Adams - Calculus A Complete Course 9th ed 2018 solutions [9th ed]

Citation preview

lOMoARcPSD|6566483

INSTRUCTOR'S SOLUTIONS MANUAL for

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL Robert A. Adams University of British Columbia

Calculus Ninth Edition Robert A. Adams University of British Columbia

Christopher Essex University of Western Ontario

dumperina

ISBN: 978-0-13-452876-2 Copyright © 2018 Pearson Canada Inc., Toronto, Ontario. All rights reserved. This work is protected by Canadian copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the Internet) will destroy the integrity of the work and is not permitted. The copyright holder grants permission to instructors who have adopted Calculus: A Complete Course, Single-Variable Calculus, or Calculus of Several Variables by Adams and Essex, to post this material online only if the use of the website is restricted by access codes to students in the instructor’s class that is using the textbook and provided the reproduced material bears this copyright notice.

Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

FOREWORD These solutions are provided for the benefit of instructors using the textbooks:

Calculus: A Complete Course (9th Edition), Single-Variable Calculus (9th Edition), and Calculus of Several Variables (9th Edition) by R. A. Adams and Chris Essex, published by Pearson Canada. For the most part, the solutions are detailed, especially in exercises on core material and techniques. Occasionally some details are omitted—for example, in exercises on applications of integration, the evaluation of the integrals encountered is not always given with the same degree of detail as the evaluation of integrals found in those exercises dealing specifically with techniques of integration. Instructors may wish to make these solutions available to their students. However, students should use such solutions with caution. It is always more beneficial for them to attempt exercises and problems on their own, before they look at solutions done by others. If they examine solutions as “study material” prior to attempting the exercises, they can lose much of the benefit that follows from diligent attempts to develop their own analytical powers. When they have tried unsuccessfully to solve a problem, then looking at a solution can give them a “hint” for a second attempt. Separate Student Solutions Manuals for the books are available for students. They contain the solutions to the even-numbered exercises only. November, 2016. R. A. Adams [email protected]

Chris Essex [email protected]

Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

CONTENTS

Solutions for Chapter P

1

Solutions for Chapter 1

23

Solutions for Chapter 2

39

Solutions for Chapter 3

81

Solutions for Chapter 4

108

Solutions for Chapter 5

177

Solutions for Chapter 6

213

Solutions for Chapter 7

267

Solutions for Chapter 8

316

Solutions for Chapter 9

351

Solutions for Chapter 10

392

Solutions for Chapter 11

420

Solutions for Chapter 12

448

Solutions for Chapter 13

491

Solutions for Chapter 14

538

Solutions for Chapter 15

579

Solutions for Chapter 16

610

Solutions for Chapter 17

637

Solutions for Chapter 18

644

Solutions for Chapter 18-cosv9

671

Solutions for Appendices

683

NOTE: “Solutions for Chapter 18-cosv9” is only needed by users of Calculus of Several Variables (9th Edition), which includes extra material in Sections 18.2 and 18.5 that is found in Calculus: a Complete Course and in Single-Variable Calculus in Sections 7.9 and 3.7 respectively. Solutions for Chapter 18-cosv9 contains only the solutions for the two Sections 18.2 and 18.9 in the Several Variables book. All other Sections are in “Solutions for Chapter 18.” It should also be noted that some of the material in Chapter 18 is beyond the scope of most students in single-variable calculus courses as it requires the use of multivariable functions and partial derivatives.

Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

CHAPTER P.

SECTION P.1 (PAGE 10)

PRELIMINARIES

19.

Given: 1=.2 x/ < 3. CASE I. If x < 2, then 1 < 3.2 x/ D 6 3x, so 3x < 5 and x < 5=3. This case has solutions x < 5=3. CASE II. If x > 2, then 1 > 3.2 x/ D 6 3x, so 3x > 5 and x > 5=3. This case has solutions x > 2. Solution: . 1; 5=3/ [ .2; 1/.

20.

Given: .x C 1/=x  2. CASE I. If x > 0, then x C 1  2x, so x  1. CASE II. If x < 0, then x C 1  2x, so x  1. (not possible) Solution: .0; 1.

21.

Given: x 2 2x  0. Then x.x 2/  0. This is only possible if x  0 and x  2. Solution: Œ0; 2.

Section P.1 Real Numbers and the Real Line (page 10) 1.

2 D 0:22222222


D 0:2 9

2.

1 D 0:09090909


D 0:09 11

3. If x D 0:121212


, then 100x D 12:121212


D 12 C x. Thus 99x D 12 and x D 12=99 D 4=33. 4. If x D 3:277777


, then 10x 32 D 0:77777


and 100x 320 D 7 C .10x 32/, or 90x D 295. Thus x D 295=90 D 59=18. 5.

1=7 D 0:142857142857


D 0:142857

22.

Given 6x 2 5x  1, then .2x 1/.3x 1/  0, so either x  1=2 and x  1=3, or x  1=3 and x  1=2. The latter combination is not possible. The solution set is Œ1=3; 1=2.

23.

Given x 3 > 4x, we have x.x 2 4/ > 0. This is possible if x < 0 and x 2 < 4, or if x > 0 and x 2 > 4. The possibilities are, therefore, 2 < x < 0 or 2 < x < 1. Solution: . 2; 0/ [ .2; 1/.

2=7 D 0:285714285714


D 0:285714 3=7 D 0:428571428571


D 0:428571

4=7 D 0:571428571428


D 0:571428 note the same cyclic order of the repeating digits 5=7 D 0:714285714285


D 0:714285

24.

6=7 D 0:857142857142


D 0:857142

6. Two different decimal expansions can represent the same number. For instance, both 0:999999


D 0:9 and 1:000000


D 1:0 represent the number 1. 7.

25.

x  0 and x  5 define the interval Œ0; 5.

8. x < 2 and x  3 define the interval Π3; 2/. 5 or x
10. x  11.

6 defines the union . 1; 6/ [ . 5; 1/.

1 defines the interval . 1; 1.

2x > 4, then x
0. Solution: .0; 1/ 2

18. If x < 9, then jxj < 3 and . 3; 3/

3 < x < 3. Solution:

2 3 < . x 1 xC1 CASE I. If x > 1 then .x 1/.x C 1/ > 0, so that 3.x C 1/ < 2.x 1/. Thus x < 5. There are no solutions in this case. CASE II. If 1 < x < 1, then .x 1/.x C 1/ < 0, so 3.x C 1/ > 2.x 1/. Thus x > 5. In this case all numbers in . 1; 1/ are solutions. CASE III. If x < 1, then .x 1/.x C 1/ > 0, so that 3.x C 1/ < 2.x 1/. Thus x < 5. All numbers x < 5 are solutions. Solutions: . 1; 5/ [ . 1; 1/.

Given:

12. x < 4 or x  2 defines the interval . 1; 1/, that is, the whole real line. 13. If

4 x 1C . 2 x CASE I. If x > 0, then x 2  2x C 8, so that x 2 2x 8  0, or .x 4/.x C 2/  0. This is possible for x > 0 only if x  4. CASE II. If x < 0, then we must have .x 4/.x C 2/  0, which is possible for x < 0 only if x  2. Solution: Œ 2; 0/ [ Œ4; 1/. Given:

26.

2 defines the interval . 2; 1/.

x>

Given x 2 x  2, then x 2 x 2  0 so .x 2/.x C1/  0. This is possible if x  2 and x  1 or if x  2 and x  1. The latter situation is not possible. The solution set is Œ 1; 2.

3j D 7, then x

t j D 1, then 1

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

3 D ˙7, so x D

4 or x D 10. 9=2 or

t D ˙1, so t D 0 or t D 2. 1 or 17, and

1

lOMoARcPSD|6566483

SECTION P.1 (PAGE 10)

ADAMS and ESSEX: CALCULUS 9

ˇs ˇ s ˇ ˇ 32. If ˇ 1ˇ D 1, then 1 D ˙1, so s D 0 or s D 4. 2 2 33. If jxj < 2, then x is in . 2; 2/.

4.

From A.0:5; 3/ to B.2; 3/, x D 2 y D 3 3 D 0. jABj D 1:5.

5.

Starting point: . 2; 3/. Increments x D 4, y D New position is . 2 C 4; 3 C . 7//, that is, .2; 4/.

6.

Arrival point: . 2; 2/. Increments x D 5, y D 1. Starting point was . 2 . 5/; 2 1/, that is, .3; 3/.

7.

x 2 C y 2 D 1 represents a circle of radius 1 centred at the origin. p x 2 C y 2 D 2 represents a circle of radius 2 centred at the origin.

34. If jxj  2, then x is in Œ 2; 2. 35. If js

1j  2, then 1

36. If jt C 2j < 1, then . 3; 1/.

2  s  1 C 2, so s is in Œ 1; 3. 2

2 C 1, so t is in

1 < t
jx 3j says that the distance from x to 1 is greater than the distance from x to 3, so x must be to the right of the point half-way between 1 and 3. Thus x > 1.

jxj  jx C yj Apply this inequality with x D a ja

bj  jaj

jyj: b and y D b to get jbj:

ˇ ˇ ˇ ˇ Similarly, ja bj D jb aj  jbj jaj. Since ˇjaj jbjˇ is equal to either jaj jbj or jbj jaj, depending on the sizes of a and b, we have ja

ˇ ˇ bj  ˇjaj

ˇ ˇ jbjˇ:

Section P.2 Cartesian Coordinates in the Plane (page 16) 1.

From A.0; 3/ to B.4; 0/, xp D 4 0 D 4 and y D 0 3 D 3. jABj D 42 C . 3/2 D 5.

2. From A. 1; 2/ to B.4; 10/, xpD 4 . 1/ D 5 and y D 10 2 D 12. jABj D 52 C . 12/2 D 13. 3. From A.3; 2/ to B. 1; 2/, x and pD 1 3 D 4p y D 2 2 D 4. jABj D . 4/2 C . 4/2 D 4 2. 2

11.

x 2 C y 2  1 represents points inside and on the circle of radius 1 centred at the origin.

y  x 2 represents all points lying on or above the parabola y D x 2 .

12. y < x 2 represents all points lying below the parabola y D x2. The vertical line through . 2; 5=3/ is x D 2; the horizontal line through that point is y D 5=3. p p The vertical line through . 2; 1:3/ is x D 2; the horizontal line through that point is y D 1:3.

15.

Line through . 1; 1/ with slope m D 1 is y D 1C1.xC1/, or y D x C 2.

16.

Line through . 2; 2/ with slope m D 1=2 is y D 2 C .1=2/.x C 2/, or x 2y D 6.

17.

Line through .0; b/ with slope m D 2 is y D b C 2x.

1/,

45. The triangle inequality jx C yj  jxj C jyj implies that

7.

10. x 2 C y 2 D 0 represents the origin.

14.

43. jaj D a if and only if a  0. It is false if a < 0. .x

9.

13.

42. jx 3j < 2jxj , x 2 6x C 9 D .x 3/2 < 4x 2 , 3x 2 C 6x 9 > 0 , 3.x C 3/.x 1/ > 0. This inequality holds if x < 3 or x > 1.

44. The equation jx 1j D 1 x holds if jx 1j D that is, if x 1  0, or, equivalently, if x  1.

8.

0:5 D 1:5 and

18. Line through .a; 0/ with slope m D or y D 2a 2x.

2 is y D 0 2.x a/,

19.

At x D 2, the height of the line 2x C 3y D 6 is y D .6 4/=3 D 2=3. Thus .2; 1/ lies above the line.

20.

At x D 3, the height of the line x 4y D 7 is y D .3 7/=4 D 1. Thus .3; 1/ lies on the line.

21.

The line through .0; 0/ and .2; 3/ has slope m D .3 0/=.2 0/ D 3=2 and equation y D .3=2/x or 3x 2y D 0.

22.

The line through . 2; 1/ and .2; 2/ has slope m D . 2 1/=.2 C 2/ D 3=4 and equation y D 1 .3=4/.x C 2/ or 3x C 4y D 2.

23.

The line through .4; 1/ and . 2; 3/ has slope m D .3 1/=. 2 4/ D or x C 3y D 7.

24. 25.

1=3 and equation y D 1

The line through . 2; 0/ and .0; 2/ has slope m D .2 0/=.0 C 2/ D 1 and equation y D 2 C x. p If m D 2 and p b D 2, then the line has equation y D 2x C 2.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

1 .x 4/ 3

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION P.2 (PAGE 16)

y

26. If m D 1=2 and b D 3, then the line has equation y D .1=2/x 3, or x C 2y D 6. 1:5x 27.

3x C 4y D 12 has x-intercept a D 12=3 D 4 and yintercept b D 12=4 D 3. Its slope is b=a D 3=4. y

2y D

3

x

Fig. P.2-30 3x C 4y D 12

31.

line through .2; 1/ parallel to y D x C 2 is y D x perpendicular to y D x C 2 is y D x C 3.

32.

line through . 2; 2/ parallel to 2xCy D 4 is 2xCy D line perpendicular to 2x C y D 4 is x 2y D 6.

33.

We have

x Fig. P.2-27

3x C 4y D 6 2x 3y D 13

28. x C 2y D 4 has x-intercept a D 4 and y-intercept b D 4=2 D 2. Its slope is b=a D 2=. 4/ D 1=2. y

2;

6x C 8y D 12 6x 9y D 39:

Subtracting these equations gives 17y D 51, so y D 3 and x D .13 9/=2 D 2. The intersection point is .2; 3/. 34.

x x C 2y D

÷

1; line

We have 2x C y D 8 5x 7y D 1

4

÷

14x C 7y D 56 5x 7y D 1:

Adding these equations gives 19x D 57, so x D 3 and y D 8 2x D 2. The intersection point is .3; 2/. 35.

If a ¤ 0 and b ¤ 0, then .x=a/ C .y=b/ D 1 represents a straight line that is neither horizontal nor vertical, and does not pass through the origin. Putting y D 0 we get x=a D 1, so the x-intercept of this line is x D a; putting x D 0 gives y=b D 1, so the y-intercept is y D b.

36.

The line .x=2/ .y=3/ D 1 has x-intercept a D 2, and y-intercept b D 3. y

Fig. P.2-28

29.

p

p p p 2x 3y D 2 has x-intercept a D 2= 2 D 2 p and y-intercept 2= 3. Its slope is p b pD b=a D 2= 6 D 2=3. y

2

p

2x

p

x

x 2

3y D 2

x

y D1 3

3

Fig. P.2-29 Fig. P.2-36 30. 1:5x 2y D 3 has x-intercept a D 3=1:5 D 2 and y-intercept b D 3=. 2/ D 3=2. Its slope is b=a D 3=4.

37. The line through .2; 1/ and .3; 1/ has slope m D . 1 1/=.3 2/ D 2 and equation y D 1 2.x 2/ D 5 2x. Its y-intercept is 5.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

3

lOMoARcPSD|6566483

SECTION P.2 (PAGE 16)

ADAMS and ESSEX: CALCULUS 9

38. The line through . 2; 5/ and .k; 1/ has x-intercept 3, so also passes through .3; 0/. Its slope m satisfies 1 k Thus k

3D

0 5 0 DmD D 3 3C2

43.

B D .1; 3/; C D . 3; 2/ p p jABj D .1 2/2 C .3 C 1/2 D 17 p p p p jAC j D . 3 2/2 C .2 C 1/2 D 34 D 2 17 p p jBC j D . 3 1/2 C .2 3/2 D 17: p Since jABj D jBC j and jAC j D 2jABj, triangle ABC is an isosceles right-angled triangle with right angle at B. Thus ABCD is a square if D is displaced from C by the same amount A is from B, that is, by increments x D 2 1 D 1 and y D 1 3 D 4. Thus D D . 3 C 1; 2 C . 4// D . 2; 2/.

44.

If M D .xm ; ym / is the midpoint of P1 P2 , then the displacement of M from P1 equals the displacement of P2 from M :

1:

1, and so k D 2.

39. C D Ax C B. If C D 5; 000 when x D 10; 000 and C D 6; 000 when x D 15; 000, then 10; 000A C B D 5; 000 15; 000A C B D 6; 000 Subtracting these equations gives 5; 000A D 1; 000, so A D 1=5. From the first equation, 2; 000 C B D 5; 000, so B D 3; 000. The cost of printing 100,000 pamphlets is $100; 000=5 C 3; 000 D $23; 000.

xm

40ı and 40ı is the same temperature on both the Fahrenheit and Celsius scales. C 40

xq

C DF

-20 -30

46. 10 20 30 40 50 60 70 80 F C D

5 .F 9

47.

A D .2; 1/; B D .6; 4/; C D .5; 3/ p p jABj D .6 2/2 C .4 1/2 D 25 D 5 p p jAC j D .5 2/2 C . 3 1/2 D 25 D 5 p p p jBC j D .6 5/2 C .4 C 3/2 D 50 D 5 2: Since jABj D jAC j, triangle ABC is isosceles. p A D .0; 0/; B D .1; 3/; C D .2; 0/ q p p jABj D .1 0/2 C . 3 0/2 D 4 D 2 p p jAC j D .2 0/2 C .0 0/2 D 4 D 2 q p p 3/2 D 4 D 2: jBC j D .2 1/2 C .0 Since jABj D jAC j D jBC j, triangle ABC is equilateral.

4

x1 D 2.x2

xq /;

.x C X/=2 D 2;

Fig. P.2-40

42.

y1 D y2

ym :

yq

y1 D 2.y2

yq /:

Let the coordinates of P be .x; 0/ and those of Q be .X; 2X/. If the midpoint of PQ is .2; 1/, then

32/

-40 . 40; 40/ -50

41.

ym

Thus xq D .x1 C 2x2 /=3 and yq D .y1 C 2y2 /=3.

10 -50 -40 -30 -20 -10 -10

xm ;

If Q D .xq ; yq / is the point on P1 P2 that is two thirds of the way from P1 to P2 , then the displacement of Q from P1 equals twice the displacement of P2 from Q:

30 20

x1 D x2

Thus xm D .x1 C x2 /=2 and ym D .y1 C y2 /=2. 45.

40.

A D .2; 1/;

48.

.0

2X/=2 D 1:

The second equation implies that X D 1, and the second then implies that x D 5. Thus P is .5; 0/. p .x 2/2 C y 2 D 4 says that the distance of .x; y/ from .2; 0/ is 4, so the equation represents a circle of radius 4 centred at .2; 0/. p p .x 2/2 C y 2 D x 2 C .y 2/2 says that .x; y/ is equidistant from .2; 0/ and .0; 2/. Thus .x; y/ must lie on the line that is the right bisector of the line from .2; 0/ to .0; 2/. A simpler equation for this line is x D y.

49.

The line 2x C ky D 3 has slope m D 2=k. This line is perpendicular to 4x C y D 1, which has slope 4, provided m D 1=4, that is, provided k D 8. The line is parallel to 4x C y D 1 if m D 4, that is, if k D 1=2.

50.

For any value of k, the coordinates of the point of intersection of x C 2y D 3 and 2x 3y D 1 will also satisfy the equation .x C 2y

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

3/ C k.2x

3y C 1/ D 0

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION P.3 (PAGE 22)

because they cause both expressions in parentheses to be 0. The equation above is linear in x and y, and so represents a straight line for any choice of k. This line will pass through .1; 2/ provided 1 C 4 3 C k.2 6 C 1/ D 0, that is, if k D 2=3. Therefore, the line through the point of intersection of the two given lines and through the point .1; 2/ has equation x C 2y

2 3 C .2x 3

3y C 1/ D 0;

13.

Together, x 2 C y 2 > 1 and x 2 C y 2 < 4 represent annulus (washer-shaped region) consisting of all points that are outside the circle of radius 1 centred at the origin and inside the circle of radius 2 centred at the origin.

14.

Together, x 2 C y 2  4 and .x C 2/2 C y 2  4 represent the region consisting of all points that are inside or on both the circle of radius 2 centred at the origin and the circle of radius 2 centred at . 2; 0/.

15.

Together, x 2 Cy 2 < 2x and x 2 Cy 2 < 2y (or, equivalently, .x 1/2 C y 2 < 1 and x 2 C .y 1/2 < 1) represent the region consisting of all points that are inside both the circle of radius 1 centred at .1; 0/ and the circle of radius 1 centred at .0; 1/.

16.

x2 C y2 4x C 2y > 4 can be rewritten .x 2/2 C .y C 1/2 > 9. This equation, taken together with x C y > 1, represents all points that lie both outside the circle of radius 3 centred at .2; 1/ and above the line x C y D 1.

17.

The interior of the circle with centre . 1; 2/ and radius p 6 is given by .x C 1/2 C .y 2/2 < 6, or 2 x C y 2 C 2x 4y < 1.

or, on simplification, x D 1.

Section P.3 Graphs of Quadratic Equations (page 22) 1.

x 2 C y 2 D 16

2. x 2 C .y

2/2 D 4, or x 2 C y 2

4y D 0

3. .x C 2/2 C y 2 D 9, or x 2 C y 2 C 4y D 5 4. .x 5.

x2 C y2 x

6.

3/2 C .y C 4/2 D 25, or x 2 C y 2

2

6x C 8y D 0.

2x D 3

2x C 1 C y 2 D 4

.x 1/2 C y 2 D 4 centre: .1; 0/; radius 2.

18. The exterior of the circle with centre .2; 3/ and radius 4 is given by .x 2/2 C .y C 3/2 > 16, or x 2 C y 2 4x C 6y > 3.

x 2 C y 2 C 4y D 0

19.

2

2

x C y C 4y C 4 D 4 2

7.

x2 C y2

x

2

2x C 4y D 4

2x C 1 C y 2 C 4y C 4 D 9

.x 1/2 C .y C 2/2 D 9 centre: .1; 2/; radius 3. 8.

x2 C y2

x2

2x

yC1D0

2x C 1 C y 2

yC

1 4

D


2 .x 1/2 C y 12 D 41 centre: .1; 1=2/; radius 1=2.

1 4

2

2

x1 1/2 C .y

3/2 < 10

20.

x C y > 4;

21.

The parabola with focus .0; 4/ and directrix y D equation x 2 D 16y.

22.

The parabola with focus .0; 1=2/ and directrix y D 1=2 has equation x 2 D 2y.

23.

The parabola with focus .2; 0/ and directrix x D equation y 2 D 8x.

24.

The parabola with focus . 1; 0/ and directrix x D 1 has equation y 2 D 4x.

25.

y D x 2 =2 has focus .0; 1=2/ and directrix y D

2

x C .y C 2/ D 4 centre: .0; 2/; radius 2.

x 2 C y 2 < 2;

.x

2 has

1=2.

y

9. x 2 C y 2 > 1 represents all points lying outside the circle of radius 1 centred at the origin.

yDx 2 =2

.0;1=2/

10. x 2 C y 2 < 4 represents the open disk consisting of all points lying inside the circle of radius 2 centred at the origin. 11.

4 has

x yD 1=2

.x C 1/2 C y 2  4 represents the closed disk consisting of all points lying inside or on the circle of radius 2 centred at the point . 1; 0/.

12. x 2 C .y 2/2  4 represents the closed disk consisting of all points lying inside or on the circle of radius 2 centred at the point .0; 2/.

Fig. P.3-25 26.

yD

x 2 has focus .0; 1=4/ and directrix y D 1=4.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

5

lOMoARcPSD|6566483

SECTION P.3 (PAGE 22)

ADAMS and ESSEX: CALCULUS 9

y

y

Version (c) yD1=4

y D x2

x

.3; 3/ Version (b)

.0; 1=4/ yD x 2

Fig. P.3-26

x 4 Version (d) .4; 2/

27.

xD

Version (a) 3

y 2 =4 has focus . 1; 0/ and directrix x D 1. y

Fig. P.3-29 a) has equation y D x 2

xD1 . 1;0/

b) has equation y D .x

x

c) has equation y D .x d) has equation y D .x

xD y 2 =4

30. Fig. P.3-27

28. x D y 2 =16 has focus .4; 0/ and directrix x D

4.

y

31. .4;0/

32. x

33.

xD 4 xDy 2 =16

34. Fig. P.3-28

6

8x C 16.

3/2 C 3 or y D x 2 4/2

2, or y D x 2

6x C 12. 8x C 14.

b) If y D mx is shifted vertically by amount y1 , the equation y D mx C y1 results. If .a; b/ satisfies this equation, then b D ma C y1 , and so y1 D b ma. Thus the shifted equation is y D mx C b ma D m.x a/ C b, the same equation obtained in part (a). p y D .x=3/ C 1 p 4y D x C 1 p y D .3x=2/ C 1 p .y=2/ D 4x C 1 x 2 shifted down 1, left 1 gives y D

35.

yD1

36.

x2 C y2 D .x C 4/2 C .y

38.

4/2 or y D x 2

a) If y D mx is shifted to the right by amount x1 , the equation y D m.x x1 / results. If .a; b/ satisfies this equation, then b D m.a x1 /, and so x1 D a .b=m/. Thus the shifted equation is y D m.x a C .b=m// D m.x a/ C b.

37. y D .x y D .x 29.

3.

.x C 1/2 .

5 shifted up 2, left 4 gives 2/2 D 5.

1/2 2/2

1 shifted down 1, right 1 gives 2. p p y D x shifted down 2, left 4 gives y D x C 4

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

2.

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION P.3 (PAGE 22)

y

39. y D x 2 C 3, y D 3x C 1. Subtracting these equations gives x 2 3x C 2 D 0, or .x 1/.x 2/ D 0. Thus x D 1 or x D 2. The corresponding values of y are 4 and 7. The intersection points are .1; 4/ and .2; 7/.

9x 2 C16y 2 D144

x

40. y D x 2 6, y D 4x x 2 . Subtracting these equations gives 2x 2 4x 6 D 0, or 2.x 3/.x C 1/ D 0. Thus x D 3 or x D 1. The corresponding values of y are 3 and 5. The intersection points are .3; 3/ and . 1; 5/. 41. x 2 C y 2 D 25, 3x C 4y D 0. The second equation says that y D 3x=4. Substituting this into the first equation gives 25x 2 =16 D 25, so x D ˙4. If x D 4, then the second equation gives y D 3; if x D 4, then y D 3. The intersection points are .4; 3/ and . 4; 3/. Note that having found values for x, we substituted them into the linear equation rather than the quadratic equation to find the corresponding values of y. Had we substituted into the quadratic equation we would have got more solutions (four points in all), but two of them would have failed to satisfy 3x C 4y D 12. When solving systems of nonlinear equations you should always verify that the solutions you find do satisfy the given equations. 42. 2x 2 C 2y 2 D 5, xy D 1. The second equation says that y D 1=x. Substituting this into the first equation gives 2x 2 C .2=x 2 / D 5, or 2x 4 5x 2 C 2 D 0. This equation factors to p .2x 2 1/.x 2 p2/ D 0, so its solutions are x D ˙1= 2 and x D ˙ 2. The corresponding values of y are givenpby p y D 1=x. p Therefore, p the p intersection p points are .1= 2; 2/, . 1= 2; 2/, . 2; 1= 2/, and p p 2; 1= 2/. .

Fig. P.3-44

45.

3/2

.y C 2/2 D 1 is an ellipse with centre at 9 4 .3; 2/, major axis between .0; 2/ and .6; 2/ and minor axis between .3; 4/ and .3; 0/. .x

C

y

x

.3; 2/

.x 3/2 .yC2/2 C D1 9 4

Fig. P.3-45

46.

43. .x 2 =4/ C y 2 D 1 is an ellipse with major axis between . 2; 0/ and .2; 0/ and minor axis between .0; 1/ and .0; 1/.

.y C 1/2 D 4 is an ellipse with centre at 4 .1; 1/, major axis between .1; 5/ and .1; 3/ and minor axis between . 1; 1/ and .3; 1/. .x

1/2 C

y

.x 1/2 C

.yC1/2 D4 4

y x2 2 4 Cy D1

x .1; 1/ x

Fig. P.3-46 Fig. P.3-43 44. 9x 2 C 16y 2 D 144 is an ellipse with major axis between . 4; 0/ and .4; 0/ and minor axis between .0; 3/ and .0; 3/.

47.

.x 2 =4/ y 2 D 1 is a hyperbola with centre at the origin and passing through .˙2; 0/. Its asymptotes are y D ˙x=2.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

7

lOMoARcPSD|6566483

SECTION P.3 (PAGE 22)

ADAMS and ESSEX: CALCULUS 9

y

y x2 4

y 2 D1

yD x=2

.x

x 1/.y C 2/ D 1 yD

x yDx=2

2

xD1

Fig. P.3-47

Fig. P.3-50 51.

48. x 2 y 2 D 1 is a rectangular hyperbola with centre at the origin and passing through .0; ˙1/. Its asymptotes are y D ˙x. y x2 y2 D 1

yDx

a) Replacing x with x replaces a graph with its reflection across the y-axis. b) Replacing y with y replaces a graph with its reflection across the x-axis.

52.

Replacing x with x and y with y reflects the graph in both axes. This is equivalent to rotating the graph 180ı about the origin.

53.

jxj C jyj D 1. In the first quadrant the equation is x C y D 1. In the second quadrant the equation is x C y D 1. In the third quadrant the equation is x y D 1. In the fourth quadrant the equation is x y D 1.

x yD x

y Fig. P.3-48 1 jxj C jyj D 1 1

49. xy D 4 is a rectangular hyperbola with centre at the origin and passing through .2; 2/ and . 2; 2/. Its asymptotes are the coordinate axes.

x

1

y

1

Fig. P.3-53 x

Section P.4 Functions and Their Graphs (page 32)

xyD 4

Fig. P.3-49

50. .x 1/.y C 2/ D 1 is a rectangular hyperbola with centre at .1; 2/ and passing through .2; 1/ and .0; 3/. Its asymptotes are x D 1 and y D 2.

8

1.

f .x/ D 1 C x 2 ; domain R, range Œ1; 1/

2.

f .x/ D 1

3.

G.x/ D

4.

F .x/ D 1=.x 1/; domain . 1; 1/ [ .1; 1/, range . 1; 0/ [ .0; 1/

p

p

8

x; domain Œ0; 1/, range . 1; 1 2x; domain . 1; 4, range Œ0; 1/

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION P.4 (PAGE 32)

t ; domain . 1; 2/, range R. (The equation 2 t y D h.t / can be squared and rewritten as t 2 C y 2 t 2y 2 D 0, a quadratic equation in t having real solutions for every real value of y. Thus the range of h contains all real numbers.)

1 p ; domain Œ2; 3/ [ .3; 1/, range 1 x 2 . 1; 0/ [ .0; 1/. The equation y D g.x/ can be solved for x D 2 .1 .1=y//2 so has a real solution provided y ¤ 0.

b) is the graph of x 2 if x < 1.

y

x/, which is positive

d) is the graph of x 3 x 4 , which is positive if 0 < x < 1 and behaves like x 3 near 0. 9. x 0 ˙0:5 ˙1 ˙1:5 ˙2

y graph (i)

x 3 D x 2 .1

c) is the graph of x x 4 , which is positive if 0 < x < 1 and behaves like x near 0.

6. g.x/ D

7.

x/2 , which is positive for x > 0.

a) is the graph of x.1

5. h.t / D p

graph (ii)

f .x/ D x 4 0 0:0625 1 5:0625 16

y x y

y D x4

x y

graph (iii)

graph (iv)

x

x x

Fig. P.4-7 Graph (ii) is the graph of a function because vertical lines can meet the graph only once. Graphs (i), (iii), and (iv) do not have this property, so are not graphs of functions.

Fig. P.4-9 10.

8. y

y

graph (a)

graph (b)

x

f .x/ D x 2=3

0 ˙0:5 ˙1 ˙1:5 ˙2

0 0:62996 1 1:3104 1:5874 y

x

y

y

graph (c)

yDx

x

2=3

graph (d) x Fig. P.4-10 11.

x Fig. P.4-8

x

f .x/ D x 2 C 1 is even: f . x/ D f .x/

12. f .x/ D x 3 C x is odd: f . x/ D f .x/ x is odd: f . x/ D f .x/ 13. f .x/ D 2 x 1

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

9

lOMoARcPSD|6566483

SECTION P.4 (PAGE 32)

1

14.

f .x/ D

x2

15.

f .x/ D

x

16. f .x/ D f. 4

17.

1

1 2

ADAMS and ESSEX: CALCULUS 9

26.

is even: f . x/ D f .x/ is odd about .2; 0/: f .2

y

x/ D

f .2 C x/

1 is odd about . 4; 0/: xC4 x/ D f . 4 C x/

f .x/ D x 2

x

18. f .x/ D x 3 2 is odd about .0; 2/: f . x/ C 2 D .f .x/ C 2/ 3

yD.x 1/2 C1

6x is even about x D 3: f .3 x/ D f .3 C x/ 27.

y

3

19. f .x/ D jx j D jxj is even: f . x/ D f .x/ 20. f .x/ D jx C 1j is even about x D 1: f . 1 x/ D f . 1 C x/ p 21. f .x/ D 2x has no symmetry. p 22. f .x/ D .x 1/2 is even about x D 1: f .1 x/ D f .1 C x/ 23.

yD1 x 3

x

28.

y

y yD x 2

x

x yD.xC2/3

24.

29.

y

y yD1 x 2 p yD xC1 x

x

25.

30. y

y

yD.x 1/2 p yD xC1

x x

10

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION P.4 (PAGE 32)

31.

36. y

y x

xD2

x yD jxj yD

32.

1 2 x

37.

y

y x yD xC1

yDjxj 1

yD1 x x xD 1

33.

38.

y

y xD1 yDjx 2j x yD 1 x yD 1 x

x

34. y

39.

yD1Cjx 2j

y

y .1;3/ yDf .x/C2 2

.2;2/

.1;1/ yDf .x/ x

2

x

x

Fig. P.4.39(a)

35.

Fig. P.4.39(b)

y

40. y

y .1;3/ yDf .x/C2

xD 2 2

.2;2/

x 1 x 1 yD

yDf .x/ 1 x .2; 1/

2 xC2

Fig. P.4.40(a)

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

Fig. P.4.40(b)

11

lOMoARcPSD|6566483

SECTION P.4 (PAGE 32)

ADAMS and ESSEX: CALCULUS 9

y 0.8 0.6

41. y

y D 0:68 x C 2 yD 2 x C 2x C 3

0.4 0.2

. 1;1/ yDf .xC2/ x

2

-5

-4

-3

-2

-1 -0.2

1y D2 0:18 3

4

x

-0.4 -0.6

42. y

.2;1/

1

-0.8 -1.0 Fig. P.4-47

yDf .x 1/ 3

x

48.

Range is approximately . 1; 0:1 [ Œ2:9; 1/. y

43. 4

y

3 2 2 yD f .x/ .1; 1/

1

x

-4

-3 -2 -1

1 -1

44. y

-2

2

3 4 5 x 2 yD x.x C 2/

6 x

-3

. 1;1/ yDf . x/

-4 x

2

-5 Fig. P.4-48

45. y

49. y yDf .4 x/

5

.3;1/

4 4 x

2

3 2 46. y

1 -5

. 1;1/

-4

-3

.1;1/

1

Range is approximately Œ 0:18; 0:68.

2

y D x4 Fig. P.4-49

x

12

-1 -1

yD1 f .1 x/

47.

-2

3

4

x

6x 3 C 9x 2

1

Apparent symmetry about x D 1:5. This can be confirmed by calculating f .3 x/, which turns out to be equal to f .x/.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION P.5 (PAGE 38)

Apparent symmetry about . 2; 2/. This can be confirmed by calculating shifting the graph right by 2 (replace x with x 2) and then down 2 (subtract 2). The result is 5x=.1 C x 2 /, which is odd.

50. y yD

3 2

2

2

2x C x 2x C x 2

53.

1

-5

-4

-3

-2

-1

1

2

3

4

Section P.5 Combining Functions to Make New Functions (page 38)

Apparent symmetry about x D 1. This can be confirmed by calculating f .2 x/, which turns out to be equal to f .x/.

1.

51. y 4 yDx 1 x 1 x 2

3

yD

2 1 -2

-1

1

2

3

4

-1

5

xC3

2.

Fig. P.4-51 Apparent symmetry about .2; 1/, and about the lines y D x 1 and y D 3 x. These can be confirmed by noting that f .x/ D 1 C

1

x 2 so the graph is that of 1=x shifted right 2 units and up one. yD

2x 2 C 3x C 4x C 5

y

x2

p f .x/ D x, g.x/ D x 1. D.f / D R, D.g/ D Œ1; 1/. D.f C g/ D D.f g/ D D.fg/ D D.g=f / D Œ1; 1/, D.f =g/ D .1; 1/. p .f C g/.x/ D x C x 1 p .f g/.x/ D x x 1 p .fg/.x/ D x x 1 p .f =g/.x/ D x= x 1 p .g=f /.x/ D . 1 x/=x

x

6

yD

-2

52.

f .x/,

x

-1 Fig. P.4-50

-3

If f is both even and odd the f .x/ D f . x/ D so f .x/ D 0 identically.

,

p p f .x/ D 1 x, g.x/ D 1 C x. D.f / D . 1; 1, D.g/ D Œ 1; 1/. D.f C g/ D D.f g/ D D.fg/ D Œ 1; 1, D.f =g/ D . 1;p1, D.g=fp/ D Œ 1; 1/. .f C g/.x/ D 1 x C 1 C x p p .f g/.x/ D 1 x 1Cx p 2 .fg/.x/ D 1 x p .f =g/.x/ D .1 x/=.1 C x/ p .g=f /.x/ D .1 C x/=.1 x/

3. y

5 4

yDx

3

x2

x

2 yDx

1 -7

-6

-5

-4

-3

-2

-1

1

2

x

-1 -2

yD

x2

Fig. P.4-52

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

13

lOMoARcPSD|6566483

SECTION P.5 (PAGE 38)

ADAMS and ESSEX: CALCULUS 9

4.

8.

y yD

x 1

-2

-1

x

1 y D x3

x

f .x/ D 2=x, g.x/ D x=.1 x/. f ı f .x/ D 2=.2=x/ D xI D.f ı f / D fx W x ¤ 0g f ı g.x/ D 2=.x=.1 x// D 2.1 x/=xI D.f ı g/ D fx W x ¤ 0; 1g g ı f .x/ D .2=x/=.1 .2=x// D 2=.x 2/I D.g ı f / D fx W x ¤ 0; 2g g ı g.x/ D .x=.1 x//=.1 .x=.1 x/// D x=.1 2x/I D.g ı g/ D fx W x ¤ 1=2; 1g

-1 y D x3

-2

9.

5. y

y D x C jxj

y D jxj

p f .x/ D 1=.1 x/, g.x/ D x 1. f ı f .x/ D 1=.1 .1=.1 x/// D .x 1/=xI D.f ı f / D fx W x ¤ 0; 1g p x 1/I f ı g.x/ D 1=.1 D.f ı g/ D fx W x  1; x ¤ 2g p p g ı f .x/ D .1=.1 x// 1 D x=.1 x/I D.g ı f / D Œ0; 1/ q p g ı g.x/ D x 1 1I D.g ı g/ D Œ2; 1/

y D x D jxj x

yDx

6. y 4

y D jxj C jx

3

2j

y D jxj

2

f .x/ y D jx

1 -2

-1

10. f .x/ D .x C 1/=.x 1/ D 1 C 2=.x 1/, g.x/ D sgn .x/. f ı f .x/ D 1 C 2=.1 C .2=.x 1/ 1// D xI D.f ı f / D fx W x ¤ 1g sgn x C 1 f ı g.x/ D D 0I D.f ı g/ D . 1; 0/ sgn x 1   n xC1 1 if x < 1 or x > 1 g ı f .x/ D sgn D I 1 if 1 < x < 1 x 1 D.g ı f / D fx W x ¤ 1; 1g g ı g.x/ D sgn .sgn .x// D sgn .x/I D.g ı g/ D fx W x ¤ 0g

1

2

3

4

2j

5

11. 12. 13. 14. 15. 16.

x

-1

7.

x

2

xp 4 x 2x 3 C 3 .x C 1/=x 1=.x C 1/2

g.x/

f ı g.x/

xC1 xC4 x2 x 1=3 1=.x 1/ x 1

.x C 1/2 x jxj 2x C 3 x 1=x 2

f .x/ D x C 5, g.x/ D x 2 3. f ı g.0/ D f . 3/ D 2; g.f .0// D g.5/ D 22 f .g.x// D f .x 2

3/ D x 2 C 2

g ı f .x/ D g.f .x// D g.x C 5/ D .x C 5/2 f ı f . 5/ D f .0/ D 5; g.g.2// D g.1/ D f .f .x// D f .x C 5/ D x C 10 g ı g.x/ D g.g.x// D .x 2

14

3/2

3

3 2

17.

p y D x. p y D 2 C px: previous graph is raised 2 units. y D 2 C 3p C x: previous graph is shiftend left 3 units. y D 1=.2 C 3 C x/: previous graph turned upside down and shrunk vertically.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION P.5 (PAGE 38)

y

21. y

y D2C y D2C

p

xC3

y D 1=.2 C

p

p

.1=2;1/ yDf .2x/

x

yD

x C 3/

x

1

p

x 22. y

x yDf .x=3/

Fig. P.5-17

18. yD

p

1

23.

y

2x

y

y D 2x yD p

y D 2x

1 1

6 x

3

. 2;2/

1 yD1Cf . x=2/

2x

x

x yD p

24.

1 1

2x

y

1

yD2f ..x 1/=2/

yD1

2x

Fig. P.5-18

1

5

x

19. y

25. y .1;2/

y D f .x/ .1; 1/

yD2f .x/

82

x

2

x

26.

20.

-2y

y

y D g.x/ .1; 1/ 2 x yD .1=2/f .x/

x

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

15

lOMoARcPSD|6566483

SECTION P.5 (PAGE 38)

27.

ADAMS and ESSEX: CALCULUS 9

F .x/ D Ax C B (a) F ı F .x/ D F .x/ ) A.Ax C B/ C B D Ax C B ) AŒ.A 1/x C B D 0 Thus, either A D 0 or A D 1 and B D 0. (b) F ı F .x/ D x ) A.Ax C B/ C B D x ) .A2 1/x C .A C 1/B D 0 Thus, either A D 1 or A D 1 and B D 0

28. bxc D 0 for 0  x < 1; dxe D 0 for

35.

a)

Let E.x/ D 12 Œf .x/ C f . x/. Then E. x/ D 21 Œf . x/ C f .x/ D E.x/. Hence, E.x/ is even. Let O.x/ D 12 Œf .x/ f . x/. Then O. x/ D 12 Œf . x/ f .x/ D O.x/ and O.x/ is odd. E.x/ C O.x/

D 12 Œf .x/ C f . x/ C 12 Œf .x/

1  x < 0.

f . x/

D f .x/:

29. bxc D dxe for all integers x.

Hence, f .x/ is the sum of an even function and an odd function.

30. d xe D bxc is true for all real x; if x D n C y where n is an integer and 0  y < 1, then x D n y, so that d xe D n and bxc D n. 31. y y D x bxc

b) If f .x/ D E1 .x/ C O1 .x/ where E1 is even and O1 is odd, then E1 .x/ C O1 .x/ D f .x/ D E.x/ C O.x/: Thus E1 .x/ E.x/ D O.x/ O1 .x/. The left side of this equation is an even function and the right side is an odd function. Hence both sides are both even and odd, and are therefore identically 0 by Exercise 36. Hence E1 D E and O1 D O. This shows that f can be written in only one way as the sum of an even function and an odd function.

x 32. f .x/ is called the integer part of x because jf .x/j is the largest integer that does not exceed x; i.e. jxj D jf .x/j C y, where 0  y < 1. y

Section P.6 Polynomials and Rational Functions (page 45) x y D f .x/

1.

x 2 7x C 10 D .x C 5/.x C 2/ The roots are 5 and 2.

2.

x 2 3x 10 D .x 5/.x C 2/ The roots are 5 and 2.

34. f even , f . x/ D f .x/ f odd , f . x/ D f .x/ f even and odd ) f .x/ D ) f .x/ D 0

16

f .x/ ) 2f .x/ D 0

4

8

D

1 ˙ i.

If x C 2x C 2 D 0, then x D 2 The roots are 1 C i and 1 i . x 2 C 2x C 2 D .x C 1 i /.x C 1 C i /.

4.

Rather than use the quadratic formula this time, let us complete the square. x2

6x C 13 D x 2 D .x D .x

f ı g. x/ D f .g. x// D f . g.x// D f .g.x// D f ı g.x/ .fg/. x/ D f . x/g. x/ D f .x/Œ g.x/ D f .x/g.x/ D .fg/.x/: The others are similar.

p

3. Fig. P.5-32 33. If f is even and g is odd, then: f 2 , g 2 , f ı g, g ı f , and f ı f are all even. fg, f =g, g=f , and g ı g are odd, and f C g is neither even nor odd. Here are two typical verifications:

2˙

2

The roots are 3 C 2i and 3

6x C 9 C 4

3/2 C 22 3 2i /.x

3 C 2i /:

2i .

5.

16x 4 8x 2 C 1 D .4x 2 1/2 D .2x 1/2 .2x C 1/2 . There are two double roots: 1=2 and 1=2.

6.

x 4 C 6x 3 C 9x 2 D x 2 .x 2 C 6x C 9/ D x 2 .x C 3/2 . There are two double roots, 0 and 3.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

7.

SECTION P.6 (PAGE 45)

x 3 C 1 D .x C 1/.x 2 x C 1/. One root is 1. The other two are the solutions of x 2 x C 1 D 0, namely xD

1˙

p

1

4

p 3 1 ˙ i: 2 2

D

2

15.

The denominator is x 3 C x 2 D x 2 .x C 1/ which is zero only if x D 0 or x D 1. Thus the rational function is defined for all real numbers except 0 and 1.

16.

The denominator is x 2 Cx 1, which is a quadratic polynomial whose roots can p be found with the quadratic formula. They are x D . 1 ˙ 1 C 4/=2. Hence the given rational p 5/=2 function is p defined for all real numbers except . 1 and . 1 C 5/=2.

We have p

3 i 2

1 2

x 3 C 1 D .x C 1/ x

8. x 4 1 D .x 2 1/.x 2 C 1/ D .x The roots are 1, 1, i , and i .

!

p ! 1 3 C i : 2 2

x

1/.x C 1/.x

10.

x

x

4

16x C 16 D .x

The roots are 1, 2, 11.

1/.x

4

2, 2i , and

2i .

x 5 C x 3 C 8x 2 C 8 D .x 2 C 1/.x 3 C 8/

i /.x C i /.x 2

D .x C 2/.x

18.

16/

1/.x 2 4/.x 4 C 4/ 1/.x 2/.x C 2/.x

D .x D .x

2i /.x C 2i /:

19.

p p 20. x Cx C8x C8 D .xC2/.x i /.xCi /.x aC 3 i /.x a 3 i /: 12.

x9

3

4x 7

2

x 6 C 4x 4 D x 4 .x 5 4

D x .x

x2

3

1/.x

4

D x .x

4x 3 C 4/

2

4/

2/.x C 2/.x 2 C x C 1/:

1/.x

Seven of the nine roots are: 0 (with multiplicity 4), 1, 2, and 2. The other two roots are solutions of x 2 C x C 1 D 0, namely xD

1˙

p 2

1

4

D

The required factorization of x 9 x 4 .x 1/.x 2/.xC2/ x

Following the method of Example 6, we calculate .x 2 bxCa2 /.x 2 CbxCa2 / D x 4 Ca4 C.2a2 b 2 /x 2 D x 4 Cx 2 C1

! 3 i : 2

p

13. The denominator is x 2 C 2x C 2 D .x C 1/2 C 1 which is never 0. Thus the rational function is defined for all real numbers. 14.

x4 C x2 x.x 3 C x 2 C 1/ x 3 x C x 2 D 2 Cx C1 x3 C x2 C 1 3 .x C x 2 C 1/ C x 2 C 1 x C x 2 DxC x3 C x2 C 1 2 2x xC1 : D x 1C 3 x C x2 C 1

x3

22.

provided a D 1 and b 2 D 1 C 2a2 D 1, so b D ˙1. Thus P .x/ D .x 2 x C 1/.x 2 C x C 1/.

x 6 C 4x 4 is 1 2

x3 x 3 C 2x 2 C 3x 2x 2 3x D C 2x C 3 x 2 C 2x C 3 x.x 2 C 2x C 3/ 2x 2 3x D x 2 C 2x C 3 2.x 2 C 2x C 3/ 4x 6 C 3x Dx x 2 C 2x C 3 xC6 Dx 2C 2 : x C 2x C 3

4 2 As in Example p 6, we wantpa D 4, so a D 2 and a D 2, b D ˙ 2a D ˙2. Thus P .x/ D .x 2 2x C 2/.x 2 C 2x C 2/.

3 1 ˙ i: 2 2

p ! 1 3 C i x 2 2

x2

3

21.

p

4x 7

1 x 3 2x C 2x 1 D 2 x2 2 x.x 2 2/ C 2x 1 D x2 2 2x 1 DxC 2 : x 2

x 2 C 5x C 3 5x x2 D x 2 C 5x C 3 x 2 C 5x C 3 5x 3 : D1C 2 x C 5x C 3

2x C 4/

Three of the five roots are 2, i and i . The remain2 ing two are 2x C 4 D 0, namely p solutions of x p 2 ˙ 4 16 xD D 1 ˙ 3 i . We have 2 5

x3 x2

i /.x C i /.

9. x 6 3x 4 C 3x 2 1 D .x 2 1/3 D .x 1/3 .x C 1/3 . The roots are 1 and 1, each with multiplicity 3. 5

17.

The denominator is x 3 x D x.x 1/.x C 1/ which is zero if x D 0, 1, or 1. Thus the rational function is defined for all real numbers except 0, 1, and 1.

23.

24.

Let P .x/ D an x n C an 1 x n 1 C


C a1 x C a0 , where n  1. By the Factor Theorem, x 1 is a factor of P .x/ if and only if P .1/ D 0, that is, if and only if an C an 1 C


C a1 C a0 D 0.

Let P .x/ D an x n C an 1 x n 1 C


C a1 x C a0 , where n  1. By the Factor Theorem, x C 1 is a factor of P .x/ if and only if P . 1/ D 0, that is, if and only if a0 a1 C a2 a3 C


C . 1/n an D 0. This condition says that the sum of the coefficients of even powers is equal to the sum of coefficients of odd powers.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

17

lOMoARcPSD|6566483

SECTION P.6 (PAGE 45)

ADAMS and ESSEX: CALCULUS 9

25. Let P .x/ D an x n C an 1 x n 1 C


C a1 x C a0 , where the coefficients ak , 0  k  n are all real numbers, so that ak D ak . Using the facts about conjugates of sums and products mentioned in the statement of the problem, we see that if z D x C iy, where x and y are real, then P .z/ D an z n C an n

D an z C an D P .z/:

1z

n 1

1z

n 1

4.

sin

5.

cos

6.

sin

C


C a1 z C a0

C


C a1 z C a0

If z is a root of P , then P .z/ D P .z/ D 0 D 0, and z is also a root of P . 26. By the previous exercise, z D u iv is also a root of P . Therefore P .x/ has two linear factors x u iv and x u C iv. The product of these factors is the real quadratic factor .x u/2 i 2 v 2 D x 2 2ux C u2 C v 2 , which must also be a factor of P .x/. 27.

By the previous exercise P .x/ D 2ux C u2 C v 2 .x

x2

u

P .x/ iv/.x

u C iv/

D Q1 .x/;

where Q1 , being a quotient of two polynomials with real coefficients, must also have real coefficients. If z D u C iv is a root of P having multiplicity m > 1, then it must also be a root of Q1 (of multiplicity m 1), and so, therefore, z must be a root of Q1 , as must be the real quadratic x 2 2ux C u2 C v 2 . Thus .x 2

P .x/ D 2 2ux C u2 C v 2 /2 x

Q1 .x/ D Q2 .x/; 2ux C u2 C v 2

where Q2 is a polynomial with real coefficients. We can continue in this way until we get

.x 2

7.

P .x/ D Qm .x/; 2ux C u2 C v 2 /m



7 12



1.

cos

2. tan

3. sin

18



3 4



 D cos 

3 D 4

tan

 2 D sin  3

 D 4

3 D 4

cos

 D 4

 11 D sin 12 12  D sin 3 4     cos sin D sin cos 3 4 3 4 p !     3 1 1 1 D p p 2 2 2 2 p 3 1 D p 2 2

  cos. C x/ D cos 2 . x/   D cos . x/ D cos.

8.

sin.2

9.

sin

10.



x/ D

3 2

x



3 Cx cos 2

!

x/ D

cos x

sin x     D sin  x 2   D sin x  2  D sin x 2 D cos x D cos

3 cos x 2

sin

3 sin x 2

D . 1/. sin x/ D sin x

1 p 2 11.

1

C

  2  5 D cos 12 3 4 2  2  D cos cos C sin sin 3 4 3 4 p !     1 3 1 1 C D p p 2 2 2 2 p 3 1 D p 2 2

where Qm no longer has z (or z) as a root. Thus z and z must have the same multiplicity as roots of P .

Section P.7 The Trigonometric Functions (page 57)



 4 3     D sin cos C cos sin 4 3 p 4 3 p 1 1 3 1 1C 3 D p p Cp D 22 2 2 2 2

D sin

cos x sin x C cos x sin x sin2 x C cos2 x D cos x sin x 1 D cos x sin x

tan x C cot x D

p  3  D sin D 3 3 2

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

12.

 cos x sin x tan x cot x cos x sin x  D  cos x sin x tan x C cot x C cos x sin x ! 2 sin x cos2 x cos x sin x ! D sin2 x C cos2 x cos x sin x 

D sin2 x 13.

cos4 x

sin4 x D .cos2 x

16.

17.



.cos x sin x/2 cos x sin x D cos x C sin x .cos x C sin x/.cos x sin x/ cos2 x 2 sin x cos x C sin2 x D cos2 x sin2 x 1 sin.2x/ D cos.2x/ D sec.2x/ tan.2x/

D 2 sin x cos x C sin x.1 D 3 sin x

sin

x has period 4. 2 y

sin2 x D cos.2x/

sin 3x D sin.2x C x/ D sin 2x cos x C cos 2x sin x

1



Fig. P.7-20

21.

sin x has period 2. y

y D sin.x/

1

1

D 2 cos3 x

D 4 cos3 x

2

sin x/ C sin x

1

Fig. P.7-21

22.

cos

x has period 4. 2 y 1 1

x

1

Fig. P.7-22 2

2 sin3 x

23. y

4 sin x

2

2.1

5 3

3

cos x

4 x

3

2 sin x/

2

1/ cos x

2 x

1

cos 3x D cos.2x C x/ D cos 2x cos x sin 2x sin x D .2 cos2 x

2 x

Fig. P.7-19

sin2 x/.cos2 x C sin2 x/

2

18.

=2

.1 cos x/.1 C cos x/ D 1 cos2 x D sin2 x implies 1 cos x sin x D . Now sin x 1 C cos x   x 1 cos 2 1 cos x  x 2 D sin x sin 2  2  x  1 1 2 sin2 2 D x x 2 sin cos 2 2 x sin 2 D tan x D x 2 cos 2 x  2 sin2 x  1 cos x  2x  D tan2 D 1 C cos x 2 2 cos2 2

D 2 sin x.1

y D cos.2x/

1

20.

D cos x

15.

y

cos2 x

2

14.

SECTION P.7 (PAGE 57)

 y D 2 cos x

 3

1 



2 sin2 x cos x cos2 x/ cos x

3 cos x

x -1 -2 -3

19. cos 2x has period .

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

19

lOMoARcPSD|6566483

SECTION P.7 (PAGE 57)

ADAMS and ESSEX: CALCULUS 9

24.

  y D 1 C sin x C 4

y 2

29.

sin x D cos x D

1 ; 2p

 0 for

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

Section 1.4 Continuity 1.

SECTION 1.4 (PAGE 87)

(page 87)

8.

g is continuous at x D 2, discontinuous at x D 1; 0; 1, and 2. It is left continuous at x D 0 and right continuous at x D 1. y .1; 2/ 2 y D g.x/

. 1; 1/

-2

1

lim f .x/ D

x! 1

1

2

x

Fig. 1.4-1 2. g has removable discontinuities at x D 1 and x D 2. Redefine g. 1/ D 1 and g.2/ D 0 to make g continuous at those points.

D f . 1/ D

4. Function f is discontinuous at x D 1; 2; 3; 4, and 5. f is left continuous at x D 4 and right continuous at x D 2 and x D 5. y

lim f .x/:

x! 1C

12. C.t / is discontinuous only at the integers. It is continuous on the left at the integers, but not on the right. 13.

1 15.

4

5

6

x 16.

-1

Fig. 1.4-4 5. f cannot be redefined at x D 1 to become continuous there because limx!1 f .x/ .D 1/ does not exist. (1 is not a real number.) 6. sgn x is not defined at x D 0, so cannot be either continuous or discontinuous there. (Functions can be continuous or discontinuous only at points in their domains!)  x if x < 0 is continuous everywhere on the 7. f .x/ D x 2 if x  0 real line, even at x D 0 where its left and right limits are both 0, which is f .0/.

The least integer function dxe is continuous everywhere on

right continuous.

2

3

lim x 2 D

x! 1C

R except at the integers, where it is left continuous but not

14.

y D f .x/

2

1¤1

 2 if x ¤ 0 is continuous everywhere except f .x/ D 1=x 0 if x D 0 at x D 0, where it is neither left nor right continuous since it does not have a real limit there.  2 if x  1 is continuous everywhere ex10. f .x/ D x 0:987 if x > 1 cept at x D 1, where it is left continuous but not right continuous because 0:987 ¤ 1. Close, as they say, but no cigar. 11.

3. g has no absolute maximum value on Œ 2; 2. It takes on every positive real value less than 2, but does not take the value 2. It has absolute minimum value 0 on that interval, assuming this value at the three points x D 2, x D 1, and x D 1.

1

lim x D

x! 1

9.

-1

3



x if x < 1 is continuous everywhere on the x 2 if x  1 real line except at x D 1 where it is right continuous, but not left continuous.

f .x/ D

17.

x2 4 Since D x C 2 for x ¤ 2, we can define the x 2 function to be 2 C 2 D 4 at x D 2 to make it continuous there. The continuous extension is x C 2.

1 C t3 .1 C t /.1 t C t 2 / 1 t C t2 D D for 2 1 t .1 C t /.1 t / 1 t t ¤ 1, we can define the function to be 3=2 at t D 1 to make it continuous there. The continuous extension is 1 t C t2 . 1 t Since

.t 2/.t 3/ t 2 t 2 5t C 6 D D for t ¤ 3, 2 t t 6 .t C 2/.t 3/ t C2 we can define the function to be 1=5 at t D 3 to make it t 2 continuous there. The continuous extension is . t C2 Since p p p xC 2 x2 2 .x 2/.x C 2/ p p D p D x4 4 2/.x C 2/.x 2 C 2/ .x C 2/.x 2 C 2/ p.x for x p ¤ 2, we can define the function to be 1=4 at there. The continuous x D 2 to make it continuous p xC 2 extension is p . (Note: cancelling the .x C 2/.x 2 C 2/ p 2 factors provides a further continuous extension to xC p 2. xD Since

limx!2C f .x/ D k 4 and limx!2 f .x/ D 4 D f .2/. Thus f will be continuous at x D 2 if k 4 D 4, that is, if k D 8.

18. limx!3 g.x/ D 3 m and limx!3C g.x/ D 1 3m D g.3/. Thus g will be continuous at x D 3 if 3 m D 1 3m, that is, if m D 1.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

31

lOMoARcPSD|6566483

SECTION 1.4 (PAGE 87)

ADAMS and ESSEX: CALCULUS 9

19. x 2 has no maximum value on 1 < x < 1; it takes all positive real values less than 1, but it does not take the value 1. It does have a minimum value, namely 0 taken on at x D 0. 20. The Max-Min Theorem says that a continuous function defined on a closed, finite interval must have maximum and minimum values. It does not say that other functions cannot have such values. The Heaviside function is not continuous on Œ 1; 1 (because it is discontinuous at x D 0), but it still has maximum and minimum values. Do not confuse a theorem with its converse. 21.

Let the numbers be x and y, where x  0, y  0, and x C y D 8. If P is the product of the numbers, then P D xy D x.8

x/ D 8x

x 2 D 16

.x

f .x/ D

29.

f .x/ D x 3 C x 1, f .0/ D 1, f .1/ D 1. Since f is continuous and changes sign between 0 and 1, it must be zero at some point between 0 and 1 by IVT.

30.

f .x/ D x 3 15x C 1 is continuous everywhere. f . 4/ D 3; f . 3/ D 19; f .1/ D 13; f .4/ D 5: Because of the sign changes f has a zero between 4 and 3, another zero between 3 and 1, and another between 1 and 4.

31.

F .x/ D .x a/2 .x b/2 C x. Without loss of generality, we can assume that a < b. Being a polynomial, F is continuous on Œa; b. Also F .a/ D a and F .b/ D b. Since a < 21 .a C b/ < b, the Intermediate-Value Theorem guarantees that there is an x in .a; b/ such that F .x/ D .a C b/=2.

32.

Let g.x/ D f .x/ x. Since 0  f .x/  1 if 0  x  1, therefore, g.0/  0 and g.1/  0. If g.0/ D 0 let c D 0, or if g.1/ D 0 let c D 1. (In either case f .c/ D c.) Otherwise, g.0/ > 0 and g.1/ < 0, and, by IVT, there exists c in .0; 1/ such that g.c/ D 0, i.e., f .c/ D c.

33.

The domain of an even function is symmetric about the y-axis. Since f is continuous on the right at x D 0, therefore it must be defined on an interval Œ0; h for some h > 0. Being even, f must therefore be defined on Œ h; h. If x D y, then

4/2 :

Therefore P  16, so P is bounded. Clearly P D 16 if x D y D 4, so the largest value of P is 16. 22. Let the numbers be x and y, where x  0, y  0, and x C y D 8. If S is the sum of their squares then S D x 2 C y 2 D x 2 C .8 D 2x 2

x/2 4/2 C 32:

16x C 64 D 2.x

Since 0  x  8, the maximum value of S occurs at x D 0 or x D 8, and is 64. The minimum value occurs at x D 4 and is 32.

23. Since T D 100 30x C 3x 2 D 3.x 5/2 C 25, T will be minimum when x D 5. Five programmers should be assigned, and the project will be completed in 25 days. 24.

lim f .x/ D lim f . y/ D lim f .y/ D f .0/:

x!0

If x desks are shipped, the shipping cost per desk is C D

245x

30x 2 C x 3 D x2 x D .x

34.

This cost is minimized if x D 15. The manufacturer should send 15 desks in each shipment, and the shipping cost will then be $20 per desk. x2

x!0C

x!0

x, we obtain

t!0C

D

t!0C

f .t /

f .0/ D f . 0/ D f .0/:

Therefore f is continuous at 0 and f .0/ D 0.

.x 1/.x C 1/ x2 1 D f .x/ D 2 x 4 .x 2/.x C 2/ f D 0 at x D ˙1. f is not defined at x D ˙2. f .x/ > 0 on . 1; 2/, . 1; 1/, and .2; 1/. f .x/ < 0 on . 2; 1/ and .1; 2/.

32

f odd , f . x/ D f .x/ f continuous on the right , lim f .x/ D f .0/

Therefore, letting t D

.x

26. f .x/ D x 2 C 4x C 3 D .x C 1/.x C 3/ f .x/ > 0 on . 1; 3/ and . 1; 1/ f .x/ < 0 on . 3; 1/. 27.

y!0C

lim f .x/ D lim f . t / D lim

1/.x C 1/ 25. f .x/ D D x x f D 0 at x D ˙1. f is not defined at 0. f .x/ > 0 on . 1; 0/ and .1; 1/. f .x/ < 0 on . 1; 1/ and .0; 1/. 1

y!0C

Thus, f is continuous on the left at x D 0. Being continuous on both sides, it is therefore continuous.

30x C 245 15/2 C 20:

.x C 2/.x 1/ x2 C x 2 D 3 x x3 f .x/ > 0 on . 2; 0/ and .1; 1/ f .x/ < 0 on . 1; 2/ and .0; 1/.

28.

35.

max 1:593 at

0:831, min

0:756 at 0:629

36.

max 0:133 at x D 1:437; min

0:232 at x D

1:805

37. max 10:333 at x D 3; min 4:762 at x D 1:260 38.

max 1:510 at x D 0:465; min 0 at x D 0 and x D 1

39.

root x D 0:682

40.

root x D 0:739

41. roots x D

0:637 and x D 1:410

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

42. roots x D

SECTION 1.5 (PAGE 92)

0:7244919590 and x D 1:220744085

43. fsolve gives an approximation to the single real root to 10 significant figures; solve gives the three roots (including a complex conjugate pair) in exact form involving the quan p 1=3 ; evalf(solve) gives approximations tity 108 C 12 69 to the three roots using 10 significant figures for the real and imaginary parts.

Section 1.5 The Formal Definition of Limit (page 92) 1.

We require 39:9  L  40:1. Thus

10. We need 1 0:05  1=.x C 1/  1 C 0:05, or 1:0526  x C 1  0:9524. This will occur if 0:0476  x  0:0526. In this case we can take ı D 0:0476. 11.

x!1

Proof: Let  > 0 be given. Then j.3x C 1/ 4j <  holds if 3jx 1j < , and so if jx 1j < ı D =3. This confirms the limit. 12. To be proved: lim .5 x!2

Let  > 0 be given. p Then jx 2 jx 0j D jxj < ı D .

14.

2. Since 1.2% of 8,000 is 96, we require the edge length x of the cube to satisfy 7904  x 3  8096. It is sufficient that 19:920  x  20:079. The edge of the cube must be within 0.079 cm of 20 cm.

4.

5.

6.

0:02  2x 1  3 C 0:02 3:98  2x  4:02 1:99  x  2:01

2

1 4x 2 D 2. 2x x!1=2 1 Proof: Let  > 0 be given. Then if x ¤ 1=2 we have

provided jx 16.

1  2 C 0:01 x 1 1 x 2:01 1:99 0:5025  x  0:4975

We need 0:03  .3x C 1/ 7  0:03, which is equivalent to 0:01  x 2  0:01 Thus ı D 0:01 will do. p 8. We need 0:01  2x C 3 3  0:01. Thus p 2:99  2x C 3  3:01 8:9401  2x C 3  9:0601 2:97005  x  3:03005 0:02995  x 3  0:03005:

Here ı D 0:02995 will do. 9. We need 8 0:2  x 3  8:2, or 1:9832  x  2:0165. Thus, we need 0:0168  x 2  0:0165. Here ı D 0:0165 will do.

lim

ˇ ˇ 2ˇˇ D j.1 C 2x/ 1 2j

2j D j2x

< ı D =2.

ˇ ˇ 1j D 2 ˇˇx

ˇ 1 ˇˇ 0 be given. For x ¤ 2 we have To be proved:

lim

ˇ 2 ˇ x C 2x ˇ ˇ xC2

7.

2j < 

2j < ı D .

To be proved:

ˇ ˇ 1 4x 2 ˇ ˇ 1 2x

0:01 

3

x 2 D 0. 1 C x2 Proof: Let  > 0 be given. Then ˇ ˇ ˇ ˇ x 2 ˇ D jx 2j  jx ˇ 0 ˇ ˇ 1 C x2 1 C x2

To be proved: lim

provided jx

15.

4 0:1  x 2  4 C 0:1 1:9749  x  2:0024 p 1 0:1  x  1:1 0:81  x  1:21

0j <  holds if

x!2

ı

The temperature should be kept between 12 C and 20 C.

3

To be proved: lim x 2 D 0. x!0

39:9  39:6 C 0:025T  40:1 0:3  0:025T  0:5 12  T  20:

3.

2x/ D 1.

Proof: Let  > 0 be given. Then j.5 2x/ 1j <  holds if j2x 4j < , and so if jx 2j < ı D =2. This confirms the limit. 13.

ı

To be proved: lim .3x C 1/ D 4.

ˇ ˇ . 2/ˇˇ D jx C 2j < 

provided jx C 2j < ı D . This completes the proof.

17.

1 1 D . xC1 2 Proof: Let  > 0 be given. We have ˇ ˇ ˇ ˇ ˇ 1 1 ˇˇ ˇˇ 1 x ˇˇ jx 1j ˇ D ˇ x C 1 2 ˇ ˇ 2.x C 1/ ˇ D 2jx C 1j : To be proved: lim

x!1

If jx 1j < 1, then 0 < x < 2 and 1 < x C 1 < 3, so that jx C 1j > 1. Let ı D min.1; 2/. If jx 1j < ı, then ˇ ˇ ˇ 1 1 ˇˇ 2 jx 1j ˇ ˇ x C 1 2 ˇ D 2jx C 1j < 2 D :

This establishes the required limit.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

33

lOMoARcPSD|6566483

SECTION 1.5 (PAGE 92)

ADAMS and ESSEX: CALCULUS 9

1 xC1 D . x2 1 2 Proof: Let  > 0 be given. If x ¤ 1, we have ˇ  ˇ ˇ  ˇ ˇ xC1 jx C 1j 1 ˇˇ ˇˇ 1 1 ˇˇ ˇ D D : ˇ x2 1 2 ˇ ˇx 1 2 ˇ 2jx 1j

18. To be proved:

lim

If jx C 1j < 1, then 2 < x < 0, so 3 < x 1 < 1 and jx 1j > 1. Ler ı D min.1; 2/. If 0 < jx . 1/j < ı then jx 1j > 1 and jx C 1j < 2. Thus ˇ  ˇ ˇ xC1 1 ˇˇ jx C 1j 2 ˇ D < D : ˇ x2 1 2 ˇ 2jx 1j 2

This completes the required proof. p 19. To be proved: lim x D 1.

Proof: Let  > 0 be given. We have ˇ ˇ ˇ x 1 ˇ p ˇ  jx j x 1j D ˇˇ p x C 1ˇ

26.

28.

3

20. To be proved: lim x D 8. x!2

Proof: Let  > 0 be given. We have jx 3 8j D jx 2jjx 2 C2xC4j. If jx 2j < 1, then 1 < x < 3 and x 2 < 9. Therefore jx 2 C 2x C 4j  9 C 2  3 C 4 D 19. If jx 2j < ı D min.1; =19/, then jx 3

  19 D : 2jjx 2 C 2x C 4j < 19

8j D jx

29.

implies jf .x/

ı x 1 > 1=B, be given. We have x 1 that is, if 1 ı < x < 1, where ı D 1=B:. This completes the proof. To be proved: limx!1 p

23. We say that limx!a f .x/ D 1 if the following condition holds: for every number B > 0 there exists a number ı > 0, depending on B, such that 0 < jx 24.

aj < ı

implies f .x/
0 there exists a number R > 0, depending on B, such that x>R

34

implies f .x/ > B:

1 x2

C1

D 0. Proof: Let  > 0

provided x > R, where R D 1=. This completes the proof. 30.

Lj < :

implies f .x/ > B:

ˇ ˇ ˇ ˇ 1 ˇp 1 ˇD p 1 < 0 there exists a number R > 0, depending on , such that

ı 0, depending on , such that a

B:

1 To be proved: limx!1C D 1. Proof: Let B > 0 x 1 1 > B if 0 < x 1 < 1=B, that be given. We have x 1 is, if 1 < x < 1 C ı, where ı D 1=B. This completes the proof.

This completes the proof. 21.

implies f .x/
0 there exists a number ı > 0, depending on B, such that a

1j < 

1j < ı D . This completes the proof.

We say that limx!aC f .x/ D 1 if the following condition holds: for every number B > 0 there exists a number ı > 0, depending on R, such that a B if x > R where R D B . This completes the proof. To be proved: if lim f .x/ D L and lim f .x/ D M , then x!a x!a L D M. Proof: Suppose L ¤ M . Let  D jL M j=3. Then  > 0. Since lim f .x/ D L, there exists ı1 > 0 such that x!a

jf .x/ Lj <  if jx aj < ı1 . Since lim f .x/ D M , there x!a

exists ı2 > 0 such that jf .x/ M j <  if jx Let ı D min.ı1 ; ı2 /. If jx aj < ı, then 3 D jL

aj < ı2 .

M j D j.f .x/ M / C .L f .x/j  jf .x/ M j C jf .x/ Lj <  C  D 2:

This implies that 3 < 2, a contradiction. Thus the original assumption that L ¤ M must be incorrect. Therefore L D M.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 1 (PAGE 93)

32. To be proved: if lim g.x/ D M , then there exists ı > 0

This completes the proof.

x!a

such that if 0 < jx aj < ı, then jg.x/j < 1 C jM j. Proof: Taking  D 1 in the definition of limit, we obtain a number ı > 0 such that if 0 < jx aj < ı, then jg.x/ M j < 1. It follows from this latter inequality that

36.

To be proved: if lim f .x/ D L and lim f .x/ D M ¤ 0, x!a

x!a

L f .x/ D . x!a g.x/ M Proof: By Exercises 33 and 35 we have

then lim

jg.x/j D j.g.x/ M /CM j  jG.x/ M jCjM j < 1CjM j: lim

x!a

f .x/ 1 1 L D lim f .x/  DL D : x!a g.x/ g.x/ M M

33. To be proved: if lim f .x/ D L and lim g.x/ D M , then x!a

x!a

lim f .x/g.x/ D LM .

x!a

37. To be proved: if f is continuous at L and lim g.x/ D L,

Proof: Let  > 0 be given. Since lim f .x/ D L, there

x!c

x!a

then lim f .g.x// D f .L/.

exists ı1 > 0 such that jf .x/ Lj < =.2.1 C jM j// if 0 < jx aj < ı1 . Since lim g.x/ D M , there ex-

x!c

Proof: Let  > 0 be given. Since f is continuous at L, there exists a number > 0 such that if jy Lj < , then jf .y/ f .L/j < . Since limx!c g.x/ D L, there exists ı > 0 such that if 0 < jx cj < ı, then jg.x/ Lj < . Taking y D g.x/, it follows that if 0 < jx cj < ı, then jf .g.x// f .L/j < , so that limx!c f .g.x// D f .L/.

x!a

ists ı2 > 0 such that jg.x/ M j < =.2.1 C jLj// if 0 < jx aj < ı2 . By Exercise 32, there exists ı3 > 0 such that jg.x/j < 1 C jM j if 0 < jx aj < ı3 . Let ı D min.ı1 ; ı2 ; ı3 /. If jx aj < ı, then jf .x/g.x/

LM D jf .x/g.x/ Lg.x/ C Lg.x/ LM j D j.f .x/ L/g.x/ C L.g.x/ M /j  j.f .x/ L/g.x/j C jL.g.x/ M /j D jf .x/ Ljjg.x/j C jLjjg.x/ M j   < .1 C jM j/ C jLj 2.1 C jM j/ 2.1 C jLj/    C D : 2 2

38.

Thus lim f .x/g.x/ D LM . x!a

34. To be proved: if lim g.x/ D M where M ¤ 0, then

jg.x/

x!a

there exists ı > 0 such that if 0 < jx aj < ı, then jg.x/j > jM j=2. Proof: By the definition of limit, there exists ı > 0 such that if 0 < jx aj < ı, then jg.x/ M j < jM j=2 (since jM j=2 is a positive number). This latter inequality implies that jM j D jg.x/C.M g.x//j  jg.x/jCjg.x/ M j < jg.x/jC

It follows that jg.x/j > jM j required.

jM j : 2

Lj D jg.x/  jg.x/  jh.x/ D jh.x/  jh.x/   < C 3 3

Review Exercises 1 (page 93) 1.

The average rate of change of x 3 over Œ1; 3 is 33 3

x!a

2

there exists ı1 > 0 such that jg.x/ M j < jM j =2 if 0 < jx aj < ı1 . By Exercise 34, there exists ı2 > 0 such that jg.x/j > jM j=2 if 0 < jx aj < ı3 . Let ı D min.ı1 ; ı2 /. If 0 < jx aj < ı, then ˇ ˇ ˇ 1 1 ˇˇ jM g.x/j jM j2 2 ˇ D < D : ˇ g.x/ M ˇ jM jjg.x/j 2 jM j2

f .x/ C f .x/ Lj f .x/j C jf .x/ Lj f .x/j C jf .x/ Lj L C L f .x/j C jf .x/ Lj Lj C jf .x/ Lj C jf .x/ Lj  C D : 3

Thus limx!a g.x/ D L.

.jM j=2/ D jM j=2, as

35. To be proved: if lim g.x/ D M where M ¤ 0, then x!a 1 1 D . lim x!a g.x/ M Proof: Let  > 0 be given. Since lim g.x/ D M ¤ 0,

To be proved: if f .x/  g.x/  h.x/ in an open interval containing x D a (say, for a ı1 < x < a C ı1 , where ı1 > 0), and if limx!a f .x/ D limx!a h.x/ D L, then also limx!a g.x/ D L. Proof: Let  > 0 be given. Since limx!a f .x/ D L, there exists ı2 > 0 such that if 0 < jx aj < ı2 , then jf .x/ Lj < =3. Since limx!a h.x/ D L, there exists ı3 > 0 such that if 0 < jx aj < ı3 , then jh.x/ Lj < =3. Let ı D min.ı1 ; ı2 ; ı3 /. If 0 < jx aj < ı, then

2.

13 26 D D 13: 1 2

The average rate of change of 1=x over Œ 2; 1 is .1=. 1// .1=. 2// D 1 . 2/

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

1=2 D 1

1 : 2

35

lOMoARcPSD|6566483

REVIEW EXERCISES 1 (PAGE 93)

ADAMS and ESSEX: CALCULUS 9

3. The rate of change of x 3 at x D 2 is lim

h!0

.2 C h/3 h

23

8 C 12h C 6h2 C h3 8 h h!0 D lim .12 C 6h C h2 / D 12:

18.

h!0

19.

lim

h!0



1 3=2



5.

lim .x

4x C 7/ D 1

x!1

22 x2 D D 6. lim 2 x!2 1 x 1 22

3=2 is

20. 21.

23.

25.

lim

9.

10.

2/.x C 2/ xC2 D lim D x!2 x 2/.x 3/ 3

x2

4 xC2 .x 2/.x C 2/ D lim D lim x!2 x 2 x!2 x x!2 4x C 4 .x 2/2 2 does not exist. The denominator approaches 0 (from both sides) while the numerator does not. lim

x2

lim

x2

x!2

xC2 4 D lim D x!2 x 4x C 4 2

1

lim

x! 2C

26.

4 27.

lim p

x2

h!0

15.

lim

h x C 3h

p

x!0C

16.

lim

x!0

p

p

x

x

29.

36

x4

x!1 x 2

lim p

4

D lim

x!1

1 x

lim p

D1

x2

1 D p D2 1=4

1 x

lim sin x does not exist; sin x takes the values

1 and 1

in any interval .R; 1/, and limits, if they exist, must be unique. cos x D 0 by the squeeze theorem, since lim x!1 x 1 cos x 1   for all x > 0 x x x and limx!1 . 1=x/ D limx!1 .1=x/ D 0. 1 D 0 by the squeeze theorem, since x 1 jxj  x sin  jxj for all x ¤ 0 x and limx!0 . jxj/ D limx!0 jxj D 0. lim x sin

1 does not exist; sin.1=x 2 / takes the values 1 x2 and 1 in any interval . ı; ı/, where ı > 0, and limits, if they exist, must be unique. p lim Œx C x 2 4x C 1 lim sin

x! 1

x2 x

.x 2 p x2

4x C 1/

4x C 1 4x 1

p jxj 1 .4=x/ C .1=x 2 / xŒ4 .1=x/ D lim p x! 1 x C x 1 .4=x/ C .1=x 2 / 4 .1=x/ D lim p D 2: x! 1 1 C 1 .4=x/ C .1=x 2 / Note how we have used jxj D x (in the second last line), because x ! 1. p lim Œx C x 2 4x C 1 D 1 C 1 D 1 D lim

x! 1

30. p

1

1

x2 D1 .4=x 2 /

x2

D lim

p p h. x C 3h C x/ .x C 3h/ x h!0 x p p p 2 x x C 3h C x D D lim 3 3 h!0

does not exist because

fined for x < 0.

lim

x! 1

D lim

x2 D 0 x2

x .1=x 2 / x3 1 D lim D 2 x! 1 1 C .4=x 2 / 1 x C4

x!0

x!3

14.

1 3

x!0

28.

x 2 x2 4 D lim D 1 x! 2C C 4x C 4 xC2 p 4 x 2 x 1 D lim 12. lim p D x!4 .2 C x!4 x 4 4 x/.x 4/ p p 2 x 9 .x 3/.x C 3/. x C 3/ 13. lim p p D lim x!3 x!3 x 3 x 3 p p p D lim .x C 3/. x C 3/ D 12 3 11.

x 2 is not de-

.1=x 2 / 1 D .1=x/ .1=x 2 /

lim

x!1

4 3

8.

x

2x C 100 .2=x/ C .100=x 2 / D lim D0 2 1 x C3 x! 1 1 C .3=x 2 /

x!1=2

x2 does not exist. The denominator approaches 0 x!1 1 x2 (from both sides) while the numerator does not.

p

lim

x!0C

4C7D4

4 .x D lim x!2 .x 5x C 6

x!1 3x 2

x!

22.

x2 D lim x!1 3 x 1

1

lim

x!

2 C 2h 3 3 D lim h h!0 2.3 C 2h 3/ D lim 3/h h!0 3.2h 4 4 D lim D : 3/ 9 h!0 3.2h

lim

x2

x 2 does not exist because

x

fined for x > 1. p lim x x2 D 0

2

7.

x!2 x 2

p

x!1

24. 2

lim

x!1

D lim

4. The rate of change of 1=x at x D 1 .3=2/ C h h

17.

x

x!1

x

x2

is not de-

31.

f .x/ D x 3 4x 2 C 1 is continuous on the whole real line and so is discontinuous nowhere.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 1 (PAGE 94)

x is continuous everywhere on its domain, xC1 which consists of all real numbers except x D 1. It is discontinuous nowhere.  x 2 if x > 2 is defined everywhere and disf .x/ D x if x  2 continuous at x D 2, where it is, however, left continuous since limx!2 f .x/ D 2 D f .2/.  x 2 if x > 1 is defined and continuous evf .x/ D x if x  1 erywhere, and so discontinuous nowhere. Observe that limx!1 f .x/ D 1 D limx!1C f .x/. n 1 if x  1 is defined everywhere f .x/ D H.x 1/ D 0 if x < 1 and discontinuous at x D 1 where it is, however, right continuous. n 1 if 3  x  3 is defined f .x/ D H.9 x 2 / D 0 if x < 3 or x > 3 everywhere and discontinuous at x D ˙3. It is right continuous at 3 and left continuous at 3.

32. f .x/ D

33.

34.

35.

36.

37. f .x/ D jxj C jx C 1j is defined and continuous everywhere. It is discontinuous nowhere. n jxj=jx C 1j if x ¤ 1 38. f .x/ D is defined everywhere 1 if x D 1 and discontinuous at x D 1 where it is neither left nor right continuous since limx! 1 f .x/ D 1, while f . 1/ D 1.

Challenging Problems 1 1.

2.

lim

3.

.c C h/3 h h!0 lim

p

.a2

c3

D lim

4.

h!0

b 2 /=3,

2

2

2

If c D C ab C then 3c D a C ab C b , so the average rate ofpchange over Œa; b is the instantaneous rate of change at .a2 C ab C b 2 /=3. p 2 Claim: .a C ab C b 2 /=3 > .a C b/=2. Proof: Since a2 2ab C b 2 D .a b/2 > 0, we have 2

2

2

4a C 4ab C 4b > 3a C 6ab C 3b s

2

a2 C ab C b 2 a2 C 2ab C b 2 > D 3 4

a2 C ab C b 2 aCb > : 3 2



aCb 2

2

jx C 1j

1j

j5 jx

2xj jx 5j j3x

1j D 1

D lim

For x near 3 we have j5 jx 5j D 5 x, and j3x lim

x!0

x .1

x/

.x C 1/

D

1 : 2

2xj D 2x 5, jx 2j D x 7j D 3x 7. Thus 2x 5 .x 2j D lim x!3 5 7j x .3x x 3 D lim D x!3 4.3 x/

x 1=3 x!64 x 1=2

D lim

y!2

5.

x and jx C 1j D x C 1.

2,

2/ 7/ 1 : 4

Let y D x 1=6 . Then we have lim

y2 4 D lim 3 8 y!2 y

4 1 y C2 D D : y 2 C 2y C 4 12 3

Use a b D

have

a2

4 8 .y 2/.y C 2/ D lim y!2 .y 2/.y 2 C 2y C 4/

a3 b 3 to handle the denominator. We C ab C b 2

p 3Cx 2 lim p x!1 3 7 C x 2 .7 C x/2=3 C 2.7 C x/1=3 C 4 3Cx 4 D lim p  x!1 .7 C x/ 8 3CxC2 .7 C x/2=3 C 2.7 C x/1=3 C 4 4C4C4 D 3: D lim p D x!1 2C2 3CxC2 6.

3c 2 h C 3ch2 C h3 D 3c 2 : h

jx

x!3

Let 0 < a < b. The average rate of change of x 3 over Œa; b is b 3 a3 D b 2 C ab C a2 : b a The instantaneous rate of change of x at x D c is

x

x!0

(page 94)

3

For x near 0 we have jx Thus

1C

p

p

1Ca , r .a/ D . a a a) lima!0 r .a/ does not exist. Observe that the right limit is 1 and the left limit is 1.

rC .a/ D

1Ca

1

b) From the following table it appears that lima!0 rC .a/ D 1=2, the solution of the linear equation 2x 1 D 0 which results from setting a D 0 in the quadratic equation ax 2 C 2x 1 D 0. a

rC .a/

1 0:1 0:1 0:01 0:01 0:001 0:001

0:41421 0:48810 0:51317 0:49876 0:50126 0:49988 0:50013

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

37

lOMoARcPSD|6566483

CHALLENGING PROBLEMS 1 (PAGE 94)

p 1Ca 1 a!0 a .1 C a/ 1 D lim p a!0 a. 1 C a C 1/ 1 1 D : D lim p a!0 2 1CaC1

8.

c) lim rC .a/ D lim a!0

7.

ADAMS and ESSEX: CALCULUS 9

TRUE or FALSE

a) If limx!a f .x/ exists and limx!a   g.x/ does not exist, then limx!a f .x/ C g.x/ does not exist.   TRUE, because if limx!a f .x/ C g.x/ were to exist then   lim g.x/ D lim f .x/ C g.x/ f .x/ x!a x!a   lim f .x/ D lim f .x/ C g.x/

9.

x!a

x!a

would also exist. b) If neither  limx!a f .x/  nor limx!a g.x/ exists, then limx!a f .x/ C g.x/ does not exist. FALSE. Neither limx!0 1=x nor limx!0 . 1=x/ exist,  but limx!0 .1=x/ C . 1=x/ D limx!0 0 D 0 exists. c) If f is continuous at a, then so is jf j. TRUE. For any two real numbers u and v we have ˇ ˇ ˇjuj

This follows from juj D ju jvj D jv

ˇ ˇ jvjˇ  ju

v C vj  ju u C uj  jv

Now we have ˇ ˇ ˇjf .x/j

vj:

b) If the domain of the continuous function f is an open interval, the range of f can be any interval (open, closed, half open, finite, or infinite). n x2 1 1 if 1 < x < 1 D . f .x/ D 2 1 if x < 1 or x > 1 jx 1j f is continuous wherever it is defined, that is at all points except x D ˙1. f has left and right limits 1 and 1, respectively, at x D 1, and has left and right limits 1 and 1, respectively, at x D 1. It is not, however, discontinuous at any point, since 1 and 1 are not in its domain. 1 1 1
D D 1
2 . 1 2 1 x x2 x C x x 12 4 4 4 Observe that f .x/  f .1=2/ D 4 for all x in .0; 1/.

10. f .x/ D 11.

Suppose f is continuous on Œ0; 1 and f .0/ D f .1/.

a) To be proved: f .a/ D f .a C 12 / for some a in Œ0; 21 . Proof: If f .1=2/ D f .0/ we can take a D 0 and be done. If not, let g.x/ D f .x C 21 /

vj C jvj; and uj C juj D ju vj C juj:

ˇ ˇ jf .a/jˇ  jf .x/

f .a/j

so the left side approaches zero whenever the right side does. This happens when x ! a by the continuity of f at a. d) If jf j is continuous at a, then so  is f . 1 if x < 0 is FALSE. The function f .x/ D 1 if x  0 discontinuous at x D 0, but jf .x/j D 1 everywhere, and so is continuous at x D 0. e) If f .x/ < g.x/ in an interval around a and if limx!a f .x/ D L and limx!a g.x/ D M both exist, then L < M .  2 if x ¤ 0 and let FALSE. Let g.x/ D x 1 if x D 0 f .x/ D g.x/. Then f .x/ < g.x/ for all x, but limx!0 f .x/ D 0 D limx!0 g.x/. (Note: under the given conditions, it is TRUE that L  M , but not necessarily true that L < M .) 38

a) To be proved: if f is a continuous function defined on a closed interval Œa; b, then the range of f is a closed interval. Proof: By the Max-Min Theorem there exist numbers u and v in Œa; b such that f .u/  f .x/  f .v/ for all x in Œa; b. By the Intermediate-Value Theorem, f .x/ takes on all values between f .u/ and f .v/ at values of x between u and v, and hence at points of Œa; b. Thus the range of f is Œf .u/; f .v/, a closed interval.

f .x/:

Then g.0/ ¤ 0 and g.1=2/ D f .1/

f .1=2/ D f .0/

f .1=2/ D

g.0/:

Since g is continuous and has opposite signs at x D 0 and x D 1=2, the Intermediate-Value Theorem assures us that there exists a between 0 and 1/2 such that g.a/ D 0, that is, f .a/ D f .a C 12 /. b) To be proved: if n > 2 is an integer, then f .a/ D f .a C n1 / for some a in Œ0; 1 n1 . Proof: Let g.x/ D f .x C n1 / f .x/. Consider the numbers x D 0, x D 1=n, x D 2=n, : : : , x D .n 1/=n. If g.x/ D 0 for any of these numbers, then we can let a be that number. Otherwise, g.x/ ¤ 0 at any of these numbers. Suppose that the values of g at all these numbers has the same sign (say positive). Then we have f .1/ > f . n n 1 / >


> f . n2 / >

1 n

> f .0/;

which is a contradiction, since f .0/ D f .1/. Therefore there exists j in the set f0; 1; 2; : : : ; n 1g such that g.j=n/ and g..j C 1/=n/ have opposite sign. By the Intermediate-Value Theorem, g.a/ D 0 for some a between j=n and .j C 1/=n, which we had to prove.

Copyright © 2018 Pearson Canada Inc.

Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

CHAPTER 2.

SECTION 2.1 (PAGE 100)

DIFFERENTIATION

7.

Slope of y D

x C 1 at x D 3 is p p 4Ch 2 4ChC2
p m D lim h h!0 4ChC2 4Ch 4 p D lim
h!0 h hChC2 1 1 D : D lim p 4 h!0 4ChC2

Section 2.1 Tangent Lines and Their Slopes (page 100) 1.

Slope of y D 3x m D lim

h!0

1 at .1; 2/ is

3.1 C h/

1 .3  1 h

1/

D lim

h!0

3h D 3: h

The tangent line is y 2 D 3.x 1/, or y D 3x 1. (The tangent to a straight line at any point on it is the same straight line.)

Tangent line is y 8.

2.2 C h/2

5 .2.22 / h h!0 8 C 8h C 2h2 8 D lim h h!0 D lim .8 C 2h/ D 8

5/

3 D 8.x

4. The slope of y D 6

x

6

m D lim

h!0

. 2 C h/ h2

3h

D lim

h!0

2/ or y D 8x

x 2 at x D

h

h

13.

9.

Slope of y D

. 2 C h/2

4

h/ D 3:

h!0

The tangent line at . 2; 4/ is y D 3x C 10. 5. Slope of y D x 3 C 8 at x D

. 2 C h/3 C 8 . 8 C 8/ h h!0 8 C 12h 6h2 C h3 C 8 D lim h h!0
D lim 12 6h C h2 D 12

6. The slope of y D

0

m D lim

h!0

0 D 12.x C 2/ or y D 12x C 24.

1 1 h h2 C 1

1 D lim

The tangent line at .0; 1/ is y D 1.

h!0

h D 0: h2 C 1

1D

1 .x 4

1 .x 54

9/, or

2/,

p

D lim p h!0 5

1 at .0; 1/ is x2 C 1 !

1 3

5 x 2 at x D 1 is p 5 .1 C h/2 2 m D lim h h!0 5 .1 C h/2 4  D lim p h!0 h 5 .1 C h/2 C 2

10. The slope of y D

h!0

Tangent line is y

5.

2x at x D 2 is xC2

Tangent line is y or x 4y D 2.

2 is

m D lim

4y D

2.2 C h/ 1 2ChC2 m D lim h h!0 4 C 2h 2 h 2 D lim h.2 C h C 2/ h!0 1 h D : D lim 4 h!0 h.4 C h/

2 is

D lim .3

3/, or x

The tangent line at .9; 13 / is y D 1 y D 12 54 x.

h!0

Tangent line is y

1 .x 4

! 1 1 1 m D lim p h!0 h 9Ch 3 p p 3 9Ch 3C 9Ch D lim p
p h!0 3h 9 C h 3C 9Ch 9 9 h D lim p p h!0 3h 9 C h.3 C 9 C h/ 1 1 D : D 3.3/.6/ 54

5 at .2; 3/ is

m D lim

2D

1 The slope of y D p at x D 9 is x

2. Since y D x=2 is a straight line, its tangent at any point .a; a=2/ on it is the same line y D x=2. 3. Slope of y D 2x 2

p

2

h

.1 C

h/2

The tangent line at .1; 2/ is y D 2 y D 52 12 x. 11.

C2 1 2 .x

D

1 2

1/, or

Slope of y D x 2 at x D x0 is m D lim

h!0

.x0 C h/2 h

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

x02

D lim

h!0

2x0 h C h2 D 2x0 : h

39

lOMoARcPSD|6566483

SECTION 2.1 (PAGE 100)

ADAMS and ESSEX: CALCULUS 9

Tangent line is y x02 D 2x0 .x or y D 2x0 x x02 .

x0 /,

19.

1 at .a; a1 / is 12. The slope of y D x ! 1 1 1 a a h 1 . C D D lim m D lim a a2 h!0 h.a C h/.a/ h!0 h a C h 1 1 1 The tangent line at .a; / is y D .x a/, or a a a2 x 2 . yD a a2 p j0 C hj 0 1 D lim does not 13. Since limh!0 h h!0 jhjsgn .h/ p exist (and is not 1 or 1), the graph of f .x/ D jxj has no tangent at x D 0. 14.

The slope of f .x/ D .x m D lim

h!0

.1 C h

0

1=3

D lim h h!0

Slope of y D x 3 at x D a is .a C h/3 a3 h h!0 a3 C 3a2 h C 3ah2 C h3 D lim h h!0 D lim .3a2 C 3ah C h2 / D 3a2

m D lim

a3

h!0

b) We have m D 3 if 3a2 D 3, i.e., if a D ˙1. Lines of slope 3 tangent to y D x 3 are y D 1 C 3.x 1/ and y D 1 C 3.x C 1/, or y D 3x 2 and y D 3x C 2.

20.

The slope of y D x 3

3x at x D a is

i 1h .a C h/3 3.a C h/ .a3 3a/ h!0 h 1h 3 a C 3a2 h C 3ah2 C h3 3a 3h D lim h!0 h D lim Œ3a2 C 3ah C h2 3 D 3a2 3:

m D lim

1/4=3 at x D 1 is

1/4=3 h

a)

D 0:

a3 C 3a

h!0

The graph of f has a tangent line with slope 0 at x D 1. Since f .1/ D 0, the tangent has equation y D 0 15.

The slope of f .x/ D .x C 2/3=5 at x D m D lim

h!0

. 2 C h C 2/3=5 h

0

At points where the tangent line is parallel to the x-axis, the slope is zero, so such points must satisfy 3a2 3 D 0. Thus, a D ˙1. Hence, the tangent line is parallel to the x-axis at the points .1; 2/ and . 1; 2/.

2 is

D lim h

2=5

h!0

The graph of f has vertical tangent x D

D 1:

2 at x D

21.

f .0 C h/ h h!0C f .0 C h/ lim h h!0

f .0/

p

h D1 h p h D1 h

D lim

h!0C

f .0/

D lim h!0

m D lim

The tangent at x D a is parallel to the line y D 2x C 5 if 3a2 1 D 2, that is, if a D ˙1. The corresponding points on the curve are . 1; 1/ and .1; 1/. 22.

h!0

1 h

.x02

1/

2x0 h C h2 D 2x0 : h h!0

D lim

1 a D lim a .a C h/ D h!0 ah.a C h/

1 : a2

The tangent at x D a is perpendicular to the line y D 4x 3 if 1=a2 D 1=4, that is, if a D ˙2. The corresponding points on the curve are . 2; 1=2/ and .2; 1=2/. 23.

The slope of the curve y D x 2 at x D a is m D lim

h!0

If m D 3, then x0 D 32 . The tangent line with slope m D 3 at . 32 ; 54 / is y D 45 3.x C 23 /, that is, 13 y D 3x 4 .

40

The slope of the curve y D 1=x at x D a is 1 a C h m D lim h h!0

1 at x D x0 is Œ.x0 C h/2

.a C h/3

h!0

Thus the graph of f has a vertical tangent x D 0.

18. The slope of y D x 2

x C 1 at x D a is

.a C h/ C 1 .a3 a C 1/ h h!0 3a2 h C 3ah2 C a3 h D lim h h!0 D lim .3a2 C 3ah C h2 1/ D 3a2 1:

m D lim

2.

16. The slope of f .x/ D jx 2 1j at x D 1 is j.1 C h/2 1j j1 1j j2h C h2 j m D limh!0 D lim , h h h!0 which does not exist, and is not 1 or 1. The graph of f has no tangent at x D 1. p x if x  0 p 17. If f .x/ D , then x if x < 0 lim

The slope of the curve y D x 3

.a C h/2 h

a2

D lim .2a C h/ D 2a: h!0

The normal at x D a has slope y

a2 D

1 .x 2a

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

a/;

or

1=.2a/, and has equation x 1 C y D C a2 : 2a 2

i

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 2.1 (PAGE 100)

y

This is the line x C y D k if 2a D 1, and so k D .1=2/ C .1=2/2 D 3=4. 24.

The curves y D kx 2 and y D k.x The slope of y D kx 2 at x D 1 is k.1 C h/2 m1 D lim h h!0 The slope of y D k.x m2 D lim

k.2

h!0

k

2

2/2 intersect at .1; k/.

1 -3

-2

-1

D lim .2 C h/k D 2k:

-2

2/2 at x D 1 is k

D lim . 2 C h/k D h!0

x

2

-1

h!0

.1 C h//2 h

1 1j

x

-3 Fig. 2.1-27

2k:

The two curves intersect at right angles if 2k D 1=. 2k/, that is, if 4k 2 D 1, which is satisfied if k D ˙1=2.

y D jx 2

28.

25. Horizontal tangents at .0; 0/, .3; 108/, and .5; 0/. y .3; 108/

Horizontal tangent at .a; 2/ and . a; 2/ for all a > 1. No tangents at .1; 2/ and . 1; 2/. y y D jx C 1j jx 1j 2 1

100 80

-3

-2

-1

60

1

2

x

-1

40 y D x 3 .5

20 -1

1

2

-2

x/2

3

4

5

-3 Fig. 2.1-28

x

-20 Fig. 2.1-25 29. 26. Horizontal tangent at . 1; 8/ and .2; 19/. y

Horizontal tangent at .0; 1/. The tangents at .˙1; 0/ are vertical. y y D .x 2

20 . 1; 8/ 10 -2

-1

y D 2x 3

3x 2

1

2

1/1=3 2 1

12x C 1 3

-3

x

-2

-1

1

2

x

-1

-10 -2 -20

.2; 19/ -3 Fig. 2.1-29

-30 Fig. 2.1-26

27.

Horizontal tangent at . 1=2; 5=4/. No tangents at . 1; 1/ and .1; 1/.

30.

Horizontal tangent at .0; 1/. No tangents at . 1; 0/ and .1; 0/.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

41

lOMoARcPSD|6566483

SECTION 2.1 (PAGE 100)

ADAMS and ESSEX: CALCULUS 9

y

2. y

y D ..x 2

1/2 /1=3 2

1 y D g 0 .x/ -1

-2

1

2

x

x

Fig. 2.1-30 31. The graph of the function f .x/ D x 2=3 (see Figure 2.1.7 in the text) has a cusp at the origin O, so does not have a tangent line there. However, the angle between OP and the positive y-axis does ! 0 as P approaches 0 along the graph. Thus the answer is NO.

3. y

y D h0 .x/ x

32. The slope of P .x/ at x D a is m D lim

h!0

P .a C h/ h

P .a/

:

Since P .a C h/ D a0 C a1 h C a2 h2 C


C an hn and P .a/ D a0 , the slope is a0 C a1 h C a2 h2 C


C an hn a0 h h!0 D lim a1 C a2 h C


C an hn 1 D a1 :

m D lim

4. y

h!0

Thus the line y D `.x/ D m.x a/ C b is tangent to y D P .x/ at x D a if and only if m D a1 and b D a0 , that is, if and only if

x

P .x/ `.x/ D a2 .x a/2 C a3 .x a/3 C


C an .x a/n h i D .x a/2 a2 C a3 .x a/ C


C an .x a/n 2 D .x

y D k 0 .x/

a/2 Q.x/

where Q is a polynomial.

Section 2.2 The Derivative

(page 107)

5.

Assuming the tick marks are spaced 1 unit apart, the function f is differentiable on the intervals . 2; 1/, . 1; 1/, and .1; 2/.

6.

Assuming the tick marks are spaced 1 unit apart, the function g is differentiable on the intervals . 2; 1/, . 1; 0/, .0; 1/, and .1; 2/.

7.

y D f .x/ has its minimum at x D 3=2 where f 0 .x/ D 0

1. y 0

y D f .x/

x

42

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 2.2 (PAGE 107)

y y D f .x/ D 3x

x

2

y D f .x/ D jx 3

y 1j

1

x

x y y D f 0 .x/

y y D f 0 .x/

x x

Fig. 2.2-7

Fig. 2.2-9

8. y D f .x/ has horizontal tangents at the points near 1=2 and 3=2 where f 0 .x/ D 0 y

10. y D f .x/ is constant on the intervals . 1; 2/, . 1; 1/, and .2; 1/. It is not differentiable at x D ˙2 and x D ˙1. y y D f .x/ D jx 2 1j jx 2 4j

x x y D f .x/ D x 3

3x 2 C 2x C 1 y y D f 0 .x/

y

x x y D f 0 .x/ Fig. 2.2-10 Fig. 2.2-8 11.

y D x2 y 0 D lim

h!0

9. y D f .x/ fails to be differentiable at x D 1, x D 0, and x D 1. It has horizontal tangents at two points, one between 1 and 0 and the other between 0 and 1.

3x .x C h/2

2xh C h2 h h!0 dy D .2x 3/ dx D lim

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

3.x C h/ .x 2 h 3h D 2x 3

3x/

43

lOMoARcPSD|6566483

SECTION 2.2 (PAGE 107)

12.

f .x/ D 1 C 4x f 0 .x/ D lim

h!0

D lim

h!0

df .x/ D .4

13.

5x 2

1 C 4.x C h/

4h

ADAMS and ESSEX: CALCULUS 9

10xh h 10x/ dx

17. 5.x C h/2 h 5h2

D4

.1 C 4x

5x 2 /

10x

2 D lim p p h!0 2.t C h/ C 1 C 2t C 1 1 D p 2t C 1 1 dF .t / D p dt 2t C 1

f .x/ D x 3

.x C h/3 x 3 h h!0 3x 2 h C 3xh2 C h3 D lim D 3x 2 h h!0 df .x/ D 3x 2 dx f 0 .x/ D lim

18.

f .x/ D 0

14.

15.

16.

1 3 C 4t   ds 1 1 1 D lim dt 3 C 4t h!0 h 3 C 4.t C h/ 4 3 C 4t 3 4t 4h D D lim .3 C 4t /2 h!0 h.3 C 4t /Œ3 C .4t C h/ 4 dt ds D .3 C 4t /2

2 x g.x/ D 2Cx 2 .x C h/ 2 x 2CxCh 2Cx 0 g .x/ D lim h h!0 .2 x h/.2 C x/ .2 C x C h/.2 D lim h.2 C x C h/.2 C x/ h!0 4 D .2 C x/2 4 dx dg.x/ D .2 C x/2

y 0 D lim

x 1h

1 .x C h/3 .x C h/ . 13 x 3 h 3  1 2 x h C xh2 C 31 h3 h D lim h!0 h D lim .x 2 C xh C 31 h2 1/ D x 2 1 h!0

h!0

dy D .x 2

44

1/ dx

i x/

3 4

p

f .x/ D lim

2 3 4

h!0

3 D lim h!0 4

sD

y D 31 x 3

p

2t C 1 p p 2.t C h/ C 1 2t C 1 F 0 .t / D lim h h!0 2t C 2h C 1 2t 1  D lim p p h!0 h 2.t C h/ C 1 C 2t C 1 F .t / D

D df .x/ D

19.

yDxC

x p 2 "

3

3 4

.x C h/ h 2

p h. 2

p

2

x

2Cx p .x C h/ C 2 x

h

x/

#

p 8 2 x 3 p dx 8 2 x

1 x

1 1 x x C h x y D lim h h!0   x x h D lim 1 C h.x C h/x h!0 1 1 D 1 C lim D1 x2 h!0 .x C h/x   1 dx dy D 1 x2 0

x/

20.

xChC

s 1Cs   dz 1 sCh s D lim ds 1Cs h!0 h 1 C s C h .s C h/.1 C s/ s.1 C s C h/ 1 D lim D h.1 C s/.1 C s C h/ .1 C s/2 h!0 1 dz D ds .1 C s/2 zD

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

F .x/ D p

21.

SECTION 2.2 (PAGE 107)

1

25.

1 C x2

1 1 p p 2 1 C x2 1 C .x C h/ F 0 .x/ D lim h h!0 p 26. p 1 C x2 1 C .x C h/2 D lim p p h!0 h 1 C .x C h/2 1 C x 2 1 C x 2 1 x 2 2hx h2 p  27. D lim p p p h!0 h 1 C .x C h/2 1 C x 2 1 C x 2 C 1 C .x C h/2 2x x D 2 3=2 2.1 C x / .1 C x 2 /3=2 x dF .x/ D dx .1 C x 2 /3=2 D

22.

yD

1 x2



28.

23.

yD p

p

2x f .x/

1

ˇ ˇ ˇ 2x/ˇ ˇ

1 p 1Cx

t2 3 t2 C 3   t2 3 1 .t C h/2 3 f 0 .t / D lim .t C h/2 C 3 t 2 C 3 h!0 h 2 Œ.t C h/ 3.t 2 C 3/ .t 2 3/Œ.t C h/2 C 3 D lim h.t 2 C 3/Œ.t C h/2 C 3 h!0 12t h C 6h2 12t D lim D 2 2 2 .t C 3/2 h!0 h.t C 3/Œ.t C h/ C 3 12t df .t / D 2 dt .t C 3/2

f .1/

x 1 0:71000 0:97010 0:99700 0:99970

d 3 .x dx

1CxCh y 0 .x/ D lim h h!0 p p 1Cx 1CxCh D lim p p h!0 h 1 C x C h 1 C x 1Cx 1 x h p D lim p
p p h!0 h 1 C x C h 1 C x 1CxC 1CxCh 1 p D lim p
p p h!0 1CxCh 1Cx 1CxC 1CxCh 1 D 2.1 C x/3=2 1 dy D dx 2.1 C x/3=2 24.

y D x3

0:9 0:99 0:999 0:9999

1 1Cx

h.x/ D jx 2 C 3x C 2j fails to be differentiable where x 2 C 3x C 2 D 0, that is, at x D 2 and x D 1. Note: both of these are single zeros of x 2 C 3x C 2. If they were higher order zeros (i.e. if .x C 2/n or .x C 1/n were a factor of x 2 C 3x C 2 for some integer n  2) then h would be differentiable at the corresponding point.

x



1 1 1 2 x2 h!0 h .x C h/ 2 2 x .x C h/ 2 D lim D x3 h!0 hx 2 .x C h/2 2 dx dy D x3 y 0 D lim

Since f .x/ D x sgn x D jxj, for x ¤ 0, f will become continuous at x D 0 if we define f .0/ D 0. However, f will still not be differentiable at x D 0 since jxj is not differentiable at x D 0.  2 Since g.x/ D x 2 sgn x D xjxj D x 2 if x > 0 , g will x if x < 0 become continuous and differentiable at x D 0 if we define g.0/ D 0.

xD1

x

f .x/

x 1 1:31000 1:03010 1:00300 1:00030

1:1 1:01 1:001 1:0001

D lim

h!0

f .1/

.1 C h/3

2.1 C h/ h

. 1/

h C 3h2 C h3 h h!0 D lim 1 C 3h C h2 D 1

D lim

h!0

29.

f .x/ D 1=x f .x/

x

f .2/

x 2 0:26316 0:25126 0:25013 0:25001

1:9 1:99 1:999 1:9999

x 2:1 2:01 2:001 2:0001

f .x/

f .2/

x 2 0:23810 0:24876 0:24988 0:24999

1 2 2 .2 C h/ 2Ch f .2/ D lim D lim h h!0 h!0 h.2 C h/2 1 1 D lim D .2 C h/2 4 h!0 0

f .t / D

30.

The slope of y D 5 C 4x x 2 at x D 2 is ˇ 5 C 4.2 C h/ .2 C h/2 dy ˇˇ D lim ˇ ˇ dx h h!0

9

xD2

D lim

h!0

h2 D 0: h

Thus, the tangent line at x D 2 has the equation y D 9.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

45

lOMoARcPSD|6566483

SECTION 2.2 (PAGE 107)

31. y D

ADAMS and ESSEX: CALCULUS 9

p

x C 6. Slope at .3; 3/ is p 1 9Ch 3 9Ch 9 p D lim m D lim
D : h 6 h!0 h h!0 9ChC3

Tangent line is y

32. The slope of y D ˇ dy ˇˇ ˇ dt ˇ

tD 2

3D

1 .x 6

t t2

2

3/, or x

at t D

6y D

2 and y D

 2Ch 1 2 h!0 h . 2 C h/2

D lim

D lim

h!0

44.

3 : 2

45.

yD

2 t2 C t

0

17x

35. g .t / D 22t 36. 37. 38. 39. 40.

41.

42.

43.

21

46

xDa

D

1 . a2

1 D a2 .x a

a/,

Slope at t D a is

18

46.

Therefore x D

9 4

18 D 0 2/ D 0:

or x D 2.

dy The slope of y D x 2 is m1 D D 2x. dx 9 9 , m1 D . At x D 2, m1 D 4. At x D 4 2 The slope of x C 4y D 18, i.e. y D 14 x C 18 4 , is m2 D 41 . Thus, at x D 2, the product of these slopes is .4/. 41 / D 1. So, the curve and line intersect at right angles at that point.

for all t

tD4

The intersection points of y D x 2 and x C 4y D 18 satisfy 4x 2 C x .4x C 9/.x

for x ¤ 0

1 dy D x 2=3 for x ¤ 0 dx 3 dy 1 4=3 D x for x ¤ 0 dx 3 d 2:25 t D 2:25t 3:25 for t > 0 dt d 119=4 119 115=4 s D s for s > 0 ds 4 ˇ ˇ 1 d p ˇˇ 1 ˇˇ D : sˇ D p ˇ ˇ ds 6 2 sˇ sD9 sD9   1 1 0 1 ; F D 16 F .x/ D ; F 0 .x/ D x x2 4 ˇ 2 5=3 ˇˇ 1 0 f .8/ D x D ˇ ˇ 3 48 xD8 ˇ ˇ ˇ ˇ dy ˇ 1 1 ˇ D t 3=4 ˇ D p ˇ ˇ dt ˇ 4 8 2 tD4

ˇ 1 ˇˇ ˇ x2 ˇ

Normal has slope a , and equation y 1 or y D a2 x a3 C a

2 2 .a C h/2 C .a C h/ a2 C a m D lim h h!0 2.a2 C a a2 2ah h2 a h/ D lim hŒ.a C h/2 C a C h.a2 C a/ h!0 4a 2h 2 D lim 2 h!0 Œ.a C h/ C a C h.a2 C a/ 4a C 2 D .a2 C a/2 2.2a C 1/ 2 .t a/ Tangent line is y D 2 a C a .a2 C a/2

34. f 0 .x/ D

1 at x D a is Slope of y D x 2

Thus, the tangent line has the equation y D 1 23 .t C 2/, that is, y D 32 t 4. 33.

1 D p : 2 x0

Thus, the equation of the tangent line is x C x0 1 p y D x0 C p .x x0 /, that is, y D p . 2 x0 2 x0

1 is

D

x at x D x0 is

xDx0

15.

2

p

ˇ dy ˇˇ ˇ dx ˇ

 . 1/

2 C h C Œ. 2 C h/2 hŒ. 2 C h/2 2

The slope of y D

47.

Let theˇ point of tangency be .a; a2 /. Slope of tangent is d 2 ˇˇ x ˇ D 2a dx ˇ xDa

This is the slope from .a; a2 / to .1; 3/, so a2 C 3 D 2a, and a 1 a2 C 3 D 2a2

2a

a2 2a 3 D 0 a D 3 or 1 The two tangent lines are (for a D 3): y 9 D 6.x 3/ or 6x 9 (for a D 1): y 1 D 2.x C 1/ or y D

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

2x

1

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 2.2 (PAGE 107)

p p 4a2 4b D a ˙ a2 b. 2 p If b < a2 , i.e. a2 b > 0, then t D a ˙ a2 b has two real solutions. Therefore, there will be two distinct tangent lines passing through .a; b/ with equations  p y D b C 2 a ˙ a2 b .x a/. If b D a2 , then t D a. There will be only one tangent line with slope 2a and equation y D b C 2a.x a/. If b > a2 , then a2 b < 0. There will be no real solution for t . Thus, there will be no tangent line.

y

Hence t D .a;a2 /

y D x2

x

2a ˙

.1; 3/

Fig. 2.2-47

48. The slope of y D

1 at x D a is x ˇ dy ˇˇ ˇ dx ˇ

xDa

or y D

Suppose f is odd: f . x/ D f .x/. Then f . x C h/ f . x/ f 0 . x/ D lim h h!0 f .x h/ f .x/ D lim h h!0 .let h D k/ f .x C k/ f .x/ D lim D f 0 .x/ k k!0 Thus f 0 is even. Now suppose f is even: f . x/ D f .x/. Then f . x C h/ f . x/ f 0 . x/ D lim h h!0 f .x h/ f .x/ D lim h h!0 f .x C k/ f .x/ D lim k k!0 D f 0 .x/ so f 0 is odd.

52.

Let f .x/ D x

1 : a2

1 1 D 2, or a D ˙ p . Therea2 2 equations of the are   two straight lines   p 1 1 2 x p and y D 2 2 xCp , 2 p 2 2x ˙ 2 2.

If the slope is fore, the p yD 2

D

51.

2, then

p 49. Let the point of tangency be ˇ .a; a/ d p ˇˇ 1 Slope of tangent is xˇ D p ˇ dx 2 a xDa p a 0 1 , so a C 2 D 2a, and a D 2. Thus p D aC2 2 a 1 The required slope is p . 2 2 y

p .a; a/ p yD x x

2

Fig. 2.2-49 50. If aˇ line is tangent to y D x 2 at .t; t 2 /, then its slope is dy ˇˇ D 2t . If this line also passes through .a; b/, then ˇ dx ˇ xDt its slope satisfies t2 b D 2t; t a

that is t 2

2at C b D 0:

f 0 .x/ D lim

n

. Then .x C h/

n

x

n

h   1 1 1 D lim .x C h/n x n h!0 h n x .x C h/n D lim n h!0 hx .x C h/n x .x C h/  D lim h!0 hx n ..x C h/n  x n 1 C x n 2 .x C h/ C


C .x C h/n h!0

D

1  nx n x 2n

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

1

D

nx

.nC1/

1



:

47

lOMoARcPSD|6566483

SECTION 2.2 (PAGE 107)

53.

ADAMS and ESSEX: CALCULUS 9

f .x/ D x 1=3

If f 0 .aC/ is finite, call the half-line with equation y D f .a/ C f 0 .aC/.x a/, (x  a), the right tangent line to the graph of f at x D a. Similarly, if f 0 .a / is finite, call the half-line y D f .a/ C f 0 .a /.x a/, (x  a), the left tangent line. If f 0 .aC/ D 1 (or 1), the right tangent line is the half-line x D a, y  f .a/ (or x D a, y  f .a/). If f 0 .a / D 1 (or 1), the right tangent line is the half-line x D a, y  f .a/ (or x D a, y  f .a/). The graph has a tangent line at x D a if and only if f 0 .aC/ D f 0 .a /. (This includes the possibility that both quantities may be C1 or both may be 1.) In this case the right and left tangents are two opposite halves of the same straight line. For f .x/ D x 2=3 , f 0 .x/ D 32 x 1=3 : At .0; 0/, we have f 0 .0C/ D C1 and f 0 .0 / D 1. In this case both left and right tangents are the positive y-axis, and the curve does not have a tangent line at the origin. For f .x/ D jxj, we have

.x C h/1=3 x 1=3 h h!0 .x C h/1=3 x 1=3 D lim h h!0 .x C h/2=3 C .x C h/1=3 x 1=3 C x 2=3  .x C h/2=3 C .x C h/1=3 x 1=3 C x 2=3 xCh x D lim h!0 hŒ.x C h/2=3 C .x C h/1=3 x 1=3 C x 2=3  1 D lim h!0 .x C h/2=3 C .x C h/1=3 x 1=3 C x 2=3 1 1 D D x 2=3 3 3x 2=3

f 0 .x/ D lim

54. Let f .x/ D x 1=n . Then .x C h/1=n f 0 .x/ D lim h h!0 a b D lim n bn a!b a D lim

a!b

1 D nb n

55.

an 1

x 1=n

1 C an 2 b C an 1 D x .1=n/ 1 : n 1

(let x C h D an , x D b n )

3 b2

C


C bn

d n .x C h/n x n x D lim dx h h!0  n.n 1/ n 2 2 1 n n n 1 hC x h x C x D lim 1 12 h!0 h  n.n 1/.n 2/ n 3 3 C x h C


C hn x n 123  n.n 1/ n 2 x h D lim nx n 1 C h 12 h!0 ! n.n 1/.n 2/ n 3 2 n 1 C x h C


C h 123 D nx n

f 0 .x/ D sgn .x/ D

1

1

1.

y D 3x 2

2.

y D 4x 1=2

3.

f .x/ D Ax 2 C Bx C C;

4.

f .x/ D

5.

zD

6.

y D x 45

7.

8.

f .a/

h!0C

48

f .a/

1

Section 2.3 Differentiation Rules (page 115)

9. f .a C h/ h f .a C h/ 0 f .a / D lim h h!0

if x > 0 if x < 0.

1

At .0; 0/, f 0 .0C/ D 1, and f 0 .0 / D 1. In this case the right tangent is y D x, .x  0/, and the left tangent is y D x, .x  0/. There is no tangent line.

56. Let f 0 .aC/ D lim

n

10.

5x

7;

5 ; x

y 0 D 2x

2 6 C 2 x3 x

s5

s3 15

2;

45

5:

1=2

C 5x

2

f 0 .x/ D 2Ax C B: f 0 .x/ D

dz 1 D s4 dx 3

;

x

y 0 D 6x

18 x4

4 x3

1 2 s : 5

y 0 D 45x 44 C 45x

46

g.t / D t 1=3 C 2t 1=4 C 3t 1=5 1 3 1 g 0 .t / D t 2=3 C t 3=4 C t 4=5 3 2 5 p 2 3 2=3 2 p D 3t yD3 t 2t 3=2 t3 dy D 2t 1=3 C 3t 5=2 dt 3 5 3=5 u D x 5=3 x 5 3 du D x 2=3 C x 8=5 dx

F .x/ D .3x F 0 .x/ D 3.1

2/.1 5x/ 5x/ C .3x 2/. 5/ D 13

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

30x

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

11.

yD

p

 x 5

3p x 2

5 y0 D p 2 x

1

12. g.t / D 13.

14. 15.

2t

3

yD

4 3

x

f .t / D

20.

21.

22.

x 3=2

1 5=2 x 3

2 .2t

3/2

24. 2x C 5 .x 2 C 5x/2

4 x/2

.3

 2

25.

t  2 . / D 2 .2  t / .2  t /2 2

f .x/ D

23.

5 3=2 x 6

g 0 .t / D

;

y0 D

;

y2

1

f 0 .x/ D

19.

p D5 x

1 x 2 C 5x 1 y0 D .2x C 5/ D .x 2 C 5x/2

16. g.y/ D

18.



yD

f 0 .t / D

17.

x2 3

x

SECTION 2.3 (PAGE 115)

g 0 .y/ D

;

4x 2

1

x3 3x

4

Dx

C 4x

3

2

.1

4y y 2 /2

4 x 4x 2 3 D x4

26.

p u u 3 D u 1=2 3u 2 u2 p 1 3=2 12 u u g 0 .u/ D u C 6u 3 D 2 2u3 g.u/ D

27.

p 2 C t C t2 D 2t 1=2 C t C t 3=2 p t 1 3p dy 3t 2 C t 2 D t 3=2 C p C p tD dt 2 2 t 2t t yD

x 1 z D 2=3 D x 1=3 x dz 1 2=3 2 D x C x dx 3 3

x 5=3

x3 4 xC1 .x C 1/.3x 2 / .x 3 0 f .x/ D .x C 1/2 3 2x C 3x 2 C 4 D .x C 1/2 f .x/ D

3 4x 3 C 4x .3 C 4x/. 4/ .3 f 0 .x/ D .3 C 4x/2 24 D .3 C 4x/2

4/.1/

ax C b cx C d .cx C d /a .ax C b/c f 0 .x/ D .cx C d /2 ad bc D .cx C d /2 f .x/ D

t 2 C 7t 8 t2 t C 1 2 .t t C 1/.2t C 7/ .t 2 C 7t F 0 .t / D .t 2 t C 1/2 2 8t C 18t 1 D .t 2 t C 1/2 F .t / D

8/.2t

1/

f .x/ D .1 C x/.1 C 2x/.1 C 3x/.1 C 4x/ f 0 .x/ D .1 C 2x/.1 C 3x/.1 C 4x/ C 2.1 C x/.1 C 3x/.1 C 4x/ C 3.1 C x/.1 C 2x/.1 C 4x/ C 4.1 C x/.1 C 2x/.1 C 3x/ OR f .x/ D Œ.1 C x/.1 C 4x/ Œ.1 C 2x/.1 C 3x/ D 1 C 10x C 25x 2 C 10x 2 .1 C 5x/ C 24x 4

2=3

D

p 1 .1 C t /. p / 2 t p 2 t/

D .1 C 5x C 4x 2 /.1 C 5x C 6x 2 /

0

D 1 C 10x C 35x 2 C 50x 3 C 24x 4

f .x/ D 10 C 70x C 150x 2 C 96x 3

xC2 3x 5=3

28.

f .r/ D .r

2

f 0 .r/ D . 2r

f .x/ D

4x/.4/

Cr

or f .r/ D 0

f .r/ D

t 2 C 2t t2 1 2 .t 1/.2t C 2/ .t 2 C 2t /.2t / z0 D .t 2 1/2 2.t 2 C t C 1/ D .t 2 1/2 zD

p 1C t p 1 t p 1 .1 t/ p ds 2 t D dt .1 1 p D p t .1 t /2 sD

29.

3

3

2

1

4

3r

C .r

2Cr

4/.r 2 C r 3 C 1/

2

Cr 3

/.r 2 C r 3 C 1/

Cr 2

3

Cr

4

4/.2r C 3r 2 /

3

Cr

2r 3r C 1 8r p y D .x 2 C 4/. x C 1/.5x 2=3 2/ p y 0 D 2x. x C 1/.5x 2=3 2/ 1 C p .x 2 C 4/.5x 2=3 2/ 2 x p 10 C x 1=3 .x 2 C 4/. x C 1/ 3 r

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

4r 2

4r 3

12r 2

49

lOMoARcPSD|6566483

SECTION 2.3 (PAGE 115)

30.

yD D y0 D

D D

31.

32.

yD

ADAMS and ESSEX: CALCULUS 9

.x 2 C 1/.x 3 C 2/ .x 2 C 2/.x 3 C 1/ x 5 C x 3 C 2x 2 C 2 x 5 C 2x 3 C x 2 C 2 .x 5 C 2x 3 C x 2 C 2/.5x 4 C 3x 2 C 4x/ .x 5 C 2x 3 C x 2 C 2/2 5 .x C x 3 C 2x 2 C 2/.5x 4 C 6x 2 C 2x/ .x 5 C 2x 3 C x 2 C 2/2 2x 7 3x 6 3x 4 6x 2 C 4x .x 5 C 2x 3 C x 2 C 2/2 7 2x 3x 6 3x 4 6x 2 C 4x .x 2 C 2/2 .x 3 C 1/2

36.

37.

1 3x C 1 .6x 2 C 2x C 1/.6x C 1/ .3x 2 C x/.12x C 2/ y0 D .6x 2 C 2x C 1/2 6x C 1 D .6x 2 C 2x C 1/2 2x C

p . x

33.

34.

35.

d dx

d dx



39. 1 p x



ˇˇ ˇ ˇ ˇ

ˇ f .x/ ˇˇ ˇ x2 ˇ

xD2

D

xD2

ˇˇ d  2 x f .x/ ˇˇ dx

50

ˇ f .x/.2x/ x 2 f 0 .x/ ˇˇ D ˇ ˇ Œf .x/2 4f .2/ 4f 0 .2/ D Œf .2/2

ˇ x 2 f 0 .x/ 2xf .x/ ˇˇ D ˇ ˇ x4

!

40.

xD2

18 14 1 .4 C f .2//f 0 .2/ f .2/.4 C f 0 .2// D D D .4 C f .2//2 62 9 ˇ  2   ˇ x 4 d d 8 ˇ j D 1 ˇ xD 2 dx x 2 C 4 dx x2 C 4 ˇ xD 2 ˇ ˇ 8 ˇ D 2 .2x/ˇ ˇ .x C 4/2 D

p #ˇ t .1 C t / ˇˇ ˇ ˇ 5 t ˇ " #tD4 d t C t 3=2 ˇˇ D ˇ ˇ dt 5 t

d dt

"

32 D 64

1 2

tD4 ˇ t /.1 C 23 t 1=2 / .t C t 3=2 /. 1/ ˇˇ ˇ ˇ .5 t /2

.5

tD4

.12/. 1/ D D 16 .1/2 p x f .x/ D xC1 p 1 .x C 1/ p x.1/ 2 x f 0 .x/ D .x C 1/2 p 3 p 2 1 2 2 f 0 .2/ D p D 9 18 2

ˇ ˇ d Œ.1 C t /.1 C 2t /.1 C 3t /.1 C 4t /ˇˇ dt tD0 D .1/.1 C 2t /.1 C 3t /.1 C 4t / C .1 C t /.2/.1 C 3t /.1 C 4t /C ˇ ˇ .1 C t /.1 C 2t /.3/.1 C 4t / C .1 C t /.1 C 2t /.1 C 3t /.4/ˇˇ tD0

D 1 C 2 C 3 C 4 D 10

xD2

4 D 4

1

41. y D

2 p , y0 D 4 x

3

2 p 2 4 x

 3



4 p

2 x



8 D4 . 1/2 2 Tangent line has the equation y D 2 C 4.x 1/ or y D 4x 6 Slope of tangent at .1; 2/ is m D

xD2

4 1 4f 0 .2/ 4f .2/ D D D 16 16 4

xD2

ˇˇ ˇ ˇ ˇ

xD2 ˇ .x 2 C f .x//f 0 .x/ f .x/.2x C f 0 .x// ˇˇ D ˇ ˇ .x 2 C f .x//2

D

.3 C 2x/. 1

x2 f .x/

f .x/ x 2 C f .x/

.1/.4/

4x C 3x 2 / .2 x 2x 2 C x 3 /.2/ .3 C 2x/2 2 .2 x/.1 x / D 2x 3=2 .3 C 2x/ ! 1 4x 3 C 5x 2 12x 7 C 1 p .3 C 2x/2 x



38.

2

1/.2 x/.1 x / p x.3 C 2x/ ! 1 2 x 2x 2 C x 3 D 1 p
3 C 2x x ! 1 3=2 2 x 2x 2 C x 3 x C 1 f 0 .x/ D 2 3 C 2x f .x/ D



xD 2

3x 2 C x D 2 6x C 2x C 1

x

d dx

 ˇˇ D 2xf .x/ C x 2 f 0 .x/ ˇˇ D 4f .2/ C 4f 0 .2/ D 20

42. xD2

For y D

xC1 we calculate x 1

y0 D

.x

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

1/.1/ .x

.x C 1/.1/ D 1/2

2 .x

1/2

:

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 2.3 (PAGE 115)

y

At x D 2 we have y D 3 and y 0 D 2. Thus, the equation of the tangent line is y D 3 2.x 2/, or y D 2x C 7. The normal line is y D 3 C 12 .x 2/, or y D 12 x C 2.



b

a;

1 1 0 , y D1 x x2 1 For horizontal tangent: 0 D y 0 D 1 so x 2 D 1 and x2 x D ˙1 The tangent is horizontal at .1; 2/ and at . 1; 2/

43. y D x C

44. If y D x 2 .4 0

y D 2x.4

Fig. 2.3-47 2

2

3

x / C x . 2x/ D 8x

4x D 4x.2

2

x /:

48.

Thus 2x C 1 D 0 and x D

1 2

The tangent is horizontal only at



xD1

2x C 1 . .x 2 C x C 1/2

49.

˙ 12 ,

or .x C 2/ D

xD1

1 : 2

D

1. Hence, the

3a2 C 4 D .a

2/2 .a C 1/:

The two tangent lines to y D x 3 passing through .2; 8/ correspond to a D 2 and a D 1, so their equations are y D 12x 16 and y D 3x C 2.

1 .x C 2/.1/ .x C 1/.1/ D : .x C 2/2 .x C 2/2 50.

The tangent to y D x 2 =.x mD

1 4:

D

ˇ ˇ ˇ ˇ

3=2 ˇ

The tangent to y D x 3 at .a; a3 / has equation y D a3 C 3a2 .x a/, or y D 3a2 x 2a3 . This line passes through .2; 8/ if 8 D 6a2 2a3 or, equivalently, if a3 3a2 C 4 D 0. Since .2; 8/ lies on y D x 3 , a D 2 must be a solution of this equation. In fact it must be a double root; .a 2/2 must be a factor of a3 3a2 C 4. Dividing by this factor, we find that the other factor is a C 1, that is, a3

2

1 x 2


The product of the slopes is .2/ 12 D two curves intersect at right angles.

 1 4 ; . 2 3

In order to be parallel to y D 4x, the tangent line must have slope equal to 4, i.e.,

.x

1/ at .a; a2 =.a ˇ 1/2x x 2 .1/ ˇˇ a2 D ˇ 2 ˇ .x 1/ .a xDa

1// has slope 2a : 1/2

The equation of the tangent is 3 2

5 . 2

and x D or At x D Hence x C 2 D y D 1, and at x D 52 , y D 3. Hence, the is
parallel to y D 4x at the points
tangent 5 3 ; 1 and ; 3 . 2 2 47.

ˇ dy ˇˇ ˇ dx ˇ

xC1 , then xC2

1 D 4; .x C 2/2

1 Since p D y D x 2 ) x 5=2 D 1, therefore x D 1 at x theˇ intersection point. The slope of y D x 2 at x D 1 is ˇ 1 2x ˇˇ D 2. The slope of y D p at x D 1 is x xD1

2x C 1 2 .x C x C 1/2

For horizontal tangent we want 0 D y 0 D

y0 D



x 2 /, then

1 , y0 D 2 x CxC1

46. If y D

1 a

x

The slope of a horizontal line must be zero, so p 4x.2 x 2 / D 0, which impliesp that x D 0 or x D ˙ 2. At x D 0; y D 0 and at x D ˙ 2; y D 4. Hence, there are two horizontal lines that are tangent to the curve. Their equations are y D 0 and y D 4. 45. y D

1 x

yD

3 , 2

Let the point of tangency be .a; a1 /. The slope of the tanb a1 1 2 gent is D . Thus b a1 D a1 and a D . a2 0 a b b2 b2 so has equation y D b x. Tangent has slope 4 4

y

a2 a

1

D

a2 .a

2a .x 1/2

a/:

This line passes through .2; 0/ provided 0

a2 a

1

D

a2 .a

2a .2 1/2

a/;

or, upon simplification, 3a2 4a D 0. Thus we can have either a D 0 or a D 4=3. There are two tangents through .2; 0/. Their equations are y D 0 and y D 8x C 16.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

51

lOMoARcPSD|6566483

SECTION 2.3 (PAGE 115)

51.

ADAMS and ESSEX: CALCULUS 9

p p f .x C h/ f .x/ h f .x C h/ f .x/ 1 p D lim p h h!0 f .x C h/ C f .x/

d p f .x/ D lim dx h!0

Proof: The case n D 2 is just the Product Rule. Assume the formula holds for n D k for some integer k > 2. Using the Product Rule and this hypothesis we calculate .f1 f2


fk fkC1 /0 D Œ.f1 f2


fk /fkC1 0 0 D .f1 f2


fk /0 fkC1 C .f1 f2


fk /fkC1

f 0 .x/ D p 2 f .x/ 2x x d p 2 x C1D p D p dx 2 x2 C 1 x2 C 1  3 52. f .x/ D jx 3 j D x 3 if x  0 . Therefore f is differenx if x < 0 tiable everywhere except possibly at x D 0, However, f .0 C h/ h h!0C f .0 C h/ lim h h!0 lim

f .0/

D .f10 f2


fk C f1 f20


fk C


C f1 f2


fk0 /fkC1 0 C .f1 f2


fk /fkC1

D f10 f2


fk fkC1 C f1 f20


fk fkC1 C


0 C f1 f2


fk0 fkC1 C f1 f2


fk fkC1

D lim h2 D 0

so the formula is also true for n D k C 1. The formula is therefore for all integers n  2 by induction.

D lim . h2 / D 0:

Section 2.4 The Chain Rule

h!0C

f .0/

h!0

Thus f 0 .0/ exists and equals 0. We have  2 f 0 .x/ D 3x 2 3x

1. 2.

if x  0 if x < 0.

n d n=2 D x .n=2/ 1 for n D 1, 2, 3, : : : . x 53. To be proved: dx 2 Proof: It is already known that the case n D 1 is true: the derivative of x 1=2 is .1=2/x 1=2 . Assume that the formula is valid for n D k for some positive integer k: d k=2 k x D x .k=2/ dx 2

3.

4.

5. 1

:

6. d .kC1/=2 d 1=2 k=2 x D x x dx dx 1 k D x 1=2 x k=2 C x 1=2 x .k=2/ 2 2

f .x/ D .4

x 2 /10

f 0 .x/ D 10.4 d p dy D 1 dx dx  F .t / D 2 C

x 2 /9 . 2x/ D

1

D

k C 1 .kC1/=2 x 2

1

:

d 1 d n=2 x D dx dx x m=2 1 m D m x .m=2/ 1 x 2 n m .m=2/ 1 x D x .n=2/ D 2 2

7.

z D .1 C x 2=3 /3=2

y D 8.

:

.f1 f2


fn /0 D f10 f2


fn C f1 f20


fn C


C f1 f2


fn0

2t 2 /

1=3

.1 C x 2=3 /1=2

y D j1

3 2 .1

x 2 j;

3=2

2t 2 /

5=2

y0 D

. 4t / D 6t .1 2xsgn .1

f .t / D j2 C t 3 j f 0 .t / D Œsgn .2 C t 3 /.3t 2 / D

11.

54. To be proved:

/Dx

4x 3 12 . 4/ D .5 4x/2 .5 4x/2

y D .1

y0 D

10.

1=3

3 5

0

9.

1

yD

x 2 /9

20x.4

3x 6x D p 3x 2 D p 2 2 1 3x 1 3x 2  10 3 t     30 3 11 3 11 3 D 2 C 10 2 C t t2 t2 t

z 0 D 32 .1 C x 2=3 /1=2 . 23 x

Thus the formula is also true for n D k C 1. Therefore it is true for all positive integers n by induction. For negative n D m (where m > 0) we have

52

y D .2x C 3/6 ; y 0 D 6.2x C 3/5 2 D 12.2x C 3/5  x 99 yD 1 3     x 98 1 x 98 D 33 1 y 0 D 99 1 3 3 3

F 0 .t / D

Then, by the Product Rule and this hypothesis,

(page 120)

y D 4x C j4x 1j y 0 D 4 C 4.sgn .4x 1//  8 if x > 14 D 0 if x < 14

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

2t 2 /

x2/ D

5=2

2x 3 2x j1 x 2 j

3t 2 .2 C t 3 / j2 C t 3 j

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

12.

y D .2 C jxj3 /1=3

y 0 D 31 .2 C jxj3 /

SECTION 2.4 (PAGE 120)

18. y

.3jxj2 /sgn .x/   x D jxj2 .2 C jxj3 / 2=3 D xjxj.2 C jxj3 / jxj

13.

f .x/ D 1 C

D

D

17.

2 3

r

 z D uC dz D du



3  2 p 2 3x C 4 2 C 3x C 4 r

f .x/ D 4 1 C

16.

yD4xCj4x 1j

slope 0

p

0

15.

slope 8

2=3

1 yD p 2 C 3x C 4   3 1 p y0 D   2 p 2 3x C 4 2 C 3x C 4 D

14.

2=3

x

2 3

r

x

x

2 1

2 3

3



!4

!3

1C

r

1 2 x

r 2

3

3

x !3

!  1 2 3

5=3

u 1    8=3  5 1 1 uC 1 3 u 1 .u 1/2   8=3  5 1 1 1 u C 3 .u 1/2 u 1

p x5 3 C x6 yD .4 C x 2 /3     p 3x 5 1 2 3 4 6 C x5 p 3 C x y0 D .4 C x / 5x .4 C x 2 /6 3 C x6 ! h i p x 5 3 C x 6 3.4 C x 2 /2 .2x/ h i .4 C x 2 / 5x 4 .3 C x 6 / C 3x 10 x 5 .3 C x 6 /.6x/ p D .4 C x 2 /4 3 C x 6 60x 4 3x 6 C 32x 10 C 2x 12 D p .4 C x 2 /4 3 C x 6

21=3

d 1=4 d x D dx dx

q

20.

d 3=4 d x D dx dx

q

21.

3 1 d 3=2 d p 3 x D p .3x 2 / D x 1=2 x D dx dx 2 2 x3

22.

d f .2t C 3/ D 2f 0 .2t C 3/ dt

23.

d f .5x dx

24.

25.

26.

27.

28.

29.

t

x

19.

y

yDj2Ct 3 j



1 4 ;1

p

1 1 1 x D pp  p D x 4 2 x 2 x

p 1 x xD p p 2 x x

x 2 / D .5

2x/f 0 .5x

 p

3=4

x xC p 2 x



D

3 x 4

1=4

x2/

  2       3 2 2 2 2 d D3 f f0 f dx x x x x2     2 6 0 2 2 D f f 2 x x x 2f 0 .x/ d p f 0 .x/ 3 C 2f .x/ D p D p dx 2 3 C 2f .x/ 3 C 2f .x/ p p d 2 f . 3 C 2t / D f 0 . 3 C 2t/ p dt 2 3 C 2t p 1 0 D p f . 3 C 2t / 3 C 2t

p p d 1 f .3 C 2 x/ D p f 0 .3 C 2 x/ dx x    d f 2f 3f .x/ dt      0 D f 2f 3f .x/
2f 0 3f .x/
3f 0 .x/      D 6f 0 .x/f 0 3f .x/ f 0 2f 3f .x/

 d  f 2 3f .4 5t / dx    D f 0 2 3f .4 5t / 3f 0 .4 5t / . 5/   D 15f 0 .4 5t /f 0 2 3f .4 5t /

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

53

lOMoARcPSD|6566483

SECTION 2.4 (PAGE 120)

30.

31.

32.

!ˇ x 2 1 ˇˇ ˇ x2 C 1 ˇ xD 2 p x 2 x 2 1.2x/ ˇˇ .x C 1/ p 2 ˇ x 1 D ˇ ˇ .x 2 C 1/2 xD   p 2 p 3. 4/ .5/ 2 3 D p D 25 25 3 p

d dx

ˇ ˇ ˇ 7ˇ ˇ

d p 3t dt

3

tD3

D p 2 3t

1 f .x/ D p 2x C 1 0

f .4/ D 33.

ADAMS and ESSEX: CALCULUS 9

ˇ ˇ 1 ˇ ˇ .2x C 1/3=2 ˇ

ˇ ˇ ˇ ˇ 7ˇ

xD4

tD3

34.

xD 1

The tangent p line at . 1; 23=2 / has equation 3=2 2.x C 1/. yD2

2

38.

b The slope of y D .ax C b/8 at x D is a ˇ ˇ ˇ dy ˇˇ 7ˇ D 8a.ax C b/ ˇ D 1024ab 7 : ˇ ˇ dx ˇ xDb=a

3 D p 2 2

y D .2b/8 D 256b 8 is D

 256b 8 C 1024ab 7 x

y D 210 ab 7 x

1 27

y D .x 3 C 9/17=2 ˇ ˇ ˇ ˇ 17 3 0ˇ 15=2 2ˇ yˇ D .x C 9/ 3x ˇ ˇ ˇ 2

xD 2

D

40.

Given that f .x/ D .x f 0 .x/ D m.x

2.1 C x/.2 C x/.3 C x/3 .4 C x/4 C

D .x

3.1 C x/.2 C x/2 .3 C x/2 .4 C x/4 C

F 0 .0/ D .22 /.33 /.44 / C 2.1/.2/.33 /.44 /C D 4.22
33
44 / D 110; 592

0

y D

D

36. The slope of y D ˇ dy ˇˇ ˇ dx ˇ

1=2  6

ˇ ˇ ˇ D p ˇ 2 1 C 2x 2 ˇ 4x

xD2

D

.x

.x

b/n C n.x n 1

b/

.mx

a/m .x mb C nx

b/n

1

na/:

mb C nx

na D 0;

n m aC b: mCn mCn

This point lies lies between a and b. 41. x.x 4 C 2x 2

2/=.x 2 C 1/5=2

42.

4.7x 4

43.

857; 592

44.

5=8

45.

The Chain Rule does not enable you to calculate the derivatives of jxj2 and jx 2 j at x D 0 directly as a composition of two functions, one of which is jxj, because jxj is not differentiable at x D 0. However, jxj2 D x 2 and jx 2 j D x 2 , so both functions are differentiable at x D 0 and have derivative 0 there.

46.

It may happen that k D g.x C h/ g.x/ D 0 for values of h arbitrarily close to 0 so that the division by k in the “proof” is not justified.

4 : 3

Thus, the equation of the tangent line at .2; 3/ is y D 3 C 43 .x 2/, or y D 43 x C 31 :

54

a/

xD

1 C 2x 2 at x D 2 is

xD2

1

m 1

b/n then

which is equivalent to

   1=2  7 5 6 x C .3x/ 2    3=2   1 .3x/5 2 5.3x/4 3  1 2    3=2  15 6 1 .3x/4 .3x/5 2 2    1=2  7  x C .3x/5 2 p

a/m

mx

3.1/.22 /.32 /.44 / C 4.1/.22 /.33 /.43 / 

a/m .x

If x ¤ a and x ¤ b, then f 0 .x/ D 0 if and only if

4.1 C x/.2 C x/2 .3 C x/3 .4 C x/3

2

 b , or a

Slope of y D 1=.x 2 x Cˇ3/3=2 at x D 2 is ˇ 3 5 3 2 D .x xC3/ 5=2 .2x 1/ˇˇ .9 5=2 /. 5/ D 2 2 162 xD 2 1 The tangent line at . 2; / has equation 27 5 1 C .x C 2/. yD 27 162

F 0 .x/ D .2 C x/2 .3 C x/3 .4 C x/4 C

35.

3  28 b 8 .

b and a

39.

17 .12/ D 102 2

F .x/ D .1 C x/.2 C x/2 .3 C x/3 .4 C x/4

  y D x C .3x/5

xDb=a

The equation of the tangent line at x D y

D

xD 2

37. Slope of y D .1 C x 2=3 /3=2ˇ at x D 1 is   p 2 1=3 ˇˇ 3 .1 C x 2=3 /1=2 x D 2 ˇ ˇ 2 3

49x 2 C 54/=x 7

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 2.5 (PAGE 126)

Section 2.5 Derivatives of Trigonometric Functions (page 126)

24.

1.

d d 1 csc x D D dx dx sin x

cos x D sin2 x

2.

d d cos x cot x D D dx dx sin x

cos2 x sin2 x D sin2 x

y0 D

3. y D cos 3x; 4. y D sin

x ; 5

y0 D

y D cot.4

25. csc 2 x

1 x cos : 5 5

y 0 D 3 csc2 .4

3x/;

26.

3 sin 3x 27. 28.

y 0 D a sec ax tan ax:

6. y D sec ax; 7.

csc x cot x

y 0 D  sec2 x

5. y D tan x;

23.

29.

3x/

 x 1  x d sin D cos dx 3 3 3 9. f .x/ D cos.s rx/; f 0 .x/ D r sin.s 8.

rx/

30.

0

10. y D sin.Ax C B/;

y D A cos.Ax C B/

d sin.x 2 / D 2x cos.x 2 / dx p p d 1 12. cos. x/ D p sin. x/ dx 2 x p sin x 13. y D 1 C cos x; y 0 D p 2 1 C cos x 11.

14.

15.

16.

31. 32.

d sin.2 cos x/ D cos.2 cos x/. 2 sin x/ dx D 2 sin x cos.2 cos x/

f .x/ D cos.x C sin x/ f 0 .x/ D .1 C cos x/ sin.x C sin x/

33.

g. / D tan. sin  /

34.

g 0 . / D .sin  C  cos  / sec2 . sin  / 17.

u D sin3 .x=2/;

18. y D sec.1=x/; 19.

u0 D y0 D

F .t / D sin at cos at 0

F .t / D a cos at cos at . D a cos 2at / 20.

21. 22.

3 cos.x=2/ sin2 .x=2/ 2

.1=x 2 / sec.1=x/ tan.1=x/ 1 sin 2at / 2 a sin at sin at

.D

35. 36.

sin a cos b a cos b cos a C b sin a sin b G 0 . / D : cos2 b  d  sin.2x/ cos.2x/ D 2 cos.2x/ C 2 sin.2x/ dx d d .cos2 x sin2 x/ D cos.2x/ dx dx D 2 sin.2x/ D 4 sin x cos x G. / D

d .tan x C cot x/ D sec2 x csc2 x dx d .sec x csc x/ D sec x tan x C csc x cot x dx d .tan x x/ D sec2 x 1 D tan2 x dx d d tan.3x/ cot.3x/ D .1/ D 0 dx dx d .t cos t sin t / D cos t t sin t cos t D t sin t dt d .t sin t C cos t / D sin t C t cos t sin t D t cos t dt d .1 C cos x/.cos x/ sin.x/. sin x/ sin x D dx 1 C cos x .1 C cos x/2 1 cos x C 1 D D 2 .1 C cos x/ 1 C cos x d cos x .1 C sin x/. sin x/ cos.x/.cos x/ D dx 1 C sin x .1 C sin x/2 sin x 1 1 D D .1 C sin x/2 1 C sin x

d 2 x cos.3x/ D 2x cos.3x/ 3x 2 sin.3x/ dx p g.t / D .sin t /=t t cos t sin t 1  g 0 .t / D p t2 2 .sin t /=t t cos t sin t D p 2t 3=2 sin t v D sec.x 2 / tan.x 2 /

v 0 D 2x sec.x 2 / tan2 .x 2 / C 2x sec3 .x 2 / p sin x zD p 1 C cos x p p p p p p .1 C cos x/.cos x=2 x/ .sin x/. sin x=2 x/ p 2 z0 D .1 C cos x/ p 1 1 C cos x D p p p D p 2 x.1 C cos x/2 2 x.1 C cos x/

d sin.cos.tan t // D dt

.sec2 t /.sin.tan t // cos.cos.tan t //

f .s/ D cos.s C cos.s C cos s// f 0 .s/ D Œsin.s C cos.s C cos s//  Œ1 .sin.s C cos s//.1 sin s/

37. Differentiate both sides of sin.2x/ D 2 sin x cos x and divide by 2 to get cos.2x/ D cos2 x sin2 x.

sin2 x and

38.

Differentiate both sides of cos.2x/ D cos2 x divide by 2 to get sin.2x/ D 2 sin x cos x.

39.

Slope of y D sin x at .; 0/ is cos  D 1. Therefore the tangent and normal lines to y D sin x at .; 0/ have equations y D .x / and y D x , respectively.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

55

lOMoARcPSD|6566483

SECTION 2.5 (PAGE 126)

ADAMS and ESSEX: CALCULUS 9

40. The slope of y D tan.2x/ at .0; 0/ is 2 sec2 .0/ D 2. Therefore the tangent and normal lines to y D tan.2x/ at .0; 0/ have equations y D 2x and y D x=2, respectively. p 41. Thepslope of y D 2 cos.x=4/ at .; 1/ is . 2=4/ sin.=4/ Dp 1=4. Therefore the tangent and normal lines to y D 2 cos.x=4/ at .; 1/ have equations y D 1 .x /=4 and y D 1 C 4.x /, respectively.

42. The slope of y Dpcos2 x at .=3; 1=4/ is sin.2=3/ D 3=2. Therefore the tangent and normal lines to y D tan.2x/ at .0; 0/ have equations p y D .1=4/ . 3=2/.x .=3// and p y D .1=4/ C .2= 3/.x .=3//, respectively.  x   x   is y 0 D cos . 43. Slope of y D sin.x ı / D sin 180 180 180 At x D 45 the tangent line has equation 1  yD p C p .x 45/. 2 180 2  x  44. For y D sec .x ı / D sec we have 180  x   x   dy D sec tan : dx 180 180 180

p  p  3 .2 3/ D : 180 90 90 Thus, the normal line has slope p and has equation  3 90 yD2 p .x 60/.  3

49.

y D x C sin x has a horizontal tangent at x D  because dy=dx D 1 C cos x D 0 there.

50.

y D 2x C sin x has no horizontal tangents because dy=dx D 2 C cos x  1 everywhere.

51.

y D x C 2 sin x has horizontal tangents at x D 2=3 and x D 4=3 because dy=dx D 1 C 2 cos x D 0 at those points.

52.

y D x C 2 cos x has horizontal tangents at x D =6 and x D 5=6 because dy=dx D 1 2 sin x D 0 at those points.

53. 54.

2 sin.2x/ t an.2x/ D lim D12D2 x!0 x 2x cos.2x/

lim sec.1 C cos x/ D sec.1

x!

55. 56.

57. 58.

At x D 60 the slope is

45. The slope of y D tan x at x D a is sec2 a. The tangent there is parallel to y D 2x if sec2 a D 2, or p cos a D ˙1= 2. The only solutions in . =2; =2/ are a D ˙=4. The corresponding points on the graph are .=4; 1/ and . =4; 1/.

lim

x!0

 x 2 cos x D 12  1 D 1 x!0 x!0 sin x       cos2 x sin x 2 lim cos D cos  D 1 D lim cos  x!0 x!0 x2 x   1 2 sin2 .h=2/ 1 sin.h=2/ 2 1 cos h D D lim D lim lim h2 h2 h=2 2 h!0 h!0 2 h!0 lim x 2 csc x cot x D lim

f will be differentiable at x D 0 if 2 sin 0 C 3 cos 0 D b; and ˇ ˇ d D a: .2 sin x C 3 cos x/ˇˇ dx xD0

Thus we need b D 3 and a D 2. 59.

There are infinitely many lines through the origin that are tangent to y D cos x. The two with largest slope are shown in the figure. y

46. The slope of y D tan.2x/ at x D a is 2 sec2 .2a/. The tangent there is normal to y D x=8 if 2 sec2 .2a/ D 8, or cos.2a/ D ˙1=2. The only solutions in . =4; =4/ are a D points on the graph are p ˙=6. The corresponding p .=6; 3/ and . =6; 3/. 47.

48.



56



2 x

y D cos x

d sin x D cos x D 0 at odd multiples of =2. dx d cos x D sin x D 0 at multiples of . dx d sec x D sec x tan x D 0 at multiples of . dx d csc x D csc x cot x D 0 at odd multiples of =2. dx Thus each of these functions has horizontal tangents at infinitely many points on its graph. d tan x D sec2 x D 0 nowhere. dx d cot x D csc2 x D 0 nowhere. dx Thus neither of these functions has a horizontal tangent.

1/ D sec 0 D 1

Fig. 2.5-59 The tangent to y D cos x at x D a has equation y D cos a .sin a/.x a/. This line passes through the origin if cos a D a sin a. We use a calculator with a “solve” function to find solutions of this equation near a D  and a D 2 as suggested in the figure. The solutions are a  2:798386 and a  6:121250. The slopes of the corresponding tangents are given by sin a, so they are 0:336508 and 0:161228 to six decimal places. 60. 61.

1 p

2 C 3.2 3=2

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

4 C 3/=

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

62.

SECTION 2.6 (PAGE 131)

a) As suggested by the figure in the problem, the square of the length of chord AP is .1 cos  /2 C.0 sin  /2 , and the square of the length of arc AP is  2 . Hence

5.

y D x 1=3 x 1=3 1 1 y 0 D x 2=3 C x 4=3 3 3 2 5=3 4 7=3 00 y D x x 9 9 10 8=3 28 10=3 x C x y 000 D 27 27

6.

y D x 10 C 2x 8

.1 C cos  /2 C sin2  <  2 ; and, since squares cannot be negative, each term in the sum on the left is less than  2 . Therefore 0  j1

cos  j < j j;

0

y D 10x C 16x

0  j sin  j < j j: 7.

Since lim!0 j j D 0, the squeeze theorem implies that lim 1 cos  D 0; lim sin  D 0: !0

!0

From the first of these, lim!0 cos  D 1. b) Using the result of (a) and the addition formulas for cosine and sine we obtain 8. lim cos.0 C h/ D lim .cos 0 cos h

h!0

h!0

sin 0 sin h/ D cos 0

lim sin.0 C h/ D lim .sin 0 cos h C cos 0 sin h/ D sin 0 :

h!0

h!0

9. This says that cosine and sine are continuous at any point 0 .

y D .3

1.

0

2.

3.

14.3

2x/6

y 00 D 168.3

2x/5

000

1680.3

D

12.

2x/4

1 x 1 y 0 D 2x C 2 x

y 000 D

6 D 6.x .x 1/2 0 y D 12.x 1/ 3 yD

y 000 D

1/

144.x

p ax C b a y0 D p 2 ax C b yD

2 x3

y 00 D 2

y D x2

y 00 D 36.x

4.

2x/

y D

y

x 1 xC1 2 0 y D .x C 1/2

6 x4

1/

13.

2

2

1/

a2 4.ax C b/3=2 3a3 D 8.ax C b/5=2

y 000

y 000 D sec x tan3 x C 5 sec3 x tan x

y D cos.x 2 /

y 00 D

2x sin.x 2 /

y 000 D

2 sin.x 2 /

4x 2 cos.x 2 /

12x cos.x 2 / C 8x 3 sin.x 2 /

sin x x cos x sin x 0 y D x x2 2 .2 x / sin x 2 cos x y 00 D x3 x2 2 2 3.x 2/ sin x .6 x / cos x C y 000 D 3 4 x x yD

1 Dx x f 0 .x/ D x 2 f .x/ D

.4/

y 00 D

y 00 D sec x tan2 x C sec3 x

y D sec x y 0 D sec x tan x

f 000 .x/ D

5

y 000

y 000 D 2 sec4 x C 4 sec2 x tan2 x

f 00 .x/ D 2x

4

4 .x C 1/3 12 D .x C 1/4

y 00 D

y 00 D 2 sec2 x tan x

y D tan x

y0 D

7

y 000 D 720x 7 C 672x 5

yD

y D sec x

11.

y 00 D 90x 8 C 112x 6

7

p y D .x 2 C 3/ x D x 5=2 C 3x 1=2 5 3 y 0 D x 3=2 C x 1=2 2 2 15 1=2 3 3=2 y 00 D x x 4 4 15 1=2 9 5=2 x C x y 000 D 8 8

0

10.

Section 2.6 Higher-Order Derivatives (page 131)

9

1

3

3Šx

4

5

f .x/ D 4Šx Guess: f .n/ .x/ D . 1/n nŠx .nC1/ ./ Proof: (*) is valid for n D 1 (and 2, 3, 4). .kC1/ Assume f .k/ .x/ D . 1/k kŠx for   some k  1 .kC1/ k Then f .x/ D . 1/ kŠ .k C 1/ x .kC1/ 1

D . 1/kC1 .k C 1/Šx ..kC1/C1/ which is (*) for n D k C 1. Therefore, (*) holds for n D 1; 2; 3; : : : by induction.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

57

lOMoARcPSD|6566483

SECTION 2.6 (PAGE 131)

14.

1 Dx 2 x2 f 0 .x/ D 2x 3 f 00 .x/ D 2. 3/x 4 D 3Šx 4 f .3/ .x/ D 2. 3/. 4/x 5 D 4Šx Conjecture:

ADAMS and ESSEX: CALCULUS 9

Thus, the formula is also true for n D k C 1. Hence, it is true for n  2 by induction.

f .x/ D

f .n/ .x/ D . 1/n .n C 1/Šx

17.

5

.nC2/

f 00 .x/ D 2b 2 .a C bx/

for n D 1; 2; 3; : : :

f .kC1/ .x/ D

D . 1/kC1 .k C 2/Šx

Œ.kC1/C2

.kC2/ 1

15.

D .2 2 x f 0 .x/ D C.2 x/ 2 f .x/ D

f 00 .x/ D 2.2

x/

x/

18. f .x/ D x 2=3 f 0 .x/ D 32 x f 00 .x/ D 32 . f 000 .x/ D 32 . Conjecture:

1

3

D .k C 1/Š.2 x/ : Thus (*) holds for n D k C 1 if it holds for k. Therefore, (*) holds for n D 1; 2; 3; : : : by induction. p 16. f .x/ D x D x 1=2 f 0 .x/ D 21 x 1=2 f 00 .x/ D 12 . 12 /x 3=2 f 000 .x/ D 21 . 12 /. 32 /x 5=2 f .4/ .x/ D 21 . 12 /. 32 /. 25 /x 7=2 Conjecture:
3
5


.2n 2n

3/

x

.2n 1/=2

11


3
5


.2k 2k

3/

x


4
7



.3k 3k

5/

x

.3k 2/=3

d .k/ f .x/ dx   .3k 2/ 1
4
7



.3k 5/
D 2. 1/k 1 x Œ.3k 2/=3 3 3k 1
4
7



.3k 5/Œ3.k C 1/ 5 Œ3.kC1/ x D 2. 1/.kC1/ 1 3. k C 1/

:

19. f 00 .x/ D

:

d .k/ f .kC1/ .x/ D f .x/ dx   .2k 1/ 1
3
5


.2k 3/
D . 1/k 1 x Œ.2k 1/=2 2 2k 1
3
5


.2k 3/Œ2.k C 1/ 3 Œ2.kC1/ x D . 1/.kC1/ 1 2kC1

1

2=3

:

Thus, the formula is also true for n D k C 1. Hence, it is true for n  2 by induction.

Then

58

11

f .kC1/ .x/ D

.n  2/:

.2k 1/=2

1 4=3 3 /x 1 4 7=3 3 /. 3 /x

Then,

Proof: Evidently, the above formula holds for n D 2; 3 and 4. Assume that it holds for n D k, i.e. f .k/ .x/ D . 1/k

.b/

1=3

f .k/ .x/ D 2. 1/k

..kC1/C1/

11

1

1
4
7



.3n 5/ .3n 2/=3 f .n/ .x/ D 2. 1/n 1 x for 3n n  2. Proof: Evidently, the above formula holds for n D 2 and 3. Assume that it holds for n D k, i.e.

f 000 .x/ D C3Š.2 x/ 4 Guess: f .n/ .x/ D nŠ.2 x/ .nC1/ ./ Proof: (*) holds for n D 1; 2; 3. Assume f .k/ .x/ D kŠ.2 x/ .kC1/ (i.e., (*) holds for n D k)   Then f .kC1/ .x/ D kŠ .k C 1/.2 x/ .kC1/ 1 . 1/

f .n/ .x/ D . 1/n

3

D . 1/kC1 .k C 1/Šb kC1 .a C bx/..kC1/C1/ So (*) holds for n D k C 1 if it holds for n D k. Therefore, (*) holds for n D 1; 2; 3; 4; : : : by induction.

:

Thus, the formula is also true for n D k C 1. Hence it is true for n D 1; 2; 3; : : : by induction. 1

1

f 000 .x/ D 3Šb 3 .a C bx/ 4 Guess: f .n/ .x/ D . 1/n nŠb n .a C bx/ .nC1/ ./ Proof: (*) holds for n D 1; 2; 3 Assume (*) holds for n D k: f .k/ .x/ D . 1/k kŠb k .a C bx/ .kC1/ Then   f .kC1/ .x/ D . 1/k kŠb k .k C 1/ .a C bx/ .kC1/

Proof: Evidently, the above formula holds for n D 1; 2 and 3. Assume it holds for n D k, i.e., f .k/ .x/ D . 1/k .k C 1/Šx .kC2/ : Then d .k/ f .x/ dx D . 1/k .k C 1/ŠŒ. 1/.k C 2/x

1 D .a C bx/ a C bx f 0 .x/ D b.a C bx/ 2 f .x/ D

f .x/ D cos.ax/ f 0 .x/ D a sin.ax/

a2 cos.ax/

f 000 .x/ D a3 sin.ax/

f .4/ .x/ D a4 cos.ax/ D a4 f .x/ It follows that f .n/ .x/ D a4 f .n 4/ .x/ for n  4, and 1

1=2

:

8 n a cos.ax/ ˆ < n a sin.ax/ .n/ f .x/ D n ˆ : a cos.ax/ n a sin.ax/

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

if if if if

n D 4k n D 4k C 1 .k D 0; 1; 2; : : :/ n D 4k C 2 n D 4k C 3

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 2.6 (PAGE 131)

Differentiating any of these four formulas produces the one for the next higher value of n, so induction confirms the overall formula. 20.

f .x/ D x cos x f 0 .x/ D cos x x sin x f 00 .x/ D 2 sin x x cos x f 000 .x/ D 3 cos x C x sin x

The pattern suggests that  nŠjxj .nC1/ sgn x f .n/ .x/ D nŠjxj .nC1/

23.

f .4/ .x/ D 4 sin x C x cos x This suggests the formula (for k D 0; 1; 2; : : :) 8 n sin x C x cos x ˆ < n cos x x sin x .n/ f .x/ D ˆ : n sin x x cos x n cos x C x sin x

if if if if

n D 4k n D 4k C 1 n D 4k C 2 n D 4k C 3

Differentiating any of these four formulas produces the one for the next higher value of n, so induction confirms the overall formula. 21.

f .x/ D x sin.ax/ f 0 .x/ D sin.ax/ C ax cos.ax/

f 00 .x/ D 2a cos.ax/ 000

f .x/ D

2

3a sin.ax/

a2 x sin.ax/ a3 x cos.ax/

f 4/ .x/ D 4a3 cos.ax/ C a4 x sin.ax/ This suggests the formula 8 nan 1 cos.ax/ C an x sin.ax/ ˆ ˆ < n 1 na sin.ax/ C an x cos.ax/ f .n/ .x/ D n 1 ˆ na cos.ax/ an x sin.ax/ ˆ : nan 1 sin.ax/ an x cos.ax/

if if if if

n D 4k n D 4k C 1 n D 4k C 2 n D 4k C 3

for k D 0; 1; 2; : : :. Differentiating any of these four formulas produces the one for the next higher value of n, so induction confirms the overall formula.

22. f .x/ D

1 D jxj jxj

1

. Recall that

f 0 .x/ D

jxj

24.

sgn x:

25.

f 00 .x/ D 2jxj

3

3Šjxj

.sgn x/ D 1:

.x/ D 4Šjxj

.sgn x/2 D 2jxj

5

4

:

D k 2 y.2 sec2 .kx/

2

Thus we can calculate successive derivatives of f using the product rule where necessary, but will get only one nonzero term in each case:

f

If y D sec.kx/, then y 0 D k sec.kx/ tan.kx/ and y 00 D k 2 .sec2 .kx/ tan2 .kx/ C sec3 .kx//

d sgn x D 0 and dx

.4/

If y D tan.kx/, then y 0 D k sec2 .kx/ and

D 2k 2 .1 C tan2 .kx// tan.kx/ D 2k 2 y.1 C y 2 /:

If x ¤ 0 we have

f .3/ .x/ D

Differentiating this formula leads to the same formula with n replaced by n C 1 so the formula is valid for all n  1 by induction. p f .x/ D 1 3x D .1 3x/1=2 1 f 0 .x/ D . 3/.1 3x/ 1=2 2  1 1 f 00 .x/ D . 3/2 .1 3x/ 3=2 2 2    1 3 1 . 3/3 .1 3x/ 5=2 f 000 .x/ D 2 2 2     1 1 3 5 f .4/ .x/ D . 3/4 .1 3x/7=2 2 2 2 2 1  3  5 


 .2n 3/ n 3 Guess: f .n/ .x/ D 2n .2n 1/=2 .1 3x/ ./ Proof: (*) is valid for n D 2; 3; 4; (but not n D 1) Assume (*) holds for n D k for some integer k  2 1  3  5  : : :  .2k 3/ k i.e., f .k/ .x/ D 3 2k .1 3x/ .2k 1/=2 1  3  5 


 .2k 3/ k .kC1/ Then f .x/ D 3 2k   2.k 1/ .1 3x/ .2k 1/=2 1 . 3/ 2   1  3  5 


2.k C 1/ 1 D 3kC1 2kC1 .1 3x/ .2.kC1/ 1/=2 Thus (*) holds for n D k C 1 if it holds for n D k. Therefore, (*) holds for n D 2; 3; 4; : : : by induction. y 00 D 2k 2 sec 2 .kx/t an.kx/

d jxj D sgn x, so dx 2

if n is odd if n is even

sgn x

3

26.

1/ D k 2 y.2y 2

1/:

To be proved: if f .x/ D sin.ax C b/, then  . 1/k an sin.ax C b/ if n D 2k f .n/ .x/ D . 1/k an cos.ax C b/ if n D 2k C 1 for k D 0; 1; 2; : : : Proof: The formula works for k D 0 (n D 2  0 D 0 and n D 2  0 C 1 D 1):  f .0/ .x/ D f .x/ D . 1/0 a0 sin.ax C b/ D sin.ax C b/ f .1/ .x/ D f 0 .x/ D . 1/0 a1 cos.ax C b/ D a cos.ax C b/

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

59

lOMoARcPSD|6566483

SECTION 2.6 (PAGE 131)

ADAMS and ESSEX: CALCULUS 9

Section 2.7 Using Differentials and Derivatives (page 137)

Now assume the formula holds for some k  0. If n D 2.k C 1/, then f .n/ .x/ D

d .n f dx

1/

d .2kC1/ f .x/ dx   d D . 1/k a2kC1 cos.ax C b/ dx

.x/ D

D . 1/kC1 a2kC2 sin.ax C b/

and if n D 2.k C 1/ C 1 D 2k C 3, then  d . 1/kC1 a2kC2 sin.ax C b/ f .n/ .x/ D dx

1.

2.

3.

D . 1/kC1 a2kC3 cos.ax C b/:

27.

0:01 1 dx D D 0:0025. y  dy D x2 22 If x D 2:01, then y  0:5 0:0025 D 0:4975. 3 dx 3 f .x/  df .x/ D p D .0:08/ D 0:06 4 2 3x C 1 f .1:08/  f .1/ C 0:06 D 2:06.

 t  1 1 h.t /  dh.t / D sin dt .1/ D . 4 4 4 10 40   1 1 1 D :  h.2/ h 2C 10 40 40 s 1 1 u  du D sec2 ds D .2/. 0:04/ D 0:04. 4 4 4 If s D  0:06, then u  1 0:04  0:96.

Thus the formula also holds for k C 1. Therefore it holds for all positive integers k by induction.

4.

If y D tan x, then

5.

If y D x 2 , then y  dy D 2x dx. If dx D .2=100/x, then y  .4=100/x 2 D .4=100/y, so y increases by about 4%.

6.

If y D 1=x, then y  dy D . 1=x 2 / dx. If dx D .2=100/x, then y  . 2=100/=x D . 2=100/y, so y decreases by about 2%.

7.

If y D 1=x 2 , then y  dy D . 2=x 3 / dx. If dx D .2=100/x, then y  . 4=100/=x 2 D . 4=100/y, so y decreases by about 4%.

8.

If y D x 3 , then y  dy D 3x 2 dx. If dx D .2=100/x, then y  .6=100/x 3 D .6=100/y, so y increases by about 6%. p p x, then dy  dy D .1=2 If y D p x/ dx. If x D .2=100/x, then y  .1=100/ x D .1=100/y, so y increases by about 1%.

y 0 D sec2 x D 1 C tan2 x D 1 C y 2 D P2 .y/; where P2 is a polynomial of degree 2. Assume that y .n/ D PnC1 .y/ where PnC1 is a polynomial of degree n C 1. The derivative of any polynomial is a polynomial of one lower degree, so y .nC1/ D

dy d PnC1 .y/ D Pn .y/ D Pn .y/.1Cy 2 / D PnC2 .y/; dx dx

a polynomial of degree n C 2. By induction, .d=dx/n tan x D PnC1 .tan x/, a polynomial of degree n C 1 in tan x. 28.

.fg/00 D .f 0 g C fg 0 / D f 00 g C f 0 g 0 C f 0 g 0 C fg 00 D f 00 g C 2f 0 g 0 C fg 00

29.

.fg/.3/ D

d .fg/00 dx d D Œf 00 g C 2f 0 g 0 C fg 00  dx D f .3/ g C f 00 g 0 C 2f 00 g 0 C 2f 0 g 00 C f 0 g 00 C fg .3/ .3/

.4/

.fg/

00 0

0 00

C 3f 0 g .3/ C f 0 g .3/ C fg .4/

D f .4/ g C 4f .3/ g 0 C 6f 00 g 00 C 4f 0 g .3/ C fg .4/ : nŠ f .n 2/ g 00 .fg/.n/ D f .n/ g C nf .n 1/g 0 C 2Š.n 2/Š nŠ f .n 3/ g .3/ C


C nf 0 g .n 1/ C fg .n/ C 3Š.n 3/Š n X nŠ f .n k/ g .k/ : D kŠ.n k/Š

60

10. If y D x 2=3 , then y  dy D . 2=3/x 5=3 dx. If dx D .2=100/x, then y  . 4=300/x 2=3 D . 4=300/y, so y decreases by about 1.33%. 11.

.3/

D f g C 3f g C 3f g C fg : d .fg/.3/ D dx d D Œf .3/ g C 3f 00 g 0 C 3f 0 g 00 C fg .3/  dx D f .4/ g C f .3/ g 0 C 3f .3/ g 0 C 3f 00 g 00 C 3f 00 g 00

kD0

9.

If V D 34  r 3 , then V  d V D 4 r 2 dr. If r increases by 2%, then dr D 2r=100 and V  8 r 3 =100. Therefore V =V  6=100. The volume increases by about 6%.

12. If V is the volume and x is the edge length of the cube then V D x 3 . Thus V  d V D 3x 2 x. If V D .6=100/V , then 6x 3 =100  3x 2 dx, so dx  .2=100/x. The edge of the cube decreases by about 2%. 13.

14.

Rate change of Area A with respect to side s, where dA A D s 2 , is D 2s: When s D 4 ft, the area is changing ds 2 at rate 8 ft /ft. p p If A D s 2 , then s D A and ds=dA D 1=.2 A/. If A D 16 m2 , then the side is changing at rate ds=dA D 1=8 m/m2 .

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

15.

SECTION 2.7 (PAGE 137)

The diameter D and area A of a circle are related by p D D 2 A=. The rate of change of diameter with respect p to area is dD=dA D 1=.A/ units per square unit.

The flow rate will increase by 10% if the radius is increased by about 2.5%. 23.

F D k=r 2 implies that dF=dr D 2k=r 3 . Since dF=dr D 1 pound/mi when r D 4; 000 mi, we have 2k D 4; 0003 . If r D 8; 000, we have dF=dr D .4; 000=8; 000/3 D 1=8. At r D 8; 000 mi F decreases with respect to r at a rate of 1/8 pounds/mi.

24.

If price = $p, then revenue is $R D 4; 000p

16. Since A D D 2 =4, the rate of change of area with respect to diameter is dA=dD D D=2 square units per unit. 17.

Rate of change of V D

4 3  r with respect to radius r is 3

dV D 4 r 2 . When r D 2 m, this rate of change is 16 dr 3 m /m.

a) Sensitivity of R to p is dR=dp D 4; 000 20p. If p D 100, 200, and 300, this sensitivity is 2,000 $/$, 0 $/$, and 2; 000 $/$ respectively.

18. Let A be the area of a square, s be its side length and L be its diagonal. Then, L2 D s 2 C s 2 D 2s 2 and dA D L. Thus, the rate of change of A D s 2 D 12 L2 , so dL the area of a square with respect to its diagonal L is L.

b) The distributor should charge $200. This maximizes the revenue. 25.

19. If the radius of the circle is r then C D 2 r and A D  r 2. r p p A Thus C D 2 D 2  A.  Rate of p change of C with respect to A is  dC 1 D p D . dA r A 20. Let s be the side length and V be the volume of a cube. ds D 31 V 2=3 . Hence, Then V D s 3 ) s D V 1=3 and dV the rate of change of the side length of a cube with respect to its volume V is 31 V 2=3 . 21.

ˇ ˇ ˇ ˇ ˇ

tD5

D

700.20

ˇ ˇ ˇ t /ˇ ˇ

tD5

D

26.

ˇ ˇ ˇ ˇ ˇ

tD15

D

700.20

ˇ ˇ ˇ t /ˇ ˇ

tD15

b) To maximize daily profit, production should be 800 sheets/day.

10; 500:

D

27.

3; 500:

b) Average rate of change between t D 5 and t D 15 is 350  .25 V .5/ D 5 10

225/

D

80; 000 n2 C 4n C n 100 n 80; 000 dC C4C : D dn n2 50 dC (a) n D 100; D 2. Thus, the marginal cost of dn production is $2.

C D

82 dC D  9:11. Thus, the marginal cost dn 9 of production is approximately $9.11.

(b) n D 300;

Water is draining out at 3,500 L/min at that time.

V .15/ 15

Daily profit if production is x sheets per day is $P .x/ where P .x/ D 8x 0:005x 2 1; 000: a) Marginal profit P 0 .x/ D 8 0:01x. This is positive if x < 800 and negative if x > 800.

Water is draining out at 10,500 L/min at that time. At t D 15, water volume is changing at rate dV dt

0:5x 2 if x units are

b) C.101/ C.100/ D 43; 299:50 43; 000 D $299:50 which is approximately C 0 .100/.

a) At t D 5, water volume is changing at rate dV dt

Cost is $C.x/ D 8; 000 C 400x manufactured. a) Marginal cost if x D 100 is C 0 .100/ D 400 100 D $300.

t /2 L at t min.

Volume in tank is V .t / D 350.20

10p 2 .

7; 000:

The average rate of draining is 7,000 L/min over that interval. 22. Flow rate F D kr 4 , so F  4kr 3 r. If F D F=10, then kr 4 F D D 0:025r: r  3 40kr 40kr 3

28.

Daily profit P D 13x

C x D 13x

10x

20

x2 1000

x2 D 3x 20 1000 Graph of P is a parabola opening downward. P will be maximum where the slope is zero: 0D

dP D3 dx

2x so x D 1500 1000

Should extract 1500 tonnes of ore per day to maximize profit.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

61

lOMoARcPSD|6566483

SECTION 2.7 (PAGE 137)

ADAMS and ESSEX: CALCULUS 9

29. One of the components comprising C.x/ is usually a fixed cost, $S, for setting up the manufacturing operation. On a per item basis, this fixed cost $S=x, decreases as the number x of items produced increases, especially when x is small. However, for large x other components of the total cost may increase on a per unit basis, for instance labour costs when overtime is required or maintenance costs for machinery when it is over used. C.x/ Let the average cost be A.x/ D . The minimal avx erage cost occurs at point where the graph of A.x/ has a horizontal tangent: 0D

r

30. If y D Cp

If f .x/ D cos x C .x 2 =2/, then f 0 .x/ D x sin x > 0 for x > 0. By the MVT, if x > 0, then f .x/ f .0/ D f 0 .c/.x 0/ for some c > 0, so f .x/ > f .0/ D 1. Thus cos x C .x 2 =2/ > 1 and cos x > 1 .x 2 =2/ for x > 0. Since both sides of the inequality are even functions, it must hold for x < 0 as well.

5.

Let f .x/ D tan x. If 0 < x < =2, then by the MVT f .x/ f .0/ D f 0 .c/.x 0/ for some c in .0; =2/. Thus tan x D x sec2 c > x, since secc > 1.

6.

dA xC 0 .x/ C.x/ : D dx x2

C.x/ D A.x/. x Thus the marginal cost C 0 .x/ equals the average cost at the minimizing value of x.

Hence, xC 0 .x/

4.

C.x/ D 0 ) C 0 .x/ D

7.

f .x/ f .0/ D f 0 .c/ x 0 .1 x/r 1 D r.1 C c/r ) x

p Cp

r

. r/Cp

r 1

D r:

f .x/ D x 2 ; b2 b

.1 C c/r

1

r 1

rx.1 C c/

>1

.since r

< rx

f .b/ a2 D a b

f .a/ a bCa )cD 2

c>0 1Cc >1

.1 C c/r

1

r 1

2. If f .x/ D

1 , and f 0 .x/ D x

f .2/ 2 where c D

p

f .1/ 1 D 1 2

rx.1 C c/

1 then x2 1D

1 D 2

1 D f 0 .c/ c2

0. 8.

If f .x/ D x 3 12x C 1, then f 0 .x/ D 3.x 2 4/. The critical points of f are x D ˙2. f is increasing on . 1; 2/ and .2; 1/ where f 0 .x/ > 0, and is decreasing on . 2; 2/ where f 0 .x/ < 0.

9.

If f .x/ D x 2 4, then f 0 .x/ D 2x. The critical point of f is x D 0. f is increasing on .0; 1/ and decreasing on . 1; 0/.

2 lies between 1 and 2.

3. f .x/ D x 3 3x C 1, f 0 .x/ D 3x 2 3, a D 2, b D 2 f .b/ f .a/ f .2/ f . 2/ D b a 4 8 6 C 1 . 8 C 6 C 1/ D 4 4 D D1 4 0 2 f .c/ D 3c 3 2 3c 2 3 D 1 ) 3c 2 D 4 ) c D ˙ p 3 (Both points will be in . 2; 2/.)

62

1 < 0/;

.since x < 0/:

Hence, .1 C x/r < 1 C rx. If x > 0, then

f 0 .x/ D 2x

D f 0 .c/ D 2c

1

for some c between 0 and x. Thus, .1 C x/r D 1 C rx.1 C c/r 1 . If 1  x < 0, then c < 0 and 0 < 1 C c < 1. Hence

Section 2.8 The Mean-Value Theorem (page 144)

bCa D

Let f .x/ D .1 C x/r where 0 < r < 1. Thus, f 0 .x/ D r.1 C x/r 1 . By the Mean-Value Theorem, for x  1, and x ¤ 0,

, then the elasticity of y is p dy D y dp

1.

Let f .x/ D .1 C x/r 1 rx where r > 1. Then f 0 .x/ D r.1 C x/r 1 r. If 1  x < 0 then f 0 .x/ < 0; if x > 0, then f 0 .x/ > 0. Thus f .x/ > f .0/ D 0 if 1  x < 0 or x > 0. Thus .1 C x/r > 1 C rx if 1  x < 0 or x > 0.

10. If y D 1 x x 5 , then y 0 D 1 5x 4 < 0 for all x. Thus y has no critical points and is decreasing on the whole real line. 11.

If y D x 3 C 6x 2 , then y 0 D 3x 2 C 12x D 3x.x C 4/. The critical points of y are x D 0 and x D 4. y is increasing on . 1; 4/ and .0; 1/ where y 0 > 0, and is decreasing on . 4; 0/ where y 0 < 0.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 2.8 (PAGE 144)

12. If f .x/ D x 2 C 2x C 2 then f 0 .x/ D 2x C 2 D 2.x C 1/. Evidently, f 0 .x/ > 0 if x > 1 and f 0 .x/ < 0 if x < 1. Therefore, f is increasing on . 1; 1/ and decreasing on . 1; 1/.

22.

There is no guarantee that the MVT applications for f and g yield the same c.

23.

CPs x D 0:535898 and x D 7:464102

13. f .x/ D x 3 4x C 1 f 0 .x/ D 3x 2 4

24.

CPs x D

1:366025 and x D 0:366025

25.

CPs x D

0:518784 and x D 0

26.

CP x D 0:521350

27.

If x1 < x2 < : : : < xn belong to I , and f .xi / D 0, .1  i  n/, then there exists yi in .xi ; xiC1 / such that f 0 .yi / D 0, .1  i  n 1/ by MVT.

28.

For x ¤ 0, we have f 0 .x/ D 2x sin.1=x/ cos.1=x/ which has no limit as x ! 0. However, f 0 .0/ D limh!0 f .h/= h D limh!0 h sin.1= h/ D 0 does exist even though f 0 cannot be continuous at 0.

29.

If f 0 exists on Œa; b and f 0 .a/ ¤ f 0 .b/, let us assume, without loss of generality, that f 0 .a/ > k > f 0 .b/. If g.x/ D f .x/ kx on Œa; b, then g is continuous on Œa; b because f , having a derivative, must be continuous there. By the Max-Min Theorem, g must have a maximum value (and a minimum value) on that interval. Suppose the maximum value occurs at c. Since g 0 .a/ > 0 we must have c > a; since g 0 .b/ < 0 we must have c < b. By Theorem 14, we must have g 0 .c/ D 0 and so f 0 .c/ D k. Thus f 0 takes on the (arbitrary) intermediate value k.  2 f .x/ D x C 2x sin.1=x/ if x ¤ 0 0 if x D 0.

2 f 0 .x/ > 0 if jxj > p 3 2 0 f .x/ < 0 if jxj < p 3

2 2 p / and . p ; 1/: 3 3 2 2 f is decreasing on . p ; p /: 3 3 f is increasing on . 1;

14.

If f .x/ D x 3 C 4x C 1, then f 0 .x/ D 3x 2 C 4. Since f 0 .x/ > 0 for all real x, hence f .x/ is increasing on the whole real line, i.e., on . 1; 1/.

15.

f .x/ D .x 2 4/2 f 0 .x/ D 2x2.x 2 4/ D 4x.x 2/.x C 2/ f 0 .x/ > 0 if x > 2 or 2 < x < 0 f 0 .x/ < 0 if x < 2 or 0 < x < 2 f is increasing on . 2; 0/ and .2; 1/. f is decreasing on . 1; 2/ and .0; 2/. 2x 1 then f 0 .x/ D . Evidently, x2 C 1 .x 2 C 1/2 f 0 .x/ > 0 if x < 0 and f 0 .x/ < 0 if x > 0. Therefore, f is increasing on . 1; 0/ and decreasing on .0; 1/.

16. If f .x/ D

17.

f .x/ D x 3 .5 0

2

f .x/ D 3x .5 2

D x .5

x/2

30. 2

3

x/ C 2x .5

x/.15

x/. 1/

f .0 C h/ f .0/ h h!0 h C 2h2 sin.1= h/ D lim h h!0 D lim .1 C 2h sin.1= h/ D 1;

a) f 0 .0/ D lim

5x/

D 5x 2 .5 x/.3 x/ f .x/ > 0 if x < 0, 0 < x < 3, or x > 5 f 0 .x/ < 0 if 3 < x < 5 f is increasing on . 1; 3/ and .5; 1/. f is decreasing on .3; 5/. 0

h!0

because j2h sin.1= h/j  2jhj ! 0 as h ! 0. b) For x ¤ 0, we have

18. If f .x/ D x 2 sin x, then f 0 .x/ D 1 2 cos x D 0 at x D ˙=3 C 2nfor n D 0; ˙1; ˙2; : : :. f is decreasing on . =3 C 2n;  C 2n/. f is increasing on .=3 C 2n; =3 C 2.n C 1// for integers n.

f 0 .x/ D 1 C 4x sin.1=x/

There are numbers x arbitrarily close to 0 where f 0 .x/ D 1; namely, the numbers x D ˙1=.2n/, where n D 1, 2, 3, : : : . Since f 0 .x/ is continuous at every x ¤ 0, it is negative in a small interval about every such number. Thus f cannot be increasing on any interval containing x D 0.

19. If f .x/ D x C sin x, then f 0 .x/ D 1 C cos x  0 f 0 .x/ D 0 only at isolated points x D ˙; ˙3; :::. Hence f is increasing everywhere. 20. If f .x/ D x C 2 sin x, then f 0 .x/ D 1 C 2 cos x > 0 if cos x > 1=2. Thus f is increasing on the intervals . .4=3/ C 2n; .4=3/ C 2n/ where n is any integer. 21.

f .x/ D x 3 is increasing on . 1; 0/ and .0; 1/ because f 0 .x/ D 3x 2 > 0 there. But f .x1 / < f .0/ D 0 < f .x2 / whenever x1 < 0 < x2 , so f is also increasing on intervals containing the origin.

2 cos.1=x/:

31.

Let a, b, and c be three points in I where f vanishes; that is, f .a/ D f .b/ D f .c/ D 0. Suppose a < b < c. By the Mean-Value Theorem, there exist points r in .a; b/ and s in .b; c/ such that f 0 .r/ D f 0 .s/ D 0. By the MeanValue Theorem applied to f 0 on Œr; s, there is some point t in .r; s/ (and therefore in I ) such that f 00 .t / D 0.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

63

lOMoARcPSD|6566483

SECTION 2.8 (PAGE 144)

ADAMS and ESSEX: CALCULUS 9

32. If f .n/ exists on interval I and f vanishes at n C 1 distinct points of I , then f .n/ vanishes at at least one point of I . Proof: True for n D 2 by Exercise 8. Assume true for n D k. (Induction hypothesis) Suppose n D k C 1, i.e., f vanishes at k C 2 points of I and f .kC1/ exists. By Exercise 7, f 0 vanishes at k C 1 points of I . By the induction hypothesis, f .kC1/ D .f 0 /.k/ vanishes at a point of I so the statement is true for n D k C 1. Therefore the statement is true for all n  2 by induction. (case n D 1 is just MVT.)

3.

x 2 C xy D y 3 Differentiate with respect to x: 2x C y C xy 0 D 3y 2 y 0 2x C y y0 D 2 3y x

4.

x 3 y C xy 5 D 2 3x 2 y C x 3 y 0 C y 5 C 5xy 4 y 0 D 0 3x 2 y y 5 y0 D 3 x C 5xy 4

5.

x 2 y 3 D 2x y 2xy 3 C 3x 2 y 2 y 0 D 2 2 2xy 3 y0 D 3x 2 y 2 C 1

33. Given that f .0/ D f .1/ D 0 and f .2/ D 1: a) By MVT, f 0 .a/ D

1 f .0/ D 0 2

f .2/ 2

0 1 D 0 2

6.

for some a in .0; 2/. b) By MVT, for some r in .0; 1/, f 0 .r/ D

f .1/ 1

7.

f .0/ 0 D 0 1

0 D 0: 0

Also, for some s in .1; 2/, f 0 .s/ D

f .2/ 2

f .1/ 1 D 1 2

0 D 1: 1

8.

Then, by MVT applied to f 0 on the interval Œr; s, for some b in .r; s/, 1 f 0 .s/ f 0 .r/ D s r s 1 1 D > s r 2

f 00 .b/ D

since s

0 r

r < 2.

c) Since f 00 .x/ exists on Œ0; 2, therefore f 0 .x/ is continuous there. Since f 0 .r/ D 0 and f 0 .s/ D 1, and since 0 < 71 < 1, the Intermediate-Value Theorem assures us that f 0 .c/ D 71 for some c between r and s.

Section 2.9 Implicit Differentiation (page 149) 1.

xy x C 2y D 1 Differentiate with respect to x: y C xy 0 1 C 2y 0 D 0 1 y Thus y 0 D 2Cx

2. x 3 C y 3 D 1

3x 2 C 3y 2 y 0 D 0, so y 0 D

64

9.

x2 . y2

x 2 C 4.y 2x C 8.y

y0

1/2 D 4 1/y 0 D 0, so y 0 D

x 4.1

y/

x2 x2 C y x y D C1D xCy y y Thus xy y 2 D x 3 Cx 2 y Cxy Cy 2 , or x 3 Cx 2 y C2y 2 D 0 Differentiate with respect to x: 3x 2 C 2xy C x 2 y 0 C 4yy 0 D 0 3x 2 C 2xy y0 D x 2 C 4y p x x C y D 8 xy 1 p xCyCx p .1 C y 0 / D y xy 0 2 xCy p 2.x C y/ C x.1 C y 0 /p D 2 x C y.y C xy 0 / 3x C 2y C 2y x C y y0 D p x C 2x x C y 2x 2 C 3y 2 D 5 4x C 6yy 0 D 0

At .1; 1/: 4 C 6y 0 D 0, y 0 D 2 Tangent line: y 1 D .x 3

2 3 1/ or 2x C 3y D 5

10. x 2 y 3 x 3 y 2 D 12 2xy 3 C 3x 2 y 2 y 0 3x 2 y 2 2x 3 yy 0 D 0 At . 1; 2/: 16 C 12y 0 12 C 4y 0 D 0, so the slope is 28 7 12 C 16 D D : y0 D 12 C 4 16 4 Thus, the equation of the tangent line is y D 2 C 74 .x C 1/, or 7x 4y C 15 D 0: 11.

x  y 3 C D2 y x x 4 C y 4 D 2x 3 y 4x 3 C 4y 3 y 0 D 6x 2 y C 2x 3 y 0 at . 1; 1/: 4 4y 0 D 6 2y 0 2y 0 D 2, y 0 D 1 Tangent line: y C 1 D 1.x C 1/ or y D x.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

12. x C 2y C 1 D

SECTION 2.9 (PAGE 149)

y2

19.

1 1/2yy 0 y 2 .1/ 1 C 2y D .x 1/2 At .2; 1/ we have 1 C 2y 0 D 2y 0 1 so y 0 D Thus, the equation of the tangent is y D 1 21 .x 2/, or x C 2y D 0: p 13. 2x C y p 2 sin.xy/ D =2 2 C y0 2 cos.xy/.y C xy 0 / D 0 At .=4; 1/: 2 C y 0 .1 C .=4/y 0 / D 0, so 0 y D 4=.4 /. The tangent has equation 4

yD1 14.

3x 2



1 2.

20.

 : 4

x

 2 1 C .x C /: 2 4. 1/

17.

21.

y 00 D D 22.

0

y 00 D

8.y 0 /2 D 8y

1 4y

1C

Ax 2 C By 2 D C

x2 D 16y 3

4y 2 x 2 D 16y 3

1 : 4y 3

Ax By

2A C 2B.y 0 /2 C 2Byy 00 D 0. Thus,

A

B.y 0 /2 D By D

2 C 8.y 0 /2 C 8yy 00 D 0.

x y

x2 y2

2Ax C 2Byy 0 D 0 ) y 0 D

0

18. x 2 C 4y 2 D 4; 2x C 8yy 0 D 0; x Thus, y 0 D and 4y

0 2

1 C .y / D y y a2 y2 C x2 D y3 y3

y 00 D

y 1 y C xy D 1 C y ) y D 1 x y 0 C y 0 C xy 00 D y 00 2.y 1/ 2y 0 D Therefore, y 00 D 1 x .1 x/2

2

x 2 C y 2 D a2 2x C 2yy 0 D 0 so x C yy 0 D 0 and y 0 D 1 C y 0 y 0 C yy 00 D 0 so

xy D x C y 0

x 3 3xy C y 3 D 1 3x 2 3y 3xy 0 C 3y 2 y 0 D 0 6x 3y 0 3y 0 3xy 00 C 6y.y 0 /2 C 3y 2 y 00 D 0 Thus

2

17 x D x y 2  y i .xy 0 y/ 2xy x 2 y 0 sin D : 2 x p x y2 3 .3y 0 1/ D 6 9y 0 , At .3; 1/: 2 p 9 p so y 0 D .108 3/=.162 3 3/. The tangent has equation p 3 108 p .x 3/: y D1C 162 3 3

16. cos h

6x

y x2 y2 x 2x C 2y 0 2y.y 0 /2 y 00 D y2 x "    2 # 2 y x2 y x2 D 2 xC y y x y2 x y2 x   2 4xy 2xy D 2 D : y x .y 2 x/2 .x y 2 /3

x sin.xy y 2 / D x 2 1 sin.xy y 2 / C x.cos.xy y 2 //.y C xy 0 2yy 0 / D 2x. At .1; 1/: 0 C .1/.1/.1 y 0 / D 2, so y 0 D 1. The tangent has equation y D 1 .x 1/, or y D 2 x.  y 

2yy 0 C 3y 2 y 0 D 1 ) y 0 D

y0 D

tan.xy 2 / D .2=/xy .sec2 .xy 2 //.y 2 C 2xyy 0 / D .2=/.y C xy 0 /. At . ; 1=2/: 2..1=4/ y 0 / D .1=/ 2y 0 , so y 0 D . 2/=.4. 1//. The tangent has equation yD

15.

4



y2 C y3 D x

1 3x 2 3y 2 2y 6x 2.y 0 /2 2yy 00 C 6y.y 0 /2 C 3y 2 y 00 D 0 .1 3x 2 /2 .2 6y/ 0 2 .2 6y/.y / 6x .3y 2 2y/2 y 00 D D 3y 2 2y 3y 2 2y .2 6y/.1 3x 2 /2 6x D 2 3 2 .3y 2y/ 3y 2y

x

.x

0

x3

23.

Maple gives 0 for the value.

24.

Maple gives the slope as

25.

Maple gives the value

26.

Maple gives the value

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

A

B



Ax By

2

By A.By 2 C Ax 2 / D B 2y3

AC : B 2y3

206 . 55 26. 855; 000 . 371; 293

65

lOMoARcPSD|6566483

SECTION 2.9 (PAGE 149)

27.

ADAMS and ESSEX: CALCULUS 9

Ellipse: x 2 C 2y 2 D 2 2x C 4yy 0 D 0

30.

x Slope of ellipse: D 2y Hyperbola: 2x 2 2y 2 D 1 4x 4yy 0 D 0 x 0 Slope of hyperbola: yH D  2 y 2 x C 2y D 2 At intersection points 2x 2 2y 2 D 1 1 3x 2 D 3 so x 2 D 1, y 2 D 2 x2 x x 0 0 D 1 D yH D Thus yE 2y y 2y 2 Therefore the curves intersect at right angles. 0 yE

2x C 4yy 0 C y C xy 0 D 0

Similarly, the slope of the hyperbola .x; y/ satisfies 2x A2

2y 0 y D 0; B2

y2 D 1 at B2

x2 A2

or y 0 D

B 2x : A2 y

2

2

2 2

Section 2.10 Antiderivatives and Initial-Value Problems (page 155) 1.

b CB A a x D
2 . y2 B 2 b2 a A2 Since a2 b 2 D A2 C B 2 , therefore B 2 C b 2 D a2 A2 , x2 A2 a2 and 2 D 2 2 . Thus, the product of the slope of the y B b two curves at .x; y/ is

3. 4. 5.

Thus,

b2 x B 2 x
D a2 y A2 y

b 2 B 2 A2 a2
D a2 A2 B 2 b 2

1:

Therefore, the curves intersect at right angles.

6. 7.

8. 9.

29. If z D tan.x=2/, then 1 C tan2 .x=2/ dx 1 C z 2 dx 1 dx D D : 1 D sec2 .x=2/ 2 dz 2 dz 2 dz Thus dx=dz D 2=.1 C z 2 /. Also cos x D 2 cos2 .x=2/ D

2 1 C z2

1D

1D

2

1 z2 1 C z2

sin x D 2 sin.x=2/ cos.x=2/ D

66

10. 11.

sec2 .x=2/

2x C y : 4y C x

the only solution is x D 0, y D 0, and these values do not satisfy the original equation. There are no points on the given curve.

2.

If the point .x; y/ is an intersection of the two curves, then y2 x2 y2 x2 C D 2 2 b A2 B 2   a 1 1 1 1 2 2 D y : C x A2 a2 B2 b2 2

y0 D

1 7 y 7 x C xy C y 2 C y 2 D 0; or .x C /2 C y 2 D 0; 4 4 2 4

b2 x : a2 y

i.e. y 0 D

)

However, since x 2 C 2y 2 C xy D 0 can be written

x2 y2 28. The slope of the ellipse 2 C 2 D 1 is found from a b 2x 2y C 2 y 0 D 0; a2 b

x x y D C 1 , xy y 2 D x 2 C xy C xy C y 2 xCy y , x 2 C 2y 2 C xy D 0 Differentiate with respect to x:

1

12.

2 tan.x=2/ 2z : D 2 1 C tan .x=2/ 1 C z2

13.

Z

Z

Z

Z

Z

5 dx D 5x C C x 2 dx D 31 x 3 C C p

x dx D

x 12 dx D x 3 dx D

2 3=2 x CC 3 1 13 13 x

CC

1 4 x CC 4

Z

.x C cos x/ dx D

Z

1 C cos3 x dx D cos2 x

x2 C sin x C C 2 Z Z tan x cos x dx D sin x dx D cos x C C

Z

Z

Z

.a2

Z

.sec2 xCcos x/ dx D tan xCsin xCC

x 2 / dx D a2 x

1 3 x CC 3

.A C Bx C C x 2 / dx D Ax C .2x 1=2 C 3x 1=3 dx D

B 2 C 3 x C x CK 2 3

4 3=2 9 4=3 x C x CC 3 4

Z

Z 6.x 1/ dx D .6x 1=3 6x 4=3 / dx x 4=3 D 9x 2=3 C 18x 1=3 C C  Z  3 x2 1 4 1 3 1 2 x C x 1 dx D x x C x 3 2 12 6 2

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

x CC

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

14.

105

Z

SECTION 2.10

.1 C t 2 C t 4 C t 6 / dt

29.



30.

Given that

D 105.t C 13 t 3 C 15 t 5 C 17 t 7 / C C

D 105t C 35t 3 C 21t 5 C 15t 7 C C Z 1 15. cos.2x/ dx D sin.2x/ C C 2 Z x  x  16. sin dx D 2 cos CC 2 2 Z 1 dx D CC 17. .1 C x/2 1Cx Z 18. sec.1 x/ tan.1 x/ dx D sec.1 19.

Z

p

20. Since

21.

Z

p

23. 24. 25. 26.

27.

Z

Z

Z

Z

(

tan2 x dx D

p Z

2x x2 C 1

sin x cos x dx D cos2 x dx D sin2 x dx D 0

y Dx

2

Thus y D

1 2 x 2

y.0/ D 3

Z

then y D

Z

33.

2

interval . 1; 0/.

x

3

/ dx D

x

34.

1 . 2

x

yD

Z

For



C 12 x

2

2

CC

9=7

dx D

9=7

4;

2=7 7 2x 7 2

C C.

cos x dx D sin x C C C D

3 2

y 0 D sin.2x/ , we have y.=2/ D 1 Z

1 cos.2x/ C C 2 1 1 1 1D cos  C C D C C ÷ C D 2 2 2  1 yD 1 cos.2x/ (for all x): 2

yD

35.

1/ C C so C D 32 . . 1/ C 1 1 3 C which is valid on the x 2x 2 2 1

Z

y0 D x y.1/ D

 1 CC D CC ÷ 6 2 3 (for all x): y D sin x C 2

3

1



2 D sin

2x C 3 for all x. y0 D x 2 x y. 1/ D 0;

C 5 which is valid on the whole real

Also, 4 D y.1/ D C C , so C D 12 . Hence, 7 1 2=7 y D 2x 2 , which is valid in the interval .0; 1/.  0 y D cos x , we have For y.=6/ D 2

1 cos.2x/ C C 4

cos.2x/ x sin.2x/ dx D CC 2 2 4 1 ) y D x 2 2x C C 2 ) 3 D 0 C C therefore C D 3

and 0 D y. 1/ D Hence, y.x/ D

1 sin.2x/ dx D 2

C C and 5 D y.0/ D C .

Given that

xCC

1

 .x

1/ dx D tan x

x sin.2x/ 1 C cos.2x/ dx D C CC 2 2 4

28. Given that Z

Z

3 4=3 4x

32.

then y D

p dx D 2 x 2 C 1 C C:

.sec2 x

3 4=3 4x

y 0 D x 1=3 y.0/ D 5;

Since y 0 D Ax 2 C Bx C C we have B A y D x 3 C x 2 C C x C D. Since y.1/ D 1, therefore 3 2 A B A B C, 1 D y.1/ D C C C C D. Thus D D 1 3 2 3 2 and B A y D .x 3 1/ C .x 2 1/ C C.x 1/ C 1 for all x 3 2

cos.x 2 / C C

d p 2 x 22. Since x C1D p , therefore 2 dx x C1 Z

x 1=3 dx D

15

31. x/ C C

p 4 dx D 8 x C 1 C C: xC1

2x sin.x 2 / dx D



Hence, y.x/ D line.

d p 1 xC1D p , therefore dx 2 xC1 Z

Z

then y D

1 .2x C 3/3=2 C C 3

2x C 3 dx D

p y 0 D 3 x ) y D 2x 3=2 C C y.4/ D 1 ) 1 D 16 C C so C D Thus y D 2x 3=2 15 for x > 0.

(PAGE 155)

sin.2x/ dx D



y 0 D sec2 x ; we have y.0/ D 1 Z y D sec2 x dx D tan x C C

For

1 D tan 0 C C D C ÷ C D 1 y D tan x C 1 (for =2 < x < =2/.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

67

lOMoARcPSD|6566483

SECTION 2.10 (PAGE 155)

36. For



ADAMS and ESSEX: CALCULUS 9

8 00 < y D cos x 41. For y.0/ D 0 we have : 0 y .0/ D 1

y 0 D sec2 x , we have y./ D 1 yD

Z

sec2 x dx D tan x C C

y0 D

1 D tan  C C D C ÷ C D 1 y D tan x C 1 (for =2 < x < 3=2/:

then y 0 D

Z

4

x 0

dx D

Since 2 D y .1/ D and y 0 D 13 x 3 C yD

Z 

1 3 7 3.

1 x 3 3

1 x 3 3

7 3

C 7 3



C C.

dx D 61 x

2

C 73 x C D;

C D, so that D D 23 . Hence, which is valid in the interval

3 , 2

43.

1 4 x x C C1 . 4 0 Since y .0/ D 0, therefore 0 D 0 0 C C1 , and 1 y 0 D x 4 x. 4 1 5 1 2 x x C C2 . Thus y D 20 2 Since y.0/ D 8, we have 8 D 0 0 C C2 . 1 5 1 2 Hence y D x x C 8 for all x. 20 2 40. Given that 8 < y 00 D 5x 2 3x 1=2 y 0 .1/ D 2 : y.1/ D 0; Z we have y 0 D 5x 2 3x 1=2 dx D 53 x 3 6x 1=2 C C . 1, therefore y 0 D

5 3 6x 1=2 C 19 3 ,

Also, 2 D y 0 .1/ D y 0 D 35 x 3 yD

Z 

5 3 x 3

6x

1=2

C

6 C C so that C D

19 . Thus, 3

and

19 3



dx D

4x

C

19 x C D: 3

5 4 C 19 Finally, 0 D y.1/ D 12 3 C D so that D D 5 4 11 3=2 Hence, y.x/ D 12 x 4x C 19 3 x 4 .

68

y0 D

Z

yD

x3 6

.x C sin x/ dx D

sin x C x C 2:

B . Then y 0 D A x Thus, for all x ¤ 0, Let y D Ax C

x 2 y 00 C xy 0

yD

11 4 .

2B C Ax x

B 2B , and y 00 D 3 . x2 x B x

B D 0: x

Ax

We will also have y.1/ D 2 and y 0 .1/ D 4 provided A C B D 2;

and A

B D 4:

These equations have solution A D 3, B D 1, so the initial value problem has solution y D 3x .1=x/. 44.

Let r1 and r2 be distinct rational roots of the equation ar.r 1/ C br C c D 0 Let y D Ax r1 C Bx r2 .x > 0/ Then y 0 D Ar1 x r1 1 C Br2 x r2 1 , and y 00 D Ar1 .r1 1/x r1 2 C Br2 .r2 1/x r2 2 . Thus ax 2 y 00 C bxy 0 C cy 1/x r1

r1 1

3=2

C2 D 1

8 00 < y D x C sin x we have For y.0/ D 2 : 0 y .0/ D 0

D ax 2 .Ar1 .r1

5 4 x 12

÷

x2 cos x C C1 2 0 D 0 cos 0 C C1 ÷ C1 D 1  Z  2 x x3 yD cos x C 1 dx D sin x C x C C2 2 6 2 D 0 sin 0 C 0 C C2 ÷ C2 D 2

C C , therefore C D 37 , Thus

and 1 D y.1/ D 16 C y.x/ D 16 x 2 C 37 x .0; 1/.

39. Since y 00 D x 3

0 D cos 0 C 0 C C2 y D 1 C x cos x:

42.

8 < y 00 D x 4 y 0 .1/ D 2 : y.1/ D 1;

cos x dx D sin x C C1

1 D sin 0 C C1 ÷ C1 D 1 Z y D .sin x C 1/ dx D cos x C x C C2

37. Since y 00 D 2, therefore y 0 D 2x C C1 . Since y 0 .0/ D 5, therefore 5 D 0 C C1 , and y 0 D 2x C 5. Thus y D x 2 C 5x C C2 . Since y.0/ D 3, therefore 3 D 0C0CC2 , and C2 D 3. Finally, y D x 2 C 5x 3, for all x. 38. Given that

Z

2

C Br2 .r2 r2 1

1/x r2

C bx.Ar1 x C Br2 x / C c.Ax   r1 D A ar1 .r1 1/ C br1 C c x  C B.ar2 .r2 1/ C br2 C c x r2

D 0x r1 C 0x r2  0 .x > 0/

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

r1

2

C Bx r2 /

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

8 < 4x 2 y 00 C 4xy 0 45. y.4/ D 2 : 0 y .4/ D 2

yD0

Auxilary Equation: 4r.r 4r 2

./

SECTION 2.11

) a D 4; b D 4; c D

1/ C 4r

1

d) never accelerating to the left e) particle is speeding up for t > 2

1D0

1D0 1 r D˙ 2 By #31, y D Ax 1=2 C Bx 1=2 solves ./ for x > 0. A B 3=2 Now y 0 D x 1=2 x 2 2 Substitute the initial conditions: 2 D 2A C 2D

A 4

B 2 B 16

)1 D A C )

f) slowing down for t < 2 g) the acceleration is 2 at all times h) average velocity over 0  t  4 is

B : 4

2.

2.

d) The point is accelerating to the left if a < 0, i.e., for all t . e) The particle is speeding up if v and a have the same sign, i.e., for t > 52 . f) The particle is slowing down if v and a have opposite sign, i.e., for t < 52 .

Since this equation must hold for all x > 0, we must have

g) Since a D

6D0

r2 r 6 D 0 .r 3/.r C 2/ D 0: There are two roots: r1 D 2, and r2 D 3. Thus the differential equation has solutions of the form y D Ax 2 C Bx 3 . Then y 0 D 2Ax 3 C 3Bx 2 . Since 1 D y.1/ D A C B and 1 D y 0 .1/ D 2A C 3B, therefore A D 52 and B D 35 . Hence, y D 25 x 2 C 53 x 3 .

Section 2.11 Velocity and Acceleration (page 162) 1.

2t , a D

D0

c) The point is accelerating to the right if a > 0, but a D 2 at all t ; hence, the point never accelerates to the right.

1/x r 2  6x r D 0 Œr.r 1/ 6x r D 0:

1/

t 2, v D 5

3

b) The point is moving to the left if v < 0, i.e., when t > 25 .

Let y D x r ; y 0 D rx r 1 ; y 00 D r.r 1/x r 2 . Substituting these expressions into the differential equation we obtain

r.r

x D 4 C 5t

16 C 3 4

a) The point is moving to the right if v > 0, i.e., when t < 25 .

B 7 , so B D 18, A D . 2 2 7 1=2 x C 18x 1=2 (for x > 0). Thus y D 2 46. Consider 8 < x 2 y 00 6y D 0 y.1/ D 1 : 0 y .1/ D 1: Hence 9 D

x 2 Œr.r

16 x.0/ D 0

x.4/ 4

B 4

8DA

(PAGE 162)

dx dv xDt 4t C 3, v D D 2t 4, a D D2 dt dt a) particle is moving: to the right for t > 2 2

b) to the left for t < 2 c) particle is always accelerating to the right

3.

2 at all t , a D

2 at t D

5 2

when v D 0.

h) The average velocity over Œ0; 4 is 8 4 x.4/ x.0/ D D 1. 4 4 dx dv x D t 3 4t C 1, v D D 3t 2 4, a D D 6t dt dt p a) particlepmoving: to the right for t < 2= 3 or t > 2= 3, b) to the left for

p p 2= 3 < t < 2= 3

c) particle is accelerating: to the right for t > 0 d) to the left for t < 0

p e) particle p is speeding up for t > 2= 3 or for 2= 3 < t < 0 p f) particle is slowing down for t < 2= 3 or for p 0 < t < 2= 3 p g) velocity is zeropat t D ˙2= 3. Acceleration at these times is ˙12= 3. h) average velocity on Œ0; 4 is

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

43

44C1 4 0

1

D 12

69

lOMoARcPSD|6566483

SECTION 2.11 (PAGE 162)

4.

ADAMS and ESSEX: CALCULUS 9

.t 2 C 1/.1/ .t /.2t / 1 t2 t ; vD D 2 ; 2 2 C1 .t C 1/ .t C 1/2 2t .t 2 3/ .t 2 C 1/2 . 2t / .1 t 2 /.2/.t 2 C 1/.2t / aD D 2 : 2 4 .t C 1/ .t C 1/3

xD

7.

D D t 2 , D in metres, t in seconds dD D 2t velocity v D dt Aircraft becomes airborne if 500 200; 000 D m/s: v D 200 km/h D 3600 9 250 Time for aircraft to become airborne is t D s, that is, 9 about 27:8 s. Distance travelled during takeoff run is t 2  771:6 metres.

8.

Let y.t / be the height of the projectile t seconds after it is fired upward from ground level with initial speed v0 . Then

t2

a) The point is moving to the right if v > 0, i.e., when 1 t 2 > 0, or 1 < t < 1. b) The point is moving to the left if v < 0, i.e., when t < 1 or t > 1. c) The point is accelerating to the right if a > 0, i.e., whenp2t .t 2 p3/ > 0, that is, when t > 3 or 3 < t < 0.

y 00 .t / D

d) The point p to the left if a < 0, i.e., for p is accelerating t< 3 or 0 < t < 3.

Two antidifferentiations give

2. 2/ D At t D 1, a D .2/3

yD

2. 2/ 1 D . .2/3 2

1 . 2

9.

tD1

Ball strikes the ground when y D 0, .t > 0/, i.e., 0 D t .9:8 4:9t / so t D 2. Velocity at t D 2 is 9:8 9:8.2/ D 9:8 m/s. Ball strikes the ground travelling at 9.8 m/s (downward).

hD

6. Given that y D 100 2t 4:9t , the time t at which the ball reaches the ground is the positive root of the equation y D 0, i.e., 100 2t 4:9t 2 D 0, namely, tD

2C

p

4 C 4.4:9/.100/  4:318 s: 9:8

100 The average velocity of the ball is D 23:16 m=s. 4:318 Since 23:159 D v D 2 9:8t , then t ' 2:159 s.

70

v2 v0 g v02
2 C v0
D 0: 2 g g 2g

An initial speed of 2v0 means the maximum height will be 4v02 =2g D 4h. To get a maximum height of 2h an p initial speed of 2v0 is required. 10. To get to 3h metres above Mars, the ball would have to be thrown upward with speed q p p vM D 6gM h D 6gM v02 =.2g/ D v0 3gM =g: Since gM D 3:72 and g D 9:80, we have vM  1:067v0 m/s.

11. 2

1:86t /:

The height of the ball after t seconds is y.t / D .g=2/t 2 C v0 t m if its initial speed was v0 m/s. Maximum height h occurs when dy=dt D 0, that is, at t D v0 =g. Hence

2

4:9t metres (t in seconds) dy D 9:8 9:8t velocity v D dt dv acceleration a D D 9:8 dt The acceleration is 9:8 m/s2 downward at all times. Ball is at maximum height ˇwhen v D 0, i.e., at t D 1. ˇ Thus maximum height is y ˇ D 9:8 4:9 D 4:9 metres.

1:86t 2 C v0 t D t .49

The time taken to fall back to ground level on Mars would be t D 49=1:86  26:3 s.

h) The average velocity over Œ0; 4 is 4 0 x.4/ x.0/ 1 D 17 D . 4 4 17 5. y D 9:8t

4:9t /:

Since the projectile returns to the ground at t D 10 s, we have y.10/ D 0, so v0 D 49 m/s. On Mars, the acceleration of gravity is 3.72 m/s2 rather than 9.8 m/s2 , so the height of the projectile would be

f) The particle is slowing down if v and a have opposite p 3 < t < 1, or 0 < t < 1 or sign,pi.e., for t > 3. 1, a D

4:9t 2 C v0 t D t .v0

yD

e) The particle is speeding p up if v and a have the same sign, i.e.,pfor t < 3, or 1 < t < 0 or 1 < t < 3.

g) v D 0 at t D ˙1. At t D

9:8; y 0 .0/ D v0 ; y.0/ D 0:

If the cliff is h ft high, then the height of the rock t seconds after it falls is y D h p 16t 2 ft. The rock hits the ground (y D 0) at time p t D h=16 s. Its speedpat that time is v D 32t D 8 h D 160 ft/s. Thus h D 20, and the cliff is h D 400 ft high.

12. If the cliff is h ft high, then the height of the rock t seconds after it is thrown down is y D h 32t 16t 2 ft. The rock hits the ground (y D 0) at time p 1p 32 C 322 C 64h D 1C tD 16 C h s: 32 4

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 2.11

v

Its speed at that time is vD

32

32t D

p 8 16 C h D

(PAGE 162)

.4; 96/

160 ft/s:

Solving this equation for h gives the height of the cliff as 384 ft.

t

13. Let x.t / be the distance travelled by the train in the t seconds after the brakes are applied. Since d 2 x=dt 2 D 1=6 m/s2 and since the initial speed is v0 D 60 km/h D 100=6 m/s, we have 1 2 100 x.t / D t C t: 12 6 The speed of the train at time t is v.t / D .t =6/ C .100=6/ m/s, so it takes the train 100 s to come to a stop. In that time it travels x.100/ D 1002 =12 C 1002 =6 D 1002 =12  833 metres. 14.

15.

x D At 2 C Bt C C; v D 2At C B. The average velocity over Œt1 ; t2  is x.t2 / x.t1 / t2 t1 At22 C Bt1 C C At12 Bt1 C D t2 t1 A.t22 t12 / C B.t2 t1 / D .t2 t1 / A.t2 C t1 /.t2 t1 / C B.t2 t1 / D .t2 t1 / D A.t2 C t1 / C B. The velocity atthe midpoint of Œt1 ; t2  is    instantaneous t2 C t1 t 2 C t1 D 2A C B D A.t2 C t1 / C B. v 2 2 Hence, the average velocity over the interval is equal to the instantaneous velocity at the midpoint. 8 < t2 0t 2 s D 4t 4 2 jhj ! 1 ˇ ˇ ˇ h jhj jhj

f .0 C h/

76

h/

D lim

h!0

27a3 27 C 2 8 4a

27 27a3 C 2 8 4a f .0/ D 1:

Thus f .0/ D 1.

jhj

h

jhj

D lim

h!0

0 D 0: h

 3a : 2

 x

 a

3a 2



D 0:

If a ¤ 0, the x-axis is another tangent to y D x 3 that passes through .a; 0/. The number of tangents to y D x 3 that pass through .x0 ; y0 / is

f .x/ f .x/

h!0

h/

This line passes through .a; 0/ because

6. Given that f .0/ D k, f .0/ ¤ 0, and f .x C y/ D f .x/f .y/, we have

f .x C h/ f .x/ D lim h h!0 f .x/f .h/ D lim h h!0

f .x h

The tangent to y D x 3 at x D 3a=2 has equation

0

0

f .0 2h

yD

as h ! 0. Therefore f 0 .0/ does not exist.

f .0/ D 0 or

f .x/

c) If f .x/ D jxj, then f 0 .0/ does not exist, but

:

5. If h ¤ 0, then

÷

lim

and

:

are always equal if either exists.

f is differentiable elsewhere, including at x D 1 where its derivative is 2.

f .0/ D f .0C0/ D f .0/f .0/

f .x/

h/

D f .x/f 0 .0/ D kf .x/:

three, if x0 ¤ 0 and y0 is between 0 and x03 ;

two, if x0 ¤ 0 and either y0 D 0 or y0 D x03 ;

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 2 (PAGE 164)

The curve y D Ax 2 C Bx C C has slope m D 2Aa C B at .a; Aa2 C Ba C C /. Thus a D .m B/=.2A/, and the tangent has equation

one, otherwise. This is the number of distinct real solutions b of the cubic equation 2b 3 3b 2 x0 C y0 D 0, which states that the tangent to y D x 3 at .b; b 3 / passes through .x0 ; y0 /.

y D Aa2 C Ba C C C m.x

10. By symmetry, any line tangent to both curves must pass through the origin. y y D x 2 C 4x C 1

where f .m/ D C 14. x

a/

B/2 B.m B/ D mx C C CC 4A 2A 2 .m B/ .m B/2 D mx C C C 4A 2A D mx C f .m/; .m

B/2 =.4A/.

.m

Parabola y D x 2 has tangent y D 2ax a2 at .a; a2 /. Parabola y D Ax 2 C Bx C C has tangent Ab 2 C C

y D .2Ab C B/x x 2 C 4x

yD

1

at .b; Ab 2 C Bb C C /. These two tangents coincide if 2Ab C B D 2a

Fig. C-2-10

Ab

2

The tangent to y D x C 4x C 1 at x D a has equation y D a2 C 4a C 1 C .2a C 4/.x D .2a C 4/x

.a2

a/

11.

The slope of y D x 2 at x D a is 2a. The slope of the line from .0; b/ to .a; a2 / is .a2 b/=a. This line is normal to y D x 2 if either a D 0 or 2a..a2 b/=a/ D 1, that is, if a D 0 or 2a2 D 2b 1. There are three real solutions for a if b > 1=2 and only one (a D 0) if b  1=2. 2

2

12. The point Q D .a; a / on y D x that is closest to P D .3; 0/ is such that PQ is normal to y D x 2 at Q. Since PQ has slope a2 =.a 3/ and y D x 2 has slope 2a at Q, we require a2 1 D ; a 3 2a which simplifies to 2a3 C a 3 D 0. Observe that a D 1 is a solution of this cubic equation. Since the slope of y D 2x 3 C x 3 is 6x 2 C 1, which is always positive, the cubic equation can have only one real solution. Thus Q D .1; 1/ is the point on y D x 2 that is closest p to P . The distance from P to the curve is jPQj D 5 units. 13. The curve y D x 2 has slope m D 2a at .a; a2 /. The tangent there has equation y D a2 C m.x

a/ D mx

m2 : 4

2

./

2

C Da :

The two curves have one (or more) common tangents if ./ has real solutions for a and b. Eliminating a between the two equations leads to

1/;

which passes through the origin if a D ˙1. The two common tangents are y D 6x and y D 2x.

m.m B/ 2A

.2Ab C B/2 D 4Ab 2

4C;

or, on simplification, 4A.A

1/b 2 C 4ABb C .B 2 C 4C / D 0:

This quadratic equation in b has discriminant D D 16A2 B 2 16A.A 1/.B 2 C4C / D 16A.B 2 4.A 1/C /: There are five cases to consider: CASE I. If A D 1, B ¤ 0, then ./ gives bD

B 2 C 4C ; 4B

aD

B2

4C : 4B

There is a single common tangent in this case. CASE II. If A D 1, B D 0, then ./ forces C D 0, which is not allowed. There is no common tangent in this case. CASE III. If A ¤ 1 but B 2 D 4.A bD

B 2.A

1/

1/C , then

D a:

There is a single common tangent, and since the points of tangency on the two curves coincide, the two curves are tangent to each other. CASE IV. If A ¤ 1 and B 2 4.A 1/C < 0, there are no real solutions for b, so there can be no common tangents.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

77

lOMoARcPSD|6566483

CHALLENGING PROBLEMS 2 (PAGE 164)

ADAMS and ESSEX: CALCULUS 9

b) The tangent to y D x 4

CASE V. If A ¤ 1 and B 2 4.A 1/C > 0, there are two distinct real solutions for b, and hence two common tangent lines. y y

y D a4

D 4a.a

2x 2 at x D a has equation

2a2 C .4a3

2

4a/.x

a/

3a4 C 2a2 :

1/x

Similarly, the tangent at x D b has equation x

y D 4b.b 2

x

one common two common tangent tangents tangent curves y

1/x

These tangents are the same line (and hence a double tangent) if 4a.a2

y

1/ D 4b.b 2

3a4 C 2a2 D

x

x

Fig. C-2-14 a) The tangent to y D x 3 at .a; a3 / has equation y D 3a2 x

0 D a2 C b 2

2a3 :

x

.x

2

3a x C 2a D 0

a/2 .x C 2a/ D 0:

b) The slope of y D x 3 at x D 2a is 3. 2a/2 D 12a2 , which is four times the slope at x D a.

17.

a) The tangent to y D f .x/ D ax 4 C bx 3 C cx 2 C dx C e

3

c) If the tangent to y D x at x D a were also tangent at x D b, then the slope at b would be four times that at a and the slope at a would be four times that at b. This is clearly impossible. d) No line can be tangent to the graph of a cubic polynomial P .x/ at two distinct points a and b, because if there was such a double tangent y D L.x/, then .x a/2 .x b/2 would be a factor of the cubic polynomial P .x/ L.x/, and cubic polynomials do not have factors that are 4th degree polynomials. 16.

a) y D x 4 2x 2 has horizontal tangents at points x satisfying 4x 3 4x D 0, that is, at x D 0 and x D ˙1. The horizontal tangents are y D 0 and y D 1. Note that y D 1 is a double tangent; it is tangent at the two points .˙1; 1/.

78

b/2 > 0:

c) If y D Ax C B is a double tangent to y D x 4 2x 2 C x, then y D .A 1/x C B is a double tangent to y D x 4 2x 2 . By (b) we must have A 1 D 0 and B D 1. Thus the only double tangent to y D x 4 2x 2 C x is y D x 1.

3

The tangent also intersects y D x 3 at .b; b 3 /, where b D 2a.

2ab D .a

Thus y D 1 is the only double tangent to y D x 4 2x 2 .

For intersections of this line with y D x 3 we solve 3

1/

3b 4 C 2b 2 :

The second equation says that either a2 D b 2 or 3.a2 C b 2 / D 2; the first equation says that a3 b 3 D a b, or, equivalently, a2 C ab C b 2 D 1. If a2 D b 2 , then a D b (a D b is not allowed). Thus a2 D b 2 D 1 and the two points are .˙1; 1/ as discovered in part (a). If a2 Cb 2 D 2=3, then ab D 1=3. This is not possible since it implies that

no common tangent

15.

3b 4 C 2b 2 :

at x D p has equation y D .4ap 3 C3bp 2 C2cpCd /x 3ap 4 2bp 3 cp 2 Ce: This line meets y D f .x/ at x D p (a double root), and xD

2ap

b˙

p

b2

4ac 2a

4abp

8a2 p 2

:

These two latter roots are equal (and hence correspond to a double tangent) if the expression under the square root is 0, that is, if 8a2 p 2 C 4abp C 4ac

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

b 2 D 0:

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 2 (PAGE 164)

This quadratic has two real solutions for p provided its discriminant is positive, that is, provided 16a2 b 2

4.8a2 /.4ac

b 2 / > 0:

This condition simplifies to 3b 2 > 8ac: For example, for y D x 4 2x 2 C x 1, we have a D 1, b D 0, and c D 2, so 3b 2 D 0 > 16 D 8ac, and the curve has a double tangent. b) From the discussion above, the second point of tangency is

so the formula above is true for n D 1. Assume it is true for n D k, where k is a positive integer. Then    d kC1 d k k cos.ax/ D a cos ax C dx 2 dx kC1    k k Da a sin ax C 2   .k C 1/ kC1 Da cos ax C : 2 Thus the formula holds for n D 1; 2; 3; : : : by induction. b) Claim:

qD

2ap b D 2a

p

b : 2a

The slope of PQ is f .q/ q

b3 f .p/ D p

4abc C 8a2 d : 8a2

Calculating f 0 ..p C q/=2/ leads to the same expression, so the double tangent PQ is parallel to the tangent at the point horizontally midway between P and Q. c) The inflection points are the real zeros of f 00 .x/ D 2.6ax 2 C 3bx C c/: This equation has distinct real roots provided 9b 2 > 24ac, that is, 3b 2 > 8ac. The roots are rD sD

p 9b 2 p12a 3b C 9b 2 12a

3b

b3 f .r/ D r

so the formula above is true for n D 1. Assume it is true for n D k, where k is a positive integer. Then    d kC1 k d k a sin ax C sin.ax/ D dx 2 dx kC1    k k D a a cos ax C 2   .k C 1/ kC1 Da sin ax C : 2

c) Note that

4abc C 8a2 d ; 8a2

so this line is also parallel to the double tangent.

18.

  d sin.ax/ D a cos.ax/ D a sin ax C ; dx 2

:

The slope of the line joining these inflection points is f .s/ s

Proof: For n D 1 we have

Thus the formula holds for n D 1; 2; 3; : : : by induction.

24ac 24ac

 n  dn sin.ax/ D an sin ax C . n dx 2

 dn n  n cos.ax/ D a cos ax C . dx n 2 Proof: For n D 1 we have

d .cos4 x C sin4 x/ D 4 cos3 x sin x C 4 sin3 x cos x dx D 4 sin x cos x.cos2 sin2 x/ D 2 sin.2x/ cos.2x/   D sin.4x/ D cos 4x C : 2 It now follows from part (a) that

a) Claim:

d cos.ax/ D dx

  a sin.ax/ D a cos ax C ; 2

dn .cos4 x C sin4 x/ D 4n dx n

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

1

 n  cos 4x C : 2

79

lOMoARcPSD|6566483

CHALLENGING PROBLEMS 2 (PAGE 164)

19.

ADAMS and ESSEX: CALCULUS 9

v (m/s)

d) The upward acceleration in Œ0; 3 was 39:2=3  13:07 m/s2 .

.3; 39:2/

40

e) The maximum height achieved by the rocket is the distance it fell from t D 7 to t D 15. This is the area under the t -axis and above the graph of v on that interval, that is,

30 20 10 t (s) -10

2

4

6

8

10

12

14

12

.15; 1/

7 2

.49/ C

49 C 1 .15 2

12/ D 197:5 m:

-20 -30

f) During the time interval Œ0; 7, the rocket rose a distance equal to the area under the velocity graph and above the t -axis, that is,

-40 .12; 49/ Fig. C-2-19

1 .7 2

a) The fuel lasted for 3 seconds. b) Maximum height was reached at t D 7 s. c) The parachute was deployed at t D 12 s.

80

0/.39:2/ D 137:2 m:

Therefore the height of the tower was 197:5 137:2 D 60:3 m.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

CHAPTER 3. FUNCTIONS

TRANSCENDENTAL

Section 3.1 Inverse Functions 1.

SECTION 3.1 (PAGE 171)

f .x1 / D f .x2 / , x12 D x22 ; .x1  0; x2  0/ , x1 D x2 Thus f is one-to-one. Let y D f 1 .x/. Then x D f .y/ p D y 2 .y  0/. p x and f 1 .x/ D x. therefore y D D.f / D . 1; 0 D R.f 1 /, D.f 1 / D Œ0; 1/ D R.f /.

(page 171)

f .x/ D x 1 f .x1 / D f .x2 / ) x1 1 D x2 1 ) x1 D x2 . Thus f is one-to-one. Let y D f 1 .x/. Then x D f .y/ D y 1 and y D x C 1. Thus f 1 .x/ D x C 1. D.f / D D.f 1 / D R D R.f / D R.f 1 /.

2. f .x/ D 2x 1. If f .x1 / D f .x2 /, then 2x1 1 D 2x2 1. Thus 2.x1 x2 / D 0 and x1 D x2 . Hence, f is one-toone. Let y D f 1 .x/. Thus x D f .y/ D 2y 1, so y D 21 .x C 1/. Thus f 1 .x/ D 12 .x C 1/. D.f / D R.f 1 / D . 1; 1/. R.f / D D.f 1 / D . 1; 1/.

8.

f .x/ D .1 2x/3 . If f .x1 / D f .x2 /, then .1 2x1 /3 D .1 2x2 /3 and x1 D x2 . Thus, f is one-toone. 1 3 Let y D f p .x/. Then x D f .y/ D .1 2y/ p so y D 21 .1 3 x/. Thus, f 1 .x/ D 12 .1 3 x/. D.f / D R.f 1 / D . 1; 1/. R.f / D D.f 1 / D . 1; 1/.

9.

f .x/ D

p 3. f .x/ D x 1 p p f .x1 / D f .x2 / , x1 1 D x2 1; .x1 ; x2  1/ , x1 1 D x2 1 D 0 , x1 D x2 Thus f is one-to-one.pLet y D f 1 .x/. Then x D f .y/ D y 1, and y D 1 C x 2 . Thus f 1 .x/ D 1 C x 2 , .x  0/. D.f / D R.f 1 / D Œ1; 1/, R.f / D D.f 1 / D Œ0; 1/. p x 1 for x  1. 4. f .x/ D p p If f .x1 / D f .x2 /, then x1 1 D x2 1 and x1 1 D x2 1. Thus x1 D x2 and f is one-to-one. p y 1 so Let y D f 1 .x/. Then x D f .y/ D 2 2 1 x D y 1 and y D x C 1. Thus, f .x/ D x 2 C 1. D.f / D R.f 1 / D Œ1; 1/. R.f / D D.f 1 / D . 1; 0. 5. f .x/ D x 3 f .x1 / D f .x2 / , x13 D x23

) .x1 x2 /.x12 C x1 x2 C x22 / D 0 ) x1 D x2 Thus f is one-to-one. Let y D f 1 .x/. Then x D f .y/ D y 3 so y D x 1=3 . Thus f 1 .x/ D x 1=3 . D.f / D D.f 1 / D R D R.f / D R.f 1 /.

p 6. f .x/ pD 1 C 3 x. p If f .x1 / D f .x2 /, then 1 C 3 x 1 D 1 C 3 x 2 so x1 D x2 . Thus, f is oneto-one. p Let y D f 1 .x/ so that x D f .y/ D 1 C 3 y. Thus 3 1 3 y D .x 1/ and f .x/ D .x 1/ . D.f / D R.f 1 / D . 1; 1/. R.f / D D.f 1 / D . 1; 1/.

f .x/ D x 2 ; .x  0/

7.

1 : D.f / D fx W x ¤ 1g D R.f xC1 1 1 D f .x1 / D f .x2 / , x1 C 1 x2 C 1 , x2 C 1 D x1 C 1 , x2 D x1 Thus f is one-to-one; Let y D f 1 .x/. 1 Then x D f .y/ D y C1 1 1 and y D f 1 .x/ D 1. so y C 1 D x x 1 D.f / D fx W x ¤ 0g D R.f /.

1

/.

x . If f .x1 / D f .x2 /, then 10. f .x/ D 1Cx x1 x2 D . Hence x1 .1 C x2 / D x2 .1 C x1 / 1 C x1 1 C x2 and, on simplification, x1 D x2 . Thus, f is one-to-one. y Let y D f 1 .x/. Then x D f .y/ D and 1Cy x D f 1 .x/. x.1 C y/ D y. Thus y D 1 x D.f / D R.f 1 / D . 1; 1/ [ . 1; 1/. R.f / D D.f 1 / D . 1; 1/ [ .1; 1/. 11.

1 2x : D.f / D fx W x ¤ 1g D R.f 1 / 1Cx 1 2x2 1 2x1 D f .x1 / D f .x2 / , 1 C x1 1 C x2 , 1 C x2 2x1 2x1 x2 D 1 C x1 2x2 2x1 x2 , 3x2 D 3x1 , x1 D x2 Thus f is one-to-one. Let y D f 1 .x/. 1 2y Then x D f .y/ D 1Cy so x C xy D 1 2y 1 x . and f 1 .x/ D y D 2Cx 1 D.f / D fx W x ¤ 2g D R.f /. f .x/ D

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

81

lOMoARcPSD|6566483

SECTION 3.1 (PAGE 171)

ADAMS and ESSEX: CALCULUS 9

x . If f .x1 / D f .x2 /, then x2 C 1 x1 x2 q D q : ./ 2 x1 C 1 x22 C 1

12. f .x/ D p

20.

Thus x12 .x22 C 1/ D x22 .x12 C 1/ and x12 D x22 . From (*), x1 and x2 must have the same sign. Hence, x1 D x2 and f is one-to-one. y , and Let y D f 1 .x/. Then x D f .y/ D p 2 y C1 x2 x 2 .y 2 C 1/ D y 2 . Hence y 2 D . Since f .y/ and y 1 x2 x , so have the same sign, we must have y D p 1 x2 x . f 1 .x/ D p 1 x2 1 D.f / D R.f / D . 1; 1/. R.f / D D.f 1 / D . 1; 1/.

21.

13. g.x/ D f .x/ 2 Let y D g 1 .x/. Then x D g.y/ D f .y/ 2, so f .y/ D x C 2 and g 1 .x/ D y D f 1 .x C 2/. 14.

15.

f .x/ D x 2 C 1 if x  0, and f .x/ D x C 1 if x < 0. If f .x1 / D f .x2 / then if x1  0 and x2  0 then x12 C 1 D x22 C 1 so x1 D x2 ; if x1  0 and x2 < 0 then x12 C 1 D x2 C 1 so x2 D x12 (not possible); if x1 < 0 and x2  0 then x1 D x22 (not possible); if x1 < 0 and x2 < 0 then x1 C 1 D x2 C 1 so x1 D x2 . Therefore f is Let y D f 1 .x/. Then  one-to-one. y 2 C 1 if y  0 x D f .y/ D y C 1  if y < 0. p 1 x 1 if x  1 Thus f .x/ D y D x 1 if x < 1. y

h.x/ D f .2x/. Let y D h 1 .x/. Then x D h.y/ D f .2y/ and 2y D f 1 .x/. Thus h 1 .x/ D y D 21 f 1 .x/.

1

y D f .x/

k.x/ D 3f .x/. Let y D k 1 .x/. Then x and x D k.y/ D 3f .y/, so f .y/ D 3  x 1 1 . k .x/ D y D f 3

16. m.x/ D f .x 2/. Let y D m 1 .x/. Then x D m.y/ D f .y 2/, and y 2 D f 1 .x/. Hence m 1 .x/ D y D f 1 .x/ C 2. 17.

1 C f .x/ . Let y D s 1 .x/. 1 f .x/ 1 C f .y/ Then x D s.y/ D . Solving for f .y/ we obtain 1 f .y/   x 1 x 1 . Hence s 1 .x/ D y D f 1 . f .y/ D xC1 xC1 s.x/ D

1 . Let y D p 1 .x/. p.x/ D 1 C f .x/ 1 1 so f .y/ D Then x D p.y/ D 1 C f .y/ x  1 1 . and p 1 .x/ D y D f 1 x

x

Fig. 3.1-21 22.

g.x/ D x 3 if x  0, and g.x/ D x 1=3 if x < 0. Suppose f .x1 / D f .x2 /. If x1  0 and x2  0 then x13 D x23 so x1 D x2 . Similarly, x1 D x2 if both are negative. If x1 and x2 have opposite sign, then so do g.x1 / and g.x2 /. Therefore g is Let y D g 1 .x/. Then  one-to-one. 3 y if y  0 x D g.y/ D y 1=3 if y < 0. 1=3 if x  0 Thus g 1 .x/ D y D x 3 x if x < 0.

23.

If x1 and x2 are both positive or both negative, and h.x1 / D h.x2 /, then x12 D x22 so x1 D x2 . If x1 and x2 have opposite sign, then h.x1 / and h.x2 / are on opposite sides of 1, so cannot be equal. Hence h is one-to-one. y2 C 1 if y  0 . If If y D h 1 .x/, then x D h.y/ D 2 y C 1 if y < 0 p p y  0, then y D px 1. If y < 0, then y D 1 x. Thus h 1 .x/ D px 1 if x  1 1 x if x < 1

24.

y D f 1 .x/ , x D f .y/ D y 3 C y. To find y D f 1 .2/ we solve y 3 C y D 2 for y. Evidently y D 1 is the only solution, so f 1 .2/ D 1.

1,

f .x/ 3 Let y D q 1 .x/. Then 2 f .y/ 3 and f .y/ D 2x C 3. Hence x D q.y/ D 2 1 1 q .x/ D y D f .2x C 3/.

18. q.x/ D

19. r.x/ D 1 2f .3 4x/ Let y D r 1 .x/. Then x D r.y/ D 1 f .3 3

4y/ D

f

x 2  1

4y D f

 1 and r 1 .x/ D y D 3 4

82

1

1



2f .3

1

x 2

1

x 2



 .

4y/.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 3.1 (PAGE 171)

25. g.x/ D 1 if x 3 C x D 10, that is, if x D 2. Thus g 1 .1/ D 2. 26. h.x/ D 3 if xjxj D h 1 . 3/ D 2. 27.

If y D f

4, that is, if x D

31.

2. Thus 32.

1

.x/ then x D f .y/. dy dy 1 1 Thus 1 D f 0 .y/ so D 0 D Dy 1 dx dx f .y/ y (since f 0 .x/ D 1=x).

dy dx D .2 C cos y/ dx dx ˇ dy ˇˇ 1 1 0 .g / .2/ D  0:36036: D ˇ dx ˇ 2 C cos y

1D

xD2

4x 3 , then x2 C 1

4x 2 .x 2 C 3/ .x 2 C 1/.12x 2 / 4x 3 .2x/ D : f .x/ D 2 2 .x C 1/ .x 2 C 1/2

33.

0

Since f 0 .x/ > 0 for all x, except x D 0, f must be oneto-one and so it has an inverse. 4y 3 , and If y D f 1 .x/, then x D f .y/ D 2 y C1 1 D f 0 .y/ D

.y 2 C 1/2 . Since f .1/ D 2, therefore 4y 4 C 12y 2 1 f .2/ D 1 and f

ˇ .y 2 C 1/2 ˇˇ .2/ D 4 ˇ 4y C 12y 2 ˇ

yD1

p 30. If f .x/ D x 3pC x 2 and y D f x D f .y/ D y 3 C y 2 , so, p 2yy 0 1 D y 3 C y2 C y p 2 3 C y2 0

Since f . 1/ D 2 implies that f f


1 0

If f .x/ D x sec x, then f 0 .x/ D sec x C x sec x tan x  1 for x in . =2; =2/. Thus f is increasing, and so oneto-one on that interval. Moreover, limx! .=2/C f .x/ D 1 and limx!.=2/C f .x/ D 1, so, being continuous, f has range . 1; 1/, and so f 1 has domain . 1; 1/. Since f .0/ D 0, we have f 1 .0/ D 0, and .f

.y 2 C 1/.12y 2 y 0 / 4y 3 .2yy 0 / : .y 2 C 1/2

Thus y 0 D


1 0

g.x/ D 2x C sin x ) g 0 .x/ D 2 C cos x  1 for all x. Therefore g is increasing, and so one-to-one and invertible on the whole real line. y D g 1 .x/ , x D g.y/ D 2y C sin y. For y D g 1 .2/, we need to solve 2y C sin y 2 D 0. The root is between 0 and 1; to five decimal places g 1 .2/ D y  0:68404. Also

28. f .x/ D 1 C 2x 3 Let y D f 1 .x/. Thus x D f .y/ D 1 C 2y 3 . 1 dy 1 dy D so .f 1 /0 .x/ D D 1 D 6y 2 dx dx 6y 2 6Œf 1 .x/2 29. If f .x/ D

p y D f 1 .2/ , 2 D f .y/ D y 2 =.1 C y/. We must solve p 2 2 C 2 y D y for y. There is a root between 2 and 3: f 1 .2/  2:23362 to 5 decimal places.

1

D

1 : 4

35.

.x/, then

) 1

0

y D

. 2/ D

ˇ p 3 C y 2 ˇˇ . 2/ D ˇ 3 C 2y 2 ˇ

yD 1

D

34.

p

3 C y2 : 3 C 2y 2

1, we have 2 : 5

p Note: f .x/ D x 3 C x 2 D 2 ) x 2 .3 C x 2 / D 4 4 2 ) x C 3x 4 D 0 ) .x 2 C 4/.x 2 1/ D 0. 2 Since .x C 4/ D 0 has no real solution, therefore x 2 1 D 0 and x D 1 or 1. Since it is given that f .x/ D 2, therefore x must be 1.

1 0

/ .0/ D

1 f 0 .f

1 .0/

D

1 D 1: f 0 .0/

If y D .f ı g/ 1 .x/, then x D f ı g.y/ D f .g.y//. Thus g.y/ D f 1 .x/ and y D g 1 .f 1 .x// D g 1 ı f 1 .x/. That is, .f ı g/ 1 D g 1 ı f 1 . x a bx c y a and Let y D f 1 .x/. Then x D f .y/ D by c cx a bxy cx D y a so y D . We have bx 1 cx a x a D , which simplifies to f 1 .x/ D f .x/ if bx c bx 1 f .x/ D

b.1

c/x 2 C .c 2

1/x C a..1

c/  0:

This equation is satisfied for (almost) all values of x provided that c D 1 a and b arbitrary, or c D 1 and a D b D 0. In the latter case f .x/  x, which is certainly self-inverse. For the former case, f will be self-inverse provided f is one-to-one and so has an inverse. Since, for c D 1, ab 1 f 0 .x/ D .bx 1/2 is positive (or negative) for all x ¤ Ê1=b provided ab > 1 (or ab < 1), we have that f is self-inverse in c D 1, and a and b are arbitrary except that ab ¤ 1.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

83

lOMoARcPSD|6566483

SECTION 3.1 (PAGE 171)

ADAMS and ESSEX: CALCULUS 9

36. Let f .x/ be an even function. Then f .x/ D f . x/. Hence, f is not one-to-one and it is not invertible. Therefore, it cannot be self-inverse. An odd function g.x/ may be self-inverse if its graph is symmetric about the line x D y. Examples are g.x/ D x and g.x/ D 1=x.

37. No. A function that is one-to-one on a single interval need not be either increasing or decreasing. For example, consider the function defined on Œ0; 2 by f .x/ D



if 0  x  1 if 1 < x  2.

x x

7.

log1=3 32x D log1=3

8.

43=2 D 8

9.

10

D

2x

D 3 2

2x )

p 2log4 8 D 23=2 D 2 2

1 Dx 1=x


10. Since loga x 1=.loga x/ D x 1=.loga x/ D a1 D a.

1 loga x D 1, therefore loga x

.loga b/.logb a/ D loga a D 1   12. logx x.logy y 2 / D logx .2x/ D logx x C logx 2 11.

It is one-to-one but neither increasing nor decreasing on all of Œ0; 2. 38. First we consider the case where the domain of f is a closed interval. Suppose that f is one-to-one and continuous on Œa; b, and that f .a/ < f .b/. We show that f must be increasing on Œa; b. Suppose not. Then there are numbers x1 and x2 with a  x1 < x2  b and f .x1 / > f .x2 /. If f .x1 / > f .a/, let u be a number such that u < f .x1 /, f .x2 / < u, and f .a/ < u. By the Intermediate-Value Theorem there exist numbers c1 in .a; x1 / and c2 in .x1 ; x2 / such that f .c1 / D u D f .c2 /, contradicting the one-to-oneness of f . A similar contradiction arises if f .x1 /  f .a/ because, in this case, f .x2 / < f .b/ and we can find c1 in .x1 ; x2 / and c2 in .x2 ; b/ such that f .c1 / D f .c2 /. Thus f must be increasing on Œa; b.

D 1 C logx 2 D 1 C

1 log2 x

1 D1 2

13.

.log4 16/.log4 2/ D 2 

14.

log15 75 C log15 3 D log15 225 D 2

.since 152 D 225/

15. 16.

17.

A similar argument shows that if f .a/ > f .b/, then f must be decreasing on Œa; b.

1.

log4 8 D

)

log10 .1=x/

  1 3

log6 9 C log6 4 D log6 36 D 2  2 2 4
3 2 log3 12 4 log3 6 D log3 24
34 D log3 .3 2 / D 2 loga .x 4 C 3x 2 C 2/ C loga .x 4 C 5x 2 C 6/ p 4 loga x 2 C 2     D loga .x 2 C 2/.x 2 C 1/ C loga .x 2 C 2/.x 2 C 3/ 2 log1 .x 2 C 2/

Finally, if the interval I where f is defined is not necessarily closed, the same argument shows that if Œa; b is a subinterval of I on which f is increasing (or decreasing), then f must also be increasing (or decreasing) on any intervals of either of the forms Œx1 ; b or Œa; x2 , where x1 and x2 are in I and x1  a < b  x2 . So f must be increasing (or decreasing) on the whole of I .

18.

Section 3.2 Exponential and Logarithmic Functions (page 175)

19.

p

20.

log3 5 D .log10 5/=.log10 3  1:46497

21.

22x D 5xC1 , 2x log10 2 D .x C 1/ log10 5, x D .log10 5/=.2 log10 2 log10 5/  7:21257 p p x D log10 3, x 2 D 3, 2 log p 10 x D 10.log10 3/= 2  2:17458

33 p D 33 35

2. 2

1=2 1=2

3. .x

D2

8

3

/

2

5=2

Dx

4. . 21 /x 4x=2

D 31=2 D

1=2 3=2

2

6

D loga .x 4 C 4x 2 C 3/

3

2

D2 D4 22.

2x D x D1 2

5. log5 125 D log5 53 D 3

6. If log4 . 18 / D y then 4y D 18 , or 22y D 2 2y D 3 and log4 . 18 / D y D 32 .

84

D loga .x 2 C 1/ C loga .x 2 C 3/

3

. Thus

log .1

cos x/ C log .1 C cos x/ 2 log sin x  sin2 x .1 cos x/.1 C cos x/ D log D log  sin2 x sin2 x D log 1 D 0 p p 2 yD3 p , log10 y D 2 log10 3, y D 10 2 log10 3  4:72880 

23.

logx 3 D 5, .log10 3/=.log10 x/ D 5, log10 x D .log10 3/=5, x D 10.log10 3/=5  1:24573

24.

log3 x D 5, .log10 x/=.log10 3/ D 5, log10 x D 5 log10 3, x D 105 log10 3 D 35 D 243

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

  1 1 then au D D x x x and u D  logax. 1 D loga x. Thus, loga x

25. Let u D loga

26. Let loga x D u, loga y D v. Then x D au , y D av . au x D v D au v Thus y  a x and loga D u v D loga x y 27.

1

SECTION 3.3 (PAGE 183)

. Hence, a

u

Dx

35.

h!0

1 h

D k. Thus

axCh ax h h!0 ax ah ax D lim h h!0 h a 1 D ax f 0 .0/ D ax k D kf .x/: D ax lim h h!0

f 0 .x/ D lim

loga y. 36.

Let u D loga .x y /, then au D x y and au=y D x. u D loga x, or u D y loga x. Therefore y y Thus, loga .x / D y loga x.

1

yDf

Thus .f

28. Let logb x D u, logb a D v. Thus b u D x and b v D a. Therefore x D b u D b v.u=v/ D au=v u logb x and loga x D D . v logb a 29.

ah

f .x/ D ax and f 0 .0/ D lim

.x/ ) x D f .y/ D ay dx dy )1D D kay dx dx 1 1 dy D D : ) dx kay kx 1 0

/ .x/ D 1=.kx/.

Section 3.3 The Natural Logarithm and Exponential Functions (page 183)

1.

e3 p D e3 e5

2.

ln.e 1=2 e 2=3 / D

3.

e 5 ln x D x 5

4.

e .3 ln 9/=2 D 93=2 D 27

5.

ln

log3 x 2 C log3 x 1=2 D 10

6.

x 5=2 D 310 ; so x D .310 /2=5 D 34 D 81

 2 e 2 ln cos x C ln e sin x D cos2 x C sin2 x D 1

7.

3 ln 4

8.

4 ln

9.

2 ln x C 5 ln.x

log4 .x C 4/

2 log4 .x C 1/ D

1 2

xC4 1 D 2 .x C 1/ 2 xC4 1=2 D4 D2 .x C 1/2 2x 2 C 3x 2 D 0 but we need x C 1 > 0, so x D 1=2. log4

30. First observe that log9 x D log3 x= log3 9 D 2 log3 x C log9 x D 10

1 2

log3 x. Now

log3 x 5=2 D 10

31. Note that logx 2 D 1= log2 x. Since limx!1 log2 x D 1, therefore limx!1 logx 2 D 0. 32. Note that logx .1=2/ D logx 2 D 1= log2 x. Since limx!0C log2 x D 1, therefore limx!0C logx .1=2/ D 0. 33. Note that logx 2 D 1= log2 x. Since limx!1C log2 x D 0C, therefore limx!1C logx 2 D 1. 34. Note that logx 2 D 1= log2 x. Since limx!1 log2 x D 0 , therefore limx!1 logx 2 D 1.

5=2

1 D ln e e 3x

p

p

D e 1=2 D 1 2

C

3x

D

4 ln 3 D ln

2 3

D

e

7 6

3x

64 43 D ln 34 81

x C 6 ln.x 1=3 / D 2 ln x C 2 ln x D 4 ln x  2/ D ln x 2 .x

2/5



10. ln.x 2 C 6x C 9/ D lnŒ.x C 3/2  D 2 ln.x C 3/ 11.

2xC1 D 3x .x C 1/ ln 2 D x ln 3 ln 2 ln 2 xD D ln 3 ln 2 ln.3=2/

12.

3x D 91 )

x

) 3x D 32.1

x D 2.1

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

x/

)

x/

xD

2 3

85

lOMoARcPSD|6566483

SECTION 3.3 (PAGE 183)

13.

14.

ADAMS and ESSEX: CALCULUS 9

5 1 D xC3 2x 8 x ln 2 D ln 5 .x C 3/ ln 8 D ln 5 .3x C 9/ ln 2 2x ln 2 D ln 5 9 ln 2 ln 5 9 ln 2 xD 2 ln 2 2x

2

3

2

D 4x D 22x ) x 2

y D ln ln x

34.

y D x ln x

x   1 y D ln x C x x 0

35.

3 D 2x

x 2x 3 D 0 ) .x Hence; x D 1 or 3:

3/.x C 1/ D 0

x2 2 x2 0 y D 2x ln x C x

36.

y D ln j sin xj;

37.

y D 52xC1

16. ln.x 2 x 2/ D lnŒ.x 2/.x C 1/ is defined if .x 2/.x C 1/ > 0, that is, if x < 1 or x > 2. The domain is the union . 1; 1/ [ .2; 1/.

38.

17.

ln.2x 5/ > ln.7 2x/ holds if 2x 5 > 0, 7 2x > 0, and 2x 5 > 7 2x, that is, if x > 5=2, x < 7=2, and 4x > 12 (i.e., x > 3). The solution set is the interval .3; 7=2/.

2 2 18. ln.x 2 2/  ln x holds 2  x. p if x > 2, x > 0, and x Thus we need x > 2 and x 2 x 2  0. This latter inequality says that .x 2/.x C 1/  0, so it holds for 1  x  2. The solution set of the given inequality is p . 2; 2.

39. 40. 41.

D .1

2x/e

22. y D x 2 e x=2 ; 23. y D ln.3x

42.

24.

y D ln j3x

y0 D

2j;

0

3x 2 ex y0 D 1 C ex

2

27.

yD

ex C e 2

f 0 .x/ D .2x/e x

x

y0 D

;

x

30. y D

32. y D e

86

y 0 D e x e .e

ex D1 1 C ex

31. y D e x sin x; x

ex

e

44.

2

x

2

45.

dx 1 D 3e 3t ln t C e 3t dt t

28. x D e 3t ln t; 29. y D e .e / ;

2 3

cos x;

x/

D e xCe

1 ; 1 C ex

ex .1 C e x /2

y 0 D e x .sin x C cos x/ y0 D

e

x

cos x

e

3xC8/

h.t / D t x

xt ;

y 0 D .2x

;

x

sin x

46.

3/.ln 2/2.x

g 0 .x/ D t x x t ln t C t xC1 x t h0 .t / D xt x

f .s/ D loga .bs C c/ D

1

2

3xC8/

1

x t ln x

ln.bs C c/ ln a

b .bs C c/ ln a

ln.2x C 3/ g.x/ D logx .2x C 3/ D ln x     2 1 ln x Œln.2x C 3/ 2x C 3 x 0 g .x/ D .ln x/2 2x ln x .2x C 3/ ln.2x C 3/ D x.2x C 3/.ln x/2 yDx

p

x

p

De 

x ln x

p  ln x x p C x 2 x    p 1 1 ln x C 1 Dx x p x 2  ln x 1 Given that y D , let u D ln x. Then x D e u and x  u 1 2 yD D .e u /u D e u . Hence, eu     1 dy du 2 ln x 1 ln x dy 2 : D
D . 2ue u / D dx du dx x x x p

x ln x

y D ln j sec x C tan xj

sec x tan x C sec2 x sec x C tan x D sec x p y D ln jx C x 2 a2 j 2x 1C p 2p x 2 a2 D p 1 y0 D x C x 2 a2 x 2 a2 y0 D

x

y0 D

2

g.x/ D t x x t ;

y0 D e

3 3x

y D

25. y D ln.1 C e x / 26. f .x/ D e x ;

43.

y 0 D 2xe x=2 C 12 x 2 e x=2 2/

y D 2.x

f 0 .s/ D

1

2x

2x D 2x ln x 2 cos x D cot x y0 D sin x

y 0 D 2.52xC1 / ln 5 D .2 ln 5/52xC1

19. y D e 5x ; y 0 D 5e 5x

20. y D xe x x; y 0 D e x C xe x x 21. y D 2x D xe 2x e y 0 D e 2x 2xe 2x

1 D ln x

y D x 2 ln x

ln.x=.2 x// is defined if x=.2 x/ > 0, that is, if 0 < x < 2. The domain is the interval .0; 2/.

15.

1 x ln x

y0 D

33.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

47.

48.

p y D ln. x 2 C a2 x/ x 1 p x 2 C a2 0 y D p x 2 C a2 x 1 D p x 2 C a2 x

y D .cos x/

x

cos x

53.

b)

De

x ln cos x

x

f .x/ D .x x /x D x .x



x tan x/

f 0 D xx

54.

f 00 .x/ D e ax .2a C a2 x/

Given that x x

f 000 .x/ D e ax .3a2 C a3 x/ :: : 1

2 C1

xx

.2 ln x C 1/

!

f .x/ D xe ax f 0 .x/ D e ax .1 C ax/

f .n/ .x/ D e ax .nan

2/

g.x/ D x ln g D x x ln x xx 1 0 g D x x .1 C ln x/ ln x C 0 g x   1 0 xx x g Dx x C ln x C .ln x/2 x Evidently g grows more rapidly than does f as x grows large.

.cos x/.ln x/

1 sin x ln x C cos x x

cos x

a)

ln f .x/ D x 2 ln x 1 0 f D 2x ln x C x f

e #  1 0 x ln cos x . sin x/ ln cos x C x y De cos x " # 1 .cos x/.ln x/ sin x ln x C cos x e x "

D .cos x/x .ln cos x

49.

SECTION 3.3 (PAGE 183)

:: x:

D a where a > 0, then ln a D x x

C an x/

Thus ln x D

50. Since

:: x:

ln x D a ln x:

1 ln a D ln a1=a , so x D a1=a . a

d .ax 2 C bx C c/e x D .2ax C b/e x C .ax 2 C bx C c/e x dx D Œax 2 C .2a C b/x C .b C c/e x D ŒAx 2 C Bx C C e x : 2

55.

x

Thus, differentiating .ax C bx C c/e produces another function of the same type with different constants. Any number of differentiations will do likewise. 51.

y D ex

2

y 0 D 2xe x

2

2

2

y 00 D 2e x C 4x 2 e x D 2.1 C 2x 2 /e x 2

2

2

y .4/ D 4.3 C 6x 2 /e x C 4.3x C 2x 3 /2xe x

52.

f .x/ D ln.2x C 1/ 00

2

f .x/ D . 1/2 .2x C 1/ f .4/ .x/ D

.3Š/24 .2x C 1/

2

2

2

56. f 0 .x/ D 2.2x C 1/

2

000

4

3

1

f .x/ D .2/2 .2x C 1/

Thus, if n D 1; 2; 3; : : : we have f .n/ .x/ D . 1/n 1 .n 1/Š2n .2x C 1/

4/

2

y 000 D 2.4x/e x C 2.1 C 2x 2 /2xe x D 4.3x C 2x 3 /e x D 4.3 C 12x 2 C 4x 4 /e x

f .x/ D .x 1/.x 2/.x 3/.x 4/ ln f .x/ D ln.x 1/ C ln.x 2/ C ln.x 3/ C ln.x 1 1 1 1 1 f 0 .x/ D C C C f .x/ x 1 x 2 x 3 x 4   1 1 1 1 C C C f 0 .x/ D f .x/ x 1 x 2 x 3 x 4

n

.

3

p

1 C x.1 x/1=3 .1 C 5x/4=5 1 ln F .x/ D 2 ln.1 C x/ C 13 ln.1 x/ 54 ln.1 C 5x/ 1 1 4 F 0 .x/ D F .x/ 2.1 C x/ 3.1 x/ .1 C 5x/ # "   1 1 4 23 1 1 0 4 D F .0/ D F .0/ D .1/ 2 3 1 2 3 6

F .x/ D

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

87

lOMoARcPSD|6566483

SECTION 3.3 (PAGE 183)

57.

ADAMS and ESSEX: CALCULUS 9

.x 2 1/.x 2 2/.x 2 3/ .x 2 C 1/.x 2 C 2/.x 2 C 3/ 321 1 f .2/ D D ; f .1/ D 0 567 35 2 2 ln f .x/ D ln.x 1/ C ln.x 2/ C ln.x 2

61.

f .x/ D

2

2

3/

xDa

2

ln.x C 1/ ln.x C 2/ ln.x C 3/ 1 2x 2x 2x 0 f .x/ D 2 C 2 C 2 f .x/ x 1 x 2 x 3 2x 2x 2x x2 C 1  x2 C 2 x2 C 3 1 1 1 C 2 C 2 f 0 .x/ D 2xf .x/ 2 x 1 x 2 x 3  1 1 1 x2 C 1 x2 C 2 x2 C 3   4 1 1 1 1 1 1 f 0 .2/ D C C 35 3 2 1 5 6 7 139 556 4  D D 35 105 3675 2 Since f .x/ D .x 1/g.x/ where g.1/ ¤ 0, then f 0 .x/ D 2xg.x/ C .x 2 1/g 0 .x/ and 1 . 1/. 2/ D . f 0 .1/ D 2g.1/ C 0 D 2  234 6 x2

58. Since y D x 2 e y 0 D 2xe

x2

Let the point of tangency be .a; e a /. Tangent line has slope ˇ d x ˇˇ ea 0 D e ˇ D ea : a 0 dx ˇ

Therefore, a D 1 and line has slope e. The line has equation y D ex. y

.a;e a /

y D ex x

Fig. 3.3-61 62.

, then

ˇ ˇ 1 1 ˇ The slope of y D ln x at x D a is y 0 D ˇ D : The xˇ a xDa line from .0; 0/ to .a; ln a/ is tangent to y D ln x if ln a 0 1 D a 0 a

2x 3 e

x2

D 2x.1

x/.1 C x/e

The tangent is horizontal at .0; 0/ and

x2

:

i.e., if ln a D 1, or a D e. Thus, the line is y D

  1 . ˙1; e

y

x . e

.a; ln a/ 59. f .x/ D xe f 0 .x/ D e

x x

.1

x/,

C.P. x D 1, f .1/ D

f 0 .x/ > 0 if x < 1 (f increasing) f 0 .x/ < 0 if x > 1 (f decreasing)

x

1 e

y D ln x Fig. 3.3-62

y .1;1=e/ y

Dxe

x

63. x

Let the point of tangency be .a; 2a /. Slope of the tangent is ˇ d x ˇˇ 2a 0 D 2 ˇ D 2a ln 2: a 1 dx ˇ xDa

1 1 ; a D1C . Thus a 1 D ln 2 ln 2 So the slope is 2a ln 2 D 21C.1= ln 2/ ln 2 D 2e ln 2. 1 (Note: ln 21= ln 2 D ln 2 D 1 ) 21= ln 2 D e) ln 2 The tangent line has equation y D 2e ln 2.x 1/.

Fig. 3.3-59 64. 1 D 4 then x D 14 and x y D ln 41 D ln 4. The tangent line of slope 4 is y D ln 4 C 4.x 14 /, i.e., y D 4x 1 ln 4.

60. Since y D ln x and y 0 D

88

The tangent line to y D ax which passes through the origin is tangent at the point .b; ab / where ˇ d x ˇˇ ab 0 D a ˇ D ab ln a: b 0 dx ˇ

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

xDb

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 3.3 (PAGE 183)

1 D ln a, so ab D a1= ln a D e. The line y D x b will intersect y D ax provided the slope of this tangent e line does not exceed 1, i.e., provided  1, or e ln a  1. b Thus we need a  e 1=e . Thus

70.

y

(a) If Aa CBb D 1 and Ba Ab D 0, then A D

.b; ab /

and B D Z

y D ax x

1 x DxC y y   y y xy 0 x D1 e xy .y C xy 0 / ln C e xy y x y2   1 At e; we have e   1 1 C ey 0 2 C e 2 .e e 3 y 0 / D 1 e 2 y 0 e e e 2 C 2e 2 y 0 C 1 e 2 y 0 D 1 e 2 y 0 . 1 Thus the slope is y 0 D . e2

b (b) If Aa CBb D 0 and Ba Ab D 1, then A D 2 a C b2 a and B D 2 . Thus a C b2

65. e xy ln

1 0 y y2

Z

e ax sin bx dx  1 ae ax sin bx D 2 2 a Cb

66. xe y C y 2x D ln 2 ) e y C xe y y 0 C y 0 2 D 0: At .1; ln 2/, 2 C 2y 0 C y 0 2 D 0 ) y 0 D 0. Therefore, the tangent line is y D ln 2.

68.

b . Thus a2 C b 2

a a2 C b 2

e ax cos bx dx   1 ae ax cos bx C be ax sin bx C C: D 2 2 a Cb

Fig. 3.3-64

67.

d .Ae ax cos bx C Be ax sin bx/ dx D Aae ax cos bx Abe ax sin bx C Bae ax sin bx C Bbe ax cos bx D .Aa C Bb/e ax cos bx C .Ba Ab/e ax sin bx:

f .x/ D Ax cos ln x C Bx sin ln x f 0 .x/ D A cos ln x A sin ln x C B sin ln x C B cos ln x D .A C B/ cos ln x C .B A/ sin ln x 1 If A D B D then f 0 .x/ D cos ln x. Z 2 1 1 Therefore cos ln x dx D x cos ln x C x sin ln x C C . 2 2 1 1 If B D , A D then f 0 .x/ D sin ln x. 2 Z 2 1 1 x cos ln x C C . Therefore sin ln x dx D x sin ln x 2 2 FA;B .x/ D Ae x cos x C Be x sin x d FA;B .x/ dx D Ae x cos x Ae x sin x C Be x sin x C Be x cos x D .A C B/e x cos x C .B A/e x sin x D FACB;B A .x/

71.

 be ax cos bx C C:

    1 1 d 1 1 C D ln C ln x D dx x 1=x x 2 x

1 1 C D 0: x x

1 C ln x D C (constant). Taking x D 1, we x 1 get C D ln 1 C ln 1 D 0. Thus ln D ln x. x   x 1 1 ln D ln x D ln x C ln D ln x ln y: y y y

Therefore ln

72.

73.

r rx r 1 r r d Œln.x r / r ln x D D D 0. dx xr x x x Therefore ln.x r / r ln x D C (constant). Taking x D 1, we get C D ln 1 r ln 1 D 0 0 D 0. Thus ln.x r / D r ln x.

74. Let x > 0, and F .x/ be the area bounded by y D t 2 , the t -axis, t D 0 and t D x. For h > 0, F .x C h/ F .x/ is the shaded area in the following figure. y

d FA;B .x/ D FACB;B A .x/ we have 69. Since dx 2 d d FA;B .x/ D FACB;B A .x/ D F2B; 2A .x/ a) dx 2 dx d3 x d3 d b) e cos x D F1;0 .x/ D F0; 2 .x/ 3 3 dx dx dx x x D F 2; 2 .x/ D 2e cos x 2e sin x

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

y D t2

x

xCh

t

Fig. 3.3-74

89

lOMoARcPSD|6566483

SECTION 3.3 (PAGE 183)

ADAMS and ESSEX: CALCULUS 9

c) The tangent to y D 1=t at t D 2 has slope equation is

Comparing this area with that of the two rectangles, we see that hx 2 < F .x C h/

F .x/ < h.x C h/2 :

Hence, the Newton quotient for F .x/ satisfies F .x C h/ x < h 2

F .x/

h!0C

F .x C h/ h

< .x C h/ :

F .x/

yD

1 3

F .x C h/ h

F .x/

h!0

F .x C h/ h

F .x/

< x2;

A2 D

D x2:

1

1

t

A2 2

0

lim x 3 e

x

x!1

2.

lim x

x!1



1 2



D lim

x : 9

3 1 C 4 2



D

5 : 8

4 1 C 9 3



D

7 : 18

x!1

3 x

e D lim

x!1

ex D1 x3

2e x 3 2 3e D lim x!1 e x C 5 x!1 1 C 5e

4.

x 2e x!1 x C 3e

lim

lim

x x

D lim

lim x ln x D 0 .power wins/

x!0C t

6. 2

b) If f .t / D 1=t , then f .t / D 1=t and f 00 .t / D 2=t 3 > 0 for t > 0. Thus f 0 .t / is an increasing function of t for t > 0, and so the graph of f .t / bends upward away from any of its tangent lines. (This kind of argument will be explored further in Chapter 5.)

lim

x!0C

7.

ln x D x

1

lim x.ln jxj/2 D 0

x!0

8.

D

2 0 D2 1C0

1 2=.xe x / 1 0 D D1 x!1 1 C 3=.xe x / 1C0

x x

(page 191)

x3 D 0 (exponential wins) ex

3.

5. 3

Fig. 3.3-75

90

1.

yD1=t

2

1 2

Section 3.4 Growth and Decay

a) The shaded area A in part (i) of the figure is less than the area of the rectangle (actually a square) with base from t D 1 to t D 2 and height 1=1 D 1. Since ln 2 D A < 1, we have 2 < e 1 D e; i.e., e > 2. (i) (ii) y y

A1

2 3

7 73 5 C D > 1. 8 18 72 1 Thus 3 > e D e. Combining this with the result of (a) we conclude that 2 < e < 3.

F .0/ D C D 0, therefore F .x/ D 31 x 3 . For x D 2, the area of the region is F .2/ D 83 square units.

A

3/ or y D

1=9. Its

e) ln 3 > A1 C A2 D

F .x C h/ F .x/ d F .x/ D lim D x2: dx h h!0 Z Therefore F .x/ D x 2 dx D 31 x 3 C C . Since

yD1=t

x : 4

The trapezoid bounded by x D 2, x D 3, y D 0, and y D .2=3/ .x=9/ has area

Combining these two limits, we obtain

75.

1 .x 9

A1 D

so similarly, lim

or y D 1

d) The trapezoid bounded by x D 1, x D 2, y D 0, and y D 1 .x=4/ has area

D x2:

If h < 0 and 0 < x C h < x, then .x C h/2
5 we have dx=dt D k2 x, so that x.t / D x.5/e k2 .t

5/

x.7/ D x.5/e 2k2 D 3;000

÷

 5=2 x.10/ D x.5/35k2 D x.5/ e 2k2  142:

3;000 22;918   3;000 5=2  22;918 22;918

25.

e 2k2 

so there are approximately 142 rabbits left after 10 years. 22. Let N.t / be the number of rats on the island t months after the initial population was released and before the first cull. Thus N.0/ D R and N.3/ D 2R. Since N.t / D Re k t , we have e 3k D 2, so e k D 21=3 . Hence N.5/ D Re 5k D 25=3 R. After the first 1,000 rats are killed the number remaining is 25=3 R 1;000. If this number is less than R, the number at the end of succeeding 5-year periods will decline. The minimum value of R for which this won’t happen must satisfy 25=3 R 1;000 D R, that is, R D 1;000=.25=3 1/  459:8. Thus R D 460 rats should be brought to the island initially.

dx D a bx.t /. dt This says that x.t / is increasing if it is less than a=b and decreasing if it is greater than a=b. Thus, the limiting concentration is a=b.

a) The concentration x.t / satisfies

b) The differential equation for x.t / resembles that of Exercise 21(b), except that y.x/ is replaced by x.t /, and b is replaced by b. Using the result of Exercise 21(b), we obtain, since x.0/ D 0,  a a  bt e C x.t / D x.0/ b  b a D 1 e bt : b

x.t / D x.0/e k1 t D 1;000e k1 t

x.5/ D 1;000e 5k1

26.

Let T .t / be the reading t minutes after the Thermometer is moved outdoors. Thus T .0/ D 72, T .1/ D 48. dT D k.T 20/. By Newton’s law of cooling, dt dV If V .t / D T .t / 20, then D kV , so dt kt kt V .t / D V .0/e D 52e . Also 28 D V .1/ D 52e k , so k D ln.7=13/. Thus V .5/ D 52e 5 ln.7=13/  2:354. At t D 5 the thermometer reads about T .5/ D 20 C 2:354 D 22:35 ı C. Let T .t / be the temperature of the object t minutes after its temperature was 45 ı C. Thus T .0/ D 45 and dT D k.T C 5/. Let T .40/ D 20. Also dt u.t / D T .t / C 5, so u.0/ D 50, u.40/ D 25, and dT du D D k.T C 5/ D ku. Thus, dt dt u.t / D 50e k t ;

23. f 0 .x/ D a C bf .x/.

92

a : b

b) If

in one year, if compounding is performed n times per year. For i D 9:5 and n D 12, we have

21.

 a D C e bx

25 D u.40/ D 50e 40k ; 1 25 1 1 )k D ln D ln : 40 50 40 2

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 3.5 (PAGE 199)

We wish to know t such that T .t / D 0, i.e., u.t / D 5, hence 5 D u.t / D 50e k t   5 40 ln 50   D 132:88 min: tD 1 ln 2 Hence, it will take about .132:88 more to cool to 0 ı C. 27.

30.

40/ D 92:88 minutes

Let T .t / be the temperature of the body t minutes after it was 5 ı . Thus T .0/ D 5, T .4/ D 10. Room temperature = 20ı . dT By Newton’s law of cooling (warming) D k.T 20/. dt dV If V .t / D T .t / 20 then D kV , dt so V .t / D V .0/e k t D 15e k t .   2 1 . Also 10 D V .4/ D 15e 4k , so k D ln 4 3 kt If T .t / D 15ı , then 5 D  V.t / D 15e 1   ln 1 1 3 D 4    10:838. so t D ln 2 k 3 ln 3 It will take a further 6.84 minutes to warm to 15 ı C.

31.

Ly0 y0 C .L y0 /e

k

y2 D

;

Ly0 y0 C .L y0 /e

Thus y1 .L y0 /e k D .L y1 /y0 , and y2 .L y0 /e 2k D .L y2 /y0 . Square the first equation and thus eliminate e 

.L y1 /y0 y1 .L y0 /

2

D

32.

t!1

k

:

y12 .y0 C y2 / 2y0 y1 y2 . y12 y0 y2 If y0 D 3, y1 D 5, y2 D 6, then 25.9/ 180 45 LD D  6:429. 25 18 7

29. The rate of growth of y in the logistic equation is

Since dy D dt

 k y L

L 2

2

C

kL ; 4

L 1 C Me k t L 200 D y.0/ D 1CM L 1; 000 D y.1/ D 1 C Me k 10; 000 D lim y.t / D L y.t / D

Thus 200.1 C M / D L D 10; 000, so M D 49. Also 1; 000.1 C 49e k / D L D 10; 000, so e k D 9=49 and k D ln.49=9/  1:695. 33.

L 10; 000 D  7671 cases 1 C 49.9=49/3 1 C Me 3k LkMe 3k  3; 028 cases/week: y 0 .3/ D .1 C Me 3k /2 y.3/ D

Section 3.5 The Inverse Trigonometric Functions (page 199)

Assuming L ¤ 0, L D

dy D ky 1 dt

kt

Assuming k and L are positive, but y0 is negative, we have t  > 0. The solution is therefore valid on . 1; t  /. The solution approaches 1 as t ! t  .

2k

.L y2 /y0 y2 .L y0 /

 y : L

Ly0 y0 C .L y0 /e

of the logistic equation is valid on any interval containing t D 0 and not containing any point where the denominator is zero. The denominator is zero if y0 D .y0 L/e k t , that is, if   y0 1  ln : t Dt D k y0 L

Now simplify: y0 y2 .L y1 /2 D y12 .L y0 /.L y2 / y0 y2 L2 2y1 y0 y2 LCy0 y12 y2 D y12 L2 y12 .y0 Cy2 /LCy0 y12 y2



The solution yD

28. By the solution given for the logistic equation, we have y1 D

L dy is greatest when y D . dt 2 Ly0 The solution y D is valid on the y0 C .L y0 /e k t largest interval containing t D 0 on which the denominator does not vanish. If y0 > L then y0 C .L y0 /e k t D 0 if y0 1 ln . t D t D k y0 L Then the solution is valid on .t  ; 1/. limt!t  C y.t / D 1. thus

p

3  D 2 3   2 1 D 2 3

1.

sin

1

2.

cos

1

3.

tan

1

4.

sec

1

5.

sin.sin

 4

. 1/ D p

2D

1

 4

0:7/ D 0:7

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

93

lOMoARcPSD|6566483

SECTION 3.5 (PAGE 199)

6.

1

cos.sin

q

0:7/ D

p

D tan

1

8. sin

1

7.

9.

10.

11.

cos



tan

1



 sin cos

1

 cos tan

sin.cos

14.

cos.sin

15.

cos.tan

18.

D tan

1

1



1 3/

.

1

1

17.

1

1

p

.

D p

.cos 40ı / D 50ı

1 2



q 1 q D 1

cos2 . arccos . 13 / p p 8 2 2 1 D D 9 3 3

D

D

D p 20.

1

 sec tan



1 2 1 D s  1 C tan2 tan 1

1

21.

22.

q x/ D 1 x/ D

x/ 24.

sin

1 sec.tan

2

1

sin

x/

1




x D

D p

p

1

cos.sec

94

1

25. x2

1

26.

27. p

1 x2 1 1 D 2 x jxj p ) tan.sec 1 x/ D x 2 1 sgn x p x2 1 if x  1 p D x 2 1 if x  1 1

x/ D

1 a b/ a2

1

a2

1

1

u D z 2 sec

1

x

xCp

1

f .t / D t tan

(assuming) a > 0/:

b/2

.x

f .x/ D x sin

tC 1

x x2

1

:

t t 1 C t2

.1 C z 2 /

1

.1 C z 2 / C

F .x/ D .1 C x 2 / tan

y D sin y0 D r

sin.cos x/ cos.cos 1 x p 1 x2 .by # 13/ D x 1 ) sin.sec x

a : 1 C .ax C b/2

2

.x

1

F .x/ D 2x tan

x/ D

x/ D

D p

y0 D

b a 1

0

1

r

x2

x

1

r

4x C 1/

z 2 .2z/ p .1 C .1 C z 2 /2 2 2z sgn .z/ p D 2z sec 1 .1 C z 2 / C .1 C z 2 / z 2 C 2

1C 1 ) cos. arctan x/ D p 1 C x2 x ) sin. arctan x/ D p 1 C x2

tan.cos

3 2

.ax C b/;

y D cos

1

tan. arctan x/ D x ) sec. arctan x/ D

2x

2Cx 1



2 2 3 1

.4x 2 1

9

du D 2z sec dz

x2

1 C x2 p

3



f 0 .t / D tan

x2

1

1

f .x/ D sin

2

23. cos2 .cos 1

2x

0

200/ D 200 p x/ D 1 p D 1



1

y D tan

y0 D

2  D p5 1

1

y D sin y0 D s

 3

3/ D

1

cos

19.

    sin. 0:2/ D sin 1 sin. 0:2/ 2  D C 0:2 2

1

16.



sin2 . arcsin 0:7/ p 0:49 D 0:51

1

.cos 40ı / D 90ı

12. tan.tan 13.

2 3

ADAMS and ESSEX: CALCULUS 9

1

a x

1  a 2

1

G.x/ D

1

x

1

z2/

x

xC1 h

.jxj > jaj/ a i a D p x2 jxj x 2

a2

sin 1 x sin 1 .2x/ sin

1

.2x/ p

1

sin

1

xp

2

1 4x 2 2 1 .2x/ p p 1 2 1 4x sin .2x/ 2 1 x 2 sin 1 x D  2 p p 1 x 2 1 4x 2 sin 1 .2x/

G 0 .x/ D

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

x2

1  sin

1

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

28.

sin 1 t sin  t

H.t / D

sin t 0

H .t / D

29.

30.

35.

p



1

t2 sin2 t

1

1 p .sin t / 1

D

SECTION 3.5 (PAGE 199)

1

sin

t cos t

csc t cot t sin

t2

1

t

f .x/ D .sin 1 x 2 /1=2 2x 1 f 0 .x/ D .sin 1 x 2 / 1=2 p 2 1 x4 x D p p 1 x 4 sin 1 x 2   a 1 y D cos p a2 C x 2 ! 1=2 " a 2 a2 0 .a C x 2 / y D 1 2 2 a Cx 2 asgn .x/ a2 C x 2 p y D a2 x 2 C a sin

36.

3=2

#

.2x/

37.

1 d sin 1 x D p > 0 on . 1; 1/. dx 1 x2 1 Therefore, sin is increasing. 1 d tan 1 x D > 0 on . 1; 1/: dx 1 C x2 1 Therefore tan is increasing. d 1 < 0 on . 1; 1/. cos 1 x D p dx 1 x2 1 Therefore cos is decreasing. Since the domain of sec 1 consists of two disjoint intervals . 1; 1 and Œ1; 1/, the fact that the derivative of sec 1 is positive wherever defined does not imply that sec 1 is increasing over its whole domain, only that it is increasing on each of those intervals taken independently. In fact, sec 1 . 1/ D  > 0 D sec 1 .1/ even though 1 < 1. d csc dx

1

D

31.

y0 D

p

D p

32.

x a2

a

x2 x

x2  y D a cos 1 1 0

y D

a2

"

a 1 x

D p

D

 1

1

1 a x a2

D

2

then p

1

1 x2

x, then y 0 D p

D 2 so that x D ˙

1 of the two tangentplines are   3  and y D y D C2 x 3 2



1 y

x

y D csc

1

x

. 1; =2/

38.

cot 1 x D arctan .1=x/I d 1 1 cot 1 x D D 1 x2 dx 1C 2 x y

1 1 C x2 =2

y D cot

1

x x

=2

Fig. 3.5-38

x2

p

1 x2

Fig. 3.5-37

1 1

1 x

.1;=2/

a x .a > 0/ aCx  p x 2ax x 2 .a > 0/ a  # 1=2   1 2a 2x x 2 p a a 2 2ax x 2

2ax x 2   2x x 33. tan 1 D 2 y y 1 2y 2xy 0 y 2 2xyy 0 D 2 2 y y4 4x 1C 2 y 4 4y 0 1 4 2y 0 D At .1; 2/ 2 4 16  2 8 4y 0 D 4 4y 0 ) y 0 D  1  2 At .1; 2/ the slope is  1 34. If y D sin

1

1 x2 1 p jxj x 2 1

1

Cr r

x a a

d sin dx 1 D r

xD

. If the slope is 2

3 : Thus the equations 2 p   3  C2 xC . 3 2

Remark: the domain of cot 1 can be extended to include 0 by defining, say, cot 1 0 D =2. This will make cot 1 right-continuous (but not continuous) at x D 0. It is also possible to define cot 1 in such a way that it is continuous on the whole real line, but we would then lose the identity cot 1 x D tan 1 .1=x/, which we prefer to maintain for calculation purposes.

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

95

lOMoARcPSD|6566483

SECTION 3.5 (PAGE 199)

39.

ADAMS and ESSEX: CALCULUS 9

  d 1 tan 1 x C tan 1 dx x   1 1 1 D 0 if x ¤ 0 D C 1 1 C x2 x2 1C 2 x Thus tan 1 x C cot 1 x D C1 (const. for x > 0)   At x D 1 we have C D C1 4 4  for x > 0. Thus tan 1 x C cot 1 x D 2 1 1 Also tan x C cot x D C2 for .x < 0/.   D C2 . At x D 1, we get 4 4  Thus tan 1 x C cot 1 x D for x < 0. 2 d .tan dx

1

1

x C cot

40. If g.x/ D tan.tan

1

x/ D

42.

1

1 .cos x/ D p . sin x/ 1 cos2 x n 1 if sin x > 0 D 1 if sin x < 0

sin 1 .cos x/ is continuous everywhere and differentiable everywhere except at x D n for integers n. y y D sin 1 .cos x/ =2 

Fig. 3.5-42

1

43.

.tan x/ then h is periodic with period , h0 .x/ D

y

d 1 tan 1 .tan x/ D .sec2 x/ D 1 except at odd dx 1 C tan2 x multiples of =2. tan 1 .tan x/ is continuous and differentiable everywhere except at x D .2n C 1/=2 for integers n. It is not defined at those points. y y D tan 1 .tan x/ =2

sec2 x D1 1 C tan2 x

provided that x ¤ .k C 21 / where k is an integer. h.x/ is  not defined at odd multiples of . 2 y



.=2;=2/ yDtan.tan

1



 yDtan

Fig. 3.5.40(a) d cos dx

1



x

Fig. 3.5-43

1 .tan x/

Fig. 3.5.40(b)

1 .cos x/ D p . sin x/ 2x 1 cos n 1 if sin x > 0 D 1 if sin x < 0

44.

1

cos .cos x/ is continuous everywhere and differentiable everywhere except at x D n for integers n. y y D cos 1 .cos x/ 

1 d tan 1 .cot x/ D . csc2 x/ D dx 1 C cot2 x integer multiples of .





1 except at

tan 1 .cot x/ is continuous and differentiable everywhere except at x D n for integers n. It is not defined at those points. y y D tan 1 .cot x/ =2 



x

Fig. 3.5-41

96

x

x/ x

41.

 x

sec .tan x/ 1 C x2 1 C x2 1 C Œtan.tan 1 x/2 D D 1: D 2 1Cx 1 C x2

If h.x/ D tan and

1

x/ then

2

g 0 .x/ D

d sin dx

Fig. 3.5-44

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

x

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

45. If jxj < 1 and y D tan and

tan y D p

1

p

x x2

1

SECTION 3.5 (PAGE 199)

x

1

sec2 y D 1 C .x 2 sec y D x;

x2

f 0 .x/  0 on . 1; 1/  x 1 tan Thus f .x/ D tan 1 xC1

1

D

1

x D C on . 1; 1/:

Evaluate the limit as x ! 1: lim f .x/ D tan

x! 1

Thus tan

50.

1



x 1 xC1

Since f .x/ D x



tan

tan

1

1

1

  3 D 2 4

1

xD

3 on . 1; 1/: 4

.tan x/ then sec2 x D1 1 C tan2 x

f 0 .x/ D 1

Thus sec y D x and y D sec 1p x. x2 1 If x  1 and y D  C sin 1 , then 2  y < x and sec y < 0. Therefore ! p p x2 1 x2 1 1 sin y D sin  C sin D x x 2

49.

1/ D x 2

because both x and sec y are negative. Thus y D sec 1 x in this case also. x , then y > 0 , x > 0 and 47. If y D sin 1 p 1 C x2 x sin y D p 1 C x2 1 x2 D cos2 y D 1 sin2 y D 1 2 1Cx 1 C x2 2 2 2 tan y D sec y 1 D 1 C x 1 D x2 tan y D x: x Thus y D tan 1 x and tan 1 x D sin 1 p . 1 C x2 p x2 1 48. If x  1 and y D sin 1 , then 0  y < 2 and x p x2 1 sin y D x 1 x2 1 D 2 cos2 y D 1 x2 x sec2 y D x 2 :

x

1

x

x2 1 x2 D sec2 y D 1 C 1 x2 1 x2 sin2 y D 1 cos2 y D 1 .1 x 2 / D x 2 sin y D x: x Thus y D sin 1 x and sin 1 x D tan 1 p . 1 x2 An alternative method of proof involves showing that the derivative of the left side minus the right side is 0, and both sides are 0 at x D 0. p p 46. If x  1 and y D tan 1 x 2 1, then tan y D x 2 1 and sec y D x, so that y D secp1 x. If x  1 and y D  tan 1 x 2 1, then 2 < y < 3 2 , so sec y < 0. Therefore p p tan y D tan. tan 1 x 2 1/ D x2 1

cos2 y D 1

because both x and sec y are negative. Thus y D sec in this case also.

, then y > 0 , x > 0

1D0

1 2 /

where k is an integer. Thus, f is  constant on intervals not containing odd multiples of . 2 f .0/ D 0 but f ./ D   0 D . There is no contradiction here because f 0 is not defined, so f is not constant 2 on the interval containing 0 and . if x ¤

51.

.k C

f .x/ D x 0

sin

1

.sin x/ 1

.   x  /

cos x 1 sin2 x cos x D1 j cos xj 8   < 0 if 1/ x2 1 R dx D tanh 1 x C C . 1 < x < 1/ 1 x2

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

x

y D sech x

x

1

y D csch x D

2 ex C e

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 3.6 (PAGE 205)

y

6. Let y D sinh Thus,

1

x  a

d sinh dx

Z

dy , x D a sinh y ) 1 D a.cosh y/ . dx x

1

a

D

x

1

1 a cosh y

x

1

p

dx

a2

C

x2

D sinh

1

x C C: a

.a > 0/

x , x D a Cosh y D a cosh y a dy for y  0, x  a. We have 1 D a.sinh y/ . Thus, dx 1

Fig. 3.6-8 9.

x 1 D a a sinh y 1 1 D q D p 2 2 x a2 a cosh y 1

d cosh dx

p

1

1 1 D q D p 2 C x2 2 a a 1 C sinh y

Let y D cosh

Z

y D coth

1

dx

x2

a2

D cosh

1

x C C: a

Since sech 1 x D cosh 1 .1=x/ is defined in terms of the restricted function Cosh, its domain consists of the reciprocals of numbers in Œ1; 1/, and is therefore the interval .0; 1. The range of sech 1 is the domain of Cosh, that is, Œ0; 1/. Also, d sech dx

1

.a > 0; x  a/

x dy Let y D tanh 1 , x D a tanh y ) 1 D a.sech2 y/ . a dx Thus, x 1 d tanh 1 D dx a a sech2 y a a D D 2 2 2 tanh2 x a x2 a a Z 1 x dx D tanh 1 C C: a2 x 2 a a   1 1 1 x2 1 7. a) sinh ln x D .e ln x e ln x / D x D 2 2 x 2x   2 1 1 x C1 1 ln x xC D b) cosh ln x D .e C e ln x / D 2 2 x 2x sinh ln x x2 1 c) tanh ln x D D 2 cosh ln x x C1 x 2 C 1 C .x 2 1/ cosh ln x C sinh ln x D 2 D x2 d) cosh ln x sinh ln x .x C 1/ .x 2 1/   1 xC1 1 8. The domain of coth x D ln consists of all x 2 x 1 xC1 > 0. Since satisfying jxj > 1. For such x, we have x 1 this fraction takes very large values for x close to 1 and values close to 0 for x close to 1, the range of coth 1 x consists of all real numbers except 0. d coth dx

1

d 1 cosh 1 dx x  1 1 1 D p : D s  2 2 x x 1 x2 1 1 x

xD

y

y D Sech

1

x

x

1

Fig. 3.6-9 10.

csch 1 has domain and range consisting of all real numbers x except x D 0. We have d csch dx

1

d 1 sinh 1 dx x   1 1 1 D s  2 x 2 D jxjpx 2 C 1 : 1 1C x

xD

d 1 tanh 1 dx x 1 1 1 D D 2 : 2 2 1 .1=x/ x x 1

y

y D csch

1

x

x

xD

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

Fig. 3.6-10

99

lOMoARcPSD|6566483

SECTION 3.6 (PAGE 205)

11.

fA;B .x/ D Ae kx C Be

0 fA;B .x/ D 00 fA;B .x/ D 00 Thus fA;B

kAe

kx

ADAMS and ESSEX: CALCULUS 9

kx

kBe

kx

k 2 Ae kx C k 2 Be k 2 fA;B D 0

auxiliary eqn

a/ C M sinh k.x

where A D 12 e

kx

ka

auxiliary eqn r 2 C 7r C 10 D 0 .r C 5/.r C 2/ D 0 C Be

2t

) rD

5; 2

C Be 3t

1 2;

) r D 0;

r2 D

3 2;

3/ D 0

and y D Ae

r 2 C 8r C 16 D 0

auxiliary eqn

y D Ae

4t

1; r D 3

2

2t

y 00 C 8y 0 C 16y D 0

5.

y 00

) rD

3 D 0 ) .2r C 1/.2r

C Bt e

.1=2/t

) rD

C Be .3=2/t :

4;

4

4t

2y 0 C y D 0

y 00

a/

r

6y 0 C 10y D 0

2

6r C 10 D 0

y D Ae 8.

3t

) r D3˙i

cos t C Be 3t sin t

9y 00 C 6y 0 C y D 0

9r 2 C 6r C 1 D 0 ) .3r C 1/2 D 0

Thus; r D

auxiliary eqn

y 00 C 7y 0 C 10y D 0

100

4r

M /.

Section 3.7 Second-Order Linear DEs with Constant Coefficients (page 212)

3D0

3y D 0

9.

k 2 y D 0 ) y D hL;M .x/ D L cosh k.x a/ C M sinh k.x y.a/ D y0 ) y0 D L C 0 ) L D y0 , v0 y 0 .a/ D v0 ) v0 D 0 C M k ) M D k Therefore y D hy0 ;v0 =k .x/ D y0 cosh k.x a/ C .v0 =k/ sinh k.x a/.

5t

4y 0

auxiliary eqn

D fA;B .x/

y D Ae

2

Thus; r1 D

a/

y 00

1.

4y 00 4r

k 2 y D 0 and

.L C M / and B D 12 e ka .L

t

r 2 2r C 1 D 0 ) .r 1/2 D 0 Thus; r D 1; 1; and y D Ae t C Bt e t :

hL;M .x/  M  L  kx ka e C e kxCka C e kx ka e kxCka D 2 2 ! ! L ka M ka kx L ka M ka e C e e e D e C e kx 2 2 2 2

13.

4.

7.

hence, hL;M .x/ is a solution of y 00

D Ae kx C Be

2r

y D A C Be

a/

a/ C M k 2 sinh k.x

3y D 0

r 2 C 2r D 0

auxiliary eqn

6.

D k hL;M .x/

2

y 00 C 2y 0 D 0

3.

12. Since

2

2y 0

y D Ae

00 gC;D .x/ D k 2 C cosh kx C k 2 D sinh kx 00 k 2 gC;D D 0 Thus gC;D cosh kx C sinh kx D e kx cosh kx sinh kx D e kx Thus fA;B .x/ D .A C B/ cosh kx C .A B/ sinh kx, that is, fA;B .x/ D gACB;A B .x/, and c D gC;D .x/ D .e kx C e kx / C .e kx e kx /, 2 2 that is gC;D .x/ D f.C CD/=2;.C D/=2 .x/.

00 hL;M .x/ D Lk 2 cosh k.x

r

kx

gC;D .x/ D C cosh kx C D sinh kx 0 gC;D .x/ D kC cosh kx C kD sinh kx

hL;M .x/ D L cosh k.x

y 00

2.

1 3;

1 3;

and y D Ae

.1=3/t

C Bt e

.1=3/t

:

y 00 C 2y 0 C 5y D 0

r 2 C 2r C 5 D 0 ) r D 1 ˙ 2i y D Ae t cos 2t C Be t sin 2t

10. For y 00 4y 0 C 5y D 0 the auxiliary equation is r 2 4r C 5 D 0, which has roots r D 2 ˙ i . Thus, the general solution of the DE is y D Ae 2t cos t C Be 2t sin t . 11.

For y 00 C 2y 0 C 3y D 0 the auxiliary equation isp r 2 C 2r C 3 D 0, which has solutions r D 1 ˙ 2i . Thus the general solution of the givenpequation is p y D Ae t cos. 2t / C Be t sin. 2t /.

12. Given that y 00 C y 0 C y D 0, hence r 2 C r C 1 D 0. Since a D 1, b D 1 and c D 1, the discriminant is D D b 2 4ac D 3 < 0 and .b=2a/ D 21 and p ! D 3=2. Thus, the general solution is p   p  3 3 t C Be .1=2/t sin t . y D Ae .1=2/t cos 2 2

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 3.7 (PAGE 212)

8 00 < 2y C 5y 0 3y D 0 13. y.0/ D 1 : 0 y .0/ D 0 The DE has auxiliary equation 2r 2 C 5y 3 D 0, with roots r D 12 and r D 3. Thus y D Ae t=2 C Be 3t . A Now 1 D y.0/ D A C B, and 0 D y 0 .0/ D 3B. 2 Thus B D 1=7 and A D 6=7. The solution is 6 1 y D e t=2 C e 3t . 7 7 14.

00

0

Given that y C 10y C 25y D 0, hence r 2 C 10r C 25 D 0 ) .r C 5/2 D 0 ) r D y D Ae

y0 D

5t

5e

C Bt e

5t

5. Thus,

0

.A C Bt / C Be

2 D y .1/ D

5

5e

C Be 5

5t

0

2t

cos t C Be

y D . 2Ae

2t

C Be

2t

.Ae

1/e

2t

4ac < 0

t!1

5

;

y.t / D Ae r1 t C Bt e r2 t :

5.t 1/

.

sin t

/ cos t

4ac < b

therefore r1 and r2 are negative. The general solution is

5

2t

4ac < b 2

Case 2: If D D b 2 4ac D 0 then the two equal roots r1 D r2 D b=.2a/ are negative. The general solution is

8 00 < y C 4y 0 C 5y D 0 y.0/ D 2 : 0 y .0/ D 0 The auxiliary equation for the DE is r 2 C 4r C 5 D 0, which has roots r D 2 ˙ i . Thus y D Ae

b2 p ˙ b2 p b ˙ b2

If t ! 1, then e r1 t ! 0 and e r2 t ! 0. Thus, lim y.t / D 0.

:

5

we have A D 2e and B D 2e . Thus, y D 2e 5 e 5t C 2t e 5 e 5t D 2.t 15.

Since

y.t / D Ae r1 t C Be r2 t :

.A C B/ C Be

5

Given that a > 0, b > 0 and c > 0: Case 1: If D D b 2 4ac > 0 then the two roots are p b ˙ b 2 4ac : r1;2 D 2a

5t

Since 0 D y.1/ D Ae

17.

C 2Be

2t

/ sin t:

Now 2 D y.0/ D A ) A D 2, and 2 D y 0 .0/ D 2A C B ) B D 6. Therefore y D e 2t .2 cos t C 6 sin t /.

If t ! 1, then e r1 t ! 0 and e r2 t ! 0 at a faster rate than Bt ! 1. Thus, lim y.t / D 0. t!1

Case 3: If D D b 2 y D Ae

.b=2a/t

e .1C/t e t lim y .t / D lim !0 !0  e tCh e t D t lim h h!0   d t Dt e D t et dt which is, along with e t , a solution of the CASE II DE y 00 2y 0 C y D 0.

cos.!t / C Be

.b=2a/t

sin.!t /

p 4ac b 2 . If t ! 1, then the amplitude of 2a .b=2a/t both terms Ae ! 0 and Be .b=2a/t ! 0. Thus, lim y.t / D 0. where ! D t!1

18. The auxiliary equation ar 2 C br C c D 0 has roots r1 D

16. The auxiliary equation r 2 .2 C /r C .1 C / factors to .r 1 /.r 1/ D 0 and so has roots r D 1 C  and r D 1. Thus the DE y 00 .2 C /y 0 C .1 C /y D 0 has general solution y D Ae .1C/t C Be t . The function e .1C/t e t is of this form with A D B D 1=. y .t / D  We have, substituting  D h=t ,

4ac < 0 then the general solution is

b

p

2a

D

;

r2 D

p bC D ; 2a

where D D p b 2 4ac. Note that a.r2 r1 / D D D .2ar1 C b/. If y D e r1 t u, then y 0 D e r1 t .u0 Cr1 u/, and y 00 D e r1 t .u00 C2r1 u0 Cr12 u/. Substituting these expressions into the DE ay 00 C by 0 C cy D 0, and simplifying, we obtain e r1 t .au00 C 2ar1 u0 C bu0 / D 0; or, more simply, u00 .r2 r1 /u0 D 0. Putting v D u0 reduces this equation to first order: v 0 D .r2

r1 /v;

which has general solution v D C e .r2 r1 /t : Hence Z u D C e .r2 r1 /t dt D Be .r2 r1 /t C A; and y D e r1 t u D Ae r1 t C Be r2 t .

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

101

lOMoARcPSD|6566483

SECTION 3.7 (PAGE 212)

ADAMS and ESSEX: CALCULUS 9

by the previous problem. Therefore ay 00 C by 0 C cy D 0 has general solution

19. If y D A cos !t C B sin !t then y 00 C ! 2 y D

A! 2 cos !t

B! 2 sin !t y D Ae k t cos.!t / C Be k t sin.!t /:

2

C ! .A cos !t C B sin !t / D 0

for all t . So y is a solution of (†). 20. If f .t / is any solution of .†/ then f 00 .t / D all t . Thus,

! 2 f .t / for

24.

2  2 i d h 2 ! f .t / C f 0 .t / dt D 2! 2 f .t /f 0 .t / C 2f 0 .t /f 00 .t / D 2! 2 f .t /f 0 .t /

y 0 .0/ D

If g.t / satisfies .†/ and also g.0/ D g 0 .0/ D 0, then by Exercise 20,  2  2 ! 2 g.t / C g 0 .t /  2  2 D ! 2 g.0/ C g 0 .0/ D 0:

25.

Since a sum of squares cannot vanish unless each term vanishes, g.t / D 0 for all t . 22. If f .t / is any solution of .†/, let g.t / D f .t / A cos !t B sin !t where A D f .0/ and B! D f 0 .0/. Then g is also solution of .†/. Also g.0/ D f .0/ A D 0 and g 0 .0/ D f 0 .0/ B! D 0. Thus, g.t / D 0 for all t by Exercise 24, and therefore f .x/ D A cos !t C B sin !t . Thus, it is proved that every solution of .†/ is of this form.

26.

4ac b 2 b and ! 2 D which is 2a 4a2 kt positive for Case III. If y D e u, then

8 00 < y C 100y D 0 y.0/ D 0 : 0 y .0/ D 3 y D A cos.10t / C B sin.10t / A D y.0/ D 0; 10B D y 0 .0/ D 3 3 yD sin.10t / 10     y D A cos !.t c/ C B sin !.t c/

D A cos !t C B sin !t where A D A cos.!c/ B sin.!c/ and B D A sin.!c/ C B cos.!c/ 27.

For y 00 C y D 0, we have y D A sin t C B cos t . Since, y.2/ D 3 D A sin 2 C B cos 2 y 0 .2/ D 4 D A cos 2 B sin 2;

Substituting into ay 00 C by 0 C cy D 0 leads to   0 D e k t au00 C .2ka C b/u0 C .ak 2 C bk C c/u   D e k t au00 C 0 C ..b 2 =.4a/ .b 2 =.2a/ C c/u   D a e k t u00 C ! 2 u : 00

therefore A D 3 sin 2 4 cos 2 B D 4 sin 2 C 3 cos 2: Thus,

2

Thus u satisfies u C ! u D 0, which has general solution u D A cos.!t / C B sin.!t /

102

5 2:

(easy tocalculate y 00 C ! 2 y D 0)  y D A cos.!t / cos.!c/ C sin.!t / sin.!c/   C B sin.!t / cos.!c/ cos.!t / sin.!c/   D A cos.!c/ B sin.!c/ cos !t   C A sin.!c/ C B cos.!c/ sin !t

23. We are given that k D

  y 0 D e k t u0 C ku   y 00 D e k t u00 C 2ku0 C k 2 u :

5)B D

Thus, y D 2 cos 2t 25 sin 2t . circular frequency = ! D 2, frequency = 1 ! D  0:318 2  2 period = D   3:14 !q amplitude = .2/2 C . 52 /2 ' 3:20

2! 2 f .t /f 0 .t / D 0

 2  2 for all t . Thus, ! 2 f .t / C f 0 .t / is constant. (This can be interpreted as a conservation of energy statement.) 21.

Because y 00 C 4y D 0, therefore y D A cos 2t C B sin 2t . Now y.0/ D 2 ) A D 2;

y D .3 sin 2 D 3 cos.t

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

4 cos 2/ sin t C .4 sin 2 C 3 cos 2/ cos t 2/ 4 sin.t 2/:

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 3.7 (PAGE 212)

8 < y 00 C ! 2 y D 0 28. y.a/ D A : 0 y .a/ D B   B   y D A cos !.t a/ C sin !.t a/ ! 29. From Example 9, the spring constant is k D 9  104 gm=sec2 . For a frequency of 10 Hz (i.e., a circular frequency ! D 20 rad=sec.), a mass m satisfying p k=m D 20 should be used. So,

32.

y D e k t ŒA cosh !.t t0 /B sinh !.t  A !.t t0 / A !.t t0 / e C e D ek t 2 2  B !.t t0 / B !.t t0 / e C e 2 2 D A1 e .kC!/t C B1 e .k

9  104 k D D 22:8 gm: mD 400 2 400 2

therefore, y D A cos 20 t C B sin 20 t and 1)AD

1 2 1 y 0 .0/ D 2 ) B D D : 20 10

33.

t0 /B sin !.t

t0 /

D e k t ŒA cos !t cos !t0 C A sin !t sin !t0 C B sin !t cos !t0 B cos !t sin !t0  D e k t ŒA1 cos !t C B1 sin !t ;

where A1 D A cos !t0 B sin !t0 and B1 D A sin !t0 C B cos !t0 . Under the conditions of this problem we know that e k t cos !t and e k t sin !t are independent solutions of ay 00 C by 0 C cy D 0, so our function y must also be a solution, and, since it involves two arbitrary constants, it is a general solution.

!/t

8 00 < y C 2y 0 C 5y D 0 y.3/ D 2 : 0 y .3/ D 0

The DE has auxiliary equation r 2 C 2r C 5 D 0 with roots r D 1 ˙ 2i . By the second previous problem, a general solution can be expressed in the form y D e t ŒA cos 2.t 3/ C B sin 2.t 3/ for which

1 Thus, y D cos 20 t C sin 20 t , with y in cm and t 10 in r second, gives the displacement at time t . The amplitude 1 2 is . 1/2 C . /  1:0005 cm. 10 k ! , !2 D (k = spring const, m = mass) 30. Frequency D 2 m Since the spring does not change, ! 2 m D k (constant) For m D 400 gm, ! D 2.24/ (frequency = 24 Hz) 4 2 .24/2 .400/ If m D 900 gm, then ! 2 D 900 2  24  2 D 32. so ! D 3 32 Thus frequency = D 16 Hz 2 4 2 .24/2 400 For m D 100 gm, ! D 100 ! so ! D 96 and frequency = D 48 Hz. 2 31. Using the addition identities for cosine and sine, y D e k t ŒA cos !.t

t0 /

where A1 D .A=2/e !t0 C .B=2/e !t0 and B1 D .A=2/e !t0 .B=2/e !t0 . Under the conditions of this problem we know that Rr D k ˙ ! are the two real roots of the auxiliary equation ar 2 C br C c D 0, so e .k˙!/t are independent solutions of ay 00 C by 0 C cy D 0, and our function y must also be a solution. Since it involves two arbitrary constants, it is a general solution.

The motion is determined by 8 < y 00 C 400 2 y D 0 y.0/ D 1 : 0 y .0/ D 2 y.0/ D

Expanding the hyperbolic functions in terms of exponentials,

y0 D

e t ŒA cos 2.t 3/ C B sin 2.t 3/ C e t Œ 2A sin 2.t 3/ C 2B cos 2.t 3/:

The initial conditions give 2 D y.3/ D e 0

0 D y .3/ D

Thus A D 2e 3 and B D solution

e

A 3

.A C 2B/

A=2 D

y D e 3 t Œ2 cos 2.t

34.

3

3/

e 3 . The IVP has

sin 2.t

3/:

8 00 < y C 4y 0 C 3y D 0 y.3/ D 1 : 0 y .3/ D 0

The DE has auxiliary equation r 2 C 4r C 3 D 0 with roots r D 2 C 1 D 1 and r D 2 1 D 3 (i.e. k ˙ !, where k D 2 and ! D 1). By the second previous problem, a general solution can be expressed in the form y D e 2t ŒA cosh.t 3/ C B sinh.t 3/ for which y0 D

2e

2t

Ce

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

ŒA cosh.t

2t

ŒA sinh.t

3/ C B sinh.t

3/ C B cosh.t

3/ 3/:

103

lOMoARcPSD|6566483

SECTION 3.7 (PAGE 212)

ADAMS and ESSEX: CALCULUS 9

The initial conditions give 1 D y.3/ D e 0

0 D y .3/ D

6

e

The conditions for stopping the motion are met at t D 2; the mass remains at rest thereafter. Thus 84 1 < 5 cos t C 5 if 0  t   2 1 x.t / D 5 cos t 5 if  < t  2 :1 if t > 2 5

A 6

. 2A C B/

Thus A D e 6 and B D 2A D 2e 6 . The IVP has solution y D e6

2t

Œcosh.t

3/ C 2 sinh.t

3/:

Review Exercises 3 (page 213) 1.

35. Let u.x/ D c Also u0 .x/ D u00 .x/ D

k 2 y.x/. Then u.0/ D c k 2 a. k 2 y 0 .x/, so u0 .0/ D k 2 b. We have k 2 y 00 .x/ D

 k2 c

 k 2 y.x/ D

k 2 u.x/

This IVP for the equation of simple harmonic motion has solution u.x/ D .c

k 2 a/ cos.kx/

xD0

2.

kb sin.kx/

so that  1  y.x/ D 2 c u.x/ k   c D 2 c .c k 2 a/ cos.kx/ C kb sin.kx/ k b c D 2 .1 cos.kx/ C a cos.kx/ C sin.kx/: k k 36. Since x 0 .0/ D 0 and x.0/ D 1 > 1=5, the motion will be governed by x 00 D x C .1=5/ until such time t > 0 when x 0 .t / D 0 again. Let u D x .1=5/. Then u00 D x 00 D .x 1=5/ D u, u.0/ D 4=5, and u0 .0/ D x 0 .0/ D 0. This simple harmonic motion initial-value problem has solution u.t / D .4=5/ cos t . Thus x.t / D .4=5/ cos t C .1=4/ and x 0 .t / D u0 .t / D .4=5/ sin t . These formulas remain valid until t D  when x 0 .t / becomes 0 again. Note that x./ D .4=5/ C .1=5/ D .3=5/.

104

f .x/ D sec2 x tan x ) f 0 .x/ D 2 sec2 x tan2 x C sec4 x > 0 for x in . =2; =2/, so f is increasing and therefore one-to-one and invertible there. The domain of f 1 is . 1; 1/, the range of f . Since f .=4/ D 2, therefore f 1 .2/ D =4, and 1 0

/ .2/ D

.f

3.

lim f .x/ D

x!˙1

4.

5. 6.

lim

1 f 0 .f

x!˙1

x ex2

1 .2//

D

1 1 D : f 0 .=4/ 8

D 0:

2

Observe f 0 .x/ D e x .1 2x 2 / is positive if x 2 < 1=2 2 and ispnegative p if x > 1=2. Thus f is increasing p on 2; 1= 2/ and is decreasing on . 1; 1= 2/ and . 1= p on .1= 2; 1/. p p The maxpand min values p of f are 1= 2e (at x D 1= 2) and 1= 2e (at x D 1= 2). y D e where

x

sin x, .0  x  2/ has a horizontal tangent

dy D e x .cos x sin x/: dx This occurs if tan x D 1, p The pso x D =4 or x D 5=4. points are .=4; e =4 = 2/ and .5=4; e 5=4 = 2/. 0D

7.

Since x./ < .1=5/, the motion for t >  will be governed by x 00 D x .1=5/ until such time t >  when x 0 .t / D 0 again.

Let v D x C .1=5/. Then v 00 D x 00 D .x C 1=5/ D v, v./ D .3=5/ C .1=5/ D .2=5/, and v 0 ./ D x 0 ./ D 0. Thius initial-value problem has solution v.t / D .2=5/ cos.t / D .2=5/ cos t , so that x.t / D .2=5/ cos t .1=5/ and x 0 .t / D .2=5/ sin t . These formulas remain valid for t   until t D 2 when x 0 becomes 0 again. We have x.2/ D .2=5/ .1=5/ D 1=5 and x 0 .2/ D 0.

f .x/ D 3x C x 3 ) f 0 .x/ D 3.1 C x 2 / > 0 for all x, so f is increasing and therefore one-to-one and invertible. Since f .0/ D 0, therefore f 1 .0/ D 0, and ˇ ˇ 1 1 1 d ˇ 1 .f /.x/ˇ D 0 D : D 0 ˇ dx f .f 1 .0// f .0/ 3

If f 0 .x/ D x for all x, then d f .x/ f 0 .x/ xf .x/ D D 0: 2 =2 2 x dx e e x =2 2

Thus f .x/=e x =2 D C (constant) for all x. Since f .2/ D 3, we have C D 3=e 2 and 2 2 f .x/ D .3=e 2 /e x =2 D 3e .x =2/ 2 . 8.

Let the length, radius, and volume of the clay cylinder at time t be `, r, and V , respectively. Then V D  r 2 `, and dr d` dV D 2 r` C r2 : dt dt dt

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

lOMoARcPSD|6566483

INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 3 (PAGE 213)

Since d V =dt D 0 and d `=dt D k` for some constant k > 0, we have 2 r`

dr D dt

k r 2 `;

)

dr D dt

13.

kr : 2

y D aa C aa .1 C ln a/.x

a) An investment of $P at r% compounded continuously grows to $P e rT =100 in T years. This will be $2P provided e rT =100 D 2, that is, rT D 100 ln 2. If T D 5, then r D 20 ln 2  13:86%.

14.

b) Since the doubling time is T D 100 ln 2=r, we have

If r D 13:863% and r D T 

0:5%, then

100 ln 2 . 0:5/  0:1803 years: 13:8632

The doubling time will increase by about 66 days. ˇ ah 1 a0Ch a0 d x ˇˇ 10. a) lim D lim D a ˇ D ln a. h h dx ˇ h!0 h!0 xD0   Putting h D 1=n, we get lim n a1=n 1 D ln a. n!1

11.

ln 2 ln x D is satisfied if x D 2 or x D 4 (because x 2 ln 4 D 2 ln 2).

1 ln b DmD : b b Thus ln b D 1, and b D e.

15.

b) Using the technique described in the exercise, we calculate   10 210 21=2 1  0:69338183   11 211 21=2 1  0:69326449

Thus ln 2  0:693. 2  2 d  f .x/ D f 0 .x/ dx  2 ) 2f .x/f 0 .x/ D f 0 .x/

a)

b) The line y D mx through the origin intersects the curve y D ln x at .b; ln b/ if m D .ln b/=b. The same line intersects y D ln x at a different point .x; ln x/ if .ln x/=x D m D .ln b/=b. This equation will have only one solution x D b if the line y D mx intersects the curve y D ln x only once, at x D b, that is, if the line is tangent to the curve at x D b. In this case m is the slope of y D ln x at x D b, so

100 ln 2 r: r2

dT T  r D dr

Let the rate be r%. The interest paid by account A is 1; 000.r=100/ D 10r. The interest paid by account B is 1; 000.e r=100 1/. This is $10 more than account A pays, so 1; 000.e r=100

If y D cos

1

x, then x D cos y and 0  y  . Thus

tan y D sgn x

0

Thus cos

sec2

x D tan

y

1

1 D sgn x

p .. 1

1  D x p 2

2

12. If f .x/ D .ln x/=x, then f .x/ D .1 ln x/=x . Thus f 0 .x/ > 0 if ln x < 1 (i.e., x < e) and f 0 .x/ < 0 if ln x > 1 (i.e., x > e). Since f is increasing to the left of e and decreasing to the right, it has a maximum value f .e/ D 1=e at x D e. Thus, if x > 0 and x ¤ e, then 17.

Putting x D  we obtain .ln /= < 1=e. Thus ln. e / D e ln  <  D  ln e D ln e  ; and  e < e  follows because ln is increasing.

1

p

Since cot x D 1= tan x, cot

ln x 1 < : x e

1/ D 10r C 10:

A TI-85 solve routine gives r  13:8165%. 16.

) f 0 .x/ D 0 or f 0 .x/ D 2f .x/: Since f .x/ is given to be nonconstant, we have f 0 .x/ D 2f .x/. Thus f .x/ D f .0/e 2x D e 2x .

a/:

This line passes through the origin if 0 D aa Œ1 a.1 C ln a/, that is, if .1 C ln a/a D 1. Observe that a D 1 solves this equation. Therefore the slope of the line is 11 .1 C ln 1/ D 1, and the line is y D x.

That is, r is decreasing at a rate proportional to itself. 9.

y D x x D e x ln x ) y 0 D x x .1 C ln x/. The tangent to y D x x at x D a has equation

r

1 x2

1D

p

x2

1 x

:

x 2 /=x/. 1

x D tan

1

.1=x/.

1 csc 1 x D sin 1 cos 1 x 1 .1=x/2  tan 1 D 2 1=x p  1 D x 2 1: sgn xtan 2  cos 1 x D sin 1 x. 2 If y D cot

1

x, then x D cot y and 0 < y < =2. Thus

p p csc y D sgn x 1 C cot2 y D sgn x 1 C x 2 sgn x : sin y D p 1 C x2

Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage ([email protected])

105

lOMoARcPSD|6566483

REVIEW EXERCISES 3 (PAGE 213)

Thus cot

1

x D sin

1

ADAMS and ESSEX: CALCULUS 9

sgn x p D sgn xsin 1 C x2

1

p

1 1 C x2

:

21.

1 . x 18. Let T .t / be the temperature of the milk t minutes after it is removed from the refrigerator. Let U.t / D T .t / 20. By Newton’s law, csc

1

x D sin

ex > 1 C x C

1

U 0 .t / D kU.t /

g.t / D e t

U.t / D U.0/e k t :

)

kD

If T .s/ D 18, then U.s/ D sk D ln.2=15/, and

1 12

2, so

ln.8=15/: 2D

1.

T .0/ D 96 ) U.0/ D 96

T .20/ D 40 ) U.20/ D 40

ln y D x ln x ln.ln y/ D ln x C ln.ln x/ g.y/ ln.ln y/ x.ln x C ln.ln x// D lim lim x!1 y!1 ln y x ln x   ln.ln x/ : D lim 1 C x!1 ln x

R R ) 60

R D .96

R ) 40

R D .96

R/e 10k R/e 20k :

p Now ln x