Algebra and Trigonometry, Complete Solutions Manual [4 ed.]

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CONTENTS ¥

CHAPTER

PROLOGUE: Principles of Problem Solving 1

P PREREQUISITES P.1

Modeling the Real World with Algebra 3

P.2 P.3

Real Numbers 4 Integer Exponents and Scientific Notation 9

P.4

Rational Exponents and Radicals 14

P.5

Algebraic Expressions 18

P.6

Factoring 22

P.7

Rational Expressions 27

P.8

Solving Basic Equations 34

P.9

Modeling with Equations 39

3

Chapter P Review 45 Chapter P Test 51 ¥

CHAPTER

FOCUS ON MODELING: Making the Best Decisions 54

1 EQUATIONS AND GRAPHS 1.1 1.2

The Coordinate Plane 57 Graphs of Equations in Two Variables; Circles 65

1.3 1.4

Lines 79 Solving Quadratic Equations 90

1.5

Complex Numbers 98

1.6

Solving Other Types of Equations 101

1.7

Solving Inequalities 110

1.8

Solving Absolute Value Equations and Inequalities 129

1.9

Solving Equations and Inequalities Graphically 131

1.10

57

Modeling Variation 139 Chapter 1 Review 143 Chapter 1 Test 161 iii

iv

Contents

¥

CHAPTER

FOCUS ON MODELING: Fitting Lines to Data 165

2 FUNCTIONS 2.1 2.2

Functions 169 Graphs of Functions 178

2.3

Getting Information from the Graph of a Function 190

2.4

Average Rate of Change of a Function 201

2.5 2.6 2.7

Linear Functions and Models 206 Transformations of Functions 212 Combining Functions 226

2.8

One-to-One Functions and Their Inverses 234 Chapter 2 Review 243

169

Chapter 2 Test 255 ¥

CHAPTER

FOCUS ON MODELING: Modeling with Functions 259

3 POLYNOMIAL AND RATIONAL FUNCTIONS 3.1

Quadratic Functions and Models 267

3.2

Polynomial Functions and Their Graphs 276

3.3

Dividing Polynomials 291

3.4

Real Zeros of Polynomials 301

3.5

Complex Zeros and the Fundamental Theorem of Algebra 334

3.6

Rational Functions 344 Chapter 3 Review 377

267

Chapter 3 Test 395 ¥

CHAPTER

FOCUS ON MODELING: Fitting Polynomial Curves to Data 398

4 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 4.1

Exponential Functions 401

4.2

The Natural Exponential Function 409

4.3

Logarithmic Functions 414

4.4

Laws of Logarithms 422

4.5

Exponential and Logarithmic Equations 426

401

Contents

4.6

Modeling with Exponential Functions 433

4.7

Logarithmic Scales 438 Chapter 4 Review 440 Chapter 4 Test 448

¥

CHAPTER

FOCUS ON MODELING: Fitting Exponential and Power Curves to Data 450

5 TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE APPROACH 5.1

Angle Measure 455

5.2

Trigonometry of Right Triangles 459

5.3

Trigonometric Functions of Angles 464

5.4

Inverse Trigonometric Functions and Right Triangles 468

5.5 5.6

The Law of Sines 471 The Law of Cosines 476 Chapter 5 Review 481

455

Chapter 5 Test 486 ¥

CHAPTER

FOCUS ON MODELING: Surveying 536

6 TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH 6.1 6.2

The Unit Circle 491 Trigonometric Functions of Real Numbers 495

6.3

Trigonometric Graphs 500

6.4

More Trigonometric Graphs 511

6.5

Inverse Trigonometric Functions and Their Graphs 519

6.6

Modeling Harmonic Motion 521

491

Chapter 6 Review 527 Chapter 6 Test 534 ¥

CHAPTER

FOCUS ON MODELING: Fitting Sinusoidal Curves to Data 487

7 ANALYTIC TRIGONOMETRY 7.1

Trigonometric Identities 541

7.2 7.3

Addition and Subtraction Formulas 549 Double-Angle, Half-Angle, and Product-Sum Formulas 556

541

v

vi

Contents

7.4

Basic Trigonometric Equations 567

7.5

More Trigonometric Equations 571 Chapter 7 Review 578 Chapter 7 Test 584

¥

CHAPTER

FOCUS ON MODELING: Traveling and Standing Waves 586

8 POLAR COORDINATES AND PARAMETRIC EQUATIONS 8.1 8.2

Polar Coordinates 589 Graphs of Polar Equations 593

8.3

Polar Form of Complex Numbers; De Moivre’s Theorem 600

8.4

Plane Curves and Parametric Equations 612

589

Chapter 8 Review 623 Chapter 8 Test 630 ¥

CHAPTER

FOCUS ON MODELING: The Path of a Projectile 631

9 VECTORS IN TWO AND THREE DIMENSIONS 9.1 9.2 9.3

Vectors in Two Dimensions 635 The Dot Product 641 Three-Dimensional Coordinate Geometry 644

9.4 9.5 9.6

Vectors in Three Dimensions 646 The Cross Product 649 Equations of Lines and Planes 652

635

Chapter 9 Review 654 Chapter 9 Test 658 ¥

CHAPTER

FOCUS ON MODELING: Vector Fields 659

10 SYSTEMS OF EQUATIONS AND INEQUALITIES 10.1

Systems of Linear Equations in Two Variables 663

10.2

Systems of Linear Equations in Several Variables 670

10.3 10.4

Partial Fractions 678 Systems of Nonlinear Equations 689

10.5

Systems of Inequalities 696

663

Contents

vii

Chapter 10 Review 709 Chapter 10 Test 717 ¥

CHAPTER

FOCUS ON MODELING: Linear Programming 720

11 MATRICES AND DETERMINANTS 11.1

Matrices and Systems of Linear Equations 729

11.2

The Algebra of Matrices 740

11.3

Inverses of Matrices and Matrix Equations 748

11.4

Determinants and Cramer’s Rule 758 Chapter 11 Review 772

729

Chapter 11 Test 782 ¥

CHAPTER

FOCUS ON MODELING: Computer Graphics 785

12 CONIC SECTIONS 12.1 12.2

Parabolas 789 Ellipses 794

12.3

Hyperbolas 803

12.4 12.5 12.6

Shifted Conics 810 Rotation of Axes 822 Polar Equations of Conics 834

789

Chapter 12 Review 842 Chapter 12 Test 856 ¥

CHAPTER

FOCUS ON MODELING: Conics in Architecture 858

13 SEQUENCES AND SERIES 13.1

Sequences and Summation Notation 861

13.2

Arithmetic Sequences 866

13.3

Geometric Sequences 871

13.4 13.5 13.6

Mathematics of Finance 879 Mathematical Induction 883 The Binomial Theorem 892 Chapter 13 Review 896

861

viii

Contents

Chapter 13 Test 903 ¥

CHAPTER

FOCUS ON MODELING: Modeling with Recursive Sequences 904

14 COUNTING AND PROBABILITY 14.1

Counting 907

14.2

Probability 914

14.3

Binomial Probability 922

14.4

Expected Value 927

907

Chapter 14 Review 929 Chapter 14 Test 935 ¥

FOCUS ON MODELING: The Monte Carlo Method 936

APPENDIXES A

Geometry Review 939

B

Calculations and Significant Figures 940

C

Graphing with a Graphing Calculator 941

939

PROLOGUE: Principles of Problem Solving 1 1 distance ; the ascent takes h, the descent takes h, and the rate 15 r 1 1 1 1 1 2  h. Thus we have     0, which is impossible. So the car cannot go total trip should take 30 15 15 r 15 r fast enough to average 30 mi/h for the 2-mile trip.

1. Let r be the rate of the descent. We use the formula time 

2. Let us start with a given price P. After a discount of 40%, the price decreases to 06P. After a discount of 20%, the price decreases to 08P, and after another 20% discount, it becomes 08 08P  064P. Since 06P  064P, a 40% discount is better. 3. We continue the pattern. Three parallel cuts produce 10 pieces. Thus, each new cut produces an additional 3 pieces. Since the first cut produces 4 pieces, we get the formula f n  4  3 n  1, n  1. Since f 142  4  3 141  427, we see that 142 parallel cuts produce 427 pieces. 4. By placing two amoebas into the vessel, we skip the first simple division which took 3 minutes. Thus when we place two amoebas into the vessel, it will take 60  3  57 minutes for the vessel to be full of amoebas. 5. The statement is false. Here is one particular counterexample:

First half Second half Entire season

Player A

Player B

1 1 hit in 99 at-bats: average  99 1 hit in 1 at-bat: average  11

0 hit in 1 at-bat: average  01

2 2 hits in 100 at-bats: average  100

98 hits in 99 at-bats: average  98 99

99 99 hits in 100 at-bats: average  100

6. Method 1: After the exchanges, the volume of liquid in the pitcher and in the cup is the same as it was to begin with. Thus, any coffee in the pitcher of cream must be replacing an equal amount of cream that has ended up in the coffee cup. Method 2: Alternatively, look at the drawing of the spoonful of coffee and cream

cream

mixture being returned to the pitcher of cream. Suppose it is possible to separate the cream and the coffee, as shown. Then you can see that the coffee going into the

coffee

cream occupies the same volume as the cream that was left in the coffee. Method 3 (an algebraic approach): Suppose the cup of coffee has y spoonfuls of coffee. When one spoonful of cream 1 coffee y cream  and  . is added to the coffee cup, the resulting mixture has the following ratios: mixture y1 mixture y1 1 So, when we remove a spoonful of the mixture and put it into the pitcher of cream, we are really removing of a y1 y spoonful of coffee. Thus the amount of cream left in the mixture (cream in the coffee) is spoonful of cream and y 1 1 y 1  of a spoonful. This is the same as the amount of coffee we added to the cream. y 1 y1 7. Let r be the radius of the earth in feet. Then the circumference (length of the ribbon) is 2r. When we increase the radius by 1 foot, the new radius is r  1, so the new circumference is 2 r  1. Thus you need 2 r  1  2r  2 extra feet of ribbon. 1

2

Principles of Problem Solving

8. The north pole is such a point. And there are others: Consider a point a1 near the south pole such that the parallel passing through a1 forms a circle C1 with circumference exactly one mile. Any point P1 exactly one mile north of the circle C1 along a meridian is a point satisfying the conditions in the problem: starting at P1 she walks one mile south to the point a1 on the circle C1 , then one mile east along C1 returning to the point a1 , then north for one mile to P1 . That’s not all. If a point a2 (or a3 , a4 , a5 ,   ) is chosen near the south pole so that the parallel passing through it forms a circle C2 (C3 , C4 , C5 ,   ) with a circumference of exactly 12 mile ( 13 mi, 14 mi, 15 mi,   ), then the point P2 (P3 , P4 , P5 ,   ) one mile north of a2 (a3 , a4 , a5 ,   ) along a meridian satisfies the conditions of the problem: she walks one mile south from P2 (P3 , P4 , P5 ,   ) arriving at a2 ( a3 , a4 , a5 ,   ) along the circle C2 (C3 , C4 , C5 ,   ), walks east along the circle for one mile thus traversing the circle twice (three times, four times, five times,   ) returning to a2 (a3 , a4 , a5 ,   ), and then walks north one mile to P2 ( P3 , P4 , P5 ,   ).

P

PREREQUISITES

P.1

MODELING THE REAL WORLD WITH ALGEBRA

1. Using this model, we find that if S  12, L  4S  4 12  48. Thus, 12 sheep have 48 legs. 2. If each gallon of gas costs $350, then x gallons of gas costs $35x. Thus, C  35x. 3. If x  $120 and T  006x, then T  006 120  72. The sales tax is $720.

4. If x  62,000 and T  0005x, then T  0005 62,000  310. The wage tax is $310.

5. If   70, t  35, and d  t, then d  70  35  245. The car has traveled 245 miles.   6. V  r 2 h   32 5  45  1414 in3 240 N   30 miles/gallon G 8 175 175 G  7 gallons (b) 25  G 25   9. (a) V  95S  95 4 km3  38 km3

8. (a) T  70  0003h  70  0003 1500  655 F

7. (a) M 

(b) 64  70  0003h  0003h  6  h  2000 ft

  10. (a) P  006s 3  006 123  1037 hp

(b) 19 km3  95S  S  2 km3

11. (a) Depth (ft)

(b) 75  006s 3  s 3  125 so s  5 knots

Pressure (lb/in2 ) 045 0  147  147

0

045 10  147  192

10

045 20  147  237

20

(b) We know that P  30 and we want to find d, so we solve the equation 30  147  045d  153  045d 

153  340. Thus, if the pressure is 30 lb/in2 , the depth 045 is 34 ft. d

045 30  147  282

30

045 40  147  327

40

045 50  147  372

50

045 60  147  417

60 12. (a) Population

(b) We solve the equation 40x  120,000 

Water use (gal)

0

0

1000

40 1000  40,000

2000 3000 4000 5000

x

120,000  3000. Thus, the population is about 3000. 40

40 2000  80,000

40 3000  120,000 40 4000  160,000 40 5000  200,000

13. The number N of cents in q quarters is N  25q. ab 14. The average A of two numbers, a and b, is A  . 2 15. The cost C of purchasing x gallons of gas at $350 a gallon is C  35x.

16. The amount T of a 15% tip on a restaurant bill of x dollars is T  015x. 17. The distance d in miles that a car travels in t hours at 60 mi/h is d  60t.

3

4

CHAPTER P Prerequisites

18. The speed r of a boat that travels d miles in 3 hours is r 

d . 3

19. (a) $12  3 $1  $12  $3  $15

(b) The cost C, in dollars, of a pizza with n toppings is C  12  n.

(c) Using the model C  12  n with C  16, we get 16  12  n  n  4. So the pizza has four toppings.

20. (a) 3 30  280 010  90  28  $118         daily days cost miles (b) The cost is    , so C  30n  01m. rental rented per mile driven (c) We have C  140 and n  3. Substituting, we get 140  30 3  01m  140  90  01m  50  01m  m  500. So the rental was driven 500 miles. 21. (a) (i) For an all-electric car, the energy cost of driving x miles is Ce  004x.

(ii) For an average gasoline powered car, the energy cost of driving x miles is C g  012x.

(b) (i) The cost of driving 10,000 miles with an all-electric car is Ce  004 10,000  $400.

(ii) The cost of driving 10,000 miles with a gasoline powered car is C g  012 10,000  $1200.

22. (a) If the width is 20, then the length is 40, so the volume is 20  20  40  16,000 in3 . (b) In terms of width, V  x  x  2x  2x 3 .

4a  3b  2c  d 4a  3b  2c  1d  0 f  . abcd  f abcd  f (b) Using a  2  3  6, b  4, c  3  3  9, and d  f  0 in the formula from part (a), we find the GPA to be 463429 54   284. 649 19

23. (a) The GPA is

P.2

THE REAL NUMBERS

1. (a) The natural numbers are 1 2 3   .

(b) The numbers     3 2 1 0 are integers but not natural numbers. p 5 , 1729 . (c) Any irreducible fraction with q  1 is rational but is not an integer. Examples: 32 ,  12 23 q   p (d) Any number which cannot be expressed as a ratio of two integers is irrational. Examples are 2, 3, , and e. q

2. (a) ab  ba; Commutative Property of Multiplication

(b) a  b  c  a  b  c; Associative Property of Addition (c) a b  c  ab  ac; Distributive Property

3. The set of numbers between but not including 2 and 7 can be written as (a) x  2  x  7 in interval notation, or (b) 2 7 in interval notation. 4. The symbol x stands for the absolute value of the number x. If x is not 0, then the sign of x is always positive.

5. The distance between a and b on the real line is d a b  b  a. So the distance between 5 and 2 is 2  5  7. a c ad  bc   . b d bd (b) No, the sum of two irrational numbers can be irrational (    2) or rational (    0).

6. (a) Yes, the sum of two rational numbers is rational:

7. (a) No: a  b   b  a  b  a in general.

(b) No; by the Distributive Property, 2 a  5  2a  2 5  2a  10  2a  10.

8. (a) Yes, absolute values (such as the distance between two different numbers) are always positive. (b) Yes, b  a  a  b.

SECTION P.2 The Real Numbers

 10. (a) Natural number: 16  4  (b) Integers: 500, 16,  20 5  4

9. (a) Natural number: 100 (b) Integers: 0, 100, 8 (c) Rational numbers: 15, 0, 52 , 271, 314, 100, 8  (d) Irrational numbers: 7, 

(c) Rational numbers: 13, 13333   , 534, 500, 1 23 ,  20 16, 246 579 ,  5  (d) Irrational number: 5

11. Commutative Property of addition

12. Commutative Property of multiplication

13. Associative Property of addition

14. Distributive Property

15. Distributive Property

16. Distributive Property

17. Commutative Property of multiplication

18. Distributive Property

19. x  3  3  x

20. 7 3x  7  3 x

21. 4 A  B  4A  4B

22. 5x  5y  5 x  y

23. 3 x  y  3x  3y

24. a  b 8  8a  8b   26. 43 6y  43 6 y  8y

25. 4 2m  4  2 m  8m 27.  52 2x  4y   52 2x  52 4y  5x  10y 3  4  9  8  17 29. (a) 10 15 30 30 30 5  4  9 (b) 14  15  20 20 20

28. 3a b  c  2d  3ab  3ac  6ad

9 1 30. (a) 23  35  10 15  15  15 15 4 35 (b) 1  58  16  24 24  24  24  24

  31. (a) 23 6  32  23  6  23  32  4  1  3       1 5  4  13  1  13 (b) 3  14 1  45  12 4  4 5 5 4 5 20

2 32. (a) 2  3  2  32  23  12  3  13  93  13  83 2

33. (a) 2  3  6 and 2  72  7, so 3  72

34. (a) 3  23  2 and 3  067  201, so 23  067

2

3 2  1 2  1 2  1 45 9 (b) 15 23  51 21  51 21  10 10  12  3  3    10 15 10 5 10 5

(b) 6  7

(b) 23  067

(c) 35  72

(c) 067  067

35. (a) False

36. (a) False:

(b) True

(b) False

 3  173205  17325.

37. (a) True

(b) False

38. (a) True

(b) True

39. (a) x  0

(b) t  4

40. (a) y  0

(b) z  1

(c) a  

(d) 5  x  13

(e)  p  3  5 41. (a) A  B  1 2 3 4 5 6 7 8 (b) A  B  2 4 6

5

(c) b  8

(d) 0    17

(e) y    2 42. (a) B  C  2 4 6 7 8 9 10 (b) B  C  8

6

CHAPTER P Prerequisites

43. (a) A  C  1 2 3 4 5 6 7 8 9 10

44. (a) A  B  C  1 2 3 4 5 6 7 8 9 10 (b) A  B  C  ∅

(b) A  C  7

46. (a) A  C  x  1  x  5

45. (a) B  C  x  x  5 (b) B  C  x  1  x  4

(b) A  B  x  2  x  4 48. 2 8]  x  2  x  8

47. 3 0  x  3  x  0 _3

0

8

_6

2

1

53. x  1  x   1]

54. 1  x  2  x  [1 2] 1

1

55. 2  x  1  x  2 1]

_5

1

58. 5  x  2  x  5 2

_1

_5

(b) 3 5]

1

63. [4 6]  [0 8  [0 6] 0

(b) 2 0]

60. (a) [0 2

_1

0

64. [4 6]  [0 8  [4 8 6

65.  4  4  _4

2

62. 2 0  1   1 0

61. 2 0  1 1  2 1 _2

2

56. x  5  x  [5 

57. x  1  x  1 

59. (a) [3 5]

1

_ _2

52.  1  x  x  1

51. [2   x  x  2

_2

8

    50. 6  12  x  6  x   12

49. [2 8  x  2  x  8 2

2

_4

8

66.  6]  2 10  2 6] 4

2

6

SECTION P.2 The Real Numbers

67. (a) 100  100 (b) 73  73 69. (a) 6  4  6  4  2  2

        68. (a)  5  5   5  5  5  5, since 5  5. (b) 10    10  , since 10  .

70. (a) 2  12  2  12  10  10

1  1  1 (b) 1 1

71. (a) 2  6  12  12      (b)   13 15  5  5

73. 2  3  5  5 75. (a) 17  2  15

(b) 21  3  21  3  24  24        3   12 55   67  67 (c)  10  11 8    40  40    40   40

(b) 1  1  1  1  1  1  1  0  1        1 1 72. (a)  6 24    4   4        5 (b)  712 127    5   1  1 74. 25  15  4  4

         7 1    49  5    54    18   18 76. (a)  15   21   105 105   105   35  35 (b) 38  57  38  57  19  19.

(c) 26  18  26  18  08  08.

77. (a) Let x  0777   . So 10x  77777     x  07777     9x  7. Thus, x  79 .

13 (b) Let x  02888   . So 100x  288888     10x  28888     90x  26. Thus, x  26 90  45 . 19 (c) Let x  0575757   . So 100x  575757     x  05757     99x  57. Thus, x  57 99  33 .

78. (a) Let x  52323   . So 100x  5232323     1x  52323     99x  518. Thus, x  518 99 .

62 (b) Let x  13777   . So 100x  1377777     10x  137777     90x  124. Thus, x  124 90  45 .

1057 (c) Let x  213535   . So 1000x  21353535     10x  213535     990x  2114. Thus, x  2114 990  495 .

      2  1, so 1  2  2  1.

79.   3, so   3    3.

80.

81. a  b, so a  b   a  b  b  a.

82. a  b  a  b  a  b  b  a  2b

83. (a) a is negative because a is positive.

(b) bc is positive because the product of two negative numbers is positive. (c) a  ba  b is positive because it is the sum of two positive numbers.

(d) ab  ac is negative: each summand is the product of a positive number and a negative number, and the sum of two negative numbers is negative. 84. (a) b is positive because b is negative.

(b) a  bc is positive because it is the sum of two positive numbers.

(c) c  a  c  a is negative because c and a are both negative. (d) ab2 is positive because both a and b2 are positive.

85. Distributive Property

7

8

CHAPTER P Prerequisites

86. Day

TO

TG

TO  TG

TO  TG 

Sunday

68

77

9

9

Monday

72

75

3

3

Tuesday

74

74

0

0

Wednesday

80

75

5

5

Thursday

77

69

8

8

Friday

71

70

1

1

Saturday

70

71

1

1

TO  TG gives more information because it tells us which city had the higher temperature. 87. (a) When L  60, x  8, and y  6, we have L  2 x  y  60  2 8  6  60  28  88. Because 88  108 the post office will accept this package. When L  48, x  24, and y  24, we have L  2 x  y  48  2 24  24  48  96  144, and since 144  108, the post office will not accept this package. (b) If x  y  9, then L  2 9  9  108  L  36  108  L  72. So the length can be as long as 72 in.  6 ft.

m2 m1 m m1n2  m2n1 m1 and y  be rational numbers. Then x  y   2  , n1 n2 n1 n2 n1 n2 m m n  m 2 n1 m m m m m , and x  y  1  2  1 2 . This shows that the sum, difference, and product xy 1  2  1 2 n1 n2 n1 n2 n1 n2 n1n2 of two rational numbers are again rational numbers. However the product of two irrational numbers is not necessarily   irrational; for example, 2  2  2, which is rational. Also, the sum of two irrational numbers is not necessarily irrational;     for example, 2   2  0 which is rational.

88. Let x 

       89. 12  2 is irrational. If it were rational, then by Exercise 6(a), the sum 12  2   12  2 would be rational, but

this is not the case.  Similarly, 12  2 is irrational. (a) Following the hint, suppose that r  t  q, a rational number. Then by Exercise 6(a), the sum of the two rational numbers r  t and r is rational. But r  t  r  t, which we know to be irrational. This is a contradiction, and hence our original premise—that r  t is rational—was false. a (b) r is rational, so r  for some integers a and b. Let us assume that rt  q, a rational number. Then by definition, b c a c bc q  for some integers c and d. But then rt  q  t  , whence t  , implying that t is rational. Once again d b d ad we have arrived at a contradiction, and we conclude that the product of a rational number and an irrational number is irrational.

90. x

1

2

10

100

1000

1 x

1

1 2

1 10

1 100

1 1000

As x gets large, the fraction 1x gets small. Mathematically, we say that 1x goes to zero. x 1 x

1

05

01

001

0001

1

1 05  2

1 01  10

1 001  100

1 0001  1000

As x gets small, the fraction 1x gets large. Mathematically, we say that 1x goes to infinity.

SECTION P.3 Integer Exponents and Scientific Notation

91. (a) Construct the number

 2 on the number line by transferring

Ï2

the length of the hypotenuse of a right triangle with legs of length 1 and 1.

_1

0

1

(b) Construct a right triangle with legs of length 1 and 2. By the Pythagorean Theorem, the length of the hypotenuse is   12  22  5. Then transfer the length of the hypotenuse to the number line.

 (c) Construct a right triangle with legs of length 2 and 2  [construct 2 as in part (a)]. By the Pythagorean Theorem,    2  the length of the hypotenuse is 2  22  6. Then transfer the length of the hypotenuse to the number line.

1 Ï2

Ï5

_1

0

2

3

2 Ï5

3

2

3

1

1 Ï6 Ï2

_1

0

Ï2

1 1

Ï2

Ï6

92. (a) Subtraction is not commutative. For example, 5  1  1  5. (b) Division is not commutative. For example, 5  1  1  5.

(c) Putting on your socks and putting on your shoes are not commutative. If you put on your socks first, then your shoes, the result is not the same as if you proceed the other way around. (d) Putting on your hat and putting on your coat are commutative. They can be done in either order, with the same result. (e) Washing laundry and drying it are not commutative. (f) Answers will vary. (g) Answers will vary. 93. Answers will vary. 94. (a) If x  2 and y  3, then x  y  2  3  5  5 and x  y  2  3  5. If x  2 and y  3, then x  y  5  5 and x  y  5. If x  2 and y  3, then x  y  2  3  1 and x  y  5. In each case, x  y  x  y and the Triangle Inequality is satisfied. (b) Case 0: If either x or y is 0, the result is equality, trivially.   xy Case 1: If x and y have the same sign, then x  y    x  y

 if x and y are positive   x  y. if x and y are negative 

Case 2: If x and y have opposite signs, then suppose without loss of generality that x  0 and y  0. Then x  y  x  y  x  y.

P.3

INTEGER EXPONENTS AND SCIENTIFIC NOTATION

1. Using exponential notation we can write the product 5  5  5  5  5  5 as 56 .

2. Yes, there is a difference: 54  5 5 5 5  625, while 54   5  5  5  5  625. 3. In the expression 34 , the number 3 is called the base and the number 4 is called the exponent.

4. When we multiply two powers with the same base, we add the exponents. So 34  35  39 . 35 5. When we divide two powers with the same base, we subtract the exponents. So 2  33 . 3  2 6. When we raise a power to a new power, we multiply the exponents. So 34  38 .

9

10

CHAPTER P Prerequisites

1 7. (a) 21  2

(b) 23 

1 8

(c)

 1 1 2 2

1 (d) 3  23  8 2

8. Scientists express very large or very small numbers using scientific notation. In scientific notation, 8,300,000 is 83  106

and 00000327 is 327  105 .  2  2 9 3 2   . 9. (a) No, 3 2 4  3 10. (a) No, x 2  x 23  x 6 .

  (b) Yes, 54  625 and 54   54  625. 3   3 (b) No, 2x 4  23  x 4  8x 12 .

11. (a) 26  64

(b) 26  64

12. (a) 53  125

(b) 53  125

13. (a)

 0 1 5  21  3 2

(b)

23 1 1  3  8 30 2

1 1 14. (a) 23  20   3   8 2

(b) 23  20  23  8

15. (a) 53  5  54  625

(b) 32  30  32  9

16. (a) 38  35  313  1,594,323

(b) 60  6  6

17. (a) 54  52  52  25

(b)

1 1 18. (a) 33  31  34  4  81 3

(b)

19. (a) x 2 x 3  x 23  x 5 20. (a) y 5  y 2  y 52  y 7

107  103  1000 104

54  53  125 5 3  (b) x 2  13 x 23  x 6

 2 12 33 27 1  33   5 25 52  2 2 2 2 5 2 (c) 52   4 5 52  2 1 (c)  42  16 4   33 27 2 3 (c)   3 3 8 2  3 (c) 22  26  64 (c)

 2 (c) 54  58  390,625 32 1 1 (c) 4  2  9 3 3

72 1 1 (c) 5  3  343 7 7 (c) t 3 t 5  t 35  t 2

(b) 8x2  82 x 2  64x 2

(c) x 4 x 3  x 43  x

1 21. (a) x 5  x 3  x 53  x 2  2 x y 10 y 0 (c)  y 1007  y 3 y7

(b) 2 4 5  245  1 

1 

1 22. (a) y 2  y 5  y 25  y 3  3 y

1 x6 1 (b) z 5 z 3 z 4  z 534  z 2  2 (c) 10  x 610  4 z x x 3  3  3  (b) a 2 a 4  a 24  a 6  a 63  a 18

a 9 a 2  a 921  a 6 a   (c) 2x2 5x 6  22 x 2  5x 6  20x 26  20x 8

23. (a)

4  4  4  z2 z4 z 24 z6  31  2  z 62  z 4 (b) 2a 3 a 2  2a 32  2a 5  24 a 54  16a 20 3 1 z z z z 3    2z 3  33 z 23  2z 3  54z 63  54z 9 (c) 3z 2    25. (a) 3x 2 y 2x 3  3  2x 23 y  6x 5 y    (b) 2a 2 b1 3a 2 b2  2  3a 22 b12  6b 24. (a)

SECTION P.3 Integer Exponents and Scientific Notation

(c)

  2 4y 2 x 4 y  4y 2 x 42 y 2  4x 8 y 22  4x 8 y 4

   4x 3 y 2 7y 5  4  7x 3 y 25  28x 3 y 7    (b) 9y 2 z 2 3y 3 z  9  3y 23 z 21  27yz 3

26. (a)

(c)

27. (a)

  2 8x 7 y 2 22  8x 7 y 2 32x 7 y 2 8x 7 y 2 12 x 3 y   6 2  32x 76 y 22  32x 2  32 2 x y x y 1 x3 y 2 2  2x 2 y 3 3y  22 x 22 y 32  3y  12x 4 y 7

x 2 y 1 x7  x 25 y 1  x 7 y 1  5 y x 3  x 23 y 3 x 6 y3 x2 y   (c) 3 3 27 3

(b)

28. (a)

  2 5x 4 y 3 8x 3  5x 4 y 3  82 x 32  5  82 x 46 y 3  320x 2 y 3

y 2 z 3 y 1  2 3  3 1 y y z yz 2  3 2 6 4 a b a6 a b (c)   b3 b6 b10 (b)

1

29. (a)



x 3 y3

30. (a)



x 2 y 4

1  3 3 x y 3  2  b6 a3 (b) a 2 b2  a 23 b23 a 32  a 6 b6 a 6  12 a 2  3  2 3 8 2y x (c)  x 22 y 22  23 y 33 x 23  x 4 y 4  8y 9 x 6  8x 46 y 49  10 13 y 2 x2 x y 3





y4 x2

3





x2 y4

3

x6  12 y

1  3 x9 2x 3 y 4  y 2 23 x 33 y 43  14 8y 3  2  1 1 b 2a 1 (c)  23 a 13 b23 b12 22 a 22  2 2 b 2a 32ab8 (b)



y2

3x 2 y 5 x y3  3 9x 3 y 2 2  2  y 32 y6 2x 3 2x 3 y 1    (b) y2 y3 22 x 32 4x 6 1  2  x 4 y5 3x 3 y 1 (c)  y 11 x 21 32 x 32 y 22  2 2 9 x y

31. (a)

1 a 3 b4 a2 32. (a) 2 5 1  12  21 a 35 b41  14 a 2 b3  3 2a b 4b

11

12

CHAPTER P Prerequisites



x2 y 5x 4

2





5x 4 x2 y

2





5x 2 y

2

25x 4 y2 1       1 y3 y 2 y2 2y 1 z 2  (c)  2 2 4 yz z 3z 9z 18z 3 (b)



 b3 3a 1  31 a 1 b31  3 3a b 1  r 5 sq 8 s3 q 1r 1 s 2  1 1 2  q 81 r 51 s 12  7 4 (b) 5 8 r sq q r s q r

33. (a)



34. (a)



s 2 t 4 5s 1 t

(b)



x y 2 z 3 x 2 y 3 z 4

2

 s 2212 t 4212 52 

3

25t 10 s6

 x 323 y 2333 z 3343 

35. (a) 69,300,000  693  107

x 3 y 15 z3

36. (a) 129,540,000  12954  108

(b) 7,200,000,000,000  72  1012

(b) 7,259,000,000  7259  109

(c) 0000028536  28536  105

(c) 00000000014  14  109

(d) 00001213  1213  104

(d) 00007029  7029  104

37. (a) 319  105  319,000

38. (a) 71  1014  710,000,000,000,000

(b) 2721  108  272,100,000

(b) 6  1012  6,000,000,000,000

(c) 2670  108  000000002670

(c) 855  103  000855

(d) 9999  109  0000000009999

(d) 6257  1010  00000000006257

39. (a) 5,900,000,000,000 mi 59  1012 mi

(b) 00000000000004 cm  4  1013 cm

(c) 33 billion billion molecules  33  109  109  33  1019 molecules

40. (a) 93,000,000 mi  93  107 mi

(b) 0000000000000000000000053 g  53  1023 g

(c) 5,970,000,000,000,000,000,000,000 kg  597  1024 kg    41. 72  109 1806  1012  72  1806  109  1012  130  1021  13  1020    42. 1062  1024 861  1019  1062  861  1024  1019  914  1043

1295643 1295643  109     109176  01429  1019  1429  1019 43.  3610  2511 3610  1017 2511  106     731  10 16341  1028 731 16341  1028 731  16341   101289  63  1038 44.  00000000019 19 19  109    162  105 1582  102 162  1582 00000162 001582     105283  0074  1012 45.   594621  58 594621000 00058 594621  108 58  103  74  1014

SECTION P.3 Integer Exponents and Scientific Notation

 9 3542  106 8774796 35429  1054  105448  319  104  10102  319  10106 46.   12  12  1048 27510376710 4 505 505  10         47. 1050  1010   1050 , whereas 10101  10100   10100 10  1  9  10100  1050 . So 1010 is closer to 1050 than 10100 is to 10101 .

48. (a) b5 is negative since a negative number raised to an odd power is negative.

(b) b10 is positive since a negative number raised to an even power is positive. (c) ab2 c3 we have positive negative2 negative3  positive positive negative which is negative. (d) Since b  a is negative, b  a3  negative3 which is negative.

(e) Since b  a is negative, b  a4  negative4 which is positive. (f)

a 3 c3 negative positive3 negative3 positive negative which is negative.    6 6 positive positive positive b c negative6 negative6

49. Since one light year is 59  1012 miles, Centauri is about 43  59  1012  254  1013 miles away or 25,400,000,000,000 miles away. 93  107 mi t st  s  500 s  8 13 min. s 186 000    103 liters   3 14 2  133  1021 liters 51. Volume  average depth area  37  10 m 36  10 m m3

50. 93  107 mi  186 000

52. Each person’s share is equal to

1674  1013 national debt   $52,900. population 3164  108

53. The number of molecules is equal to         liters molecules 602  1023 3   5  10  3  10   403  1027 volume  224 liters 224 m3 54. (a) BMI  703

W H2

Person

Weight

Height

Result

Brian

295 lb

5 ft 10 in.  70 in.

4232

obese

Linda

105 lb

5 ft 6 in.  66 in.

1695

underweight

Larry

220 lb

6 ft 4 in.  76 in.

2678

overweight

Helen

110 lb

5 ft 2 in.  62 in.

2012

normal

(b) Answers will vary. 55. Year

Total interest

1

$15208

2

30879

3

47026

4

63664

5

80808

56. Since 106  103  103 it would take 1000 days  274 years to spend the million dollars.

Since 109  103  106 it would take 106  1,000,000 days  273972 years to spend the billion dollars.

13

14

CHAPTER P Prerequisites

57. (a)

185  95



 18 5  25  32 9

(b) 206  056  20  056  106  1,000,000 am 58. (a) We wish to prove that n  a mn for positive integers m  n. By definition, a m factors

   a  a    a ak  a  a      a . Thus,  . Because m  n, m  n  0, so we can write    an a  a       a am

k factors

n factors

n factors

mn factors

mn factors

         am a  a    a  a  a    a a  a    a  a mn .   an a  a      a 1    n factors

(b) We wish to prove that

 a n b



an for positive integers m  n. By definition, bn

n factors

 a n b

   a a a an a  a    a        n. b b b b b    b     b n factors

n factors bn

 a n

 n . By definition, and using 59. (a) We wish to prove that b a  a n n 1 1 b   a n  n  n . a b a b bn 1 n a n bm a n a (b) We wish to prove that m  n . By definition, m   1 b a b bm

P.4

the result from Exercise 58(b),

bm 1 bm  n.  n a 1 a

RATIONAL EXPONENTS AND RADICALS

 1. Using exponential notation we can write 3 5 as 513 .  2. Using radicals we can write 512 as 5. 2  12  2    5212  5 and 5  512  5122  5. 3. No. 52  52  3  12 4. 412  23  8; 43  6412  8 5. Because the denominator is of the form

    a, we multiply numerator and denominator by a: 1  1  3  33 . 3

3

3

6. 513  523  51  5  7. No. If a is negative, then 4a 2  2a.    8. No. For example, if a  2, then a 2  4  8  2 2, but a  2  0.  1 3 9.   312 10. 72  723 3 1  1    1 3 11. 423  42  3 16 12. 1032  1032  103  103  1 1 5 3 13. 5  535 14. 215  232     8 23

SECTION P.4 Rational Exponents and Radicals

15. a 25  17.

 3

y 4  y 43

19. (a) (b) (c) 21. (a) (b) (c) 23. (a) (b) (c) 25. (a) (b) (c)

27. 29. 31. 33. 35. 37. 39. 40. 41. 42. 43. 44. 45. 46.

 4

 5 2 a

  16  42  4   4 4 16  24  2    1 1 4 4 4 1   16 2 2    3 3 3 16  3 2  23  6 3 2     18 18 2 2     81 9 3 81    2 3 3 27 33   4 2 22     7 28  7  28  196  14   48 48   16  4   3 3     4 24 4 54  4 24  54  4 1296  6   216 216   36  6   6 6    3 2 3 32  3 64  4    1 1 1 4 1 4 1   4   4 4 64 256 4 256

1 1 16.   52  x 52 x x5 1 1 18. y 53  53   3 5 y y   2 20. (a) 64  8  8   3 (b) 3 64  43  4   5 (c) 5 32  25  2

   3 22. (a) 2 3 81  2 3  33  6 3 3    2 3 12 3  22  (b)   5 5 25    3 2 18 2  32 (c)   2 49 7 7       24. (a) 12 24  12  24  288  2  122  12 2   54 54  (b)    93 6 6       (c) 3 15 3 75  3 15  75  3 1125  3 125  9  5 3 9    1 1 1 1  26. (a) 5 5  5 8 4 32 2    1 (b) 6 6 128  6 64  2 2    3 4 1 1 3 4 3 1      (c)  3 3 108 27 3 108 27  15  5 28. x 10  x 10  x2    3 3 3 30. 8a 5  23 a 3 a 2  2a a 2 13   32. 3 x 3 y 6  x 3 y 6  x y2 12   34. x 4 y 4  x 4 y 4  x 2 y2

x 4  x     5 32y 6  5 25 y 6  2 5 y 6  2y 5 y   4 4 16x 8  24 x 8  2x 2  13   3 3 x y  x3 y 13  x 3 y      2 4 4 4 36r 2 t 4  6rt 2  6 r t 2 36. 48a 7 b4  24 a 4 b4  3a 3  2 ab 3a 3    13     3 4   64x 6  8 x 3   2 x 38. 4 x 4 y 2 z 2  x 4 4 y 2 z 2  x 4 y 2 z 2          32  18  16  2  9  2  42  2  32  2  4 2  3 2  7 2          75  48  25  3  16  3  52  3  42  3  5 3  4 3  9 3          125  45  25  5  9  5  52  5  32  5  5 5  3 5  2 5        3 3 3 54  3 16  2  33  23  2  3 3 2  2 3 2  3 2        9a 3  a  32 a 2  a  a  3a a  a  3a  1 a        2   16x  x 5  42 x  x 2 x  4 x  x 2 x  x 2  4 x        3 4 3 3 x  3 8x  x 3 x  23 x  x 3 x  2 3 x  x  2 3 x         3 2y 4  3 2y  3 2y  y 3  3 2y  3 2y y 3  3 2y  y  1 3 2y

15

16

CHAPTER P Prerequisites

       81x 2  81  81 x 2  1  81 x 2  1  9 x 2  1        48. 36x 2  36y 2  36 x 2  y 2  36 x 2  y 2  6 x 2  y 2

47.

49. (a) 1614  2

(b) 12513  5

50. (a) 2713  3

(b) 813  2

2  51. (a) 3225  3215  22  4

 12  12 4 9 3   9 4 2  32  3 125 25 5 (b)   64 8 512

52. (a) 12523  52  25 53. (a) 523  513  52313  51  5

1 1 (c) 912  12  3 9  13 1 1 (c)   8 2  34  3 16 2 8 (c)   81 3 27

(b)

(c) 2743  34 

 335 (b) 25  33525  5 3 3

(c)

1 81

  3 3 4  4133  4

  10 723 1 1 54. (a) 327  3127  327127  32  9 (b) 53  72353  (c) 5 6  61510  7 36 7     55. When x  3, y  4, z  1 we have x 2  y 2  32  42  9  16  25  5.      4 56. When x  3, y  4, z  1 we have 4 x 3  14y  2z  4 33  14 4  2 1  4 27  56  2  4 81  34  3. 57. When x  3, y  4, z  1 we have

 23  23  23  113 9x23  2y23  z 23  9  323  2  423  123  33  32  22  1  9  4  1  14.

1 . 58. When x  3, y  4, z  1 we have x y2z  3  42  1  122  144

59. (a) x 34 x 54  x 3454  x 2

(b) y 23 y 43  y 2343  y 2

60. (a) r 16r 56  r 1656  r

(b) a 35 a 310  a 35310  a 910

43 23 61. (a)  432313  53 13

(b)

x 34 x 74 62. (a)  x 347454  x 54 x 54 23   823 a 623 b3223  4a 4 b 63. (a) 8a 6 b32 64. (a)

23  64a 6 b3  6423 a 623 b323  16a 4 b2

(b)

 3 a 54 2a 34 a 14 2

 2y 43

 23 a 5234314  8a 134

y 23

y 73

4  4y 832373  4y 13   3 y

(b) 4a 6 b8 32  432 a 632 b832  8a 9 b12

34  (b) 168 z 32  1634 834 z 3234  86 z 98

23 13   1 1 8y 3  823 y 323  2 (b) u 4  6  u 413  613  43 2 4y u  35  x3 66. (a) x 5 y 13  x 535 y 1335  15 y 12  15    32t 54 (b) 4r 8 t 12  412r 812 t 1212 3215 t 5415  2r 4 t 14 12 t 14  r 4 t 0  r 4 65. (a)

67. (a)



x 23 y 12



x 2 y 3

16

 x 23216 y 12316 

1 x

SECTION P.4 Rational Exponents and Radicals

(b)



x 12 y 2 2y 14

4 

 813 y 3413313 z 613  

(b)

8y 34 y3 z6

 9

73.



 x 2 y 8 24 y 1 2x 1 y 2 y 1  241 x 21 y 8121 

13



71.

 x 124 y 24 214 y 144 412 x 212 y 412 y 212

 1614 x 814 y 4144314 

68. (a)

x 8 y 4 16y 43



12

14



69.

4x 2 y 4 y2

x 3  x 32 x 5  x 59

6

y5

  3

 y 2  y 56  y 23  y 5623  y 32

     75. 5 3 x 2 4 x  5  2x 1314  10x 712  4 7  x 4 77.   x4  x 4 3 x   16u 3  16u 2 4u 79.   2 u 5 4   xy x 14 y 14 81.   1614 x 1214 y 1214  4 2 16x y 13    83. 3 y y  y 112  y 3213  y 12   1 6 6 1 85. (a)       6 6 6 6     3 3 2 6 (b)    2 2 2 2  9 234 948 9  14  32  (c)  4 2 2 2 2   1 5x 5x 1 87. (a)       5x 5x 5x 5x     x x 5 5x (b)     5 5 5 5  25 1 x x 25 1 (c) 5 3  35  25  x x x x   3 2 3 2 1 1 x x      89. (a)  3 3 x x x 3 x2   6 6 1 1 x x  (b)      6 5 6 5 6 x x x x   7 4 7 4 1 x x 1      (c)  7 3 7 3 7 4 x x x x

70.



x y4 8

2y 43 x2 y 34 z 2 2

x 5  x 52

1 1 72.   35  x 35 5 3 x x   4 74. b3 b  b3412  b54      3 2 76. 2 a a  2a 1223  2a 76  3  8x 2 78.   2x 23  x 12  2x 16  2 6 x x   2 4 3 3y 3 27y 3 54x y 80.   x 2x 5 y x3  a3b 82.   a 3234 b1224  a 34 4 3 2 a b   12  84. s s 3  s 132  s 54    12 3 12 12 3 86. (a)       4 3 3 3 3 3      12 12 5 60 2 15 (b)      5 5 5 5 5  8 513 835 8  23  13  (c)  3 2 5 5 5 5     s s 3t 3st    88. (a) 3t 3t 3t 3t a a b23 ab23 (b)     6 2 b b13 b23 b 25 25 1 1 c c (c) 35  35  25  c c c c   3 3 1 1 x x  90. (a)      3 2 3 2 3 x x x x   4 4 1 1 x x  (b)      4 3 4 3 4 x x x x

 3 2 x 1 1 1 1     (c)      3 4 3 3 3 2 3 3 x x x x x x x x   3 2 3 2 x x    2 3 x x x3

17

18

CHAPTER P Prerequisites

91. (a) Since 12  13 , 212  213 .  12  13  12  13 (b) 12  212 and 12  213 . Since  12   13 , we have 12  12 .  112  112 92. (a) We find a common root: 714  7312  73  343112 ; 413  4412  44  256112 . So 714  413 .  16  16    2516 ; 3  312  336  33  2716 . So (b) We find a common root: 3 5  513  526  52   3 5  3. 1 mile  0215 mi. Thus the distance you can see is given 93. First convert 1135 feet to miles. This gives 1135 ft  1135  5280 feet    by D  2r h  h 2  2 3960 0215  02152  17028  413 miles.   94. (a) Using f  04 and substituting d  65, we obtain s  30 f d  30  04  65  28 mi/h.    (b) Using f  05 and substituting s  50, we find d. This gives s  30 f d  50  30  05 d  50  15d  2500  15d  d  500 3  167 feet.

95. (a) Substituting, we get 030 60038 340012 3 65013  18038 58313 866  1822162598  1418. Since this value is less than 16, the sailboat qualifies for the race. (b) Solve for A when L  65and V  600. Substituting, we get 030 65  038A12  3 60013  16 

195  038A12  2530  16  038A12  580  16  038A12  2180  A12  5738  A  32920. Thus, the largest possible sail is 3292 ft2 .

7523  005012  17707 ft/s. 24123  0040 (b) Since the volume of the flow is V  A, the canal discharge is 17707  75  13280 ft3 s.

96. (a) Substituting the given values we get V  1486 97. (a) n

1

2

5

10

100

21n

211  2

212  1414

215  1149

2110  1072

21100  1007

So when n gets large, 21n decreases toward 1. (b) n  1n 1 2

1  11 1 2

2  05

 12 1 2

 0707

 1n So when n gets large, 12 increases toward 1.

P.5

5  15 1 2

 0871

10  110 1  0933 2

100  1100 1  0993 2

ALGEBRAIC EXPRESSIONS

 1. (a) 2x 3  12 x  3 is a polynomial. (The constant term is not an integer, but all exponents are integers.)  (b) x 2  12  3 x  x 2  12  3x 12 is not a polynomial because the exponent 12 is not an integer. (c)

1 is not a polynomial. (It is the reciprocal of the polynomial x 2  4x  7.) x 2  4x  7

(d) x 5  7x 2  x  100 is a polynomial.  3 (e) 8x 6  5x 3  7x  3 is not a polynomial. (It is the cube root of the polynomial 8x 6  5x 3  7x  3.)   (f) 3x 4  5x 2  15x is a polynomial. (Some coefficients are not integers, but all exponents are integers.)

SECTION P.5 Algebraic Expressions

2. To add polynomials we add like terms. So     3x 2  2x  4  8x 2  x  1  3  8 x 2  2  1 x  4  1  11x 2  x  5.

19

3. To subtract polynomials we subtract like terms. So     2x 3  9x 2  x  10  x 3  x 2  6x  8  2  1 x 3  9  1 x 2  1  6 x  10  8  x 3  8x 2  5x  2.

4. We use FOIL to multiply two polynomials:x  2 x  3  x  x  x  3  2  x  2  3  x 2  5x  6. 5. The Special Product Formula for the “square of a sum” is A  B2  A2  2AB  B 2 . So 2x  32  2x2  2 2x 3  32  4x 2  12x  9.

6. The Special Product Formula for the “product of the sum and difference of terms” is A  B A  B  A2  B 2 . So 5  x 5  x  52  x 2  25  x 2 .

7. (a) No, x  52  x 2  10x  25  x 2  25.

(b) Yes, if a  0, then x  a2  x 2  2ax  a 2 .

8. (a) Yes, x  5 x  5  x 2  5x  5x  25  x 2  25.

(b) Yes, if a  0, then x  a x  a  x 2  ax  ax  a 2  x 2  a 2 .

9. Binomial, terms 5x 3 and 6, degree 3 11. Monomial, term 8, degree 0 13. Four terms, terms x, x 2 , x 3 , and x 4 , degree 4

10. Trinomial, terms 2x 2 , 5x, and 3, degree 2 12. Monomial, term 12 x 7 , degree 7   14. Binomial, terms 2x and  3, degree 1

15. 6x  3  3x  7  6x  3x  3  7  9x  4

16. 3  7x  11  4x  7x  4x  3  11  11x  8       17. 2x 2  5x  x 2  8x  3  2x 2  x 2  [5x  8x]  3  x 2  3x  3       18. 2x 2  3x  1  3x 2  5x  4  2x 2  3x 2  3x  5x  1  4  x 2  2x  3

19. 3 x  1  4 x  2  3x  3  4x  8  7x  5

20. 8 2x  5  7 x  9  16x  40  7x  63  9x  103       21. 5x 3  4x 2  3x  x 2  7x  2  5x 3  4x 2  x 2  3x  7x  2  5x 3  3x 2  10x  2     22. 4 x 2  3x  5  3 x 2  2x  1  4x 2  12x  20  3x 2  6x  3  x 2  6x  17 23. 2x x  1  2x 2  2x

25. x 2 x  3  x 3  3x 2 27. 2 2  5t  t t  10  4  10t  t 2  10t  t 2  4   29. r r 2  9  3r 2 2r  1  r 3  9r  6r 3  3r 2  7r 3  3r 2  9r

  31. x 2 2x 2  x  1  2x 4  x 3  x 2

33. x  3 x  5  x 2  5x  3x  15  x 2  2x  15

24. 3y 2y  5  6y 2  15y   26. y y 2  2  y 3  2y

28. 5 3t  4  2t t  3  2t 2  21t  20 30.  3   9  2 2 2  2   4  5 3  4 2

  32. 3x 3 x 4  4x 2  5  3x 7  12x 5  15x 3

34. 4  x 2  x  8  4x  2x  x 2  x 2  6x  8

35. s  6 2s  3  2s 2  3s  12s  18  2s 2  15s  18 36. 2t  3 t  1  2t 2  2t  3t  3  2t 2  t  3 37. 3t  2 7t  4  21t 2  12t  14t  8  21t 2  26t  8 38. 4s  1 2s  5  8s 2  18s  5 39. 3x  5 2x  1  6x 2  10x  3x  5  6x 2  7x  5 40. 7y  3 2y  1  14y 2  13y  3

20

CHAPTER P Prerequisites

41. x  3y 2x  y  2x 2  5x y  3y 2

42. 4x  5y 3x  y  12x 2  19x y  5y 2

43. 2r  5s 3r  2s  6r 2  19rs  10s 2

44. 6u  5 u  2  6u 2  7u  10 2

45. 5x  12  25x 2  10x  1

46. 2  7y2  49y 2  28y  4

47. 3y  12  3y2  2 3y 1  12  9y 2  6y  1

48. 2y  52  2y2  2 2y 5  52  4y 2  20y  25

49. 2u  2  4u 2  4u   2

50. x  3y2  x 2  6x y  9y 2

51. 2x  3y2  4x 2  12x y  9y 2  2 53. x 2  1  x 4  2x 2  1

52. r  2s2  r 2  4rs  4s 2 2  54. 2  y 3  y 6  4y 3  4

57. 3x  4 3x  4  3x2  42  9x 2  16

58. 2y  5 2y  5  4y 2  25

55. x  6 x  6  x 2  36

56. 5  y 5  y  25  y 2

60. 2u   2u    4u 2   2 59. x  3y x  3y  x 2  3y2  x 2  9y 2           61. x 2 x 2  x 4 62. y 2 y  2  y 2   63. y  23  y 3  3y 2 2  3y 22  23  y 3  6y 2  12y  8

64. x  33  x 3  3x 2 3  3x 32  33  x 3  9x 2  27x  27   65. 1  2r3  13  3 12 2r  3 1 2r2  2r3  8r 3  12r 2  6r  1   66. 3  2y3  33  3 32 2y  3 3 2y2  2y3  8y 3  36y 2  54y  27   67. x  2 x 2  2x  3  x 3  2x 2  3x  2x 2  4x  6  x 3  4x 2  7x  6   68. x  1 2x 2  x  1  2x 3  x 2  x  2x 2  x  1  2x 3  x 2  1   69. 2x  5 x 2  x  1  2x 3  2x 2  2x  5x 2  5x  5  2x 3  7x 2  7x  5   70. 1  2x x 2  3x  1  x 2  3x  1  2x 3  6x 2  2x  2x 3  5x 2  x  1  2       x x x x x x x xx   73. y 13 y 23  y 53  y 1323  y 1353  y 2  y 71.

75.



x 2  y2

2



  x  1 x  x 2  x    74. x 14 2x 34  x 14  2x  x

72. x 32

 2  2  x 2  y 2  2x 2 y 2  x 4  y 4  2x 2 y 2

  1 2 1 c  2  c2  2 c c    77. x 2  a 2 x 2  a 2  x 4  a 4

76.

    ab a  b  a  b2    81. 1  x 23 1  x 23  1  x 43

79.

   78. x 12  y 12 x 12  y 12  x  y     80. h2  1  1 h2  1  1  h2 82. 1  b2 1  b2  b4  2b2  1

    2 83. x  1  x 2 x  1  x 2  x  12  x 2  x 2  2x  1  x 4  x 4  x 2  2x  1       x  2  x 2  x 4  3x 2  4 84. x  2  x 2 85. 2x  y  3 2x  y  3  2x  y2  32  4x 2  4x y  y 2  9

SECTION P.5 Algebraic Expressions

21

86. x  y  z x  y  z  x 2  y 2  z 2  2yz       87. (a) RHS  12 a  b2  a 2  b2  12 a 2  b2  2ab  a 2  b2  12 2ab  ab  LHS    2  2  2  2  2  2 (b) LHS  a 2  b2  a 2  b2  a 2  b2  2a 2 b2  a 2  b2  2a 2 b2  4a 2 b2  RHS

   88. LHS  a 2  b2 c2  d 2  a 2 c2  a 2 d 2  b2 c2  b2 d 2      a 2 c2  b2 d 2  2abcd  a 2 d 2  b2 c2  2abcd  ac  bd2  ad  bc2  RHS

89. (a) The height of the box is x, its width is 6  2x, and its length is 10  2x. Since Volume  height  width  length, we have V  x 6  2x 10  2x.   (b) V  x 60  32x  4x 2  60x  32x 2  4x 3 , degree 3.     (c) When x  1, the volume is V  60 1  32 12  4 13  32, and when x  2, the volume is     V  60 2  32 22  4 23  24. 90. (a) The width is the width of the lot minus the setbacks of 10 feet each. Thus width  x  20 and length  y  20. Since Area  width  length, we get A  x  20 y  20. (b) A  x  20 y  20  x y  20x  20y  400

(c) For the 100  400 lot, the building envelope has A  100  20 400  20  80 380  30,400. For the 200  200, lot the building envelope has A  200  20 200  20  180 180  32,400. The 200  200 lot has a larger building envelope.   91. (a) A  2000 1  r3  2000 1  3r  3r 2  r 3  2000  6000r  6000r 2  2000r 3 , degree 3. (b) Remember that % means divide by 100, so 2%  002. Interest rate r

2%

3%

45%

6%

10%

Amount A $212242 $218545 $228233 $238203 $266200     92. (a) P  R  C  50x  005x 2  50  30x  01x 2  50x  005x 2  50  30x  01x 2  005x 2  20x  50.   (b) The profit on 10 calculators is P  005 102  20 10  50  $155. The profit on 20 calculators is   P  005 202  20 20  50  $370 . 93. (a) When x  1, x  52  1  52  36 and x 2  25  12  25  26. (b) x  52  x 2  10x  25

94. (a) The degree of the product is the sum of the degrees of the original polynomials. (b) The degree of the sum could be lower than either of the degrees of the original polynomials, but is at most the largest of the degrees of the original polynomials.    (c) Product: 2x 3  x  3 2x 3  x  7  4x 6  2x 4  14x 3  2x 4  x 2  7x  6x 3  3x  21  4x 6  4x 4  20x 3  x 2  10x  21     Sum: 2x 3  x  3  2x 3  x  7  4.

22

CHAPTER P Prerequisites

P.6

FACTORING

1. The polynomial 2x 5  6x 4  4x 3 has three terms: 2x 5 , 6x 4 , and 4x 3 .   2. The factor 2x 3 is common to each term, so 2x 5  6x 4  4x 3  2x 3 x 2  3x  2 . [In fact, the polynomial can be factored further as 2x 3 x  2 x  1.]

3. To factor the trinomial x 2  7x  10 we look for two integers whose product is 10 and whose sum is 7. These integers are 5 and 2, so the trinomial factors as x  5 x  2. 4. The greatest common factor in the expression 4 x  12  x x  12 is x  12 , and the expression factors as 4 x  12  x x  12  x  12 4  x.

5. The Special Factoring Formula for the “difference of squares” is A2  B 2  A  B A  B. So 4x 2  25  2x  5 2x  5.

6. The Special Factoring Formula for a “perfect square” is A2  2AB  B 2  A  B2 . So x 2  10x  25  x  52 . 7. 5a  20  5 a  4   9. 2x 3  x  x 2x 2  1

11. 2x 2 y  6x y 2  3x y  x y 2x  6y  3 13. y y  6  9 y  6  y  6 y  9

8. 3b  12  3 b  4  3 b  4   10. 3x 4  6x 3  x 2  x 2 3x 2  6x  1   12. 7x 4 y 2  14x y 3  21x y 4  7x y 2 x 3  2y  3y 2

14. z  22  5 z  2  z  2 [z  2  5]  z  2 z  3 15. x 2  8x  7  x  7 x  1

16. x 2  4x  5  x  5 x  1

17. x 2  2x  15  x  5 x  3

18. 2x 2  5x  7  x  1 2x  7

19. 3x 2  16x  5  3x  1 x  5

20. 5x 2  7x  6  5x  3 x  2

21. 3x  22  8 3x  2  12  [3x  2  2] [3x  2  6]  3x  4 3x  8

22. 2 a  b2  5 a  b  3  [a  b  3] [2 a  b  1]  a  b  3 2a  2b  1 23. x 2  25  x  5 x  5

24. 9  y 2  3  y 3  y

25. 49  4z 2  7  2z 7  2z

26. 9a 2  16  3a  4 3a  4

28. a 2  36b2  a  6b a  6b 27. 16y 2  z 2  4y  z 4y  z    29. x  32  y 2  x  3  y x  3  y  x  y  3 x  y  3    30. x 2  y  52  x  y  5 x  y  5  x  y  5 x  y  5 31. x 2  10x  25  x  52

32. 9  6y  y 2  3  y2

33. z 2  12z  36  z  62

34. 2  16  64    82

35. 4t 2  20t  25  2t  52

36. 16a 2  24a  9  4a  32

37. 9u 2  6u   2  3u  2   39. x 3  27  x  3 x 2  3x  9

38. x 2  10x y  25y 2  x  5y2   40. y 3  64  y  4 y 2  4y  16

SECTION P.6 Factoring

  41. 8a 3  1  2a  1 4a 2  2a  1   43. 27x 3  y 3  3x  y 9x 2  3x y  y 2

   3  45. u 3   6  u 3   2  u   2 u 2  u 2   4 47. 48. 49. 50. 51. 52. 53. 54.

23

  42. 8  273  2  3 4  6  92   44. 1  1000y 3  1  10y 1  10y  100y 2

   46. 8r 3  64t 6  2r  4t 2 4r 2  8r t 2  16t 4   x 3  4x 2  x  4  x 2 x  4  1 x  4  x  4 x 2  1   3x 3  x 2  6x  2  x 2 3x  1  2 3x  1  3x  1 x 2  2   5x 3  x 2  5x  1  x 2 5x  1  5x  1  x 2  1 5x  1   18x 3  9x 2  2x  1  9x 2 2x  1  2x  1  9x 2  1 2x  1   x 3  x 2  x  1  x 2 x  1  1 x  1  x  1 x 2  1   x 5  x 4  x  1  x 4 x  1  1 x  1  x  1 x 4  1    x 52  x 12  x 12 x 2  1  x x  1 x  1    1  12 12 32 12 2 3x 3  4x  x    4x x x 3  x 1  x x

55. Start by factoring out the power of x with the smallest exponent, that is, x 32 . So   1  x2 . x 32  2x 12  x 12  x 32 1  2x  x 2  x 32   56. x  172  x  132  x  132 x  12  1  x  132 [x  1  1] [x  1  1]  x  132 x  2 x

   12 57. Start by factoring out the power of x 2  1 with the smallest exponent, that is, x 2  1 . So 

12  12  12    x2  3 x2  1 x2  1  2    2 x2  1  x2  1 . x2  1

2x  1 58. x 12 x  112  x 12 x  112  x 12 x  112 [x  1  x]    x x 1 59. 2x 13 x  223  5x 43 x  213  x 13 x  213 [2 x  2  5x]  x 13 x  213 2x  4  5x  3x  4 3 x  x 13 x  213 3x  4   3 x 2    54 14 14     60. 3x 12 x 2  1 3 x 2  1  x 2 1  x 32 x 2  1  x 12 x 2  1    4 2    14  14   x  1 2x 2  3 3x 2  3  x 2  x 12 x 2  1 2x 2  3   x 12 x 2  1  x   61. 12x 3  18x  6x 2x 2  3 62. 30x 3  15x 4  15x 3 2  x

63. 6y 4  15y 3  3y 3 2y  5

64. 5ab  8abc  ab 5  8c

65. x 2  2x  8  x  4 x  2

66. x 2  14x  48  x  8 x  6

24

CHAPTER P Prerequisites

67. y 2  8y  15  y  3 y  5

68. z 2  6z  16  z  2 z  8

69. 2x 2  5x  3  2x  3 x  1   71. 9x 2  36x  45  9 x 2  4x  5  9 x  5 x  1

70. 2x 2  7x  4  2x  1 x  4

73. 6x 2  5x  6  3x  2 2x  3

74. 6  5t  6t 2  3  2t 2  3t

75. x 2  36  x  6 x  6

76. 4x 2  25  2x  5 2x  5

77. 49  4y 2  7  2y 7  2y

78. 4t 2  9s 2  2t  3s 2t  3s

79. t 2  6t  9  t  32

80. x 2  10x  25  x  52

81. 4x 2  4x y  y 2  2x  y2   83. t 3  1  t  1 t 2  t  1

82. r 2  6rs  9s 2  r  3s2

72. 8x 2  10x  3  4x  3 2x  1

  84. x 3  27  x 3  33  x  3 x 2  3x  9     85. 8x 3  125  2x3  53  2x  5 2x2  2x 5  52  2x  5 4x 2  10x  25     86. 125  27y 3  53  3y3  5  3y 52  5 3y  3y2  3y  5 9y 2  15y  25   87. x 3  2x 2  x  x x 2  2x  1  x x  12   88. 3x 3  27x  3x x 2  9  3x x  3 x  3

  89. x 4  2x 3  3x 2  x 2 x 2  2x  3  x 2 x  1 x  3

  90. 35  54  23  3 32  5  2  3 3  1   2

  91. x 4 y 3  x 2 y 5  x 2 y 3 x 2  y 2  x 2 y 3 x  y x  y

92. 18y 3 x 2  2x y 4  2x y 3 9x  y     3    2    x 2  x 2 2y  2y2  x 2  2y x 4  2x 2 y  4y 2 93. x 6  8y 3  x 2  2y3  x 2  2y

    3     2    3a  b2 9a 2  3ab2  b4 94. 27a 3  b6  3a3  b2  3a  b2 3a2  3a b2  b2 95.

y 3  3y 2  4y  12

     y 3  3y 2  4y  12  y 2 y  3  4 y  3  y  3 y 2  4

 y  3 y  2 y  2 (factor by grouping)   96. y 3  y 2  y  1  y 2 y  1  1 y  1  y 2  1 y  1

      97. 3x 3  x 2  12x  4  3x 3  12x  x 2  4  3x x 2  4  x 2  4  3x  1 x 2  4  3x  1 x  2 x  2 (factor by grouping)

  98. 9x 3  18x 2  x  2  9x 2 x  2  x  2  9x 2  1 x  2  3x  1 3x  1 x  2

99. a  b2  a  b2  [a  b  a  b] [a  b  a  b]  2b 2a  4ab

SECTION P.6 Factoring

100.

25

            1 1 1 1 2 1 2 1  1 1  1 1  1  1 x x x x x x      1 1 1 2 4 1 1 1   1 1 2  x x x x x x

       101. x 2 x 2  1  9 x 2  1  x 2  1 x 2  9  x  1 x  1 x  3 x  3        102. a 2  1 b2  4 a 2  1  a 2  1 b2  4  a  1 a  1 b  2 b  2

103. x  1 x  22  x  12 x  2  x  1 x  2 [x  2  x  1]  3 x  1 x  2

  104. x  13 x  2 x  12 x 2  x 3 x  1  x x  1 x  12  2 x  1 x  x 2  x x  1 [x  1  x]2  x x  1 12  x x  1

    105. y 4 y  23  y 5 y  24  y 4 y  23 1  y y  2  y 4 y  23 y 2  2y  1  y 4 y  23 y  12 106. n x  y  n  1 y  x  n x  y  n  1 x  y  x  y [n  n  1]  x  y 107. Start by factoring y 2  7y  10, and then substitute a 2  1 for y. This gives

 2            a 2  1  7 a 2  1  10  a 2  1  2 a 2  1  5  a 2  1 a 2  4  a  1 a  1 a  2 a  2

 2            108. a 2  2a  2 a 2  2a  3  a 2  2a  3 a 2  2a  1  a 2  2a  3 a 2  2a  1  a  1 a  3 a  12

109. 3x 2 4x  122  x 3 2 4x  12 4  x 2 4x  12 [3 4x  12  x 2 4]  4x 2 x  3 12x  36  8x  4x 2 x  3 20x  36  16x 2 x  3 5x  9    4  5  4   110. 5 x 2  4 2x x  24  x 2  4 4 x  23  2 x 2  4 x  23 5 x x  2  x 2  4 2

   4   4   2 x 2  4 x  23 5x 2  10x  2x 2  8  2 x 2  4 x  23 7x 2  10x  8

     111. 3 2x  12 2 x  312  2x  13 12 x  312  2x  12 x  312 6 x  3  2x  1 12      2x  12 x  312 6x  18  x  12  2x  12 x  312 7x  35 2 112. 13 x  623 2x  32  x  613 2 2x  3 2  13 x  623 2x  3 [2x  3  3 x  6 4]  13 x  623 2x  3 [2x  3  12x  72]  13 x  623 2x  3 14x  69 113.



1 2    13 43  43   43   1 x2  3  3 x  3 x 2  3  23 x 2  x 2  3  23 x 2 x 2  3  x2  3 x2  3  43 3 x2  3

114. 12 x 12 3x  412  32 x 12 3x  412  12 x 12 3x  412 [3x  4  3x]  12 x 12 3x  412 6x  4  x 12 3x  412 3x  2

26

CHAPTER P Prerequisites

115. The volume of the shell is the difference between the volumes of the outside cylinder (with radius R) and the inside cylinder   R r  h  R  r. The (with radius r). Thus V   R 2 h  r 2 h   R 2  r 2 h   R  r R  r  h  2  2 R r R r average radius is and 2  is the average circumference (length of the rectangular box), h is the height, and 2 2 R r  h  R  r  2  average radius  R  r is the thickness of the rectangular box. Thus V   R 2 h  r 2 h  2  2 height  thickness R h

length rl

h thickness

116. (a) Mowed portion  field  habitat

(b) Using the difference of squares, we get b2  b  2x2  [b  b  2x] [b  b  x]  2x 2b  2x  4x b  x.

117. (a) 5282  5272  528  527 528  527  1 1055  1055 (b) 1222  1202  122  120 122  120  2 242  484

(c) 10202  10102  1020  1010 1020  1010  10 2030  20,300

118. (a) 501  499  500  1 500  1  5002  1  250,000  1  249,999 (b) 79  61  70  9 70  9  702  92  4900  81  4819

(c) 2007  1993  2000  7 2000  7  20002  72  4,000,000  49  3,999,951

     119. (a) A4  B 4  A2  B 2 A2  B 2  A  B A  B A2  B 2    A6  B 6  A3  B 3 A3  B 3 (difference of squares)      A  B A2  AB  B 2 A  B A2  AB  B 2 (difference and sum of cubes) (b) 124  74  20,736  2,401  18,335; 126  76  2,985,984  117,649  2,868,335   (c) 18,335  124  74  12  7 12  7 122  72  5 19 144  49  5 19 193    2,868,335  126  76  12  7 12  7 122  12 7  72 122  12 7  72  5 19 144  84  49 144  84  49  5 19 277 109

120. (a) A  1 A  1  A2  A  A  1  A2  1   A  1 A2  A  1  A3  A2  A  A2  A  1  A3  1   A  1 A3  A2  A  1  A4  A3  A2  A  A3  A2  A  1   (b) We conjecture that A5  1  A  1 A4  A3  A2  A  1 . Expanding the right-hand side, we have   A  1 A4  A3  A2  A  1  A5  A4  A3  A2  A  A4  A3  A2  A  1  A5  1, verifying our   conjecture. Generally, An  1  A  1 An1  An2      A  1 for any positive integer n.

SECTION P.7 Rational Expressions

121. (a)

A2  A  1

A  1 

A  1



A  1

A2  A A2

A3  A2  A  1

A  1



A2  A  1

A3  A2  A

A3

 1

 1

27

A  1

A3  A2  A  1

A 4  A3  A2  A A4

 1  (b) Based on the pattern in part (a), we suspect that A5  1  A  1 A4  A3  A2  A  1 . Check: 

A 4  A 3  A2  A  1



A  1

A4  A3  A2  A  1

A 5  A 4  A 3  A2  A

A5  1   The general pattern is An  1  A  1 An1  An2      A2  A  1 , where n is a positive integer.

P.7

RATIONAL EXPRESSIONS

3x is a rational expression. x2  1  x 1 (b) is not a rational expression. A rational expression must be a polynomial divided by a polynomial, and the 2x  3  numerator of the expression is x  1, which is not a polynomial.

1. (a)

x3  x xx 2  1  is a rational expression. x 3 x 3 2. To simplify a rational expression we cancel factors that are common to the numerator and denominator. So, the expression x  1x  2 x 1 simplifies to . x  3x  2 x 3 3. To multiply two rational expressions we multiply their numerators together and multiply their denominators together. So 2 x 2x 2x  is the same as .  2 x 1 x 3 x  1  x  3 x  4x  3 1 2 x 4. (a)   has three terms. x x  1 x  12 (c)

(b) The least common denominator of all the terms is x x  12 . (c)

2 x 2x x  1 x x 1 x  12 x  12  2x x  1  x 2       x x  1 x  12 x  1 x x  12 x x  12 x  12 

x 2  2x  1  2x 2  2x  x 2

x x  12 x x  1 x . 5. (a) Yes. Cancelling x  1, we have  x 1 x  12



2x 2  1

x x  12

  (b) No; x  52  x 2  10x  25  x 2  25, so x  5  x 2  10x  25  x 2  25. 3a 3 a a 6. (a) Yes,   1 . 3 3 3 3 (b) No. We cannot “separate” the denominator in this way; only the numerator, as in part (a). (See also Exercise 101.)

7. The domain of 4x 2  10x  3 is all real numbers.

8. The domain of x 4  x 3  9x is all real numbers.

28

CHAPTER P Prerequisites

9. Since x  3  0 we have x  3. Domain: x  x  3 11. Since x  3  0, x  3. Domain; x  x  3

10. Since 3t  6  0 we have t  2. Domain: t  t  2 12. Since x  1  0, x  1. Domain; x  x  1

13. x 2  x  2  x  1 x  2  0  x  1 or 2, so the domain is x  x  1 2.

14. 2x  0 and x  1  0  x  0 and x  1, so the domain is x  x  0.   4 x2  1 2x  1 4 x  1 x  1 x 1 5 x  3 2x  1 5 x  3 2x  1  16.    15. 2 5 x  3  2 x  3 2 x  3 12 x  2 x  1 12 x  2 x  1 3 x  2 10 x  3

17. 19.

1 x 2 x 2   2 x 2 x  2 x  2 x 4

x 2 x 2  5x  6 x  2 x  3   2 x 5  5  3 x x x  8x  15

18.

x2  x  2 x 2 x  2 x  1   2 x 1 x  1 x  1 x 1

x 2  x  12 x 4 x  4 x  3 20. 2   x 2  2  3 x x x  5x  6

y2  y y 2  3y  18 y y  1 y y6 y  6 y  3 22.     y1 2y  1 y  1 y  1 2y  1 y  3 y2  1 2y 2  7y  3   x 2x 2  x  6 x 2x  3 x  2 x 2x  3 2x 3  x 2  6x    23. 2 2x  3  3  2  3  2 2x x 2x x 2x  7x  6

21.

1  x2  x  1 1  x  x  1 1  x 1  x      x3  1 x2  x  1 x  1 x 2  x  1 x  1 x 2  x  1 4x 1 x 2 x 2 4x     25. 2 4 x  2 x  2 x  2 16x x  4 16x 24.

26. 27.

x 2  25 x  4 x 5 x  5 x  5 x  4     2 x 4 x  4 x  4 x  5 x  16 x  5

x 2  2x  15 x  5 x 3 x  5 x  3 x  5    x 2 x 2 x  5 x  5 x  2 x 2  25

x 2  2x x 2  2x t 3 29. 2  t 9 28.

30. 31. 32. 33. 34.

3 3x  x  1 x  1 1x x  3 x  1  x  3       x 3 x 1 x 1 1x x  3 x  1 3 3x t 3 1 t  3 t  3  2  2 t  9 t  3 t  3 t2  9 t 9

x2  x  6 x 2 x  1 x3  x2 x  3 x  2   2  x 2 x x  2 x  3 x  1 x  2x x  2x  3

x 4 x 2  7x  12 x 2  5x  6 x  3 x  4 x  2 x  3     x 1 x  1 x  2 x  3 x  3 x 2  3x  2 x 2  6x  9

x 2  2x y  y 2 2x 2  x y  y 2 2x  y x  y x  y x  y 2x  y    2  x  2y x  y x  y x  2y x  y x 2  y2 x  x y  2y 2

x 3 x 2  7x  12 x  3 2x 2  7x  15 x 3 x 5 x  5 2x  3  2  2     2 2x  3 2x  3 x  3 x  4 2x  3 x  4 4x  9 2x  7x  15 4x  9 x 2  7x  12 2x  1

2x 2  x  15 x3



2x  1 6x 2  x  2 x 3 1    x 3 x  3 2x  5 2x  1 3x  2 2x  5 3x  2

x3 x 2  2x  1 x 3 x  1 x  1 x 1     x 2 x  1 x x 1 x x  1 x x 2  2x  1 2x 2  3x  2 x 2 x2  x  2 x  2 2x  1 x  1 x  2 2x 2  3x  2 x2  1      36. x 1 x  1 x  1 x  2 2x  1 x2  1 2x 2  5x  2 2x 2  5x  2 x2  x  2 35.

SECTION P.7 Rational Expressions

29

x 1 x xy    z y z yz x y x z xz 38. x    yz z 1 y y x 3 1 x 4 1    39. 1  x 3 x 3 x 3 x 3 3x  2 2 x  1 3x  2  2x  2 x 4 3x  2 2    40. x 1 x 1 x 1 x 1 x 1 2 x 3 1 2 x  5 x  3  2x  10 3x  7   41.    x 5 x 3 x  5 x  3 x  5 x  3 x  5 x  3 x  5 x  3 1 x 1 1 x 1 x 1x 1 2x   42.    x 1 x 1 x  1 x  1 x  1 x  1 x  1 x  1 x  1 x  1 3 1 3 x  2 x 1 3x  6  x  1 2x  5 43.      x 1 x 2 x  1 x  2 x  1 x  2 x  1 x  2 x  1 x  2 37.

3 x x  6 3 x  4 x 2  6x  3x  12 x 2  3x  12 x      x 4 x 6 x  4 x  6 x  4 x  6 x  4 x  6 x  4 x  6 5 3 5 2x  3 3 10x  15  3 10x  18 2 5x  9 45.       2x  3 2x  32 2x  32 2x  32 2x  32 2x  32 2x  32 x x 2 2 x  1 3x  2 x  2x  2  46.     x 1 x  1 x  1 x  12 x  12 x  12 x  12 44.

47. u  1 

u u 2  2u  1  u u 2  3u  1 u u  1 u  1     u1 u1 u1 u1 u1

4 3 2b2 3ab 4a 2 2b2  3ab  4a 2 2  2  2 2  2 2  2 2   2 ab b a a b a b a b a 2 b2 x 1 1 1 1 1 x 2x  1  2  2  49. 2  2  2  2 x x  1 x x x x x x  1 x x  1 x x  1 48.

1 1 x2 1  2  3  3  x x x x 1 2  51. x  3 x 2  7x  12

x 1 x2  x  1  3  3 x x x3 1 2 2 x  4 1     x  3 x  3 x  4 x  3 x  4 x  3 x  4 2x  8  1 2x  7   x  3 x  4 x  3 x  4 x x 1 x 1 x 2    52. 2   x  2 x  2 x  2 x  2 x  2 x  2 x  2 x 4 x 2 2x  2 2 x  1   x  2 x  2 x  2 x  2 1 1 1 1 x 3 1 x 2 53.       x  3 x2  9 x  3 x  3 x  3 x  3 x  3 x  3 x  3 x  3 x  3 x 2 x 2 54. 2    x  1 x  2 x  1 x  4 x  x  2 x 2  5x  4

50.

2 x  2 x 2  4x  2x  4 x 2  6x  4 x x  4    x  1 x  2 x  4 x  1 x  2 x  4 x  1 x  2 x  4 x  1 x  2 x  4 3 4 3 4 2 x  1 3x 4 2x  2  3x  4 5x  6 2 2        55.    x x  1 x2  x x x  1 x x  1 x x  1 x x  1 x x  1 x x  1 x x  1 x 2 x 2 1 1 56. 2      x  2 x  3 x  2 x 3  3  2 x x x x 6 1 x  3 2 x  2 x  x  3  2x  4 2x  1 x      x  3 x  2 x  3 x  2 x  3 x  2 x  3 x  2 x  3 x  2 

30

CHAPTER P Prerequisites

1 1 1  2   x  2 x  1 x  3 x  1 x  2x  3 x 3  x  2 x 3x 2 5     x  3 x  2 x  1 x  3 x  2 x  1 x  3 x  2 x  1 x  3 x  2 x  1 2 2 1 3 3 1   58.  2   x  1 x  12 x  1 x  12 x  1 x  1 x 1 2 x  1 3 x  1 x  1 x  1    2 2 x  1 x  1 x  1 x  1 x  1 x  12

57.

1

x 2  3x  2

x2  1

2x  2 3x  3 x 2  1  2x  2  3x  3 x2  x  4     x  1 x  12 x  1 x  12 x  1 x  12 x  1 x  12 x  1 x  12   x 1  1x 1 1 x 1   59. 1 x   1 2 1  2x  2 x x x   y 1  2y 1  2y y 2     60. 3 3 3y 1 y 1 

y

61.

62.

63.

64.

65.

66.

y

  1 1  2 1  x 1 x 3 x  2  1 x 2 x 2     1 1 x 1 x  2  1 1 x  2 1  x 2 x 2 1 1 c1  c11  c 1 c11 c2 1 c1   1 1 1 1  1  3  x x  2 x  1 x  3  x  1 x 1 x 3 x 1 x 3    x 1 x  1 x  3 x  1 x  1 x  1 x  3 x  1 x  1 x  3 2  x  1 x  3   x 3 x 2  x 2  2x  3  x 2  2x  8 5 x  4 x  1  x  3 x  1  x  2 x  4   x 3 x  4 x  3 x  1 x  4 x  3 x  1 x  4 x  3 x  1   x x xy x  x x 2 y  1 x2 y  x2 y y  y  y   x y 2  y 2  y 2 x  1 y xy y  x x    y y 2 xy x  x 2 2 y y  x x x  x yy     x  x x y2  x 2 x x  y2 y x y y  y y

y x x 2  y2  x 2 y2 x 2  y2 xy y x xy 67.  2  x y. An alternative method is to multiply the  2   2 1 1 xy 1 y  x2 y  x  x2 y2 x 2 y2

numerator and denominator by the common denominator of both the numerator and denominator, in this case x 2 y 2 :     y x y x   x y x 2  y2 x 2 y2 x 3 y  x y3 y x y x    2   x y. 1 1 1 1 x 2 y2 y  x2 y2  x 2   x2 y2 x2 y2

SECTION P.7 Rational Expressions





x x 2  y2 x y2 y xy y x y2 x 3  x y2  x y2 x3  x   x   68. x  x   2   2 y y xy x 2 2 2 2 2 2 2 x y x y x y x y x  y2   y x y x 1 x2 1 y2  2  2 2 2 2 2 y2  x 2 xy yx y  x y  x x y x y x y x y       x y 1 2 y2 2 y 2 y  x 1 1 y  x xy y x x   xy xy x y   1 1  yx y2  x 2 x 2  y 2 x 2 y2 y  x y  x x2 y2     .    Alternatively, 1 1 1 x y y  x xy x  y 1 x 2 y2 x y2  x 2 y  x y

x 2  69. 1 x 

y 2

1 1 1 1   y x  y  x x  y x y x y x y x  y 70.     1 1 1 x y  y xy x x  y xy xy x 1  y 1

 71. 1 

72. 1 

1 1 1 x

x y  y2  x 2  x y x 2  2x y  y 2 x  y2   xy xy xy

1

1 1 1 1x

x 1x 1 x   x 1 x 1 1x

1

x 2x 1 2x  3 x 1 1x   1 x 2 x 2 x 2 1  x  1

1 1  1 1  x  h 1  x  1  x  1  x  h   73. h h 1  x 1  x  h 1  x 1  x  h 74. In calculus it is necessary to eliminate the h in the denominator, and we do this by rationalizing the numerator: 1 1       x  x h x  x h x  x  h 1 x x h              .    h h x x h x  x h h x x h x  x h x x h x  x h

  1 1 2  x 2  2xh  h 2  2 x 2 2 2 x  x  h 2x  h x  h x    75. h hx 2 x  h2 hx 2 x  h2 x 2 x  h2   x  h3  7 x  h  x 3  7x x 3  3x 2 h  3xh 2  h 3  7x  7h  x 3  7x 3x 2 h  3xh 2  h 3  7h   76. h h h  

h 3x 2  3xh  h 2  7 h

 3x 2  3xh  h 2  7

  2      x2 x2 1 1  x2 1 x  77. 1    1     2 2 2 2 1x 1x 1x 1x 1  x2 1  x2       1 2 2x 3 1 1 1 1 3 6 6 78. 1  x  3  1x  3   1x  2   x 6  12  6 6 6 4x 4x 16x 16x 16x   2   1 1  x3  3   x 3  3  4x 4x

31

32

79.

CHAPTER P Prerequisites

3 x  22 x  32  x  23 2 x  3 x  34

 

x  22 x  3 [3 x  3  x  2 2] x  34 x  22 3x  9  2x  4



x  22 x  13

x  33 x  33   x  63 2x x  6  4x 2 2x x  64  x 2 4 x  63 2x 2  12x  4x 2 12x  2x 2 2x 6  x 80.     8 8 5 x  6 x  6 x  6 x  65 x  65 x 2 2 1  x12  x 1  x12 1  x12 [2 1  x  x]   1x 1x 1  x32   12  12 12   1  x2 1  x2 1  x2  x2  x2 1  x2 1 82.   32 2 2 1x 1x 1  x2 81.

83.

84. 85.

86.

87. 88.

89. 90. 91.

92.

3 1  x13  x 1  x23

1  x23 [3 1  x  x]

2x  3  1  x23 1  x43   7  3x12 7  3x  32 x 7  32 x 7  3x12  32 x 7  3x12   7  3x 7  3x 7  3x32    5 3 1 1 5 3 5 3        25  3 22 5 3 5 3 5 3     2 5 3  3 63 5        4  5  6  3 5 2 5 2 5 2 5            2 2 2 2  7 2  7 7  2 2 2 2 7          27 5 5 2 7 2 7 2 7   1 1 x 1 x 1     x 1 x 1 x 1 x 1        3 y y y 3y y 3 y y y         3 y 3y 3 y 3 y 3 y       2 x  y x  y x y  2 x  y 2 x  y    2 x  y 2 x 2 y        xy x y x y x y    1 5 1 5 1 5 15 4            3 3 1 5 3 1 5 3 1 5       3 5 3 5 3 5 35 2 1              2 2 3 5 3 5 2 3 5 2 3 5 1  x23



      r 2 r 2 r 2 r 2    93.     5 5 r 2 5 r 2       x  x  h x  x h x  x h x  x h            94.   h x x h h x x h x  x h h x x h x  x h h 1            h x x h x  x h x x h x  x h    x2  1  x x2  1  x x2  1  x2 1 95. x 2  1  x     2 2 2 1 x 1x x 1x x 1x

SECTION P.7 Rational Expressions

96.



x 1

97. (a) R 



x



x 1 1



33

  x x 1 x x 1x 1       x 1 x x 1 x x 1 x

1 R R R1 R2 1   1 2  1 1 1 1 R1 R2 R2  R1   R1 R2 R1 R2

(b) Substituting R1  10 ohms and R2  20 ohms gives R  98. (a) The average cost A 

200 10 20  67 ohms.  30 20  10

500  6x  001x 2 Cost  . number of shirts x

(b) x

10

20

50

100

200

500

1000

Average cost

$5610

$3120

$1650

$1200

$1050

$1200

$1650

99. x x2  9 x 3

280

290

295

299

2999

3

3001

301

305

310

320

580

590

595

599

5999

?

6001

601

605

610

620

From the table, we see that the expression

x2  9 approaches 6 as x approaches 3. We simplify the expression: x 3

x2  9 x  3 x  3   x  3, x  3. Clearly as x approaches 3, x  3 approaches 6. This explains the result in the x 3 x 3 table. 2 2 100. No, squaring  changes its value by a factor of  . x x 101. Answers will vary. Algebraic Error 1 1 1   a b ab

a  b2  a 2  b2  a 2  b2  a  b ab  b a 1 a  ab b am mn  a an 102. (a)

Counterexample 1 1 1   2 2 22

1  32  12  32  52  122  5  12 26  6 2 1  1 11 35  352 32

5 a a 5a    1  , so the statement is true. 5 5 5 5

(b) This statement is false. For example, take x  5 and y  2. Then LHS  RHS 

x 5 5  , and 2  . y 2 2

(c) This statement is false. For example, take x  0 and y  1. Then LHS  RHS 

1 1 1 1   , and 0  . 1y 11 2 2

x 1 51 6    2, while y1 21 3 0 x   0, while xy 01

34

CHAPTER P Prerequisites

(d) This statement is false. For example, take x  1 and y  1. Then LHS  2

a 

2

  1  2, while 1

b 2a 2 RHS    1, and 2  1. 2b 2     a  1 1 a a  a  1 a  1  . (e) This statement is true: b b b b b 2 2 2 1 (f) This statement is false. For example, take x  2. Then LHS     , while 4x 42 6 3 1 2 1 2 3 1 3 RHS      , and  . 2 x 2 2 2 3 2 103. (a) 1 9 99 999 9999 x 1 3 2 10 100 1000 10,000 1 x 2 3333 25 2011 20001 2000001 200000001 x 1 It appears that the smallest possible value of x  is 2. x   1 1  2x  (b) Because x  0, we can multiply both sides by x and preserve the inequality: x   2  x x  x x

x 2  1  2x  x 2  2x  1  0  x  12  0. The last statement is true for all x  0, and because each step is 1 reversible, we have shown that x   2 for all x  0. x

P.8

SOLVING BASIC EQUATIONS

1. Substituting x  3 in the equation 4x  2  10 makes the equation true, so the number 3 is a solution of the equation. 2. Subtracting 4 from both sides of the given equation, 3x  4  10, we obtain 3x  4  4  10  4  3x  6. Multiplying

by 13 , we have 13 3x  13 6  x  2, so the solution is x  2. x  2x  10 is equivalent to 52 x  10  0, so it is a linear equation. 3. (a) 2 2 2 (b)  2x  1 is not linear because it contains the term , a multiple of the reciprocal of the variable. x x (c) x  7  5  3x  4x  2  0, so it is linear. 4. (a) x x  1  6  x 2  x  6 is not linear because it contains the square of the variable.  (b) x  2  x is not linear because it contains the square root of x  2.

(c) 3x 2  2x  1  0 is not linear because it contains a multiple of the square of the variable.

5. (a) This is true: If a  b, then a  x  b  x.

(b) This is false, because the number could be zero. However, it is true that multiplying each side of an equation by a nonzero number always gives an equivalent equation.

(c) This is false. For example, 5  5 is false, but 52  52 is true.

6. To solve the equation x 3  125 we take the cube root of each side. So the solution is x 

 3 125  5.

7. (a) When x  2, LHS  4 2  7  8  7  1 and RHS  9 2  3  18  3  21. Since LHS  RHS, x  2 is not a solution.

(b) When x  2, LHS  4 2  7  8  7  15 and RHS  9 2  3  18  3  15. Since LHS  RHS, x  2 is a solution.

8. (a) When x  1, LHS  2  5 1  2  5  7 and RHS  8  1  7. Since LHS  RHS, x  1 is a solution. (b) When x  1, LHS  2  5 1  2  5  3 and RHS  8  1  9. Since LHS  RHS, x  1 is not a solution.

SECTION P.8 Solving Basic Equations

35

9. (a) When x  2, LHS  1  [2  3  2]  1  [2  1]  1  1  0 and RHS  4 2  6  2  8  8  0. Since LHS  RHS, x  2 is a solution. (b) When x  4 LHS  1  [2  3  4]  1  [2  1]  1  3  2 and RHS  4 4  6  4  16  10  6. Since LHS  RHS, x  4 is not a solution.

1 1  12   12  12  1 and RHS  1. Since LHS  RHS, x  2 is a solution. 24 2 1 is not defined, so x  4 is not a solution. (b) When x  4 the expression 44

10. (a) When x  2, LHS  12 

11. (a) When x  1, LHS  2 113  3  2 1  3  2  3  5. Since LHS  1, x  1 is not a solution. (b) When x  8 LHS  2 813  3  2 2  3  4  3  1  RHS. So x  8 is a solution.

432 23 8    4 and RHS  4  8  4. Since LHS  RHS, x  4 is a solution. 46 2 2  32 23 832 292 When x  8, LHS     272 and RHS  8  8  0. Since LHS  RHS, x  8 is not a 86 2 2 solution. 0a a a When x  0, LHS     RHS. So x  0 is a solution. 0b b b ba ba When x  b, LHS   is not defined, so x  b is not a solution. bb 0  2   b b2 b2 b b2 b b When x  , LHS   14 b2     0  RHS. So x  is a solution. b 2 2 2 4 2 4 2  2   1 1 1 1 b2 1  14 b2  2  1  , so x  is not a solution. b When x  , LHS  b b b 4 b b

12. (a) When x  4, LHS  (b)

13. (a) (b) 14. (a) (b)

15. 5x  6  14  5x  20  x  4

16. 3x  4  7  3x  3  x  1

17. 7  2x  15  2x  8  x  4

18. 4x  95  1  4x  96  x  24

19. 12 x  7  3  12 x  4  x  8

20. 2  13 x  4  13 x  6  x  18

21. 3x  3  5x  3  0  8x  x  0

22. 2x  3  5  2x  4x  2  x  12

23. 7x  1  4  2x  9x  3  x  13

24. 1  x  x  4  3  2x  x   32

25. x  3  4x  3  5x  x  35

26. 2x  3  7  3x  5x  4  x  45

27. x3  1  53 x  7  x  3  5x  21  4x  24  x  6

3 x  3  4x  10  3x  30  x  40 28. 25 x  1  10

29. 2 1  x  3 1  2x  5  2  2x  3  6x  5  2  2x  8  6x  6  8x  x   34

30. 5 x  3  9  2 x  2  1  5x  15  9  2x  4  1  5x  24  2x  3  7x  21  x  3   31. 4 y  12  y  6 5  y  4y  2  y  30  6y  3y  2  30  6y  9y  32  y  32 9

32. r  2 [1  3 2r  4]  61  r  2 1  6r  12  61  r  2 6r  11  61  r  12r  22  61  13r  39  r 3 33. x  13 x  12 x  5  0  6x  2x  3x  30  0 (multiply both sides by 6)  x  30 y1 34. 23 y  12 y  3   8y  6 y  3  3 y  1  8y  6y  18  3y  3  14y  18  3y  3  11y  21 4  y  21 11

36

CHAPTER P Prerequisites

x 1 x 1   6x  8x  2x  x  1  24x  7x  1  24x  1  17x  x  17 2 4 x 1 1 5x    18x  15x  2 x  1  1  3x  2x  1  x  1 36. 3x  2 3 6 35. 2x 

37. x  1 x  2  x  2 x  3  x 2  x  2  x 2  5x  6  x  2  5x  6  6x  8  x  43 38. x x  1  x  32  x 2  x  x 2  6x  9  x  6x  9  5x  9  x   95

39. x  1 4x  5  2x  32  4x 2  x  5  4x 2  12x  9  x  5  12x  9  13x  14  x  14 13 40. t  42  t  42  32  t 2  8t  16  t 2  8t  16  32  16t  32  t  2 4 1   1  3  4  3x (multiply both sides by the LCD, 3x)  1  3x  x   13 41. x 3x 6 2 42.  5   4  2  5x  6  4x  4  9x   49  x x x 2x  1 4 43.   5 2x  1  4 x  2  10x  5  4x  8  6x  13  x  13 6 x 2 5 2x  7  23  2x  7 3  2 2x  4 (cross multiply)  6x  21  4x  8  2x  29  x  29 44. 2 2x  4 3 2   2 t  1  3 t  6 [multiply both sides by the LCD, t  1 t  6]  2t  2  3t  18  20  t 45. t 6 t 1 6 5 46.   6 x  4  5 x  3  6x  24  5x  15  x  39 x 3 x 4 3 1 47. 1   3 6  3x  3  2 [multiply both sides by 6 x  1]  18  3x  3  2  3x  15  2  x 1 2 3x  3 3x  13  x  13 3 48.

49.

50.

51.

52. 53.

54.

5 12x  5  2   12x  5 x  2x 6x  3  5 6x  3  12x 2  5x  12x 2  6x  30x  15  6x  3 x 12x 2  5x  12x 2  24x  15  19x  15  x   15 19

1 1 10 1     10 z  1  5 z  1  2 z  1  10 10z [multiply both sides by 10z z  1]  z 2z 5z z1 3 z 3 z  1  100z  3z  3  100z  3  97z  97

4 15 1    0  3  t  4 3  t  15  0  3  t  12  4t  15  0  3t  30  0  3t  30 3  t 3  t 9  t2  t  10 x 1 2   x  2 2x  4  2 [multiply both sides by 2 x  2]  x  4x  8  2  3x  6  x  2. 2x  4 x 2 But substituting x  2 into the original equation does not work, since we cannot divide by 0. Thus there is no solution. 1 5 2   x  3  5  2 x  3  x  2  2x  6  x  4  x  3 x2  9 x 3 1 6x  12 3  3 x  x  4  6x  12 (multiply both sides by x x  4]  3x  7x  16  4x  16   2 x 4 x x  4x  x  4. But substituting x  4 into the original equation does not work, since we cannot divide by 0. Thus, there is no solution. 1 2 1   2  2x  1  2 x  1  1  1. This is an identity for x  0 and x   12 , so the solutions are x 2x  1 2x  x all real numbers except 0 and  12 .

55. x 2  25  x  5

56. 3x 2  48  x 2  16  x  4  57. 5x 2  15  x 2  3  x   3

SECTION P.8 Solving Basic Equations

37

  58. x 2  1000  x   1000  10 10

  59. 8x 2  64  0  x 2  8  0  x 2  8  x   8  2 2   60. 5x 2  125  0  5 x 2  25  0  x 2  25  x  5 61. x 2  16  0  x 2  16 which has no real solution.

62. 6x 2  100  0  6x 2  100  x 2   50 3 , which has no real solution.   2 63. x  3  5  x  3   5  x  3  5    4 7 64. 3x  42  7  3x  4   7  3x  4  7  x  3 65. x 3  27  x  2713  3

66. x 5  32  0  x 5  32  x  3215  2      67. 0  x 4  16  x 2  4 x 2  4  x 2  4 x  2 x  2 x 2  4  0 has no real solution. If x  2  0, then x  2. If x  2  0, then x  2. The solutions are 2.   16 27 2716 3 27 6 6 68. 64x  27  x  x    16   64 64 2 64

69. x 4  64  0  x 4  64 which has no real solution.

70. x  13  8  0  x  13  8  x  1  813  2  x  1. 14   8114  x  2  3. So x  2  3, then x  1. If 71. x  24  81  0  x  24  81  x  24 x  2  3, then x  5. The solutions are 5 and 1.

72. x  14  16  0  x  14  16, which has no real solution.

73. 3 x  33  375  x  33  125  x  3  12513  5  x  3  5  8   74. 4 x  25  1  x  25  14  x  2  5 14  x  2  5 14  75. 3 x  5  x  53  125 3  3   14 76. x 43  16  0  x 43  16  24  x 43  24  212  x 4  212  x   212  23  8 15   23  8 77. 2x 53  64  0  2x 53  64  x 53  32  x  3235  25 32  32  78. 6x 23  216  0  6x 23  216  x 23  36  62  x 23  62  x  63  216 944  313 302 161 80. 836  095x  997  095x  161  x   169 095 582  506 81. 215x  463  x  119  115x  582  x  119 195  059 82. 395  x  232x  200  195  332x  x  332

79. 302x  148  1092  302x  944  x 

4497  4366 103 84. 214 x  406  227  011x  214x  86684  227  011x  225x  109584  x  48704  487 026x  194 85.  176  026x  194  176 303  244x  026x  194  533  429x  455x  727  303  244x 727  160 x 455 173x 320 86.  151  173x  151 212  x  173x  320  151x  022x  320  x   1455 212  x 022 83. 316 x  463  419 x  724  316x  1463  419x  3034  4497  103x  x 

38

CHAPTER P Prerequisites

87. r 

89. P V  n RT  R 

88. d  r T H  T 

PV nT

91. P  2l  2  2  P  2l    92.

93. 94. 95. 96. 97. 98.

d rH mM Fr 2 90. F  G 2  m  GM r

12 12 M M r

P  2l 2

1 1 1    R1 R2  R R2  R R1 (multiply both sides by the LCD, R R1 R2 ). Thus R1 R2  R R1  R R2  R R1 R2 R R2 . R1 R2  R  R R2  R1  R2  R  3V 3V r  V  13 r 2 h  r 2  h h  mM mM mM 2 r  G F  G 2 r  G F F r  3V 3V 3 r  V  43 r 3  r 3  4 4  2 2 2 2 2 2 a  b  c  b  c  a  b   c2  a 2        i 2 i i 2 A i A A A  1    1   i  100  100 A  P 1  1 100 P 100 100 P 100 P P     a 2 x  a  1  a  1 x  a 2 x  a  1 x   a  1  a 2  a  1 x  a  1  a 2  a  1 x  a  1

a  1 x  2 a a1 2d  b ax  b  2  ax  b  2 cx  d  ax  b  2cx  2d  ax  2cx  2d  b  a  2c x  2d  b  x  99. cx  d a  2c a1 a1 b1 100.    a a  1  a a  1  b b  1  a 2  a  a 2  a  b2  b  2a  b2  b  b  b  a a  12 b2  b

8  25 0032 250  25   000055. So the beam shrinks 10,000 10,000 000055  12025  0007 m, so when it dries it will be 12025  0007  12018 m long. 0032  25 (b) Substituting S  000050 we get 000050   5  0032  25  75  0032  10,000 75   234375. So the water content should be 234375 kg/m3 . 0032 3150 102. Substituting C  3600 we get 3600  450  375x  3150  375x  x   840. So the toy manufacturer can 375 manufacture 840 toy trucks. 101. (a) The shrinkage factor when   250 is S 

103. (a) Solving for  when P  10,000 we get 10,000  156 3   3  64102    86 km/h.

(b) Solving for  when P  50,000 we get 50,000  156 3   3  320513    147 km/h.

104. Substituting F  300 we get 300  03x 34  1000  103  x 34  x 14  10  x  104  10,000 lb. 105. (a) 3 0  k  5  k 0  k  1  k  5  k  1  2k  6  k  3

(b) 3 1  k  5  k 1  k  1  3  k  5  k  k  1  k  2  1  k  3

(c) 3 2  k  5  k 2  k  1  6  k  5  2k  k  1  k  1  k  1. x  2 is a solution for every value of k. That is, x  2 is a solution to every member of this family of equations.

106. When we multiplied by x, we introduced x  0 as a solution. When we divided by x  1, we are really dividing by 0, since x  1  x  1  0.

SECTION P.9 Modeling with Equations

P.9

39

MODELING WITH EQUATIONS

1. An equation modeling a real-world situation can be used to help us understand a real-world problem using mathematical methods. We translate real-world ideas into the language of algebra to construct our model, and translate our mathematical results back into real-world ideas in order to interpret our findings. 2. In the formula I  Pr t for simple interest, P stands for principal, r for interest rate, and t for time (in years). 3. (a) A square of side x has area A  x 2 .

(b) A rectangle of length l and width  has area A  l. (c) A circle of radius r has area A  r 2 .

5  16 ounces 4. Balsamic vinegar contains 5% acetic acid, so a 32 ounce bottle of balsamic vinegar contains 32  5%  32  100

of acetic acid.

1 wall 1  . x hours x d rt d d rt d 6. Solving d  rt for r, we find   r  . Solving d  rt for t, we find  t  . t t t r r r 7. If n is the first integer, then n  1 is the middle integer, and n  2 is the third integer. So the sum of the three consecutive integers is n  n  1  n  2  3n  3. 5. A painter paints a wall in x hours, so the fraction of the wall she paints in one hour is

8. If n is the middle integer, then n  1 is the first integer, and n  1 is the third integer. So the sum of the three consecutive integers is n  1  n  n  1  3n. 9. If n is the first even integer, then n  2 is the second even integer and n  4 is the third. So the sum of three consecutive even integers is n  n  2  n  4  3n  6.

10. If n is the first integer, then the next integer is n  1. The sum of their squares is   n 2  n  12  n 2  n 2  2n  1  2n 2  2n  1.

11. If s is the third test score, then since the other test scores are 78 and 82, the average of the three test scores is 160  s 78  82  s  . 3 3 12. If q is the fourth quiz score, then since the other quiz scores are 8, 8, and 8, the average of the four quiz scores is 24  q 888q  . 4 4 13. If x dollars are invested at 2 12 % simple interest, then the first year you will receive 0025x dollars in interest. 14. If n is the number of months the apartment is rented, and each month the rent is $795, then the total rent paid is 795n. 15. Since  is the width of the rectangle, the length is four times the width, or 4. Then area  length  width  4    42 ft2

16. Since  is the width of the rectangle, the length is   4. Then perimeter  2  length  2  width  2   4  2   4  8 ft distance d 17. If d is the given distance, in miles, and distance  rate  time, we have time   . rate 55 1h 18. Since distance  rate time we have distance  s  45 min  34 s mi. 60 min 19. If x is the quantity of pure water added, the mixture will contain 25 oz of salt and 3  x gallons of water. Thus the 25 . concentration is 3x 20. If p is the number of pennies in the purse, then the number of nickels is 2p, the number of dimes is 4  2 p, and the number of quarters is 2 p  4  2 p  4 p  4. Thus the value (in cents) of the change in the purse is 1  p  5  2 p  10  4  2 p  25  4 p  4  p  10 p  40  20 p  100 p  100  131 p  140.

40

CHAPTER P Prerequisites

21. If d is the number of days and m the number of miles, then the cost of a rental is C  65d  020m. In this case, d  3 80  400. Thus, and C  275, so we solve for m: 275  65  3  020m  275  195  02m  02m  80  m  02 Michael drove 400 miles. 22. If m is the number of messages, then a monthly cell phone bill (above $10) is B  10  010 m  1000. In this case, 285  285  B  385 and we solve for m: 385  10  010 m  1000  010 m  1000  285  m  1000  01 m  1285. Thus, Miriam sent 1285 text messages in June. 23. If x is Linh’s score on her final exam, then because the final counts twice as much as each midterm, her average score 228  2x 114  x 114  x 82  75  71  2x   . For her to average 80%, we must have  80%  08  is 3 100  200 500 250 250 114  x  250 08  200  x  86. So Linh scored 86% on her final exam. 24. Six students scored 100 and three students scored 60. Let x be the average score of the remaining 25  6  3  16 students. 6 100  3 60  16x  084  780  16x  084 2500  2100 Because the overall average is 84%  084, we have 25 100  16x  1320  x  1320 16  825. Thus, the remaining 16 students’ average score was 825%. 25. Let m be the amount invested at 4 12 %. Then 12,000  m is the amount invested at 4%.

Since the total interest is equal to the interest earned at 4 12 % plus the interest earned at 4%, we have

525  0045m  004 12,000  m  525  0045m  480  004m  45  0005m  m 

45  9000. Thus 0005

$9000 is invested at 4 12 %, and $12,000  9000  $3000 is invested at 4%. 26. Let m be the amount invested at 5 12 %. Then 4000  m is the total amount invested. Thus

4 12 % of the total investment  interest earned at 4%  interest earned at 5 12 %

So 0045 4000  m  004 4000  0055m  180  0045m  160  0055m  20  001m  m 

20  2000. 001

Thus $2,000 needs to be invested at 5 12 %. 2625  0075 or 75%. 27. Using the formula I  Prt and solving for r, we get 26250  3500  r  1  r  3500   28. If $1000 is invested at an interest rate a%, then 2000 is invested at a  12 %, so, remembering that a is expressed as a 1

a 2 a  1  2000   1  10a  20a  10  30a  10. Since the total interest 100 100 is $190, we have 190  30a  10  180  30a  a  6. Thus, the $1000 is invested at 6% interest.

percentage, the total interest is I  1000 

29. Let x be her monthly salary. Since her annual salary  12  monthly salary  Christmas bonus we have 97,300  12x  8,500  88,800  12x  x  7,400. Her monthly salary is $7,400. 30. Let s be the husband’s annual salary. Then her annual salary is 115s. Since husband’s annual salarywife’s annual salary  total annual income, we have s  115s  69,875  215s  69,875  s  32,500. Thus the husband’s annual salary is $32,500. 31. Let x be the overtime hours Helen works. Since gross pay  regular salary  overtime pay, we obtain the equation 90  8. Thus Helen 35250  750  35  750  15  x  35250  26250  1125x  90  1125x  x  1125 worked 8 hours of overtime. 32. Let x be the hours the assistant worked. Then 2x is the hours the plumber worked. Since the labor charge is equal to the plumber’s labor plus the assistant’s labor, we have 4025  45 2x  25x  4025  90x  25x  4025  115x  x  4025 115  35. Thus the assistant works for 35 hours, and the plumber works for 2  35  70 hours.

SECTION P.9 Modeling with Equations

41

33. All ages are in terms of the daughter’s age 7 years ago. Let y be age of the daughter 7 years ago. Then 11y is the age of the movie star 7 years ago. Today, the daughter is y  7, and the movie star is 11y  7. But the movie star is also 4 times his daughter’s age today. So 4 y  7  11y  7  4y  28  11y  7  21  7y  y  3. Thus the movie star’s age today is 11 3  7  40 years. 34. Let h be number of home runs Babe Ruth hit. Then h  41 is the number of home runs that Hank Aaron hit. So 1469  h  h  41  1428  2h  h  714. Thus Babe Ruth hit 714 home runs. 35. Let p be the number of pennies. Then p is the number of nickels and p is the number of dimes. So the value of the coins in the purse is the value of the pennies plus the value of the nickels plus the value of the dimes. Thus 144  001 p  005 p  010 p  144  016 p  p  144 016  9. So the purse contains 9 pennies, 9 nickels, and 9 dimes. 36. Let q be the number of quarters. Then 2q is the number of dimes, and 2q  5 is the number of nickels. Thus 300  value of the nickels value of the dimes value of the quarters. So 300  005 2q  5  010 2q  025q  300  010q  025  020q  025q  275  055q  q  275 055  5.

Thus Mary has 5 quarters, 2 5  10 dimes, and 2 5  5  15 nickels.

37. Let l be the length of the garden. Since area  width  length, we obtain the equation 1125  25l  l  1125 25  45 ft. So the garden is 45 feet long. 38. Let  be the width of the pasture. Then the length of the pasture is 2. Since area length width we have 115,200   2  22  2  57,600    240. Thus the width of the pasture is 240 feet. x

39. Let x be the length of a side of the square plot. As shown in the figure, area of the plot  area of the building  area of the parking lot. Thus,

x 2  60 40  12,000  2,400  12,000  14,400  x  120. So the plot of

x

land measures 120 feet by 120 feet.

60 40

40. Let  be the width of the building lot. Then the length of the building lot is 5. Since a half-acre is 12  43,560  21,780 and area is length times width, we have 21,780   5  52  2  4,356    66. Thus the width of the building lot is 66 feet and the length of the building lot is 5 66  330 feet. base1  base2 height. Putting in the known quantities, we have 2    y  2y 120  y  32 y 2  y 2  80  y   80  4 5. Since length is positive, y  4 5  894 inches. 2

41. The figure is a trapezoid, so its area is

x

42. First we write a formula for the area of the figure in terms of x. Region A has dimensions 10 cm and x cm and region B has dimensions 6 cm and x cm. So the shaded region has area 10  x  6  x  16x cm2 . We are given that this is equal to 144 cm2 , so 144  16x  x  144 16  9 cm.

10 cm

A

6 cm B

x

43. Let x be the width of the strip. Then the length of the mat is 20  2x, and the width of the mat is 15  2x. Now the perimeter is twice the length plus twice the width, so 102  2 20  2x  2 15  2x  102  40  4x  30  4x  102  70  8x  32  8x  x  4. Thus the strip of mat is 4 inches wide. 44. Let x be the width of the strip. Then the width of the poster is 100  2x and its length is 140  2x. The perimeter of the printed area is 2 100  2 140  480, and the perimeter of the poster is 2 100  2x  2 140  2x. Now we use the

fact that the perimeter of the poster is 1 12 times the perimeter of the printed area: 2 100  2x  2 140  2x  32  480  480  8x  720  8x  240  x  30. The blank strip is thus 30 cm wide.

42

CHAPTER P Prerequisites

45. Let x be the length of the man’s shadow, in meters. Using similar triangles,

x 10  x   20  2x  6x  4x  20  6 2

x  5. Thus the man’s shadow is 5 meters long. 46. Let x be the height of the tall tree. Here we use the property that corresponding sides in similar triangles are proportional. The base of the similar triangles starts at x-5

x 5 150   15 25 25 x  5  15 150  25x  125  2250  25x  2375  x  95. Thus the eye level of the woodcutter, 5 feet. Thus we obtain the proportion

5

tree is 95 feet tall.

25

15

125

47. Let x be the amount (in mL) of 60% acid solution to be used. Then 300  x mL of 30% solution would have to be used to yield a total of 300 mL of solution. 60% acid

30% acid

Mixture

x

300  x

300

Rate (% acid)

060

030

050

Value

060x

030 300  x

050 300

mL

60  200. 03 So 200 mL of 60% acid solution must be mixed with 100 mL of 30% solution to get 300 mL of 50% acid solution.

Thus the total amount of pure acid used is 060x  030 300  x  050 300  03x  90  150  x 

48. The amount of pure acid in the original solution is 300 50%  150. Let x be the number of mL of pure acid added. Then 150  x the final volume of solution is 300  x. Because its concentration is to be 60%, we must have  60%  06  300  x 30 150  x  06 300  x  150  x  180  06x  04x  30  x   75. Thus, 75 mL of pure acid must be 04 added. 49. Let x be the number of grams of silver added. The weight of the rings is 5  18 g  90 g.

Grams

Pure silver

Mixture

90

x

90  x

090

0

075

090 90

0x

075 90  x

Rate (% gold) Value

5 rings

So 090 90  0x  075 90  x  81  675  075x  075x  135  x  135 075  18. Thus 18 grams of silver

must be added to get the required mixture.

50. Let x be the number of liters of water to be boiled off. The result will contain 6  x liters.

Liters Concentration Amount

Original

Water

Final

6

x

6x

120

0

200

120 6

0

200 6  x

So 120 6  0  200 6  x  720  1200  200x  200x  480  x  24. Thus 24 liters need to be boiled off.

SECTION P.9 Modeling with Equations

43

51. Let x be the number of liters of coolant removed and replaced by water. 60% antifreeze

60% antifreeze (removed)

Water

Mixture

Liters

36

x

x

36

Rate (% antifreeze)

060

060

0

050

060 36

060x

0x

050 36

Value

036  06. Thus 06 liters 06

So 060 36  060x  0x  050 36  216  06x  18  06x  036  x  must be removed and replaced by water.

52. Let x be the number of gallons of 2% bleach removed from the tank. This is also the number of gallons of pure bleach added to make the 5% mixture. Original 2%

Pure bleach

5% mixture

100  x

x

100

002

1

005

002 100  x

1x

005  100

Gallons Concentration Bleach

So 002 100  x  x  005  100  2  002x  x  5  098x  3  x  306. Thus 306 gallons need to removed and replaced with pure bleach. 53. Let c be the concentration of fruit juice in the cheaper brand. The new mixture that Jill makes will consist of 650 mL of the original fruit punch and 100 mL of the cheaper fruit punch. Original Fruit Punch

Cheaper Fruit Punch

Mixture

mL

650

100

750

Concentration

050

c

048

050  650

100c

048  750

Juice

So 050  650  100c  048  750  325  100c  360  100c  35  c  035. Thus the cheaper brand is only 35% fruit juice. 54. Let x be the number of ounces of $300oz tea Then 80  x is the number of ounces of $275oz tea.

Pounds

$300 tea

$275 tea

Mixture

x

80  x

80

Rate (cost per ounce)

300

275

290

Value

300x

275 80  x

290 80

So 300x  275 80  x  290 80  300x  220  275x  232  025x  12  x  48. The mixture uses 48 ounces of $300oz tea and 80  48  32 ounces of $275oz tea.

55. Let t be the time in minutes it would take Candy and Tim if they work together. Candy delivers the papers at a rate of 1 1 70 of the job per minute, while Tim delivers the paper at a rate of 80 of the job per minute. The sum of the fractions of the

job that each can do individually in one minute equals the fraction of the job they can do working together. So we have 1 1  1  560  8t  7t  560  15t  t  37 1 minutes. Since 1 of a minute is 20 seconds, it would take them  70 80 3 3 t 37 minutes 20 seconds if they worked together. 56. Let t be the time, in minutes, it takes Hilda to mow the lawn. Since Hilda is twice as fast as Stan, it takes Stan 2t minutes to 1 1 mow the lawn by himself. Thus 40   40   1  40  20  t  t  60. So it would take Stan 2 60  120 minutes t 2t to mow the lawn.

44

CHAPTER P Prerequisites

57. Let t be the time, in hours, it takes Karen to paint a house alone. Then working together, Karen and Betty can paint a house 1 1 1 3  in 23 t hours. The sum of their individual rates equals their rate working together, so  16  2   16  t t 2t t 3

6  t  9  t  3. Thus it would take Karen 3 hours to paint a house alone.

58. Let h be the time, in hours, to fill the swimming pool using Jim’s hose alone. Since Bob’s hose takes 20% less time, it uses 1 1  1  18  08  18  08h  144  18  08h  324  08h only 80% of the time, or 08h. Thus 18   18  h 08h  h  405. Jim’s hose takes 405 hours, and Bob’s hose takes 324 hours to fill the pool alone.

59. Let t be the time in hours that Wendy spent on the train. Then 11 2  t is the time in hours that Wendy spent on the bus. We construct a table: Rate By train By bus

Time

40

t

60

11  t 2

Distance 

40t

60 11 2 t



  The total distance traveled is the sum of the distances traveled by bus and by train, so 300  40t  60 11 2 t 

300  40t  330  60t  30  20t  t  30 20  15 hours. So the time spent on the train is 55  15  4 hours. 60. Let r be the speed of the slower cyclist, in mi/h. Then the speed of the faster cyclist is 2r. Rate

Time

Distance

Slower cyclist

r

2

2r

Faster cyclist

2r

2

4r

When they meet, they will have traveled a total of 90 miles, so 2r  4r  90  6r  90  r  15. The speed of the slower cyclist is 15 mi/h, while the speed of the faster cyclist is 2 15  30 mi/h.

61. Let r be the speed of the plane from Montreal to Los Angeles. Then r  020r  120r is the speed of the plane from Los Angeles to Montreal. Rate Montreal to L.A.

r

L.A. to Montreal

12r

Time 2500 r 2500 12r

Distance 2500 2500

2500 55 2500 2500 2500      r 12r 6 r 12r ,000 55  12r  2500  6  12  2500  6  66r  18,000  15,000  66r  33,000  r  3366  500. Thus the plane flew The total time is the sum of the times each way, so 9 16 

at a speed of 500 mi/h on the trip from Montreal to Los Angeles.

ft 22 62. Let x be the speed of the car in mi/h. Since a mile contains 5280 ft and an hour contains 3600 s, 1 mi/h  5280 3600 s  15 ft/s. 220 220 The truck is traveling at 50  22 15  3 ft/s. So in 6 seconds, the truck travels 6  3  440 feet. Thus the back end

of the car must travel the length of the car, the length of the truck, and the 440 feet in 6 seconds, so its speed must be 1430440  242 ft/s. Converting to mi/h, we have that the speed of the car is 242  15  55 mi/h. 6 3 3 22

63. Let x be the distance from the fulcrum to where the mother sits. Then substituting the known values into the formula given, we have 100 8  125x  800  125x  x  64. So the mother should sit 64 feet from the fulcrum.

64. Let  be the largest weight that can be hung. In this exercise, the edge of the building acts as the fulcrum, so the 240 lb man is sitting 25 feet from the fulcrum. Then substituting the known values into the formula given in Exercise 43, we have 240 25  5  6000  5    1200. Therefore, 1200 pounds is the largest weight that can be hung.

CHAPTER P

l 2  502  l  102  l 2  2500  l 2  20l  100  20l  2400  l  120.

45

l

65. Let l be the length of the lot in feet. Then the length of the diagonal is l  10. We apply the Pythagorean Theorem with the hypotenuse as the diagonal. So

Review

50

l+10

Thus the length of the lot is 120 feet.

66. Let r be the radius of the running track. The running track consists of two semicircles and two straight sections 110 yards long, so we get the equation 2r  220  440  2r  220  r  110   3503. Thus the radius of the semicircle is about 35 yards. 67. Let h be the height in feet of the structure. The structure is composed of a right cylinder with radius 10 and height 23 h and a cone with base radius 10 and height   equation 1400   102 23 h 

by

1 h. Using the formulas for the volume of a cylinder and that of a cone, we obtain the 3   1  102 1 h  1400  200 h  100 h  126  6h  h (multiply both sides 3 3 3 9

9 )  126  7h  h  18. Thus the height of the structure is 18 feet. 100

68. Let h be the height of the break, in feet. Then the portion of the bamboo above the break is 10  h. Applying the Pythagorean Theorem, we obtain

h 2  32  10  h2  h 2  9  100  20h  h 2  91  20h  h  91 20  455. Thus the break is 455 ft above the ground.

10-h

h

3

69. Pythagoras was born about 569 BC in Samos, Ionia and died about 475 BC. Euclid was born about 325 BC and died about 265 BC in Alexandria, Egypt. Archimedes was born in 287 BC in Syracuse, Sicily and died in 212 BC in Syracuse. 70. Answers will vary.

CHAPTER P REVIEW 1. (a) Since there are initially 250 tablets and she takes 2 tablets per day, the number of tablets T that are left in the bottle after she has been taking the tablets for x days is T  250  2x. (b) After 30 days, there are 250  2 30  190 tablets left.

(c) We set T  0 and solve: T  250  2x  0  250  2x  x  125. She will run out after 125 days.

2. (a) The total cost is $2 per calzone plus the $3 delivery charge, so C  2x  3. (b) 4 calzones would be 2 4  3  $11.

(c) We solve C  2x  3  15  2x  12  x  6. You can order six calzones. 3. (a) 16 is rational. It is an integer, and more precisely, a natural number. (b) 16 is rational. It is an integer, but because it is negative, it is not a natural number.  (c) 16  4 is rational. It is an integer, and more precisely, a natural number.  (d) 2 is irrational. (e) 83 is rational, but is neither a natural number nor an integer. (f)  82  4 is rational. It is an integer, but because it is negative, it is not a natural number. 4. (a) 5 is rational. It is an integer, but not a natural number. (b)  25 6 is rational, but is neither an integer nor a natural number.  (c) 25  5 is rational, a natural number, and an integer. (d) 3 is irrational.

46

CHAPTER P Prerequisites 3 (e) 24 16  2 is rational, but is neither a natural number nor an integer. (f) 1020 is rational, a natural number, and an integer.

5. Commutative Property of addition.

6. Commutative Property of multiplication.

7. Distributive Property.

8. Distributive Property.

5 4 9 3 5 2      6 3 6 6 6 2 5 4 1 5 2 (b)     6 3 6 6 6 15  12 33 15 12     11. (a) 8 5 85 21 15  5 55 15 12     (b) 8 5 8  12 84 9. (a)

7 10 7 (b) 10 30 12. (a) 7 30 (b) 7

10. (a)

9 2 25 32

13. x  [2 6  2  x  6 _2

11 21 22 1    15 30 30 30 11 21 22 43     15 30 30 30 12 30  35 55     35 7  12 12 12 30  12 6  12     35 7  35 77 

14. x  0 10]  0  x  10 6

15. x   4]  x  4

0

_2

18. x  3  x   3

17. x  5  x  [5  5

_3

  20. 0  x  12  x  0 12

19. 1  x  5  x  1 5] 5

0

  21. (a) A  B  1 0 12  1 2 3 4

22. (a) C  D  1 2]

23. (a) A  C  1 2   (b) B  D  12  1

24. (a) A  D  0 1   (b) B  C  12  1

1 1 1    3 13 6 216 216    242 242  121  11 31.   2 2

29. 21613 

33. (a) 5  3  2  2 (b) 5  3  8  8

1 2

(b) C  D  0 1]

(b) A  B  1

25. 7  10  3  3    27. 212 812  2  8  16  4

10

16. x  [2   2  x 4

_1

25 2 72 49

26. 3  9  3  9  6  6

9  8  1 28. 23  32  18  19  72 72 72  23 30. 6423  43  42  16

32.

   2 50  100  10

34. (a) 4  0  4  4 (b) 4  4  8  8

CHAPTER P

 3 7  713  5 (b) 74  745

35. (a)

 6

x 5  x 56  9  12 9 (b) x  x  x 92

37. (a)

 3 7 5  573 3   3  (b) 4 5  514  534  12  38. (a) y 3  y 3  y 32 2   2  (b) 8 y  y 18  y 14  3  2  4 a3 b b3  a 6  a 6 b2  b12 40. a 2 36. (a)

  2  3x 1 y 2  4x 6 y 2  3x 1 y 2  4  3x 61 y 22 39. 2x 3 y  12x 5 y 4

x 4 3x2 x 4  9x 2   9x 423  9x 3 3 x x3    2 3 x 3 y y4  3 x 6 y4 y2  3 x 6 y6  x 2 y2 43. 41.



6

 a 66 b212  b14

r 12 s 8 r 2 s 43  2 6  r 122 s 86  r 10 s 2 13 r s r s     2 y 2  x y 2 44. x 2 y 4  x 2  42.

52 8r 12 s 3 122 s 34  4r 52 s 7  4r  4r s7 2r 2 s 4 2  4 6 ab2 c3 a 2 b4 c6 2 a 26 b48 c6  4a 4 b12 c6  4a c 46.   2 2a 3 b4 22 a 6 b8 b12

45.

47. 78,250,000,000  7825  1010

48. 208  108  00000000208      293  106 1582  1014 000000293 1582  1014 ab 293  1582 49.   1061412   12 12 c 28064 28064  10 28064  10  165  1032 50. 80

times 60 minutes 24 hours 365 days     90 years  38  109 times minute hour day year

51. 2x 2 y  6x y 2  2x y x  3y

  52. 12x 2 y 4  3x y 5  9x 3 y 2  3x y 2 4x y 2  y 3  3x 2

53. x 2  5x  14  x  7 x  2  2      54. x 4  x 2  2  x 2  x 2  2  x 2  2 x 2  1  x 2  2 x  1 x  1 55. 3x 2  2x  1  3x  1 x  1

56. 6x 2  x  12  3x  4 2x  3

57. 4t 2  13t  12  4t  3 t  4  2 58. x 4  2x 2  1  x 2  1  [x  1 x  1]2  x  12 x  12   59. 16  4t 2  4 4  t 2  4 t  2 t  2        60. 2y 6  32y 2  2y 2 y 4  16  2y 2 y 2  4 y 2  4  2y 2 y 2  4 y  2 y  2        61. x 6  1  x 3  1 x 3  1  x  1 x 2  x  1 x  1 x 2  x  1     62. a 4 b2  ab5  ab2 a 3  b3  ab2 a  b a 2  ab  b2   63. x 3  27  x  3 x 2  3x  9     64. 3y 3  81x 3  3 y 3  27x 3  3 y  3x y 2  3x y  9x 2

Review

47

48

CHAPTER P Prerequisites

  65. 4x 3  8x 2  3x  6  4x 2 x  2  3 x  2  4x 2  3 x  2   66. 3x 3  2x 2  18x  12  x 2 3x  2  6 3x  2  3x  2 x 2  6    67. x  y2  7 x  y  6  x  y  6 x  y  1  x  y  6 x  y  1 68. a  b2  3 a  b  10  a  b  5 a  b  2

69. 2y  7 2y  7  4y 2  49

  70. 1  x 2  x  3  x 3  x  2  x  x 2  9  x 2  2  x  x 2  9  x 2  7  x   71. x 2 x  2  x x  22  x 3  2x 2  x x 2  4x  4  x 3  2x 2  x 3  4x 2  4x  2x 3  6x 2  4x   x x 2  2x  3 x 3  2x 2  3x   x 2  2x  3 72. x x               x  1 2 x  1  x  x 2 x  1  2x x  x  2x  x  2x x  x  x 73. x

74. 2x  13  2x3  3 2x2 1  3 2x 12  13  8x 3  12x 2  6x  1

x 2  2x  3 x 3 x  3 x  1   2 2x 3  3  1 2x x 2x  5x  3   t  1 t 2  t  1 t3  1 t2  t  1 76. 2   t 1 t  1 t  1 t 1 75.

77. 78.

x 2  2x  3 3x  12 3 x  3 x  3 x  1 3 x  4     x 4 x  4 x  4 x  1 x 2  8x  16 x  1

x 2  2x  15 x 2  x  12 x 1 x  5 x  3 x  1 x  1     2 2 x 4  5  1  4  3 x x x x x  6x  5 x 1

79. x  80.

81.

82.

x x  1 1 x2  x  1 1    x 1 x 1 x 1 x 1

x x2  1 x x  1 x2  1  x2  x x 1 1            2  2 2 x 1 x 1 x  1 x  1 x  1 x  1 x  1 x 2  1 x  1 x 2  1

2 x  22 x x  2 3x 1 3 2      2 2 2 x x  2 x  2 x x  2 x x  2 x x  22   2 x 2  4x  4  x 2  2x  3x 2x 2  8x  8  x 2  2x  3x 3x 2  7x  8    2 2 x x  2 x x  2 x x  22 1 1 2 1 2 1      x  2 x2  4 x2  x  2 x  2 x  2 x  2 x  2 x  1 x 1 2 x  2 x  2 x  1    x  2 x  1 x  2 x  2 x  1 x  2 x  2 x  1 x  2 

1 x 83. x 1 84. x 1 x

x 2  2x  5 x 2  x  2  x  1  2x  4  x  2 x  1 x  2 x  2 x  1 x  2

x 2  2x 1 1 x  2 1 1 2x 2x       2 x 2 2x x 2 2x x 2 2x 1 1 1   1 x 1  x x  1  x x  1  x  1  x  1 1 1 x x  1 2x  1 x  1  x   x 1 x x 1  12

CHAPTER P

85.

86. 87. 88. 89. 90.

91.

92. 93. 94. 95.

96.

  3 x  h2  5 x  h  3x 2  5x

Review

49

3x 2  6xh  3h 2  5x  5h  3x 2  5x 6xh  3h 2  5h  h h h h 6x  3h  5  6x  3h  5  h       x h x x h x x h x x  h  x h 1              h h x h x h x h x h x h x x h x   1 1 11 11     11 11 11 11    3 6 6 3 3 6      6 2 6 6 6    10 21 10 10  10 2  10  10 2     21 21 21 21     14 3 2 42  14 2 42  14 2 14 62 2         2 7 3 2 3 2 3 2 32  2     x x 2 x 2 x 2x  x x x         2 x 4 2 x 2 x 2 x 22  x     x 4 x 4 x 2 x 2 x 2     x 4 x 2 x 2 x 2 x 5 is defined whenever x  10  0  x  10, so its domain is x  x  10. x  10 2x is defined whenever x 2  9  0  x 2  9  x  3, so its domain is x  x  3 and x  3. 2 x 9   x is defined whenever x  0 (so that x is defined) and x 2  3x  4  x  1 x  4  0  x  1 and x 2  3x  4 x  4. Thus, its domain is x  x  0 and x  4.  x 3 is defined whenever x  3  0  x  3 and x 2  4x  4  x  22  0  x  2. Thus, its domain is x 2  4x  4 x  x  3. 

97. This statement is false. For example, take x  1 and y  1. Then LHS  x  y3  1  13  23  8, while RHS  x 3  y 3  13  13  1  1  2, and 8  2.    1a 1 a 1 1 a 1 a  98. This statement is true for a  1:      .    1a 1a 1 a 1 a 1  a 1  a

99. This statement is true:

12  y 12 y 12     1. y y y y

   100. This statement is false. For example, take a  1 and b  1. Then LHS  3 a  b  3 1  1  3 2, while      RHS  3 a  3 b  3 1  3 1  1  1  2, and 3 2  2.    101. This statement is false. For example, take a  1. Then LHS  a 2  12  1  1, which does not equal  a  1. The true statement is a 2  a. 1 1 1 1 1 1 5 1   , while RHS      , 102. This statement is false. For example, take x  1. Then LHS  x 4 14 5 x 4 1 4 4 1 5 and  . 5 4 103. 3x  12  24  3x  12  x  4

104. 5x  7  42  5x  49  x  49 5

50

CHAPTER P Prerequisites

105. 7x  6  4x  9  3x  15  x  5

106. 8  2x  14  x  3x  6  x  2

107. 13 x  12  2  2x  3  12  2x  15  x  15 2 6 3 108. 23 x  35  15  2x  10x  9  3  30x  40x  6  x   40 20 109. 2 x  3  4 x  5  8  5x  2x  6  4x  20  8  5x  2x  26  8  5x  3x  18  x  6

x  5 2x  5 5    3 x  5  2 2x  5  5  3x  15  4x  10  5  x  30  x  30 2 3 6 x 1 2x  1 111.   x  1 2x  1  2x  1 x  1  2x 2  3x  1  2x 2  3x  1  6x  0  x  0 x 1 2x  1

110.

112. 113.

1 7 x 3   x  3 x  2  1  x  3x  6  1  2x  7  x   x 2 x 2 2

3x 3x x x 1     x  1 x  2  x x  1  x 2  x  2  x 2  x  2  0. Since this x 1 3x  6 3 x  2 x 2 last equation is never true, there is no real solution to the original equation.

114. x  22  x  42  x  22  x  42  0  [x  2  x  4] [x  2  x  4]  0  [x  2  x  4] [x  2  x  4]  6 2x  2  0  2x  2  0  x  1. 115. x 2  144  x  12

7 116. 4x 2  49  x 2  49 4  x  2

117. x 3  27  0  x 3  27  x  3.

118. 6x 4  15  0  6x 4  15  x 4   52 . Since x 4 must be nonnegative, there is no real solution.

119. x  13  64  x  1  4  x  1  4  5.   120. x  22  2  0  x  22  2  x  2   2  x  2  2.  121. 3 x  3  x  33  27. 2  122. x 23  4  0  x 13  4  x 13  2  x  8.

123. 4x 34  500  0  4x 34  500  x 34  125  x  12543  54  625. 124. x  215  2  x  2  25  32  x  2  32  34. 125. A 

xy  2A  x  y  x  2A  y. 2

126. V  x y  yz  xz  V  y x  z  xz  V  xz  y x  z  y 

V  xz . x z

1 1 1 1 1 11 11    tJ  1   t  , J  0. t 2t 3t 2 3 6 6J  q q q q q q 128. F  k 1 2 2  r 2  k 1 2  r   k 1 2 . (In real-world applications, r represents distance, so we would take the F F r positive root.)

127. Multiply through by t: J 

129. Let x be the number of pounds of raisins. Then the number of pounds of nuts is 50  x. Pounds Rate (cost per pound)

Raisins

Nuts

Mixture

x

50  x

50

320

240

272

So 320x  240 50  x  272 50  320x  120  240x  136  08x  16  x  20. Thus the mixture uses 20 pounds of raisins and 50  20  30 pounds of nuts.

CHAPTER P

Test

51

130. Let t be the number of hours that Anthony drives. Then Helen drives for t  14 hours. Anthony Helen

Rate

Time

Distance

45

t

40

t  14

45t   40 t  14

  When they pass each other, they will have traveled a total of 160 miles. So 45t  40 t  14  160  45t  40t  10  160  85t  170  t  2. Since Anthony leaves at 2:00 P. M . and travels for 2 hours, they pass each other at 4:00 P. M . 131. Let x be the amount invested in the account earning 15% interest. Then the amount invested in the account earning 25% is 7000  x. Amount invested Interest earned

15% Account

25% Account

Total

x

7000  x

7000

0015x

0025 7000  x

12025

From the table, we see that 0015x  0025 7000  x  12025  0015x  175  0025x  12025  5475  001x  x  5475. Thus, Luc invested $5475 in the account earning 15% interest and $1525 in the account earning 25% interest. 132. The amount of interest Shania is currently earning is 6000 003  $180. If she wishes to earn a total of $300, she must earn another $120 in interest at a rate of 125% per year. If the additional amount invested is x, we have the equation 00125x  120  x  9600. Thus, Shania must invest an additional $9600 at 125% simple interest to earn a total of $300 interest per year. 133. Let t be the time it would take Abbie to paint a living room if she works alone. It would take Beth 2t hours to paint the 1 living room alone, and it would take 3t hours for Cathie to paint the living room. Thus Abbie does of the job per hour, t 1 1 1 1 1 of the job per hour, and Cathie does of the job per hour. So    1  6  3  2  6t  Beth does 2t 3t t 2t 3t 6t  11  t  11 6 . So it would Abbie 1 hour 50 minutes to paint the living room alone. 134. Let  be width of the pool. Then the length of the pool is 2, and its volume is 8  2  8464  162  8464  2  529    23. Since   0, we reject the negative value. The pool is 23 feet wide, 2 23  46 feet long, and 8 feet deep.

CHAPTER P TEST 1. (a) The cost is C  9  15x.

(b) There are four extra toppings, so x  4 and C  9  15 4  $15.

2. (a) 5 is rational. It is an integer, and more precisely, a natural number.  (b) 5 is irrational. (c)  93  3 is rational, and it is an integer.

(d) 1,000,000 is rational, and it is an integer.

3. (a) A  B  0 1 5   (b) A  B  2 0 12  1 3 5 7 4. (a) _4

2

[4 2

0

3

[0 3]

52

CHAPTER P Prerequisites

(b) 0

(c) 4  2  6  6

_4

[4 2  [0 3]  [0 2

5. (a) 26  64 (e)

2

 2  2 4 2 3   2 3 9

(b) 26  64  5 1 32 2 (f)    4 2 16

3

[4 2  [0 3]  [4 3] 1 1 (c) 26  6  64 2  8 32 9 4 3 (g)   16 216 24

6. (a) 186,000,000,000  186  1011

710 1 (d) 12  72  49 7  34 1 (h) 8134  34  33  27

(b) 00000003965  3965  107

a 3 b2 a2  b ab3 2  y4  6 (b) 2x 3 y 2 4x  2  (c) 2x 12 y 2 3x 14 y 1  2  32 x 12214 y 221  18x        (d) 20  125  4  5  25  5  2 5  5 5  3 5     2 (e) 18x 3 y 4  9  2  x 2  x  y 2  3x y 2 2x 2  1 2x 2 y  22 x 2232 y 2122  10 (f) x 3 y 12 4x y

7. (a)

8. (a) 3 x  6  4 2x  5  3x  18  8x  20  11x  2

(b) x  3 4x  5  4x 2  5x  12x  15  4x 2  7x  15        2  2 a b a b  a  b ab (c) (d) 2x  32  2x2  2 2x 3  32  4x 2  12x  9

(e) x  23  x3  3 x2 2  3 x 22  23  x 3  6x 2  12x  8   (f) x 2 x  3 x  3  x 2 x 2  9  x 4  9x 2

9. (a) 4x 2  25  2x  5 2x  5

(b) 2x 2  5x  12  2x  3 x  4

  (c) x 3  3x 2  4x  12  x 2 x  3  4 x  3  x  3 x 2  4  x  3 x  2 x  2     (d) x 4  27x  x x 3  27  x x  3 x 2  3x  9

(e) 2x  y2  10 2x  y  25  2x  y2  2 5 2x  y  52  2x  y  52   (f) x 3 y  4x y  x y x 2  4  x y x  2 x  2 10. (a) (b) (c)

x 2  3x  2 x 2 x  1 x  2   x 2 x  1 x  2 x2  x  2

x 1 2x 2  x  1 x  3 2x  1 x  1 x  3     2 2x  1 x 3 x  3 x  3 2x  1 x 9 x2

x2  4



x2 x2 x 1  x  1 x  2 x 1     x 2 x  2 x  2 x  2 x  2 x  2 x  2 x  2   2 2 x  x x 2 x 2 1    x  2 x  2 x  2 x  2 x  2

CHAPTER P

Test

53

y y x x   y2  x 2  x  y y  x y  x y  x x y x y xy      y  x (d)   1 1 xy 1 1 xy xy xy   y x y x   3  6 6 632 6 2 11. (a)  332        3 3 2 3 2 3 2 4 2 2 2       10 5  2     10 10 52 (b)   50  2 10  5 2  2 10    54 52 52 52   1 1 x 1 x 1 (c)       1x 1 x 1 x 1 x

12. (a) 4x  3  2x  7  4x  2x  7  3  2x  10  x  5.   3 (b) 8x 3  125  8x 3  3 125  2x  5  x   52 . 32   6432  x  83  512. (c) x 23  64  0  x 23  64  x 23

x 3 x   x 2x  1  x  3 2x  5  2x 2  x  2x 2  5x  6x  15  x  x  15  2x  15 2x  5 2x  1  x  15 2 .   (e) 3 x  12  18  0  3 x  12  18  x  12  6  x  1   6  x   6  1  E E 2 2 c c . (We take the positive root because c represents the speed of light, which is positive.) 13. E  mc  m m 14. Let d be the distance in km, between Bedingfield and Portsmouth. (d)

Direction

Distance

Rate

Bedingfield  Portsmouth

d

100

Portsmouth  Bedingfield

d

75

Time d 100 d 75

distance to fill in the time column of the table. We are given that the sum of the times is 35 hours. rate   d d d 1050 d   35  300   300 35  3d  4d  1050  d   150 km. Thus we get the equation 100 75 100 75 7

We have used time 

54

FOCUS ON MODELING

FOCUS ON MODELING Making the Best Decisions 

1. (a) The total cost is 



cost of



maintenance



copier

cost

 

number of months







copy cost

 

number of months



. Each month

the copy cost is 8000  003  240. Thus we get C1  5800  25n  240n  5800  265n.       rental number copy number    . Each month the copy cost is (b) In this case the cost is  cost of months cost of months

(c)

8000  006  480. Thus we get C2  95n  480n  575n. Years

n

Purchase

Rental

1

12

8,980

6,900

2

24

12,160

13,800

3

36

15,340

20,700

4

48

18,520

27,600

5

60

21,700

34,500

6

72

24,880

41,400

(d) The cost is the same when C1  C2 are equal. So 5800  265n  575n  5800  310n  n  1871 months.



daily



daily

2. (a) The cost of Plan 1 is 3  

cost

The cost of Plan 2 is 3  

cost







cost per mile



 

number of miles



  3  65  015x  195  015x.

  3  90  270.

(b) When x  400, Plan 1 costs 195  015 400  $255 and Plan 2 costs $270, so Plan 1 is cheaper. When x  800, Plan 1 costs 195  015 800  $315 and Plan 2 costs $270, so Plan 2 is cheaper.

(c) The cost is the same when 195  015x  270  015  75x  x  500. So both plans cost $270 when the businessman drives 500 miles.



3. (a) The total cost is  

(b) The revenue is 

setup cost



cost per



number



price per tire





tire

of tires

 



number of tires



. So C  8000  22x.

. So R  49x.

(c) Profit  Revenue  Cost. So P  R  C  49x  8000  22x  27x  8000.

(d) Break even is when profit is zero. Thus 27x  8000  0  27x  8000  x  2963. So they need to sell at least 297 tires to break even.

Making the Best Decisions

55

4. (a) Option 1: In this option the width is constant at 100. Let x be the increase in length. Then the additional area is   increase   100x. The cost is the sum of the costs of moving the old fence, and of installing the width   in length

new one. The cost of moving is $6  100  $600 and the cost of installation is 2  10  x  20x, so the total cost is C  600 . Substituting in the area C  20x  600. Solving for x, we get C  20x  600  20x  C  600  x  20   C  600 we have A1  100  5 C  600  5C  3,000. 20 Option 2: In this option the length is constant at 180. Let y be the increase in the width. Then the additional area is   increase   180y. The cost of moving the old fence is 6  180  $1080 and the cost of installing the new length   in width

(b)

one is 2  10  y  20x, so the total cost is C  20y  1080. Solving for y, we get C  20y  1080  20y  C  1080   C  1080 C  1080 . Substituting in the area we have A2  180  9 C  1080  9C  9,720. y 20 20 Cost, C

Area gain A1 from Option 1

$1100

2,500 ft2

Area gain A2 from Option 2 180 ft2

$1200

3,000 ft2

1,080 ft2

$1500

4,500 ft2

3,780 ft2

$2000

7,000 ft2

8,280 ft2

$2500

9,500 ft2

12,780 ft2

$3000

12,000 ft2

17,280 ft2

(c) If the farmer has only $1200, Option 1 gives him the greatest gain. If the farmer has only $2000, Option 2 gives him the greatest gain. 5. (a) Design 1 is a square and the perimeter of a square is four times the length of a side. 24  4x, so each side is x  6 feet long. Thus the area is 62  36 ft2 .

12 . Thus, the area is Design 2 is a circle with perimeter 2r and area r 2 . Thus we must solve 2r  24  r    2 12 144  458 ft2 . Design 2 gives the largest area.    

(b) In Design 1, the cost is $3 times the perimeter p, so 120  3 p and the perimeter is 40 feet. By part (a), each side is then 40  10 feet long. So the area is 102  100 ft2 . 4

In Design 2, the cost is $4 times the perimeter p. Because the perimeter is 2r, we get 120  4 2r so  2 120 15 225 15 r  . The area is r 2    716 ft2 . Design 1 gives the largest area.  8   

6. (a) Plan 1: Tomatoes every year. Profit  acres  Revenue  cost  100 1600  300  130,000. Then for n years the profit is P1  130,000n. (b) Plan 2: Soybeans followed by tomatoes. The profit for two years is Profit  acres      soybean tomato     100 1200  1600  280,000. Remember that no fertilizer is revenue revenue needed in this plan. Then for 2k years, the profit is P2  280,000k.

(c) When n  10, P1  130,000 10  1,300,000. Since 2k  10 when k  5, P2  280,000 5  1,400,000. So Plan B is more profitable.

56

FOCUS ON MODELING

7. (a) Data (GB) 1 15 2 25 3 35 4

Plan A

Plan B $25

25  5 200  $35

Plan C $40

40  5 150  $4750

$60 60  5 100  $65

25  10 200  $45

40  10 150  $55

60  10 100  $70

25  20 200  $65

40  20 150  $70

60  20 100  $80

25  15 200  $55 25  25 200  $75 25  30 200  $85

40  15 150  $6250 40  25 150  $7750

40  30 150  $85

60  15 100  $75 60  25 100  $85 60  30 100  $90

(b) For Plan A: CA  25  2 10x  10  20x  5. For Plan B: CB  40  15 10x  10  15x  25. For Plan C: CC  60  1 10x  10  10x  50. Note that these equations are valid only for x  1.

(c) If Gwendolyn uses 22 GB, Plan A costs 25  12 2  $49, Plan B costs 40  12 15  $58, and Plan C costs 60  12 1  $72. If she uses 37 GB, Plan A costs 25  27 2  $79, Plan B costs 40  27 15  $8050, and Plan C costs 60  27 1  $87. If she uses 49 GB, Plan A costs 25  39 2  $103, Plan B costs 40  39 15  $9850, and Plan C costs 60  39 1  $99.

(d) (i) We set CA  CB  20x  5  15x  25  5x  20  x  4. Plans A and B cost the same when 4 GB are used. (ii) We set CA  CC  20x  5  10x  50  10x  45  x  45. Plans A and C cost the same when 45 GB are used. (iii) We set CB  CC  15x  25  10x  50  5x  25  x  5. Plans B and C cost the same when 5 GB are used. 8. (a) In this plan, Company A gets $32 million and Company B gets $32 million. Company A’s investment is $14 million, so they make a profit of 32  14  $18 million. Company B’s investment is $26 million, so they make a profit of 32  26  $06 million. So Company A makes three times the profit that Company B does, which is not fair.

(b) The original investment is 14  26  $4 million. So after giving the original investment back, they then share the profit of $24 million. So each gets an additional $12 million. So Company A gets a total of 14  12  $26 million and Company B gets 26  12  $38 million. So even though Company B invests more, they make the same profit as Company A, which is not fair.

(c) The original investment is $4 million, so Company A gets 14 4  64  $224 million and Company B gets 26  64  $416 million. This seems the fairest. 4

1

EQUATIONS AND GRAPHS

1.1

THE COORDINATE PLANE

1. The point that is 2 units to the left of the y-axis and 4 units above the x-axis has coordinates 2 4. 2. If x is positive and y is negative, then the point x y is in Quadrant IV.  3. The distance between the points a b and c d is c  a2  d  b2 . So the distance between 1 2 and 7 10 is     7  12  10  22  62  82  36  64  100  10. 4. The point midway between a b and c d is     8 12 1  7 2  10     4 6. 2 2 2 2



 ac bd  . So the point midway between 1 2 and 7 10 is 2 2

5. A 5 1, B 1 2, C 2 6, D 6 2, E 4 1, F 2 0, G 1 3, H 2 2 6. Points A and B lie in Quadrant 1 and points E and G lie in Quadrant 3.   7. 0 5, 1 0, 1 2, and 12  23

8. 5 0, 2 0, 26 13, and 25 35

y

(0, 5)

(_1, 0) 1 (_1, _2)

9. x y  x  2

(1/2, 2/3) 1

x

10. x y  y  2

y 5l

0

y 5l

5

x

0

5

x

57

58

CHAPTER 1 Equations and Graphs

11. x y  x  4

12. x y  y  3

y

y 5l

5l

0

5

0

x

13. x y  3  x  3

14. x y  0  y  2

y

5l

5

x

5

x

y 5l

0

5

x

15. x y  x y  0

 x y  x  0 and y  0 or x  0 and y  0 y

1

0

16. x y  x y  0

 x y  x  0 and y  0 or x  0 and y  0 y

1 1

x

17. x y  x  1 and y  3

1

x

18. x y  x  2 and y  1 y

y

1l

1 1

x

1

x

SECTION 1.1 The Coordinate Plane

19. x y  1  x  1 and  2  y  2

20. x y  3  x  3 and  1  y  1

y

y

1

1 1

1

x

x

21. The two points are 0 2 and 3 0.     (a) d  3  02  0  22  32  22  9  4  13     30 02   32  1 (b) midpoint: 2 2 22. The two points are 2 1 and 2 2.     (a) d  2  22  1  22  42  32  16  9  25  5     2  2 1  2   0 12 (b) midpoint: 2 2 23. The two points are 3 3 and 5 3.     (a) d  3  52  3  32  82  62  64  36  100  10   3  5 3  3 (b) midpoint:   1 0 2 2 24. The two points are 2 3 and 4 1.      (a) d  2  42  3  12  62  22  36  4  50  2 10   2  4 3  1   1 2 (b) midpoint: 2 2 25. (a)

y

y

26. (a) (_2, 5) (6, 16)

1l

(10, 0) 2

(0, 8) 2l 1

 (b) d  0  62  8  162    62  82  100  10   0  6 8  16 (c) Midpoint:   3 12 2 2

x

 2  102  5  02    122  52  169  13     2  10 5  0 (c) Midpoint:   4 52 2 2

(b) d 

x

59

60

CHAPTER 1 Equations and Graphs y

27. (a) (_4,5)

y

28. (a)

5l (_1, 1)

0

5

1

(3,_2)

x

(_6, _3)

 3  42  2  52      72  72  49  49  98  7 2     4  3 5  2    12  32 (c) Midpoint: 2 2

(b) d 

y

29. (a)

1l

x

 1  62  1  32    52  42  41     6  1 3  1    72  1 (c) Midpoint: 2 2

(b) d 

30. (a)

y

(_6, 2) 1l

1l 1

(6, _2)

1

x

(5, 0) x

(0, _6)

  6  62  2  22  122  42     144  16  160  4 10   6  6 2  2 (c) Midpoint:   0 0 2 2

(b) d 

 0  52  6  02     52  62  25  36  61     0  5 6  0 (c) Midpoint:   52  3 2 2

(b) d 

SECTION 1.1 The Coordinate Plane

  31. d A B  1  52  3  32  42  4.   d A C  1  12  3  32  62  6. So the area is 4  6  24.

y

32. The area of a parallelogram is its base times its height. Since two sides are parallel to the x-axis, we use the length of one of these as the base. Thus, the base is   d A B  1  52  2  22  42  4. The

height is the change in the y coordinates, thus, the height

A

is 6  2  4. So the area of the parallelogram is

B

base  height  4  4  16.

1l 1

y

C

x

C

D 1l

A

D

B

1

33. From the graph, the quadrilateral ABC D has a pair of parallel sides, so ABC D is a trapezoid. The area is   b1  b2 h. From the graph we see that 2   b1  d A B  1  52  0  02  42  4;   b2  d C D  4  22  3  32  22  2; and h is the difference in y-coordinates is 3  0  3. Thus   42 the area of the trapezoid is 3  9. 2

1l A 1

we find the length of one side and square it. This gives  d Q R  5  02  1  62   52  52    25  25  50  2 50  50. So the area is

C

y

R B

Thus point A 6 7 is closer to the origin.

Q

P

1l 1

x

    6  02  7  02  62  72  36  49  85.     d 0 B  5  02  8  02  52  82  25  64  89.

35. d 0 A 

x

34. The point S must be located at 0 4. To find the area,

y

D

61

x

62

CHAPTER 1 Equations and Graphs

    6  22  3  12  42  22  16  4  20.     d E D  3  22  0  12  52  12  25  1  26.

36. d E C 

Thus point C is closer to point E.

     1  32  1  12  42  22  16  4  20  2 5.    d Q R  1  12  1  32  0  42  16  4. Thus point Q 1 3 is closer to point R.

37. d P R 

    38. (a) The distance from 7 3 to the origin is 7  02  3  02  72  32  49  9  58. The distance from     3 7 to the origin is 3  02  7  02  32  72  9  49  58. So the points are the same distance from the origin.

  (b) The distance from a b to the origin is a  02  b  02  a 2  b2 . The distance from b a to the origin is    b  02  a  02  b2  a 2  a 2  b2 . So the points are the same distance from the origin.

39. Since we do not know which pair are isosceles, we find the length of all three sides.      d A B  3  02  1  22  32  32  9  9  18  3 2.     d C B  3  42  1  32  12  42  1  16  17.     d A C  0  42  2  32  42  12  16  1  17. So sides AC and C B have the same length. 40. Since the side AB is parallel to the x-axis, we use this as the base in the formula area  12 base  height. The height is the change in the y-coordinates. Thus, the base is 2  4  6 and the height is 4  1  3. So the area is 12 6  3  9.

41. (a) Here we have A  2 2, B  3 1, and C  3 3. So     d A B  3  22  1  22  12  32  1  9  10;      d C B  3  32  1  32  62  22  36  4  40  2 10;      d A C  3  22  3  22  52  52  25  25  50  5 2.

Since [d A B]2  [d C B]2  [d A C]2 , we conclude that the triangle is a right triangle.   (b) The area of the triangle is 12  d C B  d A B  12  10  2 10  10.

    52  42  25  16  41; 11  62  3  72      16  25  41; d A C  2  62  2  72  42  52      d B C  2  112  2  32  92  12  81  1  82.

42. d A B 

Since [d A B]2  [d A C]2  [d B C]2 , we conclude that the triangle is a right triangle. The area is    1 41  41  41 2 2.

SECTION 1.1 The Coordinate Plane

63

43. We show that all sides are the same length (its a rhombus) and then show that the diagonals are equal. Here we have A  2 9, B  4 6, C  1 0, and D  5 3. So     d A B  4  22  6  92  62  32  36  9  45;     d B C  1  42  0  62  32  62  9  36  45;     d C D  5  12  3  02  62  32  36  9  45;     d D A  2  52  9  32  32  62  9  36  45. So the points form a      rhombus. Also d A C  1  22  0  92  32  92  9  81  90  3 10,      and d B D  5  42  3  62  92  32  81  9  90  3 10. Since the diagonals are equal, the rhombus is a square.

     42  82  16  64  80  4 5. 3  12  11  32       5  32  15  112  22  42  4  16  20  2 5. d B C       d A C  5  12  15  32  62  122  36  144  180  6 5. So d A B  d B C  d A C,

44. d A B 

and the points are collinear.

45. Let P  0 y be such a point. Setting the distances equal we get   0  52  y  52  0  12  y  12    25  y 2  10y  25  1  y 2  2y  1  y 2  10y  50  y 2  2y  2  12y  48  y  4. Thus, the point is P  0 4. Check:     0  52  4  52  52  12  25  1  26;     0  12  4  12  12  52  25  1  26. 



   2 3. So the length of the median CC  is d C C         18 02  2 2   92  1 . So the length of the median B B  2  8  3  2  37. The midpoint of AC is B  2 2          9  3 2  1  62  109 . The midpoint of BC is A  3  8  6  2  11  4 . So the length is d B B   2 2 2 2 2    2   11  1  4  02  145 . of the median A A is d A A  2 2

46. The midpoint of AB is C  

13 06  2 2

47. As indicated by Example 3, we must find a point S x1  y1  such that the midpoints

y

of P R and of QS are the same. Thus     4  1 2  4 x1  1 y1  1    . Setting the x-coordinates equal, 2 2 2 2

x 1 4  1  1  4  1  x1  1  x1  2Setting the 2 2 y 1 2  4  1  2  4  y1  1  y1  3. y-coordinates equal, we get 2 2 Thus S  2 3. we get

Q 1l

P

1

R

x

64

CHAPTER 1 Equations and Graphs

48. We solve the equation 6  8

2x 2x to find the x coordinate of B. This gives 6   12  2  x  x  10. Likewise, 2 2

3y  16  3  y  y  13. Thus, B  10 13. 2 y

49. (a)

C D B

1l A

1



2  7 1  7  2 2     41 24 of B D is   52  3 . 2 2

(b) The midpoint of AC is



   52  3 , the midpoint

(c) Since the they have the same midpoint, we conclude that the x

diagonals bisect each other.

   a b a0 b0    . Thus, 2 2 2 2    2 a 2  b b2 a2 a 2  b2 d C M  0  0    ; 2 2 4 4 2     2 a  a 2  b 2 2  b a2 a 2  b2 b2 d A M     a  0    ; 2 2 2 2 4 4 2     2 a  a 2  b 2 2  b a2 a 2  b2 b2 d B M  0  b    .    2 2 2 2 4 4 2

50. We have M 



51. (a) The point 5 3 is shifted to 5  3 3  2  8 5. (b) The point a b is shifted to a  3 b  2.

(c) Let x y be the point that is shifted to 3 4. Then x  3 y  2  3 4. Setting the x-coordinates equal, we get x  3  3  x  0. Setting the y-coordinates equal, we get y  2  4  y  2So the point is 0 2.

(d) A  5 1, so A  5  3 1  2  2 1; B  3 2, so B   3  3 2  2  0 4; and C  2 1, so C   2  3 1  2  5 3. 52. (a) The point 3 7 is reflected to the point 3 7. (b) The point a b is reflected to the point a b. (c) Since the point a b is the reflection of a b, the point 4 1 is the reflection of 4 1.

(d) A  3 3, so A  3 3; B  6 1, so B   6 1; and C  1 4, so C   1 4.   53. (a) d A B  32  42  25  5. (b) We want the distances from C  4 2 to D  11 26. The walking distance is 4  11  2  26  7  24  31 blocks. Straight-line distance is    4  112  2  262  72  242  625  25 blocks.

(c) The two points are on the same avenue or the same street.   3  27 7  17   15 12, which is at the intersection of 15th Street and 12th Avenue. 54. (a) The midpoint is at 2 2 (b) They each must walk 15  3  12  7  12  5  17 blocks. 55. The midpoint of the line segment is 66 45. The pressure experienced by an ocean diver at a depth of 66 feet is 45 lb/in2 .

SECTION 1.2 Graphs of Equations in Two Variables: Circles

65

2x 2x to find the x coordinate of B: 6   12  2  x  x  10. Likewise, for the y 2 2 3y coordinate of B, we have 8   16  3  y  y  13. Thus B  10 13. 2

56. We solve the equation 6 

y

57. We need to find a point S x1  y1  such that P Q RS is a parallelogram. As indicated by Example 3, this will be the case if the diagonals P R and QS bisect each other. So the midpoints of P R and QS are the same. Thus     x1  2 y1  2 0  5 3  3    . Setting the x-coordinates equal, we get 2 2 2 2

Q 1l

x 2 05  1  0  5  x1  2  x1  3. 2 2 y 2 3  3  1  3  3  y1  2  Setting the y-coordinates equal, we get 2 2 y1  2. Thus S  3 2.

1.2

1

R

x

P

GRAPHS OF EQUATIONS IN TWO VARIABLES: CIRCLES

1. If the point 2 3 is on the graph of an equation in x and y, then the equation is satisfied when we replace x by 2 and y by 3. ?

?

We check whether 2 3  2  1  6  3. This is false, so the point 2 3 is not on the graph of the equation 2y  x  1.

To complete the table, we express y in terms of x: 2y  x  1  y  12 x  1  12 x  12 . x y

x

y

2

 12

1

0

0

1 2

1

1

2

3 2

  2  12 1 0   0 12

y 1

0

1

x

1 1   2 32

2. To find the x-intercept(s) of the graph of an equation we set y equal to 0 in the equation and solve for x: 2 0  x  1  x  1, so the x-intercept of 2y  x  1 is 1. 3. To find the y-intercept(s) of the graph of an equation we set x equal to 0 in the equation and solve for y: 2y  0  1  y  12 , so the y-intercept of 2y  x  1 is 12 .

4. The graph of the equation x  12  y  22  9 is a circle with center 1 2 and radius

 9  3.

5. (a) If a graph is symmetric with respect to the x-axis and a b is on the graph, then a b is also on the graph. (b) If a graph is symmetric with respect to the y-axis and a b is on the graph, then a b is also on the graph. (c) If a graph is symmetric about the origin and a b is on the graph, then a b is also on the graph. 6. (a) The x-intercepts are 3 and 3 and the y-intercepts are 1 and 2. (b) The graph is symmetric about the y-axis.

7. Yes, this is true. If for every point x y on the graph, x y and x y are also on the graph, then x y must be on the graph as well, and so it is symmetric about the origin. 8. No, this is not necessarily the case. For example, the graph of y  x is symmetric about the origin, but not about either axis.

66

CHAPTER 1 Equations and Graphs ?

?

?

9. y  3  4x. For the point 0 3: 3  3  4 0  3  3. Yes. For 4 0: 0  3  4 4  0  13. No. For 1 1: ?

?

1  3  4 1  1  1. Yes. So the points 0 3 and 1 1 are on the graph of this equation.  ?  ?  ?  ?  1  x. For the point 2 1: 1  1  2  1  1. No. For 3 2: 2  1  3  2  4. Yes. For 0 1: ?  1  1  0. Yes. So the points 3 2 and 0 1 are on the graph of this equation.

10. y 

?

?

?

?

11. x  2y  1  0. For the point 0 0: 0  2 0  1  0  1  0. No. For 1 0: 1  2 0  1  0  1  1  0. Yes. ?

?

For 1 1: 1  2 1  1  0  1  2  1  0. Yes. So the points 1 0 and 1 1 are on the graph of this equation.          ? ? ? ? 12. y x 2  1  1. For the point 1 1: 1 12  1  1  1 2  1. No. For 1 12 : 12 12  1  1  12 2  1.      ? ? Yes. For 1 12 : 12 12  1  1  12 2  1. Yes.     So the points 1 12 and 1 12 are on the graph of this equation. ?

?

?

13. x 2  2x y  y 2  1. For the point 0 1: 02  2 0 1  12  1  1  1. Yes. For 2 1: 22  2 2 1  12  1 ?

?

?

?

 4  4  1  1  1  1. Yes. For 2 3: 22  2 2 3  32  1  4  12  9  1  1  1. Yes. So the points 0 1, 2 1, and 2 3 are on the graph of this equation. ?

?

14. 0 1: 02  12  1  0  0  1  1  0. Yes.   2  2  ? ? 1  1 : 1  1  1  0  12  12  1  0. Yes. 2 2 2 2     2  2 ? ? 3 1 : 3  12  1  0  34  14  1  0. Yes. 2 2 2     So the points 0 1, 1  1 , and 23  12 are on the graph of this equation. 2

2

16. y  2x

15. y  3x x

y

x

y

3

9

3

6

2

6

2

4

1

3

1

2

0

0

0

0

1

3

1

2

2

6

2

4

3

9

3

6

SECTION 1.2 Graphs of Equations in Two Variables: Circles

17. y  2  x x

67

18. y  2x  3 y

x

y

4

6

4

5

2

4

2

1

0

2

0

3

2

0

2

7

4

2

4

11

19. Solve for y: 2x  y  6  y  2x  6.

20. Solve for x: x  4y  8  x  4y  8.

x

y

x

y

2

10

4

0

6

2

 52

2

2

0

2

4

2

2

6

6

4 6

21. y  1  x 2

3

 32

1

 12

8

0

10

1 2

22. y  x 2  2 y

y

x

y

x

y

3

8

3

11

2

3

2

6

1

0

1

3

0

1

0

2

1

0

1

3

2

3

2

6

3

8

3

11

1l 1

x

1l 1

x

68

CHAPTER 1 Equations and Graphs

23. y  x 2  2 x

24. y  x 2  4 y

x

y

3

7

3

5

2

2

2

0

1

1

1

3

0

2

0

4

1

1

1

3

2

2

2

0

3

7

3

5

25. 9y  x 2 . To make a table, we rewrite the equation as

26. 4y  x 2 . y

y  19 x 2 . y

x

y

x

y

4

4

9

9

2

1

1

0

0

3 0

0

2

1

3

1

4

4

9

9

p

1

1

x

1 x

2

27. x  y 2  4.

2 . x

28. x y  2  y  y

x

y

12

4

5

3

0

2

3

1

4

0

3

1

0

2

5

3

12

4

y

1l 1

x

x

y

4

 12

2

1

1

2

 12  14 1 4 1 2

1

4 8 8 4 2

2

1

4

1 2

1l 1

x

SECTION 1.2 Graphs of Equations in Two Variables: Circles

29. y 



x.

30. y  2 



69

x.

y

y

x

y

x

y

0

0

0

2

1 4

1 2

1

1

2

2

1  2

3  2 2

4

4

4

2

9

5

9

3

16

4

1l 1

x

1 x

1

 31. y   9  x 2 . Since the radicand (the expression inside

32. y 

 9  x 2.

the square root) cannot be negative, we must have

Since the radicand (the expression inside the square root)

9  x 2  0  x 2  9  x  3.

cannot be negative, we must have 9  x 2  0  x 2  9

x

y

3

0   5  2 2

2 1 0 1 2 3

 x  3. x

y

3

0  5  2 2

2

3  2 2   5

1 0 2

3  2 2  5

3

0

1

0

33. y   x.

34. x  y. In the table below, we insert values of y and find y

x

y

6

6

4

4

2

2

0

0

2

the corresponding value of x. y

x

y

3

3

2

2

1

1

2

0

0

4

4

1

1

6

6

2

2

3

3

1 1

x

1l 1

x

70

CHAPTER 1 Equations and Graphs

35. y  4  x.

36. y  4  x. y

y

x

y

x

y

6

2

6

10

4

0

4

8

2

2

2

6

0

4

0

4

2

2

2

2

4

0

4

0

6

2

6

2

8

4

10

6

1l 1

x

37. x  y 3 . Since x  y 3 is solved for x in terms of y, we

1l 1

38. y  x 3  1. y

insert values for y and find the corresponding values of x in the table below. y

x

y

3

28

x

y

2

9

27

3

1

2

8

2

0

1

1

1

1

1

0

0

2

7

1

1

3

26

8

2

27

3

39. y  x 4 .

1l 1

x

40. y  16  x 4 .

y

y

x

y

3

81 16

3

65

2 1

1

0

0

1 2 3

2

0

1

15

0

16

1

1

15

16

2

0

81

3

65

1

x

1l 1

x

1

x

y

x

3l

x

3l

SECTION 1.2 Graphs of Equations in Two Variables: Circles

41. y  001x 3  x 2  5; [100 150] by [2000 2000]

71

42. y  003x 2  17x  3; [100 50] by [50 100]

2000

100

1000 50 -100

-50

50

100

150

-1000

-100

-50

50

-2000

43. y 

-50

 12x  17; [1 10] by [1 20]

44. y 

20

 4 256  x 2 ; [20 20] by [2 6] 6 4

10

2

-20 0

2

4

6

8

-10

x ; [50 50] by [02 02] 45. y  2 x  25

46. y  x 4  4x 3 ; [4 6] by [50 100] 100

0.1

-20

50

20 -0.1 -0.2

20

-2

0.2

-40

10

10

40 -4

-2

2

4

6

-50

47. y  x  6. To find x-intercepts, set y  0. This gives 0  x  6  x  6, so the x-intercept is 6. To find y-intercepts, set x  0. This gives y  0  6  y  6, so the y-intercept is 6. 48. 2x  5y  40. To find x-intercepts, set y  0. This gives 2x  5 0  40  2x  40  x  20, and the x-intercept is 20. To find y-intercepts, set x  0. This gives 2 0  5y  40  y  8, so the y-intercept is 8.   49. y  x 2  5. To find x-intercepts, set y  0. This gives 0  x 2  5  x 2  5  x   5, so the x-intercepts are  5. To find y-intercepts, set x  0. This gives y  02  5  5, so the y-intercept is 5.

50. y 2  9  x 2 . To find x-intercepts, set y  0. This gives 02  9  x 2  x 2  9  x  3, so the x-intercepts are 3. To find y-intercepts, set x  0. This gives y 2  9  02  9  y  3, so the y-intercepts are 3.

51. y  2x y  2x  1. To find x-intercepts, set y  0. This gives 0  2x 0  2x  1  2x  1  x  is 12 . To find y-intercepts, set x  0. This gives y  2 0 y  2 0  1  y  1, so the y-intercept is 1.

1 , so the x-intercept 2

72

CHAPTER 1 Equations and Graphs

52. x 2  x y  y  1. To find x-intercepts, set y  0. This gives x 2  x 0  0  1  x 2  1  x  1, so the x-intercepts are 1 and 1. To find y-intercepts, set x  0. This gives y  02  0 y  y  1  y  1, so the y-intercept is 1. 

 x  1. To find x-intercepts, set y  0. This gives 0  x  1  0  x  1  x  1, so the x-intercept is 1.  To find y-intercepts, set x  0. This gives y  0  1  y  1, so the y-intercept is 1.

53. y 

54. x y  5. To find x-intercepts, set y  0. This gives x 0  5  0  5, which is impossible, so there is no x-intercept. To find y-intercepts, set x  0. This gives 0 y  5  0  5, which is again impossible, so there is no y-intercept. 55. 4x 2  25y 2  100. To find x-intercepts, set y  0. This gives 4x 2  25 02  100  x 2  25  x  5, so the x-intercepts are 5 and 5.

To find y-intercepts, set x  0. This gives 4 02  25y 2  100  y 2  4  y  2, so the y-intercepts are 2 and 2.

56. 25x 2  y 2  100. To find x-intercepts, set y  0. This gives 25x 2  02  100  x 2  4  x  2, so the x-intercepts are 2 and 2. To find y-intercepts, set x  0. This gives 25 02  y 2  100  y 2  100, which has no solution, so there is no y-intercept. 57. y  4x  x 2 . To find x-intercepts, set y  0. This gives 0  4x  x 2  0  x 4  x  0  x or x  4, so the x-intercepts are 0 and 4. To find y-intercepts, set x  0. This gives y  4 0  02  y  0, so the y-intercept is 0. 58.

y2 x2 02 x2 x2   1. To find x-intercepts, set y  0. This gives  1  1  x 2  9  x  3, so the 9 4 9 4 9 x-intercepts are 3 and 3. To find y-intercepts, set x  0. This gives

02 y 2 y2  1  1  y 2  4  x  2, so the y-intercepts are 2 and 2. 9 4 4

59. x 4  y 2  x y  16. To find x-intercepts, set y  0. This gives x 4  02  x 0  16  x 4  16  x  2. So the x-intercepts are 2 and 2.

To find y-intercepts, set x  0. This gives 04  y 2  0 y  16  y 2  16  y  4. So the y-intercepts are 4 and 4.

60. x 2  y 3  x 2 y 2  64. To find x-intercepts, set y  0. This gives x 2  03  x 2 02  64  x 2  64  x  8. So the x-intercepts are 8 and 8. To find y-intercepts, set x  0. This gives 02  y 3  02 y 2  64  y 3  64  y  4. So the y-intercept is 4.

61. (a) y  x 3  x 2 ; [2 2] by [1 1]

(b) From the graph, it appears that the x-intercepts are 0 and 1 and the y-intercept is 0.

1.0

(c) To find x-intercepts, set y  0. This gives

0.5

0  x 3  x 2  x 2 x  1  0  x  0 or 1. So

the x-intercepts are 0 and 1. -2

-1

1 -0.5 -1.0

2

To find y-intercepts, set x  0. This gives y  03  02  0. So the y-intercept is 0.

SECTION 1.2 Graphs of Equations in Two Variables: Circles

62. (a) y  x 4  2x 3 ; [2 3] by [3 3]

73

(b) From the graph, it appears that the x-intercepts are 0 and 2 and the y-intercept is 0.

3

(c) To find x-intercepts, set y  0. This gives

2

0  x 4  2x 3  x 3 x  2  0  x  0 or 2. So

1

the x-intercepts are 0 and 2. -2

-1

1

-1

2

3

To find y-intercepts, set x  0. This gives

y  04  2 03  0. So the y-intercept is 0.

-2 -3

2 ; [5 5] by [3 1] 63. (a) y   2 x 1

(b) From the graph, it appears that there is no x-intercept and the y-intercept is 2.

1

-4

(c) To find x-intercepts, set y  0. This gives

-1

2 0 2 , which has no solution. So there is no x 1 x-intercept.

-2

To find y-intercepts, set x  0. This gives

-2

2

4

2 y 2  2. So the y-intercept is 2. 0 1

-3

x 64. (a) y  2 ; [5 5] by [2 2] x 1

(b) From the graph, it appears that the x- and y-intercepts are 0.

2 1

-4

-2

2

4

-1

(c) To find x-intercepts, set y  0. This gives x 0 2  x  0. So the x-intercept is 0. x 1 To find y-intercepts, set x  0. This gives 0 y 2  0. So the y-intercept is 0. 0 1

-2

65. (a) y 

 3

x; [5 5] by [2 2]

(b) From the graph, it appears that and the x- and y-intercepts are 0.

2

(c) To find x-intercepts, set y  0. This gives 0 

1

-4

-2

 x  0. So the x-intercept is 0.

2 -1 -2

4

To find y-intercepts, set x  0. This gives  y  3 0  0. So the y-intercept is 0.

 3

x

74

CHAPTER 1 Equations and Graphs

66. (a) y 

 3 1  x 2 ; [5 5] by [5 3]

(b) From the graph, it appears that the x-intercepts are 1 and 1 and the y-intercept is 1.

2

-4

-2

2

4

(c) To find x-intercepts, set y  0. This gives  3 0  1  x 2  1  x 2  0  x  1. So the x-intercepts are 1 and 1. To find y-intercepts, set x  0. This gives  3 y  1  02  1. So the y-intercept is 1.

-2 -4

67. x 2  y 2  9 has center 0 0 and radius 3.

68. x 2  y 2  5 has center 0 0 and radius

y

 5.

y

1l

1l 1

1

x

69. x  32  y 2  16 has center 3 0 and radius 4. y

x

70. x 2  y  22  4 has center 0 2 and radius 2. y

1l

1l 1

1

x

x

71. x  32  y  42  25 has center 3 4 and radius 5. 72. x  12  y  22  36 has center 1 2 and y

radius 6.

y

1l 1 1l 1

x

x

SECTION 1.2 Graphs of Equations in Two Variables: Circles

75

73. Using h  3, k  2, and r  5, we get x  32  y  22  52  x  32  y  22  25.

74. Using h  1, k  3, and r  3, we get x  12  y  32  32  x  12  y  32  9.

75. The equation of a circle centered at the origin is x 2  y 2  r 2 . Using the point 4 7 we solve for r 2 . This gives 42  72  r 2  16  49  65  r 2 . Thus, the equation of the circle is x 2  y 2  65.

76. Using h  1 and k  5, we get x  12  y  52  r 2  x  12  y  52  r 2 . Next, using the point 4 6, we solve for r 2 . This gives 4  12  6  52  r 2  130  r 2 . Thus, an equation of the circle is

x  12  y  52  130.

  1  5 1  9 77. The center is at the midpoint of the line segment, which is   2 5. The radius is one half the diameter, 2 2    so r  12 1  52  1  92  12 36  64  12 100  5. Thus, an equation of the circle is x  22 y  52  52  x  22  y  52  25.

  1  7 3  5 78. The center is at the midpoint of the line segment, which is   3 1. The radius is one half the 2 2   diameter, so r  12 1  72  3  52  4 2. Thus, an equation of the circle is x  32  y  12  32.

79. Since the circle is tangent to the x-axis, it must contain the point 7 0, so the radius is the change in the y-coordinates. That is, r  3  0  3. So the equation of the circle is x  72  y  32  32 , which is x  72  y  32  9. 80. Since the circle with r  5 lies in the first quadrant and is tangent to both the x-axis and the y-axis, the center of the circle is at 5 5. Therefore, the equation of the circle is x  52  y  52  25.

81. From the figure, the center of the circle is at 2 2. The radius is the change in the y-coordinates, so r  2  0  2. Thus the equation of the circle is x  22  y  22  22 , which is x  22  y  22  4.

82. From the figure, the center of the circle is at 1 1. The radius is the distance from the center to the point 2 0. Thus    r  1  22  1  02  9  1  10, and the equation of the circle is x  12  y  12  10.  2  2  2  2 2 2  4y  4  y  1   42 83. Completing the square gives x 2  y 2  2x  4y  1  0  x 2  2x  2 2 2 2

 x 2  2x  1  y 2  4y  4  1  1  4  x  12  y  22  4. Thus, the center is 1 2, and the radius is 2.  2  2  2  2 84. Completing the square gives x 2  y 2  2x  2y  2  x 2  2x  2  y 2  2y  2  2  2  2 2 2 2 2

 x 2  2x  1  y 2  2y  1  2  1  1  x  12  y  12  4. Thus, the center is 1 1, and the radius is 2.  2  2  2  2 85. Completing the square gives x 2  y 2 4x 10y 13  0  x 2 4x  4  y 2 10y  10  13 42  10 2 2 2

 x 2  4x  4  y 2  10y  25  13  4  25  x  22  y  52  16. Thus, the center is 2 5, and the radius is 4.  2  2 86. Completing the square gives x 2  y 2  6y  2  0  x 2  y 2  6y  62  2  62  x 2  y 2  6y  9  2  9   x 2  y  32  7. Thus, the circle has center 0 3 and radius 7.  2  2 87. Completing the square gives x 2  y 2  x  0  x 2  x  12  y 2  12  x 2  x  14  y 2  14  2    x  12  y 2  14 . Thus, the circle has center  12  0 and radius 12 .

 2  2  2 88. Completing the square gives x 2  y 2  2x  y  1  0  x 2  2x  22  y 2  y  12  1  1  12  2    x 2  2x  1  y 2  y  14  14  x  12  y  12  14 . Thus, the circle has center 1  12 and radius 12 .

76

CHAPTER 1 Equations and Graphs

2 2  2   2  89. Completing the square gives x 2  y 2  12 x  12 y  18  x 2  12 x  12  y 2  12 y  12  18  12  12 2 2 2 2 2  2  1  y2  1 y  1  1  1  1  2  1  x  1  x 2  12 x  16  y  14  14 . Thus, the circle has center 2 16 8 16 16 8 4 4   1   1 and radius 1 . 4 4 2         1  0  x 2  1 x  12 2  y 2  2y  2 2   1  12 2  2 2 90. Completing the square gives x 2  y 2  12 x 2y  16 2 2 2 16 2 2 2    1 1 2  x  4  y  1  1. Thus, the circle has center  4  1 and radius 1.

91. Completing the square gives x 2  y 2  4x  10y  21  92. First divide by 4, then complete the square. This gives 2  2   2 4x 2  4y 2  2x  0  x 2  y 2  12 x  0  x 2  12 x 4  21   x 2  4x  42  y 2  10y  10 2 2  2  2   __y 2  0  x 2  12 x  12  y 2  12  10 2  x  22  y  52  21  4  25  50. 2 2 2  2     1 . Thus, the circle has center  1  0 x  14  y 2  16 Thus, the circle has center 2 5 and radius 50  5 2. 4 y

and radius 14 .

y 2

2

x

1 1

x

93. Completing the square gives x 2  y 2  6x  12y  45  0 94. x 2  y 2  16x  12y  200  0  2   2  x  32  y  62  45  9  36  0. Thus, the  y 2  12y  12 x 2  16x  16 2 2 center is 3 6, and the radius is 0. This is a degenerate 2  2  circle whose graph consists only of the point 3 6.  200  16  12  2 2 y

1

x  82  y  62  200  64  36  100. Since

completing the square gives r 2  100, this is not the equation of a circle. There is no graph. 1

x

95. x-axis symmetry: y  x 4  x 2  y  x 4  x 2 , which is not the same as y  x 4  x 2 , so the graph is not symmetric with respect to the x-axis. y-axis symmetry: y  x4  x2  x 4  x 2 , so the graph is symmetric with respect to the y-axis.

Origin symmetry: y  x4  x2  y  x 4  x 2 , which is not the same as y  x 4  x 2 , so the graph is not symmetric with respect to the origin. 96. x-axis symmetry: x  y4  y2  y 4  y 2 , so the graph is symmetric with respect to the x-axis.

y-axis symmetry: x  y 4  y 2 , which is not the same as x  y 4  y 2 , so the graph is not symmetric with respect to the y-axis.

Origin symmetry: x  y4  y2  x  y 4  y 2 , which is not the same as x  y 4  y 2 , so the graph is not symmetric with respect to the origin.

SECTION 1.2 Graphs of Equations in Two Variables: Circles

77

97. x-axis symmetry: y  x 3  10x  y  x 3  10x, which is not the same as y  x 3  10x, so the graph is not symmetric with respect to the x-axis. y-axis symmetry: y  x3  10 x  y  x 3  10x, which is not the same as y  x 3  10x, so the graph is not symmetric with respect to the y-axis. Origin symmetry: y  x3  10 x  y  x 3  10x  y  x 3  10x, so the graph is symmetric with respect to the origin. 98. x-axis symmetry: y  x 2  x  y  x 2  x, which is not the same as y  x 2  x, so the graph is not symmetric with respect to the x-axis. y-axis symmetry: y  x2  x  y  x 2  x, so the graph is symmetric with respect to the y-axis. Note that x  x.

Origin symmetry: y  x2  x  y  x 2  x  y  x 2  x, which is not the same as y  x 2  x, so the graph is not symmetric with respect to the origin. 99. x-axis symmetry: x 4 y4  x 2 y2  1  x 4 y 4  x 2 y 2  1, so the graph is symmetric with respect to the x-axis. y-axis symmetry: x4 y 4  x2 y 2  1  x 4 y 4  x 2 y 2  1, so the graph is symmetric with respect to the y-axis.

Origin symmetry: x4 y4  x2 y2  1  x 4 y 4  x 2 y 2  1, so the graph is symmetric with respect to the origin. 100. x-axis symmetry: x 2 y2  x y  1  x 2 y 2  x y  1, which is not the same as x 2 y 2  x y  1, so the graph is not symmetric with respect to the x-axis. y-axis symmetry: x2 y 2  x y  1  x 2 y 2  x y  1, which is not the same as x 2 y 2  x y  1, so the graph is not symmetric with respect to the y-axis. Origin symmetry: x2 y2  x y  1  x 2 y 2  x y  1, so the graph is symmetric with respect to the origin. 102. Symmetric with respect to the x-axis.

101. Symmetric with respect to the y-axis. y

0

y

x 0

103. Symmetric with respect to the origin.

104. Symmetric with respect to the origin. y

y

0

x

x 0

x

78

CHAPTER 1 Equations and Graphs

  105. x y  x 2  y 2  1 . This is the set of points inside (and on) the circle x 2  y 2  1.

  106. x y  x 2  y 2  4 . This is the set of points outside the circle x 2  y 2  4.

y

1l

y

2l

1

x

107. Completing the square gives x 2  y 2  4y  12  0     4 2 4 2  12    x 2  y 2  4y  2 2

2

x

108. This is the top quarter of the circle of radius 3. Thus, the area is 14 9  94 . y

x 2  y  22  16. Thus, the center is 0 2, and the

radius is 4. So the circle x 2  y 2  4, with center 0 0

3l

and radius 2 sits completely inside the larger circle. Thus, the area is 42  22  16  4  12.

3

x

109. (a) The point 5 3 is shifted to 5  3 3  2  8 5. (b) The point a b is shifted to a  3 b  2.

(c) Let x y be the point that is shifted to 3 4. Then x  3 y  2  3 4. Setting the x-coordinates equal, we get x  3  3  x  0. Setting the y-coordinates equal, we get y  2  4  y  2So the point is 0 2.

(d) A  5 1, so A  5  3 1  2  2 1; B  3 2, so B   3  3 2  2  0 4; and C  2 1, so C   2  3 1  2  5 3. (b) Symmetric about the y-axis.

110. (a) Symmetric about the x-axis.

(c) Symmetric about the origin.

y

y

y

1

1

1

1

x

0

1

x

1

x

111. (a) In 1980 inflation was 14%; in 1990, it was 6%; in 1999, it was 2%. (b) Inflation exceeded 6% from 1975 to 1976 and from 1978 to 1982. (c) Between 1980 and 1985 the inflation rate generally decreased. Between 1987 and 1992 the inflation rate generally increased. (d) The highest rate was about 14% in 1980. The lowest was about 1% in 2002. 112. (a) Closest: 2 Mm. Farthest: 8 Mm.

SECTION 1.3 Lines

79

x  32 1 x  32 22 x  32  1 4 1  34  x  32  75 4 . Taking the square 25 16 25 25      5 3 5 3 5 3 5 3 root of both sides we get x  3   75 4   2  x  3  2 . So x  3  2  133 or x  3  2  733.   The distance from 133 2 to the center 0 0 is d  133  02  2  02  57689  240. The distance   from 733 2 to the center 0 0 is d  733  02  2  02  577307  760.  2  a 2  a 2  b 2 b  y 2  by   c   113. Completing the square gives x 2  y 2  ax  by  c  0  x 2  ax  2 2 2 2    a 2  b2 a 2 b 2 a 2  b2 . This equation represents a circle only when c   0. This  x  y  c  2 2 4 4 (b) When y  2 we have

a 2  b2 a 2  b2  0, and this equation represents the empty set when c   0. 4 4     a b a 2  b2 When the equation represents a circle, the center is    , and the radius is c   12 a 2  b2  4ac. 2 2 4

equation represents a point when c 

114. (a) (i) x  22  y  12  9, the center is at 2 1, and the radius is 3. x  62 

y  42  16, the center is at 6 4, and the radius is 4. The distance between centers is     2  62  1  42  42  32  16  9  25  5. Since 5  3  4, these circles intersect.

(ii) x 2  y  22  4, the center is at 0 2, and the radius is 2. x  52  y  142  9, the center is at 5 14, and the radius is 3. The distance between centers is     0  52  2  142  52  122  25  144  169  13. Since 13  2  3, these circles do not intersect.

(iii) x  32 y  12  1, the center is at 3 1, and the radius is 1. x  22 y  22  25, the center is at 2 2,     and the radius is 5. The distance between centers is 3  22  1  22  12  32  1  9  10.  Since 10  1  5, these circles intersect.

(b) If the distance d between the centers of the circles is greater than the sum r1  r2 of their radii, then the circles do not intersect, as shown in the first diagram. If d  r1  r2 , then the circles intersect at a single point, as shown in the second diagram. If d  r1  r2 , then the circles intersect at two points, as shown in the third diagram.

rª CÁ rÁ

Cª

d

CÁ rÁ

rª

Cª

d

rÁ

rª

Cª

CÁ

d

Case 1 d  r1  r2

1.3

Case 2 d  r1  r2

Case 3 d  r1  r2

LINES

1. We find the “steepness” or slope of a line passing through two points by dividing the difference in the y-coordinates of these 51 points by the difference in the x-coordinates. So the line passing through the points 0 1 and 2 5 has slope  2. 20 2. (a) The line with equation y  3x  2 has slope 3. (b) Any line parallel to this line has slope 3.

80

CHAPTER 1 Equations and Graphs

(c) Any line perpendicular to this line has slope  13 .

3. The point-slope form of the equation of the line with slope 3 passing through the point 1 2 is y  2  3 x  1.

4. For the linear equation 2x  3y  12  0, the x-intercept is 6 and the y-intercept is 4. The equation in slope-intercept form is y   23 x  4. The slope of the graph of this equation is  23 .

5. The slope of a horizontal line is 0. The equation of the horizontal line passing through 2 3 is y  3.

6. The slope of a vertical line is undefined. The equation of the vertical line passing through 2 3 is x  2. 7. (a) Yes, the graph of y  3 is a horizontal line 3 units below the x-axis.

(b) Yes, the graph of x  3 is a vertical line 3 units to the left of the y-axis.

(c) No, a line perpendicular to a horizontal line is vertical and has undefined slope. (d) Yes, a line perpendicular to a vertical line is horizontal and has slope 0. y

8.

Yes, the graphs of y  3 and x  3 are perpendicular lines.

5l x=_3

0

x

5 y=_3

y  y1 9. m  2 x2  x1 y  y1 11. m  2 x2  x1 y  y1 13. m  2 x2  x1 y  y1 15. m  2 x2  x1

y  y1 10. m  2 x2  x1 y  y1 12. m  2 x2  x1 y  y1 14. m  2 x2  x1 y  y1 16. m  2 x2  x1

02 2   2 0  1 1 1 1  2   72 5 44 0  05 3 3 5  2    6  10 4 4

1  0 1 1   30 3 3 3 3 2  1    3  5 8 8 4 4 1  3    14 3 3 2  2 0  63 y  y1 20 17. For 1 , we find two points, 1 2 and 0 0 that lie on the line. Thus the slope of 1 is m  2  2.  x2  x1 1  0 y  y1 32 For 2 , we find two points 0 2 and 2 3. Thus, the slope of 2 is m  2   12 . For 3 we find the points x2  x1 20 y  y1 1  2  3. For 4 , we find the points 2 1 and  2 2 and 3 1. Thus, the slope of 3 is m  2 x 2  x1 32 1 1 y  y1 2  1 2 2. Thus, the slope of 4 is m  2   .  x2  x1 2  2 4 4 

y

18. (a)

m=2

m=1

(b)



y

m=3

1 m=_ 2

1 m=_ 2

m=0

1 m=_ 3

1l

1l 1

x

1

x 1 m=_ _ 3

m=_1

SECTION 1.3 Lines

81

04  1. Since the y-intercept is 4, 40 the equation of the line is y  mx  b  1x  4. So y  x  4, or x  y  4  0. 04 20. We find two points on the graph, 0 4 and 2 0. So the slope is m   2. Since the y-intercept is 4, the 2  0 equation of the line is y  mx  b  2x  4, so y  2x  4  2x  y  4  0. 0  3 21. We choose the two intercepts as points, 0 3 and 2 0. So the slope is m   32 . Since the y-intercept is 3, 20 the equation of the line is y  mx  b  32 x  3, or 3x  2y  6  0. 19. First we find two points 0 4 and 4 0 that lie on the line. So the slope is m 

22. We choose the two intercepts, 0 4 and 3 0. So the slope is m  equation of the line is y  mx  b   43 x  4  4x  3y  12  0.

0  4   43 . Since the y-intercept is 4, the 3  0

23. Using y  mx  b, we have y  3x  2 or 3x  y  2  0.

24. Using y  mx  b, we have y  25 x  4  2x  5y  20  0. 25. Using the equation y  y1  m x  x1 , we get y  3  5 x  2  5x  y  7  5x  y  7  0.

26. Using the equation y  y1  m x  x1 , we get y  4  1 x  2  y  4  x  2  x  y  2  0. 27. Using the equation y  y1  m x  x1 , we get y  7  23 x  1  3y  21  2x  2  2x  3y  19  2x  3y  19  0.

28. Using the equation y  y1  m x  x1 , we get y  5   72 x  3  2y  10  7x  21  7x  2y  31  0.

5 61 y  y1   5. Substituting into y  y1  m x  x1 , we get  29. First we find the slope, which is m  2 x2  x1 12 1 y  6  5 x  1  y  6  5x  5  5x  y  11  0. 3  2 y  y1  55  1. Substituting into y  y1  m x  x1 , we get  30. First we find the slope, which is m  2 x2  x1 4  1 y  3  1 x  4  y  3  x  4  x  y  1  0. 8 3  5 y  y1   8. Substituting  31. We are given two points, 2 5 and 1 3. Thus, the slope is m  2 x2  x1 1  2 1 into y  y1  m x  x1 , we get y  5  8 [x  2]  y  8x  11 or 8x  y  11  0. 77 y  y1  0. Substituting into  32. We are given two points, 1 7 and 4 7. Thus, the slope is m  2 x2  x1 41 y  y1  m x  x1 , we get y  7  0 x  1  y  7 or y  7  0. 3 3  0 y  y1   3. Using the y-intercept,  33. We are given two points, 1 0 and 0 3. Thus, the slope is m  2 x2  x1 01 1 we have y  3x  3 or y  3x  3 or 3x  y  3  0. 60 y  y1  68  34 . Using the y-intercept  34. We are given two points, 8 0 and 0 6. Thus, the slope is m  2 x2  x1 0  8 we have y  34 x  6  3x  4y  24  0. 35. Since the equation of a line with slope 0 passing through a b is y  b, the equation of this line is y  3.

36. Since the equation of a line with undefined slope passing through a b is x  a, the equation of this line is x  1.

37. Since the equation of a line with undefined slope passing through a b is x  a, the equation of this line is x  2.

38. Since the equation of a line with slope 0 passing through a b is y  b, the equation of this line is y  1.

39. Any line parallel to y  3x  5 has slope 3. The desired line passes through 1 2, so substituting into y  y1  m x  x1 , we get y  2  3 x  1  y  3x  1 or 3x  y  1  0. 1  2. The desired line passes through 3 2, so substituting 40. Any line perpendicular to y   12 x  7 has slope  12 into y  y1  m x  x1 , we get y  2  2 [x  3]  y  2x  8 or 2x  y  8  0.

82

CHAPTER 1 Equations and Graphs

41. Since the equation of a horizontal line passing through a b is y  b, the equation of the horizontal line passing through 4 5 is y  5. 42. Any line parallel to the y-axis has undefined slope and an equation of the form x  a. Since the graph of the line passes through the point 4 5, the equation of the line is x  4. 43. Since x  2y  6  2y  x  6  y   12 x  3, the slope of this line is  12 . Thus, the line we seek is given by y  6   12 x  1  2y  12  x  1  x  2y  11  0. 44. Since 2x  3y  4  0  3y  2x  4  y   23 x  43 , the slope of this line is m   23 . Substituting m   23 and b  6 into the slope-intercept formula, the line we seek is given by y   23 x  6  2x  3y  18  0. 45. Any line parallel to x  5 has undefined slope and an equation of the form x  a. Thus, an equation of the line is x  1. 46. Any line perpendicular to y  1 has undefined slope and an equation of the form x  a. Since the graph of the line passes through the point 2 6, an equation of the line is x  2. 47. First find the slope of 2x  5y  8  0. This gives 2x  5y  8  0  5y  2x  8  y   25 x  85 . So the 1  52 . The equation of the line we seek is slope of the line that is perpendicular to 2x  5y  8  0 is m   25 y  2  52 x  1  2y  4  5x  5  5x  2y  1  0.

48. First find the slope of the line 4x  8y  1. This gives 4x  8y  1  8y  4x  1  y  12 x  18 . So the slope of the     1  2. The equation of the line we seek is y   23  2 x  12 line that is perpendicular to 4x  8y  1 is m   12  y  23  2x  1  6x  3y  1  0.

49. First find the slope of the line passing through 2 5 and 2 1. This gives m  of the line we seek is y  7  1 x  1  x  y  6  0. 50. First find the slope of the line passing through 1 1 and 5 1. This gives m  of the line that is perpendicular is m   2x  y  7  0.

(_2, 1)

2 1  1    12 , and so the slope 51 4

1  2. Thus the equation of the line we seek is y  11  2 x  2  12

y

51. (a)

4 15   1, and so the equation 2  2 4

y

52. (a)

1l 1

x 1l 1

(b) y  1  32 x  2  2y  2  3 x  2  2y  2  3x  6  3x  2y  8  0.

(4, _1)

x

(b) y  1  2 x  4  y  1  2x  8  2x  y  7  0.

SECTION 1.3 Lines

53.

b=_1 b=_3

8

b=_6

4 _4

54.

m=1.5

4

m=0.75 _4

0

_2

2

_2

4

0

2

4

m=0.25 m=0 m=_0.25

_4

_4

b=6

_8

b=3 b=1

m=_0.75

_8

m=_1.5

b=0

y  mx  3, m  0, 025, 075, 15. Each of the

y  2x  b, b  0, 1, 3, 6. They have the same

lines contains the point 0 3 because the point 0 3

slope, so they are parallel.

satisfies each equation y  mx  3. Since 0 3 is on the y-axis, they all have the same y-intercept.

m=1.5

55. 4

_2

_2

m=6 y

56.

m=0.75

2 0

2

4

83

6

8

m=2 m=1

10

m=0.25 m=0

m=0.5

m=_0.25

_12

_8

_4

x m=0

0

4 m=_0.5

m=_0.75

_4

m=_1

m=_1.5

y  m x  3, m  0, 025, 075, 15. Each of the

lines contains the point 3 0 because the point 3 0

satisfies each equation y  m x  3. Since 3 0 is on the x-axis, we could also say that they all have the same

_10 m=_6

m=_2

y  2  m x  3, m  0, 05, 1, 2, 6. Each of

the lines contains the point 3 2 because the point 3 2 satisfies each equation y  2  m x  3.

x-intercept.

57. y  3  x  x  3. So the slope is 1 and the y-intercept is 3.

58. y  23 x  2. So the slope is 23 and the y-intercept is 2. y

y

1 1

1 1

x

x

84

CHAPTER 1 Equations and Graphs

59. 2x  y  7  y  2x  7. So the slope is 2 and the y-intercept is 7.

60. 2x  5y  0  5y  2x  y  25 x. So the slope is 2 and the y-intercept is 0. 5

y

y

1l 1

x

1 1

x

61. 4x  5y  10  5y  4x  10  y   45 x  2. So

62. 3x  4y  12  4y  3x  12  y  34 x  3. So the slope is 34 and the y-intercept is 3.

the slope is  45 and the y-intercept is 2.

y

y

1l

1l 1

1

x

x

63. y  4 can also be expressed as y  0x  4. So the slope is 64. x  5 cannot be expressed in the form y  mx  b. So 0 and the y-intercept is 4.

the slope is undefined, and there is no y-intercept. This is a

y

vertical line. y

1l 1

x

1l 1

x

SECTION 1.3 Lines

85

65. x  3 cannot be expressed in the form y  mx  b. So the 66. y  2 can also be expressed as y  0x  2. So the slope slope is undefined, and there is no y-intercept. This is a

is 0 and the y-intercept is 2. y

vertical line. y

1l 1

1l 1

x

x

67. 5x  2y  10  0. To find x-intercepts, we set y  0 and

solve for x: 5x  2 0  10  0  5x  10  x  2, so

68. 6x  7y  42  0. To find x-intercepts, we set y  0 and

solve for x: 6x  7 0  42  0  6x  42  x  7, so

the x-intercept is 2.

the x-intercept is 7.

To find y-intercepts, we set x  0 and solve for y:

To find y-intercepts, we set x  0 and solve for y:

y-intercept is 5.

y-intercept is 6.

5 0  2y  10  0  2y  10  y  5, so the y

1

6 0  7y  42  0  7y  42  y  6, so the y

2 1

x

2

x

86

CHAPTER 1 Equations and Graphs

69. 12 x  13 y  1  0. To find x-intercepts, we set y  0 and

70. 13 x  15 y  2  0. To find x-intercepts, we set y  0 and

solve for x: 12 x  13 0  1  0  12 x  1  x  2,

solve for x: 13 x  15 0  2  0  13 x  2  x  6, so

so the x-intercept is 2.

the x-intercept is 6.

1 0  1 y  1  0  1 y  1  y  3, so the 2 3 3

1 0  1 y  2  0  1 y  2  y  10, so the 3 5 5

To find y-intercepts, we set x  0 and solve for y:

To find y-intercepts, we set x  0 and solve for y:

y-intercept is 3.

y-intercept is 10.

y

1

y

2 1

2

x

71. y  6x  4. To find x-intercepts, we set y  0 and solve for x: 0  6x  4  6x  4  x   23 , so the x-intercept is  23 .

To find y-intercepts, we set x  0 and solve for y: y  6 0  4  4, so the y-intercept is 4.

x

72. y  4x  10. To find x-intercepts, we set y  0 and

solve for x: 0  4x  10  4x  10  x   52 , so

the x-intercept is  52 . To find y-intercepts, we set x  0 and solve for y:

y  4 0  10  10, so the y-intercept is 10. y

y

2 1

x

1 1

x

73. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation y  2x  3 has slope 2.

The line with equation 2y  4x  5  0  2y  4x  5  y  2x  52 also has slope 2, and so the lines are parallel. 74. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation y  12 x  4 has slope 12 .

1 , and so the lines are neither The line with equation 2x  4y  1  4y  2x  1  y   12 x  14 has slope  12   12

parallel nor perpendicular. 75. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 3x  4y  4 

4y  3x  4  y  34 x  1 has slope 34 . The line with equation 4x  3y  5  3y  4x  5  y   43 x  53 has 1 , and so the lines are perpendicular. slope  43   34

76. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 2x  3y  10 

2 2 7 3y  2x  10  y  23 x  10 3 has slope 3 . The line with equation 3y  2x  7  0  3y  2x  7  y  3 x  3 also has slope 23 , and so the lines are parallel.

SECTION 1.3 Lines

87

77. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 7x  3y  2 

3y  7x  2  y  73 x  23 has slope 73 . The line with equation 9y  21x  1  9y  21x  1  y   73  19 has 1 , and so the lines are neither parallel nor perpendicular. slope  73   73

78. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 6y  2x  5 

6y  2x  5  y  13 x  56 has slope 13 . The line with equation 2y  6x  1  2y  6x  1  y  3x  12 has 1 , and so the lines are perpendicular. slope 3   13

y

79. We first plot the points to find the pairs of points that determine each side. Next we 3 1 41   , and the find the slopes of opposite sides. The slope of AB is 71 6 2 10  7 3 1 slope of DC is   . Since these slope are equal, these two sides 5  1 6 2 are parallel. The slope of AD is

C D

6 71   3, and the slope of BC is 1  1 2

B

6 10  4   3. Since these slope are equal, these two sides are parallel. 57 2 Hence ABC D is a parallelogram.

1l

A

x

1

y

80. We first plot the points to determine the perpendicular sides. Next find the slopes of the sides. The slope of AB is

4 2 3  1   , and the slope of AC is 3  3 6 3

C

8  1 9 3    . Since 9  3 6 2

   slope of AB  slope of AC  23  32  1 the sides are perpendicular, and ABC is a right triangle.

A

81. We first plot the points to find the pairs of points that determine each side. Next we 31 2 1 find the slopes of opposite sides. The slope of AB is   and the 11  1 10 5 68 2 1 slope of DC is   . Since these slope are equal, these two sides 0  10 10 5 61 5 are parallel. Slope of AD is   5, and the slope of BC is 01 1 5 38   5. Since these slope are equal, these two sides are parallel. 11  10 1

B

1l 1

x

y

C D

1l

B A

1

x

Since slope of AB  slope of AD  15  5  1, the first two sides are each perpendicular to the second two sides. So the sides form a rectangle.

82. (a) The slope of the line passing through 1 1 and 3 9 is and 6 21 is

8 91   4. The slope of the line passing through 1 1 31 2

21  1 20   4. Since the slopes are equal, the points are collinear. 61 5

88

CHAPTER 1 Equations and Graphs

(b) The slope of the line passing through 1 3 and 1 7 is 1 3 and 4 15 is

4 73   2. The slope of the line passing through 1  1 2

15  3 12  . Since the slopes are not equal, the points are not collinear. 4  1 5

83. We need the slope and the midpoint of the line AB. The midpoint of AB is



17 42  2 2



 4 1, and the slope of

2  4 6 1 1   1. The slope of the perpendicular bisector will have slope   1. Using the 71 6 m 1 point-slope form, the equation of the perpendicular bisector is y  1  1 x  4 or x  y  3  0. AB is m 

84. We find the intercepts (the length of the sides). When x  0, we have 2y  3 0  6  0  2y  6  y  3, and when y  0, we have 2 0  3x  6  0  3x  6  x  2. Thus, the area of the triangle is 12 3 2  3.

85. (a) We start with the two points a 0 and 0 b. The slope of the line that contains them is

b b0   . So the equation 0a a

b of the line containing them is y   x  b (using the slope-intercept form). Dividing by b (since b  0) gives a x x y y    1    1. b a a b y x  1  4x  3y  24  4x  3y  24  0. (b) Setting a  6 and b  8, we get  6 8

86. (a) The line tangent at 3 4 will be perpendicular to the line passing through the points 0 0 and 3 4. The slope of 4 1 3 4  0   . Thus, the slope of the tangent line will be   . Then the equation of the tangent this line is 30 3 4 43 line is y  4  34 x  3  4 y  4  3 x  3  3x  4y  25  0.

(b) Since diametrically opposite points on the circle have parallel tangent lines, the other point is 3 4. 87. (a) The slope represents an increase of 002 C every year. The T -intercept is the average surface temperature in 1950, or 15 C. (b) In 2050, t  2050  1950  100, so T  002100  15  17 degrees Celsius. 88. (a) The slope is 00417D  00417 200  834. It represents the increase in dosage for each one-year increase in the child’s age. (b) When a  0, c  834 0  1  834 mg. 89. (a)

y

(b) The slope, 4, represents the decline in number of spaces sold for

each $1 increase in rent. The y-intercept is the number of spaces at the flea market, 200, and the x-intercept is the cost per space when the

200l

manager rents no spaces, $50. 100l

20

40

60

80 100

x

SECTION 1.3 Lines y

90. (a)

89

(b) The slope is the cost per toaster oven, $6. The y-intercept, $3000, is the monthly fixed cost—the cost that is incurred no matter how many toaster ovens are produced.

10,000l

5000l

500

1000

x

91. (a)

(b) Substituting a for both F and C, we have C F

30 22

20 4

10 14

0

10

20

30

32

50

68

86

a  95 a  32   45 a  32  a  40 . Thus both scales agree at

40 .

10 5 t  t1 80  70   . So the linear 92. (a) Using n in place of x and t in place of y, we find that the slope is 2  n2  n1 168  120 48 24 5 n  168  t  80  5 n  35  t  5 n  45. equation is t  80  24 24 24

5 150  45  7625 F  76 F. (b) When n  150, the temperature is approximately given by t  24

93. (a) Using t in place of x and V in place of y, we find the slope of the line

y

(b)

using the points 0 4000 and 4 200. Thus, the slope is

4000l

200  4000 3800 m   950. Using the V -intercept, the 40 4 linear equation is V  950t  4000.

3000l 2000l

(c) The slope represents a decrease of $950 each year in the value of the 1000l

computer. The V -intercept represents the cost of the computer. (d) When t  3, the value of the computer is given by

1

V  950 3  4000  1150.

94. (a) We are given

434 change in pressure   0434. Using P for 10 feet change in depth 10

pressure and d for depth, and using the point P  15 when d  0, we have P  15  0434 d  0  P  0434d  15.

(c) The slope represents the increase in pressure per foot of descent. The y-intercept represents the pressure at the surface. (d) When P  100, then 100  0434d  15  0434d  85  d  1959 ft. Thus the pressure is 100 lb/in3 at a depth of

approximately 196 ft.

(b)

2

3

4

x

y 60l 50l 40l 30l 20l 10l 10

20

30

40

50

60 x

95. The temperature is increasing at a constant rate when the slope is positive, decreasing at a constant rate when the slope is negative, and constant when the slope is 0.

90

CHAPTER 1 Equations and Graphs

96. We label the three points A, B, and C. If the slope of the line segment AB is equal to the slope of the line segment BC, then the points A, B, and C are collinear. Using the distance formula, we find the distance between A and B, between B and C, and between A and C. If the sum of the two smaller distances equals the largest distance, the points A, B, and C are collinear. Another method: Find an equation for the line through A and B. Then check if C satisfies the equation. If so, the points are collinear.

1.4

SOLVING QUADRATIC EQUATIONS

1. (a) The Quadratic Formula states that x 

b 

(b) In the equation 12 x 2  x  4  0, a      1  12  4 12 4   x  2 12

 b2  4ac . 2a

1 , b  1, and c  4. So, the solution of the equation is 2

13  2 or 4. 1

2. (a) To solve the equation x 2  4x  5  0 by factoring, we write x 2  4x  5  x  5 x  1  0 and use the Zero-Product Property to get x  5 or x  1.  2 (b) To solve by completing the square, we add 5 to both sides to get x 2  4x  5, and then add  42 to both sides to get x 2  4x  4  5  4  x  22  9  x  2  3  x  5 or x  1.

(c) To solve using the Quadratic Formula, we substitute a  1, b  4, and c  5, obtaining    4  42  4 1 5 4  36   2  3  x  5 or x  1. x 2 1 2 3. For the quadratic equation ax 2  bx  c  0 the discriminant is D  b2  4ac. If D  0, the equation has two real solutions; if D  0, the equation has one real solution; and if D  0, the equation has no real solution.

4. There are many possibilities. For example, x 2  1 has two solutions, x 2  0 has one solution, and x 2  1 has no solution. 5. x 2  8x  15  0  x  3 x  5  0  x  3  0 or x  5  0. Thus, x  3 or x  5. 6. x 2  5x  6  0  x  3 x  2  0  x  3  0 or x  2  0. Thus, x  3 or x  2.

7. x 2  x  6  x 2  x  6  0  x  2 x  3  0  x  2  0 or x  3  0. Thus, x  2 or x  3.

8. x 2  4x  21  x 2  4x  21  0  x  3 x  7  0  x  3  0 or x  7  0. Thus, x  3 or x  7.

9. 5x 2  9x  2  0  5x  1 x  2  0  5x  1  0 or x  2  0. Thus, x   15 or x  2.

10. 6x 2  x  12  0  3x  4 2x  3  0  3x  4  0 or 2x  3  0. Thus, x   43 or x  32 . 11. 2s 2  5s  3  2s 2  5s  3  0  2s  1 s  3  0  2s  1  0 or s  3  0. Thus, s   12 or s  3.

12. 4y 2  9y  28  4y 2  9y  28  0  4y  7 y  4  0  4y  7  0 or y  4  0. Thus, y   74 or y  4.

13. 12z 2  44z  45  12z 2  44z  45  0  6z  5 2z  9  0  6z  5  0 or 2z  9  0. Thus, z   56 or z  92 . 14. 42  4  3  42  4  3  0  2  1 2  3  0  2  1  0 or 2  3  0. If 2  1  0, then    12 ; if 2  3  0, then   32 .

15. x 2  5 x  100  x 2  5x  500  x 2  5x  500  0  x  25 x  20  0  x  25  0 or x  20  0. Thus, x  25 or x  20. 16. 6x x  1  21  x  6x 2  6x  21  x  6x 2  5x  21  0  2x  3 3x  7  0  2x  3  0 or 3x  7  0. If 2x  3  0, then x   32 ; if 3x  7  0, then x  73 .

SECTION 1.4 Solving Quadratic Equations

91

  17. x 2  8x  1  0  x 2  8x  1  x 2  8x  16  1  16  x  42  15  x  4   15  x  4  15.   18. x 2  6x  2  0  x 2  6x  2  x 2  6x  9  2  9  x  32  11  x  3   11  x  3  11.   19. x 2  6x  11  0  x 2  6x  11  x 2  6x  9  11  9  x  32  20  x  3  2 5  x  3  2 5. 2  3 3 20. x 2  3x  74  0  x 2  3x  74  x 2  3x  94  74  94  x  32  16 4  4  x  2  2  x   2  2  x  12 or x   72 .

2  3  1  x  12  1  x   12  1. So 21. x 2  x  34  0  x 2  x   x 2  x  14  34  14  x  12 4 x   12  1   32 or x   12  1  12 .     25  x  5 2  21  x  5   21   21   1  22. x 2  5x  1  0  x 2  5x  1  x 2  5x  25 4 4 2 4 2 4 2 

x  52  221 .

23. x 2  22x  21  0  x 2  22x  21  x 2  22x  112  21  112  21  121  x  112  100  x  11  10  x  11  10. Thus, x  1 or x  21.

24. x 2  18x  19  x 2  18x  92  19  92  19  81  x  92  100  x  9  10  x  9  10, so x  1 or x  19.  12 25. 5x 2  10x  7  0  x 2  2x  75  0  x 2  2x  75  x 2  2x  1  75  1  x  12  12  x  1   5 5 

 x  1  2 515

26. 2x 2  16x  5  0  x 2  8x  52  0  x 2  8x   52  x 2  8x  16   52  16  x  42  27 2    3 6 x  4   27 2  x  4  2 .   49 7 2  17  27. 2x 2  7x  4  0  x 2  72 x  2  0  x 2  72 x  2  x 2  72 x  49 16  2  16  x  4 16   17 7 17 7 x  4   16  x   4  4 .    25  x  5 2  153  x  5   153 28. 4x 2  5x  8  0  x 2  54 x  2  0  x 2  54 x  2  x 2  54 x  25  2  64 64 8 64 8 64 

 x   58  3 817 .

29. x 2  8x  12  0  x  2 x  6  0  x  2 or x  6.

30. x 2  3x  18  0  x  3 x  6  0  x  3 or x  6.

31. x 2  8x  20  0  x  10 x  2  0  x  10 or x  2.

32. 10x 2  9x  7  0  5x  7 2x  1  0  x   75 or x  12 . 33. 2x 2  x  3  0  x  1 2x  3  0  x  1  0 or 2x  3  0. If x  1  0, then x  1; if 2x  3  0, then x   32 .

34. 3x 2  7x  4  0  3x  4 x  1  0  3x  4  0 or x  1  0. Thus, x   43 or x  1.

 35. 3x 2  6x  5  0  x 2  2x  53  0  x 2  2x  53  x 2  2x  1  53  1  x  12  83  x  1   83  

x  1  2 3 6 .

 0  36. x 2  6x  1       6  62  4 1 1  6  36  4 6  32 64 2 b  b2  4ac      3  2 2. x 2a 2 1 2 2 2 37. x 2  43 x  49  0  9x 2  12x  4  0  3x  22  0  x  23 .

92

CHAPTER 1 Equations and Graphs



6  b  b2  4ac  38. 2x 2 3x  12  0  4x 2 6x 1  0  x  2a

   6  52 3  13 62  4 4 1   . 2 4 8 4

39. 4x 2  16x  9  0  2x  1 2x  9  0  2x  1  0 or 2x  9  0. If 2x  1  0, then x  12 ; if 2x  9  0, then x   92 . x 2  4x  1  0        4  42  4 1 1  b  b2  4ac 4  16  4 4  12 42 3 x      2  3. 2a 2 1 2 2 2     3  32  4 1 3 3  9  12 3  3   . Since the 41. 2  3   1  2  3  3  0    2 1 2 2 discriminant is less than 0, the equation has no real solution.     2  4ac  5  52  4 1 3 5  25  12 5  13 b  b 2    . 42. 3  5z  z  0  z  2a 2 1 2 2 40. 0



43. 10y 2  16y  5  0        16  162  4 10 5 b  b2  4ac 16  256  200 16  56 8  14 x     . 2a 2 10 20 20 10 44. 25x 2  70x  49  0  5x  72  0  5x  7  0  5x  7  x   75 .     2   2 22  4 3 2 2  4  24 2  20 b  b  4ac    . Since the 45. 3x 2  2x  2  0  x  2a 2 3 6 6 discriminant is less than 0, the equation has no real solution.     2  4ac  7  72  4 5 5 7  49  100 7  51 b  b 2    . 46. 5x  7x  5  x  2a 2 5 10 10 Since the discriminant is less than 0, the equation has no real solution. 47. x 2  0011x  0064  0     0011  00112  4 1 0064 0011  0000121  0256 0011  0506 x   . 2 1 2 2 0011  0506 0011  0506 Thus, x   0259 or x   0248. 2 2 48. x 2  2450x  1500  0      2450  24502  4 1 1500 2450  60025  6 2450  00025 2450  0050 x    . Thus, 2 1 2 2 2 2450  0050 2450  0050 x  1250 or x   1200. 2 2 49. x 2  2450x  1501  0      2450  24502  4 1 1501 2450  60025  6004 2450  00015 x   . 2 1 2 2 Thus, there is no real solution. 50. x 2  1800x  0810  0      1800  18002  4 1 0810 1800  324  324 1800  0 x     0900. Thus the only 2 1 2 2 solution is x  0900.

SECTION 1.4 Solving Quadratic Equations

51. h 

93

1 gt 2   t  1 gt 2   t  h 0 0 2 2

 0. Using the Quadratic Formula,       0    0 2  4 12 g h  0   02  2gh   t .  g 2 12 g

n n  1  2S  n 2  n  n 2  n  2S  0. 2   1  12  4 1 2S 1  1  8S n  . 2 1 2

52. S 

Using the Quadratic Formula,

53. A  2x 2  4xh  2x 2  4xh  A  0. Using the Quadratic Formula,       4h  4 4h 2  2A  4h  4h2  4 2 A 4h  16h 2  8A 4h  2 4h 2  2A    x  2 2 4 4 4     2 2h  4h 2  2A 2h  4h 2  2A   4 2 54. A  2r 2  2r h  2r 2  2r h  A  0. Using the Quadratic Formula,     2h  2h2  4 2 A 2h  4 2 h 2  8 A h   2 h 2  2 A   . r 2 2 4 2 1 1 1    c s  b  c s  a  s  a s  b  cs  bc  cs  ac  s 2  as  bs  ab  55. sa s b c s 2  a  b  2c s  ab  ac  bc  0. Using the Quadratic Formula,   a  b  2c  a  b  2c2  4 1 ab  ac  bc s  2 1   a  b  2c  a 2  b2  4c2  2ab  4ac  4bc  4ab  4ac  4bc  2   a  b  2c  a 2  b2  4c2  2ab  2     2 4 2 1 4 1  2  r 2 1  r   r 2 1  r  2  r 1  r  2r 2  4 1  r  r  r 2  2r 2  4  4r 56.  r 1r r 1r r r     5  52  4 1 4 5  25  16 5  41 2  r  5r  4  0. Using the Quadratic Formula, r    . 2 1 2 2 57. D  b2  4ac  62  4 1 1  32. Since D is positive, this equation has two real solutions.

58. x 2  6x  9  x 2  6x  9, so D  b2  4ac  62  4 1 9  36  36  0. Since D  0, this equation has one real solution. 59. D  b2  4ac  2202  4 1 121  484  484  0. Since D  0, this equation has one real solution. 60. D  b2  4ac  2212  4 1 121  48841  484  00441. Since D  0, this equation has two real solutions.   61. D  b2  4ac  52  4 4 13 8  25  26  1. Since D is negative, this equation has no real solution.

62. D  b2  4ac  r2  4 1 s  r 2  4s. Since D is positive, this equation has two real solutions. 1 63. a 2 x 2  2ax  1  0  ax  12  0  ax  1  0. So ax  1  0 then ax  1  x   . a 64. ax 2  2a  1 x  a  1  0  [ax  a  1] x  1  0  ax  a  1  0 or x  1  0. If ax  a  1  0, a1 then x  ; if x  1  0, then x  1. a

94

CHAPTER 1 Equations and Graphs

65. We want to find the values of k that make the discriminant 0. Thus k 2  4 4 25  0  k 2  400  k  20

66. We want to find the values of k that make the discriminant 0. Thus D  362  4 k k  0  4k 2  362  2k  36  k  18.

67. Let n be one number. Then the other number must be 55  nsince n  55  n  55. Because

the product is 684, we have n 55  n  684  55n  n 2  684  n 2  55n  684  0 

   55 552 41684 5517  72  36 or n   55 30252736  552 289  5517 21 2 2 . So n  2 2 5517 38 n  2  2  19. In either case, the two numbers are 19 and 36.

68. Let n be one even number. Then the next even number is n  2. Thus we get the equation n 2  n  22  1252    n 2  n 2  4n  4  1252  0  2n 2  4n  1248  2 n 2  2n  624  2 n  24 n  26. So n  24 or n  26. Thus the consecutive even integers are 24 and 26 or 26 and 24.

69. Let  be the width of the garden in feet. Then the length is   10. Thus 875     10  2  10  875  0    35   25  0. So   35  0 in which case   35 which is not possible, or   25  0 and so   25. Thus the width is 25 feet and the length is 35 feet. 70. Let  be the width of the bedroom. Then its length is   7. Since area is length times width, we have

228    7   2  7  2  7  228  0    19   12  0    19  0 or   12  0. Thus   19 or   12. Since the width must be positive, the width is 12 feet.

71. Let  be the width of the garden in feet. We use the perimeter to express the length l of the garden in terms of width. Since the perimeter is twice the width plus twice the length, we have 200  2  2l  2l  200  2  l  100  . Using the formula for area, we have 2400   100    100  2  2  100  2400  0    40   60  0. So   40  0    40, or   60  0    60. If   40, then l  100  40  60. And if   60, then l  100  60  40. So the length is 60 feet and the width is 40 feet.

72. First we write a formula for the area of the figure in terms of x. Region A has dimensions 14 in. and x in. and region B has dimensions 13  x in. and x in. So

the area of the figure is 14  x  [13  x x]  14x  13x  x 2  x 2  27x. We

are given that this is equal to 160 in2 , so 160  x 2  27x  x 2  27x  160  0

x A

14 in. 13 in. B

x

 x  32 x  5  x  32 or x  5. x must be positive, so x  5 in.

73. The shaded area is the sum of the area of a rectangle and the area of a triangle. So A  y 1  12 y y  12 y 2  y. We

are given that the area is 1200 cm2 , so 1200  12 y 2  y  y 2  2y  2400  0  y  50 y  48  0. y is positive, so y  48 cm.

1 x 300  x  30x  1 x 2  1 x 2  30x  1250  0. 74. Setting P  1250 and solving for x, we have 1250  10 10 10    1 1250   30  302  4 10 30  900  500 30  20     . Thus Using the Quadratic Formula, x  1 02 02 2 10

30  20 30  20  50 or x   250. Since he must have 0  x  200, he should make 50 ovens per week. 02 02 75. Let x be the length of one side of the cardboard, so we start with a piece of cardboard x by x. When 4 inches are x

removed from each side, the base of the box is x  8 by x  8. Since the volume is 100 in3 , we get 4 x  82  100 

x 2  16x  64  25  x 2  16x  39  0  x  3 x  13  0. So x  3 or x  13. But x  3 is not possible, since then the length of the base would be 3  8  5 and all lengths must be positive. Thus x  13, and the piece of cardboard is 13 inches by 13 inches.

SECTION 1.4 Solving Quadratic Equations

95

76. Let r be the radius of the can. Now using the formula V  r 2 h with V  40 cm3 and h  10, we solve for r. Thus 40  r 2 10  4  r 2  r  2. Since r represents radius, r  0. Thus r  2, and the diameter is 4 cm.

77. Let  be the width of the lot in feet. Then the length is   6. Using the Pythagorean Theorem, we have

2    62  1742  2  2  12  36  30,276  22  12  30240  0  2  6  15120  0    126   120  0. So either   126  0 in which case   126 which is not possible, or   120  0 in which case   120. Thus the width is 120 feet and the length is 126 feet.

78. Let h be the height of the flagpole, in feet. Then the length of each guy wire is h  5. Since the distance between the points where the wires are fixed to the ground is equal to one guy wire, the triangle is equilateral, and the flagpole is the perpendicular bisector of the base. Thus from the Pythagorean Theorem, we get   1 h  5 2  h 2  h  52  h 2  10h  25  4h 2  4h 2  40h  100  h 2  30h  75  0  2

     30 302 4175 3 . Since h  3020 3  0, we reject it. Thus  30 900300  30 2 1200  3020 h 21 2 2 2  3  15  103  3232 ft  32 ft 4 in. the height is h  3020 2

79. Let x be the rate, in mi/h, at which the salesman drove between Ajax and Barrington. Direction

Distance

Rate

Ajax  Barrington

120

x

Barrington  Collins

150

x  10

Time 120 x 150 x  10

distance to fill in the “Time” column of the table. Since the second part of the trip rate 1 hour) more than the first, we can use the time column to get the equation 120  1  150  took 6 minutes (or 10 x 10 x  10 120 10 x  10  x x  10  150 10x  1200x  12,000  x 2  10x  1500x  x 2  290x  12,000  0  We have used the equation time 

   290 2902 4112,000 290 84,10048,000 290 36,100    290190  145  95. Hence, the salesman x 2 2 2 2

drove either 50 mi/h or 240 mi/h between Ajax and Barrington. (The first choice seems more likely!) 80. Let x be the rate, in mi/h, at which Kiran drove from Tortula to Cactus. Direction

Distance

Rate

Tortula  Cactus

250

x

Cactus  Dry Junction

360

x  10

Time 250 x 360 x  10

distance to fill in the time column of the table. We are given that the sum of rate 360 250   11  250 x  10  360x  the times is 11 hours. Thus we get the equation x x  10 11x x  10  250x  2500  360x  11x 2  110x  11x 2  500x  2500  0      500  5002  4 11 2500 500  250,000  110,000 500  360,000 500  600    . Hence, x 2 11 22 22 22 Kiran drove either 454 mi/h (impossible) or 50 mi/h between Tortula and Cactus. We have used time 

96

CHAPTER 1 Equations and Graphs

81. Let r be the rowing rate in km/h of the crew in still water. Then their rate upstream was r  3 km/h, and their rate downstream was r  3 km/h. Direction

Distance

Rate

Upstream

6

r 3

Downstream

6

r 3

Time 6 r 3 6 r 3

Since the time to row upstream plus the time to row downstream was 2 hours 40 minutes  83 hour, we get the equation

6 8 6    6 3 r  3  6 3 r  3  8 r  3 r  3  18r  54  18r  54  8r 2  72  r 3 r 3 3   0  8r 2  36r  72  4 2r 2  9r  18  4 2r  3 r  6  2r  3  0 or r  6  0. If 2r  3  0, then r   32 ,

which is impossible because the rowing rate is positive. If r  6  0, then r  6. So the rate of the rowing crew in still water is 6 km/h.

82. Let r be the speed of the southbound boat. Then r  3 is the speed of the eastbound boat. In two hours the southbound boat has traveled 2r miles and the eastbound boat has traveled 2 r  3  2r  6 miles. Since they are traveling is directions with are 90 apart, we can use the Pythagorean Theorem to get 2r 2  2r  62  302  4r 2  4r 2  24r  36  900    8r 2  24r  864  0  8 r 2  3r  108  0  8 r  12 r  9  0. So r  12 or r  9. Since speed is positive, the speed of the southbound boat is 9 mi/h.

83. Using h 0  288, we solve 0  16t 2  288, for t  0. So 0  16t 2  288  16t 2  288  t 2  18     t   18  3 2. Thus it takes 3 2  424 seconds for the ball the hit the ground.

84. (a) Using h 0  96, half the distance is 48, so we solve the equation 48  16t 2  96  48  16t 2  3  t 2    t   3. Since t  0, it takes 3  1732 s.  (b) The ball hits the ground when h  0, so we solve the equation 0  16t 2  96  16t 2  96  t 2  6  t   6.  Since t  0, it takes 6  2449 s. 85. We are given  o  40 ft/s.

  (a) Setting h  24, we have 24  16t 2  40t  16t 2  40t  24  0  8 2t 2  5t  3  0  8 2t  3 t  1  0  t  1 or t  1 12 . Therefore, the ball reaches 24 feet in 1 second (ascending) and again after 1 12 seconds (descending).

(b) Setting h  48, we have 48  16t 2  40t  16t 2  40t  48  0  2t 2  5t  6  0    5  25  48 5  23 t  . However, since the discriminant D  0, there is no real solution, and hence the ball 4 4 never reaches a height of 48 feet. (c) The greatest height h is reached only once. So h  16t 2  40t  16t 2  40t  h  0 has only one solution. Thus D  402  4 16 h  0  1600  64h  0  h  25. So the greatest height reached by the ball is 25 feet.

(d) Setting h  25, we have 25  16t 2  40t  16t 2  40t  25  0  4t  52  0  t  1 14 . Thus the ball reaches the highest point of its path after 1 14 seconds. (e) Setting h  0 (ground level), we have 0  16t 2  40t  2t 2  5t  0  t 2t  5  0  t  0 (start) or t  2 12 . So the ball hits the ground in 2 12 s.

86. If the maximum height is 100 feet, then the discriminant of the equation, 16t 2   o t  100  0, must equal zero. So

0  b2  4ac   o 2  4 16 100   o2  6400   o  80. Since  o  80 does not make sense, we must have  o  80 ft/s.

SECTION 1.4 Solving Quadratic Equations

97

  87. (a) The fish population on January 1, 2002 corresponds to t  0, so F  1000 30  17 0  02  30 000. To find

when the population will again reach this value, we set F  30 000, giving   30000  1000 30  17t  t 2  30000  17000t  1000t 2  0  17000t  1000t 2  1000t 17  t  t  0 or

t  17. Thus the fish population will again be the same 17 years later, that is, on January 1, 2019.   (b) Setting F  0, we have 0  1000 30  17t  t 2  t 2  17t  30  0    17  289  120 17  409 17  2022 t    . Thus t  1612 or t  18612. Since 2 2 2 t  0 is inadmissible, it follows that the fish in the lake will have died out 18612 years after January 1, 2002, that is on August 12, 2020. 88. Let y be the circumference of the circle, so 360  y is the perimeter of the square. Use the circumference to find the  2 radius, r, in terms of y: y  2r  r  y 2. Thus the area of the circle is  y 2  y 2  4. Now if the  2 perimeter of the square is 360  y, the length of each side is 14 360  y  and the area of the square is 14 360  y .  2      Setting these areas equal, we obtain y 2  4  14 360  y  y 2   14 360  y  2y  360   y          2   y  360 . Therefore, y  360  2    1691. Thus one wire is 1691 in. long and the other is 1909 in. long. 89. Let  be the uniform width of the lawn. With  cut off each end, the area of the factory is 240  2 180  2. Since

the lawn and the factory are equal in size this area, is 12  240  180. So 21,600  43,200  480  360  42    0  42  840  21,600  4 2  210  5400  4   30   180    30 or   180. Since 180 ft is too wide, the width of the lawn is 30 ft, and the factory is 120 ft by 180 ft.

2   2 90. Let h be the height the ladder reaches (in feet). Using the Pythagorean Theorem we have 7 12  h 2  19 12   2  2  2  2  15 225 1296  h 2  39  h 2  39  15  1521 2 4 4 2 4  4  4  324. So h  324  18 feet. 91. Let t be the time, in hours it takes Irene to wash all the windows. Then it takes Henry t  32 hours to wash all the windows, and the sum of the fraction of the job per hour they can do individually equals the fraction of the 1 1 1 4 9  9  job they can do together. Since 1 hour 48 minutes  1  48 60  1  5  5 , we have t  3 t  2

5

2 1   59  9 2t  3  2 9t  5t 2t  3  18t  27  18t  10t 2  15t  10t 2  21t  27  0 t 2t  3    21  212  4 10 27 21  441  1080 21  39 21  39 9 t    . So t    2 10 20 20 20 10 21  39 or t   3. Since t  0 is impossible, all the windows are washed by Irene alone in 3 hours and by Henry alone in 20 3  32  4 12 hours.

92. Let t be the time, in hours, it takes Kay to deliver all the flyers alone. Then it takes Lynn t  1 hours to deliver all the flyers 1 1 1 1 1   14 04t 04t alone, and it takes the group 04t hours to do it together. Thus 14   04t  1 t t 1 04t t t 1 4t t 4  10  t t  1  4 t  1  4t  10 t  1  t 2  t  4t  4  4t  10t  10  t 2  t  6  0  t 1 t  3 t  2  0. So t  3 or t  2. Since t  2 is impossible, it takes Kay 3 hours to deliver all the flyers alone.

98

CHAPTER 1 Equations and Graphs

93. Let x be the distance from the center of the earth to the dead spot (in thousands of miles). Now setting 0012K K 0012K K  2   K 239  x2  0012K x 2  F  0, we have 0   2  x x 239  x2 239  x2

57121  478x  x 2  0012x 2  0988x 2  478x  57121  0. Using the Quadratic Formula, we obtain

   478 4782 4098857121 2741808  47852362  241903  26499. x  478 228484225742192  4781976 20988 1976 1976

So either x  241903  26499  268 or x  241903  26499  215. Since 268 is greater than the distance from the earth to the moon, we reject it; thus x  215,000 miles.

94. If we have x 2  9x  20  x  4 x  5  0, then x  4 or x  5, so the roots are 4 and 5. The product is 4  5  20, and

the sum is 4  5  9. If we have x 2  2x  8  x  4 x  2  0, then x  4 or x  2, so the roots are 4 and 2. The

product is 42  8, and the sum is 42  2. Lastly, if we have x 2 4x 2  0, then using the Quadratic Formula,       4  42  4 1 2 4  8 4  2 2 we have x     2  2. The roots are 2  2 and 2  2. The 2 1 2 2             product is 2  2  2  2  4  2  2, and the sum is 2  2  2  2  4. In general, if x  r1

and x  r2 are roots, then x 2  bx  c  x  r1  x  r2   x 2  r1 x  r2 x  r1r2  x 2  r1  r2  x  r1r2 . Equating the coefficients, we get c  r1r2 and b   r1  r2 . 95. Let x equal the original length of the reed in cubits. Then x  1 is the piece that fits 60 times along the length of the field, that is, the length is 60 x  1. The width is 30x. Then converting cubits to ninda, we have 2 2 375  60 x  1  30x  12  25 2 x x  1  30  x  x  x  x  30  0  x  6 x  5  0. So x  6 or 12 x  5. Since x must be positive, the original length of the reed is 6 cubits.

1.5

COMPLEX NUMBERS

1. The imaginary number i has the property that i 2  1. 2. For the complex number 3  4i the real part is 3 and the imaginary part is 4. 3. (a) The complex conjugate of 3  4i is 3  4i  3  4i.   (b) 3  4i 3  4i  32  42  25

4. If 3  4i is a solution of a quadratic equation with real coefficients, then 3  4i  3  4i is also a solution of the equation. 5. Yes, every real number a is a complex number of the form a  0i. 6. Yes. For any complex number z, z  z  a  bi  a  bi  a  bi  a  bi  2a, which is a real number. 7. 5  7i: real part 5, imaginary part 7. 2  5i 9.   23  53 i: real part  23 , imaginary part  53 . 3

8. 6  4i: real part 6, imaginary part 4. 10.

4  7i  2  72 i: real part 2, imaginary part 72 . 2

13.  23 i: real part 0, imaginary part  23 .     15. 3  4  3  2i: real part 3, imaginary part 2.

12.  12 : real part  12 , imaginary part 0.   14. i 3: real part 0, imaginary part 3.    16. 2  5  2  i 5: real part 2, imaginary part  5.

17. 3  2i  5i  3  2  5 i  3  7i

18. 3i  2  3i  2  [3  3] i  2  6i

19. 5  3i  4  7i  5  4  3  7 i  1  10i

20. 3  4i2  5i  3  2[4  5] i  59i     22. 3  2i 5  13 i  3  5 2  13 i  2 73 i

11. 3: real part 3, imaginary part 0.

21. 6  6i  9  i  6  9  6  1 i  3  5i       23. 7  12 i  5  32 i  7  5   12  32 i  2  2i

SECTION 1.5 Complex Numbers

24. 4  i  2  5i  4  i  2  5i  4  2  1  5 i  6  6i

25. 12  8i  7  4i  12  8i  7  4i  12  7  8  4 i  19  4i 26. 6i  4  i  6i  4  i  4  6  1 i  4  7i

27. 4 1  2i  4  8i

28. 2 3  4i  6  8i

29. 7  i 4  2i  28  14i  4i  2i 2  28  2  14  4 i  30  10i

30. 5  3i 1  i  5  5i  3i  3i 2  5  3  5  3 i  8  2i

31. 6  5i 2  3i  12  18i  10i  15i 2  12  15  18  10 i  27  8i 32. 2  i 3  7i  6  14i  3i  7i 2  6  7  14  3 i  1  17i

33. 2  5i 2  5i  22  5i2  4  25 1  29

34. 3  7i 3  7i  32  7i2  58

35. 2  5i2  22  5i2  2 2 5i  4  25  20i  21  20i

36. 3  7i2  32  7i2  2 3 7i  40  42i 1 1 i i i 37.    2   i i i i 1 i 1 1i 1i 1i 1 1 1 1i      2  2i 38.  1i 1i 1i 11 2 1  i2 39. 40.

2  3i 1  2i 2  4i  3i  6i 2 8i 2  3i 2  6  4  3 i     or 85  15 i  1  2i 1  2i 1  2i 14 5 1  4i 2

5i 5  i 3  4i 15  20i  3i  4i 2 11  23i 15  4  20  3 i 23      11  25  25 i 3  4i 3  4i 3  4i 9  16 25 9  16i 2

10i 1  2i 10i  20i 2 5 4  2i 20  10i 10i      4  2i  1  2i 1  2i 1  2i 14 5 1  4i 2 1 2  3i 2  3i 2  3i 1 2  3i 2  3i      13  42. 2  3i1  13 2  3i 2  3i 2  3i 49 13 4  9i 2

41.

43.

4  6i 3i 12i  18i 2 18 12 18  12i 4  6i      i  2  43 i  3i 3i 3i 9 9 9 9i 2

3  5i 15i 45i  75i 2 75 45 75  45i 3  5i      i  13  15 i  15i 15i 15i 225 225 225 225i 2 1 1 1i 1 1i 1i 1  i 1i 1i 1         i   45. 1i 1i 1i 1i 1i 1i 2 2 1  i2 1  i2 44.

3  i  6i  2i 2 5  5i 2  i 10  5i  10i  5i 2 10  5  5  10 i 1  2i 3  i      2i 2i 2i 2i 5 4  i2 15  5i 5  15  5  5i  3  i 5  5 48. i 10  i 2  15  1 47. i 3  i 2 i  i

46.

 2 49. 3i5  35 i 2 i  243 12 i  243i  250 51. i 1000  i 4  1250  1

50. 2i4  24 i 4  16 1  16

 250 52. i 1002  i 4 i 2  1i 2  1     9 53. 49  49 1  7i 54. 81 16  4 i         55. 3 12  i 3  2i 3  6i 2  6 56. 13 27  13  3i 3  3i                 57. 3  5 1  1  3  i 5 1  i  3  3i  i 5  i 2 5  3  5  3  5 i

99

100

CHAPTER 1 Equations and Graphs

               58. 3  4 6  8  3  2i 6  2i 2  18  2i 6  2i 6  4i 2 2            3 2  4 2  2 6  2 6 i   2  4i 6      2 1i 2 2  2i 2 2  8 2  59.     1  2 1i 2 1i 2      2 i 2 2i 2 i 2 6i 36 60.         i 2    2 1 2i 2 9 i 2  3i i 2 i 2 61. x 2  49  0  x 2  49  x  7i



62. 3x 2  1  0  3x 2  1  x 2   13  x   33 i     2  4ac  1  12  4 1 2 1  7 b  b    12  27 i 63. x 2  x  2  0  x  2a 2 1 2    2  22  4 1 2 2  4  8 2  4 2  2i 64. x 2  2x  2  0  x      1  i 2 1 2 2 2    3  19 3  32  4 1 7    32  219 i 65. x 2  3x  7  0  x  2 1 2     6  62  4 1 10 6  36  40 6  4 6  2i     3i 66. x 2  6x  10  0  x  2 1 2 2 2       1  12  4 1 1 1  1  4 1  3 1  i 3 2 67. x  x  1  0  x       12  23 i 2 1 2 2 2       3  32  4 1 3 3  9  12 3  3 3i 3 2 68. x  3x  3  0  x      32  23 i 2 1 2 2 2     2  22  4 2 1 2 48 2  4 2  2i 69. 2x 2  2x  1  0  x      12  12 i 2 2 4 4 4 

    3 32 413 3 3 912 3 3 3i 3   3  3 i 70. t  3   0  t 2  3t  3  0  t     21 2 2 2 2 2 t 2 71. 6x  12x  7  0        12 122 467 144168 6  12  2i 6  1  6 i  12 12  1212 24  122i x 26 12 12 12 6 72. x 2  12 x  1  0 

x

   2 1  12   4 1 1 2 2 1



 12 

 2

1 4 4



 12 

  15 4

2

   12  12 i 15    14  415 i 2

73. z    3  4i  5  2i  3  4i  5  2i  8  2i 74. z    3  4i  5  2i  8  2i  8  2i 75. z  z  3  4i 3  4i  32  42  25

76. z    3  4i 5  2i  15  6i  20i  8i 2  23  14i

77. LHS  z    a  bi  c  di  a  bi  c  di  a  c  b  d i  a  c  b  d i. RHS  z    a  bi  c  di  a  c  b  d i  a  c  b  d i. Since LHS  RHS, this proves the statement.

78. LHS  z  a  bi c  di  ac  adi  bci  bdi 2  ac  bd  ad  bc i  ac  bd  ad  bc i. RHS  z    a  bi  c  di  a  bi c  di  ac  adi  bci  bdi 2  ac  bd  ad  bc i. Since LHS  RHS, this proves the statement.

SECTION 1.6 Solving Other Types of Equations

101

   2 79. LHS  z2  a  bi  a  bi2  a 2  2abi  b2 i 2  a 2  b2  2abi.     RHS  z 2  a  bi2  a 2  2abi  b2 i 2  a 2  b2  2abi  a 2  b2  2abi. Since LHS  RHS, this proves the statement.

80. z  a  bi  a  bi  a  bi  z. 81. z  z  a  bi  a  bi  a  bi  a  bi  2a, which is a real number.

82. z  z  a  bi  a  bi  a  bi  a  bi  a  bi  a  bi  2bi, which is a pure imaginary number.

83. z  z  a  bi  a  bi  a  bi  a  bi  a 2  b2 i 2  a 2  b2 , which is a real number.

84. Suppose z  z. Then we have a  bi  a  bi  a  bi  a  bi  0  2bi  b  0, so z is real. Now if z is real, then z  a  0i(where a is real). Since z  a  0i, we have z  z.  b  b2  4ac . Since both solutions are imaginary, 85. Using the Quadratic Formula, the solutions to the equation are x  2a   b 4ac  b2 we have b2  4ac  0  4ac  b2  0, so the solutions are x   i, where 4ac  b2 is a real number. 2a 2a Thus the solutions are complex conjugates of each other. 86. i  i, i 5  i 4  i  i, i 9  i 8  i  i;

i 2  1, i 6  i 4  i 2  1, i 10  i 8  i 2  1;

i 3  i, i 7  i 4  i 3  i, i 11  i 8  i 3  i; i 4  1, i 8  i 4  i 4  1, i 12  i 8  i 4  1. Because i 4  1, we have i n  i r , where r is the remainder when n is divided by 4, that is, n  4  k  r , where k is an integer and 0  r  4. Since 4446  4  1111  2, we must have i 4446  i 2  1.

1.6

SOLVING OTHER TYPES OF EQUATIONS

Note: In cases where both sides of an equation are squared, the implication symbol  is sometimes used loosely. For example,  2  x  x  1 “” x  x  12 is valid only for positive x. In these cases, inadmissible solutions are identified later in the

solution.

1. (a) To solve the equation x 3  4x 2  0 we factor the left-hand side: x 2 x  4  0, as above. (b) The solutions of the equation x 2 x  4  0 are x  0 and x  4.   2. (a) Isolating the radical in 2x  x  0, we obtain 2x  x.  2 2x  x2  2x  x 2 . (b) Now square both sides:

(c) Solving the resulting quadratic equation, we find 2x  x 2  x 2  2x  x x  2  0, so the solutions are x  0 and x  2.  (d) We substitute these possible solutions into the original equation: 2  0  0  0, so x  0 is a solution, but  2  2  2  4  0, so x  2 is not a solution. The only real solution is x  0.

3. The equation x  12  5 x  1  6  0 is of quadratic type. To solve the equation we set W  x  1. The resulting quadratic equation is W 2  5W  6  0  W  3 W  2  0  W  2 or W  3  x  1  2 or x  1  3  x  1 or x  2. You can verify that these are both solutions to the original equation.

4. The equation x 6  7x 3  8  0 is of quadratic type. To solve the equation we set W  x 3 . The resulting quadratic equation is W 2  7W  8  0. 5. x 2  x  0  x x  1  0  x  0 or x  1  0. Thus, the two real solutions are 0 and 1.

6. 3x 3  6x 2  0  3x 2 x  2  0  x  0 or x  2  0. Thus, the two real solutions are 0 and 2.   7. x 3  25x  x 3  25x  0  x x 2  25  0  x x  5 x  5  0  x  0 or x  5  0 or x  5  0. The three real solutions are 5, 0, and 5.

102

CHAPTER 1 Equations and Graphs

   8. x 5  5x 3  x 5  5x 3  0  x 3 x 2  5  0  x  0 or x 2  5  0. The solutions are 0 and  5.    9. x 5  3x 2  0  x 2 x 3  3  0  x  0 or x 3  3  0. The solutions are 0 and 3 3.      10. 6x 5  24x  0  6x x 4  4  0  6x x 2  2 x 2  2  0. Thus, x  0, or x 2  2  0 (which has no solution), or  x 2  2  0. The solutions are 0 and  2.    11. 0  4z 5  10z 2  2z 2 2z 3  5 . If 2z 2  0, then z  0. If 2z 3  5  0, then 2z 3  5  z  3 52 . The solutions are 0  and 3 52 .     2  3 2 . The solutions are 0 12. 0  125t 10  2t 7  t 7 125t 3  2 . If t 7  0, then t  0. If 125t 3  2  0, then t  3 125 5  3

and 52 .

    13. 0  x 5  8x 2  x 2 x 3  8  x 2 x  2 x 2  2x  4  x 2  0, x  2  0, or x 2  2x  4  0. If x 2  0, then

x  0; if x  2  0, then x  2, and x 2  2x  4  0 has no real solution. Thus the solutions are x  0 and x  2.   14. 0  x 4  64x  x x 3  64  x  0 or x 3  64  0. If x 3  64  0, then x 3  64  x  4. The solutions are 0 and 4.

  15. 0  x 3  5x 2  6x  x x 2  5x  6  x x  2 x  3  x  0, x  2  0, or x  3  0. Thus x  0, or x  2, or x  3. The solutions are x  0, x  2, and x  3.   16. 0  x 4  x 3  6x 2  x 2 x 2  x  6  x 2 x  3 x  2. Thus either x 2  0, so x  0,or x  3, or x  2. The solutions are 0, 3, and 2.

  17. 0  x 4  4x 3  2x 2  x 2 x 2  4x  2 . So either x 2  0  x  0, or using the Quadratic Formula on x 2  4x  2  0,       42 412 8  42 2  2  2. The solutions are 0, 2  2, and  4 2 168  4 we have x  4 21 2 2  2  2.   18. 0  y 5  8y 4  4y 3  y 3 y 2  8y  4 . If y 3  0, then y  0. If y 2  8y  4  0, then using the Quadratic Formula, we    8  82  4 1 4    8  48 have y    4  2 3. Thus, the three solutions are 0, 4  2 3, and 4  2 3. 2 1 2 19. 3x  54  3x  53  0. Let y  3x  5. The equation becomes y 4  y 3  0      y y 3  1  y y  1 y 2  y  1  0. If y  0, then 3x  5  0  x   53 . If y  1  0, then 3x  5  1  0  x   43 . If y 2  y  1  0, then 3x  52  3x  5  1  0  9x 2  33x  31  0. The discriminant is b2  4ac  332  4 9 31  27  0, so this case gives no real solution. The solutions are x   53 and x   43 .

20. x  54  16 x  52  0. Let y  x  5. The equation becomes y 4  16y 2  y 2 y  4 y  4  0. If y 2  0, then x  5  0 and x  5. If y  4  0, then x  5  4  0 and x  1. If y  4  0, then x  5  4  0 and x  9. Thus, the solutions are 9, 5, and 1.   21. 0  x 3  5x 2  2x  10  x 2 x  5  2 x  5  x  5 x 2  2 . If x  5  0, then x  5. If x 2  2  0, then   x 2  2  x   2. The solutions are 5 and  2.   22. 0  2x 3  x 2  18x  9  x 2 2x  1  9 2x  1  2x  1 x 2  9  2x  1 x  3 x  3. The solutions are  12 , 3, and 3.

  23. x 3  x 2  x  1  x 2  1  0  x 3  2x 2  x  2  x 2 x  2  x  2  x  2 x 2  1 . Since x 2  1  0 has no real solution, the only solution comes from x  2  0  x  2.

SECTION 1.6 Solving Other Types of Equations

103

  24. 7x 3  x  1  x 3  3x 2  x  0  6x 3  3x 2  2x  1  3x 2 2x  1  2x  1  2x  1 3x 2  1  2x  1  0  or 3x 2  1  0. If 2x  1  0, then x  12 . If 3x 2  1  0, then 3x 2  1  x 2  13  x   13 . The solutions are 12  and  13 .

  4 4  3  z  1 z   z  1 3  z 2  z  4  3z  3  z 2  2z  1  0  z  12  0. The z1 z1 solution is z  1. We must check the original equation to make sure this value of z does not result in a zero denominator.   10 10  15  3m  m  5  15  m  5 3m  10  15m  75  3m 2  15m  3m 2  85  0  26. m5 m5  m   85 3 . Verifying that neither of these values of m results in a zero denominator in the original equation, we see that   85 . and the solutions are  85 3 3 25. z 

27.

    1 5 1 1 1 5    4 x  1 x  2   4 x  1 x  2  x 1 x 2 4 x 1 x 2 4

4 x  2  4 x  1  5 x  1 x  2  4x  8  4x  4  5x 2  5x  10  5x 2  3x  14  0  7 5x  7 x  2  0. If 5x  7  0, then x   75 ; if x  2  0, then x  2. The solutions are  and 2. 5   10 12 12 10 28.   4  0  x x  3   4  0  x  3 10  12x  4x x  3  0  x x 3 x x 3

10x  30  12x  4x 2  12x  0  4x 2  14x  30  0. Using the Quadratic Formula, we have     14  142  4 4 30 14  196  480 14  676 14  26    . So the solutions are 5 and  32 . x 2 4 8 8 8

29.

30.

x2  50  x 2  50 x  100  50x  5000  x 2  50x  5000  0  x  100 x  50  0  x  100  0 x  100 or x  50  0. Thus x  100 or x  50. The solutions are 100 and 50. 2x x2  1

 1  2x  x 2  1  x 2  2x  1  x  12  0, so x  1. This is indeed a solution to the original equation.

1 2 1   x  1 x  2  1  2 x  2  x  1  x 2  3x  2  1  2x  4  x  1  x 1 x 2 x  1 x  2   x 2  2  0  x   2. We verify that these are both solutions to the original equation.

31. 1 

32.

33.

2 1 x    x x  3  2 x  3  1  x 2  3x  2x  6  1  x 2  5x  5  0. Using the Quadratic x 3 x  3 x2  9    5  52  4 1 5 53 5 Formula, x   . We verify that both are solutions to the original equation. 2 2 x x 1   1  x x  3  x  1 2x  7  2x  7 x  3  x 2  3x  2x 2  9x  7  2x 2  13x  21 2x  7 x  3  3x 2  19x  28  0  3x  7 x  4  0. Thus either 3x  7  0, so x   73 , or x  4. The solutions are  73 and 4.

34.

2 1  2 x 1 x

 0  x 2  2 x  1  0  x 2  2x  2  0      2  22  4 1 2 2 48 2  4   . Since the radicand is negative, there is no real solution. x 2 1 2 2

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 x  x2 x x2  2    5x  x 2  2  5x 3x  4  x 2  2  15x 2  20x  0  14x 2  20x  2 4 x 3x  4 3 x       20  202  4 14 2 20  400  112 20  512 20  16 2 5  4 2    . The  x  2 14 28 28 28 7  5  4 2 . solutions are 7     3 1   3  1x x 4  x 2  4x x  3x  1  2x 2  4x  2x 2  7x  1  0. Using the Quadratic  x  x 2 x 36. 4 4 2 x 2 x     7  72  4 2 1 7  57 7  57  . Both are admissible, so the solutions are . Formula, we find x  2 2 4 4   2 37. 5  4x  3  52  4x  3  25  4x  3  4x  28  x  7 is a potential solution. Substituting into the   original equation, we get 5  4 7  3  5  25, which is true, so the solution is x  7.   2 38. 8x  1  3  8x  1  32  8x  1  9  x  54 . Substituting into the original equation, we get     8 54  1  3  9  3, which is true, so the solution is x  54 . x  2x 35.  5x  3  4x

39.

40.

41.

42.

43.

44.



   2  2 2x  1  3x  5  2x  1  3x  5  2x  1  3x  5  x  4. Substituting into the original equation,     we get 2 4  1  3 4  5  7  7, which is true, so the solution is x  4. 2   2   3  x  x2  1  3x  x 2  1  3  x  x 2  1  x 2  x  2  0  x  1 x  2  0      x  1 or x  2. Substituting into the original equation, we get 3  1  12  1  2  2, which is true,   and 3  2  22  1, which is also true. So the solutions are x  1 and x  2.   2 x 2  x  x  2  x 2  x  2  x 2  x 2  x  2  x  1 x  2  0  x  1 or x  2. Substituting     into the original equation, we get 1  2  1  1  1, which is false, and 2  2  2  4  2, which is true. So x  2 is the only real solution.   2 4  6x  2x  4  6x  2x2  4  6x  4x 2  2x 2  3x  2  x  2 2x  1  0  x  2   or x  12 . Substituting into the original equation, we get 4  6 2  2 2  16  4, which is false, and       4  6 12  2 12  1  1, which is true. So x  12 is the only real solution.   2x  1  1  x  2x  1  x  1  2x  1  x  12  2x  1  x 2  2x  1  0  x 2  4x  x x  4. Potential solutions are x  0 and x  4  x  4. These are only potential solutions since squaring is not a reversible operation. We must check each potential solution in the original equation.   Checking x  0: 2 0  1  1  0  1  1  0 is false.   Checking x  4: 2 4  1  1  4  9  1  4  3  1  4 is true. The only solution is x  4.   x  9  3x  0  x  9  3x  x 2  9  3x  0  x 2  3x  9. Using the Quadratic Formula to find the potential    3  32  4 1 9 3  45 3  3 5 solutions, we have x    . Substituting each of these solutions into the 2 1 2 2 





5 is a solution, but x  33 5 is not. Thus x  33 5 is the only solution. original equation, we see that x  33 2 2 2   2  2 2 x  1  x  6x  9  x  1  x 2  7x  10  0  45. x  x  1  3  x  3  x  1  x  3  x  2 x  5  0. Potential solutions are x  2 and x  5. We must check each potential solution in the original   equation. Checking x  2: 2  2  1  3, which is false, so x  2 is not a solution. Checking x  5: 5  5  1  3  5  2  3, which is true, so x  5 is the only solution.

SECTION 1.6 Solving Other Types of Equations

46.

105

   2 3  x  2  1  x  3  x  1  x  3  x  1  x2  3  x  x 2  2x  1  x 2  3x  2  0. Using   3  17 3  32  4 1 2  . Substituting the Quadratic Formula to find the potential solutions, we have x  2 1 2 



each of these solutions into the original equation, we see that x  32 17 is a solution, but x  32 17 is not. Thus 

x  32 17 is the only solution.    2   2   47. 3x  1  2  x  1  3x  1  2  x  1  3x  1  4  4 x  1  x  1  2x  4  4 x  1     2 x  2  2 x  1  x  22  2 x  1  x 2  4x  4  4 x  1  x 2  8x  0  x x  8  0  x  0 or x  8. Substituting each of these solutions into the original equation, we see that x  0 is not a solution but x  8 is a solution. Thus, x  8 is the only solution.    2    1  x  1  x  22  1  x  1  x  2 1  x 1  x  4  48. 1  x  1  x  2      2  2 1  x 1  x  4  1  x 1  x  1  1  x 1  x  1  1  x 2  1  x 2  0, so x  0. We verify that this is a solution to the original equation. 49. x 4  4x 2  3  0. Let y  x 2 . Then the equation becomes y 2  4y  3  0  y  1 y  3  0, so y  1 or y  3. If  y  1, then x 2  1  x  1, and if y  3, then x 2  3  x   3. 50. x 4  5x 2  6  0. Let y  x 2 . Then the equation becomes y 2  5y  6  0  y  2 y  3  0, so y  2 or y  3. If   y  2, then x 2  2  x   2, and if y  3, then x 2  3  x   3.

51. 2x 4  4x 2  1  0. The LHS is the sum of two nonnegative numbers and a positive number, so 2x 4  4x 2  1  1  0. This equation has no real solution.      52. 0  x 6  2x 3  3  x 3  3 x 3  1 . If x 3  3  0, then x 3  3  x  3 3, or if x 3  1  x  1. Thus x  3 3  or x  1. The solutions are 3 3 and 1.    53. 0  x 6  26x 3  27  x 3  27 x 3  1 . If x 3  27  0  x 3  27, so x  3. If x 3  1  0  x 3  1, so x  1. The solutions are 3 and 1.

   54. x 8  15x 4  16  0  x 8  15x 4  16  x 4  16 x 4  1 . If x 4  16  0, then x 4  16 which is impossible (for real numbers). If x 4  1  0  x 4  1, so x  1. The solutions are 1 and 1.

55. 0  x  52  3 x  5  10  [x  5  5] [x  5  2]  x x  7  x  0 or x  7. The solutions are 0 and 7.     x 1 x 1 2 x 1 56. Let   . Then 0   3 becomes 0  2  4  3    1   3. Now if   1  0, 4 x x x x 1 x 1 x 1 x 1 then 10  1  x  1  x  x   12 , and if   3  0, then 3  0  3 x x x x  x  1  3x  x   14 . The solutions are  12 and  14 .  2   1 1 1 . Then  8  0 becomes 2  2  8  0    4   2  0. So   4  0 2 57. Let   x 1 x 1 x 1 1    4, and   2  0    2. When   4, we have  4  1  4x  4  3  4x  x   34 . When x 1 1   2, we have  2  1  2x  2  3  2x  x   32 . Solutions are  34 and  32 . x 1  2 x 4x x 58. Let   . Then  4 becomes 2  4  4  0  2  4  4    22 . Now if  x 2 x 2 x 2 x x 20  2  x  2x  4  x  4. The solution is 4.   2  0, then x 2 x 2

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59. Let u  x 23 . Then 0  x 43  5x 23  6 becomes u 2  5u  6  0  u  3 u  2  0  u  3  0 or u  2  0.  If u  3  0, then x 23  3  0  x 23  3  x  332  3 3. If u  2  0, then x 23  2  0  x 23  2     x  232  2 2. The solutions are 3 3 and 2 2.      60. Let u  4 x; then0  x  3 4 x  4  u 2  3u  4  u  4 u  1. So u  4  4 x  4  0  4 x  4     x  44  256, or u  1  4 x  1  0  4 x  1. However, 4 x is the positive fourth root, so this cannot equal 1. The only solution is 256.    61. 4 x  112  5 x  132  x  152  0  x  1 4  5 x  1  x  12  0         x  1 4  5x  5  x 2  2x  1  0  x  1 x 2  3x  0  x  1  x x  3  0  x  1 or x  0 or x  3. The solutions are 1, 0, and 3.

62. Let u  x  4; then 0  2 x  473  x  443  x  413  2u 73  u 43  u 13  u 13 2u  1 u  1. So

u  x  4  0  x  4, or 2u  1  2 x  4  1  2x  7  0  2x  7  x  72 , or u  1  x  4  1  x  5  0

 x  5. The solutions are 4, 72 ,and 5.

  63. x 32  10x 12  25x 12  0  x 12 x 2  10x  25  0  x 12 x  52  0. Now x 12   0, so the only solution is x  5.

  64. x 12  x 12  6x 32  0  x 32 x 2  x  6  0  x 32 x  2 x  3  0. Now x 12  0, and furthermore the original equation cannot have a negative solution. Thus, the only solution is x  3.

65. Let u  x 16 . (We choose the exponent 16 because the LCD of 2, 3, and 6 is 6.) Then x 12  3x 13  3x 16  9    x 36  3x 26  3x 16  9  u 3  3u 2  3u  9  0  u 3  3u 2  3u  9  u 2 u  3  3 u  3  u  3 u 2  3 . So u  3  0 or u 2  3  0. If u  3  0, then x 16  3  0  x 16  3  x  36  729. If u 2  3  0, then

x 13  3  0  x 13  3  x  33  27. The solutions are 729 and 27.    66. Let u  x. Then 0  x  5 x  6 becomes u 2  5u  6  u  3 u  2  0. If u  3  0, then x  3  0     x  3  x  9. If u  2  0, then x  2  0  x  2  x  4. The solutions are 9 and 4. 67.

4 4 1  2   0  1  4x  4x 2  0  1  2x2  0  1  2x  0  2x  1  x   12 . The solution is  12 . x x3 x

x4 we get, 0  1  4x 2  x 4 . Substituting u  x 2 , we get 0  1  4u  u 2 , and 4      4 42 411  4 2164  42 12  422 3  2  3. Substituting using the Quadratic Formula, we get u  21        back, we have x 2  2  3, and since 2  3 and 2  3 are both positive we have x   2  3 or x   2  3.         Thus the solutions are  2  3, 2  3,  2  3, and 2  3.    x  5  x  5. Squaring both sides, we get x  5  x  25  x  5  25  x. Squaring both sides again, we 69.

68. 0  4x 4  16x 2  4. Multiplying by

get x  5  25  x2  x  5  625  50x  x 2  0  x 2  51x  620  x  20 x  31. Potential solutions are x  20 and x  31. We must check each potential solution in the original equation.    Checking x  20: 20  5  20  5  25  20  5  5  20  5, which is true, and hence x  20 is a solution.    Checking x  31: 36  31  5  37  5, which is false, and hence x  31 is not a 31  5  31  5  solution. The only real solution is x  20.    3 70. 4x 2  4x  x  4x 2  4x  x 3  0  x 3  4x 2  4x  x x 2  4x  4  x x  22 . So x  0 or x  2. The solutions are 0 and 2.

SECTION 1.6 Solving Other Types of Equations

   71. x 2 x  3  x  332  0  x 2 x  3  x  332  0  x  3





107

    x 2  x  3  0  x  3 x 2  x  3 . 

If x  312  0, then x  3  0  x  3. If x 2  x  3  0, then using the Quadratic Formula x  12 13 . The 

solutions are 3 and 12 13 .   2 2 72. Let u  11  x 2 . By definition of u we require it to be nonnegative. Now 11  x 2    1  u   1. 2 u 11  x

Multiplying both sides by u we obtain u 2  2  u  0  u 2  u  2  u  2 u  1. So u  2 or u  1. But since u   must be nonnegative, we only have u  2  11  x 2  2  11  x 2  4  x 2  7  x   7. The solutions are   7.     73. x  x  2  2. Squaring both sides, we get x  x  2  4  x  2  4  x. Squaring both sides again, we get

74.

75.

76.

77.

x  2  4  x2  16  8x  x 2  0  x 2  9x  14  0  x  7 x  2. If x  7  0, then x  7. If x  2  0, then x  2. So x  2 is a solution but x  7 is not, since it does not satisfy the original equation.         1  x  2x  1  5  x. We square both sides to get 1  x  2x  1  5  x      2   x  2x  1  4  x  16  8 x  x  2x  1  16  8 x. Again, squaring both sides, we obtain     2 2x  1  16  8 x  256  256 x  64x  62x  255  256 x. We could continue squaring both sides until  we found possible solutions; however, consider the last equation. Since we are working with real numbers, for x to be  defined, we must have x  0. Then 62x  255  0 while 256 x  0, so there is no solution.     0  x 4  5ax 2  4a 2  a  x 2 4a  x 2 . Since a is positive, a  x 2  0  x 2  a  x  a. Again, since a is    positive, 4a  x 2  0  x 2  4a  x  2 a. Thus the four solutions are  a and 2 a.   b 0  a 3 x 3  b3  ax  b a 2 x 2  abx  b2 . So ax  b  0  ax  b  x   or a       ab  ab2  4 a 2 b2 ab  3a 2 b2 b    , but this gives no real solution. Thus, the solution is x   . x a 2 a2 2a 2     x  a  x  a  2 x  6. Squaring both sides, we have             x  a  x  a  2 x  6  2x  2 x  a x  a  2x  12  2 x  a x  a  12 x a 2 x  a      x a x  a  6. Squaring both sides again we have x  a x  a  36  x 2  a 2  36  x 2  a 2  36    x   a 2  36. Checking these answers, we see that x   a 2  36 is not a solution (for example, try substituting  a  8), but x  a 2  36 is a solution.

78. Let   x 16 . Then x 13  2 and x 12  3 , and so

       0  3  a2  b  ab  2   a  b   a  2  b   a  3 x  b 6 x  a . So 6 x  a  0     a  6 x  x  a 6 is one solution. Setting the first factor equal to zero, we have 3 x  b  0  3 x  b  x  b3 .  However, the original equation includes the term b 6 x, and we cannot take the sixth root of a negative number, so this is not a solution. The only solution is x  a 6 .

900 . After 5 people 79. Let x be the number of people originally intended to take the trip. Then originally, the cost of the trip is x   900 900 4500 cancel, there are now x  5 people, each paying  2. Thus 900  x  5  2  900  900  2x   10 x x x 4500  0  2x  10   0  2x 2  10x  4500  2x  100 x  45. Thus either 2x  100  0, so x  50, or x x  45  0, x  45. Since the number of people on the trip must be positive, originally 50 people intended to take the trip.

108

CHAPTER 1 Equations and Graphs

120,000 . If one person joins the group, then there would n   120,000 120,000 be n  1 members in the group, and each person would pay  6000. So n  1  6000  120,000 n n     n  n  120,000  6000 n  1  120,000  20  n n  1  20n  n 2  19n  20  20n   6000 n 6000

80. Let n be the number of people in the group, so each person now pays

0  n 2  n  20  n  4 n  5. Thus n  4 or n  5. Since n must be positive, there are now 4 friends in the group.

  t and substituting, we have 500  3t  10 t  140    5  1105 500  3u 2  10u  140  0  3u 2  10u  360  u  . Since u  t, we must have u  0. So 3   5  1105  9414  t  8862. So it will take 89 days for the fish population to reach 500. t u 3

81. We want to solve for t when P  500. Letting u 

82. Let d be the distance from the lens to the object. Then the distance from the lens to the image is d  4. So substituting 1 1 1   . Now we multiply by the F  48, x  d, and y  d  4, and then solving for x, we have 48 d d 4 LCD, 48d d  4, to get d d  4  48 d  4  48d  d 2  4d  96d  192  0  d 2  136d  192  136  104 . So d  16 or d  12. Since d  4 must also be positive, the object is 12 cm from the lens. d 2 83. Let x be the height of the pile in feet. Then the diameter is 3x and the radius is 32 x feet. Since the volume of the cone is     3x 2 3x 3 4000 4000 1000 ft3 , we have  1000  x 3  x  3  752 feet. x  1000  3 2 4 3 3 84. Let r be the radius of the tank, in feet. The volume of the spherical tank is 43 r 3 and is also 750  01337  100275. So 4 r 3  100275  r 3  23938  r  288 feet. 3

  85. Let r be the radius of the larger sphere, in mm. Equating the volumes, we have 43 r 3  43  23  33  43   r 3  23  33  44  r 3  99  r  3 99  463. Therefore, the radius of the larger sphere is about 463 mm. 86. We have that the volume is 180 ft3 , so x x  4 x  9  180  x 3  5x 2  36x  180  x 3  5x 2  36x  180  0    x 2 x  5  36 x  5  0  x  5 x 2  36  0  x  5 x  6 x  6  0  x  6 is the only positive solution. So the box is 2 feet by 6 feet by 15 feet.

87. Let x be the length, in miles, of the abandoned road to be used. Then the length of the abandoned road not used  is 40  x, and the length of the new road is 102  40  x2 miles, by the Pythagorean Theorem. Since the  cost of the road is cost per mile  number of miles, we have 100,000x  200,000 x 2  80x  1700  6,800,000   2 x 2  80x  1700  68  x. Squaring both sides, we get 4x 2  320x  6800  4624  136x  x 2  

1  18488  x  136 3x 2  184x  2176  0  x  184 3385626112 6 6 3 or x  16. Since 45 3 is longer than the existing  road, 16 miles of the abandoned road should be used. A completely new road would have length 102  402 (let x  0)  and would cost 1700  200,000  83 million dollars. So no, it would not be cheaper.

SECTION 1.6 Solving Other Types of Equations

109

88. Let x be the distance, in feet, that he goes on the boardwalk before veering off onto the sand. The distance along the boardwalk from where he started to the point on the boardwalk closest  to the umbrella is 7502  2102  720 ft. Thus the distance that he walks on the sand is    720  x2  2102  518,400  1440x  x 2  44,100  x 2  1440x  562,500. Along boardwalk Across sand



Distance

Rate

x

4

x 2  1440x  562,500

2



Time x 4

x 2  1440x  562,500 2

Since 4 minutes 45 seconds  285 seconds, we equate the time it takes to walk along the boardwalk and across the sand   x 2  1440x  562,500 x  1140  x  2 x 2  1440x  562,500. Squaring both to the total time to get 285   4 2  sides, we get 1140  x2  4 x 2  1440x  562,500  1,299,600  2280x  x 2  4x 2  5760x  2,250,000    0  3x 2  3480x  950,400  3 x 2  1160x  316,800  3 x  720 x  440. So x  720  0  x  720, and x  440  0  x  440. Checking x  720, the distance across the sand is

210 210 feet. So 720 4  2  180  105  285 seconds. Checking x  440, the distance across the sand is  350 720  4402  2102  350 feet. So 440 4  2  110  175  285 seconds. Since both solutions are less than or equal

to 720 feet, we have two solutions: he walks 440 feet down the boardwalk and then heads towards his umbrella, or he walks 720 feet down the boardwalk and then heads toward his umbrella.

89. Let x be the length of the hypotenuse of the triangle, in feet. Then one of the other sides has length x  7 feet, and since the perimeter is 392 feet, the remaining side

must have length 392  x  x  7  399  2x. From the Pythagorean Theorem,

x-7

x

we get x  72  399  2x2  x 2  4x 2  1610x  159250  0. Using the

Quadratic Formula, we get   2 44159250 x  1610 161024  16108 44100  1610210 , and so x  2275 or x  175. But if x  2275, then the 8 side of length x  7 combined with the hypotenuse already exceeds the perimeter of 392 feet, and so we must have x  175. Thus the other sides have length 175  7  168 and 399  2 175  49. The lot has sides of length 49 feet, 168 feet, and 175 feet. 90. Let h be the height of the screens in inches. The width of the smaller screen is h  7 inches, and the width of the bigger  screen is 18h inches. The diagonal measure of the smaller screen is h 2  h  72 , and the diagonal measure of the     larger screen is h 2  18h2  424h 2  206h. Thus h 2  h  72  3  206h  h 2  h  72  206h  3. Squaring both sides gives h 2  h 2  14h  49  424h 2  1236h  9  0  224h 2  2636h  40. Applying

  2636 26362 422440 2636 10532496  2636  3245 . So  the Quadratic Formula, we obtain h  2224 448 448

2636  3245  1313. Thus, the screens are approximately 131 inches high. 448   d d 1 2  1 2  1 3  0  . Letting   d, we have 3  14  1090 91. Since the total time is 3 s, we have 3  1090 4 4 1090  545  591054 . Since   0, we have d    1151, so d  13256. The well  22  545  6540  0    4 is 1326 ft deep. h

110

CHAPTER 1 Equations and Graphs

  92. (a) Method 1: Let u  x, so u 2  x. Thus x  x  2  0 becomes u 2  u  2  0  u  2 u  1  0. So u  2   or u  1. If u  2, then x  2  x  4. If u  1, then x  1  x  1. So the possible solutions are 4 and   1. Checking x  4 we have 4  4  2  4  2  2  0. Checking x  1 we have 1  1  2  1  1  2  0. The only solution is 4.   Method 2: x  x  2  0  x  2  x  x 2  4x  4  x  x 2  5x  4  0  x  4 x  1  0. So the possible solutions are 4 and 1. Checking will result in the same solution. 1 1 12 10 (b) Method 1: Let u  , so u 2   1  0 becomes 12u 2  10u  1  0. Using . Thus  2 2 x 3 x 3 x  3 x  3      102 4121 52  102 13  5 13 . If u  5 13 , the Quadratic Formula, we have u  10 212  10 24 24 12 12      12 5 13   1 5  13 5 13 12   then  x 3    5  13. So x  2  13. 12 5 13 5 13 x 3 12       12 5 13  5  1 13 13 , then 12  513   x 3   5  13. So If u  5 12 12 5 13 5 13 x 3 12  x  2  13.  The solutions are 2  13.   10 12  1  0  x  32   Method 2: Multiplying by the LCD, x  32 , we get x  32 x 3 x  32 12  10 x  3  x  32  0  12  10x  30  x 2  6x  9  0  x 2  4x  9  0. Using the Quadratic    2 13  2  13. The solutions are 2  13. Formula, we have u  4 4 419  42 52  42 2 22

1.7

SOLVING INEQUALITIES

1. (a) If x  5, then x  3  5  3  x  3  2. (b) If x  5, then 3  x  3  5  3x  15.

(c) If x  2, then 3  x  3  2  3x  6.

(d) If x  2, then x  2.

2. To solve the nonlinear inequality

x 1  0 we x 2

Interval

 1

1 2

2 

first observe that the numbers 1 and 2 are zeros

Sign of x  1







of the numerator and denominator. These numbers divide the real line into the three intervals  1, 1 2, and 2 .

Sign of x  2













The endpoint 1 satisfies the inequality, because

Sign of x  1  x  2

21 1  1  0  0, but 2 fails to satisfy the inequality because is not 1  2 22

defined. Thus, referring to the table, we see that the solution of the inequality is [1 2. 3. (a) No. For example, if x  2, then x x  1  2 1  2  0. (b) No. For example, if x  2, then x x  1  2 3  6.

4. (a) To solve 3x  7, start by dividing both sides of the inequality by 3.

(b) To solve 5x  2  1, start by adding 2 to both sides of the inequality.

SECTION 1.7 Solving Inequalities

5. x

2  3x  13

6. x

1  2x  5x

5 17  13 ; no

11  25; yes

1

5  13 ; no

5 1

3  5; yes

0

2  0; no

2 3 5 6

1  5 3

0

1  2

1 47  7

0

1 ; no 3 1 ; yes 3 1 ; yes 3 1 ; yes 3 1 ; yes 3 1 ; yes 3

 13  10 3 ; no  23  25 6 ; no

1  5

5 13   The elements 56 , 1, 5, 3, and 5 satisfy the inequality. 7.

1  0; yes

2 3 5 6

1  5; no 347  1118; no

3

5  15; no

5

9  25; no

The elements 5, 1, and 0 satisfy the inequality. 8.

x 5 1 0 2 3 5 6

1  5 3 5

1  2x  4  7

x

1  6  7; no

5 1 0 2 3 5 6

1  5 3 5

Graph:

2 3 5 6

1  2  7; no

2  73  2; no

2  13 6  2; no 2  2  2; no

1  5

1  047  7; no

1  2  7; yes

2  076  2; yes

2  0  2; yes

3

1  6  7; yes

2  2  2; yes

5 The elements

 5, 3, and 5 satisfy the inequality.

10. x 5 1

1 is undefined; no 0 3  1 ; no 2 2 6  1 ; no 5 2 1  12 ; no 045  12 ; yes 1  1 ; yes 3 2 1  1 ; yes 5 2

x2  2  4 27  4; no

3  4; yes

0

2  4; yes

2 3 5 6

22  4; yes 9 97  4; yes 36

1  5

3  4; yes 7  4; no

3

11  4; no

5

27  4; no

The elements 1, 0, 23 , 56 , and 1 satisfy the inequality. 12. 2x  8  x  4. Interval: [4  Graph:

6 5

2  3  2; no

0

1   83  7; no 1   73  7; no

 The elements 5, 1, 5, 3, and 5 satisfy the inequality.   11. 5x  6  x  65 . Interval:  65

2  4  2; no

1

1  4  7; no

1 1 x  2 1  5  12 ; yes 1  12 ; yes

2  8  2; no

5

The elements 3 and 5 satisfy the inequality. 9.

2  3  x  2

x

1  14  7; no

4

111

112

CHAPTER 1 Equations and Graphs

13. 2x  5  3  2x  8  x  4 Interval: 4  Graph:

Graph:

4

15. 2  3x  8  3x  2  8  x  2 Interval:  2 Graph:

_2

17. 2x  1  0  2x  1  x   12   Interval:   12 Graph:

1

__ 2

19. 1  4x  5  2x  6x  4  x  23   Interval:  23 Graph:

2 _ 3

21. 12 x  23  2  12 x  83  x  16 3   16 Interval: 3   Graph:

16 __ 3

 x  1 Interval:  1]

_1

25. 2  x  5  4  3  x  1 Interval: [3 1 Graph:

_3

_1

27. 6  3x  7  8  1  3x  15  13  x  5   Interval: 13  5 Graph:

1 _ 3

_2

16. 1  5  2x  2x  5  1  x  2 Interval:  2 Graph:

2

18. 0  5  2x  2x  5  x  52   Interval:  52 Graph:

5 _ 2

20. 5  3x  2  9x  6x  3  x   12   Interval:   12 Graph:

1

__ 2

22. 23  12 x  16  x (multiply both sides by 6)  4  3x  1  6x  3  9x  13  x   Interval:  13 Graph:

23. 4  3x   1  8x  4  3x  1  8x  5x  5

Graph:

14. 3x  11  5  3x  6  x  2 Interval:  2

5

1 _ 3

24. 2 7x  3  12x  16  14x  6  12x  16  2x  22  x  11 Interval:  11] Graph:

11

26. 5  3x  4  14  9  3x  18  3  x  6 Interval: [3 6] Graph:

3

6

28. 8  5x  4  5  4  5x  9   45  x  95   Interval:  45  95 Graph:

4

_ _5_

9 __ 5

SECTION 1.7 Solving Inequalities

29. 2  8  2x  1  10  2x  9  5  x  92

30. 3  3x  7  12  10  3x   13 2 

 92  x  5   Interval: 92  5 Graph:

31.

9 _ 2

13  10 3 x 6   13 Interval:  10 3  6

Graph:

5

2x  3 1 2    8  2x  3  2 (multiply each 3 12 6

5 __ 2

10

13

_ _3_

_ _6_

4  3x 1 1    (multiply each expression by 20) 2 5 4 10  4 4  3x  5  10  16  12x  5 

32. 

5 expression by 12)  11  2x  5  11 2 x  2   Interval: 52  11 2

Graph:

113

11 11 13 26  12x  11  13 6  x  12  12  x  6   13 Interval: 11 12  6

11 __ 2

Graph:

11 __ 12

13 __ 6

33. x  2 x  3  0. The expression on the left of the inequality changes sign where x  2 and where x  3. Thus we must check the intervals in the following table.  2

2 3

3 

Sign of x  2







Sign of x  3













Interval

Sign of x  2 x  3

From the table, the solution set is x  2  x  3. Interval: 2 3. Graph:

_2

3

34. x  5 x  4  0. The expression on the left of the inequality changes sign when x  5 and x  4. Thus we must check the intervals in the following table.  4

4 5

5 

Sign of x  5







Sign of x  4













Interval

Sign of x  5 x  4

From the table, the solution set is x  x  4 or 5  x. Interval:  4]  [5 . Graph:

_4

5

35. x 2x  7  0. The expression on the left of the inequality changes sign where x  0 and where x   72 . Thus we must check the intervals in the following table. Interval Sign of x Sign of 2x  7

Sign of x 2x  7

    72

   72  0

0 



















From the table, the solution set is   x  x   72 or 0  x .   Interval:   72  [0 . Graph:

7

__ 2

0

114

CHAPTER 1 Equations and Graphs

36. x 2  3x  0. The expression on the left of the inequality changes sign when x  0 and x  23 . Thus we must check the intervals in the following table.   0 23



 0

Sign of x



Sign of 2  3x













Sign of x 2  3x





From the table, the solution set is   x  x  0 or 23  x .   Interval:  0]  23   .

2 3

Interval



Graph:

0

2 _ 3

37. x 2  3x  18  0  x  3 x  6  0. The expression on the left of the inequality changes sign where x  6 and where x  3. Thus we must check the intervals in the following table.  3

3 6

6 

Sign of x  3







Sign of x  6













Interval

Sign of x  3 x  6

From the table, the solution set is x  3  x  6. Interval: [3 6]. Graph:

_3

6

38. x 2  5x  6  0  x  3 x  2  0. The expression on the left of the inequality changes sign when x  3 and x  2. Thus we must check the intervals in the following table.  3

3 2

2 

Sign of x  3







Sign of x  2













Interval

Sign of x  3 x  2

From the table, the solution set is x  x  3 or  2  x.

Interval:  3  2 . Graph:

_3

_2

39. 2x 2  x  1  2x 2  x  1  0  x  1 2x  1  0. The expression on the left of the inequality changes sign where x  1 and where x  12 . Thus we must check the intervals in the following table. Interval

 1

  1 12



Sign of x  1



Sign of 2x  1













Sign of x  1 2x  1





1 2



From the table, the solution set is   x  x  1 or 12  x .   Interval:  1]  12   . Graph:

_1

1 _ 2

40. x 2  x  2  x 2  x  2  0  x  1 x  2  0. The expression on the left of the inequality changes sign when x  1 and x  2. Thus we must check the intervals in the following table.  1

1 2

2 

Sign of x  1







Sign of x  2













Interval

Sign of x  1 x  2

From the table, the solution set is x  1  x  2. Interval: 1 2. Graph:

_1

2

SECTION 1.7 Solving Inequalities

115

41. 3x 2  3x  2x 2  4  x 2  3x  4  0  x  1 x  4  0. The expression on the left of the inequality changes sign where x  1 and where x  4. Thus we must check the intervals in the following table.  1

1 4

4 

Sign of x  1







Sign of x  4













Interval

Sign of x  1 x  4

From the table, the solution set is x  1  x  4. Interval: 1 4. Graph:

_1

4

42. 5x 2  3x  3x 2  2  2x 2  3x  2  0  2x  1 x  2  0. The expression on the left of the inequality changes sign when x  12 and x  2. Thus we must check the intervals in the following table. Interval

 2

  2 12



Sign of 2x  1



Sign of x  2













Sign of 2x  1 x  2





1 2



From the table, the solution set is   x  x  2 or 12  x .   Interval:  2]  12   . Graph:

_2

1 _ 2

43. x 2  3 x  6  x 2  3x  18  0  x  3 x  6  0. The expression on the left of the inequality changes sign where x  6 and where x  3. Thus we must check the intervals in the following table.  3

3 6

6 

Sign of x  3







Sign of x  6













Interval

Sign of x  3 x  6

From the table, the solution set is x  x  3 or 6  x. Interval:  3  6 . Graph:

_3

6

44. x 2  2x  3  x 2  2x  3  0  x  3 x  1  0. The expression on the left of the inequality changes sign when x  3 and x  1. Thus we must check the intervals in the following table.  3

3 1

1 

Sign of x  3







Sign of x  1













Interval

Sign of x  3 x  1

From the table, the solution set is x  x  3 or 1  x. Interval:  3  1 . Graph:

_3

1

45. x 2  4  x 2  4  0  x  2 x  2  0. The expression on the left of the inequality changes sign where x  2 and where x  2. Thus we must check the intervals in the following table.  2

2 2

2 

Sign of x  2







Sign of x  2













Interval

Sign of x  2 x  2

From the table, the solution set is x  2  x  2. Interval: 2 2. Graph:

_2

2

116

CHAPTER 1 Equations and Graphs

46. x 2  9  x 2  9  0  x  3 x  3  0. The expression on the left of the inequality changes sign when x  3 and x  3. Thus we must check the intervals in the following table.  3

3 3

3 

Sign of x  3







Sign of x  3













Interval

Sign of x  3 x  3

From the table, the solution set is x  x  3 or 3  x. Interval:  3]  [3 . Graph:

_3

3

47. x  2 x  1 x  3  0. The expression on the left of the inequality changes sign when x  2, x  1, and x  3. Thus we must check the intervals in the following table.  2

2 1

1 3

3 

Sign of x  2









Sign of x  1









Sign of x  3

















Interval

Sign of x  2 x  1 x  3

From the table, the solution set is x  x  2 or 1  x  3. Interval:  2][1 3]. Graph:

_2

1

3

48. x  5 x  2 x  1  0. The expression on the left of the inequality changes sign when x  5, x  2, and x  1. Thus we must check the intervals in the following table.  1

1 2

2 5

5 

Sign of x  5









Sign of x  2









Sign of x  1

















Interval

Sign of x  5 x  2 x  1

From the table, the solution set is x  1  x  2 or 5  x. Interval: 1 25 . Graph:

_1

2

5

49. x  4 x  22  0. Note that x  22  0 for all x  2, so the expression on the left of the original inequality changes sign only when x  4. We check the intervals in the following table. Interval Sign of x  4

Sign of x  22

Sign of x  4 x  22

 2

2 4

4 



















From the table, the solution set is x  x  2 and x  4. We exclude the

endpoint 2 since the original expression cannot be 0. Interval:  2  2 4. Graph:

_2

4

SECTION 1.7 Solving Inequalities

117

50. x  32 x  1  0. Note that x  32  0 for all x  3, so the expression on the left of the original inequality changes sign only when x  1. We check the intervals in the following table. Interval Sign of x  32

 3

3 1

1 



















Sign of x  1

Sign of x  32 x  1

From the table, the solution set is x  x  1. (The endpoint 3 is already excluded.) Interval: 1 . Graph:

_1

51. x  22 x  3 x  1  0. Note that x  22  0 for all x, so the expression on the left of the original inequality changes sign only when x  1 and x  3. We check the intervals in the following table. Interval Sign of x  22

 1

1 2

2 3

3 









Sign of x  3









Sign of x  1

















Sign of x  22 x  3 x  1

From the table, the solution set is x  1  x  3. Interval: [1 3]. Graph:

_1

3

  52. x 2 x 2  1  0  x 2 x  1 x  1  0. The expression on the left of the inequality changes sign when x  1 and x  0. Thus we must check the intervals in the following table. Interval Sign of x 2

 1

1 0

0 1

1 









Sign of x  1









Sign of x  1   Sign of x 2 x 2  1

















From the table, the solution set is x  x  1, x  0, or 1  x. (The endpoint 0 is included since the original expression is allowed to be 0.) Interval:  1]  0  [1 . Graph:

_1

0

1

  53. x 3  4x  0  x x 2  4  0  x x  2 x  2  0. The expression on the left of the inequality changes sign where x  0, x  2 and where x  4. Thus we must check the intervals in the following table. Interval

 2

2 0

0 2

2 

Sign of x









Sign of x  2









Sign of x  2

















Sign of x x  2 x  2

From the table, the solution set is x  2  x  0 or x  2. Interval: 2 02 . Graph:

_2

0

2

118

CHAPTER 1 Equations and Graphs

  54. 16x  x 3  0  x 3  16x  x x 2  16  x x  4 x  4. The expression on the left of the inequality changes sign when x  4, x  0, and x  4. Thus we must check the intervals in the following table.  4

4 0

0 4

4 

Sign of x  4









Sign of x









Sign of x  4

















Interval

Sign of x x  4 x  4

From the table, the solution set is x  4  x  0 or 4  x. Interval: [4 0]  [4 . Graph:

55.

 3

  3 12



Sign of x  3



Sign of 2x  1







Sign of







x 3 2x  1





1 2

4



From the table, the solution set is   x  x  3 or x  12 . Since the denominator

cannot equal 0, x  12 .   Interval:  3]  12   . Graph:

_3

1 _ 2

4x  0. The expression on the left of the inequality changes sign when x  4 and x  4. Thus we must check the x 4 intervals in the following table.  4

4 4

4 

Sign of 4  x







Sign of x  4







Sign of







Interval

57.

0

x 3  0. The expression on the left of the inequality changes sign where x  3 and where x  12 . Thus we must 2x  1 check the intervals in the following table. Interval

56.

_4

4x x 4

From the table, the solution set is x  x  4 or x  4. Interval:  4  4 . Graph:

_4

4

4x  0. The expression on the left of the inequality changes sign where x  4. Thus we must check the intervals in x 4 the following table. Interval

 4

4 4

4 

Sign of 4  x







Sign of x  4







Sign of







4x x 4

From the table, the solution set is x  x  4 or x  4. Interval:  4  4 . Graph:

_4

4

SECTION 1.7 Solving Inequalities

119

x 1 x 1 2 x  3 3x  5 x 1 0 20   0  The expression on the left of the inequality x 3 x 3 x 3 x 3 x 3 changes sign when x  53 and x  3. Thus we must check the intervals in the following table.

58. 2 

   53

Interval Sign of 3x  5



Sign of x  3

3x  5 Sign of x 3

59.



53 3

3 

















From the table, the solution set is   x  x  53 or 3  x   .   Interval:  53  3 . Graph:

5 _ 3

3

2x  1 2x  1 3 x  5 x  16 2x  1 3 3 0  0  0. The expression on the left of the inequality x 5 x 5 x 5 x 5 x 5 changes sign where x  16 and where x  5. Thus we must check the intervals in the following table.  5

5 16

16 

Sign of x  16







Sign of x  5













Interval

x  16 Sign of x 5 60.



From the table, the solution set is x  x  5 or x  16. Since the denominator cannot equal 0, we must have x  5. Interval:  5  [16 . Graph:

5

16

3x 3x 3x 2x 3x 1 1 0  0  0. The expression on the left of the inequality changes 3x 3x 3x 3x 3x sign when x  0 and x  3. Thus we must check the intervals in the following table. Since the denominator cannot equal 0, we must  0

0 3

3 

Sign of 3  x







Sign of 2x













Interval

Sign of

61.

2x 3x

have x  3. The solution set is x  0  x  3. Interval: [0 3. Graph:

0

3

4 4 xx 4  x2 4 2  x 2  x  x  x 0  0 0  0. The expression on the left of the x x x x x x inequality changes sign where x  0, where x  2, and where x  2. Thus we must check the intervals in the following table.  2

2 0

0 2

2 

Sign of 2  x









Sign of x









Sign of 2  x









Sign of









Interval

2  x 2  x x

From the table, the solution set is x  2  x  0 or 2  x. Interval: 2 02 . Graph:

_2

0

2

120

62.

CHAPTER 1 Equations and Graphs

x x 3x x  1 2x  3x 2 x 2  3x x  3x   3x  0   0 0  0. The expression on x 1 x 1 x 1 x 1 x 1 x 1 the left of the inequality changes sign whenx  0, x   23 , and x  1. Thus we must check the intervals in the following table.  1

Interval

63. 1 

_1

2

__ 3

   23  0

0 





Sign of x



Sign of 2  3x









Sign of x  1









Sign of









2  x 2  x x

From the table, the solution set is

Graph:

  1  23





   x  x  1 or  23  x  0 . Interval:  1   23  0 .

0

2 2 2 2 x x  1 2x 2 x  1 x 2  x  2x  2x  2  1  0   0 0 x 1 x x 1 x x x  1 x x  1 x x  1 x x  1

x2  x  2 x  2 x  1 0  0. The expression on the left of the inequality changes sign where x  2, where x x  1 x x  1 x  1, where x  0, and where x  1. Thus we must check the intervals in the following table.  2

2 1

1 0

0 1

1 

Sign of x  2











Sign of x  1











Sign of x











Sign of x  1





















Interval

x  2 x  1 Sign of x x  1

Since x  1 and x  0 yield undefined expressions, we cannot include them in the solution. From the table, the solution set is x  2  x  1 or 0  x  1. Interval: [2 1  0 1]. Graph:

_2

_1

0

1

SECTION 1.7 Solving Inequalities

64.

121

4 3 4 3x 4 x  1 x x  1 3x  4x  4  x 2  x 3  1  1  0   0 0 x 1 x x 1 x x x  1 x x  1 x x  1 x x  1 4  x2 2  x 2  x 0  0 The expression on the left of the inequality changes sign when x  2, x  2, x x  1 x x  1 x  0, and x  1. Thus we must check the intervals in the following table.  2

2 0

0 1

1 2

2 

Sign of 2  x











Sign of 2  x











Sign of x











Sign of x  1





















Interval

2  x 2  x Sign of x x  1

Since x  0 and x  1 give undefined expressions, we cannot include them in the solution. From the table, the solution set is x  2  x  0 or 1  x  2. Interval: [2 0  1 2]. Graph:

65.

_2

0

1

2

6 6 6 6x 6 x  1 x x  1 6  1  10   0 x 1 x x 1 x x x  1 x x  1 x x  1 6x  6x  6  x 2  x x 2  x  6 x  3 x  2 0 0  0. The x x  1 x x  1 x x  1 expression on the left of the inequality changes sign where x  3, where x  2, where x  0, and where x  1. Thus we must check the intervals in the following table.  2

2 0

0 1

1 3

3 

Sign of x  3











Sign of x  2











Sign of x











Sign of x  1





















Interval

x  3 x  2 Sign of x x  1

From the table, the solution set is x  2  x  0 or 1  x  3. The points x  0 and x  1 are excluded from the solution set because they make the denominator zero. Interval: [2 0  1 3]. Graph:

_2

0

1

3

122

66.

CHAPTER 1 Equations and Graphs

5 x 5 x x  1 25 4 2 x  1 x 2  x  10  8x  8 x  4  4 0   0 0 2 x 1 2 x 1 2 x  1 2 x  1 2 x  1 2 x  1 x 2  7x  18 x  9 x  2 0  0. The expression on the left of the inequality changes sign when x  9, x  2, 2 x  1 2 x  1 and x  1. Thus we must check the intervals in the following table.  2

2 1

1 9

9 

Sign of x  9









Sign of x  2









Sign of x  1

















Interval

x  9 x  2 Sign of 2 x  1

From the table, the solution set is x  2  x  1 or 9  x. The point x  1 is excluded from the solution set because it makes the expression undefined. Interval: [2 1  [9 . Graph:

67.

_1 _2

9

x 1 x 2 x 1 x 2 x  2 x  2 x  1 x  3    0  0 x 3 x 2 x 3 x 2 x  3 x  2 x  2 x  3 x 2  4  x 2  2x  3 2x  1 0  0. The expression on the left of the inequality x  3 x  2 x  3 x  2 changes sign where x   12 , where x  3, and where x  2. Thus we must check the intervals in the following table.  3

Interval

1

__ 2

2 







Sign of x  3









Sign of x  2

















From the table, the solution set is

_3

   12  2

Sign of 2x  1

2x  1 Sign of x  3 x  2

Graph:

  3  12

2





   x  3  x   12 or 2  x . Interval: 3  12  2 .

SECTION 1.7 Solving Inequalities

68.

123

1 x 2 1 x 1 x 2x 1 2x  3  0  0 0  0. The x 1 x 2 x  1 x  2 x  1 x  2 x  1 x  2 x  1 x  2 expression on the left of the inequality changes sign when x   32 , x  1, and x  2. Thus we must check the intervals in the following table.  2

Interval

  2  32

   32  1

1 

Sign of 2x  3







Sign of x  1









Sign of x  2









Sign of









2x  3 x  1 x  2



  From the table, the solution set is x  x  2 or  32  x  1 . The points x  2 and x  1 are   excluded from the solution because the expression is undefined at those values. Interval:  2   32  1 . Graph:

69.

_2

x  1 x  2 x  22

3

_1

__ 2

 0. Note that x  22  0 for all x. The expression on the left of the original inequality changes sign

when x  2 and x  1. We check the intervals in the following table.  2

2 1

1 2

2 

Sign of x  1









Sign of x  2

























Interval

Sign of x  22 Sign of

x  1 x  2 x  22

From the table, and recalling that the point x  2 is excluded from the solution because the expression is

undefined at those values, the solution set is x  x  2 or x  1 and x  2. Interval:  2]  [1 2  2 . Graph:

_2

1

2

124

70.

CHAPTER 1 Equations and Graphs

2x  1 x  32  0. Note that x  32  0 for all x  3. The expression on the left of the inequality changes sign x 4

when x  12 and x  4. We check the intervals in the following table.     13  12 Interval 2 Sign of 2x  1

Sign of x  32 Sign of x  4

3 4

4 

























2x  1 x  32 Sign of     x 4   From the table, the solution set is x  x  3 and 12  x  4 . We exclude the endpoint 3 because the original expression

  cannot be 0. Interval: 12  3  3 4. Graph:

1 2

3

4

  71. x 4  x 2  x 4  x 2  0  x 2 x 2  1  0  x 2 x  1 x  1  0. The expression on the left of the inequality changes sign where x  0, where x  1, and where x  1. Thus we must check the intervals in the following table.  1

1 0

0 1

1 









Sign of x  1









Sign of x  1

















Interval Sign of x 2

Sign of x 2 x  1 x  1

From the table, the solution set is x  x  1 or 1  x. Interval:  1  1 . Graph:

_1

1

    72. x 5  x 2  x 5  x 2  0  x 2 x 3  1  0  x 2 x  1 x 2  x  1  0. The expression on the left of the inequality   1  12  4 1 1 1  3 2  . changes sign when x  0 and x  1. But the solution of x  x  1  0 are x  2 1 2 Since these are not real solutions. The expression x 2  x  1 does not change signs, so we must check the intervals in the

following table.

 0

0 1

Sign of x 2

1 







Sign of x  1



















Interval

Sign of x 2  x  1





Sign of x 2 x  1 x 2  x  1

From the table, the solution set is x  1  x. Interval: 1 . Graph:

1

SECTION 1.7 Solving Inequalities

73. For

125

 16  9x 2 to be defined as a real number we must have 16  9x 2  0  4  3x 4  3x  0. The expression in the

inequality changes sign at x  43 and x   43 . Interval Sign of 4  3x

    43

   43  43















Sign of 4  3x

Sign of 4  3x 4  3x







4 3



Thus  43  x  43 .  74. For 3x 2  5x  2 to be defined as a real number we must have 3x 2  5x  2  0  3x  2 x  1  0. The

expression on the left of the inequality changes sign whenx  23 and x  1. Thus we must check the intervals in the following table.     21  23 Interval 1  3 Sign of 3x  2







Sign of x  1













Sign of 3x  2 x  1

Thus x  23 or 1  x. 12  1 to be defined as a real number we must have x 2  5x  14  0  x  7 x  2  0. The 75. For x 2  5x  14 expression in the inequality changes sign at x  7 and x  2.

 2

2 7

7 

Sign of x  7







Sign of x  2













Interval

Sign of x  7 x  2

Thus x  2 or 7  x, and the solution set is  2  7 .  1x 1x to be defined as a real number we must have  0. The expression on the left of the inequality changes 76. For 4 2x 2x sign when x  1 and x  2. Thus we must check the intervals in the following table.  2

2 1

1 

Sign of 1  x







Sign of 2  x







Sign of







Interval

1x 2x

Thus 2  x  1. Note that x  2 has been excluded from the solution set because the expression is undefined at that

value.

77. a bx  c  bc (where a, b, c  0)  bx  c 

bc 1 bc  bx  cx  a a b

78. We have a  bx  c  2a, where a,b, c  0  a  c  bx  2a  c 



 bc c c c c c   x   . a a b a b

ac 2a  c x . b b

126

CHAPTER 1 Equations and Graphs

79. Inserting the relationship C  59 F  32, we have 20  C  30  20  59 F  32  30  36  F  32  54  68  F  86. 80. Inserting the relationship F  95 C  32, we have 50  F  95  50  95 C  32  95  18  95 C  63  10  C  35. 81. Let x be the average number of miles driven per day. Each day the cost of Plan A is 30  010x, and the cost of Plan B is

50. Plan B saves money when 50  30  010x  20  01x  200  x. So Plan B saves money when you average more than 200 miles a day.

82. Let m be the number of minutes of long-distance calls placed per month. Then under Plan A, the cost will be 25  005m, and under Plan B, the cost will be 5  012m. To determine when Plan B is advantageous, we must solve

25  005m  5  012m  20  007m  2857  m. So Plan B is advantageous if a person places fewer than 286 minutes of long-distance calls during a month.

83. We need to solve 6400  035m  2200  7100 for m. So 6400  035m  2200  7100  4200  035m  4900  12,000  m  14,000. She plans on driving between 12,000 and 14,000 miles. h , where T is the temperature in  C, and h is the height in meters. 100 (b) Solving the expression in part (a) for h, we get h  100 20  T . So 0  h  5000  0  100 20  T   5000  0  20  T  50  20  T  30  20  T  30. Thus the range of temperature is from 20 C down to 30 C.

84. (a) T  20 

85. (a) Let x be the number of $3 increases. Then the number of seats sold is 120  x. So P  200  3x

 3x  P  200  x  13 P  200. Substituting for x we have that the number of seats sold is

120  x  120  13 P  200   13 P  560 3 .

(b) 90   13 P  560 3  115  270  360  P  200  345  270  P  560  345  290  P  215  290  P  215. Putting this into standard order, we have 215  P  290. So the ticket prices are between $215 and $290. c . Since the 65 c  303  scale’s accuracy is 003 lb, and the scale shows 3 lb, we have 3  003  x  3  003  297  65 650 297  c  650 303  19305  c  19695. Since the customer paid $1950, he could have been over- or

86. If the customer buys x pounds of coffee at $650 per pound, then his cost c will be 650x. Thus x 

undercharged by as much as 195 cents.

87. 00004 

4,000,000  001. Since d 2  0 and d  0, we can multiply each expression by d 2 to obtain d2

00004d 2  4,000,000  001d 2 . Solving each pair, we have 00004d 2  4,000,000  d 2  10,000,000,000

 d  100,000 (recall that d represents distance, so it is always nonnegative). Solving 4,000,000  001d 2 

400,000,000  d 2  20,000  d. Putting these together, we have 20,000  d  100,000.

SECTION 1.7 Solving Inequalities

88.

127

  600,000  500  600,000  500 x 2  300 (Note that x 2  300  300  0, so we can multiply both sides by the 2 x  300

denominator and not worry that we might be multiplying both sides by a negative number or by zero.) 1200  x 2  300  0  x 2  900  0  x  30 x  30. The expression in the inequality changes sign at x  30 and x  30. However, since x represents distance, we must have x  0.

0 30

30 

Sign of x  30





Sign of x  30









Interval

Sign of x  30 x  30

So x  30 and you must stand at least 30 meters from the center of the fire.   89. 128  16t  16t 2  32  16t 2  16t  96  0  16 t 2  t  6  0  16 t  3 t  2  0. The expression on the left of the inequality changes sign at x  2, at t  3, and at t  2. However, t  0, so the only endpoint is t  3. 0 3

3 

Sign of 16





Sign of t  3





Sign of t  2









Interval

Sign of 16 t  3 t  2 So 0  t  3.

90. Solve 30  10  09  001 2 for 10    75. We have 30  10  09  001 2  001 2  09  20  0  01  4 01  5  0. The possible endpoints are 01  4  0  01  4    40 and 01  5  0  01  5    50.

10 40

40 50

50 75

Sign of 01  4







Sign of 01  5













Interval

Sign of 01  4 01  5 Thus he must drive between 40 and 50 mi/h.

  2 1  2    240  0  1   3   80  0. The expression in the inequality changes sign at  20 20 20   60 and   80. However, since  represents the speed, we must have   0.

91. 240   

Interval 1  3 Sign of 20

Sign of   80   1   3   80 Sign of 20

So Kerry must drive between 0 and 60 mi/h.

0 60

60 













128

CHAPTER 1 Equations and Graphs

  92. Solve 2400  20x  2000  8x  00025x 2  2400  20x  2000  8x  00025x 2  00025x 2  12x  4400  0  00025x  1 x  4400  0. The expression on the left of the inequality changes sign whenx  400 and x  4400.

Since the manufacturer can only sell positive units, we check the intervals in the following table. 0 400

400 4400

4400 

Sign of 00025x  1







Sign of x  4400













Interval

Sign of 00025x  1 x  4400

So the manufacturer must sell between 400 and 4400 units to enjoy a profit of at least $2400. 93. Let x be the length of the garden and  its width. Using the fact that the perimeter is 120 ft, we must have 2x  2  120    60  x. Now since the area must be at least 800 ft2 , we have 800  x 60  x  800  60x  x 2 

x 2  60x  800  0  x  20 x  40  0. The expression in the inequality changes sign at x  20 and x  40.

However, since x represents length, we must have x  0.

0 20

20 40

40 

Sign of x  20







Sign of x  40













Interval

Sign of x  20 x  40 The length of the garden should be between 20 and 40 feet. 94. Case 1: a  b  0 a b0

We have a  a  a  b, since a  0, and b  a  b  b, since b  0. So a 2  a  b  b2 , that is

a 2  b2 . Continuing, we have a  a 2  a  b2 , since a  0 and b2  a  b2  b, since b2  0. So

a 3  ab2  b3 . Thus a  b  0  a 3  b3 . So a  b  0  a n  bn , if n is even, and a n  b, if n is odd. Case 2: 0  a  b

We have a  a  a  b, since a  0, and b  a  b  b, since b  0. So a 2  a  b  b2 . Thus 0 

a  b  a 2  b2 . Likewise, a 2  a  a 2  b and b  a 2  b  b2 , thus a 3  b3 . So 0  a  b  a n  bn , for all positive integers n.

Case 3: a  0  b

If n is odd, then a n  bn , because a n is negative and bn is positive. If n is even, then we could have

either a n  bn or a n  bn . For example, 1  2 and 12  22 , but 3  2 and 32  22 .

95. The rule we want to apply here is “a  b  ac  bc if c  0 and a  b  ac  bc if c  0 ”. Thus we cannot simply

Case 1: x  0

3 , we must consider two cases. x Multiplying both sides by x, we have x  3. Together with our initial condition, we have 0  x  3.

Case 2: x  0

Multiplying both sides by x, we have x  3. But x  0 and x  3 have no elements in common, so this

multiply by x, since we don’t yet know if x is positive or negative, so in solving 1 

gives no additional solution. Hence, the only solutions are 0  x  3. 96. a  b, so by Rule 1, a  c  b  c. Using Rule 1 again, b  c  b  d, and so by transitivity, a  c  b  d. 97.

a c a c ad ad  , so by Rule 3, d  d   c. Adding a to both sides, we have  a  c  a. Rewriting the left-hand b d b d b b ad ab a b  d a ac side as   and dividing both sides by b  d gives  . b b b b bd cb c b  d ac c Similarly, a  c  c  , so  . d d bd d

SECTION 1.8 Solving Absolute Value Equations and Inequalities

1.8

129

SOLVING ABSOLUTE VALUE EQUATIONS AND INEQUALITIES

1. The equation x  3 has the two solutions 3 and 3. 2. (a) The solution of the inequality x  3 is the interval [3 3].

(b) The solution of the inequality x  3 is a union of two intervals  3]  [3 .

3. (a) The set of all points on the real line whose distance from zero is less than 3 can be described by the absolute value inequality x  3.

(b) The set of all points on the real line whose distance from zero is greater than 3 can be described by the absolute value inequality x  3.

4. (a) 2x  1  5 is equivalent to the two equations 2x  1  5 and 2x  1  5. (b) 3x  2  8 is equivalent to 8  3x  2  8.

5. 5x  20  5x  20  x  4.

6. 3x  10  3x  10  x   10 3.

7. 5 x  3  28  5 x  25  x  5  x  5.

8. 12 x  7  2  12 x  9  x  18  x  18.

9. x  3  2 is equivalent to x  3  2  x  3  2  x  1 or x  5.

10. 2x  3  7 is equivalent to either 2x  3  7  2x  10  x  5; or 2x  3  7  2x  4  x  2. The two solutions are x  5 and x  2. 11. x  4  05 is equivalent to x  4  05  x  4  05  x  45 or x  35.

12. x  4  3. Since the absolute value is always nonnegative, there is no solution.

13. 2x  3  11 is equivalent to either 2x  3  11  2x  14  x  7; or 2x  3  11  2x  8  x  4. The two solutions are x  7 and x  4.

14. 2  x  11 is equivalent to either 2  x  11  x  9; or 2  x  11  x  13. The two solutions are x  9 and x  13.

15. 4  3x  6  1   3x  6  3  3x  6  3, which is equivalent to either 3x  6  3  3x  3  x  1; or 3x  6  3  3x  9  x  3. The two solutions are x  1 and x  3.

16. 5  2x  6  14  5  2x  8 which is equivalent to either 5  2x  8  2x  3  x   32 ; or 5  2x  8 

3 13 2x  13  x  13 2 . The two solutions are x   2 and x  2 . 17. 3 x  5  6  15  3 x  5  9  x  5  3, which is equivalent to either x  5  3  x  2; or x  5  3  x  8. The two solutions are x  2 and x  8.

18. 20  2x  4  15  2x  4  5. Since the absolute value is always nonnegative, there is no solution.             19. 8  5  13 x  56   33  5  13 x  56   25   13 x  56   5, which is equivalent to either 13 x  56  5  13 x  35 6 

1 5 1 25 25 25 35 x  35 2 ; or 3 x  6  5  3 x   6  x   2 . The two solutions are x   2 and x  2 .         3 9 20.  35 x  2  12  4   35 x  2  92 which is equivalent to either 35 x  2  92  35 x  52  x  25 6 ; or 5 x  2   2  3 x   13  x   65 . The two solutions are x  25 and x   65 . 5 2 6 6 6

21. x  1  3x  2, which is equivalent to either x  1  3x  2  2x  3  x   32 ; or x  1   3x  2  x  1  3x  2  4x  1  x   14 . The two solutions are x   32 and x   14 .

22. x  3  2x  1 is equivalent to either x  3  2x  1  x  2  x  2; or x  3   2x  1  x  3  2x  1  3x  4  x   43 . The two solutions are x  2 and x   43 .

23. x  5  5  x  5. Interval: [5 5].

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24. 2x  20  20  2x  20  10  x  10. Interval: [10 10].

    25. 2x  7 is equivalent to 2x  7  x  72 ; or 2x  7  x   72 . Interval:   72  72   .

26. 12 x  1  x  2 is equivalent to x  2 or x  2. Interval:  2]  [2 .

27. x  4  10 is equivalent to 10  x  4  10  6  x  14. Interval: [6 14].

28. x  3  9 is equivalent to x  3  9  x  6; or x  3  9  x  12. Interval:  6  12 . 29. x  1  1 is equivalent to x  1  1  x  0; or x  1  1  x  2. Interval:  2]  [0 .

30. x  4  0 is equivalent to x  4  0  x  4  0  x  4. The only solution is x  4.

31. 2x  1  3 is equivalent to 2x  1  3  2x  4  x  2; or 2x  1  3  2x  2  x  1. Interval:  2]  [1 .

32. 3x  2  7 is equivalent to 3x  2  7  3x  5  x   53 ; or 3x  2  7  3x  9  x  3. Interval:     53  3 .

33. 2x  3  04  04  2x  3  04  26  2x  34  13  x  17. Interval: [13 17].   34. 5x  2  6  6  5x  2  6  4  5x  8   45  x  85 . Interval:  45  85 .   x  2   2  2  x  2  2  6  x  2  6  4  x  8. Interval: 4 8.  35.  3  3     x  1   4   1 x  1  4  1 x  1  4  x  1  8 which is equivalent to either x  1  8  x  7; or 36.   2 2 2 x  1  8  x  9. Interval:  9]  [7 .

37. x  6  0001  0001  x  6  0001  6001  x  5999. Interval: 6001 5999. 38. x  a  d  d  x  a  d  a  d  x  a  d. Interval: a  d a  d.

39. 4 x  2  3  13  4 x  2  16  x  2  4  4  x  2  4  6  x  2. Interval: 6 2.

40. 3  2x  4  1   2x  4  2  2x  4  2 which is equivalent to either 2x  4  2  2x  2  x  1; or 2x  4  2  2x  6  x  3. Interval:  3]  [1 .

41. 8  2x  1  6   2x  1  2  2x  1  2  2  2x  1  2  1  2x  3   12  x  32 .   Interval:  12  32 .

42. 7 x  2  5  4  7 x  2  1  x  2   17 . Since the absolute value is always nonnegative, the inequality is true for all real numbers. In interval notation, we have  .     1     43. 12 4x  13   56  4x  13   53 , which is equivalent to either 4x  13  53  4x  43  x  13 ; or 4x    53  3     4x  2  x   12 . Interval:   12  13   .             44. 2  12 x  3  3  51  2  12 x  3  48   12 x  3  24  24  12 x  3  24  27  12 x  21  54  x  42. Interval: [54 42].

45. 1  x  4. If x  0, then this is equivalent to 1  x  4. If x  0, then this is equivalent to 1  x  4  1  x  4  4  x  1. Interval: [4 1]  [1 4].

46. 0  x  5  12 . For x  5, this is equivalent to  12  x  5  12  92  x  11 2 . Since x  5 is excluded, the solution is     9  5  5 11 . 2 2

47.

1 13  2  1  2 x  7 (x  7)  x  7  12   12  x  7  12   15 2  x   2 and x  7. x  7     13 Interval:  15 2  7  7  2 .

SECTION 1.9 Solving Equations and Inequalities Graphically

48.

131

1  5  15  2x  3, since 2x  3  0, provided 2x  3  0  x  32 . Now for x  32 , we have 2x  3

1  2x  3 is equivalent to either 1  2x  3  16  2x  8  x; or 2x  3   1  2x  14  x  7 . 5 5 5 5 5 5 5     7 8 Interval:  5  5   .

49. x  3

50. x  2

51. x  7  5

52. x  2  4

53. x  2

54. x  1

55. x  3

56. x  4

57. (a) Let x be the thickness of the laminate. Then x  0020  0003.

(b) x  0020  0003  0003  x  0020  0003  0017  x  0023.    h  682    2  2  h  682  2  58  h  682  58  624  h  740. Thus 95% of the adult males are 58.  29  29 between 624 in and 740 in.

59. x  1 is the distance between x and 1; x  3 is the distance between x and 3. So x  1  x  3 represents those points closer to 1 than to 3, and the solution is x  2, since 2 is the point halfway between 1 and 3. If a  b, then the ab . solution to x  a  x  b is x  2

1.9

SOLVING EQUATIONS AND INEQUALITIES GRAPHICALLY

1. The solutions of the equation x 2  2x  3  0 are the x-intercepts of the graph of y  x 2  2x  3. 2. The solutions of the inequality x 2  2x  3  0 are the x-coordinates of the points on the graph of y  x 2  2x  3 that lie above the x-axis. 3. (a) From the graph, it appears that the graph of y  x 4  3x 3  x 2  3x has x-intercepts 1, 0, 1, and 3, so the solutions to the equation x 4  3x 3  x 2  3x  0 are x  1, x  0, x  1, and x  3.

(b) From the graph, we see that where 1  x  0 or 1  x  3, the graph lies below the x-axis. Thus, the inequality x 4  3x 3  x 2  3x  0 is satisfied for x  1  x  0 or 1  x  3  [1 0]  [1 3].

4. (a) The graphs of y  5x  x 2 and y  4 intersect at x  1 and at x  4, so the equation 5x  x 2  4 has solutions x  1 and x  4. (b) The graph of y  5x  x 2 lies strictly above the graph of y  4 when 1  x  4, so the inequality 5x  x 2  4 is satisfied for those values of x, that is, for x  1  x  4  1 4.

5. Algebraically: x  4  5x  12  16  4x  x  4. Graphically: We graph the two equations y1  x  4 and y2  5x  12 in the viewing rectangle [6 4] by

[10 2]. Zooming in, we see that the solution is x  4.

6. Algebraically: 12 x  3  6  2x  9  32 x  x  6. Graphically: We graph the two equations y1  12 x  3 and y2  6  2x in the viewing rectangle [10 5] by

[10 5]. Zooming in, we see that the solution is x  6. 5

-6

-4

-2

2 -5

4 -10

-5

5 -5

-10

-10

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CHAPTER 1 Equations and Graphs

7. Algebraically:

1 2   7  2x x 2x



1 2  x 2x



 2x 7

5.  4  1  14x  x  14

1 2  x 2x and y2  7 in the viewing rectangle [2 2] by [2 8].

Graphically: We graph the two equations y1 

Zooming in, we see that the solution is x  036.

5

6 5 4    8. Algebraically: x  2 2x 2x  4     6 4 5 2x x  2   2x x  2  x  2 2x 2x  4 2x 4  x  2 6  x 5  8x  6x  12  5x  12  3x  4  x. Graphically: We graph the two equations 6 4 5  and y2  in the viewing y1  x  2 2x 2x  4 rectangle [5 5] by [10 10]. Zooming in, we see that there is only one solution at x  4. 10

-2

-1

1

2 -4

-2

2

4

-10

9. Algebraically: x 2  32  0  x 2  32    x   32  4 2. Graphically: We graph the equation y1  x 2  32 and determine where this curve intersects the x-axis. We use the viewing rectangle [10 10] by [5 5]. Zooming in, we see that solutions are x  566 and x  566.

 10. Algebraically: x 3  16  0  x 3  16  x  2 3 2. Graphically: We graph the equation y  x 3  16 and

determine where this curve intersects the x-axis. We use the viewing rectangle [5 5] by [5 5]. Zooming in, we see that the solution is x  252. 4

4

2

2 -10

-5

-2

5

10

-4

-2

11. Algebraically: x 2  9  0  x 2  9, which has no real solution.

Graphically: We graph the equation y  x 2  9 and see that this curve does not intersect the x-axis. We use the viewing rectangle [5 5] by [5 30]. 30 20

12. Algebraically: x 2  3  2x  x 2  2x  3  0     2  22  4 1 3 2  8  . x 2 1 2 1 Because the discriminant is negative, there is no real solution. Graphically: We graph the two equations y1  x 2  3 and y2  2x in the viewing rectangle [4 6] by [6 12], and see that the two curves do not intersect.

10 -2

4

-4

-4

-4

2

-2

10 2

5

4 -4

-2

2 -5

4

6

SECTION 1.9 Solving Equations and Inequalities Graphically

13. Algebraically: 16x 4  625  x 4  625 16  x   52  25. Graphically: We graph the two equations y1  16x 4 and

y2  625 in the viewing rectangle [5 5] by [610 640].

Zooming in, we see that solutions are x  25. 640

14. Algebraically: 2x 5  243  0  2x 5  243  x 5  243 2   5 5 243 3 x  2  2 16. Graphically: We graph the equation y  2x 5  243 and

determine where this curve intersects the x-axis. We use the viewing rectangle [5 10] by [5 5]. Zooming in, we see that the solution is x  261. 4

630

2

620 -4

-2

133

0

2

-5

4

5

-2

10

-4

15. Algebraically: x  54  80  0  x  54  80     x  5   4 80  2 4 5  x  5  2 4 5.

Graphically: We graph the equation y1  x  54  80

and determine where this curve intersects the x-axis. We use the viewing rectangle [1 9] by [5 5]. Zooming in, we see that solutions are x  201 and x  799.

32 16. Algebraically: 6 x  25  64  x  25  64 6  3    2 5 2 5  x  2  5 32 3  3 81  x  2  3 81.

Graphically: We graph the two equations y1  6 x  25

and y2  64 in the viewing rectangle [5 5] by [50 70]. Zooming in, we see that the solution is x  039 70

4 2

60 -2

2

4

6

8

-4

-4

-2

0

2

4

17. We graph y  x 2  7x  12 in the viewing rectangle [0 6] 18. We graph y  x 2  075x  0125 in the viewing by [01 01]. The solutions appear to be exactly x  3

and x  4. [In fact x 2  7x  12  x  3 x  4.]

rectangle [2 2] by [01 01]. The solutions are x  025 and x  050. 0.1

0.1

0.0 2 -0.1

4

6

-2

-1

1 -0.1

2

134

CHAPTER 1 Equations and Graphs

19. We graph y  x 3  6x 2  11x  6 in the viewing

20. Since 16x 3  16x 2  x  1  16x 3  16x 2  x  1  0, we graph y  16x 3  16x 2  x  1 in the viewing

rectangle [1 4] by [01 01]. The solutions are x  100, x  200,and x  300.

rectangle [2 2] by [01 01]. The solutions are: x  100, x  025, and x  025.

0.1

0.1 -1

1

2

3

4 -2

-0.1

-1

1

2

-0.1

21. We first graph y  x 



x  1 in the viewing rectangle [1 5] by [01 01] and

find that the solution is near 16. Zooming in, we see that solutions is x  162.

0.1

-1

1

2

3

4

5

-0.1

 1  x2     1  x  1  x 2  0Since x is only defined

22. 1 



x

1

0.01

for x  0, we start with the viewing rectangle [1 5] by [1 1]. In this rectangle, there

appears to be an exact solution at x  0 and

another solution between x  2 and x  25. We then use the viewing rectangle [23 235] by

0.00 -1

1

2

3

4

5

-1

2.32

2.34

-0.01

[001 001], and isolate the second solution as x  2314. Thus the solutions are x  0 and x  231. 23. We graph y  x 13  x in the viewing rectangle [3 3] by [1 1]. The solutions

1

are x  1, x  0, and x  1, as can be verified by substitution.

-3

-2

-1

1

2

3

-1

24. Since x 12 is defined only for x  0, we start by

1

graphing y  x 12  x 13  x in the viewing

0.01

rectangle [1 5] by [1 1] We see a solution at x  0 and another one between x  3 and x  35. We then use the viewing rectangle

[33 34] by [001 001], and isolate the second solution as x  331. Thus, the solutions are x  0 and x  331.

0.00 -1 -1

1

2

3

4

5

3.35 -0.01

3.40

SECTION 1.9 Solving Equations and Inequalities Graphically

 2x  1  1 and y  x in the viewing rectangle [3 6] by [0 6]  and see that the only solution to the equation 2x  1  1  x is x  4, which can

25. We graph y 

135

6 4

be verified by substitution.

2 -2

0

2

 3  x  2 and y  1  x in the viewing rectangle [7 4] by  [2 8] and see that the only solution to the equation 3  x  2  1  x is

4

6

26. We graph y 

5

x  356, which can be verified by substitution.

-6

-4

-2

27. We graph y  2x 4  4x 2  1 in the viewing rectangle [2 2] by [5 40] and see

2

4

40

that the equation 2x 4  4x 2  1  0 has no solution.

30 20 10 -2

-1

1

2

1

2

28. We graph y  x 6  2x 3  3 in the viewing rectangle [2 2] by [5 15] and see

that the equation x 6  2x 3  3  0 has solutions x  1 and x  144, which can

10

be verified by substitution.

-2

29. x 3  2x 2  x  1  0, so we start by graphing

-1

100

the function y  x 3  2x 2  x  1 in the viewing

1

rectangle [10 10] by [100 100]. There

appear to be two solutions, one near x  0 and

-10

another one between x  2 and x  3. We then use the viewing rectangle [1 5] by [1 1] and

-5

5

10

-100

-1

1

2

3

4

5

-1

zoom in on the only solution, x  255. 30. x 4  8x 2  2  0. We start by graphing the

10

function y  x 4  8x 2  2 in the viewing

1

rectangle [10 10] by [10 10]. There appear to be four solutions between x  3 and x  3. We then use the viewing rectangle [5 5] by [1 1], and zoom to find the four solutions x  278, x  051, x  051, and x  278.

-10

-5

5 -10

10

-4

-2

2 -1

4

136

CHAPTER 1 Equations and Graphs

31. x x  1 x  2  16 x 

10

x x  1 x  2  16 x  0. We start by graphing the function y  x x  1 x  2  16 x in the

-4

viewing rectangle [5 5] by [10 10]. There

1

-2

appear to be three solutions. We then use the

2

4

-2

-1

-10

1

2

-1

viewing rectangle [25 25] by [1 1] and zoom into the solutions at x  205, x  000, and x  105. 32. x 4  16  x 3 . We start by graphing the functions y1  x 4 and y2  16  x 3 in the viewing rectangle [10 10] by [5 40]. There appears to be two solutions, one near x  2 and another one near x  2. We then use the viewing rectangle [24 22] by [27 29], and zoom in to find the solution at x  231. We then use the viewing rectangle [17 18] by [95 105], and zoom in to find the solution at x  179. 40

29

30 28

20 10 -10

-5

5

10

-2.4

-2.3

27 -2.2

10.4 10.2 10.0 9.8 9.6 1.70

1.75

33. We graph y  x 2 and y  3x  10 in the viewing rectangle [4 7] by [5 30].

1.80

30

The solution to the inequality is [2 5].

20 10 -4

-2

2

34. Since 05x 2  0875x  025  05x 2  0875x  025  0, we graph

4

6

4

y  05x 2  0875x  025 in the viewing rectangle [3 1] by [5 5]. Thus the

2

solution to the inequality is [2 025].

-3

-2

-1

1

-2 -4

35. Since x 3  11x  6x 2  6  x 3  6x 2  11x  6  0, we graph

y  x 3  6x 2  11x  6 in the viewing rectangle [0 5] by [5 5]. The solution

set is  10]  [20 30].

4 2 0 -2 -4

2

4

SECTION 1.9 Solving Equations and Inequalities Graphically

36. Since 16x 3  24x 2  9x  1 

0.01

4

16x 3  24x 2  9x  1  0, we graph

2

y  16x 3  24x 2  9x  1 in the viewing

rectangle [3 1] by [5 5]. From this rectangle,

137

0.00 -3

-2

-1

we see that x  1 is an x-intercept, but it is

1

-2

-1.0

-0.5

-4

unclear what is occurring between x  05 and

-0.01

x  0. We then use the viewing rectangle [1 0] by [001 001]. It shows y  0 at x  025. Thus in interval notation, the solution is 1 025  025 . 37. Since x 13  x  x 13  x  0, we graph y  x 13  x

  05x 2  1  2 x  05x 2  1  2 x  0, we  graph y  05x 2  1  2 x in the viewing rectangle

38. Since

in the viewing rectangle [3 3] by [1 1]. From this, we find that the solution set is 1 0  1 .

[1 1] by [1 1]. We locate the x-intercepts at

x  0535. Thus in interval notation, the solution is approximately  0535]  [0535 .

1

1 -3

-2

-1

1

2

3

-1 -1.0

-0.5

0.5

1.0

-1

39. Since x  12  x  12  x  12  x  12  0, we graph y  x  12  x  12 in the viewing

40. Since x  12  x 3  x  12  x 3  0, we graph

rectangle [2 2] by [5 5]. The solution set is  0.

y  x  12  x 3 in the viewing rectangle [4 4] by

[1 1]. The x-intercept is close to x  2. Using a trace function, we obtain x  2148. Thus the solution is [2148 .

4 2

1 -2

-1

-2

1

2

-4 -4

-2

2

4

-1

41. We graph the equations y  3x 2  3x and y  2x 2  4 in the viewing rectangle

[2 6] by [5 50]. We see that the two curves intersect at x  1 and at x  4,

40

the inequality 3x 2  3x  2x 2  4 has the solution set 1 4.

20

and that the first curve is lower than the second for 1  x  4. Thus, we see that

-2

2

4

6

138

CHAPTER 1 Equations and Graphs

42. We graph the equations y  5x 2  3x and y  3x 2  2 in the viewing rectangle

20

[3 2] by [5 20]. We see that the two curves intersect at x  2 and at x  12 , which can be verified by substitution. The first curve is larger than the second for x  2 and for x  12 , so the solution set of the inequality 5x 2  3x  3x 2  2 is    2]  12   .

10

-3

-2

-1

1

2

43. We graph the equation y  x  22 x  3 x  1 in the viewing rectangle

[2 4] by [15 5] and see that the inequality x  22 x  3 x  1  0 has the solution set [1 3].

-2

-1

1

2

3

4

-10

  44. We graph the equation y  x 2 x 2  1 in the viewing rectangle [2 2] by   [1 1] and see that the inequality x 2 x 2  1  0 has the solution set  1]  0  [1 .

1

-2

-1

1

2

2

3

-1

45. To solve 5  3x  8x  20 by drawing the graph of a single equation, we isolate all terms on the left-hand side: 5  3x  8x  20  5  3x  8x  20  8x  20  8x  20  11x  25  0 or 11x  25  0. We graph y  11x  25, and see that the solution is x  227, as in Example 2.

1

-1

1 -1

46. Graphing y  x 3  6x 2  9x and y 



x in the viewing rectangle [001 002] 0.2

by [005 02], we see that x  0 and x  001 are solutions of the equation  x 3  6x 2  9x  x.

0.1

-0.01

0.01

0.02

SECTION 1.10 Modeling Variation

139

(c) We graph the equations y  15,000 and

47. (a) We graph the equation y  10x  05x 2  0001x 3  5000 in the viewing

y  10x  05x 2  0001x 3  5000 in the viewing

rectangle [0 600] by [30000 20000].

rectangle [250 450] by [11000 17000]. We use a zoom or trace function on a graphing calculator, and find that

20000

the company’s profits are greater than $15,000 for 279  x  400.

0 200

400

600 16000

-20000

14000

(b) From the graph it appears that

12000

0  10x  005x 2  0001x 3  5000 for

300

400

100  x  500, and so 101 cooktops must be produced

to begin to make a profit. 48. (a)

(b) Using a zoom or trace function, we find that y  10 for x  667. We  x 2  000036. So for could estimate this since if x  100, then 5280    x 2   15x. Solving 15x  10 we x  100 we have 15x  5280

15 10 5 0 0

50

100

get 15  100 or x  100 15  667 mi.

49. Answers will vary. 50. Calculators perform operations in the following order: exponents are applied before division and division is applied before addition. Therefore, Y_1=x^1/3 is interpreted as y  interpreted as y 

x x1  , which is the equation of a line. Likewise, Y_2=x/x+4 is 3 3

x  4  1  4  5. Instead, enter the following: Y_1=x^(1/3), Y_2=x/(x+4). x

1.10 MODELING VARIATION 1. If the quantities x and y are related by the equation y  3x then we say that y is directly proportional to x, and the constant of proportionality is 3. 3 2. If the quantities x and y are related by the equation y  then we say that y is inversely proportional to x, and the constant x of proportionality is 3. x 3. If the quantities x, y, and z are related by the equation z  3 then we say that z is directly proportional to x and inversely y proportional to y. 4. Because z is jointly proportional to x and y, we must have z  kx y. Substituting the given values, we get 10  k 4 5  20k  k  12 . Thus, x, y, and z are related by the equation z  12 x y. 5. (a) In the equation y  3x, y is directly proportional to x. (b) In the equation y  3x  1, y is not proportional to x. 3 , y is not proportional to x. 6. (a) In the equation y  x 1 3 (b) In the equation y  , y is inversely proportional to x. x

140

CHAPTER 1 Equations and Graphs

7. T  kx, where k is constant.

8. P  k, where k is constant.

k , where k is constant. z ks 11. y  , where k is constant. t 9.  

10.   kmn, where k is constant.

 13. z  k y, where k is constant.

k , where k is constant. T kx 2 14. A  3 , where k is constant. t

15. V  klh, where k is constant.

16. S  kr 2 2 , where k is constant.

12. P 

k P2t 2  , where k is constant. 18. A  k x y, where k is constant. 3 b 19. Since y is directly proportional to x, y  kx. Since y  42 when x  6, we have 42  k 6  k  7. So y  7x. k k 24 20.  is inversely proportional to t, so   . Since   3 when t  8, we have 3   k  24, so   . t 8 t k 21 k . 21. A varies inversely as r, so A  . Since A  7 when r  3, we have 7   k  21. So A  r 3 r 17. R 

1 . So P  1 T . 22. P is directly proportional to T , so P  kT . Since P  20 when T  300, we have 20  k 300  k  15 15 kx . Since A  42 when x  7 and t  3, we 23. Since A is directly proportional to x and inversely proportional to t, A  t 18x k 7  k  18. Therefore, A  . have 42  3 t 24. S  kpq. Since S  180 when p  4 and q  5, we have 180  k 4 5  180  20k  k  9. So S  9 pq. k k  k  360. 25. Since W is inversely proportional to the square of r, W  2 . Since W  10 when r  6, we have 10  r 62 360 So W  2 . r xy xy 2 3 26. t  k . Since t  25 when x  2, y  3, and r  12, we have 25  k  k  50. So t  50 . r 12 r 27. Since C is jointly proportional to l, , and h, we have C  klh. Since C  128 when l    h  2, we have 128  k 2 2 2  128  8k  k  16. Therefore, C  16lh.  2 28. H  kl 2 2 . Since H  36 when l  2 and   13 , we have 36  k 22 13  36  49 k  k  81. So H  81l 2 2 .

k k k 275  29. R   . Since R  25 when x  121, 25    k  275. Thus, R   . 11 x x 121 a 2 2 abc abc . Since M  128 when a  d and b  c  2, we have 128  k  4k  k  32. So M  32 . 30. M  k d a d

x3 31. (a) z  k 2 y

  x3 27 k 2 , so z changes by a factor of 27 (b) If we replace x with 3x and y with 2y, then z  k  4. 4 y 2y2 3x3

x2 32. (a) z  k 4 y

  x2 9 9. k 4 , so z changes by a factor of 16 (b) If we replace x with 3x and y with 2y, then z  k  16 y 2y4 3x2

33. (a) z  kx 3 y 5

(b) If we replace x with 3x and y with 2y, then z  k 3x3 2y5  864kx 3 y 5 , so z changes by a factor of 864.

SECTION 1.10 Modeling Variation

141

k 34. (a) z  2 3 x y (b) If we replace x with 3x and y with 2y, then z 

k 3x2 2y3



1 k 1 . , so z changes by a factor of 72 72 x 2 y 3

35. (a) The force F needed is F  kx.

(b) Since F  30 N when x  9 cm and the spring’s natural length is 5 cm, we have 30  k 9  5  k  75. (c) From part (b), we have F  75x. Substituting x  11  5  6 into F  75x gives F  75 6  45 N.

36. (a) C  kpm

(b) Since C  60,000 when p  120 and m  4000, we get 60,000  k 120 4000  k  18 . So C  18 pm.

(c) Substituting p  92 and m  5000, we get C  18 92 5000  $57,500. 37. (a) P  ks 3 .

(b) Since P  96 when s  20, we get 96  k  203  k  0012. So P  0012s 3 .

(c) Substituting x  30, we get P  0012  303  324 watts.

38. (a) The power P is directly proportional to the cube of the speed s, so P  ks 3 .

80  2  008. (b) Because P  80 when s  10, we have 80  k 103  k  1000 25 2 and s  15, we have P  2 153  270 hp. (c) Substituting k  25 25

39. D  ks 2 . Since D  150 when s  40, we have 150  k 402 , so k  009375. Thus, D  009375s 2 . If D  200, then 200  009375s 2  s 2  21333, so s  46 mi/h (for safety reasons we round down).   40. L  ks 2 A. Since L  1700 when s  50 and A  500, we have 1700  k 502 500  k  000136. Thus   L  000136s 2 A. When A  600 and s  40 we get the lift is L  000136 402 600  13056 lb.

41. F  k As 2 . Since F  220 when A  40 and s  5. Solving for k we have 220  k 40 52  220  1000k   k  022. Now when A  28 and F  175 we get 175  0220 28 s 2  284090  s 2 so s  284090  533 mi/h. 42. (a) T 2  kd 3

3  (b) Substituting T  365 and d  93  106 , we get 3652  k  93  106  k  166  1019 .  3 (c) T 2  166  1019 279  109  360  109  T  600  104 . Hence the period of Neptune is 6.00104 days 164 years.

43. (a) P 

kT . V

(b) Substituting P  332, T  400, and V  100, we get 332  P

k 400  k  83. Thus k  83 and the equation is 100

83T . V

(c) Substituting T  500 and V  80, we have P 

83 500  51875 kPa. Hence the pressure of the sample of gas is 80

about 519 kPa. s 2 r (b) For the first car we have 1  1600 and s1  60 and for the second car we have 2  2500. Since the forces are equal

44. (a) F  k

we have k

2500  s22 16  602 1600  602 k   s22 , so s2  48 mi/h. r r 25

142

CHAPTER 1 Equations and Graphs

k 45. (a) The loudness L is inversely proportional to the square of the distance d, so L  2 . d k (b) Substituting d  10 and L  70, we have 70  2  k  7000. 10   1 k k , so the loudness is changed by a factor of 14 .  (c) Substituting 2d for d, we have L  4 d2 2d2   k k , so the loudness is changed by a factor of 4. (d) Substituting 12 d for d, we have L   2  4 d2 1d 2

46. (a) The power P is jointly proportional to the area A and the cube of the velocity , so P  k A 3 .   (b) Substituting 2 for  and 12 A for A, we have P  k 12 A 23  4k A 3 , so the power is changed by a factor of 4.  3 (c) Substituting 12  for  and 3A for A, we have P  k 3A 12   38 Ak 3 , so the power is changed by a factor of 38 .

47. (a) R 

kL d2

k 12 7  0002916.  k  2400 00052 7 3 4375 (c) Substituting L  3 and d  0008, we have R    137 .  2400 00082 32 (b) Since R  140 when L  12 and d  0005, we get 140 

k 3L

3 kL  , so the resistance is changed by a factor of 34 . 4 d2 2d2 48. Let S be the final size of the cabbage, in pounds, let N be the amount of nutrients it receives, in ounces, and let c be the N number of other cabbages around it. Then S  k . When N  20 and c  12, we have S  30, so substituting, we have c   N 20 30  k 12  k  18. Thus S  18 . When N  10 and c  5, the final size is S  18 10 5  36 lb. c 4  E k60004 49. (a) For the sun, E S  k60004 and for earth E E  k3004 . Thus S   6000  204  160,000. So the sun 300 4 EE k300 produces 160,000 times the radiation energy per unit area than the Earth. (d) If we substitute 2d for d and 3L for L, then R 

(b) The surface area of the sun is 4 435,0002 and the surface area of the Earth is 4 3,9602 . So the sun has   4 435,0002 435,000 2  times the surface area of the Earth. Thus the total radiation emitted by the sun is 3,960 4 3,9602   435,000 2 160,000   1,930,670,340 times the total radiation emitted by the Earth. 3,960 50. Let V be the value of a building lot on Galiano Island, A the area of the lot, and q the quantity of the water produced. Since V is jointly proportional to the area and water quantity, we have V  k Aq. When A  200  300  60,000 and q  10, we have V  $48 000, so 48,000  k 60,000 10  k  008. Thus V  008Aq. Now when A  400  400  160,000 and q  4, the value is V  008 160,000 4  $51,200.  51. (a) Let T and l be the period and the length of the pendulum, respectively. Then T  k l.

 T2 T2 2T 2 (b) T  k l  T 2  k 2 l  l  2 . If the period is doubled, the new length is  4 2  4l. So we would 2 k k k quadruple the length l to double the period T . 52. Let H be the heat experienced by a hiker at a campfire, let A be the amount of wood, and let d be the distance from A A campfire. So H  k 3 . When the hiker is 20 feet from the fire, the heat experienced is H  k 3 , and when the amount d 20  2A A 2A  k 3  d 3  16 000  d  20 3 2  252 feet. of wood is doubled, the heat experienced is H  k 3 . So k 8 000 d d

CHAPTER 1

53. (a) Since f is inversely proportional to L, we have f 

Review

143

k , where k is a positive constant. L

k k  12   12 f . So the frequency of the vibration is cut in half. 2L L 54. (a) Since r is jointly proportional to x and P  x, we have r  kx P  x, where k is a positive constant. (b) If we replace L by 2L we have

(b) When 10 people are infected the rate is r  k10 5000  10  49,900k. When 1000 people are infected the rate is r  k  1000  5000  1000  4,000,000k. So the rate is much higher when 1000 people are infected. Comparing 1000 people infected 4,000,000k these rates, we find that   80. So the infection rate when 1000 people are infected 10 people infected 49,900k is about 80 times as large as when 10 people are infected. (c) When the entire population is infected the rate is r  k 5000 5000  5000  0. This makes sense since there are no more people who can be infected.

L 25  1026 14 . 55. Using B  k 2 with k  0080, L  25  1026 , and d  24  1019 , we have B  0080  2  347  10 d 24  1019

The star’s apparent brightness is about 347  1014 Wm2 .  L L L 2 56. First, we solve B  k 2 for d: d  k  d  k because d is positive. Substituting k  0080, L  58  1030 , and B B d  58  1030 16 B  82  10 , we find d  0080  238  1022 , so the star is approximately 238  1022 m from earth. 82  1016 57. Examples include radioactive decay and exponential growth in biology.

CHAPTER 1 REVIEW 1. (a)

y

Q

(b) The distance from P to Q is  d P Q  5  22  12  02    49  144  193     3 5  2 12  0    6 . (c) The midpoint is 2 2 2

1 1

(d) The line has slope m 

P

x

12  0   12 7 , and has 5  2

12 24 equation y  0   12 7 x  2  y   7 x  7  12x  7y  24  0. Q

(e) The radius of this circle was found in part (b). It is  r  d P Q  193. So an equation is  2 193  x  22  y 2  193. x  22  y  02  Q

y

y

2

P 2

x

P

1 1

P

x

144

CHAPTER 1 Equations and Graphs

2. (a)

y 1 1

x

P

(b) The distance from P to Q is  d P Q  2  72  11  12     25  100  125  5 5     9 2  7 11  1    6 . (c) The midpoint is 2 2 2

Q

10 11  1   2, and 27 5 its equation is y  11  2 x  2 

(d) The line has slope m 

y  11  2x  4  y  2x  15.

(e) The radius of this circle was found in part (b). It is  r  d P, Q  5 5. So an equation is   2 x  72  y  12  5 5  x  72  y  12  125.

y

1

y

1

x

P

2 2

x

P

Q Q

y

3. (a)

P

4

x

4 Q

(d) The line has slope m 

(c) The midpoint is

16 8 2  14   6  4 10 5

and equation y  2   85 x  6 

8 38 y  2   85 x  48 5  y  5x  5 . y

P

(b) The distance from P to Q is  d P Q  6  42  [2  14]2     100  256  356  2 89 

6  4 2  14  2 2

x

4 Q

 1 6.

(e) The radius of this circle was found in part (b). It is  r  d P Q  2 89. So an equation is   2 [x  6]2  y  22  2 89  x  62  y  22  356.

4



P

y

4 4 Q

x

CHAPTER 1 y

4. (a)

2 2

Q

P

145

(b) The distance from P to Q is  d P Q  [5  3]2  [2  6]2     64  16  80  4 5.

x

(c) The midpoint is 2  6  48  12 , and 5  3

(d) The line has slope m 

Review

has equation y  2  12 x  5  y  2  12 x  52  y  12 x  92 . y



5  3 2  6  2 2



 1 4.

(e) The radius of this circle was found in part (b). It is  r  d P Q  4 5. So an equation is  2   2 x  52  y  2  4 5  x  52  y  22  80. y

2 Q

2

P

x

2 Q

y

5.

6. x y  x  4 or y  2

2

x

P

y

1 1

x

1 1

x

   7. d A C  74 and 4  12  4  32  4  12  4  32     d B C  5  12  3  32  5  12  3  32  72. Therefore, B is closer to C. 8. The circle with center at 2 5 and radius

 2  2 has equation x  22  y  52  2  x  22  y  52  2.

9. The center is C  5 1, and the point P  0 0 is on the circle. The radius of the circle is    r  d P C  0  52  0  12 = 0  52  0  12  26. Thus, the equation of the circle is x  52  y  12  26.

    1 11 21 38    , and the radius is 12 of the distance from P to Q, or 10. The midpoint of segment P Q is 2 2 2 2    r  12  d P, Q  12 2  12  3  82  12 2  12  3  82  r  12 34. Thus the equation is 2  2  x  12  y  11  17 2 2 .

146

CHAPTER 1 Equations and Graphs

    11. (a) x 2  y 2  2x  6y  9  0  x 2  2x  y 2  6y  9      x 2  2x  1  y 2  6y  9  9  1  9 

(b) The circle has center 1 3 and radius 1. y

x  12  y  32  1, an equation of a circle.

1l 1

12. (a) 2x 2  2y 2  2x  8y  12  x 2  x  y 2  4y  14      x 2  x  14  y 2  4y  4  14  14  4  

2 x  12  y  22  92 , an equation of a circle.

x

  (b) The circle has center 12  2 

and radius 3 2 2 . y 1l

1

x

    13. (a) x 2  y 2  72  12x  x 2  12x  y 2  72  x 2  12x  36  y 2  72  36  x  62  y 2  36. Since the left side of this equation must be greater than or equal to zero, this equation has no graph.

14. (a) x 2  y 2  6x  10y  34  0  x 2  6x  y 2  10y  34      x 2  6x  9  y 2  10y  25  34  9  25  x  32  y  52  0, an equation of a point.

(b) This is the equation of the point 3 5. y

1l 1

x

CHAPTER 1

15. y  2  3x

y

x

y

x

y

2

8

2

3

0

2

0

1

 12

0

2 3

17.

16. 2x  y  1  0  y  2x  1

1

0

1

x

y x   1  y  72 x  7 2 7

18. y

x

2

14

4

5

0

7

0

0

2

0

4

5

2 x

20. 8x  y 2  0  y 2  8x

y

y

x

y

3

7

8

8

1

15

2

4

0

16

0

0

1

15

3

7

1

x

1

x

1

x

1

y

1

x

 22. y   1  x 2

y

x

y

x

0

0

1

1

1

2

4 9

1

2

y

3

1

y

x



y

y

y

2

21. x 

147

y x   0  5x  4y  0 4 5

x

19. y  16  x 2

Review

1 2

1 1

x

y

y 0

  23

0

1

1

0

1

1

x

148

CHAPTER 1 Equations and Graphs

23. y  9  x 2 (a) x-axis symmetry: replacing y by y gives y  9  x 2 , which is not the same as the original equation, so the graph is not symmetric about the x-axis. y-axis symmetry: replacing x by x gives y  9  x2  9  x 2 , which is the same as the original equation, so the graph is symmetric about the y-axis. Origin symmetry: replacing x by x and y by y gives y  9  x2  y  9  x 2 , which is not the same as the original equation, so the graph is not symmetric about the origin. (b) To find x-intercepts, we set y  0 and solve for x: 0  9  x 2  x 2  9  x  3, so the x-intercepts are 3 and 3. To find y-intercepts, we set x  0 and solve for y: y  9  02  9, so the y-intercept is 9.

24. 6x  y 2  36 (a) x-axis symmetry: replacing y by y gives 6x  y2  36  6x  y 2  36, which is the same as the original equation, so the graph is symmetric about the x-axis. y-axis symmetry: replacing x by x gives 6 x  y 2  36  6x  y 2  36, which is not the same as the original equation, so the graph is not symmetric about the y-axis. Origin symmetry: replacing x by x and y by y gives 6 x  y2  36  6x  y 2  36, which is not the same as the original equation, so the graph is not symmetric about the origin. (b) To find x-intercepts, we set y  0 and solve for x: 6x  02  36  x  6, so the x-intercept is 6.

To find y-intercepts, we set x  0 and solve for y: 6 0  y 2  36  y  6, so the y-intercepts are 6 and 6.

25. x 2  y  12  1

 2 (a) x-axis symmetry: replacing y by y gives x 2  y  1  1  x 2  y  12  1, so the graph is not symmetric about the x-axis. y-axis symmetry: replacing x by x gives x2  y  12  1  x 2  y  12  1, so the graph is symmetric about the y-axis.  2 Origin symmetry: replacing x by x and y by y gives x2  y  1  1  x 2  y  12  1, so the graph is not symmetric about the origin.

(b) To find x-intercepts, we set y  0 and solve for x: x 2  0  12  1  x 2  0, so the x-intercept is 0.

To find y-intercepts, we set x  0 and solve for y: 02  y  12  1  y  1  1  y  0 or 2, so the y-intercepts are 0 and 2.

26. x 4  16  y (a) x-axis symmetry: replacing y by y gives x 4  16  y  x 4  16  y, so the graph is not symmetric about the x-axis. y-axis symmetry: replacing x by x gives x4  16  y  x 4  16  y, so the graph is symmetric about the y-axis. Origin symmetry: replacing x by x and y by y gives x4  16  y  x 4  16  y, so the graph is not symmetric about the origin.

(b) To find x-intercepts, we set y  0 and solve for x: x 4  16  0  x 4  16  x  2, so the x-intercepts are 2 and 2. To find y-intercepts, we set x  0 and solve for y: 04  16  y  y  16, so the y-intercept is 16.

27. 9x 2  16y 2  144 (a) x-axis symmetry: replacing y by y gives 9x 2  16 y2  144  9x 2  16y 2  144, so the graph is symmetric about the x-axis. y-axis symmetry: replacing x by x gives 9 x2  16y 2  144  9x 2  16y 2  144, so the graph is symmetric about the y-axis. Origin symmetry: replacing x by x and y by y gives 9 x2  16 y2  144  9x 2  16y 2  144, so the graph is symmetric about the origin.

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149

(b) To find x-intercepts, we set y  0 and solve for x: 9x 2  16 02  144  9x 2  144  x  4, so the x-intercepts are 4 and 4. To find y-intercepts, we set x  0 and solve for y: 9 02  16y 2  144  16y 2  144, so there is no y-intercept.

4 x 4 (a) x-axis symmetry: replacing y by y gives y  , which is different from the original equation, so the graph is not x symmetric about the x-axis. 4 y-axis symmetry: replacing x by x gives y  , which is different from the original equation, so the graph is not x symmetric about the y-axis. 4 4  y  , so the graph is symmetric about the Origin symmetry: replacing x by x and y by y gives y  x x origin. 4 (b) To find x-intercepts, we set y  0 and solve for x: 0  has no solution, so there is no x-intercept. x To find y-intercepts, we set x  0 and solve for y. But we cannot substitute x  0, so there is no y-intercept.

28. y 

29. x 2  4x y  y 2  1 (a) x-axis symmetry: replacing y by y gives x 2  4x y  y2  1, which is different from the original equation, so the graph is not symmetric about the x-axis. y-axis symmetry: replacing x by x gives x2  4 x y  y 2  1, which is different from the original equation, so the graph is not symmetric about the y-axis. Origin symmetry: replacing x by x and y by y gives x2  4 x y  y2  1  x 2  4x y  y 2  1, so the graph is symmetric about the origin. (b) To find x-intercepts, we set y  0 and solve for x: x 2  4x 0  02  1  x 2  1  x  1, so the x-intercepts are 1 and 1. To find y-intercepts, we set x  0 and solve for y: 02  4 0 y  y 2  1  y 2  1  y  1, so the y-intercepts are 1 and 1.

30. x 3  x y 2  5 (a) x-axis symmetry: replacing y by y gives x 3  x y2  5  x 3  x y 2  5, so the graph is symmetric about the x-axis. y-axis symmetry: replacing x by x gives x3  x y 2  5, which is different from the original equation, so the graph is not symmetric about the y-axis. Origin symmetry: replacing x by x and y by y gives x3  x y2  5, which is different from the original equation, so the graph is not symmetric about the origin.   (b) To find x-intercepts, we set y  0 and solve for x: x 3  x 02  5  x 3  5  x  3 5, so the x-intercept is 3 5. To find y-intercepts, we set x  0 and solve for y: 03  0y 2  5 has no solution, so there is no y-intercept.

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CHAPTER 1 Equations and Graphs

31. (a) We graph y  x 2  6x in the viewing rectangle [10 10] by [10 10].

32. (a) We graph y 

 5  x in the viewing rectangle

[10 6] by [1 5].

10

4 2

-10

-5

5

10 -10

-10

(b) From the graph, we see that the x-intercepts are 0

-5

5

(b) From the graph, we see that the x-intercept is 5 and

and 6 and the y-intercept is 0.

the y-intercept is approximately 224.

33. (a) We graph y  x 3  4x 2  5x in the viewing rectangle [4 8] by [30 20]. 20

x2 x2  y2  1  y2  1   34. (a) We graph 4 4  x2 in the viewing rectangle [3 3] by y  1 4 [2 2].

-4

-2

2

4

6

8

2 1

-20 -3

(b) From the graph, we see that the x-intercepts are 1,

-2

-1 -1

1

2

3

-2

0, and 5 and the y-intercept is 0.

(b) From the graph, we see that the x-intercepts are 2 and 2 and the y-intercepts are 1 and 1.

35. (a) The line that has slope 2 and y-intercept 6 has the slope-intercept equation

(c)

y

y  2x  6. (b) An equation of the line in general form is 2x  y  6  0.

1 1

x

CHAPTER 1

36. (a) The line that has slope  12 and passes through the point 6 3 has

Review

151

y

(c)

equation y  3   12 x  6  y  3   12 x  6  y   12 x. (b)  12 x  3  y  3  x  6  2y  6  x  2y  0. 1 1

37. (a) The line that passes through the points 1 6 and 2 4 has slope

(c)

4  6 2 m  , so y  6  23 [x  1]  y  6  23 x  23 2  1 3

x

y 1 1

 y  23 x  16 3 .

x

(b) y  23 x  16 3  3y  2x  16  2x  3y  16  0.

38. (a) The line that has x-intercept 4 and y-intercept 12 passes through the points

(c)

y

12  0  3 and the equation is 04 y  0  3 x  4  y  3x  12.

4 0 and 0 12, so m 

(b) y  3x  12  3x  y  12  0. 2

39. (a) The vertical line that passes through the point 3 2 has equation x  3.

(c)

1

x

1

x

y

(b) x  3  x  3  0. 1

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CHAPTER 1 Equations and Graphs

40. (a) The horizontal line with y-intercept 5 has equation y  5.

(c)

y

(b) y  5  y  5  0. 1 1

41. (a) 2x  5y  10  5y  2x  10  y  25 x  2, so the given line has slope (c) m  25 . Thus, an equation of the line passing through 1 1 parallel to this line is y  1  25 x  1  y  25 x  35 .

y

1

(b) y  25 x  35  5y  2x  3  2x  5y  3  0.

42. (a) The line containing 2 4 and 4 4 has slope

x

(c)

1

x

1

x

y

8 4  4   4, and the line passing through the origin with 42 2 this slope has equation y  4x. m

1

(b) y  4x  4x  y  0.

43. (a) The line y  12 x  10 has slope 12 , so a line perpendicular to this one has

(c)

y

1  2. In particular, the line passing through the origin slope  12

perpendicular to the given line has equation y  2x. (b) y  2x  2x  y  0.

1 1

x

CHAPTER 1

44. (a) x  3y  16  0  3y  x  16  y  13 x  16 3 , so the given line has

(c)

Review

153

y

slope 13 . The line passing through 1 7 perpendicular to the given line has equation y  7  

1 x  1  y  7  3 x  1  y  3x  10. 13

(b) y  3x  10  3x  y  10  0.

x 1 1

45. The line with equation y   13 x  1 has slope  13 . The line with equation 9y  3x  3  0  9y  3x  3  y   13 x  13 also has slope  13 , so the lines are parallel.

46. The line with equation 5x  8y  3  8y  5x  3  y  58 x  38 has slope 58 . The line with equation 10y  16x  1  1 has slope  8   1 , so the lines are perpendicular. 10y  16x  1  y   85 x  10 5 58

47. (a) The slope represents a stretch of 03 inches for each one-pound increase in weight. The s-intercept represents the length of the unstretched spring. (b) When   5, s  03 5  25  15  25  40 inches.

48. (a) We use the information to find two points, 0 60000 and 3 70500. Then the slope is 10,500 70,500  60,000   3,500. So S  3,500t  60,000. m 30 3 (b) The slope represents an annual salary increase of $3500, and the S-intercept represents her initial salary. (c) When t  12, her salary will be S  3500 12  60,000  42,000  60,000  $102,000. 49. x 2  9x  14  0  x  7 x  2  0  x  7 or x  2.

50. x 2  24x  144  0  x  122  0  x  12  0  x  12.

51. 2x 2  x  1  2x 2  x  1  0  2x  1 x  1  0. So either 2x  1  0  2x  1  x  x  1. 52. 3x 2  5x  2  0  3x  1 x  2  0  x  13 or x  2.   53. 0  4x 3  25x  x 4x 2  25  x 2x  5 2x  5  0. So either x  0 or 2x  5  0  2x

1 ; or x  1  0  2

 5  x  52 ; or

2x  5  0  2x  5  x   52 .

   54. x 3  2x 2  5x  10  0  x 2 x  2  5 x  2  0  x  2 x 2  5  0  x  2 or x   5.

55. 3x 2  4x  1  0           2 2 7 4 42 431 b b2 4ac 4 1612 42 7 2 7 . 4 28       x 2a 23 6 6 6 6 3     2 419 3 3 2 56. x 2  3x  9  0  x  b 2ab 4ac   3 2936  3 227 , which are not real numbers. 21 There is no real solution. 2 1   3  x  1  2 x  3 x x  1  x  1  2x  3x 2  3x  0  3x 2  6x  1  57. x x 1          2 3 6 6 62 431 b b2 4ac 6 3612 6 24 62 6 3 6 . x       2a 6 6 6 6 3 23 58.

1 8 x   2  x x  2  x  2  8  x 2  2x  x  2  8  x 2  3x  10  0  x  2 x  5  0 x 2 x 2 x 4  x  2 or x  5. However, since x  2 makes the expression undefined, we reject this solution. Hence the only solution is x  5.

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CHAPTER 1 Equations and Graphs

     59. x 4  8x 2  9  0  x 2  9 x 2  1  0  x  3 x  3 x 2  1  0  x  3  0  x  3, or x  3  0  x  3, however x 2  1  0 has no real solution. The solutions are x  3.

  60. x  4 x  32. Let u  x. Then u 2  4u  32  u 2  4u  32  0  u  8 u  4  0 So either u  8  0 or   u  4  0. If u  8  0, then u  8  x  8  x  64. If u  4  0, then u  4  x  4, which has no real solution. So the only solution is x  64.    61. x 12  2x 12  x 32  0  x 12 1  2x  x 2  0  x 12 1  x2  0. Since x 12  1 x is never 0, the only solution comes from 1  x2  0  1  x  0  x  1.

      2 62. 1  x  2 1  x  15  0. Let u  1  x, then the equation becomes u 2  2u  15  0  u  5 u  3  0    u  5  0 or u  3  0. If u  5  0, then u  5  1  x  5  x  4  x  16. If u  3  0, then u  3    1  x  3  x  4, which has no real solution. So the only solution is x  16. 63. x  7  4  x  7  4  x  7  4, so x  11 or x  3. 64. 2x  5  9 is equivalent to 2x  5  9  2x  5  9  x 

59 . So x  2 or x  7. 2

65. (a) 2  3i  1  4i  2  1  3  4 i  3  i

(b) 2  i 3  2i  6  4i  3i  2i 2  6  i  2  8  i

66. (a) 3  6i  6  4i  3  6i  6  4i  3  6  6  4 i  3  2i   (b) 4i 2  12 i  8i  2i 2  8i  2  2  8i 8  8i  2 4  2i 4  2i 2  i 8  8i  2i 2 6  8i       65  85 i 2i 2i 2i 41 5 4  i2      (b) 1  1 1  1  1  i 1  i  1  i  i  i 2  1  1  2

67. (a)

8  3i 4  3i 32  12i  9i 2 41  12i 8  3i 32  12i  9 12      41  25  25 i 2 4  3i 4  3i 4  3i 16  9 25 16  9i     (b) 10  40  i 10  2i 10  20i 2  20

68. (a)

69. x 2  16  0  x 2  16  x  4i   70. x 2  12  x   12  2 3i b  71. x 2  6x  10  0  x 

72. 2x 2  3x  2  0  x 

   6  62  4 1 10 6  36  40 b2  4ac    3  i 2a 2 1 2

 3 

 32  4 2 2 2 2



3

  7 7 3   i 4 4 4

   73. x 4  256  0  x 2  16 x 2  16  0  x  4 or x  4i   74. x 3  2x 2  4x  8  0  x  2 x 2  4  x  2 or x  2i

CHAPTER 1

Review

155

75. Let r be the rate the woman runs in mi/h. Then she cycles at r  8 mi/h. Rate Cycle

r 8

Run

r

Time

Distance

4 r 8 25 r

4 25

25 4   1. Multiplying by 2r r  8, we r 8 r get 4 2r   25 2 r  8  2r r  8  8r  5r  40  2r 2  16r  0  2r 2  3r  40  Since the total time of the workout is 1 hour, we have

   3 32 4240 3 9320 3 329 . Since r  0, we reject the negative value. She runs at r    22 4 4  3 329  378 mi/h. r 4 x2  1500  20x  x 2  x 2  20x  1500  0  x  30 x  50  0. So 76. Substituting 75 for d, we have 75  x 

20 x  30 or x  50. The speed of the car was 30 mi/h.

77. Let x be the length of one side in cm. Then 28  x is the length of the other side. Using the Pythagorean Theorem, we   have x 2  28  x2  202  x 2  784  56x  x 2  400  2x 2  56x  384  0  2 x 2  28x  192  0 

2 x  12 x  16  0. So x  12 or x  16. If x  12, then the other side is 28  12  16. Similarly, if x  16, then the other side is 12. The sides are 12 cm and 16 cm. 80 and the total amount of fencing material is 78. Let l be length of each garden plot. The width of each plot is then l     480 80 4 l  6  88. Thus 4l   88  4l 2  480  88l  4l 2  88l  480  0  4 l 2  22l  120  0  l l 4 l  10 l  12  0. So l  10 or l  12. If l  10 ft, then the width of each plot is 80 10  8 ft. If l  12 ft, then the

width of each plot is 80 12  667 ft. Both solutions are possible.

80. 12  x  7x  12  8x  32  x.   Interval:  32

79. 3x  2  11  3x  9  x  3. Interval: 3 . Graph:

-3

Graph:

81. 3  x  2x  7  10  3x  10 3 x   Interval: 10 3  Graph:

3 2

82. 1  2x  5  3  6  2x  2  3  x  1 Interval: 3 1]. Graph:

10 3

_3

_1

83. x 2  4x  12  0  x  2 x  6  0. The expression on the left of the inequality changes sign where x  2 and where x  6. Thus we must check the intervals in the following table.  6

6 2

2 

Sign of x  2







Sign of x  6













Interval

Sign of x  2 x  6

Interval:  6  2   Graph:

_6

2

156

CHAPTER 1 Equations and Graphs

84. x 2  1  x 2  1  0  x  1 x  1  0. The expression on the left of the inequality changes sign when x  1 and x  1. Thus we must check the intervals in the following table. Interval: [1 1]  1

1 1

1 

Sign of x  1







Sign of x  1













Interval

Sign of x  1 x  1

85.

Graph:

_1

1

2x  5 2x  5 x 1 x 4 2x  5 1 1  0  0  0. The expression on the left of the inequality x 1 x 1 x 1 x 1 x 1 changes sign where x  1 and where x  4. Thus we must check the intervals in the following table. We exclude x  1, since the expression is not

 4

4 1

1 

Sign of x  4







defined at this value. Thus the solution is [4 1.

Sign of x  1







Graph:

Sign of







Interval

x 4 x 1

_4

_1

86. 2x 2  x  3  2x 2  x  3  0  2x  3 x  1  0. The expression on the left of the inequality changes sign when 1 and 32 . Thus we must check the intervals in the following table.  1

Interval



 3 2

Sign of 2x  3



Sign of x  1













Sign of 2x  3 x  1

87.

  1 32 



  Interval:  1]  32   Graph:

3 2

_1

x 4 x 4 0  0. The expression on the left of the inequality changes sign where x  2, where x  2, 2  2 x  2 x x 4 and where x  4. Thus we must check the intervals in the following table.  2

2 2

2 4

4 

Sign of x  4









Sign of x  2









Interval

Sign of x  2     x 4 Sign of     x  2 x  2 Since the expression is not defined when x  2we exclude these values and the solution is  2  2 4]. Graph:

_2

2

4

CHAPTER 1

88.

Review

157

5 5 5  0  0 2  0. The 0 2 x  1 x  2 x  2 x x  1  4 x  1 x  1 x  4 expression on the left of the inequality changes sign when 2 1and 2. Thus we must check the intervals in the following table. 5

x 3  x 2  4x  4

 2

2 1

1 2

2 

Sign of x  1









Sign of x  2









Sign of x  2

















Interval

5 Sign of x  1 x  2 x  2 Interval:  2  1 2 Graph:

_2

1

2

90. x  4  002  002  x  4  002 

89. x  5  3  3  x  5  3  2  x  8. Interval: [2 8] Graph:

2

398  x  402 Interval: 398 402

8

Graph:

3.98

4.02

91. 2x  1  1 is equivalent to 2x  1  1 or 2x  1  1. Case 1: 2x  1  1  2x  0  x  0. Case 2: 2x  1  1  2x  2  x  1. Interval:  1]  [0 . Graph:

_1

0

92. x  1 is the distance between x and 1 on the number line, and x  3 is the distance between x and 3. We want those points that are closer to 1 than to 3. Since 2 is midway between 1 and 3, we get x   2 as the solution. Graph: 2

93. (a) For

 24  x  3x 2 to define a real number, we must have 24  x  3x 2  0  8  3x 3  x  0. The expression

on the left of the inequality changes sign where 8  3x  0  3x  8  x  83 ; or where x  3. Thus we must check the intervals in the following table.   Interval: 3 83 .     8 3 83 Interval  3 3 Graph: Sign of 8  3x    Sign of 3  x

Sign of 8  3x 3  x













_3

8 3

158

CHAPTER 1 Equations and Graphs

(b) For  4

1 x  x4

    to define a real number we must have x  x 4  0  x 1  x 3  0  x 1  x 1  x  x 2  0.

The expression on the left of the inequality changes sign where x  0; or where x  1; or where 1  x  x 2  0    12 411 x  1 21  1 214 which is imaginary. We check the intervals in the following table. Interval: 0 1. Interval

 0

0 1

1 

Sign of x







Sign of 1  x







  Sign of x 1  x 1  x  x 2













Sign of 1  x  x 2

6 9  r3   94. We have 8  43 r 3  12   

 3

6 r  

 3

9 . Thus r  

Graph: 0

 3

6  

1

  3 9 . 

95. From the graph, we see that the graphs of y  x 2  4x and y  x  6 intersect at x  1 and x  6, so these are the solutions of the equation x 2  4x  x  6.

96. From the graph, we see that the graph of y  x 2  4x crosses the x-axis at x  0 and x  4, so these are the solutions of the equation x 2  4x  0.

97. From the graph, we see that the graph of y  x 2  4x lies below the graph of y  x  6 for 1  x  6, so the inequality x 2  4x  x  6 is satisfied on the interval [1 6].

98. From the graph, we see that the graph of y  x 2  4x lies above the graph of y  x  6 for   x  1 and 6  x  , so the inequality x 2  4x  x  6 is satisfied on the intervals  1] and [6 .

99. From the graph, we see that the graph of y  x 2  4x lies above the x-axis for x  0 and for x  4, so the inequality x 2  4x  0 is satisfied on the intervals  0] and [4 .

100. From the graph, we see that the graph of y  x 2  4x lies below the x-axis for 0  x  4, so the inequality x 2  4x  0 is satisfied on the interval [0 4].   101. x 2  4x  2x  7. We graph the equations y1  x 2  4x 102. x  4  x 2  5. We graph the equations y1  x  4 and y2  2x  7 in the viewing rectangle [10 10] by

[5 25]. Using a zoom or trace function, we get the solutions x  1 and x  7.

-10

-5

and y2  x 2  5 in the viewing rectangle [4 5] by [0 10]. Using a zoom or trace function, we get the solutions x  250 and x  276.

20

10

10

5

5

10

-4

-2

0

2

4

CHAPTER 1

Review

159

103. x 4  9x 2  x  9. We graph the equations y1  x 4  9x 2 104. x  3  5  2. We graph the equations and y2  x  9 in the viewing rectangle [5 5] by

y1  x  3  5 and y2  2 in the viewing rectangle

[25 10]. Using a zoom or trace function, we get the

[20 20] by [0 10]. Using Zoom and/or Trace, we get the

solutions x  272, x  115, x  100, and x  287.

solutions x  10, x  6, x  0, and x  4.

10 -4

10

-2 -10

2

5

4

-20

-20

105. 4x  3  x 2 . We graph the equations y1  4x  3 and

y2  x 2 in the viewing rectangle [5 5] by [0 15]. Using

-10

0

10

20

106. x 3  4x 2  5x  2. We graph the equations

y1  x 3  4x 2  5x and y2  2 in the viewing rectangle

a zoom or trace function, we find the points of intersection

[10 10] by [5 5]. We find that the point of intersection

are at x  1 and x  3. Since we want 4x  3  x 2 , the

is at x  507. Since we want x 3  4x 2  5x  2, the solution is the interval 507 .

solution is the interval [1 3]. 15

4

10

2

5 -4

-2

-10 0

2

-5

-2

5

10

-4

4

    108. x 2  16  10  0. We graph the equation     y1  x 4  4x 2 and y2  12 x  1 in the viewing rectangle y  x 2  16  10 in the viewing rectangle [10 10] by [5 5] by [5 5]. We find the points of intersection are [10 10]. Using a zoom or trace function, we find that the at x  185, x  060, x  045, and x  200. Since x-intercepts are x  510 and x  245. Since we   we want x 4  4x 2  12 x  1, the solution is   want x 2  16  10  0, the solution is approximately 185 060  045 200.  510]  [245 245]  [510 .

107. x 4  4x 2  12 x  1. We graph the equations

4

10

2 -4

-2

-2

2

4 -10

-5

5

-4 -10

10

160

CHAPTER 1 Equations and Graphs

109. Here the center is at 0 0, and the circle passes through the point 5 12, so the radius is    r  5  02  12  02  25  144  169  13. The equation of the circle is x 2  y 2  132 

x 2  y 2  169. The line shown is the tangent that passes through the point 5 12, so it is perpendicular to the line 12 12  0   . The slope of the line we seek is through the points 0 0 and 5 12. This line has slope m 1  5  0 5 5 1 1 5 x  5  y  12  5 x  25  m2    . Thus, an equation of the tangent line is y  12  12  12 12 m1 125 12

5 x  169  5x  12y  169  0. y  12 12 110. Because the circle is tangent to the x-axis at the point 5 0 and tangent to the y-axis at the point 0 5, the center is at

5 5 and the radius is 5. Thus an equation is x  52  y  52  52  x  52  y  52  25. The slope of 4 4 51    , so an equation of the line we seek is the line passing through the points 8 1 and 5 5 is m  58 3 3 y  1   43 x  8  4x  3y  35  0.

111. Since M varies directly as z we have M  kz. Substituting M  120 when z  15, we find 120  k 15  k  8. Therefore, M  8z. k k  k  192. 112. Since z is inversely proportional to y, we have z  . Substituting z  12 when y  16, we find 12  y 16 192 Therefore z  . y k 113. (a) The intensity I varies inversely as the square of the distance d, so I  2 . d k (b) Substituting I  1000 when d  8, we get 1000   k  64,000. 82 64,000 64,000 (c) From parts (a) and (b), we have I  . Substituting d  20, we get I   160 candles. d2 202 114. Let f be the frequency of the string and l be the length of the string. Since the frequency is inversely proportional to the k k 5280 length, we have f  . Substituting l  12 when k  440, we find 440   k  5280. Therefore f  . For l 12 l 5280  l  5280 f  660, we must have 660  660  8. So the string needs to be shortened to 8 inches. l 115. Let  be the terminal velocity of the parachutist in mi/h and  be his weight in pounds. Since the terminal velocity is  directly proportional to the square root of the weight, we have   k . Substituting   9 when   160, we solve   9 for k. This gives 9  k 160  k    0712. Thus   0712 . When   240, the terminal velocity is 160    0712 240  11 mi/h. 116. Let r be the maximum range of the baseball and  be the velocity of the baseball. Since the maximum range is directly

proportional to the square of the velocity, we have r  l 2 . Substituting   60 and r  242, we find 242  k 602  k  00672. If   70, then we have a maximum range of r  00672 702  3294 feet.

CHAPTER 1

Test

161

CHAPTER 1 TEST

y

1. (a)

There are several ways to determine the coordinates of S. The diagonals of a

S

P

square have equal length and are perpendicular. The diagonal P R is horizontal R

and has length is 6 units, so the diagonal QS is vertical and also has length 6. Thus, the coordinates of S are 3 6.    (b) The length of P Q is 0  32  3  02  18  3 2. So the area of   2 P Q RS is 3 2  18.

1 1

Q

x

(b) The x-intercept occurs when y  0, so 0  x 2  4  x 2  4  x  2. The

y

2. (a)

y-intercept occurs when x  0, so y  4.

(c) x-axis symmetry: y  x 2  4  y  x 2  4, which is not the same as the original equation, so the graph is not symmetric with respect to the x-axis.

1 1

x

y-axis symmetry: y  x2  4  y  x 2  4, which is the same as the

original equation, so the graph is symmetric with respect to the y-axis.

Origin symmetry: y  x2  4  y  x 2  4, which is not the same

as the original equation, so the graph is not symmetric with respect to the origin.

y

3. (a)

Q

P

1 1

x

(b) The distance between P and Q is    d P Q  3  52  1  62  64  25  89.     3  5 1  6 (c) The midpoint is   1 72 . 2 2 16 5 5 (d) The slope of the line is   . 3  5 8 8   (e) The perpendicular bisector of P Q contains the midpoint, 1 72 , and it slope is

1   8 . Hence the equation the negative reciprocal of 58 . Thus the slope is  58 5

is y  72   85 x  1  y   85 x  85  72   85 x  51 10 . That is, y   85 x  51 10 .    (f) The center of the circle is the midpoint, 1 72 , and the length of the radius is 12 89 . Thus the equation of the circle 2   2 2   whose diameter is P Q is x  12  y  72  12 89  x  12  y  72  89 4 .

162

CHAPTER 1 Equations and Graphs

4. (a) x 2  y 2  25  52 has center 0 0 (b) x  22  y  12  9  32 has center 2 1 and radius 3.

and radius 5.

y

y

(c) x 2  6x  y 2  2y  6  0 

x 2  6x  9  y 2  2y  1  4 

x  32  y  12  4  22 has

center 3 1 and radius 2. y 1 1

x

1 1

x 1 1

5. (a) x  4  y 2 . To test for symmetry about the x-axis, we replace y with y:

x

y

x  4  y2  x  4  y 2 , so the graph is symmetric about the x-axis.

To test for symmetry about the y-axis, we replace x with x:

x  4  y 2 is different from the original equation, so the graph is not

1

symmetric about the y-axis.

1

x

1

x

For symmetry about the origin, we replace x with x and y with y:

x  4  y2  x  4  y 2 , which is different from the original

equation, so the graph is not symmetric about the origin.

To find x-intercepts, we set y  0 and solve for x: x  4  02  4, so the x-intercept is 4.

To find y-intercepts, we set x  0 and solve for y:: 0  4  y 2  y 2  4  y  2, so the y-intercepts are 2 and 2.

(b) y  x  2. To test for symmetry about the x-axis, we replace y with y:

y

y  x  2 is different from the original equation, so the graph is not symmetric about the x-axis.

To test for symmetry about the y-axis, we replace x with x:

y  x  2  x  2 is different from the original equation, so the

graph is not symmetric about the y-axis.

To test for symmetry about the origin, we replace x with x and y with

y: y  x  2  y   x  2, which is different from the original

equation, so the graph is not symmetric about the origin.

To find x-intercepts, we set y  0 and solve for x: 0  x  2  x  2  0  x  2, so the x-intercept is 2.

To find y-intercepts, we set x  0 and solve for y: y  0  2  2  2, so the y-intercept is 2.

1

CHAPTER 1

6. (a) To find the x-intercept, we set y  0 and solve for x: 3x  5 0  15

Test

163

y

(b)

 3x  15  x  5, so the x-intercept is 5.

To find the y-intercept, we set x  0 and solve for y: 3 0  5y  15  5y  15  y  3, so the y-intercept is 3.

1

(c) 3x  5y  15  5y  3x  15  y  35 x  3.

x

1

(d) From part (c), the slope is 35 . (e) The slope of any line perpendicular to the given line is the negative 1  5. reciprocal of its slope, that is,  35 3

7. (a) 3x  y  10  0  y  3x  10, so the slope of the line we seek is 3. Using the point-slope, y  6  3 x  3  y  6  3x  9  3x  y  3  0. x y (b) Using the intercept form we get   1  2x  3y  12  2x  3y  12  0. 6 4 8. (a) When x  100 we have T  008 100  4  8  4  4, so the

(b)

T

temperature at one meter is 4 C.

(c) The slope represents an increase of 008 C for each one-centimeter

5

increase in depth, the x-intercept is the depth at which the temperature is 0 C, and the T -intercept is the temperature at ground level. 20

40

60

80

100 120 x

_5

9. (a) x 2  x  12  0  x  4 x  3  0. So x  4 or x  3.      4  16  8 4  8 4  2 2 2  2 4  42  4 2 1 2     . (b) 2x  4x  1  0  x  2 2 4 4 4 2    2 (c) 3  x  3  x  3  x  x  3  3  x2  3  x  x 2  6x  9  3  x 

x 2  5x  6  x  2 x  3  0. Thus, x  2 and x  3 are potential solutions. Checking in the original equation, we see that only x  3 is valid.

(d) x 12  3x 14  2  0. Let u  x 14 , then we have u 2  3u  2  0  u  2 u  1  0. So either u  2  0 or

u  1  0. If u  2  0, then u  2  x 14  2  x  24  16. If u  1  0, then u  1  x 14  1  x  1. So x  1 or x  16.     (e) x 4  3x 2  2  0  x 2  1 x 2  2  0. So x 2  1  0  x  1 or x 2  2  0  x   2. Thus the   solutions are x  1, x  1, x   2, and x  2.

10 10 10 2 (f) 3 x  4  10  0  3 x  4  10  x  4  10 3  x  4   3  x  4  3 . So x  4  3  3 or 22 2 22 x  4  10 3  3 . Thus the solutions are x  3 and x  3 .

10. (a) 3  2i  4  3i  3  4  2i  3i  7  i

(b) 3  2i  4  3i  3  4  2i  3i  1  5i

(c) 3  2i 4  3i  3  4  3  3i  2i  4  2i  3i  12  9i  8i  6i 2  12  i  6 1  18  i

3  2i 4  3i 12  17i  6i 2 6 17 12  17i  6 3  2i      i  4  3i 4  3i 4  3i 16  9 25 25 16  9i 2  24 (e) i 48  i 2  124  1 (d)

164

CHAPTER 1 Equations and Graphs

(f)

        2   2  2 8  2  2  8  2 2  2 8  2  4  2i  4i  2  6  2i

4  11. Using the Quadratic Formula, 2x 2  4x  3  0  x 

   42  4 2 3 4  8   1  22 i. 2 2 4

12. Let  be the width of the parcel of land. Then   70 is the length of the parcel of land. Then 2    702  1302 

2  2  140  4900  16,900  22  140  12,000  0  2  70  6000  0    50   120  0. So   50 or   120. Since   0, the width is   50 ft and the length is   70  120 ft.

13. (a) 4  5  3x  17  9  3x  12  3  x  4. Expressing in standard form we have: 4  x  3. Interval: [4 3. Graph:

_4

3

(b) x x  1 x  2  0. The expression on the left of the inequality changes sign when x  0, x  1, and x  2. Thus we must check the intervals in the following table.  2

2 0

0 1

1 

Sign of x









Sign of x  1









Sign of x  2

















Interval

Sign of x x  1 x  2

From the table, the solution set is x  2  x  0 or 1  x. Interval: 2 0  1 . Graph:

_2

0

1

(c) x  4  3 is equivalent to 3  x  4  3  1  x  7. Interval: 1 7. Graph: (d)

1

7

2x  3 2x  3 2x  3 x 1 x 4 1 1  0  0  0. The expression on the left of the inequality x 1 x 1 x 1 x 1 x 1 changes sign where x  4 and where x  1. Thus we must check the intervals in the following table. Interval

 1

1 4

4 

Sign of x  4







Sign of x  1    x 4 Sign of    x 1 Since x  1 makes the expression in the inequality undefined, we exclude this value. Interval: 1 4]. Graph:

_1

4

14. 5  59 F  32  10  9  F  32  18  41  F  50. Thus the medicine is to be stored at a temperature between 41 F and 50 F.

Fitting Lines to Data

165

 15. For 6x  x 2 to be defined as a real number 6x  x 2  0  x 6  x  0. The expression on the left of the inequality changes sign when x  0 and x  6. Thus we must check the intervals in the following table.  0

0 6

6 

Sign of x







Sign of 6  x











Interval

Sign of x 6  x   From the table, we see that 6x  x 2 is defined when 0  x  6.

16. (a) x 3  9x  1  0. We graph the equation

y  x 3  9x  1 in the viewing rectangle [5 5]

(b) x 2  1  x  1. We graph the equations

y1  x 2  1 and y2  x  1 in the viewing

by [10 10]. We find that the points of

rectangle [5 5] by [5 10]. We find that the

intersection occur at x  294, 011, 305.

points of intersection occur at x  1 and x  2. Since we want x 2  1  x  1, the solution is

10

the interval [1 2].

10 -4

-2

2

4 5

-10 -4

-2

2

4

-5

17. (a) M  k

h 2 L

  4 62

h 2 (b) Substituting   4, h  6, L  12, and M  4800, we have 4800  k  k  400. Thus M  400 . 12 L   3 102  12,000. So the beam can support 12,000 pounds. (c) Now if L  10,   3, and h  10, then M  400 10

FOCUS ON MODELING Fitting Lines to Data 1. (a)

y

(b) Using a graphing calculator, we obtain the regression line y  18807x  8265.

180

(c) Using x  58 in the equation y  18807x  8265, we get y  18807 58  8265  1917 cm.

160

140 40

50

Femur length (cm)

x

166

FOCUS ON MODELING

2. (a)

y 800

(b) Using a graphing calculator, we obtain the regression line y  164163x  62183. (c) Using x  95 in the equation

600

y  164163x  62183, we get

y  164163 95  62183  938 cans. 400

50

60

70

90 x

80

High temperature (°F)

3. (a)

y 100

(b) Using a graphing calculator, we obtain the regression

80

(c) Using x  18 in the equation y  6451x  01523,

line y  6451x  01523. we get y  6451 18  01523  116 years.

60 40 20 0

2

4

6

8

10

12

14

16

18

20 x

Diameter (in.)

4. (a)

y 400

(b) Letting x  0 correspond to 1990, we obtain the regression line y  18446x  3522.

390

(c) Using x  21 in the equation y  18446x  3522,

380

we get y  18446 21  3522  3909 ppm CO2 ,

370

slightly lower than the measured value.

360 350 1990

1995

2000 2005 Year

2010

x

Fitting Lines to Data

5. (a)

y

167

(b) Using a graphing calculator, we obtain the regression line y  4857x  22097.

200

(c) Using x  100 F in the equation

y  4857x  22097, we get y  265 chirps per

minute. 100

0

50

60

70

80

90

x

Temperature (°F) y

6. (a)

(b) Using a graphing calculator, we obtain the regression line y  01275x  7929.

8

(c) Using x  30 in the regression line equation, we get

6

y  01275 30  7929  410 million km2 .

4 2 0

10

x

20

Years since 1986

7. (a)

y

(b) Using a graphing calculator, we obtain the regression line y  0168x  1989.

20

(c) Using the regression line equation y  0168x  1989, we get y  813% when

x  70%.

10

0

20

40

60

80

100 x

Flow rate (%) 8. (a)

y

(b) Using a graphing calculator, we obtain y  39018x  4197.

100

(c) The correlation coefficient is r  098, so linear model is appropriate for x between 80 dB and 104 dB.

50

(d) Substituting x  94 into the regression equation, we get y  39018 94  4197  53. So the intelligibility is about 53%.

0

80

90

100

Noise level (dB)

110 x

168

FOCUS ON MODELING

9. (a)

y 80

(b) Using a graphing calculator, we obtain y  027083x  4629. (c) We substitute x  2006 in the model y  027083x  4629 to get y  804, that is, a life

70

expectancy of 804 years.

(d) The life expectancy of a child born in the US in 2006

60

was 777 years, considerably less than our estimate in part (b). 1920

1940

1960

1980

2000

x

Year 10. (a)

y

(c) Year

x

Height (m)

1972

0

564

1976

4

564

1980

8

578

1984

12

575

1988

16

590

1992

20

587

1996

24

592

2000

28

590

2004

32

595

2008

36

596

(b) Using a graphing calculator, we obtain the regression line y  5664  000929x.

6.0

5.9 5.8 5.7 5.6 0

20

30

x

Years since 1972

The regression line provides a good model. (d) The regression line predicts the winning pole vault height in 2012 to be y  000929 2012  1972  5664  604 meters.

11. Students should find a fairly strong correlation between shoe size and height. 12. Results will depend on student surveys in each class.

10

2

FUNCTIONS

2.1

FUNCTIONS

1. If f x  x 3  1, then (a) the value of f at x  1 is f 1  13  1  0. (b) the value of f at x  2 is f 2  23  1  9.

(c) the net change in the value of f between x  1 and x  2 is f 2  f 1  9  0  9.

2. For a function f , the set of all possible inputs is called the domain of f , and the set of all possible outputs is called the range of f . x 5 3. (a) f x  x 2  3x and g x  have 5 in their domain because they are defined when x  5. However, x    h x  x  10 is undefined when x  5 because 5  10  5, so 5 is not in the domain of h. 0 55 (b) f 5  52  3 5  25  15  10 and g 5    0. 5 5 4. (a) Verbal: “Subtract 4, then square and add 3.” (b) Numerical: x

f x

0

19

2

7

4

3

6

7

5. A function f is a rule that assigns to each element x in a set A exactly one element called f x in a set B. Table (i) defines y as a function of x, but table (ii) does not, because f 1 is not uniquely defined. 6. (a) Yes, it is possible that f 1  f 2  5. [For instance, let f x  5 for all x.]

(b) No, it is not possible to have f 1  5 and f 1  6. A function assigns each value of x in its domain exactly one value of f x.

7. Multiplying x by 3 gives 3x, then subtracting 5 gives f x  3x  5. 8. Squaring x gives x 2 , then adding two gives f x  x 2  2. 9. Subtracting 1 gives x  1, then squaring gives f x  x  12 . 10. Adding 1 gives x  1, taking the square root gives



x  1, then dividing by 6 gives f x 



x 1 . 6

11. f x  2x  3: Multiply by 2, then add 3.

12. g x 

x 2 : Add 2, then divide by 3. 3

13. h x  5 x  1: Add 1, then multiply by 5.

14. k x 

x2  4 : Square, then subtract 4, then divide by 3. 3 169

170

CHAPTER 2 Functions

15. Machine diagram for f x 



16. Machine diagram for f x 

x  1.

1

subtract 1, then take square root

0

2

subtract 1, then take square root

1

5

subtract 1, then take square root

2

17. f x  2 x  12

3 . x 2

3

subtract 2, take reciprocal, multiply by 3

3

_1

subtract 2, take reciprocal, multiply by 3

_1

1

subtract 2, take reciprocal, multiply by 3

_3

18. g x  2x  3

x

f x

x

g x

1

2 1  12  8

3

2 3  3  3

0 1 2 3

2 12  2

2

2 1  12  0

0

2 2  12  2

1

2 3  12  8

3

2 2  3  1

2 0  3  3 2 1  3  5 2 3  3  9

19. f x  x 2  6; f 3  32  6  9  6  3; f 3  32  6  9  6  3; f 0  02  6  6;    2 f 12  12  6  14  6   23 4. 20. f x  x 3  2x; f 2  23  2 2  8  4  12; f 1  13  2 1  1  2  3;    3   f 0  03  2 0  0; f 12  12  2 12  18  1  98 . 1  2 2 1  2 2 5 1  2x ; f 2   1; f 2   ; f 3 3 3 3 1  2a 1  2 a  1 3  2a 1  2 a  ; f a  1   . f a  3 3 3 3

21. f x 

  1 2



  1  2 12 3

 0; f a 

1  2a ; 3

22  4 8 8 a2  4 x2  4 x2  4 22  4 x2  4 ; h 2   ; h 2   ; h a  ; h x   ; 5 5 5 5 5 5 5 5    2 x 4 a 2  4a  8   x 4 a  22  4  ;h x   . h a  2  5 5 5 5

22. h x 

23. f x  x 2  2x; f 0  02  2 0  0; f 3  32  2 3  9  6  15; f 3  32  2 3  9  6  3;    2   1 1 2 1 1 2 f a  a 2  2 a  a 2  2a; f x  x2  2 x  x 2  2x; f   2  . a a a a a   1 1  1  1  2; h 2  2  1  5 ; h 1  1  1  1  2  5 ; ; h 1  1  1 2 2 2 1 x 2 2 2 2   1 1 1 1 1 ;h   h x  1  x  1    x. 1 x 1 x x x x

24. h x  x 

SECTION 2.1 Functions

171

 

  1 1 1 1  2 1 1 1  1 1x 1 1 2    2  ; ; g 2     ; g 1  , which is undefined; g  3 1 1x 1  2 3 3 1  1 2 3 1 2 2    1  x2  1 2  x2 1  a 1a 1  a  1 1a1 2a  2    g a   ; g a  1    ;g x 1  . 1  a 1a 1  a  1 1a1 a 1  x2  1 x2

25. g x 

2  2 22 02 a2 t 2 ; g 2   0; g 2  , which is undefined; g 0   1; g a  ; t 2 2  2 22 02 a2   a2  2  2 a3 a2 a12  .  2 ; g a  1  g a2  2  2 a12 a1 a 22 a 4

26. g t 

27. k x  x 2  2x  3; k 0  02  2 0  3  3; k 2  22  2 2  3  5; k 2   22  2 2  3  3;  2      2   2 2 2  3  1  2 2; k a  2   a  22  2 a  2  3  a 2  6a  5; k  2     k x   x2  2 x  3  x 2  2x  3; k x 2   x 2  2 x 2  3  x 4  2x 2  3. 28. k x  2x 3  3x 2 ; k 0  2 03  3 02  0; k 3  2 33  3 32  27; k 3  2 33  3 32  81;  3    2  3    2 a 3  3a 2 ; k x  2 x3  3 x2  2x 3  3x 2 ; k 12  2 12  3 12   12 ; k a2  2 a2  3 a2  4  3  2   k x 3  2 x 3  3 x 3  2x 9  3x 6 .

29. f x  2 x  1; f 2  2 2  1  2 3  6; f 0  2 0  1  2 1  2;         f 12  2  12  1  2 12  1; f 2  2 2  1  2 1  2; f x  1  2 x  1  1  2 x;            f x 2  2  2  x 2  2  1  2 x 2  1  2x 2  2 (since x 2  1  0 ).

2 1 x 2 1 ; f 2    1; f 1    1; f x is not defined at x  0; x 2 2 1 1       x 2  1x 5 5 x x2 1 .   1; f x 2  2  2  1 since x 2  0, x  0; f   f 5  x 5 5 x 1x x x

30. f x 

31. Since 2  0, we have f 2  22  4. Since 1  0, we have f 1  12  1. Since 0  0, we have f 0  0  1  1. Since 1  0, we have f 1  1  1  2. Since 2  0, we have f 2  2  1  3.

32. Since 3  2, we have f 3  5. Since 0  2, we have f 0  5. Since 2  2, we have f 2  5. Since 3  2, we have f 3  2 3  3  3. Since 5  2, we have f 5  2 5  3  7.

33. Since 4  1, we have f 4  42  2 4  16  8  8. Since  32  1, we have      2 f  32   32  2  32  94  3   34 . Since 1  1, we have f 1  12  2 1  1  2  1. Since 1  0  1, we have f 0  0. Since 25  1, we have f 25  1.

34. Since 5  0, we have f 5  3 5  15. Since 0  0  2, we have f 0  0  1  1. Since 0  1  2, we have f 1  1  1  2. Since 0  2  2, we have f 2  2  1  3. Since 5  2, we have f 5  5  22  9.

35. f x  2  x  22  1  x 2  4x  4  1  x 2  4x  5; f x  f 2  x 2  1  22  1  x 2  1  4  1  x 2  6.

36. f 2x  3 2x  1  6x  1; 2 f x  2 3x  1  6x  2.    2 37. f x 2  x 2  4; f x  [x  4]2  x 2  8x  16. x  x  f x 6x  18 3 2x  6 38. f 6  18  2x  18;    2x  6 3 3 3 3 3 39. f x  3x  2, so f 1  3 1  2  1 and f 5  3 5  2  13. Thus, the net change is f 5  f 1  13  1  12. 40. f x  4  5x, so f 3  4  5 3  11 and f 5  4  5 5  21. Thus, the net change is f 5  f 3  21  11  10.

172

CHAPTER 2 Functions

41. g t  1  t 2 , so g 2  1  22  1  4  3 and g 5  1  52  24. Thus, the net change is g 5  g 2  24  3  21.

42. h t  t 2 5, so h 3  32 5  14 and h 6  62 5  41. Thus, the net change is h 6h 3  4114  27.

43. f a  5  2a; f a  h  5  2 a  h  5  2a  2h; 5  2a  2h  5  2a 5  2a  2h  5  2a 2h f a  h  f a     2. h h h h 44. f a  3a 2  2; f a  h  3 a  h2  2  3a 2  6ah  3h 2  2;     2  6ah  3h 2  2  3a 2  2 3a f a  h  f a 6ah  3h 2    6a  3h h h h f a  h  f a 55 45. f a  5; f a  h  5;   0. h h 1 1 46. f a  ; f a  h  ; a1 ah1 a1 ah1 1 1   f a  h  f a  1  h  1  1 a  h  1 a a a  ah1 a1  h h h h 1 a  1 a  h  1   . h a  1 a  h  1 a ah 47. f a  ; f a  h  ; a1 ah1 a a  h  1 a  h a  1 a ah   f a  h  f a a  h  1 a  1 a  h  1 a  1 a  h  1 a  1   h h h   a  h a  1  a a  h  1 2  a  ah  h  a 2  ah  a a a  h  1 a  1   h h a  h  1 a  1 1  a  h  1 a  1 2 a  h 2a ; f a  h  ; 48. f a  a1 ah1 2a a  h  1 2a  2h a  1 2a 2 a  h   f a  h  f a a  h  1 a  1 a  h  1 a  1 a  h  1 a  1   h h h 2 a  h a  1  2a a  h  1 2a 2  2ah  2a  2h  2a 2  2ah  2a a  h  1 a  1   h h a  h  1 a  1 2h 2   h a  h  1 a  1 a  h  1 a  1 49. f a  3  5a  4a 2 ;

  f a  h  3  5 a  h  4 a  h2  3  5a  5h  4 a 2  2ah  h 2

 3  5a  5h  4a 2  8ah  4h 2 ;     2  8ah  4h 2  3  5a  4a 2 3  5a  5h  4a f a  h  f a  h h 3  5a  5h  4a 2  8ah  4h 2  3  5a  4a 2 5h  8ah  4h 2   h h h 5  8a  4h  5  8a  4h.  h

SECTION 2.1 Functions

173

50. f a  a 3 ; f a  h  a  h3  a 3  3a 2 h  3ah 2  h 3 ;     a 3  3a 2 h  3ah 2  h 3  a 3 f a  h  f a 3a 2 h  3ah 2  h 3   h h h   2 2 h 3a  3ah  h  3a 2  3ah  h 2 .  h 51. f x  3x. Since there is no restriction, the domain is all real numbers,  . Since every real number y is three times the real number 13 y, the range is all real numbers  .

52. f x  5x 2  4. Since there is no restriction, the domain is all real numbers,  . Since 5x 2  0 for all x, 5x 2  4  4 for all x, so the range is [4 .

53. f x  3x, 2  x  6. The domain is [2 6], f 2  3 2  6, and f 6  3 6  18, so the range is [6 18]. 54. f x  5x 2  4, 0  x  2. The domain is [0 2], f 0  5 02  4  4, and f 2  5 22  4  24, so the range is [4 24]. 1 . Since the denominator cannot equal 0 we have x  3  0  x  3. Thus the domain is x  x  3. In x 3 interval notation, the domain is  3  3 .

55. f x 

1 . Since the denominator cannot equal 0, we have 3x  6  0  3x  6  x  2. In interval notation, the 3x  6 domain is  2  2 .

56. f x 

x 2 57. f x  2 . Since the denominator cannot equal 0 we have x 2  1  0  x 2  1  x  1. Thus the domain is x 1 x  x  1. In interval notation, the domain is  1  1 1  1 .

x4 . Since the denominator cannot equal 0, x 2  x  6  0  x  3 x  2  0  x  3 or x  2. 58. f x  2 x x 6 In interval notation, the domain is  3  3 2  2 .  59. f x  x  1. We must have x  1  0  x  1. Thus, the domain is [1 .  60. g x  x 2  9. The argument of the square root is positive for all x, so the domain is  .  61. f t  3 t  1. Since the odd root is defined for all real numbers, the domain is the set of real numbers,  .  62. g x  7  3x. For the square root to be defined, we must have 7  3x  0  7  3x  73  x. Thus the domain is    73 .  63. f x  1  2x. Since the square root is defined as a real number only for nonnegative numbers, we require that   1  2x  0  x  12 . So the domain is x  x  12 . In interval notation, the domain is  12 .  64. g x  x 2  4. We must have x 2  4  0  x  2 x  2  0. We make a table:  2

2 2

2 

Sign of x  2







Sign of x  2













Sign of x  2 x  2

Thus the domain is  2]  [2 .  2x . We require 2  x  0, and the denominator cannot equal 0. Now 2  x  0  x  2, and 3  x  0 65. g x  3x  x  3. Thus the domain is x  x  2 and x  3, which can be expressed in interval notation as [2 3  3 .

174

CHAPTER 2 Functions



x . We must have x  0 for the numerator and 2x 2  x  1  0 for the denominator. So 2x 2  x  1  0 2x 2  x  1      2x  1 x  1  0  2x  1  0 or x  1  0  x  12 or x  1. Thus the domain is 0 12  12   .

66. g x 

67. g x  a table:

 4

x 2  6x. Since the input to an even root must be nonnegative, we have x 2  6x  0  x x  6  0. We make  0

0 6

6 

Sign of x







Sign of x  6













Sign of x x  6

Thus the domain is  0]  [6 . 68. g x 



x 2  2x  8. We must have x 2  2x  8  0  x  4 x  2  0. We make a table:  2

2 4

4 

Sign of x  4







Sign of x  2













Sign of x  4 x  2

Thus the domain is  2]  [4 .

3 . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have x 4 x  4  0  x  4. Thus the domain is 4 .

69. f x  

x2 70. f x   . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 6x 6  x  0  6  x. Thus the domain is  6. x  12 71. f x   . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 2x  1   2x  1  0  x  12 . Thus the domain is 12   . x 72. f x   . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 4 9  x2 9  x 2  0  3  x 3  x  0. We make a table:

 3

3 3

3 

Sign of 3  x







Sign of 3  x













Interval

Thus the domain is 3 3.

Sign of x  4 x  2

SECTION 2.1 Functions

73. To evaluate f x, divide the input by 3 and add 23 to the result. 2 x (a) f x   3 3

175

y

(c)

(b) x

f x

2

4 3

4

2

6

8 3 10 3

8

1l 1

x

1

x

74. To evaluate g x, subtract 4 from the input and multiply the result by 34 . (a) g x  x  4  34  34 x  4

y

(c)

(b) x

g x

2

 32

4

0

6

3 2

8

3

1l

75. Let T x be the amount of sales tax charged in Lemon County on a purchase of x dollars. To find the tax, take 8% of the purchase price. (a) T x  008x

(c)

y

(b) x

T x

2

016

4

032

6

048

8

064

1l

1

x

76. Let V d be the volume of a sphere of diameter d. To find the volume, take the cube of the diameter, then multiply by  and divide by 6. 3 (a) V d  d 3  6   6d

(c)

y

(b) x

f x

2

4  42 3 32  335 3

4 6 8

36  113

256  268 3

10l 1

x

176

CHAPTER 2 Functions

  1 if x is rational 77. f x   5 if x is irrational

The domain of f is all real numbers, since every real number is either rational or

irrational; and the range of f is 1 5.   1 if x is rational 78. f x  The domain of f is all real numbers, since every real number is either rational or  5x if x is irrational

irrational. If x is irrational, then 5x is also irrational, and so the range of f is x  x  1 or x is irrational.     0 2  50 and V 20  50 1  20 2  0. (c) 79. (a) V 0  50 1  20 20 x V x (b) V 0  50 represents the volume of the full tank at time t  0, and 0 50 V 20  0 represents the volume of the empty tank twenty minutes 5 28125 later. 10 125 (d) The net change in V as t changes from 0 minutes to 20 minutes is 15 3125 V 20  V 0  0  50  50 gallons. 20 0 80. (a) S 2  4 22  16  5027, S 3  4 32  36  11310.

(b) S 2 represents the surface area of a sphere of radius 2, and S 3 represents the surface area of a sphere of radius 3.   05c2 075c2 81. (a) L 05c  10 1   866 m, L  10 1   661 m, and 075c c2 c2  09c2 L 09c  10 1   436 m. c2 (b) It will appear to get shorter.   13  7 104 20  2 mm, 82. (a) R 1   (b) 5 1  4 104 x R x  04 1 2 13  7 10 R 10   166 mm, and 04 1  4 10 10 166  100 148 13  7 10004  148 mm. R 100  04 200 144 1  4 100 (c) The net change in R as x changes from 10 to 100 is R 100  R 10  148  166  018 mm.   83. (a)  01  18500 025  012  4440,    04  18500 025  042  1665.

(b) They tell us that the blood flows much faster (about 275 times faster) 01 cm from the center than 01 cm from the edge.

(d) The net change in V as r changes from 01 cm to 05 cm is V 05  V 01  0  4440  4440 cms.   84. (a) D 01  2 3960 01  012  79201  281 miles   D 02  2 3960 02  022  158404  398 miles

500

141

1000

139

(c) r

 r

0

4625

01

4440

02

3885

03

2960

04

1665

05

0

SECTION 2.1 Functions

  (b) 1135 feet  1135 miles  0215 miles. D  2 3960 0215  02152  1702846  413 miles 0215 5280   (c) D 7  2 3960 7  72  55489  2356 miles

177

(d) The net change in D as h changes from 1135 ft (or 0215 mi) to 7 mi is D 7 D 0215  2356413  1943 miles.

85. (a) Since 0  5,000  10,000 we have T 5,000  0. Since 10,000  12,000  20,000 we have T 12,000  008 12,000  960. Since 20,000  25,000 we have T 25,000  1600  015 25,000  5350. (b) There is no tax on $5000, a tax of $960 on $12,000 income, and a tax of $5350 on $25,000.

86. (a) C 75  75  15  $90; C 90  90  15  $105; C 100  $100; and C 105  $105. (b) The total price of the books purchased, including shipping.

  75x if 0  x  2 87. (a) T x   150  50 x  2 if x  2

(b) T 2  75 2  150; T 3  150  50 3  2  200; and T 5  150  50 5  2  300. (c) The total cost of the lodgings.

    15 40  x if 0  x  40 88. (a) F x  0 if 40  x  65    15 x  65 if x  65

(b) F 30  15 40  10  15  10  $150; F 50  $0; and F 75  15 75  65 15  10  $150. (c) The fines for violating the speed limits on the freeway.

89. We assume the grass grows linearly. h

0

90.

T

0

91.

t

T 60

50 0

5

10

t

W

W

W

W

t

178

CHAPTER 2 Functions P

92.

1000 900 Population (×1000) 800 700 600 0

1980

1990

2000

2010

t

Year

93. Answers will vary. 94. Answers will vary. 95. Answers will vary.

2.2

GRAPHS OF FUNCTIONS

1. To graph the function f we plot the points x f x in a coordinate plane. To graph f x  x 2  2, we plot the     points x x 2  2 . So, the point 3 32  2  3 7 is

on the graph of f . The height of the graph of f above the

x

f x

x y

2

2

2 2

1

1

0

2

1

1

2

2

x-axis when x  3 is 7.

y 1

1 1

0

0 2

1

x

1 1 2 2

2. If f 4  10 then the point 4 10 is on the graph of f . 3. If the point 3 7 is on the graph of f , then f 3  7. 4. (a) f x  x 2 is a power function with an even exponent. It has graph IV. (b) f x  x 3 is a power function with an odd exponent. It has graph II.  (c) f x  x is a root function. It has graph I. (d) f x  x is an absolute value function. It has graph III.

5.

6.

y

x

f x  x  2

6

4

2

8

4

2

1

6

2

0

0

4

0

2

1

2

2

4

2

0

4

6

3

2

6

8

4

4

x

1 1

x

f x  4  2x

y

1 1

x

SECTION 2.2 Graphs of Functions

7. x

f x  x  3, 3  x  3

3

6

2

5

0

3

2

1

3

0

x

f x  x 2

4

16

3

9

2

4

1

1

0

0

11.

x

1

2

x

1

g x   x  12

5

16

3

4

2 1 0

0

15

1

1

2

05

3

0

4

05

5

1

10.

y

1

x

x 3 , 2 0x 5

f x 

x

1

1

9.

8.

y

y

f x  x 2  4

5

21

4

12

3

5

2

0

1

3

0

4

y

16

3

4

1

2

1

0

1

0

1

0

1

1

4

1

4

3

16

3

16

13. x

r x  3x 4

3

_5

y

x

r x  1  x 4

243

3

80

2

48

2

15

1

3

1

0

0

0

0

1

1

3

1

0

2

48

2

15

3

243

3

80

0

1

x

10

_1 0

x

1

y

14.

100

x

1

5

x

1

y

1

g x  x 2  2x  1

0

x

1

x

_1

y

1

x

12.

5

179

10 _1

0 _10

1

x

180

CHAPTER 2 Functions

15. x

g x  x 3  8

3

16.

y

x

g x  x  13

35

2

27

2

16

1

8

1

9

0

1

0

8

1

0

1

7

2

1

2

0

3

8

3

19

4

27

5

17. k x 

x

y

 3 x

27

3

8

2

1

1

0

18.

3

8

2

1

1

0

0

0

1

1

1

1

8

2

8

2

27

3

27

3

f x  1 



1 0 _1

x

x

10

20.

y

f x 

x

10 0



1 0 _1

y

1

2

0

1

2

3

1

4

3

6

2

1

9

4

11

3

0 2

16

5

18

4

27

5

38

6

25

10

6

21. x

1 C t  2 t

2

1 4

1

 12  14

0

x

20

y

22. x 3

1 4 16 

2

1 _1 0

1

t

C t   1 2

1

 32

2

1



 12

2

0

1

1 4 1 2

16 4

1

1

1

2

2

1 4

 12  13

x

10

x 2

0

1

x

1

y

 k x   3 x

x 27

19. x

x

1

y

1 t 1

x

10

y

2 0

2

t

SECTION 2.2 Graphs of Functions

23. x

H x  2x

5

10

4

8

3

6

2

4

1 0

24.

y

x

H x  x  1

5

4

4

3

3

2

2

1

2

1

0

0

0

1

1

2

1 x

1

25. x

G x  x  x

5

0

2

0

0

0

1 1

x

G x  x  x

5

10

2

4

1

2

0

0

2

2

4

1

0

5

10

3

0

x

f x 

x

f x  2x  2

5

12

2

8

0

2

1

0

2

2

5

8

28.

y

1 1

x

3

1

2

1

1

1

0

29. f x  8x  x 2

(a) [5 5] by [5 5]

1

x

1

27.

y

26.

y

1

x

1

x

y

1

y

x x

1 x

1

undefined

1

1

2

1

3

1

(b) [10 10] by [10 10] 10

4 2 -4

-2

-2 -4

2

4

181

-10

-5

5 -10

10

182

CHAPTER 2 Functions

(c) [2 10] by [5 20]

(d) [10 10] by [100 100]

20

100

10 -10 -2

2

4

6

8

-5

10

5

10

5

10

5

10

-100

The viewing rectangle in part (c) produces the most appropriate graph of the equation. 30. g x  x 2  x  20

(b) [10 10] by [10 10]

(a) [2 2] by [5 5]

10

4 2 -2

-1

1

-2

2

-10

-5

-4

-10

(c) [7 7] by [25 20]

(d) [10 10] by [100 100] 20

100

10 -6 -4 -2 -10

2

4

6

-10

-5

-20

-100

The viewing rectangle in part (c) produces the most appropriate graph of the equation. 31. h x  x 3  5x  4

(b) [3 3] by [10 10]

(a) [2 2] by [2 2] 2

10

1 -2

-1

1

-1

2

-3

-2

-1

-2

3

(d) [10 10] by [10 10] 5

-2

2

-10

(c) [3 3] by [10 5]

-3

1

-1 -5

10

1

2

3 -10

-10

The viewing rectangle in part (c) produces the most appropriate graph of the equation.

-5

5 -10

10

SECTION 2.2 Graphs of Functions 1 x4  x2  2 32. k x  32

(b) [2 2] by [2 2]

(a) [1 1] by [1 1]

1

2 1

-1.0

-0.5

0.5

1.0

-2

-1

-1

-1

1

2

5

10

-2

(c) [5 5] by [5 5]

(d) [10 10] by [10 10] 10

4 2 -4

-2

2

-2

4

-10

-5

-4

-10

The viewing rectangle in part (d) produces the most appropriate graph of the equation.   0 if x  2 33. f x   1 if x  2

 1 if x  1 34. f x   x  1 if x  1

y

y

1 2 1

x

 3 if x  2 35. f x   x  1 if x  2

1

x

1

x

  1  x if x  2 36. f x  5 if x  2

y

y

1

1 1

x

183

184

CHAPTER 2 Functions

 x if x  0 37. f x   x  1 if x  0

  2x  3 if x  1 38. f x   3  x if x  1

y

y

1

1 x

1

    1 if x  1 39. f x  1 if 1  x  1    1 if x  1

1

    1 if x  1 40. f x  x if 1  x  1    1 if x  1

y

y

2 2

  2 if x  1 41. f x   x 2 if x  1

1 x

1

x

1

x

1

x

  1  x 2 if x  2 42. f x  x if x  2 y

y

1

1 1

  0 if x  2 43. f x   3 if x  2

x

x

  x 2 if x  1 44. f x   1 if x  1

y

1

y

1 1

x

SECTION 2.2 Graphs of Functions

  if x  2  4 45. f x  x 2 if 2  x  2    x  6 if x  2

185

  if x  0   x 46. f x  9  x 2 if 0  x  3    x  3 if x  3

y

y

1 x

1

1 x

1

  x  2 if x  1 47. f x   x2 if x  1

5

-6 -4 -2

2

4

6

-5

  2x  x 2 if x  1 48. f x   x  13 if x  1

The first graph shows the output of a typical graphing device. However, the actual graph

of this function is also shown, and its difference from the graphing device’s version should be noted. y 1

2

-3 -2 -1 -2

    2 if x  2 49. f x  x if 2  x  2    2 if x  2

1

1

2

x

3

  if x  1  1 50. f x  1  x if 1  x  2    2 if x  2

51. The curves in parts (a) and (c) are graphs of a function of x, by the Vertical Line Test. 52. The curves in parts (b) and (c) are graphs of functions of x, by the Vertical Line Test.

53. The given curve is the graph of a function of x, by the Vertical Line Test. Domain: [3 2]. Range: [2 2]. 54. No, the given curve is not the graph of a function of x, by the Vertical Line Test. 55. No, the given curve is not the graph of a function of x, by the Vertical Line Test. 56. The given curve is the graph of a function of x, by the Vertical Line Test. Domain: [3 2]. Range: 2  0 3].

57. Solving for y in terms of x gives 3x  5y  7  y  35 x  75 . This defines y as a function of x.

58. Solving for y in terms of x gives 3x 2  y  5  y  3x 2  5. This defines y as a function of x.  59. Solving for y in terms of x gives x  y 2  y   x. The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x.

186

CHAPTER 2 Functions

  60. Solving for y in terms of x gives x 2  y  12  4  y  12  4  x 2  y  1   4  x 2  y  1  4  x 2 . The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x.  61. Solving for y in terms of x gives 2x  4y 2  3  4y 2  2x  3  y   12 2x  3. The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x.  62. Solving for y in terms of x gives 2x 2  4y 2  3  4y 2  2x 2  3  y   12 2x 2  3. The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x. 63. Solving for y in terms of x using the Quadratic Formula gives 2x y  5y 2  4  5y 2  2x y  4  0      2x  2x2  4 5 4 2x  4x 2  80 x  x 2  20   . The last equation gives two values of y for a y 2 5 10 5 given value of x. Thus, this equation does not define y as a function of x.  64. Solving for y in terms of x gives y  5  x  y  x  52 . This defines y as a function of x. 65. Solving for y in terms of x gives 2 x  y  0  y  2 x. This defines y as a function of x.

66. Solving for y in terms of x gives 2x  y  0  y  2x. Since a  a, the last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x.  67. Solving for y in terms of x gives x  y 3  y  3 x. This defines y as a function of x.  68. Solving for y in terms of x gives x  y 4  y   4 x. The last equation gives two values of y for any positive value of x. Thus, this equation does not define y as a function of x.

69. (a) f x  x 2  c, for c  0, 2, 4, and 6. c=4

10 8

c=2

(b) f x  x 2  c, for c  0, 2, 4 , and 6. c=_2

c=6

6

4 2

c=_6

4

c=0

0 -2

-2

c=0

8

6

-4

c=_4

10

2

2

-4

4

0 -2

-2

-4

-4

-6

-6

-8

-8

-10

-10

2

4

(c) The graphs in part (a) are obtained by shifting the graph of f x  x 2 upward c units, c  0. The graphs in part (b) are obtained by shifting the graph of f x  x 2 downward c units.

70. (a) f x  x  c2 , for c  0, 1, 2, and 3.

(b) f x  x  c2 , for c  0, 1, 2, and 3.

c=0 c=1

10 8

c=_1 c=0

c=2

c=_2

c=3

c=_3

8

6

6

4 2 -4

-2

0 -2

10

4 2

2

4

-4

-2

0 -2

-4

-4

-6

-6

-8

-8

-10

-10

2

4

(c) The graphs in part (a) are obtained by shifting the graph of y  x 2 to the right 1, 2, and 3 units, while the graphs in part (b) are obtained by shifting the graph of y  x 2 to the left 1, 2, and 3 units.

SECTION 2.2 Graphs of Functions

71. (a) f x  x  c3 , for c  0, 2, 4, and 6. c=2

10

-8

-4

(b) f x  x  c3 , for c  0, 2, 4, and 6.

c=6

c=_4

8

8

6

6

4

4

2

2

-2

0 -2

4

8

-8

-4

-4

-4

-6

-6

-8

c=0

187

c=0

4

8

-8

c=4

c=_6

-10

c=_2

(c) The graphs in part (a) are obtained by shifting the graph of f x  x 3 to the right c units, c  0. The graphs in part (b) are obtained by shifting the graph of f x  x 3 to the left c units, c  0.

72. (a) f x  cx 2 , for c  1, 12 , 2, and 4. c=1

c=4

(b) f x  cx 2 , for c  1, 1,  12 , and 2.

c=2

10

10

1 c=_ 2

6 4

8

c=1

8

6 4 2

2 -4

-4

0 -2

-2

2

0 -2

-2

4

2

4 1

-4

-4

c=__ 2

-6

-6

-8

-8

-10

c=_4

-10

c=_2

(c) As c increases, the graph of f x  cx 2 is stretched vertically. As c decreases, the graph of f is flattened. When c  0, the graph is reflected about the x-axis. 73. (a) f x  x c , for c  12 , 14 , and 16 .

(b) f x  x c , for c  1, 13 , and 15 .

3

2

1

1

1

c= _ 5

1 6

1

c= _

0

1

2

3

-1

4

-1

-2



(c) Graphs of even roots are similar to y  x, graphs of odd roots are similar to y   y  c x becomes steeper near x  0 and flatter when x  1. 1 74. (a) f x  n , for n  1 and 3. x

 3

x. As c increases, the graph of

1 (b) f x  n , for n  2 and 4. x 3

3 2

n=3

-2

-1

0

2

n=1

1

-3

1

c= _ 3 c= _ 4

-1

c=1

2 1 c= _ 2

1

2

n=4

1

3

-3

-2

-1

0

-1

-1

-2

-2

-3

-3

n=2 1

2

3

188

CHAPTER 2 Functions

(c) As n increases, the graphs of y  1x n go to zero faster for x large. Also, as n increases and x goes to 0, the graphs of y  1x n go to infinity faster. The graphs of y  1x n for n odd are similar to each other. Likewise, the graphs for n even are similar to each other.

75. The slope of the line segment joining the points 2 1 and 4 6 is m 

6  1   76 . Using the point-slope form, 4  2

we have y  1   76 x  2  y   76 x  73  1  y   76 x  43 . Thus the function is f x   76 x  43 for 2  x  4. 76. The slope of the line containing the points 3 2 and 6 3 is m 

2  3 5   59 . Using the point-slope equation 3  6 9

5 1 5 1 of the line, we have y  3  59 x  6  y  59 x  10 3  3  9 x  3 . Thus the function is f x  9 x  3 , for 3  x  6.

 77. First solve the circle for y: x 2  y 2  9  y 2  9  x 2  y   9  x 2 . Since we seek the top half of the circle, we   choose y  9  x 2 . So the function is f x  9  x 2 , 3  x  3.

 78. First solve the circle for y: x 2  y 2  9  y 2  9  x 2  y   9  x 2 . Since we seek the bottom half of the circle,   we choose y   9  x 2 . So the function is f x   9  x 2 , 3  x  3. 05 for 10  r  100. As the balloon r2 is inflated, the skin gets thinner, as we would expect.

79. We graph T r 

80. We graph P   141 3 for 1    10. As wind speed increases, so does power output, as expected. P

T

20,000

0.004 0.003 10,000

0.002 0.001 0

50

r

  600  010x if 0  x  300 81. (a) E x   3600  006 x  300 if 300  x

0

(b)

5

10

E 60 50 40 30 20 10 0

100

200

300

400

500

600

x

SECTION 2.2 Graphs of Functions

  200      220    240 82. C x             400

189

C

4.00

if 0  x  1

if 1  x  11

if 11  x  12

3.00 2.00

if 19  x  2 0

1

2

x

P

  049      070 83. P x    091     112

1.00

if 0  x  1

0.80

if 1  x  2 if 2  x  3

0.60

if 3  x  35

0.40 0.20 0

1

2

3

4

x

84. The graph of x  y 2 is not the graph of a function because both 1 1 and 1 1 satisfy the equation x  y 2 . The graph of x  y 3 is the graph of a function because x  y 3  x 13  y. If n is even, then both 1 1 and 1 1 satisfies the equation x  y n , so the graph of x  y n is not the graph of a function. When n is odd, y  x 1n is defined for all real

numbers, and since y  x 1n  x  y n , the graph of x  y n is the graph of a function.

85. Answers will vary. Some examples are almost anything we purchase based on weight, volume, length, or time, for example gasoline. Although the amount delivered by the pump is continuous, the amount we pay is rounded to the penny. An example involving time would be the cost of a telephone call.

86.

y

y

1

y

1 1

f x  [[x]]

x

1 1

x

g x  [[2x]] 1 The graph of k x  [[nx]] is a step function whose steps are each wide. n

1

h x  [[3x]]

x

190

CHAPTER 2 Functions

    87. (a) The graphs of f x  x 2  x  6 and g x  x 2  x  6 are shown in the viewing rectangle [10 10] by [10 10].

10

-10

10

-5

5

10

-10

-5

5

-10

10

-10

For those values of x where f x  0 , the graphs of f and g coincide, and for those values of x where f x  0, the graph of g is obtained from that of f by reflecting the part below the x-axis about the x-axis.     (b) The graphs of f x  x 4  6x 2 and g x  x 4  6x 2  are shown in the viewing rectangle [5 5] by [10 15]. 10

-4

-2

10

2

4

-4

-10

-2

2

4

-10

For those values of x where f x  0 , the graphs of f and g coincide, and for those values of x where f x  0, the graph of g is obtained from that of f by reflecting the part below the x-axis above the x-axis. (c) In general, if g x   f x, then for those values of x where f x  0, the graphs of f and g coincide, and for those values of x where f x  0, the graph of g is obtained from that of f by reflecting the part below the x-axis above the x-axis. y

y

x

y  f x

2.3

x

y  g x

GETTING INFORMATION FROM THE GRAPH OF A FUNCTION

1. To find a function value f a from the graph of f we find the height of the graph above the x-axis at x  a. From the graph of f we see that f 3  4 and f 1  0. The net change in f between x  1 and x  3 is f 3  f 1  4  0  4. 2. The domain of the function f is all the x-values of the points on the graph, and the range is all the corresponding y-values. From the graph of f we see that the domain of f is the interval   and the range of f is the interval  7]. 3. (a) If f is increasing on an interval, then the y-values of the points on the graph rise as the x-values increase. From the graph of f we see that f is increasing on the intervals  2 and 4 5. (b) If f is decreasing on an interval, then y-values of the points on the graph fall as the x-values increase. From the graph of f we see that f is decreasing on the intervals 2 4 and 5 .

SECTION 2.3 Getting Information from the Graph of a Function

191

4. (a) A function value f a is a local maximum value of f if f a is the largest value of f on some interval containing a. From the graph of f we see that there are two local maximum values of f : one maximum is 7, and it occurs when x  2; the other maximum is 6, and it occurs when x  5.

(b) A function value f a is a local minimum value of f if f a is the smallest value of f on some interval containing a. From the graph of f we see that there is one local minimum value of f . The minimum value is 2, and it occurs when x  4.

5. The solutions of the equation f x  0 are the x-intercepts of the graph of f . The solution of the inequality f x  0 is the set of x-values at which the graph of f is on or above the x-axis. From the graph of f we find that the solutions of the equation f x  0 are x  1 and x  7, and the solution of the inequality f x  0 is the interval [1 7]. 6. (a) To solve the equation 2x  1  x  4 graphically we graph the

y

functions f x  2x  1 and g x  x  4 on the same set of axes and determine the values of x at which the graphs of f and g intersect. From the graph, we see that the solution is x  1.

1 0

1

x

(b) To solve the inequality 2x  1  x  4 graphically we graph the functions f x  2x  1 and g x  x  4 on the same set of axes and find the values of x at which the graph of g is higher than the graph of f . From the graphs in part (a) we see that the solution of the inequality is  1. 7. (a) h 2  1, h 0  1, h 2  3, and h 3  4. (b) Domain: [3 4]. Range: [1 4].

(c) h 3  3, h 2  3, and h 4  3, so h x  3 when x  3, x  2, or x  4.

(d) The graph of h lies below or on the horizontal line y  3 when 3  x  2 or x  4, so h x  3 for those values of x. (e) The net change in h between x  3 and x  3 is h 3  h 3  4  3  1.

8. (a) g 4  3, g 2  2, g 0  2, g 2  1, and g 4  0. (b) Domain: [4 4]. Range: [2 3].

(c) g 4  3. [Note that g 2  1 not 3.]

(d) It appears that g x  0 for 1  x  18 and for x  4; that is, for x  1  x  18  4. (e) g 1  0 and g 2  1, so the net change between x  1 and x  2 is 1  0  1.

9. (a) f 0  3  12  g 0. So f 0 is larger.

(b) f 3  1  25  g 3. So g 3 is larger. (c) f x  g x for x  2 and x  2.

(d) f x  g x for 4  x  2 and 2  x  3; that is, on the intervals [4 2] and [2 3]. (e) f x  g x for 2  x  2; that is, on the interval 2 2.

10. (a) The graph of g is higher than the graph of f at x  6, so g 6 is larger.

(b) The graph of f is higher than the graph of g at x  3, so f 3 is larger.

(c) The graphs of f and g intersect at x  2, x  5, and x  7, so f x  g x for these values of x. (d) f x  g x for 1  x  2 and approximately 5  x  7; that is, on [1 2] and [5 7].

(e) f x  g x for 2  x  5 and approximately 7  x  8; that is, on [2 5 and 7 8].

192

CHAPTER 2 Functions y

11. (a)

y

12. (a)

2 0

x

1

2 0

x

1

(b) Domain:  ; Range:  

13. (a)

y

(b) Domain:  ; Range  

14. (a)

y

1 0

1 0

1

x

1

x

(b) Domain: 1 4; Range 4 2 (b) Domain: [2 5]; Range [4 3]

15. (a)

y

16. (a)

y

2 0

1

x

2 0

x

1

(b) Domain: [3 3]; Range: [1 8] (b) Domain: [3 3]; Range [6 3]

18. (a)

17. (a) 10

-2

5

2

4

-5 -6

-4

-2

2

(b) Domain:  ; Range: [1 

-10

(b) Domain:  ; Range:  2]

SECTION 2.3 Getting Information from the Graph of a Function

19. (a)

193

20. (a) 2

2

1

1 2

4

6

-3 -2 -1 -1

-1

(b) Domain: [1 ; Range: [0 

1

2

3

(b) Domain: [2 ; Range: [0 

21. (a)

22. (a) 5

-8 -6 -4 -2

5

2 4 6 8

-8 -6 -4 -2

-5

-5

(b) Domain: [4 4]; Range: [0 4]

(b) Domain: [5 5]; Range: [5 0]

y

23.

y

24. y=4-x

_2x+3

y=x-2

1l

1l 1

2 4 6 8

3x-7 1

x

x

(a) From the graph, we see that x  2  4  x when

(a) From the graph, we see that 2x  3  3x  7 when

(b) From the graph, we see x  2  4  x when x  3.

(b) From the graph, we see that 2x  3  3x  7 when

x  3.

x  2. x  2.

194

CHAPTER 2 Functions y

25.

y

26.

1l

1

x y=_x@l

y=x@l y=3-4x 1l

y=2-xl 1

x

(a) From the graph, we see that x 2  2  x when

(a) From the graph, we see that x 2  3  4x when

(b) From the graph, we see that x 2  2  x when

(b) From the graph, we see that x 2  3  4x when

x  2 or x  1.

x  1 or x  3.

2  x  1.

1  x  3.

28.

27.

20

10

10 -4

-2

-10

2

-20

-10

-30

-20

(a) We graph y  x 3  3x 2 (black) and

2

4

6

(a) We graph y  5x 2  x 3 (black) and

y  x 2  3x  7 (gray). From the graph, we see

y  x 2  3x  4 (gray). From the graph, we see

that the graphs intersect at x  432, x  112,

that the graphs intersect at x  058, x  129, and

and x  144.

x  529.

(b) From the graph, we see that

(b) From the graph, we see that

x 3  3x 2  x 2  3x  7 on approximately

5x 2  x 3  x 2  3x  4 on approximately

[432 112] and [144 . 29.

[058 129] and [529 .

30. 4 2

3 2

-2

-1

1

1 -2

(a) We graph y  16x 3  16x 2 (black) and y  x  1 (gray). From the graph, we see that the graphs intersect at x  1, x   14 , and x  14 . (b) From the graph, we see that 16x 3  16x 2  x  1 on     1  14 and 14   .

-1

0

(a) We graph y  1 

1



2

3

4

5

x (black) and y 



x2  1

(gray). From the graph, we see that the solutions are x  0 and x  231.

(b) From the graph, we see that 1  approximately 0 231.



x



x 2  1 on

SECTION 2.3 Getting Information from the Graph of a Function

195

31. (a) The domain is [1 4] and the range is [1 3]. (b) The function is increasing on 1 1 and 2 4 and decreasing on 1 2.

32. (a) The domain is [2 3] and the range is [2 3]. (b) The function is increasing on 0 1 and decreasing on 2 0 and 1 3.

33. (a) The domain is [3 3] and the range is [2 2]. (b) The function is increasing on 2 1 and 1 2 and decreasing on 3 2, 1 1, and 2 3. 34. (a) The domain is [2 2] and the range is [2 2]. (b) The function is increasing on 1 1 and decreasing on 2 1 and 1 2. 35. (a) f x  x 2  5x is graphed in the viewing rectangle [2 7] by [10 10].

36. (a) f x  x 3  4x is graphed in the viewing rectangle [10 10] by [10 10].

10

10

-2

2

4

6

-10

-5

-10

5

10

-10

(b) The domain is   and the range is [625 .

(c) The function is increasing on 25 . It is decreasing on  25.

37. (a) f x  2x 3  3x 2  12x is graphed in the viewing rectangle [3 5] by [25 20].

(b) The domain and range are  . (c) The function is increasing on  115 and 115 . It is decreasing on 115 115.

38. (a) f x  x 4  16x 2 is graphed in the viewing rectangle [10 10] by [70 10].

20 -10

10 -2

-10

-5

5

10

-20 2

4

-40 -60

-20

(b) The domain and range are  .

(b) The domain is   and the range is [64 .

(c) The function is increasing on  1 and 2 .

(c) The function is increasing on 283 0 and

It is decreasing on 1 2.

283 . It is decreasing on  283 and 0 283.

196

CHAPTER 2 Functions

39. (a) f x  x 3  2x 2  x  2 is graphed in the viewing rectangle [5 5] by [3 3].

40. (a) f x  x 4  4x 3  2x 2  4x  3 is graphed in the viewing rectangle [3 5] by [5 5]. 4

2

2 -4

-2

2

4

-2

-2

2

-2

4

-4

(b) The domain and range are  .

(b) The domain is   and the range is [4 .

(c) The function is increasing on  155 and

(c) The function is increasing on 04 1 and 24 .

022 . It is decreasing on 155 022.

41. (a) f x  x 25 is graphed in the viewing rectangle [10 10] by [5 5].

It is decreasing on  04 and 1 24.

42. (a) f x  4  x 23 is graphed in the viewing rectangle [10 10] by [10 10].

10

4 2 -10

-5

-2

5

10

-10

-5

-4

5

10

-10

(b) The domain is   and the range is [0 .

(b) The domain is   and the range is  4].

(c) The function is increasing on 0 . It is decreasing

(c) The function is increasing on  0. It is

on  0.

decreasing on 0 .

43. (a) Local maximum: 2 at x  0. Local minimum: 1 at x  2 and 0 at x  2.

(b) The function is increasing on 2 0 and 2  and decreasing on  2 and 0 2.

44. (a) Local maximum: 2 at x  2 and 1 at x  2. Local minimum: 1 at x  0.

(b) The function is increasing on  2 and 0 2 and decreasing on 2 0 and 2 .

45. (a) Local maximum: 0 at x  0 and 1 at x  3. Local minimum: 2 at x  2 and 1 at x  1.

(b) The function is increasing on 2 0 and 1 3 and decreasing on  2, 0 1, and 3 .

46. (a) Local maximum: 3 at x  2 and 2 at x  1. Local minimum: 0 at x  1 and 1 at x  2.

(b) The function is increasing on  2, 1 1, and 2  and decreasing on 2 1 and 1 2.

47. (a) In the first graph, we see that f x  x 3  x has a local minimum and a local maximum. Smaller x- and y-ranges show that f x has a local maximum of about 038 when x  058 and a local minimum of about 038 when x  058. 5

0.5

0.50 -0.3

0.4 -5

-0.4

5 -5

-0.60

-0.55

0.3 -0.50

-0.5

(b) The function is increasing on  058 and 058  and decreasing on 058 058.

0.55

0.60

SECTION 2.3 Getting Information from the Graph of a Function

197

48. (a) In the first graph, we see that f x  3  x  x 2  x 3 has a local minimum and a local maximum. Smaller x- and y-ranges show that f x has a local maximum of about 400 when x  100 and a local minimum of about 281 when x  033. 4 2

-2

0

2

-0.40

-0.35

2.9

4.1

2.8

4.0

2.7 -0.30

3.9 0.9

1.0

1.1

(b) The function is increasing on 033 100 and decreasing on  033 and 100 .

49. (a) In the first graph, we see that g x  x 4  2x 3  11x 2 has two local minimums and a local maximum. The local maximum is g x  0 when x  0. Smaller x- and y-ranges show that local minima are g x  1361 when x  171 and g x  7332 when x  321. -5

5

-1.75

-1.70

-1.65 -13.4

-73.0

3.1

-50

-13.6

-73.5

-100

-13.8

-74.0

3.2

3.3

(b) The function is increasing on 171 0 and 321  and decreasing on  171 and 0 321.

50. (a) In the first graph, we see that g x  x 5  8x 3  20x has two local minimums and two local maximums. The local maximums are g x  787 when x  193 and g x  1302 when x  104. Smaller x- and y-ranges show that local minimums are g x  1302 when x  104 and g x  787 when x  193. Notice that since g x is odd, the local maxima and minima are related. 20

-2.0

-1.8 -7.8

13.1 13.0

-5

-7.9

5

12.9 -20

1.0

-8.0 -1.2

-1.0

1.2

7.90 -12.8 7.85 -13.0 -13.2

7.80 1.90

1.95

2.00

(b) The function is increasing on  193, 104 104, and 193  and decreasing on 193 104 and 104 193.

198

CHAPTER 2 Functions

 51. (a) In the first graph, we see that U x  x 6  x has only a local maximum. Smaller x- and y-ranges show that U x has a local maximum of about 566 when x  400. 10

5.70

5

5.65 5.60 3.9

5

4.0

4.1

(b) The function is increasing on  400 and decreasing on 400 6.  52. (a) In the first viewing rectangle below, we see that U x  x x  x 2 has only a local maximum. Smaller x- and y-ranges show that U x has a local maximum of about 032 when x  075. 1.0

0.40

0.5

0.35 0.30

0.0 0.0

0.5

1.0

0.7

0.8

0.9

(b) The function is increasing on 0 075 and decreasing on 075 1. 1  x2 has a local minimum and a local maximum. Smaller x- and y-ranges x3 show that V x has a local maximum of about 038 when x  173 and a local minimum of about 038 when x  173.

53. (a) In the first graph, we see that V x 

2

0.40

1.6

1.7

1.8

-0.30 0.35 -5

-0.35

5 -1.8

-2

-1.7

0.30 -1.6

-0.40

(b) The function is increasing on  173 and 173  and decreasing on 173 0 and 0 173. 1 has only a local maximum. Smaller x- and x2  x  1 y-ranges show that V x has a local maximum of about 133 when x  050.

54. (a) In the first viewing rectangle below, we see that V x 

2

1.40 1.35

-5

5 -0.6

-0.5

1.30 -0.4

(b) The function is increasing on  050 and decreasing on 050 . 55. (a) At 6 A . M . the graph shows that the power consumption is about 500 megawatts. Since t  18 represents 6 P. M ., the graph shows that the power consumption at 6 P. M . is about 725 megawatts. (b) The power consumption is lowest between 3 A . M . and 4 A . M ..

SECTION 2.3 Getting Information from the Graph of a Function

199

(c) The power consumption is highest just before 12 noon. (d) The net change in power consumption from 9 A . M . to 7 P. M . is P 19  P 9  690  790  100 megawatts. 56. (a) The first noticeable movements occurred at time t  5 seconds. (b) It seemed to end at time t  30 seconds.

(c) Maximum intensity was reached at t  17 seconds.

57. (a) This person appears to be gaining weight steadily until the age of 21 when this person’s weight gain slows down. The person continues to gain weight until the age of 30, at which point this person experiences a sudden weight loss. Weight gain resumes around the age of 32, and the person dies at about age 68. Thus, the person’s weight W is increasing on 0 30 and 32 68 and decreasing on 30 32. (b) The sudden weight loss could be due to a number of reasons, among them major illness, a weight loss program, etc. (c) The net change in the person’s weight from age 10 to age 20 is W 20  W 10  150  50  100 lb. 58. (a) Measuring in hours since midnight, the salesman’s distance from home D is increasing on 8 9, 10 12, and 15 17, constant on 9 10, 12 13, and 17 18, and decreasing on 13 15 and 18 19. (b) The salesman travels away from home and stops to make a sales call between 9 A . M . and 10 A . M ., and then travels further from home for a sales call between 12 noon and 1 P. M . Next he travels along a route that takes him closer to home before taking him further away from home. He then makes a final sales call between 5 P. M . and 6 P. M . and then returns home. (c) The net change in the distance D from noon to 1 P. M . is D 1 P. M .  D noon  0. 59. (a) The function W is increasing on 0 150 and 300  and decreasing on 150 300. (b) W has a local maximum at x  150 and a local minimum at x  300.

(c) The net change in the depth W from 100 days to 300 days is W 300  W 100  25  75  50 ft.

60. (a) The function P is increasing on 0 25 and decreasing on 25 50. (b) The maximum population was 50,000, and it was attained at x  25 years, which represents the year 1975. (c) The net change in the population P from 1970 to 1990 is P 40  P 20  40  40  0.

61. Runner A won the race. All runners finished the race. Runner B fell, but got up and finished the race. 62. (a)

F

63. (a)

80

E 400

70 60

300

50 40

200

30

100

20 10 0

1

2

3

4

5

6

7

8

9 10

(b) As the distance x increases, the gravitational attraction F decreases. The rate of decrease is rapid at first, and slows as the distance increases.

x

0

50

100 150 200 250 300

(b) As the temperature T increases, the energy E increases. The rate of increase gets larger as the temperature increases.

T

200

CHAPTER 2 Functions

64. In the first graph, we see the general location of the minimum of V  99987  006426T  00085043T 2  00000679T 3 is around T  4. In the second graph, we isolate the minimum, and from this graph, we see that the minimum volume of 1 kg of water occurs at T  396 C. 1005

999.76

1000

999.75

995

999.74 0

20

3.5

4.0

4.5

10 . In the second graph, we isolate the  5 minimum, and from this graph, we see that energy is minimized when   75 mi/h.

65. In the first graph, we see the general location of the minimum of E   273 3

10000

4700 4650

5000 4600 6

8

10

7.4

7.5

7.6

66. In the first graph, we see the general location of the maximum of  r   32 1  r  r 2 is around r  07 cm. In the second graph, we isolate the maximum, and from this graph we see that at the maximum velocity is approximately 047 when r  067 cm. 1.0

0.50 0.48

0.5

0.46 0.0 0.6

0.8

1.0

67. (a) f x is always increasing, and f x  0 for all x. y

0

0.60

0.65

0.70

(b) f x is always decreasing, and f x  0 for all x. y

x

0

x

SECTION 2.4 Average Rate of Change of a Function

(c) f x is always increasing, and f x  0 for all x.

201

(d) f x is always decreasing, and f x  0 for all x.

y

y

x 0

0

x

68. Numerous answers are possible. 69. (a) If x  a is a local maximum of f x then

f a  f x  0 for all x around x  a. So  2  2 g a  g x and thus g a  g x.

(c) Let f x  x 4  x 2  6x  9. From the graph, we

Similarly, if x  b is a local minimum of f x, then f x  f b  0 for all x around x  b. So 2  2  g x  g b and thus g x  g b.

see that f x has a minimum at x  1. Thus g x also has a minimum at x  1 and this minimum   value is g 1  14  12  6 1  9  5. 10 5

(b) Using the distance formula,   2 g x  x  32  x 2  0   x 4  x 2  6x  9

2.4

0 0

2

AVERAGE RATE OF CHANGE OF A FUNCTION

100 miles  50 mi/h. 2 hours f b  f a 2. The average rate of change of a function f between x  a and x  b is average rate of change  . ba

1. If you travel 100 miles in two hours then your average speed for the trip is average speed 

3. The average rate of change of the function f x  x 2 between x  1 and x  5 is

25  1 24 52  12 f 1 average rate of change  f 5    6.  51 4 4 4 4. (a) The average rate of change of a function f between x  a and x  b is the slope of the secant line between a f a and b f b. (b) The average rate of change of the linear function f x  3x  5 between any two points is 3.

5. (a) Yes, the average rate of change of a function between x  a and x  b is the slope of the secant line through a f a f b  f a . and b f b; that is, ba (b) Yes, the average rate of change of a linear function y  mx  b is the same (namely m) for all intervals.

6. (a) No, the average rate of change of an increasing function is positive over any interval.

(b) No, just because the average rate of change of a function between x  a and x  b is negative, it does not follow

that the function is decreasing on that interval. For example, f x  x 2 has negative average rate of change between x  2 and x  1, but f is increasing for 0  x  1.

7. (a) The net change is f 4  f 1  5  3  2.

(b) We use the points 1 3 and 4 5, so the average rate of change is

2 53  . 41 3

202

CHAPTER 2 Functions

8. (a) The net change is f 5  f 1  2  4  2. (b) We use the points 1 4 and 5 2, so the average rate of change is 9. (a) The net change is f 5  f 0  2  6  4. (b) We use the points 0 6 and 5 2, so the average rate of change is 10. (a) The net change is f 5  f 1  4  0  4.

24 2 1   . 51 4 2 26 4  . 50 5

(b) We use the points 1 0 and 5 4, so the average rate of change is

4 2 40   . 5  1 6 3

11. (a) The net change is f 3  f 2  [3 3  2]  [3 2  2]  7  4  3. 3 f 3  f 2   3. (b) The average rate of change is 32 1     1 12. (a) The net change is r 6  r 3  3  3 6  3  13 3  1  2  1.

1 r 6  r 3  . 63 3     3 13. (a) The net change is h 1  h 4  1  2   4  32  12  11 2  5. 5 h 1  h 4   1. (b) The average rate of change is 1  4 5     14. (a) The net change is g 2  g 3  2  23 2  2  23 3  23  4   10 3 . (b) The average rate of change is

 10 g 2  g 3 3  2.  2  3 5 3     15. (a) The net change is h 6  h 3  2 62  6  2 32  3  66  15  51. (b) The average rate of change is

h 6  h 3 51   17. 63 3     16. (a) The net change is f 0  f 2  1  3 02  1  3 22  1  11  12. (b) The average rate of change is

f 0  f 2 12   6. 0  2 2       17. (a) The net change is f 10  f 0  103  4 102  03  4 02  600  0  600. (b) The average rate of change is

f 10  f 0 600   60. 10  0 10     18. (a) The net change is g 2  g 2  24  23  22  24  23  22  12  28  16. (b) The average rate of change is

g 2  g 2 16   4. 2  2 4     19. (a) The net change is f 3  h  f 3  5 3  h2  5 32  45  30h  5h 2  45  5h 2  30h. (b) The average rate of change is

f 3  h  f 3 5h 2  30h  5h  30.  h 3  h  3       20. (a) The net change is f 2  h f 2  1  3 2  h2  1  3 22  3h 2  12h  11 11  3h 2 12h. (b) The average rate of change is

(b) The average rate of change is

3h 2  12h f 2  h  f 2  3h  12.  h 2  h  2

SECTION 2.4 Average Rate of Change of a Function

203

1a 1 1   . a 1 a 1a 1a 1 g a  g 1  a   . (b) The average rate of change is a1 a1 a a  1 a

21. (a) The net change is g a  g 1 

2 2h 2   . h1 01 h1 2h  g h  g 0 h  1  2h   2 . (b) The average rate of change is  h0 h h h  1 h1

22. (a) The net change is g h  g 0 

23. (a) The net change is f a  h  f a 

2 2 2h   . ah a a a  h

(b) The average rate of change is 2h  2h 2 f a  h  f a a a  h   .  h ah a  h a a  h a  h  a 24. (a) The net change is f a  h  f a 

  a  h  a.

(b) The average rate of change is     f a  h  f a ah a ah a h 1 a  h  a             . h a  h  a ah a h ah a h ah a ah a

25. (a) The average rate of change is     1 a  h  3  1 a  3 1a  1h  3  1a  3 1h 1 f a  h  f a 2 2 2 2  2  2  .  h h h 2 a  h  a (b) The slope of the line f x  12 x  3 is 12 , which is also the average rate of change.

26. (a) The average rate of change is [4 a  h  2]  [4a  2] g a  h  g a 4a  4h  2  4a  2 4h    4.  h h h a  h  a (b) The slope of the line g x  4x  2 is 4, which is also the average rate of change. 27. The function f has a greater average rate of change between x  0 and x  1. The function g has a greater average rate of change between x  1 and x  2. The functions f and g have the same average rate of change between x  0 and x  15. 28. The average rate of change of f is constant, that of g increases, and that of h decreases. 29. The average rate of change is

50  75 25 1 W 200  W 100     ft/day. 200  100 200  100 100 4

P 40  P 20 40  40 0    0. 40  20 40  20 20 (b) The population increased and decreased the same amount during the 20 years.

30. (a) The average rate of change is

735 1,591  856   245 persons/yr. 2001  1998 3 657 826  1 483   3285 persons/yr. (b) The average rate of change of population is 2004  2002 2 (c) The population was increasing from 1997 to 2001.

31. (a) The average rate of change of population is

(d) The population was decreasing from 2001 to 2006.

204

CHAPTER 2 Functions

400 100 800  400    476 m/s. 152  68 84 21 400 1,600  1,200   268 m/s. (b) The average speed is 412  263 149 (c) Lap Length of time to run lap

32. (a) The average speed is

Average speed of lap.

1

32

625 m/s

2

36

556 m/s

3

40

500 m/s

4

44

455 m/s

5

51

392 m/s

6

60

333 m/s

7

72

278 m/s

8

77

260 m/s

The man is slowing down throughout the run.

635  495 140   14 players/yr. 2013  2003 10 18 513  495   18 players/yr. (b) The average rate of change of sales is 2004  2003 1 103 410  513   103 players/yr. (c) The average rate of change of sales is 2005  2004 1 (d) Year DVD players sold Change in sales from previous year

33. (a) The average rate of change of sales is

2003

495

—

2004

513

18

2005

410

103

2006

402

8

2007

520

118

2008

580

60

2009

631

51

2010

719

88

2011

624

95

2012

582

42

2013

635

53

Sales increased most quickly between 2006 and 2007, and decreased most quickly between 2004 and 2005.

SECTION 2.4 Average Rate of Change of a Function

205

34. Year

Number of books

1980

420

1981

460

1982

500

1985

620

1990

820

1992

900

1995

1020

1997

1100

1998

1140

1999

1180

2000

1220

35. The average rate of change of the temperature of the soup over the first 20 minutes is 119  200 81 T 20  T 0    405 F/min. Over the next 20 minutes, it is 20  0 20  0 20 T 40  T 20 89  119 30    15 F/min. The first 20 minutes had a higher average rate of change of 40  20 40  20 20 temperature (in absolute value).

36. (a) (i) Between 1860 and 1890, the average rate of change was about 84 farms per year. (ii) Between 1950 and 1970, the average rate of change was about 131 farms per year.

4570  2040 y 1890  y 1860   84, a gain of 1890  1860 30 2780  5390 y 1970  y 1950   131, a loss of 1970  1950 20

(b) From the graph, it appears that the steepest rate of decline was during the period from 1950 to 1960.

d 10  d 0 100   10. 10  0 10 (b) Skier A gets a great start, but slows at the end of the race. Skier B maintains a steady pace. Runner C is slow at the beginning, but accelerates down the hill.

37. (a) For all three skiers, the average rate of change is

38. (a) Skater B won the race, because he travels 500 meters before Skater A. A 10  A 0 200  0 (b) Skater A’s average speed during the first 10 seconds is   20 ms. 10  0 10 B 10  B 0 100  0 Skater B’s average speed during the first 10 seconds is   10 ms. 10  0 10 500  395 A 40  A 25   7 ms. (c) Skater A’s average speed during his last 15 seconds is 40  25 15 B 35  B 20 500  200 Skater B’s average speed during his last 15 seconds is   20 ms. 35  20 15

206

CHAPTER 2 Functions

39. t a

t b

3

35

3

31

3

301

3

3001

3

30001

Average Speed 

f b  f a ba

16 352  16 32  104 35  3

16 312  16 32  976 31  3

16 3012  16 32  9616 301  3

16 30012  16 32  96016 3001  3

16 300012  16 32  960016 30001  3

From the table it appears that the average speed approaches 96 fts as the time intervals get smaller and smaller. It seems reasonable to say that the speed of the object is 96 fts at the instant t  3.

2.5

LINEAR FUNCTIONS AND MODELS

1. If f is a function with constant rate of change, then (a) f is a linear function of the form f x  ax  b. (b) The graph of f is a line.

2. If f x  5x  7, then (a) The rate of change of f is 5.

(b) The graph of f is a line with slope 5 and y-intercept 7.

3. From the graph, we see that y 2  50 and y 0  20, so the slope of the graph is 50  20 y 2  y 0   15 galmin. m 20 2

4. From Exercise 3, we see that the pool is being filled at the rate of 15 gallons per minute. 5. If a linear function has positive rate of change, its graph slopes upward.

6. f x  3 is a linear function because it is of the form f x  ax  b, with a  0 and b  3. Its slope (and hence its rate of change) is 0. 7. f x  3  13 x  13 x  3 is linear with a  13 and b  3. 8. f x  2  4x  4x  2 is linear with a  4 and b  2. 9. f x  x 4  x  4x  x 2 is not of the form f x  ax  b for constants a and b, so it is not linear.  10. f x  x  1 is not linear. 11. f x 

x 1  15 x  15 is linear with a  15 and b  15 . 5

12. f x 

3 2x  3  2  is not linear. x x

13. f x  x  12  x 2  2x  1 is not of the form f x  ax  b for constants a and b, so it is not linear. 14. f x  12 3x  1  32 x  12 is linear with a  32 and b   12 .

SECTION 2.5 Linear Functions and Models y

15. x

f x  2x  5

1

7

0

5

1

3

2

1

3

1

4

3

2 0

x

1

The slope of the graph of f x  2x  5 is 2. y

16. x

g x  4  2x

1

6

0

4

1

2

2

0

3

2

4

4

2 0

1

x

The slope of the graph of g x  4  2x  2x  4 is 2. 17.

r t   23 t  2

t 1

y

267

0

2

2

1

133

0

2

067

3

0

4

067

2

t

The slope of the graph of r t   23 t  2 is  23 . 18. t 2 1

h t  12  34 t 2 125

0

05

1

025

2

1

3

175

The slope of the graph of h t  12  34 t is  34 .

y

1 0

1

t

207

208

CHAPTER 2 Functions

19. (a)

f

g

20. (a)

1

0

0

_1

1

1

z

x

1

_10

(b) The graph of f x  2x  6 has slope 2.

(b) The graph of g z  3z  9 has slope 3.

(c) f x  2x  6 has rate of change 2.

(c) g z  3z  9 has rate of change 3.

h

21. (a)

s

22. (a)

1

1 _1

_10 0

0

w

t

1

_1

10

(b) The graph of h t  05t  2 has slope 05.

(b) The graph of s   02  6 has slope 02.

(c) h t  05t  2 has rate of change 05.

(c) s   02  6 has rate of change 02.

23. (a)

v

A

24. (a)

10 _2 0 _10

1 2

t

10 (b) The graph of  t   10 3 t  20 has slope  3 . 10 (c)  t   10 3 t  20 has rate of change  3 .

_1

0 _1

1

r

(b) The graph of A r   23 r  1 has slope  23 . (c) A r   23 r  1 has rate of change  23 .

SECTION 2.5 Linear Functions and Models f

25. (a)

26. (a)

209

g 1 0

x

1

1 _1

0 _1

t

1

(b) The graph of f t   32 t  2 has slope  32 .

(b) The graph of g x  54 x  10 has slope 54 .

(c) f t   32 t  2 has rate of change  32 .

(c) g x  54 x  10 has rate of change 54 .

27. The linear function f with rate of change 3 and initial value 1 has equation f x  3x  1.

28. The linear function g with rate of change 12 and initial value 100 has equation g x  12x  100. 29. The linear function h with slope 12 and y-intercept 3 has equation h x  12 x  3.

30. The linear function k with slope  45 and y-intercept 2 has equation k x   45 x  2. 31. (a) From the table, we see that for every increase of 2 in the value of x, f x increases by 3. Thus, the rate of change of f is 32 . (b) When x  0, f x  7, so b  7. From part (a), a  32 , and so f x  32 x  7. 32. (a) From the table, we see that f 3  11 and f 0  2. Thus, when x increases by 3, f x decreases by 9, and so the rate of change of f is 3. (b) When x  0, f x  2, so b  2. From part (a), a  3, and so f x  3x  2.

33. (a) From the graph, we see that f 0  3 and f 1  4, so the rate of change of f is (b) From part (a), a  1, and f 0  b  3, so f x  x  3. 34. (a) From the graph, we see that f 0  4 and f 2  0, so the rate of change of f is (b) From part (a), a  2, and f 0  b  4, so f x  2x  4. 35. (a) From the graph, we see that f 0  2 and f 4  0, so the rate of change of f is (b) From part (a), a   12 , and f 0  b  2, so f x   12 x  2.

43  1. 10 04  2. 20 02 1  . 40 2

36. (a) From the graph, we see that f 0  1 and f 2  0, so the rate of change of f is (b) From part (a), a  12 , and f 0  b  1, so f x  12 x  1. 37.

f

Increasing the value of a makes the graph of f steeper. In other words, it

a=2

increases the rate of change of f .

a=1 a= 21

1 01

0  1 1  . 20 2

t

210

CHAPTER 2 Functions f

38.

Increasing the value of b moves the graph of f upward, but does not affect

b=2

the rate of change of f . b=1 b= 21

1

0

39. (a)

t

1

T 38,000

(b) The slope of T x  150x  32,000 is the value of a, 150. (c) The amount of trash is changing at a rate equal to the slope of the graph, 150 thousand tons per year.

36,000 34,000 32,000

0

40. (a)

10

20

30

x

f 900

(b) The slope of the graph of f x  200  32x is 32.

700

(c) Ore is being produced at a rate equal to the slope of the graph, 32 thousand tons per year.

800 600 500 400 300 200 100

0

5

10

15

20

25 x

41. (a) Let V t  at  b represent the volume of hydrogen. The balloon is being filled at the rate of 05 ft3 s, so a  05, and initially it contains 2 ft3 , so b  2. Thus, V t  05t  2.

(b) We solve V t  15  05t  2  15  05t  13  t  26. Thus, it takes 26 seconds to fill the balloon. 42. (a) Let V t  at  b represent the volume of water. The pool is being filled at the rate of 10 galmin, so a  10, and initially it contains 300 gal, so b  300. Thus, V t  10t  300. (b) We solve V t  1300  10t  300  1300  10t  1000  t  100. Thus, it takes 100 minutes to fill the pool.

1 . The ramp 43. (a) Let H x  ax  b represent the height of the ramp. The maximum rise is 1 inch per 12 inches, so a  12 1 x. starts on the ground, so b  0. Thus, H x  12 1 150  125. Thus, the ramp reaches a height of 125 inches. (b) We find H 150  12

1200  0075, or 75%. 15,000 500 Brianna descends 500 vertical feet over 10,000 feet, so the grade of her road is  005, or 5%. 10,000

44. Meilin descends 1200 vertical feet over 15,000 feet, so the grade of her road is

45. (a) From the graph, we see that the slope of Jari’s trip is steeper than that of Jade. Thus, Jari is traveling faster.

SECTION 2.5 Linear Functions and Models

211

  7 70 7  miles per minute or 60  70 mih. (b) The points 0 0 and 6 7 are on Jari’s graph, so her speed is 60 6 6 16  10 The points 0 10 and 6 16 are on Jade’s graph, so her speed is 60   60 mih. 60 1 hmin  1 mi/min and Jari’s speed is (c) t is measured in minutes, so Jade’s speed is 60 mih  60 1 hmin  7 mi/min. Thus, Jade’s distance is modeled by f t  1 t  0  10  t  10 and Jari’s 70 mih  60 6

distance is modeled by g t  76 t  0  0  76 t.

46. (a) Let d t represent the distance traveled. When t  0, d  0, and when

(b)

40  0 t  50, d  40. Thus, the slope of the graph is  08. The 50  0 y-intercept is 0, so d t  08t.

(c) Jacqueline’s speed is equal to the slope of the graph of d, that is, 08 mimin or 08 60  48 mih.

d 110 100 90 80 70 60 50 40 30 20 10 0

20

40

60

80 100 120 t

6 , so if we take 0 0 as the starting point, the 47. Let x be the horizontal distance and y the elevation. The slope is  100

6 x. We have descended 1000 ft, so we substitute y  1000 and solve for x: 1000   6 x  elevation is y   100 100 1 16,667  316 mi. x  16,667 ft. Converting to miles, the horizontal distance is 5280

48. (a)

D

(b) The slope of the graph of D x  20  024x is 024.

30

(c) The rate of sedimentation is equal to the slope of the graph, 024 cmyr or 24 mmyr.

20 10

0

10

20

30

40

50

x

480 miles at a cost of $380, and in June she drove 800 miles at a cost of

C 600

$460. Thus, the points 480 380 and 800 460 are on the graph, so the

500

49. (a) Let C x  ax  b be the cost of driving x miles. In May Lynn drove

slope is a 

1 460  380  . We use the point 480 380 to find the 800  480 4

(b)

400 300

value of b: 380  14 480  b  b  260. Thus, C x  14 x  260.

200

(c) The rate at which her cost increases is equal to the slope of the line, that is

100

1 . So her cost increases by $025 for every additional mile she drives. 4

0

200 400 600 800 1000 1200 1400 x

The slope of the graph of C x  14 x  260 is the value of a, 14 .

212

CHAPTER 2 Functions

50. (a) Let C x  ax  b be the cost of producing x chairs in one day. The first (b) day, it cost $2200 to produce 100 chairs, and the other day it cost $4800 to produce 300 chairs.. Thus, the points 100 2200 and 300 4800 are on 4800  2200  13. We use the point 300  100 100 2200 to find the value of b: 2200  13 100  b  b  900. Thus,

the graph, so the slope is a  C x  13x  900.

(c) The rate at which the factory’s cost increases is equal to the slope of the line, that is $13chair.

C 10,000 9000 8000 7000 6000 5000 4000 3000 2000 1000 0

100 200 300 400 500 600 x

The slope of the graph of C x  13x  900 is the value of a, 13. f x2   f x1  ax  ax1 ax2  b  ax1  b   2 . x2  x1 x2  x1 x 2  x1 a x2  x1  ax  ax1   a. (b) Factoring the numerator and cancelling, the average rate of change is 2 x2  x1 x2  x1

51. (a) By definition, the average rate of change between x1 and x2 is

f x  f a  c. x a (b) Multiplying the equation in part (a) by x  a, we obtain f x  f a  c x  a. Rearranging and adding f a to both sides, we have f x  cx   f a  ca, as desired. Because this equation is of the form f x  Ax  B with constants A  c and B  f a  ca, it represents a linear function with slope c and y-intercept f a  ca.

52. (a) The rate of change between any two points is c. In particular, between a and x, the rate of change is

2.6

TRANSFORMATIONS OF FUNCTIONS

1. (a) The graph of y  f x  3 is obtained from the graph of y  f x by shifting upward 3 units. (b) The graph of y  f x  3 is obtained from the graph of y  f x by shifting left 3 units.

2. (a) The graph of y  f x  3 is obtained from the graph of y  f x by shifting downward 3 units. (b) The graph of y  f x  3 is obtained from the graph of y  f x by shifting right 3 units.

3. (a) The graph of y   f x is obtained from the graph of y  f x by reflecting in the x-axis. (b) The graph of y  f x is obtained from the graph of y  f xby reflecting in the y-axis.

4. (a) The graph of f x  2 is obtained from that of y  f x by shifting upward 2 units, so it has graph II.

(b) The graph of f x  3 is obtained from that of y  f x by shifting to the left 3 units, so it has graph I.

(c) The graph of f x  2 is obtained from that of y  f x by shifting to the right 2 units, so it has graph III. (d) The graph of f x  4 is obtained from that of y  f x by shifting downward 4 units, so it has graph IV.

5. If f is an even function, then f x  f x and the graph of f is symmetric about the y-axis. 6. If f is an odd function, then f x   f x and the graph of f is symmetric about the origin. 7. (a) The graph of y  f x  1 can be obtained by shifting the graph of y  f x downward 1 unit.

(b) The graph of y  f x  2 can be obtained by shifting the graph of y  f x to the right 2 units.

8. (a) The graph of y  f x  4 can be obtained by shifting the graph of y  f x to the left 5 units. (b) The graph of y  f x  4 can be obtained by shifting the graph of y  f x upward 4 units.

9. (a) The graph of y  f x can be obtained by reflecting the graph of y  f x in the y-axis.

(b) The graph of y  3 f x can be obtained by stretching the graph of y  f x vertically by a factor of 3.

SECTION 2.6 Transformations of Functions

213

10. (a) The graph of y   f x can be obtained by reflecting the graph of y  f x about the x-axis.

(b) The graph of y  13 f x can be obtained by shrinking the graph of y  f x vertically by a factor of 13 .

11. (a) The graph of y  f x  5  2 can be obtained by shifting the graph of y  f x to the right 5 units and upward 2 units. (b) The graph of y  f x  1  1 can be obtained by shifting the graph of y  f x to the left 1 unit and downward 1 unit. 12. (a) The graph of y  f x  3  2 can be obtained by shifting the graph of y  f x to the left 3 units and upward 2 units. (b) The graph of y  f x  7  3 can be obtained by shifting the graph of y  f x to the right 7 units and downward 3 units. 13. (a) The graph of y   f x  5 can be obtained by reflecting the graph of y  f x in the x-axis, then shifting the resulting graph upward 5 units. (b) The graph of y  3 f x  5 can be obtained by stretching the graph of y  f x vertically by a factor of 3, then shifting the resulting graph downward 5 units. 14. (a) The graph of y  1  f x can be obtained by reflect the graph of y  f x about the x-axis, then reflecting about the y-axis, then shifting upward 1 unit. (b) The graph of y  2  15 f x can be obtained by shrinking the graph of y  f x vertically by a factor of 15 , then reflecting about the x-axis, then shifting upward 2 units. 15. (a) The graph of y  2 f x  5  1 can be obtained by shifting the graph of y  f x to the left 5 units, stretching vertically by a factor of 2, then shifting downward 1 unit. (b) The graph of y  14 f x  3  5 can be obtained by shifting the graph of y  f x to the right 3 units, shrinking vertically by a factor of 14 , then shifting upward 5 units. 16. (a) The graph of y  13 f x  2  5 can be obtained by shifting the graph of y  f x to the right 2 units, shrinking vertically by a factor of 13 , then shifting upward 5 units. (b) The graph of y  4 f x  1  3 can be obtained by shifting the graph of y  f x to the left 1 unit, stretching vertically by a factor of 4, then shifting upward 3 units. 17. (a) The graph of y  f 4x can be obtained by shrinking the graph of y  f x horizontally by a factor of 14 .   (b) The graph of y  f 14 x can be obtained by stretching the graph of y  f x horizontally by a factor of 4.

18. (a) The graph of y  f 2x  1 can be obtained by shrinking the graph of y  f x horizontally by a factor of 12 , then shifting it downward 1 unit.   (b) The graph of y  2 f 12 x can be obtained by stretching the graph of y  f x horizontally by a factor of 2 and stretching it vertically by a factor of 2.

19. (a) The graph of g x  x  22 is obtained by shifting the graph of f x to the left 2 units. (b) The graph of g x  x 2  2 is obtained by shifting the graph of f x upward 2 units.

20. (a) The graph of g x  x  43 is obtained by shifting the graph of f x to the right 4 units. (b) The graph of g x  x 3  4 is obtained by shifting the graph of f x downward 4 units.

21. (a) The graph of g x  x  2  2 is obtained by shifting the graph of f x to the left 2 units and downward 2 units.

(b) The graph of g x  g x  x  2  2 is obtained from by shifting the graph of f x to the right 2 units and upward 2 units.

214

CHAPTER 2 Functions

 22. (a) The graph of g x   x  1 is obtained by reflecting the graph of f x in the x-axis, then shifting the resulting graph upward 1 unit.  (b) The graph of g x  x  1 is obtained by reflecting the graph of f x in the y-axis, then shifting the resulting graph upward 1 unit. 23. (a)

y

y=x@+1

y

(b)

y=x@

y=x@ 1 1

(c)

y=(x-1)@

1 1

x

y

x

y

(d)

y=(x-1)@+3

y=x@

y=x@

1

1 1

1

x

x

y=_x@

24. (a)

y

(b)

y

y=Ïx+1

y=Ïx y=Ïx-2

1 1

(c)

y

1

x

(d)

y=Ïx+2+2

y=Ïx

1 1

y=Ïx

1

y

y=Ïx

1 x

x

1

x y=_Ïx+1

25. The graph of y  x  1 is obtained from that of y  x by shifting to the left 1 unit, so it has graph II. 26. y  x  1 is obtained from that of y  x by shifting to the right 1 unit, so it has graph IV. 27. The graph of y  x  1 is obtained from that of y  x by shifting downward 1 unit, so it has graph I.

SECTION 2.6 Transformations of Functions

215

28. The graph of y   x is obtained from that of y  x by reflecting in the x-axis, so it has graph III.

29. f x  x 2  3. Shift the graph of y  x 2 upward 3 units. 30. f x  x 2  4. Shift the graph of y  x 2 downward y

4 units.

y y=x@+3 y=x@ y=x@ 1 0

1 0

y=x@-4

x

1

31. f x  x  1. Shift the graph of y  x downward 1 unit.

32. f x 



x  1. Shift the graph of y  y

y=| x |

x upward 1 unit.

y=Ïx

1 y=| x |-1

1



y=Ïx+1

y

1

x

1

1

x

x

33. f x  x  52 . Shift the graph of y  x 2 to the right 5 units.

34. f x  x  12 . Shift the graph of y  x 2 to the left 1 unit.

y

y y=x@ y=(x-5)@ y=(x+1)@

5 1

x

y=x@ 1 1

x

216

CHAPTER 2 Functions

35. f x  x  2. Shift the graph of y  x to the left 2 units.

y

36. f x  4 units.



x  4. Shift the graph of y 

y=Ïx

y=| x |

y=Ïx-4

1

1

1

x

37. f x  x 3 . Reflect the graph of y  x 3 in the x-axis. y

1

x

38. f x   x. Reflect the graph of y  x in the x-axis. y

y=| x |

y=x#

2

1 1

x y=_x#

39. y 

x to the right

y

y=| x+2 |

1



x y=_| x |

  4 x. Reflect the graph of y  4 x in the y-axis. y

40. y 

  3 x. Reflect the graph of y  3 x in the y-axis. y

2 y=Îx 4 y=Ï_x

4 y=Ïx

2

1

x y=Î_x

10

x

SECTION 2.6 Transformations of Functions

41. y  14 x 2 . Shrink the graph of y  x 2 vertically by a factor of 14 .

  42. y  5 x. Stretch the graph of y  x vertically by a factor of 5, then reflect it in the x-axis. y

y y=x@

y=Ïx

2 1

x

y=41 x@ y=_5Ïx

1 1

x

43. y  3 x. Stretch the graph of y  x vertically by a factor of 3.

44. y  12 x. Shrink the graph of y  x vertically by a factor of 12 .

y

y y=3 | x | y=| x | 1 y=| x |

1 1

y=21 | x | 1

3 units and upward 5 units.

46. y 



x  4  3. Shift the graph of y 

y=x@

x to the left

y=Ïx

1 1 y=Ïx+4-3

y=(x-3)@+5 1



4 units and downward 3 units. y

y

5

x

x

45. y  x  32  5. Shift the graph of y  x 2 to the right

x

217

x

218

CHAPTER 2 Functions

47. y  3  12 x  12 . Shift the graph of y  x 2 to the right one unit, shrink vertically by a factor of 12 , reflect in the x-axis, then shift upward 3 units.

48. y  2 



x  1. Shift the graph of y 

x to the left

1 unit, reflect the result in the x-axis, then shift upward

2 units. y

y

y=Ïx

y=x@ 1

2 1

y=3-21 (x-1)@



x

1

x y=2-Ïx+1

49. y  x  2  2. Shift the graph of y  x to the left 2 units and upward 2 units.

y

50. y  2  x. Reflect the graph of y  x in the x-axis, then shift upward 2 units.

y

y=| x+2 |+2

y=| x | 1 y=| x |

1

1

x

y=2-| x | 1

x

  51. y  12 x  4  3. Shrink the graph of y  x vertically by a factor of 12 , then shift the result to the left 4 units and downward 3 units.

52. y  3  2 x  12 . Stretch the graph of y  x 2 vertically by a factor of 2, reflect the result in the x-axis, then shift

the result to the right 1 unit and upward 3 units. y

y

y=x@ y=Ïx 1 1

1 4

x

x

y=3-2(x-1)@

1

y=2Ïx+4-3

53. y  f x  3. When f x  x 2 , y  x 2  3. 55. y  f x  2. When f x 



x, y 



x  2.

57. y  f x  2  5. When f x  x, y  x  2  5.

54. y  f x  5. When f x  x 3 , y  x 3  5. 56. y  f x  1. When f x 

 3

x, y 

58. y   f x  4  3. When f x  x, y   x  4  3.

 3

x  1.

SECTION 2.6 Transformations of Functions

59. y  f x  1. When f x 

 4

x, y 

 4 x  1.

219

60. y   f x  2. When f x  x 2 , y   x  22 . 62. y  12 f x  1  3. When f x  x,

61. y  2 f x  3  2. When f x  x 2 , y  2 x  32  2.

y  12 x  1  3.

63. g x  f x  2  x  22  x 2  4x  4

64. g x  f x  3  x 3  3

65. g x  f x  1  2  x  1  2  67. g x   f x  2   x  2

66. g x  2 f x  2 x 68. g x   f x  2  1   x  22  1  x 2  4x  3

69. (a) y  f x  4 is graph #3.

70. (a) y  13 f x is graph #2.

(b) y  f x  3 is graph #1.

(b) y   f x  4 is graph #3.

(c) y  2 f x  6 is graph #2.

(c) y  f x  5  3 is graph #1.

(d) y   f 2x is graph #4.

(d) y  f x is graph #4.

71. (a) y  f x  2

(b) y  f x  2

y

1

y

y

1 1

1 1

x

(d) y   f x  3

(e) y  f x

y

x

x

y

1

x

(c) y  g x  2

y

1 1

x

1

x

1

x

1

1

(b) y  g x

x

y

1 1

1

(f) y  12 f x  1

y

1

72. (a) y  g x  1

(c) y  2 f x

y

1 1

x

220

CHAPTER 2 Functions

(d) y  g x  2

(e) y  g x

y

1

(f) y  2g x

y

1 1

1 1

x

y

1

1

x

74. (a) y  h 3x

1

  (b) y  h 13 x

y y=h(3x)

  76. y  14 x

x

y

1 1

x

2

x

For part (b), shift the graph in (a) to the left 5 units; for part (c), shift the graph

8 (d)

4

2

4

6

in (a) to the left 5 units, and stretch it vertically by a factor of 2; for part (d), shift

(c)

the graph in (a) to the left 5 units, stretch it vertically by a factor of 2, and then shift

(b)

it upward 4 units.

(a) -4

1 y=h( _ 3 x)

3

1

-2 0

y

x

y

75. y  [[2x]]

x

y=h(x)

y=h(x)

1

-6

x

y

1

-8

1

x

  (b) y  g 12 x

73. (a) y  g 2x

77.

y

8

SECTION 2.6 Transformations of Functions

78. (a)

-8

-6

-4

6

For (b), reflect the graph in (a) in the x-axis; for (c), stretch the graph in (a)

4

vertically by a factor of 3 and reflect in the x-axis; for (d), shift the graph in (a) to

2

the right 5 units, stretch it vertically by a factor of 3, and reflect it in the x-axis. The

-2

2

4

6

8

(c)

-4 -6

79.

order in which each operation is applied to the graph in (a) is not important to obtain the graphs in part (c) and (d).

-2 (b)

221

(c)

(d)

For part (b), shrink the graph in (a) vertically by a factor of 13 ; for part (c), shrink

4 (a) (b)

the graph in (a) vertically by a factor of 13 , and reflect it in the x-axis; for part (d),

2

shift the graph in (a) to the right 4 units, shrink vertically by a factor of 13 , and then -4

-2

2 -2

4

(c)

6 (d)

reflect it in the x-axis.

-4

80. (b)

4

For (b), shift the graph in (a) to the left 3 units; for (c), shift the graph in (a) to the

2 (a)

left 3 units and shrink it vertically by a factor of 12 ; for (d), shift the graph in (a) to the left 3 units, shrink it vertically by a factor of 12 , and then shift it downward

(c) -6

-4

2

4

-2

6

3 units. The order in which each operation is applied to the graph in (a) is not important to sketch (c), while it is important in (d).

(d) -4

 (b) y  f 2x  2 2x  2x2   4x  4x 2

 81. (a) y  f x  2x  x 2

-4

-2



1x 2



     2  2 12 x  12 x

(c) y  f   x  14 x 2

4

4

4

2

2

2

-2

2

4

-4

-2

-4

-2

2

4

-4

-2

-4

-2

2

4

-4

The graph in part (b) is obtained by horizontally shrinking the graph in part (a) by a factor of 12 (so the graph is half as wide). The graph in part (c) is obtained by horizontally stretching the graph in part (a) by a factor of 2 (so the graph is twice as wide).    82. (a) y  f x  2x  x 2 (b) y  f x  2 x  x2 (c) y   f x   2 x  x2    2x  x 2   2x  x 2

-4

-2

4

4

4

2

2

2

-2 -4

2

4

-4

-2

-2 -4

2

4

-4

-2

-2 -4

2

4

222

CHAPTER 2 Functions

      2 1 (e) y  f  2 x  2  12 x   12 x   x  14 x 2

 (d) y  f 2x  2 2x  2x2     2x  x 2  4x  4x 2

-4

-2

4

4

2

2 2

-2

4

-4

-4

-2

2

-2

4

-4

The graph in part (b) is obtained by reflecting the graph in part (a) in the y-axis. The graph in part (c) is obtained by rotating the graph in part (a) through 180 about the origin [or by reflecting the graph in part (a) first in the x-axis and then in the y-axis]. The graph in part (d) is obtained by reflecting the graph in part (a) in the y-axis and then horizontally shrinking the graph by a factor of 12 (so the graph is half as wide). The graph in part (e) is obtained by reflecting the graph in part (a) in the y-axis and then horizontally stretching the graph by a factor of 2 (so the graph is twice as wide).

83. f x  x 4 . f x  x4  x 4  f x. Thus f x is even.

84. f x  x 3 . f x  x3  x 3   f x. Thus f x is odd.

y

y

2 1

x

2 1

x

85. f x  x 2  x. f x  x2  x  x 2  x. Thus 86. f x  x 4  4x 2 . f x  f x. Also, f x   f x, so f x is

neither odd nor even.

f x  x4  4 x2  x 4  4x 2  f x. Thus f x is even.

y

y

1 1

x

1 1

x

SECTION 2.6 Transformations of Functions

87. f x  x 3  x.

88. f x  3x 3  2x 2  1.

f x  3 x3  2 x2  1  3x 3  2x 2  1.

f x  x3  x  x 3  x     x 3  x   f x.

Thus f x  f x. Also f x   f x, so f x is neither odd nor even.

Thus f x is odd.

y

y

1

1

 3

x. f x  1 

  3 x  1  3 x. Thus

f x  f x. Also f x   f x, so f x is

neither odd nor even.

x

1

x

1

89. f x  1 

223

90. f x  x  1x. f x  x  1 x  x  1x   x  1x   f x.

y

Thus f x is odd. y

1 x

1

1 x

1

91. (a) Even

(b) Odd y

y

1

1 1

x

1

x

224

CHAPTER 2 Functions

92. (a) Even

(b) Odd y

y

1

1 1

1

x

93. Since f x  x 2  4  0, for 2  x  2, the graph of

y  g x is found by sketching the graph of y  f x for

x  2 and x  2, then reflecting in the x-axis the part of

    94. g x  x 4  4x 2 

x

y g

4

the graph of y  f x for 2  x  2.

2 _3

95. (a) f x  4x  x 2

    (b) f x  4x  x 2 

y

1 1

96. (a) f x  x 3

_1

    (b) g x  x 3 

1 1

x

y

1

x

y

3 x

1

1

x

1

x

y

1

SECTION 2.6 Transformations of Functions

97. (a) Luisa drops to a height of 200 feet, bounces up and down, then settles at

(b)

350 feet.

225

y (ft) 500

(c) To obtain the graph of H from that of h, we shift downward 100 feet. Thus, H t  h t  100.

0

98. (a) Miyuki swims two and a half laps, slowing down with each successive lap.

(b)

In the first 30 seconds she swims 50 meters, so her average speed is

4

t (s)

80 60

50  167 ms. 30

40 20

(c) Here Miyuki swims 60 meters in 30 seconds, so her average speed is 60  2 ms. 30

0

100

200

t

This graph is obtained by stretching the original graph vertically by a factor of 12. 99. (a) The trip to the park corresponds to the first piece of the graph. The class travels 800 feet in 10 minutes, so their average speed is 800 10  80 ftmin. The second (horizontal) piece of the graph stretches from t  10 to t  30, so the class

spends 20 minutes at the park. The park is 800 feet from the school. y 600

(b)

(c)

800

400

600 400

200 0

y 1000

200 20

40

60 t

0

20

40

60 t

The new graph is obtained by shrinking the original

This graph is obtained by shifting the original graph

graph vertically by a factor of 050. The new average

to the right 10 minutes. The class leaves ten minutes

speed is 40 ftmin, and the new park is 400 ft from

later than it did in the original scenario.

the school. 100. To obtain the graph of g x  x  22  5 from that of f x  x  22 , we shift to the right 4 units and upward 5 units.

101. To obtain the graph of g x from that of f x, we reflect the graph about the y-axis, then reflect about the x-axis, then shift upward 6 units. 102. f even implies f x  f x  g even implies g x  g x; f odd implies f x   f x  and g odd implies g x  g x If f and g are both even, then  f  g x  f x  g x  f x  g x   f  g x and f  g is even. If f and g are both odd, then  f  g x  f x  g x   f x  g x    f  g x and f  g is odd. If f odd and g even, then  f  g x  f x  g x   f x  g x, which is neither odd nor even.

103. f even implies f x  f x; g even implies g x  g x; f odd implies f x   f x; and g odd implies g x  g x. If f and g are both even, then  f g x  f x  g x  f x  g x   f g x. Thus f g is even. If f and g are both odd, then  f g x  f x  g x   f x  g x  f x  g x   f g x. Thus f g is even If f if odd and g is even, then  f g x  f x  g x  f x  g x   f x  g x    f g x. Thus f g is odd.

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104. f x  x n is even when n is an even integer and f x  x n is odd when n is an odd integer. These names were chosen because polynomials with only terms with odd powers are odd functions, and polynomials with only terms with even powers are even functions.

2.7

COMBINING FUNCTIONS

1. From the graphs of f and g in the figure, we find  f  g 2  f 2  g 2  3  5  8,   3 f 2 f  .  f  g 2  f 2  g 2  3  5  2,  f g 2  f 2 g 2  3  5  15, and 2  g g 2 5

2. By definition, f  g x  f g x. So, if g 2  5 and f 5  12, then f  g 2  f g 2  f 5  12. 3. If the rule of the function f is “add one” and the rule of the function g is “multiply by 2” then the rule of f  g is “multiply by 2, then add one” and the rule of g  f is “add one, then multiply by 2.” 4. We can express the functions in Exercise 3 algebraically as f x  x  1, g x  2x,  f  g x  2x  1, and g  f  x  2 x  1.

5. (a) The function  f  g x is defined for all values of x that are in the domains of both f and g. (b) The function  f g x is defined for all values of x that are in the domains of both f and g.

(c) The function  f g x is defined for all values of x that are in the domains of both f and g, and g x is not equal to 0. 6. The composition  f  g x is defined for all values of x for which x is in the domain of g and g x is in the domain of f .

7. f x  x has domain  . g x  2x has domain  . The intersection of the domains of f and g is  .  f  g x  x  2x  3x, and the domain is  .  f  g x  x  2x  x, and the domain is  .   1 x f  , and the domain is  0  0 .  f g x  x 2x  2x 2 , and the domain is  . x  g 2x 2  8. f x  x has domain  . g x  x has domain [0 . The intersection of the domains of f and g is [0 .    f  g x  x  x, and the domain is [0 .  f  g x  x  x, and the domain is [0 .     x f  f g x  x x  x 32 , and the domain is [0 . x    x, and the domain is 0 . g x 9. f x  x 2  x and g x  x 2 each have domain  . The intersection of the domains of f and g is  .

 f  g x  2x 2  x, and the domain is  .  f  g x  x, and the domain is  .   x2  x f 1  1  , and the domain is  0  0 . x   f g x  x 4  x 3 , and the domain is  . 2 g x x

10. f x  3  x 2 and g x  x 2  4 each have domain  . The intersection of the domains of f and g is  .

 f  g x  1, and the domain is  .  f  g x  2x 2  7, and the domain is  .      3  x2 3  x2 f  ,  f g x  3  x 2 x 2  4  x 4  7x 2  12, and the domain is  . x  2 g x  2 x  2 x 4 and the domain is  2  2 2  2 .

11. f x  5  x and g x  x 2  3x each have domain  . The intersection of the domains of f and g is  .    f  g x  5  x  x 2  3x  x 2  4x  5, and the domain is  .    f  g x  5  x  x 2  3x  x 2  2x  5, and the domain is  .    f g x  5  x x 2  3x  x 3  8x 2  15x, and the domain is  .   f 5x 5x , and the domain is  0  0 3  3 .  x  2 g x x  3 x  3x

SECTION 2.7 Combining Functions

227

12. f x  x 2  2x has domain  . g x  3x 2  1 has domain  . The intersection of the domains of f and g is  .    f  g x  x 2  2x  3x 2  1  4x 2  2x  1, and the domain is  .    f  g x  x 2  2x  3x 2  1  2x 2  2x  1, and the domain is  .     f g x  x 2  2x 3x 2  1  3x 4  6x 3  x 2  2x, and the domain is  .       x 2  2x f , 3x 2  1  0  x   33 , and the domain is x  x   33 . x  2 g 3x  1

  13. f x  25  x 2 , has domain [5 5]. g x  x  3, has domain [3 . The intersection of the domains of f and g is [3 5].    f  g x  25  x 2  x  3 , and the domain is [3 5].    f  g x  25  x 2  x  3, and the domain is [3 5].    f g x  25  x 2 x  3, and the domain is [3 5].    25  x 2 f , and the domain is 3 5]. x  g x 3

  14. f x  16  x 2 has domain [4 4]. g x  x 2  1 has domain  1]  [1 . The intersection of the domains of f and g is [4 1]  [1 4].    f  g x  16  x 2  x 2  1, and the domain is [4 1]  [1 4].    f  g x  16  x 2  x 2  1, and the domain is [4 1]  [1 4].     f g x  16  x 2 x 2  1 , and the domain is [4 1]  [1 4].    f 16  x 2 , and the domain is [4 1  1 4]. x  g x2  1

2 4 has domain x  0. g x  , has domain x  4. The intersection of the domains of f and g is x x 4 x  x  0 4; in interval notation, this is  4  4 0  0 . 2 4 2 4 2 3x  4    , and the domain is  4  4 0  0 .  f  g x   x x 4 x x 4 x x  4 4 2 x  4 2  , and the domain is  4  4 0  0 .  f  g x   x x 4 x x  4 4 8 2  , and the domain is  4  4 0  0 .  f g x   x x 4 x x  4 2   f x 4 x , and the domain is  4  4 0  0 .  x  4 g 2x x 4

15. f x 

228

CHAPTER 2 Functions

x 2 has domain x  1. g x  has domain x  1 The intersection of the domains of f and g is x 1 x 1 x  x  1; in interval notation, this is  1  1 . 2 x x 2   , and the domain is  1  1 .  f  g x  x 1 x 1 x 1 2 x 2x   , and the domain is  1  1 .  f  g x  x 1 x 1 x 1 2 x 2x   , and the domain is  1  1 .  f g x  x 1 x 1 x  12 2   2 f 1 x  x  x  x , so x  0 as well. Thus the domain is  1  1 0  0 . g x 1     17. f x  x  3  x. The domain of x is [0 , and the domain of 3  x is  3]. Thus, the domain of f is  3]  [0   [0 3].     1x 1x 18. f x  x  4  . The domain of x  4 is [4 , and the domain of is  0  0 1]. Thus, the x x domain of f is [4    0  0 1]  [4 0  0 1].

16. f x 

1 . Since 14 is an even root and the denominator can not equal 0, x  3  0  x  3 x  314 So the domain is 3 .   1 x 3 20. k x  . The domain of x  3 is [3 , and the domain of is x  1. Since x  1 is  1  1 , x 1 x 1 the domain is [3    1  1   [3 1  1 . 19. h x  x  314 

y

21.

y

22.

f

f+g

0

2

f

x

-4

24. 5

25.

f+g

40

f+g

g

-2

26.

f+g

g

0

2

4

3

f

20

f+g 2

3

f

-4

2

-2

2

-40

1

4

f

1

-20

g

1 0

g

x

f

4

4

f+g

g

0

23.

2

2 27. f x  2x  3 and  g x  4  x .

(a) f g 0  f 4  02  f 4  2 4  3  5 (b) g  f 0  g 2 0  3  g 3  4  32  5

28. (a) f  f 2  f 2 2  3  f 1  2 1  3  1   (b) g g 3  g 4  32  g 5  4  52  21

g -4

-2

00

2

4

SECTION 2.7 Combining Functions

  29. (a)  f  g 2  f g 2  f 4  22  f 0  2 0  3  3

(b) g  f  2  g  f 2  g 2 2  3  g 7  4  72  45

30. (a)  f  f  1  f  f 1  f 2 1  3  f 5  2 5  3  13   (b) g  g 1  g g 1  g 4  12  g 3  4  32  5

    31. (a)  f  g x  f g x  f 4  x 2  2 4  x 2  3  8  2x 2  3  5  2x 2   (b) g  f  x  g  f x  g 2x  3  4  2x  32  4  4x 2  12x  9  4x 2  12x  5

32. (a)  f  f  x  f  f x  f 3x  5  3 3x  5  5  9x  15  5  9x  20   2    (b) g  g x  g g x  g 2  x 2  2  2  x 2  2  4  4x 2  x 4  x 4  4x 2  2 33. f g 2  f 5  4

34. f 0  0, so g  f 0  g 0  3.

35. g  f  4  g  f 4  g 2  5

36. g 0  3, so  f  g 0  f 3  0.

37. g  g 2  g g 2  g 1  4

38. f 4  2, so  f  f  4  f 2  2.

39. From the table, g 2  5 and f 5  6, so f g 2  6. 40. From the table, f 2  3 and g 3  6, so g  f 2  6. 41. From the table, f 1  2 and f 2  3, so f  f 1  3. 42. From the table, g 2  5 and g 5  1, so g g 2  1. 43. From the table, g 6  4 and f 4  1, so  f  g 6  1. 44. From the table, f 2  3 and g 3  6, so g  f  2  6. 45. From the table, f 5  6 and f 6  3, so  f  f  5  3. 46. From the table, g 2  5 and g 5  1, so g  g 5  1. 47. f x  2x  3, has domain  ; g x  4x  1, has domain  .  f  g x  f 4x  1  2 4x  1  3  8x  1, and the domain is  . g  f  x  g 2x  3  4 2x  3  1  8x  11, and the domain is  .  f  f  x  f 2x  3  2 2x  3  3  4x  9, and the domain is  . g  g x  g 4x  1  4 4x  1  1  16x  5, and the domain is  . x 48. f x  6x  5 has domain  . g x  has domain  . 2 x  x  6  5  3x  5, and the domain is  .  f  g x  f 2 2 6x  5  3x  52 , and the domain is  . g  f  x  g 6x  5  2  f  f  x  f 6x  5  6 6x  5  5  36x  35, and the domain is  . x x  x  2  , and the domain is  . g  g x  g 2 2 4 49. f x  x 2 , has domain  ; g x  x  1, has domain  .

 f  g x  f x  1  x  12  x 2  2x  1, and the domain is  .     g  f  x  g x 2  x 2  1  x 2  1, and the domain is  .    2  f  f  x  f x 2  x 2  x 4 , and the domain is  .

g  g x  g x  1  x  1  1  x  2, and the domain is  .

229

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CHAPTER 2 Functions

 50. f x  x 3  2 has domain  . g x  3 x has domain  .      3  f  g x  f 3 x  3 x  2  x  2, and the domain is  .    3 g  f  x  g x 3  2  x 3  2 and the domain is  .   3   f  f  x  f x 3  2  x 3  2  2  x 9  6x 6  12x 3  8  2  x 9  6x 6  12x 3  10, and the domain is  .

g  g x  g

13     3 x  3 3 x  x 13  x 19 , and the domain is  .

1 , has domain x  x  0; g x  2x  4, has domain  . x 1 .  f  g x is defined for 2x  4  0  x  2. So the domain is  f  g x  f 2x  4  2x  4 x  x  2   2  2 .     1 2 1 2  4   4, the domain is x  x  0   0  0 . g  f  x  g x x x   1 1     x.  f  f  x is defined whenever both f x and f  f x are defined; that is,  f  f  x  f 1 x x whenever x  x  0   0  0 . g  g x  g 2x  4  2 2x  4  4  4x  8  4  4x  12, and the domain is  .

51. f x 

 52. f x  x 2 has domain  . g x  x  3 has domain [3 .    2 x 3  x  3  x  3, and the domain is [3 .  f  g x  f      g  f  x  g x 2  x 2  3. For the domain we must have x 2  3  x   3 or x  3. Thus the domain is        3  3  .    2  f  f  x  f x 2  x 2  x 4 , and the domain is  .     x  3  3. For the domain we must have x  3  3  x  3  9  x  12, so the g  g x  g x  3  domain is [12 . 53. f x  x, has domain  ; g x  2x  3, has domain    f  g x  f 2x  4  2x  3, and the domain is  . g  f  x  g x  2 x  3, and the domain is  .  f  f  x  f x  x  x, and the domain is  . g  g x  g 2x  3  2 2x  3  3  4x  6  3  4x  9. Domain is  . 54. f x  x  4 has domain  . g x  x  4 has domain  .  f  g x  f x  4  x  4  4, and the domain is  . g  f  x  g x  4  x  4  4  x, and the domain is  .  f  f  x  f x  4  x  4  4  x  8, and the domain is  . g  g x  g x  4  x  4  4  x  4  4 (x  4  4 is always positive). The domain is  .

SECTION 2.7 Combining Functions

231

x , has domain x  x  1; g x  2x  1, has domain   x 1 2x  1 2x  1 , and the domain is x  x  0   0  0 .   f  g x  f 2x  1  2x 2x  1  1     2x x x 2 1   1, and the domain is x  x  1   1  1  g  f  x  g x 1 x 1 x 1 x   x x x 1 x x   x 1    .  f  f  x is defined whenever both f x and  f  f  x  f x 1 x  1 x  x  1 2x 1 1 x 1     f  f x are defined; that is, whenever x  1 and 2x  1  0  x   12 , which is  1  1  12   12   .

55. f x 

g  g x  g 2x  1  2 2x  1  1  4x  2  1  4x  3, and the domain is  . 1 56. f x   has domain x  x  0  g x  x 2  4x has domain  . x   1 .  f  g x is defined whenever 0  x 2  4x  x x  4. The product of two  f  g x  f x 2  4x   2 x  4x numbers is positive either when both numbers are negative or when both numbers are positive. So the domain of f  g is x  x  0 and x  4  x  x  0 and x  4 which is  0  4 .       1 2 1 1 4 1      . g  f  x is defined whenever both f x and g  f x are 4  g  f  x  g  x x x x x defined, that is, whenever x  0. So the domain of g  f is 0 .   1 1    x 14 .  f  f  x is defined whenever both f x and f  f x are defined, that is,  f  f  x  f  x 1  x whenever x  0. So the domain of f  f is 0 .    2   g  g x  g x 2  4x  x 2  4x  4 x 2  4x  x 4  8x 3  16x 2  4x 2  16x  x 4  8x 3  12x 2  16x,

and the domain is  .

1 x , has domain x  x  1; g x  has domain x  x  0. x 1 x   1 1 1 1    1 x .  f  g x is defined whenever both g x and f g x are    f  g x  f 1 1 x x 1  1 x x

57. f x 

x

defined, so the domain is x  x  1 0.   1 x 1 x  x  . g  f  x is defined whenever both f x and g  f x are defined, so the g  f  x  g x 1 x x1

domain is x  x  1 0.   x x x x     xx1  .  f  f  x is defined whenever both f x and  f  f  x  f x x 1  1 2x 1 x  1 x1  1 x1   f  f x are defined, so the domain is x  x  1  12 .   1 1  1  x. g  g x is defined whenever both g x and g g x are defined, so the domain is g  g x  g x x

x  x  0.

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CHAPTER 2 Functions

x 2 has domain x  x  0; g x  has domain x  x  2. x x 2   x 2x  4 2  x  .  f  g x is defined whenever both g x and f g x are defined; that  f  g x  f x 2 x x 2 is, whenever x  0 and x  2. So the domain is x  x  0 2. 2   1 2 2 x   . g  f  x is defined whenever both f x and g  f x are defined;  g  f  x  g 2 x 2  2x 1x 2 x that is, whenever x  0 and x  1. So the domain is x  x  0 1.   2 2   x.  f  f  x is defined whenever both f x and f  f x are defined; that is, whenever  f  f  x  f 2 x x x  0. So the domain is x  x  0. x   x x x x   x 2   . g  g x is defined whenever both g x and g  g x  g x 2 x  2  2 3x 4 x 2 x 2   g g x are defined; that is whenever x  2 and x   43 . So the domain is x  x  2  43 .    x 1  x 11  f  g  h x  f g h x  f g x  1  f    3 g  h x  g x 2  2  x 2  2  x 6  6x 4  12x 2  8.   1 .  f  g  h x  f x 6  6x 4  12x 2  8  6 4 x  6x  12x 2  8       4 x 5  x 5 1  f  g  h x  f g h x  f g x  f       3 3 3   x x x 3 .  f  g  h x  f    . x   g  h x  g 3 3 3 x 1 x 1 x 1

58. f x 

59. 60.

61. 62.

For Exercises 63–72, many answers are possible. 63. F x  x  95 . Let f x  x 5 and g x  x  9, then F x   f  g x.   64. F x  x  1. If f x  x  1 and g x  x, then F x   f  g x.

x2 x 65. G x  2 and g x  x 2 , then G x   f  g x. . Let f x  x 4 x 4 1 1 . If f x  and g x  x  3, then G x   f  g x. 66. G x  x 3 x     67. H x  1  x 3 . Let f x  x and g x  1  x 3 , then H x   f  g x.     68. H x  1  x. If f x  1  x and g x  x, then H x   f  g x.

1 1 69. F x  2 . Let f x  , g x  x  1, and h x  x 2 , then F x   f  g  h x. x x 1     70. F x  3 x  1. If g x  x  1 and h x  x, then g  h x  x  1, and if f x  3 x, then F x   f  g  h x.   9  71. G x  4  3 x . Let f x  x 9 , g x  4  x, and h x  3 x, then G x   f  g  h x.

  2 2 72. G x    2 . If g x  3  x and h x  x, then g  h x  3  x, and if f x  x 2 , then 3 x G x   f  g  h x.

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233

73. Yes. If f x  m 1 x  b1 and g x  m 2 x  b2 , then  f  g x  f m 2 x  b2   m 1 m 2 x  b2   b1  m 1 m 2 x  m 1 b2  b1, which is a linear function, because it is of the form y  mx  b. The slope is m 1 m 2 . 74. g x  2x  1 and h x  4x 2  4x  7.

Method 1: Notice that 2x  12  4x 2  4x  1. We see that adding 6 to this quantity gives

2x  12  6  4x 2  4x  1  6  4x 2  4x  7, which is h x. So let f x  x 2  6, and we have

 f  g x  2x  12  6  h x. Method 2: Since g x is linear and h x is a second degree polynomial, f x must be a second degree polynomial, that is, f x  ax 2  bx  c for some a, b, and c. Thus f g x  f 2x  1  a 2x  12  b 2x  1  c 

4ax 2  4ax  a  2bx  b  c  4ax 2  4a  2b x  a  b  c  4x 2  4x  7. Comparing this with f g x, we

have 4a  4 (the x 2 coefficients), 4a  2b  4 (the x coefficients), and a  b  c  7 (the constant terms)  a  1 and

2a  b  2 and a  b  c  7  a  1, b  0 c  6. Thus f x  x 2  6.

f x  3x  5 and h x  3x 2  3x  2. Note since f x is linear and h x is quadratic, g x must also be quadratic. We can then use trial and error to find g x.

Another method is the following: We wish to find g so that  f  g x  h x. Thus f g x  3x 2  3x  2  3 g x  5  3x 2  3x  2  3 g x  3x 2  3x  3  g x  x 2  x  1.

75. The price per sticker is 015  0000002x and the number sold is x, so the revenue is R x  015  0000002x x  015x  0000002x 2 .

76. As found in Exercise 75, the revenue is R x  015x  0000002x 2 , and the cost is 0095x  00000005x 2 , so the profit   is P x  015x  0000002x 2  0095x  00000005x 2  0055x  00000015x 2 .

77. (a) Because the ripple travels at a speed of 60 cm/s, the distance traveled in t seconds is the radius, so g t  60t. (b) The area of a circle is r 2 , so f r  r 2 .

(c) f  g   g t2   60t2  3600t 2 cm2 . This function represents the area of the ripple as a function of time.

78. (a) Let f t be the radius of the spherical balloon in centimeters. Since the radius is increasing at a rate of 1 cm/s, the radius is f t  t after t seconds. (b) The volume of the balloon can be written as g r  43 r 3 .

(c) g  f  43  t3  43 t 3 . g  f represents the volume as a function of time. 79. Let r be the radius of the spherical balloon in centimeters. Since the radius is increasing at a rate of 2 cm/s, the radius is r  2t   after t seconds. Therefore, the surface area of the balloon can be written as S  4r 2  4 2t2  4 4t 2  16t 2 .

80. (a) f x  080x

(b) g x  x  50

(c)  f  g x  f x  50  080 x  50  080x  40. f  g represents applying the $50 coupon, then the 20% discount. g  f  x  g 080x  080x  50. g  f represents applying the 20% discount, then the $50 coupon. So applying the 20% discount, then the $50 coupon gives the lower price.

81. (a) f x  090x

(b) g x  x  100

(c)  f  g x  f x  100  090 x  100  090x  90. f  g represents applying the $100 coupon, then the 10% discount. g  f  x  g 090x  090x  100. g  f represents applying the 10% discount, then the $100 coupon. So applying the 10% discount, then the $100 coupon gives the lower price. 82. Let t be the time since the plane flew over the radar station. (a) Let s be the distance in miles between the plane and the radar station, and let d be the horizontal distance that the plane  has flown. Using the Pythagorean theorem, s  f d  1  d 2 .

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CHAPTER 2 Functions

(b) Since distance  rate  time, we have d  g t  350t.   (c) s t   f  g t  f 350t  1  350t2  1  122,500t 2 .

83. A x  105x A  A x  A A x  A 105x  105 105x  1052 x.     A  A  A x  A A  A x  A 1052 x  105 1052 x  1053 x.     A  A  A  A x  A A  A  A x  A 1053 x  105 1053 x  1054 x. A represents the amount in

the account after 1 year; A  A represents the amount in the account after 2 years; A  A  A represents the amount in the account after 3 years; and A  A  A  A represents the amount in the account after 4 years. We can see that if we compose n copies of A, we get 105n x.

84. If g x is even, then h x  f g x  f g x  h x. So yes, h is always an even function.

If g x is odd, then h is not necessarily an odd function. For example, if we let f x  x  1 and g x  x 3 , g is an odd   function, but h x   f  g x  f x 3  x 3  1 is not an odd function.

If g x is odd and f is also odd, then h x   f  g x  f g x  f g x   f g x    f  g x  h x. So in this case, h is also an odd function. If g x is odd and f is even, then h x   f  g x  f g x  f g x  f g x   f  g x  h x, so in this case, h is an even function.

2.8

ONE-TO-ONE FUNCTIONS AND THEIR INVERSES

1. A function f is one-to-one if different inputs produce different outputs. You can tell from the graph that a function is one-to-one by using the Horizontal Line Test. 2. (a) For a function to have an inverse, it must be one-to-one. f x  x 2 is not one-to-one, so it does not have an inverse. However g x  x 3 is one-to-one, so it has an inverse.  (b) The inverse of g x  x 3 is g1 x  3 x.

3. (a) Proceeding backward through the description of f , we can describe f 1 as follows: “Take the third root, subtract 5, then divide by 3.”  3 x 5 (b) f x  3x  53 and f 1 x  . 3 4. Yes, the graph of f is one-to-one, so f has an inverse. Because f 4  1, f 1 1  4, and because f 5  3, f 1 3  5.

5. If the point 3 4 is on the graph of f , then the point 4 3 is on the graph of f 1 . [This is another way of saying that f 3  4  f 1 4  3.] 6. (a) False. For instance, if f x  x, then f 1 x  x, but

1 1   f 1 x. f x x

(b) This is true, by definition. 7. By the Horizontal Line Test, f is not one-to-one. 9. By the Horizontal Line Test, f is one-to-one. 11. By the Horizontal Line Test, f is not one-to-one.

8. By the Horizontal Line Test, f is one-to-one. 10. By the Horizontal Line Test, f is not one-to-one. 12. By the Horizontal Line Test, f is one-to-one.

13. f x  2x  4. If x1  x2 , then 2x1  2x2 and 2x1  4  2x2  4. So f is a one-to-one function. 14. f x  3x  2. If x1  x2 , then 3x1  3x2 and 3x1  2  3x2  2. So f is a one-to-one function.

SECTION 2.8 One-to-One Functions and Their Inverses

235

   15. g x  x. If x1  x2 , then x1  x2 because two different numbers cannot have the same square root. Therefore, g is a one-to-one function. 16. g x  x. Because every number and its negative have the same absolute value (for example, 1  1  1), g is not a one-to-one function. 17. h x  x 2  2x. Because h 0  0 and h 2  2  2 2  0 we have h 0  h 2. So f is not a one-to-one function. 18. h x  x 3  8. If x1  x2 , then x13  x23 and x13  8  x23  8. So f is a one-to-one function.

19. f x  x 4  5. Every nonzero number and its negative have the same fourth power. For example, 14  1  14 , so f 1  f 1. Thus f is not a one-to-one function.

20. f x  x 4  5, 0  x  2. If x1  x2 , then x14  x24 because two different positive numbers cannot have the same fourth power. Thus, x14  5  x24  5. So f is a one-to-one function.

21. r t  t 6  3, 0  t  5. If t1  t2 , then t16  t26 because two different positive numbers cannot have the same sixth power. Thus, t16  3  t26  3. So r is a one-to-one function.

22. r t  t 4  1. Every nonzero number and its negative have the same fourth power. For example, 14  1  14 , so r 1  r 1. Thus r is not a one-to-one function. 1 1 1 23. f x  2 . Every nonzero number and its negative have the same square. For example, 1 , so 2 x 1 12 f 1  f 1. Thus f is not a one-to-one function. 1 1 1 24. f x  . If x1  x2 , then  . So f is a one-to-one function. x x1 x2 25. (a) f 2  7. Since f is one-to-one, f 1 7  2.

(b) f 1 3  1. Since f is one-to-one, f 1  3.

26. (a) f 5  18. Since f is one-to-one, f 1 18  5. (b) f 1 4  2. Since f is one-to-one, f 2  4.

27. f x  5  2x. Since f is one-to-one and f 1  5  2 1  3, then f 1 3  1. (Find 1 by solving the equation 5  2x  3.)

28. To find g1 5, we find the x value such that g x  5; that is, we solve the equation g x  x 2  4x  5. Now

x 2  4x  5  x 2  4x  5  0  x  1 x  5  0  x  1 or x  5. Since the domain of g is [2 , x  1 is

the only value where g x  5. Therefore, g1 5  1. 29. (a) Because f 6  2, f 1 2  6.

(b) Because f 2  5, f 1 5  2.

(c) Because f 0  6, f 1 6  0.

30. (a) Because g 4  2, g1 2  4.

(b) Because g 7  5, g1 5  7.

(c) Because g 8  6, g1 6  8.

31. From the table, f 4  5, so f 1 5  4.

32. From the table, f 5  0, so f 1 0  5.   34. f f 1 6  6 33. f 1  f 1  1   35. From the table, f 6  1, so f 1 1  6. Also, f 2  6, so f 1 6  1. Thus, f 1 f 1 1  f 1 6  1.   36. From the table, f 5  0, so f 1 0  5. Also, f 4  5, so f 1 5  4. Thus, f 1 f 1 0  f 1 5  4. 37. f g x  f x  6  x  6  6  x for all x. g  f x  g x  6  x  6  6  x for all x. Thus f and g are inverses of each other. x  x  3  x for all x. 38. f g x  f 3 3 3x  x for all x. Thus f and g are inverses of each other. g  f x  g 3x  3

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CHAPTER 2 Functions

 x 4  4  x  4  4  x for all x. 3 3x  4  4 g  f x  g 3x  4   x for all x. Thus f and g are inverses of each other. 3     2x 2x 40. f g x  f 25  2  2  x  x for all x. 5 5 2  2  5x 5x g  f x  g 2  5x    x for all x. Thus f and g are inverses of each other. 5 5   1 1   x for all x  0. Since f x  g x, we also have g  f x  x for all x   0. Thus f and 41. f g x  f x 1x g are inverses of each other.      5 42. f g x  f 5 x  5 x  x for all x.    5 g  f x  g x 5  x 5  x for all x. Thus f and g are inverses of each other.

39. f g x  f



x 4 3



3





  2 x 9  x  9  9  x  9  9  x for all x  9.      g  f x  g x 2  9  x 2  9  9  x 2  x for all x  0. Thus f and g are inverses of each other.   3  44. f g x  f x  113  x  113  1  x  1  1  x for all x. 3    g  f x  g x 3  1  x  113  1  x  1  1  x for all x. Thus f and g are inverses of each other.   1 1   x for all x  0. 1   45. f g x  f 1 x 1 1 x   1 1   1  x  1  1  x for all x  1. Thus f and g are inverses of each other.  g  f x  g 1 x 1 x 1   2     4  x2  4  4  x 2  4  4  x 2  x 2  x, for all 0  x  2. (Note that the last 46. f g x  f 43. f g x  f

equality is possible since x  0.)    2    2 g  f x  g 4x  4 4  x 2  4  4  x 2  x 2  x, for all 0  x  2. (Again, the last equality

is possible since x  0.) Thus f and g are inverses of each other.   2x2  2 2x  2  2 x  1 4x 2x  2 x1   2x2   x for all x  1. 47. f g x  f x 1 2x  2  2 x  1 4  2 x1    2 x2  2  4x 2 x  2  2 x  2 x 2 x2  x2   x for all x  2. Thus f and g are inverses of  g  f x  g x 2 x  2  1 x  2 4 1 x2

each other.





54x  5 5  4x  5 1  3x 19x 13x     x for all x  13 . 54x 3 5  4x  4 1  3x 19 3 13x  4     5  4 x5 19x 5 3x  4  4 x  5 x 5 3x4    g  f x  g   for all x   43 . Thus f and g are inverses x5 3x  4 3x  4  3  5 19 x 1  3 3x4

48. f g x  f

5  4x 1  3x



of each other.

49. f x  3x  5. y  3x  5  3x  y  5  x  13 y  5  13 y  53 . So f 1 x  13 x  53 .

50. f x  7  5x. y  7  5x  5x  7  y  x  15 7  y   15 y  75 . So f 1 x   15 x  75 .

SECTION 2.8 One-to-One Functions and Their Inverses



51. f x  5  4x 3 . y  5  4x 3  4x 3  5  y  x 3  14 5  y  x  3 14 5  y. So f 1 x  3 14 5  x.   52. f x  3x 3  8. y  3x 3  8  3x 3  y  8  x 3  13 y  83  x  3 13 y  83 . So f 1 x  3 13 x  8.

1 1 1 1 1 .y  x  2   x   2. So f 1 x   2. x 2 x 2 y y x x 2 x 2 .y  y x  2  x  2  x y  2y  x  2  x y  x  2  2y  x y  1  2 y  1 f x  x 2 x 2 2 y  1 2 x  1 x  . So f 1 x  . y1 x 1 x 4y x . y  y x  4  x  x y  4y  x  x  x y  4y  x 1  y  4y  x  . So f x  x 4 x 4 1y 4x f 1 x  . 1x 3x 2y 3x .y  y x  2  3x  x y  2y  3x  x y  3x  2y  x y  3  2y  x  . So f x  x 2 x 2 y3 2x f 1 x  . x 3 2x  5 2x  5 .y  y x  7  2x  5  x y  7y  2x  5  x y  2x  7y  5  x y  2  7y  5 f x  x 7 x 7 7y  5 7x  5 x  . So f 1 x  . y2 x 2 4x  2 4x  2 .y  y 3x  1  4x  2  3x y  y  4x  2  4x  3x y  y  2  x 4  3y  y  2 f x  3x  1 3x  1 y2 x 2 x  . So f 1 x  . 4  3y 4  3x 2x  3 2x  3 f x  .y  y 1  5x  2x  3  y  5x y  2x  3  2x  5x y  y  3  x 2  5y  y  3 1  5x 1  5x y3 x 3 x  . So f 1 x  . 5y  2 5x  2 3  4x y3 3  4x .y  y 8x  1  3  4x  8x y  y  3  4x  4x 2y  1  y  3  x  . f x  8x  1 8x  1 4 2y  1 x 3 So f 1 x  . 4 2x  1   f x  4  x 2 , x  0. y  4  x 2  x 2  4  y  x  4  y. So f 1 x  4  x, x  4. [Note that x  0  f x  4.]  2 2 2     1  y  14  f x  x 2  x  x 2  x  14  14  x  12  14 , x   . y  x  12  14  x  12 2    x  12  y  14  x  y  14  12 , y   14 . So f 1 x  x  14  12 , x   14 . (Note that x   12 , so that  2 2   x  12  0, and hence x  12  y  14  x  12  y  14 . Also, since x   12 , y  x  12  14   14 so that  y  14  0, and hence y  14 is defined.)   f x  x 6 , x  0. y  x 6  x  6 y for x  0. The range of f is y  y  0, so f 1 x  6 x, x  0. 1 1 1 1 1 f x  2 , x  0. y  2  x 2   x   . The range of f is y  y  0, so f 1 x   , x  0. y y x x x

53. f x 

54.

55.

56.

57.

58.

59.

60.

61.

62.

63. 64.

237



  2  x3 2  x3 .y  5y  2  x 3  x 3  2  5y  x  3 2  5y. Thus, f 1 x  3 2  5x. 5 5  7  7     66. f x  x 5  6 . y  x 5  6  7 y  x 5  6  x 5  7 y  6  x  5 7 y  6. Thus, f 1 x  5 7 x  6. 65. f x 

238

CHAPTER 2 Functions

67. f x 

  5  8x. Note that the range of f (and thus the domain of f 1 ) is [0 . y  5  8x  y 2  5  8x 

y2  5 x2  5 . Thus, f 1 x  , x  0. 8x  y 2  5  x  8 8

68. f x  2 

   3  x. The range of f is [2 . y  2  3  x  y  2  3  x  y  22  3  x 

x  y  22  3. Thus, f 1 x  x  22  3, x  2.

69. f x  2 

 3

x. y  2 

 3

x  y2

 3

x  x  y  23 . Thus, f 1 x  x  23 .

   4  x 2 , 0  x  2. The range of f is [0 2]. y  4  x 2  y 2  4  x 2  x 2  4  y 2  x  4  y 2 .  Thus, f 1 x  4  x 2 , 0  x  2.

70. f x 

71. (a), (b) f x  3x  6

72. (a), (b) f x  16  x 2 , x  0

y

y

f f Ð!

1

f

x

1

f Ð!

2 2

(c) f x  3x  6. y  3x  6  3x  y  6  x  13 y  6. So f 1 x  13 x  6.

x

(c) f x  16  x 2 , x  0. y  16  x 2   x 2  16  y  x  16  y. So  f 1 x  16  x, x  16. (Note: x  0  f x  16  x 2  16.)

73. (a), (b) f x 

 y

74. (a), (b) f x  x 3  1

x 1

y

f f Ð!

f Ð!

1 1

f

x

1 x

1

(c) f x 



x  1, x  1. y 



x  1, y  0

 y 2  x  1  x  y 2  1 and y  0. So f 1 x  x 2  1, x  0.

(c) f x  x 3  1  y  x 3  1  x 3  y  1    x  3 y  1. So f 1 x  3 x  1.

SECTION 2.8 One-to-One Functions and Their Inverses

75. f x  x 3  x. Using a graphing device and the

Horizontal Line Test, we see that f is not a one-to-one function. For example, f 0  0  f 1.

-2

76. f x  x 3  x. Using a graphing device and the

Horizontal Line Test, we see that f is a one-to-one function.

-2

2

x  12 . Using a graphing device and the x 6 Horizontal Line Test, we see that f is a one-to-one

77. f x 

function.

239

78. f x 



2

x 3  4x  1. Using a graphing device and the

Horizontal Line Test, we see that f is not a one-to-one function. For example, f 0  1  f 2.

10 2

-20

20 -2

-10

79. f x  x  x  6. Using a graphing device and the Horizontal Line Test, we see that f is not a one-to-one function. For example f 0  6  f 2.

0

2

80. f x  x  x. Using a graphing device and the

Horizontal Line Test, we see that f is a one-to-one function. 20

10

-10

-5

10

5 -20

-10

81. (a) y  f x  2  x  x  y  2. So

82. (a) y  f x  2  12 x  12 x  2  y  x  4  2y.

f 1 x  x  2.

So f 1 x  4  2x.

(b)

(b) 10 5

-5

5

-10

10

-5 -10

240

CHAPTER 2 Functions

83. (a) y  g x 



x  3, y  0  x  3  y 2 , y  0

 x  y 2  3, y  0. So g1 x  x 2  3, x  0.

84. (a) y  g x  x 2  1, x  0  x 2  y  1, x  0    x  y  1. So g1 x  x  1.

(b)

(b) 10

10

5

5

5

10 5

85. If we restrict the domain of f x to [0 , then y  4  x 2  x 2  4  y  x   positive square root). So f 1 x  4  x.

10

 4  y (since x  0, we take the

 If we restrict the domain of f x to  0], then y  4  x 2  x 2  4  y  x   4  y (since x  0, we take the  negative square root). So f 1 x   4  x.

86. If we restrict the domain of g x to [1 , then y  x  12  x  1    root)  x  1  y. So g 1 x  1  x.



87. If we restrict the domain of h x to [2 , then y  x  22  x  2    root)  x  2  y. So h 1 x  2  x.



y (since x  1 we take the positive square

 If we restrict the domain of g x to  1], then y  x  12  x  1   y (since x  1 we take the negative square   root)  x  1  y. So g 1 x  1  x.

y (since x  2, we take the positive square

 If we restrict the domain of h x to  2], then y  x  22  x  2   y (since x  2, we take the negative   square root)  x  2  y. So h 1 x  2  x.

   x  3 if x  3  0  x  3 88. k x  x  3   x 3 if x  3  0  x  3

If we restrict the domain of k x to [3 , then y  x  3  x  3  y. So k 1 x  3  x.

If we restrict the domain of k x to  3], then y   x  3  y  x  3  x  3  y. So k 1 x  3  x.

89.

y f Ð!

90.

y

1

f

1

f

x 1

f Ð! 1

x

SECTION 2.8 One-to-One Functions and Their Inverses y

91. (a)

241

y

92. (a)

1 0

1

(b) Yes, the graph is unchanged upon reflection about the line y  x. (c) y 

1 1 1  x  , so f 1 x  . x y x

1

x

0

1

x

(b) Yes, the graph is unchanged upon reflection about the line y  x.

x 3  y x  1  x  3  x 1 y3 x y  1  y  3  x  . Thus, y1

(c) y 

f 1 x 

x 3 . x 1

93. (a) The price of a pizza with no toppings (corresponding to the y-intercept) is $16, and the cost of each additional topping (the rate of change of cost with respect to number of toppings) is $150. Thus, f n  16  15n. (b) p  f n  16  15n  p  16  15n  n  23  p  16. Thus, n  f 1  p  23  p  16. This function represents the number of toppings on a pizza that costs x dollars.

(c) f 1 25  23 25  16  23 9  6. Thus, a $25 pizza has 6 toppings. 94. (a) f x  500  80x.

p  500 1 p  500 . So x  f 1  p  . f 80 80 represents the number of hours the investigator spends on a case for x dollars. 720 1220  500 (c) f 1 1220    9. If the investigator charges $1220, he spent 9 hours investigating the case. 80 80 (b) p  f x  500  80x. p  500  80x  80x  p  500  x 

       t 2 V t 2 t 2 V t 95. (a) V  f t  100 1   1   , 0  t  40. V  100 1   1 40 40 100 40 40 100    t V 1  t  40  4 V . Since t  40, we must have t  f 1 V   40  4 V . f 1 represents time that 40 10 has elapsed since the tank started to leak.  (b) f 1 15  40  4 15  245 minutes. In 245 minutes the tank has drained to just 15 gallons of water.     96. (a)   g r  18,500 025  r 2 .   18,500 025  r 2    4625  18,500r 2  18,500r 2  4625      4625   4625   4625   1 2 1 r  r  . Since r represents a distance, r  0, so g   . g  18,500 18,500 18,500 represents the radial distance from the center of the vein at which the blood has velocity .  4625  30 (b) g 1 30   0498 cm. The velocity is 30 cms at a distance of 0498 cm from the center of the artery 18,500 or vein.

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CHAPTER 2 Functions

97. (a) D  f  p  3 p  150. D  3 p  150  3 p  150  D  p  50  13 D. So f 1 D  50  13 D. f 1 D represents the price that is associated with demand D. (b) f 1 30  50  13 30  40. So when the demand is 30 units, the price per unit is $40. 98. (a) F  g C  95 C  32. F  95 C  32  95 C  F  32  C  59 F  32. So g1 F  59 F  32. g1 F represents the Celsius temperature that corresponds to the Fahrenheit temperature of F. (b) F 1 86  59 86  32  59 54  30. So 86 Fahrenheit is the same as 30 Celsius. 99. (a) f 1 U   102396U .

(b) U  f x  09766x. U  09766x  x  10240U . So f 1 U   10240U . f 1 U  represents the value of U US dollars in Canadian dollars.

(c) f 1 12,250  10240 12,250  12,54352. So $12,250 in US currency is worth $12,54352 in Canadian currency.

  01x, if 0  x  20,000 100. (a) f x   2000  02 x  20,000 if x  20,000 (b) We will find the inverse of each piece of the function f .

f 1 x  01x. T  01x  x  10T . So f 11 T   10T . f 2 x  2000  02 x  20,000  02x  2000. T  02x  2000  02x  T  2000  x  5T  10,000. So f 21 T   5T  10,000.

  10T , if 0  T  2000 Since f 0  0 and f 20,000  2000 we have f 1 T    5T  10,000 if T  2000

This represents the

taxpayer’s income.

(c) f 1 10,000  5 10,000  10,000  60,000. The required income is 60,000. 101. (a) f x  085x.

(b) g x  x  1000.

(c) H x   f  g x  f x  1000  085 x  1000  085x  850.

(d) P  H x  085x  850. P  085x  850  085x  P  850  x  1176P  1000. So

H 1 P  1176P  1000. The function H 1 represents the original sticker price for a given discounted price P.

(e) H 1 13,000  1176 13,000  1000  16,288. So the original price of the car is $16,288 when the discounted price ($1000 rebate, then 15% off) is $13,000. 102. f x  mx  b. Notice that f x1   f x2   mx1  b  mx2  b  mx1  mx2 . We can conclude that x1  x2 if and only if m  0. Therefore f is one-to-one if and only if m  0. If m  0, f x  mx  b  y  mx  b  mx  y  b x 

yb x b . So, f 1 x  . m m

2x  1 is “multiply by 2, add 1, and then divide by 5 ”. So the reverse is “multiply by 5, subtract 1, and then 5   5x  1  2  1 5x  1  1 5x 5x  1 5x  1 2 . Check: f  f 1 x  f    x divide by 2 ” or f 1 x  2 2 5 5 5   2x  1   5 1 2x  1 2x  1  1 2x 5 and f 1  f x  f 1     x. 5 2 2 2

103. (a) f x 

CHAPTER 2

Review

243

1 1   3 is “take the negative reciprocal and add 3 ”. Since the reverse of “take the negative x x reciprocal” is “take the negative reciprocal ”, f 1 x is “subtract 3 and take the negative reciprocal ”, that is,     1 x 3 1 1 1 1 . Check: f  f x  f  3  3  x  3  x and f x   3 1 1 x 3 x 3 1 x 3   1 1 1 x     x. f 1  f x  f 1 3   1  1 1 x 1  3 3 x x  (c) f x  x 3  2 is “cube, add 2, and then take the square root”. So the reverse is “square, subtract 2, then take     3 the cube root ” or f 1 x  x 2  2. Domain for f x is  3 2  ; domain for f 1 x is [0 . Check:    3    3 2 3 2 1 x 2  x  2  2  x 2  2  2  x 2  x (on the appropriate domain) and f  f x  f    2   3 3 3 1 1 3 f  f x  f x 2  x 3  2  2  x 3  2  2  x 3  x (on the appropriate domain). (b) f x  3 

(d) f x  2x  53 is “double, subtract 5, and then cube”. So the reverse is “take the cube root, add  3 x 5 Domain for both f x and f 1 x is  . Check: 5, and divide by 2” or f 1 x  2     3  3 3   3   3 x 5 x 5 3 f  f 1 x  f  2  5  3 x  5  5  3 x  x 3  x and 2 2  x   2x  53  5 2x 2x  5  5 1 1 3 f    x.  f x  f 2x  5  2 2 2 In a function like f x  3x  2, the variable occurs only once and it easy to see how to reverse the operations step by step. But in f x  x 3  2x  6, you apply two different operations to the variable x (cubing and multiplying by 2) and then add 6, so it is not possible to reverse the operations step by step.

104. f I x  f x; therefore f  I  f . I  f x  f x; therefore I  f  f . By definition, f  f 1 x  x  I x; therefore f  f 1  I . Similarly, f 1  f x  x  I x; therefore f 1  f  I . 105. (a) We find g 1 x: y  2x  1  2x  y  1  x  12 y  1. So g1 x  12 x  1. Thus     2  f x  h  g1 x  h 12 x  1  4 12 x  1  4 12 x  1  7  x 2  2x  1  2x  2  7  x 2  6.

(b) f  g  h  f 1  f  g  f 1  h  I  g  f 1  h  g  f 1  h. Note that we compose with f 1 on the left on each side of the equation. We find f 1 : y  3x  5  3x  y  5  x  13 y  5. So f 1 x  13 x  5.        Thus g x  f 1  h x  f 1 3x 2  3x  2  13 3x 2  3x  2  5  13 3x 2  3x  3  x 2  x  1.

CHAPTER 2 REVIEW 1. “Square, then subtract 5” can be represented by the function f x  x 2  5. 2. “Divide by 2, then add 9” can be represented by the function g x 

x  9. 2

3. f x  3 x  10: “Add 10, then multiply by 3.”  4. f x  6x  10: “Multiply by 6, then subtract 10, then take the square root.”

244

CHAPTER 2 Functions

5. g x  x 2  4x

6. h x  3x 2  2x  5 x

g x

x

h x

5

2

3

0

0

1

4

1

3

0

5

2

4

1

0

3

3

2

11

1

7. C x  5000  30x  0001x 2 (a) C 1000  5000  30 1000  0001 10002  $34,000 and

C 10,000  5000  30 10,000  0001 10,0002  $205,000.

(b) From part (a), we see that the total cost of printing 1000 copies of the book is $34,000 and the total cost of printing 10,000 copies is $205,000. (c) C 0  5000  30 0  0001 02  $5000. This represents the fixed costs associated with getting the print run ready.

(d) The net change in C as x changes from 1000 to 10,000 is C 10,000  C 1000  205,000  34,000  $171,000, and 171,000 C 10,000  C 1000   $19copy. the average rate of change is 10,000  1000 9000 8. E x  400  003x (a) E 2000  400  003 2000  $460 and E 15 000  400  003 15,000  $850. (b) From part (a), we see that if Reynalda sells $2000 worth of goods, she makes $460, and if she sells $15,000 worth of goods, she makes $850. (c) E 0  400  003 0  $400 is Reynalda’s base weekly salary.

(d) The net change in E as x changes from 2000 to 15,000 is E 15,000  E 2000  850  460  $390, and the average 390 E 15,000  E 2000   $003 per dollar. rate of change is 15,000  2000 13,000 (e) Because the value of goods sold x is multiplied by 003 or 3%, we see that Reynalda earns a percentage of 3% on the goods that she sells. 9. f x  x 2  4x  6; f 0  02  4 0  6  6; f 2  22  4 2  6  2;

f 2  22  4 2  6  18; f a  a2  4 a  6  a 2  4a  6; f a  a2  4 a  6  a 2  4a  6;

f x  1  x  12 4 x  16  x 2 2x 14x 46  x 2 2x 3; f 2x  2x2 4 2x6  4x 2 8x 6.     10. f x  4  3x  6; f 5  4  15  6  1; f 9  4  27  6  4  21;        f a  2  4  3a  6  6  4  3a; f x  4  3 x  6  4  3x  6; f x 2  4  3x 2  6.

11. By the Vertical Line Test, figures (b) and (c) are graphs of functions. By the Horizontal Line Test, figure (c) is the graph of a one-to-one function. 12. (a) f 2  1 and f 2  2.

(b) The net change in f from 2 to 2 is f 2  f 2  2  1  3, and the average rate of change is f 2  f 2 3  . 2  2 4 (c) The domain of f is [4 5] and the range of f is [4 4]. (d) f is increasing on 4 2 and 1 4; f is decreasing on 2 1 and 4 5. (e) f has local maximum values of 1 (at x  2) and 4 (at x  4).

(f) f is not a one-to-one, for example, f 2  1  f 0. There are many more examples. 13. Domain: We must have x  3  0  x  3. In interval notation, the domain is [3 .  Range: For x in the domain of f , we have x  3  x  3  0  x  3  0  f x  0. So the range is [0 .

CHAPTER 2

Review

245

  14. F t  t 2  2t  5  t 2  2t  1  5  1  t  12  4. Therefore F t  4 for all t. Since there are no restrictions on t, the domain of F is  , and the range is [4 .

15. f x  7x  15. The domain is all real numbers,  . 16. f x  17. f x 

  2x  1 . Then 2x  1  0  x  12 . So the domain of f is x  x  12 . 2x  1 

x  4. We require x  4  0  x  4. Thus the domain is [4 .

18. f x  3x  

19. f x 

2 . The domain of f is the set of x where x  1  0  x  1. So the domain is 1 . x 1

1 1 1   . The denominators cannot equal 0, therefore the domain is x  x  0 1 2. x x 1 x 2

2x 2  5x  3 2x 2  5x  3 . The domain of g is the set of all x where the denominator is not 0. So the  2 2x  1 x  3 2x  5x  3   domain is x  2x  1  0 and x  3  0  x  x   12 and x  3 .

20. g x 

21. h x 

  4  x  x 2  1. We require the expression inside the radicals be nonnegative. So 4  x  0  4  x; also

x 2  1  0  x  1 x  1  0. We make a table:

 1

1 1

1 

Sign of x  1







Sign of x  1













Interval

Sign of x  1 x  1

Thus the domain is  4]   1]  [1    1]  [1 4].

 3 2x  1 . Since we have an odd root, the domain is the set of all x where the denominator is not 0. Now 22. f x   3 2x  2   3 2x  2  0  3 2x  2  2x  8  x  4. Thus the domain of f is x  x  4. 23. f x  1  2x

24. f x  13 x  5, 2  x  8

y

y

1

1 1

x

1

x

246

CHAPTER 2 Functions

25. f x  3x 2

26. f x   14 x 2

y

y

1

1

0

x

5 0

27. f x  2x 2  1

1

x

28. f x   x  14

y

y 5 0

1

x

1 0

29. f x  1 



1

x

30. f x  1 

x y



x 2 y

1 0

2

x

1 0

1

x

31. f x  12 x 3

32. f x  y

y

1

2 0

 3 x

1

x

0

2

x

CHAPTER 2

33. f x   x

34. f x  x  1

y

Review

247

y

1 0

1

x

1 0

1 35. f x   2 x

36. f x  y

1

x

1 x  13

y

1 0

1

x 5 0

1

x

    x if x  0 38. f x  x 2 if 0  x  2    1 if x  2

  1  x if x  0 37. f x  1 if x  0

y

y

1 1

x

1 1

x

 39. x  y 2  14  y 2  14  x  y   14  x, so the original equation does not define y as a function of x. 40. 3x 



y 8



y  3x  8  y  3x  82 , so the original equation defines y as a function of x.

 13 41. x 3  y 3  27  y 3  x 3  27  y  x 3  27 , so the original equation defines y as a function of x (since the cube root function is one-to-one).

 42. 2x  y 4  16  y 4  2x  16  y   4 2x  16, so the original equation does not define y as a function of x.

248

CHAPTER 2 Functions

43. f x  6x 3  15x 2  4x  1

(ii) [8 8] by [8 8]

(i) [2 2] by [2 2]

2 5

-2

-5

2

5 -5

-2

(iii) [4 4] by [12 12]

(iv) [100 100] by [100 100] 100

10

-4

-2

2

4

-100

-10

100 -100

From the graphs, we see that the viewing rectangle in (iii) produces the most appropriate graph.

44. f x 

 100  x 3

(i) [4 4] by [4 4]

(ii) [10 10] by [10 10] 10

-4

-2

2

4

-10

10 -10

(iii) [10 10] by [10 40]

(iv) [100 100] by [100 100]

40

100

20 -100 -10

10

100 -100

From the graphs, we see that the viewing rectangle in (iii) produces the most appropriate graph of f .

CHAPTER 2

45. (a) We graph f x 

 9  x 2 in the viewing rectangle

[4 4] by [1 4].

Review

 46. (a) We graph f x   x 2  3 in the viewing rectangle [5 5] by [6 1].

4 -5

5

2

-4

-2

2

-5

4

(b) From the graph, the domain of f is [3 3] and the

(b) From the graph, the domain of f is  173]  [173  and the range of f is

range of f is [0 3].

 0].

47. (a) We graph f x 



x 3  4x  1 in the viewing

rectangle [5 5] by [1 5].

48. (a) We graph f x  x 4  x 3  x 2  3x  6 in the viewing rectangle [3 4] by [20 100]. 100

4

50

2

-5

5

(b) From the graph, the domain of f is approximately [211 025]  [186  and the range of f is

[0 .

49. f x  x 3  4x 2 is graphed in the viewing rectangle

[5 5] by [20 10]. f x is increasing on  0 and 267 . It is decreasing on 0 267.

-2

2

4

(b) From the graph, the domain of f is   and the range of f is approximately [710 .

    50. f x  x 4  16 is graphed in the viewing rectangle

[5 5] by [5 20]. f x is increasing on 2 0 and 2 . It is decreasing on  2 and 0 2. 20

-5

5

-20

10

-5

5

f 8  f 4 4   1. 84 4

51. The net change is f 8  f 4  8  12  4 and the average rate of change is

52. The net change is g 30  g 10  30  5  35 and the average rate of change is 53. The net change is f 2  f 1  6  2  4 and the average rate of change is

35 7 g 30  g 10   . 30  10 20 4

4 f 2  f 1  . 2  1 3

54. The net change is f 3  f 1  1  5  6 and the average rate of change is

6 f 3  f 1   3. 31 2

249

250

CHAPTER 2 Functions

    55. The net change is f 4  f 1  42  2 4  12  2 1  8  1  9 and the average rate of change is 9 f 4  f 1   3. 41 3

56. The net change is g a  h  g a  a  h  12  a  12  2ah  2h  h 2 and the average rate of change is g a  h  g a 2ah  2h  h 2   2a  2  h. aha h

57. f x  2  3x2  9x 2  12x  4 is not linear. It cannot be expressed in the form f x  ax  b with constant a and b. 58. g x 

x 3  15 x  35 is linear with a  15 and b  35 . 5 y

59. (a)

y

60. (a)

1 0

1

1

x

0

(b) The slope of the graph is the value of a in the

1

x

(b) The slope of the graph is the value of a in the

equation f x  ax  b  3x  2; that is, 3.

equation f x  ax  b   12 x  3; that is,  12 .

(c) The rate of change is the slope of the graph, 3.

(c) The rate of change is the slope of the graph,  12 .

61. The linear function with rate of change 2 and initial value 3 has a  2 and b  3, so f x  2x  3. 62. The linear function whose graph has slope 12 and y-intercept 1 has a  12 and b  1, so f x  12 x  1. 63. Between x  0 and x  1, the rate of change is f x  2x  3. 64. Between x  0 and x  2, the rate of change is is f x   14 x  6.

53 f 1  f 0   2. At x  0, f x  3. Thus, an equation is 10 1 f 2  f 0 55  6    14 . At x  0, f x  6. Thus, an equation 20 2

65. The points 0 4 and 8 0 lie on the graph, so the rate of change is y   12 x  4.

04 1   . At x  0, y  4. Thus, an equation is 80 2

66. The points 0 4 and 2 0 lie on the graph, so the rate of change is is y  2x  4.

0  4  2. At x  0, y  4. Thus, an equation 20

67. P t  3000  200t  01t 2 (a) P 10  3000  200 10  01 102  5010 represents the population in its 10th year (that is, in 1995), and

P 20  3000  200 20  01 202  7040 represents its population in its 20th year (in 2005). P 20  P 10 7040  5010 2030 (b) The average rate of change is    203 peopleyear. This represents the 20  10 10 10 average yearly change in population between 1995 and 2005.

CHAPTER 2

Review

251

68. D t  3500  15t 2 (a) D 0  3500  15 02  $3500 represents the amount deposited in 1995 and D 15  3500  15 152  $6875 represents the amount deposited in 2010. 13500 (b) Solving the equation D t  17,000, we get 17,000  3500  15t 2  15t 2  13,500  t 2   900  15 t  30, so thirty years after 1995 (that is, in the year 2025) she will deposit $17,000. 6875  3500 D 15  D 0   $225year. This represents the average annual (c) The average rate of change is 15  0 15 increase in contributions between 1995 and 2010. 69. f x  12 x  6 (a) The average rate of change of f between x  0 and x  2 is     1 2  6  1 0  6 5  6 1 f 2  f 0 2 2    , and the average rate of change of f between x  15 20 2 2 2 and x  50 is     1 50  6  1 15  6 19  32 f 50  f 15 1 2 2    . 50  15 35 35 2 (b) The rates of change are the same. (c) Yes, f is a linear function with rate of change 12 . 70. f x  8  3x

[8  3 2]  [8  3 0] 28 f 2  f 0    3, 20 2 2 and the average rate of change of f between x  15 and x  50 is [8  3 50]  [8  3 15] f 50  f 15 142  37    3. 50  15 35 35 (b) The rates of change are the same. (a) The average rate of change of f between x  0 and x  2 is

(c) Yes, f is a linear function with rate of change 3. 71. (a) y  f x  8. Shift the graph of f x upward 8 units.

(b) y  f x  8. Shift the graph of f x to the left 8 units.

(c) y  1  2 f x. Stretch the graph of f x vertically by a factor of 2, then shift it upward 1 unit. (d) y  f x  2  2. Shift the graph of f x to the right 2 units, then downward 2 units. (e) y  f x. Reflect the graph of f x about the y-axis.

(f) y   f x. Reflect the graph of f x first about the y-axis, then reflect about the x-axis.

(g) y   f x. Reflect the graph of f x about the x-axis.

(h) y  f 1 x. Reflect the graph of f x about the line y  x.

72. (a) y  f x  2

(b) y   f x

y

1

(c) y  3  f x

y

1 1

x

y

1 1

x

1

x

252

CHAPTER 2 Functions

(d) y  12 f x  1

(e) y  f 1 x

y

(f) y  f x

y

1

1 1

1 1

x

y

1

x

x

73. (a) f x  2x 5  3x 2  2. f x  2 x5  3 x2  2  2x 5  3x 2  2. Since f x  f x, f is not even.  f x  2x 5  3x 2  2. Since  f x  f x, f is not odd.   (b) f x  x 3  x 7 . f x  x3  x7   x 3  x 7   f x, hence f is odd.

1  x2 1  x2 1  x2 . f    f x. Since f x  f x, f is even. x 1  x2 1  x2 1  x2 1 1 1 1 . f x  .  f x   . Since f x  f x , f is not even, and since  (d) f x  x 2 2x x 2 x  2 f x   f x, f is not odd. (c) f x 

74. (a) This function is odd. (b) This function is neither even nor odd. (c) This function is even. (d) This function is neither even nor odd.     75. g x  2x 2  4x  5  2 x 2  2x  5  2 x 2  2x  1  5  2  2 x  12  7. So the local minimum value 7 when x  1.

 2     76. f x  1  x  x 2   x 2  x  1   x 2  x  14  1  14   x  12  54 . So the local maximum value is 54 when x   12 .

77. f x  33  16x  25x 3 . In the first viewing rectangle, [2 2] by [4 8], we see that f x has a local maximum and a local minimum. In the next viewing rectangle, [04 05] by [378 380], we isolate the local maximum value as approximately 379 when x  046. In the last viewing rectangle, [05 04] by [280 282], we isolate the local minimum value as 281 when x  046. 3.80

2.82

3.79

2.81

5

-2

2

3.78 0.40

0.45

0.50

-0.50

-0.45

2.80 -0.40

CHAPTER 2

Review

253

78. f x  x 23 6  x13 . In the first viewing rectangle, [10 10] by [10 10], we see that f x has a local maximum and a local minimum. The local minimum is 0 at x  0 (and is easily verified). In the next viewing rectangle, [395 405] by [316 318], we isolate the local maximum value as approximately 3175 when x  400. 10

3.18 3.17

-10

10 3.16 3.95

-10

4.00

4.05

    79. h t  16t 2  48t  32  16 t 2  3t  32  16 t 2  3t  94  32  36   2   16 t 2  3t  94  68  16 t  32  68 The stone reaches a maximum height of 68 feet.

  1500  12x  00004x 2  00004 x 2  30,000x    1500  00004 x 2  30,000x  225,000,000  1500  90,000  00004 x  15,0002  88,500

80. P x



The maximum profit occurs when 15,000 units are sold, and the maximum profit is $88,500. 81. f x  x  2, g x  x 2

82. f x  x 2  1, g x  3  x 2 5

5

-4

-2

2

-2

2

83. f x  x 2  3x 2 and g x 4  3x. (a)  f  g x  x 2  3x  2  4  3x  x 2  6x  6   (b)  f  g x  x 2  3x  2  4  3x  x 2  2   (c)  f g x  x 2  3x  2 4  3x  4x 2  12x  8  3x 3  9x 2  6x  3x 3  13x 2  18x  8   x 2  3x  2 f , x  43 (d) x  g 4  3x

(e)  f  g x  f 4  3x  4  3x2  3 4  3x  2  16  24x  9x 2  12  9x  2  9x 2  15x  6     (f) g  f  x  g x 2  3x  2  4  3 x 2  3x  2  3x 2  9x  2  84. f x  1  x 2 and g x  x  1. (Remember that the proper domains must apply.)    2 (a)  f  g x  f x 1 1 x 1 1x 1 x      (b) g  f  x  g 1  x 2  1  x 2  1  x 2  x   (c)  f  g 2  f g 2  f 2  1  f 1  1  12  2.   (d)  f  f  2  f  f 2  f 1  22  f 5  1  52  26. (e)  f  g  f  x  f g  f  x  f x  1  x2  1  x 2 . Note that g  f  x  x by part (b).  (f) g  f  g x  g  f  g x  g x  x  1. Note that  f  g x  x by part (a).

254

CHAPTER 2 Functions

85. f x  3x  1 and g x  2x  x 2 .      f  g x  f 2x  x 2  3 2x  x 2  1  3x 2  6x  1, and the domain is  .

g  f  x  g 3x  1  2 3x  1  3x  12  6x  2  9x 2  6x  1  9x 2  12x  3 , and the domain is    f  f  x  f 3x  1  3 3x  1  1  9x  4, and the domain is  .     2  g  g x  g 2x  x 2  2 2x  x 2  2x  x 2  4x  2x 2  4x 2  4x 3  x 4  x 4  4x 3  6x 2  4x, and domain is  .

2 , has domain x  x  4. x, has domain x  x  0. g x  x 4    2 2  .  f  g x is defined whenever both g x and f g x are defined; that is,  f  g x  f x 4 x 4 2 2 whenever x  4 and  0. Now  0  x  4  0  x  4. So the domain of f  g is 4 . x 4 x 4   2 . g  f  x is defined whenever both f x and g  f x are defined; that is, whenever g  f  x  g x   x 4   x  0 and x  4  0. Now x  4  0  x  16. So the domain of g  f is [0 16  16 .    x  x  x 14 .  f  f  x is defined whenever both f x and f  f x are defined; that is,  f  f  x  f whenever x  0. So the domain of f  f is [0 .   2 x 4 2 x  4 2    g  g x is defined whenever both g x and  g  g x  g 2 x 4 2  4 x  4 9  2x 4 x 4 g g x are defined; that is, whenever x  4 and 9  2x  0. Now 9  2x  0  2x  9  x  92 . So the domain of   g  g is x  x  92  4 .

86. f x 



  1  x, g x  1  x 2 and h x  1  x.            2   f 1 12 x x  f  g  h x  f g h x  f g 1  x  f 1  1  x      2          f x  2 x  1  x  2 x  1  2 x  x  1  x  1  x

87. f x 

88. If h x 



x and g x  1  x , then g  h x  g

   1  f  g  h x  f 1  x     T x. 1 x

   1 x  1  x. If f x   , then x

89. f x  3  x 3 . If x1  x2 , then x13  x23 (unequal numbers have unequal cubes), and therefore 3  x13  3  x23 . Thus f is a one-to-one function.   90. g x  2  2x  x 2  x 2  2x  1  1  x  12  1. Since g 0  2  g 2 , as is true for all pairs of numbers equidistant from 1, g is not a one-to-one function.

1 . Since the fourth powers of a number and its negative are equal, h is not one-to-one. For example, x4 1 1  1 and h 1   1, so h 1  h 1. h 1  4 1 14      92. r x  2  x  3. If x1  x2 , then x1  3  x2  3, so x1  3  x2  3 and 2  x1  3  2  x2  3. Thus r is one-to-one. 91. h x 

CHAPTER 2

93. p x  33  16x  25x 3 . Using a graphing device and the Horizontal Line Test, we see that p is not a one-to-one

Test

255

94. q x  33  16x  25x 3 . Using a graphing device and the Horizontal Line Test, we see that q is a one-to-one

function.

function. 10

10

-5

5

-5

5

-10

-10

95. f x  3x  2  y  3x  2  3x  y  2  x  13 y  2. So f 1 x  13 x  2. 2x  1 2x  1 y  2x  1  3y  2x  3y  1  x  12 3y  1  So f 1 x  12 3x  1. 96. f x  3 3    97. f x  x  13  y  x  13  x  1  3 y  x  3 y  1. So f 1 x  3 x  1.    98. f x  1  5 x  2. y  1  5 x  2  y  1  5 x  2  x  2  y  15  x  2  y  15 . So f 1 x  2  x  15 .

99. The graph passes the Horizontal Line Test, so f has an inverse. Because f 1  0, f 1 0  1, and because f 3  4, f 1 4  3.

100. The graph fails the Horizontal Line Test, so f does not have an inverse. 101. (a), (b) f x  x 2  4, x  0

102.

y

f

  (a) If x1  x2 , then 4 x1  4 x2 , and so   1  4 x1  1  4 x2 . Therefore, f is a one-to-one function.

f Ð!

y

(b), (c)

1 1

f Ð!

x

f 1

(c) f x  x 2  4, x  0  y  x 2  4, y  4   x 2  y  4  x  y  4. So  f 1 x  x  4, x  4.

x

1

(d) f x  1 

 4

x. y  1 

 4

x

 4

x  y 1

 x  y  14 . So f 1 x  x  14 ,

x  1. Note that the domain of f is [0  , so  y  1  4 x  1. Hence, the domain of f 1 is

[1 .

CHAPTER 2 TEST 1. By the Vertical Line Test, figures (a) and (b) are graphs of functions. By the Horizontal Line Test, only figure (a) is the graph of a one-to-one function.

256

CHAPTER 2 Functions

     0 2 2 a2 a2  0; f 2   ; f a  2   . 2. (a) f 0  01 21 3 a21 a3  x (b) f x  . Our restrictions are that the input to the radical is nonnegative and that the denominator must not be 0. x 1 Thus, x  0 and x  1  0  x  1. (The second restriction is made irrelevant by the first.) In interval notation, the domain is [0 .   10 2    f 10  f 2 3 10  11 2 10  1 2  1 (c) The average rate of change is   . 10  2 10  2 264

3. (a) “Subtract 2, then cube the result” can be expressed

y

(c)

algebraically as f x  x  23 . (b) x

f x

1

27

0

8

1

1

2

0

3

1

2

1 x

4 8 (d) We know that f has an inverse because it passes the Horizontal Line Test. A verbal description for f 1 is, “Take the cube root, then add 2.”    (e) y  x  23  3 y  x  2  x  3 y  2. Thus, a formula for f 1 is f 1 x  3 x  2. 4. (a) f has a local minimum value of 4 at x  1 and local maximum values of 1 at x  4 and 4 at x  3. (b) f is increasing on  4 and 1 3 and decreasing on 4 1 and 3 .

5. R x  500x 2  3000x

(a) R 2  500 22  3000 2  $4000 represents their total sales revenue when their price is $2 per bar and

R 4  500 42  3000 4  $4000 represents their total

sales revenue when their price is $4 per bar

(c) The maximum revenue is $4500, and it is achieved at a price of x  $3.

(b)

R 5000 4000 3000 2000 1000 0

1

2

3

4

5

x

      6. The net change is f 2  h  f 2  2  h2  2 2  h  22  2 2  4  h 2  4h  4  2h  0  2h  h 2 and the average rate of change is

2h  h 2 f 2  h  f 2   2  h. 2h2 h

CHAPTER 2

7. (a) f x  x  52  x 2  10x  25 is not linear because it cannot be

Test

257

y

(b)

expressed in the form f x  ax  b for constants a and b. g x  1  5x is linear.

y=g(x)

y=f(x)

(c) g x has rate of change 5. 10 0

8. (a) f x  x 3

1

x

(b) g x  x  13  2. To obtain the graph of g,

y

shift the graph of f to the right 1 unit and downward 2 units. y

1 1

x 1 x

1

9. (a) y  f x  3  2. Shift the graph of f x to the right 3 units, then shift the graph upward 2 units. (b) y  f x. Reflect the graph of f x about the y-axis.

10. (a) f 2  1  2  1  2  3 (since 2  1 ).

(b)

y

f 1  1  1  0 (since 1  1 ).

1 1

x

11. f x  x 2  x  1; g x  x  3.  (a)  f  g x  f x  g x  x 2  x  1  x  3  x 2  2x  2   (b)  f  g x  f x  g x  x 2  x  1  x  3  x 2  4

(c)  f  g x  f g x  f x  3  x  32  x  3  1  x 2  6x  9  x  3  1  x 2  5x  7     (d) g  f  x  g  f x  g x 2  x  1  x 2  x  1  3  x 2  x  2

(e) f g 2  f 1  12  1  1  1. [We have used the fact that g 2  2  3  1.] (f) g  f 2  g 7  7  3  4. [We have used the fact that f 2  22  2  1  7.]

(g) g  g  g x  g g g x  g g x  3  g x  6  x  6  3  x  9. [We have used the fact that g x  3  x  3  3  x  6.]

258

CHAPTER 2 Functions

12. (a) f x  x 3  1 is one-to-one because each real number has a unique cube.

(b) g x  x  1 is not one-to-one because, for example, g 2  g 0  1.

1   1 2 2 x the Inverse Function Property,

13. f g x  

1 1  x for all x  0, and g  f x   2  x  2  2  x for all x  2. Thus, by 1 1 x x 2 f and g are inverse functions.

x 3 x 3 5y  3 . y   2x  5 y  x  3  x 2y  1  5x  3  x   . Thus, 2x  5 2x  5 2y  1 5x  3 f 1 x   . 2x  1    15. (a) f x  3  x, x  3  y  3  x  (b) f x  3  x, x  3 and f 1 x  3  x 2 ,

14. f x 

y 2  3  x  x  3  y 2 . Thus

x 0

y

f 1 x  3  x 2 , x  0.

f 1 x

1

f Ð!

16. The domain of f is [0 6], and the range of f is [1 7]. 17. The graph passes through the points 0 1 and 4 3, so f 0  1 and f 4  3. 18. The graph of f x  2 can be obtained by shifting the graph of f x to the right

y

2 units. The graph of f x  2 can be obtained by shifting the graph of f x

y=f(x)+2

upward 2 units.

y=f(x-2) 1

f 1

19. The net change of f between x  2 and x  6 is f 6  f 2  7  2  5 and the average rate of change is f 6  f 2 5  . 62 4 20. Because f 0  1, f 1 1  0. Because f 4  3, f 1 3  4. 21.

y f fÐ! 1

1

x

x

Modeling with Functions

259

22. (a) f x  3x 4  14x 2  5x  3. The graph is shown in the viewing rectangle [10 10] by [30 10].

-10

10 -20

(b) No, by the Horizontal Line Test. (c) The local maximum is approximately 255 when x  018, as shown in the first viewing rectangle [015 025] by [26 25]. One local minimum is approximately 2718 when x  161, as shown in the second viewing rectangle [165 155] by [275 27]. The other local minimum is approximately 1193 when x  143, as shown is the viewing rectangle [14 15] by [12 119]. 0.15 -2.50

0.20

0.25

-1.65

-1.60

-2.55

-1.55 -27.0 -27.2

1.40 -11.90

1.45

1.50

-11.95

-27.4 -2.60

-12.00

(d) Using the graph in part (a) and the local minimum, 2718, found in part (c), we see that the range is [2718 .

(e) Using the information from part (c) and the graph in part (a), f x is increasing on the intervals 161 018 and 143  and decreasing on the intervals  161 and 018 143.

FOCUS ON MODELING Modeling with Functions 1. Let  be the width of the building lot. Then the length of the lot is 3. So the area of the building lot is A   32 ,   0. 2. Let  be the width of the poster. Then the length of the poster is   10. So the area of the poster is A      10  2  10.

3. Let  be the width of the base of the rectangle. Then the height of the rectangle is 12 . Thus the volume of the box is given by the function V   12 3 ,   0. 4. Let r be the radius of the cylinder. Then the height of the cylinder is 4r. Since for a cylinder V  r 2 h, the volume of the cylinder is given by the function V r   r 2 4r  4r 3 .

5. Let P be the perimeter of the rectangle and y be the length of the other side. Since P  2x  2y and the perimeter is 20, we have 2x  2y  20  x  y  10  y  10  x. Since area is A  x y, substituting gives A x  x 10  x  10x  x 2 , and since A must be positive, the domain is 0  x  10.

6. Let A be the area and y be the length of the other side. Then A  x y  16  y  P  2x  2 

32 16  2x  , where x  0. x x

16 . Substituting into P  2x  2y gives x

260

FOCUS ON MODELING

7.

Let h be the height of an altitude of the equilateral triangle whose side has length x, x

x

h 1 _x 2

as shown in the diagram. Thus the area is given by A  12 xh. By the Pythagorean   2 Theorem, h 2  12 x  x 2  h 2  14 x 2  x 2  h 2  34 x 2  h  23 x.

Substituting into the area of a triangle, we get    A x  12 xh  12 x 23 x  43 x 2 , x  0.

8. Let d represent the length of any side of a cube. Then the surface area is S  6d 2 , and the volume is V  d 3  d    2 Substituting for d gives S V   6 3 V  6V 23 , V  0. A r  9. We solve for r in the formula for the area of a circle. This gives A  r 2  r 2  r A 







 3

V.

A , so the model is 

A , A  0. 

C . Substituting for r 10. Let r be the radius of a circle. Then the area is A  r 2 , and the circumference is C  2r  r  2  2 C C2 , C  0. gives A C    2 4 60 11. Let h be the height of the box in feet. The volume of the box is V  60. Then x 2 h  60  h  2 . x The surface area, S, of the box is the sum of the area of the 4 sides and the area of the base and top. Thus   240 240 60 S  4xh  2x 2  4x  2x 2   2x 2 , so the model is S x   2x 2 , x  0. x x x2 5 12 5d   5 L  d  12L  5d  7L  L  . The model is L d  57 d. L L d 7

12. By similar triangles,

dª

13.

Let d1 be the distance traveled south by the first ship and d2 be the distance traveled east by the second ship. The first ship travels south for t hours at 5 mi/h, so

dÁ

d1  15t and, similarly, d2  20t. Since the ships are traveling at right angles to

D

each other, we can apply the Pythagorean Theorem to get    D t  d12  d22  15t2  20t2  225t 2  400t 2  25t.

14. Let n be one of the numbers. Then the other number is 60  n, so the product is given by the function P n  n 60  n  60n  n 2 .

Let b be the length of the base, l be the length of the equal sides, and h be the

15. l

h

b

l

height in centimeters. Since the perimeter is 8, 2l  b  8  2l  8  b   2 l  12 8  b. By the Pythagorean Theorem, h 2  12 b  l 2   h  l 2  14 b2 . Therefore the area of the triangle is  b1 A  12  b  h  12  b l 2  14 b2  8  b2  14 b2 2 4   b b b   64  16b  b2  b2  64  16b   4 4  b  b 4  b 4 4 4  so the model is A b  b 4  b, 0  b  4.

Modeling with Functions

261

16. Let x be the length of the shorter leg of the right triangle. Then the length of the other triangle is 2x. Since it is a right    triangle, the length of the hypotenuse is x 2  2x2  5x 2  5 x (since x  0 ). Thus the perimeter of the triangle is     P x  x  2x  5 x  3  5 x.

 2 2 17. Let  be the length of the rectangle. By the Pythagorean Theorem, 12   h 2  102   h 2  102  4     2  4 100  h 2    2 100  h 2 (since   0 ). Therefore, the area of the rectangle is A  h  2h 100  h 2 ,  so the model is A h  2h 100  h 2 , 0  h  10.

18. Using the formula for the volume of a cone, V  13 r 2 h, we substitute V  100 and solve for h. Thus 100  13 r 2 h  h r 

300 . r 2

19. (a) We complete the table. First number

Second number

Product

1 2 3 4 5 6 7 8 9 10 11

18 17 16 15 14 13 12 11 10 9 8

18 34 48 60 70 78 84 88 90 90 88

(b) Let x be one number: then 19  x is the other number, and so the product, p, is p x  x 19  x  19x  x 2 .   (c) p x  19x  x 2   x 2  19x   2   2  19   x 2  19x  19 2 2   x  952  9025

So the product is maximized when the numbers are both 95.

From the table we conclude that the numbers is still increasing, the numbers whose product is a maximum should both be 95.

20. Let the positive numbers be x and y. Since their sum is 100, we have x  y  100  y  100  x. We wish to minimize

the sum of squares, which is S  x 2  y 2  x 2  100  x2 . So S x  x 2  100  x2  x 2  10,000  200x      x 2  2x 2  200x  10,000  2 x 2  100x  10,000  2 x 2  100x  2500  10,000  5000  2 x  502  5000.

Thus the minimum sum of squares occurs when x  50. Then y  100  50  50. Therefore both numbers are 50.

262

FOCUS ON MODELING

21. (a) Let x be the width of the field (in feet) and l be the length of the field (in feet). Since the farmer has 2400 ft of fencing we must have 2x  l  2400. 2000

200

400

200

Area=2000(200)=400,000 1000

1000

700

1000

700 Area=400(1000)=400,000

Area=1000(700)=700,000

Width

Length

Area

200

2000

400,000

300

1800

540,000

400

1600

640,000

500

1400

700,000

600

1200

720,000

700

1000

700,000

800

800

640,000

It appears that the field of largest area is about 600 ft  1200 ft.

(b) Let x be the width of the field (in feet) and l be the length of the field (in feet). Since the farmer has 2400 ft of fencing we must have 2x  l  2400  l  2400  2x. The area of the fenced-in field is given by   A x  l  x  2400  2x x  2x 2  2400x  2 x 2  1200x .     (c) The area is A x  2 x 2  1200x  6002  2 6002  2 x  6002  720,000. So the maximum area occurs when x  600 feet and l  2400  2 600  1200 feet.

22. (a) Let  be the width of the rectangular area (in feet) and l be the length of the field (in feet). Since the farmer has 750 feet of fencing, we must have 5  2l  750  2l  750  5  l  52 150  . Thus the total area of the four   pens is A   l    52  150     52 2  150 .       (b) We complete the square to get A    52 2  150   52 2  150  752  52  752   52   752  140625. Therefore, the largest possible total area of the four pens is 14,0625 square feet.

23. (a) Let x be the length of the fence along the road. If the area is 1200, we have 1200  x width, so the width of the garden   1200 7200 1200 . Then the cost of the fence is given by the function C x  5 x  3 x  2   8x  . is x x x (b) We graph the function y  C x in the viewing

(c) We graph the function y  C x and y  600 in

cost is minimized when x  30 ft. Then the

From this we get that the cost is at most $600

rectangle [0 75]  [0 800]. From this we get the width is 1200 30  40 ft. So the length is 30 ft and

the width is 40 ft.

the viewing rectangle [10 65]  [450 650].

when 15  x  60. So the range of lengths he can fence along the road is 15 feet to 60 feet.

600 500 500 0 0

50

20

40

60

Modeling with Functions

24. (a) Let x be the length of wire in cm that is bent into a square. So 10  x is the length of wire in 10  x x and , and the area cm that is bent into the second square. The width of each square is 4 4   2  x 2 x2 100  20x  x 2 10  x and . Thus the sum of the areas is of each square is   4 16 4 16

x2 100  20x  x 2 100  20x  2x 2    18 x 2  54 x  25 4 . 16 16 16     1 x 2  10x  25  1 x 2  10x  25  25  (b) We complete the square. A x  18 x 2  54 x  25  4 8 4 8 4 25  1 x  52  25 So the minimum area is 25 cm2 when each piece is 5 cm long. 8 8 8 8 A x 

25. (a) Let h be the height in feet of the straight portion of the window. The circumference of the semicircle is C  12 x. Since the perimeter of the window is 30 feet, we have x  2h  12 x  30. Solving for h, we get 2h  30  x  12 x  h  15  12 x  14 x. The area of the window is  2   A x  xh  12  12 x  x 15  12 x  14 x  18 x 2  15x  12 x 2  18 x 2 .   120 (b) A x  15x  18   4 x 2   18   4 x 2  x 4   2  2  450 120 450 60 60   x   18   4 x    18   4 x 2  4 4 4 4 4 The area is maximized when x 

60  840, and hence h  15  12 840  14  840  420. 4

26. (a) The height of the box is x, the width of the box is 12  2x, and the length of the

box is 20  2x. Therefore, the volume of the box is

(b) We graph the function y  V x in the viewing rectangle [0 6]  [200 270].

250

V x  x 12  2x 20  2x  4x 3  64x 2  240x, 0  x  6

200 0

(c) From the graph, the volume of the box with the largest volume is 262682 in3 when x  2427.

5

From the calculator we get that the volume of the box is greater than 200 in3 for 1174  x  3898 (accurate to 3 decimal places).

27. (a) Let x be the length of one side of the base and let h be the height of the box in feet. Since the volume of 12 the box is V  x 2 h  12, we have x 2 h  12  h  2 . The surface area, A, of the box is sum of the x area of the four sides and the area of the base. Thus the surface area of the box is given by the formula   12 48 A x  4xh  x 2  4x  x2   x 2 , x  0. 2 x x

263

264

FOCUS ON MODELING

(b) The function y  A x is shown in the first viewing rectangle below. In the second viewing rectangle, we isolate the minimum, and we see that the amount of material is minimized when x (the length and width) is 288 ft. Then the 12 height is h  2  144 ft. x 26

50

25 24

0 0

2

4

3.0

28. Let A, B, C, and D be the vertices of a rectangle with base AB on the x-axis and its other two vertices C and D above the x-axis and lying on the parabola y  8  x 2 . Let C have the coordinates x y, x  0. By symmetry, the coordinates of D must be x y. So the width of the rectangle is 2x, and the length is y  8  x 2 . Thus the area of the rectangle is   A x  length  width  2x 8  x 2  16x  2x 3 . The graphs of A x below show that the area is maximized when

x  163. Hence the maximum area occurs when the width is 326 and the length is 533. y

y=8-x@

D

A

C

x

18

10

17 16

0

x

B

20

0

2

4

1.5

2.0

29. (a) Let  be the width of the pen and l be the length in meters. We use the area to establish a relationship between 100  and l. Since the area is 100 m2 , we have l    100  l  . So the amount of fencing used is    200  22 100  2  . F  2l  2  2   (b) Using a graphing device, we first graph F in the viewing rectangle [0 40] by [0 100], and locate the approximate location of the minimum value. In the second viewing rectangle, [8 12] by [39 41], we see that the minimum value of F occurs when   10. Therefore the pen should be a square with side 10 m. 100

41

50

40 39

0 0

20

40

8

10

12

Modeling with Functions

30. (a) Let t1 represent the time, in hours, spent walking, and let t2 represent the time spent rowing. Since the distance walked is x and the walking

265

(b) We graph y  T x. Using the zoom

function, we see that T is minimized when x  613. He should land at a point

speed is 5 mi/h, the time spent walking is t1  15 x. By the Pythagorean Theorem, the distance rowed is   d  22  7  x2  x 2  14x  53, and so the time spent  rowing is t2  12  x 2  14x  53. Thus the total time is  T x  12 x 2  14x  53  15 x.

613 miles from point B. 4 2 0 0

31. (a) Let x be the distance from point B to C, in miles. Then the distance from A to C is  flying from A to C then C to D is f x  14 x 2  25  10 12  x.



2

4

6

x 2  25, and the energy used in

(b) By using a graphing device, the energy expenditure is minimized when the distance from B to C is about 51 miles. 200

169.1

100

169.0 168.9

0 0

5

10

5.0

5.1

5.2

 32. (a) Using the Pythagorean Theorem, we have that the height of the upper triangles is 25  x 2 and the height of the lower   triangles is 144  x 2 . So the area of the each of the upper triangles is 12 x 25  x 2 , and the area of the each of the  lower triangles is 12 x 144  x 2 . Since there are two upper triangles and two lower triangles, we get that the total area          25  x 2  144  x 2 . is A x  2  12 x 25  x 2  2  12 x 144  x 2  x    (b) The function y  A x  x 25  x 2  144  x 2 is shown in the first viewing rectangle below. In the second viewing rectangle, we isolate the maximum, and we see that the area of the kite is maximized when x  4615. So the length of the horizontal crosspiece must be 2  4615  923. The length of the vertical crosspiece is   52  46152  122  46152  1300. 100

60.1

50

60.0

0 0

2

4

59.9 4.60

4.62

4.64

3

POLYNOMIAL AND RATIONAL FUNCTIONS

3.1

QUADRATIC FUNCTIONS AND MODELS

1. To put the quadratic function f x  ax 2  bx  c in standard form we complete the square. 2. The quadratic function f x  a x  h2  k is in standard form. (a) The graph of f is a parabola with vertex h k. (b) If a  0 the graph of f opens upward. In this case f h  k is the minimum value of f .

(c) If a  0 the graph of f opens downward. In this case f h  k is the maximum value of f .

3. The graph of f x  3 x  22  6 is a parabola that opens upward, with its vertex at 2 6, and f 2  6 is the minimum value of f . 4. The graph of f x  3 x  22  6 is a parabola that opens downward, with its vertex at 2 6, and f 2  6 is the maximum value of f . 5. (a) The vertex is 3 4, the x-intercepts are 1 and 5, and it appears that the y-intercept is approximately 5.

6. (a) The vertex is 2 8, the x-intercepts are 6 and 2, and the y-intercept is 6.

(b) Maximum value of f : 4

(b) Maximum value of f : 8

(c) Domain  , range:  4]

(c) Domain:  , range:  8]

7. (a) The vertex is 1 3, the x-intercepts are

approximately 02 and 22, and the y-intercept is

8. (a) The vertex is 1 4, the x-intercepts are

1.

approximately 22 and 02, and the y-intercept is 1.

(b) Minimum value of f : 3

(b) Minimum value of f : 4

(c) Domain:  , range: [3 

(c) Domain:  , range: [4 

9. (a) f x  x 2  2x  3  x  12  1  3  x  12  2

(c)

y

(b) The vertex is at 1 2. x-intercepts: y  0  0  x  12  2  x  12  2. This has

no real solution, so there is no x-intercept.

y-intercept: x  0  y  0  12  2  3. The y-intercept is 3. (d) Domain:  , range: [2 

1 1

x

267

268

CHAPTER 3 Polynomial and Rational Functions

10. (a) f x  x 2  4x  1  x  22  4  1  x  22  5

y

(c)

(b) The vertex is at 2 5.

x-intercepts: y  0  0  x  22  5  x  22  5      x  2   5  x  2  5. So x  2  5 or x  2  5.   The x-intercepts are 2  5 and 2  5.

1 1

y-intercept: x  0  y  0  22  5  1. The y-intercept is 1.

x

(d) Domain:  , range: [5 

11. (a) f x  x 2  6x  x 2  6x  x 2  6x  9  9  x  32  9

(c)

(b) The vertex is at 3 9.

y

1

x-intercepts: y  0  0  x 2  6x  x x  6. So x  0 or x  6.

x

1

The x-intercepts are 0 and 6.

y-intercept: x  0  y  0. The y-intercept is 0. (d) Domain:  , range: [9 

12. (a) f x  x 2  8x  x  42  16

y

(c)

(b) The vertex is at 4 16.

x-intercepts: y  0  0  x 2  8x  x x  8. So x  0 or

x  8. The x-intercepts are 0 and 8.

3

y-intercept: x  0  y  0. The y-intercept is 0.

1

x

1

x

(d) Domain:  , range: [16 

  13. (a) f x  3x 2  6x  3 x 2  2x  3 x  12  3

(c)

y

(b) The vertex is at 1 3.

x-intercepts: y  0  0  3 x  12  3  x  12  1  x  2 or 0. The x-intercepts are 2 and 0.

y-intercept: x  0  y  3 02  6 0  0. The y-intercept is 0.

(d) Domain:  , range: [3 

1

SECTION 3.1 Quadratic Functions and Models





14. (a) f x  x 2  10x   x 2  10x   x  52  25 (b) The vertex is at 5 25.

(c)

x-intercepts: y  0  0   x  52  25  x  52  25 

269

y

10

x  0 or 10. The x-intercepts are 0 and 10.

y-intercept: x  0  y  02  10 0  0. The y-intercept is 0.

x

1

(d) Domain:  , range:  25]

15. (a) f x  x 2  4x  3  x  22  1

y

(c)

(b) The vertex is at 2 1.

x-intercepts: y  0  0  x 2  4x  3  x  1 x  3. So x  1 or x  3. The x-intercepts are 1 and 3.

1

y-intercept: x  0  y  3. The y-intercept is 3.

1

x

1

x

(d) Domain:  , range: [1 

16. (a) f x  x 2  2x  2  x  12  1

y

(c)

(b) The vertex is at 1 1. x-intercepts: y  0  x  12  1  0  x  12  1. Since

this last equation has no real solution, there is no x-intercept.

1

y-intercept: x  0  y  2. The y-intercept is 2. (d) Domain:  , range: [1 

17. (a) f x  x 2  6x  4   x  32  13

(c)

y

(b) The vertex is at 3 13. x-intercepts: y  0  0   x  32  13  x  32  13     x  3   13  x  3  13. The x-intercepts are 3  13 and  3  13. y-intercept: x  0  y  4. The y-intercept is 4. (d) Domain:  , range:  13]

2 1

x

270

CHAPTER 3 Polynomial and Rational Functions

18. (a) f x  x 2  4x  4   x  22  8

y

(c)

(b) The vertex is at 2 8. x-intercepts: y  0  0  x 2  4x  4  0  x 2  4x  4.

Using the Quadratic Formula,

2

     2 22 2  4 42 414 4 32    2  2 2. x 21 2 2

1

x

  The x-intercepts are 2  2 2 and 2  2 2.

y-intercept: x  0  y  4. The y-intercept is 4.

(d) Domain:  , range:  8]

19. (a) f x  2x 2  4x  3  2 x  12  1

y

(c)

(b) The vertex is at 1 1. x-intercepts: y  0  0  2x 2  4x  3  2 x  12  1 

2 x  12  1. Since this last equation has no real solution, there is

no x-intercept.

y-intercept: x  0  y  3. The y-intercept is 3.

1

(d) Domain:  , range: [1 

20. (a) f x  3x 2  6x  2  3 x  12  1

1

x

y

(c)

(b) The vertex is at 1 1. x-intercepts: y  0  0  3 x  12  1  0  x  12  13     x  1   13  x  1  13 . The x-intercepts are 1  13 and  1  13 .

2 1

x

y-intercept: x  0  y  2. The y-intercept is 2.

(d) Domain:  , range:  1]

21. (a) f x  2x 2  20x  57  2 x  52  7

(c)

y

(b) The vertex is at 5 7. x-intercepts: y  0  0  2x 2  20x  57  2 x  52  7 

2 x  52  7. Since this last equation has no real solution, there is

no x-intercept.

y-intercept: x  0  y  57. The y-intercept is 57. (d) Domain:  , range: [7 

2 1

x

SECTION 3.1 Quadratic Functions and Models

 22. (a) f x  2x 2  12x  10  2 x 2  6x  9  18  10  2 x  32  8



y

(c)

(b) The vertex is at 3 8. The x-intercepts are 5 and 1 and the

2

y-intercept is 10.

0

(d) Domain:  , range: [8 

  23. (a) f x  4x 2  12x  1  4 x 2  3  1 2 2    4 x  32  9  1  4 x  32  10   (b) The vertex is at  32  10 .

1

x

y

(c)

10

 2 2  x-intercepts: y  0  0  4 x  32  10  x  32  52     x  32   52  x   32  52   32  210 . The x-intercepts are

 32 

271

5

1

  10 10 3 2 and  2  2 .

_3

_2

x

_1

y-intercept: x  0  y  4 02  12 0  1  1. The y-intercept

is 1.

(d) Domain:  , range:  10]

  24. (a) f x  3x 2  2x  2  3 x 2  23 x  2 2 2    3 x  13  13  2  3 x  13  73   (b) The vertex is at  13   73 . 

x-intercepts: y  0  0  3 x  13 



2



 73  x  13

(c)

2

10

 79  

x  13   37  x   13  37 . The x-intercepts are  13  37 and

  13  37 .

y-intercept: x  0  y  3 02  2 0  2  2. The y-intercept

is 2.

  (d) Domain:  , range:  73  

y 20

1

x

272

CHAPTER 3 Polynomial and Rational Functions

  25. (a) f x  x 2  2x  1  x 2  2x  1    x 2  2x  1  1  1  x  12  2 y

(b)

  26. (a) f x  x 2  8x  8  x 2  8x  16  8  16  x  42  8

(b)

2

1 1

 3 x  12  2 y

x

1

x

(c) The minimum value is f 1  2.   27. (a) f x  3x 2  6x  1  3 x 2  2x  1    3 x 2  2x  1  1  3 (b)

y

(c) The minimum value is f 4  8.   28. (a) f x  5x 2  30x  4  5 x 2  6x  4    5 x 2  6x  9  4  45  5 x  32  41

(b)

y 10 1

x

1 1

x

(c) The minimum value is f 1  2.   29. (a) f x  x 2  3x  3   x 2  3x  3     x 2  3x  94  3  94 2    x  32  21 4 (b)

y

(c) The minimum value is f 3  41.

  30. (a) f x  1  6x  x 2   x 2  6x  1     x 2  6x  9  1  9   x  32  10

(b)

y

2

1 1

  (c) The maximum value is f  32  21 4.

x

1

(c) The maximum value is f 3  10.

x

SECTION 3.1 Quadratic Functions and Models





31. (a) g x  3x 2  12x  13  3 x 2  4x  13    3 x 2  4x  4  13  12  3 x  22  1

(b)

y





32. (a) g x  2x 2  8x  11  2 x 2  4x  11    2 x 2  4x  4  11  8  2 x  22  3

y

(b)

2 x

1

2 1

(c) The minimum value is g 2  1. 

(c) The minimum value is g 2  3.



  34. (a) h x  3  4x  4x 2  4 x 2  x  3    4 x 2  x  14  3  1 2   4 x  12  4

33. (a) h x  1  x  x 2   x 2  x  1     x 2  x  14  1  14 2    x  12  54 (b)

x

y

(b)

y

1

1 1

x

1

    (c) The maximum value is h  12  54 . (c) The maximum value is h  12  4.       35. f x  2x 2  4x  1  2x 2  4x  1  2 x 2  2x  1  2 x 2  2x  1  2  1  2 x  12  3.

Therefore, the minimum value is f 1  3.     36. f x  3  4x  x 2  x 2  4x  3   x 2  4x  3   x 2  4x  4  4  3   x  22  7. Therefore, the maximum value is f 2  7.

  37. f t  3  80t  20t 2  20t 2  80t  3  20 t 2  4t  4  80  3  20 t  22  77. Therefore, the maximum value is f 2  77.   38. f x  6x 2  24x  100  6 x 2  4x  4  24  100  6 x  22  124.

Therefore, the minimum value is f 2  124.     39. f s  s 2  12s  16  s 2  12s  16  s 2  12s  036  16  036  s  062  1564. Therefore, the minimum value is f 06  1564.

x

273

274

CHAPTER 3 Polynomial and Rational Functions

  2     5625  10 x  15 40. g x  100x 2  1500x  100 x 2  15x  100 x 2  15x  225  5625. 4 2   Therefore, the minimum value is g 15 2  5625.     41. h x  12 x 2  2x  6  12 x 2  4x  6  12 x 2  4x  4  6  2  12 x  22  8. Therefore, the minimum value is h 2  8.

    x2  2x  7   13 x 2  6x  7   13 x 2  6x  9  7  3   13 x  32  10. 3 Therefore, the maximum value is f 3  10.     43. f x  3  x  12 x 2   12 x 2  2x  3   12 x 2  2x  1  3  12   12 x  1  72 . Therefore, the maximum

42. f x  

value is f 1  72 .

    44. g x  2x x  4  7  2x 2  8x  7  2 x 2  4x  7  2 x 2  4x  4  7  8  2 x  22  1. Therefore, the minimum value is g 2  1.

45. (a) The graph of f x  x 2  179x  321 is shown. The minimum value is f x  401. -1.0

-0.9

(b) f x  x 2  179x  321   2  2   321  179  x 2  179x  179 2 2  x  08952  4011025

-0.8 -3.9

Therefore, the exact minimum of f x is 4011025.

-4.0 -4.1

46. (a) The graph of f x  1  x 

 2 2x is

shown. The maximum value is f x  118. 1.180



Therefore, the exact maximum of f x is 88 2 .

1.175 1.170 0.30

     (b) f x  1  x  2 x 2   2 x 2  22 x  1      2    1  82   2 x 2  22 x  42  2      2 x  42  88 2

0.35

0.40

47. The vertex is 2 3, so the parabola has equation y  a x  22  3. Substituting the point 3 1, we have 1  a 3  22  3  a  4, so f x  4 x  22  3.

48. The vertex is 1 5, so the parabola has equation y  a x  12  5. Substituting the point 3 7, we have

7  a 3  12  5  a  3, so f x  3 x  12  5.   49. Substituting t  x 2 , we have f t  3  4t  t 2   t 2  4t  4  4  3   t  22  7. Thus, the maximum value  is 7, when t  2 (or x  2).   50. Substituting t  x 3 , we have f t  2  16t  4t 2  4 t 2  4t  4  16  2  4 t  22  14. Thus, the minimum  value is 14, when t  2 (or x   3 2).

SECTION 3.1 Quadratic Functions and Models

275

  2  2   2    16 54  16 t  54  25. Thus the maximum 51. y  f t  40t  16t 2  16 t 2  52  16 t 2  52 t  54   height attained by the ball is f 54  25 feet.         32 x 2  x  5   2 x 2  25 x  5   2 x 2  25 x  25 2  2 25 2  5 52. (a) We complete the square: y   400 25 2 25 2 4 25 4   2 x  25 2  65 , so the maximum height of the ball is 65  8125 ft.  y   25 4 8 8

2 x 2  x  5. Using the Quadratic (b) The ball hits the ground when its vertical displacement y is 0, that is, when 0   25    2 5  1  1  4  25 25  5 65  . Taking the positive root, we find that x  163 ft. Formula, we find x  4 4  25     53. R x  80x  04x 2  04 x 2  200x  04 x 2  200x  10,000  4,000  04 x  1002  4,000. So

revenue is maximized at $4,000 when 100 units are sold.     54. P x  0001x 2  3x  1800  0001 x 2  3000x  1800  0001 x 2  3000x  2 250 000  1800 

2250  0001 x  15002  450. The vendor’s maximum profit occurs when he sells 1500 cans and the profit is $450.     1 n 2   1 n 2  60n   1 n 2  60n  900  10   1 n  302  10. Since the maximum of 55. E n  23 n  90 90 90 90 the function occurs when n  30, the viewer should watch the commercial 30 times for maximum effectiveness.     56. C t  006t  00002t 2  00002 t 2  300t  00002 t 2  300t  22,500  45  00002 t  1502  45. The maximum concentration of 45 mg/L occurs after 150 minutes.

57. A n  n 900  9n  9n 2  900n is a quadratic function with a  9 and b  900, so by the formula, the maximum 900 b   50 trees, and because a  0, this gives a maximum value. or minimum value occurs at n   2a 2 9 58. A n  700  n 10  001n  001n 2  10n  001 700 n  7000  001n 2  3n  7000. This is a quadratic b 3 function with a  001 and b  3, so the maximum (a  0) occurs at x     150. Since n  150 2a 2 001 is the number of additional vines that should be planted, the total number of vines that maximizes grape production is 700  150  850 vines. 59. The area of the fenced-in field is given by A x  2400  2x x  2x 2  2400x. Thus, by the formula in this section, 2400 b   600. The maximum area occurs when x  600 feet the maximum or minimum value occurs at x   2a 2 2 and l  2400  2 600  1200 feet. 60. The total area of the four pens is A   52  150     52 2  375. Thus, by the formula, the maximum or 375 b      75. Therefore, the largest possible total area of the four pens is minimum value occurs at    2a 2  52   A 75   52 752  375 75  14,0625 square feet.

15 b   84 ft 61. A x  15x  18   4 x 2 , so by the formula, the maximum area occurs when x     2a 2 18   4 and h  15  12 840  14  840  42 ft.

 54 b    5. The   62. A x  18 x 2  54 x  25 , so by the formula, the maximum or minimum area occurs where x   4 2a 2 18   25 2 minimum area is 18 52  54 5  25 4  8 cm when each piece is 5 cm long.

276

CHAPTER 3 Polynomial and Rational Functions

63. (a) The area of the corral is A x  x 1200  x  1200x  x 2  x 2  1200x.

(b) A is a quadratic function with a  1 and b  1200, so by the formula, it has a maximum or minimum at 1200 b   600, and because a  0, this gives a maximum value. The desired dimensions are 600 ft by x  2a 2 1 600 ft.

64. (a) The dimensions of the gutter are x inches and 30  x  x  30  2x inches, so the cross sectional area is A  x 30  2x  30x  2x 2 .

(b) Since A is a quadratic function with a  2 and b  30, the maximum occurs at x  

b 30   75 inches. 2a 2 2

(c) The maximum cross section is A 75  2 752  30 75  1125 in2 .

65. (a) To model the revenue, we need to find the total attendance. Let x be the ticket price. Then the amount by which the ticket price is lowered is 10  x, and we are given that for every dollar it is lowered, the attendance increases by 3000; that is, the increase in attendance is 3000 10  x. Thus, the attendance is 27,000  3000 10  x, and since each spectator pays $x, the revenue is R x  x [27,000  3000 10  x]  3000x 2  57,000x.

(b) Since R is a quadratic function with a  3000 and b  57,000, the maximum occurs at 57,000 b   95; that is, when admission is $950. x  2a 2 3000 (c) We solve R x  0 for x: 3000x 2  57,000x  0  3000x x  19  0  x  0 or x  19. Thus, if admission is $19, nobody will attend and no revenue will be generated. 66. (a) Let x be the price per feeder. Then the amount by which the price is increased is x  10, and we are given that for every dollar increase, sales decrease by 2; that is, the change in sales is 2 x  10, so the total number sold is 20  2 x  10  40  2x. The profit per feeder is equal to the sale price minus the cost, that is, x  6. Multiplying the number of feeders sold by the profit per feeder sold, we find the profit to be P x  40  2x x  6  2x 2  52x  240.

(b) Using the formula, profit is maximized when x  

b 52   13; that is, when the society charges $13 per 2a 2 2

feeder. The maximum weekly profit is P 13  2 132  52 13  240  $98.

y

67. Because f x  x  m x  n  0 when x  m or x  n, those are its x-intercepts. By symmetry, we expect that the vertex is halfway between mn these values; that is, at x  . We obtain the graph shown at right. 2

Expanding, we see that f x  x 2  m  n x  mn, a quadratic

function with a  1 and b   m  n. Because a  0, the minimum value occurs at x  

3.2

y=(x-a)(x-b)

a

0

a+b 2

b

x

b m n  , the x-value of the vertex, as expected. 2a 2

POLYNOMIAL FUNCTIONS AND THEIR GRAPHS

1. Graph I cannot be that of a polynomial because it is not smooth (it has a cusp.) Graph II could be that of a polynomial function, because it is smooth and continuous. Graph III could not be that of a polynomial function because it has a break. Graph IV could not be that of a polynomial function because it is not smooth. 2. (a) y  x 3  8x 2  2x  15 has odd degree and a positive leading coefficient, so y   as x   and y   as x  .

SECTION 3.2 Polynomial Functions and Their Graphs

277

(b) y  2x 4  12x  100 has even degree and a negative leading coefficient, so y   as x   and y   as x  . 3. (a) If c is a zero of the polynomial P, then P c  0.

(b) If c is a zero of the polynomial P, then x  c is a factor of P x.

(c) If c is a zero of the polynomial P, then c is an x-intercept of the graph of P.

4. (a) This is impossible. If P has degree n  3, it has at most n  1  2 local extrema.

(b) This is possible. For example, y  x 3 has degree 3 and no local maxima or minima.

(c) This is possible. For example, P x  x 4 has one local maximum and no local minima.

5. (a) P x  x 2  4

(b) Q x  x  42

y

y

1 1

(_2, 0)

x

(2, 0)

4

(0, _4)

(c) P x  2x 2  3

(0, 16)

1

(d) P x   x  22

y

x

(4, 0)

(_2, 0)

y 1 0

1

x

(0, _4)

1 0

6. (a) P x  x 4  16

(0, 3) x

1

(b) P x   x  54

y

y 100 0

4 (_2, 0)

1

(2, 0)

x

(0, _625) (0, _16)

1

x

278

CHAPTER 3 Polynomial and Rational Functions

(c) P x  5x 4  5

(d) P x  x  54

y

y

(0, 5)

1

(_1, 0)

(1, 0)

0

1

(0, 625)

x

100 0

7. (a) P x  x 3  8

(5, 0) x

1

(b) Q x  x 3  27

y

y

1 1

(2, 0)

x (0, 27)

(0, _8)

4

(3, 0)

x

1

(c) R x   x  23

(_2, 0)

(d) S x  12 x  13  4

y

y

1 x

1

(0, _72) (_1, 0)

(0, _8)

8. (a) P x  x  35

1

(b) Q x  2 x  35  64

y

y (0, 422)

(0, 243)

(_3, 0)

x

1

(_1, 0)

50 1x

100 1x

SECTION 3.2 Polynomial Functions and Their Graphs

(c) R x   12 x  25

279

(d) S x   12 x  25  16

y

y (0, 16) (0, 32)

2

(2, 0) 1

x

4

(4, 0)

1

x

  9. (a) P x  x x 2  4  x 3  4x has odd degree and a positive leading coefficient, so y   as x   and y   as x  .

(b) This corresponds to graph III.   10. (a) Q x  x 2 x 2  4  x 4  4x 2 has even degree and a negative leading coefficient, so y   as x   and y   as x  .

(b) This corresponds to graph I. 11. (a) R x  x 5  5x 3  4x has odd degree and a negative leading coefficient, so y   as x   and y   as x  . (b) This corresponds to graph V.

12. (a) S x  12 x 6  2x 4 has even degree and a positive leading coefficient, so y   as x   and y   as x  . (b) This corresponds to graph II.

13. (a) T x  x 4  2x 3 has even degree and a positive leading coefficient, so y   as x   and y   as x  . (b) This corresponds to graph VI.

14. (a) U x  x 3  2x 2 has odd degree and a negative leading coefficient, so y   as x   and y   as x  . (b) This corresponds to graph IV.

15. P x  x  1 x  2

16. P x  2  x x  5

y

1 (_2, 0)

1

(1, 0)

(0, _2)

y 20

10 (0, 10) x (_5, 0)

1

(2, 0)

x

280

CHAPTER 3 Polynomial and Rational Functions

17. P x  x x  3 x  2

18. P x  x x  3 x  2

y

y

1

(_2, 0) (_2, 0) 1

1

1

(3, 0)

(0, 0)

x

(3, 0)

x

20. P x  x  3 x  2 3x  2

y

y

(0, 12)

(0, 3) 1

(_1, 0)

(3, 0)

x

19. P x   2x  1 x  1 x  3

(_3, 0)

(0, 0)

(_2, 0)

1

4

1

(1/2, 0) x

21. P x  x  2 x  1 x  2 x  3

( _23 , 0)

22. P x  x x  1 x  1 2  x y

y

(_1, 0)

1 0

(0, 0) 1 (1, 0)

(2, 0)

x

(0, 12) 5

(2, 0)

0

(_2, 0)

(3, 0) x

1

(_1, 0)

24. P x  15 x x  52

23. P x  2x x  22 y

y

1 (0, 0)

(2, 0) 1

2 (0, 0) x

1

(5, 0)

x

SECTION 3.2 Polynomial Functions and Their Graphs

25. P x  x  1 x  12 2x  3

26. P x   x  12 x  13 x  2

y

y

1

(_1, 0) (_1, 0) 2 0 (_2, 0)

(1, 0)

0

(2, 0)

1

x

(3/2, 0) 1

x

(0, _2)

(0, _6)

1 x  22 x  32 27. P x  12

28. P x  x  12 x  23

y

y

(0, 8) 2

1

(0, 3)

x

(1, 0)

(_2, 0)

1 1

(_2, 0)

x

(3, 0)

29. P x  x 3 x  2 x  32

30. P x  x  32 x  12

y

10

y

1 (0, 0)

(_2, 0)

(3, 0)

x

(0, 9) 2 1

(_1, 0)

31. P x  x 3  x 2  6x  x x  2 x  3

2 (0, 0) 1

y

4

(_4, 0) (3, 0)

x

32. P x  x 3  2x 2  8x  x x  2 x  4

y

(_2, 0)

(3, 0)

x

(0, 0)

1

(2, 0)

x

281

282

CHAPTER 3 Polynomial and Rational Functions

33. P x  x 3  x 2  12x  x x  3 x  4 y

  34. P x  2x 3  x 2  x  x 2x 2  x  1  x 2x  1 x  1 y

4 (0, 0) 1

(_3, 0)

(4, 0)

1 x

( _21 , 0)

(0, 0)

1

(_1, 0)

35. P x  x 4  3x 3  2x 2  x 2 x  1 x  2 y

1

36. P x  x 5  9x 3  x 3 x  3 x  3 y

10

(_3, 0)

(3, 0)

(0, 0) (1, 0) 1

(0, 0)

(2, 0)

y

y

(_2, 0)

(1, 0) 1

(0, _1)

x

38. P x  x 3  3x 2  4x  12  x  3 x  2 x  2

(_3, 0) (_1, 0)

1

x

37. P x  x 3  x 2  x  1  x  1 x  12

1

x

2

(2, 0) 1

x

(0, _12)

x

SECTION 3.2 Polynomial Functions and Their Graphs

 2 40. P x  18 2x 4  3x 3  16x  24  2  18 x  22 2x  32 x 2  2x  4

39. P x  2x 3  x 2  18x  9  x  3 2x  1 x  3 y

y

40 (0, 9)

(_3, 0)

1

( _21 , 0)

(3, 0)

x (0, 72) 20 1

(_ _32 , 0)

41. P x  x 4  2x 3  8x  16    x  22 x 2  2x  4

x

42. P x  x 4  2x 3  8x  16    x  2 x  2 x 2  2x  4

y

y

(_2, 0)

10

(2, 0)

10 (2, 0)

x

1 (0, _16)

(0, 16) x

1 (2, 0)

  43. P x  x 4  3x 2  4  x  2 x  2 x 2  1 y

44. P x  x 6  2x 3    2 2 1  x 3  1  x  12 x 2  x  1 y

(_2, 0)

1

(2, 0) 1

x

(0, _4) 1

(0, 1)

1 (1, 0)

x

283

284

CHAPTER 3 Polynomial and Rational Functions

45. P x  3x 3  x 2  5x  1; Q x  3x 3 . Since P has odd degree and positive leading coefficient, it has the following end behavior: y   as x   and y   as x  . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, we see that the graphs of P and Q have different intercepts. 100

10

50 -3

-2

-1

P 1

5

Q

-50

2

3

-1.5 -1

-0.5 -5

-100

-10

P

Q

0.5

1

1.5

46. P x   18 x 3  14 x 2  12x; Q x   18 x 3 . Since P has odd degree and negative leading coefficient, it has the following end behavior: y   as x   and y   as x  . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different and seem (wrongly) to have different end behavior. 60 40

1000 P

Q

Q

500 -6

-30 -20 -10 -500

10

20

P

20

-4

-2

30

0 -20

2

4

6

-40 -60

-1000

-80

47. P x  x 4  7x 2  5x  5; Q x  x 4 . Since P has even degree and positive leading coefficient, it has the following end behavior: y   as x   and y   as x  . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different and we see that they have different intercepts. 600 10

500 Q

400

P

300

P

Q

-3

200

-2

-1

-4

-2 0 -100

1

2

3

-10

100 -6

0

2

4

6

-20

48. P x  x 5  2x 2  x; Q x  x 5 . Since P has odd degree and negative leading coefficient, it has the following end behavior: y   as x   and y   as x  . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different. 3

200 Q

P

-3

-2

Q

100 -1 0 -100 -200

1

2

3

P

-1

2 1 0 -1 -2 -3

1

SECTION 3.2 Polynomial Functions and Their Graphs

285

49. P x  x 11  9x 9 ; Q x  x 11 . Since P has odd degree and positive leading coefficient, it has the following end behavior: y   as x   and y   as x  . On a large viewing rectangle, the graphs of P and Q look like they have the same end behavior. On a small viewing rectangle, the graphs of P and Q look very different and seem (wrongly) to have different end behavior. 600000 400000

Q

P

100

Q

-6

-4

-2 0 -200000

P

50

200000 2

4

6

0

-1 P

-400000

1

-50

Q

-100

-600000

50. P x  2x 2  x 12 ; Q x  x 12 . Since P has even degree and negative leading coefficient, it has the following end behavior: y   as x   and y   as x  . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different. 200 -3

-2

-1 0 -200

1

2

2

3

-400

P Q

-800

0

-1

-600 P

Q

51. (a) x-intercepts at 0 and 4, y-intercept at 0. (b) Local maximum at 2 4, no local minimum.

53. (a) x-intercepts at 2 and 1, y-intercept at 1. (b) Local maximum at 1 0, local minimum at 1 2.

55. y  x 2  8x, [4 12] by [50 30]

No local minimum. Local maximum at 4 16. Domain:  , range:  16].

-4

52. (a) x-intercepts at 0 and 45, y-intercept at 0. (b) Local maximum at 0 0, local minimum at 3 3.

54. (a) x-intercepts at 0 and 4, y-intercept at 0. (b) No local maximum, local minimum at 3 3.

56. y  x 3  3x 2 , [2 5] by [10 10]

Local minimum at 2 4. Local maximum at 0 0. Domain:  , range:  . 10

20

-40

-2

-1000 -1200

-4 -2 -20

1

2 4 6 8 10 12 -2

2 -10

4

286

CHAPTER 3 Polynomial and Rational Functions

57. y  x 3  12x  9, [5 5] by [30 30]

Local maximum at 2 25. Local minimum at 2 7. Domain:  , range:  .

58. y  2x 3  3x 2  12x  32, [5 5] by [60 30] Local minimum at 2 52. Local maximum at 1 25.

Domain:  , range:  .

20 20 -4

-2

2

4

-4

-2 -20

-20

2

4

-40 -60

59. y  x 4  4x 3 , [5 5] by [30 30]

Local minimum at 3 27. No local maximum.

Domain:  , range: [27 .

60. y  x 4  18x 2  32, [5 5] by [100 100]

Local minima at 3 49 and 3 49. Local maximum at 0 32.

Domain:  , range: [49 .

20

-4

-2

100 2

4 -4

-20

-2

2

4

-100

61. y  3x 5  5x 3  3, [3 3] by [5 10]

Local maximum at 1 5. Local minimum at 1 1. Domain:  , range:  .

62. y  x 5  5x 2  6, [3 3] by [5 10]

Local minimum at 126 124. Local maximum at 0 6. Domain:  , range:  .

10

10

5

-3

-2

-1

1

2

3

-3

-2

-1

1

2

3

-5

63. y  2x 2  3x  5 has one local maximum at 075 613.

64. y  x 3  12x has no local maximum or minimum. 20

10

10

5

-4

-2

-4 2

-5

4

-2 -10 -20

2

4

SECTION 3.2 Polynomial Functions and Their Graphs

287

65. y  x 3  x 2  x has one local maximum at 033 019 66. y  6x 3  3x  1 has no local maximum or minimum. and one local minimum at 100 100.

20 10

2 -3 -3

-2

-1

1

2

-2

3

-1 -10

67. y  x 4  5x 2  4 has one local maximum at 0 4 and two local minima at 158 225 and 158 225. 10

3

68. y  12x 5  375x 4  7x 3  15x 2  18x has two local

maxima at 050 465 and 297 1210 and two local minima at 140 2744 and 140 254. 20

5

-2

2

-20

-2

-3

1

-1

1

2

3

-4

-2

2

-5

-20

69. y  x  25  32 has no maximum or minimum.

 3 70. y  x 2  2 has one local minimum at 0 8.

60

10

40 20 -3 -2

2

4

-2

6

-1

1

2

3

-10

71. y  x 8  3x 4  x has one local maximum at 044 033 and two local minima at 109 115 and 112 336.

72. y  13 x 7  17x 2  7 has one local maximum at 0 7 and one local minimum at 171 2846. 20

4 2 -3

-2

-1 -2 -4

-3 1

2

3

-2

-1 -20

1

2

3

288

CHAPTER 3 Polynomial and Rational Functions

73. y  cx 3 ; c  1, 2, 5, 12 . Increasing the value of c stretches 74. P x  x  c4 ; c  1, 0, 1, 2. Increasing the value of c shifts the graph to the right.

the graph vertically. c=5

100

c=0

c=1

c=_1

-4

-2

2

c=2

1

4

0

-2

1 c=_ 2

2

-1

-100

c=2

c=1

2

75. P x  x 4  c; c  1, 0, 1, and 2. Increasing the value of c moves the graph up.

76. P x  x 3  cx; c  2, 0, 2, 4. Increasing the value of c makes the “bumps” in the graph flatter.

c=2 c=1 c=0 c=_1

6 4

c=0 10

c=_2

c=2

c=_4

2 0

-2

0

-1

2

1

-2 -10

77. P x  x 4  cx; c  0, 1, 8, and 27. Increasing the value 78. P x  x c ; c  1, 3, 5, 7. The larger c gets, the flatter the of c causes a deeper dip in the graph, in the fourth

graph is near the origin, and the steeper it is away from the

quadrant, and moves the positive x-intercept to the right.

origin.

c=1

10

c=5

c=3

c=27

20

c=0

c=7

c=1 -4

-2

0

2 c=8

4

-2

-1

0

1

2

-20

-10 -40

79. (a)

y

y=x#-2x@-x+2 y=_x@+5x+2

5 1

x

(b) The two graphs appear to intersect at 3 points. (c) x 3  2x 2  x  2  x 2  5x  2  x 3  x 2  6x  0    x x 2  x  6  0  x x  3 x  2  0. Then either x  0, x  3, or x  2. If x  0, then y  2; if x  3 then

y  8 if x  2, then y  12. Hence the points where the

two graphs intersect are 0 2, 3 8, and 2 12.

80. Graph 1 belongs to y  x 4 . Graph 2 belongs to y  x 2 . Graph 3 belongs to y  x 6 . Graph 4 belongs to y  x 3 . Graph 5 belongs to y  x 5 .

SECTION 3.2 Polynomial Functions and Their Graphs

289

81. (a) Let P x be a polynomial containing only odd powers of x. Then each term of P x can be written as C x 2n1 , for some constant C and integer n. Since C x2n1  C x 2n1 , each term of P x is an odd function. Thus by part (a), P x is an odd function. (b) Let P x be a polynomial containing only even powers of x. Then each term of P x can be written as C x 2n , for some constant C and integer n. Since C x2n  C x 2n , each term of P x is an even function. Thus by part (b), P x is an even function. (c) Since P x contains both even and odd powers of x, we can write it in the form P x  R x  Q x, where R x contains all the even-powered terms in P x and Q x contains all the odd-powered terms. By part (d), Q x is an odd function, and by part (e), R x is an even function. Thus, since neither Q x nor R x are constantly 0 (by assumption), by part (c), P x  R x  Q x is neither even nor odd.     (d) P x  x 5  6x 3  x 2  2x  5  x 5  6x 3  2x  x 2  5  PO x  PE x where PO x  x 5  6x 3  2x and PE x  x 2  5. Since PO x contains only odd powers of x, it is an odd function, and since PE x contains only even powers of x, it is an even function.

82. (a) From the graph, P x  x 3  4x  x x  2 x  2 has three x-intercepts, one local maximum, and one local minimum.

  (b) From the graph, Q x  x 3  4x  x x 2  4 has one x-intercept and no local maximum or minimum.

10

-5

10

5

-5

-10

5 -10

  (c) For the x-intercepts of P x  x 3  ax, we solve x 3  ax  0. Then we have x x 2  a  0  x  0 or x 2  a. If         x 2  a, then x   a. So P has 3 x-intercepts. Since P x  x x 2  a  x x  a x  a , by part (c) of problem 67, P has 2 local extrema. For the x-intercepts of Q x  x 3  ax, we solve x 3  ax  0. Then we   have x x 2  a  0  x  0 or x 2  a. The equation x 2  a has no real solutions because a  0. So Q has 1

x-intercept. We now show that Q is always increasing and hence has no extrema. Ifx1  x2 , then ax1  ax2 (because

a  0) and x13  x23 . So we have x13  ax1  x23  ax2 , and hence Q x1   Q x2 . Thus Q is increasing, that is, its graph always rises, and so it has no local extrema. 83. (a) P x  x  1 x  3 x  4. Local maximum at 18 21.

Local minimum at 36 06.

(b) Since Q x  P x  5, each point on the graph of Q has y-coordinate 5 units more than the corresponding point on the graph of P. Thus Q has a local maximum at 18 71 and a local minimum at 35 44.

10 10 5 -10

5

290

CHAPTER 3 Polynomial and Rational Functions

84. (a) P x  x  2 x  4 x  5 has one local maximum and one local minimum.

(b) Since P a  P b  0, and P x  0 for a  x  b (see the table

below), the graph of P must first rise and then fall on the interval a b, and so P must have at least one local maximum between a and b. Using similar reasoning, the fact that P b  P c  0 and P x  0 for

10

b  x  c shows that P must have at least one local minimum between b and c. Thus P has at least two local extrema.

5

 a a b b c c 

Interval

-10

Sign of x  a









Sign of x  b









Sign of x  c

















Sign of x  a x  b x  c

85. Since the polynomial shown has five zeros, it has at least five factors, and so the degree of the polynomial is greater than or equal to 5. 86. No, it is not possible. Clearly a polynomial must have a local minimum between any two local maxima. 87. P x  8x  03x 2  00013x 3  372 4000

(a) For the firm to break even, P x  0. From the graph, we see that P x  0 when x  252. Of course, the firm cannot produce

fractions of a blender, so the manufacturer must produce at least 26 blenders a year.

2000

(b) No, the profit does not increase indefinitely. The largest profit is approximately $ 327622, which occurs when the firm produces

0 0

200

166 blenders per year.

88. P t  120t  04t 4  1000

(a) A maximum population of approximately 1380 is attained after 422 months. (b) The rabbit population disappears after approximately 842 months.

1000 500 0 0

5

10

89. (a) The length of the bottom is 40  2x, the width of the bottom is 20  2x, and the height is x, so the volume of the box is V  x 20  2x 40  2x  4x 3  120x 2  800x.

(c) Using the domain from part (b), we graph V in the viewing rectangle [0 10] by [0 1600]. The maximum volume is V  15396 when x  423.

(b) Since the height and width must be positive, we must have x  0 and 20  2x  0, and so the domain of V is 0  x  10.

1000

0 0

5

10

SECTION 3.3 Dividing Polynomials

90. (a) Let h be the height of the box. Then the total length of all 12 edges is 8x  4h  144 in. Thus, 8x  4h  144 

2x  h  36  h  36  2x. The volume of the box is equal to   area of baseheight  x 2 36  2x  2x 3 36x 2 .

291

(c) Using the domain from part (b), we graph V in the viewing rectangle [0 18] by [0 2000]. The maximum volume is V  1728 in3 when x  12 in. 2000

Therefore, the volume of the box is

V x  2x 3  36x 2  2x 2 18  x. (b) Since the length of the base is x, we must have x  0. Likewise, the height must be positive so 36  2x  0 

1000 0 0

10

x  18. Putting these together, we get that the domain of V

is 0  x  18.

The graph of y  x 100 is close to the x-axis for x  1, but passes through the

y

91. (_1, 1) y=x$ y=x%

points 1 1 and 1 1. The graph of y  x 101 behaves similarly except that the

(1, 1)

1

y-values are negative for negative values of x, and it passes through 1 1

y=x@

instead of 1 1. 1

x

y=x#

(1, _1)

92. No, it is impossible. The end behavior of a third degree polynomial is the same as that of y  kx 3 , and for this function, the values of y go off in opposite directions as x   and x  . But for a function with just one extremum, the values of y would head off in the same direction (either both up or both down) on either side of the extremum. An nth-degree polynomial can have n  1 extrema or n  3 extrema or n  5 extrema, and so on (decreasing by 2). A polynomial that has six local extrema must be of degree 7 or higher. For example, P x  x  1 x  2 x  3 x  4 x  5 x  6 x  7 has six local extrema.

3.3

DIVIDING POLYNOMIALS

1. If we divide the polynomial P by the factor x  c, and we obtain the equation P x  x  c Q x  R x, then we say that x  c is the divisor, Q x is the quotient, and R x is the remainder. 2. (a) If we divide the polynomial P x by the factor x  c, and we obtain a remainder of 0, then we know that c is a factor of P. (b) If we divide the polynomial P x by the factor x  c, and we obtain a remainder of k, then we know that P c  k.

292

CHAPTER 3 Polynomial and Rational Functions

3.

2

2

5

7

4

2

2

1

9

4.

P x  D x

x 2

5. 2x  1

 2x  1 

3 3

Thus, the quotient is 2x  1 and the remainder is 9, and 2x 2  5x  7

4

5

1

12

12

28

3

7

29

Thus, the quotient is 3x 2  3x  7 and the remainder is 29, and

9 . x 2

2x  12 4x 2  3x  7

9

3x 3  9x 2  5x  1 P x  D x x 4   29  3x 2  3x  7  . x 4

2x 2  3x

6. 3x  4

4x 2  2x

6x 3  x 2  12x  5

6x 3  8x 2

9x 2  12x

x  7

x  12  15 2

9x 2  12x 5

Thus, the quotient is 2x  12 and the remainder is  15 2 ,   15 P x 4x 2  3x  7  2 . and   2x  12  D x 2x  1 2x  1

7. x2  4

2x 2  x 

1

x 2  3x  1

 8x 2

x 3  x 2 x 3

2x 3  6x 2  17x 

2x 5  0x 4 

3x  5

x 3  2x 2

6x 4  18x 3  6x 2

 4x

17x 3  8x 2 

 4

43x 2  14x  5

43x 2  129x  43

Thus, the quotient is 2x 2  x  1 and the remainder is   2x 4  x 3  9x 2 P x 2  x  1  4x  4 .   2x D x x2  4 x2  4

3x

17x 3  51x 2  17x

4x  4

4x  4, and

43

x 3  2x 2 

2x 5  6x 4  2x 3 6x 4 

x 2  4x x2

 P x 6x 3  x 2  12x  5  2 5   2x  3x  . D x 3x  4 3x  4

8.

2x 4  x 3  9x 2 2x 4

Thus, the quotient is 2x 2  3x and the remainder is 5, and

115x  48

Thus, the quotient is 2x 3  6x 2  17x  43 and the remainder is 115x  48 and

2x 5  x 3  2x 2  3x  5 P x  D x x 2  3x  1   115x  48 3  2x  6x 2  17x  43  2 x  3x  1

SECTION 3.3 Dividing Polynomials

x 2 

9. x 1

x  3

293

10. Long division:

x 3  0x 2  2x  6

x 3  5x 2  15x  35

x 3  x 2

x 3

x 2  2x

x2  x

x 4  2x 3  0x 2  10x 

0

x 3  3x 3 5x 3

3x  6

5x 3  15x 2

3x  3

 10x

15x 2  10x

9

15x 2  45x

Thus, the quotient is x 2  x  3 and the remainder is 9,   so P x  x 3  2x  6  x  1  x 2  x  3  9.

35x 35x  105 105 P x  x 4  2x 3  10x    x  3  x 3  5x 2  15x  35  105

Synthetic division:

3

1 1

2

0

10

0

3

15

45

105

5

15

35

105

Thus, the quotient is x 3  5x 2  15x  35 and the

remainder is 105, as above.

x2

11. 2x  3

 1

2x 2 

12.

2x 3  3x 2  2x

2x  1

2x 3  3x 2

4x 3

x  4

4x 3  2x 2

 7x  9

2x 2  7x

2x

2x 2  x

2x  3

8x  9

3 Thus, the quotient is x 2  1 and the remainder is 3, and   P x  2x 3  3x 2  2x  2x  3 x 2  1  3.

8x  4 5 Thus

  P x  4x 3  7x  9  2x  1  2x 2  x  4  5.

294

CHAPTER 3 Polynomial and Rational Functions

13.

4x 2  2x  1

2x 2  1

8x 4  4x 3  6x 2 8x 4

9x 3  6x 2 

14. 3x 2  3x  1

 4x 2

4x 3  2x 2 4x 3

3x  2

27x 5  9x 4  0x 3  3x 2  0x  3 27x 5  27x 4  9x 3

18x 4  9x 3  3x 2



2x 2 

18x 4  18x 3  6x 2

2x

9x 3  3x 2

2x

2x 2

9x 3  9x 2  3x

 1

6x 2  3x  3

2x  1

6x 2  6x  2

Thus, the quotient is 4x 2  2x  1 and the remainder is 2x  1, and

P x  27x 5  9x 4  3x 2  3      3x 2  3x  1  9x 3  6x 2  3x  2  3x  5

     2x 2  1  4x 2  2x  1  2x  1

15.

x  1

x2 

16.

x 2  3x  7

x 2

3x  5

Thus

P x  8x 4  4x 3  6x 2

x 3

x 2  2x

x  2

x 3  2x 2  x 3  3x 2

x  1

x 2  x

x  7

x 2  3x

x  2 5

2x  1 2x  6

Thus, the quotient is x  1 and the remainder is 5.

5 Thus, the quotient is x 2  x  2 and the remainder is 5.

2x 2

17. 2x  1

18.

 1

4x 3  2x 2  2x  3

4x 3  2x 2

2x  3 2x  1  2 Thus, the quotient is 2x 2  1 and the remainder is 2.

3x  6

1 x2  1 x  2 3 3 3 x 3  3x 2  4x  3

x 3  2x 2

x 2  4x x 2  2x 2x  3 2x  4 1

Thus, the quotient is 13 x 2  13 x  23 and the remainder is 1.

SECTION 3.3 Dividing Polynomials

19. x2  x  3

x 

1

20.

x 3  0x 2  2x  1

x 2  5x  1

x 3  x 2  3x

x2  x  1 x2 

x 2  2x 

9

x 4  3x 3 

0x 2 

2x 3 

x2 

x 4  5x 3 

x  2

x2

x

2x 3  10x 2 

x  3

2x

9x 2 

2

295

x  2

9x 2  45x  9

Thus, the quotient is x  1 and the remainder is 2.

44x  11 Thus, the quotient is x 2  2x  9 and the remainder is 44x  11.

21. 2x 2  0x  5

3x 

1

22.

6x 3  2x 2  22x  0 6x 3

3 3x 2  7x

 15x

2x 2 

2x 2

9x 2 

x  5

9x 2  21x

7x  0

20x  5

 5

Thus, the quotient is 3 and the remainder is 20x  5.

7x  5 Thus, the quotient is 3x  1 and the remainder is 7x  5. x4

23. x2  1

24.

 1

x 6  0x 5  x 4  0x 3  x 2  0x  1

x6

 x4

 x2

0

x2

4x 2  6x  8

1 x3  x2  5 x  7 2 2 4 2x 5  7x 4  0x 3  0x 2  0x  13

2x 5  3x 4  4x 3

4x 4  4x 3  0x 2

 1

4x 4  6x 3  8x 2

 1

10x 3  8x 2  0x

0

10x 3  15x 2  20x

Thus, the quotient is x 4  1 and the remainder is 0.

7x 2  20x  13

7x 2  21 2 x  14

19 x  1 2

Thus, the quotient is 12 x 3  x 2  52 x  74 and the remainder is 19 2 x  1. 25. The synthetic division table for this problem takes the following form.

26. The synthetic division table for this problem takes the following form.

3

2 2

5

3

6

3

1

6

Thus, the quotient is 2x  1 and the remainder is 6.

1

1 1

1

4

1

2

2

6

Thus, the quotient is x  2 and the remainder is 6.

296

CHAPTER 3 Polynomial and Rational Functions

27. The synthetic division table for this problem takes the following form.

28. The synthetic division table for this problem takes the following form.

3

1

3

1

0

3

2

2

2

2

4

Thus, the quotient is 3x  2 and the remainder is 2. 29. Since x  2  x  2, the synthetic division table for this problem takes the following form. 2

1

2

2

1

2

0

4

0

2

3

1

0

8

2

3

9

3

3

1

1

Thus, the quotient is x 2  3x  1 and the remainder is 1. 33. Since x 5  3x 3  6  x 5  0x 4  3x 3  0x 2  0x  6, the synthetic division table for this problem takes the

3

1

12

9

1

15

15

30

3

6

31

following form. 2

1

1 2

2

6

10

1

1

3

5

12

0

3

0

0

6

1

1

4

4

4

1

4

4

4

2

Thus, the quotient is x 4  x 3  4x 2  4x  4 and the

1

2

1

Thus, the quotient is x 3  x 2  3x  5 and the remainder is 12.

34. The synthetic division table for this problem takes the following form. 3

1

13

Thus, the quotient is 3x 2  3x  6 and the remainder is 31.

following form. 1

16

8

following form.

table for this problem takes the following form.

1

8

32. The synthetic division table for this problem takes the

x 3  8x  2  x 3  0x 2  8x  2, the synthetic division 1

3

30. The synthetic division table for this problem takes the

3

31. Since x  3  x  3 and

0

Thus, the quotient is 4x  8 and the remainder is 13.

5

Thus, the quotient is x 2  2 and the remainder is 3.

3

4

1

9

27

27

3

18

27

1

6

9

0

Thus, the quotient is x 2  6x  9 and the remainder is 0.

remainder is 2.

35. The synthetic division table for this problem takes the following form.

36. The synthetic division table for this problem takes the following form.

1 2

2 2

3

2

1

1

2

0

4

0

1

Thus, the quotient is 2x 2  4x and the remainder is 1.

 23

6 6

10

5

1

1

4

4

6

1

 23 1 3

 29 7 9

Thus, the quotient is 6x 3  6x 2  x  13 and the remainder is 79 .

SECTION 3.3 Dividing Polynomials

37. Since x 3  27  x 3  0x 2  0x  27, the synthetic

division table for this problem takes the following form. 3

1 1

0

0

27

3

9

27

3

9

0

4

5

4

8

8

3

1

3

7

6

1

2

10

6

5

3

12

30

2

0

7

2

0

0

0

0

7

5

40

0

2

9

1

1

5

10

6

  Therefore, by the Remainder Theorem, P 12  6. 1

1

1

5

1

2

3

2

3

2

Therefore, by the Remainder Theorem, P 1  2.

2

21

9

200

22

11

220

1

20

20

36

14

35

35

35

497

5

5

71

483

Therefore, by the Remainder Theorem, P 11  20. 46. P x  6x 5  10x 3  x  1, c  2 2

6 6

0

10

0

1

1

12

24

68

136

274

12

34

68

137

273

Therefore, by the Remainder Theorem, P 2  273. 48. P x  2x 6  7x 5  40x 4  7x 2  10x  112, c  3

 x 7  0x 6  0x 5  0x 4  0x 3  3x 2  0x  1 c3

1

16

8

2

47. P x  x 7  3x 2  1

1

8

4

is 0.

11

Therefore, by the Remainder Theorem, P 7  483.

3

4

2

44. P x  2x 3  21x 2  9x  200, c  11

45. P x  5x 4  30x 3  40x 2  36x  14, c  7 5

2

1

Therefore, by the Remainder Theorem, P 2  7.

7

16

Thus, the quotient is x 3  2x 2  4x  8 and the remainder

1

43. P x  x 3  2x 2  7, c  2 1

0

42. P x  x 3  x 2  x  5, c  1

Therefore, by the Remainder Theorem, P 2  12.

2

0

2

41. P x  x 3  3x 2  7x  6, c  2 1

0

1 2

Therefore, by the Remainder Theorem, P 1  3.

2

1

40. P x  2x 2  9x  1, c  12

12

4

following form.

1

39. P x  4x 2  12x  5 c  1 1

38. The synthetic division table for this problem takes the

2

Thus, the quotient is x 2  3x  9 and the remainder is 0.

297

0

0

0

0

3

0

1

3

9

27

81

243

720

2160

3

9

27

81

240

720

2159

Therefore by the Remainder Theorem, P 3  2159.

3

2 2

7

40

0

7

10

112

6

39

3

9

6

12

13

1

3

2

4

100

Therefore, by the Remainder Theorem, P 3  100.

298

CHAPTER 3 Polynomial and Rational Functions

49. P x  3x 3  4x 2  2x  1, c  23 2 3

3 3

4

2

2

4

6

2

50. P x  x 3  x  1, c  14 1 4

1 4 3 7 3

1

0

1

1 4 1 4

1

1 16 15  16

1  15 64 49 64

  Therefore, by the Remainder Theorem, P 14  49 64 .

 

Therefore, by the Remainder Theorem, P 23  73 . 51. P x  x 3  2x 2  3x  8, c  01 01

1 1

2

3

8

01

021

0279

21

279

8279

Therefore, by the Remainder Theorem, P 01  8279.

52. (a) P x  6x 7  40x 6  16x 5  200x 4  60x 3  69x 2  13x  139, c  7 7

6 6

40

16

200

60

69

13

139

42

14

210

70

70

7

140

2

30

10

10

1

20

1

Therefore, by the Remainder Theorem, P 7  1. (b) P 7  6 77  40 76  16 75  200 74  60 73  69 72  13 7  139  6 823,543  40 117,649  16 16,807  200 2401  60 343  69 49  13 7  139  1

which agrees with the value obtained by synthetic division, but requires more work. 53. P x  x 3  3x 2  3x  1, c  1 1

1 1

54. P x  x 3  2x 2  3x  10, c  2

3

3

1

1

2

1

2

1

0

Since the remainder is 0, x  1 is a factor. 55. P x  2x 3  7x 2  6x  5, c  12 1 2

2 2

2

1 1

2

3

10

2

8

10

4

5

0

Since the remainder is 0, x  2 is a factor. 56. P x  x 4  3x 3  16x 2  27x  63, c  3, 3

7

6

5

1

4

5

8

10

0

Since the remainder is 0, x  12 is a factor.

3

1 1

3

16

27

63

3

18

6

63

6

2

21

0

Since the remainder is 0, x  3 is a factor. We next show that x  3 is also a factor by using synthetic division on the quotient of the above synthetic division, x 3  6x 2  2x  21. 3

1 1

6

2

21

3

9

21

3

7

0

Since the remainder is 0, x  3 is a factor.

SECTION 3.3 Dividing Polynomials

57. P x  x 3  2x 2  9x  18, c  2 1

2

58. P x  x 3  5x 2  2x  10, c  5

2

9

18

2

0

18

0

9

0

1

5

Hence, the zeros are 3, 2, and 3. .

1

5

2

10

5

0

10

0

2

0

Since the remainder is 0, we know that 5 is a zero and        P x  x  5 x 2  2  x  5 x  2 x  2 .  Hence, the zeros are 5 and  2. 60. P x  3x 4  x 3  21x 2  11x  6, c  2, 13

59. x 3  x 2  11x  15, c  3 1

1 1

Since the remainder is 0, we know that 2 is a zero and   P x  x  2 x 2  9  x  2 x  3 x  3.

3

1

11

15

3

6

15

2

5

0

3

2

3

Since the remainder is 0, we know that 3 is a zero and   P x  x  3 x 2  2x  5 . Now x 2  2x  5  0   2  22  4 1 5  1  6. Hence, when x  2 1  the zeros are 3 and 1  6.

1

21

11

6

6

14

14

6

7

7

3

0

Since the remainder is 0, we know that 2 is a zero and   P x  x  2 3x 3  7x 2  7x  3 . 1 3

3

7

7

3

3

1

2

3

6

9

0

Since the remainder is 0, we know that 13 is a zero and    P x  x  2 x  13 3x 2  6x  9    3 x  2 x  13 x  1 x  3. Hence, the zeros are 2, 13 , 1, and 3.

61. P x  3x 4  8x 3  14x 2  31x  6, c  2, 3 3

2

3

8

14

31

6

6

28

28

6

14

14

3

0

62. P x  2x 4  13x 3  7x 2  37x  15, c  1, 3 2

1

2

13

7

37

15

2

15

22

15

15

22

15

0

Since the remainder is 0, we know that 2 is a zero and   P x  x  2 3x 3  14x 2  14x  3 . 14

14

3

Since the remainder is 0, we know that 1 is a zero and   P x  x  1 2x 3  15x 2  22x  15 . 15

22

15

9

15

3

6

27

15

5

1

0

9

5

0

3

3 3

299

Since the remainder is 0, we know that 3 is a zero and   P x  x  2 x  3 3x 2  5x  1 . Now

3x 2  5x  1  0 when    5  52  4 3 1 5  37  , and x 2 3 6  5  37 . hence, the zeros are 2, 3, and 6

3

2 2

The remainder is 0, so 3 is a zero and   P x  x  1 x  3 2x 2  9x  5

 x  1 x  3 2x  1 x  5.

Hence, the zeros are 1, 3,  12 , and 5.

300

CHAPTER 3 Polynomial and Rational Functions

63. Since the zeros are x  1, x  1, and x  3, the factors are x  1, x  1, and x  3. Thus P x  x  1 x  1 x  3  x 3  3x 2  x  3.

64. Since the zeros are x  2, x  0, x  2, and x  4, the factors are x  2, x, x  2, and x  4.

Thus P x  c x  2 x x  2 x  4. If we let c  1, then P x  x 4  4x 3  4x 2  16x.

65. Since the zeros are x  1, x  1, x  3, and x  5, the factors are x  1, x  1, x  3, and x  5. Thus P x  x  1 x  1 x  3 x  5  x 4  8x 3  14x 2  8x  15.

66. Since the zeros are x  2, x  1, x  0, x  1, and x  2, the factors are x  2, x  1, x, x  1, and x  2. Thus P x  c x  2 x  1 x x  1 x  2. If we let c  1, then P x  x 5  5x 3  4x.

67. Since the zeros of the polynomial are 2, 0, 1, and 3, it follows that P x  C x  2 x x  1 x  3  C x 4 2C x 3 

5C x 2  6C x. Since the coefficient of x 3 is to be 4, 2C  4, so C  2. Therefore, P x  2x 4  4x 3  10x 2  12x is the polynomial.   68. Since the zeros of the polynomial are 1, 0, 2, and 12 , it follows that P x  C x  1 x x  2 x  12  C x 4 

3 C x 3  3 C x 2 C x. Since the coefficient of x 3 is to be 3,  3 C  3, so C  2. Therefore, P x  2x 4 3x 3 3x 2 2x 2 2 2

is the polynomial.

  2 and integer coefficients, the fourth zero must be  2, otherwise the       constant term would be irrational. Thus, P x  C x  1 x  1 x  2 x  2  C x 4  3C x 2  2C. Requiring

69. Since the polynomial degree 4 and zeros 1, 1, and

that the constant term be 6 gives C  3, so P x  3x 4  9x 2  6.   70. Since the polynomial degree 5 and zeros 2, 1, 2, and 5 and integer coefficients, the fourth zero must be  5,      otherwise the constant term would be irrational. Thus, P x  C x  2 x  1 x  2 x  5 x  5  C x 5  C x 4  9C x 3  9C x 2  20C x  20C. Requiring that the constant term be 40 gives C  2, so P x  2x 5  2x 4  18x 3  18x 2  40x  40.

71. The y-intercept is 2 and the zeros of the polynomial are 1, 1, and 2.   It follows that P x  C x  1 x  1 x  2  C x 3  2x 2  x  2 . Since P 0  2 we have   2  C 03  2 02  0  2  2  2C  C  1 and P x  x  1 x  1 x  2  x 3  2x 2  x  2.

72. The y-intercept is 4 and the zeros of the polynomial are 1 and 2 with 2 being degree two.     It follows that P x  C x  1 x  22  C x 3  3x 2  4 . Since P 0  4 we have 4  C 03  3 02  4  4  4C  C  1 and P x  x 3  3x 2  4.

73. The y-intercept is 4 and the zeros of the polynomial are 2 and 1 both being degree two.   It follows that P x  C x  22 x  12  C x 4  2x 3  3x 2  4x  4 . Since P 0  4 we have   4  C 04  2 03  3 02  4 0  4  4  4C  C  1. Thus P x  x  22 x  12  x 4  2x 3  3x 2  4x  4.

74. The y-intercept is 2 and the zeros of the polynomial are 2, 1, and 1 with 1 being degree two.   It follows that P x  C x  2 x  1 x  12  C x 4  x 3  3x 2  x  2 . Since P 0  2 we have   4  C 04  03  3 02  0  2  2  2C so C  1 and P x  x 4  x 3  3x 2  x  2.

75. A. By the Remainder Theorem, the remainder when P x  6x 1000  17x 562  12x  26 is divided by x  1 is P 1  6 11000  17 1562  12 1  26  6  17  12  26  3.

B. If x  1 is a factor of Q x  x 567  3x 400  x 9  2, then Q 1 must equal 0. Q 1  1567  3 1400  19  2  1  3  1  2  1  0, so x  1 is not a factor.

SECTION 3.4 Real Zeros of Polynomials

301



    76. R x  x 5  2x 4  3x 3  2x 2  3x  4  x 4  2x 3  3x 2  2x  3 x  4  x 3  2x 2  3x  2 x  3 x  4      x 2  2x  3 x  2 x  3 x  4  [x  2 x  3] x  2 x  3 x  4

So to calculate R 3, we start with 3, then subtract 2, multiply by 3, add 3, multiply by 3, subtract 2, multiply by 3, add 3, multiply by 3, and add 4, to get 157.

3.4

REAL ZEROS OF POLYNOMIALS

1. If the polynomial function P x  an x n  an1 x n1     a1 x  a0 has integer coefficients, then the only numbers that p could possibly be rational zeros of P are all of the form , where p is a factor of the constant coefficient a0 and q is a factor q of the leading coefficient an . The possible rational zeros of P x  6x 3  5x 2  19x  10 are 1,  12 ,  13 ,  16 , 2,  23 , 5,  52 ,  53 ,  56 , 10, and  10 3.

2. Using Descartes’ Rule of Signs, we can tell that the polynomial P x  x 5  3x 4  2x 3  x 2  8x  8 has 1, 3 or 5 positive real zeros and no negative real zero. 3. This is true. If c is a real zero of the polynomial P, then P x  x  c Q x, and any other zero of P x is also a zero of Q x  P x  x  c. 4. False. Consider the polynomial P x  x  1 x  2 x  4  x 3  x 2  10x  8. An upper bound is 3, but 3 is not a lower bound.

5. P x  x 3  4x 2  3 has possible rational zeros 1 and 3.

6. Q x  x 4  3x 3  6x  8 has possible rational zeros 1, 2, 4, 8.

7. R x  2x 5  3x 3  4x 2  8 has possible rational zeros 1, 2, 4, 8,  12 .

8. S x  6x 4  x 2  2x  12 has possible rational zeros 1, 2, 3, 4, 6, 12,  12 ,  32 ,  13 ,  23 ,  43 ,  16 .

9. T x  4x 4  2x 2  7 has possible rational zeros 1, 7,  12 ,  72 ,  14 ,  74 .

1 . 10. U x  12x 5  6x 3  2x  8 has possible rational zeros 1, 2, 4, 8  12 ,  13 ,  23 ,  43 ,  83 ,  14 ,  16 ,  12

11. (a) P x  5x 3  x 2  5x  1 has possible rational zeros 1,  15 . (b) From the graph, the actual zeros are 1, 15 , and 1.

12. (a) P x  3x 3  4x 2  x  2 has possible rational zeros 1, 2,  13 ,  23 . (b) From the graph, the actual zeros are 1 and 23 .

13. (a) P x  2x 4  9x 3  9x 2  x  3 has possible rational zeros 1, 3,  12 ,  32 . (b) From the graph, the actual zeros are  12 , 1, and 3.

14. (a) P x  4x 4  x 3  4x  1 has possible rational zeros 1,  12 ,  14 . (b) From the graph, the actual zeros are 14 and 1.

15. P x  x 3  2x 2  13x  10. The possible rational zeros are 1, 2, 5, 10. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  x 3  2x 2  13x  10 has 1 variation in sign and hence P has 1 negative real zero. 1

2 2

1

13

10

2

3

10

10 0  x  1 is a zero.  P x  x 3  2x 2  13x  10  x  1 x 2  3x  10  x  5 x  1 x  2. Therefore, the zeros are 5, 1, and 2. 

3

302

CHAPTER 3 Polynomial and Rational Functions

16. P x  x 3  4x 2  19x  14. The possible rational zeros are 1, 7, 14. P x has 1 variation in sign and hence 1 positive real zero. P x  x 3  4x 2  19x  14 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1

1

4

19

14

1

3

22

1

1

4

19

14

1

5

14

 x  1 is not a zero. 1 5 14 0  x  1 is a zero.   P x  x 3  4x 2  19x  14  x  1 x 2  5x  14  x  2 x  1 x  7. Therefore, the zeros are 2, 1, and 3

22

36

7.

17. P x  x 3  3x 2  4. The possible rational zeros are 1, 2, 4. P x has 1 variation in sign and hence 1 positive real zero.P x  x 3  3x 2  4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1

3

0

4

1

4

4

1

4 4 0  x  1 is a zero.  P x  x 3  3x 2  4  x  1 x 2  4x  4  x  1 x  22 . Therefore, the zeros are 2 and 1. 

18. P x  x 3  3x  2. The possible rational zeros are 1, 2. P x has 1 variation in sign and hence 1 positive real zero.P x  x 3  3x  2 has 2 variations in sign and P has hence 0 or 2 negative real zeros. 1

1 1

0

3

1

1

1

2

1

2

2 2

0

3

2

2

4

2

 x  1 is not a zero. 1 2 1 0  x  2 is a zero.   P x  x 3  3x  2  x  2 x 2  2x  1  x  2 x  12 . Therefore, the zeros are 2 and 1. 4

19. P x  x 3  6x 2  12x  8. The possible rational zeros are 1, 2, 4, 8. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x  x 3  6x 2  12x  8 has no variations in sign and hence P has no negative real zero. 1

1 1

6

12

8

1

5

7

7

1

1

2

6

12

8

2

8

8

 x  1 is not a zero. 1 4 4 0  x  2 is a zero.   P x  x 3  6x 2  12x  8  x  2 x 2  4x  4  x  23 . Therefore, the only zero is x  2. 5

20. P x  x 3  12x 2  48x  64. The possible rational zeros are 1, 2, 4, 8, 16, 32, 64. P x has 0 variations in sign and hence no positive real zero. P x  x 3  12x 2  48x  64 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1

1 1

12

48

64

1

11

37

11

37

27

2  x  1 is not a zero. 4

1 1

1 1

12

48

64

4

32

64

8

16

12

48

64

2

20

56

10

28

6

 x  2 is not a zero.

0  x  4 is a zero.  P x  x 3  12x 2  48x  64  x  4 x 2  8x  16  x  43 . Therefore, the only zero is x  4. 

SECTION 3.4 Real Zeros of Polynomials

303

21. P x  x 3  19x  30. The possible rational zeros are 1, 2, 3, 5, 6, 15, 30. P x has 1 variations in sign and hence 1 positive real zero. P x  x 3  19x  30 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1 1

0

19

30

1

1

18

1

18

12

1

2  x  1 is not a zero. 1

4

1

12

48

64

4

32

64

8

16

1

0

19

30

2

4

30

2

15

0

 x  2 is a zero.

0  x  4 is a zero.  P x  x 3  19x  30  x  2 x 2  2x  15  x  3 x  2 x  5. Therefore, the zeros are 3, 2, and 5. 

22. P x  x 3  11x 2  8x  20. The possible rational zeros are 1, 2, 4, 5, 10, 20. P x has 1 variation in sign

and hence 1 positive real zeros. P x  x 3  11x 2  8x  20 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1

11

8

20

1

12

20

1 12 20 0  x  1 is a zero.   P x  x 3  11x 2  8x  20  x  1 x 2  12x  20  x  10 x  2 x  1. Therefore, the zeros are 10, 2,

and 1.

23. P x  x 3  3x 2  x  3. The possible rational zeros are 1, 3. P x has 1 variation in sign and hence 1 positive real zero. P x  x 3  3x 2  x  4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1

3

1

3

1

2

3

2

3 

0

1 

 x  1 is a zero.

So P x  x 3  3x 2  x  3  x  1 x 2  2x  3  x  1 x  3 x  1. Therefore, the zeros are 1, 3, and 1.

24. P x  x 3  4x 2  11x  30. The possible rational zeros are 1, 2, 3, 5, 10, 15, 30. P x has 2 variations

in sign and hence 0 or 2 positive real zeros. P x  x 3  4x 2  11x  30 has 1 variation in sign and hence P has 1 negative real zero. 1

1 1

4

11

30

1

3

14

2

1

4

11

30

2

4

30

1 2 15 0  x  2 is a zero.  So P x  x 3  4x 2  11x  30  x  2 x 2  2x  15  x  2 x  5 x  3. Therefore, the zeros are 3, 2, and 5.

3

14

16



304

CHAPTER 3 Polynomial and Rational Functions

25. Method 1: P x  x 4  5x 2  4The possible rational zeros are 1, 2, 4. P x has 1 variation in sign and hence 1 positive real zero. P x  x 4  5x 2  4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1

0

5

0

4

1

1

4

4

1 1 4 4 0  x  1 is a zero.   Thus P x  x 4  5x 2  4  x  1 x 3  x 2  4x  4 . Continuing with the quotient we have: 1

1

1

4

4

1

0

4

1 0 4 0  x  1 is a zero.   P x  x 4  5x 2  4  x  1 x  1 x 2  4  x  1 x  1 x  2 x  2. Therefore, the zeros are 1, 2.

Method 2: Substituting u  x 2 , the polynomial becomes P u  u 2  5u  4, which factors:    u 2  5u  4  u  1 u  4  x 2  1 x 2  4 , so either x 2  1 or x 2  4. If x 2  1, then x  1; if x 2  4, then x  2. Therefore, the zeros are 1 and 2.

26. P x  x 4  2x 3  3x 2  8x  4. Using synthetic division, we see that x  1 is a factor of P x: 1

1 1

2

3

8

4

1

1

4

4

1

4

4

0

 x  1 is a zero.

We continue by factoring the quotient, and we see that x  1 is again a factor: 1

1

1

1

4

4

1

0

4

0

4 0  x  1 is a zero.   P x  x 4  2x 3  3x 2  8x  4  x  1 x  1 x 2  4  x  12 x  2 x  2

Therefore, the zeros are 1 and 2.

27. P x  x 4  6x 3  7x 2  6x  8. The possible rational zeros are 1, 2, 4, 8. P x has 1 variation in sign and hence 1 positive real zero. P x  x 4  6x 3  7x 2  6x  8 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1

1 1

6

7

6

8

1

7

14

8

7

14

8

0

 x  1 is a zero

  and there are no other positive zeros. Thus P x  x 4  6x 3  7x 2  6x  8  x  1 x 3  7x 2  14x  8 . Continuing

by factoring the quotient, we have:

1

1 1

7

14

8

1

6

8

6

8

 x  1 is a zero.  So P x  x 4  6x 3  7x 2  6x  8  x  1 x  1 x 2  6x  8  x  1 x  1 x  2 x  4. Therefore, the zeros are 4, 2, and 1.

0



SECTION 3.4 Real Zeros of Polynomials

305

28. P x  x 4  x 3  23x 2  3x  90. The possible rational zeros are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. Since P x has 2 variations in sign, P has 0 or 2 positive real zeros. Since P x  x 4  x 3  23x 2  3x  90 has 2 variations in sign, P has 0 or 2 negative real zeros. 1

1

1

23

3

90

1

0

23

26

0

23

26

64

1 3

1

1

21

45

3

12

27

1

2

1 1

5

1

23

3

90

2

2

42

90

1

21

45

0

1

21

45

5

30

45

 x  2 is a zero.

1 6 9 0  x  5 is a zero. 4 9 72   2 2 P x  x  2 x  5 x  6x  9  x  2 x  5 x  3 . Therefore, the zeros are 3, 2, and 5. 1

29. P x  4x 4  37x 2  9 has possible rational zeros 1, 3, 9,  12 ,  32 ,  92 ,  14 ,  34 ,  94 . Since P x has 2 variations in sign, there are 0 or 2 positive real zeros, and since P x  4x 4  37x 2  36 has 2 variations in sign, P has 0 or 2 negative real zeros. 1

4 4

0

37

0

9

4

4

33

33

4

33

33

24 1 2

4

3

4

4 4

12

1

3

2

7

3

14

6

0

Because P is even, we conclude that the zeros are 3 and  12 and

0

37

0

9

12

36

3

9

12

1

3

0

 x  3 is a zero.

 x  12 is a zero.

P x  4x 4  37x 2  9  x  3 2x  1 2x  1 x  3. Note: Since P x has only even terms, factoring by substitution also works. Let x 2  u; then    P u  4u 2  37u  9  4u  1 u  9  4x 2  1 x 2  3 , which gives the same results.

30. P x  6x 4 23x 3 13x 2 32x 16. The possible rational zeros are 1, 2, 4, 8, 16,  12 ,  13 ,  23 ,  43 ,  83 ,  16 3 ,  16 . Since P x has 2 variations in sign, P has 0 or 2 positive real zeros. Since P x  6x 4  23x 3  13x 2  32x  16 has 2 variations in sign, P has 0 or 2 negative real zeros. 1

6 6

23

13

32

16

6

17

30

2

17

30

2

18

4

6

2

6

23

13

32

16

24

4

36

16

6

6

23

13

32

16

12

22

70

76

11

35

38

60

9 4 0  x  4 is a zero   P x  6x 4  23x 3  13x 2  32x  16  x  4 6x 3  x 2  9x  4 . We continue:  12

1

6

1

9

4

3

1

4

6 2 8 0  x   12 is a zero   Thus, P x  x  4 2x  1 3x 2  x  4  x  1 2x  1 3x  4 x  4 and the rational zeros are 1,  12 , 43 ,

and 4.

306

CHAPTER 3 Polynomial and Rational Functions

31. P x  3x 4  10x 3  9x 2  40x  12. The possible rational zeros are 1, 2, 3, 4, 6, 12. P x has 3 variations in sign and hence 1, 3, or 5 positive real zeros. P x  3x 4  10x 3  9x 2  40x  12 has 1 variation in sign and hence P has 1 negative real zero. 1 3 10

9

3 3

40 12

2 3 10

24

6

7 16

40 12

9

12

8 34

24  x  1 is not a zero. 3 4 17 6 0  x  2 is a zero.   Thus P x  3x 4  10x 3  9x 2  40x  12  x  2 3x 3  4x 2  17x  6 . Continuing by factoring the quotient, we 7 16

24

have

3

3

4

17

6

9

15

6

3 5 2 0  x  3 is a zero.   Thus P x  x  3 x  2 3x 2  5x  2  x  3 x  2 3x  1 x  2. Therefore, the zeros are 2, 13 , 2, and 3.

32. P x  2x 3  7x 2  4x  4. The possible rational zeros are 1, 2, 4,  12 . Since P x has 1 variation in sign, P has 1 positive real zero. Since P x  2x 3  7x 2  4x  4 has 2 variations in sign, P has 0 or 2 negative real zeros. 1

2

7

4

4

2

9

13

1 2

2

7

4

4

1

4

4

9 13 9  x  1 is an upper bound. 2 8 8 0  x  12 is a zero.        P x  x  12 2x 2  8x  8  2 x  12 x 2  4x  4  2 x  12 x  22 . Therefore, the zeros are 2 and 12 . 2



33. Factoring by grouping can be applied to this exercise.

  4x 3  4x 2  x  1  4x 2 x  1  x  1  x  1 4x 2  1  x  1 2x  1 2x  1. Therefore, the zeros are 1 and  12 .

  34. We use factoring by grouping: P x  2x 3  3x 2  2x  3  2x x 2  1      3 x 2  1  x 2  1 2x  3  x  1 x  1 2x  3. Therefore, the zeros are 32 and 1. 35. P x  4x 3  7x  3. The possible rational zeros are 1, 3,  12 ,  32 ,  14 ,  34 . Since P x has 2 variations in sign, there are 0 or 2 positive zeros. Since P x  4x 3  7x  3 has 1 variation in sign, there is 1 negative zero. 1 2

4

0

7

3

2

1

3

4 2 6 0  x  12 is a zero.     P x  x  12 4x 2  2x  6  2x  1 2x 2  x  3  2x  1 x  1 2x  3  0. Thus, the zeros are  32 , 

1 , and 1. 2

SECTION 3.4 Real Zeros of Polynomials

307

1 . Since P x has 2 36. P x  12x 3  25x 2  x  2. The possible rational zeros are 1, 2,  12 ,  13 ,  23 ,  14 ,  16 ,  12

variations in sign, P has 0 or 2 positive real zeros, and since P x  12x 3  25x 2  x  2 has 1 variations in sign, P has 1 negative real zero. 1

12

25

1

2

12

13

12

2

12

25

1

2

24

2

2

12 1 1 0  x  2 is a zero. 10   P x  12x 3  25x 2  x  2  x  2 12x 2  x  1  4x  1 3x  1 x  2. Therefore, the zeros are  14 , 13 , 12

13

12

and 2.

37. P x  24x 3  10x 2  13x  6. The possible rational zeros are 1, 2, 3, 6,  12 ,  32 ,  13 ,  23 ,  14 ,  34 ,  16 ,  18 , 1 ,  1 . P x has 1 variation in sign and hence 1 positive real zero. P x  24x 3  10x 2  13x  6 has 2  38 ,  12 24

variations in sign, so P has 0 or 2 negative real zeros. 1 24

10 13 6 24

14 1

24 14 3 24

2 24

48

1 7  x  1 is not a zero.

10 13

10 13

24 38 6 24

6

72 186 519

6

76 126 63 132  x  2 is not a zero.

10 13

6

144 804 4746

24 62 173 525  x  3 is not a zero.  12

24 24

24 134 791 4752  x  6 is not a zero.

10

13

6

12

1

6

2

12

0

 x   12 is a zero.

  Thus P x  24x 3  10x 2  13x  6  2x  1 12x 2  x  6  3x  2 2x  1 4x  3 has zeros  23 ,  12 , and 3. 4

1 ,  3 ,  3 . P x has 2 38. P x  12x 3  20x 2  x  3. The possible rational zeros are 1, 2, 3,  12 ,  13 ,  14 ,  16 ,  12 2 4

variations in sign and hence 0 or 2 positive real zeros. P x  12x 3  20x 2  x  3 has 1 variations in sign and hence P has 1 negative real zero. 1

12 12

3

12 12

20

1

3

12

8

7

8

7

4

20

1

3

36

48

147

16

49

150

12

2  x  1 is not a zero.

12 1 2

 x  3 is not a zero.

12 12

20

1

3

24

8

18

4

9

21

 x  2 is not a zero.

20

1

3

6

7

3

14

6

0

 x  12 is a zero.

308

CHAPTER 3 Polynomial and Rational Functions

   Thus, P x  12x 3  20x 2  x  3  x  12 12x 2  14x  6 . Continuing: 3 2

12 12

14

6

18

6

4

0

 x  32 is a zero.

Thus, P x  2x  1 2x  3 3x  1 has zeros 12 , 32 , and  13 .

39. P x  2x 4  7x 3  3x 2  8x  4. The possible rational zeros are 1, 2, 4,  12 . P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x  2x 4  7x 3  3x 2  8x  4 has 1 variation in sign and hence P has 1 negative real zero. 1

2

7

3

8

4

2

5

2

6

2

6

2

1 2

2

7

3

8

4

1

3

0

4

 x  1 is not a zero. 2 6 0 8 0  x  12 is a zero.    Thus P x  2x 4  7x 3  3x 2  8x  4  x  12 2x 3  6x 2  8 . Continuing by factoring the quotient, we have: 2

5

2

2

6

0

8

4

4

8

2 2 4 0  x  2 is a zero.         P x  x  12 x  2 2x 2  2x  4  2 x  12 x  2 x 2  x  2  2 x  12 x  22 x  1. Thus, the 



zeros are 12 , 2, and 1.

40. P x  6x 4  7x 3  12x 2  3x  2. The possible rational zeros are 1, 2,  12 ,  13 ,  23 ,  16 . Since P x has 2 variations in sign, P has 0 or 2 positive real zeros. Since P x  6x 4  7x 3  12x 2  3x  2 has 2 variations in sign, P

has 0 or 2 negative real zeros.

7

12

3

2

12

10

4

2

6 5 2   P x  6x 4  7x 3  12x 2  3x  2  x  2 6x 3  5x 2  2x  1 .

1

0

1

6 6

7

12

3

2

6

1

13

10

1

13

10

8

2

6

 x  2 is a zero.

SECTION 3.4 Real Zeros of Polynomials

309

Continuing by factoring the quotient, we first note that the possible rational zeros are 1,  12 ,  13 ,  16 . We have: 1 2

6

5

2

1

3

4

1

0  x  12 is a zero.        P x  x  2 x  12 6x 2  8x  2  2 x  2 x  12 3x 2  4x  1  2 x  2 x  12 x  1 3x  1. 6

8

2



Therefore, the zeros are 1,  13 , 12 , and 2.

41. P x  x 5  3x 4  9x 3  31x 2  36. The possible rational zeros are 1, 2, 3, 4, 6, 8, 9, 12, 18. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  x 5  3x 4  9x 3  31x 2  36 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1

1 1

3

9

31

0

36

1

4

5

36

36

4

5

36 36 0  x  1 is a zero.   So P x  x 5  3x 4  9x 3  31x 2  36  x  1 x 4  4x 3  5x 2  36x  36 . Continuing by factoring the quotient, we have:

1

1 1

4

5

36

36

1

5

0

36

1

0

36

72

3

1 1

2

1 1

4

5

36

36

3

21

48

36

7

16

4

5

36

36

2

12

14

44

6

7

22

80

12 0  x  3 is a zero.  So P x  x  1 x  3 x 3  7x 2  16x  12 . Since we have 2 positive zeros, there are no more positive zeros, so 

we continue by factoring the quotient with possible negative zeros. 1

1

7

16

12

1

6

10

6

10

1

2

1

7

16

12

2

10

12

2

1 5 6 0  x  2 is a zero.  Then P x  x  1 x  3 x  2 x 2  5x  6  x  1 x  3 x  22 x  3. Thus, the zeros are 1, 3, 2, 

and 3.

42. P x  x 5  4x 4  3x 3  22x 2  4x  24 has possible rational zeros 1, 2, 3, 4, 6, 8, 12, 24. Since P x has 3 variations in sign, there are 1 or 3 positive real zeros. Since P x  x 5  4x 4  3x 3  22x 2  4x  24 has 2 variations in sign, P has 0 or 2 negative real zeros. 1

1 1

4

3

22

4

24

1

3

6

16

12

3 6 16 12 12   P x  x  2 x 4  2x 3  7x 2  8x  12

2

1 1

4

3

22

4

24

2

4

14

16

24

2

7

8

12

0

 x  2 is a zero.

310

CHAPTER 3 Polynomial and Rational Functions

2

1

2

7

8

12

1

2

0

14

12

0

7

6

0

  P x  x  22 x 3  7x  6 2

1

0

7

6

2

4

6

 x  2 is a zero again.

1

3

0

7

6

3

9

6

1 3 2 0  x  3 is a zero. 2 3 12   P x  x  22 x  3 x 2  3x  2  x  22 x  3 x  1 x  2  0. Therefore, the zeros are 1, 2, and 3. 1

43. P x  3x 5  14x 4  14x 3  36x 2  43x  10 has possible rational zeros 1, 2, 5, 10,  13 ,  23 ,  53 ,  10 3 . Since P x has 2 variations in sign, there are 0 or 2 positive real zeros. Since P x  3x 5  14x 4  14x 3  36x 2  43x  10 has 3 variations in sign, P has 1 or 3 negative real zeros. 1

3

14

14

36

43

10

3

11

25

11

54

2

3

11 25 11 54 64   P x  x  2 3x 4  8x 3  30x 2  24x  5 2

3

8

30

24

5

6

4

68

184

3

2

3 3

5

14

14

36

43

10

6

16

60

48

10

8

30

24

5

0

3

8

30

24

5

15

35

25

5

 x  2 is a zero.

34 92 189 3 7 5 1 0  x  5 is a zero.   P x  x  2 x  5 3x 3  7x 2  5x  1 . Since 3x 3  7x 2  5x  1 has no variation in sign, there are no more

positive zeros.

1

3

7

5

1

3

4

1

3 4 1 0  x  1 is a zero.   P x  x  2 x  5 x  1 3x 2  4x  1  x  2 x  5 x  1 x  1 3x  1. Therefore, the zeros are 1,

 13 , 2, and 5.

SECTION 3.4 Real Zeros of Polynomials

311

44. P x  2x 6  3x 5  13x 4  29x 3  27x 2  32x  12 has possible rational zeros 1, 2, 3, 4,

6, 12,  12 ,  32 . Since P x has 5 variations in sign, there are 1 or 3 or 5 positive real zeros. Since P x  2x 6  3x 5  13x 4  29x 3  27x 2  32x  12 has 3 variations in sign, P has 1 or 3 negative real zeros. 1

2

2

3

13

29

2

2

1

1

14

13

29

4

2

1

11

3

2

2

27

32

12

14

15

12

20

15

12

20

8

27

32

12

22

14

26

12

7

13

0  x  2 is a zero.   P x  x  2 2x 5  x 4  11x 3  7x 2  13x  6 . We continue with the quotient: 2

2

6

1

11

7

13

6

4

10

2

10

6

2 5 1 5 3 0  x  2 is a zero again.   P x  x  22 2x 4  5x 3  x 2  5x  3 . We continue with the quotient, first noting 2 is no longer a possible rational

solution:

3

2 2

5

1

5

3

6

22

42

94

11

21

47

91

 x  3 is an upper bound.

We know that there is at least 1 more positive zero. 1 2

2

5

1

5

3

1

2

1

3

2 6 2 6 0  x  12 is a zero.   P x  x  22 x  12 2x 3  6x 2  2x  6 . We can factor 2x 3  6x 2  2x  6 by grouping;         2x 3  6x 2  2x  6  2x 3  6x 2  2x  6  2x  6 x 2  1 . So P x  2 x  22 x  12 x  3 x 2  1 . 

Since x 2  1 has no real zeros, the zeros of P are 3, 2, and 12 .

45. P x  3x 3  5x 2  2x  4. The possible rational zeros are 1, 2, 4,  13 ,  23 ,  43 . P x has 1 variation in sign and hence 1 positive real zero. P x  3x 3  5x 2  2x  4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1 3 5 2 4 5 2 4 1 3 3 3

2

0

3

2

4

2 0 4 3 2 4 0  x  1 is a zero.   So P x  x  1 3x 2  2x  4 . Using the Quadratic Formula on the second factor, we have    2 434  13 13 . Therefore, the zeros of P are 1 and 13 13 . x  2 223

312

CHAPTER 3 Polynomial and Rational Functions

46. P x  3x 4  5x 3  16x 2  7x  15. The possible rational zeros are 1, 3, 5, 15,  13 ,  53 . P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  3x 4  5x 3  16x 2  7x  15 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

3

5

16

7

15

3

3

5

2 18 11 4   So P x  x  3 3x 3  4x 2  4x  5 . Continuing:

3

2

18

11 3

3

1

3

4

4

5

3

1

5

16

7

15

9

12

12

15

4

4

5

0

 x  3 is a zero.

3 1 5 0  x  1 is a zero.     Thus, P x  x  3 3x 3  4x 2  4x  5  x  1 x  3 3x 2  x  5 . Using the Quadratic Formula on the last    2 435 factor, we have x  1 123  16 61 . Therefore, the zeros of P are 1, 3, and 16 61 .

47. P x  x 4  6x 3  4x 2  15x  4. The possible rational zeros are 1, 2, 4. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  x 4  6x 3  4x 2  15x  4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1 1 6 4 15 4 4 15 4 2 1 6 1

1

5

1

14

5

1

14

18

4

1

1

6

4

15

4

4

8

16

4

2

8

8

14

4

4

7

18

2 4 1 0  x  4 is a zero.  So P x  x  4 x 3  2x 2  4x  1 . Continuing by factoring the quotient, we have: 4

1 1

2

1



4

1

4

8

16

2

4

15

1



 x  4 is an upper bound. 

1 1

2

4

1

1

3

1

3

1

0

 x  1 is a zero.

So P x  x  4 x  1 x 2  3x  1 . Using the Quadratic Formula on the third factor, we have:    3 32 411 3 13 . Therefore, the zeros are 4, 1, and 3 13 .  x 2 2 21

SECTION 3.4 Real Zeros of Polynomials

313

48. P x  x 4  2x 3  2x 2  3x  2. The possible rational zeros are 1, 2. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  x 4  2x 3  2x 2  3x  2 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1





1 1

2

2

3

2

1

3

1

2

3

1

2

0

 x  1 is a zero.

P x  x  1 x 3  3x 2  x  2 . Continuing with the quotient: 1 1 3 1 2 1 4

1 1

5

3

1 2

2 1

1

1 2

3

1 2

2 2

2

1 4 5

3  x  1 is an 1 2 1 1 1 1 1 0  x  2 is a zero. upper bound.   So P x  x  1 x  2 x 2  x  1 . Using the Quadratic Formula on the third factor, we have    1  12  4 1 1 1  5 5. x  . Therefore, the zeros are 1, 2, and 1 2 2 1 2

49. P x  x 4  7x 3  14x 2  3x  9. The possible rational zeros are 1, 3, 9. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x  x 4  7x 3  14x 2  3x  4 has 1 variation in sign and hence P has 1 negative real zero. 1

1

7

14

3

9

1

6

8

5

1

3

7

14

3

9

3

12

6

9

1

6 8 5 4 1 4 2 3 0  x  3 is a zero.   So P x  x  3 x 3  4x 2  2x  3 . Since the constant term of the second term is 3, 9 are no longer possible zeros.

Continuing by factoring the quotient, we have:





3

1 1

4

2

3

3

3

3

1

1

0

 x  3 is a zero again.

So P x  x  32 x 2  x  1 . Using the Quadratic Formula on the second factor, we have:    1 12 411 1 5 . Therefore, the zeros are 3 and 1 5 .  x 2 2 21

314

CHAPTER 3 Polynomial and Rational Functions

50. P x  x 5  4x 4  x 3  10x 2  2x  4. The possible rational zeros are 1, 2, 4. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x  x 5  4x 4  x 3  10x 2  2x  4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1 1

4

1

10

2

4

1

3

4

6

8

2

4 6 8 4   So P x  x  2 x 4  2x 3  5x 2  2 . 2

1

3

is 2, 4 are no longer possible zeros.

1

4

1

10

2

4

2

4

10

0

4

2

5

0

2

0

 x  2 is a zero.

Since the constant term of the second factor

Continuing by factoring the quotient, we have:

1

2

5

0

2

2

0

10

20

1

0

1

1

2

5

0

2

1

3

2

2

5 10 18 1 3 2 2 0  x  1 is a zero.   So P x  x  2 x  1 x 3  3x 2  2x  2 . Continuing by factoring the quotient, we have: 1

1

3

2

2

1

4

2

1

4 2 0  x  1 is a zero again.  So P x  x  2 x  12 x 2  4x  2 . Using the Quadratic Formula on the second factor, we have: 

     4 42 412 4 8  42 2  2  2. Therefore, the zeros are 1, 2, and 2  2.  x 21 2 2

51. P x  4x 3  6x 2  1. The possible rational zeros are 1,  12 ,  14 . P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  4x 3  6x 2  1 has 1 variation in sign and hence P has 1 negative real zero. 1

So P x 



4

6

0

1

4

2

2

1 2

4

6

0

1

2

2

1

4 2 2 1 4 4 2 0  x  12 is a zero.   x  12 4x 2  4x  2 . Using the Quadratic Formula on the second factor, we have:

     4 42 442 4 48  44 3  1 3 . Therefore, the zeros are 1 and 1 3 .  x 8 8 2 2 2 24

SECTION 3.4 Real Zeros of Polynomials

315

52. P x  3x 3  5x 2  8x  2. The possible rational zeros are 1, 2,  13 ,  23 . P x has 1 variation in sign and hence 1 positive real zero. P x  3x 3  5x 2  8x  2 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1 3

3

5

3 3

5

8

2

3

2

10

2

10

12

8

2

 43

1 3

2 3

3

 46 9

5

8

2

6

2

12

3

1

6

14

3

5

8

2

2

3

10

 26 3

 28 9

 28 3

4

2

3

2

 20 3

Thus we have tried all the positive rational zeros, so we try the negative zeros. 1

3 3

5

8

3

8

0

8

0

2

 13

2

3

2

5

8

2

1

2

2

3

5

8

2

6

22

28

3

11

14

30

3 6 6 0  x   13 is a zero.       So P x  x  13 3x 2  6x  6  3 x  13 x 2  2x  2 . Using the Quadratic Formula on the second factor, we      2 22 412  22 12  222 3  1  3. Therefore, the zeros are  13 and 1  3. have: x  21

53. P x  2x 4  15x 3  17x 2  3x  1. The possible rational zeros are 1,  12 . P x has 1 variation in sign and hence 1 positive real zero. P x  2x 4  15x 3  17x 2  3x  1 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1 2

2 2  12

15

17

3

1

8

16

25

25 2 31 2

2

1 31 4 27 4

 x  12 is an upper bound.

15

17

3

1

1

7

5

1

14 10 2 0  x   12 is a zero.       So P x  x  12 2x 3  14x 2  10x  2  2 x  12 x 3  7x 2  5x  1 . 2

1

1

7

5

1

1

6

1

1 6 1 0  x  1 is a zero.       So P x  x  12 2x 3  14x 2  10x  2  2 x  12 x  1 x 2  6x  1 Using the Quadratic Formula on the 

   6 62 411 10  3  10. Therefore, the zeros are 1,  1 , third factor, we have x   62 40  62 21 2 2

and 3 

 10.

316

CHAPTER 3 Polynomial and Rational Functions

54. P x  4x 5  18x 4  6x 3  91x 2  60x  9. The possible rational zeros are 1, 3, 9,  12 ,  32 ,  92 ,  14 ,  34 ,  94 .

P x has 4 variations in sign and hence 0 or 2 or 4 positive real zeros. P x  4x 5  18x 4  6x 3  91x 2  60x  9 has 1 variation in sign and hence P has 1 negative real zero. 1

4

18

6

91

60

9

4

14

20

71

1

4

14

3

4

18

6

91

60

9

12

18

72

57

9

20 71 11 10 4 6 24 19 3 0  x  3 is a zero.   So P x  x  3 4x 4  6x 3  24x 2  19x  3 . Continuing by factoring the quotient, we have: 4

3

6

24

19

3

12

18

18

3

4 6 6 1 0  x  3 is a zero again.   So P x  x  32 4x 3  6x 2  6x  1 . Continuing by factoring the quotient, we have: 3

4

6

6

1

12

54

144

1 2

4

6

6

1

2

4

1

48 1445  x  3 is an upper bound. 4 8 2 0  x  12 is a zero.       So P x  x  32 x  12 4x 2  8x  2  2 x  32 x  12 2x 2  4x  1 . Using the Quadratic Formula on     2 421 the second factor, we have x  4 422  44 24  1  26 . Therefore, the zeros are 12 , 3, and 1  26 . 4

18

55. (a) P x  x 3  3x 2  4x  12 has possible rational zeros 1, 2, 3,

(b)

y

4, 6, 12. 1

2

1

3

1 1 1

4

12

1

2

6

2

6

6

3

4

12

2

2

12

1

6

0

2 x

1

 x  2 is a zero.

  So P x  x  2 x 2  x  6  x  2 x  2 x  3. The real zeros of P are 2, 2, and 3.

56. (a) P x  x 3  2x 2  5x  6 has possible rational zeros 1, 2, 3,

(b)

y

6. 1

1 1

2

1 1

2

5

6

1

3

2

3

2

8

2 1

2

5

6

2

8

6

4

3

0

x

 x  2 is a zero.

    So P x  x  2 x 2  4x  3   x  2 x 2  4x  3   x  2 x  1 x  3. The real zeros of P are 2, 1, and 3.

SECTION 3.4 Real Zeros of Polynomials

57. (a) P x  2x 3  7x 2  4x  4 has possible rational zeros 1, 2, 4,

317

y

(b)

 12 .

1 2 7

4

4

2 2 7

2 5 1

4

4

4 6 4

2 5 1 3

2 3 2 0  x  2 is a zero.   So P x  x  2 2x 2  3x  2 . Continuing: 2

2 2

3

2

4

2

1

0

1 x

1

 x  2 is a zero again.

Thus P x  x  22 2x  1. The real zeros of P are 2 and  12 . 58. (a) P x  3x 3  17x 2  21x  9 has possible rational zeros 1, 3,

y

(b)

2

9,  13 ,  23 . 3

1

1

17

21

9

3

20

41

3

20

41

32

1 3

3

17

3

x

 x  1 is an upper bound.

21

9

1

6

9

18

27

0

 x  13 is a zero.

        So P x  x  13 3x 2  18x  27  3 x  13 x 2  6x  9  3 x  13 x  32 . The real zeros of P are 3 and 13 .

59. (a) P x  x 4  5x 3  6x 2  4x  8 has possible rational zeros 1,

(b)

y

2, 4, 8. 1

2

1

5

6

4

8

1

4

2

6

1

4

2

6

2

1

5

6

4

8

2

6

0

8

3

0

4

0

1

2 1

x

 x  2 is a zero.

  So P x  x  2 x 3  3x 2  4 and the possible rational zeros are restricted to 1, 2, 4. 2

1

3

0

4

2

2

4

1 1 2 0  x  2 is a zero again.  P x  x  22 x 2  x  2  x  22 x  2 x  1  x  23 x  1. So the real zeros of P are 1 and 2. 

318

CHAPTER 3 Polynomial and Rational Functions

60. (a) P x  x 4  10x 2  8x  8 has possible rational zeros 1, 2,

y

(b)

4, 8. 1 1

0 10

8 8

1 1

9 17

1 1 4 1

0

9 17 10

8 8

2 4 12 40

9 8

0 10

2 1

6 20 32

1 2

5

8

1

4 16 24 64 6 16 72  x  4 is an upper bound.

1 4

0

10

8

2

4

12

8

1 9 1 7 1 2   So P x  x  2 x 3  2x 2  6x  4 . Continuing, we have:

6

4

0

1

x

1

0

10

8

8

1

1

9

1

1

2

1

2

1

2

6

2

8

4

2

8  x  2 is a zero.

4

4

 x  2 is a zero again.   P x  x  22 x 2  4x  2 . Using the Quadratic Formula on the second factor, we have      2 412 8  42 2  2  2. So the real zeros of P are 2 and 2  2.  4 x  4 421 2 2 1

0

61. (a) P x  x 5  x 4  5x 3  x 2  8x  4 has possible rational zeros

(b)

y

1, 2, 4. 1

2

1 1

1

1

5

1

8

4

1

0

5

4

4

1

0

5

4

4

8

1

5

1

8

4

2

2

6

10

4

1 1

1

3 5 2 0  x  2 is a zero.   So P x  x  2 x 4  x 3  3x 2  5x  2 , and the possible rational zeros are restricted to 1, 2. 2

1

1 2

6

6

2

1

3

3

1

0

1

1

3

3

1

1

2

1

3

5

2

 x  2 is a zero again.  So P x  x  22 x 3  3x 2  3x  1 , and the possible rational zeros are restricted to 1. 

1

2 1 0  x  1 is a zero.  So P x  x  22 x  1 x 2  2x  1  x  22 x  13 ., and the real zeros of P are 1 and 2. 

x

SECTION 3.4 Real Zeros of Polynomials

62. (a) P x  x 5  x 4  6x 3  14x 2  11x  3 has possible rational zeros

(b)

319

y

1, 3. 1

1

1

6

14

11

3

1

0

6

8

3

1

0

6 8 3 0   So P x  x  1 x 4  6x 2  8x  3 : 1

1 1

0

6

8

3

1

1

5

3

 x  1 is a zero.

10 1

x

1

5 3 0  x  1 is a zero again.   2 3 2 So P x  x  1 x  x  5x  3 . 1

1

1

5

3

1

2

3

1 2 3 0  x  1 is a zero again.  So P x  x  13 x 2  2x  3  x  14 x  3, and the real zeros of P are 1 and 3. 

63. P x  x 3  x 2  x  3. Since P x has 1 variation in sign, P has 1 positive real zero. Since P x  x 3  x 2  x  3 has 2 variations in sign, P has 2 or 0 negative real zeros. Thus, P has 1 or 3 real zeros. 64. P x  2x 3  x 2  4x  7. Since P x has 3 variations in signs, P has 3 or 1 positive real zeros. Since P x  2x 3  x 2  4x  7 has no variation in sign, there is no negative real zero. Thus, P has 1 or 3 real zeros. 65. P x  2x 6  5x 4  x 3  5x  1. Since P x has 1 variation in sign, P has 1 positive real zero. Since P x  2x 6  5x 4  x 3  5x  1 has 1 variation in sign, P has 1 negative real zero. Therefore, P has 2 real zeros. 66. P x  x 4  x 3  x 2  x  12. Since P x has no variations in sign, P has no positive real zeros. Since P x  x 4  x 3  x 2  x  12 has 4 variations in sign, P has 4, 2, or 0 negative real zeros. Therefore, P x has 0, 2, or 4 real zeros. 67. P x  x 5  4x 3  x 2  6x. Since P x has 2 variations in sign, P has 2 or 0 positive real zeros. Since P x  x 5  4x 3  x 2  6x has no variation in sign, P has no negative real zero. Therefore, P has a total of 1 or 3 real zeros (since x  0 is a zero, but is neither positive nor negative). 68. P x  x 8  x 5  x 4  x 3  x 2  x  1. Since P x has 6 variations in sign, the polynomial has 6, 4, 2, or 0 positive real zeros. Since P x has no variation in sign, the polynomial has no negative real zeros. Therefore, P has 6, 4, 2, or 0 real zeros. 69. P x  2x 3  5x 2  x  2; a  3, b  1 3

2 2

1

2 2

5

1

2

6

3

12

1

4

14

5

1

2

2

7

8

7

8

6

Therefore a  3 and b  1 are lower and upper bounds.

alternating signs  lower bound.

all nonnegative  upper bound.

320

CHAPTER 3 Polynomial and Rational Functions

70. P x  x 4  2x 3  9x 2  2x  8; a  3, b  5 1

3

5

2

9

2

8

3

15

18

48

1

5

6

16

56

1

2

9

2

8

5

15

30

160

3

6

32

168

1

Alternating signs  lower bound.

All nonnegative  upper bound.

Therefore a  3 and b  5 are lower and upper bounds.

71. P x  8x 3  10x 2  39x  9; a  3, b  2 8

3

8 2

10

39

9

24

42

9

14

3

0

8 8

10

39

16

52

26

26

13

35

alternating signs  lower bound.

9 all nonnegative  upper bound.

Therefore a  3 and b  2 are lower and upper bounds. Note that x  3 is also a zero.

72. P x  3x 4  17x 3  24x 2  9x  1; a  0, b  6 0

3 3

6

3

17

24

9

1

0

0

0

0

17

24

9

1

Alternating signs  lower bound.

17

24

9

1

18

6

180

1026

1

30

171

1027

3

All nonnegative  upper bound.

Therefore a  0 and b  6 are lower and upper bounds. Note that because P x alternates in sign, by Descartes’ Rule of Signs, 0 is automatically a lower bound.

73. P x  x 4  2x 3  3x 2  5x  1; a  2, b  1 2

1 1

1

1 1

2

3

5

1

2

0

6

2

0

3

1

1

2

3

5

1

1 3

3

6

11

6

11

10

Therefore a  2 and b  1 are lower and upper bounds.

Alternating signs  lower bound.

All nonnegative  upper bound.

SECTION 3.4 Real Zeros of Polynomials

74. P x  x 4  3x 3  4x 2  2x  7; a  4, b  2 1

4

1 2

3

4

2

7

4

4

0

8

1

0

2

1

1 1

3

4

2

7

2

10

12

20

5

6

10

13

Alternating signs  lower bound.

All nonnegative  upper bound.

Therefore a  4 and b  2 are lower and upper bounds.

75. P x  2x 4  6x 3  x 2  2x  3; a  1, b  3 2

1

2 3

6

1

2 8

2

2

3

8

9

11

9

11

14

6

1

2

3

6

0

3

3

0

1

1

6

2

Alternating signs  lower bound.

All nonnegative  upper bound.

Therefore a  1 and b  3 are lower and upper bounds.

76. P x  3x 4  5x 3  2x 2  x  1; a  1, b  2 3

1

2

5

2

1

1

3

8

6

5

3

8

6

5

4

3

5

2

1

1

6

2

0

2

1

0

1

1

3

Therefore a  1 and b  2 are lower and upper bounds.

Alternating signs  lower bound.

All nonnegative  upper bound.

77. P x  x 3  3x 2  4 and use the Upper and Lower Bounds Theorem: 1

1

3

0

4

1

4

4

1

4

4

0

1

3

0

4

3

0

0

0

0

4

3

1

alternating signs  lower bound.

all nonnegative  upper bound.

Therefore 1 is a lower bound (and a zero) and 3 is an upper bound. (There are many possible solutions.)

78. P x  2x 3  3x 2  8x  12 and using the Upper and Lower Bounds Theorem: 2

2 2 3

3

8

12

4

14

12

7

6

0

2

3

8

6

9

3

3

1

15

2 (There are many possible solutions.)

Alternating signs  x  2 is a lower bound (and a zero). 12 All nonnegative  x  3 is an upper bound.

321

322

CHAPTER 3 Polynomial and Rational Functions

79. P x  x 4  2x 3  x 2  9x  2. 1

1

1

2

1

9

2

1

1

0

9

1

0

9

7

3

1

1 1

1 1

2

1

9

2

3

3

12

9

1

4

3

11

1 1

2

2

1

9

2

1

3

4

13

3

4

13

15

2

1

9

2

2

0

2

14

0

1

7

12

all positive  upper bound.

alternating signs  lower bound.

Therefore 1 is a lower bound and 3 is an upper bound. (There are many possible solutions.)

80. Set P x  x 5  x 4  1. 1

1

1

0

0

0

1

1

0

0

0

0

0

0

0

0

1

1 1

1 1

All nonnegative  x  1 is an upper bound.

1

0

0

0

1

1

2

2

2

2

2

2

2

2

1

(There are many possible solutions.)

81. P x  2x 4  3x 3  4x 2  3x  2. 1

  P x  x  1 2x 3  5x 2  x  2

1

2 2

2

Alternating signs  x  1 is a lower bound.

3

4

3

2

2

5

1

2

5

1

2

0

5

1

2

2

3

2

 x  1 is a zero.

2 3 2 0  x  1 is a zero.   P x  x  1 x  1 2x 2  3x  2  x  1 x  1 2x  1 x  2. Therefore, the zeros are 2, 12 , 1.

SECTION 3.4 Real Zeros of Polynomials

323

82. P x  2x 4  15x 3  31x 2  20x  4. The possible rational zeros are 1, 2, 4,  12 . Since all of the coefficients are positive, there are no positive zeros. Since P x  2x 4  15x 3  31x 2  20x  4 has 4 variations in sign, there are 0, 2, or 4 negative real zeros. 1 2 15

31

20

4

2 2 15

2 13 18 2 2 13

18

2

31

20

4 22 18 4 2 11

2

9

2

  P x  x  2 2x 3  11x 2  9x  2 : 2 2 11

9

2

4 2 11

4 14 10 2

7

4

9

 12 2 11

2

8 12 12 2

5 12

3

0  x  2 is a zero.

9

2

1 5 2

3 14

2 10

4

0 x   12 is a zero.

    P x  x  2 2x  1 x 2  5x  2 . Now if x 2  5x  2  0, then x  5 25412  52 17 . Thus, the zeros 2 

are 2,  12 , and 52 17 .

83. Method 1: P x  4x 4  21x 2  5 has 2 variations in sign, so by Descartes’ rule of signs there are either 2 or 0 positive zeros. If we replace x with x, the function does not change, so there are either 2 or 0 negative zeros. Possible rational zeros are 1,  12 ,  14 , 5,  52 ,  54 By inspection, 1 and 5 are not zeros, so we must look for non-integer solutions: 1 2

4

0

4

21

0

5

2

1

10

5

2

20

10

0

 x  12 is a zero.

   P x  x  12 4x 3  2x 2  20x  10 , continuing with the quotient, we have:  12

4 4



P x  x  12



2

20

10

2

0

10

0

20

0

 x   12 is a zero.

    1  2 4x  20  0. If 4x 2  20  0, then x   5. Thus the zeros are  12   5. x 2

Method 2: Substituting u  x 2 , the equation becomes 4u 2  21u  5  0, which factors:     4u 2  21u  5  4u  1 u  5  4x 2  1 x 2  5 . Then either we have x 2  5, so that x   5, or we have   x 2  14 , so that x   14   12 . Thus the zeros are  12   5.

324

CHAPTER 3 Polynomial and Rational Functions

  84. P x  6x 4  7x 3  8x 2  5x  x 6x 3  7x 2  8x  5 . So x  0 is a zero. Continuing with the quotient,

1 Q x  6x 3  7x 2  8x  5. The possible rational zeros are 1, 5,  12 ,  52 ,  ,  53 ,  16 ,  56 . Since Q x has 2 3

variations in sign, there are 0 or 2 positive real zeros. Since Q x  6x 4  7x 3  8x 2  5x has 1 variation in sign, there is 1 negative real zero. 6

1

6

7

8

5

6

1

9

1

9

4 1 2

6

7

8

5

30

115

535

6

23

107

540

8

5

5

6

7 3

2

5

6

4

10

0

All positive  upper bound.

 x  12 is a zero.

  P x  x 2x  1 3x 2  2x  5  x 2x  1 3x  5 x  1. Therefore, the zeros are 0, 1, 12 and 53 .

85. P x  x 5  7x 4  9x 3  23x 2  50x  24. The possible rational zeros are 1, 2, 3, 4, 6, 8, 12, 24. P x has 4 variations in sign and hence 0, 2, or 4 positive real zeros. P x  x 5  7x 4  9x 3  23x 2  50x  24 has 1 variation in sign, and hence P has 1 negative real zero. 1

1 1

7

9

23

50

24

1

6

3

26

24

6

3

26

24

0

 x  1 is a zero.

  P x  x  1 x 4  6x 3  3x 2  26x  24 ; continuing with the quotient, we try 1 again. 1

1 1

6

3

26

24

1

5

2

24

5

2

24

0

 x  1 is a zero again.

  P x  x  12 x 3  5x 2  2x  24 ; continuing with the quotient, we start by trying 1 again. 1

1 1

5

2

24

1

4

6

4

6

18

2

1 1

5

2

24

2

6

16

3

8

8

3

1 1

5

2

24

3

6

24

2

8

0

 x  3 is a zero.

  P x  x  12 x  3 x 2  2x  8  x  12 x  3 x  4 x  2. Therefore, the zeros are 2, 1, 3 4.

SECTION 3.4 Real Zeros of Polynomials

325

1 86. P x  8x 5  14x 4  22x 3  57x 2  35x  6. The possible rational zeros are 1, 2, 3, 6,  , 2 1  32 ,  14 ,  34 ,  ,  38 . Since P x has 4 variations in sign, there are 0, 2, or 4 positive real zeros. Since 8 P x  8x 5  14x 4  22x 3  57x 2  35x  6 has 1 variation in sign, there is 1 negative real zero. 1

8

14

22

57

35

6

8 8

22

22

0

57

92

0

57

92

98

2

8

14 16

8

30

57

35

6

60

76

38

6

38

19

3

0

22

 x  2 is a zero.

  P x  x  2 8x 4  30x 3  38x 2  19x  3 . All the other real zeros are positive. 8

1

8

30

38

19

3

8

22

16

3

22

16

3

0

  P x  x  2 x  1 8x 3  22x 2  16x  3 . 1

8 8

Since f

  1 2

22

16

3

8

14

2

14

2

1

 x  1 is a zero.

1 2

8 8

22

16

4

9

18

7

3 7 2 1 2

 0  f 1, there must be a zero between 12 and 1. We try 34 : 3 4

8 8

22

16

3

6

12

3

16

4

0

 x  34 is a zero.

    P x  x  2 x  1 4x  3 2x 2  4x  1 . Now, 2x 2  4x  1  0 when x  4 16421  22 2 . Thus, the 22 

zeros are 1, 34 , 2, and 22 2 .

87. P x  x 3  x  2. The only possible rational zeros of P x are 1 and 2. 1

1 1

0

1

1

1

0

1

0

2

2

2

1 1

0

1

2

2

4

6

2

3

4

1

1 1

0

1

1

1

0

1

0

2

2

Since the row that contains 1 alternates between nonnegative and nonpositive, 1 is a lower bound and there is no need to try 2. Therefore, P x does not have any rational zeros.

326

CHAPTER 3 Polynomial and Rational Functions

88. P x  2x 4  x 3  x  2. The only possible rational zeros of P x are 1, 2,  12 . 1 2

2

1

0

1 0

2 2

1

2

1

2

0

0

0

1

1 2 5 2

1

0

1

2

2

3

3

2

3

3

2

4

 12

All nonnegative  x  12 is an upper bound.

Alternating signs  x  1 is a lower bound. 2 2

1

0

1

2

1

1

2

1

 12 1 2

 14 7 4

Therefore, there is no rational zero.

89. P x  3x 3  x 2  6x  12 has possible rational zeros 1, 2, 3, 4, 6, 12,  13 ,  23 ,  43 . 3 1

3

1

6

12

2

4

8

2

3

5

4

20

1

3

4

2

14

2

3

7

8

4

1 3 2 3 4 3  13  23  43

all positive  x  2 is an upper bound alternating signs  x  2 is a lower bound

3

1

6

12

3

0

6

10

3

1

 16 3

3

3

2

3

2

 16 3

3

3

4

3

5

2 3

76 9 28 3 124 9 44 3 100 9

Therefore, there is no rational zero.

90. P x  x 50 5x 25 x 2 1. The only possible rational zeros of P x are 1. Since P 1  150 5 125 12 1  4 and P 1  150  5 125  12  1  6, P x does not have a rational zero.

91. P x  x 3  3x 2  4x  12, [4 4] by [15 15]. The

possible rational zeros are 1, 2, 3, 4, 6, 12. By

observing the graph of P, the rational zeros are x  2, 2, 3.

92. P x  x 4  5x 2  4, [4 4] by [30 30].

The possible rational solutions are 1, 2, 4.

By observing the graph of the equation, the solutions of the given equation are x  1, 2.

10

-4

-2

20

2 -10

4

-4

-2

2 -20

4

SECTION 3.4 Real Zeros of Polynomials

93. P x  2x 4  5x 3  14x 2  5x  12, [2 5] by

327

94. P x  3x 3  8x 2  5x  2, [3 3] by [10 10]

[40 40]. The possible rational zeros are 1, 2, 3,

The possible rational solutions are 1, 2,  13 ,  23 . By

4, 6, 12,  12 ,  32 . By observing the graph of P, the

observing the graph of the equation, the only real solution

zeros are  32 , 1, 1, 4.

of the given equation is x  2. 10

40 20 -2

2

-20

-2

4

2 -10

-40

95. x 4  x  4  0. Possible rational solutions are 1, 2, 4. 1

1 1

1

0

0

1

4

1

1

1

0

1

1

0

4

0

2

1

1

0 1

1

1

2

1

1

1

2

2

1

1

4

2

0

0

1

4

2

4

8

14

2

4

7

10

1

4

0

 x  2 is an upper bound.

1

0 2

4

8

18

1

2

4

9

14

 x  2 is a lower bound.

Therefore, we graph the function P x  x 4  x  4 in the viewing rectangle [2 2] by [5 20] and see there are two

solutions. In the viewing rectangle [13 125] by [01 01], we find the solution x  128. In the viewing rectangle

[1516] by [01 01], we find the solution x  153. Thus the solutions are x  128, 153. 20 10 -1.30 -2

-1

1

-1.28

0.1

0.1

0.0

0.0

-1.26

2

1.55 -0.1

1.60

-0.1

96. 2x 3  8x 2  9x  9  0. Possible rational solutions are 1, 3, 9,  12 ,  32 ,  92 . 1

2 2

8

9

9

2

6

3

6

3

6

3

2 2

8

9

9

6

6

9

2

3

0

 x  3 is a zero.

We graph P x  2x 3  8x 2  9x  9 in the viewing rectangle [4 6] by

[40 40]. It appears that the equation has no other real solution. We can factor   2x 3  8x 2  9x  9  x  3 2x 2  2x  3 . Since the quotient is a quadratic expression, we can use the Quadratic Formula to locate the other possible  2 423 , which are not real. So the only solution is x  3. solutions: x  2 222

40 20 -4

-2 -20 -40

2

4

6

328

CHAPTER 3 Polynomial and Rational Functions

97. 400x 4  400x 3  1096x 2  588x  909  0. 1 4 4 1096 588 4

2 4

4 1096 588

8 296 884

8

296 884

025

4 12

4 1096 588

909

4 8 2 4

909

8

8

4 4

296

3 4

592 008 004

24 2608 4040 1304

4 1096 12

901

4

8

909

202 4949  x  2 is an upper bound. 588

909

24 3912

135

1304

45 14409  x  3 is a lower bound.

Therefore, we graph the function P x  400x 4  400x 3  1096x 2  588x  909 in the viewing rectangle [3 2] by [10 40]. There appear to be two solutions. In the viewing rectangle [16 14] by [01 01], we find the solution x  150. In the viewing rectangle [08 12] by [0 1], we see that the graph comes close but does not go through the x-axis. Thus there is no solution here. Therefore, the only solution is x  150. 40

0.1

20

1.0 0.5

0.0 -1.6

-1.5 0.0

-3

-2

-1

1

2

0.8

-0.1

0.9

1.0

1.1

1.2

98. x 5  2x 4  096x 3  5x 2  10x  48  0. Since all the coefficients are positive, there is no positive solution. So x  0 is an upper bound. 2 1

2 096 2

1

5

10

48

3 1

0 192 616 768

0 096

308

384 288

2 096 3

5

10

48

3 1188 2064 9192

1 1 396

688 3064 8712  x  3 is a lower bound.

Therefore, we graph P x  x 5  2x 4  096x 3  5x 2  10x  48 in the viewing rectangle [3 0] by [10 5] and see that there are three possible solutions. In the viewing rectangle [175 17] by [01 01], we find the solution x  171. In the viewing rectangle [125 115] by [01 01], we find the solution x  120. In the viewing

rectangle [085 075] by [01 01], we find the solution x  080. So the solutions are x  171, 120, 080. 0.1

0.1

0.1

0 -2

0.0 -1.74 -10

-1.72

0.0 -1.25

-0.1

-1.20

0.0 -0.85

-0.1

-0.80 -0.1

SECTION 3.4 Real Zeros of Polynomials



329



99. Let r be the radius of the silo. The volume of the hemispherical roof is 12 43 r 3  23 r 3 . The volume of the cylindrical   section is  r 2 30  30r 2 . Because the total volume of the silo is 15,000 ft3 , we get the following equation: 2 r 3  30r 2  15000  2 r 3  30r 2  15000  0  r 3  45r 2  22500  0. Using a graphing device, we 3 3

first graph the polynomial in the viewing rectangle [0 15] by [10000 10000]. The solution, r  1128 ft., is shown in the viewing rectangle [112 114] by [1 1]. 1

10000

0

0 11.3

11.4

10

-1

-10000

100. Given that x is the length of a side of the rectangle, we have that the length of the diagonal is x  10, and the length of   the other side of the rectangle is x  102  x 2 . Hence x x  102  x 2  5000  x 2 20x  100  25,000,000  2x 3  10x 2  2,500,000  0  x 3  5x 2  1,250,000  0. The first viewing rectangle, [0 120] by [100 500],

shows there is one solution. The second viewing rectangle, [106 1061] by [01 01], shows the solution is x  10608.

Therefore, the dimensions of the rectangle are 47 ft by 106 ft.

0.1

400 200

0.0 106.05

106.10

0 50

101. h t  1160t  1241t 2  620t 3

 158t 4  020t 5  001t 6 is shown in the viewing rectangle

100

-0.1

(a) It started to snow again. (b) No, h t  4. (c) The function h t is shown in the viewing rectangle [6 65] by

[0 10] by [0 6].

[0 05]. The x-intercept of the function is a little less than 65, which means that the snow melted just before midnight on Saturday night.

5

0.4 0.2

0 0

5

10 0.0 6.0

6.2

6.4

330

CHAPTER 3 Polynomial and Rational Functions

102. The volume of the box is V  1500  x 20  2x 40  2x  4x 3  120x 2  800x    4x 3  120x 2  800x  1500  4 x 3  30x 2  200x  375  0. Clearly, we must have 20  2x  0, and so 0  x  10. 5

1

30

200

375

5

125

375

1 25 75 0  x  5 is a zero.   x 3  30x 2  200x  375  x  5 x 2  25x  75  0. Using the Quadratic Formula, we find the other zeros: 







 252 325  2552 13 . Since 2552 13  10, the two answers are: x  height  5 cm, x  25 6254175 2 

width  20  2 5  10 cm, and length  40  2 5  30 cm; and x  height  2552 13  349 cm,         width  20  25  5 13  5 13  5  1303 cm, and length  40  25  5 13  15  5 13  3303 cm. 103. Let r be the radius of the cone and cylinder and let h be the height of the cone. r

Since the height and diameter are equal, we get h  2r. So the volume of the

cylinder is V1  r 2  cylinder height  20r 2 , and the volume of the cone is 500 , it follows V2  13 r 2 h  13 r 2 2r  23 r 3 . Since the total volume is 3 500  r 3  30r 2  250  0. By Descartes’ Rule of that 23 r 3  20r 2  3 Signs, there is 1 positive zero. Since r is between 276 and 2765 (see the table),

the radius should be 276 m (correct to two decimals).

r 3  30r 2  250

1

219

2

122

3

47

27

1162

276

233

277

144

2765

144

28

715

 104. (a) Let x be the length, in ft, of each side of the base and let h be the height. The volume of the box is V  2 2  hx 2 ,    and so hx 2  2 2. The length of the diagonal on the base is x 2  x 2  2x 2 , and hence the length of the diagonal  between opposite corners is 2x 2  h 2  x  1. Squaring both sides of the equation, we have 2x 2  h 2  x 2  2x  1     x 2  2x  1 x 2   h 2  x 2  2x  1  h  x 2  2x  1. Therefore, 2 2  hx 2    x 2  2x  1 x 4  8  x 6  2x 5  x 4  8  0. (b) We graph y  x 6  2x 5  x 4  8 in the viewing rectangle [0 5] by [10 10], and we see that there are two solutions. In the second viewing rectangle, [14 15] by [1 1], we see the solution x  145. The third viewing rectangle, [225 235] by [1 1], shows the solution x  231. 1

10

0

0

2 -10

1

0 1.45

4 -1

1.50

2.30

2.35

-1

 If x  width  length  145 ft, then height  x 2  2x  1  134 ft, and if x  width  length  231 ft, then  height  x 2  2x  1  053 ft.

SECTION 3.4 Real Zeros of Polynomials

331

105. Let b be the width of the base, and let l be the length of the box. Then the length plus girth is l  4b  108, and the volume

is V  lb2  2200. Solving the first equation for l and substituting this value into the second equation yields l  108  4b    V  108  4b b2  2200  4b3  108b2  2200  0  4 b3  27b2  550  0. Now P b  b3  27b2  550 has two variations in sign, so there are 0 or 2 positive real zeros. We also observe that since l  0 b  27, so b  27 is an upper bound. Thus the possible positive rational real zeros are 1 2 3 10 11 22 25. 1

1 1

27

0

550

1

26

26

26

26

524

5

2

1 1

1

27

0

550

5

110

550

1

22

27

0

550

2

50

100

25

50

450

0  b  5 is a zero.     22 924  2230397 . The positive P b  b  5 b2  22b  110 . The other zeros are b  22 48441110  2 2 2 110

answer from this factor is b  2620Thus we have two possible solutions, b  5 or b  2620. If b  5, then

l  108  4 5  88; if b  2620, then l  108  4 2620  320. Thus the length of the box is either 88 in. or 320 in.

106. (a) An odd-degree polynomial must have a real zero. The end behavior of such a polynomial requires that the graph of the polynomial heads off in opposite directions as x   and x  . Thus the graph must cross the x-axis. (b) There are many possibilities one of which is P x  x 4  1.      (c) P x  x x  2 x  2  x 3  2x.          (d) P x  x  2 x  2 x  3 x  3  x 4  5x 2  6. If a polynomial with integer coefficients has no real zeros, then the polynomial must have even degree.

a for x we have 3    a 3 a 2 a x 3  ax 2  bx  c  X  c a X  b X  3 3 3   a2 2a a3 a2 ab 3 2 2  X  aX  X a X  X  bX  c 3 27 3 9 3

107. (a) Substituting X 

a3 a3 ab a2 2a 2 X  a X2  X  bX  c 3 27 3 9 3     2a 2 a3 ab a2 a3 b X    c  X 3  a  a X 2    3 3 27 9 3     ab 4a 3  X 3  b  a2 X   c 27 3  X3  a X2 

  (b) x 3 6x 2  9x  4  0. Setting a  6, b  9, and c  4, we have: X 3  9  62 X  32  18  4  X 3 27X  18.

332

CHAPTER 3 Polynomial and Rational Functions

        2 3   3 2 3 2 2 22 3 33         3 1  3 1  1  1  2 108. (a) Using the cubic formula, x  2 4 27 2 4 27 2

1

0

3

2

2

4

2

1 2 1 0  So x  2 x 2  2x  1  x  2 x  12  0  x  2, 1. Using the methods from this section, we have 

1

1

0

1

3

2

1

1

2

1

2

0  So x 3  3x  2  x  1 x 2  x  2  x  12 x  2  0  x  2, 1. 

Since this factors easily, the factoring method was easier.

(b) Using the cubic formula,         2 3 3  54 3  54 54 542 27 273   x       2 4 27 2 4 27       3 54 3 54 3 3 2 2 2 2  2  27  27  2  27  27  27  27  3  3  6 6

1 1

0

27

54

6

36

54

6

9

0  x 3  27x  54  x  6 x 2  6x  9  x  6 x  32  0  x  3, 6. 

Using methods from this section, 1

1 1

0

27

54

1

1

26

2

1

0

27

54

2

4

46

3

1 2 23 8  So x 3  27x  54  x  3 x 2  3x  18  x  6 x  32  0  x  3 6. 1

26



28

1 1

Since this factors easily, the factoring method was easier.

(c) Using the cubic formula,         2 3 3 3 4 3 33 4 4 42   x       2 4 27 2 4 27         3 3 3 3 2  4  1  2  4  1  2  5  2  5 

0

27

54

3

9

54

3

18

0

SECTION 3.4 Real Zeros of Polynomials

333

Using a graphing calculator, we see that P x  x 3  3x  4 has one zero. Using methods from this section, P x has possible rational zeros 1, 2, 4. 1

1 1

0

3

4

1

1

4

1

4

4

1

1

0

3

4

1

1

4

 1 is an upper bound. 1 1 4 0  x  1 is a zero.   P x  x 3  3x  4  x  1 x 2  x  4 . Using the Quadratic Formula we have

  1 12 414  1 215 which is not a real number. Since it is not easy to see that x  2

    3 3 2  5  2  5  1, we see that the factoring method was much easier.

109. (a) Since z  b, we have z  b  0. Since all the coefficients of Q x are nonnegative, and since z  0, we have Q z  0 (being a sum of positive terms). Thus, P z  z  b  Q z  r  0, since the sum of a positive number and a nonnegative number. (b) In part (a), we showed that if b satisfies the conditions of the first part of the Upper and Lower Bounds Theorem and z  b, then P z  0. This means that no real zero of P can be larger than b, so b is an upper bound for the real zeros. (c) Suppose b is a negative lower bound for the real zeros of P x. Then clearly b is an upper bound for P1 x  P x. Thus, as in Part (a), we can write P1 x  x  b  Q x  r, where r  0 and the coefficients of Q are all nonnegative, and P x  P1 x  x  b  Q x  r  x  b  [Q x]  r . Since the coefficients of Q x are all nonnegative, the coefficients of Q x will be alternately nonpositive and nonnegative, which proves the second part of the Upper and Lower Bounds Theorem. 110. P x  x 5  x 4  x 3  5x 2  12x  6 has possible rational zeros 1, 2, 3, 6. Since P x has 1 variation in sign,

there is 1 positive real zero. Since P x  x 5  x 4  x 3  5x 2  12x  6 has 4 variations in sign, there are 0, 2, or 4 negative real zeros. 1

1

1

1

5

12

6

1

0

1

6

18

0

1

6

18

24

1

2

1

3 1 1 1 5 12 6 3

1

1

5

12

6

2

2

2

6

36

1

1

3

18

42

1 1 1 1 5 12 6 2 1

6

5 10 18 48  3 is an upper bound. 1 2 1 6   P x  x  1 x 4  2x 3  x 2  6x  6 , continuing with the quotient we have

6

1

6 15

1

30 54

1

2

1

1 1

2

1

6

6

1

3

4

10

3

4

10

4

6 0  x  1 is a zero.

 1 is a lower bound.

Therefore, there is 1 rational zero, namely 1. Since there are 1, 3 or 5 real zeros, and we found 1 rational zero, there must

be 0, 2 or 4 irrational zeros. However, since 1 zero must be positive, there cannot be 0 irrational zeros. Therefore, there is exactly 1 rational zero and 2 or 4 irrational zeros.

334

CHAPTER 3 Polynomial and Rational Functions

3.5

COMPLEX ZEROS AND THE FUNDAMENTAL THEOREM OF ALGEBRA

1. The polynomial P x  5x 2 x  43 x  7 has degree 6. It has zeros 0, 4, and 7. The zero 0 has multiplicity 2, and the zero 4 has multiplicity 3. 2. (a) If a is a zero of the polynomial P, then x  a must be a factor of P x.

(b) If a is a zero of multiplicity m of the polynomial P, then x  am must be a factor of P x when we factor P completely.

3. A polynomial of degree n  1 has exactly n zeros, if a zero of multiplicity m is counted m times. 4. If the polynomial function P has real coefficients and if a  bi is a zero of P, then a  bi is also a zero of P. So if 3  i is a zero of P, then 3  i is also a zero of P. 5. P x  x 4  1 (a) True, P has degree 4, so by the Zeros Theorem it has four (not necessarily distinct) complex zeros. (b) True, by the Complete Factorization Theorem, this is true. (c) False, the fourth power of any real number is nonnegative, so P x  1 for all real x and P has no real zeros. 6. P x  x 3  x

  (a) False. P x  x x 2  1 , and x 2  1 has no real zeros. (b) True, P 0  0.

(c) False, P x cannot be factored into linear factors with real coefficients because x 2  1 has no real zeros.   7. (a) x 4  4x 2  0  x 2 x 2  4  0. So x  0 or x 2  4  0. If x 2  4  0 then x 2  4  x  2i. Therefore, the solutions are x  0 and 2i.

(b) To get the complete factorization, we factor the remaining quadratic factor P x  x 2 x  4  x 2 x  2i x  2i.   8. (a) x 5  9x 3  0  x 3 x 2  9  0. So x  0 or x 2  9  0. If x 2  9  0 then x  3i. Therefore, the zeros of P are x  0, 3i.

(b) Since 3i and 3i are the zeros from x 2  9  0, x  3i and x  3i are the factors of x 2  9. Thus the complete   factorization is P x  x 3 x 2  9  x 3 x  3i x  3i.

  9. (a) x 3  2x 2  2x  0  x x 2  2x  2  0. So x  0 or x 2  2x  2  0. If x 2  2x  2  0 then   2 22 412  2 2 4  22i x 2 2  1  i. Therefore, the solutions are x  0, 1  i.

(b) Since 1  i and 1  i are zeros, x  1  i  x  1  i and x  1  i  x  1  i are the factors of x 2  2x  2.   Thus the complete factorization is P x  x x 2  2x  2  x x  1  i x  1  i.

  10. (a) x 3  x 2  x  0 x x 2  x  1  0. So x  0 or x 2  x  1  0. If x 2  x  1  0 then

    1 12 411  12 3   12  i 23 . Therefore, the zeros of P are x  0,  12  i 23 . x 21   (b) The zeros of x 2  x  1  0 are  12  i 23 and  12  i 23 , so factoring we get x 2  x               1  x   12  i 23 x   12  i 23  x  12  i 23 x  12  i 23 . Thus the complete factorization is        P x  x x 2  x  1  x x  12  i 23 x  12  i 23 .

SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra

335

 2 11. (a) x 4  2x 2  1  0  x 2  1  0  x 2  1  0  x 2  1  x  i. Therefore the zeros of P are x  i. (b) Since i and i are zeros, x  i and x  i are the factors of x 2  1. Thus the complete factorization is  2 P x  x 2  1  [x  i x  i]2  x  i2 x  i2 .

    12. (a) x 4  x 2  2  0  x 2  2 x 2  1  0. So x 2  2  0 or x 2  1  0. If x 2  2  0 then x 2  2  x   2.  And if x 2  1  0 then x 2  1  x  i. Therefore, the zeros of P are x   2, i. (b) To get the complete factorization, we factor the quadratic factors to get         P x  x 2  2 x 2  1  x  2 x  2 x  i x  i.

     13. (a) x 4  16  0  0  x 2  4 x 2  4  x  2 x  2 x 2  4 . So x  2 or x 2  4  0. If x 2  4  0 then x 2  4  x  2i. Therefore the zeros of P are x  2, 2i.

(b) Since i and i are zeros, x  i and x  i are the factors of x 2  1. Thus the complete factorization is   P x  x  2 x  2 x 2  4  x  2 x  2 x  2i x  2i.

 2  14. (a) x 4  6x 2  9  0  x 2  3  0  x 2  3. So x  i 3 are the only zeros of P (each of multiplicity 2). (b) To get the complete factorization, we factor the quadratic factor to get  2   2    2   2  x i 3 . P x  x 2  3  x  i 3 x  i 3  x i 3

  15. (a) x 3  8  0  x  2 x 2  2x  4  0. So x  2 or x 2  2x  4  0. If x 2  2x  4  0 then

   2 22 414 2 12 22i 3  1i 3. Therefore, the zeros of P are x  2, 1  i 3.   x 2 2 2

        (b) Since 1  i 3 and 1  i 3 are the zeros from the x 2  2x  4  0, x  1  i 3 and x  1  i 3 are the factors of x 2  2x  4. Thus the complete factorization is           P x  x  2 x 2  2x  4  x  2 x  1  i 3 x  1i 3       x  2 x  1  i 3 x  1  i 3

  16. (a) x 3  8  0  x  2 x 2  2x  4  0. So x  2 or x 2  2x  4  0. If x 2  2x  4  0 then

   2 22 414 3  1i 3. Therefore, the zeros of P are x  2, 1  i 3.  22 12  22i x 2 2

        (b) Since 1  i 3 and 1  i 3 are the zeros from x 2  2x  4  0, x  1  i 3 and x  1  i 3 are the factors of x 2  2x  4. Thus the complete factorization is           P x  x  2 x 2  2x  4  x  2 x  1  i 3 x  1  i 3       x  2 x  1  i 3 x  1  i 3

       17. (a) x 6  1  0  0  x 3  1 x 3  1  x  1 x 2  x  1 x  1 x 2  x  1 . Clearly, x  1 are solutions. 







If x 2  x  1  0, then x  1 1411  12 3   12  23 so x   12  i 23 . And if x 2  x  1  0, then 2 



 1 2 3  12  x  1 1411 2

    3 1  i 3 . Therefore, the zeros of P are x  1,  1  i 3 , 1  i 3 .  2 2 2 2 2 2 2

336

CHAPTER 3 Polynomial and Rational Functions   3 and  1  i 3 , so x 2  x  1 factors as 2 2 2      3 1 x  2 i 2 x  12  i 23 . Similarly,since   the zeros of x 2  x  1  0 are 12  i 23 and 12  i 23 , so x 2  x  1 factors as              1 x  12  i 23  x  12  i 23 x  12  i 23 . Thus the complete  i 23 x 2

(b) The zeros of x 2  x  1  0 are  12  i         x   12  i 23  x   12  i 23

factorization is

    P x  x  1 x 2  x  1 x  1 x 2  x  1

          x  1 x  1 x  12  i 23 x  12  i 23 x  12  i 23 x  12  i 23        18. (a) x 6  7x 3  8  0  0  x 3  8 x 3  1  x  2 x 2  2x  4 x  1 x 2  x  1 . Clearly, x  1 and     x  2 are solutions. If x 2  2x  4  0, then x  2 4414  22 12   22  2 23 so x  1  i 3. If 2 







 1 2 3  12  23  12  i 23 . Therefore, the zeros of P are x  1, 2, x 2  x  1  0, then x  1 1411 2   1  i 3, 12  i 23 .      (b) From Exercise 10, x 2  2x  4  x  1  i 3 x  1  i 3 and from Exercise 11,      x 2  x  1  x  12  i 23 x  12  i 23 . Thus the complete factorization is     P x  x  2 x 2  2x  4 x  1 x 2  x  1

          x  2 x  1 x  1  i 3 x  1  i 3 x  12  i 23 x  12  i 23

19. P x  x 2  25  x  5i x  5i. The zeros of P are 5i and 5i, both multiplicity 1.

20. P x  4x 2  9  2x  3i 2x  3i. The zeros of P are 32 i and  32 i, both multiplicity 1.

  2 22 412 2 21. Q x  x  2x  2. Using the Quadratic Formula x   22 4  22i  1  i. So 21 2

Q x  x  1  i x  1  i. The zeros of Q are 1  i (multiplicity 1) and 1  i (multiplicity 1).   22. Q x  x 2  8x  17  x 2  8x  16  1  x  42  1  [x  4  i] [x  4  i]  x  4  i x  4  i. The zeros of Q are 4  i and 4  i, both multiplicity 1.   23. P x  x 3  4x  x x 2  4  x x  2i x  2i. The zeros of P are 0, 2i, and 2i (all multiplicity 1).   24. P x  x 3  x 2  x  x x 2  x  1 . Using the Quadratic Formula, we have

     1 12 411  1 2 3  12  i 23 . The zeros of P are 0, 12  i 23 , and 12  i 23 , all of multiplicity 1. And x 21      P x  x x  12  i 23 x  12  i 23 .

     25. Q x  x 4  1  x 2  1 x 2  1  x  1 x  1 x 2  1  x  1 x  1 x  i x  i. The zeros of Q are

1, 1, i, and i (all of multiplicity 1).      26. Q x  x 4  625  x 2  25 x 2  25  x  5 x  5 x 2  25  x  5 x  5 x  5i x  5i. The zeros

of Q are 5, 5, 5i, and 5i, all multiplicity 1.    27. P x  16x 4  81  4x 2  9 4x 2  9  2x  3 2x  3 2x  3i 2x  3i. The zeros of P are 32 ,  32 , 32 i, and  32 i (all of multiplicity 1).

SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra



 28. P x  x 3  64  x  4 x 2  4x  16 .

337

Using the Quadratic Formula, we have

   3  2  2i 3. The zeros of P are 4, 2  2i 3, and 2  2i 3, all of x  4 164116  42 48  44i 2 2

     multiplicity 1, and P x  x  4 x  2  2i 3 x  2  2i 3 .   29. P x  x 3  x 2  9x  9  x 2 x  1  9 x  1  x  1 x 2  9  x  1 x  3i x  3i. The zeros of P are 1, 3i, and 3i (all of multiplicity 1).        30. P x  x 6  729  x 3  27 x 3  27  x  3 x 2  3x  9 x  3 x 2  3x  9 . Using the 

 27 2 . Using the     3 9419 3 27 27 3  9 we have x    2  2  32  3 2 3 i 2 2      3 2 3 i,  32  3 2 3 i, 32  3 2 3 i, and 32  3 2 3 i, all multiplicity 1. And          3 2 3 i x  32  3 2 3 i x  32  3 2 3 i x  32  3 2 3 i . 

Quadratic Formula on x 2  3x  9 we have x  3 9419  32 27   32  2

Quadratic Formula on x 2  3x The zeros of P are 3, 3,  32  P x  x  3 x  3 x  32

 2 31. Q x  x 4  2x 2  1  x 2  1  x  i2 x  i2 . The zeros of Q are i and i (both of multiplicity 2).

 2   2    2   2   x  i 5 . The zeros of Q are i 5  x i 5 32. Q x  x 4  10x 2  25  x 2  5  x  i 5 x  i 5  and i 5, both of multiplicity 2.    33. P x  x 4  3x 2  4  x 2  1 x 2  4  x  1 x  1 x  2i x  2i. The zeros of P are 1, 1, 2i, and 2i (all of multiplicity 1).

         34. P x  x 5  7x 3  x 3 x 2  7  x 3 x  i 7 x  i 7 . The zeros of P are 0 (multiplicity 3), i 7 and i 7, both of multiplicity 1.

   2   2  2  x  i 3 . The zeros of P are 0 35. P x  x 5  6x 3  9x  x x 4  6x 2  9  x x 2  3  x x  i 3   (multiplicity 1), i 3 (multiplicity 2), and i 3 (multiplicity 2).   2 2 36. P x  x 6  16x 3  64  x 3  8  x  22 x 2  2x  4 . Using the Quadratic Formula, on x 2  2x  4 we have           2 22 414 2 12 22i 3  1  i 3, so P x  x  22 x  1  i 3 2 x  1  i 3 2 .   x 2 2 2

  The zeros of P are 2, 1  i 3, and 1  i 3, all multiplicity 2.

37. Since 1  i and 1  i are conjugates, the factorization of the polynomial must be   P x  a x  [1  i] x  [1  i]  a x 2  2x  2 . If we let a  1, we get P x  x 2  2x  2.   38. Since 1  i 2 and 1  i 2 are conjugates, the factorization of the polynomial must be           P x  c x  1  i 2 x  1  i 2  c x 2  2x  3 . If we let c  1, we get P x  x 2  2x  3.

39. Since 2i and 2i are conjugates, the factorization of the polynomial must be     Q x  b x  3 x  2i x  2i]  b x  3 x 2  4  b x 3  3x 2  4x  12 . If we let b  1, we get Q x  x 3  3x 2  4x  12.

40. Since i is a zero, by the Conjugate Roots Theorem, i is also a zero. So the factorization of the polynomial must be     Q x  b x  0 x  i x  i  bx x 2  1  b x 3  x . If we let b  1, we get Q x  x 3  x.

41. Since i is a zero, by the Conjugate Roots Theorem, i is also a zero. So the factorization of the polynomial must be   P x  a x  2 x  i x  i  a x 3  2x 2  x  2 . If we let a  1, we get P x  x 3  2x 2  x  2.

338

CHAPTER 3 Polynomial and Rational Functions

42. Since 1  i is a zero, by the Conjugate Roots Theorem, 1  i is also a zero. So the factorization of the polynomial must be     Q x  a x  3 x  [1  i] x  [1  i]  a x  3 x 2  2x  2  a x 3  x 2  4x  6 . If we let a  1, we get Q x  x 3  x 2  4x  6.

43. Since the zeros are 1  2i and 1 (with multiplicity 2), by the Conjugate Roots Theorem, the other zero is 1  2i. So a factorization is R x  c x  [1  2i] x  [1  2i] x  12  c [x  1]  2i [x  1]  2i x  12           c [x  1]2  [2i]2 x 2  2x  1  c x 2  2x  1  4 x 2  2x  1  c x 2  2x  5 x 2  2x  1      c x 4  2x 3  x 2  2x 3  4x 2  2x  5x 2  10x  5  c x 4  4x 3  10x 2  12x  5 If we let c  1 we get R x  x 4  4x 3  10x 2  12x  5.

44. Since S x has zeros 2i and 3i, by the Conjugate Roots Theorem, the other zeros of S x are 2i and 3i. So a factorization of S x is         S x  C x  2i x  2i x  3i x  3i  C x 2  4i 2 x 2  9i 2  C x 2  4 x 2  9  C x 4  13x 2  36 If we let C  1, we get S x  x 4  13x 2  36.

45. Since the zeros are i and 1  i, by the Conjugate Roots Theorem, the other zeros are i and 1  i. So a factorization is T x  C x  i x  i x  [1  i] x  [1  i]          C x 2  i 2 [x  1]  i [x  1]  i  C x 2  1 x 2  2x  1  i 2  C x 2  1 x 2  2x  2      C x 4  2x 3  2x 2  x 2  2x  2  C x 4  2x 3  3x 2  2x  2  C x 4  2C x 3  3C x 2  2C x  2C Since the constant coefficient is 12, it follows that 2C  12  C  6, and so   T x  6 x 4  2x 3  3x 2  2x  2  6x 4  12x 3  18x 2  12x  12.

46. Since U x has zeros 12 , 1 (with multiplicity two), and i, by the Conjugate Roots Theorem, the other zero is i. So a

factorization of U x is        U x  c x  12 x  12 x  i x  i  12 c 2x  1 x 2  2x  1 x 2  1  12 c 2x 5  3x 4  2x 3  2x 2  1 Since the leading coefficient is 4, we have 4  12 c 2  c. Thus we have   U x  12 4 2x 5  3x 4  2x 3  2x 2  1  4x 5  6x 4  4x 3  4x 2  2.

  47. P x  x 3  2x 2  4x  8  x 2 x  2  4 x  2  x  2 x 2  4  x  2 x  2i x  2i. Thus the zeros are 2 and 2i.

48. P x  x 3  7x 2  17x  15. We start by trying the possible rational factors of the polynomial: 1

1

7

17

15

1

6

11

3

1

7

17

15

3

12

15

1 4 5 0  x  3 is a zero. 4  So P x  x  3 x 2  4x  5 . Using the Quadratic Formula on the second factor, we have 1





6



11

x  4 16415  4 2 4  42i 2 2  2  i. Thus the zeros are 3, 2  i.

SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra

339

49. P x  x 3  2x 2  2x  1. By inspection, P 1  1  2  2  1  0, and hence x  1 is a zero. 1

1 1





2

2

1

1

1

1

1

1

0

Thus P x  x  1 x 2  x  1 . So x  1 or x 2  x  1  0.     1i2 3 . Hence, the zeros are 1 and 1i2 3 . Using the Quadratic Formula, we have x  1 1411 2 50. P x  x 3  7x 2  18x  18 has possible rational zeros 1, 2, 3, 6, 9, 18. Since all of the coefficients are positive, there are no positive real zeros. 1 1

7 18

18

2 1

1 6 12

7

18

18

3 1

2 10 16

7

18

18

3 12 18

1 5 8 2 1 4 6 0  x  3 is a zero. 6 12 6   So P x  x  3 x 2  4x  6 . Using the Quadratic Formula on the second factor, we have      2  2  i 2. Thus the zeros are 3, 2  i 2. x  4 16416  42 8  42i 2 2 1

51. P x  x 3  3x 2  3x  2. 3

3

2

2

2

2

1 1 1  Thus P x  x  2 x 2  x  1 . So x  2 or x 2  x  1  0

0

2



1







Using the Quadratic Formula we have x  1 1411  1i2 3 . Hence, the zeros are 2, and 1i2 3 . 2 52. P x  x 3  x  6 has possible zeros 1, 2, 3. 1

1

0

1

6

1

1

0

2

1

0

1

6

2

4

6

1 2 3 0  x  2 is a zero. 0 6     2  1  i 2. Thus the P x  x  2 x 2  2x  3 . Now x 2  2x  3 has zeros x  2 4413  22i 2 2  zeros are 2, 1  i 2. 1

1



53. P x  2x 3  7x 2  12x  9 has possible rational zeros 1, 3, 9,  12 ,  32 ,  92 . Since all coefficients are positive, there are no positive real zeros. 1

2 2

7

12

9

2

5

7

5

7

2

2

2

There is a zero between 1 and 2.

 32

2

2

7

12

9

3

6

9

7

12

9

4

6

12

3

6

3

2 4 6 0  x   32 is a zero.      P x  x  32 2x 2  4x  6  2 x  32 x 2  2x  3 . Now x 2  2x  3 has zeros     22 2 x  2 4431   1  i 2. Hence, the zeros are  32 and 1  i 2. 2 2 

340

CHAPTER 3 Polynomial and Rational Functions

54. Using synthetic division, we see that x  3 is a factor of the polynomial: 1

2

8

9

9

2

6

3

2

3

8

9

9

6

6

9

2 2 3 0  x  3 is a zero.   So P x  2x 3  8x 2  9x  9  x  3 2x 2  2x  3 . Using the Quadratic Formula, we find the other two solutions:     2  20 2  4  4 3 2   12  25 i. Thus the zeros are 3, 12  25 i. x 2 2 4 2

3

6

6

55. P x  x 4  x 3  7x 2  9x  18. Since P x has one change in sign, we are guaranteed a positive zero, and since P x  x 4  x 3  7x 2  9x  18, there are 1 or 3 negative zeros. 1

1

1

7

9

18

1

2

9

18

1 2 9 18 0  Therefore, P x  x  1 x 3  2x 2  9x  18 . Continuing with the quotient, we try negative zeros. 

1

1

2

9

18

1

1

8

2

1

2

9

18

2

0

18

1 0 9 0 1 8 10  P x  x  1 x  2 x 2  9  x  1 x  2 x  3i x  3i. Therefore,the zeros are 1, 2, and 3i. 1



56. P x  x 4  2x 3  2x 2  2x  3 has possible zeros 1, 3. 1

1

2

2

2

3

1

1

3

5

3

1

2

2

2

3

3

3

3

3

1 1 1 1 0  x  3 is a zero. 5 8  P x  x  3 x 3  x 2  x  1 . If we factor the second factor by grouping, we get   x 3  x 2  x  1  x 2 x  1  1 x  1  x  1 x 2  1 . So we have   P x  x  3 x  1 x 2  1  x  3 x  1 x  i x  i. Thus the zeros are 3, 1, i, and i. 1 

1

3

57. We see a pattern and use it to factor by grouping. This gives   P x  x 5  x 4  7x 3  7x 2  12x  12  x 4 x  1  7x 2 x  1  12 x  1  x  1 x 4  7x 2  12          x  1 x 2  3 x 2  4  x  1 x  i 3 x  i 3 x  2i x  2i

 Therefore,the zeros are 1, i 3, and 2i.

           58. P x  x 5  x 3  8x 2  8  x 3 x 2  1  8 x 2  1  x 2  1 x 3  8  x 2  1 x  2 x 2  2x  4

(factoring a sum of cubes). So x  2, or x 2  1  0. If x 2  1  0, then x 2  1  x  i. If x 2  2x  4  0, then     12 x  2 4414  1   1  i 3. Thus, the zeros are 2, i, 1  i 3. 2 2

SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra

341

59. P x  x 4  6x 3  13x 2  24x  36 has possible rational zeros 1, 2, 3, 4, 6, 9, 12, 18. P x has 4 variations in sign and P x has no variation in sign. 1 1 6 13 24 1 5 1 5

36

8 16

8 16

Continuing:

2 1 6 13 24 2 8

20

1 4

36

3 1 6 13 24

10 28

5 14

3 9

8

1 3

36

12 36

4 12

0  x  3 is a zero.

3 1 3 4 12 3 0

12

1 0 4 0  x  3 is a zero.  P x  x  32 x 2  4  x  32 x  2i x  2i. Therefore,the zeros are 3 (multiplicity 2) and 2i. 

60. P x  x 4  x 2  2x  2 has possible rational zeros 1, 2. 1 1 0 1 2 2 1

1 1

0 1 2

1 0 2

2

1 1 1 0

1 0 2

1

2

1 2 2

1 1

0 2 4 1 is an upper bound. 1 1 0 2 0 1 2 2 0   P x  x  12 x 2  2x  2 . Using the Quadratic Formula on x 2  2x  2, we have x  2 248  22i 2  1  i. Thus, the zeros of P x are 1 (multiplicity 2) and 1  i. 

61. P x  4x 4  4x 3  5x 2  4x  1 has possible rational zeros 1,  12 ,  14 . Since there is no variation in sign, all real zeros (if there are any) are negative. 1

4

4

5

4

1

4

0

5

1

 12

4 0 5 1 2   1 3 P x  x  2 4x  2x 2  4x  2 . Continuing: 

 12

4 4

4 4

2

4

2

2

0

2

0

4

0

4

5

4

1

2

1

2

1

2

4

2

0

 x   12 is a zero.

 x   12 is a zero again.

 2   P x  x  12 4x 2  4 . Thus, the zeros of P x are  12 (multiplicity 2) and i.

62. P x  4x 4  2x 3  2x 2  3x  1 has possible rational zeros 1,  12 ,  14 . P has one variation in sign, so P has one positive real zero. 4

2

2

3

1

4

6

4

1

4 6 4   P x  x  1 4x 3  6x 2  4x  1 . Continuing:

1

0

 12

4

1

1

4

6

4

1

4

2

2

 1 is a zero.

6

4

1

2

2

1

2 2 3 4 4 2 0   12 is a zero.   P x  x  1 x  12 4x 2  4x  2 . Using the Quadratic Formula on 4x 2  4x  2, we find 4





x  4 81632   12  12 i. Thus, P has zeros 1,  12 ,  12  12 i.

342

CHAPTER 3 Polynomial and Rational Functions

63. P x  x 5  3x 4  12x 3  28x 2  27x  9 has possible rational zeros 1, 3, 9. P x has 4 variations in sign and P x has 1 variation in sign.

1

1

1 1

1

3

12

28

27

9

1

2

10

18

9

2

10

18

9

0

2

10

18

9

1

1

9

9

 x  1 is a zero. 1

1

1

9

9

1

0

9

1

1 9 9 0  x  1 is a zero. 1 0 9 0  x  1 is a zero.   P x  x  13 x 2  9  x  13 x  3i x  3i. Therefore,the zeros are 1 (multiplicity 3) and 3i 64. P x  x 5  2x 4  2x 3  4x 2  x  2 has possible rational zeros 1, 2. 1

1

2

2

4

1

2

1

1

1

3

2

2

1

2

2

4

1

2

2

0

4

0

2

1

1 1 3 2 4 1 0 2 0 1 0  x  2 is a zero.   2  P x  x  2 x 4  2x 2  1  x  2 x 2  1  x  2 x  i2 x  i2 . Thus, the zeros of P x are 2, i.

  65. (a) P x  x 3  5x 2  4x  20  x 2 x  5  4 x  5  x  5 x 2  4 (b) P x  x  5 x  2i x  2i

66. (a) P x  x 3  2x  4 1

1 1

0

2

4

1

1

1

1 

1

2

5  P x  x 3  2x  4  x  2 x 2  2x  2

1 1

0

2

4

2

4

4

2

2

0

(b) P x  x  2 x  1  i x  1  i

     67. (a) P x  x 4  8x 2  9  x 2  1 x 2  9  x  1 x  1 x 2  9 (b) P x  x  1 x  1 x  3i x  3i

 2 68. (a) P x  x 4  8x 2  16  x 2  4 (b) P x  x  2i2 x  2i2

       69. (a) P x  x 6  64  x 3  8 x 3  8  x  2 x 2  2x  4 x  2 x 2  2x  4          (b) P x  x  2 x  2 x  1  i 3 x  1  i 3 x  1  i 3 x  1  i 3        70. (a) P x  x 5  16x  x x 4  16  x x 2  4 x 2  4  x x  2 x  2 x 2  4 (b) P x  x x  2 x  2 x  2i x  2i

SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra

343



 71. (a) x 4  2x 3  11x 2  12x  x x 3  2x 2  11x  12  0. We first find the bounds for our viewing rectangle.

5

1

2

11

12

5

1

3

4

32

 x  5 is an upper bound.

4

1

6

13

50

 x  4 is a lower bound.

-20 -40

We graph P x  x 4  2x 3  11x 2  12x in the viewing rectangle [4 5] by [50 10] and see that it has 4 real solutions. Since this matches the degree of P x, P x has no nonreal solution. (b) x 4  2x 3  11x 2  12x  5  0. We use the same bounds for our viewing rectangle, [4 5] by [50 10], and see that R x  x 4  2x 3  11x 2  12x  5 has

(c) x 4  2x 3  11x 2  12x  40  0. We graph

T x  x 4  2x 3  11x 2  12x  40 in the viewing

rectangle [4 5] by [10 50], and see that T has no

2 real solutions. Since the degree of R x is 4, R x

real solution. Since the degree of T is 4, T must have

must have 2 nonreal solutions.

4 nonreal solutions.

5 -20

40 20

-40 5

72. (a) 2x  4i  1  2x  1  4i  x  12  2i. (b) x 2  i x  0  x x  i  0  x  0, i.

(c) x 2  2i x  1  0  x  i2  0  x  i.

(d) i x 2  2x  i  0. Using the Quadratic Formula, we get           2 22 4ii 2  1 2  1  2 i  1  2 i. x  22i 8  22 i 2i 2i 73. (a) P x  x 2  1  i x  2  2i. So P 2i  2i2  1  i 2i  2  2i  4  2i  2  2  2i  0, and P 1  i  1  i2  1  i 1  i  2  2i  1  2i  1  1  1  2  2i  0.

Therefore, 2i and 1  i are solutions of the equation x 2  1  i x  2  2i  0. However, P 2i  2i2  1  i 2i  2  2i  4  2i  2  2  2i  4  4i, and

P 1  i  1  i2  1  i 1  i  2  2i  2  2i. Since, P 2i  0 and P 1  i  0, 2i and 1  i are not solutions. (b) This does not violate the Conjugate Roots Theorem because the coefficients of the polynomial P x are not all real. 74. (a) Because i and 1  i are zeros, i and 1  i are also zeros. Thus,

   P x  C x  i x  i x  [1  i] x  [1  i]  C x 2  1 x 2  2x  2      C x 4  2x 3  2x 2  x 2  2x  2  C x 4  2x 3  3x 2  2x  2

Because C  1, the polynomial is P x  x 4  2x 3  3x 2  2x  2.

344

CHAPTER 3 Polynomial and Rational Functions

(b) Because i and 1  i are zeros,

    P x  C x  i x  [i  1]  C x 2  xi  x  xi  1  i  C x 2  1  2i x  1  i

Because C  1, the polynomial is P x  x 2  1  2i x  1  i.

75. Because P has real coefficients, the imaginary zeros come in pairs: a  bi (by the Conjugate Roots Theorem), where b  0. Thus there must be an even number of nonreal zeros. Since P is of odd degree, it has an odd number of zeros (counting multiplicity). It follows that P has at least one real zero.    76. x 4  1  0  x 2  1 x 2  1  0  x  1 x  1 x  i x  i  0  x  1, i. So there are four fourth

roots of 1, two that are real and two that are nonreal. Consider P x  x n  1, where n is even. P has one change in sign

so P has exactly one real positive zero, namely x  1. Since P x  P x, P also has exactly one real negative zero,

namely x  1. Thus P must have n  2 complex roots. As a result, x n  1 has two real n th zeros and n  2 complex roots.    3 . So there is one real cube zero of unity and two nonreal x 3  1  0  x  1 x 2  x  1  0  x  1, 1i 2

roots. Now consider Q x  x k  1, where k is odd. Since Q has one change in sign, Q has exactly one real positive zero,

namely x  1. But Q x  x k  1 has no change in sign, so there is no negative real zero. As a result, x k  1 has one real kth zero and k  1 nonreal roots.

3.6

RATIONAL FUNCTIONS

1. If the rational function y  r x has the vertical asymptote x  2, then as x  2 , either y   or y  . 2. If the rational function y  r x has the horizontal asymptote y  2, then y  2 as x  . 3. The function r x 

x  1 x  2 has x-intercepts 1 and 2. x  2 x  3

4. The function r has y-intercept 13 . 5. The function r has vertical asymptotes x  2 and x  3. 6. The function r has horizontal asymptote y  1. x2  x x x  1 x for x  1.     2 x  2 x  1 2x  4 x  1 2x  4 (a) False. r does not have vertical asymptote x  1. It has a “hole” at 1 16 .

7. r x 

(b) True, r has vertical asymptote x  2.

(c) False, r has a horizontal asymptote y  12 but not a horizontal asymptote y  1.

(d) True, r has horizontal asymptote y  12 .

x2  x crosses its 8. True, the graph of a rational function may cross a horizontal asymptote. For example, r x  2 x x 1   horizontal asymptote y  1 at the point 12  1 . 9. r x 

x x 2

SECTION 3.6 Rational Functions

(a) x

r x

x

x

r x

15

3

25

5

10

125

10

0833

19

19

21

21

50

1042

50

0962

199

199

201

201

100

1020

100

0980

1999

1999

2001

2001

1000

1002

1000

0998

x

r x

r x

(b) r x   as x  2 and r x   as x  2 .

(c) r has horizontal asymptote y  1. 4x  1 10. r x  x 2 (a) x

r x

x

r x

x

r x

x

r x

15

14

25

22

10

5125

10

325

19

86

21

94

50

4188

50

3827

199

896

201

904

100

4092

100

3912

1999

8996

2001

9004

1000

4009

1000

3991

(b) r x   as x  2 and r x   as x  2 .

(c) r has horizontal asymptote y  4. 3x  10 11. r x  x  22 (a) x

r x

r x

x

r x

x

r x

x

15

22

25

10

10

03125

10

02778

19

430

21

370

50

00608

50

00592

39,700

100

00302

100

00298

3,997,000

1000

00030

1000

00030

x

r x

x

r x

199 1999

40,300 4,003,000

201 2001

(b) r x   as x  2 and r x   as x  2 . (c) r has horizontal asymptote y  0.

12. r x  (a)

3x 2  1

x  22 x

r x

x

r x

15

31

25

79

10

4703

10

209

19

1183

21

1423

50

3256

50

2774

131,203

100

3124

100

2884

13,012,003

1000

3012

1000

2988

199 1999

128,803 12,988,003

201 2001

(b) r x   as x  2 and r x   as x  2 .

(c) r has horizontal asymptote y  3. In Exercises 13–20, let f x 

1 . x

345

346

CHAPTER 3 Polynomial and Rational Functions

1  f x  1. From this form we see that the graph of r is obtained x 1 from the graph of f by shifting 1 unit to the right. Thus r has vertical asymptote

y

x  1 and horizontal asymptote y  0. The domain of r is  1  1  and

1

13. r x 

x

1

its range is  0  0 .

1  f x  4. From this form we see that the graph of r is obtained x 4 from the graph of f by shifting 4 units to the left. Thus r has vertical asymptote

y

x  4 and horizontal asymptote y  0. The domain of r is  4  4 

1

14. r x 

1

and its range is  0  0 .

15. s x 

  3 1 3  3 f x  1. From this form we see that the graph x 1 x 1

x

y

of s is obtained from the graph of f by shifting 1 unit to the left and stretching

3

vertically by a factor of 3. Thus s has vertical asymptote x  1 and horizontal

1

asymptote y  0. The domain of s is  1  1  and its range is

x

 0  0 .

16. s x 

  2 1  2  2 f x  2. From this form we see that the x 2 x 2

y

graph of s is obtained from the graph of f by shifting 2 units to the right, stretching 1

vertically by a factor of 2, and then reflecting about the x-axis. Thus s has vertical

x

1

asymptote x  2 and horizontal asymptote y  0. The domain of s is  2  2  and its range is  0  0 .

1 2x  3 2  f x  2  2 (see the long x 2 x 2 division at right). From this form we see that the graph of t is

17. t x 

obtained from the graph of f by shifting 2 units to the right and 2 units vertically. Thus t has vertical asymptote x  2 and horizontal asymptote y  2. The domain of t is

 2  2  and its range is  2  2 .

y

2 x 2

2x  3 2x  2 1

1 1

x

SECTION 3.6 Rational Functions



9 3x  3 1  3 39 x 2 x 2 x 2  9 f x  2  3

18. t x 



y

3 x 2

3x  3 3x  6

From this form we see that the graph of t is obtained from the

347

5

9

graph of f by shifting 2 units to the left, stretching vertically

x

1

by a factor of 9, reflecting about the x-axis, and then shifting 3 units vertically. Thus t has vertical asymptote x  2 and horizontal asymptote y  3. The domain of t is

 2  2  and its range is  3  3 .

1 x 2 1   f x  3  1 (see the long x 3 x 3 division at right). From this form we see that the graph of r is

19. r x 

y

1 x 3

x  2

obtained from the graph of f by shifting 3 units to the left,

x  3

reflect about the x-axis, and then shifting vertically 1 unit.

1

1 1

Thus r has vertical asymptote x  3 and horizontal

x

asymptote y  1. The domain of r is  3  3  and its range is  1  1 .

1 2x  9  2 2 x 4 x 4   f x  4  2

20. t x 



1 x 4



y

2 x 4

2x  9

From this form we see that the graph of t is obtained from the

2x  8

graph of f by shifting 4 units to the right, reflecting about the

1

x-axis, and then shifting 2 units vertically. Thus t has vertical

1 1

x

asymptote x  4 and horizontal asymptote y  2. The domain of r is  4  4  and its range is  2  2 .

x 1 . When x  0, we have r 0   14 , so the y-intercept is  14 . The numerator is 0 when x  1, so the x 4 x-intercept is 1.

21. r x 

3x . When x  0, we have s 0  0, so the y-intercept is 0. The numerator is zero when 3x  0 or x  0, so x 5 the x-intercept is 0.

22. s x 

23. t x 

2 x2  x  2 . When x  0, we have t 0   13 , so the y-intercept is 13 . The numerator is 0 when x 6 6

x 2  x  2  x  2 x  1  0 or when x  2 or x  1, so the x-intercepts are 2 and 1.

2 2 24. r x  2   12 , so the y-intercept is  12 . The numerator is never zero, so . When x  0, we have r 0  4 x  3x  4 there is no x-intercept. 25. r x 

x2  9 . Since 0 is not in the domain of r x, there is no y-intercept. The numerator is 0 when x2

x 2  9  x  3 x  3  0 or when x  3, so the x-intercepts are 3.

348

CHAPTER 3 Polynomial and Rational Functions

x3  8 26. r x  2 . When x  0, we have r 0  84  2, so the y-intercept is 2. The x-intercept occurs when x 3  8  0  x 4    x  2 x 2  2x  4  0  x  2 or x  1  i 3, which has only one real solution, so the x-intercept is 2. 27. From the graph, the x-intercept is 3, the y-intercept is 3, the vertical asymptote is x  2, and the horizontal asymptote is y  2.

28. From the graph, the x-intercept is 0, the y-intercept is 0, the horizontal asymptote is y  0, and the vertical asymptotes are x  1 and x  2.

29. From the graph, the x-intercepts are 1 and 1, the y-intercept is about 14 , the vertical asymptotes are x  2 and x  2, and the horizontal asymptote is y  1. 30. From the graph, the x-intercepts are 2, the y-intercept is 6, the horizontal asymptote is y  2, and there are no vertical asymptotes 5 has a vertical asymptote where x  2  0  x  2, and y  0 is a horizontal asymptote because the degree x 2 of the denominator is greater than that of the numerator.

31. r x 

2x  3 has are vertical asymptotes where x 2  1  0  x  1 or x  1, and y  0 is a horizontal asymptote 32. r x  2 x 1 because the degree of the denominator is greater than that of the numerator. 3x  1 has no vertical asymptote since 4x 2  1  0 for all x. y  0 is a horizontal asymptote because the degree 4x 2  1 of the denominator is greater than that of the numerator.

33. r x 

   3x 2  5x has vertical asymptotes where x 4  1  x 2  1 x 2  1  0  x  1, and y  0 is a horizontal 4 x 1 asymptote because the degree of the denominator is greater than that of the numerator.

34. r x 

35. s x 

6x 2  1 has vertical asymptotes where 2x 2  x  1  0  x  1 2x  1  0  x  1 or x  12 , and 2x 2  x  1

horizontal asymptote y  62  3. 36. s x 

8x 2  1

4x 2  2x  6

has vertical asymptotes where 4x 2  2x  6  0  2 2x  3 x  1  0  x   32 or x  1,

and horizontal asymptote y  84  2. 37. r x 

x  1 2x  3 has vertical asymptotes where x  2 4x  7  0  x   74 or x  2, and horizontal x  2 4x  7

asymptote y  38. r x 

x  3 x  2 has vertical asymptotes where 5x  1 2x  3  0  x   15 or x  32 , and horizontal 5x  1 2x  3

asymptote y  39. r x 

1 12  . 14 2

1 11  . 52 10

6x 3  2 6x 3  2   . Because the quadratic in the denominator has no real zero, r has vertical  2x 3  5x 2  6x x 2x 2  5x  6

asymptote x  0 and horizontal asymptote y  62  3.

SECTION 3.6 Rational Functions

349

5x 3 5x 3 5x 2    40. r x  3  . Because the denominator has no real zero, r has no vertical x  2x 2  5x x x 2  2x  5 x 2  2x  5 asymptote. r has horizontal asymptote y  51  5.

x2  2 . A vertical asymptote occurs when x  1  0  x  1. There is no horizontal asymptote because the degree x 1 of the numerator is greater than the degree of the denominator.

41. y 

x 2 x  3 x 3  3x 2  . Because the degree of the numerator is greater than the degree of the denominator, 2 x  2 x  2 x 4 the function has no horizontal asymptote. Two vertical asymptotes occur at x  2 and x  2. By using long division, we 4x  12 see that r x  x  3  2 so y  x  3 is a slant asymptote. x 4

42. r x 

4x  4 . When x  0, y  2, so the y-intercept is 2. When y  0, x 2 4x  4  0  x  1, so the x-intercept is 1. Since the degree of the numerator and

y

43. y 

denominator are the same, the horizontal asymptote is y  41  4. A vertical 4x  4  , and as asymptote occurs when x  2. As x  2 , y  x 2 4x  4 x  2 , y   . The domain is x  x  2 and the range is x 2 y  y  4. 44. r x 

4 1

2 x  3 2x  6  . When x  0, we have y  2, so the y-intercept 6x  3 3 2x  1

x

y

is 2. When y  0, we have x  3  0  x  3, so the x-intercept is 3. A

vertical asymptote occurs when 2x  1  0  x  12 . Because the degree of the denominator and the numerator are the same, the horizontal asymptote is     2   1 . The domain is x  x  1 and the range is y  y   1 . y  6 3 2 3





3 x 2  4x  4  1 3x 2  12x  13 1 45. r x  . When x  0,  3 x 2  4x  4 x 2  4x  4 x  22

1 1

x

y

1 13 y  13 is positive 4 , so the y-intercept is 4 . There is no x-intercept since x  22

on its domain. There is a vertical asymptote at x  2. The horizontal asymptote is y  3. The domain is x  x  2 and the range is y  y  3.

1 1

x

350

CHAPTER 3 Polynomial and Rational Functions

  2 x 2  4x  4  1 2x 2  8x  9 1 46. r x  . When   2  2 2 x  4x  4 x  4x  4 x  22 x  0, y   94 , so the y-intercept is  94 . There is no x-intercept since

1

x  22

y

is

1 0

positive on its domain. There is a vertical asymptote at x  2. The horizontal

x

1

asymptote is y  2. The domain is x  x  2 and the range is y  y  2.

   x 2  8x  16  2 x 2  8x  18 2 47. r x  2 . When   1  2 x  8x  16 x  8x  16 x  42 x  0, y   98 , and so the y-intercept is  98 . There is no x-intercept since 2

is positive on its domain. There is a vertical asymptote at x  4. The x  42 horizontal asymptote is y  1. The domain is x  x  4 and the range is

y

1 0

1

x

y  y  1.





1 2x 2  4x  2  2 1 x 2  2x  3 1 2 48. r x  2 . is true. When    2 x  12 2x  4x  2 2x 2  4x  2

y

x  0, we have y  32 , so the y-intercept is 2. There is no x-intercept since 1 is positive on its domain. There is a vertical asymptote at x  1. The x  12

horizontal asymptote is y  12 . The domain is x  x  1 and the range is   y  y  12 .

49. s x 

4x  8 8 . When x  0, y   2, so the y-intercept is 2. x  4 x  1 4 1

1 0

1

x

y

When y  0, 4x  8  0  x  2, so the x-intercept is 2. The vertical asymptotes are x  1 and x  4, and because the degree of the numerator is less than the

degree of the denominator, the horizontal asymptote is y  0. The domain is x  x  1 4 and the range is R.

1 1

x

SECTION 3.6 Rational Functions

6 6  1, so the y-intercept is 1. . When x  0, y  50. s x  2 6 x  5x  6 Since the numerator is never zero, there is no x-intercept. The vertical asymptotes occur when x 2  5x  6  x  1 x  6  x  1 and x  6, and because the degree of the numerator is less less than the degree of the denominator, the

351

y

1

1

x

horizontal asymptote is y  0. The domain is x  x  1 6 and the range is y  y  05 or y  0.

2x  4 2 x  2 51. s x  2 . When x  0, y  2, so the y-intercept is   1 x  2 x x x 2 2. When y  0, we have 2x  4  0  x  2, so the x-intercept is 2. A vertical

y

asymptote occurs when x  1 x  2  0  x  1 and x  2. Because the

1

degree of the denominator is greater than the degree of the numerator, the

horizontal asymptote is y  0. The domain is x  x  2 1 and the range is

1

x

1

x

y  y  02 or y  2.

52. s x 

x 2 2 , so the y-intercept is  23 When . When x  0, y  3 x  3 x  1

y

y  0, we have x  2  0  x  2, so the x-intercept is 2. A vertical

asymptote occurs when x  3 x  1  0  x  3 and x  1. Because the

1

degree of the denominator is greater than the degree of the numerator, the

horizontal asymptote is y  0. The domain is x  x  3 1 and the range is R.

53. r x 

x  1 x  2 . When x  0, y  23 , so the y-intercept is 23 . When x  1 x  3

y

y  0, x  1 x  2  0  x  2, 1, so, the x-intercepts are 2 and 1. The

vertical asymptotes are x  1 and x  3, and because the degree of the

numerator and denominator are the same the horizontal asymptote is y  11  1. The domain is x  x  1 3 and the range is R.

2x 2  10x  12 2 x  1 x  6 2 1 6  . When x  0, y   2, x  2 x  3 2 3 x2  x  6 so the y-intercept is 2. When y  0, 2 x  1 x  6  0  x  6, 1, so the

54. r x 

x-intercepts are 6 and 1. Vertical asymptotes occur when x  2 x  3  0

 x  3 or x  2. Because the degree of the numerator and denominator are the same the horizontal asymptote is y  21  2. The domain is x  x  3 2 and the range is R.

1 x

1

y

2 1

x

352

CHAPTER 3 Polynomial and Rational Functions

2x 2  2x  4 2 x  2 x  1 . Vertical asymptotes occur at x  0  x x  1 x2  x and x  1. Since x cannot equal zero, there is no y-intercept. When y  0, we

y

55. r x 

have x  2 or 1, so the x-intercepts are 2 and 1. Because the degree of the

5

denominator and numerator are the same, the horizontal asymptote is y  21  2. The domain is x  x  1 0 and the range is y  y  2 or y  184.





3 x2  2 3x 2  6  56. r x  2 . When x  0, y  2, so the y-intercept x  3 x  1 x  2x  3

1

x

y

is 2. Since the numerator can never equal zero, there is no x-intercept. Vertical

asymptotes occur when x  1, 3. Because the degree of the numerator and

denominator are the same, the horizontal asymptote is.y  31  3. The domain is x  x  1 3 and the range is y  y  18 or y  25.

57. s x 

x 2  2x  1 x  12  . Since x  0 is not in the domain of s x, x 3  3x 2 x 2 x  3

there is no y-intercept. The x-intercept occurs when y  0 

4 1

x

1

x

y 1

x 2  2x  1  x  12  0  x  1, so the x-intercept is 1. Vertical asymptotes

occur when x  0, 3. Since the degree of the numerator is less than the degree of

the denominator, the horizontal asymptote is y  0. The domain is x  x  0 3 and the range is R.

x2  x  6 x  3 x  2 . The x-intercept occurs when y  0   2 x x  3 x  3x x  3 x  2  0  x  2, 3, so the x-intercepts are 2 and 3. There is no

y

58. y 

y-intercept because y is undefined when x  0. The vertical asymptotes are x  0

2

and x  3. Because the degree of the numerator and denominator are the same,

1

the horizontal asymptotes is y  11  1. The domain is x  x  3 0 and the range is R.

x 2  2x  1 x  12 59. r x  2   x  2x  1 x  12



 x 1 2 . When x  0, y  1, so the x 1

x

y

y-intercept is 1. When y  0, x  1, so the x-intercept is 1. A vertical asymptote

occurs at x  1  0  x  1. Because the degree of the numerator and

denominator are the same the horizontal asymptote is y  11  1. The domain is x  x  1 and the range is y  y  0.

1 1

x

SECTION 3.6 Rational Functions

4x 2 4x 2 60. r x  2 . When x  0, we have y  0, so the  x  3 x  1 x  2x  3 graph passes through the origin. Vertical asymptotes occur at x  1 and x  3.

353

y

Because the degree of the denominator and numerator are the same, the horizontal

asymptote is y  41  4. The domain is x  x  1 3 and the range is y  y  0 or y  29.

1

  5 x2  1 5x 2  5 5 61. r x  2 . When x  0, we have y  , so the  2 4 x  4x  4 x  2

2

x

2

x

y

y-intercept is 54 . Since x 2  1  0 for all real x, y never equals zero, and there is

no x-intercept. The vertical asymptote is x  2. Because the degree of the

denominator and numerator are the same, the horizontal asymptote occurs at y  51  5. The domain is x  x  2 and the range is y  y  10.

x 2 x  1 x3  x2  3 . When x  0, we have y  0, so the 62. r x  3 x  3x  2 x  3x  2

2

y

y-intercept is 0. When y  0, we have x 2 x  1  0, so the x-intercepts are 0

and 1. Vertical asymptotes occur when x 3  3x  2  0. Since x 3  3x  2  0

when x  2, we can factor x  2 x  12  0, so the vertical asymptotes occur

at x  2 and x  1. Because the degree of the denominator and numerator are

the same, the horizontal asymptote is y  11  1. The domain is x  x  1 2 and the range is R.

63. r x 

x 2  4x  5 x 5 x  5 x  1 for x  1. When x  0, y  52 ,   2 x 2 x  2 x  1 x x 2

1 x

1

y

so the y-intercept is 52 . When y  0, x  5, so the x-intercept is 5. Vertical asymptotes occur when x  2  0  x  2. Because the degree of the

denominator and numerator are the same, the horizontal asymptote is y  1. The

domain is x  x  2 1 and the range is y  y  1 2.

(1, 2) 1 0

1

x

354

CHAPTER 3 Polynomial and Rational Functions

64. r x 

x 2  3x  10 x 2 x  2 x  5   for x  1 x  3 x  5 x  1 x  3 x  5 x  1 x  3

y

x  5. When x  0, we have y  23 , so the y-intercept is 23 . When y  0, we have x  2, so the x-intercept is 2. Vertical asymptotes occur when

1

x  1 x  3  0  x  1 or 3, so the vertical asymptotes occur at x  1 and x  3. Because the degree of the denominator is greater than that of the

numerator, the horizontal asymptote is y  0. The domain is x  x  5 1 3

0

(_5, _ 327 )

1

x

and the range is R.

x 2  2x  3 x  3 x  1   x  3 for x  1. When x  0, x 1 x 1 y  3, so the y-intercept is 3. When y  0, x  3, so the x-intercept is 3.

y

65. r x 

There are no asymptotes. The domain is x  x  1 and the range is

1

y  y  4.

0

1

x

(_1, _4)

x 3  2x 2  3x x x  1 x  3   x x  1 for x  3. When x  0, x 3 x 3 we have y  0, so the y-intercept is 0. When y  0, x  1 or 0, so the

y

66. r x 

(3, 12)

x-intercepts are 1 and 0. There are no asymptotes. The domain is x  x  3 and   the range is y  y   14 .

1 0

x 3  5x 2  3x  9 . We use synthetic division to check whether the x 1 denominator divides the numerator: 1 1 Thus, r x 



 x 2  6x  9 x  1

5

3

9

1

6

9

6

9

0

 x 2  6x  9  x  32 for x  1.

x 1 When x  0, y  9, so the y-intercept is 9. When y  0, x  3, so the x-intercept is 3. There are no asymptotes. The domain is x  x  1 and the range is

y  y  0.

x

y

67. r x 

1

1

(_1, 16)

2 0

1

x

SECTION 3.6 Rational Functions

x 1 x 2  4x  5 x  1 x  5  for x  5. x  0 is 68. r x  3  x x  2 x  5 x x  2 x  7x 2  10x not in the domain of r , so there is no y-intercept. When y  0, x  1, so the

355

y

x-intercept is 1. There are vertical asymptotes at x  2 and x  0. Because the

degree of the denominator is less than the degree of the numerator, y  0 is a horizontal asymptote. The domain is x  x  5 2 0 and the range is y  y  0134 or y  1866.

69. r x 

x2 . When x  0, y  0, so the graph passes through the origin. There x 2

1 0

(_5, _ 327 )

1 x

y

is a vertical asymptote when x  2  0  x  2, with y   as x  2 , and y   as x  2 . Because the degree of the numerator is greater than the

degree of the denominator, there is no horizontal asymptotes. By using long

2

4 division, we see that y  x  2  , so y  x  2 is a slant asymptote. x 2 x x  2 x 2  2x  . When x  0, we have y  0, so the graph passes x 1 x 1 through the origin. Also, when y  0, we have x  0 or 2, so the x-intercepts are

70. r x 

2

x

1

x

y

2 and 0. The vertical asymptote is x  1. There is no horizontal asymptote, and the line y  x  3 is a slant asymptote because by long division, we have y  x 3

2 . x 1

x 2  2x  8 x  4 x  2  . The vertical asymptote is x  0, thus, x x there is no y-intercept. If y  0, then x  4 x  2  0  x  2, 4, so the

71. r x 

x-intercepts are 2 and 4. Because the degree of the numerator is greater than the

degree of the denominator, there are no horizontal asymptotes. By using long

1

y

2 2

8 division, we see that y  x  2  , so y  x  2 is a slant asymptote. x

x 3  x 3x  x 2  . When x  0, we have y  0, so the graph passes 2x  2 2 x  1 through the origin. Also, when y  0, we have x  0 or x  3, so the x-intercepts are 0 and 3. The vertical asymptote is x  1. There is no horizontal asymptote, and

72. r x 

the line y   12 x  1 is a slant asymptote because by long division we have 1 y   12 x  1  . x 1

x

y

1 1

x

356

CHAPTER 3 Polynomial and Rational Functions

73. r x 

x 2  5x  4 x  4 x  1  . When x  0, y   43 , so the y-intercept x 3 x 3

y

is  43 . When y  0, x  4 x  1  0  x  4, 1, so the two x-intercepts are 4 and 1. A vertical asymptote occurs when x  3, with y   as x  3 , and y   as x  3 . Using long division, we see that 28 y  x 8 , so y  x  8 is a slant asymptote. x 3

x3  4 x3  4 . When x  0, we have  2x  1 x  1 2x 2  x  1  04  4, so the y-intercept is 4. Since x 3  4  0  x   3 4, y 001  the x-intercept is x   3 4. There are vertical asymptotes where

74. r x 

5 x

2

y

1 1

x

2x  1 x  1  0  x  12 or x  1. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. By long division, we have y  12 x  14  slant asymptote.

75. r x 

3 x  15 4 4 , so the line y  1 x  1 is a 2 4 2x 2  x  1

x3  x2 x 2 x  1  . When x  0, y  0, so the graph passes 2 x  2 x  2 x 4

through the origin. Moreover, when y  0, we have x 2 x  1  0  x  0, 1, so the x-intercepts are 0 and 1. Vertical asymptotes occur when x  2; as

x  2 , y   and as x  2 , y  . Because the degree of the numerator is greater than the degree of the denominator, there is no horizontal 4x  4 asymptote. Using long division, we see that y  x  1  2 , so y  x  1 is a x 4 slant asymptote.

  2x x 2  1 2x 3  2x 76. r x  2  . When x  0, we have y  0, so the graph x  1 x  1 x 1

passes through the origin. Also, note that x 2  1  0, for all real x, so the only x-intercept is 0. There are two vertical asymptotes at x  1 and x  1. There is no horizontal asymptote, and the line y  2x is a slant asymptote because by long 4x division, we have y  2x  2 . x 1

y

2 1

x

1

x

y

2

SECTION 3.6 Rational Functions

77. f x 

357

2x 2  6x  6 , g x  2x. f has vertical asymptote x  3. x 3 50 -10

-5

5 -20

20

-20

78. f x 

-50

x 3  6x 2  5 , g x  x  4. f has vertical asymptotes x  0 and x  2. x 2  2x 20

20

-5

-20

5

20

-20

79. f x 

-20

x 3  2x 2  16 , g x  x 2 . f has vertical asymptote x  2. x 2 50

50

-10

10

5 -50

80. f x 

x 4  2x 3  2x x  12

-4

, g x  1  x 2 . f has vertical asymptote x  1.

-2

2

4

-4

-2

2

-5

-5

-10

-10

4

2x 2  5x has vertical asymptote x  15, x-intercepts 0 and 25, y-intercept 0, local maximum 39 104, 2x  3 12 . From the graph, we see that the end and local minimum 09 06. Using long division, we get f x  x  4  2x  3 behavior of f x is like the end behavior of g x  x  4.

81. f x 

x  4

2x  3

2x 2  5x

50

20

2x 2  3x  8x

-10

 8x  12 12

10

-20

20

-20 -50

358

CHAPTER 3 Polynomial and Rational Functions

x 4  3x 3  x 2  3x  3 has vertical asymptotes are x  0, x  3, x-intercept 082, and no y-intercept. The x 2  3x local minima are 080 263 and 338 1476. The local maximum is 256 488. By using long division, we see that 3 f x  x 2  1  2 . From the second graph, we see that the end behavior of f x is the same as the end behavior x  3x 2 of g x  x  1.

82. f x 

x2

x 2  3x

 1

20

x 4  3x 3  x 2  3x  3

50

x 4  3x 3

0x 3  x 2  3x

-5

x 2  3x

5

-5 -50

-20

3

5

x5 83. f x  3 has vertical asymptote x  1, x-intercept 0, y-intercept 0, and local minimum 14 31. x 1 x2 Thus y  x 2  3 . From the graph we see that the end behavior of f x is like the end behavior of g x  x 2 . x 1 x2 x3  1

10

x5

10

x5  x2 x2

-5

5

-5

5

-10

-10

Graph of f 84. f x 

x4

Graph of f and g

has vertical asymptotes x  141, x-intercept 0, and y-intercept 0. The local maximum is 0 0. The

x2  2

4 . From the second graph, local minima are 2 8 and 2 8. By using long division, we see that f x  x 2  2  2 x 2 2 we see that the end behavior of f x is the same as the end behavior of g x  x  2. x2

x2  2

2



20

x 4  0x 3  0x 2  0x  0

x4

50

 2x 2 2x 2 2x 2

 4

-5

5 -5

4

5

x 4  3x 3  6 has vertical asymptote x  3, x-intercepts 16 and 27, y-intercept 2, local maxima 04 18 x 3 6 and 24 38, and local minima 06 23 and 34 543. Thus y  x 3  . From the graphs, we see that the end x 3 behavior of f x is like the end behavior of g x  x 3 .

85. f x 

x3

x 3

x 4  3x 3  6

100

100

x 4  3x 3

6

-5

5 -100

-5

5 -100

SECTION 3.6 Rational Functions

359

4  x2  x4 x 4  x 2  4 has vertical asymptotes x  1, x-intercepts 16, and y-intercept 4. The local  x  1 x  1 x2  1 maximum is 0 4 and there is no local minimum. 6 Thus y  x 2  2 . From the graphs, we see that the end behavior of f x is like the end behavior of g x  x 2 . x 1 x 2

86. r x 

x2  1

20

x 4  0x 3  x 2  0x  4 x 4

 x2

0

20

-5

 4

5

-5

5

-20

-20

88. (a)

87. (a)

10

2000 0

0 0

20

40

0

3000 3000t  3000  . So as t  , we t 1 t 1 have p t  3000.

(b) p t 

5t 89. c t  2 t 1

10

20

30t (b) c t  2 . Since the degree of the denominator t 2 is larger than the degree of the numerator, c t  0 as t  .

(a) The highest concentration of drug is 250 mg/L, and it is reached 1 hour after the drug is administered. (b) The concentration of the drug in the bloodstream goes to 0.

2

(c) From the first viewing rectangle, we see that an approximate solution

0 0

10

20

5t is near t  15. Thus we graph y  2 and y  03 in the t 1 viewing rectangle [14 18] by [0 05]. So it takes about 1661 hours for the concentration to drop below 03 mg/L. 0.4 0.2 0.0 14

  64  106  2   90. Substituting for R and g, we have h   . The vertical 2 98 64  106   2 asymptote is   11,000, and it represents the escape velocity from the earth’s gravitational pull: 11,000 m/s  1900 mi/h.

16

18

2e+7 1e+7 0 0

10000

360

CHAPTER 3 Polynomial and Rational Functions

91. P   P0



s0 s0  



 P   440



332 332  



4000

If the speed of the train approaches the speed of sound, the pitch of the whistle becomes very loud. This would be experienced as a “sonic boom”— an effect

2000

seldom heard with trains. 0 0

92. (a)

1 1 1 1 1 1 xF xF 1        y . Using x y F y F x y xF xF

200

F  55, we get y 

100

55x . Since y  0, we use the viewing x  55 rectangle [0 1000] by [0 250].

200

400

0

(b) y approaches 55 millimeters.

0

500

1000

(c) y approaches . 2x 1 . Vertical asymptote x  3 and horizontal asymptote y  2: r x  . x 3 x 3 x 4 Vertical asymptotes x  1 and x  1, horizontal asymptote 0, and x-intercept 4: q x  . Of course, x  1 x  1

93. Vertical asymptote x  3: p x 

other answers are possible.

x 6  10 94. r x  4 has no x-intercept since the numerator has no real roots. Likewise, r x has no vertical asymptotes, x  8x 2  15 since the denominator has no real roots. Since the degree of the numerator is two greater than the degree of the denominator, r x has no horizontal or slant asymptotes. 1 1 95. (a) Let f x  2 . Then r x   f x  2. From this form we see that x x  22 the graph of r is obtained from the graph of f by shifting 2 units to the right. Thus

y

r has vertical asymptote x  2 and horizontal asymptote y  0.

1 x

1

3 2x 2  4x  5 2  3 f x  1  2. From this form we see that the graph of s is obtained from 2 x  2x  1 x  12 the graph of f by shifting 1 unit to the left, stretching vertically by a factor of 3, and shifting 2 units vertically. Thus r has vertical asymptote x  1 and horizontal asymptote y  2.

(b) s x 

y

2 x 2  2x  1

2x 2 2x 2



4x



5



4x



2 3 1 1

x

SECTION 3.7 Polynomial and Rational Inequalities

361

2  3x 2 12x  14 (c) Using long division, we see that p x  2  3  2 which cannot be graphed by transforming x  4x  4 x  4x  4 1 f x  2 . Using long division on q we have: x 3

x 2  4x  4

3x 2 3x 2



0x



2



12x



12



12x



14

3

x 2  4x  4

3x 2 3x 2



12x



0



12x



12 12

12x  3x 2 12 So q x  2  3   12 f x  2  3. From this form we see that the graph of q is obtained x  4x  4 x  22 from the graph of f by shifting 2 units to the right, stretching vertically by a factor of 12, and then shifting 3 units vertically down. Thus the vertical asymptote is x  2 and the horizontal asymptote is y  3. We show y  p x just 1 to verify that we cannot obtain p x from y  2 . x

y 1 1

y

1

1

x

y  q x

3.7

x

y  p x

POLYNOMIAL AND RATIONAL INEQUALITIES

1. To solve a polynomial inequality, we factor the polynomial into irreducible factors and find all the real zeros of the polynomial. Then we find the intervals determined by the real zeros and use test points in each interval to find the sign of the polynomial on that interval.  2

2 0

0 1

1 

x









x 2









x 1

















Sign of

P x  x x  2 x  1

From the table, we see that P x  0 on the intervals [2 0] and [1 .

362

CHAPTER 3 Polynomial and Rational Functions

2. To solve a rational inequality, we factor the numerator and the denominator into irreducible factors. The cut points are the real zeros of the numerator and the real zeros denominator. Then we find the intervals determined by the cut points, and we use test points to find the sign of the rational function on each interval. Sign of

 4

4 2

2 1

1 3

3 

x 2











x 1











x 3



















x 4

   x  2 x  1 r x     x  3 x  4 From the table, we see that r x  0 on the intervals  4, [2 1], and 3 .

3. The inequality x  3 x  5 2x  5  0 already has all terms on one side and the polynomial is factored. The intervals     determined by the zeros 3, 5, and  52 are  5, 5  52 ,  52  3 , and 3 . We make a sign diagram:     5  52  52  3 Sign of  5 3  x 3









x 5









2x  5













P x  x  3 x  5 2x  5



   None of the endpoints satisfies the inequality. The solution is  5   52  3 .

4. The inequality x  1 x  2 x  3 x  4  0 already has all terms on one side and the polynomial is factored. The intervals determined by the zeros 1, 2, 3, and 4 are  4, 4 2, 2 1, 1 3, and 3 . We make a sign diagram: Sign of

 4

4 2

2 1

1 3

3 

x 1











x 2











x 3











x 4





















P x  x  1 x  2 x  3 x  4

All of the endpoints satisfy the inequality. The solution is  4]  [2 1]  [3 .

5. The inequality x  52 x  3 x  1  0 already has all terms on one side and the polynomial is factored. The intervals determined by the zeros 5, 3, and 1 are  5, 5 3, 3 1, and 1 . We make a sign diagram:  5

5 3

3 1

1 









x 3









x 1

















Sign of x  52

P x  x  52 x  3 x  1

None of the endpoints satisfies the inequality. The solution is  5  5 3  1 .

SECTION 3.7 Polynomial and Rational Inequalities

363

6. The inequality 2x  74 x  13 x  1  0 already has all terms on one side and the polynomial is factored. The     intervals determined by the zeros 72 , 1, and 1 are  1, 1 1, 1 72 , and 72   . We make a sign diagram:     7 1 72 Sign of  1 1 1 2 2x  74 x  13 x 1

























P x  2x  74 x  13 x  1         All of the endpoints satisfy the inequality [note that P 72  0]. The solution is [1 1]  72 .

7. We start by moving all terms to one side and factoring: x 3  4x 2  4x  16  x 3  4x 2  4x  16  x  4 x  2 x  2  0. The intervals determined by the zeros 4, 2, and 2 are  4, 4 2, 2 2, and 2 . We make a sign diagram:  4

4 2

2 2

2 

x 4









x 2









x 2

















Sign of

P x  x  4 x  2 x  2

All of the endpoints satisfy the inequality. The solution is [4 2]  [2 .

8. We start by moving all terms to one side and factoring: 2x 3  18x  x 2  9  2x 3  x 2  18x      9  x  3 2x  1 x  3  0. The intervals determined by the zeros 3, 12 , and 3 are  3, 3 12 , 12  3 , and 3 . We make a sign diagram:

  3 12





13 2

3 





Sign of

 3

x 3



2x  1









x 3











 





P x  x  3 2x  1 x  3

None of the endpoints satisfies the inequality. The solution is  3 





13 . 2

  9. We start by moving all terms to one side and factoring: 2x 3 x 2  918x  2x 3 x 2 18x 9  2x  1 x 2  9  0.   Note that x 2  9  0 for all x, so the sign of P x  2x  1 x 2  9 is negative where 2x  1 is negative and positive   where 2x  1 is positive. The endpoint x  12 does not satisfy the inequality, so the solution is  12 .

364

CHAPTER 3 Polynomial and Rational Functions

10. We start by moving all terms to one side and factoring: x 4  3x 3  x  3  x 4  3x 3  x  3  0. The possible rational zeros of P x  x 4  3x 3  x  3 are 1 and 3.

1 1 3 1 3 1

4

3

1 4 3 0    Thus, P x  x  1 x 3  4x 2  4x  3  x  1 x  3 x 2  x  1 . The last factor is positive everywhere, so 

we test the intervals  3, 3 1, and 1 . Sign of

 3

3 1

1 

x 3







x 1

















P x  x  1 x  3 x 2  x  1

Neither of the endpoints satisfies the inequality. The solution is 3 1.

11. All the terms are on the left size. We factor, using the substitution t  x 2 : x 4  7x 2  18  0       t 2  7t  18  t  2 t  9  0  x 2  2 x 2  9  0  x 2  2 x  3 x  3  0. The first factor is positive everywhere, so we test  3, 3 3, and 3 : Sign of

 3

3 3

3 

x 3



















x 3





P x  x 2  2 x  3 x  3

Neither of the endpoints satisfies the inequality. The solution is 3 3.

12. All the terms are on the left size. We factor, using the substitution t  x 2 : 4x 4  25x 2  36  0     4t 2  25t  36  4t  9 t  4  0  4x 2  9 x 2  4  2x  3 2x  3 x  2 x  2  0. The zeros are        32 and 2, so we test  2, 2  32 ,  32  32 , 32  2 , and 2 :       32 2  32  32  32 Sign of  2 2  2 x 2

x  32 x  32 x 2 P x















































      All of the endpoints satisfy the inequality. 2  32  32  2 .

SECTION 3.7 Polynomial and Rational Inequalities

365

13. All the terms are on the left size. To factor, note that the possible rational zeros of P x  x 3  x 2  17x  15 are 1, 3, 5, 15. 1 1 1 17 1

15

2 15

1 2 15 0  Thus, P x  x  1 x 2  2x  15  x  5 x  1 x  3. The zeros are 5, 1, and 3, so we test  5, 

5 1, 1 3, and 3 :

 5

5 1

1 3

3 

x 5









x 1









x 3

















Sign of

P x

All of the endpoints satisfy the inequality P x  0, so the solution is [5 1]  [3 .

14. All the terms are on the left size. To factor, note that the possible rational zeros of P x  x 4  3x 3  3x 2  3x  4 are 1, 2, 4. 1 1 3 3 3 4 1

4 1

4

1 4

1 4 

0







Thus, P x  x  1 x 3  4x 2  x  4  x  1 x  4 x 2  1 . The last factor is positive everywhere, so we test  4, 4 1, and 1 :

Sign of

 4

4 1

1 

x 4







x 1













P x

None of the endpoints satisfies P x  0, so the solution is 4 1.

3 3 3    15. We start by moving all terms to one side and factoring: x 1  x 2  7 1  x 2  x  7 1  x 2  0 

P x  x  7 1  x3 1  x3  0. The intervals determined by the zeros 1, 1, and 7 are  1, 1 1, 1 7, and 7 . We make a sign diagram: Sign of

 1

1 1

1 7

7 

x 7

































1  x3 1  x3 P x

None of the endpoints satisfies the inequality. The solution is  1  1 7.

366

CHAPTER 3 Polynomial and Rational Functions

16. We start by moving all terms to one side and factoring: x 2 7  6x  1  6x 3  7x 2  1  0. The possible rational zeros of P x  6x 3  7x 2  1 are 1,  12 ,  13 ,  16 .

1 6

7 0 1 6 1

1

6 1 1 0   Thus, P x  x  1 6x 2  x  1  1  x 2x  1 3x  1. The intervals determined by the zeros  13 , 12 , and 1         13 ,  13  12 , 12  1 , and 1 . We make a sign diagram:       11   13  13  12 Sign of 1  2 3x  1









2x  1









1x













P x

    All of the endpoints satisfy the inequality P x  0. The solution is  13  12  [1 .

x 1  0. Since all nonzero terms are already on one side of the inequality symbol and there is no factoring x  10 needed, we find the intervals determined by the cut points 1 and 10. These are  1, 1 10, and 10 . We make a sign diagram:

17. r x 

Sign of

 1

1 10

10 

x 1







x  10













r x

The cut point 1 does not satisfy the inequality, and the cut point 10 is not in the domain of r. Thus, the solution is 1 10.

3x  7  0. Since all nonzero terms are already on one side of the inequality symbol and there is no factoring x 2     needed, we find the intervals determined by the cut points 73 and 2. These are  2, 2 73 , and 73   . We

18. r x 

make a sign diagram:

  2 73



 2

3x  7



x 2













r x





7 3

Sign of



  The cut point 73 satisfies equality, but the cut point 2 is not in the domain of r. Thus, the solution is 2 73 .

SECTION 3.7 Polynomial and Rational Inequalities

19. r x 

2x  5

x 2  2x  35 2x  5

 0.

367

Since all nonzero terms are already on one side, we factor:

2x  5 . Thus, the cut points are 7,  52 , and 5. The intervals determined by these x  7 x  5     points are  7, 7  52 ,  52  5 , and 5 . We make a sign diagram:     7  52  52  5 Sign of  7 5 

r x 

x 2  2x  35



2x  5









x 7









x 5

















r x

The cut point  52 satisfies equality, but the cut points 7 and 5 are not in the domain of r. Thus, the solution is   7  52  5 .

4x 2  25 4x 2  25 2x  5 2x  5 .  0. Since all nonzero terms are already on one side, we factor: r x  2  2 x  3 x  3 x 9 x 9       Thus, the cut points are  52 and 3. The intervals determined by these points are  3, 3  52 ,  52  52 , 52  3 ,

20. r x 

and 3 . We make a sign diagram:

  3  52

   52  52











x 3











x 3





















Sign of

 3

2x  5



2x  5

r x









53 2

3 





    The cut points  52 satisfy equality, but the cut points 3 are not in the domain of r. Thus, the solution is 3  52  52  3 . x  0. Since all nonzero terms are already on one side but the denominator cannot be factored, we x 2  2x  2   2  22  4 1 2 2  1  3. Thus, the cut points are use the Quadratic Formula: x  2x  2  0  x  2            1  3, 0, and 1  3. The intervals determined by these points are  1  3 , 1  3 0 , 0 3  1 ,   and 3  1  . We make a sign diagram:             1  3 1  3 0 0 3  1 Sign of 3  1 

21. r x 

x

x 2  2x  2

r x



















    The cut point 0 satisfies equality, but the cut points 1  3 are not in the domain of r. Thus, the solution is        1  3  0 3  1 .

368

CHAPTER 3 Polynomial and Rational Functions

x 1

 0. Since all nonzero terms are already on one side but the denominator cannot be factored, we    4  42  4 2 1 2 2 use the Quadratic Formula: 2x 2  4x  1  0  x   . Thus, the cut points are 2 2 2          1, 1  22 , and 1  22 . The intervals determined by these points are  1, 1 1  22 , 1  22  1  22 ,    and 1  22   . We make a sign diagram:           1 1  22 1  22  1  22 1  22   Sign of  1

22. r x 

2x 2  4x  1

x 1

 x  1  22  x  1  22

r x





























        None of the cut points satisfies the strict inequality, so the solution is 1 1  22  1  22   .

  x 2  2x  3 x  3 x  1   0. The intervals determined by the cut points are  3, 3  23 , 23. r x  2 3x  2 x  3 3x  7x  6   2  3  1 , 1 3, and 3 .     3  23  23  1 Sign of  3 1 3 3  x 3











3x  2











x 1











x 3

















r x

    None of the cut points satisfies the strict inequality, so the solution is  3   23  1  3 .

x 1 x 1   0. The second factor in the denominator is positive for all x, so the intervals   24. r x  3 x 1 x  1 x 2  x  1 determined by the cut points are  1, 1 1, and 1 . Sign of

 1

1 1

1 

x 1







x 1













Sign of

 4

4 3

3 

x 4







x 3













r x

The cut point 1 satisfies equality, but 1 is not in the domain of r. Thus, the solution is  1  [1 .     x x 2  6x  9  3 x 2  6x  9 x 3  3x 2  9x  27 x  3 x  32 25. r x     0. The second factor in the x 4 x 4 x 4 numerator is positive for all x, so the intervals determined by the cut points are  4, 4 3, and 3 .

r x

The cut point 3 satisfies equality, but 4 is not in the domain of r. Thus, the solution is 4 3].

SECTION 3.7 Polynomial and Rational Inequalities

369

x 2  16 x  4 x  4 x  4 x  4      0. The last factor in the denominator is positive for  26. r x  4   2 x  4 x2  4 x  16 x  2 x  2 x 2  4 all x, so the intervals determined by the cut points are  4, 4 2, 2 2, 2 4, and 4 . Sign of

 4

4 2

2 2

2 4

4 

x 4











x 2











x 2











x 4





















r x

None of the cut points satisfy the strict inequality. Thus, the solution is 4 2  2 4.

x 3 x  3  2x  5 x 3 1 1  0 0 2x  5 2x  5 2x  5      x  8 r x   0. The intervals determined by the cut points are  8, 8  52 , and  52   . 2x  5     8  52  52   Sign of  8

27. We start by moving all terms to one side and simplifying:

 x  8







2x  5













r x

  The cut point 8 satisfies equality, but  52 is not in the domain of r. Thus, the solution is 8  52 .

1 1 2 1 1 2       0 x x 1 x 2 x x 1 x 2 x 2  3x  2  x 2  2x  2x 2  2x x  1 x  2  x x  2  2x x  1  0   0   x x  1 x  2 x x  1 x  2     3x  2 r x   0. The intervals determined by the cut points are  2, 2 1, 1  23 ,  23  0 , x x  1 x  2 and 0 .     1  23  23  0 Sign of  2 2 1 0 

28. We start by moving all terms to one side and simplifying:

x 2











x 1











3x  2











x



















r x

   None of the cut points satisfies the strict inequality. Thus, the solution is 2 1   23  0 .

370

CHAPTER 3 Polynomial and Rational Functions

3 1 3 1  2   0 1x x 1x x 2x 2  6x  3 2x 2  6x  3 2 1  x x  x  3 1  x  0  r x    0. The numerator is 0 when x 1  x x 1  x x x  1       6  62  4 2 3 3 3 3 3 x  , so the intervals determined by the cut points are  0, 0 , 2 2 2 2          3 3 3 3 3 3  1 , 1 , and  . 2 2 2             3 3 3 3 3 3 3 3 0 1 1  Sign of  0 2 2 2 2

29. We start by moving all terms to one side and simplifying: 2 

x











x 1





















2x 2  6x  3

r x

      3 3 satisfy equality, but 0 and 1 are not in the domain of r, so the solution is The cut points 2       3 3 3 3  1 . 0 2 2

1 1 2x   2  x 3 x 2 x x 2 1 1 2x x  2 x  1  x  3 x  1  2x x  3    0   0  x 3 x 2 x  2 x  1 x  2 x  1 x  3     3x  1  0. The intervals determined by the cut points are  2, 2  13 ,  13  1 , 1 3, and x  2 x  1 x  3 3 .     2  13  13  1 Sign of  2 1 3 3 

30. We start by moving all terms to one side and factoring:

x 2











3x  1











x 1











x 3





















r x

The cut point  13 satisfies equality, but 2, 1, and 3 are not in the domain of r, so the solution is    2   13  1  3 .

SECTION 3.7 Polynomial and Rational Inequalities

371

x  12  0. The numerator is nonnegative for all x, but note that x  1 fails to satisfy the strict inequality. x  1 x  2 The intervals determined by the cut points are  2, 2 1, and 1 .

31. r x 

 2

2 1

1 







x 2







x 1











 (except at x  1)

Sign of x  12

r x

The cut points 2 and 1 are not in the domain of r, so the solution is  2  1 1  1 . 32. r x  1 .

x 2  2x  1 x  12  0. The intervals determined by the cut points are  1, 1 1, and  x 3  3x 2  3x  1 x  13 Sign of x  13

 1

1 1

1 



















x  12

r x

The cut point 1 is not in the domain of r and the cut point 1 satisfies the inequality. Thus, the solution is  1]  1. 6 6 6 6   1  1  0 x 1 x x 1 x 6x  6 x  1  x x  1 x2  x  6 x  2 x  3 0  0  r x    0. The intervals determined by the x x  1 x x  1 x x  1 cut points are  2, 2 0, 0 1, 1 3, and 3 .

33. We start by moving all terms to one side and factoring:

Sign of

 2

2 0

0 1

1 3

3 

x 2











x











x 1











x 3





















r x (note negative sign)

The cut points 0 and 1 are inadmissible in the original inequality, so the solution is [2 0  1 3]. 5 x 5 x  4  4  0 2 x 1 2 x 1 x 2  7x  18 x x  1  5 2  4 2 x  1 x  2 x  9 0 0  0. The intervals determined by the cut 2 x  1 2 x  1 2 x  1 points are  2, 2 1, 1 9, and 9 .

34. We start by moving all terms to one side and factoring:

Sign of

 2

2 1

1 9

9 

x 2









x 1









x 9

















r x

The cut point 1 is inadmissible in the original inequality, so the solution is [2 1  [9 .

372

CHAPTER 3 Polynomial and Rational Functions

x 1 x 2 x 1 x 2     0  35. We start by moving all terms to one side and factoring: x 3 x 2 x 3 x 2   2 x  12 x  2 x  2  x  1 x  3  0  r x   0. The intervals determined by the cut points are x  3 x  2 x  3 x  2      3, 3  12 ,  12  2 , and 2 .     3  12  12  2 Sign of  3 2  x 3

x  12 x 2

r x (note negative sign)



























    The cut points fail to satisfy the strict inequality, so the solution is 3  12  2 .



1 1 1 1 1 1      0 x 1 x 2 x 3 x 1 x 2 x 3 x 2  6x  7 x  2 x  3  x  1 x  3  x  1 x  2  0  r x   0. The numerator is x  1 x  2 x  3 x  1 x  2 x  3      6  62  4 1 7  3  2, so the intervals determined by the cut points are  3  2 , 0 when x  2 1          3  2 3 , 3 2, 2 3  2 , 3  2 1 , and 1 .              3  2 3  2 3 2 3  2 3  2 1 Sign of 3 2 1 

36. We start by moving all terms to one side and factoring:

x 2  6x  7













x 3













x 2













x 1



















r x

         The cut points 3, 2, and 1 are not in the domain, so the solution is  3  2  3 2  3  2 1 .

37. The graph of f lies above that of g where f x  g x; that is, where x 2  3x  10  x 2  3x  10  0  x  2 x  5  0. We make a sign diagram:  2

2 5

5 

x 2







x 5













Sign of

x  2 x  5

Thus, the graph of f lies above the graph of g on  2 and 5 .

SECTION 3.7 Polynomial and Rational Inequalities

1 1 1 1 x  1  x    0 0 x x 1 x x 1 x x  1

38. The graph of f lies above that of g where f x  g x; that is, where 

1 . We make a sign diagram: x x  1  0

0 1

1 







x 1  1  x x  1 Thus, the graph of f lies above the graph of g on 0 1.









Sign of x

39. The graph of f lies above that of g where f x  g x; that is, where 4x  r x 

2x  1 2x  1  0. We make a sign diagram: x       12  12  0 Sign of

1 4x 2  1 1  4x   0  0 x x x

  0 12







x









2x  1









       Thus, the graph of f lies above the graph of g on  12  0 and 12   .







1 2

2x  1

r x

373



x3  x2  2 2 2 40. The graph of f lies above that of g where f x  g x; that is, where x 2  x   x 2  x   0  0 x x x   x  1 x 2  2x  2  0. The second factor in the numerator is positive for all x. We make a sign diagram:  r x  x  0

0 1

1 

x







x 1













Sign of

r x

Thus, the graph of f lies above the graph of g on  0 and 1 .

41. f x 

 6  x  x 2 is defined where 6  x  x 2   x  2 x  3  0. We make a sign diagram: Sign of

 2

2 3

3 

x 2







x 3













 x  2 x  3 Thus, the domain of f is [2 3].

374

CHAPTER 3 Polynomial and Rational Functions

42. g x 



5x 5x is defined where  0 and 5  x  0. We make a sign diagram: 5x 5x  5

5 5

5 







5x   5x   5x The cut point 5 is permissible, and so the domain of g is [5 5.



Sign of 5x

43. h x 

 4



  x 4  1 is defined where x 4  1  x  1 x  1 x 2  1  0. The last factor is positive for all x. We make a

sign diagram:

 1

1 1

1 

x 1







x 1







  x  1 x  1 x 2  1













Sign of

x2  1

Thus, the domain of h is  1]  [1 .

44. f x  

1

x 4  5x 2  4 make a sign diagram:

   is defined where x 4  5x 2  4  x 2  4 x 2  1  x  2 x  2 x  1 x  1  0. We  2

2 1

1 1

1 2

2 

x 2











x 1











x 1











x 2





















Sign of

x 4  5x 2  4

Thus, the domain of h is  2  1 1  2 .

46.

45.

-4

20

20

10

10

-2

2

4

-10

From the graph, we see that x 3  2x 2  5x  6  0 on [2 1]  [3 .

-4

-2

-10

2

-20

From the graph, we see that 2x 3  x 2  8x  4  0 on    2]   12  2 .

SECTION 3.7 Polynomial and Rational Inequalities

375

48.

47.

-2

-1

4

4

2

2 1

-2

2

-2

-4

-1 -2

1

2

3

4

-4

From the graph, we see that 2x 3  3x  1  0 on

From the graph, we see that x 4  4x 3  8x  0 on

approximately  137  037 1.

approximately  124  0 2  324 .

49.

50. 20

40

10 20 -2 -1

0

1

1

2

-10

  From the graph, we see that 5x 4  8x 3 on 0 85 . 51.

-1

2

From the graph, we see that x 5  x 3  x 2  6x on

approximately [131 0]  [151 .

    1  x2  x 4 x x  1   1  x2 1  x2  4 x x  1   4 x x  1  0   0     x x x

1  2x  x 2  4x 2  4x 3x 2  2x  1 1  x 3x  1 0  0  r x   0. The domain of r is 0 , and    x x x  both 3x  1 and x are positive there. 1  x  0 for x  1, so the solution is 0 1]. 7x  8 52. 23 x 13 x  212  12 x 23 x  212  0  16 x  212 x 13 [4 x  2  3x]  0  r x    0.  63 x x 2 Note that the domain of r is 2 . We make a sign diagram with cut points  87 and 0:     2  87  87  0 Sign of 0  7x  8  3 x  x 2



















   Neither cut point is a solution. The solution is  87  0 .





r x

53. We want to solve P x  x  a x  b x  c x  d  0, where a  b  c  d. We make a sign diagram with cut points a, b, c, and d:  a

a b

b c

c d

d 

x a











x b











x c











x d





















Sign of

P x

Each cut points satisfies equality, so the solution is  a]  [b c]  [d .

376

CHAPTER 3 Polynomial and Rational Functions

[x  a] x  b x 2  a  b x  ab x  a x  b    0. Note that x c x c x  c 0  a  c, so c  a  0. We make a sign diagram with cut points c, a, and b:

54. Factoring the numerator, we have r x  Sign of

x  c x b

x  a P x

 c

c a

a b

b 

































The cut points a and b satisfy equality, but c is not in the domain of P. Thus, the solution is  c  [a b]. 500,000 500,000 55. We want to solve the inequality T x  300  2  300  2  300  0  x  400 x  400     2 500,000  300 x 2  400 300 3800 300x 2  380,000 3 x  0   0  . Because x represents distance, it is x 2  400 x 2  400 x 2  400  3800  356. The positive, and the denominator is positive for all x. Thus, the inequality holds where x 2  3800 3 x  3 temperature is below 300  C at distances greater than 356 meters from the center of the fire.

2  175  0   2  25  4375  0. Using the Quadratic Formula, we find 56. We want to solve d   175    25    25  252  4 1 4375 25  18,125 25  18,125   . Because  is positive, we have    548, so 2 1 2 2 Kerry can travel at up to 548 mih. 88x 57. We graph N x   x 2 and y  40 in the viewing rectangle 60 17  17 20 40

[0 100] by [0 60], and see that N x  40 for approximately 95  x  423.

20

Thus, cars can travel at between 95 and 423 mih.

0 0

50

100

20000

We graph S x  8x  08x 2  0002x 3  4000 and y  12,000 in the viewing

10000

approximately 183  x  344. Thus, profits are above $12,000 when between 183

58.

rectangle [0 400] by [10000 20000], and see that S x  12,000 for and 344 units are produced.

0 100 200 300 400 -10000

CHAPTER 3

Review

377

CHAPTER 3 REVIEW   1. (a) f x  x 2  6x  2  x 2  6x  2    x 2  6x  9  2  9

  2. (a) f x  2x 2  8x  4  2 x 2  4x  4    2 x 2  4x  4  4  8  2 x  22  4

 x  32  7

y

(b)

y

(b)

1 _1

  3. (a) f x  1  10x  x 2   x 2  10x  1     x 2  10x  25  1  25   x  52  26

(b)

1

0 x

y

0

1

x

  4. (a) f x  2x 2  12x  2 x 2  6x    2 x 2  6x  9  18  2 x  32  18

(b)

y

10 0

1 x

5 0

1

x

 2     5. f x  x 2  3x  1   x 2  3x  1   x 2  3x  94  1  94   x  32  54 has the maximum value 54 when x  32 .

    6. f x  3x 2  18x  5  3 x 2  6x  5  3 x 2  6x  9  5  27  3 x  32  22 has the minimum value 22 when x  3.

    7. We write the height function in standard form: h t  16t 2  48t  32  16 t 2  3t  32  16 t 2  3t  94  2  32  36  16 t  32  68. The stone reaches a maximum height of 68 ft. 8. We write the profit function in standard form:

  P x  1500  12x  0004x 2  0004 x 2  3000x  1500    0004 x 2  3000x  15002  1500  0004 15002  0004 x  15002  7500

Thus, the maximum profit of $7500 is achieved when 1500 units are sold.

378

CHAPTER 3 Polynomial and Rational Functions

9. P x  x 3  64

  10. P x  2x 3  16  2 x  2 x 2  2x  4

y

y

5

(0, 64)

(2, 0)

x

1 (0, _16)

10

(4, 0) 1

11. P x  2 x  14  32

x

12. P x  81  x  34 y

y

5

25

(1, 0) x

(_3, 0)

(0, 0)

(6, 0) 1

x

(0, _30)

13. P x  32  x  15

14. P x  3 x  25  96

y

y

(0, 31) 10

(_1, 0)

1

x

20 (0, 0)

1

15. (a) P x  x  3 x  1 x  5  x 3 7x 2 7x15 has odd degree and a positive leading coefficient, so

  16. (a) P x   x  5 x 2  9 x  2

 x 4  3x 3  19x 2  27x  90

y   as x   and y   as x  . (b)

x

has even degree and a negative leading coefficient, so

y

y   as x   and y   as x  . (b)

y 20

10 1

1 x

x

CHAPTER 3

17. (a) P x   x  12 x  4 x  22  x 5  2x 4  11x 3  8x 2  20x  16

379

  18. (a) P x  x 2 x 2  4 x 2  9  x 6  13x 4  36x 2 has even degree and a positive leading coefficient, so y   as x   and y   as x  .

has odd degree and a negative leading coefficient, so y   as x   and y   as x  .

Review



y

(b)

y

(b)

100

20 x

1 5

_5

x

_100

19. (a) P x  x 3 x  22 . The zeros of P are 0 and 2, with multiplicities 3 and 2, respectively.

(b) We sketch the graph using the guidelines on page 295.

20. (a) P x  x x  13 x  12 . The zeros of P are 1, 0, and 1, with multiplicities 3, 1, and 2, respectively.

(b) We sketch the graph using the guidelines on page 295.

y

y

1

1

x

0.1 1

21. P x  x 3  4x  1. x-intercepts: 21, 03, and 19. y-intercept: 1. Local maximum is 12 41. Local

minimum is 12 21. y   as x  ; y   as x  .

22. P x  2x 3  6x 2  2. x-intercepts: 05, 07, and 29. y-intercept: 2. Local maximum is 2 6. Local

minimum is 0 2. y   as x   and y   as x  .

4 2 -2

-2 -4

5 2

x

-2

2 -5

4

380

CHAPTER 3 Polynomial and Rational Functions

23. P x  3x 4  4x 3  10x  1. x-intercepts: 01 and 21. y-intercept: 1. Local maximum is 14 145.

24. P x  x 5  x 4  7x 3  x 2  6x  3. x-intercepts: 30, 13, and 19. y-intercept: 3. Local maxima are

24 332 and 05 50. Local minima are 06 06

There is no local maximum. y   as x  .

and 16 16. y   as x   and y   as x  .

20 10 -1-10 -20

1

2

40

3

20 -4

25. (a) Use the Pythagorean Theorem and solving for y 2 we have, x 2  y 2  102  y 2  100  x 2 . Substituting   we get S  138x 100  x 2  1380x  138x 3 .

-2

2

-20

26. (a) The area of the four sides is 2x 2  2x y  1200 

600  x 2 2x y  1200  2x 2  y  . Substituting x   600  x 2  600x  x 3 . we get V  x 2 y  x 2 x

(b) Domain is [0 10]. (c)

(b) 5000

4000 2000 0 0

5

0

10

0

(d) The strongest beam has width 58 inches.

27.

x 2  5x  2 x 3 1 1

5

2

3

6

2

4

2x 3  x 2  3x  4 x 5

2

1

3

4

10

55

290

11

58

294

1

5

6

10

5

5

Using synthetic division, we see that Q x  3x  5 and R x  5.

30. 2

3 3

R x  4.

5

3x 2  x  5 x 2 2

Using synthetic division, we see that Q x  x  2 and

29.

20

(c) V is maximized when x  1414, y  2828.

28. 3

10

x 3  2x  4 x 7 7

1 1

0

2

4

7

49

329

7

47

325

Using synthetic division, we see that

Using synthetic division, we see that

Q x  2x 2  11x  58 and R x  294.

Q x  x 2  7x  47 and R x  325.

CHAPTER 3

31.

x 4  8x 2  2x  7 x 5

32.

1

5

1

33.

0

8

2

7

5

25

85

415

5

17

83

422

Review

2x 4  3x 3  12 x 4 4

2 2

3

0

0

12

8

20

80

320

5

20

80

308

Using synthetic division, we see that

Using synthetic division, we see that

Q x  x 3  5x 2  17x  83 and R x  422.

Q x  2x 3  5x 2  20x  80 and R x  308.

2x 3  x 2  8x  15 x 2  2x  1

34. 2x 

x 2  2x  1

2x 3 

x 4  2x 2  7x x2  x  3

x2 

3 x 2  8x  15

2x 3  4x 2  2x

381

x2  x  3

x 

4

x 4  0x 3  2x 2  7x  0

x 4  x 3  3x 3

3x 2  6x  15

x 3  5x 2  7x

3x 2  6x  3

x 3  x 2  3x

4x 2  4x  0

12

4x 2  4x  12

Therefore, Q x  2x  3, and R x  12.

12 Therefore, Q x  x 2  x  4, and R x  12.

35. P x  2x 3  9x 2  7x  13; find P 5. 5

2 2

9

7

13

10

5

10

1

2

3

36. Q x  x 4  4x 3  7x 2  10x  15; find Q 3 3

1 1

Therefore, P 5  3.

4

7

10

15

3

3

12

6

1

4

2

21

By the Remainder Theorem, we have Q 3  21.

37. The remainder when dividing P x  x 500 6x 101 x 2 2x 4 by x 1 is P 1  1500 6 1201 12 2 14  8.

38. The remainder when dividing P x  x 101  x 4  2 by x  1 is P 1  1101  14  2  0. 39. 12 is a zero of P x  2x 4  x 3  5x 2  10x  4 if   P 12  0. 1 2

2 2

Since P

  1 2

1

5

10

4

1

1

2

4

2

4

8

0

 0, 12 is a zero of the polynomial.

40. x  4 is a factor of

P x  x 5  4x 4  7x 3  23x 2  23x  12 if P 4  0. 4

1 1

4

7

23

23

12

4

0

28

20

12

0

7

5

3

0

Since P 4  0, x  4 is a factor of the polynomial.

41. (a) P x  x 5  6x 3  x 2  2x  18 has possible rational zeros 1, 2, 3, 6, 9, 18.

(b) Since P x has 2 variations in sign, there are either 0 or 2 positive real zeros. Since P x  x 5 6x 3  x 2 2x 18 has 3 variations in sign, there are 1 or 3 negative real zeros.

42. (a) P x  6x 4  3x 3  x 2  3x  4 has possible rational zeros 1, 2, 4,  12 ,  13 ,  23 ,  43 ,  16 .

(b) Since P x has no variations in sign, there are no positive real zeros. Since P x  6x 4  3x 3  x 2  3x  4 has 4 variations in sign, there are 0, 2, or 4 negative real zeros.

382

CHAPTER 3 Polynomial and Rational Functions

43. (a) P x  3x 7  x 5  5x 4  x 3  8 has possible rational zeros 1, 2, 4, 8,  13 ,  23 ,  43 ,  83 .

(b) Since P x has 2 variations in sign, there are either 0 or 2 positive real zeros. Since P x  3x 7  x 5 5x 4  x 3 8 has 3 variations in sign, there are 1 or 3 negative real zeros.

44. (a) P x  6x 10  2x 8  5x 3  2x 2  12 has possible rational zeros 1, 2, 3, 4, 6, 12,  12 ,  13 ,  16 ,  23 ,  32 ,  43 .

(b) Since P x has 2 variations in sign, there are either 0 or 2 positive real zeros. Since P x  6x 10  2x 8  5x 3  2x 2  12 has 2 variations in sign, there are 0 or 2 negative real zeros.

  45. (a) P x  x 3  16x  x x 2  16

46. (a) P x  x 3  3x 2  4x

 x x  1 x  4

 x x  4 x  4

has zeros 0, 1, and 4 (all of multiplicity 1).

has zeros 4, 0, 4 (all of multiplicity 1).

y

(b)

y

(b)

2

5

  47. (a) P x  x 4  x 3  2x 2  x 2 x 2  x  2  x 2 x  2 x  1

   48. (a) P x  x 4  5x 2  4  x 2  4 x 2  1  x  2 x  2 x  1 x  1

Thus, the zeros are 1, 1, 2, 2 (all of multiplicity

The zeros are 0 (multiplicity 2), 2 (multiplicity 1),

1).

and 1 (multiplicity 1).

(b)

(b)

y

x

1

x

1

y

5

1 1

x

1

x

CHAPTER 3

Review

383

49. (a) P x  x 4  2x 3  7x 2  8x  12. The possible rational zeros are 1, 2, 3, 4, 6, 12. P has 2 variations in sign, so it has either 2 or 0 positive real zeros. 1

1 1

2

7

8

12

1

1

8

0

1

8

0

12

1

2

2



2

7

8

12

2

0

14

12

0

7

6

0



 x  2 is a root.

P x  x 4  2x 3  7x 2  8x  12  x  2 x 3  7x  6 . Continuing: 2 1 0 6

6

2

4

3

9

6

1 2 2 10

1 3

2

0

4

3 1 0 7 6

(b)

y

so x  3 is a root and

  P x  x  2 x  3 x 2  3x  2

5

 x  2 x  3 x  1 x  2

x

1

Therefore the real roots are 2, 1, 2, and 3 (all of multiplicity 1).

50. (a) P x  x 4  2x 3  2x 2  8x  8      x 2 x 2  2x  2  4 x 2  2x  2     x 2  4 x 2  2x  2    x  2 x  2 x 2  2x  2

(b)

y

5 x

1

The quadratic is irreducible, so the real roots are 2 (each of multiplicity 1).

51. (a) P x  2x 4  x 3  2x 2  3x  2. The possible rational roots are 1, 2,  12 . P has one variation in sign, and hence 1 positive real root. P x has 3 variations in sign and hence either 3 or 1 negative real roots. 1

2 2

1

2

3

2

2

3

5

2

5

2

0

3 

 x  1 is a zero. 

P x  2x 4  x 3  2x 2  3x  2  x  1 2x 3  3x 2  5x  2 . Continuing: 1 2

3

5

2

2 1 4 2  12 2

1

4 2

3

5

2

4

3 5

y

2

4 2 14 2 1 7 12

2

1 1 2 2

2 2

(b)

0  x   12 is a zero.

5 1

x

   P x  x  1 x  12 2x 2  2x  4 . The quadratic is irreducible, so the real zeros are 1 and  12 (each of

multiplicity 1).

384

CHAPTER 3 Polynomial and Rational Functions

52. (a) P x  9x 5  21x 4  10x 3  6x 2  3x  1. The possible rational zeros are 1,  13 ,  19 . P has 3 variations in sign, hence 3 or 1 positive real roots. P x has 2 variations in sign, hence 2 or 0 real negative roots. 1 9 21

10

6 3 1

9 12 2

4

1

9 12 2 4 1 0  x  1 is a zero.   P x  x  1 9x 4  12x 3  2x 2  4x  1 . Continuing: 1 9 12 2

4

1

9 3 5 1 9 3 5 1   P x  x  12 9x 3  3x 2  5x  1 .

0  x  1 is a zero again. (b)

y

Continuing:

4

1 9 3 5 1 9

9

6

6

1

1 1

0  x  1 is a zero yet again.   P x  x  13 9x 2  6x  1

x

 x  13 3x  12

So the real zeros of P are 1 (multiplicity 3) and  13 (multiplicity 2).

  53. Because it has degree 3 and zeros  12 , 2, and 3, we can write P x  C x  12 x  2 x  3 

C x 3  92 C x 2  72 C x  3C. In order that the constant coefficient be 12, we must have 3C  12  C  4, so

P x  4x 3  18x 2  14x  12. 54. Because it has degree 4 and zeros 4, 4, and 3i, 3i must also be a zero. Thus,   P x  C x  42 x 2  9  C x 4  8C x 3  25C x 2  72C x  144C.

The coefficient of x 2 is 25  25C, so C  1 and hence P x  x 4  8x 3  25x 2  72x  144.

55. No, there is no polynomial of degree 4 with integer coefficients that has zeros i, 2i, 3i and 4i. Since the imaginary zeros of polynomial equations with real coefficients come in complex conjugate pairs, there would have to be 8 zeros, which is impossible for a polynomial of degree 4.    56. P x  3x 4  5x 2  2  3x 2  2 x 2  1 . Since 3x 2  2  0 and x 2  1  0 have no real zeros, it follows that 3x 4  5x 2  2 has no real zeros.

57. P x  x 3  x 2  x  1 has possible rational zeros 1. 1

1

1

1

1

1

0

1

1 0 1 0  x  1 is a zero.  So P x  x  1 x 2  1 . Therefore, the zeros are 1 and i.   58. P x  x 3  8  x  2 x 2  2x  4  0, so 2 is a zero. Using the Quadratic Formula, the other zeros are    2  22  4 1 4  1  12  1  3i. x 2 2 

CHAPTER 3

Review

385

59. P x  x 3  3x 2  13x  15 has possible rational zeros 1, 3, 5, 15. 1

1 1

3

13

15

1

2

15

2 

15

0 

 x  1 is a zero.

So P x  x 3  3x 2  13x  15  x  1 x 2  2x  15  x  1 x  5 x  3. Therefore, the zeros are 3, 1, and 5.

60. P x  2x 3  5x 2  6x  9 has possible rational zeros 1, 3, 9,  12 ,  32 ,  92 . Since there is one variation in sign, there is a positive real zero. 1 2 5 6 9 2

7

3 2

1

5 6 9

3 2 5 6 9 2

6 33 81

3 12

9

2 11 27 72  x  3 is an upper bound 2 8 6 0  x  32 is a zero.   So P x  2x 3  5x 2  6x  9  2x  3 x 2  4x  3  2x  3 x  3 x  1. Therefore, the zeros are 3, 1 2 7

1 8

and 32 .

61. P x  x 4  6x 3  17x 2  28x  20 has possible rational zeros 1, 2, 4, 5, 10, 20. Since all of the coefficients are positive, there are no positive real zeros. 1

1

6

17

28

20

1

5

12

16

2

1

17

28

20

8

18

20

9 10 0  x  2 is a zero.  P x  x 4  6x 3  17x 2  28x  20  x  2 x 3  4x 2  9x  10 . Continuing with the quotient, we have 1

5

12

16

1

6 2

4

2

1

4

1



9

10

2

4

10

2

5

4

0  x  2 is a zero.   Thus P x  x 4  6x 3  17x 2  28x  20  x  22 x 2  2x  5 . Now x 2  2x  5  0 when 

x  2 4451  24i  1  2i. Thus, the zeros are 2 (multiplicity 2) and 1  2i. 2 2

62. P x  x 4  7x 3  9x 2  17x  20 has possible rational zeros 1, 2, 4, 5, 10, 20. 1 1 7 1

9 17 20 2 1 7 8

17

0

9 17 20

2 18

54

1 9 27

37

1 1

74

7

9 17 20

1 6

3

20

54  x  2 is an 1 6 3 20 0  x  1 is a zero. upper bound.   So P x  x 4  7x 3  9x 2  17x  20  x  1 x 3  6x 2  3x  20 . Continuing with the quotient, we have 1 8 17

0 20

1 1

6

3 20

1 5 1

2 1

2

5 2 18

6

3 20

2 8 1

10

4 5 10

4 1

6

3 20

4 8

20

1 2 5 0  x  4 is a zero.   So P x  x 4  7x 3  9x 2  17x  20  x  1 x  4 x 2  2x  5 . Now using the Quadratic Formula on     2  2 6 2  4  4 1 5   1  6. Thus, the zeros are 4, 1, and 1  6. x 2  2x  5 we have: x  2 2

386

CHAPTER 3 Polynomial and Rational Functions

63. P x  x 5  3x 4  x 3  11x 2  12x  4 has possible rational zeros 1, 2, 4. 1 1 3 1 11 12 1 2 3 1 2 3

4

8 4

8 4 

0  x  1 is a zero. 

P x  x 5  3x 4  x 3  11x 2  12x  4  x  1 x 4  2x 3  3x 2  8x  4 . Continuing with the quotient, we have 1 1 2 3

8 4

1 1 4

4

1 1 4 4 0  x  1 is a zero.     x 5  3x 4  x 3  11x 2  12x  4  x  12 x 3  x 2  4x  4  x  13 x 2  4  x  13 x  2 x  2

Therefore, the zeros are 1 (multiplicity 3), 2, and 2.      64. P x  x 4  81  x 2  9 x 2  9  x  3 x  3 x 2  9  x  3 x  3 x  3i x  3i. Thus, the zeros are 3, 3i.

       65. P x  x 6  64  x 3  8 x 3  8  x  2 x 2  2x  4 x  2 x 2  2x  4 . Now using the Quadratic

Formula to find the zeros of x 2  2x  4, we have   3  1  i 3, and using the Quadratic Formula to find the zeros of x 2  2x  4, we have x  2 4441  22i 2 2      22i 3  1  i 3. Therefore, the zeros are 2, 2, 1  i 3, and 1  i 3.  x  2 4441 2 2

1 . 66. P x  18x 3  3x 2  4x  1 has possible rational zeros 1,  12 ,  13 ,  16 ,  19 ,  18

1 18

1 18 2

3 4 1 18 21 17

3 4 1 9

6

1

18 12 2 0  x  12 is a zero. 18 21 17 16  x  1 is an upper bound.   So P x  18x 3  3x 2  4x  1  2x  1 9x 2  6x  1  2x  1 3x  12 . Thus the zeros are 12 and  13

(multiplicity 2).

  67. P x  6x 4  18x 3  6x 2  30x  36  6 x 4  3x 3  x 2  5x  6 has possible rational zeros 1, 2, 3, 6. 1 6 18

6 30

6 12 6 12

36

6 36

6 36 

0  x  1 is a zero.     6 x  1 x 3  2x 2  x  6 .

So P x  6x 4  18x 3  6x 2  30x  36  x  1 6x 3  12x 2  6x  36 Continuing with the quotient we have 1 1 2 1 6

2 1 2 1 6

1 1 2 1 1 2 8

2

0 2



0 1 8

3

3

6

1 2 0  x  3 is a zero.  So P x  6x 4  18x 3  6x 2  30x  36  6 x  1 x  3 x 2  x  2 . Now x 2  x  2  0 when 

1

3 1 2 1 6 1





7 , and so the zeros are 1, 3, and 1i 7 . x  1 1412  1i 2 2 2     68. P x  x 4  15x 2  54  x 2  9 x 2  6 . If x 2  9, then x  3i. If x 2  6, then x  i 6. Therefore, the  zeros are 3i and i 6.

CHAPTER 3

69. 2x 2  5x  3  2x 2  5x  3  0. The solutions are x  05, 3.

2

Review

387

70. Let P x  x 3  x 2  14x  24. The solutions to P x  0 are x  3, 2, and 4.

-5

4

5 -20

-5 -40

71. x 4  3x 3  3x 2  9x  2  0 has solutions x  024, 424.

72. x 5  x  3  x 5  x  3  0. We graph

P x  x 5  x  3. The only real solution is 134. 10

5

-2

2

-50 -10

73. P x  x 3  2x  4 1

1

0

2

4

1

1

1

2

1

0

2

4

2

4

4

1

1 1 5 1 2 2 0   P x  x 3  2x  4  x  2 x 2  2x  2 . Since x 2  2x  2  0 has no real solution, the only real zero of P is

x  2.

   74. P x  x 4  3x 2  4  x 2  1 x 2  4 . Since x 2  4  0 has no real solution, the only real zeros of P are x  1 and x  1.

3 . The vertical asymptote is x  4. Because the x 4 denominator has higher degree than the numerator, the horizontal

75. (a) r x 

asymptote is y  0. When x  0, y  34 , so the y-intercept is 34 . There is no x-intercept because the numerator is never 0. The domain of r is  4  4  and its range is  0  0 .   1 3 1 (b) If f x  , then r x  3  3 f x  4, so we x x 4 x 4 obtain the graph of r by shifting the graph of f to the left 4 units and stretching vertically by a factor of 3.

y

1 1

x

388

CHAPTER 3 Polynomial and Rational Functions

76. (a) r x  

1 . The vertical asymptote is x  5 and the horizontal x 5

y

asymptote is y  0. When x  0, y  15 , so the y-intercept is 15 .

There is no x-intercept. The domain of r is  5  5  and its range is  0  0 .

  1 1 1    f x  5. (b) If f x  , then r x   x x 5 x 5

1 x

1

Thus, we obtain the graph of r by shifting the graph of f to the right 5 units and reflecting in the x-axis. 3x  4 . The vertical asymptote is x  1. Because the x 1 denominator has the same degree as the numerator, the horizontal

77. (a) r x 

y

asymptote is y  31  3. When x  0, y  4, so the y-intercept is 4. When y  0, 3x  4  0  x  43 , so the x-intercept is 43 . The domain of r is  1  1  and its range is  3  3 .

1 , then x 3 x  1  1 1 r x  3  3  f x  1. Thus, we x 1 x 1 obtain the graph of r by shifting the graph of f to the right 1 unit,

(b) If f x 

1 x

1

reflecting in the x-axis, and shifting upward 3 units. 78. (a) r x 

2x  5 . The vertical asymptote is x  2 and the horizontal x 2

y

asymptote is y  2. When x  0, y  52 , so the y-intercept is 52 .

When y  0, x   52 , so the x-intercept is  52 . The domain of r is  2  2  and its range is  2  2 .

1

1 , then x 2 x  2  1 1 r x  2  2  f x  2. Thus, we x 2 x 2 obtain the graph of r by shifting the graph of f to the left 2 units and

(b) If f x 

1x

_1

upward 2 units. 3x  12 12 . When x  0, we have r 0   12, so the y-intercept is x 1 1 12. Since y  0, when 3x  12  0  x  4, the x-intercept is 4. The vertical

79. r x 

y

asymptote is x  1. Because the denominator has the same degree as the

numerator, the horizontal asymptote is y  31  3. The domain of r is  1  1  and its range is  3  3 .

5 1

x

CHAPTER 3

80. r x 

1 x  22

1 1 . When x  0, we have r 0  2  , so the y-intercept is 14 . 4 2

Review

389

y

Since the numerator is 1, y never equals zero and there is no x-intercept. There is a vertical asymptote at x  2. The horizontal asymptote is y  0 because the

degree of the denominator is greater than the degree of the numerator. The domain of r is  2  2  and its range is 0 . 1

x 2 x 2 1 . When x  0, we have r 0  2 81. r x  2  8  4 , x  2 x  4 x  2x  8

1

x

1

x

y

so the y-intercept is 14 . When y  0, we have x  2  0  x  2, so the

x-intercept is 2. There are vertical asymptotes at x  2 and x  4. The domain

1

of r is  2  2 4  4  and its range is  .

82. r x 

x 3  27 27 . When x  0, we have r 0  27 4 , so the y-intercept is y  4 . x 4

y

When y  0, we have x 3  27  0  x 3  27  x  3. Thus the x-intercept

is x  3. The vertical asymptote is x  4. Because the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. By long division, we have r x  x 2  4x  16 

37 . So the end behavior of y is x 4

like the end behavior of g x  x 2  4x  16. The domain of r is

20 x

2

 4  4  and its range is  .

83. r x 

x2  9 x  3 x  3  . When x  0, we have r 0  9 1 , so the 2x 2  1 2x 2  1

y-intercept is 9. When y  0, we have x 2  9  0  x  3 so the x-intercepts

are 3 and 3. Since 2x 2  1  0, the denominator is never zero so there are no

vertical asymptotes. The horizontal asymptote is at y  12 because the degree of the denominator and numerator are the same. The domain of r is   and its   range is 9 12 .

y 1 1

x

390

CHAPTER 3 Polynomial and Rational Functions

84. r x 

2x 2  6x  7 7 . When x  0, we have r 0  7 4  4 , so the y-intercept x 4

y

is y  74 . We use the Quadratic Formula to find the x-intercepts:

   6 62 427 6 92  3 23 . Thus the x-intercepts are x  22 4 2

x  39 and x  09. The vertical asymptote is x  4. Because the degree of the

numerator is greater than the degree of the denominator, there is no horizontal

2

1 , so the slant asymptote. By long division, we have r x  2x  2  x 4 asymptote is r x  2x  2. The domain of r is  4  4  and its range is

x

1

approximately  717]  [1283 .

x 2  5x  14 x  7 x  2   x  7 for x  2. The x-intercept is x 2 x 2 7 and the y-intercept is 7, there are no asymptotes, the domain is x  x  2, and

y

85. r x 

(2, 9)

the range is y  y  9.

1 0

x 3  3x 2  10x

  x x 2  3x  10

x x  5 x  2    x x  5 x 2 x 2 x 2 for x  2. The x-intercepts are 0 and 5 and the y-intercept is 0, there are no   asymptotes, the domain is x  x  2, and the range is y  y   25 4 .

86. r x 

1

x

y (_2, 14)

2 0

x 2  3x  18 x 6 x  6 x  3 for x  3. The x-intercept is 87. r x  2   x 5  5  3 x x x  8x  15

6 and the y-intercept is  65 , the vertical asymptote is x  5, the horizontal   asymptote is y  1, the domain is x  x  3 5, and the range is y  y  1  92 .

1

x

y

2 2

(3, _ 92 )

x

CHAPTER 3

x 2  2x  15 x 3 x  3 x  5 88. r x  3  for  x  1 x  2 x  5 x  1 x  2 x  4x 2  7x  10

1

x  1 and x  2, the horizontal asymptote is y  0, the domain is   x  x  5 1 2, and the range is y  y  19 or y  1 (You can use a

asymptote at x  3 and a horizontal asymptote at

0

(_5, _ 27 )

graphing calculator to find the range.)

x 3 . From the graph we see that the 2x  6 x-intercept is 3, the y-intercept is 05, there is a vertical

391

y

x  5. The x-intercept is 3 and the y-intercept is 32 , the vertical asymptotes are

89. r x 

Review

1

x

2x  7 90. r x  2 . From the graph we see that the x 9 x-intercept is 35, the y-intercept is 078, there is a

horizontal asymptote at y  0 and no vertical asymptote,

y  05, and there is no local extremum.

the local minimum is 111 090, and the local maximum is 811 012.

5

-10 -8 -6 -4 -2

2

4

-20

20

-5 -1

x3  8 91. r x  2 . From the graph we see that the x-intercept is 2, the x x 2 y-intercept is 4, there are vertical asymptotes at x  1 and x  2, there is no

10

horizontal asymptote, the local maximum is 0425 3599, and the local minimum is 4216 7175. By using long division, we see that

-5

10  x f x  x  1  2 , so f has a slant asymptote of y  x  1. x x 2

5 -10

2x 3  x 2 . From the graph we see that the x-intercepts are 0 and 12 , the x 1 y-intercept is 0, there is a vertical asymptote at x  1, the local maximum is

92. r x 

50

0 0, and the local minima are 157 1790 and 032 003. Using long division, we see that r x  2x 2  3x  3 

3 . So the end x 1

-5

behavior of r is the same as the end behavior of g x  2x 2  3x  3.

5

93. 2x 2  x  3  2x 2  x  3  0  P x  x  1 2x  3  0. The cut points occur where x  1  0 and where 2x  3  0; that is, at x  1 and x  32 . We make a sign diagram:   1 32 Sign of  1



x 1





2x  3









 



P x

 Both endpoints satisfy the inequality. The solution is  1]  32   .



3 2



392

CHAPTER 3 Polynomial and Rational Functions

94. x 3  3x 2  4x  12  0  x 2 x  3  4 x  3  0  P x  x  2 x  2 x  3  0. The cut points are 2, 2, and 3. We make a sign diagram:  2

2 2

2 3

3 

x 2









x 2









x 3

















Sign of

P x

All endpoints satisfy the inequality. The solution is  2]  [2 3].

     95. x 4  7x 2  18  0  x 2  9 x 2  2  0  P x  x  3 x  3 x 2  2  0. The last factor is positive for all x. We make a sign diagram:

Sign of

 3

3 3

3 

x 3







x 3













P x

Neither endpoint satisfies the strict inequality. The solution is 3 3.

        96. x 8  17x 4  16  0  x 4  16 x 4  1  0  x 2  4 x 2  4 x 2  1 x 2  1  0     P x  x 2  4 x 2  1 x  4 x  4 x  1 x  1  0. The first two factors are positive for all x. We make a sign diagram:

Sign of

 4

4 1

1 1

1 4

4 

x 4











x 1











x 1











x 4





















P x

None of the endpoints satisfies the strict inequality. The solution is  2  1 1  2 .

97.

5 x 3  x 2  4x  4

0



5  

  0  r x 

x x2  4  x2  4

5  0. We make a sign diagram: x  1 x  2 x  2

 2

2 1

1 2

2 

x 2









x 1









x 2

















Sign of

r x

None of the endpoints is in the domain of r . The solution is  2  1 2.

CHAPTER 3

98.

Review

393

2 3x  1 2 3 3x  1  2 x  2 9x  3  2x  4 7x  1 3x  1    0 0  0  r x   0. We x 2 3 x 2 3 3 x  2 3 x  2 3 x  2 make a sign diagram:     1 2 17 Sign of  2 7 x 2







7x  1













r x

  The cut point 17 satisfies equality, but 2 is not in the domain of r. The solution is 2 17 . 99.

1 2 3 1 2 3 x x  3  2x x  2  3 x  2 x  3       0  0 x 2 x 3 x x 2 x 3 x x x  2 x  3 4x  18 2 9  2x  0  r x   0. We make a sign diagram: x x  2 x  3 x x  2 x  3     9 2 92 Sign of  3 3 0 0 2 2 x 3











x











x 2











9  2x





















r x

  The cut point 92 satisfies equality, but the other cut points are not in the domain of r . The solution is 3 0  2 92 . 100.

1 3 4 1 3 4 x x  3  3x x  2  4 x  2 x  3       0  0 x 2 x 3 x x 2 x 3 x x x  2 x  3 7x  24 r x   0. We make a sign diagram: x x  2 x  3       24  24 Sign of 2 0 0 3 3  7 7  2 7x  24











x 2











x











x 3





















r x

  24 The cut point  24 7 satisfies equality, but the other cut points are not in the domain of r. The solution is  7  2  0 3. 101. f x 

  24  x  3x 2  x  3 8  3x is defined where r x  x  3 8  3x  0. We make a sign diagram:     8 3 83 Sign of  3 3 x 3







8  3x











r x

   Both cut points satisfy equality, so the domain of f is 3 83 .

394

CHAPTER 3 Polynomial and Rational Functions

102. g x   4

  1 1   is defined where r x  x 1  x x 2  x  1  0.       4 4 x  x4 x 1  x3 x 1  x x 2  x  1 1

(Equality is excluded because the denominator cannot be 0.) The last factor of r x is positive for all x. We make a sign diagram:  0

0 1

1 

x







1x



















Sign of

x2  x  1

r x

Neither cut point satisfies the strict inequality, so the domain of g is 0 1. 104.

103. 40

20 20 -2 -3

-2

-1

1

2

3

-1

1

2

3

-20

From the graph, we see that x 4  x 3  5x 2  4x  5 on

From the graph, we see that x 5  4x 4  7x 3  12x  2  0

on approximately 106 017  191 .

approximately [074 195].

105. (a) We use synthetic division to show that 1 is a zero of P x  2x 4  5x 3  x  4: 2

1

2

5

0

1

4

2

3

3

4

3

3

4

0   Thus, 1 is a zero and P x  x  1 2x 3  3x 2  3x  4 .

(b) P x has no change of sign, and hence no positive real zeros. But P x  x  1 Q x, so Q cannot have a positive real zero either. 106. We want to find the solutions of x 4  x 2  24x  6x 3  20  P x  x 4  6x 3  x 2  24x  20  0. The possible rational zeros of P are 1, 2, 4, 5, 10, 20. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x  x 4  6x 3  x 2  24x  20 has 1 variation in sign, and so P has 1 negative real zeros. 1

1

6

1

24

20

1

1

5

4

20

5

4

20

0

2  x  1 is a zero.

1

5

4

20

1

2

6

20

3

10

0

 P x  x  1 x  2 x 2  3x  10  x  1 x  2 x  2 x  5, and 

y

 x  2 is a zero.

so the zeros of P (and hence the x-coordinates of the points of intersection) are 2, 1, 2, and 5. From the original equation, the coordinates of the points of intersection are 2 28, 1 26, 2 68, 5 770.

100 1

x

CHAPTER 3

Test

395

CHAPTER 3 TEST y

1. f x  x 2  x  6    x2  x  6    x 2  x  14  6  14 2   x  12  25 4

2. g x  2x 2  6x  3 is a quadratic

function with a  2 and b  6, so it has a maximum or minimum value where b 6 3 x     . Because 2a 2 2 2

1 1

x

a  0, this gives the minimum value  2     g  32  2  32 6  32 3   32 .

    3. (a) We write the function in standard form: h x  10x001x 2  001 x 2  1000x  001 x 2  1000x  5002    001 5002  001 x  5002  2500. Thus, the maximum height reached by the cannonball is 2500 feet. (b) By the symmetry of the parabola, we see that the cannonball’s height will be 0 again (and thus it will splash into the water) when x  1000 ft. y

4. f x   x  23  27 has y-intercept y  23  27  19 and x-intercept where  x  23  27  x  1.

10

(0, 19) (1, 0)

x

1

5. (a)

2

1 1

0

4

2

5

2

4

0

4

2

0

2

(b) 2x 2  1

9

x 3  2x 2



2x 5  4x 4  x 3  2x 5

Therefore, the quotient is

4x 4

Q x  x 3  2x 2  2, and the remainder is

4x 4

 x3

1 2 x 2  0x 

7

 x2  2x 2 x2

R x  9.

x2



7

 12 15 2

Therefore, the quotient is Q x  x 3  2x 2  12 and the remainder is R x  15 2 . 6. (a) Possible rational zeros are: 1, 3,  12 ,  32 . (b)

1

2

5

4

3

2

7

3

2 7 3 0  x  1 is a zero.   P x  x  1 2x 2  7x  3  x  1 2x  1 x  3    2 x  1 x  12 x  3

(c) The zeros of P are x  1, 3, 12 .

y

2 1

x

396

CHAPTER 3 Polynomial and Rational Functions

7. P x  x 3  x 2  4x  6. Possible rational zeros are: 1, 2, 3, 6. 1 1 1 4 1

0

2 1 1 4

6

2

4

3 1 1 4 6

6

2

3

4

6

6

1

0 4 10 1 1 2 10 1 2 2 0  x  3 is a zero.   So P x  x  3 x 2  2x  2 . Using the Quadratic Formula on the second factor, we have    2  4 2  2 1 2  22  4 1 2    1  i. So zeros of P x are 3, 1  i, and 1  i. x 2 1 2 2 8. P x  x 4  2x 3  5x 2  8x  4. The possible rational zeros of P are: 1, 2, and 4. Since there are four changes in sign, P has 4, 2, or 0 positive real zeros. 1

1

2

5

8

4

1

1

4

4

1 1 4 4 0  So P x  x  1 x 3  x 2  4x  4 . Factoring the second factor by grouping, we have     P x  x  1 x 2 x  1  4 x  1  x  1 x 2  4 x  1  x  12 x  2i x  2i. 

9. Since 3i is a zero of P x, 3i is also a zero of P x. And since 1 is a zero of multiplicity 2,    P x  x  12 x  3i x  3i  x 2  2x  1 x 2  9  x 4  2x 3  10x 2  18x  9.

10. P x  2x 4  7x 3  x 2  18x  3. (a) Since P x has 4 variations in sign, P x can have 4, 2, or 0 positive real zeros. Since

P x  2x 4  7x 3  x 2  18x  3 has no variations in sign, there are no negative real zeros. (b)

4

2 2

7

1

18

3

8

4

20

8

1

5

2

11

Since the last row contains no negative entry, 4 is an upper bound for the real zeros of P x. 1

2 2

1

18

3

1

9

10

28

9

10

28

31

7

Since the last row alternates in sign, 1 is a lower bound for the real zeros of P x. (c) Using the upper and lower limit from part (b), we graph P x in the viewing rectangle [1 4] by

(d) Local minimum 282 7031.

[1 1]. The two real zeros are 017 and 393.

50

1 -2

2

4

-50 2

4

-1

x2  x  6 2x  1 x 3  27 x 3  9x x 3  6x 2  9x , u x  . 11. r x  2 , s x  2 , t x  , and  x  2 x 2 x 3 x x 2 x 4 x  25

CHAPTER 3

Test

397

(a) r x has the horizontal asymptote y  0 because the degree of the denominator is greater than the degree of the

numerator. u x has the horizontal asymptote y  11  1 because the degrees of the numerator and the denominator are the same.

(b) The degree of the numerator of s x is one more than the degree of the denominator, so s has a slant asymptote.   x x 2  6x  9 x x  32   x x  3 (c) The denominator of s x is never 0, so s has no vertical asymptote.  x  x 3 x 3 for x  3, so  has no vertical asymptote.

y

(d) From part (c),  has a “hole” at 3 0. 2x  1 2x  1 (e) r x  2 , so r has vertical asymptotes at x  1  x  1 x  2 x x 2 and x  2. y  0 is a horizontal asymptote because the degree of the numerator is less than the degree of the denominator.

1

x2  x  6 x  3 x  2 (f) u x   . When x  0, we have 2 x  5 x  5 x  25

2

x

6  6 , so the y-intercept is y  6 . When y  0, we have x  3 or u x  25 25 25 x  2, so the x-intercepts are 3 and 2. The vertical asymptotes are x  5 and

x  5. The horizontal asymptote occurs at y  11  1 because the degree of the denominator and numerator are the same. x 2  2x  5

(g) x 2

x 3  0x 2  9x  x 3  2x 2

y

0

2x 2  9x 2x 2  4x 5x  0 5x  10 10

Thus P x  x 2  2x  5 and t x 

12. x 

10 2

x

x 3  9x have the same end behavior. x 2

6x 6  x  x 2x  5 2x 2  4x  6 6x x  1 3  x 0 x 0 0   r x. We 2x  5 2x  5 2x  5 2x  5 x5 2

make a sign diagram:

  1 52





53 2

3 





Sign of

 1

x 1



3x

























x  52

r x



  The cut point 52 is excluded so that the denominator is not 0. The solution is  1]  52  3 .

398

FOCUS ON MODELING

1 13. f x   is defined where 4  2x  x 2  0. Using the Quadratic Formula to solve x 2  2x  4  0, we 4  2x  x 2   2  22  4 1 4   1  5. The radicand is positive between these two roots, so the domain of have x  2 1     f is 1  5 1  5 .

P x  x 4  4x 3  8x. From the graph, the x-intercepts are approximately

14. (a) 10

124, 0, 2, and 324, P has local maximum P 1  5, and P has local minima P 073  P 273  4.

-2

2

4

(b) From part (a), P x  0 on approximately  124]  [0 2]  [324 .

FOCUS ON MODELING Fitting Polynomial Curves to Data

1. (a) Using a graphing calculator, we obtain the quadratic

2. (a) Using a graphing calculator, we obtain the quadratic

polynomial

polynomial

y  0275428x 2  197485x  2735523 (where

y  02783333x 2  184655x  166732 (where

miles are measured in thousands). (b)

y

(b)

100

y 120

80 60

80

40

40

20 0

plants/acre are measured in thousands).

25

30

35

40

45

50 x

Pressure (lb/in@)

(c) Moving the cursor along the path of the polynomial, we find that 3585 lb/in2 gives the longest tire life.

0

10

20

30

40

50

60 x

Density (thousand plants/acre)

(c) Moving the cursor along the path of the polynomial, we find that yield when 37,000 plants are planted per acre is about 135 bushels/acre.

Fitting Polynomial Curves to Data

3. (a) Using a graphing calculator, we obtain the cubic

y 60

4. (a)

polynomial

50

y  000203709x 3  0104522x 2

40

 1966206x  145576.

(b)

399

30

y

20 10

20

0

1

10

2

x

3

Time (s) A quadratic model seems appropriate. 35 x

(b) Using a graphing calculator, we obtain the quadratic

(c) Moving the cursor along the path of the polynomial,

(c) Moving the cursor along the path of the polynomial,

0

5

10

15

20

25

30

Seconds

polynomial y  160x 2  518429x  420714.

we find that the subjects could name about

we find that the ball is 20 ft. above the ground 03

43 vegetables in 40 seconds.

seconds and 29 seconds after it is thrown upward.

(d) Moving the cursor along the path of the polynomial,

(d) Again, moving the cursor along the path of the

we find that the subjects could name 5 vegetables in

polynomial, we find that the maximum height is

about 20 seconds.

462 ft.

5. (a) Using a graphing calculator, we obtain the quadratic polynomial y  00120536x 2  0490357x  496571. (c) Moving the cursor along the path of the polynomial, we find that the tank should drain in 190 minutes.

(b)

y 6 5 4 3 2 1 0

10

Time (min)

20 x

4

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

4.1

EXPONENTIAL FUNCTIONS

1 , f 0  50  1, f 2  52  25, 1. The function f x  5x is an exponential function with base 5; f 2  52  25 and f 6  56  15,625.

2. (a) f x  2x is an exponential function with base 2. It has graph III.  x (b) f x  2x  12 is an exponential function with base 12 . It has graph I.

(c) The graph of f x  2x is obtained from the graph of 2x by reflecting about the x-axis. Thus, it has graph II.

(d) The graph of f x  2x is obtained from the graph of 2x by reflecting about the x- and y-axes. Thus, it has graph IV. 3. (a) To obtain the graph of g x  2x  1 we start with the graph of f x  2x and shift it downward 1 unit. 4.

5.

6.

7. 8.

(b) To obtain the graph of h x  2x1 we start with the graph of f x  2x and shift it to the right 1 unit. nt  for compound interest, the letters P, r, n, and t stand for principal, interest rate per In the formula A t  P 1  nr year, number of times interest is compounded per year, and number of years, respectively, and A t stands for the amount after t years. So if $100 is invested at an interest rate of 6% compounded quarterly, then the amount after 2 years is  4  2 100 1  006  11265. 4  x The exponential function f x  12 has the horizontal asymptote y  0. This means that as x  , we have  x 1  0. 2  x The exponential function f x  12  3 has the horizontal asymptote y  3. This means that as x  , we have  x 1  3  3. 2      5  22195, f 2  0063, f 03  1516. f x  4x ; f 12  412  4  2, f     f x  3x1 ; f 12  0577, f 25  5196, f 1  0111, f 14  0439.

   x1   9. g x  13 ; g 12  0192, g 2  0070, g 35  15588, g 14  1552.

       3x ; f  12  0650, f 6  8281, f 3  0075, f 43  3160. 10. f x  43

11.

f x  2x

y

12.

g x  8x

x

y

x

y

4

2

2

1 16 1 4

1

1 64 1 8

0

1

0

1

1

8

2

64

2

4

4

16

1 1

x

y

1 1

x

401

402

13.

CHAPTER 4 Exponential and Logarithmic Functions

 x f x  13

y

x

y

x

y

2

9

5

0620921323

1

3

1

090

0

1

1

1 3 1 9

2 15.

14. h x  11x

y

1 x

1

g x  3 13x

0

1

5

161051

259374246  x 16. h x  2 14

y

y

x

y

2

1775

32

2308

2

1

1

8

30

0

2

1

1 2 1 8 1 32 1 128

1

39

2

507

3

6591

3

4

8568

4

1 1

x

2

x

1

10

x

0

1

y

2 1

x

 x 18. f x  3x and g x  13

17. f x  2x and g x  2x y

y

g

f=g

f

1

1

x

1

x

1

 x 20. f x  34 and g x  15x .

19. f x  4x and g x  7x . y

y

g f

f

2

5 1

x

21. From the graph, f 2  a 2  9, so a  3. Thus f x  3x .

g

0

1

x

SECTION 4.1 Exponential Functions

403

22. From the graph, f 1  a 1  15 , so a  5. Thus f x  5x .   1 , so a  1 . Thus f x  1 x . 23. From the graph, f 2  a 2  16 4 4  x 24. From the graph, f 3  a 3  8, so a  12 . Thus f x  12 . 25. The graph of f x  5x1 is obtained from that of y  5x by shifting 1 unit to the left, so it has graph II. 26. The graph of f x  5x  1 is obtained from that of y  5x by shifting 1 unit upward, so it has graph I.

27. g x  2x  3. The graph of g is obtained by shifting the graph of y  2x downward 3 units. Domain:  .

Range: 3 . Asymptote: y  3.

 x 28. h x  4  12 . The graph of h is obtained by shifting  x the graph of y  12 upward 4 units. Domain:  . Range: 4 . Asymptote: y  4.

y

y

1

(_1, 6)

1 (1, _1)

x 1 x

1

29. The graph of f x  3x is obtained by reflecting the graph of y  3x about the x-axis. Domain:  . Range:  0. Asymptote: y  0. y

30. The graph of f x  10x is obtained by reflecting the graph of y  10x about the y-axis. Domain:  . Range: 0 . Asymptote: y  0. y

1 x

1 (1, _3)

1 1

x

404

CHAPTER 4 Exponential and Logarithmic Functions

31. f x  10x3 . The graph of f is obtained by shifting the graph of y  10x to the left 3 units. Domain:  .

Range: 0 . Asymptote: y  0.

y

Range: 0 . Asymptote: y  0.

(3, 1)

1 x

_1

33. y  5x  1. The graph of y is obtained by reflecting the

graph of y  5x about the x-axis and then shifting upward 1 unit. Domain:  . Range: 1 . Asymptote: y  1.

graph of y  2x to the right 3 units. Domain:  . y

100

(_3, 1)

32. g x  2x3 . The graph of g is obtained by shifting the

y

x

1

34. h x  6  3x . The graph of h is obtained by reflecting

the graph of y  3x about the x-axis and shifting upward 6 units. Domain:  . Range:  6. Asymptote: y  6.

y

(1, 3) 1 x

1

(0, 2) 1 1

x

 x 35. y  2  13 . The graph is obtained by reflecting the  x graph of y  13 about the x-axis and shifting upward 2 units. Domain:  . Range:  2. Asymptote: y  2.

36. y  5x  3. The graph is obtained by reflecting the

graph of y  5x about the y-axis, then shifting downward

3 units. Domain:  . Range: 3 . Asymptote: y  3. y

y

1 0

1

x

1 0

1

x

SECTION 4.1 Exponential Functions

37. h x  2x4  1. The graph of h is obtained by shifting

the graph of y  2x to the right 4 units and upward 1 unit.

Domain:  . Range: 1 . Asymptote: y  1. y

405

38. y  3  10x1  10x1  3. The graph of y is

obtained by reflecting the graph of y  10x about the

y-axis, then shifting to the right 1 unit and upward 3 units.

Domain:  . Range:  3. Asymptote: y  3. y

2

(1, 2)

(4, 2)

1

1 1

1

x

x

 x 39. g x  1  3x  3x  1. The graph of g is obtained 40. y  3  15  5x  3. The graph of y is obtained by reflecting the graph of y  3x about the x- and y-axes

and then shifting upward 1 unit. Domain:  . Range:  1. Asymptote: y  1.

by reflecting the graph of y  5x about the x- and y-axes and then shifting upward 3 unit. Domain:  . Range:  3. Asymptote: y  3.

y

y

1 (_1, _2)

41. (a)

1

(0, 2)

1

x

x

1

y

42. (a)

y

g f

1

1 1

x

  (b) Since g x  3 2x  3 f x and f x  0, the

height of the graph of g x is always three times the height of the graph of f x  2x , so the graph of g

is steeper than the graph of f .

1

x

 x2 (b) f x  9x2  32  32x2  3x  g x. So f x  g x, and the graphs are the same.

406

CHAPTER 4 Exponential and Logarithmic Functions

43.

f x  x 3

x

y

g x  3x

0

0

1

1

1

3

2

8

9

3

27

27

4

64

81

6

216

729

8

512

6561

10

1000

59,049

x

g(x)=3

f(x)=x#

20 0

x

1

From the graph and the table, we see that the graph of g ultimately increases much more quickly than the graph of f . 44.

f x  x 4

x

y

g x  4x

0

0

1

1

1

3

2

16

16

3

81

64

4

256

256

6

1296

4096

8

4096

65,536

10

10,000

1,048,576

g(x)=4

x

f(x)=x$

100 0

x

1

From the graph and the table, we see that the graph of g ultimately increases much more quickly than the graph of f . 45. (a) From the graphs below, we see that the graph of f ultimately increases much more quickly than the  of g.  graph   (i) [0 5] by [0 20] (ii) [0 25] by 0 107 (iii) [0 50] by 0 108 20

10,000,000

f

8,000,000 6,000,000

10

g

f

1

2

4,000,000

3

4

80,000,000

f

g

40,000,000 20,000,000

0

5

100,000,000

60,000,000

2,000,000

0

g

10

20

0

20

40

(b) From the graphs in parts (a)(i) and (a)(ii), we see that the approximate solutions are x  12 and x  224. 46. (a) (i) [4 4] by [0 20]

  (iii) [0 20] by 0 105

(ii) [0 10] by [0 5000]

20 g

f

4000

f

g

100,000

80,000 60,000

10

f

g

2000

40,000 20,000 -4

-3

-2

-1 0

1

2

3

4

0

2

4

6

8

10

0

10

(b) From the graphs in parts (i) and (ii), we see that the solutions of 3x  x 4 are x  080, x  152 and x  717.

20

SECTION 4.1 Exponential Functions

47.

c=2 c=1

5

48.

10

c=0.5

4

c=2

c=1

8

3 c=4

c=4

c=0.5

6 c=0.25

2

4

1

c=0.25

2 -3

-2

-1

0 -1

1

2

3 0

-2

The larger the value of c, the more rapidly the graph of shifted horizontally 1 unit. This is because of our choice of c; each c in this exercise is of the form 2k . So f x  2k  2x  2xk . 2

2

4

The larger the value of c, the more rapidly the graph of  x f x  2cx increases. In general, f x  2cx  2c ;  x so, for example, f x  22x  22  4x .

f x  c2x increases. Also notice that the graphs are just

49. y  10xx

407

50. y  x2x 2 1 -4 -4

-2

0

2

(a) From the graph, we see that the function is increasing

on 144  and decreasing on  144.

(b) From the graph, we see that the range is

(b) From the graph, we see that the range is

approximately 0 178].

approximately [053 .

10xh  10x f x  h  f x 51. f x  10x , so   10x

52. f x  3x1 , so

2

(a) From the graph, we see that the function is increasing

on  050 and decreasing on 050 .

h

-2

4

h



 10h  1 . h

3xh1  3x1 f x  h  f x   3x1 h h



 3h  1 . h

53. (a) After 1 hour, there are 1500  2  3000 bacteria. After 2 hours, there are 1500  2  2  6000 bacteria. After 3 hours, there are 1500  2  2  2  12,000 bacteria. We see that after t hours, there are N t  1500  2t bacteria. (b) After 24 hours, there are N 24  1500  224  25,165,824,000 bacteria.

54. (a) Because the population doubles every year, there are N t  320  2t mice after t years. (b) After 8 years, there are approximately N 8  320  28  81,920 mice.

408

CHAPTER 4 Exponential and Logarithmic Functions

55. Using the formula A t  P 1  ik with P  5000,

004 per month, and k  12  number 12 of years, we fill in the table:

i  4% per year 

56. Using the formula A t  P 1  ik with P  5000, rate per year i per month, and k  12  5  60 months, 12 we fill in the table:

Amount

Rate per year

Amount

1

$520371

1%

$525625

2

$541571

2%

$552539

3

$563636

3%

$580808

4

$586599

4%

$610498

5

$610498

5%

$641679

6

$635371

6%

$674425

Time (years)

2t  57. P  10,000, r  003, and n  2. So A t  10,000 1  003  10,000  10152t . 2

(a) A 5  10000  101510  11,60541, and so the value of the investment is $11,60541.

(b) A 10  10000  101520  13,46855, and so the value of the investment is $13,46855. (c) A 15  10000  101530  15,63080, and so the value of the investment is $15,63080.

365t  . 58. P  2500, r  0025, and n  365. So A t  2500 1  0025 365 365  2  (a) A 2  2500 1  0025  262817, and so the value of the investment is $262817. 365 365  3   269470, and so the value of the investment is $269470. (b) A 3  2500 1  0025 365 365  6   290457, and so the value of the investment is $290457. (c) A 6  2500 1  0025 365

4t  . 59. P  500, r  00375, and n  4. So A t  500 1  00375 4 4   51902, and so the value of the investment is $51902. (a) A 1  500 1  00375 4 8   53875, and so the value of the investment is $53875. (b) A 2  500 1  00375 4 40   72623, and so the value of the investment is $72623. (c) A 10  500 1  00375 4 4t  . 60. P  4,000, r  00575, and n  4. So A t  4000 1  00575 4 16   502616, and so the amount due is $502616. (a) A 4  4000 1  00575 4 24   563410, and so the amount due is $563410. (b) A 6  4000 1  00575 4 32   631558, and so the amount due is $631558. (c) A 8  4000 1  00575 4

23   P 10456  10000  13023P  P  767896. 61. We must solve for P in the equation 10000  P 1  009 2 Thus, the present value is $7,67896.

 125 62. We must solve for P in the equation 100000  P 1  008  P 10066760  100000  14898P  12 P  $67,12104.

SECTION 4.2 The Natural Exponential Function

409

   r n 008 12 63. rAPY  1   1. Here r  008 and n  12, so rAPY  1   1  1006666712  1  0083000. n 12 Thus, the annual percentage yield is about 83%.    r n 0055 4  1. Here r  0055 and n  4, so rAPY  1   1  0056145. Thus, the annual 64. rAPY  1  n 4 percentage yield is about 561%.

65. (a) In this case the payment is $1 million. (b) In this case the total pay is 2  22  23      230  230 cents  $10,737,41824. Since this is much more than method (a), method (b) is more profitable. 66. Since f 40  240  1,099,511,627,776, it would take a sheet of paper 4 inches by 1,099,511,627,776 inches. Since there are 12 inches in a foot and 5,280 feet in a mile, 1,099,511,627,776 inches  174 million miles. So the dimensions of the sheet of paper required are 4 inches by about 174 million miles.

4.2

THE NATURAL EXPONENTIAL FUNCTION

1. The function f x  e x is called the natural exponential function. The number e is approximately equal to 271828. 2. In the formula A t  Pert for continuously compound interest, the letters P, r, and t stand for principal, interest rate per year, and number of years respectively, and A t stands for amount after t years. So, if $100 is invested at an interest rate of 6% compounded continuously, then the amount after 2 years is A 2  100  e0062  $11275. 3. h x  e x ; h 1  2718, h   23141, h 3  0050, h

  2  4113

  4. h x  e3x ; h 13  0368, h 15  0011, h 1  20086, h   12,391648 5.

f x  15e x x

y

6. g x  4e13x

y

x

y

2

020

3

1087

1

055

2

779

05

091

1

558

0

15

05

247

1

408

2

1108

2 0

1

x

0

4

1

287

2

205

3

147

y

2 0

1

x

410

CHAPTER 4 Exponential and Logarithmic Functions

7. g x  2  e x . The graph of g is obtained from the graph of y  e x by shifting it upward 2 units. Domain:  . Range: 2 . Asymptote: y  2. y

8. h x  ex  3. The graph of h is obtained from the graph of y  e x by reflecting it about the y-axis and

shifting it downward 3 units. Domain:  . Range: 3 . Asymptote: y  3. y

5

5 0

1

0

x

9. y  e x . The graph of y  e x is obtained from the

graph of y  e x by reflecting it about the x-axis. Domain:  . Range:  0. Asymptote: y  0. y

1 1

1

10. f x  1  e x . The graph of f x  1  e x is obtained

by reflecting the graph of y  e x about the x-axis and then shifting upward 1 unit. Domain:  . Range:  1. Asymptote: y  1. y

x

1 1

11. y  ex  1. The graph of y  ex  1 is obtained from the graph of y  e x by reflecting it about the y-axis then shifting downward 1 unit. Domain:  . Range: 1 . Asymptote: y  1. y

x

x

12. f x  ex . The graph of f x  ex is obtained

by reflecting the graph of y  e x about the y-axis and then about the x-axis. Domain:  . Range:  0. Asymptote: y  0.

y 1 1

1 1

x

x

SECTION 4.2 The Natural Exponential Function

13. y  e x2 . The graph of y  e x2 is obtained from the

graph of y  e x by shifting it to the right 2 units. Domain:  . Range: 0 . Asymptote: y  0. y

411

14. f x  e x3  4. The graph of f x  e x3  4 is obtained by shifting the graph of y  e x to the right

3 units, and then upward 4 units. Domain:  . Range: 4 . Asymptote: y  4. y

(3, 5) 1

(2, 1) x

1

1 x

1

15. h x  e x1  3. The graph of h is obtained from the graph of y  e x by shifting it to the left 1 unit and

downward 3 units. Domain:  . Range: 3 .

Asymptote: y  3.

y

16. g x  e x1  2. The graph of g is obtained by shifting the graph of y  e x to the right 1 unit, reflecting it about the x-axis, then shifting it downward 2 units.

Domain:  . Range:  2. Asymptote: y  2.

y 1

1

1

(0, e-3)

(1, _3)

x

1

x

(_1, _2)

y

17. (a)

y

18. (a)

1

1 y=_ 2 e¨

y=sinh x

y=cosh x 1 y=_ 2 eШ

1

1

1 y=_ 2 e¨

1

x

ex  e x ex  ex  2 2 e x  ex   cosh x 2

(b) cosh x 

x

1 y=_ _ 2 eШ

ex  e x ex  ex  2 2 ex  e x e x  ex     sinh x 2 2

(b) sinh x 

412

CHAPTER 4 Exponential and Logarithmic Functions a=1

19. (a)

a=1.5

10 a=0.5

20. 4

a=2

8

2

6 4

0

2 -4

-2

0

0 2

4

 a  xa e (b) As a increases the curve y   exa 2 flattens out and the y intercept increases.

21. g x  x x . Notice that g x is only defined for x  0. The graph of g x is shown in the viewing rectangle

[0 15] by [0 15]. From the graph, we see that there is a local minimum of about 069 when x  037.

20

40

x  From the graph, we see that y  1  1x approaches e as x get large.

22. g x  e x  e2x . The graph of g x is shown in the

viewing rectangle [1 2] by [1 6]. From the graph, we see that there is a local minimum of about 189 when x  023. 5

1

0 0

1

-1

1

2

23. D t  50e02t . So when t  3 we have D 3  50e023  274 milligrams.

24. m t  13e0015t (a) m 0  13 kg.

(b) m 45  13e001545  13e0675  6619 kg. Thus the mass of the radioactive substance after 45 days is about 66 kg.

  25.  t  180 1  e02t

  (a)  0  180 1  e0  180 1  1  0.   (b)  5  180 1  e025  180 0632  11376 ft/s. So the

velocity after 5 s is about 1138 ft/s.    10  180 1  e0210  180 0865  1557 ft/s. So the

(c)

y 200

100

velocity after 10 s is about 1557 ft/s. (d) The terminal velocity is 180 ft/s.

5

x

SECTION 4.2 The Natural Exponential Function

  26. (a) Q 5  15 1  e0045  15 01813  27345. Thus

(c) 20

approximately 27 lb of salt are in the barrel after 5 minutes.   (b) Q 10  15 1  e00410  15 03297  4946Thus

10

approximately 49 lb of salt are in the barrel after 10 minutes.

0 0

(d) The amount of salt approaches 15 lb. This is to be expected, since

100

200

50 gal  03 lb/gal  15 lb. 1200 1  11e02t 1200 1200  100. (a) P 0   1  11 1  11e020 1200 1200 1200 (b) P 10   482. P 20   999. P 30   1168. 1  11e0210 1  11e0220 1  11e0230 1200  1200. The graph shown confirms this. (c) As t   we have e02t  0, so P t  10

27. P t 

28. n t 

5600 05  275e0044t

(b) 10000

(a) n 0  5600 28  200 (c) From the graph, we see that n t approaches about 11,200 as t gets large.

0 0

29. P t 

732 61  59e002t

(a) In the year 2200, t  2200  2000  200, and the population is

732  1179 billion. In 61  59e002200 732  1197 billion. 2300, t  300, and P 300  61  59e002300 predicted to be P 200 

(c) As t increases, the denominator approaches 61, so according to this model, the world population approaches 732 61  12 billion people. 30. D t 

(b)

200

P 14 12 10 8 6 4 2 0

100

200

300

400

500 t

54 54 . So D 20   1600 ft. 1  29e001t 1  29e00120

31. Using the formula A t  Pert with P  7000 and r  3%  003, we fill in the table:

32. Using the formula A t  Pert with P  7000 and t  10 years, we fill in the table:

Time (years)

Amount

Rate per year

1

$721318

1%

$773620

2

$743286

2%

$854982

3

$765922

3%

$944901

4

$789248

4%

$10,44277

5

$813284

5%

$11,54105

6

$838052

6%

$12,75483

Amount

413

414

CHAPTER 4 Exponential and Logarithmic Functions

33. We use the formula A t  Pert with P  2000 and r  35%  0035. (a) A 2  2000e0035  2  $214502

(b) A 42  2000e0035  4  $230055 (c) A 12  2000e0035  12  $304392

34. We use the formula A t  Pert with P  3500 and r  625%  00625. (a) A 3  3500e00625  3  $422181 (b) A 6  3500e00625  6  $509247 (c) A 9  3500e00625  9  $614269 35. (a) Using the formula A t  P 1  ik with P  600, i  25% per year  0025, and k  10, we calculate A 10  600 102510  $76805.

  0025 20  $76922. semiannually and k  10  2  20, so A  600 1  (b) Here i  0025 10 2 2   0025 40  $76982. quarterly and k  10  4  40, so A 10  600 1  (c) Here i  25% per year  0025 4 4 (d) Using the formula A t  Pert with P  600, r  25%  0025, and t  10, we have A 10  600e0025  10  $77042.

36. We use the formula A t  Pert with P  8000 and t  12.

(a) If r  2%  002, then A 12  8000e002  12  $10,16999.

(b) If r  3%  003, then A 12  8000e003  12  $11,46664.

(c) If r  45%  0045, then A 12  8000e0045  12  $13,72805. (d) If r  7%  007, then A 12  8000e007  12  $18,53094.

2   10252. 37. Investment 1: After 1 year, a $100 investment grows to A 1  100 1  0025 2 4   10227. Investment 2: After 1 year, a $100 investment grows to A 1  100 1  00225 4

Investment 3: After 1 year, a $100 investment grows to A 1  100e002  10202. We see that Investment 1 yields the highest return. 2   10519. 38. Investment 1: After 1 year, a $100 investment grows to A 1  100 1  005125 2 Investment 2: After 1 year, a $100 investment grows to A 1  100e005  10512. We see that Investment 1 yields a higher return.

39. (a) A t  Pert  5000e009t (b) 20000

0 0

10

20

(c) A t  25,000 when t  1788 years.

4.3

LOGARITHMIC FUNCTIONS

1. log x is the exponent to which the base 10 must be raised in order to get x. x log x

103

102

101

100

101

102

103

1012

3

2

1

0

1

2

3

12

SECTION 4.3 Logarithmic Functions

415

2. The function f x  log9 x is the logarithm function with base 9. So f 9  log9 91  1, f 1  log9 90  0,   f 19  log9 91  1, f 81  log9 92  2, and f 3  log9 912  12 . 3. (a) 53  125, so log5 125  3.

(b) log5 25  2, so 52  25.

4. (a) f x  log2 x is a logarithmic function with base 2. It has graph III.

(b) The graph of f x  log2 x is obtained from that of y  log2 x by reflecting about the y-axis. It has graph II.

(c) f x   log2 x is obtained from that of y  log2 x by reflecting about the x-axis. It has graph I.

(d) f x   log2 x is obtained from that of y  log2 x by reflecting about the x- and y-axes. It has graph IV.

5. The natural logarithmic function f x  ln x has the vertical asymptote x  0. 6. The natural logarithmic function f x  ln x  1 has the vertical asymptote x  1. 7.

8.

Logarithmic form

Logarithmic form

Exponential form

log8 8  1

81  8

log4 64  3

823  4

log4 8  32

log8 64  2

log8 4  23

log8 512  3

log8 81  1

1  2 log8 64

9. (a) 34  81 (b) 30  1 13. (a) 3x  5 (b) 72  3y

82  64 83  512

81  18 1 82  64

10. (a) 51  15 (b) 43  64 14. (a) 61  z (b) 102t  3

Exponential form 43  64

log4 2  12

412  2

1  2 log4 16 log4 21   12 1  5 log4 32 2

1 42  16 412  12

11. (a) 813  2 (b) 102  001 15. (a) e3y  5

432  8

1 452  32 1 12. (a) 53  125

(b) 823  4 16. (a) e2  x  1

(b) e1  t  1

(b) e4  x  1

17. (a) log10 10,000  4   1  2 (b) log5 25

18. (a) log6 36  2   1  1 (b) log10 10

19. (a) log8 81  1   (b) log2 18  3

20. (a) log4 0125   32

21. (a) log4 70  x

22. (a) log3 10  2x

23. (a) ln 2  x

24. (a) ln 05  x  1

(b) log3   5 25. (a) log2 2  1 (b) log5 1  log5 50  0

(b) log10 01  4x 26. (a) log3 37  7 (b) log4 64  log4 43  3

(c) log6 65  5 (c) log5 125  3   1  log 33  3 29. (a) log3 27 3  (b) log10 10  log10 1012  12   (c) log5 02  log5 15  log5 51  1

31. (a) 3log3 5  5

(b) ln y  3

(b) log7 343  3

(b) ln t  05x

27. (a) log6 36  log6 62  2 28. (a) log2 32  log2 25  5 (b) log9 81  log9 92  2

(b) log8 817  17

(c) log7 710  10

(c) log6 1  log6 60  0

30. (a) log5 125  log5 53  3 (b) log49 7  log49 4912  12  12  (c) log9 3  log9 312  log9 912  14   32. (a) eln 3  3

(b) 5log5 27  27

1 (b) eln1  

(c) eln 10  10

(c) 10log 13  13

416

CHAPTER 4 Exponential and Logarithmic Functions

 12  2  log4 212  log4 412  14    1 (b) log4 12  log4 21  log4 412   12  3 (c) log4 8  log4 23  log4 412  log4 432  32

33. (a) log8 025  log8 823   23

34. (a) log4

(b) ln e4  4   1  ln e1  1 (c) ln e

36. (a) log3 x  2  x  32  19

35. (a) log4 x  3  x  43  64 (b) log10 001  x  10x  001  x  2

(b) log5 125  x  5x  125  x  3

37. (a) ln x  3  x  e3

38. (a) ln x  1  x  e1  1e   (b) ln 1e  x  x  ln e1   ln e  1

(b) ln e2  x  x  2 ln e  2   1  x  7x  1  x  2 39. (a) log7 49 49

40. (a) x  log4 2  log4 412  12

  41. (a) log2 12  x  2x  12  x  1

42. (a) logx 1000  3  x 3  1000  x  10

43. (a) logx 16  4  x 4  16  x  2

44. (a) logx 6  12  x 12  6  x  36

(b) log4 x  2  x  42  16

(b) log2 x  5  25  x  x  32

(b) logx 25  2  x 2  25  x  5

1 (b) log10 x  3  103  x  x  1000

(b) logx 8  32  x 32  8  x  823  4

(b) logx 3  13  x 13  3  x  27

45. (a) log 2  03010

46. (a) log 50  16990  (b) log 2  01505    (c) log 3 2  06276

(b) log 352  15465   (c) log 23  01761

48. (a) ln 27  32958

47. (a) ln 5  16094

(b) ln 739  20001

(b) ln 253  32308    (c) ln 1  3  10051

(c) ln 546  40000

y

49. x 1 33 1 32 1 3

f x 3

x

1

_3

1 43 1 42 1 4

0

2

_1

1

2

3

4

_2

1

1

0

3

1

32

2

y

50.

2

5

6

x

g x 3

2 1 0

2

_1

2

3

4

_2

1

_3

_4

1

0

_4

_5

4

1

_5

42

2

f x  log3 x

1

g x  log4 x

5

6

x

SECTION 4.3 Logarithmic Functions y

51. x 1 103 1 102 1 10

f x

y

52.

2

x

1

2

g x

2

_3

1 103 1 102 1 10

1

0

_4

1

1

_4

10

2

_5

10

2

_5

102

4

102

3

6

0

4

1

_1

2

3

4

5

6

x

_2

f x  2 log x

417

1

2

0

1

_1

1

2

3

4

5

6

x

_2

0

_3

g x  log x  1

53. Since the point 5 1 is on the graph, we have 1  loga 5  a 1  5. Thus the function is y  log5 x.     54. Since the point 12  1 is on the graph, we have 1  loga 12  a 1  12  a  2. Thus the function is y  log2 x.   55. Since the point 3 12 is on the graph, we have 12  loga 3  a 12  3  a  9. Thus the function is y  log9 x.

56. Since the point 9 2 is on the graph, we have 2  loga 9  a 2  9  a  3. Thus the function is y  log3 x.

57. The graph of f x  2  ln x is obtained from that of y  ln x by shifting it upward 2 units, as in graph I.

58. The graph of f x  ln x  2 is obtained from that of y  ln x by shifting it to the right 2 units, as in graph II. 59. The graph of y  log4 x is obtained from that of y  4x by reflecting it in the line y  x. y

60. The graph of y  log3 x is obtained from that of y  3x by reflecting it in the line y  x. y

y=4¨

y=3¨

y=x

y=log£ x

y=log¢ x

1

1 x

1

61. The graph of g x  log5 x is obtained from that of y  log5 x by reflecting it about the y-axis.

Domain:  0. Range:  . Vertical asymptote: x  0.

y=x

1

x

62. The graph of f x   log10 x is obtained from that of y  log10 x by reflecting it about the x-axis.

Domain: 0 . Range:  . Vertical asymptote: x  0.

y

y

1

1 1

x

1

x

418

CHAPTER 4 Exponential and Logarithmic Functions

63. The graph of f x  log2 x  4 is obtained from that of 64. The graph of g x  ln x  2 is obtained from that of

y  ln x by shifting to the left 2 units. Domain: 2 .

y  log2 x by shifting it to the right 4 units.

Range:  . Vertical asymptote: x  2.

Domain: 4 . Range:  .

y

Vertical asymptote: x  4. y

1 1

1

x

x

1

65. The graph of h x  ln x  5 is obtained from that of

y  ln x by shifting to the left 5 units. Domain: 5 .

Range:  . Vertical asymptote: x  5. y

66. The graph of g x  log6 x  3 is obtained from that of y  log6 x by shifting to the right 3 units.

Domain: 3 . Range:  . Vertical asymptote: x  3. y

1

1 1

x

67. The graph of y  2  log3 x is obtained from that of

y  log3 x by shifting upward 2 units. Domain: 0 .

Range:  . Vertical asymptote: x  0. y

x

1

68. The graph of y  1  log10 x is obtained from that of y  log10 x by reflecting it about the x-axis, and then

shifting it upward 1 unit. Domain: 0 .

Range:  . Vertical asymptote: x  0. y

(1, 2) 1

(1, 1)

1 1

x

1

x

SECTION 4.3 Logarithmic Functions

419

69. The graph of y  log3 x  1  2 is obtained from that of 70. The graph of y  1  ln x is obtained from that of y  log3 x by shifting to the right 1 unit and then

downward 2 units. Domain: 1 . Range:  . Vertical asymptote: x  1. 1

y

y  ln x by reflecting it about the y-axis and then shifting

it upward 1 unit. Domain:  0. Range:  . Vertical asymptote: x  0.

y

x

1

(_1, 1) 1

(2, _2)

1

71. The graph of y  ln x is obtained from that of y  ln x

by reflecting the part of the graph for 0  x  1 about the

x

  ln x if x  0 72. Note that y  ln x   ln x if x  0

The graph

of y  ln x is obtained by combining the graph of

x-axis. Domain: 0 . Range: [0 .

y  ln x and its reflection about the y-axis.

Vertical asymptote: x  0.

Domain:  0  0 . Range:  . Vertical

y

asymptote: x  0.

y 1 x

1 1 1

x

73. f x  log10 x  3. We require that x  3  0  x  3, so the domain is 3 . 74. f x  log5 8  2x. Then we must have 8  2x  0  8  2x  4  x, and so the domain is  4.   75. g x  log3 x 2  1 . We require that x 2  1  0  x 2  1  x  1 or x  1, so the domain is  1  1 .   76. g x  ln x  x 2 . Then we must have x  x 2  0 

x 1  x  0. Using the methods from Chapter 1 with the

endpoints 0 and 1, we get the table at right. Thus the domain is 0 1.

Interval

 0

0 1

1 

Sign of x







Sign of 1  x













Sign of x 1  x

77. h x  ln x  ln 2  x. We require that x  0 and 2  x  0  x  0 and x  2  0  x  2, so the domain is 0 2.  78. h x  x  2  log5 10  x . Then we must have x  2  0 and 10  x  0  x  2 and 10  x  2  x  10. So the domain is [2 10.

420

CHAPTER 4 Exponential and Logarithmic Functions

  79. y  log10 1  x 2 has domain 1 1, vertical

asymptotes x  1 and x  1, and local maximum y  0 at x  0.

  80. y  ln x 2  x  ln x x  1 has domain

 0  1 , vertical asymptotes x  0 and x  1, and no local maximum or minimum. 10

-2

2 -10

10

-2

-10

81. y  x  ln x has domain 0 , vertical asymptote x  0, and no local maximum or minimum.

82. y  x ln x2 has domain 0 , no vertical asymptote, local minimum y  0 at x  1, and local maximum y  054 at x  014. 10

2 5 -5 0

ln x has domain 0 , vertical asymptote x  0, x horizontal asymptote y  0, and local maximum y  037

83. y 

at x  272.

2

4

84. y  x log10 x  10 has domain 10 , vertical

asymptote x  10, and local minimum y  362 at x  587.

5 0 10

20 -10

10

-2 -5

85. f x  2x and g x  x  1 both have domain  , so  f  g x  f g x  2gx  2x1 with domain   and g  f  x  g  f x  2x  1 with domain  . 2 86. f x  3x and g x  x 2  1 both have domain  , so  f  g x  f g x  3x 1 with domain    x 2  1  32x  1 with domain  . and g  f  x  g  f x  3

87. f x  log2 x has domain 0  and g x  x  2 has domain  , so  f  g x  f g x  log2 x  2 with domain 2  and g  f  x  g  f x  log2 x  2 with domain 0 .   88. f x  log x has domain 0  and g x  x 2 has domain  , so  f  g x  f g x  log x 2 with domain  0  0  and g  f  x  g  f x  log x2 with domain 0 .

SECTION 4.3 Logarithmic Functions

89. The graph of g x 



x grows faster than the graph of

90. (a)

f x  ln x.

6 4 2

0

91. (a)

6

g

4

f

g

2

f

0

10

20

20

30

(b) From the graph, we see that the solution to the  equation x  1  ln 1  x is x  1350.

30

c=4 c=3 c=2 c=1

2

10

92. (a)

c=4

6

c=3

4

c=2

1 2

0

421

10

20

30

0

(b) Notice that f x  log cx  log c  log x, so

c=1 10

20

30

(b) As c increases, the graph of f x  c log x

as c increases, the graph of f x  log cx is

stretches vertically by a factor of c. shifted upward log c units.   93. (a) f x  log2 log10 x . Since the domain of log2 x is the positive real numbers, we have: log10 x  0  x  100  1. Thus the domain of f x is 1 .   y x (b) y  log2 log10 x  2 y  log10 x  102  x. Thus f 1 x  102 .

94. (a) f x  ln ln ln x. We must have ln ln x  0  ln x  1  x  e. So the domain of f is e . y

(b) y  ln ln ln x  e y  ln ln x  ee  ln x  ee 2x 2x .y  y  y2x  2x x 12 1  2x y  y  2x  y2x  2x 1  y  2x  1 y   y  x  log2 . Thus 1y   x f 1 x  log2 . 1x

95. (a) f x 





ey

x

e  x. Thus the inverse function is f 1 x  ee .

(b)

x  0. Solving this using the methods from 1x Chapter 1, we start with the endpoints, 0 and 1. Interval

 0

0 1

1 

Sign of x







Sign of 1  x x Sign of 1x













Thus the domain of f 1 x is 0 1.

I  2500 ln 07  89169 moles/liter. I0   D  8267 ln 073  2602 years. 97. Using D  073D0 we have A  8267 ln D0 log N50 log 20,000 98. Substituting N  1,000,000 we get t  3 3  4286 hours. log 2 log 2 ln 2 ln 2  116 years. When r  7% we have t   99 years. And when r  8% we have 99. When r  6% we have t  006 007 ln 2  87 years. t 008 96. Using I  07I0 we have C  2500 ln

422

CHAPTER 4 Exponential and Logarithmic Functions

100. Using k

 025 and substituting C  09C0 we have   C  025 ln 1  09  025 ln 01  058 hours. t  025 ln 1  C0 log 2  1005 log 40 log 2AW     532. Using A  100 and 101. Using A  100 and W  5 we find the ID to be log 2 log 2 log 2 log 2  10010 log 20 523 log 2AW     432. So the smaller icon is  123 times W  10 we find the ID to be log 2 log 2 log 2 432 harder.

102. (a) Since 2 feet  24 inches, the height of the graph is 224  1677216 inches. Now, since there are 12 inches per foot and ,216 5280 feet per mile, there are 12 5280  63,360 inches per mile. So the height of the graph is 1,677 63360  2648, or about 265 miles.   (b) Since log2 224  24, we must be about 224 inches  265 miles to the right of the origin before the height of the

graph of y  log2 x reaches 24 inches or 2 feet.   103. log log 10100  log 100  2      log log log 10googol  log log googol  log log 10100  log 100  2

104. Notice that loga x is increasing for a  1. So we have log4 17  log4 16  log4 42  2. Also, we have log5 24  log5 25  log5 52  2. Thus, log5 24  2  log4 17.

105. The numbers between 1000 and 9999 (inclusive) each have 4 digits, while log 1000  3 and log 10,000  4. Since     log x  3 for all integers x where 1000  x  10,000, the number of digits is log x  1. Likewise, if x is an integer       where 10n1  x  10n , then x has n digits and log x  n  1. Since log x  n  1  n  log x  1, the number   of digits in x is log x  1.

4.4

LAWS OF LOGARITHMS

1. The logarithm of a product of two numbers is the same as the sum of the logarithms of these numbers. So log5 25  125  log5 25  log5 125  2  3  5.

2. The logarithm of a quotient of two numbers is the same as the difference of the logarithms of these numbers. So   25  log 25  log 125  2  3  1. log5 125 5 5

3. The logarithm of a number raised to a power is the same as the power times the logarithm of the number. So   log5 2510  10  log5 25  10  2  20.     x2 y  log x 2 y  log z  log x 2  log y  log z  2 log x  log y  log z 4. log z     x2 y 5. 2 log x  log y  log z  log x 2  log y  log z  log x 2 y  log z  log . z

6. (a) Most calculators can find logarithms with base 10 and base e. To find logarithms with different bases we use the change 1079 log 12 of base formula. To find log7 12, we write log7 12    1277. log 7 0845 2485 ln 12 (b) Yes, the result is the same: log7 12    1277. ln 7 1946 7. (a) False. log A  B  log A  log B. (b) True. log AB  log A  log B.

SECTION 4.4 Laws of Logarithms

A  log A  log B. B log A  log A  log B. (b) False. log B

8. (a) True. log

9. log 50  log 200  log 50  200  4 60  2 11. log2 60  log2 15  log2 15

10. log6 9  log6 24  log6 9  24  3 12. log3 135  log3 45  log3 135 45  1

13. 14 log3 81  14 4  1    15. log5 5  log5 512  12

17.

18. 19. 20. 21. 22.

14.  13 log3 27   13 3  1   16. log5  1  log5 532   32 125   2 6 log2 6  log2 15  log2 20  log2 15  log2 20  log2 5  20  log2 8  log2 23  3     100  log3 19  log3 32  2 log3 100  log3 18  log3 50  log3 18  50  100 log4 16100  log4 42  log4 4200  200  33 log2 833  log2 23  log2 299  99   log log 1010,000  log 10,000 log 10  log 10,000  1  log 10,000  log 104  4 log 10  4     200  ln e200 ln e  ln e200  200 ln e  200 ln ln ee

23. log3 8x  log3 8  log3 x

24. log6 7r  log6 7  log6 r

25. log3 2x y  log3 2  log3 x  log3 y

26. log5 4st  log5 4  log5 s  log5 t  28. log t 5  log t 52  52 log t  30. ln ab  12 ln ab  12 ln a  ln b     32. log3 x y  log3 x  log3 y  log3 x  12 log3 y r  ln r  ln 3  ln s 34. ln 3s   s5  5 log2 s  log2 7  2 log2 t 36. log2 7t 2

27. ln a 3  3 ln a

  29. log2 x y10  10 log2 x y  10 log2 x  log2 y   31. log2 AB 2  log2 A  log2 B 2  log2 A  2 log2 B

2x  log3 2  log3 x  log3 y y   3x 2  log5 3  2 log5 x  3 log5 y log5 y3  y3 3x 5  12  52 log3 x  log3 y log3 38. log   3 log y  12 log 2  12 log x y 2x     3 4 x y  log x 3 y 4  log z 6  3 log x  4 log y  6 log z log z6       x2 2  log  log yz 3  2 loga x  loga y  3 loga z  2 loga x  loga y  3 loga z loga x a a yz 3    ln x 4  2  12 ln x 4  2    3 log x 2  4  13 log x 2  4      y y 1  ln x  2 ln  ln x  12 ln y  ln z ln x z z   3x 2 2  ln x  110  ln 3  2 ln x  10 ln x  1 ln  ln 3x x  110

33. log3 35.

37. 39.

40. 41. 42. 43. 44.

423

424

CHAPTER 4 Exponential and Logarithmic Functions

  x 2  y 2  14 log x 2  y 2    x  log x  log 3 1  x  log x  13 log 1  x log  3 1x        2   x2  4 x2  4 1 1 2 2 3 log    2  2 log   2  2 log x  4  log x  1 x  7 x2  1 x3  7 x2  1 x3  7         12 log x 2  4  log x 2  1  2 log x 3  7                log x y z  12 log x y z  12 log x  log y z  12 log x  12 log y z     12 log x  12 log y  12 log z  12 log x  14 log y  18 log z   log4 6  2 log4 7  log4 6  log4 72  log4 6  72  log4 294

45. log 46.

47.

48.

49.

 4



 50. 12 log2 5  2 log2 7  log2 5  log2 49  log2 495

51. 2 log x  3 log x  1  log x 2  log x  13  log

x2 x  13

 8x 2 52. 3 ln 2  2 ln x  12 ln x  4  ln 23  ln x 2  ln x  4  ln  x 4    3 1 53. 4 log x  3 log x 2  1  2 log x  1  log x 4  log x 2  1  log x  12     x4 x 4 x  12 2   log x  1  log  log  3 2 3 2 x 1 x 1

  x2  1 x  1 x  1  log5  log5 x  1 54. log5 x 2  1  log5 x  1  log5 x 1 x 1   a 2  b2 55. ln a  b  ln a  b  2 ln c  ln a  b a  b  ln c2  ln c2  2   x y2 x 2 y4 x y2 56. 2 log5 x  2 log5 y  3 log5 z  2 log5 3  log5  log5 6 3 z z z    2 x4  3  13 log x  2  12 log  57. 13 log x  23  12 log x 4  log x 2  x  6 2 x2  x  6 12  x2 x 2 x  2 x2 x4  log x  2  log  log  2  log  log  log x x 3 x  3 x  2 x  3 x  2 [x  3 x  2]2   bd c 58. loga b  c loga d  r loga s  loga bd c  loga s r  loga r s log 5 log 2  2321928 60. log5 2   0430677 59. log2 5  log 2 log 5

61. log3 16 

log 16  2523719 log 3

62. log6 92 

log 92  2523658 log 6

63. log7 261 

log 261  0493008 log 7

64. log6 532 

log 532  3503061 log 6

65. log4 125 

log 125  3482892 log 4

66. log12 25 

log 25  0368743 log 12

SECTION 4.4 Laws of Logarithms

67. log3 x 

ln x 1 1 loge x   ln x. The graph of y  ln x is loge 3 ln 3 ln 3 ln 3

425

2

shown in the viewing rectangle [1 4] by [3 2]. 2

4

-2

68. Note that logc x 



 1 ln x (by the change of base formula). So ln c

a=2 a=e a=5 a=10

2

the graph of y  logc x is obtained from the graph of y  ln x by

1 depending ln c on whether ln c  1 or ln c  1. All of the graphs pass through 1 0 either shrinking or stretching vertically by a factor of

0

2

4

-2

because logc 1  0 for all c.

   log 5 log 7 log 7 70. log2 5 log5 7     log2 7 log 2 log 5 log 2        1 x  x2  1 x  x2  1 1       ln  ln  x2  x2  1 x  x2  1 x  x2  1 x  x2  1     ln x  x 2  1

1 ln e  ln 10 ln 10     71.  ln x  x 2  1  ln 69. log e 

80 P0 . Substituting P0  80, t  24, and c  03 we have P   305. So the t  1c 24  103 student should get a score of 30.   c c 73. (a) log P  log c  k log W  log P  log c  log W k  log P  log  P  k. Wk W 8000 8000 (b) Using k  21 and c  8000, when W  2 we have P  21  1866 and when W  10 we have P  21  64. 2 10   74. (a) log S  log c  k log A  log S  log c  log Ak  log S  log c Ak  S  c Ak . 72. From Example 5(a), P 

(b) If A  2A0 when k  3 we get S  c 2A0 3  c  23  A30  8  c A30 . Thus doubling the area increases the species eightfold.

75. (a) M  25 log BB0   25 log B  25logB0 .

(b) Suppose B1 and B2 are the brightness of two stars such that B1  B2 and let M1 and M2 be their respective magnitudes. Since log is an increasing function, we have log B1  log B2 . Then log B1  log B2  log B1  log B0  log B2  log B0  log B1 B0   log B2 B0   25 log B1 B0   25 log B2 B0   M1  M2 . Thus the brighter star has less magnitudes.

(c) Let B1 be the brightness of the star Albiero. Then 100B1 is the brightness of Betelgeuse, and its magnitude is     M  25log 100B1 B0   25 log 100  log B1 B0   25 2  log B1 B0   5  25 log B1 B0   5  magnitude of Albiero

76. (a) False; log xy  log x  log y  log x  log y.

(b) False; log2 x  log2 y  log2 xy  log2 x  y.   (c) True; the equation is an identity: log5 ab2  log5 a  log5 b2  log5 a  2 log5 b. (d) True; the equation is an identity: log 2z  z log 2.

(e) False; log P  log Q  log P Q  log P log Q. (f) False; log a  log b  log ab  log a  log b.

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 x (g) False; x log2 7  log2 7x  log2 7 .

(h) True; the equation is an identity. loga a a  a loga a  a  1  a.

(i) False; log x  y  log x  log y. For example, 0  log 3  2  log 3  log 2. (j) True; the equation is an identity:  ln 1A   ln A1  1  ln A  ln A.

77. The error is on the first line: log 01  0, so 2 log 01  log 01.

78. Let f x  x 2 . Then f 2x  2x2  4x 2  4 f x. Now the graph of f 2x is the same as the graph of f shrunk

horizontally by a factor of 12 , whereas the graph of 4 f x is the same as the graph of f x stretched vertically by a factor of 4. Let g x  e x . Then g x  2  e x2  e2 e x  e2 g x. This shows that a horizontal shift of 2 units to the right is the same as a vertical stretch by a factor of e2 .

Let h x  ln x. Then h 2x  ln 2x  ln 2  ln x  ln 2  h x. This shows that a horizontal shrinking by a factor of 12 is the same as a vertical shift upward by ln 2.

4.5

EXPONENTIAL AND LOGARITHMIC EQUATIONS

1. (a) First we isolate e x to get the equivalent equation e x  25.

(b) Next, we take the natural logarithm of each side to get the equivalent equation x  ln 25.

(c) Now we use a calculator to find x  3219.

2. (a) First we combine the logarithms to get the equivalent equation log 3 x  2  log x.

(b) Next, we write each side in exponential form to get the equivalent equation 3 x  2  x. (c) Now we find x  3.

3. Because the function 5x is one-to-one, 5x1  125  5x1  53  x  1  3  x  4. 2

4. Because the function e x is one-to-one, e x  e9  x 2  9  x  3.

5. Because the function 5x is one-to-one, 52x3  1  52x3  50  2x  3  0  x  32 . 1  2x  3  1  x  1. 6. Because the function 10x is one-to-one, 102x3  10

7. Because the function 7x is one-to-one, 72x3  765x  2x  3  6  5x  3x  9  x  3. 8. Because the function e x is one-to-one, e12x  e3x5  1  2x  3x  5  5x  6  x  65 .

2 2 9. Because the function 6x is one-to-one, 6x 1  61x  x 2  1  1  x 2  2x 2  2  x  1.

2 2 10. Because the function 10x is one-to-one, 102x 3  109x  2x 2  3  9  x 2  3x 2  12  x  2.

11. (a) 10x  25  log 10x  log 25  x log 10  log 25  x  log 25  2 log 5 (b) x  1397940

12. (a) 10x  4  log 10x  log 4  x  log 4  x   log 4 (b) x  0602060

13. (a) e5x  10  ln e5x  ln 10  5x ln e  ln 10  5x  ln 10  x   15 ln 10 (b) x  0460517

14. (a) e04x  8  ln e04x  ln 8  04x  ln 8  x  52 ln 8 (b) x  5198604

3 log 3 15. (a) 21x  3  log 21x  log 3  1  x log 2  log 3  1  x  log log 2  x  1  log 2 (b) x  0584963

SECTION 4.5 Exponential and Logarithmic Equations

427

 1 5 log 5 log 5 16. (a) 32x1  5  log 32x1  log 5  2x  1 log 3  log 5  2x  1  log log 3  2x  1  log 3  x  2 1  log 3 (b) x  1232487   10 17. (a) 3e x  10  e x  10 3  x  ln 3 (b) x  1203973

    17 1 17 18. (a) 2e12x  17  e12x  17 2  12x  ln 2  x  12 ln 2 (b) x  0178339

10 10  12t 10  12t ln 41  ln 10  12t  ln 3  t  ln 3  19. (a) 300 102512t  1000  41 40 3 40 3 ln 41 12 ln 41 40 40 (b) x  4063202  10t ln 5  5  10t ln 11 20. (a) 10 137510t  50  11 8 8  ln 5  t  10 ln 11 8

(b) t  0505391

21. (a) e14x  2  1  4x  ln 2  4x  1  ln 2  x  14 1  ln 2 (b) x  0076713 22. (a) e35x  16  3  5x  ln 16  5x  ln 16  3  x   15 ln 16  3 (b) x  0045482

ln 15 5 ln 15 ln 15  7x  5  x   23. (a) 257x  15  5  7x ln 2  ln 15  5  7x  ln 2 ln 2 7 7 ln 2 (b) x  0156158 34 24. (a) 23x  34  log 23x  log 34  3x log 2  log 34  x  3log log 2 (b) x  1695821 x  log 01 log 3  log 01  x  14log 25. (a) 3x14  01  log 3x14  log 01  3 14 (b) x  29342646 x log 2 log 5  log 2  x   100 26. (a) 5x100  2  log 5x100  log 2   log 5 100 (b) x  43067656   27. (a) 4 1  105x  9  1  105x  94  105x  54  5x  log 54  x  15 log 54

(b) x  0019382   ln 45 ln 45 28. (a) 2 5  3x1  100  5  3x1  50  3x1  45  x  1 ln 3  ln 45  x  1  x  1 ln 3 ln 3 (b) x  2464974 29. (a) 8  e14x  20  e14x  12  1  4x  ln 12  x 

1  ln 12 4

(b) x  0371227 30. (a) 1  e4x1  20  e4x1  19  4x  1  ln 19  x 

ln 19  1 4

(b) x  0486110 ln 50 3  2x ln 2  ln 50  ln 3  x  31. (a) 4x  212x  50  22x  212x  50  22x 1  2  50  22x  50 3 ln 4 (b) x  2029447

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CHAPTER 4 Exponential and Logarithmic Functions

32. (a) 125x  53x1  200  53x  53x1  200  53x 1  5  200  53x  100 3  3x ln 5  ln 100  ln 3  100 ln 3 x 3 ln 5 (b) x  0726249 33. (a) 5x  4x1  log 5x  log 4x1  x log 5  x  1 log 4  x log 4  log 4  x log 5  x log 4  log 4  log 4 x log 5  log 4  log 4  x  log 54 (b) x  6212567 34. (a) 101x  6x  log 101x  log 6x  1  x  x log 6  1  x log 6  x  1  x log 6  1  x  log 161 (b) x  0562382 35. (a) 23x1  3x2  log 23x1  log 3x2  3x  1 log 2  x  2 log 3  3x log 2  log 2  x log 3  2 log 3  3x log 2  x log 3   log 2  2 log 3  x 3 log 2  log 3   log 2  2 log 3    log 2  32 log 18 log 22 log 3    x   3 log 2log 3   log 23 3 log 8 3

(b) x  2946865

x  x  36. (a) 7x2  51x  log 7x2  log 51x  log 7  1  x log 5  log 7  log 5  x log 5  2  2  x  log 7  x log 5  log 5  x 12 log 7  log 5  log 5  x  1 log 5 2 2 log 7log 5 (b) x  0623235 50  4  50  4  4ex  46  4ex  115  ex  ln 115  x  x   ln 115 1  ex (b) x  2442347

37. (a)

10  2  10  2  2ex  8  2ex  4  ex  ln 4  x  x   ln 4 1  ex (b) x  1386294    39. e2x  3e x  2  0  e x  1 e x  2  0  e x  1  0 or e x  2  0. If e x  1  0, then e x  1  x  ln 1  0. If 38. (a)

e x  2  0, then e x  2  x  ln 2  06931. So the solutions are x  0 and x  06931.    40. e2x  e x  6  0  e x  3 e x  2  0  e x  2  0 (impossible) or e x  3  0. If e x  3  0, then e x  3  x  ln 3  10986. So the only solution is x  10986.    41. e4x  4e2x  21  0  e2x  7 e2x  3  0  e2x  7 or e2x  3. Now e2x  7 has no solution, since e2x  0

for all x. But we can solve e2x  3  2x  ln 3  x  12 ln 3  05493. So the only solution is x  05493.    42. 34x  32x  6  0  32x  3 32x  2  0  32x  3 or 32x  2. The latter equation has no solution, so we solve

32x  3  2x  1  x  12 .        43. 2x  10 2x  3  0  2x 22x  10  3  2x  0  2x 2x  5 2x  2  0. The first two factors are positive

everywhere, so we solve 2x  2  0  x  1.      44. e x  15ex  8  0  ex e2x  15  8e x  0  ex e x  5 e x  3  0. The first factor is positive everywhere,

so the solutions occur where e x  5 or e x  3; that is, x  ln 3  10986 or x  ln 5  16094.   45. x 2 2x  2x  0  2x x 2  1  0  2x  0 (never) or x 2  1  0. If x 2  1  0, then x 2  1  x  1. So the only solutions are x  1.

SECTION 4.5 Exponential and Logarithmic Equations

429



     46. x 2 10x  x10x  2 10x  x 2 10x  x10x  2 10x  0  10x x 2  x  2  0  10x  0(never) or x 2  x  2  0. If x 2  x  2  0, then x  2 x  1  0  x  2, 1. So the only solutions are x  2, 1.

47. 4x 3 e3x  3x 4 e3x  0  x 3 e3x 4  3x  0  x  0 or e3x  0 (never) or 4  3x  0. If 4  3x  0, then

3x  4  x  43 . So the solutions are x  0 and x  43 .   48. x 2 e x  xe x  e x  0  e x x 2  x  1  0  e x  0 (impossible) or x 2  x  1  0. If x 2  x  1  0, then 



5 . So the solutions are x  1 5 . x  1 2 2

49. log x  log x  1  log 4x  log [x x  1]  log 4x  x 2  x  4x  x 2  5x  0  x x  5  0  x  0 or x  5. So the possible solutions are x  0 and x  5. However, when x  0, log x is undefined. Thus the only solution is x  5.   50. log5 x  log5 x  1  log5 20  log5 x 2  x  log5 20  x 2  x  20  x 2  x  20  0  x  5 x  4  0

 x  5 or x  4. Since log5 5 is undefined, the only solution is x  4.   51. 2 log x  log 2  log 3x  4  log x 2  log 6x  8  x 2  6x  8  x 2  6x  8  0  x  4 x  2  0  x  4 or x  2. Thus the solutions are x  4 and x  2.      52. ln x  12  ln 2  2 ln x  ln 2 x  12  ln x 2  2x  1  x 2  x 2  2x  1  0  x  12  0  x  1. Thus the only solution is x  1.

53. log2 3  log2 x  log2 5  log2 x  2  log2 3x  log2 5x  10  3x  5x  10  2x  10  x  5

54. log4 x  2  log4 3  log4 5  log4 2x  3  log4 3 x  2  log4 5 2x  3  3x  6  10x  15  7x  21 x 3

55. ln x  10  x  e10  22,026

56. ln 2  x  1  2  x  e1  x  e  2  07183 57. log x  2  x  102  001

58. log x  4  3  x  4  103  1000  x  1004

59. log 3x  5  2  3x  5  102  100  3x  95  x  95 3  316667

60. log3 2  x  3  2  x  33  27  x  25  x  25

61. 4  log 3  x  3  log 3  x  1  3  x  10  x  7   62. log2 x 2  x  2  2  x 2  x  2  22  4  x 2  x  6  0  x  3 x  2  0  x  3 or x  2. Thus the solutions are x  3 and x  2.

63. log2 x  log2 x  3  2  log2 [x x  3]  2  x 2  3x  22  x 2  3x  4  0  x  4 x  1  x  1 or x  4. Since log 1  3  log 4 is undefined, the only solution is x  4. 64. log x  log x  3  1  log [x x  3]  1  x 2  3x  10  x 2  3x  10  0  x  2 x  5  0  x  2 or x  5. Since log 2 is undefined, the only solution is x  5.

65. log9 x  5  log9 x  3  1  log9 [x  5 x  3]  1  x  5 x  3  91  x 2  2x  24  0  x  6 x  4  0  x  6 or4. However, x  4 is inadmissible, so x  6 is the only solution.

66. ln x  1  ln x  2  1  ln [x  1 x  2]  1  x 2  x  2  e  x 2  x  2  e  0  







x  1 142e  1 2 94e . Since x  1  0 when x  1 2 94e , the only solution is x  1 2 94e  17290 2   x 1 x 1 2  52  x  1  25x  25  24x  26  x  13 67. log5 x  1  log5 x  1  2  log5 12 x 1 x 1 x  15 x  15 x  15 2  32   9  x  15  9 x  1  68. log3 x  15  log3 x  1  2  log3 x 1 x 1 x 1 x  15  9x  9  8x  24  x  3

430

CHAPTER 4 Exponential and Logarithmic Functions

69. ln x  3  x  ln x  x  3  0. Let f x  ln x  x  3. We need to solve the equation f x  0. From the graph of f , we get x  221.

70. log x  x 2  2  log x  x 2  2  0. Let

f x  log x  x 2  2. We need to solve the equation

f x  0. From the graph of f , we get x  001 or

x  147.

2 0 2

4 0 1

2

-2

71. x 3  x  log10 x  1  x 3  x  log10 x  1  0.

Let f x  x 3  x  log10 x  1. We need to solve the

equation f x  0. From the graph of f , we get x  0 or x  114.

    72. x  ln 4  x 2  x  ln 4  x 2  0. Let   f x  x  ln 4  x 2 . We need to solve the equation f x  0. From the graph of f , we get x  196 or

x  106.

2

-1

1

2 -2

2

-2

73. e x  x  e x  x  0. Let f x  e x  x. We need to

solve the equation f x  0. From the graph of f , we get x  057.

74. 2x  x  1  2x  x  1  0. Let

f x  2x  x  1. We need to solve the equation

f x  0. From the graph of f , we get x  138.

2

2

0 -1.0

-0.5

-1 -2

   75. 4x  x  4x  x  0. Let f x  4x  x.

We need to solve the equation f x  0. From the graph of f , we get x  036.

1

2

-2 2

2

76. e x  2  x 3  x  e x  2  x 3  x  0. Let 2

f x  e x  2  x 3  x. We need to solve the equation

f x  0. From the graph of f , we get x  089 or

x  071.

2

2 0 1

2 -2

2

-2 -2

SECTION 4.5 Exponential and Logarithmic Equations

431

  2 1  log 22 log5 x  log 1  4  log5 x   12  x  512  1  04472 77. 22 log5 x  16 2 2 16  5 log5 x   78. log2 log3 x  4  log3 x  24  16  x  316  43,046,721   79. log x  2  log 9  x  1  log [x  2 9  x]  1  log x 2  11x  18  1  x 2  11x  18  101  0  x 2  11x  28  0  x  7 x  4. Also, since the domain of a logarithm is positive we must have

0  x 2  11x  18  0  x  2 9  x. Using the methods from Chapter 1 with the endpoints 2, 4, 7, 9 for the intervals, we make the following table:  2

2 4

4 7

7 9

9 

Sign of x  7











Sign of x  4











Sign of x  2











Sign of 9  x































Interval

Sign of x  7 x  4 Sign of x  2 9  x

Thus the solution is 2 4  7 9.

80. 3  log2 x  4  23  x  24  8  x  16.

81. 2  10x  5  log 2  x  log 5  03010  x  06990. Hence the solution to the inequality is approximately the interval 03010 06990.        82. x 2 e x  2e x  0  e x x 2  2  0  e x x  2 x  2  0. We use the methods of Chapter 1 with the endpoints    2 and 2, noting that e x  0 for all x. We make a table: Interval Sign of e x    Sign of x  2    Sign of x  2      Sign of e x x  2 x  2

  Thus  2  x  2.

     2

     2 2

  2 

























  ln y . 83. To find the inverse of f x  22x , we set y  f x and solve for x. y  22x  ln y  ln 22x  2x ln 2  x  2 ln 2 ln x ln x Interchange x and y: y  . Thus, f 1 x  . 2 ln 2 2 ln 2   84. To find the inverse of f x  3x1 , we set y  f x and solve for x. y  3x1  ln y  ln 3x1  x  1 ln 3  x 1

ln y ln y ln x ln x x   1. Interchange x and y: y   1. Thus, f 1 x   1. ln 3 ln 3 ln 3 ln 3

85. To find the inverse of f x  log2 x  1, we set y  f x and solve for x. y  log2 x  1  2 y  2log2 x1  x  1  x  2 y  1. Interchange x and y: y  2x  1. Thus, f 1 x  2x  1.

86. To find the inverse of f x  log 3x, we set y  f x and solve for x. y  log 3x  10 y  3x  x  x and y: y 

10x 10x . Thus, f 1 x  . 3 3

10 y . Interchange 3

432

CHAPTER 4 Exponential and Logarithmic Functions

87. log x  3  log x  log 3  log x  3  log 3x  x  3  3x  2x  3  x  32   88. log x3  3 log x  log x3  3 log x  0  log x log x2  3  log x  0 or log x2  3  0. Now log x  0     x  1. Also log x2  3  0  log x2  3  log x   3  x  10 3 , so x  10 3  539574 or 





x  10 3  00185. Thus the solutions to the equation are x  1, x  10 3  539574 and x  10 3  00185.     0085 43 89. (a) A 3  5000 1   5000 10212512  643509. Thus the amount after 3 years is $6,43509. 4     0085 4t  5000 1021254t  2  1021254t  log 2  4t log 102125  (b) 10000  5000 1  4 log 2  824 years. Thus the investment will double in about 824 years. t 4 log 102125

90. (a) A 2  6500e0062  $732873

91.

92.

93. 94.

95.

96.

    006t  ln 16  006t  t  1 ln 16  346. So the investment doubles in (b) 8000  6500e006t  16 13  e 13 13 006 1 about 3 2 years.     0075 4t 8000  5000 1   5000 1018754t  16  1018754t  log 16  4t log 101875  4 log 16 t  633 years. The investment will increase to $8000 in approximately 6 years and 4 months. 4 log 101875   00975 2t log 125 5000  4000 1   2344. So it  125  1048752t  log 125  2t log 104875  t  2 2 log 104875 1 takes about 2 years to save $5000. 3 ln 2 2  e0085t  ln 2  0085t  t   815 years. Thus the investment will double in about 815 years. 0085     r 24 r 8 r r 143577  1000 1   8 143577   8 143577  1   143577  1  1 2 2 2  2 r  2 8 143577  1  00925. Thus the rate was about 925%.   ln 3 15e0087t  5  e0087t  13  0087t  ln 13   ln 3  t   126277. So only 5 grams remain after 0087 approximately 13 days.   We want to solve for t in the equation 80 e02t  1  70 (when motion is downwards, the velocity is negative). Then       ln 18 80 e02t  1  70  e02t  1   78  e02t  18  02t  ln 18  t   104 seconds. Thus the 02 velocity is 70 ft/s after about 10 seconds. 10  7337, so there are approximately 7337 fish after 3 years. 1  4e083 10 08t  1  e08t  025  08t  ln 025  (b) We solve for t.  5  1  4e08t  10 5  2  4e 1  4e08t ln 025  173. So the population will reach 5000 fish in about 1 year and 9 months. t 08

97. (a) P 3 

98. (a) I  10e000830  10e024  787. So at 30 ft the intensity is 787 lumens.   ln 12 (b) 5  10e0008x  e0008x  12  0008x  ln 12  x   866. So the intensity drops to 25 lumens 0008 at 866 ft.

SECTION 4.6 Modeling with Exponential Functions

99. (a) ln



P P0





433

P h   ehk  P  P0 ehk . Substituting k  7 and P0  100 we get P  100eh7 . k P0

(b) When h  4 we have P  100e47  5647 kPa.   T  20 T  20  011t   e011t  T  20  200e011t  T  20  200e011t 100. (a) ln 200 200 (b) When t  20 we have T  20  200e01120  20  200e22  422 F.     13t5  13 I  1  e13t5  e13t5  1  13 I   13 t  ln 1  13 I  1  e 101. (a) I  60 13 60 60 5 60   5 13 t   13 ln 1  60 I .   5 ln 1  13 2  0218 seconds. (b) Substituting I  2, we have t   13 60 102. (a) P  M  Cekt  Cekt  M  P  ekt      1 MP MP  t   ln kt  ln C k C

MP  C

(c)

(b) P t  20  14e0024t . Substituting M  20, C  14, k  0024,   1 MP , we have and P  12 into t   ln k C   1 20  12 t  ln  2332. So it takes about 23 months. 0024 14

20 10 0 0

100

200

103. Since 91  9, 92  81, and 93  729, the solution of 9x  20 must be between 1 and 2 (because 20 is between 9 and 81), whereas the solution to 9x  100 must be between 2 and 3 (because 100 is between 81 and 729).   1 log x  1, so 1 log x  101 for all x  0. So 104. Notice that log x 1 log x  log x x 1 log x  5 has no solution, and x 1 log x  k has a solution only when k  10.

10

This is verified by the graph of f x  x 1 log x .

0 0

5

10

  105. (a) x  1logx1  100 x  1  log x  1logx1  log 100 x  1     2 log x  1 log x  1  log 100  log x  1  log x  1  log x  1  2  0     log x  1  2 log x  1  1  0. Thus either log x  1  2  x  101 or log x  1  1  x  11 10 .         (b) log2 x  log4 x  log8 x  11  log2 x  log2 x  log2 3 x  11  log2 x x 3 x  11  log2 x 116  11 6  11 6 log2 x  11  log2 x  6  x  2  64

 2      ln 3 or 2x  1, which (c) 4x  2x1  3  2x  2 2x  3  0  2x  3 2x  1  0  either 2x  3  x  ln 2 ln 3 has no real solution. So x  is the only real solution. ln 2

4.6

MODELING WITH EXPONENTIAL FUNCTIONS

1. (a) Here n 0  10 and a  15 hours, so n t  10  2t15  10  22t3 . (b) After 35 hours, there will be n 35  10  22353  106  108 bacteria.

434

CHAPTER 4 Exponential and Logarithmic Functions

  2t 3 ln 1000 ln 2  ln 1000  t   149, (c) n t  10  22t3  10,000  22t3  1000  ln 22t3  ln 1000  3 2 ln 2 so the bacteria count will reach 10,000 in about 149 hours. 2. (a) Here n 0  25 and a  5 hours, so n t  25  2t5 . (b) After 18 hours, there will be n 18  25  2185  303 bacteria.   t (c) n t  25  2t5  1,000,000  2t5  40,000  ln 2t5  ln 40,000  ln 2  ln 40,000  5 5 ln 40,000  764, so the bacteria count will reach 1,000,000 in about 764 hours. t ln 2 3. (a) A model for the squirrel population is n t  n 0  2t6 . We are given

n

(c)

1,000,000

that n 30  100,000, so n 0  2306  100,000 

800,000

100,000  3125. Initially, there were approximately 25 3125 squirrels. n0 

600,000 400,000 200,000

(b) In 10 years, we will have t  40, so the population will be n 40  3125  2406  317,480 squirrels.

4. (a) A model for the bird population is n t  n 0  2t10 . We are given

0

20

30

40

50 t (years)

n

(c)

that n 25  13,000, so n 0  22510  13,000 

60,000

n0 

40,000

50,000

13,000  2298. Initially, there were approximately 2300 birds. 252

30,000

(b) In 5 years, we will have t  30, so the population will be

20,000 10,000

approximately n 30  2298  23010  18,384  18,400 birds.

5. (a) r  008 and n 0  18,000. Thus the population is given by the

10

10

0

(d)

20

30

40 t (years)

n 40,000

formula n t  18,000e008t .

30,000

(b) t  2021  2013  8. Then we have

20,000

n 8  18,000e0088  18,000e064  34,100. Thus there should

10,000

be about 34,100 foxes in the region by the year 2021.

1

2

3

4

5

6

7

8 t (years)

(c) Solving n t  25,000, we get 18,000e008t  25,000    1 ln 25  ln 18  41, so the fox 18e008t  25  ln 18e008t  ln 25  ln 18  008t  ln 25  t  008 population will reach 25,000 after about 41 years.

6. (a) Taking t  0 in the year 2010 and measuring n in millions, we have n 0  12 and r  0012. Therefore, an exponential model is

n t  12e0012t .

(b) In 2015, t  5, so the population was approximately

n 5  12e00125  12742, or 12,742,000 fish.   (c) Solving n t  14, we get 12e0012t  14  ln 12e0012t  ln 14

1 ln 14  ln 12  128, so the  ln 12  0012t  ln 14  t  0012 fish population will reach 14 million after about 128 years.

(d)

n (millions) 15 10 5 0

2020

2030 year

SECTION 4.6 Modeling with Exponential Functions

435

7. n t  n 0 ert ; n 0  110 million, t  2036  2011  25.

(a) r  003; n 25  110,000,000e00325  110,000,000e075  232,870,000. Thus at a 3% growth rate, the projected population will be approximately 233 million people by the year 2036.

(b) r  002; n 25  110,000,000e00225  110,000,000e050  181,359,340. Thus at a 2% growth rate, the projected population will be approximately 181 million people by the year 2036. 8. (a) In this case, a model for the bacteria population is n t  n 0 ert  22e012t , so after 24 hours the population is approximately n 24  22e01224  392 bacteria.

(b) In this case, a model is n t  n 0 ert  22e005t , so after 24 hours the population is approximately n 24  22e00524  73 bacteria.

9. (a) The doubling time is 18 years and the initial population is 112,000, so

(c)

a model is n t  112,000  2t18 . (b) We need to find the relative growth rate r. Since the population is 2  112,000  224,000 when t  18, we have 224,000  112,000e18r  2  e18r  ln 2  18r  r  ln182  00385. Thus, a model is

n 800,000 700,000 600,000 500,000 400,000 300,000 200,000 100,000 0

10 20 30 40 n t  112,000e00385t . t18 t18 (d) Using the model in part (a), we solve the equation n t  112,000  2  500,000  2  125 28  125 18 ln 28 t 125  3885. Therefore, it takes about 3885 years for the ln 2t18  ln 125 28  18 ln 2  ln 28  t  ln 2 population to reach 500,000.

10. (a) The doubling time is 25 years and the initial population is 350,000, so a model is n t  350,000  2t25 . (b) r  ln252  00277, so a model is n t  350,000e00277t .

(c)

t

n 2,000,000 1,000,000

(d) We solve the equation n t  350,000  2t25  2,000,000  t25  ln 40  t ln 2  ln 40  2t25  40 7  ln 2 7 7 25

0

20

40

60

t

25 ln 40 7  6286. Therefore, it takes about 629 years for the t

ln 2 population to reach 2,000,000.

11. (a) The deer population in 2010 was 20,000. (b) Using the model n t  20,000ert and the point 4 31000, we have 31,000  20,000e4r  155  e4r  4r  ln 155  r  14 ln 155  01096. Thus n t  20,000e01096t

(c) n 8  20,000e010968  48,218, so the projected deer population in 2018 is about 48,000. ln 5  1468. Thus, it takes about 147 years (d) 100,000  20,000e01096t  5  e01096t  01096t  ln 5  t  01096 for the deer population to reach 100,000. 12. (a) From the graph, we see that the initial bullfrog population was 100. (b) We use a model of the form n t  n 0 ert with n 0  100. Because we know that the population was 225 at t  2, we

04055t . solve n 2  225  100e2r  225  r  12 ln 225 100  04055. Thus, a model is n t  100e (c) The estimated population after 15 years is n 15  100e0405515  43,800 frogs.

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CHAPTER 4 Exponential and Logarithmic Functions

(d) The population will reach 75,000 when n t  100e04055t  75,000  e04055t  750  t 

ln 750  1632. So it 04055

will take about 163 years for the population to reach 75,000. 13. (a) Using the formula n t  n 0 ert with n 0  8600 and n 1  10000, we solve for r, giving 10000  n 1  8600er   50 r 01508t .  50 43  e  r  ln 43  01508. Thus n t  8600e (b) n 2  8600e015082  11627. Thus the number of bacteria after two hours is about 11,600. ln 2 (c) 17200  8600e01508t  2  e01508t  01508t  ln 2  t   4596. Thus the number of bacteria will 01508 double in about 46 hours.

14. (a) Using n t  n 0 ert with n 2  400 and n 6  25,600, we have n 0 e2r  400 and n 0 e6r  25,600. Dividing the

n e6r 25,600 second equation by the first gives 0 2r   64  e4r  64  4r  ln 64  r  14 ln 64  104. Thus the 400 n0e relative rate of growth is about 104%.

(b) Since r  14 ln 64  12 ln 8, we have from part (a) n t  n 0 e



1 2

 ln 8 t

. Since n 2  400, we have 400  n 0 eln 8 

400 n 0  ln 8  400 8  50. So the initial size of the culture was 50. e (c) Substituting n 0  50 and r  104, we have n t  n 0 ert  50e104t .

(d) n 45  50e10445  50e468  53885, so the size after 45 hours is approximately 5400. ln 1000  664. Hence the population will (e) n t  50,000  50e104t  e104t  1000  104t  ln 1000  t  104 reach 50,000 after roughly 6 hours 40 minutes. 15. (a) Calculating dates relative to 1990 gives n 0  2976 and n 10  3387. Then n 10  2976e10r  3387 

1 0012936t million e10r  3387 2976  11381  10r  ln 11381  r  10 ln 11381  0012936. Thus n t  2976e people. ln 2  5358, so the population (b) 2 2976  2976e0012936t  2  e0012936t  ln 2  0012936t  t  0012936 doubles in about 54 years.

(c) t  2010  1990  20, so our model gives the 2010 population as n 20  2976e001293620  3855 million. The actual population was estimated at 3696 million in 2009. 16. (a) 2n 0  n 0 e002t  2  e002t  002t  ln 2  t  50 ln 2  3466. So at the current growth rate, it will take approximately 3466 years for the population to double. (b) 3n 0  n 0 e002t  3  e002t  002t  ln 3  t  50 ln 3  5493. So at the current growth rate, it will take approximately 5493 years for the population to double. 17. (a) Because the half-life is 1600 years and the sample weighs 22 mg initially, a suitable model is m t  22  2t1600 . ln 2 ln 2 (b) From the formula for radioactive decay, we have m t  m 0 ert , where m 0  22 and r    0000433. h 1600 Thus, the amount after t years is given by m t  22e0000433t . (c) m 4000  22e00004334000  389, so the amount after 4000 years is about 4 mg.

9  e0000433t  (d) We have to solve for t in the equation 18  22 e0000433t . This gives 18  22e0000433t  11   9   ln 11 9 t  0000433t  ln 11  4634, so it takes about 463 years. 0000433

18. (a) Because the half-life is 30 years and the sample weighs 10 g initially, a suitable model is m t  10  2t30 . ln 2 ln 2 (b) Using m t  m 0 ert with m 0  10 and h  30, we have r    00231. Thus m t  10e00231t . h 30

SECTION 4.6 Modeling with Exponential Functions

437

(c) m 80  10e0023180  16 grams.

19.

20.

21.

22.

   ln 5 (d) 2  10e00231t  15  e00231t  ln 15  00231t  t   70 years. 00231 ln 2 , so m t  50e[ln 228]t . We need to solve for t in the By the formula in the text, m t  m 0 ert where r  h     28 ln 2 32 32 equation 32  50e[ln 228]t . This gives e[ln 228]t  32 50   28 t  ln 50  t   ln 2  ln 50  1803, so it takes about 18 years. ln 2 From the formula for radioactive decay, we have m t  m 0 ert , where r  . Since h  30, we have h ln 2  00231 and m t  m 0 e00231t . In this exercise we have to solve for t in the equation 005m 0  m 0 e00231t r 30 ln 005  1297. So it will take about 130 s.  e00231t  005  00231t  ln 005  t  00231 ln 2 , in other words m t  m 0 e[ln 2 h ]t . In By the formula for radioactive decay, we have m t  m 0 ert , where r  h ln 2 48 this exercise we have to solve for h in the equation 200  250e[ln 2 h ]48  08  e[ln 2 h ]48  ln 08   h ln 2  48  1491 hours. So the half-life is approximately 149 hours. h  ln 08 ln 2 From the formula for radioactive decay, we have m t  m 0 ert , where r  . In other words, m t  m 0 e[ln 2 h ]t . h (a) Using m 3  058m 0 , we have to solve for h in the equation 058m 0  m 3  m 0 e[ln 2 h ]3 . 3 ln 2 3 ln 2 Then 058m 0  m 0 e[3 ln 2 h ]  e[3 ln 2 h ]  058    ln 058  h    38 days. Thus the h ln 058 half-life of radon-222 is about 38 days. (b) Here we have to solve for t in the equation 02m  m e[ln 2382]t . So we have 02m  m e[ln 2382]t  0

0

0

0

382 ln 02 ln 2 t  ln 02  t    887. So it takes roughly 9 days for a sample of 02  e[ln 2382]t   382 ln 2 Radon-222 to decay to 20% of its original mass. ln 2 23. By the formula in the text, m t  m 0 e[ln 2 h ]t , so we have 065  1  e[ln 25730]t  ln 065   t 5730 5730 ln 065  3561. Thus the artifact is about 3560 years old. t  ln 2 ln 2 ln 2 . Since h  5730, r   0000121 24. From the formula for radioactive decay, we have m t  m 0 ert where r  h 5730 and m t  m 0 e0000121t . We need to solve for t in the equation 059m 0  m 0 e0000121t  e0000121t  059  ln 059  43606. So the mummy was buried about 4360 years ago. 0000121t  ln 059  t  0000121 25. (a) T 0  65  145e0050  65  145  210 F.

(b) T 10  65  145e00510  1529. Thus the temperature after 10 minutes is about 153 F.

(c) 100  65  145e005t  35  145e005t  02414  e005t  ln 02414  005t  t  

ln 02414  284. 005

Thus the temperature will be 100 F in about 28 minutes. 26. (a) We use Newton’s Law of Cooling: T t  Ts  D0 ekt with k  01947, Ts  60, and D0  986  60  386 . So T t  60  386e01947t .

  12 12 (b) Solve T t  72. So 72  60  386e01947t  386e01947t  12  e01947t   01947t  ln 386 386   1 12 t  ln  600, and the time of death was about 6 hours ago. 01947 386

438

CHAPTER 4 Exponential and Logarithmic Functions

kt 27. Using Newton’s Law of Cooling, T t  Ts  D0 ekt with Ts  75 and D0  185  75  110. So T t  75  110e  . 15 (a) Since T 30  150, we have T 30  75  110e30k  150  110e30k  75  e30k  15 22  30k  ln 22   1 ln 15 . Thus we have T 45  75  110e4530 ln1522  1369, and so the temperature of the turkey  k   30 22

after 45 minutes is about 137 F.

25 5  (b) The temperature will be 100 F when 75  110et30 ln1522  100  et30 ln1522   22 110   5       ln 22 t 15 5   1161. So the temperature will be 100 F after about 2 hours.  ln 22  ln 22  t  30 30 ln 15 22

28. We use Newton’s Law of Cooling: T t  Ts  D0 ekt , with Ts  20 and

D0  100  20  80. So T t  20  80ekt . Since T 15  75, we have   11 20  80e15k  75  80e15k  55  e15k  11 16  15k  ln 16    1 ln 11 . Thus T 25  20  80e2515ln1116  628, and so the k   15 16 temperature after another 10 min is 63 C. The function

100 80 60 0

20

T t  20  80e115ln1116t is shown in the viewing rectangle [0 30] by

[50 100].

4.7

LOGARITHMIC SCALES

    1. (a) pH   log H   log 50  103  23     (b) pH   log H   log 32  104  35     (c) pH   log H   log 50  109  83     2. pH   log H   log 31  108  75 and the substance is basic.

    3. (a) pH   log H  30  H  103 M     (b) pH   log H  65  H  1065  32  107 M     4. (a) pH   log H  46  H  1046 M  25  105 M     (b) pH   log H  73  H  1073 M  50  108 M         5. 40  107  H  16  105  log 40  107  log H  log 16  105       log 40  107  pH   log 16  105  64  pH  48. Therefore the range of pH readings for cheese is approximately 48 to 64.

  6. 28  pH  38  28  pH  38  1028  10pH  1038  158  103  H  158  104 . The   range of H is 158  104 to 158  103 .       7. (a) For the California wine, we have pH   log H  log H  32  H  1032  63  104 M. For the       Italian wine, pH   log H  log H  29  H  1029  13  103 M. (b) The California wine has lower hydrogen ion concentration.

SECTION 4.7 Logarithmic Scales

439

      8. (a) pH   log H  log H  55  H  1055  32  106 M.

(b) As pH increases, hydrogen ion concentration decreases. So as Marco gets better, the pH of his saliva will increase and its hydrogen ion concentration will decrease. (c) The saliva was more acidic when he was sick.

3125 I with S  104 and I  3125, so M  log 4  55. S 10 I I (b) M  log  10 M   I  S  10 M . We have M  48 and S  104 , so I  104  1048  63. S S

9. (a) M  log

I 721 with S  104 and I  721, so M  log10 4  59. S 10 (b) I  104  1058  631.

10. (a) M  log

11. Let I0 be the intensity of the smaller earthquake and I1 the intensity of the larger earthquake. Then I1  20I0 . 20I0 I I  log 20  log I0  log S. Then Notice that M0  log 0  log I0  log S and M1  log 1  log S S S M1  M0  log 20  log I0  log S  log I0  log S  log 20  13. Therefore the magnitude is 13 times larger.

I 12. Let the subscript S represent the San Francisco earthquake and J the Japan earthquake. Then we have M S  log S  83 S 83 I I 10  I S  S  1083 and M J  log J  49  I J  S  1049 . So S  49  1034  25119, and so the San Francisco S IJ 10 earthquake was 2500 times more intense than the Japan earthquake.

I 13. Let the subscript J represent the Japan earthquake and S represent the San Francisco earthquake. Then M J  log J  91 S 91 I I S  10  1008  63, and hence the Japan  I J  S  1091 and M S  log S  83  I S  S  1083 . So J  S IS S  1083 earthquake was about six times more intense than the San Francisco earthquake. 14. Let the subscript N represent the Northridge, California earthquake and K the Kobe, Japan earthquake. Then I I I 1072 M N  log N  68  I N  S  1068 and M K  log K  72  I K  S  1072 . So K  68  1004  251, and S S IN 10 so the Kobe, Japan earthquake was 25 times more intense than the Northridge, California earthquake.     20  105 I  10 log  10 log 2  107  10 log 2  log 107  10 log 2  7  73. Therefore the 12 I0 10  10 intensity level was 73 dB.

15.   10 log

16.   10 log

32  102 I  10 log  10 log 32  1010  105 dB. I0 10  1012

I I  70  10 log  log I  12  7  log I  5, so the intensity was 105 wattsm2 . I0 10  1012   I 18. 98  10 log 12  log I  1012  98  log I  98  log 1012  22  I  1022  63  103 . So the 10 intensity was 63  103 wattsm2 . 17.   10 log

19. (a) The intensity is 31  105 Wm2 , so   10 log

  I 31  105  10 log  10 log 31  107  75 dB. 12 I0 10  10

I  log I  12  9  I  103 Wm2 . 1012 103 Ie   323. (c) The ratio of the intensities is Is 31  105 (b) Here   90 dB, so 90  10 log

440

CHAPTER 4 Exponential and Logarithmic Functions

  IM  20. Let the subscript M represent the power mower and C the rock concert. Then 106  10 log 1012       IC 12  120   log I  10 log I M  1012  106  I M  1012  10106 . Also 120  10 log C 1012

I 1012 IC  1012  10120 . So C  106  1014  2512, and so the ratio of intensity is roughly 25. IM 10     k I k k k 21. (a) 1  10 log 1 and I1  2  1  10 log 2  10 log  2 log d1  10 log  20 log d1 . Similarly, I0 I0 I0 d d I0 1

1

k  20 log d2 . Substituting the expression for 1 gives 2  10 log I0   k d  20 log d1  20 log d1  20 log d2  1  20 log d1  20 log d2  1  20 log 1 . 2  10 log I0 d2 d 2  120  20 log 02  106, and so the (b) 1  120, d1  2, and d2  10. Then 2  1  20 log 1  120  20 log 10 d2 intensity level at 10 m is approximately 106 dB.

CHAPTER 4 REVIEW   2  9739, f 25  55902   7  18775, f 55  135765 2. f x  3  2x ; f 22  0653, f

1. f x  5x ; f 15  0089, f

3. g x  4e x2 ; g 07  0269, g 1  1472, g   12527   3  26888, g 36  174098 4. g x  74 e x1 ; g 2  0644, g 5. f x  3x2 . domain  , range 0 , asymptote y  0.

y

6. f x  2x1 . Domain  , range 0 , asymptote y  0.

1

1 x

1

7. g x  3  2x . Domain  , range 3 , asymptote y  3.

y

y

x

1

8. g x  5x  5. Domain  , range 5 , asymptote y  5.

y

1 1

1 1

x

x

CHAPTER 4

9. F x  e x1  1. Domain  , range 1 , asymptote y  1.

y

Review

10. G x  e x1  2. Domain  , range  2, asymptote y  2.

y 1 x

1

1 x

1

11. f x  log3 x  1. Domain 1 , range  , asymptote x  1.

y

12. g x  log x. Domain  0, range  , asymptote x  0.

y

1

1 x

1

13. f x  2  log2 x. Domain 0 , range  , asymptote x  0.

1

x

14. f x  3  log5 x  4. Domain 4 , range  , asymptote x  4. y

y (1, 2) 1 1

x

(_3, 3) 1 1

x

441

442

CHAPTER 4 Exponential and Logarithmic Functions

15. g x  2 ln x. Domain 0 , range  , asymptote x  0.

  16. g x  ln x 2 . Domain

x  x  0   0  0 , range  ,

y

asymptote x  0.

y

1 x

1

1 1

x

2

17. f x  10x  log 1  2x. Since log u is defined only for u  0, we require 1  2x  0  2x  1  x  12 , and so   the domain is  12 .   18. g x  log 2  x  x 2 . We must have 2  x  x 2  0 (since log y is defined only for y  0)  x 2  x  2  0  x  2 x  1  0. The endpoints of the intervals are 2 and 1.

 1

1 2

2 

Sign of x  2







Sign of x  1













 2

2 2

2 

Sign of x  2







Sign of x  2













Interval

Sign of x  2 x  1

Thus the domain is 1 2.   19. h x  ln x 2  4 . We must have x 2  4  0 ( since ln y is defined only for y  0)  x 2  4  0  x  2 x  2  0. The endpoints of the intervals are 2 and 2. Interval

Sign of x  2 x  2

Thus the domain is  2  2 .

20. k x  ln x. We must have x  0. So x  0  x  0 or x  0. Since x  0  x  0, the domain is x  0 or x  0 which is equivalent to x  0. In interval notation,  0  0 . 21. log2 1024  10  210  1024

22. log6 37  x  6x  37

23. log x  y  10 y  x

24. ln c  17  e17  c

25. 26  64  log2 64  6

26. 4912  17  log49 17   12

27. 10x  74  log10 74  x  log 74  x   29. log2 128  log2 27  7

28. ek  m  ln m  k   30. log8 1  log8 80  0

31. 10log 45  45   33. ln e6  6

32. log 0000001  log 106  6   34. log4 8  log4 432  32

CHAPTER 4 1  log 33  3 35. log3 27 3  37. log5 5  log5 512  12

Review

443

36. 2log2 13  13 2  38. e2 ln 7  eln 7  72  49    40. log3 243  log3 352  52

39. log 25  log 4  log 25  4  log 102  2    23 3 41. log2 1623  log2 24  log2 292  92 42. log5 250  log5 2  log5 250 2  log5 125  log5 5  3   43. log8 6  log8 3  log8 2  log8 63  2  log8 4  log8 823  23   44. log10 log10 10100  log10 100  log10 102  2   45. log AB 2 C 3  log A  2 log B  3 log C       46. log2 x x 2  1  log2 x  log2 x 2  1  log2 x  12 log2 x 2  1            1 x2  1 x2  1 1 ln x 2  1  ln x 2  1 1 ln x  1 x  1  ln x 2  1  ln  47. ln  2 2 2 x2  1 x2  1     12 ln x  1  ln x  1  ln x 2  1         4x 3 3  log y 2 x  15  log 4  3 log x  2 log y  5 log x  1  log 4x 48. log y 2 x  15        x 2 1  5x3/2   log5 x 2 1  5x3/2  log5 x x 2  1  2 log5 x  32 log5 1  5x  12 log5 x 3  x 49. log5 x3  x     2 log5 x  32 log5 1  5x  12 log5 x  log5 x 2  1    2 log5 x  32 log5 1  5x  12 log5 x  log5 x  1  log5 x  1    3 4     x  12 50. ln  13 ln x 4  12  ln x  16  12 ln x  3  x  16 x  3   51. log 6  4 log 2  log 6  log 24  log 6  24  log 96       52. log x  log x 2 y  3 log y  log x  x 2 y  y 3  log x 3 y 4      2 x  y3/2 3 2 2 3 / 2 2 2  log2  53. 2 log2 x  y  2 log2 x  y  log2 x  y  log2 x  y 2 x 2  y2   2 x  1 54. log5 2  log5 x  1  13 log5 3x  7  log5 [2 x  1]  log5 3x  713  log5  3 3x  7      x2  4 1 2 2 55. log x  2  log x  2  2 log x  4  log [x  2 x  2]  log x  4  log  x2  4       5  5  56. 12 ln x  4  5 ln x 2  4x  12 ln x  4 x 2  4x  ln x  4 x 2  4x 57. 32x7  27  32x7  33  2x  7  3  2x  10  x  5

1  54x  53  4  x  3  x  7 58. 54x  125     59. 23x5  7  log2 23x5  log2 7  3x  5  log2 7  x  13 log2 7  5 . Using the Change of Base Formula, we 7 1 have log2 7  log log 2  2807, so x  3 2807  5  260.     60. 1063x  18  log10 1063x  log10 18  6  3x  log10 18  x  13 6  log10 18  158

444

CHAPTER 4 Exponential and Logarithmic Functions

61. 41x  32x5  log 41x  log 32x5  1  x log 4  2x  5 log 3  log 4  5 log 3  2x log 3  x log 4  log 4  5 log 3  115 x log 3  log 4  log 4  5 log 3  x  2 log 3  log 4   62. e3x4  10  ln e3x4  ln 10  34 x  ln 10  x  43 ln 10  307

  63. x 2 e2x  2xe2x  8e2x  e2x x 2  2x  8  0  x 2  2x  8  0 (since e2x  0)  x  4 x  2  0  x  4 or x  2

   64. 32x  3x  6  0  3x  3 3x  2  0  3x  3 or 3x  2  x  1 (the second equation has no solution)

65. log x  log x  1  log 12  log x x  1  log 12  x x  1  12  x 2  x  12  0  x  3 x  4  0  x  4 or 3. Since log 4 is undefined, the only solution is x  3. 66. ln x  2  ln 3  ln 5x  7  ln 3 x  2  ln 5x  7  3 x  2  5x  7  2x  1  x  12 , but since ln x  2 is undefined for x  12 , there is no solution.

67. log2 1  x  4  1  x  24  x  1  16  15

  68. ln 2x  3  1  0  ln 2x  3  1  2x  3  e1  x  12 1e  3  168 69. log3 x  8  log3 x  2  log3 x x  8  2  x x  8  9  x 2  8x  9  0  x  9 x  1  0  x  1 or 9. We reject 1 because it does not satisfy the original equation, so the only solution is x  9. 70. log8 x  5  log8 x  2  1  log8 x 3

x 5 x 5 1  8  x  5  8 x  2  x  5  8x  16  7x  21 x 2 x 2

3 log 063 2x log 5  log 063  x    0430618 71. 52x /3  063  3 2 log 5   log 7  2602452 72. 23x5  7  3x  5 log 2  log 7  x  13 5  log 2 73. 52x1  34x1  2x  1 log 5  4x  1 log 3  2x log 5  log 5  4x log 3  log 3  log 3  log 5  2303600 x 2 log 5  4 log 3   log 3  log 5  x  4 log 3  2 log 5 1 ln 10000  0614023 74. e15k  10000  15k  ln 10000  k   15

75. y  e xx2 . Vertical asymptote x  2, horizontal asymptote y  272, no maximum or minimum. 10

76. y  10x  5x . No vertical asymptote, horizontal

asymptote y  0, local minimum of about 013 at x  052.

2 5 1 -20

0

20 -2

2

CHAPTER 4



 77. y  log x 3  x . Vertical asymptotes x  1, x  0,

x  1, no horizontal asymptote, local maximum of about

041 when x  058.

Review

445

78. y  2x 2  ln x. Vertical asymptote x  0, no horizontal asymptote, local minimum of about 119 at x  050. 100 50

-1

1

2

0 0

5

10

-2

79. 3 log x  6  2x. We graph y  3 log x and y  6  2x in 80. 4  x 2  e2x . From the graphs, we see that the solutions the same viewing rectangle. The solution occurs where the two graphs intersect. From the graphs, we see that the

are x  064 and x  2.

5

solution is x  242. 10 -2 5

2

10

-10

81. ln x  x  2We graph the function f x  ln x  x  2, and we see that the graph lies above the x-axis for

016  x  315. So the approximate solution of the given inequality is 016  x  315.

82. e x  4x 2  e x  4x 2  0. We graph the function

f x  e x  4x 2 , and we see that the graph lies below the

x-axis for  041  071 431.

2

5 -10

0 5 -2

83. f x  e x  3ex  4x. We graph the function f x,

and we see that the function is increasing on  0 and 110  and that it is decreasing on 0 110.

-2

84. The line has x-intercept at x  e0  1. When x  ea ,

y  ln ea  a. Therefore, using the point-slope equation,

a a0 we have y  0  a x  1  y  a x  1. e 1 e 1

2 -5

85. log4 15 

log 15  1953445 log 4

  log 3 4  0147839 86. log7 34  log 7

446

CHAPTER 4 Exponential and Logarithmic Functions

87. log9 028 

log 028  0579352 log 9

88. log100 250 

log 250  1198970 log 100

89. Notice that log4 258  log4 256  log4 44  4 and so log4 258  4. Also log5 620  log5 625  log5 54  4 and so log5 620  4. Then log4 258  4  log5 620 and so log4 258 is larger.     x x 90. f x  23 . Then y  23  log2 y  3x  log3 log2 y  x, and so the inverse function is f 1 x  log3 log2 x . Since log3 y is defined only when y  0, we have log2 x  0  x  1. Therefore the domain is 1 , and the range is  .  r nt . 91. P  12,000, r  010, and t  3. Then A  P 1     n 23

(a) For n  2, A  12,000 1  010  12,000 1056  $16,08115. 2 123   $16,17818. (b) For n  12, A  12,000 1  010 12 3653   $16,19764. (c) For n  365, A  12,000 1  010 365

(d) For n  , A  Pert  12,000e0103  $16,19831.

92. P  5000, r  0085, and n  2.  215 (a) For t  15, A  5000 1  0085  5000  104253  $566498. 2

7 (b) We want to find t such that A  7000. Then A  5000  104252t  7000  104252t  7000 5000  5        log 75 log 75 7 t   404, and so the investment will amount to $7000 2t log 10425  log 5  2t  log 10425 2 log 10425 after approximately 4 years.

(c) In this case, we solve n t  n 0 ert for t when n 0  5000, r  0085, and n t  7000: 7000  5000e0085t  7 7  e0085t  ln 7  0085t  t  ln 5  396, so the investment will grow to $7000 in just under 4 years. 5 5 0085  r nt

with P  100,000, r  0052, n  365, and A  100,000  10,000  110,000,  365t 365t    and solve for t: 110,000  100,000 1  0052   11  1  0052  log 11  365t log 1  0052 365 365 365

93. We use the formula A  P 1 

t

n

log 11   1833. The account will accumulate $10,000 in interest in approximately 18 years.  365 log 1  0052 365

94. We solve n t  n 0 ert for n t  2n 0 and r  0045: 2n 0  n 0 e0045t  ln 2  0045t  t 

ln 2  15403. The 0045

retirement savings plan will double in about 154 years.   00425 365 95. After one year, a principal P will grow to the amount A  P 1   P 104341. The formula for simple 365 interest is A  P 1  r. Comparing, we see that 1  r  104341, so r  004341. Thus the annual percentage yield is 4341%.   0032 12 96. A  P 1   P 103247  P 1  r  r  3247% 12 97. (a) Using the model n t  n 0 ert , with n 0  30 and r  015, we have the formula n t  30e015t . (b) n 4  30e0154  55.

    015t  015t  ln 50  t  1 ln 50  1876. So the stray cat population will (c) 500  30e015t  50 3 e 3 3 015 reach 500 in about 19 years.

CHAPTER 4

Review

447

98. Using the model n t  n 0 ert , with n 0  10000 and n 1  25000, we have 25000  n 1  10000er1  er  52  r  ln 52  0916. So n t  10000e0916t . (a) Here we must solve the equation n t  20000 for t. So n t  10000e0916t  20000  e0916t  2  ln 2  0756. Thus the doubling period is about 45 minutes. 0916t  ln 2  t  0916 (b) n 3  10000e09163  156250, so the population after 3 hours is about 156,250.

99. (a) From the formula for radioactive decay, we have m t  10ert , where r   



5

the amount remaining is m 1000  10  e  ln 2 2710 Therefore the amount remaining is about 997 mg. 



ln 2 . So after 1000 years 27  105

   1000  10eln 2 27102  10eln 2270  997. 







5 5 (b) We solve for t in the equation 7  10e ln 2 2710 t . We have 7  10e ln 2 2710 t     5 ln 2 ln 07  27  105  138,93475. Thus it takes about t t  07  e ln 2 2710 t  ln 07   ln 2 27  105 139,000 years.

ln 2 . So m t  m 0 e[ln 2 h ]t . h (a) Using m 8  033m 0 , we solve for h. We have 033m 0  m 8  m 0 eln 2 h  033  e8 ln 2 h  8 ln 2 8 ln 2   ln 033  h    5002. So the half-life of this element is roughly 5 days. h ln 033 (b) m 12  m 0 e[ln 25]12  019m 0 , so about 19% of the original mass remains.

100. From the formula for radioactive decay, we have m t  m 0 ert , where r 

ln 2  00004359 and n t  150  e00004359t . 1590 (b) n 1000  150  e000043591000  9700, and so the amount remaining is about 9700 mg.

101. (a) From the formula for radioactive decay, r 

(c) Find t so that 50  150  e00004359t . We have 50  150  e00004359t  13  e00004359t    1 ln 13  2520. Thus only 50 mg remain after about 2520 years. t  00004359

102. From the formula for radioactive decay, we have m t  m 0 ert , where r 

ln 2 ln 2 . Since h  4, we have r   0173 h 4

and m t  m 0 e0173t . (a) Using m 20  0375, we solve for m 0 . We have 0375  m 20  m 0 e017320  003125m 0  0375  0375  12. So the initial mass of the sample was about 12 g. m0  003125 (b) m t  12e0173t . (c) m 3  12e01733  7135. So there are about 71 g remaining after 3 days.

(d) Here we solve m t  015 for t: 015  12e0173t  00125  e0173t  0173t  ln 00125  ln 00125 t  253. So it will take about 25 days until only 15% of the substance remains. 0173 103. (a) Using n 0  1500 and n 5  3200 in the formula n t  n 0 ert , we have 3200  n 5  1500e5r  e5r  32 15      1 32 01515t . 5r  ln 32 15  r  5 ln 15  01515. Thus n t  1500  e

(b) We have t  2020  2009  11 so n 11  1500e0151511  7940. Thus in 2009 the bird population should be about 7940.

448

CHAPTER 4 Exponential and Logarithmic Functions

104. We use Newton’s Law of Cooling: T t  Ts  D0 ekt with k  00341, Ts  60 and D0  190  60  130.

3  So 90  T t  60  130e00341t  90  60  130e00341t  130e00341t  30  e00341t  13   3  t   ln 313  430, so the engine cools to 90 F in about 43 minutes. 00341t  ln 13 00341       105. H  13  108 M. Then pH   log H   log 13  108  79, and so fresh egg whites are basic.

    106. pH  19   log H . Then H  1019  126  102 M.

107. Let I0 be the intensity of the smaller earthquake and I1 be the intensity of the larger earthquake. Then I1  35I0 . Since     I I M  log , we have M0  log 0  65 and S S       35I0 I I1  log  log 35  log 0  log 35  M0  log 35  65  804. So the magnitude on the M1  log S S S Richter scale of the larger earthquake is approximately 80.     I I 108. Let the subscript J represent the jackhammer and W the whispering:  J  132  10 log J  log J  132 I0 I0 I I I 10132  J  10132 . Similarly W  1028 . So J   10104  251  1010 , and so the ratio of intensities is I0 I0 IW 1028

251  1010 .

CHAPTER 4 TEST y

1. (a)

y

(b)

10

4

8 6 (0, 5) 4

(_2, 0)

_4 _2 0 _2

2 _4 _2 0

2

2

4

6

8 x

f x  2x  4 has domain  , range 4 ,

and horizontal asymptote y  4.

(0, 1) 2

4

6

4. (a) 10log 36  36

(b) ln e3  3  12  (c) log3 27  log3 33  log3 332  32   3 (d) log2 80  log2 10  log2 80 10  log2 8  log2 2  3

(e) log8 4  log8 823  23 (f) log6 4  log6 9  log6 4  9  log6 62  2

x

_4

g x  log3 x  3 has domain 3 , range

 , and vertical asymptote x  3.

  2. (a) f t  ln 2t  3 is defined where 2t  3  0  2t  3  t  32 , so its domain is 32   .   (b) g x  log x 2  1 is defined where x 2  1  0  x  1, so its domain is  1  1 .

3. (a) 62x  25  log6 62x  log6 25  2x  log6 25

8

(b) ln A  3  eln A  e3  A  e3

CHAPTER 4

5. (a) log



x y3 z2



Test

449

 log x  log y 3  log z 2  log x  3 log y  2 log z

     x 12 x 1 x  ln  12 ln x  12 ln y (b) ln  ln y y 2 y        1 x 2 x 2 3     log   13 log x  2  4 log x  log x 2  4 (c) log 3 x4 x2  4 x4 x2  4   1  3 log x  2  43 log x  13 log x 2  4 

  6. (a) log a  2 log b  log a  log b2  log ab2

  x 2  25 x  5 x  5  ln  ln x  5 (b) ln x 2  25  ln x  5  ln x 5 x 5 (c) log2 3  3 log2 x  12 log2 x  1  log2 3  log2 x 3  log2 x  112  log2

 3 x 1 x3

7. (a) 34x  3100  4x  100  x  25

2 (b) e3x2  e x  3x  2  x 2  x 2  3x  2  0  x  1 x  2  0  x  1 or x  2

(c) 5x10  1  7  5x10  6  log5 5x10  log5 6  x10  log5 6  x  10 log5 6  1113

(d) 10x3  62x  log 10x3  log 62x  x  3  x 

3 log6 10  539 2  log6 10

    log6 62x  log6 10 x  3  2x  2  log6 10 x  3 log6 10 log6 10

8. (a) log 2x  3  2x  103  x  12 1000  500

(b) log x  1  log 2  log 5x  log 2 x  1  log 5x  2x  2  5x  3x  2  x  23

(c) 5 ln 3  x  4  ln 3  x  45  3  x  e45  x  3  e45  0774

(d) log2 x  2  log2 x  1  2  log2 x  2 x  1  2  x  2 x  1  22  4  x 2  x  2  4 

x 2  x  6  0  x  3 x  2  0  x  3 or 2. However, x  3 does not satisfy the original equation, so the only solution is x  2.

9. Using the Change of Base Formula, we have log12 27 

log 27  1326. log 12

10. (a) From the formula for population growth, we have 8000  1000er1

y

(d)

 8  er  r  ln 8  207944. Thus n t  1000e207944t .

(b) n 15  1000e20794415  22,600 (c) 15,000  1000e207944t  15  e207944t  ln 15  207944t  ln 15  13. Thus the population will reach 15,000 after t 207944 approximately 13 hours. 12t  11. (a) A t  12,000 1  0056 , where t is in years. 12

50,000

  365t  0056 3653 . So A  12,000 1   $14,19506. (b) A t  12,000 1  0056 3 365 365

1

x

450

FOCUS ON MODELING

  (c) A t  12,000e0056t . So 20,000  12,000e0056t  5  3e0056t  ln 5  ln 3e0056t  ln 5  ln 3  0056t  1 ln 5  ln 3  912. Thus, the amount will grow to $20,000 in approximately 912 years. t  0056

12. (a) The initial mass is m 0  3 and the half-life is h  10, so using the formula m t  m 0 2t h , we have m t  3  2t10 .

(b) Using the radioactive decay model with m 0  3 and r  lnh2  ln102 , we have m t  3e[ln 210]t  3e00693t . (c) After 1 minute  60 seconds, the amount remaining is m 60  3e0069360  0047 g.     106 106 106 00693t 6 00693t 00693t  ln e  00693t  ln   10  e   ln (d) We solve 3e 3 3 3   1 106 ln  215, so there is 1 g of 91 Kr remaining after about 215 seconds, or 36 minutes. t  00693 3

  IJ IJ 13. Let the subscripts J and P represent the two earthquakes. Then we have M J  log  64  1064   S S   I I I 1064 S 1064 S  I J . Similarly, M P  log P  31  1031  P  1031 S  I P . So J  31  1033  19953, S S IP 10 S and so the Japan earthquake was about 1995 times more intense than the Pennsylvania earthquake.

FOCUS ON MODELING Fitting Exponential and Power Curves to Data

1. (a)

y

(b) Using a graphing calculator, we obtain the model y  abt , where a  33349260  1015 and

300

b  10198444, and y is the population (in millions) in

250

the year t.

200

(c) Substituting t  2020 into the model of part (b), we get y  ab2020  5776 million.

150

(d) According to the model, the population in 1965 was

100

y  ab1965  1960 million.

50 0 1780

1820

1860

1900

Year

1940

1980

2020 x

Fitting Exponential and Power Curves to Data y

2. (a)

451

(b) We let t represent the time (in seconds) and y the

5

distance fallen (in meters). Using a calculator, we obtain the power model: y  49622t 20027 .

4

(c) When t  3 the model predicts that y  44792 m.

3 2 1

0

0.2

0.4

0.6

0.8

1

x

Time (s)

3. (a) Yes. (b) Year t

Health Expenditures E ($bn)

ln E

ln E

1970

743

430811

1980

2558

554440

1985

4446

609718

1987

5191

625210

5

1990

7243

658521

4

1992

8579

675449

3

1994

9727

688008

2

1996

10818

698638

1998

12089

709747

2000

13772

722781

2002

16380

740123

2005

20354

761845

2008

24117

778809

2010

25990

786288

2012

27934

793501

7 6

1 0

10

20

30

40

x

Year since 1970

Yes, the scatter plot appears to be roughly linear.

(c) Let t be the number of years elapsed since 1970 . Then ln E  47473926  008213193t, where E is expenditure in billions of dollars. (d) E  e47473926008213193t  11528330e008213193t

(e) In 2020 we have t  2020  1970  50, so the estimated 2020 health-care expenditures are 11528330e00821319350  70023 billion dollars.

452

FOCUS ON MODELING

4. (a)

y

(b) Let t be the time (in hours) and y be amount of

5

iodine-131 (in grams). Using a calculator, we obtain the exponential model y  abt , where a  479246 and

4.8

b  099642.

4.6

(c) To find the half-life of iodine-131, we must find the time when the sample has decayed to half its original

4.4

mass. Setting y  240 g, we get

4.2

240  479246  099642t 

4 3.8 0

10

20

30

40

ln 240  ln 479246  t ln 099642  ln 240  ln 479246  1928 h. t ln 099642

50 x

Time (h)

5. (a) Using a graphing calculator, we find that

y

(b)

I0  227586444 and k  01062398.

14

(c) We solve 015  227586444e01062398x for x: 015  227586444e01062398x 

12 10

0006590902  e01062398x  5022065  01062398x  x  4727. So light

8 6

intensity drops below 015 lumens below around

4

4727 feet.

2 0

10

20

30

40

x

Depth (ft)

6. (a) Using a graphing calculator, we find the power function model y  4970030t 015437 and the exponential   model y  4482418  099317t .

(c) The power function, y  4970030t 015437 , seems to provide a better model.

(b)

y 80 70 60 50 40 30

exponential power

20 10 0

20

40

60

80

Time (hours)

100

120 x

Fitting Exponential and Power Curves to Data

7. (a) Let A be the area of the cave and S the number of species of bat. Using a graphing calculator, we obtain

S

(b)

7

the power function model S  014A064 .

6 5

(c) According to the model, there are

4

S  014 205064  4 species of bat living in the El

3

Sapo cave.

2 1

0

100

200

300

500 A

400

Area (m@ ) The model fits the data reasonably well.

8. Let x be the reduction in emissions (in percent), and y be the cost (in

y

dollars). First we make a scatter plot of the data. A linear model does

600

not appear appropriate, so we try an exponential model. Using a calculator, we get the model y  ab x , where a  2414 and

500 400

b  105452. This model is graphed on the scatter plot.

300 200 100 0 40

50

60

70

80

90

Reduction in emissions (%)

9. (a)

y

(b)

1.2 1 0.8 0.6 0.4 0.2 0

2

4

6

8

10

12

14

16

18 x

x

y

ln x

ln y

2

008

069315

252573

4

012

138629

212026

6

018

179176

171480

8

025

207944

138629

10

036

230259

102165

12

052

248491

065393

14

073

263906

031471

16

106

277259

005827

100 x

453

454

FOCUS ON MODELING ln y

ln y

(b) (cont’d)

2

0

4

6

8

10

12

14

16

18

_1

_1

_2

_2

y

2

2.5

3 ln x

(b)

800 700 600 500 400 300 200 100

(b) (cont’d)

1.5

(d) y  a  b x where a  0057697 and b  1200236.

(c) The exponential function.

0

1

_3

_3

10. (a)

0.5

0

x

20

40

60

100 x

80

ln y

ln y

8

8

7

7

6

6

5

5

4

4

3

3

2

2

1

1

0

20

40

(c) The power function is more appropriate.

60

80

100 x

y

ln x

ln y

10 20 30 40 50 60 70 80 90

29 82 151 235 330 430 546 669 797

230259 299573 340120 368888 391202 409434 424850 438203 449981

336730 440672 501728 545959 579909 606379 630262 650578 668085

1

2

3

4

5

ln x

(d) y  ax n where a  0893421326 and n  150983.

11. (a) Using the Logistic command on a TI-83 we get y  c  500855793.

0

x

c where a  4910976596, b  04981144989, and 1  aebx

c cN c c c  aebt  1  we solve for t. So N   1aebt  bt bt N N N 1  ae 1  ae cN 1 bt  e  bt  ln c  N   ln a N  t  [ln a N  ln c  N ]. Substituting the values for a, b, and aN b 1 c, with N  400 we have t  04981144989 ln 1964390638  ln 100855793  1058 days.

(b) Using the model N 

5

TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE APPROACH

5.1

ANGLE MEASURE

1. (a) The radian measure of an angle  is the length of the arc that subtends the angle in a circle of radius 1.  . (b) To convert degrees to radians we multiply by 180

(c) To convert radians to degrees we multiply by 180  . 2. (a) If a central angle  is drawn in a circle of radius r, the length of the arc subtended by  is s  r . (b) The area of the sector with central angle  is A  12 r 2 .

3. (a) The angular speed of the point is  

 . t

s (b) The linear speed of the point is   . t (c) The linear speed  and the angular speed ome are related by the equation   r.

4. No, if the common angular speed is , Object A has linear speed 2, while Object B has angular speed 5. Object B has greater linear speed.  rad   rad  0262 rad 5. 15  15  180  12

 rad   rad  0628 rad 6. 36  36  180  5

 rad  3 rad  0942 rad 7. 54  54  180  10

 rad  5 rad  1309 rad 8. 75  75  180  12

 rad    rad  0785 rad 9. 45  45  180  4

 rad    rad  0524 rad 10. 30  30  180  6

 rad  5 rad  1745 rad 11. 100  100  180  9

 rad  10 rad  3491 rad 12. 200  200  180  9

 rad  50 rad  17453 rad 13. 1000  1000  180  9

 rad  20 rad  62832 rad 14. 3600  3600  180 

 rad   7 rad  1222 rad 15. 70  70  180  18

 rad   5 rad  2618 rad 16. 150  150  180  6

  17. 53  53  180   300

  18. 34  34  180   135

  19. 56  56  180   150

 20.  32   32  180   270

 540  21. 3  3  180     1719

 360  22. 2  2  180      1146

 216  23. 12  12  180      688

 612  24. 34  34  180     1948

    180  18 25. 10  10

  5  180  50 26. 518  18







   2  180  24  13 180  27.  215 28.  13  15 12   12    195 29. 50 is coterminal with 50  360  410 , 50  720  770 , 50  360  310 , and 50  720  670 . (Other answers are possible.)

30. 135 is coterminal with 135  360  495 , 135  720  855 , 135  360  225 , and 135  720  585 . (Other answers are possible.) 31. 34 is coterminal with 34  2  114 , 34  4  194 , 34  2   54 , and 34  4   134 . (Other answers are possible.) 11 13 32. 116 is coterminal with 116  2  236 , 116  4  356 , 116  2    6 , and 6  4   6 . (Other answers are possible.)

455

456

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

 7  15  9  17 33.   4 is coterminal with  4  2  4 ,  4  4  4 ,  4  2   4 , and  4  4   4 . (Other answers are possible.)

34. 45 is coterminal with 45  360  315 , 45  720  675 , 45  360  405 , and 45  720  765 . (Other answers are possible.) 35. Since 430  70  360 , the angles are coterminal.

36. Since 330  30   360 , the angles are coterminal.

37. Since 176  56  126  2; the angles are coterminal. 38. Since 323  113  213  7 is not a multiple of 2, the angles are not coterminal. 39. Since 875  155  720  2  360 , the angles are coterminal. 40. Since 340  50  290 is not a multiple of 360 , the angles are not coterminal.

41. Since 400  360  40 , the angles 400 and 40 are coterminal. 42. Since 375  360  15 , the angles 375 and 15 are coterminal.

43. Since 780  2  360  60 , the angles 780 and 60 are coterminal.

44. Since 100  260  360 is a multiple of 360 , the angles 100 and 260 are coterminal. 45. Since 800  3  360  280 , the angles 800 and 280 are coterminal.

46. Since 1270  190  1080  3  360 is a multiple of 360 , the angles 1270 and 190 are coterminal.

47. Since 196  2  76 , the angles 196 and 76 are coterminal. 5  48. Since  53  2   3 , the angles  3 and 3 are coterminal.

49. Since 25  12  2  , the angles 25 and  are coterminal.

50. Since 10  2  3717, the angles 10 and 10  2 are coterminal.

17  51. Since 174  2  2   4 , the angles 4 and 4 are coterminal. 52. Since 512  32  24  12  2, the angles 512 and 32 are coterminal.

53. Using the formula s  r, s  56  9  152 .

 7 35   5 5  122. 54. Using the formula s  r, the length of the arc is s  140  180 9 9  s 10 180 55.     2 rad  2   1146 r 5  8 s 56. Solving for r, we have r  , so the radius of the circle is r   4.  2 57. Solving for s, we have s  r , so the length of the arc is 5  3  15 cm.  8 58. Solving for s, we have s  r   12  40    838 m. 180 3  14 14 180 s rad    891 . 59. Solving for , we have    r 9 9  s 15 5 5 180 60. Solving for , we have     rad    955 . r 9 3 3  15 18 s   573 m 61. r    56  s 20 72 62. r     2292 cm   50 180     32  4  128  4468 63. (a) A  12 r 2   12  82  80  180  9 9

(b) A  12 r 2   12  102  05  25   2A 2  12 , so the radius is  586. 64. (a) A  12 r 2   r   07

SECTION 5.1 Angle Measure

(b) r 



2A  



457

2  12   303 150  180 

   2 65.   23 rad and r  10 m, so A  12 r 2   12 102 23  100 3  1047 m .     1 2 1 2 29  29  456 ft2 .  29 66.   145  145  36 and r  6 ft, so A  2 r   2 6 36 2 180  7 1 7 2 rad. Thus, A  12 r 2   70   r   67. A  70 m2 and   140  140  180 9 2 9   6 5 9   76 m. r  2  70  7    4 6 12 5  1 5  2 2 2 68. A  20 m and   12 , so A  2 r   20  12 r  r  2  20    55 m. 5 

69. r  80 mi and A  1600 mi2 , so A  12 r 2   1600  12 802     12 rad.  600 . Thus, the area of the sector is 70. The area of the circle is r 2  600 m2 , so r   900 1 600 3  2865 m2 . A  12 r 2    2   71. Referring to the figure, we have AC  3  1  4, BC  1  2  3, and AB  2  3  5. Since

2

AB 2  AC 2  BC 2 , then by the Pythagorean Theorem, the

triangle is a right triangle. Therefore,    2 and  2 A  12 r 2   12  12   2  4 ft .

3 A

3

B 2

1 1 C

s 1 72. The triangle is equilateral, so 1   3 rad. To find  2 , we use the formula  2  r  1  1 rad. Thus,  1  2   3  1  0047 rad, or approximately 27 .

73. Between 1:00 P. M . and 1:00 P. M ., the minute hand traverses three-quarters of a complete revolution, or 34 2  32 rad, while the hour hand moves three-quarters of the way from 12 to 1, which is itself one-twelfth of a revolution. So the hour   1 2   rad. hand traverses 34 12 8

74. Between 1:00 P. M . and 6:45 P. M ., the minute hand traverses five complete revolutions plus three-quarters of a revolution;

that is, 5 2  34 2  232 rad, while the hour hand moves through five-twelfths of a revolution, plus three-quarters of   5 2  3 1 2  23 rad. the way from 6 to 7; that is, 12 4 12 24

75. The circumference of each wheel is d  28 in. If the wheels revolve 10,000 times, the distance traveled is 1 ft 1 mi 10,000  28 in.    1388 mi. 12 in. 5280 ft 76. Since the diameter is 30 in., we have r  15 in. In one revolution, the arc length (distance traveled) is s  r  2  1 rev 15  30 in. The total distance traveled is 1 mi  5280 ft/mi  12 in/ft  63,360 in.  63,360 in.   67227 rev. 30 in Therefore the car wheel will make approximately 672 revolutions.  rad   rad. 77. We find the measure of the angle in degrees and then convert to radians.   405 255  15 and 15 180  12

Then using the formula s  r , we have s  roughly 1037 mi.  rad  78.   35  30  5  5  180 

 12  3960  330  1036725 and so the distance between the two cities is

 rad. Then using the formula s  r , the length of the arc is 3

s 3  3960  110  345575. So the distance between the two cities is roughly 346 mi.

458

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

1 of its orbit which is 2 rad. Then s  r  2  93,000,000  1,600,9113, so the 79. In one day, the earth travels 365 365 365 distance traveled is approximately 16 million miles.

80. Since the sun is so far away, we can assume that the rays of the sun are parallel when striking the earth. Thus, the angle 500 s 180  500 formed at the center of the earth is also   72 . So r     72  3980 mi, and the circumference  72  180  2  180  500  25,000 mi. 72   1   1   rad   rad. Then s  r    3960  1152, and so a 81. The central angle is 1 minute  60 60 180 10,800 10,800 nautical mile is approximately 1152 mi.     219,900 ft2 . 82. The area is A  12 r 2   12  3002  280  180  is c  2r 

83. The area is equal to the area of the large sector (with radius 34 in.) minus the area of the small sector (with radius 14 in.)      1131 in.2 . Thus, A  12 r12   12 r22   12 342  142 135  180  84. The area available to the cow is shown in the diagram. Its area is the sum of four quarter-circles:   A  14  1002  502  402  302  3750

 11,781 ft2

50 ft 30 ft

20 ft 50 ft

100 ft

60 ft 50 ft

40 ft

45  2 rad  90 rad/min. 1 min 45  2  16  1440 in./min  45239 in./min. (b) The linear speed is   1 1000  2 rad  2000 rad/min. 86. (a) The angular speed is   1 min

85. (a) The angular speed is  

(b) The linear speed is   87.  

6 1000  2  12 50  ft/s  524 ft/s. 60 3

8  2  2 32   6702 ft/s. 15 15

600  2  11 12 ft  1 mi  60 min  125 mi/h  393 mi/h. 1 min 5280 ft 1 hr 1 day 1  2  3960   103957 mi/h. 89. 23 h 56 min 4 s  239344 hr. So the linear speed is 1 day 239344 hr

88.  

50 mi/h 1h 5280 ft linear speed     2200 rad/min. radius 2 ft 60 min 1 mi angular speed 2200 rad/min (b) The rate of revolution is   350 rev/min. 2 2 100  2  020 m 2 91.     209 m/s. 60 s 3 40  2  4 linear speed of pedal   160 rad/min. 92. (a) The angular speed is   radius of wheel sprocket 2 (b) The linear speed of the bicycle is   angular speed  radius  160 rad/min  13 in  2080 in/min  62 mi/h. 90. (a) The radius is 2 ft, so the angular speed is  

SECTION 5.2 Trigonometry of Right Triangles

459

93. (a) The circumference of the opening is the length of the arc subtended by the angle  on the flat piece of paper, that is, C  s  r   6  53  10  314 cm.

10 C   5 cm. 2 2  (c) By the Pythagorean Theorem, h 2  62  52  11, so h  11  33 cm.  (d) The volume of a cone is V  13 r 2 h. In this case V  13   52  11  868 cm3 . (b) Solving for r, we find r 

94. (a) With an arbitrary angle , the circumference of the opening is   3 92 C 2 2  , h  6  r  36  2 , and C  6, r  2      92 9 2 9 V  13 r 2 h   2 36  2  2  2 42  2 . 3   

(b) 100 50 0 0

2

4

6

(c) The volume seems to be maximized for   513 rad or about 293 .

95. Answers will vary, although of course everybody prefers radians.

5.2 1. (a)

TRIGONOMETRY OF RIGHT TRIANGLES (b) sin  

opposite adjacent

adjacent opposite opposite , cos   , and tan   . hypotenuse hypotenuse adjacent

(c) The trigonometric ratios do not depend on the size of the triangle because all right triangles with angle  are similar.

¬

hypotenuse

2. The reciprocal identities state that csc  

1 1 1 , sec   , and cot   . sin  cos  tan 

3. sin   45 , cos   35 , tan   43 , csc   54 , sec   53 , cot   34

7 , cos   24 , tan   7 , csc   25 , sec   25 , cot   24 4. sin   25 7 7 25 24 24   9 2 5. The remaining side is obtained by the Pythagorean Theorem: 41  402  81  9. Then sin   40 41 , cos   41 , 41 41 9 tan   40 9 , csc   40 , sec   9 , cot   40

6. The hypotenuse is obtained by the Pythagorean Theorem: 17 17 8 tan   15 8 , csc   15 , sec   8 , cot   15

  8 82  152  289  17. Then sin   15 17 , cos   17 ,

7. The remaining side is obtained by the Pythagorean Theorem: 





  32  22  13. Then sin   2



 2 1313 , 13

cos   3  3 1313 , tan   23 , csc   213 , sec   313 , cot   32 13    8. The remaining side is obtained by the Pythagorean Theorem: 82  72  15. Then sin   78 , cos   815 , 





tan   7  7 1515 , csc   87 , sec   8  8 1515 , cot   715 15 15   2 2 9. c  5  3  34 

(a) sin   cos   3  3 3434 34   2 2 10. b  7  4  33

(b) tan   cot   35



(c) sec   csc   534

460

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

(a) sin   cos   47

(b) tan   cot   4

(c) sec   csc   7

33

33

11. (a) sin 22  037461

(b) cot 23  235585

12. (a) cos 37  079864

(b) csc 48  134563

13. (a) sec 13  102630

(b) tan 51  123490

14. (a) csc 10  575877 x , we have x  25 sin 30  25  12  25 15. Since sin 30  2. 25  12 12 12 , we have x   1  12 2. 16. Since sin 45   x sin 45 

(b) sin 46  071934

2

  x 17. Since sin 60  , we have x  13 sin 60  13  23  132 3 . 13  4 4 4  1  4 3. 18. Since tan 30  , we have x  x tan 30  3

19. 20. 21. 22. 23.

12 12  1651658. Since tan 36  , we have x  x tan 36 25 25 Since sin 53  , we have x   3130339. x sin 53 y x  cos   x  28 cos , and  sin   y  28 sin . 28 28 4 4 x  tan   x  4 tan , and  cos   y   4 sec . 4 y cos    24. cos   12 tan   56 . Then the third side is x  52  62  61. 13 . The third side is     y  132  122  25  5. The other five ratios are The other five ratios are sin   5 6161 , cos   6 6161 , 



csc   561 , sec   661 , and cot   65 . Ï61 ¬

5 , tan   5 , csc   13 , sec   13 , and sin   13 12 5 12

cot   12 5. 13

5

5

¬ 12

6

25. cot   1. Then the third side is r 

  12  12  2. 

The other five ratios are sin   1  22 , 2   2 1 cos     2 , tan   1, csc   2, and 2  sec   2.

Ï2

¬ 1

1

26. tan  

  3. The third side is r  12  3  2. The other 

five ratios are sin   23 , cos   12 , csc   2 , 3 sec   2, and cot   1 . 3

2

Ï3

¬ 1

SECTION 5.2 Trigonometry of Right Triangles

27. csc   11 6 . The third side is x 

    112  62  85. The 28. cot   53 . The third side is x  52  32  34. The 





6 , cos   85 , other five ratios are sin   11 11

other five ratios are sin   3 3434 , cos   5 3434 ,

tan   6 8585 , sec   118585 , and cot   685 .

tan   35 , csc   334 and sec   534 .







11





Ï34

6

¬

¬ Ï85     1  3  1 3 29. sin   cos 6 6 2 2 2

30. sin 30 csc 30  sin 30 

461

1 1 sin 30



3

5



31. sin 30 cos 60  sin 60 cos 30  12  12  23  23  14  34  1   2  2 32. sin 60 2  cos 60 2  23  12  34  14  1

 2  12  34    3  1  1   sin  cos  2  34. sin  cos 3 4 4 3 2  2 2  18 4  2 3  33. cos 30 2  sin 30 2 

  2 3 2

1  1 4 2

2     2 2   1  12   31  18 3  1  18 3  2 3  1   2 2 1 2 3 4

 2      2  1 2  3  2  2  1  1 2 1  sin  35. cos  4 6 2 2 2 4 2 2 4 2   2  2       3 3  2 36. sin 3 tan   12  2  94  2 6  csc 4 2  3  2

37. This is an isosceles right triangle, so the other leg has length 16 tan 45  16, the hypotenuse has length  16  16 2  2263, and the other angle is 90  45  45 .  sin 45 100  10352, and the other angle is 38. The other leg has length 100 tan 75  2679, the hypotenuse has length sin 75    90  75  15 . 35  5685, and the other angle is 39. The other leg has length 35 tan 52  4480, the hypotenuse has length cos 52    90  52  38 .

40. The adjacent leg has length 1000 cos 68  37461, the opposite leg has length 1000 sin 68  92718, and the other angle is 90  68  22 .

 41. The adjacent leg has length 335 cos  8  3095, the opposite leg has length 335 sin 8  1282, and the other angle is     3 . 2 8 8

42. The opposite leg has length 723 tan  6  4174, the hypotenuse has length

723  8348, and the other angle is cos  6

    . 2 6 3

43. The adjacent leg has length     3 . 2 5 10

106 106  14590, the hypotenuse has length   18034, and the other angle is tan  sin 5 5

44. The adjacent leg has length 425 cos 38  16264, the opposite leg has length 425 sin 38  39265, and the other angle is   3   . 2 8 8

462

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

1  045 cos   2  089, tan   1 , csc   224, sec   224  112, cot   200. 45. sin   224 224 2 2 064     083, csc 40  156, sin 40  064, cos 40  077, tan 40  46. 077 sec 40  131, cot 40  120. 40¡

47. x 

100 100   2309 tan 60 tan 30

48. Let d be the length of the base of the 60 triangle. Then tan 60  dx 

85 85 x   d  981. tan 30 tan 30

85 85 85 d   49075, and so tan 30  d tan 60 dx

50 h 50 h h   57735  sin 65   x   637 h sin 60 x sin 65 5 5 h h    10, and so tan 60  50. Let h be the hypotenuse of the top triangle. Then sin 30  h sin 30 x 10 h   58. x tan 60 tan 60 y x , so x  y sin   10 sin  tan . 10 From the diagram, sin   and tan   51. y 10 49. Let h be the length of the shared side. Then sin 60 

¬

¹ -¬ 2

y

x

¬

a b 1 d  a  sin , tan    b  tan , cos    c  sec , cos    d  cos  1 1 c 1 h   h  5280  tan 11  1026 ft. 53. Let h be the height, in feet, of the Empire State Building. Then tan 11  5280 35,000 54. (a) Let r be the distance, in feet, between the plane and the Gateway Arch. Therefore, sin 22   r 35,000 r  93,431 ft. sin 22 (b) Let x be the distance, also in feet, between a point on the ground directly below the plane and the Gateway Arch. Then 35,000 35,000 x   86,628 ft. tan 22  x tan 22 h 55. (a) Let h be the distance, in miles, that the beam has diverged. Then tan 05   240,000 h  240,000  tan 05  2100 mi. 52. sin  

56. 57. 58. 59.

(b) Since the deflection is about 2100 mi whereas the radius of the moon is about 1000 mi, the beam will not strike the moon. 200 200 x   471 ft. Let x be the distance, in feet, of the ship from the base of the lighthouse. Then tan 23  x tan 23 h  h  20 sin 72  19 ft. Let h represent the height, in feet, that the ladder reaches on the building. Then sin 72  20 h  h  600 sin 65  544 ft. Let h be the height, in feet, of the communication tower. Then sin 65  600 h Let h be the height, in feet, of the kite above the ground. Then sin 50   h  450 sin 50  345 ft. 450

SECTION 5.2 Trigonometry of Right Triangles

60.

hÁ

18¡ x 14¡

hª

463

Let h 1 be the height of the flagpole above elevation and let h 2 be the height below, h as shown in the figure. So tan 18  1  h 1  x tan 18 . Similarly, x h 2  x tan 14 . Since the flagpole is 60 feet tall, we have h 1  h 2  60, so x tan 18  tan 14   60  x 

60  1045 ft. tan 18  tan 14

h 61. Let h 1 be the height of the window in feet and h 2 be the height from the window to the top of the tower. Then tan 25  1 325 h  h 1  325  tan 25  152 ft. Also, tan 39  2  h 2  325  tan 39  263 ft. Therefore, the height of the window 325 is approximately 152 ft and the height of the tower is approximately 152  263  415 ft. 35¡

52¡

62.

52¡

5150

dÁ

Let d1 be the distance, in feet, between a point directly below the plane and one car, and d2 be the distance, in feet, between the same point and the other car. Then

dª

35¡

tan 52 

5150 5150 5150  d1   402362 ft, and tan 35   d1 tan 52 d2

5150  735496 ft. So the distance between the two cars is about d1  d2  402362  735496  11,379 ft. tan 35 63. Let d1 be the distance, in feet, between a point directly below the plane and one car, and d2 be the distance, in feet, between d d the same point and the other car. Then tan 52  1  d1  5150  tan 52  65917 ft. Also, tan 38  2  5150 5150 d2  5150  tan 38  40236 ft. So in this case, the distance between the two cars is about 2570 ft. d2 

64. Let x be the distance, in feet, between a point directly below the balloon and the first mile post. Let h be the height, h h in feet, of the balloon. Then tan 22  and tan 20  . So h  x tan 22  x  5280 tan 20  x x  5280 5280  tan 20  47,977 ft. Therefore h  47,9769  tan 22  19,384 ft  37 mi. x tan 22  tan 20 65. Let x be the horizontal distance, in feet, between a point on the ground directly below the top of the mountain and h the point on the plain closest to the mountain. Let h be the height, in feet, of the mountain. Then tan 35  x h 1000  tan 32    and tan 32  . So h  x tan 35  x  1000 tan 32  x   82942. Thus x  1000 tan 35  tan 32 h  82942  tan 35  5808 ft. Since the angle of elevation from the observer is 45 , the distance from the

66.

observer is h, as shown in the figure. Thus, the length of the leg in the smaller right h 75¡ h

ã600ã

h  600  h tan 75  h  600  h 600 tan 75  473 m. 600 tan 75  h 1  tan 75   h  1  tan 75 triangle is 600  h. Then tan 75 

600-h

67. Let d be the distance, in miles, from the earth to the sun.

Then sec 8985 

d  240,000

d  240,000  sec 8985  917 million miles.  68. (a) s  r    rs  6155 3960  15543 rad  8905

(b) Let d represent the distance, in miles, from the center of the earth to the moon. Since cos   d

3960 3960   239,9615. So the distance AC is 239,9615  3960  236,000 mi. cos  cos 8905

3960 , we have d

464

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

r 69. Let r represent the radius, in miles, of the earth. Then sin 60276  r600  r  600 sin 60276  r  sin 60276 600 sin 60276  r 1  sin 60276   r  600 1sin 60276  3960099. So the earth’s radius is about 3960 mi.

,000 , we have 70. Let d represent the distance, in miles, from the earth to Alpha Centauri. Since sin 0000211  93,000 d 93,000,000 13 d  sin 0000211  25,253,590,022,410. So the distance from the earth to Alpha Centauri is about 253  10 mi.

d  d, so d  sin 463  0723 AU. 1 72. If two triangles are similar, then their corresponding angles are equal and their corresponding sides are proportional. That       is, if triangle ABC is similar to triangle A B  C  then AB  r  A B  , AC  r  A C  , and BC  r  B  C  . Thus when we express any trigonometric ratio of these lengths as a fraction, the factor r cancels out.

71. Let d be the distance, in AU, between Venus and the sun. Then sin 463 

5.3

TRIGONOMETRIC FUNCTIONS OF ANGLES

1. If the angle  is in standard position and P x y is a point on the terminal side of , and r is the distance from the origin to x y y P, then sin   , cos   , and tan   . r r x 2. The sign of a trigonometric function of  depends on the quadrant in which the terminal angle of  lies. For example, if  is in quadrant II, sin  is positive. In quadrant III, cos  is negative. In quadrant IV, sin  is negative. 3. (a) If  is in standard position, then the reference angle  is the acute angle formed by the terminal side of  and the x-axis. So the reference angle for   100 is   80 and that for   190 is   10 .

(b) If  is any angle, the value of a trigonometric function of  is the same, except possibly for sign, as the value of the trigonometric function of . So sin 100  sin 80 and sin 190   sin 10 .

4. The area A of a triangle with sides of lengths a and b and with included angle  is given by the formula A  12 ab sin . So the area of the triangle with sides 4 and 7 and included angle   30 is 12 4 7 sin 30  7.

5. (a) The reference angle for 120 is 180  120  60 .

6. (a) The reference angle for 175 is 180  175  5 .

(b) The reference angle for 200 is 200  180  20 .

(b) The reference angle for 310 is 360  310  50 .

(c) The reference angle for 285 is 360  285  75 .

(c) The reference angle for 730 is 730  720  10 .

7. (a) The reference angle for 225 is 225  180  45 . (b) The reference angle for 810 is 810  720  90 . (c) The reference angle for 105 is 180  105  75 .

 is   7  3 . 9. (a) The reference angle for 710 10 10

8. (a) The reference angle for 99 is 180  99  81 . (b) The reference angle for 199 is 199  180  19 .

(c) The reference angle for 359 is 360  359  1 . 10. (a) The reference angle for 56 is   56   6.

(b) The reference angle for 98 is 98     8.

(b) The reference angle for 109 is 109     9.

(c) The reference angle for 103 is 103  3   3.

(c) The reference angle for 237 is 237  3  27 .

11. (a) The reference angle for 57 is   57  27 .

12. (a) The reference angle for 23 is 23  2  03.

(b) The reference angle for 14 is 14    04.

(b) The reference angle for 23 is   23  084.

(c) The reference angle for 14 is 14 because 14   2.

(c) The reference angle for 10 is 10  10  0.

 13. cos 150   cos 30   23



14. sin 240   sin 60   23

SECTION 5.3 Trigonometric Functions of Angles  15. tan 330   tan 30   33  17. cot 120   cot 60  33

16. sin 30    sin 30   12 

18. csc 300   csc 60   2 3 3

 3

19. csc 630   csc 90  sin190  1

20. cot 210  cot 30  tan130 

21. cos 570   cos 30   23

22. sec 120   sec 60   cos160  2



 23. tan 750  tan 30  1  33 3

25. sin 32   sin  2  1    27. tan  43   tan  3  3     29. csc  56   csc  6  2

465

24. cos 660  cos 60  12 1 26. cos 43   cos  3  2    3 28. cos  116  cos   6 2 

2 3 30. sec 76   sec  6  3

1 31. sec 173  sec  3  cos   2 3   1 33. cot  4   cot  4  tan   1 4

32. csc 54   csc  4 

 1  2 sin  4 

2 1 34. cos 74  cos  4  2  2

1 35. tan 52  tan  36. sin 116   sin  2 which is undefined. 6  2 37. Since sin   0 and cos   0,  is in quadrant III. 38. Since both tan  and sin  are negative,  is in quadrant IV. sin   0  sin   0 (since cos   0). Since sin   0 and cos   0,  is 39. sec   0  cos   0. Also tan   0  cos  in quadrant IV. 40. Since csc   0  sin   0 and cos   0,  is in quadrant II.   sin  1  cos2  2  . 41. Since sin  is negative in quadrant III, sin    1  cos  and we have tan   cos  cos    1  sin2  cos   because cos   0 in quadrant II. 42. cot   sin  sin   43. cos2   sin2   1  cos   1  sin2  because cos   0 in quadrant IV. 1 1 44. sec    because all trigonometric functions are positive in quadrant I. cos  1  sin2   45. sec2   1  tan2   sec    1  tan2  because sec   0 in quadrant II.  46. csc2   1  cot2   csc    1  cot2  because csc   0 in quadrant III.   47. sin    45 . Then x   52  42   9  3, since  is in quadrant IV. Thus, cos    35 , tan    43 , csc    54 ,

sec   53 , and cot    34 .  48. tan   43 , so r  42  32  5. Since  is in quadrant III, x  3 and y  4, and so sin    45 , cos    35 , csc    54 , sec    53 , and cot   34 .       2 7 95 7 7 49. cos   12 , so sin    1  12   1295 , tan    795 , csc    129595 , sec   12 , and cot    7 95      9 145 8 145 8 9 9 50. cot    9 , so r  82  92  145. Thus, sin       145 , cos   145 , tan    8 , csc    145 9 ,  and sec   145 8 .

51. csc   2. Then sin   12 and x   and cot   3.

145

     22  12  3. So sin   12 , cos   23 , tan   1  33 , sec   2  2 3 3 , 3 3

466

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

   52. cot   14 . Then tan   4 and r  42  12  17. So sin    4 , cos    1 , tan   4, csc    417 , and 17 17  sec    17.       7  7 5 , sec    7 , 53. cos    27 . Then y  72  22  45  3 5, and so sin   3 7 5 , tan    3 2 5 , csc    15 2 3 5

 2  2 5. and cot     15 3 5

54. tan   4. Then r  cot    14 .

    42  12  17, and so sin   4 , cos    1 , csc   417 , sec    17, and 17

17

  2  3 and 2 sin   2 sin   3. 55. If    , then sin 2  sin 3 3 2 3     2   sin  2  3 and sin  2  sin  2  0890. 56. If    , then sin 3 3 4 9

57. The area is 12 7 9 sin 72  300.

58. The area is 12 10 22 sin 10  191.  59. The area is 12 102 sin 60  25 3  433. 

60. The area is 12 132 sin 60  1694 3  732.

  1 32  1154, or 61. Let the angle be . Then A  12 x y sin   16  12 5 7 sin   sin   32 35    sin 35 approximately 661 .    48 1 1 5  2 2 62. Let the lengths of the equal sides be x. Then A  2 x sin   24  2 x sin 6  x   96  4 6  98 cm. 5  sin 6

  4 . For the triangle defined by the two sides, 63. For the sector defined by the two sides, A1  12 r 2   12  22  120  180  3   A2  12 ab sin   12  2  2  sin 120  2 sin 60  3. Thus the area of the region is A1  A2  43  3  246.

64. The area of the entire circle is r 2    122  144, the area of the sector is 12 r 2   12  122   3  24,  1 1  2 and the area of the triangle is 2 ab sin   2  12  sin 3  36 3, so the area of the shaded region is   A1  A2  A3  144  24  36 3  120  36 3  4393.

65. (a) tan  

5280 ft h , so h  tan   1 mile   5280 tan  ft. 1 mile 1 mile

(b) 

20

60

80

85

h

1922

9145

29,944

60,351

66. (a) Let the depth of the water be h. Then the cross-sectional area of the gutter is a trapezoid whose height is h  10 sin . The bases are 10 and 10  2 10 cos   10  20 cos . Thus, the area is b  b2 10  10  20 cos  A   1 h   10 sin   100 sin   100 sin  cos . 2 2 (b)

(c) 200

130.0

100

129.5

0 0

1

129.0 1.00

1.05

1.10

From the graph, the largest area is achieved when   1047 rad  60 .

SECTION 5.3 Trigonometric Functions of Angles

67. (a) From the figure in the text, we express depth and width in terms of .

467

(b)

width depth and cos   , we have depth  20 sin  Since sin   20 20 and width  20 cos . Thus, the cross-section area of the beam is A   depth width  20 cos  20 sin   400 cos  sin .

200

0 0

(c) The beam with the largest cross-sectional area is the square beam,   10 2 by 10 2 (about 1414 by 1414).

1

68. Using depth  20 sin  and width  20 cos , from Exercise 65, we have strength  k width depth2 . Thus S   k 20 cos  20 sin 2  8000k cos  sin2 .

69. (a) On Earth, the range is R 

 122 sin   02 sin 2 3  9 3  3897 ft and the height is  g 32 4

122 sin2   2 sin2  6  9  05625 ft. H 0  2g 2  32 16

2 2  122 sin  3  23982 ft and H  12 sin 6  3462 ft (b) On the moon, R  52 2  52   2000 70. Substituting, t   5 10  158 s. 16 sin 30

71. (a) W  302  038 cot   065 csc 

10

(b) From the graph, it appears that W has its minimum value at about   0946  542 .

5 0 0

72. (a) We label the lengths L 1 and L 2 as shown in the

¬ 6 Lª

9 6 figure. Since sin   and cos   , we L1 L2 ¬

have L 1  9 csc  and L 2  6 sec . Thus L   L 1  L 2  9 csc   6 sec .

(c) The minimum value of L  is 2107. (d) The minimum value of L  is the shortest distance the pipe must pass through.

LÁ ¬

2

(b) 40 20 0 0.0

0.5

1.0

1.5

9

73. We have sin   k sin , where   594 and k  133. Substituting, sin 594  133 sin  sin 594  06472. Using a calculator, we find that   sin1 06472  403 , so  sin   133   4  2  4 403   2 594   424 . 74. sin 4  00697564737 but sin 4  07568024953. Your partner has found sin 4 instead of sin 4 radians. O P O P adj    O P. Since QS is tangent to the circle at R, O R Q is a right triangle. Then 75. cos   O R hyp 1 R Q O Q hyp opp  R Q and sec    O Q. Since  SO Q is a right angle S O Q is a right   tan   O R O R adj adj O S S R adj hyp  O S and cot    S R. Summarizing, we have   triangle and  O S R  . Then csc   O R O R opp opp sin   P R, cos   O P, tan   R Q, sec   O Q, csc   O S, and cot   S R.

468

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

  76. (a) sin2   cos2   1  sin2   cos2  

1 1 1  tan2   1  sec2  2 cos  cos2    1 1 (b) sin2   cos2   1  sin2   cos2   1  1  cot2   csc2  sin2  sin2 

77. Let x represent the desired real number, in radians. We wish to find the smallest nonzero solution of the equation sin x  sin 180x  . (Since x is in radians, 180x  is in degrees.) Now if a is in degrees with 0  a  180 and sin a  b, then either a  b or 180  a  b. The graph shows that the first positive solution occurs when

y 1

180x y=sin ¹ y=sin x

0

x

2

_1 180x 180x  180, so we have 180   x.   180 180  3088. You can verify that the value of sin is the same whether Solving this equation, we get x  180   180   your calculator is set to radian mode or degree mode.

0

5.4

INVERSE TRIGONOMETRIC FUNCTIONS AND TRIANGLES

1. For a function to have an inverse, it must be one-to-one. To define the inverse sine function we restrict the domain of the   sine function to the interval   2 2 .   2. (a) The function sin1 has domain [1 1] and range   2 2 . (b) The function cos1 has domain [1 1] and range [0 ].   (c) The function tan1 has domain R and range   2 2 .

8  sin1 4 6  cos1 3 (b)   cos1 10 3. (a)   sin1 10 5 5   5 , we let   cos1 5 and 4. To find sin cos1 13 13

complete the right triangle as shown. We find that   5  12 . sin cos1 13 13

(c)   tan1 68  tan1 43

13 ¬

12

5

 3  3

5. (a) sin1 1   2

(b) cos1 0   2

(c) tan1

6. (a) sin1 0  0    7. (a) sin1  22    4

(b) cos1 1      (b) cos1  22  34

(c) tan1 0  0

   8. (a) sin1  23    3 7. sin1 030  030469 13. tan1 3  124905

  (b) cos1  12  23

(c) tan1 1    4

   (c) tan1  3    3

  10. cos1 02  177215 11. cos1 13  123096

14. tan1 4  132582 15. cos1 3 is undefined.

12. sin1 65  098511

16. sin1 2 is undefined.

6  3 , so   sin1 3  369 . 17. sin   10 5 5

7 , so   tan1 7  213 . 18. tan   18 18

9 , so   tan1 9  347 . 19. tan   13 13

3 1 3  254 . 20. sin   30 7 70  7 , so   sin

21. sin   47 , so   sin1 47  348 .

22. cos   89 , so   cos1 98  273 .

SECTION 5.4 Inverse Trigonometric Functions and Triangles

469

  23. We use sin1 to find one solution in the interval 90  90 . sin   23    sin1 23  418 . Another solution with  between 0 and 180 is obtained by taking the supplement of the angle: 180  418  1382 . So the solutions of the equation with  between 0 and 180 are approximately   418 and   1382 .   24. One solution is given by   cos1 43  414 . This is the only solution, because cos x is one-to-one on 0 180 .

    25. One solution is given by   cos1  25  1136 . This is the only solution, because cos x is one-to-one on 0 180 .   26. tan1 20  871 , so the only solution in 0 180 is approximately 180  871  929 .   27. tan1 5  787 . This is the only solution on 0 180 .

    28. One solution is   sin1 45  531 . Another solution on 0 180 is approximately 180  531  1269 .   29. To find cos sin1 45 , first let   sin1 54 . Then  is the number in the interval     2  2 whose sine is 45 . We draw a right triangle with  as one of its acute

5

angles, with opposite side 4 and hypotenuse 5. The remaining leg of the triangle is

¬

found by the Pythagorean Theorem to be 3. From the figure we get   cos sin1 54  sin   35 .

4

3

    Another method: By the cancellation properties of inverse functions, sin sin1 54 is exactly 45 . To find cos sin1 45 , we

first write the cosine function in terms of the sine function. Let u  sin1 45 . Since 0  u   2 , cos u is positive, and since       2   9 3 cos2 u  sin2 u  1, we can write cos u  1  sin2 u  1  sin2 sin1 45  1  45  1  16 25  25  5 .   Therefore, cos sin1 54  35 .   30. To find cos tan1 54 , we draw a right triangle with angle , opposite side 4, and   adjacent side 3. From the figure we see that cos tan1 34  cos   35 .

  12 , we draw a right triangle with 31. To find sec sin1 13

angle , opposite side 12, and hypotenuse 13. From the   12  sec   13 . figure we see that sec sin1 13 5 13 12

¬

5

5 ¬

4

3

  7 , we draw a right triangle with 32. To find csc cos1 25

angle , opposite side 7, and hypotenuse 25. From the   7  csc   25 . figure we see that csc cos1 25 24 25

¬

24

7

470

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

    1 2 , we draw a right triangle with angle , we draw a right triangle with angle 33. To find tan sin1 12 34. To find cot sin 13 3 , opposite side 12, and hypotenuse 13. From the figure   12  tan   12 . we see that tan sin1 13 5 13

¬

, opposite side 2, and hypotenuse 3. From the figure we    see that cot sin1 32  cot   25 .

5

12

3

2

¬ Ï5

  35. We want to find cos sin1 x . Let   sin1 x, so sin   x. We sketch a right triangle with an acute angle , opposite side x, and hypotenuse 1. By the  Pythagorean Theorem, the remaining leg is 1  x 2 . From the figure we have    cos sin1 x  cos   1  x 2 .

1

x

¬ Ï1-x@

Another method: Let u  sin1 x. We need to find cos u in terms of x. To do so, we write cosine in terms of sine. Note     1 x. Now cos u  1  sin2 u is positive because u lies in the interval     . that   2  u  2 because u  sin 2 2  Substituting u  sin1 x and using the cancellation property sinsin1 x  x gives cossin1 x  1  x 2 .   36. We want to find sin tan1 x . Let   tan1 x, so tan   x. We sketch a right Ïx@+1

triangle with an acute angle , opposite side x, and adjacent side 1. By the  Pythagorean Theorem, the remaining leg is x 2  1. From the figure we have   x . sin tan1 x  sin    x2  1   37. We want to find tan sin1 x . Let   sin1 x, so sin   x. We sketch a right

triangle with an acute angle , opposite side x, and hypotenuse 1. By the  Pythagorean Theorem, the remaining leg is 1  x 2 . From the figure we have   x . tan sin1 x  tan    1  x2   38. We want to find cos tan1 x . Let   tan1 x, so tan   x. We sketch a right

¬

x

1

1

x

¬ Ï1-x@

Ïx@+1

x triangle with an acute angle , opposite side x, and adjacent side 1. By the  ¬ Pythagorean Theorem, the remaining leg is x 2  1. From the figure we have 1   1 1 cos tan x  cos    . x2  1 39. Let  represent the angle of elevation of the ladder. Let h represent the height, in feet, that the ladder reaches on the

6  03    cos1 03  1266 rad  725 . By the Pythagorean Theorem, h 2  62  202  building. Then cos   20   h  400  36  364  19 ft.

96  08    tan1 08  0675  387 . 40. Let  be the angle of elevation of the sun. Then tan   120 41. (a) Solving tan   h2 for h, we have h  2 tan .

(b) Solving tan   h2 for  we have   tan1 h2.

42. (a) Solving tan   50s. Solving for , we have   tan1 50s.

5 1 5  682 . (b) Set s  20 ft to get tan   50 20  2 . Solving for , we have   tan 2

SECTION 5.5 The Law of Sines

471

43. (a) Solving sin   h680 for  we have   sin1 h680.    (b) Set h  500 to get   sin1 500 680  0826 rad  473 .

3960 . Solving for , we get 44. (a) Since the radius of the earth is 3960 miles, we have the relationship cos   h3960   3960 .   cos1 h3960

(b) The arc length is s  radius included angle  3960 2  7920.   3960 (c) s  7920 cos1 h3960     3960 (d) When h  100 we have s  7920 cos1 1003960  7920 cos1 3960 4060  17615 miles.       3960 3960 3960 2450 1 (e) When s  2450 we have 2450  7920 cos1 h3960  2450 7920  cos h3960  h3960  cos 7920      2450 h  3960  3960 sec 2450 7920  h  3960 sec 7920  3960  1973 mi.     1 1 1 1  sin  sin1 08102  541 45. (a)   sin 7 tan 10 2  3  1 tan 10     1 1  . For n  3,   sin1  483  322 . For n  4, (b) For n  2,   sin1 5 tan 15 7 tan 15   1 1 1  245 . n  0 and n  1 are outside of the domain for   15 , because  3732   sin 9 tan 15 tan 15 1 and  1244, neither of which is in the domain of sin1 . 3 tan 15 1 46. Let   sec1 x. Then sec   x, as shown in the figure. Then cos   , so x     x 1 1 . Thus, sec1 x  cos1 , x  1.   cos1 x x In particular, sec1 2  cos1 12   3.

1 Let   csc1 x. Then csc   x, as shown in the figure. Then sin   , so x     1 1   sin1 . Thus, csc1 x  sin1 , x  1. x x In particular, csc1 3  sin1 31  0340.

1 Let   cot1 x. Then cot   x, as shown in the figure. Then tan   , so x     1 1   tan1 . Thus, cot1 x  tan1 , x  1. x x In particular, cot1 4  tan1 41  0245.

5.5

THE LAW OF SINES

sin B sin C sin A   . a b c 2. (a) The Law of Sines can be used to solve triangles in cases ASA or SSA.

1. In triangle ABC with sides a, b, and c the Law of Sines states that

(b) The Law of Sines can give ambiguous solutions in case SSA. 376 sin 57  31875. 3.  C  180  984  246  57 . x  sin 984

¬

1

x

1

º

1 

x

472

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

17 sin 1144 4.  C  180  375  281  1144 . x   254. sin 375 267 sin 52 5.  C  180  52  70  58 . x   248. sin 58 6. sin  

563 sin 67  0646. Then   sin1 0646  403 . 802

7. sin C 

36 sin 120  0693   C  sin1 0693  44 . 45

185 sin 50  1449. 8.  C  180  102  28  50 . x  sin 102 65 sin 46 65 sin 20 9.  C  180  46  20  114 . Then a   51 and b   24.  sin 114 sin 114 2 sin 100 2 sin 30  257 and a   131. 10.  B  180  30  100  50 . Then c   sin 50 sin 50 12 sin 44 11.  B  68 , so  A  180  68  68  44 and a   899. sin 68 12. sin B 

34 sin 80 65 sin 69  0515, so  B  sin1 0515  31 . Then  C  180 80 31  69 and c   62. 65 sin 80

13.  C  180  50  68  62 . Then 230 sin 50 230 sin 68 a  200 and b   242.  sin 62 sin 62

14.  C  180  110  23  47 . Then 50 sin 23 50 sin 110 a  267 and b   642.  sin 47 sin 47 C

C

50¡

A

68¡ 230

23¡

A

110¡ 50

B

B

15.  B  180  30  65  85 . Then 10 sin 65 10 sin 30  50 and c   9. a  sin 85 sin 85

16.  C  180  95  22  63 . Then 420 sin 63 420 sin 95  11169 and c   9990. b  sin 22 sin 22

C 10

65¡

30¡

A

C 420 22¡

A

B

17.  A  180  51  29  100 . Then 44 sin 51 44 sin 100  89 and c   71. a  sin 29 sin 29

95¡

18.  A  180  100  10  70 . Then 115 sin 10 115 sin 70  1097 and b   203. a  sin 100 sin 100 C

C

A

44 A

10¡

100¡

51¡

29¡

B

B

115

B

SECTION 5.5 The Law of Sines

473

15 sin 110  0503   B  sin1 0503  30 . Then 19. Since  A  90 there is only one triangle. sin B  28 28 sin 40  C  180  110  30  40 , and so c   19. Thus  B  30 ,  C  40 , and c  19. sin 110 40 sin 37 20. sin C   0802   C1  sin1 0822  534 or  C2  180  534  1266 . 30 30 sin 896 If  C1  534 , then  B1  180  37  534  896 and b1   498. sin 37 30 sin 164 If  C2  1266 , then  B2  180  37  1266  164 and b2   141. sin 37 Thus, one triangle has  B1  896 ,  C1  534 , and b1  498; the other has  B2  164 ,  C2  1266 , and b2  141. 21.  A  125 is the largest angle, but since side a is not the longest side, there can be no such triangle. 45 sin 38 22. sin B   0660   B1  sin1 0660  413 or  B2  180  413  1387 . 42 42 sin 1007  67. If  B1  413 , then  A1  180  38  413  1007 and a1  sin 38  42 sin 33 If  B2  1387 , then  A2  180  38  1387  33 and a2   39. sin 38   Thus, one triangle has  A1  1007 ,  B1  413 , and a1  67; the other has  A2  33 ,  B2  1387 , and a2  39. 30 sin 25 23. sin C   0507   C1  sin1 0507  3047 or  C2  180  3947  14953 . 25 25 sin 12453  4873. If  C1  3047 , then  A1  180  25  3047  12453 and a1  sin 25  25 sin 547 If  C2  14953 , then  A2  180  25  14953  547 and a2   564. sin 25 Thus, one triangle has  A1  125 ,  C1  30 , and a1  49; the other has  A2  5 ,  C2  150 , and a2  56. 100 sin 30 24. sin B   23   B1  sin1 32  418 or  B2  180  418  1382 . 75 75 sin 1082  1425. If  B1  418 , then  C1  180  30  418  1082 and c1  sin 30 75 sin 118 If  B2  1382 , then  C2  180  30  1382  118 and c2   307. sin 30 Thus, one triangle has  B1  418 ,  C1  1082 , and c1  1425; the other has  B2  1382 ,  C2  118 , and c2  307. 100 sin 50  1532. Since sin   1 for all , there can be no such angle B, and thus no such triangle. 25. sin B  50  80 sin 135 26. sin B   0566   B1  sin1 0566  344 or  B2  180  344  1456 . 100 100 sin 106 If  B1  344 , then  C  180  135  344  106 and c   259. sin 135 If  B2  180  344  1456 , then  A   B2  135  1456  180 , so there is no such triangle. Thus, the only possible triangle is  B  344 ,  C  106 , and c  259. 26 sin 29 27. sin A   0840   A1  sin1 0840  572 or  A2  180  572  1228 . 15 15 sin 938  309. If  A1  572 , then  B1  180  29  572  938 and b1  sin 29 15 sin 281  146. If  A2  1228 , then  B2  180  29  1228  282 and b2  sin 29 Thus, one triangle has  A1  572 ,  B1  938 , and b1  309; the other has  A2  1228 ,  B2  282 , and b2  146.

474

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

82 sin 58  0953, so  C1  sin1 0953  724 or  C2  180  724  1076 . 73 73 sin 496  656. If  C1  724  then  A1  180  58  724  496 and a1  sin 58 73 sin 144 If  C2  1076 , then  A2  180  58  1076  144 and a2   214. sin 58   Thus, one triangle has  A1  496 ,  C1  724 , and a1  656; the other has  A2  144 ,  C2  1076 , and a2  214.

28. sin C 

sin B 28 sin 30 sin 30   sin B   07, so 20 28 20  B  sin1 07  44427 . Since BC D is isosceles,  B   B DC  44427 . Thus,  BC D  180  2  B  91146  911 .

29. (a) From ABC and the Law of Sines we get

(b) From ABC we get  BC A  180   A   B  180  30  44427  105573 . Hence  DC A   BC A   BC D  105573  91146  144 . 30. By symmetry,  DC B  25 , so  A  180  25  50  105 . Then by the Law of Sines, AD 

12 sin 25  525. sin 105

31. (a) Let a be the distance from satellite to the tracking station A in miles. Then the subtended angle at the satellite is 50 sin 842  C  180  93  842  28 , and so a   1018 mi. sin 28 (b) Let d be the distance above the ground in miles. Then d  10183 sin 87  1017 mi. 32. (a) Let x be the distance from the plane to point A. Then x 5

sin 48  377 mi. sin 100

(b) Let h be the height of the plane. Then sin 32 

sin 48 x sin 48       AB sin 180  32  48  sin 100

h  h  377 sin 32  200 mi. x

AC AB AB sin 52   AC  , so substituting we 33.  C  180  82  52  46 , so by the Law of Sines,   sin 52 sin 46 sin 46 200 sin 52  219 ft. have AC  sin 46 312 sin 486  0444   ABC  sin1 0444  264 , and so  BC A  180  486  264  105 . 527 527 sin 105  6785 ft. Then the distance between A and B is AB  sin 486

34. sin  ABC 

35.

We draw a diagram. A is the position of the tourist and C is the top of the tower.

C

 B  90  56  844 and so  C  180  292  844  664 . Thus, by 105 sin 292

5.6¡

the Law of Sines, the length of the tower is BC 

29.2¡ B

C 165

The situation is illustrated in the diagram. 180

67¡ B

 559 m.

A

105

36.

sin 664

A

D

AC  165 sin 67  1519 ft, so using the Pythagorean Theorem  we can calculate B A  1652  15192  644 ft and  AD  1802  15192  966 ft. Thus the anchor points are B A  AD  644  966  161 ft apart.

SECTION 5.5 The Law of Sines

475

37. The angle subtended by the top of the tree and the sun’s rays is  A  180  90  52  38 . Thus the height of the tree 215 sin 30 is h   175 ft. sin 38 C

38.

Let x be the length of the wire, as shown in the figure. Since   12 , other angles in ABC are   90  58  148 , and   180  12  148   20 .



x

Œ

Thus,

100 sin 148 x   x  100   155 m. sin 148 sin 20 sin 20

º B 100

A 58¡

39. Call the balloon’s position R. Then in P Q R, we see that  P  62  32  30 , and  Q  180  71  32  141 . Q R P Q sin 30 Therefore,  R  180  30  141  9 . So by the Law of Sines,   Q R  60   192 m.   sin 30 sin 9 sin 9 B

40. º

30 C

Label the diagram as shown, and let the hill’s angle of elevation be . Then applying the Law of Sines to ABC,

sin  sin 8   120 30

sin   4 sin 8  055669    sin1 055669  338 . But from AB D,

 B AD   B    8     90 , so   90  8  338  482 . 8¡

A

Œ

120

D

41. Let d be the distance from the earth to Venus, and let  be the angle formed by sun, Venus, and earth. By the Law sin 394 sin    0878, so either   sin1 0878  614 or   180  sin1 0878  1186 . of Sines, 1 0723 0723 d   d  1119 AU; in the second case, In the first case, sin 180  394  614  sin 394 d 0723  d  0427 AU.  sin 180  394  1186  sin 394 sin 60 b sin 60 sin B  or sin B  . Similarly, applying the Law of b c c   sin B sin 120 r sin 120 r b Sines to BC D gives  or sin B  . Since sin 120  sin 60 , we have   r cd cd c cd   c sin D sin 60 b sin 60 b  (). Similarly, from ADC and the Law of Sines we have  or sin D  , and r cd b d d a sin 120 b sin 60 a sin 120 b d from B DC we have sin D  . Thus,    . Combining this with cd d cd a cd b c d cd b b ab r a b    1. Solving for r, we find  1      (), we get   r a cd cd cd r a a b ab ab . r ab

42. (a) Applying the Law of Sines to ABC, we get

476

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

43  12 cm. 43 (c) If a  b, then r is infinite, and so the face is a flat disk.

(b) r 

43. By the area formula from Section 5.3, the area of ABC is A  12 ab sin C. Because we are given a and the three sin A a sin B sin B  b  . Thus, angles, we need to find b in terms of these quantities. By the Law of Sines, b a sin A   2 a sin B sin C a sin B A  12 ab sin C  12 a sin C  . sin A 2 sin A 44. By the area formula from Section 5.3,

1 ab sin C sin C Area of  ABC  12  , because a and b are the same for both     Area of  A B C sin C  ab sin C 2

triangles. C

45. b

a

B

C b

B

A

a  b: One solution

a

C b

a»

B

B

b  a  b sin A: Two solutions

A

C b

a B

a  b sin A: One solution

A

a  b sin A: No solution

 A  30 , b  100, sin A  1 . If a  b  100 then there is one triangle. If 100  a  100 sin 30  50, then there are 2

two possible triangles. If a  50, then there is one (right) triangle. And if a  50, then no triangle is possible.

5.6

THE LAW OF COSINES

1. For triangle ABC with sides a, b, and c the Law of Cosines states c2  a 2  b2  2ab cos C.

2. The Law of Cosines is required to solve triangles in cases SSS and SAS.

 3. x 2  212  422  2  21  42  cos 39  441  1764  1764 cos 39  834115 and so x  834115  289.  4. x 2  152  182  2  15  18  cos 108  225  324  540 cos 108  715869 and so x  715869  268.  5. x 2  252  252  2  25  25  cos 140  625  625  1250 cos 140  2207556 and so x  2207556  47.  6. x 2  22  82  2  2  8  cos 88  4  64  32 cos 88  66883 and so x  66883  82. 7. 37832  68012  42152  2  6801  4215  cos . Then cos     cos1 0867  2989 . 8. 15462  6012  12252  2  601  1225  cos . Then cos     cos1 0359  111 .

37832  68012  42152  0867  2  6801  4215

15462  6012  12252  0359  2  601  1225

 9. x 2  242  302  2  24  30  cos 30  576  900  1440 cos 30  228923 and so x  228923  15.

156 202  102  122   065    cos1 065  13054 . 2  10  12 240  11. c2  102  182  2  10  18  cos 120  100  324  360 cos 120  604 and so c  604  24576. Then 18 sin 120 sin A   0634295   A  sin1 0634295  394 , and  B  180  120  394  206 . 24576

10. 202  102  122  2  10  12  cos . Then cos  

12. 122  402  442  2  40  44  cos B  cos B  sin A 

122  402  442  0964   B  cos1 0964  15 . Then 2  40  44

40 sin 155  0891   A  sin1 0891  63 , and so  C  180  15  63  102 . 12

SECTION 5.6 The Law of Cosines

477

 4 sin 53  0983 13. c2  32  42  2  3  4  cos 53  9  16  24 cos 53  10556  c  10556  32. Then sin B  325   B  sin1 0983  79 and  A  180  53  79  48 . 14. a 2  602  302  2  60  30  cos 70  3600  900  3600 cos 70  326873  30 sin 70  a  326873  572. Then sin C   0493  572  C  sin1 0493  295 , and  B  180  70  295  805 .

2 252 222 15. 202  252  222  2  25  22  cos A  cos A  20 22522  0644   A  cos1 0644  50 . Then  sin B  25 sin20499  0956   B  sin1 0956  73 , and so  C  180  50  73  57 . 2 122 162  078125  16. 102  122  162  2  12  16  cos A  cos A  10 21216   A  cos1 078125  386 . Then sin B  12 sin 386  0749  10

 B  sin1 0749  485 , and so  C  180  386  485  929 .

sin 40  0833   C  sin1 0833  564 or  C  180  564  1236 . 17. sin C  162125 1 2 sin 836  193. If  C1  564 , then  A1  180  40  564  836 and a1  125sin 40 125 sin 164  549.        If C2  1236 , then A2  180  40  1236  164 and a2  sin 40

Thus, one triangle has  A  836 ,  C  564 , and a  193; the other has  A  164 ,  C  1236 , and a  549.

52  1024. Since sin   1 for all , there is no such  A, and hence there is no such triangle. 18. sin A  65 sin 50

55  1065. Since sin   1 for all , there is no such  B, and hence there is no such triangle. 19. sin B  65 sin 50 sin 61  1094 and c  735 sin 83  1241. 20.  A  180  61  83  36 . Then b  735 sin 36 sin 36 sin 35  2. 21.  B  180  35  85  60 . Then x  3sin 60

 22. x 2  10  18  2  10  18  cos 40  100  324  360 cos 40  148224 and so x  148224  122. 

sin 30 23. x  50 sin 100  254

2 102 112 1 0932  213 .  205 24. 42  102  112  2  10  11  cos . Then cos   4 21011 220  0932    cos

25. b2  1102  1382  2 110 138  cos 38  12,100  19,044  30,360 cos 38  72200 and so b  850. Therefore, 2 852 1282 using the Law of Cosines again, we have cos   1102110138

   8915 .

40  0803    sin1 0803  535 or   180  535  1265 , but 535 doesn’t fit the 26. sin   10 sin 8 picture, so   1265 .

27. x 2  382  482  2  38  48  cos 30  1444  2304  3648 cos 30  588739 and so x  243. sin 98  11808. 28.  A  180  98  25  57 . Then x  1000 sin 57

 18, so by Heron’s Formula the area is 29. The semiperimeter is s  91215 2   A  18 18  9 18  12 18  15  2916  54.  52 , so by Heron’s Formula the area is 30. The semiperimeter is s  122 2        15 A  52 52  1 52  2 52  2  15 16  4  0968.

 12, so by Heron’s Formula the area is 31. The semiperimeter is s  789 2    A  12 12  7 12  8 12  9  720  12 5  268.

32. The semiperimeter is s  11100101  106, so by Heron’s Formula the area is 2    A  106 106  11 106  100 106  101  302,100  10 3021  550.

478

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

33. The semiperimeter is s  346  13 2 2 , so by Heron’s Formula the area is        13  3 13  4 13  6  455  455  533. A  13 2 2 2 2 16 4  6, so by Heron’s Formula the 34. Both of the smaller triangles have the same area. The semiperimeter of each is s  255 2     area of each is A  6 6  2 6  5 6  5  24  2 6, so the shaded area is 4 6  980. 35. We draw a diagonal connecting the vertices adjacent to the 100 angle. This forms two triangles. Consider the triangle with sides of length 5 and 6 containing the 100 angle. The area of this triangle is A1  12 5 6 sin 100  1477. To use Heron’s Formula to find the area of the second triangle, we need to find the length of the diagonal using the Law of Cosines: c2  a 2  b2  2ab cos C  52  62  2  5  6 cos 100  71419  c  845. Thus the second triangle has semiperimeter  8  7  845  117255 and area A2  117255 117255  8 117255  7 117255  845  2600. The area s 2 of the quadrilateral is the sum of the areas of the two triangles: A  A1  A2  1477  2600  4077. 36. We draw a line segment with length x bisecting the 60 angle to create two triangles. By the Law of Cosines,    32  42  x 2  2  4x cos 30  x 2  4 3x  7  0. Using the Quadratic Formula, we find x  2 3  5. The minus sign provides the correct length of about 123 (the other solution is about 57, which corresponds to a convex quadrilateral   342 3 5 with the same side lengths), so the semiperimeter of each triangle is s  and the total area of the figure 2      is A  2 s s  3 s  4 s  2 3  5  246. 37.

Label the centers of the circles A, B, and C, as in the figure. By the Law of C

6

5

6

B 5

4

4 A

Cosines, cos A 

92  102  112 AB 2  AC 2  BC 2   13   A  7053 . 2 AB AC 2 9 10

Now, by the Law of Sines,

sin 7053 sin B sin C   . So 11 AC AB

 1 085710  5899 and sin B  10 11 sin 7053  085710  B  sin 9 sin 7053  077139  C  sin1 077139  5048 . The area of sin C  11

ABC is 12 AB AC sin A  12 9 10 sin 7053   42426.   2  7053  9848. Similarly, the areas of sectors B and C   The area of sector A is given by S A   R 2  4 360 360 are S B  12870 and SC  15859. Thus, the area enclosed between the circles is A  ABC  S A  S B  SC  A  42426  9848  12870  15859  385 cm2 .

 38. By the Law of Cosines we have a 2  62  42  2 6 4 cos 45  52  24 2,     b2  62 x 2 2 6 x cos 30  x 2 6 3x36, and c2  x 2 42 2 x 4 cos 30  45   x 2 2 2  6 x16.        By the Pythagorean Theorem, a 2  b2  c2 , so we have 52  24 2  x 2  6 3x  36  x 2  2 2  6 x  16       12 3  2          26  12 2  3 3x  18  2  6 x  8  36  12 2  3 3  2  6 x, so x     . 3 3 2 6 39. Let c be the distance across the lake, in miles. Then c2  2822  3562  2 282 356  cos 403  5313  c  230 mi.

SECTION 5.6 The Law of Cosines D

40. Suppose ABC D is a parallelogram with AB  DC  5, AD  BC  3, and  A  50 (see the figure). Since opposite angles are equal in a

parallelogram, it follows that  C  50 , and

260  B   D  360  100  260 . Thus,  B   D   130 .

479

C 3

50¡l

A

B

5

2  By the Law of Cosines, AC 2  32  52  2 3 5  cos 130  AC  9  25  30  cos 130  73. Similarly,  B D  32  52  2 3 5  cos 50  38. 41. In half an hour, the faster car travels 25 miles while the slower car travels 15 miles. The distance between them is given by the Law of Cosines: d 2  252  152  2 25 15  cos 65    d  252  152  2 25 15  cos 65  5 25  9  30  cos 65  231 mi. 42. Let x be the car’s distance from its original position. Since the car travels

x

at a constant speed of 40 miles per hour, it must have traveled 40 miles

º 45¡

east, and then 20 miles northeast (which is 45 east of “due north”). From 40

the diagram, we see that    135 , so   x  202  402  2 20 40  cos 135  10 4  16  16  cos 135  560 mi. 43. The pilot travels a distance of 625  15  9375 miles in her original

direction and 625  2  1250 miles in the new direction. Since she makes a course correction of 10 to the right, the included angle is

20

10¡

937.5

1250

d

180  10  170 . From the figure, we use the Law of Cosines to get

the expression d 2  93752  12502  2 9375 1250  cos 170  4,749,54942, so d  2179 miles. Thus, the pilot’s distance from her original position is approximately 2179 miles.

44. Let d be the distance between the two boats in miles. After one hour, the boats have traveled distances of 30 miles and 26 miles. Also, the angle subtended by their directions is 180  50  70  60 . Then  d 2  302  262  2  30  26  cos 60  796  d  796  282. Thus the distance between the two boats is about 28 miles. 45. (a) The angle subtended at Egg Island is 100 . Thus using the Law of

Forrest Island

Cosines, the distance from Forrest Island to the fisherman’s home port is x 2  302  502  2  30  50  cos 100  900  2500  3000 cos 100  3920945

10¡

 and so x  3920945  6262 miles.

S 182 E.

x

80¡

(b) Let  be the angle shown in the figure. Using the Law of Sines, 50 sin 100  07863    sin1 07863  518 . Then sin   6262   90  20  518  182 . Thus the bearing to his home port is

50

Egg Island

20¡ 30

 ¬

70¡ Home Port

480

CHAPTER 5 Trigonometric Functions: Right Triangle Approach B

46. (a) In 30 minutes the pilot flies 100 miles due east, so using the Law of Cosines we have x 2  1002  3002  2  100  300  cos 40

300

 1002 1  9  6 cos 40   1002 5404. Thus, x 

 1002 5404  2325, and so the pilot is 2325 miles from

A

his destination.

(b) Using the Law of Sines, sin  

50¡ 40¡

x ¬

100

300 sin 40  0829  2325

  sin1 0829  56 . However, since   90 , the angle we seek is

180  56  124 . Hence the bearing is 124  90  34 , that is, N 34 E.

47. The largest angle is the one opposite the longest side; call this angle . Then by the Law of Cosines, 442  362  222  2 36 22  cos   cos  

362  222  442  009848    cos1 009848  96 . 2 36 22

48. Let  be the angle formed by the cables. The two tugboats and the barge form a triangle: the side opposite  has a length of 120 ft and the other two sides have lengths of 212 and 230 ft. Therefore, 1202  2122  2302  2 212 230  cos   cos  

2122  2302  1202  cos   08557    cos1 08557  31 . 2 212 230

49. Let d be the distance between the kites. Then d 2  3802  4202  2 380 420  cos 30   d  3802  4202  2 380 420  cos 30  211 ft. 50. Let x be the length of the wire and  the angle opposite x, as shown in the figure. Since the mountain is inclined 32 , we must have

125

x

  180  90  32   122 . Thus,  x  552  1252  2 55 125  cos 122  161 ft.

¬ 55 32¡

51. Solution 1: From the figure, we see that   106 and sin 74 

3400 b



3400  3537. Thus, x 2  8002  35372  2 800 3537 cos 106 sin 74  x  8002  35372  2 800 3537 cos 106  x  3835 ft.

b

Solution 2: Notice that tan 74 

3400 a



a

Pythagorean Theorem, x 2  a  8002  34002 . So  x  9749  8002  34002  3835 ft.



x

3400  9749. By the tan 74



b

3400

74¡

800

52. Let the woman be at point A, the first landmark (at 62 ) be at point B, and the other landmark be at point C. We 1150 1150 1150  AB    2450. Similarly, cos 54  want to find the length BC. Now, cos 62  AB cos 62 AC 1150  1956. Therefore, by the Law of Cosines, BC 2  AB 2  AC 2  2 AB AC  cos 43  AC  cos 54  BC  24502  19562  2 2450 1956  cos 43  BC  1679. Thus, the two landmarks are roughly 1679 feet apart.

CHAPTER 5

Review

481

 112  148  190 abc   225. Thus, 53. By Heron’s formula, A  s s  a s  b s  c, where s  2 2  A  225 225  112 225  148 225  190  82777 ft2 . Since the land value is $20 per square foot, the value of the lot is approximately 82777  20  $165,554. sin 465 sin B 54. Having found a  132 using the Law of Cosines, we use the Law of Sines to find  B:   105 132  105 sin 465 sin B   0577. Now there are two angles  B between 0 and 180 which have sin B  0577, namely 132  B  352 and  B  1448 . But we must choose  B , since otherwise  A   B  180 . 1 2 1 sin 465 180 sin 465 sin C   sin C   0989, so either  C  815 Using the Law of Sines again, 180 132 132 or  C  985 . In this case we must choose  C  985 so that the sum of the angles in the triangle is  A   B   C  465  352  985  180 . (The fact that the angles do not sum to exactly 180 , and the discrepancies between these results and those of Example 3, are due to roundoff error.) The method in this exercise is slightly easier computationally, but the method in Example 3 is more reliable. 55. In any ABC, the Law of Cosines gives a 2  b2 c2 2bccos A, b2  a 2 c2 2accos B, and c2  a 2 b2 2abcos C. Adding the second and third equations gives b2  a 2  c2  2ac  cos B

c2  a 2  b2  2ab  cos C

b2  c2  2a 2  b2  c2  2a c cos B  b cos C Thus 2a 2  2a c cos B  b cos C  0, and so 2a a  c cos B  b cos C  0. Since a  0 we must have a  c cos B  b cos C  0  a  b cos C  c cos B . The other laws follow from the symmetry of a, b, and c.

CHAPTER 5 REVIEW     052 rad 1. (a) 30  30  180 6

  7  183 rad 2. (a) 105  105  180 12

  5  262 rad (b) 150  150  180 6

  2  126 rad (b) 72  72  180 5

     035 rad (c) 20  20  180 9

   9  707 rad (c) 405  405  180 4

   5  393 rad (d) 225  225  180 4

   7  550 rad (d) 315  315  180 4

 3. (a) 56 rad  56  180   150

 4. (a)  53 rad   53  180   300

 180  (b)   9 rad   9    20

 (b) 109 rad  109  180   200

 (c)  43 rad   43  180   240

900  (c) 5 rad  5  180      2865

720  (d) 4 rad  4  180     2292

 (d) 113 rad  113  180   660

5. r  10 m,   25 rad. Then s  r  10  25  4  126 m. 6. s  7 cm, r  25 cm. Then  

s 7  28 rad  1604  25 r

  5 rad. Then r  s  25  18  90  286 ft. 7. s  25 ft,   50  50  180  18 5    13 rad. Then r  s  13  18  18 m. 8. s  13 m,   130  130  180 18 13 

482

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

9. Since the diameter is 28 in, r  14 in. In one revolution, the arc length (distance traveled) is s  r  2  14  28 in. The total distance traveled is 60 mi/h  05 h  30 mi  30 mi  5280 ft/mi  12 in./ft  1,900,800 in. The number of 1 rev revolution is 1,900,800 in   216087 rev. Therefore the car wheel will make approximately 21,609 revolutions. 28 in. 10. r  3960 miles, s  2450 miles. Then  

2450 s   35448 and so the angle is   0619 rad  0619  180 r 3960

approximately 354 . 11. r  5 m,   2 rad. Then A  12 r 2   12  52  2  25 m2 .     18,151 ft2 12. A  12 r 2   12 2002 52  180 

2A 2  125    250 625  04 rad  229 r2 252   2A 100  11  42 m. 14. A  50 m2 and   116 rad. Thus, r   13. A  125 ft2 , r  25 ft. Then  

6

150  2 rad  300 rad/min  9425 rad/min. The linear speed is 1 min

15. The angular speed is   

150  2  8  2400 in./min  75398 in./min  6283 ftmin. 1

 rad  7000 rad/min  21,9911 rad/min. 16. (a) The angular speed of the engine is e  35002 1 min

e  7000rad/min  09  (b) To find the angular speed  of the wheels, we calculate g       77778 rad/min  24,4346 rad/min.

(c) The speed of the car is the angular speed of the wheels times their radius: 77778 rad  11 in  60 min  1 mile  2545 mi/h. min 1 hr 63,360 in.

    52  72  74. Then sin   5 , cos   7 , tan   57 , csc   574 , sec   774 , and cot   75 . 74 74     3 , cos   91 , tan   3 , csc   10 , sec   10 , and cot   91 . 18. x  102  32  91. Then sin   10 10 3 3

17. r 

91

x 19.  cos 40 5

20. cos 35 



x  5 cos 40  383, and 5y  sin 40



91

y  5 sin 40  321.

2  x  cos235  244, and tan 35  2y  y  2 tan 35  140. x

1 2924  311.  sin 20  x  sin120  292, and xy  cos 20  y  cosx20  09397 x x 22. cos 30   x  4 cos 30  346, and sin 30  xy  y  x sin 30  346  05  173. 4

21.

23. A  90  20  70 , a  3 cos 20  2819, and b  3 sin 20  1026. A

C

b

24. C  90  60  30 , cos 60  20a 

a  20 cos 60   40, and tan 60  b20  b  20 tan 60  3464. C

70¡ 3

20¡

a b B A

30¡

a 60¡ 20

B

CHAPTER 5

25. c 

 7  02960  170 , 252  72  24, A  sin1 24

 and C  sin1 24 25  12870  737 . 25 A

c

26. b 

Review

483

 5  03948  226 , 122  52  13, A  sin1 13

 and C  sin1 12 13  11760  674 .

C

b

7 B

A

12

C 5 B

1 1 1 1 a   cot , sin    b   csc  a tan  b sin  h 28. Let h be the height of the tower in meters. Then tan 2881   h  1000 tan 2881  550 m. 1000

27. tan  

One side of the hexagon together with radial line segments through its endpoints

29. x

8

forms a triangle with two sides of length 8 m and subtended angle 60 . Let x be the

60¡

length of one such side (in meters). By the Law of Cosines,

8

x 2  82  82  2  8  8  cos 60  64

x  8. Thus the perimeter of the



hexagon is 6x  6  8  48 m.

y

30. As the crankshaft moves in its circular pattern, point Q is

y

determined by the angle , namely it has coordinates Q 2 cos , 2 sin . We split the triangle into two right triangles O Q R and P Q R, as shown in the figure. Let h be the height of

the piston. We consider two cases, 0    180 and

180    360 . If 0    180 , then h is the sum of O R and R P. Using the Pythagorean Theorem, we find R P 

 82  2 cos 2 , while

8

h

h

8 R ¬ O 2

Q(2 cos ¬, 2 sin ¬) x

Q(2 cos ¬, 2 sin ¬)

¬

O 2 R

x

O R is the y-coordinate of the point Q, 2 sin . Thus  h  64  4 cos2   2 sin .

 If 180    360 , then h is the difference between R P and R O. Again, R P  64  4 cos2  and O R is the  y-coordinate of the point Q, 2 sin . Thus h  64  4 cos2   2 sin . Since sin   0 for 180    360 , this also  reduces to h  64  4 cos2   2 sin .  Since we get the same result in both cases, the height of the piston in terms of  is h  64  4 cos2   2 sin .  r 31. Let r represent the radius, in miles, of the moon. Then tan  ,   0518  r  r  236,900tan 0259 2 r  AB 236,900  tan 0259  r 1  tan 0259   236,900  tan 0259  r   1076 and so the radius of the moon is 1  tan 0259 roughly 1076 miles. 32. Let d1 represent the horizontal distance from a point directly below the plane to the closer ship in feet, and d2 represent the 35,000 35,000 35,000  d1  , and similarly tan 40   horizontal distance to the other ship in feet. Then tan 52  d1 tan 52 d2 35,000 35,000 35,000 . So the distance between the two ships is d2  d1    14,400 ft. d2  tan 40 tan 40 tan 52   34. csc 94  csc  33. sin 315   sin 45   1   22 4  2 2

35. tan 135   tan 45  1



3 36. cos 56   cos  6  2

484

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

   1   3 37. cot  223  cot 23  cot    3 3 3



38. sin 405  sin 45  1  22 2



39. cos 585  cos 225   cos 45   1   22

40. sec 223  sec 43   sec  3  2

2 3 2 41. csc 83  csc 23  csc  3  3  3

2 3 42. sec 136  sec  6  3

2





 43. cot 390   cot 30    cot 30   3 44. tan 234  tan 34   tan  4  1   5 12 13 13 45. r  52  122  169  13. Then sin   12 13 , cos    13 , tan    5 , csc   12 , sec    5 , and 5 . cot    12

46. If  is in standard position, then the terminal point of  on the unit circle is simply cos  sin . Since the terminal point is    given as  23  12 , sin   12 .     47. y  3x  1  0  y  3x  1, so the slope of the line is m  3. Then tan   m  3    60 .   48. 4y  2x  1  0  y  12 x  14 . The slope of the line is m  12 . Then tan   m  12 and r  12  22  5. So   sin    1 , cos    2 , tan   12 , csc    5, sec    25 , and cot   2. 5 5   sin  1  cos2  49. Since sin  is positive in quadrant II, sin   1  cos2  and we have tan    . cos  cos  1 1 50. sec     (because cos   0 in quadrant III). cos   1  sin2  51. tan2  

sin2  sin2   cos2  1  sin2 

52. csc2  cos2  

1

sin2 

 cos2  

1  sin2  sin2 



1

1

sin2      53. tan   37 , sec   43 . Then cos   34 and sin   tan   cos   47 , csc   4  4 7 7 , and cot   3  3 7 7 . 7 7 9 sin   41 41 9 40 9 40 54. sec   41 40 , csc    9 . Then sin    41 , cos   41 , tan   cos   40   40 , and cot    9 . 41   55. sin   35 . Since cos   0,  is in quadrant II. Thus, x   52  32   16  4 and so cos    45 , tan    34 , csc   53 , sec    54 , cot    43 . 5 56. sec    13 5 and tan   0. Then cos    13 , and  must be in quadrant III  sin   0. Therefore,

  25   12 , csc    13 , tan   sin   12 , and cot   5 . sin    1  cos2    1  169 13 12 5 12 cos   4  cos    45   2 since 57. tan    12 . sec2   1  tan2   1  14  54  cos2   5 5   sin  1 1 1 2 1   . Therefore,   2  sin    2 cos    2   cos   0 in quadrant II. But tan   5 5 cos     5 1 2 1 sin   cos            5 . 5 5 5  1 1 1 sin   1 sin   1 sin  2 3      3. 58. sin   12 for  in quadrant I. Then tan   sec     2 cos  cos  cos  2 1  sin2  1 1 2

59. By the Pythagorean Theorem, sin2   cos2   1 for any angle . 



3 5 10 5 5  60. cos    23 and  2    . Then   6  2  6  3 . So sin 2  sin 3   sin 3   2 . 

61. sin1 23   3



62. tan1 33   6

CHAPTER 5

Review

485

63. Let u  sin1 25 and so sin u  25 . Then from the triangle, 64. Let u  cos1 83 then cos u  38 . From the triangle, we      tan sin1 25  tan u  2 . have sin cos1 38  sin u  855 . 21

u

5

8

Ï21

u

2

65. Let   tan1 x

 tan   x. Then from the   x . triangle, we have sin tan1 x  sin    1  x2 Ï1+x@ ¬

3

66. Let   sin1 x. Then sin   x. From the triangle, we   1 have sec sin1 x  sec    . 1  x2

x

1

x

¬

1

Ï1-x@

x  x    tan1 2 2  10 sin 30      532. 69.  B  180  30  80  70 , and so by the Law of Sines, x  sin 70 2 sin 45  146 70. x  sin 105 67. cos  

x  x    cos1 3 3

Ï55

68. tan  

71. x 2  1002  2102  2  100  210  cos 40  21,926133  x  14807 20 sin 60 72. sin B   0247   B  sin1 0247  1433 . Then  C  180  60  1433 10567 , and so 70 70 sin 10567 x  7782. sin 60  73. x 2  22  82  2 2 8 cos 120  84  x  84  917 6 sin 3121 4 sin 110  0626   B  3879 . Then  C  180 110 3879  3121 , and so x   33. 74. sin B  6 sin 110   sin 25 23 sin 25 sin  23 sin 25   sin      sin1  541 or 75. By the Law of Sines, 23 12 12 12   180  541  1259 .   sin  sin 80 4 sin 80 4 sin 80 76. By the Law of Sines,   sin      sin1  520 . 4 5 5 5 77. By the Law of Cosines, 1202  1002  852  2 100 85 cos , so cos     cos1 016618  804 .

1202  1002  852  016618. Thus, 2 100 85

sin 10 5 sin 10 sin A   sin A   02894, so A  sin1 02894  168 5 3 3 or A  180  168  1632 . Therefore,   180  10  168  1532 or   180  10  1632  68 .

78. We first use the Law of Sines to find  A:

79. After 2 hours the ships have traveled distances d1  40 mi and d2  56 mi. The subtended angle is 180  32  42  106 . Let d be the distance between the two ships in miles. Then by the Law of Cosines, d 2  402  562  2 40 56 cos 106  5970855  d  773 miles.

486

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

80. Let h represent the height of the building in feet, and x the horizontal distance from the building to point B. Then h h h tan 241  and tan 302   x  h cot 302 . Substituting for x gives tan 241   x  600 x h cot 302  600  600  tan 241 h  tan 241 h cot 302  600  h   1160 ft. 1  tan 241 cot 302 81. Let d be the distance, in miles, between the points A and B . Then by the Law of Cosines, d 2  322  562  2 32 56 cos 42  14966  d  39 mi.

120 sin 689 82.  C  180  423  689  688 . Then b   12008 miles. Let d be the shortest distance, in miles, to sin 688 the shore. Then d  b sin A  12008 sin 423  808 miles.

83. A  12 ab sin   12 8 14 sin 35  3212  568 abc   95. Thus, 84. By Heron’s Formula, A  s s  a s  b s  c, where s  2 2  A  95 95  5 95  6 95  8  1498.

CHAPTER 5 TEST   11 rad. 135  135     3 rad. 1. 330  330  180 6 180 4 180 234   2. 43 rad  43  180   240 . 13 rad  13       745

3. (a) The angular speed is 

120  2 rad  240 rad/min  75398 rad/min. 1 min

(b) The linear speed is 120  2  16  3840 ft/min 1  12,0637 ft/min

 



4. (a) sin 405  sin 45  1  22 2



(b) tan 150   tan 30  1  33 3 (c) sec 53  sec  3 2 (d) csc 52  csc  2 1

 137 mi/h

   2 13  3   2 2 26  6 13 . 5. r  32  22  13. Then tan   sin        3 39 13 3 13 b a  a  24 sin . Also, cos    b  24 cos . 6. sin   24 24   7. cos    13 and  is in quadrant III, so r  3, x  1, and y   32  12  2 2. Then

  1 23  43 2. 3  2   csc   1   4 2 2 2 2 tan  1 sin  1 5 13 13 1 5 , tan    5 . Then sec   8. sin   13    tan      . 12 cos  cos  sin  sin  12 5 12   9. sec2   1  tan2   tan    sec2   1. Thus, tan    sec2   1 since tan   0 in quadrant II.

tan  cot   csc   tan  

h  h  6 tan 73  196 ft. 6 x  x 11. (a) tan      tan1 4 4   3 3 (b) cos      cos1 x x 10. tan 73 

Surveying 9 so tan u  9 . From the triangle, r  12. Let u  tan1 40 40   40 9 1 cos tan 40  cos u  41 .

 92  402  41. So

41 u

487

9

40

13. By the Law of Cosines, x 2  102  122  2 10 12  cos 48  8409  x  91.

230 sin 69 14.  C  180  52  69  59 . Then by the Law of Sines, x   2505. sin 59 h x h  h  50 tan 20 and tan 28   x  h  50 tan 28 15. Let h be the height of the shorter altitude. Then tan 20  50 50  x  50 tan 28  h  50 tan 28  50 tan 20  84. 15 sin 108 16. Let  A and  X be the other angles in the triangle. Then sin A   0509   A  3063 . Then 28 28 sin 4137  X  180  108  3063  4137 , and so x   195. sin 108 17. By the Law of Cosines, 92  82 62 2 8 6 cos   cos  

82  62  92  01979, so   cos1 01979  786 . 2 8 6

18. We find the length of the third side x using the Law of Cosines: x 2  52  72  2 5 7 cos 75  5588  x  7475. sin 75 5 sin 75 sin    sin    06461, so   sin1 06461  402 . Therefore, by the Law of Sines, 5 7475 7475   50  72 . A triangle  1 r  r sin   1  102 sin 72 . Thus, the area of the 19. (a) A sector  12 r 2   12  102  72  180 180 2  2     153 m2 . shaded region is A shaded  A sector  A triangle  50 72 180  sin 72

(b) The shaded region is bounded by two pieces: one piece is part of the triangle, the other is part of the circle.   The first part has length l  102  102  2 10 10  cos 72  10 2  2  cos 72 . The second has length   4. Thus, the perimeter of the shaded region is p  l  s  102  2 cos 72  4  243 m. s  10  72  180

20. (a) If  is the angle opposite the longest side, then by the Law of Cosines cos  

92  132  202  06410. Therefore, 2 9 20

  cos1 06410  1299 .

(b) From part (a),   1299 , so the area of the triangle is A  12 9 13 sin 1299  449 units2 . Another way to find  9  13  20 abc   21. Thus, the area is to use Heron’s Formula: A  s s  a s  b s  c, where s  2 2   A  21 21  20 21  13 21  9  2016  449 units2 . 21. Label the figure as shown. Now    85  75  10 , so by the Law of Sines, 100 x  sin 75 sin 10



x  100 

sin 75 h . Now sin 85  sin 10 x

sin 75  h  x sin 85  100   sin 85  554. sin 10

º



x 75¡ 100

h

85¡

FOCUS ON MODELING Surveying 1. Let x be the distance between the church and City Hall. To apply the Law of Sines to the triangle with vertices at City Hall, the church, and the first bridge, we first need the measure of the angle at the first bridge, which is 180  25  30  125 . x 086 086 sin 125 Then   x   14089. So the distance between the church and City Hall is about sin 125 sin 30 sin 30 141 miles.

488

FOCUS ON MODELING

2. To find the distance z between the fire hall and the school, we use the distance found in the text between the bank and the cliff. To find z we first need to find the length of the edges labeled x and y.

bank 50¡

In the bank-second bridge-cliff triangle, the third angle is

1.55

155 sin 50 180  50  60  70 , so x   126. 

sin 70 In the second bridge-school-cliff triangle, the third angle is

60¡

cliff

126 sin 45  109. sin 55 Finally, in the school-fire hall-cliff triangle, the third angle is

80¡

180  80  55  45 , so y 

180  45  80  55 , so z 

109 sin 80

 131. sin 55 Thus, the fire hall and the school are about 131 miles apart.

fire hall

55¡

z

70¡

x

second bridge

45¡

80¡ y 45¡

55¡

school

3. First notice that  D BC  180  20  95  65 and  D AC  180  60  45  75 . AC 20 20 sin 45   AC   146 . From BC D we get From AC D we get   sin 45 sin 75 sin 75 BC 20 20 sin 95 BC     220. By applying the Law of Cosines to ABC we get sin 95 sin 65 sin 65  AB2  AC2 BC2 2 AC BC cos 40  1462 2202 2146220cos 40  205, so AB  205  143 m. Therefore, the distance between A and B is approximately 143 m.

4. Let h represent the height in meters of the cliff, and d the horizontal distance to the cliff. The third horizontal 200 sin 516 h angle is 180  694  516  59 and so d    182857. Then tan 331   sin 59 d   h  d tan 331  182857 tan 331  1192 m.

5. (a) In ABC,  B  180  , so  C  180    180      . By the Law of Sines,

AB BC  sin  sin   

d sin  sin   . sin    sin    d sin  h h (b) From part (a) we know that BC  . But sin   . Therefore,  BC  BC sin    sin  d sin  h d sin  sin  BC   h  . sin    sin  sin    800 sin 25 sin 29 d sin  sin   (c) h   2350 ft sin    sin 4  BC  AB

6. Let the surveyor be at point A, the first landmark (with angle of depression 42 ) be at point B, and the other landmark 2430 2430 2430  36316. Similarly, sin 39  be at point C. We want to find BC. Now sin 42   AB  AB AC sin 42 2430  38613. Therefore, by the Law of Cosines, BC2  AB2  AC2  2 AB AC cos 68   AC  sin 39  BC  363162  386132  2 36316 38613 cos 68  4194. Thus, the two landmarks are approximately 4194 ft apart.

Surveying

489

7. We start by labeling the edges and calculating the remaining angles, as shown in the first figure. Using the Law of Sines, a 150 150 sin 29 b 150 150 sin 91 we find the following:   a   8397,   b   17318, sin 29 sin 60 sin 60 sin 91 sin 60 sin 60   c 17318 17318 sin 32 d 17318 17318 sin 61  c   9190,  e   15167, sin 32 sin 87 sin 87 sin 61 sin 87 sin 87 15167 15167 sin 41 f 15167 15167 sin 88 e  e   12804,   f   19504,      sin 41 sin 51 sin 51 sin 88 sin 51 sin 51 19504 19504 sin 50 h 19504 19504 sin 38 g  g  14950, and  h   12015. Note that      sin 50 sin 92 sin 92 sin 38 sin 92 sin 92 we used two decimal places throughout our calculations. Our results are shown (to one decimal place) in the second figure.

29¡ 150

c

61¡

87¡ 88¡

91¡

60¡ a

32¡ 41¡

128.0

51¡ 38¡

d

b

91.9

e

g

f 50¡ h

92¡

150

173.2

84

151.7

195.0

149.5

120.2

8. Answers will vary. Measurements from the Great Trigonometric Survey were used to calculate the height of Mount Everest to be exactly 29,000 ft, but in order to make it clear that the figure was considered accurate to within a foot, the height was published as 29,002 ft. The accepted figure today is 29,029 ft.

6

TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH

6.1

THE UNIT CIRCLE

1. (a) The unit circle is the circle centered at 0 0 with radius 1. (b) The equation of the unit circle is x 2  y 2  1.

(c) (i) Since 12  02  1, the point is P 1 0.

(ii) P 0 1

(iii) P 1 0

(iv) P 0 1

2. (a) If we mark off a distance t along the unit circle, starting at 1 0 and moving in a counterclockwise direction, we arrive at the terminal point determined by t.  (b) The terminal points determined by  2 , ,  2 , 2 are 0 1, 1 0, 0 1, and 1 0, respectively.    2  2 9  16  1, P 3   4 lies on the unit circle. 3. Since 35   45  25 25 5 5 2      7 2  576  49  1, P  24   7 lies on the unit circle.   25 4. Since  24 25 625 25 25 25

   2   2  9  7  1, P 3   7 lies on the unit circle. 5. Since 34   47  16 16 4 4    2   2  24  1, P  5   2 6 lies on the unit circle. 6. Since  57   2 7 6  25  7 7 49 49

     2  2 7. Since  35  23  59  49  1, P  35  23 lies on the unit circle.   2  2   11  5 lies on the unit circle. 25  1, P 8. Since 611  56  11  36 36 6 6  2 9  y 2  16  y   4 . Since P x y is in quadrant III, y is negative, so the point is 9.  35  y 2  1  y 2  1  25 25 5   P  35   45 .   7 2  1  x 2  1  49  x 2  576  x   24 . Since P is in quadrant IV, x is positive, so the point is 10. x 2   25 625 625 25   24 7 P 25   25 .   2  1  x 2  1  19  x 2  89  x   2 3 2 . Since P is in quadrant II, x is negative, so the point is 11. x 2  13    P  2 3 2  13 .   2 4  y 2  21  y   21 . Since P is in quadrant I, y is positive, so the point is 12. 25  y 2  1  y 2  1  25 25 5    21 2 P 5 5 .   2 4  x 2  45  x   3 5 . Since P x y is in quadrant IV, x is positive, so the point is 13. x 2   27  1  x 2  1  49 7 49    3 5 2 P 7 7 .   2 14.  23  y 2  1  y 2  1  49  y 2  59  y   35 . Since P is in quadrant II, y is positive, so the point is    P  23  35 .    2 5 25  y 2  144  y   12 . Since its y-coordinate is negative, the point is P 5   12 .  y 2  1  y 2  1  169 15. 13 169 13 13 13 491

492

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

  2  9  x 2  16  x   4 . Since its x-coordinate is positive, the point is P 4   3 . 16. x 2   35  1  x 2  1  25 25 5 5 5

   2   17. x 2  23  1  x 2  1  49  x 2  59  x   35 . Since its x-coordinate is negative, the point is P  35  23 .    2 5  x 2  20  x   2 5 . Since its x-coordinate is positive, the point is  1  x 2  1  25 18. x 2   55 25 5     2 5 5 P 5  5 .

   2 19.  32  y 2  1  y 2  1  29  y 2  79  y   37 . Since P lies below the x-axis, its y-coordinate is negative, so     the point is P  32   37 .

  2 4  y 2  21  y   21 . Since P lies above the x-axis, its y-coordinate is positive, so 20.  25  y 2  1  y 2  1  25 25 5    the point is P  25  521 .

21.

22. t

Terminal Point

0

1 0    2 2 2  2

 4

 2 3 4



t

Terminal Point

t

Terminal Point



1 0      22   22

0 

1 0   3 1 2 2    1 3 2 2

5 4 3 2 7 4

0 1

     22  22 1 0

2

0 1



  2 2 2  2

1 0

6

 3

 2 2 3 5 6

 23. t  0, so t  4 corresponds to P x y  1 0. y

t=0 P(1, 0) t=4¹ x

3 25. t   2 , so t  2 corresponds to P x y  0 1. y ¹ Q(0, 1) t= 2

t

Terminal Point



1 0     23   12     12   23

7 6 4 3 3 2 5 3 11 6

0 1     12  23     23  12 1 0



1 0

2

24. t  0, so t  3 corresponds to P x y  1 0. y

P(_1, 0)

t=_3¹

Q(1, 0)

t=0

x

5 26. t   2 , so t  2 corresponds to P x y  0 1. y P(0, 1)

5¹

t= 2

¹

t= 2 x

3¹

t= 2

P(0, _1)

0 1

  1 3 2  2  31 2 2

x

SECTION 6.1 The Unit Circle

27. t    6 corresponds to P x y  y

 31 . 2 2



493

7 28. t   6 and t  6 corresponds to    P x y   23   12 . y

¹

t=_ 6

7¹ t= 6

x

(

)

P Ï3 , _21 2

x 1 P(_ Ï3 2 , _ 2)

5 29. t   4 and t  4 corresponds to     P x y   22   22 .

4 30. t   3 and t  3 corresponds to    P x y   12   23 .

y

y

4¹

t= 3 x

5¹

t= 4

P(_ Ï2 , _ Ï2 ) 2 2

x P (_ 21 , _ Ï3 2)

   5 corresponds to P x y  1   3 . 32. t   and t  3 3 2 2

7 31. t   6 and t   6 corresponds to    P x y   23  12 .

y

y

P(_ 2 , 2 ) Ï3 1

x

x

5¹

t= 3

P ( 21 , _ Ï3 2 )

7¹

t=_ 6

7 33. t   4 and t   4 corresponds to    P x y  22  22 .

4 34. t   3 and t   3 corresponds to    P x y   12  23 .

y

7¹

t=_ 4

P( Ï2 , Ï2) 2 2

P (_ 21 , Ï3 2)

y

4¹

t=_ 3 x

x

494

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

3 35. t   4 and t   4 corresponds to     P x y   22   22 . y

11 36. t   6 and t  6 corresponds to   P x y  23   12 .

¹ 4

(

P _ Ï2 , _ Ï2 2 2

y

Q(Ï2 , Ï2 2 2 )

¹ 6

x

)

11¹

3¹ t=_ 4

37. (a) t  43     3 (b) t  2  53   3   7  (c) t     6   6 (d) t  35    036 39. (a) t    57  27

t= 6

38. (a) t  9  9  0   (b) t     54   4 (c) t    56   6

(d) t  4    086 40. (a) t  115  2   5

(b) t    79  29

(b) t  97    27

(c) t    3  0142

(c) t  2  6  0283

(d) t  2  5  1283

(d) t  7  2  0717

41. (a) t  2  116   6   3 1 (b) P 2   2   43. (a) t     43   3    3 1 (b) P  2  2 45. (a) t    23   3    3 1 (b) P  2   2

42. (a) t    23   3    3 1 (b) P  2  2

44. (a) t  2  53   3    3 1 (b) P 2   2

46. (a) t  76     6    3 1 (b) P  2  2

47. (a) t  134  3   4     2 2 (b) P  2   2

48. (a) t  136  2   6   3 1 (b) P 2  2

51. (a) t  4  113   3    3 1 (b) P 2  2 53. (a) t  163  5   3    3 1 (b) P  2   2

52. (a) t  316  5   6    3 1 (b) P  2   2   54. (a) t  10   414   4    2 2 (b) P 2   2 56. t  25 

P 08 06

57. t  11  05 09

58. t  42 

P 05 09

49. (a) t  7  416   6    3 1 (b) P  2  2

55. t  1  05 08

50. (a) t  174  4   4    2 2 (b) P 2  2

Q(Ï3 ,1 2 2) x

(

P Ï3 , _ 21 2

)

SECTION 6.2 Trigonometric Functions of Real Numbers





59. Let Q x y  35  45 be the terminal point determined by t. y

¹-t

¹+t

2¹+t

(b) Q( 35 , 45)

(c)

t

(d) x

_t

   60. Let Q x y  34  47 be the terminal point determined by t. y ¹-t 4¹+t t-¹

(a)

495

    t determines the point P x y   35  45 .   t determines the point P x y  35   45 .     t determines the point P x y   35   45 .   2  t determines the point P x y  35  45 .

Q( 34 , Ï7 4) t _t

x

   (a) t determines the point P x y  34   47 .    (b) 4  t determines the point P x y  34  47 .    (c)   t determines the point P x y   34  47 .    (d) t   determines the point P x y   34   47 .

61. The distances P Q and P R are equal because they both subtend arcs of length  3 . Since P x y is  a point on the unit circle, x 2  y 2  1. Now d P Q  x  x2  y  y2  2y and    d R S  x  02  y  12  x 2  y 2  2y  1  2  2y (using the fact that x 2  y 2  1). Setting these equal  gives 2y  2  2y  4y 2  2  2y  4y 2  2y  2  0  2 2y  1 y  1  0. So y  1 or y  12 . Since  2 P is in quadrant I, y  12 is the only viable solution. Again using x 2  y 2  1 we have x 2  12  1  x 2  34     x   23 . Again, since P is in quadrant I the coordinates must be 23  12 . 62. P is the reflection of Q about the line y  x. Since Q is the point Q

6.2



3 1 2 2



   it follows that P is the point P 12  23 .

TRIGONOMETRIC FUNCTIONS OF REAL NUMBERS

1. If Px y is the terminal point on the unit circle determined by t, then sin t  y, cos t  x, and tan t  yx. 2. If Px y is on the unit circle, then x 2  y 2  1. So for all t we have sin2 t  cos2 t  1.

496

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

3.

4. t

sin t

cos t

t

sin t

cos t

t

sin t

cos t

0

0

1

0

0

1

0

1 2  3 2

 3 2 1 2



 12   23

1

 23

1

0

7 6 4 3 3 2 5 3 11 6

1

0

  23  12

1 2  3 2

2

0

1

 4



(b) cos 54 

0



  22





0



(c) cos 74   1  13. (a) cos   3  2   (b) sec   3 2    3   (c) sin   3 2

17. (a) csc 76  2  23  (b) sec   6  3    (c) cot  56  3 (b) cos 14  1 (c) tan 15  0

  22

1

3

1

  22

2 2 3 5 6

 3 2 1 2

0



0

  22

 2 2

0

1 

 3 3   22   22  2 2

21. (a) sin 13  0

6

1

(b) cos 176   23

9. (a) cos 34 



 2 2

2

(c) tan 76 

 2 2

2 3 4 5 4 3 2 7 4

5. (a) sin 76   12

 2 2

6. (a) sin 53   23 (b) cos 113  12  (c) tan 53   3 

10. (a) sin 34  22



 12

  23

1



7. (a) sin 114  22    2   (b) sin   4 2 

(c) sin 54   22 

11. (a) sin 73  23



(b) sin 54   22

(b) csc 73  2 3 3

(c) sin 74   22   14. (a) tan   4  1    (b) csc   4  2   (c) cot   4  1

(c) cot 73  33  3  15. (a) cos   6  2    2 3   (b) csc   3 3    (c) tan  6   33



 18. (a) sec 34   2   (b) cos  23   12    (c) tan  76   33





19. (a) sin 43   23 

(b) sec 116  2 3 3    3   (c) cot   3 3

24. t   2  sin t  1, cos t  0, csc t  1, cot t  0, tan t and sec t are undefined. 25. t    sin t  0, cos t  1, tan t  0, sec t  1, csc t and cot t are undefined.

26. t  32  sin t  1, cos t  0, csc t  1, cot t  0, tan t and sec t are undefined. 4  2  2 9  16  1. So sin t   4 , cos t   3 , and tan t   5  4 . 27.  35   45  25 25 5 5 3  35 

3   2   2  28.  12  23  14  34  1. So sin t  23 , cos t   12 , and tan t  21   3. 2

 12



8. (a) cos 196   23    (b) cos  76   23  3  (c) cos   6  2  12. (a) csc 54   2  (b) sec 54   2 (c) tan 54  1    2 16. (a) sin   4  2    (b) sec   4  2    (c) cot   6  3 

20. (a) csc 23  2 3 3   (b) sec  53  2   (c) cos 103   12

  22. (a) sin 252  sin  2  24  1   (b) cos 252  cos  2  24  0   (c) cot 252  cot  2  24  0

23. t  0  sin t  0, cos t  1, tan t  0, sec t  1, csc t and cot t are undefined.



SECTION 6.2 Trigonometric Functions of Real Numbers

29.

30.

31.

32.

33.

34.

35.

36.

497

 2 2   2   2   13  2 3 2  19  89  1. So sin t  2 3 2 , cos t   13 , and tan t  31  2 2. 3    2   2  256 2 6 2 6 1 1 24 1   5  25  25  1. So sin t   5 , cos t  5 , and tan t  1  2 6. 5 5  13    2   2 7   13 . 13 13 6  67  713  36 49  49  1. So sin t  7 , cos t   7 , and tan t  6 6 7 9  2  2 41 40 9 81 9 40 9  41  1600 41 1681  1681  1. So sin t  41 , cos t  41 , and tan t  40  40 . 41 12 2  2  5 25  144  1. So sin t   12 , cos t   5 , and tan t  13  12 .   12   13 13 169 169 13 13 5 5  13  2 5     2   2 5 5  20  1. So sin t  2 5 , cos t  5 , and tan t  5  2.  2 5 5  25 5 25 5 5 5 5 21 2  2  29 441 21 20  20  21  400   21 29 29 841  841  1. So sin t  29 , cos t   29 , and tan t  20 .  20 29 7  2   24 7 2  576  49  1. So sin t   7 , cos t  24 , and tan t   25   7 .   25 25 625 625 25 25 24 24 25

37. (a) 08

38. (a) 07

(b) 084147 41. (a) 10

(b) 069671 42. (a) 36

(b) 102964

(b) 360210

39. (a) 09

40. (a) 03

(b) 093204

(b) 028366

43. (a) 06

44. (a) 09

(b) 057482

(b) 088345

45. sin t  cos t. Since sin t is positive in quadrant II and cos t is negative in quadrant II, their product is negative. 46. tan t  sec t is negative in quadrant IV because tan t is negative and cos t is positive in quadrant IV. 47.

1 tan t  sin t  tan t   sin t  tan t  tan t  sin t  tan2 t  sin t. Since tan2 t is always positive and sin t is negative in cot t cot t quadrant III, the expression is negative in quadrant III.

48. cos t  sec t is positive in any quadrant, since cos t  sec t  cos t  49. Quadrant II  53. sin t  1  cos2 t

sin t sin t  cos t 1  sin2 t  57. sec t   1  tan2 t 55. tan t 

 59. tan t  sec2 t  1 61. tan2 t 

sin2 t sin2 t  cos2 t 1  sin2 t

50. Quadrant III

1  1, provided cos t  0. cos t

51. Quadrant II  54. cos t  1  sin2 t  1  cos2 t 56. tan t   cos t  58. csc t   1  cot2 t

52. Quadrant II

 1 sec2 t  1   sec t sec2 t (sin t  0 and sec t  0 in quadrant IV)

  2 60. sin t   1  cos t   1 

62. sec2 t  sin2 t 

  1 1 2t   1  cos 1 2 cos t cos2 t

498

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

  63. sin t   45 and the terminal point of t is in quadrant IV, so the terminal point determined by t is P x  45 . Since P is    2 9 3 on the unit circle, x 2   45  1. Solving for x gives x   1  16 25   25   5 . Since the terminal point is   in quadrant IV, x  35 . Thus the terminal point is P 35   45 . Thus, cos t  35 , tan t   43 , csc t   54 , sec t  53 , cot t   34 .

  7 and the terminal point of t lies in quadrant III, so the terminal point determined by t is P  7  y . Since 64. cos t   25 25     7 2  y 2  1. Solving for y gives x   1  49   576   24 . Since the terminal P is on the unit circle,  25 625 625 25   24 7 24 24 25 point is in quadrant III, y   25 . Thus the terminal point is P  25   25 . Thus, sin t   25 , tan t  24 7 , csc t   24 , 7 sec t   25 7 , cot t  24 .

  65. sec t  3 and the terminal point of t lies in quadrant IV. Thus, cos t  13 and the terminal point determined by t is P 13  y .     2 Since P is on the unit circle, 13  y 2  1. Solving for y gives y   1  19   89   2 3 2 . Since the terminal      point is in quadrant IV, y   2 3 2 . Thus the terminal point is P 13   2 3 2 . Therefore, sin t   2 3 2 , cos t  13 ,    3   3 2 , cot t    1   2. tan t  2 2, csc t    4 4 2 2

2 2

 2 1 1  17 . 66. tan t  14 and the terminal point of t lies in quadrant III. Since sec2 t  tan2 t 1 we have sec2 t  14 1  16 16    17 17 Thus sec t   17 16   4 . Since sec t  0 in quadrant III we have sec t   4 , so      4 17 1 4 1   1   1717 .       . Since tan t  cos t  sin t we have sin t  14  4 cos t  17 17 sec t 17 17  417        Thus, the terminal point determined by t is P  4 1717   1717 . Therefore, sin t   1717 , cos t   4 1717 , csc t   17, 

sec t   417 , cot t  4.

  12 2  1  169 , so because 2 2 2 67. tan t   12 5 and sin t  0, so t is in quadrant II. Since sec t  tan t  1 we have sec t   5 25    1 13 5 5 secant is negative in quadrant II, sec t   169 25   5 . Thus, cos t  sec t   13 and we have P  13  y . Since      5  12 . Thus, the terminal point determined by t is P  5  12 , and so  tan t  cos t  sin t we have sin t   12 5 13 13 13 13 5 13 13 5 sin t  12 13 , cos t   13 , csc t  12 , sec t   5 , cot t   12 .

  68. csc t  5 and cos t  0, so t is in quadrant II. Thus, sin t  15 and the terminal point determined by t is P x 15 . Since    2 1   2 6 . Since the terminal point is in P is on the unit circle, x 2  15  1. Solving for x gives x   1  25 5       2 6 2 6 1 1 1   6, quadrant II, x   5 and the terminal point is P  5  5 . Thus, sin t  5 , cos t   2 5 6 , tan t    12 2 6   5 6 5 sec t      12 , cot t  2 6. 2 6

SECTION 6.2 Trigonometric Functions of Real Numbers

499

 69. sin t   14 , sec t  0, so t is in quadrant III. So the terminal point determined by t is P x  14 . Since P is on the unit     2 1   15   15 . Since the terminal point is in quadrant III, circle, x 2   14  1. Solving for x gives x   1  16 16 4       15 . Thus, the terminal point determined by t is P  415   14 , and so cos t   415 , tan t  1  1515 , x  15 4   4 15 4 csc t  4, sec t      15 , cot t  15. 

15

70. tan t  4 and the terminal point of t lies in quadrant II. Since sec2 t  tan2 t 1 we have sec2 t  42 1  161  17.    1 17 Thus sec t   17. Since sec t  0, we have sec t   17and cos t   1   . Since tan t  cos t  sin t  17 sec t 17         we have sin t  4  1717  4 1717 . Thus, the terminal point determined by t is P  1717  4 1717 . Thus,     sin t  4 1717 , cos t   1717 , csc t  417 , sec t   17, cot t   14 .

71. f x  x2 sin x  x 2 sin x   f x, so f is odd.

72. f x  x2 cos 2 x  x 2 cos 2x  f x, so f is even.

73. f x  sin x cos x   sin x cos x   f x, so f is odd.

74. f x  sin x  cos x   sin x  cos x which is neither f x nor  f x, so f is neither even nor odd. 75. f x  x cos x  x cos x  f x, so f is even.

76. f x  x sin3 x  x [sin x]3  x  sin x3  x sin3 x  f x, so f is even.

77. f x  x3  cos x  x 3  cos x which is neither f x nor  f x, so f is neither even nor odd. 78. f x  cos sin x  cos  sin x  cos sin x  f x, so f is even. 79. t

0

025

y t 4 283

050 075 100 125 0

80. (a) B 6  80  7 sin  2  87 mmHG  (b) B 105  80  7 sin 105 12  827 mmHG

283 4 283

(c) B 12  80  7 sin   80 mmHG 

7 3  (d) B 20  80  7 sin 20 12  80  2  739 mmHG

81. (a) I 01  08e03 sin 1  0499 A

82.

(b) I 05  08e15 sin 5  0171 A

t

0

1

2

4

6

8

12

H t 175 1504 100 386 100 1503 588

83. Notice that if P t  x y, then P t    x y. Thus, (a) sin t    y and sin t  y. Therefore, sin t     sin t.

(b) cos t    x and cos t  x. Therefore, cos t     cos t. y y sin t sin t       tan t. (c) tan t    cos t   x x cos t

84. To prove that AO B  C DO, first note that O B  O D  1 and  O AB   OC D   2 . Now  C O D   AO B      C O D     AO B   AB O. Since we know two angles and one side to be  2 2 equal, the triangles are (SAA) congruent. Thus AB  OC and O A  C D, so if B has coordinates x y, then D has coordinates y x. Therefore,   (a) sin t   2  x  cos t     (b) cos t   2  y, and sin t  y. Therefore, cos t  2   sin t.   x cos t x (c) tan t   2  y   y   sin t   cot t

500

6.3

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

TRIGONOMETRIC GRAPHS

1. If a function f is periodic with period p, then f t  p  f t for every t. The trigonometric functions y  sin x and cos x are periodic, with period 2 and amplitude 1. y

y

1

1

2¹

2¹ x

¹

_1

x

¹

_1

2. To obtain the graph of y  5  sin x, we start with the graph of y  sin x, then shift it 5 units upward. To obtain the graph of y   cos x, we start with the graph of y  cos x, then reflect it in the x-axis. 3. The sine and cosine curves y  a sin kx and y  a cos kx, k  0, have amplitude a and period 2k. The sine curve y  3 sin 2x has amplitude 3  3 and period 22  . 4. The sine curve y  a sin k x  b has amplitude a, period 2k, and horizontal shift b. The sine curve   2  y  4 sin 3 x   6 has amplitude 4  4, period 3 , and horizontal shift 6 . 6. f x  2  cos x

5. f x  2  sin x

y 1

y

2 ¹

2¹

¹

2¹

x

¹

2¹

x

x _2

7. f x   sin x

8. f x  2  cos x

y

y

1 ¹

2¹

x 1

SECTION 6.3 Trigonometric Graphs

9. f x  2  sin x

10. f x  1  cos x

y

y

¹

2¹

1

x

_1

11. g x  3 cos x

12. g x  2 sin x

y

2

¹

2¹

x

¹

2¹

x

¹

2¹

¹

2¹

y

2 ¹

2¹

x

13. g x   12 sin x

14. g x   23 cos x y

y

1

1 ¹

2¹

x

x

15. g x  3  3 cos x

16. g x  4  2 sin x

y

y

2

2 ¹

2¹

x

x

501

502

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

17. h x  cos x

18. h x  sin x

y

1

y

1 ¹

2¹

x

19. y  cos 2x has amplitude 1 and period .

¹

2¹

x

20. y   sin 2x has amplitude 1 and period .

y

y

1

1 ¹

2¹

x

21. y   sin 3x has amplitude 1 and period 23 .

¹

2¹

x

22. y  cos 4x has amplitude 1 and period 12 . y

y

1

1

¹

2¹

x

1

1 2

x

_1

23. y  2 cos 3x has amplitude 2 and period 23 . y 2

_1

24. y  3 sin 6x has amplitude 3 and period  3. y

1

x

3

¹ 6

_2 _3

x

SECTION 6.3 Trigonometric Graphs

25. y  10 sin 12 x has amplitude 10 and period 4.

26. y  5 cos 14 x has amplitude 5 and period 8.

y

y

10

5 2¹

4¹

x

27. y   13 cos 13 x has amplitude 13 and period 6. y

1 2

4¹

8¹

x

28. y  4 sin 2x has amplitude 4 and period . y

4 x

3¹

_

¹

2¹

x

1 2

29. y  2 sin 2x has amplitude 2 and period 1. y

2

30. y  3 sin x has amplitude 3 and period 2. y

2 1

2

x

31. y  1  12 cos x has amplitude 12 and period 2. y

1

2

x

32. y  2  cos 4x has amplitude 1 and period 12 . y 1

1

_1 1

2

x

2

x

503

504

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

  33. y  cos x   2 has amplitude 1, period 2, and horizontal shift  2.

  34. y  2 sin x   3 has amplitude 2, period 2, and horizontal shift  3.

y

y

1

2 4¹ 3

¹

2¹

¹ 3

x

  35. y  2 sin x   6 has amplitude 2, period 2, and horizontal shift  6.

7¹ 3

¹

2¹

x

  36. y  3 cos x   4 has amplitude 3, period 2, and horizontal shift   4. y

y

3

2 ¹ 6

7¹ 6

¹

13¹ 6

2¹

_

¹ 4

3¹ 4

x

  37. y  4 sin 2 x   2 has amplitude 4, period , and horizontal shift   2.

¹

y

y 1

¹

2¹

x

    39. y  5 cos 3x   4  5 cos 3 x  12 has amplitude 5, . period 23 , and horizontal shift 12

4¹

x

period 3, and horizontal shift  4. y 2

x

15¹ 4

  2  40. y  2 sin 23 x   6  2 sin 3 x  4 has amplitude 2,

y

3¹ 4

7¹ 4

¹

_4

5

5¹ 12

x

horizontal shift   4.

4

¹ 2

2¹

  38. y  sin 12 x   4 has amplitude 1, period 4, and

_2

¹ 12

7¹ 4

¹ 4

7¹ 4

13¹ 4

x

SECTION 6.3 Trigonometric Graphs

 1 1    41. y  12  12 cos 2x   3  2  2 cos 2 x  6 has amplitude 12 , period , and horizontal shift  6. y

505

    42. y  1  cos 3x   2  1  cos 3 x  6 has

amplitude 1, period 23 , and horizontal shift   6. y

1

2

1 2

1

¹ 6

2¹ 3

7¹ 6

x

  43. y  3 cos  x  12 has amplitude 3, period 2, and horizontal shift  12 .

¹

¹ 6

_6

x

¹ 2

44. y  3  2 sin 3 x  1 has amplitude 2, period 23 , and horizontal shift 1.

y

y 3

3 1

1 2

_2

x

3 2

1

2¹ -1 3

_1

x

      45. y  sin 3x    sin 3 x   3 has amplitude 1, period 46. y  cos 2  x  cos x  2 has amplitude 1, period 2 , and horizontal shift   2, and horizontal shift  2. 3

3

y

y

1

1

¹

_3

¹ 3

x

¹ 2

3¹ 2

5¹ 2

47. (a) This function has amplitude a  4, period 2k  2, and horizontal shift b  0 as a sine curve. (b) y  a sin k x  b  4 sin x

48. (a) This curve has amplitude a  2, period 2k  , and horizontal shift b  0 as a cosine curve. (b) y  a cos k x  b  2 cos 2x

49. (a) This curve has amplitude a  32 , period 2k  23 , and horizontal shift b  0 as a cosine curve. (b) y  a cos k x  b  32 cos 3x

50. (a) This curve has amplitude a  3, period 2k  4, and horizontal shift b  0 as a sine curve. (b) y  3 sin 12 x

51. (a) This curve has amplitude a  12 , period 2k  , and horizontal shift b    3 as a cosine curve.   1  (b) y   2 cos 2 x  3

1 , period 2  , and horizontal shift b   3 as a cosine curve. 52. (a) This curve has amplitude a  10 k 4   1 cos 2 x  3 (b) y   10 4

x

506

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

53. (a) This curve has amplitude a  4, period 2k  32 , and horizontal shift b   12 as a sine curve.   (b) y  4 sin 43 x  12 54. (a) This curve has amplitude a  5, period 2k  1, and horizontal shift b   14 as a sine curve.   (b) y  5 sin 2 x  14 55. f x  cos 100x, [01 01] by [15 15]

56. f x  3 sin 120x, [01 01] by [4 4]

1

-0.1

0.1

-0.1

0.1

-1

57. f x  sin

x , [250 250] by [15 15] 40

58. f x  cos

x , [0 500] by [1 1] 80 1

1

0 -200

200

200

400

-1 -1

59. y  tan 25x, [02 02] by [3 3]

60. y  csc 40x, [01 01] by [10 10] 10

2

-0.2

0.2

-0.1

0.1

-2 -10

61. y  sin2 20x, [05 05] by [02 12]

62. y 

 tan 10x, [02 02] by [1 4] 4

1.0

2

0.5

-0.5

0.5

-0.2

0.2

SECTION 6.3 Trigonometric Graphs

63. f x  x, g x  sin x

64. f x  sin x, g x  sin 2x 6

g

f+g

2 -4

-2

2

f

4

-6

507

0 -2

2

4

f

g 1

6

-6

-4

-4

f+g

-2

0

2

4

6

-1

-6

-2

65. f x  sin 3x, g x  cos 12 x

_5

66. f x  05 sin 5x, g x   cos 2x.

2

f+g

2

1

g

1

0 _1 _2

5

f+g f

_2

2 _1

f

g

_2

67. y  x 2 sin x is a sine curve that lies between the graphs of y  x 2 and y  x 2 .

68. y  x cos x is a cosine curve that lies between the graphs of y  x and y  x.

200

5

-10

-5

10

5 -5

-200

69. y 



x sin 5x is a sine curve that lies between the   graphs of y  x and y   x. 2

cos 2x is a cosine curve that lies between the 1  x2 1 1 and y   . graphs of y  1  x2 1  x2

70. y 

1 2 -2

4 -5

5 -1

508

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

71. y  cos 3x cos 21x is a cosine curve that lies between the graphs of y  cos 3x and y   cos 3x.

72. y  sin 2x sin 10x is a sine curve that lies between the graphs of y  sin 2x and y   sin 2x.

1

1

-0.5

0.5

-0.5

-1

0.5 -1

73. y  sin x  sin 2x. The period is 2, so we graph the

function over one period,  . Maximum value 176 when x  094  2n, minimum value 176 when x  094  2n, n any integer.

74. y  x  2 sin x, 0  x  2. Maximum value 697 when x  524, minimum value 068 when x  105.

5

2 0 2 -2

4

6

2 -2

75. y  2 sin x  sin2 x. The period is 2, so we graph the

function over one period,  . Maximum value 300 when x  157  2n, minimum value 100 when x  157  2n, n any integer.

cos x . The period is 2, so we graph the function 2  sin x over one period. Maximum value 058 when

76. y 

x  576  2n (exact value x  116  2n); Minimum value 058 when x  367  2n (exact value x  76  2n) for any integer n.

2

1 -2

2 0

-2

2

4

6

-1

77. cos x  04, x  [0 ]. The solution is x  116.

78. tan x  2, x  [0 ]. The solution is x  111.

1 2 0 1

2

3 0

-1

0

1

2

3

SECTION 6.3 Trigonometric Graphs

79. csc x  3, x  [0 ]. The solutions are x  034, 280.

509

80. cos x  x, x  [0 ]. The solution is x  074.

4

2

2 0 0

1 0

81. f x 

1

2

2

3

3

1  cos x x

(a) Since f x  is odd.

1  cos x 1  cos x    f x, the function x x

(c) 1

(b) The function is undefined at x  0, so the x-intercepts occur when

-20

1  cos x  0, x  0  cos x  1, x  0  x  2, 4, 6, 

20 -1

(d) As x  , f x  0. (e) As x  0, f x  0.

82. f x 

sin 4x 2x

(a) f x 

sin 4x  sin 4x   f x, so the function is even. 2 x 2x

(b) The x-intercepts occur when sin 4x  0, x  0  4x  n x  14 n, n any nonzero integer.

(c) 2

 1

(d) As x  , f x  0.

-2

(e) As x  0, f x  2.

2

2  20 seconds. 10 (b) Since h 0  3 and h 10  3, the wave height is 3  3  6 feet.

83. (a) The period of the wave is

84. (a) The period of the vibration is (b) Since each vibration takes

1 2  second. 880 440

1 of a second, there are 440 vibrations 440

(c)

v 1

per second. 0

_1

0.002 0.004 0.006 0.008

t

510

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

85. (a) The period of p is

1 2  minute. 160 80

p

(c)

160

(b) Since each period represents a heart beat, there are 80 heart beats per

140

minute.

120 115 100 90 80

(d) The maximum (or systolic) is 115  25  140 and the minimum (or

diastolic) is 115  25  90. The read would be 14090 which higher than normal.

0

86. (a) The period of R Leonis is

2  312 days. 156

2 t (s)

1

b

(c)

10 8

(b) The maximum brightness is 79  21  10; the minimum brightness

6

is 79  21  58.

4 2 0

87. (a) y  sin

  x . This graph looks like a sine function

which has been stretched horizontally (stretched

more for larger values of x). It is defined only for x  0, so it is neither even nor odd.

200

400

600

t

  (b) y  sin x 2 . This graph looks like a graph of sin x which

has been shrunk for x  1 (shrunk more for larger values of x) and stretched for x  1. It is an even function, whereas

sin x is odd.

1

1

0 200

400

-5

-1

5 -1

88. (a) This function is periodic with period 2. (b) This function is not periodic. f 25  f 45.

(c) This function is not periodic; its graph on the interval [2 0] is different from its graph on the interval [0 2].

(d) This function is periodic with period 3. 89. (a) The graph of y  sin x is shown in the viewing rectangle [628 628] by [05 15]. This

(b) The graph of y  sin x is shown in the viewing rectangle

[10 10] by [15 15]. The function is not periodic. Note that while sin x  2  sin x for many values of x, it is     false for x  2 0. For example sin   2  sin 2  1   3  while sin   2  2  sin 2  1.

function is periodic with period.

1

1

-5

5 -10

10 -1

SECTION 6.4 More Trigonometric Graphs

(c) The graph of y  2cos x is shown in the viewing rectangle [10 10] by [1 3]. This function is periodic with period  2.

511

(d) The graph of y  x  [[x]] is shown in the viewing rectangle [75 75] by [05 15]. This function is periodic with period1. Be sure to turn off “connected” mode when graphing functions with gaps in their graph.

2 1 -10

10 -5

5

90. From the graph we see that the amplitude is 5 and the period is 2. In Exercises 25 and 26 in Section 7.2, we will prove the reduction formulas sin x     sin x and cos x     cos x. So starting with y  5 sin x we may also represent the curve by y  5 sin x  b where b  2n, where n is an integer; y  5 sin x  b where b  2n  1 , where   n is an integer. Then starting with y  5 cos x   2 we may also represent the curve by y  5 cos x  b where  b  2  2n, where n is an integer; y  5 cos x  b where b  2  2n  1 , where n is an integer.

6.4

MORE TRIGONOMETRIC GRAPHS

1. The trigonometric function y  tan x has period  and asymptotes x   2  n, n an integer.

2. The trigonometric function y  csc x has period 2 and asymptotes x  n, n an integer. y

y 10

5

¹/2 x

_¹/2

_10

_¹

0

¹ x

_5

   3 3. f x  tan x   4 corresponds to Graph II. f is undefined at x  4 and x  4 , and Graph II has the shape of a graph of a tangent function. 3 4. f x  sec 2x corresponds to Graph III. f is undefined at x   4 and x  4 , and Graph III has the shape of a graph of a secant function.

5. f x  cot 4x corresponds to Graph VI.

6. f x   tan x corresponds to Graph I.

7. f x  2 sec x corresponds to Graph IV.

8. f x  1  csc x corresponds to Graph V.

512

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

9. y  4 tan x has period .

_2¹

10. y  3 tan x has period .

y 10

_2¹

0

_¹

_¹

0

x

¹

_10

¹

x

¹

x

_10

11. y   32 tan x has period .

12. y  34 tan x has period . y 5

_2¹

y 10

y 4 2

_¹

0

x

¹

_2¹

0

_¹

_2 _4

_5

13. y  2 cot x has period .

14. y  2 cot x has period .

y

y 10

4

5 _2¹

0

_¹

x

¹

_¹

x

¹

_5 _10

15. y  2 csc x has period 2.

16. y  12 csc x has period 2.

y 5

_¹

y 1

¹

x

_¹

¹

x

SECTION 6.4 More Trigonometric Graphs

17. y  3 sec x has period 2.

18. y  3 sec x has period 2.

y

y

5

5

_¹

¹

19. y  tan 3x has period  3. y

y

4

4 2 ¹

0

_¹/2

x

¹/2

_¹

_2

_4

_4

0

1

2

3 x

_1

0

_1/2

_5

_20

_10

23. y  2 cot 3x has period 13 .

24. y  3 cot 2x has period 12 .

_1/3

y 10

y 10

5

5

0

1/2

1 x

1/2

1 x

5

_10

_2/3

x

y 10

10 _1

¹ ¹/2

22. y  3 tan 4x has period 14 .

y 20

_2

0

_¹/2

_2

21. y  5 tan x has period 1.

_3

¹

20. y  tan 4x has period  4.

2 _¹

x

_¹

x

1/3

2/3

1 x

_1

_1/2

0

_5

_5

_10

_10

513

514

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

   25. y  tan  4 x has period   4.

     26. y  cot  2 x has period   2.

4

2

y

y

2

2

_2

2

x

_1

1

   28. y  2 tan  2 x has period   2.

27. y  2 tan 3x has period 13 .

2

y

y

4 _0.5

4 0.5

x

29. y  csc 4x has period 24   2.

_1

x

¹

x

y

2

10

_¹ 2

¹ 2

_¹

x

2  . 2

32. y  12 sec 4x has period 12 . y 3

y

2

2 _¹

1

30. y  5 csc 3x has period 23 .

y

31. y  sec 2x has period

x

1

¹

x

_1

_1/2

0 1/8 _1 _2 _3

1/2

1 x

SECTION 6.4 More Trigonometric Graphs

2 4 33. y  5 csc 32 x has period 3  . 3

34. y  5 sec 2x has period 22   1. y

2

y

10

10 _1

1

_0.5

x

  35. y  tan x   4 has period .

0.5

  36. y  tan x   4 has period .

y 4

y 4

2 _2¹

_¹

2

0 ¹/4

2¹ x

¹

_2

_2¹

_¹

  38. y  2 cot x   3 has period . 4

0

y

2

2

_2¹

¹ 3¹/4

_¹

2¹ x

0 ¹/3

_2

_2

_4

_4

  39. y  csc x   4 has period 2.

¹

2¹

  40. y  sec x   4 has period 2.

y 4

y

2 _¹

2

0 ¹/4

_2

2¹ x

¹

_¹

¹

x

¹

x

_4

  41. y  12 sec x   6 has period 2.

  42. y  3 csc x   2 has period 2. y

y

6

1 _¹

x

_4

y 4

_2¹

2¹

_2

  37. y  cot x   4 has period .

_¹

¹ 3¹/4

0

_4

_2¹

x

¹

x

_¹

x

515

516

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

   43. y  tan 2 x   3 has period 2 .

   44. y  cot 2x   4 has period 2 . y 4

y 4

2

2 _¹/2

0¹/12

_2

¹/2 x

_¹

_¹/2

46. y  4 tan 4x  2 has period  4. y 10

y 10

5

5 0 ¹/6 ¹/2

¹ x

0 ¹/8

_¹/2

_5

_5

_10

_10

     47. y  cot 2x   2  cot 2 x  4 has period 2 . y

_¹

y

1 ¹ x

_0.5

    1 49. y  2 csc x   3  2 csc  x  3 has period y

0.5

y 10

_ 16¹ 3

1

x

x

    1 2 50. y  3 sec 14 x   6  3 sec 4 x  3 has period 8. 5

4 1

¹/2 x

48. y  12 tan x    12 tan  x  1 has period    1.

2

2  2. 

x

_4

   45. y  5 cot 3x   2 has period 3 .

_¹/2

¹

_2

_4

_¹

¹/2

0 ¹/8

_ 16¹ 3

_ 4¹ 3

_5

_10

0

8¹ 3

20¹ 3

x

SECTION 6.4 More Trigonometric Graphs

  51. y  sec 2 x   4 has period .

  52. y  csc 2 x   2 has period .

y 5

_¹

_¹/2

0

y 5

¹ x

¹/2

0

_¹

_5

x

¹

_5

    2 53. y  5 sec 3x   2  5 sec 3 x  6 has period 3 . y

  54. y  12 sec 2x    12 sec 2 x  12 has period 2  1. 2

y

10 _¹/2

¹/2

1

x

_0.5

  2  55. y  tan 23 x   6  tan 3 x  4 has period    23  32 .

0.5

x

     56. y  tan 12 x   4 has period 1  2. 2

y

y

2 2

5¹

_¹

¹

x

  57. y  3 sec  x  12 has period 2  2. y

x

    2 58. y  sec 3x   2  sec 3 x  6 has period 3 . y

2

6 1

3¹ 4

_4

1

x

_¹/2

¹/2

x

517

518

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

     59. y  2 tan 2x   3  2 tan 2 x  6 has period 2 . y

60. y  2 cot 3x  3  2 cot 3 x  1 has period 13 . y 4

4 _¹/3

¹/6

2 2¹/3 x

0

_1

1 x

_2 _4

61. (a) d t  3 tan t, so d 015  153,

d

(b)

d 025  300, and d 045  1894.

10

(c) d   as t  12 .

0

  t , so S 2  1039, S 6  0, 62. (a) S t  6 cot 12 S 8  346, and S 1175  9154.

(b)

0.1

0.2

0.3

0.4

0.5 t

S 10

(c) From the graph, it appears that S t  6 at

approximately t  3 and t  9, corresponding to 9 A . M . and 3 P. M .

(d) As t  12 , the sun approaches the horizon and the

0

10

t

man’s shadow grows longer and longer.

63. (a) If f is periodic with period p, then by the definition of a period, f x  p  f x for all x in the domain of f . 1 1 1  for all f x  0. Thus, is also periodic with period p. Therefore, f x  p f x f 1 also has period 2. Similarly, since cos x has (b) Since sin x has period 2, it follows from part (a) that csc x  sin x 1 also has period 2. period 2, we conclude sec x  cos x 64. If f and g are periodic with period p, then f x  p  f x for all x and g x  p  g x for all x. Thus f x f x  p  for all x [unless g x is undefined, in which case both are undefined.] But consider f x  sin x g x  p g x f x sin x (period 2) and g x  cos x (period 2). Their quotient is   tan x, whose period is . g x cos x 65. The graph of y   cot x is the same as the graph of y  tan x shifted  2 units to the right, and the graph of y  csc x is the  same as the graph of y  sec x shifted 2 units to the right.

SECTION 6.5 Inverse Trigonometric Functions and Their Graphs

6.5

INVERSE TRIGONOMETRIC FUNCTIONS AND THEIR GRAPHS

  1. (a) To define the inverse sine function we restrict the domain of sine to the interval   2  2 . On this interval the sine function is one-to-one and its inverse function sin1 is defined by sin1 x  y  sin y  x. For example,  1 sin1 21   6 because sin 6  2 . (b) To define the inverse cosine function we restrict the domain of cosine to the interval [0 ]. On this interval the

cosine function is one-to-one and its inverse function cos1 is defined by cos1 x  y  cos y  x. For example,

 1 cos1 12   3 because cos 3  2 .     1 sin    ; 2. The cancellation property sin1 sin x  x is valid for x in the interval   2  2 . Therefore, (i) sin 3 3      1 sin      . (ii) sin1 sin 103  103 because 103   ; and (iii) sin 2 4 4        3. (a) sin1 1  2 because sin 2  1 and 2 lies in  2  2 .      3   (b) sin1 23   3 because sin 3  2 and 3 lies in  2  2 .

(c) sin1 2 is undefined because there is no real number x such that sin x  2. 

4. (a) sin1 1    2

(b) sin1 22   4

5. (a) cos1 1     6. (a) cos1 22   4

(b) cos1 21   3

7. (a) tan1 1    4 8. (a) tan1 0  0   9. (a) cos1  12  23

10. (a) cos1 0   2

(b) cos1 1  0  3  3    1  3   (b) tan 3    (b) sin1  22    4 (b) tan1

(b) sin1 0  0

11. sin1 23  072973   13. cos1  37  201371

(c) sin1 2 is undefined.    (c) cos1  23  56    (c) cos1  22  34 

(c) tan1 33   6    3 1  3   (c) tan 6

(c) tan1 1   4   (c) sin1  12    6

  12. sin1  89  109491   14. cos1 49  111024

15. cos1 092761  275876

16. sin1 013844  013889

17. tan1 10  147113

18. tan1 26  153235

19. tan1 123456  088998

20. cos1 123456 is undefined because 123456  1.

21. sin1 025713  026005   23. sin sin1 14  14   25. tan tan1 5  5    27. sin sin1 32 is undefined because 32  1.    29. cos cos1  15   15      4 31. sin1 sin  4     33. sin1 sin 34 4

22. tan1 025713  025168   24. cos cos1 32  23   26. sin sin1 5 is undefined because 5  1.    28. tan tan1 32  32    30. sin sin1  34   34      4 32. cos1 cos  4     34 34. cos1 cos 34

519

520

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

    56 35. cos1 cos 56

    36. sin1 sin 56 6

    56 37. cos1 cos 76

     38. sin1 sin 76 6

     4 39. tan1 tan  4

     40. tan1 tan   3 3

       2  41. tan1 tan 23    3 because tan  3  tan 3 and  3 lies in  2  2 .    42. sin1 sin 114  4    3 43. tan sin1 12  tan   6 3

  44. cos sin1 0  cos 0  1

   1 45. cos sin1 23  cos  3  2

   46. tan sin1 22  tan  4 1

     2 47. sin tan1 1  sin   4  2

       3 48. sin tan1  3  sin   3  2

From the graph of y  sin1 x  cos1 x, it appears that y  157. We suspect

49. (a)

that the actual value is  2. 2

-1

(b) To show that sin1 x  cos1 x   2 , start with the identity   sin a   2   cos a and take arcsin of both sides to obtain

1  cos a. Now let a  cos1 x. Then a 2  sin    1  cos cos1 x cos1 x    sin1 x   sin1 x, so 2  sin

1

sin1 x  cos1 x   2.

From the graph of y  tan1 x  tan1 1x , it appears that y  157 for x  0

50. (a) 2

-2

and y  157 for x  0. We suspect that the actual values are   2.

2

-2

(b) To show that tan1 x  tan1 x1   2 for x  0, we start with the identity    a and take arccot of both sides to obtain tan a  cot  2 cot1 tan a    a. Now use the identity cot1 x  tan1 x1 to write   2 1 1 , we have tan1 tan1 a   x 2  a. Substituting a  tan    1     tan1 1  tan1 1    tan1 1  tan1 1 x x 2 1x 2 tan tan

1x

. For the case x  0, simply note that tan1 x   tan1 x, so for positive x, tan1 x  tan1 x1    2    1   tan1 x  tan1 x1    tan1 x  tan1 x 2.

SECTION 6.6 Modeling Harmonic Motion

521

  51. The domain of f x  sin sin1 x is the same as that of sin1 x, [1 1], and the graph of f is the same as that of y  x on [1 1].

The domain of g x  sin1 sin x is the same as that of sin x,  , because for all x, the value of sin x lies within

 the domain of sin1 x. g x  sin1 sin x  x for   2  x  2 . Because the graph of y  sin x is symmetric about the  3 line x   2 , we can obtain the part of the graph of g for 2  x  2 by reflecting the graph of y  x about this vertical line. The graph of g is periodic with period 2. y

y

¹/2

1

y=g(x)

y=f(x) _1

1

x

_3¹/2

_¹

_¹/2

_1

6.6

¹/2

¹

3¹/2 x

_¹/2

MODELING HARMONIC MOTION

1. (a) Because y  0 at time t  0, y  a sin t is an appropriate model.

(b) Because y  a at time t  0, y  a cos t is an appropriate model.

2. (a) Because y  0 at time t  0, y  kect sin t is an appropriate model.

(b) Because y  a at time t  0, y  kect cos t is an appropriate model.

2 , and the phase 3. (a) For an object in harmonic motion modeled by y  A sin kt  b the amplitude is A, the period is k   b is b. To find the horizontal shift, we factor k to get y  A sin k t  . From this form of the equation we see that the k b horizontal shift is . k (b) For an object in harmonic motion modeled by y  5 sin 4t   the amplitude is 5, the period is  2 , the phase is , and the horizontal shift is  4.

  4. Objects A and B are in harmonic motion modeled by y  3 sin 2t   and y  3 sin 2t   2 . The phase of A is  and  the phase of B is  2 . The phase difference is 2 , so the objects are moving out of phase. 6. y  3 cos 12 t

5. y  2 sin 3t 1 (a) Amplitude 2, period 23 , frequency period  23 .

(b)

2  4, frequency (a) Amplitude 3, period 12

y

1 1 period  4 .

2

(b) ¹ 3

2¹ 3

y 3

x 2¹

4¹ x

522

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

7. y   cos 03t

8. y  24 sin 36t

  20 , frequency 3 . (a) Amplitude 1, period 203 3 20 y

(b)

  5 , frequency 9 . (a) Amplitude 24, period 236 9 5 y

(b)

2.4

1

5¹ 18

5¹ 9

x

20¹ x 3

10¹ 3

_1

    3t    025 cos 9. y  025 cos 15t   3  2 3   025 cos 32 t  29

2  4 , frequency 3 . (a) Amplitude 025, period 32 3 4

(b)

10. y   32 sin 02t  14   32 sin 02 t  7   10, frequency 1 . (a) Amplitude 32 , period 202 10 y

(b)

1.5

y 1 -7 2¹ 9

8¹ 9

x

14¹ 9

5¹-7

x

10¹-7

_1.5

_1

    11. y  5 cos 23 t  34  5 cos 23 t  98

12. y  16 sin t  18

2  3, frequency 1 . (a) Amplitude 5, period 23 3 y

(b)

(a) Amplitude 16, period 2, frequency 21 . (b)

y 1.6

5

¹

3¹

1.8+2 9

_8

3¹ 9 2 _8

3¹ _ 9 8

x

1.8

1.8+2 1.8+¹

13. The amplitude is a  10 cm, the period is 2k  3 s, and f 0  0, so f t  10 sin 23 t. 14. The amplitude is 24 ft, the period is 2k  2 min, and f 0  0, so f t  24 sin t. 5 Hz, and f 0  0, so f t  6 sin 10t. 15. The amplitude is 6 in., the frequency is 2k  

16. The amplitude is 12 m, the frequency is 2k  05 Hz, and f 0  0, so f t  12 sin t. 17. The amplitude is 60 ft, the period is 2k  05 min, and f 0  60, so f t  60 cos 4t. 18. The amplitude is 35 cm, the period is 2k  8 s, and f 0  35, so f t  35 cos  4 t. 19. The amplitude is 24 m, the frequency is 2k  750 Hz, and f 0  24, so f t  24 cos 1500t. 20. The amplitude is 625 in., the frequency is 2k  60 Hz, and f 0  625, so f t  625 cos 120t.

x

1.8+2¹

SECTION 6.6 Modeling Harmonic Motion

21. (a) k  2, c  15, and f  3    6, so we have

22. (a) k  15, c  025, and f  06    12, so we

y  2e15t cos 6t. (b)

have y  15e025t cos 12t.

y 2

(b)

23. (a) k  100, c  005, and p  4     2 , so we have y  100e005t cos  2 t. y 100

24. (a) k  075, c  3, and p  3    23 , so we have (b)

y 0.75

x

25. (a) k  7, c  10, and p   6    12, so we have

26. (a) k  1, c  1, and p  1    2, so we have y  et sin 2t. y

(b)

y

x

¹/5

y  7e10t sin 12t.

0.5

1

27. (a) k  03, c  02, and f  20    40, so we

x

1

x

¹/30

28. (a) k  12, c  001, and f  8    16, so we

have y  03e02t sin 40t. (b)

x

5

y  075e3t cos 23 t.

4

(b)

y 15

x

1

(b)

have y  12e001t sin 16t.

y

(b)

y 10

0.2 0.2

523

0.4

0.6

0.8

x

    29. y  5 sin 2t   2 has amplitude 5, period , phase 2 , and horizontal shift 4 .     30. y  10 sin t   3 has amplitude 10, period 2, phase 3 , and horizontal shift 3 .

31. y  100 sin 5t   has amplitude 100, period 25 , phase , and horizontal shift   5.    2 32. y  50 sin 12 t   5 has amplitude 50, period 4, phase  5 , and horizontal shift  5 .

1

2

x

524

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

    33. y  20 sin 2 t   4 has amplitude 20, period , phase 2 , and horizontal shift 4 .    has amplitude 8, period  , phase   , and horizontal shift   . 34. y  8 sin 4 t  12 2 3 12     5 35. y1  10 sin 3t   ; y  10 sin 3t  2 2 2 5 (a) y1 has phase  2 and y2 has phase 2 .

y 10

(d)

5 (b) The phase difference is  2  2  2.

yÁ, yª

5

(c) Because the phase difference is a multiple of 2, the curves are in

_¹

phase.

0

¹ x

_5 _10

    36. y1  15 sin 2t   3 ; y2  15 sin 2t  6  (a) y1 has phase  3 and y2 has phase 6 .

y

(d)

  (b) The phase difference is  3  6  6.

10 _¹

(c) Because the phase difference is not a multiple of 2, the curves are out of phase.

yª

yÁ

0

¹ x

_10

    80 sin 5t   ; y  80 sin 5t    37. y1  80 sin 5 t  10 2 2 3  (a) y1 has phase  2 and y2 has phase 3 .

y 80

(d)

  (b) The phase difference is  2  3  6.

(c) Because the phase difference is not a multiple of 2, the curves are

_¹

out of phase.

yª

0

yÁ

¹ x

_80

    3  20 sin 2t  3 38. y1  20 sin 2 t   2  20 sin 2t  ; y2  20 sin 2 t  2 (a) y1 has phase  and y2 has phase 3.

y 20

(d)

(b) The phase difference is   3  2. (c) Because the phase difference is a multiple of 2, the curves are in phase.

yÁ, yª _¹

0

_20

¹ x

SECTION 6.6 Modeling Harmonic Motion

    40. y  a sin 2 915  107 t . The period is

39. y  02 cos 20t  8  (a) The frequency is 20 2  10 cycles/min.

1 2     107  109  108 and the 915 2 915  107   2 915  107  915  107 . frequency is 2

y

(b)

525

8.4 8.2 8 7.8

0

0.2

0.4

t

(c) Since y  02 cos 20t  8  02 1  8  82 and when t  0, y  82, the maximum displacement is 82 m.

41. p t  115  25 sin 160t

1 2  1  00125, frequency  80. (a) Amplitude 25, period 160  80 period

(b)

(c) The period decreases and the frequency increases.

P 140 120 100 80 0

0.01

0.02

t

42. (a) When t  0, , 2,    we have cos 2t  1 so y is maximized at y  8900.

(b) The length of time between successive periods of maximum population is the length of a period which is 2    314 years. 2

2 2 5  43. The graph resembles a sine wave with an amplitude of 5, a period of 25 , and no phase shift. Therefore, a  5,    5, and a formula is d t  5 sin 5t. 44. From the graph we see that the amplitude is 6 feet and the period is 12 hours. Also, the sine curve is shifted 6 hours (exactly half its period) to the right, which is equivalent to reflecting the curve about the t-axis. Thus, the equation is    t  6 sin  t. y   6 sin 212 6

 1 cycle/hour  1  12 45. a  21, f  12    6 . So, 2   y  21 sin 6 t (assuming the tide is at mean level and rising when t  0).

y

20

0

2

4

6

8

10

12 t

_20

2  1    2. So y  2 cos 2t.  47. Since the mass travels from its highest point (compressed spring) to its lowest point in 12 s, it completes half a period in 12 s.

46. a  2,

So, 12 one period  12 s



1  2  1 2  2



  2. Also, a  5. So y  5 cos 2t.

526

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

48. (a) m  10 g, k  3, a  5 cm. Then f t  5 cos





3 10 t .

 (b) The function f t  a cos kmt describes an object oscillating in simple harmonic motion, so by  comparing it with the general equation y  a cos t we see that   km. This means the frequency is   1  km k    . f  2 2 2 m  k (c) If the mass is increased, then the denominator of increases, so overall the frequency decreases. If the frequency m has decreased, then by definition the oscillations are slower.  k increases. Thus, overall the frequency increases. If (d) If a stiffer spring is used, k is larger and so the numerator of m the frequency has increased, then by definition the oscillations are faster. 49. Since the Ferris wheel has a radius of 10 m and the bottom of the wheel is 1 m above the ground, the minimum height is 2  , and so y  11  10 sin   t , where t 1 m and the maximum height is 21 m. Then a  10 and  20 s    10 10  is in seconds. 50. Let f t represent the measure of the angle  at time t. The amplitude of this motion is 10. Since the period is 2s, we have 2  2    . Thus f t  10 sin t.    2  51. a  02,  10    . Then y  38  02 sin  5t .  5   1 2  2    . So R t  20  15 sin t , where R is in millions of miles and t is in days. 52. a  15, 2 54 54 54 53. The amplitude is 12 100  80  10 mmHG, the period is 24 hours, and the phase shift is 8 hours, so   f t  10 sin 12 t  8  90. 54. E 0  310, frequency is 100 

  100 2



  200. Then, E t  310 cos 200t. The maximum voltage

310 produced occurs when cos 2t  1, and hence is E max  310 V. The rms voltage is   219 V. 2 55. (a) The maximum voltage is the amplitude, that is, Vmax  a  45 V.

(b) From the graph we see that 4 cycles are completed every 01 seconds, or equivalently, 40 cycles are completed every second, so f  40.   f  40. (c) The number of revolutions per second of the armature is the frequency, that is, 2  (d) a  45, f   40    80. Then V t  45 cos 80t. 2   1130 56. (a) As the car approaches, the perceived frequency is f  500 1130110  5539 Hz. As it moves away, the perceived   1130 frequency is f  500 1130110  4556 Hz.      1130 1130 (b) The frequency is  500 1130110 , so   1000 1130110  11078 (approaching) or 9113 (receding). 2 Thus, models are y  A sin 11078t and A sin 9113t.   1    . Since f 0  0, f t  e09t sin t. 57. k  1, c  09, and 2 2  58. k  6, c  28, and  2    4. 2 (a) Since f 0  6, f t  6e28t cos 4t.

CHAPTER 6 1 (b) The amplitude of the vibration is 05 when 6e28t  05  e28t  12 1  ln 12  089 s. t   12 ln 12 28

59.

kect  4  ectct3  4 kect3

60. (a)

3 kec0 5  06 kec2





e2c  5

e3c  4 



2c  ln 5

3c  ln 4 



1  28t  ln 12

Review

527



c  13 ln 4  046.

c  12 ln 5  080.

(b) f t  3e08t cos t. If the frequency is 165 cycles per second, then

  165 2



  330. Thus,

f t  3e08t cos 330t. 61. (a) For fan A, the amplitude is 1, the frequency is 100, and the phase is 0, so

  100    200 and an equation is 2

y  sin 200t.

For fan B, the amplitude is 1, the frequency is 100, and the phase is  34 , so again   200 and an equation is   y  sin 200t  34 .

(b) The phase difference is 34 , so the fans are out of phase. If fan A were rotated 34 counterclockwise, the phase difference would become 0 and the fans would be in phase.   62. (a) E I  50 sin 120t, so the voltage phase is 0. E I I  50 sin 120t  54 , so the voltage phase is 54 .

(b) The phase difference is 54 , so the generators are out of phase. If the armature in the second generator were rotated 34 , then the phase difference would become 54  34  2, and the generators would produce voltage in phase.

1  63. From left to right: at t   6 , sin t  2 and the values are increasing; at t  2 , sin t  1 and the values reach a maximum;

at t  56 , sin t  12 and the values are decreasing; at t  , sin t  0 and the values are decreasing; at t  76 , sin t   12 and the values are decreasing; at t  32 , sin t  1 and the values reach a minimum, and at t  116 , sin t   12 and the values are increasing.

64. From left to right: new moon, waxing crescent moon, waxing gibbous moon, full moon, waning gibbous moon, third quarter moon, waning crescent moon, new moon. Tides and werewolf sightings are in phase with the lunar cycle; paying rent and cellphone bills are out of phase.

CHAPTER 6 REVIEW    2  2   1. (a) Since  23  12  34  14  1, the point P  23  12 lies on the unit circle. 

(b) sin t  12 , cos t   23 , tan t 

1  2   3 . 3  23

   2  2 9  1  1, the point P 3   4 lies on the unit circle. 2. (a) Since 35   45  25 2 5 5 4 (b) sin t   45 , cos t  35 , tan t  35   43 . 5

528

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

3. t  23

4. t  53

(a) t    23   3    3 1 (b) P  2  2

   (c) sin t  23 , cos t   12 , tan t   3, csc t  2 3 3 , 

  (c) sin t   23 , cos t  12 , tan t   3, 



csc t   2 3 3 , sec t  2, and cot t   33 .

sec t  2, and cot t   33 . 5. t   114

6. t   76

  (a) t  3   114   4     (b) P  22   22

   (c) sin t   22 , cos t   22 , tan t  1, csc t   2,  sec t   2, and cot t  1. 

2 7. (a) sin 34  sin  4  2

(a) t  2  53   3    3 1 (b) P 2   2

(a) t  76     6    3 1 (b) P  2  2





(c) sin t  12 , cos t   23 , tan t   33 , csc t  2,   sec t   2 3 3 , and cot t   3.

 8. (a) tan  3  3    (b) tan   3  3



2 (b) cos 34   cos  4  2

10. (a) cos  5  080902   (b) cos  5  080902

9. (a) sin 11  089121 (b) cos 11  045360 11. (a) cos 92  cos  2 0

12. (a) sin  7  043388

(b) sec 92 is undefined

(b) csc  7  230476

13. (a) tan 52 is undefined

14. (a) sin 2  0

(b) cot 52  cot  2 0

(b) csc 2is undefined.



15. (a) tan 56   33  (b) cot 56   3

1 16. (a) cos  3  2 1 (b) sin  6  2

sin t tan t cos t  sin t  sin t  17. cos t cos t cos2 t 1  sin2 t

sin2 t 1  cos2 t 1 1    cos t cos2 t cos t cos3 t sin t sin t sin t    (because t is in quadrant IV, cos t is positive). 19. tan t  2 cos t  1  sin t 1  sin2 t 18. tan2 t  sec t  tan2 t 

20. sec t 

1 1 1     (because t is in quadrant II, cos t is negative). 2 cos t  1  sin t  1  sin2 t 5

5 , cos t   12 . Then tan t  13   5 , csc t  13 , sec t   13 , and cot t   12 . 21. sin t  13 13 12 5 12 5  12 13

22. sin t   12 , cos t  0. Since sin t is negative and cos t is positive, t is in quadrant IV. Thus, t determines the terminal point       2 3 3 , csc t  2, sec t  , cot t   3. P 23   12 , and cos t  23 , tan t   3 3

CHAPTER 6

Review

529

  1 cos t 5 2  2 5 . Now cot t  , so 2 . Since csc t  sin t , we know sin t  5 5 sin t      1 1 1       5   5. cos t  sin t  cot t  2 5 5   12   55 , and tan t     2 while sec t  5 1 5 cos t 2  5

23. cot t   12 , csc t 

24. cos t   35 , tan t  0. Since cos t and tan t are both negative, t is in quadrant II. Thus, t determines the terminal point   5 P  35 , 45 , and sin t  45 , tan t   43 , csc t  54 , sec t   , cot t   34 . 3  2 1 sin t   sin t  14 cos t. Thus, cos2 t  sin2 t  1  cos2 t  14 cos t  1  25. tan t  14 , so cot t  4 and cos t 4 

17 17 2 cos2 t  16 17  sec t  16 . Because cosine and secant are negative in Quadrant III, we have sec t   4 , and thus, 

sec t  cot t  4  417 .   8 , so csc t   17 and cos2 t  1  sin2 t  1   8 2  225  sec2 t  289 . Because cosine and secant 26. sin t   17 8 17 289 225 17 17 119 are positive in Quadrant IV, sec t  17 15 , and so csc t  sec t   8  15   120 .

27. cos t  35 , so because the terminal point is in Quadrant I, sin t  45 . Thus, tan t  43 and sec t  53 , so tan t  sec t  43  53  3. 28. sin2 t  cos2 t  1 for any value of t. 29. y  10 cos 12 x

30. y  4 sin 2x

2 (a) This function has amplitude 10, period 1  4,

(a) This function has amplitude 4, period 22   1, and phase shift 0.

2

and phase shift 0.

y

(b)

y

(b)

4 10 2¹

4¹

0.5

x

31. y   sin 12 x 2  4, and (a) This function has amplitude 1, period 12

(a) This function has amplitude 2, period 2, and phase shift  4.

(b)

y

x

  32. y  2 sin x   4

phase shift 0. (b)

1

y

2

1

2¹ 2¹

4¹

x

¹/4

¹

x

530

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

33. y  3 sin 2x  2  3 sin 2 x  1 (a) This function has amplitude 3, period 22  , and

(a) This function has amplitude 1, period 22  , and phase shift  2.

phase shift 1. (b)

  34. y  cos 2 x   2

y

y

(b)

3 x 1-¹/2

1

1+¹/2

1

¹/2

       cos  x  1 35. y   cos  x  2 6 2 3

(a) This function has amplitude 1, period 2 2  4, and phase shift  13 . y

(b)

x

¹

    36. y  10 sin 2x   2  10 sin 2 x  4

(a) This function has amplitude 10, period 22  , and phase shift  4.

(b)

y

1

_13/3

11/3 x

_1/3

10 x ¹/2

¹

37. From the graph we see that the amplitude is 5, the period is  2 , and there is no phase shift. Therefore, the function is y  5 sin 4x. 38. From the graph we see that the amplitude is 2, the period is 4 since 14 of the period has been completed at 1, 2, and there is 2  4k   no phase shift. Thus, 2 . Therefore, the function is y  2 sin 2 x. k 39. From the graph we see that the amplitude is 12 , the period is 1, and there is a phase shift of  13 . Therefore, the function is   y  12 sin 2 x  13 .

40. From the graph we see that the amplitude is 4, the period is 43 . Thus, 2k  43  k  32 . The phase shift is   3.   3  Therefore, the function is y  4 sin 2 x  3 . 41. y  3 tan x has period .

42. y  tan x has period    1.

y

y

10

_¹

¹

2

x _1

1

x

CHAPTER 6

  43. y  2 cot x   2 has period .

Review





1 44. y  sec 12 x   2  sec 2 x   has period

y

2    4. 1 2

4

y

x

¹

2 _2¹

  45. y  4 csc 2x    4 csc 2 x   2 has period 2  . 2

2¹

x

5¹/6

x

  46. y  tan x   6 has period . y

y

2 5

_¹

_¹/6 ¹

x

   1  47. y  tan 12 x   8  tan 2 x  4 has period 1  2. 2

1 48. y  4 sec 4x has period 24   2. y

y

10 2 ¹/4

5¹/4

x

_0.5 _0.25

0.25

x

0.5

       1 1  3 50. cos1  12  23 51. sin1 sin 136   52. cos 6 2         . 53. y  100 sin 8 t  16  100 sin 8t  2 has amplitude 100, period 4 , phase  2 , and horizontal shift  16     3 has amplitude 80, period 2 , phase 3 , and horizontal shift  .  80 sin 3t  54. y  80 sin 3 t   2 2 3 2 2

49. sin1 1   2

      3 5 55. y1  25 sin 3 t   2  25 sin 3t  2 ; y2  10 sin 3t  2 (a) y1 has phase 32 and y2 has phase 52 .

y

(d)

20

(b) The phase difference is 32  52  . (c) Because the phase difference is not a multiple of 2, the curves are out of phase.

_¹

0 _20

yÁ

yª ¹ x

531

532

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

      56. y1  50 sin 10t   2 ; y2  50 sin 10 t  20  50 sin 10t  2  (a) y1 has phase  2 and y2 has phase 2 .

y

(d)

yÁ, yª

40

 (b) The phase difference is  2  2  0.

(c) Because the phase difference is a multiple of 2, the curves are in

0

_¹/2

phase.

¹/2 x

_40

58. (a) y  sin cos x

57. (a) y  cos x

1

1 -5 -5

5 -1

5

(b) This function has period .

(b) This function has period 2.

(c) This function is even.

(c) This function is even.

  59. (a) y  cos 201x

60. (a) y  1  2cos x 4 1 2

-50

50 -1

-10

10

(b) This function is not periodic.

(b) This function has period 2.

(c) This function is neither even nor odd.

(c) This function is even.

61. (a) y  x cos 3x

62. (a) y 



x sin 3x (x  0)

5

0 -5

5

5

-5

(b) This function is not periodic.

(b) This function is not periodic.

(c) This function is even.

(c) This function is neither even nor odd.

10

CHAPTER 6

63. y  x sin x is a sine function whose graph lies between those of y  x and y  x.

Review

533

64. y  2x cos 4x is a cosine function whose graph lies between the graphs of y  2x and y  2x .

10

5

-10

10

-2

-10

2 -5

65. y  x  sin 4x is the sum of the two functions y  x and y  sin 4x.

66. y  sin2 x  cos2 x is the sum of the two functions

y  sin2 x and y  cos2 x. Note that sin2 x  cos2 x  1

for all x. 2

1 -2

2 -2 -5

67. We graph y  f x  cos x  sin 2x in the viewing

rectangle [0 2]  [2 2] and see that the function has local maxima of approximately f 063  176 and

5

68. We graph y  f x  cos x  sin2 x in the viewing

rectangle [0 2]  [2 2] and see that the function has local maxima of approximately

f 414  037 and local minima of approximately

f 105  f 524  125 and local minima of

periodic with period 2.

periodic with period 2.

f 251  176 and f 528  037. The function is

f 0  f 2  1 and f   1. The function is 2

2

0

0 2

4

2

6

4

6

-2

-2

69. We want to find solutions to sin x  03 in the interval [0 2], so we plot the functions y  sin x and

y  03 and look for their intersection. We see that the

graphs intersect at x  0305 and at x  2837.

70. We want to find solutions to cos 3x  x in the interval [0 ], so we plot the functions y  cos 3x and y  x and

look for their intersection. We see that the graphs intersect at x  0390. 1

1 0

0 2 -1

4

1

6 -1

2

3

534

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

71. y1  cos sin x, y2  sin cos x (a)

1

yÁ

(b) y1 has period , while y2 has period 2. (c) sin cos x  cos sin x for all x.

-10

0

10

-1

yª

72. The amplitude is a  50 cm. The frequency is 8 Hz, so   8 2  16. Since the mass is at its maximum displacement when t  0, the motion follows a cosine curve. So a function describing the motion of P is f t  50 cos 16t. 73. The amplitude is 12 100  50 cm, the frequency is 4 Hz, so   4 2  8. Since the mass is at its lowest point when t  0, a function describing the distance of the mass from its rest position is f t  50 cos 8t. (b)

74. (a) The initial amplitude is 16 cm and the frequency is 14 Hz, so

20

a function describing the motion is y  16e072t cos 28t. (c) When t  10 s, y  16e72 cos 28  00119 cm.

0 2 -20

CHAPTER 6 TEST 

1. Since P x y lies on the unit circle, x 2  y 2  1 y   1  quadrant. Therefore y is negative



y   56 .

  2 11 6

 5   25 36   6 . But P x y lies in the fourth

 2 9 , and so x   3 . From the diagram, 2. Since P is on the unit circle, x 2  y 2  1  x 2  1  y 2 . Thus, x 2  1  45  25 5   x is clearly negative, so x   35 . Therefore, P is the point  35  45 . (a) sin t  45

(b) cos t   35

4

(c) tan t  53   43 5

(d) sec t   53  13   22 4   3 (d) csc 2  1

3. (a) sin 76  05    (c) tan  53  3

(b) cos

sin t sin t   . But t is in quadrant II cos t  1  sin2 t sin t . Thus, tan t    1  sin2 t

4. tan t 

8 , t in quadrant III 5. cos t   17

quadrant III) 1 





cos t is negative, so we choose the negative square root.

tan t  cot t  csc t  1 

1 1 2 .   1  15   15 64  1  289 17

1  (since t is in  1  cos2 t

CHAPTER 6

6. y  5 cos 4x (a) This function has amplitude5, period 24   2, phase 0, and horizontal shift 0.

535

1  7. y  2 sin 12 x   6  sin 2 x  3

2 (a) This function has amplitude2, period 1  4, 2

 phase  6 , and horizontal shift 3 .

y

(b)

Test





5

y 2

(b)

0

¹/2 x

¹/4

¹/3

   9. y  tan 2 x   4 has period 2 .

8. y   csc 2x has period 22  . y

y

2 _¹/2

2 ¹/2

_¹/2

¹/2

x

   (b) cos1  23  56

10. (a) tan1 1   4

13¹/3 x

7¹/3

x

(c) tan1 tan 3  0

  (d) sin sin1 21  12

2 11. From the graph, we see that the amplitude is 2 and the phase shift is   3 . Also, the period is , so k     k  2  2. Thus, the function is y  2 sin 2 x   3 .

    12. y1  30 sin 6t   2 ; y2  30 sin 6t  3  (a) y1 has phase  2 and y2 has phase 3 .

y 30

(d)

  (b) The phase difference is  2  3  6.

(c) Because the phase difference is not a multiple of 2, the curves are out of phase.

_¹/2

0

yª



yÁ

¹/2 x

_30

13. y  (a)

cos x 1  x2

(b) The function is even. 1.0

(c) The function has a minimum value of approximately 011 when x  254 and a maximum value of 1 when x  0.

0.5

-10

10

14. The amplitude is 12 10  5 cm and the frequency is 2 Hz. Assuming that the mass is at its rest position and moving upward when t  0, a function describing the distance of the mass from its rest position is f t  5 sin 4t.

536

FOCUS ON MODELING

(b)

15. (a) The initial amplitude is 16 in. and the frequency

20

is 12 Hz, so a function describing the motion is y  16e01t cos 24t.

0 0.5

1.0

-20

FOCUS ON MODELING Fitting Sinusoidal Curves to Data

1. (a) See the graph in part (c). (b) Using the method of Example 1, we find the vertical shift b  12 maximum value  minimum value  12 21  21  0, the amplitude 2  2 6  0  12 (so a  12 maximum value  minimum value  12 21  21  21, the period     05236), and the phase shift c  0. Thus, our model is y  21 cos 6 t. (c)

y

(d) Using the SinReg command on the TI-83, we find

2

y  2048714222 sin 05030795477t  1551856108

1

 00089616507.

y 0

2

4

6

8

10

12

2

14 x

_1

1

_2

0

The curve fits the data quite well.

2

4

6

8

10

12

14 x

_1 _2

 (e) Our model from part (d) is equivalent to y  205 cos 050t  155   2  001  205 cos 050t  002  001. This is the same as the function in part (b), correct to one decimal place. 

2. (a) See the graph in part (c). (b) Using the method of Example 1, we find the vertical shift b  12 maximum value  minimum value  12 200  95  1475, the amplitude 2  2 275  125  300 (so a  12 maximum value  minimum value  12 200  95  525, the period    002094), and the phase shift c  25. Thus, our model is y  525 cos 002094 t  25  1475.

Fitting Sinusoidal Curves to Data y

(c)

(d) Using the SinReg command on the TI-83, we find

240

y  4970329025 sin 00200800412x  2093611012

200

 1491294508.

y

160

240

120

200

80

160

40 0

537

120

400 x

200

80

The curve fits the data quite well.

40 0

200

400

x

  (e) Our model from part (d) is equivalent to y  4970 cos 002t  209   2  14913  4970 cos 002t  052  14913. This is close but not identical to the function in part (b).

3. (a) See the graph in part (c). (b) Using the method of Example 1, we find the vertical shift b  12 maximum value  minimum value  12 251  10  1305, the amplitude 2  2 15  09  12 (so a  12 maximum value  minimum value  12 251  10  1205, the period    5236), and the phase shift c  03. Thus, our model is y  1205 cos 5236 t  03  1305. (c)

y 30

(d) Using the SinReg command on the TI-83, we find y  1171905062 sin 5048853286t  02388957877

10

0

 1296070536.

y 30

20

20

1

x

10

The curve fits the data fairly well. 0

1

x

  (e) Our model from part (d) is equivalent to y  1172 cos 505t  024   2 1296  1172 cos 505t  1331296. This is close but not identical to the function in part (b). 4. (a) See the graph in part (c). (b) Using the method of Example 1, we find the vertical shift b  12 maximum value  minimum value  12 063  010  0265, the amplitude 2  2 55  25  6 (so a  12 maximum value  minimum value  12 063  010  0365, the period    1047), and the phase shift c  05. Thus, our model is y  0365 cos 1047 t  05  0265.

538

FOCUS ON MODELING

(c)

y

(d) Using the SinReg command on the TI-83, we find

0.6

y  0327038879 sin 1021164911t  2118186963

0.4

0.6

0.2 0

 02896017397.

y

2

4

6

0.4

x

0.2

_0.2

The curve fits the data reasonably well. 0

2

4

6

x

_0.2

  (e) Our model from part (d) is equivalent to 033 cos 102t  212   2  029  033 cos 102t  055  029. This is the same as the function in part (b), correct to one decimal place.

5. (a) See the graph in part (c). (b) Let t be the time since midnight. We find a function of the form y  a sin  t  c  b. a  12 374  366  04.   026. b  1 374  366  37. Because the maximum value occurs at t  16, we The period is 24 and so   224 2 get c  16. Thus the function is y  04 cos 026 t  16  37. y

(c)

37

36 0

10

20

x

Time (d) Using the SinReg command on the TI-83 we obtain the function y  a sin bt  c  d, where a  04, b  026, c  264, and d  370. Thus we get the model y  04 sin 026t  264  370.

6. (a) See the graph in part (c). (b) Let t be the time years. We find a function of the form y  a sin  t  c  b, where y is the owl population.

  052. b  1 80  20  50. Because the a  12 80  20  30. The period is 2 9  3  12 and so   212 2 values start at the middle we have c  0. Thus the function is y  30 sin 052t  50.

Fitting Sinusoidal Curves to Data

539

y

(c)

80 60 40 20

0

2

4

6

8

10

12 x

Year (d) Using the SinReg command on the TI–83 we find that for the function y  a sin bt  c  d, where a  258, b  052, c  002, and d  506. Thus we get the model y  258 sin 052t  002  506.

7. (a) See the graph in part (c). (b) Let t be the time since 1985. We find a function of the form y  a sin  t  c  b. a  12 63  22  205. The   052. b  1 63  22  425. The average value occurs in the ninth year, periodis 2 15  9  12 and so   220 2 so c  6. Thus, our model is y  205 sin 052 t  6  425. y

(c)

60

40

20

0

2

4

6

8

10

12

14

x

Year since 1985 (d) Using the SinReg command on the TI-83, we find that for the function y  a sin bt  c  d, where a  178, b  052, c  311, and d  424. Thus we get the model y  178 sin 052t  311  424.

8. (a) See the graph in part (c). (b) Let t be the number of years since 1968. We find a function of the form y  a cos  t  c  b. We have

a  12 157  3  77. The number of sunspots varies over a cycle of approximately 11 years (as the problem   057. Also, we have b  1 157  3  80. The first maximum value seems to occur at indicates), so   211 2

approximately c  1. Thus, our model is y  77 cos 057 t  1  80, where y is the average daily sunspot count and t is the number of years since 1968.

540

FOCUS ON MODELING

(c)

y 160 140 120

(c)

100 80 60

(d)

40 20 0 1970

1980

1990

2000

2010

Year

(d) Using the SinReg command on the TI-83, we find the function y  a sin bt  c  d, where a  a  5527740309, b  05772889486, c  1032016596, and d  6397239705. Thus we get the model y  5527sin 058t  103  640. Converting this to a cosine function for comparison, we have   y  5527cos 058t  103   2  640  5527 cos 058t  054  640. This model is less accurate for the peaks around 1980 and 1990, but more accurate for the early 2000s.

7

ANALYTIC TRIGONOMETRY

7.1

TRIGONOMETRIC IDENTITIES

1. An equation is called an identity if it is valid for all values of the variable. The equation 2x  x  x is an algebraic identity and the equation sin2 x  cos2 x  1 is a trigonometric identity.

2. The fact that cos x has the same value as cos x for any x can be expressed as the identity cos x  cos x. sin t 3. cos t tan t  cos t   sin t cos t 1 4. cos t csc t  cos t   cot t sin t 1 5. sin  sec   sin    tan  cos  1 1 sin     sec  6. tan  csc   cos  sin  cos  sin2 x 1 sin2 x  1  cos2 x 7. tan2 x  sec2 x      1 2 2 2 cos x cos x cos x cos2 x 1 sec x cos x  sin x  tan x 8.  1 csc x cos x sin x cos u sin2 u  cos2 u 1 9. sin u  cot u cos u  sin u   cos u    csc u sin u sin u sin u     sin2   cos2   sin2   1 10. cos2  1  tan2   cos2  1  cos2 

1  cos  1  cos2  sin2  sin  sec   cos   cos      tan  11. sin  sin  sin  cos  sin  cos  cos  cos  cos  cot  cos  sin    sec   12.  2 1 csc   sin  cos2  1  sin   sin  sin  1 sin x  sin x sec x cos x 1 13.  cos x tan x sin x 1 cos x  cos x sec x cos x  sin x 14.  cos x cot x cos x sin x sin t sin t  tan t tan t   cos t  1 15.  sin t tan t tan t cos t cos A 1  cot A  1  cot A sin A  sin A  sin A cot A  sin A  sin A  sin A  cos A 16. csc A sin A   17. cos3 x  sin2 x cos x  cos x cos2 x  sin2 x  cos x

541

542

CHAPTER 7 Analytic Trigonometry

   18. sin4   cos4   cos2   sin2   cos2  sin2   cos2   cos2   sin2  19.

sec2 x  1 tan2 x sin2 x    cos2 x  sin2 x. sec2 x sec2 x cos2 x 1 sec2 x  1 1  1  cos2 x  sin2 x Another method: sec2 x sec2 x

cos2 x 1  2 2 sec x  cos x cos x  1  cos x  sin x  sin x  cos x 20. sin x tan x sin x sin x cos x

21.

22.

23.

1  cos y cos y 1  cos y 1  cos y 1  cos y    cos y   1 cos y  1 1  sec y 1 cos y  1 1 cos y cos y 1  sin y 1  sin y 1  sin y  sin y   1 1 1  csc y 1 1  sin y sin y sin y 1  sin u cos u 1  2 sin u  sin2 u  cos2 u 1  2 sin u  1 1  sin u2  cos2 u     cos u 1  sin u cos u 1  sin u cos u 1  sin u cos u 1  sin u 2  2 sin u 2 1  sin u 2     2 sec u cos u 1  sin u cos u 1  sin u cos u

24. Note that because sin2 t  1  cos2 t  1  cos t 1  cos t, we can write sin t 1  cos t 1 cos t  csc t     cot t. 1  cos t sin t sin t sin t

25.

cos x  sec x  tan x

cos x 1  sin2 x cos2 x 1  sin x 1  sin x    1  sin x  1 sin x 1  sin x 1  sin x 1  sin x  cos x cos x

26. Because tan A   tan A,

27.

28.

1  cos t sin t  . Thus, 1  cos t sin t

cot A  1 tan A 1  tan A cot A  1     cot A 1  tan A 1  tan A tan A tan A 1  tan A

1 1 1  sin   1  sin  2     2 sec2  2 1  sin  1  sin  cos2  1  sin  2  tan2 x 1  1  tan2 x 1 1  tan2 x 1 sec2 x 1  1  1  1 2 2 2 2 2 sec x sec x sec x sec x sec x sec2 x 1 1 11  cos2 x  sec2 x sec2 x

SECTION 7.1 Trigonometric Identities

cos x 29. (a)  sec x sin x

1  sin2 x cos2 x   sin x sin x

cos x

1  sin x cos x sin2 x 1   csc x  sin x  sin x sin x

sin y tan y 1  cos2 y sin2 y cos y 30. (a)    1 csc y cos y cos y sin y 

(b) We graph each side of the equation and see that the cos x and y  csc x  sin x are graphs of y  sec x sin x identical, confirming that the equation is an identity.

cos2 y 1   sec y  cos y cos y cos y

(b) We graph each side of the equation and see that the tan x graphs of y  and y  sec x  cos x are csc x identical, confirming that the equation is an identity.

1

1

-5

5

-5

-1

sin  sin   sin    sin  tan  cos  1 cos u sec u  cos u 33. tan u cos u sin y tan y sin2 y cos y  35.  1 csc y cos y sin y 31.

5 -1

cos   cos  sin 

32.

tan x sin x   cos x  sin x sec x cos x

cot u  cot u

34.

cot x sec x cos x 1    sin x  1 csc x sin x cos x



543

1  cos2 y 1 1 1   cos y   cos y cos y cos y sec y

1  sin2  1 cos2     sin   csc   sin  sin  sin  sin  37. cos x  sin x  cos x   sin x  cos x  sin x 36.

38. cot  cos   sin    39. tan   cot  

 cos2   sin2  1 cos  cos   sin      csc  sin  sin  sin 

cos  sin2   cos2  1 sin      sec  csc  cos  sin  cos  sin  cos  sin 

40. sin x  cos x2  sin2 x  2 sin x cos x  cos2 x  1  2 sin x cos x 1 41. 1  cos  1  cos   1  cos2   sin2   csc2  cos x sin x 42.   cos2 x  sin2 x  1 sec x csc x 1 1 43.  sec2 y  1  tan2 y  2 cos2 y 1  sin y 44. csc x  sin x 

1 1  sin2 x cos2 x  sin x    cos x cot x sin x sin x sin x

    45. tan x  cot x2  tan2 x  2 tan x cot x  cot2 x  tan2 x  2  cot2 x  tan2 x  1  cot2 x  1  sec2 x  csc2 x     46. tan2 x  cot2 x  sec2 x  1  csc2 x  1  sec2 x  csc2 x  2  2   47. 1  sin2 t  cos2 t  4 sin2 t cos2 t  2 cos2 t  4 sin2 t cos2 t  4 cos2 t cos2 t  sin2 t  4 cos2 t

544

48.

CHAPTER 7 Analytic Trigonometry

2 sin x cos x sin x  cos x2  1

 

2 sin x cos x sin2 x  cos2 x  2 sin x cos x  1 2 sin x cos x 1 2 sin x cos x  1  1



2 sin x cos x   2 sin x cos x  sin2 x  cos2 x  1

sin2 x 1 cos2 x    csc x sin x sin x sin x     cos2 t 1 2 t  cos2 t 50. cot2 t  cos2 t   cos  1  cos2 t csc2 t  1  cot2 t cos2 t 2 2 sin t sin t

49. csc x cos2 x  sin x 

sin x  cos x sin x  cos x sin x  cos x sin2 x  cos2 x sin x  cos x2    sin x  cos x sin x  cos x sin x  cos x sin x  cos x sin x  cos x sin x  cos x2 2   2 52. sin x  cos x4  sin x  cos x2  sin2 x  2 sin x cos x  cos2 x  1  2 sin x cos x2

51.

sin x  cos x2 sin2 x  cos2 x



1 1  cos t  cos t cos t 1  cos2 t sec t  cos t cos t cos t    sin2 t 53.   1 1 sec t cos t 1 cos t cos t cos2 x cos x cos x 1 54. cot x  csc x cos x  1  cot x cos x  cot x  csc x cos x  csc x     sin x sin x sin x sin x cos2 x  1  sin2 x     sin x sin x sin x   55. cos2 x  sin2 x  cos2 x  1  cos2 x  2 cos2 x  1   56. 2 cos2 x  1  2 1  sin2 x  1  2  2 sin2 x  1  1  2 sin2 x  2  2    57. sin4   cos4   sin2   cos2   sin2   cos2  sin2   cos2   sin2   cos2       cos2 x 2 2 2  sin2 x  cos2 x  1 58. 1  cos x 1  cot x  sin x 1  sin2 x

sin2 t  2 sin t cos t  cos2 t sin2 t  cos2 t 2 sin t cos t 1 sin t  cos t2      2  2  sec t csc t sin t cos t sin t cos t sin t cos t sin t cos t sin t cos t   1 cos t 1 1 1 sin t    2  sec2 t  csc2 t 60. sec t csc t tan t  cot t  cos t sin t cos t sin t cos2 t sin t

59.

61.

62.

63.

64.

65.

sin2 u sin2 u 1  2 2 2 1 cos2 u  cos2 u  cos u  cos u  sin u   2 2 2 2 2 1  tan u cos u  sin2 u cos2 u  sin2 u sin u sin u cos u 1 1 cos2 u cos2 u 2 2 1  sec x 1  sec x 1    1  cos2 x  1 1  tan2 x sec2 x sec2 x 1 1 1 1   sec x  csc x cos x sin x cos x sin x  sin x cos x  sin x  cos x  sin x  cos x   cos x cos x sin x cos x sin x sin x tan x  cot x sin2 x  cos2 x   cos x sin x cos x sin x sin x  cos x sin x  cos x sin x  cos x cos x sin x   cos x sin x   sin x  cos x 1 1 sin x  cos x sec x  csc x sin x  cos x  cos x sin x cos x sin x 1  cos x sin x 1  cos x 1  cos x sin x sin x 1  2 cos x  cos2 x  sin2 x       sin x 1  cos x sin x 1  cos x 1  cos x sin x sin x 1  cos x 2  2 cos x 2 1  cos x    2 csc x sin x 1  cos x sin x 1  cos x 1  tan2 u

1

SECTION 7.1 Trigonometric Identities

cos x cos x 1 1   csc x  cot x sin x  sin x sin x  sin x cos x  cos x 1  cos x  cos x  cot x  sin x 66. 1 1 sec x  1 sin x cos x sin x 1  cos x sin x 1 1 cos x cos x  sin2 u cos2 u sin2 u  sin2 u 67. tan2 u  sin2 u    1  cos2 u  tan2 u sin2 u 2 2 2 cos u cos u cos u      4 4 2 2 2 68. sec x  tan x  sec x  tan x sec x  tan2 x  1 sec2 x  tan2 x  sec2 x  tan2 x

sin x 1 1  tan x cos x  cos x  cos x  sin x  69. sin x cos x 1  tan x cos x  sin x 1 cos x

sin2   1 1 sin   csc   cos2  cos  sin   sin  70.    cos  cos  sin   cos  cos   cot  cos  sin   1 1  sin  cos   sin  sin  1 sec x  tan x  sec x  tan x 2 sec x 2 sec x 1    2 sec x   71. sec x  tan x sec x  tan x 1 sec x  tan x sec x  tan x sec2 x  tan2 x sin  

72.

cos2 t  tan2 t  1 sin2 t



 sin2 t  tan2 t sin2 t

 1 

1 sin2 t   1  sec2 t  tan2 t cos2 t sin2 t

1  sin x 1  2 sin x  sin2 x  1  2 sin x  sin2 x 1  sin x 1  sin x2  1  sin x2    1  sin x 1  sin x 1  sin x 1  sin x 1  sin2 x 4 sin x 1 sin x    4 tan x sec x 4 cos x cos x cos2 x sin x sin x cos y  cos x sin y sin y  tan x  tan y cos x cos y cos x cos y  cos x 74. cos y  cos x sin y  sin x cos y cot x  cot y  sin x sin y sin x sin y    sin x cos y  cos x sin y sin x sin y sin x sin y    tan x tan y cos x cos y cos x sin y  sin x cos y cos x cos y   sin x  cos x sin2 x  sin x cos x  cos2 x sin3 x  cos3 x   sin2  sin x cos x  cos2 x  1  sin x cos x 75. sin x  cos x sin x  cos x tan   cot  1 tan   cot  1 sin  cos  76.     cos  sin  cos  sin  tan   cot  tan   cot  tan   cot  tan2   cot2   cos  sin  sin  cos    sin  cos  sin2   cos2 

73.

77.

1  cos  1  cos  1  cos2  sin2  sin  1  cos       sin  sin  1  cos  sin  1  cos  sin  1  cos  1  cos 

sin x  1 sin x  1 sin2 x  1  cos2 x sin x  1     2 sin x  1 sin x  1 sin x  1 sin x  1 sin x  12 1 sin  sin  cos   tan    79. 1 sin   cos  sin   cos  1  tan  cos  sin A sin A 1  cos A cos A sin A 1  cos A sin A 1  cos A 80.  cot A   cot A    cot A   1  cos A 1  cos A 1  cos A sin A 1  cos2 A sin2 A 1 cos A cos A 1      csc A sin A sin A sin A sin A

78.

545

546

81.

CHAPTER 7 Analytic Trigonometry

sec x sec x  tan x sec x sec x  tan x sec x sec x sec x  tan x     sec x sec x  tan x  2 2 sec x  tan x sec x  tan x sec x  tan x 1 sec x  tan x

82. sec   tan   sec   tan   83. 84.

85.

sec   tan  sec2   tan2  1   sec   tan  sec   tan  sec   tan 

cos  1  sin  cos  1  sin  sin  cos  1  sin  1 cos       sec   tan    1  sin  1  sin  1  sin  cos  cos  cos2  1  sin2  tan  sin  tan  sin  tan   sin  tan  sin  tan   sin  tan  sin  tan   sin        2 2 tan   sin  tan   sin  tan   sin  tan   sin  sin2  sec2   1 tan  sin  tan   sin  tan   sin    2 2 tan  sin  tan  sin  1  sin x 1  sin x 1  2 sin x  sin2 x 1 2 sin x sin2 x 1  2 sin x  sin2 x 1  sin x        1  sin x 1  sin x 1  sin x cos2 x cos2 x cos2 x cos2 x 1  sin2 x  sec2 x  2 sec x tan x  tan2 x  sec x  tan x2

86.

1  sin x 1  sin x 1  sin x 1  sin x2 1  sin x2      1  sin x 1  sin x 1  sin x cos2 x 1  sin2 x



 1  sin x 2  tan x  sec x2 cos x

cos x 1  cos x 1  cos x 1  cos x 1  cos2 x sin2 x 1       sin x sin x sin x sin x 1  cos x sin x 1  cos x sin x 1  cos x 1 1 1    1 1 cos x csc x  cot x  1  cos x sin x sin x sin x

87. csc x  cot x 

1 sin u  1 sin u  sin u tan u  sin u sec u  1  cos u  cos u   88. 1 sin u sec u  1 sin u tan u  sin u 1  sin u cos u cos u sin  sin  x sin    tan  (since cos   0 for 0      89. x  sin ; then   2 ). 2 2 2 cos  cos  1x 1  sin        90. x  tan ; then 1  x 2  1  tan2   1  sec2   1  sec2   sec (since sec   0 for 0     2 ). 91. x  sec ; then



x2  1 

    sec2   1  tan2   1  1  tan2   tan 

(since tan   0 for 0     2)

92. x  2 tan ; then 1 1 1 1          2 2 2 2 2 4 tan   2 sec2  x 4x 4 tan   4 1  tan  2 tan 2 4  2 tan 2 1 1 cos2  cos    18 cot2  cos    2 8 8 tan   sec  sin2         93. x  3 sin ; then 9  x 2  9  3 sin 2  9  9 sin2   9 1  sin2   3 cos2   3 cos 

(since

cos   0 for 0     2 ).

94. x  5 sec ; then



x 2  25  x

tan   0 for 0     2 ).

 5 sec 2  25 5 sec 



   25 sec2   1 5 sec 

sin   5 tan2  tan  cos   sin  (since    1 5 sec  sec  cos 

SECTION 7.1 Trigonometric Identities

95. f x  cos2 x  sin2 x, g x  1  2 sin2 x. From the graph, f x  g x this appears to be an identity. Proof:

1

f x  cos2 x  sin2 x  cos2 x  sin2 x  2 sin2 x  1  2 sin2 x  g x. Since

f x  g x for all x, this is an identity.

-5

5 -1

96. f x  tan x 1  sin x, g x 

sin x cos x . From the graph, f x  g x does 1  sin x

not appear to be an identity. In order to show this, let x   4 . Then,      21 . However,   f 4  1  1  1 2

2

1

-5

1  1       1 1 g 4   2  2     g   22 . Since f  4  4 , this is not an 1 21 2  1  2 2

5 -1

identity.

97. f x  sin x  cos x2 , g x  1. From the graph, f x  g x does not

appear to be an identity. In order to show this, we can set x   4 . Then we have        2 2 2     1  1  2  2  2  1  g  f  4  4 . Since 2 2 2   f 4  g 4 , this is not an identity.

1

-5

5 -1

98. f x  cos4 x  sin4 x, g x  2 cos2 x  1. From the graph, f x  g x appears to be an identity. In order to prove this, simplify the expression f x:    f x  cos4 x  sin4 x  cos2 x  sin2 x cos2 x  sin2 x      cos2 x  sin2 x 1  2 cos2 x  cos2 x  sin2 x    2 cos2 x  cos2 x  sin2 x  2 cos2 x  1  g x

1

-5

5 -1

Since f x  g x for all x, this is an identity.

99. sin x sin y  cos x cos y sin x sin y  cos x cos y  sin x sin y2  cos x cos y2      1  cos2 x sin2 y  cos2 x 1  sin2 y  sin2 y  cos2 x

1  cos2 x  sin2 x  2 cos x  sin x  cos x sin x 1  cos x  sin x 1  cos x  sin x 1  cos x  sin x   1  cos x  sin x 1  cos x  sin x 1  cos x  sin x 1  cos2 x  sin2 x  2 cos x 2  2 cos x  2 sin x  2 sin x cos x 1  sin x 1  cos x  sin x 1  cos x    cos x 1  cos x cos x 2 cos2 x  2 cos x 4  4  2   4 cos x sin x sin x  cos2 x 1 101. tan x  cot x4      sec4 x csc4 x cos x sin x sin x cos x sin x cos x       cos   1 1 sin   cos    sin  1   cos  1  102. sin   tan  cos   cot   sin   cos  sin  cos  sin      1 1  cos  1  sin  1   cos   1 sin   1 cos  sin 

100.

547

548

CHAPTER 7 Analytic Trigonometry

  sin y  csc y sin2 y  sin y csc y  csc2 y sin3 y  csc3 y   sin2 y  csc2 y  1 103. sin y  csc y sin y  csc y

   104. sin6   cos6   sin2   cos2  sin4   sin2  cos2   cos4       sin2  1  cos2   sin2  cos2   cos2  1  sin2   sin2   cos2   3 sin2  cos2   1  3 sin2  cos2 

   sin x    ln sin x  ln sin x  ln cos x  ln sin x 105. ln tan x sin x  ln tan x  ln sin x  ln  cos x     1    2 ln sin x  ln sec x  2 ln sin x  ln  cos x  106. ln tan x  ln cot x  ln tan x cot x  ln 1  0

2 2 2 2 2 2 2 2 107. esin x etan x  e1cos x esec x1  e1cos xsec x1  esec x e cos x

108. e x2 lnsin x  e x e2 lnsin x  e x elnsin x elnsin x  e x sin2 x 2 2 109. This is an identity: LHS  esin xcos x  e1  RHS.

110.

x  1  x  x  x  12  x 2  x  1  0, which has no real solution. x 1

 111. Squaring both sides, we have sin2 x  1  sin2 x  2 sin2 x  1  sin x  0, which has solutions x  k, k an integer. 2

112. xeln x  x  x 2  x 3 is true for all x  0.

  113. LHS  R cos  sin 2  R sin  sin 2  R cos 2  R sin 2 cos2   sin2   R cos 2  R sin 2  R cos 2  RHS

114. (a) x  y2  x 2  2x y  y 2 is an identity.

(b) x 2  y 2  1 is not an identity. For example, 12  12  1. (c) x y  z  x y  xz is the Distributive Law, an identity.

(d) t 2  cos2 t  t  cos t t  cos t (difference of squares).

 (e) sin t  cos t  1 is not an identity. For example, sin  4  cos 4 

 2.

(f) x 2  tan2 x  0 is not an identity.

 115. (a) Choose x   2 . Then sin 2x  sin   0, whereas 2 sin x  2 sin 2  2.

    1 1 2 (b) Choose x   4 and y  4 . Then sin x  y  sin 2  1, whereas sin x  sin y  sin 4  sin 4  2  2  2 . Since these are not equal, the equation is not an identity.  2  2 2 2 (c) Choose    2  2  4  1. 4 . Then sec   csc  

1 1 1  1 whereas (d) Choose x   4 . Then sin x  cos x  sin   cos   1 1 2   4 4 2 2    csc x  sec x  csc  4  sec 4  2  2. Since these are not equal, the equation is not an identity.

SECTION 7.2 Addition and Subtraction Formulas

549

116. No. All this proves is that f x  g x for x in the range of the viewing rectangle. It does not prove that these functions

x4 x6 x2   and g x  cos x. In the first viewing rectangle 2 24 720 the graphs of these two functions appear identical. However, when the domain is expanded in the second viewing rectangle, you can see that these two functions are not identical. are equal for all values of x. For example, let f x  1 

1 -10 -2

10 -5

2 -1

-10

117. Answers will vary.  118. Label a the side opposite , b the side opposite u, and c the hypotenuse. Since u     2  , we must have u    2   a     2  u. Next we express all six trigonometric function for each angle: cos u  c  sin   cos u  sin 2  u ,     b a sin u  bc  cos   sin u  cos  2  u , tan u  a  cot   tan u  cot 2  u , cot u  b  tan         c cot u  tan 2  u , sec u  ac  csc   sec u  csc  2  u , and csc u  b  sec   csc u  sec 2  u .

7.2

ADDITION AND SUBTRACTION FORMULAS

1. If we know the values of the sine and cosine of x and y we can find the value of sin x  y using the addition formula for sine, sin x  y  sin x cos y  cos x sin y. 2. If we know the values of the sine and cosine of x and y we can find the value of cos x  y using the subtraction formula for cosine, cos x  y  cos x cos y  sin x sin y. 





  6 2 4      2 3 2 6 2 1        sin 15  sin 45  30   sin 45 cos 30  cos 45 sin 30  2  2  2  2  4      6 cos 105  cos 60  45   cos 60 cos 45  sin 60 sin 45  12  22  23  22  2 4       2 cos 195   cos 15   cos 45  30    cos 45 cos 30  sin 45 sin 30    22  23  22  12   6 4   3   tan 30  1  3 3 tan 45 3      tan 15  tan 45  30     2 3 3 1  tan 45 tan 30 3 3 11 3   3 1 tan 45  tan 30 33  3        tan 165   tan 15   tan 45  30      32   1  tan 45 tan 30 3 3 1  1  33       3 6 2 2 1 2  7      sin 19 12   sin 12   sin 4  3   sin 4 cos 3  cos 4 sin 3   2  2  2  2   4           cos 5   cos      cos  cos   sin  sin    2  3  2  1  2 6 cos 17 12 12 4 6 4 6 4 6 2 2 2 2 4

3. sin 75  sin 45  30   sin 45 cos 30  cos 45 sin 30  22  23  22  12  4. 5. 6. 7.

8. 9. 10.

  tan  tan      1 3  3 4   11. tan  12   tan 12   tan 3  4     32   1  tan  1 3 3 tan 4           sin 5   sin        sin  cos   cos  sin    2  3  2  1   6 2 12. sin  512 12 4 6 4 6 4 6 2 2 2 2 4        3 6 2 2 2         1 13. cos 11 12   cos 12   cos 3  4   cos 3 cos 4  sin 3 sin 4   2  2  2  2   4

550

CHAPTER 7 Analytic Trigonometry

  3  1  3 3 3 7  5     14. tan 12   tan 12   tan 4  6    2  3   3 1  tan  tan 3  3 4 6 11 3





 tan  4  tan 6



15. sin 18 cos 27  cos 18 sin 27  sin 18  27   sin 45  1  22 2 16. cos 10 cos 80  sin 10 sin 80  cos 10  80   cos 90  0     sin 3 sin 2  cos 3  2  cos 7  cos   1 17. cos 37 cos 221 7 7 21 21 21 3 2 18. 19. 20. 21. 22. 23. 24. 25. 26.

  tan  tan 18 9



    tan   3  tan 318  tan 18    9 6 3 1  tan 18 tan 9

 tan 73  tan 13  tan 73  13   tan 60  3 1  tan 73 tan 13          13 13  2 1  cos 10 cos 13 15 cos  5  sin 15 sin  5  cos 15   5 15  cos 3   2       sin  2  u   sin 2 cos u  cos 2 sin u  1  cos u  0  sin u  cos u  cot u  tan   u    2 cos 2 cos u  sin  0  cos u  1  sin u sin u cos 2  u 2 sin u       cos 2  u cos 2 cos u  sin 2 sin u 0  cos u  1  sin u sin u   cot  2  u  sin   u  sin  cos u  cos  sin u  1  cos u  0  sin u  cos u  tan u 2 2 2   1 1 1 1   sec 2  u   cos u  sin  sin u  0  cos u  1  sin u  sin u  csc u cos cos   u 2 2 2   1 1 1 1    csc 2  u   cos u  cos  sin u  1  cos u  0  sin u  cos u  sec u sin sin   u 2 2 2     cos x sin   0  sin x  1  cos x   cos x  sin x cos sin x   2 2 2     cos x   2  cos x cos 2  sin x sin 2  0  cos x  1  sin x  sin x

27. sin x    sin x cos   cos x sin   1  sin x  0  cos x   sin x 28. cos x    cos x cos   sin x sin   1  cos x  0  sin x   cos x tan x  0 tan x  tan    tan x 29. tan x    1  tan x tan  1  tan x  0      sin x   sin x cos   cos x 2  2  cos x sin 2  30. tan x   2  cos x    cos x cos   sin x sin   sin x   cot x 2 2 2    cos x  cos  sin x  1  cos x  0  sin x  cos x and 31. LHS  sin   x  sin 2 2  2  cos x  cos  sin x  1  cos x  0  sin x  cos x. Therefore, LHS  RHS.  x  sin RHS  sin  2 2 2         1 cos x  3 sin x  3 sin x  1 cos x  0 32. cos x    sin x  3 6 2 2 2 2   tan x  tan   3  tan x 3 33. tan x    3  1  tan x tan   1  3 tan x 3  tan x  tan   4  tan x  1  tan x  1  34. tan x   4 1  tan x tan  1  tan x  1 tan x  1 4

35. sin x  y  sin x  y  sin x cos y  cos x sin y  sin x cos y  cos x sin y  2 cos x sin y 36. cos x  y  cos x  y  cos x cos y  sin x sin y  cos x cos y  sin x sin y  2 cos x cos y 1 1 1 1 1  tan x tan y cot x cot y  1 cot x cot y cot x cot y 37. cot x  y      1 1 tan x  y tan x  tan y cot x cot y cot y  cot x  cot x cot y 1 1 1 1  tan x tan y cot x cot y  1 1 cot x cot y cot x cot y    38. cot x  y   1 1 tan x  y tan x  tan y cot x cot y cot x  cot y  cot x cot y

SECTION 7.2 Addition and Subtraction Formulas

39. tan x  tan y 

551

sin y sin x cos y  cos x sin y sin x  y sin x    cos x cos y cos x cos y cos x cos y

40. 1  tan x tan y  1 

cos x cos y  sin x sin y cos x  y sin x sin y   cos x cos y cos x cos y cos x cos y

41.

sin x  y sin x cos y  cos x sin y tan x  tan y tan x  tan y cos x cos y    1  tan x tan y cos x cos y  sin x sin y cos x  y 1  tan x tan y cos x cos y

42.

sin x cos y  cos x sin y  sin x cos y  cos x sin y 2 cos x sin y sin x  y  sin x  y    tan y cos x  y  cos x  y cos x cos y  sin x sin y  cos x cos y  sin x sin y 2 cos x cos y

43. cos x  y cos x  y  cos x cos y  sin x sin y cos x cos y  sin x sin y  cos2 x cos2 y  sin2 x sin2 y      cos2 x 1  sin2 y  1  cos2 x sin2 y  cos2 x  sin2 y cos2 x  sin2 y cos2 x  sin2 y  cos2 x  sin2 y

44. cos x  y cos y  sin x  y sin y  cos x cos y  sin x sin y cos y  sin x cos y  cos x sin y sin y    cos2 y cos x  sin2 y cos x  cos2 y  sin2 y cos x  cos x 45. sin x  y  z  sin x  y  z  sin x  y cos z  cos x  y sin z

 cos z sin x cos y  cos x sin y  sin z cos x cos y  sin x sin y

 sin x cos y cos z  cos x sin y cos z  cos x cos y sin z  sin x sin y sin z 46. The addition formula for the tangent function can be written as tan A  tan B  tan A  B 1  tan A tan B. Also note that tan A   tan A. Using these facts, we get   tan x  y  tan y  z  tan z  x  tan x  y  y  z 1  tan x  y tan y  z  tan z  x    tan x  z 1  tan x  y tan y  z  tan z  x  tan x  z  tan z  x  tan x  y tan y  z tan x  z

 tan x  z  tan x  z  tan x  y tan y  z tan x  z  0  tan x  y tan y  z tan x  z  tan x  y tan y  z tan z  x 47. We want to write cossin1 x  tan1 y in terms of x and y only. We let   sin1 x and   tan1 y and sketch triangles with angles  and  such  that sin   x and tan   y. From the triangles, we have cos   1  x 2 ,

1 ¬

x

Ï1+y@ ú

1

y

Ï1-x@ 1 y cos    , and sin    . tan   y sin   x 1  y2 1  y2 From the subtraction formula for cosine we have     1  x2  x y 1 y  cos sin1 x  tan1 y  cos     cos  cos   sin  sin   1  x 2   x    1  y2 1  y2 1  y2

552

CHAPTER 7 Analytic Trigonometry

x 48. Let   sin1 x and   cos1 y. From the triangles, tan    and 1  x2  1  y2 , so using the addition formula for tangent, we have tan   y  x 1  y2     y 1  x2  tan sin1 x  cos1 y  tan     1  y2 x 1   y 1  x2   x y  1  x 2 1  y2    y 1  x 2  x 1  y2 1 , 49. Let   tan1 x and   tan1 y. From the triangles, cos    1  x2 x 1 y sin    , cos    , and sin    , so using the 2 2 1y 1  y2 1x subtraction formula for sine, we have   sin tan1 x  tan1 y  sin     sin  cos   cos  sin 

1

x

1

¬

ú

Ï1-x@

sin   x

Ï1-y@ y

cos   y

Ï1+x@

Ï1+y@

x

ú

¬

y

1

1

tan   x

tan   y

x 1 1 y xy       1  y2 1  y2 1  x2 1  x2 1  x 2 1  y2

 50. Let   sin1 x and   cos1 y. From the triangles, cos   1  x 2 and  sin   1  y 2 , so using the addition formula for sine, we have   sin sin1 x  cos1 y  sin     sin  cos   cos  sin       x y  1  x 2  1  y2  x y  1  x 2 1  y2

1

1

x

ú

¬

Ï1-x@

Ï1-y@

y

cos   y

sin   x

1 1   , so the addition formula for sine gives 51. We know that cos1 21   3 and tan 4          3 2 2 6 2.      1 1 1 1 sin cos 2  tan 1  sin 3  4  sin 3 cos  4  cos 3 sin 4  2  2  2  2  4   1 3   , so the addition formula for cosine gives 52. We know that sin1 23   3 and cot 6         3     sin  sin   1  3  3  1  0. 1 1  cos  cot 3  cos  cos cos sin 2 3 6 3 6 3 6 2 2 2 2

53. We sketch triangles such that   sin1 43 and   cos1 31 . From the  triangles, we have tan   3 and tan   2 2, so the subtraction formula for 7

4

tangent gives  3  2 2   tan   tan  7  tan sin1 34  cos1 13   1  tan  tan  1  3  2 2 7  3  2 14   76 2

¬ Ï7

sin   34

3

3

ú

2Ï2

1

cos   13

SECTION 7.2 Addition and Subtraction Formulas

553

54. We sketch triangles such that   cos1 23 and   tan1 12 . From the 

triangles, we have sin   35 , cos   2 , and sin   1 , so the addition 5 5

3

formula for sine gives    sin cos1 32  tan1 12  sin  cos   cos  sin   35  2  23  1 5 5

¬

remaining sides using the Pythagorean Theorem. To find cos   ,

x

ú 1

(3, _4)

cos     cos  cos   sin  sin       4 3  35  12   45 23   3 10

y 3

have sin     sin  cos   cos  sin          4 3 10 10 3 15 10      5 10 5 10 50  3 10  10

4

y

¬ x

ú

3Ï10

5

Ï10

10



sin    1010

tan   43

(12, 5)

13 ¬

(_2Ï5, Ï5)

x

12

y

y

2Ï2

x 3

(_1, _2Ï2)

(_Ï15, 1) 1

4 Ï15

sin   14

cos    13

    2  3  12  4  2. Thus, sin   12 and cos   2 3    56 , so     3 sin x  cos x  k sin x    2 sin x  56 .

59. k 



A2  B 2 

  12  12  2 and  satisfies sin   1 , cos   1     4 . Thus, 2 2    sin x  cos x  k sin x    2 sin x   4 .

60. k 



A2  B 2 

x

 cos    2 5 5

¬ 1

ú

2Ï5

5 sin   13

have

y 5

Ï5

5

58. Using the addition formula for tangent and the triangles shown, we

x

(3Ï10, _Ï10)

(_3, _4)

y

x

 tan    3

cos   35

56. Using the subtraction formula for sine and the triangles shown, we

   1 2   2 tan   tan  15 tan        1  1  tan  tan   12 2  15  2 30  1   15  2 2

2

Ï3

4

sketched:

sin     sin  cos   cos  sin        5 2 5 12 2 5 5     13 5 13 5 65

y

(_1, Ï3)

5

we use the addition formula for sine and the triangles we have

57. Using the addition formula for sine and the triangles shown, we have

tan   12

y 3

1

2

2

cos   23

55. As in Example 7, we sketch the angles  and  in standard position with terminal sides in the appropriate quadrants and find the

Ï5 ú

¬

 2  10  2 5  23   15 3 5

Ï5

ú x

554

CHAPTER 7 Analytic Trigonometry

   5   1 and cos    5  1    7 , so 52  52  50  5 2. Thus, sin     4 5 2 2 5 2 2    7  5 sin 2x  cos 2x  k sin 2x    5 2 sin 2x  4 .

61. k 



A2  B 2 



    2  32  3 3  36  6 and  satisfies sin   3 6 3  23 , cos   36  12     3 . Thus,      1 3 sin x  3 3 cos x  k sin x    6 sin x   3  6 sin  x  3 .

62. k 



A2  B 2 

 63. (a) g x  cos 2x  3 sin 2x    2  k  12  3  4  2, and  satisfies

(b) This is a sine curve with amplitude 2, period , and . phase shift  12 y

 sin   12 , cos   23     6 . Thus, we can

2 1

write

    g x  k sin 2x    2 sin 2x   6  2 sin 2 x  12 .

_2¹

0

_¹

¹

2¹ x

_1 _2

64. (a) f x  sin x  cos x  k 

  12  12  2, and

 satisfies sin   cos   1     4 . Thus, 2

(b) This is a sine curve with amplitude

 2, period 2,

and phase shift   4. y

we can write

2

   f x  k sin x    2 sin x   4 .

1 _2¹

_¹

0

¹

_1 _2

65. f x  cos x. Now cos x  h  cos x cos x cos h  sin x sin h  cos x f x  h  f x   h h h      cos x 1  cos h  sin h sin x 1  cos h sin h    cos x  sin x h h h 66. g x  sin x. Now sin x  h  sin x sin x cos h  cos x sin h  sin x g x  h  g x   h h h     sin h cos x  sin x 1  cos h sin h 1  cos h   cos x  sin x h h h

2¹ x

SECTION 7.2 Addition and Subtraction Formulas

    2 67. (a) y  sin2 x   4  sin x  4 . From the graph we see that the value of y seems to always be equal to 1.

555

1

-5

5

       2  sin x cos   cos x sin  2 2 (b) y  sin2 x   4  sin x  4  sin x cos 4  cos x sin 4 4 4    2  2 1 1 1 2 2   sin x  cos x   sin x  cos x  2 sin x  cos x  sin x  cos x 2 2      12 sin2 x  2 sin x cos x  cos2 x  sin2 x  2 sin x cos x  cos2 x  12 [1  2 sin x cos x  1  2 sin x cos x]  12  2  1

68. (a) y   12 [cos x    cos x  ]. The graph of y appears to be the same as 1

that of cos x.

-5

5 -1

(b) y   12 [cos x    cos x  ]

  12 [cos x cos   sin x sin   cos x cos   sin x sin ]

  12 [ cos x  0   cos x  0]   12 2 cos x  cos x

   69. If      2 , then     2 . Now if we let y  x  , then sin x    cos x    2  sin y    cos y   2  sin y   sin y  0. Therefore, sin x    cos x    0. 70. Let  A and  B be the two angles shown in the diagram. Then

180    A  B, 90    A, and 90    B. Subtracting the second and third equation from the first, we get

180  90  90    A  B    A    B 

    . Then

6 Œ

4

3

A

B

º

4



4  3 8  9 tan   tan  17  6 4 4 3  12 112  2  17 12  6 . 1  tan  tan  1 6  4 1 2     xy tan u  tan   tan1  tan1 tan u    u    tan1 x  tan1 y 71. tan1 1  xy 1  tan u tan    cot  cot u  1 x 1x  1  1 72. cot u      0, so tan1 x tan1  u   cot1 cot u    cot1 0  . cot   cot u x  1x x 2

tan   tan    

y y and tan   . Thus, m  tan . x x tan 2  tan 1 (b) tan   tan 2   1   . From part (a), we have m 1  tan 1 and m 2  tan 2 . Then by 1  tan  2 tan  1 m  m1 substitution, tan   2 . 1  m1m2

73. (a) By definition, m 

556

CHAPTER 7 Analytic Trigonometry

(c) Let  be the unknown angle as in part (b). Since m 1  13 and m 2  12 , 1  1 1 m  m1 tan   2  2 3   67  17    tan1 17  0142 rad  81 . 1  m1m2 1  13 12 6

(d) From part (b), we have cot   have 0 

1  m1m2 . If the two lines are perpendicular then   90 and so cot   0. Thus we m2  m1

1  m1m2  0  1  m 1 m 2  m 1 m 2  1  m 2  1m 1 . Thus m 2 is the negative reciprocal of m 1 . m2  m1 1

1

tan A  tan B 3  2  1. Thus A  B   , so A  B  C       . 74. Clearly C   4 . Now tan A  B  1  tan A tan B  4 4 4 2 1  13  12   A2  B 2  52  52  5 2. Therefore, B 5 1 sin        and 2 2 5 2 2 A B 1 A   . Thus,    cos    4. 2 A2  B 2

75. (a) y  f 1 t  f 2 t  5 sin t  5 cos t  5 sin t  cos t

(b) k 

10

-5

5



-10

76. (a) f t  C sin t  C sin t    C sin t  C sin t cos   cos t sin   C 1  cos  sin t  C sin  cos t  A sin t  B cos t

where A  C 1  cos  and B  C sin .     (b) In this case, f t  10 1  cos  3 sin t  10 sin 3 cos t  15 sin t  5 3 cos t. Thus      2    3 and sin   5 3  1 , so    . Therefore, k  152  5 3  10 3 and  has cos   15 2 2 6 10 3 10 3    . f t  10 3 sin t   6          77. sin s  t  cos 2  s  t  cos  2  s  t  cos 2  s cos t  sin 2  s sin t  sin s cos t  cos s sin t. The last equality comes from again applying the cofunction identities. sin s  t sin s cos t  cos s sin t  cos s  t cos s cos t  sin s sin t 1 sin s  sin s cos t  cos s sin t cos s cos t cos s    sin s 1 cos s cos t  sin s sin t 1 cos s cos t cos s

78. tan s  t 

7.3

sin t cos t  tan s  tan t sin t 1  tan s tan t  cos t

DOUBLE-ANGLE, HALF-ANGLE, AND PRODUCT-SUM FORMULAS

1. If we know the values of sin x and cos x, we can find the value of sin 2x using the Double-Angle Formula for Sine: sin 2x  2 sin x cos x.

x x 2. If we know the value of cos x and the quadrant in which lies, we can find the value of sin using the Half-Angle Formula 2 2  x 1  cos x . for Sine: sin   2 2

SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas





557



5 , x in quadrant I  cos x  12 and tan x  5 . Thus, sin 2x  2 sin x cos x  2 5 12  120 , 3. sin x  13 13 12 13 13 169 120  2  2 sin 2x 169 5 119 120 169 120 cos 2x  cos2 x  sin2 x  12  13  14425 13 169  169 , and tan 2x  cos 2x  119  169  119  119 . 169

  4. tan x   43 . Then sin x  45 and cos x   35 (x is in quadrant II). Thus, sin 2x  2 sin x cos x  2  45  35   24 25 , 24  2  2 7 , and tan 2x  sin 2x   25  24  25  24 . cos 2x  cos2 x  sin2 x   35  45  916   7 25 25 25 7 cos 2x 7 25

  5. cos x  45 . Then sin x   35 (csc x  0) and tan x   34 . Thus, sin 2x  2 sin x cos x  2  35  45   24 25 , 24  2  2 7 , and tan 2x  sin 2x   25   24  25   24 .  cos 2x  cos2 x  sin2 x  45   35  169 7 25 25 25 7 7 cos 2x

25  6. csc x  4. Then sin x  14 , cos x   415 , and tan x   1 (tan x  0). Thus, 15       2  2 7 sin 2x  2 sin x cos x  2  14  415   815 , cos 2x  cos2 x  sin2 x   415  14  151 16  8 , and     15 sin 2x  78   815  87   715 . tan 2x  cos 2x 8

   7. sin x   35 . Then, cos x   45 and tan x  34 (x is in quadrant III). Thus, sin 2x  2 sin x cos x  2  35  45  24 25 , 24  2  2 7 , and tan 2x  sin 2x  25  24  25  24 .  cos 2x  cos2 x  sin2 x   45   35  169 7 25 25 25 7 7 cos 2x 25   8. sec x  2. Then cos x  12 , sin x   23 , and tan x   3 (x is in quadrant IV). Thus,          2 1   3 , cos 2x  cos2 x  sin2 x  1 2   3 sin 2x  2 sin x cos x  2  23   12 , and 2 2 2 2    23 sin 2x tan 2x    3. 1 cos 2x 2

9. tan x   13 and cos x  0, so sin x  0. Thus, sin x   1 and cos x  3 . Thus, 10 10        6   3 , cos 2x  cos2 x  sin2 x  3 2   1 2  8  4 , 3   10 sin 2x  2 sin x cos x  2  1 5 10 5 10 10 10 10 and tan 2x 

3 sin 2x  45   35  54   34 . cos 2x 5

10. cot x  23 .

Then tan x  32 , sin x  3 (sin x  0), and cos x  2 . Thus, 13 13   2  2   5 2  12 , cos 2x  cos2 x  sin2 x  2  3  49 sin 2x  2 sin x cos x  2 3 13 13   13 , and 13 13 13 13 12   sin 2x  135  12   12   13 tan 2x  13 5 5. cos 2x  13

 2  1  cos 2x 2 11. sin4 x  sin2 x   14  12 cos 2x  14 cos2 2x 2 1  cos 4x  14  12 cos 2x  14   14  12 cos 2x  18  18 cos 4x  38  12 cos 2x  18 cos 4x 2   12 34  cos 2x  14 cos 4x

558

CHAPTER 7 Analytic Trigonometry

 2  1  cos 2x 2 12. cos4 x  cos2 x   14  12 cos 2x  14 cos2 2x 2 1  cos 4x  14  12 cos 2x  14   14  12 cos 2x  18  18 cos 4x  38  12 cos 2x  18 cos 4x 2   12 34  cos 2x  14 cos 4x

13. We use the result of Example 4 to get       1 1  cos 2x  cos 4x  cos 2x cos 4x. cos2 x sin4 x  sin2 x cos2 x sin2 x  18  18 cos 4x  12  12 cos 2x  16 14. Using Example 4, we have   1 1  cos 4x  cos 2x  cos 2x cos 4x. cos4 x sin2 x  cos2 x cos2 x sin2 x  12 1  cos 2x  18 1  cos 4x  16  2 15. Since sin4 x cos4 x  sin2 x cos2 x we can use the result of Example 4 to get 

sin4 x cos4 x 

 1  1 cos 4x 2  1  1 cos 4x  1 cos2 4x 8 8 64 32 64

1  1 cos 4x  1  1 1  cos 8x  1  1 cos 4x  1  1 cos 8x  64 32 64 2 64 32 128 128   3 1 1 1 3 1  128  32 cos 4x  128 cos 8x  32 4  cos 4x  4 cos 8x

16. Using the result of Exercise 12, we have  2    cos6 x  cos2 x  cos2 x  38  12 cos 2x  18 cos 4x 12  12 cos 2x

3  1 cos 2x  1 cos 4x  3 cos 2x  1 cos2 2x  1 cos 2x cos 4x  16 4 16 16 4 16

1 cos 4x 1 1  cos 4x   , the last expression is equal to Because 14 cos2 2x   4 2 8 8 3 1 1 3 1 1 1 16  4 cos 2x  16 cos 4x  16 cos 2x  8  8 cos 4x  16 cos 2x cos 4x 5  7 cos 2x  3 cos 4x  1 cos 2x cos 4x  1 5  7 cos 2x  3 cos 4x  cos 2x cos 4x  16 16 16 16 16            17. sin 15  12 1  cos 30   12 1  23  14 2  3  12 2  3

  1  23 1  cos 30   2 3 18. tan 15   1 sin 30 2  2   1  1  cos 45   2  21 19. tan 225   2 sin 45 2             20. sin 75  12 1  cos 150   12 1   23  14 2  3  12 2  3

        21. cos 165   12 1  cos 330    12 1  cos 30    12 1  23   12 2  3         1 1    22. cos 1125   2 1  cos 225    2 1  cos 45    12 1  22   12 2  2 

2  1  cos  4  1 2  21   23. tan   8

sin 4 

24. cos 38  cos 12  34

2 2







1  cos 34  2





1  22  2

root because 38 is in quadrant I, so cos 38  0.



   2 2  12 2  2. Note that we have chosen the positive 4

SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas

559

          1  25. cos 12  2 1  cos 6  12 1  23  12 2  3   1  cos 56 1  23 5   2 3 26. tan 12  5  1 sin 6 2

         27. sin 98   12 1  cos 94   12 1  22   12 2  2. We have chosen the negative root because 98 is in quadrant III, so sin 98  0.

 28. sin 11 12 

         1 1  cos 11  1 1  3  1 2  3. We have chosen the positive root because 11 is in 2 6 2 2 2 12

 quadrant II, so sin 11 12  0.

2 tan 7  tan 14 1  tan2 7 2 tan 7 (b)  tan 14 1  tan2 7

29. (a) 2 sin 18 cos 18  sin 36

30. (a)

(b) 2 sin 3 cos 3  sin 6

   sin2  cos  2 2   (b) 2 sin cos  sin  2 2  1  cos 30 34. (a)  sin 15 2  1  cos 8 (b)  sin 4 2

31. (a) cos2 34  sin2 34  cos 68

32. (a) cos2

(b) cos2 5  sin2 5  cos 10 sin 8 8  tan  tan 4  1  cos 8 2 4 1  cos 4  tan  tan 2 (b) sin 4 2

33. (a)

35. sin x  x  sin x cos x  cos x sin x  2 sin x cos x

36. tan x  x 

2 tan x tan x  tan x  1  tan x tan x 1  tan2 x

37. sin x  35 . Since x is in quadrant I, cos x  45 and x2 is also in quadrant I. Thus,          10 x 1 1 4 1 x 1  sin 2  2 1  cos x  2 1  5   10 , cos 2  2 1  cos x  12 1  45  3



 3 1010 , and 10

10

 sin x2  1  310  13 . tan x2  x 10 cos 2

38. cos x   45 . Since x is in quadrant III, sin x   35 and tan x  34 . Also, since 180  x  270 ,      90  x2  135 and so x2 is in quadrant II. Thus, sin x2  12 1  cos x  12 1  45  3  3 1010 , 10       sin x2 10  3. x 1  3  1 cos 2   2 1  cos x   12 1  45   1   1010 , and tan x2  x 10 10 cos 2 

2 2 . Since 90  x  180 , we have 39. csc x  3. Then, sin x  13 and since x is in quadrant II, cos x   3          45  x2  90 and so x2 is in quadrant I. Thus, sin x2  12 1  cos x  12 1  2 3 2  16 3  2 2 , cos x2 



1 1  cos x  2

     x       1 1  2 2  1 3  2 2 , and tan x  sin 2  322  3  2 2. x 2 3 6 2 32 2 cos 2

560

CHAPTER 7 Analytic Trigonometry   2 and cos x  2   2 2 , since x is in quadrant I. Also, since 0  x  90 ,        0  x2  45 and so x2 is also in quadrant I. Thus, sin x2  12 1  cos x  12 1  22  12 2  2,          1  22 1  cos x 2 x 1 1 1 x    2  1. cos 2  2 1  cos x  2 1  2  2 2  2, and tan 2  2 sin x 2

40. tan x  1. Then sin x 



41. sec x  32 . Then cos x  23 and since x is in quadrant IV, sin x   35 . Since 270  x  360 , we have      135  x2  180 and so x2 is in quadrant II. Thus, sin x2  12 1  cos x  12 1  23  1  66 , 6    x      sin 30 2  1  6   1   5 . cos x2   12 1  cos x   12 1  23   5   , and tan x2  x 5 6 6  5 5 6 cos 2 42. cot x  5.

Then, cos x   5

26

and sin x   1 (csc x  0). 2

Since cot x  0 and

csc x  0, it follows that x is in quadrant III. Thus 180  x  270 and so 90  x2  135 .       26 , 1 1  cos x  1 1  5 Thus x2 is in quadrant II. sin x2   12 265 2 2 13 26

     1  5   26 26 , and tan x  1  cos x  cos x2   12 1  cos x   12 1  5   12 265  5  26. 13 2 1 26 sin x  2

  43. To write sin 2 tan1 x as an algebraic expression in x, we let   tan1 x

x and sketch a suitable triangle. We see that sin    and 1  x2 1 cos    , so using the double-angle formula for sine, we have 1  x2   x 1 2x sin 2 tan1 x  sin 2  2 sin  cos   2     . 2 2 1  x2 1x 1x

Ï1+x@ ¬

x

1

  44. To write tan 2 cos1 x as an algebraic expression in x, we let   cos1 x 1  1  x2 , so using the and sketch a suitable triangle. We see that tan   ¬ x x double-angle formula for tangent, we have  1  x2   2   2 tan  2 1  x2 2x 1  x 2 x 1  tan 2 cos x  tan 2   . 2    1  tan2  2x 2  1 1  x2 1  x2 x 1 1 x2 x

Ï1-x@

      1  cos cos1 x    1x  . Because cos1 45. Using the half-angle formula for sine, we have sin 12 cos1 x   2 2     1  x   1 1 1 1 1 1 . has range [0 ], 2 cos x lies in 0 2 and so sin 2 cos x is positive. Thus, sin 2 cos x  2

    46. Using a double-angle formula for cosine, we have cos 2 sin1 x  1  2 sin2 sin1 x  1  2x 2 .

SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas 7 and find that sin   24 . Thus, using 47. We sketch a triangle with   cos1 25 25 the double-angle formula for sine,   7  sin 2  2 sin  cos   2  24  7  336 . sin 2 cos1 25 25 25 625

25

¬ 7

24

7 cos   25

12 48. We sketch a triangle with   tan1 12 5 and find that sin   13 . Thus, using a

¬

13

double-angle formula for cosine,   2  2   1  2 12  cos 2  1  2 sin   119 cos 2 tan1 12 5 13 169 .

5

12

tan   12 5

49. Rewriting the given expression and using a double-angle formula for cosine, we have   1 8 1 1      sec 2 sin1 14   2  . 1 1 1 2 1 7 cos 2 sin 4 1  2 sin sin 4 12 1

4  50. We sketch a triangle with   cos1 23 and find that sin   35 . Thus, using a  5 3

 sin  5 1 1 2 1  .  tan 2 cos 3  tan 2   2 1  cos  5 1 3 



2

Ï5

 2 7. 51. Using a double-angle formula for cosine, we have cos 2  1  2 sin2   1  2  35  25

cos   23

y ¬

52. To evaluate sin 2, we first sketch the angle  in standard position with

terminal side in quadrant IV and find the remaining side using the Pythagorean Theorem. Using the half-angle formula for sine, we have    1  12 1  cos   13   26 . Because 2 lies in  sin   26 2 2 2

¬

3

half-angle formula for tangent,

12 13

x

5

(12, _5)



quadrant II, where sine is positive, we take the positive value. Thus, sin 2  2626 . 53. To evaluate sin 2, we first sketch the angle  in standard position with terminal side in quadrant II and find the remaining side using the Pythagorean Theorem. Using the double-angle formula for sine, we have      sin 2  2 sin  cos   2 17  4 7 3   8493 .

54. To evaluate tan 2, we first sketch the angle  in standard position with terminal

1

y

(_4Ï3, 1)

y

(3, 4) 5

Using the double-angle formula for tangent, we have 2 tan  tan 2   1  tan2 

24  2   . 7 1 4 3

55. sin 2x cos 3x  12 [sin 2x  3x  sin 2x  3x]  12 sin 5x  sin x 56. sin x sin 5x  12 [cos x  5x  cos x  5x]  12 cos 4x  cos 6x 57. cos x sin 4x  12 [sin 4x  x  sin 4x  x]  12 sin 5x  sin 3x 58. cos 5x cos 3x  12 [cos 5x  3x  cos 5x  3x]  12 cos 8x  cos 2x

x

4Ï3

side in quadrant I and find the remaining side using the Pythagorean Theorem. 2  43

¬

7

¬

3

4 x

561

562

CHAPTER 7 Analytic Trigonometry

59. 3 cos 4x cos 7x  3  12 [cos 4x  7x  cos 4x  7x]  32 cos 11x  cos 3x        3x x 60. 11 sin x2 cos x4  11  12 sin x2  x4  sin x2  x4  11 2 sin 4  sin 4     5x  3x 5x  3x 61. sin 5x  sin 3x  2 sin cos  2 sin 4x cos x 2 2       x  4x x  4x 5x 3x 5x 3x 62. sin x  sin 4x  2 cos sin  2 cos sin   2 cos sin 2 2 2 2 2 2     4x  6x 4x  6x sin  2 sin 5x sin x  2 sin 5x sin x 63. cos 4x  cos 6x  2 sin 2 2     11x 7x 9x  2x 9x  2x 64. cos 9x  cos 2x  2 cos cos  2 cos cos 2 2 2 2       9x 5x 9x 5x 2x  7x 2x  7x sin  2 cos sin   2 cos sin 65. sin 2x  sin 7x  2 cos 2 2 2 2 2 2       7x x 7x x 3x  4x 3x  4x cos  2 sin cos   2 sin cos 66. sin 3x  sin 4x  2 sin 2 2 2 2 2 2   67. 2 sin 525 sin 975  2  12 cos 525  975   cos 525  975   cos 45   cos 150       cos 45  cos 150  22  23  12 2 3       68. 3 cos 375 cos 75  32 cos 45  cos 30   32 22  23  34 2 3     21 69. cos 375 sin 75  12 sin 45  sin 30   12 22  12  14          75  15 75  15 cos  2 sin 45 cos 30  2  22  23  26 70. sin 75  sin 15  2 sin 2 2           255  195 255  195 sin  2 sin 225 sin 30  2  22 12  22 71. cos 255  cos 195  2 sin 2 2          3 6 2   cos 5  2 cos 1   5 cos 1   5   2 cos  72. cos 12 12 2 12 12 2 12 12 4 cos 6  2  2  2  2 73. cos2 5x  sin2 5x  cos 2  5x  cos 10x 74. sin 8x  sin 2  4x  2 sin 4x cos 4x

75. sin x  cos x2  sin2 x  2 sin x cos x  cos2 x  1  2 sin x cos x  1  sin 2x    76. cos4 x  sin4 x  cos2 x  sin2 x cos2 x  sin2 x  cos2 x  sin2 x  cos 2x 77.

78. 79. 80. 81. 82.

2 tan x sin x 2 tan x  2 cos2 x  2 sin x cos x  sin 2x cos x 1  tan2 x sec2 x   1  1  2 sin2 x 2 sin2 x 1  cos 2x    tan x sin 2x 2 sin x cos x 2 sin x cos x   x   x  1  cos x 1  cos x  cos x  cos2 x sin2 x 1  cos x tan  cos x tan   cos x    sin x 2 2 sin x sin x sin x sin x x  1  cos x 1 2  cos x tan  csc x    2 sin x sin x sin x 2 sin 2x cos 2x 2 2 sin x cos x cos 2x sin 4x    4 cos x cos 2x sin x sin x sin x 1  sin 2x 1  2 sin x cos x 1  1  1  12 csc x sec x sin 2x 2 sin x cos x 2 sin x cos x

SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas

563

2 2 tan x  cot x 2 2 tan x  cot x    2 2 cos x sin x tan x  cot x tan x  cot x tan x  cot x tan x  cot x  cos x sin x 2 sin x cos x 2 sin x cos x   2 sin x cos x  sin 2x   cos x sin x cos x sin x sin2 x  cos2 x  cos x sin x 2 sin x cos x sin 2x    tan x  84. 1  cos 2x 1  2 cos2 x  1

83.

1  tan2 x 1  2 tan x 2 tan x 1  tan2 x    3   86. 4 sin6 x  cos6 x  4 sin2 x  cos2 x  3 sin4 x cos2 x  sin2 x cos4 x     4 1  3 sin2 x cos2 x sin2 x  cos2 x  4  12 sin2 x cos2 x 85. cot 2x 

1  tan 2x

 4  3 2 sin x cos x2  4  3 sin2 2x

  2 tan x  tan x 2 tan x  tan x 1  tan2 x 2x tan 2x  tan x 1  tan 87. tan 3x  tan 2x  x    2 tan x 1  tan 2x tan x 1  tan2 x  2 tan x tan x 1 tan x 1  tan2 x 3 tan x  tan3 x  1  3 tan2 x 88.

sin 3x  cos 3x sin 2x cos x  cos 2x sin x  cos 2x cos x  sin 2x sin x  cos x  sin x cos x  sin x   cos2 x  sin2 x cos x  sin x sin 2x cos x  sin x  cos 2x cos x  sin x   sin 2x  cos x  sin x cos x  sin x  sin 2x  cos x  sin x2  1  4 sin x cos x

89. 90. 91.

2 sin 3x cos 2x sin 3x sin x  sin 5x    tan 3x cos x  cos 5x 2 cos 3x cos 2x cos 3x

2 sin 5x cos 2x cos 2x sin 3x  sin 7x    cot 2x cos 3x  cos 7x 2 sin 5x sin 2x sin 2x sin 10x 2 sin 5x cos 5x cos 5x   sin 9x  sin x 2 sin 5x cos 4x cos 4x

sin x  sin 5x  sin 3x 2 sin 3x cos 2x  sin 3x sin 3x 2 cos 2x  1 sin x  sin 3x  sin 5x     tan 3x cos x  cos 3x  cos 5x cos x  cos 5x  cos 3x 2 cos 3x cos 2x  cos 3x cos 3x 2 cos 2x  1       xy xy xy   cos 2 sin sin sin x  sin y xy 2 2 2      tan    93. xy xy xy cos x  cos y 2 cos 2 cos cos 2 2 2         2x xyxy xyxy 2y 2 cos 2 cos sin sin sin y sin x  y  sin x  y 2 2 2 2           tan y 94. x yx y 2y xyxy 2x cos x  y  cos x  y cos y cos cos 2 cos 2 cos 2 2 2 2   x   2y  x  . Then 95. Let y  2 4 2   x 1  cos x    1  cos 2y 2   1   sin x  1  sin x  tan2   tan2 y   2 4 1  cos 2y 1   sin x 1  sin x 1  cos x   2 92.

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CHAPTER 7 Analytic Trigonometry

           96. 1  cos 4x 2  tan2 x  cot2 x  1  1  2 sin2 2x tan2 x  1  cot2 x  1  2 sin2 2x sec2 x  csc2 x       2 4 cos2 x sin2 x sec2 x  csc2 x  8 sin2 x  cos2 x  8   130  110 130  110 sin  2 cos 120 sin 10  2  12 sin 10   sin 10 97. sin 130  sin 110  2 cos 2 2   200   200 100 100 sin  2 sin 150 sin 50  98. cos 100  cos 200  2 sin 2 2    2 12  sin 50   sin 50       45  15 45  15 cos  2 sin 30 cos 15  2  12  cos 15 99. sin 45  sin 15  2 sin 2 2  cos 15  sin 90  15   sin 75 (applying the cofunction identity) 87  33 87  33 cos  2 cos 60 cos 27 2 2  2  12 cos 27  cos 27  sin 90  27   sin 63

100. cos 87  cos 33  2 cos

sin x  sin 2x  sin 3x  sin 4x  sin 5x sin x  sin 5x  sin 2x  sin 4x  sin 3x  cos x  cos 2x  cos 3x  cos 4x  cos 5x cos x  cos 5x  cos 2x  cos 4x  cos 3x sin 3x 2 cos 2x  2 cos x  1 2 sin 3x cos 2x  2 sin 3x cos x  sin 3x   tan 3x  2 cos 3x cos 2x  2 cos 3x cos x  cos 3x cos 3x 2 cos 2x  2 cos x  1     102. n  1: sin 21 x  2 sin x cos x  21 sin x cos 20 x     n  2: sin 22 x  sin 4x  2 sin 2x cos 2x  2 2 sin x cos x cos 2x  4 sin x cos x cos 2x  22 sin x cos x cos 21 x   n  3: sin 23 x  sin 8x  2 sin 4x cos 4x  2 4 sin x cos x cos 2x cos 4x  8 sin x cos x cos 2x cos 4x    23 sin x cos x cos 2x cos 22 x

101.

In general, for n  0 we have         sin 2n x  sin 2 2n1 x  2 sin 2n1 x cos 2n1 x       2 2n1 sin x cos x cos 2x cos 4x cos 8x    cos 2n2 x cos 2n1 x      2n sin x cos x cos 2x cos 4x cos 8x    cos 2n2 x cos 2n1 x

103. With u  sin1 x for 0  x  1, we can write     cos1 1  2x 2  cos1 1  2 sin2 u  cos1 cos 2u  2u  2 sin1 x.       x2  1 1 1 cos 2u  2u  2 tan1 1 , we can write cos1  cos 104. With u  tan1 x x x2  1 105. (a) f x 

sin 3x cos 3x  sin x cos x

cos 3x sin 3x  sin x cos x sin 3x cos x  cos 3x sin x sin 3x  x   sin x cos x sin x cos x 2 sin x cos x sin 2x  2  sin x cos x sin x cos x for all x for which the function is defined.

(b) f x 

2 1

-5

5

The function appears to have a constant value of 2 wherever it is defined.

SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas

106. (a) f x  cos 2x  2 sin2 x

565

(b) f x  cos 2x  2 sin2 x    cos2 x  sin2 x  2 sin2 x  cos2 x  sin2 x  1

1

-5

5

The function appears to have a constant value of 1.

(b) By a sum-to-product formula,

107. (a) y  sin 6x  sin 7x 2

-5

5 -2

y  sin 6x  sin 7x     6x  7x 6x  7x  2 sin cos 2 2     1  2 sin 13 2 x cos  2 x

(c) We graph y  sin 6x  sin 7x,   y  2 cos 12 x , and   y  2 cos 12 x . 2

1  2 sin 13 2 x cos 2 x

-5

5 -2

The graph of y  f x lies

between the other two graphs.

 1 3 108. From Example 2, we have cos 3x  4 cos3 x  3 cos x. If 3x   3 , then cos 3  2  4 cos x  3 cos x  1  8 cos3 x  6 cos x  8 cos3 x  6 cos x  1  0. Substituting y  cos x gives 8y 3  6y  1  0.

 2 109. (a) cos 4x  cos 2x  2x  2 cos2 2x  1  2 2 cos2 x  1  1  8 cos4 x  8 cos2 x  1. Thus the desired polynomial is P t  8t 4  8t 2  1.

  (b) cos 5x  cos 4x  x  cos 4x cos x  sin 4x sin x  cos x 8 cos4 x  8 cos2 x  1  2 sin 2x cos 2x sin x    8 cos5 x  8 cos3 x  cos x  4 sin x cos x 2 cos2 x  1 sin x [from part (a)]    8 cos5 x  8 cos3 x  cos x  4 cos x 2 cos2 x  1 sin2 x     8 cos5 x  8 cos3 x  cos x  4 cos x 2 cos2 x  1 1  cos2 x  8 cos5 x  8 cos3 x  cos x  8 cos5 x  12 cos3 x  4 cos x  16 cos5 x  20 cos3 x  5 cos x

Thus, the desired polynomial is P t  16t 5  20t 3  5t.

566

CHAPTER 7 Analytic Trigonometry

110. Let c1 and c2 be the lengths of the segments shown in the figure. By the Law of

C b b sin 2x c  or c  . Also by the x sin 2x sin B sin B b a x s c1 s  and Law of Sines applied to BC D and AC D we have sin B sin x B A cÁ cª D s c2 s sin x s sin x  . So c1  and c2  . Since c  c1  c2 , we have sin A sin x sin B  sin A  s sin x s sin x 1 b a b sin 2x 1    s sin x  . By applying the Law of Sines to ABC,  sin B sin B sin A sin B sin A sin B sin A     b b sin 2x 1 b sin x b 1  . Substituting we have  s sin x  s 1   sin A a sin B sin B sin B a sin B sin B a         b b ab 2ab cos x b  2b sin x cos x  sin x 1   2b cos x  s 1  s s  . b sin 2x  s sin x 1  a a a a ab

Sines applied to ABC, we have

111. Using

a

product-to-sum

formula,   RHS  4 sin A sin B sin C  4 sin A 12 [cos B  C  cos B  C]  2 sin A cos B  C  2 sin A cos B  C.

Using another product-to-sum formula, this is equal to     2 12 [sin A  B  C  sin A  B  C]  2 12 [sin A  B  C  sin A  B  C]

 sin A  B  C  sin A  B  C  sin A  B  C  sin A  B  C Now A  B  C  , so A  B  C    2C, A  B  C    2B, and A  B  C  2A  . Thus our expression simplifies to sin A  B  C  sin A  B  C  sin A  B  C  sin A  B  C  sin   2C  sin   2B  0  sin 2A    sin 2C  sin 2B  sin 2A  LHS

112. (a) The length of the base of the inscribed rectangle twice the length of the adjacent side which is 2 5 cos  and the length of the opposite side is 5 sin . Thus the area of the rectangle is modeled by A   2 5 cos  5 sin   25 2 sin  cos   25 sin 2.

  (b) The function y  sin u is maximized when u   2 . So 2  2    4 . Thus the maximum cross-sectional area is   A 4  25 sin 2 4  25 cm2 .       5 2  354 cm. (c) The length of the base is 2 5 cos   5 2  707 cm and the width of the rectangle is 5 sin 4 4 2

113. (a) In both logs the length of the adjacent side is 20 cos  and the length of the opposite side is 20 sin . Thus the cross-sectional area of the beam is modeled by A   20 cos  20 sin   400 sin  cos   200 2 sin  cos   200 sin 2.

  (b) The function y  sin u is maximized when u   2 . So 2  2    4 . Thus the maximum cross-sectional area is   A 4  200 sin 2 4  200.

114. We first label the figure as shown. Because the sheet of paper is folded over,  E AC   C AB  . Thus  BC A   AC E  90  . It follows that  EC D  180   BC A   AC E  180  90    90    2. Also,

from the figure we see that BC  L sin  and C E  L sin , so

A ¬ L

E

DC  EC cos 2  L sin  cos 2. Thus 6  D B  DC 

C B  L sin  cos 2  L sin   L sin  1  cos 2  L sin   2 cos2 . So L

3 6  . 2 sin  cos2  sin  cos2 

D

C

B

SECTION 7.4 Basic Trigonometric Equations

115. (a) y  f 1 t  f 2 t  cos 11t  cos 13t

567

(c) We graph y  cos 11t  cos 13t, y  2 cos t, and y  2 cos t.

2

2 -5

5 -5

-2

5 -2

(b) Using the identity

y cos   cos y  2  cos 2



The graph of f lies between the graphs

  y cos , we have 2     11t  13t 11t  13t f t  cos 11t  cos 13t  2  cos cos 2 2  2  cos 12t  cos t  2 cos 12t cos t 



116. (a) f 1  770 Hz and f 2  1209 Hz, so

of y  2 cos t and y  2 cos t. Thus,

the loudness of the sound varies between y  2 cos t.

(c) 2

y  sin 2  770t  sin 2  1209t  sin 1540t  sin 2418t

0 0.005

(b) Using a sum-to-product formula, we have 1540t  2418t 1540t  2418t cos 2 2  2 sin 1979t cos 439t

-2

y  2 sin

117. We find the area of ABC in two different ways. First, let AB be the base and C D

C

be the height. Since  B OC  2 we see that C D  sin 2. So the area is

1 base height  1  2  sin 2  sin 2. On the other hand, in ABC we see 2 2  that C is a right angle. So BC  2 sin  and AC  2 cos , and the area is 1 base height  1  2 sin  2 cos   2 sin  cos . Equating the two 2 2

1

A

¬

1

O

D

B

expressions for the area of ABC, we get sin 2  2 sin  cos .

7.4

BASIC TRIGONOMETRIC EQUATIONS

1. Because the trigonometric functions are periodic, if a basic trigonometric equation has one solution, it has infinitely many solutions. 2. The basic equation sin x  2 has no solution (because the sine function has range [1 1]), whereas the basic equation sin x  03 has infinitely many solutions. 3. We can find some of the solutions of sin x  03 graphically by graphing y  sin x and y  03. The solutions shown are x  97, x  60, x  34, x  03, x  28, x  66, and x  91. 4. (a) To find one solution of sin x  03 in the interval [0 2, we take sin1 to get x  sin1 03  030 The other solution in this interval is x    sin1 03  284.

(b) To find all solutions, we add multiples of 2 to the solutions in [0 2. The solutions are x  030  2k and x  284  2k.

568

CHAPTER 7 Analytic Trigonometry y 1

5. Because sine has period 2, we first find the solutions in the interval  [0 2. From the unit circle shown, we see that sin   23 in quadrants I 2 and II, so the solutions are    3 and   3 . We get all solutions of the

5¹

¬= 3

equation by adding integer multiples of 2 to these solutions: 2   3  2k and   3  2k for any integer k.

y=Ï3/2

_1

0

¹

¬= 3

1

x



6. The sine function is negative in quadrants III and IV, so solutions of sin    22 on the interval [0 2 are   54 and

  74 . Adding integer multiples of 2 to these solutions gives all solutions:   54  2k, 74  2k for any integer k. 7. The cosine function is negative in quadrants II and III, so the solution of cos   1 on the interval [0 2 is   . Adding integer multiples of 2 to this solution gives all solutions:     2k  2k  1  for any integer k. 

8. The cosine function is positive in quadrants I and IV, so the solutions of cos   23 on the interval [0 2 are    6 and 11   116 . Adding integer multiples of 2 to these solutions gives all solutions:    6  2k, 6  2k for any integer k.

9. The cosine function is positive in quadrants I and IV, so the solutions of cos   14 on the interval [0 2 are   cos1 41  132 and   2  cos1 41  497. Adding integer multiples of 2 to these solutions gives all solutions:   132  2k, 497  2k for any integer k.

10. The sine function is negative in quadrants III and IV, so the solutions of sin   03 on the interval [0 2 are

    sin1 03  345 and   2  sin1 03  598. Adding integer multiples of 2 to these solutions gives all solutions,   345  2k and   598  2k for any integer k.

11. The sine function is negative in quadrants III and IV, so the solutions of sin   045 on the interval [0 2 are

    sin1 045  361 and   2  sin1 045  582. Adding integer multiples of 2 to these solutions gives all solutions,   361  2k, 582  2k for any integer k.

12. The cosine function is positive in quadrants I and IV, so the solutions of cos   032 on the interval [0 2 are

  cos1 032  125 and   2  cos1 032  504. Adding integer multiples of 2 to these solutions gives all solutions,   125  2k, 504  2k for any integer k.    13. We first find one solution by taking tan1 of each side of the equation:   tan1  3    3 . By definition, this is the    only solution in the interval  2  2 . Since tangent has period , we get all solutions of the equation by adding integer multiples of :     3  k for any integer k.

14. One solution of tan   1 is   tan1 1   4 . Adding integer multiples of  to this solution gives all solutions:    4  k for any integer k. 15. One solution of tan   5 is   tan1 5  137. Adding integer multiples of  to this solution gives all solutions:   137  k for any integer k.   16. One solution of tan    13 is   tan1  13  032. Adding integer multiples of  to this solution gives all solutions:   032  k for any integer k.

    17. One solution of cos    23 is   cos1  23  56 and another is   2  56  76 . All solutions are

  56  2k, 76  2k for any integer k. Specific solutions include   56  2   76 ,   76  2   56 ,   56 ,   76 ,   56  2  176 , and   76  2  196 .

 18. One solution of cos   12 is   cos1 12   3 and another is   2  3     53  2k for any integer k. Specific solutions include    53 ,   3, 3,

5 . All solutions are     2k and 3 3 5 , 7 , and 11 . 3 3 3

SECTION 7.4 Basic Trigonometric Equations

569

   3  19. One solution of sin   22 is   sin1 22   4 and another is     4  4 . All solutions are   4  2k and 3 9 11   34  2k for any integer k. Specific solutions include    74 ,  54 ,  4 , 4 , 4 , and 4 .    20. One solution of sin    23 is     sin1 23  43 and another is   2  sin1 23  53 . All solutions are 4 5 10 11   43  2k and   53  2k for any integer k. Specific solutions include    23 ,   3 , 3 , 3 , 3 , and 3 . 21. One solution of cos   028 is   cos1 028  129 and another is   2  cos1 028  500. All solutions are

  129  2k and   500  2k for any integer k. Specific solutions include   500, 129, 129, 500, 757, and 1128.

22. One solution of tan   25 is   tan1 25  119. All solutions are   119  k for any integer k. Specific solutions include   509, 195, 119, 433, 747, and 1061.

23. One solution of tan   10 is   tan1 10  147. All solutions are   147  k for any integer k. Specific solutions include   775, 461, 147, 167, 481, and 795.

24. One solution of sin   09 is     sin1 09  426 and another is   2  sin1 09  516. All solutions are   426  2k and   516  2k for any integer k. Specific solutions include   202, 112, 426, 516, 1054, and 1144. 25. cos   1  0  cos   1. In the interval [0 2 the only solution is   Thus the solutions are   2k  1  for any integer k. 26. sin   1  0  sin   1. In the interval [0 2 the only solution is   32 . Therefore, the solutions are   32  2k for any integer k.   27. 2 sin   1  0  2 sin   1  sin    1 The solutions in the interval [0 2 are   54 , 74 . Thus the

2 5  7  solutions are   4  2k, 4  2k for any integer k.   7 28. 2 cos   1  0  2 cos   1  cos   1 . The solutions in the interval [0 2 are    4 , 4 . Thus the solutions 2 7 are    4  2k, 4  2k for any integer k. 29. 5 sin   1  0  sin   15 . The solutions in the interval [0 2 are   sin1 15  020 and     sin1 51  294.

Thus the solutions are   020  2k, 294  2k for any integer k.

  30. 4 cos   1  0  cos    14 . The solutions in the interval [0 2 are   cos1  14  182 and     2  cos1  14  446. Thus the solutions are   182  2k, 446  2k for any integer k. 

   31. 3 tan2   1  0  tan2   13  tan    33 . The solutions in the interval   2  2 are    6 , so all solutions are     6  k, 6  k for any integer k.

32. cot   1  0  cot   1. The solution in the interval 0  is   34 . Thus, the solutions are   34  k for any integer k.

3 5 7  33. 2 cos2   1  0  cos2   12  cos    1     4 , 4 , 4 , 4 in [0 2. Thus, the solutions are   4  k, 2

3  k for any integer k. 4



2 4 5  34. 4 sin2   3  0  sin2   34  sin    23     3 , 3 , 3 , 3 in [0 2. Thus, the solutions are   3  k, 2  k for any integer k. 3   35. tan2   4  0  tan2   4  tan   2    tan1 2  111 or   tan1 2  111 in   2  2 . Thus, the

solutions are   111  k, 111  k for any integer k.

36. 9 sin2   1  0  sin2   19  sin    13    sin1 31  034,     sin1 31  280,     sin1 13  348, or   2  sin1 13  594 on [0 2. Thus, the solutions are   034  k, 280  k for any integer k.

570

CHAPTER 7 Analytic Trigonometry

 3 5 7 37. sec2   2  0  sec2   2  sec    2. In the interval [0 2 the solutions are    4 , 4 , 4 , 4 . Thus, the solutions are   2k  1  4 for any integer k. 5 7 11 38. csc2   4  0  csc2   4  csc   2. In the interval [0 2 the solutions are    6 , 6 , 6 , 6 . So the 5 solutions are    6  k, 6  k for any integer k.   39. tan2   4 2 cos   1  0  tan2   4 or 2 cos   1. From Exercise 35, we know that the first equation has   solutions   111  k, 111  k for any integer k. 2 cos   1  cos    12 has solutions cos1  12  23

and 43 on [0 2, so all solutions are   23  2k and 43  2k for any integer k. Thus, the original equation has solutions   111  k, 111  k, 23  2k, and 43  2k for any integer k.   1 . The first equation has solution   tan1 2  111 40. tan   2 16 sin2   1  0  tan   2 or sin2   16   1 1 1  025, on   2  2 . The second equation is equivalent to sin    4 , which has solutions   sin 4     sin1 41  289,     sin1 41  339, and   2  sin1 41  603 on [0 2.

Thus, the original equation has solutions   111  k, 025  2k, 289  2k, 339  2k, and 603  2k for any integer k. 5 41. 4 cos2   4 cos   1  0  2 cos   12  0  2 cos   1  0  cos   12     3  2k, 3  2k for any

integer k.

42. 2 sin2   sin   1  0  2 sin   1 sin   1  0  2 sin   1  0 or sin   1  0. Since 2 sin   1  0  7 2 sin    12    76 , 116 in [0 2 and sin   1     2 in [0 2. Thus the solutions are   6  2k, 11  2k,   2k for any integer k. 6 2

43. 3 sin2   7 sin   2  0  3 sin   1 sin   2  0  3 sin   1  0 or sin   2  0. Since sin   1,

sin   2  0 has no solution. Thus 3 sin   1  0  sin   13    033984 and     033984  280176 are the solutions in [0 2, and all solutions are   033984  2k, 280176  2k for any integer k.    44. tan4   13 tan2   36  0  tan2   4 tan2   9  0  tan   2 or tan   3    110715, 110715,   124905, 124905 in   2  2 . Since the period for tangent is , the solutions are   124905  k, 110715  k for any integer k.

45. 2 cos2   7 cos   3  0  2 cos   1 cos   3  0  cos   12 or cos   3 (which is inadmissible)     3, 5 . Therefore, the solutions are     2k, 5  2k for any integer k. 3 3 3

46. sin2   sin   2  0  sin   2 sin   1  0  sin   2 (inadmissible) or sin   1. Thus, the solutions are   32  2k for any integer k.

47. cos2   cos   6  0  cos   2 cos x  3  0  cos x  2 or cos x  3, neither of which has a solution. Thus, the original equation has no solution. 48. 2 sin2   5 sin   12  0  sin x  4 2 sin x  3  0  sin x  4 or sin x  32 , neither of which has a solution. Thus, the original equation has no solution. 49. sin2   2 sin   3  sin2   2 sin   3  0  sin   3 sin   1  0  sin   3  0 or sin   1  0. Since

sin   1 for all , there is no solution for sin   3  0. Hence sin   1  0  sin   1    32  2k for any integer k.   50. 3 tan3   tan   3 tan3   tan   0  tan  3 tan2   1  0  tan   0 or 3 tan2   1  0. Now tan   0 

5   k and 3 tan2   1  0  tan2   13  tan    1     6  k, 6  k. Thus the solutions are   k, 3

  k, 5  k for any integer k. 6 6

SECTION 7.5 More Trigonometric Equations

571

7 11 51. cos  2 sin   1  0  cos   0 or sin    12     2  k, 6  2k, 6  2k for any integer k.     52. sec  2 cos   2  0  sec   0 or 2 cos   2  0. Since sec   1, sec   0 has no solution. Thus    7  7 2 cos   2  0  2 cos   2  cos   22     4  4 in [0 2. Thus   4  2k, 4  2k for any

integer k.

53. cos  sin   2 cos   0  cos  sin   2  0  cos   0 or sin   2  0. Since sin   1 for all , there is no 3  solution for sin   2  0. Hence, cos   0     2  2k, 2  2k    2  k for any integer k.

54. tan  sin   sin   0  sin  tan   1  0  sin   0 or tan   1  0. Now sin   0 when   k and tan   1  0  tan   1    34  k. Thus, the solutions are   k, 34  k for any integer k.

  55. 3 tan  sin   2 tan   0  tan  3 sin   2  0  tan   0 or sin   23 . tan   0 has solution   0 on   2 2 and sin   23 has solutions   sin1 23  073 and     sin1 32  241 on [0 2, so the original equation has solutions   k,   073  2k, 241  2k for any integer k.

3 56. 4 cos  sin   3 cos   0  cos  4 sin   3  0  cos   0 or sin    34 . cos   0 has solutions    2 , 2 on [0 2, while sin    34 has solutions     sin1 43  399 and   2  sin1 43  544 on [0 2. Thus, the

original equation has solutions    2  k,   399  2k, 544  2k for any integer k.  sin 70 sin 70  07065   133  sin 2  57. We substitute  1  70 and 1  133 into Snell’s Law to get 2 sin  2 133 2  4495 . sin 1 1  0658, so we substitute 2  90 into Snell’s Law to get  0658 58. The index of refraction from glass to air is 152 sin 90   sin 1  0658  1  411 .

59. (a) F  12 1  cos   0  cos   1    0

(b) F  12 1  cos   025  1  cos   05  cos   05    60 or 120 (c) F  12 1  cos   05  1  cos   1  cos   0    90 or 270

(d) F  12 1  cos   1  1  cos   2  cos   1    180 60. Statement A is true: every identity is an equation. However, Statement B is false: not every equation is an identity. The difference between an identity and an equation is that an identity is true for all values in the domain, whereas an equation may be true only for certain values in the domain and false for others. For example, x  0 is an equation but not an identity, because it is true for only one value of x.

7.5

MORE TRIGONOMETRIC EQUATIONS

1. Using a Pythagorean identity, we calculate sin x  sin2 x  cos2 x  1  sin x  1  1  sin x  0, whose solutions are x  k for any integer k.

2. Using a double-angle formula, we we see that the equation sin x  sin 2x  0 is equivalent to the equation sin x  2 sin x cos x  0. Factoring the left-hand side as sin x 1  2 cos x, we see that solving this equation is equivalent to solving the two basic equations sin x  0 and 1  2 cos x  0.   3. 2 cos2   sin   1  2 1  sin2   sin   1  0  2 sin2   sin   1  0  2 sin2   sin   1  0. From Exercise 7.4.42, the solutions are   76  2k, 116  2k,  2  2k for any integer k.

4. sin2   4  2 cos2   sin2   cos2   cos2   4  1  cos2   4  cos2   3 Since cos   1 for all , it follows that the original equation has no solution.

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5. tan2   2 sec   2  sec2   1  2 sec   2  sec2   2 sec   3  0  sec   3 sec   1  0  sec   3 or sec   1. If sec   3, then cos   13 , which has solutions   cos1 31  123 and   2  cos1 31  505

on [0 2. If sec   1, then cos   1, which has solution    on [0 2. Thus, solutions are   2k  1 ,   123  2k, 505  2k for any integer k.

6. csc2   cot   3  1  cot2   cot   3  cot2   cot   2  0  cot   2 cot   1  0  cot   2 or   cot   1. If cot   2, then tan   12 , which has solution   tan1 12  046 on   2  2 , and if cot   1, then     tan   1, which has solution     4 on  2  2 . Thus, solutions are    4  k,   046  k for any integer k. 7. 2 sin 2  3 sin   0  2 2 sin  cos   3 sin   0  sin  4 cos   3  0  sin   0 or cos   34 . The first equation has solutions   0,  on [0 2, and the second has solutions   cos1 34  072 and   2  cos1 34  556 on [0 2. Thus, solutions are   k,   072  2k, 556  2k for any integer k.

8. 3 sin 2  2 sin   0  3 2 sin  cos   2 sin   0  2 sin  3 cos   1  0  sin   0 or cos   13 . The first equation has solutions   0,  on [0 2, and the second has solutions   cos1 13  123 and   2  cos1 13  505 on [0 2. Thus, solutions are   k,   123  2k, 505  2k for any integer k.

9. cos 2  3 sin   1  1  2 sin2   3 sin   1  2 sin2   3 sin   2  0  sin   2 2 sin   1  0

 sin   2 or sin   12 . The first equation has no solution and the second has solutions   sin1 21   6 and 5   5  1 1     sin 2  6 on [0 2, so the original equation has solutions   6  2k, 6  2k for any integer k. 

10. cos 2  cos2   12  2 cos2   1  cos2   12  cos2   12  cos    22 . Thus, the solutions are    4  k, 3  k for any integer k. 4   11. 2 sin2   cos   1  2 1  cos2   cos   1  0  2 cos2   cos   1  0  2 cos   1 cos   1  0  5 2 cos   1  0 or cos   1  0  cos   12 or cos   1     3  2k, 3  2k, 2k  1  for any integer k.

12. tan   3 cot   0 

3 cos  sin2   3 cos2  sin2   cos2   4 cos2  sin    0  0  0 cos  sin  cos  sin  cos  sin 

1  4 cos2  2  0  1  4 cos2   0  4 cos2   1  cos    12     3  k, 3  k for any integer k. cos  sin  13. sin   1  cos   sin   cos   1. Squaring both sides, we have sin2   cos2   2 sin  cos   1  sin 2  0,

3  which has solutions   0,  2 , , 2 in [0 2. Checking in the original equation, we see that only   2 and    are valid. (The extraneous solutions were introduced by squaring both sides.) Thus, the solutions are   2k  1 ,  2  2k for any integer k.

14. Square both sides of cos   sin   1 to get cos2   sin2   2 sin  cos   1  sin 2  0, which has solutions   0,  , , 3 on [0 2. Checking in the original equation, we see that only   0 and   3 are valid. Thus, the solutions 2 2 2 are   2k, 3  2k for any integer k. 2

1 sin  1   sin cos   1. Squaring both sides, we have sin2 cos2 2 sin  cos   1 cos  cos  3  sin 2  0, which has solutions   0,  2 , , 2 on [0 2. Checking in the original equation, we see that only   0

15. tan 1  sec  

is valid. Thus, the solutions are   2k for any integer k.   16. 2 tan   sec2   4  2 tan   1  tan2   4  tan2   2 tan   3  0  tan   3 tan   1  0    tan   3 or tan   1. The first equation has the solution   tan1 3  125 on   2  2 , and the second has     solution  4 on  2  2 . Thus, the solutions are   125  k,   4  k for any integer k.

SECTION 7.5 More Trigonometric Equations

573

5  2 5 2 17. (a) 2 cos 3  1  cos 3  12  3   3 , 3 for 3 in [0 2. Thus, solutions are 9  3 k, 9  3 k for any integer k. 5 7 11 13 17 (b) We take k  0, 1, 2 in the expressions in part (a) to obtain the solutions    9 , 9 , 9 , 9 , 9 , 9 in [0 2.

5  5 18. (a) 2 sin 2  1  sin 2  12  2   6 , 6 for 2 in [0 . Thus, solutions are 12  k, 12  k for any integer k.

 , 5 , 13 , 17 in [0 2. (b) Take k  0, 1 in the expressions in part (a) to obtain the solutions   12 12 12 12

2 19. (a) 2 cos 2  1  0  cos 2   12  2  23  2k, 43  2k     3  k, 3  k for any integer k. 2 4 5 (b) The solutions in [0 2 are  3, 3 , 3 , 3 .

  2 k, 11  2 k for any integer k. 20. (a) 2 sin 3  1  0  2 sin 3  1  sin 3   12 , which has solutions   718 3 18 3

 , 11 , 19 , 23 , 31 , 35 . (b) The solutions in [0 2 are   718 18 18 18 18 18    1 k for any integer k. 21. (a) 3 tan 3  1  0  tan 3   1  3  56  k    518 3 3  , 11 , 17 , 23 , 29 , 35 . (b) The solutions in [0 2 are 518 18 18 18 18 18

5  1 5 1 22. (a) sec 4  2  0  sec 4  2  4   3  2k, 3  2k    12  2 k, 12  2 k for any integer k.

 , 5 , 7 , 11 , 13 , 17 , 19 , 23 . (b) The solutions in [0 2 are 12 12 12 12 12 12 12 12

23. (a) cos 2  1  0  cos 2  1  2  2k    4k for any integer k.

(b) The only solution in [0 2 is   0.   24. (a) tan 4  3  0  tan 4   3  4  23  k    83  4k for any integer k. (b) There is no solution in [0 2.    25. (a) 2 sin 3  3  0  2 sin 3   3  sin 3   23  3  43  2k, 53  2k    4  6k, 5  6k for any integer k. (b) There is no solution in [0 2. 26. (a) sec 2  cos 2  cos2 2  1  cos 2  1  2  k    2k for any integer k. (b) The only solution in [0 2 is   0.   1 27. (a) sin 2  3 cos 2  tan 2  3    12 tan1 3  062 on   4  4 . Thus, solutions are   062  2 k for any integer k.

(b) The solutions in [0 2 are   062, 219, 376, 533.  1  5 sin 3  1  5 sin2 3  sin 3   55 , which has solutions 28. (a) csc 3  5 sin 3  sin 3       5 5  089,     1 sin1 5  120, and 1 1 1 1  015,      3 sin 5 3  3 sin 5 3 3 5       23  13 sin1 55  194 on 0 23 . Thus, solutions are   015  13 k, 089  13 k for any integer k. (b) The solutions in [0 2 are   015, 089, 120, 194, 224, 298, 329, 403, 434, 508, 539, 613.

29. (a) 1  2 sin   cos 2  1  2 sin   1  2 sin2   2 sin2   2 sin   0  2 sin  sin   1  0  sin   0 or  sin   1    0, , or  2 in [0 2. Thus, the solutions are   k,   2  2k for any integer k. (b) The solutions in [0 2 are   0,  2 , .

30. (a) tan 3  1  sec 3  tan 3  12  sec2 3  tan2 3  2 tan 3  1  sec2 3  sec2 3  2 tan 3  sec2 3  2 tan 3  0  3  k for any integer k. Because squaring both sides is an operation that can introduce extraneous solutions, we must check each of the possible solutions in the original equation, and we see that only   23 k are valid solutions.

(b) The solutions in [0 2 are   0, 23 , 43 .

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  31. (a) 3 tan3   3 tan2   tan   1  0  tan   1 3 tan2   1  0  tan   1 or 3 tan2   1  tan   1 or  5 tan    1     6  k, 4  k, 6  k for any integer k. 3

 5 7 5 11 (b) The solutions in [0 2 are    6, 4, 6 , 6 , 4 , 6 .

32. (a) 4 sin  cos   2 sin   2 cos   1  0  2 sin   1 2 cos   1  0  2 sin   1  0 or 2 cos   1  0  5 2 4 sin   12 or cos    12     6  2k, 6  2k, 3  2k, 3  2k for any integer k.

5 2 4 (b) The solutions in [0 2 are    6, 6 , 3 , 3 .

33. (a) 2 sin  tan   tan   1  2 sin   2 sin  tan   tan   2 sin   1  0  2 sin   1 tan   1  0 

5 3 2 sin   1  0 or tan   1  0  sin   12 or tan   1     6  2k, 6  2k, 4  k for any integer k.

3 5 7 (b) The solutions in [0 2 are  6, 4 , 6 , 4 .

cos  sin  1 sin  cos2   cos    sin    sin .  cos  cos  sin  sin  cos2  Multiplying both sides by the common denominator cos2  sin  gives sin2   cos4   sin2  cos2     sin2   cos4   1  cos2  cos2   sin2   cos4   cos2   cos4   sin2   cos2   sin2   1  sin2 

34. (a) sec  tan   cos  cot   sin  

3  2 sin2   1  sin    1     4  k, 4  k for any integer k. Since we multiplied the original 2

equation by cos2  sin  (which could be zero) we must check to see if we have introduced extraneous solutions. However, each of the values of  does indeed satisfy the original equation. 3 5 7 (b) The solutions in [0 2 are    4, 4 , 4 , 4 .

(b) f x  3 cos x  1; g x  cos x  1. f x  g x when

35. (a) 4

3 cos x  1  cos x  1  2 cos x  2  cos x  1  x    2k  2k  1 . The points of intersection are

2

2k  1  2 for any integer k.

-6

-4

-2

2

4

6

-2

The points of intersection are approximately 314 2. (b) f x  sin 2x  1; gx  2 sin 2x  1. f x  g x when

36. (a)

sin 2x  1  2 sin 2x  1  sin 2x  0  x  12 k. The points of   intersection are 12 k 1 for any integer k.

2

-6

-4

-2

2

4

6

The points of intersection are approximately 628 1, 471 1, 314 1, 157 1, and 0 1.

SECTION 7.5 More Trigonometric Equations

575

  3. f x  g x when tan x  3      x 3  k. The intersection points are 3  k 3 for any integer k.

(b) f x  tan x; g x 

37. (a) 10

-1

1 -10

The point of intersection is approximately 104 173. (b) f x  sin x  1; g x  cos x. f x  g x when sin x  1  cos x 

38. (a)

sin x  12  cos2 x  sin2 x  2 sin x  1  cos2 x 

-6

-4

-2

2

4

6

sin2 x  2 sin x  1  cos2 x  0  2 sin2 x  2 sin x  0 

2 sin x sin x  1  0  sin x  0 or sin x  1  x  k,  2  2k. However, x  k is not a solution when k is even. (The extraneous

-2

solutions were introduced by squaring both sides.) So the solutions are

The points of intersection are approximately 471 0, 314 1, 157 0, and 314 1.

x  2k  1 ,  2  2k, and the intersection points are   2k 1,   2  2k 0 for any integer k.

3 5 7 9 11 13 15 39. cos  cos 3  sin  sin 3  0  cos   3  0  cos 4  0  4   2 , 2 , 2 , 2 , 2 , 2 , 2 , 2 in 3 5 7 9 11 13 15 [0 8     8 , 8 , 8 , 8 , 8 , 8 , 8 , 8 in [0 2. 5 40. cos  cos 2  sin  sin 2  12  cos   2  12  cos   12  cos   12     3 , 3 in [0 2. 





2 41. sin 2 cos   cos 2 sin   23  sin 2    23  sin   23     3 , 3 in [0 2.

42. sin 3 cos   cos 3 sin   0  sin 3    0  sin 2  0  2  0, , 2, 3, 4 in [0 4    0,  2 , , 3 in [0 2. 2

7 43. sin 2  cos   0  2 sin  cos   cos   0  cos  2 sin   1  0  cos   0 or sin    12     2, 6 , 3 , 11 in [0 2. 2 6

 sin   sin   0   sin   0  sin   sin  1  cos   0 (and cos   1    )  2 1  cos  3 sin   cos   0  sin   0 or cos   0    0,  2 , 2 in [0 2 (   is inadmissible).

44. tan

45. cos 2  cos   2  2 cos2   1  cos   2  0  2 cos2   cos   3  0  2 cos   3 cos   1  0 

2 cos   3  0 or cos   1  0  cos    32 (which is impossible) or cos   1    0 in [0 2.   cos  sin  cos  sin    8 sin  cos     sin  cos   8 sin  cos   sin  cos   46. tan   cot   4 sin 2  cos  sin  cos  sin  sin2   cos2   8 sin2  cos2   1  2 2 sin  cos 2  sin 22  12  sin 2   1 . Therefore, 2   4  k 2

k 3 k  3 5 7 or 2  34  k     8  2 or   8  2 . Thus on the interval [0 2 the solutions are   8 , 8 , 8 , 8 , 9 , 11 , 13 , 15 . 8 8 8 8

47. cos 2  cos2   0  2 cos2   1  cos2   0  cos2   1    k for any integer k. On [0 2, the solutions are   0, .

576

CHAPTER 7 Analytic Trigonometry 

48. 2 sin2   2  cos 2  2 sin2   2  1  2 sin2   4 sin2   3  sin    23    23  k, 43  k for any 2 4 5 integer k. On [0 2, the solutions are  3, 3 , 3 , 3 .

  49. cos 2  cos 4  0  cos 2  2 cos2 2  1  0  cos 2  1 2 cos 2  1  0. The first factor has zeros at

2 4 5  2   0,  and the second has zeros at    3 , 3 , 3 , 3 . Thus, solutions of the original equation are are   0, 3 , 3 , , 43 , 53 in [0 2.

50. sin 3  sin 6  0  sin 3  2 sin 3 cos 3  0  sin 3 1  2 cos 3  0. The first factor has zeros at   0,  3,

2 , , 4 , 5 and the second has zeros at  , 5 , 7 , 11 , 13 , 17 , so these are all solutions of the original equation 3 3 3 9 9 9 9 9 9

in [0 2.      1  cos   . Squaring both sides, we have 51. cos   sin   2 sin 2  cos   sin   2  2 cos2   sin2   2 sin  cos   1  cos   1  2 sin  cos   1  cos   either 2 sin   1 or cos   0     6,  , 5 , 3 in [0 2. Of these, only  and 3 satisfy the original equation. 2 6 2 6 2

52. Square both sides of sin   cos   12 to get sin2   cos2   2 sin  cos   14  1  sin 2  14  sin 2  34 . 1 1 3  115,   1 sin1 3  357, and Thus, the possible solutions in [0 2 are   12 sin1 34  042,  2  2 sin 4 2 4 3  1 sin1 3  429. We find that only valid solutions to the original equation on [0 2 are   115, 357. 2 2 4

53. sin   sin 3  0  2 sin 2 cos   0  2 sin 2 cos   0  sin 2  0 or cos   0  2  k or   k  2    12 k for any integer k.

54. cos 5  cos 7  0  2 sin 6 sin   0  sin 6 sin   0  sin 6  0 or sin   0  6  k or   k    16 k for any integer k.

55. cos 4  cos 2  cos   2 cos 3 cos   cos   cos  2 cos 3  1  0  cos   0 or cos 3  12     2 or

5 7 11 13 17   2 5 2 3   3  2k, 3  2k, 3  2k, 3  2k, 3  2k, 3  2k    2  k, 9  3 k, 9  3 k for any integer k.

56. sin 5  sin 3  cos 4  2 cos 4 sin   cos 4  cos 4 2 sin   1  0  cos 4  0 or sin   12   5  1  5 4   2  k or   6  2k, 6  2k    8  4 k, 6  2k, 6  2k for any integer k. 58. cos x 

57. sin 2x  x

x 3

1

-1

1 1

-1

The three solutions are x  0 and x  095.

-4

-2

2 -1

The three solutions are x  117, 266, and 294.

SECTION 7.5 More Trigonometric Equations

59. 2sin x  x

60. sin x  x 3 1

2 -1 2

1 -1

4

The only solution is x  192. 61.

577

The three solutions are x  0 and x  093.   62. cos x  12 e x  ex

cos x  x2 1  x2

1

1

-1

-1

1

The two solutions are x  071.

1

The only solution is x  0.

x  2x tan u  tan  63. With u  tan1 x and   tan1 2x, we have u     4  tan u    1  1  tan u tan   1  1  x 2x  1    3  17 3  32  4 2 1 2 2 x  2x  1  2x  2x  3x  1  0  x   . Because tan x is not one-to-one, 2 2 4 we must check both roots, and find that only

 173 is a solution to the original equation. 4

64. With u  sin1 x and   cos1 x, we have 2u      cos 2u    1    cos 2u cos   sin 2u sin   1  1  2 sin2 u cos   2 sin u cos u sin   1. Referring to the diagram, this becomes     1  2x 2 x  2x 1  x 2 1  x 2  1  x  2x 3  2x  2x 3  1  x  1.

22002 sin 2  5000  151250 sin 2  32 sin 2  003308  2  189442 or 2  180  189442  17810558 . If 2  189442 , then   094721 , and if 2  17810558 , then   8905279 .

65. We substitute  0  2200 and R   5000 and solve for . So 5000 

66. Since 4e3t  0, we have 0  4e3t sin 2t  0  sin 2t  2t  0, , 2,     t  0, 12 , 1, 32 ,   .       67. (a) 10  12  283 sin 23 t  80  283 sin 23 t  80  2  sin 23 t  80  070671. Now

sin   070671 and   078484. If 23 t  80  078484  t  80  456  t  344. Now in the interval [0 2, we have     078484  392644 and   2  078484  549834. If 23 t  80  392644 

t  80  2281  t  3081. And if 23 t  80  549834  t  80  3194  t  3994 3994  365  344. So according to this model, there should be 10 hours of sunshine on the 34th day (February 3) and on the 308th day (November 4).   (b) Since L t  12  283 sin 23 t  80  10 for t  [34 308], the number of days with more than 10 hours of daylight is 308  34  1  275 days.

578

CHAPTER 7 Analytic Trigonometry

   . The part of the belt touching the larger 2 2 pulley has length 2    R     R and similarly the part

68. (a) First note that  

touching the smaller belt has length    r. To calculate a and b, we write cot

a b       a  R cot and b  r cot , so 2 R r 2 2

R

a Œ

¬l

Œ

b

R

r r

 the length of the straight parts of the belt is 2a  2b  2 R  r cot . Thus, the total length of the belt is 2     L  and so     2 cot   L     R     r  2 R  r cot  R  r     2 cot 2 2 2 R r L   .   2 cot  2 R r  L 2778 (b) We plot   2 cot and      45113 in the same 2 R r 242  121 10 viewing rectangle. The solution is   1047 rad  60 . 5 0 0

2 y

69. sin cos x is a function of a function, that is, a composition of trigonometric functions (see Section 2.6). Most of the other equations involve sums, products,

1

differences, or quotients of trigonometric functions. sin cos x  0  cos x  0 or cos x  . However, since cos x  1, the only

solution is cos x  0  x   2  k. The graph of f x  sin cos x is shown.

_2¹

0

_¹

_1

CHAPTER 7 REVIEW 

sin  cos   1. sin  cot   tan   sin  sin  cos 



 cos  

sin2  cos2   sin2  1    sec  cos  cos  cos 

2. sec   1 sec   1  sec2   1  tan2    3. cos2 x csc x  csc x  1  sin2 x csc x  csc x  csc x  sin2 x csc x  csc x   sin2 x  4.

1

1  sin2 x



1  sec2 x  1  tan2 x cos2 x

5.

cos2 x  tan2 x

6.

1 1  cos x 1  cos2 x sin2 x 1  sec x   1  1  cos x  1  cos x    sec x sec x 1  cos x 1  cos x 1  cos x

7.

sin2 x

cos2 x  1  sin x

1   sin x sin x



cos2 x sin2 x



tan2 x sin2 x

 cot2 x 

1  cot2 x  sec2 x cos2 x

cos x cos x cos x   sin x 1 1 sec x  tan x  1  sin x cos x cos x cos x

8. 1  tan x 1  cot x  1  cot x  tan x  tan x cot x  2  cot x  tan x  2  2

1 cos2 x  sin2 x 2  2  sec x csc x cos x sin x cos x sin x



sin x cos x  sin x cos x



¹

x

CHAPTER 7

579

sin2 x  cos2 x  sin2 x  1 cos2 x sin2 x  2   2  cos x 2 sin2 x  cos2 x sin x 1 2 10. tan x  cot x      sec x csc x2  csc2 x sec2 x cos x sin x cos x sin x cos x sin x 9. sin2 x cot2 x  cos2 x tan2 x  sin2 x 

11. 12. 13. 14. 15. 16.

cos2 x

Review

 cos2 x 

2 sin x cos x 2 sin x cos x 2 sin x sin 2x   tan x   1  cos 2x 2 cos x 1  2 cos2 x  1 2 cos2 x cos x cos y  sin x sin y cos x cos y sin x sin y cos x  y     cos x sin y cos x sin y cos x sin y cos x sin y x 1  cos x csc x  tan  csc x   csc x  csc x  cot x  cot x 2 sin x x sin x 1  cos x 1  cos x 1 1  tan x tan  1   1 1 2 cos x sin x cos x cos x cos 2x 2 sin x cos x 2 cos2 x  1 sin 2x     2 cos x  2 cos x  sin x cos x sin x cos x  tan x  tan   1  tan x 4 tan x   4  1  tan x tan   1  tan x 4

cos y sin x   cot y  tan x sin y cos x

1

1  sec x cos x

1  sec x cos x

1 1 1  1 cos x sec x  1 1  cos x x cos x cos x 17.    tan   1 1 sin x sec x cos x sin x 2 sin x sin x cos x cos x

  18. cos x  cos y2  sin x  sin y2  cos2 x  2 cos x cos y  cos2 y  sin2 x  2 sin x sin y  sin2 y  cos2 x  sin2 x    sin2 y  cos2 y  2 cos x cos y  sin x sin y  2  2 cos x  y   x x x 2 x x x x x x x x 19. cos  sin  1  sin x  cos2  2 sin cos  sin2  sin2  cos2  2 sin cos  1  sin 2  2 2 2 2 2 2 2 2 2 2 2     3x  7x 3x  7x 2 sin sin 2 sin 5x sin 2x sin 2x cos 3x  cos 7x 2 sin 5x sin 2x 2 2          tan 2x 20. 3x  7x 3x  7x sin 3x  sin 7x 2 sin 5x cos 2x 2 sin 5x cos 2x cos 2x cos 2 sin 2 2     x  y  x  y x  y  x  y 2 sin cos 2 sin x cos y sin x sin x  y  sin x  y 2 2        tan x 21. x  y  x  y x  y  x  y cos x  y  cos x  y 2 cos x cos y cos x cos 2 cos 2 2   1 22. sin x  y sin x  y  2 cos x  y  x  y  cos x  y  x  y  12 cos 2y  cos 2x       12 1  2 sin2 y  1  2 sin2 x  12 2 sin2 x  2 sin2 y  sin2 x  sin2 y

2  23. (a) f x  1  cos x2  sin x2 , g x  sin x

5 -1

prove this, expand f x and simplify, using the double-angle formula for sine:

1

-5

(b) The graphs suggest that f x  g x is an identity. To

2  f x  1  cos x2  sin x2    1  cos2 x2  2 cos x2 sin x2  sin2 2x    1  2 cos x2 sin x2  cos2 x2  sin2 2x  1  sin x  1  sin x  g x

580

CHAPTER 7 Analytic Trigonometry

24. (a) f x  sin x  cos x, g x  2

 sin2 x  cos2 x

-5

5

(b) The graphs suggest that f x  g x in general. For example, choose x   6 and evaluate the functions:     1 3 f 6  2  2  12 3 , whereas   1 3  g  6  4  4  1  1, so f x  g x.

-2

25. (a) f x  tan x tan x2 , g x 

1 cos x

(b) The graphs suggest that f x  g x in general. For   example, choose x   3 and evaluate: f 3  tan 3

5

-5

tan  6 

5

   1 3  1  1, whereas g  3  1  2, so 3

2

f x  g x.

-5

26. (a) f x  1  8 sin2 x  8 sin4 x, g x  cos 4x 1

-5

5

(b) The graphs suggest that f x  g x is an identity. To

show this, expand g x by using double-angle identities: g x  cos 4x  cos 2 2x  1  2 sin2 2x    1  2 2 sin x cos x2  1  2 4 sin2 x cos2 x    1  8 sin2 x  1  sin2 x

-1

 1  8 sin2 x  8 sin4 x  f x

27. (a) f x  2 sin2 3x  cos 6x

(b) The graph suggests that f x  1 for all x. To prove this, we use the double angle formula to note that

1

-5

5

cos 6x  cos 2 3x  1  2 sin2 3x, so   f x  2 sin2 3x  1  2 sin2 3x  1.

-1

x 28. (a) f x  sin x cot , g x  cos x 2

(b) Proof: f x  sin x cot x2  sin x 

2

-5

Conjecture: f x  g x  1

5

   2 sin x2 cos x2 





 cos x2 sin x2

cos x2 sin x2



Now subtract and add 1, so f x  2 x cos2  1  1  cos x  1  g x  1. 2

 2 cos2 x2

CHAPTER 7

Review

581

29. 4 sin   3  0  4 sin   3  sin   34    sin1 34  08481 or     sin1 34  22935.     30. 5 cos   3  0  5 cos   3  cos    35    cos1  35  22143 or   2  cos1  35  40689.

31. cos x sin x  sin x  0  sin x cos x  1  0  sin x  0 or cos x  1  x  0,  or x  0. Therefore, the solutions are x  0 and . 5 32. sin x  2 sin2 x  0  sin x 1  2 sin x  0  sin x  0 or sin x  12  x  0,  or x   6 , 6 . Therefore, the 5 solutions in [0 2 are x  0,  6 , 6 , .

33. 2 sin2 x  5 sin x  2  0  2 sin x  1 sin x  2  0  sin x  12 or sin x  2 (which is inadmissible)  x   6, 5 . Thus, the solutions in [0 2 are x   and 5 . 6 6 6 34. sin x  cos x  tan x  1  sin x cos x  cos2 x  sin x   cos x  sin x cos x  sin x  cos2 x  cos x  0  sin x cos x  1  cos x cos x  1  0  sin x  cos x cos x  1  0  sin x  cos x or cos x  1  tan x  1 or 5  5 cos x  1  x   4 , 4 or x  0. Therefore, the solutions in [0 2 are x  0, 4 , 4 .

35. 2 cos2 x  7 cos x  3  0  2 cos x  1 cos x  3  0  cos x  12 or cos x  3 (which is inadmissible)  x   3, 5 . Therefore, the solutions in [0 2 are x   , 5 . 3 3 3   36. 4 sin2 x  2 cos2 x  3  2 sin2 x  2 sin2 x  cos2 x  3  0  2 sin2 x  2  3  0  2 sin2 x  1  sin x   1 . 2

3 5 7 So the solutions in [0 2 are x   4, 4 , 4 , 4 .

37. Note that x   is not a solution because the denominator is zero. 4 cos x  2  cos x   12  x  23 , 43 in [0 2.

1  cos x  3  1  cos x  3  3 cos x  1  cos x

38. sin x  cos 2x  sin x  1  2 sin2 x  2 sin2 x  sin x  1  0  2 sin x  1 sin x  1  0  sin x  12 or 5 3  3 5 sin x  1  x   6 , 6 or x  2 . Thus, the solutions in [0 2 are x  6 , 2 , 6 .    39. Factor by grouping: tan3 x  tan2 x  3 tan x  3  0  tan x  1 tan2 x  3  0  tan x  1 or tan x   3  2 4 5  2 3 4 5 7 x  34 , 74 or x   3 , 3 , 3 , 3 . Therefore, the solutions in [0 2 are x  3 , 3 , 4 , 3 , 3 , 4 .   40. cos 2x csc2 x  2 cos 2x  cos 2x csc2 x  2 cos 2x  0  cos 2x csc2 x  2  0  cos 2x  0 or csc2 x  2 

cos 2x  0or sin2 x  12  cos 2x  0 or sin x   1 . 2

3 5 7  3 5 7 For cos 2x  0, the solutions in [0 4 are 2x   2 , 2 , 2 , 2  the solutions in [0 2 are x  4 , 4 , 4 , 4 . 3 5 7 For sin x   1 , the solutions in [0 2 are x   4, 4 , 4 , 4 . 2

3 5 7 Thus, the solutions of the equation in [0 2 are x   4, 4 , 4 , 4 . 1  cos x 1 4 sin x cos x   1cos x 4 sin2 x cos x  1  4 sin2 x cos x cos x  0 41. tan 12 x 2 sin 2x  csc x  sin x sin x   3  5 7 11  cos x 4 sin2 x  1  0  cos x  0 or sin x   12  x   2 , 2 or x  6 , 6 , 6 , 6 . Thus, the solutions in

 5 7 3 11 [0 2 are x   6, 2, 6 , 6 , 2 , 6 . 42. cos 3xcos 2xcos x  0  cos 2x cos xsin 2x sin xcos 2xcos x  0  cos 2x cos xcos 2xsin 2x sin xcos x  0    cos 2x cos x  1  2 sin2 x cos x  cos x  0  cos 2x cos x  1  cos x 1  2 sin2 x  0 

cos 2x cos x  1  cos x cos 2x  0  cos 2x cos x  1  cos x  0  cos 2x 2 cos x  1  0  cos 2x  0 or

3 5 7 2 4  cos x   12  2x   2 , 2 , 2 , 2 (in [0 4) or x  3 , 3 (in [0 2). Thus, the solutions in [0 2 are x  4 , 2 , 3 , 5 , 4 , 7 . 3 4 4 3 4

582

CHAPTER 7 Analytic Trigonometry

     1 sin x   3  sin x  1  3 cos x  3 cos x  sin x  1  23 cos x  12 sin x  12 3 cos x cos x  sin x  1  cos x     1  x     , 5  x   , 3 . However, x  3 is  cos  cos x  sin 6 6 2 6 2 6 3 3 6 2 2 inadmissible because sec 32 is undefined. Thus, the only solution in [0 2 is x   . 6   sin x 2  0  2 cos x  3 sin x  0 (cos x  0)  2 1  sin2 x  3 sin x  0  44. 2 cos x  3 tan x  0  2 cos x  3 cos x 2 sin2 x  3 sin x  2  0  2 sin x  1 sin x  2  0  sin x  12 or sin x  2 (which has no solution)  x   6,

43. tan x  sec x 

5 . 6

45. We graph f x  cos x and g x  x 2  1 in the viewing 46. We graph f x  esin x and g x  x in the viewing rectangle [0 65] by [2 2]. The two functions intersect

rectangle [0 65] by [1 3]. The two functions intersect

at only one point, x  118.

at only one point, x  222.

2 2 0 2

4

6 0 2

-2

4

6

4002 sin2   sin2   08  sin   08944    634 64 4002 sin2   2500 sin2   2500. Therefore it is impossible for the projectile to reach a height of 3000 ft. (b) 64 (c) The function M   2500 sin2  is maximized when sin2   1, so   90 . The projectile will travel the highest when it is shot straight up. k 48. Since e02t  0 we have f t  e02t sin 4t  0  sin 4t  0  4t  k  t  where k is any integer. Thus 4 the shock absorber is at equilibrium position every quarter second.      1  cos 30 2 3 2 3   49. Since 15 is in quadrant I, cos 15    . 2 4 2

47. (a) 2000 

 







2 , which is Another method: cos 15  cos 45  30   cos 45 cos 30  sin 45 sin 30  22 23  22 12  6 4   equal to 12 2  3.        3 5 1  1  cos 2 3 2 3 2 6   is in quadrant I, sin 5    . 50. Since 512 12 2 2 4 2       sin       sin  cos   cos  sin   1 3  1 1  31   6 2 , which is Another method: sin 512 4 6 4 6 4 6 2 2 4 2 2 2 2   1 equal to 2 2  3.

51. tan  8 

1  1    1  cos  2 4   1  1 2 21  1 2 sin 4  2

 cos   sin 2     sin   1 52. 2 sin 12 12 12 6 2



53. sin 5 cos 40  cos 5 sin 40  sin 5  40   sin 45  1  22 2    tan 66  tan 6 54.  tan 66  6   tan 60  3 1  tan 66 tan 6

CHAPTER 7

Review

583

    2   cos 2   cos   1  2 55. cos2   sin 8 8 8 4 2 2     3 sin   sin  cos   cos  sin   sin       sin   1  2 56. 12 cos 12 2 12 6 12 6 12 6 12 4 2 2       57. We use a product-to-sum formula: cos 375 cos 75  12 cos 45  cos 30   12 22  23  14 2 3 .

    1 1  cos 45 675  225 675  225 cos  2 cos 45 cos 225  2    2 2 2 2      1  cos 45  1  1  22 2

58. cos 675  cos 225  2 cos



2





In Exercises 59–64, x and y are in quadrant I, so we know that sec x  32  cos x  23 , so sin x  35 and tan x  25 . 



1  2. Also, csc y  3  sin y  13 , and so cos y  2 3 2 , and tan y   4 2 2      59. sin x  y  sin x cos y  cos x sin y  35  2 3 2  23  13  29 1  10 .

      60. cos x  y  cos x cos y  sin x sin y  23  2 3 2  35  13  19 4 2  5 .

        5  2 5  2  2 2 5  2 8  10   tan x  tan y 8  2 5  61. tan x  y     23 2   4    2   4     2 2 1  tan x tan y 8 8  10 8  10 1  25 1  25 4 4 



62. sin 2x  2 sin x cos x  2  35  23  4 9 5 .         1  2 2  y 1  cos y 32 2 3 63. cos    (since cosine is positive in quadrant I) 2 2 2 6    1  232 1  cos y 32 2 y  32 2  64. tan  1 2 sin y 1 3



65. We sketch a triangle such that   cos1 73 . We see that tan   2 310 , and the double-angle formula for tangent gives    4 10 2  2 310 2 tan  12 10 3 tan 2  .     2  31 1  tan2  1  40 9 1  2 310

7 ¬

sin     sin  cos   cos  sin  5  4  12  63  35  13 5 13 65

  67. The double-angle formula for tangent gives tan 2 tan1 x 

3

cos   37

5 . From the 66. We sketch triangles such that   tan1 43 and   cos1 13

triangles, we see that sin   35 , cos   45 , and sin   12 13 , so the addition formula for sine gives

2Ï10

5

¬

3

tan   34

  2 tan tan1 x 2x    . 1  tan2 tan1 x 1  x2

4

13 12

5 cos   13

ú

5

584

CHAPTER 7 Analytic Trigonometry

 68. Let   sin1 x and   cos1 y. From the triangles, cos   1  x 2 and  sin   1  y 2 , so using the addition formula for cosine, we have   cos     cos  cos   sin  sin   1  x 2  y  x 1  y 2    y 1  x 2  x 1  y2 



10 10    tan1 x x   10 , for x  0. Since the road sign can first be seen when   2 , (b)   tan1 x   10 10 x  we have 2  tan1  2864 ft. Thus, the sign can first be x tan 2

69. (a) tan  

1

1

x

ú

¬

Ï1-x@

y

cos   y

sin   x 2 1 0 20

seen at a height of 2864 ft.

40

70. (a) Let  be the angle formed by the top of the tower, the car, and the base of the

40

building, and let  be the angle formed by base of the tower, the car, and the base   420 420 of the building, as shown in the diagram. Then tan      tan1 x x   380 380 and tan      tan1 . Thus, x x     420 380       tan1  tan1 . x x (b) We graph  x and find that  is maximized when x  400.

Ï1-y@

380

¬ 

º x

0.055 0.050 0.045 300

400

500

CHAPTER 7 TEST 1. tan  sin   cos  

sin  sin2  cos2  1 sin   cos      sec  cos  cos  cos  cos 

sin x 1  cos x tan x 1  cos x tan x 1  cos x 1  cos x 1 tan x cos x      csc x 1  sec x   2. 2 1  cos x 1  cos x 1  cos x sin x cos x 1  cos2 x sin x 2 tan x 2 tan x 2 sin x  cos2 x  2 sin x cos x  sin 2x   2 2 cos x 1  tan x sec x   x  1  cos x  sin x  1  cos x 4. sin x tan 2 sin x 3.

5. 2 sin2 3x  1  cos 2 3x  1  cos 6x

  6. cos 4x  1  2 sin2 2x  1  2 2 sin x cos x2  1  8 sin2 x 1  sin2 x  1  8 sin2 x  8 sin4 x

CHAPTER 7

Test

585



2    x 2  x  1  cos x 1  cos x 1  cos x 1  cos x 1  cos2 x  cos   2  1  sin x 7. sin   2 2 2 2 2 2 4  x  x  x   x 2 x  x  x  Another method: sin  sin2  cos  2 sin cos  cos2  1  2 sin  1  sin x 2 2 2 2 2 2 2 sin  2 sin  2 sin  x 2 sin  sin  sin    tan  (because     8.       2 2 2 cos  cos  cos  4  x2 4  4 sin  2 1  sin  4  2 sin 2  cos   0 for   2  2)

9. (a) sin 8 cos 22  cos 8 sin 22  sin 8  22   sin 30  12

      (b) sin 75  sin 45  30   sin 45 cos 30  cos 45 sin 30  22  23  22  12  14 6 2           1   3   1  cos 150 2 3 2 3 2   Another method: Since 75 is in quadrant I, sin 75     , 2 2 4 2    which is equal to 14 6 2 .      3    1  cos 1  2 3 6 2     12 2  3 (c) sin 12  2 2 4

 is in quadrant I, Another method: Since 12          sin     sin  cos   cos  sin   3 2  1 2  1 6  2 , which is equal to sin 12 3 4 3 4 3 4 2 2 2 2 4   1 2  3. 2 





52  102 5 . 10. From the figures, we have cos     cos  cos   sin  sin   2  35  1  23  2  15 5 5 3 5

11. sin 3x cos 5x  12 [sin 3x  5x  sin 3x  5x]  12 sin 8x  sin 2x     2x  5x 7x 3x 2x  5x sin  2 cos sin 12. sin 2x  sin 5x  2 cos 2 2 2 2

  1   35 1  35 1  cos   53   2.    13. sin    45 . Since  is in quadrant III, cos    35 . Then tan  4 4 2 sin  4 5 5 14. 3 sin   1  0  3 sin   1  sin   13    sin1 13  034 or     sin1 31  280 on [0 2. 15. 2 cos   1 sin   1  0  cos   12 or sin   1. The first equation has solutions    3  105 and 5     3  524 on [0 2, while the second has the solution   2  157. 16. 2 cos2   5 cos   2  0  2 cos   1 cos   2  0  cos    12 or cos   2 (which is impossible). So in the interval [0 2, the solutions are   23  209, 43  419.

3 17. sin 2  cos   0  2 sin  cos   cos   0  cos  2 sin   1  0  cos   0 or sin   12     2 , 2 or 5   5 3   6 , 6 . Therefore, the solutions in [0 2 are   6  052, 2  157, 6  262, 2  471.

18. 5 cos 2  2  cos 2  25  2  cos1 04  1159279. The solutions in [0 4 are 2  1159279, 2  1159279, 2  1159279, 4  1159279  2  1159279, 5123906, 7442465, 11407091    057964, 256195, 372123, 570355 in [0 2.   19. 2 cos2 x  cos 2x  0  2 cos2 x  2 cos2 x  1  0  cos x   12 . The solutions in [0 4 are x   3  105, x  23  209, x  43  419, and x  53  524.

586

FOCUS ON MODELING

20. 2 tan

x  2

 csc x  0  2

x  53  524.



1  cos x sin x



1  0  cos x  12 . The solutions in [0 4 are x   3  105 and sin x



9 so tan u  9 . From the triangle, cos u  40 , so using a 21. Let u  tan1 40 40 41  2  1  1519 double-angle formula for cosine, cos 2u  2 cos2 u  1  2 40 41 1681 .

41 u

22. We sketch triangles such that   cos1 x and   tan1 y. From the  y 1 triangles, we have sin   1  x 2 , sin    , and cos    , 2 1y 1  y2

1 ¬

sin     sin  cos   cos  sin 

 1 y  1  x2   x   2 1y 1  y2

40

Ï1-x@

Ï1+y@ ú

x

so the addition formula for sine gives

9

cos   x

y

1

tan   y

 1  x2  x y  1  y2

FOCUS ON MODELING Traveling and Standing Waves       1. (a) Substituting x  0, we get y 0 t  5 sin 2  0   2 t  5 sin  2 t  5 sin 2 t. (b)

y

6

t=0

0.8

1.6

4 2

x

¹ ¹/2

_2

3¹/2

_4 t=0.4

_6

1.2

     (c) We express the function in the standard form y x t  A sin k x  t: y x t  5 sin 2x   2 t  5 sin 2 x  4 t . Comparing this to the standard form, we see that the velocity of the wave is    4. 2. (a) y  02 sin 1047x  0524t  02 sin 1047x  0524t  2 02 sin 1047x cos 0524t  04 sin 1047x cos 0524t.  m  3m. So the nodes are at 3, 6, 9, 12,   . The nodes occur when 1047x  m  x  1047 (b)

y

t=0

0.4

t=1

0.2

_0.2 _0.4

t=2 1

2

3

4

t=3l t=4

5

6

x

t=5 t=6

Note that when t  3, cos 0524  3  0001, so y x 3  00004. Thus, the graph of y x 3 cannot be distinguished from the x-axis in the diagram. Yes, this is a standing wave.   068. Since   6, we have 3. From the graph, we see that the amplitude is A  27 and the period is 92, so k  292   6  410, so the equation we seek is y x t  27 sin 068x  410t. k  292

Traveling and Standing Waves

587

4. (a) We are given A  5, period 23 , and   05. Since the period is 23 , we have k  223  3. Thus, expressing the function in standard form, we have y x t  5 sin 3 x  05t  5 sin 3x  15t. (b)

y

t=0

1

2

4 2

_2

x

¹ ¹/2

3¹/2

_4 t=0.5

1.5

5. From the graphs, we see that the amplitude is A  06. The nodes occur at x  0, 1, 2, 3. Since sin x  0 when x  k (k any integer), we have   . Then since the frequency is 2, we get 20  2    40. Thus, an equation for this model is f x t  06 sin x cos 40t. 6. From the graph, we see that the amplitude is A  7. Now sin x  0 when x  k (k an integer). So for k  1, we must   1 have   2      2. Then since the period is 4, we have 2  4    2 . Thus, an equation for this model is f x t  7 sin 2x cos 12 t.

7. (a) The first standing wave has   1, the second has   2, the third has   3, and the fourth has   4.

(b)  is equal to the number of nodes minus 1. The first string has two nodes and   1; the second string has three nodes and   2, and so forth. Thus, the next two values of  would be   5 and   6, as sketched below.

(c) Since the frequency is 2, we have 440  2    880.

(d) The first standing wave has equation y  sin x cos 880t, the second has equation y  sin 2x cos 880t, the third has equation y  sin 3x cos 880t, and the fourth has equation y  sin 4x cos 880t.

8. (a) The nodes of the tube occur when cos 12 x  0 and 0  x  377. So 12 x  2k  1  2  x  2k  1 . Thus, the nodes are at x  , 3, 5, 7, 9, and 11. We stop there since 13  377. Note that the endpoints of the tube (x  0 and x  377) are not nodes.  (b) In the function y  A cos x cos t, the frequency is 2. In this case, the frequency is 50 2  25 Hz.

8

POLAR COORDINATES AND PARAMETRIC EQUATIONS

8.1

POLAR COORDINATES

1. We can describe the location of a point in the plane using different coordinate systems. The point P shown in the figure has   rectangular coordinates 1 1 and polar coordinates 2  4 .

2. (a) If a point P in the plane has polar coordinates r  then it has rectangular coordinates x y where x  r cos  and y  r sin . y (b) If P has rectangular coordinates x y then it has polar coordinates r  where r 2  x 2  y 2 and tan   . x       7 3. Yes; both 2  3 1 in Cartesian coordinates. 6 and 2 6 correspond to the point 4. No; adding a multiple of 2 to  gives the same point, as does adding an odd multiple of  and reversing the sign of r .

5.

O

8.

6.

(4, ¹4 )

O

(1, 0)

7.

(6, _ 7¹ 6 ) O

¹ 4

7¹

_ 6 O

9. 2¹ _ 3

(_2, 4¹ 3) 4¹ 3

(3, _ 2¹ 3 )

(3, ¹2 )

      5 13. 1 76 has polar coordinates 1  6 or 1  6 .

O

O

1

_ 17¹ 6

      12. 2 34 has polar coordinates 2 114 or 2 74 .

(2, 3¹ 4 ) O

1

(_1, 7¹ 6 )

(_5, _ 17¹ 6 )

O

Answers to Exercises 11–16 will vary.       5 or 3 3 11. 3  has polar coordinates 3 2 2 2

O

10.

1

      2 5 14. 2   3 has polar coordinates 2 3 or 2 3 .

(_2, _ ¹3 ) O

1

589

590

CHAPTER 8 Polar Coordinates and Parametric Equations

15. 5 0 has polar coordinates 5  or 5 2.

16. 3 1 has polar coordinates 3 1  2 or 3 1  . (3, 1)

(_5, 0)

O

1

O

1

  17. Q has coordinates 4 34 .   3   19. Q has coordinates 4   4  4 4 .     21. P has coordinates 4  234  4  4 .       4 5  4  . 23. P has coordinates 4 101 4 4 4 25.

26. 27.

28.

    18. R has coordinates 4  34  4 54 .       20. P has coordinates 4 134  4 54  4  4 .     22. Q has coordinates 4 234  4 4  4 34 .       4 7 . 24. S has coordinates 4 103 4 4  P  3 3 in rectangular coordinates, so r 2  x 2  y 2  32  32  18 and we can take r  3 2.    y 3 tan     1, so since P is in quadrant 2 we take   34 . Thus, polar coordinates for P are 3 2 34 . x 3   Q  0 3 in rectangular coordinates, so r  3 and   32 . Polar coordinates for Q are 3 32 .      Here r  5 and    23 , so x  r cos   5 cos  23   52 and y  r sin   5 sin  23   5 2 3 . R has    rectangular coordinates  52   5 2 3 .      r  2 and   56 , so S has rectangular coordinates r cos  r sin   2 cos 56  2 sin 56   3 1 .

     4  3  23 and y  r sin   4 sin   4  1  2. Thus, the rectangular . So x  r cos   4 cos 29. r   4  6 6 2 6 2    coordinates are 2 3 2 .     30. r   6 23 . So x  r cos   6 cos 23  6  12  3 and y  r sin   6 sin 23  6  23  3 3. Thus, the    rectangular coordinates are 3 3 3 .        2   2 cos   2  1  1, and 31. r   4 . So x  r cos   4 2     1    y  r sin   2 sin    1. Thus, the rectangular coordinates are 1 1.  2  4 2   32. r   1 52 . So x  r cos   1 cos 52  1  0  0 and y  r sin   1 sin 52  1  1  1. Thus, the

rectangular coordinates are 0 1.

33. r   5 5. So x  r cos   5 cos 5  5, and y  r sin   5 sin 5  0. Thus, the rectangular coordinates are 5 0. 34. r   0 13. So x y  0 0 because r  0.        35. r   6 2 116 . So x  r cos   6 2 cos 116  3 6 and y  r sin   6 2 sin 116  3 2. Thus, the     rectangular coordinates are 3 6 3 2 .          3  53 . So x  r cos   3 cos  53  23 and y  r sin   3 sin  53  32 . Thus, the 36. r     rectangular coordinates are 23  32 .

SECTION 8.1 Polar Coordinates

591

 1 y  1, so, since 37. x y  1 1. Since r 2  x 2  y 2 , we have r 2  12  12  2, so r  2. Now tan    x 1   the point is in the second quadrant,   34 . Thus, polar coordinates are 2 34 .      2 38. x y  3 3 3 . Since r 2  x 2  y 2 , we have r 2  3 3  32  36, so r  6. Now   y 1 3   , so, since the point is in the fourth quadrant,   116 . Thus, polar coordinates are 6 116 . tan      3 3 x 3   2  2    y 8 8 . Since r 2  x 2  y 2 , we have r 2  8  8  16, so r  4. Now tan    8  1, so, 39. x y  8 x   . . Thus, polar coordinates are 4 since the point is in the first quadrant,    4 4     2   2    2 2 2 2 40. x y   6  2 . Since r  x  y , we have r   6   2  8, so r  2 2. Now      2 y tan      1 , so, since the point is in the third quadrant,   76 . Thus, polar coordinates are 2 2 76 . 3 x  6 y 41. x y  3 4. Since r 2  x 2  y 2 , we have r 2  32  42  25, so r  5. Now tan    43 , so, since the point is in x   the first quadrant,   tan1 34 . Thus, polar coordinates are 5 tan1 43 .

 2 y  2, and 5. Now, tan    x 1 since the point is in the fourth quadrant,   2  tan1 2 (since we need 0    2). Thus, polar coordinates are   5 2  tan1 2 .   y 43. x y  6 0. r 2  62  36, so r  6. Now tan    0, so since the point is on the negative x-axis,   . x Thus, polar coordinates are 6 .       44. x y  0  3 . r  3 and since the point is on the negative y-axis,   32 . Thus, polar coordinates are 3 32 . 42. x y  1 2. Since r 2  x 2  y 2 , we have r 2  12  22  2, so r 

45. x  y  r cos   r sin   tan   1, and so    4.

  46. x 2  y 2  9. By substitution, r cos 2  r sin 2  9  r 2 cos2   sin2   9  r 2  9  r  3.

47. y  x 2 . We substitute and then solve for r: r sin   r cos 2  r 2 cos2   sin   r cos2   sin   tan  sec . r cos2  5  5 csc . 48. y  5. By substitution, r sin   5  r  sin  4  4 sec . 49. x  4. We substitute and then solve for r: r cos   4  r  cos    50. x 2  y 2  1. By substitution, r cos 2  r sin 2  1  r 2 cos2   sin2   1  r 2 cos 2  1  r2 

1  sec 2. cos 2

51. r  7. But r 2  x 2  y 2 , so x 2  y 2  r 2  49. Hence, the equivalent equation in rectangular coordinates is x 2  y 2  49. 52. r  3  x 2  y 2  r 2  9, so an equivalent equation in rectangular coordinates is x 2  y 2  9.

53.     2  cos   0, so an equivalent equation in rectangular coordinates is x  0. y 54.     tan   0   0  y  0. x 55. r cos   6. But x  r cos , and so x  6 is an equivalent rectangular equation. 2 56. r  2 csc   r   r sin   2. But r sin   y, so y  2 is an equivalent rectangular equation. sin 

592

CHAPTER 8 Polar Coordinates and Parametric Equations

57. r  4 sin   r 2  4r sin . Thus, x 2  y 2  4y is an equivalent rectangular equation. Completing the square, it can be written as x 2  y  22  4.

  58. r  6 cos   r 2  6r cos . By substitution, x 2  y 2  6x  x 2  6x  9  y 2  9  x  32  y 2  9.

59. r  1  cos . If we multiply both sides of this equation by r we get r 2  r  r cos . Thus r 2  r cos   r, and squaring  2  2 both sides gives r 2  r cos   r 2 , or x 2  y 2  x  x 2  y 2 in rectangular coordinates.

60. r  3 1  sin   3  3 sin   r 2  3r  3r sin   r 2  3r sin   3r. Squaring both sides gives   2  2  r 2  3r sin   9r 2 , or x 2  y 2  3y  9 x 2  y 2 in rectangular coordinates.

61. r  1  2 sin . If we multiply both sides of this equation by r we get r 2  r  2r sin . Thus r 2  2r sin   r, and  2  2 squaring both sides gives r 2  2r sin   r 2 , or x 2  y 2  2y  x 2  y 2 in rectangular coordinates. 62. r  2  cos   r 2  2r  r cos   r 2  r cos   2r  2   x 2  y2  x  4 x 2  y2 1 sin   cos  y  x  1.

63. r 



 2  2  r 2  r cos   2r2  r 2  r cos   4r 2 

r sin   cos   1  r sin   r cos   1, and since r cos   x and r sin   y, we get

1  r 1  sin   1  r  r sin   1. Thus r  1  r sin , and squaring both sides gives 1  sin  r 2  1  r sin 2  x 2  y 2  1  y2  1  2y  y 2  x 2  2y  1  0.

64. r 

4  r 1  2 sin   4  r  2r sin   4. Thus r  4  2r sin . Squaring both sides, we get 1  2 sin  r 2  4  2r sin 2 . Substituting, x 2  y 2  4  2y2  x 2  y 2  16  16y  4y 2  x 2  3y 2  16y  16  0.

65. r 

2  r 1  cos   2  r r cos   2  r  2 r cos . Squaring both sides, we get r 2  2  r cos 2 . 1  cos  Substituting, x 2  y 2  2  x2  4  4x  x 2  y 2  4x  4. y y 67. r 2  tan . Substituting r 2  x 2  y 2 and tan   , we get x 2  y 2  . x x 2  68. r 2  sin 2  2 sin  cos   r 4  2r 2 sin  cos   2 r cos  r sin . By substitution, x 2  y 2  2x y  66. r 

x 4  2x 2 y 2  y 4  2x y  0.

   y 2 2 2 2 69. sec   2  cos   12      3  tan    3  x   3  y   3x  y  3x  y  3x  0. y 70. cos 2  1 means that 2  0    0  tan   0   0  y  0 x 71. (a) In rectangular coordinates, the points r1  1  and r2  2  are x1  y1   r1 cos  1  r1 sin 1  and x2  y2   r2 cos 2  r2 sin 2 . Then, the distance between the points is   D  x1  x2 2  y1  y2 2  r1 cos 1  r2 cos 2 2  r1 sin 1  r2 sin 2 2       r12 cos2  1  sin2 1  r22 cos2 2  sin2 2  2r1r2 cos 1 cos 2  sin 1 sin 2  

 r12  r22  2r1r2 cos 2  1 

SECTION 8.2 Graphs of Polar Equations

593

    (b) The distance between the points 3 34 and 1 76 is D

      340 32  12  2 3 1 cos 76  34  9  1  6 cos 512

72. (a) Because streets are laid out in a grid, rectangular coordinates are more appropriate when giving directions to a taxi driver. (b) Descartes’ famous declaration cogito ergo sum does not apply to pigeons, so polar coordinates are probably more useful in this case.

8.2

GRAPHS OF POLAR EQUATIONS

1. To plot points in polar coordinates we use a grid consisting of circles centered at the pole and rays emanating from the pole. 2. (a) To graph a polar equation r  f  we plot all the points r  that satisfy the equation.

(b) The graph of the polar equation r  3 is a circle with radius 3 centered at the pole. The graph of the polar equation   4 is a line passing through the pole with slope 1.

3

O

3. VI

4. III

5. II

O

6. IV

7. I

8. V

9. Polar axis: 2  sin   2  sin   r, so the graph is not symmetric about the polar axis. Pole: 2  sin     2  sin  cos   cos  sin   2   sin   2  sin   r, so the graph is not symmetric about the pole.  Line    2 : 2  sin     2  sin  cos   cos  sin   2  sin   r, so the graph is symmetric about   2 . 10. Polar axis: 4  8 cos   4  8 cos   r, so the graph is symmetric about the polar axis. Pole: 4  8 cos     4  8 cos  cos   sin  sin   4  8  cos   r, so the graph is not symmetric about the pole. Line    2 : 4  8 cos     4  8 cos  cos   sin  sin   4  8  cos   r, so the graph is not symmetric about    2.

11. Polar axis: 3 sec   3 sec   r, so the graph is symmetric about the polar axis. 3 1 3 Pole: 3 sec        3 sec   r , so the graph is not symmetric cos    cos  cos   sin  sin   cos  about the pole. 1 3 3 Line    2 : 3 sec     cos     cos  cos   sin  sin    cos   3 sec   r, so the graph is not symmetric about    2.

594

CHAPTER 8 Polar Coordinates and Parametric Equations

12. Polar axis: 5 cos  csc   5 cos   csc   r, so the graph is not symmetric about the polar axis. 1 Pole: 5 cos    csc     5 cos    sin    1 1  5 cos  cos   sin  sin   5  cos   5 cos  csc   r , sin  cos   cos  sin   sin  so the graph is symmetric about the pole. 1 Line    2 : 5 cos    csc     5 cos    sin    1 1  5 cos  cos   sin  sin   5  cos   5 cos  csc   r , sin  cos   cos  sin  sin   so the graph is not symmetric about   2 . 4 4   r, so the graph is not symmetric about the polar axis. 3  2 sin  3  2 sin  4 4 4 4 Pole:     r, so the graph is not 3  2 sin    3  2 sin  cos   cos  sin  3  2  sin  3  2 sin  symmetric about the pole. 4 4 4 Line    2 : 3  2 sin     3  2 sin  cos   cos  sin   3  2 sin   r, so the graph is symmetric about   2.

13. Polar axis:

5 5   r , so the graph is symmetric about the polar axis. 1  3 cos  1  3 cos  5 5 5 5 Pole:     r, so the graph is not 1  3 cos    1  3 cos  cos   sin  sin  1  3  cos  1  3 cos  symmetric about the pole. 5 5 5 Line    2 : 1  3 cos     1  3 cos  cos   sin  sin   1  3 cos   r, so the graph is not symmetric about    2.

14. Polar axis:

15. Polar axis: 4 cos 2   4 cos 2  r 2 , so the graph is symmetric about the polar axis. Pole: r2  r 2 , so the graph is symmetric about the pole.

 2 Line    2 : 4 cos 2     4 cos 2  2  4 cos 2  4 cos 2  r , so the graph is symmetric about   2 .

16. Polar axis: 9 sin   9 sin   r 2 , so the graph is not symmetric about the polar axis. Pole: r2  r 2 , so the graph is symmetric about the pole.

 2 Line    2 : 9 sin     9 sin  cos   cos  sin   9 sin   r , so the graph is symmetric about   2 .

17. r  2  r 2  4  x 2  y 2  4 is an equation of a circle with radius 2 centered at the origin.

O

1

18. r  1  r 2  1  x 2  y 2  1 is an equation of a circle with radius 1 centered at the origin.

O

1

SECTION 8.2 Graphs of Polar Equations

19.     2  cos   0  x  0 is an equation of a vertical line.

O

1

O

21. r  6 sin   r 2  6r sin   x 2  y 2  6y 

x 2  y  32  9, a circle of radius 3 centered at 0 3.

O

1

22. r  cos   r 2  r cos   x 2  y 2  x   2 x  12  y 2  14 , a circle of radius 12 centered at   10 . 2

O

1

23. r  2 cos . Circle.

1

24. r  3 sin . Circle. (3, ¹2 ) O

1

O

25. r  2  2 cos . Cardioid.

595

   y 20.   56  tan    33    33  y   33 x, a x  3 line with slope  3 passing through the origin.

1

26. r  1  sin . Cardioid.

1

1

596

CHAPTER 8 Polar Coordinates and Parametric Equations

27. r  3 1  sin . Cardioid.

28. r  cos   1. Cardioid. 1 1

29. r  sin 2

30. r  2 cos 3

O

O

1

31. r   cos 5

1

32. r  sin 4 1

1

_1

_1

1

O

1

_1

_1

33. r  2 sin 5

O

34. r  3 cos 4

(5, ¹2 )

(_3, 0)

O

(_3, 3¹ 2 )

(_3, ¹)

O

(_3, ¹2 )

35. r 

 3  2 sin 

_2

_1 O

1

2

36. r  2  sin 

_1 _2 _3 _4

O

1

SECTION 8.2 Graphs of Polar Equations

37. r 

 3  cos 

38. r  1  2 cos 

2 1 _1

O

1

2

1

O

3

_1 _2

 39. r  2  2 2 cos 

40. r  3  6 sin 

(9, ¹2 )

(2, ¹2 ) (2-2Ï2, 0) (2+2Ï2, ¹)

(_3, 3¹ 2 )

O

(2, 3¹ 2 )

41. r 2  cos 2

(3, ¹)

(3, 0)

O

42. r 2  4 sin 2

O

O

1

1

44. r  1,   0

43. r  ,   0

O

10 0.2

45. r  2  sec 

46. r  sin  tan  2 O

1

O

1

597

598

CHAPTER 8 Polar Coordinates and Parametric Equations

47. r  cos

   ,   [0 4] 2

48. r  sin



 8 ,   [0 10] 5

1

1

-1

1

-1

-1

49. r  1  2 sin

1 -1

   ,   [0 4] 2

50. r 

 1  08 sin2 ,   [0 2] 1

2

-1 -2

1 -1

51. r  1  sin n. There are n loops. 2

-2

2

2

-2

-2

n1 52. r  1  c sin 2 2

-2

2 -2

2

2

-2

2

2

-2

2

2

-2

2

-2

-2

-2

-2

n2

n3

n4

n5

2

2

2

2

-2

2 -2

-2

2 -2

-2

2 -2

-2

2 -2

c  03 c  06 c1 c  15 c2 3  As c increases, the graph becomes “pinched” along the line   4 , eventually growing more “leaves” with that line as their axis.     1   1 , and so on. is IV, since the graph must contain the points 0 0     53. The graph of r  sin 2 2 2 1 54. The graph of r   is I, since as  increase, r decreases (  0). So this is a spiral into the origin.  5 7  5 7 55. The graph of r   sin  is III, since for    2  2  2     the values of r are also 2  2  2    . Thus the graph must cross the vertical axis at an infinite number of points. 56. The graph of r  1  3 cos 3 is II, since when   0 r  1  3  4 and when    r  1  3  2, so there should be two intercepts on the positive-axis.

SECTION 8.2 Graphs of Polar Equations

57.



x 2  y2

3

 3  4x 2 y 2  r 2  4 r cos 2 r sin 2 

r 6  4r 4 cos2  sin2   r 2  4 cos2  sin2   r  2 cos  sin   sin 2. The equation is r  sin 2, a rose.

58.



x 2  y2

3

O

1

2 2   3   x 2  y 2  r 2  r cos 2  r sin 2 

 2   2 r 6  r 2 cos2   r 2 sin2   r 6  r 2 cos2   sin2  

 2 r 6  r 4 cos2   sin2   r 2  cos 22  r   cos 2. The equations

1

r  cos 2 and r   cos 2 have the same graph, a rose.

59.

 2  x 2  y 2  r 2  r cos 2  r sin 2    r 4  r 2 cos2   r 2 sin2   r 4  r 2 cos2   sin2   

x 2  y2

2

r 2  cos2   sin2   cos 2. The graph is r 2  cos 2, a leminiscate.

O

1

 2  2 60. x 2  y 2  x 2  y 2  x  r 2  r 2  r cos   r 2  [r r  cos ]2  r 2  r 2 r  cos 2  1  r  cos 2  1  r  cos   r  cos   1.

Both r  cos   1 and r  cos   1 give the same graph. To see this, replace 

1

with   : cos     1   cos   1   cos   1. This is a cardioid.

61. (a) r  a cos   b sin   r 2  ar cos   br sin  

(b) r  2 sin   2 cos  has center 1 1   and radius 12 22  22  2.

x 2  y 2  ax  by  x 2  ax  y 2  by  0 

x 2  ax  14 a 2  y 2  by  14 b2  14 a 2  14 b2     2  2 x  12 a  y  12 b  14 a 2  b2 . Thus, in rectangular    coordinates the center is 12 a 12 b and the radius is 12 a 2  b2 .

(b) r  tan  sec  

62. (a)

y  x 2.

5

-2

2

O

sin  cos 



1 cos 





yr xr2

1

ry y  2  2 1 x x

599

600

CHAPTER 8 Polar Coordinates and Parametric Equations

64. (a)

63. (a) 5000

-5000

5000

5000

-5000

-5000

5000

-5000

At   0, the satellite is at the “rightmost” point in

The satellite starts out at the same point, 5625 0,

its orbit, 5625 0. As  increases, it travels

but its orbit decays. It narrowly misses the earth at its

counterclockwise. Note that it is moving fastest

first perigee (closest approach to the earth), and

when   .

crashes during its second orbit.

(b) The satellite is closest to earth when   . Its height above the earth’s surface at this point is

(b) The satellite crashes when   837 rad  480 .

22,500 4  cos 3960  45003960  540 mi.     65. The graphs of r  1  sin    6 and r  1  sin   3 have

2

 , respectively. Similarly, the graph of r  f    is the 3

1

the same shape as r  1  sin , rotated through angles of  6 and graph of r  f  rotated by the angle .

-2

2 -1

66. The circle r  2 (in polar coordinates) has rectangular coordinate equation x 2  y 2  4. The polar coordinate

form is simpler. The graph of the equation r  sin 2 is a four-leafed rose. Multiplying both sides by r 2 , we get 32  r 3  r 2 2 cos  sin   2 r cos  r sin  which is x 2  y 2  2x y in rectangular form. The polar form is definitely simpler.

67. y  2  r sin   2  r  2 csc . The rectangular coordinate system gives the simpler equation here. It is easier to study lines in rectangular coordinates.

8.3

POLAR FORM OF COMPLEX NUMBERS; DE MOIVRE’S THEOREM

1. A complex number z  a  bi has two parts: a is the real part and b is the imaginary part. To graph a  bi we graph the ordered pair a b in the complex plane.  2. (a) The modulus of z is r  a 2  b2 and an argument of z is an angle  satisfying tan   ba.

(b) We can express z in polar form as z  r cos   i sin  where r is the modulus of z and  is the argument of z.    3. (a) The complex number z  1  i in polar form is z  2 cos 34  i sin 34 .    (b) The complex number z  2 cos  6  i sin 6 in rectangular form is z  3  i.    (c) The complex number z can be expressed in rectangular form as 1  i or in polar form as 2 cos  4  i sin 4 .

SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem

4. A nonzero complex number has n different nth roots. The number 16 has four

Im

fourth roots. These roots are 2, 2, 2i, and 2i. In the complex plane, these

2i

roots all lie on a circle of radius 2.

_2

2

0 _2i

5. 4i 

 02  42  4

6. 3i 

 09 3

Im

Im 4i

1 0

1 1

0

Re

1

Re

_3i

7. 2 

 402

8. 6  6 Im

Im

1 _2

9. 5  2i 

0

1 1

0

Re

  52  22  29

10. 7  3i 

Im

6

1

Re

  49  9  58 Im

5+2i 1 0

1 1

Re

0

1

Re 7-3i

     11.  3  i   3  1  2

        12. 1  33 i   1  13  43  2 3 3

Im

Im

0

1

Ï3+i

1 1

0

Re _1- Ï3 i 3

1

Re

Re

601

602

CHAPTER 8 Polar Coordinates and Parametric Equations

   3  4i   9 16  13.  25  25  1 5 

     2  i 2     14.    12  12  1   2

Im

Im

1

3+4i 5

0

1

Im 2z

0

0

Re

15. z  1  i, 2z  2  2i, z  1  i, 12 z  12  12 i

1

1

_Ï2+iÏ2 2

16. z  1 

1

Re

   3i, 2z  2  2 3i, z  1  3i,

 1z  1  3i 2 2 2

Im

z

2z

1z 2

1

z

Re

_z

1z 1 2

0

1

Re

_z

17. z  8  2i, z  8  2i

18. z  5  6i, z  5  6i

Im

z

z

1 0

Im

1

1 0

Re

1

Re

z

z

19. z 1  2  i, z 2  2  i, z 1  z 2  2  i  2  i  4, z 1 z 2  2  i 2  i  4  i 2  5

20. z 1  1  i, z 2  2  3i,

z 1  z 2  1  i  2  3i  1  2i,

z 1 z 2  1  i 2  3i  2  3i  2i  3i 2  1  5i

Im

Im 1 0

zÁzª

zª 1

zÁzª zÁ+zª

zÁ

Re

zÁ 1 0

1

zÁ+zª zª

Re

SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem

22. z  a  bi  a  1 b  1

21. z  a  bi  a  0 b  0

Im

Im

1

1 0

23. z  z  3

1

0

Re

24. z  z  1

Im

25. z  z  2

1

0

Re

26. z  2  z  5

Im

1

Re

Im

1

0

1

0

Re

28. z  a  bi  a  b

Im

1

Re

1

Re

Im

1

1 0

Re

1

1

27. z  a  bi  a  b  2

1

Im

1 0

603

1

0

Re

29. 1 i. Then tan   11  1 with  in quadrant I     4 , and r 

     12  12  2. Hence, 1  i  2 cos  4  i sin 4 .

7 30. 1  i. Then tan   1 1  1 with  in quadrant IV    4 , and r     1  i  2 cos 74  i sin 74 .

2  1 with  in quadrant II    3 , and r  31. 2  2i. Then tan   2 4    2  2i  2 2 cos 34  i sin 34 .

  12  12  2. Hence,

  22  22  2 2. Hence,

604

CHAPTER 8 Polar Coordinates and Parametric Equations

   32.  2  2i. Then tan   2  1 with  in quadrant III    54 , and r   2     5   2  2i  2 cos 4  i sin 54 .

    3 with  in quadrant III    7 , and r  33.  3  i. Then tan   1 3 6  3    7  7   3  i  2 cos 6  i sin 6 .

  2   2  2   2  2. Hence,   2  3  12  2. Hence,

   34. 5  5 3i. Then tan   553   3 with  in quadrant II    23 , and r     5  5 3i  10 cos 23  i sin 23 .



  2 52  5 3  10. Hence,

    2    3 with  in quadrant IV    11 , and r  35. 2 3  2i. Then tan   2 2 3  22  4. Hence, 3 6 2 3    11  2 3  2i  4 cos 6  i sin 116 .     2   36. 3  3 3i. Then tan   3 3 3  3 with  in quadrant I     , and r  32  3 3  6. Hence, 3    3  3 3i  6 cos  3  i sin 3 .    37. 2i. Then tan  is undefined,    2 , and r  2. Hence, 2i  2 cos 2  i sin 2 .   38. 5i. Then tan  is undefined,   32 , and r  5. Hence, 5i  5 cos 32  i sin 32 .

39. 3. Then tan   0,   , and r  3. Hence, 3  3 cos   i sin .     40. 2. Then tan   0,   0, and r  2. Hence, 2  2 cos 0  i sin 0.

    41.  6  2i. Then tan   2   33 with  in quadrant II    56 , and r   6       6  2i  2 2 cos 56  i sin 56 .

  2  2   6  2  2 2. Hence,

    42.  5  15i. Then tan    15  3 with  in quadrant III    43 , and r   5      Hence,  5  15i  2 5 cos 43  i sin 43 .

43. 4  3i. Then tan     4  3i  5 cos tan1

  2   2   5   15  2 5.

3 with  in quadrant I    tan1 3  06435, and r  4 4    3  i sin tan1 3 . 4 4

44. 3  2i. Then tan   23 with  in quadrant I    tan1 23  05880, and r        3  2i  13 cos tan1 23  i sin tan1 23 .

 42  32  5. Hence,

  32  22  13. Hence,

     2     3 with  in quadrant IV    11 , and r  4 3  42  8. 3  i  4 3  4i. Then tan   4 3 6 4 3     11  3  i  8 cos 6  i sin 116 . Hence, 4         46. i 2  6i  6  2i. Then tan   2  33 with  in quadrant I    tan1 33   6 , and 6        2  2    6  2  2 2. Hence, i 2  6i  2 2 cos  r 6  i sin 6 .   3  1 with  in quadrant II    3 , and r  32  32  3 2. Hence, 47. 3 1  i  3  3i. Then tan   3 4    3 1  i  3 2 cos 34  i sin 34 .

45. 4

SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem

605

  2  1 with  in quadrant II    3 , and r  22  22  2 2. Hence, 48. 2i 1  i  2  2i. Then tan   2 4    2i 1  i  2 2 cos 34  i sin 34 .

            6 cos   i sin   49. z 1  3 cos  3  i sin 3 , z 2  2 cos 6  i sin 6 , z 1 z 2  3  2 cos 3  6  i sin 3  6 2 2      i sin       3 cos   i sin   z 1 z 2  32 cos   3 6 3 6 2 6 6    3 cos 54  i sin 54 , z 2  2 cos   i sin ,          z1 z2  3  2 cos 54    i sin 54    2 3 cos  4  i sin 4 ,          z 1 z 2  23 cos 54    i sin 54    23 cos  4  i sin 4

50. z 1 

      2 cos 53  i sin 53 , z 2  2 2 cos 32  i sin 32 ,           4 cos 76  i sin 76 , 2  2 2 cos 53  32  i sin 53  32 z1 z2          z 1 z 2  2 cos 53  32  i sin 53  32  12 cos  6  i sin 6

51. z 1



2 2

     , z z  cos 3    i sin 3    cos 13  i sin 13 ,  i sin 52. z 1  cos 34  i sin 34 , z 2  cos  1 2 3 3 4 3 4 3 12 12     3   3   5  5  z 1 z 2  cos 4  3  i sin 4  3  cos 12  i sin 12 53. z 1  4 cos 120  i sin 120 , z 2  2 cos 30  i sin 30 ,   z 1 z 2  4  2 cos 120  30   i sin 120  30   8 cos 150  i sin 150 ,   z 1 z 2  42 cos 120  30   i sin 120  30   2 cos 90  i sin 90 

  54. z 1  2 cos 75  i sin 75 , z 2  3 2 cos 60  i sin 60 ,     z 1 z 2  2  3 2  cos 75  60   i sin 75  60   6 cos 135  i sin 135 ,    z 1 z 2  2 cos 75  60   i sin 75  60   13 cos 15  i sin 15  3 2

55. z 1  4 cos 200  i sin 200 , z 2  25 cos 150  i sin 150 ,   z 1 z 2  4  25 cos 200  150   i sin 200  150   100 cos 350  i sin 350 , 4 cos 200  150   i sin 200  150   4 cos 50  i sin 50  z 1 z 2  25 25

56. z 1  45 cos 25  i sin 25 , z 2  15 cos 155  i sin 155 ,   4 cos 180  i sin 180 , z 1 z 2  45  15  cos 25  155   i sin 25  155   25             z 1 z 2  45 15 cos 25  155   i sin 25  155   4 cos 130   i sin 130   4 cos 130  i sin 130    3  i, so tan  1  1 with 1 in quadrant I  1   6 , and r1  3  1  2. 3    z 2  1  3i, so tan 2  3 with 2 in quadrant I  2   3 , and r1  1  3  2.      Hence, z 1  2 cos  6  i sin 6 and z 2  2 cos 3  i sin 3 .        4 cos   i sin  , Thus, z 1 z 2  2  2 cos  6  3  i sin 6  3 2 2      i sin       cos     i sin   , and 1z  1 cos     i sin   . z 1 z 2  22 cos   1 6 3 6 3 6 6 2 6 6

57. z 1 

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CHAPTER 8 Polar Coordinates and Parametric Equations

   2  2i, so tan 1  1 with  1 in quadrant IV   1  74 , and r1  2  2  2.   z 2  1  i, so tan 2  1 with 2 in quadrant IV  2  74 , and r2  1  1  2.      Hence, z 1  2 cos 74  i sin 74 and z 2  2 cos 74  i sin 74 .             Thus, z 1 z 2  2  2 cos 74  74  i sin 74  74  2 2 cos 72  i sin 72  2 2 cos 32  i sin 32 ,       z 1 z 2  2 cos 74  74  i sin 74  74  2 cos 0  i sin 0, and 2        1z 1  12 cos  74  i sin  74  12 cos  4  i sin 4 .

58. z 1 

     1 with  1 in quadrant IV   1  11 , and r1  12  4  4. 59. z 1  2 3  2i, so tan 1  2 6 2 3 3   z 2  1  i, so tan 2  1 with  2 in quadrant II  2  34 , and r2  1  1  2.      Hence, z 1  4 cos 116  i sin 116 and z 2  2 cos 34  i sin 34 .            i sin 7 , Thus, z 1 z 2  4  2 cos 116  34  i sin 116  34  4 2 cos 712 12       4    13 z 1 z 2   cos 116  34  i sin 116  34  2 2 cos 13 12  i sin 12 , and 2       1   14 cos  1z 1  4 cos  116  i sin  116 6  i sin 6 .   60. z 1   2i, so  1  32 , and r1  2.    z 2  3  3 3i, so tan  2  3 with 2 in quadrant IV  2  43 , and r2  9  27  6.      Hence, z 1  2 cos 32  i sin 32 and z 2  6 cos 43  i sin 43 .          Thus, z 1 z 2   6 2 cos 56  i sin 56 , 2  6 cos 32  43  i sin 32  43          z 1 z 2  62 cos 32  43  i sin 32  43  62 cos  6  i sin 6 , and         1z 1  1 cos  32  i sin  32  22 cos  2  i sin 2 . 2

  61. z 1  5  5i, so tan 1  55  1with 1 in quadrant I  1   4 , and r1  25  25  5 2. z 2  4, so  2  0, and r2  4.    Hence, z 1  5 2 cos  4  i sin 4 and z 2  4 cos 0  i sin 0.            i sin  , z z  5 2 cos   i sin  , and  0  i sin  0  20 2 cos Thus, z 1 z 2  5 2  4 cos  1 2 4 4 4 4 4 4 4  1 cos     i sin     2 cos     i sin   . 1z 1   4 4 10 4 4 5 2

      3 with  1 in quadrant III   1  11 , and r1  48  16  8. 62. z 1  4 3  4i, so tan 1  4 3 6 4 3

, and r2  8. z 2  8i, so  2     2   Hence, z 1  8 cos 116  i sin 116 and z 2  8 cos  2  i sin 2 .       11     64 cos  Thus, z 1 z 2  8  8 cos 116   2  i sin 6 2 3  i sin 3 ,       11    z 1 z 2  88 cos 116    cos 43  i sin 43 , and 1z 1  18 cos  2  i sin 6 2 6  i sin 6 .

SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem

607

63. z 1  20, so 1  , and r1  20.   z 2  3  i, so tan  2  1 with 2 in quadrant I  2   6 , and r2  3  1  2. 3   Hence, z 1  20 cos   i sin  and z 2  2 cos  6  i sin 6 .          40 cos 7  i sin 7 , Thus, z 1 z 2  20  2 cos    6  i sin   6 6 6          10 cos 5  i sin 5 , and z 1 z 2  20 2 cos   6  i sin   6 6 6 1 [cos   i sin ]  1 cos   i sin . 1z 1  20 20

 64. z 1  3  4i, so tan 1  43 with 1 in quadrant I  1  tan1 43 , and r1  9  16  5.   z 2  2  2i, so tan 2  1 with 2 in quadrant IV  2  74 , and r2  4  4  2 2.         Hence, z 1  5 cos tan1 43  i sin tan1 43  5 cos 0927  i sin 0927 and z 2  2 2 cos 74  i sin 74 .        Thus, z 1 z 2  5  2 2 cos tan1 43  74  i sin tan1 34  74  10 2 cos 0142  i sin 0142,            5 cos tan1 34  74  i sin tan1 34  74  5 4 2 cos tan1 43  74  i sin tan1 43  74 z 1 z 2   22

 5 4 2 [cos 457  i sin 457], and      1z 1  15 cos  tan1 43  i sin  tan1 43  15 [cos 0927  i sin 0927].

      6  65.  3  i  2 cos 56  i sin 56 , so  3  i  26 cos 306  i sin 306  64 cos 5  i sin 5  64. 66. 1  i 

     10    cos 704  i sin 704  32 cos 32  i sin 32  32i. 2 cos 74  i sin 74 , so 1  i10  2

          5  67.  2  2i  2 cos 54  i sin 54 , so  2  2i  25 cos 254  i sin 254  16 2  16 2i.

  7     7 68. 1  i  2 cos  cos 74  i sin 74  8  8i. 2 4  i sin 4 , so 1  i     1  1  1 and tan   1     . Thus 2  2 i  cos   i sin  . Therefore, 69. r  2 2 4 2 2 4 4  12      2 2   cos 12  2  2 i 4  i sin 12 4  cos 3  i sin 3  1.     1 with  in quadrant IV    11 . Thus 3  i  2 cos 11  i sin 11 , so 70. r  3  1  2 and tan    6 6 6 3       10  10         3   i sin 110  1 cos    i sin    1 1 1 cos 110 3i  12 6 6 1024 3 3 1024 2  2 i  2048 1  3i .      71. r  4  4  4 2 and tan   1 with  in quadrant IV    74 . Thus 2  2i  2 2 cos 74  i sin 74 , so   8 2  2i8  2 2 cos 14  i sin 14  4096 1  0i  4096.    72. r  14  34  1 and tan   3 with  in quadrant III     43 . Thus  12  23 i  cos 43  i sin 43 , so  15   12  23 i  cos 603  i sin 603  cos 20  i sin 20  1.      73. r  1  1  2 and tan   1 with  in quadrant III    54 . Thus 1  i  2 cos 54  i sin 54 , so     7      cos 354  i sin 354  8 2 cos 34  i sin 34  8 2 1  i 1  8 1  i. 2 1  i7  2

2

        74. r  9  3  2 3 and tan   33 with  in quadrant I     6 . Thus 3  3i  2 3 cos 6  i sin 6 , so         4   3  3i  144 cos 23  i sin 23  144  12  23 i  72 1  3i .

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CHAPTER 8 Polar Coordinates and Parametric Equations

  2  1     . Thus 2 3  2i  4 cos   i sin  , so 75. r  12  4  4 and tan    6 6 6 2 3 3        5  5    1 1 2 3  2i cos 56  i sin 56  1024  23  12 i  2048  3i  14

     1  1  2 and tan   1 with  in quadrant IV    74 . Thus 1  i  2 cos 74  i sin 74 , so   1 cos 56  i sin 56  1 cos 14  i sin 14  1 . 1  i8  16 4 4 16 16

76. r 

 4  1     . Thus 77. r  48  16  8 and tan    6 4 3 3     4 3  4i  8 cos  6  i sin 6 . So,      12    6  2k 6  2k  i sin for k  0, 1.  8 cos 4 3  4i 2 2     i sin   and Thus the two roots are 0  2 2 cos 12 12    13  13  1  2 2 cos 12  i sin 12 .

Im

1 0 wÁ

 4  1     . Thus 78. r  48  16  8 and tan    6 4 3 3     4 3  4i  8 cos  6  i sin 6 . So      13   6  2k 6  2k  i sin for k  0, 1, 2.  2 cos 4 3  4i 3 3    i sin  , Thus the three roots are 0  2 cos 18 18     13  13    i sin 25 . 1  2 cos 18  i sin 8 , and 2  2 cos 25 18 18   79. 81i  81 cos 32  i sin 32 . Thus,      32  2k 32  2k  i sin for k  0, 1, 81i14  8114 cos 4 4   2, 3. The four roots are 0  3 cos 38  i sin 38 ,     1  3 cos 78  i sin 78 , 2  3 cos 118  i sin 118 , and   3  3 cos 158  i sin 158 .   2k 2k cos  i sin for k  0, 1, 2, 3, 5 5   4. Thus the five roots are 0  2 cos 0  i sin 0, 1  2 cos 25  i sin 25 ,     2  2 cos 45  i sin 45 , 3  2 cos 65  i sin 65 , and   4  2 cos 85  i sin 85 .

wü 1

Re

Im

wÁ 1 0

wü 1

Re

wª

wü

Im

wÁ

1 0

1

Re

w£ wª

80. 32  32 cos 0  i sin 0. Thus, 3215

Im

wÁ wª

1 0

w£

wü 1

w¢

Re

SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem

2k 2k  i sin , for k  0, 1, 2, 3, 4, 5, 6, 8 8 7. So the eight roots are 0  cos 0  i sin 0  1,

81. 1  cos 0  i sin 0. Thus, 118  cos 1  cos 3  cos 5  cos 7  cos

    i sin   2  i 2 ,   cos 2 4 4 2 2   3  i sin 3   2  i 2 ,   4 4 4 2 2   5  i sin 5   2  i 2 ,   6 4 4 2 2   7  i sin 7  2  i 2 . 4 4 2 2

Im

wª

w£

  i sin   i, 2 2

cos 32  i sin 32  i, and

wÁ wü 1

Re

w¦

w°

   2 cos  4  i sin 4 . So    13   4  2k  4  2k cos  i sin for k  0, 1, 2 1  i13  3 3    i sin  , 2. Thus the three roots are 0  216 cos 12 12       i sin 9 , and   216 cos 17  i sin 17 . 1  216 cos 912 2 12 12 12

1

0

w¢

cos   i sin   1,

609

w§

82. 1  i 

Im

wÁ

1 wü 0

1

Re

wª

 13  cos 83. i  cos  2  i sin 2 , so i



2  2k 3



 i sin



2  2k 3 



for

Im

3  1 k  0, 1, 2. Thus the three roots are 0  cos  6  i sin 6  2  2 i,

1

wÁ



wü

1  cos 56  i sin 56   23  12 i, and 2  cos 32  i sin 32  i.

0

1

Re

wª

 15  cos 84. i  cos  2  i sin 2 , so i



2  2k 5



 i sin



2  2k 5

  i sin  , k  0, 1, 2, 3, 4. Thus the five roots are 0  cos 10 10



for

Im

1

 9 9 13 13 1  cos  2  i sin 2 , 2  cos 10  i sin 10 , 3  cos 10  i sin 10 , and

wü

wª 0

 17 4  cos 17 10  i sin 10 .



  2k 4





  2k  i sin 4 



2 2  k  0, 1, 2, 3. So the four roots are 0  cos  4  i sin 4  2  i 2 ,



for

    1  cos 34  i sin 34   22  i 22 , 2  cos 54  i sin 54   22  i 22 ,   and 3  cos 74  i sin 74  22  i 22 .

1

Re

1

Re

w¢

w£

85. 1  cos   i sin . Then 114  cos

wÁ

Im

wÁ

1

0 wª

wü

w£

610

CHAPTER 8 Polar Coordinates and Parametric Equations    162 1  3  32 and tan   1616 3  3 with  in quadrant III       43 . Thus 16  16 3i  32 cos 43  i sin 43 . So        15 43  2k 43  2k 16  16 3i  i sin for  3215 cos 5 5     i sin 4 , k  0, 1, 2, 3, 4. The five roots are 0  2 cos 415 15       i sin 16 , 1  2 cos 23  i sin 23 , 2  2 cos 16 15 15     22  22  28   3  2 cos 15  i sin 15 , and 4  2 cos 15  i sin 28 15 .

86. r 







Im

wÁ

wü 1

wª

0

1

Re

w¢ w£



87. z 4  1  0  z  114  z  22  22 i, z   22  22 i (from Exercise 85)      , then, z  i 18  cos 2  2k  i sin 2  2k for  i sin 88. z 8  i  0  x  i 18 . Since i  cos  2 2 8 8

  i sin  , cos 5  i sin 5 , cos 9  i sin 9 , k  0, 1, 2, 3, 4, 5, 6, 7. Thus there are eight solutions: z  cos 16 16 16 16 16 16   i sin 13 , cos 17  i sin 17 , cos 21  i sin 21 , cos 25  i sin 25 , and cos 29  i sin 29 . cos 13 16 16 16 16 16 16 16 16 16 16   13      89. z 3  4 3  4i  0  z  4 3  4i . Since 4 3  4i  8 cos  6  i sin 6 ,      13   6  2k 6  2k  i sin , for k  0, 1, 2. Thus the three roots are  813 cos 4 3  4i 3 3         i sin  , z  2 cos 13  i sin 13 , and z  2 cos 25  i sin 25 . z  2 cos 18 18 18 8 18 18

90. z 6  1  0  z  116 . Since 1  cos 0  i sin 0 z  116  cos 2k6  i sin 2k6 for k  0, 1, 2, 3, 4, 5. Thus there are 



six solutions: z  1, 12  23 i,  12  23 i.

   2 cos 54  i sin 54 , 91. z 3  1  i  z  1  i13 . Since 1  i       54  2k 54  2k  i sin for k  0, 1, 2. Thus the three solutions to this z  1  i13  216 cos 3 3         i sin 5 , 216 cos 13  i sin 13 , and 216 cos 21  i sin 21 . equation are z  216 cos 512 12 12 12 12 12

2k 2k 92. z 3  1  0  z  113 . Since 1  cos 0  i sin 0 z  113  cos  i sin for k  0, 1, 2. Thus the three 3 3 



solutions to this equation are z  cos 0  i sin 0, cos 23  i sin 23 , and cos 43  i sin 43 or z  1,  12  23 i,  12  23 i.     i  i2  4 1 1 i  5 1 5 2   i 93. z  i z  1  0  z  2 1 2 2   i  9 i  i 2  4 1 2   2i or i 94. z 2  i z  2  0  z  2 2    2i  2i2  4 1 2 2i  4 95. z 2  2i z  2  0  z   i 1 2 2 96. z 2  1  i z  i  z  i z  1, so z  1 or i. 97. 1 zk z1

1 cos 0  i sin 0, so by De Moivre’s Theorem, its n roots are      2k 2k 0  2k 0  2k  i sin  cos  i sin for k  0 1 2    n  1. So z 0  1,  11n cos n n n n 2 4 2 4  i sin  , z 2  cos  i sin  2 , and so on.  cos n n n n 

SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem

611

 n 98. From Exercise 97, k  1 for 0  k  n  1. Multiplying both sides by s n and noting that s n  z, we have  n n  s n k  s n  sk  z for 0  k  n  1, showing that 1 s s2      sn1 are nth roots of z. 



99. The cube roots of 1 are 0  1, 1  cos 23  i sin 23   12  23 i, and 2  cos 43  i sin 43  12  23 i, so their       sum is 0  1  2  1   12  23 i  12  23 i  0.

The fourth roots of 1 are 0  1, 1  i, 2  1, and 3  i, so their sum is 0  1  2  3  1  i  1  i  0.

The fifth roots of 1 are 0  1, 1  cos 25  i sin 25 , 2  cos 45  i sin 45 , 3  cos 65  i sin 65 , and 4  cos 85  i sin 85 , so their sum is         1  cos 25  i sin 25  cos 45  i sin 45  cos 65  i sin 65  cos 85  i sin 85       5 5 1 1  1  2 cos 25  2 cos 65 (most terms cancel)  1  2 0 4  4   4  4 



3 3  1 2 2 1 2 The sixth roots of 1 are 0  1, 1  cos  3  i sin 3  2  2 i,   cos 3  i sin 3   2  2 i,

  3 3 5 5 1 5 2 i, and   cos 3  i sin 3  2  2 i, so their sum is      3 3 1 2 i  2  2 i  0.     2 2 2 2  3 3 2 3 The eight roots of 1 are 0  1, 1  cos  4  i sin 4  2  2 i,   i,   cos 4  i sin 4   2  2 i,     4  1, 5  cos 54  i sin 54   22  22 i, 6  i, and 7  cos 74  i sin 74  22  22 i, so their sum is               1  22  22 i  i   22  22  1   22  22 i  i  22  22 i  0.

3  1, 4  cos 43  i sin 43   12     1     1  12  23 i    23 i  1   12  2

It seems that the sum of any set of nth roots is 0.    To prove this, factor n  1  1  1 1  1  2  3      n1 . Since this is 0 and 1  0, we must have

1  1  2  3      n1  0.

100. The cube roots of 1 are 0  1, 1  cos 23  i sin 23 , and 2  cos 43  i sin 43 , so their product is    0  1  2  1 cos 23  i sin 23 cos 43  i sin 43  cos 2  i sin 2  1. The fourth roots of 1 are 0  1, 1  i, 2  1, and 3  i, so their product is 0  1  2  3  1  i  1  i  i 2  1.

The fifth roots of 1 are 0  1, 1  cos 25  i sin 25 , 2  cos 45  i sin 45 , 3  cos 65  i sin 65 , and 4  cos 85  i sin 85 , so their product is      1 cos 25  i sin 25 cos 45  i sin 45 cos 65  i sin 65 cos 85  i sin 85  cos 4  i sin 4  1. 

3  2 2 1 2 3 The sixth roots of 1 are 0  1, 1  cos  3  i sin 3 ,   cos 3  i sin 3   2  2 i,   1, 



4  cos 43  i sin 43   12  23 i, and 5  cos 53  i sin 53  12  23 i, so their product is         cos 2  i sin 2 1 cos 4  i sin 4 5  i sin 5  cos 5  i sin 5  1. cos 1 cos   i sin 3 3 3 3 3 3 3 3

 3 3 2 3 The eight roots of 1 are 0  1, 1  cos  4  i sin 4 ,   i,   cos 4  i sin 4 , 4  1, 5  cos 54  i sin 54 , 6  i, 7  cos 74  i sin 74 , so their product is          i cos 3  i sin 3 1 cos 5  i sin 5 i cos 7  i sin 7  i2   i sin 1 cos  4 4 4 4 4 4 4 4

cos 2  i sin 2  1.

612

CHAPTER 8 Polar Coordinates and Parametric Equations

The product of the nth roots of 1 is 1 if n is even and 1 if n is odd. m m  1 . The proof requires the fact that the sum of the first m integers is 2 2 2 2k 2k Let   cos  i sin . Then k  cos  i sin for k  0 1 2     n  1. The argument of the n n n n product of the n roots of unity can be found by adding the arguments of each k . So the argument of the product is 2 n  2  2 n  1  2 2 1  2 2  2 3  [0  1  2  3      n  2  n  1].        0 n n n n n n 2 n  1 n   n  1 . Thus the product of the n roots of Since this is the sum of the first n  1 integers, this sum is n 2 unity is cos n  1   i sin n  1   1 if n is even and 1 if n is odd. 101.

z1 r cos 1  i sin 1  cos 2  i sin 2 r cos 1 cos 2  i 2 sin 1 cos 2  i sin 1 cos 2  i sin 2 cos 1  1  1  z2 r2 cos 2  i sin 2  cos 2  i sin 2 r2 cos2 2  i 2 sin2  2  r1  cos 1  2   i sin  1  2   r2

8.4

PLANE CURVES AND PARAMETRIC EQUATIONS

1. (a) The parametric equations x  f t and y  g t give the coordinates of a point x y   f t  g t for appropriate values of t. The variable t is called a parameter.     (b) When t  0 the object is at 0 02  0 0 and when t  1 the object is at 1 12  1 1. (c) If we eliminate the parameter in part (b) we get the equation y  x 2 . We see from this equation that the path of the moving object is a parabola.

2. (a) It is true that the same curve can be described by parametric equations in many different ways. For instance, the  parabola y  x 2 can be represented by x  t, y  t 2 or by x  t, y  t 2 or by x   t, y  t. (b) The parametric equations x  2t, y  2t2 model the position of a

y

moving object at time t. When t  0 the object is at 0 0, and when t  1 the object is at 2 4.

t=1 [Ex. 2(b)]

(c) If we eliminate the parameter we get the equation y  x 2 , which is the same equation as in Exercise 1(b). So the objects in Exercises 1(b) and 2(b) move along the same path, but traverse the path differently.

1

t=1 [Ex. 1(b)]

0

1

2

t=0 [1(b) and 2(b)]

3. (a) x  2t, y  t  6

y

y

1 1 2

x  2y  12  0.

x

4. (a) x  6t  4, y  3t, t  0

1

(b) Since x  2t, t 

3

x

x

x x and so y   6  2 2

y (b) Since y  3t, t  and so 3 y  4  2y  4  x  2y  4  0, y  0. x 6 3

SECTION 8.4 Plane Curves and Parametric Equations

613

 2 6. (a) x  2t  1, y  t  12

5. (a) x  t 2 , y  t  2, 2  t  4 y

y

1

1 x

4

t  y  2, we have x  t 2

(b) Since y  t  2  

x  y  22 , and since 2  t  4, we have

4  x  16.

7. (a) x 

1

2 2   (b) Since y  t  12 , 4y  4 t  12  2t  12 ,

and since x  2t  1, we have x 2  2t  12  4y 

 t, y  1  t  t  0 y

1

x

y  14 x 2 .

8. (a) x  t 2 , y  t 4  1 y

x

20 10

_10

x

1

(b) Since x 

 t, we have x 2  t, and so y  1  x 2

with x  0.

9. (a) x 

1 , y t 1 t

x  0.

10. (a) x  t  1, y  y

1 1

(b) Since x 

(b) Since x  t 2 , we have x 2  t 4 and so y  x 2  1,

x

1 1 1 we have t  and so y   1. t x x

t t 1

y

1 1

x

(b) Since x  t  1, we have t  x  1, so y 

x 1 . x

614

CHAPTER 8 Polar Coordinates and Parametric Equations

11. (a) x  4t 2 , y  8t 3

12. (a) x  t, y  1  t

y

y

1 x

1

1 x

1

(b) Since y  8t 3



have y 2  x 3 .

 3 y 2  64t 6  4t 2  x 3 , we

13. (a) x  2 sin t, y  2 cos t, 0  t  

(b) Since x  t, we have y  1  x, where x  0.

14. (a) x  2 cos t, y  3 sin t, 0  t  2

y

y

1

1 x

1

(b) x 2  2 sin t2  4 sin2 t and y 2  4 cos2 t. Hence, x 2  y 2  4 sin2 t  4 cos2 t  4  x 2  y 2  4,

where x  0.

15. (a) x  sin2 t, y  sin4 t

(b) We have cos t  x2 and sin t  y3, so

x22  y32  cos2 t  sin2 t  1  1 x 2  1 y 2  1. 4 9

16. (a) x  sin2 t, y  cos t

y

x

1

y 1

1

1

1

x

(b) Since x  sin2 t we have x 2  sin4 t and so y  x 2 . But since 0  sin2 t  1 we only get the part of this parabola for which 0  x  1.

x

(b) Since y  cos t, we have y 2  cos2 t and so

x  y 2  sin2 t  cos2 t  1 or x  1  y 2 (actually  y   1  x), where 0  x  1.

SECTION 8.4 Plane Curves and Parametric Equations

17. (a) x  cos t, y  cos 2t

18. (a) x  cos 2t, y  sin 2t

y 1

y 1

1

x

(b) Since x  cos t we have x 2  cos2 t, so

2x 2  1  2 cos2 t  1  cos 2t  y. Hence, the

rectangular equation is y  2x 2  1, 1  x  1.

19. (a) x  sec t, y  tan t, 0  t   2  x  1 and y  0.

1

x

(b) x 2  cos2 2t and y 2  sin2 2t. Then

x 2  y 2  cos2 2t  sin2 2t  1  x 2  y 2  1.

20. (a) x  cot t, y  csc t, 0  t   so y  1. y

y

1 1

1

x

x

1

(b) x 2  sec2 t, y 2  tan2 t, and

y 2  1  tan2 t  1  sec2 t  x 2 . Therefore, y 2  1  x 2  x 2  y 2  1, x  1, y  0.

21. (a) x  tan t, y  cot t, 0  t   2  x  0.

(b) x 2  cot2 t, y 2  csc2 t, and so

x 2  1  cot2 t  1  csc2 t  y 2 . Therefore,

y 2  x 2  1, with y  1. This is the top half of the

hyperbola y 2  x 2  1.

22. (a) x  et , y  et , so y  0.

y

y

1

1 1

x

(b) x y  tan t  cot t  1, so y  1x for x  0.

1

(b) x y  et et  1, so y  1x, x  0.

x

615

616

CHAPTER 8 Polar Coordinates and Parametric Equations

24. (a) x  sec t, y  tan2 t, 0  t   2 , so y  0 and x  1.

23. (a) x  e2t , y  et , so y  0. y

y

2 1 _1 0

1

2

3

4

5

7 x

6

2

 2 (b) x  e2t  et  y 2 , y  0.

x

1

(b) x 2  sec2 t  1  tan2 t, so x 2  1  y, x  1, y  0.

25. (a) x  cos2 t, y  sin2 t, so 0  x  1 and 0  y  1. y

26. (a) x  cos3 t, y  sin3 t, 0  t  2 y

1

1

1

x

(b) x  y  cos2 t  sin2 t  1. Hence, the equation is x  y  1 for 0  x  1 and 0  y  1.

1

x

(b) x 23  cos2 t and y 23  sin2 t, and so x 23  y 23  1.

27. x  3 cos t, y  3 sin t. The radius of the circle is 3, the position at time 0 is x 0  y 0  3 cos 0 3 sin 0  3 0 and the orientation is counterclockwise (because x is decreasing and y is increasing initially). x y  3 0 again when t  2, so it takes 2 units of time to complete one revolution. 28. x  2 sin t, y  2 cos t. The radius is 2, the position at time 0 is 0 2, the orientation is clockwise (because x is increasing and y is decreasing initially), and it takes 2 units of time to complete one revolution. 29. x  sin 2t, y  cos 2t. The radius of the circle is 1, the position at time 0 is x 0  y 0  sin 0 cos 0  0 1 and the orientation is clockwise (because x is increasing and y is decreasing initially). x y  0 1 again when t  , so it takes  units of time to complete one revolution. 30. x  4 cos 3t, y  4 sin 3t. The radius is 4, the position at time 0 is 4 0, the orientation is counterclockwise, and it takes 2 units of time to complete one revolution. 3

31. Since the line passes through the point 4 1 and has slope 12 , parametric equations for the line are x  4  t, y  1  12 t.

32. Since the line passes through the point 10 20 and has slope 2, parametric equations for the line are x  10  t, y  20  2t. 87  1. Thus, parametric equations for the line are 33. Since the line passes through the points 6 7 and 7 8, its slope is 76 x  6  t, y  7  t. 70 7 34. Since the line passes through the points 0 0 and 12 7, its slope is  . Thus, parametric equations for the line 12  0 12 7 t. are x  t, y  12 35. Since cos2 t  sin2 t  1, we have a 2 cos2 t  a 2 sin2 t  a 2 . If we let x  a cos t and y  a sin t, then x 2  y 2  a 2 . Hence, parametric equations for the circle are x  a cos t, y  a sin t.

SECTION 8.4 Plane Curves and Parametric Equations

617

a 2 cos2 t b2 sin2 t x2 y2   1. If we let x  a cos t and y  b sin t, then   1. a2 b2 a2 b2 Hence, parametric equations for the ellipse are x  a cos t, y  b sin t.

36. Since cos2 t  sin2 t  1, we have

37. x   0 cos  t, y   0 sin  t  16t 2 . From the equation for x, t 

x . Substituting into the equation for y gives  0 cos 

 2 x 16x 2 x . Thus the equation is of the form y  c1 x  c2 x 2 ,  16  x tan   2  0 cos   0 cos   0 cos2  where c1 and c2 are constants, so its graph is a parabola. y   0 sin 

38.  0  2048 fts and   30 . (a) The bullet will hit the ground when y  0. Since y   0 sin  t  16t 2 , we have 0  2048  sin 30  t  16t 2  0  1024t  16t 2

 16t t  64  0  t  0 or t  64. Hence, the bullet will hit the ground after 64 seconds.  (b) After 64 seconds, the bullet will hit the ground at x  2048  cos 30   64  65,536 3  113,5117 ft  215 miles. (c) The maximum height attained by the bullet is the maximum value of y. Since     y  1024t  16t 2  16 t 2  64t  16 t 2  64t  1024  16,384  16 t  322  16,384, y is maximized when t  32, and therefore the maximum height is 16,384 ft  31 miles. 40. x  2 sin t, y  cos 4t

39. x  sin t, y  2 cos 3t 2

1 -1

-2

1

2 -1

-2

41. x  3 sin 5t, y  5 cos 3t

42. x  sin 4t, y  cos 3t 5

1

-1

1 -1

-5

43. x  sin cos t, y  cos t 32 , 0  t  2

44. x  2 cos t  cos 2t, y  2 sin t  sin 2t 2

1

-1

1 -1

-2

2 -2

618

CHAPTER 8 Polar Coordinates and Parametric Equations

45. (a) r  212 , 0    4  x  2t12 cos t, y  2t12 sin t

46. (a) r  sin   2 cos   x  sin t  2 cos t cos t, y  sin t  2 cos t sin t

(b)

(b)

2 2

-2

2 -2

47. (a) r 

1

4 cos t 4 sin t 4 x  ,y 2  cos  2  cos t 2  cos t

(b)

2

48. (a) r  2sin   x  2sin t cos t, y  2sin t sin t (b)

2

2

-2

2

4

-2

-2

2

 2   49. x  t 3  2t, y  t 2  t is Graph III, since y  t 2  t  t 2  t  14  14  t  12  14 , and so y   14 on this curve, while x is unbounded.

50. x  sin 3t, y  sin 4t is Graph IV, since when t  0 and t   the curve passes through 0 0. Thus this curve must pass through 0 0 twice. 51. x  t  sin 2t, y  t  sin 3t is Graph II, since the values of x and y oscillate about their values on the line x  t, y  t  y  x. 52. x  sin t  sin t, y  cos t  cos t is Graph I, since this curve does not pass through the point 0 0. 53. (a) It is apparent that x  O Q and

(b) The curve is graphed with a  3 and b  2.

y  Q P  ST . From the diagram,

x  O Q  a cos  and y  ST   b sin .

2

Thus, parametric equations are x  a cos  and y  b sin .

-4 y

a

b O

-2

2

4

-2 S ¬ T

(c) To eliminate  we rearrange: sin   yb

P Q

x

sin2   yb2 and cos   xa





cos2   xa2 . Adding the two equations:

sin2   cos2   1  x 2 a 2  y 2 b2 . As

indicated in part (b), the curve is an ellipse.

SECTION 8.4 Plane Curves and Parametric Equations

619

54. (a) A has coordinates a cos  a sin . Since O A is perpendicular to AB, 2

O AB is a right triangle and B has coordinates a sec  0. It follows that P has coordinates a sec  b sin . Thus, the parametric equations

0

are x  a sec , y  b sin .

5

(b) The right half of the curve is graphed with a  3 and b  2.

55. (a) If we modify Figure 8 so that PC  b, then by the

10

-2

(b) 5

same reasoning as in Example 6, we see that x  OT   P Q  a  b sin  and y  T C  C Q  a  b cos .

We graph the case where a  3 and b  2.

-20

20

56. From the figure, we see that x  OT   P Q  a  b sin  and y  T C  C Q  a  b cos . When a  1 and b  2, the parametric equations become x    2 sin , y  1  2 cos . The curve is graphed for 0    4. y

(b)

C(a¬,a)

b

P(x, y) O

¬

a¬

-10

Q

T

10

x

x2 y y2  sec2   2 . Since . Also, y  b sec   sec   2 b a b 2 2 2 2 x y y x tan2   sec2   1, we have 2  2  1  2  2  1, which is the equation of a hyperbola. a b b a

57. x  a tan   tan  

x a

tan2  



58. Substituting the given values for x and y into the equation we derived in Exercise 57, we get   2   2 a t b t 1   t  1  t  1. Thus the points on this curve satisfy the equation, which is that of a hyperbola. 2 b a2 y2 x2 However, this hyperbola is only the part of 2  2  1 for which x  0 and y  0. b a 59. x  t cos t, y  t sin t, t  0 t

x

y

t

0

0

0

5 4 3 2 7 4

 4

 2 3 4



  2 8

  2 8

0



  38 2

2 3 2 8



0

2

x

y

   58 2  58 2 0  32   7 2  7 2 8 8

2

0

60. x  sin t, y  sin 2t

y

y

1

1

1 1

x

x

620

CHAPTER 8 Polar Coordinates and Parametric Equations

61. x 

3t 3t 2 , y  , t  1 1  t3 1  t3 t

09

x

y

996 897

075 389 292 05

171 086 0

t 2

y

x

y

067 133

t 11

x

y

t

997 1097

4

x

y

25 045 113

125 393

492

45 015 067

3

032 096

15

189

284

5

012 060

4

018 074

2

086

171

6

008 050

0

0

05

133 067

5

012 060

25

051

128

7

006 043

1

15

6

008 050

3

035

104

8

005 038

15

103 154

35

025

087

15

1

019 076

1

x

As t  1 we have x   and y  . As t  1 we have x   and y  . As t   we have x  0 and y  0 . As t   we have x  0 and y  0 . 62. x  cot t, y  2 sin2 t, 0  t  

y

2

1

1

x

63. (a) We first note that the center of circle C (the small circle) has coordinates

y

[a  b] cos  [a  b] sin . Now the arc P Q has the same length as the arc

a ab a , and so         . Thus the b b b x-coordinate of P is the x-coordinate of the center of circle C plus   ab b cos     b cos  , and the y-coordinate of P is the b   ab y-coordinate of the center of circle C minus b  sin     b sin  . b   ab So x  a  b cos   b cos  and b   ab y  a  b sin   b sin  . b P  Q, so b  a   

(b) If a  4b, b 

a , and x  34 a cos   14 a cos 3, y  34 a sin   14 a sin 3. 4

¬ ú P»

¬

Q

(a, 0) P x

y

a

From Example 2 in Section 7.3, cos 3  4 cos3   3 cos . Similarly, one can prove that sin 3  3 sin   4 sin3 . Substituting, we get   x  34 a cos   14 a 4 cos3   3 cos   a cos3    y  34 a sin   14 a 3 sin   4 sin3   a sin3 . Thus,

x 23  y 23  a 23 cos2   a 23 sin2   a 23 , so x 23  y 23  a 23 .

a x

SECTION 8.4 Plane Curves and Parametric Equations

621

64. The coordinates of C are [a  b] cos  [a  b] sin . Let  be  OC P as shown in the figure. Then the arcs traced a out by rolling the circle along the outside are equal, that is, b  a    . P is displaced from C by amounts b equal to the legs of the right triangle C PT , where C P is the hypotenuse. Since OC Q is a right triangle, it follows that    a    ab  OC Q    . Thus,               . So 2 2 b 2 b 2          ab ab   a  b cos   b sin   x  a  b cos   b sin b 2 2 b   ab   a  b cos   b cos b and          ab ab    a  b sin   b cos  y  a  b sin   b cos b 2 2 b   ab   a  b sin   b sin b Therefore, parametric equations for the epicycloid are     ab ab x  a  b cos   b cos  and y  a  b sin   b sin . b b

y

C a ¬

ú

Œ

T

P Q

x

65. A polar equation for the circle is r  2a sin . Thus the coordinates of Q are x  r cos   2a sin  cos  and

y  r sin   2a sin2 . The coordinates of R are x  2a cot  and y  2a. Since P is the midpoint of Q R, we use the   midpoint formula to get x  a sin  cos   cot  and y  a 1  sin2  .

66. (a) C  2a cot  2a, so the x-coordinate of P is x  2a cot . Let B  0 2a. Then  O AB is a right angle and  O B A  , so   O A  2a sin  and A  2a sin  cos  2a sin2  . Thus, the

5

y-coordinate of P is y  2a sin2 .

(b) The curve is graphed with a  1.

-10

10

ay  cos   67. We use the equation for y from Example 6 and solve for . Thus for 0    , y  a 1  cos   a        ay ay ay   cos1 . Substituting into the equation for x, we get x  a cos1  sin cos1 . a a a             2ay  y 2 2ay  y 2 ay ay ay 2 1 1 However, sin cos  1 . Thus, x  a cos  ,  a a a a a      y 2ay  y 2  x 2ay  y 2  x ay and we have  cos1  1  cos  a a a a    2ay  y 2  x . y  a 1  cos a

622

CHAPTER 8 Polar Coordinates and Parametric Equations

68. (a) 2

5

2

-2

-2

2

2

-5

-5

5

-2

-2

5 5 -5

-5

R  05 R1 R3 (b) The graph with R  5 seems to most closely resemble the profile of the engine housing.

R5

69. (a) In the figure, since O Q and QT are perpendicular and OT and T D are perpendicular, the angles formed by their intersections are equal, that is,    DT Q. Now the coordinates of T are cos  sin . Since T D is the length of the string that has been unwound from the circle, it must also have arc length , so T D  . Thus the x-displacement from T to D is   sin  while the y-displacement from T to D is   cos . So the coordinates of D are x  cos    sin  and y  sin    cos . (b) y 10

T 1 ¬

D Q

1

-20

20

x

-10

70. (a) It takes 2 units of time. The parametric equations of the particle that moves twice as fast around the circle are x  sin 2t, y  cos 2t.

(b) From the first table, we see that the particle travels counterclockwise. If we want the particle to travel in a clockwise direction, then we want the second table to apply. Possible parametric equations for clockwise traversal of the unit circle are x   sin t, y  cos t. y

t

x  sin t

0

0



1

2

0

0





0

1

1

0

2

0

1

 71. C: x  t, y  t 2 ; D: x  t, y  t, t  0 (a) For C, x  t, y  t 2  y  x 2 .  For D, x  t, y  t  y  x 2 . For F, x  et equation.

 

t

1

3 2

For E, x  sin t

¹

1 t= 2

y  cos t

t=0

t=¹

1

3¹ t= 2

E: x  sin t, y  1  cos2 t

x 2  sin2 t  1  cos2 t  y and so y  x 2 .

x

2



x

y

0

1

1

0

0

3 2

1

1

0

2

0

1

F: x  et , y  e2t

x 2  e2t  y and so y  x 2 . Therefore, the points on all four curves satisfy the same rectangular

CHAPTER 8

Review

623

(b) Curve C is the entire parabola y  x 2 . Curve D is the right half of the parabola because t  0 and so x  0. Curve E is the portion of the parabola for 1  x  1. Curve F is the portion of the parabola where x  0, since et  0 for all t. y

y

y

y

1

1

1

1

0

1

x

C

0

1

0

x

D

1

0

x

E

1

x

F

CHAPTER 8 REVIEW (12, ¹6)

1. (a)

O

(8, _ 3¹ 4)

¹/6



 3 (b) x  12 cos  6  12  2  6 3, y  12 sin   12  1  6. Thus, the rectangular 6

2

   coordinates of P are 6 3 6 .

3. (a)

(

7¹ _3, 4

7¹ 4

)

O

2. (a)

      (b) x  8 cos  34  8  22  4 2,       y  8 sin  34  8  22  4 2. Thus, the     rectangular coordinates of P are 4 2 4 2 . 2¹ 3

4. (a) O

O

 

 (b) x  3 cos 74  3 22   3 2 2 ,     y  3 sin 74  3  22  3 2 2 . Thus, the     rectangular coordinates of P are  3 2 2  3 2 2 .

5. (a)

(_Ï3, 2¹ 3 )      (b) x   3 cos 23   3  12  23 ,     y   3 sin 23   3 23   32 . Thus, the   rectangular coordinates of P are 23   32 . 6. (a)

(4Ï3, _ 5¹ 3) _ 5¹ 3

O

       (b) x  4 3 cos  53  4 3 12  2 3,       y  4 3 sin  53  4 3 23  6. Thus, the    rectangular coordinates of P are 2 3 6 .

_ 3¹ 4

(_6Ï2, _ ¹4 ) O

_¹ 4

  2      6, (b) x  6 2 cos   4  6 2 2        2  6. Thus, y  6 2 sin   4  6 2  2 the rectangular coordinates of P are 6 6.

624

CHAPTER 8 Polar Coordinates and Parametric Equations y

7. (a)

y

8. (a) P

P

1

2 x

2

(b) r 

   82  82  128  8 2 and   tan1 88 .

Since P is in quadrant I,    4 . Polar coordinates    for P are 8 2  4 .    (c) 8 2 54

y

9. (a)

1 x

  2  2    2  (b) r  6  8  2 2 and   tan1

    6  tan1  3    . Since x is 3  2

negative and y is positive,   23 . Polar    coordinates for P are 2 2 23 .    (c) 2 2 53 y

10. (a)

2 2

x

P

P

1 x

1

(b) r 

  2   2  6 2  6 2  144  12 and

(b) r 



  tan1 62   4 . Since P is in quadrant III, 6 2     54 . Polar coordinates for P are 12 54 .   (c) 12  4

y

11. (a)



3  tan1 3   . Since P is in   tan1  3 6 3 3

quadrant I,    6 . Polar coordinates for P are   6 6 .   (c) 6 76 12. (a)

P

1 x

32  

x

1 1



y 1

P

(b) r 

  2  3 3  32  36  6 and

 2   3  12  2 3 and

  tan1 33 . Since P is in quadrant II,   56 .    Polar coordinates for P are 2 3 56 .    (c) 2 3   6

(b) r 

   42  42  32  4 2 and

7   tan1 4 4 . Since P is in quadrant IV,   4 .    Polar coordinates for P are 4 2 74 .    (c) 4 2 34

CHAPTER 8

13. (a) x  y  4  r cos   r sin   4 

4 r cos   sin   4  r  cos   sin 

14. (a) x y  1  r cos  r sin   1  r 2 cos  sin   1  r 2 

(b) The rectangular equation is easier to graph. y

1

1

15. (a) x 2  y 2  4x  4y  r 2  4r cos   4r sin   r 2  r 4 cos   4 sin   r  4 cos   4 sin 

x

1

x

1

2   2 16. (a) x 2  y 2  2x y  r 2  2 r cos  r sin   r 4  2r 2 cos  sin   r 2  2 cos  sin  

r 2  sin 2

(b) The polar equation is easier to graph.

(b) The polar equation is easier to graph.

O

1

O

17. (a)

1

18. (a)

2 O

(b) r  3  3 cos   r 2  3r  3r cos , which gives  x 2  y 2  3 x 2  y 2  3x   x 2  3x  y 2  3 x 2  y 2 . Squaring both sides 2    gives x 2  3x  y 2  9 x 2  y 2 .

625

2 1  or cos  sin  sin 2

r 2  2 csc 2.

(b) The rectangular equation is easier to graph. y

Review

1

(b) r  3 sin   r 2  3r sin , so x 2  y 2  3y   2   x 2  y 2  3y  94  94  x 2  y  32  94 .

626

CHAPTER 8 Polar Coordinates and Parametric Equations

19. (a)

20. (a)

O

O

2

2

(b) r  4 cos 3  4 cos 2 cos   sin 2 sin       4 cos  cos2   sin2   2 sin2  cos     r  4 cos3   3 sin2  cos     r 4  4r 3 cos3   3 sin2  cos  , which gives

(b) r  2 sin 2  r  2  2 sin  cos  

r 3  4r 2 sin  cos    32 r2  4 r sin  r cos  and so, since x  r cos  and y  r sin , we get 3  x 2  y 2  16x 2 y 2 .



x 2  y2

2

 4x 3  12x y 2 .

22. (a)

21. (a)

O

1

O

1 1   cos 2 cos2   sin2    r 2 cos2   sin2   1 

(b) r 2  sec 2 

1

(b) r 2  4 sin 2  8 sin  cos   2  r 4  8r 2 sin  cos , so x 2  y 2  8x y.

r 2 cos2   r 2 sin2   1 

r cos 2  r sin 2  1  x 2  y 2  1. (b) r  sin   cos   r 2  r sin   r cos , so x 2  y 2  y  x    2  2   x 2  x  14  y 2  y  14  12  x  12  y  12 

23. (a)

O

 1. 2

1

(b) r 

24. (a)

O

1

 4  2r  r cos   4, so 2 x 2  y 2  x  4 2  cos 

 2 x 2  y 2  4  x. Squaring both sides gives   4 x 2  y 2  4  x2  16  8x  x 2 



3x 2  8x  4y 2  16.

CHAPTER 8

25. r  cos 3,   [0 3].

26. r  sin 94,   [0 8]

1

1

-1

1

-1

-1

27. r  1  4 cos 3,   [0 6].

28. r   sin ,   [3 3]. This graph is not bounded.

5

20

5

-20

20

-20

-5

30. (a)

Im

Im 2 2

4+4i 1 1

(b) 4  4i has r 

  16  16  4 2, and

(b) 10i has r  10 and   32 .   (c) 10i  10 cos 32  i sin 32

32. (a)

Im

Re

_10i

Re

  tan1 44   4 (in quadrant I).    (c) 4  4i  4 2 cos  4  i sin 4

31. (a)

Im 1+Ï3 i 1

5+3i 1 1

(b) 5  3i. Then r 

627

1

-1

29. (a)

Review

1

Re

  25  9  34, and

  tan1 53 .       (c) 5  3i  34 cos tan1 53  i sin tan1 53

Re

  3i has r  1  3  2, and    tan1 3   3 (since  is in quadrant I).    (c) 1  3i  2 cos  3  i sin 3

(b) 1 

628

CHAPTER 8 Polar Coordinates and Parametric Equations

33. (a)

34. (a)

Im

_1+i

5

1 _20

_1

(b) 1  i has r 

Im

5

Re

Re

  1 with 1  1  2 and tan   1

 in quadrant II    34 .    (c) 1  i  2 cos 34  i sin 34

(b) 20 has r  20 and   . (c) 20  20 cos   i sin 

   1  3  2 and tan   1 3   3 with  in    quadrant III    53 . Therefore, 1  3i  2 cos 53  i sin 53 , and so             4 1  3i  24 cos 203  i sin 203  16 cos 23  i sin 23  16  12  i 23  8 1  i 3 .

35. 1 

 3i has r



     11  2 and    2 cos  36. 1  i has r  4 . Thus, 1  i  4  i sin 4 , and so  8 2 cos 2  i sin 2  16 1  i  0  16. 1  i8 

37.

      3  i has r  3  1  2 and tan   1 with  in quadrant I     6 . Therefore, 3  i  2 cos 6  i sin 6 , 3 and so         4 1 cos 2  i sin 2  1  1  i 3 3i  24 cos 46  i sin 46  16 3 3 16 2 2       1 1  i 3   1 1  i 3  32 32

    32 3 1   38. 12  23 i has r  14  34  1 and   tan1 12  3 (since  is in quadrant I). So 2  2 i  cos 3  i sin 3 . Thus  20        1  3i 20 cos 20  i sin 20  cos 2  i sin 2   1  3 i  1 1  i 3  1 2 2 3 3 3 3 2 2 2

  39. 16i has r  16 and   32 . Thus, 16i  16 cos 32  i sin 32 and so      3  4k 3  4k 12 12 cos  i sin for k  0, 1. Thus  16 16i 4 4       4  1  i 1  2 2 1  i and the roots are 0  4 cos 34  i sin 34 2 2      1  4 cos 74  i sin 74  4 1  i 1  2 2 1  i. 2

2

       40. 4  4 3i has r  16  48  8 and   tan1 4 4 3   3 . Therefore, 4  4 3i  8 cos 3  i sin 3 .        13    6k   6k Thus 4  4 3i  i sin for k  0, 1, 2. Thus the three roots are  3 8 cos 9 9       ,   2 cos 7  i sin 7 , and   2 cos 13  i sin 13 .  i sin 0  2 cos  1 2 9 9 9 9 9 9

  2k 2k 41. 1  cos 0  i sin 0. Then 116  1 cos  i sin for k  0, 1, 2, 3, 4, 5. Thus the six roots are 6 6      3  1 2 2   1  i 3 , 0  1 cos 0  i sin 0  1, 1  1 cos  3  i sin 3  2  i 2 , 2  1 cos 3  i sin 3 2 2       3  1 cos   i sin   1, 4  1 cos 43  i sin 43   12  i 23 , and 5  1 cos 53  i sin 53  12  i 23 .

CHAPTER 8  18  cos 42. i  cos  2  i sin 2 . Then i



2  2k 8



 i sin



2  2k 8



 cos



  4k 16



 i sin



Review

  4k 16

629



for

  i sin  ,   cos 5  i sin 5 ,   cos 9  i sin 9 , k  0, 1, 2, 3, 4, 5, 6, 7. Thus the eight roots are 0  cos 16 1 2 16 16 16 16 16

 13 17 17 21 21 25 25 3  cos 13 16  i sin 16 , 4  cos 16  i sin 16 , 5  cos 16  i sin 6 , 6  cos 16  i sin 16 , and  29 7  cos 29 16  i sin 16 . y

43. (a)

y

44. (a)

1 1 x

1 x

1

(b) x  1  t 2 , y  1  t  t  y  1. Substituting for t gives x  1  y  12  x  2y  y 2 in

(b) x  t 2  1, y  t 2  1

y

t 2  y  1.

Substituting for t 2 gives x  y  1  1 y  x  2 where x  1 and y  1.

rectangular coordinates. 45. (a)





y

46. (a)

1 1

x

1

1

(b) x  1  cos t  cos t  x  1, and y  1  sin t  sin t  1  y. Since cos2 t  sin2 t  1, it follows that x  12  1  y2  1 

x  12  y  12  1. Since t is restricted by

  0t   2 , 1  cos 0  x  1  cos 2 1  x  2, and similarly, 0  y  1. (This is the

lower right quarter of the circle.)

(b) x 

x

1 2 1  2   x  2, y  2 . Substituting for t t t

1 gives y  2 x  22 . Since t is restricted by t 1 1 0  t  2, we have  , so x  52 and y  12 . t 2 The rectangular coordinate equation is y  2 x  22 , x  52 . 48. x  sin t  cos 2t, y  cos t  sin 3t

47. x  cos 2t, y  sin 3t 1

-1

1

1 -1

-1

1 -1

49. The coordinates of Q are x  cos  and y  sin . The coordinates of R are x  1 and y  tan . Hence, the midpoint P   1  cos  sin   tan  1  cos  sin   tan   , so parametric equations for the curve are x  and y  . is 2 2 2 2

630

CHAPTER 8 Polar Coordinates and Parametric Equations

CHAPTER 8 TEST         1. (a) x  8 cos 54  8  22  4 2, y  8 sin 54  8  22  4 2. So the point has rectangular coordinates     4 2 4 2 .     (b) P  6 2 3 in rectangular coordinates. So tan   263 and the reference angle is    6 . Since P is in   2  quadrant II, we have   56 . Next, r 2  62  2 3  36  12  48, so r  4 3. Thus, polar coordinates for       the point are 4 3 56 or 4 3 116 . (b) r  8 cos 

2. (a)



x 2  8x  y 2  0

O

x  42  y 2  16

1

r 2  8r cos  

x 2  y 2  8x



x 2  8x  16  y 2  16





The curve is a circle.

3.

The curve is a limaçon.

O

4. (a)

2

1+Ï3 i 1

1

    3i has r  1  3  2 and   tan1 3  3 . So, in     trigonometric form, 1  3i  2 cos  3  i sin 3 .     (c) z  1  3i  2 cos  3  i sin 3   z 9  29 cos 93  i sin 93  512 cos 3  i sin 3

(b) 1 

Im

Re

 512 1  i 0  512

      i sin 7 and z  2 cos 5  i sin 5 . 5. z 1  4 cos 712 2 12 12 12      7  5 7  5  i sin  8 cos   i sin  Then z 1 z 2  4  2 cos 12 12        7  5 7  5 3   2  i sin z 1 z 2  42 cos  i sin  2 cos  6 6 2  12 12

 8 and 1i 2





 3  i.

The Path of a Projectile

   6. 27i has r  27 and    2 , so 27i  27 cos 2  i sin 2 . Thus,        2k  2k 3 2 2 13  i sin  27 cos 27i 3 3        4k   4k  i sin  3 cos 6 6

Im wÁ

wü 1 0

for k  0, 1, 2. Thus, the three roots are      3  1 i  3 3  i ,   3  i sin 0  3 cos  6 6 2 2 2         1  3 cos 56  i sin 56  3  23  12 i  32  3  i , and   2  3 cos 96  i sin 96  3i. 7. (a) x  3 sin t  3, y  2 cos t, 0  t  . From the

work of part (b), we see that this is the half-ellipse

Squaring both sides gives y

y  2 cos t 

1 1

x

1

Re

wª

(b) x  3 sin t  3  x  3  3 sin t 

shown.

631

x 3  sin t. 3

x  32  sin2 t. Similarly, 9

y  cos t, and squaring both sides gives 2

y2  cos2 t. Since sin2 t  cos2 t  1, it follows that 4 y2 x  32   1. Since 0  t  , sin t  0, so 9 4 3 sin t  0  3 sin t  3  3, and so x  3. Thus the curve consists of only the right half of the ellipse.

8. We start at the point 3 5. Because the line has slope 2, for every 1 unit we move to the right, we must move up 2 units. Therefore, parametric equations are x  3  t, y  5  2t. 9. (a) x  3 sin 2t, y  3 cos 2t. The radius is 3 and the position at time t  0 is 3 sin 2 0  3 cos 2 0  0 3. Initially x is increasing and y is decreasing, so the motion is clockwise. At time t   the object is back at 0 3, so it takes  units of time to complete one revolution. (b) If the speed is doubled, the time to complete one revolution is halved, to  2 . Parametric equations modeling this motion are x  3 sin 4t, y  3 cos 4t.   (c) x 2  y 2  3 sin 2t2  3 cos 2t2  9 sin2 2t  cos2 2t  9, so an equation in rectangular coordinates is x 2  y 2  9.

(d) In polar coordinates, an equation is r  3.

FOCUS ON MODELING The Path of a Projectile x . Substituting this value for t into the equation for y, we get  0 cos     2 x x g x 2 . This shows y   0 sin  t  12 gt 2  y   0 sin   y  tan  x  2  12 g  0 cos   0 cos  2 cos2 

1. From x   0 cos  t, we get t 

0

that y is a quadratic function of x, so its graph is a parabola as long as   90 . When   90 , the path of the projectile is a straight line up (then down).

632

FOCUS ON MODELING

2. (a) Applying the given values, we get x   0 cos  t  15t and  y  4   0 sin  t  12 gt 2  4  15 3t  16t 2 as the parametric equations for the path of the baseball.

y 10

0

10

x

20

(b) The baseball will hit the ground when y  0  4  2598t  16t 2 . Using the Quadratic Formula,  2598 25982 4164 t   177 seconds (since t must be positive). So the baseball travels 32 x  15  177  265 ft before hitting the ground after 177 s.

3. (a) We find the time at which the projectile hits the ground using the equation t  2 0 gsin  , where  0  1000 and g  16. Since the rocket is fired 5 from the vertical axis and  is measured from the horizontal, we have sin 85  6226, so the rocket is in the air for 6226 seconds.   90  5  85 . Then t  200032

(b) Substituting the given values into y   0 sin  t  12 gt 2 , we get

(d)

y  1000 sin 85  t  16t 2  996t  16t 2 . Then

y  f t  at 2  bt  c, where a  16, b  996, and c  0. So y is a

quadratic function whose maximum value is attained at

b   996  31125, and the maximum value is t   2a 216

f 31125  996 31125  16 311252  15,500 (see Section 3.1 for a

guide to finding the maximum value of a quadratic function). Thus, the

y 15,000 10,000 5000

0

2000

4000

6000 x

rocket reaches a maximum height of 15,500 feet. (c) We use the equation x   0 cos  t to find the horizontal distance traveled after t seconds. Since the rocket hits the ground after

6226 seconds, substituting into the equation for horizontal distance gives x  1000 cos 85  6226  5426. Thus, the rocket travels a horizontal

distance of 5426 feet.

4. (a) The missile hits the ground when t  x   0 cos 

2 0 sin  , so the missile travels g

2 2 0 sin   0 sin 2 meters. Substituting  0  330 ms, g g

3302 sin 2  sin 2  08999. 98 So 2  6415    3208 or 2  11585    5793 . x  10 km, and g  98 m/s2 , we get 10000 

y 4000 3000 2000 1000 0

4000

8000

x

(b) The missile fired at 3208 will hit the target in 357 seconds, while the missile fired at 5793 reaches the target in 571 seconds.

The Path of a Projectile

633

5. We use the equation of the parabola from Exercise 1 and find its vertex:   2 2 sin  cos x g g x2  y   2 x2  0  y  tan  x  2 g 2 0 cos2  2 0 cos2       2 2 sin  cos x 2 sin  cos  2  02 sin  cos   2 g g 0 0 2     y   2   2 x  g g g 2 0 cos2  2 0 cos2   2    02 sin  cos   02 sin  cos   02 sin2   02 sin2  g x y 2 . Thus the vertex is at  , so the maximum  g 2g g 2g 2 0 cos2   2 sin2  . height is 0 2g 6. Since the horizontal component of the projectile’s velocity has been reduced by , the parametric equations become x   0 cos    t, y   0 sin  t  12 gt 2 . y

7. In Exercise 6 we derived the equations x   0 cos    t,

y   0 sin  t  12 gt 2 . We plot the graphs for the given values of

10 ¬=60¡

 0 , , and  in the figure to the right. The projectile will be blown

8

backwards if the horizontal component of its velocity is less than the

6

speed of the wind, that is, 32 cos   24  cos   34    414 .

4 ¬=45¡

¬=40¡ ¬=30¡

2

_14 _12 _10 _8 _6 _4 _2 0

The optimal firing angle appears to be between 15 and 30 . We

y

graph the trajectory for   20 ,   23 , and   25 . The solution appears to be close to 23 .

3

2

4 x

¬=15¡ ¬=5¡

¬=25¡ ¬=23¡

2

¬=20¡ 1

0

1

2

3

4

x

8. (a) Answers will vary. (b) Both projectiles land at the same spot. The projectile fired with   30 flies lower and lands first.    2   2 o  o o  0 . Thus, when  0 t  49 t 2  t (c) When   90 , we have y   0 t  49t 2  49 t 2  49 49 98 196 is doubled the maximum height of the ball increases by a factor of 4.

9

VECTORS IN TWO AND THREE DIMENSIONS

9.1

VECTORS IN TWO DIMENSIONS

1. (a) The vector u has initial point A and terminal point B.

(b) The vector u has initial point 2 1 and terminal point 4 3. In component form we write u  2 2 and v  3 6. Then 2u  4 4 and u  v  1 8.

v

u

u+ v

2u

u

2. (a) The length of a vector w  a1  a 2  is w     u  22  22  8  2 2.

 a12  a22 , so the length of the vector u in Figure II is

(b) If we know the length w and direction  of a vector w then we can express the vector in component form as w  w cos  w sin .

3. 2u  2 2 3  4 6

4. v   3 4  3 4

y

y

2u v

u 1

1

v

u 1

x

5. u  v  2 3  3 4  2  3 3  4  1 7 y

1

_v

x

6. u  v  2 3  3 4  2  3 3  4  5 1 y

v u+v

_v

u 1

1

x

u 1 u-v

1

x

635

636

CHAPTER 9 Vectors in Two and Three Dimensions

7. v  2u  3 4  2 2 3  3  2 2  4  2 3

8. 2u  v  2 2 3  3 4  2 2  3 2 3  4  1 10

 7 2 y

y

v

2u+v _2u

v

1

2u

x

1

1

v-2u

x

1

In Solutions 9–18, v represents the vector with initial point P and terminal point Q.

9. P 2 1, Q 5 4. v  5  2 4  1  3 3

10. P 2 1, Q 3 4. v  3  2 4  1  5 3

11. P 1 2, Q 4 1. v  4  1 1  2  3 1

12. P 3 1, Q 1 2. v  1  3 2  1  2 3

13. P 3 2, Q 8 9. v  8  3 9  2  5 7

14. P 1 1, Q 9 9. v  9  1 9  1  8 8

15. P 5 3, Q 1 0. v  1  5 0  3  4 3

16. P 1 3, Q 6 1. v  6  1  1  3  5 4

17. P 1 1, Q 1 1.

18. P 8 6, Q 1 1.

v  1  1  1  1  0 2 y

19.

v  1  8  1  6  7 5

(6, 7)

y

20.

(3, 5)

u

u (4, 3)

(4, 3)

1

1 x

1

The terminal point is 4  2 3  4  6 7. 21.

y

The terminal point is 4  1 3  2  3 5. y

22.

(4, 3) 1 1

(4, 3)

u u

(_4, 2) (8, 0) x

The terminal point is 4  4 3  3  8 0.

x

1

1 1

x

The terminal point is 4  8 3  1  4 2.

SECTION 9.1 Vectors in Two Dimensions y

23.

637

y

24. (_3, 5)

(2, 3)

u u 1

u

(_3, 5)

u

1

x

1

y

u

x

u

1

25.

u

(2, 3)

26.

(_3, 5)

y (2, 3)

(2, 3)

u u

(_3, 5)

u

1 1

1 1

x

u

x

u

27. u  1 4  i  4j

28. u  2 10  2i  10j

29. u  3 0  3i

30. u  0 5  5j

31. u  2 7, v  3 1. 2u  2  2 7  4 14; 3v  3  3 1  9 3; u  v  2 7  3 1  5 8; 3u  4v  6 21  12 4  6 17

32. u  2 5  v  2 8. 2u  2  2 5  4 10; 3v  3  2 8  6 24; u  v  2 5  2 8  0 3; 3u  4v  6 15  8 32  14 47

33. u  0 1, v  2 0. 2u  20 1  0 2; 3v  32 0  6 0; uv  0 12 0  2 1; 3u  4v  0 3  8 0  8 3

34. u  i, v  2j. 2u  2i; 3v  3 2j  6j; u  v  i  2j; 3u  4v  3i  8j

35. u  2i, v  3i  2j. 2u  2  2i  4i; 3v  3 3i  2j  9i  6j; u  v  2i  3i  2j  5i  2j; 3u  4v  3  2i  4 3i  2j  6i  8j

36. u  i  j, v  i  j. 2u  2i  2j 3v  3i  3j u  v  i  j  i  j  2i; 3u  4v  3 i  j  4 i  j  i  7j       37. u  2i  j v  3i  2j. Then u  22  12  5; v  32  22  13; 2u  4i  2j; 2u  42  22  2 5;    2    1 v  3 ij;  1 v  3  12  12 13; uv  5ij; u  v  52  12  26; uv  2ij3i2j  i3j; 2  2 2 2     u  v  12  32  10; u  v  5  13       38. u  2i  3j v  i  2j. Then u  4  9  13; v  1  4  5; 2u  4i  6j; 2u  16  36  2 13;      1 v  1 i  j;  1 v  1  1  1 5; u  v  i  j; u  v  1  1  2; u  v  3i  5j;   2 2 2 4 2     u  v  9  25  34; u  v  13  5     39. u  10 1, v  2 2. Then u  102  12  101; v  22  22  2 2; 2u  20 2;          2u  202  22  404  2 101; 12 v  1 1;  12 v  12  12  2; u  v  8 3;       u  v  82  32  73; u  v  12 1; u  v  122  12  145; u  v  101  2 2

638

CHAPTER 9 Vectors in Two and Three Dimensions

    40. u  6 6, v  2 1. Then u  36  36  6 2; v  4  1  5; 2u  12 12;             2u  144  144  12 2; 12 v  1  12 ;  12 v  1  14  12 5; u  v  8 5; u  v  64  25  89;     u  v  4 7; u  v  16  49  65; u  v  6 2  5 In Solutions 41–46, x represents the horizontal component and y the vertical component.   41. v  40, direction   30 . x  40 cos 30  20 3 and y  40 sin 30  20. Thus, v  xi  yj  20 3i  20j.   42. v  50, direction   120 . x  50 cos 120  25 and y  50 sin 120  25 3. Thus, v  xi  yj  25i  25 3j. 43. v  1, direction   225 . x  cos 225   1 and y  sin 225   1 . Thus,   v  xi  yj   1 i  1 j   22 i  22 j. 2 2

2

2

44. v  800, direction   125 . x  800 cos 125  45886 and y  800 sin 125  65532. Thus, v  xi  yj  800 cos 125  i  800 sin 125  j  45886i  65532j. 45. v  4, direction   10 . x  4 cos 10  394 and y  4 sin 10  069. Thus, v  xi  yj  4 cos 10  i  4 sin 10  j  394i  069j.      46. v  3, direction   300 . x  3 cos 300  23 and y  3 sin 300   32 . Thus, v  xi  yj  23 i  32 j.    47. v  3 4. The magnitude is v  32  42  5. The direction is , where tan   43    tan1 43  5313 .

     48. v   22   22 . The magnitude is v  12  12  1. The direction is , where tan   1 with  in quadrant III    180  tan1 1  225 .

  5 with  in 49. v  12 5. The magnitude is v  122  52  169  13. The direction is , where tan    12   5  15738 . quadrant II      tan1  12 50. v  40 9. The magnitude is v  9  1268 .   tan1 40

 9 with  in quadrant I  1600  81  41. The direction is , where tan   40

  2   3j. The magnitude is v  12  3  2. The direction is , where tan   3 with  in quadrant I     tan1 3  60 .   52. v  i  j. The magnitude is v  1  1  2. The direction is , where tan   1 with  in quadrant I 

51. v  i 

  tan1 1  45 .

 3  2598, y  30 sin 30  15. So the horizontal component of 2  force is 15 3 lb and the vertical component is 15 lb.

53. v  30, direction   30 . x  30 cos 30  30 

54. v  500, direction   70 . x  500 cos 70  17101, y  500 sin 70  46985. So the east component of the velocity is 17101 mi/h and the north component is 46985 mi/h. 55. The flow of the river can be represented by the vector v  3j and the swimmer can be represented by the vector u  2i. Therefore the true velocity is u  v  2i  3j. 56. If the current is 12 mi/h, it can be represented by the vector v  12j. If the swimmer heads in a direction  north of east, his velocity relative to the water is u  2 cos  i  2 sin  j and so his velocity relative to land is u  v  2 cos  i  2 sin   12 j. We want the y-component of his velocity to be 0, so we calculate 2 sin   12  0  sin   35    sin1 53  369 . Therefore, he should swim 369 north of east, or N 531 E.

SECTION 9.1 Vectors in Two Dimensions

u

57. The speed of the airplane is 300 mi/h, so its velocity relative to the air is v  300 cos  i  300 sin  j. The wind has velocity w  30j, so

639

w

¬

v

the true course of the airplane is given by

u  v+w  300 cos  i  300 sin   30 j. We want the y-component of the airplane’s velocity to be 0, so we

1    574 . Therefore, the airplane should head in the direction 18574 (or solve 300 sin   30  0  sin   10 S 8426 W). 58. The ocean currents can be represented by the vector v  3i and the salmon can be represented by the vector       u  5 2 2 i  5 2 2 j. Therefore u  v  5 2 2  3 i  5 2 2 j represents the true velocity of the fish.

59. (a) The velocity of the wind is 40j.

(b) The velocity of the jet relative to the air is 425i. (c) The true velocity of the jet is v  425i  40j  425 40.    40  54   is (d) The true speed of the jet is v  4252  402  427 mi/h, and the true direction is   tan1 425 N 846 E.

   55 3 60. (a) The velocity of the wind is w  55 cos 60  55 sin 60   55 2 2 .     (b) The velocity of the jet is w  765 cos 45  765 sin 45   7652 2  7652 2 .      765 2  55 3  765 2  56844 58857. (c) The true velocity of the jet is v  w  55 2  2 2 2  (d) The true speed of the jet is w  v  568442  588572  818 mih. The direction of the vector w  v is       tan1 58857 56844  46 . Thus the true direction of the jet is approximately N 44 E.   61. If the direction of the plane is N 30 W, the airplane’s velocity is u  ux  u y where ux  765 cos 60  3825,

and u y  765 sin 60  66251. If the direction of the wind is N 30 E, the wind velocity is   w  x   y where x  55 cos 60  275, and  y  55 sin 60  4763. Thus, the actual flight path is v  u  w  3825  275 66251  4763  355 71014, and so the true speed is    v  3552  710142  794 mih, and the true direction is   tan1  71014  1166 so  is N 266 W. 355

62. Let v be the velocity vector of the jet and let  be the direction of this vector. Thus v  765 cos  i  765 sin  j. If w      55 3 j. is the velocity vector of the wind then the true course of the jet is v  w  765 cos   55 2 i  765 sin   2 To achieve a course of true north, the east-west component (the i component) of the jet’s velocity vector must be 0. That   275 1  275  921 . Thus the pilot should head his plane in the is 765 cos   55 2  0  cos    765    cos 765

direction N 21 W.

63. (a) The velocity of the river is represented by the vector r  10 0.

(b) Since the boater direction is 60 from the shore at 20 mih, the velocity of the boat is represented by the vector    b  20 cos 60  20 sin 60   10 10 3  10 1732. (c) w  r  b  10  10 0  1732  20 1732  (d) The true speed of the boat is w  202  17322  265 mih, and the true direction is    409  N 491 E.   tan1 1732 20

64. Let w be the velocity vector of the water, let v be the velocity vector of the boat, and let  be the direction of v. Then v  20 cos  i  20 sin  j and w  10i. The true course of the boat is v  w  20 cos   10 i  20 sin  j. To achieve a course of true north, the east-west component of the boat’s velocity vector must be 0. Thus, 20 cos   10  0    10   12    cos1  12  120 . Thus the boater should head her boat in the direction N 30 W. cos   20

640

CHAPTER 9 Vectors in Two and Three Dimensions

  65. (a) Let b  bx  b y represent the velocity of the boat relative to the water. Then b  24 cos 18  24 sin 18   228 74.   (b) Let w  x   y represent the velocity of the water. Then w  0  where  is the speed of the water. So the true velocity of the boat is b  w  24 cos 18  24 sin 18  . For the direction to be due east, we must have 24 sin 18    0    742 mih. Therefore, the true speed of the water is 74 mi/h. Since b  w  24 cos 18  0, the true speed of the boat is b  w  24 cos 18  228 mih. 66. Let w represent the velocity of the woman and l the velocity of the ocean liner. Then w  2 0, and l  0 25, and so  r  2  0 0  25. Hence, relative to the water, the woman’s speed is r  4  625  2508 mih, and her direction 25  9457 or approximately N 457 W. is   tan1 2

67. F1  2 5 and F2  3 8. (a) F1  F2  2  3 5  8  5 3

(b) The additional force required is F3  0 0  5 3  5 3.

68. F1  3 7, F2  4 2, and F3  7 9. (a) F1  F2  F3  3  4  7 7  2  9  0 0 (b) No additional force is required.

69. F1  4i  j, F2  3i  7j, F3  8i  3j, and F4  i  j. (a) F1  F2  F3  F4  4  3  8  1 i  1  7  3  1 j  0i  4j  4j (b) The additional force required is F5  0i  0j  0i  4j  4j.

70. F1  i  j, F2  i  j, and F3  2i  j. (a) F1  F2  F3  1  1  2 i  1  1  1 j  0i  j  j

(b) The additional force required is F4  0i  0j  0i  j  j.

71. F1  10 cos 60  10 sin 60  

      5 5 3 , F2  8 cos 30  8 sin 30   4 3 4 , and

F3  6 cos 20  6 sin 20   5638 2052.    (a) F1  F2  F3  5  4 3  5638 5 3  4  2052  757 1061.

(b) The additional force required is F4  0 0  757 1061  757 1061. 72. F1  3 1, F2  1 2, F3  2 1, and F4  0 4. (a) F1  F2  F3  F4  3  1  2  0 1  2  1  4  2 4 (b) The additional force required is F5  0 0  2 4  2 4.

73. From the figure we see that T1   T1  cos 50 i  T1  sin 50 j and T2  T2  cos 30 i  T2  sin 30 j. Since T1  T2  100j we get  T1  cos 50  T2  cos 30  0 and T1  sin 50  T2  sin 30  100. From the first cos 50 cos 50 sin 30 , and substituting into the second equation gives T1  sin 50  T1   100 equation, T2   T1   cos 30 cos 30  T1  sin 50 cos 30  cos 50 sin 30   100 cos 30  T1  sin 50  30   100 cos 30  cos 30 T1   100  879385. sin 80 cos 30 Similarly, solving for T1  in the first equation gives T1   T2  and substituting gives cos 50 cos 30 sin 50 T2   T2  sin 30  100  T2  cos 30 sin 50  cos 50 sin 30   100 cos 50  cos 50 100 cos 50 T2    652704. Thus, T1  879416 cos 50  i  879416 sin 50  j  565i  674j and sin 80 T2  652704 cos 30  i  652704 sin 30  j  565i  326j.

SECTION 9.2 The Dot Product

641

74. From the figure we see that T1   T1  cos 223 i  T1  sin 223 j and T2  T2  cos 415 i  T2  sin 415 j. Since T1  T2  18 278j, we get  T1  cos 223  T2  cos 415  0 and T1  sin 223  T2  sin 415  18 278. From the cos 223 first equation, T2   T1  , and substituting into the second equation gives cos 415  cos 223 sin 415 T1  sin 223  T1   18 278  cos 415    T1  sin 223 cos 415  cos 223 sin 415   18 278 cos 415  cos 415 T1  sin 223  415   18,278 cos 415  T1   18,278  15 257. sin 638 cos 415 Similarly, solving for T1  in the first equation gives T1   T2  and substituting gives cos 223 cos 415 sin 223 T2   T2  sin 415  18,278  T2  sin 223 cos 415  sin 415 cos 223   18,278 cos 223 cos 223 18,278 cos 223  T2    18,847. sin 638 Thus, T1  15,257 cos 223  i  15,257 sin 223  j  14,116i  5,789j and T2  18,847 cos 415  i  18,847 sin 415  j  14,116i  12,488j. 75. When we add two (or more vectors), the resultant vector can be found by first placing the initial point of the second vector at the terminal point of the first vector. The resultant vector can then found by using the new terminal point of the second vector and the initial point of the first vector. When the n vectors are placed head to tail in the plane so that they form a polygon, the initial point and the terminal point are the same. Thus the sum of these n vectors is the zero vector.

9.2

THE DOT PRODUCT

1. The dot product of u  a1  a2  and v  b1  b2  is defined by u  v  a1 a2  b1 b2 . The dot product of two vectors is a real number, or scalar, not a vector. uv . So if u  v  0, the vectors are perpendicular. 2. The angle  satisfies cos   u v u 3. (a) The component of u along v is the scalar u cos  and can be uv expressed in terms of the dot product as . v projv u ¬   v uv compv u v. (b) The projection of u onto v is the vector projv u  v2 4. The work done by a force F in moving an object along a vector D is W  F  D.         5. (a) u  v  2 0  1 1  2  0  2 6. (a) u  v  1 3   3 1   3  3  0 uv uv 2  1    45 (b) cos     (b) cos    0    90 2 2 2 u v u v

7. (a) u  v  2 7  3 1  6  7  13 uv   13    56 (b) cos   53 10 u v 9. (a) u  v  3 2  1 2  3  4  1 uv (b) cos     1    97 13 5 u v      11. (a) u  v  0 5  1  3  0  5 3  5 3   uv (b) cos    5523  23    30 u v

8. (a) u  v  6 6  1 1  6  6  12 uv  12  1    180 (b) cos   6 2 2 u v 10. (a) u  v  2 1  3 2  6  2  4 uv (b) cos     4    60 5 13 u v 12. (a) u  v  1 1  1 1  1  1  0 uv   0  0    90 (b) cos   2 2 u v

642

CHAPTER 9 Vectors in Two and Three Dimensions

13. (a) u  v  i  3j  4i  j  4  3  1 uv   1    856 (b) cos   10 17 u v

14. (a) u  v  3i  4j  2i  j  6  4  10 uv    10 (b) cos   5 5 u v     1534   cos1 10 5 5

15. u  v  12  12  0  vectors are orthogonal

16. u  v  0  0  0  vectors are orthogonal

17. u  v  8  12  4  0  vectors are not orthogonal

18. u  v  0  0  0  vectors are orthogonal

19. u  v  24  24  0  vectors are orthogonal

20. u  v  4  0  4  vectors are not orthogonal

21. u  v  u  w  2 1  1 3  2 1  3 4

22. u  v  w  2 1  [1 3  3 4]

23649 23. u  v  u  v  [2 1  1 3]  [2 1  1 3]  3 2  1 4  3  8  5 25. 27. 29.

30.

31.

32.

33.

34.

 2 1  4 1  8  1  9 24. u  v u  w  2 1  1 3 2 1  3 4  1 10  10

 3  5  12  24 uv 12 uv 2 2 2   x  26. x    2 v v 5 5 1 2 uv uv 0  24 56  0 28 x  24 28. x   5   v v 1 10     2 4  1 1 uv 1 1  1 1. v  (a) u1  projv u  12  12 v2 (b) u2  u  u1  2 4  1 1  3 3. We resolve the vector u as u1  u2 , where u1  1 1 and u2  3 3.     7 4  2 1 uv 2 1  2 2 1  4 2 v  (a) u1  projv u  22  12 v2 (b) u2  u  u1  7 4  4 2  3 6. We resolve the vector u as u1  u2 , where u1  4 2 and u2  3 6.       1 2  1 3 uv 1 1 3   1  3 1 v  3   (a) u1  projv u  2 2 2 v2 12  32         (b) u2  u  u1  1 2   12  32  32  12 . We resolve the vector u as u1  u2 , where u1   12  32 and u2  32  12 .     11 3  3 2 uv 3 2  3 3 2  9 6 (a) u1  projv u  v  v2 32  22 (b) u2  u  u1  11 3  9 6  2 3. We resolve the vector u as u1  u2 , where u1  9 6 and u2  2 3.       2 9  3 4 uv 24 3 4  65 3 4   18 v  (a) u1  projv u  5 5 2 3 2 v 3  4       18 24 21 18 24 (b) u2  u  u1  2 9   5  5  28 5  5 . We resolve the vector u as u1  u2 , where u1   5  5 and   21 u2  28 5  5 .       1 1  2 1 uv 2 1  15 2 1  25   15 v  (a) u1  projv u  2 2 v 22  1         (b) u2  u  u1  1 1  25   15  35  65 . We resolve the vector u as u1  u2 , where u1  25   15 and u2  35  65 .

35. W  F  d  4 5  3 8  28

36. W  F  d  400 50  201 0  80 400

37. W  F  d  10 3  4 5  25

38. W  F  d  4 20  5 15  280

39. Let u  u1  u2  and v  v1  v2 . Then u  v  u 1  u 2    1   2   u 1  1  u 2  2   1 u 1   2 u 2   1   2   u 1  u 2   v  u

SECTION 9.2 The Dot Product

643

40. Let u  u1  u2  and v  v1  v2 . Then cu  v  c u 1  u 2    1   2   cu 1  cu 2    1   2   cu 1  1  cu 2  2  c u 1  1  u 2  2   c u  v  u 1 c 1  u 2 c 2  u 1  u 2   c 1  c 2   u  cv

41. Let u  u 1  u 2 , v   1   2 , and w  1  2 . Then u  v  w  u 1  u 2    1   2   1  2   u 1   1  u 2   2   1  2 

 u 1 1   1 1  u 2 2   2 2  u 1 1  u 2 2   1 1   2 2  u 1  u 2   1  2    1   2   1  2   u  w  v  w

42. Let u  u 1  u 2  and v   1   2 . Then u  v  u  v  u 1  u 2    1   2   u 1  u 2    1   2   u 1   1  u 2   2   u 1   1  u 2   2  2  2      u 21  u 22   12   22  u2  v2  u 21   12  u 22   22  u 21  u 22   12   22  43. We use the definition that projv u    projv u  u  projv u  





uv



v. Then v2           uv uv uv uv uv v u v  v v v  u  v2 v2 v2 v2 v2 

u  v2 v2



u  v2 v4

v2 

Thus u and u  projv u are orthogonal.   uv u  v v  v v=  u  v. 44. v  projv u  v  v2 v2

u  v2 v2



u  v2 v2

0

45. W  F  d  4 7  4 0  16 ft-lb

46. The displacement of the object is D  11 13  2 5  9 8. Hence, the work done is W  F  D  2 8  9 8  18  64  82 ft-lb.

47. The distance vector is D  200 0 and the force vector is F  50 cos 30  50 sin 30 . Hence, the work done is W  F  D  200 0  50 cos 30  50 sin 30   200  50 cos 30  8660 ft-lb.

48. W  F  d, and in this problem F  0 2500, d  500 cos 12  500 sin 12  Thus,

W  0 2500  500 cos 12  500 sin 12   0  2500 1040  260,000 ft-lb. This is the work done by gravity; the

car does (positive) work in overcoming the force of gravity. So the work done by the car is 260,000 ft-lb. 490  28218, and thus the weight of the car 49. (a) Since the force parallel to the driveway is 490  w sin 10  w  sin 10 is about 2822 lb. (b) The force exerted against the driveway is 28218 cos 10  2779 lb.

50. Since the weight of the car is 2755 lb, the force exerted perpendicular to the earth is 2755 lb. Resolving this into a force u perpendicular to the driveway gives u  2766 cos 65  1164 lb. Thus, a force of about 1164 lb is required.

51. Since the force required parallel to the plane is 80 lb and the weight of the package is 200 lb, it follows that 80  200 sin ,   80  2358 , and so the angle of inclination is where  is the angle of inclination of the plane. Then   sin1 200 approximately 236 .

52. Let R represent the force exerted by the rope and d the force causing the cart to roll down the ramp. Gravity acting on the cart exerts a force w of 40 lb directly downward. So the magnitude of the part of that force causing the cart to roll down the ramp is d  40 sin 15 . The angle between R and d is 45 , so the magnitude of the force holding the cart up is R cos 45 . Equating these two, we have R cos 45  40 sin 15 , so R 

40 sin 15  1464 lb. cos 45

644

CHAPTER 9 Vectors in Two and Three Dimensions

53. (a) 2 0  4 2  8, so Q 0 2 lies on L. 2 2  4 1  4  4  8,

(c) From the graph, we can see that u  w is

so R 2 1 lies on L.  (b) u  Q P  0 2  3 4  3 2.  v  Q R  0 2  2 1  2 1.   3 2  2 1 uv 2 1 w  projv u  v 2 2 v 2  2  1 8   85 2 1  16 5 5

9.3

orthogonal to v (and thus to L). Thus, the

distance from P to L is u  w. y

P

L

u Q

u-w

w v

R x

0

THREE-DIMENSIONAL COORDINATE GEOMETRY

z

1. In a three-dimensional coordinate system the three mutually perpendicular axes are called the x-axis, the y-axis, and the z-axis. The point P has coordinates 5 2 3. The equation of the plane passing through P and parallel to the x z-plane is y  2.

0

P

y

x

2. The distance between the point P x1  y1  z 1  and Q x2  y2  z 2  is given by the formula  d P Q  x2  x1 2  y2  y1 2  z 2  z 1 2 . The distance between the point P in the figure and the   origin is 5  02  2  02  3  02  38. The equation of the sphere centered at P with radius 3 is x  52  y  22  z  32  9.

z

3. (a)

z

4. (a) 0

P(3, 1, 0)

x

y

P(5, 0, 10) Q(3, _6, 7)

Q(_1, 2, _5)

 1  32  2  12  5  02   42

0

(b) d P Q 

x

(b) d P Q 

x

 3  52  6  02  7  102  7

SECTION 9.3 Three-Dimensional Coordinate Geometry

5. (a)

645

6. (a) Q(8, _7, 4)

Q(_12, 3, 0) P(_2, _1, 0) x

0

z 0

y

 (b) d P Q  12  22  3  12  0  02   2 29 7. x  4 is a plane parallel to the yz-plane.

P(5, _4, _6)

 (b) d P Q  8  52  7  42  4  62   118 8. y  2 is a plane parallel to the xz-plane.

z

z

0

y

4

_2

x

0

y

x

9. z  8 is a plane parallel to the x y-plane.

10. y  1 is a plane parallel to the xz-plane.

z 8

z

0

x

y

_1

0

y

x

 2 11. A sphere with radius r  5 and center C 2 5 3 has equation x  22  y  5  z  32  52 , or x  22  y  52  z  32  25.

12. A sphere with radius r  3 and center C 1 4 7 has equation x  12  y  42  z  72  9.  13. A sphere with radius r  6 and center C 3 1 0 has equation x  32  y  12  z 2  6.  14. A sphere with radius r  11 and center C 10 0 1 has equation x  102  y 2  z  12  11. 15. We complete the squares in x, y, and z: x 2  y 2  z 2  10x  2y  8z  9        x 2  10x  25  y 2  2y  1  z 2  8z  16  9  25  1  16  x  52  y  12  z  42  51. This  is an equation of a sphere with center 5 1 4 and radius 51.

646

CHAPTER 9 Vectors in Two and Three Dimensions

16. We complete the squares in x, y, and z: x 2  y 2  z 2  4x  6y  2z  10        x 2  4x  4  y 2  6y  9  z 2  2z  1  10  4  9  1  x  22  y  32  z  12  24. This is an  equation of a sphere with center 2 3 1 and radius 2 6.     17. We complete the squares in x, y, and z: x 2  y 2  z 2  12x  2y  x 2  12x  36  y 2  2y  1  z 2  36  1   x  62  y  12  z 2  37. This is an equation of a sphere with center 6 1 0 and radius 37.     18. We complete the squares in x, y, and z: x 2  y 2  z 2  14y  6z  x 2  y 2  14y  49  z 2  6z  9  49  9   x 2  y  72  z  32  58. This is an equation of a sphere with center 0 7 3 and radius 58.

19. (a) To find the trace in the yz-plane, we set x  0: 0  12  y  22  z  102  100  y  22  z  102  99.  This represents a circle with center 0 2 10 and radius 3 11. (b) We set x  4 and find 4  12  y  22  z  102  100  y  22  z  102  75. This represents a circle  with center 4 2 10 and radius 5 3.

20. (a) To find the trace in the xz-plane, we set y  0: x 2  0  42  z  32  144  x 2  z  32  128. This  represents a circle with center 0 0 3 and radius 8 2.

(b) We set z  2 and find x 2  y  42  2  32  144  x 2  y  42  119. This is a circle with center  0 4 2 and radius 119.

21. With the origin at its center, an equation of the tank is x 2  y 2  z 2  25. The metal circle is the trace in the plane z  4, so its equation is x 2  y 2  42  25 or x 2  y 2  9. Therefore, its radius is 3.

22. (a) With the origin at the center of the sphere, an equation is x 2  y 2  z 2  4.

(b) Six inches of the buoy are submerged, so the waterline is 2  12 or 32 ft below the center of the sphere. Therefore, an  2 equation of the circle is x 2  y 2   32  4  x 2  y 2  74 .

23. We use the Distance Formula to write the condition that X x y z is equidistant from P 0 0 0 and Q 0 3 0. (It is easier to equate the squares of the distances.) X P2  X Q2  x  02 y  02 z  02  x  02 y  32 z  02

 y 2  y  32  y 2  y 2  6y  9  6y  9  y  32 . This is an equation of a plane parallel to the xz-plane.   24. Using the Distance Formula (squaring both sides first), we have x 2  y  32  z 2  2 x 2  y 2  z 2 

x 2  y  32  z 2  2x 2  2y 2  2z 2  x 2  2y 2  y  32  z 2  0  x 2  y 2  6y  9  z 2  0. Completing

the square in y, we have x 2  y  32  z 2  18. This is an equation of the sphere with center 0 3 0 and radius   18  3 2.

9.4

VECTORS IN THREE DIMENSIONS

1. A vector v  a1  a2  a3  in three dimensions can be written in terms of the unit vectors i, j, and k as v  a1 i  a2 j  a3 k.  The magnitude of the vector v is v  a12  a22  a32 . So 4 2 4  4i  2 j  4k and 7j  24k  0 7 24. uv 2. The angle  between the vectors u and v satisfies cos   . So if u and v are perpendicular then u  v  0. If u v u  4 5 6 and v  3 0 2 then u  v  4  3  5  0  6  2  0, so u and v are perpendicular. 3. The vector with initial point P 1 1 0 and terminal point Q 0 2 5 is v  0  1 2  1  5  0  1 1 5. 4. The vector with initial point P 1 2 1 and terminal point Q 3 1 2 is v  3  1 1  2 2  1  2 3 3. 5. The vector with initial point P 6 1 0 and terminal point Q 0 3 0 is v  0  6 3  1  0  0  6 2 0. 6. The vector with initial point P 1 1 1 and terminal point Q 0 0 1 is v  0  1 0  1  1  1  1 1 0.

SECTION 9.4 Vectors in Three Dimensions

647

7. If the vector v  3 4 2 has initial point P 2 0 1, its terminal point is 2  3 0  4 1  2  5 4 1.

8. If the vector v  0 0 1 has initial point P 0 1 1, its terminal point is 0  0 1  0 1  1  0 1 0.

9. If the vector v  2 0 2 has initial point P 3 0 3, its terminal point is 3  2 0  0 3  2  1 0 1.

10. If the vector v  23 5 12 has initial point P 6 4 2, its terminal point is 6  23 4  5 2  12  17 1 14.   11. 2 1 2  22  12  22  3 12. 5 0 12  52  02  122  13      2      2 2 2 14.  1 6 2 2   12  62  2 2  3 5 13. 3 5 4  3  5  4  5 2

15. If u  2 7 3 and v  0 4 1, then u  v  2  0 7  4 3  1  2 3 2, u  v  2  0 7  4 3  1  2 11 4, and     3u  12 v  3 2  12 0  3 7  12 4  3 3  12 1  6 23 19 2 .

16. If u  0 1 3 and v  4 2 0, then u  v  0  4 1  2 3  0  4 3 3, u  v  0  4 1  2 3  0  4 1 3, and   3u  12 v  3 0  12 4  3 1  12 2  3 3  12 0  2 2 9.

17. If u  i  j and v  j  2k, then u  v  i  j  j  2k  i  2k, u  v  i  j  j  2k  i  2j  2k, and 3u  12 v  3 i  j  12 j  2k  3i  72 j  k.

18. If u  a 2b 3c and v  4a b 2c, then u  v  a  4a 2b  b 3c  2c  3a 3b c, u      b 10c . v  a  4a 2b  b 3c  2c  5a b 5c, and 3u  12 v  3a  12 4a  6b  12 b 9c  12 2c  5a 11 2 19. 12 0 2  12i  2k 21. 3 3 0  3i  3j 23. (a) 2u  3v  2 0 2 1  3 1 1 0  3 1 2

20. 0 3 5  3j  5k   22. a 13 a 4  ai  13 aj  4k

24. (a) 2u  3v  2 3 1 0  3 3 0 5  3 2 15     25. u  v  2 5 0  12  1 10  2 12  5 1  0 10  4     26. u  v  3 0 4  2 4 12  3 2  0 4  4 12  4       27. u  v= 6i  4j  2k  56 i  32 j  k  6 56  4 32  2 1  1       28. u  v  3j  2k  56 i  53 j  0 56  3  53  2 0  5

(b) 2u  3v  3i  j  2k (b) 2u  3v  3i  2j  15k

29. 4 2 4  1 2 2  4 1  2 2  4 2  0, so the vectors are perpendicular.

30. 4j  k  i  2j  9k  0 1  4 2  1 9  1, so the vectors are not perpendicular.

31. 03 12 09  10 5 10  03 10  12 5  09 10  12, so the vectors are not perpendicular.

32. x 2x 3x  5 7 3  x 5  2x 7  3x 3  0, so the vectors are perpendicular (regardless of the value of x). 2 2 1  1 2 2 2 1  2 2  1 2 4 uv      , so 33. cos    u v 2 2 1 1 2 2 9 22  22  12 12  22  22     cos1  49  1164 .

4 0 2  2 1 0 4 2 4 uv     , so   cos1 45  369 . u v 4 0 2 2 1 0 5 2 2 2 2 4  2 2  1     uv 1 2  1 3 28 j  k  i  2j  3k  35. cos   , so   cos1  2828  1009 .    u v j  k i  2j  3k 28 12  12 12  22  32

34. cos  

648

CHAPTER 9 Vectors in Two and Three Dimensions

1 4  2 3 uv 2 i  2j  2k  4i  3k     , so   cos1 23  482 . u v i  2j  2k 4i  3k 3 2 2 2 2 2 1  2  2 4  3   3 , 37. The length of the vector 3i  4j  5k is 32  42  52  5 2, so by definition, its direction angles satisfy cos    36. cos  

 4 , and cos   1 . Thus,   cos1  3  65 ,   cos1 2 2  56 , and   cos1 1  45 . cos    5 5 2 2 5 2 2

5 2

  2 , and cos    1    cos1 1  66 , 38. i  2j  k  12  22  12  6, so cos   1 , cos    6 6 6 6     2 1 1  1    cos    145 , and   cos    114 . 6 6  1 2  73 , 39. 2 3 6  22  32  62  7, so cos   27 , cos   37 , and cos   6 7    cos 7     cos1 73  65 , and   cos1  67  149 .  2 1 2  48 , 40. 2 1 2  22  12  22  3, so cos   23 , cos   1 3 , and cos   3    cos 3     1 2 1  1   3  109 , and   cos   cos 3  48 .

2 2 2 2 41. We are given that    3 ,   3 , and  is acute. Using the property of direction cosines cos   cos   cos   1,       2 2 2  2 we have cos   cos2   cos 23  1  12  cos2    12  1  cos2   12  cos    1 . Because 3 2

 is acute, we have   cos1 1  45 . 2

 2   2 1 2 2 2 2  2  22  14  cos    12 . 42.   23 and    4  cos   cos 3  cos 4  1  cos   1   2 Because  is acute, we have   cos1 21  60 .

43. We are given that   60 ,   50 , and  is obtuse, so cos2 60  cos2 50  cos2   1     cos2   1  cos2 60  cos2 50  0337. Because  is obtuse,   cos1  0337  125 .

44.   75 and   15  cos2 75  cos2   cos2 15  1  cos2   1  cos2 75  cos2 15 . Using double-angle     1  cos 30 1  cos 150   0. Therefore, cos   0, and so identities, we can write the right-hand side as 1  2 2   90 .

45. Here cos2   cos2   cos2 20  cos2 45  138  1, so there is no angle  satisfying the property of direction cosines cos2   cos2   cos2   1.

46. Here cos2   cos2   cos2 150  cos2 25  157  1, so there is no angle  satisfying cos2   cos2   cos2   1. 47. (a) We solve v  au  6 4 8  a 3 2 4  a  2. Therefore, the vectors are parallel and v  2u. (b) v  au  12 8 16  a 9 6 12  a   43 , so the vectors are parallel and v   43 u.

(c) v  au  2i  2j  2k  a i  j  k has no solution, so the vectors are not parallel.   1  1 1 1 48. (a) If v has magnitude m, then  v  v  m  1. Therefore, v is a unit vector. m m m m    (b) 1 2 2  12  22  22  3, so 13 1 2 2  13   23  23 is a unit vector in the same direction as 1 2 2.

6 8 10 

  62  82  102  10 2, so

the same direction as 6 8 10.   6 5 9  62  52  92  142, so  1

direction as 6 5 9.

     1 6 8 10   3 2  2 2   2 is a unit vector in 10 5 2 10 2

     142  9 142 is a unit vector in the same 6 5 9  3 71142  5 142 142 142

SECTION 9.5 The Cross Product

649

49. (a) The second and third forces are F2  24j and F3  25k. Therefore, F1  F2  F3  F4  0  7i  24j  25k  F4  0  F4  7i  24j  25k.   (b) F4   72  242  252  25 2

    50. (a) The side lengths are AB  0  12  1  02  2, BC  0  12  1  02  2,     C D  1  02  1  02  2, and D A  1  12  0  12  0  12  2. Because the side lengths are equal, the tetrahedron is regular.

(b) We calculate

    1 1 1  1 1 1     1 E A  E B 1 4 2 2 2 2 2 2    , so the central angle is  cos  AE B         4 3 3 1  1  1 1  1  1  E A  E B  4 4 4 4 4 4   cos1 13  1095 . 



51. (a) r  u  r  v  0  x  2 y  2 z  2  x  2  y  2  z  0  0  x  2 x  2 

y  2 y  2  z  2 z  0  x 2  4  y 2  4  z 2  2z  0  x 2  y 2  z  12  4  4  1  9.

(b) The sphere with equation x 2  y 2  z  12  9 has center 0 0 1 and radius 3.

(x, y, z)

(c) The diagram shows the plane determined by u, v, and r, along with the trace of the sphere in that plane. We see that the equation

r-v

r  u  r  v  0 states the fact that lines from the ends of a diameter of a circle to any point on its surface meet at right angles.

r-u

r v

(0, 0, 1)

u

(d) Let u  0 1 3 and v  2 1 4. Then r  u  r  v  0  x y  1 z  3  x  2 y  1 z  4  0  x x  2  y  1 y  1  z  3 z  4 2  x  12  y 2  z  72  1  12  1  49 4 

9.5

 0  x 2  2x  y 2  1  z 2  7x  12  0    1 , an equation of a circle with center 1 0 7 and radius 1 . 4 2 2

THE CROSS PRODUCT

1. The cross product of the vectors u  a1  a2  a3  and v  b1  b2  b3  is the vector    i j k      u  v   a1 a2 a3   a2 b3  a3 b2  i  a3 b1  a1 b3  j  a1 b2  a2 b1  k. So the cross product of u  1 0 1 and    b1 b2 b3     i j k      v  2 3 0 is u  v   1 0 1   3i  2j  3k.   2 3 0 2. The cross product of two vectors u and v is perpendicular to u and to v. Thus if both vectors u and v lie inside a plane, the vector u  v is perpendicular to the plane.

   i j k      3. u  v   1 0 3   9i  6j  3k   2 3 0

  i j k   4. u  v   0 4 1   1 1 2

      7i  j  4k   

650

CHAPTER 9 Vectors in Two and Three Dimensions

    i j k  k      6. u  v   2 3 4   0 6 2 8   0  1  1 1   9 3 12  6 4 3     i j k i j k       8. u  v   3 1 0   i  3j  9k 1 1 1   4i  7j  3k     0 3 1  3 0 4     i j k       9. (a) u  v   1 1 1   0 2 2 is perpendicular to both u and v.    1 1 1      0 2 2 uv   0 22  22 is a unit vector perpendicular to both u and v. (b) u  v 22  22    i j k      10. (a) u  v   2 5 3   1 11 19 is perpendicular to both u and v.    3 2 1      5. u  v         7. u  v    

i

j

    1 11 19 uv 483  11 483   19 483 is a unit vector perpendicular to both u and v.  483  483 483 u  v 12  112  192    i j k     11. (a) u  v   12 1 23   14 7 0 is perpendicular to both u and v.    6 12 6      14 7 0 uv (b)   2 5 5  55  0 is a unit vector perpendicular to both u and v. u  v 142  72    i j k     12. (a) u  v   0 3 5   6 5 3 is perpendicular to both u and v.    1 0 2 

(b)

     6 5 3 uv  3 3570   1470  3 7070 is a unit vector perpendicular to both u and v.  u  v 62  52  32    14. u  v  u v sin   4 5 sin 30  10 13. u  v  u v sin   6 12 sin 60  3 2 3

(b)

15. u  v  u v sin   10 10 sin 90  100

16. u  v  u v sin   012 125 sin 75  0145

  17. P Q  P R  1 1 1  2 0 0  0 2 2 is perpendicular to the plane passing through P, Q, and R.   18. P Q  P R  2 2 2  1 3 1  4 0 4 is perpendicular to the plane passing through P, Q, and R.   19. P Q  P R  1 1 5  1 1 5  10 10 0 is perpendicular to the plane passing through P, Q, and R.   20. P Q  P R  3 2 5  5 0 6  12 43 10 is perpendicular to the plane passing through P, Q, and R.  21. The area of the parallelogram determined by u  3 2 1 and v  1 2 3 is u  v  4 8 4  4 6.  22. The area of the parallelogram determined by u  0 3 2 and v  5 6 0 is u  v  12 10 15  469.     9   5 14 . 23. The area of the parallelogram determined by u  2i  j  4k and v  12 i  2j  32 k is u  v    13  5  2 2 2  24. The area of the parallelogram determined by u  i  j  k and v  i  j  k is u  v  0 2 2  2 2.

SECTION 9.5 The Cross Product

651

  25. The area of triangle P Q R is one-half the area of the parallelogram determined by P Q and P R, that is,       1 1  1 2 2 2 2  P Q  P R   2 1 1 1  1 3 3  2 6  2  4  14.

  26. The area of triangle P Q R is one-half the area of the parallelogram determined by P Q and P R, that is,     1  P Q  P R   1 2 1 1  6 1 0  1 12  62  82  101 .  2 2 2 2

  27. The area of triangle P Q R is one-half the area of the parallelogram determined by P Q and P R, that is,     1  P Q  P R   1 6 6 0  6 0 6  1 362  362  362  18 3.  2 2 2

  28. The area of triangle P Q R is one-half the area of the parallelogram determined by P Q and P R, that is,     1  P Q  P R   1 4 2 12  0 6 0  1 722  02  242  12 10.  2 2 2 29. (a) u  v  w  1 2 3  3 2 1  0 8 10  1 2 3  12 30 24  0 (b) Because their scalar triple product is 0, the vectors are coplanar.

30. (a) u  v  w  3 0 4  1 1 1  7 4 0  3 0 4  4 7 3  0 (b) Because their scalar triple product is 0, the vectors are coplanar.

31. (a) u  v  w  2 3 2  1 4 0  3 1 3  2 3 2  12 3 11  55

(b) Because their scalar triple product is nonzero, the vectors are not coplanar. The volume of the parallelepiped that they determine is u  v  w  55.

32. (a) u  v  w  1 1 0  1 0 1  0 1 1  1 1 0  1 1 1  0 (b) Because their scalar triple product is 0, the vectors are coplanar.

33. (a) u  v  w  1 1 1  0 1 1  1 1 1  1 1 1  2 1 1  2

(b) Because their scalar triple product is nonzero, the vectors are not coplanar. The volume of the parallelepiped that they determine is u  v  w  2.

34. (a) u  v  w  2 2 3  3 1 1  6 0 0  2 2 3  0 6 6  6

(b) Because their scalar triple product is nonzero, the vectors are not coplanar. The volume of the parallelepiped that they determine is u  v  w  6.

35. (a) We have u  120 cm, v  150 cm, w  300 cm, the angle between v and w is 90  30  60 , and the angle between u and v  w is 0 (because u is perpendicular to both v and w). Therefore,  u  v  w  u v  w cos 0  120 150  300  sin 60   2,700,000 3  4,676,537.

4,676,537 cm3  4677 liters. 1000 cm3 L 36. (a) From Exercise 9.4.48, the object with vertices A 1 0 0, B 0 1 0, C 0 0 1, and D 1 1 1 is a Rubik’s    Tetrahedron. The volume of the parallelepiped determined by the vectors AB, AC, and AD is calculated as       AB  AC  AD   1 1 0  1 0 1  0 1 1  1 1 0  1 1 1  2. Therefore, the volume of (b) The capacity in liters is approximately

the Rubik’s Tetrahedron is 16 2  13 .

37. (a) u  v  w  0 1 1  1 0 1  1 1 0  0 1 1  1 1 1  2, u  w  v  0 1 1  1 1 0  1 0 1  0 1 1  1 1 1  2, v  u  w  1 0 1  0 1 1  1 1 0  1 0 1  1 1 1  2, v  w  u  1 0 1  1 1 0  0 1 1  1 0 1  1 1 1  2, w  u  v  1 1 0  0 1 1  1 0 1  1 1 0  1 1 1  2, and w  v  u  1 1 0  1 0 1  0 1 1  1 1 0  1 1 1  2. (b) It appears that u  v  w  v  w  u  w  u  v  u  w  v  v  u  w  w  v  u.

(c) We know that the absolute values of the six scalar triple products must be equal because they all represent the volume of the parallelepiped determined by u, v, and w. The fact that a  b   b  a completes the proof.

652

CHAPTER 9 Vectors in Two and Three Dimensions

9.6

EQUATIONS OF LINES AND PLANES

1. A line in space is described algebraically by using parametric equations. The line that passes through the point P x0  y0  z 0  and is parallel to the vector v  a b c is described by the equations x  x0  at, y  y0  bt, z  z 0  ct. 2. The plane containing the point P x0  y0  z 0  and having the normal vector n  a b c is described algebraically by the equation a x  x0   b y  y0   c z  z 0   0. 3. The line passing through P 1 0 2 parallel to v  3 2 3 has parametric equations x  1  3t, y  2t, z  2  3t. 4. The line passing through P 0 5 3 parallel to v  2 0 4 has parametric equations x  2t, y  5, z  3  4t. 5. x  3, y  2  4t, z  1  2t

6. x  4t, y  3t, z  5t

7. x  1  2t, y  0, z  2  5t

8. x  1  t, y  1  t, z  1  t

9. We first find a vector determined by P 1 3 2 and Q 2 1 1: v  2  1 1  3  1  2  1 4 3. Now we use v and the point 1 3 2 to find parametric equations: x  1  t, y  3  4t, z  2  3t where t is any real number. 10. We first find a vector determined by P 2 1 2 and Q 0 1 3: v  0  2 1  1  3  2  2 2 1. Now we use v and the point 2 1 2 to find parametric equations: x  2  2t, y  1  2t, z  2  t. 11. A vector determined by P 1 1 0 and Q 0 2 2 is 1 1 2, so parametric equations are x  1  t, y  1  t, z  2t. 12. A vector determined by P 3 3 3 and Q 7 0 0 is 4 3 3, so parametric equations are x  3  4t, y  3  3t, z  3  3t. 13. A vector determined by P 3 7 5 and Q 7 3 5 is 4 4 0, so parametric equations are x  3  4t, y  7  4t, z  5. 14. A vector determined by P 12 16 18 and Q 12 6 0 is 0 22 18, so parametric equations are x  12, y  16  22t, z  18  18t. 15. (a) An equation of the plane with normal vector

16. (a) An equation of the plane with normal vector

n  1 1 1 that passes through P 0 2 3 is

n  3 2 0 that passes through P 1 2 7 is

x  y  z  5.

3x  2y  7.

1 x  0  1 y  2  1 [z  3]  0 or

(b) Setting y  z  0, we find x  5, so the x-intercept

is 5. Similarly, the y-intercept is 5 and the z-intercept

is 5.

x

(b) Setting y  z  0, we find 3x  7, so the x-intercept is 73 . Similarly, the y-intercept is 72 . If we set

x  y  0, we have 3 0  2 0  7, which has no

z

solution, and so there is no z-intercept. 0

5

3 x  1  2 y  2  0 z  7  0 or

z

5 y

_5

7 3

x

0

7 2

y

SECTION 9.6 Equations of Lines and Planes

17. (a) 3 x  2  12 z  8  0  6x  z  4 (b) x-intercept 23 , no y-intercept, z-intercept 4 z

18. (a) 23 x  6  13 y  0  z  3  0  2 x  6  y  3 z  3  0  2x  y  3z  3 (b) x-intercept 32 , y-intercept 3, z-intercept 1 z

0

2 3

x

19. (a) 3x  y  2  2 z  3  0  3x  y  2z  8 (b) x-intercept  83 , y-intercept 8, z-intercept 4 z

_3

y

_4

y

0

3 2

_1

_4

x

20. (a) x  1  4y  0  x  4y  1 (b) x-intercept 1, y-intercept 14 , no z-intercept z

_83 0

x

653

8

0

y

1

1 4

y

x

    21. The vector P Q  P R  1 1 2  1 2 1  5 3 1 is perpendicular to both P Q and P R and is therefore perpendicular to the plane through P, Q, and R. Using the formula for an equation of a plane with the point P, we have 5 x  6  3 y  2  z  1  0  5x  3y  z  35.     22. The vector P Q  P R  2 2 2  1 3 1  4 0 4 is perpendicular to both P Q and P R and is therefore perpendicular to the plane through P, Q, and R. Using the formula for an equation of a plane, we have 4 x  3  4 z  5  0  x  z  2.         1   23. P Q  P R  1 3  2  1  13  6  83  8 0 , so an equation is 83 x  3  8 y  13  0  x  3y  2.   24. P Q  P R  2 2 2  2 4 4  0 4 4, so an equation is 4 y  4  4 z  2  0  y  z  2.   25. P Q  P R  3 1 1  6 1 1  2 3 9, so an equation is 2 x  6  3 y  1  9 z  1  0  2x  3y  9z  0   26. P Q  P R  2 2 2  2 0 4  8 12 4, so an equation is 8 x  2  12y  4z  0  2x  3y  z  4.

27. The line passes through 0 0 4 and 2 5 0. A vector determined by these two points is v  2  0 5  0 0  4  2 5 4. Now we use v and the point 0 0 4 to find parametric equations: x  2t, y  5t, z  4  4t, where t is any real number. 28. The line passes through 2 0 0 and 0 0 10, so a direction vector is v  2 0 10 and equations are x  2  2t, y  0, z  10t, where t is any real number.

29. The line passes through 2 1 5 and is parallel to j, so equations are x  2, y  1  t, z  5, where t is any real number.

30. The line passes through 3 0 2 and is parallel to j, so equations are x  3, y  t, z  2, where t is any real number.   31. The plane passes through P 1 0 0, Q 0 3 0, and R 0 0 4. The vector P Q P R  1 3 01 0 4  12 4 3   is perpendicular to both P Q and P R and is therefore perpendicular to the plane through P, Q, and R. Using the formula for an equation of a plane, we have 12 x  1  4y  3z  0  12x  4y  3z  12.

32. This plane has the same normal vector as x  2y  4z  6, and because it contains the origin, its equation is x  2y  4z  0.

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33. If the point x y z is equidistant from P 3 2 5 and Q 1 1 4, then it satisfies

x  32 y  22 z  52  x  12 y  12 z  42  6x 94y 410z 25  2x 12y 18z 16  8x  6y  2z  20  4x  3y  z  10.

34. Taking t  0 and t  1 in the equation of the given line, we see that the points Q 1 2 0 and R 0 3 3 lie on the   plane. P Q  P R  1 2 6  2 3 3  12 9 1, so an equation is 12 x  2  9y  z  6  0  12x  9y  z  30. 35. (a) To find the point of intersection, we substitute the parametric equations of the line into the equation of the plane: 5 2  t  2 3t  2 5  t  1  10  5t  6t  10  2t  1  t  1. (b) The parameter value t  1 corresponds to the point 3 3 4.

36. (a) If the line is perpendicular to the plane, then it is parallel to the plane’s normal vector, so v is parallel to n. (b) If the line is parallel to the plane, then it is perpendicular to the plane’s normal vector, so v is perpendicular to n. (c) The plane’s normal vector is n  1 1 4. Line 1 is parallel to 2 2 8  2 1 1 4, so, as in part (a), it is perpendicular to the given plane. Line 2 is parallel to v  2 2 1, and v  n  2  2  4  0, so as in part (b), it is parallel to the given plane. 37. (a) Setting t  0 in the equation for Line 1 gives the point P 1 0 6. Setting t  1 gives Q 0 3 1. If we set t  1 in the equation for Line 2, we have the point P  1 0 6  P, and if we set t  12 , we get Q  0 3 1  Q. (b) Setting t  0 in the equation for Line 1 gives the point 0 3 5. But if a point on Line 4 has x-coordinate 0, it must have x  8  2t  0  t  4. But this parameter value gives the point 0 3 2 on Line 2, so the two lines are not the same.

CHAPTER 9 REVIEW 1. u  2 3, v  8 1. u 

  22  32  13, u  v  2  8 3  1  6 4,

u  v  2  8 3  1  10 2, 2u  2 2  2 3  4 6, and 3u  2v  3 2  2 8  3 3  2 1  22 7.   2. u  5 2, v  3 0. u  52  22  29, u  v  5  3 2  0  2 2,

u  v  5  3  2  0  8 2, 2u  2 5  2 2  10 4, and 3u  2v  3 5  2 3  3 2  2 0  21 6.   3. u  2i  j, v  i  2j. u  22  12  5, u  v  2  1 i  1  2 j  3i  j, u  v  2  1 i  1  2 j  i  3j, 2u  4i  2j, and 3u  2v  3 2i  j  2 i  2j  4i  7j.  4. u  3j, v  i  2j. u  32  3, u  v  3j  i  2j  i  5j, u  v  3j  i  2j  i  j, 2u  6j, and 3u  2v  3 3j  2 i  2j  2i  5j. 5. The vector with initial point P 0 3 and terminal point Q 3 1 is 3  0 1  3  3 4.

6. If the vector 5i  8j is placed in the plane with its initial point at P 5 6, its terminal point is 5  5 6  8  10 2.      2    2 3 2 7. u  2 2 3 has length 2  2 3  4. Its direction is given by tan     3 with  in quadrant II, 2    so     tan1  3  120 .   8. v  2i  5j has length 22  52  29. Its direction is given by tan    52 with  in quadrant IV, so     2  tan1  52  2918 .    9. u  u cos  u sin   20 cos 60  20 sin 60   10 10 3 .

10. u  u cos  u sin   135 cos 125  135 sin 125   135 cos 125  135 sin 125   77 111

CHAPTER 9

Review

655

11. (a) The force exerted by the first tugboat can be expressed in component form as   u  20  104 cos 40  20  104 sin 40  15321 12856, and that of the second tugboat is   34  104 cos 15   34  104 sin 15   32841 8800. Therefore, the resultant force is

12.

13. 14. 15. 16. 17. 18.

w  u  v  15321  32841 12856  8800  48162 4056.  (b) The magnitude of the resultant force is 481622  40562  48,332 lb. Its direction is given by 4056  0084, so   tan1 0084  48 or N 852 E. tan   48,162 (a) The velocity of the airplane is the sum of its velocity relative to the air, which is       u  600 cos 30  600 sin 30   300 3 300 , and the wind velocity v  50 cos 120  50 sin 120   25 25 3 .     Thus, its velocity is w  300 3  25 300  25 3 .  2    2 300 3  25  300  25 3  602 mi/h, and its direction is given by (b) The true speed of the airplane is w   300  25 3 tan    0694, so   tan1 0694  348 , or N 552 E.  300 3  25  u  4 3, v  9 8. u  42  32  5, u  u  42  32  25, and u  v  4 9  3 8  60.  u  5 12, v  10 4. u  52  122  13, u  u  52  122  169, and u  v  5 10  12 4  2.   u  2i  2j, v  i  j. u  22  22  2 2, u  u  22  22  8, and u  v  2 1  2 1  0.  u  10j, v  5i  3j. u  102  10, u  u  102  100, and u  v  10 3  30. u  v  4 2  3 6  4 3  2 6  0, so the vectors are perpendicular. u  v  5 3  2 6  5 2  3 6  8  0, so the vectors are not perpendicular. The angle between them is given by   uv 340 8 340  775 .  cos   , so   cos1 85   u v 85 52  32 22  62

19. u  v  2i  j  i  3j  2 1  1 3  5, so the vectors are not perpendicular. The angle between them is given by   uv 2 5  cos   , so   cos1 22  45 .   u v 2 22  12 12  32

20. u  v  i  j  i  j  1 1  1 1  0, so the vectors are perpendicular.

 3 6  1 1 17 37 uv   .  21. (a) u  3 1, v  6 1. The component of u along v is v 37 62  12     uv 17 6 1  102   17 v  (b) projv u  37 37 37 v2       17 and u  u  proj u  3 1  102   17  9  54 .   (c) u1  projv u  102 2 v 37 37 37 37 37 37

22. (a) u  8 6, v  20 20. The component of u along v is (b) projv u 



uv v2



v

40 20 20  1 1 800

 uv 8 20  6 20   2.   v 2 2 20  20

(c) u1  projv u  1 1 and u2  u  projv u  8 6  1 1  7 7. 23. (a) u  i  2j, v  4i  9j. The component of u along v is (b) projv u 



uv v2



56 126 v  14 97 4i  9j   97 i  97 j

 1 4  2 9 14 97 uv   .  v 97 42  92

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  126 j and u  u  proj u  i  2j   56 i  126 j  153 i  68 j. (c) u1  projv u   56 i  2 v 97 97 97 97 97 97 24. (a) u  2i  4j, v  10j. The component of u along v is 



4 10 uv    4. v 102

40 10j  4j v 100 v2 (c) u1  projv u  4j and u2  u  projv u  2i  4j  4j  2i.

(b) projv u 

uv

z

25.

z

26.

P(0, 2, 4)

Q(3, _2, 3) P(1, 0, 2)

0

y

0

x

The distance between P and Q is  3  12  2  02  3  22  3.

x

Q(1, 3, 0)

y

The distance between P and Q is   1  02  3  22  0  42  3 2.

27. The sphere with radius r  6 and center C 0 0 0 has equation x  02  y  02  z  02  62  x 2  y 2  z 2  36.  2 28. The sphere with radius r  2 and center C 1 2 4 has equation x  12  y  2  z  42  22  x  12  y  22  z  42  4.

      29. We complete the squares to find x 2  y 2 z 2 2x 6y 4z  2  x 2  2x  1  y 2  6y  9  z 2  4z  4  2

1  9  4 x  12  y  32  z  22  16, an equation of the sphere with center 1 3 2 and radius 4.     30. We complete the squares to find x 2  y 2  z 2  4y  4z  x 2  y 2  4y  4  z 2  4z  4  4  4   x 2  y  22  z  22  8, an equation of the sphere with center 0 2 2 and radius 2 2.  31. u  4 2 4 and v  2 3 1, so u  42  22  42  6, u  v  4  2 2  3 4  1  6 1 3,   u  v  4  2 2  3 4  1  2 5 5, and 34 u  2v  34 4 2 4  2 2 3 1  1  15 2 5 .  32. u  6i  8k and v  i  j  k, so u  62  82  10, u  v  7i  j  7k, u  v  5i  j  9k, and 3 u  2v  3 6i  8k  2 i  j  k  5 i  2j  8k. 4 4 2

33. (a) u  v  3 2 4  3 1 2  3 3  2 1  4 2  1

(b) u  v  0, so the vectors are not perpendicular. The angle between them is given by     406 1 uv 406  928 .   , so   cos1  406  cos   u v 406 2 2 2 2 2 2 3  2  4 3  1  2     34. (a) u  v  2 6 5  1  12  1  2 1  6  12  5 1  0 (b) u  v  0, so the vectors are perpendicular.

35. (a) u  v  2i  j  4k  3i  2j  k  2 3  1 2  4 1  0 (b) u  v  0, so the vectors are perpendicular.

36. (a) u  v  j  k  i  j  0 1  1 1  1 0  1 (b) u  v  0, so the vectors are not perpendicular. The angle between them is given by cos     cos1 21  60 .

1 1 uv     , so u v 2 2 2

CHAPTER 9

Review

657

37. (a) u  v  1 1 3  5 0 2  2  0 i  2  15 j  0  5 k  2 17 5      2 17 5 uv 318  17 318   5 318 .    159 (b) A unit vector perpendicular to u and v is 318 318 u  v 22  172  52 38. (a) u  v  2 3 0  0 4 1  3 2 8      3 2 8 uv (b) A unit vector perpendicular to u and v is    3 7777  2 7777  8 7777 . u  v 32  22  82 39. (a) u  v  i  j  2j  k  1  0 i  1  0 j  2  0 k  i  j  2k    uv i  j  2k (b) A unit vector perpendicular to u and v is   66 i  66 j  36 k. u  v 12  12  22

40. (a) u  v  i  j  k  i  j  k  1  1 i  1  1 j  1  1 k  2j  2k   uv 2j  2k (b) A unit vector perpendicular to u and v is   22 j  22 k.  u  v 22  22   41. The area of triangle P Q R is one-half the area of the parallelogram determined by P Q and P R, that is,     1  P Q  P R   1 1  6 i  2  8 j  6  4 k  1 52  102  102  1 225  15 .  2 2 2 2 2

42. The area of the parallelogram determined by u  4 1 1 and v  1 2 2 is the absolute value of their cross product:   A  u  v  0 9 9  92  92  9 2.

43. The volume of the parallelepiped determined by u  2i  j, v  2j  k, and w  3i  j  k is the absolute value of their scalar triple product: V  u  v  w  2i  j  3i  3j  6k  6  3  9. 44. This parallelepiped is determined by the vectors u  0 2 2, v  3 1 1, and w  1 4 1, so its volume is V  u  v  w  0 2 2  5 4 11  8  22  14.

45. The line that passes through P 2 0 6 and is parallel to v  3 1 0 has parametric equations x  2  3t, y  t, z  6.

46. The line that passes through P 5 2 8 and is parallel to v  2i  j  5k has parametric equations x  5  2t, y  2  t, z  8  5t.

47. A vector determined by P 6 2 3 and Q 4 1 2 is 2 3 1, so parametric equations are x  6  2t, y  2  3t, z  3  t. 48. A vector determined by P 1 0 0 and Q 3 4 2 is 2 4 2 or 1 2 1, so parametric equations are x  1  t, y  2t, z  t.

49. Using the formula for an equation of a plane, the plane with normal vector n  2 3 5 passing through P 2 1 1 has equation 2 x  2  3 y  1  5 z  1  0  2x  3y  5z  2.

50. Using the formula for an equation of a plane, the plane with normal vector n  i  2j  7k passing through P 2 5 2 has equation  [x  2]  2 y  5  7 z  2  0  x  2y  7z  6.

51. The plane passes through P 1 1 1, Q 3 4 2, and R 6 1 0.    i j k        P Q  P R  2 5 1  5 2 1   2 5 1   7i  7 j  21k  i  j  3k, so an equation is    5 2 1  1 x  1  1 y  1  3 z  1  0  x  y  3z  5.

  52. The plane passes through P 4 0 0, Q 0 3 0, and R 0 0 5. P Q  P R  4 3 0  4 0 5  15 20 12, so an equation is 15 x  4  20y  12z  0  15x  20y  12z  60.

53. The line passes through the points 2 0 0 and 0 0 4. A vector determined by these points is 2 0 4 or 1 0 2, so parametric equations are x  2  t, y  0, z  2t.

658

CHAPTER 9 Vectors in Two and Three Dimensions

54. We are given P 5 3 0. We find two points on the given line by substituting t  0 and t  1 in its parametric equations,   finding Q 2 0 6 and R 4 4 6. P Q  P R  3 3 6  1 1 6  24 12 6, so an equation is 24 x  5  12 y  3  6z  0  4x  2y  z  14.

CHAPTER 9 TEST 1. (a)

y

(b) u  3  3 i  [9  1] j  6i  10j   (c) u  62  102  2 34

(_3, 9)

u 1 1

(3, _1)

x

2. (a) u  3v  1 3  3 6 2  1  3 6  3  3 2  19 3   (b) u  v  1 3  6 2  5 5  52  52  5 2 (c) u  v  1 3  6 2  1 6  3 2  0

(d) Because u  v  0, u and v are perpendicular.

  2 4 3  42  8. Its direction is given by (b) The length of u is u 

y

3. (a) (_4Ï3, 4)

 4   3 with  in quadrant II, so 3 4 3     180  tan1 33  150 .

tan  

u 1 1

x

4. (a) The river’s current can be represented by the vector u  8 0 and the motorboat’s velocity relative to the water by the       vector v  12 cos 60  12 sin 60   6 6 3 . Thus, the true velocity is w  u  v  14 6 3 .     2 6 13 2  0742, so (b) The true speed is w  14  6 3  174 mi/h. The direction is given by tan   14   tan1 0742  366 or N 534 E.    uv 338 3 5  2 1 338  cos1 2  45 .  , so   cos1 26 5. (a) cos     2 u v 26 32  22 52  12  uv 13 26 (b) The component of u along v is .   v 2 26   uv 5 1 v  13 (c) projv u  26 5i  j  2 i  2 j v2   6. The work is W  F  d  3i  5j  7  2 i  13  2 j  3i  5j  5i  15j  3 5  5 15  90.  7. (a) The distance between P 4 3 1 and Q 6 1 3 is d  6  42  1  32  [3  1]2  6.

Vector Fields

659

(b) An equation is x  42  y  32  [z  1]2  62  x  42  y  32  z  12  36. (c) u  6  4 1  3 3  1  2 4 4  2i  4j  4k

8. u  i  j  2k, v  3i  2j  k, and w  j  5k. (a) 2u  3v  [2 1  3 3] i  [2 1  3 2] j  [2 2  3 1] k  11i  4j  k   (b) u  12  12  22  6 (c) u  v  1 3  1 2  2 1  1    i j k       (d) u  v   1 1 2   3i  7j  5k    3 2 1     i j k        (e) v  w   3 2 1   9i  15j  3k  92  152  32  3 35    0 1 5 

(f) u  v  w  i  j  2k  9i  15j  3k  1 9  1 15  2 3  18     84 1 uv , so   cos1  8484  96 .    (g) cos   u v 84 6 32  22  12

9. A vector perpendicular to both u  j  2k and v  i  2j  3k is u  v  0 1 2  1 2 3  7 2 1, so two unit           7 2 1 uv vectors perpendicular to u and v are   7186  96   186 and  7186   96  186 . u  v 72  22  12

10. (a) A vector perpendicular to the plane that contains the points P 1 0 0, Q 2 0 1, and R 1 4 3 is   P Q  P R  1 0 1  0 4 3  4 3 4. (b) An equation of the plane is 4 x  1  3y  4z  0  4x  3y  4z  4.

  (c) The area of triangle P Q R is half the area of the parallelogram determined by P Q and P R, that is,     1  P Q  P R   1 42  32  42  41 .  2 2 2

 11. A vector determined by the two points is P Q  2 1 2, so parametric equations are x  2  2t, y  4  t, z  7  2t.

FOCUS ON MODELING Vector Fields 1. F x y  12 i  12 j

 All vectors point in the same direction and have length 22 . y

0

x

2. F x y  i  xj

The vectors lie on lines with slope x. y

0

x

660

FOCUS ON MODELING

3. F x y  yi  12 j

4. F x y  x  y i  xj

The vectors point to the left for y  0 and to the right for

The vectors form a spiral pattern. y

y  0. y

0 0

x

x

yi  xj 5. F x y   x 2  y2

yi  xj 6. F x y   x 2  y2

All the vectors F x y are unit vectors tangent to circles  centered at the origin with radius x 2  y 2 .

yi  xj The length of the vector  is 1. x 2  y2 y

0

y

x

7. F x y z  j

All vectors in this field are parallel to the y-axis and have length 1.

0

x

8. F x y z  j  k All vectors are parallel to 0 1 1 and have length z

z

x

0

x y

0

y

 2.

Vector Fields

9. F x y z  zj

At each point x y z, F x y z is a vector of length z. For z  0, all point in the direction of the positive y-axis

661

10. F x y z  yk

At each point x y z, F x y z is a vector of length y. For y  0, all point in the direction of the positive z-axis

while for z  0, all are in the direction of the negative

while for y  0, all are in the direction of the negative

y-axis.

z-axis. z z

x

0

y

0

y

x

11. F x y  y x corresponds to graph II, because in the first quadrant all the vectors have positive x- and y-components, in the second quadrant all vectors have positive x-components and negative y-components, in the third quadrant all vectors have negative x- and y-components, and in the fourth quadrant all vectors have negative x-components and positive y-components.

12. F x y  1 sin y corresponds to graph IV because the x-component of each vector is constant, the vectors are independent of x (vectors along horizontal lines are identical), and the vector field appears to repeat the same pattern vertically.

13. F x y  x  2 x  1 corresponds to graph I because the vectors are independent of y (vectors along vertical lines are identical) and, as we move to the right, both the x- and the y-components get larger.

14. F x y  y 1x corresponds to graph III. As in Problem 11, all the vectors in the first quadrant have positive x- and y-components, in the second quadrant all vectors have positive x-components and negative y-components, in the third quadrant all vectors have negative x- and y-components, and in the fourth quadrant all vectors have negative x-components and positive y-components. Also, the vectors become longer as we approach the y-axis.

15. F x y z  i  2j  3k corresponds to graph IV, since all vectors have identical length and direction. 16. F x y z  i  2j  zk corresponds to graph I, since the horizontal vector components remain constant, but the vectors above the x y-plane point generally upward while the vectors below the x y-plane point generally downward.

17. F x y z  xi  yj  3k corresponds to graph III; the projection of each vector onto the x y-plane is xi  yj, which points away from the origin, and the vectors point generally upward because their z-components are all 3.

18. F x y z  xi  yj  zk corresponds to graph II; each vector F x y z has the same length and direction as the position vector of the point x y z, and therefore the vectors all point directly away from the origin.

662

FOCUS ON MODELING y 5

19.

4

(a)

3 2 (b)

1

_5 _4 _3 _2 _1 0 _1 (c) _2 _3 _4 _5

1

2

3

4

5 x

10

SYSTEMS OF EQUATIONS AND INEQUALITIES

10.1 SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 1. The given system is a system of two equations in the two variables x and y. To check if 5 1 is a solution of this system, we check if x  5 and y  1 satisfy each equation in the system. The only solution of the given system is 2 1.

2. A system of equations in two variables can be solved by the substitution method, the elimination method, or the graphical method. 3. A system of two linear equations in two variables can have one solution, no solution, or infinitely many solutions. 4. For the given system, the graph of the first equation is the same as the graph of the second equation, so the system has  x t where t is any real number. Some of the infinitely many solutions. We express these solutions by writing y 1t

solutions of this system are 1 0, 3 4, and 5 4.  x  y 1 Solving the first equation for x, we get x  y  1, and substituting this into the second equation gives 5. 4x  3y  18

4 y  1  3y  18  7y  4  18  7y  14  y  2. Substituting for y we get x  y  1  2  1  3. Thus, the solution is 3 2.  3x  y  1 6. Solving the first equation for y, we get y  1  3x, and substituting this into the second equation gives 5x  2y  1

5x  2 1  3x  1  x  2  1  x  1. Substituting for x we get y  1  3 1  2. Thus, the solution is 1 2.  x  y2 7. Solving the first equation for x, we get x  y  2, and substituting this into the second equation gives 2x  3y  9

2 y  2  3y  9  5y  4  9  5y  5  y  1. Substituting for y we get x  y  2  1  2  3. Thus, the solution is 3 1.  2x  y  7 Solving the first equation for y, we get y  7  2x, and substituting this into the second equation gives 8. x  2y  2

x  2 7  2x  2  x  14  4x  2  3x  12  x  4. Substituting for x we get y  7  2x  7  2 4  1. Thus, the solution is 4 1.  3x  4y  10 9. Adding the two equations, we get 4x  8  x  2, and substituting into the first equation in the x  4y  2

original system gives 3 2  4y  10  4y  4  y  1. Thus, the solution is 2 1.   2x  5y  15 4x  10y  30 10. Multiplying the first equation by 2 gives the system 4x  y  21 4x  y  21

Subtracting the second

equation from the first gives 9y  9  y  1, and substituting into the original second equation gives 4x  1  21  4x  20  x  5. Thus, the solution is 5 1.   3x  2y  13 6x  4y  26 11. Multiplying the first equation by 2 gives the system Adding, we get y  2, 6x  5y  28 6x  5y  28

and substituting into the first equation in the original system gives 3x  2 2  13  3x  9  x  3. The solution is 3 2. 663

664

12.

CHAPTER 10 Systems of Equations and Inequalities



2x  5y  18

Multiplying the first equation by 3 and the second by 2 gives the 3x  4y  19  6x  15y  54 Subtracting the equations gives 23y  92  y  4. Substituting this value into the first system 6x  8y  38 equation in the original system gives 2x  5 4  18  2x  2  x  1. Thus, the solution is 1 4.

13.



2x  y  1

By inspection of the graph, it appears that 2 3 is the solution to the system. We check this in both

x  2y  8

equations to verify that it is a solution. 2 2  3  4  3  1 and 2  2 3  2  6  8. Since both equations are satisfied, the solution is 2 3.

14.



xy2

2x  y  5

By inspection of the graph, it appears that 3 1 is the solution to the system. We check this in both

equations to verify that it is a solution. 3  1  2 and 2 3  1  6  1  5. Since both equations are satisfied, the solution is 3 1.

15.



xy4

16.

2x  y  2



2x  y  4

3x  y  6

The solution is x  2, y  0.

The solution is x  2, y  2. y

y

1 1

17.



1

x

2x  3y  12

x  32 y  4 The lines are parallel, so there is no intersection and hence no solution.

1

18.



x

2x  6y  0

3x  9y  18

The lines are parallel, so there is no intersection and hence no solution.

y

y

1 1

x

1 1

x

SECTION 10.1 Systems of Linear Equations in Two Variables

  x  1 y  5 2 19.  2x  y  10

20.

There are infinitely many solutions. y



12x  15y  18

2x  52 y  3 There are infinitely many solutions. The lines are the same. y

1 1

21.



xy4

x  y  0

665

x

1 1

x

Adding the two equations gives 2y  4  y  2. Substituting for y in the first equation gives

x  2  4  x  2. Hence, the solution is 2 2.  x  y3 Subtracting the first equation from the second equation gives 4y  4  y  1. Substituting, we have 22. x  3y  7 x  1  3  x  4. Hence, the solution is 4 1.  2x  3y  9 23. Adding the two equations gives 6x  18  x  3. Substituting for x in the second equation gives 4x  3y  9 4 3  3y  9  12  3y  9  3y  3  x  1. Hence, the solution is 3 1.  3x  2y  0 24. Adding the two equations gives 2x  8  x  4. Substituting for x in the second equation gives x  2y  8

3 4  2y  0  12  2y  0  y  6. Hence, the solution is 4 6.  x  3y  5 Solving the first equation for x gives x  3y  5. Substituting for x in the second equation gives 25. 2x  y  3

2 3y  5  y  3  6y  10  y  3  7y  7  y  1. Then x  3 1  5  2. Hence, the solution is 2 1.  x  y 7 26. Adding 3 times the first equation to the second equation gives 5x  20  x  4. So 2x  3y  1 27. 28. 29. 30.

4  y  7  y  3, and the solution is 4 3. x  y  2  y  x  2. Substituting for y into 4x  3y  3 gives 4x  3 x  2  3  4x  3x  6  3  x  3, and so y  3  2  5. Hence, the solution is 3 5. 9x  y  6  y  9x  6. Substituting for y into 4x  3y  28 gives 4x  3 9x  6  28  23x  46  x  2, and so y  9 2  6  12. Thus, the solution is 2 12. x 2y  7  x  72y. Substituting for x into 5x  y  2 gives 5 7  2y y  2  3510y  y  2  11y  33  y  3, and so x  7  2 3  1. Hence, the solution is 1 3. 4x  12y  0  x  3y. Substituting for x into 12x  4y  160 gives 12 3y  4y  160  40y  160  y  4, and so x  3 4  12. Therefore, the solution is 12 4.

31.  13 x  16 y  1  2x  y  6  y  2x  6. Substituting for y into 23 x  16 y  3 gives 23 x  16 2x  6  3  4x  2x  6  18  2x  12  x  6, and so y  2 6  6  6. Hence, the solution is 6 6.   32. 34 x  12 y  5  y  10  32 x. Substituting for y into  14 x  32 y  1 gives  14 x  32 10  32 x  1  x  60  9x  4  8x  64  x  8, and so y  10  32 8  2. Hence, the solution is 8 2.

666

CHAPTER 10 Systems of Equations and Inequalities

  33. 12 x  13 y  2  x  23 y  4  x  4  23 y. Substituting for x into 15 x  23 y  8 gives 15 4  23 y  23 y  8 

4  2 y  10 y  8  12  2y  10y  120  y  9, and so x  4  2 9  10. Hence, the solution is 10 9. 5 15 15 3

34. 02x 02y  18  x  y9. Substituting for x into 03x 05y  33 gives 03 y  905y  33  02y  06  y  3, and so x  3  9  6. Hence, the solution is 6 3.   3x  2y  8 3x  2y  8 Subtracting the second Multiplying the second equation by 3 gives the system 35. 3x  6y  0 x  2y  0 equation from the first gives 8y  8  y  1. Substituting into the first equation we get 3x  2 1  8  3x  6  x  2. Thus, the solution is 2 1.  4x  2y  16 36. Adding the first equation to 4 times the second equation gives 22y  264  y  12, so x  5y  70

4x  2 12  16  x  10, and the solution is 10 12.  x  4y  8 Adding 3 times the first equation to the second equation gives 0  22, which is never true. Thus, 37. 3x  12y  2 the system has no solution.  3x  5y  2 38. Adding 3 times the first equation to the second equation gives 0  12, which is false. Therefore, 9x  15y  6

there is no solution to this system.  2x  6y  10 39. Adding 3 times the first equation to 2 times the second equation gives 0  0. Writing the equation 3x  9y  15

in slope-intercept form, we have 2x  6y  10  6y  2x  10  y  13 x  53 , so the solutions are all pairs of the   form x 13 x  53 where x is a real number.  2x  3y  8 40. Adding 7 times the first equation to 1 times the second equation gives 0  59, which is false. 14x  21y  3 Therefore, there is no solution to this system.  6x  4y  12 41. Adding 3 times the first equation to 2 times the second equation gives 0  0. Writing the equation in 9x  6y  18

slope-intercept form, we have 6x  4y  12  4y  6x  12  y   32 x  3, so the solutions are all pairs of the form   x  32 x  3 where x is a real number.  25x  75y  100 1 times the first equation to 1 times the second equation gives 0  0, which is always 42. Adding 25 10 10x  30y  40 true. We now put the equation in slope-intercept form. We have x  3y  4  3y  x  4  y  13 x  43 , so the   solutions are all pairs of the form x 13 x  43 where x is a real number.  8s  3t  3 43. Adding 2 times the first equation to 3 times the second equation gives s  3, so 5s  2t  1 8 3  3t  3  24  3t  3  t  7. Thus, the solution is 3 7.  u  30  5 Adding 3 times the first equation to the second equation gives 10  10    1, so 44. 3u  80  5 u  30 1  5  u  25. Thus, the solution is u   25 1.

SECTION 10.1 Systems of Linear Equations in Two Variables

667

  1x  3y  3 2 5 45.  5 x  2y  10 3

Adding 10 times the first equation to 3 times the second equation gives 0  0. Writing the equation

  3x  1y  1 2 3 2 46.  2x  1 y   1 2 2

Adding 6 times the first equation to 4 times the second equation gives x  5  x  5. So

in slope-intercept form, we have 12 x  35 y  3  35 y   12 x  3  y   56 x  5, so the solutions are all pairs of the   form x  56 x  5 where x is a real number.

9 5  2y  3  y  21. Thus, the solution is 5 21.

47.



04x  12y  14 12x 

Adding 30 times the first equation to 1 times the second equation gives 41y  410  y  10, so

5y  10

12x  5 10  10  12x  60  x  5. Thus, the solution is 5 10.

48.



26x  10y  4

06x  12y 

Adding 3 times the first equation to 25 times the second equation gives 63x  63  x  1, so

3

26 1  10y  4  10y  30  y  3. Thus, the solution is 1 3.   1x  1y  2 3 4 49.  8x  6y  10

Adding 24 times the first equation to the second equation gives 0  58, which is never true. Thus,

the system has no solution.

  1 x  1y  4 10 2 50.  2x  10y  80

Adding 20 times the first equation to the second equation gives 0  0, which is always true. We

now put the equation in slope-intercept form. We have 2x  10y  80  10y  2x  80  y  15 x  8, so the   solutions are all pairs of the form x 15 x  8 where x is a real number.

51.



021x  317y  951

52.

235x  117y  589

The solution is approximately 387 274.



1872x  1491y  1233

621x  1292y  1782

The solution is approximately 071 172.

5

-5

5

5 -5

-5

5 -5

668

53.

CHAPTER 10 Systems of Equations and Inequalities



2371x  6552y  13,591

54.

9815x  992y  618,555

The solution is approximately 6100 2000.



435x  912y 

0

132x  455y  994

The solution is approximately 285 136.

5

20 -5

5

0

-5

50

1 , a  1. So 55. Subtracting the first equation from the second, we get ay  y  1  y a  1  1  y  a1     1 1 1 1 1 0x   . Thus, the solution is   . x a1 1a a1 a1 a1 a , a  b. So x  56. Subtracting the first equation from a times the second, we get a  b y  a  y  ab   b a b . Hence, the solution is  . x  ba ba ab



a ab



1

  ab 1 57. Subtracting b times the first equation from a times the second, we get a 2  b2 y  a  b  y  2 ,  ab a  b2   a 1 1 b 1  1  ax  x  . Thus, the solution is  . a 2  b2  0. So ax  ab ab ab ab ab 1 b 58. Subtracting a times the first equation from the second, we get b b  a y  1  y  . So ax  0 b b  a b b  a     1 1 1 1 1 . Hence, the solution is     2 . x  2 a b  a a b  a b b  a a  ab b  ab 59. Let the two numbers be x and y. Then



x  y  34 x  y  10

Adding these two equations gives 2x  44  x  22. So

22  y  34  y  12. Therefore, the two numbers are 22 and 12. 60. Let x be the larger number and y be the other number. This gives



x  y  2 x  y x  6  2y





two equations gives y  6, so x  6  2 6  18. Therefore, the two numbers are 18 and 6. 61. Let d be the number of dimes and q be the number of quarters. This gives



d 

x  3y  0

x  2y  6

q  14

010d  025q  275

Adding these

Subtracting the

first equation from 10 times the second gives 15q  135  q  9. So d  9  14  d  5. Thus, the number of dimes is 5 and the number of quarters is 9. 62. Let c be the number of children and a be the number of adults. This gives



c

a  2200

150c  400a  5050

Subtracting 3 times

the first equation from 2 times the second gives 5a  3500  a  700, so c  700  2200  c  1500. Therefore, the number of children admitted was 1500 and the number of adults was 700.

SECTION 10.1 Systems of Linear Equations in Two Variables

63. Let r be the amount of regular gas sold and p the amount of premium gas sold. Then



r 

p  280

220r  300 p  680

669

Subtracting

the second equation from three times the first equation gives 3r  22  3 280  680  08r  160  r  200. Substituting this value of r into the original first equation gives 200  p  280  p  80. Thus, 200 gallons of regular gas and 80 gallons of premium were sold. 64. Let s be the number of boxes of regular strawberries sold and d the number of deluxe strawberries sold. Then  s  d  135 Subtracting 7 times the first equation from the second equation gives 10d  7d  1110  7 135 7s  10d  1110

 3d  165  d  55. Substituting this value of d into the original first equation gives s  55  135  s  80. Thus, 80 boxes of standard strawberries and 55 boxes of deluxe strawberries were sold.  2x  2y  180 65. Let x be the speed of the plane in still air and y be the speed of the wind. This gives Subtracting 12x  12y  180

6 times the first equation from 10 times the second gives 24x  2880  x  120, so 2 120  2y  180  2y  60  y  30. Therefore, the speed of the plane is 120 mi/h and the wind speed is 30 mi/h.  Downriver: x  y  20 66. Let x be speed of the boat in still water and y be speed of the river flow. Adding Upriver: 25x  25y  20

5 times the first equation to 2 times the second gives 10x  140  x  14, so 14  y  20  y  6. Therefore, the boat’s speed is 14 mi/h and the current in the river flows at 6 mi/h.  012a  020b  32 Subtracting 250 times the 67. Let a and b be the number of grams of food A and food B. Then 100a  50b  22,000 first equation from the second, we get 70a  14,000  a  200, so 012 200  020b  32  020b  8  b  40. Thus, she should use 200 grams of food A and 40 grams of food B.

68. Let x be the number of pounds of Kenyan coffee and y be the number of pounds of Sri Lankan coffee. This gives  350x  560y  1155 Adding 10 times the first equation and 35 times the second gives 21y  105  y  05, x  y 3 so x  25  3  x  25. Thus, 25 pounds of Kenyan coffee and 05 pounds of Sri Lankan coffee should be mixed.

69. Let x and y be the sulfuric acid concentrations in the first and second containers.  300x  600y  900 015 Subtracting the first equation from 3 times the second gives 900y  90  y  010, so 100x  500y  600 0125 100x  500 010  75  x  025. Thus, the concentrations of sulfuric acid are 25% in the first container and 10% in the second.

70. Let x and y be the number of milliliters of the two brine solutions.  Quantity: x  y  1000 Subtracting the first equation from 20 times the second gives 3y  1800 Concentrations: 005x  020y  014  y  600, so x  600  1000  x  400. Therefore, 400 milliliters of the 5% solution and 600 milliliters of the 20% solution should be mixed.  Total invested: x  y  20,000 71. Let x be the amount invested at 5% and y the amount invested at 8%. Interest earned: 005x  008y  1180

Subtracting 5 times the first equation from 100 times the second gives 3y  18,000  y  6,000, so x  6,000  20,000  x  14,000. She invests $14,000 at 5% and $6,000 at 8%.

670

CHAPTER 10 Systems of Equations and Inequalities

72. Let x be the amount invested at 6% and y the amount invested at 10%. The ratio of the amounts invested gives x  2y. Then the interest earned is 006x  010y  3520  6x  10y  352,000. Substituting gives 6 2y  10y  352,000  22y  352,000  y  16,000. Then x  2 16,000  32,000. Thus, he invests $32,000 at 6% and $16,000 at 10%. 73. Let x be the length of time John drives and y be the length of time Mary drives. Then y  x  025, so x  y  025, and multiplying by 40, we get 40x  40y  10. Comparing the distances, we get 60x  40y  35, or 60x  40y  35. This  40x  40y  10 gives the system Adding, we get 20x  45  x  225, so y  225  025  25. Thus, John 60x  40y  35 drives for 2 14 hours and Mary drives for 2 12 hours.

74. Let x be the cycling speed and y be the running speed. (Remember to divide by 60 to convert minutes to decimal hours.)  05x  05y  125 We have Subtracting 2 times the first equation from 5 times the second, we get 275x  55  075x  02y  16 x  20, so 20  y  25  y  5. Thus, the cycling speed is 20 mi/h and the running speed is 5 mi/h.  xy7 75. Let x be the tens digit and y be the ones digit of the number. Adding 9 times the first 10y  x  27  10x  y

equation to the second gives 18x  36  x  2, so 2  y  7  y  5. Thus, the number is 25. 76. First let us find the intersection point of the two lines. The y-coordinate of the intersection point is the height of the triangle.  y  2x  4 We have Adding 2 times the first equation to the second gives 3y  12, so the triangle has height 4. y  4x  20 Furthermore, y  2x  4 intersects the x-axis at x  2, and y  4x  20 intersects the x-axis at x  5. Thus the base has

length 5  2  3. Therefore, the area of the triangle is A  12 bh  12  3  4  6.   77. n  5, so nk1 xk  1  2  3  5  7  18, nk1 yk  3  5  6  6  9  29, n k1 xk yk  1 3  2 5  3 6  5 6  7 9  124, and n 2 2 2 2 2 2 k1 xk  1  2  3  5  7  88. Thus we get the system  18a  5b  29 Subtracting 18 times the first equation from 5 times the 88a  18b  124 second, we get 116a  98  a  0845. Then

b  15 [18 0845  29]  2758. So the regression line is y  0845x  2758.

y 10 8 6 4 2 0

1

2 3 4 5 6 7 x

10.2 SYSTEMS OF LINEAR EQUATIONS IN SEVERAL VARIABLES 1. If we add 2 times the first equation to the second equation, the second equation becomes x  3z  1. 2. To eliminate x from the third equation, we add 3 times the first equation to the third equation. The third equation becomes 4y  5z  4.  3. The equation 6x  3y  12 z  0 is linear. 4. The equation x 2  y 2  z 2  4 is not linear, since it contains squares of variables.     x y  3y  z  5 is not a linear system, since the first equation contains a product of variables. In fact 5. The system x  y 2  5z  0    2x  yz  3 both the second and the third equation are not linear.

SECTION 10.2 Systems of Linear Equations in Several Variables

    x  2y  3z  10 6. The system 2x  5y  2    y  2z  4

    x  3y  z  0 7. yz  3    z  2

671

is linear.

Substituting z  2 into the second equation gives y  2  3  y  1. Substituting z  2

and y  1 into the first equation gives x  3 1  2  0  x  5. Thus, the solution is 5 1 2.     3x  3y  z  0 8. Substituting z  3 into the second equation gives y  4 3  10  y  2. Substituting z  3 y  4z  10    z 3 and y  2 into the first equation gives 3x  3 2  3  0  x  3. Thus, the solution is 3 2 3.     x  2y  z  7 Solving we get 2z  6  z  3. Substituting z  3 into the second equation gives 9. y  3z  9    2z  6

y  3 3  9  y  0. Substituting z  3 and y  0 into the first equation gives x  2 0  3  7  x  4. Thus, the solution is 4 0 3.     x  2y  3z  10 Solving we get 3z  12  z  4. Substituting z  4 into the second equation gives 10. 2y  z  2    3z  12

2y  4  2  y  3. Substituting z  4 and y  3 into the first equation gives x  2 3  3 4  10  x  4. Thus, the solution is 4 3 4.     2x  y  6z  5 11. Solving we get 2z  1  z   12 . Substituting z   12 into the second equation gives y  4z  0    2z  1     y  4  12  0  y  2. Substituting z   12 and y  2 into the first equation gives 2x  2  6  12  5  x  5.   Thus, the solution is 5 2  12 .

12.

    4x    

2y 

3z  10

z  6

1z  2

2

Solving we get 12 z  4  z  8 . Substituting z  8 into the second equation gives

2y  8  6  2y  2  y  1. Substituting z  8 into the first equation gives 4x  3 8  10  4x  14    x   72 . Thus, the solution is  72  1 8 .

    3x  y  z  4 Add the third equation to the second equation: y  z  1    x  2y  z  1     3x  y  z  4 Or, add the first equation to three times the second equation: 4y  7z  4    x  2y  z  1

    3x  y  z  4 13. x  y  2z  0    x  2y  z  1

672

CHAPTER 10 Systems of Equations and Inequalities

  3   5x  2y  3z  14. 10x  3y  z  20    x  3y  z  8

  3   5x  2y  3z  Add twice the first equation to the second equation: y  5z  14    x  3y  z  8     5x  2y  3z  3 Or, add 10 times the third equation to the second equation: 27y  11z  60    x  3y  z  8     2x  y  3z  5 Add 3 times the first equation to the third equation: 2x  3y  z  13    8y  8z  8   5   2x  y  3z  Or, add 3 times the second equation to the third equation: 2x  3y  z  13    14y  4z  32

    2x  y  3z  5 15. 2x  3y  z  13    6x  5y  z  7

    x  3y  2z  1 Add 2 times the first equation to 3 times the third equation: y  z  1    2x  z 1     x  3y  2z  1 Or, add 2 times the second equation to the third equation: y  z  1    3z  3

    x  3y  2z  1 16. y  z  1    2y  z  1

    x  y  z  4    x  y  z  4 17. 2y  z  1  2y  z  1      x  y  2z  5  3z  9

Eq. 1  Eq. 3

So z  3 and 2y  3  1  2y  2  y  1. Thus, x  1  3  4  x  2. So the solution is 2 1 3.

      x  y  z  0 x  y  z  0 18. y  2z  2  y  2z  2     x  y  z  2  2y  2z  2

1  Eq. 1  Eq. 3

   x  y  z  0  y  2z  2    6z  6

2  Eq. 2  Eq. 3

So z  1 and y  2 1  2  y  0. Then x  0  1  0  x  1. So the solution is 1 0 1.

      x  2y  z  6  x  2y  z  6 19. y  3z  16  y  3z  16      x  3y  2z  14 5y  3z  20    x  2y  z  6 y  3z  16   12z  60 5  Eq. 2  Eq. 3

Eq. 1  1  Eq. 3



So z  5 and y  3 5  16  y  1. Then x  2 1  5  6  x  1. So the solution is 1 1 5.

      x  2y  3z  10  x  2y  3z  10 20. 3y  z  7  3y  z  7     x  y  z  3y  4z  17 7

1  Eq. 1Eq. 3

   x  2y  3z  10  3y  z  7   5z  10

Eq. 2  1  Eq. 3

So z  2 and 3y  2  7  y  3. Then x  2 3  3 2  10  x  2. So the solution is 2 3 2.

SECTION 10.2 Systems of Linear Equations in Several Variables

       x y z 4 x  y  z  4 21. x  3y  3z  10  2y  2z  6      2x  y  z  3  y  3z  5    x  y  z  4 y  3z  5    4z  4 2  Eq. 2  Eq. 3

1  Eq. 1  Eq. 2 2  Eq. 1  1  Eq. 3

   x  y  z  4  y  3z  5    2y  2z  6

Eq. 3

673



Eq. 2

So z  1 and y  3 1  8  y  2. Then x  2  1  4  x  1. So the solution is 1 2 1.       x  y  z  0 x  y  z  0    x  y  z  0   22. x  2y  5z  3  y  2z  1 y  2z  1 13  Eq. 1  13  Eq. 2          3x  y  5z  10 4  Eq. 2  Eq. 3 6 4y  3z  6 3  Eq. 1  Eq. 3 So z  2 and y  2 2  1  y  3. Then x  3  2  0  x  1. So the solution is 1 3 2.        4z  1 4z  1  4z  1    x  x  x 23. 2x  y  6z  4  y  2z  2 y  2z  2 2  Eq. 1  Eq. 2          2x  3y  2z  8  3y  6z  6 2  Eq. 1  Eq. 3 12z  12 3  Eq. 2  Eq. 3

So z  1 and y  2 1  2  y  0. Then x  4 1  1  x  5. So the solution is 5 0 1.       2 2     x  y  2z   x  y  2z  2  x  y  2z  24. y  6z  11 3x  y  5z  8  y  6z  11 3  Eq. 1  Eq. 3          2x  y  2z  7  4y  z  23z  46 4  Eq. 2  Eq. 2 2 2  Eq. 1  Eq. 2

So z  2 and y  6 2  11  y  1. Then x  1  2  2  x  1. So the solution is 1 1 2.       2x  4y  z  2 2x  4y  z  2      x  2y  3z  4  25. 5z  10 x  2y  3z  4  Eq. 1  2  Eq. 2  14y  5z  4 Eq. 2  Eq. 3        14y  5z  4 3  Eq. 1  2  Eq. 3   3x  y  z  1 5z  10 Eq. 2  Eq. 3 So z  2 and 14y  5 2  4  y  1.        2x  y  2x  y  z  8 26. 3y x  y  z  3        2x  4z  18 y

Then x  2 1  3 2  4  x  0. So the solution is 0 1 2.    z  8   2x  y  z  8  z  2 Eq. 1  2  Eq. 2  3y  z  2    8z  32 Eq. 2  3  Eq. 3  3z  10 Eq. 1  Eq. 3

So z  4 and 3y  4  2  y  2. Then 2x  2  4  8  x  1. So the solution is 1 2 4.     Eq. 2 2y  4z  1     2x  y  2z  1 27. 2x  y  2z  1  2y  4z  1 Eq. 1      4x  2y   0 4z  2 2Eq. 2  Eq. 3       So z   12 and 2y  4  12  1  y  12 . Then 2x  12  2  12  1  x  14 . So the solution is 14  12   12 .      Eq. 2 y  z  1 2    z 2   6x  2y  z   6x  2y  28.   6x  2y  z  2 y  z  1 Eq. 1 y z  1        x  y  3z  2  21z  14 4  Eq. 2  Eq. 3 4y  17z  10 Eq. 2  6  Eq. 3     So z  23 and y  23  1  y   13 . Then 6x  2  13  23  2  x  13 . So the solution is 13   13  23 .     x  2y  z  1    x  2y  z  1  29. 2x  3y  4z  3  Since 0  1 is false, this system is inconsistent. y  2z  5 2  Eq. 1  Eq. 2       3x  6y  3z  4 0  1 3  Eq. 1  Eq. 3

674

CHAPTER 10 Systems of Equations and Inequalities

       2z  0  2z  0 Eq. 2    x  x  2y  5z  4  x 30. 2y  3z  4 x  2z  0  x  2y  5z  4 Eq. 1         2y  3z  2  4x  2y  11z  2  4x  2y  11z  2    2z  0  x Since 0  6 is false, this system is inconsistent. 2y  3z  4    0  6 Eq. 2  Eq. 3

Eq. 1  Eq. 2



4  Eq. 1  Eq. 3

       3 Eq. 2  3     x  2y  x  2y  2x  3y  z  1 31. y  z  5 Eq. 2  2  Eq. 1  x  2y  3  2x  3y  z  1 Eq. 1         x  3y  z  4   x  3y  z  4 yz 1 Eq. 3  Eq. 1    3   x  2y Since 0  4 is false, this system is inconsistent. y  z  5    0  4 Eq. 2  Eq. 3     5    x  2y  3z  5  x  2y  3z  32. 2x  y  z  5  5y  5z  5      4x  3y  7z  5  5y  5z  5

Eq. 2  2  Eq. 1 2  Eq. 2  Eq. 3

  5   x  2y  3z   5y  5z  5    0  10

Since 0  10 is false, this system is inconsistent.       x  y  z  0 x  y  z  0    x  y  z  0   33. y  2z  3 x  2y  3z  3  y  2z  3 Eq. 2  Eq. 1          2x  3y  4z  3  00 y  2z  3 2  Eq. 1  Eq. 3

Eq. 3  Eq. 2

Eq. 2  Eq. 3

So z  t and y  2t  3  y  2t  3. Then x  2t  3  t  0  x  t  3. So the solutions are t  3 2t  3 t, where t is any real number.           x  2y  z  3  x  2y  z  3  x  2y  z  3 34. 2x  5y  6z  7  y  4z  1 Eq. 2  2  Eq. 1  y  4z  1         2x  3y  2z  5  0  0 Eq. 2  Eq. 3 y  4z  1 Eq. 3  2  Eq. 1

So z  t and we have y  4t  1  y  4t  1. Substituting into the first equation, we have x  2 4t  1  t  3  x  7t  1. So the solutions are 7t  1 4t  1 t, where t is any real number.       x  3y  2z  0 x  3y  2z  0      x  3y  2z  0  35. 2x 6y  8z  4 6y  8z  4 Eq. 2  2  Eq. 1   4z  4           4x  6y 0  0 Eq. 2  Eq. 3 4 6y  8z  4 Eq. 3  4  Eq. 1   So z  t and 6y  8t  4  6y  8t  4  y  43 t  23 . Then x  3 43 t  23  2t  0  x  2t  2. So the   solutions are 2t  2 43 t  23  t , where t is any real number.           2x  4y  z  3  x  2y  2z  0 Eq. 3  x  2y  2z  0 36. x  2y  4z  6  2x  4y  z  3 Eq. 1  3z  3 2  Eq. 1  Eq. 2         x  2y  2z  0  x  2y  4z  6 Eq. 2  6z  6 Eq. 3  Eq. 1     x  2y  2z  0 3z  3    0  0 2  Eq. 2  Eq. 3

So z  1 and y  t, and substituting into the first equation we have x  2t  2 1  0  x  2t  2. Thus, the solutions are 2t  2 t 1, where t is any real number.

SECTION 10.2 Systems of Linear Equations in Several Variables

37.

  x    

 z  2 

y  2z

6

 3

  x    

 z  2 

y  2z

675

6

 3

    2y  2z  2  8 x  2y  z  2 Eq. 3  Eq. 1         2x  y  3z  2  0 y  z  6  12 Eq. 4  2  Eq. 1     x  z  2  6 x  z  2  6        y  2z  y  2z  3  3  1 Eq. 3   z    1 2z  2  2 Eq. 3  2  Eq. 2   2       3z  6  9 Eq. 4  Eq. 2 6  12 3  Eq. 3  2  Eq. 4

So   2 and z  2  1  z  1. Then y  2 1  3  y  1 and x  1  2 2  6  x  1. Thus, the solution is 1 1 1 2.

    x y z 0 x y z 0        x  y  2z  2  0  z 0 38.   2x  2y  3z  4  1  z  2  1         2x  3y  4z  5  2 y z 1   x y z 0     yz 1  z 0       1 Eq. 4  Eq. 3

Eq. 2  Eq. 1 Eq. 3  2  Eq. 1 Eq. 4  Eq. 3

  x y z 0     yz 1   z 0     z  2  1

Eq. 4 Eq. 2



Eq. 3

So   1 and z  1  0  z  1. Then y  1  1  1  y  1 and x  1  1  1  0  x  1. So the solution is 1 1 1 1.

39. Let x be the amount invested at 4%, y the amount invested at 5%, and z the amount invested at 6%. We set   x  y  z  100,000   Total money: up a model and get the following equations: Annual income: 004x  005y  006z  0051 100,000    Equal amounts: xy        x  y  z  100,000  x  y  z  100,000   4x  5y  6z  510,000 y  2z  110,000 Eq. 2  4  Eq. 1       x  y   0 2y  z  100,000 Eq. 3  Eq. 1     x  y  z  100,000 y  2z  110,000    3z  120,000 2  Eq. 2  Eq. 3

So z  40,000 and y  2 40,000  110,000  y  30,000. Since x  y, x  30,000. Mark should invest $30,000 in short-term bonds, $30,000 in intermediate-term bonds, and $40,000 in long-term bonds.

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CHAPTER 10 Systems of Equations and Inequalities

40. Let x be the amount invested at 3%, y the amount invested at 5 12 %, and z the amount invested at 9%. We   x  y  z  50,000   Total investment: set up a model and get the following equations: Annual income: 003x  0055y  009z  2540    Twice as much: x  2z        x  y  z  50,000  x  y  z  50,000  6x  11y  18z  508,000 200  Eq. 2  5x  7z  42,000 Eq. 2  11  Eq. 1        x x  2z  0  2z  0     x  y  z  50,000 5x  7z  42,000    3z  42,000 Eq. 2  5  Eq. 3

So z  14,000 and x  2z  28,000. Since x  y  z  50,000, we have y  50,000  14,000  28,000  8000. Cyndee should invest $28,000 in the least risky account, $8,000 in the intermediate account, and 14,000 in the highest-yielding account.

41. Let x, y, and z be the number of acres of land planted with corn, wheat, and soybeans. We set up a model and   x  y  z  1200   Total acres: Substituting 2x for y, we get get the following equations: Market demand: 2x  y    Total cost: 45x  60y  50z  63,750       x  2x  z  1200  z  1200  z  1200      3x  3x  2x  y 2x  y  0  2x  y  0        45x  60 2x  50z  63,750  165x   50z  63,750 15x  3750 Eq. 3  50  Eq. 1

So 15x  3,750  x  250 and y  2 250  500. Substituting into the original equation, we have 250  500  z  1200  z  450. Thus the farmer should plant 250 acres of corn, 500 acres of wheat, and 450 acres of soybeans.

42. Let a, b, and c be the number of gallons of Regular, Performance Plus, and Premium gas sold. The information provided     b c  6500 a b c  6500   a   gives the following system: 300a  320b  330c  20,050  02b  03c  550 Eq. 2  3  Eq. 1        a  3c  0 b  4c  6500 Eq. 1  Eq. 3   b c  6500  a  02b  03c  550    5c  7500 10  Eq. 2  2  Eq. 3

Thus, c  1500, so 02b  03 1500  550  b  500 and a  500  1500  6500  a  4500. The gas station sold 4500 gallons of Regular, 500 gallons of Performance Plus, and 1500 gallons of Premium gas.

43. Let a, b, and c be the number of ounces of Type A, Type B, and Type C pellets used. The     2a  3b  c  9 requirements for the different vitamins gives the following system: 3a  b  3c  14     8a  5b  7c  32     2a  3b  c  9 Equations 2 and 3 are inconsistent, so there is no solution. 7b  3c  1 2  Eq. 2  3  Eq. 1    7b  3c  4 Eq. 3  4  Eq. 1

SECTION 10.2 Systems of Linear Equations in Several Variables

677

44. Let a, b, and c represent the number of servings of toast, cottage cheese, and fruit, respectively. Then    2c  6   2a the given dietary requirements give rise to the following system: a  5b  11     100a  120b  60c  460     2a  2c  6  c 6     2a 10b  2c  16 10b  2c  16 2  Eq. 2  Eq. 1        16c  32 Eq. 3  12  Eq. 2 120b  40c  160 Eq. 3  50  Eq. 1

Thus, c  2, so 10b  2 2  16  10b  20  b  2 and 2a  2 2  6  a  1. Nicole should eat one serving of toast and two each of cottage cheese and fruit.

45. Let a, b, and c represent the number of Midnight Mango, Tropical Torrent, and Pineapple Power smoothies sold. The        8a  6b  2c  820  8a  6b  2c  820 given information leads to the system 3a  5b  8c  690  22b  58c  3060 8  Eq. 2  3  Eq. 1        3a  3b  4c  450 2b  4c  240 Eq. 2  Eq. 3     8a  6b  2c  820 22b  58c  3060    14c  420 Eq. 2  11  Eq. 3

Thus, c  30, so 22b  58 30  3060  22b  1320  b  60 and 8a  6 60  2 30  820  a  50. Thus, The Juice Company sold 50 Midnight Mango, 60 Tropical Torrent, and 30 Pineapple Power smoothies on that particular day.

46. Let a, b, and c represent the number of days required at each plant. The given information leads        8a  10b  14c  110  8a  10b  14c  110 to the system 16a  12b  10c  150  8b  18c  70 2  Eq. 1  1  Eq. 2        10a  18b  6c  114 84b  2c  162 5  Eq. 2  8  Eq. 3     8a  10b  14c  110 8b  18c  70    382c  1146 21  Eq. 2  2  Eq. 3

Thus, c  3, so 8b  18 3  70  8b  16  b  2 and 8a  10 2  14 3  110  a  6. Thus, Factory A should be scheduled for 6 days, Factory B for 2 days, and Factory C for 3 days.

47. Let a, b, and c be the number of shares of Stock A, Stock B, and Stock C in the investor’s portfolio. Since the total value remains unchanged, we get the following system:     10a  25b  29c  74,000     10a  25b  29c  74,000 12a  20b  32c  74,000  50b  14c  74,000 6  Eq. 1  5  Eq. 2       16a  15b  32c  74,000  125b  72c  222,000 8  Eq. 1  5  Eq. 3     10a  25b  29c  74,000   

50b  14c 

74,000

74c  74,000

5  Eq. 2  2  Eq. 3

So c  1,000. Back-substituting we have 50b  14 1000  74,000  50b  60,000  b  1,200 . And finally 10a  25 1200  29 1000  74,000 10a  30,000  29,000  74,000  10a  15,000  a  1,500. Thus the portfolio consists of 1,500 shares of Stock A, 1,200 shares of Stock B, and 1,000 shares of Stock C.

678

CHAPTER 10 Systems of Equations and Inequalities

    I1  I2  I3  0 48. 16I1  8I2 4  24I2  16I3  4 16  Eq. 1  Eq. 2        8I2  4I3  5 8I2  4I3  5   I2  I3  0   I1  24I2  16I3  4    28I3  19 Eq. 2  3  Eq. 3   19 2 2 19 11 So I3  19 28  068 and 24I2  16 28  4  I2  7  029. Then I1  7  28  0  I1  28  039.    

I 1  I2  I3  0

x  x1 y0  y1 z  z1 49. (a) We begin by substituting 0 , , and 0 into the left-hand side of the first equation: 2 2       2   x0  x1 y0  y1 z  z1  b1  c1 0  12 a1 x0  b1 y0  c1 z 0   a1 x1  b1 y1  c1 z 1  a1 2 2 2    12 d1  d1  d1

Thus the given ordered triple satisfies the first equation. We can show that it satisfies the second and the third in exactly the same way. Thus it is a solution of the system.

(b) We have shown in part (a) that if the system has two different solutions, we can find a third one by averaging the two solutions. But then we can find a fourth and a fifth solution by averaging the new one with each of the previous two. Then we can find four more by repeating this process with these new solutions, and so on. Clearly this process can continue indefinitely, so there are infinitely many solutions.

10.3 PARTIAL FRACTIONS 1. (iii): r x  2. (ii): r x  3. 5.

7. 8.

4 x x  22



B C A   x x  2 x  22

Bx  C A 2x  8     2 x 1 x 4 x  1 x 2  4

B A 1   x 1 x 2 x  1 x  2 x 2  3x  5

x  22 x  4



B C A   x  2 x  22 x 4

x A B x 4. 2    x  1 x  4 x  1 x  4 x  3x  4 1 B 1 C D A 6. 4  3   2  3  x x 1 x  x3 x x  1 x x

Bx  C x2 A    2 x 3 x 4 x  3 x 2  4

B Cx  D 1 1 A 1        2  2 x 1 x 1 x  1 x2  1 x4  1 x 1 x  1 x  1 x 2  1

Ax  B Cx  D x 3  4x 2  2     2 9.  2 x  1 x2  2 x2  1 x 2

10.

B A Cx  D x4  x2  1 Ex  F   2  x  2  2 2 2 2 x x  4 x2  4 x x 4

B C x3  x  1 A D Ex  F Gx  H   2   2  x  2x  5  2 x  2x  5 2x  52 2x  53 x 2  2x  5 x 2x  53 x 2  2x  5        12. Since x 3  1 x 2  1  x  1 x 2  x  1 x  1 x  1  x  12 x  1 x 2  x  1 , we have 11.



B Dx  E C 1 A 1      2  .    x  1 x  12 x 1 x3  1 x2  1 x x 1 x  12 x  1 x 2  x  1

SECTION 10.3 Partial Fractions

13.

B 2 A  . Multiplying by x  1 x  1, we get 2  A x  1  B x  1   x 1 x 1 x  1 x  1  AB 0 Adding we get 2A  2  A  1. Now A  B  0  B  A, so 2  Ax  A  Bx  B. Thus AB 2 B  1. Thus, the required partial fraction decomposition is

14.

2 1 1  .  x 1 x 1 x  1 x  1

B A 2x  . Multiplying by x  1 x  1, we get 2x  A x  1  B x  1   x 1 x 1 x  1 x  1  AB 2 2x  Ax  A  Bx  B. Thus Adding we 2A  2  A  1. Since A  B  0  B  A, B  1. The AB 0 required partial fraction decomposition is

15.

1 2x 1  .  x 1 x 1 x  1 x  1

5 B A  . Multiplying by x  1 x  4, we get 5  A x  4  B x  1   x 1 x 4 x  1 x  4  AB 0 5  Ax  4A  Bx  B. Thus Now A  B  0  B  A, so substituting,we get 4A  A  5  4A  B  5 5A  5  A  1and B  1. The required partial fraction decomposition is

16.

1 1 5  .  x 1 x 4 x  1 x  4

A B x 6   . x x  3 x x 3

Multiplying by x x  3we get x  6  A x  3  x B   AB1 x  6  Ax  3A  Bx  A  B x  3A. Thus Now 3A  6  A  2, and 2  B  1  B  1. 3A 6 The required partial fraction decomposition is

17.

679

x 6 2 1   . x x  3 x x 3

B 12 A 12  . Multiplying by x  3 x  3, we get 12  A x  3  B x  3    x 3 x 3 x  3 x  3 x2  9   A B  0 AB 0 12  Ax  3A  Bx  3B. Thus  Adding, we get 2A  4  A  2. So 2  B  0 3A  3B  12 AB 4 2 2 12   .  B  2. The required partial fraction decomposition is 2 x  3 x  3 x 9

18.

x  12 A B x  12   . Multiplying by x x  4, we get  x x  4 x x 4 x 2  4x x 12  A x  4 Bx  Ax 4A  Bx  A  B x 4A. Thus we must solve the system 2 3 x  12 .   gives 4A  12  A  3, and 3  B  1  B  2. Thus 2 x x 4 x  4x

19.



AB 4A

4 B 4 A  . Multiplying by x 2  4, we get   x 2 x 2 x  2 x  2 x2  4   A B 0 AB 0 4  A x  2  B x  2  A  B x  2A  2B, and so  2A  2B  4 AB 2 1 1 4  .   A  1, and B  1. Therefore, 2 x 2 x 2 x 4

1

This

 12

Adding we get 2A  2

680

20.

CHAPTER 10 Systems of Equations and Inequalities

2x  1

x2  x  2 and so





B 2x  1 A  . Thus, 2x  1  A x  1  B x  2  A  B x  A  2B,  x 2 x 1 x  2 x  1

A B 2

A  2B  1

Adding the two equations, we get 3B  3  B  1. Thus A  1  2  A  1. Therefore,

2x  1 1 1  .  x 2 x 1 x2  x  2 21.

x  14 B x  14 A  . Hence, x  14  A x  2  B x  4  A  B x  2A  4B,   x 4 x 2 x  4 x  2 x 2  2x  8   A B  1 2A  2B  2 and so  Adding, we get 3A  9  A  3. So 3  B  1  B  2. 2A  4B  14 A  2B  7 2 3 x  14  .  Therefore, 2 x 4 x 2 x  2x  8

22.

23.

A B 8x  3 8x  3   . Hence, 8x  3  A 2x  1  Bx  2A  B x  A, giving  x 2x  1 x 2x  1 2x 2  x  2A  B  8 2 3 8x  3 .   So A  3  A  3, and 2 3  B  8  B  2. Therefore, 2 x 2x  1 2x  x A  3 B x A x  . Hence,   4x  3 2x  1 4x  3 2x  1 8x 2  10x  3 x  A 2x  1  B 4x  3  2A  4B x  A  3B, and so Adding, we get 2B  1  B   12 , and A  32 . Therefore,

24.



2A  4B  1

A  3B  0





3 2

2A  4B  1

2A  6B  0

1 x 2  .  4x  3 2x  1 8x 2  10x  3

A B C 7x  3 7x  3    . Hence,  x x  3 x  1 x x 3 x 1 x 3  2x 2  3x       7x  3  A x  3 x  1  Bx x  1  C x x  3  A x 2  2x  3  B x 2  x  C x 2  3x

Thus 

   

 A  B  C x 2  2A  B  3C x  3A AB C

0

2A  B  3C  7    3A  3

1 B C 0

2  B  3C  7

Coefficients of x 2 Coefficients of x Constant terms

So 3A  3  A  1. Substituting, the system reduces to

Adding these two equations, we get 3  4C  7  C  1. Thus 1  B  1  0  B  2.

7x  3 2 1 1 Therefore, 3  .   x x 3 x 1 x  2x 2  3x 25.

B C A 9x 2  9x  6 9x 2  9x  6   . Thus,   x  2 x  2 2x  1 x  2 x  2 2x  1 2x 3  x 2  8x  4 9x 2  9x  6  A x  2 2x  1  B x  2 2x  1  C x  2 x  2        A 2x 2  3x  2  B 2x 2  5x  2  C x 2  4  2A  2B  C x 2  3A  5B x  2A  2B  4C

SECTION 10.3 Partial Fractions

681

    Coefficients of x 2    2A  2B  C  9  2A  2B  C  9 This leads to the system 16B  3C  45  3A  5B  9 Coefficients of x        2A  2B  4C  6 4B  3C  15 Constant terms   9   2A  2B  C  Hence, 15C  15  C  1; 16B  3  45  B  3; and 2A  6  1  9  A  2. 16B  3C  45    15C  15 Therefore,

26.

9x 2  9x  6

2x 3  x 2  8x  4



3 1 2   . x  2 x  2 2x  1

B C 3x 2  3x  27 A 3x 2  3x  27      . Thus,  2 x  2 2x  3 x  3 x  2 2x  3 x  3 x  2 2x  3x  9

3x 2  3x  27  A 2x  3 x  3  B x  2 x  3  C x  2 2x  3        A 2x 2  3x  9  B x 2  5x  6  C 2x 2  x  6

    2A  B  2C  3 So 3A  5B  C  3    9A  6B  6C  27

 2A  B  2C x 2  3A  5B  C x  9A  6B  6C     Coefficients of x 2    2A  B  2C  3  2A  B  2C  3 7B  4C  3  7B  6C  3 Coefficients of x        21B  6C  27 24C  24 Constant terms

Hence, 24C  24  C  1; then 7B  4  3  B  1; and 2A  1  2  3  A  3. Therefore,

27.

1 1 3 3x 2  3x  27     . 2 x  2 2x  3 x  3 x  2 2x  3x  9

x2  1 B x2  1 C A . Hence,    2  x x 1 x3  x2 x 2 x  1 x

x 2  1  Ax x  1  B x  1  C x 2  A  C x 2  A  B x  B, and so B  1; A  1  0  A  1; and

1 x2  1 1 2  2  . 1  C  1  C  2. Therefore, 3  2 x x  1 x x x 28.

3x 2  5x  13 B C A 3x 2  5x  13       . Thus, 2 2 3x  2 x  2 3x  2 x  4x  4 3x  2 x  2 x  22

3x 2  5x  13  A x  22  B 3x  2 x  2  C 3x  2      A x 2  4x  4  B 3x 2  4x  4  C 3x  2

 A  3B x 2  4A  4B  3C x  4A  4B  2C     A  3B  3 Coefficients of x 2     A  3B This leads to the following system: 4A  4B  3C  5 Coefficients of x  8B      4A  4B  2C  13  Constant terms 8B    3   A  3B   Hence, 8C  9  C  98 ; 8B  3 98  8B  27  8B  3C  17 8  17  B     8C  9



3

 3C  17

 5C  8

109 ; and 64

135 109 9   2 64 64  8 327  3  A   135 . Therefore, 3x  5x  13  A  3 109  A   . 64 64 64 3x  2 x  2 x  22 3x  2 x  22

682

29.

CHAPTER 10 Systems of Equations and Inequalities

B A  . Hence, 2x  A 2x  3  B  2Ax  3A  B. So 2A  2  2x  3 2x  32 2x 3 1 A  1; and 3 1  B  0  B  3. Therefore, 2 .   2x  3 2x  32 4x  12x  9 2x

4x 2  12x  9



2x

2x  32



B A   . Hence, x  4  A 2x  5  B  2Ax  5A  B, and so 30. 2 2x  5 2x  5 2x  52 x 4

  A  12 and 5 12  B  4  B   32 . Therefore,

31.

x 4

1

3



2A



1

5A  B  4



2 2   . 2x  5 2x  52 2x  52

B 4x 2  x  2 C D A 4x 2  x  2 . Hence,  3   2  3  4 3 x x 2 x  2x x x  2 x x 4x 2  x  2  Ax 2 x  2  Bx x  2  C x  2  Dx 3  A  D x 3  2A  B x 2  2B  C x  2C So 2C  2  C  1; 2B  1  1  B  0; 2A  0  4  A  2; and 2  D  0  D  2. Therefore, 1 2 2 4x 2  x  2 .   3  4 3 x x  2 x  2x x

32.

x 3  2x 2  4x  3 B A C D   2  3  4 . Hence, x 3  2x 2  4x  3  Ax 3  Bx 2  C x  D. Thus A  1; B  2; 4 x x x x x C  4; and D  3. Therefore,

33.

10x 2  27x  14 x  13 x  2



2 1 4 3 x 3  2x 2  4x  3   2  3  4. x x4 x x x

B C D A    . Thus, x  2 x  1 x  12 x  13

10x 2  27x  14  A x  13  B x  2 x  12  C x  2 x  1  D x  2

       A x 3  3x 2  3x  1  B x  2 x 2  2x  1  C x 2  x  2  D x  2        A x 3  3x 2  3x  1  B x 3  3x  2  C x 2  x  2  D x  2

 A  B x 3  3A  C x 2  3A  3B  C  D x  A  2B  2C  2D

SECTION 10.3 Partial Fractions

683

which leads to the system   A B     3A  C  3A  3B  C  D     A  2B  2C  2D

  A B     3B  C  3C  D     3C  8D

0



 10 27



 14

  A B  0     2 3B  C  10 Coefficients of x    3B  2C  D  17 Coefficients of x     3B  5C  7D  15 Constant terms Coefficients of x 3

  A B      10 3B  C    7 3C  D      2 9D 

0



0

 10  

7 9

Hence, 9D  9  D  1, 3C  1  7  C  2, 3B  2  10  B  4, and A  4  0  A  4. Therefore, 10x 2  27x  14 x  13 x  2

34.



4 1 4 2  .   x  2 x  1 x  12 x  13

2x 2  5x  1 B C 2x 2  5x  1 2x 2  5x  1 D A        . Thus,  x  1 x  1 x  12 x 4  2x 3  2x  1 x  1 x 3  x 2  x  1 x  13 x  1 x  13 2x 2  5x  1  A x  13  B x  1 x  12  C x  1 x  1  D x  1        A x 3  3x 2  3x  1  B x  1 x 2  2x  1  C x 2  1  D x  1        A x 3  3x 2  3x  1  B x 3  x 2  x  1  C x 2  1  D x  1

 A  B x 3  3A  B  C x 2  3A  B  D x  A  B  C  D

 AB  0     3ABC 2 which leads to the system  3AB D  5     ABCD 1     A B A B  0  0        2B C  2B C 2 2    2C D  1 2C D  1         2C5D  5 6D  6

  A B     2B C Coefficients of x 2   2B C D Coefficients of x     2B3C4D Constant terms Coefficients of x 3

 0 2  3



 2

Hence, 6D  6  D  1, 2C  1  1  C  0, 2B  0  2

2x 2  5x  1 1 1 1  B  1, and A  1  0  A  1. Therefore, 4   .  x  1 x  1 x  13 x  2x 3  2x  1

684

35.

CHAPTER 10 Systems of Equations and Inequalities

3x 3  22x 2  53x  41 x  22 x  32



B D A C    . Thus, x  2 x  22 x  3 x  32

3x 3  22x 2  53x  41  A x  2 x  32  B x  32  C x  22 x  3  D x  22      A x 3  8x 2  21x  18  B x 2  6x  9

    C x 3  7x 2  16x  12  D x 2  4x  4

 A  C x 3  8A  B  7C  D x 2

 21A  6B  16C  4D x  18A  9B  12C  4D     A  C  3 A  C  3 Coefficients of x 3         8A  B  7C  D  22 2 B  C  D  2 Coefficients of x  so we must solve the system   6B  5C  4D  10 21A  6B  16C  4D  53 Coefficients of x         9B  6C  4D  13 18A  9B  12C  4D  41 Constant terms     A  C  3 A C  3         B  C  D  2 B  C  D  2 Hence, D  1, C  2  2  C  0, B  0  1  2     C  2D  2 C  2D  2         3C  5D  5 D  1  B  1, and A  0  3  A  3. Therefore,

36.

3x 3  22x 2  53x  41 x  22 x  32



1 3 1   . x  2 x  22 x  32

B D 3x 2  12x  20 A C 3x 2  12x  20 3x 2  12x  20      . Thus,    2 2 x  22 2 x  2 x  2 2 x 4  8x 2  16  2  2  22 x x x x 4

3x 2  12x  20  A x  2 x  22  B x  22  C x  22 x  2  D x  22          A x 3  2x 2  4x  8  B x 2  4x  4  C x 3  2x 2  4x  8  D x 2  4x  4

 A  C x 3  2A  B  2C  D x 2  4A  4B  4C  4D x  8A  4B  8C  4D     A  C  0 Coefficients of x 3 A  C  0         2A  B  2C  D  2 B  4C  D  3 3 Coefficients of x  which leads to the system   4A  4B  4C  4D  12 6B  8C  2D  6 Coefficients of x         4B  16C  12D  4 8A  4B  8C  4D  20 Constant terms     A  C  0 A  C  0         B  4C  D  3 B  4C  D  3  Hence, 48D  48  D  1,    16C  8D  24 16C  8D  24         32C  32D  0 48D  48

16C  8  24  C  1; B  4  1  3  B  2, and A  1  0  A  1. Therefore, 3x 2  12x  20 2 1 1 1    .  x  2 x  22 x  2 x  22 x 4  8x 2  16

37.

  Bx  C x 3 A x 3    2   2 . Hence, x  3  A x 2  3  Bx 2  C x  A  B x 2  C x  3A. So 3 x x  3x x x 3 x 3 x 3 x 1 1 3A  3  A  1; C  1; and 1  B  0  B  1. Therefore, 3   2 . x x  3x x 3

SECTION 10.3 Partial Fractions

38.

685

3x 2  2x  8 3x 2  2x  8 Ax  B C   . Thus,   2  x 1 x 2  2 x  1 x 3  x 2  2x  2 x 2   3x 2  2x  8  Ax  B x  1  C x 2  2  A  C x 2  A  B x  B  2C, which leads to the system   2   C3 Coefficients of x 2 A  C  3 Coefficients of x   A   A  B  2 Coefficients of x  B C1 Coefficients of x       B  2C  8 Constant terms 2C  6 Constant terms Hence, 2C  6  C  3, B  3  1  B  2; and A  3  3  A  0. Therefore, 3x 2  2x  8 2 3 .  2  x 3  x 2  2x  2 x 2 x 1

Ax  B Cx  D 2x 3  7x  5    2 . Thus, 39.  2 2 2 x x 2 x 1 x x 2 x 1     2x 3  7x  5  Ax  B x 2  1  C x  D x 2  x  2

 Ax 3  Ax  Bx 2  B  C x 3  C x 2  2C x  Dx 2  Dx  2D

 A  C x 3  B  C  D x 2  A  2C  D x  B  2D

We must solve the system  A C       B C  D  A  2C  D      B  2D 

2 0 7 5

    A C  A C  2         2 BC  D BC D 0 Coefficients of x     C  D C  D  5 Coefficients of x         2D  C  D  5 Constant terms Coefficients of x 3

2 0 5 10

Hence, 2D  10  D  5, C  5  5  C  0, B  0  5  0  B  5, and A  0  2  A  2. Therefore, 

40.

2x  5 5 2x 3  7x  5    . x2  x  2 x2  1 x2  x  2 x2  1

x2  x  1 Ax  B Cx  D x2  x  1      2 . Thus, 2x 4  3x 2  1 2x 2  1 x 1 2x 2  1 x 2  1     x 2  x  1  Ax  B x 2  1  C x  D 2x 2  1  Ax 3  Ax  Bx 2  B  2C x 3  2Dx 2  C x  D  A  2C x 3  B  2D x 2  A  C x  B  D

which leads to the system   A  2C 0     B  2D  1   C 1 A    B  D1

  A  2C     2 B  2D Coefficients of x   C Coefficients of x     D Constant terms Coefficients of x 3

 

0 1

 1 

0

Hence, D  0, C  1, B  0  1  B  1, and A  2  0  A  2. Therefore,

x2  x  1 2x  1 x  2  2 . 4 2 2x  3x  1 2x  1 x  1

686

41.

CHAPTER 10 Systems of Equations and Inequalities

Bx  C x4  x3  x2  x  1 A Dx  E   2   2 2 . Hence, x 2 x  1 x2  1 x x 1

 2   x 4  x 3  x 2  x  1  A x 2  1  Bx  C x x 2  1  x Dx  E       A x 4  2x 2  1  Bx 2  C x x 2  1  Dx 2  E x    A x 4  2x 2  1  Bx 4  Bx 2  C x 3  C x  Dx 2  E x  A  B x 4  C x 3  2A  B  D x 2  C  E x  A

So A  1, 1  B  1  B  0; C  1; 2  0  D  1  D  1; and 1  E  1  E  2. Therefore, x4  x3  x2  x  1 1 1 x 2   2   2 2 . x x 1 x2  1 x x2  1

42.

Ax  B Cx  D 2x 2  x  8   2  2 2 . Thus, x 4 x2  4 x2  4   2x 2  x  8  Ax  B x 2  4  C x  D  Ax 3  4Ax  Bx 2  4B  C x  D  Ax 3  Bx 2  4A  C x  4B  D

2x 2  x  8 1 2 and so A  0, B  2, 0  C  1  C  1, and 8  D  8  D  0. Therefore,   2  2 2 . x 4 x2  4 x2  4

43. We must first get a proper rational function. Using long division, we find that

x 5  2x 4  x 3  x  5  x2  x 3  2x 2  x  2

2 2x 2  x  5 2  2x  x  5   x 2  A  Bx  C . Hence,  x x 2 x 3  2x 2  x  2 x2  1 x  2 x 2  1   2x 2  x  5  A x 2  1  Bx  C x  2  Ax 2  A  Bx 2  C x  2Bx  2C

 A  B x 2  C  2B x  A  2C

Equating coefficients, we get the system       2 Coefficients of x 2  2  2    A  B A  B A  B 2B  C  1 Coefficients of x  2B  C  1  2B  C  1       A    2C  5 Constant terms B  2C  3 5C  5

Therefore, 5C  5  C  1, 2B  1  1  B  1, and A  1  2  A  3, so x 1 3 x 5  2x 4  x 3  x  5  2  x2  . 3 2 x 2 x 1 x  2x  x  2

SECTION 10.3 Partial Fractions

44.

687

x 5  3x 4  3x 3  4x 2  4x  12 x 5  3x 4  3x 3  4x 2  4x  12 . We use long division to get a proper rational    x 4  4x 3  6x 2  8x  8 x  22 x 2  2

function:

x 4  4x 3  6x 2  8x  8

x 

1

x 5  3x 4  3x 3  4x 2  4x  12 x 5  4x 4  6x 3  8x 2  8x x 4  3x 3  4x 2  4x  12 x 4  4x 3  6x 2  8x  8 x 3  2x 2  4x  4

B x 5  3x 4  3x 3  4x 2  4x  12 Cx  D x 3  2x 2  4x  4 A  , so  2  x  1       x 1 2 2 2 2 2 x  2 x 2 x  2 x  2 x  2 x  2 x  2     x 3  2x 2  4x  4  A x  2 x 2  2  B x 2  2  C x  D x  22

Thus,

       A x 3  2x 2  2x  4  B x 2  2  C x  D x 2  4x  4

 Ax 3  2Ax 2  2Ax  4A  Bx 2  2B  C x 3  4C x 2  4C x  Dx 2  4Dx  4D

 A  C x 3  2A  B  4C  D x 2  2A  4C  4D x  4A  2B  4D   A  C  1 Coefficients of x 3     2A  B  4C  D  2 Coefficients of x 2  which leads to the system  2A  4C  4D  4 Coefficients of x     4A  2B  4D  4 Constant terms     A  C  1 A  C 1         B  2C  D  0 2  Eq. 1  Eq. 2 B  2C  D  0     2C  4D  2 B  3D  2 Eq. 2  Eq. 3 Eq. 2  Eq. 3         2B  8C  4D  12 2  Eq. 3  Eq. 4 8C  2D  8 2  Eq. 3  Eq. 4   A  C 1     B  2C  D  0 Then D  0; 2C  2  C  1; B  2  0  B  2; and A  1  1  2C  4D  2     18D  0 4  Eq. 3  Eq. 4  A  0. Therefore,

45.

x x 5  3x 4  3x 3  4x 2  4x  12 2  2 .  x 1   2 2 2 x 2 x  2 x  2 x  2

A B ax  b   . Hence, ax  b  A x  1  B x  1  A  B x  A  B. x 1 x 1 x2  1  AB a ab . So Adding, we get 2A  a  b  A  2 AB b Substituting, we get B  a  A 

ab ab ab ab 2a   . Therefore, A  and B  . 2 2 2 2 2

688

CHAPTER 10 Systems of Equations and Inequalities

ax 3  bx 2 Ax  B Cx  D 46.   2  2 2 . Hence, 2 x  1 x 1 x2  1   ax 3  bx 2  Ax  B x 2  1  C x  D  Ax 3  Ax  Bx 2  B  C x  D  Ax 3  Bx 2  A  C x  B  D

and so A  a, B  b, a  C  0  C  a, and b  D  0  D  b. Therefore, A  a, B  b, C  a, and D  b. 1 x is already a partial fraction decomposition. The denominator in the first term is a  47. (a) The expression 2 x 1 x 1 quadratic which cannot be factored and the degree of the numerator is less than 2. The denominator of the second term is linear and the numerator is a constant. x can be decomposed further, since the numerator and denominator both have linear factors. (b) The term x  12 B A x   . Hence, x  A x  1  B  Ax  A  B. So A  1, B  1, and x 1 x  12 x  12 x 1 1   . x  1 x  12 x  12

2 1  is already a partial fraction decomposition, since each numerator is constant. x  1 x  12 x 2 (d) The expression  2 is already a partial fraction decomposition, since the denominator is the square of a quadratic x2  1 which cannot be factored,and the degree of the numerator is less than 2. (c) The expression

48. Combining the terms, we have     2 x2  1 1 x 2  2x  1 1 1 1 x  1 2      x  1 x  12 x 1 x  12 x  1 x  12 x  1 x  12 x  1 

2x 2  2  x  1  x 2  2x  1 x  12 x  1 3x 2  x



3x 2  x

x  12 x  1

B C A  , and so  x  1 x  12 x 1     3x 2  x  A x  1 x  1  B x  1  C x  12  A x 2  1  B x  1  C x 2  2x  1

Now to find the partial fraction decomposition, we have

 Ax 2  A  Bx  B  C x 2  2C x    C 3   A which result in the system B  2C  1    A  B  C  0

   A   

 C

3

4C 

4

x  12 x  1



 C  A  C x 2  B  2C x  A  B  C   Coefficients of x 2  C 3  A  Coefficients of x  B  2C  1    Constant terms B  2C  3 Eq. 1  Eq. 3

so 4C  4  C  1, B  2 1  1  B  1, and A  1  3  A  2.

B  2C  1

1  Eq. 2  Eq. 3

Therefore, we get back the same expression:

3x 2  x

x  12 x  1



1 1 2  .  2 x  1 x  1 x 1

SECTION 10.4 Systems of Nonlinear Equations

689

10.4 SYSTEMS OF NONLINEAR EQUATIONS 1. The solutions of the system are the points of intersection of the two graphs, namely 2 2 and 4 8. 2. For 2 2: 2y  x 2  2 2  22  0 and y  x  2  2  4, so 2 2 is a solution.

For 4 8: 2y  x 2  2 8  42  0 and y  x  8  4  4, so 4 8 is a solution.  y  x2 Substituting y  x 2 into the second equation gives x 2  x  12  3. y  x  12

0  x 2  x  12  x  4 x  3  x  4 or x  3. So since y  x 2 , the solutions are 3 9 and 4 16.   x 2  y 2  25 Substituting for y in the first equation gives x 2  2x2  25  5x 2  25  x 2  5  x   5 4. y  2x           5 2 5 and When x  5 then y  2 5, and when x   5 then y  2  5  2 5. Thus the solutions are      5 2 5 .

5.



x 2  y2  8 x  y0

Solving the second equation for y gives y  x, and substituting this into the first equation gives

x 2  x2  8  2x 2  8  x  2. So since y  x, the solutions are 2 2 and 2 2.  x2  y  9 Solving the first equation for y, we get y  9  x 2 . Substituting this into the second equation gives 6. x  y 3  0   x  9  x 2  3  0  x 2  x  6  0  x  3 x  2  0  x  3 or x  2. If x  3, then y  9  32  0,

and if x  2, then y  9  22  5. Thus the solutions are 3 0 and 2 5.  x  y2  0 7. Solving the first equation for x gives x  y 2 , and substituting this into the second equation gives 2x  5y 2  75   2 y 2  5y 2  75  3y 2  75  y 2  25  y  5. So since x  y 2 , the solutions are 25 5 and 25 5. 8.



x2  y  1

Solving the first equation for y, we get y  x 2  1. Substituting this into the second equation gives 2x 2  3y  17   2x 2  3 x 2  1  17  2x 2  3x 2  3  17  5x 2  20  x 2  4  x  2. If x  2, then y  22  1  3,

and if x  2, then y  22  1  3. Thus the solutions are 2 3 and 2 3.  x 2  2y  1 9. Subtracting the first equation from the second equation gives 7y  28  y  4. Substituting y  4 into x 2  5y  29

the first equation of the original system gives x 2  2 4  1  x 2  9  x  3. The solutions are 3 4 and 3 4.  3x 2  4y  17 10. Multiplying the first equation by 2 and the second by 3 gives the system 2x 2  5y  2  6x 2  8y  34 Adding we get 7y  28  y  4. Substituting this value into the second equation gives 6x 2  15y  6       2x 2  5 4  2  2x 2  22  x 2  11  x   11. Thus the solutions are 11 4 and  11 4 .

690

11.

CHAPTER 10 Systems of Equations and Inequalities



3x 2  y 2  11

Multiplying the first equation by 4 gives the system

x 2  4y 2  8



12x 2  4y 2  44 x 2  4y 2  8

Adding the equations

gives 13x 2  52  x  2. Substituting into the first equation we get 3 4  y 2  11  y  1. Thus, the solutions are 2 1, 2 1, 2 1, and 2 1.

12.





2x 2  4y  13

Multiplying the second equation by 2 gives the system

x 2  y 2  72

2x 2  4y  13

2x 2  2y 2  7

Subtracting the

equations gives 4y  2y 2  6  y 2  2y  3  0  y  3 y  1  0  y  3, y  1. If y  3, then   5 2 2x 2  4 3  13  x 2  25 . If y  1, then 2x 2  4 1  13  x 2  92  x   3 2 2 . Hence, the  x   2 2       5 2 3 2 solutions are  2  3 and  2  1 .

13.



14.



x  y2  3  0

Adding the two equations gives 2x 2  x  1  0. Using the Quadratic Formula we have 2x 2  y 2  4  0   1  1  4 2 1 1  9 1  3 1  3 1  3 x   . So x   1 or x   12 . Substituting x  1 2 2 4 4 4 4  into the first equation gives 1  y 2  3  0  y 2  2  y   2. Substituting x  12 into the first equation gives        1  y 2  3  0  y 2  7  y   7 . Thus the solutions are 1  2 and 1   7 . 2 2 2 2 2 x 2  y2  1

2x 2  y 2  x  3

Subtracting the first equation from the second equation gives x 2  x  2 

x 2  x  2  0  x  2 x  1  0  x  2, x  1. Solving the first equation for y 2 we have y 2  x 2  1. When  x  1, y 2  12  1  0 so y  0 and when x  2, y 2  22  1  3 so y   3. Thus, the solutions are 1 0,       2  3 , and 2 3 .

15.



x2  y 

8

x  2y  6

By inspection of the graph, it appears that 2 4 is a solution, but is difficult to get accurate values

for the other point. Multiplying the first equation by 2 gives the system



2x 2  2y  16 x  2y  6

Adding the equations gives

2x 2  x  10  2x 2  x  10  0  2x  5 x  2  0. So x   52 or x  2. If x   52 , then  52  2y  6    2y   72  y  74 , and if x  2, then 2  2y  6  2y  8  y  4. Hence, the solutions are  52  74 and

2 4.

16.



x  y 2  4

x  y

2

By inspection of the graph, it appears that 0 2 and 5 3 are solutions to the system. We check

each point in both equations to verify that it is a solution. For 0 2: 0  22  4 and 0  2  2. For 5 3: 5  32  5  9  4 and 5  3  2. Thus, the solutions are 0 2 and 5 3.

SECTION 10.4 Systems of Nonlinear Equations

17.



x2 

y0

x 3  2x  y  0

691

By inspection of the graph, it appears that 2 4, 0 0, and 1 1 are solutions to the system.

We check each point in both equations to verify that it is a solution. For 2 4: 22  4  4  4  0 and 23  2 2  4  8  4  4  0.

For 0 0: 02  0  0 and 03  2 0  0  0.

For 1 1: 12  1  1  1  0 and 13  2 1  1  1  2  1  0. Thus, the solutions are 2 4, 0 0, and 1 1.  x 2  y 2  4x 18. By inspection of the graph, it appears that 0 0 is a solution, but is difficult to get accurate values for x  y2

the other points. Substituting for y 2 we have x 2  x  4x  x 2  3x  0  x x  3  0. So x  0 or x  3. If x  0,        then y 2  0 so y  0. And is x  3 then y 2  3 so y   3. Hence, the solutions are 0 0, 3  3 , and 3 3 .  y  x 2  4x Subtracting the second equation from the first equation gives x 2  4x  4x  16  19. y  4x  16

x 2  8x  16  0  x  42  0  x  4. Substituting this value for x into either of the original equations gives y  0. Therefore, the solution is 4 0.   x  y2  0 x  y2 20. Substituting for y in the Solving the first equation for x and the second equation for y gives 2 yx  0 y  x2   first equation gives x  x 4  x x 3  1  0  x  0, x  1. Thus, the solutions are 0 0 and 1 1.  x  2y  2 21. Now x  2y  2  x  2y  2. Substituting for x gives y 2  x 2  2x  4  y 2  x 2  2x  4

y 2  2y  22  2 2y  2  4  y 2  4y 2  8y  4  4y  4  4  y 2  4y  4  0  y  22  0  y  2. Since x  2y  2, we have x  2 2  2  2. Thus, the solution is 2 2.  y  4  x2 22. Setting the two equations equal, we get 4  x 2  x 2  4  2x 2  8  x  2. Therefore, the solutions y  x2  4 are 2 0 and 2 0.  xy 4 23. Now x  y  4  x  4  y. Substituting for x gives x y  12  4  y y  12  y 2  4y  12  0 x y  12  y  6 y  2  0  y  6, y  2. Since x  4  y, the solutions are 2 6 and 6 2.  x y  24 24 Since x  0 is not a solution, from the first equation we get y  24. . Substituting into the 2 2 x 2x  y  4  0  2    24  2x 4  4x 2  576  x 4  2x 2  288  0  x 2  18 x 2  16  0. second equation, we get 2x 2  4  x

Since x 2  18 cannot be 0 if x is real, we have x 2  16  0  x  4. When x  4, we have y  24 4  6 and when 24  6. Thus the solutions are 4 6 and 4 6. x  4, we have y  4  x 2 y  16 16 16 . Substituting for x 2 gives  4y  16  0  4y 2  16y  16  0 25. Now x 2 y  16  x 2  y y x 2  4y  16  0  y 2  4y  4  0  y  22  0  y  2. Therefore, x 2  has no solution.

16  8, which has no real solution, and so the system 2

692

26.

27.

CHAPTER 10 Systems of Equations and Inequalities



x



y 0  Solving the first equation for x, we get x   y. Substituting for x gives y 2  4x 2  12   2  y 2  4  y  12  y 2  4y  12  0  y  6 y  2  0  y  6, y  2. Since x   2 is not a real    solution, the only solution is  6 6 . 

x 2  y2  9

x 2  y2  1

  Adding the equations gives 2x 2  10  x 2  5  x   5. Now x   5  y 2  9  5  4 

y  2, and so the solutions are

28.



x 2  2y 2  2

          5 2 , 5 2 ,  5 2 , and  5 2 .

Multiplying the first equation by 2 gives the system

2x 2  3y  15

equations gives 4y 2  3y  11  4y 2  3y  11  0  y 

3 



2x 2  4y 2  4

2x 2  3y  15

Subtracting the two

 9  4 4 11 which is not a real number. 2 4

Therefore, there are no real solutions.

29.



2x 2  8y 3  19

4x 2  16y 3  34

Multiplying the first equation by 2 gives the system



4x 2  16y 3  38 4x 2  16y 3  34

Adding the two

equations gives 8x 2  72  x  3, and then substituting into the first equation we have 2 9  8y 3  19      y 3   18  y   12 . Therefore, the solutions are 3  12 and 3  12 . 30.



x 4  y 3  15

3x 4  5y 3  53

Multiplying the first equation by 3 gives the system



3x 4  3y 3  45

3x 4  5y 3  53

Subtracting the equations

gives 8y 3  8  y 3  1  y  1, and then x 4  1  15  x  2. Therefore, the solutions are 2 1 and 2 1.   

2 3  1 x y 31. 4 7     1 x y

1 1 If we let u  and   ,the system is equivalent to x y

equation by 4 gives the system



4u  6  2

4u  7  1



2u  3  1

4u  7  1

Multiplying the first

Adding the equations gives   3, and then substituting into the first

  equation gives 2u  9  1  u  5. Thus, the solution is 15  13 .

 4 6 7    2  4  2 x y 32.  1  2 0   2 x y4

  4u  6  7 1 1 2 , and multiplying the If we let u  2 and   4 , the system is equivalent to  u  2  0 x y

  4u  6  7 2 second equation by 3, gives  3u  6  0

Adding the equations gives 7u  72  u  12 , and   14 . Therefore,

            1 1 2 2 , 2  2 ,  2 2 , x 2   2  x   2, and y 4   4  y   2. Thus, the solutions are   u   and  2  2 .

SECTION 10.4 Systems of Nonlinear Equations

33.



y  x 2  8x

34.

y  2x  16

The solutions are 8 0 and 2 20.



y  x 2  4x

2x  y  2





y  x 2  4x

y  2x  2

The solutions are approximately 035 130 and 565 930.

20 10 -10

10 -20

35.



x 2  y 2  25





5

 y   25  x 2

y   13 x  23 The solutions are 451 217 and 491 097. x  3y  2

36.



x 2  y 2  17

x 2  2x  y 2  13





 y   17  x 2  y   13  2x  x 2

The solutions are approximately 2 361.

5

5

-5

5

-5

-5

5 -5

   2  y2 x y   18  2x 2  1 37.  9 18  y  x 2  6x  2  y  x 2  6x  2

38.



x 2  y2  3

y  x 2  2x  8





 y   x2  3

y  x 2  2x  8

The solutions are approximately 222 140,

The solutions are 123 387 and 035 421.

188 072, 345 299, and 465 431.

5

5

-5

5 -5

5

-5 -5

39.



  

 4 32  x 4 y  2  x 2  2x  y  0  y  x 2  2x x 4  16y 4  32

The solutions are 230 070 and 048 119.

40.



y  e x  ex y  5  x2

119 359 and 119 359. 5

2

-2

The solution are approximately

2 -2

-2

2

693

694

CHAPTER 10 Systems of Equations and Inequalities

  log x  log y  3 2 41.  2 log x  log y  0

 Adding the two equations gives 3 log x  32  log x  12  x  10. Substituting into the

second equation we get 2 log 1012  log y  0  log 10  log y  0  log y  1  y  10. Thus, the solution is   10 10 .

42.



2x  2 y  10

4x  4 y  68





2x  2 y  10

22x  22y  68

If we let u  2x and   2 y , the system becomes



u    10

u 2   2  68

Solving the first equation for u, and substituting this into the second equation gives u    10  u  10  , so 10  2   2  68  100  20   2   2  68   2  10  16  0    8   2  0    2 or   8. If   2, then u  8, and so y  1 and x  3. If   8, then u  2, and so y  3 and x  1. Thus, the solutions are 1 3 and 3 1.  xy 3 43. Solving the first equation for x gives x  3  y and using the hint, x 3  y 3  387  3 x  y 3  387     x  y x 2  x y  y 2  387. Next, substituting for x, we get 3 3  y2  y 3  y  y 2  387  9  6y  y 2  3y  y 2  y 2  129  3y 2  9y  9  129  y  8 y  5  0  y  8 or y  5. If y  8, then

x  3  8  5, and if y  5, then x  3  5  8. Thus the solutions are 5 8 and 8 5.  x2  xy  1 44. Adding the equations gives x 2 x yx y y 2  4  x 2 2x y y 2  4  x  y2  4  x  y  2. x y  y2  3 If x  y  2, then from the first equation we get x x  y  1  x  2  1  x  12 , and so y  2  12  32 . If   x  y  2, then from the first equation we get x x  y  1  x  2  1  x   12 , and so y  2   12   32 .     Thus the solutions are 12  32 and  12   32 . 45. Let  and l be the lengths of the sides, in cm. Then we have the system



l  180

2l  2  54

We solve the second equation

for  giving,   27  l, and substitute into the first equation to get l 27  l  180  l 2  27l  180  0 

l  15 l  12  0  l  15 or l  12. If l  15, then   27  15  12, and if l  12, then   27  12  15. Therefore, the dimensions of the rectangle are 12 cm by 15 cm.

46. Let b be the length of the base of the triangle, in feet, and h be the height of the triangle,  1 bh  84  168 2 . By substitution, The first equation gives b  in feet. Then  b2  h 2  252  625 h 

    168 2  h 2  625  h 4  625h 2  1682  0  h 2  49 h 2  576  0  h  7 or h  24. Thus, the lengths of h

the other two sides are 7 ft and 24 ft.

47. Let l and  be the length and width, respectively, of the rectangle. Then, the system of equations is  2l  2  70  Solving the first equation for l, we have l  35  , and substituting into the second gives l 2  2  25  l 2  2  25  l 2  2  625  35  2  2  625  1225  70  2  2  625  22  70  600  0    15   20  0    15 or   20. So the dimensions of the rectangle are 15 and 20.

SECTION 10.4 Systems of Nonlinear Equations

695

48. Let  be the width and l be the length of the rectangle, in inches. From the figure, the diagonals of the rectangle are  l  160 1602 160 simply diameters of the circle. Then, . By substitution, 2  l 2  400    2 2 2 l l   l  20  400        l 4  400l 2  1602  0  l 2  80 l 2  320  0  l  80  4 5 or l  320  8 5. Therefore, the dimensions   of the rectangle are 4 5 in. and 8 5 in..

  y  1x 2 49. At the points where the rocket path and the hillside meet, we have Substituting for y in the second  y  x 2  401x   801  0  x  0, x  801 . When x  0, the rocket has x  0  x x  equation gives 12 x  x 2  401x  x 2  801 2 2 2     801 1 801 801 801 not left the pad. When x  2 , then y  2 2  4 . So the rocket lands at the point 801 2  4 . The distance from     the base of the hill is

801 2  2

801 2  44777 meters. 4

50. Let x be the circumference and y be length of the stove pipe. Using the circumference we can determine  x 2 1 2 x . Thus the volume is  x y. So the system is given by y  the radius, 2r  x  r  2 2 4   x y  1200 1 1 1 2 x y x x y  x 1200  600  Substituting for x y in the second equation gives 1  4 4 4 x 2 y  600 4 1200 600 600 1200   . Thus the dimensions of the sheet metal are 2  63 in and  1910 in. x  2. So y  x 2  

696

CHAPTER 10 Systems of Equations and Inequalities

51. The point P is at an intersection of the circle of radius 26 centered at A 22 32

y

40 and the circle of radius 20 centered at B 28 20. We have the system A  B x  222  y  322  262 20  2 2 2 x  28  y  20  20  0 _20 20 40 x x 2  44x  484  y 2  64y  1024  676  2 2 x  56x  784  y  40y  400  400 _20  x 2  44x  y 2  64y  832 Subtracting the two equations, we get 12x  24y  48  x  2y  4, x 2  56x  y 2  40y  784

which is the equation of a line. Solving for x, we have x  2y  4. Substituting into the first equation gives

2y  42 44 2y  4 y 2 64y  832  4y 2 16y1688y176 y 2 64y  832  5y 2 168y192  832   2 451024  5y 2  168y  1024  0. Using the Quadratic Formula, we have y  168 16825  16810 7744  16888 10

 y  8 or y  2560. Since the y-coordinate of the point P must be less than that of point A, we have y  8. Then x  2 8  4  12. So the coordinates of P are 12 8.

To solve graphically, we must solve each equation for y. This gives x  222  y  322  262    y  322  262  x  222  y  32   676  x  222  y  32  676  x  222 . We use the function  y  32  676  x  222 because the intersection we at interested in is below the point A. Likewise, solving the second  equation for y, we would get the function y  20  400  x  282 . In a three-dimensional situation, you would need a minimum of three satellites, since a point on the earth can be uniquely specified as the intersection of three spheres centered at the satellites.

52. The graphs of y  x 2 and y  x  k for various values of k are shown. If we solve  y  x2 the system we get x 2  x  k  0. Using the Quadratic Formula, y  x k   1  1  4k . So there is no solution if 1  4k is undefined, that we have x  2

y k=2 1

k=_ 4 1

1

x

k=_3

is, if 1  4k  0  k   14 . There is exactly one solution if 1  4k  0  k   14 , and there are two solutions if 1  4k  0  k   14 .

10.5 SYSTEMS OF INEQUALITIES 1. If the point 2 3 is a solution of an inequality in x and y, then the inequality is satisfied when we replace x by 2 and y by 3. Because 4 2  2 3  8  6  2  1, the point 2 3 is a solution of the inequality 4x  2y  1.

SECTION 10.5 Systems of Inequalities

697

y

2. To graph an inequality we first graph the corresponding equation. So to graph y  x  1, we first graph the equation y  x  1. To decide which

side is the graph of the inequality we use test points. Test Point

Inequality y  x  1 ?

0 0

001X

Part of graph

201

Not part of graph

?

0 2

1

x

1

y=x+1

Conclusion

3. If the point 2 3 is a solution of a system of inequalities in x and y, then each inequality is satisfied when we replace x by 2 and y by 3. Because 2 2  4 3  16  17 and 6 2  5 3  27  29, the point 2 3 is a solution of the given system. 4. (a)

y

(b) x-y=0



xy0 xy2

(c)

y



xy0 xy2

xy0 xy2

y

y

x-y=0

1

(d)



x-y=0

x-y=0

x+y=2 x

1

1

1

x+y=2 1

5.

x

1

x+y=2 x

1

x+y=2 1

6. Test Point Inequality x  5y  3 ?

1 2 1  5 2  3 X ?

Conclusion Solution

1 2

1  5 2  3 X

Solution

1 2

1  5 2  3 

Not a solution

8  5 1  3 

Not a solution

8 1

? ?

7. Test Point System

0 0

1 2

1 1

3 1



Test Point Inequality 3x  2y  2 2 1 1 3 1 3 0 1

?

3 2  2 1  2 X ?

3 1  2 3  2  ?

2x  y  3

 ?  3 0  2 0  5X ?  2 0  0  3  ?  3 1  2 2  5X ?  2 1  2  3X  ?  3 1  2 1  5 X ?  2 1  1  3X  ?  3 3  2 1  5  ?  2 3  1  3X

Conclusion

0 0

Solution

1 3

Solution

3 0

Not a solution

1 2

Not a solution Solution

3 0  2 1  2 X

Solution

?

Test Point System

Not a solution

Solution

3 1  2 3  2 X

8. 3x  2y  5

Conclusion

 



x  2y  4

4x  3y  11

Conclusion

?

0  2 0  4 

?  4 0  3 0  11   ?  1  2 3  4 X ?  4 1  3 3  11 X  ?  3  2 0  4  ?  4 3  3 0  11   ?  1  2 2  4 X ?  4 1  3 2  11 

Not a solution

Solution

Not a solution

Not a solution

x

698

CHAPTER 10 Systems of Equations and Inequalities

9. y  2x. The test point

10. y  3x. The test point

1 0 satisfies the

11. y  2. The test point

1 0 satisfies the

inequality.

inequality.

2 0 satisfies the

inequality.

y

y

12. x  1. The test point

0 3 satisfies the

inequality. y

y y=3x

1

1

x

1

x

1

1

x=_1

y=2

1

x

1

x

1

y=_2x

13. x  2. The test point

14. y  1. The test point

0 0 satisfies the

0 2 satisfies the

inequality.

inequality.

x=2

1 1

1

inequality.

1

1

point 0 6 satisfies the

inequality. y

10

x

_x@+y=5 1

21. x 2  y 2  9. The test point 0 4 satisfies the inequality.

1

22. x 2  y  22  4. The test point 0 1 satisfies the inequality.

x@+(y-2)@=4

1

y=x@+1

1

x

y

x@+y@=9 1

point 0 2 satisfies the

y

3x-y-9=0

y

20. y  x 2  1. The test

inequality.

2 2

x

x y=1-x

1

19. x 2  y  5. The test

y

y

1

x

1

x

point 0 0 satisfies the

inequality.

y y=x-3

y=1

18. 3x  y  9  0. The test

point 0 0 satisfies the

inequality.

y

1

17. 2x  y  4. The test

point 0 0 satisfies the

inequality.

1

x

16. y  1  x. The test

point 0 0 satisfies the

y

y

2x-y=_4

15. y  x  3. The test

x 1 1

x

x

SECTION 10.5 Systems of Inequalities

23. 3x  2y  18

699

24. 4x  3y  9 8

4 -2 0

2

4

6

8

4

10 12

-4

-4

-8

-2 0

2

4

6

8

-4

-12

25. 5x  2y  8

26. 5x  3y  15 8

4

4 0 -2

0 -4

2

4

6

5

-4

-8

-8

27. The boundary is a solid curve, so we have the inequality y  12 x  1. We take the test point 0 2 and verify that it satisfies the inequality: 2  12 0  1. 28. The boundary is a solid curve, so we have the inequality y  x 2  2. We take the test point 0 0 and verify that it satisfies the inequality: 0  02  2.

29. The boundary is a broken curve, so we have the inequality x 2  y 2  4. We take the test point 0 4 and verify that it satisfies the inequality: 02  42  4.

30. The boundary is a solid curve, so we have the inequality y  x 3  4x. We take the test point 1 1 and verify that it satisfies the inequality: 1  13  4 1.

31.



xy4 yx

The vertices occur where



xy4 yx

y

Substituting, we have

y=x 1

2x  4  x  2. Since y  x, the vertex is 2 2, and the solution set is not

32.



2x  3y  12



2x  3y  12

3x  y  21

The vertices occur where



2x  3y  12 3x  y  21

x+y=4



and adding the two equations gives 11x  75  x  75 11 .

9x  3y  63   132150   18  y   6 . Therefore, the Then 2 75 11  3y  12  3y  11 11 11   75 6 vertex is 11   11 , and the solution set is not bounded. The test point 0 5 satisfies each inequality.

x

1

bounded. The test point 0 1 satisfies each inequality.

y 2x+3y=12 1 1 3x-y=21

x

700

CHAPTER 10 Systems of Equations and Inequalities

  y  1x  2 4 33.  y  2x  5

  y  1x  2 4 The vertex occurs where  y  2x  5

y

Substituting for y y=41 x+2

gives 14 x  2  2x  5  74 x  7  x  4, so y  3. Hence, the vertex is 4 3,

1

34.



xy0

4  y  2x

The vertices occur where



y x

4  y  2x

y

set is not bounded. The test point 3 0 satisfies each inequality.

One vertex occurs where



y=2x-5

Substituting for y

gives 4  x  2x  x  4, so y  4. Hence, the vertex is 4 4, and the solution

    y  2x  8 35. y   12 x  5    x  0, y  0

x

1

and the solution is not bounded. The test point 0 1 satisfies each inequality.

y  2x  8

y   12 x  5

4+y=2x 2 x-y=0

x

2

y

Substituting for

y=_2x+8 1

y=_ 2 x+5

y gives 2x  8   12 x  5   32 x  3  x  2, so y  2 2  8  4.  y  2x  8 Hence, this vertex is 2 4. Another vertex occurs where  y0

x

1 1

2x  8  0  x  4; this vertex is 4 0. Another occurs where  y   12 x  5  y  5; this gives the vertex 0 5. The origin is another x 0 vertex, and the solution set is bounded. The test point 1 1 satisfies each inequality.     4x  3y  18 36. 2x  y  8    x  0, y  0

One vertex occurs where



4x  3y  18 2x  y  8

y

Subtracting twice 2x+y=8

the second equation from the first gives y  2, so 2x  2  8  x  3 and the

4x+3y=18

1

vertex is 3 2. Other vertices occur at the x-intercept of 2x  y  8 and the

x

1

y-intercept of 4x  3y  18; these are 4 0 and 0 6, respectively. The origin is

another vertex, and the solution set is bounded. The test point 1 1 satisfies each inequality.   x     y 37.   3x  5y    3x  2y

 0  0  15

From the graph, the points 3 0, 0 3 and 0 0 are vertices,

y 3x+2y=9

 9

and the fourth vertex occurs where the lines 3x  5y  15 and 3x  2y  9

intersect. Subtracting these two equations gives 3y  6  y  2, and so x  53 .   Thus, the fourth vertex is 53  2 , and the solution set is bounded. The test point 1 1 satisfies each inequality.

3x+5y=15

1 1

x

SECTION 10.5 Systems of Inequalities

38.

   

y

x 2 y  12

   2x  4y  8

From the graph, the vertices occur at 2 1 and 28 12.

39.

y  9  x2

x  0, y  0

y=12

6

2x-4y=8 x

6

The solution set is not bounded. The test point 6 0 satisfies each inequality.



x=2

y

From the graph, the vertices occur at 0 0, 3 0, and 0 9.

y=9-x@

The solution set is bounded. The test point 1 1 satisfies each inequality. 1

y=0

x

1 x=0

 2   y  x 40. y  4   x  0

y x=0

From the graph, the vertices occur at 0 0, 0 2, and 2 4. The

y=4

solution set is bounded. The test point 1 3 satisfies each inequality. y=x@

1 1

41.



y  9  x2 y  x 3

The vertices occur where



y  9  x2

y  x 3

Substituting for y

y=9-x@

y y=x+3

gives 9  x 2  x  3  x 2  x  6  0  x  2 x  3  0  x  3, x  2. Therefore, the vertices are 3 0 and 2 5, and the solution set is

1

42.

y  x2

xy6

The vertices occur where



y  x2

xy6

x

1

bounded. The test point 0 4 satisfies each inequality.



y

Substituting for y y=x@

gives x 2  x  6  x 2  x  6  0  x  3 x  2  0  x  3, x  2. Since y  x 2 , the vertices are 3 9 and 2 4, and the solution set is not bounded. The test point 0 7 satisfies each inequality.

43.



x 2  y2  4 xy0

The vertices occur where



x 2  y2  4 xy0

x

x+y=6

2 1 y

Since x  y  0

  x  y, substituting for x gives y 2  y 2  4  y 2  2  y   2, and         x   2. Therefore, the vertices are  2  2 and 2 2 , and the

solution set is bounded. The test point 1 0 satisfies each inequality.

x

x-y=0

1 1 x@+y@=4

x

701

702

44.

CHAPTER 10 Systems of Equations and Inequalities

     

y

x 0

y 0  x  y  10     2 x  y2  9

From the graph, the vertices are 0 3, 0 10, 3 0, and

1

10 0. The solution set is bounded. The test point 4 1 satisfies each inequality.

45.

 

x2  y  0

2x 2  y  12 2x 2  2y  0

2x 2  y  12

The vertices occur where



x2  y  0

2x 2  y  12

46.

x2  y  8

x@+y@=9

y



x@-y=0

Subtracting the equations gives 3y  12  y  4, and

The vertices occur where



2x 2  y  4 x2  y  8

y

Adding the

2 4 and 2 4. The solution set is not bounded. The test point 0 6 satisfies each inequality.

47.

x 2  y2  9

2x  y 2  1

The vertices occur where



x 2  y2  9

2x  y 2  1

Subtracting the

equations gives x 2  2x  8  x 2  2x  8  x  2 x  4  0. Therefore,

2 x@-y=8

point 0 0 satisfies each inequality.

48.

x 2  y2  4

x 2  2y  1

The vertices occur where

2x@+y=4

(_2, _4)

1

y

2x+y@=1

x (2, _4)

x@+y@=9

(_2, Ï5)

      the vertices are 2  5 and 2 5 . The solution set is bounded. The test



x

1

equations gives 2x 2  x 2  12  x 2  4  x  2. Therefore, the vertices are



2x@+y=12

1

The test point 0 1 satisfies each inequality.

2x 2  y  4

x

y=0

1

x  2. Thus, the vertices are 2 4 and 2 4, and the solution set is bounded.



x+y=10

x=0

1 x

1

(_2, _Ï5)



x 2  y2  4

x 2  2y  1

equations gives y 2  2y  3  y 2  2y  3  y  3 y  1  0. However,

     y  3 is extraneous, and the vertices are  3 1 and 3 1 . The solution

set is bounded. The test point 0 2 satisfies each inequality.

y

Subtracting the

x@-2y=1

(_Ï3, 1)

(Ï3, 1)

1 1

x x@+y@=4

SECTION 10.5 Systems of Inequalities

    x  2y  14 49. 3x  y  0    x y 2

We find the vertices of the region by solving pairs of the

corresponding equations: 

3x  y  0 xy2







x  2y  14 x y 2

3x  y 

0

2x  y  2





x  2y  14

3y  12

x+2y=14

3x-y=0 1

 y  4 and x  6.

 x  1 and y  3. Therefore, the vertices

are 6 4 and 1 3, and the solution set is not bounded. The test point 0 4 satisfies each inequality.   y  x 6   50. 3x  2y  12 To find where the line y  x  6 intersects the lines    x  2y  2

point 4 2 satisfies each inequality.  x  0, y  0 51. The points of intersection are 0 7, 0 0, 7 0, 5 2, x  5, x  y  7 and 5 0. However, the point 7 0 is not in the solution set. Therefore, the

x-y=2

y 3x+2y=12 y=x+6

52.

x  0, y  0

y  4, 2x  y  8

y x=5 x=0 x+y=7

1

y

2x+y=8 y=4

x=0 1 1

point 1 1 satisfies each inequality.

53.

x  2y  12    x 1 0

x

1

y=0

vertices are 0 4, 2 4, 4 0, and 0 0. The solution set is bounded. The test

y  x 1

y

We find the vertices of the region by solving pairs of the

corresponding equations. Using x  1 and substituting for x in the line y  x  1 gives the point 1 0. Substituting for x in the line x  2y  12 gives    y  x 1 13 the point 1 2 .  x  y  1 and y  1  2y  12  x  2y  12   10 13 3y  13  y  13 3 and x  3 . So the vertices are 1 0, 1 2 , and   10  13 , and none of these vertices is in the solution set. The solution set is 3 3 bounded. The test point 0 2 satisfies each inequality.

x

2

The points of intersection are 0 8, 0 4, 4 0, 2 4,

and 3 2. However, the point 0 8 is not in the solution set. Therefore, the

   

2

x-2y=2

vertices are 0 7, 0 0, 5 0, and 5 2, and the solution set is bounded. The test point 1 1 satisfies each inequality.

x

1

3x  2y  12 and x  2y  2, we substitute for y: 3x  2 x  6  12  x  0 and y  6; x  2 x  6  2  x  14 and y  8. Next, adding the equations 3x  2y  12 and x  2y  2 gives 4x  14  x  72 , so these lines   intersect at the point 72  34 . Since the vertex 14 8 is not part of the solution   set, the vertices are 0 6 and 72  34 , and the solution set is not bounded. The test



y

y=0

x

x+2y=12

x+1=0 y=x+1

1 1

x

703

704

CHAPTER 10 Systems of Equations and Inequalities

    x  y  12 54. y  12 x  6    3x  y  6

y

Graphing these inequalities, we see that there are no points 3x+y=6

that satisfy all three, and hence the solution set is empty.

x+y=12 2

y=21 x-6

55.



x 2  y2  8

x  2, y  0

   The intersection points are 2 2, 2 0, and 2 2 0 .

x@+y@=8

x

2 1 2

y

x=2

1

However, since 2 2 is not part of the solution set, the vertices are 2 2, 2 0,    and 2 2 0 . The solution set is bounded. The test point 21 1 satisfies each

y=0

x

1

inequality.

 2   x  y  0 56. xy6    xy6

x+y=6

y

x@-y=0

Adding the equations x  y  6 and x  y  6 yields 2x  12 2

 x  6. So these curves intersect at the point 6 0. To find where x 2  y  0

1

x-y=6

and x  y  6 intersect, we solve the first for y, giving y  x 2 , and then substitute

x

into the second equation to get x  x 2  6  x 2  x  6  0  x  3 x  2  0  x  3 or x  2. When x  3, we have

y  9, and when x  2, we have y  4, so the points of intersection are 3 9 and 2 4. Substituting y  x 2 into the

equation x  y  6 gives x  x 2  6, which has no solution. Thus the vertices (which are not in the solution set) are 6 0, 3 9, and 2 4. The solution set is not bounded. The test point 1 0 satisfies each inequality.

57.



x 2  y2  9

x  y  0, x  0

x+y=0 y

Substituting x  0 into the equations x 2  y 2  9 and

x  y  0 gives the vertices 0 3 and 0 0. To find the points of intersection

for the equations x 2  y 2  9 and x  y  0, we solve for x  y and substitute

1

x=0 1

x

 into the first equation. This gives y2  y 2  9  y   3 2 2 . The points     x@+y@=9 0 3and 3 2 2   3 2 2 lie away from the solution set, so the vertices are 0 0,     0 3, and  3 2 2  3 2 2 . Note that the vertices are not solutions in this case. The solution set is bounded. The test point

0 1 satisfies each inequality.

SECTION 10.5 Systems of Inequalities

58.

   

y  x3

y  2x  4   x  y  0

y

The curves y  x 3 and x  y  0 intersect when

y=2x+4

x 3  x  0  x  0  y  0. The lines x  y  0 and y  2x  4 intersect

y=x#

1

when x  2x  4  3x  4  x   43  y  43 . To find where y  x 3 and y  2x  4 intersect, we substitute for y and get x 3  2x  4    x 3  2x  4  0  x  2 x 2  2x  2  0  x  2 (the other factor has no

x

1 x+y=0

real solution). When x  2 we have y  8Thus, the vertices of the region are   0 0,  43  43 , and 2 8. The solution set is bounded. The test point 0 2 satisfies each inequality.

    x  2y  14 59. 3x  y  0    x y2

14  x  6x  x  2 and y  6. The lines 3x  y  0 and x  y  2 intersect where 3x  2  x  x  1 and y  3. The lines x  2y  14 and x  y  2 intersect where 14  2y  2  y  y  4 and x  6. Thus, the vertices of the region are 1 3, 2 6, and 6 4. The solution set is bounded. The test point 1 0 satisfies each inequality.

    x  2y  14 60. 3x  y  0    x y2

y

The lines x  2y  14 and 3x  y  0 intersect when

(2, 6) 2 (_1, _3)

(2, 6) 2 (_1, _3)

The lines x  y  12 and y  12 x  6 intersect where

(6, 4)

2

x

y (2, 12)

12  x  12 x  6  x  12 and y  0. The lines y  12 x  6 and y  2x  6 intersect where 12 x  6  2x  6  x  8 and y  10. The lines x  y  12

and y  2x  6 intersect where 12  x  2x  6  x  2 and y  12. The solution set is not bounded. The test point 0 8 satisfies each inequality.

x

y

The vertices are 1 3, 2 6, and 6 4. In this case the

solution set is not bounded. The test point 3 0 satisfies each inequality.

    x  y  12 61. y  12 x  6    y  2x  6

(6, 4)

2

2 2 (_8, _10)

(12, 0)

x

705

706

62.

CHAPTER 10 Systems of Equations and Inequalities

   

y  x 1

x  2y  12

   x 1 0

The lines y  x  1 and x  2y  12 intersect where

(_1, 132 )

y

( 103 , 133 )

10 y  1  12  2y  y  13 3 and x  3 . The lines y  x  1 and x  1  0 intersect at 1 0, and the lines x  2y  12 and x  1  0 intersect where

12  2y  1  y  13 2 and x  1. Thus, the vertices of the region are     10 13 1 0, 1 13 2 , and 3  3 . The solution set is bounded. The test point

1

(_1, 0)

    3x  y  5 30x  10y  50        x  2y  5  10x  20y  50  63.   x  6y  9 10x  60y  90         x  0, y  0 x  0 y  0

y

   y  x 3 65. y  2x  6   y 8

Using a graphing calculator, we find the region shown. The

vertices are 3 0, 1 8, and 11 8.

    x  y  12 66. 2x  y  24    x  y  6

Using the graphing calculator we find the region shown. The

(3, 1)

1

(9, 0)

1 y

  7 The vertices are 1 5, 17 4  4 , and 8 1. The solution set is bounded. The test point 4 3 satisfies each inequality.

(0, 5)

(1, 2)

The vertices are 0 5, 1 2, 3 1, and 9 0. The solution set is not bounded. The test point 1 3 satisfies each inequality.

  x y6     4x  7y  39 64.  x  5y  13     x  0 y  0

x

1

0 2 satisfies each inequality.

1

x

(1, 5)

( 174 , 74 )

(8, 1) x

1

10 8 6 4 2 0 -2 -4

10

10

vertices are 3 9, 6 12, and 12 0.

67.



y  6x  x 2

xy4

0

Using a graphing calculator, we find the region shown. The

vertices are 06 34 and 64 24.

10

10 8 6 4 2 -4 -2 -2 0 2 -4

4

6

8 10

SECTION 10.5 Systems of Inequalities

68.

      

y  x3

2x  y  0

707

12 10 8 6 4 2

Using a graphing calculator, we find the region shown. The

y  2x  6

vertices are 0 0, 22 103 and 15 3. -4

-2

0 -2

2

4

y

69. (a) Let x and y be the numbers of acres of potatoes and corn, respectively.   x  y  500     90x  50y  40,000 A system describing the possibilities is  30x  80y  30,000     x  0 y  0

(0, 375) (200, 300) (375, 125)

100

x ( 4000 9 , 0)

100

(b) Because the point 300 180 lies in the feasible region, the farmer can plant 300 acres of potatoes and 180 acres of corn. (c) Because the point 150 325 lies outside the feasible region, the farmer cannot plant this combination of crops. y

70. (a) Let x and y be the numbers of acres of cauliflower and cabbage, respectively.   x  y  300     70x  35y  17,500 A system describing the possibilities is  25x  55y  12,000     x  0 y  0

(0, 2400 11 ) (150, 150)

(200, 100)

50 50

(250, 0)

(b) Because the point 155 115 lies in the feasible region, the farmer can plant 155 acres of cauliflower and 115 acres of cabbage. (c) Because the point 115 175 lies outside the feasible region, the farmer cannot plant this combination of crops. 71. Let x be the number of fiction books published in a year and y the number of nonfiction books. Then the following system of inequalities holds:     x  0, y  0 From the graph, we see that the vertices are 50 50, 80 20 x  y  100    y  20, x  y and 20 20.

72. Let x be the number of chairs made and y the number of tables. Then the following     2x  3y  12 system of inequalities holds: 2x  y  8 The intersection points are 0 4,    x  0, y  0 0 8, 6 0, 4 0, 0 0, and 3 2. Since the points 0 8 and 6 0 are not in the solution set, the vertices are 0 4, 0 0, 4 0, and 3 2.

y

120 100 80

x+y=100 x=y

60 40

y=20

20

x

20 40 60 80 100

y 2x+y=8

2x+3y=12

1 1

x

x

708

CHAPTER 10 Systems of Equations and Inequalities y 200

73. Let x be the number of Standard Blend packages and y be the number of Deluxe Blend packages. Since there are 16 ounces per pound, we get the following system

3 x+38 y=90 4

of inequalities:      

1 x+58 y=80 4

100

x 0

y 0 1 x  5 y  80   4 8   3 3 y  90 x  4 8

200 x

100

From the graph, we see that the vertices are 0 0, 120 0, 70 100 and 0 128. y

74. Let x be the amount of fish and y the amount of beef (in ounces) in each can. Then the following system of inequalities holds:     12x  6y  60 3x  9y  45    x  0, y  0

12x+6y=60 3x+9y=45

2

x

2

From the graph, the vertices are 15 0, 3 4, and 0 10.

75. x  2y  4, x  y  1, x  3y  9, x  3. Method 1: We shade the solution to each inequality with lines perpendicular to the boundary. As you can see, as the number of inequalities in the system increases, it gets harder to locate the region where all of the shaded parts overlap. Method 2: Here, if a region is shaded then it fails to satisfy at least one inequality. As a result, the region that is left unshaded satisfies each inequality, and is the solution to the system of inequalities. In this case, this method makes it easier to identify the solution set.   To finish, we find the vertices of the solution set. The line x  3 intersects the line x  2y  4 at 3 12 and the line

x  3y  9 at 3 2. To find where the lines x  y  1 and x  2y  4 intersect, we add the two equations, which gives

3y  5  y  53 , and x  23 . To find where the lines x  y  1 and x  3y  9 intersect, we add the two equations,       5 3 1 2 5 3 5 which gives 4y  10  y  10 4  2 , and x  2 . The vertices are 3 2 , 3 2, 3  3 , and 2  2 , and the solution set is bounded.

_x+y=1

y

1

x+3y=9 x

1 x=3

Method 1

Method 2

Solution Set

x+2y=4

CHAPTER 10

Review

CHAPTER 10 REVIEW   3x  y  5 1.  2x  y  5

y

Adding, we get 5x  10  x  2. So 2 2  y  5  y  1. 1

Thus, the solution is 2 1.

  y  2x  6 2.  y  x  3

x

1

x

y

Subtracting the second equation from the first, we get 1

0  3x  3  x  1. So y   1  3  4. Thus, the solution is 1 4.

   2x  7y  28  2x  7y  28 3.    2x  7y  28 y  27 x  4

y

Since these equations represent the

2

same line, any point on this line will satisfy the system. Thus the solution are   x 27 x  4 , where x is any real number.

  6x  8y  15 6x  8y  15 4.   3 x  2y  4 6x  8y  16 2

2

x

y

Adding gives 0  1 which is

false. Hence, there is no solution. The lines are parallel.

    2x  y  1 5. x  3y  10    3x  4y  15

1

1 1

x

1

x

y

Solving the first equation for y, we get y  2x  1.

Substituting into the second equation gives x  3 2x  1  10  5x  7      7 12 x   75 . So y    75  1  12 5 Checking the point  5  5 in the third     ? 21 48 equation we have 3  75  4 12 5  15 but  5  5  15. Thus, there is no

solution, and the lines do not intersect at one point.

1

709

710

CHAPTER 10 Systems of Equations and Inequalities

    2x  5y  9 6. x  3y  1    7x  2y  14

y

Adding the first equation to twice the second equation gives

11y  11  y  1. Substituting back into the second equation, we get

1

x  3 1  1  x  4. Checking point 4 1 in the third equation gives

7 4  2 1  26  14. Thus there is no solution, and the lines do not intersect at

x

1

one point.

  y  x 2  2x 7. y 6x

Substituting for y gives 6  x  x 2  2x  x 2  x  6  0. Factoring, we have x  2 x  3  0.

Thus x  2 or 3. If x  2, then y  8, and if x  3, then y  3. Thus the solutions are 3 3 and 2 8.

  x 2  y2  8 8.  y  x 2

Substituting for y in the first equation gives x 2  x  22  8 

  2x 2  4x  4  0  2 x 2  2x  2  0. Using the Quadratic Formula, we have          2  2 3 2  4  8   1  3. If x  1  3. then y  1  3  2  1  3, and if x  1  3, x 2 2             then y  1  3  2  1  3. Thus, the solutions are 1  3 1  3 and 1  3 1  3 .

 4   3x   6 y 9. 8   x 4 y

16 8 Adding twice the first equation to the second gives 7x  16  x  16 7 . So 7  y  4 

  16 14 16y  56  28y  12y  56  y   14 3 . Thus, the solution is 7   3 .

10.

 

x 2  y 2  10

 x 2  2y 2  7y  0

Subtracting the first equation from the second gives y 2  7y  10 

 y 2  7y  10  0  y  2 y  5  0  y  2, y  5. If y  2, then x 2  4  10  x 2  6  x   6, and if      y  5, then x 2  25  10  x 2  15, which leads to no real solution. Thus the solutions are 6 2 and  6 2 .

  32x   032x  043y  0 y  43 11.   7x  12y  341   y  7x  341 12 The solution is approximately 2141 1593. 20

22

24

    12x  32y  660  y  6 x  1102 3 12.   7137x  3931y  20,000  y   7137 x  20,000 3931 3931 The solution is approximately 6104 10573. 55

-10

-100

-15

-105

-20

-110

60

65

CHAPTER 10

   x  y 2  10  y  x  10 13.  1 y  12   y  22 x  12 x  22

  y  5x  x 14.  y  x5  5

The solutions are 1194 139 and 1207 144.

Review

711

The solutions are approximately

145 135, 1 6, and 151 1293.

5 10 0 10

15 -2

-5

15.

   

x  2y  z 

8

   x 

4x z 9       2x  y  z  8 

  8   x  2y  z   8y  3z  23 8y  3z  23    3y  z  8 z  5 2y  z 

2

8

Therefore, z  5, 8y 3 5  23

 y  1, and x  2 1  5  8  x  1. Hence, the solution is 1 1 5.       x  y  3z  4 x  y  3z  4       x  y  3z  4 16. 4x  2y  z  11  6y  13z  27  6y  13z  27        5x  y  z  16   3z  9 6y  16z  36

Therefore, z  3,

6y  13 3  27  y  2, and x  2  3 3  4  x  3. Hence, the solution is 3 2 3.           x  y  2z  6  x  y  2z  6  x  y  2z  6 17. 2x Therefore, 3z  6  z  2, 2y  2  0   5z  12  2y  z  0  2y  z  0        x  2y  3z  9   4y  z  6 3z  6 y  1, and x  1  2 2  6  x  1. Hence, the solution is 1 1 2.       x  2y  3z  1 x  2y  3z  1     x  2y  3z  1     18. y  4z  1 y  4z  1 x  3y  z  0          2x 28z  0 6y  4z  6  6z  6

 x  3, and so the solution is 3 1 0.           x  2y  3z  1  x  2y  3z  1  x  2y  3z  1 19. 2x  y  z  3  3y  5z  1  3y  5z  1        2x  7y  11z  2   6y  10z  1 0  1

system has no solution.   x  y z 2      2x  3z 5 20.    4  9  x  2y     x  y  2z  3  5   x  y z  2      2y  5z  2  1  13z  12  11      14  28

Thus, z  0, y  1, and x  2 1  3 0  1

which is impossible. Therefore, the

    x  y z  2 x  y z  2           2y  5z  2  1 2y  5z  2  1     3y  z  3  7 13z  12  11           z  2  3 z  2  3 Therefore, 14  28    2, 13z12 2  11  z  1; 2y5 12 2  1

 2y  1  1  y  0; and x  1  2  2  x  1. So the solution is 1 0 1 2.

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CHAPTER 10 Systems of Equations and Inequalities

   x  3y  z  4  x  3y  z  4 21.   4x  y  15z  5  y  z  1

Thus, the system has infinitely many solutions given by z  t,

y  t  1  y  1  t, and x  3 1  t  t  4  x  1  4t. Therefore, the solutions are 1  4t 1  t t, where t is any real number.

      2x  3y  4z  3 2x  3y  4z  3 3       2x  3y  4z  22. 4x  5y  9z  13  y z 7  y z 7        2x    7z  0 3y  3z  3 0  24

Since this last equation is impossible, the system is inconsistent and has no solution.

      x  z  2 x  z    2 x  z    2                2x  y y  2z  4  8 y  2z  4  8  2  12    23.    5z  13  20 3y  z    4 3y  z    4                x  yz z  4  20 y  8  10    z  2  x    y  2z  4    8  20  z   60 , 5z  13  40 Therefore, 33  120     40 11 11 11 ,  5z  13  20      33  120         40  8  y  48 ; and x   60   40  2  x  2 . Hence, the solution is y  2  60  4  11 11 11 11 11 11   2  48   60   40 . 11 11 11 11       x  4y  z  8 x  4y  z  8       x  4y  z  8 24. 2x  6y  z  9  2y  3z  7  2y  3z  7        x  6y  4z  15   6y  9z  21 00

Thus, the system has infinitely many solutions. Letting z  t, we find 2y  3t  7  y  72  32 t, and     x  4 72  32 t  t  8  x  6  5t. Therefore, the solutions are 6  5t 72  32 t t , where t is any real number.

25. Let k and s be be Kieran’s and Siobhan’s ages. From the given information, we have the system  k  s 4 Subtracting the first equation from the second gives s  22  s  4  2s  18  s  9, so k  s  22 k  22  9  13. Thus, Kieran is 13 and Siobhan is 9.

26. Let x be the amount in the 6% account and y the amount in the 7% account. The system is

 

y  2x  006x  007y  600

Substituting gives 006x  007 2x  600  02x  600  x  3000, so y  2 3000  6000. Hence, the man has $3000 invested at 6% and $6000 invested at 7%.

CHAPTER 10

Review

713

27. Let n be the number of nickels, d the number of dimes, and q the number of quarter in the piggy bank. We get the following   50   n d  q Since 10d  25n, we have d  52 n, so substituting into the first equation system: 5n  10d  25z  560    10d  5 5n we get n  52 n  q  50  72 n  q  50  q  50  72 n. Now substituting this into the second equation we have     115 115 5n  10 52 n  25 50  72 n  560  5n  25n  1250  175 2 n  560  1250  2 n  560  2 n  650 

n  12. Then d  52 12  30 and q  50  n  d  50  12  30  8. Thus the piggy bank contains 12 nickels, 30 dimes, and 8 quarters. 28. Let c, s, and p be the number of each kind of salmon caught.          c  s  c  s  p  25  c  s  p  25 c  s p3  c  s  p  3  2s         c  2s   0 3s c  2s

The given information leads to the system    p  25   c  s  p  25  2 p  22  2s  2 p  22  p  4, so    4 p  16  p  25

2s  2 4  22  s  7 and c  7  4  25  c  14. Tornie caught 14 coho, 7 sockeye, and 4 pink salmon.

29.

3x  1

B 3x  1 A  .Thus, 3x  1  A x  3  B x  5  x A  B  3A  5B,  x 5 x 3 x  5 x  3    3A  3B  9  AB 3 Adding, we have 8B  8  B  1, and A  2. Hence,  and so   3A  5B  1 3A  5B  1 x 2  2x  15



3x  1 1 2  .  x 5 x 3 x 2  2x  15

30.

  A B C 8 8 8 2  4  Bx x  2        . Then 8  A x  x x  2 x  2 x x 2 x 2 x 3  4x x x2  4

C x x  2  x 2 A  B  C  x 2B  2C  4A. Thus, 4A  8  A  2, 2B  2C  0  B  C, and 1 1 2 8  .   2  2B  0  B  1, so C  1. Therefore, 3 x x 2 x 2 x  4x 31.

2x  4

x x  12



B C A   . Then 2x  4  A x  12  Bx x  1  C x  Ax 2  2Ax  A  Bx 2  x x 1 x  12

Bx  C x  x 2 A  B  x 2A  B  C  A. So A  4, 4  B  0  B  4, and 8  4  C  2  C  2. 2x  4 4 2 4 Therefore,    . 2 x x  1 x  12 x x  1 32.

x 6

x 6 Ax  B C  . Thus,  2  2  x 2 x 4 x  4 x  2   x  6  Ax  B x  2  C x 2  4  Ax 2  2Ax  Bx  2B  C x 2  4C

x 3  2x 2  4x  8

 x 2 A  C  x 2A  B  2B  4C       A  C0  C0  C0     A A and we get the system 2A  B 1  B  2C  1  B  2C  1          2B  4C  6 2B  4C  6 8C  8

x 6 x 1 1 Thus, 8C  8  C  1, B  2  1  B  1, and A  1  0  A  1. So 3 .  2  2 x  2x  4x  8 x 4 x 2

714

33.

CHAPTER 10 Systems of Equations and Inequalities

  2x  1 A Bx  C 2x  1     2 . Then 2x 1  A x 2  1 Bx  C x  Ax 2  A Bx 2 C x  A  B x 2  3 2 x x x x x 1 x 1 x 2 1 2x  1   2 . C x  A. So A  1, C  2, and A  B  0 gives us B  1. Thus 3 x x x x 1

     B Cx  D 5x 2  3x  10 A   2  . 34. Since x 4 x 2 2  x 2  1 x 2  2  x  1 x  1 x 2  2 , we have 4 2 x  1 x  1 x x 2 x 2 Thus       5x 2  3x  10  x  1 x 2  2 A  x  1 x 2  2 B  x 2  1 C x  D  Ax 3  Ax 2  2Ax  2A  Bx 3  Bx 2  2Bx  2B  C x 3  Dx 2  C x  D

 A  B  C x 3  A  B  D x 2  2A  2B  C x  2A  2B  D     A BC  0 Coefficients of x 3 A B C  0          A B  2 2B  C  D  5 D 5 Coefficients of x  This leads to the system    3C  3 2A  2B  C  3 Coefficients of x          2A  2B  4B  C  D  13  D  10 Constant terms   A B C  0      2B  C  D  5 Thus C  1, 3 1  3D  3  D  0, 2B  1  0  5  B  3, and  C  1      3C  3D  3 A  3  1  0  A  2. Thus,

35.

3 x 2 5x 2  3x  10   2  . x 1 x 1 x4  x2  2 x 2

Cx  D Ax  B 3x 2  x  6   2 2 . Thus x 2  22 x 2 x2  2   3x 2  x  6  x 2  2 Ax  B  C x  D  Ax 3  Bx 2  2Ax  2B  C x  D

 x 3 A  x 2 B  x 2A  C  2B  D     A A  0         B B  3  This leads to the system   C C  1   2A       D 2B D6

3x 2  x  6 x 3 Thus   2 2 . 2 2 2 x  2 x 2 x 2

 

0 3

 1 

0

CHAPTER 10

Review

715

Bx  C x2  x  1 A Dx  E   2  2 . Thus x xx 2  12 x 1 x2  1  2   x 2  x  1  x 2  1 A  x x 2  1 Bx  C  x Dx  E  Ax 4  2Ax 2  A  Bx 4  C x 3  Bx 2  C x  Dx 2  E x 36.

 x 4 A  B  x 3 C  x 2 2A  B  D  x C  E  A   AB  0      C  0   Thus A  1, B  1, C  0, D  0, and E  1, so This leads to the system 2A  B D  1     C E 1     A 1 x2  x  1 x 1 1    2 2 . x 2 xx 2  12 x 1 x 1

  2x  3y  7 37.  x  2y  0

By inspection of the graph, it appears that 2 1 is the solution to the system. We check this in both

equations to verify that it is the solution. 2 2  3 1  4  3  7 and 2  2 1  2  2  0. Since both equations are satisfied, the solution is indeed 2 1.

  3x  y  8 38.  y  x 2  5x

By inspection of the graph, it appears that 2 14 and 4 4 are solutions to the system.

We check each possible solution in both equations to verify that it is a solution: 3 2  14  6  14  8 and

22  5 2  4  10  14; also 3 4  4  12  4  8 and 42  5 4  16  20  4. Since both points satisfy both equations, the solutions are 2 14 and 4 4.

39.

 

x2  y  2

 x 2  3x  y  0

By inspection of the graph, it appears that 2 2 is a solution to the system, but is difficult

to get accurate values for the other point. Adding the equations, we get 2x 2  3x  2  2x 2  3x  2  0   2 2x  1 x  2  0. So 2x  1  0  x   12 or x  2. If x   12 , then  12  y  2  y  74 . If x  2, then   22  y  2  y  2. Thus, the solutions are  12  74 and 2 2. 40.

 

x  y  2  x 2  y 2  4y  4

By inspection of the graph, it appears that 2 0 and 2 4 are solutions to the system. We

check each possible solution in both equations to verify that it is a solution: 2  0  2 and 22  02  4 0  4;

also 2  4  2 and 22  42  4 4  4  16  16  4. Since both points satisfy both equations, the solutions are 2 0 and 2 4.

41. The boundary is a solid curve, so we have the inequality x  y 2  4. We take the test point 0 0 and verify that it satisfies the inequality: 0  02  4.

42. The boundary is a solid curve, so we have the inequality x 2  y 2  8. We take the test point 0 3 and verify that it satisfies the inequality: 02  32  8.

716

CHAPTER 10 Systems of Equations and Inequalities

44. y  x 2  3

43. 3x  y  6

y

45. x 2  y 2  9

y

46. x  y 2  4

y

y

1 1 1

1

x

y

y

1

x

0

1

x

y 4

1 1

  x 2  y2  9 51.  xy0

    y  2x 50. y  2x    y  1x  2 2

y

1

1

4

x

y

The vertices occur where y  x. By substitution, 1

x 2  x 2  9  x   3 , and so y   3 . Therefore, the vertices are 2 2     3 3 3 3    and     and the solution set is bounded. The test point 2

2

x

1

   x  y  2 49. y  x  2    x 3

 y  x 1 48.  x 2  y2  1

3

1

x

x

1

  y  x 2  3x 47.  y  1x  1 0

1

2

x

1

2

1 1 satisfies each inequality.

  y  x2  4 52.  y  20

y

The vertices occur where y  x 2  4 and y  2. By

substitution, x 2  4  20  x 2  16  x  4 and y  20. Thus, the vertices

(_4, 20)

(4, 20)

are 4 20 and the solution set is bounded. The test point 0 10 satisfies each inequality.

2 2

    x  0, y  0 53. x  2y  12    y  x 4

The intersection points are 4 0, 0 4,





y

4  16 , 0 6, 3 3

0 0, and 12 0. Since the points 4 0 and 0 6 are not in the solution set,   the vertices are 0 4, 43  16 3 , 12 0, and 0 0. The solution set is bounded. The test point 1 1 satisfies each inequality.

x

y=x+4

x+2y=12 1 1

x

x

CHAPTER 10

54.

      

x4 x  y  24

y

Test

717

x=4

The lines x  y  24 and x  2y  12 intersect at the

x  2y  12

point 20 4. The lines x  4 and x  2y  12 intersect at the point 4 4, but this vertex does not satisfy the other inequality. The lines x  y  24 and x  4 intersect at the point 4 20. Hence, the vertices are 4 20 and 20 4. The solution set is not bounded. The test point 12 12 satisfies each inequality.

x+y=24 1 1

x=2y+12 x

       x  y  z  a  x  y  z  a ab ac ab ac ,z , and x   a 55. x yz b  2z  a  b Thus, y    2 2 2 2    x yz c  2y  ac   bc bc ac ab . The solution is   . x 2 2 2 2        ax  by  cz  a  b  c  ax  by  cz  a  b  c 56. bx  by  cz  a  b  c  0  bx  by  cz  Subtracting the second equation from c c      cx  cy  cz   c x yz  1 the first gives a  b x  a  b  x  1Subtracting the third equation from the second, b  c x  b  c y  0  y  x  1. So 1  1  z  1  z  1, and the solution is 1 1 1.

57. Solving the second equation for y, we have y  kx. Substituting for yin the first equation gives us 12 x  kx  12  1  k x  12  x  . Substituting for y in the third equation gives us kx  x  2k  k1 2k 12 2k . These points of intersection are the same when the x-values are equal. Thus,  k  1 x  2k  x  k 1 k1 k 1    12 k  1  2k k  1  12k  12  2k 2  2k  0  2k 2  10k  12  2 k 2  5k  6  2 k  3 k  2. Hence, k  2 or k  3.

58. The system will have infinitely many solutions when the system has solutions other than 0 0 0. So we solve the system with x  0, y  0, and z  0:     kx  y  z0    kx  y  z  0 kx  y  z  0        2y  k  3 z  0 x  2y  kz  0  2y  k  3 z  0          x   y  3k  1 z  0  3z  0 [k  3  2 3k  1] z  0 Since z  0, we must have k  3  2 3k  1  5k  1  0  k  15 .

CHAPTER 10 TEST 1. (a) The system is linear.   x  3y  7 (b) Multiplying the first equation by 5 and then adding gives 13y  39  y  3. So  5x  2y  4 x  3 3  7  x  2. Thus, the solution is 2 3.

2. (a) The system is nonlinear.

718

CHAPTER 10 Systems of Equations and Inequalities

   6x  y 2  10  6x  y 2  10 (b)   3x  y  5  y 2  2y  0

Thus y 2  2y  y y  2  0, so either y  0 or y  2. If y  0, then

  3x  5  x  53 and if y  2, then 3x  2  5  3x  3  x  1. Thus the solutions are 53  0 and 1 2.

3. (a) The system is nonlinear.   x 2  y 2  100 Substituting y  3x into the first equation gives x 2  3x2  100  x 2  10  x   10. (b) y  3x         If x   10, then y  3 10, and if x  10, then y  3 10. We can verify that  10 3 10 and    10 3 10 are valid solutions to the first equation in the given system.   x  2y  1 4.  y  x 3  2x 2

2

The solutions are approximately 055 078, 043 029, and 212 056. -2

2 -2

    x  2y  z  3     x  2y  z  3 5. (a) x  3y  2z  3  y z 0      2x  3y  z  8  y  3z  2

Eq. 1  Eq. 2 2  Eq. 1  Eq. 3

    x  2y  z  3  y z 0    2z  2

 Eq. 2  Eq. 3

z  1, so y  1  0  y  1 and z  2 1  1  3  x  2. Thus, the solution is 2 1 1.

(b) The system is neither inconsistent nor dependent.

       x  y  9z  3  x  y  9z  3 6. (a) x  4z  7  y  13z  10      3x  y  z  5  2y  26z  14

Eq. 1  Eq. 2 3  Eq. 1  Eq. 3

The last equation cannot be satisfied, so the system has no solution.

(b) The system is inconsistent.

    2x  y  z  0     2x  y  z  0 7. (a) 3x  2y  3z  1  7y  9z  2      x  4y  5z  1  7y  9z  2

    x  y  9z  3  y  13z  10    0 6

3  Eq. 1  2  Eq. 2

Eq. 1  2  Eq. 3





    2x  y  z  0  7y  9z  2    00

2  Eq. 2  Eq. 3

Eq. 2  Eq. 3

Letting z  t, we have 7y  9t  2  y  27  97 t, so 2x  27  97 t  t  0  x  17  17 t. The solutions are   1  1 t 2  9 t t where t is any real number. 7 7 7 7

(b) The system is dependent.

       x  y  2z  8  x  y  2z  8 8. (a) 2x  y  20  3y  4z  4      2x  2y  5z  15  z  1

2  Eq. 1  Eq. 2 2  Eq. 1  Eq. 3

x  0  2 1  8  x  10. Thus, the solution is 10 0 1.

(b) The system is neither inconsistent nor dependent.

 z  1, so 3y  4 1  4  y  0 and

CHAPTER 10

Test

719

9. Let  be the speed of the wind and a the speed of the airplane in still air, in kilometers per hour. Then the speed of the of the plane flying against the wind is a   and the speed of the plane flying with the wind is a  . Using distance  rate  time,    600  25 a    240  a   Adding the two equations, we get 600  2a  a  300. we get the system  50  300   360  a   a   60

So 360  300      60Thus the speed of the airplane in still air is 300 km/h and the speed of the wind is 60 km/h.

10. Let x, y, and z represent the price in dollars for coffee, juice, and donuts respectively. Then the system of equations is       2x  y  2z  625 Anne 2x  y  2z  625       2x  y  2z  625 x  3z  375 Barry  y  4z  125  y  4z  125        3x  y  4z  925   Cathy y  5z  200 z  075

Thus, z  075, y  4 075  125  y  175, and 2x  175  2 075  625  x  15. Thus coffee costs $150, juice costs $175, and donuts cost $075.

11. 3x  4y  6

12. x 2  y  3

y

y

1 1

x 1 x

1

    2x  y  8 13. x  y  2    x  2y  4

y

From the graph, the points 4 0 and 0 2 are vertices. The

third vertex occurs where the lines 2x  y  8 and x  y  2 intersect. Adding these two equations gives 3x  6  x  2, and so y  8  2 2  4 . Thus, the

third vertex is 2 4.

1 x

1

  x2  5  y  0 14.  y  5  2x

Substituting y  5  2x into the first equation gives

y x@-5+y=0

y=5+2x

x 2  5  5  2x  0  x 2  2x  0  x x  2  0  x  0 or x  2. If

x  0, then y  5  2 0  5, and if x  2, then y  5  2 2  1. Thus, the

vertices are 0 5 and 2 1.

1 1

15.

4x  1

x  12 x  2



B C A  . Thus,  x  1 x  12 x 2

x

    4x  1  A x  1 x  2  B x  2  C x  12  A x 2  x  2  B x  2  C x 2  2x  1  A  C x 2  A  B  2C x  2A  2B  C

720

FOCUS ON MODELING

which leads to the system of equations       A  C  0 A  C  0  C 0      A A  B  2C  4  B  3C  4  B  3C  4        2A  2B  C  1   2B  3C  1 9C  9

Therefore, 9C  9  C  1, B  3 1  4  B  1, and A  1  0  A  1. Therefore, 4x  1

x  12 x  2

16.



1 1 1  .  x  1 x  12 x 2

2x  3 Bx  C 2x  3 A      2 . Then x x 3  3x x x 3 x2  3   2x  3  A x 2  3  Bx  C x  Ax 2  3A  Bx 2  C x  A  B x 2  C x  3A.

x 2 1 2x  3   2 . So 3A  3  A  1, C  2 and A  B  0 gives us B  1. Thus 3 x x x x 3

FOCUS ON MODELING Linear Programming

2.

1. Vertex 0 2 0 5 4 0

M  200  x  y

200  0  2  198 200  0  5  195 200  4  0  196

Thus, the maximum value is 198 and the minimum value is 195.

    x  0, y  0 3. 2x  y  10    2x  4y  28

Vertex 1 0   1 1 2 2 2 2 4 0

N  12 x  14 y  40 1 1  1 0  40  405 4  2  1 1  1 1  40  40375 2 2 4 2 1 2  1 2  40  415 2 4 1 4  1 0  40  42 2 4

Thus, the maximum value is 42 and the minimum value is 40375.

The objective function is P  140  x  3y. From the graph,

y

2x+y=10

the vertices are 0 0, 5 0, 2 6, and 0 7. Vertex 0 0 5 0 2 6 0 7

P  140  x  3y

140  0  3 0  140 140  5  3 0  135 140  2  3 6  156 140  0  3 7  161

Thus the maximum value is 161, and the minimum value is 135.

2x+4y=28 1 1

x

Linear Programming

  x  0, y  0      x  10, y  20 4.  xy5      x  2y  18

The objective function is Q  70x  82y. From the graph,

y

x+2y=18

the vertices are at 0 9, 0 5, 5 0, 10 0, and 10 4. Note that the restriction y  20 is irrelevant, superseded by x  2y  18 and x  0. Vertex 0 9 0 5 5 0 10 0 10 4

721

Q  70x  82y

x+y=5

1

70 0  82 9  738

x=10

x

1

70 0  82 5  410 70 5  82 0  350

70 10  82 0  700

70 10  82 4  1028

Thus, the maximum value of Q is 1028 and the minimum value is 350.

5. Let t be the number of tables made daily and c be the number of chairs made daily. Then the data given can be summarized by the following table: Tables t

Chairs c

Available time

2h

3h

108 h

Finishing

1h

1 h 2

20 h

Profit

$35

$20

Carpentry

    2t  3c  108 Thus we wish to maximize the total profit P  35t  20c subject to the constraints t  12 c  20    t  0, c  0 From the graph, the vertices occur at 0 0, 20 0, 0 36, and 3 34.

y

Vertex 0 0 20 0 0 36 3 34

P  35t  20c

35 0  20 0 

0

1

t+2 c=20

35 20  20 0  700

35 0  20 36  720 35 3  20 34  785

2t+3c=108 10 10

Hence, 3 tables and 34 chairs should be produced daily for a maximum profit of $785.

x

722

FOCUS ON MODELING

6. Let c be the number of colonial homes built and r the number of ranch homes built. Since there are 100 lots available, c  r  100. From the capital restriction, we get 30,000c  40,000r  3,600,000, or 3c  4r  360. Thus, we wish to     c  0, r  0 maximize the profit P  4000c  8000r subject to the constraints c  r  100    3c  4r  360 From the graph, the vertices occur at 0 0, 100 0, 40 60, and 0 90.

y

Vertex 0 0 100 0 40 60 0 90

P  4000c  8000r

4000 0  8000 0 

c+r=100

0

4000 100  8000 0  400,000

4000 40  8000 60  640,000

3c+4r=360

20

4000 0  8000 90  720,000

x

20

Therefore, he should build no colonial style and 90 ranch style houses for a maximum profit of $720,000. Note that ten of the lots will be left vacant.

7. Let x be the number of crates of oranges and y the number of crates of grapefruit. Then the data given can be summarized by the following table: Oranges

Grapefruit

Available

Volume

4 ft3

6 ft3

300 ft3

Weight

80 lb

100 lb

5600 lb

Profit

$250

$400

In addition, x  y. Thus we wish to maximize the total profit P  25x  4y subject to the constraints     x  0, y  0, x  y 4x  6y  300    80x  100y  5600 From the graph, the vertices occur at 0 0, 30 30, 45 20, and 70 0.

y

Vertex 0 0 30 30 45 20 70 0

80x+100y=5600

P  25x  4y

25 0  4 0 

0

25 30  4 30  195

25 45  4 20  1925 25 70  4 0  175

x=y

4x+6y=300 10 10

Thus, she should carry 30 crates of oranges and 30 crates of grapefruit for a maximum profit of $195.

x

Linear Programming y

8. Let x be the daily production of standard calculators and y the daily production of     x  100, y  80 scientific calculators. Then the inequalities x  200, y  170 describe the    x  y  200

180 120 90

100 100 100 170 200 170 200 80

C  5x  7y

30 60 90 120 150 180 x

Vertex 100 100

5 200  7 170  2190

200 170

5 100  7 170  1690

100 170

5 200  7 80  1560

200 80

5 120  7 80  1160

x+y=200

30

(b) Maximize the objective function P  2x  5y:

5 100  7 100  1200

120 80

y=80

60

200 170, 200 80, and 120 80.

Vertex

x=200

150

constraints. From the graph, the vertices occur at 100 100, 100 170,

(a) Minimize the objective function C  5x  7y:

x=100 y=170

723

P  2x  5y

2 100  5 100  300 2 100  5 170  650 2 200  5 170  450 2 200  5 80 

0

2 120  5 80  160

120 80

So to minimize cost, they should produce

So to maximize profit, they should produce

120 standard and 80 scientific calculators.

100 standard and 170 scientific calculators.

9. Let x be the number of stereo sets shipped from Long Beach to Santa Monica and y the number of stereo sets shipped from Long Beach to El Toro. Thus, 15  x sets must be shipped to Santa Monica from Pasadena and 19  y sets to El Toro from Pasadena. Thus, x  0, y  0, 15  x  0, 19  y  0, x  y  24, and 15  x  19  y  18. Simplifying, we get   x  0, y  0      x  15, y  19 the constraints  x  y  24      x  y  16 The objective function is the cost C  5x  6y  4 15  x  55 19  y  x  05y  1645, which we wish to

minimize. From the graph, the vertices occur at 0 16, 0 19, 5 19, 15 9, and 15 1. Vertex 0 16 0 19 5 19 15 9 15 1

C  x  05y  1645

y

x=15

0  05 16  1645  1725 0  05 19  1645  174 5  05 19  1645  179 15  05 9  1645  184 15  05 1  1645  180

10 x+y=16 10

x+y=24 x

The minimum cost is $17250 and occurs when x  0 and y  16. Hence, no stereo should be shipped from Long Beach to Santa Monica, 16 from Long Beach to El Toro, 15 from Pasadena to Santa Monica, and 3 from Pasadena to El Toro.

724

FOCUS ON MODELING

10. Let x be the number of sheets shipped from the east side store to customer A and y be the number of sheets shipped from the east side store to customer B. Then 50  x sheets must be shipped to customer A from the west side store and 70  y sheets must be shipped to customer B from the west side store. Thus, we obtain the constraints     x  0, y  0 x  0, y  0          x  50, y  70  x  50, y  70     x  y  80  x  y  80        x  y  75  50  x  70  y  45

The objective function is the cost C  05x  06y  04 50  x  055 70  y  01x  005y  585, which we wish to minimize. From the graph, the vertices occur at 5 70, 10 70, 50 30, and 50 25. y

Vertex 5 70 10 70 50 30 50 25

x=70

C  01x  005y  585

01 5  005 70  585  625

y=70

01 10  005 70  585  63

x+y=75

01 50  005 30  585  65

x+y=80

10

01 50  005 25  585  6475

x

10

Therefore, the minimum cost is $6250 and occurs when x  5 and y  70. So 5 sheets should be shipped from the east side store to customer A, 70 sheets from the east side store to customer B, 45 sheets from the west side store to customer A, and no sheet from the west side store to customer B.

11. Let x be the number of bags of standard mixtures and y be the number of bags of deluxe mixtures. Then the data can be summarized by the following table: Standard

Deluxe

Available

Cashews

100 g

150 g

15 kg

Peanuts

200 g

50 g

20 kg

Selling price

$195

$220

Thus the total revenue, which we want to maximize, is given by R  195x  225yWe have the constraints        x  0, y  0, x  y  x  0, y  0, x  y  01x  015y  15 10x  15y  1500      02x  005y  20  20x  5y  2000

From the graph, the vertices occur at 0 0, 60 60, 90 40, and 100 0.

y

Vertex 0 0 60 60 90 40 100 0

R  195x  225y

196 0  225 0  0

10x+15y=1500

195 60  225 60  252

195 90  225 40  2655 195 100  225 0  195

x=y 20x+5y=2000

10 10

x

Hence, he should pack 90 bags of standard and 40 bags of deluxe mixture for a maximum revenue of $26550.

Linear Programming

725

12. Let x be the quantity of type I food and y the quantity of type II food, in ounces. Then the data can be summarized by the following table: Type I

Type II

Required

Fat

8g

12 g

24 g

Carbohydrate

12 g

12 g

36 g

Protein

2g

1g

4g

Cost

$020

$030

    x  0, y  0, Also, the total amount of food must be no more than 5 oz. Thus, the constraints are x  y  5, 8x  12y  24    12x  12y  36, 2x  y  4

The objective function is the cost C  02x  03y, which we wish to minimize. From the graph, the vertices occur at 1 2, 0 4, 0 5, 5 0, and 3 0. Vertex 1 2 0 4 0 5 5 0 3 0

y

C  02x  03y

x+y=5

02 1  03 2  08 02 0  03 4  12 02 0  03 5  15

2x+y=4 12x+12y=36

02 5  03 0  10 02 3  03 0  06

8x+12y=24 1 x

1

Hence, the rabbits should be fed 3 oz of type I food and no type II food, for a minimum cost of $060.

13. Let x be the amount in municipal bonds and y the amount in bank certificates, both in dollars. Then 12000  x  y is the amount in high-risk bonds. So our constraints can be stated as     x  0, y  0, x  3y     x  0, y  0, x  3y  x  y  12,000 12,000  x  y  0      12,000  x  y  2000  x  y  10,000

From the graph, the vertices occur at 7500 2500, 10000 0, 12000 0, and 9000 3000. The objective function is P  007x  008y  012 12000  x  y  1440  005x  004y, which we wish to maximize. y

Vertex 7500 2500 10000 0

1440  005 10,000  004 0  940

12000 0 9000 3000

x+y=12,000

P  1440  005x  004y

1440  005 7500  004 2500  965

1440  005 12,000  004 0  840

1440  005 9000  004 3000  870

x+y=10,000

x=3y

2000 2000

x

Hence, she should invest $7500 in municipal bonds, $2500 in bank certificates, and the remaining $2000 in high-risk bonds for a maximum yield of $965.

726

FOCUS ON MODELING

14. The only change that needs to be made to the constraints in Exercise 13 is that the 2000 in the last inequality becomes 3000. Then we have     x  0, y  0, x  3y     x  0, y  0, x  3y  x  y  12,000 12,000  x  y  0      12,000  x  y  3000  x  y  9000

From the graph, the vertices occur at 6750 2250, 9000 0, 12000 0, and 9000 3000. The objective function is Y  007x  008y  012 12000  x  y  1440  005x  004y, which we wish to maximize. y

Vertex 6750 2250

1440  005 9000  004 0  990

9000 0 12000 0 9000 3000

x+y=12,000

Y  1440  005x  004y

1440  005 6750  004 2250  10125 1440  005 12,000  004 0  840

1440  005 9000  004 3000  870

x=3y

x+y=9000 2000

x

2000

Hence, she should invest $6750 in municipal bonds, $2250 in bank certificates, and $3000 in high-risk bonds for a maximum yield of $101250, which is an increase of $4750, over her yield in Exercise 13.

15. Let g be the number of games published and e be the number of educational programs published. Then the number of utility programs published is 36  g  e. Hence we wish to maximize profit, P  5000g  8000e  6000 36  g  e  216,000  1000g  2000e, subject to the constraints        g  4, e  0  g  4, e  0 36  g  e  0  g  e  36      36  g  e  2e  g  3e  36.

  From the graph, the vertices are at 4 32 3 , 4 32, and 36 0. The objective function is P  216,000  1000g  2000e. e

Vertex   4 32 3 4 32 36 0

g=4

P  216,000  1000g  2000e   216,000  1000 4  2000 32 3  233,33333 216,000  1000 4  2000 32  276,000 216,000  1000 36  2000 0  180,000

g+e=36 g+3e=36

10 10

g

So, they should publish 4 games, 32 educational programs, and no utility program for a maximum profit of $276,000 annually.

Linear Programming

    x  0, x  y 16. x  2y  12    x  y  10 (a)

y

(b) x+y=10

y

x=y P=40 P=36 P=32 P=28

(5, 5) x+2y=12

1 1

1

x

1

(c) These lines will first touch the feasible region at the top vertex, 5 5. (d) Vertex 0 0 10 0 5 5 8 2

P  x  4y

0  4 0  0

10  4 0  10 5  4 5  25 8  4 2  16

The maximum value of the objective function occurs at the vertex 5 5.

x

727

11

MATRICES AND DETERMINANTS

11.1 MATRICES AND SYSTEMS OF LINEAR EQUATIONS 1. A system of linear equations with infinitely many solutions is called dependent. A system of linear equations with no solution is called inconsistent.     1 1 1 1  x  y  z  1    2. x  2z  3    1 0 2 3     2y  z  3 0 2 1 3

3. (a) The leading variables are x and y. (b) The system is dependent.

(c) The solution of the system is x  3  t y  5  2t z  t. 

   x  2    The solution is 2 1 3. 0 1 y1    z3 1 3   1 2  x z2  The solution is z  t, y  1  t, x  2  t. 1 1  yz1 0 0    0 2 2  x   This system is inconsistent and has no solution.  0 1 y 1    03 0 3

1 0 0 2

 4. (a)  0 0  1   (b)  0 0  1  (c)  0 0

1 0 0 1 0 0 1 0

5. 3  2 

3



6. 2  4 1 1 2

7. 2  1



8. 3  1 

1

9. 1  3 0

1 1

10. 2  2



   11.   2 1 0 1  1 0 1 3

 12.  

13. (a) Yes, this matrix is in row-echelon form.

14. (a) Yes, this matrix is in row-echelon form.

(b) Yes, this matrix is in reduced row-echelon form.   x  3 (c) y  5

0

3 2

1 1

3

 7  3

(b) No, this matrix not in reduced row-echelon form. The entry above the leading 1 in the second row is not 0.   x  3y  3 (c)  y 5

729

730

CHAPTER 11 Matrices and Determinants

16. (a) Yes, the matrix is in row-echelon form.

15. (a) Yes, this matrix is in row-echelon form.

(b) Yes, the matrix is in reduced row-echelon form.    7z  0  x (c) y  3z  0    01

(b) No, this matrix is not in reduced row-echelon form, since the leading 1 in the second row does not have a zero above it.     x  2y  8z  0 (c) y  3z  2    00 17. (a) No, this matrix is not in row-echelon form, since the row of zeros is not at the bottom.

18. (a) Yes, the matrix is in row-echelon form. (b) Yes, the matrix is in reduced row-echelon form.   1  x (c) y 2    z3

(b) No, this matrix is not in reduced row-echelon form.   0  x (c) 00    y  5z  1 19. (a) Yes, this matrix is in row-echelon form.

20. (a) No, this matrix is not in row-echelon form, since the fourth column has the leading 1 of two rows.

(b) Yes, this matrix is in reduced row-echelon form.   x  3y    0      z  2  0 (c)  01      00

(b) No, this matrix is not in reduced row-echelon form.   x  3y   0      y  4 0 (c)  u2       0

Notice that this system has no solution.



1

1

2

3

1

1



5



2



1 3

 21.  

0



 4  1 2 1 1 2 3

 22.   10 3 1 3

3

1 3

5



2 1



 1 1   2 1 1 1



 1 20   1 8

   23.   2 3 1 13  6 5 1 7  24.  0 0

3R1  R2  R2 

   

2R2  R3  R3 

1

2

0

4

7

0



3



 4  1 2 1 1

   

2R1  R2  R2 

3R1  R3  R3 

1

5 2 3

 0 1 5 14   1 3 1 8



2

 2 3  0 8

5



 1 13   8 8



1 3

2 1



 0  0

1 3

 1 1   0 3 3 1

SECTION 11.1 Matrices and Systems of Linear Equations

    x  2y  4z  3 25. (a) y  2z  7    z2

(b) y  2 2  7  y  3, so x  2 3  4 2  3 

(b) y  3 1  5  y  2, so x  2  3 1  8 

x  1. The solution is 1 3 2.

  x  2y  3z         y  2z  27. (a)  z  2       

x  3. The solution is 3 2 1.

  x  2z  2  5      y  3z  1 28. (a)  z  0        1

7 5 5 3

(b) z  1  0  z  1, so y  3 1  1 

(b) z  2 3  5  z  1, so y  2 1  5 

y  2 and x  2 1  2 1  5  x  5. The

y  3 and x  2 3  3 1  3  7  x  7. The

solution is 5 2 1 1.

solution is 7 3 1 3. 

1 2 1 1

 29.  0 1



 1 2 5  1 3 8



1 2 1 1

 0  0

R3  R1  R3



731

    x  y  3z  8 26. (a) y  3z  5    z  1



 1 2 5  3 2 7



1 2

1

1

2

 0  0

R3  3R2  R3



1



 5 . Thus, 4z  8  0 4 8

z  2; y  2 2  5  y  1; and x  2 1  2  1  x  1. Therefore, the solution is 1 1 2.



1 1 6 3



   30.  1 1 3 3 1 2 4 7

R2  R1  R2





1 1

6 3



   0 0 3 0    1 2 4 7

 13 R2



R3  R1  R3



1 1

6 3



  0 0 1 0   0 1 2 4



1 1

6 3



   0 1 2 4 .   0 0 1 0

R3  R2



Thus, z  0; y  2z  y  0  y  4; and x  y  6z  3  x  4  0  3  x  1. Therefore, the solution is 1 4 0. 

1

1

1 2



   31.   2 3 2 4  4 1 3 1

R2  2R1  R2  R3  4R1  R3



1

1

1

2



   0 5 0 0    0 3 7 7

R3  35 R2  R3





1

1

1

2



   0 5 0 0 .   0 0 7 7

Thus, 7z  7  z  1; 5y  0  y  0; and x  0  1  2  x  1. Therefore, the solution is 1 0 1. 

1

1 1

4



   32.   1 2 3 17  2 1 0 7

R2  R1  R2  R3  2R1  R3



1

1

1

4



   0 3 4 21    0 3 2 15



1 1 1

4



   0 3 4 21 .   0 0 2 6

R3  R2  R3  R3  2R1  R3

Thus, 2z  6  z  3; 3y  4 3  21  y  3; and x  3  3  4  x  2. The solution is 2 3 3. 

1

 33.  1 2  1  0  0

2 1 2





1

2 1 2

  R2  R1  R2  0 2 2 2  0   R3  2R1  R3 0 5 1 1 1 1 3  2 1 2  1 1 1 . Thus, 6z  6  z  1; y  1  1  0 6 6 0

1

solution is 1 0 1.

   

 12 R2 



1

2 1 2

 0 1  0 5

1 1



 1  1

R3  5R2  R3



y  0; and x  2 0  1  2  x  1. Therefore, the

732

CHAPTER 11 Matrices and Determinants



0 2

1

4



   34.  1 1 0 4 3 3 1 10

R1  R2





1 1

0

4



  0 2 1 4   3 3 1 10



1 1

0



4

   0 2 1 4 .   0 0 1 2

R3  3R1  R3



Thus, z  2  z  2; 2y  2  4  y  1; and x  1  4  x  3. The solution is 3 1 2 

1 2 1



9

   35.   2 0 1 2  3 5 2 22



1

2 1

  0 4  0 1

R2  2R1  R2  R3  3R1  R3

9



 1 20   5 5



1

2 1

9



   0 4 1 20 .   0 0 19 0

4R3  R2  R3 

Thus, 19x3  0  x3  0; 4x2  20  x2  5; and x1  2 5  9  x1  1. Therefore, the solution is 1 5 0. 

2

1 0

7



   36.   2 1 1 6  3 2 4 11



2

1 0

7



   0 2 1 1    0 7 8 1

R2  R1  R2  2R3  3R1  R3



2

1 0



7

   0 2 1 1 . Then, 9x3  9    0 0 9 9

2R3  7R2  R3 

x3  1; 2x2  1  1  x2  1; and 2x1  1  7  x1  3. Therefore, the solution is x1  x2  x3   3 1 1.



2 3 1 13



   37.   1 2 5 6  5 1 1 49



2 3

1 13



   0 1 11 25    0 13 3 33

2R2  R1  R2  2R3  5R1  R3



2 3

 0  0

R3  13R2  R3



1

13



 25  . 0 146 292 1 11

Thus, 146z  292  z  2; y  11 2  25  y  3; and 2x  3  3  2  13  x  10. Therefore, the solution is 10 3 2. 

10 10 20

60



   38.   15 20 30 25  5 30 10 45

2R2  3R1  R2  2R3  R1  R3



10 10 20

60



   0 10 120 230    0 70 40 150



10 10

R3  7R2  R3  

20

60



 120 230  . 0 880 1760

  0 10

0

Thus, 880z  1760  z  2; 10y  120 2  230  y  1; and 10x  10  40  60  x  1. Therefore, the solution is 1 1 2. 

1 1

1 2



   39.   0 1 3 1  2 1 5 0

R3  2R1  R3





1

1

1

2



   0 1 3 1    0 1 3 4

R3  R2  R3





1 1

1

3



   0 1 3 1 . The third row of the   0 0 0 3

matrix states 0  3, which is impossible. Hence, the system is inconsistent, and there is no solution. 

1

0

3 3



   40.   2 1 2 5  0 1 8 8



1

0

3

3



   0 1 8 1    0 1 8 8

R2  2R1  R2



 41.  

   

2 3 9 5



 R R 1 2 2   3 1 4 3   1 0 3 2 1  R R R  3 2 3   0 1 5 3    0 0 1 5 3 0 1

0

3

   

1

0

3

2



1 0

1

0

3

3



   0 1 8 1 .   0 0 0 7

R3  R2  R3



The system is inconsistent, and there is no solution. 





3

2



 R2  2R1  R2    1 R2 3   0 3 15 9  2 3 9 5       R3  3R1  R3 3 1 4 3 0 1 5 3  0 3 2  1 5 3 . Therefore, this system has infinitely many solutions, given by x  3t  2 0 0 0

 x  2  3t, and y  5t  3  y  3  5t. Hence, the solutions are 2  3t 3  5t t, where t is any real number.

SECTION 11.1 Matrices and Systems of Linear Equations





1 2 5 3   42.  2 6 11 1   3 16 20 26







1 2 5 3   0 2 1 7   0 10 5 35

R2  2R1  R2  R3  3R1  R3

R3  5R2  R3  

733



1 2

5 3  2 1 7  . 0 0 0

  0 0

The system is dependent; there are infinitely many solutions, given by 2y  t  7  2y  t  7  y  12 t  72 ; and     x  2 12 t  72  5t  3  x  t  7  5t  3  x  4t  10. The solutions are 4t  10 12 t  72  t , where t is any real number. 

1 1

3

3



   43.   4 8 32 24  2 3 11 4



1 1

3

3



   0 4 20 12    0 1 5 2

R2  4R1  R2  R3  2R1  R3

 14 R2

 

1 1

3

3



 1 5 3  . The third row of the 0 0 5

  0 R3  R2  R3 

0

matrix states 0  5, which is impossible. Hence, the system is inconsistent, and there is no solution. 

 44.  

2

6 2 12

1 3

2

3

2

1





1 3 1

6





1 3 1

  1 R1   R2  R1  R2      0 2 10     1 3 2 10   R3  R1  R3  6 1 3 2 6 0

6





1 3 1 6

 R  3R  R  3 2 3  4    0 0 3 12 0 0 1



 0 1 4 . 0 0 0

The system is dependent; there are infinitely many solutions, given by z  4, and x  3y  z  6, so x  3t  4  6  x  3t  2. The solutions are 3t  2 t 4, where t is any real number. 

1

4 2 3



   45.   2 1 5 12  8 5 11 30



1

4 2 3



   0 9 9 18    0 27 27 54

R2  2R1  R2  R3  8R1  R3

R3  3R2  R3





1

4 2 3

  0 9  0 0



 9 18  . 0 0

Therefore, this system has infinitely many solutions, given by 9y  9t  18  y  2  t, and x  4 2  t  2t  3  x  5  2t. Hence, the solutions are 5  2t 2  t t, where t is any real number. 

3

2 3 10





1 1 1 5





1 1 1 5

  R R   R2  3R1  R2  1 2     0 5 0 25  46.   1 1 1 5    3 2 3 10  R3  R1  R3 1 4 1 20 1 4 1 20 0 5 0 25   1 1 1 5    0 5 0 25 . The system is dependent; there are infinitely many solutions, given by 5s   0 0 0 0 r  5  t  5  r  t. Hence, the solutions are t 5 t, where t is any real number. 

2

1 2 12



  1 47.  1 6    1  2 3 3 18 3 2

R1  R2  R1



1 12 1

6



   2 1 2 12    3 32 3 18

R2  2R1  R2  R3  3R1  R3



1 12 1 6

 0 0  0 0

   

R3  R2  R3



 25  s  5, and



 0 0 . 0 0

Therefore, this system has infinitely many solutions, given by x  12 s  t  6  x  6  12 s  t. Hence, the solutions are   6  12 s  t s t , where s and t are any real numbers. 

 0 1 5 7   R R 2  1 48.   3 2 0 12   3 0 10 80



 3 2 0 12   3  R1  R3  0 1 5 7  R    3 0 10 80

Therefore, the system is inconsistent and has no solution.



 3 2 0 12   3  2R2  R3  0 1 5 7  R    0 2 10 68



 3 2 0 12    0 1 5 7 .   0 0 0 82

734

CHAPTER 11 Matrices and Determinants



4 3

1 8



   49.   2 1 3 4  1 1 2 3   1 1 2 3    0 1 1 2    0 1 7 20



1 1

2

3



   2 1 3 4    4 3 1 8  1 1 2 3  R3  R2  R3  0 1 1 2   0 0 6 18 R1  R3





1 1

2

3



   0 1 1 2   0 1 7 20

R2  2R1  R2  R3  4R1  R3



R2 

 . Therefore, 6z  18  z  3; y  3  2  

y  1; and x  1  2 3  3  x  2. Hence, the solution is 2 1 3.



2 3

5

14





1

1 1

3





1

1 1 3

 2  4R1  R2    R1  R3     4 1 2 17  R   0 5 50.    4 1 2 17   R3  2R1  R3  R1  2 3 5 14 0 5 1 1 1 3





1

1 1 3

0

0

 R R R  3 2 3  2 5     0 5

7 20



 2 5  . 5 25

Thus 5z  25  z  5; 5y  2 5  5  5y  15  y  3; and x  3  5  3  x  1. Hence, the solution is 1 3 5. 

2 1

3

9



   51.   1 0 7 10  3 2 1 4



2 1

3

9



   0 1 11 29    0 2 22 34

2R2  R1  R2  R3  3R2  R3



 52.  

4 1 36 24 1 2

9

1

6

2



 3  6

   

4R2  R1  R2  R3  2R2  R3

4 1 36 24



 0 9 72 36   0 3 24 12

3

9



   0 1 11 29 .   0 0 0 24

2R2  R3  R3 

Therefore, the system is inconsistent and there is no solution. 

2 1

 19 R2

 3R3  R2  R3

   

4 1 36 24



 1 8 4  . Therefore, 0 0 0

0 0

the system is dependent. Let z  t. Then y  8t  4  y  4  8t and 4x  y  36t  24 

x   14 [24  4  8t  36t]  5  7t. The solutions are 5  7t 4  8t t where t is any real number.



1

2 3

5



   53.   2 4 6 10  3 7 2 13



1 2

3 5

  0 0 12  0 1 7

R2  2R1  R2  R3  3R1  R3



 0  2



1 2

3 5

 0 1 7  0 0 12

R2  R3





 2 . Therefore, 0

12z  0  z  0; y  7 0  2  y  2; and x  2 2  3 0  5  x  9. Hence, the solution is 9 2 0. 

3 1 0 2



   54.   4 3 1 4  2 5 1 0

3R2  4R1  R2  2R3  R2  R3



3

1 0

2



   0 13 3 20    0 13 3 4

R2  R3  R3



Therefore, the system is inconsistent and there is no solution. 

1 1

6

0

1

 55.  1 1

8



 5  3 14 4

R2  R1  R2  R3  R1  R3



1 1

6

1

5

 0  0

8



 3   4 20 12



3

1 0

2



   0 13 3 20 .   0 0 0 16

R3  4R2  R3





1 1

 0  0

6

8



 1 5 3  . Therefore, 0 0 0

the system is dependent. Let z  t. Then y 5t  3  y  35t and x  y 6t  8  x  83  5t6t  5t. The solutions are 5  t 3  5t t, where t is any real number.

SECTION 11.1 Matrices and Systems of Linear Equations



 56.      

3 1



2 1  1 7   1 3 2 1  1 3 2 1  0 2 1 1   0 10 7 11 4 2







735



1 3 2 1 1 3 2 1    R2  1 R3 2  4R1  R2   4 2 1 7  R 4   0 10 7 11    R3  3R1  R3    3 1 2 1 0 8 4 4   1 3 2 1   R3  5R2  R3  0 2 1 1    . Thus 2z  6  z  3; 2y  3  1  2y  2  0 0 2 6 R1  R3

 R1

y  1; and x  3 1  2 3  1  x  2. Hence, the solution is 2 1 3.     1 2 1 3 3 1 2 1 3 3     R2  3R1  R2  3 4 1 1 9   3 4 1 1 9    R1   R R1  R3 57.   3      1 1 1 1 0   1 1 1 1 0  R4  2R1  R4     2 1 4 2 3 2 1 4 2 3     1 2 1 3 3 1 2 1 3 3      0 1 2 4 9  R  3R  R  0 1 2 4 1 9   323   2 R2       0 3 0 4 3  R4  5R2  R4  0 0 6 8 24      0 5 6 8 9 0 0 4 12 36   1 2 1 3 3    0 1 2 4 9    . Therefore, 20  60    3; 6z  24  24  z  0 0 6 8 24    0 0 0 20 60



1 2 1

 0 2    0 3  0 5

3 3



 4 8 18    0 4 3   6 8

9

3R4  2R3  R4 

 0. Then y  12  9  y  3 and

x  6  9  3  x  0. Hence, the solution is 0 3 0 3.     1 1 1 1 6 1 1 1 1 6     R2  R4  R2 R2  2R1  R2  2 0 1 3  0 2 3 1 4  8 R3  R4  R3     R3 R1  R3 58.       1  1 1 0 4 10   0 2 1 5 16  R4  3R1  R4 2 R4     3 5 1 1 20 0 2 2 2 2     1 1 1 1 6 1 1 1 1 6     0 1 1 1 0 1 1 1 1 1 R4  R2     5 R  3 R  R 4 3 4     . R3  R4  0 0 5 1 2    0 0 5 1 2      0 0 3 7 14 0 0 0 32 64



1 1 1 1

 0 0   0 0  0 1

5 3 1



 2    7 14   1 1 1

Thus 32  64    2; 5z  2  2  5z  0  z  0; y  0  2  1  y  3; and x  3  0  2  6  x  1. Hence the solution is 1 3 0 2.       1 1 2 1 2 1 1 2 1 2 1 1 2 1 2        0 3 1 2 2 0 3 1 2 2 0 3 1 2 2 R3  R1  R3       R  R2  R4 4           59.         1 1 0 3 2  R4  3R1  R4  0 0 2 4 4   0 0 2 4 4        0 3 7 1 1 0 0 6 3 3 3 0 1 2 5   1 1 2 1 2   0 3 1 2 2   R4  3R3  R4 . Therefore, 9  9    1; 2z  4 1  4  z  0. Then    0 0 2 4 4    0 0 0 9 9 3y  0  2 1  2  y  0and x  0  2 0  1  2  x  1. Hence, the solution is 1 0 0 1.

6

736

CHAPTER 11 Matrices and Determinants



1 3 2

1

2



   1 2 0 2 10    60.    0 0 1 5 15    3 0 2 1 3  1 3  0 1  R4  14R3  R4   0 0  0 0



R2  R1  R2  R4  3R1  R4

2

1

2

3

2

1 3

 0   0  0 

2

1 2



 1 2 3 8    0 1 5 15   9 4 2

R4  9R2  R4



3



1 3

 0   0  0

2

1 2



 1 2 3 8    0 1 5 15  

0 14 25 75

 8   . Thus, 45  135    3; z  5 3  15  z  0; 1 5 15   0 45 135

y  2 0  3 3  8  y  1; and x  3 1  2 0  3  2  x  2. Hence, the solution is 2 1 0 3.



1 1

0 1 0



   61.   3 0 1 2 0  1 4 1 2 0



1 1

0

1 0



   0 3 1 1 0    0 3 1 1 0

R2  3R1  R2  R3  R1  R3

R3  R2  R3





1 1

 0  0

0

1 0



 3 1 1 0  . 0 0 0 0

Therefore, the system has infinitely many solutions, given by 3y  s  t  0  y  13 s  t and x  13 s  t  t  0    x  13 s  2t. So the solutions are 13 s  2t  13 s  t  s t , where s and t are any real numbers. 

 2 1 2 1 5   R1  R2   62.   1 1 4 1 3   R1 3 2 1 0 0  1 1 4 1 3  R3  R2  R3    0 1 10 1 11 0 0 1 2 2



1 1 4 1 3

  2 1 2 1  3 2 1 0 



 5  0

R2  2R1  R2  R3  3R1  R3



1 1 4

 0  0

1 3



 1 10 1 11   1 11 3 9

 . Thus, the system has infinitely many solutions, given by z  2t  2  

z  2  2t; y  10 2  2t  t  11  y  31  19t; and x  31  19t  4 2  2t  t  3  x  20  12t. Hence, the solutions are 20  12t 31  19t 2  2t t, where t is any real number.



1

0

1 1

4





1

0

1 1

4





1 0

1 1

4



       0 1 1 0 4   0 1 1 0 4   0 1 1 0 4  R3  R1  R3       R  2 R  R 3 2 3  63.        1 2 3 1 12  R4  2R1  R4  0 2 2 0 8    0 0 0 0 0        2 0 2 5 1 0 0 4 3 9 0 0 4 3 9   1 0 1 1 4    0 1 1 0 4    R3  R4  . Therefore, 4z  3t  9  4z  9  3t  z  94  34 t. Then we have   0 0 4 3 9    0 0 0 0 0     3 9 3 7 7 y  94  34 t  4  y  7 4  4 t and x  4  4 t  t  4  x  4  4 t. Hence, the solutions are   7  7 t  7  3 t 9  3 t t , where t is any real number. 4 4 4 4 4 4

SECTION 11.1 Matrices and Systems of Linear Equations



0  R2  R3  R2 0 R4  R3  R4    1 2 0 0 4 12  2 R3  2 0 2 5 6  1 0 0 2 6   0 R2  2R3  R2  0 1 1 2   R2  R3  0 0 2 9 18  0 0 2 9 18

   64.   

0 1 1 2



3 2

0 1



      



0 1 1 2 0    1 2 0 3 12  R1  R3      R  1 0 0 2 2  R1  R2 6  0 0 2 9 18  1 0 0 2 6   0 1 1 2 0  R4  R3  R4    0 0 2 9 18  0 0 0 0 0





737



1 0 0 2 6    0 2 0 5 18        0 1 1 2 0   0 0 2 9 18

   .  

  Thus, the system has infinitely many solutions given by 2z  9t  18  z  92 t  9; y  92 t  9  2t  0    y  52 t  9; and x  2t  6  x  2t  6. So the solutions are 2t  6 52 t  9 92 t  9 t , where t is any real number. 

075 375 295 40875

 65. Using rref on the matrix   095 875 0 

 3375  , we find that the solution is x  125, y  025, z  075. 125 015 275 36625 131 272 371 139534

 66. Using rref on the matrix   021 0 0

z  381.

42 31



49 27 52

  6  67. Using rref on the matrix   35  0

373

0 42

04



 45   , we find that the solution is x  12, y  34, z  52, 0 67 32 3488   31 48 52 766 0

0

9

0 145

  0 27 0 43 1187  68. Using rref on the matrix   0 31 42 0 721  73 54 0 0 1327   32.



 134322  , we find that the solution is x  371, y  172, 234 456 213984



  13.





   , we find that the solution is x  13, y  07, z  12,  

69. Let x, y, z represent the number of VitaMax, Vitron, and VitaPlus pills taken daily. The matrix representation for the system of equations is   1       5 10 15 50 1 2 3 10 1 2 3 10 1 2 3 10 5 R1   1 2   2  3R1  R2     3  R2  R3   15 20 0 50  5R  3 4 0 10  R    0 2 9 20  R   1       0 2 9 20 . R3  2R1  R3  5 R3 10 10 10 50 2 2 2 10 0 2 4 10 0 0 5 10 Thus, 5z  10  z  2; 2y  18  20  y  1; and x  2  6  10  x  2. Hence, he should take 2 VitaMax, 1 Vitron, and 2 VitaPlus pills daily.

738

CHAPTER 11 Matrices and Determinants

70. Let x be the quantity, in mL, of 10% acid, y the quantity of 20% acid, and z the quantity of 40% acid. Then     1 1 1 100  10R2  01x  02y  04z  18   R2 R1  R2   Writing this equation in matrix form, we get  x  y  z  100 01 02 04 18       R   3 R1  R3   x  4z  0 1 0 4 0     1 1 1 100 1 1 1 100   R3  R2  R3   0 1 3   80    0 1 3 80   . Thus 2z  20  z  10; y  30  80  y  50; and R3 0 1 5 100 0 0 2 20

x  50  10  100  x  40. So he should mix together 40 mL of 10% acid, 50 mL of 20% acid, and 10 mL of 40% acid.

71. Let x, y, and z represent the distance, in miles, of the run, swim, and cycle parts of the race respectively. Then, since distance , we get the following equations from the three contestants’ race times: time  speed        y x z       10    4    20   25  2x  5y  z  50 x  y  z 3  which has the following matrix representation: 4x  5y  2z  90 75 6 15       x    y    z   175  8x  40y  3z  210 

15

2

3

5 1

50



   4 5 2 90    8 40 3 210

40

R2  2R1  R2  R3  4R1  R3



2

5

1

50



   0 5 0 10    0 20 1 10



2

5

1

50



   0 5 0 10 .   0 0 1 30

R3  4R2  R3



Thus, z  30  z  30; 5y  10  y  2; and 2x  10  30  50  x  5. So the race has a 5 mile run, 2 mile swim, and 30 mile cycle.

72. Let a, b, and c be the number of students in classrooms A, B, and C, respectively, where a, b, c  0. Then, the     a  b  c  100     a  b  c  100 1a  1b  system of equations is By substitution, a  52 a  32 a  100  a  20; b  52 a 2 5       c  3b  3a 1b  1c 5 3 5 2 b  52 20  50; and c  32 20  30. So there are 20 students in classroom A, 50 students in classroom B, and 30 students in classroom C.

73. Let t be the number of tables produced, c the number of chairs, and a the number of armoires. Then, the system of equations   1      t  2c  2a  600  2 t  c  a  300 1 3 and a matrix representation is is 2 t  2 c  a  400  t  3c  2a  800      2t  3c  4a  1180  t  3 c  2a  590 

2

1 2 2

600



   1 3 2 800    2 3 4 1180

R2  R1  R2  R3  2R1  R3



1

2 2 600



   0 1 0 200    0 1 0 20

R3  R2  R3





1 2 2 600



   0 1 0 200 .   0 0 0 180

The third row states 0  180, which is impossible, and so the system is inconsistent. Therefore, it is impossible to use all of the available labor-hours.

SECTION 11.1 Matrices and Systems of Linear Equations

739

74. The number of cars entering each intersection must equal the number of cars leaving that intersection. This leads   200  180  x  z      x  70  20   to the following equations: Simplifying and writing this in matrix form, we get     200  y  30     y  z  400  200       1 0 1 0 380 1 0 1 0 380 1 0 1 0 380        1 0 0 1 50  R  R  R  0 0 1 1 430   0 1 0 1 170  2 1 2        R  R 2 3    .     0 1 0 1 170  R4  R3  R4  0 1 0 1 170    0 0 1 1 430        0 1 1 0 600 0 0 1 1 430 0 0 1 1 430

Therefore, z  t  430  z  430  t; y  t  170  y  170  t; and x  430  t  380  x  t  50. Since x, y, z,   0, it follows that 50  t  430, and so the solutions are t  50 170  t 430  t t, where 50  t  430.

75. Line containing the points 0 0 and 1 12: Using the general form of a line, y  ax  b, we substitute for x and y and solve for a and b. The point 0 0 gives 0  a 0  b  b  0; the point 1 12 gives 12  a 1  b  a  12. Since a  12 and b  0, the equation of the line is y  12x. Quadratic containing the points 0 0, 1 12, and 3 6: Using the general form of a quadratic, y  ax 2  bx  c, we substitute for x and y and solve for a, b, and c. The point 0 0 gives 0  a 02  b 0  c  c  0; the point 1 12

gives 12  a 12  b 1  c  a  b  12; the point 3 6 gives 6  a 32  b 3  c  9a  3b  6. Subtracting the third equation from 3 times the third gives 6a  30  a  5. So a  b  12  b  12  a  b  17. Since a  5, b  17, and c  0, the equation of the quadratic is y  5x 2  17x.

Cubic containing the points 0 0, 1 12, 2 40, and 3 6: Using the general form of a cubic, y  ax 3  bx 2  cx  d,

we substitute for x and y and solve for a, b, c, and d. The point 0 0 gives 0  a 03  b 02  c 0  d  d  0; the point the point 1 12 gives 12  a 13  b 12  c 1  d  a  b  c  d  12; the point 2 40 gives

40  a 23  b 22  c 2  d  8a  4b  2c  d  40; the point 3 6 gives 6  a 33  b 32  c 3  d      a  b  c  12 27a  9b  3c  d  6. Since d  0, the system reduces to 8a  4b  2c  40 which has representation    27a  9b  3c  6 

1 1 1 12





1

1

1

12



 12 R2



1 1 1 12





1 1

1 12



   R  3R  R     2  8R1  R2  3 2 3   8 4 2 40  R   0 4 6 56    0 2 3 28  2 3 28    0 2     .   R3 1  27R1  R3  6 R3 0 18 24 318 0 3 4 53 0 0 1 22 27 9 3 6

So c  22 and back-substituting we have 2b  3 22  28  b  47 and a  47  22  0  a  13. So the

cubic is y  13x 3  47x 2  22x. Fourth-degree polynomial containing the points 0 0, 1 12, 2 40, 3 6, and 1 14: Using the general form of a

fourth-degree polynomial, y  ax 4  bx 3  cx 2  dx  e, we substitute for x and y and solve for a, b, c, d, and e. The point

0 0 gives 0  a 04 b 03 c 02 d 0e  e  0; the point 1 12 gives 12  a 14 b 13 c 12 d 1e;

the point 2 40 gives 40  a 24 b 23 c 22 d 2e; the point 3 6 gives 6  a 34 b 33 c 32 d 3e; the point 1 14 gives 14  a 14  b 13  c 12  d 1  e.

740

CHAPTER 11 Matrices and Determinants

Because the first equation is e  0, we eliminate e from the other equations to get      a  b  c  d  12 1 1 1 1 12 1 1 1 1         16a  8b  4c  2d  40  16 8 4 2 40  R2  16R1  R2  0 8 12 14   R3 81R1  R3         6 6  81a  27b  9c  3d    81 27 9 3  R4  R1  R4  0 54 72 78    a  b  c  d  14 1 1 1 1 14 0 2 0 2      1 1 1 1 12 1 1 1 1 12 1 1        0   1 0 1 13  R3  8R2  R3  0 1 0 1 13  R4  6R3  R4  0 1         0 8 12 14 152  R4  54R2  R4  0 0 12 6 48    0 0      0 54 72 78 966 0 0 72 24 264 0 0

So d  2. Then 12c  6 2  48  c  3and b  2  13  b  11.

Finally, a  11  3  2  12  a  4. So the fourth-degree polynomial

12



  1 R4  R2 2 152   R2  R3   966   R3  R4 26  1 1 12  0 1 13   . 12 6 48   0 12 24

40

containing these points is y  4x 4  11x 3  3x 2  2x.

20

-2

2

4

-20

11.2 THE ALGEBRA OF MATRICES 1. We can add (or subtract) two matrices only if they have the same dimension. 2. (a) We can multiply two matrices only if the number of columns in the first matrix is the same as the number of rows in the second matrix. (b) If A is a 3  3 matrix and B is a 4  3 matrix, then (ii) B A and (iii) A A are defined, but (i) AB and (iv) B B are not.

3. (i) A  A and (ii) 2A exist for all matrices A, but (iii) A  A is not defined when A is not square.

4. The entry in the first row, second column is a12  3 3  1 2  2 1  9; the entry in the second row, third column is a23  1 2  2 1  0 0  0; the entry in the third row, first column is a31  1 1  3 3  2 2  4, and      3 1 2 1 3 2 4 9 7          so on:   1 2 0   3 2 1    7 7 0  1 3 2 2 1 0 4 5 5 5. The matrices have different dimensions, so they cannot be equal.  6. Because 14  025, ln 1  0, 2  4, and 3  62 , the corresponding entries are equal, so the matrices are equal.

7. All corresponding entries must be equal, so a  5 and b  3.

8. All corresponding entries must be equal, so a  5 and b  7.             2 6 1 3 1 3 0 1 1 2 1 1 2 0 2       9.  10.  5 3 6 2 1 5 1 1 0 1 3 2 0 2 2         1 1 0 1 1 1 2 3 6               11. 3  12. 2   1 0 1    2 1  is undefined because these  4 1    12 3  0 1 1 3 1 1 0 3 0 matrices have incompatible dimensions.

SECTION 11.2 The Algebra of Matrices









2 6 1 2      3 6  is undefined because these 13.  1 3    2 4 2 0

  14.  6 3 4 

matrices have incompatible dimensions.

















2 1 2

2 3





   1 2   6  7 6 0

1 2 3 2 

741





7



   5  1 1    1 4 2 2 1 7 10 7 1 7 1 2         1  2 5   4 6  1  2 1   1  12  1 17. 2X  A  B  X  2 B  A  .    2 2 1 2 3 7 2 4 1 3

15. 

1 2



1 2

3



5

2

1

 16.  0



18. 3X  B  C  3X  C  B. but C is a 3  2 matrix and B is a 2  2 matrix, so C  B is impossible. Thus, there is no solution. 19. 2 B  X  D. Since B is a 2  2 matrix, B  X is defined only when X is a 2  2 matrix, so 2 B  X is a 2  2 matrix. But D is a 3  2 matrix. Thus, there is no solution. 20. 5 X  C  D  X  C  15 D            10 20 2 3 2 4 2 3 4 7          1         X  15 D  C   30 20     1 0    6 4    1 0    7 4 . 5 10 0 0 2 2 0 0 2 2 2 21. 15 X  D  C  X  D  5C       2 3 10 20 10 15           X  5C  D  5  1 0    30 20    5 0 0 2 10 0 0 10 22. 2A  B  3X  2A  B  3X 





10 20





0

5



        30 20    25 20 .      10 0 10 10

X   13 2A  B              1  8 12   2 5  1  6 7   2  73  1   4 6   2 5  2       1 1 3 3 3 1 1 1 3 3 7 2 6 3 7 3 3

In Solutions 23–36, the matrices A, B, C, D, E, F, G, and H are defined as follows:       0 2 5 2  52 3 12 5    A C  B 1 1 3 0 2 3 0 7       1 1 0 0 5 3 10          E  F  G 2 0 1 0  6 1 0 0 0 0 1 5 2 2       0 3 12 5 2  52 5 2 5    23. (a) B  C   1 1 3 0 2 3 1 1 0

(b) B  F is undefined because B 2  3 and F 3  3 don’t have the same dimensions.       0 3 12 5 1 3 5 2  52    24. (a) C  B   0 2 3 1 1 3 1 3 6

D



7 3





3

1

H 

2 1

 

742

CHAPTER 11 Matrices and Determinants

     0 2  52 3 12 5 14 8 30   6   (b) 2C  6B  2  0 2 3 1 1 3 6 10 24 



2 5

25. (a) 5A  5 

0

7







10 25 0

35

 

(b) C  5A is undefined because C 2  3 and A 2  2 don’t have the same dimensions.       2  52 0 13  72 15 3 12 5  2   26. (a) 3B  2C  3  1 1 3 0 2 3 3 1 3

(b) 2H  D is undefined because 2H 2  2 and D 1  2 don’t have the same dimensions.

27. (a) AD is undefined because A 2  2 and D 1  2 have incompatible dimensions.     2 5     14 14 (b) D A  7 3  0 7 28. (a) D H 







7 3 

3

1

2 1







27 4



(b) H D is undefined because H and D have incompatible dimensions.      3 1 4 7 2 5    29. (a) AH   2 1 14 7 0 7      3 1 2 5 6 8    (b)  2 1 0 7 4 17

30. (a) BC is undefined because B 2  3 and C 2  3 have incompatible dimensions.    1 0 0    1  3 12 5 3 2 5   0 1 0     (b) B F    1 1 3  1 1 3 0 0 1 

 31. (a) G F    

 (b) G E   

5 3 10 6

1

5

2

5 3 6

1

5

2



  0  2  10   0  2

1 0 0





5 3 10

   0 1 0  6 0 0 1 5    1 1       2  8  0 1

32. (a) B 2 is undefined because B 2  3 is not square.      1 0 0 1 0 0 1 0 0          (b) F 2    0 1 0  0 1 0    0 1 0  0 0 1 0 0 1 0 0 1 

33. (a) A2  

2 5 0

7

 

2 5 0

7







4 45 0

49

 

1 2



 0  2

SECTION 11.2 The Algebra of Matrices



(b) A3  

2 5 0

34. (a) D A B  (b) D AB  

35. (a) AB E  

7  

  

0 

7

0



 2 7 3  0 7



2 5

7 3 

2 5 0

2 5



 

3 1

2 5

0  

7 3





 

4 45 0

49

 

2 5 0

7 

1 5   3 2   14 14 





8 335 0 

343

743

 

1 5   2   28 21 28

1 1 3    1 6 5   5 3 12 5  7 3    28 21 28  1 1 3 7 7 21 7      1   1   1 5     1 6 5 13 2  2     2     1 3   7 7 21   7 0 0 7



1 1 3







(b) AH E is undefined because the dimensions of AH 2  2 and E 3  1 are incompatible.         5 2 5     0 2  52 3 12 5    7 3    38 11 35 36. (a) D BDC  D B  C  7 3  1 1 3 0 2 3 1 1 0 (b) B F  F E is undefined because the dimensions 2  3  3  3  2  3 and 3  3  3  1  3  1 are incompatible.

In Solutions 37–42, the matrices A, B, and C are defined as follows:       12 01 03 11 24     02 02 01    C  B A  0 05   09 01 04  11 21 21 05 21 07 03 05   156 562    37. AB   128 088  38. B has 2 columns and A has 3 rows, so B A is undefined.  109 097     035 003 033   019 029   39. BC   40. C B    055 105 105  027 325 241 431 446   06 094 004   41. B and C have different dimensions, so B  C is undefined. 42. A2   112 192   01  041 095 131     x 2        2y  2 x 2y 2 2  . Thus we must solve the system So x  2 and 2y  2  y  1. Since 43.   4  2x 4 6 2x 6y      6  6y these values for x and y also satisfy the last two equations, the solution is x  2, y  1.

  3x        x y 6 9 x y 3x 3y 3y  . Since 3   , we must solve the system 44. 3   y x 9 6 y x 3y 3x 3x      3y 











x  2 and 3y  9  y  3. Thus, the solution is x  2, y  3.



6

 9 

6

 9

So 3x  6 

744

CHAPTER 11 Matrices and Determinants



45. 2 

x

y





2 4





x



y



2x



2y

 . Since 2   , Thus we must solve the xy xy 2 6 xy xy 2 x  y 2 x  y   2x  2      2y  4 So x  1 and y  2. Since these values for x and y also satisfy the last two equations, the system  2 x  y  2      2 x  y  6

solution is x  1, y  2.           x y y x 4 4 x y yx 4 4     . Thus we must solve the system 46.  y x x y 6 6 x  y x  y 6 6   xy 4      y  x  4 Adding the first equation to the last equation gives 2x  10  x  5 and 5  y  6  y  1. So the   x  y  6     xy 6

solution is x  5, y  1.        2x  5y  7 2 5 x 7      . 47. written as a matrix equation is   3x  2y  4 3 2 y 4     6x  y  z  12 48. 2x  z 7    y  2z  4





2 2 2 2 

1



x





12

0

0







    written as a matrix equation is   1 0 1 0    0 3 1 1  

   written as a matrix equation is   





        1   y    7 . 4 1 2 z

 written as a matrix equation is  2

    3x1  2x2  x3  x4  0 49. x1  x3 5    3x2  x3 x4  4   x y z      4x  2y  z 50.   x  y  5z     x  y  z

6 1

1 1

3 2 1

1





1





x1 x2 x3 x4 2

 x    4 2 1   2     y    2 1 1 5  z 1 1 1 2 1







    0       5 .   4

   .  

   0    . ABC is undefined because the dimensions of A (2  4) 1 7 9 2 , and C    1  2 12 4 0   2      3  3 21 27 6  1 7 9 2   . B AC is undefined because and B (1  4) are not compatible. AC B   2 2 14 18 4

51. A  

1 0 6 1

, B 



the dimensions of B (1  4) and A (2  4) are not compatible. BC A is undefined because the dimensions of C (4  1) and A (2  4) are not compatible. C AB is undefined because the dimensions of C (4  1) and A (2  4) are not compatible. C B A is undefined because the dimensions of B (1  4) and A (2  4) are not compatible.

SECTION 11.2 The Algebra of Matrices



52. (a) Let A  

a b c d 





 and B  

e f g h







. Then A  B  





A2  

a b



AB  

a b

c d

c d

cg d h

cg d h

a  e2  b  f  c  g

c  g a  e  d  h c  g

 

a b

 

e f g h

A2  AB  B A  B 2  













c d







cg d h



ae b f ae b f   A  B2   

ae b f

ac  cd bc  d 2

ae  bg a f  bh ce  dg c f  dh

, and



a  e b  f   b  f  d  h

a 2  bc ab  bd

c  g b  f   d  h2









; B 2  

; B A  

e f g h e f g h

 

 

e f g h a b c d

;





e2  f g e f  f h







ae  c f be  d f



 

eg  gh f g  h 2

ag  ch bg  dh

a 2  bc  ae  bg  ae  c f  e2  f g ab  bd  e f  f h  a f  bh  be  d f

ac  cd  eg  gh  ce  dg  ag  ch bc  d 2  f g  h 2  c f  dh  bg  dh a 2  2ae  e2  b c  g  f c  g

745



;

. Then  

a b  f   e b  f   b d  h  f d  h

c b  f   g b  f   d 2  2dh  h 2   a  e b  f   b  f  d  h a  e2  b  f  c  g   A  B2  c  g a  e  d  h c  g c  g b  f   d  h2 c a  e  g a  e  d c  g  h c  g

(b) No. From part (a), A  B2  A  B A  B  A2  AB  B A  B 2  A2  2AB  B 2 unless AB  B A which is not true in general, as we saw in Example 5.      5 075 010 0 4          53. (a) AB    025 070 070   20    22  7 0 020 030 10

(b) Five members of the group have no postsecondary education, 22 have 1 to 4 years, and seven have more than 4 years.      075 020 005 80 96          54. (a) AB    060 030 010   170    103  040 030 030 40 95

(b) 96 students slept less than 4 hours, 103 slept 4 to 7 hours, and 95 slept more than 7 hours.      50 20 15 350 35375          55. (a) AB    40 75 20   575    65625  35 60 100 425 89250 (b) The total revenue for Monday is the 1 1th entry of the product matrix, $35375.

(c) The total revenue is the sum of the three entries in the product matrix, $190250.      4000 1000 3500      $4690 $1690 $13,210 56. (a) B A  $090 $080 $110  400 300 200   700 500 9000

 

746

CHAPTER 11 Matrices and Determinants

(b) The entries in the product matrix represent the total food sales in Santa Monica, Long Beach, and Anaheim, respectively. 

12 10

 57. (a) AB    4 8

0



$1000

$500





$32,000 $18,000



        4 20    $2000 $1200    $42,000 $26,800  9 12 $1500 $1000 $44,000 $26,800

(b) The daily profit in January from the Biloxi plant is the 2 1 matrix entry, namely $42,000. (c) The total daily profit from all three plants in February was $18,000  $26,800  $26,800  $71,600.

58. (a) AB 





2000 2500



     3000 1500       105,000 58,000 6 10 14 28   2500 1000    1000 500

(b) That day they canned 105,000 ounces of tomato sauce and 58,000 ounces of tomato paste. 

120 50 60



010





9700



         59. (a) AC    40 25 30   050    4650  60 30 20 100 4100

Amy’s stand sold $97 worth of produce on Saturday, Beth’s stand

sold $4650 worth, and Chad’s stand sold $41 worth.      100 60 30 010 7000          (b) BC    35 20 20   050    3350  Amy’s stand sold $70 worth of produce on Sunday, Beth’s stand 60 25 30 100 4850

sold $3350 worth, and Chad’s stand sold $4850 worth.       220 110 90 100 60 30 120 50 60            (c) A  B    40 25 30    35 20 20    75 45 50  120 55 50 60 25 30 60 30 20

This represents the melons, squash, and

tomatoes they sold during the weekend.           120 50 60 100 60 30 010 220 110 90 010 16700                        (d) A  B C   40 25 30    35 20 20   050    75 45 50   050    8000 60 30 20 60 25 30 100 120 55 50 100 8950

   

During the weekend, Amy’s stand sold $167 worth, Beth’s stand sold $80 worth, and Chad’s stand sold $8950 worth of       9700 7000 16700            produce. Notice that A  B C  AC  BC    4650    3350    8000 . 4100 4850 8950 

     60. (a)      

1 0 1 0 1 1



 0 3 0 1 2 1   1 2 0 0 3 0   1 3 2 3 2 0  0 3 0 0 2 1  1 2 0 1 3 1



     (b)      

2 1 2 1 2 2



 1 3 1 2 3 2   2 3 1 1 3 1   2 3 3 3 3 1  1 3 1 1 3 2  2 3 1 2 3 2



     (c)      

2 3 2 3 2 2



 3 0 3 2 1 2   2 1 3 3 0 3   2 0 1 0 1 3  3 0 3 3 1 2  2 1 3 2 0 2

SECTION 11.2 The Algebra of Matrices



     (d)      

0 0 0 0 0 0

747



 0 3 0 0 3 0   0 3 0 0 3 0   0 3 3 3 3 0  0 3 0 0 3 0  0 3 0 0 3 0

(e)

61. Suppose A is n  m and B is i  j. If the product AB is defined, then m  i. If the product B A is defined, then j  n. Thus if both products are defined and if A is n  m, then B must be m  n.



62. A  

1 1 0 1

























0 1 0 1 0 1 0 1 0 1       1 1 1 3 1 n 1 4   . Therefore, it seems that An   . A4  A  A3   0 1 0 1 0 1 0 1













1 1 1 1 1 1 1 2 1 2 1 3 ; A2           ; A3  A  A2   ; 0 1



1 1







1 1

2 2

1 1



2 2

4 4



        ; A3  A  A2   ; 1 1 1 1 1 1 2 2 2 2 4 4        n1 2n1 1 1 4 4 8 8 2   . From this pattern, we see that An   . A4  A  A3   1 1 4 4 8 8 2n1 2n1

63. A  

1 1

; A2  

1 1

748

CHAPTER 11 Matrices and Determinants

64. Let A

 



a b c d



  



a b c d



.

For the first matrix, we have A2

a 2  bc ab  bd

ac  cd bc  d 2







a 2  bc b a  d

c a  d bc  d 2



 

a b



c d 



  

4 0  . So A2   0 9

  a 2  bc  4      b a  d  0

 c a  d  0      bc  d 2  9

If a  d  0, then a  d, so 4  a 2  bc  d2  bc  d 2  bc  9, which is a contradiction. Thus a  d  0. Since b a  d  0 and c a  d  0, we must have b  0 and c  0. So the first equation becomes a 2  4  a  2, and the

fourth equation becomes d 2  9  d  3.           4 0 2 0 2 0 2 0 2 0  are A1   , A2   , A3   , and A4   . Thus the square roots of  0 9 0 3 0 3 0 3 0 3   a 2  bc  1        b a  d  5 1 5  For the second matrix, we have A2   Since a  d  0 and c a  d  0, we must have  c a  d  0 0 9      bc  d 2  9     a2  1 a  1     c  0. The equations then simplify into the system b a  d  5  b a  d  5       d2  9 d  3 We consider the four possible values of a and d. If a  1 and d  3, then b a  d  5  b 4  5  b  54 . If a  1 and d  3, then b a  d  5  b 2  5  b   52 . If a  1 and d  3, then b a  d  5  b 2  5    1 5  are b  52 . If a  1 and d  3, then b a  d  5  b 4  5  b   54 . Thus, the square roots of  0 9         1 54 1  52 1 52 1  54 , A2   , A3   , and A4   . A1   0 3 0 3 0 3 0 3

11.3 INVERSES OF MATRICES AND MATRIX EQUATIONS 

1. (a) The matrix I  

1 0 0 1



 is called an identity matrix.

(b) If A is a 2  2 matrix then A  I  A and I  A  A.

(c) If A and B are 2  2 matrices with AB  I then B is the inverse of A. 

2. (a) 

5 3 3 2

 

x y







4 3

 

      d b 2 3 2 3 1 1     . (b) The inverse of A is A1  ad  bc c a 5  2  3  3 3 5 3 5

SECTION 11.3 Inverses of Matrices and Matrix Equations



(c) The solution of the matrix equation is X  A1 B  

2 3 5

3

(d) The solution of the system is x  1, y  3. 

4 1

3. A  





2 1

 



4





3

1 3

749



.



; B   . 7 2 7 4           1 0 4 1 2 1 1 0 2 1 4 1 .    and B A     AB   0 1 7 2 7 4 0 1 7 4 7 2 

4. A  







7 3 2 . ; B   2

2 3

2 1      7 3 2 3 1 0 2 2   . BA   2 1 4 7 0 1 

 5. A   

4 7

1

3 1

1

4





8 3



AB  

4



     0 ; B   2 1 1 . 1 3 2 1 0 1     8 3 4 1 3 1 1         BA     2 1 1   1 4 0   0 1 0 1 1 3 2 0 

3 2

4





9 10 8



       6. A    1 1 6 ; B   12 14 11 . 1 1 2 1 12  12 2 2     9 10 8 3 2 4 1       1 1 6    0 BA   12 14 11     1 1  12 2 1 12 0 2 2

2 3 4 7



 AB    0 0











8 3

4

9 10 8





7 3 2    1 0  and  2

2 1

1

3 1

1

4

1 3

  0   2 2 1

0 1





1 0 0



     1 1     0 1 0  and 0 0 1 0 1

 1 0  0 1 

3 2

4



   AB    1 1 6   12 2 1 12  12  0 0  1 0 . 0 1



1 0 0



     14 11     0 1 0  and 1 1 0 0 1 2 2

         2 4 7 4 1 2 1 0 1 2 1 . Then, A A1          A1  7. A   14  12 3 7 3 2  32 72 0 1 3 2  32 27      1 2 7 4 1 0  . and A1 A   3 7   2 2 3 2 0 1 

7 4



750

CHAPTER 11 Matrices and Determinants



1

 8. B   

3 2



 2 2  We begin with a 3  6 matrix whose left half is B and whose right half is I3 . 2 1 0       1 3 2 1 0 0 1 3 2 1 0 0 1 3 2 1 0 0   R  2R  R   2R  5R  R   2 3  0 2 2 0 1 0  313  0 2 2 0 1 0  3 0 2 2 0 1 0        2 1 0 0 0 1 0 5 4 2 0 1 0 0 2 4 5 2     0 1 3 2 1 0 0 1 0 1 1  32 1  R1  3R2  R1  R R R   2 R2 1 3 1 0 1 1 0 1  0 1 0 2 2   0 1  R2     2 R3  R2  12 R3 0 0 1 2 52 1 0 0 1 2 52 1          1 1 1 1 1 1 1 0 0 1 1 1 1 3 2 1 0 0           0 1 0 2 2 1 . Then B 1   2 2 1 ; B 1 B   2 2 1   0 2 2    0 1 0 ;          2 52 1 2 52 1 0 0 1 2 52 1 2 1 0 0 0 1      1 3 2 1 1 1 1 0 0          and B B 1    0 2 2   2 2 1    0 1 0 . 2 1 0 2 52 1 0 0 1   1 1 2  and verify that A1 A  A A1  I . 9. Using a calculator, we find A1   3 2 2 2   1 1 0   1 1  10. Using a calculator, we find B 1    33 31 3  and verify that B B  B B  I3 . 13 12 1  1     3 5 3 5 3 5 1      11.  9  10 2 3 2 3 2 3 

12. 

3 4



13. 

2

7 9

0

1 

5

5 13

      9 4 9 4 9 4 1        27  28 7 3 7 3 7 3

1 

    13 5 13 5 1     26  25 5 2 5 2

    5 4  53  43 1      14.  35  32 8 7  83  73 8 5    1 4 3 6 3 1  , which is not defined, and so there is no inverse.   15.  24  24 8 6 8 4 



1 16.  2



17. 

4

7

1

1 1 3  

5 4

04 12 03

06

   4  13 12 1    1 2  53 5 15 32 2

1 

1



    06 12 1 2 1     024  036 03 04  12 23

SECTION 11.3 Inverses of Matrices and Matrix Equations



4  18.  3 1  1  0  0

2 3 1 0 0







4 2 3 1 0 0   R1  R3  R1  0 6 1 3 4 0     3R3 R2  R3 0 2 1 1 0 4     0 1 0 0 1 1 0 0 3 2 5 1 0 1  R1  R3  R1     0 1 6 R2   0 6 0 6 6 1 3 4 0  6 6  R2  R3  R2     0 1 3 2 6 0 0 1 3 2 6 0 0   3 2 5    Therefore, the inverse matrix is  1 1 1  . 3 2 6



 3 2 0 1 0  0 1 0 0 1

2 4

4R2  3R1  R2  4R3  R1  R3

1 1 0 0

 19.   1 1 1 0 1 0 1 4 0 0 0 1  6  R1  5R3  R1 0   R2  R3  R2 0

   



2 4

1

1 0 0





4 0 4 0 0 4    0 6 1 3 4 0    0 0 2 6 4 12  0 3 2 5  0 1 1 1  . 1 3 2 6





6 0

5

751 1 4 R1

 1 2 R3

1 4 0



    3R3  2R2  R3  0 6 1 1 2 0   0 6 1 1 2 0       3R1  2R2  R1 0 4 1 1 0 2 0 0 1 5 4 6    0 0 24 24 30 1 0 0 4 4 5 1    6 R1  0 1 0 1 1 1 .  6 0 6 6 6     1 6 R2 , R3 0 1 5 4 6 0 0 1 5 4 6   4 4 5    Therefore, the inverse matrix is   1 1 1 . 5 4 6



5

7 4 1 0 0

2R2  R1  R2  2R3  R1  R3





1

0 1 1 0 1

  R3  R1  R3    3 1 3 0 1 0  20.   3 1 3 0 1 0    R1  R3 6 7 5 0 0 1 5 7 4 1 0 0   1 0 1 1 0 1   1  R3  R1 R3  7R2  R3  0 1 0 3 1 3  R     R 2 , R3 0 0 1 27 7 26   26 7 25   Therefore, the inverse matrix is  3  3 1 . 27 7 26



 1 2 3 1 0 0    21.   4 5 1 0 1 0  1 1 10 0 0 1

R2  4R1  R2  R3  R1  R3



   

   



1

0

1 1 0

  0 1 0  0 7 1  1 0 0 26 7 25  0 1 0 3 1 3 . 0 0 1 27 7 26 R2  3R1  R2  R3  5R1  R3

 1 2 3 1 0 0    0 3 13 4 1 0    0 3 13 1 0 1

R3  R2  R3



Since the left half of the last row consists entirely of zeros, there is no inverse matrix.



1



 3 1 3   6 0 5

 1 2 3 1 0 0    0 3 13 4 1 0 .   0 0 0 3 1 1

752

CHAPTER 11 Matrices and Determinants



2 1 0 1 0 0

 22.  1 1 4 0 1 0 2 1 2 0 0 1  1 0 4 1   0 1 8 1  0 0 2 1

   



1 1 4 0 1 0



  2 1 0 1 0 0   2 1 2 0 0 1   1 0 1 0  R1  2R3  R1   0 1  2 0   R2  4R3  R2  0 1 0 0   1 1 2    Therefore, the inverse matrix is  3 2 4  .  12 0 12



0 2 2 1 0 0

R1  R2







1 2 3 0 0 1

   R R 1 3   23.   3 1 3 0 1 0    3 0 1 2 3 0 0 1  1 0 1 1 0 1  R1  R3  R1  0 1 0 3 1 3   R2  3R3  R2 0 2 2 1 0 0   1 0 1 1 0 1   1  R3  R1  0 1 0 3 1 3  R    0 0 1 72 1 3    92 1 4    3 1 3 .   7 1 3 2



3 2 0 1 0 0

 24.  5 2  1  0  0





1

R2  2R1  R2  R3  2R1  R3

0 1 1 2

   



 0 3 2 4   2 1 0 1



 1 3 0 1 0  2 2 1 0 0    



1

1

4 0

  0 1 8  0 1 6  1  R2 0  1  2 R3 0



1 0



 1 2 0   0 2 1

R1  R2  R1  R3  R2  R3

0 0 1 1

1 0

0 1  12

3 0 0

0 0

1 0 1



 2 4  . 1 0 2

3

1 2

2

1



   0 7 6 0 1 3    0 2 2 1 0 0   1 0 1 1 0 1   1 R3  2R2  R3  0 1 0 3 1 3  2 R3     0 0 2 7 2 6  1 0 0  92 1 4  0 1 0 3 1 3  . Therefore, the inverse matrix is 7 0 0 1 1 3 2

0 0 1 0 1



R2  3R1  R2





1



 R2  5R1  R2    R R R  1 3 1     0 1 1 5 1 5  1 1 0 1 0     5 1 1 0 1 0   R3  2R1  R3  2 2 0 0 0 1 0 2 0 2 0 3 2 0 0 0 1     0 0 1 0 1 1 0 0 1 0 1 1 1     2 R3 3      1 1 5 1 5  0 1 0 1 0  2 . Therefore, the inverse matrix is  1 R2  R3  R2  0 0 1 6 1 13 6 0 2 12 2 13 2

R3  2R2  R3



0 1



 0  32  . 13 1 2

SECTION 11.3 Inverses of Matrices and Matrix Equations







1 2 0 3 1 0 0 0 1     0 1 1 1 0 1 0 0  R R R 0 3 2 3      25.     0 1 0 1 0 0 1 0  R4  R1  R4 0    1 2 0 2 0 0 0 1 0   1 2 0 3 1 0 0 0    0 1 1 1 0 1 0 0  R  2R  R  121    R2  R3  R2  0 0 1 0 0 1 1 0    0 0 0 1 1 0 0 1   1 0 0 1 1 0 2 0    0 1 0 0 1 0 1 1    R1  R1  R4    0 0 1 0 0 1 1 0     0 0 0 1 1 0 0 1   0 0 2 1    1 0 1 1     .  0 1 1 0    1 0 0 1 

1 0  0 1  26.  1 1  1 1

1 0 1 0 0 0



 0 1 0 1 0 0 4  R1  R4 R   R3  R1  R3 1 0 0 0 1 0  1 1 0 0 0 1

Therefore, there is no inverse matrix. 

 27.  



   29.    

3

2

3

 3  0 2

0 1 1

1 4 1 0 2

0

1



2  3

1

1 0 0 0

1

  0 2 0 0   33.   0 0 4 0   0 0 0 7



3 1 2

2



1 0 0 0



  0 1 0 0    2   0 0 1 0    4 0 0 0 17

1

1

0

0 1

0

1 0

0 0 0

 1 0 0   0 1 1 0  

0 1 1

0 0 1

1 0 2 1 1 2 0 1 0 1 0 0 0 1 1 0 0

R3  R4

0

0 0 1 1 0

0

0 1

1

1 0 0 0



0

0





5

2

5

1



3

 28.      30.   

1

1



1 0  0 1   0 0  0 0

1

 0  0 1 2 0 1

1

3

1

2 3 1 0 0 0



1 0

1

  2 5 0 0   32.   4 2 3 0   5 1 2 1 0

0

1

0

1

0

1



  



0 0 0

2 5  15



1 3 25  25  2 1  5 5  1  5  35

 1 16  16    1 1 0 0    4  2    1 1 3 3   3 13 1 2 56 6 6 5 6

1

0

1 0

0

0 0



  1  2 0 0   5 5   16  2 1 0   15 3  15  1 2 1  37 15 15 3 



 1 0 0  . 0 1 1 1 1 0   0 0 1 1 0 1

2  25

0

1

 0  0 0 3 0 2

1

1



 1 1    0 1 

3 1



 34.  



 0 1  2R3  R1  R   R  2  R4  R2 1 1 0  0 0 1  0 2 1  0 1 1  . Therefore, the inverse matrix is 1 1 0   0 0 1 0

 0 1 0 1 0 0 4  R2  R4 R   R3  R2  R3 0 0 1 0 1 0   0 1 1 0 0 1

   0 1 0 1    2    2 2 1 2    1 1 1 0 

3

1

 0 1   0 0  0 0  1 0  0 1   0 0  0 0





0

1 0 

4 3 3 

1 1  72 6   1 1   0  2 6  1 0 0 3

1

1 0  0 1   0 1  0 1

   1 1 3 1 2 1 3 3

 0   4 1 2   2 2 0 0 1

1 7 3    31.  0 2 1   0 0 3 

1



2



   0 0



 0   0  13 1 2

753

754

CHAPTER 11 Matrices and Determinants

In Solutions 35–38, the matrices A and B are defined as follows:   1 0 2    A  0 2 1  4 2 1 



 14

3 3 4 4   7 23 3  35. A1 B     16  16  16  7 1 5 8 8 8



7 3 4

 22 2 37. B AB 1    7 7 50 7

26 7

 

16  7  37 7



 B  

2 1 2 0

3

1

0



 1  2

0

0

1

25 7

13 7

22 7



  1 5 2  36. AB 1    7 7 7  



4 8 13 7 7   7 1 8 17 6  38. B AB    7  7 7  31 9  10 7 7 7

       3x  5y  4 3 5 x 4      . Using the inverse from Exercise 11, 39. is equivalent to the matrix equation   2x  3y  0 2 3 y 0        x 3 5 4 12       . Therefore, x  12 and y  8. y 2 3 0 8        3x  4y  10 3 4 x 10       is equivalent to the matrix equation  40.  7x  9y  20 7 9 y 20        x 9 4 10 10     . Therefore, x  10 and y  10. y 7 3 20 10

       2x  5y  2 2 5 x 2     . Using the inverse from 41. is equivalent to the matrix equation   5x  13y  20 5 13 y 20        x 13 5 2 126   . Therefore, x  126 and y  50. Exercise 13,     y 5 2 20 50        7x  4y  0 7 4 x 0       42. is equivalent to the matrix equation   8x  5y  100 8 5 y 100        0  400 x  53  43 3 . Therefore, x   400 and y   700 .     3 3  83  73  700 100 y 3

       2 4 1 x 7   2x  4y  z  7          43. x  y  z  0 is equivalent to the matrix equation   1 1 1   y    0 . Using the inverse from    x  4y  2 1 4 0 z 2        x 4 4 5 7 38               Exercise 19,  y    1 1 1   0    9 . Therefore, x  38, y  9, and z  47. z 5 4 6 2 47

SECTION 11.3 Inverses of Matrices and Matrix Equations

   5x  7y  4z  1 5    44. 3x  y  3z  1 is equivalent to the matrix equation  3    6x  7y  5z  1 6        x 26 7 25 1 8         y    3 1   1    1 . Therefore, x  8, 3        z 27 7 26 1 8    2y  2z  12 0     45. 3x  y  3z  2 is equivalent to the matrix equation  3    x  2y  3z  8 1 

x





 92 1

4



12





20









7 4 x 1     y  1  1 3      7 5 z 1 y  1, and z  8. 2 2



x





12



        1 3   y    2 . Using the inverse from 8 2 3 z

             Exercise 23,   y    3 1 3   2    10 . Therefore, x  20, y  10, and z  16. 7 z 1 3 8 16 2        0 x  2y  3  0 1 2 0 3 x                 yz 1 1  0 1 1 1  y  46. is equivalent to the matrix equation        2  0 1 0 1  z   y  2           x  2y 3   2  3 1 2 0 2        1 0 0 2 1 0 x         y   1 0 1 1   1   5           Therefore, x  1, y  5, z  1, and   3.        z   0 1 1 0   2   1         3 1 0 0 1 3  47. Using a calculator, we get the result 3 2 1.

48. Using a calculator, we get the result 1 2 3.

49. Using a calculator, we get the result 3 2 2.

50. Using a calculator, we get the result 6 12 24.

51. Using a calculator, we get the result 8 1 0 3.

755



52. Using a calculator, we get the result 8 4 2 1.   53. This has the form M X  C, so M 1 M X  M 1 C and M 1 M X  M 1 M X  X.    1   3 2 3 2 3 2 1 1      . Since X  M 1 C, we get Now M    98 4 3 4 3 4 3        x y z 3 2 1 0 1 7 2 3     . u   4 3 2 1 3 10 3 5      39 3 6  39  92 1 4 2          54. Using the inverse matrix from Exercise 23, we see that   3 1 3   6 12    15 30 . 7 33 1 3 0 0 33 2 2      39 39 x u    2     Hence,   y     15 30 . 33 z  33 2  1       a a a a 1 1 a a 1 1 1          55.  2a 1 1 a 2  a 2 2a 2 a a a a a a

756

CHAPTER 11 Matrices and Determinants



a 0 0 0 1 0 0 0



2 x



  0 b 0 0 0 1 0 0   56.   0 0 c 0 0 0 1 0   0 0 0 d 0 0 0 1  a 0 0 0  0 b 0 0  Thus the matrix  0 0 c 0  0 0 0 d 57. 

x x2

1 

1 a R1 1 b R2



1 0 0 0 1a

0

0

0

  0 1 0 0 0 1b 0 0     1  0 1c 0 c R3 0 0 1 0 0 1 d R4 0 0 0 1 0 0 0 1d    1a 0 0 0     0 1b 0  0     .  has inverse    0  0 1c 0    0 0 0 1d



   .  

      1 1x x 2 x x 2 x 1 1     .  2 2x  x 2 x 2 x 2 x 2 1x 2x 2

The inverse does not exist when x  0.



58.  

e x e2x

e2x 1

e3x

ex

1 

    x e2x e3x e2x e 1   1 . The inverse exists for all x.  4x 2 e  e4x e2x e x e2x e3x

0 1 0 0





1

ex

0

1

0 0



  12 e2x R2   0 2e2x 0 e x 1 0      1 2 R3 0 0 2 0 0 1    1 1 ex 1 0 0 0 1 0 0 2 2  R  ex R  R   1 ex  1 e2x 0  1 x  1 e2x 0 . 1 2 1     0 1 0 2 e  2 2 2 1 1 0 0 0 0 1 0 0 2 2   1 1 ex 0 2  2  1 ex  1 e2x 0 . The inverse exists for all x. Therefore, the inverse matrix is   2  2 1 0 0 2

 x 2x 59.   e e 0 0  x 1 e 0  0 1 0  0 0 1

 0 0 1 0  2 0 0 1

1

R2  e x R1  R2



   1  x 1  1 1  1 1 1   x 1  x 1   x2  x 1  x2  . 60.    x 1 x   1   x 2  1  x x x x x x x x 1 x 1 x x The inverse exists for x  0, 1. 

x



1

3 1 3 1 0 0







4 2 4

0 1 0





1 1 1 1 1 0



  R3  R1  R3     1  R2  R1  3 1 3 1 0 0 R    61. (a)   4 2 4 0 1 0       3 1 3 1 0 0  R1  R2 3 2 4 0 0 1 0 1 1 1 0 1 0 1 1 1 0 1     R1  12 R2  R1 1 1 1 1 1 0 1 0 1 1  12 0     R R R  12 R2 R2  3R1  R2 3 0 1 3 1  0 2 0 4 3 0    0 1 0 2        2 1 R3  2 R2  R3 0 0 1 1  32 1 0 1 1 1 0 1     0 1 1 1 0 0 0 1 1     3 3   0 1 0 2 0 0 . Therefore, the inverse of the matrix is  2 .  2 2 3 3 0 0 1 1 2 1 2 1 1

SECTION 11.3 Inverses of Matrices and Matrix Equations













757



A 0 1 1 10 1        3    2   14    1 . (b)  B 0        2 C 1  32 1 13 2 Therefore, he should feed the rats 1 oz of food A, 1 oz of food B, and 2 oz of food C.        A 0 1 1 9 2        3       (c)    0   12   0   B   2 . 2 3 C 1 2 1 10 1 Therefore, he should feed the rats 2 oz of food A, no food B, and 1 oz of food C.        2 7 A 0 1 1        3       (d)  0   4    2 .  B    2 2 1  32 1 11 7 C Since A  0, there is no combination of foods giving the required supply.



3 1 4 1 0 0

 62.  4 2 6 0 1 0 3 2 5 0 0 1  1 1 2 1   0 2 2 4  0 1 1 1

   



 3 0   0 1

3 1 4

1 0 0



  1  R2  1 1 2 1 1 0  R    0 1 1 1 0 1  1 1 2 1 1 0  R3  12 R2  R3  0 2 2 4 3 0   0 0 0 1  32 1

R2  R1  R2  R3  R1  R3

1 0





1 1 2 1 1 0



  3 1 4 1 0 0   0 1 1 1 0 1 

R2  3R1  R2



 . 

Since the inverse matrix does not exist, it would not be possible to use matrix inversion in the solutions of parts (b), (c), and (d).

       9 11 8 x 740   9x  11y  8z  740          63. (a) 13x  15y  16z  1204 (b)   13 15 16   y    1204     8x  7y  14z  828 8 7 14 z 828     9 11 8 1 0 0 9 11 8 1 0 0   9R2  13R1  R2   R3  31R2  R3    0 8 40 13 9 0  20 (c)  13 15 16 0 1 0      9R3  8R1  R3 8 7 14 0 0 1 0 25 62 8 0 9     9 11 8 1 0 0 9 11 8 1 0 0 1   R R    252 R2 2 3 0    8 40 13 9 0     0 252 0 243 279 180  63 R3  2R2  R3 0 252 0 243 279 180 0 8 40 13 9 0     9 11 8 1 0 0 9 11 8 1 0 0 1    R  11R  8R  R  R3 27 5 27 5 31 31 1 2 3 1 0 1    2520 7  0  28    0 1 0  28 28  7   28 1 25 0 0 2520 1305 1125 360 0 0 1  29 56 56  7     63  63 7 7 9 1 9 0 0 1 0 0 4 4 4 4 1     31  5   0 1 0  27  0 1 0  27 31  5 . 9 R1    7  7  28 28 28 28 25  1 25  1 0 0 1  29 0 0 1  29 7 7 56 56 56 56

758

CHAPTER 11 Matrices and Determinants





7 7 1 4 4   27 31  5 . (You could also use a calculator to find the inverse.) Therefore, Thus, the inverse of the matrix is    28 28 7  25  1  29 7 56 56



  7 7 1 x 4 4     y     27 31  5    28 28 7 25  1  29 z 7 56 56



   740 16       1204    28 . She earns $16 on a standard phone, $28 on a deluxe phone, and     828 36

$36 on a super deluxe phone.



64. No. Consider the following counterexample: A  

0 1 0 0





 and B  

0 2 0 0



. Then, AB  O, but neither A  O nor 

B  O. There are infinitely many matrices for which A2  O. One example is A  

0 1 0 0



. Then, A2  O, but A  O.

11.4 DETERMINANTS AND CRAMER’S RULE 1. True. det A is defined only for a square matrix A. 2. True. det A is a number not a matrix. 3. True. If det A  0 then A is not invertible.      2 1   2 4  1 3  11 4. (a)    3 4    1 0 2     (b)  3 2 1   1 [2 4  1 3]  0 [3 4  1 0]  2 [3 3  2 0]  8  3  2 9  7    0 3 4    2 0  has determinant D  2 3  0 0  6. 5. The matrix  0 3   0 1  has determinant D  0 0  1 2  2. 6. The matrix  2 0   3   1  has determinant D  3  2  1 1  0. 7. The matrix  2 2 3 1  23   02 04  has determinant D  02 08  04 04  0. 8. The matrix  04 08   4 5  has determinant D  4 1  5 0  4. 9. The matrix  0 1   2 1  has determinant D  2 2  1 3  1. 10. The matrix  3 2   11. The matrix 2 5 does not have a determinant because it is not square.

SECTION 11.4 Determinants and Cramer’s Rule



12. The matrix 



3 0

1 13. The matrix  2



14. The matrix 

759



 does not have a determinant because it is not square. 

1 8  has determinant D  1  1  1  1  1  1  1 . 2 2 8 4 8 8 1 12

22 14 05

10 



 has determinant D  22 10  05 14  22  07  29. 1 0 12



   In Solutions 15–20, A    3 5 2 . 0 0 4

15. M11  5  4  0  2  20, A11  12 M11  20

16. M33  1  5  3  0  5, A33  16 M33  5

17. M12  3  4  0  2  12, A12  13 M12  12

18. M13  3  0  0  5  0, A13  14 M13  0

20. M32  1  2  3  12  72 , A32  15 M32   72 19. M23  1  0  0  0  0, A23  15 M23  0     2 1 0      2 4    2 6  4  4. Since M  0, the . Therefore, expanding by the first column, M  2  21. M   0 2 4      1 3  0 1 3

matrix has an inverse.   1 2 5    22. M    2 3 2 . Therefore, 3 5 3              2 2   3 2   2 3    2   9  10  2 6  6  5 10  9  19  24  5  0, and so   5 M  1        3 3  5 3  3 5

the matrix does not have an inverse.   30 0 20    23. M    0 10 20 . Therefore, expanding by the first row, 40 0 10          0 10   10 20    20    30 100  0  20 0  400  3000  8000  5000, and so M 1 exists. M  30     0 10  0   40 

     2  32 12      2  32    2 4  1       1 4  2  8  3  1  5  4, and the  24. M   2 4 0 . Therefore, M   1   1  2 2 2 2 4    2 1 2 1 2 matrix has an inverse.    1 3 7    3 7   M 25. M   . Therefore, expanding by the second row,  2  2 0 8 2 2 0 2 2 M  0, the matrix does not have an inverse.

    1 3  8   0 2 

     2 6  14  16  0. Since  

760

CHAPTER 11 Matrices and Determinants





   2 4  M . Therefore, expanding by the first row,   1 6 4   1 3 0 3

0 1 0

 26. M   2 1

has an inverse. 

1 3 3 0



2 0 2



  0 27. M     1 1     M  1    

1

     6  4  2. Since M  0, the matrix  

 2 0 1 . Therefore, expanding by the third row,  0 0 2 6 4 1    1 3 3       3 3 0             1 3 3 3 3 3     6  6  4  4, and so M 1 exists.        2  4  1  1 2 0 1 0 2 0         1 4 2 0 6 4  1 6 4 6 4 1

  3 4 0 4 28. M    0 1 6 0 1 0 2 0 and so M 1 exists.

   1 2 2  2 0 2         . Therefore, M  1  4 0 4   2  3 4 4         0 1 0  1 6 0

   1 2 1      29.  2 2 1   6. The matrix has an inverse.   1 2 2

         1 2  2 2  2   6     3 4  4 4   

  10 20 31   30.  10 11 45   20 40 50

     6  16  2  2  92,  

      1080. The matrix has an inverse.   

        7  1 3 2 5   1 10 2      3 9 11 5   2 18 18 13      31.    0. The matrix has no inverse.   12. The matrix has an inverse. 32.   2 6  3 30 4 24  0 31          5 15 10 39   1 10 2 10     4 3 2   8 6 24  33.   20 15 3    12 9 6

 0   2 35. M   2  3  0   M  1  2  3

  10   1    0. The matrix has no inverse. 27   1 

   0 4 6      1 1 3      1 2 3      0 1 7   0 6   2  1 3   6  3  0 7

  3 5 10  2   2 2 26 3  34.   6 9 16 45    8 12 20 36

        8. The matrix has an inverse.    

 0 0 4 6   2 1 1 3  , by replacing R3 with R3 R2 . Then, expanding by the third row, 0 0 1 0  3 0 1 7   1   6 2  0  3  1  18.  0

SECTION 11.4 Determinants and Cramer’s Rule



  36. M    

2

3 1 7



 6 2 3  .  7 7 0 5 3 12 4 0 4

761

     2 0 1 7   2 3 1 7          4 0 2 3    4 6 2 3 ,    M Then       7 0 5  7 7 0 5  7      3 0 4 0  3 12 4 0 

by replacing C2 with C2  3C3 . So expanding about the second column,       2 1 7           2 1    4 2     7 7  22  3  5  1183.   3 M  7  4 2 3   7 7      3 4  3 4   3 4 0     1 2 3 4 5       1 2 3 4   1 2 3 0 2 4 6 8         0 2 4 6     1 2       60  2  120. 37. M   0 0 3 6 9 , so M  5    5  4  0 2 4   20  3     0 0 3 6 0 2   0 0 0 4 8   0 0 3     0 0 0 4 0 0 0 0 5           2 1 6 4  2 1 6 4   0 1 6 4         7 2 2 5   11 2 2 5   7 2 2 5        38. M   . Then, M   , by replacing C1 with C1  2C2 . So   4 2 10 8   0 2 10 8   4 2 10 8            6 1 1 4  8 1 1 4 6 1 1 4             1 6 0   1 6 0   1 6 4   1 6 4              1 6           1 6   104  M  11  2 10 8   8  2 2 5   11  2 10 0   8  2 2 13   88     2 10           2 10  2 10 0   1 1 8  2 10 8   1 1 4  88  2  104  2  32   4 1 0    39. B    2 1 1  4  0 3           4 1 4 0 1 0   6  12  4  2   1   1 (a) B  2       4 0 4 3 0 3          4 1 4 1   4  6  2  3 (b) B  1      2 1  4 0 (c) Yes, as expected, the results agree.

40. If we expand along the first row of each submatrix, we see that the determinant is 210  1024.               2x  y  9    2 9   9 1   2 1       25.     41. Then D     10, and D y     5, Dx      x  2y  8  8 2 1 2 1 8    Dy  Dx  10 25 Hence, x     2, y   5, and so the solution is 2 5. D D 5 5               6x  12y  33    6 33   33 12   6 12    9, and  D y      6, Dx      12. 42. Then D         4x  7y  20  20 7  4 7  4 20       Dy  Dx  3 9 12 Hence, x   and y   2, and so the solution is 32  2 .   D D 6 2 6

     

762

CHAPTER 11 Matrices and Determinants

              1 3  3 6   1 6     8.       Then, D      12, and D y     20, Dx    3 1 1 2 3 2    Dy  Dx  8 12 Hence, x   04,and so the solution is 06 04.  20  06, y   D D 20        1   1 1  1   1x  1y  1  2     1 1 1 1 2 3 2 3 3        44. Then, D   1    6 , Dx    3   3 , and D y   1 1 1 3  1x  1 y  3  4 2  2 6   4 6  4 6 2   1  Dy  Dx  1 Hence, x   31  2, y   1  6, and so the solution is 2 6. D D     x  6y  3 43.  3x  2y  1

6

     1.  

6

               04 04   04 12   04 12    08.   32, and  D y      08, Dx    Then, D         12 32   32 16   12 16     Dy  Dx  32 08 Hence, x   4, y   1,and so the solution is 4 1.   D D 08 08               10x  17y  21    10 21   21 17   10 17    30.   12, and  D y      30, Dx    46.  Then, D         20x  31y  39  20 39   39 31   20 31        Dx  2 , y  D y  30  1, and so the solution is 2  1 . Hence, x    12 30 5 5 D D 30     x  y  2z  0 47. Then expanding by the second row, 3x  z  11    x  2y  0   04x  12y  04 45.  12x  16y  32

  1 1   D   3 0   1 2   1 0      D y    3 11    1 0

Therefore, x  44 11     5x  3y  z 48. 4y  6z    7x  10y

  2    1  1   3   2  0   2    1   1   11   1  0

       0 1 2         1 1   1 2   2   12  1  11, Dx    11 0 1   11    1       0  1 2   2 0    0 2 0       1 1 0        1 1    2   11.   22, and Dz    3 0 11   11       0  1 2     1 2 0 

     44,  

11  4, y  22 11  2, z  11  1, and so the solution is 4 2 1.        5 3 1   6        5 3  0 4     28  426  398,   6 Then D   0 4 6   1   22     7 10   7 10     7 10 0   13            5 6 3 1  6 1               22 4   0 22     6 3    272  126  398,  D y    0 22 6   1  6  Dx    22 4 6   1          13 10   13 10   7 13       13 10 0   7 13 0           5 3 6          5 3  5 5   6 6   428  1562  1990.        6    154  642  796, and Dz    0 4 22   4    22   7 10   7 13   7 13     7 10 13 

796 1990 Therefore, x  398 398  1, y  398  2, and z  398  5, and so the solution is 1 2 5.

SECTION 11.4 Determinants and Cramer’s Rule

    2x1  3x2  5x3  1 49. x 1  x 2  x3  2    2x2  x3  8

Then, expanding by the third row,     2 3 5      2 5   D   1 1 1   2   1 1   0 2 1

     2 3     1 1

     6  1  7,  

        1 3 5            2 1  2 1    1 1      3  30  20  7,  Dx    2 1 1      5   3 1         8 2 8 1  2 1  8 2 1

    2 1 5       2 5     Dx    1 2 1   8  2     1 1   0 8 1

 2 3 1      Dx    1 1 2 3   0 2 8

     2 1     1 2

     24  3  21, and  

           2 3 2 1    6  8  14.   8   2       1 1 1 2  

21 14 Thus, x1  7 7  1, x2  7  3, x 3  7  2, and so the solution is 1 3 2.

   c 2   2a 50. a  2b  c  9    3a  5b  2c  22

       2 0 1        1 2  2 1      18  1  19,   1 Then D   1 2 1   2     3 5 5 2    3 5 2        2 0 1        9 2  2 1      18  1  19,   1 Da    9 2 1   2      22 5  5 2    22 5 2 

        2 2 1         1 9  1 1   9 1      1   2 Db    1 9 1   2       3 22 3 2  22 2     3 22 2 

  2 0 2   Dc    1 2 9   3 5 22

     2 9    2     5 22  

    1 2   2   3 5 

     2  2  0.  

Hence, a  1, b  5, and c  0, and so the solution is 1 5 0.

     80  10  5  95, and  

763

764

CHAPTER 11 Matrices and Determinants

   

1x  1y  1z  3 5 2 51.  23 x  25 y  32 z     x  45 y  z 



7   10  10x  6y  15z  21 11  20x  12y  45z  33 10    5x  4y  5z  9 9 5

Then

         10 6 15           20 12   20 45   12 45      15   6   2400  1950  300  750, D   20 12 45   10       5 4  5 5    4 5     5 4 5           21 6 15           33 12   33 45   12 45      5040  1440  3600  0,   15    6 Dx    33 12 45   21        9 4   9 5  4 5     9 4 5           10 21 15              20 33   20 45   33 45    D y    20 33 45   10    2400  6825  5175  750, and   15    21          5 9 5 5    9 5    5 9 5          10 6 21           20 12   20 33   12 33      2400  2070  420  750.   21    6 Dz    20 12 33   10       5 4  5 9    4 9     5 4 9 

Therefore, x  0, y  1, z  1, and so the solution   is 0 1 1.         5       2 1 0   2x  y 5 3 0 3     24  35  11,  1 52. 5x Then D   5 0 3   2   3z  19      0 7 4 7    0 4 7 4y  7z  17        5 1 0         19 3  0 3     60  82  22,   1 Dx    19 0 3   5      17 7  4 7    17 4 7        2 5 0          5 3  19 3    D y    5 19 3   2    164  175  11, and   5       0 7  17 7     0 17 7         2 1 5         2 1  2 5     52  85  33. Thus, x  2, y  1, and z  3, and so the   17  Dz    5 0 19   4     5 0  5 19     0 4 17 

solution is 2 1 3.        0 3 5  3y  5z  4         2 0  2 1      12  70  58,   5 53. 2x Then D   2 0 1   3   z  10      4 7 4 0    4x  7y 4 7 0  0         0 4 5  4 3 5            4 5  4 5       216, and   378,  D y    2 10 1   4  Dx    10 0 1   7        10 1   10 1      4 0 0  0 7 0       0 3 4       0 4 3 4     120  56  176.   7 Dz    2 0 10   4      2 10   0 10    4 7 0   108 88 189 108 88 Thus, x  189 29 , y   29 , and z  29 , and so the solution is 29   29  29 .

SECTION 11.4 Determinants and Cramer’s Rule

    2 5 0   2x  5y  4       54. Then D   1 1 1   x  y z8      3x 3 0 5  5z  0      2  4 5 0           4 5      220,  D y    1 Dx    8 1 1   5    8 1    3 0 0 5     2 5 4      5 4     Dz    1 1 8   3   1 8   3 0 0

         1 1   1 1    10  40  50,   5 2     3 5 0 5

     0       1 1   8 1     80  32  48, and   4 8 1   2     3 5 0 5  0 5 4

       132. Thus, x  22 , y  24 , and z   66 , and so the solution is 22  24   66 .  5 25 25 5 25 25 

  x y z0      2z    0 Then 55.  y z 0     x  2z 1         1 1 1 1 1 1 1 2 0 1          2 0 0 1      1 0   D     1  0 1 0   1  2 0 1    2   0 1 1 0       2 0   1 2 0 1 2 0   1 0 2 0  2 0  1 1  1 1  2 1  4,       0 1 1 1  1 1 1      0 0 0 1 1 1       Dx      1  0 0 1   1 1   0 1 1 0   1 1      1 1 0    1 0 2 0

        Dy       

      Dz       

   1 1 1 1     2 0 0 1   1  2 0  0 0 1 0    0 1  1 1 2 0    1 1 0 1 1 1 1     2 0 0 1 2 0 1  1   0 1 0 0   0 1 0  1 0 1 0

      D        y

1 0

   Dy  D



         2 1  0 1    1    1     1 0  1 0  

    1 1    2   2 1 

     

     2,  

     1       1 1  0 1    1  2 1  1,     1   1    2  1 0   1 0   0

           1 1 0 1    1 1  2 1  1, and  2   1       1 0 1 0  

   1 0 1 1 1    2 0 0 0  2 0 0  1   0 1 1 0    0 1 1  1 0 2 1 1 1

765

       1 1    2 2  4. Hence, we have x  Dx   2  1 ,   2     D 4 2 1 1    

  Dz  D  1 1 1 1 4  ,z  , and    1, and the solution is 12  14  14  1 .   D D 4 4 4 4 4

766

CHAPTER 11 Matrices and Determinants

          xy1   1 1 0 0    0 1 0 1 1 0                 yz 2  0 1 1 0         1 1 1 0     1    1  1  2, 56. Then D     1  0 1 1   1  0 1 1   1        0 1 1 1 0 0 1 1 z    3              1 0 1  0 0 1       x  4  1 0 0 1          1 1 0 0        2 1 0 1 1 0             2 1 1 0 3 1 1 1 1 1         1  3  2,           Dx      1 0 1 1   1 3 1 1       1   2 3 0 1 1 4 1 0 1 0 1                 4 0 1 0 0 1   4 0 0 1          1 1 0 0      0 1 0 2 1 0                 0 2 1 0 1 0 3 1  1 1                   3  1  4,  Dy        1  2  1  1 1        0 1 1 3 1 1   0 3 1 1 1 1 4 1  0 1          1 0 1 4 0 1    1 4 0 1           1 1 1 0     1 1 0 1 2 0            0 1 2 0         1 1 3 1     1  1  0,   1 Dz      1  0 3 1   1  1 2 0   1      0 0 3 1 1 2 4 1       0 3 1 0 4 1    1 0 4 1           1 1 0 1       1 0 1  1 1 2              0 1 1 2           0 1 1 2 1 3     4  2  6.   1   1  D      1  0 1 3   1  1 1 2          0 0 1 3 1 3 1 3 0 4             0 1 3 0 0 4    1 0 0 4     Dy  Dx  Dz  D  2 4 0 6 Hence, the solution is x   1, y     2, z    0, and     3. D D D D 2 2 2 2

 0 0 1  1  57. Area    6 2 1 2 3 8 1

      6 2 1     2  3 8 

      1 48  6  1 42  21  2 2 

y

       3 5     1     2 2  

  12 [5  2  6  10]   12 [3  16]  12 19  19 2

(6, 2)

1 (0, 0)

   1 0 1       5 1 1 1    58. Area    3 5 1    1  2 2  2 1  2 2 1 

(3, 8)

x

1

y (_2, 2)

(3, 5)

1 (1, 0)

x

SECTION 11.4 Determinants and Cramer’s Rule

    1 3 1           2 1  9 1 1 1    3 59. Area    2 9 1    1    2 2  5 1  6 1   5 6 1    12 [1 9  6  3 2  5  1 12  45]

    2 9    1    5 6 

     

y (_1, 3)

    2 5 1           7 1  2 1 1 1    5 60. Area    7 2 1    2    2 2  3 1  4 1   3 4 1 

       62.      

  a 0 0 0 0      0 b 0 0 0     0 0 c 0 0 a    0 0 0 d 0      0 0 0 0 e

  a a a a a      0 a a a a     0 0 a a a   a    0 0 0 a a     0 0 0 0 a

  x 12 13   63.  0 x  1 23  0 0 x 2  x 1 1   64.  1 1 x  x 1 x

   a a a a a a a     0 a a a 2   a  0 a a  0 0 a a  0 0 a  0 0 0 a

     1 x     x x

     1 0    0  1   0 1  

 a b x a   66.  x x  b x  0 1 1

x

    7 2    1    3 4 

     

(_2, 5)

     d 0    abc    0 e  

     abcde  

      a a   a3     0 a 

     a5  

y (7, 2)

1

x

1 (3, _4)

       x  12    0  x  2  x x  1  0  x  0, 1, or 2   0  x  2      0 x 1   

     1 x  x   1 x  

  1 0 x   65.  x 2 1 0   x 0 1

   b 0 0 0 c 0 0    0 c 0 0    ab  0 d 0 0 0 d 0   0 0 e  0 0 0 e

2

(5, _6)

  12 [2 2  4  5 7  3  28  6]   12 [2 6  5 4  34]   12 12  20  34  12 66  33

       61.      

(2, 9)

1

  12 [15  3 3  57]   12 63  63 2

767

     1 1     x 1

       x 0  x  x 2  1  x  x 2  2x  1  0  x  12  0  x  1  

    2   x 1  x   0  1  x 2  0  x 2  1  x  1     x 0 

      a x a    1    x x  

     a b      x x b

     1 [ax  x x  a]  a x  b  bx  

 ax  x 2  ax  ax  ab  bx  x 2  ax  bx  ab  x  a x  b  0  x  a or x  b

768

CHAPTER 11 Matrices and Determinants

  1 x x2   67.  1 y y 2   1 z z2

68.

   

      y y2    1    z z2  

     x x2    1    z z2 

     x x2    1    y y2 

         yz 2  y 2 z  xz 2  x 2 z  x y 2  x y 2  

 yz 2  y 2 z  x z 2  x 2 z  x y 2  x y 2  x yz  x yz  x yz  xz 2  y 2 z  yz 2  x 2 y  x 2 z  zy 2  x yz        z x y  xz  y 2  yz  x x y  xz  y 2  yz  z  x x y  x z  y 2  yz    z  x x y  z  y y  z  z  x x  y y  z

x  2y  6z  5

3x  6y  5z  8    2x  6y  9z  7 (a) If x  1, y  0, and z  1, then x 2y 6z  12 06 1  5, 3x 6y 5z  3 16 05 1  8, and 2x  6y  9z  2 1  6 0  9 1  7. Therefore, x  1, y  0, z  1 is a solution of the system.        1 0 6  1 2 6 1 2 6           . Then, M   3 6 5    3 0 5  (replacing C2 with C2  2C1 ), so (b) M   3 6 5            2 2 9  2 6 9 2 6 9      1 6   2 5  18  46.  M  2    3 5       1 2 6 x 5           (c) We can write the system as a matrix equation:    3 6 5   y   8  or M X  B. Since M  0, M has 2 6 9 z 7 an inverse. If we multiply both sides of the matrix equation by M 1 , then we get a unique solution for X, given by X  M 1 B. Thus, the equation has no other solution.

(d) Yes, since M  0.

   1  1 69. (a) If three points lie on a line then the area of the “triangle” they determine is 0, that is,  2 Q    2 

 a1 b1 1   a2 b2 1   0   a3 b3 1 

Q  0. If the points are not collinear, then the point form a triangle, and the area of the triangle determined by these

points is nonzero. If Q  0, then  12 Q   12 0  0, so the “triangle” has no area, and the points are collinear.         6 4 1           y   2 10   6 4   6 4      (6, 13) (b) (i)  2 10 1           6 13   6 13   2 10    6 13 1  (2, 10)  26  60  78  24  60  8  34  104  68  0

Thus, these points are collinear.

(_6, 4) 1 1

x

SECTION 11.4 Determinants and Cramer’s Rule

   5 10 1        2 6  (ii)  2 6 1      15 2   15 2 1 

      5 10      15 2

        5 10         2 6

769

y (_5, 10) (2, 6)

 4  90  10  150  30  20  94  140  50  4

These points are not collinear. Note that this is difficult to determine from the diagram.

1 2

(15, _2)

x

   x y 1      70. (a) Let M   x1 y1 1 . Then, expanding by the third column,    x2 y2 1               x1 y1   x y   x y      x1 y2  x2 y1   x y2  x2 y  x y1  x1 y M         x2 y2   x2 y2   x1 y1   x1 y2  x2 y1  x y2  x2 y  x y1  x1 y  x2 y  x1 y  x y2  x y1  x1 y2  x2 y1

 x2  x1  y  y2  y1  x  x1 y2  x2 y1 So M  0  x2  x1  y  y2  y1  x  x1 y2  x2 y1  0  x2  x1  y  y2  y1  x  x1 y2  x2 y1  y  y1 x y  y1  y x  x1  x 1 2  1 2  x2  x1  y  y2  y1  x  x1 y2  x1 y1  x1 y1  x2 y1  y  2 x2  x1 x2  x1 x2  x1 y  y1 y  y1 y 2 x  x1   y1  y  y1  2 x  x1 , which is the “two-point” form of the equation for the line x2  x1 x2  x1 passing through the points x1  y1  and x2  y2 . (b) Using the result of part (a), the line has equation       x y 1       20 50     x y   20 50 1   0         10 25   10 25    10 25 1 

      x y      20 50

   0  

500  500  25x  10y  50x  20y  0  25x  30y  1000  0  5x  6y  200  0.

71. (a) Let x be the amount of apples, y the amount of peaches, and z the amount of pears (in pounds).   x  y z  18   We get the model 075x  090y  060z  1380    075x  090y  060z  180

770

CHAPTER 11 Matrices and Determinants

          1 1 1          075 090   075 060   090 060      0  090  135  045, 1 1 (b) D   075 090 060   1         075 090   075 060   090 060     075 090 060            18 1 1          1380 090   1380 060   090 060      0  72  108  36, 1 1 Dx    1380 090 060   18         180 090   180 060   090 060     180 090 060            1 18 1          075 1380   075 060   1380 060       1   18    D y    075 1380 060   1            075 180   075 060   180 060     075 180 060   72  162  117  27, and           1 1 18          075 090   075 1380   090 1380       18   1 Dz    075 090 1380   1         075 090   075 180   090 180     075 090 180 

 108  117  243  18.   Dy  Dx  Dz  36 27 18 So x   8; y   6; and z   4.    D D D 045 045 045 Thus, Muriel buys 8 pounds of apples, 6 pounds of peaches, and 4 pounds of pears.

72. (a) Using the points 10 25, 15 3375, and 40 40,we substitute for x and y and get the system     100a  10b  c  25 225a  15b  c  3375    1600a  40b  c  40   100 10 1   (b) D   225 15 1   1600 40 1

      225 15  1    1600 40  

      100 10 1     1600 40

     100 10  1    225 15 

     

 9000  24,000  4000  16,000  1500  2250  15,000  12,000  750  3750,          25 10 1          25  25 10   3375 15    10   1 1 Da    3375 15 1   1        40   3375 15   40 40   40    40 40 1 

 1350  600  1000  400  375  3375  750  600  375  1875,          100 25 1           100 25  100 25   225 3375     1 1 Db    225 3375 1   1         225 3375   1600 40    1600 40    1600 40 1

 9000  54,000  4000  40,000  3375  5625  45,000  36,000  2250  11,250, and

SECTION 11.4 Determinants and Cramer’s Rule

  100 10 25   Dc    225 15 3375   1600 40 40

      225 15    25      1600 40  

     100 10    3375      1600 40 

     100 10    40      225 15 

 25  9,000  24,000  3375  4,000  16,000  40  1,500  2,250

771

     

 25  15,000  3375  12,000  40  750  375,000  405,000  30,000  0.

Thus, a 

Db  Dc  Da  1875 11,250 0     005, b   3, and c   0. Thus, the model is D D D 3750 3,750 3,750

y  005x 2  3x.

73. Using the determinant formula for the area of a triangle, we have     1000 2000 1            1000 2000  5000 4000  1 1  1 Area    5000 4000 1    1     2 2   2000 6000  2000 6000   2000 6000 1 

     1000 2000  1    5000 4000 

  12  22,000,000  2,000,000  6,000,000   12  14,000,000  7,000,000

     

Thus, the area is 7,000,000 ft2 .

74. (a) The coordinates of the vertices of the surrounding rectangle are a1  b1 , a2  b1 , a2  b3 , and a1  b3 . The area of the surrounding rectangle is given by a2  a1   b3  b1   a2 b3  a1 b1  a2 b1  a1 b3  a1 b1  a2 b3  a1 b3  a2 b1 . (b) The area of the three blue triangles are as follows:

Area of  a1  b1   a2  b1   a2  b2 : 12 a2  a1   b2  b1   12 a2 b2  a1 b1  a2 b1  a1 b2  Area of  a2  b2   a2  b3   a3  b3 : 12 a2  a3   b3  b2   12 a2 b3  a3 b2  a2 b2  a3 b3  Area of  a1  b  a1  b3   a3  b3 : 12 a3  a1   b3  b1   12 a3 b3  a1 b1  a3 b1  a1 b3 . Thus the sum of the areas of the blue triangles, B, is

B  12 a2 b2  a1 b1  a2 b1  a1 b2   12 a2 b3  a3 b2  a2 b2  a3 b3   12 a3 b3  a1 b1  a3 b1  a1 b3   12 a1 b1  a1 b1  a2 b2  a2 b3  a3 b2  a3 b3   12 a1 b2  a1 b3  a2 b1  a2 b2  a3 b1  a3 b3 

 a1 b1  12 a2 b3  a3 b2   12 a1 b2  a1 b3  a2 b1  a3 b1  So the area of the red triangle A is the area of the rectangle minus the sum of the areas of the blue triangles, that is,   A  a1 b1  a2 b3  a1 b3  a2 b1   a1 b1  12 a2 b3  a3 b2   12 a1 b2  a1 b3  a2 b1  a3 b1   a1 b1  a2 b3  a1 b3  a2 b1  a1 b1  12 a2 b3  a3 b2   12 a1 b2  a1 b3  a2 b1  a3 b1 

 12 a1 b2  a2 b3  a3 b1   12 a1 b3  a2 b1  a3 b2    a b 1  1 1    (c) We first find Q   a2 b2 1  by expanding about the third column.    a3 b3 1               a1 b1   a1 b1   a2 b2    1   1   a2 b3  a3 b2  a1 b3  a3 b1   a1 b2  a2 b1 Q  1        a3 b3   a2 b2   a3 b3   a1 b2  a2 b3  a3 b1  a1 b3  a2 b1  a3 b2

So 12 Q  12 a1 b2  a2 b3  a3 b1   12 a1 b3  a2 b1  a3 b2 , the area of the red triangle. Since 12 Q is not always positive, the area is  12 Q.

75. (a) If A is a matrix with a row or column consisting entirely of zeros, then if we expand the determinant by this row or       column, we get A  0   A1 j   0   A2 j       0   Anj   0.

772

CHAPTER 11 Matrices and Determinants

(b) Use the principle that if matrix B is a square matrix obtained from A by adding a multiple of one row to another, or a multiple of one column to another, then A  B. If we let B be the matrix obtained by subtracting the two rows (or columns) that are the same, then matrix B will have a row or column that consists entirely of zeros. So B  0  A  0.

(c) Again use the principle that if matrix B is a square matrix obtained from A by adding a multiple of one row to another, or a multiple of one column to another, then A  B. If we let B be the matrix obtained by subtracting the proper multiple of the row (or column) from the other similar row (or column), then matrix B will have a row or column that consists entirely of zeros. So B  0  A  0.

75. Gaussian elimination is superior, since it takes much longer to evaluate six 5  5 determinants than it does to perform one five-equation Gaussian elimination.

CHAPTER 11 REVIEW 1. (a) 2  3

2. (a) 2  3

(b) Yes, this matrix is in row-echelon form.

(b) Yes, this matrix is in row-echelon form.

(c) No, this matrix is not in reduced row-echelon form, since the leading 1 in the second row does not have a

(c) Yes, this matrix is in reduced row-echelon form.  x  6 (d) y 0

0 above it.   x  2y  5 (d)  y 3

3. (a) 3  4

(b) Yes, this matrix is in row-echelon form. (c) Yes, this matrix is in reduced row-echelon form.    8z  0  x (d) y  5z  1    0 0

5. (a) 3  4 (b) No, this matrix is not in row-echelon form. The leading 1 in the second row is not to the left of the one above it. (c) No, this matrix is not in reduced row-echelon form.   y  3z  4   (d) x  y 7    x  2y  z  2

4. (a) 3  4 (b) No, this matrix is not in row-echelon form, since the leading 1 in the second row is not to the left of the one above it. (c) Since this matrix is not in row-echelon form, it is not in reduced row-echelon form.     x  3y  6z  2 (d) 2x  y 5    z0

6. (a) 4  4

(b) No, this matrix is not in row-echelon form. The leading 1 in the fourth row is not to the left of the one above it. (c) No, this matrix is not in reduced row-echelon form.   x  8y  6z  4      y  3z  5 (d)  2z  7     x  y  z  0

CHAPTER 11



1

2 2



6



1 1 0 1

  R R  1   2 7.   1 1 0 1    1 2 1 3 7 2

2 2 1 3





1 1 0 1

 R2  R1  R2    0 6  R3  2R1  R3  7 0

3 2 3 3





Review

773

1 1 0 1



 R R R  3 2 3  7    0 9 0

3 2 0 1

Thus, z  2, 3y  2 2  7  3y  3  y  1, and x  1  1  x  0, and so the solution is 0 1 2.



1 1 1 2

 8.  1 0



 1 3 6  2 3 5

R2  R1  R2





1 1 1 2

 0  0



 2 2 4  2 3 5



1 1 1 2

 0  0

R3  R2  R3



1 2

3 2



   9.   2 1 1 2  2 7 11 9



1 2

3 2



   0 3 5 6    0 3 5 5

R2  2R1  R2  R3  2R1  R3



 2 2 4 . Thus, z  1; 2y  2 1  4 0 1 1

 y  1; and x  1  1  2  x  2, and so the solution is 2 1 1.





1 2

 0  0

R3  R2  R3



3 2

0

1 1 1

2



   10.  1 1 3 6 3 1 5 10  R1  R2  R1



R2  R1  R2  R3  3R1  R3

1 0 2 4





1 1 1 2

 0  0



 2 2 4  2 2 4

R3  R2  R3





1 1 1 2

0

 0  0



 2 2 4  0 0 0



 6 . 1

3 5

The last row corresponds to the equation 0  1, which is always false. Thus, there is no solution.



 7 . 2



 0  0

1 2 R2





1 1 1 2

 1 1 2  0 0 0

   0 1 1 2 . Let z  t. Then y  t  2  y  2  t and x  2t  4  x  4  2t, and so the   0 0 0 0

solutions are 4  2t 2  t t, where t is any real number.



1

1

1

1

0



   1 1 4 1 1    11.    1 2 0 4 7    2 2 3 4 3  1 1 1   0 1 4 R3  3R2  R3     0 0 13  0 0 1



1

1

  0 2     0 3 R4  2R1  R4  0 0   1 0 1   0 5 6   R3  R4     0 12 11    2 3 0 R2  R1  R2 R3  R1  R3

1

1

0



 5 2 1    1 3 7   1 2 3 1

1

1

1

4

5



 6   0 1 2 3   0 13 12 11

1

1

1

1

0



   0 1 4 5 6       0 3 1 3 7    0 0 1 2 3   1 1 1 1 0    6  R4  13R3  R4  0 1 4 5 .    0 0 1 2 3    0 0 0 14 28

R3  R2  R3 

0



Therefore, 14  28    2, z  2 2  3  z  1, y  4 1  5 2  6  y  0, and x  0  1  2  0  x  1. So the solution is 1 0 1 2.

774

CHAPTER 11 Matrices and Determinants



1 0 3

0 1

  0 1 0 4 5  12.  0 2 1 1 0  2 1 5 4 4  1 0  0 1  R3  R4   0 0  0 0  1 0 3 0 1   0 1 0 4 5    0 0 1 0 1  0 0 0 1 1 

1 1 3 2

 13.  2 1 1 2 3 0 4 4  1 1 3   0 1 5  3 0 0 0

   

      



3

0

1

1 1       



 5   0 1  9 10

0 4

0 1





1 0  0 1   0 0  0 0

 R2  4R4  R2

  23   0

3

R4  3R3  R4

R1  3R3  R1



1 0

  0 1 0 4   0 2 1 1  0 1 1 4 

R4  2R1  R4



1 1

0 0 0 0 1 0 0 1 3

2



 5   0  6

1 0



0 1

3

0

1



   0 1 0 4 5      0 0 1 9 10    0 0 1 0 1

R3  2R2  R3  R4  R2  R4

3

1 0



   0 1 0 4 5  R3      1  0 0 1 0 1  9 R4   0 0 0 9 9  2  1  . Therefore, the solution is 2 1 1 1. 1   1



   0 3 5 2    0 3 5 2  4 4 1 0 3 3  R1  R2  R1 5 2  0 1     3 3 0 0 0 0







R2  2R1  R2  R3  3R1  R3

2



R3  R2  R3





1 1

 0  0

3

2



 3 5 2   0 0 0

1 3 R2



 . The system is dependent, so let z  t: y  5 t   2  3 3 

  y  53 t  23 and x  43 t  43  x   43 t  43 . So the solution is  43 t  43  53 t  23  t , where t is any real number.



1 1

0

1

 14.  1 1 2 3 1 3 2 1   1 1 0 1   0 1 1 1   0 0 0 0

   

R2  R1  R2  R3  R1  R3

   

R1  R2  R1



1 1

0

1



1 1 0 1



1     3  R2  R3 0 2 2 2 R   2 R4     0 2 2 2   0 2 2 2 0 0 0 0  1 0 1 2  0 1 1 1 . Since the system is dependent, let z  t. Then y  t  1  0 0 0 0

y  t  1 and x  t  2  x  t  2. So, the solution is t  2 t  1 t where t is any real number.



1 1

1 1 0





1 1

1 1 0



1 2 R4



1 1

1 1 0



2  3R1  R2 1  R2  R1      R  R   0 2 4 2 2 0 1 2 1 1 3 1 1 1 2   1 0 1 0 1  . Since the system is dependent, Let z  s and   t. Then y  2s  t  1  y  2s  t  1 and 0 1 2 1 1

15. 

x  s  1  x  s  1. So the solution is s  1 2s  t  1 s t, where s and t are any real numbers.



 1 1 3    16.  2 1 6 1 2 9

R2  2R1  R2  R3  R1  R3



 1 1 3   0 3 0   0 1 6

R2  3R3  R2





 1 1 3    0 0 18 . Since the second row corresponds   0 1 6

to the equation 0  18, which is always false, this system has no solution.

CHAPTER 11





1 1 1 0   17.  3 2 1 6    1 4 3 3

R2  3R1  R2  R3  R1  R3





1 1 1 0    0 5 4 6    0 5 4 3

R3  R2  R3





Review

775



1 1 1 0    0 5 4 6 . The last row of this   0 0 0 3

matrix corresponds to the equation 0  3, which is always false. Hence there is no solution.       1 2 3 2 1 2 3 2 1 2 3 2       R2  2R1  R2 3  R2  R3  0 5 11 3  R     18.      0 5 11 3 . Since the third  2 1 5 1   R3  4R1  R3 0 5 11 2 0 0 0 1 4 3 1 6

row corresponds to the equation 0  1, which is always false, this system has no solution.       1 0 0 1 1 1 0 0 1 1 1 1 1 1 2       R2  R1  R2  1 1 1 1 0   0 1 1 2 1   1 1 1 1 0       R1  12 R3  R3  R1  R3 19.         0 1 1 2 1   2 0 0 2 2    1 1 1 1 2  R4  2R1  R4       2 4 4 2 6 0 4 4 4 4 2 4 4 2 6     1 0 0 1 1 1 0 0 1 1     R1  R3  R1  0 1 1 2 1   0 1 1 2 1   14 R3 R3  R2  R3     R2  2R3  R2          R4  4R2  R4  0 0 0 4 0   121 R4 0 0 0 0 0 R4  R3  R4     0 0 0 12 0 0 0 0 1 0   1 0 0 0 1    0 1 1 0 1     . This system is dependent. Let z  t, so y  t  1  y  t  1; x  1  x  1. So the 0 0 0 1 0   0 0 0 0 0 solution is 1 t  1 t 0, where t is any real number.    1 1 2 3 0 1 1 2 3 0   R  3R  R   313  0 1 1 1 1 20.  0 1 1 1 1    3 2 7 10 2 0 1 1 1 2

   

R3  R2  R3





 1 1 2 3 0    0 1 1 1 1 . Since the   0 0 0 0 1

third row corresponds to the equation 0  1, which is always false, this system has no solution.

21. A 3  3 and B 2  3 have different dimensions, so they are not equal.       25 1 5 1 5 e0     B, so A and B are equal. 22. A   0 21 log 1 12 0 12 In Solutions 23–34, the matrices A, B, C, D, E, F, and G are defined as follows:

A 

1



2 0 1 4



   D  0 1  2 0





B 

E 

2 1

 12

1

 

1 2 4 2 1 0



 



4 0 2



 C  

   F   1 1 0  7 5 0

23. A  B is not defined because the matrix dimensions 1  3 and 2  3 are not compatible.

3



G



1 2

 2 32   2 1 5



776

CHAPTER 11 Matrices and Determinants



 24. C  D   

1 2

3





1

4





 12 1

       2 32     0 1    2 2 1 2 0 4      1 3 1 4  2     3     25. 2C  3D  2  2 2   3  0 1   2 1 2 0

 

5  2 

1

1 6





3 12





4 18

       4 3    0 3    4 4 2 6 0 2



 0  2

26. 5B  2C is not defined because the matrix dimensions 2  3 and 3  2 are not compatible.      27. G A  5 2 0 1  10 0 5

28. AG is undefined because the matrix dimensions 1  3 and 1  1 are not compatible.          1 3  1 3  4 2  11 2 2 2 7     1 2 4  10  1 2 4 2 3 3 11         29. BC   30. C B   2 2   2 2  8  1 2 2 1 0  1  92 2 1 0 4 3 8 2 1 2 1        1 3     2 14 4 0 2 4 0 2 2       1 2 4  30 22 2  2 3    3 3    1 1 0     31. B F   32. FC   1 1 0        2  2 2  2 1 0 9 1 4 27 57 2 1 7 5 0 7 5 0 2 2            1 3 3 7  1 4  1 11      2  2 2   2 2 1 2 1 15 3 1 3              33. C  D E   2 2    0 1   1  2 2   4 2  2  12 1 1 2 1 2 0 0 1  12 1            4 0 2 1 6 1 4 4 0 2 0 2 12 12                     34. F 2C  D   4 2  1 1 0   4 3    0 1    1 1 0   4 4     7 5 0 4 2 2 0 7 5 0 6 2 20 34 In Solutions 35–44, the matrices A and B are defined as follows:     1 4 1 3 0 3       B A  1 1 0   2 1 2  2 0 2 1 6 0     27 0 21 6 42 24       35. AB 2   36. A2 B    20 5 13   3 37 22  5 22 7 3 42 27     19 14 26 8  32 4 4     7 1 11  7  37. A1 B A   38. B AB 1    3  3   2 1  4  3  35 13 18 80 7 16 3 3 2   1   39. AB  12 40. B A  12 41. A1   3     1 1  1   1  44. A  B A  4  43. A B A  4 42. A 3           2 5 3 52 1 0 3 52 2 5 1 0    and B A     . 45. AB   2 6 1 1 0 1 1 1 2 6 0 1

   

CHAPTER 11

















Review



777



2 1 3 3 2 52 2 52 1 0 0 3 2 1 3 1 0 0   2     2       1 1 2    0 1 0  and B A   1 1 2   2 2 1    0 1 0 . 46. AB   2 2 1           0 1 1 1 1 1 0 0 1 1 1 1 0 1 1 0 0 1 

In Solutions 47–52, A  

2 1 3 2





, B  

1 2 2

4





, and C  



0 1 3

.

2 4 0 

47. A  3X  B  3X  B  A  X  13 B  A. Thus, X  13 

1 2 4

2







2 1 3 2

48. 12 X  2B  A  X  2B  2A  X  2A  2B  2 A  B.         2 1 1 2 3 1 6 2    2   . X  2  3 2 2 4 1 6 2 12





  1  3

1 3 5

2



.

Thus,

49. 2 X  A  3B  X  A  32 B  X  A  32 B. Thus,           3 3 7 2 1 2 2 1 2 1  2 .   3   2 X  2 3 6 0 8 2 4 3 2 3 2 50. 2X  C  5A  2X  5A  C, but the difference 5A  C is not defined because the dimensions of 5A and C are not the same. 51. AX  C  A1 AX  X  A1 C. Now          2 1 2 1 0 1 3 2 1 2 2 6 1     . Thus, X  A1 C   . A1  4  3 3 2 3 2 2 4 0 3 2 4 5 9 52. AX  B  A1 AX  X  A1 B. From Exercise 65,        2 1 2 1 1 2 4 8 1 1 . Thus, X  A B     . A  3 2 3 2 2 4 7 14 

1 4



2

53. D  

2 9





. Then D  1 9  2 4  1, and so D 1   

9 4 2

1



.









3 1 3 2 4 .  8 . Then D  2 3  1 2  8, and so D 1   1  54. D   8 1 1  1 2 1 3 8 4



55. D  



2

4 12 2

6



. Then D  4 6  2 12  0, and so D has no inverse.



    1 2 . Then D  2  56. D   1 1 2    3 2 0 3 2 2 4 0

     1 2    4    0 2 

     2 2  6  4 2  0, and so D has no inverse.  

778

57.

58.

59.

60.

CHAPTER 11 Matrices and Determinants





        3 0    2 3    4  12  9  1. So D 1 exists.   1 . Then, D  1  D 2 3 0        2 3   4 2  4 2 1       1 3 1 1 1 0 1 3 1 1 1 0 3 0 1 1 0 0   R2  2R1  R2     R2 1  R2  R1     0 9 2 2 3 0  3  2 3 0 0 1 0  R        2 3 0 0 1 0   2R3 R3  4R1  R3  4 2 1 0 0 1 0 14 3 4 4 1 4 2 1 0 0 1       1 3 1 1 1 0 1 3 1 1 1 0 1 3 1 1 1 0     R3  R2   R3  9R2  R3 3  R2  R3   0 27 6 6 9 0  R          0 27 6 6 9 0  1 R3  0 1 0 2 1 2   R1  3R2  R1 3 0 28 6 8 8 2 0 1 0 2 1 2 0 9 2 2 3 0       3 2 3 1 0 1 5 4 6 1 0 0 3 2 3 1       2 R3 0 1 0 . Thus, D 1     2 1 2  2 1 2  0 1 0 2 1 2   .     R1  R3  R1 8 6 9 0 0 2 16 12 18 0 0 1 8 6 9          1 2 3 1 0 0 1 2 3       2 4 2 5    4 5   1  4  6  1. So D 1 exists.  2 4 5 0 1 0   3   2 . Then, D  1  D 2 4 5          2 5 2 6 5 6 2 5 6 0 0 1 2 5 6     1 2 3 1 0 0 1 2 3 1 0 0   R  R   R  3R  R R2  2R1  R2 2 3 1 3 1       0 0 1 2 1 0    0 1 0 2 0 1   R3  2R1  R3 0 1 0 2 0 1 0 0 1 2 1 0       1 3 2 1 2 0 5 3 0 1 0 0 1 3 2       1 1  2R2  R1    0 1 0 2 0 1  R       0 1 0 2 0 1 . Thus, D   2 0 1 . 2 1 0 0 0 1 2 1 0 0 0 1 2 1 0       1 0 0 1 1 0 0 1 1 0 0 0 1         2 R2   2 0 2 0 2 0 2 0 2 0 2 0 1 0 0 1 3 3   R3     1     D  . Thus, D   0 3 3   2   3   24 and D exists.  0 0 3 3   1 0 4     0 0 3 3 0 0 1 0 4 R4 0 0 4 0 0 0 4 0 0 0 4 0 0 0 1       1 0 0  14 1 0 0 1 1 0 0 0 1 0 0 0 1 0 0  14       R1  R4  R1  0 1 0 1  0 1 0 1 0 1 0 0  0 1 0 0 0 1 0 1        R2 R4  R2 2 2 4 2 4 1     . Therefore, D   .  0 0 1 1  0 0 1 1 0 0 1 0  0 0 1 0 0 0 1 1  R3  R4  R3 3 3 4  3 4      1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 14 4 4   1 0 1 0   0 1 0 1   D . Thus, 1 1 1 2   1 2 1 2           1 0 1   0 1 1                  1 2 1 1  1 2 1 1           0  1  0  1  0.  D   1 1 2    1 1 2           1 2 1 2  1 2 2 1    2 1 2 1 2 2 3

0 1

Hence, D 1 does not exist.

   

CHAPTER 11



61. 

 

12 5 x

 

5 2   

x y

2





5



 

10 17 10





. If we let A  







65



12 5 5 2



, then A1 







Review

779



2 5 2 5 1   , and so 24  25 5 12 5 12

. Therefore, the solution is 65 154. 5 12 17 154        6 5 x 1 6 5     . , then 62.  If we let A   8 7 y 1 8 7              7 5 7 5 1 7 5 7 5 x 6 1 2 , and so     2 2     .  2    1  A1  2 42  40 8 6 4 3 4 3 1 8 6 y 7 y

Therefore, the solution is 6 7.            1 2 1 5 2 1 5 1 0 0 1 2 2 0 1 0 2 1 5 x 3     R R        1 2          1  63.   1 2 2   y    4 . Let A   1 2 2 . Then  1 2 2 0 1 0    2 1 5 1 0 0  1 1 0 3 1 0 3 0 0 1 1 0 3 0 0 1 1 0 3 z 6     1 2 2 0 1 0 1 2 2 0 1 0   R  2R  R   R1  2R2  R1 R2  2R1  R2 2 3 2        0 3 1 1 2 0    0 1 1 1 0 2   R3  R1  R3 R3  R3  2R2 0 2 1 0 1 1 0 2 1 0 1 1       1 0 4 2 1 4 1 0 4 2 1 4 1 0 0 6 3 8     R1  4R3  R1   R3  0 1 1 1 0 2      0 1 0 1 1 1 .      0 1 1 1 0 2  R2   R3  R2 0 0 1 2 1 3 0 0 1 2 1 3 0 0 1 2 1 3          1 1 6 3 8 x 6 3 8  12 3                1   1  Hence, A1    1 1 1  and  y    1 1 1   4    12 , and so the solution is 1 1 2 1 3 z 2 1 3 6 12   1 1  1 .  12 12 12          2 0 3 x 5 2 0 3 2 0 3 1 0 0          R R 1 2         64.   1 1 6   y    0 . Let A   1 1 6 . Then  1 1 6 0 1 0   3 1 1 z 5 3 1 1 3 1 1 0 0 1       1 1 6 0 1 0 1 1 6 0 1 0 1 1 6 0 1 0       2  2R1  R2 2  2R3  R2 2 0 3 1 0 0 R  0 2 9 1 2 0  R           0 2 9 1 2 0  R3  3R1  R3 3 1 1 0 0 1 0 4 17 0 3 1 0 0 1 2 1 1     1 1 0 12 5 6 1 1 0 12 5 6  1    R1  9R3  R1 1  R2  R1  0 2 0 17 7 9   2 R2  0 1 0 17  7  9  R           2 2 2 R2  6R3  2R2 0 0 1 2 1 1 0 0 1 2 1 1            7 3 3 7 3 3 x 10 1 0 0 72  32  32 2 2 2  2 2 2  5           0 1 0 17  7  9 . Hence, A1   17  7  9  and  y    17  7  9   0    20 , and    2    2  2 2 2  2 2  2 2   z 5 0 0 1 2 1 1 2 1 1 2 1 1 5 so the solution is 10 20 5.

65. (a) The i jth entry of A represents how many pounds of vegetable j were sold on day i, and the ith entry of B represents the price of vegetable i.

780

CHAPTER 11 Matrices and Determinants







    685   100    . The jth entry of AB represents the total revenue on day j. (b) AB    14 12 16  410 050 25 16 30



150

66. (a) We are given that x  y  18, and since the total is $600, 20x  50y  600.      20 50 x 600      (b)  1 1 y 18     1 5 1 50  1 3 , so     30 (c) A1  1 2 20 1  50 1 1 20 30 3          1 1 5 18 5   20  30 10 600  600 3  3   . Brodie received 10    30 X  A1 B   30 1 2 1 600  2 18 20  12 8 18 30 3 30 3 $20 bills and 8 $50 bills.

         2 13     60  78  18.   208  210  2, and  D y         6 30    2  1 and y  18  9 , and so the solution is 1  9 . Therefore, x  10 5 10 5 5 5

  2 7 67. D    6 16

      13 7   32  42  10, Dx        30 16

     12 11    108  77  185, Dx   68. D    9  7        12 140   Dy      240  980  740. Therefore, x    7 20 

     140 11    1260  220  1480, and    9  20

740  1480 185  8 and y  185  4, and so the solution is 8 4.

       2 1 5         2 1   1 7      195  39  156,   3 69. D   1 7 0   5      1 7   5 4    5 4 3        0 1 5         0 1   9 7     495  27  522,   3 Dx    9 7 0   5     9 7  9 4     9 4 3         2 0 5           2 0  1 9    D y    1 9 0   5    180  54  126, and   3        1 9   5 9     5 9 3         2 1 0         2 1   2 1      117  117  234.  9 Dz    1 7 9   9      1 7  5 4    5 4 9 

  522   87 , y  126  21 , and z  234  3 , and so the solution is  87  21  3 . Therefore, x  156 26 156 26 156 2 26 26 2

CHAPTER 11

Review

781

       3 4 1         3 1   1 4      52  11  41,   1 70. D   1 0 4   4      1 4  2 5   2 1 5        10 4 1         10 1   20 4      880  20  860,   1 Dx    20 0 4   4      20 4   30 5     30 1 5           3 10 1           10 1      10 1   20 4     660  80  40  540, and  D y    1 20 4   3    2   1          20 4   30 5   30 5     2 30 5         3 4 10         3 10   1 20      40  50  10.   1 Dz    1 0 20   4      1 20   2 30     2 1 30    860 540 10 860 540 10 Therefore, x  860 41  41 , y   41 , and z  41 , and so the solution is 41   41  41 .

      1 3 1               3 1   1 3   1 3  1 1     1 4  8  10  11.   71. The area is   3 1 1     2 2 2  2 2   2 2   3 1    2 2 1        5 2 1               1 5   5 2   5 2  1 1     1 21  3  27  51 .   72. The area is   1 5 1     2 2 2 2  4 1   4 1   1 5    4 1 1 

73. Let x be the amount invested in Bank A, y the amount invested in Bank B, and z the amount invested in Bank C.     x  y z  60,000 y z  60,000     x  We get the following system: 002x  0025y 003z  1575  2x  25y 3z  157,500 which      2x   2x  2z  y y 2z  0 

1

1 1

60,000

 has matrix representation   2 25 3 157,500 2 1 2 0    1 1 1 60,000 1    R1 R2  R1  0 1 0 40,000  0   R3 05R2  R3  0 05 1 37,500 0

   



 1 0 40,000   0 1 17,500

$2500 in Bank A, $40,000 in Bank B, and $17,500 in Bank C.

1

1

1

60,000



  R 1R 2  0 05 1 3 3 37,500    0 3 0 120,000   1 0 1 2500   R1  R3  R1  0 1 0 40,000 . Thus, she invests    0 0 1 17,500

R2  2R1  R2  R3  2R1  R3

0 1 20,000



782

CHAPTER 11 Matrices and Determinants

74. Let x be the amount of haddock, y the amount of sea bass, and z the amount of red snapper, in pounds. Our system     1 1 1 560 1 1 1 560   R R R   R  125R  R 2 3 2 3 1 3  0 has the matrix representation  075 0 255   125 075 200 575    125 0 200 320 125 0 200 320       1 1 1 560 1 1 1 560 1 1 1 560   4 R2   R  125R  R   4 R3 3 2 3  0 075 0   3 3 255  0 340  340      0 1    0 1 0   0 125 075 380 0 125 075 380 0 0 075 45       1 1 1 560 1 1 0 500 1 0 0 160     R R R   1  R3  R1 1 2 1  0 1 0 340  R         0 1 0 340   0 1 0 340 . Thus, he caught 160 lb of 0 0 1 60 0 0 1 60 0 0 1 60 haddock, 340 lb of sea bass, and 60 lb of red snapper.

CHAPTER 11 TEST 

1 8 0

0



4



   1.   0 1 7 10  is in row-echelon form, but not reduced row-echelon form because the 1 in the second row does not have a 0 0 0 0 0 above it. 

0 0



1 0 0



1 0 0

0

 0 0 2 5  2.   0 1 2 7  1 0 3 0 3. 

0 0 1



    is in neither row-echelon nor reduced row-echelon form.  

 is in reduced row-echelon form. 3



   4.   0 1 0 2  is in reduced row-echelon form. 0 0 1 32

      1 1 2 0 1 1 2 0   x  y  2z  0     R R R 2 1 2    5. 2x  4y  5z  5 has the matrix representation   2 4 5 5    0 2 1 5     0 2 3 5 2y  3z  5 0 2 3 5     1 1 2 0 1 1 2 0    1 R3   R3  R2  R3 5     2   0 2 1 5    0 2 1 5 . Thus z  0, 2y  0  5  y  2 , and 0 0 2 0 0 0 1 0   x  52  2 0  0  x  52 . Thus, the solution is 52  52  0 .

CHAPTER 11

    2x  3y  z  3 2 3 1 3     6. has the matrix representation  x  2y  2z  1 1 2 2 1       4x  y  5z  4 4 1 5 4    1 2 2 1 1 2 2 1    R2  2R1  R2 3  R2  R3  0 7 3 5  R   0     7 3 5 R3  4R1  R3 0 7 3 8 0 0 0 3

R1  R2







3R3  4R2  R3 

1 2 0

3



  0 3 1 2   0 0 7 7

1 7 R3





1 2 0

3

783

1 2 2 1    2 3 1 3    4 1 5 4

 . 

Since the last row corresponds to the equation 0  3, this system has no solution.      3 1 2 0 3   x  2y   R R R 3 1 3  7. has the matrix representation  3y  z  2  0 3 1 2      x  2y  z  2 1 2 1 2 

Test







1

2 0

3



  0 3 1 2   0 4 1 5

   0 3 1 2 .   0 0 1 1

Thus z  1, 3y  1  2  y  1, and x  2 1  3  x  1. Hence, the solution is 1 1 1.       1 3 1 0 1 3 1 0   x  3y  z  0     R2  3R1  R2  0 5 1 1    8. has the matrix representation  3x  4y  2z  1    3 4 2 1  R3 R  R  1 3   x  2y 0 5 1 1  1 1 2 0 1       1 3 1 0 1 3 1 0 1 0  25  35 1      R  3R  R  R3  R2  R3 1 1  1 1 . 1 2 1  0 5 1 1   5 R2        0 1  5 5    0 1  5 5  0 0 0 0 0 0 0 0 0 0 0 0

Since this system is dependent, let z  t. Then y  15 t  15  y  15 t  15 and x  25 t   35  x  25 t  35 . Thus, the   solution is 25 t  35  15 t  15  t . 

In Solutions 9–16, A  





2 4





1 0 4



    , B   1 1 , and C   1 1 2 .     2 4 3 0 0 1 3 2 3

9. A  B is undefined because A is 2  2 and B is 3  2, so they have incompatible dimensions.

10. AB is undefined because A is 2  2 and B is 3  2, so they have incompatible dimensions.            2 4  2 4 12 22 6 12 6 10    2 3                  11. B A  3B    1 1  2 4  3  1 1    0 1    3 3    3 2  3 0 3 0 6 9 9 0 3 9          1 0 4 2 4  14 4  36 58    2 3   2 3     1 1      3 3      0 3  12. C B A   1 1 2    2 4   2 4   0 1 3 3 0 8 1 18 28       3 2 3 4 3 2  1 2    A1    13. A   8  6 2 2 2 4 1 1

14. B 1 does not exist because B is not a square matrix.

15. det B is not defined because B is not a square matrix.

784

CHAPTER 11 Matrices and Determinants

      1 0 4       1 1 1 2    4 16. det C   1 1 2   1     0 1 1 3    0 1 3   4x  3y  10 17. (a) The system  3x  2y  30    4 3 (b) We have D    3 2

     1  4  3  



is equivalent to the matrix equation 

4 3 3 2

 

x y







10 30



.

            2 3 x 2 3 10 70   4 2  3 3  1. So D 1    and       .  3 4 y 3 4 30 90 

Therefore, x  70 and y  90.

 1 4 1   18. A   0 2 0  1 0 1

     1 1    2   1 1  

   1 4 0             1 0   0, B   0 2 0   2    2. Since A  0, A does not have an inverse, and          3 1   3 0 1  

 since B  0, B does have an inverse.   

1 0 0 1 2 0



  0 2 0 0 1 0   0 0 1 3 6 1    z  14   2x 19. 3x  y  5z  0    4x  2y  3z  2

1 2 R2





1 4 0 1 0 0



 0 2 0 0 1 0  3 0 1 0 0 1

1 0 0 1 2 0







4 0 1 0 0

1 2 0

R1  2R2  R1  R3  6R2  R3



     0 1 0 0 1 0 . Therefore, B 1   0 1 0 .     2 2 0 0 1 3 6 1 3 6 1

    2 0 1      1 5   Then D   3 1 5   2   2 3   4 2 3

    14 0 1      1 5   Dx    0 1 5   14   2 3    2 2 3 

       3 1     26  10  36,   1    4 2 

       0 1     182  2  180,   1    4 2 

       2 14 1            2 14   14 1    D y    3 0 5   3    120  300  180, and  5        4 2   2 3     4 2 3    2 0 14   Dz    3 1 0   4 2 2



  0 2 0 0 1 0   0 12 1 3 0 1

R3  3R1  R3



1

      1 0    2    2 2  

       3 1     4  140  144.   14     2 2 

180 144 Therefore, x  180 36  5, y  36  5, z  36  4, and so the solution is 5 5 4.

Computer Graphics

785

20. Let x and y represent the number of pounds of almonds and walnuts respectively. Then the problem is modeled by the system

of equations



x 

  1 1   Then D  , so det D   475 345  475 345 

y 3

475x  345y  1191

  3 1 det Dx     1191 345

1

1



       1 3   1035  1191  156, and det D y       475 1191

     345  475  13,  

     1191  1425  234. Then  

   D y  234 Dx  156 x    12 and y   18, so she bought 12 pounds of almonds and 18 pounds of walnuts. D D 13 13

FOCUS ON MODELING Computer Graphics 

1. The data matrix D  

0 1 1 0 0 0 1 1

square.



 represents the gray

Reflection using T   

TD 

y

2 1

0



1

0

0 1

 

1

0

0 1



:

0 1 1 0 0 0 1 1

y







0 1

1

0 0 1 1

2 1

x

2

_1

1

0

1 2

x

_1



Expansion with c  2 using T  

2 0



: 0 1      2 0 0 1 1 0 0 2 2 0    TD  0 1 0 0 1 1 0 0 1 1 y

2

_1

Shearing with c  1 using T  

1 1



: 0 1      1 1 0 1 1 0 0 1 2 1    TD  0 1 0 0 1 1 0 0 1 1 y

2

1

0



1

1

2

x

0 _1

1

2

0

x

 

786

FOCUS ON MODELING



2. The data matrix D  

0 1 1 0 0 0 1 1

square.





 represents the

Reflection in y-axis using T1   

T1 D  

y

2

1

0

0 1

 

1

0 1 1 0 0 0 1 1 y

1

_1

Expansion in y-direction with c  3 using T2   T2 D  

0 3



0 1 1 0 0 0 1 1

_1



3. (a) T  

1 15

0 1 1 0 0

0

1 1

1







:

2

x

2

_1

1 0

0 1  



3

0



1 0

y







0 1 1 0 0 0 3 3

1 0

0 3  



:

0

Shear in y-direction with c  1 using T3  



y

2

2

1

1

x

1 0

: 1 1      1 0 0 1 1 0 0 1 1 0    T3 D   1 1 0 0 1 1 0 1 2 1 3

1

x



3

0

1

_1

0

1

x



 is a shear in the x-direction. 0 1     1 15 1 15 1   (b) T 1   1 0 1 0 1

(c) T 1 is a leftward shear in the x-direction.

  (d) The result is the original matrix. Algebraically, T 1 T D  T 1 T D  I D  D where I is the 2  2 identity           1 15 1 15 0 1 1 0 1 15 0 1 25 15 0 1 1 0          matrix:  0 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 1 1

 

Computer Graphics



4. (a) T  

1 15 0 1

x-direction.



 is an expansion by a factor of 3 in the



(b) S  

1 0 0 2

y-direction.

 is an expansion by a factor of 2 in the

y

y

2

2

1

1

0

1

3

2

x

0

_1



3 0



0 3 3 0

(c) T S D  



0 1

1

 

1 0

0 2 

 

0 1 1 0 0 0 1 1





  

3 0 0 1

 

0 1 1 0 0 0 2 2



y



2 1



0

This corresponds to expansion by a factor of 3 in the x-direction and a factor of 2 in the y-direction.      3 0 1 0 3 0    (d) W  T S   0 1 0 2 0 2      3 0 0 1 1 0 0 3 3 0   . It is the same as T S D. (e) W D   0 2 0 0 1 1 0 0 2 2



5. (a) D  

0 1 1 4 4 1 1 6 6 0 0

0 0 4 4 5 5   075 0 , (b) T   0 1   075 0 0  TD  0 1 0   1 025 , (c) S   0 1   1 025 0  SD   0 1 0

x

3

2

_1

 

0 0 2 2

787



7 7 8 8 0

1

3

2

_1

 

1 1 4 4 1 1 6 6 0 0 0 4 4 5 5 7 7 8 8 0

1 1 4 4 1 1 6 6 0 0 0 4 4 5 5 7 7 8 8 0







0 0





0 1 2 5 525 225 275 775 8 2 0



0 075 075 3 3 075 075 45 45 0 0 4

0 0 4 4 5

4 5 5

5

7

7

7

7

8

8 8 0

8 0

 

 

x

788

FOCUS ON MODELING



6. (a) The data matrix D  

0 1 2 1 0 0 0 0 2 4 4 0

figure at right.



 represents the

y

4 3 2 1



(b) T  

1

1



, 0 1    1 1 0 1 2 1 0 0   TD  0 1 0 0 2 4 4 0   0 1 4 5 4 0   0 0 2 4 4 0

The transformation is a reflection about the x-axis and a shear in the x-direction.      1 1 1 0 1 1    (c) T   0 1 0 1 0 1

0

1

2

3

4

5

x

1

2

3

4

5

x

y 0 _1 _2 _3 _4

12

CONIC SECTIONS

12.1 PARABOLAS 1. A parabola is the set of all points in the plane equidistant from a fixed point called the focus and a fixed line called the directrix of the parabola. 2. The graph of the equation x 2  4 py is a parabola with focus F 0 p and directrix y   p. So the graph of x 2  12y is a parabola with focus F 0 3 and directrix y  3.

3. The graph of the equation y 2  4 px is a parabola with focus F  p 0 and directrix x   p. So the graph of y 2  12x is a parabola with focus F 3 0 and directrix x  3. 4. (a) From top to bottom: focus 0 3, vertex 0 0, directrix y  3. (b) From left to right: directrix x  3, vertex 0 0, focus 3 0.

5. y 2  2x is Graph III, which opens to the right and is not as wide as the graph for Exercise 5.

6. y 2   14 x is Graph V, the only graph that opens downward.

7. x 2  6y is Graph II, which opens downward and is narrower than the graph for Exercise 6. 8. 2x 2  y is Graph I, the only graph that opens to the right.

9. y 2  8x  0 is Graph VI, which opens to the right and is wider than the graph for Exercise 1.

10. 12y  x 2  0 is Graph IV, which opens downward and is wider than the graph for Exercise 3. 11. (a) x 2  8y, so 4 p  8  p  2. The focus is 0 2,

the directrix is y  2, and the focal diameter is 8.

12. (a) x 2  4y, so 4 p  4  p  1. The focus is

0 1, the directrix is y  1, and the focal diameter is 4.

y

(b)

y

(b)

1 1 1

13. (a) y 2  24x, so 4 p  24  p  6. The focus is

6 0, the directrix is x  6, and the focal diameter is 24.

(b)

1

x

x

14. (a) y 2  16x, so 4 p  16  p  4. The focus is 4 0, the directrix is x  4, and the focal diameter is 16.

(b)

y

y 2 2 1

1

x

x

789

790

CHAPTER 12 Conic Sections

15. (a) y   18 x 2  x 2  8y, so 4 p  8  p  2. The focus is 0 2, the directrix is y  2, and the focal diameter is 8.

16. (a) x  2y 2  y 2  12 x, so 4 p  12  p  18 . The   focus is 18  0 , the directrix is x   18 , and the focal diameter is 12 .

y

(b)

y

(b)

1

1 x

1

1

17. (a) x  2y 2  y 2   12 x, so 4 p   12  p   18 .   The focus is  18  0 , the directrix is x  18 , and the focal diameter is 12 .

18. (a) y  14 x 2  x 2  4y, so 4 p  4  p  1. The focus is 0 1, the directrix is y  1, and the focal diameter is 4.

y

(b) y

(b)

x

1

1

x

1

  19. (a) 5y  x 2 , so 4 p  5  p  54 . The focus is 0 54 , the directrix is y   54 , and the focal diameter is 5.

(b)

1

y

  20. (a) 9x  y 2 , so 4 p  9  p  94 . The focus is 94  0 , the directrix is x   94 , and the focal diameter is 9.

(b)

y

2 1 1

x

x

1

x

SECTION 12.1 Parabolas

21. (a) x 2  12y  0  x 2  12y, so 4 p  12 

p  3. The focus is 0 3, the directrix is y  3,

and the focal diameter is 12.

22. (a) x  15 y 2  0  y 2  5x, so 4 p  5    p   54 . The focus is  54  0 , the directrix is x  54 , and the focal diameter is 5.

y

(b)

y

(b) 1 x

2

1 1

23. (a) 5x  3y 2  0  y 2   53 x. Then 4 p   53    5 . The focus is  5  0 , the directrix is p   12 12 5 , and the focal diameter is 5 . x  12 3

(b)

24. (a) 8x 2  12y  0  x 2   32 y. Then 4 p   32    p   38 . The focus is 0  38 , the directrix is y  38 , and the focal diameter is 32 . y

(b)

y

1 1

1 1

x

x

x

25. x 2  16y

26. x 2  8y 1 -5

-5

-2

5

27. y 2   13 x

5

28. 8y 2  x 1

2

5

-2 -1

-2

10

791

792

CHAPTER 12 Conic Sections

29. 4x  y 2  0

30. x  2y 2  0 2

-2

-1

1

2

4

-2

31. Since the focus is 0 6, p  6  4 p  24. Hence, an equation of the parabola is x 2  24y.   32. Since the focus is 0  14 , p   14  4 p  1. So an equation of the parabola is x 2  y.

33. Since the focus is 8 0, p  8  4 p  32. Hence, an equation of the parabola is y 2  32x.

34. Since the focus is 5 0, p  5  4 p  20. Hence, an equation of the parabola is y 2  20x.   35. Since the focus is 0  34 , p   34  4 p  3. Hence, an equation of the parabola is x 2  3y.   1  0 , p   1  4 p   1 . Hence, an equation of the parabola is y 2   1 x. 36. Since the focus is  12 12 3 3

37. Since the directrix is x  4, p  4  4 p  16. Hence, an equation of the parabola is y 2  16x.

38. Since the directrix is y  12 , p   12  4 p  2. Hence, an equation of the parabola is x 2  2y.

1 , p   1  4 p   2 . Hence, an equation of the parabola is x 2   2 y. 39. Since the directrix is y  10 10 5 5

40. Since the directrix is x   18 , p  18  4 p  12 . Hence, an equation of the parabola is y 2  12 x.

1 , p   1  4 p   1 . Hence, an equation of the parabola is y 2   1 x. 41. Since the directrix is x  20 20 5 5

42. Since the directrix is y  5, p  5  4 p  20. Hence, an equation of the parabola is x 2  20y.

43. The focus is on the positive x-axis, so the parabola opens horizontally with 2 p  2  4 p  4. So an equation of the parabola is y 2  4x.

44. The focus is on the negative y-axis, so the parabola opens vertically with 2 p  6  4 p  12. So an equation of the parabola is x 2  12y.

45. The parabola opens downward with focus 10 units from 0 0, so p  10  4 p  40 and an equation of the parabola is x 2  40y.

46. Since the parabola opens upward with focus 5 units from the vertex, the focus is 5 0. So p  5  4 p  20. Thus an equation of the parabola is x 2  20y.

47. The directrix has y-intercept 6, and so p  6  4 p  24. Therefore, an equation of the parabola is x 2  24y. 48. Since the focal diameter is 8 and the focus is on the negative y-axis, 4 p  8. So the equation is x 2  8y.

49. p  2  4 p  8. Since the parabola opens upward, its equation is x 2  8y.

50. The directrix is x  2, and so p  2  4 p  8. Since the parabola opens to the left, its equation is y 2  8x. 51. p  4  4 p  16. Since the parabola opens to the left, its equation is y 2  16x.

52. p  3  4 p  12. Since the parabola opens downward, its equation is x 2  12y.

53. The focal diameter is 4 p  32  32  3. Since the parabola opens to the left, its equation is y 2  3x. 54. The focal diameter is 4 p  2 5  10. Since the parabola opens upward, its equation is x 2  10y.

55. The equation of the parabola has the form y 2  4 px. Since the parabola passes through the point 4 2, 22  4 p 4  4 p  1, and so an equation is y 2  x.

56. Since the directrix is x   p, we have p2  16, so p  4, and an equation is y 2  4 px or y 2  16x.

SECTION 12.1 Parabolas

793

 57. The area of the shaded region is width  height  4 p  p  8, and so p2  2  p   2 (because the parabola opens   downward). Therefore, an equation is x 2  4 py  4 2y  x 2  4 2y. 58. The focus is 0 p. Since the line has slope 12 , an equation of the line is y  12 x  p. Therefore, the point where the line intersects the parabola has y-coordinate 12 2  p  p  1. The parabola’s equation is of the form x 2  4 py,  1  5 (since p  0). Hence, an equation of the parabola is so 22  4 p  p  1  p2  p  1  0  p  2   5  1 y. x2  2

59. (a) A parabola with directrix y   p has equation x 2  4 py. If the directrix   is y  12 , then p   12 , so an equation is x 2  4  12 y  x 2  2y.

(b)

1

-4

-2

4 x@=_32y x@=_16y

-3

x 2  4y. If the directrix is y  4, then p  4, so an equation is

-4 x@=_4y x@=_2y

-5

x 2  4 4 y  x 2  16y. If the directrix is y  8, then p  8, so

an equation is x 2  4 8 y  x 2  32y.

As the directrix moves further from the vertex, the parabolas get flatter. (b)

10

focal diameter is 4 p  1, an equation is x 2  y. If the focal diameter is

y=x@

2y=x@

8

4y=x@

6

4 p  2, an equation is x 2  2y. If the focal diameter is 4 p  4, an

4

equation is x 2  4y. If the focal diameter is 4 p  8, an equation is x 2  8y.

2

-2

If the directrix is y  1, then p  1, so an equation is x 2  4 1 y 

60. (a) If the focal diameter of a parabola is 4 p, it has equation x 2  4 py. If the

0 -1

8y=x@

2 -4

-2

0

2

4

As the focal diameter increases, the parabolas get flatter.

61. (a) Since the focal diameter is 12 cm, 4 p  12. Hence, the parabola has equation y 2  12x.

(b) At a point 20 cm horizontally from the vertex, the parabola passes through the point 20 y, and hence from part (a),   y 2  12 20  y 2  240  y  4 15. Thus, C D  8 15  31 cm.

62. The equation of the parabola has the form x 2  4 py. From the diagram, the parabola passes through the point 10 1, and so 102  4 p 1  4 p  100 and p  25. Therefore, the receiver is 25 ft from the vertex.

63. With the vertex at the origin, the top of one tower will be at the point 300 150. Inserting this point into the equation x 2  4 py gives 3002  4 p 150  90000  600 p  p  150. So an equation of the parabolic part of the cables is

x 2  4 150 y  x 2  600y.

64. The equation of the parabola has the form x 2  4 py. From the diagram, the parabola passes through the point 100 379, and so 1002  4 p 379  1516 p  10000 and p  65963. Therefore, the receiver is about 65963 inches  55 feet from the vertex.

65. Many answers are possible: satellite dish TV antennas, sound surveillance equipment, solar collectors for hot water heating or electricity generation, bridge pillars, etc. 66. Yes. If a cone intersects a plane that is parallel to a line on the cone, the resulting curve is a parabola, as shown in the text.

794

CHAPTER 12 Conic Sections

12.2 ELLIPSES 1. An ellipse is the set of all points in the plane for which the sum of the distances from two fixed points F1 and F2 is constant. The points F1 and F2 are called the foci of the ellipse. y2 x2 2. The graph of the equation 2  2  a b  where c  a 2  b2 . So the graph of

1 with a  b  0 is an ellipse with vertices a 0 and a 0 and foci c 0, x2 y2  2  1 is an ellipse with vertices 5 0 and 5 0 and foci 3 0 and 2 5 4

3 0.

x2 y2 3. The graph of the equation 2  2  1 with a  b  0 is an ellipse with vertices 0 a and 0 a and foci 0 c, b a  x2 y2 where c  a 2  b2 . So the graph of 2  2  1 is an ellipse with vertices 0 5 and 0 5 and foci 0 3 and 4 5 0 3. 4. (a) From left to right: vertex 5 0, focus 3 0, focus 3 0, vertex 5 0. (b) From top to bottom: vertex 0 5, focus 0 3, focus 0 3, vertex 0 5. 5.

y2 x2   1 is Graph II. The major axis is horizontal and the vertices are 4 0. 16 4

y2 6. x 2   1 is Graph IV. The major axis is vertical and the vertices are 0 3. 9 7. 4x 2  y 2  4 is Graph I. The major axis is vertical and the vertices are 0 2. 8. 16x 2  25y 2  400 is Graph III. The major axis is horizontal and the vertices are 5 0. 9.

y2 x2   1. 25 9

(c)

(a) This ellipse has a  5, b  3, and so c2  a 2  b2  16  c  4. The vertices c are 5 0, the foci are 4 0, and the eccentricity is e   45  08. a

y

1 1

x

1

x

(b) The length of the major axis is 2a  10, and the length of the minor axis is 2b  6.

10.

y2 x2  1 16 25 (a) This ellipse has a  5, b  4, and so c2  25  16  9  c  3. The vertices are 0 5, the foci are 0 3, and the eccentricity is e  ac  35  06.

(b) The length of the major axis is 2a  10, and the length of the minor axis is 2b  8.

(c)

y

1

SECTION 12.2 Ellipses

11.

y2 x2  1 36 81

(c)

 (a) This ellipse has a  9, b  6, and so c2  81  36  45  c  3 5. The    vertices are 0 9, the foci are 0 3 5 , and the eccentricity is

795

y

2 x

2

 e  ac  35 .

(b) The length of the major axis is 2a  18 and the length of the minor axis is 2b  12.

12.

x2  y2  1 4

(c)

 (a) This ellipse has a  2, b  1, and so c2  4  1  3  c  3. The vertices     are 2 0, the foci are  3 0 , and the eccentricity is e  ac  23 .

y

1 1

x

(b) The length of the major axis is 2a  4 and the length of the minor axis is 2b  2.

13.

y2 x2  1 49 25

(c)

 (a) This ellipse has a  7, b  5, and so c2  49  25  24  c  2 6. The    vertices are 7 0, the foci are 2 6 0 , and the eccentricity is

y

2 2

x

2

x

 e  ac  2 7 6 .

(b) The length of the major axis is 2a  14 and the length of the minor axis is 2b  10.

14.

x2 y2  1 9 64

(c)

 (a) This ellipse has a  8, b  3, and so c2  64  9  55  c  55. The    vertices are 0 8, the foci are 0  55 , and the eccentricity is

y

2

 e  ac  855 .

(b) The length of the major axis is 2a  16 and the length of the minor axis is 2b  6.

15. 9x 2  4y 2  36 

y2 x2  1 4 9

 (a) This ellipse has a  3, b  2, and so c2  9  4  5  c  5. The vertices     are 0 3, the foci are 0  5 , and the eccentricity is e  ac  35 .

(b) The length of the major axis is 2a  6, and the length of the minor axis is 2b  4.

(c)

y

1 1

x

796

CHAPTER 12 Conic Sections

16. 4x 2  25y 2  100 

y2 x2  1 25 4

(c)

 (a) This ellipse has a  5, b  2, and so c2  25  4  21  c  21. The    vertices are 5 0, the foci are  21 0 , and the eccentricity is

y

1 1

x

1

x

1

x

2

x

 e  ac  521 .

(b) The length of the major axis is 2a  10, and the length of the minor axis is 2b  4.

y2 x2  1 17. x 2  4y 2  16  16 4

(c)

 (a) This ellipse has a  4, b  2, and so c2  16  4  12  c  2 3.The    vertices are 4 0, the foci are 2 3 0 , and the eccentricity is

y

1

  e  ac  2 4 3  23 .

(b) The length of the major axis is 2a  8, and the length of the minor axis is 2b  4.

18. 4x 2  y 2  16 

y2 x2  1 4 16

(c)

 (a) This ellipse has a  4, b  2, and so c2  16  4  12  c  2 3. The    vertices are 0 4, the foci are 0 2 3 , and the eccentricity is

y

1

  e  ac  2 4 3  23 .

(b) The length of the major axis is 2a  8, and the length of the minor axis is 2b  4.

19. 16x 2  25y 2  1600 

y2 x2  1 100 64

(c)

(a) This ellipse has a  10, b  8, and so c2  100  64  36  c  6. The

y

2

vertices are 10 0, the foci are 6 0, and the eccentricity is e  ac  35 .

(b) The length of the major axis is 2a  20 and the length of the minor axis is 2b  16.

y2 x2  1 49 2   (a) This ellipse has a  7, b  2, and so c2  49  2  47  c  47. The    vertices are 7 0, the foci are  47 0 , and the eccentricity is

20. 2x 2  49y 2  98 

 e  ac  747 .

(b) The length of the major axis is 2a  14 and the length of the minor axis is  2b  2 2.

(c)

y

1 1

x

SECTION 12.2 Ellipses

y2 x2  1 3 9   (a) This ellipse has a  3, b  3, and so c2  9  3  6  c  6. The    vertices are 0 3, the foci are 0  6 , and the eccentricity is

21. 3x 2  y 2  9 

(c)

797

y

1 1

x

1

x

1

x

1

x

 e  ac  36 .

(b) The length of the major axis is 2a  6 and the length of the minor axis is  2b  2 3. x2 y2 22. x 2  3y 2  9   1 9 3   (a) This ellipse has a  3, b  3, and so c2  9  3  6  c  6. The    vertices are 3 0, the foci are  6 0 , and the eccentricity is

(c)

y

1

 e  ac  36 .

(b) The length of the major axis is 2a  6 and the length of the minor axis is  2b  2 3. y2 x2  1 2 4   (a) This ellipse has a  2, b  2, and so c2  4  2  2  c  2. The    vertices are 0 2, the foci are 0  2 , and the eccentricity is

23. 2x 2  y 2  4 

(c)

y

1

 e  ac  22 .

(b) The length of the major axis is 2a  4 and the length of the minor axis is  2b  2 2. y2 x2  1 4 3  (a) This ellipse has a  2, b  3, and so c2  4  3  c  1. The vertices are

24. 3x 2  4y 2  12 

(c)

y

1

2 0, the foci are 1 0, and the eccentricity is e  ac  12 .

(b) The length of the major axis is 2a  4 and the length of the minor axis is  2b  2 3. y2 x2  1 1 25. x 2  4y 2  1  1 4

 (a) This ellipse has a  1, b  12 , and so c2  1  14  34  c  23 . The vertices    are 1 0, the foci are  23  0 , and the eccentricity is   3 e  ac  32 1  2 .

(b) The length of the major axis is 2a  2, and the length of the minor axis is 2b  1.

(c)

y 1

1

x

798

CHAPTER 12 Conic Sections

26. 9x 2  4y 2  1 

y2 x2  1 19 14

(c)

 5  c  5 . The vertices are (a) This ellipse has a  12 , and so c2  14  19  36 6        56 5 1 0  2 , the foci are 0  6 , and the eccentricity is e  ac  12  35 .

y 0.5

0.5 x

(b) The length of the major axis is 2a  1, and the length of the minor axis is 2b  23 .

27. x 2  4  2y 2  x 2  2y 2  4 

y2 x2  1 4 2

(c)

  2, and so c2  4  2  2  c  2. The    vertices are 2 0, the foci are  2 0 , and the eccentricity is

y

1

(a) This ellipse has a  2, b 

x

1

 e  ac  22 .

(b) The length of the major axis is 2a  4, and the length of the minor axis is  2b  2 2. x2 y2 1 28. y 2  1  2x 2  2x 2  y 2  1  1  1

(c)

y 1

2   2 (a) This ellipse has a  1, b  2 , and so c2  1  12  12  c  22 . The    vertices are 0 1, the foci are 0  22 , and the eccentricity is   e  ac  11 2  22 .

1

x

(b) The length of the major axis is 2a  2, and the length of the minor axis is  2b  2. 29. This ellipse has a horizontal major axis with a  5 and b  4, so an equation is

x2 52



y2 42

1

y2 x2   1. 25 16

y2 x2 y2 x2   1. 30. This ellipse has a vertical major axis with a  5 and b  2. Thus an equation is 2  2  1  4 25 2 5  31. This ellipse has a vertical major axis with c  2 and b  2. So a 2  c2  b2  22  22  8  a  2 2. So an equation is

x2

y2 y2 x2   1.   1     2 4 8 22 2 2

32. This ellipse has a vertical major axis with a  4 and c  3. So c2  a 2  b2  9  16  b2  b2  7. Thus an equation is

x2 y2 y2 x2  2 1   1. 7 7 16 4

x2 y2  2  1. Substituting 2 16 b 36 36 36 3 4 36 64 1 2  48. Thus, an the point 8 6 into the equation, we get 256  2  1  2  1  4  2   b  4 3 b b b y2 x2   1. equation of the ellipse is 256 48

33. This ellipse has a horizontal major axis with a  16, so an equation of the ellipse is of the form

SECTION 12.2 Ellipses

799

x2 y2 34. This ellipse has a vertical major axis with b  2, so an equation of the ellipse is of the form 2  2  1. Substituting the 2 a 4 16 4 4 1 3 4 4  . Thus, an equation of point 1 2 into the equation, we get 14  2  1  2  1   2   a 2  4 4 3 3 a a a x2 y2 x2 3y 2 the ellipse is  1   1. 4 163 4 16 35.

x2 y2 y2 x2 4x 2  1 1  y 2  20   25 20 20 25 5  4x 2 . y   20  5

y2 y2 1  1  x 2  y 2  12  12x 2  12 12  y   12  12x 2 .

36. x 2 

4 2

5 -2 -5

-1

-2

1

2

5 -4 -5

 37. 6x 2  y 2  36  y 2  36  6x 2  y   36  6x 2 .

x2 38. x 2  2y 2  8  2y 2  8  x 2  y 2  4   2  x2 y  4 . 2

5

-10

10

2

-5 -2

2 -2

39. The foci are 4 0, and the vertices are 5 0. Thus, c  4 and a  5, and so b2  25  16  9. Therefore, an equation of the ellipse is

y2 x2   1. 25 9

40. The foci are 0 3 and the vertices are 0 5. Thus, c  3 and a  5, and so c2  a 2  b2  9  25  b2  b2  25  9  16. Therefore, an equation of the ellipse is

y2 x2   1. 16 25

41. The foci are 1 0 and the vertices are 2 0. Thus, c  1 and a  2, so c2  a 2  b2  1  4  b2  b2  4  1  3. Therefore, an equation of the ellipse is

y2 x2   1. 4 3

42. The foci are 0 2 and the vertices are 0 3. Thus, c  2 and a  3, so c2  a 2  b2  4  9  b2  b2  9  4  5. Therefore, an equation of the ellipse is

x2 y2   1. 5 9

    43. The foci are 0  10 and the vertices are 0 7. Thus, c  10 and a  7, so c2  a 2  b2  10  49  b2  b2  49  10  39. Therefore, an equation of the ellipse is

y2 x2   1. 39 49

800

CHAPTER 12 Conic Sections

    44. The foci are  15 0 and the vertices are 6 0. Thus, c  15 and a  6, so c2  a 2  b2  15  36  b2 

y2 x2   1. 36 21 45. The length of the major axis is 2a  4  a  2, the length of the minor axis is 2b  2  b  1, and the foci are on the b2  36  15  21. Therefore, an equation of the ellipse is

y2  1. 4 46. The length of the major axis is 2a  6  a  3, the length of the minor axis is 2b  4  b  2, and the foci are on the y-axis. Therefore, an equation of the ellipse is x 2 

x-axis. Therefore, an equation of the ellipse is

x2 y2   1. 9 4

47. The foci are 0 2, and the length of the minor axis is 2b  6  b  3. Thus, a 2  4  9  13. Since the foci are on the y-axis, an equation is

x2 y2   1. 9 13

48. The foci are 5 0, and the length of the major axis is 2a  12  a  6. Thus, c2  a 2  b2  25  36  b2 

y2 x2   1. 36 11 49. The endpoints of the major axis are 10 0  a  10, and the distance between the foci is 2c  6  c  3. Therefore, b2  36  25  11. Since the foci are on the x-axis, an equation is

x2 y2   1. 100 91 50. Since the endpoints of the minor axis are 0 3, we have b  3. The distance between the foci is 2c  8, so c  4. Thus, b2  100  9  91, and so an equation of the ellipse is

y2 x2   1. 25 9 51. The length of the major axis is 10, so 2a  10  a  5, and the foci are on the x-axis, so the form of the equation is  2   5 y2 4 x2 5 4 4 22  2  1. Since the ellipse passes through  2 1  5 2 , we have 1 2   25 25 25 b2 5 b b b y2 x2   1. b2  5, and so an equation is 25 5 52. The length of the minor axis is 10, so 2b  10  b  5, and the foci are on the y-axis, so the form of the equation is  2  2    5 40 x 2 y2 5 40 4 40 5 40 , we have 1 1 2   2  1. Since the ellipse passes through   25 a 25 25 a 2 5 a2 a 2 2 y x   1.  a 2  50, and so an equation is 25 50 a 2  b2  c2  9  16  25, and an equation of the ellipse is

2  6 and 53. The eccentricity is 13 , so e  13 , and the foci are 0 2, so c  2. Thus, e  ac  a  ce  13

y2 x2   1. b2  a 2  c2  36  4  32. The major axis lies on the y-axis, so an equation is 32 36   54. The eccentricity is e  075  34 and the foci are 15 0  32  0 , so c  32 . Thus, a  ce  32 34  2 and y2 x2  7  1. b2  a 2  c2  4  94  74 . The major axis lies on the x-axis, so an equation is 4

4   3 55. Since the length of the major axis is 2a  4, we have a  2. The eccentricity is 2  ac  2c , so c  3. Then y2 b2  a 2  c2  4  3  1, and since the foci are on the y-axis, an equation of the ellipse is x 2   1.

4

  56. The eccentricity is e  35 and the major axis has length 2a  12, so a  6. Thus, c  ae  2 5 and y2 x2   1. b2  a 2  c2  36  20  16. The foci are on the x-axis, so an equation of the ellipse is

36

16

SECTION 12.2 Ellipses

  4x 2  y 2  4 57.  4x 2  9y 2  36

y (0, 2)

Subtracting the first equation from the second gives 1

8y 2  32  y 2  4  y  2. Substituting y  2 in the first equation gives

(0, _2)

  y2 1  9x 2  16y 2  144  144x 2  256y 2  2304 9   2 2  16x  9y  144  144x 2  81y 2  1296 y2 1 16

y

(_ 125 , 125)

Dividing the first equation by 100 gives x 2  y2

( 125 , 125)

1

x

1

Adding gives 175y 2  1008  y   12 5 . Substituting for y gives  2 1296 12  144  9x 2  144  2304 9x 2  16 12 5 25  25  x   5 , and so the   12 four points of intersection are  12 5  5 .

   100x 2  25y 2  100 59. y2   x 2  1 9

x

1

4x 2  22  4  x  0, and so the points of intersection are 0 2.

 2 x     16 58. 2   x  9

801

(

12 12 _5 ,_5

)

(

12 12 5,_5

)

y

y2  1. 4

1

(_1, 0)

y2

(1, 0) 1

 0 Subtracting this equation from the second equation gives 9 4   1  1 y 2  0  y  0. Substituting y  0 in the second equation gives 9 4

x

x 2  02  1  x  1, and so the points of intersection are 1 0.

   25x 2  144y 2  3600  3600x 2  20,736y 2  518,400 60.   144x 2  25y 2  3600  3600x 2  625y 2  90,000

y

Subtracting

3600  the second equation from the first, we have 20,111y 2  428,400, so y 2  169  2 60 2 y   60  3600  x   60 13 . Substituting for y gives 25x  144 13 13 , and so   60 the four points of intersection are  60 13   13 .

10

(_ 6013 , 6013 )

( 6013 , 6013 ) 10 x

(

_ 60 , _ 60 13 13

(

)

60 , _ 60 13 13

)

y2 x2   1 has a  4 and b  2. Thus, an equation of the ancillary circle is 61. (a) The ellipse x 2  4y 2  16  16 4 x 2  y 2  4.

(b) If s t is a point on the ancillary circle, then s 2  t 2  4  4s 2  4t 2  16  2s2  4 t2  16, which implies that 2s t is a point on the ellipse.

1 62. (a) x 2  ky 2  100  ky 2  100  x 2  y   100  x 2 . For the top k 1 half, we graph y  100  x 2 for k  4, 10, 25, and 50. k (b) This family of ellipses have common major axes and vertices, and the eccentricity increases as k increases.

k=4 2 1 k=50 -10-8 -6 -4 -2 0

k=10 k=25 2 4 6 8 10

802

63.

CHAPTER 12 Conic Sections

y2 x2   1 is an ellipse for k  0. Then a 2  4  k, b2  k, and so c2  4  k  k  4  c  2. Therefore, all of k 4k the ellipses’ foci are 0 2 regardless of the value of k.

64. The foci are c 0, where c2  a 2  b2 . The endpoints of one latus rectum are the points c k, and the length is 2k.   2 a 2  c2 2 2 2 2 2 2 b k k c a c c  k2  . Since Substituting this point into the equation, we get 2  2  1  2  1  2  2 a b b a a2 a  2b2 b4 b2 b2 b2  a 2  c2 , the last equation becomes k 2  2  k  . Thus, the length of the latus rectum is 2k  2  . a a a a 65. Using the perihelion, a  c  147,000,000, while using the aphelion, a  c  153,000,000. Adding, we have 2  2  2a  300,000,000  a  150,000,000. So b2  a 2 c2  150  106  3  106  22,4911012  224911016 .

y2 x2   1. 16 22500  10 22491  1016 c 66. Using the eccentricity, e  025   c  025a. Using the length of the minor axis, 2b  10,000,000,000  a 18 b  5  109 . Since a 2  c2  b2 , a 2  025a2  25  1018  15 a 2  25  1018  a 2  80 3  10    16     5 5 9 9 9  5  109 . Since the Sun is at one focus of the ellipse, a  80 3  10  4 3  10 . Then c  025 4 3  10 3    the distance from Pluto to the Sun at perihelion is a  c  4 53  109  53  109  3 53  109  387  109 km; the    distance from Pluto to the Sun at aphelion is a  c  4 53  109  53  109  5 53  109  645  109 km. Thus, an equation of the orbit is

67. Using the perilune, a  c  1075  68  1143, and using the apolune, a  c  1075  195  1270. Adding, we get

2a  2413  a  12065. So c  1270  12065  c  635. Therefore, b2  120652  6352  1,451,610. Since

y2 x2   1. 1,455,642 1,451,610 68. Placing the origin at the center of the sheet of plywood and letting the x-axis be the long central axis, we have 2a  8, so  that a  4, and 2b  4, so that b  2. So c2  a 2  b2  42  22  12  c  2 3  346. So the tacks should be located 2 346  692 feet apart and the string should be 2a  8 feet long. a 2  1,455,642, an equation of Apollo 11’s orbit is

69. From the diagram, a  40 and b  20, and so an equation of the ellipse whose top half is the window is

x2 y2   1. 1600 400

252 h2 Since the ellipse passes through the point 25 h, by substituting, we have   1  625  4y 2  1600  1600 400   5 39 975   1561 in. Therefore, the window is approximately 156 inches high at the specified point. y 2 2 70. Have each friend hold one end of the string on the blackboard. These fixed points will be the foci. Then, keeping the string taut with the chalk, draw the ellipse. 71. We start with the flashlight perpendicular to the wall; this shape is a circle. As the angle of elevation increases, the shape of the light changes to an ellipse. When the flashlight is angled so that the outer edge of the light cone is parallel to the wall, the shape of the light is a parabola. Finally, as the angle of elevation increases further, the shape of the light is hyperbolic. 72. The shape drawn on the paper is almost, but not quite, an ellipse. For example, when the bottle has radius 1 unit and the compass legs are set 1 unit apart, then it can be shown that an equation of the resulting curve is 1  y 2  2 cos x. The graph of this curve differs very slightly from the ellipse with the same major and minor axis. This example shows that in mathematics, things are not always as they appear to be.

SECTION 12.3 Hyperbolas

803

12.3 HYPERBOLAS 1. A hyperbola is the set of all points in the plane for which the difference of the distances from two fixed point F1 and F2 is constant. The points F1 and F2 are called the foci of the hyperbola. y2 x2 2. The graph of the equation 2  2  1 with a  0, b  0 is a hyperbola with horizontal transverse axis, vertices a 0 a b  x2 y2 and a 0 and foci c 0, where c  a 2  b2 . So the graph of 2  2  1 is a hyperbola with vertices 4 0 and 4 3 4 0 and foci 5 0 and 5 0. y2 x2 3. The graph of the equation 2  2  1 with a  0, b  0 is a hyperbola with vertical transverse axis, vertices 0 a a b  y2 x2 and 0 a and foci 0 c, where c  a 2  b2 . So the graph of 2  2  1 is a hyperbola with vertices 0 4 and 4 3 0 4 and foci 0 5 and 0 5. 4. (a) From left to right: focus 5 0, vertex 4 0, asymptote y   34 x, asymptote y  34 x, vertex 4 0, focus 5 0.

(b) From top to bottom: focus 0 5, vertex 0 4, asymptote y   43 x, asymptote y  43 x, vertex 0 4, focus 0 5. x2  y 2  1 is Graph III, which opens horizontally and has vertices at 2 0. 5. 4 x2 6. y 2   1 is Graph IV, which opens vertically and has vertices at 0 1. 9 7. 16y 2  x 2  144 is Graph II, which pens vertically and has vertices at 0 3. 8. 9x 2  25y 2  225 is Graph I, which opens horizontally and has vertices at 5 0.

9.

y2 x2  1 4 16

(c)

 (a) The hyperbola has a  2, b  4, and c2  16  4  c  2 5. The vertices    are 2 0, the foci are 2 5 0 , and the asymptotes are y   42 x 

y

1 1

x

1

x

1

x

y  2x.

(b) The transverse axis has length 2a  4. 10.

x2 y2  1 9 16

(c)

(a) The hyperbola has a  3, b  4, and c2  9  16  25  c  5. The vertices

y

1

are 0 3, the foci are 0 5, and the asymptotes are y   34 x.

(b) The transverse axis has length 2a  6.

11.

x2 y2  1 36 4  (a) The hyperbola has a  6, b  2, and c2  36  4  40  c  2 10. The    vertices are 0 6, the foci are 0 2 10 , and the asymptotes are y  3x.

(b) The transverse axis has length 2a  12.

(c)

y

2

804

12.

CHAPTER 12 Conic Sections

y2 x2  1 9 64

(c)

 (a) The hyperbola has a  3, b  8, and c2  9  64  73  c  73. The    vertices are 3 0, the foci are  73 0 , and the asymptotes are y   83 x.

y

5 2

x

1

x

(b) The transverse axis has length 2a  6.

13.

x2 y2  1 1 25

(c)

y 1

 (a) The hyperbola has a  1, b  5, and c2  1  25  26  c  26. The    vertices are 0 1, the foci are 0  26 , and the asymptotes are y   15 x.

(b) The transverse axis has length 2a  2.

14.

y2 x2  1 2 1

(c)

  (a) The hyperbola has a  2, b  1, and c2  2  1  3  c  3. The       vertices are  2 0 , the foci are  3 0 , and the asymptotes are

y

1 1

x

1

x

 y   22 x.

 (b) The transverse axis has length 2a  2 2. 15. x 2  y 2  1

(c)

 (a) The hyperbola has a  1, b  1, and c2  1  1  2  c  2. The vertices    are 1 0, the foci are  2 0 , and the asymptotes are y  x.

y

1

(b) The transverse axis has length 2a  2.

16.

y2 x2  1 16 12

(c)

 (a) The hyperbola has a  4, b  2 3, and c2  16  12  28       c  28  2 7. The vertices are 4 0, the foci are 2 7 0 , and the

y

2 2

x

1

x

 asymptotes are y   23 x.

(b) The transverse axis has length 2a  8. 17. 9x 2  4y 2  36 

y2 x2  1 4 9

 (a) The hyperbola has a  2, b  3, and c2  4  9  13  c  13. The    vertices are 2 0, the foci are  13 0 , and the asymptotes are y   32 x.

(b) The transverse axis has length 2a  4.

(c)

y

1

SECTION 12.3 Hyperbolas

18. 25y 2  9x 2  225 

x2 y2  1 9 25

(c)

 (a) The hyperbola has a  3, b  5, and c2  25  9  34  c  34. The    vertices are 0 3, the foci are 0  34 , and the asymptotes are y   35 x.

805

y

2 2

x

2

x

1

x

1

x

1

x

(b) The transverse axis has length 2a  6.

19. 4y 2  9x 2  144 

x2 y2  1 36 16

(c)

 (a) The hyperbola has a  6, b  4, and c2  a 2  b2  52  c  2 13. The    vertices are 0 6, the foci are 0 2 13 , and the asymptotes are

y

2

y   32 x.

(b) The transverse axis has length 2a  12.

x2 y2  1 20. y 2  25x 2  100  100 4

(c)

 (a) The hyperbola has a  10, b  2, and c2  100  4  104  c  2 26. The    vertices are 0 10, the foci are 0 2 26 , and the asymptotes are

y

5

y  5x.

(b) The transverse axis has length 2a  20.

y2 x2  1 8 2    (a) The hyperbola has a  2 2, b  2, and c2  8  2  10  c  10. The       vertices are 2 2 0 , the foci are  10 0 , and the asymptotes are

21. x 2  4y 2  8  0 

(c)

y

1

 y   2 x   12 x. 8

 (b) The transverse axis has length 2a  4 2.

y2 x2  1 3 9   (a) The hyperbola has a  3, b  3, and c2  3  9  12  c  2 3. The       vertices are 0  3 , the foci are 0 2 3 , and the asymptotes are

22. 3y 2  x 2  9  0 

 y   33 x.

 (b) The transverse axis has length 2a  2 3.

(c)

y

1

806

CHAPTER 12 Conic Sections

x2 y2  1 23. x 2  y 2  4  0  y 2  x 2  4  4 4

y

(c)

 (a) The hyperbola has a  2, b  2, and c2  4  4  8  2 2. The vertices are    0 2, the foci are 0 2 2 , and the asymptotes are y  x.

1 1

x

1

x

(b) The transverse axis has length 2a  4.

x2 y2  1 24. x 2  3y 2  12  0  4 12  (a) The hyperbola has a  2, b  2 3, and c2  4  12  16  c  4. The

y

(c)

1

vertices are 0 2, the foci are 0 4, and the asymptotes are  2 x   3 x. y  3 2 3

(b) The transverse axis has length 2a  4. y2 25. 4y 2  x 2  1  1  x 2  1

y

(c)

1

4

 (a) The hyperbola has a  12 , b  1, and c2  14  1  54  c  25 . The      vertices are 0  12 , the foci are 0  25 , and the asymptotes are

1

x

1

x

1 y   12 1 x   2 x.

(b) The transverse axis has length 2a  1. 26. 9x 2  16y 2  1 

y2 x2  1 19 116

y

(c)

1

1  25  c  5 . The (a) The hyperbola has a  13 , b  14 , and c2  19  16 144 12     1 5 vertices are  3  0 , the foci are  12  0 , and the asymptotes are 3 y   14 13 x   4 x.

(b) The transverse axis has length 2a  23 . 27. From the graph, the foci are 4 0, and the vertices are 2 0, so c  4 and a  2. Thus, b2  16  4  12, and since the vertices are on the x-axis, an equation of the hyperbola is

x2 y2   1. 4 12

28. From the graph, the foci are 0 13 and the vertices are 0 12, so c  13 and a  12. Then b2  c2  a 2  169  144  25, and since the vertices are on the y-axis, an equation of the hyperbola is

y2 x2   1. 144 25

29. From the graph, the vertices are 0 4, the foci are on the y-axis, and the hyperbola passes through the point 3 5. So

y2 x2 9 25 32 52  2  1. Substituting the point 3 5, we have  2 1 1  2  16 16 16 b b b 9 x2 y2 9  2  b2  16. Thus, an equation of the hyperbola is   1. 16 16 16 b

the equation is of the form

SECTION 12.3 Hyperbolas

807

    x2 y2 30. The vertices are 2 3 0 , so a  2 3, so an equation of the hyperbola is of the form   2  2  1. Substituting b 2 3 the point 4 4 into the equation, we get hyperbola is

16 16 16 16 16 4  1 2   b2  48. Thus, an equation of the 1 2  12 b2 12 12 b b

y2 x2   1. 12 48

a 3 31. From the graph, the vertices are 0 3, so a  3. Since the asymptotes are y  3x   x, we have  3  b  1. b b y2 x2 y2 Since the vertices are on the x-axis, an equation is 2  2  1   x 2  1. 9 3 1 b 1 3 b 32. The vertices are 3 0, so a  3. Since the asymptotes are y   12 x   x, we have   b  . Since the a 3 2 2 4y 2 y2 x2 x2   1. 1 vertices are on the x-axis, an equation is 2  9 9 3 322 33. x 2  2y 2  8  2y 2  x 2  8  y 2  12 x 2  4   y   12 x 2  4

34. 3y 2  4x 2  24  3y 2  4x 2  24  y 2  43 x 2  8   y   43 x 2  8 10

5

-5

5

-10

10

-5

35.

-10

x2 y2 x2 y2 x2  1   1  y2  2 2 6 2 6 3  x2 2 y 3 5

-5

36.

y2 x2   1  16x 2  25y 2  1600  100 64 2 25y 2  16x 2  1600  y 2  16 25 x  64   2 y   16 25 x  64

20

5 -20

20

-5 -20

37. The foci are 5 0 and the vertices are 3 0, so c  5 and a  3. Then b2  25  9  16, and since the vertices are on the x-axis, an equation of the hyperbola is

x2 y2   1. 9 16

38. The foci are 0 10 and the vertices are 0 8, so c  10 and a  8. Then b2  c2  a 2  100  64  36, and since the vertices are on the y-axis, an equation of the hyperbola is

x2 y2   1. 64 36

39. The foci are 0 2 and the vertices are 0 1, so c  2 and a  1. Then b2  4  1  3, and since the vertices are on x2  1. the y-axis, an equation is y 2  3

808

CHAPTER 12 Conic Sections

40. The foci are 6 0 and the vertices are 2 0, so c  6 and a  2. Then b2  c2  a 2  36  4  32, and since the vertices are on the x-axis, an equation is

y2 x2   1. 4 32

b b 41. The vertices are 1 0 and the asymptotes are y  5x, so a  1. The asymptotes are y   x, so  5  b  5. a 1 2 y Therefore, an equation of the hyperbola is x 2   1. 25 6 a 42. The vertices are 0 6, so a  6. The asymptotes are y   13 x   x   13  b  18. Since the vertices are on b b y2 x2 the y-axis, an equation of the hyperbola is   1. 36 324 43. The vertices are 0 6, so a  6. Since the vertices are on the y-axis, the hyperbola has an equation of the form

x2 25 81 25 45 y2  2  1. Since the hyperbola passes through the point 5 9, we have  2 1 2   b2  20. 36 36 36 b b b x2 y2   1. Thus, an equation is 36 20

44. The vertices are 2 0, so a  2. Since the vertices are on the x-axis, the hyperbola has an equation of the form    y2 30 30 9 5 x2  2  1. Since the hyperbola passes through the point 3 30 , we have  2  1   2  b2  24. 4 4 4 b b b y2 x2   1. Thus, an equation is 4 24 45. The asymptotes of the hyperbola are y  x, so b  a. Since the hyperbola passes through the point 5 3, its foci are on x2 y2 25 9 the x-axis, and its equation has the form, 2  2  1, so it follows that 2  2  1  a 2  16  b2 . Therefore, an a a a a x2 y2 equation of the hyperbola is   1. 16 16

46. The asymptotes of the hyperbola are y  x, so b  a. Since the hyperbola passes through the point 1 2, its foci are on

y2 x2 4 1 the y-axis, and its equation has the form 2  2  1, so it follows that 2  2  1  a 2  3. Therefore, an equation a a a a y2 x2 of the hyperbola is   1. 3 3

x2 42 12 y2 47. The foci are 0 3, so c  3 and an equation is 2  2  1. The hyperbola passes through 1 4, so 2  2  1  a  b b       a 2 2 2 2 2 2 2 2 4 2 2 2 16b  a  a b  16b  9  b  9  b b  b  8b  9  0  b  1 b  9  0, Thus, b2  1, a 2  8, and an equation is

y2  x 2  1. 8

       x2 y2  1. The hyperbola passes through 4 18 , 48. The foci are  10 0 , so c  10 and an equation is 2  b a     18 16 2 2 2 2 2 so 2  2  1  16b  18a  a b  16b  18 10  b2  10  b2 b2  b4  24b2  180  0  a b    y2 x2 b2  30 b2  6  0. Thus, b2  6, a 2  4, and an equation is   1. 4 6

49. The foci are 5 0, and the length of the transverse axis is 6, so c  5 and 2a  6  a  3. Thus, b2  25  9  16, and an equation is

y2 x2   1. 9 16

SECTION 12.3 Hyperbolas

809

50. The foci are 0 1, and the length of the transverse axis is 1, so c  1 and 2a  1  a  12 . Then y2 4x 2 x2 b2  c2  a 2  1  14  34 , and since the foci are on the y-axis, an equation is   1  4y 2   1. 14 34 3   y2 x2   1 has a  5 and b  5. Thus, the asymptotes are y  x, and their 5 5 slopes are m 1  1 and m 2  1. Since m 1  m 2  1, the asymptotes are perpendicular.

51. (a) The hyperbola x 2  y 2  5 

(b) Since the asymptotes are perpendicular, they must have slopes 1, so a  b. Therefore, c2  2a 2  a 2 

c2 , and 2

x2 y2 c2 since the vertices are on the x-axis, an equation is 1  1  1  x 2  y 2  . 2 2 2 2c 2c y2 x2 52. The hyperbolas 2  2  1 and a b

x2 y2   1 are conjugate to each other. a2 b2

y2 x2 (a) x 2  4y 2  16  0    1 and 4y 2  x 2  16  0  16 4 y2 x2   1. So the hyperbolas are conjugate to each other. 16 4 (b) They both have the same asymptotes, y   12 x.

y

1 1

x

(c) The two general conjugate hyperbolas both have asymptotes b y   x. a 53.

  x  c2  y 2  x  c2  y 2  2a. Let us consider the positive case only. Then   x  c2  y 2  2a  x  c2  y 2 , and squaring both sides gives x 2  2cx  c2  y 2  4a 2    4a x  c2  y 2  x 2  2cx  c2  y 2  4a x  c2  y 2  4cx  4a 2 . Dividing by 4 and squaring both sides   gives a 2 x 2  2cx  c2  y 2  c2 x 2  2a 2 cx  a 4  a 2 x 2  2a 2 cx  a 2 c2  a 2 y 2  c2 x 2  2a 2 cx  a 4

 a 2 x 2  a 2 c2  a 2 y 2  c2 x 2  a 4 . Rearranging the order, we have c2 x 2  a 2 x 2  a 2 y 2  a 2 c2  a 4      c2  a 2 x 2  a 2 y 2  a 2 c2  a 2 . The negative case gives the same result. 54. (a) The hyperbola

x2 y2   1 has a  3, b  4, so c2  9  16  25, and c  5. Therefore, the foci are F1 5 0 and 9 16

F2 5 0.

  25 2569 25 16 (b) Substituting P 5 16 3 into an equation of the hyperbola, we get 9  16  9  9  1, so P lies on the hyperbola.   1 34 2 2 (c) d P F1   5  52  163  02  16 3 , and d P F2   5  5  163  0  3 900  256  3 .

16 (d) d P F2   d P F1   34 3  3  6  2 3  2a.

55. (a) From the equation, we have a 2  k and b2  16  k. Thus, c2  a 2  b2  k  16  k  16  c  4. Thus the foci of the family of hyperbolas are 0 4.

810

CHAPTER 12 Conic Sections

   2 y2 x kx 2 x2 (b)   1  y2  k 1  . For y k k 16  k 16  k 16  k  kx 2 , k  1, 4, 8, 12. As k the top branch, we graph y  k  16  k

k=12 8

k=8

6

increases, the asymptotes get steeper and the vertices move further apart.

-10

4

k=4

2

k=1

0

10

56. d AB  500  2c  c  250. (a) Since t  2640 and   980, we have d  d P A  d P B  t  980 fts  2640 s  2,587,200 ft  490mi.

(b) c  250, 2a  490  a  245 and the foci are on the y-axis. Then b2  2502  2452  2475. Hence, an equation is x2 y2   1. 60,025 2475

2502 x2 (c) Since P is due east of A, c  250 is the y-coordinate of P. Therefore, P is at x 250, and so  1 2 2475 245   2502 x 2  2475  1  10205. Then x  101, and so P is approximately 101 miles from A. 2452 57. Since the asymptotes are perpendicular, a  b. Also, since the sun is a focus and the closest distance is 2  109 , it follows   2  109 and that c  a  2  109 . Now c2  a 2  b2  2a 2 , and so c  2a. Thus, 2a  a  2  109  a   21 4  1018 x2 y2 a 2  b2   1   23  1019 . Therefore, an equation of the hyperbola is 23  1019 23  1019 32 2 x 2  y 2  23  1019 .

58. (a) These equally spaced concentric circles can be used as a kind of measure where we count the number of rings. In the case of the red dots, the sum of the number of wave crests from each center is a constant, in this case 17. As you move out one wave crest from the left stone you move in one wave crest from the right stone. Therefore this satisfies the geometric definition of an ellipse. (b) Similarly, in the case of the blue dots, the difference of the number of wave crests from each center is a constant. As you move out one wave crest from the left center you also move out one wave crest from the right stone. Therefore this satisfies the geometric definition of a hyperbola. 59. Some possible answers are: as cross-sections of nuclear power plant cooling towers, or as reflectors for camouflaging the location of secret installations. 60. The wall is parallel to the axis of the cone of light coming from the top of the shade, so the intersection of the wall and the cone of light is a hyperbola. In the case of the flashlight, hold it parallel to the ground to form a hyperbola.

12.4 SHIFTED CONICS 1. (a) If we replace x by x  3 the graph of the equation is shifted to the right by 3 units. If we replace x by x  3 the graph is shifted to the left by 3 units. (b) If we replace y by y  1 the graph of the equation is shifted upward by 1 unit. If we replace y by y  1 the graph is shifted downward by 1 unit. 2. x 2  12y, from top to bottom: focus 0 3, vertex 0 0, directrix y  3. x  32  12 y  1, from top to bottom: focus 3 4, vertex 3 1, directrix y  2.

SECTION 12.4 Shifted Conics

3.

4.

5.

811

x2 y2 x  32 y  12   1, from left to right: vertex 0, focus 0, focus 0, vertex 0.   1, from 5 3 3 5 52 42 52 42 left to right: vertex 2 1, focus 0 1, focus 6 1, vertex 8 1.

x2 y2  2  1, from left to right: focus 5 0, vertex 4 0, asymptote y   34 x, asymptote y  34 x, vertex 4 0, 2 4 3 x  32 y  12 focus 5 0.   1, from left to right: focus 2 1, vertex 1 1, asymptote y   34 x  13 4 , 42 32 asymptote y  34 x  54 , vertex 7 1, focus 8 1.

y  12 x  22  1 9 4

y

(c)

x2 y2   1 by shifting it 2 units to 9 4   the right and 1 unit upward. So a  3, b  2, and c  9  4  5. The

(a) This ellipse is obtained from the ellipse

1 x

1

center is 2 1, the vertices are 2  3 1  1 1 and 5 1, and the foci are    2  5 1 .

(b) The length of the major axis is 2a  6 and the length of the minor axis is 2b  4.

6.

x  32  y  32  1 16

(c)

x2  y 2  1 by shifting to the right (a) This ellipse is obtained from the ellipse 16   3 units and downward 3 units. So a  4, b  1, and c  16  1  15. The center is 3 3, the vertices are 3  4 3  1 3 and 7 3, and the    foci are 3  15 3 .

y

1 x

1

(b) The length of the major axis is 2a  8 and the length of the minor axis is 2b  2.

7.

x2 y  52  1 9 25 x2 y2 (a) This ellipse is obtained from the ellipse   1 by shifting it 5 units 9 25  downward. So a  5, b  3, and c  25  9  4. The center is 0 5, the vertices are 0 5  5  0 10 and 0 0, and the foci are 0 5  4  0 9 and 0 1.

(b) The length of the major axis is 2a  10 and the length of the minor axis is 2b  6.

(c)

y

1 1

x

812

CHAPTER 12 Conic Sections

y  22 1 8. x 2  4

(c)

y

1

y2  1 by shifting it 2 units (a) This ellipse is obtained from the ellipse x 2  4   downward. So a  2, b  1, and c  4  1  3. The center is 0 2, the

x

1

vertices are 0 2  2  0 4 and 0 0, and the foci are          0 2  3  0 2  3 and 0 2  3 .

(b) The length of the major axis is 2a  4 and the length of the minor axis is 2b  2.

9.

y  12 x  52  1 16 4

y

(c)

y2 x2   1 by shifting it 5 units to 16 4   the left and 1 units upward. So a  4, b  2, and c  16  4  2 3. The

(a) This ellipse is obtained from the ellipse

1 1

x

center is 5 1, the vertices are 5  4 1  9 1 and 1 1, and the          foci are 5  2 3 1  5  2 3 1 and 5  2 3 1 .

(b) The length of the major axis is 2a  8 and the length of the minor axis is 2b  4.

10.

y  1 x  12  1 36 64

(c)

y2 x2   1 by shifting it 1 unit to the 36 64   left and 1 unit downward. So a  8, b  6, and c  64  36  2 7. The center is 1 1, the vertices are 1 1  8  1 9 and 1 7, and          the foci are 1 1  2 7  1 1  2 7 and 1 2 7  1 .

(a) This ellipse is obtained from the ellipse

y

2 2

x

(b) The length of the major axis is 2a  16 and the length of the minor axis is 2b  12.

11. 4x 2  25y 2  50y  75  4x 2  25 y  12  25  75 

x2 y  12  1 25 4

y2 x2   1 by shifting it 1 unit 25 4   upward. So a  5, b  2, and c  25  4  21. The center is 0 1, the

(c)

y

(a) This ellipse is obtained from the ellipse

vertices are 5 1  5 1 and 5 1, and the foci are         21 1 .  21 1   21 1 and

(b) The length of the major axis is 2a  10 and the length of the minor axis is 2b  4.

1 1

x

SECTION 12.4 Shifted Conics

12. 9x 2  54x  y 2  2y  46  0  9 x  32  81  y  12  1  46  0 

(c)

x  32 y  12  1 4 36

813

y

1

x2

x

2

y2

 1 by shifting it 3 units to 4 36   the right and 1 unit downward. So a  6, b  2, and c  36  4  4 2. The

(a) This ellipse is obtained from the ellipse



center is 3 1, the vertices are 3 1  6  3 7 and 3 5, and the foci          are 3 1  4 2  3 1  4 2 and 3 4 2  1 .

(b) The length of the major axis is 2a  12 and the length of the minor axis is 2b  4.

13. x  32  8 y  1

(b)

(a) This parabola is obtained from the parabola x 2  8y by shifting it 3 units to the

y

1

right and 1 unit down. So 4 p  8  p  2. The vertex is 3 1, the focus is 3 1  2  3 1, and the directrix is y  1  2  3.

14. y  12  16 x  3

(b)

(a) This parabola is obtained from the parabola y 2  16x by shifting it 3 units to

x

1

x

y

2

the right and 1 unit down. So 4 p  16  p  4. The vertex is 3 1, the focus is 3  4 1  7 1, and the directrix is x  3  4  1.

15. y  52  6x  12  6 x  2

1

y

(b)

1 1

(a) This parabola is obtained from the parabola y 2  6x by shifting to the right

x

2 units and down 5 units. So 4 p  6  p   32 . The vertex is 2 5, the     focus is 2  32  5  12  5 , and the directrix is x  2  32  72 .

  16. y 2  16x  8  16 x  12

(a) This parabola is obtained from the parabola y 2  16x by shifting to the right 12   unit. So 4 p  16  p  4. The vertex is 12  0 , the focus is     1  4 0  9  0 , and the directrix is x  1  4   7 . 2 2 2 2

(b)

y

2 1

x

814

CHAPTER 12 Conic Sections

17. 2 x  12  y  x  12  12 y

y

(b)

(a) This parabola is obtained from the parabola x 2  12 y by shifting it 1 unit to the   right. So 4 p  12  p  18 . The vertex is 1 0, the focus is 1 18 , and the directrix is y   18 .

1 x

1

2 2   18. 4 x  12  y  x  12   14 y

y

(b)

1

(a) This parabola is obtained from the parabola x 2   14 y by shifting it 12 unit to   1 . The vertex is  1  0 , the focus is the left, so 4 p   14  p   16 2     1 1 1 1 1  1 .  2  0  16   2   16 , and the directrix is y  0  16 16

19. y 2  6y  12x  33  0  y  32  9  12x  33  0  y  32  12 x  2

1

(b)

x

y

(a) This parabola is obtained from the parabola y 2  12x by shifting it 2 units to the right and 3 units upward. So 4 p  12  p  3. The vertex is 2 3, the

2

focus is 2  3 3  5 3, and the directrix is x  2  3  1.

20. x 2  2x  20y  41  0  x  12  1  20y  41  0  x  12  20 y  2

x

1

(b)

y

(a) This parabola is obtained from the parabola x 2  20y by shifting it 1 unit to the left and 2 units upward, so 4 p  20  p  5. The vertex is 1 2, the focus

1

is 1 2  5  1 7, and the directrix is y  2  5  3.

21.

y  32 x  12  1 9 16 x2 y2   1 by shifting it 9 16  1 unit to the left and 3 units up. So a  3, b  4, and c  9  16  5. The

(b)

1

x

1

x

y

(a) This hyperbola is obtained from the hyperbola

center is 1 3, the vertices are 1  3 3  4 3 and 2 3, the foci are 1  5 3  6 3 and 4 3, and the asymptotes are

y  3   43 x  1  y   43 x  1  3  y  43 x  13 3 and y   43 x  53 .

1

SECTION 12.4 Shifted Conics

22. x  82  y  62  1

(b)

y

(a) This hyperbola is obtained from the hyperbola x 2  y 2  1 by shifting to the   right 8 units and downward 6 units. So a  1, b  1, and c  1  1  2.

10

The center is 8 6, the vertices are 8  1 6  7 6 and 9 6, the    foci are 8  2 6 , and the asymptotes are y  6   x  8 

20

x

_10

_20

y  x  14 and y  x  2.

x  12 1 23. y 2  4

815

y

(b)

x2 (a) This hyperbola is obtained from the hyperbola y 2   1 by shifting it 1 unit 4   to the left. So a  1, b  2, and c  1  4  5. The center is 1 0, the

1 x

1

vertices are 1 1  1 1 and 1 1, the foci are          1  5  1  5 and 1 5 , and the asymptotes are y   12 x  1  y  12 x  12 and y   12 x  12 .

24.

y  12  x  32  1 25

y

(b)

y2  x 2  1 by shifting to the 25   left 3 units and upward 1 unit. So a  5, b  1, and c  25  1  26. The

(a) This hyperbola is obtained from the hyperbola

5 1

x

center is 3 1, the vertices are 3 1  5  3 4 and 3 6, the foci    are 3 1  26 , and the asymptotes are y  1  5 x  3  y  5x  16 and y  5x  14.

25.

y  12 x  12  1 9 4

y

(b)

x 2 y2   1 by shifting it 1 unit 9 4   to the left and 1 unit downward, so a  3, b  2, and c  9  4  13. The center is 1 1, the vertices are 1  3 1  4 1 and 2 1, the         foci are 1  13 1  1  13 1 and 13  1 1 , and the

1

(a) This hyperbola is obtained from the hyperbola

1

x

5

x

asymptotes are y  1   23 x  1  y  23 x  13 and y   23 x  53 .

26.

x2 y  22  1 36 64 x2 y2   1 by shifting it (a) This hyperbola is obtained from the hyperbola 36 64  2 unit downward. So a  6, b  8, and c  36  64  10. The center is 0 2, the vertices are 0 2  6  0 8 and 0 4, the foci are

0 2  10  0 12 and 0 8, and the asymptotes are y  2   34 x  y  34 x  2 and y   34 x  2.

(b)

y 5

816

CHAPTER 12 Conic Sections

27. 36x 2  72x  4y 2  32y  116  0  36 x  12  36  4 y  42  64  116  0 

y

(b) x  12 y  42  1 36 4

y2 x 2   1 by shifting it 1 unit 36 4   to the left and 4 units upward. So a  6, b  2, and c  36  4  2 10.

2

(a) This hyperbola is obtained from the hyperbola

1

x

The center is 1 4, the vertices are 1 4  6  1 2 and 1 10,          the foci are 1 4  2 10  1 2 10  4 and 1 4  2 10 , and

the asymptotes are y  4  3 x  1  y  3x  7 and y  3x  1. 28. 25x 2  9y 2  54y  306  25x 2  9 y  32  81  306 

x 2 y  32  1 9 25

(b)

x2 y2   1 by shifting it 9 25   3 units downward. So a  3, b  5, and c  9  25  34. The center is 0 3, the vertices are 3 3  3 3 and 3 3, the foci are         34 3 , and the asymptotes are  34 3   34 3 and

y

2

(a) This hyperbola is obtained from the hyperbola

1

x

y  3   53 x  y  53 x  3 and y   53 x  3.

29. This is a parabola that opens down with its vertex at 0 4, so its equation is of the form x 2  a y  4. Since 1 0 is a point on this parabola, we have 12  a 0  4  1  4a  a   14 . Thus, an equation is x 2   14 y  4.

30. This is a parabola that opens to the right with its vertex at 6 0, so its equation is of the form y 2  4 p x  6, with

p  0. Since the distance from the vertex to the directrix is p  6  12  6, an equation is y 2  4 6 x  6 

y 2  24 x  6

31. This is an ellipse with the major axis parallel to the x-axis, with one vertex at 0 0, the other vertex at 10 0, and one   focus at 8 0. The center is at 010 2  0  5 0, a  5, and c  3 (the distance from one focus to the center). So b2  a 2  c2  25  9  16. Thus, an equation is

y2 x  52   1. 25 16

32. This is an ellipse with the major axis parallel to the y-axis. From the graph the center is 2 3, with a  3 and b  2. Thus, an equation is

x  22 y  32   1. 4 9

33. This is a hyperbola with center 0 1 and vertices 0 0 and 0 2. Since a is the distance form the center to a vertex, a we have a  1. The slope of the given asymptote is 1, so  1  b  1. Thus, an equation of the hyperbola is b 2 2  1  x  1. y 34. From the graph, the vertices are 2 0 and 6 0. The center is the midpoint between the vertices, so the center is   26 00   4 0. Since a is the distance from the center to a vertex, a  2. Since the vertices are on the x-axis, the 2 2 y2 x  42 42 0  42  2  1. Since the point 0 4 lies on the hyperbola, we have  2 1 4 4 b b 2 2 2 y 3y 2 16 16 16 16  4  4 x x  2  1  3  2  b2  . Thus, an equation of the hyperbola is  16  1    1. 4 3 4 4 16 b b

equation is of the form

3

SECTION 12.4 Shifted Conics

817

35. The ellipse with center C 2 3, vertices V1 8 3 and V2 12 3, and foci F1 4 3 and F2 8 3 has

x  22 y  32   1. The distance between the vertices 2 a b2 is 2a  12  8  20, so a  10. Also, the distance from the center to each focus is c  2  4  6, so a horizontal major axis, so its equation has the form

b2  a 2  c2  100  36  64. Thus, an equation is

y  32 x  22   1. 100 64

36. The ellipse with vertices V1 1 4 and V2 1 6 and foci F1 1 3 and F2 1 5 is centered midway between the   vertices; that is, it has C 1 12 4  6  1 1. The distance between the vertices is 2a  6  4  10, so a  5.

Also, the distance from the center to each focus is c  1  3  4, so b2  a 2  c2  25  16  9. Thus, an equation is x  12 y  12   1. 9 25

37. The hyperbola with center C 1 4, vertices V1 1 3 and V2 1 11, and foci F1 1 5 and F2 1 13 has y  42 x  12   1. The distance between the vertices a2 b2 is 2a  11  3  14, so a  7. Also, the distance from the center to each focus is c  4  5  9, so a vertical transverse axis, so its equation has the form

b2  c2  a 2  81  49  32. Thus, an equation is

x  12 y  42   1. 49 32

38. The hyperbola with vertices V1 1 1 and V2 5 1, and foci F1 4 1 and F2 8 1 is centered midway between   the vertices; that is, it has C 12 1  5  1  2 1. It has a horizontal transverse axis, so its equation has the form

y  12 x  22   1. The distance between the vertices is 2a  5  1  6, so a  3. Also, the distance from the 2 a b2 x  22 y  12   1. center to each focus is c  2  4  6, so b2  c2  a 2  36  9  27. Thus, an equation is 9 27

39. The parabola with vertex V 3 5 and directrix y  2 has an equation of the form x  32  4 p y  5. The distance from the vertex to the directrix is p  5  2  3, so an equation is x  32  12 y  5.

40. The parabola with focus F 1 3 and directrix x  3 has vertex midway between the focus and the directrix; that is, it has   V 12 3  1  3  2 3. The distance from the vertex to the directrix is p  2  3  1. (Since the directrix is to the right of the focus, p is negative.) Thus, an equation is y  32  4 1 x  2  4 x  2.

41. The hyperbola with foci F1 1 5 and F2 1 5 that passes through the point 1 4 is centered midway between the foci;   y2 x  12 that is, it has C 1 12 5  5  1 0. It has a vertical transverse axis, so its equation has the form 2   1. a b2 The point 1 4 lies on the transverse axis, so it is a vertex and we have a  4  0  4. Also, the distance from the center to each focus is c  5  0  5, so b2  c2  a 2  25  16  9. Thus, an equation is

y2 x  12   1. 16 9

42. The hyperbola with foci F1 2 2 and F2 4 2 that passes through the point 3 2 is centered midway between   the foci; that is, it has C 12 2  4  2  1 2. It has a horizontal transverse axis, so its equation has the form

y  22 x  12   1. The point 3 2 lies on the transverse axis, so it is a vertex and we have a  3  1  2. 2 a b2 Also, the distance from the center to each focus is c  4  1  3, so b2  c2  a 2  9  4  5. Thus, an equation is x  12 y  22   1. 4 5

818

CHAPTER 12 Conic Sections

43. The ellipse with foci F1 1 4 and F2 5 4 that passes through the point 3 1 is centered midway between the foci;   y  42 x  32 that is, it has C 12 1  5  4  3 4, and so its equation has the form   1. The distance from 2 a b2 the center to each focus is c  3  1  2, so c2  a 2  b2  a 2  b2  4. Substituting the point x y  3 1 into the

3  32 1  42   1  b2  25. From above, we have a 2  b2  4  29. Thus, an 2 a b2 x  32 y  42 equation of the ellipse is   1. 29 25 44. The ellipse with foci F1 3 4 and F2 3 4 and x-intercepts 0 and 6 is centered midway between the foci; that is, it has   x  32 y 2 C 3 12 4  4  3 0, and so its equation has the form  2  1. Because 0 is an x-intercept, we substitute b2 a 2 2 0 3  2  1  b  3. We also have c  4  0  4, so a 2  b2  c2  25 and an 0 0 into this equation, obtaining b2 a y2 x  32 equation is   1. 9 25 45. The parabola that passes through the point 6 1, with vertex V 1 2 and horizontal axis of symmetry has an equation of equation of the ellipse, we have

1 . the form y  22  4 p x  1. Substituting the point 6 1 into this equation, we have 1  22  4 p 6  1  p  28 Thus, an equation is y  22  17 x  1.

46. Because 6 2 is lower than V 4 1, the parabola opens downward, so an equation is x  42  4 p y  1.

Substituting the point 6 2, we have 6  42  4 p 2  1  p  1, so an equation is x  42  4 y  1. y

47. y 2  4 x  2y  y 2  8y  4x  y 2  8y  16  4x  16 

y  42  4 x  4. This is a parabola with 4 p  4  p  1. The vertex is

4 4, the focus is 4  1 4  3 4, and the directrix is x  4  1  5. 1 x

1

  48. 9x 2  36x  4y 2  0  9 x 2  4x  4  36  4y 2  0 

y

y2 x  22   1. This is an ellipse with a  3, 4 9      b  2, and c  9  4  5. The center is 2 0, the foci are 2  5 , the

9 x  22  4y 2  36 

1 x

1

vertices are 2 3, the length of the major axis is 2a  6, and the length of the minor axis is 2b  4.

    49. x 2  5y 2  2x  20y  44  x 2  2x  1  5 y 2  4y  4  44  1  20 y  22 x  12  x  12  5 y  22  25    1. This is a hyperbola 25 5    with a  5, b  5, and c  25  5  30. The center is 1 2, the foci are    1  30 2 , the vertices are 1  5 2  4 2 and 6 2, and the asymptotes 







are y  2   55 x  1  y   55 x  1  2  y   55 x  2  55 and 



y  55 x  2  55 .

y

1 2

x

SECTION 12.4 Shifted Conics y

50. x 2  6x  12y  9  0  x 2  6x  9  12y  x  32  12y. This is a

parabola with 4 p  12  p  3. The vertex is 3 0, the focus is 3 3,

_1

1

x

and the directrix is y  3.

y

51. 4x 2  25y 2  24x  250y  561  0      4 x 2  6x  9  25 y 2  10y  25  561  36  625  4 x  32  25 y  52  100 

x  32

1 x

1

y  52

  1. This is an ellipse 25 4   with a  5, b  2, and c  25  4  21. The center is 3 5, the foci are    3  21 5 , the vertices are 3  5 5  2 5 and 8 5, the length

of the major axis is 2a  10, and the length of the minor axis is 2b  4.   52. 2x 2  y 2  2y  1  2x 2  y 2  2y  1  1  1  2x 2  y  12  2 

y

  y  12  1. This is an ellipse with a  2, b  1, and c  2  1  1. 2 The center is 0 1, the foci are 0 1  1  0 0 and 0 2, the vertices are     0 1  2 , the length of the major axis is 2a  2 2, and the length of the minor

1

x2 

1

x

axis is 2b  2.

  53. 16x 2  9y 2  96x  288  0  16 x 2  6x  9y 2  288  0    16 x 2  6x  9  9y 2  144  288  16 x  32  9y 2  144 

y

2 x

1

y2 x  32   1. This is a hyperbola with a  4, b  3, and 16 9  c  16  9  5. The center is 3 0, the foci are 3 5, the vertices are

3 4, and the asymptotes are y   43 x  3  y  43 x  4 and y  4  43 x.   54. 4x 2  4x  8y  9  0  4 x 2  x  8y  9  0 

y

   2 4 x 2  x  14  8y  9  1  4 x  12  8 y  1  

2 x  12  2 y  1. This is a parabola with 4 p  2  p  12 . The vertex is     1  1 , the focus is 1  3 , and the directrix is y   1  1  1 . 2 2 2 2 2   55. x 2  16  4 y 2  2x  x 2  8x  4y 2  16  0    x 2  8x  16  4y 2  16  16  4y 2  x  42  y   12 x  4.

Thus, the conic is degenerate, and its graph is the pair of lines y  12 x  4 and y   12 x  4.

1 1

x

y

1 1

x

819

820

CHAPTER 12 Conic Sections y

56. x 2  y 2  10 x  y  1  x 2  10x  y 2  10y  1      x 2  10x  25  y 2  10y  25  1  25  25  x  52  y  52  1.   This is a hyperbola with a  1, b  1, and c  1  1  2. The center is 5 5,    the foci are 5  2 5 , the vertices are 5  1 5  4 5 and 6 5, and the

1 x

1

asymptotes are y  5   x  5  y  x and y  x  10.     57. 3x 2  4y 2  6x  24y  39  0  3 x 2  2x  4 y 2  6y  39      3 x 2  2x  1  4 y 2  6y  9  39  3  36 

y

(1, 3)

1 x

1

3 x  12  4 y  32  0  x  1 and y  3. This is a degenerate conic whose

graph is the point 1 3.

      58. x 2  4y 2  20x  40y  300  0  x 2  20x  4 y 2  10y  300  x 2  20x  100    4 y 2  10y  25  300  100  100  x  102  4 y  52  100. Since u 2   2  0 for all u,   R, there is no x y such that x  102  4 y  52  100. So there is no solution, and the graph is empty.

59. 2x 2  4x  y  5  0  y  2x 2  4x  5.

-2

2

4

-5 -10







 60. 4x 2  9y 2  36y  0  4x 2  9 y 2  4y  0  4x 2  9 y 2  4y  4  36    9 y 2  4y  4  36  4x 2  y 2  4y  4  4  49 x 2    y  2   4  49 x 2  y  2  4  49 x 2 61. 9x 2  36  y 2  36x  6y  x 2  36x  36  y 2  6y    9x 2  36x  45  y 2  6y  9  9 x 2  4x  5  y  32      y  3   9 x 2  4x  5  y  3  3 x 2  4x  5

5

-4

-2

2

2

2 y  1   x  2  y  1   12 x  2  y  1  12 x  2.So

conic.

-5

4

4

-10

  62. x 2  4y 2  4x  8y  0  x 2  4x  4 y 2  2y  0      x 2  4x  4  4 y 2  2y  1  0  4 y  12  x  22 

y  1  12 x  2  12 x  2 and y  1  12 x  2   12 x. This is a degenerate

2

SECTION 12.4 Shifted Conics









63. 4x 2  y 2  4 x  2y  F  0  4 x 2  x  y 2  8y      4 x 2  x  14  y 2  8y  16  16  1  F  4 x  (a) For an ellipse, 17  F  0  F  17.

821

 F   1 2  y  12  17  F 2

(b) For a single point, 17  F  0  F  17. (c) For the empty set, 17  F  0  F  17.

64. The parabola x 2  y  100  x 2   y  100 has 4 p  1  p   14 . The vertex is 0 100 and so the focus is       399 0 100  14  0 399 4 . Thus, one vertex of the ellipse is 0 100, and one focus is 0 4 . Since the second focus of   401 the ellipse is 0 0, the second vertex is 0  14 . So 2a  100  14  401 4  a  8 . Since 2c is the distance between 399 4012 3992 2 2 2 the foci of the ellipse, 2c  399 4 , c  8 , and then b  a  c  64  64  25. The center of the ellipse is  2   2 y  399 x x2 8y  3992 8 and so its equation is 0 399     1. 2  1, which simplifies to 8 25 25 160,801 401 8

65. (a) x 2  4 p y  p, for

p  2,  32 , 1,  12 , 12 , 1, 32 , 2. 3

1 2

p= 2

p= 1

2

0

4

8

-4 3

1

_2

(c) The parabolas become narrower as the vertex moves toward the origin.

-8 p= _ 2

focus of x 2  4 py is at 0 p, so this point is also shifted  p units at the origin.

4 -4

x 2  4 py vertically  p units so that the vertex is at 0  p. The

vertically to the point 0 p  p  0 0. Thus, the focus is located

8

-8

(b) The graph of x 2  4 p y  p is obtained by shifting the graph of

p= _1 _2

66. Since 0 0 and 1600 0 are both points on the parabola, the x-coordinate of the vertex is 800. And since the highest point it reaches is 3200, the y-coordinate of the vertex is 3200. Thus the vertex is 800 3200, and the equation is of the form x  8002  4 p y  3200. Substituting the point 0 0, we get 0  8002  4 p 0  3200  640,000  12,800 p

 p  50. So an equation is x  8002  4 50 y  3200  x  8002  200 y  3200. y

67. Since the height of the satellite above the earth varies between 140 and 440, the length of the major axis is 2a  140  2 3960  440  8500  a  4250. Since the center of the earth is at one focus, we have a  c  earth radius  140  3960  140  4100 

c  a  4100  4250  4100  150. Thus, the center of the ellipse is 150 0. So b2  a 2  c2  42502  1502  18,062,500  22500  18,040,000. Hence,

an equation is

410

Path of satellite Earth

3960 Center of ellipse is (_150, 0)

140 x

y2 x  1502   1. 18,062,500 18,040,000

68. (a) We assume that 0 1 is the focus closer to the vertex 0 0, as shown in the figure in the text. Then the center of the ellipse is 0 a and 1  a  c. So c  a  1 and a  12  a 2  b2  a 2  2a  1  a 2  b2  b2  2a  1.

x2 x2 y  a2 y  22    1. If we choose  1. If we choose a  2, then we get 2a  1 3 4 a2 x2 y  52   1. (Answers will vary, depending on your choices of a  1.) a  5, then we get 9 25 Thus, the equation is

822

CHAPTER 12 Conic Sections

(b) Since a vertex is at 0 0 and a focus is at 0 1, we must have c  a  1 (a  0), and the center of the hyperbola is 0 a. So c  a  1 and a  12  a 2  b2  a 2  2a  1  a 2  b2  b2  2a  1. Thus the equation x2 x2 y  a2 y  22  1. If we let a  2, then we get   1. If we let a  5, then we get  2a  1 4 5 a2 x2 y  52   1. (Answers will vary, depending on your choices of a.) 25 11

is

(c) Since the vertex is at 0 0 and the focus is at 0 1, we must have

y

p  1. So x  02  4 1 y  0  x 2  4y, and there is no other

possible parabola.

(d) Graphs will vary, depending on the choices of a in parts (a) and (b). (e) The ellipses are inside the parabola and the hyperbolas are outside the

1 1

parabola. All touch at the origin.

x

12.5 ROTATION OF AXES 1. If the x- and y-axes are rotated through an acute angle  to produce the new X- and Y -axes, then the x y-coordinates x y and the XY -coordinates X Y  of a point P in the plane are related by the formulas x  X cos   Y sin , y  X sin   Y cos , X  x cos   y sin , and Y  x sin   y cos . 2. (a) In general, the graph of the equation Ax 2  Bx y  C y 2  Dx  E y  F  0 is a conic section.

AC . B (c) The discriminant of this equation is B 2  4AC. If the discriminant is 0 the graph is a parabola, if it is positive the graph is an ellipse, and if it is negative the graph is a hyperbola.  2 and 3. x y  1 1,   45 . Then X  x cos   y sin   1  1  1  1  2 2   Y  x sin   y cos   1  1  1  1  0. Therefore, the XY -coordinates of the given point are X Y   2 0 . (b) To eliminate the x y-term from this equation we rotate the axes through an angle  that satisfies cot 2 

2

2

    4. x y  2 1,   30 . Then X  x cos   y sin   2  23  1  12  12 1  2 3 and     Y  x sin   y cos   2  12  1  23  12 2  3 . Therefore, the XY -coordinates of the given point are        X Y   12 1  2 3  12 2  3 .      3  3 ,   60 . Then X  x cos   y sin   3  12  3  23  0 and    Y  x sin   y cos   3  23  3  12  2 3. Therefore, the XY -coordinates of the given point are    X Y   0 2 3 .

5. x y 

6. x y  2 0,   15 . Then X  x cos   y sin   2 cos 15  0 sin 15  19319 and Y  x sin   y cos   2 sin 15  0 cos 15  05176. Therefore, the XY -coordinates of the given point are approximately X Y   19319 05176. 7. x y  0 2,   55 . Then X  x cos   y sin   0 cos 55  2 sin 55  16383 and Y  x sin   y cos   0 sin 55  2 cos 55  11472. Therefore, the XY -coordinates of the given point are approximately X Y   16383 11472.

SECTION 12.5 Rotation of Axes

823

     8. x y  2 4 2 ,   45 . Then X  x cos   y sin   2  1  4 2  1  5 and 2 2   1 1 Y  x sin   y cos    2    4 2    3. Therefore, the XY -coordinates of the given point are 2

2

X Y   5 3. 







9. x 2  3y 2  4,   60 . Then x  X cos 60  Y sin 60  12 X  23 Y and y  X sin 60  Y cos 60  23 X  12 Y .  2  2   3 23 X  12 Y  4  Substituting these values into the equation, we get 12 X  23 Y       X2 3XY 3XY 3XY 3Y 2 3X 2 Y2 X2 9 2 3Y 2 3Y 2 3 3XY   3   4  X     4 4 2 4 4 2 4 4 4 4 4 2 2   2X 2  2 3XY  4  X 2  3XY  2. 



10. y  x  12 ,   45 . Then x  X cos 45  Y sin 45  22 X  22 Y and y  X sin 45  Y cos 45  22 X  22 Y . Substituting these values into the equation, we get       2    2 X2 2 2 2 2  2 22 X 22 Y  1  22 Y  1  2 X 2 Y  2 X  2 Y 1 2    Y2  X2  2 22 X 22 Y  1   2Y  1   2 2      2 2     Y X  22 X  2X 22 Y  1   2Y  1  22 Y  0  X 2  Y 2  2XY  3 2X  2Y  2  0. 2 2   11. x 2  y 2  2y,   cos1 35 . So cos   35 and sin   45 . Then

 2  2   X cos   Y sin 2  X sin   Y cos 2  2 X sin   Y cos   35 X  45 Y  45 X  35 Y  2 45 X  35 Y

24XY 16Y 2 16X 2 24XY 9Y 2 8X 6Y 7X 2 48XY 7Y 2 8X 6Y 9X 2             0 25 25 25 25 25 25 5 5 25 25 25 5 5 7Y 2  48XY  7X 2  40X  30Y  0. 

  12. x 2  2y 2  16,   sin1 35 . So sin   35 and cos   45 . Then x  45 X  35 Y and y  35 X  45 Y . Substituting these values into the equation, we get       4 X  3 Y 2  2 3 X  4 Y 2  16  16 X 2  24 XY  9 Y 2  2 9 X 2  24 XY  16 Y 2  16  5 5 5 5 25 25 25 25 25 25 16 X 2  18 X 2  9 Y 2  32 Y 2  24 XY  48 XY  16  34 X 2  41 Y 2  24 XY  16  34X 2  41Y 2  24XY  400. 25 25 25 25 25 25 25 25 25

    13. x 2  2 3x y  y 2  4,   30 . Then x  X cos 30  Y sin 30  23 X  12 Y  12 3X  Y     and y  X sin 30  Y cos 30  12 X  23 Y  12 X  3Y . Substituting these values into the 2            1   2 1 X  3Y 3X  Y  2 3 12 3X  Y  2 X  3Y  4 equation, we get 12 2 2         2 3X  Y  2 3 3X  Y X  3Y  X  3Y  16           3X 2  2 3XY  Y 2  6X 2  4 3XY  6Y 2  X 2  2 3XY  3Y 2  16  8X 2 8Y 2  16 

X2 Y2   1. This is a hyperbola. 2 2

824

CHAPTER 12 Conic Sections

  1 X  1 Y  1 X  Y  14. x y  x  y,    4 . Therefore, x  X cos 4  Y sin 4  2 2 2  1 X  1 Y  1 X  Y . Substituting into the equation gives and y  X sin  4  Y cos 4  2 2 2       1 1 1 1  X  Y   X  Y    X  Y    X  Y   1 X 2  Y 2  2 X  X 2  Y 2  2 2X  2 2 2 2 2 2   2 X 2 Y2   1. This is a hyperbola. 2 2

15. (a) x y  8  0x 2  x y  0y 2  8. So A  0, B  1, and C  0, and so the

(c)

discriminant is B 2  4AC  12  4 0 0  1. Since the discriminant is

1

positive, the equation represents a hyperbola. (b) cot 2  

y



x

1

AC  0  2  90    45 . Therefore, B   2 2 2 X  2 Y . After substitution, the original     2 2 2 2 Y 2 X  2 Y 8 X2 Y2

x  22 X  22 Y and y   equation becomes 22 X 

 X  Y  X  Y  8   1. This is a hyperbola with a  4, b  4, and c  4 2. Hence, the vertices 2 16 16    are V 4 0 and the foci are F 4 2 0 .

16. (a) x y  4  0  0x 2  x y  0y 2  4  0. So A  0, B  1, and C  0,

y

(c)

and so the discriminant is B 2  4AC  12  4 0 0  1. Since the

1

discriminant is positive, the equation represents a hyperbola.





x  22 X  22 Y and y   equation becomes 22 X 

x

1

AC  0  2  90    45 . Therefore, (b) cot 2  B   2 2 2 X  2 Y . After substitution, the original     2 2 2 2 Y 2 X  2 Y 40 Y2 X2

  X  Y  X  Y   4    1. This is a hyperbola with a  2 2, b  2 2, and c  4. Hence, the 2 8 8    vertices are V 0 2 2 and the foci are F 0 4.

  17. (a) x 2  2 3x y  y 2  2  0. So A  1, B  2 3, and C  1, and so the   2 discriminant is B 2  4AC  2 3  4 1 1  0. Since the discriminant is positive, the equation represents a hyperbola.

(c)

y

1 1

11 AC 1      2  60    30 . Therefore, (b) cot 2  B 2 3 3 



x  23 X  12 Y and y  12 X  23 Y . After substitution, the original equation becomes     2  2    3 1 1 X  3Y  1 X  3Y  2 3 23 X  12 Y 20 2 X  2Y 2 2 2 2       3 X 2  3 XY  1 Y 2  3 3X 2  2XY  3Y 2  14 X 2  23 XY  34 Y 2  2  0  4 2 4 2          X 2 34  32  14  XY  23  3  23  Y 2 14  32  34  2  2X 2  2Y 2  2  Y 2  X 2  1.

x

SECTION 12.5 Rotation of Axes

  18. (a) 13x 2  6 3x y  7y 2  16. Then A  13, B  6 3, and C  7, and so   2 the discriminant is B 2  4AC  6 3  4 13 7  256. Since the

(c)

825

y

1

discriminant is negative, the equation represents an ellipse.

1

x

13  7 AC 1      2  60    30 . Therefore, (b) cot 2  B 6 3 2 



x  23 X  12 Y and y  12 X  23 Y . After substitution, the original equation becomes    2  2     1 X  3Y  7 1 X  3Y 13 23 X  12 Y  6 3 23 X  12 Y  16  2 2 2 2           13 3X 2  2 3XY  Y 2  3 3 3X 2  2XY  3Y 2  74 X 2  2 3XY  3Y 2  16  4 2          9  7  XY  13 3  6 3  7 3  Y 2 13  9  21  16  16X 2  4Y 2  16  X 2 39  4 2 4 2 2 2 4 2 4

  Y2  1. This is an ellipse with a  2, b  1, and c  4  1  3. Thus, the vertices are V 0 2 and the 4    foci are F 0  3 `. X2 

19. (a) 11x 2  24x y  4y 2  20  0. So A  11, B  24, and C  4, and so the discriminant is B 2  4AC  242  4 11 4  0. Since the

discriminant is positive, the equation represents a hyperbola.

AC 11  4 7 7 . Therefore,     cos 2   25 B 24 24   cos   1725  35 and sin   1725  45 . Hence, 2 2

(c)

y

1 1

(b) cot 2 

3X  45 Y and y  45 X  5 equation becomes 2   11 35 X  45 Y  24 35 X    11 9X 2  24XY  16Y 2  25 x

3 Y . After substitution, the original 5



7 , we have Since cos 2   25

2  10626 , so   53 .

 2  4 45 X  35 Y  20  0      24 12X 2  7XY  12Y 2  4 16X 2  24XY  9Y 2  20  0  25 25

4Y 5

4 X  3Y 5 5

x



X 2 99  288  64  XY 264  168  96  Y 2 176  288  36  500  125X 2  500Y 2  500  1 X 2  Y 2  1. 4

826

CHAPTER 12 Conic Sections

  20. (a) 21x 2  10 3x y  31y 2  144. Then A  21, B  10 3, and C  31,   2 and so the discriminant is B 2  4AC  10 3  4 21 31  2304.

(c)

y

1

Since the discriminant is negative, the equation represents an ellipse.

1

x

21  31 AC 1  (b) cot 2       2  120    60 . B 10 3 3 



Therefore, x  12 X  23 Y and y  23 X  12 Y . After substitution, the original equation becomes  2       2   3 3 1 1 21 12 X  23 Y  10 3 12 X  23 Y  144  2 X  2 Y  31 2 X  2 Y           21 X 2  2 3XY  3Y 2  5 3 2 2  144  3X 2  2XY  3Y 2  31 4 2 4 3X  2 3XY  Y          15  93  XY  21 3  10 3  31 3  Y 2 63  15  31  144  36X 2  16Y 2  144  X 2 21  4 2 4 2 2 2 4 2 4  1 X 2  1 Y 2  1. This is an ellipse with a  3, b  2, and c  9  4  5. 4 9 21. (a)

  2 3x  3x y  3. So A  3, B  3, and C  0, and so the discriminant   3 0  9. Since the discriminant is positive, is B 2  4AC  32  4

(c)

y

1

the equation represents a hyperbola.

x

1

1 AC    2  60    30 . Therefore, (b) cot 2  B 3 



x  23 X  12 Y and y  12 X  23 Y . After substitution, the equation   2     1 X  3Y  3 becomes 3 23 X  12 Y  3 23 X  12 Y 2 2        3 3  4 3X 2  2 3XY  Y 2  4 3X 2  2XY  3Y 2  3              3  3 3  3  3 3 X2  3 Y 2  3  3 X2   6  Y2 1 Y 2  1. This  X 2 3 4 3  3 4 3  XY 6 4 4 4 4 2 2 2 2 3    is a hyperbola with a  2 and b  2 3. 3

22. (a) 153x 2  192x y  97y 2  225. Then A  153, B  192, and C  97, and so the discriminant is B 2  4AC  1922  4 153 97  22500.

Since the discriminant is negative, the equation represents an ellipse.

AC 153  97 56 56 (b) cot 2     cos 2  . Therefore, B 192 192 200   156200 4 3  16  12 cos   156200 2 20  5 and sin   2 20  5 

(c)

y

1

  cos1 54  369 . Substituting gives  2      3 X  4 Y  97 3 X  4 Y 2  225  153 45 X  35 Y  192 45 X  35 Y 5 5 5 5       153 16X 2  24XY  9Y 2  192 12X 2  7XY  12Y 2  97 9X 2  24XY  16Y 2  225  25 25 25

1

X 2 2448  2304  873  XY 3672  1344  2328  Y 2 1377  2304  1552  5625    5625X 2  625Y 2  5625  X 2  19 Y 2  1. This is an ellipse with a  3, b  1, and c  9  1  2 2.

x

SECTION 12.5 Rotation of Axes

23. (a) x 2  2x y  y 2  x  y  0. So A  1, B  2, and C  1, and so the

y

(c)

discriminant is B 2  4AC  22  4 1 1  0. Since the discriminant is

1

zero, the equation represents a parabola.

1

AC (b) cot 2   0  2  90    45 . Therefore, B 





827

x



x  22 X  22 Y and y  22 X  22 Y . After substitution, the original equation becomes  2       2 X  2Y 2 X  2Y 2 X  2Y  2 2 2 2 2 2 2

 2       2 2  22 X  22 Y  22 X  22 Y  0  2 X 2 Y    1 X 2  XY  1 Y 2  X 2  Y 2  1 X 2  XY  Y 2  2Y  0  2X 2  2Y  0  X 2  2 Y . This is a 2 2 2 2





  1 . parabola with 4 p  1 and hence the focus is F 0  2

4 2

24. (a) 25x 2  120x y  144y 2  156x  65y  0. Then A  25, B  120, and

(c)

y

C  144, and so the discriminant is

B 2  4AC  1202  4 25 144  0. Since the discriminant is zero,

the equation represents a parabola.

AC 25  144 119    cos 2  119 169 . Therefore, B 120 120   1119169 5 . Hence, cos   1119169  12  13 2 13 and sin   2

(b) cot 2 

2 2

x

Since cos 2  119 169 , we have

12X 5Y 5X 12Y 2  452 , so   23 . x  and y   . Substituting gives 13 13 13 13          5Y 2 5Y 12Y 12Y 2 5Y 5X 5X 12X 12X 12X     144     120  156 25 13 13 13 13 13 13 13 13 13 13     Y2  12Y X2  5X 65  0 25  122  120  12  5  144  52  25  52  120 5 12  144  122  13 13 169 169 Y X 156  12  65  5  156  5  65  12  0  169Y 2  169X  0  X  Y 2 . This is a parabola with 13 13 4 p  1.

828

CHAPTER 12 Conic Sections

   25. (a) 2 3x 2  6x y  3x  3y  0. So A  2 3, B  6, and C  0, and    so the discriminant is B 2  4AC  62  4 2 3 0  36. Since the

y

(c)

2

discriminant is positive, the equation represents a hyperbola.  AC 2 3 1 (b) cot 2       2  120    60 . Therefore, B 6 3 

2

x



x  12 X  23 Y and y  23 X  12 Y , and substituting gives  2             3 3 3 1 1 1 2 3 12 X  23 Y  6 12 X  23 Y 2 X  2Y  3 2 X  2 Y  3 2 X  2Y  0            3 X 2  23XY  3Y 2  3 3X 2  2XY  3Y 2  3 X  3Y  3 3X  Y  0  2 2 2 2             X 2 23  3 2 3  X 23  3 2 3  XY 3  3  Y 2 3 2 3  3 2 3  Y  32  32  0        3X 2  2 3X  3 3Y 2  0  X 2  2X  3Y 2  0  3Y 2  X 2  2X  1  1    X  12  3Y 2  1. This is a hyperbola with a  1, b  33 , c  1  13  2 , and C 1 0. 3

26. (a) 9x 2  24x y  16y 2  100 x  y  1. Then A  9, B  24, and

C  16, and so the discriminant is B 2  4AC  242  4 9 16  0. Since the discriminant is zero, the equation represents a parabola.

(c)

y 1 1

x

AC 9  16 7 7    369 . Now    cos 2  25 B 24 24   4 and sin   1725  3 , and so cos   1725  2 5 2 5

(b) cot 2 

x  45 X  35 Y , y  35 X  45 Y . By substitution,  2        3 X  4 Y  16 3 X  4 Y 2  100 4 X  3 Y  3 X  4 Y  1  9 45 X  35 Y  24 45 X  35 Y 5 5 5 5 5 5 5 5         9 2 2  24 12X 2  7XY  12Y 2  16 9X 2  24XY  16Y 2  100 1 X  7 Y  1 25 16X  24XY  9Y 25 25 5 5  625Y 2  500X  3500Y  2500  5Y 2  28Y  4X  20          196 196 96 24 14 2  4 X  24 . This is a 5 Y 2  28 5 Y  25  4X  20  5  4X  5  4 X  5  Y  5 5 5   14 . parabola with 4 p  45 and V  24   5 5

27. (a) 52x 2  72x y  73y 2  40x  30y  75. So A  52, B  72, and C  73, and so the discriminant is

B 2  4AC  722  4 52 73  10,000. Since the discriminant is decidedly negative, the equation represents an ellipse.

SECTION 12.5 Rotation of Axes

829

52  73 7 AC    . Therefore, as in Exercise 19(b), we get cos   35 , sin   45 , and B 72 24 x  35 X  45 Y , y  45 X  35 Y . By substitution,  2      4 X  3 Y  73 4 X  3 Y 2 52 35 X  45 Y  72 35 X  45 Y 5 5 5 5      40 35 X  45 Y  30 45 X  35 Y  75        52 9X 2  24XY  16Y 2  72 12X 2  7XY  12Y 2  73 16X 2  24XY  9Y 2 25 25 25

(b) cot 2 

 24X  32Y  24X  18Y  75  468X 2  832Y 2  864X 2  864Y 2  1168X 2  657Y 2  1250Y  1875  2500X 2  625Y 2  1250Y  1875    100X 2  25Y 2  50Y  75  X 2  14 Y  12  1. This is an ellipse with a  2, b  1, c  4  1  3, and center C 0 1.

(c)

y

1 x

1

  7 , we have 2  cos1  7  10626 and so   53 . Since cos 2   25 25

28. (a) 7x  24y2  49x 2  336x y  576y 2  600x  175y  25. So A  49, B  336, and C  576, and so the

discriminant is B 2  4AC  3362  4 49 576  0. Since the discriminant is zero, the equation represents a parabola.  AC 49  576 527 1527625 7 and (b) cot 2  . Therefore, cos    25    cos 2   527 625 2 B 336 336  7 24 24 7  24 sin   1527625 2 25 . Substituting x  25 X  25 Y and y  25 X  25 Y gives        7 X  24 Y 2  336 7 X  24 Y 24 X  7 Y  576 24 X  7 Y 2 49 25 25 25 25 25 25 25 25     7 X  24 Y  175 24 X  7 Y  25   600 25 25 25 25       49 2 2  336 168X 2  527XY  168Y 2  576 576X 2  336XY  49Y 2 625 49X  336XY  576Y 625 625

 168X  576Y  168X  49Y  25  X 2 2401  56,448  331,776  XY 16,464  177,072  193,536  Y 2 28,224  56,448  28,224

 6252 Y  15,625    1   Y  1 . This is a parabola with 390,625X 2  390,625Y  15,625  25X 2  25Y  1  X 2  Y  25 25   1 4 p  1 and vertex 0 25 . (c)

y

1 1

x

  1  527  14748 and so   74 . Since cos 2   527 625 , we have 2  cos 625

830

CHAPTER 12 Conic Sections

29. (a) The discriminant is B 2  4AC  42  4 2 2  0. Since the discriminant is 0, the equation represents a parabola. (b) 2x 2  4x y  2y 2  5x  5  0  2y 2  4x y  2x 2  5x  5    2 y 2  2x y  2x 2  5x  5    2 y 2  2x y  x 2  2x 2  5x  5  2x 2  2 y  x2  5x  5    y  x2  52 x  52  y  x   52 x  52  y  x  52 x  52

5

5

30. (a) The discriminant is B 2  4AC  22  4 1 3  8  0. Since the discriminant is negative, the equation represents an ellipse.   (b) x 2  2x y  3y 2  8  3y 2  2x y  8  x 2  3 y 2  23 x y  8  x 2 





 3 y 2  23 x y  19 x 2  8  x 2  13 x 2  3 y  13 x  2 y  13 x  83  29 x 2  y  13 x   83  29 x 2   y  13 x  83  29 x 2 

2

5

 8  23 x 2 

-5

5 -5

31. (a) The discriminant is B 2  4AC  102  4 6 3  28  0. Since the discriminant is positive, the equation represents a hyperbola. (b) 6x 2  10x y  3y 2  6y  36  3y 2  10x y  6y  36  6x 2    3y 2  2 5x  3 y  36  6x 2  y 2  2 53 x  1 y  12  2x 2     2  2 y 2  2 53 x  1 y  53 x  1  53 x  1  12  2x 2    2 10 2 2 y  53 x  1  25 9 x  3 x  1  12  2x 

10

-10

10 -10



 2 y  53 x  1  79 x 2  10 3 x  13     y  53 x  1   79 x 2  10 3 x  13   y   53 x  1  79 x 2  10 3 x  13

32. (a) The discriminant is B 2  4AC  62  4 9 1  0. Since the discriminant is 0, the equation represents a parabola. (b) 9x 2  6x y  y 2  6x  2y  0  y 2  6x y  2y  9x 2  6x 

5

y 2  2 3x  1 y  9x 2  6x 

y 2  2 3x  1 y  3x  12  3x  12  9x 2  6x   2 y  3x  1  1  y  3x  1  1  y  3x  1  1. So y  3x or y  3x  2. This is a degenerate conic.

-2

2 -5

SECTION 12.5 Rotation of Axes

831

33. (a) 7x 2  48x y  7y 2  200x  150y  600  0. Then A  7, B  48, and C  7, and so the discriminant is

B 2  4AC  482  4 7 7  0. Since the discriminant is positive, the equation represents a hyperbola. We 7 AC   cos   45 and sin   35 . now find the equation in terms of XY -coordinates. We have cot 2  B 24 Therefore, x  45 X  35 Y and y  35 X  45 Y , and substitution gives

2           3 X  4 Y  7 3 X  4 Y 2  200 4 X  3 Y  150 3 X  4 Y  600  0 7 45 X  35 Y  48 45 X  35 Y 5 5 5 5 5 5 5 5       7 16X 2  24XY  9Y 2  48 12X 2  7XY  12Y 2  7 9X 2  24XY  16Y 2  25 25 25

 160X  120Y  90X  120Y  600  0  112X 2  168XY  63Y 2  576X 2  336XY  576Y 2  63X 2  168XY  112Y 2  6250X  15,000  0    25X 2  25Y 2  250X  600  0  25 X 2  10X  25  25Y 2  600  625  X  52  Y 2  1. This is a   hyperbola with a  1, b  1, c  1  1  2, and center C 5 0.

(b) In the XY -plane, the center is C 5 0, the vertices are V 5  1 0  V1 4 0 and V2 6 0, and the foci      are F 5  2 0 . In the x y-plane, the center is C 45  5  35  0 35  5  45  0  C 4 3, the vertices are         12 4 3 3 4 24 18 V1 45  4  35  0 35  4  45  0  V1 16 5  5 and V2 5  6  5  0 5  6  5  0  V2 5  5 , and the foci are         F1 4  45 2 3  35 2 and F2 4  45 2 3  35 2 . (c) In the XY -plane, the equations of the asymptotes are Y  X  5 and Y  X  5. In the x y-plane, these equations

become x  35  y  45  x  45  y  35  5  7x  y  25  0. Similarly, x  35  y  45  x  45  y  35  5  x  7y  25  0.

       34. (a) 2 2 x  y2  7x  9y  2 2x 2  4 2x y  2 2y 2  7x  9y. Therefore, A  2 2, B  4 2, and C  2 2,  X Y X Y AC  x  y  2X. Thus the  0  2  90    45 . Thus x   , y   and so cot 2  B 2 2     2 16X  2Y  8X 2  16X  2Y  4X 2  8X  Y  4 X 2  2X  1  Y  4 equation becomes 2 2 2X   2  X  12  14 Y  4. This is a parabola.

  1 . Thus, in XY -coordinates the vertex is V 1 4 and the focus is F 1  63 . In x y-coordinates, (b) 4 p  14  p  16 16         5 2 3 2 79 2 47 2 V and F 32   32 . 2  2 (c) The directrix is Y   65 16 . Thus

 65 x  y   or y  x  6516 2 .  16 2

35. We use the hint and eliminate Y by adding: x  X cos   Y sin   x cos   X cos2   Y sin  cos  and

y  X sin   Y cos   y sin   X sin2   Y sin  cos , and adding these two equations gives x cos   y   sin   X cos2   sin2   x cos   y sin   X. In a similar manner, we eliminate X by subtracting:

x  X cos   Y sin   x sin   X cos  sin   Y sin2  and y  X sin   Y cos     y cos   X sin  cos   Y cos2 , so x sin   y cos   Y cos2   sin2   x sin   y cos   Y . Thus, X  x cos   y sin  and Y  x sin   y cos .

832

36.

CHAPTER 12 Conic Sections



x



  y  1. Squaring both sides gives x  2 x y  y  1  2 x y  1  x  y, and squaring both sides again

gives 4x y  1  x  y2  1  x  y  x  x 2  x y  y  x y  y 2  4x y  x 2  y 2  2x y  2x  2y  1  AC x 2  y 2  2x y  2x  2y  1  0. Then A  1, B  2, and C  1, and so cot 2   0  2  90  B 







  45 . Therefore, x  22 X  22 Y and y  22 X  22 Y , and substituting gives  2       2 X  2Y 2 X  2Y 2 X  2Y  2 2 2 2 2 2 2  2         22 X  22 Y  2 22 X  22 Y  2 22 X  22 Y  1  0           1 X 2  2XY  Y 2  X 2  Y 2  1 X 2  2XY  Y 2  2 2X  1  0  X 2 1  1  1  XY 1  1  2 2 2 2         1 . This is a parabola with Y 2 12  1  12  2 2X  1  0  2Y 2  2 2X  1  Y 2  2X  12  2 X   2 2    1 4 p  2 and vertex V   0 . However, in the original equation we must have x  0 and y  0, so we get only the 2 2 part of the parabola that lies in the first quadrant. 

37. (a) Z  

x











X cos   sin  , Z    , and R   .

cos      x cos   sin  X X cos   Y sin Y      . Equating the entries in this Thus Z  R Z      y sin  cos  Y X sin   Y cos  y



Y



sin 



matrix equation gives the first pair of rotation of axes formulas. Now     cos  sin  cos  sin  1 1      and so Z   R 1 Z  R  cos2   sin2   sin  cos   sin  cos         X cos  sin  x x cos   y sin        . Equating the entries in this matrix equation gives Y  sin  cos  y x sin   y cos 

the second pair of rotation of axes formulas.    cos 1  sin 1 cos 2  sin 2   (b) R1 R2   sin 1 cos 1 sin 2 cos 2        cos 1 cos 2  sin 1 sin 2  cos 1 sin 2  sin 1 cos 2 cos 1  2  sin 1  2       sin 1  2 cos 1  2 sin 1 cos 2  cos 1 sin 2  sin 1 sin 2  cos 1 cos 2   38. (a) Using A  A cos2   B sin  cos   C sin2 , B   2 C  a sin  cos   B cos2   sin2  , and C   A sin2   B sin  cos   C cos2 , we first expand the terms.   2   2 B  2 C  a sin  cos   B cos2   sin2 

   2  4 C  A2 sin2  cos2   4B C  A sin  cos  cos2   sin2   B 2 cos2   sin2   4 C  A2 sin2  cos2   4B C  A sin  cos3   4B C  A sin3  cos 

 B 2 cos4   2B 2 sin2  cos2   B 2 sin4 

SECTION 12.5 Rotation of Axes

833



  4A C   4 A cos2   B sin  cos   C sin2  A sin2   B sin  cos   C cos2 

 4A2 sin2  cos2   4AB sin  cos3   4AC cos4   4AB sin3  cos   4B 2 sin2  cos2 

 4BC sin  cos3   4AC sin4   4BC sin3  cos   4C 2 sin2  cos2 

Since there are many terms in this expansion, we find the coefficients of like trignometric terms.   2 B Term 4A C  Sum cos4 

B2

sin  cos3 

4B C  A

sin2  cos2  sin3  cos  sin4  So 2

B   4A C  

4 C  A2  2B 2 4B C  A B2





4AC

B 2  4AC

4AB  4BC

4BC  4AB  4AB  4BC  0   8AC  2B 2  2 B 2  4AC

4A2  4B 2  4C 2 4AB  4BC 4AC

4BC  AB  4AB  4BC  0 B 2  4AC

     B 2  4AC cos4   2 B 2  4AC sin2  cos2   B 2  4AC sin4 

    2 cos4   2 sin2  cos2   sin4   B 2  4AC cos2   sin2      B 2  4AC 12  B 2  4AC 

B 2  4AC

So B 2  4AC  B 2  4A C  .

(b) Using A  A cos2   B sin  cos   C sin2  and C   A sin2   B sin  cos   C cos2 , we have A  C   A cos2   B sin  cos   C sin2   A sin2   B sin  cos   C cos2       A sin2   cos2   C sin2   cos2   AC

(c) Since F   F, F is also invariant under rotation. 39. Let P be the point x1  y1  and Q be the point x2  y2  and let P  X 1  Y1  and Q  X 2  Y2  be the images of P and Q under the rotation of . So X 1  x1 cos   y1 sin  Y1  x1 sin   y1 cos , X 2  x2 cos   y2 sin and    Y2  x2 sin   y2 cos . Thus d P   Q   X 2  X 1 2  Y2  Y1 2 , where X 2  X 1 2 

 2  2 x2 cos   y2 sin   x1 cos   y1 sin   x2  x1  cos   y2  y1  sin 

 x2  x1 2 cos2   x2  x1  y2  y1  sin  cos   y2  y1 2 sin2  and Y2  Y1 2 

 2  2 x2 sin   y2 cos   x1 sin   y1 cos    x2  x1  sin   y2  y1  cos 

 x2  x1 2 sin2   x2  x1  y2  y1  sin  cos   y2  y1 2 cos2  So

X 2  X 1 2  Y2  Y1 2  x2  x1 2 cos2   x2  x1  y2  y1  sin  cos   y2  y1 2 sin2   x2  x1 2 sin2   x2  x1  y2  y1  sin  cos   y2  y1 2 cos2 

 x2  x1 2 cos2   y2  y1 2 sin2   x2  x1 2 sin2   y2  y1 2 cos2       x2  x1 2 cos2   sin2   y2  y1 2 sin2   cos2   x2  x1 2  y2  y1 2

834

CHAPTER 12 Conic Sections

    Putting these equations together gives d P   Q   X 2  X 1 2  Y2  Y1 2  x2  x1 2  y2  y1 2  d P Q.

12.6 POLAR EQUATIONS OF CONICS 1. All conics can be described geometrically using a fixed point F called the focus and a fixed line  called the directrix. For a distance from P to F fixed positive number e the set of all points P satisfying  e is a conic section. If e  1 the conic is a distance from P to  parabola, if e  1 the conic is an ellipse, and if e  1 the conic is a hyperbola. The number e is called the eccentricity of the conic. ed ed 2. The polar equation of a conic with eccentricity e has one of the following forms: r  or r  . 1  e cos  1  e sin 

2 3 3  3. Substituting e  23 and d  3 into the general equation of a conic with vertical directrix, we get r  1  23 cos 

r

6 . 3  2 cos 

4 3 3  4. Substituting e  43 and d  3 into the general equation of a conic with vertical directrix, we get r  1  43 cos 

r

12 . 3  4 cos 

5. Substituting e  1 and d  2 into the general equation of a conic with horizontal directrix, we get r  r

2 . 1  sin 

12  1  sin 

1 4 2 6. Substituting e  12 and d  4 into the general equation of a conic with horizontal directrix, we get r   1  12 sin  4 . r 2  sin  20 45 r  . 7. r  5 sec   r cos   5  x  5. So d  5 and e  4 gives r  1  4 cos  1  4 cos  12 06  2 r  . 8. r  2 csc   r sin   2  y  2. So d  2 and e  06 gives r  1  06 sin  1  06 sin  9. Since this is a parabola whose focus is at the origin and vertex at 5 2, the directrix must be y  10. So d  10 and 10 1  10  . e  1 gives r  1  sin  1  sin  d P F 2 2 10. Since the vertex is at 2 0 we have d P F  2. Now, since  e we get  04  d P    5. d P  d P  04 04  7 28 The directrix is x  7, so substituting e  04 and d  7 we get r  r  . 1  04 cos  1  04 cos  6 is Graph II. The eccentricity is 1, so this is a parabola. When   0, we have r  3 and when    11. r  2 , we 1  cos  have r  6. 2 1 12. r  is Graph III. r  , so e  12 and this is an ellipse. When   0, r  2, and when   , r  23 . 2  cos  1  1 cos  2

3 13. r  is Graph VI. e  2, so this is a hyperbola. When   0, r  3, and when   , r  3. 1  2 sin  5

14. r 

5 3 is Graph I. r  , so e  1 and this is a parabola. When   0, r  53 and when   , r  53 . 3  3 sin  1  sin 

SECTION 12.6 Polar Equations of Conics

835

4 12 , so e  23 and this is an ellipse. When   0, r  4, and when   , r  4. is Graph IV. r  2 3  2 sin  1  3 sin 

15. r 

6 12 is Graph V. r  , so e  32 and this is a hyperbola. When   0, r  12 5 , and when   , 2  3 cos  1  32 cos  we have r  12.

16. r 

4 has e  1 and d  4, so 1  sin  it represents a parabola.

17. (a) The equation r 

3

18. (a) The equation r 

3 2  has e  1 2  2 sin  1  sin 

and d  32 , so it represents a parabola. V(3/4, ¹/2) _4

_8

_4 y=_4

O

4 V(2, 3¹/2)

ed , 1  e sin  the directrix is parallel to the polar axis and has   equation y  4. The vertex is 2 32 .

e  1 and d  53 , so it represents a parabola. x=5/3

O

O

4

ed , 1  e sin  the directrix is parallel to the polar axis and has   equation y  d  32 . The vertex is 34   2 .

(b) Because the equation is of the form r 

5

5 3  has 3  3 cos  1  cos 

1

y=3/2 2

8

(b) Because the equation is of the form r 

19. (a) The equation r 

_2

2

V(5/6, 0)

3

ed , 1  e cos  the directrix is parallel to the polar axis and has   equation x  d  53 . The vertex is 56  0 .

(b) Because the equation is of the form r 

2

20. (a) The equation r 

2 5  has 5  5 cos  1  cos 

e  1 and d  25 , so it represents a parabola. x=_2/5

1 V(1/5, ¹) O

1

ed , 1  e cos  the directrix is parallel to the polar axis and has   equation x  d   25 . The vertex is 15   .

(b) Because the equation is of the form r 

836

CHAPTER 12 Conic Sections

21. (a) The equation r 

2 4 has e  12  1, so it represents an  1 2  cos  1  2 cos 

x=_4

ellipse.

1

ed (b) Because the equation is of the form r  with d  4, the directrix is 1  e cos    vertical and has equation x  4. Thus, the vertices are V1 4 0 and V2 43   .

Vª(4/3, ¹)

1

O

VÁ(4, 0)

(c) The length of the major axis is 2a  V1 V2   4  43  16 3 and the center is at the   midpoint of V1 V2 , 43  0 . The minor axis has length 2b where 2  2  b2  a 2  c2  a 2  ae2  83  83  12  16 3 , so   8 3 2b  2  16 3  3  462.

22. (a) The equation r 

6 2 has e  23  1, so it represents an  2 3  2 sin  1  3 sin 

VÁ(6, ¹/2)

ellipse. ed with d  3, the directrix is 1  e sin    horizontal and has equation y  3. Thus, the vertices are V1 6  2 and   V2 65  32 .

(b) Because the equation is of the form r 

1 O Vª(6/5, 3¹/2)

1 y=_3

(c) The length of the major axis is 2a  V1 V2   6  65  36 5 and the center is at the    midpoint of V1 V2 , 12 5  2 . The minor axis has length 2b where   2  18  2 2  36 , so b2  a 2  c2  a 2  ae2  18  5 5 3 5   12 5 2b  2  36 5  5  537.

23. (a) The equation r 

3 12 has e  34  1, so it represents an  4  3 sin  1  34 sin 

ellipse. ed (b) Because the equation is of the form r  with d  4, the directrix is 1  e sin     horizontal and has equation y  4. Thus, the vertices are V1 12 7  2 and   V2 12 32 .

96 (c) The length of the major axis is 2a  V1 V2   12 7  12  7 and the center is at   3 the midpoint of V1 V2 , 36 7  2 . The minor axis has length 2b where

  2  48  3 2  144 , so b2  a 2  c2  a 2  ae2  48  7 7 4 7   24 7 2b  2  144 7  7  907.

VÁ(12/7, ¹/2) O 1 4 8 Vª(12, 3¹/2)

y=4

SECTION 12.6 Polar Equations of Conics 9

24. (a) The equation r 

18 2 has e  34  1, so it represents an  4  3 cos  1  34 cos 

x=6

ellipse.

VÁ(18/7, 0)

ed (b) Because the equation is of the form r  with d  6, the directrix is 1  e cos     0 and V2 18 . vertical and has equation x  6. Thus, the vertices are V1 18 7

Vª(18, ¹)

2 O

5

144 (c) The length of the major axis is 2a  V1 V2   18 7  18  7 and the center is at   the midpoint of V1 V2 , 54 7   . The minor axis has length 2b where

  2  72  3 2  324 , so b2  a 2  c2  a 2  ae2  72  7 7 4 7   36 7 2b  2  324 7  7  1361.

8 has e  2  1, so it represents a hyperbola. 1  2 cos  ed with d  4, the transverse axis is (b) Because the equation has the form r  1  cos    horizontal and the directrix has equation x  4. The vertices are V1 83  0 and

x=4

25. (a) The equation r 

VÁ(8/3, 0)

Vª(8, 0)

O

V2 8   8 0.

  (c) The center is the midpoint of V1 V2 , 16 3  0 . To sketch the central box and the

asymptotes, we find a and b. The length of the transverse axis is 2a  16 3 , and so  2  2 a  83 , and b2  c2  a 2  ae2  a 2  83  2  83  64 3 , so   8 3 b  64 3  3  462.

10 has e  4  1, so it represents a hyperbola. 1  4 sin  ed with d  52 , the transverse axis (b) Because the equation has the form r  1  cos 

26. (a) The equation r 

VÁ(2, 3¹/2)

is vertical and the directrix has equation y   52 . The vertices are

       10 3 3 V1  10 3  2  3  2 and V2 2 2 .   (c) The center is the midpoint of V1 V2 , 83  32 . To sketch the central box and the

asymptotes, we find a and b. The length of the transverse axis is 2a  43 , and so  2  2 a  23 , and b2  c2  a 2  ae2  a 2  23  4  23  20 3 , so  b  20 3  258.

y=_5/2 Vª(10/3, 3¹/2)

837

838

CHAPTER 12 Conic Sections

27. (a) The equation r 

10 20 has e  32  1, so it represents a  2  3 sin  1  32 sin 

VÁ(4, 3¹/2) O

hyperbola. (b) Because the equation has the form r 

y=_20/3

ed with d  20 3 , the transverse axis 1  cos 

is vertical and the directrix has equation y   20 3 . The vertices are       3 3 V1 20  2  20 2 and V2 4 2 .   (c) The center is the midpoint of V1 V2 , 12 32 . To sketch the central box and the

Vª(20, 3¹/2)

asymptotes, we find a and b. The length of the transverse axis is 2a  16, and so 2  a  8, and b2  c2  a 2  ae2  a 2  8  32  82  80, so   b  80  4 5  894.

28. (a) The equation r 

3 6  has e  72  1, so it represents a 7 2  7 cos  1  2 cos 

hyperbola. ed with d  67 , the transverse axis (b) Because the equation has the form r  1  cos    is horizontal and the directrix has equation x  67 . The vertices are V1 23  0 and     V2  65    65  0 .   (c) The center is the midpoint of V1 V2 , 14 15  0 . To sketch the central box and the asymptotes, we find a and b. The length of the transverse axis is 2a     4 , and b2  c2  a 2  ae2  a 2  4  7 2  4 2  a  15 15 2 15   b  45  2 5 5  089.

29. (a) r 

4  e  3, so the conic is a hyperbola. 1  3 cos 

(b) The vertices occur where   0 and   . Now   0  r 

x=6/7 1 O VÁ(2/3, 0)

Vª(6/5, 0)

8 , and so  15 4 , so 5

4  1, 1  3 cos 0

4 4 and     r    2. Thus the vertices are 1 0 and 1  3 cos  2 2 .

(_2, ¹)

(1, 0)

8

8 3 r   e  1, so the conic is a parabola. 3  3 cos  1  cos    8 8 4 (b) Substituting   0, we have r    . Thus the vertex is 43  0 . 3  3 cos 0 6 3

30. (a) r 

1

( 43 , 0)

SECTION 12.6 Polar Equations of Conics

2  e  1, so the conic is a parabola. 1  cos  2 (b) Substituting   , we have r   22  1. Thus the vertex is 1 . 1  cos 

31. (a) r 

(1, ¹)

1

10

32. (a) r 

10 3 r   e  23 , so the conic is an ellipse. 3  2 sin  1  23 sin 

(10, ¹2 )

3  (b) The vertices occur where    2 and   2 . Now   2  10 10 10 10  10 and   32  r   2. Thus,   r 3  3  2 sin  1 5 3  2 sin 2 2     3 . the vertices are 10  and 2 2 2

33. (a) r 

1

(2, 3¹ 2 )

1 6 6 2 r   e  12 , so the conic is an ellipse. 2  sin  1  12 sin 

(2, ¹2)

3  (b) The vertices occur where    2 and   2 . Now   2 

1

6 6 6 6   2 and   32  r    6. Thus, the  3  2  sin 2 3 1 2  sin 2     vertices are 2 2 and 6 32 .

r

(6, 3¹ 2 )

5

34. (a) r 

5 2  e  32 , so the conic is a hyperbola. r  2  3 sin  1  32 sin 

1

3  (b) The vertices occur where    2 and   2 . Now   2  5 5 5  5 and   32  r    55  1. Thus, r 3 2  3 sin  1 2  3 sin 2 2      3  the vertices are 5 2 and 1 2 .

(

¹ _5, 2

)

(1, 3¹ 2)

7

7 2 r   e  52 , so the conic is a hyperbola. 2  5 sin  1  52 sin  7 3 7 7 (b) The vertices occur where    2 and   2 . r  2  5 sin   3   3 and 2   7 7  1. Thus, the vertices are  7   and    32  r  7 3 2 2  5 sin 32   1 32 .

35. (a) r 

(_ 73 , ¹2 )

1

(1, 3¹ 2)

8

36. (a) r 

8 3 r   e  13 , so the conic is an ellipse. 3  cos  1  13 cos 

(b) The vertices occur where   0 and   . Now   0  8 8 r  84  2 and     r   82  4. Thus, the vertices 3  cos 0 3  cos  are 2 0 and 4 .

(4, ¹)

(2, 0) 1

839

840

CHAPTER 12 Conic Sections 1

37. (a) r 

1 4  e  34 , so the  4  3 cos  1  34 cos 

conic is an ellipse. The vertices occur where   0 and   . Now   0  r 

(b) If the ellipse is rotated through  3 , the equation of the resulting conic is r 

1  1 and 4  3 cos 0

1 .  4  3 cos    3

1.0

1  17 . Thus, the vertices  r  4  3 cos    are 1 0 and 17   . We have d  13 , so the

0.5

directrix is x   13 .

x=-1/3

0.5

O

1

2

38. (a) r 

2 5   e  35 , so the 5  3 sin  1  35 sin 

conic is an ellipse. The vertices occur where    2 and   32 . Now    2 r  and   32  r 

2 1 5  3 sin  2

(b) If the ellipse is rotated through   23 , the equation 2 .  of the resulting conic is r  5  3 sin   23 0.5

2

 14 . Thus, the 5  3 sin 32

    1 3 2 vertices are 1  2 and 4  2 . We have d  3 , so the directrix is y   23 .

1

O y=_2/3

-1.0

-0.5 -0.5

SECTION 12.6 Polar Equations of Conics

2  e  1, so the conic is a parabola. 1  sin  2 Substituting    2 , we have r  t 1  sin   1, 2   so the vertex is 1  . Because d  2, the directrix 2

39. (a) r 

is y  2.

(b) If the ellipse is rotated through     4 , the equation of the resulting conic is r

2 .  1  sin    4 2

y=2 O

-8

-6

-4

-2

2 -2 -4 -6 -8

9 9 2   e  1, so the conic 2  2 cos  1  cos  is a parabola. Substituting   0, we have   9 9  , so the vertex is 94  0 . r 2  2 cos 0 4

40. (a) r 

Because d  92 , the directrix is x  92 .

(b) If the ellipse is rotated through    56 , the equation of the resulting conic is r

9 .  2  2 cos   56

x=9/2

O

10

5 1

5 -5

41. The ellipse is nearly circular when e is close to 0 and becomes more elongated as

4

e  1 . At e  1, the curve becomes a parabola.

e=0.4 e=0.6

2

2 -2

4 e=0.8

e=1

-4

d where d  12 , d  2, and 1  sin  d  10. As d increases, the parabolas get flatter while the vertex moves further

42. (a) Shown are the graphs of the conics r 

10 d=10

from the focus at the origin.

-10

10 d=2

-10

1

d=2

841

842

CHAPTER 12 Conic Sections

e where e  05, e  1, and 1  e sin  e  10. As e increases, the conic changes from an ellipse to a parabola, and

(b) Shown are the graphs of the conics r 

finally to a hyperbola. The vertex gets closer to the directrix (shown as a

2

e=10 -2

2

e=0.5 e=1 -2

dashed line in the figure).

  ed we need to show that ed  a 1  e2 . 1  e cos  e2 d 2 From the proof of the Equivalent Description of Conics we have a 2   2 . Since the conic is an ellipse, e  1 1  e2   and so the quantities a, d, and 1  e2 are all positive. Thus we can take the square roots of both sides and maintain     2d 2 a 1  e2 ed e ed equality. Thus a 2   r  .  ed  a 1  e2 . As a result, r  2  a  1  e cos  1  e cos  1  e2 1  e2

43. (a) Since the polar form of an ellipse with directrix x  d is r 

(b) Since 2a  299  108 we have a  1495  108 , so a polar equation for the earth’s orbit (using e  0017) is   1495  108 1  00172 149  108  . r 1  0017 cos  1  0017 cos 

44. (a) Using the form of the equation from Exercise 27, the perihelion distance occurs when    and the aphelion distance occurs when   0. Since the focus is at the origin, the perihelion distance is     2 a 1  e a 1  e2 a 1  e 1  e a 1  e 1  e   a 1  e and the aphelion distance is   a 1  e. 1e 1e 1e 1e (b) Given e  0017 and 2a  299  108 we have a  1495  108 . Thus the perihelion distance is

1495  108 [1  0017]  1468  108 km and the aphelion distance is 1495  108 1  0017  1520  108 km.

45. From Exercise 44, we know that at perihelion r  443  109  a 1  e and at aphelion r  737  109  a 1  e. 737  109 1e a 1  e  1664   1664 1  e  1  e   9 a 1  e 1e 443  10 0664 1664  1  e  1664  0664  2664e  e   025. 2664 Dividing these equations gives

46. Since the focus is at the origin, the distance from the focus to any point on the conic is the absolute value of the r-coordinate of that point, which we can obtain from the polar equation. 47. The r-coordinate of the satellite will be its distance from the focus (the center of the earth). From the r-coordinate we can easily calculate the height of the satellite.

CHAPTER 12 REVIEW 1. (a) y 2  4x. This is a parabola with 4 p  4  p  1. The vertex is 0 0, the

(b)

y

focus is 1 0, and the directrix is x  1.

1 1

x

CHAPTER 12 1 y 2  y 2  12x. This is a parabola with 4 p  12  p  3. The vertex 2. (a) x  12 is 0 0, the focus is 3 0, and the directrix is x  3.

(b)

Review

843

y

2 x

1

3. (a) 18 x 2  y  x 2  8y. This is a parabola with 4 p  8  p  2. The vertex is 0 0, the focus is 0 2, and the directrix is y  2.

y

(b)

1

4. (a) x 2  8y. This is a parabola with 4 p  8  p  2. The vertex is 0 0,

_1

7. (a) y  22  4 x  2. This is a parabola with 4 p  4  p  1. The vertex is

(b)

x

1

x

y

vertex is 0 0, the focus is 0 2, and the directrix is y  2.

(b)

1

1

(b)

6. (a) 2x  y 2  0  y 2  2x. This is a parabola with 4 p  2  p  12 . The vertex   is 0 0, the focus is 12  0 , and the directrix is x   12 .

x

y

(b)

the focus is 0 2, and the directrix is y  2.

5. (a) x 2  8y  0  x 2  8y. This is a parabola with 4 p  8  p  2. The

1

y

1 x

1

y

2 2, the focus is 1 2, and the directrix is x  3.

1 1

x

844

CHAPTER 12 Conic Sections

8. (a) x  32  20 y  2. This is a parabola with 4 p  20  p  5. The

(b)

9. (a) 12 y  32  x  0  y  32  2x. This is a parabola with 4 p  2    p   12 . The vertex is 0 3, the focus is  12  3 , and the directrix is x  12 .

(b)

y

1

vertex is 3 2, the focus is 3 7, and the directrix is y  3.

x

5

y

1 1

10. (a) 2 x  12  y  x  12  12 y. This is a parabola with 4 p  12  p  18 .   The vertex is 1 0, the focus is 1 18 , and the directrix is y   18 .

x

y

(b)

1 1 x

11. (a) 12 x 2  2x  2y  4  x 2  4x  4y  8  x 2  4x  4  4y  8 

y

(b)

x  22  4 y  3. This is a parabola with 4 p  4  p  1. The vertex is 2 3, the focus is 2 3  1  2 2, and the directrix is y  3  1  4.

12. (a) x 2  3 x  y  x 2  3x  3y  x 2  3x  94  3y  94   2   x  32  3 y  34 . This is a parabola with 4 p  3  p  34 . The vertex       is 32   34 , the focus is 32   34  34  32  0 , and the directrix is

1 1

(b)

y   34  34   32 .

13. (a)

 y2 x2   1. This is an ellipse with a  5, b  3, and c  25  9  4. The 9 25 center is 0 0, the vertices are 0 5, and the foci are 0 4.

(b) The length of the major axis is 2a  10 and the length of the minor axis is 2b  6.

y

1

(c)

x

1

x

y

1 1

x

CHAPTER 12

14. (a)

  y2 x2   1. This is an ellipse with a  7, b  3, and c  49  9  2 10. 49 9    The center is 0 0, the vertices are 7 0, and the foci are 2 10 0 .

Review

845

y

(c)

1 x

1

(b) The length of the major axis is 2a  14 and the length of the minor axis is 2b  6.

15. (a)

  y2 x2   1. This is an ellipse with a  7, b  2, and c  49  4  3 5. 49 4    The center is 0 0, the vertices are 7 0, and the foci are 3 5 0 .

y

(c)

1 x

1

(b) The length of the major axis is 2a  14 and the length of the minor axis is 2b  4.

16. (a)

  y2 x2   1. This is an ellipse with a  6, b  2, and c  36  4  4 2. 4 36    The center is 0 0, the vertices are 0 6, and the foci are 0 4 2 .

y

(c)

1 x

1

(b) The length of the major axis is 2a  12 and the length of the minor axis is 2b  4.

y2 x2   1. This is an ellipse with a  4, b  2, and 17. (a) x 2  4y 2  16  16 4   c  16  4  2 3. The center is 0 0, the vertices are 4 0, and the foci    are 2 3 0 .

y

(c)

1 x

1

(b) The length of the major axis is 2a  8 and the length of the minor axis is 2b  4.

y2 x2   1. This is an ellipse with a  12 , b  13 , and 19 14     c  14  19  16 5. The center is 0 0, the vertices are 0  12 , and the    foci are 0  16 5 .

18. (a) 9x 2  4y 2  1 

y 0.5

(c)

0.5 x

(b) The length of the major axis is 2a  1 and the length of the minor axis is 2b  23 .

19. (a)

y2 x  32   1This is an ellipse with a  4, b  3, and 9 16   c  16  9  7. The center is 3 0, the vertices are 3 4, and the foci    are 3  7 .

(b) The length of the major axis is 2a  8 and the length of the minor axis is 2b  6.

(c)

y

1 1

x

846

CHAPTER 12 Conic Sections

20. (a)

y  32 x  22   1. This is an ellipse with a  5, b  4, and 25 16  c  25  16  3. The center is 2 3, the vertices are

(c)

y 1 x

1

2  5 3  3 3 and 7 3, and the foci are 2  3 3  1 3 and 5 3.

(b) The length of the major axis is 2a  10 and the length of the minor axis is 2b  8.

21. (a)

x  22 y  32   1. This is an ellipse with a  6, b  3, and 9 36   c  36  9  3 3. The center is 2 3, the vertices are    2 3  6  2 9 and 2 3, and the foci are 2 3  3 3 .

y

(c)

1 x

1

(b) The length of the major axis is 2a  12 and the length of the minor axis is 2b  6.

22. (a)

 x2 y  52   1. This is an ellipse with a  5, b  3, and 3 25   c  25  3  22. The center is 0 5, the vertices are    0 5  5  0 10 and 0 0, and the foci are 0 5  22 .

y 1

(c)

1

x

(b) The length of the major axis is 2a  10 and the length of the minor axis is  2b  2 3.   23. (a) 4x 2  9y 2  36y  4x 2  9 y 2  4y  4  36  4x 2  9 y  22  36

y

(c)

x2 y  22   1. This is an ellipse with a  3, b  2, and 9 4   c  9  4  5. The center is 0 2, the vertices are 3 2, and the foci    are  5 2 .



1 1

x

(b) The length of the major axis is 2a  6 and the length of the minor axis is 2b  4.

24. (a) 2x 2  y 2  2  4 x  y  2x 2  4x  y 2  4y  2      2 x 2  2x  1  y 2  4y  4  2  2  4  2 x  12  y  22  8  x  12 y  22    1. This is an ellipse with a  2 2, b  2, and 4 8     c  8  4  2. The center is 1 2, the vertices are 1 2  2 2 , and

the foci are 1 2  2  1 0 and 1 4.  (b) The length of the major axis is 2a  4 2 and the length of the minor axis is 2b  4.

(c)

y

1 1

x

CHAPTER 12

y2 y2 x2 x2  1   0. This is a hyperbola with a  4, b  3, and 9 16 16 9   c  16  9  25  5. The center is 0 0, the vertices are 0 4, the foci

25. (a) 

(b)

 y2 x2   1. This is a hyperbola with a  7, b  4 2, and 49 32  c  49  32  9. The center is 0 0, the vertices are 7 0, the foci are

(b)

28. (a)

y2 x2   1. This is a hyperbola with a  2, b  7, and 4 49   c  4  49  53. The center is 0 0, the vertices are 2 0, the foci are     53 0 , and the asymptotes are y   72 x.

x2 y2   1. This is a hyperbola with a  5, b  2, and 25 4   c  25  4  29. The center is 0 0, the vertices are 0 5, the foci are    0  29 , and the asymptotes are y   52 x.

 y2 x2   1. This is a hyperbola with a  4, b  2 2, 29. (a) x 2  2y 2  16  16 8    and c  16  8  24  2 6. The center is 0 0, the vertices are 4 0,     the foci are 2 6 0 , and the asymptotes are y   2 4 2 x  y   1 x.

y

2

y   12 x.

x

2

x

1

x

1

x

1

x

1

x

y

(b)

y

2

(b)

y

2

(b)

y

1

2

2

x y   1. This is a hyperbola 30. (a) x 2  4y 2  16  0  4y 2  x 2  16  4 16   with a  2, b  4, and c  4  16  2 5. The center is 0 0, the vertices    are 0 2, the foci are 0 2 5 , and the asymptotes are y   24 x 

1

2



9 0, and the asymptotes are y   4 7 2 x.

27. (a)

847

1

are 0 5, and the asymptotes are y   43 x.

26. (a)

Review

(b)

y

1

848

31. (a)

CHAPTER 12 Conic Sections

y2 x  42   1. This is a hyperbola with a  4, b  4 and 16 16   c  16  16  4 2. The center is 4 0, the vertices are 4  4 0    which are 8 0 and 0 0, the foci are 4  4 2 0 , and the asymptotes

y

(b)

1

1 x

are y   x  4.

32. (a)

  y  22 x  22   1. This is a hyperbola with a  2 2, b  2 2 and 8 8     c  8  8  4. The center is 2 2, the vertices are 2  2 2 2 , the

(b)

y 1

x

1

foci are 2  4 2  2 2 and 6 2, and the asymptotes are y  2   x  2  y  x and y  x  4.

33. (a)

x  12 y  32   1. This is a hyperbola with a  2, b  6, and 4 36   c  4  36  2 10. The center is 1 3, the vertices are    1 3  2  1 1 and 1 5, the foci are 1 3  2 10 , and the

(b)

1

1 8 asymptotes are y  3   13 x  1  y  13 x  10 3 and y   3 x  3 .

34. (a)

 x2 y  32   1. This is a hyperbola with a  3, b  4, and 3 16      c  3  16  19. The center is 0 3, the vertices are 0 3  3 , the     foci are 0 3  19 , and the asymptotes are y  3   43 x 

(b)

2

x

2

x

y

1

  y  43 x  3 and y   43 x  3.

    35. (a) 9y 2  18y  x 2  6x  18  9 y 2  2y  1  x 2  6x  9  9  9  18

y

y

(b)

x  32 y  12  9 y  12  x  32  18    1. This is a 2 18     hyperbola with a  2, b  3 2, and c  2  18  2 5.The center is       3 1, the vertices are 3 1  2 , the foci are 3 1  2 5 , and

the asymptotes are y  1   13 x  3  y  13 x and y   13 x  2.   36. (a) y 2  x 2  6y  y 2  6y  9  x 2  9  y  32  x 2  9 

x2 y  32   1. This is a hyperbola with a  3, b  3, and 9 9   c  9  9  3 2. The center is 0 3, the vertices are 0 3  3  0 6    and 0 0, the foci are 0 3  3 2 , and the asymptotes are y  3  x  y  x  3 and y  x  3.

1 1

(b)

x

y

1 1

x

CHAPTER 12

Review

849

37. This is a parabola that opens to the right with its vertex at 0 0 and the focus at 2 0. So p  2, and the equation is y 2  4 2 x  y 2  8x.

38. This is an ellipse with the center at 0 0, a  12, and b  5. The equation is then

x2 y2 y2 x2   1.  2 1 2 144 25 12 5

39. From the graph, the center is 0 0, and the vertices are 0 4 and 0 4. Since a is the distance from the center to a

vertex, we have a  4. Because one focus is 0 5, we have c  5, and since c2  a 2  b2 , we have 25  16  b2 

b2  9. Thus an equation of the hyperbola is

x2 y2   1. 16 9

40. This is a parabola that opens to the left with its vertex at 4 4, so its equation is of the form y  42  4 p x  4 with

p  0. Since 0 0 is a point on this hyperbola, we must have 0  42  4 p 0  4  16  16 p  p  1. Thus the

equation is y  42  4 x  4.

41. From the graph, the center of the ellipse is 4 2, and so a  4 and b  2. The equation is

y  22 x  42  1 42 22

x  42 y  22   1. 16 4 42. From the graph, the center is at 1 0, and the vertices are 0 0 and 2 0. Since a is the distance form the center to a b vertex, a  1. From the graph, the slope of one of the asymptotes is  1  b  1. Thus an equation of the hyperbola is a x  12  y 2  1. 43.

x2 x2 y 1   y  1  x 2  12 y  1. This is a parabola with 12 12 4 p  12  p  3. The vertex is 0 1, the focus is 0 1  3  0 2, and

y 1

the directrix is y  1  3  4.

  y2 y x2    12x 2  y 2  12y  0  12x 2  y 2  12y  36  36  12 144 12 2 2  x y  6   1. This is an ellipse with a  6, b  3, and 3 36      c  36  3  33. The center is 0 6, the foci are 0 6  33 , and the

y

x2 y2   1. This is a hyperbola with a  12, b  12, 45. x 2  y 2  144  0  144 144      and c  144  144  12 2. The center is 0 0, the foci are 0 12 2 , the

y

44.

vertices are 0 6  6  0 0 and 0 12.

vertices are 0 12, and the asymptotes are y  x.

x

1

1 1

x

4 4

x

850

CHAPTER 12 Conic Sections

  x  32  y 2  1. This is a 46. x 2  6x  9y 2  x 2  6x  9  9y 2  9  9   hyperbola with a  3, b  1, and c  9  1  10. The center is 3 0, the    foci are 3  10 0 , the vertices are 3  3 0  6 0 and 0 0, and the

y

1 1

x

asymptotes are y   13 x  3  y  13 x  1 and y   13 x  1.

    47. 4x 2  y 2  8 x  y  4 x 2  2x  y 2  8y  0      4 x 2  2x  1  y 2  8y  16  4  16  4 x  12  y  42  20 

y

  x  12 y  42   1. This is an ellipse with a  2 5, b  5, and 5 20      c  20  5  15. The center is 1 4, the foci are 1 4  15 , and the    vertices are 1 4  2 5 .

3 3

  48. 3x 2  6 x  y  10  3x 2  6x  6y  10  3 x 2  2x  1  6y  10  3     2  2 y  13 .   1  3 x  12  6y  13  3 x  12  6 y  13 x 6 6   1 13 This is a parabola with 4 p  2  p  2 . The vertex is 1  6 the focus is       1  1  10  1  5 , and the directrix is y   13  1   8 . 1  13  6 2 6 3 6 2 3

y

1 x

1

y

49. x  y 2  16y  x  64  y 2  16y  64  y  82  x  64. This is a parabola with 4 p  1  p  14 . The vertex is 64 8, the focus is     64  14  8   255  8 , and the directrix is x  64  14   257 4 4 .

2 10

  50. 2x 2  4  4x  y 2  y 2  2x 2  4x  4  y 2  2 x 2  2x  1  4  2   y2  x  12  1. This is a hyperbola with a  2, 2      b  1, and c  2  1  3. The center is 1 0, the foci are 1  3 , the     vertices are 1  2 , and the asymptotes are y   2 x  1      y  2 x  2 and y   2 x  2. y 2  2 x  12  2 

x

x

y

1 1

x

CHAPTER 12

    51. 2x 2  12x  y 2  6y  26  0  2 x 2  6x  y 2  6y  26      2 x 2  6x  9  y 2  6y  9  26  18  9  2 x  32  y  32  1 

x  32 1 2

c

Review

851

y x

1

_1



 y  32  1. This is an ellipse with a  1, b  22 , and

     1  12  22 . The center is 3 3, the foci are 3 3  22 , and the

vertices are 3 3  1  3 4 and 3 2.

    52. 36x 2  4y 2  36x  8y  31  36 x 2  x  4 y 2  2y  31      36 x 2  x  14  4 y 2  2y  1  31  9  4 

y

1



2 2 y  12  36 x  12  4 y  12  36  x  12   1. This is a 9

1

x

    hyperbola with a  1, b  3, and c  1  9  10. The center is 12  1 , the          foci are 12  10 1 , the vertices are 12  1 1   12  1 and 32  1 ,   and the asymptotes are y  1  3 x  12  y  3x  52 and y  3x  12 .

        1 25 5 2 1 2   75 . 2 53. 9x 2 8y 2 15x8y27  0  9 x 2  53 x  25 36 8 y  y  4  27 4 2  9 x  6 8 y  2 4 However, since the left-hand side of the equation is greater than or equal to 0, there is no point that satisfies this equation. The graph is empty.   y 54. x 2  4y 2  4x  8  x 2  4x  4  4y 2  8  4  x  22  4y 2  12    y2 x  22   1. This is an ellipse with a  2 3, b  3, and 12 3  c  12  3  3. The center is 2 0, the foci are 2  3 0  1 0 and    5 0, and the vertices are 2  2 3 0 .

1 1

x

55. The parabola has focus 0 1 and directrix y  1. Therefore, p  1 and so 4 p  4. Since the focus is on the y-axis and the vertex is 0 0, an equation of the parabola is x 2  4y.

56. The parabola with vertex at the origin and focus F 5 0 has p  5, so 4 p  20. The focus is on the x-axis, so an equation is y 2  20x.

57. The ellipse with center at the origin and with x-intercepts 2 and y-intercepts 5 has a vertical major axis, a  5, and b  2, so an equation is

x2 y2   1. 4 25

58. The hyperbola has vertices 0 2 and asymptotes y   12 x. Therefore, a  2, and the foci are on the y-axis. Since the slopes of the asymptotes are  12  

a y2 x2  b  2a  4, an equation of the hyperbola is   1. b 4 16

59. The ellipse has center C 0 4, foci F1 0 0 and F2 0 8, and major axis of length 10. Then 2c  8  0  c  4. Also,

since the length of the major axis is 10, 2a  10  a  5. Therefore, b2  a 2  c2  25  16  9. Since the foci are on

the y-axis, the vertices are on the y-axis, and an equation of the ellipse is

x2 y  42   1. 9 25

852

CHAPTER 12 Conic Sections

60. The hyperbola has center C 2 4, foci F1 2 7 and F2 2 1, and vertices V1 2 6 and V2 2 2. Thus, 2a  6  2  4  a  2. Also, 2c  7  1  6  c  3. So b2  9  4  5. Since the hyperbola has center C 2 4, its equation is y  42 x  22   1. 4 5

61. The ellipse has foci F1 1 1 and F2 1 3, and one vertex is on the x-axis. Thus, 2c  3  1  2  c  1, and so the

center of the ellipse is C 1 2. Also, since one vertex is on the x-axis, a  2  0  2, and thus b2  4  1  3. So an

equation of the ellipse is

x  12 y  22   1. 3 4

62. The parabola has vertex V 5 5 and directrix the y-axis. Therefore,  p  0  5  p  5  4 p  20. Since the parabola opens to the right, its equation is y  52  20 x  5.

63. The ellipse has vertices V1 7 12 and V2 7 8 and passes through the point P 1 8. Thus, 2a  12  8  20    8  12 x  72 y  22  7 2. Thus an equation of the ellipse has the form  1. a  10, and the center is 7  2 2 100 b

1  72 8  22  1  3600  36b2  100b2  64b2  3600  b2  225  4 . 100 b2 4 x  72 y  22 y  22 x  72  1   1. Therefore, an equation of the ellipse is 2254 100 225 100 Since the point P 1 8 is on the ellipse,

64. The parabola has vertex V 1 0, horizontal axis of symmetry, and crosses the y-axis where y  2. Since the parabola has a horizontal axis of symmetry and V 1 0, its equation is of the form y 2  4 p x  1. Also, since the parabola crosses

the y-axis where y  2, it passes through the point 0 2. Substituting this point gives 22  4 p 0  1  4 p  4.

Therefore, an equation of the parabola is y 2  4 x  1.

65. The length of the major axis is 2a  186,000,000  a  93,000,000. The eccentricity is e  ca  0017, and so c  0017 93,000,000  1,581,000. (a) The earth is closest to the sun when the distance is a  c  93,000,000  1,581,000  91,419,000.

(b) The earth is furthest from the sun when the distance is a  c  93,000,000  1,581,000  94,581,000. y

66. We sketch the LORAN station on the y-axis and place the x-axis halfway between them as

A

x2 y2 suggested in the exercise. This gives us the general form 2  2  1. Since the ship is 80 miles a b closer to A than to B we have 2a  80  a  40Since the foci are 0 150, we have c  150.

150

x

Thus b2  c2  a 2  1502  402  20900. So this places the ship on the hyperbola given by the

150

x2 y2 1600 y2 225 y2   1. When x  40, we get  1   1600 20,900 1600 20,900 1600 209 y  415. (Note that y  0, since A is on the positive y-axis.) Thus, the ship’s position is

equation

B

approximately 40 415.

67. (a) The graphs of

x2 y2   1 for k  1, 2, 4, and 8 are shown in 16  k 2 k2

the figure.   (b) c2  16  k 2  k 2  16 

c  4. Since the center is 0 0,

the foci of each of the ellipses are 4 0.

y k=8

4

40

2 1 1 1

x

CHAPTER 12

68. (a) The graphs of y  kx 2 for k  12 , 1, 2, and 4 are shown in the figure.     1 1 1 y. Thus the foci are 0 . (b) y  kx 2  x 2  y  4 k 4k 4k

10

k=4

Review

853

k=2

8

k=1

6 4

(c) As k increases, the focus gets closer to the vertex.

1

k=2

2 0

-2

2

69. (a) x 2  4x y  y 2  1. Then A  1, B  4, and C  1, so the discriminant is 42  4 1 1  12. Since the discriminant is positive, the equation represents a hyperbola.     11 AC   0  2  90    45 . Therefore, x  22 X  22 Y and y  22 X  22 Y . (b) cot 2  B 4 Substituting into the original equation gives  2       2   2 X  2Y 2 X  2Y 2 X  2Y  2 X  2Y  4 1 2 2 2 2 2 2 2 2       y 1 X 2  2XY  Y 2  2 X 2  XY  XY  Y 2  1 X 2  2XY  Y 2  1  (c) 2 2 3X 2  Y 2  1  3X 2  Y 2  1. This is a hyperbola with a  1 , b  1, and 3    1 2 1 c  3  1   . Therefore, the hyperbola has vertices V    0 and foci 3 3   2 F    0 , in XY -coordinates.

1

x

1

3

70. (a) 5x 2  6x y  5y 2  8x  8y  8  0. Then A  5, B  6, and C  5, so the discriminant is 62  4 5 5  64. Since the discriminant is negative, the equation represents an ellipse.     AC (b) cot 2   0  2  90    45 . Therefore, x  22 X  22 Y and y  22 X  22 Y . Substituting B into the original equation gives  2        2 X  2Y 5 22 X  22 Y  6 22 X  22 Y 2 2  2         5 22 X  22 Y  8 22 X  22 Y  8 22 X  22 Y  8  0            5 X 2  2XY  Y 2  3 X 2  Y 2  5 X 2  2XY  Y 2  4 2X  4 2Y  4 2X  4 2Y  8  0  2 2  5X 2  3X 2  3Y 2  5Y 2  8 2Y  8  0     X2 Y  12   1. This ellipse has 2X 2  8 Y 2  2Y  12  8  4  6 32     a  6, b  32, and c  6  32  3 2 2 . Therefore, the vertices are       V  6 1 and the foci are F  3 2 2  1 .

(c)

y

1 1

   71. (a) 7x 2  6 3x y  13y 2  4 3x  4y  0. Then A  7, B  6 3, and C  13, so the discriminant is   2 6 3  4 7 13  256. Since the discriminant is negative, the equation represents an ellipse.

x

854

CHAPTER 12 Conic Sections   7  13 AC 1      2  60    30 . Therefore, x  23 X  12 Y and y  12 X  23 Y . B 6 3 3 Substituting into the original equation gives   2     1 X  3Y 7 23 X  12 Y  6 3 23 X  12 Y 2 2  2         13 12 X  23 Y  4 3 23 X  12 Y  4 12 X  23 Y  0         7 3X 2  2 3XY  Y 2  3 3 3X 2  3XY  XY  3Y 2 4 2      2 2  6X  2 3Y  2X  2 3Y  0   13 4 X  2 3XY  3Y     y 9  13  8X  Y 2 7  9  39  0  4X 2  8X  16Y 2  0  (c)  X 2 21 4 2 4 4 2 4 1   4 X 2  2X  1  16Y 2  4  X  12  4Y 2  1. This ellipse has a  1,   x 1 b  12 , and c  1  14  12 3. Therefore, the vertices are    V 1  1 0  V1 0 0 and V2 2 0 and the foci are F 1  12 3 0 .

(b) cot 2 

72. (a) 9x 2  24x y  16y 2  25. Then A  9, B  24, and C  16, so the discriminant is 242  4 9 16  0. Since the discriminant is zero, the equation represents a parabola.

(b) cot 2 

AC 9  16 7 7     cos 2   , so cos   B 24 24 25



1725  35 , sin   2

   53 , and thus x  35 X  45 Y and y  45 X  35 Y . Substituting,



1725  45 2 y

(c)

9x 2  24x y  16y 2  25   2      4 X  3 Y  16 4 X  3 Y 2  25  9 35 X  45 Y  24 35 X  45 Y 5 5 5 5

1 x

1

25X 2  25  X 2  1  X  1. Thus the graph is a degenerate conic that consists of two lines. Converting back to x y-coordinates, we see that X  35 x  45 y, so 35 x  45 y  1  3x  4y  5.

73. 5x 2  3y 2  60  3y 2  60  5x 2  y 2  20  53 x 2 . This conic is an ellipse.

5

-5

5 -5

 74. 9x 2  12y 2  36  0  12y 2  9x 2  36  y 2  34 x 2  3  y   34 x 2  3. This conic is a hyperbola.

5

-5

5 -5

CHAPTER 12

75. 6x  y 2  12y  30  y 2  12y  30  6x  y 2  12y  36  66  6x    y  62  66  6x  y  6   66  6x  y  6  66  6x. This conic is

Review

20

a parabola.

10

-10

10

76. 52x 2  72x y  73y 2  100  73y 2  72x y  52x 2  100  0. Using the quadratic formula,

1

   72x 72x2 473 52x 2 100 y  146



 72x

10,000x 2 29,200 146

 10 2  36 73 x  73 73  25x

-2

2



-1

7325x 2

 72x20146

This conic is an ellipse.

1  e  1. Therefore, this is a 1  cos  parabola.

2  e  23 . Therefore, this is an 3  2 sin  ellipse.

77. (a) r 

78. (a) r 

(b)

(b)

( 25 , ¹2) 1

( 21 , ¹)

_1

1

(2, 3¹ 2 ) 4  e  2. Therefore, this is a 1  2 sin  hyperbola.

12  e  4. Therefore, this is a 1  4 cos  hyperbola.

79. (a) r 

80. (a) r 

(b)

(b)

(_4, 3¹ 2) ( 43 , ¹2) 1

(_4, 0)

( 125 , ¹)

1

855

856

CHAPTER 12 Conic Sections

CHAPTER 12 TEST 1. x 2  12y. This is a parabola with 4 p  12  p  3. The focus is 0 3 and the directrix is y  3.

2.

y _1

  y2 x2   1. This is an ellipse with a  4, b  2, and c  16  4  2 3. The 16 4    vertices are 4 0, the foci are 2 3 0 , the length of the major axis is

y

 x2 y2   1. This is a hyperbola with a  3, b  4, and c  9  16  5. The 9 16

y

vertices are 0 3, the foci are 0 5, and the asymptotes are y   34 x.

x

1

x

1

x

1

2a  8, and the length of the minor axis is 2b  4.

3.

1

1

4. The parabola with vertex 0 0 and focus 4 0 has p  4, so 4 p  16. The focus lies to the right of the vertex, so an equation is y 2  16x.

5. The ellipse with foci 3 0 and vertices 4 0 has a  4 and c  3, so c2  a 2  b2  9  16  b2  b2  16  9  7. Thus, an equation is

x2 y2   1. 16 7

6. The hyperbola has foci 0 5 and asymptotes y   34 x. Since the foci are 0 5, c  5, the foci are on a 3 a  a  34 b. Then the y-axis, and the center is 0 0. Also, since y   34 x   x, it follows that  b b 4  2 3 2 2 c2  52  25  a 2  b2  34 b  b2  25 16 b  b  16, and by substitution, a  4 4  3. Therefore, an equation of the hyperbola is

x2 y2   1. 9 16

7. This is a parabola that opens to the left with its vertex at 0 0. So its equation is of the form y 2  4 px with p  0.   Substituting the point 4 2, we have 22  4 p 4  4  16 p  p   14 . So an equation is y 2  4  14 x  y 2  x.

8. This is an ellipse tangent to the x-axis at 0 0 and with one vertex at the point 4 3. The center is 0 3, and a  4 and b  3. Thus the equation is

x2 y  32   1. 16 9

CHAPTER 12

Test

857

9. This a hyperbola with a horizontal transverse axis, vertices at 1 0 and 3 0, and foci at 0 0 and 4 0. Thus the center x  22 y2 1 is 2 0, and a  3  2  1 and c  4  2  2. Thus b2  22  12  3. So an equation is  2 3 1 y2  1. x  22  3     y 10. 16x 2  36y 2  96x  36y  9  0  16 x 2  6x  36 y 2  y  9      16 x 2  6x  9  36 y 2  y  14  9  144  9  1  2 2 y  12  3 x 16 x  32  36 y  12  144    1. This is an 9 4     ellipse with a  3, b  2, and c  9  4  5. The center is 3  12 , the       vertices are 3  3  12  0  12 and 6  12 , and the foci are          h  c k  3  5  12  3  5  12 and 3  5  12 . 

2

x

1

    11. 9x 2  8y 2  36x  64y  164  9 x 2  4x  8 y 2  8y  164      9 x 2  4x  4  8 y 2  8y  16  164  36  128 

y

y  42 x  22   1. This is a hyperbola 9 x  22  8 y  42  72  8 9  with a  2 2, b  3, and c2  a 2  b2  17. The center is 2 4, the foci are       2  17 4 , the vertices are 2  2 2 4 , and the asymptotes are 







1 1

x



y  4   3 4 2 x  2  y  3 4 2 x  4  3 2 2 and y   3 4 2 x  4  3 2 2 .

12. 2x  y 2  8y  8  0  y 2  8y  16  2x  8  16  y  42  2 x  4. This is a parabola with 4 p  2  p   12 . The vertex is 4 4, the focus is     4  12  4  72  4 , and the directrix is x  4  12  92 .

y 1 1

x

13. The ellipse with center 2 0, foci 2 3 and major axis of length 8 has a horizontal major axis with 2a  8  a  4.

y2 x  22   1. 7 16 14. The parabola has focus 2 4 and directrix the x-axis (y  0). Therefore, 2 p  4  0  4  p  2  4 p  8, and the Also, c  3, so b2  a 2  c2  16  9  7. Thus, an equation is

vertex is 2 4  p  2 2. Hence, an equation of the parabola is x  22  8 y  2  x 2  4x  4  8y  16  x 2  4x  8y  20  0.

15. We place the vertex of the parabola at the origin, so the parabola contains the points 3 3, and the equation is of the form   y 2  4 px. Substituting the point 3 3, we get 32  4 p 3  9  12 p  p  34 . So the focus is 34  0 , and we should place the light bulb 34 inch from the vertex.

16. (a) 5x 2  4x y  2y 2  18. Then A  5, B  4, and C  2, so the discriminant is 42  4 5 2  24. Since the discriminant is negative, the equation represents an ellipse.

858

FOCUS ON MODELING

  52 3 AC   . Thus, cos 2  35 and so cos   135 (b) cot 2   255, 2 B 4 4       135 5 sin    5 . It follows that x  2 5 5 X  55 Y and y  55 X  2 5 5 Y . By 2  2      2      5 X  2 5Y  2 5 X  2 5Y  18 substitution, 5 2 5 5 X  55 Y  4 2 5 5 X  55 Y 5 5 5 5      4X 2  4XY  Y 2  45 2X 2  4XY  XY  2Y 2  25 X 2  4XY  4Y 2  18 

      2 2 8 2 1  8  4  18  6X 2  Y 2  18  X  Y  1. This is an X 2 4  85  25  XY 4  12 5  5 Y 5 5 3 18   ellipse with a  3 2 and b  3.    y (c) (d) In XY -coordinates, the vertices are V 0 3 2 . Therefore, in 

1 x

1



x y-coordinates, the vertices are x   3 2 and y  6 2  5 5       3 2 6 2 3 2 6 2  V1     , and x   and y    5 5 5 5      2 . V2 3 2  6 5

Since cos 2  35 we have

5

2  cos1 35  5313 , so   27 . 17. (a) Since the focus of this conic is the origin and the directrix is x  2, the equation has the form

ed . Subsituting e  12 and d  2 we 1  e cos  1 2 get r  r  . 1 2  cos  1  2 cos 

(b) r 

3 r  2  sin  1

3 2 . So e  12 and the 1 sin  2

conic is an ellipse.

r

O

O

1

1

FOCUS ON MODELING Conics in Architecture 1. Answers will vary. 2. (a) The difference P F2   P A  c is a constant (the length of the string). Also, P F1   P A  d is constant. Subtracting these two equations, we find that P F2   P F1   c  d is a constant. This is the definition of a hyperbola. (b) Reflect the entire apparatus through a line perpendicular to F1 F2 .

4. As the lids are twisted more, the vertices of the hyperbolic cross sections get closer together.   5. (a) The tangent line passes though the point a a 2 , so an equation is y  a 2  m x  a.

Conics in Architecture

  (b) Because the tangent line intersects the parabola at only the one point a a 2 , the system



y  a 2  m x  a y  x2

859

has

only one solution, namely x  a, y  a 2 .   y  a 2  m x  a y  a 2  m x  a   a 2  m x  a  x 2  x 2  mx  am  a 2  0. This quadratic (c) 2 yx y  x2   has discriminant m2  4 1 am  a 2  m 2  4am  4a 2  m  2a2 . Setting this equal to 0, we find m  2a. (d) An equation of the tangent line is y  a 2  2a x  a  y  a 2  2ax  2a 2  y  2ax  a 2 .

6. (a) F1 and F2 are points of tangency of the spheres to the plane, and the vertical line through Q 1 and Q 2 is also tangent to each sphere. Since P lies outside of both spheres, we see that P F1  P Q 1 and P F2  P Q 2 . (b) Each sphere is tangent to the enclosing cylinder along a circle in a vertical plane. The distance between these two planes is constant, so P Q 1  P Q 2  Q 1 Q 2 is constant for any choice of P.

(c) Combining the results of parts (a) and (b), we see that P F1  P F2  P Q 1  P Q 2 is constant.

(d) By definition, an ellipse is a curve consisting of all points whose distances to two fixed points F1 and F2 add up to a constant. That is the case here.

13

SEQUENCES AND SERIES

13.1 SEQUENCES AND SUMMATION NOTATION 1. A sequence is a function whose domain is the natural numbers. 2. The nth partial sum of a sequence is the sum of the first n terms of the sequence. So for the sequence an  n 2 the fourth partial sum is S4  12  22  32  42  30.

3. an  n  3. Then a1  1  3  2, a2  2  3  1, a3  3  3  0, a4  4  3  1, and a100  100  3  97.

4. an  2n  1. Then a1  2 1  1  1, a2  2 2  1  3, a3  2 3  1  5, a4  2 4  1  7, and a100  2 100  1  199. 1 1 1 1 1 1 1 1 1 . Then a1   , a2   , a3   , a4   , and 5. an  2n  1 2 1  1 3 2 2  1 5 2 3  1 7 2 4  1 9 1 1 a100   . 2 100  1 201

6. an  n 2  1. Then a1  12  1  0, a2  22  1  3, a3  32  1  8, a4  42  1  15, and a100  1002  1  9999.

7. an  5n . Then a1  51  5, a2  52  25, a3  53  125, a4  54  625, and a100  5100  79  1069 .  n  1  2  3   1 , a  1 4  1 , and . Then a1  1   13 , a2  1  19 , a3  1   27 8. an  1 4 3 3 3 3 3 81  100  3100  19  1048 . a100  1 3

1 1 1 1n 11 12 13 14 . Then a1   1, a2   , a3    , a4   , and 2 2 2 2 4 9 16 n 1 2 3 42 1 1100 .  a100  2 10,000 100 1 1 1 1 1 1 1 1 1 1 , and a100  .  1, a2   , a3   , a4    10. an  2 . Then a1  2 2 2 2 2 4 9 16 10,000 n 1 2 3 4 100 9. an 

11. an  1  1n . Then a1  1  11  0, a2  1  12  2, a3  1  13  0, a4  1  14  2, and a100  1  1100  2.

1 2 3 4 1n1 n 12  1 13  2 14  3 15  4 . Then a1   , a2    , a3   , a4    , and n1 11 2 21 3 31 4 41 5 100 1101  100  . a100  101 101 13. an  n n . Then a1  11  1, a2  22  4, a3  33  27, a4  44  256, and a100  100100  10200 . 12. an 

14. an  3. Then a1  3, a2  3, a3  3, a4  3, and a100  3.   15. an  2 an1  3 and a1  4. Then a2  2 4  3  14, a3  2 14  3  34, a4  2 34  3  74, and a5  2 74  3  154.

1 2 2 1 1 a 24 4  4, a3    , a4  3   , and a5  9   . 16. an  n1 and a1  24. Then a2  6 6 6 3 6 9 6 54 17. an  2an1  1 and a1  1. Then a2  2 1  1  3, a3  2 3  1  7, a4  2 7  1  15, and a5  2 15  1  31. 1 1 1 1 2 1 3 1 5  , a3  and a1  1. Then a2   , a4   , and a5   . 18. an  1  an1 11 2 3 5 8 1 1 1 2 1 3 2

3

5

861

862

CHAPTER 13 Sequences and Series

19. an  an1  an2 , a1  1, and a2  2. Then a3  2  1  3, a4  3  2  5, and a5  5  3  8.

20. an  an1  an2  an3 and a1  1, a2  1, and a3  1. Then a4  1  1  1  3 and a5  3  1  1  5. 21. (a) a1  7, a2  11, a3  15, a4  19, a5  23, a6  27, a7  31, a8  35, a9  39, a10  43 (b)

22. (a) a1  2, a2  6, a3  12, a4  20, a5  30,

a6  42, a7  56, a8  72, a9  90, a10  110

(b) 40

100

20

50

0

0 0

5

10

12 12 23. (a) a1  12 1  12, a2  2  6, a3  3  4, 12 12 12 a4  12 4  3, a5  5 , a6  6  2, a7  7 , 3 12 4 12 6 a8  12 8  2 , a9  9  3 , a10  10  5

0

5

10

24. (a) a1  6, a2  2, a3  6, a4  2, a5  6, a6  2, a7  6, a8  2, a9  6, a10  2

(b)

(b)

5 10 0

5

0

5

10

0 0

5

10

25. (a) a1  2, a2  05, a3  2, a4  05, a5  2,

a6  05, a7  2, a8  05, a9  2, a10  05

(b)

26. (a) a1  1, a2  3, a3  2, a4  1, a5  3,

a6  2, a7  1, a8  3, a9  2, a10  1

(b) 2 1 0 0

5

10

4 2 0 -2 -4

5

10

27. 2, 4, 6, 8,   . All are multiples of 2, so a1  2, a2  2  2, a3  3  2, a4  4  2,   . Thus an  2n.

28. 1, 3, 5, 7,   . All are odd numbers, so a1  2  1, a2  2  2  1, a3  3  2  1, a4  4  2  1,   . Thus an  2n  1. 29. 2, 4, 8, 16,   . All are powers of 2, so a1  2, a2  22 , a3  23 , a4  24 ,   . Thus an  2n .

1 2 1 , 1 ,   . The denominators are all powers of 3, and the terms alternate in sign. Thus a  1 , a  1 , 30.  13 , 19 ,  27 1 2 81 31 32 3 4 n 1 1 1 a3  , a4  ,   . So an  . 3n 33 34 31. 2, 3, 8, 13,   . The difference between any two consecutive terms is 5, so a1  5 1  7, a2  5 2  7, a3  5 3  7, a4  5 4  7,   . Thus, an  5n  7.

32. 7, 4, 1, 2,   . The difference between any two consecutive terms is 3, so a1  3 1  10, a2  3 2  10, a3  3 3  10, a4  3 4  10,   . Thus, an  3n  10.

33. 5, 25, 125, 625,   . These terms are powers of 5, and the terms alternate in sign. So a1  12  51 , a2  13  52 , a3  14  53 , a4  15  54 ,   . Thus an  1n1  5n .

34. 3, 03, 003, 0003,   . The ratio of any two consecutive terms is 10, so a1  3  100 , a2  3  101 , a3  3  102 ,  n1 1 a4  3  103 ,   . Thus an  3  10 .

SECTION 13.1 Sequences and Summation Notation

863

7 , 9 ,   . We consider the numerator separately from the denominator. The numerators of the terms differ 35. 1, 34 , 59 , 16 25 2 1  1 2 2  1 2 3  1 2 4  1 by 2, and the denominators are perfect squares. So a1  , a2  , a3  , a4  , 2 2 2 1 2 3 42 2 5  1 2n  1 ,   . Thus an  . a5  2 5 n2

12 22 32 36. 34 , 45 , 56 , 67 ,   . Both the numerator and the denominator increase by 1, so a1  , a2  , a3  , 13 23 33 42 n2 a4  ,   . Thus an  . 43 n3 37. 0, 2, 0, 2, 0, 2,   . These terms alternate between 0 and 2. So a1  1  1, a2  1  1, a3  1  1, a4  1  1, a5  1  1, a6  1  1,    Thus an  1  1n . n1 38. 1, 12 , 3, 14 , 5, 16 ,   . So a1  1, a2  21 , a3  31 , a4  41 ,   . Thus an  n 1 .

39. a1  1, a2  3, a3  5, a4  7,   . Therefore, an  2n  1. So S1  1, S2  1  3  4, S3  1  3  5  9, S4  1  3  5  7  16, S5  1  3  5  7  9  25, and S6  1  3  5  7  9  11  36. 40. a1  12 , a2  22 , a3  32 , a4  42 ,   . Therefore, an  n 2 . So S1  12  1, S2  122  5, S3  532  59  14, S4  14  42  14  16  30, S5  30  52  30  25  55, and S6  55  62  55  36  91.

1 1 1 1 1 4 1 1 41. a1  13 , a2  2 , a3  3 , a4  4 ,   . Therefore, an  n . So S1  13 , S2  13  2  , S3  13  2  3  13 27 , 3 9 3 3 3 3 3 3 1 1 1 1 1 1 1 1 1 1 1 1 , and S5  13  2  3  4  5  121 , S6  13  2  3  4  5  6  364 S4  13  2  3  4  40 81 243 729 . 3 3 3 3 3 3 3 3 3 3 3 3 42. a1  1, a2  1, a3  1, a4  1,   . Therefore, an  1n . So S1  1, S2  1  1  0, S3  0  1  1, S4  1  1  0, S5  0  1  1, and S6  1  1  0. 2 2 2 2 2 2 2 2 43. an  n . So S1  23 , S2   2  89 , S3  23  2  3  26 , and S4  23  2  3  4  80 27 81 . Therefore, 3 3 3 3 3 3 3 3 n 3 1 Sn  . 3n       1 1  . So S1  12  13 , S2  12  13  13  14  12   13  13  14  12  14 , n1 n2           1 1 S3  2  3  13  14  14  15  12   13  13   14  14  15  12  15 , and               S4  12  13  13  14  14  15  15  16  12   13  13   14  14   15  15  16  12  16 . Therefore,           1  1 1  1  1       1  1 1 1 1 Sn  12  13  13  14      n1  n2 2 3 3 n1 n1  n2  2  n2 .

44. an 

                 n  n  1. So S1  1  2  1  2, S2  1 2  2  3  1   2  2  3  1  3,                    1 2  2 3  3  4  1   2  2   3  3  4  1  4, S3              S4  1 2  2 3  3 4  4 5               1  2 2   3 3   4 4  51 5

45. an 

Therefore,          1 2  2  3   n  n  1 Sn                 1   2  2   3  3     n  n  n  1  1  n  1

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CHAPTER 13 Sequences and Series

 k  log k  log k  1. So S1  log 1  log 2   log 2, S2   log 2  log 2  log 3   log 3, k 1 S3   log 2  log 2  log 3  log 3  log 4   log 2  log 2   log 3  log 3  log 4   log 4, and

46. an  log



S4   log 2  log 2  log 3  log 3  log 4  log 4  log 5   log 2  log 2   log 3  log 3   log 4  log 4  log 5   log 5

Therefore, Sn   log n  1. 4 47. k1 k  1  2  3  4  10 4 48. k1 k 2  1  22  32  42  1  4  9  16  30 3

1 1 1 6 3 2 11 k1 k  1  2  3  6  6  6  6  j 1 2 3 4 99 100  1  1  1  1  1      1  1  0 50. 100 j1 1  1  1  1  1      1  1 49.

51.

52.

8



i i 1 1  1

12



02020202 8

i 4 10  10  10  10  10  10  10  10  10  10  90 53. k1 2k1  20  21  22  23  24  1  2  4  8  16  31  54. i31 i2i  1  21  2  22  3  23  2  8  24  34

5

55. 385 56. 15,550 57. 46,438 58. 0153146 59. 22 60. 0688172 4 3 3 3 3 3 61. k1 k  1  2  3  4  1  8  27  64         4 j 1 0 1 2 3 3 2 15 62.     0   j1 j 1 2 3 4 5 3 2 5          63. 6k0 k  4  4  5  6  7  8  9  10  64. 9k6 k k  3  6  9  7  10  8  11  9  12  54  70  88  108  k 3 4 5 100 65. 100 k3 x  x  x  x      x n j1 x j  12 x  13 x 2  14 x 3      1n1 x n  x  x 2  x 3      1n1 x n 66. j1 1  67. 2  4  6      50  25 k1 2k 10 68. 2  5  8      29  k1 3k  1  2 69. 12  22  32      102  10 k1 k 70.

71. 72. 73. 74. 75.

 1 1 1 1 1 1k       100 k2 k ln k 2 ln 2 3 ln 3 4 ln 4 5 ln 5 100 ln 100 999 1 1 1 1 1       k1 12 23 34 999  1000 k k  1      1 2 3 n  k  2  2      2  nk1 2 12 2 3 n k  k 1  x  x 2  x 3      x 100  100 k0 x  k1  k  x k1 1  2x  3x 2  4x 3  5x 4      100x 99  100 k1 1           2, 2 2, 2 2 2, 2 2 2 2,   . We simplify each term in an attempt to determine a formula for an . So a1  212 ,       a2  2  212  232  234 , a3  2  234  274  278 , a4  2  278  2158  21516 ,   . Thus

n n an  22 12 .

76. G 1  1, G 2  1, G 3  2, G 4  3, G 5  5, G 6  8, G 7  13, G 8  21, G 9  34, G 10  55 77. (a) A1  $2004, A2  $200801, A3  $201202, A4  $201605, A5  $202008, A6  $202412

SECTION 13.1 Sequences and Summation Notation

865

(b) Since 3 years is 36 months, we get A36  $214916. 78. (a) I1  0, I2  $050, I3  $150, I4  $301, I5  $503, I6  $755 (b) Since 5 years is 60 months, we have I60  $97700.

79. (a) P1  35,700, P2  36,414, P3  37,142, P4  37,885, P5  38,643 (b) Since 2014 is 10 years after 2004, P10  42,665.

80. (a) The amount she owes at the end of the month, An , is the amount she owes at the beginning of the month, An1 , plus the interest, 0005An1 , minus the $200 she repay her uncle. Thus An  An1  0005An1  200  An  1005An1  200.

(b) A1  9850, A2  969925, A3  954774, A4  939548, A5  924246, A6  908867. Thus she owes her uncle $908867 after six months.

81. (a) The number of catfish at the end of the month, Pn , is the population at the start of the month, Pn1 , plus the increase in population, 008Pn1 , minus the 300 catfish harvested. Thus Pn  Pn1  008Pn1  300  Pn  108Pn1  300. (b) P1  5100, P2  5208, P3  5325, P4  5451, P5  5587, P6  5734, P7  5892, P8  6064, P9  6249, P10  6449, P11  6665, P12  6898. Thus there should be 6898 catfish in the pond at the end of 12 months.

82. (a) Let n be the years since 2002. So P0  $240,000. Each month the median price of a house in Orange County increases to 106 the previous months. Thus Pn  106n P0 . (b) Since 2010  2002  8. Thus P8  382,52353. Thus in 2010, the median price of a house in Orange County should be $382,524.

83. (a) Let Sn be his salary in the nth year. Then S1  $30,000. Since his salary increase by 2000 each year, Sn  Sn1  2000. Thus S1  $30,000 and Sn  Sn1  2000.

(b) S5  S4  2000  S3  2000  2000  S2  2000  4000  S1  2000  6000  $38,000. 84. (a) Let n be the days after she starts, so C0  4. And each day the concentration increases by 10%. Thus after n days the brine solution is Cn  110Cn1 with C0  4. (b) C8  110C7  110 110C6       1108  C0  1108  4  86. Thus the brine solution is 86 g/L of salt.

85. Let Fn be the number of pairs of rabbits in the nth month. Clearly F1  F2  1. In the nth month each pair that is two or more months old (that is, Fn2 pairs) will add a pair of offspring to the Fn1 pairs already present. Thus Fn  Fn1  Fn2 . So Fn is the Fibonacci sequence. 86. (a) an  n 2 . Then a1  12  1, a2  22  4, a3  32  9, a4  42  16. (b) an  n 2  n  1 n  2 n  3 n  4, a1  12  1  1 1  2 1  3 1  4  1  0 1 2 3  1, a2  22  2  1 2  2 2  3 2  4  4  1  0 1 2  4, a3  32  3  1 3  2 3  3 3  4  9  2  1  0 1  9,

a4  42  4  1 4  2 4  3 4  4  16  3  2  1  0  16.

Hence, the sequences agree in the first four terms. However, for the second sequence, a5  52  5  1 5  2 5  3 5  4  25  4  3  2  1  49, and for the first sequence, a5  52  25, and thus the sequences disagree from the fifth term on. (c) an  n 2  n  1 n  2 n  3 n  4 n  5 n  6 agrees with an  n 2 in the first six terms only. (d) an  2n and bn  2n  n  1 n  2 n  3 n  4.

866

CHAPTER 13 Sequences and Series

a  n if an is even 2 87. an1  3an  1if an is odd

With a1  11, we have a2  34, a3  17, a4  52, a5  26, a6  13, a7  40,

a8  20, a9  10, a10  5, a11  16, a12  8, a13  4, a14  2, a15  1, a16  4, a17  2, a18  1,    (with 4, 2, 1 repeating). So a3n1  4, a3n2  2, and a3n  1, for n  5. With a1  25, we have a2  76, a3  38, a4  19, a5  58, a6  29, a7  88, a8  44, a9  22, a10  11, a11  34, a12  17, a13  52, a14  26, a15  13, a16  40, a17  20, a18  10, a19  5, a20  16, a21  8, a22  4, a23  2, a24  1, a25  4, a26  2, a27  1,    (with 4, 2, 1 repeating). So a3n1  4, a3n2  2, and a3n3  1 for n  7. We conjecture that the sequence will always return to the numbers 4, 2, 1 repeating.

88. an  anan1  anan2 , a1  1, and a2  1. So a3  a31  a31  a2  a2  1  1  2, a4  a42  a41  a2  a3  1  2  3, a5  a53  a52  a2  a3  1  2  3, a6  a63  a63  a3  a3  2  2  4, a7  a74  a73  a3  a4  2  3  5, a8  a85  a84  a3  a4  2  3  5, a9  a95  a95  a4  a4  3  3  6, and a10  a106  a105  a4  a5  3  3  6. The definition of an depends on the values of certain preceding terms. So an is the sum of two preceding terms whose choice depends on the values of an1 and an2 (not on n  1 and n  2).

13.2 ARITHMETIC SEQUENCES 1. An arithmetic sequence is sequence where the difference between successive terms is constant. 2. The sequence an  a  n  1 d is an arithmetic sequence where a is the first term and d is the common difference. So, for the arithmetic sequence an  2  5 n  1 the first term is 2 and the common difference is 5. 3. True. The nth partial sum of an arithmetic sequence is the average of the first and last terms times n.

4. True. If we know the first and second terms of an arithmetic sequence then we can find any other term.

5. (a) a1  7  3 1  1  7, a2  7  3 2  1  10, a3  7  3 3  1  13, a4  7  3 4  1  16,

6. (a) a1  10  20 1  1  10, a2  10  20 2  1  10, a3  10  20 3  1  30,

a5  7  3 5  1  19

a4  10  20 4  1  50,

(b) The common difference is 3. (c)

a5  10  20 5  1  70

an 20

(b) The common difference is 20.

15

(c)

10

60

5 0

an

40 1

2

3

4

5 n

20 0

1

2

3

4

5 n

SECTION 13.2 Arithmetic Sequences

7. (a) a1  6  4 1  1  6,

8. (a) a1  10  4 1  1  10,

a2  6  4 2  1  10,

a2  10  4 2  1  6,

a3  6  4 3  1  14,

a3  10  4 3  1  2,

a4  6  4 4  1  18, a5  6  4 5  1  22 (b) The common difference is 4. (c)

a4  10  4 4  1  2, a5  10  4 5  1  6

(b) The common difference is 4.

an 0

(c) 1

2

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3

4

5 n

an 10

_10

5

_20

0 _5

3 1

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4

5 n

_10

9. (a) a1  52  1  1  52 , a2  52  2  1  32 , a3  52  3  1  12 , a4  52  4  1   12 ,

a3  12 3  1  1, a4  12 4  1  32 ,

a5  52  5  1   32

a5  12 5  1  2 (b) The common difference is 12 .

(b) The common difference is 1. (c)

10. (a) a1  12 1  1  0, a2  12 2  1  12 ,

an

(c)

2

2 4 0 _2

an

1

2

3

1 5 n 0

1

2

3

4

5 n

11. a  9, d  4, an  a  d n  1  9  4 n  1. So a10  9  4 10  1  45. 12. a  5, d  4, an  a  d n  1  5  4 n  1. So a10  5  4 10  1  31.

13. a  07, d  02, an  a  d n  1  07  02 n  1. So a10  07  02 10  1  25. 14. a  14, d   32 , an  a  d n  1  14  32 n  1. So a10  14  32 10  1  12 .

15. a  52 , d   12 , an  a  d n  1  52  12 n  1. So a10  52  12 10  1  2.        16. a  3, d  3, an  a  d n  1  3  3 n  1. So a10  3  3 10  1  10 3.

17. a4  a3  a3  a2  a2  a1  6. The sequence is arithmetic with common difference 6.

18. a4  a3  a3  a2  a2  a1  12. The sequence is arithmetic with common difference 12. 19. Since a3  a2  7 and a4  a3  6, the terms of the sequence do not have a common difference. This sequence is not arithmetic. 20. a4  a3  a3  a2  a2  a1  32. The sequence is arithmetic with common difference 32.

21. Since a2  a1  4  2  2 and a4  a3  16  8  8, the terms of the sequence do not have a common difference. This sequence is not arithmetic. 22. a4  a3  a3  a2  a2  a1  2. This sequence is arithmetic with common difference 2. 3 23. a4  a3   32  0   32 , a3  a2  0    32 , a2  a1  32  3   32 . This sequence is arithmetic with common 2 difference  32 . 8 4 24. a4  a3  ln 16  ln 8  ln 16 8  ln 2, a3  a2  ln 8  ln 4  ln 4  ln 2, a2  a1  ln 4  ln 2  ln 2  ln 2. This sequence is arithmetic with common difference ln 2.

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CHAPTER 13 Sequences and Series

25. a4  a3  77  60  17, a3  a2  60  43  17, 4  a1  43  26  17. This sequence is arithmetic with common difference 17. 1 and a  a  1  1   1 , the terms of the sequence do not have a common difference. 26. Since a4  a3  15  14   20 3 2 4 3 12 This sequence is not arithmetic.

27. a1  4  7 1  11, a2  4  7 2  18, a3  4  7 3  25, a4  4  7 4  32, a5  4  7 5  39. This sequence is arithmetic, the common difference is d  7 and an  4  7n  4  7n  7  7  11  7 n  1. 28. a1  4  21  6, a2  4  22  8, a3  4  23  12, a4  4  24  20, a5  4  25  36. Since a4  a3  8 and a3  a2  4, the terms of the sequence do not have a common difference. This sequence is not arithmetic. 1 1 1 1 1 1 1 1 1 1  , a2   , a3   , a4   , a5   . Since 29. a1  1  2 1 3 1  2 2 5 1  2 3 7 1  2 4 9 1  2 5 11

2 and a  a  1  1   2 , the terms of the sequence do not have a common difference. This a4  a3  19  17   63 3 2 7 3 21 sequence is not arithmetic.

30. a1  1  12  32 , a2  1  22  2, a3  1  32  52 , a4  1  42  3, a5  1  52  72 . This sequence is arithmetic, the common difference is d  12 and an  1  n2  1  12 n  12  12  32  12 n  1. 31. a1  6 1  10  4, a2  6 2  10  2, a3  6 3  10  8, a4  6 4  10  14, a5  6 5  10  20. This sequence is arithmetic, the common difference is d  6 and an  6n  10  6n  6  6  10  4  6 n  1. 32. a1  3  11 1  2, a2  3  12 2  5, a3  3  13 3  0, a4  3  14 4  7,

a5  3  15 5  2. Since a4  a3  7 and a3  a2  5, the terms of the sequence do not have a common difference. This sequence is not arithmetic.

33. 4, 10, 16, 22,   . Then d  a2  a1  10  4  6, a5  a4  6  22  6  28, an  4  6 n  1, and a100  4  6 99  598.

34. 1, 11, 23, 35,   . Then d  a2  a1  11  1  12, a5  a4  12  35  12  47, an  1  12 n  1, and a100  1  12 99  1187.

35. 29, 11, 7, 25,   . Then d  a2  a1  11  29  18, a5  a4  18  25  18  43, an  29  18 n  1, and a100  29  18 99  1753. 36. 64, 49, 34, 19,   . Then d  a2  a1  49  64  15, a5  a4  15  19  15  4, an  64  15 n  1, and a100  64  15 99  1421. 37. 4, 9, 14, 19,   . Then d  a2  a1  9  4  5, a5  a4  5  19  5  24, an  4  5 n  1, and a100  4  5 99  499.

38. 11, 8, 5, 2,   . Then d  a2  a1  8  11  3, a5  a4  3  2  3  1, an  11  3 n  1, and a100  11  3 99  286.

39. 12, 8, 4, 0,   . Then d  a2  a1  8  12  4, a5  a4  4  0  4  4, an  12  4 n  1, and a100  12  4 99  384.

8 10 7 1 1 8 1 19 7 1 40. 76 , 53 , 13 6 , 3 ,   . Then d  a2  a1  6  6  2 , a5  a4  2  3  2  6 , an  6  2 n  1, and  152 a100  76  12 99  7297 6 3 .

41. 25, 265, 28, 295,   . Then d  a2  a1  265  25  15, a5  a4  15  295  15  31, an  25  15 n  1, and a100  25  15 99  1735.

42. 15, 123, 96, 69,   . Then d  a2  a1  123  15  27, a5  a4  27  69  27  42, an  15  27 n  1, and a100  15  27 99  2523.

43. 2, 2  s, 2  2s, 2  3s,   . Then d  a2  a1  2  s  2  s, a5  a4  s  2  3s  s  2  4s, an  2  n  1 s, and a100  2  99s.

44. t, t  3, t  6, t  9,   . Then d  a2  a1  t  3  t  3, a5  a4  3  t  9  3  t  12, an  t  3 n  1, and a100  t  3 99  t  297.

SECTION 13.2 Arithmetic Sequences

869

45. a50  1000 and d  6. Thus, a50  a1  d 50  1  1000  a1  6 50  1  a1  1000  294  706 and a2  706  6  712. 46. a100  750 and d  20. Thus, 750  a1  20 99  a1  1230 and a5  1230  20 4  1150.

  1 . Thus, a  1  8 1   5 and 47. a14  23 and a9  14 , so 23  a1  13d and 14  a1  8d  23  14  5d  d  12 1 4 12 12

5  1 n  1. an   12 12 48. a12  118 and a8  146, so 118  a1  11d and 146  a1  7d  118  146  4d  d  7. Thus, a1  146  7 7  195 and an  195  7 n  1.

49. a1  25 and d  18, so an  601  25  18 n  1  601  33. Thus, 601 is the 33rd term of the sequence.

50. a1  3500 and d  15, so an  2795  2795  3500  15 n  1  n  48. Thus, 2795 is the 48th term of the sequence.

51. a  3, d  5, n  20. Then S20  20 2 2  3  19  5  1010. 52. a  10, d  8, n  30. Then S30  30 2 [2  10  29 8]  3180.

53. a  40, d  14, n  15. Then S15  15 2 [2 40  14  14]  870. 25 54. a  2, d  23, n  25. Then S25  2 2 2  24  23  6850.

55. a1  55, d  12, n  10. Then S10  10 2 2  55  9  12  1090. 56. a2  8, a5  95, n  15. Thus a2  a  d  8 and a5  a  4d  95. Subtracting the first equation from the second gives 3d  15  d  05. Substituting for d in the first equation gives a  05  8  a  75. Thus S15  15 2 2  75  14  05  165.

57. 1  5  9      401 is a partial sum of an arithmetic series with a  1 and d  5  1  4. The last term is

401  an  1  4 n  1, so n  1  100  n  101. So the partial sum is S101  101 2 1  401  101  201  20,301.     3 3 58. 3   2  0  2  3      30 is a partial sum of an arithmetic sequence with a  3 and d   32  3  32 . The

621 last term is 30  an  3  32 n  1, so 22  n 1  n  23. So the partial sum is S23  23 2 3  30  2  3105. 59. 250  233  216      97 is a partial sum of an arithmetic sequence with a  250 and d  233  250  17. The last

10 term is 97  an  250  17 n  1, so n  1  97250 17  9  n  10. So the partial sum is S10  2 250  97  1735. 60. 89  85  81      13 is a partial sum of an arithmetic sequence with a  89 and d  85  89  4. The last term is 20 13  an  89  4 n  1, so n  1  1389 4  19  n  20. So the partial sum is S20  2 89  13  1020. 61. 07  27  47      567 is a partial sum of an arithmetic sequence with a  07 and d  27  07  2. The last term

is 567  an  07  2 n  1  28  n  1  n  29. So the partial sum is S29  29 2 07  567  8323. 62. 10  99  98      01 is a partial sum of an arithmetic sequence with a  10 and d  01. The last term

is 01  an  10  01 n  1, so 99  n  1  n  100. So the partial sum is S100  100 2 10  01  505. 10 63. k0 3  025k is a partial sum of an arithmetic sequence with a  3  025  0  3 and d  025. The last term is a11  3  025  10  55. So the partial sum is S11  11 2 3  55  4675. 20 64. n0 1  2n is a partial sum of an arithmetic sequence where a  1  2  0  1, d  2, and the last term is a21  1  2  20  39. So the partial sum is S21  21 2 1  39  399.

65. We have an arithmetic sequence with a  5 and d  2. We seek n such that 2700  Sn 

n [2a  n  1 d]. Solving for 2

n [10  2 n  1]  5400  10n  2n 2  2n  n 2  4n  2700  0  n  50 n  54  0  2 n  50 or n  54. Since n is a positive integer, 50 terms of the sequence must be added to get 2700. n 66. a  12 and d  8, so Sn  2700  2700  [2  12  n  1  8]  4n 2  8n  2700  0  4 n  27 n  25  0. 2 Thus, 2700 is the sum of the first 25 terms of the sequence. n, we have 2700 

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CHAPTER 13 Sequences and Series

67. Let x denote the length of the side between the length of the other two sides. Then the lengths of the three sides of the triangle are x  a, x, and x  a, for some a  0. Since x  a is the longest side, it is the hypotenuse, and by the Pythagorean Theorem, we know that x  a2  x 2  x  a2  x 2  2ax  a 2  x 2  x 2  2ax  a 2  x 2  4ax  0  x x  4a  0  x  4a (x  0 is not a possible solution). Thus, the lengths of the three sides are x  a  4a  a  3a, x  4a, and x  a  4a  a  5a. The lengths 3a, 4a, 5a are proportional to 3, 4, 5, and so the triangle is similar to a 3-4-5 triangle.

68. P  10110  10210  10310      101910  101231910 . Now, 1  2  3      19 is an arithmetic series with 1  19 a  1, d  1, and n  19. Thus, 1  2  3      19  S19  19  190, and so P  1019010  1019 . 2 69. The sequence 1, 35 , 37 , 13 ,    is harmonic if 1, 53 , 73 , 3,    forms an arithmetic sequence. Since 53  1  73  53  3  73  23 , the sequence of reciprocals is arithmetic and thus the original sequence is harmonic. 70. The two original numbers are 3 and 5. Thus, the reciprocals are 13 and 15 , and their average is     1 1  1  1 5  3  4 . Therefore, the harmonic mean is 15 . 2 3 5 2 15 15 15 4

71. The diminishing values of the computer form an arithmetic sequence with a1  12,500 and common difference d  1875. Thus the value of the computer after 6 years is a7  12,500  7  1 1875  $1250. 72. The number of poles in a layer can be viewed as an arithmetic sequence, where a1  25 and the common difference is 1. The number of poles in the first 12 layers is S12  12 2 [2 25  11 1]  6  39  234.

73. The increasing values of the man’s salary form an arithmetic sequence with a1  30,000 and common difference d  2300. Then his total earnings for a ten-year period are S10  10 2 [2 30,000  9 2300]  403,500. Thus his total earnings for the 10 year period are $403,500.

74. The number of cars that can park in a row can be viewed as an arithmetic sequence, where a1  20 and the common difference is 2. Thus the number of cars that can park in the 21 rows is S21  21 2 [2 20  20 2]  105  80  840. 75. The number of seats in the nth row is given by the nth term of an arithmetic sequence with a1  15 and common n difference d  3. We need to find n such that Sn  870. So we solve 870  Sn  [2 15  n  1 3] for n. We have 2 n 2 2 2 870  27  3n  1740  3n  27n  3n  27n  1740  0  n  9n  580  0  x  20 x  29  0  2 n  20 or n  29. Since the number of rows is positive, the theater must have 20 rows. 76. The sequence is 16, 48, 80,   . This is an arithmetic sequence with a  16 and d  48  16  32. (a) The total distance after 6 seconds is S6  62 32  5  32  3  192  576 ft. (b) The total distance after n seconds is Sn  n2 [32  32 n  1]  16n 2 ft.

77. The number of gifts on the 12th day is 1  2  3  4      12. Since a2  a1  a3  a2  a4  a3      1, the number of gifts on the 12th day is the partial sum of an arithmetic sequence with a  1 and d  1. So the sum is   1  12  6  13  78. S12  12 2 78. (a) We want an arithmetic sequence with4 terms, so let a1  10 and a4  18. Since the sequence is arithmetic,   8  46 are the a4  a1  3d  18  10  8  3d  8  d  83 . Therefore, a2  10  83  38 and a  10  2 3 3 3 3 two arithmetic means between 10 and 18.

(b) We want an arithmetic sequence with5 terms, so let a1  10 and a5  18. Since the sequence is arithmetic, a5  a1  4d  18  10  8  4d  8  d  2. Therefore, a2  10  2  12, a3  10  2 2  14, and a4  10  3 2  16 are the three arithmetic means between 10 and 18.

SECTION 13.3 Geometric Sequences

871

(c) We want an arithmetic sequence with6 terms, with the starting dosage a1  100 and the final dosage a6  300. Since the sequence is arithmetic, a6  a1  5d  300  100  200  5d  200  d  40. Therefore, a2  140, a3  180, a4  220, a5  260, and a6  300. The patient should take 140 mg, then 180 mg, then 220 mg, then 260 mg, and finally arrive at 300 mg.

13.3 GEOMETRIC SEQUENCES 1. A geometric sequence is a sequence where the ratio between successive terms is constant. 2. The sequence an  ar n1 is a geometric sequence where a is the first term and r is the common ratio. So, for the geometric sequence an  2 5n1 the first term is 2 and the common ratio is 5. 3. True. If we know the first and second terms of a geometric sequence then we can find all other terms. 4. (a) The nth partial sum of a geometric sequence an  ar n1 is given by Sn  a

1  rn . 1r

 k1  a  ar  ar 2  ar 3     is is an infinite geometric series. If r  1, then this series (b) The series  k1 ar converges and its sum is S  a 1  r. If r  1 the series diverges.

5. (a) a1  7 30  7, a2  7 31  21,

6. (a) a1  6 050  6, a2  6 051  3,

a3  7 32  63, a4  7 33  189,

a3  6 052  15, a4  6 053  075,

a5  7 34  567

a5  6 054  0375

(b) The common ratio is 3. (c)

(b) The common ratio is 05.

an

an

(c)

600

6 4

400

2

200

0 _2

500 300

2

3

4

5 n

_4

100

0

1

2

3

4

5 n

 0  1 7. (a) a1  52  12  52 , a2  52  12   54 ,

 2  3 5 , a3  52  12  58 , a4  52  12   16  4 5 a5  52  12  32

8. (a) a1  30  1, a2  31  3, a3  32  9, a4  33  27, a5  34  81

(b) The common ratio is 3. (c)

60

an

40

2

0

an 80

(b) The common ratio is  12 . (c)

1

20

1

2

3

4

5 n

_2

9. a  7, r  4. So an  ar n1  7 4n1 and a4  7  43  448. 10. a  3, r  2. So an  ar n1  3 2n1 and a4  3 23  24.  n1  3 5 . 11. a  52 , r   12 . So an  ar n1  52  12 and a4  52   12   16

0

1

2

3

4

5 n

872

CHAPTER 13 Sequences and Series

12. a 

    n1  n   3 3, r  3. So an  ar n1  3 3  3 and a4  3  3  9.

a2 a a 6 12 24  2, and 4   2. Since these ratios are the same, the sequence is geometric with the   2, 3  a1 3 a2 6 a3 12 common ratio 2. a 31 48 93 a  16 and 3   . Since these ratios are not the same, this is not a geometric sequence. 14. 2  a1 3 a2 48 16 a 1536 768 384 1 a 1 a 1 15. 2   , 3   , and 4   . Since these ratios are the same, the sequence is geometric with a1 3072 2 a2 1536 2 a3 768 2 13.

the common ratio 12 . 16.

1 144 a2  ,  a1 432 3

a3 1 a 1 48 16   , and 4    . Since these ratios are the same, the sequence is  a2 144 3 a3 48 3

geometric with the common ratio  13 . 17.

1 a 1 a 1 32 34 38 a2  , 3   , and 4   . Since these ratios are the same, the sequence is geometric with the  a1 3 2 a2 32 2 a3 34 2

common ratio 12 . 18.

1 9 a2  ,  a1 27 3 common ratio  13 .

a3 1 3   , and  a2 9 3

a4 1   . Since these ratios are the same, the sequence is geometric with the a3 3

19.

2 a 4 a2 13 15  and 4   . Since these ratios are not the same, this is not a geometric sequence.  a1 12 3 a3 14 5

20.

a2 e4 a e6 a e8  2  e2 , 3  4  e2 , and 4  6  e2 . Since these ratios are the same, the sequence is geometric with the a1 a2 a3 e e e

common ratio e2 . a 11 121 a  11, 3   11, and 21. 2  a1 10 a2 11 with the common ratio 11.

a4 1331  11. Since these ratios are the same, the sequence is geometric  a3 121

1

22.

1 a2 a 1 3  41  and 4  81  . Since these ratios are not the same, this is not a geometric sequence. a1 2 a3 4 2

6

23. a1  2 31  6, a2  2 32  18, a3  2 33  54, a4  2 34  162, and a5  2 35  486. This sequence is geometric, the common ratio is r  3, and an  a1 r n1  6 3n1 .

24. a1  4  31  7, a2  4  32  13, a3  4  33  31, a4  4  34  85, and a5  4  35  247. Since a2 a3 31  13 7 and a  13 are different ratios, this is not a geometric sequence. a1 2 1 1 1 1 1 1 1 1 1 25. a1  , a2  2  , a3  3  , a4  4  , and a5  5  . This sequence is geometric, the 4 16 64 256 1024 4 4 4 4  n1 common ratio is r  14 and an  a1r n1  14 14 . 26. a1  11 21  2, a2  12 22  4, a3  13 23  8, a4  14 24  16, and

a5  15 25  32. This sequence is geometric, the common ratio is r  2, and an  a1r n1  2 2n1 .         27. Since ln a b  b ln a, we have a1  ln 50  ln 1  0, a2  ln 51  ln 5, a3  ln 52  2 ln 5, a4  ln 53  3 ln 5,   a5  ln 54  4 ln 5. Since a1  0 and a2  0, this sequence is not geometric. a a 4 27 28. a1  11  1, a2  22  4, a3  33  27, a4  44  256, and a5  55  3125. Since 2   4 and 3  are a1 1 a2 4 different, this is not a geometric sequence.

SECTION 13.3 Geometric Sequences

873

a 29. 2, 6, 18, 54,   . Then r  2  62  3, a5  a4  3  54 3  162, and an  2  3n1 . a1 14  n1 a2 3 28 56 2 2 56 2 112 2 30. 7, 14 . 3 , 9 , 27 ,   . Then r  a  7  3 , a5  a4  3  27  3  81 , and an  7 3 1 009 a  03, a5  a4  03  00081 03  000243, and 31. 03, 009, 0027, 00081,   . Then r  2  a1 03 an  03 03n1 .

  n1      a 2   2, a5  a4  2  2 2  2  4, and an  2, 2, 2 2,   . Then r  2  2 . a1 1       1 ,   . Then r  a2  12   1 , a  a   1   1  1  1 , a  144  1 n1 . 33. 144, 12, 1,  12 n 5 4 12 12 12 12 144 12 a1 144   a 2 1 2 1 , and a  8 1 n1 . 34. 8, 2,  12 ,  18 ,   . Then r   14 , a5  a4    18  14   32  n 4 a1 8 4   a 353 35. 3, 353 , 373 , 27,   . Then r  2   323 , a5  a4  323  27  323  3113 , and a1 3 n1   3  32n23  32n13 . an  3 323

32. 1,

t2  n1 a2 t t4 t t5 t2 t3 t4 t t   , and an  t 36. t, , , ,   . Then r   2  , a5  a4   . 2 4 8 a1 t 2 2 8 2 16 2

n1  a s 27 37. 1, s 27 , s 47 , s 67 ,   . Then r  2   s 27 , a5  a4 s 27  s 67 s 27  s 87 , and an  s 27  s 2n27 . a1 1 a 5c1 38. 5, 5c1 , 52c1 , 53c1 ,   . Then r  2   5c , a5  a4  a1 5  n1  5  5cnc  5cnc1 . an  5 5c  3 2 a 6 39. a1  15, a2  6. Thus r  2   and a4  a1r 41  15 25  a1 15 5 12 1 1 1 40. a1  , a   . Thus r   6, so a6  65  648. 12 2 2 112 12

41. a3  

5c  53c1  5c  54c1 , and

24 . 25

1 a a r5 9 and a6  9. Thus, 6  1 2  r 3  r 3   27, so r  3. Therefore, a1 32   13  3 a3 13 a1r

1 and a   1 3  1 . a1   27 2 27 9

a7 8 2 81 32 329 a 12 . Thus,  , so r  . Therefore, a4  r 3 a1  a1  34    r3  r3  9 a4 12 27 3 827 2 r  n1 81 2 and the nth term is a1 r n  . 2 3

42. a4  12 and a7 

9 a 9216 a 18  512  r  8. Therefore, a1  23   and 43. a3  18 and a6  9216. Thus, r 3  6  a3 18 64 32 r 9 an   8n1 . 32 a6 729256 a 54 27 3 3 44. a3  54 and a6  729  384 and   r   . Thus, a1  23  256 . Thus, r  a  54 512 8 r 382 3   a2  384  38  144. a 729  1728, a2  1728 075  1296, and a3  1296 075  972. 45. r  075 and a4  729, so a1  34  r 0753

874

CHAPTER 13 Sequences and Series

 6 a 18 1 . 46. r  16 and a3  18, so a1  23   648 and a7  648 16  2 72 r 16  n1 47. a  1536 and r  12 , so an  ar n1  6  1536 12  1536 21n  log2 6  log2 1536  1  n  n  1  log2 1536 6  1  log2 256  9. Thus, 6 is the ninth term.

n5  468,750  48. a2  30 and a5  3750, so r 3  3750 30  125 and r  5. Thus, an  468,750  3750  5

,750 n  5  log5 468 3750  log5 125  n  8. Therefore, the eighth term of the sequence is 468,750.

1  26  5 63  315. 49. a  5, r  2, n  6. Then S6  5 12     1 1 4   80 2 3 2 81  80 . 50. a  23 , r  13 , n  4. Then S4   3 81 2 3 1 1 3

3

a ar 5 a 2 51. a3  28, a6  224, n  6. So 6  2  r 3 . So we have r 3  6  224 28  8, and hence r  2. Since a3  a  r , we a3 a3 ar a 28 1  26  7 63  441. get a  23  2  7. So S6  7 12 r 2 012 a 000096 a 52. a2  012, a5  000096, n  4. So r 3  5   0008  r  02, and thus a1  2   06. a2 012 r 02 Therefore, S4  06

1  024  07488. 1  02

a 53. 1  3  9      2187 is a partial sum of a geometric sequence, where a  1 and r  2  31  3. Then the last term is a1 2187  an  1  3n1  n  1  log3 2187  7  n  8. So the partial sum is S8  1

1  38  3280. 13

1 1 is a partial sum of a geometric sequence for which a  1 and r  a2   2   1 . The last 54. 1  12  14  18      512 2 a1 1  10  n1 1   12 1 1    341 term is an , where an   1 2 , so n  10. So the partial sum is S10  1 512 . 512 1  1 2

30 a  2. The 55. 15  30  60      960 is a partial sum of a geometric sequence for which a  15 and r  2  a1 15 last term is an  960  15 2n1  n  7, so the partial sum is S7  15

1  27  645. 1  2

a 1 2560 56. 5120  2560  1280      20 is a partial sum of a geometric sequence for which a  5120 and r  2   . a1 5120 2  9  n1 1  12  n  9, so the partial sum is S9  5120  10,220. The last term is an  20  5120 12 1  12 a 57. 125  125  125      12,500,000 is a partial sum of a geometric sequence for which a  125 and r  2  10. The a1 1  108  13,888,88875. 1  10 a 1 58. 10,800  1080  108      0000108 is a partial sum of a geometric sequence for which a  10,800 and r  2  . a1 10  9 1  n1 1  10 1 The last term is an  0000108  10,800 10  n  9, so the partial sum is S9  10,800 11,999999988. 1 1  10 last term is an  12,500,000  125  10n1  n  8, so the partial sum is S8  125

SECTION 13.3 Geometric Sequences

59.

5 

k1

 k1

3 12

3

 5 1  12 1  12

93  16

1  26  105 1  2 k1  5  k1 1  23 5  211 63. 3 23 3  2 27 1 3 k1

61.

6 

5 2k1  5

875

 5  k1 1   32 55    60. 8  32 8 3 2 k1 1  2 5 

1  56  39,060 15 k1  6  k1 1  32 6     1330 64. 64 32  64 k1 1  32

62.

6 

10  5k1  10

1     is an infinite geometric series with a  1 and r  1 . Therefore, it is convergent with sum 65. 1  13  19  27 3 3 a 1    . S  1r 2 1 1 3

66. 1  12  14  18     is an infinite geometric series with a  1 and r   12 . Therefore, it is convergent with sum S

2 1 1     . 3 1 3 1  2 2

1     is an infinite geometric series with a  1 and r   1 . Therefore, it is convergent with sum 67. 1  13  19  27 3

S

3 a 1   .  1 1r 4 1  3

4  8     is an infinite geometric series with a  2 and r  2 . Therefore, it is convergent with sum 68. 25  25 125 5 5 2 5

2

2  53  . 2 3 1 5 5  2  3 69. 1  32  32  32     is an infinite geometric series with a  1 and r  32 1 1 1 1 1 70. 6  8  10  12     is an infinite geometric series with a  6 and r  3 3 3 3 3 S

 1. Therefore, the series diverges. 1 1  . Therefore, it is convergent with 2 9 3

1

sum S 

1 9 1 a 6 3    .   6 1 1r 648 3 8 1 9

71. 3  32  34  38     is an infinite geometric series with a  3 and r   12 . Therefore, it is convergent with sum 3    2. S 1   12

72. 1  1  1  1     is an infinite geometric series with a  1 and r  1. Because r  1  1  1, the series diverges. 73. 3  3 11  3 112  3 113     is an infinite geometric series with a  3 and r  11  1. Therefore, the series diverges. 10

100 3 3 10 3 . Therefore, it is convergent  74.  100 9  3  1  10     is an infinite geometric series with a   9 and r  100 10  9  100  100 a 9 10 1000 9   with sum S     100 9  13   117 . 13 3 1r 1   10 10 1  1     is an infinite geometric series with a  1 and r  1 . Therefore, the sum of the series is 75. 1  12   4 2 2 2 2 2

S

1 2

1  1 2

 1  2  1.  21

876

CHAPTER 13 Sequences and Series

    76. 1  2  2  2 2  4     is an infinite geometric series with a  1 and r   2. Because r  2  1, the series diverges. 7  7  7     is an infinite geometric series with a  7 and r  1 . Thus 77. 0777     10 100 1000 10 10 7 7 a  10 1  . 0777     1r 9 1 10

53  53 53 53 78. 02535353     02  1000 100,000  10,000,000     is an infinite geometric series (after the first term) with a  1000 53

1 . Thus 005353     1000  53  100  53 , and so 02535353     2  53  2  99  53  251 . and r  100 1000 99 990 1 10 990 990 990 1  100

3  3 3 3 1 79. 0030303     100 10,000  1,000,000     is an infinite geometric series with a  100 and r  100 . Thus 3

0030303    

1 3 a  1001   . 1r 99 33 1  100

25 25 80. 211252525     211  1025 ,000  1,000,000  100,000,000     is an infinite geometric series (after 25

25 10,000 1 , and so  the first term) with a  1025 ,000 and r  100 . Thus 000252525     1 9900 1  100 211252525    

25 211  99  25 20,914 10,457 211     . 100 9900 9900 9900 4950

112  112 112 112 81. 0112  0112112112     1000 1,000,000  1,000,000,000     is an infinite geometric series with a  1000 and 112

112 1 . Thus 0112112112     a  10001  . r  1000 1r 999 1  1000

123  123 123 123 1 82. 0123123123     1000 1,000,000  1,000,000,000     is an infinite geometric series with a  1000 and r  1000 .

Thus 0123123123    

123 41 123 1000  .  1 999 333 1  1000

a 83. Since we have 5 terms, let us denote a1  5 and a5  80. Also, 5  r 4 because the sequence is geometric, and so a1 r 4  80 5  16  r  2 . If r  2, the three geometric means are a2  10, a3  20, and a4  40. (If r  2, the three geometric means are a2  10, a3  20, and a4  40, but these are not between 5 and 80.)

84. The sum is given by         a  b  a 2  2b  a 3  3b      a 10  10b  a  a 2  a 3      a 10  b  2b  3b      10b  a

1  a 10 10 1  a 10   b  10b  a   55b 1a 2 1a

a a3 85. (a) 5 3 5 3    is neither arithmetic nor geometric because a2  a1  a3  a2 and 2   . a1 a2 (b) 13  1 53  73     is arithmetic with d  23 , so the next term is a5  73  23  3.     (c) 3 3 3 3 9    is geometric with r  3, so the next term is a5  9 3.

(d) 3  32  0 32     is arithmetic with d  32 , so the next term is a5  32  32  3. 86. (a) 1 1 1 1    is geometric with r  1, so the next term is a5  1.    1 (b) 5 3 5 6 5 1    is geometric with r  516 , so a5  1  516   . 6 5 a (c) 2 1 12  2 is neither arithmetic nor geometric because a2  a1  a3  a2 and 2  a1

a3 . a2

SECTION 13.3 Geometric Sequences

877

(d) x  1 x x  1 x  2 is arithmetic with d  1, so a5  x  2  1  x  3. 87. (a) The value at the end of the year is equal to the value at beginning less the depreciation, so Vn  Vn1  02Vn1  08Vn1 with V1  160,000. Thus Vn  160,000  08n1 .

(b) Vn  100,000  08n1  160,000  100,000  08n1  0625  n  1 log 08  log 0625  log 0625  211. Thus it will depreciate to below $100,000 during the fourth year. n1 log 08 88. Let an denote the number of ancestors a person has n generations back. Then a1  2, a2  4, a3  8,   . Since 4  8      2, this is a geometric sequence with r  2. Therefore, a  2  214  215  32,768. 15 2 4 89. Since the ball is dropped from a height of 80 feet, a  80 . Also since the ball rebounds three-fourths of the distance  n fallen, r  34 . So on the nth bounce, the ball attains a height of an  80 34 . Hence, on the fifth bounce, the ball goes  5 a5  80 34  80243 1024  19 ft high.

90. a  5000, r  108. After 1 hour, there are 5000  108  5400, after 2 hours, 5400  108  5832, after 3 hours, 5832  108  629856, after 4 hours, 629856  108  68024448, and after 5 hours, 68024448  108  7347 bacteria. After n hours, the number of bacteria is an  5000  108n .

91. Let an be the amount of water remaining at the nth stage. We start with 5 gallons, so a  5. When 1 gallon (that is, 15 of the mixture) is removed, 45 of the mixture (and hence 45 of the water in the mixture) remains. Thus, a1  5  45  a2  5  45  45    ,  n  3 and in general, an  5 45 . The amount of water remaining after 3 repetitions is a3  5 45  64 25 , and after  5 5 repetitions it is a5  5 45  1024 625 .

92. Let aC denote the term of the geometric series that is the frequency of middle C. Then aC  256 and aC1  512. Since aC 256 this is a geometric sequence, r  512 256  2, and so aC2  r 2  22  64.

93. Let an be the height the ball reaches on the nth bounce. From the given information, an is the geometric sequence  n an  9  13 . (Notice that the ball hits the ground for the fifth time after the fourth bounce.)  2  3  4 (a) a0  9, a1  9  13  3, a2  9  13  1, a3  9  13  13 , and a4  9  13  19 . The total distance traveled is 8 a0  2a1  2a2  2a3  2a4  9  2  3  2  1  2  13  2  19  161 9  17 9 ft. (b) The total distance traveled at the instant the ball hits the ground for the nth time is  2  3  4  n1 Dn  9  2  9  13  2  9  13  2  9  13  2  9  13      2  9  13   2  3  4  n1  1 1 1 1 9  2 9  9  3  9  3  9  3  9  3      9  13



 n   n3   n  1  13   1 1  2 9   9  18   9  27 1   3 3 1  13

1  2n  2n  1. At the end of 30 days, she 12 will have S30  230  1  1,073,741,823 cents  $10,737,41823. To become a billionaire, we want 2n  1  1011 or 11 approximately 2n  1011 . So log 2n  log 1011  n   365. Thus it will take 37 days. log 2

94. We have a geometric sequence with a  1 and r  2. Then Sn  1

878

CHAPTER 13 Sequences and Series

95. Let a1  1 be the man with 7 wives. Also, let a2  7 (the wives), a3  7a2  72 (the sacks), a4  7a3  73 (the cats), and a5  7a4  74 (the kits). The total is a1  a2  a3  a4  a5  1  7  72  73  74 , which is a partial sum of a geometric sequence with a  1 and r  7. Thus, the number in the party is S5  1 

96. (a) (b)

10

 k1

1 k1 50 2

 50 

 k1 1  k1 50 2



 10 1  12 1  12

1  75  2801. 17

 100 1  00009765  999023 mg

50  100 mg 1  12

97. Let an be the height the ball reaches on the nth bounce. We have a0  1 and an  12 an1 . Since the total distance d traveled includes the bounce up as well and the distance down, we have  2    3  4 d  a0  2  a1  2  a2      1  2 12  2 12  2 12  2 12       i  2  3  1 1  1  1  12  12  12      1  1 3 2 1 1  i0 2

Thus the total distance traveled is about 3 m.  2 98. The time required for the ball to stop bouncing is t  1  1  1     which is an infinite geometric series with 2 2      2 2 1 2 2 21     2  2. Thus the  a  1 and r  1 . The sum of this series is t     2 21 21 21 21 1  1 2  time required for the ball to stop is 2  2 341 s. 99. (a) If a square has side x, then by the Pythagorean Theorem the length of the side of the square formed by    x2 x x2 x 2  x 2 joining the midpoints is,    . In our case, x  1 and the side of the   2 2 4 4 2  2 1 first inscribed square is  , the side of the second inscribed square is 1  1  1 , the side of the 2 2 2 2  3 third inscribed square is 1 , and so on. Since this pattern continues, the total area of all the squares is 2  2  4  6  2  3 1  1  1      1  12  12  12       2. A  12  1 2 2 2 1  12  2  3 (b) As in part (a), the sides of the squares are 1, 1 , 1 , 1 ,   . Thus the sum of the perimeters is 2 2 2  2  3 1 1 1 4      , which is an infinite geometric series with a  4 and r  1 . Thus S  414  4  2 2 2 2      21 4 2 424 2 4 2 4     8  4 2.  the sum of the perimeters is S  21 21 21 21 1  1 2

 2 2 100. Let An be the area of the disks of paper placed at the nth stage. Then A1   R 2 , A2  2   12 R   2R ,  2 1 1 2 2 2 2 A3  4   14 R   4 R ,   . We see from this pattern that the total area is A   R  2  R  4  R     . Thus,  R2  2 R 2 . the total area, A, is an infinite geometric series with a1   R 2 and r  12 . So A  1  12

SECTION 13.4 Mathematics of Finance

879

101. Let an denote the area colored blue at nth stage. Since only the middle squares are colored blue, an  19  area remaining yellow at the n  1 th stage. Also, the area remaining yellow at the nth stage is 89 of the area    2  3 remaining yellow at the preceding stage. So a1  19 , a2  19 89 , a3  19 89 , a4  19 89 ,   . Thus the total area    2  3 colored blue A  19  19 89  19 89  19 89     is an infinite geometric series with a  19 and r  89 . So the total area is A 

1 9

 1. 1  89

102. Let a1 , a2 , a3 ,    be a geometric sequence with common ratio r. Thus a2  a1 r, a3  a1  r 2 ,   , an  a1  r n1 . Hence,    2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 n1 1 1    ,      ,        , and so a2 a1  r a1 r a3 a1 r 2 a1 r an a1 r n1 a1 r a1  r 2 a1  r n1 1 1 1 1 , , ,    is a geometric sequence with common ratio . a1 a2 a3 r 103. a1  a2  a3     is a geometric sequence with common ratio r . Thus a2  a1 r, a3  a1  r 2 ,   , an  a1  r n1 .     Hence log a2  log a1r  log a1  log r, log a3  log a1  r 2  log a1  log r 2  log a1  2 log r,   ,     log an  log a1  r n1  log a1  log r n1  log a1  n  1 log r , and so log a1  log a2  log a3     is an arithmetic

sequence with common difference log r. 104. Since a1  a2  a3     is an arithmetic sequence with common difference d, the terms can be expressed as a2  a1  d, 2  a3  a1  2d,   , an  a1  n  1 d. So 10a2  10a1 d  10a1  10d , 10a3  10a1 2d  10a1  10d ,   , n1  , and so 10a1  10a2  10a3     is a geometric sequence with common ratio r  10d . 10an  10a1 n1d  10a1  10d

13.4 MATHEMATICS OF FINANCE 1. An annuity is a sum of money that is paid in regular equal payments. The amount of an annuity is the sum of all the individual payments together with all the interest. 2. The present value of an annuity is the amount that must be invested now at interest rate i per time period in order to provide n payments each of amount R. 1  00610  1 1  in  1  1000  $13,18079. i 006 008 1006666724  1 1  in  1 R  500, n  24, i   00066667. So A f  R  500  $12,96659. 12 i 00066667 1  01220  1 1  in  1 n  20, R  $5000, i  012. So A f  R  5000  $360,26221. i 012 006 1  in  1 10320  1 R  500, n  20, i   003. So A f  R  500  $13,43519. 2 i 003 008 1  00216  1 1  in  1 n  16, R  $300, i   002. So A f  R  300  $5,59179. 4 i 002 1  00540  1 1  in  1 n  40, R  2000, i  005. So A f  R  2000  $241,59955. i 005 iAf 010 0025 5000 A f  5000, n  4  2  8, i   0025. So R    $57234. n 4 1  i  1 10258  1 iAf 006 0005 2000  0005, n  8. Then R    $24566. A f  $2000, i  12 1  in  1 1  00058  1

3. n  10, R  $1000, i  006. So A f  R 4. 5. 6. 7. 8. 9. 10.

880

CHAPTER 13 Sequences and Series

1  104520 009 1  1  in  0045. So A p  R  1000  $13,00794. 2 i 0045  30 1  1  008 008 1  1  in 12 12. R  300, n  30, i  . So A p  R  300  $813265. 008 12 i 11. R  1000, n  20, i 

12

13. R  $200, n  20, i 

1  1  004520 009 1  1  in  0045. So A p  R  200  $260159. 2 i 0045

14. A p  50,000, n  10 2  20, i  15. A p  $12 000, i 

i Ap 008 004 50,000  004. So R    $367909. n 2 1  1  i 1  10420

i Ap 0105 000875 12000  000875, n  48. Then R    $30724. n 12 1  1  i 1  1  00087548

009  00075. Over a 30 year period, n  30 12  360, and the monthly payment is 12 i Ap 00075 80,000 R   $64370. Over a 15 year period, n  15  12  180, and so the monthly 1  1  in 1  10075360 80,000  00075  $81141. payment is R  1  10075180

16. A p  80,000, i 

i Ap 008 0006667 100,000  0006667, n  360. Then R    $73376. n 12 1  1  i 1  1  0006667360 Therefore, the total amount paid on this loan over the 30 year period is 360 73376  $264,15360.

17. A p  $100,000, i 

i Ap 006 0005 200,000  0005, n  180. Then R    $168771. Therefore, 12 1  1  in 1  1  0005180 the total amount paid on this loan over the 15 year period is 180 168771  $303,78780.

18. A p  $200,000, i 

1  1005360 1  1  in 006  0005. So A p  R  3500  $583,770.65. 12 i 0005 Therefore, Dr. Gupta can afford a loan of $583,770.65.

19. R  3500, n  12 30  360, i 

009 1  10075360 1  1  in  00075. So A p  R  650  $80,78321. 12 i 00075 Therefore, the couple can afford a loan of $80,78321.

20. R  650, n  12 30  360, i 

008  000667. The amount borrowed is 12 1  10066736 1  1  in Ap  R  220  $7,02060. So she purchased the car for i 000667 $7,02060  $2000  $902060.

21. R  220, n  12 3  36, i 

22. R  $30, i 

1  1  000833312 1  1  in 010  0008333, n  12. Then A p  R  30  $34124. 12 i 0008333

00975  0008125. 23. A p  100,000, n  360, i  12 i Ap 0008125 100,000 (a) R    $85915. 1  1  in 1  1  0008125360

(b) The total amount that will be paid over the 30 year period is 360 85915  $309,29400. (c) R  $85915, i 

00975 1  0008125360  1  0008125, n  360. So A f  85915  $1,841,51929. 12 0008125

SECTION 13.4 Mathematics of Finance

24. (a) For the 30-year mortgage, A p  300,000, n  12 30  360, and i  0065 12 , so R30 

i Ap  1  1  in

0065 300,000 12 360  $189620.  1  1  0065 12

For the 15-year mortgage, A p  300,000, n  12 15  180, and i  00575 12 , so R15 

i Ap  1  1  in

monthly payment.

00575 300,000 12 180  $249123. The 15-year mortgage has the larger  1  1  00575 12

(b) The total amount of the 30-year mortgage is 360 189620  $682,63200, and the total amount of the 15-year mortgage is 180 249123  $448,42140. The 15-year mortgage has the lower total payment.

25. A p  $640, R  $32, n  24. We want to solve the equation R 

i Ap 1  1  in

x . So we can express R for the interest rate i. Let x be the interest rate, then i  12 x  640 12 as a function of x by R x  . We graph R x and y  32 in  x 24 1 1 12 the rectangle [012 022]  [30 34]. The x-coordinate of the intersection is about

34 32 30 0.15

0.20

01816, which corresponds to an interest rate of 1816%.

26. A p  $12,500, R  $420, n  36. We want to solve for the interest rate using the

i AP x . So we can . Let x be the interest rate, then i  n 12 1  1  i x  12,500 12 . We graph express R as a function of x as follows: R x   x 36 1 1 12 R x and y  420 in the rectangle [012 013] by [415 425]. The x-coordinate of equation R 

425 420 415 0.120

0.125

the intersection is about 01280, which corresponds to an interest rate of 1280%.

27. A p  $18999, R  $1050, n  20. We want to solve the equation

i Ap x . So for the interest rate i. Let x be the interest rate, then i  1  1  in 12 x  18999 12 we can express R as a function of x by R x  . We graph  x 20 1 1 12 R x and y  1050 in the rectangle [010 018]  [10 11]. The x-coordinate of R

the intersection is about 01168, which corresponds to an interest rate of 1168%.

11.0 10.5 10.0 0.10

0.15

0.130

881

882

CHAPTER 13 Sequences and Series

28. A p  $2000  $200  $1800, R  $88, n  24. We want to solve for the interest

89

i AP x . . Let x be the interest rate, then i  12 1  1  in x  1800 12 . So we can express R as a function of x as follows: R x   x 24 1 1 12 [014 [87 We graph R x and y  88 in the viewing rectangle 016] by 89]. The rate using the equation R 

88 87 0.14

0.15

0.16

x-coordinate of the intersection is about 01584, which corresponds to an interest

rate of 1584%.

R . The present value of an annuity is the sum of 1  ik the present values of each of the payments of R dollars, as shown in the time line.

29. (a) The present value of the kth payment is PV  R 1  ik  Time

1

2

3

4

Payment

R

R

R

R

¤¤¤

n-1

n

R

R

Present value R/(1+i) R/(1+i)@ R/(1+i)# R/(1+i)$ R/(1+i)n-1 R/(1+i)n

(b)

R R R R      2 3 1i 1  in 1  i 1  i      2  n1  R 1 R 1 1 R R       1i 1i 1i 1i 1i 1i 1i 1 R 1  rn and r  . Since Sn  a , we have This is a geometric series with a  1i 1i 1r  n 1  1  1  1  in 1  1  in 1  1  in R 1  i    R   R  Ap  . R 1 1 1i i 1  i  1 1 1  i 1  1i 1i Ap



R . The amount of money to be invested now (A p ) to ensure an 1  ik annuity in perpetuity is the (infinite) sum of the present values of each of the payments, as shown in the time line.

30. (a) The present value of the kth payment is P V  Time

1

2

3

4

Payment

R

R

R

R

Present value R/(1+i) R/(1+i)@ R/(1+i)# R/(1+i)$ R/(1+i)n-1 R/(1+i)n

¤¤¤

n-1

n

R

R

¤¤¤

SECTION 13.5 Mathematical Induction

883

R R R R R  and         . This is an infinite geometric series with a  n 2 3 1i 1i 1  i 1  i 1  i R 1 1  i  R  1i  R. . Therefore, A p  r 1 1i 1i i i 1 1i 5000 R (c) Using the result from part (b), we have R  5000 and i  010. Then A p    $50,000. i 010 (d) We are given two different time periods: the interest is compounded quarterly, while the annuity is paid yearly. In order to use the formula in part (b), we need to find the effective annual interest rate produced by 8% interest compounded quarterly, which is 1  inn  1. Now i  8% and it is compounded quarterly, n  4, so the effective

(b) A p 

annual yield is 1  0024  1  008243216. Thus by the formula in part (b), the amount that must be invested is 3000 A p  008243216  $36,39356.

31. (a) Using the hint, we calculate the present value of the remaining 240 payments with R  72417, i  00075, and n  240. Since A p  R

1  10075240 1  1  in  72417  80,48784, they still owe $80,48784 on their i 00075

mortgage. (b) On their next payment, 00075 80,48784  $60366 is interest and $72417  60366  $12051 goes toward the principal.

13.5 MATHEMATICAL INDUCTION 1. Mathematical induction is a method of proving that a statement P n is true for all natural numbers n. In Step 1 we prove that P 1 is true. 2. (b) We prove “if P kis true then P k  1 is true.” 3. Let P n denote the statement 2  4  6      2n  n n  1. Step 1: P 1 is the statement that 2  1 1  1, which is true. Step 2: Assume that P k is true; that is, 2  4  6      2k  k k  1. We want to use this to show that P k  1 is true. Now 2  4  6      2k  2 k  1  k k  1  2 k  1

induction hypothesis

 k  1 k  2  k  1 [k  1  1] Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 4. Let P n denote the statement 1  4  7  10      3n  2 

n 3n  1 . 2

1 [3 1  1] 12  , which is true. 2 2 k 3k  1 Step 2: Assume that P k is true; that is. 1  4  7      3k  2  . We want to use this to show that 2 P k  1 is true. Now k 3k  1  3k  1 induction hypothesis 1  4  7  10      3k  2  [3 k  1  2]  2 2 2 3k  k  6k  2 3k  5k  2 k 3k  1 6k  2 k  1 3k  2 k  1 [3 k  1  1]       2 2 2 2 2 2 Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. Step 1: P 1 is the statement that 1 

884

CHAPTER 13 Sequences and Series

n 3n  7 . 2 1  3  1  7 Step 1: We need to show that P 1 is true. But P 1 says that 5  , which is true. 2 k 3k  7 Step 2: Assume that P k is true; that is, 5  8  11      3k  2  . We want to use this to show that 2 P k  1 is true. Now

5. Let P n denote the statement 5  8  11      3n  2 

5  8  11      3k  2  [3 k  1  2] 

k 3k  7  3k  5 2

induction hypothesis

3k 2  7k 6k  10 3k 2  13k  10   2 2 2 3k  10 k  1 k  1 [3 k  1  7]   2 2 

Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 6. Let P n denote the statement 12  22  32      n 2  Step 1: P 1 is the statement that 12 

n n  1 2n  1 . 6

123 , which is true. 6

Step 2: Assume that P k is true; that is, 12  22  32      k 2 

k k  1 2k  1 . We want to use this to show that 6

P k  1 is true. Now, k k  1 2k  1  k  12 induction hypothesis 6     2k 2  k  6k  6 k 2k  1  6 k  1  k  1  k  1 6 6   2k 2  7k  6 k  1 k  2 2k  3   k  1 6 6

12  22  32      k 2  k  12 



k  1 [k  1  1] [2 k  1  1]  6

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 7. Let P n denote the statement 1  2  2  3  3  4      n n  1  Step 1: P 1 is the statement that 1  2 

n n  1 n  2 . 3

1  1  1  1  2 , which is true. 3

Step 2: Assume that P k is true; that is, 1  2  2  3  3  4      k k  1 

k k  1 k  2 . We want to use this to 3

show that P k  1 is true. Now 1  2  2  3  3  4      k k  1  k  1 [k  1  1] k k  1 k  2  k  1 k  2 3 k k  1 k  2 3 k  1 k  2 k  1 k  2 k  3    3 3 3 

Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

induction hypothesis

SECTION 13.5 Mathematical Induction

8. Let P n denote the statement 1  3  2  4  3  5      n n  2  Step 1: P 1 is the statement that 1  3 

885

n n  1 2n  7 . 6

129 , which is true. 6

Step 2: Assume that P k is true; that is, 1  3  2  4  3  5      k k  2 

k k  1 2k  7 . We want to use this to 6

show that P k  1 is true. Now 1  3  2  4  3  5      k k  2  k  1 [k  1  2] k k  1 2k  7  k  1 k  3 6     2k 2  7k  6k  18 k 2k  7 6 k  3   k  1  k  1 6 6 6 



induction hypothesis

k  1 [k  1  1] [2 k  1  7] 6

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

9. Let P n denote the statement 13  23  33      n 3 

n 2 n  12 . 4

12  1  12 , which is clearly true. 4 k 2 k  12 Step 2: Assume that P k is true; that is, 13  23  33      k 3  . We want to use this to show that 4 P k  1 is true. Now

Step 1: P 1 is the statement that 13 

13  23  33      k 3  k  13   

k 2 k  12  k  13 4   k  12 k 2  4 k  1 4

k  12 k  22

induction hypothesis



  k  12 k 2  4k  4 4

k  12 [k  1  1]2

 4 4 Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

  10. Let P n denote the statement 13  33  53      2n  13  n 2 2n 2  1 .   Step 1: P 1 is the statement that 13  12 2  12  1 , which is clearly true.   Step 2: Assume that P k is true; that is, 13  33  53      2k  13  k 2 2k 2  1 . We want to use this to show that P k  1 is true. Now

  13  33  53      2k  13  2k  13  k 2 2k 2  1  2k  13

induction hypothesis

 2k 4  k 2  8k 3  12k 2  6k  1  2k 4  8k 3  11k 2  6k  1       k 2  2k  1 2k 2  4k  1  k  12 2 k  12  1

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

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CHAPTER 13 Sequences and Series

11. Let P n denote the statement 23  43  63      2n3  2n 2 n  12 . Step 1: P 1 is true since 23  2 12 1  12  2  4  8. Step 2: Assume that P k is true; that is, 23  43  63      2k3  2k 2 k  12 . We want to use this to show that P k  1 is true. Now 23  43  63      2k3  [2 k  1]3  2k 2 k  12  [2 k  1]3

induction hypothesis

   2k 2 k  12  8 k  1 k  12  k  12 2k 2  8k  8  2 k  12 k  22  2 k  12 [k  1  1]2

Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

1 1 1 n 1       . 12 23 34 n n  1 n1 1  12 , which is clearly true. Step 1: P 1 is the statement that 12 1 1 1 k 1     . We want to use this to show that Step 2: Assume that P k is true; that is 12 23 34 k k  1 k 1 P k  1 is true. Now

12. Let P n denote the statement

1 1 1 1 k 1        12 23 k k  1 k  1 k  2 k  1 k  1 k  2  

induction hypothesis

k k  2  1 k 2  2k  1 k  12   k  1 k  2 k  1 k  2 k  1 k  2 k 1 k 1  k 2 k  1  1

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.   13. Let P n denote the statement 1  2  2  22  3  23  4  24      n  2n  2 1  n  1 2n . Step 1: P 1 is the statement that 1  2  2 [1  0], which is clearly true.

  Step 2: Assume that P k is true; that is, 1  2  2  22  3  23  4  24      k  2k  2 1  k  1 2k . We want to use

this to show that P k  1 is true. Now

1  2  2  22  3  23  4  24      k  2k  k  1  2k1    2 1  k  1 2k  k  1  2k1

induction hypothesis

     2 1  k  1  2k  k  1  2k  2 1  2k  2k      2 1  k  2k1  2 1  [k  1  1] 2k1

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 14. Let P n denote the statement 1  2  22      2n1  2n  1.

Step 1: P 1 is the statement that 1  21  1, which is clearly true.

Step 2: Assume that P k is true; that is, 1  2  22      2k1  2k  1. We want to use this to show that P k  1 is true. Now

1  2  22      2k1  2k  2k  1  2k  2  2k  1  2k1  1 Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

induction hypothesis

SECTION 13.5 Mathematical Induction

887

15. Let P n denote the statement n 2  n is divisible by 2.

Step 1: P 1 is the statement that 12  1  2 is divisible by 2, which is clearly true.

Step 2: Assume that P k is true; that is, k 2  k is divisible by 2. Now     k  12  k  1  k 2  2k  1  k  1  k 2  k  2k  2  k 2  k  2 k  1. By the induction hypothesis, k 2  k is divisible by 2, and clearly 2 k  1 is divisible by 2. Thus, the sum is divisible by 2, so P k  1 is true. Therefore, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

16. Let P n denote the statement that 5n  1 is divisible by 4.

Step 1: P 1 is the statement that 51  1  4 is divisible by 4, which is clearly true.

Step 2: Assume that P k is true; that is, 5k  1 is divisible by 4. We want to use this to show that P k  1 is true. Now,     5k1  1  5  5k  1  5  5k  5  4  5 5k  1  4 which is divisible by 4 since 5 5k  1 is divisible by 4 by the induction hypothesis. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

17. Let P n denote the statement that n 2  n  41 is odd.

Step 1: P 1 is the statement that 12  1  41  41 is odd, which is clearly true.

Step 2: Assume that P k is true; that is, k 2  k  41 is odd. We want to use this to show that P k  1 is true. Now,   k  12  k  1  41  k 2  2k  1  k  1  41  k 2  k  41  2k, which is also odd because k 2  k  41 is odd by the induction hypothesis, 2k is always even, and an odd number plus an even number is always odd. Therefore, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 18. Let P n denote the statement that n 3  n  3 is divisible by 3.

Step 1: P 1 is the statement that 13  1  3  3 is divisible by 3, which is true.

Step 2: Assume that P k is true; that is, k 3  k  3 is divisible by 3. We want to use this to show that P k  1 is true.     Now, k  13  k  1  3  k 3  3k 2  3k  1  k  1  3  k 3  k  3  3k 2  3k  k 3  k  3  3 k 2  k ,   which is divisible by 3, since k 3  k  3 is divisible by 3 by the induction hypothesis, and 3 k 2  k is divisible by 3. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

19. Let P n denote the statement that 8n  3n is divisible by 5.

Step 1: P 1 is the statement that 81  31  5 is divisible by 5, which is clearly true.

Step 2: Assume that P k is true; that is, 8k  3k is divisible by 5. We want to use this to show that P k  1 is true. Now,   8k1  3k1  8  8k  3  3k  8  8k  8  5  3k  8  8k  3k  5  3k , which is divisible by 5 because 8k  3k

is divisible by 5 by our induction hypothesis, and 5  3k is divisible by 5. Thus P k  1 follows from P k. So by the

Principle of Mathematical Induction, P n is true for all n.

20. Let P n denote the statement that 32n  1 is divisible by 8.

Step 1: P 1 is the statement that 32  1  8 is divisible by 8, which is clearly true.

Step 2: Assume that P k is true; that is, 32k  1 is divisible by 8. We want to use this to show that P k  1 is true. Now,   32k1  1  9  32k  1  9  32k  9  8  9 32k  1  8, which is divisible by 8, since 32k  1 is divisible by 8 by the induction hypothesis. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true

for all n.

888

CHAPTER 13 Sequences and Series

21. Let P n denote the statement n  2n . Step 1: P 1 is the statement that 1  21  2, which is clearly true.

Step 2: Assume that P k is true; that is, k  2k . We want to use this to show that P k  1 is true. Adding 1 to both sides

of P k we have k  1  2k  1. Since 1  2k for k  1, we have 2k  1  2k  2k  2  2k  2k1 . Thus k  1  2k1 ,

which is exactly P k  1. Therefore, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

22. Let P n denote the statement n  12  2n 2 , for all n  3.

Step 1: P 3 is the statement that 3  12  2  32 or 16  18, which is true.

Step 2: Assume that P k is true; that is, k  12  2k 2 , k  3. We want to use this to show that P k  1 is true. Now   k  22  k 2  4k  4  k 2  2k  1  2k  3  k  12  2k  3  2k 2  2k  3

induction hypothesis

 2k 2  2k  3  2k  1

because 2k  1  0 for k  3

 2k 2  4k  2  2 k  12 Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n  3. 23. Let P n denote the statement 1  xn  1  nx, if x  1.

Step 1: P 1 is the statement that 1  x1  1  1x, which is clearly true.

Step 2: Assume that P k is true; that is, 1  xk  1  kx. Now, 1  xk1  1  x 1  xk  1  x 1  kx,

by the induction hypothesis. Since 1  x 1  kx  1  k  1 x  kx 2  1  k  1 x (since kx 2  0), we have

1  xk1  1  k  1 x, which is P k  1. Thus P k  1 follows from P k. So the Principle of Mathematical Induction, P n is true for all n. 24. Let P n denote the statement 100n  n 2 , for all n  100.

Step 1: P 100 is the statement that 100 100  1002 , which is true.

Step 2: Assume that P k is true; that is, 100k  k 2 . We want to use this to show that P k  1 is true. Now 100 k  1  100k  100  k 2  100  k 2  2k  1  k  12

induction hypothesis because 2k  1  100 for k  100

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n  100. 25. Let P n be the statement that an  5  3n1 .

Step 1: P 1 is the statement that a1  5  30  5, which is true.

Step 2: Assume that P k is true; that is, ak  5  3k1 . We want to use this to show that P k  1 is true. Now,     ak1  3ak  3  5  3k1 , by the induction hypothesis. Therefore, ak1  3  5  3k1  5  3k , which is exactly P k  1. Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

26. an1  3an  8 and a1  4. Then a2  3  4  8  4, a3  3  4  8  4, a4  3  4  8  4,   , and the conjecture is that an  4. Let P n denote the statement that an  4. Step 1: P 1 is the statement that a1  4, which is true. Step 2: Assume that P k is true; that is, ak  4. We want to use this to show that P k  1 is true. Now, ak1  3  ak  8  3  4  8  4, by the induction hypothesis. This is exactly P k  1, so by the Principle of Mathematical Induction, P n is true for all n.

SECTION 13.5 Mathematical Induction

889

27. Let P n be the statement that x  y is a factor of x n  y n for all natural numbers n. Step 1: P 1 is the statement that x  y is a factor of x 1  y 1 , which is clearly true.

Step 2: Assume that P k is true; that is, x  y is a factor of x k  y k . We want to use this to show that P k  1 is true.   Now, x k1  y k1  x k1  x k y  x k y  y k1  x k x  y  x k  y k y, for which x  y is a factor because x  y   is a factor of x k x  y, and x  y is a factor of x k  y k y, by the induction hypothesis. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

28. Let P n be the statement that x  y is a factor of x 2n1  y 2n1 .

Step 1: P 1 is the statement that x  y is a factor of x 1  y 1 , which is clearly true.

Step 2: Assume that P k is true; that is, x  y is a factor of x 2k1  y 2k1 . We want to use this to show that P k  1 is true. Now x 2k11  y 2k11  x 2k1  y 2k1  x 2k1  x 2k1 y 2  x 2k1 y 2  y 2k1      x 2k1 x 2  y 2  x 2k1  y 2k1 y 2

for which x  y is a factor. This is because x  y is a factor of x 2  y 2  x  y x  y and x  y is a factor of

x 2k1  y 2k1 by our induction hypothesis. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n .

29. Let P n denote the statement that F3n is even for all natural numbers n. Step 1: P 1 is the statement that F3 is even. Since F3  F2  F1  1  1  2, this statement is true. Step 2: Assume that P k is true; that is, F3k is even. We want to use this to show that P k  1 is true. Now, F3k1  F3k3  F3k2  F3k1  F3k1  F3k  F3k1  F3k  2  F3k1 , which is even because F3k is even by the induction hypothesis, and 2  F3k1 is even. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 30. Let P n denote the statement that F1  F2  F3      Fn  Fn2  1. Step 1: P 1 is the statement that F1  F3  1. But F1  1  2  1  F3  1, which is true. Step 2: Assume that P k is true; that is, F1  F2  F3      Fk  Fk2  1. We want to use this to show that P k  1   is true. Now Fk12  1  Fk3  1  Fk2  Fk1  1  Fk2  1  Fk1  F1  F2  F3      Fk  Fk1 by the induction hypothesis. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 31. Let P n denote the statement that F12  F22  F32      Fn2  Fn  Fn1 . Step 1: P 1 is the statement that F12  F1  F2 or 12  1  1, which is true. Step 2: Assume that P k is true, that is, F12  F22  F32      Fk2  Fk  Fk1 . We want to use this to show that P k  1 is true. Now 2 2  Fk  Fk1  Fk1 F12  F22  F32      Fk2  Fk1

   Fk1 Fk  Fk1  Fk1  Fk2

induction hypothesis

by definition of the Fibonacci sequence

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CHAPTER 13 Sequences and Series

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

32. Let P n denote the statement that F1  F3      F2n1  F2n . Step 1: P 1 is the statement that F1  F2 , which is true since F1  1 and F2  1. Step 2: Assume that P k is true; that is, F1  F3      F2k1  F2k , for some k  1. We want to use this to show that P k  1 is true; that is, F1  F3      F2k11  F2k1 . Now F1  F3      F2k1  F2k1  F2k  F2k1

induction hypothesis

 F2k2

definition of F2k2

 F2k1 Therefore, P k  1 is true. So by the Principle of Mathematical Induction, P n is true for all n. 

33. Let P n denote the statement  

Step 1: Since 

1 1 1 0



2

 

1 1 1 0

1 1 1 0

n



1 1

 

1 0

k1 

Fn1

Fn

Fn

Fn1





2 1

k



  1 1 1 0 

Step 2: Assume that P k is true; that is,  





1 1 1 0

1 1

 

.





F2 F1

Fk1

Fk

Fk Fk1

 

1 1



 

Fk1  Fk Fk1

 

1 1 1 0

Fk  Fk1

F3 F2





1 0

k 







 Fk







, it follows that P 2 is true.

. We show that P k  1 follows from this. Now,

Fk1

Fk

Fk Fk1 



 

Fk2 Fk1 Fk1

Fk

1 1 1 0

 

 

induction hypothesis

by definition of the Fibonacci sequence

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n  2.

34. Let a1  1 and an1 

1 Fn , for n  1 . Let P n be the statement that an  , for all n  1. 1  an Fn1

Step 1: P 1 is the statement that a1 

F1 F , which is true since a1  1 and 1  11  1. F2 F2

Step 2: Assume that P k is true; that is, ak  ak1 

1 (definition of ak1 )  1  ak

Fk . We want to use this to show that P k  1 is true. Now Fk1

1 Fk 1 Fk1

(induction hypothesis) 

Fk1 F  k1 (definition of Fk2 ). Fk  Fk1 Fk2

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n  100.

SECTION 13.5 Mathematical Induction

891

35. Since F1  1, F2  1, F3  2, F4  3, F5  5, F6  8, F7  13,    our conjecture is that Fn  n, for all n  5. Let P n denote the statement that Fn  n.

Step 1: P 5 is the statement that F5  5  5, which is clearly true.

Step 2: Assume that P k is true; that is, Fk  k, for some k  5. We want to use this to show that P k  1 is true. Now, Fk1  Fk  Fk1  k  Fk1 (by the induction hypothesis) k  1 (because Fk1  1). Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n  5.

36. Since 100  10  103 , 100  11  113 , 100  12  123 ,   , our conjecture is that 100n  n 3 , for all natural numbers n  10. Let P n denote the statement that 100n  n 3 , for n  10.

Step 1: P 10 is the statement that 100  10  1,000  103  1,000, which is true.

Step 2: Assume that P k is true; that is, 100k  k 3 . We want to use this to show that P k  1 is true. Now 100 k  1  100k  100  k 3  100

induction hypothesis

 k3  k2

because k  10

 k 3  3k 2  3k  1  k  13 Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n  10. 37. (a) P n  n 2  n  11 is prime for all n. This is false as the case for n  11 demonstrates: P 11  112  11  11  121, which is not prime since 112  121.

(b) n 2  n, for all n  2. This is true. Let P n denote the statement that n 2  n. Step 1: P 2 is the statement that 22  4  2, which is clearly true.

Step 2: Assume that P k is true; that is, k 2  k. We want to use this to show that P k  1

is true. Now k  12  k 2  2k  1. Using the induction hypothesis (to replace k 2 ), we have

k 2  2k  1  k  2k  1  3k  1  k  1, since k  2. Therefore, k  12  k  1, which is exactly P k  1. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

(c) 22n1  1 is divisible by 3, for all n  1 . This is true. Let P n denote the statement that 22n1  1 is divisible by 3. Step 1: P 1 is the statement that 23  1  9 is divisible by 3, which is clearly true.

Step 2: Assume that P k is true; that is, 22k1  1 is divisible by 3. We want to use this to show that P k  1 is   true. Now, 22k11  1  22k3  1  4  22k1  1  3  1 22k1  1  3  22k1  22k1  1 , which is

divisible by 3 since 22k1  1 is divisible by 3 by the induction hypothesis, and 3  22k1 is clearly divisible by 3. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

(d) The statement n 3  n  12 for all n  2 is false. The statement fails when n  2: 23  8  2  12  9.

(e) n 3  n is divisible by 3, for all n  2. This is true. Let P n denote the statement that n 3  n is divisible by 3. Step 1: P 2 is the statement that 23  2  6 is divisible by 3, which is clearly true.

Step 2: Assume that P k is true; that is, k 3 k is divisible by 3. We want to use this to show that P k  1 is true. Now     k  13  k  1  k 3  3k 2  3k  1  k  1  k 3  3k 2  2k  k 3  k  3k 2  2k  k  k 3  k  3 k 2  k .   The term k 3  k is divisible by 3 by our induction hypothesis, and the term 3 k 2  k is clearly divisible by 3. Thus k  13  k  1 is divisible by 3, which is exactly P k  1. So by the Principle of Mathematical Induction, P n is true for all n.

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CHAPTER 13 Sequences and Series

(f) n 3  6n 2  11n is divisible by 6, for all n  1. This is true. Let P n denote the statement that n 3  6n 2  11n is divisible by 6. Step 1: P 1 is the statement that 13  6 12  11 1  6 is divisible by 6, which is clearly true.

Step 2: Assume that P k is true; that is, k 3  6k 2  11k is divisible by 6. We show that P k  1 is then also true. Now

k  13  6 k  12  11 k  1  k 3  3k 2  3k  1  6k 2  12k  6  11k  11    k 3  3k 2  2k  6  k 3  6k 2  11k  3k 2  9k  6        k 3  6k 2  11k  3 k 2  3k  2  k 3  6k 2  11k  3 k  1 k  2

In this last expression, the first term is divisible by 6 by our induction hypothesis. The second term is also divisible by 6. To see this, notice that k  1 and k  2 are consecutive natural numbers, and so one of them must be even (divisible by 2). Since 3 also appears in this second term, it follows that this term is divisible by 2 and 3 and so is divisible by 6. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

38. The induction step fails when k  2, that is, P 2 does not follow from P 1. If there are only two cats, Midnight and

Sparky, and we remove Sparky, then only Midnight remains. So at this point, we still know only that Midnight is black. Now removing Midnight and putting Sparky back leaves Midnight alone. So the induction hypothesis does not allow us to

conclude that Sparky is black.

13.6 THE BINOMIAL THEOREM 1. An algebraic expression of the form a  b, which consists of a sum of two terms, is called a binomial. 2. We can find the coefficients in the expansion of a  bn from the nth row of Pascal’s Triangle. So a  b4  1a 4  4a 3 b  6a 2 b2  4ab3  1b4 .

    n n! 4 4!  . So   4. k k! n  k! 3 3!1!     4 4 4 3 4. To expand a  bn we can use the Binomial Theorem. Using this theorem we find a  b4  a  a b 0 1       4 4 4 4 2 2 ab3  b . a b  3 4 2

3. The binomial coefficients can be calculated directly using the formula

5. x  y6  x 6  6x 5 y  15x 4 y 2  20x 3 y 3  15x 2 y 4  6x y 5  y 6

6. 2x  14  2x4  4 2x3  6 2x2  4  2x  1  16x 4  32x 3  24x 2  8x  1  2   3  4  1 1 1 1 4 1 1 4 4 3 2  x  4x   6x  4x   x 4  4x 2  6  2  4 7. x  x x x x x x x 8. x  y5  x 5  5x 4 y  10x 3 y 2  10x 2 y 3  5x y 4  y 5

9. x  15  x 5  5x 4  10x 3  10x 2  5x  1   6       a  b  a 3  6a 2 a b  15a 2 b  20a ab b  15ab2  6 ab2 b  b3 10.     a 3  6a 2 ab  15a 2 b  20ab ab  15ab2  6b2 ab  b3

or a 3  6a 52 b12  15a 2 b  20a 32 b32  15ab2  6a 12 b52  b3 .  5  5  4  3  2 11. x 2 y  1  x 2 y 5 x 2 y 10 x 2 y  10 x 2 y 5x 2 y 1  x 10 y 5 5x 8 y 4  10x 6 y 3 10x 4 y 2 5x 2 y 1      6 12. 1  2  16  6  15  2  15  14  2  20  13  2 2  15  12  4  6  1  4 2  23      1  6 2  30  40 2  60  24 2  8  99  70 2

SECTION 13.6 The Binomial Theorem

893

13. 2x  3y3  2x3  3 2x2 3y  3  2x 3y2  3y3  8x 3  36x 2 y  54x y 2  27y 3 3   2  3 14. 1  x 3  13  3  12  x 3  3  1 x 3  x 3  1  3x 3  3x 6  x 9    5  4  3  2     1  5 1 1  1 1 1  x x2  x2 x 15.  5 x  10 x  10 x x 5 x x x x x x 1 5 10 10  5  72  2  12  5x  x 52 x x x x  x 2  x 3   x 4  x 5 x 5 x 1 x5 16. 2   25 5 24 10 23 10 22 5 2   3240x 20x 2 5x 3  58 x 4  32 2 2 2 2 2 2   6  5  4! 6! 6   15 17.  4! 2! 2  1  4! 4   8  7  6  5! 8 8!   8  7  56 18.  3! 5! 3  2  1  5! 3   100  99  98! 100 100! 19.   4950  98! 2! 98!  2  1 98   10  9  8  7  6  5! 10 10! 20.   3  2  7  6  252  5! 5! 5  4  3  2  1  5! 5    3  2!  4  3  2! 3! 4! 3 4   18 21.  1! 2! 2! 2! 1  2!  2  1  2! 1 2    5! 5  4  3! 5  4  3! 5 5 5! 22.     10  10  100  2! 3! 3! 2! 2  1  3! 3!  2  1 2 3             23. 50  51  52  53  54  55  1  15  25  32             5 5! 5! 5! 5 5 5 5 5 5! 24.     1  0. Notice that the first and sixth       1 1! 4! 2! 3! 3! 2! 4! 1! 0 1 2 3 4 5 terms cancel, as do the second and fifth terms and the third and fourth terms.           25. x  2y4  40 x 4  41 x 3  2y  42 x 2  4y 2  43 x  8y 3  44 16y 4  x 4  8x 3 y  24x 2 y 2  32x y 3  16y 4             26. 1  x5  50 15  51 14 x  52 13 x 2  53 12 x 3  54 1 x 4  55 x 5  1  5x  10x 2  10x 3  5x 4  x 5            2    3    4    5    6 1 6 6 4 1 6 6 6 5 1 6 3 1 6 2 1 6 6 1 1  27. 1       1 1  1 1 1 1 x x x x x x x 2 0 1 3 4 5 6 15 6 20 15 6 1 1  2  3  4  5  6 x x x x x x 4        2     3    4 4 4 4 2 4 3 2 28. 2A  B  0 2A  1 2A B  2 2A2 B 2  43 2A B 2  44 B 2 29.

30.

31. 32.

 16A4  32A3 B 2  24A2 B 4  8AB 6  B 8   20   19 The first three terms in the expansion of x  2y20 are 20  x 20 , 20  2y  40x 19 y, and 0 x 1 x 20 18 2 18 2 2 x  2y  760x y .  30    12 30    12 29 The first four terms in the expansion of x 12  1 are 30  x 15 , 30 x x 1  30x 292 , 0 1    12 27 3 30  12 28 2 x x 1  435x 14 , and 30 1  4060x 272 . 2 3 25    23  13 24   253 The last two terms in the expansion of a 23  a 13 are 25  a  25a 263 , and 25  a 253 . 24 a 25 a       40   1 40 40 , 40 x 39 1 The first three terms in the expansion of x  x are 40  x  40x 38 , and 0 1 x x   40 38 1 2  780x 36 . x 2 x

894

CHAPTER 13 Sequences and Series

 18 33. The middle term in the expansion of x 2  1 occurs when both terms are raised to the 9th power. So this term is   18 2 9 9 x 1  48,620x 18 . 9   16 4 16 16 34. The fifth term in the expansion of ab  120 is 20 4 ab 1  4845a b .   2 23 2 23 35. The 24th term in the expansion of a  b25 is 25 23 a b  300a b .   3 27 3 27 36. The 28th term in the expansion of A  B30 is 30 27 A B  4060A B .   1 99 99 37. The 100th term in the expansion of 1  y100 is 100 99 1  y  100y .  25      2 24 1 1   25x 47 . x is 25 38. The second term in the expansion of x 2  1 x x

39. The term that contains x 4 in the expansion of x  2y10 has exponent r  4. So this term is 10 4 104  13,440x 4 y 6 . 4 x  2y  12    r 40. The rth term in the expansion of 2y is 12 2 y 12r . The term that contains y 3 occurs when 12  r  3  r     9 3 2 y  3520 2y 3 . r  9 . Therefore, the term is 12 9   r  2 12r 12 r 242r 41. The rth term is 12  r a b . Thus the term that contains b8 occurs where 24  2r  8  r  8. So r a b   8 8 8 8 the term is 12 8 a b  495a b .  8r   8r   8r   xr 1  r8 8r  8r  r8 8r  x 2r8 . So the term that does not contain x occurs 42. The rth term is r8 8xr 2x 2 x 2  4   1 when 2r  8  0  r  4. Thus, the term is 84 8x4  17,920. 2x 43. x 4  4x 3 y  6x 2 y 2  4x y 3  y 4  x  y4

44. x  15  5 x  14  10 x  13  10 x  12  5 x  1  1  [x  1  1]5  x 5         45. 8a 3  12a 2 b  6ab2  b3  30 2a3  31 2a2 b  32 2ab2  33 b3  2a  b3  4    4    3    2     46. x 8  4x 6 y  6x 4 y 2  4x 2 y 3  y 4  40 x 2  41 x 2 y  42 x 2 y 2  43 x 2 y 3  44 y 4  x 2  y   h 3x 2  3xh  h 2 x 3  3x 2 h  3xh 2  h 3  x 3 3x 2 h  3xh 2  h 3 x  h3  x 3     3x 2  3xh  h 2 47. h h h h 4 4 4 3       x  1 x h  42 x 2 h 2  43 xh 3  44 h 4  x 4 x 4  4x 3 h  6x 2 h 2  4xh 3  h 4  x 4 x  h4  x 4 0   48. h h h   3  6x 2 h  4xh 2  h 3 3 2 2 3 4 h 4x 4x h  6x h  4xh  h   4x 3  6x 2 h  4xh 2  h 3  h h   100 49. 101100  1  001100 . Now the first term in the expansion is 100  1, the second term is 0 1 100 98 100 99 2 1 1 001  1, and the third term is 2 1 001  0495. Now each term is nonnegative, so 101100  1  001100  1  1  0495  2. Thus 101100  2. n  n  n  n  n! n! n! n!   1.   1. Therefore, 50.    . 0!n! 1  n! n!0! n!  1 n 0 0 n   n  n n  1! n n n! n n  1! n!    n.    n. Therefore, 51.  1! n  1! 1 n  1! 1 n1 1 n  1! 1! n  1! 1 n   n    n. 1 n1

SECTION 13.6 The Binomial Theorem

52.

n  r





n! n!   r! n  r ! n  r! r! 

n



n n r



895

for 0  r  n.

n! n! .  r  1! [n  r  1]! r! n  r! n! n! r  n! n  r  1  n! (b)    r  1! [n  r  1]! r! n  r ! r  r  1! n  r  1! r! n  r  1 n  r ! r  n! n  r  1  n!   r! n  r  1! r! n  r  1! Thus a common denominator is r! n  r  1!.

53. (a)

n r 1



r



(c) Therefore, using the results of parts (a) and (b),     r  n! n! n n n! n  r  1  n!      r! n  r  1! r! n  r  1! r 1 r r  1! [n  r  1]! r! n  r!

  n! r  n  r  1 n! n  1 r  n!  n  r  1  n! n1 n  1!     r! n  r  1! r! n  r  1! r! n  1  r ! r ! n  1  r! r n  54. Let P n be the proposition that is an integer for the number n, 0  r  n. r n  0 Step 1: Suppose n  0. If 0  r  n, then r  0, and so   1, which is obviously an integer. Therefore, P 0 r 0 is true.   k1 Step 2: Suppose that P k is true. We want to use this to show that P k  1 must also be true; that is, is an r       k k k 1  by the key property of binomial coefficients integer for 0  r  k  1. But we know that  r 1 r r     k k (see Exercise 49). Furthermore, and are both integers by the induction hypothesis. Since the sum of two r 1 r   k 1 integers is always an integer, must be an integer. Thus, P k  1 is true if P k is true. So by the Principal of r n Mathematical induction, is an integer for all n  0, 0  r  n. r 

55. By the Binomial Theorem, the volume of a cube of side x  2 inches is         3 3 3 2 3 3 3 x  x 2  x 22  2  x 3  3  2x 2  3  4x  8  x 3  6x 2  12x  8. The volume x  23  0 1 2 3 of a cube of side x inches is x 3 , so the difference in volumes is x 3  6x 2  12x  8  x 3  6x 2  12x  8 cubic inches.  n pr q nr . In this case, n r p  09, q  01, n  5, and r  3, so the probability that the archer hits the target exactly 3 times in 5 attempts is     5 5 3 53  P 093 012  00729. 09 01 2 53

56. By the Binomial Theorem, the coefficient of pr in the expansion of  p  qn is



57. Notice that 100!101  100!100  100! and 101!100  101  100!100  101100  100!100 . Now

100!  1  2  3  4      99  100 and 101100  101  101  101      101. Thus each of these last two expressions consists of 100 factors multiplied together, and since each factor in the product for 101100 is larger than each factor in the product for

100!, it follows that 100!  101100 . Thus 100!100  100!  100!100  101100 . So 100!101  101!100 .

896

CHAPTER 13 Sequences and Series

58.

11 2 1214 13318 1  4  6  4  1  16 1  5  10  10  5  1  32

Conjecture: The sum is 2n . Proof: 2n  1  1n 

n

n

11  1n1 

n 

12  1n2     

n 

1n  10 1 2 n n  n  n  n       0 1 2 n n n  n  n  n n 0 n 1 59. 0  0  1  1  1 1  1 1n1  12 1n2      1n 10 0 1 2 n n n  n  n  n        1n        1k k n 0 1 2 0

10  1n 

CHAPTER 13 REVIEW 1 4 9 16 100 n2 12 22 32 42 102 . Then a1   , a2   , a3   , a4   , and a10   . n1 11 2 21 3 31 4 41 5 10  1 11 8 2n 21 22 23 24 . Then a1  11  2, a2  12  2, a3  13   , a4  14  4, and an  1n n 1 2 3 3 4 210 1024 512 a10  110   . 10 10 5 1 2 1n  1 11  1 12  1 13  1 an   , a3  . Then a1   0, a2    0, 3 3 3 8 4 n 1 2 33 1 2 1 14  1 110  1  , and a .    a4  10 3 3 64 32 500 4 10 n n  1 1 1  1 2 2  1 3 3  1 4 4  1 . Then a1   1, a2   3, a3   6, a4   10, and an  2 2 2 2 2 10 10  1 a10   55. 2 654 2n! 2  1! 2  2! 2  3! an  n . Then a1  1  1, a2  2  3, a3  3   15, 2 n! 8 2  1! 2  2! 2  3! 8765 2  4! 2  10!  105, and a10  10   654,729,075. a4  4 16 2  4!  2  10!       n1 11 21 3! 31 4! an   3, a3   6, . Then a1   1, a2    2! 1! 2! 2! 2 2 2 2     41 5! 10  1 11!  10, and a10   55. a4    2! 3! 2! 9! 2 2 an  an1  2n  1 and a1  1. Then a2  a1  4  1  4, a3  a2  6  1  9, a4  a3  8  1  16, a5  a4  10  1  25, a6  a5  12  1  36, and a7  a6  14  1  49. 1 1 1 1 1 a a a a a a ,a  4  ,a  5  , and an  n1 and a1  1. Then a2  1  , a3  2  , a4  3  n 2 2 3 6 4 24 5 5 120 6 6 720 1 a . a7  6  7 5040 an  an1  2an2 , a1  1 and a2  3. Then a3  a2  2a1  5, a4  a3  2a2  11, a5  a4  2a3  21, a6  a5  2a4  43, and a7  a6  2a5  85.

1. an  2.

3.

4.

5.

6.

7.

8.

9.

CHAPTER 13

Review

897

12       3an1 and a1  3  312 . Then a2  3a1  3 3  3  312  334 ,        3a2  3  334  374  378 , a4  3a3  3  378  3158  31516 ,        3a4  3  31516  33116  33132 , a6  3a5  3  33132  36332  36364 ,     3a6  3  36364  312764  3127128 .

10. an  a3 a5 a7

11. (a) a1  2 1  5  7, a2  2 2  5  9,

a3  2 3  5  11, a4  2 4  5  13,

a5  2 5  5  15 an

(b)

5 5 5 5 5 5 12. (a) a1  1  , a2  2  , a3  3  , 2 4 8 2 2 2 5 5 5 5 , a5  5  a4  4  16 32 2 2 an

(b)

14 12 10 8 6 4 2 0

2 1

1

2

3

4

0

5 n

1

2

3

4

5 n

(c) S5  7  9  11  13  15  55

5 5 5 5 5 155 (c) S5       2 4 8 16 32 32

(d) This sequence is arithmetic with common difference

(d) This sequence is geometric with common ratio 12 .

2. 31 3 32 9 33 27 , 13. (a) a1  2  , a2  3  , a3  4  4 8 16 2 2 2

7 1 2  , a2  4   3, 2 2 2 5 3 3 4 5 a3  4   , a4  4   2, a5  4   2 2 2 2 2

14. (a) a1  4 

34 81 35 243 , a5  6  a4  5  32 64 2 2 (b)

(b)

an

3

4

2

2

0

(c) S5 

an

1 1

2

3

4

5 n

633 3 9 27 81 243      4 8 16 32 64 64

(d) This sequence is geometric with common ratio 32 .

0

1

2

3

4

5 n

5 3 25 7 (c) S5   3   2   2 2 2 2 (d) This sequence is arithmetic with common difference  12 .

15. 5 55 6 65   . Since 55  5  6  55  65  6  05, this is an arithmetic sequence with a1  5 and d  05. Then a5  a4  05  7.            16. 2 2 2 3 2 4 2   . Since 2 2  2  3 2  2 2  4 2  3 2  2, this is an arithmetic sequence with       a1  2 and d  2. Then a5  a4  2  4 2  2  5 2. 17. t  3 t  2 t  1 t   . Since t  2  t  3  t  1  t  2  t  t  1  1, this is an arithmetic sequence with a1  t  3 and d  1. Then a5  a4  1  t  1.       4  2, this is a geometric sequence with a  2 and r  2. Then 18. 2 2 2 2 4   . Since 2  2 2 2   1 2 2 2   a5  a4  r  4  2  4 2. 1 t2 t 1 1 1 19. t 3  t 2  t 1   . Since 3  2  , this is a geometric sequence with a1  t 3 and r  . Then a5  a4  r  1   . t t t t t t

898

CHAPTER 13 Sequences and Series

3 2  20. 1  32  2  52    . Since  32  1  2  32 , and 2  , the series is neither arithmetic nor geometric. 1  32 1

1

2

21. 34  12  13  29    . Since 23  31  91  23 , this is a geometric sequence with a1  34 and r  23 . Then 4 2 2 4 a5  a4  r  9  3  27 .

2

3

1 1 2 1 1 1 a a  and 22. a 1  2    . Since  1 a a a 1 a 1 1 1 a5  a4  r  2   3 . a a a 6i 12 23. 3 6i 12 24i   . Since  2i,  3 6i ratio r  2i.

1 1 , this is a geometric sequence with a1  a and r  . Then a a

2 2i 24i  2  2i,  2i, this is a geometric sequence with common i 12 i

a 2  2i 24. The sequence 2, 2  2i, 4i, 4  4i, 8,    is geometric (where i 2  1), since 2   1  i, a1 2 4i 2  2i 8  8i a 1 1 i i 4i 4  4i a3     1  i, 4     1     1  2  1  1  i,  a2 2  2i 2  2i 2  2i 8 a3 4i i i i i 8 4  4i 32  32i 8 a5     1  i. Thus the common ratio is i  1 and the first term is 2. So the  a4 4  4i 4  4i 4  4i 32 nth term is an  a1r n1  2 1  in1 .

25. a6  17  a  5d and a4  11  a  3d. Then, a6  a4  17  11  a  5d  a  3d  6  6  2d  d  3. Substituting into 11  a  3d gives 11  a  3  3, and so a  2. Thus a2  a  2  1 d  2  3  5. 26. a20  96 and d  5. Then 96  a20  a  19  5  a  95  a  96  95  1. Therefore, an  1  5 n  1.  2 27. a3  9 and r  32 . Then a5  a3  r 2  9  32  81 4.

a5 3 28. a2  10 and a5  1250 27 . Then r  a  2  n1 an  ar n1  6 53 .

1250 27  125  r  5 and a  a  a2  10  6. Therefore, 1 27 3 5 10 r 3

29. (a) An  32,000  105n1

(b) A1  $32,000, A2  32,000  1051  $33,600, A3  32,000  1052  $35,280, A4  32,000  1053  $37,044, A5  32,000  1054  $38,89620, A6  32,000  1055  $40,84101, A7  32,000  1056  $42,88306,

A8  32,000  1057  $45,02721

30. (a) An  35,000  1,200 n  1 (b) A8  35,000  1,200 7  $43,400. The salary for this teacher is higher for the first 6 years is higher, but from the 7th year on, the salary of the teacher in Exercise 29 is higher. 31. Let an be the number of bacteria in the dish at the end of 5n seconds. So a0  3, a1  3  2, a2  3  22 , a3  3  23 ,   . Then, clearly, an is a geometric sequence with r  2 and a  3. Thus at the end of 60  5 12 seconds, the number of bacteria is a12  3  212  12,288.

32. Let d be the common difference in the arithmetic sequence a1 , a2 , a3 ,   , so that an  a1  n  1 d, n  1, 2, 3,   , and let e be the common difference for b1 , b2 , b3 ,   , so that bn  b1  n  1 e. Then     an  bn  a1  n  1 a  b1  n  1 e  a1  b1   n  1 d  e, n  1, 2, 3,   . Thus a1  b1  a2  b2     is an arithmetic sequence with first term a1  b1 and common difference d  e.

CHAPTER 13

Review

899

33. Suppose that the common ratio in the sequence a1  a2  a3     is r . Also, suppose that the common ratio in the sequence b1  b2  b3     is s. Then an  a1r n1 and bn  b1 s n1 , n  1 2 3   . Thus an bn  a1r n1 b1 s n1  a1 b1  rsn1 . So the sequence a1 b1  a2 b2  a3 b3     is geometric with first term a1 b1 and common ratio rs. 34. (a) Yes. If the common difference is d, then an  a1  n  1 d. So an  2  a1  2  n  1 d, and thus the sequence a1  2 a2  2 a3  2    is an arithmetic sequence with the common difference d, but with the first term a1  2. (b) Yes. If the common ratio is r, then an  a1  r n1 . So 5an  5a1   r n1 , and the sequence 5a1  5a2  5a3     is also geometric, with common ratio r, but with the first term 5a1 .

35. (a) 6 x 12    is arithmetic if x  6  12  x  2x  18  x  9.  12 x  x 2  72  x  6 2. (b) 6 x 12    is geometric if  6 x 36. (a) 2 x y 17    is arithmetic. Therefore, 15  17  2  a4  a1  a  3d  a  3d. So d  5, and hence, x  a  d  2  5  7 and y  a  2d  2  2  5  12.  13 a4 ar 3 3  r  17  r   . So (b) 2 x y 17    is geometric. Therefore, 17 2 2 a1 a   2  13  23 17 13 17 23 1713 and y  a  ar 2  2  2  2  213 1723 . x  a2  ar  2 17 3 2 2 2 37.

38.

6

2 2 2 2 2 k3 k  1  3  1  4  1  5  1  6  1  16  25  36  49  126

4

2i

21

22

23

24

4

6

8

210  140  126  120

596

 i 1 2i  1  2  1  1  2  2  1  2  3  1  2  4  1  2  3  5  7  105 105 6 k1 0 1 2 3 4 5 39. k1 k  1 2  2  2  3  2  4  2  5  2  6  2  7  2  2  6  16  40  96  224  384 5 40. m1 3m2  31  30  31  32  33  13  1  3  9  27  121 3 10 2 2 2 2 2 2 2 2 2 2 2 41. k1 k  1  0  1  2  3  4  5  6  7  8  9 42. 43. 44.

100

1 1 1 1 1 1 1 1 j2 j  1  1  2  3  4  5      98  99

50

k1

10

3k

2k1

3 32 33 34 349 350  2  3  4  5      50  51 2 2 2 2 2 2

2 n 2 1 2 2 2 3 2 9 2 10 n1 n 2  1  2  2  2  3  2      9  2  10  2

45. 3  6  9  12      99  3 1  3 2  3 3      3 33   2 46. 12  22  33      1002  100 k1 k

33

k1 3k

47. 1  23  2  24  3  25  4  26      100  2102

48.

 1 212  2 222  3 232  4 242      100 21002  k2  100 k1 k  2

 1 1 1 1 1       999 k1 k k  1 12 23 34 999  1000

49. 1  09  092      095 is a geometric series with a  1 and r 

09  09. Thus, the sum of the series is 1

1  096 1  0531441   468559. 1  09 01 50. 3  37  44      10 is an arithmetic series with a  3 and d  07. Then 10  an  3  07 n  1  07 n  1  7 S6 

143  n  11. So the sum of the series is S11  11 2 3  10  2  715.          51. 5  2 5  3 5      100 5 is an arithmetic series with a  5 and d  5. Then 100 5  an  5  5 n  1         n  100. So the sum is S100  100 5  100 5  50 101 5  5050 5. 2

900

CHAPTER 13 Sequences and Series

52. 13  23  1  43      33 is an arithmetic series with a  13 and d  13 . Then an  33  13  13 n  1  n  99. So the   2  99  1  99  100  1650. sum is S99  99 2 3 2 3 3 6 53. n0 3  4n is a geometric series with a  3, r  4, and n  7. Therefore, the sum of the series is S7  3 

54.

8

  1  47  35 1  47  9831. 1  4

k2 is a geometric series with a  7, r  512 , and n  9. Thus, the sum of the series is k0 7  5   1  592 1  5 1  592  5  55 1  592 7 S9  7   7

15 1  5  1 5 1  5    74 1  625 5  5  3125  5467  1092 5  79088

4  8     is a geometric series with a  1 and r   2 . Therefore, it is convergent with sum 55. 1  25  25 125 5 5 1 a   .  S 1r 7 1  2 5

56. 01  001  0001  00001     is an infinite geometric series with a  01 and r  01. Therefore, it is convergent with 1 01  . sum S  1  01 9

57. 5  5 101  5 1012  5 1013     is an infinite geometric series with a  5 and r  101. Because r  101  1, the series diverges. 1 1 1 58. 1  12  13  32     is an infinite geometric series with a  1 and r   . Thus, it is convergent with sum 3 3 3     1 3  S  12 3  3 . 31 1  1 3

 2

59. 1  98  98 diverges.

 3  98     is an infinite geometric series with a  1 and r   98 . Because r  98  1, the series

    60. a  ab2  ab4  ab6     is an infinite geometric series with first term a and common ratio b2 . Because b  1, b2   1, a . 1  b2 61. We have an arithmetic sequence with a  7 and d  3. Then n n n Sn  325  [2a  n  1 d]  [14  3 n  1]  11  3n  650  3n 2  11n  3n  50 n  13  0  2 2 2 is inadmissible). Thus, 13 terms must be added. n  13 (because n   50 3 and so the series is convergent with sum S 

62. We have a geometric series with S3  52 and r  3. Then 52  S3  a  3a  9a  13a  a  4, and so the first term is 4.   1  215  2 215  1  65,534, and so the total 63. This is a geometric sequence with a  2 and r  2. Then S15  2  12 number of ancestors is 65,534.   10816  1  12,500 10816  1  $30,32428. 64. R  1000, i  008, and n  16. Thus, A  1000 108  1 65. A  10,000, i  003, and n  4. Thus, 10,000  R

10,000  003 1034  1 R  $239027. 003 1034  1

009  00075. 12 60,000  00075 (a) If the period is 30 years, n  360 and R   $48277. 1  10075360

66. A  60,000 and i 

CHAPTER 13

(b) If the period is 15 years, n  180 and R 

Review

901

60,000  00075  $60856. 1  10075180

67. Let P n denote the statement that 1  4  7      3n  2 

n 3n  1 . 2

1 [3 1  1] 12  , which is true. 2 2 k 3k  1 Step 2: Assume that P k is true; that is, 1  4  7      3k  2  . We want to use this to show that 2 P k  1 is true. Now Step 1: P 1 is the statement that 1 

1  4  7  10      3k  2  [3 k  1  2]  

k 3k  1  3k  1 2

induction hypothesis

3k 2  k  6k  2 k 3k  1 6k  2   2 2 2

3k 2  5k  2 k  1 3k  2  2 2 k  1 [3 k  1  1]  2



Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

1 1 1 n 1      .  13 35 57 2n  1 2n  1 2n  1 1 1 Step 1: P 1 is the statement that  , which is true. 13 211 1 1 1 1 k Step 2: Assume that P k is true; that is,    . We want to use this to  13 35 57 2k  1 2k  1 2k  1 show that P k  1 is true. Now 1 1 1 1 1       13 35 57 2k  1 2k  1 2k  1 2k  3

68. Let P n denote the statement that

  

k 1  2k  1 2k  1 2k  3

k 2k  3  1 2k 2  3k  1  2k  1 2k  3 2k  1 2k  3

k 1 k 1 k  1 2k  1   2k  3 2 k  1  1 2k  1 2k  3

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

induction hypothesis

902

CHAPTER 13 Sequences and Series

      1 1 1 1 1 1    1   n  1. 69. Let P n denote the statement that 1  1 2 3 n

Step 1: P 1 is the statement that 1  11  1  1, which is clearly true.       1 1 1 1 1 1    1   k  1. We want to use this to Step 2: Assume that P k is true; that is, 1  1 2 3 k show that P k  1 is true. Now        1 1 1 1 1 1 1    1  1 1 1 2 3 k k 1         1 1 1 1 1 1 1    1  1  1 1 2 3 k k 1   1  k  1 1  induction hypothesis k1  k  1  1

Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

70. Let P n denote the statement that 7n  1 is divisible by 6.

Step 1: P 1 is the statement that 71  1  6 is divisible by 6, which is clearly true.

Step 2: Assume that P k is true; that is, 7k  1 is divisible by 6. We want to use this to show that P k  1 is true. Now   7k1  1  7  7k  1  7  7k  7  6  7 7k  1  6, which is divisible by 6. This is because 7k  1 is divisible by

6 by the induction hypothesis, and clearly 6 is divisible by 6. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 71. an1  3an  4 and a1  4. Let P n denote the statement that an  2  3n  2. Step 1: P 1 is the statement that a1  2  31  2  4, which is clearly true.

Step 2: Assume that P k is true; that is, ak  2  3k  2. We want to use this to show that P k  1 is true. Now ak1  3ak  4    3 2  3k  2  4

definition of ak1 induction hypothesis

 2  3k1  6  4  2  3k1  2

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

72. Let P n denote the statement that F4n is divisible by 3. Step 1: Show that P 1 is true, but P 1 is true since F4  3 is divisible by 3. Step 2: Assume that P k is true; that is, F4k is divisible by 3. We want to use this to show that P k  1 is true.     Now, F4k1  F4k4  F4k2  F4k3  F4k  F4k1  F4k1  F4k2  F4k  F4k1  F4k1    F4k  F4k1  2  F4k  3  F4k1 , which is divisible by 3 because F4k is divisible by 3 by our induction hypothesis, and 3  F4k1 is clearly divisible by 3. Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.    5! 54 54 5 5 5! 73.     10  10  100  2! 3! 3! 2! 2 2 2 3     10! 10  9 10  9  8  7 10 10 10! 74.     45  210  255   2! 8! 6! 4! 2 432 2 6                  75. 5k0 5k  50  51  52  53  54  55  2 0!5!5!  1!5!4!  2!5!3!  2 1  5  10  32   8                   2 80 88  2 81 87  2 82 86  2 83 85  84 84  2  2  82  2  282  2  562  702  12,870 76. 8k0 8k 8k         77. A  B3  30 A3  31 A2 B  32 AB 2  33 B 3  A3  3A2 B  3AB 2  B 3

CHAPTER 13

Test

903

                   78. x  25  50 x 5  51 x 4 2  52 x 3 22  53 x 2 23  54 x 24  55 25

 x 5  10x 4  40x 3  80x 2  80x  32 6                79. 1  x 2  60 16  61 15 x 2  62 14 x 4  63 13 x 6  64 12 x 8  65 x 10  66 x 12 80. 81. 82.

83.

 1  6x 2  15x 4  20x 6  15x 8  6x 10  x 12           2x  y4  40 2x4  41 2x3 y  42 2x2 y 2  43  2x y 3  44 y 4  16x 4  32x 3 y  24x 2 y 2  8x y 3  y 4   3 19 3 19 The 20th term is 22 19 a b  1540a b .  20    23 20    23 19 The first three terms in the expansion of b23  b13 b b are 20  b403 , 20 0 1    18  2   b23 b13  20b373 , and 20 b13  190b343 . 2   r 10r . The term that contains A6 occurs when r  6. Thus, the The rth term in the expansion of A  3B10 is 10 r A 3B 10 6 term is 6 A 3B4  210A6 81B 4  17,010A6 B 4 .

CHAPTER 13 TEST 1. an  2n 2  n  a1  1, a2  6, a3  15, a4  28, a5  45, a6  66, and S6  1  6  15  28  45  66  161.

2. an1  3an  n, a1  2  a2  3 2  1  5, a3  3 5  2  13, a4  3 13  3  36, a5  3 36  4  104, and a6  3 104  5  307.

3. (a) The common difference is d  5  2  3. (b) an  2  n  1 3

(c) a35  2  3 35  1  104

3  1. 4. (a) The common ratio is r  12 4  n1 (b) an  a1r n1  12 14  101 (c) a10  12 14  38  65,3536 4

1

1 1.  r  15 , so a5  ra4  25 5. (a) a1  25, a4  15 . Then r 3  5  25 125  8 1  15 97,656 58  1   (b) S8  25 1 12,500 3125 1 5

6. (a) a1  10 and a10  2, so 9d  8  d   89 and a100  a1  99d  10  88  78.    10 2  10  9  8 [2a (b) S10  10  60   1 d]  10 2 2 9

7. Let the common ratio for the geometric series a1  a2  a3     be r, so that an  a1r n1 , n  1, 2, 3,   . Then 2    n1  an2  a1r n1  a12 r 2 . Therefore, the sequence a12 , a22 , a32 ,    is geometric with common ratio r 2 .            5  2  1  12  1  22  1  32  1  42  1  52  0  3  8  15  24  50 8. (a) n1 1  n  (b) 6n3 1n 2n2  13 232  14 242  15 252  16 262  2  4  8  16  10 2

3

9

9. (a) The geometric sum 13  22  23  24      210 has a  13 , r  23 , and n  10. So 3 3 3 3   10 1  3 1  1024  58,025 . S10  13  123  123 3 59,049 59,049

904

FOCUS ON MODELING 1  1  1     has a  1 and r  212  1 . Thus, (b) The infinite geometric series 1  12 2 2 232 2     2 2 21 1    S    2  2. 11 2

21

21

21

n n  1 2n  1 . 6 123 , which is true. Step 1: Show that P 1 is true. But P 1 says that 12  6 k k  1 2k  1 . We want to use this to show that Step 2: Assume that P k is true; that is, 12  22  32      k 2  6 P k  1 is true. Now

10. Let P n denote the statement that 12  22  32      n 2 

12  22  32      k 2  k  12 

k k  1 2k  1  k  12 6

induction hypothesis

  k  1 2k 2  k  6k  6 k  1

k k  1 2k  1  6 k  12  6 6     2 2 k  1 2k  7k  6 k  1 2k  k  6k  6   6 6 k  1 k  2 2k  3 k  1 [k  1  1] [2 k  1  1]   6 6 

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.  5    2    3    4    5     11. 2x  y 2  50 2x5  51 2x4 y 2  52 2x3 y 2  53 2x2 y 2  54 2x y 2  55 y 2  32x 5  80x 4 y 2  80x 3 y 4  40x 2 y 6  10x y 8  y 10

  3 7 3 3 12. 10 3 3x 2  120  27x 128  414,720x

13. (a) Each week he gains 24% in weight, that is, 024an . Thus, an1  an  024an  124an for n  1.

a0 is given to be 085 lb. Then a0  085, a1  124 085, a2  124 124 085  1242 085,   a3  124 1242 085  1243 085, and so on. So we can see that an  085 124n .

(b) a6  124a5  124 124a4       1246 a0  1246 085  31 lb (c) The sequence a1  a2  a3     is geometric with common ratio 124.

FOCUS ON MODELING Modeling with Recursive Sequences 00365  00001. Thus the amount in the account at 365 the end of the nth day is An  10001An1 with A0  $275,000. (b) A0  $275,000, A1  10001A0  10001  275,000  $275,02750,

1. (a) Since there are 365 days in a year, the interest earned per day is

A2  10001A1  10001 10001A0   100012 A0  $275,05500,

A3  10001A2  100013 A0  $275,08251, A4  100014 A0  $275,11002, A5  100015 A0  $275,13753,

A6  100016 A0  $275,16504, A7  100017 A0  $275,19256 (c) An  10001n  275,000 2. (a) Tn  Tn1  15 with T1  5.

(b) T1  5, T2  T1  15  5  15  65, T3  T2  15  5  15  15  5  2  15  80, T4  T3  15  5  2  15  15  5  3  15  95, T5  T4  15  5  3  15  15  5  4  15  110, T6  T5  15  5  4  15  15  5  5  15  125

Modeling with Recursive Sequences

905

(c) This is an arithmetic sequence with Tn  5  15 n  1.

(d) Tn  65  5  15n  15  615  15n  n  41So Sheila swims 65 minutes on the 41st day.

(e) Using the partial sum of an arithmetic sequence, she swims for 30 2 [2 5  30  1 15]  15  535  8025 minutes  13 hours 225 minutes. 003  00025. Thus the amount in the account at the 3. (a) Since there are 12 months in a year, the interest earned per day is 12 end of the nth month is An  10025An1  100 with A0  $100. (b) A0  $100, A1  10025A0  100  10025  100  100  $20025, A2  10025A1  100  10025 10025  100  100  100  100252  100  10025  100  100  $30075,   A3  10025A2  100  10025 100252  100  10025  100  100  100  100253  100  100252  100  10025  100  100  $40150,   A4  10025A3  100  10025 100253  100  100252  100  10025  100  100  100  100254  100  100253  100  100252  100  10025  100  100  $50251

(c) An  10025n  100      100252  100  10025  100  100, the partial sum of a geometric series, so 10025n1  1 1  10025n1 An  100   100  . 1  10025 00025

1002561  1  $658083. 00025 4. (a) The amount An of pollutants in the lake in the nth year is 30% of the amount from the preceding year (030An1 ) plus the amount discharged that year (2400 tons). Thus An  030An1  2400. (d) Since 5 years is 60 months, we have A60  100 

(b) A0  2400, A1  030 2400  2400  3120,

A2  030 [030 2400  2400]  2400  0302 2400  2400 2400  2400  3336,   A3  030 0302 2400  2400 2400  2400  2400  0033 2400  0302 2400  2400 2400  2400  34008,   A4  030 0033 2400  0302 2400  2400 2400  2400  2400

 0034 2400  0033 2400  0302 2400  2400 2400  2400  34202

(c) An is the partial sum of a geometric series, so An  2400 

1  030n1

 2400  

 1  030  34286 1  030n1

1  030n1 070

(e) 4000 2000

1  0307 0  34278 tons. The sum of a geometric 0 070 1  34286 tons. series, is A  2400  070   5. (a) Un  105Un1  010 105Un1  110 105 Un1  1155Un1 with U0  5000. (b) U0  $5000, U1  1155U0  $5775, (d) A6  2400 

U2  1155U1  1155 1155  5000  11552  5000  $667013,   U3  1155U2  1155 11552  5000  11553  5000  $770399,   U4  1155U3  1155 11553  5000  11554  5000  $889811.

(c) Using the pattern we found in part (b), we have Un  1155n  5000.

10

20

906

FOCUS ON MODELING

(d) U10  115510  5000  $21,12467. 6. (a) In the nth year since Victoria’s initial deposit the amount Vn in her CD is the amount from the preceding year (Vn1 ), plus the 5% interest earned on that amount (005Vn1 ), plus $500 times the number of years since her initial deposit (500n). Thus Vn  105Vn1  500n. (b) Using the sequence mode on a calculator and scrolling down, we see that U31  435,52342 and U32  503,02955, while V 31  455,67849 and V 32  494,46242. Thus, Ursula’s savings surpass Victoria’s savings in the 32nd year.

14

COUNTING AND PROBABILITY

14.1 COUNTING 1. The Fundamental Counting Principle says that if one event can occur in m ways and a second event can occurs in n ways, then the two events can occur in order in m  n ways. So if you have two choices for shoes and three choices for hats, then the number of different shoe-hat combinations you can wear is 2  3  6. 2. The number of ways of arranging r objects from n objects in order is called the number of permutations of n objects taken r n! . at a time, and is given by the formula P n r   n  r! 3. The number of ways of choosing r objects from n objects is called the number of combinations of n objects taken r at a n! . time, and is given by the formula Cn r  r! n  r! 4. (a) False. When counting combinations order does not matter. (b) True. When counting permutations order matters. (c) False. For a set of n distinct objects, the number of different combinations of these objects is less than the number of different permutations. (d) True. If we have a set with 5 distinct objects then the number of different ways of choosing 2 members of this set is the same as the number of ways of choosing 3 members. 9! 9!  9  8  72  7! 9  2!

5. P 8 3 

8! 8!  8  7  6  336  5! 8  3!

6. P 9 2 

7. P 11 4 

11! 11!  11  10  9  8  7920  7! 11  4!

8. P 10 5 

10! 10!  30,240  5! 10  5!

10. P 99 3 

99! 99!   941,094 99  3! 96!

9. P 100 1  11. C 8 3 

100! 100!   100 100  1! 99!

8! 876 8!    56 3! 8  3! 3! 5! 321

13. C 11 4 

11  10  9  8 11!   330 4! 7! 4321

15. C 100 1 

100!  100 1  100 1! 99!

12. C 9 2 

9! 9! 98    36 2! 9  2! 2! 7! 21

14. C 10 5 

10! 10!   252 5! 10  5! 5! 5!

16. C 99 3 

99! 99  98  97   156,849 3! 96! 321

17. By the Fundamental Counting Principle, the number of possible single-scoop ice cream cones is     number of ways to number of ways to   4  3  12. choose the flavor choose the type of cone 18. By the Fundamental Counting Principle, the possible number of 3-letter words is       number of ways to number of ways to number of ways to   . choose the 1st letter choose the 2nd letter choose the 3rd letter

(a) Since repetitions are allowed, we have 26 choices for each letter. Thus, there are 26  26  26  17,576 words.

(b) Since repetitions are not allowed, we have 26 choices for the 1st letter, 25 choices for the 2nd letter, and 24 choices for the 3rd letter. Thus there are 26  25  24  15,600 words.

907

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CHAPTER 14 Counting and Probability

19. (a) By the Fundamental Counting Principle, the possible number of ways 8 horses can complete a race, assuming no ties in any position, is       number of ways to number of ways to number of ways to     87654321 choose the 1st finisher choose the 2nd finisher choose the 8th finisher  8!  40,320

20. 21.

22.

23.

24.

25.

(b) By the Fundamental Counting Principle, the possible number of ways the first, second, and third place can be decided,       number of ways to number of ways to number of ways to assuming no ties, is    8  7  6  336. choose the 1st finisher choose the 2nd finisher choose the 3th finisher Since there are four choices for each of the five questions, by the Fundamental Counting Principle there are 4  4  4  4  4  1024 different ways the test can be completed. The number of possible seven-digit phone numbers is       number of ways to number of ways to number of ways to  . Since the first digit cannot be a 0 or a 1, there  choose the 2nd digit choose the 7th digit choose the 1st digit are only 8 digits to choose from, while there are 10 digits to choose from for the other six digits in the phone number. Thus the number of possible seven-digit phone numbers is 8  10  10  10  10  10  10  8,000,000. Since a runner can only finish once, there are no repetitions. And since we are assuming that there is no tie, the number of       number of ways to number of ways to number of ways to different finishes is      5  4  3  2  1  120. choose the 1st runner choose the 2nd runner choose the 5th runner Since there are 4 main courses, there are 6 ways to choose a main course. Likewise, there are 5 drinks and 3 desserts so there are 5 ways to choose a drink and 3 ways to choose a dessert. So the number of different meals consisting of a main course, a       number of ways to number of ways to number of ways to drink, and a dessert is    4 5 3  60. choose the main course choose a drink choose a dessert By the Fundamental Counting Principle, the number of different routes from town A to town D via towns B and C is       number of routes number of routes number of routes   4 5 6  120.  from C to D from A to B from B to C The number of possible sequences of heads and tails when a coin is flipped 5 times is       number of possible number of possible number of possible      2 2 2 2 2 outcomes on the 1st flip outcomes on the 2nd flip outcomes on the 5th flip  25  32

Here there are only two choices, heads or tails, for each flip. 26. Since each die has six different faces, the number of different outcomes when rolling a red and a white die is 6  6  36. 27. Since there are six different faces on each die, the number of possible outcomes when a red die and a blue die and a white die  are rolled is      number of possible number of possible number of possible    6 6 6  63  216. outcomes on the red die outcomes on the blue die outcomes on the white die 28. The number of possible skirt-blouse-shoe outfits is       number of ways number of ways number of ways    5 8 12  480. to choose a skirt to choose a blouse to choose shoes 29. The number of different California license plates possible is          number of ways to number of ways number of ways    9 263 103  158,184,000. choose a nonzero digit to choose 3 letters to choose 3 digits 30. The number of possible ID numbers consisting of one letter followed by 3 digits is 26 10 10 10  26,000. 31. Since successive numbers cannot be the same, the number of possible choices for the second number in the combination is only 59. The third number in the combination cannot be the same as the second in the combination, but it can be the same as the first number, so the number of possible choices for the third number in the combination is also 59. So the number of possible combinations consisting of a number in the clockwise direction, a number in the counterclockwise direction, and then a number in the clockwise direction is 60 59 59  208,860.

SECTION 14.1 Counting

909

32. The number of possible license plates of two letters followed by three digits is        number of ways to number of ways   262 103  676,000. Since 676,000  8,000,000, there will not be choose 2 letters to choose 3 digits enough different license plates for the state’s 8 million registered cars. 33. Since a student can hold only one office, the number of ways that a president, a vice-president and a secretary can be chosen from  a class of 30 students  is    number of ways number of ways to number of ways    30 29 28  24,360. to choose a president choose a vice-president to choose a secretary 34. The number of ways a chairman, vice-chairman, and a secretary can be chosen if the chairman must be a Democrat and the vice-chairman must be a Republican is      number of ways to choose

 

a Democratic chairman

from the 10 Democrats

number of ways to choose

    a Republican vice-chairman    from the 10 Republicans

number of ways to choose a secretary from

   10 7 15  1050.

the remaining 15 members

35. We have seven choices for the first digit and 10 choices for each of the other 8 digits. Thus, the number of Social Security numbers is 7  108  700,000,000. 36. The number of possible ways to arrange three girls and four boys is     number of ways to number of ways   3!  4!  144. arrange 3 girls to arrange 4 boys 37. (a) The number of ways to select 5 of the 8 objects is C 8 5 

8!  56. 5! 3!

(b) A set with 8 elements has 28  256 subsets.

38. We may choose any subset of the 8 available brochures. There are 28  256 ways to do this. 39. Each subset of toppings constitutes a different way a hamburger can be ordered. Since a set with 10 elements has 210  1024 subsets, there are 1024 different ways to order a hamburger. 40. We consider a set of 20 objects (the shoppers in the mall) and a subset that corresponds to those shoppers that enter the store. Since a set of 20 objects has 220  1,048,576 subsets, there are 1,048,576 outcomes to their decisions. 41. (a) The number of ways to seat ten people in a row of ten chairs is 10!  3,628,800. (b) The number of ways to choose six out of ten people and seat them in six chairs is C 10 6  6! 

10!  6! 4!

10!  151,200. 4! 42. The number of ways of selecting 3 objects in order (a 3-letter word) from 6 distinct objects (the 6 letters) assuming that the letters cannot be repeated is P 6 3  6  5  4  120. 6! 

43. In selecting these officers, order is important and repetition is not allowed, so the number of ways of choosing 3 officers from 15 students is P 15 3  2730. 44. The number of ways of selecting 3 objects in order (a 3-digit number) from 4 distinct objects (the 4 digits) with no repetition of the digits is P 4 3  4  3  2  24. 45. Since the order of finish is important, we want the number of permutations of 8 objects (the contestants) taken three at a 8! 8!  8  7  6  336.  time, which is P 8 3  5! 8  3!

46. The number of ways of ordering 8 pieces in order (without repeats) is P 8 8  8!  40,320.

47. The number of ways of ordering 9 distinct objects (the contestants) is P 9 9  9!  362,880. Here a runner cannot finish more than once, so no repetitions are allowed, and order is important. 48. The number of ways of ordering three of the five distinct flags is P 5 3  60.

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CHAPTER 14 Counting and Probability

49. The number of ways of ordering 1000 distinct objects (the contestants) taking 3 at a time is P 1000 3  1000  999  998  997,002,000. We are assuming that a person cannot win more than once, that is, there are no repetitions. 50. In selecting these officers, order is important and repetition is not allowed, so the number of ways of choosing 4 officers from 30 students is P 30 4  657,720. 51. We first place Jack in the first seat, and then seat the remaining four students. Thus the number of these arrangements is     number of ways to number of ways to seat   P 1 1  P 4 4  1! 4!  24. seat Jack in the first seat the remaining four students 52. We start by first placing Jack in the middle seat, and then we place the remaining four students in the remaining four seats.     number of ways number of ways to seat Thus the number of a these arrangements is   1  P 4 4  1! 4!  24. the remaining four students to seat Jack 53. Here we have 6 objects, of which 2 are blue marbles and 4 are red marbles. Thus the number of distinguishable permutations 6  5  4! 6!   15. is 2! 4! 2  4! 54. Here we have 14 objects (the 14 balls) of which 5 are red balls, 2 are white balls, and 7 are blue balls. So the number of 14!  72,072. distinguishable permutations is 5! 2! 7! 55. The number of distinguishable permutations of 12 objects (the 12 coins), from like groups of size 4 (the pennies), of size 3 12!  277,200. (the nickels), of size 2 (the dimes) and of size 3 (the quarters) is 4! 3! 2! 3! 56. The word ELEEMOSYNARY has 12 letters of which 3 are E, 2 are Y, and the remaining letters are distinct. So we wish to find the number of distinguishable permutations of 12 objects (the 12 letters) from like groups of sizes 3 and 2 and seven 12! like groups of size 1. We get  39,916,800. 3! 2! 1! 1! 1! 1! 1! 1! 57. The number of distinguishable permutations of 12 objects (the 12 ice cream cones) from like groups of size 3 (the vanilla cones), of size 2 (the chocolate cones), of size 4 (the strawberry cones), and of size 5 (the butterscotch cones) is 14!  2,522,520. 3! 2! 4! 5! 58. This is the number of distinguishable permutations of 7 objects (the students) from like groups of size 3 (the ones who stay in the 3-person room), size 2 (the ones who stay in the 2-person room), size 1 (the one who stays in the 1-person room), and 7! size 1 (the one who sleeps in the car). This number is  420. 3! 2! 1! 1! 59. The number of distinguishable permutations of 8 objects (the 8 cleaning tasks) from like groups of size 5, 2, and 1 workers, 8! respectively is  168. 5! 2! 1! 60. The number of distinguishable permutations of 30 objects (the students) from like groups of sizes 8, 11, and 11 is 30!  4,128 ,840,588,600. 8! 11! 11! 61. Here we are interested in the number of ways of choosing three objects (the three members of the committee) from a set of 25!  2300. 25 objects (the 25 members). The number of combinations of 25 objects taken three at a time is C 25 3  3! 22! 6! 62. We want the number of ways of choosing a group of three from a group of six. This number is C 6 3   20. 3! 3! 12! 63. We want the number of ways of choosing a group of three from a group of 12. This number is C 12 3   220. 3! 9! 64. We want the number of ways of choosing a group of six people from a group of ten people. The number of combinations of 10!  210. ten objects (people) taken six at a time is C 10 6  6! 4!

SECTION 14.1 Counting

911

65. We want the number of ways of choosing a group (the 5-card hand) where order of selection is not important. The number 52!  2,598,960. of combinations of 52 objects (the 52 cards) taken 5 at a time is C 52 5  5! 47! 66. Since order is not important in a 7-card hand, the number of combinations of 52 objects (the 52 cards) taken 7 at a time is 52!  133,784,560. C 52 7  7! 45! 67. The order of selection is not important, hence we must calculate the number of combinations of 10 objects (the 10 questions) 10! taken 7 at a time, this gives C 10 7   120. 7! 3! 68. In this exercise, we assume that the pizza toppings cannot be repeated, so we are interested in the number of ways to select a 16!  560. subset of 3 toppings from a set of 16 toppings. The number of ways this can occur is C 16 3  3! 13! 69. We assume that the order in which he plays the pieces in the recital is not important, so the number of combinations of 12!  495. 12 objects (the 12 pieces) taken 8 at a time is C 12 8  8! 4! 70. The order in which the skirts are selected is not important and no skirt is repeated. So the number of combinations of 8!  56. eight skirts taken five at a time is C 8 5  5! 3! 71. The order in which the pants are selected is not important and no pair of pants is repeated, so the number of combinations of 10! ten pairs of pants taken three at a time is C 10 3   120. 3! 7! 72. (a) Since Jack must go on the field trip, we first pick Jack to go on the field trip, and then select the six other students from 29! the remaining 29 students. Since C 29 6   475,020, there are 475,020ways to select the students to go on 6! 23! the field trip with Jack. (b) We first take Jack out of the class of 30 students and select the 7 students from the remaining 29 students. Thus there 29! are C 29 7   1,560,780 ways to pick the 7 students for the field trip. 7! 22! (c) We are interested only in the group of seven students taken from the class of 30 students, not the order in which they are 30!  2,035,800. picked. Thus the number is C 30 7  7! 23! 73. Since the order in which the numbers are selected is not important, the number of combinations of 49 numbers taken 6 at a 49!  13,983,816. time is C 49 6  6! 43! 74. The number of distinguishable permutations of 13 objects (the total number of blocks he must travel) which can be 13! partitioned into like groups of size 8 (the east blocks) and of size 5 (the north blocks) is  1,287. 8! 5! 20!  15,504. 75. (a) The number of ways of choosing 5 students from the 20 students is C 20,5  5! 15! 12! (b) The number of ways of choosing 5 students for the committee from the 12 females is C 12 5   792. 5! 7! (c) We use the Fundamental Counting Principle to count the number of possible committees with 3 females and 2 males. Thus, we get     number of ways to choose the number of ways to choose the   C 12 3  C 8 2  220 28  6160. 3 females from the 12 females 2 males from the 8 males 76. We pick the two men from the group of ten men, and we pick the two women from the group of ten women. So the number of ways 2 men and 2 women can be chosen is     number of ways to pick number of ways to pick   C 10 2  C 10 2  45 45  2025. 2 of the 10 men 2 of the 10 women

912

CHAPTER 14 Counting and Probability

77. The number of ways the committee can be chosen is       number of ways to number of ways to number of ways to    C 20 1  C 19 1  C 18 4 choose 1 president choose 1 vice-president choose 4 other members  20  19  3060  1,162,800 78. (a) We choose 2 girls from the girls on the camping trip and the 4 boys from the boys on the camping trip. Thus the number of ways to pick the 6 to gather firewood is     number of ways to number of ways to   C 9 2  C 16 4  36 1820  65,520. choose 2 of 9 girls choose 4 of 16 boys (b) Method 1: We consider the number of ways of selecting the group of 6 from the 25 campers and subtract off the groups that contain no girls and those that contain one girls. Thus the number groups that contain at least two girls is C 25 6  C 9 0  C 16 6  C 9 1  C 16 5  177,100  1 8008  9 4368  129,780. Method 2: In the method we construct all the groups that are possible. Thus the number of groups is         groups with 2 girls and 4 boys



groups with 3 girls and 3 boys



groups with 4 girls and 2 boys



groups with 5 girls and 1 boy

 groups with 6 girls

 C 9 2  C 16 4  C 9 3  C 16 3  C 9 4  C 16 2  C 9 5  C 16 1  C 9 6  C 16 0  36 1820  84 560  126 120  126 16  84 1

 65,520  47,040  15,120  2,016  84  129,780. 79. The number of ways the committee can be chosen is         number of ways to number of ways to number of ways to number of ways to    choose 2 of 6 freshmen choose 3 of 8 sophomores choose 4 of 12 juniors choose 5 of 10 seniors  C 6 2  C 8 3  C 12 4  C 10 5  15  56  495  252  104,781,600 80. The leading and supporting roles are different (order counts), while the extra roles are not (order doesn’t count). Also, the male roles must be filled by the male actors and the female roles filled by the female actresses. Thus, number of ways the actors and actresses can be chosen is         number of ways the number of ways the     number of ways to number of ways to leading and the supporting  leading and the supporting         choose 5 of 8 male extras  choose 3 of 10 female extras  actors can be chosen

actresses can be chosen

 P 10 2  P 12 2  C 8 5  C 10 3  90 132 56 120  79,833,600.

81. We choose 3 forwards from the forwards, 2 defensemen from the defensemen, and the goalie from the two goalies. Thus the number of ways to pick the 6 starting players is       number of ways to number of ways to number of ways to    C 12 3  C 6 2  C 2 1 pick 3 of 12 forwards pick 2 of 6 defensemen pick 1 of 2 goalies  220 15 2  6600 82. To order a pizza, we must make several choices. First the size (4 choices), the type of crust (2 choices), and then the toppings. Since there are 14 toppings, the number of possible choices is the number of subsets of the 14 toppings, that is 214 choices. So by the Fundamental Counting Principle, the number of possible pizzas is 4 2  214  131,072. 83. We count the total number of committees and subtract the number that contain both Barry and Harry. The total number of committees possible is C 10 4 and the number that contain both Barry and Harry is C 8 2, so the number of possible committees is C 10 4  C 8 2  210  28  182.

SECTION 14.1 Counting

913

84. Method 1: We consider the number of 5-member committees that can be formed from the 26 interested people and subtract the numbers of those that contain no teacher and those that contain no students. Thus the number of committees is C 26 5  C 12 5  C 14 5  65,780  792  2002  62,986. Method 2: Here we construct all the committees that are possible. Thus the number of committees is         committees with



1 student and 4 teachers

committees with

2 students and 3 teachers



committees with

3 students and 2 teachers



committees with

4 students and 1 teacher

 C 14 1  C 12 4  C 14 2  C 12 3  C 14 3  C 12 2  C 14 4  C 12 1  14 495  91 220  364 66  1001 12  6,930  20,020  24,024  12,012  62,986 85. Since the two algebra books must be next to each other, we first consider them as one object. So we now have four objects to arrange and there are 4! ways to arrange these four objects. Now there are 2 ways to arrange the two algebra books. Thus the number of ways that 5 mathematics books may be placed on a shelf if the two algebra books are to be next to each other is 2  4!  48.

86. We treat John and Jane as one object and Mike and Molly as one object, so we need the number of ways of permuting 8 objects. We then multiply this by the number of ways of arranging John and Jane within their group and arranging Mike and Molly within their group. Thus the number of possible arrangements is P 8 8  P 2 2  P 2 2  8!  2!  2!  161,280. 87. (a) To find the number of ways the men and women can be seated we first select and place a man in the first seat and then     select 1 of arrange the arrange the other 7 people. Thus we get   C 4 1  P 7 7  4  7!  20,160. the 4 men remaining 7 people (b) To find the number of ways the men and women can be seated we first select and place a woman in the first and last seats and then arrange the other 6 people. Thus we get     arrange 2 of arrange the   P 4 2  P 6 6  12  6!  8,640. the 4 women remaining 6 people 88. (a) We treat the women as one object and find the number of ways of permuting 5 objects. We then permute the 4 women. Thus the number of arrangements is P 5 5  P 4 4  5!  4!  2880. (b) There are two ways to solve this exercise. Method 1: The number of ways the men and women can be seated is       number of ways the number of ways the number of ways to select    P 4 4  P 4 4  C 2 1 men can be arranged women can be arranged the gender of the first seat  4!  4!  2  1152

Method 2: There are 8 choices for the first seat, 4 for the second, 3 for the third, 3 for the fourth, 2 for the fifth, 2 for the sixth, 1 for the seventh, and 1 for the eighth. Thus the number of ways of seating the men and women in the required fashion is 8  4  3  3  2  2  1  1  1152.

89. The number of ways the top finalist can be chosen is       

number of ways to choose the 6

  

semifinalists from the 30

number of ways to choose the 2

number of ways



     to choose the winner  C 30 6  C 6 2  C 2 1

finalists from the 6

from the 2 finalists

 593,775 15 2  17,813,250

90. There are many different possibilities here, so we consider the complement where no professor is chosen for the delegates and subtract this number from the way to select 3 people from the group of 8 people, which is C 8 3. If the professor cannot to be selected, then we must select 3 people from a group of 5 and this can be done in C 5 3 ways. Thus the number of delegations that contain a professor is C 8 3  C 5 3  56  10  46. 91. Since there are 26 letters, the possible number of combinations of the first and the last initials is 26 26  676. Since 677  676, there must be at least two people that have the same first and last initials in any group of 677 people.

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CHAPTER 14 Counting and Probability

92. When two objects are chosen from ten objects, it determines a unique set of eight objects, those not chosen. So choosing two objects from ten objects is the same a choosing eight objects from ten objects. In general, every subset of r objects chosen from a set of n objects determines a corresponding set of n  r objects, namely, those not chosen. Therefore, the total number of combinations for each type are equal. 93. We are only interested in selecting a set of three marbles to give to Luke and a set of two marbles to give to Mark, not the order in which we hand out the marbles. Since both C 10 3  C 7 2 and C 10 2  C 8 3 count the number of ways this can be done, these numbers must be equal. (Calculating these values shows that they are indeed equal.) In general, if we wish to find two distinct sets of k and r objects selected from n objects (k  r  n), then we can either first select the k objects from the n objects and then select the r objects from the n  k remaining objects, or we can first select the r objects         n n r n nk from the n objects and then the k objects from the n  r remaining objects. Thus    . r k k r 94. (a) x  y5  x  y x  y x  y x  y x  y

 x  y x  y x  y x x  x y  yx  yy

 x  y x  y x x x  x x y  x yx  x yy  yx x  yx y  yyx  yyy

 x  y x x x x  x x x y  x x yx  x x yy  x yx x  x yx y  x yyx  x yyy  yx x x  yx x y  yx yx  yx yy  yyx x  yyx y  yyyx  yyyy

 x x x x x  x x x x y  x x x yx  x x x yy  x x yx x  x x yx y  x x yyx  x x yyy x yx x x  x yx x y  x yx yx  x yx yy  x yyx x  x yyx y  x yyyx  x yyyy yx x x x  yx x x y  yx x yx  yx x yy  yx yx x  yx yx y  yx yyx  yx yyy yyx x x  yyx x y  yyx yx  yyx yy  yyyx x  yyyx y  yyyyx  yyyyy (b) There are ten terms that contain two x’s and three y’s. Their sum is x x yyy  x yx yy  x yyx y  x yyyx  yx x yy  yx yx y  yx yyx  yyx x y  yyx yx  yyyx x (c) To count the number of terms with two x’s, we must count the number of ways to pick two of the five positions to contain an x. This number is C 5 2.   n (d) In the Binomial Theorem, the coefficient is the number of ways of picking r positions in a term with n factors to r contain an x. By definition, this is C n r .

14.2 PROBABILITY 1. The set of all possible outcomes of an experiment is called the sample space. A subset of the sample space is called an event. The sample space for the experiment of tossing two coins is S  H H H T T H T T , and the event “getting at least 3 n E  . one head” is E  H H H T T H. The probability of getting at least one head is P E  n S 4 2. (a) The probability of E or F occurring is P E  F  P E  P F  P E  F.

(b) If the events E and F have no outcome in common (that is, the intersection of E and F is empty), then the events are called mutually exclusive. So in drawing a card from a deck, the event E of “getting a heart” and the event F of “getting a spade” are mutually exclusive.

(c) If E and F are mutually exclusive, then the probability of E or F is PE  F  P E  P F. n E  F . So in tossing a die, the conditional 3. The conditional probability of E given that F occurs is P E  F  n F probability of the event E “getting a six” given that that the event F “getting an even number” has occurred is 1 P E  F  . 3

SECTION 14.2 Probability

915

4. (a) The probability of E and F occurring is PE  F  P E P F  E.

(b) If the occurrence of E does not affect the probability of the occurrence F, then the events are called independent. So in tossing a coin twice, the event E of “getting heads on the first toss” and the event F of “getting heads on the second toss” are independent.

(c) If E and F are independent events, then the probability of E and F is PE  F  P E P F. 5. (a) S  1 2 3 4 5 6 (b) E  2 4 6 (c) E  5 6

6. (a) There are two possible outcomes of the coin toss and 52 possible outcomes of drawing a card, so n S  2  52  104. (b) H A H A H A H A

(c) T J  T Q T K  T J  T Q T K  T J  T Q T K  T J  T Q T K 

(d) H 2 H 3 H4 H5 H6 H7 H8 H9 H10 H J  H Q H K  H A

7. Let H stand for head and T for tails. (a) The sample space is S  H H H T T H T T .

1 (b) Let E be the event of getting exactly two heads, so E  H H. Then P E  nE nS  4 .

3 (c) Let F be the event of getting at least one head. Then F  H H H T T H, and P F  nF nS  4 . 2 1 (d) Let G be the event of getting exactly one head, that is, G  H T T H . Then P G  nG nS  4  2 . 8. Let H stand for heads and T for tails; the numbers 1, 2,   , 6 are the faces of the die. (a) S  H 1 H2 H3 H 4 H5 H6 T 1 T 2 T 3 T 4 T 5 T 6

3 1 (b) Let E be the event of getting heads and rolling an even number. So E  H2 H4 H6, and P E  nE nS  12  4 .

(c) Let F be the event of getting heads and rolling a number greater than 4. So F  H5 H6, and 2 1 P F  nF nS  12  6 .

3 1 (d) Let G be the event of getting tails and rolling an odd number. So G  T 1 T 3 T 5, and P G  nG nS  12  4 . 1 9. (a) Let E be the event of rolling a six. Then P E  nE nS  6 . 3 1 (b) Let F be the event of rolling an even number. Then F  2 4 6. So P F  nF nS  6  2 . 1 (c) Let G be the event of rolling a number greater than 5. Since 6 is the only face greater than 5, P G  nG nS  6 . 2 1 10. (a) Let E be the event of rolling a two or a three. Then P E  nE nS  6  3 . 3 1 (b) Let F be the event of rolling an odd number. So F  1 3 5, and P F  nF nS  6  2 . 1 (c) Let G be the event of rolling a number divisible by 3. So G  3 6, and P G  nG nS  3 . 4 1 11. (a) Let E be the event of choosing a king. Since a deck has four kings, P E  nE nS  52  13 . (b) Let F be the event of choosing a face card. Since there are three face cards per suit and four suits, 12 3 P F  nF nS  52  13 .

  3  10 . (c) Let F be the event of choosing a face card. Then P F   1  P F  1  13 13

13 1 12. (a) Let E be the event of choosing a heart. Since there are 13 hearts, P E  nE nS  52  4 . 26 1 (b) Let F be the event of choosing a heart or a spade. Since there are 13 hearts and 13 spades, P F  nE nS  52  2 . (c) Let G be the event of choosing a heart, a diamond or a spade. Since there are 13 cards in each suit, 39 3 P G  nG nS  52  4 .

916

CHAPTER 14 Counting and Probability

5 13. (a) Let E be the event of selecting a red ball. Since the jar contains five red balls, P E  nE nS  8 . (b) Let F be the event of selecting a yellow ball. Since there is only one yellow ball,   1 7 P F   1  P F  1  nF nS  1  8  8 .

0 (c) Let G be the event of selecting a black ball. Since there are no black balls in the jar, P G  nG nS  8  0.

14. (a) Let E be the event of selecting a white or a yellow ball. Since there are two white balls and one yellow ball,   3 5 P E   1  P E  1  nE nS  1  8  8 .

(b) Let F be the event of selecting a red, a white or a yellow ball. Since all the types of balls are in the jar, P E  1. (c) Let G be the event of selecting a white ball. Since there are 2 white balls,   2 6 3 P E   1  P E  1  nE nS  1  8  8  4 .

1287 C 13 5   0000495. C 52 5 2,598,960 (b) Let E be the event of choosing five cards of the same suit. Since there are four suits and 13 cards in each suit, 5,148 4  C 13 5   000198. n E  4  C 13 5. Also, n S  C 52 5. Therefore, P E  C 52 5 2,598,960 (c) Let E be the event of dealing five face cards. Since there are 3 face cards for each suit and 4 suits, 792 C 12 5   0000305. P E  C 52 5 2,598,960 (d) Let E be the event of dealing a royal flush (ace, king, queen, jack, and 10 of the same suit). Since there is only one such 4 4   153908  106 . sequence for each suit, there are only 4 royal flushes, so P E  C 52 5 2,598,960

15. (a) Let E be the event of dealing five hearts. Since there are 13 hearts, P E 

16. (a) Let E be the event of choosing three defective CDs. Since there are four defective CDs, n E  C 4 3. Also, C 4 3 4 1 n E     0018. n S  C 12 3. Therefore, P E  n S C 12 3 220 55 (b) Let E be the event of choosing three functioning CDs. Since there are eight functioning CDs, n E  C 8 3, so C 8 3 56 14 n E     0255. P E  n S C 12 3 220 55 17. (a) Let E be the event of choosing two red balls. Since there are three red balls, n E  C 3 2. Also, n S  C 8 2. C 3 2 3 n E    0107. Therefore, P E  n S C 8 2 28 (b) Let E be the event of choosing two white balls. Since there are five white balls, n E  C 5 2, so C 5 2 10 5 n E     0357. P E  n S C 8 2 28 14 3 . 18. (a) Let E be the event of choosing a T. Since 3 of the 16 letters are Ts, P E  16

(b) Let F be the event of choosing a vowel. Since there are 6 vowels,   (c) Let F be the event of choosing a vowel. Then P F   1  38 

6  3. P F  16 8 5. 8

19. (a) Let E be the probability that at least one card is a spade. The number of hands that do not contain a spade is the number 7411 C 39 5   0778. of possible 5-card hands using the other three suits, that is, C 39 5. Thus, P E  1  C 52 5 9520 (b) Let E be the probability that at least one card is a face card. The number of hands that do not contain a face card is the number of possible 5-card hands using the cards of the deck that are not face cards, that is, C 40 5. Thus, 6221 C 40 5   0747. P E  1  C 52 5 8330

SECTION 14.2 Probability

917

20. (a) S  1 1  1 2  1 3  1 4  1 5  1 6  2 1  2 2  2 3  2 4  2 5  2 6  3 1  3 2  3 3  3 4  3 5  3 6  4 1  4 2  4 3  4 4  4 5  4 6  5 1  5 2  5 3  5 4  5 5  5 6  6 1  6 2  6 3  6 4  6 5  6 6 6  1. (b) Let E be the event of getting a sum of 7. Then E  1 6  2 5  3 4  4 3  5 2  6 1, and P E  36 6

4  1. (c) Let F be the event of getting a sum of 9. Then F  3 6  4 5  5 4  6 3, and P F  36 9

6  1. (d) Let E be the event that the two dice show the same number. Then P E  36 6 (e) Let E be the event that the two dice show different numbers. Then E  is the event that the two dice show the same   number. Thus, P E  1  P E   1  16  56 . 5 (f) Let E be the event of getting a sum of 9 or higher. Then P E  10 36  18 .

3 21. (a) Let E be the event that the spinner stops on red. Since 12 of the regions are red, P E  12 16  4 .

(b) Let F be the event that the spinner stops on an even number. Since 8 of the regions are even-numbered, 8  1. P F  16 2

4  1. (c) Since 4 of the even-numbered regions are red, P E  F  P E  P F  P E  F  34  12  16 4  1. 22. (a) Let E be the event that the spinner stops on blue. Since only 4 of the regions are blue, P E  16 4

8  1. (b) Let F be the event that the spinner stops on an odd number. Since 8 of the regions are odd-numbered, P F  16 2

(c) Since none of the odd-numbered regions are blue, P E  F  P E  P F  14  12  34 .

23. (a) Yes, the events are mutually exclusive since the number cannot be both even and odd. So P E  F  P E  P F  36  36  1. (b) No, the events are not mutually exclusive since 6 is both even and greater than 4. So P E  F  P E  P F  P E  F  36  26  16  23 . 24. (a) No, the events are not mutually exclusive since 4 is greater than 3 and also less than 5. So P E  F  P E  P F  P E  F  36  46  16  1. (b) Yes, the events are mutually exclusive since there are only 2 numbers less than 3, namely 1 and 2, but they are not divisible by 3. So P E  F  P E  P F  26  26  23 . 25. (a) No, the events E and F are not mutually exclusive since the jack, queen, and king of spades are both face cards and 12 3 11 spades. So P E  F  P E  P F  P E  F  13 52  52  52  26 . (b) Yes, the events E and F are mutually exclusive since the card cannot be both a heart and a spade. So 13 1 P E  F  P E  P F  13 52  52  2 .

26. (a) No, events E and F are not mutually exclusive since a king can be a club. So 4 1 4 P E  F  P E  P F  P E  F  13 52  52  52  13 .

(b) No, events E and F are not mutually exclusive since an ace can be a spade. So 4  13  1  4 . P E  F  P E  P F  P E  F  52 52 52 13

27. (a) Let E denote a roll of five and F a roll greater than three. Using the formula for conditional probability, we have 1 n E  F  . P E  F  n F 3 (b) Let E denote a roll of three and F an odd roll. Using the formula for conditional probability, we have 1 n E  F  . P E  F  n F 3

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CHAPTER 14 Counting and Probability

4 1 n E  F   . n F 12 3 n E  F 1 (b) Let E denote drawing a king and F drawing a spade. Then P E  F   . n F 13 n E  F 1 (c) Let E denote drawing a spade and F drawing a king. Then P E  F   . n F 4

28. (a) Let E denote drawing a queen and F drawing a face card. Then P E  F 

29. Let E denote the spinner stopping on an even number and F the spinner stopping on red. Then n E  F 4 1 P E  F    . n F 12 3 30. Let E denote the spinner stopping on a number divisible by 3 and F the spinner stopping on blue. Then n E  F 1 P E  F   . n F 4 31. (a) There is only one red ball numbered 3 and one green ball numbered 3. If the ball drawn is numbered 3, then the probability it is red is 12 . (b) There is only one ball numbered 7 and it is green, so if the ball drawn is numbered 7, then the probability it is green is 1. (c) There are two even-numbered red balls and three even-numbered green balls, so if the ball is even-numbered, then the probability it is a red ball is 25 . (d) There are five red balls, and two are even-numbered, so if the ball drawn is red, then the probability it is even-numbered is 25 . 32. (a) If the first ball drawn is red, there are four red balls and seven green balls remaining, so the probability that the second 4 . ball is red is 11 5. (b) If the first ball drawn is green, there are five red balls and six green balls remaining, so the required probability is 11

(c) If the first ball drawn is odd-numbered, then there are six odd-numbered and five even-numbered balls remaining, so the 5 . required probability is 11

(d) If the first ball drawn is even-numbered, then there are seven odd-numbered and four even-numbered balls remaining, 4 . so the required probability is 11

33. (a) Let E be the event of drawing a black ball first. Because the jar contains seven black balls and three white balls, the 7 . The probability of the second ball being white is 3  1 , so the probability of the first ball being black is P E  10 9 3 7 1 7 probability of the intersection is 10  3  30 .

7  2  7 . (b) Here the probability of the second ball being black is 69  23 , so the probability of the intersection is 10 3 15

34. (a) Let E be the event of drawing a red sock. Since three pairs are red, the drawer contains six red socks, and so 6 1 P E  nE nS  18  3 . 5 (b) Let F be the event of drawing another red sock. Since there are 17 socks left of which 5 are red, P F  nF nS  17 . 5  5 . (c) In this case, the probability is P E  F  P E  P F  13  17 51 4 and the probability of the second being a king is 4 , so the 35. (a) The probability of the first card being an ace is 52 51 4  4  4 . probability of the intersection is 52 51 663

4 and the probability of the second being an ace is 3 , so the (b) The probability of the first card being an ace is 52 51 4  3  1 . probability of the intersection is 52 51 221

36. (a) Yes, the first roll does not influence the outcome of the second roll.

   1  1 . (b) The probability of getting a six on both rolls is P E  F  P E  P F  16 6 36

SECTION 14.2 Probability

919

37. Let E be the event of getting a 1 on the first roll, and let F be the event of getting an even number on the second roll. Since 1 . these events are independent, P E  F  P E  P F  16  36  12 38. (a) Yes, they are independent. The toss of the coin does not influence the roll of the die.

(b) The probability of getting a tail is 12 and the probability of getting an even number is 36  12 , so the probability of the intersection is 12  12  14 . 39. (a) Yes. What happens on spinner A does not influence what happens on spinner B.

   2  1. (b) The probability that A stops on red and B stops on yellow is P E  F  P E  P F  24 8 8

40. (a) Let E A and E B be the event that the respective spinners stop on a purple region. Since these events are independent,     1 . P E A  E B  P E A  P E B   14  28  16 (b) Let FA and FB be the event that the respective spinners stop on a blue region. Since these events are independent,     1. P FA  FB  P FA  P FB   14  18  32

41. (a) Let B and G stand for “boy”and “girl”. Then S  B B B B G B B B BG B B B BG B B B BG GG B B G BG B G B BG BGG B BG BG B BGG BGGG G BGG GG BG GGG B GGGG 1. (b) Let E be the event that the couple has only boys. Then E  B B B B and P E  16

(c) Let F be the event that the couple has 2 boys and 2 girls. Then

6  3. F  GG B B G BG B G B BG BGG B BG BG B BGG, so P F  16 8

2  1. (d) Let G be the event that the couple has 4 children of the same sex. Then G  B B B B GGGG, and P G  16 8 (e) Let H be the event that the couple has at least 2 girls. Then H  is the event that the couple has fewer than two girls.     5  11 . Thus, H   B B B B G B B B BG B B B BG B B B BG, so n H   5, and P H  1  P H   1  16 16

42. Let E be the event that a 13-card bridge hand consists of all cards from the same suit. Since there are exactly 4 such hands 4  63  1012 . (one for each suit), P E  C 52 13 43. Let E be the event that the ball lands in an odd numbered slot. Since there are 18 odd numbers between 1 and 36

9 P E  18 38  19 . 44. (a) Let E be the event that the toddler arranges the word FRENCH. Since the letters are distinct, there are P 6 6 ways of 1 1   00014. arranging the blocks of which only one spells the word FRENCH. Thus P E  P 6 6 720 (b) Let E be the event that the toddler arranges the letters in alphabetical order. Since there are P 6 6 ways of arranging 1 1   00014. the blocks of which only one is in alphabetical order, P E  P 6 6 720 45. Let E be the event of picking the 6 winning numbers. Since there is only one way to pick these, 1 1   715  108 . P E  C 49 6 13,983,816 46. Let E be the event that no women are chosen. The number of ways that no women are chosen is the same as the number of C 11 6  000078. ways that only men are chosen, which is C 11 6. Thus P E  C 30 6

47. The sample space consist of all possible True-False combinations, so n S  210 . Let E be the event that the student 1 1 . answers all 10 questions correctly. Since there is only one way to answer all 10 questions correctly, P E  10  1024 2 48. Let E be the event that the batch will be discarded. Thus, E is the event that at least one defective bulb is found. It is easier to find E  , the event that no defective bulbs are found. Since there are 10 bulbs in the batch of which 8 are non-defective,     C 8 3 C 8 3 . Thus P E  1  P E   1   1  04667  05333. P E  C 10 3 C 10 3

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CHAPTER 14 Counting and Probability

49. (a) Let E be the event that the monkey types “Hamlet” as his first word. Since “Hamlet” contains 6 letters and there are 48 1 typewriter keys, P E  6  818  1011 . 48 (b) Let F be the event that the monkey types “to be or not to be” as his first words. Since this phrase has 18 characters 1 (including the blanks), P F  18  547  1031 . 48 1 1  00014. 50. (a) Let E be the event that the monkey arranges the 6 blocks to spell HAMLET. Then P E   720 6! (b) The probability that the monkey arranges the 6 blocks to spell HAMLET three consecutive times is the probability of   1 3  268  109 . three independent events E, and hence is equal to [P E]3  720

51. Let E be the event that the toddler will arrange the 8 blocks to spell TRIANGLE or INTEGRAL. The number of ways of arranging these blocks is the number of distinguishable permutations of 8 blocks. Since no two blocks are the same, the 2  496  105 number of distinguishable permutations is 8!. Two of these arrangements result in event E, so P E  8! or 00000496. 52. Let E be the event that you predict the correct order for the horses to finish the race. Since there are eight horses, there are P 8 8  8! ways that the horses could finish, with only one being the correct order. Thus, 1 1   248  105 . P E  P 8 8 40,320 53. (a) Let E be the event that the pea is tall. Since tall is dominant, E  T T T t t T . So P E  34 .   (b) E  is the event that the pea is short. So P E   1  P E  1  34  14 .

(a) Let E be the event that the offspring will be tall. Since only offspring

54.

Parent 2

Parent 1

t

t

T

Tt

Tt

t

tt

tt

with genotype Tt will be tall, P E  24  12 . (b) E  is the event that the offspring will not be tall (thus, the offspring is   short). So P E   1  P E  1  12  12 .

55. Let E be the event that the player wins on spin 1, and let F be the event that the player wins on spin 2. What happens on the first spin does not influence what happens on the second spin, so the events are independent. Thus,

1  1  1 . P E  F  P E  P F  38 38 1444 56. Let E be the event that the committee is all male and F the event it is all female. The sample space is the set of all ways that 5 people can be chosen from the group of 14. These events are mutually exclusive, so C 8 5 6  56 31 C 6 5    . P E  F  P E  P F  C 14 5 C 14 5 2002 1001 57. Let E, F and G denote the events of rolling two ones on the first, second, and third rolls, respectively, of a pair of dice. The 1  1  1  1  214  105 . events are independent, so P E  F  G  P E  P F  P G  36 36 36 363 58. Let E be the event that a player has exactly 5 winning numbers and F be the event that a player has all 6 winning numbers. These events are mutually exclusive. For a players to have exactly 5 winning numbers means that the player has 5 of the 6 winning numbers and 1 number that was not selected in the lottery. So n E  C 6 5  C 43 1. Thus, 1 C 6 5  C 43 1   00000185. P at least 5 winning numbers  P E  F  P E  P F  C 49 6 C 49 6

59. Let E be the event that the marble is red and F be the event that the number is odd-numbered. Then E  is the event that the marble is blue, and F  is the event that the marble is even-numbered. 6  3 (a) P E  16 8 8  1 (b) P F  16 2

SECTION 14.2 Probability

921

6  8  3  11 (c) P E  F  P E  P F  P E  F  16 16 16 16         8  5  13 . (d) P E   F   P E   P F   P E   F   10  16 16 16 16

60. The number of ways a set of six numbers can be selected from a group of 49 numbers is C 49 6. Since the games are  2 1 independent, the probability of winning the lottery two times in a row is  511  1015 . C 49 6    61. The probability of getting 2 red balls by picking from jar B is 57 46  10 21 . The probability of getting 2 red balls by    5  15 . The probability of getting 2 red balls after putting all balls in one jar is picking one ball from each jar is 37 7 49    8 7  4 . Hence, picking both balls from jar B gives the greatest probability. 14 13 13 1 , and is the same for the second and third wheels. The events are 62. (a) The probability that the first wheel has a bar is 11 1  1  1  1 . independent, and so the probability of getting 3 bars is 11 11 11 1331

(b) The probability of getting a number on the first wheel is 10 11 , the probability of getting the same number on the second 1 , and the probability of getting the same number on the third wheel is 1 . Thus, the probability of getting wheel is 11 11 1 1 10 the same number on each wheel is 10 11  11  11  1331 . (c) We use the complement, no bar, to determine the probability of at least one bar. The probability that the first wheel does

not have a bar is 10 11 , and is the same for the second and third wheels. Since the events are independent, the probability 10 10 103 1000 1000 331 of getting no bar is 10 11  11  11  113  1331 , and so P at least one bar  1  1331  1331 .

63. Let E be the event that she opens the lock within an hour. The number of combinations she can try in one hour is 10  60  600. The number of possible combinations is P 40 3 assuming that no number can be repeated. Thus 600 5 600    0010 . P E  P 40 3 59,280 494 64. (a) Let E be the event that the curriculum committee consists of 2 women and 4 men. So number of committees with 2 women and 4 men C 8 2  C 10 4 28  210 490 P E     . number of ways to select 6 member committee C 18 6 18564 1547 (b) Let F be the event that curriculum committee consists of two or fewer women. Then F is the event that the committee has no woman or one woman or two women. Then P F  

1  210  8  252  28  210 C 8 0  C 10 6 C 8 1  C 10 5 C 8 2  C 10 4    C 18 6 C 18 6 C 18 6 18,564 193 8106  18,564 442

  249 (c) So F  is the event the at curriculum committee has more that two women. The P F   1  P F  1  193 442  442 .

65. Let E be the event that Paul stands next to Phyllis. To find n E we treat Paul and Phyllis as one object and find the number of ways to arrange the 19 objects and then multiply the result by the number of ways to arrange Paul and Phyllis. 2 19!  2!   010. So n E  19!  2!. The sample space is all the ways that 20 people can be arranged. Thus P E  20! 20 66. Let E be the event that the monkey arranges the 6 blocks to spell BUBBLE. The number of ways of arranging these blocks is the number of distinguishable permutations of 6 blocks. Since there are three blocks labeled B, the number of distinguishable 1 1 6! 3!  . permutations is . Only one of these arrangements spells the word BUBBLE. Thus P E   6! 3! 6! 120 3!

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CHAPTER 14 Counting and Probability

67. Let E be the event that the monkey arranges the 11 blocks to spell PROBABILITY. The number of ways of arranging these blocks is the number of distinguishable permutations of 11 blocks. Since there are two blocks labeled B and two 11! . Only one of these arrangements spells the word blocks labeled I, the number of distinguishable permutations is 2! 2! 1 1 2! 2! PROBABILITY. Thus P E   .  11! 11! 9,979,200 2! 2! 68. (a) Because the events are independent, the probability that the family has two boys given that the oldest child is a boy is 12 . (b) There are four equally likely possibilities: boy-boy, boy-girl, girl-boy, and girl-girl. In three cases, at least one of the children is a boy, and in one of those cases both children are boys. Thus, the probability that the family has two boys given that one of the children is a boy is 13 .

14.3 BINOMIAL PROBABILITY 1. A binomial experiment is one in which there are exactly two outcomes. One outcome is called success and the other is called failure. 2. If a binomial experiment has probability p of success then the probability of failure is 1  p. The probability of getting 3. 4. 5. 6. 7. 8.

exactly r successes in n trials of this experiment is C n r pr 1  pnr .    P 2 successes in 5  C 5 2  072 033  013230    P 3 successes in 5  C 5 3  073 032  030870    P 0 success in 5  C 5 0  070 035  000243    P 5 successes in 5  C 5 5  075 030  016807    P 1 success in 5  C 5 1  071 034  002835    P 1 failure in 5  C 5 1  031 074  036015. Note that exactly one failure is the same as exactly 4 successes,    and P 4 successes in 5  C 5 4  074 031  036015.

9. P at least 4 successes  P 4 successes  P 5 successes  C 5 4 074 031  C 5 5 075 030  036015  016807  052822 10. P at least 3 successes  P 3 successes  P 4 successes  P 5 successes

 C 5 3 073 032  C 5 4 074 031  C 5 5 075 030  030870  036015  016807  083692

11. P at most 1 failure  P 0 failure  P 1 failure  P 5 successes  P 4 successes  C 5 5 075 030  C 5 4 074 031  016807  036015  052822 12. P at most 2 failures  P 0 failure  P 1 failure  P 2 failure

 P 5 successes  P 4 successes  P 3 successes

 C 5 5 075 030  C 5 4 074 031  C 5 3 073 032  016807  036015  030870  083692

SECTION 14.3 Binomial Probability

13. P at least 2 successes  P 2 successes  P 3 successes  P 4 successes  P 5 successes  013230  030870  036015  016807  096922 14. P at most 3 failures  P 0 failure  P 1 failure  P 2 failures  P 3 failures

 P 5 successes  P 4 successes  P 3 successes  P 2 successes  016807  036015  030870  013230  096922 (b)

15. (a) Outcome

Probability

1

02

2

02

3

02

4

02

5

02

Outcome

Probability

1

05

2

03

3

01

4

01

5

0

16. (a)

0.2

0

1

2

3

4

5

(b)

17. (a)

0.1

0

1

2

3

4

3

4

(b) r

Probability

0

4

1 16 1 4 3 8 1 4 1 16

r

Probability

0

00776

1

02592

2

03456

3

02304

4

00768

5

00102

1 2 3

18. (a)

1 16

0

(b)

1

2

0.5

0

1

2

3

4

5

923

924

CHAPTER 14 Counting and Probability

19. (a)

(b) r

Probability

0

02097

1

03670

2

02753

3

01147

4

00287

5

00043

6

000036

7

0000013

r

Probability

0

0000001

1

0000054

2

0001215

3

0014580

4

0098415

5

0354294

6

0531441

20. (a)

0.3

(b)

0

1

0

1

2

3

4

5

6

7

0.5

2

3

4

5

6

 2  4 5 21. Here “success” is “face is 4” and P face is 4  16 . Then P 2 successes in 6  C 6 2  16  020094. 6 22. Here P success  08 and P failure   02.   (a) P 0 success in 7  C 7 0  080 027  00000128    (b) P 7 successes in 7  C 7 7  087 020  0209715

(c) P he hits target more than once  1  [P 0 success in 7  P 1 success in 7]         1  C 7 0  080 027  C 7 1  081 026  099963

(d) P at least 5 successes  P 5 successes  P 6 successes  P 7 successes           C 7 5  085 022  C 7 6  086 021  C 7 7  087 020  085197

   23. P 4 successes in 10  C 10 4  044 066  025082

24. The complement is that none of the raccoons had rabies.

   P at least 1 had rabies  1  P none had rabies  1  C 4 0  010 094  1  06561  03439

   25. (a) P 5 in 10  C 10 5  0455 0555  023403

(b) P at least 3  1  P at most 2  P 0 in 10  P 1 in 10  P 2 in 10            1  C 10 0  0450 05510  C 10 1  0451 0559  C 10 2  0452 0558  090044

   26. (a) P 12 in 15  C 15 12  0112 093  331695  1010

SECTION 14.3 Binomial Probability

925

(b) P at least 12  P 12 in 15  P 13 in 15  P 14 in 15  P 15 in 15        C 15 12  0112 093  C 15 13  0113 092       C 15 14  0114 091  C 15 15  0115 090  340336  1010

27. (a) The complement of at least 1 germinating is no seed germinating, so    P at least 1 germinates  1  P 0 germinates  1  C 4 0  0750 0254  099609.

(b) P at least 2 germinate  P 2 germinates  P 3 germinates  P 4 germinates           C 4 2  0752 0252  C 4 3  0753 0251  C 4 4  0754 0250

 094922    (c) P 4 germinates  C 4 4  0754 0250  031641

28. (a) P at least 3 boys  P 3 boys  P 4 boys  P 5 boys           C 5 3  053 052  C 5 4  054 051  C 5 5  055 050  05

(b) P at least 4 girls  P 4 girls  P 5 girls  P 6 girls  P 7 girls              C 7 4  054 053  C 7 5  055 052  C 7 6  056 051  C 7 7  057 050  05

   29. (a) P all 10 are boys  C 10 10  05210 0480  00014456.    (b) P all 10 are girls  C 10 0  0520 04810  000064925.    (c) P 5 in 10 are boys  C 10 5  0525 0485  024413.    30. (a) P 2 in 12  C 12 2  022 0810  028347. (b) The complement of “at least 3” is “at most 2”, so

P at least 3 in 12  1  P at most 2 in 12  1  [P 0 in 12  P 1 in 12  P 2 in 12]            1  C 12 0  020 0812  C 12 1  021 0811  C 12 2  022 0810

 044165    31. (a) P 3 in 3  C 3 3  00053 09950  0000000125.

(b) The complement of “one or more bulbs is defective” is “none of the bulbs is defective.” So    P at least 1 defective  1  P none is defective  1  C 3 0  00050 09953  0014925.

   32. P at least 1 in 10  1  P 0 in 10  1  C 10 0  0050 09510  040126

33. The complement of “2 or more workers call in sick” is “0 or 1 worker calls in sick.” So P 2 or more  1  [P 0 in 8  P 1 in 8]         1  C 8 0  0040 0968  C 8 1  0041 0967  0038147

   34. P 3 in 5 favor  C 5 3  063 042  03456

   35. (a) P 6 in 6  C 6 6  0756 0250  017798    (b) P 0 in 6  C 6 0  0750 0256  000024414    (c) P 3 in 6  C 6 3  0753 0253  013184

926

CHAPTER 14 Counting and Probability

(d) P at least 2 seasick  1  P at most 1 seasick  1  [P 6 in 6 OK  P 5 in 6 OK]         1  C 6 6  0756 0250  C 6 5  0755 0251  046606 36. (a) P machine breaks  P at least 1 component fails  1  P 0 component fails     1  C 4 0  0010 0994  0039404    (b) P 0 component fails  C 4 0  0010 0994  0960596    (c) P 3 components fail  C 4 3  0013 0991  000000396 37. (a) The complement of “at least one gets the disease” is “none gets the disease.” Then    P at least 1 gets the disease  1  P 0 gets the disease  1  C 4 0  0250 0754  068359. (b) P at least 3 get the disease  P 3 get the disease  P 4 get the disease        C 4 3  0253 0751  C 4 4  0254 0750  005078

38. There are 52 cards in the deck of which 13 belong to any one suit, so P heart  P spade  P diamond 025.    P club

(a) P 3 in 3 are hearts  C 3 3  0253 0750  0015625    (b) P 2 in 3 are spades  C 3 2  0252 0751  0140625    (c) P 0 in 3 are diamonds  C 3 0  0250 0753  0421875    (d) P at least 1 is a club  1  P none is a club  1  C 3 0  0250 0753  0578125

39. Fred (a nonsmoker) is already in the room, concerns the remaining 4 participants assigned to the room.  exercise   sothis (a) P 1 in 4 is a smoker  C 4 1  031

073  04116

   (b) P at least 1 smoker  1  P 0 in 4 is a smoker  1  C 4 0  030 074  07599

40. (a) P 2 or more  1  [P 0 in 100  P 1 in 100]         1  C 100 0  0020 098100  C 100 1  0021 09899  059673

(b) Since P at least 1 interested  1  P 0 interested and 09835  0507, the telephone consultant needs to make at least 35 calls to ensure at least a 05 probability of reaching one or more interested parties.

41. (a) P 8 or more recover  P 0 dies  P 1 dies  P 2 die  C 10 0 060 0410  C 10 1 061 049  C 10 2 062 048  00123 (b) Yes, the drug appears to be effective. 42. (a) P 5 or more hits  P 5 hits  P 6 hits  P 7 hits  P 8 hits  C 8 5 065 043  C 8 6 066 042  C 8 7 067 041  C 8 8 068 040  0594 (b) No, the coaching does not appear to have made any difference.

SECTION 14.4 Expected Value

927

0.3

43. (a) Number of heads

Probability

0

0003906

1

003125

2

0109375

3

021875

4

0273475

5

021875

6

0109375

7

003125

8

0003906

0.25 0.2 Probability

0.15 0.1 0.05 0

2

4

6

8

Number of heads

If n  8, then 4 heads has the greatest probability of occurring. If the coin is flipped 100 times, then 50 heads has the greatest probability of occurring. 0.3

(b) Number of heads

Probability

0

0001953

1

0017578

2

0070313

3

0164063

4

0246094

5

0246094

6

0164063

7

0070313

8

0017578

9

0001953

0.25 0.2 Probability

0.15 0.1 0.05 0

2

4

6

8

Number of heads

If n  9, then 4 and 5 heads are the most likely outcomes. If the coin is flipped 101 times, then 50 and 51 heads are the most likely outcomes.

14.4 EXPECTED VALUE 1. If a game gives payoffs of $10 and $100 with probabilities 09 and 01, respectively, then the expected value of this game is E  10  09  100  01  $19. 2. If you played the game in Exercise 1 many times then you would expect your average payoff per game to be about $19.     3. Mike gets $2 with probability 12 and $1 with probability 12 . Thus, E  2 12  1 12  15, and so his expected winnings are $150 per game.

    4. The probability that Jane gets $10 is 16 , and the probability that she loses $1 is 56 . Thus E  10 16  1 56  0833, and so her expectation is $0833.

1 , the expected value of this game is 5. Since the probability of drawing the ace of spades is 52     1  1 51  49  094. So your expected winnings are $094 per game. E  100 52 52 52     6. The expected value of this game is E  3 12  2 12  52  25. So Tim’s expected winnings are $250 per game.

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    7. Since the probability that Carol rolls a six is 16 , the expected value of this game is E  3 16 050 56  55 6  09167. So Carol expects to win $092 per game.  2  14 , the probability that Albert gets one tail and one head is 8. The probability that Albert gets two tails is 12  2  2 C 2 1 12  12 , and the probability that Albert gets two heads is 12  14 . If Albert gets two heads, he will receive

$4, if he get one head and one tail, he will get $2 $1  $1, and if he get two tails, he will lose $2. Thus the expected value       1 of this game is E  4 14  1  2 14  1. So Albert’s expected winnings are $1 per game. 2

9. Since the probability that the die shows an even number equals the probability that that die shows an odd number, the     expected value of this game is E  2 12  2 12  0. So Tom should expect to break even after playing this game many times.

10. Since there are 4 aces, 12 face cards and only one 8 of clubs, the expected value of this game is       4  26 12  13 1  $1425. E  104 52 52 52

11. Since it costs $050 to play, if you get a silver dollar, you win only 1  050  $050. Thus the expected value of this game     2  050 8  030. So your expected winnings are $030 per game. In other words, you is E  050 10 10 should expect to lose $030 per game.

8  7  56 , and the probability of not choosing 2 white 12. The probability of choosing 2 white balls (that is, no black ball) is 10 9 90       56 34 34  3111. 0 balls (at least one black) is 1  90  90 . Therefore, the expected value of this game is E  5 56 90 90

Thus, John’s expected winnings are $311 per game.

    1  1 37   2  00526. 13. You can either win $35 or lose $1, so the expected value of this game is E  35 38 38 38 Thus the expected value is $00526 per game. 1 . After the first prize winner is selected, then 2  106 1 1 P winning the second prize  . Similarly, P winning the third prize  . So the expected 2  106  1 2  106  2             1 1 1 5  10  104  $0555. value of this game is E  106 2  106 2  106  1 2  106  2 (b) Since we expect to win $0555 on the average per game, if we pay $100, then our net outcome is a loss of $0445 per game. Hence, it is not worth playing, because on average you will lose $0445 per game.

14. (a) We have P winning the first prize 

15. By the rules of the game, a player can win $10 or $5, break even, or lose $100. Thus the expected value of this game is         10  5 10  100 2 78  050. So the expected winnings per game are $050.  E  10 100 0 100 100 100

16. Since the safe has a six digit combination, there are 106 possible combinations to the safe, of which only one is correct. The     1  106  1  0. expected value of this game is E  106  1  1 106 106 17. If the stock goes up to $20, she expects to make $20  $5  $15. And if the stock falls to $1, then she has lost $5  $1  $4. So the expected value of her profit is E  15 01  4 09  21. Thus, her expected profit per share is $210, that is, she should expect to lose $210 per share. She did not make a wise investment.  3 1 18. Since the wheels of the slot machine are independent, the probability that you get three watermelons is 11 . So the     expected value of this game is E  475 13  025 1  13  $0246. 11

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19. There are C 49 6 ways to select a group of six numbers from the group of 49 numbers, of which only one is a winning      1 1 set. Thus the expected value of this game is E  106  1  1 1   $093. C 49 6 C 49 6 20. (a) Since the life insurance policy costs $25 per year, we have the expected value E   7500  25 00003  25 09997  2275. (b) The expected yearly income is 450,000 2275  $10,237,500.

21. The expected number is E  2 015  3 045  4 030  5 010  335 hours of TV. 22. The expected number is 005 3  015 2  045 1  035 0  09 foreign languages.

23. The expected number is 3 030  2 045  1 015  0 010  195 times in any given week. 24. (a) Number of girls

Probability

0

1 8 3 8 3 8 1 8

1 2 3

(b) The expected number of girls is         0 18  1 38  2 38  3 18  15.

  1 12  51 1   27 , and so the game is not fair. 25. (a) The expected value is 52 52 2 104   1 x  51 1  0  x  51  $2550. (b) The game is fair with payout x, where 52 52 2 2

26. (a) The expected value is 13 20  23 10  0, and so the game is fair.

3070 10 27. (a) The expected value is 16  16 30  35 36 2  36   9 , and so the game is not fair. 1 x  35 2  0  x  $70. (b) The game is fair with payout x, where 36 36

1 28. (a) The expected value is 16  12 10  11 12 1   12 , and so the game is not fair.

(b) The game is fair with payout x, where 16  12 x  11 12 1  0  x  $11.   1  1  1 600  1  1  1  1 1   23 , and so the game is not fair. 29. (a) The expected value is 52 6 2 52 6 2 624   1 1 1 1 1 (b) The game is fair with payout x, where 52  6  2 x  1  52  6  12  0  x  $623.   30. (a) The expected value is 28 1  68 12   18 , and so the game is not fair.   (b) The game is fair with payout x, where 28 x  68 12  0  28 x  38  x  $150.

31. If you win, you win $1 million minus the price of the stamp. If you lose, you lose only the price of the stamp (currently 44 cents). So the expected value of this game is 999,99956  expect to lose 39 cents on each entry, and so it’s not worth it.

1 20  106  1    039. Thus, you 044 20  106 20  106

CHAPTER 14 REVIEW 1. The number of possible outcomes is       number of outcomes number of outcomes number of ways    2 6 52  624. when a coin is tossed a die is rolled to draw a card 2. (a) The number of 3-digit numbers that can be formed using the digits 1–6 if a digit can be used any number of times is 6 6 6  216.

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(b) The number of 3-digit numbers that can be formed using the digits 1–6 if a digit can be used only once is 6 5 4  120. 3. (a) Order is not important, and there are no repetitions, so the number of different two-element subsets is 54 5!   10. C 5 2  2! 3! 2 5!  20. 3! 4. Since the order in which the people are chosen is not important and a person cannot be bumped more than once (no (b) Order is important, and there are no repetitions, so the number of different two-letter words is P 5 2  repetitions), the number of ways that 7 passengers can be bumped is C 120 7  59488  1010 .

5. You earn a score of 70% by answering exactly 7 of the 10 questions correctly. The number of different ways to answer the 10! questions correctly is C 10 7   120. 7! 3! 6. There 2 ways to answer each of the 10 true-false questions and 4 ways to answer each of the 5 multiple choice questions. So    the number of ways that this test can be completed is 210 45  1,048,576. 7. You must choose two of the ten questions to omit, and the number of ways of choosing these two questions is 10!  45. C 10 2  2! 8! 8. Since the order of the scoops of ice cream is not important and the scoops cannot be repeated, the number of ways to have a banana split is C 15 4  1365.

9. The maximum number of employees using this security system is       number of choices number of choices number of choices    26 26 26  17,576. for the first letter for the second letter for the third letter 10. Since there are n! ways to arrange a group of size n and 5!  120, there are 5 students in this class.

11. We could count the number of ways of choosing 7 of the flips to be heads; equivalently we could count the number of ways of 10!  120. choosing 3 of the flips to be tails. Thus, the number of different ways this can occur is C 10 7  C 10 3  3! 7! 12. The number ways to form a license plate consisting of 2 letters followed by 3 numbers is 26 26 10 10 10  676,000. Since there are fewer possible license plates than 700,000, there must be fewer than 700,000 licensed cars in the Yukon. x! x!  10   20  x x  1  20 13. Let x be the number of people in the group. Then C x 2  10  2! x  2! x  2!  x 2  x  20  0  x  5 x  4  0  x  5 or x  4. So there are 5 people in this group.

14. Each topping corresponds to a subset of a set with n elements. Since a set with n elements has 2n subsets and 211  2048, there are 11 toppings that the pizza parlor offers. 15. A letter can be represented by a sequence of length 1, a sequence of length 2, or a sequence of length 3. Since each symbol is either a dot or a dash, the possible number of letters is       number of letters number of letters number of letters    23  22  2  14. using 3 symbols using 2 symbols using 1 symbol

16. Since the nucleotides can be repeated, the number of possible words of length n is 4n . Since 42  16  20 and 43  64, the minimum length of word needed is 3.

17. (a) Since we cannot choose a major and a minor in the same subject, the number of ways a student can select a major and a minor is P 16 2  16  15  240. (b) Again, since we cannot have repetitions and the order of selection is important, the number of ways to select a major, a first minor, and a second minor is P 16 3  16  15  14  3360.

(c) When we select a major and 2 minors, the order in which we choose the minors is not important. Thus the number of     number of ways number of ways to ways to select a major and 2 minors is   16  C 15 2  16  105  1680. to select a major select two minors

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18. (a) Solution 1: Since the leftmost digit of a three-digit number cannot be zero, there are 9 choices for this first digit and 10 choices for each of the other two digits. Thus, the number of three-digit numbers is 9 10 10  900. Solution 2: Since there are 999 numbers between 1 and 999, of which the numbers between 1 and 99 do not have three digits, there are 999  99  900 three-digit numbers. (b) There are 1001 numbers from 0–1000. From part (a), there are 900 three-digit numbers. Therefore the probability that 900  0899. the number chosen is a three-digit number is P E  1001

19. Because the letters are distinct, the number of anagrams of the word RANDOM is 6!  720. 20. Because two letters are the same, the number of anagrams of the word BLOB is

4!  12. 2!

21. Because three letters are the same, the number of anagrams of the word BUBBLE is

6!  120. 3!

22. Because there are two sets of four indistinguishable letters (I, S) and one set of two indistinguishable letters (P), the number 11!  34,650. of anagrams of the word MISSISSIPPI is 4! 4! 2! 23. (a) The possible number of committees is C 18 7  31,824.

(b) Since we must select the 4 men from the group of 10 men and the 3 women from the group of 8 women, the possible     number of ways to number of ways to number of committees is   C 10 4  C 8 3  210  56  11,760. choose 4 of 10 men choose 3 of 8 women (c) We remove Susie from the group of 18so the possible number of committees is C 17 7  19,448.

(d) The possible number of committees is       possible number of possible number of possible number of   committees with 5 women committees with 6 women committees with 7 women  C 8 5  C 10 2  C 8 6  C 10 1  C 8 7  C 10 0  56  45  28  10  8  1  2808

(e) Since the committee is to have 7 members, “at most two men” is the same as “at least five women,” which we found in part (d). So the number is also 2808. (f) We select the specific offices first, then complete the committee from the remaining members of the group. So the number of possible committees is     number of ways to choose number of ways to choose    P 18 3  C 15 4  4896  1365  6,683,040.  a chairman, a vice-chairman,   4 other members and a secretary

24. Method 1: We choose the 5 states first and then one of the two senators from each state. Thus the number of committees is C 50 5  25  67,800,320. Method 2: We choose one of 100 senators, then choose one of the remaining 98 senators (deleting the chosen senator and the other senator form that state), then choose one of the remaining 96 senators, continuing this way until the 5 senators are chosen. Finally, we need to divide by the number of ways to arrange the 5 senators. Thus the number of committees is 10098969492  67,800,320. 5! 2 25. (a) The probability that the ball is red is 10 15  3 .

8 . (b) The probability that the ball is even numbered is 15 2 . (c) The probability that the ball is white and an odd number is 15 7 5 12 4 (d) The probability that the ball is red or odd numbered is P red  P odd  P red  odd  10 15  15  15  15  5 .

26. Let Rn denote the event that the nth ball is red and let Wn denote the event that the nth ball is white. 9 3 (a) P both balls are red  P R1  R2   P R1   P R2  R1   10 15  14  7 .

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(b) Solution 1: The probability that one is white and that the other is red is C 10 1  C 5 1 10 number of ways to select one white and one red   . number of ways to select two balls C 15 2 21 Solution 2: P one white and one red  P W1  R2   P R1  W2   P W1   P R2  W1   P R1   P W2  R1 

5  10  10  5  10  15 14 15 14 21 (c) Solution 1: Let E be the event “at least one is red”. Then E  is the event “both are white”.   5  4  2 . Thus P E  1  2  19 . P E   P W1  W2   P W1   P W2  W1   15 14 21 21 21

9 19 Solution 2: P at least one is red  P one red and one white  P both red  10 21  21  21 (from (a) and (b)). (d) Since 5 of the 15 balls are both red and even-numbered, the probability that both balls are red and even-numbered is 5 4 2 15  14  21 .

2  1  1 . (e) Since 2 of the 15 balls are both white and odd-numbered, the probability that both are white is 15 14 105

27. (a) S  H H H H H T H T H H T T T H H T H T T T H T T T . (b) P H H H  18

(c) P 2 or more heads  P exactly 2 heads  P 3 heads  38  18  48  12 (d) P tails on the first toss  48  12

28. The probability that you select a mathematics book is

4 2 number of ways to select a mathematics book    04. number of ways to select a book 10 5

29. Since rolling a die and selecting a card is independent, 4  1. P both show a six  P die shows a six  P card is a six  16  52 78 4  1 30. (a) P ace  52 13 (b) Let E be the event the card chosen is an ace, and let F be the event the card chosen is a jack. Then 4  4  2. P E  F  P E  P F  52 52 13 (c) Let E be the event the card chosen is an ace, and let F be the event the card chosen is a spade. Then 4  13  1  4 . P E  F  P E  P F  P E  F  52 52 52 13 (d) Let E be the event the card chosen is an ace, and let F be the event the card chosen is a red card. Then n E  F 2  1. P E  F   52 26 n S 1  1  1  1 . 31. (a) Since these events are independent, the probability of getting the ace of spades, a six, and a head is 52 6 2 624 1 1 1 (b) The probability of getting a spade, a six, and a head is 13 52  6  2  48 .

3 1 3 (c) The probability of getting a face card, a number greater than 3, and a head is 12 52  6  2  52 .

32. (a) The probability the first die shows some number is 1, and the probability the second die shows the same number is 16 . So the probability each die shows the same number is 1  16  16 .

(b) By part (a), the event of showing the same number has probability of 16 , and the complement of this event is that the dice show different numbers. Thus the probability that the dice show different numbers is 1  16  56 . 33. (a) Since there are four kings in a standard deck, P 4 kings 

C 4 4 1 1  52515049   369  106 . C 52 4 270,725 4321 13121110

C 13 4 4321 11  000264. (b) Since there are 13 spades in a standard deck, P 4 spades   52515049  4165 C 52 4 4321

CHAPTER 14

(c) Since there are 26 red cards and 26 black cards, P all same color 

Review

933

2  26252423 2  C 26 4 4321 92  011044.  52515049  833 C 52 4 4321

34. In the numbers game lottery, there are 1000 possible “winning” numbers. 1 . (a) The probability that John wins $500 is 1000 (b) There are P 3 3  6 ways to arrange the digits 1, 5, 9. However, if John wins only $50, it means that his number 159 5  1 . was not the winning number. Thus the probability is 1000 200

35. She knows the first digit and must arrange the other four digits. Since only one of the P 4 4  24 arrangements is correct, 1 . the probability that she guesses correctly is 24

36. The number of different pizzas is the number of subsets of the set of 12 toppings, that is, 212  4096. The number of pizzas with anchovies is the number of ways of choosing anchovies and then choosing a subset of the 11 remaining toppings, that 1 is, 1  211  2048. Thus, P getting anchovies  2048 4096  2 . Note that this makes intuitive sense: for each pizza combination without anchovies there is a corresponding one with anchovies, so half will have anchovies and half will not.

37. (a) Since there are only two colors of socks, any 3 socks must contain a matching pair. (b) Method 1: If the two socks drawn form a matching pair then they are either both red or both blue. So     C 20 2 C 30 2 choosing a both red or   051. P P  P both red  P both blue  C 50 2 C 50 2 matching pair both blue Method 2: The complement of choosing a matching pair is choosing one sock of each color. So C 20 1  C 30 1  1  049  051. P choosing a matching pair  1  P different colors  1  C 50 2 38. (a) number of codes  choices for 1st digit  choices for 2nd digit      choices for 5th digit  10  10  10  10  10  105  100,000

(b) Since there are five numbers (0, 1, 6, 8 and 9) that can be read upside down, we have number of codes  choices for 1st digit  choices for 2nd digit      choices for 5th digit  55  3125.

55 n E 1  5  . n S 32 10 (d) Suppose a zip code is turned upside down. Then the middle digit remains the middle digit, so it must be a digit that reads the same when turned upside down, that is, a 0, 1 or 8. Also, the last digit becomes the first digit, and the next to last digit becomes the second digit. Thus, once the first two digits are chosen, the last two are determined. Therefore, the number of zip codes that read the same upside down as right side up is number of codes  choices for 1st digit  choices for 2nd digit      choices for 5th digit  5  5  3  1  1  75. (c) Let E be the event that a zip code can be read upside down. Then by parts (a) and (b), P E 

39. (a) Order is important, and repeats are possible. Thus there are 10 choices for each digit. So the number of different Zip+4 codes is 10  10      10  109 .

(b) If a Zip+4 code is to be a palindrome, the first 5 digits can be chosen arbitrarily. But once chosen, the last 4 digits are determined. Since there are 10 ways to choose each of the first 5 digits, there are 105 palindromes. 5 (c) By parts (a) and (b), the probability that a randomly chosen Zip+4 code is a palindrome is 109  104 .

10

40. (a) Using the rule for the number of distinguishable combinations, the number of divisors of N is 7  1 2  1 5  1  144.

(b) An even divisor of N must contain 2 as a factor. Thus we place a 2 as one of the factors and count the number of distinguishable combinations of M  26 32 55 . So using the rule for the number of distinguishable combinations, the number of even divisors of N is 6  1 2  1 5  1  126.

934

CHAPTER 14 Counting and Probability

(c) A divisor is a multiple of 6 if 2 is a factor and 3 is a factor. Thus we place a 2 as one of the factors and a 3 as one of the factors. Then we count the number of distinguishable combinations of K  26 31 55 . So using the rule for the number of distinguishable combinations, the number of even divisors of N is 6  1 1  1 5  1  84.

126 7 (d) Let E be the event that the divisor is even. Then using parts (a) and (b), P E  nE nS  144  8 . 4  1 41. (a) P king  52 13

8  2 (b) P king or ace  52 13

4 1 number of kings   . number of face cards 12 3 number of kings 4 1 (d) The probability that the card is a king given that it is not an ace is   . number of non-aces 48 12 (c) The probability that the card is a king given that it is a face card is

42. (a)

(b) (c) (d)



   1 3 12 0 1 Because each card is replaced, the probability that all three cards are kings is C 3 3 .  13 13 2197 4 4 4 1 1 Another method:We can calculate the probability as    3  . 52 52 52 2197 13  2  1 1 36 12 The probability that exactly two cards are jacks is C 3 2 .  13 13 2197  0  3 10 1000 3  . The probability that none of the cards is a face card is C 3 0 13 13 2197 “At least one of the cards is a face card” is the complement of the event in part (c), so its probability is 1197 1000  . 1 2197 2197

 4  4 5 43. (a) P 4 sixes in 8 rolls  C 8 4  16  0026048. 6

(b) There are three even numbers on a die and three odds numbers, so P even  P odd  05. Thus P 2 or more evens in 8 rolls  1  P fewer than 2 evens  1  [P 0 evens  P 1 even]         1  C 8 0  050 058  C 8 1  051 057  096484

   44. (a) P 5 in 5 are white flesh  C 5 5  035 070  000243    (b) P 0 in 5 are white flesh  C 5 0  030 075  016807    (c) P 2 in 5 are white flesh  C 5 2  032 073  03087

(d) P 3 or more are red flesh  P 3 in 5 are red flesh  P 4 in 5 are red flesh  P 5 in 5 are red flesh           C 5 3  032 073  C 5 4  031 074  C 5 5  030 075  083692

45. (a) The probability that nine or more patients would have recovered without the drug is C 12 9 0659 0353  C 12 10 06510 0352  C 12 11 06511 0351  C 12 12 06512 0350  0347.

(b) No, the drug does not appear to be effective.

CHAPTER 14

Test

935

46. The probabilities are as follows: Zero heads: C 4 0 070 034  00081 One heads: C 4 1 071 033  00756

Two heads: C 4 2 072 032  02646

Three heads: C 4 3 073 031  04116

Four heads: C 4 4 074 030  02401

Outcome (heads)

Probability

0

00081

1

00756

2

02646

3

04116

4

02401

47. There are 36 possible outcomes in rolling two dice and 6 ways in which both dice show the same numbers, namely, 1 1,     6  1 30  0. 2 2, 3 3, 4 4, 5 5, and 6 6. So the expected value of this game is E  5 36 36 1 , 48. Using the same logic as in Exercise 32(a), the probability that all three dice show the same number is 1  16  16  36 1  35 . Thus, the expected value of this game is while the probability they are not all the same is 1  36 36     1 35 30 E  5 36  1 36   36  083. So John’s expected winnings per game are $083, that is, he expects to

lose $083 per game.

49. Since Mary makes a guess as to the order of ratification of the 13 original states, the number of such guesses 1 is P 13 13  13!, while the probability that she guesses the correct order is . Thus the expected value is 13!     13!  1 1  0  000016. So Mary’s expected winnings are $000016. E  1,000,000 13! 13! 50. The expected number of times Liam goes jogging in any given week is 04 3  01 2  02 1  03 0  16.

CHAPTER 14 TEST 1. The order is fixed, but for each grandchild they have three choices of pictures. Thus, the number of possibilities is 3  3  3  3  81. 2. There are 4 main courses, 3 desserts, and 6 drinks to choose from, so the total number of possibilities is 4  3  6  72. 3. (a) If repetition is allowed, then each letter can be chosen in 26 ways and each digit in 10 ways, so the number of possible passwords is 264  103  456,976,000.

(b) If repetition is not allowed, then the first letter can be chosen in 26 ways, the second in 25 ways, the third in 24 ways, and the fourth in 23 ways. The first digit can be chosen in 10 ways, the second in 9 ways, and the third in 8 ways. Thus, in this case the total number of possible passwords is 26  25  24  23  10  9  8  258,336,000. 4. (a) Order is important in the arrangement, therefore the number of ways to arrange P 30 4  657,720.

(b) Here we are interested in the group of books to be taken on vacation so order is not important, therefore the number of ways to choose these books is C 30 4  27,405.

5. There are two choices to be made: choose a road to travel from Ajax to Barrie, and then choose a different road from Barrie to Ajax. Since there are 4 roads joining the two cities, we need the number of permutations of 4 objects (the roads) taken 2 at a time (the road there and the road back). This number is P 4 2  4  3  12. 6. A customer must choose a size of pizza and must make a choice of toppings. There are 4 sizes of pizza, and each choice of toppings from the 14 available corresponds to a subset of the 14 objects. Since a set with 14 objects has 214 subsets, the number of different pizzas this parlor offers is 4  214  65,536. 7. (a) We want the number of ways of arranging 4 distinct objects (the letters L, O, V, E). This is the number of permutations of 4 objects taken 4 at a time. Therefore, the number of anagrams of the word LOVE is P 4 4  4!  24.

936

FOCUS ON MODELING

(b) We want the number of distinguishable permutations of 6 objects (the letters K, I, S, S, E, S) consisting of three like groups of size 1 and a like group of size 3 (the S’s). Therefore, the number of different anagrams of the word KISSES is 6! 6!   120. 1! 1! 1! 3! 3! 8. We choose the officers first. Here order is important, because the officers are different. Thus there are P 30 3 ways to do this. Next we choose the other 5 members from the remaining 27 members. Here order is not important, so there are C 27 5 ways to do this. Therefore the number of ways that the board of directors can be chosen is 27!  1,966,582,800. P 30 3  C 27 5  30  29  28  5! 22! 9. One card is drawn from a deck. 1 (a) Since there are 26 red cards, the probability that the card is red is 26 52  2 . 4  1 . (b) Since there are 4 kings, the probability that the card is a king is 52 13

2  1. (c) Since there are 2 red kings, the probability that the card is a red king is 52 26

10. Let R be the event that the ball chosen is red. Let E be the event that the ball chosen is even-numbered. 5  03846. (a) Since 5 of the 13 balls are red, P R  13 6  4615. (b) Since 6 of the 13 balls are even-numbered, P E  13

5  6  2  9  06923. (c) P R or E  P R  P E  P R  E  13 13 13 13

11. Let E be the event of choosing 3 men. Then P E 

number of ways to choose 3 men C 5 3 n E    0022. n S number of ways to choose 3 people C 15 3

12. Two dice are rolled. Let E be the event of getting doubles. Since a double may occur in 6 ways, P E 

n E 6 1   . n S 36 6

13. There are 4 students and 12 astrological signs. Let E be the event that at least 2 have the same astrological sign. Then E  is the event that no 2 have the same astrological sign. It is easier to find E  . So   number of ways to assign 4 different astrological signs 12  11  10  9 P 12 4 55    . P E  number of ways to assign 4 astrological signs 12  12  12  12 96 124   41 Therefore, P E  1  P E   1  55 96  96  0427.    14. (a) P 6 heads in 10 tosses  C 10 6  0556 0454  023837. (b) “Fewer than 3 heads” is the same as “0, 1, or 2 heads.” So

P fewer than 3 heads  P 0 head in 10  P 1 head in 10  P 2 heads in 10           C 10 0  0550 04510  C 10 1  0551 0459  C 10 2  0552 0458  002739

4  1 , the probability 15. A deck of cards contains 4 aces, 12 face cards, and 36 other cards. So the probability of an ace is 52 13 3 , and the probability of a non-ace, non-face card is 36  9 . Thus the expected value of this game of a face card is 12  13 52 13  52     1  1 3  5 9  85  0654, that is, about $065. is E  10 13 13 13 13

FOCUS ON MODELING The Monte Carlo Method 1. (a) You should find that with the switching strategy, you win about 90% of the time. The more games you play, the closer to 90% your winning ratio will be.

The Monte Carlo Method

937

1 , since there are ten doors and only (b) The probability that the contestant has selected the winning door to begin with is 10 9 . If the contestant switches, he exchanges a one is a winner. So the probability that he has selected a losing door is 10 1 , and the probability that he losing door for a winning door (and vice versa), so the probability that he loses is now 10 9 . wins is now 10

2. (a) You should find that you get a combination consisting of one head and one tail about 50% of the time. (b) The possible gender combinations are B B BG G B GG. Thus, the probability of having one child of each sex is 2  1. 4 2

3. (a) You should find that player A wins about 78 of the time. That is, if you play this game 80 times, player A should win approximately 70 times. (b) The game will end when either player A gets one more head or player B gets three more tails. Each toss is independent, and both heads and tails have probability 12 , so we obtain the following probabilities. Outcome

Probability

H

1 2 1  1  1 2 2 4 1  1  1  1 2 2 2 8 1  1  1  1 2 2 2 8

TH TTH TTT

Since Player A wins for any outcome that ends in heads, the probability that he wins is 12  14  18  78 . 4. (a) If you simulate 80 World Series with coin tosses, you should expect the series to end in 4 games about 10 times, in 5 games about 20 times, in 6 games about 25 times, and in 7 games about 25 times. (b) We first calculate the number of ways that the series can end with team A winning. (Note that a team must win the final game plus three of the preceding games to win the series.) To win in 4 games, team A must win 4 games right off the bat, and there is only 1 way this can happen. To win in 5 games, team A must win the final game plus 3 of the first 4 games, so this can happen in C 4 3  4 ways. To win in 6 games, team A must win the final game plus 3 of the first 54  10 ways. To win in 7 games, team A must win the final game plus 3 5 games, so this can happen in C 5 3  21 654 of the first 6 games, so this can happen in C 6 3   20 ways. By symmetry, it is also true for team B that 321 they can win in 4 games 1 way, in 5 games 4 ways, in 6 games 10 ways, and in 7 games 20 ways. The probability that any particular team wins a given game is 12 ; this fact, together with the assumption that the games are independent allows us to calculate the probabilities in the following table. Series

Number of ways

4 games

2

5 games

8

6 games

20

7 games

40

Probability  2  12  12  12  8  12  12  12  12  20  12  12  12  12  12  40  12  12  12  12  12  12

  12    12    12    12 

1 8 1 4 5 16 5 16

5  6  5  7  5 13  58. Thus, on average, we expect a World Series to end in (c) The expected value is 18  4  14  5  16 16 16

about 58 games.

5. With 1000 trials, you are likely to obtain an estimate for  that is between 31 and 32.

938

FOCUS ON MODELING

6. Modify the TI-83 program in Problem 5 to the following: PROGRAM:PROB6 :0P :For(N,1,1000) :randX:randY :P+(X2 Y)P :End :Disp “PROBABILITY IS APPROX”,P/1000 You should find that the probability is very close to 13 . 7. (a) We can use the following TI-83 program to model this experiment. It is a minor modification of the one given in Problem 5. PROGRAM:PROB7 :0P :For(N,1,1000) :randX:randY :P+((X+Y)1)P :End :Disp “PROBABILITY IS APPROX”,P/1000 You should find that the probability is very close to 12 . (b) Following the hint, the points in the square for which x  y  1 are the ones that lie below the line x  y  1. This triangle has area 12 (it takes up half the square), so the probability that x  y  1 is 12 .

APPENDIXES A

GEOMETRY REVIEW

1. Congruent by ASA 2. Congruent by SSS 3. Not necessarily congruent 4. Congruent by SAS 5. Similar 6. Similar 7. Similar 8. Not similar x 5  x  125 9.  6 150 36 y 10.   y  30 y 25 y 21 x 92 7 y , so   x  6. 11.  2 32 4 7 214 2 x 12.  x 4 x 8 c ac x  x  13. a ab ab b c ac 14.  x  a a x a b

C

15. From the diagram, we see that because AC  E F and BC  E D, a   and b  . Thus, ABC  AE D  E B F.

F

D

A

16.

b»

º

E

Œ

b

B

(a) Because AB  F G, a  , and so ADG  GC F.

C G Œ

a

º

(b) Because a  , b  180  a  90    b, and similarly a   a.

F

Thus, ADG  F E B.

a»

FE AD  , so AD  E B  DG  F E. DG EB FE AD  , so AD  E B  DG  F E  D E2 , and (d) By similarity, DG EB the result follows. (c) By similarity,

A

a

D

 82  62  10  18. x  732  552  48   19. x  22  12  3

E

b

B

17. x 

939

940

APPENDIXES

 20. x 2  3x2  202  10x 2  400  x 2  40  x  2 10

21. x 2  x  22  582  2x 2  4x  4  3364  2x 2  4x  3360  2 x  42 x  40  0  x  40 22. x 2  172  x  12  x 2  289  x 2  2x  1  2x  288  x  144

23. 52  122  169  132 , so the triangle is a right triangle.

24. 152  202  625  252 , so the triangle is a right triangle.

25. 82  102  164  122 , so the triangle is not a right triangle. 26. 62  172  325  182 , so the triangle is not a right triangle. 27. 482  552  5329  732 , so the triangle is a right triangle.

28. 132  842  7225  852 , so the triangle is a right triangle.

29. Let the other leg have length x. Then 112  x 2  x  12  121  x 2  x 2  2x  1  2x  122  x  61 cm. 30. Let the width of the rectangle be x. Then x 2  x  12  1692  2x 2  2x  1  28,561 

2x 2  2x  28,560  2 x  120 x  119  0  x  119 ft. Then the length is x  1  120 ft, and the dimensions of the rectangles are 119 ft by 120 ft.

31. If the quadrilateral were a rectangle, we would have 172  212  272 . But this is false, so it is not a rectangle.  32. AB  202  152  25. The area of the triangle can be written as 12 BC C A  12 15 20  150, or as 1 AB h  1 25 h  150  1 25 h  h  12. 2 2 2

  33. The diagonal of the left face of the box is 32  42  5, so the length of the desired diagonal is 52  122  13. 2  34. (a) a 2  b2  m 2  n 2  2mn2  m 4  n 4  2m 2 n 2  4m 2 n 2 (b) m n a b c 2   m 4  n 4  2m 2 n 2  m 2  n 2  c2 2 1 3 4 5 so a b c is a Pythagorean triple.

35.

8

h a

B

d

b

CALCULATIONS AND SIGNIFICANT FIGURES

1. 327  01834  309

1

8 6 10

3

2

5 12 13

4

1

15 8 17

4

2

12 16 20

4

3

7 24 25

5

1

24 10 26

5

2

21 20 29

5

3

16 30 34

5

4

9 40 41

24 h 8 h  and  . Thus, 24a  8b  b  3a a d b d h 24  4a  d, and so   h  6. a 4a

By similarity, 24

3

2. 10268  267  1294

APPENDIX C Graphing with a Graphing Calculator

3. 2836  501375  14,220

4.

201,186  3841 5238

5. 1363  252

6.

 4273  2067

7. 33 64275  66787  33 70954  2300

701 8. 127105  759

    9. 510  103 124  107 6007  106  380

10.

   1361107 47717105  5066 1281876

11. The circumference is 2r  2 527  331 ft and the area is r 2   5272  873 ft2 . 12. The volume is 13 r 2 h  13  42672 523  997 cm3 . 2 m m 13. The force is F  G 1 2 2  667428  1011  11,4262  266  1012 N. 57,200 r

14. The Sun and the Earth are 150  1011 m apart, with masses 19891  1030 kg and 5972  1024 kg respectively.    m m 198911030 59721024 (a) The force is F  G 1 2 2  667428  1011   352  1022 N.  2 1501011 r   lb  793  1021 lb. (b) The force is 352  1022 N  352  1022 N 0225 1N

C

GRAPHING WITH A GRAPHING CALCULATOR

1. y  x 4  2

(b) [0 4] by [0 4]

(a) [2 2] by [2 2] 2

4 3

1 -2

-1

-1

1

2

2 1 0 0

-2

(c) [8 8] by [4 40]

1

2

3

4

20

40

(d) [40 40] by [80 800] 40

800

30

600

20

400

10

200

-8 -6 -4 -2

2 4 6 8

-40

The viewing rectangle in part (c) produces the most appropriate graph of the equation.

-20

941

942

APPENDIXES

2. y  x 2  7x  6

(b) [0 10] by [20 100]

(a) [5 5] by [5 5]

100

4 2

50 -4

-2

2

-2

4 0

-4

5

(c) [15 8] by [20 100]

10

(d) [10 3] by [100 20] 100 -10 -8 -6 -4 -2 50

2

-50

-14-12-10-8 -6 -4 -2

2 4 6 8

-100

The viewing rectangle in part (c) produces the most appropriate graph of the equation.

3. y  100  x 2

(a) [4 4] by [4 4]

(b) [10 10] by [10 10] 4

10

2 -4

-2

2

-2

4

-10

-5

-4

10

2

4

-10

(c) [15 15] by [30 110]

(d) [4 4] by [30 110]

100

100

50

50

-15 -10 -5

5

5

10 15

-4

The viewing rectangle in part (c) produces the most appropriate graph of the equation.

-2

APPENDIX C Graphing with a Graphing Calculator

4. y  2x 2  1000

(b) [10 10] by [100 100]

(a) [10 10] by [10 10] 10

-10

-5

100

5

10

-10

-5

-10

5

10

-100

(c) [10 10] by [1000 1000]

(d) [25 25] by [1200 200]

1000 -20

-10

10

20

-500 -10

-5

5

10 -1000

-1000

The viewing rectangle in part (d) produces the most appropriate graph of the equation. 5. y  10  25x  x 3

(a) [4 4] by [4 4]

(b) [10 10] by [10 10] 4

10

2 -4

-2

-2

2

4

-10

-5

5

-4

10

-10

(c) [20 20] by [100 100]

(d) [100 100] by [200 200]

100

200 100

-20

-10

10

20

-100

100

-100

-100

-200

The viewing rectangle in part (c) produces the most appropriate graph of the equation.  6. y  8x  x 2 (a) [4 4] by [4 4]

(b) [5 5] by [0 100]

4

100

2 50 -4

-2

-2 -4

2

4 -4

-2

0

2

4

943

944

APPENDIXES

(c) [10 10] by [10 40]

(d) [2 10] by [2 6]

40

6 4

20

-10

2

-5

5

-2 -2

10

2

4

6

8

10

From the graphs we see that the viewing rectangle in (d) produces the most appropriate graph of the equation. Note: Squaring both sides yields the equation y 2  8x  x 2  x  42  y 2  16. Since this gives a circle, the original equation represents the top half of a circle. 7. y  100x 2 , [2 2] by [10 400]

8. y  100x 2 , [2 2] by [400 10] -2

400

-1

0

1

2

300 200

-100

100

-200 -300

-2

-1

0

1

2

9. y  4  6x  x 2 , [4 10] by [10 20]

-400

10. y  03x 2  17x  3, [15 10] by [10 20]

20

20

10

10

-4 -2 -10

2

4

6

8 10

 4 256  x 2 . We require that 256  x 2  0   4 16  x  16, so we graph y  256  x 2 in the

11. y 

-15 -10

12. y 

-5 -10

5

10

 12x  17, [0 10] by [0 20] 20

viewing rectangle [20 20] by [1 5].

10 4 0 0

2

-20

-10

10

10

20

13. y  001x 3  x 2  5, [50 150] by [2000 2000]

14. y  x x  6 x  9, [10 10] by [250 150]

2000

100 -10 100

-2000

5

-5 -100 -200

5

10

APPENDIX C Graphing with a Graphing Calculator

1 , [2 4] by [8 8] 15. y  2 x  2x

945

x , [10 10] by [02 02] 16. y  2 x  25 0.2

5

0.1 -2

-1

1

2

3

4

-10

-5

-5

-0.1

5

10

-0.2

    18. y  2x  x 2  5, [10 10] by [10 10]

17. y  1  x  1, [3 5] by [1 5]

10

4 2

-10 -2

2

-5

4

rectangle [4 4] by [1 3], there is no point of intersection. You can verify this by zooming in.

20. Although the graphs of y 

 49  x 2 and

y  15 41  3x appear to intersect in the viewing rectangle [8 8] by [1 8], there is no point of intersection. You can verify this by zooming in. 8

3

6

2

4

1 -2

10

-10

19. Although the graphs of y  3x 2  6x  12 and  7 x 2 appear to intersect in the viewing y  7  12

-4

5

2 2

-1

4

-8 -6 -4 -2

2 4 6 8

21. The graphs of y  6  4x  x 2 and y  3x  18 appear to 22. The graphs of y  x 3  4x and y  x  5 appear to have have two points of intersection in the viewing rectangle

one point of intersection in the viewing rectangle [4 4]

[6 2] by [5 20]. You can verify that x  4 and

by [15 15]. The solution is x  2627.

x  3 are exact solutions.

10

20 -4

10

-2

2 -10

-6

-4

-2

2

4

946

APPENDIXES

 23. x 2  y 2  9  y 2  9  x 2  y   9  x 2 . So we 24. y  12  x 2  1  y  12  1  x 2      graph the functions y1  9  x 2 and y2   9  x 2 in y  1   1  x 2  y  1  1  x 2 . So we graph   the viewing rectangle [6 6] by [4 4]. the functions y1  1  1  x 2 and y2  1  1  x 2 in the viewing rectangle [3 3] by [1 3].

4 2

-6

-4

-2 -2

3 2

4

2

6

1

-4 -3

1  4x 2  25. 4x 2  2y 2  1  2y 2  1  4x 2  y 2  2  1  4x 2 y . So we graph the functions 2   1  4x 2 1  4x 2 and y2   in the viewing y1  2 2

-2

-1 -1

[5 5].

4

-4

-2

-2 -4

-0.5

3

2

0.5

0.5

2

 26. y 2  9x 2  1  y 2  1  9x 2  y   1  9x 2 . So  we graph the functions y1  1  9x 2 and  y2   1  9x 2 in the viewing rectangle [5 5] by

rectangle [12 12] by [08 08].

-1.0 -0.5

1

1.0

2

4