Student solutions manual : Algebra and trigonometry, fourth edition, James Stewart, Lothar Redlin, Saleem Watson (Compressed) [4 ed.] 1305118154, 9781305118157

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Algebra and Trigonometry FOURTH EDITION James Stewart McMaster University and University of Toronto

The Pennsylvania State University

California State University, Long Beach

Prepared by

Andy Bulman-Fleming eo » CENGAGE ** Learning’ Australia + Brazil * Mexico » Singapore * United Kingdom + United States

CENGAGE Learning © 2016 Cengage Learning

ISBN: 978-1-305-11815-7

WCN: 01-100-101 Cengage Learning ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher.

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ty. tt

CONTENTS

CHAPTER

m

PROLOGUE: Principles of Problem Solving

P_

PREREQUISITES

P.1 P.2

3

Modeling the Real World with Algebra RealNumbers

Integer Exponents and Scientific Notation

P.4

Rational Exponents and Radicals

P.5

Algebraic Expressions

P.6

Factoring

P.7

Rational Expressions

P.8

Solving Basic Equations

P.9

Modeling with Equations

ChapterPTest

8

11

16

19 22

25 28

FOCUS ON MODELING: Making the Best Decisions

33

TheCoordinate Plane

1.2

Graphs of Equations in Two Variables; Circles

1.3

Lines

1.4

Solving Quadratic Equations

33 38

45

ComplexNumbers

50

55

1.6

Solving Other Types of Equations

1.7

Solving Inequalities

1.8

Solving Absolute Value Equations and Inequalities

1.9

Solving Equations and Inequalities Graphically

1.10

29

EQUATIONS AND GRAPHS

1.1.

1.5

6

13

Chapter P Review

CHAPTER 1

3

3

P.3

w

1

Modeling Variation Chapter 1Review Chapter 1Test

56

60 70 71

74 77 -

86

© 2016 Cengage Learning. AlJ Rights Reserved. May not be scanned, copied or duplicated, or posted to a.publicly accessible website, in whole or in part.

iv

Contents

FOCUS ON MODELING: Fitting Lines to Data

CHAPTER 2

89

FUNCTIONS

91

2.1

Functions

2.2

Graphs of Functions

2.3

Getting Information from the Graph of aFunction

2.4

Average Rate of Change ofa Function

2.5

Linear Functions and Models = 111

2.6

Transformations of Functions

2.7

Combining Functions

2.8

One-to-One Functions and Their Inverses

91

Chapter2Review

Chapter 2 Test

95

108

114

121 125

130

137

FOCUS ON MODELING: Modeling with Functions

CHAPTER 3

3.1 3.2 3.3 3.4 3.5 3.6

102

138

POLYNOMIAL AND RATIONAL FUNCTIONS Quadratic Functions and Models =143 Polynomial Functions and TheirGraphs Dividing Polynomials

148

156

Real Zeros of Polynomials

160

Complex Zeros and the Fundamental Theorem of Algebra Rational Functions

Chapter 3Review Chapter 3 Test

4.1 4.2 4.3 4.4 4.5

176

181

198 208

FOCUS ON MODELING: Fitting Polynomial Curves toData

CHAPTER 4

143

EXPONENTIAL AND LOGARITHMIC FUNCTIONS Exponential Functions

Laws of Logarithms

213

213

The Natural Exponential Function Logarithmic Functions

210

217

220

225

Exponential and Logarithmic Equations ©227

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or inpart.

i]

Contents

4.6

Modeling with Exponential Functions

4.7

Logarithmic Scales

231

233

Chapter 4 Review Chapter 4 Test

234 239

FOCUS ON MODELING: Fitting Exponential and Power Curves to Data

240

CHAPTER 5 TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE APPROACH

5.1 5.2 5.3 5.4

Angle Measure

Trigonometric Functions of Angles

245 247

Inverse Trigonometric Functions and Right Triangles

5.5

The Law of Sines

5.6

The Law of Cosines

250

252

Chapter5Review

254 257

259

FOCUS ON MODELING: Surveying

287

CHAPTER 6 TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH 6.1

The Unit Circle

6.2

Trigonometric Functions of RealNumbers

6.3 6.4 6.5 6.6

263

263

Trigonometric Graphs

265

267

More Trigonometric Graphs

274

Inverse Trigonometric Functions‘and Their Graphs Modeling Harmonic Motion Chapter 6 Review Chapter 6 Test

243

243

Trigonometry of Right Triangles

Chapter 5 Test

v

277

279

282 286

FOCUS ON MODELING: Fitting Sinusoidal Curves to Data

260

CHAPTER 7 ANALYTIC TRIGONOMETRY

291

7.1

Trigonometric Identities

291

7.2

Addition and Subtraction Formulas

7.3

Double-Angle, Half-Angle, and Product-Sum Formulas

295 298

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

vi

Contents

7.4

Basic Trigonometric Equations

304

7.5

More Trigonometric Equations

306

Chapter 7 Review

Chapter 7 Test

309

312

FOCUS ON MODELING: Traveling and Standing Waves

CHAPTER 8

POLAR COORDINATES AND PARAMETRIC EQUATIONS

8.1

Polar Coordinates

8.2

Graphs of Polar Equations

8.3

Polar Form of Complex Numbers; De Moivre’s Theorem

8.4

Plane Curves and Parametric Equations

Chapter 8 Test

317

334 337 338

CHAPTER 9 VECTORS IN TWO AND THREE DIMENSIONS 9.1

Vectors inTwo Dimensions

9.2

The Dot Product

9.3

Three-Dimensional Coordinate Geometry

341

341

344

Vectors in Three Dimensions

346

347

9.5

The Cross Product

9.6

Equations of Lines and Planes

349

Chapter9 Review

350

352

354

FOCUS ON MODELING: Vector Fields

CHAPTER 10

321

327

FOCUS ON MODELING: The Path ofaProjectile

Chapter 9 Test

315

315

Chapter 8 Review

98

313

355

SYSTEMS OF EQUATIONS AND INEQUALITIES

10.1 10.2

Systems of Linear Equations in Two Variables

10.3

Partial Fractions

10.4

Systems of Nonlinear Equations

10.5

Systems of Inequalities

Systems of Linear Equations in Several Variables

357 ;

357 361

365 370

374

© 2016 Cengage Learning. All Rights Reserved, May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in-part.

, 1729 q 23

‘

(d) Any number which cannot be expressed as a ratio P are a2, xf3! a, and e. Pp q of two integers is irrational. Examples p

. The set of numbers between but not including 2 and 7 can be written as (a) {x | 2 < x < 7} in interval notation, or (b) (2, 7) in interval notation.

5. The distance between a and b on the real line is d (a, b) = |b — a]. So the distance between —5 and 2 is |2 — (—5)| = 7.

7. (a) No:a—b=—(b—a)

#b —a in general.

(b) No; by the Distributive Property, —2(a — 5) = —2a + —2 (—5) = —2a + 10 4 —2a — 10. . (a) Natural number: 100 (b) Integers: 0, 100, —8 (c) Rational numbers: —1.5, 0, 5, 2.71. 3.14, 100; —8

(d) Irrational numbers: /7, —7

11. Commutative Property of addition

13. Associative Property of addition

15. Distributive Property

17. Commutative Property of multiplication

19, x+3=3+x

21.4(4+

23. 3@+y)

25. 4(2m) = (4-2)m = 8m

27.

=3x+3y

—3 (2x — 4y) = —3 (2x) + 3 Gy) = —Sx + 10y

B) =44A+4B

29.) H+Bp=% 0 1

Tres 5

-+

Sleell

wp

4.559

(b) 4+ 5 = 39 + 20 = 20 S16

(Bees 7 (c) 3.5= 4 37. (a) True _ (b) False

35. (a) False (b) True

Sh} (a)x >0

41. (a)

(b)t n

(d) -S 0, x > —3. Domain; {x | x > —3}

13.x2

—x

—-2=(x +1) (x —2)

#0

x

$—1 or2, so the

domain is {x | x 4 —1, 2}. 5,

DET DOF

_ 5 (x — 3) (2x + 1) ws 2x +1

10 (x — 3)? ie: x7 +5x+6

23. 25. PA 29.

31. ap

S(x —3)-2(x-3) _

&+2)@+3)_x+2

x248x+4+15

(x+5)(x+3)

2x3 — x? — 6x D2 Gs

x (2x? -x -6) eS) (2)

4x

x27

pytear

x45

4x

x2-4 y-1

x—2

‘abba

(x-2)(¥+2)

x42

PUNY SEAT Sepa

(G-1O4+) > yl

x (2x +3)(x—2) x (2x +3) x ea eee ees |te

1

y2—4 1) 1k XG —2)~+2y, 16x 40-2) x?42x-15 x-5 (x+5)(e-3)(@-5) x =3 SS eR a KES SG PD) eee t-—3

¢t+3

249

2-9

(t — 3) (t¢ +3)

x24 3x42

x43 4x2—9

1

(249)¢-3)¢4+3) 1249 x2 4604-9

ODO

HG

43)

x247x412 — x43

2x7 47x -15 _

2x247x—15

x247x412

x3

35.

Peaicama OF

2(x -3)

4x2—9 :

sry

xo

xt 42e41

x

Tay teie

x

PCENGTH =

(x +1)x

43)

xe 1

x+3

(x+5)(QQx-3)

(2x —3)Qx+3)

(4+3)44)

x +5

(2x4+3)044)

—, oi. Co aa

574241

3

39; 41.

pee

ee:

z yz yz 1

x+3

Se Sie SOR 1 2

Es)

Ses.

1

x+4

WRN Vie ST. x —3

POSS

245)

+ FSSA Ze 10. og

(x +5)(x—3)

&+5)@—3)

3x'+-7

@+5)@—3) '

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

O ae a

SECTION P.7 Rational Expressions

43

3

ae,

ee te

.3 (x +2)

"

x+1

2 (4-1). e $2). 9 1)

eas

3.

§(Qx —3)

3

Deere, 23)4 | (22 —3) uo

u+1

I Co

1

x2

u+l

a

x24x

1

+ Oana

uo

2x +5

PD?)

«10x -15-3

On ay

(wu+1)(u+1)

47. u+1+——

je

2)

Oe

ED

10x -18

8)

+2)

2(Sx —9)

Oxe ae Oe aye

uw + Qut lu

ur +3utl

u+1

u+1

Fie eh

7

l

eel x 25-41 ae ee SeaRoie Sa Sai Weare Ceyeed ei LR Pe

x20

x(x+1)

$2 Gk ieee ee)

x2 (x + 1)

ee 1 _ 2 +4) eee oe exes 12. x+3 °° 4+3)04+4). “643044 i

et hor

i)

—1 6430449)

2x +7

43) (44) See 3) G +4)

1

1

x4+3

x2-9

aE

I See,

ee

x+3

1 ee

ee x

x-1l

ee

x

l

1

x-1

x(e—-1)

>

I

=

6

eter)

1

l ee

x —2 Lt

&—-3)0@+4+3).

&-3)04+3)

Bae as rt

x(e—-1)

x(Qe—1)

Ry 243074 Sx 6.

x@-1)

1

x2 — 2x -—3 (x +2041). 2g x—3 i"

~ &%-3)@4+2041 :

(-—3)@4+3))

AO x2-x

ee ox $2

Fe ee ea Coes,

(x—3)(x +3)

x(x —1)

x (x — 1)

}

&=3)04+1) —(x +2)

@&-3)@42)@41

eu

x—3

=x = 2

@&-3)@42)@41)

ae

—5

&-3)@4+2)@41)

4

Yo2° x(i-2) 1-2 ue, Bee

x 4-2) 1+) "xt2) _(@+2)+1_ x43

. = ‘iho X+2

1

pda a

+2)(1

4 I - 27) Se .

il

(x —1)(« +3) (¢ +1) 2 ~ (x —1)(«-+3)

x

4 oe

x41

1 1 Ree WeAPOuG sre eeely,

x+1

(+3)1). °y

2(x +1)

~~ @=-)@4D@ +3) &-1I +1) @ +3)

9

nen

65.

~ (+2)-1

a3

xy

=

J, heme

y

2 od ( 4) x? — y? =

gYaa aii x2. y2 x2y2

Ah OS

aS

a

_xty-x?

=

xy2—y2

x2—y2

cay

x?(Qy-1)

=-y2 (x - 1)

—x2y2

Pe

ae

xy ee

xy.

An alternative method is to multiply the

numerator and denominator by the common denominator of both the numerator and denominator, in this case x*y?:

eA s I ta ‘) x2y2 na

gta x2

y2

ae ae erie ye —x2 p22

er)

©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

18

CHAPTERP Prerequisites

1 69.

x~2—y

2

—

ee Pe

y?

oe

y? - x2y2

x2y2 hey? a xe

ee

Beat

xy

; te Gal

1

x2

xy

1

=x) O +x) xy SE SSRs x?y? (y +x) xy

Ee

Fae rd

wey? &|Yitelx

(5+5)

x2y2 ~~ xy24+x2y0

xy"

xy

ba ae reli-k amie

ee

xy(ytx)

oxy

es

FogCSRS Pgx-1IRS x-1 dkSat A l—x el Pe x

1

1

“3. l+x+h

1+x _ (+x)-C+x+h)

h 1

A

a

1

(1+x)(+x+h)

1

Gem

2 4 PE w+ hy? Caribe

: We

__

h(+x)(i+x+h) pie “

i+( set| na 7=3)

Ges

=|

rae

G+

(@ — 3)*

1-x2

aC

_ (+2)? Gx

K/d see

3) (x + 2) (2)]

(«= 3)4

-9-2x-4) | @ +2)? @ -13)

(x — 3)

2

Ee) ace (ltx)7'/*(2(+x)-x]

1+x

thy?

eat eae} 1—x2

Dileg

Bs fore

x +h

pape ls: x2

(aes) Lee

ai 2a)

oe

1+x

Vi@reaye

x +2

~ (4x3?

ei 3(1+x)!3 —x (1 +x)72/9 oiGiha he? Bees 2x +3 aye Se (+x) (1+x)2/3 '

ge,

de. as TS SO ee ees NS W/E Swe Sana See ae ae

Fe:

2

Va-yi_2(v2-v7)_2(v2-v7)

“LEME hl telTibiPEA) ee

fo

'W8.4-a/9.

oi; eS 3

ON

er

AQAA eee

Vr+V2_ 93,7 fe 5

la

5

ae

aay,

Pee 3 (1+¥5)

Vrtv2 yr-v2 r—-2ee ee Nea 5

95. /x2+1—-x= es 97. (a)

a8

y(V3~= V9) _ yw3—yy9

Oe ee 3(1+¥5)

TS 1+V5

PN 3

ed

_2(v7-v2)

R=

rial

5s (yr- V2)

chit ves Saree bye 1+xoe

2+ l+x 1

1

vie

=

x? +1-— x?

vx?+

Ve ae l+x

R, Ro

Vi 1 Ge: Ree Ro +R ay ad ape eeeBo eeke oe R, Ro

Ri

RP

Rink

I 10)

(20

CO} (b): Spbstihuting Ry = 1¢.olimas anid tp =: 20 oni eres tae

200

= — ae

*¥ 6.7 ohms.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION P8 Solving Basic Equations

99.

ae

3.05 | 3.10 | 3.20

Ey | | [09| [99 [com[an[ots [em .

From the table, we see that the expression ~ ae se Raiire 6 as x approaches 3. We simplify the expression:

x*—9 = er 3)G43) x—3 table.

x-—3

.

x

+3, x # 3. Clearly as x approaches 3, x + 3 approaches 6. This explains the result in the

101. Answers will vary.

af 2 $45 (a+b)? 4a2+b2

Rene

Va2+b24at+b

| V524+122 45412 2-6 "46

ait!

103. (a) 9999 10,000

1 It appears that the smallest possible value of x + —is 2. ss f : (b) Because x > 0, we can multiply both sides by x and preserve the inequality: x24]

Sy

x*—2x+1

>0Sek-

1 x + — > 24x eX

l («+ =)>2xo os

1)? > 0. The last statement is true for all x > 0, and because each step is

1 reversible, we have shown that x + — > 2 forall x > 0. x

P8

SOLVING BASIC EQUATIONS

1. Substituting x = 3 in the equation 4x — 2 = 10 makes the equation true, so the number 3 is a solution of the equation.

3. (a) ;+ 2x = 10 is equivalent to 3x — 10 =0, so it is a linear equation.

: 22 (b) —x — 2x = 1 is not linear because it contains the term =8 multiple of the reciprocal of the variable.

:: } b

aot 2.80 | 2.90 2

19

(c) x +7 =5 —3x

Kg

& 4x —2 = 0, so it is linear.

5. (a) This is true: Ifa = b, thena+x =b+x.

(b) This is false, because the number could be zero. However, it is true that multiplying each side of an equation by a nonzero number always gives an equivalent equation. (c) This is false. For example, —5 = S is false, but (—5)* = 5? is true.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER P Prerequisites

20

- (a) When x = —2, LHS = 4(—2)

+ 7 = —8 +7 = —1 and RHS = 9 (—2)

—3 = —18 —3 = —21. Since LHS # RHS;

x = —2 is nota solution.

(b) When x = 2, LHS = 4(—2)

+ 7 = 8+7

= 15 and RHS = 9 (2) —3 = 18 —3 = 15. Since LHS = RHS, x =2isa

solution.

. (a)

t

Whenx = 2, LHS = 1—{2—-G—(Q))]=1-—[2—1]=1—-—1=0 and RHS = 4(2) —6+(2)) =8 —8 =0, Since

LHS = RHS, x = 2 is a solution.

(b) When x = 4 LHS = 1—[2 — (3 — (4))] = 1-[2 — (-1)] = 1—3 = —2 and RHS = 4 (4) — (6 + (4)) = 16-10 =6. Since LHS: 4 RHS, x = 4 is not a solution.

11. (a) When x = —1, LHS = 2 (—1)!/3 —3 = 2(-1) —3 = —2 —3 = —5. Since LHS $ 1,x= —1 is not a solution.

(b) When x = 8 LHS me 13.

—3=2(2)-3=4-—3=1=RHS. Sox = 8isasolution.

0.23

ies

(a) When x= 0, LHS =—— ae e ; = RHS. So x = 0 isa solution. al

os

(b) When x = b, LHS = ga.

15. 19. 23, a7, 29, 31. 33. 35,

Si

6 = 14.45 5x = 20 Sia

,

ea “ is not defined, so x = b is not a solution. 4

17.7 -—2x = 15@2x

=-8

ox =-4

$x+7=30 5x =-40x =-8

21. -3x —3 = 5x -3 0 = 8 @ x =0

Ix+1=4-2x

25.-x+3=4x

99x

=3x

= 4

o3=Sxrex=Z

F—-l=}x+7ex-3=5x+2l 4x =-M ox =-6 2(1—x) =3(1+2x) +5492-2x =3 + 6x +5

2—-2x =8 + 6x

4(y—$) —y =6(5 -—») 2 4y -2-y = 30 -6y

-6 = 8x

x = -}

3-2 = 30 Sy Ny =

y= ZB

x — 4x — 5x —5 =0 © 6x — 2x — 3x — 30 = 0 (multiply both sides by 6) = x = 30 x x+1 Ble x at

= 6x 8x —2x+x4+1= 2x eIxtl=Mxel=lkexr=zh

ade (x — 1) (@ +2) =(@ -2)(e@ -3) 39. (x — 1) (4x +5) = (2x —3)?

x?

+4 -2=x*

—Sx +6

4x? +x —5 = 4x? - 12x49

x-2=-Sx

+6

6x

=8 ex

=F

x -S=-12x 49S 13x = lox

1 4 41. -= as +1=>3=4-+3x (multiply both sides by the LCD, 3x) = —1 = 3x =x

=

=-}

x

43.

2x—-1 4 az) = 5S

2 45. ee

47.

(Ar — 1) = 4

Le

es

3

1

1

x+1

2

3x43

+2)

Ae

10x

— 5

ox £8 > Or

ox

=

B3

[multiply both sides by the LCD, (t — 1) (¢ + 6)] @ 2t —-2 =3t+ 18@—20=t

= 3 (6) — (3x +3) = 2 [multiply both sides by 6 (x + 1)] @ 18—3x —3 =2@—-3x+15=206

3x =-l3e@x=4 1 1 49. See Zz

v4

1

10

ee

et => 10(z2+1)-—S(z4+1)- 2+

3(z+ 1) = 1002 6 3z4+3= 100z = 3 = 972

51. 5

1) = 10(10z) [multiply both sides by 10z (z + 1)] ©

Z

& ir

]

: ri-—2= rye => x —2 (2x — 4) = 2 [multiply both sides by 2 (« — 2)]) @x

:

—44 +8 =2 4 -3x =-6@x

=2..

x — x— But substituting x = 2 into the original equation does not work, since we cannot divide by 0. Thus there is no solution. 3 1 6x+12 53. = 3 (x) = (x +4) + 6x + 12 (multiply both sides byx (x + 4)] = 3x = 7x + 16 = —4x = 16

x4+4° x

x24 4x

= x = —4. But substituting x = —4 into the original equation does not work, since we cannot divide by 0. Thus, there is no solution.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

‘

\

SECTION P8 Solving Basic Equations

21

§5. x7 = 25x =+5 57. 5x? = 15 @x* =3 x =4V3 59. 8x* — 64 =0 x7 -8 =06 x2 =8 ex =4vV8 = 422 61. x* + 16 = 0 © x2 = —16 which has no real solution.

63. (x —3 =Sex-3=4/5 ox =34 S55: 65. x3 = 27@ x =27!/3 =3 67. 0=x*-16=

(x? +4) (x? -4) = (x? +4) (x — 2) (x + 2) .x* +4 = 0 has no real solution. If x — 2 = 0, then x = 2.

Ifx + 2 = 0, then x = —2. The solutions are +2.

69. x+ + 64 = 0 y* = 80> y = +80 = +445. Since length is positive, y = 4./5 = 8.94 inches.

2

43. Let x be the width of the strip. Then the length of the mat is 20 + 2x, and the width of the mat is 15 + 2x. Now the

perimeter is twice the length plus twice the width, so 102 = 2 (20 + 2x)-+ 2 (15 + 2x) = 102 = 40 + 4x + 30+ 4x @ 102 = 70 + 8x + } => eo-+ i =—©> }

6 +t =9 >t =3. Thus it would take Karen 3 hours to paint a house alone.

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HI

CHAPTERP =Review

25

59. Let t be the time in hours that Wendy spent on the train. Then i — t is the time in hours that Wendy spent on the bus. We construct a table:

By train | 40

t

40t

Bybus | 60 | +--+ | 60(4-+ The total distance traveled is the sum of the distances traveled by bus and by train, so 300 = 40t + 60 (4 - t)S

300 = 40r + 330 — 60t © —30 = -20t

et = = = 1.5 hours. So the time spent on the train is 5.5 — 1.5 = 4 hours.

61. Let r be the speed of the plane from Montreal to Los Angeles. Thenr + 0.20r = 1.20r is the speed of the plane from Los Angeles to Montreal.

Montreal to L.A. L.A. to Montreal

:

The total time is the sum of the times each way, so 9} =

55+ 1.2r=2500 -6- 1.2

aie

2500

2500

2500

2500 |2500 55 _ 2500 _2500 r

| ae

6

i

iz

1.2r

+2500 -6 66r = 18,000 + 15,000 66r=33,000 r = 33:99 — 500. Thus the plane flew

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

63. Let x be the distance from the fulcrum to where the mother sits. Then substituting the known values into the formula given, we have 100(8) = 125x

=

800 = 125x / = 120. Thus the length of the lot is 120 feet.

:

67. Let h be the height in feet of the structure. The structure is composed of a right cylinder with radius 10 and height zh anda = cone with base radius 10 and height xh. Using the formulas for the volume of a cylinder and that of a cone, we obtain the

equation 1400m = m (10) ($h) + $7 (10)? ($2) 1400m = 20074+ 100) 126 = 6h + h (multiply both sides by a) © 126 = 7h = h = 18. Thus the height of the structure is 18 feet. Ls 69. Pythagoras was born about 569 BC in Samos, Ionia and died about 475 BC. Euclid was born about 325 BC and died about 265 BC in Alexandria, Egypt. Archimedes was born in 287 BC in Syracuse, Sicily and died in 212 BC in Syracuse. °

CHAPTER P REVIEW Jip (a) Since there are initially 250 tablets and she takes 2 tablets per day, the number of tablets 7 that are left in the bottle after she has been taking the tablets for x days is T = 250 — 2x. (b) After 30 days, there are 250— 2 (30)= 190 tablets left.

(c) We set T = 0 and solve:

T = 250 — 2x = 0

250= 2x

x = 125. She will run out after 125 days.

3. (a) 16 is rational. It is an integer, and more precisely, a natural number. (b) —16 is rational. It is an integer, but because it is negative, it is not a natural number.

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26

CHAPTER P Prerequisites

(c)

16 = 4 is rational. It is an integer, and more precisely, a natural number.

(d) V2 is irrational.

(e) $ is rational, but is neither a natural number nor an integer.

(f) -3 = —4 is rational. It is an integer, but because it is negative, it is not a natural number. 5. Commutative Property of addition.

7. Distributive Property.

Op eee Ae O38 ne -+2=2>4+-=-=2 Ure has iso

9. (a)

Shar

Blo2

)

2 Spee

Pade

4,

1

Ona

Oe

6

ee

11/@)

15)

5)eae

Os

13. x € [-2,6) 9 -2 a

cy are v/

‘

1

EQUATIONS AND GRAPHS

1.1

THE COORDINATE PLANE

1. The point that is 2 units to the left of the y-axis and 4 units above the x-axis has coordinates (—2, 4).

3. The distance between the points (a, b) and (c, d) is \/ (ce — a)’ +(d—- b)?. So the distance between (1, 2) and (7, 10) is

J

(7-1)? + (10 — 2)? = 62 + 82 = 36+ 64 = V/100 = 10.

5. A(5, 1), B (1,2), C (—2, 6),

D(—6, 2), E (—4, —1), F (—2, 0), G (—1,,=3), HQ, —2)

7. (0, 5), (=1,0),(1, -2), and (5,3)

9. {(x,y)|x > 2}

YA

@(0,5)

(-1, oy Ho(1/2, 2/3) 1

11. {@,y) |x = —4}

x

13. {(x,y) |—3 landy < 3}

= {(x, y) |x Oorx > Oandy < 0}

19. {(x, y) |-l (x — 2)? + (y — 5)? =25. 79. Since the circle is tangent to the x-axis, it must contain the point (7, 0), so the radius is the change in the y-coordinates.

That is, r = |—3 — 0| = 3. So the equation ofthe circle is (x — 7)? + (y — (—3))* = 32, which is (x — 7)? + (y + 3)? =9. 81. From the figure, the center of the circle is at (—2, 2). The radius is the change in the y-coordinates, sor = |2 — 0| = 2.

Thus the equation ofthe circle is (x — (—2))? + (y — 2)? = 22, which is (x + 2)? + (y — 2)? =4. . Completing the square gives x2 +y2 —2x+4y+1=0

x2

-2x 4 (2)

2

2

+y? +4y+ (3) =-l+

(2)

2

2

+ (3)

ox*—-2x +14)? 44y4+4=-141446 (x — 1)? + (y +2)? = 4. Thus, the center is (1, —2), and the radius is 2. 85. Completing the square gives x2 + y2—4x +10y+13 a x?

7

2

4x4 (53) +y?+10y+ (4?) = -13+(3)

2

2

+ (+2)

ex? —4x +44 y? + 10y +25 = -13 +4425 © (x — 2)? + (yp+5)? = 16. Thus, the center is (2, —5), and the radius is 4.

87. Completing the square gives x? + y2 +x =Oexr2txt

ey + y? =: OF exrztxt

1 +y=

es

(x+f. Hy = y? = t. Thus, the circle has center (-3, 0)and radius 5.

89. Completing the square givesx? +y*—5x+4y = jcox det (VP) ay24dy4 (2) = Orie x?

sett

Vr t+pt+eK = 4 +e+%

(3.-}) and radius ;:

etre):

= 5 = G eS (—A)os (yet) = i Thus, the circle has center

~

oT: Completing the square gives x? + y* +4x —10y =21

44x +(4) +)?- 1oy + (=}2)" =21+(3) + (=}2)" o> +2)? + (y-5)? = 21 44-425 = 50.

93. Completing the square gives x* + y? + 6x — 12y +45 =0

(x +3)? + (y — 6)? = —45 +9 + 36 = 0. Thus, the center is (—3, 6), and the radius is 0. This is a degenerate

circle whose graph consists only of the point (—3, 6).

Thus, the circle has center (—2, 5) and radius V/50 = 5V2.

. x-axis symmetry: (—y) = xt4x2@ i

—x4 — x2, which is not the same as y = Ht Hhx2, so the graph is not symmetric

with respect to the x-axis. y-axis symmetry: y = cxyt+ (—x)? = x44 x, so the graph is symmetric with respect to the y-axis. Origin symmetry: (—y) = (-x)4 + (—x)? &-y=

x44 x2, which is not the same as y = x4 + x2) so the graph is not

symmetric with respect to the origin.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Ad

CHAPTER 1 Equations and Graphs

97. x-axis symmetry: (—y) = x? + 10x y = —x? — 10x, which is not the same as y = x3 + 10x, so the graph is not symmetric with respect to the x-axis.

y-axis symmetry:

y = (—x)? + 10 (—x) ot 3) =—¢c+

e

“ 2

24 . This equation represents a circle only when —c + .

2

-

> 0. This 2

b2

= 0, and this equation represents the empty set when —c + sam

When the equation represents a circle, the center is

“a7

5) and the radius is ,/—c +

4

2

(5) + (5)

4

b2

equation represents a point when —c + tne

1.3

b

2

?

.

113. Completing the square gives x? +2 +ax+by+c=0

45 b

< 0.

= 5Vae+ b+ — 4ac.

LINES

1. We find the “steepness” or slope of a line passing through two points by dividing the difference in the y-coordinates of these points by the difference in the x-coordinates. So the line passing through the points (0, 1) and (2, 5) has slope Fig

Fy 2)

3. The point-slope form of the equation of the line with slope 3 passing through the point (1, 2) is y —2 = 3 (x — 1). 5. The slope of a horizontal line is 0. The equation of the horizontal line passing through (2, 3) is y = 3. 7. (a) Yes, the graph of y = —3 is a horizontal line 3 units below the x-axis. (b) Yes, the graph of x = —3 is a vertical line 3 units to the left of the y-axis.

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

5 ieee = 2,5) eee |

) 3

et 0-2

3 at aed

0 —(—1)

Habitsoo=

1

x2

—4 eee xX2—x,; O-5

AI a) x2—X]

lm

iBree) (—2

ae

7-2

5

Xj

—5-—(-re) Nope —3 eeeed 6—10 —-4

3 4

= 2-0 17. For ¢1, we find two points, (—1, 2) and (0, 0) that lie on the line. Thus the slope of £; is: m = eats === SM

sl

or £2, we find two points (0, 2) and (2, 3). Thus, the slope of &2 is m = ———— = mur lbh For €3 we find the points x2 — Xj —

wey

(2, —2) and (3, 1). Thus, the slope of £3 is m = ————

1-2) _

= Poy open

x2 — x]

Higa

3. For £4, we find the points (—2,; —1) and

a

litele MerBSO etWEVYOST davesal) esl tee, WEY

(2, —2). Thus, the slope of €4 is m = PREOS

0-4 19. First we find two points (0, 4) and (4, 0) that lie on the line. So the slope is m = as the equation of the line is y = mx +b = —1x

+ 4. Soy =—x+4,orx+y—4=0.

21. We choose the two intercepts as points, (0, —3) and (2, 0). So the slope is m = the equation of the line is y = mx

Gaol —1. Since the y-intercept is 4,

—(-3 5 ( 7 ) = 3. Since the y-intercept is —3,

+b = 3x —3, or 3x —2y -—6 =0.

23. Using y = mx + b, we have y = 3x + (—2) or 3x —y-—2=0. y — yy = m (x — x1), we get y -3 =S(x-2) 25. Using the equation

@—-5x+y=-7T

y — yj = m(x — x1), we get y —7 Il $ (x —1) 27. Using the equation

9 3y-21

5x -y-7=0.

= 2x -2@-2x+3y = 19 ©

2x —3y +19 =0. =

29. First we find the slope, which is m = De . x2 — x}

y-6=-5S(x-l)

6-1 5 EE ee bee ca ee. Substituting into y — yy = m (x — x1), we get 1—2

ad |

@y-6=-5x4+5@5x+y-11=0.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

46

CHAPTER 1 Equations and Graphs

31. We are given two points, (—2, 5) and (—1, —3). Thus, the slope is m = into y — yj, = m (x — xj), we get y —5 = —8[x — (-2)] @ y = —8

Maire Vig’ |ae xg —x, — ll or8x

—3-0 ane SRE We are given two points, (1, 0) and (0, —3). Thus, the slope is m = ee. Sian Y ha

we have y = 3x + (—3) or y = 3x —3 or3x

A|

tees

—l—(-2) +y+11=0.

si

-3 Ga

1

= —8. Substituting

; ; 3. Using the y-intercept,

a

—y—3 = 0.

35: Since the equation of a line with slope 0 passing through (a, b) is y = b, the equation of this line is y = 3. 37. Since the equation of a line with undefined slope passing through (a, b) is x = a, the equation of this line is x = 2. 39. Any line parallel to y = 3x —5 has slope 3. The desired line passes through (1, 2), so substituting into y — y = m (x — x4), we get y —2 =3(x

—1)

@ y= 3x —1lor3x

—y-1=0.

41. Since the equation of a horizontal line passing through (a, b) is y = b, the equation of the horizontal line passing through

(4,5) is7 = 5. 43. Sincex +2y

=6@2y =>-x+60y=

—4x + 3, the slope ofthis line is —5. Thus, the line we seek is given by

y—(-6) =-3 (x -1l) @2y4+12=-x+1lext2y+11=0. 45. Any line parallel to x = 5 has undefined slope and an equation of the form x = a. Thus, an equation of the line is x = —1.

47. First find the slope of 2x + 5y + 8 = 0. This gives 2x +5y+8

=0@5y

= -2x -8@

y= —3x- 8. So the

1 slope of the line that is perpendicular to 2x + 5y +8 = 0ism = —I375 = 3. The equation of the line we seek is

y—(-2)= 3

-(-1l)) @ 2p +4 =5x +5

5x -2y+1=0.

ihe

-4

.

49. First find the slope of the line passing through (2, 5) and (—2, 1). This gives m = ee: = — 4 = |, and so the equation of the line we seek is y-—7=1(x

-—1) Sx-—y+6=0.

Ss (a)

53.

b=-1

b=-6

b y = —2x +b,b = 0, +1, +3, +6. They have the same

(b) y—1=3 (« —(-2)) @2y-2=3( +26 2y —2=3x +6 3x —2y +8 =0.

slope, so they are parallel.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.3 Lines

55.

m=1.5

47

57. y =3 —x = —x +3. So the slope is —1 and the m=0.75

y-intercept is 3.

m= 0.25 m=0

~

Se —025

m=-0.75 m=-L5

y =m (x —3), m = 0, £0.25, +0.75, +1.5. Each ofthe

lines contains the point (3, 0) because the point (3, 0) satisfies each equation y = m (x — 3). Since (3, 0) is on

the x-axis, we could also say that they all have the same

sae iflaltnelobets

kek

1

x-intercept.

59.

—2x +y=7@

y=2x+7.Sotheslopeis2 andthe

y-intercept is 7.

61.4x+5y=10@5y=—4x+10G

y= —$x +2. So

the slope is -2 and the y-intercept is 2. ay

63. y = 4 can also be expressed as y = Ox + 4. So the slope is 65. x = 3 cannot be expressed in the form y = mx + b. So the 0 and the y-intercept is 4.

slope is undefined, and there is no y-intercept. This is a vertical line.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

48

CHAPTER 1 Equations and Graphs

67. 5x +2y — 10 = 0. To find x-intercepts, we set solve for x: 5x +2(0)

-10=0@5x

y= QOand

= 10@x

=2,s0

the x-intercept is 2. To find y-intercepts, we set x = 0 and solve for y: 5(0)+2y-—10=0@2y = 10 Sy =S, so the

y-intercept is 5.

69. 5x _ iy + 1 = 0. To find x-intercepts, we set y= 0 and _solve for x: 5x - 5 (0)+1=06

5x =-lox=-z,

so the x-intercept is —2. To find y-intercepts, we set x = 0 and solve for y:

5 (0) - zy +1=06

fy=ley=3,s0the

y-intercept is 3.

71. y = 6x +4. To find x-intercepts, we set y = 0 and solve forx:0=6x+4@6x

=-4@x=

—}, so the

x-intercept is — z. To find y-intercepts, we set x = 0 and solve for y: y = 6(0) + 4 = 4, so the y-intercept is 4.

73. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation y = 2x + 3 has slope 2. The line with equation

2y — 4x —-5=0@2y

=4x+5Qy=2x4+

5 also has slope 2, and so the lines are parallel.

75. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation —3x + 4y = 4 4yv=3x+40y=

3x + | has slope 3. The line with equation 4x +3y=SQ3y==—44

+5 Qy=

—fx + 3 has

slope -7 =— ya and so the lines are perpendicular.

'

77. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 7x — 3y 3y=7x -2Sey=

fx - q has slope i. The line with equation 9y + 21x

=1@9y = —-2lx -loy=

=2@ -} — 5has

slope =. x — 73, and so the lines are neither parallel nor perpendicular.

aa a

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.3 Lines

49

79. We first plot the points to find the pairs of points that determine each side. Next we

4— Va3r 1 find the slopes of opposite sides. The slope of AB is oa ea aie and the 10-7 3) 5. levees slope o f DC MCAEETD is———— = = ato = —. Since these slope are equal, these two sides are parallel. The slope of 4D is

10-4 a

T-1 EET

Bee ke —3, and the slope of BC is

6 : : ee ie —3. Since these slope are equal, these two sides are parallel.

Hence ABCD

isa parallelogram.

81. ‘We first plot the points to find the pairs of points that determine each side. Next we

bas » 3-1 2 1 find the slopes of opposite sides. The slope of AB is Tere eat aa and the 6-8 = 1 slope of DC is ——— 5 Ses . Since these slope are equal, these two sides

are parallel. Slope of AD

en

ea

11-10

1

6-1 is i

5 aa

—5, and the slope of BC is

= —5. Since these slope are equal, these two sides are parallel.

Since (slope of AB) x (slope of AD) = f x (—5S) = —1, the first two sides are each perpendicular to the second two sides. So the sides form a rectangle.

1+7 4-2 83. We need the slope and the midpoint of the line 4B. The midpoint of AB is ( + ; =) = (4, 1), and the slope of : ABism

=

—2-4 Se

-6 ; : ; ang |e pier ae —1. The slope of the perpendicular bisector will have slope Gira

‘ 1. Using the

point-slope form, the equation of the perpendicular bisector is y — 1 = 1 (x — 4) orx —y—3 =0.

85. (a) We start with the two points (a, 0) and (0, 5). The slope of the line that contains them is 0

b ' PL So the equation

b : Poy : f of the line containing them is y = ee + b (using the slope-intercept form). Dividing by b (since b ¥ 0) gives 44 x xy -_=— e-4+r-=1. b ak: oh

e+ (b) Setting a= 6 and b = —8, we get A

2 = Lendx —3y = 24 e 4x —3y — 24 =0.

87. (a) The slope represents an increase of 0.02° C every year. The T -intercept is the average surface temperature in 1950, or

16% Ce (b) In 2050, t = 2050 — 1950 = 100, so T = 0.02(100) + 15 = 17 degrees Celsius.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

50

CHAPTER 1 Equations and Graphs

89. (a)

y

(b) The slope, —4, represents the decline in number of spaces sold for each $1 increase in rent. The y-intercept is the number of spaces at the flea market, 200, and the x-intercept is the cost per space when the

manager rents no spaces, $50.

20

40

60

80.

100

x

Cane oa[eofe De

91. (a)

(b) Substituting a for both F and C, we have a=2at+32@-fa=32e

a = —40°. Thus both scales agree at

—40°. 93. (a) Using ¢ in place of x and V in place of y, we find the slope of the line

—(b)

using the points (0, 4000) and (4, 200). Thus, the slope is

200 — 4000

me

a

© —3800

ne Ue

= —950. Using the V -intercept, the

linear equation is V = —950t + 4000. (c) The slope represents a decrease of $950 each year in the value of the computer. The V -intercept represents the cost of the computer. ‘(d) When t = 3, the value of the computer is given by

V = —950 (3) + 4000 = 1150. 95. The temperature is increasing at a constant rate when the slope is positive, decreasing at a constant rate when the slope is negative, and constant when the slope is 0.

1.4

SOLVING QUADRATIC EQUATIONS xb + Vb? — 4ac_

1. (a) The Quadratic Formula states that x =

2a

(b) In the equation 5x? —-x-4=0,a=

ek

5. b = —1, and c = —4. So, the solution of the equation is.

-(-D4/-1-4(5)4. 145

se

ae

Se? Or 4.

3. For the quadratic equation ax” + bx +c = 0 the discriminant is D = b? — 4ac. If D > 0, the equation has two real solutions; if D = 0, the equation has one real solution; and if D < 0, the equation has no real solution.

5. x27 —8x +15 =06 7x2

—x

=6

x2

(x —3) (x —5) =0@ x —3 =0orx —5 =0. Thus, x = 3 orx =5. —x -6=06

(x +2) (x —3) =06x+2=0o0rx —3 =0. Thus, x = —2 orx =3.

9. 5x? — 9x —2 =0 © (5x + 1) (x — 2) =0 11. 2s? = 5s +3. 2s? — Ss —3 = 0

5x +1 =0orx —2 =0. Thus, x = —$ orx =2.

(25+ 1) (s — 3) = 0

2s +1 =0 ors —3 =0. Thus, s = —} ors =3.

13. 122? — 442 = 45 > 122? — 442 — 45 = 0 & (62 + 5) (22 —9) = 0

62 +5 = 0 or 2z—9 = 0.Thus, z =—32 orz= 3.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole orin part. Py "

é

“Sy

=

SECTION1.4 Solving Quadratic Equations

15. x2 =5(x + 100)

51

x? = 5x +.500 & x2 — 5x — 500 =0 © (x — 25) (x +20) =0 @ x —25 =0 orx +20 =0. Thus,

Kae 2) OX) = — 20:

17. x7 —8x +1=06x2

— 8% =-1

Ox? - 8x +16 =-14 16

(x 4)? = I ox —4=4JI5

x =44 V5.

19. x? — 6x —11 =0@ x? — 6x = ll @ x? — 6x $9 = 1149 6 (x -3) =U Sx -3 = H2V5 Ox =3 425. 3 2

Mex

+x-ZFsle@rtxr=Texrtxrt4y

=4+4¢(*+3)

=l>x+35

=tlex=-}+1.

So

x=-5-1=-forx=-$41=}.

23. x7 +22x +21 =O x2 +22 = -21 x+11=+10@x

=-—11+10.

@ x2 4.22% 4:11? = -214 11? = -214 121

Thus, x = —1 orx

6 (« +:11)? = 1005

= —21.

25. 5x? + 10x -7 =O ex? +2x-L=0ex?+2xratoertyr+lat+le@+PaBorti=+/¥ ex

=-1+2¥15

27. 2x7 +7xt+4=0ex+4x42s0er4+hr=-2erthir+ PB = -2+ 8 @ (x+4)

2

ti

xt} st/Yoxa=-]447

29. x? —8x +12 =06 (x -2) (x —6) =0 ex =2orx =6. 31. x? + 8x —20 =0

(x + 10) (x —2)

=0 ex

SAP 2x2 +x -3=06

(x -1) 2x +3)

=0Ex—-1=0or2x

ele

=—100rx

=2.

+3 =0. Ifx —1=0, then x = 1; if 2x +3 = 0, then

EE!

Taree

35. 3x2 + 6x —5 =O eax? 42x —$ 0x? 42x =f oxrt2xrt1 =ftlew@+l=$sx41=4/$e x=-1+2¥6.

37. x7 — Fx + J =0 9 9x? — 12x +4 =0 39. 4x2 + 16x

—9 = 0

(2x — 1) (2x +9)

Gx — 2)? =0ex =F. =0

2x —1 =0o0r2x +9 =0. If2x

x=-3.

— 1 = 0, thenx = 4; if2x +9 =0, then

|

—(-3)4 /(-3 -4() 6) 340-1

Al. > = 3(w — 1) > w* —304+3 =0>5 v0 = = — 2(1)

———_ ay

34773

= ——_. Since the 2

discriminant is less than 0, the equation has no real solution.

|

43. 10y? — l6y+5=0>

_ b+ VRP =4ac oF 2a =

~(-16) +¥(-16) - 410) (5) 16 + V256—200 16+ V56_ 8+ V4 2 (10) ii 20 Ra Ee PATH RS

~b+/b2—=4ac —@2)+/@?-43)Q@) ee SES Sig py ama

45. 3x2 +2x+2=05x= aac

-24V4—-24 ay RT -24/-20 . A Since the

ee eS ae

discriminant is less than 0, the equation has no real solution.

47. x2 —0.011x — 0.064 =0=> =

— (—0.011) + y (—0.011)? — 4 (1) (-0.064)

— (—2.450)

+/(—2.450)?— 4 (1) (1.501) _ 2.450 2450++ /6.0025 0025 — — 6.004 6.004 _ 2.450 2.450 ++ /—0.0015

2(1)

Dia LSD ae tarseee tage

Bi

Thus, there is no real solution.

Sites

Sgt? + vot ©

Sgt? + vot — h

=

0.

Using the Quadratic

Formula,

= (v9) + ,/(v0)? - 4 (5 +8)\c h) _ ne 70+ eae t=

wit OR

aeaoe:

53. A = 2x2 +. 4xh © 2x2 +4xh -—A=0. Using the Quadratic Formula,

— (4h)+ ,/(4h)?— 4(2) (—A) aie

gree

= a thgh 4 AREA 2 A)

2 (2)

2(-2nt Van? +24) SAE: 55.

1 Sta

_oy 4 Jan? 424

CTA aOR Den +

1 s+b

=-

1 c¢

ees

4

a ee eee 2

©6c(s+b)+c(s+a) =(s +a) (s +b) Secs +bce+cs

+ac=s*+as+bs+abe

s? + (a+b —2c)s + (ab — ac — bc) = 0. Using the Quadratic Formula,

—(a+b—2c)+,/(a +b —2c)* — 4(1) (ab —ac — be)

2 (1) —(a+b—2c) + Va? + b? + 4c2 + 2ab — 4ac — 4be — 4ab + 4ac + 4be

—(a+b —2c)+ Va? +b? 4 4c? — 57.

D = b* — 4ac = (-6)? — 4(1) (1) = 32. Since D is positive, this equation has two real solutions.

59. D = b* — 4ac = (2.20)? — 4(1) (1.21) = 4.84 — 4.84 = 0. Since D = 0, this equation has one real solution. 61. D = b? — 4ac = (5)? — 4 (4) (2) = 25 — 26 = —1. Since D is negative, this equation has no real solution. 63. a*x? + 2ax

+1 =06

(ax +1)* =0@ax+1=0.

Soax+1=0thenax

1 =-lex=--. a

65. We want to find the values of k that make the discriminant 0. Thus k2 — 4 (4) (25) = 0 @ k? = 400 @ k = +20. 67. Let n be one number.

Then the other number must be 55 — n,since n + (55—n)

the product is 684, we have (n)(55—n)

—(—55)+,/ (—55)?—4(1) (684)

=

684 = 55n —n?

55+./3025—2736

1 a fh al

=

55+./289

= ES

=

55.

Because

6840

n2 — 55n + 684 =

55+17

sok =

55417

05

2

= s = 19. In either case, the two numbers are 19 and 36.

Hp

69. Let w be the width of the garden in feet. Then the length is w + 10. Thus 875 = w (w + 10) © w + 10w —875=06

(w + 35) (w — 25) = 0. So w + 35 = 0 in which case w = —35, which is not possible, or w — 25 = 0 and so w = 25. Thus the width is 25 feet and the length is 35 feet. 71. Let w be the width of the garden in feet. We use the perimeter to express the length / of the garden in terms of width. Since

the perimeter is twice the width plus twice the length, we have 200 = 2w + 2/ = 2] = 200 — 2w =/ = 100 — w. Using the formula for area, we have 2400 = w (100 — w) = 100w — w? w* — 100w + 2400 = 0 Sow

—40 =0@w

= 40, or w —60

=0 Sw

(w — 40) (w — 60) = 0.

= 60. If w = 40, then / = 100 — 40 = 60. And if w = 60, then

1 = 100 — 60 = 40. So the length is. 60 feet and the width is 40 feet.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

, *

'

SECTION 1.4 Solving Quadratic Equations

53

1h The shaded area is the sum ofthe area of a rectangle and the area of a triangle. So A = y (1) + 4 (vy) () = sy? + y. We are given that the area is 1200 cm2, so 1200 = sy? + yo y* +2y — 2400 = 0

(y + 50) (vy — 48) = 0. y is positive, so

y = 48 cm.

75. Let x be the length of one side of the cardboard, so we start with a piece of cardboard x by x. When 4 inches are removed from each side, the base of the box is x — 8 by x — 8. Since the volume is 100 in?, we get 4(x — 8)? = 100

x? — 16x + 64 = 25 > x? — 16x +39 =0

© (x — 3) (x — 13) = 0. Sox =3 orx = 13. But x=3 is notpossible, since

then the length of the base would be 3 — 8 = —S, and all lengths must be positive. Thus x = 13, and the piece of cardboard is 13 inches by 13 inches.

77. Let w be the width of the lot in feet. Then the length is w + 6. Using.the Pythagorean Theorem, we have

w? + (w +6)? = (174)? & w2 + w2 + 12w + 36 = 30,276 & 2w2 + 12w — 30240 = 0 (w + 126) (w — 120) = 0. So either w + 126 = 0 in which case

w? + 6w — 151200 =06

w = —126, which is not possible, or w — 120 = 0 in

which case w = 120. Thus the width is 120 feet and the length is 126 feet.

79. Let x be the rate, in mi/h, at which the salesman drove between Ajax and Barrington. ime 120

Ajax — Barrington

Barrington We have used the equation time =

dist

Se

> Collins

150

‘to fill in the “Time” column of the table. Since the second part of the trip i

120 1 150 on took 6 minutes (or jo4 hour) more than the first, we can use the time column to get the equation ——- + — x 10° x+10

120 (10) (x + 10) + x (x + 10) = 150 (10x) = 1200x + 12,000 + x? + 10x = 1500x x? — 290x + 12,000 = 0 -(Fy

y (—290)?—4(1)(12, 200) 41) 12,000), 4 290 ./84,100—48,000 tus 48,000 _ a

— 2904190 _ 145495, Hence, the salesman

drove either 50 mi/h or 240 mi/h between Ajax and Barrington. (The first choice seems more likely!)

81. Let r be the rowing rate in km/h of the crew in still water. Then their rate upstream was r — 3 km/h, and their rate downstream was r + 3 km/h.

Upstream Downstream

Since the time to row upstream plus the time to row downstream was 2 hours 40 minutes = : hour, we get the equation

6

ON

6

ato

8

© 6(3) (r +3) +63) ¢ —3) = 8(r —3) (7 +3) © 187 +544 18 —54 = 8? -22&

0 = 8r? = 36r = 72 = 4 (2r? —9r -— 18) =4(r+3)(¢ —6) @2r+3=0orr

—6=0. If27 +3 =0, thenr = —3,

which is impossible because the rowing rate is positive. If — 6 = 0, thenr = 6. So the rate of the rowing crew in still water is 6 km/h.

83. Using ho = 288, we solve 0 = —16r? + 288, for t> 0. So 0 = —16t? + 288 = 16r7 = 288 @ t? = 18>

t = +,/18 = +32. Thus it takes 3\/2 © 4.24 seconds for the ball the hit the ground. 85. We are given vo = 40 ft/s.

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54

CHAPTER 1 Equations and Graphs

(a) Setting h= 24, we have 24 = —16r2 +401 & 161 — 40¢ +24 = 0. 8 (21?— 51 +3) =08(2t—3)(t-1) =0 Keo

aren

OL

ee 1}. Therefore, the ball reaches 24 feet in 1 second (ascending) and again after 15 seconds

(descending).

(b) Setting A = 48, we have 48 = —16r? + 40t < 16¢? — 40 + 48 = 0 5+/25—48 irr 5+ /-23 4 never reaches a height of 48 feet.

217 —S5t+6=06

. However, since the discriminant D < 0, there is no real solution, and hence the ball

(c) The greatest height h is reached only once. So h = —16t? + 40t © 16t2 — 40¢ + h = 0 has only one solution. Thus D = (—40)? — 4 (16) (h) = 0 & 1600 — 64h = 0 & h = 25. So the greatest height reached by the ball is 25 feet.

(d) Setting h = 25, we have 25 = —16t7 + 40t t (2 — 5) = 0 >t = 0 (start) or f = 25.

So the ball hits the ground in 2} s. 87.

;

(a) The fish population on January 1, 2002 corresponds to t = 0, so F = 1000 (30417(0)— (0) = 30, 000. To find when the population will again reach this value, we set F = 30, 000, giving

30000 = 1000 (30+17t — ?) = 30000 + 17000 — 100017 w = 30 or w = 180. Since 180 ftis too wide, the width of the lawn is 30 ft, and the factory is 120 ft by 180 ft.

91. Let t be the time, in hours it takes Irene to wash all the windows.

Then it takes Henry t + 3 hours to wash all

the windows, and the sum of the fraction of the job per hour they can do individually equals the fraction of the 1 1 job they can do together. Since 1 hour 48 minutes = 1 + & = 1+ t = 2, we have — + ———- = ti

|

t

t+}.

ifs

2

2

a are = 3 => 9(2t+3) +2(9t) = St (2t +.3). 18 + 27 + 18¢ = 10/7 + 15¢ x = +2.Ja. Thus the four solutions are +./a and +2/a. Lp J/x+a+t Jx —a = /2/x +6. Squaring both sides, ee lave

x+a42(J/x +a) (/x —a)+x-a =2(« + 6)

2x +2 (Vx +a) (Vx —a) = 2x +122 (Vx +a) (Vx —a) = 12

= (/x +a) (x —a) = 6. Squaring both sides again we have (x +a) (x —a) =36@ x? ~a? =36@x* =a? + 36 ex=

+a? + 36. Checking these answers, we see that x = —/a? + 36 is not a solution (for example, try substituting

a = 8), but x = Va? + 36 isa solution.

~ ©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

60

CHAPTER 1 Equations and Graphs

Len 79. Let x be the number of people originally intended to take the trip. Then originally, the cost of the trip is ——. After 5 people x : 900 900 * 4500 cancel, there are now x — 5 people, each paying —— + 2. Thus 900 = (x — 5) (= + 2)= 900 = 900 +2x — —— — 10 ‘ vit 5 a x 4500 € 0 = 2x — 10 — —— © 0 = 2x? — 10x — 4500 = (2x — 100) (x + 45). Thus either 2x — 100 = 0, sox = 50, or oo x +45 = 0, x = —45. Since the number of people on the trip must be positive, originally 50 people intended to take the trip. $1. We want to solve for t when P = 500. Letting u = \/f and substituting, we have 500 = 3¢.+ 10/7 + 140 500 = 3u2+ 10u+ 140 0 = 3u?+ 10u — 360Su =

=5 4/1105

t=u4= vt=u

:

—5 + /1105 . Since u = ./f, we must have u > 0. So 3

~9414 St =~ 88.62. So it will take 89 days for the fish population to reach 500.

——

83. Let x be the height of the pile in feet. Then the diameter is 3x and the radius is 3x feet. Since the volume of the cone is

3x

\2

1000 ft3, we have ~ (=) 3

>

3x3

x = 1000 ——~— = 1000

. Let r be the radius of the larger sphere, in mm. Pe =23433444

or

4000

4000

3r

3

x3 = —— @x = J —

4

7.52 feet.

Equating the volumes, we have tar = tn (23+ 3° 4 4°) S

=99 or = 99 © 4.63. Therefore, the radius of the larger sphere is about 4.63 mm.

87. Let x be the length, in miles, of the abandoned road to be used. Then the length of the abandoned road not used is 40 — x, and the length of the new road is ,/102 + (40 — x)? miles, by the Pythagorean Theorem.

Since the

cost of the road is cost per mile x number of miles, we have 100,000x + 200,000/x2 — 80x + 1700 = 6,800,000

© 2,/x2 — 80x +1700 = 68 — x. Squaring both sides, we get 4x2 — 320x + 6800 = 4624 — 136x +x2 © 3x2 — 184x' +2176

=O

x = Ht

A2850—26N2 = iiss =e

ae or x = 16. Since 454 is longer than the existing

road, 16 miles of the abandoned road should be used. A completely new road would have length and would cost

10? + 402 (let Se

0)

1700 x 200,000 ~ 8.3 million dollars. So no, it would not be cheaper.

89. Let x be the length of the hypotenuse of the triangle, in feet. Then one of the other sides has length x — 7 feet, and since the perimeter is 392 feet, the remaining side must have length 392 — x — (x — 7) = 399 — 2x. From the Pythagorean Theorem,

we get (x — 7)? + (399 — 2x)? = x? © 4x? — 1610x + 159250 = 0, Using the Quadratic Formula, we get

x=

1610+

Es

?

;

leit zo (159250) = 50/4410 Zs Nel 2t) and so x = 227.5-or x = 175: But ifx = 227.5, then the

side of length x — 7 combined with the hypotenuse already exceeds the perimeter of 392 feet, and so we must have x = 175. Thus the other sides have length 175 — 7 = 168 and 399 — 2 (175) = 49. The lot has sides of length 49 feet, 168 feet, and

175 feet.

|

91. Since the total time is 3 s, we have 3 = yu 4 7090" Letting w = Jd, we have 3 = Z0+ 7 © 2w? + 545w — 6540 =04w

—545+ 591.054

= SOG

is 132.6 ft deep.

1.7

= Tp

+7H-3 sr

_~

ks. Since w > 0, we have /d = w © 11.51, so d = 132.56. The well

°

SOLVING INEQUALITIES

1f: (a) Ifx 1. Interval:

(—oo, —2] U [1], 00).

33. |2x — 3] < 0.4 —0.4 < 2x —3 < 0.42.6 < 2x 9x2 — 16y2 = 144, so the

graph is symmetric about the origin. (b) To find x-intercepts, we set y = 0 and solve for x: 9x2 — 16 (0) = 144 © 9x2 = 144

x = +4,50 the x-intercepts

are —4 and 4. To find y-intercepts, we set x = 0 and solve for y: 9 (0)? — 16y? = 144 16y? = —144, so there is no y-intercept.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

80

CHAPTER 1 Equations and Graphs

29. x2 + 4xy + y? =

(a) x-axis symmetry: replacing y by —y gives x? + 4x (—y) + (—y)* = 1, which is different from the original equation, so - the graph is not symmetric about the x-axis.

y-axis symmetry: replacing x by —x gives (—x)? + 4 (—x) y + y* = 1, which is different from the original equation, so the graph is not symmetric about the y-axis. Origin symmetry: replacing x by —x andy by —y gives (—x)* +4 (—x) (—y) + (-y)? =1lex*+ 4xy +y2 =,1,'so

the graph is symmetric about the origin. (b) To find x-intercepts, we set vy= 0 and solve for x: x7 +4x(0) +0? =1e x? =1e>x

=+1,so the x-intercepts are

—1 and 1. To find y-intercepts, we set x = 0 and solve for y: 07 +4(0)y+y*=ley?=

1 =>y = +1, so the y-intercepts are

—l and 1. 31. (a) We graph y = x? — 6x in the viewing rectangle

33. (a) We graph y = x3 — 4x? — 5x in the viewing

[—10, 10] by [—10, 10].

rectangle [—4, 8] by [—30, 20].

(b) From the graph, we see that the x-intercepts are 0

(b) From the graph, we see that the x-intercepts are —1,

and 6 and the y-intercept is 0.

0, and 5 and the y-intercept is 0.

35. (a) The line that has slope 2 and y-intercept 6 has the slope-intercept equation

(c)

y=2x+ 6.

(b) An equation of the line in general form is 2x —

y + 6 = 0.

37. (a) The line that passes through the points (—1, —6) and (2, —4) has slope

Sot-6)

ae

i Leader 2gWe

a

:

ee |fendOd

ea

|

(c)

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER1

39. (a) The vertical line that passes through the point (3, —2) has equationx = 3.

Review

81

(ec)

(b) x =3 Ox -3=0.

41. (a) 2x —Sy = 10 @5y =2x-10G y= 2x — 2, so the given line has slope

(ce)

fi

= 2. Thus, an equation of the line passing through (1, 1) parallel to this line is y —1 = 2(x -1l) @ y= 2x43.

(b) y = 2x +2 Sy =2x +3. 2x — Sy +3 =0.

43. (a) The line y = 5x — 10 has slope 5, so a line perpendicular to this one has _ (c)

ya

slope ShY = —2. In particular, the line passing through the origin perpendicular to the given line has equation y = —2x. (b) y= —2x © 2x+y =0.

45. The line with equation y = ee — | has slope —}. The line with equation 9y + 3x

+3 =0@9y

= —3x -3 ©

= —4x - 4 also has slope —1 so the lines are parallel. 47. (a) The slope represents a stretch of 0.3 inches for each one-pound increase in weight. The s-intercept represents the length of the unstretched spring. (b) When w =’5, s = 0.3 (5) +2.5 = 1.54 2.5 = 4.0 inches.

49. x2 - 9x +14 =08

(x —7) (x —2) =0

Siz 2x2 +x =162x27+x-1

=08

204 ia

ca

Lyre, sorither 24 =

10

2k

=

i=

1:orx+1 wines

;

53. 0 = 4x? — 25x = x (4x? - 25) Sh

2x +5 =0

x =T ore =2.

(ree '5)Ox ¥ 5) = (0. So eitherx = 0: or 2x =) = 0 Sexe

5 Sox=

3; or

2x =-Serx =-3.

55. 3x7 +4x -—1=0> ese.

)(-1) (4)?-4(3ao —(4)4y E23), ers diy —

2 saad & eval 44/28 _ =4427 spelt 12 _ ee 44/16 ee ES ay

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82

CHAPTER 1 Equations and Graphs

1 2 57s Agar eR] =36 (x-1)4+2(x) ; xXx

o—bds/Podae

2x = 3x? —3x

6/04) 626

6

cant

x4 — 8x? 9 = 0. (x? -9) (x? +1) = 0. (x3)

6

20 = 3x27 -Ox

+15

2(34v6) 7; 33/6

=

7

6

a

:

+3) (x?+1) =0>x-3=04%x=3, ory +3

ee

x = —3, however x2 + 1 =0 has no real solution. The solutions are x= +3.

61. x71/2 — 2x 1/2 4 43/2 =~0ex-1/2 (1— 2x +x?) =0e6 x7!/2 (1 — x) = 0. Since x7!/2 - 1/./x is never 0, the only solution comes from(1 —x? =0e1-x=06x=1.

63. | 65.

67.

69.

tk ea

ae

ey

SOY

=

OL cee

(a) (2—31)+(14+4i) =(24+ 1) + (-34+4i =3+i (b) (2 +7) (3 —2i) =6 —41 +31 — 2/2 =6-i1+2=8-i

(a) 442i

442i

2+i _8+8i+2i* _8+8i-2

648i _6) 3,

ya nn on ba 2 en Cy oe ee Ce enti ye (b) (1— /=1) (1+ V=1) = -) 04+) =14i-i-i#? =14+1=2

x74+16=06x7 =-l6@x =44i

71. x7 +6x+10=06x= U3

=

x4 — 256 = 0

—b+Vb?-4ac

-64/67—4(1) (10)

2a

2(1)

-6+V36—40 |

—3+i

(x? - 16)(x? +16) = 0.x = 44 orx = t4i

75. Let r be the rate the woman runs in mi/h. Then she cycles at r + 8 mi/h.

Since the total time of the workout is 1 hour, we have

. Multiplying by 2r (r + 8), we if

get 4(2r)

+ 2.5(2)( +8) =

—3+4,/ (3)?—4(2)(—40

ae hese

2r(77°+8)

2;

© 8& + 5r +40

vi

:

=

2dr?+ 1l6r 0

;

=

277 +37

-40 >

:

s fe BEE = eee his) Since r > 0, we reject the negative value. She runs at

p = 34/329 ~ 3.78 mih. 77. Let x be the length of one side in cm. Then 28 — x is the length of the other side. Using the Pythagorean Theorem, we

havex? + (28 —x)? = 20? —1l 93x > 9x > -3. Interval:

(—3, co). ( )

81.3—x06

Review

83

(x — 2) (x + 6) > 0. The expression on the left of the inequality changes sign where x = 2 and where

x = —6. Thus we must check the intervals in the following table. Interval: (—oo, —6) U (2, 00).

Sign of x — 2

-

Sign of x + 6

+

Graph: ——-o————_—_—_——__o——_» :

0)

z

Sign of (x — 2) (x + 6)

2x.+5 2X 4-5 ax+5 x+1 x+4

-4

87.

x-4

Peres

x2 —47~

x—-4

(x — 2) (« +2)

4

< 0. The expression on the left of the inequality changes sign where x = —2, where x = 2,

and where x = 4. Thus we must check the intervals in the following table.

5

Sign of x — 4

_

-

+

Sign of x —2

25

ah

a

Sign of x +2 x= 4 Sign of ———_———_

aE

ar

AF

Soresrcess

cee F

an

Since the expression is not defined when x = +2,we exclude these values and the solution is (—o0, —2) U (2, 4]. Graph: P

89. |x —5)

=2

2

< 3-3

4

1@2x 2x

< —2 0@x

> 0. Case 2:2x+1 0 © (8 — 3x) (3 +x) > 0. The expression

on the left of the inequality changes sign where 8 — 3x =0@

-—3x

=-8 Sx

= g; or where x = —3. Thus we must

check the intervals in the following table.

Kee

eC

Sign of 8 — 3x

+

Sign of 3 + x

-

eee

+

8 Interval: ‘ [-3. §.

Graph:

Be

a

1 a real number we must have x — x4 > 0@ x (1—x°) >0ex(1-x) (b) For _———— to define Vx —x4

(1are! +x?) B0)

The expression on the left of the inequality changes sign where x = 0; or where x = 1; or where 1

x=

—1+./12—4(1) (1 2(1)

+x + x27=0>

= 1£V1=4 which is imaginary. We check the intervals in the following table. Interval: (0, 1).

Sign of x

Sign of 1 — x

1, (ca ae pane

[sinofe (ree) Sign of 1+x 4x4

95. From the graph, we see that the graphs of y = x? — 4x and y = x + 6 intersect at x = —1 and x = 6, so these are the solutions of the equation x2 — 4x = x + 6.

97. From the graph, we see that the graph of y = x” — 4x lies below the graph of y = x + 6 for —1 < x < 6, so the inequality

x? —4x < x + 6 is satisfied on the interval [—1, 6].

.

99. From the graph, we see that the graph of y = x? — 4x lies above the x-axis for x < 0 and for x > 4, so the inequality x* 2 —4x > 0 is satisfied on the intervals (—oo, 0] and [4, oo).

101. x2 — 4x = 2x +7. We graph the equations yj = x? — 4x 103. x4 — 9x2 =x —9. We graph the equations y, = x4 — 9x? and y2 = 2x +7

in the viewing rectangle [—10, 10] by

[—5, 25]. Using a zoom or trace function, we get the solutions

x = —1 andx = 7.

and y2 = x — 9 in the viewing rectangle [—5, 5] by :

[—25, 10]. Using a zoom or trace function, we get the solutions x + —2.72, x + —1.15,x = 1.00, and.x © 2.87.

a

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or inpart. b>

a

aeraty |

-

CHAPTER1

105. 4x — 3 > x”. We graph the equations yj) = 4x —3and y2 =x? in the viewing rectangle [—S, 5] by [0, 15]. Using a zoom or trace function, we find the points of intersection

107.x4—4x?

Review

< 5x — 1. We graph the equations

yy = x4 — 4x? and oS

5x — | in the viewing rectangle

[—5, 5] by [—5, 5]. We find the points of intersection are

are atx = 1 and x = 3. Since we want 4x — 3 > x2, the

atx + —1.85, x ~ —0.60, x ~ 0.45, and x = 2.00. Since

solution is the interval [1, 3].

we want x4 — 4x? < 5x — 1, the solution is

(—1.85, —0.60) U (0.45, 2.00).

109. Here the center is at (0,0), and the circle passes through the point (—5, 12), so the radius is

(—5 — 0)? + (12 —0)? = 254 144 = /169 = 13. The equation of the circle is x2 + y? = 1327 & x? 4 y? = 169. The line shown is the tangent that passes through the point (—5, 12), so it is perpendicular to the line through the points (0, 0) and (—5, 12). This line has slope my = m=

1

a,

1

12-0 a0

12 ORE The slope of the line we seek is

5 : Fy an —. Thus, an equation ofthe tangent line is y— 12 = a (x«+5)ey-12=

S12/5:oul2

x =f B oS

y= px t+lO o 5x - 12y + 169 =0. 111. Since M varies directly as z we have M = kz. Substituting M = 120 when z = 15, we find 120 = k (15) @k = 8.

. Therefore, M = 8z. k 113. (a) The intensity / varies inversely as the square of the distance d, so ] = rh k (b) Substituting 7 = 1000 when d = 8, we get 1000 = @ (c) From parts (a) and (b), we have J =

© k = 64,000.

0 64,000 > Fm Substituting d = 20, we get J = = 160 candles.

(20)?

115. Let v be the terminal velocity of the parachutist in mi/h and w be his weight in pounds. Since the terminal velocity is

directly proportional to the square root of the weight, we have v = k,/w. Substituting » = 9 when w = 160, we solve 9 , Ae} for k. This gives 9 = k./160 .

85

86

CHAPTER 1 Equations and Graphs

CHAPTER 1 TEST

1. (a)

y

There are several ways to determine the coordinates of S. The diagonals of a

square have equal length and are perpendicular. The diagonal PR is horizontal P

R

and has length is 6 units, so the diagonal QS is vertical and also has length 6.

al

Thus, the coordinates of S are (3, 6). 1

:

(b) The length of PQ is «/(0 — 3)2:4+ (3 0)? =-/18 = 3/9, So the arene es

PORS is (3v2) =

3. (a)

y

(b) The distance between P and Q is

*o (c)

¥ e

d(P, Q) = J(-3-5)* +(1—6)? = 64 +25 = V/89. 3 4 Sin | The midpoint is ( 2 “¥) = (1,5). 2 2

1

x

5

1-6

5

(d) The slope ofthe line is Tyee Fike = 3°

(e) The perpendicular bisector of

PO contains the midpoint, (1; 5),and it slope is

the negative reciprocal of 3. Thus the slope is =3) = -§. Hence the equation

isy-$=-$0-loy=—§x4+$84+f=-814+9 y=

That is,

—§x + a

(f) The center of the circle is the midpoint, (1; 5 , and the length of the radius is 51/89 . Thus the equation of the circle

whose diameter is PQis (x — IP +(y- iy =

4 ve) = (x-1+(y-J)

= ®

5. (a) x =4— y*. To test for symmetry about the x-axis, we replace y with —y:

x =4-(-y)? ox =4-— y’, so the graph is symmetric about the x-axis. To test for symmetry about the y-axis, we replace x with —x: —x = 4 — y? is different from the original equation, so the graph is not symmetric about the y-axis. For symmetry about the origin, we replace x with —x and y with —y:

—x =4-(-y)?

@ —x =4-— y?, which is different from the original

equation, so the graph is not symmetric about the origin. To find x-intercepts, we set y = 0 and solve for x:

x = 4 — 0? = 4, so the

x-intercept is 4.

To find y-intercepts, we set x = 0 and solve for y:: 0 = 4 — ” ey=4 —4. Expressing in standard form we have: —4 < x < 3.

’

,

Int nterval: 1: [[—4, )3). Graph: Pp ———0—_> “e :

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88

CHAPTER 1 Equations and Graphs

(b) x (x — 1) (x +2) > 0. The expression on the left of the inequality changes sign when x = 0, x = 1, and x = —2. Thus we must check the intervals in the following table.

Sign of x Sign of x — 1 Sign of x+2

From the table, the solution set is {x | -2 Graph:

—.———___—_.--

i

ae

0

0 0. We make a table:

P| 0,0) | 8) |600) |

oon ee ed a fsimofee 9] + | - | + Sign of x

+

+

Thus the domain is (—oo, 0] U [6, 00). &

69.

x —4> 0x

-

71. 2x

Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have

> 4. Thus the domain is (4, 00).

1)? . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have

-1>0@x>

5. Thus the domain is (3,00).

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94

CHAPTER 2 Functions

13. To evaluate f (x), divide the input by 3 and add gEto the result.

BMIO=

tee

+5

(c)

(b)

75. Let T (x) be the amount of sales tax charged in Lemon County on a purchase of x dollars. To find the tax, take 8% of the purchase price.

(a) T (x) = 0.08x

(c)

(b)

Ts f@) =

1 ifx is rational

Parana e 5 ifx is irrational

;

a

y

ater

f

The domain of / is all real numbers, since every real number is either rational or

irrational; and the range of f is {1, 5}.

the

(a) V (0) = 50 (1isor = 50 and V (20) = 50 (1a me iG

(c)

Popo| hofas| raofo |

(b) V (0) = 50 represents the volume of the full tank at time ¢ = 0, and

V (20) = 0 represents the volume of the empty tank twenty minutes later.

(d) The net change in V as ¢ changes from 0 minutes to 20 minutes is

V (20) — V (0) = 0 — 50 = —S0 gallons.

81.

0.5c)*

.75¢)?

(a) L(0.5c) = 10,/1 go _ ~ 8.66 m, L(0.75c) = 10,/1Btabodh * 6.61 m, and G

L(0.9¢e) = 10,/1 —

(0.9¢)? c2

bs

= 4.36 m.

(b) It will appear to get shorter.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Ren.

SECTION 2.2 Graphs of Functions

83. (a) v (0.1) = 18500 (0.25 = 0.17) = 4440,

(c)

v (0.4) = 18500 (0.25 « 0.4”) = 1665. (b) They tell us that the blood flows much faster (about 2.75 times faster)

0.1 cm from the center than 0.1 cm from the edge. (d) The net change in V as r changes from 0.1 cm to 0.5 cm is

V (0.5) — V (0.1) = 0 — 4440 = —4440 cm/s.

85. (a) Since 0 < 5,000 < 10,000 we have T (5,000) = 0. Since 10,000 < 12,000 < 20,000 we have T (12,000) = 0.08 (12,000) = 960. Since 20,000 < 25,000 we have T (25,000) = 1600 + 0.15 (25,000) = 5350. (b) There is no tax on $5000, a tax of $960 on $12,000 income, and a tax of $5350 on $25,000.

87. (a) T (x) =

75x

)

150+ 50(x —2)

if0 0 for all x.

y

0

y

3h

(c) f (x) is always increasing, and f (x) < 0 for all x. :

(b) f (x) is always decreasing, and f (x) > 0 for all x.

0

x

(d) f (x) is always decreasing, and f (x) < 0 for all x. *t

*

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108

CHAPTER 2 Functions

69. (a) Ifx =a isa local maximum of f (x) then

(ec) Leta

Ce)

x4 +x? — 6x +9. From the graph, we

f (a) = f (x) = 0 for all x around x = a. So

see that f (x) has a minimum at x = 1. Thus g (x)

[g (a) > [g (xf? and thus g(a) > g(x).

also has a minimum at x = | and this minimum

Similarly, if x= b is a local minimum of f (x), then

value isg(1) = V14 + 1? -6(1) +9 = V5.

i

=

4

}

=

Ff (x) = f (6) = 0 for allx around x = b. So

[e(x)]? > [g ()]° and thusg(x) > g (). (b) Using the distance formula,

g(x) = J(x — 3)? + (x2 -0)" = Vx44x2 —6x +9

~ 2.4

AVERAGE RATE OF CHANGE OF A FUNCTION

100 mil 1. If you travel 100 miles in two hours then your average speed for the trip is average speed = ira 3. The average rate of change of the function f(x)

=

x? between x

=

1 andx

=

= 50 mi/h.

5 is

) 2 eel ee fG)=7) A Se bat aaa a average rate of change = LE)-fO Shavit Ue ee

5. (a) Yes, the average rate of change of a function between x = a and x = J is the slope of the secant line through (a, f (a)) rae

and (b, f (b)); that is, fO)~ £0)

(b) Yes, the average rate of change of a linear function y = mx + 5 is the same (namely m) for all intervals. 7. (a) The net change is f (4) —

f (1) =5—3

=2.

5-3 (b) We use the points (1, 3) and (4, 5), so the average rate of change is ce

+2 es

9. (a) The net change is f (5) — f (0) = 2-6 = —4. 2-

(b) We use the points (0, 6) and (5, 2), so the average rate of change is 5

—4

;= 5%

11. (a) The net change is f (3) — f 2) = [3 G)— 2] - [3 Q) —-2] =7-4=3. (b) The average rate of change is poi

= -= 3.

13. (a) The net change is A (1) —h (—4) = [-1 + 3]= [--4) + 3]= 5 - y =—5,. h(i) —h(-4 (b) The average rate of change is ere

= = =-—1,

15. (a) The net change is h (6) — h (3) = [2(6)? - 6|= [2(3)2 — 3|66S 1S SOT (b) The average rate of change is ao)

= ~ Sly}

17. (a) The net change is f (10) — f (0) = [103eA (10) = [o°55" (0°) = 600 — 0 = 600. STO (b) The average rate of change is FU100

SOS 0

19. (a) The net change is f (3 +A) — £3) = [s(Be n)?|os[s@)7] = 45 4+ 30h + 5h2 — 45 = 5h? +30h.

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——~

SECTION 2.4 Average Rate of Change of a Function

fB+h)—f(3)

(b) The average rate of change is Arad

ere) =

5h? +30h i

109

= 5h + 30.

21. (a) The net change is g (a) —g(1) = --7= al a l—a

(b) The average rate of change is SARs EMD ie sate a-1

We eeg

a-—1

a(a—-1)

i

et a

2 2 23. (a) The net change is f (a+h) — f(a) = Tay eae ae a (b) The average rate of change is

“OS SE

(a+h)—a

aie! URAC HCE BRIT «a OP h

-—

a

ah(ath) . a(at+h)

25. (a) The average rate of change is

1

1

fatn-s@ _[2@++3|-[3a+3] (a+h)-a

h

da4dn43—fa-3 *

h

Sh 1 Wisy

foes:

(b) The slope of the line f (x) = Sx + 3 is 5 which is also the average rate of change,

27. The function f has a greater average rate of change between x = 0 and x = 1. The function g has a greater average rate of change between x = 1 and x = 2. The functions f and g have the same average rate of change between x = 0 and x = 1.5.

. W (200) — W (100) 50 — 75 h: i 29. Theaverage rate of change is te 700 100 700-100

—25 ' = 100

1 q fi/d Wwday..

1,591 — 856 735 lation isis—————_. 3001 — 1998 = —— 5 = 245 persons/yr. : 31. (a) The average rate of f chchange of f population ._. 826—1,483 (b) The average rate of change of population 7004 = 2002 t is ——_____

—657

5 = ~ = ——__

328.5328. persons/yr.

(c) The population was increasing from 1997 to 2001. (d) The population was decreasing from 2001 to 2006.

. 635 — 495 33. (a) The average rate of change of sales is mee 513 — 495 is ——___—_ (b) The average rate of change of f sales is 5004 2003

140 10 18 i

SS

14 players/yr. / * 18 players/yr.

410 — 513 —103 (c) The average rate of change of sales is 3005-2004 = —1 = —103 players/yr.

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110

CHAPTER 2 Functions

(d)

DVD players sold | Change in sales from previous year

Sales increased most quickly between 2006 and 2007, and decreased most quickly between 2004 and 2005. 35. The average rate of change of the temperature RCI se AUP =

20 —0 T (40) — T (20 oe

caw:

20 —0 89 — 119 Peay oh

=

rs

=

20

30 one

of the soup over the first 20 minutes is

-—4.05° F/min.

Over the next 20 minutes,

it is

‘ : —1.5° F/min. The first 20 minutes had a higher average rate of change of

temperature (in absolute value).

37. (a) For all three skiers, the average rate of change is

d(10)—d(0) _ 100

10-0

10

=U

(b) Skier A gets a great start, but slows at the end of the race. Skier B maintains a steady pace. Runner C is slow at the beginning, but accelerates down the hill.

39.

16 (3.5)* — 16 (3)? _

3.525 16 (3.1)? — 16 (3)? _ epee

16 (3.01)? — 16(3)? _ 3. iee3 16 (3.001)? — 16 (3)? 3.001 — 3

97.6

96.16 96.016

16 (3.0001)? — 16 (3)? __ 96.0016 3.0001 — 3 From the table it appears that the average speed approaches 96 ft/s as the time intervals get smaller and smaller. It seems reasonable to say that the speed of the object is 96 ft/s at the instant t = 3.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.5

2.5

Linear Functions and Models

LINEAR FUNCTIONS AND MODELS

1. If f is a function with constant rate of change, then

(a) f isa linear function of the form f (x) = ax + b. (b) The graph of f isa line. 3. From the graph, we see that y(2)

ed ee ODGee 20s gots

=

50 and y(0)

=

20, so the slope of the graph is

iscatfmin ee

5. Ifa linear function has positive rate of change, its graph slopes upward.

7. f(x) =3+ 3x = 5x + 3 is linear with a = + and b = 3.

9. f (x) =x (4—x) = 4x — x? is not of the form J (x) = ax + b for constants a and 5, so it is not linear.

ll. f(x) =

= 5x + & is linear with a = i and b= ;.

13. f (x) = (x +1)? = x* + 2x + 1 is not of the form f (x) = ax +b for constants a and b, $0 it is not linear.

The slope ofthe graph of r(t) = —Ft +2 is —§.

© 2016 Cengage Learning: All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

111

112

CHAPTER 2 Functions

19. (a)

21. (a)

(b) The graph of f (x) = 2x — 6 has slope 2.

(b) The graph of h (t) = —0.5t — 2 has slope —0.5.

(c) f (x) = 2x — 6 has rate of change 2.

(c) A (t) = —0.5t — 2 has rate of change —0.5.

23. (a)

25. (a)

(b) The graph of v (t) = —y — 20 has slope —.

(b) The graph of f (t) = —31 + 2 has slope -3.

(c) v (t) = —Pr — 20 has rate of change 2.

(Cee) — —31 + 2 has rate of change —3.

27. The linear function f with rate of change 3 and initial value —1 has equation f (x) = 3x — 1.

29. The linear function h with slope 4 and y-intercept 3 has equation h (x) = 5x +3.

31. (a) From the table, we see that for every increase of 2 in the value of x, f (x) increases by 3. Thus, the rate of change of f aay!

1S 5°

(b) When x = 0, f (x) = 7, so b = 7. From part (a), a = 3,and so f (x) = 3x47.

4-3 33. (a) From the graph, we see that f (0) = 3 and f (1) = 4, so the rate of change of f is pet ==): (b) From part (a), a = 1, and f (0) = 6 =3,so f (x) =x +3.

35. (a) From the graph, we see that f (0) = 2 and f (4) = 0, so the rate of change of f is ;et;oS (b) From part (a), a= -5, and) (0) ==, b ==

SOnfa(e) = —3x +2.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.5 Linear Functions and Models

oe

113

Increasing the value of a makes the graph of / steeper. In other words, it increases the rate of change of f.

39. (a)

(b) The slope of T (x) = 150x + 32,000 is the value of a, 150.

(c) The amount of trash is changing at a rate equal to the slope of the graph, 150 thousand tons per year.

41. (a) Let V (t) = at + b represent the volume of hydrogen. The balloon is being filled at the rate of 0.5 fi?/s, soa = 0.5, and initially it contains 2 ft?, so b = 2. Thus, V (t) = 0.5t +2. “(b) We solve V (t) = 15.

43. (a) Let H (x) = ax +b

0.5 +2 = 15

0.5t = 13

>t = 26. Thus, it takes 26 seconds to fill the balloon.

represent the height of the ramp. The maximum rise is 1 inch per 12 inches, soa = is The ramp

starts on the ground, so b = 0. Thus, H (x) = pe.

(b) We find H (150) = b (150) = 12.5. Thus, the ramp reaches a height of 12.5 inches.

45. (a) From the graph, we see that the slope of Jari’s trip is steeper than that of Jade. Thus, Jari is traveling faster.

7-0 7 7 ar miles per minute or 60 (Z)= 70 mi/h. 6-0 16—10 The points (0, 10) and (6, 16) are on Jade’s graph, so her speed is 60 = 60 mi/h. 6=0

(b) The points (0, 0) and (6, 7) are on Jari’s graph, so her speed is

c) ¢ is measured in minutes, so Jade’s speed is‘60 mi/h - 60} h/min =

1 mi/min and Jari’s speed is

70 mi/h - a h/min = 7 mi/min. Thus, Jade’s distance is modeled by f (t) = 1 (t —0) +10 =¢ + 10 and Jari’s

distance is modeled by g (t) = 3 (t —0) +0 = 3t. 47. Let x be the horizontal distance and y the elevation. The slope is —(t. so if we take (0, 0) as the starting point, the elevation is y = — 185: We have descended 1000 ft, so we substitute

y = —1000 and solve for x: —1000 = — 785% —

x = 16,667 ft. Converting to miles, the horizontal distance is oh (16,667) © 3.16 mi.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

114

CHAPTER 2 Functions

49. (a) Let C (x) = ax +b be the cost of driving x miles. In May Lynn drove

(b)

480 miles at a cost of $380, and in June she drove 800 miles at a cost of

$460. Thus, the points (480, 380) and (800, 460) are on the graph, so the

slope is.a =

460-380

1

800-480

4°

We use the point (480, 380) to find the

value ofb:380 = { (480) + b b = 260. Thus, C (x) = 4x + 260. (c) The rate at which her cost increases is equal to the slope of the line, that is

i. So her cost increases by $0.25 for every additional mile she drives.

0}

200 400 600 800 1000 12001400*

The slope of the graph of Co

ix + 260 is the value of a, i:

= b) b = 51. (a) By definition, the average rate of change between x, and x9 is Sf(x2) = f 1) = Sarge) TE = ad Bb, x2 — x]

: ‘ . QX2 —ax (b) Factoring the numerator and cancelling, the average rate of change is eer x2 —]

2.6

x2 —*)

x2 —%]

a (x2 —-x a at al = x2 — *]

TRANSFORMATIONS OF FUNCTIONS

1. (a) The graph of y = f (x) +3

is obtained from the graph of y = f (x) by shifting upward 3 units.

(b) The graph of y = f (x + 3) is obtained from the graph of y = f (x) by shifting Jeff 3 units. 3. (a) The graph of y = —f

(x) is obtained from the graph of y = f (x) by reflecting in the x-axis.

(b) The graph of y = f (—x) is obtained from the graph of y = f (x)by reflecting in the y-axis. 5. If f is an even function, then f (—x) = f (x) and the graph of f is symmetric about the y-axis. 7. (a) The graph of y = f (x) — 1 can be obtained by shifting the graph of y = f (x) downward 1 unit. (b) The graph of y = ‘f (x — 2) can be obtained by shifting the graph of y = f (x) to the right 2 units. 9. (a) The graph of y = f (—x) can be obtained by reflecting the graph of y = f (x) in the y-axis.

(b) The graph of y = 3 f (x) can be obtained by stretching the graph of y = f (x) vertically by a factor of 3. 11. (a) The graph of y = f (x — 5) + 2 can be obtained by shifting the graph of y = f (x) to the right 5 units and upward 2 units.

(b) The graph of y = f (x + 1) — | can be obtained by shifting the graph of y = f (x) to the left 1 unit and downward 1 unit.

:

13. (a) The graph of y = —f (x) +5 can be obtairied by reflecting the graph of y = f (x) in the x-axis, then shifting the resulting graph upward 5 units.

(b) The graph of y = 3 f (x) — 5 can be obtained by stretching the graph of y = f (x) vertically by a factor of 3, then shifting the resulting graph downward 5 units. 15. (a) The graph of y = 2 f (x + 5) — 1 can be obtained by shifting the graph of y = f (x) to the left 5 units, retdhieie vertically by a factor of 2, then shifting downward | unit.

(b) The graph of y = 7Fi (x — 3) + 5 can be obtained by shifting the graph of y = f (x) to the right 3 units, shrinking vertically by a factor of q, 4 then shifting upward 5 units. 17. (a) The graph of y = f (4x) can be obtained by shrinking the graph of y = f (x) horizontally by a factor of i: (b) The graph of y = f (4x) can be obtained by stretching the graph of y = f (x) horizontally by a factor of 4.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole orin part.

ae

SECTION 2.6 Transformations of Functions

115

19. (a) The graph of g(x) = (x + 2) is obtained by shifting the graph of f (x) to the left 2 units.

(b) The graph of g (x) = x? + 2 is obtained by shifting the graph of f (x) upward 2 units.

21. (a) The graph of g (x) = |x + 2| — 2 is obtained by shifting the graph of f (x) to the left 2 units and downward 2 units. (b) The graph of g (x) = g(x) = |x — 2| +2

is obtained from by shifting the graph of f (x) to the right 2 units and upward

2 units.

23. (a)

(b)

(d)

25. The graph of y = |x + 1| is obtained from that of y = |x| by shifting to the left 1 unit, so it has graph II.

27. The graph of y = |x| — | is obtained from that of y = |x| by shifting downward 1 unit, so it has graph I.

PANS

i 6'9) Ne x? + 3. Shift the graph of y= x? upward 3 units.

31. f (x) = |x| — 1. Shift the graph of y = |x| downward 1 unit.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

116

CHAPTER2

Functions

33. f (x) = (x — 5). Shift the graph of y = x? to the right

35. f (x) = |x + 2|. Shift the graph ofy= |x|to the left — : 2unitsis

5 units.

Gea:

a

t

S7ey (ey = —x3, Reflect the graph of y = x? inthe x-axis-

39%. y = ez, Reflect the graph of y = 4/x inthe y

-

41. y= 4x, Shrink the graph of y = x? vertically by a :

factor of§.

‘\

Cie

oily

10

i

{

ty

SP

y

(43. y = 3 |x]. Stretch the graph ofy =[xlve i factor of 3.

SECTION 2.6 Transformations of Functions

45. y = (x — 3)? +5. Shift the graph of y=x? tothe right Serene award > units.

117

47. y = 3 — 4 (x — 1)*. Shift the graph of y = x? to the right one unit, shrink vertically by a factor of 5, reflect in the x-axis, then shift upward 3 units.

49. y = |x + 2| + 2. Shift the graph ofy = |x| to the left ee

Mrward 2 units.

5l.y= 5VXx + 4 — 3. Shrink the graph of y= /X vertically by a factor of 5,then shift the result to the left 4 units and downward 3 units.

53. y = f (x) —3. When f (x) =x?, y =x? —3.

55. y = f (x +2). When f (x) = /x, y = Vx +2.

57. y = f (x +2) —5. When f (x) = |x|,y=|x+2|/—5.

59 y = f (—x) +1. When f (x) = Yx, y = Y—x +1.

61. y =2f (x — 3) —2. When f (x) = x’,

63. g(x) = f (« —2) = (x -2)? =x? 4x 44

y=2 (x —3)* —2. 65.

gx)=f@+lIt+2=[x4+1/+2

69.

(a) y = f (x —4) is graph #3.

67. g(x) =—f (« +2) =—Vx +2

(b) y = f (x) +3 is graph #1. (c) y = 2f (x + 6) is graph #2.

(d) y = —/f (2x) is graph #4.

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118

CHAPTER 2 Functions

eo

Oy =2f@)

(b) y = f(x) -2

71. (a) y = f (x —2)

i peas

on-

(d)

bP ah

a

ies

y=—-f@) +3

Ha

(e)y = f (-x)

e

a,

j

+0

zee

Ot

¢

ee

t

For part(b), shift the graph in (a) to the left 5units; for part (c),shift the of2;for in(a) to the left 5 units, and stretch itvertically by afactor the graph in (a) to the left 5units, stretchitvertically by a factor0

it upward 4 units.

re:

Pa

J

6

= '

« s

(EQN

ie

ie -

4

wv—~

fri

a\\

,

ro

cleo

a

‘cn

ae

SECTION 2.6 Transformations of Functions

119

For part (b), shrink the graph in (a) vertically by a factor of 33 for part (c), shrink the graph in (a) vertically by a factor of i and reflect it in the x-axis; for part (d), shift the graph in (a) to the right 4 units, shrink vertically by a factor of i and then reflect it in the x-axis.

(b) y= f (2x) = V2 (2x) — (2x)? = V4x — 4x2

The graph in part (b) is obtained by horizontally shrinking the graph in part (a) by a factor of 4 (so the graph is half as wide). The graph in part (c) is obtained by horizontally stretching the graph in part (a) by a factor of 2 (so the graph is twice as wide).

83. f (x) =x*. f (—x) = (—x)* =x4 = f(x). Thus f(x) is even.

85. f (x) = x* +x. f (—x) = (—x)* + (—x) = x? — x. Thus ft (—x) # f (x). Also, f (—x) # —f (x), so f (x) is neither odd nor even.

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120

CHAPTER 2 Functions

87. f (x) =x? =x.

89. f(x) =1— 3¥x. f (-x) =1- YCx) =14+ x. Thus

f (-x) = (-xy3 = (-x) = —x3 +x Se (x?wy x) Be Uy)

f (—x) # f (x). Also f (—x) # —f (x), so f (x) is neither odd nor even.

Thus f (x) is odd.

91. (a) Even

(b) Odd

93. Since f (x) = x* — 4 < 0, for —2 2, then reflecting in the x-axis the part of the graph of y = f (x) for -—2 3.

So the domain is (3, co).

21.

27 tf (eles 2x —3 and g(x)

=4—x~

(a) f(g) =f (4- ©) = $4 =24)-3=5

(b) g(f ) =8 2) —3) =8(-3) =4- (3 =-5

29. (a) (fog) (-2) = f (g(-2)) = f (4- (-2?) = f@ =2@)-3=-3

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.7 Combining Functions

(b) (go f) (—2) = 8 (Ff(—2)) = g 2 (-2) —3) = g(-7) =4 = (-7)? = -45 3 —.

(a) (fog) (x) = f(e@)) = f (4-2?) =2 (4-2?) -3 = 8 - 2x? -3 = 5 - 2%? (b) (go f)(&) =8(F (*)) = gx —3) =4—

Rx 3)? = 4 — (4x? — 12x +9) = 4x? + 12x —5

33. f (g(2)) = f (5) =4

35. (go f) (4) =2e(f 4) =g(2)=5

37. (g 0g) (—2) = g(g(—2)) =g (1) =4 39. From the table, g (2) = 5 and f (5) = 6, so f (g (2)) = 6. 4 a . From the table, f (1) = 2 and f (2) = 3, so f (f (1)) =3.

43. From the table, g (6) = 4 and f (4) = 1, so (f og) (6) = 1. 45. From the table, f (5) = 6 and f (6) = 3, so (f 0 f) (5) =3. 47. f (x) = 2x +3, has domain (—oo, 00); g (x) = 4x — 1, has domain (—00, oc).

(f 0g) (x) = f (4x — 1) =2 (4x — 1)

+3 = 8x + 1, and the domain is (—00, 00).

(go f) (x) = g (2x + 3) = 4 (2x + 3) — 1 = 8x+ 11, and the domainis (—oo, 00). (f o f) (x) = f (2x +3)

=2 (2x +3) +3 = 4x +9, and the domain is (—o00, 00).

(gog) (x) =g (4x — 1) = 4 (4x — 1) — 1 = 16x —5, and the domain is (—o0, oo). 49. f(x)= x”, has domain (—0o, 00); g(x) = x + 1, has domain (—o0, 00).

(f og) (x) = f («x+1) = (x +1)? = x? + 2x + 1, and the domain is (—00, 00). (go f) (x) =g (x?) = (x?) +1 =x? + 1, and the domain is (—00, 00). Gio) tc) =f (x?) = (x2)° = x4, and the domain is (—00, 00). (gog) (x) =g(x +1) =(*+1)4+1 Sie

=x +2, and the domainis (—00, 00),

x) — - has domain {x |x 40}; g(x) = 2x +4, has domain (—oo, 00). (f og)(x) = f(Qx+4

= soy

(f og) (x) is defined for 2x.+ 4 #4

0x

4 —2.

So the domain is

{x | x # —2} = (—oo, —2) U (—2, 00). @oN)=«(=) =2(=) +4=

-+ 4, the domainis {x | x 4 0} = (—oo, 0) U (0, 00).

x

Crop (xe ye= f (=) = at = x. (f o f) (x) is defined whenever both f (x) and f (f (x)) are defined; that is,

(7)

whenever {x | x 4 0} = (—o0, 0) U (0, ov).

(g 0g) (x) =g (2x + 4) =2(2x +4) +4 =4x

+8 +4 = 4x + 12, and the domainis (—0o, oo).

53. f (x) = |x|, has domain (—0o, 00); g (x) = 2x + 3, has domain (—00, 00)

(Uf og) (x) = f (2x + 4) = |2x + 3], and the domain is (—o0, 00). (go f) (x) = g (|x|) = 2 |x| + 3, and the domain is (—0o, 00). (fof) (x) = f (xl) = IIxl| = [x], and the domain is (—oo, 00).

(g 0g) (x) =g (2x +3) =2 (2x +3) +3 =4x +643 =4x + 9. Domainis (—00, 00).

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

123

124

CHAPTER 2 Functions

55. f (x) = ae has domain {x | x

—1}; g(x) = 2x — 1, has domain (—00, 00)

4%

2x — 1

2x — 1

Cf og) (x)= f (2x — 1) = ——————

x+1

ais

= ———., and the domain is {x | x ~ 0} = (—oo, 0) U (0, ov).

Qrai+i Pax oN) =e(5) =2(-4) aepes a —1, x+1

x

and the domainis{x | x # —1} = (—00, —1) U (-1, 00)

oy yen@=i(75)-

1

aS

-ae

soap . (f © f) (x) is defined whenever both f (x) and

x+1

f f (&)) are defined; that is, whenever x# —1 and 2x +1 #0 = x # —4, which is (—00, -1) U (-1,-}) U (-3, 00). (go g) (x) = g (2x — 1) =2 (2x — 1)

—1 = 4x —2 — 1 = 4x — 3, and the domain is (—o0, 00).

1 Sha 169) = exp has domain {x | x 4 —1};g(*) = = has domain {x | x £ 0}. Gizeoie)\ (xy

a

%

1 =~*—_=

14]

swe

oe= eet, (f © g) (x) is defined whenever both g (x) and f (g (x)) are

x($+1)

x+1

defined, so the domain is {x |x 4 —1, O}.

(go f) (x) = «(=5) =

igs x +1" (@o.f) &).is delined whenever both. /(G)landg (/G)) areal

ay |

domainis {x | x 4 —1, 0}.

(fo f)@) =a - )= HH x+1

rat

tL (+1)

_* (fc f)(x)is defined whenever both f(x)and

(34 +1)

2x

+1

Ff Uf (x)) are defined, so the domain is {x|x 4-1, —3}.

= x. (go g) (x) is defined whenever both g (x) and g (g (x)) are defined, so the domain is

@o2))=2(-) =

ae *l—|

Bel apeaiN he

59. (fogoh) (x)= f(g (h(x) = f(g @—-)) =f (Vx

-1)=vx -1-

61. (fogoh)(x) =f(g (h(x) = F (8 (V)) =f (V¥ -5) = (VE-ot For Exercises 63—72, many answers are possible.

63. F (x) = (x — 9)°. Let f (x) =x and g(x) = x — 9, then F (x) = (fog) (x). x2

65.84 (xy)=

eet)

q and g (x) = x2, then G (x) == Gio 2 )iGa):

67. H (x)=i"| Let f(x) = |x| and g(x) = 1 — x3, then H (x) = (f 0g) (x). 69527 (0s) aaaRebs Ci

~g@)=x4+1, and h (x) = x?, then F (x) = (f ogoh) (x).

71..G (x) = ifsft. Let f(x) =x°, g(x) 73. Yes. If f (x) =m

= 44x, and h(x) = Ys, then G (x) = (fogoh) (x).

x + b; andg (x) = m2x + bo, then

(f 0 g) (x) = f (m2x + b2) = my (m2x + bz) +b; = mymzx + mj, b7 + bj, which is a linear function, because it is of the form y = mx + b. The slope is mymp. 75. The price per sticker is 0.15 — 0.000002x

and the number sold is x, so the revenue

is

R (x)= (0.15 — 0.000002x) x = 0.15x — 0.000002x?. 77. (a) Because the ripple travels at a speed of 60 cm/s, the distance traveled in t seconds is the radius, so g (t) = 601.

‘

—_——

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or inae \

ir

y &

MM 0S

:

]

>

ee

SECTION 2.8 One-to-One Functions and Their Inverses

125

(b) The area ofa circle is rr, so f(r) =nr?.

(c)

fog=7(g(t))* =m (60t)* = 360071? cm2. This function represents the area ofthe ripple as a function oftime.

79. Letr

be the radius of the spherical balloon in centimeters. Since the radius is increasing at a rate of 2 cm/s, the radius isr = 2f

after t seconds. Therefore, the surface area of the balloon can be written as S = 4ar? = 42 (2)? =4r (477) = l6nt?.

81. (a) f (x) = 0.90x

(b) g (x) =x — 100 (c) (fog) (x) = f (x — 100) = 0.90 (x — 100) = 0.90x — 90. f o g represents applying the $100 coupon, then the 10% discount. (go f) (x) = g(0.90x) = 0.90x — 100. go f represents applying the 10% discount, then the

é

$100 coupon. So applying the 10% discount, then the $100 coupon gives the lower price.

83. A(x)

=

1.05x. (40 A)(x)

(A0A0A)(x)

=

A(A(x))

= A(A0A(x))

=

A(1.05x)

=

1.05(1.05x)

=

(1.05)? x.

= A ((1.05)? x) =i 1 05 [1.05)? x] = (1.05)3x.

(AooAn A)(x) = A(A0A0A(x)) =A ((1.05)3 x) = 1.05 [1.0593 x|= (1.05)4x.A represents the amount in the account after 1 year; A o A represents the amount in the account after 2 years; A o A o A represents the amount in the account after 3 years; and A o A o Ao n copies of A, we get (1.05)” x.

2.8

A represents the amount in the account after 4 years. We can see that if we compose

ONE-TO-ONE FUNCTIONS AND THEIR INVERSES

1. A function f is one-to-one if different inputs produce different outputs. You can tell from the graph that a function is

one-to-one by using the Horizontal Line Test. 3. (a) Proceeding backward through the description of f, we can describe f—! as follows: “Take the third root, subtract 5, then divide by 3.”

fx — 5

(b) f (%) = Gx +5)3 and f-! (x) = as

5. If the point (3, 4) is on the graph of f, then the point (4, 3) is on the graph of f—!. [This is another way of saying that

fQ=4e f' 4 =3] 7. By the Horizontal Line Test, f is not one-to-one.

9. By the Horizontal Line Test, f is one-to-one.

11. By the Horizontal Line Test, f is not one-to-one. 13.. f (x) = —2x +4. If x # x2, then —2x; # —2x7 and —2x,; +4 4 —2x7 +4. So f is a one-to-one function.

15. g(x) = x. Ifx; # x2, then /x; # ,/x2 because two different numbers cannot have the same square root. Therefore, g is a one-to-one function. sec)

— x? — 2x. Because h (0) = 0 and h (2) = (2) — 2 (2) = 0 we have h (0) = h (2). So f is not a one-to-one function.

ES Gf (6a pes x4 4+5, Every nonzero number and its negative have the same fourth power. For example, (—1)* = 1 = (1)*, so f (—1) = f (1). Thus / is not a one-to-one function. 21. r (t) = 19 -—3,0 0.y=4-x2

ox? =4-yoxur=/4-y.

127

So f7! (x) = /4-—x,x < 4. [Note thatx > 0=>

Sf(&) 0. y=x5eox = $y forx > 0. The range of f is {y | y > 0}, so f~! (x) = Sx, x > 0.

65. f (x)= é

&5y=2-Vexr=2-Syex=

y=

JI Sy. Thus, f~! (x) = /2—5x.

q

Fade Fis) V5 + 8x. Note that the range of f (and thus the domain of f~!) is [0, 00). y=VJV54+8x a 5 2a = ye Sex=2 wa Thus, f~! (G3) — ‘J 3 ene0;

i

%

t

oy? =54+8xO

c,

|. f6)= 2+ y= 2+ Yeery- 2= = Jrex=(y- 2)3. Thus, f~! Qs= (x —2)3,

73. (a), (b) f (x) = Vx 41 ©

: x C) (x)= BES -6.y= = 3x soetrayt6d —

Be tra1 6 6. Be i

x.Using


0.So a. fort ipe >0.

NTT: fa)=* +12 - Usinga graphing deviceandthe x-ar Horizontal Line Test,we see that fi isa one-to-one

fier

,

128

CHAPTER 2 Functions

79. f (x) = |x| — |x — 6|. Using a graphing device and the

81. (a) y = f

Horizontal Line Test, we see that f is not a one-to-one function. For example f (0) = —6 = f (—2).

@®)=24+x% Sx

=y—2. So

fol(@)=x-2. (b)

10

-10

83. (a)

y=g(x) =Vx +3, y x

>0@x+3=y",y>0

= y*—3,y >0.S0g—! (x) =x* -3,x >0.

85. If we restrict the domain of f (x) to [0, 00), then y=4-x2? x2 =4-ys>x=/4—y(sincex

> 0,

we take the positive square root). So f —l (x) = .j/g—x,

(b)

If we restrict the domain of f (x) to (—o0o, 0], then

y=4-x2

ex? =4-y>ax=—/4—y

(since x < 0,

we take the negative square root). So

f3()=-V4=x.

87. If we restrict the domain of h (x) to [—2, 00), then y = (x + 2)? > x +2= /y (since x > —2, we take the positive square

root) x

= —2+ /y. Soh! (x) = -24 x.

If we restrict the domain of h (x) to (—oo, —2], then y = (« + 2/2 =>x+2=-—.,/y (since x < —2, we take the negative

square root) x= —2— /y. Soh—! (x) = -2— x. 91. (a)

(b) Yes, the graph is unchanged upon reflection about the line y = x.

Oya

1 Jove

As

ST

y

Gee

x

93. (a) The price of a pizza with no toppings (corresponding to the y-intercept) is $16, and the cost of each additional topping (the rate of change of cost with respect to number of toppings) is $1.50. Thus, f (n) = 16 + 1.5n.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

S 7

SECTION 2.8 One-to-One Functions and Their Inverses

(b)

p = f(n) = 164+ 1.5n &

p-—16=15nOSn=

129

$(p — 16). Thus, n = igre (pie 2 (p — 16). This function

represents the number of toppings on a pizza that costs x dollars.

(c) f—! (25) = § (25 — 16) = 3 (9) = 6. Thus, a $25 pizza has 6 toppings.

a. (a) V= f(t) = 100} 1

i

;

a

t oe

tite,

C0 be Biles

vvoe

Mid

V=100}1

:

3

40

Bagh.

100 —

1

:

40

orn

Sy (V) V. Since t < 40, we must have t = f~* = 40 —4/V.

ee:

Pai.

40° ~ V 100

={ represents time that f~*

has elapsed since the tank started to leak. (b) f —! (15) = 40 — 4/15 © 24.5 minutes. In 24.5 minutes the tank has drained to just 15 gallons of water.

97. (a) D= f (p) = —3p + 150.

D= —3p + 150 & 3p = 150-—D&

p=50- 5D. So f~! (D) =50- 5D. f(D)

represents the price that is associated with demand D. (b) f ad (30) = 50 — 4 (30) = 40. So when the demand is 30 units, the price per unit is $40.

99. (a) f~! (U) = 1.02396U. (b) U = f (x) = 0.9766x. U = 0.9766x x = 1.0240U. So fia (U) = 1.0240U. pot (U) represents the value of

U US dollars in Canadian dollars. (c) f Ss (12,250) = 1.0240 (12,250) = 12,543.52. So $12,250 in US currency is worth $12,543.52 in Canadian currency.

101. (a) f (x) = 0.85x. (b) g(x) = x — 1000. (ce) H (x) =(f og)x (d)

P =

= f (x — 1000) = 0.85 (x — 1000) = 0.85x — 850.

H (x) = 0.85x — 850. P = 0.85x — 850 © 0.85x = P + 850 =x

= 1.176P + 1000. So

H-! (P) = 1.176P + 1000. The function H—! represents the original sticker price for a given discounted price P. (e) H—! (13,000) = 1.176 (13,000) + 1000 = 16,288. So the original price ofthe car is $16,288 when the discounted

price ($1000 rebate, then 15% off) is $13,000. .

103. (aap (x)= at

Re )

is “multiply by 2, add 1, and then divide by 5”. So the reverse is “multiply by 5, subtract 1, and then

. ery

5x -1 2

5x -1 Chadaeroy l(c) = ¢( 2) 2

2x + 1 and fo fx)= s-1(>= 5°) =

b—

=

—_———

(= -) i 5

;

=:

>

2

2x+1—-1 OT

oOo

2(= - -)+1

2 oe 5

S5x-1+1 5

5 oh 3 5)

. 2x a

Z

Oyo)

=3— ‘s= io + 3 is “take the negative reciprocal and add 3 ”. Since the reverse of “take the negative x x reciprocal” is “take the negative reciprocal ”, f -l (x) is “subtract 3 and take the negative reciprocal ”, that is,

f@=

a. Checks

flof@)=s7! a

fo fa!ts) = f (3)

eet ge

=3~

22 23

(1/25) a3

o3 =

ane

i

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

130

CHAPTER2

Functions

(c) f (x) = Vx3 +2

is “cube, add 2, and then take the square root”. So the reverse is “square, subtract 2, then take

the cube root ” or |ee (x)

v/x2 — 2. Domain for tf (x) is [-7/2; 00) domain for f7} (x) is [0, 00). Check: 3

fofi@e=f (Wx? _ 2)=

(vx2 — 2) +2 = Vx2 —242

= Vx2:= x (on the appropriate domain) and

;

foof(x)=f7}

:

( x3 +2) =) (v3 +2) —2= Vx3 +2—2

(d) f(x) = Qx- 5)3 is “double, subtract 5, and then cube”.

Yx +5

5, and divide by 2” or f—! (x) =

fof

@) = (2) Z

(on the appropriate domain).

So the reverse is “take the cube root, add

Domain for both f (x) and f~! (x) is (—00, 00). Check:

~ es2 Seal = (3 +5-5) = (¥x)) = VO =x and

-1 ays fof @)fl =f" (@x-5)) Ina

3

= Vx3 =x

=

V@x-S5P+5

:

Qx—5)4+5 2x |

;

:

function like f (x) = 3x — 2, the variable occurs only once and it easy to see how to reverse the operations step by

step. But in f (x) = x3 + 2x + 6, you apply two different operations to the variable x (cubing and multiplying by 2) and then add 6, so it is not possible to reverse the operations step by step.

105. (a) Wefind g~! (x): y = 2x+19 2x = y—leex = 4(y-1). Sog7!(%) = $(@-1). Thus

f)=hog (b)

fog=he

(x) =A(}@-0) =4[$@—H] +4[}@—0] +7 =x? 2x +142x-247=%7 46, fio fog=

f7!ohelog=

f~!oh eg = f-!oh. Note that we compose with f—! on the left

on each side of the equation. We find f~!: y=3x+5@3x

=y—S@x=

$(v-5). So f~! (x) = 4(x—-5).

Thus g(x) = f-! oh(x) = f7! (3x? + 3x +2) = 4 |(se? +3x +2)-5] = 5 [3x2 +3x -3] =a? +x=1.

CHAPTER 2 REVIEW 1. “Square, then subtract 5” can be represented by the function f (x) =x? —5. 3. f (x) =3 (x + 10): “Add 10, then multiply by 3.” 5. g(x) =x? —4x

7. C (x) = 5000 + 30x — 0.001x? (a) C (1000) = 5000 + 30 (1000) — 0.001 (1000)? = $34,000 and C (10,000) = 5000 + 30 (10,000) — 0.001 (10,000)? = $205,000. (b) From part (a), we see that the total cost of printing 1000 copies of the book is $34,000 and the total cost of printing 10,000 copies is $205,000.

(c) C (0) = 5000 + 30 (0) — 0.001 (0)? = $5000. This represents the fixed costs associated with getting the print runready.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or inpart. — t

;

ex

CHAPTER2

Review

131

(d) The net change in C as x changes from 1000 to 10,000 is C (10,000) — C (1000) = 205,000 — 34,000 = $171,000, and

the average rate of change is

C (10,000) — C (1000) EB 171,000 eS 10,000 — 1000 9000

iD ieopy.

9. f(x) = x*- 4x +6; f(0) = (0)? —4(0) + 6 = 6; halts = (2)? -4(2) +6 = 2; f (—2) = (-2)? —4 (-2)+6 = 18: f(a) = (a)? —4 (a) +6 = a2 — 4a +6; f (—a)= (—a)? —4 (-a) +6 = a2 +40 +6; f (41) = (@& +:1)*-4(&4:1) 4-6 = x2 42x 41-4 —446 = x? —2x 43; f (2x)= (2x)? —4 (2x) +6 = 4x?—-8x +6. 11. By the Vertical Line Test, figures ~ and (c) are graphs of functions. By the Horizontal Line Test, figure (c) is the graph of a

one-to-one function. 13. Domain: We must have x +3 > 0 —3. In interval notation, the domain is [—3, oo).

Range: Forx in the domain of f, we havex

> —3 3x+3>0@J/x+3>04

f (x) > 0. So the range is [0, 00).

15. f (x) = 7x + 15. The domain is all real numbers, (—00, 00). iempilx) = nf+4. We require x + 4 > 0 —4. Thus the domain is [—4, 00).

1 1 sae f@~)=--+o am -— any) The denominators cannot equal 0, therefore the domain is {x |x 3 x

#0, —1, —2}.

21. h(x) = /4—x + Vx? — 1. We require the expression inside the radicals be nonnegative. So 4 — x > 04 x7-1>0(x-

> x; also

1) « + 1) = 0. We make a table: Interval

Sign of x — 1 Sign of x + 1 Sign of (x — 1) (x + 1)

Thus the domain is (—oo, 4] N {(—oo, —1] U[1, o0)} = (—oo ; 2 U[1, 4].

23. f (x) =1-2x

25. f (x) = 3x?

27. f (x) =2x?-1

29. f(x) =1+ Je y

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132

CHAPTER 2 Functions

61.°f (x)= '}x?

33. f (x) =—|Ix| y

y

1

x 2

0

1

x

1 38. fx)=—-G y x

1

0

x 1

x

39. x+y? = 145 y* = 14-x => y = 4/14 —x, 50 the original equation does not define y as a function ofx. 41.3 —-p=27ey=x-27ey=

(3 _ 27)ve so the original equation defines y as a function of x (since the cube

root function is one-to-one).

43. f (x) = 6x3 — 15x? + 4x -1 (i) [-2, 2] by [-2, 2]

(iii) [—4, 4] by [—12, 12]

(ii) [—8, 8] by [—8, 8]

|

(iv) [—100, 100] by [—100, 100] 100

-100

100 -100

From the graphs, we see that the viewing rectangle in (iii) produces the most appropriate graph.

-

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or inpart.

CHAPTER2

45. (a) We graph f (x) = /9 — x2 in the viewing rectangle

Review

133

47. (a) We graph f (x) = Vx3 — 4x +1 in the viewing

[—4, 4] by [—1, 4].

rectangle [—5, 5] by [—1, 5].

(b) From the graph, the domain of f is [—3, 3] and the range of f is [0, 3].

(b) From the graph, the domain of f is approximately ;

[—2.11, 0.25] U [1.86, 00) and the range of / is

[0, oo). 49. f(x) = x3 — 4x? is graphed in the viewing rectangle [—5, 5] by [—20, 10]. f (x) is increasing on (—oo, 0) and (2.67, 00). It is decreasing on (0, 2.67).

8)-f (4 —4 51. The net change is f (8) — f (4) = 8 — 12 = —4 and the average rate of change is £O- £0 oe

Tye

platl

53. The net change is f (2) — f (—1) = 6 —2 = 4 and the average rate of change is fon

—1.

4

=a:

55. The net change is f (4) — f(1) = [” —2 (4)| - [1 = 2(1)| = 8 — (—1) = 9 and the average rate of change is

f@-f0) _9_, 4-1

3

;

S7ep (x)=: 2+ 3x)? = 9x? + 12x + 4 is not linear, It cannot be expressed in the form f (x) = ax + b with constant a and b.

59. (a)

(b) The slope of the graph is the value of a in the

equation f (x) = ax + b = 3x + 2; that is, 3. (c) The rate of change is the slope of the graph, 3.

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134

CHAPTER 2 Functions

61. The linear function with rate of change —2 and initial value 3 has a = —2 and b = 3, so f (x) = —2x +3. 63. Between x = 0 and x = 1, the rate of change is Poy

f-f£@ _ 5-3 TO ]

= 2. Atx = 0,:f (x) = 3. Thus, an equation is

s=2x 3)

; 0-4 65. The points (0, 4) and (8, 0) lie on the graph, so the rate of change is B20

1

= v5: At x = 0, y = 4. Thus, an equation is

y=-4x + 4.

67. P (t)= 3000 + 200t + 0.112 (a) P (10)= 3000 + 200 (10) + 0.1 (10)? == 5010 represents ee population in its 10th year (that is, in 1995), and P (20) = 3000 + 200 (20) + 0.1 (20) = 7040 represents its population in its 20th year (in 2005).

P (20)

— P (10

(b) The average rate of change is se

7040-5010 2030 OST a ee oe 203 people/year. This represents the

average yearly change in population between 1995 and 2005.

69.

f(x) = 4x -6 (a) The average rate of change of f between x = 0 and x = 2 is

£@-6@ _[2@-9|-[}@-§] _ and the average rate of change of f between x = 15

:

-fas) _[260-6]-[305)-6] 19-3 fo SDS 1S GC Ba ae

reat

(b) The rates of change are the same.

(c) Yes, f is a linear function with rate of change 5

Ale (a)

y= f (x) +8. Shift the graph of f (x) upward 8 units.

(b) y =f (« + 8). Shift the graph of f (x) to the left 8 units.

(c)

y=1+2/

(x). Stretch the graph of f (x) vertically-by a factor of 2, then shift it upward 1 unit.

(d) y = f (« — 2) —2. Shift the graph of f (x) to the right 2 units, then downward 2 units. (e) y = f (—x). Reflect the graph of f (x) about the y-axis.

(f) y = —/f (—x). Reflect the graph of f (x) first about the y-axis, then reflect about the x-axis. (g) y =—/f

(x). Reflect the graph of f (x) about the x-axis.

(h) y = f~! (x). Reflect the graph of f (x) about the line ee

Se

(a) f (x) = 2x5 — 3x? +2. f (—x) =2(—x)5 —3(—x)* +2 = —2x5 — 3x? +2. Since f (x) # f (—x), f is not even. —f (x) ='—2x? + 3x? — 2. Since —f (x) # f (—x), f is not odd.

(b) f (x) =x3 — x’. f (—x) = (-x)3 — (-x)? =- (3 - 2) = —f (x), hence f is odd.

PAA

Es Seen 5 1

GN te ez (=x)? _ 1 — x? ~ 14(-x)2- 14x2

(d) f(x) = Pony LUT Ne f (-x)

1

1

aoe tye

= f (x). Since f (x)= f'(—x)} F is even:

—f(x)= aaa. Since f (x) # f (—x) , f is not even, and since

# —f (x), f is not odd.

75. g(x) = 2x? +4x — 5 =2(2x? + 2x) -5=2(% +2x +1) ~5—2=2(x +1)? —7. So the local minimum value —7 when x = —1.

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a

CHAPTER2

Review

135

77. f (x) = 3.3 + 1.6x — 2.5x3. In the first viewing rectangle, [—2, 2] by [—4, 8], we see that f (x) has a local maximum

and a local minimum.

In the next viewing rectangle, [0.4, 0.5] by [3.78, 3.80], we isolate the local maximum value as

approximately 3.79 when x * 0.46. In the last viewing rectangle, [—0.5, —0.4] by [2.80, 2.82], we isolate the local minimum value as 2.81 when x + —0.46.

79. h(t) = —1617 + 481 + 32 = -16 (1?- 3) +32 = -16 AEE —3t+ a + 32'+ 36 2

* =16 (1?EEay 3)+68 = —16 (1- 3) +68 The stone reaches a maximum height of 68 feet.

81. f(x) =x +2, g(x) =x?

83. f (x) =x* — 3x +2 and g(x) = 4 — 3x.

(a) (f +8) @) =(*? —3x +2) +(4- az (b) (f - g) (x) = (x? - 3x +2) - (43x)

art's:

=x? -2

(c) (fg) (x) = (x? - 3x +2) @— 3x) = 4x? — 12x +8 ~ 3x3 + 9x? — 6x = —3x3 + 13x? — 18x +8 fi

_ x?

3x42

(d) (4)@) =a

«4!

A

(e) (f og) (x) = f (4-3x) = (43x)? —3 (4 —3x) +2 = 16 — 24x + 9x? — 124 9x +2 = 9x? — 15x +6

(f (go f)() =g (x? - 3x +2) =4-3 (x? 3x +2) =-3x2 + 9x -2 85. f (x) = 3x — 1 and g(x) = 2x— x7.

(fog)(x)=f (2x~ x’) =i3 (2x — x?) — 1 = —3x?2 + 6x — 1, and the domain is (—00, 00). (go f)(x) = g (3x — 1) = 2(3x — 1) — Gx — 1)* = 6x — 2 — 9x? + 6x — 1 = —9x? + 12x — 3, and the domain is (—00, 00)

(Uf o f) (x) = f Gx — 1) =3 (3x — 1) — 1 = 9x — 4, and the domain is (—00, 00).

(g08)(x) = g (2x— x2) =2 (2x—x2) — (2x— x2)’ = ax — 2x?— ax? +

3 — x4 = x4 4 r3 — 6x2 + dr,and

domain is (—oo, 00).

: i

87. f (x) = JI=x, g(x) =1—x? and h (x) = 14 yx. (fogoh)(x) =f (g(h(x) = f(g(1+ vx) =f (1- (1+ v%)) =F (1- (1 +2V¥ +2)) !

= f (-x —2,/x) =,/1-—(-x -2x) )=JVit2yetx = (14+ Vx) =14+ Ve

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136

CHAPTER 2 Functions

89. f(x) =3+4+ x3. If x, #X, then xf = es (unequal numbers have unequal cubes), and therefore 3 + ra ~#3+ xi Thus f is a one-to-one function.

] } ; : Y “5. Since the fourth powers of a number and its negative are equal, h is not one-to-one. 0% 1 h(-1)= ]iM Fy A SW er AN Ta

91. 163)

For example,

93% p(x) =3.3 + 1.6x — 2.5x3. Using a graphing device and the Horizontal Line Test, we see that p is not a one-to-one function.

95.

f (x) =3x -2@y =3x -23x

=y +2 ex =} (yt2). Sof! (x) = 5(+2).

if

f®=e+peyae+lPextl= Wox= V-1.80 f(x) = YF- 1.

99% The graph passes the Horizontal Line Test, so f has an inverse. Because f (1) = 0, f -1 (0) = 1, and because f (3) = 4,

101.

f- (4) =3. (a), (b) f (x) =x* -4,x 20

;

(0) f (x) =x? -4,x >0@y=x?-4,>y-4

exr=yt4ex=Vyt4.

So

fIi3@)=Vvx4+4,x 2-4.

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CHAPTER2 _ Test

137

CHAPTER 2 TEST 1. By the Vertical Line Test, figures (a) and (b) are graphs of functions. By the Horizontal Line Test, only figure (a) is the graph of a one-to-one function. 3. (a) “Subtract 2, then cube the result” can be expressed

(c)

algebraically as f (x) = (x —2)?.

(b)

(d) We know that / has an inverse because it passes the Horizontal Line Test. A verbal description for f —1 is, “Take the

cube root, then add 2.”

() y=

-2% & Jyo=x-2ox = Yp+2. Thus, a formula for f—! is f—! (x) = Y= +2.

5. R(x) = —500x2 + 3000x (a) R (2) = —500 (2) + 3000 (2) = $4000 represents their total

(b)

sales revenue when their price is $2 per bar and R (4) = —500 (4)? + 3000 (4) = $4000 represents their total

sales revenue when their price is $4 per bar (c) The maximum revenue is $4500, and it is achieved at a price OLY — o3.

a, (a)

f(x) =a+ 5)? = x? + 10x +25 is not linear because it cannot be

(b)

expressed in the form f (x) = ax + b for constants a and b. g (x) = 1 — 5x is linear.

(c) g (x) has rate of change —S.

9. (a) y = f (x — 3) +2. Shift the graph of f (x) to the right 3 units, then shift the graph upward 2 units. (b) y = f (—x). Reflect the graph of f (x) about the y-axis.

11. f(x)ax? txt lex)

=x—-3.

(@) (f +g) ) =f) +8) = (x? +x 41) + —3) =2? +24 -2

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138

FOCUS ON MODELING

(b) (f- 8)®) = f@)-g@) = (x?+x41)-@-3) =x? +4 (c) (fog) (x) = f (g(x) = f & -—3) =

—3)? +(@ -3) 4 1 =x? - 6x +9 4x -—3 41 =x? —5x $7

|

@ (of) =8(f@) =8 (x? +x 41) = (x?+441) -3 3x72 42-2 (e) f(g (2)) =f (-1l) = (-1)? + (-1) +1 = 1. [We have used the fact that g (2) = 2-3 = -1.] (f) g(f 2)) =g(7)

=7—3 =4. [We have used the fact that f (2) = 27+2+4+1=7]

(g) (gogog) (x) = g(g(g()))

= g(g @ —3))

= 2

— 6) = &* — 6) —3

= x — 9. [We have used the fact that

g(x —3)=@ —3) -3 =x -6]

A329 (2) — mers OFA =-—

1,

ake

.

1

A:

Ae

=x forallx

1 40, andg(f())= 7

x

42 =% 7242 =x forall x # —2. Thus,by

x—-2

the Inverse Function Property, f and g are inverse functions.

15. (a) f(x) = V3—-x,x 0, the minimum

:

b m+n value occurs atx = Regs a

the x-value of the vertex, as expected.

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148

CHAPTER 3 Polynomial and Rational Functions

3.2

POLYNOMIAL FUNCTIONS AND THEIR GRAPHS

1. Graph I cannot be that of a polynomial because it is not smooth (it has a cusp.) Graph II could be that of a polynomial function, because it is smooth and continuous. Graph III could not be that of a polynomial function because it has a break. Graph IV could not be that of a polynomial function because it is not smooth.

3. (a) If c is a zero of the polynomial P, then P (c) = 0.

(b) Ifc is a zero of the polynomial P, then x — c is a factor of P (x). (c) Ifc isa zero of the polynomial P, then c is an x-intercept of the graph of P.

5. (a) P(x) =x*—-4

(b) O (x) = (x — 4)?

(c) P (x) = 2x? +3

7. (a) P(x) =x3-8

(b) O (x) = —x3 +27

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole orin part.

SECTION 3.2 Polynomial Functions and Their Graphs

(c) R(x) =—(x +2)3

149

(d) S(x) = 4(x-1)3+4 y

(aera (x) =X (x? = 4)= x3 — 4x has odd degree and a positive leading coefficient, so y

00 asx

> co and y > —co

as x > —O0o. (b) This corresponds to graph III. 11. (a) R(x) = —x> + 5x3 — 4x has odd degree and a negative leading coefficient, so

y

—oo asx — oo and y — coas

x > —0d. (b) This corresponds to graph V. 13. (a) T (x) = x4 + 2x? has even degree and a positive leading coefficient, so y > oo asx > oo and y > oo asx 4 —oo. (b) This corresponds to graph VI. 15.

P(x) = (x —1)(« +2)

19. P(x) = —(2x —1) (x +1) (x +3)

17. P (x) = —x (x — 3) (x +2)

21. P (x) = (x +2) (x +1) & — 2) (« -3)

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150

CHAPTER 3 Polynomial and Rational Functions

23. P(x) = —2x (x — 2)?

27. P(x) =

(x +2)? @« -3)°

29. P(x) = x9

@+2)@ — 3) ned

33. P(x) = —x3 +x2 4 12x= y

\

35. P(x) =x4 —3x3 42x? =x2(@-1)@—2)

37. P(x) =x $22 —x -1 = (x —1Ie+ 12 '

+

#?

Ty

1

(1,0), (Lo itis

\\

SECTION 3.2 Polynomial Functions and Their Graphs

39. P(x) = 2x3 —x?2 — 18x +9

41..P (x)= x4 ~ 2x3 ~ 8x + 16

= (x — 3) (2x — 1) (x +3)

43. P (x) = x4 — 3x? ~4 =

aa

151

= (& — 2)? (x? +2x +4)

—2)(@ +2)(x? +1)

(x)= 3x° — x2 +.5x +1; Oo) — 3x3. Since P has odd degree and positive leading coefficient, it has the following end

behavior:

y > oo asx

> ooand y — —ooasx > —O.

On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, we see that the

graphs of P and Q have different intercepts.

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152

CHAPTER 3 Polynomial and Rational Functions

47. P(x) =x4 —7x? + 5x +5; O (x) = x4. Since P has even degree and positive leading coefficient, it has the following end behavior: y > oo asx > ooand y > asx

> —Ooo.

On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different and we see that they have different intercepts.

-2 -100-

49. P(x) = x!! — 9x9; Q(x) = x!!. Since P has odd degree and positive leading coefficient, it has the following end behavior:

y — oo as x — oo and y> —oo asx > —00.

On a large viewing rectangle, the graphs of P and Q look like they have the same end behavior. On a small viewing

rectangle, the graphs of P and Q look very different and seem (wrongly) to have different end behavior. 600000

51. (a) x-intercepts at 0 and 4, y-intercept at 0. (b) Local maximum at (2, 4), no local minimum.

53. (a) x-intercepts at —2 and 1, y-intercept at —1. (b) Local maximum at (1, 0), local minimum at

(—1, —2).

55, y = —x? + 8x, [—4, 12] by [—50, 30]

57. y =x? — 12x +9, [—5, 5] by [—30, 30]

No local minimum. Local maximum at (4, 16).

Local maximum at (—2, 25). Local minimum at (2, —7).

Domain: (—oo, 00), range: (—oo, 16].

Domain: (—00, 00), range: (—00, 00).

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

|

SECTION 3.2 Polynomial Functions and Their Graphs

59. y = x4 + 4x3, [—5, 5] by [—30, 30]

153

61. y = 3x° — 5x3 +3, [—3, 3] by [—5, 10]

Local minimum at (—3, —27). No local maximum.

Local maximum at (—1, 5). Local minimum at (1, 1).

Domain: (—oo, 00), range: [—27, 00).

Domain: (—00, 00), range: (—0o, 00).

20

2

4

-20

os.) — —2x? + 3x +5 has one local maximum at (0:75, 6.13).

iy

= x4 — 5x? + 4 has one local maximum at (0, 4) and

65. y =x? — x? — x has one local maximum at (—0.33, 0.19) and one local minimum at (1.00, —1.00).

69. y = (x — 2)° + 32 has.no maximum or minimum.

two local minima at (—1.58, —2.25) and (1.58, —2.25).

1. y=x8- 3x4 +x has one local maximum at (0.44, 0.33)

and two local minima at (1.09, —1.15) and

73. y =cx?;c = 1,2, 5, 5. Increasing the value of c stretches

the graph vertically.

(—1.12, —3.36).

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154

CHAPTER 3 Polynomial and Rational Functions

75. P(x) =x4+c;c =—1, 0, 1, and 2. Increasing the value

of c moves the graph up.

77. P (x) = x4 —cx;c = 0, 1, 8, and 27. Increasing the value

of c causes a deeper dip in the graph, in the fourth

quadrant, and moves the positive x-intercept to the right.

79. (a)

y

y=x-22-x4+2°

(b) The two graphs appear to intersect at 3 points. (c) x? —2x? —x 42 =-x7

y=

xr+5x+2

45x 42073

-2x7 -64 =00

x (x? -x-6) =0ex (x — 3) (x +2) =0. Then either. x =0,x =3,0rx

= —2. Ifx =0; then y = 23a

es

y = 8; ifx = —2, then y= —12. Hence the points where the two graphs intersect are (0, 2), (3, 8), and (—2, —12).

: 81. (a) Let P (x) be a polynomial containing only odd powers of x. Then each term of P (x) can be written as Cx2"t1)

some constant C and integer n. Since C (—x)?*"+! = —Cx2"+! (a), P (x) is an odd function.

for

each term of P (x) is an odd function. Thus by part

(b) Let P (x) be a polynomial containing only even powers of x. Then each term of P (x) can be written as Cx?", for some

constant C and integer n. Since C (—x)*” = Cx?", each term of P (x) is an even function. Thus by part (b),

P(x) is —

an even function.

:

(c) Since P (x) contains both even and odd powers of x, we can write it in the form P (x) = R (x) + Q (x), where R (x) ©

contains all the even-powered terms in P (x) and Q (x) contains all the odd-powered terms. By part (d), O (x) is an odd function, and by part (e), R (x) is an even function. Thus, since neither Q (x) nor R (x) are constantly 0 (by assumption), by part (c), P (x) = R (x) + Q (x) is neither even nor odd. (d) P(x) =x°+6x3 x? —2xe+5 = (x5+ 6x3 2x)+ (=x? +5) = Po (x)+ Pe (x) where Pg (x) =x5+6x3-2x

and Pg (x) = —x? +5. Since Pg (x) contains only odd powers of x, it is an odd function, and since Pg (x) contains only even powers of x, it is an even function.

SECTION 3.2 Polynomial Functions and Their Graphs

83. (a) P(x) = (x — 1) (x —3) (x —4).

155

(b) Since QO (x) = P (x) +5, each point on the graph of Q has y-coordinate

Local maximum at (1.8, 2.1).

5 units more than the corresponding point on the graph of P. Thus Q

Local minimum at (3.6, —0.6).

has a local maximum at (1.8, 7.1) and a local minimum at (3.5, 4.4).

10 10

5 -10

z

85. Since the polynomial shown has five zeros, it has at least five factors, and so the degree of the polynomial is greater than or

equal to 5. 87. P (x) = 8x + 0.3x? — 0.0013x? — 372

(a) For the firm to break even, P (x) = 0. From the graph, we see that P (x) = 0 when x © 25.2. Of course, the firm cannot produce fractions of a blender, so the manufacturer must produce at least 26 blenders a year. (b) No, the profit does not increase indefinitely. The largest profit is approximately $ 3276.22, which occurs when the firm produces

.

166 blenders per year. 89. (a) The length of the bottom is 40 — 2x, the width of the bottom

(c) Using the domain from part (b), we graph V in

is 20 — 2x, and the height is x, so the volume of the box is

the viewing rectangle [0, 10] by [0, 1600]. The

V =x (20 — 2x) (40 — 2x) = 4x3 — 120x? + 800x.

maximum volume is V © 1539.6 when x = 4.23.

(b) Since the height and width must be positive, we must have

x > 0 and 20 — 2x > 0, and so the domain of V is

1000

On x. . 2x

Thus, the quotient is x? — | and the remainder is —3, and

2x

P (x) =2x3 — 3x? — 2x = (2x— 3) (x? - 1)ta)

2

Pre: —2x -— 1

Thus, the quotient is 4x? + 2x + 1 and the remainder is —2x —1,and-

P(x) = 8x4 + 4x3 + 6x2

= (202+1): (a2 42841)4 C20

a

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in wholeorinpart.

‘*

tj

SECTION 3.3 Dividing Polynomials

15.

17,

x—]

MedDipole enBarcap7

2x4 ara Dx Yi | 4x9 + 3x3 —6 =x? + 0x4 + 3x3 + Ox? + 0x — 6,

x3 — 8x +2 =x? + 0x? — 8x + 2, the synthetic division

the synthetic division table for this problem takes the

table for this problem takes the following form.

following form.

=3}

1 ie

0

=8

2

—3

9°.

—3

ee)

1

-1

Thus, the quotient is x? —3x + 1 and the remainder is —1.

|e feWORE pm Malt ot pee

oe oe

th

i 4

eee A

ao

Sea ek

Thus, the quotient is x4 + x3 + 4x? + 4x +4 and the remainder is —2.

>]

eee © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ,

y ”

158

CHAPTER 3 Polynomial and Rational Functions

37. Since x3 — 27 = x3 + Ox? + Ox — 27, the synthetic

35. The synthetic division table for this problem takes the

division table for this problem takes the following form.

following form. 1 Paes Bs 22" | Lape86 pra Weta

Sed

Les

—-7|5

OeE—/

—2

0

0

0

0

-7

1

0

81243 = 6720

BEZIO0

(329

SE20

eRaTS9

S274!

240"

DO

14

—35

35

35

-497

-5

-5

71

—483

2 4/3

4 3.2 62

=I

SHOT

36

P (x) = 3x3 44x? —2x + 1c =F

=x? +0x® + 0x> + Ox4 + 0x3 — 3x2 + 0x -—1 3

-40

Therefore, by the Remainder Theorem, P (—7) = —483.

(49,

Orme

30 5

47. P (x) =x! — 3x71

Ore

0

45. P (x) = 5x4 + 30x3 — 40x? + 36x + 14,c = —7

Therefore, by the Remainder Theorem, P (—2) = —7.

Ome

aad:

Therefore, by the Remainder Theorem, P (2) = 12.

43. P(x) =x? 42x? -7,c =-2

Sill) oeOle

PH

41. P(x) =x? + 3x2 —-7x+6,c=2

Therefore, by the Remainder Theorem, P (—1) = —3.

1

Bay

Thus, the quotient is x? + 3x + 9 and the remainder is 0.

39. P(x) = 4x? 4+ 12x +5,c=-1

2)

UNE

SD

Thus, the quotient is 2x” + 4x and the remainder is 1.

—2 | 1

a UM

-2

1

sve

On me

Therefore, by the Remainder Theorem, P (3) — q.

Therefore by the Remainder Theorem, P (3) = 2159.

51. P (x) =x3 + 2x? —3x

-—8,c =0.1

0.1.

=—3

},,3

0.1 Liye

=8

O21

—07279

2/90

Oddo

Therefore, by the Remainder Theorem, P (0.1) = —8.279.

53. P(x) =x3 —3x2 +3x-—l,c=1

Since the remainder is 0, x — 1 is a factor.

55. P (x) = 2x3 + 7x* + 6x —5,c= 4

Since the remainder is 0, x — 4 is a factor.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

’

SECTION 3.3 Dividing Polynomials

57. P (x) =x? + 2x? —9x — 18,c= -2 —2 | 1

59. x3 —x* —1lx+15,c=3

2

-9

—18

—2

0

18

0

-9

0

1

159

3

6

Since the remainder is 0, we know that —2 is a zero and

Since the remainder is 0, we know that 3 is a zero and

P (x) =(« +2) (x? -9) = (x +2) (x +3) (x -3).

P(x) =

Hence, the zeros are —3, —2, and 3. .

— 3) (x? +2x —5). Nowx? +2x -5 =0 ok

pa 2s

when x = Sarr

=

Tame = —1+/6. Hence,

the zeros are 3 and -1+ /6.

61. P (x) = 3x4 — 8x3 — 14x? + 31x +. 6,c = —2,3 —2 | 3

-8§

3

-14

31

6

63. Since the zeros are x = —1, x = 1, and x = 3, the factors arex +1,x

—1,andx

—3.

;

Thus

—6

28:

28

—6

—14

14

3

0

P(x) =

+) @-1

@ -3) = x3 —3x? —x $3.

Since the remainder is 0, we know that —2 is a zero and

P (x) =(x +2) (3x3 — 14x? 4 14x + 3). 9 3

—5

-15

-3

—1

0

Since the remainder is 0, we know that 3 is a zero and

P (x) = (x +2)(x —3) (3x? ge 1).Now 3x2 — 5x — 1 =0 when

—(-5)+ J(-5)?-4@)(-1)

54/37

SE

i

aad

2(3)

hence, the zeros are —2, 3, and

6

5+ /37 rea

65. Since the zeros are x = —1,x = 1,x =3, and x =5, thefactors arex + 1,x —1,x —3,andx —S.

Thus P (x) = (x + 1) (x — 1) (w — 3) (x — 5) = x4 — 8x3 + 14x? 4+8x — 15. 67. Since the zeros of the polynomial are —2, 0, 1, and 3, it follows that P (x) = C (x + 2) (x) (x — 1) (x — 3) = Cx4—2Cx35Cx2 + 6Cx. Since the coefficient of x3 is to be 4, -2C = 4, so C = —2. Therefore, P (x) = —2x4 + 4x3 + 10x? — 12x

~ is the polynomial. 69. Since the polynomial degree 4 and zeros —1, 1, and \/2 and integer coefficients, the fourth zero must be —/2, otherwise the constant term would be irrational. Thus, P (x) = C (x + 1) (x — 1) (x- A) (x+ A) = Cx4 —3Cx? +2C. Requiring that the constant term be 6 givesC = 3, so P (x) = 3x4 — 9x? +6.

71. The y-intercept is 2 and the zeros of the polynomial are —1, 1, and 2.

It follows that

P(x) = C(x+1)(x-l)@—-2) = C (3 —2x2 —x +2). Since P(0) = 2 we have

2=C{ (0? 20)? - (0) +2] 22 =2C

C=

Land PQ) = & +1)

1) @—2) =

— 2x? — x +2.

t

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

160

CHAPTER 3 Polynomial and Rational Functions

73. The y-intercept is 4 and the zeros of the polynomial are —2 and | both being degree two.

It follows that P(x) = C(x +2)2(x-1)2 =C (x4 4 2x3 — 3x2 — 4x +4). Since P (0) = 4 we have

4=C[(0)' +20) - 30) -4(0) +4] 24 =4C@ C= 1. . Thus P (x) = (x +2) (x — 1)? = x4 + 2x3 — 3x? —4x +4. the A. By the Remainder Theorem, the remainder when P (x) = 6x 1000 _ 17,562 4 12x + 26 is divided by x + 1 is

P (—1) = 6 (—1)!000 — 17 (-1)956 4 12 (-1) + 26 = 6 — 17 — 12 +.26 =3. B. If x — 1 is a factor of O(x)

=

x99? — 3x400 + x9 + 2, then Q(1) must equal 0.

|

QO (1) = (1)587 — 3 (1)4 4 (1)? +2 = 1-3 4+14+2=1¥0,s0x —1 is nota factor.

3.4

REAL ZEROS OF POLYNOMIALS

. If the polynomial function P (x) = anx” + a,—,x"—! +--+ a,x + ao has integer coefficients, then the only numbers that could possibly be rational zeros of P are all of the form Ls, where p is a factor of the constant coefficient ag and q is a factor q of the leading coefficient an. The possible rational zeros of P (x) = 6x? + 5x2 — 19x — 10 are Eeil- +4, +}, +4, +2, +3, 5

5

5

10

,

SES +5, +3, +z, +10, and +.

. This is true. If c is a real zero of the polynomial P, then P (x) = (x —c) Q (x), and any other zero of P (x) is also a zero of

QO(x) = P(x)/(« —¢). . P (x) =x3 — 4x? +3 has possible rational zeros +1 and +3. . R(x) = 2x9 + 3x3 + 4x2 — 8 has possible rational zeros +1, +2, +4, +8, +}. etn Cay 4x4 — 2x2 —7 has possible rational zeros +1, +7, +4, +f, +4, +4.

11. (a) P (x) = 5x? —x? —5x +1

has possible rational zeros +1, +4.

(b) From the graph, the actual zeros are —1, z, and 1.

13. (a)

P(x) =

— 9x3 + 9x? +x —3 has possible rational zeros +1, +3, +3, +3.

(b) From the graph, the actual zeros are -3, 1, and 3.

15. Paix) x3 + 2x2 — 13x + 10. The possible rational zeros are +1, +2, +5, +10. P (x) has 2 variations in sign and hence 0 or 2 positive real zeros. P (—x) = —x? + 2x? + 13x + 10 has 1 variation in sign and hence P has 1 negative real zero.

Ne

tad Bes ge Oy

:

|

a.

eae ale

203

-10

a

0 =+x=1isazero.

‘

P(x) = x9 +2x? — 13x +10 = (x 1) (x? +3 — 10) = (x +5) @ — 1) (2). therefte, theten aa 1, and2. 17, P (x)=x? +. 3x2 — 4. The possible rational zeros are +1, +2, +4. P (x) has 1 variation in sign and hence 1 positivereal zero. P (—x)= —x> + 3x? — 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros.

1

de

4

4

eh

0

;

: ‘is

‘.

= x = 1isa zero.

rte

fh i Te

P(x) =x3 43x? —4=( -1) (x?+ 4x +4) = (x — 1) (x +2). Therefore, the zeros are —2 and 1.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, orposted to a publicly accessible website, in hho

*

os

SECTION 3.4 Real Zeros of Polynomials

161

19. P (x) =x? — 6x? + 12x —8. The possible rational zeros are +1, +2, +4, +8. P (x) has 3 variations in sign and hence | or 3 positive real zeros.

1/1

P (—x) = —x> — 6x” — 12x — 8 has no variations in sign and hence P has no negative real zero.

-6

12

-8

1

-5

7

15

7

2\ely

= )65 =>) = "his not a zero:

P(x) = x3 — 6x2 4+12x -8=

F654 1

12 5 —8

2

-8

—4

4

8 Oey t2 1S1a Zero:

(x — 2) (x? — 4x +4) = (x — 2)>. Therefore, the only zero isx = 2.

21. PN) = x? = 19x — 30. The possible rational zeros are +1, +2, +3, +5, +6, +15, +30. P (x) has 1 variations in sign and hence 1 positive real zero.

—1/1

0

-19

-1 1

-1

P (—x) = —x3 + 19x — 30 has 2 variations in sign and hence P has 0 or 2 negative real zeros.

—30

is -18

—2/

HLS -—-12

0

-19

—30

4

30

—2 =x

eg

=-1 is nota zero.

pa

1

—2

—15

0.

=> x =—21s

azero.

Tow Vigh! t64 L4ya)230 64

1

IAG

1

8

aa x? —19x —30=(x+ 2) (x? —2x-

16

0

s>x =-—4isa zero.

15) = (x + 3) (x +2) (x — 5). Therefore, the zeros are —3, —2, and 5.

23. Bias x? +3x? —x —3. The possible rational zeros are +1, +3. P (x) has 1 variation in sign and hence 1 positive real zero. P (—x) = —x? + 3x2 +x = 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros.

—-1/}1 1

3

-!

-3

-l1

-2

3

2/3

Oni=>

xi='—11s'a zero.

So L(x) = x3 4+3x2-—x-3 =(x +1) (x?+ 2x — 3)= (x + 1) (x +3) (x — 1). Therefore, the zeros are —1, —3, and 1.

25. Method 1: P (x) = x4 — 5x? + 4.The possible rational zeros are +1, +2, +4. P (x) has | variation in sign and hence 1 positive real zero. P (—x) = x4 — 5x? + 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros.

Joby

0.

5

0

“

1

-4

-4

=4

-4

1 I-11 Thus P (x) = x4 — 5x2 +4 = (x —1) (3 ei

-1]1

Ax

1 -1

0) i>

== LS 2iZero.

4).Continuing with the quotient we have:

-4

—4

0

4

PAL hI clay 18% => x =\—l 18,4 zero, Pix) = x4 —5x244= (x -1)(x +1) (x? - 4)Sb) a:1) (x + 1) (« = 2) (& +2). Therefore, the zeros are +1, +2. Method 2: Substituting

u =

uw? —5u+4=(u—1)(u-—4)

x?, the polynomial becomes P (u) =

uz — 5u + 4, which factors:

= ( 2_ 1)( 2 -4), so eitherx? = 1 or x2 = 4. Ifx? = 1, thenx = +1; if x? = 4, then

x = +2. Therefore, the zeros are +1 and +2.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

162

CHAPTER 3 Polynomial and Rational Functions

27. Pin) i= x4 4 6x3 + 7x? — 6x —8. The possible rational zeros are +1, +2, +4, +8. P (x) has 1 variation in sign and hence 1 positive real zero.

P (—x) = x4 — 6x3 + 7x? + 6x — 8 has 3 variations in sign and hence P has | or 3 negative real zeros.

Vee

Y

Ohne

Se

6

8

nel

A

8

it~

14

8

0

=x =1isazero

and there are no other positive zeros. Thus P (x) = x4 + 6x3 + 7x2 —6x —8 = (x — 1) (x? + 7x? + 14x +4 8).Continuing by factoring the quotient, we have:

1

6

8

0

=>x =-—1 1s a zero.

So P (x) = x4 + 6x3 4 7x? = 6x — 8 = (x = 1) @ +1) (x? +6 +8) = @ — 1) (+1) @ +2) ( +4). Therefore, the zeros are —4, —2, and +1.

29. 1EN 9) yi 4x4 — 37x? +9 has possible rational zeros +1, +3, +9, +3, +3, +3, +4, +3, +4. Since P (x) has 2 variations in sign, there are 0 or 2 positive real zeros, and since P (—x) = 4x4 — 37x? + 36 has 2 variations in sign, P has 0 or 2. negative real zeros. Ie

4°02

4

37

0

9

4

4

-33

—33

4

-—33

-—33

—24

3 |4

12 Zery2

2

4

0

14)

7

—37

0

9

36

—3

-9

ay

—3

Or

x == Ss a zeros

3

65.0.

=> x= }iis'a zeto.

Because P is even, we conclude that the zeros are +3 and +4 and

P (x) = 4x4 — 37x? +9 = (x +3) (2x +1) (2x — 1) (x —3). Note: Since P (x) has only even terms, factoring by substitution also works. Let x2 = u; then TENG 4u? —37u+9 = (4u — l)\u—-9)= (4x - 1)(x? - 3),which gives the same results.

S13 Pt)

3x4 — 10x3 — 9x? + 40x — 12. The possible rational zeros are +1, +2, +3, +4, +6, +12. P (x) has 3 variations

in sign and hence 1, 3, or 5 positive real zeros. P (—x) = 3x4 + 10x3 — 9x? — 40x — 12 has 1 variation in sign and hence

P has | negative real zero.

Li 3)

= 102

SS ie A

CW 3-7

-16

eee

2|3

etCon eel 24 24 =x=lisnotazero.

—10

6 ~

3

—4

~—9.'

40 °-12

tara —-17

6

ee 0 =x =2isazero.

Thus P (x) = 3x4 — 10x3 — 9x? + 40x — 12 = (x —2) (3x3 — 4x2 — 17x + 6).Continuing by factoring the quotient, we have

3 |3

3

—4 8 = 17

9

1S

5

—2

6

=6 0)

=> % = 38 isa Zero,

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.4 Real Zeros of Polynomials

163

Thus P (x) = (x — 3) (x —2) (3x? 45x — 2) = (x — 3) (x — 2) (3x — 1) (x +2). Therefore, the zeros are —2, 4, 2, and 3.

33. Factoring by grouping can be applied to this exercise.

4x3 44x? —x

—1 = 4x2 (x +1) -—(x

+1) = (+1) (4x Ss 1)= (x + 1) (2x +1) (2x — 1).

Therefore, the zeros are —1 and +4.

OS

Semiane i 4x3 —7x +3. The possible rational zeros are +1, +3, +}, +3, +4, +3. Since P (x):has 2 variations in sign,

there are 0 or 2 positive zeros. Since P (—x) = —4x3 + 7x +3 1

Coe)

Ee 2 4

Bx)

2

ioe:

5

1

-3

-6

has | variation in sign, there is 1 negative zero.

0

= x = 5isa zero.

(x_ ) (4x? +2x—- 6)= (2x — 1) (2x? +x— 3)= (2x — 1) (x — 1) (2x +3) = 0. Thus, the zeros are —3,

5,and 1.

Slee (oi 24x3 + 10x? — 13x — 6. The possible rational zeros are +1, +2, +3, +6, +}, +3, +4, +3, +4, +3, +4, +3,

+3, +4, +3. P (x) has 1 variation in sign and hence 1 positive real zero.

P (—x) = —24x3 + 10x? + 13x — 6 has 2

variations in sign, so P has 0 or 2 negative real zeros.

—1|24

10 —13 —6 —24

14-1.

24 -14

—3|24

=| 24 eee —48

1 —7 >x

10 -13

=-1 is nota zero.

24 —38

-6

—72

186 —519

24 —62

173 —525

Ol Se 6

~6|24

=> x = —3 is not a zero. 1 =$:)24

24

76 —126 63 —132 =x

10 —13

=-2 is nota zero.

—6

—144

804 —4746

24 —134

791 —4752

=> x = —6 is not a zero.

+3x4 — 9x3 — 31x? + 36. The possible rational zeros are +1, +2, +3, +4, +6, +8, +9, +12, +18. P (x) has 2 variations in sign and hence 0 or 2 positive real zeros.

P (—x) = —x> + 3x4 + 9x3 — 31x? + 36 has 3 variations in sign

and hence P has | or 3 negative real zeros.

1)

1

3.

+9.

1

4

1

4

-5

=31

0

36

-5

-36

-36

—-36

-36

Oe

—> tin sia Zer0:

So P (x) =x° + 3x4 — 9x3 — 31x2 +36 = (x -—1) (3 + 4x3 — 5x2 — 36x — 36). Continuing by factoring the quotient, we have:

fel Sivind2 65 ies hee ey

9%-3655 156 oe se Pee ae

7

Dili:

21

48

36

16

12

0

=x

Adbcap—5 bee Ou be36 Qt 126 mild ored4 1 7 =22',.—80

=3 isazero.

So P (x) = & — 1) @ — 3) (3 + Tx? + 16x + 12).Since we have 2 positive zeros, there are no more positive zeros, so

we continue by factoring the quotient with possible negative zeros.

tM

Vath

4 Lait.

l

6

ARS oar stl 10

2

ft

ty SMe 1

5

Seigh med iho oe18 6

0

=x

=-—2 isazero.

Then P (x) = (x —1) (x —3) (« +2) (x? 45x 4 6) = (x — 1) (x —3) (x +2)? (x +3). Thus, the zeros are 1, 3, —2, and —3.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.4 Real Zeros of Polynomials

165

43,6F (x) = 3x> — 14x4 — 14x3 + 36x? + 43x + 10 has possible rational zeros +1, +2, +5, +10, +4, +4, +3, +1). Since

P (x) has 2 variations in sign, there are 0 or 2 positive real zeros. Since P (—x) = —3x> — 14x4 + 14x3 +36x2 — 43x + 10 has 3 variations in sign, P has 1 or 3 negative real zeros.

feast

eid Bo

3

36 6743

eet i — 25)

-ll

-—25

i

3

D9

aes

bielAce tale

54

Tl - 54:

P(x) =(x -2) (3x4 Rx? S50 2/3

10 64

3

36 ur aay oe.1)

6

+16

-60

-48

—-§8

-30

—24

—5

—10 Onset

==

isa vero!

Awne 5)

-8

-30

—24

—5

6

—-4

-68

—184

—2

-34

-92

—189

5|3

-8

-—30

-—24

-5

15

35

25

5

5

1

3

Vy

Oya

So iska Zero

P (x) = (x —2) (x —5) (3x3 + 7x2 45x + 1).Since 3x2 + 7x? + 5x +1 has no variation in sign, there are no more

positive zeros.

peeing Cae 3

4

> 44 ae 1

0

>x=-lisa zero.

P (x) = (x —2) (x —5) (+1) (3x? + 4x4 1)= (x — 2) (x —5)(* +1) («+ 1) Gx + 1). Therefore, the zeros are —1,

—4,2,and 5. 45. P (x) = 3x3 + 5x? —2x — 4. The possible rational zeros are +1, +2, +4, +4, +3, +f. P (x) has 1 variation in sign and hence | positive real zero. P (—x) = —3x3 + 5x? + 2x — 4 has 2 variations in sign and hence P has 0 or 2 negative real

Zeros.

eta Riom So

P(x) =

cio

47.

=

SE

4

-1[3

3

2

0

PA

0

-4

5 -2

3

-4

—3

-2

4

2

—-4

0

=x

=-1 isazero.

(+1) (3x? + 2x — 4). Using the Quadratic Formula on the second factor, we have i

7

~

aS ead

ot AED Therefore, the zeros of P are —1 and ie AE

P(x) = x* — 6x3 4+ 4x2 + 15x +4. The possible rational zeros are +1, +2, +4. P (x) has 2 variations in sign and hence 0

or 2 positive real zeros. P (—x) = x4 + 6x3 + 4x2 — 15x +4 has 2 variations in sign and hence P has 0 or 2 negative real Zeros. LL

—O 1 Ly

5)

4

15

4

7-5

--1..

14

10-414,»

18

WH 1

2

oa

hae oe Z~ en

8a 16.4 -4

—1

4a

4.. i—8" oh

16

4

8) > 14 v2

18

24 0

°=>x = 418 a zero:

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166

CHAPTER 3 Polynomial and Rational Functions

So P (x) = ( — 4) (x3 ee

4)};1

—-2 ]

So P(x) =

=

-4

Ay — 1).Continuing by factoring the quotient, we have:

-1

4

ag

2

4

@-4)

-1/1

15

=x =4isan upper bound.

+1) (2 —3x-

—(—3)+,/ (-3)?—4(1)(-1

Sh G ‘ a Oe)

2 1

-4,-1

—1

3

1

-3

—1l

0

=x = -—1 is 4 zero,

1). Using the Quadratic Formula on the third factor, we have:

= ne

Therefore, the zeros are 4, —1, and =zndED)

49. P (x) = x4 —7x3 + 14x? — 3x — 9. The possible rational zeros are +1, +3, +9. P (x) has 3 variations in sign and hence 1

or 3 positive real zeros. 1/1

P (—x) = x4 + 7x3 + 14x? + 3x — 4 has 1 variation in sign and hence P has | negative real zero.

7

14 1

1

-3

-9

-6

8

5

8

5

4

-6

ee he

1

Continuing by factoring the quotient, we have:

3;}1

1 So P(x)

=

—(-1)+,/

(x -3) (x? —x(-1)?-41)(-1)

ee ey . OC

=

Rec

14

-3

-9

3

—-12

6

9

-4

2

3

0. 4=> x=

—-4

2

3

3

-3

-3

-1

-l

0

=x

34s a zero

=3 isa zero again.

1). Using the Quadratic Formula on the second factor, we have:

12 fs. Therefore, the zeros are 3 and tee

,

al Ve(b

ee 4x3 — 6x? +1. The possible rational zeros are +1, +}, +}. P (x) has 2 variations in sign and hence 0 or 2

positive real zeros. P (—x) = —4x3 — 6x? + 1 has 1 variation in sign and hence P has | negative real zero.

[|

4 SOnL (xi =

Ue

Ge

8

4

—-2

-2

-2

-2

-1

$14

4

-6

oO

2

2

-4

-2

1 -1

0

=x=}

isazero.

(x- +) (4x? — 4x - 2). Using the Quadratic Formula on the second factor, we have:

—(—4)+,/ (—4)?—4(4)(—2

= eras ALAR

— a)

;

= atAv3 = Ted3. Therefore, the zeros are :and LS

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.4 Real Zeros of Polynomials

167

53. P (x) = 2x4 + 15x3 + 17x? + 3x — 1. The possible rational zeros are +1, +}. P (x) has 1 variation in sign and hence 1 positive real zero. P (—x) = 2x4 — 15x3 + 17x? — 3x — 1 has 3 variations in sign and hence P has 1 or 3 negative real

zeros. oh1

ek

ee

2

eee

LORtEZS

2

14

3

.-1

y

2

ee

10

=> x = 4 is an upper bound.

-2

0

=x=-} isazero.

So P(x) =(x+4)(2x3 + 14x? + 10x — 2) =2(x + 4)(x? +727 + 5x -1). =

81

1

Son)

7

heey2 0) Oe)

1

—-1

-6

1

6

-l

0

= (x+ +) (2x3 + 14x* +.10x — 2) 2

third factor, we have x = poy

5.

=x

=- 1 isazero.

(x+ +)(x + 1) (x?+ 6x — 1)Using the Quadratic Formula on the

= celal = =S42/10 = —3+,/10. Therefore, the zeros are —1, ape

and —3 + /10.

5a. (ayel(x)

x? — 3x? — 4x + 12 has possible rational zeros +1, +2,+3,

(b)

4,6, 12;

2 1

-l

—2 -6

—-12 O74

= .4 = 248 a Zero:

So P (x) = (x —2) (x? —-x- 6)= (x — 2) (x +2) (x —3). The real zeros ofP are —2, 2, and 3.

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168

CHAPTER 3 Polynomial and Rational Functions

57. (a) P (x) = 2x? — 7x? + 4x + 4 has possible rational zeros +1, +2, +4,

(b)

+}. 1}2

—7

4

4

2-5 2 —5

2}2

—7

-l

-1

-3

2

4

4

4 -6

-—4

25-35

1

0

—2))

=x

0 => X12 15a. zero:

=2

isa zero again.

Thus P (x) = (x — 2) (2x + 1). The real zeros ofP are 2 and —5.

5OTi(a) ere

x4 — 5x3 + 6x? 4+ 4x — 8 has possible rational zeros +1,

(b)

y

Sh oth arte) 1)

te

5

6

4

1

-8

1

-4

2

6

-—4

2

6

-2

2

24 —6-_ 0 4.8 Lg

Se,

0

4

QO.

=> x =218 4 zero.

So P (x) = & —2) (x3 — 3x2 4 4)and the possible rational zeros are restricted to —1, +2, +4.

1-1!

-—2

0

=x

=2 isa zero again.

P (x) = (x — 2)? (x2 -x-2) = (x — 2)? (x — 2) (x +1) = (x — 2)3 (x + 1). Sothe real zeros of P are —1 and 2.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1}

I

SECTION 3.4 Real Zeros of Polynomials

61. (a) P(x)

= x9 — x4 — 5x3. 4.x2 4+ 8x +4 has possible rational zeros

169

(b)

25) joel veere

1

1

-—3..

—5

—2

0

=x

=2isazero.

So P (x) = (x -—2) i +x ~3x7—S¥'— 2, and the possible rational zeros are restricted to —1, +2.

Lei So P(x) =@-

3

1

0

=x

=2 isa zero again.

2) (3 + 3x2 43x 4 1),and the possible rational zeros are restricted to —1.

| 1

2

1

Ope

xe

— 1s a 7ero:

SOP (x) = & — 2) (x + 1) (x?+ 2x + 1)= (x - 2)2 (x + 1)3., and the real zeros of P are —1 and 2.

63. P (x) =x? —x? —x —3. Since P (x) has 1 variation in sign, P has 1 positive real zero. Since P (—x) = —x3 —x? +x —3 has 2 variations in sign, P has 2 or 0 negative real zeros. Thus, P has 1 or 3 real zeros.

One feel 6) A 2x6 4+ 5x4 — x3 — 5x —1. P(-x)= 2x6 + 5x4 4x3 + 5x —

O72 (x) = x° + 4x3 — x2 + 6x.

Since P(x) has 1 variation in sign, P has 1 positive real zero.

Since

has | variation in sign, P has | negative real zero. Therefore, P has 2 real zeros.

Since P (x) has 2 variations in sign, P has 2 or 0 positive real zeros.

Since

. P(-x) =-x°- 4x3 — x? — 6x has no variation in sign, P has no negative real zero. Therefore, P has a total of 1 or 3

real zeros (since x = 0 is a zero, but is neither positive nor negative).

69. P (x) = 2x3 + 5x2 +x —2;a=-3,b=1 soe | bee. 2

5H

si

—2

—-6

3

—-12

-1

4

—14

Diss, Dias

WS;

alternating signs > lower bound.

8 6

all nonnegative> upper bound.

Therefore a = —3 and b = | are lower and upper bounds.

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170

CHAPTER 3 Polynomial and Rational Functions

71. P(x) = 8x? + 10x? — 39x +9;a =-3,b =2

8

-—-14

3

16

D226

26

13.

8

0

alternating signs = lower bound.

35

all nonnegative => upper bound.

Therefore a = —3 and b = 2 are lower and upper bounds. Note that x = —3 is also a zero.

73. P(x) =x442x3 + 3x2 +5x -—lja=-2,b=1

1

0

3

-!1

bee 1

3

1

Alternating signs > lower bound.

6a 6

11

£10

All nonnegative= upper bound.

Therefore a = —2 and b = | are lower and upper bounds.

75. P(x) = 2x4 — 6x3 +x? -2x +3;a=-1,b=3 -1|}2

-6 2

1

a2

Oe

-8

9

—2

3

— Oy a -I11

14 _ Alternating signs => lower bound.

SM? POT 2

2

8

62>.0

3395

0.31

1

6

All nonnegative

> upper bound.

Therefore a = —1 and b = 3 are lower and upper bounds.

77. P(x) =x3 — 3x? +4

and use the Upper and Lower Bounds Theorem:

1

-4

4

3}-1.

-3

0.

4

ae”

OF

ED

0

0

4

1

0

alternating signs => lower bound.

all nonnegative = upper bound.

Therefore —1 is a lower bound (and a zero) and 3 is an upper bound. (There are many possible solutions.)

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.4 Real Zeros of Polynomials

171

79. P(x) =x4 —2x3 4x? —9x 42.

a

Sallie’ Sian Wile nae pees A Ae es a RA PTs,

EY

2,1

-2 1-9 2 JA. Gaps ron Ate he

gees 25thbax Rh Phe sys l

1

lsoud

-—3

al

3.

4°

11.

—13

15

all positive => upper bound.

alternating signs=> lower bound.

Therefore —1 is a lower bound and 3 is an upper bound. (There are many possible solutions.)

81. P (x) = 2x* + 3x3 — 4x2 — 3x +2.

DA

ps

5

ih 258)

Es)

1-2

Ole

> ct

sia Zero.

P(x)=(@- 1) (2x3 + 5x? +x -2)

2

3

-—2

Ol

=>) i=. — lis:aizero,

P(x)=(@-)N@+t+)) (2x? + 3x —2) = (x — 1) (x +1) (2x — 1) (x +2). Therefore, the zeros are —2, 5, eile 83. Method 1: P (x) = 4x4 — 21x? +5 has 2 variations in sign, so by Descartes’ rule of signs there are either 2 or 0 positive

zeros. If we replace x with (—x), the function does not change, so thereare either 2 or 0 negative zeros. Possible rational zeros are +1, +4, +4, ES, +3, +3 .By inspection, +1 and +5 are not zeros, so we must look for non-integer solutions: 1 t/4 0 +21 0

4.12

—-20.

Nn

—10

0

=> x = 5isa zero.

Nes) fe (x_ +)(4:3 + 2x? — 20x — 10),continuing with the quotient, we have:

-$|4

2-20 —2

4 1A

0 —20

10 0

=> x =—} isazero.

i (x- +) (:+ 5)(4x? _ 20) = 0. If 4x? — 20 = 0, then x = +/5. Thus the zeros are +4, +5.

Method 2: Substituting 4u2 —21lu+5=

x=

0

-10

u =

x?, the equation becomes 4u2 — 21u + 5 =

(4u —1)(u—5)

0, which factors:

= (4x? _ 1)(x? = 5). Then either we have x2 = 5, so that x= +/5, or we have

i so that x = +/1 = +h. Thus the zeros are +}, +./5.

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172

85.

CHAPTER 3 Polynomial and Rational Functions

P (x) =x — 7x4 + 9x3 + 23x? — 50x + 24. The possible rational zeros are +1, +2, +3, +4, +6, +8, +12, +24. P (x) has 4 variations in sign and hence 0, 2, or 4 positive real zeros. P (—x) = —x5 — 7x4 — 9x3 + 23x? + 50x + 24 has 1

variation in sign, and hence P has | negative real zero.

1 1

-6

lb

5

-6

3

26

3

26

—24

1

P(x) =(x—1)? (x3 at

la I,

45

-2

+2

24

a

— we

24 QO

=x

=1

Isa zero again.

Dee 24); continuing with the quotient, we start by trying | again.

4G 5-6

-5

—24

2:

18

ie

P (x) = (x — 1)? (x — 3) (x? SBE

—6.

—16

Se

8

lay

3°

-6

—24

2

8

0

=x =31S2aZeEr0,

8)= (x — 1)? (x —3) (x — 4) (x +2). Therefore, the zeros are —2, 1, 3, 4.

87. P (x) =x? —x —2. The only possible rational zeros of P (x) are +1 and +2.

1/1

0

-1

1 Le

-2

1

0

0

-—2

241

0) My

B12

-1]1

#3

4

6

2

3

4

1

0

-1l

-2

—1

1

0

-1l

0

-2

Since the row that contains —1 alternates between nonnegative and nonpositive, —1 is a lower bound and there is no need to try —2. Therefore, P (x) does not have any rational zeros.

89.

P (x) = 3x3 —x? — 6x + 12 has possible rational zeros +1, +2, +3, +4, +6, +12, +4, +3, +4. 3

=-1

-6

12

3 3

all positive > x = 2 is an upper bound

3 3

alternating signs > x = —2 is a lower bound | Wi WIR Wi) WIN WIE

Therefore, there is no rational zero.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.4 Real Zeros of Polynomials

91. P(x) =x? — 3x? — 4x + 12, [—4, 4] by [-15, 15]. The

173

93. P (x) = 2x4 — 5x3 — 14x? 4 5x 4 12, [-2, 5] by

possible rational zeros are +1, +2, +3, +4, +6, +12. By

[—40, 40]. The possible rational zeros are +1, +2, +3,

observing the graph of P, the rational zeros are x = —2, 2,

+4, +6, +12, +4, +3, By observing the graph of P, the

ae

3

zeros are —5, —1,1,4.

95. x* — x —4 = 0. Possible rational solutions are +1, +2, +4.

| ie Coe Oe

2/1

iy

al

Perk 9-7

—1

1

0

é

line s:

Oe a4

0

-1

-1

1.

-1

7

oye,

Le

=2

f= 2

|1

0

0

0

2

4

-1

a2

-4

—2}

1 1

Therefore, we graph the function P (x) = x4 —x —4

-4

8

14

7

10

0

0

-1

-4

—2

4

-8

18

—-2

4

-9

14°

=x

=2 isan upper bound.

=x =-—2isa

lower bound.

in the viewing rectangle [—2, 2] by [—5, 20] and see there are two

solutions. In the viewing rectangle [—1.3, —1.25] by [—0.1, 0.1], we find the solution x ~ —1.28. In the viewing rectangle [1.5.1.6] by [—0.1, 0.1], we find the solution x + 1.53. Thus the solutions are x + —1.28, 1.53.

97. 4.00x4 + 4.00x3 — 10.96x2 — 5.88x + 9.09 = 0. 1|4 4 —10.96

4 48

—2}4

9.09

8 —2.96

—8.84

-2.96

—8.84

0.25

4 —10.96

—5.88

9.09

—8 4-4

—5.88

87 -2.96

2}4

9.01

—5.88

8

24

26.08 40.40

13.04

20.2 49.49

412

—3|/4

5.92 -—0.08 0.04

4 —10.96

4 -10.96 —12

4

-8

9.09 = x =2

-—5.88

9.09

24 —39.12

135

13.04

—45

144.09

isan upper bound.

= x = —3 isa lower bound.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

174

CHAPTER 3 Polynomial and Rational Functions

Therefore, we graph the function P (x) = 4.00x4 + 4.00x3 — 10.96x? — 5.88x + 9.09 in the viewing rectangle [—3, 2] by [—10, 40]. There appear to be two solutions. In the viewing rectangle [—1.6, —1.4] by [—0.1, 0.1], we find the solution

x * —1.50. In the viewing rectangle [0.8, 1.2] by [0, 1], we see that the graph comes close but does not go through the

x-axis. Thus there is no solution here. Therefore, the only solution is x ~ —1.50,

99. Let r be the radius of the silo. The volume of the hemispherical roof is 5 (Far3 )= dar, The volume of the cylindrical section is 7 (-?) (30) = 30mr2. Because the total volume of the silo is 15,000 ft?, we get the following equation:

Zar? + 30mr? = 15000 > $ar> + 30nr? — 15000 = 0 > ar3 + 45mr? — 22500 = 0. Using a graphing device, we first graph the polynomial in the viewing rectangle [0, 15] by [—10000, 10000]. The solution, r + 11.28 ft., is shown in the viewing rectangle [11.2, 11.4] by [—1, 1]. 10000

-10000

101. h(t) = 11.60t — 12.411? + 6.2023

— 1.58¢4 + 0.202 — 0.0118 is shown in the viewing rectangle [0, 10] by [0, 6].

(a) It started to snow again. - (b) No, A(t) < 4.

(c) The function h (¢) is shown in the viewing rectangle [6, 6.5] by [0, 0.5]. The x-intercept of the function is a little less than 6.5, which

means that the snow melted just before midnight on Saturday night. —

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. +

SECTION 3.4 Real Zeros of Polynomials

103. Let r be the radius of the cone and cylinder and let h be the height of the cone.

175

r3 + 30r? — 250

Since the height and diameter are equal, we get h = 2r, So the volume of the

cylinder is Vj = ar? - (cylinder height) = 207r2, and the volume of the cone is 500

gQe= imr7h = iar? (2r)i— far, Since the total volume is

iy it follows

500 that ars + 20nr? = aes r> + 30r? — 250 = 0. By Descartes’ Rule of Signs, there is 1 positive zero. Since r is between 2.76 and 2.765 (see the table), the radius should be 2.76 m (correct to two decimals).

105. Let b be the width of the base, and let / be the length of the box. Then the length plus girth is / + 4b = 108, and the volume is V =1b* = 2200. Solving the first equation for /and substituting this value into the second equation yields /= 108 — 4b

= V = (108 — 4b) b* = 2200 = 46° — 10862 + 2200 =04 (° — 27b* + 550) = 0. Now P (b) = b? — 27b* +. 550 has two variations in sign, so there are 0 or 2 positive real zeros. We also observe that since / > 0,b < 27, so b = 27 is an upper bound. Thus the possible positive rational real zeros are 1, 2, 3, 10, 11, 22, 25.

EMP.

RPS Y 1

0).

550

1

-26

—26

-—26

-—26

524

Dal

des

27,

0

2 1

5 | 1

—27

0

550

5

—-110

-—550

1.

—22

-110

Ae)

—50..—100 —50

450

EIN] = aT

= a

Di

JEN e) oe (b —5) (0? — 22b— 110). The other zeros are b = S27 ithe

550

—25

siazero:

answer from this factor is b + 26.20.Thus we have two possible solutions,

b = 5 or b © 26.20.

eee The positive

If b = 5, then

1 = 108 — 4(5) = 88; if b© 26.20, then /= 108 — 4 (26.20) = 3.20. Thus the length ofthe box is either 88 in. or 3.20 in.

107. (a) Substituting X — xfor x we have

(2-4)0(r-$) 44x $)

x? + ax? + bx +c

Omar

aE

3

xX

Ee

a

2

3 2a 2 RESat

3

xe 5

27

a3

4a3—s het (bey a)x+($

3

2a 2

Sy eX 3

2

2

Bre

2

UmHae

oy. p> BE +(EE = X°+(-a+a)X 3 ll +3

2

a(x

2

9 a3

a a

5 +: a )¥

b

rox Fae 3

ab

Met 3 3

ee a7

3

+ F

5 +c

ab Sada Pe)

(b) x3 +6x2+9x +4 = 0. Setting a= 6, b = 9, andc = 4, we have: X° + (9_ 6?)K++ (32 — 18)

4) =X? ~27X4-18.

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176

CHAPTER 3 Polynomial and Rational Functions

109. (a) Since z > b, we have z — b > O. Since all the coefficients of Q (x) are nonnegative, and since z > 0, we have

QO (z) > 0 (being a sum of positive terms). Thus, P (z) = (z — b)-

Q(z) +r > 0, since the sum of a positive number

and a nonnegative number. (b) In part (a), we showed that if b satisfies the conditions of the first part of the Upper and Lower Bounds Theorem and

z > b, then P (z) > 0. This means that no real zero of P can be larger than b, so b is an upper bound for the real zeros.

(c) Suppose —b is a negative lower bound for the real zeros of P (x). Then clearly b is an upper bound for

’

P; (x) = P (—x). Thus, as in Part (a), we can write P; (x) = (x — b)-

Q(x) +r, where r > 0 and the coefficients

of Q are all nohnegative, and P (x) = P; (—x) = (-x —b)- O(—x)

+r = (x +b)

-[—Q (—x)] +r. Since the

coefficients of Q (x) are all nonnegative, the coefficients of -Q (—x) will be alternately nonpositive and nonnegative, . which proves the second part of the Upper and Lower Bounds Theorem.

3.5

COMPLEX ZEROS AND THE FUNDAMENTAL THEOREM OF ALGEBRA

1. The polynomial P (x) = 5x (x — 4)3 (x + 7) has degree 6. It has zeros 0, 4, and —7. The zero 0 has multiplicity 2, and the zero 4 has multiplicity 3. 3. A polynomial of degree n > 1 has exactly n zeros, if a zero of multiplicity m is counted m times.

5.

P(x) =x441 (a) True, P has degree 4, so by the Zeros Theorem it has four (not necessarily distinct) complex. zeros. (b) True, by the Complete Factorization Theorem, this is true. (c) False, the fourth power of any real number is nonnegative, so P (x) > 1 for all real x and P has no real zeros.

7. (a) x4 44x? =06x2 (G4+4) =0. Sox =Oorx? +4=0, Ifx? +4 =0 then x? = —4 > x = +2/. Therefore; the solutions are x = 0 and +27.

(b) To get the complete factorization, we factor the remaining quadratic factor P (x) = x2 (x+ 4)= x2 (x —2i ) (x + 2i). 9. (a) x3 = 2x? + 2x

= 0

x(x? - 2x42) = 0. Sox

—(—2)+y (-2)?—4(1)@)

=

;

(b) Since 1 — i and 1 +i are zeros, x — (1 —i)

=Oorx?

— 2x +2 =0. Ifx? — 2x+2 = 0 then

= 1 +i. Therefore, the solutions are x = 0, 1 +i.

=x

—1+iandx

—(1+i)

=x —1—i are the factors of x? — 2x +2. j

Thus the complete factorization is P (x) = x (x? —2x + 2)=x (x —1+i)(*-1-i). 2

11. (a) x4 4+2x74+1=00 (x?4 1) =0¢x74+1=0¢x*

=-1 ex

= +i. Therefore the zeros of P are x = +i.

(b) Since —i and i are zeros, x + i and x — i are the factors of x? + 1. Thus the complete factorization is

P(x) = (x2+1) =[6 +) @-DP = +2 @ 9. 13:i(a) x) —16 300 = (x? -4) (x?+4) = (x — 2) (x +2) (x?+4). Sox = +2 orx? +4 =0. Ifx2 +4 =0 then x? = —4 = x = +2i. Therefore the zeros of P are x = =

dottB

(b) Since —i and i are zeros, x + i and x — i are the factors of x2 + 1. Thus the complete factorization is

P(x) = (x -2)(¢ +2) (x?+4) = (x — 2) (x +2) (x —2i) (x +21). 15. (a) 348 = 04 (+2) (x?-2x +4) = 0. Sox = —2orx? —2x +4 =0. Ifx? —2x +4 =0 then —(—2)4,/ (—2)?-4(1)(4

x= sc

aaah nalAh = eee

= ee

= 1+i/3. Therefore, the zeros of P arex = —2,1 +i/3.

ft

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_—+-~—

=

My ge

Wigberiy 5

=

SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra

177

(b) Since 1 —i/3 and 1 +i/3 are the zeros from the x2 — 2x +4 =0,x — (1— iV3) and x — (1+iv3) are the factors

of x? — 2x + 4. Thus the complete factorization is

P(x) =(x +2) (x? a, © +4) = (x +2) E i (1i iv3)| [xBS(1+iv3)|

= (x +2) (x= +iy3) (xoy ~iy3) 17. (a) © -1=060= (x?2.1)(4 a 1)2 hee (x?txt 1)(x +1) (x? mph 1).Clearly, x= +1 are solutions.

Ifx?+x+1=0, thenx= tVIE4@)_ =1e V3 _ _1 4 V3 sg» = 1 4)¥B. Andifx? —x+1 =0,then =

yt ROO

— 1by3 _ 1473

(b) The zeros of x? +x

+1

= 0 are Ut

[ - (-}-19)] [x-(-3+792)] the zeros

of x7 —x

+1

=

BesaA)

0 are

X pan Therefore, the zeros of P are x = +1, ~harwafs, 14593,

sp¥2 and — Ne =

Ge)

so x2 + x + 1 factors as («+4 -19).

Similarly,since

1 ~ 1X8 and 4 4 132, so x2 — x + 1factorsas

nie) G9:

}: Thus the complete

factorization is

Ax!)

(x1) (x? +241) @ +0

(x?-x 41)

(@-N@+)(x+5 +7993) (r+ 5-198) (x- $4798) (x - 4-138) 19. P (x) =x* + 25 = (x — Si) (x + 5i). The zeros ofP are 5i and —5i, both multiplicity 1.

2)+,/ (2)?—4(1)(2

|

21. O(x) = x? + 2x + 2. Using the Quadratic Formula x = naa

ONE

= a2iind — saat =

eet SO

O (x) = (x + 1 —i)(« + 1+7). The zeros of O are —1 — i (multiplicity 1) and —1 + i (multiplicity 1). DS

AX

x3 44x =x (x?+ 4)= x (x — 27) (x + 27). The zeros ofP are 0, 27, and —2i (all multiplicity 1).

35. O(x) =x4-1 ( 2-1) (x? +1) hg 1) (+1) (x? +1) = (x +1) (x +1) (x -i) (x +). Thezeros of Qare 1, —1,i, and —i (all of multiplicity 1).

27. P (x) = 16x* — 81 = (4x? - 9)(4x? teye (2x — 3) (2x + 3) (2x — 3i) (2x + 3). The zeros ofP are 5,-3, 37, and —3i (all of multiplicity 1).

29. P(x) =x3 4x2 49x49 =x2

(641494) = (4+) (x? +9) = (x + 1) (x — 31) (x + 3i). The zeros ofPare

—1, 3i, and —3i (all of multiplicity 1).

SOx)

2 AeA == xf 4 2x24 Lice (x?+ 1) = (x —i)* (x +7). The zeros of Q are iand —i (both of multiplicity 2).

amex)

— x4 43x? -4=

( cle 1)(x?+4) = (x — 1) (x + 1) & — 27) (&* +27). The zeros of Pare 1, —1, 2, and —2i

(all of multiplicity 1). 2

2

2

35. P(x) = x9 + 6x3 +9x =x (x4 + 6x? +9) =x (x? +3) =x (x~iv3) (x +iv3) . The zeros ofP are0

(multiplicity 1), 73 (multiplicity 2), and —i /3 (multiplicity 2). 37. Since

1 + i and 1 — i are conjugates,

the factorization of the polynomial

must be

P(x) =a(x—[l +i]) («-[ —i}) =a (x? = 2x +2). Ifwe let a = 1, we get P (x) =x? —2x +2.

‘

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178

CHAPTER 3 Polynomial and Rational Functions

39. Since 2i and —2i

are conjugates, the factorization

of the polynomial

must be

Q(x) = b(x —3)(x—2i)(x+21] = B(x —3) (x?+4) a b (x3 — 3x2 44x — 12). If we let b = 1, we get O (x) =x? — 3x? 44x — 12. 41. Since i is a zero, by the Conjugate Roots Theorem, —i is also a zero. So the factorization of the polynomial must be P(t) =a (x —2)@

i) @ +i) =a (x3 — 2x? +x —2). If we let a = 1, we get P (x) = x9 — 2x? +x —2.

43. Since the zeros are 1 — 2i and 1 (with multiplicity 2), by the Conjugate Roots Theorem, the other zero is 1 + 27. Soa factorization is Rie)

ee

cr

~ 2i}) (x — [1 +27]

@& — 1)? =c (x — 1] + 21) (x — 1] — 28) & — 1)?

c(t - 17 - pi?) (x? - 2x +1) =¢(x? -2x+1+4) (x? - 2x +1) =c(x?-2

+5) (x? - 2x+1)

o (x4 — 2x9 4.x? 2x3 + 4x2 —2x 4+5x?-= 10x +5) (x3 — 4x3 + 10x? — 12x +5)

If we let c = 1 we get R (x) = x4 — 4x3 + 10x? — 12x +5. 45. Since the zeros are i and 1 +i, by the Conjugate Roots Theorem, the other zeros are —i and 1 — i. So a factorization is

T(x) = C@—-i (+i)

—-[1 +i] @-

Zi)

C(x? =) (e-1]-) @- 14+) =C(x?+1) (x? -2x 41-1?) = C(x? +1) (x? - 2x+2) = C (x4 = 2x3 + 2x? 4.x? - 2x +2) = C (x4 — 283 + 3x? — 2x +2) = Cx4 — 2Cx3 + 3x? — 20x +2€ Since the constant

coefficient

is 12, it follows

that 2C

=

12 =

C

=

6, and so

T (x) = 6 (x4 — 2x3 4 3x? — 2x +2) = 6x4 — 12x} + 18x? — 12x + 12. 47.

P (x) =x3 42x? 44x 48 =x2 (x +2) 44(x +2) = (x 42) (x?+4) = (x +2) (x —2i) (« + 2i). Thus the zeros are —2 and +21.

49. P (x) =x3 — 2x? + 2x — 1. By inspection, P (1) = 1 —2+2-—1=0, and hence x = 1is a zero.

1 | tl

«2

2

-!il

1

-1

1

|

1

0

Thus P (x) = (x — 1) (x? -x +1). Sox =lorx*-x+1=0. Using the Quadratic Formula, we have x =

51.

= egy Hence, the zeros are 1 and 4+!

3.

P (x) =x? = 3x? + 3x —2.

Thus P (x) = (x — 2) (x? -x +1). Sox =2o0rx?-x+1=0

Using the Quadratic Formula we have x = ied lV 63),08 EEE Hence, the zeros are 2, and 1+i¥3

|

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SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra

Sora

atx) t= 2x3 + 7x2 + 12x +9

has possible rational zeros +1, +3, +9, +}, +3, +3. Since all coefficients are positive,

there are no positive real zeros.

2

ee 9 ey Dy, ETE 33

Pie 20S

OE

=—3

-6

1-9

4

6

0

2

Roe

tO: re Ziheees Th:

9 ey N43

=> x = —} isa zero.

P(x) = (x+$) (2x7+4r +6) = 2(x +3) (x?+2x 43). Now x? + 2x +3 has zeros =

LOD)

=: eee

=—-1+i/2. Hence, the zeros are -3 and —1 +i/2.

55. P(x) =x44+x3+47x2.4 9x — 18. Since P (x) has one change in sign, we are guaranteed a positive zero, and since P (—x) = x4 — x3 + 7x? — 9x — 18, there are 1 or 3 negative zeros.

S

| 1 Rosy Pahoy eg CR i

Lakes

2g

PRP

A

Demarest 18

RIE

0

Therefore, P (x) = (x — 1) fe + 2x2 4+. 9x + 18).Continuing with the quotient, we try negative zeros. =]

pl

2

eS

ll > peels 1

1

=P)

spd

Phy

woes

8

ols,

=—2

10

1

18.

0) = 3.is\a zero:

Continuing:

3}1

—3

4 —12

Sa 05" 1

PS)

179

x =—} isa zero.

P(x)= (x+ +) (4x3 + 2x? + 4x +2). Continuing: ri1

4

a

2

4

2

aoe

AU

cae

0

4

0 - = x = —} isa zero again.

v)

Pie)

(x+ ) (4x + 4).Thus, the zeros of P (x) are -5 (multiplicity 2) and +i.

63. P (x) =x? — 3x4 + 12x3 — 28x? + 27x — 9 has possible rational zeros +1, +3, +9. P (x) has 4 variations in sign and P (—x) has | variation in sign.

I

Lop

2. es 10

1

18

=2

105°

—18

9

1

-1l

9

-1l

9

—9

1/1

-9 0)

9

-1

, =>ix% = 118 a zero;

1

9

-9

a0

9

OAs

0;

=x

=] isiazems:

P(x) =(x—-1)3 (x?+ 9)= (x — 1)3 (x —3i) (x + 3i). Therefore,the zeros are 1 (multiplicity 3) and +3i.

65.

(a) P (x) = x3 — 5x? 44x — 20 =x? (x — 5) +4

— 5) = @ -5) (x? +4)

(b) P (x)= & —5) (&« — 21) & +2i)

67.

(a) P(x) = x44 8x? —9 = (x?-1) (x2 +9) =@-)@+1) (2? +9) (b) P (x) = (x —1) (x + 1) @& — 31) (x +31)

69.

(a) P(x) = x6 — 64= (x3 -8) ( 27, either y > 00 or y > —00.

(+1) @ —2) hasas x-intercepts —1 and an 2. x-int ts —1

3.b The function r (x) = ————— G+ 2)@—3)

5. The function r has vertical asymptotes x = —2 and x = 3.

a

a: a

x+1)(x—4)

1 al le (ex+1)2x-4)

2(¢-2)

'

(a) False. r does not have vertical asymptote x = —1. It has a “hole” at (=1, 4): (b) True, r has vertical asymptote x = 2. (c) False, r has a horizontal asymptote y = 4 but not a horizontal asymptote y = 1.

(d) True, r has horizontal asymptote y = 5. CPi (ea

aay

(a)

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182

CHAPTER 3 Polynomial and Rational Functions

(b) r (x) >

—00 as x — 27 andr (x) > ocoasx > 2t.

(c) r has horizontal asymptote y = 1.

ll. r(x) (a)

a. 16 == 5 (x ='2) —10

~430 —40,300 —4,003,000

~370 ~39,700 —3,997,000

(b) r (x) —00 asx 3 27 andr (x) > —ooasx

> 2+.

(c) r has horizontal asymptote y = 0.

| In Exercises 13-20, let f (x) = -. EX 1 1357) — ne Ee

; f (x — 1). From this form we see that the graph ofr is obtained

from the graph of f by shifting 1 unit to the right. Thus r has vertical asymptote x = 1 and horizontal asymptote y = 0. The domain of r is (—oo, 1) U (1, oo) and its range is (—oo, 0) U (0, 00).

3

LS: Ses(Ge)) = ae

a

)= 3f (x + 1). From this form we see that the graph

— =3(

of s is obtained from the graph of f by shifting 1 unit to the left and stretching vertically by a factor of 3. Thus s has vertical asymptote x = —1 and horizontal

asymptote y = 0. The domain ofs is (—oo, —1) U (—1, 00) and its range is (—oo, 0) U (0, 00).

Line)

—3

2.

1

= 2+ ——5 =f

—2)+ 2 (sec the long

division at right). From this form we see that the graph of ¢ is

obtained from the graph of f by shifting2 units to the right and 2 units vertically. Thus ¢ has vertical asymptote x = 2 and horizontal asymptote y = 2. The domain of t is (—o0, 2) U (2, 00) and its range is (—0o, 2) U (2, 00).

19. re aS a

_ me

18)

1 eect Jone

division at right). From this form we see that the graph of r is obtained from the graph of f by shifting 3 units to the left, reflect about the x-axis, and then shifting vertically 1 unit. Thus r has vertical asymptote x = —3 and horizontal

asymptote y = 1. The domain ofr is (—oo, —3) U (—3, co) and its range is (—oo, 1) U (1, 00).

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SECTION 3.6

Rational Functions

x-1 oar: When x = 0, we have r (0) = —4 so the y-intercept is —j. The numerator is 0 when x =

21. Rice

183

1, so the

x-intercept is 1. 2 : x*—x-2 eae ix) = 7 4

: —2 When x = 0, we have t (0) = ce =

i. so the y-intercept is i. The numerator is 0 when

x27—-x-2= (x — 2) (x + 1) = 0 or when x = 2 orx = —1, so the x-intercepts are 2 and —1. 25hix)

x7-9

=

é

7

a

:

Since 0 is not in the domain of r (x), there is no y-intercept.

;

The numerator is 0 when

x27 -9=(x- 3) (x +3) = 0 or when x = +3, so the x-intercepts are +3.

27. From the graph, the x-intercept is 3, the y-intercept is 3, the vertical asymptote is x = 2, and the horizontal asymptote is Me 2 29. From the graph, the x-intercepts are —1 and 1, the y-intercept is about i> the vertical asymptotes are x = —2 and x = 2, and the horizontal asymptote is y = 1.

rQ)=

Sh

5 has a vertical asymptote where x — 2 = 0 = x = 2, and y = 0is a horizontal asymptote because the degree Pp4 —

of the denominator is greater than that of the numerator. 3 1 Sheed ea he A of i has no vertical asymptote since 4x” + 1 > 0 for all x. y = 0 is a horizontal asymptote because the degree x of the denominator is greater than that of the numerator.

Ox? Hil

SEOG(Ga eS ET

a

x

has vertical asymptotes where 2x“

2

+ x —1=0@

(x +1) Qx —1) =0@x

=—lorx

1 and = 5,

_—

horizontal asymptote y = 8 as

+—2)1) @x — 3) 37. r (x) = @ (ce ————-h (4x +7)

7

as vertical rtical asymptotes totes whwhere (x —2) ) (4x (4x ++7) 7) =0@x x = —Z q orx ox = 2,and riz horizontal

ee Lepy 1 asymptote y = Gade.

6x3 —2

39. ¢ (x) = ————_

2x3 + 5x24 6x

6x3 —2

= > —

ne

. Because the quadratic in the denominator has no real zero, r has vertical

~x (2x? + 5x + 6)

asymptote x = 0 and horizontal asymptote y = § es} x2 42

41 a

7 A vertical asymptote occurs when x — 1 = 0 " When x = 0, y = —2, so the y-intercept is —2. When y = 0, x 4x —-4=0 —2t, y= “ — —oo, and as = :

'

x7

es NS Aoi’ —2°,y=

— oo. The

x+2

JX. domain is

|

1

x

.Sem

{x |x # —2} and the range is

| |

fy ly #4}.

.

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184

CHAPTER 3 Polynomial and Rational Functions 3x2 =.12

3(x-Z —4x+4)4+1

13

sciiak Waar baadsypod traded + Argh ral x2 —4x+4+4

x2 —4x4+4

1

(x — 2)

=. When x = 0,

w= B. so the y-intercept is B. There is no x-intercept since Game x

is positive

—

on its domain. There is a vertical asymptote at x = 2. The horizontal asymptote is y = 3. The domain is {x | x # 2} and the range is {y | y > 3}.

47. r (x)

Me

-yar—ig

= —————_ =

Te

x2 — 8x + 16

tIe)-2

x2 — 8x + 16

(x — 4)?

. When

== OS pi -3, and so the y-intercept is —2. There is no x-intercept since Gua is positive on its domain. There is a vertical asymptote at — 4.The horizontal asymptote is y = —1. The domain is {x | x 4 4} and the range is

WLP

lb

4x —8

—8

49. sSq)— @-4@+41)

When

x =i)

= Ca

= 2, so the y-intercept is 2.

When y = 0, 4x — 8 = 0 & x = 2, so the x-intercept is 2. The vertical asymptotes are x = —1 and x = 4, and because the degree of the numerator is less than the

degree of the denominator, the horizontal asymptote is y = 0. The domain is {x |x # —1, 4} and the range is R.

Si GIES =

2A

oN)

x72

Dix (x

2)

1) (x +2)

. When x = 0, y = 2, so the y-intercept is

2. When y = 0, we have 2x — 4 = 0 x

y= z, so the y-intercept is z. When

= -2, 1, so, the x-intercepts are —2 and 1. The

vertical asymptotes are x = —1 and x = 3, and because the degree of the numerator and denominator are the same the horizontal asymptote is y = ; Nb The domain is {x |x 4 —1, 3} and the range is R.

2x2 +2x-4

55. r (x) (x) = ————_ 5 ce

=

2(¢4+2)(x-1) . Vertical asymptotes occur at x = 0 GT} ymp

and x = —1. Since x cannot equal zero, there is no y-intercept. When y = 0, we have x = —2 or 1, so the x-intercepts are —2 and 1. Because the degree of the denominator and numerator are the same, the horizontal asymptote is y = i — iM The domain is {x | x 4 —1,.0} and the range is {y | y < 2 ory > 18.4}.

mh oie) =

x2 = 2x]

an (x — 1)?

x3 = 3x2.

x2 (x -—3)

. Since x = 0 is not in the domain of s (x),

there is no y-intercept. The x-intercept occurs when x? —2x +1 =(x — 1)?

y= 0 &

=0=> x = 1, so the x-intercept is 1. Vertical asymptotes

occur when x = 0, 3. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. The domain is {x |x 4 0, 3}

and the range is R.

De ee

39. r@)=

Bo

ie

eer aeae

or

-

Re?

SO

(x + 1)?

ae.

-(F

x+1

2

7)» When: = 0, y = 1 so the

y-intercept is 1. When y = 0, x = 1, so the x-intercept is 1. A vertical asymptote occurs atx + 1 =0 O}.

Bas 61.

r(x)= pe x2 44x44

£52b1) Se

(x2)

5 4

Whenx = 0, we have y = -, so the

y-intercept is ;. Since x2 + 1 > 0 forall real x, y never equals zero, and there is no x-intercept. The vertical asymptote is x = —2. Because the degree of the

denominator and numerator are the same, the horizontal asymptote occurs at y= ; = 5. The domain is {x | x #4 —2} and the range is {y |y > 1.0}.

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185

186

CHAPTER 3 Polynomial and Rational Functions

63. r(x) =

x24 4x—-5 x2+x—-2

(x+5)(Q@—-1)_ x45 (x+2)(x-1)

x42

for x # 1. Whenx =0, y = §,

so the y-intercept is 3. When y = 0, x = —5, so the x-intercept is —5. Vertical asymptotes occur when x + 2 = 0 oo asx > 27, and

y — —co as x — 27. Because the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptotes. By using long

week division, we see that

y= x

4 ; +2+ Prac, so y = x +2 isa

slant asymptote.

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SECTION 3.6 Rational Functions

TL

(x)= e

7-2x-8

(x —-4)(x +2)

. The vertical asymptote is x = 0, thus, x ae there is no y-intercept. If y = 0, then (x — 4) (x +2) =0 > x = —2, 4, so the x-intercepts are —2 and 4. Because the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes. By using long

fae 8 é division, we see that y = x — 2 — —, so y = x — 2 isa m5

x*) +5x+4

ee

x+4)(x+4+1

slant asymptote.

. When x =0,y = —}, so the y-intercept

is —}. Wheny = 0, (x + 4) (x +1) =0@x

= —4, -1, so the two x-intercepts

are —4 and —1. A vertical asymptote occurs when x = 3, with y > oo as x > 3+, and y — —oo as x > 37. Using long division, we see that

28 y=x+8+-——,80y

J =x + 8 isaslant asymptote.

x=~—

LS TACX) =

x? +x £

x2—4

¥ (x

+1)

@ —2)(@ +2)

. When x = 0, y = 0, so the graph passes

through the origin. Moreover, when y = 0, we have x? (x+1)=0>x=0,-1,

so the x-intercepts are 0 and —1. Vertical asymptotes occur when x = +2; as x > +27, y = —oo and as x > +27, y > oo. Because the degree ofthe

numerator is greater than the degree of the denominator, there is no horizontal

,

b-vie

4x +4

asymptote. Using long division, we see that y = x +1+ aia? soy=x+lisa y&

—

slant asymptote.

2 ena)

— ena, x

g (x) = 2x. f has vertical asymptote x = —3.

Te, fix) = eae

ie — x2. f has vertical asymptote x = 2. 50

-10

10° -50

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187

188

CHAPTER 3 Polynomial and Rational Functions

2x? — 5;

S12)

eae x

has vertical asymptote x = —1.5, x-intercepts 0 and 2.5,-y-intercept 0, local maximum (—3.9, —10.4),

and local minimum (0.9, —0.6). Using long division, we get f (x) =x —4+

12 Pep

From the graph, we see that the end

behavior of / (x) is like the end behavior of g (x) = x — 4. 4 2x +3

Dy

2

50

rer

alge Sa

2x? + 3x — 8x

-20

— eh —

20

ele

-50

83. f (x) =

x

ead |

has vertical asymptote x = 1, x-intercept 0, y-intercept 0, and local minimum (1.4, 3.1). 2

= : From the graph we see that the end behavior of f (x) is like the end behavior of g (x) = x2, x31

Thus y = x? +

x2 Tot

x? x

Te x2 x2

Graph of f

S550 a)

= 2

4

3x? +6

Graph of f and g

has vertical asymptote x = 3, x-intercepts 1.6 and 2.7, y-intercept —2, local maxima (—0.4, —1.8)

—3

and (2.4, 3.8), and local minima (0.6, —2.3) and (3.4, 54.3). Thus y = x3 + oy From the graphs, we see that the end x= behavior of f (x) is like the end behavior of g (x) = x.

87. (a) 2000

0 0

b p(t) t) (b)

20

3000t 3000 = —— ee = 3000 — ——. mag

40

.

So as t > 00, we

have p (t) — 3000.

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SECTION 3.6

Rational Functions

189

(a) The highest concentration of drug is 2.50 mg/L, and it is reached 1 hour after the drug is administered. (b) The concentration of the drug in the bloodstream goes to 0.

(c) From the first viewing rectangle, we see that an approximate solution

: St is near f= 15. Thus we graph y = a: andy = 0.3 in the viewing rectangle [14, 18] by [0, 0.5]. So it takes about 16.61 hours for the concentration to drop below 0.3 mg/L.

14

16

18

SO 332 S12 2 (0) =F P (v) = 440 () : ( 4 (i—x

) © (x —7) (1—Xx ) >0e

P (x)= (x —7) (1 —x)3 (1 +x)? > 0. The intervals serene by the zeros—1, 1, and 7 are (—oo, —1), (—1, 1), (1, 7), and (7, co). We make a sign diagram:

x7

= =

=e

—

None of the endpoints satisfies the inequality. The solution is (—oo, —1) U (1, 7).

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or ss

ae

SECTION 3.7 Polynomial and Rational Inequalities

193

—1 < 0. Since all nonzero terms are already on one side of the inequality symbol and there is no factoring —10 dod we find the intervals determined by the cut points 1 and 10. These are (—0o, 1), (1, 10), and (10, 00). We make a sign diagram:

17; r(Xx)=

2x +5 ear

19. Fate) Aw) —

x? + 2x — 35 2x a

+5

>

0.

Since all nonzero terms are already on one side, we factor:

2x =

+5

. Thus, the cut points are —7, —3, and 5. The intervals determined by these

x? + 2x — 35 ~ & ENG BS)e

points are (—oo, —7), (-7, +3),Ch als and (5, 00). We rare a sign diagram:

The cut point -. satisfies equality, but the cut points —7 and.5 are not in the domain of r. Thus, the solution is

(-7, -$] UG, 00).

74te We foe) Ne ea ec

< 0. Since all nonzero terms are already on one side but the denominator cannot be factored, we

—

use the Quadratic Formula: x2 + 2x

-2 =0@x =

—2+ ae 4 (1)

(—2)

vig eus tee?) ( &

—1 + /3. Thus, the cut points are

= V3, 0, and —1 + /3. The intervals determined by these points are ess -l- v3), (-1 awl, 0), (0,V3— 1),

and (v3 —1, ouWe make a sign diagram:

pe

ee

ee

en

_ The cut point 0 satisfies equality, but the cut points —1 + iaare not in the domain of r. Thus, the solution is

(-o0, -1 = v3) U [o.vax 1).

\

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194

CHAPTER3 Polynomial and Rational Functions

pe alk 6)

x2 + 2x.—3

A

3x2 -—7x—-6

eth 3)ie. 1)

(3x +2)(x —3)

> 0. The intervals determined by the cut points are (—oo, —3), (-3, -3),

(3,1), (1,3), and @, 00).

See [Co- (3) Ch) ea ear ~ +

None of the cut points satisfies the strict inequality, so the solution is (—oo, —3) U (-3, 1) UG, co).

3

IS TALE?)

: fees b Phe

x

please Aaent Sea

2 2 (x*+6x+9)—3(x*+6x4+9

«(x2 46x+9)-3(x?+6x+9) _

=

3)2

29 @+3)" & 0; The second Eee

x+4 x+4 x+4 -; numerator is positive for all x, so the intervals determined by the cut points are (—oo, —4), (—4, 3), and (3, 00).

x+4

ae

+h

ne

tS

=

=

+

r (x)

+

a

oF

The cut point 3 satisfies equality, but —4 is not in the domain of r. Thus, the solution is (—4, 3].

x —3 simplifying x5 27. Wee start startbby moving i all lit terms to t oneside ide and and simplifying: Fe

—(x + 8)

;

x—3 => 1 GES —-1>0>0e

x —3—(2x +5) x 4 5 >0 0. The intervals determined by the cut points are (—oo, —8), (-8, -3),

The cut point —8 satisfies equality, but 3 is not in the domain of r. Thus, the solution is [-8. ~§).

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.7 Polynomial and Rational Inequalities

29. We start by moving all terms to one side and simplifying:

2(1—x)(x)

+x -3(1-

= (6) =4/(-6)? —4(2)G)' aT YP

3-73 ave 2

_

FE

34/3 5

~

< sf@2+

—2x?+6x-—3 °° 2x? -6x

peer) tA 3(LX) SUS x (1 —x) x= “So

2 + i

yaleay = rat WeDi sie Soares x (1 —x) x (x -—1)

34.3

= ae

+3

i

ee

195

ye =< 0S eles

< 0. The numerator is 0 when

-

so the intervals determined by the cut points are (—oo, 0), | 0, 3 ae :

i 54/3

2

, an

5)

» CO

f.

x-1

2x? — 6x +3

r (x) » JIBS The cut points

A

agte

;

2 satisfy equality, (0oF u(. 4 |

Sear

; : yah but 0 and 1 are not in the domain of r, so the solution is

Cea):

i

> 0. The numerator is nonnegative for all x, but note that x = | fails to satisfy the strict inequality. (x +1) (x +2) The intervals determined by the cut a saa are (—0o, Ty(—2, —1), and (—1, 00).

The cut points —2 and —1 are not in the domain ofr, so the solution is (—oo, —2) U (—1, 1) U (1, ov).

33. We start by moving all terms to one side and factoring:

6 6 arian Saale

_ -1)Sime Xm 2) (x — eee es) 2 1) >0e Riera URL =. Oe (x) = tA) x(x —1) x(x —J) x(x -1) cut points are (—oo, —2), (—2, 0), (0, 1), (1, 3), and (3, 00).

Er

6

ay

3).

6

Cre

eed

(0)

> 0. The intervals determined by the

ON ON OE

x+2

28 x-1 ;

x-—3 r (x) (note negative sign)

The cut points 0 and 1 are inadmissible in the original inequality, so the solution is [—2, 0) U (1, 3].

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196

CHAPTER 3 Polynomial and Rational Functions

35. We start by moving all terms to one side and factoring: eae

2

oe

x—1

< aay =

x+2

x—1

Ces

< 06

B29 4

Gah 2)0979) aN) (x + 3) (x —2)

et 3) A

(639) — eee < 0. The intervals determined by the cut points are (x + 3) (x — 2)

(—00, —3), (-3, -3), ci 2),and (2, 00). x+3 x+4

x—2

r (x) (note negative sign)

37. The graph of f lies above that of gwhere f (x) > g(x); that is, where x2 > 3x +10

@ x? —-3x -10 > 0@

(x + 2) (x — 5) > 0. We make a sign ear

xX +2 x—5

(x +2)

—5)

Thus, the graph of f lies above the graph of g on (—o0o, —2) and (5, 00).

1 1 39. The graph of f lies above that of g where f (x) > g(x); that is, where 4x > —-006 % x (2x + 1) (2x — 1) (G0) = > 0. We make a sign diagram: 25

4x? =]

>0o

x

Thus, the graph of f lies above the graph of g on (-3, 0)and (3;00).

41. f (x) = V6 +x — x? is defined where 6 + x — x? = — (x +2) (x —3)> 0. We make asign diagram: x+2

x—3

— (x +2) (x —3) Thus, the domain of f is [—2, 3].

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———

SECTION 3.7 Polynomial and Rational Inequalities

197

43. h(x) = Vx4 — 1 is defined where x4 — 1 = (x — 1) (x +1) (x? + 1)> 0. The last factor is positive for all x. We make a

sign diagram:

cry] aes +

(x — 1) +1) (x2 +1

fs

Thus, the domain of h is (—oo, —1] U [1], oo).

45.

47.

From the graph, we see that x? — 2x2 — 5x +6 >0on

From the graph, we see that 2x3? — 3x +1 0e8

+x0 — 4° + 4x >0e w3xt +2x 41 >0er(x)=

Je

Ji

(1 a —x)? ~ J& (4x) @ —_ - 1) am

1—x)(3x

2> 08

+1

G—x) Gx +) > 0. The domain of r is (0, 00), and

Je

both 3x + 1 and ./x are positive there. 1 — x > 0 forx < 1, so the solution is (0, 1]. 53. We want to solve P (x) = (x — a) (x — b) (x —c) (x —d) = 0, wherea < b < c < d. We make a sign diagram with cut points a, b, c, and d:

(6,6) |(4) |@.00) ~

+

+

“bh

Each cut points satisfies equality, so the solution is (—oo, a] U [b, c] U [d, 00).

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

198

CHAPTER 3 Polynomial and Rational Functions

55. We want to solve the inequality T(x)

500,000 — 300 (x? + 400) >

x? + 400

—_




ae ~ 35.6. The

temperature is below 300 °C at distances greater than 35.6 meters from the center of the fire. 88x 57. We graph WN (x) = and y = 40 in the viewing rectangle 2 17+17 (35) [0, 100] by [0, 60], and see that NV (x) > 40 for approximately 9.5 < x < 42.3. Thus, cars can travel at between 9.5 and 42.3 mi/h.

~ CHAPTER3 REVIEW 1. (a) f(x) =x? +6x4+2= (x? +6x) +2

3. (a) f(x) =1=10x =x? =-(x?+10r) +1

= (x? +6x +9) +2-9

= — (x? 4 10x +25) +1425

= (x +3)?-7

= —(x +5)? +26

(b)

y

(b)

8. f (x) =—x? +3x—1 = — (x? —3x) -1=-(x?-3r 49)-14+ §=—(x- 3) + Jhasthe maximum value 3 2

when x = 3.

7. We write the height function in standard form: h (1) = —161? + 481 + 32 = —16 (? = 31)Me

es | (??= Jae 3)rs

2

32 + 36 = —16 (1- 3) + 68. The stone reaches a maximum height of 68 ft.

9. P(x) = —x3 + 64

11. P (x) =2(x +1)4 — 32

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CHAPTER3

13. P (x) =32+(x -1)9

15. (a) P (x) = (x — 3) (x +1) (x —5) = x9 —7x247x415 y

‘

has odd degree and a positive leading coefficient, so y — ooasx > ooand y > —oo asx

(b)

17. (a) P(x) = —(x — 1) « —4) « +2)” ene

Review

> —o0.

y

19. (a) P (x) =x? (x — 2)?. The zeros of P are 0 and 2,

4 «ty? — 8x2 — 200 416

has odd degree and a negative leading coefficient, so

y > —00 as x > ooand y > coasx > —o0.

with multiplicities 3 and 2, respectively.

(b) We sketch the graph using the guidelines on

page 295.

(b)

AY 12 (68)

x3 — 4x + 1. x-intercepts: —2.1, 0.3, and 1.9.

23. P (x) = 3x4 — 4x3 — 10x — 1. x-intercepts: —0.1 and

y-intercept: 1. Local maximum is (—1.2, 4.1). Local

2.1. y-intercept: —1. Local maximum is (1.4, —14.5).

minimum is (1.2, —2.1).

There is no local maximum. y + oo as x+ +00.

as xX

y

oo asx

00; y % —00

—0O.

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199

200

CHAPTER 3 Polynomial and Rational Functions

25. (a) Use the Pythagorean Theorem and solving for y? we have, x? + y* = 10? = y? = 100 — x”. Substituting

we get S = 13.8x (100 eex?) = 1380x — 13.8x3. (b) Domain is [0, 10].

Using synthetic division, we see that QO (x) = x — 2 and

(c)

R(x) = -4.

(d) The strongest beam has width 5.8 inches.

29.

x4 — 8x2 42x +7

2x3 —x243x—4 6 ie)

ee

—5)|

2 - 1-3 2

ARE

—4

-10

55

-—290

-11

58

-—294

x+5

—5 } 1 1

0

-8

2

z

—-5

25

—-85

415

-5

17

—-83

422

Using synthetic division, we see that

Using synthetic division, we see that

QO (x) = 2x? — 11x +58 and R (x) = —294.

O (x) = x3 — 5x? + 17x — 83 and R (x) = 422.

2x3 + x2 —8x x24+2x—1

+15

ee Opa ; P (5). 35. P (x) = 2x? — 9x* — 7x + 13;find

rao x*4+2x-1

idea

| 9x3

ST

x2 "Bx15

2x3 + 4x? —

2

RE

2, G— ON ee] ae

13

sues oa

2x

6h

15

ae

Therefore; P:(3))= 3.

, 12

Therefore, O (x) = 2x — 3, and R (x) = 12.

37

The remainder when dividing P (x) = x59 46x10! —x2-2x44 by x—1 is P (1) = (1)999 +6 (1)2°! —(1)?-2 (1) +4 = 8.

Ab : is a zero of P (x) = 2x4 +.x3 — 5x? + 10x —4 if

Since P (3) —A 41. ENA

4 is a zero of the polynomial.

3) k= x? — 6x3 — x2 42x + 18 has possible rational zeros +1, +2, +3, +6, +9, +18.

(b) Since P (x) has 2 variations in sign, there are either 0 or 2 positive real zeros. Since P (—x) = —x> +6x3 —x?2-—2x +18

has 3 variations in sign, there are 1 or 3 negative real zeros.

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CHAPTER3

Review

201

43. (a) P (x) = 3x? —x° + 5x4 4x3 + 8 has possible rational zeros +1, +2, +4, +8,+4,+3,44, +8. (b) Since P (x) has 2 variations in sign, there are either 0 or 2 positive real zeros. Since

P (—x) = —3x7+x°+5x4—x3+8

has 3 variations in sign, there are 1 or 3 negative real zeros.

45. (a).P (x) =x3 — 16x =x (x?FS16)

47. (a) P(x) =x4 4x3 — 2x? =x? (x?+x -2)

=x (x —4)(x +4)

= x2 (x +2) (x —1)

has zeros —4, 0, 4 (all of multiplicity 1).

(b)

The zeros are 0 (multiplicity 2), —2 (multiplicity 1),

;

and | (multiplicity 1).

(b)

49. (a) P(x) = x4 — 2x3 — 7x2 + 8x + 12. The possible rational zeros are +1, +2, +3, +4, +6, +12. P has 2 variations in

sign, so it has either 2 or 0 positive real zeros. 1/1

2

-7 1

1

-l

-l

Sai? -8

-8

2/1

—-2.

0

Ov,

12

9) 2

0

-7

8

0

-14

-7

—6

12 —-12 Os

= 2 is 'airoot!

P (x) = x4 — 2x3 — 7x? + 8x + 12 = (x — 2) (x3 — 7x — 6). Continuing:

2|1 0 ~6 Py 1 2 —2

eee

Ss(to 7 tab

et -10

S55 Te

(b)

Oleh

ite Zee 0

so x = 3 isaroot and

P (x) = (x — 2) (x —3) (x? +3x +2) = (x —2)@

—3)(*¥ +1) @+4+2)

Therefore the real roots are —2, —1,2, and3 (all

of multiplicity 1).

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202

CHAPTER 3 Polynomial and Rational Functions

51. (a) P (x) = 2x4 4x3 4+ 2x2 — 3x — 2. The possible rational roots are +1, +2, +}. P has one variation in sign, and hence

1 positive real root. P (—x) has 3 variations in sign and hence either 3 or | negative real roots. Lfe2>

ing

P (x) = 2x4 x3 42x? — 3x —2 =

ls 2s ares he?

Re

The

2.

0

— 1) (283 43x? + 5x +2).

Continuing:

SDT

(b)

ob —2 Die

==") 1s a zero;

base

toe -1

2

21

-4

eee —4

Acme

2

2 -14

2-1

7 -12

at1 1p eee Oe -1 -1 -2 Oe?

ane

0

=x

=—} isazero.

P(x) = (x -1) (x+ +) (2x +2x + 4). The quadratic is irreducible, so the real zeros are 1 and -3 (each of multiplicity 1).

53. Because it has degree 3 and zeros -4, 2, and 3, we can write

P(x) =

C (x+ ) (x —2)(x —3)

Cx3 — 3Cx? + £Cx + 3C. In order that the constant coefficient be 12, we must have 3C =

=

12 = C = 4, so

P (x) = 4x3 — 18x? + 14x + 12. 55. No, there is no polynomial of degree 4 with integer coefficients that has zeros i, 2i, 3i and 4i. Since the imaginary zeros of polynomial equations with real coefficients come in complex conjugate pairs, there would have to be 8 zeros, which is impossible for.a polynomial of degree 4.

57. P(x) =x? —x*4+x—I1has possible rational zeros +1.

1

71°

°=1-

1.

Lee 1

Oa

=1 1

el

Om

(= x=] As alzero,

So P (x) = (x - 1) (= o 1).Therefore, the zeros are 1 and +i.

i hidail 69)

x? — 3x? — 13x + 15 has possible rational zeros +1, +3, +5, +15.

1)/1

-3 1 1

—2

-13

15

—2

—-15

-15

0:

=x) Silhsta vero:

So P (x) = x3 — 3x2 — 13x +15 = (x -1) (x? dre 15) = (x — 1) (x — 5) (x +3). Therefore, the zeros are—3, 1, and 5.

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1

tI]

CHAPTER3.

Review

203

GL Fix yi x4 + 6x3 + 17x? + 28x + 20 has possible rational zeros +1, +2, +4, +5, +10, +20. Since all of the coefficients

are positive, there are no positive real zeros. —1}

1

6 -l

1

(oy

5

ilk)

28

20

-5

-12

-—16

12

16

4

—2/

1

6

;

‘Nef

—2 1

-8

4

28

20

-18

-—20

10

0

9

=x

=-2

isazero.

x4 + 6x3 + 17x? 4+ 28x +20 = (x + 2) (x? + 4x2 49x74 10).Continuing with the quotient, we have

1

2

5

0

=x

=-2isazero.

Thus P(x) = x4 + 6x3 + 17x2 + 28x + 20 = (x +2)? (x? +2x +5). Now x2 + 2x +5 = 0 when a

pe

OD) = at. = —| + 2i. Thus, the zeros are —2 (multiplicity 2) and —1 + 2i.

63.0. (xc) — x> — 3x4 — x3 4 11x? — 12x +4 has possible rational zeros +1, +2, +4.

1)

1

1

/-3.

-1

11

—-12

4

1

—2

-3

8

—-4

—2

-3

8

—4

OS

4 —

1 5s'azero:

P(x) =x? — 3x* — x3 + 11x? —12x +4 = (x - 1) te. — 2x9 — 3x2'4-8x — 4).Continuing with the quotient, we have

1

x5 —3x4 — x3 4 1x2 - 12x44

-l

-4

4

Ole

ol 1S)a-7ero!

=(x-1) (3 — x? — 4x +4) = (x —1)3 ee -4) =O -1t) @—2) 6G +2)

Therefore, the zeros are | (multiplicity 3), —2, and 2.

65. P(x) = x5 —64= (x?ee8)(x3 " 8) 2G

(x?$e + 4)(x +2) (x? ~2x4 4). Now using the Quadratic

Formula to find the zeros of x2 + 2x + 4, we have

x= tS PhO)

= rE

= —1 +13, and using the Quadratic Formula to find the zeros of x? —2x + 4, we have

= 2ida/3 = 1+i4/3. Therefore, the zeros are 2, —2, 1 +i/73, and -1+iJ3.

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204

CHAPTER3 Polynomial and Rational Functions

67. P (x) = 6x4 — 18x? + 6x? — 30x + 36 = 6 (x4 — 3x3 4+ x2 = Sxit- 6)has possible rational zeros +1, +2, +3, +6. 1

j}-6>

-—18

6

~=30

36

6

—12

—-6

—36

WY

ree

1

oer.

So P(x) = 6x4 — 18x} + 6x? — 30x + 36 = ( - 1) (6x3 12x? — 6x — 36) ay

1) (x3 —2x? —x-6).

Continuing with the quotient we have

1;1

—-2

-1

-6

1

-1

-2

—2

-8

1 °-1

2|;1

2 1

-1

-6

2

0

-2

0

-1

-8

3;1

2 1

-1

-6

3

3

6

1

2

QO:

=> x= 318 ager:

So P(x) = 6x4 — 18x3 + 6x? — 30x + 36 = 6(x— 1) (@ =3) (x? +x +2). Nowx2 +x +2 = 0 when a

chyA) Ae)

=l4iV7 and so the zeros are 1, 3, and —ltiv7

69. 2x2 = 5x +3 & 2x? — 5x —3 = 0. The solutions are

x = —0.5, 3.

71.x4 — 3x? —3x2-—9x —2 Js0 has solutions x + —0.24,

4.24.

73. P(x) =x? -—2x -—4 1 |10

2 l

GPa Je (Gel x2 -2x-4=

-4 1

2 |) le

-l

Oe y)

SSS Rates

i

a 4

4

2

0

(x — 2) (x?+2x+ 2). Since x? + 2x +2 = 0 has no real solution, the only real zero of P is

Foch

3 . ; 75. (a) r(x) = Pew The vertical asymptote is x = —4. Because the denominator has higher degree than the numerator, the horizontal asymptote is y = 0. Whenx = 0, y = i,so the y-intercept is i. There is no x-intercept because the numerator is never 0. The domain ofr is (—oo, —4) U (—4, 00) and its range is (—o0o, 0) U (0, 00).

(b) If f(x) = Shee 9%

eas =3/ )=3/ x+4 x+4

+4, sowe

obtain the graph of r by shifting the graph of f to the left 4 units and stretching vertically by a factor of 3.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER3

x —4 77 - (a) r (x) = ——.. x-1

Review

The vertical asymptote is x = 1. Because the

denominator has the same degree as the numerator, the horizontal asymptote is y = 3 = 3. When x = 0, y = —4, so the y-intercept is —4. When

y= 0,3x

—4=008x=

;, so the x-intercept is z. The

domain of r is (—oo, 1) U (1, 00) and its range is (—o00, 3) U (3, 00).

1 (b) If f (x) = —,then 2

3(x-1)-1

rx) = SAD a1 3-1

1

=3~ f(e1), Thus,we

obtain the graph ofr by shifting the graph of f to the right 1 unit,

reflecting in the x-axis, and shifting upward 3 units.

IEA

MED Ve

SX e712 —12 TG When x = 0, we have r (0) = ee x

—12. Since y = 0, when 3x — 12

—12, so the y-intercept is

= 0 & x = 44, the x-intercept is 4. The vertical

asymptote is x = —1. Because the denominator has the same degree as the numerator, the horizontal asymptote is y = T3 — 3. The domain of r is P (—oo, —1) U (—1, 00) and its range is (—0o, 3) U (3, 00).

x—2

81. r (x) = ———_ (*) x2-—2x-8

x—2

=)

1

= ————. When x = 0, we haver (0) = = =, (x+2)(x-4) (0) vba

so the y-intercept is i: When y = 0, we have x — 2 = 0 0, the denominator is never zero so there are no vertical asymptotes. The horizontal asymptote is at y = 5 because the degree of the denominator and numerator are the same. The domain of 7 is (—0o, 00) and its range is [-9, 4).

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

205

206

CHAPTER 3 Polynomial and Rational Functions

R507) (x)=

x4 5x14. x—2

(6.47) (x -2) yg

x-—2

=x+7

forx 4 2. The x-intercept is

—7 and the y-intercept is 7, there are no asymptotes, the domain is {x |x # 2}, and the range is {y | y 4 9}.

- r(x) = ;

tea

Puy? 28x 215

_ @&+6(—-3)_ x+6 x — 5) x3)

ex 5

for x # 3. The x-intercept is

—6 and the y-intercept is — § the vertical asymptote is x = 5, the horizontal asymptote isy= 1, the domain is {x |x 4 3, 5}, and the range is {»yee, -3}.

—3 89587 (05) a— moe From the graph we see that the x-intercept is 3, the y-intercept is —0.5, there is a vertical asymptote at x = —3 and a horizontal asymptote at y = 0.5, and there is no local extremum.

210

x7 $) =E/8 SDEs lla) y— °F aa xe —x—

60 4

7 ; From the graph we see that the x-intercept is —2, the

y-intercept is —4, there are vertical asymptotes at x = —1 and x = 2, there is no horizontal asymptote, the local maximum is (0.425, —3.599), and the local

minimum is (4.216, 7.175). By using long division, we see that

Sf@)=x+1 +

10—5; $0 f has a slant asymptote of y = x + 1. x*—x—

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CHAPTER3

93. 2x7 > x +3

Review

207

2x? —x —3 > O'e P(x) = (x + 1) (2x — 3) > O. The cut points occur where x+ 1 = 0 and where

2x — 3 = 0; that is, atx = —1 andx = 3. We make a sign diagram:

Sigof n |Hee Utena ee x+1 2x —3 P(e)

Both endpoints satisfy the inequality. The solution is (—oo, —1] U [3.00).

95. x4— 7x2

18

0. Using the Quadratic Formula to solve —x* —2x+4=0, we

V4 —2x — x?

—(—2)

have x = nN

+ ./(—2)? =4(=1) (4

NT ae 2-1 fis (-1 — 3, -1+ V5).

FOCUS ON MODELING

RI

= —1+J/5. The radicand is positive between these two roots, so the domain of

Fitting Polynomial Curves to Data

1. (a) Using a graphing calculator, we obtain the quadratic

3. (a) Using a graphing calculator, we obtain the cubic

polynomial

polynomial

y = —0.275428x2 + 19.7485x — 273.5523 (where

y = 0.00203709x3 — 0.104522x2

miles are measured in thousands).

b)

2

+ 1.966206x + 1.45576.

(b)

3

3

5

3

= 2 =g

E + s=

oO

S omel

z, oO

2

Z

Pressure (Ib/in’) (c) Moving the cursor along the path of the polynomial,

we find that 35.85 Ib/in2 gives the longest tire life.

Seconds () Moving the cursor along the path of the polynomial

¢find thatthe subjects could bout hirenerarar aR eyes ae She

43 vegetables in 40 seconds.

(d) Moving the cursor along the path of the polynomial, we find that the subjects could name 5 vegetables in about 2.0 seconds.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Fitting Polynomial Curves to Data

|

‘-

5. (a) Using a graphing calculator, we obtain the quadratic

‘

(b)

polynomial

y = 0.0120536x? — 0.490357x + 4.96571. (c) Moving the cursor along the path of the polynomial,

i

_

we find that the tank should drain in 19.0 minutes.

—

(ft) Height

211

4

ht oe

4 —_EXPONENTIAL AND LOGARITHMIC FUNCTIONS 4.1

EXPONENTIAL FUNCTIONS

1. The function f (x) = 5* is an exponential function with base 5; f (—2) = 5~* = 44, f (0) =5° = 1, f (2) = S* =25, and f (6) = 5° = 15,625. 3. (a) To obtain the graph of g (x) = 2* — 1 we start with the graph of f (x) = 2* and shift it downward 1 unit.

(b) To obtain the graph of h (x) = 2*—! we start with the graph of f (x) = 2* and shift it to the right 1 unit.

5. The exponential function f(x) = (+) has the horizontal asymptote y = 0. This means that as x —> 00, we have

~

(3) — 0.

1 f(x)=4i f (3) SEs

(v5) s 22.195, f (—2) = 0.063, f (0.3) © 1.516.

l

9. g(x) = (4)" :g (5) ~ 0.192,g(v2) ~ 0.070, g(—3.5) © 15.588, g(—1.4) © 1.552. UW.

f(x*)=2

13. f(x)= (+)

: © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

213

214

CHAPTER 4 Exponential and Logarithmic Functions

19. f (x) = 4 and g(x) =7*.

21. From the graph, f (2) = a* = 9, s0 a = 3. Thus

= Oe a(x

y

23. From the graph, f (2) = a? = 74, soa = +. Thus f (x) = (4)

x

25. The graph of f (x) = 5**! is obtained from that of y = 5* by shifting 1 unit to the left, so it has graph II.

27. g(x) = 2* —3. The graph of g is obtained by shifting the

29. The graph of f (x) = —3* is obtained by reflecting the

graph of y = 2* downward 3 units. Domain: (—oo, 00).

graph of y = 3* about the x-axis. Domain: (—oo, 00).

Range: (—3, co). Asymptote:

Range: (—oo, 0). Asymptote:

y = —3.

31. f (x) = 10*+3. The graph of f is obtained by shifting the

y = 0.

33. y = 5~* +1. The graph of y is obtained by reflecting the

graph of y = 10* to the left 3 units. Domain; (—oo, oo).

graph of y = 5* about the x-axis and then shifting upward

Range: (0, co). Asymptote: y = 0.

1 unit. Domain: (—oo, oo). Range: (1, 00). Asymptote: isle

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 4.1

35. p= Qe (5) . The graph is obtained by reflecting the ee

1\* =

(3)

hard tt

; Pee -

leeds

215

37. h (x) = 2*~4 +1. The graph ofh is obtained by shifting

x

cima

Exponential Functions

1

pach besee aka

the graph of y = 2* to the right 4 units and upward 1 unit. Domain: (—oo, 00). Range: (1, 00). Asymptote: y = 1.

2 units. Domain: (—0o, 00), Range: (—oo, 2). Asymptote: y = 2.

39. g(x) =1—3-* = —3-* + 1. The graph of gis obtained

41. (a)

by reflecting the graph of y = 3* about the x- and y-axes and then shifting upward 1 unit. Domain: (—oo, 00). Range: (—oo, 1). Asymptote: y = 1.

(b) Since g (x) = 3 (2*) = 3f (x) and f (x) > 0, the height of the graph of g (x) .is always three times the height of the graph of f (x) = 2*, so the graph of g is steeper than the graph of /.

43.

From the graph and the table, we see that the graph of g ultimately increases much more quickly than the graph of f. ,

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

216

CHAPTER 4 Exponential and Logarithmic Functions

45. (a) From the graphs below, we see that the graph of f ultimately increases much more quickly than the graph of g.

(i) [0, 5] by [0, 20]

(ii) [0, 25] by [0, 107]

(iii) [0, 50] by [0,108]

47.

The larger the value of c, the more rapidly the graph of F (x) = c2* increases. Also notice that the graphs are just ; : : ae : shifted horizontally 1 unit. This is because of our choice of

c; each c in this exercise is of the form 2*. So f=

st

$6)

a) From the hh, we see that the function is increasin (a) tid & on (—oo, 0.50) and decreasing on (0.50, 00).

(b) From the graph, we see that the range is approximately (0, 1.78].

Qk 9x — oxtk

= 1h

2

LEER

LO)

ie

Oa

wie

Pe.

(2

‘)

53. (a) After 1 hour, there are 1500 - 2 = 3000 bacteria. After 2 hours, there are (1500 - 2) - 2 = 6000 bacteria. After 3 hours, there are (1500 - 2 - 2) -2 = 12,000 bacteria. We see that after ¢hours, there are N (t) = 1500 - 2! bacteria.

(b) After 24 hours, there are N (24) = 1500 - 224 = 25,165,824,000 bacteria.

55. Using the formula 4 (t) = P (1 +i)* with P = 5000, _ 0.04 i = 4% per year = 1 per month, and & = 12 - number of years, we fill in the table:

Time (years)

;

is

18

$5203.71 $5415.71 $5636.36 $5865.99 $6104.98 $6353.71

0.03 \7! = 10,000- 1.015~'. 2 57. P = 10,000, r = 0.03, and n = 2. So A (t) = 10,000 (1+ 093) (a) A (5) = 10000 - 1.015!° ~ 11,605.41, and so the value of the investment is $1 1,605.41.

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7

SECTION 4.2 The Natural Exponential Function

217

(b) A (10) = 10000 - 1.01529 = 13,468.55, and so the value of the investment is $13,468.55. (c) A (15) = 10000 - 1.01530 ~ 15,630.80, and so the value ofthe investment is $15,630.80.

4t

59. P = 500,r = 0.0375, andn = 4. So A(t) = 500 (1. 0.0575) 4

(a) 4 (1) = 500 (1- 0.0775) ~ 519.02, and so the value ofthe investment is $519.02. 0:0375 \® ~ 538.75, and so the value of the investment ; (b) 4 (2) = 500 (1+ 29775)" is $538.75. 40

(6) A (10) = 500 (1+ 29775)" ~ 726.23, and so the value of the investment is $726.23. 2(3

- 61. We must solve for P in the equation 10000 = P (1+ 092) ©) =P (1.045)® © 10000 = 1.3023P = P = 7678.96. Thus, the present value is $7,678.96.

n

0.08 \ !2

63. rapy = (1- “) — 1. Herer = 0.08 andn = 12, sorapy = (: ae =z) 0

— 1 © (1.0066667)!2 — 1 ~ 0.083000.

Thus, the annual percentage yield is about 8.3%.

65. (a) In this case the payment is $1 million.

(b) In this case the total pay is 2 + 22 + 23 +... +239 > 239 cents = $10,737,418.24. Since this is much more than method (a), method (b) is more profitable.

4.2

THENATURAL EXPONENTIAL FUNCTION

1. The function f (x) = e* is called the natural exponential function. The number e is approximately equal to 2.71828.

3. h (x) = e*; (1) © 2.718, h (4) © 23.141, A (—3) © 0.050, h (v2) = 4.113 5

fi)

15e*

7. g(x) =2+*. The graph of g is obtained from the graph

of y = e* by shifting it upward 2 units. Domain: (—oo, 00). Range: (2, oo). Asymptote:

y = 2.

y

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

218

CHAPTER 4 Exponential and Logarithmic Functions

9. y = —e”. The graph of y= —e” is obtained from the

11. y = e7* — 1. The graph of y= e~* — 1 is obtained from

graph of y = e” by reflecting it about the x-axis. Domain:

the graph of y = e” by reflecting it about the y-axis then

(—oo, 00). Range: (—oo, 0). Asymptote: y = 0.

shifting downward | unit. Domain: (—oo, 00). Range: (—1, co). Asymptote:

13. y = e*~. The graph ofy= e*~? is obtained from the graph of y = e* by shifting it to the right 2 units. Domain: (—oo, 00). Range: (0, 00). Asymptote: y = 0.

y = —1.

15. h (x) = e*+! — 3, The graph of/ is obtained from thegraph of y = e” by shifting it to the left 1 unit and downward 3 units. Domain: (—oo, 00). Range: (—3, 00). Asymptote: y = —3.

17. (a)

19. (a)

; a (b) As a increases the curve y = 2 (e/a+ e-*/4) (b) cosh (—x) = pete

EES

hE Be

flattens out and the y intercept p increases.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 4.2 The Natural Exponential Function

219

21. g(x) = x*. Notice that g (x) is only defined forx > 0. The graph of g (x) is shown in the viewing rectangle [0, 1.5] by [0, 1.5]. From the graph, we see that there is a local minimum of about 0.69 when x ~ 0.37. ~

23. D (t) = 50e~°!. So when t= 3 we have D (3) = 50e~°-2G) ~ 27.4 milligrams.

25. v(t) = 180 (1a e~ 02")

(a) » (0) = 180 (1 -e°) = 180 (= 1) 0.

()

(b) v (5) = 180 (12 4,020) ~ 180 (0.632) = 113.76 fUs. So the velocity after 5 s is about 113.8 ft/s.

0

(10) = 180 (i= e420) = 180 (0.865) = 155.7 ft/s. So the velocity after 10 s is about 155.7 ft/s.

(d) The terminal velocity is 180 ft/s.

1200 = ————_;

Reta tao = Te te020 TH

27. P (t)

= ———.~

= ——

= 100.

1200

(c) Ast

29. P(t)

©)

1200

1200 > oo we have e~?.2! -5 0, so P(t) > eer

1200. The graph shown confirms this.

S52 = ————_;;>

6.1 + 5.9e—0.02t

(a) In the year 2200, t = 2200 — 2000 = 200, and the population is Z 73.2 Of a predicted to be P (200) = 6.14 5.9e~0.02(200) ~ 11.79 billion. In :

2300, Bst‘aaa = 300, > an and P ((300) ) =a i

1200

(b)

. 2 A 8

ies

6.1 a 5.9e—0.02(300)

. *a 11.97 billion 1 .

(c) As ¢ increases, the denominator approaches 6.1, so according to this

:

4

. 0

100

200

300

400

500

model, the world population approaches i = 12 billion people.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

¢

220

CHAPTER 4 Exponential and Logarithmic Functions

31. Using the formula A (t) = Pe’’ with P = 7000 and r = 3% = 0.03, we fill in the table:

Time (years)

$7213.18 $7432.86 $7659.22 $7892.48

|,

$8132.84 $8380.52 33. We use the formula A (t) = Pe” with

P = 2000 andr = 3.5% = 0.035.

(a) A (2) = 2000e9-935 -2 = $2145.02

(b) A (42) = 2000299354 = $2300.55 (c) A (12) = 2000e9-935: 12 ~ $3043.92

35. (a) Using the formula 4 (t) = P (1+ ik with P = 600, i = 2.5% per year = 0.025, and k = 10, we calculate

A (10) = 600 (1.025)! ~ $768.05. 20

(b) Here i = 995 semiannually and k = 10 -2 = 20, so A (10) = 600 (1_ 0.925) ~ $769.22. :

40°

(c) Here i = 2.5% per year = 2-925 quarterly and k = 10 - 4 = 40, so A (10) = 600 (1+ 0.5 ) ~ $769.82. (d) Using the formula A(t) =

Pe’! with

P = 600, r = 2.5% = 0.025, and t =

10, we have

A (10) = 6009-925: 10 ~ $770.42. 2

37. Investment 1: After 1 year, a $100 investment grows to A (1) = 100 (1- 0.925) ~ 102.52. 4

_ Investment 2: After 1 year, a $100 investment grows to 4 (1) = 100 (1+ 0.0725) = 02-27:

Investment 3: After 1 year, a $100 investment grows to A (1) = 100e9-9? ~ 102.02. We see that Investment 1 yields.the highest return.

39. (a) A(t) = Pe” = 5000e9-0%

(b)

(c) A(t) = 25,000 when t © 17.88 years.

4.3

LOGARITHMIC FUNCTIONS

1. logx is the exponent to which the base 10 must be raised in order to get x.

x logx 3. (a) 5° = 125, so logs 125 = 3.

10°. 3.0

102,04y 252. be)

410%... OA

107! © 1052. 193 ie eee ee

04/2 See

(b) logs 25 = 2, so 5* = 25.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, ‘or posted to a publicly accessible website, in whole or in part.

SECTION 4.3 Logarithmic Functions

221

5. The natural logarithmic function f (x) = Inx has the vertical asymptote x = 0.

Logarithmic form

Exponential form

logg 8 = 1

g! =

logg 64 = 2 logg 4 = $

11. (a) 81/3 =2.

(b) 3° = 1

82 = 64 82/3 =4

logg 512 = 3

(b) 10-7 = 0.01

8? = 512

logg ¢= —1

Males.

logg 4A =-—2

iS; (a) 3* =5

9. (a) 34 =81

gre

A

15. (a) 2 =5

(b) 7? =3y

(b) logs (35) = -2

(bye! =1+]

21. (a) logy 70 = x

23.4(a) In2 =x

(b)logs w = 5

(b) Iny =3

29, (a) log; (x7) = log; 3-3 = -3

25.

31.

(b) log, (3) = -3

(a) logy 2 =1

27. (a) loge 36 = log, 67 = 2

(b) logs 5 1 = logs5 5° = 0

(b) logo9 81 = logy9 92 = 2

(c) logs6 6° = 5

(c) log, 7!° = 10

(a) 3! 5= 5 (b) 5logs 27 _ 97

(b) logyg V/10 = logyg 10!/? = 5

(c) 19 — 10

(c) logs 0.2 = logs (3) = logs 5~! = -1

= —42 33. (a) logg 0.25 = logg 8-7/3

19. (a) logg 81¢= —1

17. (a) logjg 10,000 = 4

35. (a) loggx =3

x= 4 = 64

(b) log;9 0.01 =x

(b) Ine* =4

10*

= 0.01

x = -2

7

= fy x

(c) In (

> —3, so the domain is (—3, 00).

0x2 > 1 =>. x 0 and2-—x 79. y = logio (1- x?) has domain (—1, 1), vertical

asymptotes x = —1 and x= 1, and local maximum y = 0

> 1,80 the domain is (—00, —1) U (1, 00).

>0@x > Oandx 10°=1.

Thus the domain of f (x) is (1, 00).

(b) y = logy (logyx) = 2” = logygx 10” =x. Thus f—! (x) = 10". gh

95. (a) f (x)= Foy ee

pee aarti

oy ty2* =2"

— y2* Sale dd — y)

y

2” PaPe p

y © x ahs = log» GS 2). Thus x

fol (8) = lows (=

x

(b) i —

> 0. Solving this using the methods from x

Chapter 1, we start with the endpoints, 0 and 1.

Sign of x

)

Sign of 1— x oe

Thus the domain of f~! (x) is (0, 1). ’ 97. Using

D D = 0.73Do we have A = —8267 In (=) = —82671n0.73 © 2602 years. :

0

In2 In2 99. Whenr = 6% we have t = ate = 11.6 years. When r= 7% we have t = a = 9.9 years. And when r = 8% we have 7

In2 008 = 8.7 years.

——

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or inpart. ‘

.

&'

wy

SECTION 4.4 Laws of Logarithms

101. Using A = 100 and W = 5 we find the ID to be W = 10 we find the ID to be*

log(2A/W) log 2

log(2A/W)

iop2

_ log(2-100/5)

_ log40

ew

~

log 2

_ log(2-100/10)

_ log20

he

~

log 2

Jog2

log2

225

= 5.32. Using A = 100 and

= 4.32. So the smaller icon is =

= 1.23 times

harder.

103. log (log1010) = log 100 = 2 log (log (log1og0cee!) )= log (log (googol)) = log (log 101%?) = log (100) = 2 105. The numbers between 1000 and 9999 (inclusive) each have 4 digits, while log 1000 = 3 and log 10,000 = 4. Since

[log x]]= 3 for all integers x where 1000 < x < 10,000, the number of digits is [[log x]]+ 1. Likewise, if x is an integer where 10”~! < x < 10", then xhas n digits and [[}og x] =n —1. Since [logx]] =n -—1 ane+ 5 |logx4 4 — log (x? ara,oleae 1/2

x4 = log (x + 2) + log —————_————

aig

o —saaae)

:

x2 x* 2 (x +2) = log (x + 2) + log———————. = log ———_———. = lo

Bt)

log 5 59. log) ~=—2 © 2.321928 082 5 = log 63. log; 2.61 = log= 2.61 ° = 0.493008

G E+)

86-3642)

x y) -

log16 61. log3 083 16= log3 me2.523719 65. logy 125 = log~= 125 7 © 3.482892

oO

| Inx Inx. The graph of y = n3 Inx is In3 In3=

log,x 67. log3 x = log.3

shown in the viewing rectangle [—1, 4] by [—3, 2].

Ine

1

Oe ees in ign HAM 71.

=

—In(x — vx? =1) =In x-

_

—

ed

=

x—Vx2-1

In

_

:

x —Vx2—-

peabae botbon Wess fda Mica —

yaad

>

etal boat

= In(x + vx? =1)

73. (a) log P = loge — klog W © log P = loge — log W* = log P = loe (5 g) oP = (b) Using & = 2.1 and c = 8000, when

000 8000 W = 2 we have P = are= 1866 and when W = 10 we have P = jonl ~ 64.

75. (a) M = —2.5log(B/Bo) = —2.5 logB + 2.5logBo. (b) Suppose B; and Bp are the brightness of two stars such that By < Bp and let M; and Mj be their respective magnitudes.

Since log is an increasing function, we have log B, < log Bz. Then log By < log By &

log B, — log Bo < log Bz — log Bo & log (B,/Bo) < log (B2/Bo)

= —2.5 log (B1/Bo) > —2.5log (B2/Bo) oo

M, > Mp. Thus the brighter star has less magnitudes.

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SECTION 4.5 Exponential and Logarithmic Equations

(c) Let By be the brightness of the star Albiero. Then 1008, is the brightness of Betelgeuse, and its magnitude is

M = —2.5log (100B)/Bo) = —2.5 [log 100 + log (B, /Bo)| = —2.5 [2 + log (B,/Bo)| = —5 — 2.5 log (Bi /Bo) = —5 + magnitude of Albiero 77. The error is on the first line: log 0.1 < 0, so 2log0.1 < log0.1.

4.5

EXPONENTIAL AND LOGARITHMIC EQUATIONS

1. (a) First we isolate e* to get the equivalent equation e* = 25. (b) Next, we take the natural logarithm of each side to get the equivalent equation x = In 25. (c) Now we use a calculator to find x + 3.219.

3. Because the function 5* is one-to-one, 5*~! = 125 5%"! =S3ox-la3ex=4. 5. Because the function 5* is one-to-one, 52*-3 = 1

52-3 = 59

2x -3 =0ex=

7. Because the function 7* is one-to-one, 72*~3 = 79+5* «> 2x —-3 =6+5x 9. Because the function 6* is one-to-one, 6)

3.

3x =-9 ex =-3.

= 61-2? 6 2 _ |] = 1 — x? © 2x* =2ex=41.

11. (a) 10° = 25 & log 10% =4og25 & x log 10 = log25 = x = log25 = 2log5

- (b) x © 1.397940 13. (a) e~>* = 10 = Ine~* = In10 © —Sx Ine = Inl0 @ —5x = Inl0x = —41n10 (b) x © —0.460517

15. (a) 2!-* =3 © log2!-* = log3 & (1 — x) log2 = log3 @ 1 —x = es epee 1 oa (b) x + —0.584963

17. (a) 3e* = 10>e*=

Wo x=In(42)

(b) x © 1.203973

41\!24 = 10 10 o> 12¢n41fh= In 108 o> 12¢ = in? In? 19. (a) 300 (1.025)!2¢ = 1000+ (33) 3 ot = —2

(b) x © 4.063202 _ 21. (a) el

=2

1 - 4x = In2 > —4x = -1 + In2 @ x = j (1—In2)

(b) x © 0.076713 Rs ane EE

2. 570) iidSin 15 RSE od?ape

Se ee a eine

eS eee Be Peuntan ge ak pln ein?

(b) x © 0.156158 25. (a) 3 x/14 =_ 0.1 = log3 x/14 =_ log0.1&

x 4

log3 ae = log 0.1

(b) x + —29.342646

> x _= 14log0.1 oR

|

27. (a) 4(1 +105") =9 > 1+ 10% = 3 10 = 3 o>Sx = log} x = § log} (b) x * 0.019382 ;

1—In12

29. (a) 8+e!-4* = 20 Se!-4#* = 12 1 -4x =InI2ex = = (b) x © —0.371227

: In

31. (a) 4% +.2142* = 50 22 412142" — 50 o> 27* (14.2) =509 2% = P 2x In2 =In50—-In3 x =a

(b) x * 2.029447

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227

228

CHAPTER 4 Exponential and Logarithmic Functions

33 (a) 5* = 4*+! & log5* = log4*t!

= x logS = (x + 1) log4 = x log4 + log4 @ x log5 — x log4 = log4@ log 4 x (log5 — log4) = log4 ex = e. log x

(b) x © 6.212567

3x log2 + log2 = x log3 — 2log3 35. (a) 23*+! — 3*-2 > Jog23*t! = log3*-2 & (3x + 1)log2 = (x — 2) log3 © 3x log2 — xlog3 = —log2 — 2log3 = x Glog2 —log3) = — (log2+2log3) © erelOf 242 Ops met

log (2 3?) 5

* = —3fog2=l0g3 = —Tog (23/3)

log 18

log :

(b) x © —2.946865 Si. (a) os

=4o50=4+4 4e%

© 46 =4e%

© 11.5=e7%

olnll.s

=—x

&x ==InI15

(b) x + —2.442347

39:

e* — 3e% +2 =0

(e — 1) (eX —2) =0 > e* —1 =0ore* —2=0. Ife* —1 =0,thene* =1

e* —2=0, thene* = 2 &x

math

e** 4 4e2%X 21

=0S

(ex +7) (c* -3) =0=> e* = —7 ore =3. Nowe

for all x. But we can solve e2* = 32x

43.

2* — 10(2-*)

=In3

= —7 has no solution, since e* > 0

x= $ In3 + 0.5493. So the only solution is x + 0.5493.

+3 = 09 2-* (2 — 10+ 3-2") = 0. 2-* (2* +5) (2* — 2) = 0. The first two factors are positive

everywhere, so we solve 2* —2=0@x

45.

x =Inl =0. If

= In2’~ 0.6931. So the solutions are x = 0 and x © 0.6931.

= 1.

x22* — 2 Sika waa 2 1)= 0 =>2° =0 (never) orx? —1 =0. Ifx? — 1 =0, thenx? = 1 x2 — 5x

=0 x(x

—5) =0>x=0

or x = 5. So the possible solutions are x = 0 and x= 5. However, when x = 0, logx is undefined. Thus the only solution ise — oe

51.

2logx = log2 +log (3x — 4) © log (x?) = log (6x — 8) x? = 6x —8 x2 -6x +8 =0 E(x -4)(x-2 =0E x =4orx = 2. Thus the solutions are x = 4 and x = 2.

53. logy 3 + logy x = logy 5 + logy (x — 2) & logy Gx) = logy (Sx — 10) 55.

Inx = 10x =e! = 22,026

She

logx = —-2 x

63.

2x =10Sx

=5

= 107? = 0.01

59. log (3x +5) =23x+5=10? 61.

3x = 5x — 10

4 —log (3 —x) =3 @ log

= 1006 3x =9 @x=

= ~ 31.6667

—x) =143-x=10@x=-7

logy x + logy (x — 3) = 2 © logs [x (x — 3)] =2 9 x2 — 3x =2* ox? - 3x —4=06 (« -4) (x41) ex =-lor x = 4. Since log (—1 — 3) = log (—4) is undefined, the only solution is x = 4.

65. logg (x — 5) + logo (x +3) = 1 & logy [(& —5) (* +3)] (x — 6) (x +4) 67.

= 1&

@ -5)@+4+3) = 91 x2

-2x -24 =08

= 0 => x = 6 or—4. However, x = —4 is inadmissible, so x = 6 is the only solution.

i =209 22> 1 = Seon 41 = 251-25 logs (+ 1)~ logs (©~ 1) = 22 logs (=27)

Mtr = 26 ox = B

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=

SECTION 4.5 Exponential and Logarithmic Equations

69. Inx

=3—x

@Inx+x—3=0.

Let

T1.x3 —x = logig(x+1)

S (x) = Inx +x — 3. We need to solve the equation

229

@ x3 —x — logig(« +1) = 0.

Let f (x) =x3 -—x—- logjg (x + 1). We need to solve the

f (x) = 0. From the graph of f, we getx ~¥ 2.21.

equation f (x) = 0. From the graph of f, we get x = 0 or

x © 1.14.

73. & = —x @e* +x =0. Let f (x) =e* +x. Weneedto

75.47-*= x 4

— /x =0. Let f (x) = 47% — JX.

solve the equation f (x) = 0. From the graph of f, we get

We need to solve the equation f (x) = 0. From the graph

x © —0.57.

of f, we getx

71. 2211985 = dg¢2logy 27/'085* = log, (ig)© 79. log (x — 2) + log (9 — x) < 1 20

< x27 -11x +280

0 < —x?2 4+ 11x -—18 60

=2

0.36.

3 = —4 #2logsx = —} ox = 5-1? = yee= 0.4472

log[(x —2)(9—x)] < 1

-

log (—x? + 11x - 18) -x? + 11x

-—18 < 10!

< (x- 7) (x — 4). Also, since the domain of a logarithm is positive we must have < (x — 2) (9 —x). Using the methods from Chapter 1 with the endpoints 2, 4, 7, 9 for the

intervals, we make the following table:

Sign of x —7

-

Sign ofx — Sign of x— 2

+

ae of 9—x

aon Sim ote OSs) Joe a

oa

+

~

ob

+

ake Ue

Le | tal + [in

Thus the solution is (2, 4) U (7, 9). 81.2 < 10%

2 = 1.02125" < log2 = 4t log 1.02125 =

log 2 Slog 1.02125 = 8.24 years. Thus the investment will double in about 8.24 years.

0.075\* 91. 8000 = 5000 (:it ) = 5000 (1.01875) € 1.6 = 1.018754 < log1.6 = 4f log 1.01875 = log 1.6 t = ——_———_ ®& 6,33 years. The investment will increase to $8000 in approximately 6 years and 4 months. 4 log 1.01875

In2 93. 2 = ¢9-085! In2 = 0.085t ot = ees

0.085

8.15 years. Thus the investment will double in about 8.15 years.

95, 15270087! — 5 ey e0.087! — } 0.087 = In(}) =-In3 oe = aaas ~ 12.6277. So only 5 grams remain after approximately 13 days.

:

975\(a)) PG): = aa

= 7.337, so there are approximately 7337 fish after 3 years.

(b) We solve for t. iw SN i

=591+4e08

— WH 2< 4e—0-8t = | ey e0.8' — 0.25 > -0.8t = In0.25@

ede) cs 1:73. So the population will reach 5000 fish in about 1 year and 9 months.

99. (a) In (=) = -; oS z = e7h/k P = Pye*/*. Substituting k = 7 and Py = 100 we get P = 100e~*/7. (b) When h = 4 we have P = 100e74/7 ~ 56.47 kPa.

101.'(a) T= $9 (1enI9/5) oo Br Sy hi 5

eS & eS = 1 - Br oe — Br =n (1— Bi) &

In(1 = $/).

(b) Substituting 7 = 2, we have t = =

In [1- a @)| = 0.218 seconds.

103. Since 9! = 9, 9? = 81, and 93 = 729, the solution of 9* = 20 must be between 1 and 2 (because 20 is between 9 and 81), whereas the solution to 9¥ = 100 must be between 2 and 3 (because 100 is between 81 and 729).

105. (a) (x — 1)!8@—-) = 100 (x — 1) 6 log («x~ ye) = log (100 (x —1)) @ [log (x — 1)] log ( — 1) = log 100 + log (x — 1) © [log (x - yp —log(x-1)-2 [log (x — 1) — 2] [log (x — 1) + 1] =0. Thus either log (x — 1) =2

(b) logy x + logy x + logg x = 11 >

logy x = 11

logy x =6

x

=0@

= 101 orlog(x—1)=-lex=H.

logy x + logy Vx + logy Y/x = 11 = logy (x./x ¥/x) = 11 & log, (24/8) =11 x = 2° = 64

(c) 4* —2°+1 = 3 & (2)? —2 (27) —3 = 0 (2* —3) (2% +1) =O either?

=3

x= as or 2F = —1, which n

In3 has no real solution. So x = > is the only real solution.

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> j 4, “S)4

.

SECTION 4.6 Modeling with Exponential Functions

4.6

231

MODELING WITH EXPONENTIAL FUNCTIONS

1. (a) Here np = 10 and a = 1.5 hours, son (t)= 10- 24/15 = 10. 24/3, (b) After 35 hours, there will be n (35)= 10 -22@5)/3= 1.06 x 108 bacteria.

(c) n(t) = 10- 27//3 = 10,000 = 22/3 = 1000 & In (27/3) = In 1000 ° ~ 181,359,340. Thus at a'2% growth rate, the projected population will be approximately 181 million people by the year 2036.

9. (a) The doubling time is 18 years and the initial population is 112,000, so

a model is n (t) = 112,000 -2!/18.

B 800,000

700,000

600,000

(b) We need to find the relative growth rate r. Since the population is

2 - 112,000 = 224,000 when t = 18, we have 224,000 = 112,000e!8”

watts/m2.

&

k k MN Nen are = 10 oe(=) ~ 21oga |= ae

k ae— 20 log d . Similarly,

1 ea k —Ii — 20 log d>. Substituting the expression for 3; 1 gives

k

[sh = 10 log ——— 20 logd; + 20logd,

— 20 logd) = 8; +20logd,

di\

— 20logdy = B; + 20log (4).

0

i

d (b) 8; = 120, d; = 2, and d) = 10. Then By = By +20log | = 120 + 20log 7 = 120+ 20 log 0.2 © 106, and so the

intensity level at 10 m is approximately 106 dB.

CHAPTER 4 REVIEW 1. f (x) =5*; f (-1.5) © 0.089, f (v2) ~ 9.739, f (2.5) © 55.902

3. g(x) = 4e¥~2; g (0.7) © 0.269, g (1) © 1.472, g (m) © 12.527 5. f (x) = 3*~?. domain (—0o, 00), range (0, co), asymptote y = 0.

7. g(x) =3+2*. Domain (—oo, 00), range (3, 00), asymptote y = 3.

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|

L]

CHAPTER4 — Review

wh TEM Ga

e*—! 4.1. Domain (—0o, 00), range (1, co),

asymptote y = 1.

235

11. f (x) = log3 (x — 1). Domain (1, 00), range (—o0, 00),

asymptote x = 1. y

13. f (x) = 2 — logy x. Domain (0, 00), range (—00, 00), asymptote x = 0.

15. g (x) = 21Inx. Domain (0, 00), range (—00, 00), asymptote x = 0. ad

ie SC

10%" + log(1 — 2x). Since logu is defined only for u > 0, we require 1 — 2x > 0} —2x > -lo&x

< 5, and so

the domain is (—co, 5):

19. h(x) =In (x? -4). We must have x* —4 > 0 (since In y is defined only for y> 0) 0@ (x —2)(x +2) > 0. The endpoints of the intervals are —2 and 2.

EE) Sign of x — 2

-

+

Sign of x + 2

aa

+

simoe-aeaD]

+ | - |+ |

Thus the domain is (—0o, —2) U (2, ov).

21. logy 1024= 10

2!9 = 1024

23. logx =y

10” =x

25. 2° = 64 = log, 64 = 6

27. 10% = 74 © logig 74 =x < log 74 =x

29. logy 128 = log, (27) = 7

31. 100845 = 45

33. In (5) =6

35. log; +4 = log; 3-3 = -3

37. logs V5 = logs 5!/2 =

1

z 23 41. log, (163) = log (24) = log, 2° = 92

39. log25+ log4= log (25 - 4) = log 10? = 2

43. logg 6 — logg 3 + logg 2 = logs ($-2) = logs 4 = logs 87/7 = 3

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236

4 nn.

CHAPTER 4 Exponential and Logarithmic Functions

log (42°C?) = logA+2logB +3logC

47. In

oe = 59954) = 5 fn?

1) -m (2 +1)] = 8 [ne

e+)

-n(2+1)]

= [in - 1) +In@ + In (x? +1)] hy i,Sa

3/2

49. logs (=) Vx3-x

= logs x? (1 — 5x)? — logs ,/x (x2 — 1) = 2logsx + 3 logs (1 — 5x) — 5 logs (x? -x) = 2 logsx + 3 logs (1 — 5x) - 5 [logs x + logs (x? = 1)

= 2logsx+ 3 logs (1 — Sx) — } [logsx+ logs (x — 1) + logs (x + 1)]

51. log6 + 4log2 = log6 + log24 = log (6-24) = log96

ievies 53. 3 logy (x — y) — 2 logy (x? + y”) = logy (x — y)3? — log (x?+?) 2 = log. (———

(? +?)

ie 55. log(x — 2) + log (x + 2) — 4 log (x? +4) = logs =2) (6 +2)1—og

57, 3-7 — 27 6 34-7 = 33 2x

4 = oa

z )

—-7=3 62x =1l0 Sx =5

59, 23*-5 =7 & log, (25) = log) 7 = 3x — 5 = logy 7

x = } (logy 7 + 5). Using the Change of Base Formula, we

tas~ 2.807, so x mol have logy 7 log = joe? © 3 (2.807 + 5) ~© 2.60.

61. 41-* = 32*+5 & log4!-* = log3**+5 & (1 —x) log4 = (2x + 5) log3 & log4 — 5log3 = 2x log3 + x log4& x (log3 + log 4) = log4 —Slog3

log 4— S5log3 @ x = ———__—_—_ ¥ - 1.15 2log3 + log 4

63. x2e2* + 2xe2* = 8e* = e* (x? +2x -8) = 0x? Ob

+2x —8 = 0 (since e* £0)

(x +4) (x -2) =0@x =-4

2

65. logx + log (x + 1) = log12 ©] log(x («+ 1)) = logl12@x@+4+1)=

12x27 +x- 12=0e0-3)@+4=00

x = —4 or 3. Since log (—4) is undefined, the only solution is x = 3.

67. logy (l—x) =4e1-—x =24x =1-16=-15 69. log; (x — 8) + logy x = 2 & log; (x (x —8))

=2

x(x

—8)

=9

x2 - 8% -9=0N ES(X-N(X+1)

=0E

x = —1 or 9. We reject —1 because it does not satisfy the original equation, so the only solution is x = 9.

71. 5-28 = 0163 ea.

—2 3

log5= lap 0.63 Soa

3 log 0.6

a ee 2 log 5

73. 5S2x+1 = 34-1 & (2x4 1)logS = (4x —1)log3 & 2xlog5 + log5 = 4xlog3 — log3 x (2log 5 — 4log3) =

—log3

—logS

log3 + log5 x = —————— © 2.303600 4log3 —2log5

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER4

75. y =

e*/@+2). Vertical asymptote x = —2, horizontal

Review

237

77. y = log (x3 = x). Vertical asymptotes x = —1, x = 0,

asymptote y © 2.72, no maximum or minimum.

x = 1, no horizontal asymptote, local maximum of about —0.41 when x + —0.58.

79. 3logx = 6 — 2x. We graph y= 3 logx and y=6—2x

in

81.Inx > x —2.We graph the function f (x) = Inx — x +2,

the same viewing rectangle. The solution occurs where the

and we see that the graph lies above the x-axis for

two graphs intersect. From the graphs, we see that the

0.16 4 > logs 620 and so log, 258 is larger. r\ant

91. P = 12,000,r =0.10, andt =3. Then A = P(1+“) en

(a) Form =2, A = 12,000 (1+ 942)"

= 12,000 (1.05%) ~ $16,081.15.

123

(b) For n= 12, A = 12,000 (1+ 210 ) ©) $16,178.18.

() Forn = 365, A = 12,000 (1+ 939

rae

~ $16,197.64.

(d) Forn = 00, A = Pe = 12,000e9-!°G) ~ $16,198.31.

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238

CHAPTER 4 Exponential and Logarithmic Functions

93. We use the formula

t

4 = P (1+ ~)" with P = 100,000, r = 0.052, n = 365, and A = 100,000 + 10,000 = 110,000, n

and solve for1:110,000 = 100,000 (1+ 9282)" i

log 1.1 satis ooltckta wt

365t

eo 1.1 = (1+ 482)

365t

> log 1.1 = 365¢ log (1+ 98%)

: F 1.833. The account will accumulate $10,000 in interest in approximately 1.8 years.

365 log (1+ 99%)

95. After one year, a principal P will grow to the amount

0.0425 \76

4 = P (:+ as)

= P (1.04341). The formula for simple

interest is A = P (1 +r). Comparing, we see that 1 +r = 1.04341, sor = 0.04341. Thus the annual percentage yield is

4.341%. 97. (a) Using the model n (t) = nge”’, with np = 30 andr = 0.15, we have the formula n (t) = 30e0-15¢,

(b) n (4) = 3009-154) x 55, 1 (c) 500 = 3029-15! «> ¥ = 29-15t 65 0.15¢ = In (2) Eh ye 015 In (2) = 18.76. So the stray cat population will reach 500 in about 19 years.

;

99, (a) From the formula for radioactive decay, we have m (t) = 10e~"', where r = —

In2

Det

10

So after 1000 years

the amount remaining is m (1000) = 10- el— n2/(2.7x10°)]-1000 _ 1gg—(In2)/(2.7x10°) _ 1gg—(n2)/270 ~ 9.97, Therefore the amount remaining is about 9.97 mg.

(b) We solve for ¢ in the equation 7 = 10e~[n2/(2.7x10°)]*

We have 7 = 10e7LIn2/(2.7%10°)]# es

0.7 a: o- [n2/2.7x10))-f co hg ge 2.7 x 105

In2

- 2.7 x 10° © 138,934.75. Thus it takes about

139,000 years.

101. (a) From the formula for radioactive decay, r =

n2 © 020004359 and n (1) = 150 - €~0-0004359", Top

(b) 7 (1000) = 150 - e~9-0004359-1000 ~ 97.00, and so the amount remaining is about 97.00 mg. (c) Find ¢ so that 50 = 150 - e~9-0004359 i

We have 50 = 150 - e~9.00043591 e” = 2 &

Sr = In (#3) er= tin (#2) ~ 0.1515. Thus n (t) = 1500 - e9-1515¢, (b) We have t = 2020 — 2009 = 11 son (11) = 1500¢9:!5!5:1! ~ 7940. Thus in 2009 the bird population should be about 7940. 105. [H*] = 1.3 x 1078 M. Then pH = — log [H+] = —log (13 x 10-8) * 7.9, and so fresh egg whites are basic. 107. Let Jp be the intensity of the smaller earthquake and /; be the intensity of the larger earthquake. Then J; = 35/p. Since I M = log (5):we have Mp = log (3) = 6.5 and I

35]

I

M, = log (+) = log (=) = log35 + log (4) = log35 + Mp = log35 + 6.5 © 8.04. So the magnitude on the Richter scale of the larger earthquake is approximately 8.0.

i ria

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* ss

CHAPTER 4 — Test

239

CHAPTER 4 TEST 1. (a)

(b)

S (x) =2™ +4 has domain (—00, 00), range (4, 00),

g (x) = log3 (x + 3) has domain (—3, oo), range

and horizontal asymptote y = 4.

(—00, 00), and vertical asymptote x = —3.

3. (a) 67% = 25 © log, 62* = log,25 2x = logs25

(b)nd=3ee™4aSBo4=e3

3

5. (a) (=) = logx + logy? — logz* = logx +3 log y— 2logz

(b) in = = in((2)'") 3 ;In (=) =4inx—Jiny (c) log 3 sa

= Lae (3575)

J [eee +2) — (4logx + log (x? +4))]

= }log (x +2) — $ logx — }log (x? +4) 7. (a) 3% = 3!% 4 4x = 100 x = 25

(b) 23*-2 =e?” 3x —2 = 72 6 x? — 3x 42=00 (x -1) (x -2) =0 Ex =lorx =2 (c) 5*/10 4.1 = 7 4 5*/19 = 6 & logs 5*/!0 = logs 6 > x/10 = logs 6 > x = 10 log, 6 © 11.13 log, 62*

(d) 10°43 = 62* & log 10°+3 = log6* oo x4+3= ce =

(logs 10) ( +3) = 2x € (2 — loge 10)x= 3 logs 10

6

l 2 — log, 10

; log 27 9. Using the Change of Base Formula, we have log), 27 = jog12~ 1.326. 12t

11. (a) A(t) = 12,000 (1se0.056) , where ¢is in years. 0.056 (b) 4 (t) = 12,000 (1a oo)

365t

So A (3)

0.056 \3650) 000(eae )

$14,195.06

(©) A(t) = 12,000e9-956, So 20,000 = 12,000e9-956 5 = 3¢9-056 & In5 = In (320.056) © In5 = In3 +.0.0561 & i

ws (In5 — In3) + 9.12. Thus, the amount will grow to $20,000 in approximately 9.12 years.

I I 13. Let the subscripts J and P represent the two earthquakes. Then we have M; = log (2) = 6.44 104 = = S

6.45 —= J,. Similarly, ia 10°*'S Mp

I

I

Srloe og (a9Ve Sl => 10) So5

I

10 Sap,Pp. Soles So Tp

10°45

~ 10345 = 1033 = 1995.3,

and so the Japan earthquake was about 1995 times more intense than the Pennsylvania earthquake.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

240

FOCUS ON MODELING

FOCUS ON MODELING _ Fitting Exponential and Power Curves to Data

1. (a)

(b) Using a graphing calculator, we obtain the model

y

nd

y =ab', where a = 3.3349260 x 107!5 and b = 1.0198444, and y is the population (in millions) in the year f. (c) Substituting ¢ = 2020 into the model of part (b), we get y = ab?920 ~ 577.6 million. (d) According to the model, the population in 1965 was

y = ab!%5 ~ 196.0 million.

millions) (in Population

0 1780

1820

1860

1900

1940

1980

2020 x

Year

3. (a) Yes.

(b) Health Expenditures E ($bn)

4.30811 5.54440 6.09718 6.25210 6.58521 6.75449 6.88008 6.98638 7.09747

7.22781 7.40123

7.61845

0

10

20

30

40

%

Year since 1970

Yes, the scatter plot appears to be roughly linear.

7.78809 7.86288 7.93501 (c) Let ¢ be the number of years elapsed since 1970 . Then In E = 4.7473926 + 0.08213193t, where E is expenditure in billions of dollars. (d) Ew

e4-7473926+0.08213193t

~

115.28330e0-08213193¢

(e) In 2020 we have t = 2020 — 1970 =

S50, so the estimated 2020 health-care expenditures are

115.28330¢9-08213193(50) ~ 7002.3 billion dollars.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Fitting Exponential and Power Curves to Data

5. (a) Using a graphing calculator, we find that

241

(b)

Ig = 22.7586444 and k = 0.1062398.

(c) We solve 0.15 = 22.7586444¢~9:1062398x for x: 0.15 = 22.7586444e—9-1062398x «, 0.006590902 = e—9-1062398x ~, —5.022065 = —0.1062398x x © 47.27. So light

€

12

>

104

Bein

— ) 1 4

intensity drops below 0.15 lumens below around

47.27 feet.

4

24 Ia a nnn ee oe 0

10

20

30

40

x

Depth (ft)

7. (a) Let A be the area of the cave and S the number of species of bat. Using a graphing calculator, we obtain

(b)

SS; df

the power function model § = 0.14.49-%.

26

6

oO

ae

(c) According to the model, there are

So oe 5 Ta Qa

S = 0.14 (205) ~ 4 species of bat living in the El

_

Sapo cave.

4

ais 2 1

0!

100

200

300

400

500 A

Area (m7) The model fits the data reasonably well.

9. (a)

y

(b) 0.69315

=2.52573

1.38629

—2.12026

1.79176

—1.71480

2.07944

—1.38629

2.30259

—1.02165

2.48491

—0.65393

2.63906

—0.31471

PRI Pee, art eGo

Sen iO

0.05827

dae 4416. 19° %

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

242

FOCUS ON MODELING

(b) (cont’d)

Iny

Iny

=]

1

=

2

St

a

(c) The exponential function.

-_

wig

(d) y =a- b* where a = 0.057697 and b = 1.200236.

11. (a) Using the Logistic command on a TI-83 we get y = SPR ad where a = 49.10976596, b = 0.4981144989, and 1 +ae—>x a

c = 500.855793.

(b) Using themodel N= se

we solve fort. SoN = as

2 ltae = © eraet = (=)-1

. thevalues for eet = —Su € —bt = In(c — N)~InaN et = =1 [InaN ~In(c—N)].Substituting

c, with N = 400 we have ¢= ggogrhgqogp (In 19643.90638 — In 100.855793) ~ 10.58days. !

\\

poe |

5

TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE APPROACH

5.1

ANGLE MEASURE

1, (a) The radian measure of an angle 6 is the length of the arc that subtends the angle in a circle of radius 1. (b) To convert degrees to radians we multiply by Th0° 180 (c) To convert radians to degrees we multiply by —.

0 . (a) The angular speed of the point is w = z (b) The linear speed of the point is v = _ (c) The linear speed v and the angular speed ome are related by the equation v = rw.

5. 15° = 15° - 7ghs rad = 7 rad © 0.262 rad 9, —45° = —45°. 180° rad = Snir§rad + —0.785 rad

7. 54° = 54°. 77h. rad = 37 rad ~ 0.942 rad 11. 100° = 100° - 180° rad = Sz rad © 1.745 rad

13. 1000° = 1000° - 5gh5 rad = 29 rad © 17.453 rad

15. —70° = —70° . 7ftz rad = — 4% rad © 1.222 rad

17, 3 = 5t . 180° — 300°

19, 5% = Sm. 180° _ 150°

21.3 = 3. 180° — 540° ~ 171.9°

23, —1.2 = —1.2. BO = — 216 _ _¢68 8°

25. 75x= 75 1802 ES = 18°

|

Prd= —4% Lees . 180° 13025_ _p40 27, —4%

. 50° is coterminal with 50° + 360° = 410°, 50° + 720° = 770°, 50° — 360° = —310°, and 50° — 720° = —670°. (Other answers are possible.)

Sh an is coterminal with a +27 = ue a +47 = Le, iz —27 = —3t, and iz —47 = aoe (Other answers are possible.)

3B —F is coterminal with —3 + 27 = Ee —¥+40r= is —F en = —2

and —3 —47 = le. (Other answers

are possible.)

ARE Since 430° — 70° = 360°, the angles are coterminal. ade Since un = = = ie = 27; the angles are coterminal. 39. Since 875° — 155° = 720° = 2 - 360°, the angles are coterminal. 41. Since 400° — 360° = 40°, the angles 400° and 40° are coterminal. 43. Since 780° — 2 - 360° = 60°, the angles 780° and 60° are coterminal. 45. Since —800° + 3 - 360° = 280°, the angles —800° and 280° are coterminal. : 197 ee 47. Since

ikis 197 : 2 == a the angles say ieand Tt 6 are coterminal.

49. Since 257% — 12 - 27 = mr, the angles 257 and 7 are coterminal. nih Since te —2:27=

i, the angles un and + are coterminal.

53. Using the formula s= Or, s= aa i 55.8

Ee,

ee Fit = 0 ws 114.6° r

5

7

Si: Solving for s, we have s = r@, so the length ofthe arc is 5 : B= 14 14 180° for 8,we have 0= =Sy ae ~ 89.1°. 59. Solving

Ch:

243 © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

244

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

Ss 15 == = ——

lr

18 = — B57

eee ae Rea OTAE ao 63. (a) A = 5770= 5 -87-80°- 7 = 32. SE = ET ~ 44.68 (b) 4 = os 5-107

-0.5 =25

65. 6 = 2F rad andr = 10m, so A = 5770 = 4 (10)? (28)=100%= 104.7 m?

67. A = 70m? and @ = 140° = 140°. =

20 Tt

7

rad. Thus, A = $770

Hheoedh 70 = 5. Sr? =

ram

wT

69. r = 80 mi andA = 1600 mi”, so A = 4770& 1600 = 5 (80)? 0 = 5 rad. 71. Referring to the figure, we have AC

BC

= 3+ 1 = 4,

=1+2=3, and AB =2+3 =5. Since

AB? = AC? + BC?, then by the Pythagorean Theorem, the

triangle is a right triangle. Therefore, ee?

a2

0 = + and

een

2

2

73. Between 1:00 P.M. and 1:00 P.M., the minute hand traverses three-quarters of a complete revolution, or 3 (27)= = rad, while the hour hand moves three-quarters of the way from 12 to 1, which is itself one-twelfth of a revolution. So the hour hand traverses 7Ge5) (2m) == 3 rad. 75. The circumference of each wheel is 7d = 287 in. If the wheels revolve 10,000 times, the distance traveled is

1 ft

1 mi

10,000 - 287r in. - Din 5280R ——— * 13.88 mi mi.

77. We find the measure of the angle in degrees and then convert to radians. 9 = 40.5°—25.5° = 15° and 15- gps rad = F rad. Then using the formula s = Or, we have s = D - 3960 = 3307 © 1036.725 and so the distance between the two cities is

roughly 1037 mi. 79. In one day, the earth travels ak of its orbit which is ae Tad, shhenis =

O71

am - 93,000,000~ 1,600,911.3, so the

distance traveled is approximately 1.6 million miles.

81. The central angle is 1 minute = (a) Pome a * Tg0> tad= T0800 rad. Then s= Or = T0800 - 3960 © 1.152, and soa nautical mile is approximately 1.152 mi. 83. The area is equal to the area of the large sector (with radius 34 in.) minus the area of the small sector (with radius 14 in.) ae) 2 2 ashe Thus, A =eg 3r?0 — 3730 =nee4 (34?p) — 14?) (135° 75>) ©~ 1131 in?.

45 . 27 rad = 907 rad/min. min 45-27-16 : :

85. (a) The angular speed is w = ——

:

;

(b) The linear speed is v = ——

87.

ivo =

:

,

= 14407 in./min © 4523.9 in./min.

8:-27-2 327 ———— = — *¥ 6.702 ft/

is

15

3

:

: 1-27 - (3960 89. 23 h 56 min 4 s = 23.9344 hr. So the linear speed is. a

1d , eae ~ 1039.57 mish.

100-27-0.20m 27 91. 9= ————_ = — © 2.09 mss.

:

60s

3

“

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 5.2 Trigonometry of Right Triangles

245

93. (a) The circumference of the opening is the length of the arc subtended by the angle 6 on the flat piece of paper, that is,

C=s=r0=6-3F = 100 314m. G 10 (b) Solving forr, we findr = — = ins oS 5 cm 2a 24 (c) By the Pythagorean Theorem, h* = 62 — 5? = 11, soh = J/11 © 3.3 cm. (d) The volume of a cone is V = imr?h. In this case V = in . 52. /11 © 86.8 cm?.

95. Answers will vary, although of course everybody prefers radians.

5.2

TRIGONOMETRY OF RIGHT TRIANGLES

1. (a @)

: opposite b) sin@ = ————., (6) hypotenuse

‘tes adjacent

adjacent cos 9 = ——_—__., hypotenuse

opposite tan @ = ———_. adjacent

(c) The trigonometric ratios do not depend on the size of the triangle because all right triangles with angle @ are similar.

hypotenuse

S.sing— 3, cos @ = 2,tan@ =

5. The remaining side is obtained by the Pythagorean Theorem: 412 — 40? 2540

tan@ = 9

_

41

SA

cscO = Zo secO=

9 , cot? = 70

7. The remaining side is obtained by the Pythagorean Theorem: cos 0 =

= /81 = 9. Then sin@ = 4. cos 8 = a

teen

13 Fe = 313 tan @ = Z,cscO = VIB sec

V3? + 22 = /13. Then sin@ = Ta =

2 ta ;

13 = VB cot @ = 53

9. c= 52 +32 = /34 (a) sina = cos

=

Fe = AX

(b) tana = cot B = 3

11. (a) sin22° ~ 0.37461

(b) cot 23° © 2.35585

13. (a) sec 13° © 1.02630

(b) tan 51° © 1.23490

(c) seca = csc 8 = “34

15. Since sin30° = =, we have x= 25sin30° = 25-4 = 28. 17. Since sin 60° = =, we have x = 13 sin60° = 13- 2 = Lee 12 ; (2 ns 19. Since tan 36° = 4 , we have x a=

21. = =cos@

1 wae

2 _ ~ 1651658.

= x = 28 cos8, and a =sind@Sy= 28sind.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

246

CHAPTERS

Trigonometric Functions: Right Triangle Approach

23. tan@ = 2. Then thethird side isx =

/52+62 = /61.

25. cot@ = 1. Then thethird side is r = V1? + 1? = V2. r

é

l

2

The other five ratios are sin@ = oie cos@ = © :hy

The other five ratios are sin@ = pe

a

Csc'o.= on sec8 = V61 and cot 8 = g

cos@ = w = ¥2 tan0 = 1, csc@ = V2, and

secO = /2. Jeli

:

6

v2

1

27. csc =11.Thethird side is x = 112 — 6 = 85. The 29.sin® +cos™ =14%3 = 143 other five ratios are sin@ = ts cos 9 =

;

tan@ = 6185 sec @ = L185 | and cot 8 = AEE 11

85 31. sin 30° cos 60° + sin 60° cos 30° = 5 5 Ww.

2

NI+

Mo Se ll

+I+

Il

BIW

2

(cos 30°)? — (sin 30°)? = (2) z (3) =3-1.}

35. (cos % +sin8)? = (2 +4) =44442(3) (2) =3+

:

37. This is an isosceles right triangle, so the other leg has length 16tan45° 16 = 16./2 © 22.63, and the other angle is 90° — 45° = 45°. sin 45°

=

16, the hypotenuse has length

39. The other leg has length 35 tan 52° ~ 44.80, the hypotenuse has length Sos 52° = 56.85, and the other angle is

90° = 522. = 138%) 41. The adjacent leg has length 33.5 cos z =~ 30.95, the opposite leg has length 33.5 sin F = 12.82, and the other angle is us

bg

Tet.

fae a

106 106 43. The adjacent leg has length aot = 145.90, the hypotenuse has length an = 180.34, and the other angle is ‘i Ey

45. sin@ ~ sy = 0.45. cos@ + “4 ~ 0.89, tan@ = fecsc 8 © 2.24, sec 0 Fe

A

—

100

100 ——_

abet

&

a

= 1.12, cot@ © 2.00.

:

ng0e nd ee

49. Leth be the length

Bt ofthe

shared side. Then sin 60° = a Sh=

eos See

57.735

= sin65° = Msers

Sy

= 63.7 oe an an see

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 5.3 Trigonometric Functions of Angles

51.

247

From the diagram, sin@ = i and tan@ = =, sox = ysin# = 10sin@ tan @. y

53. Let h be the height, in feet, of the Empire State Building. Then tan 11° = — & h = 5280 - tan 11° © 1026 ft. 55. (a) Let A be the distance, in miles, that the beam has diverged.

Then tan0.5°

h © 180° — 25° — 149.53° = 5.47° and a) = — Thus, one triangle has Z 4; © 125°, ZC, % 30°, and a, ~ 49; the other has 2 A> ~ 5°, ZC © 150°, and ay © 5.6. 100

sin 50°

25..Sitibes ae

= 1.532. Since |sin 8| < 1 for all @, there can be no such angle B, and thus no such triangle.

1!

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a]

SECTION 5.5 The Law of Sines

27. sin'A =

26 sin 29° Reyer

253

~ 0.840 ~ 122.8°, LBy © 28.2°, and

by © 14.6.

nigh 29. (a) From AABC and the Law of Sines we get — LB & sin-!0.7 © 44.427°.

Since ABCD

LBCD = 180° —2 ZB © 91.146° (b) From AABC

we get ZBCA

=

:

é

«

=

=

© sinB =

is isosceles, 4B = LBDC

30 = WeeSO 28 ==

~% 44.427°.

Thus,

91.1°.

180° — ZA — ZB &

180° — 30° — 44,427°

=

105.573°.

Hence

= ZBCA — LZBCD ® 105.573° — 91.146° = 14.4°.

LDCA

31. (a) Let a be the distance from satellite to the tracking station A in miles. Then the subtended angle at the satellite is 50 sin 84.2°

LC = 180° — 93° — 84.2° = 2.8°, and soa = = sin 2.8°

1018 mi.

(b) Let d be the distance above the ground in miles. Then d = 1018.3 sin87° ~ 1017 mi. AB 33. LC = 180° — 82° — 52° = 46°, so by the Law of Sines, _AG= = es (AG

AB|sin

52°

— ae,

so substituting we

200 sin 52°

~ 219 ff. have |AC| = ———— sin 46° 35:

ae

We draw a diagram. 4 is the position of the tourist and C is the top of the tower.

LB = 90° — 5.6° = 84.4° and so ZC = 180° — 29.2° — 84.4° = 66.4°. Thus, by

105 sin 29.2° the Law of Sines, the length of the tower is |BC] = ~~ —— = 55.9 m. sin 66.4°

56-4 B

105

A

37. The angle subtended by the top of the tree and the sun’s rays is ZA = 180° — 90° — 52° = 38°. Thus the height of the tree 215 sin 30° ish = ———— ~w 175 ft.

sin 38°

he

39. Call the balloon’s position R. Then in AP QR, we see that Z P = 62° — 32° = 30°, and ZO = 180° — 71° + 32° = 141°.

Therefore,

a

ZR = 180° (2) as— 30° one— 141

o—

‘

|OR|

rahi 9°.9°. So by the Law of Sines, —————_

=e

|PO|

66

|OR| R|

=

sin 30° =

60-

Ge ;

= 192 m.

41. Let d be the distance from the earth to Venus, and let 8 be the angle formed by sun, Venus, and earth. By the Law sinB __ sin39.4°

of Sines,

ee

ai aos7. 0 878:'80 either 8 ~ sin~! 0.878 ~ 61.4° or B © 180° — sin! 0.878 ~ 118.6°.

' dee ee SS dosing Uirst-case, —Gigge = 394° 614) d

ae Eee sin (180° — 39.4° — 118.6°)

0.723

ee —_sin39.4°

0.723 sin39.4°

eae

d =

1.119 AU; in the second case,

d & 0.427 AU.

43. By the area formula from Section 5.3, the area of AABC is A = Sab sin C. Because we are given a and the three : a : sin B angles, we need to find b in terms of these quantities. By the Law of Sines, Ra asinB\ : ales | fabsin€ = $a ( sin

yl |

A=

sin A

ice

—

pk

asin B

itl

us,

. a* sin B sinC sin C = ————__—_ aries

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

254

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

45. c

if

ere oy

ARB

a > b: One solution

b> a>

Pao B

A

bsin A: Two solutions

c

Dae ee

,a

B

A

a = bsin A: One solution

a < bsin A: No solution

LA = 30°, b = 100, sinA = 5. If a > b = 100 then there is one triangle. If 100 > a > 100sin30° = 50, then there are

two possible triangles. If a = 50, then there is one (right) triangle. And if a < 50, then no triangle is possible.

5.6

THELAW OF COSINES

1. For triangle ABC with sides a, b, and c the Law of Cosines states c? = a* + b? — 2abcosC.

3. x2 = 217 +.42? —2-21-42-cos39° = 441 + 1764 — 1764.c0s39° © 834.115 and sox © /834.115

© 28.9.

5. x2 = 257 +. 257 —2-25-25- cos 140° = 625 + 625 — 1250 cos 140° + 2207.556 and so x © »/2207.556 © 47. 37.82 Te. 37-83

=

68.012 ++ 42.152 68.01 68.015 — 2 -68.0

- 42.15 -cos cos@. Th en

cos

Oia

37.832 — 68.012 — 42.152 oe ee Sa eee Saas Olas

cos

@ = cos! 0.867 ¥ 29.89°.

9. x2 = 247 +30? —2-24-30-cos30° = 576 + 900 — 1440 cos 30° + 228.923 and so x © ./228.923 ~ 15. 11. c? = 10% + 182 —2-10- 18- cos 120° = 100 + 324 — 360.cos 120° = 604 and soc © 604 ~ 24.576. Then

sin A© —"—

18 sin 120°

~ 0.634295 «2 LA ~ sin!nail 0.634295 ~ 39.4°, oOand LB ~ 180° — 120° — 39.4° ° = 20.6°.°

53° , ©10.556~3.2. Thensin B = 4sin . ~ 10.556 ce —2.3-4-cos53° 3 = 9+ 16 — 2400s53° «9 ——— * 0.983 13. c?a = 3744?

LB ~ sin7! 0.983 © 79° and LA © 180° — 53° — 79° = 48°.

15, 20? = 25? +222 -2.25-22-cosA

> cosd = 2525 =22' ~ 0.644 > LA © cos! 0.644 ~ 50°. Then

sin B~ 258in40.9" ~ 0.956 LB ~ sin—! 0.956 © 73°, and so LC © 180° — 50° — 73° = 57°. 17. sinc = 12sin40" ~ 0.833 LC) © sin! 0.833 © 56.4° or LC2 © 180° — 56.4° © 123.6°. If LC, ~ 56.4°, then ZAy ~ 180° — 40° — 56.4° = 83.6° and a, = 1235i0 83.6" ~ 193,

If LC © 123.6°, then Zp © 180° — 40° — 123.6° = 16.4° and ay = VS8int64 ~ 54.9, Thus, one triangle has ZA ~ 83.6°, ZC *¥ 56.4°, anda ~ 193; the other has ZA © 16.4°, ZC * 123.6°, anda ~ 54.9, 195510) p= boas = 1,065. Since |sin @| < 1 for all @, there is no such ZB, and hence there is no such triangle. ap be 21. LB fate = 180° — 35° — 85° == 60°. Then xeee= AIGS38135° ~ 2. o S0 sin oO? as 9 Vo aim sm 1007 ~~ 25.4

25. b? = 110? + 1382 — 2 (110) (138) - cos 38° = 12,100 + 19,044 — 30,360 cos 38° ~ 7220.0 and so b © 85.0. Therefore, using the Law of Cosines again, we have cos 8@ = Ly hs 18

=

0

89.15°.

27. x? = 38? + 482 — 2-38-48 - cos 30° = 1444 + 2304 — 3648 cos 30° © 588.739 and so x © 24.3. 29. The semiperimeter

is s =

eels

=

18, so by Heron’s

Formula

the area

is

A = 18 (18 — 9) (18 — 12) (18 — 15) = 42916 = 54. 31. The semiperimeter

is 5

=

“tpt

=

12, so by Heron’s

Formula

the area is

A= J1202—7) 2 — 8) 2 — 9) = 7120 = 12,/5 + 26.8.

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or inpart.

er

F

;

Be

¥

f

7

’

ae”

i

SECTION 5.6 The Law of Cosines

33. The

semiperimeter

is s

=

255

2

#(#-3) (B-4) (F-6) = (8 = BS w 5.33,

SKE We draw a diagonal connecting the vertices adjacent to the 100° angle. This forms two triangles. Consider the triangle with sides of length 5 and 6 containing the 100° angle. The area of this triangle is 4; = 5 (5) (6) sin 100° + 14.77. To use Heron’s Formula to find the area of the second triangle, we need to find the length of the diagonal using the Law of Cosines: ce? =a? +b? —2abcosC = 82 + 6? —2-5-6cos 100° © 71.419 c © 8.45. Thus the second triangle has semiperimeter 8+7 + 8.45 Ss = ——,_ © 11.7255 and area Az = 4/11.7255 (11.7255 — 8) (11.7255 — 7) (11.7255 — 8.45) ~ 26.00. The area 2 of the quadrilateral is the sum of the areas of the two triangles: 4 = A, + Ap © 14.77 + 26.00 = 40.77.

“WK

Label the centers of the circles 4, B, and C, as in the figure. By the Law of

AB? + AC?—BC?

Cosines, cos A = 2B) Now, by the Law of Sines,

92 +10? —11?

dG) = XOX

sin70.53°

A 7470-53".

sinB ~ sinC

PvitneG

sin B = " sin 70.53° © 0.85710

a = t =r

AC)

SAB.

> B & sin! 0.85710 ~ 58.99° and

sinGi= 4 sin 70.53° © 0.77139 = C & sin—! 0.77139 ~ 50.48°. The area of

AABC is 5 (AB) (AC) sin A= 5 (9) (10) (sin 70.53°) © 42.426. The area of sector A is given by

0 S4 = 7R2 - Tae

1 (4) - e087 ~ 9.848. Similarly, the areas of sectors B and C

are Sg © 12.870 and Sc © 15.859. Thus, the area enclosed between the circles is 4 = AABC — S4—-Szp —-Sc> A © 42.426 — 9.848 — 12.870 — 15.859 © 3.85 cm2.

39. Let c be the distance across the lake, in miles. Then c* = 2.82? + 3.562 — 2 (2.82) (3.56) - cos40.3°

c © 2.30 mi.

41. In half an hour, the faster car travels 25 miles while the slower car travels 15 miles.

© 5.313 ©

The distance

between them is given by the Law of Cosines: d? = 252 + 152 — 2(25) (15) - cos65° => d = /252 + 152 — 2 (25) (15) - cos 65° = 5/25 + 9 — 30 - cos 65° © 23.1 mi.

43. The pilot travels a distance of 625 - 1.5 = 937.5 miles in her original direction and 625 - 2 = 1250 miles in the new direction. Since she makes a course correction of 10° to the right, the included angle is

COT

10° sceepign pce d

180° — 10° = 170°. From the figure, we use the Law of Cosines to get

the expression d? = 937.5? + 1250? — 2 (937.5) (1250) - cos 170° © 4,749,549.42, so d © 2179 miles. Thus, the pilot’s distance from her original position is approximately 2179 miles.

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256

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

45. (a) The angle subtended at Egg Island is 100°. Thus using the Law of

Forrest Island

Cosines, the distance from Forrest Island to the fisherman’s home port is

x2 = 307 +507 —2-30- 50-cos 100°

900 + 2500 — 3000 cos 100° ~ 3920.945 and so x © /3920.945

¥ 62.62 miles.

(b) Let @ be the angle shown in the figure. Using the Law of Sines,

50 sin 100° sin@ = erayT ~ 0.7863 @ © sin~! 0.7863 © 51.8°. Then ¥y = 90° — 20° — 51.8° = 18.2°. Thus the bearing to his home port is Home Port

Sul Suze lbs

47. The largest angle is the one opposite the longest side; call this angle 8. Then by the Law of Cosines,

442 = 362 + 222 — 2 (36) (22) - cos 49.

cos

Let d be the distance between the kites.

=

367 +227 — 444 2 (36) (22)

= —0,09848 = 8 ~ cos! (—0.09848) ~ 96°.

Then d* ~ 380% + 420? — 2 (380) (420) - cos30° =

3802 + 4202 — 2 (380) (420) - cos 30°

211 ft.

3400 51. Solution 1: From the figure, we see that 7 = 106° and sin 74° = Tau

3400 ao

=

S

ST. Thus, x? = 8007 + 35372 — 2 (800) (3537) cos 106°

=>

S

x = 8002 + 35372 — 2 (800) (3537) cos 106° Solution 2: Notice that tan 74°

3400

= ——-

©

a

=

3400

a=

Ki

x © 3835 ft. tan 74°

3400

= 974.9. By the

Pythagorean Theorem, x” = (a + 800)” + 34002. So

=

i.

x = ,/(974.9 + 800)” + 34002 ~ 3835 ft.

b

SR By Heron’s formula,

112 + 148 + 190

4 = 4/s (s — a) (s — b)(s —c), where s = Salas = Lived! SEAL ==(225) ies 2 2 225 (225 — 112) (225 — 148) (225 — 190) © 8277.7 fi2. Since the land value is $20 per square foot, the value of the

lot is approximately 8277.7 - 20 = $165,554.

Sos In any A ABC, the Law of Cosines gives a2 = b* +c? —2be-cos A, b* = a*+c?—2ac-cos B, and c* = a” +b? —2ab-cos C. Adding the second and third equations gives Bb? =

a* +c? —2ac-cosB

ce =

a*+b*—2ab-cosC

be +c?

=

2a*+b*+c*%-2a (ccos B + bcosC)

Thus 2a” — 2a (c cos B + bcosC) = 0, and so 2a (a — ecos B +bcosC) = 0.

Since a ~ 0 we must have a —(ccos

B + bcosC)

= 0a

= bcosC +c cos B . The other laws follow from the symmetry

of a, b, and c.

=. ——_!

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: i hin

CHAPTERS

Review

257

CHAPTER 5 REVIEW 1. (a) 30° = 30- 7, = F © 0.52 rad

3. (a) 37 rad = 5H . 180 — 599

(b) 150° = 150. 7%, = = ~ 2.62 rad

ie

(c) —20° = —20- fy = — 3 © —0.35 rad

(c)—3F rad = — 42 . 180 _ _ 940°

(d) —225° = —225 . 7%, = —S7 ~ —3.93 rad

(d) 4 rad= 4. 180 — 20 ~ 979,20

Sr

10m, @ = 22 rad. Thens

a

ee

=r0 =

180 _ _29°

2% — 4m ~ 12.6 m.

ey

vat 5 5 = 25-18 —~ Mx 286 ft

9. Since the diameter is 28 in, y = 14 in. In one revolution, the arc length (distance traveled) is s = @r = 27 - 14 = 287r in. The total distance traveled is 60 mi/h - 0.5 h = 30 mi = 30 mi- 5280 fi/mi - 12 in/ft = 1,900,800 in. The number of

:

:

1

; ; : revolution is 1,900,800 in - 58 _ = 21608.7 rev. Therefore the car wheel will make approximately 21,609 revolutions. in. 7

11. r =5 m, 0 = 2 rad. Then 4 = 5779 = 4.52. =25 2m2.

13. A = 125 f?,r = 25 ft.Then 6 = =!= aac = 38 = 0.4 rad © 22.9° 15. The angular speed is w

ate

=

3007 rad/min

= J/524+72 = J/74. Then sin@ =

942.5 rad/min.

=, cos@ = —L, tan@ = §, csc@ =

19. . =cos40°

©

x =5co0s40° ~ 3.83, and z =sin40°

LS sin20° ci 21: an

@&

x=

inahes© 2.92, and : = cos 20° Snzo>

23. A = 90° — 20°= 70°, a = 3cos20° © 2.819, and

fe

~

ed 711,026.

.

47. y—V¥3x+1=0e y = V3x — 1, 80 the slope of the line is m = /3. Then tan@ = m = V3 © @ = 60°. in@ Jl —cos?@ 49. Since sin@ is positive in quadrant II, sin@ = V1 — cos? @ and we have tan@ = ee ee! Lae cos 8 cos0 2

ras

51. tan?@= athe = a cos*?@ 1 —sin?@ 53. tand = oe = 4. Then cos 8 = : and sin@ = tan@ -cos@ = V1 csc = Ta =4 55. sin@= z. Since cos@ < 0, @ is in quadrant II. Thus, x = —/52

— 32 = —./16

7 and cot = a = 30ff

= —4 and so cos@= —%, tan@ = ~},

csc d= 3, sec@ = —}, cot? = —}.

57. tan@ =

cos@ sin

—3. sec?@ =

1+ tan?@ =

1+4

=

4 3 = cos’ 0 =e => cos@

< 0 in quadrant II. But tan@ = uly = -1 &sin@

cos 0

+cos6

Pa=

i Peg 2 Sz + (fic 4,)

2

= —1cos@

2

=

—

: =

= -}(-+)

2\

-% since

= -L.

V5

V5"

Therefore

:

ehF ype oakaBI

59. By the Pythagorean Theorem, sin? 9 + cos” @ = 1 for any angle 0.

73 _= = 61. sin~* -1 += 63. Let u = sin7! g and so sinu= E. Then from the triangle, tan (sin 1 3) = fanifee ee

5

cos

tan@ =x. Then from the

Vv1+4+x2

Ju

Vit x2

2

67. cos cos@ = x~3 = @

triangle, we have sin (tan! x) = sil G= saa

iain

5

65.Let@=tan~!x

°

ioe

x =cos~! (=) 3 10 sin 30°

69. ZB = 180° — 30° — 80° = 70°, and so by the Law ofSines, x = alas sin 70°

a2

71. x* = 1007 +210? — 2 - 100 - 210 - cos 40° ~ 21,926.133 x © 148.07 73. x* = 22 + 82 — 2 (2) (8) cos 120° = 84 @ x © V84 © 9.17

© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTERS — Test

nO 75. By the Law of Sines, =

=

atest. “-

=

sind) =

Mee —

=> 6 =

in 25° sin7! (AS)

=

259

54.1° or

6 © 180° — 54.1° = 125.9°.

1207era — 1002 ee — 852 100? + 852 — 2 (100) (85) cos 0, so cos@ = ———_— = 0.16618. Thus,

77. By the Law of Cosines, 1202

@ = cos! (0.16618) ~ 80.4°. 79. After 2 hours the ships have traveled distances d; = 40 mi and d) = 56 mi. The subtended angle is 180° — 32° — 42° = 106°. Let d be the distance between the two ships in miles. Then by the Law of Cosines,

d* = 40? + 56? — 2 (40) (56) cos 106° © 5970.855 d © 77.3 miles.

81. Let d be the distance, in miles, between the points A and B .

Then by the Law of Cosines,

d* = 3.2? + 5.62 — 2 (3.2) (5.6) cos 42° © 14.966 = d ~ 3.9 mi.

83. A = 4absin = 4 (8) (14) sin35° ~ 32.12

CHAPTER 5 TEST 1. 330° = 330 -7fq = UF rad. -135° = -135 -B, = — 3% rad. 3. (a) The angular speed is 120-2

940 1 min

5.r = /32 4+ 22 = ./13. Then

padmin © 753.98 malin

i ta.8

Pen

east

aay (vB + 3) 26 + 6/13 /iS™

ae o/s

abeos

BS

ae,

(b) The linear speed is

ixans

= 38407 ft/min

12,063.7 ft/min = 137 mi/h

7. cosO =

=. and @ is in quadrant III, sor

l tan @ cot@ + csc 8 = tan@- recy

dh

ee

= 3,x = ae wat—

—1, andy = tes 12 =

—2./2.

Then

272-3 _ he

9. sec? 9= 1 + tan? @ & tan@ = +y/sec2@— 1. Thus, tan @= —ysec? @— | since tan@ < 0 in quadrant II.

tantoled(=) = 5 =>6 ee= tan! 1. (a) tan@age Pecos

3 . REE! =~ => 8 = cos (=)

13. By the Law of Cosines, x2 = 10% + 122 — 210) (12) - cos 48° © 8.409 > x ~ 9.1, h 15. Leth be the height ofthe shorter altitude.

+h

Then tan 20° = 50 © h = 50 tan 20° and tan 28° = ar &x+h

&

= 50tan 28°

x =50tan28° —h = 50tan 28° — 50tan20° ~ 8.4. g2

62 ~ 92

17. By the Law ofCosines, 9* = 8* +6? —2 (8) (6) cos @ > cos0 = Sear ~ 0.1979, so @ © cos~! (0.1979) + 78.6°.

19. (a) A (sector)=3770 = 5-10? 72. 7%, =50- B™. A (triangle)=4 -r sin@=4. 10? sin 72°. Thus, the area of the shaded region is A (shaded) = A (sector)— A (triangle)= 50 (Rr - sin 72°) = 15.3 m2. (b) The shaded region is bounded by two pieces: one piece is part of the triangle, the other is part of the circle. The first part has length /= \/102 + 102 — 2 (10) (10) - cos 72° = 10./2 — 2- cos 72°. The second has length

s= 10-72- T80 = 4m. Thus, the perimeter ofthe shaded region is p = / +s = 10./2 — 2.cos 72° + 4a © 24.3 m.

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+ a]

260

FOCUS ON MODELING

21. Label the figure as shown. Now 23 = 85° — 75° = 10°, so by the Law of Sines, x

eee

sin 75°

pe

100

es

sin 10°

= 100-

i

sin 75°

h =x sin85° = 100-

sin 75° sin 10°

h

.Nowsin85°=—Seat

XG

&

2

sin 85° = 554.

sin 10°

FOCUS ON MODELING

Surveying

1. Let x be the distance between the church and City Hall. To apply the Law of Sines to the triangle with vertices at City Hall, the church, and the first bridge, we first need the measure of the angle at the first bridge, which is 180° — 25° — 30° = 125°. f

EG

*

0.86

sin 125° 1.41 miles.

pee.

sin 30°

IBC} sin 95°

we get

r

BC

=

sin 75°

20 sin 95°

|

Ss

sin 65°

=

20 sin 45°

20

|AC|

sin 65° ie

= 1.4089. So the distance between the church and City Hall is about

180° — 20° — 95° — 65° and DAC

sin 45°

20

—

sin 30°

=

3. First notice that ZDBC From AACD

0.86 sin 125°

AC) |AC|

4.6°. GE.

S.A sin 75°

= 22.0.

180° — 60° — 45°

=

75°.

From ABCD D we ge t

By applying the Law of Cosines to AABC

|AB|? = |AC|?+|BC|? —2|AC||BC| cos 40° © 14.6% +22.02 —2- 14.6-22.0-cos 40°

we get

205, so |AB| © /205 © 14.3 m.

Therefore, the distance between A and B is approximately 14.3 m.

BC AB 5. (a) In AABC, ZB = 180° — B, so ZC = 180° — a — (180° — B) = B — a. By the Law ofSines, Bs) = __4a sina sin(8—a)

Ch

EB ee sin(8-—a)

(b) From BCI

P

part (a) we know that |BC| = ———. gene

sin(B —a)

dsna

~ sin(B—a) (bx

sin(@—a)

eh

sin B er

But Bu sin = —g

iBC|

© |BC| = —. E

sinB

Therefore F

_ dsina sin B

sin(B — a)

dsinasinB — 800sin25° sin29° = 2350 ft sin(@—a) sin 4°

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Surveying

261

7. We start by labeling the edges and calculating the remaining angles, as shown in the first figure. Using the Law of Sines, ;

Es

lear rs Ge A

ETE = a

Ce

at ae

Wie

fe]

.

ie are aie ae 5,ay ites aes Sige NE ibe ~ 91.90, —

= —

@e@e=

sae ae ia =

& = ee

°

at

one

~ 151.67,

re © 149.50, and —

= =

eh= ae

~ 120.15. Note that

we used two decimal places throughout our calculations. Our results are shown (to one decimal place) in the second figure. 91.9

128.0

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Shoe Lan * ew

Ep tt

rel wpenartent cama

|

“te

thr ae

6

TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH

6.1

THEUNIT CIRCLE

i

a

a

ea

a

ee

ee

1. (a) The unit circle is the circle centered at (0, 0) with radius 1.

(b) The equation of the unit circle is x2 + y=,

(c) (i) Since 17 + 0? = 1, the point is P (1, 0).

(ii) P (0, 1)

(iii) P (—1, 0)

(iv) P (0, -1)

ee (3,-4) lies on the unit circle.

hve (;.-41) lies on the unit circle.

SP (-2. 3) lies on the unit circle. ey=

i ee

2 = 5 SNS

+292 Since P is in quadrant II, x is negative, so the point is

a= 4 ex=

= feo @ ll

q

a

ll

= 5=

+3. Since P (x, y) is in quadrant III, y is negative, so the point is

+35>. Since P (x, y) is in quadrant IV, x is positive, so the point is

Y = +75. Since its y-coordinate is negative, the point is P (4. -).

He

ks. Since its x-coordinate is negative, the point is P (--8, 3). 5

+97. Since P lies below the x-axis, its y-coordinate is negative, so

263 © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

264

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

25. f= F,sot= an corresponds to P (x,y) = (0, —1).

27.t = => corresponds to P (x,y)= (2. -3).

29. t= 4 andt = An corresponds to

33.¢= 4 andt= in corresponds to

35.f= 4 andt = -r corresponds to

37. (ji = 4 -w=F

39. (a) f= 1 — 58 = OR

(b) i =20-2F =F

b)f=n-B=

(?=-n-(-#) =F

(0) 7 = -3~0.142

(d) 7 =3.5 — 7 © 0.36

(d) 7 = 2m —5 © 1.283

41. (a)f=2n—-la_z

) P (8, -3) 45. (a)i=a—-2Z =F

(b) P (-3,-48) 49. (a)i=In-HE=F

(o) P (-33, 4)

43. (a) ?=—n -(-48) =F

) P (-4,8) 47. (a)? = BF -30n =F

(b) P

(-2,-2)

51.(a)i=4r—-UE

b) P (3,3)

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SECTION 6.2

53. (a) i= 19" -se =F

55.t=1

=

Trigonometric Functions of Real Numbers

265

(0.5,0.8)

) P (-3.-33) t=-1.l

Sia

=>

(0.5,-0.9)

59. Met) (ay) i= (3,¢)be the terminal point determined by r.

3

=I

(b) —t determines the point P (x, —y) = (3,-4). (c) 7 +f determines the point P (—x, —y) = (-3, ~4).

(d) 27 + ¢ determines the point P (x, y) = (3.3).

61. The distances PQ and PR are equal because they both subtend arcs of length 7 a point on the unit circle, x2 + y* =

d(R,S)=

1. Now d(P,Q)

(x — 0)? + (y— 1)? = Vx? +92? —29 41

gives2y= ./2—2y

=

=

Since P (x, y) is

V(x —x)*+ (vy —(-y))? =

2y and

= VI—-Dy (using the fact that x? + y = 1). Setting these equal

4y? =2-—2y 4y242y-2=0

2(2y-1)(¥4+ 1) =0. Soy =-lory =}. Since

P is in quadrant I, y = 4 is the only viable solution. Again using x2 + y? = | we have x2 + (3)2 =1lex?=

3 ij

iS +93, Again, since P is in quadrant I the coordinates must be (2. 3).

6.2

TRIGONOMETRIC FUNCTIONS OF REAL NUMBERS

1. If P(x, y) is the terminal point on the unit circle determined by f¢, then sint = y, cost = x, and tant = y/x.

3

5. (a) sin -E = -5 (b) cos 1 =

(c) tan 2 = 3 “419. (a) cos 32 = — /2

+" (b) cos 5%= —¥2

11.

3

(c) cos Thue FE = 15.

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>

266

(We

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

(a) csc 2 = ~2

19. (a) sin4 = —¥3

21. (a) sin13~ =0

(b) sec (-%) = 23

(b)_ sec He = a3

(b) cos 147% = 1

(c) cot (-5£) =

(c) cot (—7) =

(c) tan 157 =0

/3

23. t=0

=>

sint =0,cost = 1, tant = 0, sect = 1, csct¢ and cott are undefined.

25. t=m7

=>

sint =0,cost = —1,tant =0,sect = —1, csct and cott are undefined.

3\? + (-4) AVA = 759D + 55UANI6 = 1, Sosint aS = —%,4 cost = —z,3 and tant Sly: ae Die (-3) 5.

2

2

2

2

a2

29. (-3) + (242) = 4 4 5 =