Instructor's Solutions Manual, Single Variable for Thomas' Calculus, Twelfth Edition vol 1

INSTRUCTOR’S SOLUTIONS MANUAL SINGLE VARIABLE Collin County Community College WILLIAM ARDIS THOMAS’ CALCULUS TWELFTH E...

Author: William Ardis | Maurice D. Weir

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INSTRUCTOR’S SOLUTIONS MANUAL SINGLE VARIABLE Collin County Community College

WILLIAM ARDIS

THOMAS’ CALCULUS TWELFTH EDITION BASED ON THE ORIGINAL WORK BY

George B. Thomas, Jr. Massachusetts Institute of Technology

AS

REVISED BY

Maurice D. Weir Naval Postgraduate School

Joel Hass

University of California, Davis

This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson Addison-Wesley from electronic files supplied by the author. Copyright © 2010, 2005, 2001 Pearson Education, Inc. Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-321-60807-9 ISBN-10: 0-321-60807-0 1 2 3 4 5 6 BB 12 11 10 09

PREFACE TO THE INSTRUCTOR This Instructor's Solutions Manual contains the solutions to every exercise in the 12th Edition of THOMAS' CALCULUS by Maurice Weir and Joel Hass, including the Computer Algebra System (CAS) exercises. The corresponding Student's Solutions Manual omits the solutions to the even-numbered exercises as well as the solutions to the CAS exercises (because the CAS command templates would give them all away). In addition to including the solutions to all of the new exercises in this edition of Thomas, we have carefully revised or rewritten every solution which appeared in previous solutions manuals to ensure that each solution ì conforms exactly to the methods, procedures and steps presented in the text ì is mathematically correct ì includes all of the steps necessary so a typical calculus student can follow the logical argument and algebra ì includes a graph or figure whenever called for by the exercise, or if needed to help with the explanation ì is formatted in an appropriate style to aid in its understanding Every CAS exercise is solved in both the MAPLE and MATHEMATICA computer algebra systems. A template showing an example of the CAS commands needed to execute the solution is provided for each exercise type. Similar exercises within the text grouping require a change only in the input function or other numerical input parameters associated with the problem (such as the interval endpoints or the number of iterations). For more information about other resources available with Thomas' Calculus, visit http://pearsonhighered.com.

TABLE OF CONTENTS 1 Functions 1 1.1 1.2 1.3 1.4

Functions and Their Graphs 1 Combining Functions; Shifting and Scaling Graphs 8 Trigonometric Functions 19 Graphing with Calculators and Computers 26 Practice Exercises 30 Additional and Advanced Exercises 38

2 Limits and Continuity 43 2.1 2.2 2.3 2.4 2.5 2.6

Rates of Change and Tangents to Curves 43 Limit of a Function and Limit Laws 46 The Precise Definition of a Limit 55 One-Sided Limits 63 Continuity 67 Limits Involving Infinity; Asymptotes of Graphs 73 Practice Exercises 82 Additional and Advanced Exercises 86

3 Differentiation 93 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

Tangents and the Derivative at a Point 93 The Derivative as a Function 99 Differentiation Rules 109 The Derivative as a Rate of Change 114 Derivatives of Trigonometric Functions 120 The Chain Rule 127 Implicit Differentiation 135 Related Rates 142 Linearizations and Differentials 146 Practice Exercises 151 Additional and Advanced Exercises 162

4 Applications of Derivatives 167 4.1 4.2 4.3 4.4 4.5 4.6 4.7

Extreme Values of Functions 167 The Mean Value Theorem 179 Monotonic Functions and the First Derivative Test 188 Concavity and Curve Sketching 196 Applied Optimization 216 Newton's Method 229 Antiderivatives 233 Practice Exercises 239 Additional and Advanced Exercises 251

5 Integration 257 5.1 5.2 5.3 5.4 5.5 5.6

Area and Estimating with Finite Sums 257 Sigma Notation and Limits of Finite Sums 262 The Definite Integral 268 The Fundamental Theorem of Calculus 282 Indefinite Integrals and the Substitution Rule 290 Substitution and Area Between Curves 296 Practice Exercises 310 Additional and Advanced Exercises 320

6 Applications of Definite Integrals 327 6.1 6.2 6.3 6.4 6.5 6.6

Volumes Using Cross-Sections 327 Volumes Using Cylindrical Shells 337 Arc Lengths 347 Areas of Surfaces of Revolution 353 Work and Fluid Forces 358 Moments and Centers of Mass 365 Practice Exercises 376 Additional and Advanced Exercises 384

7 Transcendental Functions 389 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8

Inverse Functions and Their Derivatives 389 Natural Logarithms 396 Exponential Functions 403 Exponential Change and Separable Differential Equations 414 ^ Indeterminate Forms and L'Hopital's Rule 418 Inverse Trigonometric Functions 425 Hyperbolic Functions 436 Relative Rates of Growth 443 Practice Exercises 447 Additional and Advanced Exercises 458

8 Techniques of Integration 461 8.1 8.2 8.3 8.4 8.5 8.6 8.7

Integration by Parts 461 Trigonometric Integrals 471 Trigonometric Substitutions 478 Integration of Rational Functions by Partial Fractions 484 Integral Tables and Computer Algebra Systems 491 Numerical Integration 502 Improper Integrals 510 Practice Exercises 518 Additional and Advanced Exercises 528

9 First-Order Differential Equations 537 9.1 9.2 9.3 9.4 9.5

Solutions, Slope Fields and Euler's Method 537 First-Order Linear Equations 543 Applications 546 Graphical Solutions of Autonomous Equations 549 Systems of Equations and Phase Planes 557 Practice Exercises 562 Additional and Advanced Exercises 567

10 Infinite Sequences and Series 569 10.1 Sequences 569 10.2 Infinite Series 577 10.3 The Integral Test 583 10.4 Comparison Tests 590 10.5 The Ratio and Root Tests 597 10.6 Alternating Series, Absolute and Conditional Convergence 602 10.7 Power Series 608 10.8 Taylor and Maclaurin Series 617 10.9 Convergence of Taylor Series 621 10.10 The Binomial Series and Applications of Taylor Series 627 Practice Exercises 634 Additional and Advanced Exercises 642

11 Parametric Equations and Polar Coordinates 647 11.1 11.2 11.3 11.4 11.5 11.6 11.7

Parametrizations of Plane Curves 647 Calculus with Parametric Curves 654 Polar Coordinates 662 Graphing in Polar Coordinates 667 Areas and Lengths in Polar Coordinates 674 Conic Sections 679 Conics in Polar Coordinates 689 Practice Exercises 699 Additional and Advanced Exercises 709

CHAPTER 1 FUNCTIONS 1.1 FUNCTIONS AND THEIR GRAPHS 1. domain œ (_ß _); range œ [1ß _)

2. domain œ [0ß _); range œ (_ß 1]

3. domain œ Ò2ß _); y in range and y œ È5x 10 ! Ê y can be any positive real number Ê range œ Ò!ß _). 4. domain œ (_ß 0Ó Ò3, _); y in range and y œ Èx2 3x ! Ê y can be any positive real number Ê range œ Ò!ß _). 5. domain œ (_ß 3Ñ Ð3, _); y in range and y œ Ê3 t!Ê

4 3t

4 3t,

now if t 3 Ê 3 t ! Ê

4 3t

!, or if t 3

! Ê y can be any nonzero real number Ê range œ Ð_ß 0Ñ Ð!ß _).

6. domain œ (_ß %Ñ Ð4, 4Ñ Ð4, _); y in range and y œ 2

% t 4 Ê 16 Ÿ t 16 ! Ê nonzero real number Ê range œ Ð_ß

# "' 18 Ó

Ÿ

2 t2 16

2 t2 16 ,

2 t2 16

now if t % Ê t2 16 ! Ê 2

!, or if t % Ê t 16 ! Ê

2 t2 16

!, or if

! Ê y can be any

Ð!ß _).

7. (a) Not the graph of a function of x since it fails the vertical line test. (b) Is the graph of a function of x since any vertical line intersects the graph at most once. 8. (a) Not the graph of a function of x since it fails the vertical line test. (b) Not the graph of a function of x since it fails the vertical line test. #

9. base œ x; (height)# ˆ #x ‰ œ x# Ê height œ

È3 #

x; area is a(x) œ

" #

(base)(height) œ

" #

(x) Š

È3 # x‹

œ

È3 4

x# ;

perimeter is p(x) œ x x x œ 3x. 10. s œ side length Ê s# s# œ d# Ê s œ

d È2

; and area is a œ s# Ê a œ

" #

d#

11. Let D œ diagonal length of a face of the cube and j œ the length of an edge. Then j# D# œ d# and D# œ 2j# Ê 3j# œ d# Ê j œ

d È3

. The surface area is 6j# œ

6d# 3

12. The coordinates of P are ˆxß Èx‰ so the slope of the line joining P to the origin is m œ ˆx, Èx‰ œ ˆ m"# ,

#

œ 2d# and the volume is j$ œ Š d3 ‹ Èx x

œ

" Èx

$Î#

œ

(x 0). Thus,

"‰ m .

13. 2x 4y œ 5 Ê y œ "# x 54 ; L œ ÈÐx 0Ñ2 Ðy 0Ñ2 œ Éx2 Ð "# x 54 Ñ2 œ Éx2 4" x2 54 x œ É 54 x2 54 x

25 16

œ É 20x

2

20x 25 16

œ

È20x2 20x 25 4

14. y œ Èx 3 Ê y2 3 œ x; L œ ÈÐx 4Ñ2 Ðy 0Ñ2 œ ÈÐy2 3 4Ñ2 y2 œ ÈÐy2 1Ñ2 y2 œ Èy4 2y2 1 y2 œ Èy4 y2 1

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

d$ 3È 3

25 16

.

2

Chapter 1 Functions

15. The domain is a_ß _b.



18. The domain is Ð_ß !Ó.

19. The domain is a_ß !b a!ß _b.

20. The domain is a_ß !b a!ß _b.

21. The domain is a_ß 5b Ð5ß 3Ó Ò3, 5Ñ a5, _b 22. The range is Ò2, 3Ñ. 23. Neither graph passes the vertical line test (a)

(b)


Section 1.1 Functions and Their Graphs 24. Neither graph passes the vertical line test (a)

(b)

Ú xyœ" Þ Ú yœ1x Þ or or kx yk œ 1 Í Û Í Û ß ß Ü x y œ " à Ü y œ " x à 25.

x y

0 0

1 1

27. Faxb œ œ

2 0

26.

x y

0 1

1 0

2 0

" , x0 28. Gaxb œ œ x x, 0 Ÿ x

4 x2 , x Ÿ 1 x2 2x, x 1

29. (a) Line through a!ß !b and a"ß "b: y œ x; Line through a"ß "b and a#ß !b: y œ x 2 x, 0 Ÿ x Ÿ 1 f(x) œ œ x 2, 1 x Ÿ 2 Ú Ý 2, ! Ÿ x " Ý !ß " Ÿ x # (b) f(x) œ Û Ý Ý 2ß # Ÿ x $ Ü !ß $ Ÿ x Ÿ % 30. (a) Line through a!ß 2b and a#ß !b: y œ x 2 " Line through a2ß "b and a&ß !b: m œ !& # œ x #, 0 x Ÿ # f(x) œ œ " $ x &$ , # x Ÿ &

f(x) œ œ

œ "$ , so y œ "$ ax 2b " œ "$ x

$ ! ! Ð"Ñ œ " $ % #! œ #

(b) Line through a"ß !b and a!ß $b: m œ Line through a!ß $b and a#ß "b: m œ

" $

& $

$, so y œ $x $ œ #, so y œ #x $

$x $, " x Ÿ ! #x $, ! x Ÿ #


3

4

Chapter 1 Functions

31. (a) Line through a"ß "b and a!ß !b: y œ x Line through a!ß "b and a"ß "b: y œ " Line through a"ß "b and a$ß !b: m œ !" $" œ Ú x " Ÿ x ! " !xŸ" f(x) œ Û Ü "# x $# "x$

" #

(b) Line through a2ß 1b and a0ß 0b: y œ 12 x Line through a0ß 2b and a1ß 0b: y œ 2x 2 Line through a1ß 1b and a3ß 1b: y œ 1 32. (a) Line through ˆ T# ß !‰ and aTß "b: m œ faxb œ

(b)

"! TaTÎ#b

œ "# , so y œ "# ax "b " œ "# x

Ú

1 2x

faxb œ Û 2x 2 Ü 1

$ #

2 Ÿ x Ÿ 0 0xŸ1 1xŸ3

œ T# , so y œ T# ˆx T# ‰ 0 œ T# x "

!, 0 Ÿ x Ÿ T# # T T x ", # x Ÿ T

Ú A, Ý Ý Ý Aß faxb œ Û Aß Ý Ý Ý Ü Aß

! Ÿ x T# T # Ÿx T T Ÿ x $#T $T # Ÿ x Ÿ #T

33. (a) ÚxÛ œ 0 for x − [0ß 1)

(b) ÜxÝ œ 0 for x − (1ß 0]

34. ÚxÛ œ ÜxÝ only when x is an integer. 35. For any real number x, n Ÿ x Ÿ n ", where n is an integer. Now: n Ÿ x Ÿ n " Ê Ðn "Ñ Ÿ x Ÿ n. By definition: ÜxÝ œ n and ÚxÛ œ n Ê ÚxÛ œ n. So ÜxÝ œ ÚxÛ for all x − d . 36. To find f(x) you delete the decimal or fractional portion of x, leaving only the integer part.


Section 1.1 Functions and Their Graphs 37. Symmetric about the origin Dec: _ x _ Inc: nowhere

38. Symmetric about the y-axis Dec: _ x ! Inc: ! x _

39. Symmetric about the origin Dec: nowhere Inc: _ x ! !x_

40. Symmetric about the y-axis Dec: ! x _ Inc: _ x !

41. Symmetric about the y-axis Dec: _ x Ÿ ! Inc: ! x _

42. No symmetry Dec: _ x Ÿ ! Inc: nowhere


5

6

Chapter 1 Functions

43. Symmetric about the origin Dec: nowhere Inc: _ x _

44. No symmetry Dec: ! Ÿ x _ Inc: nowhere

45. No symmetry Dec: ! Ÿ x _ Inc: nowhere

46. Symmetric about the y-axis Dec: _ x Ÿ ! Inc: ! x _

47. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the function is even. 48. faxb œ x& œ

" x&

and faxb œ axb& œ

" a x b&

œ ˆ x"& ‰ œ faxb. Thus the function is odd.

49. Since faxb œ x# " œ axb# " œ faxb. The function is even. 50. Since Òfaxb œ x# xÓ Á Òfaxb œ axb# xÓ and Òfaxb œ x# xÓ Á Òfaxb œ axb# xÓ the function is neither even nor odd. 51. Since gaxb œ x$ x, gaxb œ x$ x œ ax$ xb œ gaxb. So the function is odd. 52. gaxb œ x% $x# " œ axb% $axb# " œ gaxbß thus the function is even. 53. gaxb œ

" x# "

54. gaxb œ

x x# " ;

55. hatb œ

" t ";

œ

" axb# "

œ gaxb. Thus the function is even.

gaxb œ x#x" œ gaxb. So the function is odd.

hatb œ

" t " ;

h at b œ

" " t.

Since hatb Á hatb and hatb Á hatb, the function is neither even nor odd.

56. Since l t$ | œ l atb$ |, hatb œ hatb and the function is even.


Section 1.1 Functions and Their Graphs 57. hatb œ 2t ", hatb œ 2t ". So hatb Á hatb. hatb œ 2t ", so hatb Á hatb. The function is neither even nor odd. 58. hatb œ 2l t l " and hatb œ 2l t l " œ 2l t l ". So hatb œ hatb and the function is even. 59. s œ kt Ê 25 œ kÐ75Ñ Ê k œ

" 3

Ê s œ 3" t; 60 œ 3" t Ê t œ 180

60. K œ c v# Ê 12960 œ ca18b2 Ê c œ 40 Ê K œ 40v# ; K œ 40a10b# œ 4000 joules 61. r œ 62. P œ

k s

Ê6œ

k v

k 4

Ê k œ 24 Ê r œ

Ê 14.7 œ

k 1000

24 s ;

10 œ

24 s

Ê k œ 14700 Ê P œ

Êsœ

14700 v ;

12 5

23.4 œ

14700 v

Êvœ

24500 39

¸ 628.2 in3

63. v œ f(x) œ xÐ"% 2xÑÐ22 2xÑ œ %x$ 72x# $!)x; ! x 7Þ 64. (a) Let h œ height of the triangle. Since the triangle is isosceles, AB # AB # œ 2# Ê AB œ È2Þ So, #

h# "# œ ŠÈ2‹ Ê h œ " Ê B is at a!ß "b Ê slope of AB œ " Ê The equation of AB is y œ f(x) œ x "; x − Ò!ß "Ó. (b) AÐxÑ œ 2x y œ 2xÐx "Ñ œ 2x# #x; x − Ò!ß "Ó. 65. (a) Graph h because it is an even function and rises less rapidly than does Graph g. (b) Graph f because it is an odd function. (c) Graph g because it is an even function and rises more rapidly than does Graph h. 66. (a) Graph f because it is linear. (b) Graph g because it contains a!ß "b. (c) Graph h because it is a nonlinear odd function. x #

67. (a) From the graph, (b)

x #

1

x 0:

x #

x 0:

x 2

4 x

1

Ê 4 x

x #

1

4 x

Ê x − (2ß 0) (%ß _)

1 4x 0 # 2x8 0 Ê x 2x

0 Ê

(x4)(x2) #x

0

(x4)(x2) #x

0

Ê x 4 since x is positive; 1

4 x

0 Ê

x# 2x8 2x

0 Ê

Ê x 2 since x is negative; sign of (x 4)(x 2) ïïïïïðïïïïïðïïïïî 2 % Solution interval: (#ß 0) (%ß _)


7

8

Chapter 1 Functions 3 2 x 1 x 1 3 2 x 1 x 1

68. (a) From the graph, (b) Case x 1:

Ê x − (_ß 5) (1ß 1) Ê

3(x1) x 1

2

Ê 3x 3 2x 2 Ê x 5. Thus, x − (_ß 5) solves the inequality. Case 1 x 1:

3 x 1

2 x 1

Ê

3(x1) x 1

2

Ê 3x 3 2x 2 Ê x 5 which is true if x 1. Thus, x − (1ß 1) solves the inequality. 3 Case 1 x: x1 x2 1 Ê 3x 3 2x 2 Ê x 5 which is never true if 1 x, so no solution here. In conclusion, x − (_ß 5) (1ß 1). 69. A curve symmetric about the x-axis will not pass the vertical line test because the points ax, yb and ax, yb lie on the same vertical line. The graph of the function y œ faxb œ ! is the x-axis, a horizontal line for which there is a single y-value, !, for any x. 70. price œ 40 5x, quantity œ 300 25x Ê Raxb œ a40 5xba300 25xb 71. x2 x2 œ h2 Ê x œ

h È2

œ

È2 h 2 ;

cost œ 5a2xb 10h Ê Cahb œ 10Š

È2 h 2 ‹

10h œ 5hŠÈ2 2‹

72. (a) Note that 2 mi = 10,560 ft, so there are È800# x# feet of river cable at $180 per foot and a10,560 xb feet of land cable at $100 per foot. The cost is Caxb œ 180È800# x# 100a10,560 xb. (b) Ca!b œ $"ß #!!ß !!! Ca&!!b ¸ $"ß "(&ß )"# Ca"!!!b ¸ $"ß ")'ß &"# Ca"&!!b ¸ $"ß #"#ß !!! Ca#!!!b ¸ $"ß #%$ß ($# Ca#&!!b ¸ $"ß #()ß %(* Ca$!!!b ¸ $"ß $"%ß )(! Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the point P. 1.2 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS 1. Df : _ x _, Dg : x 1 Ê Df

g

œ Dfg : x 1. Rf : _ y _, Rg : y 0, Rf g : y 1, Rfg : y 0

2. Df : x 1 0 Ê x 1, Dg : x 1 0 Ê x 1. Therefore Df Rf œ Rg : y 0, Rf g : y È2, Rfg : y 0

g

œ Dfg : x 1.

3. Df : _ x _, Dg : _ x _, DfÎg : _ x _, DgÎf : _ x _, Rf : y œ 2, Rg : y 1, RfÎg : 0 y Ÿ 2, RgÎf : "# Ÿ y _ 4. Df : _ x _, Dg : x 0 , DfÎg : x 0, DgÎf : x 0; Rf : y œ 1, Rg : y 1, RfÎg : 0 y Ÿ 1, RgÎf : 1 Ÿ y _ 5. (a) 2 (d) (x 5)# 3 œ x# 10x 22 (g) x 10

(b) 22 (e) 5 (h) (x# 3)# 3 œ x% 6x# 6

(c) x# 2 (f) 2


Section 1.2 Combining Functions; Shifting and Scaling Graphs 6. (a) "3 (d)

(b) 2

" x

(c)

(e) 0

(g) x 2

(h)

(f)

" " x 1 1

œ

x x

"

# 1

x" x#

œ

" x 1 3 4

1œ

x x1

7. af‰g‰hbaxb œ fagahaxbbb œ faga4 xbb œ fa3a4 xbb œ fa12 3xb œ a12 3xb 1 œ 13 3x 8. af‰g‰hbaxb œ fagahaxbbb œ fagax2 bb œ fa2ax2 b 1b œ fa2x2 1b œ 3a2x2 1b 4 œ 6x2 1 9. af‰g‰hbaxb œ fagahaxbbb œ fˆgˆ 1x ‰‰ œ fŠ 1 1 % ‹ œ fˆ 1 x 4x ‰ œ É 1 x 4x " œ É 15x4x" x

2

10. af‰g‰hbaxb œ fagahaxbbb œ fŠgŠÈ2 x‹‹ œ f

ŠÈ2 x‹ 2

ŠÈ2 x‹

œ fˆ $ x ‰ œ 1 2x

2 x $ x 2 3 $2 xx

8 3x 7 2x

œ

11. (a) af‰gbaxb (d) a j‰jbaxb

(b) a j‰gbaxb (e) ag‰h‰f baxb

(c) ag‰gbaxb (f) ah‰j‰f baxb

12. (a) af‰jbaxb (d) af‰f baxb

(b) ag‰hbaxb (e) a j‰g‰f baxb

(c) ah‰hbaxb (f) ag‰f‰hbaxb

g(x)

f(x)

(f ‰ g)(x)

(a)

x7

Èx

Èx 7

(b)

x2

3x

3(x 2) œ 3x 6

(c)

x#

Èx 5

Èx# 5

(d)

x x1

x x1

" x1 " x

1

13.

(e) (f)

" x

gaxb" g ax b

œ

x x (x1)

œx

x

" x

x

" lx "l .

14. (a) af‰gbaxb œ lgaxbl œ (b) af‰gbaxb œ

x x 1 x x 1 1

x x"

œ

Ê"

" g ax b

œ

x x"

Ê"

x x"

œ

" g ax b

Ê

" x"

œ

" gaxb ß so

gaxb œ x ".

(c) Since af‰gbaxb œ Ègaxb œ lxl, gaxb œ x . (d) Since af‰gbaxb œ fˆÈx‰ œ l x l, faxb œ x# . (Note that the domain of the composite is Ò!ß _Ñ.) #

The completed table is shown. Note that the absolute value sign in part (d) is optional. gaxb faxb af‰gbaxb " " lxl x" lx "l x" x# Èx

x" x

Èx #

x

15. (a) faga1bb œ fa1b œ 1 (d) gaga2bb œ ga0b œ 0

x x"

lxl lxl (b) gafa0bb œ ga2b œ 2 (e) gafa2bb œ ga1b œ 1

(c) fafa1bb œ fa0b œ 2 (f) faga1bb œ fa1b œ 0

16. (a) faga0bb œ fa1b œ 2 a1b œ 3, where ga0b œ 0 1 œ 1 (b) gafa3bb œ ga1b œ a1b œ 1, where fa3b œ 2 3 œ 1 (c) gaga1bb œ ga1b œ 1 1 œ 0, where ga1b œ a1b œ 1


9

10

Chapter 1 Functions (d) fafa2bb œ fa0b œ 2 0 œ 2, where fa2b œ 2 2 œ 0 (e) gafa0bb œ ga2b œ 2 1 œ 1, where fa0b œ 2 0 œ 2 (f) fˆgˆ "# ‰‰ œ fˆ #" ‰ œ 2 ˆ #" ‰ œ 5# , where gˆ "# ‰ œ "# 1 œ "#

17. (a) af‰gbaxb œ fagaxbb œ É 1x 1 œ É 1 x x ag‰f baxb œ gafaxbb œ

1 Èx 1

(b) Domain af‰gb: Ð_, 1Ó Ð0, _Ñ, domain ag‰f b: Ð1, _Ñ (c) Range af‰gb: Ð1, _Ñ, range ag‰f b: Ð0, _Ñ 18. (a) af‰gbaxb œ fagaxbb œ 1 2Èx x ag‰f baxb œ gafaxbb œ 1 kxk (b) Domain af‰gb: Ò0, _Ñ, domain ag‰f b: Ð_, _Ñ (c) Range af‰gb: Ð0, _Ñ, range ag‰f b: Ð_, 1Ó 19. af‰gbaxb œ x Ê fagaxbb œ x Ê

g ax b g ax b 2

œ x Ê gaxb œ agaxb 2bx œ x † gaxb 2x

Ê gaxb x † gaxb œ 2x Ê gaxb œ 1 2x x œ

2x x1

20. af‰gbaxb œ x 2 Ê fagaxbb œ x 2 Ê 2agaxbb3 4 œ x 2 Ê agaxbb3 œ 21. (a) y œ (x 7)#

(b) y œ (x 4)#

22. (a) y œ x# 3

(b) y œ x# 5

x6 2

3 x6 Ê gaxb œ É 2

23. (a) Position 4

(b) Position 1

(c) Position 2

(d) Position 3

24. (a) y œ (x 1)# 4

(b) y œ (x 2)# 3

(c) y œ (x 4)# 1

(d) y œ (x 2)#

25.

26.

27.

28.


Section 1.2 Combining Functions; Shifting and Scaling Graphs 29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.


11

12

Chapter 1 Functions

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

51.

52.


Section 1.2 Combining Functions; Shifting and Scaling Graphs 53.

54.

55. (a) domain: [0ß 2]; range: [#ß $]

(b) domain: [0ß 2]; range: [1ß 0]

(c) domain: [0ß 2]; range: [0ß 2]

(d) domain: [0ß 2]; range: [1ß 0]

(e) domain: [2ß 0]; range: [!ß 1]

(f) domain: [1ß 3]; range: [!ß "]

(g) domain: [2ß 0]; range: [!ß "]

(h) domain: [1ß 1]; range: [!ß "]


13

14

Chapter 1 Functions

56. (a) domain: [0ß 4]; range: [3ß 0]

(b) domain: [4ß 0]; range: [!ß $]

(c) domain: [4ß 0]; range: [!ß $]

(d) domain: [4ß 0]; range: ["ß %]

(e) domain: [#ß 4]; range: [3ß 0]

(f) domain: [2ß 2]; range: [3ß 0]

(g) domain: ["ß 5]; range: [3ß 0]

(h) domain: [0ß 4]; range: [0ß 3]

58. y œ a2xb# 1 œ %x# 1

57. y œ 3x# 3 59. y œ "# ˆ"

"‰ x#

œ

" #

" #x#

60. y œ 1

" axÎ$b#

œ1

61. y œ È%x 1

62. y œ 3Èx 1

# 63. y œ É% ˆ x# ‰ œ "# È16 x#

64. y œ "$ È% x#

65. y œ " a3xb$ œ " 27x$

66. y œ " ˆ x# ‰ œ "

$

* x#

x$ )


Section 1.2 Combining Functions; Shifting and Scaling Graphs 67. Let y œ È#x " œ faxb and let gaxb œ x"Î# , "Î# "Î# haxb œ ˆx " ‰ , iaxb œ È#ˆx " ‰ , and #

#

"Î# jaxb œ ’È#ˆx "# ‰ “ œ faBb. The graph of

haxb is the graph of gaxb shifted left

" #

unit; the

graph of iaxb is the graph of haxb stretched vertically by a factor of È#; and the graph of jaxb œ faxb is the graph of iaxb reflected across the x-axis. 68. Let y œ È"

x #

œ faxbÞ Let gaxb œ axb"Î# ,

haxb œ ax #b"Î# , and iaxb œ œ È"

x #

" È # a x

#b"Î#

œ faxbÞ The graph of gaxb is the

graph of y œ Èx reflected across the x-axis. The graph of haxb is the graph of gaxb shifted right two units. And the graph of iaxb is the graph of haxb compressed vertically by a factor of È#. 69. y œ faxb œ x$ . Shift faxb one unit right followed by a shift two units up to get gaxb œ ax "b3 #.

70. y œ a" Bb$ # œ Òax "b$ a#bÓ œ faxb. Let gaxb œ x$ , haxb œ ax "b$ , iaxb œ ax "b$ a#b, and jaxb œ Òax "b$ a#bÓ. The graph of haxb is the graph of gaxb shifted right one unit; the graph of iaxb is the graph of haxb shifted down two units; and the graph of faxb is the graph of iaxb reflected across the x-axis.

71. Compress the graph of faxb œ of 2 to get gaxb œ unit to get haxb œ

" #x . Then " #x ".

" x

horizontally by a factor

shift gaxb vertically down 1


15

16

Chapter 1 Functions

72. Let faxb œ œ

"

#

ŠxÎÈ#‹

" x#

and gaxb œ

"œ

# x#

"œ

" # ’Š"ÎÈ#‹B“

" # Š B# ‹

"

"Þ Since

È# ¸ "Þ%, we see that the graph of faxb stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of gaxb.

$ 73. Reflect the graph of y œ faxb œ È x across the x-axis $ to get gaxb œ Èx.

74. y œ faxb œ a#xb#Î$ œ Òa"ba#bxÓ#Î$ œ a"b#Î$ a#xb#Î$ œ a#xb#Î$ . So the graph of faxb is the graph of gaxb œ x#Î$ compressed horizontally by a factor of 2.

75.

76.

77. *x# #&y# œ ##& Ê

x# &#

y# $#

œ"

78. "'x# (y# œ ""# Ê

x# # È Š (‹

y# %#

œ"


Section 1.2 Combining Functions; Shifting and Scaling Graphs 79. $x# ay #b# œ $ Ê

x# "#

a y #b # #

ŠÈ$‹

80. ax "b# #y# œ % Ê

œ"

Ê

83.

x# "'

#

ŠÈ#‹

y# *

y a#b‘# #

ŠÈ$‹

# # 82. 'ˆx $# ‰ *ˆy "# ‰ œ &%

81. $ax "b# #ay #b# œ ' ax " b #

x a"b‘# ##

#

œ"

Ê

’xˆ $# ‰“ $#

ˆy "# ‰# #

ŠÈ'‹

œ"

œ " has its center at a!ß !b. Shiftinig 4 units

left and 3 units up gives the center at ah, kb œ a%ß $b. # x a4b‘# ay 3#3b œ " 4# a y $b # œ ". Center, C, is a%ß 3#

So the equation is Ê

ax % b # 4#

$b, and

major axis, AB, is the segment from a)ß $b to a!ß $b.

84. The ellipse

x# %

y# #&

œ " has center ah, kb œ a!ß !b.

Shifting the ellipse 3 units right and 2 units down produces an ellipse with center at ah, kb œ a$ß #b and an equation

ax 3 b# %

y a#b‘# #&

œ ". Center,

C, is a3ß #b, and AB, the segment from a$ß $b to a$ß (b is the major axis.

85. (a) (fg)(x) œ f(x)g(x) œ f(x)(g(x)) œ (fg)(x), odd (b) Š gf ‹ (x) œ

f(x) g(x)

œ

f(x) g(x)

œ Š gf ‹ (x), odd


y# # È Š #‹

œ"

17

18

Chapter 1 Functions (c) ˆ gf ‰ (x) œ (d) (e) (f) (g) (h) (i)

g(x) f(x)

œ

g(x) f(x)

œ ˆ gf ‰ (x), odd

f # (x) œ f(x)f(x) œ f(x)f(x) œ f # (x), even g# (x) œ (g(x))# œ (g(x))# œ g# (x), even (f ‰ g)(x) œ f(g(x)) œ f(g(x)) œ f(g(x)) œ (f ‰ g)(x), even (g ‰ f)(x) œ g(f(x)) œ g(f(x)) œ (g ‰ f)(x), even (f ‰ f)(x) œ f(f(x)) œ f(f(x)) œ (f ‰ f)(x), even (g ‰ g)(x) œ g(g(x)) œ g(g(x)) œ g(g(x)) œ (g ‰ g)(x), odd

86. Yes, f(x) œ 0 is both even and odd since f(x) œ 0 œ f(x) and f(x) œ 0 œ f(x). 87. (a)

(b)

(c)

(d)

88.


Section 1.3 Trigonometric Functions 1.3 TRIGONOMETRIC FUNCTIONS 1. (a) s œ r) œ (10) ˆ 451 ‰ œ 81 m radians and

51 4

1 ‰ 3. ) œ 80° Ê ) œ 80° ˆ 180° œ

41 9

2. ) œ

s r

œ

101 8

œ

51 4

1 ‰ (b) s œ r) œ (10)(110°) ˆ 180° œ

1

)

231

0

1 #

s r

œ

30 50

31 4 " È2 È" 2

sin )

0

cos )

1

tan )

0

È3

0

und.

"

und.

" È3

und.

0

1

und.

È 2

1

#

und.

È23

sec ) csc )

0

"

"

0

" und.

7. cos x œ 45 , tan x œ 34 9. sin x œ

È8 3

, tan x œ È8

"

6.

È2

3#1

)

1'

sin )

"

cos )

!

" #

tan )

und.

È 3

cot )

!

È"3

sec )

und.

#

csc )

"

È23

8. sin x œ

2 È5

10. sin x œ

12 13

13.

14.

period œ 1

13

È #3

12. cos x œ

, cos x œ

" È2

&1 ' " # È #3

È"3

"

È"3

È 3

"

È 3

2 È3

È2

È23

#

È2

#

"# È3 #

" È5

, tan x œ 12 5 È3 #

, tan x œ

" È3

period œ 41 16.

period œ 2

m

‰ ¸ 34° œ 0.6 rad or 0.6 ˆ 180° 1

11. sin x œ È"5 , cos x œ È25

15.

551 9

Ê s œ (6) ˆ 491 ‰ œ 8.4 in. (since the diameter œ 12 in. Ê radius œ 6 in.)

È #3 "#

cot )

œ

ˆ 180° ‰ œ 225° 1

4. d œ 1 meter Ê r œ 50 cm Ê ) œ 5.

1101 18

period œ 4


1 % " È2

19

20

Chapter 1 Functions

17.

18.

period œ 6

period œ 1

19.

20.

period œ 21

period œ 21

21.

22.

period œ 21

period œ 21

23. period œ 1# , symmetric about the origin

24. period œ 1, symmetric about the origin

25. period œ 4, symmetric about the s-axis

26. period œ 41, symmetric about the origin


Section 1.3 Trigonometric Functions 27. (a) Cos x and sec x are positive for x in the interval ˆ 12 , 12 ‰; and cos x and sec x are negative for x in the intervals ˆ 321 , 12 ‰ and ˆ 12 , 321 ‰. Sec x is undefined when cos x is 0. The range of sec x is (_ß 1] ["ß _); the range of cos x is ["ß 1]. (b) Sin x and csc x are positive for x in the intervals ˆ 321 , 1‰ and a!, 1b; and sin x and csc x are negative for x in the intervals a1, !b and ˆ1, 321 ‰. Csc x is undefined when sin x is 0. The range of csc x is (_ß 1] [1ß _); the range of sin x is ["ß "].

28. Since cot x œ

" tan x

, cot x is undefined when tan x œ 0

and is zero when tan x is undefined. As tan x approaches zero through positive values, cot x approaches infinity. Also, cot x approaches negative infinity as tan x approaches zero through negative values.

29. D: _ x _; R: y œ 1, 0, 1

30. D: _ x _; R: y œ 1, 0, 1

31. cos ˆx 1# ‰ œ cos x cos ˆ 1# ‰ sin x sin ˆ 1# ‰ œ (cos x)(0) (sin x)(1) œ sin x 32. cos ˆx 1# ‰ œ cos x cos ˆ 1# ‰ sin x sin ˆ 1# ‰ œ (cos x)(0) (sin x)(1) œ sin x 33. sin ˆx 1# ‰ œ sin x cos ˆ 1# ‰ cos x sin ˆ 1# ‰ œ (sin x)(0) (cos x)(1) œ cos x 34. sin ˆx 1# ‰ œ sin x cos ˆ 1# ‰ cos x sin ˆ 1# ‰ œ (sin x)(0) (cos x)(1) œ cos x 35. cos (A B) œ cos (A (B)) œ cos A cos (B) sin A sin (B) œ cos A cos B sin A (sin B) œ cos A cos B sin A sin B 36. sin (A B) œ sin (A (B)) œ sin A cos (B) cos A sin (B) œ sin A cos B cos A (sin B) œ sin A cos B cos A sin B 37. If B œ A, A B œ 0 Ê cos (A B) œ cos 0 œ 1. Also cos (A B) œ cos (A A) œ cos A cos A sin A sin A œ cos# A sin# A. Therefore, cos# A sin# A œ 1.


21

22

Chapter 1 Functions

38. If B œ 21, then cos (A 21) œ cos A cos 21 sin A sin 21 œ (cos A)(1) (sin A)(0) œ cos A and sin (A 21) œ sin A cos 21 cos A sin 21 œ (sin A)(1) (cos A)(0) œ sin A. The result agrees with the fact that the cosine and sine functions have period 21. 39. cos (1 x) œ cos 1 cos B sin 1 sin x œ (1)(cos x) (0)(sin x) œ cos x 40. sin (21 x) œ sin 21 cos (x) cos (21) sin (x) œ (0)(cos (x)) (1)(sin (x)) œ sin x 41. sin ˆ 3#1 x‰ œ sin ˆ 3#1 ‰ cos (x) cos ˆ 3#1 ‰ sin (x) œ (1)(cos x) (0)(sin (x)) œ cos x 42. cos ˆ 3#1 x‰ œ cos ˆ 3#1 ‰ cos x sin ˆ 3#1 ‰ sin x œ (0)(cos x) (1)(sin x) œ sin x œ sin ˆ 14 13 ‰ œ sin

44. cos

111 1#

45. cos

1 12

œ cos ˆ 13 14 ‰ œ cos

46. sin

51 1#

œ sin ˆ 231 14 ‰ œ sin ˆ 231 ‰ cos ˆ 14 ‰ cos ˆ 231 ‰ sin ˆ 14 ‰ œ Š

21 ‰ 3

È2

1 8

œ

1 cos ˆ 281 ‰ #

œ

1 #

49. sin#

1 1#

œ

1 cos ˆ 211# ‰ #

œ

1 #

3 4

Ê sin ) œ „

52. sin2 ) œ cos2 ) Ê

sin2 ) cos2 )

1 4

œ cos

47. cos#

51. sin2 ) œ

cos

#

È3 #

È3 2

œ

1 3

cos

cos

21 3

1 4

sin

sin

1 4

cos ˆ 14 ‰ sin

1 3

1 3

È2 È3 # ‹Š # ‹

71 1#

œ cos ˆ 14

1 4

È2 ˆ"‰ # ‹ #

43. sin

œŠ

sin 1 3

21 3

œŠ

Š

È2 ˆ "‰ # ‹ #

sin ˆ 14 ‰ œ ˆ "# ‰ Š

Š

È2 # ‹

œ

È2 È3 # ‹Š # ‹

Š

È3 È2 # ‹Š # ‹

2 È2 4

48. cos#

51 1#

œ

1‰ 1 cos ˆ 10 1# #

œ

2 È3 4

50. sin#

31 8

œ

1 cos ˆ 681 ‰ #

Ê tan2 ) œ 1 Ê tan ) œ „ 1 Ê ) œ 14 ,

31 51 71 4 , 4 , 4

cos2 ) cos2 )

œ

È3 È2 # ‹ Š # ‹

œ

Ê ) œ 13 ,

È 6 È 2 4 È 2 È 6 4 1 È 3 2È 2

œ

ˆ "# ‰ Š œ

œ

1 Š

È3 ‹ #

#

1 Š #

È2 ‹ #

È2 # ‹

œ

œ

œ

1 È 3 2È 2

2 È3 4

2 È2 4

21 41 51 3 , 3 , 3

53. sin 2) cos ) œ 0 Ê 2sin ) cos ) cos ) œ 0 Ê cos )a2sin ) 1b œ 0 Ê cos ) œ 0 or 2sin ) 1 œ 0 Ê cos ) œ 0 or sin ) œ "# Ê ) œ 12 , 321 , or ) œ 16 , 561 Ê ) œ 16 , 12 , 561 , 321 54. cos 2) cos ) œ 0 Ê 2cos2 ) 1 cos ) œ 0 Ê 2cos2 ) cos ) 1 œ 0 Ê acos ) 1ba2cos ) 1b œ 0 Ê cos ) 1 œ 0 or 2cos ) 1 œ 0 Ê cos ) œ 1 or cos ) œ "# Ê ) œ 1 or ) œ 13 , 531 Ê ) œ 13 , 1, 531 55. tan (A B) œ

sin (AB) cos (AB)

œ

sin A cos Bcos A cos B cos A cos Bsin A sin B

œ

sin A cos B cos A sin B cos A cos B cos A cos B sin A sin B cos A cos B cos A cos B cos A cos B

œ

tan Atan B 1tan A tan B

56. tan (A B) œ

sin (AB) cos (AB)

œ

sin A cos Bcos A cos B cos A cos Bsin A sin B

œ

sin A cos B cos A sin B cos A cos B cos A cos B sin A sin B cos A cos B cos A cos B cos A cos B

œ

tan Atan B 1tan A tan B

57. According to the figure in the text, we have the following: By the law of cosines, c# œ a# b# 2ab cos ) œ 1# 1# 2 cos (A B) œ 2 2 cos (A B). By distance formula, c# œ (cos A cos B)# (sin A sin B)# œ cos# A 2 cos A cos B cos# B sin# A 2 sin A sin B sin# B œ 2 2(cos A cos B sin A sin B). Thus c# œ 2 2 cos (A B) œ 2 2(cos A cos B sin A sin B) Ê cos (A B) œ cos A cos B sin A sin B.


Section 1.3 Trigonometric Functions 58. (a) cosaA Bb œ cos A cos B sin A sin B sin ) œ cosˆ 1# )‰ and cos ) œ sinˆ 1# )‰ Let ) œ A B

sinaA Bb œ cos’ 1# aA Bb“ œ cos’ˆ 1# A‰ B“ œ cos ˆ 1# A‰ cos B sin ˆ 1# A‰ sin B œ sin A cos B cos A sin B (b) cosaA Bb œ cos A cos B sin A sin B cosaA aBbb œ cos A cos aBb sin A sin aBb Ê cosaA Bb œ cos A cos aBb sin A sin aBb œ cos A cos B sin A asin Bb œ cos A cos B sin A sin B Because the cosine function is even and the sine functions is odd. 59. c# œ a# b# 2ab cos C œ 2# 3# 2(2)(3) cos (60°) œ 4 9 12 cos (60°) œ 13 12 ˆ "# ‰ œ 7. Thus, c œ È7 ¸ 2.65. 60. c# œ a# b# 2ab cos C œ 2# 3# 2(2)(3) cos (40°) œ 13 12 cos (40°). Thus, c œ È13 12 cos 40° ¸ 1.951. 61. From the figures in the text, we see that sin B œ hc . If C is an acute angle, then sin C œ hb . On the other hand, if C is obtuse (as in the figure on the right), then sin C œ sin (1 C) œ hb . Thus, in either case, h œ b sin C œ c sin B Ê ah œ ab sin C œ ac sin B. a # b # c # 2ab

By the law of cosines, cos C œ

and cos B œ

a # c # b # . 2ac

Moreover, since the sum of the

interior angles of a triangle is 1, we have sin A œ sin (1 (B C)) œ sin (B C) œ sin B cos C cos B sin C #

#

#

#

#

#

b c c b ˆ h ‰ h ‰ œ ˆ hc ‰ ’ a 2ab a2a# b# c# c# b# b œ “ ’ a 2ac “ b œ ˆ 2abc

ah bc

Ê ah œ bc sin A.

Combining our results we have ah œ ab sin C, ah œ ac sin B, and ah œ bc sin A. Dividing by abc gives h sin A sin C sin B bc œ ðóóóóóóóñóóóóóóóò a œ c œ b . law of sines 62. By the law of sines,

sin A #

œ

sin B 3

œ

È3/2 c .

By Exercise 61 we know that c œ È7. Thus sin B œ

3È 3 2È 7

¶ 0.982.

63. From the figure at the right and the law of cosines, b# œ a# 2# 2(2a) cos B œ a# 4 4a ˆ "# ‰ œ a# 2a 4. Applying the law of sines to the figure, Ê

È2/2 a

œ

È3/2 b

sin A a

œ

sin B b

Ê b œ É 3# a. Thus, combining results,

a# 2a 4 œ b# œ

3 #

a# Ê 0 œ

" #

a# 2a 4

Ê 0 œ a# 4a 8. From the quadratic formula and the fact that a 0, we have aœ

4È4# 4(1)(8) #

œ

4 È 3 4 #

¶ 1.464.

64. (a) The graphs of y œ sin x and y œ x nearly coincide when x is near the origin (when the calculator is in radians mode). (b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The curves look like intersecting straight lines near the origin when the calculator is in degree mode.


23

24

Chapter 1 Functions

65. A œ 2, B œ 21, C œ 1, D œ 1

66. A œ "# , B œ 2, C œ 1, D œ

" #

67. A œ 12 , B œ 4, C œ 0, D œ

68. A œ

L 21 ,

" 1

B œ L, C œ 0, D œ 0

69-72. Example CAS commands: Maple f := x -> A*sin((2*Pi/B)*(x-C))+D1; A:=3; C:=0; D1:=0; f_list := [seq( f(x), B=[1,3,2*Pi,5*Pi] )]; plot( f_list, x=-4*Pi..4*Pi, scaling=constrained, color=[red,blue,green,cyan], linestyle=[1,3,4,7], legend=["B=1","B=3","B=2*Pi","B=3*Pi"], title="#69 (Section 1.3)" ); Mathematica Clear[a, b, c, d, f, x] f[x_]:=a Sin[21/b (x c)] + d Plot[f[x]/.{a Ä 3, b Ä 1, c Ä 0, d Ä 0}, {x, 41, 41 }]


Section 1.3 Trigonometric Functions 69. (a) The graph stretches horizontally.

(b) The period remains the same: period œ l B l. The graph has a horizontal shift of

" #

period.

70. (a) The graph is shifted right C units.

(b) The graph is shifted left C units. (c) A shift of „ one period will produce no apparent shift. l C l œ ' 71. (a) The graph shifts upwards l D lunits for D ! (b) The graph shifts down l D lunits for D !Þ

72. (a) The graph stretches l A l units.

(b) For A !, the graph is inverted.


25

26

Chapter 1 Functions

1.4 GRAPHING WITH CALCULATORS AND COMPUTERS 1-4.

The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and has little unused space.

1. d.

2. c.

3. d.

4. b.

5-30.

For any display there are many appropriate display widows. The graphs given as answers in Exercises 530 are not unique in appearance.

5. Ò2ß 5Ó by Ò15ß 40Ó

6. Ò4ß 4Ó by Ò4ß 4Ó

7. Ò2ß 6Ó by Ò250ß 50Ó



Section 1.4 Graphing with Calculators and Computers 9. Ò4ß 4Ó by Ò5ß 5Ó






15. Ò3ß 3Ó by Ò0ß 10Ó



27

28

Chapter 1 Functions





21. Ò"!ß "!Ó by Ò'ß 'Ó

22. Ò&ß &Ó by Ò#ß #Ó

23. Ò'ß "!Ó by Ò'ß 'Ó

24. Ò$ß &Ó by Ò#ß "!Ó


Section 1.4 Graphing with Calculators and Computers 25. Ò0Þ03ß 0Þ03Ó by Ò1Þ25ß 1Þ25Ó

26. Ò0Þ1ß 0Þ1Ó by Ò3ß 3Ó

27. Ò300ß 300Ó by Ò1Þ25ß 1Þ25Ó

28. Ò50ß 50Ó by Ò0Þ1ß 0Þ1Ó

29. Ò0Þ25ß 0Þ25Ó by Ò0Þ3ß 0Þ3Ó

30. Ò0Þ15ß 0Þ15Ó by Ò0Þ02ß 0Þ05Ó

31. x# #x œ % %y y# Ê y œ # „ Èx# #x ). The lower half is produced by graphing y œ # Èx# #x ).

32. y# "'x# œ " Ê y œ „ È" "'x# . The upper branch is produced by graphing y œ È" "'x# .


29

30

Chapter 1 Functions

33.

34.

35.

36.

37.

38Þ

39.

40.

CHAPTER 1 PRACTICE EXERCISES 1. The area is A œ 1 r# and the circumference is C œ #1 r. Thus, r œ 2. The surface area is S œ %1 r# Ê r œ ˆ %S1 ‰ surface area gives S œ %1 r# œ %1 ˆ $%V1 ‰

"Î#

#Î$

C #1

#

Ê A œ 1ˆ #C1 ‰ œ

C# %1 .

$ $V . The volume is V œ %$ 1 r$ Ê r œ É %1 . Substitution into the formula for

.


Chapter 1 Practice Exercises 3. The coordinates of a point on the parabola are axß x# b. The angle of inclination ) joining this point to the origin satisfies the equation tan ) œ 4. tan ) œ

rise run

œ

h &!!

x# x

œ x. Thus the point has coordinates axß x# b œ atan )ß tan# )b.

Ê h œ &!! tan ) ft.

5.

6.

Symmetric about the origin.

Symmetric about the y-axis.

7.

8.

Neither

Symmetric about the y-axis.

9. yaxb œ axb# " œ x# " œ yaxb. Even. 10. yaxb œ axb& axb$ axb œ x& x$ x œ yaxb. Odd. 11. yaxb œ " cosaxb œ " cos x œ yaxb. Even. 12. yaxb œ secaxb tanaxb œ 13. yaxb œ

axb% " axb$ #axb

œ

x% " x$ #x

sinaxb cos# axb

œ

sin x cos# x

œ sec x tan x œ yaxb. Odd.

%

" œ xx$ # x œ yaxb. Odd.

14. yaxb œ axb sinaxb œ axb sin x œ ax sin xb œ yaxb. Odd. 15. yaxb œ x cosaxb œ x cos x. Neither even nor odd. 16. yaxb œ axbcosaxb œ x cos x œ yaxb. Odd. 17. Since f and g are odd Ê faxb œ faxb and gaxb œ gaxb. (a) af † gbaxb œ faxbgaxb œ ÒfaxbÓÒgaxbÓ œ faxbgaxb œ af † gbaxb Ê f † g is even (b) f 3 axb œ faxbfaxbfaxb œ ÒfaxbÓÒfaxbÓÒfaxbÓ œ faxb † faxb † faxb œ f 3 axb Ê f 3 is odd. (c) fasinaxbb œ fasinaxbb œ fasinaxbb Ê fasinaxbb is odd. (d) gasecaxbb œ gasecaxbb Ê gasecaxbb is even. (e) lgaxbl œ lgaxbl œ lgaxbl Ê lgl is evenÞ


31

32

Chapter 1 Functions

18. Let faa xb œ faa xb and define gaxb œ fax ab. Then gaxb œ faaxb ab œ faa xb œ faa xb œ fax ab œ gaxb Ê gaxb œ fax ab is even. 19. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) Since l x l attains all nonnegative values, the range is Ò#ß _Ñ. 20. (a) Since the square root requires " x !, the domain is Ð_ß "Ó. (b) Since È" x attains all nonnegative values, the range is Ò#ß _Ñ. 21. (a) Since the square root requires "' x# !, the domain is Ò%ß %Ó. (b) For values of x in the domain, ! Ÿ "' x# Ÿ "', so ! Ÿ È"' x# Ÿ %. The range is Ò!ß %Ó. 22. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) Since $#x attains all positive values, the range is a"ß _b. 23. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) Since #ex attains all positive values, the range is a$ß _b. 24. (a) The function is equivalent to y œ tan #x, so we require #x Á

k1 #

for odd integers k. The domain is given by x Á

k1 %

for

odd integers k. (b) Since the tangent function attains all values, the range is a_ß _b. 25. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) The sine function attains values from " to ", so # Ÿ #sina$x 1b Ÿ # and hence $ Ÿ #sina$x 1b " Ÿ ". The range is Ò3ß 1Ó. 26. (a) The function is defined for all values of x, so the domain is a_ß _b. & (b) The function is equivalent to y œ È x# , which attains all nonnegative values. The range is Ò!ß _Ñ. 27. (a) The logarithm requires x $ !, so the domain is a$ß _b. (b) The logarithm attains all real values, so the range is a_ß _b. 28. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) The cube root attains all real values, so the range is a_ß _b. 29. (a) (b) (c) (d)

Increasing because volume increases as radius increases Neither, since the greatest integer function is composed of horizontal (constant) line segments Decreasing because as the height increases, the atmospheric pressure decreases. Increasing because the kinetic (motion) energy increases as the particles velocity increases.

30. (a) Increasing on Ò2, _Ñ (c) Increasing on a_, _b

(b) Increasing on Ò1, _Ñ (d) Increasing on Ò "# , _Ñ

31. (a) The function is defined for % Ÿ x Ÿ %, so the domain is Ò%ß %Ó. (b) The function is equivalent to y œ Èl x l, % Ÿ x Ÿ %, which attains values from ! to # for x in the domain. The range is Ò!ß #Ó.


Chapter 1 Practice Exercises

33

32. (a) The function is defined for # Ÿ x Ÿ #, so the domain is Ò#ß #Ó. (b) The range is Ò"ß "Ó. 33. First piece: Line through a!ß "b and a"ß !b. m œ Second piece: Line through a"ß "b and a#ß !b. m faxb œ œ

" x, ! Ÿ x " # x, " Ÿ x Ÿ #

34. First piece: Line through a!ß !b and a2ß 5b. m œ Second piece: Line through a2ß 5b and a4ß !b. m faxb œ

10

5 2 x, 5x 2 ,

!" " "! œ " œ " " œ !# " œ "

" Ê y œ x " œ " x œ " Ê y œ ax "b " œ x # œ # x

5! 5 5 2! œ 2 Ê y œ 2x !5 5 œ 4 2 œ 2 œ 52 Ê

y œ 52 ax 2b 5 œ 52 x 10 œ 10

!Ÿx2 (Note: x œ 2 can be included on either piece.) 2ŸxŸ4

35. (a) af‰gba"b œ faga"bb œ fŠ È"" # ‹ œ fa"b œ (b) ag‰f ba#b œ gafa#bb œ gˆ "2 ‰ œ (c) af‰f baxb œ fafaxbb œ fˆ "x ‰ œ

" É "# #

" "Îx

œ

" È#Þ&

" "

œ"

or É &#

œ x, x Á !

(d) ag‰gbaxb œ gagaxbb œ gŠ Èx" # ‹ œ

"

" É Èx # #

œ

% x# È É " #È x #

$ 36. (a) af‰gba"b œ faga"bb œ fˆÈ " "‰ œ fa!b œ # ! œ # $ (b) ag‰f ba#b œ faga#bb œ ga# #b œ ga!b œ È !"œ"

(c) af‰f baxb œ fafaxbb œ fa# xb œ # a# xb œ x $ $ $ È (d) ag‰gbaxb œ gagaxbb œ gˆÈ x "‰ œ É x""

#

37. (a) af‰gbaxb œ fagaxbb œ fˆÈx #‰ œ # ˆÈx #‰ œ x, x #. ag‰f baxb œ fagaxbb œ ga# x# b œ Èa# x# b # œ È% x# (b) Domain of f‰g: Ò#ß _ÑÞ Domain of g‰f: Ò#ß #ÓÞ

(c) Range of f‰g: Ð_ß #ÓÞ Range of g‰f: Ò!ß #ÓÞ

% 38. (a) af‰gbaxb œ fagaxbb œ fŠÈ" x‹ œ ÉÈ" x œ È " x.

ag‰f baxb œ fagaxbb œ gˆÈx‰ œ É" Èx (b) Domain of f‰g: Ð_ß "ÓÞ Domain of g‰f: Ò!ß "ÓÞ 39.

y œ faxb

(c) Range of f‰g: Ò!ß _ÑÞ Range of g‰f: Ò!ß "ÓÞ y œ af‰f baxb


5x 2

34

Chapter 1 Functions

40.

41.

42.

The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis. 43.

It does not change the graph.

44.

Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis.

Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis.


Chapter 1 Practice Exercises 45.

46.

The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis.

Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis. 47.

48.

The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis. 49. (a) y œ gax 3b (c) y œ gaxb (e) y œ 5 † gaxb

" #

The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis. (b) y œ gˆx 3# ‰ 2 (d) y œ gaxb (f) y œ ga5xb

50. (a) Shift the graph of f right 5 units (b) Horizontally compress the graph of f by a factor of 4 (c) Horizontally compress the graph of f by a factor of 3 and a then reflect the graph about the y-axis (d) Horizontally compress the graph of f by a factor of 2 and then shift the graph left "# unit. (e) Horizontally stretch the graph of f by a factor of 3 and then shift the graph down 4 units. (f) Vertically stretch the graph of f by a factor of 3, then reflect the graph about the x-axis, and finally shift the graph up "4 unit. 51. Reflection of the grpah of y œ Èx about the x-axis followed by a horizontal compression by a factor of 1 2 then a shift left 2 units.


35

36

Chapter 1 Functions

52. Reflect the graph of y œ x about the x-axis, followed by a vertical compression of the graph by a factor of 3, then shift the graph up 1 unit.

53. Vertical compression of the graph of y œ

1 x2

by a

factor of 2, then shift the graph up 1 unit.

54. Reflect the graph of y œ x1Î3 about the y-axis, then compress the graph horizontally by a factor of 5.

55.

56.

period œ 1 57.

period œ 41 58.

period œ 2

period œ 4


Chapter 1 Practice Exercises 59.

60.

period œ 21

period œ 21 1 3

61. (a) sin B œ sin

œ

b c

œ

b #

Ê b œ 2 sin

1 3

œ 2Š

È3 # ‹

œ È3. By the theorem of Pythagoras,

a# b# œ c# Ê a œ Èc# b# œ È4 3 œ 1. 1 3

(b) sin B œ sin

œ

b c

œ

2 c

Ê cœ

2 sin 13

œ È23 œ Š ‹ #

4 È3

#

. Thus, a œ Èc# b# œ ÊŠ È43 ‹ (2)# œ É 34 œ

62. (a) sin A œ

a c

Ê a œ c sin A

(b) tan A œ

a b

Ê a œ b tan A

63. (a) tan B œ

b a

Ê aœ

(b) sin A œ

a c

Ê cœ

64. (a) sin A œ

a c

(c) sin A œ

a c

b tan B

œ

a sin A

È c # b # c

65. Let h œ height of vertical pole, and let b and c denote the distances of points B and C from the base of the pole, measured along the flatground, respectively. Then, tan 50° œ hc , tan 35° œ hb , and b c œ 10. Thus, h œ c tan 50° and h œ b tan 35° œ (c 10) tan 35° Ê c tan 50° œ (c 10) tan 35° Ê c (tan 50° tan 35°) œ 10 tan 35° tan 35° Ê c œ tan10 50°tan 35° Ê h œ c tan 50° œ

10 tan 35° tan 50° tan 50°tan 35°

¸ 16.98 m.

66. Let h œ height of balloon above ground. From the figure at the right, tan 40° œ ha , tan 70° œ hb , and a b œ 2. Thus, h œ b tan 70° Ê h œ (2 a) tan 70° and h œ a tan 40° Ê (2 a) tan 70° œ a tan 40° Ê a(tan 40° tan 70°) 70° œ 2 tan 70° Ê a œ tan 240°tantan 70° Ê h œ a tan 40° œ

2 tan 70° tan 40° tan 40°tan 70°

¸ 1.3 km.

67. (a)

(b) The period appears to be 41.


2 È3

.

37

38

Chapter 1 Functions (c) f(x 41) œ sin (x 41) cos ˆ x#41 ‰ œ sin (x 21) cos ˆ x# 21‰ œ sin x cos since the period of sine and cosine is 21. Thus, f(x) has period 41.

x #

68. (a)

(b) D œ (_ß 0) (!ß _); R œ [1ß 1] (c) f is not periodic. For suppose f has period p. Then f ˆ #"1 kp‰ œ f ˆ #"1 ‰ œ sin 21 œ 0 for all integers k. Choose k so large that

" #1

kp

" 1

Ê 0

" (1/21)kp

1. But then

f ˆ #"1 kp‰ œ sin Š (1/#1")kp ‹ 0 which is a contradiction. Thus f has no period, as claimed. CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES 1. There are (infinitely) many such function pairs. For example, f(x) œ 3x and g(x) œ 4x satisfy f(g(x)) œ f(4x) œ 3(4x) œ 12x œ 4(3x) œ g(3x) œ g(f(x)). 2. Yes, there are many such function pairs. For example, if g(x) œ (2x 3)$ and f(x) œ x"Î$ , then (f ‰ g)(x) œ f(g(x)) œ f a(2x 3)$ b œ a(2x 3)$ b

"Î$

œ 2x 3.

3. If f is odd and defined at x, then f(x) œ f(x). Thus g(x) œ f(x) 2 œ f(x) 2 whereas g(x) œ (f(x) 2) œ f(x) 2. Then g cannot be odd because g(x) œ g(x) Ê f(x) 2 œ f(x) 2 Ê 4 œ 0, which is a contradiction. Also, g(x) is not even unless f(x) œ 0 for all x. On the other hand, if f is even, then g(x) œ f(x) 2 is also even: g(x) œ f(x) 2 œ f(x) 2 œ g(x). 4. If g is odd and g(0) is defined, then g(0) œ g(0) œ g(0). Therefore, 2g(0) œ 0 Ê g(0) œ 0. 5. For (xß y) in the 1st quadrant, kxk kyk œ 1 x Í x y œ 1 x Í y œ 1. For (xß y) in the 2nd quadrant, kxk kyk œ x 1 Í x y œ x 1 Í y œ 2x 1. In the 3rd quadrant, kxk kyk œ x 1 Í x y œ x 1 Í y œ 2x 1. In the 4th quadrant, kxk kyk œ x 1 Í x (y) œ x 1 Í y œ 1. The graph is given at the right.


Chapter 1 Additional and Advanced Exercises 6. We use reasoning similar to Exercise 5. (1) 1st quadrant: y kyk œ x kxk Í 2y œ 2x Í y œ x. (2) 2nd quadrant: y kyk œ x kxk Í 2y œ x (x) œ 0 Í y œ 0. (3) 3rd quadrant: y kyk œ x kxk Í y (y) œ x (x) Í 0 œ 0 Ê all points in the 3rd quadrant satisfy the equation. (4) 4th quadrant: y kyk œ x kxk Í y (y) œ 2x Í 0 œ x. Combining these results we have the graph given at the right: 7. (a) sin# x cos# x œ 1 Ê sin# x œ 1 cos# x œ (1 cos x)(1 cos x) Ê (1 cos x) œ Ê

1cos x sin x

œ

sin# x 1cos x

sin x 1cos x

(b) Using the definition of the tangent function and the double angle formulas, we have #

tan ˆ x# ‰ œ

sin# ˆ x# ‰ cos# ˆ #x ‰

œ

"

x ‹‹ cos Š2 Š #

#

"cos Š2 Š #x ‹‹ #

œ

1cos x 1cos x

.

8. The angles labeled # in the accompanying figure are equal since both angles subtend arc CD. Similarly, the two angles labeled ! are equal since they both subtend arc AB. Thus, triangles AED and BEC are similar which ) b implies ab c œ 2a cos a c Ê (a c)(a c) œ b(2a cos ) b) Ê a# c# œ 2ab cos ) b# Ê c# œ a# b# 2ab cos ).

9. As in the proof of the law of sines of Section 1.3, Exercise 61, ah œ bc sin A œ ab sin C œ ac sin B Ê the area of ABC œ "# (base)(height) œ "# ah œ "# bc sin A œ "# ab sin C œ "# ac sin B. 10. As in Section 1.3, Exercise 61, (Area of ABC)# œ œ

" 4

(base)# (height)# œ

" 4

a# h # œ

" 4

a# b# sin# C

a# b# a" cos# Cb . By the law of cosines, c# œ a# b# 2ab cos C Ê cos C œ

Thus, (area of ABC)# œ œ

" 4

" 16

" 4

a# b# a" cos# Cb œ #

Š4a# b# aa# b# c# b ‹ œ

" 16

" 4

a# b# Œ" Š a

#

b# c# ‹ #ab

#

œ

a# b# 4

a# b# c# 2ab

Š"

.

aa # b # c # b 4a# b#

#

‹

ca2ab aa# b# c# bb a2ab aa# b# c# bbd

" ca(a b)# c# b ac# (a b)# bd œ 16 c((a b) c)((a b) c)(c (a b))(c (a b))d a b c a b c a b c a b c œ ˆ # ‰ ˆ # ‰ ˆ # ‰ ˆ # ‰‘ œ s(s a)(s b)(s c), where s œ a#bc .

œ

" 16

Therefore, the area of ABC equals Ès(s a)(s b)(s c) . 11. If f is even and odd, then f(x) œ f(x) and f(x) œ f(x) Ê f(x) œ f(x) for all x in the domain of f. Thus 2f(x) œ 0 Ê f(x) œ 0.


39

40

Chapter 1 Functions f(x) f((x)) œ f(x) #f(x) œ E(x) Ê E # even function. Define O(x) œ f(x) E(x) œ f(x) f(x) #f(x) œ f(x) #f(x) . Then O(x) œ f(x) #f((x)) œ f(x)# f(x) œ Š f(x) #f(x) ‹ œ O(x) Ê O is an odd function

12. (a) As suggested, let E(x) œ

f(x) f(x) #

Ê E(x) œ

is an

Ê f(x) œ E(x) O(x) is the sum of an even and an odd function. (b) Part (a) shows that f(x) œ E(x) O(x) is the sum of an even and an odd function. If also f(x) œ E" (x) O" (x), where E" is even and O" is odd, then f(x) f(x) œ 0 œ aE" (x) O" (x)b (E(x) O(x)). Thus, E(x) E" (x) œ O" (x) O(x) for all x in the domain of f (which is the same as the domain of E E" and O O" ). Now (E E" )(x) œ E(x) E" (x) œ E(x) E" (x) (since E and E" are even) œ (E E" )(x) Ê E E" is even. Likewise, (O" O)(x) œ O" (x) O(x) œ O" (x) (O(x)) (since O and O" are odd) œ (O" (x) O(x)) œ (O" O)(x) Ê O" O is odd. Therefore, E E" and O" O are both even and odd so they must be zero at each x in the domain of f by Exercise 11. That is, E" œ E and O" œ O, so the decomposition of f found in part (a) is unique. 13. y œ ax# bx c œ a Šx# ba x

b# 4a# ‹

b# 4a

c œ a ˆx

b ‰# 2a

b# 4a

c

(a) If a 0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift of the vertex toward the y-axis and upward. If a 0 the graph is a parabola that opens downward. Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward. (b) If a 0 the graph is a parabola that opens upward. If also b 0, then increasing b causes a shift of the graph downward to the left; if b 0, then decreasing b causes a shift of the graph downward and to the right. If a 0 the graph is a parabola that opens downward. If b 0, increasing b shifts the graph upward to the right. If b 0, decreasing b shifts the graph upward to the left. (c) Changing c (for fixed a and b) by ?c shifts the graph upward ?c units if ?c 0, and downward ?c units if ?c 0. 14. (a) If a 0, the graph rises to the right of the vertical line x œ b and falls to the left. If a 0, the graph falls to the right of the line x œ b and rises to the left. If a œ 0, the graph reduces to the horizontal line y œ c. As kak increases, the slope at any given point x œ x! increases in magnitude and the graph becomes steeper. As kak decreases, the slope at x! decreases in magnitude and the graph rises or falls more gradually. (b) Increasing b shifts the graph to the left; decreasing b shifts it to the right. (c) Increasing c shifts the graph upward; decreasing c shifts it downward. 15. Each of the triangles pictured has the same base b œ v?t œ v(1 sec). Moreover, the height of each triangle is the same value h. Thus "# (base)(height) œ

" #

bh

œ A" œ A# œ A$ œ á . In conclusion, the object sweeps out equal areas in each one second interval.

16. (a) Using the midpoint formula, the coordinates of P are ˆ a# 0 ß b# 0 ‰ œ ˆ #a ß b# ‰ . Thus the slope of OP œ

?y ?x

œ

b/2 a/2

œ

b a

.


Chapter 1 Additional and Advanced Exercises (b) The slope of AB œ

b 0 0 a

41

œ ba . The line segments AB and OP are perpendicular when the product #

of their slopes is " œ ˆ ba ‰ ˆ ba ‰ œ ba# . Thus, b# œ a# Ê a œ b (since both are positive). Therefore, AB is perpendicular to OP when a œ b. 17. From the figure we see that 0 Ÿ ) Ÿ cos ) œ

and AB œ AD œ 1. From trigonometry we have the following: sin ) œ sin ) cos ) .

œ CD, and tan ) œ œ We can see that: w " area ÃEB area sector DB area ÃDC Ê # aAEbaEBb "# aADb2 ) "# aADbaCDb AE AB

œ AE, tan ) œ

1 2

CD AD

EB AE

Ê "# sin ) cos ) "# a"b2 ) "# a"batan )b Ê "# sin ) cos ) "# )

" sin ) # cos )

18. af‰gbaxb œ fagaxbb œ aacx db b œ acx ad b and ag‰f baxb œ gafaxbb œ caax bb d œ acx cb d Thus af‰gbaxb œ ag‰f baxb Ê acx ad b œ acx bc d Ê ad b œ bc d. Note that fadb œ ad b and gabb œ cb d, thus af‰gbaxb œ ag‰f baxb if fadb œ gabb.


EB AB

œ EB,

42

Chapter 1 Functions

NOTES:


CHAPTER 2 LIMITS AND CONTINUITY 2.1 RATES OF CHANGE AND TANGENTS TO CURVES 1. (a)

?f ?x

œ

f(3) f(2) 3#

2. (a)

?g ?x

œ

g(1) g(1) 1 (1)

3. (a)

?h ?t

œ

h ˆ 341 ‰ h ˆ 14 ‰ 1 31 4 4

œ

?g ?t

œ

g(1) g(0) 10

(2 1) (2 1) 10

4. (a) 5.

?R ?)

œ

R(2) R(0) 20

6.

?P ?)

œ

P(2) P(1) 21

7. (a)

?y ?x

œ

?y ?x

œ

?y ?x

œ

?y ?x

œ

?y ?x

œ

?y ?x

œ

?y ?x

œ

œ œ

28 9 1

œ

œ

œ

1 1 2

œ 19

(b)

?f ?x

œ

f(1) f(") 1 (1)

œ

20 #

œ1

œ0

(b)

?g ?x

œ

g(0)g(2) 0(2)

œ

04 #

œ 2

(b)

?h ?t

œ

h ˆ 1# ‰ h ˆ 16 ‰ 11 # 6

œ

?g ?t

œ

g(1) g(1) 1 (1)

œ

1 1 1 #

È 8 1 È 1 #

œ 14 œ 12

3" #

œ

(8 16 10)(" % &) 1

â2 h b2 3 ‰ ˆ 2 2 3 ‰ h

œ

(b)

0 È3 1 3

œ

3 È 3 1

(2 1) (2 ") #1

œ0

œ1 œ22œ0

4 4h h2 3 1 h

œ

4h h2 h

œ 4 h. As h Ä 0, 4 h Ä 4 Ê at Pa2, 1b the slope is 4.

(b) y 1 œ 4ax 2b Ê y 1 œ 4x 8 Ê y œ 4x 7 8. (a)

ˆ 5 a1 h b 2 ‰ ˆ 5 1 2 ‰ h

œ

5 1 2h h2 4 h

œ

2h h2 h

œ 2 h. As h Ä 0, 2 h Ä 2 Ê at Pa1, 4b the

slope is 2. (b) y 4 œ a2bax 1b Ê y 4 œ 2x 2 Ê y œ 2x 6 9. (a)

â2 h b2 2 a 2 h b 3 ‰ ˆ 2 2 2 a 2 b 3 ‰ h

œ

4 4h h2 4 2h 3 a3b h

œ

2h h2 h

œ 2 h. As h Ä 0, 2 h Ä 2 Ê at

Pa2, 3b the slope is 2. (b) y a3b œ 2ax 2b Ê y 3 œ 2x 4 Ê y œ 2x 7. 10. (a)

â1 h b2 4 a 1 h b ‰ ˆ 1 2 4 a 1 b ‰ h

œ

1 2h h2 4 4h a3b h

œ

h2 2h h

œ h 2. As h Ä 0, h 2 Ä 2 Ê at

Pa1, 3b the slope is 2. (b) y a3b œ a2bax 1b Ê y 3 œ 2x 2 Ê y œ 2x 1. 11. (a)

a2 h b 3 2 3 h

œ

8 12h 4h2 h3 8 h

œ

12h 4h2 h3 h

œ 12 4h h2 . As h Ä 0, 12 4h h2 Ä 12, Ê at

Pa2, 8b the slope is 12. (b) y 8 œ 12ax 2b Ê y 8 œ 12x 24 Ê y œ 12x 16. 12. (a)

2 a1 h b3 ˆ 2 1 3 ‰ h

œ

2 1 3h 3h2 h3 1 h

œ

3h 3h2 h3 h

œ 3 3h h2 . As h Ä 0, 3 3h h2 Ä 3, Ê at

Pa1, 1b the slope is 3. (b) y 1 œ a3bax 1b Ê y 1 œ 3x 3 Ê y œ 3x 4. 13. (a)

a1 hb3 12a1 hb ˆ13 12a"b‰ h 2

œ

1 3h 3h2 h3 12 12h a11b h

œ

9h 3h2 h3 h

œ 9 3h h2 . As h Ä 0,

9 3h h Ä 9 Ê at Pa1, 11b the slope is 9. (b) y a11b œ a9bax 1b Ê y 11 œ 9x 9 Ê y œ 9x 2.


44

Chapter 2 Limits and Continuity

14. (a)

?y ?x

œ

a2 h b 3 3 a 2 h b 2 4 ˆ 2 3 3 a 2 b 2 4 ‰ h 2

8 12h 6h2 h3 12 12h 3h2 4 0 h

œ

œ

3h2 h3 h

œ 3h h2 . As h Ä 0,

3h h Ä 0 Ê at Pa2, 0b the slope is 0. (b) y 0 œ 0ax 2b Ê y œ 0. 15. (a)

?p ?t

Slope of PQ œ

Q

650 225 20 10 650 375 20 14 650 475 20 16.5 650 550 20 18

Q" (10ß 225) Q# (14ß 375) Q$ (16.5ß 475) Q% (18ß 550)

œ 42.5 m/sec œ 45.83 m/sec œ 50.00 m/sec œ 50.00 m/sec

(b) At t œ 20, the sportscar was traveling approximately 50 m/sec or 180 km/h. 16. (a)

Slope of PQ œ

Q Q" (5ß 20) Q# (7ß 39) Q$ (8.5ß 58) Q% (9.5ß 72)

80 20 10 5 80 39 10 7 80 58 10 8.5 80 72 10 9.5

?p ?t

œ 12 m/sec œ 13.7 m/sec œ 14.7 m/sec œ 16 m/sec

(b) Approximately 16 m/sec 17. (a)

(b)

?p ?t

œ

174 62 2004 2002

œ

112 #

œ 56 thousand dollars per year

(c) The average rate of change from 2001 to 2002 is ??pt œ

62 27 20022 2001 œ 35 thousand dollars per year. 111 62 The average rate of change from 2002 to 2003 is ??pt œ 2003 2002 œ 49 thousand dollars per year. So, the rate at which profits were changing in 2002 is approximatley "# a35 49b œ 42 thousand dollars

18. (a) F(x) œ (x 2)/(x 2) x 1.2 F(x) 4.0 ?F ?x ?F ?x ?F ?x

œ

?g ?x ?g ?x

œ

œ œ

1.1 3.4

1.01 3.04

1.001 3.004

1.0001 3.0004

1 3

4.0 (3) œ 5.0; 1.2 1 3.04 (3) œ 4.04; 1.01 1 3.!!!% (3) œ 4.!!!%; 1.0001 1

?F ?x ?F ?x

œ œ

3.4 (3) œ 4.4; 1.1 1 3.004 (3) œ 4.!!%; 1.001 1

È g(2) g(1) œ #21" ¸ 0.414213 21 È1 h" g(1 h) g(1) (1 h) 1 œ h

?g ?x

œ

g(1.5) g(1) 1.5 1

(b) The rate of change of F(x) at x œ 1 is 4. 19. (a)

œ

œ

È1.5 " 0.5

¸ 0.449489

(b) g(x) œ Èx 1h È1 h ŠÈ1 h 1‹ /h

1.1 1.04880

1.01 1.004987

1.001 1.0004998

1.0001 1.0000499

1.00001 1.000005

1.000001 1.0000005

0.4880

0.4987

0.4998

0.499

0.5

0.5

(c) The rate of change of g(x) at x œ 1 is 0.5.


per year.

Section 2.1 Rates of Change and Tangents to Curves (d) The calculator gives lim hÄ! 20. (a) i) ii) (b)

f(3) f(2) 32

œ

f(T) f(2) T#

œ

"" 3 #

1 " " T # T#

T f(T) af(T) f(2)b/aT 2b

"

œ

6

1

È1 h" h

œ "# .

œ "6

#TT T#

2 #T

œ

45

œ

2T #T(T 2)

2.1 0.476190 0.2381

œ

2T #T(2 T)

2.01 0.497512 0.2488

œ #"T , T Á 2

2.001 0.499750 0.2500

2.0001 0.4999750 0.2500

2.00001 0.499997 0.2500

2.000001 0.499999 0.2500

(c) The table indicates the rate of change is 0.25 at t œ 2. " ‰ (d) lim ˆ #T œ 4" TÄ#

NOTE: Answers will vary in Exercises 21 and 22. s 15 0 ˜s 20 15 10 ˜s 30 20 21. (a) Ò0, 1Ó: ˜ ˜t œ 1 0 œ 15 mphà Ò1, 2.5Ó: ˜t œ 2.5 1 œ 3 mphà Ò2.5, 3.5Ó: ˜t œ 3.5 2.5 œ 10 mph (b) At Pˆ "# , 7.5‰: Since the portion of the graph from t œ 0 to t œ 1 is nearly linear, the instantaneous rate of change

will be almost the same as the average rate of change, thus the instantaneous speed at t œ

" #

is

15 7.5 1 0.5

œ 15 mi/hr.

At Pa2, 20b: Since the portion of the graph from t œ 2 to t œ 2.5 is nearly linear, the instantaneous rate of change will 20 be nearly the same as the average rate of change, thus v œ 20 2.5 2 œ 0 mi/hr. For values of t less than 2, we have Slope of PQ œ

Q

?s ?t

15 20 1 2 œ 5 mi/hr 19 20 1.5 2 œ 2 mi/hr 19.9 20 1.9 2 œ 1 mi/hr

Q" (1ß 15) Q# (1.5ß 19) Q$ (1.9ß 19.9)

Thus, it appears that the instantaneous speed at t œ 2 is 0 mi/hr. At Pa3, 22b: s Q Slope of PQ œ ? ?t 35 22 43 30 22 3.5 3 23 22 3.1 3

Q" (4ß 35) Q# (3.5ß 30) Q$ (3.1ß 23)

Slope of PQ œ

Q

œ 13 mi/hr

Q" (2ß 20)

œ 16 mi/hr

Q# (2.5ß 20)

œ 10 mi/hr

Q$ (2.9ß 21.6)

20 22 2 3 œ 2 mi/hr 20 22 2.5 3 œ 4 mi/hr 21.6 22 2.9 3 œ 4 mi/hr

Thus, it appears that the instantaneous speed at t œ 3 is about 7 mi/hr. (c) It appears that the curve is increasing the fastest at t œ 3.5. Thus for Pa3.5, 30b s Q Slope of PQ œ ? Q Slope of PQ œ ?t 35 30 4 3.5 œ 10 mi/hr 34 30 3.75 3.5 œ 16 mi/hr 32 30 3.6 3.5 œ 20 mi/hr

Q" (4ß 35) Q# (3.75ß 34) Q$ (3.6ß 32)

Ã ˜t

œ

10 15 30

(b) At Pa1, 14b: Q Q" (2ß 12.2) Q# (1.5ß 13.2) Q$ (1.1ß 13.85)

¸ 1.67

gal day à

Ò0, 5Ó:

Q# (3.25ß 25) Q$ (3.4ß 28)

Slope of PQ œ

Ã ˜t

?A ?t

12.2 14 2 1 œ 1.8 gal/day 13.2 14 1.5 1 œ 1.6 gal/day 13.85 14 1.1 1 œ 1.5 gal/day

œ

3.9 15 50

¸ 2.2

gal day à Ò7,

10Ó:

?s ?t

22 30 3 3.5 œ 16 mi/hr 25 30 3.25 3.5 œ 20 mi/hr 28 30 3.4 3.5 œ 20 mi/hr

Q" (3ß 22)

Thus, it appears that the instantaneous speed at t œ 3.5 is about 20 mi/hr. 22. (a) Ò0, 3Ó:

?s ?t

Ã ˜t

Q Q" (0ß 15) Q# (0.5ß 14.6) Q$ (0.9ß 14.86)

œ

0 1.4 10 7

¸ 0.5

Slope of PQ œ

gal day ?A ?t

15 14 0 1 œ 1 gal/day 14.6 14 0.5 1 œ 1.2 gal/day 14.86 14 0.9 1 œ 1.4 gal/day

Thus, it appears that the instantaneous rate of consumption at t œ 1 is about 1.45 gal/day. At Pa4, 6b:


46

Chapter 2 Limits and Continuity Q Q" (5ß 3.9) Q# (4.5ß 4.8) Q$ (4.1ß 5.7)

Slope of PQ œ 3.9 6 54 4.8 6 4.5 4 5.7 6 4.1 4

?A ?t

Q

œ 2.1 gal/day

Q" (3ß 10)

œ 2.4 gal/day

Q# (3.5ß 7.8)

œ 3 gal/day

Q$ (3.9ß 6.3)

Slope of PQ œ

10 6 3 4 œ 4 gal/day 7.8 6 3.5 4 œ 3.6 gal/day 6.3 6 3.9 4 œ 3 gal/day

Thus, it appears that the instantaneous rate of consumption at t œ 1 is 3 gal/day. At Pa8, 1b: Q Slope of PQ œ ??At Q Slope of PQ œ Q" (9ß 0.5) Q# (8.5ß 0.7) Q$ (8.1ß 0.95)


Q" (7ß 1.4)

4.8 7.8 4.5 3.5 œ 3 gal/day 6 7.8 4 3.5 œ 3.6 gal/day 7.4 7.8 3.6 3.5 œ 4 gal/day

Q" (2.5ß 11.2)

Q# (7.5ß 1.3) Q$ (7.9ß 1.04)

?A ?t

?A ?t


Thus, it appears that the instantaneous rate of consumption at t œ 1 is 0.55 gal/day. (c) It appears that the curve (the consumption) is decreasing the fastest at t œ 3.5. Thus for Pa3.5, 7.8b s Q Slope of PQ œ ??At Q Slope of PQ œ ? ?t Q" (4.5ß 4.8) Q# (4ß 6) Q$ (3.6ß 7.4)

Q# (3ß 10) Q$ (3.4ß 8.2)

Thus, it appears that the rate of consumption at t œ 3.5 is about 4 gal/day.

11.2 7.8 2.5 3.5 œ 3.4 gal/day 10 7.8 3 3.5 œ 4.4 gal/day 8.2 7.8 3.4 3.5 œ 4 gal/day

2.2 LIMIT OF A FUNCTION AND LIMIT LAWS 1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x Ä 1. (b) 1 (c) 0 (d) 0.5 2. (a) 0 (b) 1 (c) Does not exist. As t approaches 0 from the left, f(t) approaches 1. As t approaches 0 from the right, f(t) approaches 1. There is no single number L that f(t) gets arbitrarily close to as t Ä 0. (d) 1 3. (a) True (d) False (g) True

(b) True (e) False

(c) False (f) True

4. (a) False (d) True

(b) False (e) True

(c) True

5.

x

lim x Ä 0 kx k x kx k

does not exist because

x kx k

œ

x x

œ 1 if x 0 and

approaches 1. As x approaches 0 from the right,

x kx k

x kxk

œ

x x

œ 1 if x 0. As x approaches 0 from the left,

approaches 1. There is no single number L that all the

function values get arbitrarily close to as x Ä 0. 6. As x approaches 1 from the left, the values of

" x 1

become increasingly large and negative. As x approaches 1

from the right, the values become increasingly large and positive. There is no one number L that all the function values get arbitrarily close to as x Ä 1, so lim x" 1 does not exist. xÄ1


Section 2.2 Limit of a Function and Limit Laws

47

7. Nothing can be said about f(x) because the existence of a limit as x Ä x! does not depend on how the function is defined at x! . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when x is close enough to x! . That is, the existence of a limit depends on the values of f(x) for x near x! , not on the definition of f(x) at x! itself. 8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near 0 regardless of the xÄ0

value f(0) itself. 9. No, the definition does not require that f be defined at x œ 1 in order for a limiting value to exist there. If f(1) is defined, it can be any real number, so we can conclude nothing about f(1) from lim f(x) œ 5. xÄ1

10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If lim f(x) exists, its value may be some number other than f(1) œ 5. We can conclude nothing about lim f(x), xÄ1

xÄ1

whether it exists or what its value is if it does exist, from knowing the value of f(1) alone. 11.

lim (2x 5) œ 2(7) 5 œ 14 5 œ 9

x Ä (

12. lim ax# 5x 2b œ (2)# 5(2) 2 œ 4 10 2 œ 4 xÄ#

13. lim 8(t 5)(t 7) œ 8(6 5)(6 7) œ 8 tÄ'

14.

lim ax$ 2x# 4x 8b œ (2)$ 2(2)# 4(2) 8 œ 8 8 8 8 œ 16

x Ä #

15. lim

x3

œ

x Ä # x6

17.

23 26

16. lim# 3s(2s 1) œ 3 ˆ 23 ‰ 2 ˆ 23 ‰ 1‘ œ 2 ˆ 43 1‰ œ

5 8

sÄ

$

lim 3(2x 1)# œ 3(2(1) 1)# œ 3(3)# œ 27

x Ä "

y2

18. lim

# y Ä # y 5y 6

19.

œ

œ

22 (2)# 5(#) 6

œ

4 4 10 6

œ

4 #0

œ

" 5

%

lim (5 y)%Î$ œ [5 (3)]%Î$ œ (8)%Î$ œ ˆ(8)"Î$ ‰ œ 2% œ 16

y Ä $

20. lim (2z 8)"Î$ œ (2(0) 8)"Î$ œ (8)"Î$ œ 2 zÄ!

21. lim

3

22. lim

È5h 4 2 h

h Ä ! È3h 1 1

hÄ0

œ

5 È4 2

23. lim

œ

x5

# x Ä & x 25

24.

œ

3 È3(0) 1 1

œ lim

hÄ0

œ

3 È1 1

È5h 4 2 h

†

œ

3 2

È5h 4 2 È5h 4 2

œ lim

a5h 4b 4

h Ä 0 hŠÈ5h 4 2‹

œ lim

5h

h Ä 0 hŠÈ5h 4 2‹

œ lim

5 4 x5

œ lim

x3

x Ä & (x 5)(x 5)

lim # x Ä $ x 4x 3

œ lim

œ lim

x3

1

x Ä & x5

x Ä $ (x 3)(x 1)

œ lim

œ

" 55

œ

" 10

1

œ

" 3 1

x Ä $ x 1

5

h Ä 0 È5h 4 2

œ "2


2 3

48 25.

Chapter 2 Limits and Continuity x# 3x "0 x5

lim

x Ä &

(x 5)(x 2) x5

œ lim

x Ä &

(x 5)(x 2) x2

26. lim

x# 7x "0 x#

œ lim

27. lim

t# t 2 t# 1

t Ä " (t 1)(t 1)

xÄ#

tÄ"

t# 3t 2 # t Ä " t t 2

29.

lim $ # x Ä # x 2x

lim

2x 4

5y$ 8y#

1 x 1 x x Ä 1 1

32. lim

xÄ0

33. lim

1 xc1

u% "

y# (5y 8)

œ lim

uÄ1

œ lim

4x x# x Ä % 2 Èx

œ lim

36. lim

lim

x Ä "

œ

(v 2) av# 2v 4b

Èx 3

x Ä * ˆÈ x 3 ‰ ˆ È x 3 ‰

x1 x Ä 1 Èx 3 2

37. lim

œ

Èx# 12 4 x2 xÄ2

x Ä "

œ lim

xÄ2

(x 2)(x 2)

x2 x Ä 2 È x # 5 3

lim

x Ä 2

8 16

1 ‰ x1

œ lim

xÄ1

œ lim

uÄ1

au# "b (u 1) u# u 1

"

œ lim

(x 2)(x 2)

444 (4)(8)

œ

x1

x2

œ 2

4 3

12 32

œ

3 8

xÄ%

œ lim ŠÈx 3 #‹ œ È4 2 œ 4 xÄ1

ax # 8 b *

œ lim

x Ä 1 (x 1) ŠÈx# 8 $‹

2 33

œ

(x 2) ŠÈx# 12 4‹

x Ä 2 Èx# 12 4

œ

2 1

œ lim x ˆ2 Èx‰ œ 4(2 2) œ 16

(x 1) ˆÈx 3 #‰ (x 3) 4 xÄ1

x Ä 1 È x # ) $

œ

œ

" 6

œ lim

(x 1) ŠÈx# 8 $‹

œ "3

ax# 12b 16

œ lim

x Ä 2 (x 2) ŠÈx# 12 4‹

œ

4 È16 4

ax 2b ŠÈx# 5 3‹

È x# 5 3 x2 x Ä 2

œ lim

" È9 3

œ

x Ä * Èx 3

(1 1)(1 1) 111

œ

# v Ä # (v 2) av 4b

x ˆ2 È x ‰ ˆ 2 È x ‰ 2 Èx xÄ%

œ lim

œ

v# 2v 4

œ lim

œ lim

œ lim

2

x Ä 1 ax 1bax 1b

x Ä 2 ŠÈx# 5 3‹ ŠÈx# 5 3‹

ax 2b ŠÈx# 5 3‹

œ #"

œ lim x1 œ 1

ŠÈx# 12 4‹ ŠÈx# 12 4‹

x Ä 2 (x 2) ŠÈx# 12 4‹

œ

œ

ŠÈx# 8 $‹ ŠÈx# 8 $‹

lim

(x 1)(x 1) lim x Ä 1 (x 1) ŠÈx# ) $‹

lim

5y 8

(x 1) ˆÈx 3 2‰ È ˆ x 3 #‰ ˆ È x 3 #‰ xÄ1

39. lim

40.

œ 21

œ lim

È x# 8 3 x1

œ lim

2 4

xÄ1

# v Ä # (v 2)(v 2) av 4b

x(4 x) x Ä % 2 Èx

œ

2

œ 13

1 œ lim Š ax 12x bax 1b † x ‹ œ lim

au# "b (u 1)(u 1) au# u 1b (u 1)

œ lim

Èx 3 x9

xÄ*

x

xÄ1

œ lim

35. lim

b 1b b ax c 1b c 1bax b 1b

ax

3 #

1 2 1 2

# y Ä ! 3y 16

xÄ1

œ

œ

# x Ä # x

œ lim ˆ 1 x x † ax

œ lim

t2

t Ä " t 2

œ lim

1cx x

12 11

œ

œ lim

œ lim

x Ä 1 x1

% v Ä # v 16

t2

2(x 2)

v$ 8

34. lim

xÄ#

œ lim

œ lim

$ u Ä 1 u 1

38.

t Ä " (t 2)(t 1)

# # y Ä ! y a3y 16b

x b1 1 x

œ lim (x 5) œ 2 5 œ 3

t Ä " t1

(t 2)(t 1)

œ lim

x Ä &

œ lim

# x Ä # x (x 2)

% # y Ä 0 3y 16y

31. lim

(t 2)(t 1)

œ lim

28.

30. lim

xÄ#

œ lim (x 2) œ & # œ 7

œ

œ

œ lim

" 2

ax 2b ŠÈx# 5 3‹

x Ä 2

È9 3 4

ax # 5 b 9

œ 23


Section 2.2 Limit of a Function and Limit Laws 41.

2 È x# 5 x3 x Ä 3

lim

œ

Š2 Èx# 5‹ Š2 Èx# 5‹

œ lim

(x 3) Š2 Èx# 5‹

x Ä 3

9 x#

lim

œ

x Ä 3 (x 3) Š2 Èx# 5‹

4x x Ä 4 5 È x# 9

œ lim

x Ä 4 Š5 Èx# 9‹ Š5 Èx# 9‹

a4 xb Š5 Èx# 9‹

xÄ4

x Ä 3 (x 3) Š2 Èx# 5‹

a4 xb Š5 Èx# 9‹

œ lim

42. lim

16 x#

x Ä 3 (x 3) Š2 Èx# 5‹

(3 x)(3 x)

lim

œ lim

(4 x)(4 x)

xÄ4

œ lim

3x

x Ä 3 2 È x # 5

œ

6 2 È4

œ

3 2

a4 xb Š5 Èx# 9‹ 25 ax# 9b

xÄ4

a4 xb Š5 Èx# 9‹

œ lim

4 ax # 5 b

œ lim

œ lim

xÄ4

5 È x# 9 4x

œ

5 È25 8

œ

5 4

2

43. lim a2sin x 1b œ 2sin 0 1 œ 0 1 œ 1

44. lim sin2 x œ Š lim sin x‹ œ asin 0b2 œ 02 œ 0

45. lim sec x œ

46. lim tan x œ

xÄ0

xÄ0

47. lim

xÄ0

1 x sin x 3cos x

1

lim x Ä 0 cos x œ

œ

1 0 sin 0 3cos 0

1 cos 0

œ

œ

1 1

xÄ0

œ1

100 3

œ

xÄ0

xÄ0

sin x

lim x Ä 0 cos x

œ

sin 0 cos 0

œ

0 1

œ0

1 3

48. lim ax2 1ba2 cos xb œ a02 1ba2 cos 0b œ a1ba2 1b œ a1ba1b œ 1 xÄ0

49. x Ä lim1Èx 4 cosax 1b œ x Ä lim1Èx 4 † x Ä lim1cosax 1b œ È1 4 † cos 0 œ È4 1 † 1 œ È4 1 50. lim È7 sec2 x œ É lim a7 sec2 xb œ É7 lim sec2 x œ È7 sec2 0 œ É7 a1b2 œ 2È2 xÄ0

xÄ0

xÄ0

51. (a) quotient rule (c) sum and constant multiple rules

(b) difference and power rules

52. (a) quotient rule (c) difference and constant multiple rules

(b) power and product rules

53. (a) xlim f(x) g(x) œ ’xlim f(x)“ ’ x lim g(x)“ œ (5)(2) œ 10 Äc Äc Äc (b) xlim 2f(x) g(x) œ 2 ’xlim f(x)“ ’ xlim g(x)“ œ 2(5)(2) œ 20 Äc Äc Äc (c) xlim [f(x) 3g(x)] œ xlim f(x) 3 xlim g(x) œ 5 3(2) œ 1 Äc Äc Äc lim f(x) f(x) 5 5 xÄc (d) xlim œ lim f(x) lim g(x) œ 5(2) œ 7 Ä c f(x) g(x) x

54. (a) (b) (c) (d) 55. (a) (b)

Äc

x

Äc

lim [g(x) 3] œ lim g(x) lim 3 œ $ $ œ !

xÄ%

xÄ%

xÄ%

lim xf(x) œ lim x † lim f(x) œ (4)(0) œ 0

xÄ%

xÄ%

xÄ%

#

lim [g(x)] œ ’ lim g(x)“ œ [3]# œ 9

xÄ%

#

g(x) x Ä % f(x) 1

lim

xÄ%

œ

Ä%

lim g(x)

x

lim f(x) lim 1

xÄ%

xÄ%

œ

3 01

œ3

lim [f(x) g(x)] œ lim f(x) lim g(x) œ 7 (3) œ 4

xÄb

xÄb

xÄb

lim f(x) † g(x) œ ’ lim f(x)“ ’ lim g(x)“ œ (7)(3) œ 21

xÄb

xÄb

xÄb


49

50

Chapter 2 Limits and Continuity lim 4g(x) œ ’ lim 4“ ’ lim g(x)“ œ (4)(3) œ 12

(c)

xÄb

xÄb

xÄb

xÄb

xÄb

7 3

œ 37

lim [p(x) r(x) s(x)] œ lim p(x) lim r(x) lim s(x) œ 4 0 (3) œ 1

56. (a)

x Ä #

x Ä #

x Ä #

x Ä #

lim p(x) † r(x) † s(x) œ ’ lim p(x)“ ’ lim r(x)“ ’ lim s(x)“ œ (4)(0)(3) œ 0

(b)

x Ä #

x Ä #

x Ä #

x Ä #

lim [4p(x) 5r(x)]/s(x) œ ’4 lim p(x) 5 lim r(x)“ ‚ lim s(x) œ [4(4) 5(0)]/3 œ

(c)

x Ä #

57. lim

hÄ!

x Ä #

(1 h)# 1# h

œ lim

hÄ!

(2 h)# (2)# h hÄ!

58. lim 59. lim

hÄ!

ˆ #" h ‰ ˆ "# ‰ h hÄ! È7 h È7 h hÄ!

61. lim

1 2h h# 1 h

œ lim

hÄ!

44hh# 4 h hÄ!

œ lim

[3(2 h) 4] [3(2) 4] h

60. lim

œ

xÄb

lim f(x)/g(x) œ lim f(x)/ lim g(x) œ

(d)

œ lim

2 2 h "

œ lim

h(2 h) h

œ lim

hÄ!

hÄ!

œ lim (h 4) œ 4 hÄ!

2 (2 h) 2h(# h)

hÄ!

h ŠÈ7 h È7‹

h

œ lim

œ "4

h Ä ! h(4 2h)

œ lim

(7 h) 7

h Ä ! h ŠÈ7 h È7‹

œ lim

h

h Ä ! h ŠÈ7hÈ7‹

œ lim

È3(0 h) 1 È3(0) 1 h hÄ!

œ lim

3

h Ä ! È3h 1 1

œ

œ lim

ŠÈ3h 1 "‹ ŠÈ3h 1 "‹ h ŠÈ3h 1 "‹

hÄ!

(3h 1) "

œ lim

h Ä ! h ŠÈ3h 1 1 ‹

œ lim

3h

h Ä ! h ŠÈ3h 1 "‹

3 #

63. lim È5 2x# œ È5 2(0)# œ È5 and lim È5 x# œ È5 (0)# œ È5; by the sandwich theorem, xÄ!

xÄ!

lim f(x) œ È5

xÄ!

64. lim a2 x# b œ 2 0 œ 2 and lim 2 cos x œ 2(1) œ 2; by the sandwich theorem, lim g(x) œ 2 xÄ!

65. (a)

xÄ!

lim Š1

xÄ!

x# 6‹

œ1

0 6

xÄ!

œ 1 and lim 1 œ 1; by the sandwich theorem, lim

(b) For x Á 0, y œ (x sin x)/(2 2 cos x) lies between the other two graphs in the figure, and the graphs converge as x Ä 0.

66. (a)

lim Š "#

xÄ!

lim

xÄ!

1cos x x#

x# 24 ‹

œ lim

1

xÄ! #

lim

x#

x Ä ! #4

œ

" #

x sin x

x Ä ! 22 cos x

xÄ!

0œ

" #

and lim

"

xÄ! #

œ1

œ "# ; by the sandwich theorem,

œ "# .


"

h Ä ! È 7 h È 7

" #È 7

62. lim

"6 3

œ lim (2 h) œ 2

h(h 4) h

ŠÈ7 h È7‹ ŠÈ7 h È7‹

hÄ!

x Ä #

œ3

œ lim

2h

hÄ!

œ lim

3h

hÄ! h

x Ä #

Section 2.2 Limit of a Function and Limit Laws (b) For all x Á 0, the graph of f(x) œ (1 cos x)/x# lies between the line y œ "# and the parabola yœ

" #

x# /24, and the graphs converge as x Ä 0.

67. (a) f(x) œ ax# *b/(x 3) x 3.1 f(x) 6.1 2.9 5.9

x f(x)

3.01 6.01

3.001 6.001

3.0001 6.0001

3.00001 6.00001

3.000001 6.000001

2.99 5.99

2.999 5.999

2.9999 5.9999

2.99999 5.99999

2.999999 5.999999

The estimate is lim f(x) œ 6. x Ä $

(b)

(c) f(x) œ

x# 9 x3

œ

(x 3)(x 3) x3

œ x 3 if x Á 3, and lim (x 3) œ 3 3 œ 6. x Ä $

68. (a) g(x) œ ax# #b/ Šx È2‹ x g(x)

1.4 2.81421

1.41 2.82421

1.414 2.82821

1.4142 2.828413

1.41421 2.828423

1.414213 2.828426

(b)

(c) g(x) œ

x# 2 x È2

œ

Šx È2‹ Šx È2‹ Šx È2‹

œ x È2 if x Á È2, and

69. (a) G(x) œ (x 6)/ ax# 4x 12b x 5.9 5.99 G(x) .126582 .1251564 x G(x)

6.1 .123456

6.01 .124843

5.999 .1250156 6.001 .124984

lim

x Ä È#

5.9999 .1250015 6.0001 .124998

Šx È2‹ œ È2 È2 œ 2È2.

5.99999 .1250001 6.00001 .124999

5.999999 .1250000 6.000001 .124999


51

52

Chapter 2 Limits and Continuity (b)

(c) G(x) œ

x6 ax# 4x 12b

œ

x6 (x 6)(x 2)

œ

" x#

70. (a) h(x) œ ax# 2x 3b / ax# 4x 3b x 2.9 2.99 h(x) 2.052631 2.005025 x h(x)

3.1 1.952380

3.01 1.995024

"

if x Á 6, and lim

x Ä ' x 2

œ

" ' 2

œ "8 œ 0.125.

2.999 2.000500

2.9999 2.000050

2.99999 2.000005

2.999999 2.0000005

3.001 1.999500

3.0001 1.999950

3.00001 1.999995

3.000001 1.999999

(b)

(c) h(x) œ

x# 2x 3 x# 4x 3

œ

(x 3)(x 1) (x 3)(x 1)

œ

x1 x1

71. (a) f(x) œ ax# 1b / akxk 1b x 1.1 1.01 f(x) 2.1 2.01 .9 1.9

x f(x)

.99 1.99

if x Á 3, and lim

x1

x Ä $ x1

œ

31 31

œ

4 #

œ 2.

1.001 2.001

1.0001 2.0001

1.00001 2.00001

1.000001 2.000001

.999 1.999

.9999 1.9999

.99999 1.99999

.999999 1.999999

(b)

(c) f(x) œ

x# " kx k 1

(x 1)(x 1)

1 œ (x x1)(x 1) (x 1)

œ x 1, x 0 and x Á 1 , and lim (1 x) œ 1 (1) œ 2. x Ä 1 œ 1 x, x 0 and x Á 1


Section 2.2 Limit of a Function and Limit Laws 72. (a) F(x) œ ax# 3x 2b / a2 kxkb x 2.1 2.01 F(x) 1.1 1.01 1.9 .9

x F(x)

1.99 .99

2.001 1.001

2.0001 1.0001

2.00001 1.00001

2.000001 1.000001

1.999 .999

1.9999 .9999

1.99999 .99999

1.999999 .999999

(b)

(c) F(x) œ

x# 3x 2 2 kx k

(x 2)(x 1)

œ (x 2)(x# x") 2x

73. (a) g()) œ (sin ))/) ) .1 g()) .998334

, x 0 , and lim (x 1) œ 2 1 œ 1. x Ä # œ x 1, x 0 and x Á 2

.01 .999983

.001 .999999

.0001 .999999

.00001 .999999

.000001 .999999

.1 .998334

.01 .999983

.001 .999999

.0001 .999999

.00001 .999999

.000001 .999999

74. (a) G(t) œ (1 cos t)/t# t .1 G(t) .499583

.01 .499995

.001 .499999

.0001 .5

.00001 .5

.000001 .5

.1 .499583

.01 .499995

.001 .499999

.0001 .5

.00001 .5

.000001 .5

) g()) lim g()) œ 1

)Ä!

(b)

t G(t)

lim G(t) œ 0.5

tÄ!

(b)


53

54


75. xlim f(x) exists at those points c where xlim x% œ xlim x# . Thus, c% œ c# Ê c# a1 c# b œ 0 Ê c œ 0, 1, or 1. Äc Äc Äc Moreover, lim f(x) œ lim x# œ 0 and lim f(x) œ lim f(x) œ 1. xÄ!

xÄ!

x Ä 1

xÄ1

76. Nothing can be concluded about the values of f, g, and h at x œ 2. Yes, f(2) could be 0. Since the conditions of the sandwich theorem are satisfied, lim f(x) œ 5 Á 0. xÄ#

lim f(x) lim 5 % xÄ% œ xÄlim x lim 2 œ

f(x)5 x Ä % x 2

77. 1 œ lim

xÄ%

lim f(x) 5

xÄ%

%#

xÄ%

xÄ%

78. (a) 1 œ lim

f(x) x#

lim f(x) lim f(x) œ xÄlim# x# œ xÄ %# Ê

(b) 1 œ lim

f(x) x#

œ ’ lim

x Ä # x Ä #

xÄ

#

x Ä #

79. (a) 0 œ 3 † 0 œ ’ lim

xÄ#

Ê lim f(x) 5 œ 2(1) Ê lim f(x) œ 2 5 œ 7. xÄ%

lim f(x) œ 4.

x Ä #

f(x) lim x" “ x “ ’x Ä #

œ ’ lim

x Ä #

f(x) ˆ " ‰ x “ #

Ê

lim

x Ä #

f(x) x

œ 2.

f(x) 5 x # “ ’xlim Ä#

5 (x 2)“ œ lim ’Š f(x) x # ‹ (x 2)“ œ lim [f(x) 5] œ lim f(x) 5

f(x) 5 x # “ ’xlim Ä#

(x 2)“ Ê lim f(x) œ 5 as in part (a).

xÄ#

Ê lim f(x) œ 5.

xÄ#

xÄ#

xÄ#

(b) 0 œ 4 † 0 œ ’ lim

xÄ#

80. (a) 0 œ 1 † 0 œ ’ lim

f(x) # “ ’ lim xÄ! x xÄ!

(b) 0 œ 1 † 0 œ 81. (a)

lim x sin

xÄ!

(b) 1 Ÿ sin

82. (a)

" x

’ lim f(x) # “ ’ lim xÄ! x xÄ!

" x

xÄ#

#

x“ œ ’ lim

f(x)

# xÄ! x

x“ œ

lim ’ f(x) x# xÄ!

# “ ’ lim x# “ œ lim ’ f(x) x# † x “ œ lim f(x). That is, lim f(x) œ 0.

xÄ!

† x“ œ

xÄ!

lim f(x) . xÄ! x

That is,

xÄ!

lim f(x) xÄ! x

œ 0.

œ0

Ÿ 1 for x Á 0:

x 0 Ê x Ÿ x sin

" x

Ÿ x Ê lim x sin

" x

œ 0 by the sandwich theorem;

x 0 Ê x x sin

" x

x Ê lim x sin

" x

œ 0 by the sandwich theorem.

xÄ! xÄ!

lim x# cos ˆ x"$ ‰ œ 0

xÄ!


xÄ!

Section 2.3 The Precise Definition of a Limit (b) 1 Ÿ cos ˆ x"$ ‰ Ÿ 1 for x Á 0 Ê x# Ÿ x# cos ˆ x"$ ‰ Ÿ x# Ê lim x# cos ˆ x"$ ‰ œ 0 by the sandwich xÄ!

theorem since lim x# œ 0. xÄ!

83-88. Example CAS commands: Maple: f := x -> (x^4 16)/(x 2); x0 := 2; plot( f(x), x œ x0-1..x0+1, color œ black, title œ "Section 2.2, #83(a)" ); limit( f(x), x œ x0 ); In Exercise 85, note that the standard cube root, x^(1/3), is not defined for x (surd(x+1, 3) 1)/x. Mathematica: (assigned function and values for x0 and h may vary) Clear[f, x] f[x_]:=(x3 x2 5x 3)/(x 1)2 x0= 1; h = 0.1; Plot[f[x],{x, x0 h, x0 h}] Limit[f[x], x Ä x0] 2.3 THE PRECISE DEFINITION OF A LIMIT 1. Step 1: Step 2:

kx 5k $ Ê $ x 5 $ Ê $ 5 x $ 5 $ 5 œ 7 Ê $ œ 2, or $ 5 œ 1 Ê $ œ 4. The value of $ which assures kx 5k $ Ê 1 x 7 is the smaller value, $ œ 2.

Step 1: Step 2:

kx 2k $ Ê $ x 2 $ Ê $ # x $ 2 $ 2 œ 1 Ê $ œ 1, or $ 2 œ 7 Ê $ œ 5. The value of $ which assures kx 2k $ Ê 1 x 7 is the smaller value, $ œ 1.

Step 1: Step 2:

kx (3)k $ Ê $ x $ $ Ê $ 3 x $ 3 $ 3 œ 7# Ê $ œ "# , or $ $ œ "# Ê $ œ 5# .

2.

3.

The value of $ which assures kx (3)k $ Ê 7# x "# is the smaller value, $ œ "# . 4.

Step 1:

¸x ˆ 3# ‰¸ $ Ê $ x

3 #

$ Ê $

3 #

x$

3 #

Step 2:

œ 7# Ê $ œ #, or $ 3# œ "# Ê $ œ 1. The value of $ which assures ¸x ˆ 3# ‰¸ $ Ê 7# x "# is the smaller value, $ œ ".

Step 1:

¸x "# ¸ $ Ê $ x

$

3 #

5. " #

$ Ê $

" #

x$

" #


55

56

Chapter 2 Limits and Continuity Step 2:

" or $ #" œ 47 Ê $ œ 14 . "¸ 4 ¸ The value of $ which assures x # $ Ê 9 x

$

" #

œ

4 9

Ê $œ

" 18 ,

4 7

is the smaller value, $ œ

" 18 .

6.

Step 1: Step 2:

kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3 $ $ œ 2.7591 Ê $ œ 0.2409, or $ $ œ 3.2391 Ê $ œ 0.2391. The value of $ which assures kx 3k $ Ê 2.7591 x 3.2391 is the smaller value, $ œ 0.2391.

7. Step 1: Step 2:

kx 5k $ Ê $ x 5 $ Ê $ 5 x $ 5 From the graph, $ 5 œ 4.9 Ê $ œ 0.1, or $ 5 œ 5.1 Ê $ œ 0.1; thus $ œ 0.1 in either case.

8. Step 1: Step 2:

kx (3)k $ Ê $ x 3 $ Ê $ 3 x $ 3 From the graph, $ 3 œ 3.1 Ê $ œ 0.1, or $ 3 œ 2.9 Ê $ œ 0.1; thus $ œ 0.1.

9. Step 1: Step 2:

kx 1k $ Ê $ x 1 $ Ê $ 1 x $ 1 9 7 From the graph, $ 1 œ 16 Ê $ œ 16 , or $ 1 œ 25 16 Ê $ œ

10. Step 1: Step 2:

kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3 From the graph, $ 3 œ 2.61 Ê $ œ 0.39, or $ 3 œ 3.41 Ê $ œ 0.41; thus $ œ 0.39.

11. Step 1:

kx 2k $ Ê $ x 2 $ Ê $ 2 x $ 2 From the graph, $ 2 œ È3 Ê $ œ 2 È3 ¸ 0.2679, or $ 2 œ È5 Ê $ œ È5 2 ¸ 0.2361; thus $ œ È5 2.

Step 2:

12. Step 1: Step 2:

9 16 ;

thus $ œ

kx (1)k $ Ê $ x 1 $ Ê $ 1 x $ 1 From the graph, $ 1 œ thus $ œ

È5 2 # .

È5 #

Ê $œ

È5 2 #

¸ 0.1180, or $ 1 œ

13. Step 1: Step 2:

kx (1)k $ Ê $ x 1 $ Ê $ 1 x $ 1 7 16 From the graph, $ 1 œ 16 9 Ê $ œ 9 ¸ 0.77, or $ 1 œ 25 Ê

14. Step 1:

¸x "# ¸ $ Ê $ x

Step 2:

7 16 .

From the graph, $ thus $ œ 0.00248.

" #

œ

" # 1 2.01

$ Ê $ Ê $œ

1 2

" #

x$

" #.01

" #

¸ 0.00248, or $

" #

œ

È3 #

9 25

Ê $œ

2 È3 #

œ 0.36; thus $ œ

1 1.99

Ê $œ

1 1.99

¸ 0.1340;

9 25

" #

œ 0.36.

¸ 0.00251;

15. Step 1: Step 2:

k(x 1) 5k 0.01 Ê kx 4k 0.01 Ê 0.01 x 4 0.01 Ê 3.99 x 4.01 kx 4k $ Ê $ x 4 $ Ê $ 4 x $ 4 Ê $ œ 0.01.

16. Step 1:

k(2x 2) (6)k 0.02 Ê k2x 4k 0.02 Ê 0.02 2x 4 0.02 Ê 4.02 2x 3.98 Ê 2.01 x 1.99 kx (2)k $ Ê $ x 2 $ Ê $ 2 x $ 2 Ê $ œ 0.01.

Step 2: 17. Step 1: Step 2:

¹Èx 1 "¹ 0.1 Ê 0.1 Èx 1 " 0.1 Ê 0.9 Èx 1 1.1 Ê 0.81 x 1 1.21 Ê 0.19 x 0.21 kx 0k $ Ê $ x $ . Then, $ œ !Þ"* Ê $ œ !Þ"* or $ œ !Þ#"; thus, $ œ 0.19.


Section 2.3 The Precise Definition of a Limit 18. Step 1: Step 2:

¸Èx "# ¸ 0.1 Ê 0.1 Èx "# 0.1 Ê 0.4 Èx 0.6 Ê 0.16 x 0.36 ¸x "4 ¸ $ Ê $ x 4" $ Ê $ 4" x $ 4" . Then, $

19. Step 1: Step 2:

20. Step 1: Step 2:

21. Step 1: Step 2:

22. Step 1: Step 2:

57

" 4

œ 0.16 Ê $ œ 0.09 or $

" 4

œ 0.36 Ê $ œ 0.11; thus $ œ 0.09.

¹È19 x $¹ " Ê " È19 x $ 1 Ê 2 È19 x % Ê 4 19 x 16 Ê % x 19 16 Ê 15 x 3 or 3 x 15 kx 10k $ Ê $ x 10 $ Ê $ 10 x $ 10. Then $ 10 œ 3 Ê $ œ 7, or $ 10 œ 15 Ê $ œ 5; thus $ œ 5. ¹Èx 7 4¹ 1 Ê " Èx 7 % 1 Ê 3 Èx 7 5 Ê 9 x 7 25 Ê 16 x 32 kx 23k $ Ê $ x 23 $ Ê $ 23 x $ 23. Then $ 23 œ 16 Ê $ œ 7, or $ 23 œ 32 Ê $ œ 9; thus $ œ 7. ¸ "x 4" ¸ 0.05 Ê 0.05

" x

" 4

0.05 Ê 0.2

" x

0.3 Ê

kx 4k $ Ê $ x 4 $ Ê $ 4 x $ 4. 2 2 Then $ % œ 10 3 or $ œ 3 , or $ 4 œ 5 or $ œ 1; thus $ œ 3 .

10 #

x

10 3

or

10 3

x 5.

kx# 3k !.1 Ê 0.1 x# 3 0.1 Ê 2.9 x# 3.1 Ê È2.9 x È3.1 ¹x È3¹ $ Ê $ x È3 $ Ê $ È3 x $ È3. Then $ È3 œ È2.9 Ê $ œ È3 È2.9 ¸ 0.0291, or $ È3 œ È3.1 Ê $ œ È3.1 È3 ¸ 0.0286; thus $ œ 0.0286.

23. Step 1: Step 2:

kx# 4k 0.5 Ê 0.5 x# 4 0.5 Ê 3.5 x# 4.5 Ê È3.5 kxk È4.5 Ê È4.5 x È3.5, for x near 2. kx (2)k $ Ê $ x 2 $ Ê $ # x $ 2. Then $ # œ È4.5 Ê $ œ È4.5 # ¸ 0.1213, or $ # œ È3.5 Ê $ œ # È3.5 ¸ 0.1292; thus $ œ È4.5 2 ¸ 0.12.

24. Step 1: Step 2:

25. Step 1: Step 2:

¸ "x (1)¸ 0.1 Ê 0.1

" x

1 0.1 Ê 11 10

" x

9 10 10 10 10 Ê 10 11 x 9 or 9 x 11 .

kx (1)k $ Ê $ x 1 $ Ê $ " x $ ". " 10 " Then $ " œ 10 9 Ê $ œ 9 , or $ " œ 11 Ê $ œ 11 ; thus $ œ

" 11 .

kax# 5b 11k " Ê kx# 16k 1 Ê " x# 16 1 Ê 15 x# 17 Ê È15 x È17. kx 4k $ Ê $ x 4 $ Ê $ % x $ %. Then $ % œ È15 Ê $ œ % È15 ¸ 0.1270, or $ % œ È17 Ê $ œ È17 % ¸ 0.1231; thus $ œ È17 4 ¸ 0.12.

26. Step 1: Step 2:

27. Step 1: Step 2:

¸ 120 ¸ x 5 " Ê "

120 x

&1 Ê 4

120 x

6 Ê

" 4

x 120

" 6

Ê 30 x 20 or 20 x 30.

kx 24k $ Ê $ x 24 $ Ê $ 24 x $ 24. Then $ 24 œ 20 Ê $ œ 4, or $ 24 œ 30 Ê $ œ 6; thus Ê $ œ 4. kmx 2mk 0.03 Ê 0.03 mx 2m 0.03 Ê 0.03 2m mx 0.03 2m Ê 0.03 2 0.03 m x2 m . kx 2k $ Ê $ x 2 $ Ê $ 2 x $ 2. 0.03 0.03 Then $ 2 œ 2 0.03 m Ê $ œ m , or $ 2 œ # m Ê $ œ

0.03 m .

In either case, $ œ


0.03 m .

58

Chapter 2 Limits and Continuity kmx 3mk c Ê c mx 3m c Ê c 3m mx c 3m Ê 3

28. Step 1:

kx 3k $ Ê $ x 3 $ Ê $ $ B $ $. Then $ $ œ $ mc Ê $ œ mc , or $ $ œ $ mc Ê $ œ

Step 2:

¸(mx b) ˆ m# b‰¸ - Ê c mx m# c Ê c ¸x "# ¸ $ Ê $ x "# $ Ê $ "# x $ "# .

29. Step 1: Step 2:

Then $

" #

œ

" #

c m

Ê $œ

c m,

or $

" #

œ

" #

c m

c m. m #

Ê $œ

c m

x 3


c m.

c m. " #


c m.

mx c

m #

Ê

c m

c m

x

" #

c m.

k(mx b) (m b)k 0.05 Ê 0.05 mx m 0.05 Ê 0.05 m mx 0.05 m 0.05 Ê 1 0.05 m x" m .

30. Step 1:

kx 1k $ Ê $ x 1 $ Ê $ " x $ ". 0.05 0.05 Then $ " œ " 0.05 m Ê $ œ m , or $ " œ " m Ê $ œ

Step 2:

0.05 m .


0.05 m .

31. lim (3 2x) œ 3 2(3) œ 3 xÄ3

ka3 2xb (3)k 0.02 Ê 0.02 6 2x 0.02 Ê 6.02 2x 5.98 Ê 3.01 x 2.99 or 2.99 x 3.01. 0 k x 3k $ Ê $ x 3 $ Ê $ $ x $ $ . Then $ $ œ 2.99 Ê $ œ 0.01, or $ $ œ 3.01 Ê $ œ 0.01; thus $ œ 0.01.

Step 1: Step 2:

32.

lim (3x #) œ (3)(1) 2 œ 1

x Ä 1

k(3x 2) 1k 0.03 Ê 0.03 3x 3 0.03 Ê 0.01 x 1 0.01 Ê 1.01 x 0.99. kx (1)k $ Ê $ x 1 $ Ê $ " x $ 1. Then $ " œ 1.01 Ê $ œ 0.01, or $ " œ 0.99 Ê $ œ 0.01; thus $ œ 0.01.

Step 1: Step 2:

33. lim

x# 4

x Ä # x#

34.

35.

œ lim

xÄ#

#

(x 2)(x 2) (x 2)

œ lim (x 2) œ # # œ 4, x Á 2 xÄ#

(x 2)(x 2) (x 2)

Step 1:

¹Š xx 24 ‹

Step 2:

Ê 1.95 x 2.05, x Á 2. kx 2k $ Ê $ x 2 $ Ê $ 2 x $ 2. Then $ 2 œ 1.95 Ê $ œ 0.05, or $ 2 œ 2.05 Ê $ œ 0.05; thus $ œ 0.05.

lim

x Ä &

x# 6x 5 x5

4¹ 0.05 Ê 0.05

œ lim

x Ä &

(x 5)(x 1) (x 5)

% 0.05 Ê 3.95 x 2 4.05, x Á 2

œ lim (x 1) œ 4, x Á 5. x Ä &

(x 5)(x ") (x 5)

Step 1:

# ¹Š x x 6x5 5 ‹

Step 2:

Ê 5.05 x 4.95, x Á 5. kx (5)k $ Ê $ x 5 $ Ê $ & x $ &. Then $ & œ 5.05 Ê $ œ 0.05, or $ & œ 4.95 Ê $ œ 0.05; thus $ œ 0.05.

(4)¹ 0.05 Ê 0.05

4 0.05 Ê 4.05 x 1 3.95, x Á 5

lim È1 5x œ È1 5(3) œ È16 œ 4

x Ä $

Step 1:

¹È1 5x 4¹ 0.5 Ê 0.5 È1 5x 4 0.5 Ê 3.5 È1 5x 4.5 Ê 12.25 1 5x 20.25

Step 2:

Ê 11.25 5x 19.25 Ê 3.85 x 2.25. kx (3)k $ Ê $ x 3 $ Ê $ $ x $ $. Then $ $ œ 3.85 Ê $ œ 0.85, or $ $ œ 2.25 Ê 0.75; thus $ œ 0.75.

36. lim

4

xÄ# x

Step 1:

œ

4 #

œ2

¸ 4x 2¸ 0.4 Ê 0.4

4 x

2 0.4 Ê 1.6

4 x

2.4 Ê

10 16

x 4

10 24

Ê

10 4

x


10 6

or

5 3

x 25 .

Section 2.3 The Precise Definition of a Limit Step 2:

59

kx 2k $ Ê $ x 2 $ Ê $ # x $ #. Then $ # œ 53 Ê $ œ "3 , or $ # œ #5 Ê $ œ "# ; thus $ œ 3" .

37. Step 1: Step 2:

k(9 x) 5k % Ê % 4 x % Ê % 4 x % 4 Ê % % x 4 % Ê % % x 4 %. kx 4k $ Ê $ x 4 $ Ê $ % x $ %. Then $ 4 œ % 4 Ê $ œ %, or $ % œ % % Ê $ œ %. Thus choose $ œ %.

38. Step 1:

k(3x 7) 2k % Ê % 3x 9 % Ê 9 % 3x * % Ê 3

Step 2:

39. Step 1: Step 2:

40. Step 1: Step 2:

41. Step 1: Step 2:

42. Step 1: Step 2:

43. Step 1: Step 2:

x 3 3% .

kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3. Then $ 3 œ $ 3% Ê $ œ 3% , or $ 3 œ 3 3% Ê $ œ 3% . Thus choose $ œ 3% . ¹Èx 5 2¹ % Ê % Èx 5 2 % Ê 2 % Èx 5 2 % Ê (2 %)# x 5 (2 %)# Ê (2 %)# & x (2 %)# 5. kx 9k $ Ê $ x 9 $ Ê $ 9 x $ 9. Then $ * œ %# %% * Ê $ œ %% %# , or $ * œ %# %% * Ê $ œ %% %# . Thus choose the smaller distance, $ œ %% %# . ¹È4 x 2¹ % Ê % È4 x 2 % Ê 2 % È4 x 2 % Ê (2 %)# % x (2 %)# Ê (2 %)# x 4 (2 %)# Ê (2 %)# % x (2 %)# %. k x 0k $ Ê $ x $ . Then $ œ (2 %)# 4 œ %# %% Ê $ œ %% %# , or $ œ (2 %)# 4 œ 4% %# . Thus choose the smaller distance, $ œ 4% %# . For x Á 1, kx# 1k % Ê % x# " % Ê " % x# " % Ê È1 % kxk È1 % Ê È" % x È1 % near B œ ". kx 1k $ Ê $ x 1 $ Ê $ " x $ ". Then $ " œ È1 % Ê $ œ " È1 %, or $ 1 œ È" % Ê $ œ È" % 1. Choose $ œ min š" È1 %ß È1 % "›, that is, the smaller of the two distances. For x Á 2, kx# 4k % Ê % x# 4 % Ê 4 % x# 4 % Ê È4 % kxk È4 % Ê È4 % x È4 % near B œ 2. kx (2)k $ Ê $ x 2 $ Ê $ 2 x $ 2. Then $ 2 œ È% % Ê $ œ È% % #, or $ # œ È% % Ê $ œ # È% %. Choose $ œ min šÈ% % #ß # È% %› . ¸ "x 1¸ % Ê %

" x

"% Ê "%

" x

"% Ê

" 1%

% "%,

" 1%.

x

kx 1k $ Ê $ x 1 $ Ê " $ x " $ . Then " $ œ " " % Ê $ œ " " " % œ " % % , or " $ œ " " % Ê $ œ Choose $ œ

44. Step 1:

% 3

" "%

"œ

% "%.

the smaller of the two distances.

¸ x"# "3 ¸ % Ê %

" x#

" 3

% Ê

" 3

%

" x#

" 3

% Ê

1 3% 3

" x#

1 $% 3

3 È $. Ê É 1 3 $% kxk É " 3 $% , or É " 3 $% x É "$ % for x near


Ê

3 "

%$ x#

3 "

%$ 60

Chapter 2 Limits and Continuity Step 2:

¹x È3¹ $ Ê $ x È3 $ Ê È3 $ x È3 $ . Then È3 $ œ É " 3 $% Ê $ œ È3 É " 3 $% , or È3 $ œ É " 3 $% Ê $ œ É " 3 $% È3. Choose $ œ min šÈ3 É " 3 $% ß É " 3 $% È3›.

45. Step 1: Step 2:

46. Step 1: Step 2:

47. Step 1:

#

¹Š xx 3* ‹ (6)¹ % Ê % (x 3) 6 %, x Á 3 Ê % x 3 % Ê % $ x % $. kx (3)k $ Ê $ x 3 $ Ê $ $ x $ 3. Then $ $ œ % $ Ê $ œ %, or $ $ œ % $ Ê $ œ %. Choose $ œ %. #

¹Š xx 11 ‹ 2¹ % Ê % (x 1) 2 %, x Á 1 Ê " % x " %. kx 1k $ Ê $ x 1 $ Ê " $ x " $ . Then " $ œ " % Ê $ œ %, or " $ œ " % Ê $ œ %. Choose $ œ %. x 1: l(4 2x) 2l % Ê ! 2 2x % since x 1Þ Thus, 1

% #

x !;

x 1: l(6x 4) 2l % Ê ! Ÿ 6x 6 % since x 1. Thus, " Ÿ x 1 6% . Step 2:

48. Step 1: Step 2:

kx 1k $ Ê $ x 1 $ Ê " $ x 1 $ . Then 1 $ œ " #% Ê $ œ #% , or " $ œ 1 6% Ê $ œ 6% . Choose $ œ 6% . x !: k2x 0k % Ê % 2x ! Ê #% x 0; x 0: ¸ x# !¸ % Ê ! Ÿ x #%.

k x 0k $ Ê $ x $ . Then $ œ #% Ê $ œ #% , or $ œ #% Ê $ œ #%. Choose $ œ #% .

49. By the figure, x Ÿ x sin

" x

Ÿ x for all x 0 and x x sin

then by the sandwich theorem, in either case, lim x sin xÄ!

50. By the figure, x# Ÿ x# sin

" x

" x

" x

x for x 0. Since lim (x) œ lim x œ 0, xÄ!

œ 0.

xÄ!

Ÿ x# for all x except possibly at x œ 0. Since lim ax# b œ lim x# œ 0, then

by the sandwich theorem, lim x# sin xÄ!

" x

xÄ!

œ 0.

xÄ!

51. As x approaches the value 0, the values of g(x) approach k. Thus for every number % 0, there exists a $ ! such that ! kx 0k $ Ê kg(x) kk %. 52. Write x œ h c. Then ! lx cl $ Í $ x c $ , x Á c Í $ ah cb c $ , h c Á c Í $ h $ , h Á ! Í ! lh !l $ . Thus, limfaxb œ L Í for any % !, there exists $ ! such that lfaxb Ll % whenever ! lx cl $ x Äc

Í lfah cb Ll % whenever ! lh !l $ Í lim fah cb œ L. hÄ!

53. Let f(x) œ x# . The function values do get closer to 1 as x approaches 0, but lim f(x) œ 0, not 1. The function f(x) œ x# never gets arbitrarily close to 1 for x near 0.

xÄ!


Section 2.3 The Precise Definition of a Limit

61

54. Let f(x) œ sin x, L œ "# , and x! œ 0. There exists a value of x (namely, x œ 16 ) for which ¸sin x "# ¸ % for any given

% 0. However, lim sin x œ 0, not "# . The wrong statement does not require x to be arbitrarily close to x! . As another\ xÄ!

example, let g(x) œ sin "x , L œ #" , and x! œ 0. We can choose infinitely many values of x near 0 such that sin you can see from the accompanying figure. However, lim sin xÄ!

" x

" x

œ

" #

as

fails to exist. The wrong statement does not require all

values of x arbitrarily close to x! œ 0 to lie within % 0 of L œ "# . Again you can see from the figure that there are also infinitely many values of x near 0 such that sin "x œ 0. If we choose % ¸sin "x #" ¸ % for all values of x sufficiently near x! œ 0.

#

55. kA *k Ÿ 0.01 Ê 0.01 Ÿ 1 ˆ x# ‰ 9 Ÿ 0.01 Ê 8.99 Ÿ Ê

2É 8.99 1

ŸxŸ

2É 9.01 1

1 x# 4

" 4

we cannot satisfy the inequality

Ÿ 9.01 Ê

4 1

(8.99) Ÿ x# Ÿ

4 1

(9.01)

or 3.384 Ÿ x Ÿ 3.387. To be safe, the left endpoint was rounded up and the right

endpoint was rounded down. 56. V œ RI Ê (120)(10) 51

V R

ŸRŸ

œ I Ê ¸ VR 5¸ Ÿ 0.1 Ê 0.1 Ÿ (120)(10) 49

120 R

5 Ÿ 0.1 Ê 4.9 Ÿ

120 R

Ÿ 5.1 Ê

10 49

R 1#0

10 51

Ê

Ê 23.53 Ÿ R Ÿ 24.48.

To be safe, the left endpoint was rounded up and the right endpoint was rounded down. 57. (a) $ x 1 0 Ê " $ x 1 Ê f(x) œ x. Then kf(x) 2k œ kx 2k œ 2 x 2 1 œ 1. That is, kf(x) 2k 1 "# no matter how small $ is taken when " $ x 1 Ê lim f(x) Á 2. xÄ1

(b) 0 x 1 $ Ê " x " $ Ê f(x) œ x 1. Then kf(x) 1k œ k(x 1) 1k œ kxk œ x 1. That is, kf(x) 1k 1 no matter how small $ is taken when " x " $ Ê lim f(x) Á 1. xÄ1

(c) $ x 1 ! Ê " $ x 1 Ê f(x) œ x. Then kf(x) 1.5k œ kx 1.5k œ 1.5 x 1.5 1 œ 0.5. Also, ! x 1 $ Ê 1 x " $ Ê f(x) œ x 1. Then kf(x) 1.5k œ k(x 1) 1.5k œ kx 0.5k œ x 0.5 " 0.5 œ 0.5. Thus, no matter how small $ is taken, there exists a value of x such that $ x 1 $ but kf(x) 1.5k "# Ê lim f(x) Á 1.5. xÄ1

58. (a) For 2 x 2 $ Ê h(x) œ 2 Ê kh(x) 4k œ 2. Thus for % 2, kh(x) 4k % whenever 2 x 2 $ no matter how small we choose $ 0 Ê lim h(x) Á 4. xÄ#

(b) For 2 x 2 $ Ê h(x) œ 2 Ê kh(x) 3k œ 1. Thus for % 1, kh(x) 3k % whenever 2 x 2 $ no matter how small we choose $ 0 Ê lim h(x) Á 3. xÄ#

(c) For 2 $ x 2 Ê h(x) œ x# so kh(x) 2k œ kx# 2k . No matter how small $ 0 is chosen, x# is close to 4 when x is near 2 and to the left on the real line Ê kx# 2k will be close to 2. Thus if % 1, kh(x) 2k % whenever 2 $ x 2 no mater how small we choose $ 0 Ê lim h(x) Á 2. xÄ#


62


59. (a) For 3 $ x 3 Ê f(x) 4.8 Ê kf(x) 4k 0.8. Thus for % 0.8, kf(x) 4k % whenever 3 $ x 3 no matter how small we choose $ 0 Ê lim f(x) Á 4. xÄ$

(b) For 3 x 3 $ Ê f(x) 3 Ê kf(x) 4.8k 1.8. Thus for % 1.8, kf(x) 4.8k % whenever 3 x 3 $ no matter how small we choose $ 0 Ê lim f(x) Á 4.8. xÄ$

(c) For 3 $ x 3 Ê f(x) 4.8 Ê kf(x) 3k 1.8. Again, for % 1.8, kf(x) 3k % whenever $ $ x 3 no matter how small we choose $ 0 Ê lim f(x) Á 3. xÄ$

60. (a) No matter how small we choose $ 0, for x near 1 satisfying " $ x " $ , the values of g(x) are near 1 Ê kg(x) 2k is near 1. Then, for % œ "# we have kg(x) 2k "# for some x satisfying " $ x " $ , or ! kx 1k $ Ê

lim g(x) Á 2.

x Ä 1

(b) Yes, lim g(x) œ 1 because from the graph we can find a $ ! such that kg(x) 1k % if ! kx (1)k $ . x Ä 1

61-66. Example CAS commands (values of del may vary for a specified eps): Maple: f := x -> (x^4-81)/(x-3);x0 := 3; plot( f(x), x=x0-1..x0+1, color=black, # (a) title="Section 2.3, #61(a)" ); L := limit( f(x), x=x0 ); # (b) epsilon := 0.2; # (c) plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01, color=black, linestyle=[1,3,3], title="Section 2.3, #61(c)" ); q := fsolve( abs( f(x)-L ) = epsilon, x=x0-1..x0+1 ); # (d) delta := abs(x0-q); plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" ); for eps in [0.1, 0.005, 0.001 ] do # (e) q := fsolve( abs( f(x)-L ) = eps, x=x0-1..x0+1 ); delta := abs(x0-q); head := sprintf("Section 2.3, #61(e)\n epsilon = %5f, delta = %5f\n", eps, delta ); print(plot( [f(x),L-eps,L+eps], x=x0-delta..x0+delta, color=black, linestyle=[1,3,3], title=head )); end do: Mathematica (assigned function and values for x0, eps and del may vary): Clear[f, x] y1: œ L eps; y2: œ L eps; x0 œ 1; f[x_]: œ (3x2 (7x 1)Sqrt[x] 5)/(x 1) Plot[f[x], {x, x0 0.2, x0 0.2}] L: œ Limit[f[x], x Ä x0] eps œ 0.1; del œ 0.2; Plot[{f[x], y1, y2},{x, x0 del, x0 del}, PlotRange Ä {L 2eps, L 2eps}]


Section 2.4 One-Sided Limits 2.4 ONE-SIDED LIMITS 1. (a) True (e) True (i) False

(b) True (f) True (j) False

(c) False (g) False (k) True

(d) True (h) False (l) False

2. (a) True (e) True (i) True

(b) False (f) True (j) False

(c) False (g) True (k) True

(d) True (h) True

3. (a)

lim f(x) œ

x Ä #b

2 #

" œ #, lim c f(x) œ $ # œ " xÄ#

(b) No, lim f(x) does not exist because lim b f(x) Á lim c f(x) xÄ# xÄ# xÄ# (c) lim c f(x) œ 4# 1 œ 3, lim b f(x) œ 4# " œ $ xÄ%

xÄ%

(d) Yes, lim f(x) œ 3 because 3 œ lim c f(x) œ lim b f(x) xÄ% xÄ% xÄ% 4. (a)

lim f(x) œ

x Ä #b

2 #

œ 1, lim c f(x) œ $ # œ ", f(2) œ 2 xÄ#

(b) Yes, lim f(x) œ 1 because " œ lim b f(x) œ lim c f(x) xÄ# xÄ# xÄ# (c) lim c f(x) œ 3 (1) œ 4, lim b f(x) œ 3 (1) œ 4 x Ä "

x Ä "

(d) Yes, lim f(x) œ 4 because 4 œ x Ä "

lim

x Ä "c

f(x) œ

lim

x Ä "b

f(x)

5. (a) No, lim b f(x) does not exist since sin ˆ "x ‰ does not approach any single value as x approaches 0 xÄ! (b) lim c f(x) œ lim c 0 œ 0 xÄ!

(c)

xÄ!

lim f(x) does not exist because lim b f(x) does not exist xÄ! xÄ!

6. (a) Yes, lim b g(x) œ 0 by the sandwich theorem since Èx Ÿ g(x) Ÿ Èx when x 0 xÄ! (b) No, lim c g(x) does not exist since Èx is not defined for x 0 xÄ!

(c) No, lim g(x) does not exist since lim c g(x) does not exist xÄ! xÄ! 7. (aÑ

lim f(x) œ " œ lim b f(x) xÄ1 (c) Yes, lim f(x) œ 1 since the right-hand and left-hand (b)

x Ä 1c

xÄ1

limits exist and equal 1

8. (a)

(b)

lim f(x) œ 0 œ lim c f(x) xÄ1

x Ä 1b

(c) Yes, lim f(x) œ 0 since the right-hand and left-hand xÄ1

limits exist and equal 0


63

64


9. (a) domain: 0 Ÿ x Ÿ 2 range: 0 y Ÿ 1 and y œ 2 (b) xlim f(x) exists for c belonging to Äc (0ß 1) ("ß #) (c) x œ 2 (d) x œ 0

10. (a) domain: _ x _ range: " Ÿ y Ÿ 1 (b) xlim f(x) exists for c belonging to Äc (_ß 1) ("ß ") ("ß _) (c) none (d) none

11.

x Ä !Þ&c

lim

13.

x Ä #b

14.

x Ä 1c

15.

h Ä !b

lim

2 0.5 2 È3 É 3/2 É xx É 1 œ 0.5 1 œ 1/2 œ

lim

x Ä 1b

" 1 È0 œ ! É "1 É xx # œ # œ

5‰ ˆ x x 1 ‰ ˆ 2x ˆ 2 ‰ 2(2) 5 ˆ1‰ x# x œ 2 1 Š (2)# (2) ‹ œ (2) 2 œ 1

lim

ˆ x " 1 ‰ ˆ x x 6 ‰ ˆ 3 7 x ‰ œ ˆ 1 " 1 ‰ ˆ 1 1 6 ‰ ˆ 3 7 1 ‰ œ ˆ "# ‰ ˆ 71 ‰ ˆ 27 ‰ œ 1

lim

Èh# 4h 5 È5 h

œ lim b hÄ! 16.

12.

lim

h Ä !c

(b)

18. (a)

(b)

ah# 4h 5b 5

h ŠÈh# 4h 5 È5‹

È6 È5h# 11h 6 h

œ lim c hÄ! 17. (a)

œ lim b Š hÄ!

x Ä #c

lim

x Ä 1b

lim

x Ä 1c

œ lim c Š hÄ!

6 a5h# 11h 6b

lim

lim

œ lim b hÄ!

h ŠÈ6 È5h# 11h 6‹

x Ä #b

kx 2 k x 2

(x 3)

œ

kx 2 k x2

œ

lim

lim

x Ä #c

È2x (x 1) kx 1 k

È2x (x 1) kx 1 k

œ lim b xÄ1

h(5h 11)

h ŠÈ6 È5h# 11h 6‹

(x 3)

x Ä #b

œ

œ

04 È5 È5

œ

2 È5

È5h# 11h 6 È6 È5h# 11h 6 È ‹ Š È66 ‹ h È5h# 11h 6

x Ä #b

lim

h(h 4)

h ŠÈh# 4h 5 È5‹

œ lim c hÄ!

œ

(x 3)

Èh# 4h 5 È5 È # 4h 5 È5 ‹ Š Èhh# ‹ h 4h 5 È5

(x2) (x#)

(0 11) È6 È6

11 œ 2È 6

akx 2k œ ax 2b for x 2b

(x 3) œ a(2) 3b œ 1

(x 3) ’ (x(x#2) ) “

lim

œ

x Ä #c

akx 2k œ (x 2) for x 2b

(x 3)(1) œ (2 3) œ 1

È2x (x 1) (x 1)

akx 1k œ x 1 for x 1b

œ lim b È2x œ È2 xÄ1

œ lim c xÄ1

È2x (x 1) (x 1)

akx 1k œ (x 1) for x 1b

œ lim c È2x œ È2 xÄ1


Section 2.4 One-Sided Limits 19. (a)

) Ä $b

20. (a)

t Ä %b

Ú) Û )

lim

œ1

3 3

lim at ÚtÛb œ 4 4 œ 0

21. lim

sin È2) È 2)

22. lim

sin kt t

23. lim

sin 3y 4y

)Ä!

tÄ!

yÄ!

24.

œ

tan 2x x xÄ!

25. lim

2t

27. lim

xÄ!

)Ä!

œ lim c ˆ "3 † hÄ! sin 2x ‰ ˆ cos 2x x xÄ!

œ lim

œ 2 lim

t

sin t t Ä ! ˆ cos t ‰

x csc 2x cos 5x

œ

3h ‰ sin 3h

x x cos x

2 3

lim at ÚtÛb œ 4 3 œ 1

" ‰ cos 5x

xÄ!

œ

" 3

Œ

œ

" lim

)Ä!c

(where ) œ kt) (where ) œ 3y)

3 4

" 3

œ

sin ) )

"

‹ Š lim x Ä ! cos 2x xÄ!

œ Š #" lim

†1œ

2 sin 2x #x ‹

t

Ä!

" 3

(where ) œ 3h)

œ1†2œ2

t

"

‹ Š lim cos 5x ‹ x Ä ! sin 2x xÄ!

6x# cos x

œ lim ˆ sin xxcos x

œk†1œk

tÄ!

x Ä ! sin x sin 2x

x Ä ! sin x cos x

t Ä %c

œ

œ 2 Š lim cos t‹ Œ lim" sin t œ 2 † " † " œ 2

t cos t sin t

tÄ!

xÄ!

(b)

Ú) Û )

lim

3 sin ) 4 )lim Ä! )

œ

œ Š lim

sin 2x

x Ä ! x cos 2x

œ 2 lim

xÄ!

)Ä!

" " sin 3h 3 h lim Ä !c ˆ 3h ‰

œ

œ lim

œ lim ˆ sinx2x †

sin ) )

œ k lim

sin 3y 3 4 ylim Ä ! 3y

28. lim 6x# (cot x)(csc 2x) œ lim 29. lim

k sin ) )

œ lim

3 sin 3y " 4 ylim 3y Ä!

œ

h

t Ä ! tan t

k sin kt kt

tÄ!

) Ä $c

(where x œ È2))

œ1

sin x x

xÄ!

œ lim

lim h Ä !c sin 3h

26. lim

œ lim

(b)

2x

œ lim ˆ3 cos x †

x sin x

xÄ!

x cos x ‰ sin x cos x

œ lim ˆ sinx x † xÄ!

†

œ ˆ #" † 1‰ (1) œ

2x ‰ sin 2x

" ‰ cos x

" #

œ3†"†1œ3

lim

x

x Ä ! sin x

œ lim Š sin" x ‹ † lim ˆ cos" x ‰ lim Š sin" x ‹ œ (1)(1) 1 œ 2 xÄ!

30. lim

xÄ!

xÄ!

x

x# x sin x #x

1 cos ) ) Ä ! sin 2)

31. lim

œ lim

)Ä!

)Ä!

34. lim

sin (sin h) sin h sin )

) Ä ! sin 2)

36. lim

sin 5x

x Ä ! sin 4x

œ

œ lim

)Ä!

œ lim

)Ä!

sin ) )

sin ) )

)Ä!

"# (1) œ 0

1 cos2 ) a2sin ) cos )ba1 cos )b

œ lim

)Ä!

sin2 ) a2sin ) cos )ba1 cos )b

œ lim

xÄ!

xa1 c cos xb 9x2 sin2 3x 9x2

œ lim

1 c cos x 9x

2 x Ä ! ˆ sin3x3x ‰

œ

" lim ˆ 1 9 x

Ä!

cos x ‰ x

2 lim ˆ sin3x3x ‰

xÄ!

œ

" 9 a0 b 12

œ0

œ 1 since ) œ sin h Ä 0 as h Ä 0 2) ‰ #)

5x œ lim ˆ sin sin 4x †

4x 5x

)Ä!

œ lim

" #

œ 1 since ) œ 1 cos t Ä 0 as t Ä 0

sin ) œ lim ˆ sin 2) †

xÄ!

"# ˆ sinx x ‰‰ œ 0

œ0

0 a2ba2b

xa1 cos xb sin2 3x xÄ!

sin(1 cos t) 1cos t

x

a1 cos )ba1 cos )b a2sin ) cos )ba1 cos )b

œ lim

33. lim

35. lim

" #

xÄ!

œ lim

x x cos x sin2 3x xÄ!

hÄ!

œ lim ˆ #x

sin ) a2cos )ba1 cos )b

32. lim

tÄ!

xÄ!

œ

" # )lim Ä!

† 54 ‰ œ

ˆ sin) ) †

5 4 xlim Ä!

2) ‰ sin 2)

ˆ sin5x5x †

œ

" #

4x ‰ sin 4x

†1†1œ œ

5 4

" #

†1†1œ

5 4


65

66


37. lim ) cos ) œ 0 † 1 œ 0 )Ä!

38. lim sin ) cot 2) œ lim sin ) )Ä!

)Ä!

tan 3x

39. lim

x Ä ! sin 8x

œ

3 8 xlim Ä!

40. lim

yÄ!

œ

sin 3x œ lim ˆ cos 3x †

xÄ!

" ‰ sin 8x

œ lim sin ) )Ä!

œ lim

xÄ!

3 8

†1†1†1œ

sin 3y sin 4y cos 5y y cos 4y sin 5y

yÄ!

œ lim

) cot 4) 2 2 ) Ä ! sin ) cot 2)

42. lim

)Ä!

sin ) cos ) 3) 2 ) cos sin 3)

œ lim

)Ä!

2 lim 4) cos2 4) cos ) ) Ä ! cos 2) sin 4)

œ lim

cos 4) sin 4) 2 2) 2 sin ) cos sin2 2)

" sin 8x

cos 2)

) Ä ! 2 cos )

†

8x 3x

œ

1 2

† 83 ‰

3 8

yÄ!

sin ) sin 3)

2 ) Ä ! ) cos ) cos 3)

)

œ lim

sin 4y cos 5y 3†4†5y œ lim Š siny3y ‹ Š cos 4y ‹ Š sin 5y ‹ Š 3†4†5y ‹

cos 5y ˆ 3†4 ‰ lim Š sin3y3y ‹ Š sin4y4y ‹ Š sin5y5y ‹ Š cos 4y ‹ 5 yÄ!

tan ) 2 ) Ä ! ) cot 3)

cos 2) 2sin ) cos )

sin 3x œ lim ˆ cos 3x †

ˆ cos"3x ‰ ˆ sin3x3x ‰ ˆ sin8x8x ‰ œ

sin 3y cot 5y y cot 4y

41. lim

œ

cos 2) sin 2)

œ1†1†1†1†

œ

12 5

œ lim ˆ sin) ) ‰ˆ sin3)3) ‰ˆ cos ) 3cos 3) ‰ œ a1ba1bˆ 13†1 ‰ œ 3 )Ä!

) cos 4) sin2 2) 2 2 ) Ä ! sin ) cos 2) sin 4)

œ lim

12 5

) cos 4) a2sin ) cos )b2 2 2 ) Ä ! sin ) cos 2) sin 4)

œ lim

) cos 4) ˆ4sin2 ) cos2 )‰ 2 2 ) Ä ! sin ) cos 2) sin 4)

œ lim

4) cos ) 4) cos ) œ lim ˆ sin4)4) ‰Š coscos ‹ œ lim Š sin14) ‹Š coscos ‹ œ ˆ 11 ‰Š 11†12 ‹ œ 1 2 2) 2 2) 2

)Ä!

2

)Ä!

2

4)

43. Yes. If lim b f(x) œ L œ lim c f(x), then xlim f(x) œ L. If lim b f(x) Á lim c f(x), then xlim f(x) does not exist. Äa Äa xÄa xÄa xÄa xÄa 44. Since xlim f(x) œ L if and only if lim b f(x) œ L and lim c f(x) œ L, then xlim f(x) can be found by calculating Äc Äc xÄc xÄc lim b f(x). xÄc

45. If f is an odd function of x, then f(x) œ f(x). Given lim b f(x) œ 3, then lim c f(x) œ $. xÄ! xÄ! 46. If f is an even function of x, then f(x) œ f(x). Given lim c f(x) œ 7 then xÄ#

can be said about

lim

x Ä #c

lim

x Ä #b

f(x) œ 7. However, nothing

f(x) because we don't know lim b f(x). xÄ#

47. I œ (5ß 5 $ ) Ê 5 x & $ . Also, Èx 5 % Ê x 5 %# Ê x & %# . Choose $ œ %# Ê lim Èx 5 œ 0. x Ä &b

48. I œ (% $ ß %) Ê % $ x 4. Also, È% x % Ê % x %# Ê x % %# . Choose $ œ %# Ê lim È% x œ 0. x Ä %c

49. As x Ä 0 the number x is always negative. Thus, ¹ kxxk (1)¹ % Ê ¸ xx 1¸ % Ê 0 % which is always true independent of the value of x. Hence we can choose any $ 0 with $ x ! Ê

2 ¸ x 2 ¸ 50. Since x Ä # we have x 2 and kx 2k œ x 2. Then, ¹ kxx 2 k " ¹ œ x 2 " % Ê 0 %

which is always true so long as x #. Hence we can choose any $ !, and thus # x # $ 2 Ê ¹ kxx 2k "¹ % . Thus,

x 2

lim x Ä #b kx2k

x

lim x Ä ! c kx k

œ 1.


œ 1.

Section 2.5 Continuity 51. (a) (b)

lim

x Ä %!!b

67

ÚxÛ œ 400. Just observe that if 400 x 401, then ÚxÛ œ 400. Thus if we choose $ œ ", we have for any

number % ! that 400 x 400 $ Ê lÚxÛ 400l œ l400 400l œ ! %. lim c ÚxÛ œ 399. Just observe that if 399 x 400 then ÚxÛ œ 399. Thus if we choose $ œ ", we have for any

x Ä %!!

number % ! that 400 $ x 400 Ê lÚxÛ 399l œ l399 399l œ ! %. (c) Since lim b ÚxÛ Á lim c ÚxÛ we conclude that lim ÚxÛ does not exist. x Ä %!!

x Ä %!!

52. (a)

x Ä %!!

lim f(x) œ lim b Èx œ È0 œ 0; ¸Èx 0¸ % Ê % Èx % Ê ! x %# for x positive. Choose $ œ %# xÄ! Ê lim b f(x) œ 0.

x Ä !b

xÄ!

(b)

lim f(x) œ lim c x# sin ˆ x" ‰ œ 0 by the sandwich theorem since x# Ÿ x# sin ˆ x" ‰ Ÿ x# for all x Á 0. x Ä !c xÄ! Since kx# 0k œ kx# 0k œ x# % whenever kxk È%, we choose $ œ È% and obtain ¸x# sin ˆ "x ‰ 0¸ %

if $ x 0. (c) The function f has limit 0 at x! œ 0 since both the right-hand and left-hand limits exist and equal 0. 2.5 CONTINUITY 1. No, discontinuous at x œ 2, not defined at x œ 2 2. No, discontinuous at x œ 3, " œ lim c g(x) Á g(3) œ 1.5 xÄ$ 3. Continuous on [1ß 3] 4. No, discontinuous at x œ 1, 1.5 œ lim c k(x) Á lim b k(x) œ ! xÄ" xÄ" 5. (a) Yes

(b) Yes,

(c) Yes

(d) Yes

6. (a) Yes, f(1) œ 1

lim

x Ä "b

f(x) œ 0

(b) Yes, lim f(x) œ 2 xÄ1

(c) No

(d) No

7. (a) No

(b) No

8. ["ß !) (!ß ") ("ß #) (#ß $) 9. f(2) œ 0, since lim c f(x) œ 2(2) 4 œ 0 œ lim b f(x) xÄ# xÄ# 10. f(1) should be changed to 2 œ lim f(x) xÄ1

11. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( limc f(x) œ 1 and lim b f(x) œ 0). xÄ" xÄ1 xÄ"

Removable discontinuity at x œ 0 by assigning the number lim f(x) œ 0 to be the value of f(0) rather than f(0) œ 1. xÄ!

12. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( lim c f(x) œ 2 and lim b f(x) œ 1). xÄ" xÄ1 xÄ" Removable discontinuity at x œ 2 by assigning the number lim f(x) œ 1 to be the value of f(2) rather than f(2) œ 2. xÄ#


68


13. Discontinuous only when x 2 œ 0 Ê x œ 2

14. Discontinuous only when (x 2)# œ 0 Ê x œ 2

15. Discontinuous only when x# %x $ œ ! Ê (x 3)(x 1) œ 0 Ê x œ 3 or x œ 1 16. Discontinuous only when x# 3x 10 œ 0 Ê (x 5)(x 2) œ 0 Ê x œ 5 or x œ 2 17. Continuous everywhere. ( kx 1k sin x defined for all x; limits exist and are equal to function values.) 18. Continuous everywhere. ( kxk " Á 0 for all x; limits exist and are equal to function values.) 19. Discontinuous only at x œ 0 20. Discontinuous at odd integer multiples of 1# , i.e., x = (2n ") 1# , n an integer, but continuous at all other x. 21. Discontinuous when 2x is an integer multiple of 1, i.e., 2x œ n1, n an integer Ê x œ

n1 # ,

n an integer, but

continuous at all other x. 22. Discontinuous when

1x #

is an odd integer multiple of 1# , i.e.,

1x #

œ (2n 1) 1# , n an integer Ê x œ 2n 1, n an

integer (i.e., x is an odd integer). Continuous everywhere else. 23. Discontinuous at odd integer multiples of 1# , i.e., x = (2n 1) 1# , n an integer, but continuous at all other x. 24. Continuous everywhere since x% 1 1 and " Ÿ sin x Ÿ 1 Ê 0 Ÿ sin# x Ÿ 1 Ê 1 sin# x 1; limits exist and are equal to the function values. 25. Discontinuous when 2x 3 0 or x 3# Ê continuous on the interval 3# ß _‰ . 26. Discontinuous when 3x 1 0 or x

" 3

Ê continuous on the interval 3" ß _‰ .

27. Continuous everywhere: (2x 1)"Î$ is defined for all x; limits exist and are equal to function values. 28. Continuous everywhere: (2 x)"Î& is defined for all x; limits exist and are equal to function values. 29. Continuous everywhere since lim xÄ3 30. Discontinuous at x œ 2 since

x2 x 6 x3

œ lim xÄ3

ax 3bax 2b x3

œ lim ax 2b œ 5 œ ga3b xÄ3

lim faxb does not exist while fa2b œ 4. x Ä 2

31. xlim sin (x sin x) œ sin (1 sin 1) œ sin (1 0) œ sin 1 œ 0, and function continuous at x œ 1. Ä1 32. lim sin ˆ 1# cos (tan t)‰ œ sin ˆ 1# cos (tan (0))‰ œ sin ˆ 1# cos (0)‰ œ sin ˆ 1# ‰ œ 1, and function continuous at t œ !. tÄ!

33. lim sec ay sec# y tan# y 1b œ lim sec ay sec# y sec# yb œ lim sec a(y 1) sec# yb œ sec a(" ") sec# 1b yÄ1

yÄ1

yÄ1

œ sec 0 œ 1, and function continuous at y œ ". 34. lim tan 14 cos ˆsin x"Î$ ‰‘ œ tan 14 cos (sin(0))‘ œ tan ˆ 14 cos (0)‰ œ tan ˆ 14 ‰ œ 1, and function continuous at x œ !. xÄ!


Section 2.5 Continuity 35. lim cos ’ È19 13 sec 2t “ œ cos ’ È19 13 sec 0 “ œ cos tÄ!

1 È16

œ cos

1 4

œ

È2 # ,

and function continuous at t œ !.

36. lim1 Écsc# x 5È3 tan x œ Écsc# ˆ 16 ‰ 5È3 tan ˆ 16 ‰ œ Ê4 5È3 Š È"3 ‹ œ È9 œ 3, and function continuous at xÄ

'

x œ 1' . 37. g(x) œ

x# 9 x3

(x 3)(x 3) (x 3)

œ

38. h(t) œ

t# 3t 10 t#

39. f(s) œ

s$ " s# 1

40. g(x) œ

œ

œ

œ x 3, x Á 3 Ê g(3) œ lim (x 3) œ 6 xÄ$

(t 5)(t 2) t#

as# s 1b (s 1) (s 1)(s 1)

x# 16 x# 3x 4

œ

œ t 5, t Á # Ê h(2) œ lim (t 5) œ 7 tÄ#

s# s " s1 ,

œ

(x 4)(x 4) (x 4)(x 1)

œ

x4 x1

s Á 1 Ê f(1) œ lim Š s sÄ1

#

s1 s1 ‹

4‰ , x Á 4 Ê g(4) œ lim ˆ xx 1 œ

xÄ%

œ

3 #

8 5

41. As defined, lim c f(x) œ (3)# 1 œ 8 and lim b (2a)(3) œ 6a. For f(x) to be continuous we must have xÄ$ xÄ$ 6a œ 8 Ê a œ 43 .

42. As defined,

lim

x Ä #c

g(x) œ 2 and

4b œ 2 Ê b œ "# .

lim

x Ä #b

g(x) œ b(2)# œ 4b. For g(x) to be continuous we must have

43. As defined, lim c f(x) œ 12 and lim b f(x) œ a# a2b 2a œ 2a# 2a. For f(x) to be continuous we must have xÄ# xÄ# 12 œ 2a# 2a Ê a œ 3 or a œ 2. 44. As defined, lim c g(x) œ b b1

xÄ0

0b b1

œ

b b1

œ b Ê b œ 0 or b œ 2.

45. As defined,

lim

x Ä 1 c

f(x) œ 2 and

and lim b g(x) œ a0b2 b œ b. For g(x) to be continuous we must have xÄ0

lim

x Ä 1 b

f(x) œ aa1b b œ a b, and

lim f(x) œ aa1b b œ a b and

x Ä 1c

lim f(x) œ 3. For f(x) to be continuous we must have 2 œ a b and a b œ 3 Ê a œ

x Ä 1b

5 #

and b œ "# .

46. As defined, lim c g(x) œ aa0b 2b œ 2b and lim b g(x) œ a0b2 3a b œ 3a b, and xÄ0 xÄ0 lim g(x) œ a2b2 3a b œ 4 3a b and lim b g(x) œ 3a2b 5 œ 1. For g(x) to be continuous we must xÄ0

x Ä 2c

have 2b œ 3a b and 4 3a b œ 1 Ê a œ 3# and b œ 3# .


69

70


47. The function can be extended: f(0) ¸ 2.3.

48. The function cannot be extended to be continuous at x œ 0. If f(0) ¸ 2.3, it will be continuous from the right. Or if f(0) ¸ 2.3, it will be continuous from the left.

49. The function cannot be extended to be continuous at x œ 0. If f(0) œ 1, it will be continuous from the right. Or if f(0) œ 1, it will be continuous from the left.

50. The function can be extended: f(0) ¸ 7.39.

51. f(x) is continuous on [!ß "] and f(0) 0, f(1) 0 Ê by the Intermediate Value Theorem f(x) takes on every value between f(0) and f(1) Ê the equation f(x) œ 0 has at least one solution between x œ 0 and x œ 1.

52. cos x œ x Ê (cos x) x œ 0. If x œ 1# , cos ˆ 1# ‰ ˆ 1# ‰ 0. If x œ 1# , cos ˆ 1# ‰ for some x between

1 #

and

1 #

1 #

0. Thus cos x x œ 0

according to the Intermediate Value Theorem, since the function cos x x is continuous.

53. Let f(x) œ x$ 15x 1, which is continuous on [4ß 4]. Then f(4) œ 3, f(1) œ 15, f(1) œ 13, and f(4) œ 5. By the Intermediate Value Theorem, f(x) œ 0 for some x in each of the intervals % x 1, " x 1, and " x 4. That is, x$ 15x 1 œ 0 has three solutions in [%ß 4]. Since a polynomial of degree 3 can have at most 3 solutions, these are the only solutions. 54. Without loss of generality, assume that a b. Then F(x) œ (x a)# (x b)# x is continuous for all values of x, so it is continuous on the interval [aß b]. Moreover F(a) œ a and F(b) œ b. By the Intermediate Value Theorem, since a a # b b, there is a number c between a and b such that F(x) œ a # b .


Section 2.5 Continuity 55. Answers may vary. Note that f is continuous for every value of x. (a) f(0) œ 10, f(1) œ 1$ 8(1) 10 œ 3. Since $ 1 10, by the Intermediate Value Theorem, there exists a c so that ! c 1 and f(c) œ 1. (b) f(0) œ 10, f(4) œ (4)$ 8(4) 10 œ 22. Since 22 È3 10, by the Intermediate Value Theorem, there exists a c so that 4 c 0 and f(c) œ È3. (c) f(0) œ 10, f(1000) œ (1000)$ 8(1000) 10 œ 999,992,010. Since 10 5,000,000 999,992,010, by the Intermediate Value Theorem, there exists a c so that ! c 1000 and f(c) œ 5,000,000.

56. All five statements ask for the same information because of the intermediate value property of continuous functions. (a) A root of f(x) œ x$ 3x 1 is a point c where f(c) œ 0. (b) The points where y œ x$ crosses y œ 3x 1 have the same y-coordinate, or y œ x$ œ 3x 1 Ê f(x) œ x$ 3x 1 œ 0. (c) x$ 3x œ 1 Ê x$ 3x 1 œ 0. The solutions to the equation are the roots of f(x) œ x$ 3x 1. (d) The points where y œ x$ 3x crosses y œ 1 have common y-coordinates, or y œ x$ 3x œ 1 Ê f(x) œ x$ 3x 1 œ !. (e) The solutions of x$ 3x 1 œ 0 are those points where f(x) œ x$ 3x 1 has value 0. 57. Answers may vary. For example, f(x) œ

sin (x 2) x2

is discontinuous at x œ 2 because it is not defined there.

However, the discontinuity can be removed because f has a limit (namely 1) as x Ä 2. 58. Answers may vary. For example, g(x) œ

" x1

has a discontinuity at x œ 1 because lim g(x) does not exist. x Ä "

Š lim c g(x) œ _ and lim b g(x) œ _.‹ x Ä " x Ä " 59. (a) Suppose x! is rational Ê f(x! ) œ 1. Choose % œ "# . For any $ 0 there is an irrational number x (actually infinitely many) in the interval (x! $ ß x! $ ) Ê f(x) œ 0. Then 0 kx x! k $ but kf(x) f(x! )k œ 1 "# œ %, so x lim f(x) fails to exist Ê f is discontinuous at x! rational. Äx !

On the other hand, x! irrational Ê f(x! ) œ 0 and there is a rational number x in (x! $ ß x! $ ) Ê f(x) œ 1. Again x lim f(x) fails to exist Ê f is discontinuous at x! irrational. That is, f is discontinuous at Äx !

every point. (b) f is neither right-continuous nor left-continuous at any point x! because in every interval (x! $ ß x! ) or (x! ß x! $ ) there exist both rational and irrational real numbers. Thus neither limits lim f(x) and x Ä x!

lim f(x) exist by the same arguments used in part (a).

x Ä x !

60. Yes. Both f(x) œ x and g(x) œ x g ˆ "# ‰

œ0 Ê

f(x) g(x)

" #

are continuous on [!ß "]. However

is discontinuous at x œ

f(x) g(x)

is undefined at x œ

" #

since

" #.

61. No. For instance, if f(x) œ 0, g(x) œ ÜxÝ, then h(x) œ 0 aÜxÝb œ 0 is continuous at x œ 0 and g(x) is not. 62. Let f(x) œ œ

" x1

" (x 1) 1

œ

and g(x) œ x 1. Both functions are continuous at x œ 0. The composition f ‰ g œ f(g(x)) " x

is discontinuous at x œ 0, since it is not defined there. Theorem 10 requires that f(x) be

continuous at g(0), which is not the case here since g(0) œ 1 and f is undefined at 1. 63. Yes, because of the Intermediate Value Theorem. If f(a) and f(b) did have different signs then f would have to equal zero at some point between a and b since f is continuous on [aß b].


71

72


64. Let f(x) be the new position of point x and let d(x) œ f(x) x. The displacement function d is negative if x is the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the Intermediate Value Theorem, d(x) œ 0 for some point in between. That is, f(x) œ x for some point x, which is then in its original position. 65. If f(0) œ 0 or f(1) œ 1, we are done (i.e., c œ 0 or c œ 1 in those cases). Then let f(0) œ a 0 and f(1) œ b 1 because 0 Ÿ f(x) Ÿ 1. Define g(x) œ f(x) x Ê g is continuous on [0ß 1]. Moreover, g(0) œ f(0) 0 œ a 0 and g(1) œ f(1) 1 œ b 1 0 Ê by the Intermediate Value Theorem there is a number c in (!ß ") such that g(c) œ 0 Ê f(c) c œ 0 or f(c) œ c. 66. Let % œ

kf(c)k #

0. Since f is continuous at x œ c there is a $ 0 such that kx ck $ Ê kf(x) f(c)k %

Ê f(c) % f(x) f(c) %. If f(c) 0, then % œ "# f(c) Ê " #

" #

If f(c) 0, then % œ f(c) Ê

f(c) f(x) 3 #

3 #

f(c) f(x)

f(c) Ê f(x) 0 on the interval (c $ ß c $ ). " #

f(c) Ê f(x) 0 on the interval (c $ ß c $ ).

67. By Exercises 52 in Section 2.3, we have xlim faxb œ L Í lim fac hb œ L. Äc hÄ0

Thus, faxb is continuous at x œ c Í xlim faxb œ facb Í lim fac hb œ facb. Äc hÄ0

68. By Exercise 67, it suffices to show that lim sinac hb œ sin c and lim cosac hb œ cos c. hÄ0

hÄ0

Now lim sinac hb œ lim asin cbacos hb acos cbasin hb‘ œ asin cbŠ lim cos h‹ acos cbŠ lim sin h‹ hÄ0

hÄ0

hÄ0

hÄ0

By Example 11 Section 2.2, lim cos h œ " and lim sin h œ !. So lim sinac hb œ sin c and thus faxb œ sin x is hÄ0

continuous at x œ c. Similarly,

hÄ0

hÄ0

lim cosac hb œ lim acos cbacos hb asin cbasin hb‘ œ acos cbŠ lim cos h‹ asin cbŠ lim sin h‹ œ cos c.

hÄ0

hÄ0

Thus, gaxb œ cos x is continuous at x œ c.

hÄ0

69. x ¸ 1.8794, 1.5321, 0.3473

70. x ¸ 1.4516, 0.8547, 0.4030

71. x ¸ 1.7549

72. x ¸ 1.5596

73. x ¸ 3.5156

74. x ¸ 3.9058, 3.8392, 0.0667

75. x ¸ 0.7391

76. x ¸ 1.8955, 0, 1.8955

hÄ0


Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 2.6 LIMITS INVOLVING INFINITY; ASMYPTOTES OF GRAPHS 1. (a) (c) (e) (g)

lim f(x) œ 0

(b)

xÄ2

lim

x Ä 3 c

f(x) œ 2

lim

x Ä 3 b

f(x) œ 2

(d) lim f(x) œ does not exist xÄ3

lim b f(x) œ 1

(f)

lim f(x) œ does not exist

(h) x lim f(x) œ 1 Ä_

xÄ0 xÄ0

lim f(x) œ _

x Ä 0c

(i) x Ä lim f(x) œ 0 _ 2. (a) (c)

lim f(x) œ 2

(b)

lim f(x) œ 1

(d) lim f(x) œ does not exist

xÄ4

x Ä 2c

xÄ2

lim f(x) œ _ x Ä 3 b (g) lim f(x) œ _ (e)

x Ä 3

(i)

lim f(x) œ 3

x Ä 2b

lim f(x) œ _

(f)

x Ä 3 c

lim

(h)

x Ä 0b

(k) x lim f(x) œ 0 Ä_

lim f(x) œ _

lim f(x) œ does not exist

(j)

x Ä 0c

xÄ0

(l) x Ä lim f(x) œ 1 _

Note: In these exercises we use the result

"

lim mÎn xÄ „_ x

Theorem 8 and the power rule in Theorem 1:

lim

xÄ „_

œ 0 whenever

ˆ xm"În ‰ œ

lim

(b) 3

4. (a) 1

(b) 1

5. (a)

" #

(b)

" #

6. (a)

" 8

(b)

" 8

7. (a) 53

(b)

10. 3") Ÿ 11.

lim

tÄ_

12. r Ä lim_

0. This result follows immediately from ˆ x" ‰mÎn œ Š

"

lim ‹ xÄ „_ x

mÎn

(b) 53

3 4

9. "x Ÿ

m n

xÄ „_

3. (a) 3

8. (a)

f(x) œ _

sin 2x x

Ÿ

" x

cos ) 3)

Ÿ

" 3)

2 t sin t t cos t

Ê x lim Ä_ Ê

13. (a) x lim Ä_

2x 3 5x 7

lim

) Ä _

œ lim

2 t

tÄ_

r sin r 2r 7 5 sin r

sin 2x x

œrÄ lim_

œ x lim Ä_

œ 0 by the Sandwich Theorem

cos ) 3)

œ 0 by the Sandwich Theorem

1 ˆ sint t ‰ 1 ˆ cost t ‰

œ

1 ˆ sinr r ‰ 2 7r 5 ˆ sinr r ‰ 2 3x 5 7x

3 4

œ

2 5

010 10

œ 1

œrÄ lim_

10 200

œ

(b)

" #

2 5

(same process as part (a))


œ 0mÎn œ 0.

73

74

Chapter 2 Limits and Continuity 2 Š x7$ ‹

$

2x 7 14. (a) x lim œ x lim Ä _ x$ x# x 7 Ä_ (b) 2 (same process as part (a)) " x

x"#

15. (a) x lim Ä_

x1 x# 3

œ x lim Ä_

1 x3#

16. (a) x lim Ä_

3x 7 x# 2

œ x lim Ä_

1 x2#

17. (a) x lim Ä_

7x$ x$ 3x# 6x

18. (a) x lim Ä_

x$

20. (a) x lim Ä_ 9 #

2x%

x7#

œ x lim Ä_

œ x lim Ä_

10x& x% 31 x'

19. (a) x lim Ä_

(b)

" 4x 1

3 x

œ2

œ0

(b) 0 (same process as part (a))

œ0

(b) 0 (same process as part (a)) œ(

7 1 3x x6# " x$ 4 x"$ x#

1

œ x lim Ä_

2


œ!

x"# x31' 1

10 x

œ x lim Ä_

9x% x 5x# x 6

1 "x x"# x7$


œ0


9 x"$

5 x#

x"$ x6%

œ

9 #

(same process as part (a)) 2x$ 2x 3 3x$ 3x# 5x

21. (a) x lim Ä_

2 x2# x3$

œ x lim Ä_

œ 23

3 3x x5#

(b) 23 (same process as part (a)) x % x% 7x$ 7x# 9

22. (a) x lim Ä_

" 1 7x x7# x9%

œ x lim Ä_

œ 1

(b) 1 (same process as part (a)) 8

8

3

3

3 x2 x2 É 8x 23. x lim œ Êx lim œ É 82 00 œ È4 œ 2 2x2 x œ x lim Ä_ Ä _ Ê 2 1x Ä _ 2 1x 2

24. x Ä lim Šx x1‹ _ 8x2 3 2

1 Î3

œxÄ lim _Œ

5

1

" 1x x12 8

x

3 x2

1 Î3

œ Œx Ä lim _

5

1

" 1x x12 8

3 x2

5

x

1 Î3

œ ˆ " 8 0 0 0 ‰

1 Î3

œ ˆ "8 ‰

1 Î3

œ

x2 x2 ‰ œ_ 25. x Ä lim lim œ Œx Ä lim œ ˆ 01_ Š 1x ‹ œ x Ä 0 _ x2 7x _Œ 1 7x _ 1 7x 3

1

5

1

5

5

x x x2 x2 É x 5x œ x lim 26. x lim œ Êx lim œ É 1 0 0 0 0 œ È0 œ 0 Ä _ x3 x 2 Ä _Ê 1 x12 x23 Ä _ 1 x12 x23 2

27. x lim Ä_

2Èx xc" 3x 7

29. x Ä lim _

30. x lim Ä_

œ x lim Ä_

3 xÈ 5 x È 3 xÈ 5 x È

x "x % x #x $

Š

œxÄ lim _

œ x lim Ä_

2 ‹ Š x"# ‹ x"Î# 3 7x

œ0

1 xÐ"Î&Ñ Ð"Î$Ñ 1 xÐ"Î&Ñ Ð"Î$Ñ

x x"# 1 x"

œxÄ lim _

28. x lim Ä_ " ‹ 1 Š #Î"& x

" ‹ 1 Š #Î"&

2 Èx 2 Èx

œ x lim Ä_

2 ‹" x"Î# 2 Š "Î# ‹1 x

Š

œ 1

œ1

x

œ_


" #

Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 2x&Î$ x"Î$ 7 x)Î& 3x Èx

31. x lim Ä_

3 x 5x 3 È 2x x#Î$ 4

32. x Ä lim _ 33. x lim Ä_

È x2 1 x1

34. x Ä lim _

œ x lim Ä_ œxÄ lim _

œ x lim Ä_

È x2 1 x1

1

" x#Î$

2

È x 2 1 ÎÈ x 2 a x 1 b ÎÈ x 2

œxÄ lim _

" 7 x"*Î"& x)Î& 3 " x$Î& x""Î"!

2x"Î"&

5 3x " x"Î$

È

3 œ 5# È x

4x

œ x lim Ä_

È x 2 1 ÎÈ x 2 a x 1 bÎ È x 2

œ_

È a x 2 1 bÎx 2 ax 1bÎx

œ x lim Ä_

È 1 1 Îx 2 a1 1 Îx b

È a x 2 1 bÎ x 2

È1 0 a1 0 b

œ

È 1 1 Îx 2

œxÄ lim œ x lim œ _ ax 1bÎaxb Ä _ a 1 1 Î x b

œ1

È1 0 a1 0b

a x 3 bÎ x a x 3 bÎ x a1 3 Îx b x3 35. x lim œ x lim œ x lim œ x lim œ Ä _ È4x2 25 Ä _ È4x2 25ÎÈx2 Ä _ Èa4x2 25bÎx2 Ä _ È4 25Îx2 ˆ4 3x3 ‰ÎÈx6

4 3x 36. x Ä lim œxÄ lim œxÄ lim _ Èx6 9 _ Èx6 9ÎÈx6 _ 3

"

œ_

37.

lim x Ä !b 3x

39.

lim x Ä #c x 2

41.

lim x Ä )b x8

3

2x

4

43. lim

# x Ä ( (x7)

œ _ œ _ œ_

lim "Î$ x Ä !b 3x

46. (a)

lim "Î& x Ä !b x 4

47. lim

#Î& xÄ! x

49.

51. 52.

lim

x Ä ˆ 1# ‰

œ lim

4

# x Ä ! ax"Î& b

œ x lim Ä_

ˆ 4 Îx 3 3‰ È 1 9 Îx 6

Š positive positive ‹

40.

lim x Ä $b x 3

Š negative positive ‹

42.

lim x Ä &c 2x10

positive Š positive ‹

44. lim

œ_

"

3x

"

# x Ä ! x (x1)

2

(b)

lim "Î$ x Ä !c 3x

(b)

lim "Î& x Ä !c x

48. lim

"

#Î$ xÄ! x

tan x œ _

50.

œ3

œ_

positive Š negative ‹

2

a0 3 b È1 0

" #


lim x Ä !c 2x

5

œ

œ

œ _

38.

œ_

2

ˆ4 3x3 ‰Îˆx3 ‰ Èax6 9bÎx6

positive Š positive ‹

œ_

2

45. (a)

a1 0 b È4 0

2

œ 1

œ_

Š negative negative ‹

œ _

negative Š positive †positive ‹

œ _ œ _

œ lim

"

# x Ä ! ax"Î$ b

œ_

lim sec x œ _

x Ä ˆ #1 ‰

lim (1 csc )) œ _

)Ä!

lim (2 cot )) œ _ and lim c (2 cot )) œ _, so the limit does not exist )Ä!

) Ä !b

"

œ lim b xÄ#

" (x2)(x2)

œ_

Š positive"†positive ‹

"

œ lim c xÄ#

" (x2)(x2)

œ _

Š positive†"negative ‹

53. (a)

lim # x Ä # b x 4

(b)

lim # x Ä # c x 4

(c)

lim # x Ä #b x 4

(d)

lim # x Ä #c x 4

"

œ

lim x Ä #b (x2)(x2)

"

œ _

Š positive†"negative ‹

"

œ

lim x Ä #c (x2)(x2)

"

œ_

Š negative"†negative ‹


75

76


54. (a)

lim # x Ä "b x 1

(b)

lim # x Ä "c x 1

(c)

lim # x Ä "b x 1

(d)

lim # x Ä "c x 1

x

œ lim b xÄ"

x (x1)(x1)

œ_

positive Š positive †positive ‹

x

œ lim c xÄ"

x (x1)(x1)

œ _

positive Š positive †negative ‹

x

œ

lim x Ä "b (x1)(x1)

x

œ_

negative Š positive †negative ‹

x

œ

lim x Ä "c (x1)(x1)

x

œ _

negative Š negative †negative ‹

55. (a)

lim x Ä !b #

x#

" x

œ 0 lim b xÄ!

" x

œ _

" Š negative ‹

(b)

lim x Ä !c #

x#

" x

œ 0 lim c xÄ!

" x

œ_

" Š positive ‹

(c)

lim # x Ä $È2

(d)

lim x Ä 1 #

56. (a)

x#

x#

lim x Ä #b

" x

" x

œ

x# 1 2x 4 x# 1

2#Î$ #

œ

" #

(d)

lim x Ä !c 2x 4

œ

(b) (c) (d) (e) 58. (a)

x# 3x 2 x$ 2x#

lim b

x# 3x 2 x$ 2x#

lim

xÄ#

#

x 3x 2 x$ 2x# x 3x 2 x$ 2x#

lim

œ lim c xÄ#

lim x Ä #b

(c)

x Ä 0c

(d)

x Ä "b

(e)

lim x Ä !b x(x #)

x# 1 2x 4

lim x Ä #c

œ _


œ0

(x 2)(x 1) x# (x 2)

œ _

(x 2)(x 1) x# (x 2)

œ lim b xÄ#

(x 2)(x 1) x# (x 2)

œ lim c xÄ#

œ lim

œ lim

(x 2)(x 1) x# (x 2)

œ _

xÄ!

lim

x# 3x 2 x$ 4x

lim

x# 3x 2 x$ 4x x"

2†0 #4

(x 2)(x 1) x# (x 2)

xÄ#

œ lim b xÄ#

x# 3x 2 x$ 4x

(b)

œ

œ lim

x# 3x 2 x$ 4x

lim

x Ä #b

and

œ lim b xÄ#

x 3x 2 x$ 2x#

#

xÄ!

œ lim b xÄ!

#

lim

x Ä #c

(x 1)(x 1) 2x 4

(b)

" 4

lim b

xÄ#

3 #

Š positive positive ‹

œ lim b xÄ"

xÄ!

œ 2"Î$ 2"Î$ œ 0

œ_

lim x Ä "b 2x 4

57. (a)

" #"Î$

ˆ "1 ‰ œ

(c)

x# 1

œ

xÄ#

(x 2)(x ") x(x #)(x 2)

(x 2)(x ")

œ lim c xÄ! œ lim b xÄ"

(x 2)(x ") x(x #)(x 2) (x 2)(x ") x(x #)(x 2)

œ

x1 x#

œ

" 4

,xÁ2

x1 x#

œ

" 4

,xÁ2

x1 x#

œ

" 4

,xÁ2 †negative Š negative positive†negative ‹

œ lim b xÄ#

lim x Ä #b x(x #)(x 2)

†negative Š negative positive†negative ‹

(x 1) x(x #)

œ

(x 1)

lim x Ä #b x(x #)

œ lim c xÄ! œ lim b xÄ"

œ_

(x 1) x(x #)

œ

negative Š positive †positive ‹

x"

negative Š negative †positive ‹

œ_

œ

" 8

œ_

(x 1) x(x #)

œ _

lim x Ä !c x(x #)

" #(4)

0 (1)(3)

negative Š negative †positive ‹ negative Š negative †positive ‹

œ0

so the function has no limit as x Ä 0. lim 2

59. (a)

t Ä !b

60. (a)

t Ä !b

61. (a)

x Ä !b

(c)

x Ä "b

3 ‘ t"Î$

œ _

" lim t$Î& 7‘ œ _

lim

2

lim

lim

" ’ x#Î$

2 “ (x 1)#Î$

œ_

lim

" ’ x#Î$

2 “ (x 1)#Î$

œ_

(b)

t Ä !c

(b)

t Ä !c

lim

" ’ x#Î$

2 “ (x 1)#Î$

œ_

(b)

x Ä !c

lim

" ’ x#Î$

2 “ (x 1)#Î$

œ_

(d)

x Ä "c

" t$Î&

3 ‘ t"Î$

œ_

7‘ œ _


Section 2.6 Limits Involving Infinity; Asymptotes of Graphs lim

" ’ x"Î$

1 “ (x 1)%Î$

œ_

(b)

x Ä !c

lim

" ’ x"Î$

1 “ (x 1)%Î$

œ _

(d)

x Ä "c

62. (a)

x Ä !b

(c)

x Ä "b

lim

" ’ x"Î$

1 “ (x 1)%Î$

œ _

lim

" ’ x"Î$

1 “ (x 1)%Î$

œ _

63. y œ

" x1

64. y œ

" x1

65. y œ

" #x 4

66. y œ

3 x3

67. y œ

x3 x2

68. y œ

2x x1

œ1

" x#

69. Here is one possibility.

œ#

2 x1



77

78








77. Yes. If x lim Ä_

f(x) g(x)

œ 2 then the ratio of the polynomials' leading coefficients is 2, so x Ä lim _

f(x) g(x)

œ 2 as well.

78. Yes, it can have a horizontal or oblique asymptote. 79. At most 1 horizontal asymptote: If x lim Ä_ f(x) lim x Ä _ g(x)

f(x) g(x)

œ L, then the ratio of the polynomials' leading coefficients is L, so

œ L as well.

Èx 9 Èx 4 80. x lim Š Èx 9 Èx 4‹ œ x lim ’Èx 9 Èx 4“ † ’ Èx 9 Èx 4 “ œ x lim Ä_ Ä_ Ä_ 5 Èx 5 0 œ x lim œ x lim œ 11 œ 0 9 4 Ä _ Èx 9 Èx 4 Ä_

ax 9 b a x 4 b Èx 9 Èx 4

É1 x É1 x

È 2 È 2 81. x lim Š Èx2 25 Èx2 "‹ œ x lim ’Èx2 25 Èx2 "“ † ’ Èx2 25 Èx2 " “ œ x lim Ä_ Ä_ Ä_ x 25 x "

œ x lim Ä_

26 Èx2 25 Èx2 "

œ x lim Ä_

26 x

É1 x252 É1 x12

œ

0 11

œ0

È 2 ˆx 3 ‰ ˆ x ‰ x 82. x Ä lim lim lim Š Èx2 3 x‹ œ x Ä ’Èx2 3 x“ † ’ Èxx2 33 “œxÄ _ _ _ Èx2 3 x x 3 È x2 3x 3 œxÄ lim œ lim œ lim œ 1 0 1 œ 0 2 È 3 x _ x Ä _ É1 2 È x Ä _ É1 32 1 x 3x x x x2 2

2


ˆx2 25‰ ˆx2 "‰ Èx2 25 Èx2 "

Section 2.6 Limits Involving Infinity; Asymptotes of Graphs È

ˆ4x2 ‰ ˆ4x2 3x 2‰

4x 3x 2 83. x Ä lim lim lim Š 2x È4x2 3x 2‹ œ x Ä ’2x È4x2 3x 2“ † ’ 2x “œxÄ _ _ _ 2x È4x2 3x 2 2x È4x2 3x 2 c3x b 2 c3x b 2 3 x2 Èx2 3x 2 cx œxÄ lim œxÄ lim œxÄ lim œxÄ lim _ 2x È4x2 3x 2 _ È2x É4 3 22 _ 2x É4 3 22 _ 2 É4 3 22 x x x cx x x x x2 œ 3202 œ 43 2

È 2 84. x lim Š È9x2 x 3x‹ œ œ x lim ’È9x2 x 3x“ † ’ È9x2 x 3x “ œ x lim Ä_ Ä_ Ä_ 9x x 3x

œ x lim Ä_

x È9x2 x 3x

œ x lim Ä_

xx 2 É 9x2 x

xx2 3x x

1 É9 "x 3

œ x lim Ä_

œ

1 33

ˆ9x2 x‰ ˆ9x2 ‰ È9x2 x 3x

œ "6

È 2 È 2 85. x lim Š Èx2 3x Èx2 2x‹ œ x lim ’ Èx2 3x Èx2 2x“ † ’ Èx2 3x Èx2 2x “ œ x lim Ä_ Ä_ Ä_ x 3x x 2x 5x 5 5 5 œ x lim œ x lim œ 11 œ # È 2 3 2 Ä_ È 2 Ä_ x 3x

x 2x

É1 x É1 x

Èx# x Èx# x œ lim ’Èx# x Èx# x“ † ’ Èx# x Èx# x “ œ lim 86. x lim È x# x È x# x Ä_ xÄ_ xÄ_ 2x 2 2 œ x lim œ x lim œ 11 œ 1 È # " " Ä_ È # Ä_ x x

ˆx2 3x‰ ˆx2 2x‰ Èx2 3x Èx2 2x

x x

ax # x b a x # x b È x# x È x# x

É1 x É1 x

87. For any % 0, take N œ 1. Then for all x N we have that kf(x) kk œ kk kk œ 0 %. 88. For any % 0, take N œ 1. Then for all y N we have that kf(x) kk œ kk kk œ 0 %. " x#

89. For every real number B 0, we must find a $ 0 such that for all x, 0 kx 0k $ Ê

" x#

Ê

B ! Í " x#

" x#

#

B0 Í x

" B

" ÈB

Í kxk

. Choose $ œ

" ÈB

, then 0 kxk $ Ê kxk

xÄ!

B ! Í lxl B" . Choose $ œ B" . Then ! kx 0k $ Ê lxl

" B

Ê

" lx l

" lx l

2 (x 3)#

B ! Í

2 (x 3)#

$ œ É B2 , then 0 kx 3k $ Ê

B0 Í 2 (x 3)#

(x 3) 2

#

" B

Í (x 3)#

B 0 so that lim

2

# x Ä $ (x 3)

2 B

B. Now,

x Ä ! lx l 2 (x 3)#

Now,

#

B ! Í (x 5)

Ê kx 5k

" ÈB

Ê

" (x 5)#

" B

Í kx 5k

B so that lim

"

" ÈB

# x Ä & (x 5)

. Choose $ œ

œ _.

B.

Í ! kB $k É B2 . Choose

œ _.

92. For every real number B 0, we must find a $ 0 such that for all x, 0 kx (5)k $ Ê 1 (x 5)#

"

B so that lim

91. For every real number B 0, we must find a $ 0 such that for all x, 0 kx 3k $ Ê Now,

" ÈB

B so that lim x"# œ _.

90. For every real number B 0, we must find a $ 0 such that for all x, ! kx 0k $ Ê " lx l

B. Now,

" ÈB

1 (x 5)#

B.

. Then 0 kx (5)k $

œ _.

93. (a) We say that f(x) approaches infinity as x approaches x! from the left, and write lim c f(x) œ _, if x Ä x! for every positive number B, there exists a corresponding number $ 0 such that for all x, x! $ x x! Ê f(x) B. (b) We say that f(x) approaches minus infinity as x approaches x! from the right, and write lim b f(x) œ _, x Ä x!

if for every positive number B (or negative number B) there exists a corresponding number $ 0 such that for all x, x! x x! $ Ê f(x) B.


79

80

Chapter 2 Limits and Continuity (c) We say that f(x) approaches minus infinity as x approaches x! from the left, and write lim f(x) œ _, x Ä x!

if for every positive number B (or negative number B) there exists a corresponding number $ 0 such that for all x, x! $ x x! Ê f(x) B. 94. For B 0,

" x

B 0 Í x B" . Choose $ œ B" . Then ! x $ Ê 0 x

95. For B 0,

" x

B 0 Í x" B 0 Í x

Ê B" x Ê 96. For B !,

" x#

" x

B so that lim c xÄ!

" x

" B

Ê

" x#

" x#

Ê 99. y œ

101. y œ

" x#

B ! so that lim b xÄ#

œx1

x# % x"

B so that lim b xÄ!

" x

œ _.

" B

Í x 2 B" Í x 2 B" . Choose $ œ B" . Then " x#

B 0 so that lim c xÄ#

" x#

œ _.

B Í ! x 2 B" . Choose $ œ B" . Then # x # $ Ê ! x # $ Ê ! x 2

" 1 x#

" x"

œx"

$ x"

œ _.

B Í 1 x#

" #B . Then " $ x " Ê " 1 x# B for ! x 1 and x# x"

" x

œ _.

B Í x " # B Í (x 2)

98. For B 0 and ! x 1, $

Ê

Í B" x. Choose $ œ B" . Then $ x !

2 $ x 2 Ê $ x 2 ! Ê B" x 2 0 Ê 97. For B 0,

" B

" B

Í (" x)(" x) B" . Now

$ x 1 0 Ê " x $ x near 1 Ê

"

lim # x Ä "c " x

" #B

1x #

1 since x 1. Choose Ê (" x)(" x) B" ˆ 1 # x ‰ B"

œ _.

100. y œ

x# " x1

œx"

102. y œ

x2 " #x %

œ #" x "

# x1

$ #x %


" B

Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 103. y œ

x# 1 x

105. y œ

x È 4 x#

œx

107. y œ x#Î$

" x

104. y œ

x$ 1 x#

106. y œ

" È 4 x#

œx

" x#

108. y œ sin ˆ x# 1 1 ‰

" x"Î$

109. (a) y Ä _ (see accompanying graph) (b) y Ä _ (see accompanying graph) (c) cusps at x œ „ 1 (see accompanying graph)

110. (a) y Ä 0 and a cusp at x œ 0 (see the accompanying graph) (b) y Ä 32 (see accompanying graph) (c) a vertical asymptote at x œ 1 and contains the point Š1,

3 ‹ 3 2È 4

(see accompanying graph)


81

82


CHAPTER 2 PRACTICE EXERCISES 1. At x œ 1: Ê

f(x) œ

lim

x Ä "c

lim

x Ä "b

f(x) œ 1

lim f(x) œ 1 œ f(1)

x Ä 1

Ê f is continuous at x œ 1. At x œ 0: lim c f(x) œ lim b f(x) œ 0 Ê lim f(x) œ 0. xÄ!

xÄ!

xÄ!

But f(0) œ 1 Á lim f(x) xÄ!

Ê f is discontinuous at x œ 0. If we define fa!b œ !, then the discontinuity at x œ ! is removable. At x œ 1: lim c f(x) œ 1 and lim f(x) œ 1 xÄ"

Ê lim f(x) does not exist

xÄ"

xÄ1

Ê f is discontinuous at x œ 1. 2. At x œ 1: Ê

f(x) œ 0 and

lim

x Ä "

lim

x Ä "

f(x) œ 1

lim f(x) does not exist

x Ä "

Ê f is discontinuous at x œ 1. At x œ 0: lim f(x) œ _ and lim f(x) œ _ xÄ!

Ê lim f(x) does not exist

xÄ!

xÄ!

Ê f is discontinuous at x œ 0. At x œ 1: lim f(x) œ lim f(x) œ 1 Ê lim f(x) œ 1. xÄ"

xÄ1

xÄ"

But f(1) œ 0 Á lim f(x) xÄ1

Ê f is discontinuous at x œ 1. If we define fa"b œ ", then the discontinuity at x œ " is removable. 3. (a) (b) (c) (d) (e) (f)

lim a3fatbb œ 3 lim fatb œ 3(7) œ 21

t Ä t!

t Ä t!

#

lim afatbb# œ Š lim fatb‹ œ a(b# œ 49

t Ä t!

t Ä t!

lim afatb † gatbb œ lim fatb † lim gatb œ (7)(0) œ 0

t Ä t!

t Ä t!

lim fatb t Ä t! g(t)7

Ät

t Ä t!

lim fatb

œ

Ät

t

œ

!

lim agatb 7b

t

t

!

Ät

lim fatb

Ät

t

!

Ät

lim gatb lim 7 t

!

!

œ

7 07

œ1

lim cos agatbb œ cos Š lim gatb‹ œ cos ! œ 1

t Ä t!

t Ä t!

lim kfatbk œ ¹ lim fatb¹ œ k7k œ 7

t Ä t!

t Ä t!

(g) lim afatb gatbb œ lim fatb lim gatb œ 7 0 œ 7 t Ä t!

(h)

4. (a) (b) (c) (d)

t Ä t!

lim Š " ‹ t Ä t! fatb

œ

" lim fatb

t

Ät

t Ä t!

" 7

œ

!

œ 71

lim g(x) œ lim g(x) œ È2

xÄ!

xÄ!

lim ag(x) † f(x)b œ lim g(x) † lim f(x) œ ŠÈ2‹ ˆ "# ‰ œ

xÄ!

xÄ!

xÄ!

lim af(x) g(x)b œ lim f(x) lim g(x) œ

xÄ!

"

lim x Ä ! f(x)

œ

" lim f(x)

xÄ!

xÄ!

œ

" " #

œ2

xÄ!

" #

È2 #

È2


Chapter 2 Practice Exercises (e) (f)

" #

lim ax f(x)b œ lim x lim f(x) œ 0

xÄ!

xÄ!

f(x)†cos x x 1 xÄ!

lim

xÄ!

lim f(x)† lim cos x

œ

xÄ!

xÄ!

lim x lim 1

xÄ!

xÄ!

œ

ˆ "# ‰ (1) 01

œ

83

" #

œ #"

5. Since lim x œ 0 we must have that lim (4 g(x)) œ 0. Otherwise, if lim (% g(x)) is a finite positive xÄ!

xÄ!

xÄ!

’ 4xg(x) “

’ 4xg(x) “

œ _ and lim b œ _ so the limit could not equal 1 as xÄ! x Ä 0. Similar reasoning holds if lim (4 g(x)) is a finite negative number. We conclude that lim g(x) œ 4. number, we would have lim c xÄ!

xÄ!

6. 2 œ lim

x Ä %

xÄ!

’x lim g(x)“ œ lim x † lim xÄ!

x Ä %

’ lim g(x)“ œ 4 lim xÄ!

x Ä %

(since lim g(x) is a constant) Ê lim g(x) œ xÄ!

xÄ!

2 %

x Ä %

œ #" .

’ lim g(x)“ œ 4 lim g(x) xÄ!

xÄ!

7. (a) xlim faxb œ xlim x"Î$ œ c"Î$ œ facb for every real number c Ê f is continuous on a_ß _b. Äc Äc (b) xlim gaxb œ xlim x$Î% œ c$Î% œ gacb for every nonnegative real number c Ê g is continuous on Ò!ß _Ñ. Äc Äc " c#Î$ " c"Î'

(c) xlim haxb œ xlim x#Î$ œ Äc Äc (d) xlim kaxb œ xlim x"Î' œ Äc Äc

œ hacb for every nonzero real number c Ê h is continuous on a_ß !b and a_ß _b. œ kacb for every positive real number c Ê k is continuous on a!ß _b

8. (a) - ˆˆn "# ‰1ß ˆn "# ‰1‰, where I œ the set of all integers. n−I (b) - an1ß an 1b1b, where I œ the set of all integers. n−I (c) a_ß 1b a1ß _b (d) a_ß !b a!ß _b 9.

(a)

x# 4x 4 x2

lim x Ä !c x(x 7) (b) 10. (a)

#

x# x

œ lim

1 x# (x 1)

x# x

x # a# x % a%

œ xlim Äa

13. lim

(x h)# x# h

œ lim

(x h)# x# h xÄ!

œ lim

xÄ!

x

œ _ and lim b xÄ!

" Èx

ax # a # b ax # a # b a x # a # b

hÄ!

œ lim

xÄ!

1 x# (x 1)

2 (2 x) 2x(# x)

œ lim

#

x x

"

"

œ

œ

œ

0 2(9)

œ0

, x Á 0 and x Á 1.

œ _.

, x Á 0 and x Á 1. The limit does not

1

lim # x Ä "b x (x 1)

"

# x Ä 0 x (x 1)

& % $ x Ä ! x 2x x

# x Ä " x (x 1)

" x # a#

œ xlim Äa

œ lim

œ _ Ê lim

œ lim

x2

x Ä # x(x 7)

x1

x Ä 1 1 Èx

ax# 2hx h# b x# h xÄ!

, x Á 2, and lim

# x Ä ! x (x 1)(x 1)

œ lim

ax# 2hx h# b x# h

x2

œ lim

œ _ and

x Ä 1 ˆ1 È x ‰ ˆ 1 È x ‰

, x Á 2; the limit does not exist because

x Ä # x(x 7)

x(x 1)

"

12. xlim Äa

" " x #

x(x 1)

lim # x Ä "c x (x 1)

œ lim

#

œ lim

$ # x Ä " x ax 2x 1b

1 Èx 1x

15. lim

(x 2)(x 2)

œ lim

11. lim

14. lim

œ _

$ # x Ä ! x ax 2x 1b

lim & % $ x Ä " x 2x x

hÄ!

x2 x(x 7)

x2

x Ä ! x(x 7)

x Ä # x(x 7)(x #)

œ lim

lim & % $ x Ä ! x 2x x

xÄ1

œ lim

œ _ and lim b xÄ!

x 4x 4

exist because

(x 2)(x 2)

x Ä ! x(x 7)(x 2)

lim $ # x Ä # x 5x 14x

Now lim c xÄ! (b)

œ lim

lim $ # x Ä ! x 5x 14x

œ _.

" #

" #a #

œ lim (2x h) œ 2x hÄ!

œ lim (2x h) œ h xÄ!

"

x Ä ! 4 #x

œ "4


84

Chapter 2 Limits and Continuity (# x)$ 8 x

16. lim

xÄ!

ax$ 6x# 12x 8b 8 x

œ lim

xÄ!

ˆx1Î3 1‰ x1Î3 1 œ lim ˆÈx 1‰ È x 1 xÄ1 xÄ1 œ 1 1 1 1 1 œ 23

17. lim

18.

tan 2x

œ lim

sin 2x

x Ä ! cos 2x

lim csc x œ limc x1

20.

x Ä 1c

21.

xÄ1

22.

xÄ1

ax 1bˆÈx 1‰

œ lim

2Î3 1Î3 x Ä 1 ax 1bax x 1b

Èx 1

œ lim

2Î3 1Î3 x Ä 1 x x 1

1 sin x

†

1x ‰ˆ 1x ‰ˆ 2x ‰ œ lim ˆ sin2x2x ‰ˆ cos cos 2x sin 1x 1x œ 1 † 1 † 1 †

cos 1x sin 1x

xÄ!

2 1

œ

2 1

œ_

lim sin ˆ x2 sin x‰ œ sin ˆ 12 sin 1‰ œ sin ˆ 12 ‰ œ 1

lim cos2 ax tan xb œ cos2 a1 tan 1b œ cos2 a1b œ a1b2 œ 1

23. lim xÄ0 24. lim xÄ0

8x 3sin x x

œ lim xÄ0

cos 2x 1 sin x

2x 1 œ lim ˆ cossin † x xÄ0

8 3 sinx x 1

œ

8 3 a1 b 1

œ4

cos 2x 1 ‰ cos 2x 1

œ lim xÄ0

"Î$

lim [4 g(x)]"Î$ œ 2 Ê ’ lim b 4 g(x)“ x Ä !b xÄ! lim

x Ä È&

27. lim

xÄ1

28.

ˆx2Î3 x1Î3 1‰ˆÈx 1‰ ˆÈx 1‰ax2Î3 x1Î3 1b

ˆx1Î3 4‰ˆx1Î3 4‰ ˆx1Î3 4‰ˆx1Î3 4‰ ˆx2Î3 4x1Î3 16‰ˆÈx )‰ x2Î3 16 œ lim œ lim † ˆÈx )‰ax2Î3 4x1Î3 16b È È Èx 8 x 8 x 8 x Ä 64 x Ä 64 x Ä 64 ˆx1Î3 4‰ˆÈx )‰ ax 64bˆx1Î3 4‰ˆÈx )‰ 4 4ba8 8b 8 œ lim ax 64bax2Î3 4x1Î3 16b œ lim x2Î3 4x1Î3 16 œ a16 16 16 œ 3 x Ä 64 x Ä 64

x Ä ! tan 1x

26.

xÄ!

lim

19. lim

25.

†

œ lim ax# 6x 12b œ 12

" x g(x)

3x# 1 g(x)

xÄ1

5 x#

œ

(x g(x)) œ

lim

x Ä È&

œ2 Ê " #

œ lim xÄ0

sin2 2x sin xacos 2x 1b

œ lim xÄ0

4sin x cos2 x cos 2x 1

œ0 Ê

Ê È5 lim

x Ä È5

g(x) œ

" #

Ê

lim

x Ä È5

g(x) œ

" #

È5

xÄ1

lim g(x) œ _ since lim a5 x# b œ 1

x Ä #

x Ä #

lim f(x) œ lim c x Ä "c x Ä "

lim x Ä "c

x ax # 1 b x# 1

œ

lim

x Ä "c

x ax # 1 b kx # 1 k

x œ 1, and

#

lim f(x) œ lim b xkaxx# 11k b œ lim b x Ä "b x Ä " x Ä " œ lim (x) œ (1) œ 1. Since

x ax # 1 b a x # "b

x Ä 1

lim f(x) Á lim b f(x) x Ä "c x Ä " Ê

lim f(x) does not exist, the function f cannot be

x Ä 1

extended to a continuous function at x œ 1. At x œ 1:

lim f(x) œ lim c

x Ä "c

xÄ"

x ax # 1 b kx # 1 k

x ax # 1 b kx # 1 k

œ lim c xÄ"

x ax # 1 b x# "

x ax # 1 b ax # 1 b

œ lim c (x) œ 1, and xÄ"

lim f(x) œ lim b œ lim b œ lim b x œ 1. Again lim f(x) does not exist so f xÄ1 xÄ" xÄ" xÄ1 cannot be extended to a continuous function at x œ 1 either.

x Ä "b

œ

4a0ba1b2 11

lim 4 g(x) œ 8, since 2$ œ 8. Then lim b g(x) œ 2. xÄ!

x Ä !b

œ _ Ê lim g(x) œ 0 since lim a3x# 1b œ 4

lim x Ä # Èg(x)

29. At x œ 1:

œ2 Ê

cos2 2x 1 sin xacos 2x 1b


œ0

Chapter 2 Practice Exercises 30. The discontinuity at x œ 0 of f(x) œ sin ˆ "x ‰ is nonremovable because lim sin xÄ!

" x

does not exist.

31. Yes, f does have a continuous extension to a œ 1: " define f(1) œ lim xxÈ œ 43 . % x xÄ1

32. Yes, g does have a continuous extension to a œ 1# : ) 5 g ˆ 1# ‰ œ lim1 45)cos #1 œ 4 . )Ä #

33. From the graph we see that lim h(t) Á lim h(t) tÄ! tÄ! so h cannot be extended to a continuous function at a œ 0.

34. From the graph we see that lim c k(x) Á lim b k(x) xÄ! xÄ! so k cannot be extended to a continuous function at a œ 0.

35. (a) f(1) œ 1 and f(2) œ 5 Ê f has a root between 1 and 2 by the Intermediate Value Theorem. (b), (c) root is 1.32471795724 36. (a) f(2) œ 2 and f(0) œ 2 Ê f has a root between 2 and 0 by the Intermediate Value Theorem. (b), (c) root is 1.76929235424 # $

#x $ x 37. x lim œ x lim œ Ä _ &x ( Ä _ & (x

x 39. x Ä lim _

#

%x ) $x $

#! &!

ˆ" œxÄ lim _ $x

œ

#

# &

% $x#

#

#x $ 38. x Ä lim œxÄ lim _ &x# ( _ &

) ‰ $x$

$ x# ( x#

œ

#! &!

œ!!!œ!


œ

# &

85

86

Chapter 2 Limits and Continuity "

" x# 40. x lim œ x lim œ Ä _ x # (x " Ä _ " (x x"#

! "!!

œ!

#

%

x (x x( 41. x Ä lim œxÄ lim œ _ _ x 1 _ " "x

$

x x x" 42. x lim œxÄ lim œ_ Ä _ "#x$ "#) _ "# "#) x$

sin x " sin x 43. x lim Ÿ x lim œ ! since int x Ä _ as x Ä _ Êx lim œ !. Ä _ gx h Ä _ gx h Ä _ gx h

44.

lim

)Ä_

45. x lim Ä_

cos ) " )

Ÿ lim

#

) Ä _)

x sin x #Èx x sin x

#Î$

œ ! Ê lim

)Ä_

œ x lim Ä_

cos ) " )

" sinx x È#x " sinx x

"

œ

&Î$

x x " x 46. x lim œ x lim #x œ Ä _ x#Î$ cos# x Ä _Œ " cos#Î$

œ !.

"!! "!

"! "!

œ"

œ"

x

47. (a) y œ (b) y œ

x2 4 x3

is undefined at x œ 3: lim c xx 34 œ _ and lim b xx 34 œ _, thus x œ 3 is a vertical asymptote. xÄ3 xÄ3 2

x2 x 2 x2 2x 1

2

is undefined at x œ 1: lim c xÄ1

x2 x 2 x2 2x 1

œ _ and lim b xÄ1

x2 x 2 x2 2x 1

œ _, thus x œ 1 is a vertical

asymptote. (c) y œ

x2 x ' x2 2x 8

is undefined at x œ 2 and 4: lim

x x' 2

œ lim b x Ä %

lim 2 x Ä %b x 2x 8 48. (a) y œ

1 x2 1 x2 x2 " : x lim Ä _ x2 "

x3 x4

x2 x '

2 x Ä 2 x 2x 8

x3

œ lim

x Ä 2 x4

œ 56 ; lim c x Ä %

x2 x ' x2 2x 8

œ lim c x Ä %

x3 x4

œ_

œ _. Thus x œ 4 is a vertical asymptote. 1

1

x2 œ x lim Ä _ 1

1 x2

œ

1

1

1 1

1x œ 1 and x Ä lim œ lim x2 œ _ x2 " x Ä _ 1 x12

œ

10 È1 0

2

1 1

œ 1, thus y œ 1 is a

horizontal asymptote. (b) y œ

Èx 4 Èx 4 Èx 4 : x lim Ä _ Èx 4

(c) y œ

È x2 4 È x2 4 : x lim x x Ä_

œx Ä lim _

É1 x42 1

œ

œ x lim Ä_

È1 0 1

1 È4x É1 B4

œ x lim Ä_ œ

1 1

É1 x42 1

œ

œ 1 , thus y œ 1 is a horizontal asymptote.

È1 0 1

œ 1 and x lim Ä _

thus y œ

1

2

" 3

œx Ä lim_

É1 x42

È

9

" 3

x xx

2

is a horizontal asymptote.

0.1 0.7943

œx Ä lim _

É1 x42 x

cx

0.01 0.9550

0.001 0.9931

0.0001 0.9991

1

9

x 9 0 " x2 É 9x and x Ä lim lim œ É 19 21 œ 0 œ 3, _ x Ä _ Ê 9 x12

CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES 1. (a)

x x2

œ 1, thus y œ 1 and y œ 1 are horizontal asymptotes.

x 9 x 9 0 x2 (d) y œ É 9x lim É 9x lim œ É 91 21: 21 œ 0 œ xÄ_ x Ä _ Ê 9 x12 2

È x2 4 x

0.00001 0.9999

Apparently, lim b xx œ 1 xÄ!


Chapter 2 Additional and Advanced Exercises 87 (b)

2. (a)

x ˆ "x ‰"ÎÐln xÑ Apparently,

10

100

1000

0.3679

0.3679

0.3679

"ÎÐln xÑ lim ˆ " ‰ xÄ_ x

œ 0.3678 œ

" e

(b)

3.

lim L œ lim c L! É" v Ä cc vÄc

v# c#

lim v œ L! É1 vÄcc # œ L! É1 #

c# c#

œ0

The left-hand limit was needed because the function L is undefined if v c (the rocket cannot move faster than the speed of light). 4. (a) ¹

Èx #

1¹ 0.2 Ê 0.2

Èx #

1 0.2 Ê 0.8

Èx #

1.2 Ê 1.6 Èx 2.4 Ê 2.56 x 5.76.

(b) ¹

Èx #

1¹ 0.1 Ê 0.1

Èx #

1 0.1 Ê 0.9

Èx #

1.1 Ê 1.8 Èx 2.2 Ê 3.24 x 4.84.

5. k10 (t 70) ‚ 10% 10k 0.0005 Ê k(t 70) ‚ 10% k 0.0005 Ê 0.0005 (t 70) ‚ 10% 0.0005 Ê 5 t 70 5 Ê 65° t 75° Ê Within 5° F. 6. We want to know in what interval to hold values of h to make V satisfy the inequality lV "!!!l œ l$'1h "!!!l Ÿ "!. To find out, we solve the inequality: **! l$'1h "!!!l Ÿ "! Ê "! Ÿ $'1h "!!! Ÿ "! Ê **! Ÿ $'1h Ÿ "!"! Ê $' 1 Ÿ hŸ

"!"! $'1

Ê )Þ) Ÿ h Ÿ )Þ*. where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe. The interval in which we should hold h is about )Þ* )Þ) œ !Þ" cm wide (1 mm). With stripes 1 mm wide, we can expect to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking.

7. Show lim f(x) œ lim ax# 7b œ ' œ f(1). xÄ1

xÄ1

Step 1: kax 7b 6k % Ê % x# 1 % Ê 1 % x# 1 % Ê È1 % x È1 %. #


88

Chapter 2 Limits and Continuity Step 2: kx 1k $ Ê $ x 1 $ Ê $ " x $ ". Then $ " œ È1 % or $ " œ È1 %. Choose $ œ min š1 È1 %ß È1 % 1› , then 0 kx 1k $ Ê kax# (b 6k % and lim f(x) œ 6. By the continuity test, f(x) is continuous at x œ 1. xÄ1

8. Show lim" g(x) œ lim" xÄ

xÄ

%

" 2x

œ 2 œ g ˆ 4" ‰ .

%

Step 1: ¸ #"x 2¸ % Ê % #"x # % Ê # % #"x # % Ê Step 2: ¸B "4 ¸ $ Ê $ x 4" $ Ê $ 4" x $ 4" . Then $ Choose $ œ

" 4

œ

" 4 #%

% 4(#%)

Ê $œ

" 4

" 4 #%

œ

% 4(2 %)

, or $

" 4

œ

, the smaller of the two values. Then 0 ¸x

By the continuity test, g(x) is continuous at x œ

" 4

" 4 #% Ê 4" ¸ $

" 4#%

x

" 4 #% ¸ #"x

" 4#%

.

" 4

% 4(2 %)

$œ

œ

Ê

2¸ % and lim"

.

xÄ

%

" #x

œ 2.

.

9. Show lim h(x) œ lim È2x 3 œ " œ h(2). xÄ#

xÄ#

Step 1: ¹È2x 3 1¹ % Ê % È2x 3 " % Ê " % È2x 3 " % Ê

(1 %)# $ #

x

(" %)# 3 . #

Step 2: kx 2k $ Ê $ x 2 $ or $ # x $ #. (" % )# $ Ê $œ # (" % Ñ # $ (" %Ñ# " #œ # #

Then $ # œ

#

Ê $œ

œ%

# (" %)# $ œ " (1# %) # # %# . Choose $ œ %

œ%

%# #,

%# #

, or $ # œ

(" %)# $ #

the smaller of the two values . Then,

! kx 2k $ Ê ¹È2x 3 "¹ %, so lim È2x 3 œ 1. By the continuity test, h(x) is continuous at x œ 2. xÄ#

10. Show lim F(x) œ lim È9 x œ # œ F(5). xÄ&

xÄ&

Step 1: ¹È9 x 2¹ % Ê % È9 x # % Ê 9 (2 %)# x * (# %)# . Step 2: 0 kx 5k $ Ê $ x & $ Ê $ & x $ &. Then $ & œ * (# %)# Ê $ œ (# %)# % œ %# #%, or $ & œ * (# %)# Ê $ œ % (# %)# œ %# #%. Choose $ œ %# #%, the smaller of the two values. Then, ! kx 5k $ Ê ¹È9 x #¹ %, so lim È9 x œ #. By the continuity test, F(x) is continuous at x œ 5.

xÄ&

11. Suppose L" and L# are two different limits. Without loss of generality assume L# L" . Let % œ

" 3

(L# L" ).

Since x lim f(x) œ L" there is a $" 0 such that 0 kx x! k $" Ê kf(x) L" k % Ê % f(x) L" % Äx !

Ê "3 (L# L" ) L" f(x)

" 3

(L# L" ) L" Ê 4L" L# 3f(x) 2L" L# . Likewise, x lim f(x) œ L# Ä x! so there is a $# such that 0 kx x! k $# Ê kf(x) L# k % Ê % f(x) L# % Ê "3 (L# L" ) L# f(x) 3" (L# L" ) L# Ê 2L# L" 3f(x) 4L# L" Ê L" 4L# 3f(x) 2L# L" . If $ œ min e$" ß $# f both inequalities must hold for 0 kx x! k $ : 4L" L# 3f(x) 2L" L# Ê 5(L" L# ) 0 L" L# . That is, L" L# 0 and L" L# 0, L" %L# 3f(x) 2L# L" a contradiction. 12. Suppose xlim f(x) œ L. If k œ !, then xlim kf(x) œ xlim 0 œ ! œ ! † xlim f(x) and we are done. Äc Äc Äc Äc % If k Á 0, then given any % !, there is a $ ! so that ! lx cl $ Ê lfaxb Ll l5l Ê lkllfaxb Ll % Ê lkafaxb Lb| % Ê lakfaxbb akLbl %. Thus, xlim kf(x) œ kL œ kŠxlim f(x)‹. Äc Äc


Chapter 2 Additional and Advanced Exercises 89 13. (a) Since x Ä 0 , 0 x$ x 1 Ê ax$ xb Ä 0 Ê

lim f ax$ xb œ lim c f(y) œ B where y œ x$ x. yÄ!

x Ä !b

(b) Since x Ä 0 , 1 x x$ 0 Ê ax$ xb Ä 0 Ê

(c) Since x Ä 0 , 0 x% x# 1 Ê ax# x% b Ä 0 Ê

lim f ax$ xb œ lim b f(y) œ A where y œ x$ x. yÄ!

x Ä !c

lim f ax# x% b œ lim b f(y) œ A where y œ x# x% . yÄ!

x Ä !b

(d) Since x Ä 0 , 1 x 0 Ê ! x% x# 1 Ê ax# x% b Ä 0 Ê

lim f ax# x% b œ A as in part (c).

x Ä !b

14. (a) True, because if xlim (f(x) g(x)) exists then xlim (f(x) g(x)) xlim f(x) œ xlim [(f(x) g(x)) f(x)] Äa Äa Äa Äa œ xlim g(x) exists, contrary to assumption. Äa " x

(b) False; for example take f(x) œ

and g(x) œ x" . Then neither lim f(x) nor lim g(x) exists, but xÄ!

lim (f(x) g(x)) œ lim ˆ "x x" ‰ œ lim 0 œ 0 exists.

xÄ!

xÄ!

xÄ!

xÄ!

(c) True, because g(x) œ kxk is continuous Ê g(f(x)) œ kf(x)k is continuous (it is the composite of continuous functions). 1, x Ÿ 0 Ê f(x) is discontinuous at x œ 0. However kf(x)k œ 1 is (d) False; for example let f(x) œ œ 1, x 0 continuous at x œ 0. x# "

15. Show lim f(x) œ lim

x Ä 1 x 1

x Ä 1

(x 1)(x ") (x 1)

œ lim

x Ä 1

Define the continuous extension of f(x) as F(x) œ œ

œ #, x Á 1.

x# 1 x1 ,

2

x Á " . We now prove the limit of f(x) as x Ä 1 , x œ 1

exists and has the correct value. #

Step 1: ¹ xx 1" (#)¹ % Ê %

(x 1)(x ") (x 1)

# % Ê % (x 1) # %, x Á " Ê % " x % ".

Step 2: kx (1)k $ Ê $ x 1 $ Ê $ " x $ ". Then $ " œ % " Ê $ œ %, or $ " œ % " Ê $ œ %. Choose $ œ %. Then ! kx (1)k $ #

Ê ¹ xx 1" a#b¹ % Ê

lim F(x) œ 2. Since the conditions of the continuity test are met by F(x), then f(x) has a

x Ä 1

continuous extension to F(x) at x œ 1. 16. Show lim g(x) œ lim xÄ$

xÄ$

x# 2x 3 2x 6

œ lim

xÄ$

(x 3)(x ") 2(x 3)

œ #, x Á 3. #

Define the continuous extension of g(x) as G(x) œ œ

x 2x 3 2x 6 ,

2

xÁ3 . We now prove the limit of g(x) as , xœ3

x Ä 3 exists and has the correct value. Step 1: ¹ x

#

2x 3 #x 6

2¹ % Ê %

(x 3)(x ") 2(x 3)

# % Ê %

x" #

# % , x Á $ Ê $ #% x $ #% .

Step 2: kx 3k $ Ê $ x 3 $ Ê $ $ x $ $. Then, $ $ œ $ #% Ê $ œ #%, or $ $ œ $ #% Ê $ œ #%. Choose $ œ #%. Then ! kx 3k $ Ê ¹x

#

2x 3 2x 6

2¹ % Ê lim

xÄ$

(x 3)(x ") #(x 3)

œ 2. Since the conditions of the continuity test hold for G(x),

g(x) can be continuously extended to G(x) at B œ 3. 17. (a) Let % ! be given. If x is rational, then f(x) œ x Ê kf(x) 0k œ kx 0k % Í kx 0k %; i.e., choose $ œ %. Then kx 0k $ Ê kf(x) 0k % for x rational. If x is irrational, then f(x) œ 0 Ê kf(x) 0k % Í ! % which is true no matter how close irrational x is to 0, so again we can choose $ œ %. In either case, given % ! there is a $ œ % ! such that ! kx 0k $ Ê kf(x) 0k %. Therefore, f is continuous at x œ 0. (b) Choose x œ c !. Then within any interval (c $ ß c $ ) there are both rational and irrational numbers. If c is rational, pick % œ #c . No matter how small we choose $ ! there is an irrational number x in (c $ ß c $ ) Ê kf(x) f(c)k œ k0 ck œ c

c #

œ %. That is, f is not continuous at any rational c 0. On


90

Chapter 2 Limits and Continuity the other hand, suppose c is irrational Ê f(c) œ 0. Again pick % œ #c . No matter how small we choose $ ! there is a rational number x in (c $ ß c $ ) with kx ck œ kxk

c #

œ% Í

œ % Ê f is not continuous at any irrational c 0.

If x œ c 0, repeat the argument picking % œ nonzero value x œ c. 18. (a) Let c œ

c #

kc k #

œ

c # .

x

c #

Then kf(x) f(c)k œ kx 0k

3c #.

Therefore f fails to be continuous at any

m n

be a rational number in [0ß 1] reduced to lowest terms Ê f(c) œ "n . Pick % œ

" #n

œ %. Therefore f is discontinuous at x œ c, a rational number.

" #n .

No matter how small $ ! is taken, there is an irrational number x in the interval (c $ ß c $ ) Ê kf(x) f(c)k œ ¸0 "n ¸ œ

" n

(b) Now suppose c is an irrational number Ê f(c) œ 0. Let % 0 be given. Notice that number reduced to lowest terms with denominator 2 and belonging to [0ß 1]; denominator 3 belonging to [0ß 1];

" 4

and

[0ß 1]; etc. In general, choose N so that

" N

3 4

with denominator 4 in [0ß 1];

" 3

and

" 2 3 5, 5, 5

2 3

and

" #

is the only rational

the only rationals with 4 5

with denominator 5 in

% Ê there exist only finitely many rationals in [!ß "] having

denominator Ÿ N, say r" , r# , á , rp . Let $ œ min ekc ri k : i œ 1ß á ß pf . Then the interval (c $ ß c $ ) contains no rational numbers with denominator Ÿ N. Thus, 0 kx ck $ Ê kf(x) f(c)k œ kf(x) 0k œ kf(x)k Ÿ N" % Ê f is continuous at x œ c irrational. (c) The graph looks like the markings on a typical ruler when the points (xß f(x)) on the graph of f(x) are connected to the x-axis with vertical lines.

19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the zero point, 0, on the equator Ê 0 1R represents the midnight point (at the same exact time). Suppose x" is a point on the equator “just after" noon Ê x" 1R is simultaneously “just after" midnight. It seems reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically opposite point just after midnight: That is, T(x" ) T(x" 1R) 0. At exactly the same moment in time pick x# to be a point just before midnight Ê x# 1R is just before noon. Then T(x# ) T(x# 1R) 0. Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate Value Theorem says there is a point c between 0 (noon) and 1R (simultaneously midnight) such that T(c) T(c 1R) œ 0; i.e., there is always a pair of antipodal points on the earth's equator where the temperatures are the same. #

#

# # " 20. xlim f(x)g(x) œ xlim af(x) g(x)b‹ Šxlim af(x) g(x)b‹ “ ’af(x) g(x)b af(x) g(x)b “ œ "% ’Šxlim Äc Äc % Äc Äc œ "% ˆ$# a"b# ‰ œ #.


Chapter 2 Additional and Advanced Exercises 91 21. (a) At x œ 0: lim r (a) œ lim aÄ!

œ lim

1 (" a)

aÄ!

a Ä ! a ˆ" È1 a‰

At x œ 1: (b) At x œ 0:

lim

a Ä "b

œ

r (a) œ

" È1 a a 1 " È1 0

œ lim c aÄ!

1 (" a) a ˆ" È1 a‰

" #

œ

aÄ!

1 (1 a)

lim

a Ä "b a ˆ1 È1 a‰

lim r (a) œ lim c aÄ!

a Ä !c

È1 a

œ lim Š " a

" È1 a a

œ lim c aÄ!

a

œ lim

a Ä 1 a ˆ" È1 a‰ È1 a

œ lim c Š " a aÄ!

a a ˆ 1 È 1 a ‰

" È1 a

‹ Š " È1 a ‹

œ lim c aÄ!

œ

" " È0

œ1

" È1 a

‹ Š " È1 a ‹

" œ _ (because the " È1 a " œ _ (because the " È1 a

denominator is always negative); lim b r (a) œ lim b aÄ! aÄ! is always positive). Therefore, lim r (a) does not exist.

denominator

aÄ!

At x œ 1:

lim

a Ä "b

r (a) œ

lim

a Ä "b

1 È 1 a a

œ

lim

"

a Ä 1b " È1 a

œ1

(c)

(d)

22. f(x) œ x 2 cos x Ê f(0) œ 0 2 cos 0 œ 2 0 and f(1) œ 1 2 cos (1) œ 1 # 0. Since f(x) is continuous on [1ß !], by the Intermediate Value Theorem, f(x) must take on every value between [1 #ß #]. Thus there is some number c in [1ß !] such that f(c) œ 0; i.e., c is a solution to x 2 cos x œ 0. 23. (a) The function f is bounded on D if f(x) M and f(x) Ÿ N for all x in D. This means M Ÿ f(x) Ÿ N for all x in D. Choose B to be max ekMk ß kNkf . Then kf(x)k Ÿ B. On the other hand, if kf(x)k Ÿ B, then B Ÿ f(x) Ÿ B Ê f(x) B and f(x) Ÿ B Ê f(x) is bounded on D with N œ B an upper bound and M œ B a lower bound. (b) Assume f(x) Ÿ N for all x and that L N. Let % œ L # N . Since x lim f(x) œ L there is a $ ! such that Äx !

0 kx x! k $ Ê kf(x) Lk % Í L % f(x) L % Í L Í

LN #

f(x)

3L N # .

But L N Ê

LN #

LN #

f(x) L

LN #

N Ê N f(x) contrary to the boundedness assumption

f(x) Ÿ N. This contradiction proves L Ÿ N.


92

Chapter 2 Limits and Continuity (c) Assume M Ÿ f(x) for all x and that L M. Let % œ Ê L

ML #

f(x) L

ML #

3L M #

Í

f(x)

ML # . As in part (b), 0 kx L M M, a contradiction. #

24. (a) If a b, then a b 0 Ê ka bk œ a b Ê max Öaß b× œ

ab #

ka b k #

If a Ÿ b, then a b Ÿ 0 Ê ka bk œ (a b) œ b a Ê max Öaß b× œ œ

2b #

xÄ0

ab #

sina" cos xb x

œ lim

lim b sinsinÈxx œ

xÄ0

†

sin x

xÄ0

sinasin xb x

xÄ0

sinax# xb x xÄ0

œ lim

29. lim

sinax# %b x2

œ lim

30. lim

sinˆÈx $‰ x9

28. lim

xÄ2

xÄ9

sin x " cos x

xÄ0

œ lim

ka b k #

.

sina" cos xb " cos x

lim b sinB x †

xÄ0

27. lim

œ lim

xÄ0 x

26.

ab ab 2a # # œ # œ a. ka b k ab œ a # b b # a # #

œ

œ b.

(b) Let min Öaß b× œ 25. lim œ

x! k $

sinasin xb sin x

†

" cos x x

Èx sin Èx

†

†

œ lim

sin x x

x Èx

xÄ0

sinasin xb sin x

sinax# %b x# %

† ax 2b œ lim

xÄ9

xÄ0

sinˆÈx $‰ Èx $

† lim

" cos# x

x Ä 0 xa" cos xb

" Èx $

† lim

sin x

xÄ0 x

œ " † lim

sin# x

x Ä 0 xa" cos xb

œ " † " œ ".

sinax# xb # x Ä 0 x x

† lim ax "b œ " † " œ "

sinax# %b x# %

† lim ax 2b œ " † % œ %

xÄ2

†

sina" cos xb " cos x

œ " † lim b sin"Èx † lim b Èx œ " † ! † ! œ !. x Ä 0 Š Èx ‹ x Ä 0

† ax "b œ lim

œ lim

œ lim

œ " † ˆ #! ‰ œ !.

sinax# xb # x Ä 0 x x

xÄ2

" cos x " cos x

†

œ lim

xÄ9

xÄ0

xÄ2

sinˆÈx $‰ Èx $

† lim

"

x Ä 9 Èx $

œ"†

" '

œ

" '

31. Since the highest power of x in the numerator is 1 more than the highest power of x in the denominator, there is an oblique asymptote. y œ 32. As x Ä „ _,

2x3Î2 2x 3 Èx 1 1 x

œ 2x

3 Èx 1 ,

thus the oblique asymptote is y œ 2x.

Ä 0 Ê sinˆ 1x ‰ Ä 0 Ê 1 sinˆ 1x ‰ Ä 1, thus as x Ä „ _, y œ x x sinˆ 1x ‰ œ xˆ1 sinˆ 1x ‰‰ Ä x;

thus the oblique asymptote is y œ x. 33. As x Ä „ _, x2 1 Ä x2 Ê Èx2 1 Ä Èx2 ; as x Ä _, Èx2 œ x, and as x Ä _, Èx2 œ x; thus the oblique asymptotes are y œ x and y œ x. 34. As x Ä „ _, x 2 Ä x Ê Èx2 2x œ Èxax 2b Ä Èx2 ; as x Ä _, Èx2 œ x, and as x Ä _, Èx2 œ x; asymptotes are y œ x and y œ x.


CHAPTER 3 DIFFERENTIATION 3.1 TANGENTS AND THE DERIVATIVE AT A POINT 1. P" : m" œ 1, P# : m# œ 5

2. P" : m" œ 2, P# : m# œ 0

3. P" : m" œ 5# , P# : m# œ "#

4. P" : m" œ 3, P# : m# œ 3

5. m œ lim

hÄ!

c4 (" h)# d a4 (1)# b h

a1 2h h# b1 h hÄ!

œ lim

œ lim

hÄ!

h(# h) h

œ 2;

at ("ß $): y œ $ #(x (1)) Ê y œ 2x 5, tangent line

6. m œ lim

hÄ!

c(1 h 1)# 1d c(" ")# 1d h

h#

œ lim

hÄ! h

œ lim h œ 0; at ("ß "): y œ 1 0(x 1) Ê y œ 1, hÄ!

tangent line

È 2È 1 h 2È 1 œ lim 2 1 h h 2 h hÄ! hÄ! 4(1 h) 4 œ lim œ lim È1 2h 1 h Ä ! 2h ŠÈ1 h 1‹ hÄ!

7. m œ lim

†

2È 1 h 2 2È 1 h #

œ 1;

at ("ß #): y œ 2 1(x 1) Ê y œ x 1, tangent line

8. m œ lim

hÄ!

"

( 1 h)#

( "")#

h

a2h h# b # h Ä ! h(1 h)

œ lim

1 (1 h)# # h Ä ! h(1h) 2h lim # œ 2; h Ä ! (1 h)

œ lim œ

at ("ß "): y œ 1 2(x (1)) Ê y œ 2x 3, tangent line


94

Chapter 3 Differentiation (2 h)$ (2)$ h

9. m œ lim

hÄ!

8 12h 6h# h$ 8 h

œ lim

hÄ!

œ lim a12 6h h# b œ 12; hÄ!

at (2ß 8): y œ 8 12(x (2)) Ê y œ 12x 16, tangent line

10. m œ lim

(

h

hÄ!

œ

"2 8(8)

hÄ!

hÄ!

at ˆ#ß "8 ‰ : y œ 8" Ê yœ 11. m œ lim

hÄ!

x

" #,

12 6h h# 8(2 h)$

œ lim

3 œ 16 ;

3 16

8 (# h)$ 8h(# h)$

œ lim

a12h 6h# h$ b 8h(# h)$

œ lim

hÄ!

" " # h)$ ( #)$

3 16 (x

(2))

tangent line

c(2 h)# 1d 5 h

œ lim

hÄ!

a5 4h h# b 5 h

hÄ!

at (2ß 5): y 5 œ 4(x 2), tangent line 12. m œ lim

hÄ!

c(" h) 2(1 h)# d (1) h

œ lim

hÄ!

h(4 h) h

œ lim

a1 h 2 4h 2h# b 1 h

3 (3

h h) 2

3

h

hÄ!

œ lim

hÄ!

(3 h) 3(h 1) h(h 1)

h Ä ! h(h 1)

at ($ß $): y 3 œ 2(x 3), tangent line 14. m œ lim

hÄ!

8 (2

h)#

2

h

hÄ!

(2 h)$ 8 h hÄ!

œ lim

a8 12h 6h# h$ b 8 h hÄ!

œ lim

16. m œ lim

hÄ!

c(1 h)$ 3(1 h)d 4 h

hÄ!

at ("ß %): y 4 œ 6(t 1), tangent line 17. m œ lim

hÄ!

È4 h 2 h

œ lim

hÄ!

œ "4 ; at (%ß #): y 2 œ 18. m œ lim

hÄ!

œ

" È9 3

È(8 h) 1 3 h

" 4

È4 h 2 h

†

hÄ!

h a12 6h h# b h hÄ!

a1 3h 3h# h$ 3 3hb 4 h

œ lim

œ lim

œ lim

at (2ß )): y 8 œ 12(t 2), tangent line

È4 h 2 È4 h 2

œ 3;

œ 2;

8 2 a4 4h h# b h(2 h)# hÄ!

8 2(2 h)# # h Ä ! h(2 h)

œ lim

at (2ß 2): y 2 œ 2(x 2) 15. m œ lim

2h

œ lim

h(3 2h) h

œ lim

at ("ß "): y 1 œ 3(x 1), tangent line 13. m œ lim

œ %;

œ lim

2h(4 h) h(2 h)#

œ

8 4

œ 2;

œ 12;

œ lim

hÄ!

(4 h) 4

h Ä ! h ŠÈ4 h #‹

h a6 3h h# b h

œ lim

œ 6;

h

h Ä ! h ŠÈ4 h #‹

œ

" È4 #

(x 4), tangent line

œ lim

hÄ!

È9 h 3 h

œ 6" ; at (8ß 3): y 3 œ

19. At x œ 1, y œ 5 Ê m œ lim

hÄ!

" 6

†

È9 h 3 È9 h 3

œ lim

(9 h) 9

h Ä ! h ŠÈ9 h 3‹

œ lim

h

h Ä ! h ŠÈ9 h 3‹

(x 8), tangent line

5(" h)# 5 h

œ lim

hÄ!

5 a1 2h h# b 5 h

œ lim

hÄ!

5h(2 h) h

œ 10, slope


Section 3.1 Tangents and the Derivative at a Point c1 (2 h)# d (3) h


hÄ!

" #

21. At x œ 3, y œ

Ê m œ lim

"

h) 1

(3

#"

h1 h 1


hÄ!

hÄ!

2 (2 h) 2h(2 h)

œ lim

h

hÄ!

œ lim

hÄ!

(1) h

a1 4 4h h# b 3 h

hÄ!

hÄ!

h

œ lim

œ lim

h(4 h) h

œ 4, slope

œ "4 , slope

h Ä ! 2h(2 h)

(h 1) (h ") h(h 1)

œ lim

œ lim

2h

h Ä ! h(h 1)

œ 2, slope

c(x h)# 4(x h) 1d ax# 4x 1b h hÄ! a2xh h# 4hb lim œ lim (2x h 4) œ 2x h hÄ! hÄ!

23. At a horizontal tangent the slope m œ 0 Ê 0 œ m œ lim ax# 2xh h# 4x 4h 1b ax# 4x 1b h hÄ!

œ lim

œ

4;

2x 4 œ 0 Ê x œ 2. Then f(2) œ 4 8 1 œ 5 Ê (2ß 5) is the point on the graph where there is a horizontal tangent. c(x h)$ 3(x h)d ax$ 3xb h

24. 0 œ m œ lim

hÄ!

œ lim

hÄ!

3x# h 3xh# h$ 3h h

ax$ 3x# h 3xh# h$ 3x 3hb ax$ 3xb h

œ lim

hÄ!

œ lim a3x# 3xh h# 3b œ 3x# 3; 3x# 3 œ 0 Ê x œ 1 or x œ 1. Then hÄ!

f(1) œ 2 and f(1) œ 2 Ê ("ß 2) and ("ß 2) are the points on the graph where a horizontal tangent exists. 25. 1 œ m œ lim

"

h) 1

(x

x " 1

h

hÄ!

œ lim

hÄ!

(x 1) (x h 1) h(x 1)(x h 1)

h

œ lim

h Ä ! h(x 1)(x h 1)

œ (x " 1)#

Ê (x 1)# œ 1 Ê x# 2x œ 0 Ê x(x 2) œ 0 Ê x œ 0 or x œ 2. If x œ 0, then y œ 1 and m œ 1 Ê y œ 1 (x 0) œ (x 1). If x œ 2, then y œ 1 and m œ 1 Ê y œ 1 (x 2) œ (x 3). 26.

" 4

Èx h Èx

œ m œ lim œ lim

h

y œ 2 "4 (x 4) œ

hÄ!

f(2 h) f(2) h

x 4

Èx h Èx h

hÄ!

h Ä ! h ŠÈx h Èx‹

27. lim

œ lim

h

hÄ!

œ

" #È x

. Thus,

" 4

œ

†

Èx h Èx Èx h Èx

" #Èx

(x h) x

œ lim

h Ä ! h ŠÈx h Èx‹

Ê Èx œ 2 Ê x œ 4 Ê y œ 2. The tangent line is

1.

œ lim

hÄ!

a100 4.9(# h)# b a100 4.9(2)# b h

4.9 a4 4h h# b 4.9(4) h

œ lim

hÄ!

œ lim (19.6 4.9h) œ 19.6. The minus sign indicates the object is falling downward at a speed of 19.6 m/sec. hÄ!

28. lim

hÄ!

f(10 h) f(10) h

hÄ!

1(3 h)# 1(3)# h hÄ!

f(3 h) f(3) h hÄ!

œ lim

f(2 h) f(2) h hÄ!

œ lim

29. lim

30. lim

3(10 h)# 3(10)# h

œ lim

hÄ!

41 3

3 a20h h# b h

œ lim

hÄ!

œ 60 ft/sec.

1 c9 6h h# 9d h hÄ!

œ lim

(2 h)$ 431 (2)$ h

œ lim

41 3

hÄ!

31. At ax0 , mx0 bb the slope of the tangent line is lim

hÄ!

œ lim 1(6 h) œ 61 hÄ!

c12h 6h# h$ d h

œ lim

hÄ!

amax0 hb bb am x0 bb ax 0 h b x 0

41 3

c12 6h h# d œ 161

œ lim

hÄ!

mh h

The equation of the tangent line is y am x0 bb œ max x0 b Ê y œ mx b. 32. At x œ 4, y œ

1 È4 È

œ

" #

œ lim – 22hÈ44hh † hÄ!

and m œ lim

hÄ!

2 È4 h 2 È4 h —

È4

1 h

h

"#

œ lim – hÄ!

È4

1 h

h

"#

†

2È 4 h 2È 4 h —

œ lim m œ m. hÄ!

È

œ lim Š 22hÈ44hh ‹ hÄ!

hb h œ lim È 4 a4 È œ lim È 2h 4 hŠ2 4 h‹ 2h 4 hŠ2 È4 h‹ hÄ! hÄ!


95

96

Chapter 3 Differentiation 1 1 œ lim È œ È 1 È œ 16 2 4 hŠ2 È4 h‹ 2 4Š2 4‹ hÄ!

f(0 h) f(0) h hÄ!

33. Slope at origin œ lim

h# sin ˆ "h ‰ h hÄ!

œ lim

œ lim h sin ˆ "h ‰ œ 0 Ê yes, f(x) does have a tangent at hÄ!

the origin with slope 0. g(0 h) g(0) h

34. lim

hÄ!

œ lim

hÄ!

h sin ˆ "h ‰ h

œ lim sin h" . Since lim sin hÄ!

hÄ!

" h

does not exist, f(x) has no tangent at

the origin. 35.

lim

h Ä !c

f(0 h) f(0) h

lim f(0 h)h f(0) hÄ! 36.

œ lim c hÄ!

1 0 h

œ _, and lim b hÄ!

f(0 h) f(0) h

10 h

œ lim b hÄ!

œ _ Ê yes, the graph of f has a vertical tangent at the origin.

œ _, and lim b U(0 h)h U(0) œ lim b hÄ! hÄ! does not have a vertical tangent at (!ß ") because the limit does not exist. lim

h Ä !c

œ _. Therefore,

U(0 h) U(0) h

œ lim c hÄ!

01 h

11 h

œ 0 Ê no, the graph of f

37. (a) The graph appears to have a cusp at x œ 0.

(b)

lim c

hÄ!

f(0 h) f(0) h

œ lim c hÄ!

h#Î& 0 h

œ lim c hÄ!

" h$Î&

œ _ and lim b hÄ!

" h$Î&

œ _ Ê limit does not exist

Ê the graph of y œ x#Î& does not have a vertical tangent at x œ 0.


(b)

lim

h Ä !c

f(0 h) f(0) h

œ lim c hÄ!

h%Î& 0 h

œ lim c hÄ!

" h"Î&

œ _ and lim b hÄ!

" h"Î&

œ _ Ê limit does not exist

Ê y œ x%Î& does not have a vertical tangent at x œ 0.

39. (a) The graph appears to have a vertical tangent at x œ !.

(b)

lim

hÄ!

f(0 h) f(0) h

œ lim

hÄ!

h"Î& 0 h

œ lim

"

%Î& hÄ! h

œ _ Ê y œ x"Î& has a vertical tangent at x œ 0.


Section 3.1 Tangents and the Derivative at a Point 40. (a) The graph appears to have a vertical tangent at x œ 0.

(b)

lim

hÄ!

f(0 h) f(0) h

œ lim

hÄ!

h$Î& 0 h

"

œ lim

#Î& hÄ! h

œ _ Ê the graph of y œ x$Î& has a vertical tangent at x œ 0.


(b)

lim c

hÄ!

f(0 h) f(0) h

œ lim c hÄ!

4h#Î& 2h h

œ lim c hÄ!

4 h$Î&

2 œ _ and lim b hÄ!

4 h$Î&

#œ_

Ê limit does not exist Ê the graph of y œ 4x#Î& 2x does not have a vertical tangent at x œ 0.


(b)

lim

hÄ!

f(0 h) f(0) h

œ lim

hÄ!

h&Î$ 5h#Î$ h

œ lim h#Î$ hÄ!

5 h"Î$

œ 0 lim

y œ x&Î$ 5x#Î$ does not have a vertical tangent at x œ !.

5

"Î$ hÄ! h

does not exist Ê the graph of

43. (a) The graph appears to have a vertical tangent at x œ 1 and a cusp at x œ 0.

(b) x œ 1:

lim

hÄ!

(1 h)#Î$ (1 h 1)"Î$ " h

œ lim

hÄ!

(1 h)#Î$ h"Î$ " h

œ _

Ê y œ x#Î$ (x 1)"Î$ has a vertical tangent at x œ 1;


97

98

Chapter 3 Differentiation x œ 0:

lim

hÄ!

f(0 h) f(0) h

œ lim

hÄ!

h#Î$ (h 1)"Î$ (1)"Î$ h

" œ lim ’ h"Î$

hÄ!

(h ")"Î$ h

h" “

does not exist Ê y œ x#Î$ (x 1)"Î$ does not have a vertical tangent at x œ 0. 44. (a) The graph appears to have vertical tangents at x œ 0 and x œ 1.

(b) x œ 0:

lim

hÄ!

f(0 h) f(0) h

œ lim

hÄ!

h"Î$ (h 1)"Î$ (")"Î$ h

œ _ Ê y œ x"Î$ (x 1)"Î$ has a

vertical tangent at x œ 0;

x œ 1:

lim

hÄ!

f(1 h) f(1) h

œ lim

hÄ!

(1 h)"Î$ (" h 1)"Î$ 1 h

œ _ Ê y œ x"Î$ (x 1)"Î$ has a

vertical tangent at x œ ".

45. (a) The graph appears to have a vertical tangent at x œ 0.

(b)

lim b

hÄ!

f(0 h) f(0) h

œ lim b xÄ!

Èh 0 h

œ lim

"

h Ä ! Èh

È kh k 0

f(0 h) f(0) h

œ lim c œ lim c h hÄ! hÄ! Ê y has a vertical tangent at x œ 0. lim

h Ä !c

œ _; È kh k kh k

œ lim c hÄ!

" È kh k

œ_


(b)

lim b

f(4 h) f(4) h

œ lim b hÄ!

Èk4 (4 h)k 0 h

lim

f(4 h) f(4) h

œ lim c

Èk4 (4 h)k h

hÄ!

h Ä !c

hÄ!

œ lim b hÄ!

œ lim c hÄ!

È kh k h

È kh k lhl

œ lim b hÄ!

œ lim c hÄ!

" Èh

" È kh k

œ _;

œ _

Ê y œ È% x does not have a vertical tangent at x œ 4. 47-50. Example CAS commands: Maple: f := x -> x^3 + 2*x;x0 := 0; plot( f(x), x=x0-1/2..x0+3, color=black, title="Section 3.1, #47(a)" ); q := unapply( (f(x0+h)-f(x0))/h, h );

# part (a) # part (b)


Section 3.2 The Derivative as a Function L := limit( q(h), h=0 ); # part (c) sec_lines := seq( f(x0)+q(h)*(x-x0), h=1..3 ); # part (d) tan_line := f(x0) + L*(x-x0); plot( [f(x),tan_line,sec_lines], x=x0-1/2..x0+3, color=black, linestyle=[1,2,5,6,7], title="Section 3.1, #47(d)", legend=["y=f(x)","Tangent line at x=0","Secant line (h=1)", "Secant line (h=2)","Secant line (h=3)"] ); Mathematica: (function and value for x0 may change) Clear[f, m, x, h] x0 œ p; f[x_]: œ Cos[x] 4Sin[2x] Plot[f[x], {x, x0 1, x0 3}] dq[h_]: œ (f[x0+h] f[x0])/h m œ Limit[dq[h], h Ä 0] ytan: œ f[x0] m(x x0) y1: œ f[x0] dq[1](x x0) y2: œ f[x0] dq[2](x x0) y3: œ f[x0] dq[3](x x0) Plot[{f[x], ytan, y1, y2, y3}, {x, x0 1, x0 3}] 3.2 THE DERIVATIVE AS A FUNCTION 1. Step 1: f(x) œ 4 x# and f(x h) œ 4 (x h)# f(x h) f(x) h

Step 2:

œ

c4 (x h)# d a4 x# b h

œ

a4 x# 2xh h# b 4 x# h

œ

2xh h# h

œ

h(2x h) h

œ 2x h Step 3: f w (x) œ lim (2x h) œ 2x; f w ($) œ 6, f w (0) œ 0, f w (1) œ 2 hÄ!

2. F(x) œ (x 1)# 1 and F(x h) œ (x h 1)# " Ê Fw (x) œ lim œ lim

hÄ!

hÄ!

ax# 2xh h# 2x 2h 1 1b ax# 2x 1 1b h w

w

œ lim

w

œ 2(x 1); F (1) œ 4, F (0) œ 2, F (2) œ 2 3. Step 1: g(t) œ

" t#

and g(t h) œ "

Step 2:

"

# # g(t h) g(t) œ (t h)h t h 2t h) 2t h œ h( (t h)# t# h œ (t h)# t#

Step 3: gw (t) œ lim

2t h

# # h Ä ! (t h) t

4. k(z) œ

1 z #z

and k(z h) œ

œ œ

" 2z#

œ

Œ

2t t# †t#

1 (z h) 2(z h)

(" z)(z h) lim (1 z h)z #(z h)zh hÄ!

2xh h# 2h h

œ lim (2x h 2) hÄ!

" (t h)#

œ

œ

hÄ!

c(x h 1)# 1d c(x 1)# 1d h

t# (t h)# (t h)# †t#

h

œ

2 t$

t# at# 2th h# b (t h)# †t# †h

œ

œ

2th h# (t h)# t# h

2 ; gw (1) œ 2, gw (2) œ "4 , gw ŠÈ3‹ œ 3È 3

Ê kw (z) œ lim

hÄ!

Š

"

(z h) " z #(z h) #z ‹

# z h z# zh lim z z zh 2(z h)zh hÄ!

h

œ lim

h

h Ä ! 2(z h)zh

œ lim

"

h Ä ! #(z h)z

; kw (") œ "# , kw (1) œ "# , kw ŠÈ2‹ œ 4"

5. Step 1: p()) œ È3) and p() h) œ È3() h)


99

100

Chapter 3 Differentiation

Step 2:

p() h) p()) h

œ

œ

È3() h) È3) h

3h h ŠÈ3) 3h È3)‹

Step 3: pw ()) œ lim

œ

ŠÈ3) 3h È3)‹ h

œ

3 È3) 3h È3)

3

œ

h Ä ! È3) 3h È3)

†

œ

3 È 3) È 3)

3 2È 3 )

ŠÈ3) 3h È3)‹ ŠÈ3) 3h È3)‹

; pw (1) œ

œ

(3) 3h) 3) h ŠÈ3) 3h È3)‹

, pw (3) œ "# , pw ˆ 32 ‰ œ

3 2È 3

3 #È2

È2s 2h 1 È2s 1 h hÄ!

6. r(s) œ È2s 1 and r(s h) œ È2(s h) 1 Ê rw (s) œ lim œ lim

ŠÈ2s h 1 È2s 1‹ h

hÄ!

œ lim

ŠÈ2s 2h 1 È2s 1‹

†

œ lim

2h

œ

" È2s 1

; rw (0) œ 1, rw (1) œ

2

" È3

, rw ˆ #" ‰ œ

hÄ!

6x# h 6xh# 2h$ h

hÄ!

9. s œ r(t) œ œ lim

t 2t1

Š

dr ds

œ

2 2È2s 1

2 ax$ 3x# h 3xh# h$ b 2x$ h hÄ!

2(x h)$ 2x$ h hÄ!

œ lim

œ lim

œ lim a6x# 6xh 2h# b œ 6x# hÄ!

œ lim

th 2(th)1

and r(t h) œ

(t b h)(2t b 1) c t(2t b 2h b 1) ‹ (2t b 2h b 1)(2t b 1)

h

hÄ!

œ lim

œ

2t# t 2ht h 2t# 2ht t (2t 2h 1)(2t 1)h hÄ! " " (2t 1)(2t 1) œ (2t 1)#

dv dt

œ lim

hÄ!

œ lim

10.

2 È2s 1 È2s 1

3 2 2 3 2 2 3 2 âs hb3 2as hb2 3‰ ˆs3 2s2 3‰ œ lim s 3s h 3sh h 2s h 4sh h 3 s 2s 3 h hÄ! hÄ! 2 2 3 2 hˆ3s2 3sh h2 4s h‰ lim 3s h 3sh hh 4sh h œ lim œ lim a3s2 3sh h2 4s hb œ 3s2 2s h hÄ! hÄ! hÄ!

8. r œ s3 2s2 3 Ê œ

dy dx

h a6x# 6xh 2h# b h

œ lim

œ

" È2

7. y œ f(x) œ 2x$ and f(x h) œ 2(x h)$ Ê œ lim

h Ä ! h ŠÈ2s 2h 1 È2s 1‹

h Ä ! È2s 2h 1 È2s 1

h Ä ! h ŠÈ2s 2h 1 È2s 1‹

(2s 2h 1) (2s 1)

œ lim

ŠÈ2s 2h 1 È2s 1‹

’(t h)

hÄ!

ht# h# t h h(t h)t hÄ!

œ lim

Œ

" “ ˆt " ‰ t h h

œ lim

11. p œ f(q) œ

t

" Èq 1

h

hÄ!

(t h)(2t 1) t(2t 2h 1) (2t 2h 1)(2t 1)h h

h Ä ! (2t 2h 1)(2t 1)h

œ lim

hÄ!

t# ht 1 h Ä ! (t h)t

and f(q h) œ œ lim

œ

h

t

" " t h h

t# 1 t#

" È(q h) 1

Ê

h Ä ! (2t 2h 1)(2t 1)

Š

œ lim

h(t

h)t t (t (t h)t

dp dq

h)

‹

h

hÄ!

œ1

"

œ lim

" t#

œ lim

Š È(q

hÄ!

"

h)

1

‹ Š Èq"

1

‹

h

Èq 1 Èq h 1

h Ä ! hÈ q h 1 È q 1

h

hÄ!

t ‰ Š 2(t bt bh)hb 1 ‹ ˆ 2t b 1

œ lim

ds dt

œ lim

œ lim

Èq b 1 c Èq b h b 1 Èq b h b 1 Èq b 1

Ê

œ

ˆÈ q 1 È q h 1 ‰ ˆ È q 1 È q h 1 ‰ 1) (q h 1) † ˆÈq 1 Èq h 1‰ œ lim hÈq h 1(qÈq 1 ˆÈ q 1 È q h 1 ‰ h Ä ! h Èq h 1 Èq 1 hÄ! h " lim œ lim Èq h 1 Èq 1 ˆÈq 1 Èq h 1‰ h Ä ! h È q h 1 È q 1 ˆÈ q 1 È q h 1 ‰ hÄ! " " œ È q 1 È q 1 ˆÈ q 1 È q 1 ‰ 2(q 1) Èq 1

dz dw

œ lim

œ lim œ

12.

Š È3(w " h) 2 h

hÄ!

œ lim

hÄ!

"

È3w 2 ‹

ŠÈ3w 2 È3w 3h 2‹ hÈ3w 3h 2 È3w 2

œ lim

œ lim

È3w 2 È3w 3h 2

h Ä ! hÈ3w 3h 2 È3w 2

†

ŠÈ3w2È3w3h2‹ ŠÈ3w 2 È3w 3h 2‹

3

h Ä ! È3w 3h 2 È3w 2 ŠÈ3w 2 È3w 3h 2‹

œ

œ

œ lim

(3w 2) (3w 3h 2)

h Ä ! hÈ3w 3h 2 È3w 2 ŠÈ3w 2 È3w 3h 2‹

3 È3w 2 È3w 2 ŠÈ3w 2 È3w 2‹

3 2(3w 2) È3w 2


Section 3.2 The Derivative as a Function 13. f(x) œ x œ

x(x h)# 9x x# (x h) 9(x h) x(x h)h

œ

h(x# xh 9) x(x h)h

" #x

14. k(x) œ w

hÄ!

œ œ 16.

dy dx

œ

œ

x# 9 x#

œ

9 9 (x b h) “ ’x x “

’(x h)

h

x$ 2x# h xh# 9x x$ x# h 9x 9h x(x h)h

œ

x# xh 9 h Ä ! x(x h)

and k(x h) œ

x# h xh# 9h x(x h)h

œ

œ1

9 x#

; m œ f w (3) œ 0

Š # "x h k(x h) k(x) œ lim h h hÄ! hÄ! h " " lim œ lim (2 x)(# x h) œ (2 x)# ; h Ä ! h(2 x)(2 x h) hÄ! " 2 (x h)

Ê kw (x) œ lim

$ # # $ # # $ # c(t h)$ (t h)# d at$ t# b œ lim at 3t h 3th h b h at 2th h b t t h hÄ! hÄ! # # # # $ # lim 3t h 3th hh 2th h œ lim h a3t 3th h h 2t hb œ lim a3t# 3th hÄ! hÄ! hÄ! ¸ 3t# 2t; m œ ds œ 5 dt tœ"

#

" ‹ x

œ lim

ax b hb b 3 1 c ax b hb

œ lim

b3 1x c x

h

hÄ!

œ lim

17. f(x) œ

4h

b h b 3ba1 c xb c ax b 3ba1 c x c hb a1 c x c hba1 c xb h

œ lim

8 ŠÈx 2 Èx h 2‹ hÈ x h 2 È x 2

4

†

8 È(x h) 2

ŠÈx 2 Èx h 2‹

œ

8h hÈx h 2 Èx 2 ŠÈx 2 Èx h 2‹

œ

8 Èx 2 Èx 2 ŠÈx 2 Èx 2‹

œ

œ

œ

h# 2t hb

x h 3 x2 xh 3x x 3 x2 3x xh 3h h a1 x h b a 1 x b

4 ; dy ¹ a1 xb2 dx xœ2

f(x h) f(x) h

Ê

ŠÈx 2 Èx h 2‹

œ lim

hÄ!

h Ä ! a1 x hba1 xb

and f(x h) œ

8 Èx 2

ax

œ lim

hÄ!

h Ä ! ha1 x hba1 xb

œ

f(x h) f(x) h

Ê

9 (x h)

; f w (x) œ lim

" 16

k (2) œ ds dt

x# xh 9 x(x h)

œ

(# x) (2 x h) h(2 x)(2 x h)

œ lim

15.

and f(x h) œ (x h)

9 x

œ

œ

4 a3 b 2

4 9

È(x b h) c 2 Èx c 2 8

œ

8

h

8[(x 2) (x h 2)] hÈx h 2 Èx 2 ŠÈx 2 Èx h 2‹ 8

Ê f w (x) œ lim

h Ä ! Èx h 2 Èx 2 ŠÈx 2 Èx h 2‹

4 (x 2)Èx 2

; m œ f w (6) œ

4 4È 4

œ "# Ê the equation of the tangent

line at (6ß 4) is y 4 œ "# (x 6) Ê y œ "# x $ % Ê y œ "# x (. ˆ1 È4 (z h)‰ Š1 È4 z‹

18. gw (z) œ lim

h

hÄ!

œ

h

hÄ!

(4 z h) (4 z) lim h Ä ! h ŠÈ4 z h È4 z‹ " œ "# 2È 4 3 "# z $# # Ê w

ŠÈ4 z h È4 z‹

œ lim

œ

h lim h Ä ! h ŠÈ4 z h È4 z‹

†

ŠÈ4 z h È4 z‹ ŠÈ4 z h È4 z‹ "

œ lim

h Ä ! ŠÈ4 z h È4 z‹

œ

" 2È 4 z

m œ gw (3) œ

Ê the equation of the tangent line at ($ß #) is w 2 œ "# (z 3)

Êwœ

œ "# z (# .

19. s œ f(t) œ 1 3t# and f(t h) œ 1 3(t h)# œ 1 3t# 6th 3h# Ê a1 3t# 6th 3h# b a1 3t# b h hÄ!

œ lim

20. y œ f(x) œ " œ lim

" " x

21. r œ f()) œ hÄ!

h

h

hÄ!

œ lim

x

" x

2 È4 )

œ lim (6t 3h) œ 6t Ê hÄ!

and f(x h) œ 1 œ lim

h

h Ä ! x(x h)h

œ

œ lim

Ê

dy dx "

h Ä ! x(x h)

œ

œ lim

" x#

" 3

Ê

È œ

dy dx ¹x= 3

2 È4 () h)

Ê

dr d)

œ lim

hÄ!

f(t h) f(t) h

œ6

f(x h) f(x) h hÄ!

œ lim

œ lim

Š1

x

" ‹ Š1 " ‹ h x h

hÄ!

f() h) f()) œ lim h hÄ! hÄ! È È 2È4 ) #È% ) h Š2 % ) 2 4 ) h‹ lim † È Š2 4 ) #È4 ) h‹ h Ä ! hÈ 4 ) È 4 ) h

and f() h) œ

2È 4 ) 2È 4 ) h hÈ 4 ) È 4 ) h

" xh

ds ¸ dt t=c"

ds dt

È4 c ) c h È4 c ) 2

2

h


;

101

102

Chapter 3 Differentiation 4(% )) 4(% ) h)

œ lim

h Ä ! 2hÈ4 ) È4 ) h ŠÈ4 ) È4 ) h‹

œ

2 (4 )) Š2È4 )‹

œ

" (4 ))È4 )

Ê

dr ¸ d) )œ!

œ lim

2

h Ä ! È4 ) È4 ) h ŠÈ4 ) È% ) h‹

œ

" 8

22. w œ f(z) œ z Èz and f(z h) œ (z h) Èz h Ê œ lim

Šz h Èz h‹ ˆz Èz‰

hÄ!

h

œ 1 lim

(z h) z

h Ä ! h ŠÈz h Èz‹

h Èz h Èz h hÄ!

œ lim

œ 1 lim

"

h Ä ! Èz h Èz

"

dw dz

œ lim

hÄ!

œ lim –1 hÄ!

œ"

" 2È z

Ê

f(z h) f(z) h Èz h Èz h

dw ¸ dz zœ4

œ

†

ŠÈz h Èz‹ ŠÈz h Èz‹ —

5 4

"

fazb faxb a x #b a z # b xz " z #x # 23. f w axb œ zlim œ zlim œ zlim œ zlim œ zlim œ Äx zx Ä x zx Ä x az xbaz #bax #b Ä x az xbaz #bax #b Ä x az #bax #b ˆz2 3z 4‰ ˆx2 3x 4‰

" ax #b #

fazb faxb z 3z x 3x z x 3z 3x 24. f w axb œ zlim œ zlim œ zlim œ zlim zx zx zx Ä x zx Äx Äx Äx az xbaz xb 3‘ az xbaz xb 3az xb az xb 3‘ œ 2x 3 œ zlim œ zlim œ zlim zx zx Äx Äx Äx z

2

2

2

2

x

gazb gaxb z a x "b x a z " b z x " zc" x " 25. gw axb œ zlim œ zlim œ zlim œ zlim œ zlim œ Äx zx Äx zx Ä x az xbaz "bax "b Ä x az xbaz "bax "b Ä x az "bax "b g az b g a x b 26. gw axb œ zlim œ zlim Äx zx Äx

ˆ" Èz‰ˆ" Èx‰ zx

œ zlim Äx

Èz Èx zx

†

Èz Èx Èz Èx

" a x "b #

zx " œ zlim œ zlim œ Ä x az x bˆÈ z È x ‰ Ä x Èz Èx

" #È x

27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x œ 0), then positive Ê the slope is always increasing which matches (b). 28. Note that the slope of the tangent line is never negative. For x negative, f#w (x) is positive but decreasing as x increases. When x œ 0, the slope of the tangent line to x is 0. For x 0, f#w (x) is positive and increasing. This graph matches (a). 29. f$ (x) is an oscillating function like the cosine. Everywhere that the graph of f$ has a horizontal tangent we expect f$w to be zero, and (d) matches this condition. 30. The graph matches with (c). 31. (a) f w is not defined at x œ 0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree. For example, lim c xÄ!

f(x) f(0) x0

œ slope of line joining (%ß 0) and (!ß #) œ

" #

but lim b xÄ!

line joining (0ß 2) and ("ß 2) œ 4. Since these values are not equal, f w (0) œ

f(x) f(0) x0

f(0) lim f(x)x 0 xÄ!

(b)


œ slope of

does not exist.

Section 3.2 The Derivative as a Function 32. (a)

103

(b) Shift the graph in (a) down 3 units

33.

(b) The fastest is between the 20th and 30th days; slowest is between the 40th and 50th days.

34. (a)

35. Answers may vary. In each case, draw a tangent line and estimate its slope. ‰F (a) i) slope ¸ 1.54 Ê dT ii) slope ¸ 2.86 Ê dt ¸ 1.54 hr iii) slope ¸ 0 Ê

dT dt

¸ 0‰ hrF

iv) slope ¸ 3.75

dT ‰F dt ¸ 2.86 hr ‰F Ê dT dt ¸ 3.75 hr

(b) The tangent with the steepest positive slope appears to occur at t œ 6 Ê 12 p.m. and slope ¸ 7.27 Ê The tangent with the steepest negative slope appears to occur at t œ 12 Ê 6 p.m. and ‰F slope ¸ 8.00 Ê dT dt ¸ 8.00 hr (c)

36. Answers may vary. In each case, draw a tangent line and estimate the slope. lb (a) i) slope ¸ 20.83 Ê dW ii) slope ¸ 35.00 Ê dt ¸ 20.83 month iii) slope ¸ 6.25 Ê

dW dt

dW dt

lb ¸ 35.00 month

lb ¸ 6.25 month

(b) The tangentwith the steepest positive slope appears to occur at t œ 2.7 months. and slope ¸ 7.27 lb Ê dW dt ¸ 53.13 month


dT dt

¸ 7.27‰ hrF .

104


(c)

37. Left-hand derivative: For h 0, f(0 h) œ f(h) œ h# (using y œ x# curve) Ê œ lim c hÄ!

h# 0 h

œ lim c h œ 0; hÄ!

Right-hand derivative: For h 0, f(0 h) œ f(h) œ h (using y œ x curve) Ê œ lim b hÄ! Then lim c hÄ!

h0 h

lim

h Ä !c

œ lim b 1 œ 1; hÄ!

f(0 h) f(0) h

Á lim b hÄ!

f(0 h) f(0) h

lim

h Ä !b

œ lim c 0 œ 0; hÄ!

f(1 h) f(1) h

lim

h Ä !c

Right-hand derivative: When h !, 1 h 1 Ê f(1 h) œ 2(1 h) œ 2 2h Ê

Then lim c hÄ!

(2 2h)2 h

œ lim b hÄ!

f(1 h) f(1) h

2h h

hÄ!

È1 h " h

œ lim c

lim

h Ä !b

ŠÈ1 h "‹ h

hÄ!

†

ŠÈ1 h "‹ ŠÈ1 h 1‹

œ lim c hÄ!

lim

h Ä !c

Then lim c hÄ!

(2h 1) " h f(1 h) f(1) h

40. Left-hand derivative:

lim

h Ä !c

Right-hand derivative: œ lim b hÄ! Then lim c hÄ!

h h(1 h)


f(1 h) f(") h

lim b

hÄ!

œ lim b hÄ!

f(1 h) f(1) h

f(1 h) f(1) h

Á lim b hÄ!

(1 h) " h ŠÈ1 h "‹

œ lim c hÄ!

Á lim b hÄ!

" È1 h 1

lim

h Ä !b

Ê the derivative f w (1) does not exist. (1 h) " h

œ lim c hÄ!

f(1 h) f(") h " 1h

f(1 h) f(1) h

f(" h) f(1) h

Right-hand derivative: When h 0, 1 h 1 Ê f(1 h) œ 2(1 h) 1 œ 2h 1 Ê œ lim b hÄ!

œ lim c 1 œ 1; hÄ!

Š 1 " h "‹

œ lim b hÄ!

h

œ lim b hÄ!

Š

1 (1 h) 1 h ‹

h

œ 1; f(1 h) f(1) h

Ê the derivative f w (1) does not exist.

41. f is not continuous at x œ 0 since lim faxb œ does not exist and fa0b œ 1 xÄ!

42. Left-hand derivative: Right-hand derivative: Then lim c hÄ!

g(h) g(0) h

lim

h Ä !c

g(h) g(0) h

lim

h Ä !b

œ lim b hÄ!

22 h


39. Left-hand derivative: When h 0, 1 h 1 Ê f(1 h) œ È1 h Ê œ lim c

œ lim c hÄ!


f(1 h) f(1) h

Á lim b hÄ!

f(0 h) f(0) h


38. Left-hand derivative: When h !, 1 h 1 Ê f(1 h) œ 2 Ê

œ lim b hÄ!

f(0 h) f(0) h

œ lim c hÄ!

g(h) g(0) h

œ lim b hÄ!

g(h) g(0) h

h1Î3 0 h h

2Î3

œ lim c hÄ!

0 h

1 h2Î3

œ lim b hÄ!

œ +_;

1 h1Î3

œ +_;

œ _ Ê the derivative gw (0) does not exist.


œ #" ;

f("h)f(1) h

Section 3.2 The Derivative as a Function 43. (a) The function is differentiable on its domain $ Ÿ x Ÿ 2 (it is smooth) (b) none (c) none 44. (a) The function is differentiable on its domain # Ÿ x Ÿ 3 (it is smooth) (b) none (c) none 45. (a) The function is differentiable on $ Ÿ x 0 and ! x Ÿ 3 (b) none (c) The function is neither continuous nor differentiable at x œ 0 since lim c f(x) Á lim b f(x) xÄ! xÄ! 46. (a) f is differentiable on # Ÿ x 1, " x 0, 0 x 2, and 2 x Ÿ 3 (b) f is continuous but not differentiable at x œ 1: lim f(x) œ 0 exists but there is a corner at x œ 1 since x Ä 1

œ 3 and lim b f(" h)h f(1) œ 3 Ê f w (1) does not exist hÄ! hÄ! (c) f is neither continuous nor differentiable at x œ 0 and x œ 2: at x œ 0, lim c f(x) œ 3 but lim b f(x) œ 0 Ê lim f(x) does not exist; lim c

f(1 h) f(") h

xÄ!

xÄ0

xÄ!

at x œ 2, lim f(x) exists but lim f(x) Á f(2) xÄ#

xÄ#

47. (a) f is differentiable on " Ÿ x 0 and 0 x Ÿ 2 (b) f is continuous but not differentiable at x œ 0: lim f(x) œ 0 exists but there is a cusp at x œ 0, so f(0 h) f(0) h hÄ!

f w (0) œ lim

xÄ!

does not exist

(c) none 48. (a) f is differentiable on $ Ÿ x 2, 2 x 2, and 2 x Ÿ 3 (b) f is continuous but not differentiable at x œ 2 and x œ 2: there are corners at those points (c) none 49. (a) f w (x) œ lim

hÄ!

f(x h) f(x) h

œ lim

hÄ!

(x h)# ax# b h

œ lim

hÄ!

x# 2xh h# x# h

œ lim (2x h) œ 2x hÄ!

(b)

(c) yw œ 2x is positive for x 0, yw is zero when x œ 0, yw is negative when x 0 (d) y œ x# is increasing for _ x 0 and decreasing for ! x _; the function is increasing on intervals where yw 0 and decreasing on intervals where yw 0 f(x h) f(x) h hÄ!

50. (a) f w (x) œ lim

œ lim

hÄ!

Š xc" h h

1 x ‹

œ lim

hÄ!

x (x h) x(x h)h

œ lim

"

h Ä ! x(x h)

œ

" x#


105

106


(b)

(c) yw is positive for all x Á 0, yw is never 0, yw is never negative (d) y œ "x is increasing for _ x 0 and ! x _ 51. (a) Using the alternate formula for calculating derivatives: f w (x) œ zlim Äx œ

$ $ lim z x z Ä x 3(z x)

œ

az# zx x# b lim (z x)3(z x) zÄx

œ

# # lim z zx3 x zÄx

f(z) f(x) zx

#

w

$

Š z3

œ zlim Äx

œ x Ê f (x) œ x

x$ 3 ‹

zx

#

(b)

(c) yw is positive for all x Á 0, and yw œ 0 when x œ 0; yw is never negative (d) y œ

x$ 3

is increasing for all x Á 0 (the graph is horizontal at x œ 0) because y is increasing where yw 0; y is

never decreasing

52. (a) Using the alternate form for calculating derivatives: f w (x) œ zlim Äx œ

% % lim z x z Ä x 4(z x)

œ

$ xz# x# z x$ b lim (z x) az 4(z x) zÄx

œ

f(z) f(x) zx

$ # # $ lim z xz 4 x z x zÄx

œ zlim Äx $

Œ

z% 4

x% 4

zx

w

œ x Ê f (x) œ x$

(b)

(c) yw is positive for x 0, yw is zero for x œ 0, yw is negative for x 0 (d) y œ

x% 4

is increasing on 0 x _ and decreasing on _ x 0

# # # a2(x h)# 13(x h) 5b a2x# 13x 5b œ lim 2x 4xh 2h 13x h13h 5 2x 13x 5 h hÄ! hÄ! # lim 4xh 2hh 13h œ lim (4x 2h 13) œ 4x 13, slope at x. The slope is 1 when hÄ! hÄ!

53. yw œ lim œ

4x 13 œ "

Ê 4x œ 12 Ê x œ 3 Ê y œ 2 † 3# 13 † 3 5 œ 16. Thus the tangent line is y 16 œ (1)(x 3) Ê y œ x "$ and the point of tangency is (3ß 16). 54. For the curve y œ Èx, we have yw œ lim

hÄ!

œ lim

"

h Ä ! Èx h Èx

œ

" #Èx

ŠÈx h Èx‹ h

†

ŠÈx h Èx‹ ŠÈx h Èx‹

œ lim

(x h) x

h Ä ! ŠÈx h Èx‹ h

. Suppose ˆ+ß Èa‰ is the point of tangency of such a line and ("ß !) is the point

on the line where it crosses the x-axis. Then the slope of the line is

Èa 0 a (1)

œ

Èa a1

which must also equal


Section 3.2 The Derivative as a Function " ; 2È a

using the derivative formula at x œ a Ê

exist: its point of tangency is ("ß "), its slope is

Èa a1

œ

" #È a

œ

" Ê 2a œ a 1 Ê a œ 1. #Èa " # ; and an equation of the line is

Thus such a line does y1œ

" #

(x 1)

Ê y œ "# x "# . 55. Yes; the derivative of f is f w so that f w (x! ) exists Ê f w (x! ) exists as well. 56. Yes; the derivative of 3g is 3gw so that gw (7) exists Ê 3gw (7) exists as well. 57. Yes, lim

g(t)

t Ä ! h(t)

can exist but it need not equal zero. For example, let g(t) œ mt and h(t) œ t. Then g(0) œ h(0)

œ 0, but lim

g(t)

t Ä ! h(t)

œ lim

tÄ!

mt t

œ lim m œ m, which need not be zero. tÄ!

58. (a) Suppose kf(x)k Ÿ x# for " Ÿ x Ÿ 1. Then kf(0)k Ÿ 0# Ê f(0) œ 0. Then f w (0) œ lim œ lim

hÄ!

f(h) 0 h

œ lim

hÄ!

f(h) h .

For khk Ÿ 1, h# Ÿ f(h) Ÿ h# Ê h Ÿ

hÄ!

f(h) h

f(0 h) f(0) h

Ÿ h Ê f w (0) œ lim

hÄ!

f(h) h

œ0

by the Sandwich Theorem for limits. (b) Note that for x Á 0, kf(x)k œ ¸x# sin "x ¸ œ kx# k ksin xk Ÿ kx# k † 1 œ x# (since " Ÿ sin x Ÿ 1). By part (a), f is differentiable at x œ 0 and f w (0) œ 0.

59. The graphs are shown below for h œ 1, 0.5, 0.1. The function y œ y œ Èx so that

" #È x

œ lim

hÄ!

Èx h Èx h

" 2È x

. The graphs reveal that y œ

is the derivative of the function

Èx h Èx h

gets closer to y œ

" #È x

as h gets smaller and smaller.

60. The graphs are shown below for h œ 2, 1, 0.5. The function y œ 3x# is the derivative of the function y œ x$ so that 3x# œ lim

hÄ!

(xh)$ x$ h

. The graphs reveal that y œ

(xh)$ x$ h

gets closer to y œ 3x# as h

gets smaller and smaller.


107

108


61. The graphs are the same. So we know that for f(x) œ kxk , we have f w (x) œ

kx k x

.

62. Weierstrass's nowhere differentiable continuous function.

63-68. Example CAS commands: Maple: f := x -> x^3 + x^2 - x; x0 := 1; plot( f(x), x=x0-5..x0+2, color=black, title="Section 3.2, #63(a)" ); q := unapply( (f(x+h)-f(x))/h, (x,h) ); # (b) L := limit( q(x,h), h=0 ); # (c) m := eval( L, x=x0 ); tan_line := f(x0) + m*(x-x0); plot( [f(x),tan_line], x=x0-2..x0+3, color=black, linestyle=[1,7], title="Section 3.2 #63(d)", legend=["y=f(x)","Tangent line at x=1"] ); Xvals := sort( [ x0+2^(-k) $ k=0..5, x0-2^(-k) $ k=0..5 ] ): # (e) Yvals := map( f, Xvals ): evalf[4](< convert(Xvals,Matrix) , convert(Yvals,Matrix) >); plot( L, x=x0-5..x0+3, color=black, title="Section 3.2 #63(f)" ); Mathematica: (functions and x0 may vary) (see section 2.5 re. RealOnly ): Ê kv(0)k œ l 'l œ ' m/sec and kv(')k œ l'l œ ' m/sec; Ê a(0) œ # m/sec# and a(') œ # m/sec#

(c) v œ 0 Ê ' #t œ 0 Ê t œ $. v is positive in the interval ! t $ and v is negative when $ t ' Ê the body changes direction at t œ $. 3. s œ t$ 3t# 3t, 0 Ÿ t Ÿ 3 (a) displacement œ ?s œ s(3) s(0) œ 9 m, vav œ (b) v œ

ds dt

#

?s ?t

œ

9 3

œ 3 m/sec

œ 3t 6t 3 Ê kv(0)k œ k3k œ 3 m/sec and kv(3)k œ k12k œ 12 m/sec; a œ #

#

d# s dt#

œ 6t 6

Ê a(0) œ 6 m/sec and a(3) œ 12 m/sec (c) v œ 0 Ê 3t# 6t 3 œ 0 Ê t# 2t 1 œ 0 Ê (t 1)# œ 0 Ê t œ 1. For all other values of t in the interval the velocity v is negative (the graph of v œ 3t# 6t 3 is a parabola with vertex at t œ 1 which opens downward Ê the body never changes direction).


Section 3.4 The Derivative as a Rate of Change 4. s œ

t% 4

115

t$ t# , 0 Ÿ t Ÿ $

(a) ?s œ s($) s(0) œ $

m, vav œ

?s ?t

œ

* %

$

œ

$ %

m/sec

(b) v œ t 3t 2t Ê kv(0)k œ 0 m/sec and kv($)k œ ' m/sec; a œ 3t# 6t 2 Ê a(0) œ 2 m/sec# and a($) œ "" m/sec# (c) v œ 0 Ê t$ 3t# 2t œ 0 Ê t(t 2)(t 1) œ 0 Ê t œ 0, 1, 2 Ê v œ t(t 2)(t 1) is positive in the interval for 0 t 1 and v is negative for 1 t 2 and v is positive for # t $ Ê the body changes direction at t œ 1 and at t œ #. 5. s œ

25 t#

#

* %

5t , 1 Ÿ t Ÿ 5

(a) ?s œ s(5) s(1) œ 20 m, vav œ (b) v œ

50 t$

a(5) œ

4 25

(c) v œ 0 Ê 6. s œ

25 t5

5 t#

20 4

œ 5 m/sec

Ê kv(1)k œ 45 m/sec and kv(5)k œ

" 5

m/sec; a œ

150 t%

10 t$

Ê a(1) œ 140 m/sec# and

m/sec# 50 5t t$

œ 0 Ê 50 5t œ 0 Ê t œ 10 Ê the body does not change direction in the interval

, % Ÿ t Ÿ 0

(a) ?s œ s(0) s(4) œ 20 m, vav œ 20 4 œ 5 m/sec (b) v œ a(0) (c) v œ

25 (t 5)# Ê kv(4)k œ 25 œ 25 m/sec# 0 Ê (t255)# œ 0 Ê v is

m/sec and kv(0)k œ " m/sec; a œ

50 (t5)$

Ê a(4) œ 50 m/sec# and

never 0 Ê the body never changes direction

7. s œ t$ 6t# 9t and let the positive direction be to the right on the s-axis. (a) v œ 3t# 12t 9 so that v œ 0 Ê t# 4t 3 œ (t 3)(t 1) œ 0 Ê t œ 1 or 3; a œ 6t 12 Ê a(1) œ 6 m/sec# and a(3) œ 6 m/sec# . Thus the body is motionless but being accelerated left when t œ 1, and motionless but being accelerated right when t œ 3. (b) a œ 0 Ê 6t 12 œ 0 Ê t œ 2 with speed kv(2)k œ k12 24 9k œ 3 m/sec (c) The body moves to the right or forward on 0 Ÿ t 1, and to the left or backward on 1 t 2. The positions are s(0) œ 0, s(1) œ 4 and s(2) œ 2 Ê total distance œ ks(1) s(0)k ks(2) s(1)k œ k4k k2k œ 6 m. 8. v œ t# 4t 3 Ê a œ 2t 4 (a) v œ 0 Ê t# 4t 3 œ 0 Ê t œ 1 or 3 Ê a(1) œ 2 m/sec# and a(3) œ 2 m/sec# (b) v 0 Ê (t 3)(t 1) 0 Ê 0 Ÿ t 1 or t 3 and the body is moving forward; v 0 Ê at 3bat 1b 0 Ê " t 3 and the body is moving backward (c) velocity increasing Ê a 0 Ê 2t 4 0 Ê t 2; velocity decreasing Ê a 0 Ê 2t 4 0 Ê ! Ÿ t 2 9. sm œ 1.86t# Ê vm œ 3.72t and solving 3.72t œ 27.8 Ê t ¸ 7.5 sec on Mars; sj œ 11.44t# Ê vj œ 22.88t and solving 22.88t œ 27.8 Ê t ¸ 1.2 sec on Jupiter. 10. (a) v(t) œ sw (t) œ 24 1.6t m/sec, and a(t) œ vw (t) œ sw w (t) œ 1.6 m/sec# (b) Solve v(t) œ 0 Ê 24 1.6t œ 0 Ê t œ 15 sec (c) s(15) œ 24(15) .8(15)# œ 180 m (d) Solve s(t) œ 90 Ê 24t .8t# œ 90 Ê t œ

30 „ 15È2 #

¸ 4.39 sec going up and 25.6 sec going down

(e) Twice the time it took to reach its highest point or 30 sec 11. s œ 15t "# gs t# Ê v œ 15 gs t so that v œ 0 Ê 15 gs t œ 0 Ê gs œ

15 t

. Therefore gs œ


15 20

œ

3 4

œ 0.75 m/sec#

116


12. Solving sm œ 832t 2.6t# œ 0 Ê t(832 2.6t) œ 0 Ê t œ 0 or 320 Ê 320 sec on the moon; solving se œ 832t 16t# œ 0 Ê t(832 16t) œ 0 Ê t œ 0 or 52 Ê 52 sec on the earth. Also, vm œ 832 5.2t œ 0 Ê t œ 160 and sm (160) œ 66,560 ft, the height it reaches above the moon's surface; ve œ 832 32t œ 0 Ê t œ 26 and se (26) œ 10,816 ft, the height it reaches above the earth's surface. 13. (a) s œ 179 16t# Ê v œ 32t Ê speed œ kvk œ 32t ft/sec and a œ 32 ft/sec# (b) s œ 0 Ê 179 16t# œ 0 Ê t œ É 179 16 ¸ 3.3 sec È É 179 (c) When t œ É 179 16 , v œ 32 16 œ 8 179 ¸ 107.0 ft/sec 14. (a)

lim1 v œ lim1 9.8(sin ))t œ 9.8t so we expect v œ 9.8t m/sec in free fall

)Ä

(b) a œ

#

dv dt

)Ä

#

œ 9.8 m/sec# (b) between 3 and 6 seconds: $ Ÿ t Ÿ 6 (d)

15. (a) at 2 and 7 seconds (c)

16. (a) P is moving to the left when 2 t 3 or 5 t 6; P is moving to the right when 0 t 1; P is standing still when 1 t 2 or 3 t 5 (b)

17. (a) (c) (e) (f)

190 ft/sec at 8 sec, 0 ft/sec From t œ 8 until t œ 10.8 sec, a total of 2.8 sec Greatest acceleration happens 2 sec after launch

(g) From t œ 2 to t œ 10.8 sec; during this period, a œ

(b) 2 sec (d) 10.8 sec, 90 ft/sec

v(10.8) v(2) 10.8 2

¸ 32 ft/sec#

18. (a) Forward: 0 Ÿ t 1 and 5 t 7; Backward: 1 t 5; Speeds up: 1 t 2 and 5 t 6; Slows down: 0 Ÿ t 1, 3 t 5, and 6 t 7 (b) Positive: 3 t 6; negative: 0 Ÿ t 2 and 6 t 7; zero: 2 t 3 and 7 t 9 (c) t œ 0 and 2 Ÿ t Ÿ 3 (d) 7 Ÿ t Ÿ 9 19. s œ 490t# Ê v œ 980t Ê a œ 980 (a) Solving 160 œ 490t# Ê t œ

4 7

sec. The average velocity was

s(4/7) s(0) 4/7

œ 280 cm/sec.


Section 3.4 The Derivative as a Rate of Change (b) At the 160 cm mark the balls are falling at v(4/7) œ 560 cm/sec. The acceleration at the 160 cm mark was 980 cm/sec# . 17 (c) The light was flashing at a rate of 4/7 œ 29.75 flashes per second. 20. (a)

(b)

21. C œ position, A œ velocity, and B œ acceleration. Neither A nor C can be the derivative of B because B's derivative is constant. Graph C cannot be the derivative of A either, because A has some negative slopes while C has only positive values. So, C (being the derivative of neither A nor B) must be the graph of position. Curve C has both positive and negative slopes, so its derivative, the velocity, must be A and not B. That leaves B for acceleration. 22. C œ position, B œ velocity, and A œ acceleration. Curve C cannot be the derivative of either A or B because C has only negative values while both A and B have some positive slopes. So, C represents position. Curve C has no positive slopes, so its derivative, the velocity, must be B. That leaves A for acceleration. Indeed, A is negative where B has negative slopes and positive where B has positive slopes. 23. (a) c(100) œ 11,000 Ê cav œ #

11,000 100

œ $110

(b) c(x) œ 2000 100x .1x Ê cw (x) œ 100 .2x. Marginal cost œ cw (x) Ê the marginal cost of producing 100 machines is cw (100) œ $80 (c) The cost of producing the 101st machine is c(101) c(100) œ 100 201 10 œ $79.90 24. (a) r(x) œ 20000 ˆ1 "x ‰ Ê rw (x) œ w

(b) r a"!"b œ $"Þ*'Þ (c) x lim rw (x) œ x lim Ä_ Ä_

20000 x#

20000 x#

, which is marginal revenue. rw a"!!b œ

20000 100#

œ $#Þ

œ 0. The increase in revenue as the number of items increases without bound

will approach zero. 25. b(t) œ 10' 10% t 10$ t# Ê bw (t) œ 10% (2) a10$ tb œ 10$ (10 2t) (a) bw (0) œ 10% bacteria/hr (b) bw (5) œ 0 bacteria/hr w % (c) b (10) œ 10 bacteria/hr 26. Q(t) œ 200(30 t)# œ 200 a900 60t t# b Ê Qw (t) œ 200(60 2t) Ê Qw (10) œ 8,000 gallons/min is the rate the water is running at the end of 10 min. Then

Q(10) Q(0) 10

œ 10,000 gallons/min is the average rate the water flows

during the first 10 min. The negative signs indicate water is leaving the tank.


117

118


27. (a) y œ 6 ˆ1

t ‰# 1#

œ 6 Š1

(b) The largest value of value of

dy dt

dy dt

t 6

t# 144 ‹

Ê

dy dt

œ

t 12

1

is 0 m/h when t œ 12 and the fluid level is falling the slowest at that time. The smallest

is 1 m/h, when t œ 0, and the fluid level is falling the fastest at that time.

(c) In this situation,

Ÿ 0 Ê the graph of y is

dy dt

always decreasing. As

dy dt

increases in value,

the slope of the graph of y increases from 1 to 0 over the interval 0 Ÿ t Ÿ 12.

28. (a) V œ

4 3

1 r$ Ê

(b) When r œ 2,

dV dr

dV dr

œ 41 r # Ê

dV ¸ dr r=2

œ 41(2)# œ 161 ft$ /ft

œ 161 so that when r changes by 1 unit, we expect V to change by approximately 161. Therefore

when r changes by 0.2 units V changes by approximately (161)(0.2) œ 3.21 ¸ 10.05 ft$ . Note that V(2.2) V(2) ¸ 11.09 ft$ . 29. 200 km/hr œ 55 59 m/sec œ t œ 25, D œ

10 9

#

(25) œ

500 9 m/sec, 6250 9 m

and D œ

10 # 9 t

30. s œ v! t 16t# Ê v œ v! 32t; v œ 0 Ê t œ

v! 32

Ê Vœ

20 9

t. Thus V œ

500 9

Ê

; 1900 œ v! t 16t# so that t œ

Ê v! œ È(64)(1900) œ 80È19 ft/sec and, finally,

80È19 ft sec

†

60 sec 1 min

†

60 min 1 hr

†

1 mi 5280 ft

v! 32

20 9

tœ

500 9

Ê t œ 25 sec. When

Ê 1900 œ

v!# 3#

v!# 64

¸ 238 mph.

31.

v œ 0 when t œ 6.25 sec v 0 when 0 Ÿ t 6.25 Ê body moves right (up); v 0 when 6.25 t Ÿ 12.5 Ê body moves left (down) body changes direction at t œ 6.25 sec body speeds up on (6.25ß 12.5] and slows down on [0ß 6.25) The body is moving fastest at the endpoints t œ 0 and t œ 12.5 when it is traveling 200 ft/sec. It's moving slowest at t œ 6.25 when the speed is 0. (f) When t œ 6.25 the body is s œ 625 m from the origin and farthest away. (a) (b) (c) (d) (e)


Section 3.4 The Derivative as a Rate of Change 32.

(a) v œ 0 when t œ

3 #

sec

(b) v 0 when 0 Ÿ t 1.5 Ê body moves left (down); v 0 when 1.5 t Ÿ 5 Ê body moves right (up) (c) body changes direction at t œ 3# sec (d) body speeds up on ˆ 3# ß &‘ and slows down on !ß 3# ‰ (e) body is moving fastest at t œ 5 when the speed œ kv(5)k œ 7 units/sec; it is moving slowest at t œ

3 #

when the

speed is 0 (f) When t œ 5 the body is s œ 12 units from the origin and farthest away. 33.

6 „ È15 3 6 È15 t 3


sec

(b) v 0 when

6 È15 3

Ê body moves left (down); v 0 when 0 Ÿ t

6 È15 3

or

6 È15 3

tŸ4

Ê body moves right (up) 6 „ È15 sec 3 6 È15 6 È15 Š 3 ß #‹ Š 3 ß %“

(c) body changes direction at t œ (d) body speeds up on

È15

and slows down on ’0ß 6 3

È15

‹ Š#ß 6 3

‹.

(e) The body is moving fastest at t œ 0 and t œ 4 when it is moving 7 units/sec and slowest at t œ (f) When t œ

6È15 3

the body is at position s ¸ 6.303 units and farthest from the origin.

34.


6 „ È15 3


6„È15 3

sec

119

120

Chapter 3 Differentiation È 6 È15 or 6 3 15 t Ÿ 4 Ê body is moving left (down); v 0 when 3 È 6 È15 t 6 3 15 Ê body is moving right (up) 3 È body changes direction at t œ 6 „ 3 15 sec È È È È body speeds up on Š 6 3 15 ß #‹ Š 6 3 15 ß %“ and slows down on ’!ß 6 3 15 ‹ Š#ß 6 3 15 ‹

(b) v 0 when 0 Ÿ t

(c) (d)

(e) The body is moving fastest at 7 units/sec when t œ 0 and t œ 4; it is moving slowest and stationary at t œ (f) When t œ

6 È15 3

6 „ È15 3

the position is s ¸ 10.303 units and the body is farthest from the origin.

3.5 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 1. y œ 10x 3 cos x Ê 2. y œ

3 x

5 sin x Ê

3. y œ x2 cos x Ê

dy dx

œ

dy dx

5

(cos x) œ 10 3 sin x

(sin x) œ

d dx

œ Èx sec x tan x

dy dx

5. y œ csc x 4Èx 7 Ê " x#

3 x#

d dx

3 x#

5 cos x

œ x2 asin xb 2x cos x œ x2 sin x 2x cos x

4. y œ Èx sec x 3 Ê

6. y œ x# cot x

œ 10 3

dy dx

Ê

dy dx

dy dx

œ csc x cot x

œ x#

œ x# csc# x 2x cot x

sec x 2È x

d dx

0 œ Èx sec x tan x 4 #È x

(cot x) cot x †

d dx

sec x 2È x

0 œ csc x cot x ax# b

2 x$

2 Èx

œ x# csc# x (cot x)(2x)

2 x$

2 x$

sin x 2 7. faxb œ sin x tan x Ê f w axb œ sin x sec2 x cos x tan x œ sin x sec2 x cos x cos x œ sin xasec x 1b

8. gaxb œ csc xcot x Ê gw axb œ csc xacsc2 xb acsc xcot xbcot x œ csc3 x csc x cot2 x œ csc xacsc2 x cot2 xb 9. y œ (sec x tan x)(sec x tan x) Ê #

dy dx

œ (sec x tan x)

d dx

(sec x tan x) (sec x tan x) #

d dx

(sec x tan x)

œ (sec x tan x) asec x tan x sec xb (sec x tan x) asec x tan x sec xb œ asec# x tan x sec x tan# x sec$ x sec# x tan xb asec# x tan x sec x tan# x sec$ x tan x sec# xb œ 0. ŠNote also that y œ sec# x tan# x œ atan# x 1b tan# x œ 1 Ê 10. y œ (sin x cos x) sec x Ê

œ (sin x cos x)

dy dx

d dx

sin# x cos x sin x cos# x cos x sin x cos# x

œ

" cos# x

d dx (sin x cos x) (sin x cos x) sin x x sin x cos cos cos# x x

œ sec# x

ŠNote also that y œ sin x sec x cos x sec x œ tan x 1 Ê 11. y œ œ

Ê

dy dx

œ

(1 cot x)

csc# x csc# x cot x csc# x cot x (1 cot x)#

12. y œ œ

cot x 1 cot x

cos x 1 sin x

Ê

dy dx

œ

sin x sin# x cos# x (1 sin x)#

(1 sin x)

œ

d dx

œ

œ 0.‹

(sec x) sec x

œ (sin x cos x)(sec x tan x) (sec x)(cos x sin x) œ œ

dy dx

(cot x) (cot x) (1 cot x)#

d dx

(1 cot x)

œ

dy dx

œ sec# x.‹

(1 cot x) acsc# xb (cot x) acsc# xb (1 cot x)#

csc# x (1 cot x)#

d (cos x) (cos x) dx (1 sin x) b (cos x) acos xb œ (1 sin x) a(1sin xsin (1 sin x)# x)# (1 sin x) sin x 1 " (1 sin x)# œ (1 sin x)# œ 1 sin x d dx


Section 3.5 Derivatives of Trigonometric Functions 13. y œ

4 cos x

" tan x

14. y œ

cos x x

x cos x

œ 4 sec x cot x Ê Ê

œ

dy dx

œ 4 sec x tan x csc# x

dy dx

x(sin x) (cos x)(1) x#

15. y œ x# sin x 2x cos x 2 sin x Ê

121

(cos x)(1) x(sin x) cos# x

œ

x sin x cos x x#

cos x x sin x cos# x

dy dx

œ ax# cos x (sin x)(2x)b a(2x)(sin x) (cos x)(2)b 2 cos x

dy dx

œ ax# (sin x) (cos x)(2x)b a2x cos x (sin x)(2)b 2(sin x)

œ x# cos x 2x sin x 2x sin x 2 cos x 2 cos x œ x# cos x 16. y œ x# cos x 2x sin x 2 cos x Ê

œ x# sin x 2x cos x 2x cos x 2 sin x 2 sin x œ x# sin x 17. faxb œ x3 sin x cos x Ê f w axb œ x3 sin xasin xb x3 cos xacos xb 3x2 sin x cos x œ x3 sin2 x x3 cos2 x 3x2 sin x cos x 18. gaxb œ a2 xbtan2 x Ê gw axb œ a2 xba2 tan x sec2 xb a1btan2 x œ 2a2 xbtan x sec2 x tan2 x œ 2a2 xbtan xa sec2 x tan xb 19. s œ tan t t Ê 1 csc t 1 csc t

21. s œ œ

Ê

œ sec# t 1

ds dt ds dt

20. s œ t# sec t 1 Ê

sin t 1 cos t

Ê

ds dt

œ 2t sec t tan t

(1 csc t)(csc t cot t) (" csc t)(csc t cot t) (1 csc t)#

œ

csc t cot t csc# t cot t csc t cot t csc# t cot t (1 csc t)#

22. s œ

ds dt

(1 cos t)(cos t) (sin t)(sin t) (1 cos t)#

œ

23. r œ 4 )# sin ) Ê

2 csc t cot t (1 csc t)#

œ

dr d)

œ ˆ) #

24. r œ ) sin ) cos ) Ê

dr d)

d d)

œ

cos t cos# t sin# t (1 cos t)#

œ

cos t " (1 cos t)#

œ 1 "cos t œ

" cos t 1

(sin )) (sin ))(2))‰ œ a)# cos ) 2) sin )b œ )() cos ) # sin ))

œ () cos ) (sin ))(1)) sin ) œ ) cos )

dr 25. r œ sec ) csc ) Ê d) œ (sec ))(csc ) cot )) (csc ))(sec ) tan )) " " cos ) sin ) ‰ " " # # œ ˆ cos ) ‰ ˆ sin ) ‰ ˆ sin ) ‰ ˆ sin" ) ‰ ˆ cos" ) ‰ ˆ cos ) œ sin# ) cos# ) œ sec ) csc )

26. r œ (1 sec )) sin ) Ê 27. p œ &

" cot q

œ

sin q cos q cos q

Ê

dp dq

œ

dp dq

dp dq

tan q 1 tan q

31. p œ

q sin q q2 1

Ê

Ê dp dq

œ sec# q

œ (1 csc q)(sin q) (cos q)(csc q cot q) œ (sin q 1) cot# q œ sin q csc# q

(cos q)(cos q sin q) (sin q cos q)(sin q) cos# q

cos# q cos q sin q sin# q cos q sin q cos# q

30. p œ

œ

œ (" sec )) cos ) (sin ))(sec ) tan )) œ (cos ) ") tan# ) œ cos ) sec# )

œ 5 tan q Ê

28. p œ (1 csc q) cos q Ê 29. p œ

dr d)

œ

" cos# q

œ sec# q

(1 tan q) asec# qb (tan q) asec# qb (1 tan q)#

dp dq

œ

œ

ˆq2 1‰aq cos q sin qa1bb aq sin qba2qb aq 2 1 b 2

œ

sec# q tan q sec# q tan q sec# q (1 tan q)#

œ

œ

sec# q (1 tan q)#

q3 cos q q2 sin q q cos q sin q 2q2 sin q aq 2 1 b 2

q3 cos q q2 sin q q cos q sin q aq 2 1 b 2


122

Chapter 3 Differentiation aq sec qbˆ3 sec2 q‰ a3q tan qbaq sec q tan q sec qa1bb aq sec qb2 3 2 ˆ 3q sec q q sec q 3q sec q tan q 3q sec q q sec q tan2 q sec q tan q‰ q sec3 q 3q2 sec q tan q q sec q tan2 q sec q tan q

32. p œ œ

3q tan q q sec q

Ê

dp dq

œ

œ

aq sec qb2

aq sec qb2

33. (a) y œ csc x Ê yw œ csc x cot x Ê yww œ a(csc x) acsc# xb (cot x)(csc x cot x)b œ csc$ x csc x cot# x œ (csc x) acsc# x cot# xb œ (csc x) acsc# x csc# x 1b œ 2 csc$ x csc x (b) y œ sec x Ê yw œ sec x tan x Ê yww œ (sec x) asec# xb (tan x)(sec x tan x) œ sec$ x sec x tan# x œ (sec x) asec# x tan# xb œ (sec x) asec# x sec# x 1b œ 2 sec$ x sec x 34. (a) y œ 2 sin x Ê yw œ 2 cos x Ê yww œ 2(sin x) œ 2 sin x Ê ywww œ 2 cos x Ê yÐ%Ñ œ 2 sin x (b) y œ 9 cos x Ê yw œ 9 sin x Ê yww œ 9 cos x Ê ywww œ 9(sin x) œ 9 sin x Ê yÐ%Ñ œ 9 cos x 35. y œ sin x Ê yw œ cos x Ê slope of tangent at x œ 1 is yw (1) œ cos (1) œ "; slope of tangent at x œ 0 is yw (0) œ cos (0) œ 1; and slope of tangent at x œ 3#1 is yw ˆ 3#1 ‰ œ cos 3#1

œ 0. The tangent at (1ß !) is y 0 œ 1(x 1), or y œ x 1; the tangent at (0ß 0) is y 0 œ 1(x 0), or y œ x; and the tangent at ˆ 31 ‰ # ß 1 is y œ 1.

36. y œ tan x Ê yw œ sec# x Ê slope of tangent at x œ 13 is sec# ˆ 13 ‰ œ 4; slope of tangent at x œ 0 is sec# (0) œ 1; and slope of tangent at x œ

1 3

is sec# ˆ 13 ‰ œ 4. The tangent

at ˆ 13 ß tanˆ 13 ‰‰ œ Š 13 ß È3‹ is y È3 œ 4ˆx 13 ‰ ; the tangent at (0ß 0) is y œ x; and the tangent at ˆ 13 ß tan ˆ 13 ‰‰ œ Š 13 ß È3‹ is y È3 œ 4 ˆx 13 ‰ . 37. y œ sec x Ê yw œ sec x tan x Ê slope of tangent at x œ 13 is sec ˆ 13 ‰ tan ˆ 13 ‰ œ 2È3 ; slope of tangent is sec ˆ 14 ‰ tan ˆ 14 ‰ œ È2 . The tangent at the point ˆ 1 ß sec ˆ 1 ‰‰ œ ˆ 1 ß #‰ is y 2 œ #È3 ˆx 1 ‰ ;

at x œ

1 4

3

3

3

the tangent at the point

3

ˆ 14 ß sec ˆ 14 ‰‰

œ

Š 14 ß È2‹

is y È2

œ È2 ˆx 14 ‰ .

38. y œ 1 cos x Ê yw œ sin x Ê slope of tangent at È

x œ 13 is sin ˆ 13 ‰ œ #3 ; slope of tangent at x œ ‰ œ 1. The tangent at the point is sin ˆ 31 #

31 #

ˆ 13 ß " cos ˆ 13 ‰‰ œ ˆ 13 ß 3# ‰ È

is y 3# œ #3 ˆx 13 ‰ ; the tangent at the point ˆ 3#1 ß " cos ˆ 3#1 ‰‰ œ ˆ 3#1 ß 1‰ is y 1 œ x 3#1 39. Yes, y œ x sin x Ê yw œ " cos x; horizontal tangent occurs where 1 cos x œ 0 Ê cos x œ 1 Ê x œ 1


Section 3.5 Derivatives of Trigonometric Functions 40. No, y œ 2x sin x Ê yw œ 2 cos x; horizontal tangent occurs where 2 cos x œ 0 Ê cos x œ #. But there are no x-values for which cos x œ #. 41. No, y œ x cot x Ê yw œ 1 csc# x; horizontal tangent occurs where 1 csc# x œ 0 Ê csc# x œ 1. But there are no x-values for which csc# x œ 1. 42. Yes, y œ x 2 cos x Ê yw œ 1 2 sin x; horizontal tangent occurs where 1 2 sin x œ 0 Ê 1 œ 2 sin x Ê "# œ sin x Ê x œ 16 or x œ 561 43. We want all points on the curve where the tangent line has slope 2. Thus, y œ tan x Ê yw œ sec# x so that yw œ 2 Ê sec# x œ 2 Ê sec x œ „ È2 Ê x œ „ 14 . Then the tangent line at ˆ 14 ß "‰ has equation y 1 œ 2 ˆx 14 ‰ ; the tangent line at ˆ 14 ß "‰ has equation y 1 œ 2 ˆx 14 ‰ .

44. We want all points on the curve y œ cot x where the tangent line has slope 1. Thus y œ cot x Ê yw œ csc# x so that yw œ 1 Ê csc# x œ 1 Ê csc# x œ 1 Ê csc x œ „ 1 Ê x œ 1# . The tangent line at ˆ 1# ß !‰ is y œ x 12 .

2 cos x ‰ 45. y œ 4 cot x 2 csc x Ê yw œ csc# x 2 csc x cot x œ ˆ sin" x ‰ ˆ 1 sin x

(a) When x œ 1# , then yw œ 1; the tangent line is y œ x w

1 #

2.

(b) To find the location of the horizontal tangent set y œ 0 Ê 1 2 cos x œ 0 Ê x œ then y œ % È3 is the horizontal tangent. 46. y œ 1 È2 csc x cot x Ê yw œ È2 csc x cot x csc# x œ ˆ sin" x ‰ Š

1 3

È2 cos x 1 ‹ sin x

(a) If x œ 14 , then yw œ 4; the tangent line is y œ 4x 1 4. (b) To find the location of the horizontal tangent set yw œ 0 Ê È2 cos x 1 œ 0 Ê x œ xœ

31 4 ,

radians. When x œ 13 ,

31 4

radians. When

then y œ 2 is the horizontal tangent.

47. lim sin ˆ "x #" ‰ œ sin ˆ #" #" ‰ œ sin 0 œ 0 xÄ2

48.

lim

x Ä 16

49. lim1 )Ä

6

È1 cos (1 csc x) œ É1 cos ˆ1 csc ˆ 1 ‰‰ œ È1 cos a1 † a2bb œ È2

sin ) "# ) 16

6

œ

d d) asin )b¹)œ 1 6

œ cos )¹

)œ 16

œ cosˆ 16 ‰ œ

È3 2


123

124

Chapter 3 Differentiation tan ) 1 ) 14

50. lim1 )Ä

4

œ

d d) atan )b¹)œ 1

œ sec2 )¹

4

œ sec2 ˆ 14 ‰ œ 2

)œ 14

1 ‰ ‘ ˆ 1 ‰ ‘ ˆ 1 ‰‘ œ sec 1 œ 1 51. lim sec cos x 1 tan ˆ 4 sec x 1 œ sec 1 1 tan 4 sec 0 1 œ œ sec 1 tan 4

xÄ!

x ‰ ˆ 1 tan 0 ‰ ˆ 1‰ 52. lim sin ˆ tan1xtan 2 sec x œ sin tan 0 2 sec 0 œ sin # œ 1

xÄ!

53. lim tan ˆ1 tÄ!

sin t ‰ t

œ tan Š1 lim

tÄ!

1) ‰ 54. lim cos ˆ sin ) œ cos Š1 lim

)

sin t t ‹

‹ ) Ä ! sin )

)Ä!

œ tan (1 1) œ 0 "

œ cos Œ1 †

lim

sin ) )

)Ä!

" œ cos ˆ1 † 1 ‰ œ 1

dv da ˆ1‰ 55. s œ 2 2 sin t Ê v œ ds dt œ 2 cos t Ê a œ dt œ 2 sin t Ê j œ dt œ 2 cos t. Therefore, velocity œ v 4 œ È2 m/sec; speed œ ¸v ˆ 1 ‰¸ œ È2 m/sec; acceleration œ a ˆ 1 ‰ œ È2 m/sec# ; jerk œ j ˆ 1 ‰ œ È2 m/sec$ . 4

56. s œ sin t cos t Ê v œ velocity œ

v ˆ 14 ‰

58.

œ cos t sin t Ê a œ

ds dt

œ 0 m/sec; speed œ

sin# 3x x# xÄ!

57. lim f(x) œ lim xÄ!

4

¸v ˆ 14 ‰¸

4

œ sin t cos t Ê j œ

dv dt

œ 0 m/sec; acceleration œ

œ cos t sin t. Therefore œ È2 m/sec# ; jerk œ j ˆ 1 ‰ œ 0 m/sec$ .

da dt

a ˆ 14 ‰

4

œ lim 9 ˆ sin3x3x ‰ ˆ sin3x3x ‰ œ 9 so that f is continuous at x œ 0 Ê lim f(x) œ f(0) Ê 9 œ c. xÄ!

xÄ!

lim g(x) œ lim c (x b) œ b and lim b g(x) œ lim b cos x œ 1 so that g is continuous at x œ 0 Ê lim c g(x) xÄ! xÄ! xÄ! xÄ! œ lim b g(x) Ê b œ 1. Now g is not differentiable at x œ 0: At x œ 0, the left-hand derivative is

x Ä !c

xÄ!

d dx

(x b)¸ x = 0 œ 1, but the right-hand derivative is

(cos x)¸ x=0 œ sin 0 œ 0. The left- and right-hand

d dx

derivatives can never agree at x œ 0, so g is not differentiable at x œ 0 for any value of b (including b œ 1). 59.

d*** dx***

d% dx%

(cos x) œ sin x because

(cos x) œ cos x Ê the derivative of cos x any number of times that is a

multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999 œ 249 † 4 3 Ê œ

$

d dx$

#%*†%

d ’ dx #%* % (cos x)“ œ †

60. (a) y œ sec x œ Ê

d dx

" cos x

$

d dx$

Ê

d*** dx***

(cos x)

(cos x) œ sin x.

dy dx

œ

(cos x)(0) (1)(sin x) (cos x)#

œ

(sin x)(0) (1)(cos x) (sin x)#

œ

sin x cos# x

sin x ‰ œ ˆ cos" x ‰ ˆ cos x œ sec x tan x

(sec x) œ sec x tan x

(b) y œ csc x œ Ê

d dx

d dx

Ê

dy dx

cos x sin# x

œ

" ‰ ˆ cos x ‰ œ ˆ sin x sin x œ csc x cot x

(csc x) œ csc x cot x

(c) y œ cot x œ Ê

" sin x

cos x sin x

Ê

dy dx #

œ

(sin x)(sin x) (cos x)(cos x) (sin x)#

œ

sin# xcos# x sin# x

œ

" sin# x

œ csc# x

(cot x) œ csc x

61. (a) t œ 0 Ä x œ 10 cosa0b œ 10 cm; t œ cm (b) t œ 0 Ä v œ 10 sina0b œ 0 sec ;tœ

1 3 1 3

Ä x œ 10 cosˆ 13 ‰ œ 5 cm; t œ 341 Ä x œ 10 cosˆ 341 ‰ œ 5È2 cm cm cm Ä v œ 10 sinˆ 13 ‰ œ 5È3 sec ; t œ 341 Ä v œ 10 sinˆ 341 ‰ œ 5È2 sec

62. (a) t œ 0 Ä x œ 3 cosa0b 4 sina0b œ 3 ft; t œ

1 2

Ä x œ 3 cosˆ 12 ‰ 4 sinˆ 12 ‰ œ 4 ft;

t œ 1 Ä x œ 3 cosa1b 4 sina1b œ 3 ft ft (b) t œ 0 Ä v œ 3 sina0b 4 cosa0b œ 4 sec ;tœ t œ 1 Ä v œ 3 sina1b 4 cosa1b œ 4

1 2

Ä v œ 3 sinˆ 12 ‰ 4 cosˆ 12 ‰ œ 3

ft sec ;

ft sec


Section 3.5 Derivatives of Trigonometric Functions

125

63.

As h takes on the values of 1, 0.5, 0.3 and 0.1 the corresponding dashed curves of y œ closer and closer to the black curve y œ cos x because

d dx

(sin x) œ

is true as h takes on the values of 1, 0.5, 0.3 and 0.1.

sin x lim sin (x h) h hÄ!

sin (x h) sin x h

get

œ cos x. The same

64.

cos (x h) cos x h cos x lim cos (x h) œ sin x. h hÄ!

As h takes on the values of 1, 0.5, 0.3, and 0.1 the corresponding dashed curves of y œ

get

closer and closer to the black curve y œ sin x because

The

d dx

(cos x) œ

same is true as h takes on the values of 1, 0.5, 0.3, and 0.1. 65. (a)

The dashed curves of y œ

sinax hb sinax hb #h

are closer to the black curve y œ cos x than the corresponding dashed

curves in Exercise 63 illustrating that the centered difference quotient is a better approximation of the derivative of this function. (b)

The dashed curves of y œ

cosax hb cosax hb #h

are closer to the black curve y œ sin x than the corresponding dashed

curves in Exercise 64 illustrating that the centered difference quotient is a better approximation of the derivative of this function. 66. lim

hÄ!

k0 h k k 0 h k 2h

œ lim

xÄ!

k h k k hk 2h

œ lim 0 œ 0 Ê the limits of the centered difference quotient exists even hÄ!

though the derivative of f(x) œ kxk does not exist at x œ 0.


126


67. y œ tan x Ê yw œ sec# x, so the smallest value yw œ sec# x takes on is yw œ 1 when x œ 0; yw has no maximum value since sec# x has no largest value on ˆ 1# ß 1# ‰ ; yw is never negative since sec# x 1.

68. y œ cot x Ê yw œ csc# x so yw has no smallest value since csc# x has no minimum value on (!ß 1); the largest value of yw is 1, when x œ 1# ; the slope is never positive since the largest value yw œ csc2 x takes on is 1. 69. y œ

sin x x appears to cross the y-axis at y œ 1, since lim sin x œ 1; y œ sinx2x appears to cross the y-axis xÄ! x

at y œ 2, since lim sinx2x œ 2; y œ sinx4x appears to xÄ! cross the y-axis at y œ 4, since lim sinx4x œ 4. xÄ!

However, none of these graphs actually cross the y-axis since x œ 0 is not in the domain of the functions. Also, lim

xÄ!

sin 5x x

sin (3x) x

œ 5, lim

xÄ!

œ k Ê the graphs of y œ yœ

sin kx x

œ 3, and lim

sin kx x

xÄ!

yœ

sin 5x x ,

sin (3x) , x

and

approach 5, 3, and k, respectively, as

x Ä 0. However, the graphs do not actually cross the y-axis. 70. (a)

sin h h

h 1 0.01 0.001 0.0001 lim

hÄ!

sin h° h

ˆ sinh h ‰ ˆ 180 ‰ 1 .99994923 1 1 1

.017452406 .017453292 .017453292 .017453292 œ lim

xÄ!

1 ‰ sin ˆh† 180 h

œ lim

1

180

hÄ!

1 ‰ sin ˆh† 180 1 180 †h

œ lim

)Ä!

1 sin )

180

)

œ

1 180

() œ h †

1 180 )

(converting to radians) (b)

cos h1 h

h 1 0.01 0.001 0.0001 lim

hÄ!

0.0001523 0.0000015 0.0000001 0

cos h1 h

(c) In degrees,

œ 0, whether h is measured in degrees or radians.

d dx

(sin x) œ lim

hÄ!

œ lim ˆsin x † hÄ!

cos h 1 ‰ h

sin (x h) sin x h

lim ˆcos x † hÄ!

1 ‰ œ (sin x)(0) (cos x) ˆ 180 œ

1 180

œ lim

hÄ!

sin h ‰ h

(sin x cos h cos x sin h) sin x h

œ (sin x) † lim ˆ cos hh 1 ‰ (cos x) † lim ˆ sinh h ‰ hÄ!

hÄ!

cos x


Section 3.6 The Chain Rule (d) In degrees, œ lim

hÄ!

d dx

hÄ!

(cos x)(cos h 1) sin x sin h h

œ (cos x) lim ˆ hÄ!

(e)

d# dx# d# dx#

cos (x h) cos x h

(cos x) œ lim cos h 1 ‰ h

œ lim

hÄ!

œ lim ˆcos x † hÄ!

127

(cos x cos h sin x sin h) cos x h

cos h 1 ‰ h

lim ˆsin x † hÄ!

sin h ‰ h

1 ‰ 1 (sin x) lim ˆ sinh h ‰ œ (cos x)(0) (sin x) ˆ 180 œ 180 sin x

hÄ!

1 ‰# ˆ 180

d$ dx$

(sin x) œ

d dx

1 ˆ 180

(cos x) œ

d dx

1 1 ‰# ˆ 180 sin x‰ œ ˆ 180 cos x;

cos x‰ œ

sin x;

(sin x) œ d$ dx$

d dx

#

$

1 ‰ 1 ‰ sin x‹ œ ˆ 180 cos x; Š ˆ 180

(cos x) œ

d dx

#

$

1 ‰ 1 ‰ cos x‹ œ ˆ 180 sin x Š ˆ 180

3.6 THE CHAIN RULE 1. f(u) œ 6u 9 Ê f w (u) œ 6 Ê f w (g(x)) œ 6; g(x) œ

" #

x% Ê gw (x) œ 2x$ ; therefore

dy dx

œ f w (g(x))gw (x) œ 6 † 2x$ œ 12x$

2. f(u) œ 2u$ Ê f w (u) œ 6u# Ê f w (g(x)) œ 6(8x 1)# ; g(x) œ 8x 1 Ê gw (x) œ 8; therefore œ 6(8x 1)# † 8 œ 48(8x 1)#

dy dx

œ f w (g(x))gw (x)

3. f(u) œ sin u Ê f w (u) œ cos u Ê f w (g(x)) œ cos (3x 1); g(x) œ 3x 1 Ê gw (x) œ 3; therefore

œ f w (g(x))gw (x)

dy dx

œ (cos (3x 1))(3) œ 3 cos (3x 1) 4. f(u) œ cos u Ê f w (u) œ sin u Ê f w (g(x)) œ sin ˆ 3x ‰ ; g(x) œ ‰ œ "3 sin ˆ 3x ‰ œ sin ˆ 3x ‰ † ˆ " 3

x 3

Ê gw (x) œ "3 ; therefore

dy dx

œ f w (g(x))gw (x)

5. f(u) œ cos u Ê f w (u) œ sin u Ê f w (g(x)) œ sin (sin x); g(x) œ sin x Ê gw (x) œ cos x; therefore dy dx

œ f w (g(x))gw (x) œ (sin (sin x)) cos x

6. f(u) œ sin u Ê f w (u) œ cos u Ê f w (g(x)) œ cos (x cos x); g(x) œ x cos x Ê gw (x) œ 1 sin x; therefore dy dx

œ f w (g(x))gw (x) œ (cos (x cos x))(1 sin x)

7. f(u) œ tan u Ê f w (u) œ sec# u Ê f w (g(x)) œ sec# (10x 5); g(x) œ 10x 5 Ê gw (x) œ 10; therefore dy dx

œ f w (g(x))gw (x) œ asec# (10x 5)b (10) œ 10 sec# (10x 5)

8. f(u) œ sec u Ê f w (u) œ sec u tan u Ê f w (g(x)) œ sec ax# 7xb tan ax# 7xb ; g(x) œ x# 7x Ê gw (x) œ 2x 7; therefore

dy dx

œ f w (g(x))gw (x) œ (2x 7) sec ax# 7xb tan ax# 7xb

9. With u œ (2x 1), y œ u& :

dy dx

œ

dy du du dx

œ 5u% † 2 œ 10(2x 1)%

10. With u œ (4 3x), y œ u* :

dy dx

œ

dy du du dx

œ 9u) † (3) œ 27(4 3x))

11. With u œ ˆ1 x7 ‰ , y œ u( : 12. With u œ ˆ x# 1‰ , y œ u"! :

dy dx

œ

dy dx

œ

#

13. With u œ Š x8 x "x ‹ , y œ u% : 14. With u œ 3x2 4x 6, y œ u1Î2 :

) œ 7u) † ˆ "7 ‰ œ ˆ" x7 ‰

dy du du dx dy du du dx

dy dx

dy dx

œ œ

œ 10u"" † ˆ "# ‰ œ 5 ˆ x# 1‰

dy du du dx

dy du du dx

œ 4u$ † ˆ x4 1

"‰ x#

""

#

$

œ 4 Š x8 x "x ‹ ˆ x4 1

œ "# u1Î2 † a6x 4b œ

3x 2 È3x2 4x6


"‰ x#

128


15. With u œ tan x, y œ sec u: 16. With u œ 1 "x , y œ cot u: 17. With u œ sin x, y œ u$ :

œ

dy dx dy dx

œ

dy du du dx

dy dx

œ

dy du du dx

18. With u œ cos x, y œ 5u% :

dy dx

œ

19. p œ È3 t œ (3 t)"Î# Ê 3 20. q œ È 2r r# œ a2r r# b

21. s œ œ

4 31 4 1

sin 3t

4 51

dp dt

"Î3

œ

œ acsc# ub ˆ x"# ‰ œ x"# csc# ˆ1 x" ‰

œ 3u# cos x œ 3 asin# xb (cos x)

dy du du dx

Ê

cos 5t Ê

œ (sec u tan u) asec# xb œ (sec (tan x) tan (tan x)) sec# x

dy du du dx

" #

dq dr

œ a20u& b (sin x) œ 20 acos& xb (sin x)

(3 t)"Î# † œ

" 3

d dt

a2r r# b

(3 t) œ "# (3 t)"Î# œ

2 Î3

†

d dr

a2r r# b œ

ds dt

œ

cos 3t †

d dt

(3t)

ds dt

œ cos ˆ 3#1t ‰ †

d dt

ˆ 3#1t ‰ sin ˆ 3#1t ‰ †

4 31

4 51

(sin 5t) †

d dt

" 3

" 2È 3 t

a2r r# b

(5t) œ

4 1

2 Î3

2 2r 3a2rr# b2Î3

(2 2r) œ

cos 3t

4 1

sin 5t

(cos 3t sin 5t)

22. s œ sin ˆ 3#1t ‰ cos ˆ 3#1t ‰ Ê œ 321 ˆcos 3#1t sin 3#1t ‰ 23. r œ (csc ) cot ))" Ê

dr d)

24. r œ 6(sec ) tan ))3Î2 Ê

œ (csc ) cot ))# dr d)

d d)

d dt

ˆ 3#1t ‰ œ

(csc ) cot )) œ

œ 6 † 3# asec ) tan )b1Î#

d d) asec

31 2

cos ˆ 3#1t ‰

csc ) cot ) csc# ) (csc ) cot ))#

œ

31 2

sin ˆ 3#1t ‰

csc ) (cot ) csc )) (csc ) cot ))#

œ

csc ) csc ) cot )

) tan )b œ 9Èsec ) tan )asec ) tan ) sec2 )b

d d # d % % # # 25. y œ x# sin% x x cos# x Ê dy xb cos# x † dx œ x dx asin xb sin x † dx ax b x dx acos d d œ x# ˆ4 sin$ x dx (sin x)‰ 2x sin% x x ˆ2 cos$ x † dx (cos x)‰ cos# x

d dx

(x)

œ x# a4 sin$ x cos xb 2x sin% x xa a2 cos$ xb (sin x)b cos# x œ 4x# sin$ x cos x 2x sin% x 2x sin x cos$ x cos# x d ˆ"‰ x d $ $ asin& xb sin& x † dx x 3 dx acos xb cos x † œ "x a5 sin' x cos xb asin& xb ˆ x"# ‰ 3x a a3 cos# xb (sin x)b acos$ xb ˆ 3" ‰

26. y œ

" x

sin& x

x 3

cos$ x Ê yw œ

œ 5x sin' x cos x

" x#

" d x dx

sin& x x cos# x sin x

" 3

ˆ x3 ‰

cos$ x

" " ‰" 7 d ( ' ˆ Ê dy 21 (3x 2) 4 #x# dx œ 21 (3x 2) † dx (3x 2) 7 " ‰# ˆ " ‰ " ' ' ˆ # 21 (3x 2) † 3 (1) 4 #x# x$ œ (3x 2) $ x Š4 "# ‹

27. y œ œ

d dx

(1) ˆ4

" ‰# #x #

†

d dx

ˆ4

" ‰ #x#

#x

% 28. y œ (5 2x)$ "8 ˆ x2 1‰ Ê

dy dx

$ $ œ 3(5 2x)% (2) 84 ˆ x2 1‰ ˆ x2# ‰ œ 6(5 2x)% ˆ x"# ‰ ˆ x2 1‰

$

œ

6 (5 2x)%

Š 2x 1‹ x#

29. y œ (4x 3)% (x 1)$ Ê %

dy dx

œ (4x 3)% (3)(x 1)% †

%

$

$

d dx

(x 1) (x 1)$ (4)(4x 3)$ † %

œ (4x 3) (3)(x 1) (1) (x 1) (4)(4x 3) (4) œ 3(4x 3) (x 1) œ

$

(4x 3) (x 1)%

c3(4x 3) 16(x 1)d œ

%

d dx $

(4x 3)

16(4x 3) (x 1)$

$

(4x 3) (4x 7) (x 1)%


Section 3.6 The Chain Rule '

30. y œ (2x 5)" ax# 5xb Ê &

2 ax# 5xb (2x 5)#

œ 6 ax# 5xb

dy dx

&

'

œ (2x 5)" (6) ax# 5xb (2x 5) ax# 5xb (1)(2x 5)# (2)

'

d ˆ d 31. h(x) œ x tan ˆ2Èx‰ 7 Ê hw (x) œ x dx tan ˆ2x"Î# ‰‰ tan ˆ2x"Î# ‰ † dx (x) 0 d ˆ "Î# ‰ " # ˆ "Î# ‰ "Î# ‰ #ˆ È ‰ ˆ ˆ † dx 2x œ x sec 2x tan 2x œ x sec 2 x † È tan 2Èx‰ œ Èx sec# ˆ2Èx‰ tan ˆ2Èx‰ x

d ˆ d 32. k(x) œ x# sec ˆ "x ‰ Ê kw (x) œ x# dx sec x" ‰ sec ˆ x" ‰ † dx ax# b œ x# sec ˆ x" ‰ tan ˆ x" ‰ † œ x# sec ˆ "x ‰ tan ˆ x" ‰ † ˆ x"# ‰ 2x sec ˆ x" ‰ œ 2x sec ˆ x" ‰ sec ˆ x" ‰ tan ˆ x" ‰

33. faxb œ È7 x sec x Ê f w axb œ "# a7 x sec xb1Î2 ax † asec x tan xb asec xb † "b œ 34. gaxb œ œ

tan 3x ax 7 b 4

ax 7b4 ˆsec2 3x†$‰ atan 3xb4ax 7b3 †1 ax 7b4 ‘2

Ê gw axb œ

x sec x tan x sec x #È7x sec x

ax 7b3 ˆ$ax 7bsec2 3x 4tan 3x‰ ax 7 b 8

ˆ$ax 7bsec2 3x 4tan 3x‰ ax 7 b5 #

35. f()) œ ˆ 1 sincos) ) ‰ Ê f w ()) œ 2 ˆ 1 sincos) ) ‰ † œ

œ

ˆ x" ‰ 2x sec ˆ x" ‰

d dx

(2 sin )) acos ) cos# ) sin# )b (1 cos ))$

sin 3t ‰ 36. g(t) œ ˆ 1 3 2t

"

(2 sin )) (cos ) 1) (1 cos ))$

œ

3 2t 1 sin 3t

œ

37. r œ sin a)# b cos (2)) Ê

d d)

Ê gw (t) œ

ˆ 1 sincos) ) ‰ œ œ

†

(1 cos ))(cos )) (sin ))(sin )) (1 cos ))#

2 sin ) (1 cos ))#

a1 sin 3tba2b a3 2tba3 cos 3tb a1 sin 3tb2

œ sin a)# b (sin 2))

dr d)

2 sin ) 1 cos )

d d)

œ

2 2sin 3t 9 cos 3t 6t cos 3t a1 sin 3tb2

(2)) cos (2)) acos a)# bb †

d d)

a) # b

œ sin a)# b (sin 2))(2) (cos 2)) acos a)# bb (2)) œ 2 sin a)# b sin (#)) 2) cos (2)) cos a)# b 38. r œ Šsec È)‹ tan ˆ ") ‰ Ê

dr d)

œ )"# sec È) sec# ˆ ") ‰ 39. q œ sin Š Ètt 1 ‹ Ê œ cos Š Ètt 1 ‹ †

2

t1

dq dt

41. y œ sin# (1t 2) Ê

" #È )

tan ˆ ") ‰ sec È) tan È) œ Šsec È)‹ ”

œ cos Š Ètt 1 ‹ †

dq dt

Èt 1

40. q œ cot ˆ sint t ‰ Ê

œ Šsec È)‹ ˆ sec# ") ‰ ˆ )"# ‰ tan ˆ ") ‰ Šsec È) tan È)‹ Š

È

t t

1

d dt

Èt 1 (1)t †

d dt

sec# ˆ )" ‰ )# •

ˆÈ t 1 ‰

ˆÈ t 1 ‰

#

1) t œ cos Š Ètt 1 ‹ Š 2(t ‹ œ Š 2(tt1)2$Î# ‹ cos Š Ètt 1 ‹ 2(t 1)$Î#

œ csc# ˆ sint t ‰ †

d dt

ˆ sint t ‰ œ ˆcsc# ˆ sint t ‰‰ ˆ t cos tt# sin t ‰

œ 2 sin (1t 2) †

dy dt

Š Ètt 1 ‹ œ cos Š Ètt 1 ‹ †

tan È) tan ˆ ") ‰ #È )

" ‹ #È )

d dt

sin (1t 2) œ 2 sin (1t 2) † cos (1t 2) †

d dt

(1t 2)

œ 21 sin (1t 2) cos (1t 2) 42. y œ sec# 1t Ê

dy dt

œ (2 sec 1t) †

43. y œ (1 cos 2t)% Ê 44. y œ ˆ1 cot ˆ #t ‰‰ œ

csc# ˆ #t ‰ $ ˆ1 cot ˆ t ‰‰ #

#

dy dt

Ê

d dt

(sec 1t) œ (2 sec 1t)(sec 1t tan 1t) †

œ 4(1 cos 2t)& † dy dt

œ 2 ˆ1 cot ˆ #t ‰‰

d dt $

d dt

(1t) œ 21 sec# 1t tan 1t

(1 cos 2t) œ 4(1 cos 2t)& (sin 2t) † †

dˆ dt 1

cot ˆ #t ‰‰ œ 2 ˆ1 cot ˆ #t ‰‰

$

d dt

(2t) œ

8 sin 2t (1 cos 2t)&

† ˆcsc# ˆ #t ‰‰ †


dˆt‰ dt #

129

130


45. y œ at tan tb10 Ê 46. y œ ˆt3Î4 sin t‰ œ

4 Î3

œ t1 asin tb4Î3 Ê

dy dt

œ t1 ˆ 34 ‰asin tb1Î3 cost t2 asin tb4Î3 œ

4asin tb1Î3 cos t 3t

asin tb4Î3 t2

asin tb1Î3 a4t cos t 3cos tb 3t2 3

2

47. y œ Š t3 t 4t ‹ Ê 4‰ 48. y œ ˆ 3t 5t 2

œ

œ 10at tan tb9 at † sec2 t 1 † tan tb œ 10 t9 tan9 tat sec2 t tan tb œ 10 t10 tan9 t sec2 t 10 t9 tan10 t

dy dt

5

2

2

dy dt

œ 3Š t3 t 4t ‹ †

dy dt

4‰ œ 5ˆ 3t 5t 2

Ê

ˆt3 4t‰a2tb t2 ˆ3t2 4‰

6

at3 4tb2

†

a5t 2b†3 a3t 4b†5 a5t 2b2

œ

3t4 at3 4tb2

†

2t4 8t2 3t4 4t2 at3 4tb2 6

2‰ œ 5ˆ 5t † 3t 4

œ

15t 6 15t 20 a5t 2b2

3t4 ˆt4 4t2 ‰ t4 at2 4b4

3t2 ˆt2 4‰ at2 4b4

œ 6

5t 2b œ 5 aa3t † 4 b6

26 a5t 2b2

130a5t 2b4 a3t 4b6

49. y œ sin acos (2t 5)b Ê

dy dt

œ cos (cos (2t 5)) †

d dt

cos (2t 5) œ cos (cos (2t 5)) † (sin (2t 5)) †

d dt

(2t 5)

œ 2 cos (cos (2t 5))(sin (2t 5)) ˆ ˆ t ‰‰ † 50. y œ cos ˆ5 sin ˆ 3t ‰‰ Ê dy dt œ sin 5 sin 3 œ 53 sin ˆ5 sin ˆ 3t ‰‰ ˆcos ˆ 3t ‰‰

d dt

ˆ5 sin ˆ 3t ‰‰ œ sin ˆ5 sin ˆ 3t ‰‰ ˆ5 cos ˆ 3t ‰‰ †

$

dy % ˆ t ‰‘# d † dt 1 dt œ 3 1 tan 1# " ‘ % ˆ t ‰‘# $ˆ t ‰ #ˆ t ‰ tan 1# tan 1# sec 1# † 1# œ 1

51. y œ 1 tan% ˆ 1t# ‰‘ Ê œ 12 1 52. y œ

" 6

$

c1 cos# (7t)d Ê "Î#

Ê

œ "# a1 cos at# bb

"Î#

53. y œ a1 cos at# bb

dy dt

2 cos ŒÉ1 Èt É1 Èt†2Èt

# tan% ˆ 1t# ‰‘ tan$ ˆ 1t# ‰ sec# ˆ 1t# ‰‘

#

d dt

tan ˆ 1t# ‰‘

#

c1 cos# (7t)d † 2 cos (7t)(sin (7t))(7) œ 7 c1 cos# (7t)d (cos (7t) sin (7t))

3 6 " #

# tan% ˆ 1t# ‰‘ œ 3 1 tan% ˆ 1t# ‰‘ 4 tan$ ˆ 1t# ‰ †

a1 cos at# bb

"Î#

†

d dt

a1 cos at# bb œ

#

" #

a1 cos at# bb

"Î#

ˆsin at# b †

d dt

a t# b ‰

dy dt

œ 4 cos ŒÉ1 Èt †

d dt

"

ŒÉ1 Èt œ 4 cos ŒÉ1 Èt †

# É 1 È t

†

d dt

ˆ1 Èt‰

cos ŒÉ1 Èt

œ

55. y œ tan2 asin3 tb Ê

œ

ˆ 3t ‰

at b asin at# bb † 2t œ È1t sin cos at# b

54. y œ 4 sin ŒÉ1 Èt Ê œ

œ

dy dt

d dt

Ét tÈt

dy dt

56. y œ cos4 asec2 3tb Ê

œ 2 tanasin3 tb † sec2 asin3 tb † a3sin2 t † acos tbb œ 6 tanasin3 tbsec2 asin3 tbsin2 t cos t dy dt

œ 4 cos3 asec2 a3tbbasinasec2 a3tbb † 2 aseca3tbbaseca3tb tana3tb † 3bb

œ 24 cos3 asec2 a3tbbsinasec2 a3tbbsec2 a3tb tana3tb 4

57. y œ 3ta2t2 5b Ê

dy dt

3

58. y œ Ê3t É2 È1 t Ê œ

" #Ê3tÉ2È1t

3

4

3

3

œ 3t † 4a2t2 5b a4tb 3 † a2t2 5b œ 3a2t2 5b ’16t2 2t2 5“ œ 3a2t2 5b a18t2 5b

dy dt

" #É2È1t

1 Î2

œ "# Œ3t É2 È1 t †

1 #È 1 t

œ

" #Ê3tÉ2È1t

" Œ3 # Š2 È1 t‹

12È1tÉ2È1t " 4È1tÉ2È1t

œ

1 Î2

" # a1

tb1Î2 a1b

12È1tÉ2È1t " 8È1tÉ2È1tÊ3tÉ2È1t


Section 3.6 The Chain Rule $ # # 59. y œ ˆ1 "x ‰ Ê yw œ 3 ˆ1 x" ‰ ˆ x"# ‰ œ x3# ˆ1 x" ‰ Ê yww œ ˆ x3# ‰ † # œ ˆ x3# ‰ ˆ2 ˆ1 x" ‰ ˆ x"# ‰‰ ˆ x6$ ‰ ˆ1 x" ‰ œ œ x6$ ˆ1 x" ‰ ˆ1 2x ‰

60. y œ ˆ1 Èx‰ Ê yww œ œ

" #

œ

" #x

61. y œ

" #

"

Ê yw œ ˆ1 Èx‰

œ

ˆ1 x" ‰

ˆ "# x"Î# ‰ œ

" #

6 x$

ˆ1 x" ‰# œ

ˆ1 x" ‰# ˆ1 x" ‰# †

ˆ1 Èx‰$ Š

ˆ x3# ‰

ˆ1 x" ‰ ˆ x" 1 x" ‰

ˆ1 Èx‰# x"Î#

#

" #È x

$

x" ˆ1 Èx‰ “ œ

" #

1‹ œ

" #x

" #

x" ˆ1 Èx‰

ˆ1 Èx‰$ Š 3#

$

"# x"Î# ˆ1 Èx‰ 1‘

" ‹ #Èx

cot (3x 1) Ê yw œ 9" csc# (3x 1)(3) œ 3" csc# (3x 1) Ê yww œ ˆ 32 ‰ (csc (3x 1) † 2 3

d dx

# $ ’ˆ1 Èx‰ ˆ "# x$Î# ‰ x"Î# (2) ˆ1 Èx‰ ˆ "# x"Î# ‰“

$Î# ˆ 1 Èx‰ ’ " # x

" 9

#

6 x%

d dx 6 x$

131

csc (3x 1)(csc (3x 1) cot (3x 1) †

d dx

csc (3x 1))

d dx

#

(3x 1)) œ 2 csc (3x 1) cot (3x 1)

62. y œ 9 tan ˆ x3 ‰ Ê yw œ 9 ˆsec# ˆ x3 ‰‰ ˆ 3" ‰ œ 3 sec# ˆ x3 ‰ Ê yww œ 3 † 2 sec ˆ x3 ‰ ˆsec ˆ x3 ‰ tan ˆ x3 ‰‰ ˆ "3 ‰ œ 2 sec# ˆ 3x ‰ tan ˆ 3x ‰ 63. y œ xa2x 1b4 Ê yw œ x † 4a2x 1b3 a2b 1 † a2x 1b4 œ a2x 1b3 a8x a2x 1bb œ a2x 1b3 a10x 1b Ê yww œ a2x 1b3 a10b 3a2x 1b2 a2ba10x 1b œ 2a2x 1b2 a5a2x 1b 3a10x 1bb œ 2a2x 1b2 a40x 8b œ 16a2x 1b2 a5x 1b 5

4

5

4

4

64. y œ x2 ax3 1b Ê yw œ x2 † 5ax3 1b a3x2 b 2xax3 1b œ xax3 1b ’15x3 2ax3 1b“ œ ax3 1b a17x4 2xb 4

3

3

Ê yww œ ax3 1b a68x3 2b 4ax3 1b a3x2 ba17x4 2xb œ 2ax3 1b ’ax3 1ba34x3 1b 6x2 a17x4 2xb“ 3

œ 2ax3 1b a136x6 47x3 1b 65. g(x) œ Èx Ê gw (x) œ

" #È x

Ê g(1) œ 1 and gw (1) œ

therefore, (f ‰ g)w (1) œ f w (g(1)) † gw (1) œ 5 †

" #

œ

" u#

; f(u) œ u& 1 Ê f w (u) œ 5u% Ê f w (g(1)) œ f w (1) œ 5;

5 #

66. g(x) œ (1 x)" Ê gw (x) œ (1 x)# (1) œ Ê f w (u) œ

" #

" (1x)#

Ê g(1) œ

" # w

and gw (1) œ

" 4

; f(u) œ 1

Ê f w (g(1)) œ f w ˆ #" ‰ œ 4; therefore, (f ‰ g)w (1) œ f (g(1))gw (1) œ 4 †

67. g(x) œ 5Èx Ê gw (x) œ Ê f w (g(1)) œ f w (5) œ

5 Ê g(1) œ 5 and gw (1) œ #5 #È x 1 1 # ˆ1‰ 10 csc # œ 10 ; therefore,

" 4

" u

œ1

1‰ ; f(u) œ cot ˆ 110u ‰ Ê f w (u) œ csc# ˆ 110u ‰ ˆ 10 œ

1 10

csc# ˆ 110u ‰

1 (f ‰ g)w (1) œ f w (g(1))gw (1) œ 10 † 5# = 14

68. g(x) œ 1x Ê gw (x) œ 1 Ê g ˆ "4 ‰ œ 14 and gw ˆ 4" ‰ œ 1; f(u) œ u sec# u Ê f w (u) œ 1 2 sec u † sec u tan u œ 1 2 sec# u tan u Ê f w ˆg ˆ "4 ‰‰ œ f w ˆ 14 ‰ œ 1 2 sec# 14 tan 14 œ 5; therefore, (f ‰ g)w ˆ 4" ‰ œ f w ˆg ˆ 4" ‰‰ gw ˆ 4" ‰ œ 51 69. g(x) œ 10x# x 1 Ê gw (x) œ 20x 1 Ê g(0) œ 1 and gw (0) œ 1; f(u) œ œ

2u# 2 au # 1 b #

Ê f w (u) œ

au# 1b(2) (2u)(2u) au # 1 b #

Ê f w (g(0)) œ f w (1) œ 0; therefore, (f ‰ g)w (0) œ f w (g(0))gw (0) œ 0 † 1 œ 0

" 2 w x# 1 Ê g (x) œ x$ Ê g(1) œ 0 and 4(u 1) 1 ‰ (u 1)(1) (u 1)(1) 2 ˆ uu œ 2(u(u1)(2) 1 † (u 1)# 1)$ œ (u 1)$ w w w

70. g(x) œ œ

2u u # 1

#

1‰ 1‰ gw (1) œ 2; f(u) œ ˆ uu Ê f w (u) œ 2 ˆ uu 1 1

Ê f w (g(1)) œ f w (0) œ 4; therefore,

(f ‰ g) (1) œ f (g(1))g (1) œ (4)(2) œ 8


d du

1‰ ˆ uu 1

132


71. y œ fagaxbb, f w a3b œ 1, gw a2b œ 5, ga2b œ 3 Ê y w œ f w agaxbbg w axb Ê y w ¹

x œ2

œ a1b † 5 œ 5 72. r œ sinafatbb, fa0b œ 13 , f w a0b œ 4 Ê 73. (a) y œ 2f(x) Ê

dy dx

œ 2f w (x) Ê

(b) y œ f(x) g(x) Ê (c) y œ f(x) † g(x) Ê

dr dt

œ cosafatbb † f w atb Ê

œ f w (x) gw (x) Ê

dy dx

dy dx

œ

g(x)f w (x) f(x)gw (x) [g(x)]#

(e) y œ f(g(x)) Ê

dy dx

œ f w (g(x))gw (x) Ê

(d) y œ

f(x) g(x)

Ê

" #

Ê

dy dx ¹ x=2

dy dx

œ

(g) y œ (g(x))# Ê

dy dx

œ 2(g(x))$ † gw (x) Ê

Ê

dy dx ¹ x=2

œ

" #

(f(x))"Î# † f w (x) œ

"Î#

Ê

" # # "Î#

dy dx

a(f(2))# (g(2)) b

74. (a) y œ 5f(x) g(x) Ê (b) y œ f(x)(g(x))$ Ê

f (x) #Èf(x)

œ

dy dx ¹ x=3

dy dx ¹ x=2

œ

" 3

f (2) #Èf(2)

dy dx ¹ x=1

ˆ "3 ‰ #È 8

œ

œ

37 6

(3) œ 1 œ

" 6È 8

" 1 #È 2

œ

œ 2(g(3))$ gw (3) œ 2(4)$ † 5 œ "Î#

œ

È2 24

5 3#

a2f(x) † f w (x) 2g(x) † gw (x)b " #

a8# 2# b

"Î#

ˆ2 † 8 †

" 3

2 † 2 † (3)‰ œ 3È517

œ 5f w (1) gw (1) œ 5 ˆ 3" ‰ ˆ 38 ‰ œ 1

œ f(x) a3(g(x))# gw (x)b (g(x))$ f w (x) Ê

dy dx

(2) ˆ "3 ‰ (8)(3) ##

w

a2f(2)f w (2) 2g(2)gw (2)b œ

œ 5f w (x) gw (x) Ê

dy dx

Ê

a(f(x))# (g(x))# b

œ

œ f(3)gw (3) g(3)f w (3) œ 3 † 5 (4)(21) œ 15 81

œ f w (g(2))gw (2) œ f w (2)(3) œ w

(f) y œ (f(x))"Î# Ê

(h) y œ a(f(x))# (g(x))# b

dy dx ¹ x=3

g(2)f w (2) f(2)gw (2) [g(2)]#

œ

dy dx ¹ x=2

2 3

œ f w (3) gw (3) œ 21 5

dy dx ¹ x=3

œ f(x)gw (x) g(x)f w (x) Ê

dy dx

œ cosafa0bb † f w a0b œ cosˆ 13 ‰ † 4 œ ˆ "# ‰ † 4 œ 2

dr dt ¹tœ0

œ 2f w (2) œ 2 ˆ "3 ‰ œ

dy dx ¹ x=2

œ f w aga2bbgw a2b œ f w a3b † 5

dy dx ¹ x = 0

œ $f(0)(g(0))# gw (0) (g(0))$ f w (0)

œ 3(1)(1)# ˆ 3" ‰ (1)$ (5) œ 6 (c) y œ œ

f(x) g(x) 1

Ê

(g(x) 1)f w (x) f(x) gw (x) (g(x) 1)#

œ

dy dx

(4") ˆ "3 ‰(3) ˆ 83 ‰ (41)#

Ê

dy dx ¹ x = 1

(g(1) 1)f w (1) f(1)gw (1) (g(1) 1)#

œ

œ1

(d) y œ f(g(x)) Ê

dy dx

œ f w (g(x))gw (x) Ê

dy dx ¹ x = 0

œ f w (g(0))gw (0) œ f w (1) ˆ "3 ‰ œ ˆ "3 ‰ ˆ 3" ‰ œ 9"

(e) y œ g(f(x)) Ê

dy dx

œ gw (f(x))f w (x) Ê

dy dx ¹ x = 0

œ gw (f(0))f w (0) œ gw (1)(5) œ ˆ 83 ‰ (5) œ 40 3

(f) y œ ax"" f(x)b

#

Ê

œ 2 ax"" f(x)b

dy dx

$

a11x"! f w (x)b Ê

dy dx ¹ x=1

œ 2(1 f(1))$ a11 f w (1)b

" ‰ œ 2(1 3)$ ˆ11 "3 ‰ œ ˆ 42$ ‰ ˆ 32 3 œ 3

(g) y œ f(x g(x)) Ê

dy dx

œ f w (x g(x)) a1 gw (x)b Ê

dy dx ¹ x = 0

œ f w (0 g(0)) a1 gw (0)b œ f w (1) ˆ1 "3 ‰

œ ˆ "3 ‰ ˆ 43 ‰ œ 49 75.

ds dt

œ

ds d)

†

d) dt :

s œ cos ) Ê

76.

dy dt

œ

dy dx

†

dx dt :

y œ x# 7x 5 Ê

77. With y œ x, we should get (a) y œ (b)

u 5

7 Ê

y œ 1 "u Ê " œ " u# † (x 1)#

dy du dy du

œ

œ

" 5

œ sin ) Ê

ds d)

dy dx

dy dx

œ 2x 7 Ê

œ

dy dx ¹ x = 1

œ 9 so that

dy dt

œ

ds dt

dy dx

œ †

dx dt

ds d)

†

d) dt

œ9†

œ 1†5œ 5 " 3

œ3

œ 1 for both (a) and (b):

; u œ 5x 35 Ê " u#

œ sin ˆ 3#1 ‰ œ 1 so that

ds ¸ d) ) = 321

; u œ (x 1)

" a(x 1) " b#

†

" (x 1)#

"

du dx

œ 5; therefore,

Ê

du dx

œ (x 1)# †

dy dy dx œ du #

†

œ (x 1) (1) œ

" (x 1)#

du " dx œ 5 " (x 1)# ;

† 5 œ 1, as expected therefore

dy dx

œ

œ 1, again as expected


dy du

†

du dx

Section 3.6 The Chain Rule 78. With y œ x$Î# , we should get (a) y œ u$ Ê

dy du

as expected. (b) y œ Èu Ê

œ

dy dx

3 #

x"Î# for both (a) and (b):

œ 3u# ; u œ Èx Ê

dy du

œ

" #È u

du dx

; u œ x$ Ê

du dx

" #È x

; therefore,

dy dx

œ

dy du

†

du dx

œ 3u# †

œ 3x# ; therefore,

dy dx

œ

dy du

†

du dx

œ

œ

" #Èu

" #È x

# œ 3 ˆÈx‰ †

† 3x# œ

" #È x $

" #Èx

† 3x# œ

3 #

œ

3 #

Èx,

x"Î# ,

again as expected. 2

2

1‰ 1‰ 1‰ 79. y œ ˆ xx and x œ 0 Ê y œ ˆ 00 œ a1b2 œ 1. yw œ 2ˆ xx 1 1 1 †

yw ¹

xœ0

œ

4 a0 1 b a0 1 b 3

œ

4 13

ax 1b†1 ax 1b†1 ax 1 b 2

1b 2 œ 2 aaxx 1 b ax 1 b 2 œ

œ 4 Ê y 1 œ 4ax 0b Ê y œ 4x 1

80. y œ Èx2 x 7 and x œ 2 Ê y œ Éa2b2 a2b 7 œ È9 œ 3. y w œ "# ax2 x 7b y w¹

xœ2

œ

2 a2 b 1 2 É a2 b 2 a 2 b 7

81. y œ 2 tan ˆ 14x ‰ Ê (a)

dy dx ¹ x = 1

œ

1 #

dy dx

4 ax 1 b ax 1 b 3

œ

3 6

œ

" #

œ ˆ2 sec#

1 Î2

a2x 1b œ

2x 1 2 È x 2 x 7

Ê y 3 œ "# ax 2b Ê y œ "# x 2 1x ‰ ˆ 1 ‰ 4 4

œ

1 #

sec#

1x 4

sec# ˆ 14 ‰ œ 1 Ê slope of tangent is 2; thus, y(1) œ 2 tan ˆ 14 ‰ œ 2 and yw (1) œ 1 Ê tangent line is

given by y 2 œ 1(x 1) Ê y œ 1x 2 1 (b) yw œ 1# sec# ˆ 14x ‰ and the smallest value the secant function can have in # x 2 is 1 Ê the minimum value of yw is 1# and that occurs when 1# œ 1# sec# ˆ 14x ‰ Ê 1 œ sec# ˆ 14x ‰ Ê „ 1 œ sec ˆ 14x ‰ Ê x œ 0. 82. (a) y œ sin 2x Ê yw œ 2 cos 2x Ê yw (0) œ 2 cos (0) œ 2 Ê tangent to y œ sin 2x at the origin is y œ 2x; y œ sin ˆ x# ‰ Ê yw œ "# cos ˆ x# ‰ Ê yw (0) œ "# cos 0 œ "# Ê tangent to y œ sin ˆ x# ‰ at the origin is y œ "# x. The tangents are perpendicular to each other at the origin since the product of their slopes is 1. (b) y œ sin (mx) Ê yw œ m cos (mx) Ê yw (0) œ m cos 0 œ m; y œ sin ˆ mx ‰ Ê yw œ m" cos ˆ mx ‰ Ê yw (0) œ m" cos (0) œ m" . Since m † ˆ m" ‰ œ 1, the tangent lines are perpendicular at the origin.

(c) y œ sin (mx) Ê yw œ m cos (mx). The largest value cos (mx) can attain is 1 at x œ 0 Ê the largest value yw can attain is kmk because kyw k œ km cos (mx)k œ kmk kcos mxk Ÿ kmk † 1 œ kmk . Also, y œ sin ˆ mx ‰ ˆ x ‰¸ Ÿ ¸ m" ¸ ¸cos ˆ mx ‰¸ Ÿ km" k Ê the largest value yw can attain is ¸ m" ¸ . Ê yw œ m" cos ˆ mx ‰ Ê kyw k œ ¸ " m cos m (d) y œ sin (mx) Ê yw œ m cos (mx) Ê yw (0) œ m Ê slope of curve at the origin is m. Also, sin (mx) completes m periods on [0ß 21]. Therefore the slope of the curve y œ sin (mx) at the origin is the same as the number of periods it completes on [0ß 21]. In particular, for large m, we can think of “compressing" the graph of y œ sin x horizontally which gives more periods completed on [0ß 21], but also increases the slope of the graph at the origin. 83. s œ A cos (21bt) Ê v œ

ds dt

œ A sin (21bt)(21b) œ 21bA sin (21bt). If we replace b with 2b to double the

frequency, the velocity formula gives v œ 41bA sin (41bt) Ê doubling the frequency causes the velocity to # # double. Also v œ #1bA sin (21bt) Ê a œ dv dt œ 41 b A cos (21bt). If we replace b with 2b in the

acceleration formula, we get a œ 161# b# A cos (41bt) Ê doubling the frequency causes the acceleration to $ $ quadruple. Finally, a œ 41# b# A cos (21bt) Ê j œ da dt œ 81 b A sin (21bt). If we replace b with 2b in the jerk formula, we get j œ 641$ b$ A sin (41bt) Ê doubling the frequency multiplies the jerk by a factor of 8. 21 21 21 ‰ 84. (a) y œ 37 sin 365 (x 101)‘ 25 Ê yw œ 37 cos 365 (x 101)‘ ˆ 365 œ

741 365

21 cos 365 (x 101)‘ .

The temperature is increasing the fastest when yw is as large as possible. The largest value of 21 21 cos 365 (x 101)‘ is 1 and occurs when 365 (x 101) œ 0 Ê x œ 101 Ê on day 101 of the year

( µ April 11), the temperature is increasing the fastest.


133

134


(b) yw (101) œ

741 365

21 cos 365 (101 101)‘ œ

85. s œ (" 4t)"Î# Ê v œ v œ 2(" 4t)

"Î#

ds dt

œ

Ê aœ

dv dt

86. We need to show a œ œ

k 2È s

† kÈs œ

87. v proportional to œ 2sk$Î# † dx dt

88. Let

k Ès

kT 2

k #

cos (0) œ

741 365

¸ 0.64 °F/day

(1 4t)"Î# (4) œ 2(1 4t)"Î# Ê v(6) œ 2(" % † 6)"Î# œ " #

œ † 2(1 4t)

is constant: a œ

dv dt

œ

$Î#

dv ds

(4) œ 4(1 4t)

†

ds dt

and

dv ds

œ

" Ès

Ê vœ

k Ès

for some constant k Ê

dv ds

m/sec;

Ê a(6) œ 4(1 4 † 6)$Î# œ 14#5 m/sec#

ˆkÈs‰ œ

k 2È s

œ 2sk$Î# . Thus, a œ

#

œ f(x). Then, a œ dT dL

d ds

$Î#

2 5

Ê aœ

dv ds

†

†

ds dt

ds dt

œ

dv ds

†v

which is a constant.

œ k# ˆ s"# ‰ Ê acceleration is a constant times

89. T œ 21É Lg Ê œ

#

dv dt

" #

741 365

dv dt

œ 21 †

œ

" #É Lg

dv dx

†

dx dt

œ

†

" g

œ

1 gÉ Lg

dv dx

† f(x) œ œ

1 ÈgL

d dx

" s#

dT du

œ

dv ds

œ

dv ds

†v

so a is inversely proportional to s# .

ˆ dx ‰ dt † f(x) œ

. Therefore,

dv dt

œ

d dx

(f(x)) † f(x) œ f w (x)f(x), as required.

dT dL

†

dL du

œ

1 ÈgL

† kL œ

1 kÈ L Èg

œ

" #

† 21kÉ Lg

, as required.

90. No. The chain rule says that when g is differentiable at 0 and f is differentiable at g(0), then f ‰ g is differentiable at 0. But the chain rule says nothing about what happens when g is not differentiable at 0 so there is no contradiction. 91. As h Ä 0, the graph of y œ

sin 2(xh)sin 2x h

approaches the graph of y œ 2 cos 2x because lim

hÄ!

sin 2(xh)sin 2x h

œ

d dx

(sin 2x) œ 2 cos 2x.

92. As h Ä 0, the graph of y œ

cos c(x h)# dcos ax# b h #

approaches the graph of y œ 2x sin ax b because cos c(x h)# dcos ax# b h hÄ!

lim

œ

d dx

ccos ax# bd œ 2x sin ax# b.


Section 3.7 Implicit Differentiation 93. (a)

(b)

df dt

œ 1.27324 sin 2t 0.42444 sin 6t 0.2546 sin 10t 0.18186 sin 14t

(c) The curve of y œ

approximates y œ

df dt

dg dt

the best when t is not 1, 1# , 0, 1# , nor 1.

94. (a)

(b)

dh dt

œ 2.5464 cos (2t) 2.5464 cos (6t) 2.5465 cos (10t) 2.54646 cos (14t) 2.54646 cos (18t)

(c)

3.7 IMPLICIT DIFFERENTIATION 1. x# y xy# œ 6: Step 1:

Šx#

Step 2:

x#

dy dx

Step 3:

dy dx dy dx

ax# 2xyb œ 2xy y#

Step 4:

dy dx

œ

y † 2x‹ Šx † 2y

2xy

dy dx

dy dx

y# † 1‹ œ 0

œ 2xy y#

2xy y# x# 2xy

2. x$ y$ œ 18xy Ê 3x# 3y#

dy dx

œ 18y 18x

dy dx

Ê a3y# 18xb

dy dx

œ 18y 3x# Ê

dy dx

œ

6y x# y# 6x

3. 2xy y# œ x y: Step 1:

Š2x

Step 2:

2x

dy dx

dy dx

2y‹ 2y

2y

dy dx

dy dx

dy dx

œ1

dy dx

œ 1 2y


135

136

Chapter 3 Differentiation dy dx dy dx

Step 3: Step 4:

(2x 2y 1) œ " 2y œ

1 2y 2x 2y 1

4. x$ xy y$ œ 1 Ê 3x# y x

dy dx

3y#

œ 0 Ê a3y# xb

dy dx

dy dx

œ y 3x# Ê

dy dx

y 3x# 3y# x

œ

5. x# (x y)# œ x# y# : Step 1:

x# ’2(x y) Š1

Step 2:

2x# (x y)

Step 3:

dy dx dy dx

Step 4:

dy dx

dy dx ‹“

2y

(x y)# (2x) œ 2x 2y

dy dx

œ 2x 2x# (x y) 2x(x y)#

c2x# (x y) 2yd œ 2x c1 x(x y) (x y)# d œ

œ

2x c1 x(x y) (x y)# d 2x# (x y) 2y

œ

dy dx

x c1 x(x y) (x y)# d y x# (x y)

œ

x a1 x# xy x# 2xy y# b x# y x$ y

x 2x$ 3x# y xy# x# y x$ y

6. (3xy 7)# œ 6y Ê 2(3xy 7) † Š3x Ê

dy dx

dy dx

3y‹ œ 6

[6x(3xy 7) 6] œ 6y(3xy 7) Ê (x 1) (x 1) (x 1)#

dy dx

dy dx

Ê 2(3xy 7)(3x)

y(3xy 7) œ x(3xy 7) 1 œ

dy dx

6

dy dx

œ 6y(3xy 7)

#

3xy 7y 1 3x# y 7x

7. y# œ

x" x1

Ê 2y

8. x3 œ

2x y x 3y

Ê x4 3x3 y œ 2x y Ê 4x3 9x2 y 3x3 y w œ 2 y w Ê a3x3 1by w œ 2 4x3 9x2 y

Ê yw œ

œ

dy dx

œ

Ê

2 (x 1)#

dy dx

œ

" y(x 1)#

2 4x3 9x2 y 3x3 1

9. x œ tan y Ê 1 œ asec# yb

dy dx

Ê

dy dx

œ

" sec# y

œ cos# y

dy dy dy # # # 10. xy œ cot axyb Ê x dy dx y œ csc (xy)Šx dx y‹ Ê x dx x csc (xy) dx œ y csc (xy) y

Ê

dy dx x

x csc# (xy)‘ œ y csc# (xy) "‘ Ê

11. x tan (xy) œ ! Ê 1 csec# (xy)d Šy x œ

1 x sec# (xy)

y x

œ

cos# (xy) x

y x

œ

dy dx ‹

œ

y csc# (xy) "‘ x" csc# (xy)‘

œ yx

œ 0 Ê x sec# (xy)

dy dx

œ 1 y sec# (xy) Ê

dy dx

œ

" y sec# (xy) x sec# (xy)

cos# (xy) y x

12. x4 sin y œ x3 y2 Ê 4x3 (cos y)

dy dx

œ 3x2 y2 x3 † 2y

13. y sin Š "y ‹ œ 1 xy Ê y ’cos Š y" ‹ † (1) dy dx

dy dx

’ "y cos Š "y ‹ sin Š y" ‹ x“ œ y Ê

" y# dy dx

†

dy dx “

œ

dy dx

Ê acos y 2x3 yb

sin Š y" ‹ †

dy dx

œ x

y "y cos Š "y ‹ sin Š "y ‹ x

œ

dy dx

dy dx

œ 3x2 y2 4x3 Ê

dy dx

œ

3x2 y2 4x3 cos y 2x3 y

y Ê y #

y sin Š "y ‹ cos Š "y ‹ xy

14. x cosa2x 3yb œ y sin x Ê x sina2x 3yba2 3y w b cosa2x 3yb œ y cos x y w sin x Ê 2x sina2x 3yb 3x y w sina2x 3yb cosa2x 3yb œ y cos x y w sin x Ê cosa2x 3yb 2x sina2x 3yb y cos x œ asin x 3x sina2x 3ybby w Ê y w œ 15. )"Î# r"Î# œ 1 Ê 16. r 2È) œ

3 #

" #

)"Î# "# r"Î# †

)#Î$ 34 )$Î% Ê

dr d)

dr d)

œ0 Ê

dr d)

" ’ #È “œ r

)"Î# œ )"Î$ )"Î% Ê

" #È )

dr d)

Ê

dr d)

œ

2È r 2È )

cosa2x 3yb 2x sina2x 3yb y cos x sin x 3x sina2x 3yb Èr

œÈ

)

œ )"Î# )"Î$ )"Î%


Section 3.7 Implicit Differentiation 17. sin (r )) œ

" #

Ê [cos (r ))] ˆr )

18. cos r cot ) œ r ) Ê (sin r)

dr d)

œ0Ê

dr ‰ d)

dr d)

[) cos (r ))] œ r cos (r )) Ê

csc# ) œ r )

Ê

dr d)

19. x# y# œ 1 Ê 2x 2yyw œ 0 Ê 2yyw œ 2x Ê y(1) xyw y#

Ê yww œ

œ

20. x#Î$ y#Î$ œ 1 Ê

2 3

y x Š xy ‹ y#

since yw œ xy Ê

x"Î$ 23 y"Î$

œ0 Ê

dy dx

dy dx

d# y dx#

œ

" 3

2x 2 2y

21. y# œ x# 2x Ê 2yyw œ 2x 2 Ê yw œ

d# y dx#

œ yww œ

d# y dx#

Ê

y# (x 1)# y$

ww

œy œ

y # x # y$

d# y dx#

d dx

,

y# a" y# b y$

œ

dr d)

#

csc ) œ rsin r)

ayw b œ

œ

dy dx

œ )r , cos (r )) Á 0

d dx

Š xy ‹

" y$

œ yx

"Î$ "Î$

"Î$

œ ˆ yx ‰

y (x 1)yw y#

; then yww œ

" y1

œ

y (x 1) Š x y 1 ‹ y#

œ (y 1)" ; then yww œ (y 1)# † yw

" (y 1)$

œ yww œ

23. 2Èy œ x y Ê y"Î# yw œ 1 yw Ê yw ˆy"Î# 1‰ œ 1 Ê

dy dx

œ yw œ

" y "Î# 1

Èy Èy 1

œ

; we can

" #

differentiate the equation yw ˆy"Î# 1‰ œ 1 again to find yww : yw ˆ y$Î# yw ‰ ˆy"Î# 1‰ yww œ 0 Ê ˆy"Î# 1‰ yww œ

" #

w # $Î#

cy d y

Ê

d# y dx#

" #

œ yww œ

#

Œy

" $Î# "Î# 1 y

ay "Î# 1b

" $ 2y$Î# ay "Î# 1b

œ

œ

" $ # ˆ1 È y ‰

24. xy y# œ 1 Ê xyw y 2yyw œ 0 Ê xyw 2yyw œ y Ê yw (x 2y) œ y Ê yw œ œ

(x 2y)yw y(1 2yw ) (x 2y)#

œ

2y(x 2y) 2y# (x 2y)$

œ

y

œ

y

(x 2y) ’ (x 2y) “ y ’1 2 Š (x 2y) ‹“ (x 2y)#

2y# 2xy (x 2y)$

œ

œ

"

(x 2y)

y (x2y)

;

d# y dx#

œ yww

cy(x 2y) y(x 2y) 2y# d (x 2y)#

2y(x y) (x 2y)$ #

25. x$ y$ œ 16 Ê 3x# 3y# yw œ 0 Ê 3y# yw œ 3x# Ê yw œ xy# ; we differentiate y# yw œ x# to find yww : # ww

w

w

w #

# ww

y y y c2y † y d œ 2x Ê y y œ 2x 2y cy d œ

2xy$ 2x% y&

Ê

d# y dx# ¹ (2ß2)

œ

32 32 32

ww

Ê y œ

2x 2y Š y#

26. xy y# œ 1 Ê xyw y 2yyw œ 0 Ê yw (x 2y) œ y Ê yw œ since yw k (0

ß

1)

œ "# we obtain yww k (0ß

1)

œ

27. y# x# œ y% 2x at (#ß ") and (#ß 1) Ê 2y Ê

dy dx

a2y 4y$ b œ 2 2x Ê

#

x# ‹ y#

œ

2x y#

2x% y$

œ 2

(2) ˆ "# ‰ (1)(0) 4

dy dx

œ

x" #y $ y

Ê

;

"Î$ x"Î$ †ˆ "3 y #Î$ ‰ Œ y"Î$ y"Î$ ˆ "3 x #Î$ ‰ x x#Î$

22. y# 2x œ 1 2y Ê 2y † yw 2 œ 2yw Ê yw (2y 2) œ 2 Ê yw œ œ (y 1)# (y 1)" Ê

d# y dx#

23 y"Î$ ‘ œ 23 x"Î$ Ê yw œ

x1 y

œ

[sin r )] œ r csc# ) Ê

œ yw œ xy ; now to find

x"Î$ †ˆ 3" y #Î$ ‰ yw y"Î$ ˆ "3 x #Î$ ‰ œ x#Î$ "Î$ y " "Î$ %Î$ " x œ 3x %Î$ 3y"Î$ x#Î$ 3 y

x#Î$ y"Î$

r cos (r )) ) cos (r ))

œ

dy dx

Differentiating again, yww œ Ê

dr d)

dr d)

y (x2y)

Ê yww œ

(x 2y) ayw b (y) a1 2yw b (x 2y)#

œ 4"

dy dx 2x dy dx ¹ ( 2ß1)

œ 4y$

dy dx

2 Ê 2y

œ 1 and

dy dx ¹ ( 2ß 1)

dy dx

4y$

dy dx

œ 2 2x

œ1


;

137

138

Chapter 3 Differentiation #

28. ax# y# b œ (x y)# at ("ß !) and ("ß 1) Ê 2 ax# y# b Š2x 2y Ê and

dy dx

c2y ax# y# b (x y)d œ 2x ax# y# b (x y) Ê

dy dx ¹ (1ßc1)

dy dx

dy dx ‹

œ

œ 2(x y) Š1

2x ax# y# b (x y) 2y ax# y# b (x y)

dy dx ‹

Ê

dy dx ¹ (1ß0)

œ 1

œ1

29. x# xy y# œ 1 Ê 2x y xyw 2yyw œ 0 Ê (x 2y)yw œ 2x y Ê yw œ (a) the slope of the tangent line m œ yw k (2 3) œ ß

Ê the tangent line is y 3 œ

7 4

(b) the normal line is y 3 œ 47 (x 2) Ê y œ 47 x

7 4

2x y 2y x

;

(x 2) Ê y œ

7 4

x

" #

29 7

30. x# y# œ 25 Ê 2x 2yyw œ 0 Ê yw œ xy ; (a) the slope of the tangent line m œ yw k (3

4)

ß

œ xy ¹

(3ß 4)

œ


3 4

3 4

(x 3) Ê y œ

3 4

x

25 4

(b) the normal line is y 4 œ 43 (x 3) Ê y œ 43 x 31. x# y# œ 9 Ê 2xy# 2x# yyw œ 0 Ê x# yyw œ xy# Ê yw œ yx ; (a) the slope of the tangent line m œ yw k ( 1ß3) œ yx ¸ ( 1ß3) œ 3 Ê the tangent line is y 3 œ 3(x 1) Ê y œ 3x 6 (b) the normal line is y 3 œ "3 (x 1) Ê y œ 3" x

8 3

32. y# 2x 4y " œ ! Ê 2yyw 2 4yw œ 0 Ê 2(y 2)yw œ 2 Ê yw œ

" y#

;

(a) the slope of the tangent line m œ yw k ( 2ß1) œ 1 Ê the tangent line is y 1 œ 1(x 2) Ê y œ x 1 (b) the normal line is y 1 œ 1(x 2) Ê y œ x 3 33. 6x# 3xy 2y# 17y 6 œ 0 Ê 12x 3y 3xyw 4yyw 17yw œ 0 Ê yw (3x 4y 17) œ 12x 3y 12x 3y Ê yw œ 3x 4y 17 ; (a) the slope of the tangent line m œ yw k ( 1ß0) œ Ê yœ

6 7

x

"2x 3y 3x 4y 17 ¹ ( 1ß0)

œ

6 7


6 7

(x 1)

6 7

(b) the normal line is y 0 œ 76 (x 1) Ê y œ 76 x

7 6

34. x# È3xy 2y# œ 5 Ê 2x È3xyw È3y 4yyw œ 0 Ê yw Š4y È3x‹ œ È3y 2x Ê yw œ (a) the slope of the tangent line m œ yw k ŠÈ3 2‹ œ ß

È3y 2x ¹ 4y È3x ŠÈ3ß2‹

œ 0 Ê the tangent line is y œ 2

(b) the normal line is x œ È3 35. 2xy 1 sin y œ 21 Ê 2xyw 2y 1(cos y)yw œ 0 Ê yw (2x 1 cos y) œ 2y Ê yw œ (a) the slope of the tangent line m œ yw k ˆ1 12 ‰ œ ß

2y 2x 1 cos y ¹ ˆ1ß 1 ‰ 2

y

1 #

œ 1# (x 1) Ê y œ 1# x 1

(b) the normal line is y

1 #

œ

2 1

(x 1) Ê y œ

2 1

x

2 1

2y 2x 1 cos y

œ 1# Ê the tangent line is 1 #

36. x sin 2y œ y cos 2x Ê x(cos 2y)2yw sin 2y œ 2y sin 2x yw cos 2x Ê yw (2x cos 2y cos 2x) œ sin 2y 2y sin 2x Ê yw œ

sin 2y 2y sin 2x cos 2x 2x cos 2y

(a) the slope of the tangent line m œ yw k ˆ 14

ß

1‰ 2

œ

; sin 2y 2y sin 2x cos 2x 2x cos 2y ¹ ˆ 1 ß 1 ‰ 4 2

y

1 #

œ 2 ˆx 14 ‰ Ê y œ 2x

(b) the normal line is y

1 #

œ "# ˆx 14 ‰ Ê y œ "# x

œ

1 1 #

œ 2 Ê the tangent line is

51 8


;

È3y 2x 4y È3x

;

Section 3.7 Implicit Differentiation 37. y œ 2 sin (1x y) Ê yw œ 2 [cos (1x y)] † a1 yw b Ê yw [1 2 cos (1x y)] œ 21 cos (1x y) Ê yw œ (a) the slope of the tangent line m œ yw k (1 0) œ ß

21 cos (1x y) 1 2 cos (1x y) ¹(1ß0)

y 0 œ 21(x 1) Ê y œ 21x 21 (b) the normal line is y 0 œ #"1 (x 1) Ê y œ 2x1

139

21 cos (1x y) 1 # cos (1x y)

;

œ 21 Ê the tangent line is

" #1

38. x# cos# y sin y œ 0 Ê x# (2 cos y)(sin y)yw 2x cos# y yw cos y œ 0 Ê yw c2x# cos y sin y cos yd 2x cos# y 2x# cos y sin y cos y

œ 2x cos# y Ê yw œ

;

(a) the slope of the tangent line m œ yw k (0 1) œ ß

2x cos# y 2x# cos y sin y cos y ¹ (0ß1)

œ 0 Ê the tangent line is y œ 1

(b) the normal line is x œ 0 39. Solving x# xy y# œ 7 and y œ 0 Ê x# œ 7 Ê x œ „ È7 Ê ŠÈ7ß !‹ and ŠÈ7ß !‹ are the points where the curve crosses the x-axis. Now x# xy y# œ 7 Ê 2x y xyw 2yyw œ 0 Ê (x 2y)yw œ 2x y 2 È 7 y 2x y È È Ê yw œ 2x x 2y Ê m œ x 2y Ê the slope at Š 7ß !‹ is m œ È7 œ 2 and the slope at Š 7ß !‹ is È

m œ 2È77 œ 2. Since the slope is 2 in each case, the corresponding tangents must be parallel. 40. xy 2x y œ 0 Ê x

dy dx

y2

dy dx

œ0 Ê

dy dx

œ

y2 1x

; the slope of the line 2x y œ 0 is 2. In order to be

parallel, the normal lines must also have slope of 2. Since a normal is perpendicular to a tangent, the slope of 2 " the tangent is "# . Therefore, y1 x œ # Ê 2y 4 œ 1 x Ê x œ 3 2y. Substituting in the original equation,

y(3 2y) 2(3 2y) y œ 0 Ê y# 4y 3 œ 0 Ê y œ 3 or y œ 1. If y œ 3, then x œ 3 and y 3 œ 2(x 3) Ê y œ 2x 3. If y œ 1, then x œ 1 and y 1 œ 2(x 1) Ê y œ 2x 3.

41. y% œ y# x# Ê 4y$ yw œ 2yyw 2x Ê 2 a2y$ yb yw œ 2x Ê yw œ y x2y$ ; the slope of the tangent line at È3 " È È È Š 43 ß #3 ‹ is y x2y$ ¹ È3 È3 œ È3 4 6È3 œ " 4 3 œ # " 3 œ 1; the slope of the tangent line at Š 43 ß #" ‹ # 4 Œ

is

x y2y$ ¹

Œ

È3 4

ß

œ

1 2

È3 " #

4

28

4

ß

#

2

œ

2È 3 42

8

œ È3

42. y# (2 x) œ x$ Ê 2yyw (2 x) y# (1) œ 3x# Ê yw œ œ

4 #

y# 3x# 2y(2 x)

; the slope of the tangent line is m œ

œ 2 Ê the tangent line is y 1 œ 2(x 1) Ê y œ 2x 1; the normal line is y 1 œ "# (x 1) Ê y œ "# x

43. y% 4y# œ x% 9x# Ê 4y$ yw 8yyw œ 4x$ 18x Ê yw a4y$ 8yb œ 4x$ 18x Ê yw œ œ

y# 3x# 2y(2 x) ¹ (1ß1)

x a2x# 9b y a2y# 4b

œ m; (3ß 2): m œ

(3)(18 9) 2(8 4)

œ 27 8 ; ($ß #): m œ

27 8

; (3ß #): m œ

27 8

ß

5 4

(b) yw œ 0 Ê

and yw k (2 4) œ ß

#

3y x y# 3x

4 5

œ

2x$ 9x 2y$ 4y

; (3ß #): m œ 27 8

44. x$ y$ 9xy œ 0 Ê 3x# 3y# yw 9xyw 9y œ 0 Ê yw a3y# 9xb œ 9y 3x# Ê yw œ (a) yw k (4 2) œ

4x$ 18x 4y$ 8y

3 #

9y 3x# 3y# 9x

œ

3y x# y# 3x

;

œ 0 Ê 3y x# œ 0 Ê y œ

x# 3

#

$

#

Ê x$ Š x3 ‹ 9x Š x3 ‹ œ 0 Ê x' 54x$ œ 0

Ê x$ ax$ 54b œ 0 Ê x œ 0 or x œ $È54 œ 3 $È2 Ê there is a horizontal tangent at x œ 3 $È2 . To find the corresponding y-value, we will use part (c). (c)

dx dy

œ0 Ê

y# 3x 3y x#

$

œ 0 Ê y# 3x œ 0 Ê y œ „ È3x ; y œ È3x Ê x$ ŠÈ3x‹ 9xÈ3x œ 0

3 3 Ê x$ 6È3 x$Î# œ 0 Ê x$Î# Šx$Î# 6È3‹ œ 0 Ê x$Î# œ 0 or x$Î# œ 6È3 Ê x œ 0 or x œ È 108 œ 3 È 4.

Since the equation x$ y$ 9xy œ 0 is symmetric in x and y, the graph is symmetric about the line y œ x. That is, if


140

Chapter 3 Differentiation (aß b) is a point on the folium, then so is (bß a). Moreover, if yw k (a b) œ m, then yw k (b a) œ ß

ß

" m

. Thus, if the folium has a

horizontal tangent at (aß b), it has a vertical tangent at (bß a) so one might expect that with a horizontal tangent at 3 3 3 3 3 xœÈ 54 and a vertical tangent at x œ 3 $È4, the points of tangency are ŠÈ 54ß 3 È 4‹ and Š3 È 4ß È 54‹, respectively. One can check that these points do satisfy the equation x$ y$ 9xy œ 0. 45. x# 2xy 3y# œ 0 Ê 2x 2xyw 2y 6yyw œ 0 Ê yw (2x 6y) œ 2x 2y Ê yw œ xy 3y x ¹ (1ß1)

line m œ yw k (1 1) œ ß

xy 3y x

Ê the slope of the tangent

œ 1 Ê the equation of the normal line at (1ß 1) is y 1 œ 1(x 1) Ê y œ x 2. To find

where the normal line intersects the curve we substitute into its equation: x# 2x(2 x) 3(2 x)# œ 0 Ê x# 4x 2x# 3 a4 4x x# b œ 0 Ê 4x# 16x 12 œ 0 Ê x# 4x 3 œ 0 Ê (x 3)(x 1) œ 0 Ê x œ 3 and y œ x 2 œ 1. Therefore, the normal to the curve at (1ß 1) intersects the curve at the point (3ß 1). Note that it also intersects the curve at (1ß 1). 46. Let p and q be integers with q 0 and suppose that y œ Èxp œ xpÎq . Then yq œ xp . Since p and q are integers and q

d q dx ay b xp 1ap pÎqb œ qp

assuming y is a differentiable function of x, œ

p q

†

xp

axpÎq b

1 q

47. y# œ x Ê

1

œ

p q

†

dy dx

œ

" #y

xp 1 xp pÎq

œ

p q

†

œ

p1 Ê qyq 1 dy Ê dx œ px

d p dx ax b

dy dx

pxp qyq

œ

1

œ

1

p q

xp yq

†

1 1

a p Îq b 1

†x

. If a normal is drawn from (aß 0) to (x" ß y" ) on the curve its slope satisfies

Ê y" œ 2y" (x" a) or a œ x" "# . Since x" 0 on the curve, we must have that a

" #

y" 0 x" a

œ 2y"

. By symmetry, the two Èx

Èx

points on the parabola are ˆx" ß Èx" ‰ and ˆx" ß Èx" ‰ . For the normal to be perpendicular, Š x" "a ‹ Š a x"" ‹ œ 1 Ê

x" (a x" )#

œ 1 Ê x" œ (a x" )# Ê x" œ ˆx"

" #

#

2x w 48. 2x# 3y# œ 5 Ê 4x 6yyw œ 0 Ê yw œ 2x 3y Ê y k (1 1) œ 3y ¹ 3x# 2y

Ê yw k (1 1) œ ß

3x# 2y ¹ (1ß1)

œ

and yw k (1

3 #

and y" œ „ #" . Therefore, ˆ 4" ß „ #" ‰ and a œ

œ 23 and yw k (1

(1ß1)

ß

y# œ x$ Ê 2yyw œ 3x# Ê yw œ

" 4

x" ‰ Ê x " œ

ß

1)

œ

3x# 2y ¹ (1ß 1)

ß

1)

œ 2x 3y ¹

(1ß 1)

œ

2 3

; also,

œ #3 . Therefore the

tangents to the curves are perpendicular at (1ß 1) and (1ß 1) (i.e., the curves are orthogonal at these two points of intersection). 49. (a) x2 y2 œ 4, x2 œ 3y2 Ê a3y2 b y2 œ 4 Ê y2 œ 1 Ê y œ „ 1. If y œ 1 Ê x2 a1b2 œ 4 Ê x2 œ 3 Ê x œ „ È3. If y œ 1 Ê x2 a1b2 œ 4 Ê x2 œ 3 Ê x œ „ È3. x2 y2 œ 4 Ê 2x 2y dy dx œ 0 Ê m1 œ At ŠÈ3ß 1‹: m1 œ

œ

dy dx

È3 1

dy dx

œ

At ŠÈ3ß 1‹: m1 œ

dy dx

œ


dy dx

1

È3 2

Ê x œ 1 Š

dy Ê 1 œ 23 y dx Ê m2 œ

At Š 14 ß

È3 2 ‹:

At Š 14 ß

m1 œ

È3 2 ‹:

dy dx

m1 œ

dy dx

È3 2 2 ‹

œ

œ dy dx

œ

œ

œ È3 and m2 œ

dy dx

œ

ŠÈ3‹ a 1 b

œ È3 and m2 œ 3 4

È3 3

œ

È 3 3 a1 b

œ

dy dx

Êyœ „

È 33

œ

Ê m1 † m2 œ

È3 3

ŠÈ3‹ 3 a 1 b

È3 2 .

œ

dy dx

Ê m1 † m2 œ ŠÈ3‹Š

œ

If y œ

œ 14 . x œ 1 y2 Ê 1 œ 2y dy dx Ê m1 œ

x 3y

È3 3 ‹

œ 1

È ŠÈ3‹Š 33 ‹

Ê m1 † m2 œ ŠÈ3‹Š È3 3 È3 2 dy dx

œ 1

È3 3 ‹

Ê m1 † m2 œ ŠÈ3‹Š Êxœ1Š

2

È3 2 ‹

œ 1

È3 3 ‹

œ 1

œ 41 . If

1 œ 2y and x œ 13 y2

3 2y

1 2ŠÈ3Î2‹

œ

È3 3 a1 b

œ

dy dx

(b) x œ 1 y2 , x œ 13 y2 Ê ˆ 13 y2 ‰ œ 1 y2 Ê y2 œ yœ

dy dx

È3 3 a 1 b

œ È3 and m2 œ

ŠÈ3‹

œ

œ xy and x2 œ 3y2 Ê 2x œ 6y dy dx Ê m2 œ

œ È3 and m2 œ

È a13b


dy dx

œ È13 and m2 œ

1 2ŠÈ3Î2‹

œ

1 È3

dy dx

and m2 œ

œ

dy dx

3 2ŠÈ3Î2‹

œ

œ

3 2ŠÈ3Î2‹

3 È3

Ê m1 † m2 œ Š È13 ‹Š È33 ‹ œ 1

œ È33 Ê m1 † m2 œ Š È13 ‹Š È33 ‹ œ 1


3 4

.

Section 3.7 Implicit Differentiation 50. y œ 13 x b, y2 œ x3 Ê Ê

x4 4

dy dx

dy œ 13 and 2y dx œ 3x2 Ê

51. xy$ x# y œ 6 Ê x Š3y#

dy dx ‹

a4 b 2 2

3x2 2y

2

Ê ˆ 13 ‰Š 3x 2y ‹ œ 1 Ê

y$ x#

dy dx

2xy œ 0 Ê

$

#

#

x œ 3xy y$ 2xy ; thus

dx dy

a0b2 2

x2 2

appears to equal

"

dy dx

dx dy

dy dx

2

œ ! and ˆ 13 ‰Š 3x 2y ‹ œ 1 is

a3xy# x# b œ y$ 2xy Ê

x# y Š2x

dx dy ‹

œ0 Ê

dx dy

2

2

œ y Ê Š x2 ‹ œ x3

œ 8. At a4ß 8b, y œ 31 x b Ê 8 œ 31 a4b b Ê b œ

y 2xy $ # # $ œ 3xy # x# ; also, xy x y œ 6 Ê x a3y b y dx dy

œ

œ x3 Ê x4 4x3 œ 0 Ê x3 ax 4b œ 0 Ê x œ 0 or x œ 4. If x œ 0 Ê y œ

indeterminant at a0, 0b. If x œ 4 Ê y œ

Ê

dy dx

dy dx

œ

28 3 . y$ 2xy 3xy# x#

ay$ 2xyb œ 3xy# x#

. The two different treatments view the graphs as functions

symmetric across the line y œ x, so their slopes are reciprocals of one another at the corresponding points (aß b) and (bß a). 52. x$ y# œ sin# y Ê 3x# 2y œ

3x# 2 sin y cos y 2y

appears to equal

dy dx

œ (2 sin y)(cos y)

; also, x$ y# œ sin# y Ê 3x# "

dy dx

dx dy

dy dx

Ê

dy dx

(2y 2 sin y cos y) œ 3x# Ê

2y œ 2 sin y cos y Ê

dx dy

œ

2 sin y cos y 2y 3x#

dy dx

œ

3x# 2y 2 sin y cos y

; thus

dx dy

. The two different treatments view the graphs as functions symmetric across the line

y œ x so their slopes are reciprocals of one another at the corresponding points (aß b) and (bß a). 53-60. Example CAS commands: Maple: q1 := x^3-x*y+y^3 = 7; pt := [x=2,y=1]; p1 := implicitplot( q1, x=-3..3, y=-3..3 ): p1; eval( q1, pt ); q2 := implicitdiff( q1, y, x ); m := eval( q2, pt ); tan_line := y = 1 + m*(x-2); p2 := implicitplot( tan_line, x=-5..5, y=-5..5, color=green ): p3 := pointplot( eval([x,y],pt), color=blue ): display( [p1,p2,p3], ="Section 3.7 #57(c)" ); Mathematica: (functions and x0 may vary): Note use of double equal sign (logic statement) in definition of eqn and tanline. x^4 - 8*x^2 + 4*x + 2; domain := x=-20/25..64/25; plot( f(x), domain, color=black, title="Section 4.1 #81(a)" ); Df := D(f); plot( Df(x), domain, color=black, title="Section 4.1 # 81(b)" ) StatPt := fsolve( Df(x)=0, domain ) SingPt := NULL; EndPt := op(rhs(domain)); Pts :=evalf([EndPt,StatPt,SingPt]); Values := [seq( f(x), x=Pts )]; Maximum value is 2.7608 and occurs at x=2.56 (right endpoint). %

Minimum value $ is -6.2680 and occurs at x=1.86081 (singular point). Mathematica: (functions may vary) (see section 2.5 re. RealsOnly ): x^3-4*x^2+3*x; a := 0; b := 4; F := unapply( int(f(t),t=a..x), x ); # (a) p1 := plot( [f(x),F(x)], x=a..b, legend=["y = f(x)","y = F(x)"], title="#67(a) (Section 5.4)" ): p1; dF := D(F); # (b) q1 := solve( dF(x)=0, x ); pts1 := [ seq( [x,f(x)], x=remove(has,evalf([q1]),I) ) ]; p2 := plot( pts1, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where F '(x)=0" ): display( [p1,p2], title="81(b) (Section 5.4)" ); incr := solve( dF(x)>0, x ); # (c) decr := solve( dF(x) x^2; f := x -> sqrt(1-x^2); F := unapply( int( f(t), t=a..u(x) ), x ); dF := D(F); # (b) cp := solve( dF(x)=0, x ); solve( dF(x)>0, x ); solve( dF(x)29. :

V œ 'a 1R# (x) dx œ 'c1 1 a3x% b dx œ 1 'c1 9x) dx b

1

1

#

"

œ 1 cx* d " œ 21

(b) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x a3x% b dx œ 21 † 3'0 x& dx œ 21 † 3 ’ x6 “ œ 1 b

1

1

'

!

Note: The lower limit of integration is 0 rather than 1. (c) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ 21'c1 (1 x) a3x% b dx œ 21 ’ 3x5 b

"

1

&

(d) A+=2/< 7/>29. :

" x' 2 “ "

œ 21 ˆ 35 "# ‰ ˆ 35 "# ‰‘ œ

121 5

R(x) œ 3, r(x) œ 3 3x% œ 3 a1 x% b Ê V œ 'a 1 cR# (x) r# (x)d dx œ 'c1 1 ’9 9 a1 x% b “ dx b

1

œ 91 'c1 c1 a1 2x% x) bd dx œ 91 'c1 a2x% x) b dx œ 91 ’ 2x5 1

1

8. (a) A+=2/< 7/>29. : R(x) œ

, r(x) œ

4 x$

" #

&

" x* 9 “ "

b

21†13 5

œ

261 5

2

(b) =2/66 7/>29. :

V œ 21'1 x ˆ x4$ "# ‰ dx œ 21 ’4x" 2

(c) =2/66 7/>29. :

" #

# x# 4 “"

16 5

4" ‰ œ

1 20

(2 10 64 5) œ

b

2

x

(d) A+=2/< 7/>29. :

# x# 4 “"

571 #0

œ 21 ˆ 4# 1‰ ˆ4 4" ‰‘ œ 21 ˆ 54 ‰ œ

shell ‰ shell V œ 21'a ˆ radius Š height ‹ dx œ 21'1 (2 x) ˆ x4$ "# ‰ dx œ 21'1 ˆ x8$ 4 x

œ 181 25 "9 ‘ œ

# # # & Ê V œ 'a 1cR# (x) r# (x)d dx œ '1 1 ’ˆ x4$ ‰ ˆ "# ‰ “ dx œ 1 16 x4 ‘ " 5 x

"‰ " ˆ 16 " ‰‘ œ 1 ˆ 10 œ 1 ˆ 5†16 32 # 5 4

œ 21 ’ x4#

#

2

4 x#

51 #

1 x# ‰ dx

œ 21 (1 2 2 1) ˆ4 4 1 4" ‰‘ œ

31 #

V œ 'a 1cR# (x) r# (x)d dx b

# œ 1 '1 ’ˆ 7# ‰ ˆ4 2

dx

œ

491 4

161'1 a1 2x$ x' b dx

œ

491 4

161 ’x x#

œ

491 4 491 4 491 4

161 ˆ2 4" 5†"3# ‰ ˆ1 1 5" ‰‘ " 161 ˆ 4" 160 5" ‰

œ œ 9.

4 ‰# x$ “

2

161 160

# x & 5 “"

(40 1 32) œ

491 4

711 10

œ

1031 20

(a) .3=5 7/>29. :

# V œ 1 '1 ŠÈx 1‹ dx œ 1'1 (x 1) dx œ 1 ’ x# x“

#

5

5

‰ ˆ" ‰‘ œ 1 ˆ 24 ‰ œ 1 ˆ 25 # 5 # 1 # 4 œ 81

& "

(b) A+=2/< 7/>29. :

R(y) œ 5, r(y) œ y# 1 Ê V œ 'c 1 cR# (y) r# (y)d dy œ 1 'c2 ’25 ay# 1b “ dy d

2

œ 1'c2 a25 y% 2y# 1b dy œ 1 'c2 a24 y% 2y# b dy œ 1 ’24y 2

2

#

y& 5

32 y$ “

# #

œ 21 ˆ24 † 2


32 5

2 3

† 8‰

378

Chapter 6 Applications of Definite Integrals œ 321 ˆ3

2 5

"3 ‰ œ

321 15

(45 6 5) œ

10881 15

(c) .3=5 7/>29. : R(y) œ 5 ay# 1b œ 4 y#

Ê V œ 'c 1R# (y) dy œ 'c2 1 a4 y# b dy d

2

#

œ 1 'c2 a16 8y# y% b dy 2

œ 1 ’16y

8y$ 3

œ 641 ˆ1

2 3

# y& 5 “ #

"5 ‰ œ

œ 21 ˆ32

641 15

64 3

(15 10 3) œ

32 ‰ 5 5121 15

10. (a) =2/66 7/>29. :

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y Šy d

4

œ 21'0 Šy# 4

œ

21 1#

† 64 œ

y$ 4‹

$

dy œ 21 ’ y3

321 3

%

y% 16 “ !

y# 4‹

dy

œ 21 ˆ 64 3

64 ‰ 4

(b) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ˆ2Èx x‰ dx œ 21'0 ˆ2x$Î# x# ‰ dx œ 21 ’ 45 x&Î# b

4

œ 21 ˆ 45 † 32

64 ‰ 3

œ

4

1281 15

% x$ 3 “!

(c) =2/66 7/>29. :

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21(4 x) ˆ2Èx x‰ dx œ 21'0 ˆ8x"Î# 4x 2x$Î# x# ‰ dx b

4

$Î# œ 21 ’ 16 2x# 54 x&Î# 3 x

œ 641 ˆ1 45 ‰ œ

641 5

4

% x$ 3 “!

œ 21 ˆ 16 3 † 8 32

(d) =2/66 7/>29. :

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(4 y) Šy d

4

œ 21'0 Š4y 2y# 4

y$ 4‹

dy œ 21 ’2y# 32 y$

y# 4‹

% y% 16 “ !

4 5

† 32

64 ‰ 3

œ 641 ˆ 34 1

dy œ 21'0 Š4y y# y# 4

œ 21 ˆ32

2 3

4 5

y$ 4‹

32 ‰

dy

† 64 16‰ œ 321 ˆ2

8 3

1‰ œ

321 3

11. .3=5 7/>29. : R(x) œ tan x, a œ 0, b œ

1 3

Ê V œ 1 '0 tan# x dx œ 1'0 asec# x 1b dx œ 1[tan x x]! 1Î3

12. .3=5 7/>29. :

1Î3

1Î$

V œ 1'0 (2 sin x)# dx œ 1 '0 a4 4 sin x sin# xb dx œ 1'0 ˆ4 4 sin x 1

1

œ 1 4x 4 cos x

x #

sin 2x ‘ 1 4 !

1

œ 1 ˆ41 4

1 #

0‰ (0 4 0 0)‘ œ

œ

1cos 2x ‰ dx # 9 1 1 ˆ # 8‰ œ 1#

1 Š3È31‹ 3

(91 16)

13. (a) .3=5 7/>29. :

V œ 1'0 ax# 2xb dx œ 1'0 ax% 4x$ 4x# b dx œ 1 ’ x5 x% 34 x$ “ œ 1 ˆ 32 5 16 2

œ

161 15

2

#

(6 15 10) œ

#

&

!

161 15

32 ‰ 3

(b) A+=2/< 7/>29. :

V œ '0 1’1# ax# 2x "b “ dx œ '0 1 dx '0 1 ax "b% dx œ #1 ’1 2

2

#

2

(c) =2/66 7/>29. :

# ax"b& & “!

œ #1 1 †

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ 21'0 (2 x) c ax# 2xbd dx œ 21'0 (2 x) a2x x# b dx b

2

2


# &

œ

)1 &

Chapter 6 Practice Exercises œ 21'0 a4x 2x# 2x# x$ b dx œ 21'0 ax$ 4x# 4xb dx œ 21 ’ x4 34 x$ 2x# “ œ 21 ˆ4 2

œ

21 3

2

#

%

81 3

(36 32) œ

32 3

!

8‰

(d) A+=2/< 7/>29. :

V œ 1 '0 c2 ax# 2xbd dx 1'0 2# dx œ 1'0 ’4 4 ax# 2xb ax# 2xb “ dx 81 2

2

#

2

#

œ 1'0 a4 4x# 8x x% 4x$ 4x# b dx 81 œ 1'0 ax% 4x$ 8x 4b dx 81 2

2

#

&

‰ œ 1 ’ x5 x% 4x# 4x“ 81 œ 1 ˆ 32 5 16 16 8 81 œ !

1 5

(32 40) 81 œ

721 5

401 5

œ

321 5

14. .3=5 7/>29. :

V œ 21'0 4 tan# x dx œ 81'0 asec# x 1b dx œ 81[tan x x]! 1Î4

1Î4

1Î%

œ 21(4 1)

15. The material removed from the sphere consists of a cylinder and two "caps." From the diagram, the height of the cylinder #

is 2h, where h# ŠÈ$‹ œ ## , i.e. h œ ". Thus #

Vcyl œ a#hb1ŠÈ$‹ œ '1 ft$ . To get the volume of a cap, use the disk method and x# y# œ ## : Vcap œ '" 1x# dy 2

œ '" 1a% y# bdy œ 1’%y 2

œ 1ˆ8 83 ‰ ˆ% 3" ‰‘ œ

# y3 3 “"

&1 3

Vremoved œ Vcyl #Vcap œ '1

ft$ . Therefore, "!1 3

œ

#)1 3

ft$ .

16. We rotate the region enclosed by the curve y œ É12 ˆ1

4x# ‰ 121

and the x-axis around the x-axis. To find the

11Î2

volume we use the .3=5 method: V œ 'a 1R# (x) dx œ 'c11Î2 1 ŠÉ12 ˆ1 b

œ 121'c11Î2 Š1 11Î2

4x# 121 ‹

œ 1321 ˆ1 "3 ‰ œ 17. y œ x"Î#

x$Î# 3

Ê

dx œ 121 ’x

2641 3

dy dx

œ

""Î# 4x$ 363 “ ""Î#

ˆ4

2 3

18. x œ y#Î$ Ê

" #

#

x"Î# "# x"Î# Ê Š dy dx ‹ œ

œ '1

8

† 8‰ ˆ2 23 ‰‘ œ

È9y#Î$ 4 3y"Î$

œ

" 4

" 3

5 12

12 Š1

4x# 121 ‹

dx

#

4

ˆ2

14 ‰ 3

œ

4y #Î$ 9

" #

ˆx"Î# x"Î# ‰ dx œ

" #

2x"Î# 23 x$Î# ‘ % "

10 3 dx Ê L œ '1 Ê1 Š dy ‹ dy œ '1 É1 #

8

8

4 9y#Î$

dy

'18 È9y#Î$ 4 ˆy"Î$ ‰ dy; u œ 9y#Î$ 4 Ê du œ 6y"Î$ dy; y œ 1 Ê u œ 13,

y œ 8 Ê u œ 40d Ä L œ 19. y œ

" #

#

dy œ

11Î2

ˆ x" 2 x‰ Ê L œ ' É1 4" ˆ x" 2 x‰ dx 1 4

y"Î$ Ê Š dx dy ‹ œ

2 3

11Î2

$

4

dx dy

dx œ 1 '

œ 881 ¸ 276 in$

4

" #

#

4 ‰ 11 ˆ 4 ‰ ˆ 11 ‰ “ œ 1321 ’1 ˆ 363 œ 241 ’ 11 Š 4 ‹“ 2 363 #

# Ê L œ '1 É 4" ˆ x" 2 x‰ dx œ '1 É 4" ax"Î# x"Î# b dx œ '1

œ

4x# ‰ 121 ‹

x'Î& 58 x%Î& Ê

dy dx

" 18

œ

'1340 u"Î# du œ 18" 32 u$Î# ‘ %! œ #"7 40$Î# 13$Î# ‘ ¸ 7.634 "$ " #

#

dy x"Î& "# x"Î& Ê Š dx ‹ œ

" 4

ˆx#Î& 2 x#Î& ‰

# Ê L œ '1 É1 "4 ax#Î& 2 x#Î& b dx Ê L œ '1 É 4" ax#Î& 2 x#Î& b dx œ ' É 4" ax"Î& x"Î& b dx 32

32

32

1


379

380

Chapter 6 Applications of Definite Integrals œ '1

32

œ

" 48

20. x œ

" #

ˆx"Î& x"Î& ‰ dx œ

(1260 450) œ

" 1#

y$

" y

Ê

" % œ '1 É 16 y 2

" #

œ

1710 48

" 5 # 6 285 8

$#

x'Î& 45 x%Î& ‘ " œ

" y#

#

dx dy

œ

" 4

" y%

dy œ '1 ÊŠ 4" y#

y#

2

8 œ ˆ 12 "# ‰ ˆ 1"# 1‰ œ

7 1#

" #

21. S œ 'a 21y Ê1 Š dy dx ‹ dx;

dy dx

œ

#

b

" 16

Ê Š dx dy ‹ œ

œ

" y# ‹

#

" #

ˆ 65 † 2'

y%

" #

" y%

5 4

† 2% ‰ ˆ 56 54 ‰‘ œ

" #

ˆ 315 6

" % Ê L œ '1 Ê1 Š 16 y 2

dy œ '1 Š 4" y# 2

" y# ‹

dy œ ’ 1"# y$ y" “

" #

75 ‰ 4

" y% ‹

dy

# "

13 12 #

" È2x 1

Ê Š dy dx ‹ œ

" #x 1

Ê S œ '0 21È2x 1 É1 3

" #x 1

dx

2 È ' Èx 1 dx œ 2È21 2 (x 1)$Î# ‘ $ œ 2È21 † 2 (8 1) œ œ 21'0 È2x 1 É 2x 2x1 dx œ 2 21 0 3 3 ! 3

3

22. S œ 'a 21y Ê1 Š dy dx ‹ dx; #

b

œ

1 6

dy dx

% ' œ x# Ê Š dy dx ‹ œ x Ê S œ 0 21 † #

1

x$ 3

È1 x% dx œ

1 6

281È2 3

'01 È1 x% a4x$ b dx

'01 È1 x% d a1 x% b œ 16 ’ 32 a1 x% b$Î# “ " œ 19 ’2È2 1“ !

23. S œ 'c 21x Ê1 Š dx dy ‹ dy; #

d

dx dy

ˆ "# ‰ (4 2y) È4y y#

œ

œ

2y È4y y#

#

Ê 1 Š dx dy ‹ œ

4y y# 4 4y y# 4y y#

œ

4 4y y#

Ê S œ '1 21 È4y y# É 4y 4 y# dy œ 41'1 dx œ 41 2

2

24. S œ 'c 21x Ê1 Š dx dy ‹ dy; #

d

œ 1'2 È4y 1 dy œ 6

1 4

dx dy

œ

1 2È y

#

Ê 1 Š dx dy ‹ œ 1

32 (4y 1)$Î# ‘ ' œ #

1 6

(125 27) œ

1 6

" 4y

œ

4y 1 4y

(98) œ

Ê S œ '2 21Èy † 6

È4y 1 È4y

dy

491 3

25. The equipment alone: the force required to lift the equipment is equal to its weight Ê F" (x) œ 100 N.

The work done is W" œ 'a F" (x) dx œ '0 100 dx œ [100x]%! ! œ 4000 J; the rope alone: the force required b

40

to lift the rope is equal to the weight of the rope paid out at elevation x Ê F# (x) œ 0.8(40 x). The work done is W# œ 'a F# (x) dx œ '0 0.8(40 x) dx œ 0.8 ’40x b

40

the total work is W œ W" W# œ 4000 640 œ 4640 J

%! x# # “!

œ 0.8 Š40#

40# # ‹

œ

(0.8)(1600) #

œ 640 J;

26. The force required to lift the water is equal to the water's weight, which varies steadily from 8 † 800 lb to 8 † 400 lb over the 4750 ft elevation. When the truck is x ft off the base of Mt. Washington, the water weight is x‰ F(x) œ 8 † 800 † ˆ 2†24750 œ (6400) ˆ1 †4750

œ '0

4750

6400 ˆ1

x ‰ 9500

dx œ 6400 ’x

œ 22,800,000 ft † lb

x ‰ 9500

lb. The work done is W œ 'a F(x) dx

%(&! x# 2†9500 “ !

b

œ 6400 Š4750

4750# 4†4750 ‹

œ ˆ 34 ‰ (6400)(4750)

27. Force constant: F œ kx Ê 20 œ k † 1 Ê k œ 20 lb/ft; the work to stretch the spring 1 ft is W œ '0 kx dx œ k'0 x dx œ ’20 x# “ œ 10 ft † lb; the work to stretch the spring an additional foot is 1

1

#

"

! #

W œ '1 kx dx œ k '1 x dx œ 20 ’ x# “ œ 20 ˆ 4# "# ‰ œ 20 ˆ 3# ‰ œ 30 ft † lb 2

2

#

"


Chapter 6 Practice Exercises 28. Force constant: F œ kx Ê 200 œ k(0.8) Ê k œ 250 N/m; the 300 N force stretches the spring x œ œ

300 250

381

F k

œ 1.2 m; the work required to stretch the spring that far is then W œ '0 F(x) dx œ '0 250x dx 1Þ2

1Þ2

œ [125x# ]!"Þ# œ 125(1.2)# œ 180 J 29. We imagine the water divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [0ß 8]. The typical slab between the planes at y and y ?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆ 54 y‰ ?y œ 25161 y# ?y ft$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ 62.4 ?V œ

(62.4)(25) 16

1y# ?y lb. The distance through which F(y)

must act to lift this slab to the level 6 ft above the top is about (6 8 y) ft, so the work done lifting the slab is about ?W œ

(62.4)(25) 16

1y# (14 y) ?y ft † lb. The work done

lifting all the slabs from y œ 0 to y œ 8 to the level 6 ft above the top is approximately 8

W¸! !

(62.4)(25) 16

1y# (14 y) ?y ft † lb so the work to pump the water is the limit of these Riemann sums as the norm of

the partition goes to zero: W œ '0

8

$ œ (62.4) ˆ 25161 ‰ Š 14 3 †8

8% 4‹

(62.4)(25) (16)

(62.4)(25)1 16

1y# (14 y) dy œ

'08 a14y# y$ b dy œ (62.4) ˆ 25161 ‰ ’ 143 y$ y4 “ ) %

!

¸ 418,208.81 ft † lb

30. The same as in Exercise 29, but change the distance through which F(y) must act to (8 y) rather than (6 8 y). Also change the upper limit of integration from 8 to 5. The integral is:W œ '0

5

œ (62.4) ˆ 25161 ‰'0 a8y# y$ b dy œ (62.4) ˆ 25161 ‰ ’ 83 y$ 5

& y% 4 “!

(62.4)(25)1 16

y# (8 y) dy

œ (62.4) ˆ 25161 ‰ Š 83 † 5$

5% 4‹

31. The tank's cross section looks like the figure in Exercise 29 with right edge given by x œ #

slab has volume ?V œ 1(radius)# (thickness) œ 1 ˆ #y ‰ ?y œ F(y) œ 60 †

1 4

y œ y# . A typical horizontal

y# ?y. The force required to lift thisslab is its weight:

y# ?y. The distance through which F(y) must act is (2 10 y) ft, so the work to pump the liquid is

W œ 60'0 1a12 ybŠ y4 ‹dy œ 151 ’ 12y 3 10

22,5001 ft†lb 275 ft†lb/sec

1 4

5 10

¸ 54,241.56 ft † lb

#

$

"! y% 4 “!

œ 22,5001 ft † lb; the time needed to empty the tank is

¸ 257 sec

32. A typical horizontal slab has volume about ?V œ (20)(2x)?y œ (20) ˆ2È16 y# ‰ ?y and the force required to lift this slab is its weight F(y) œ (57)(20) ˆ2È16 y# ‰ ?y. The distance through which F(y) must act is (6 4 y) ft, so the work to pump the olive oil from the half-full tank is W œ 57'c4 (10 y)(20) ˆ2È16 y# ‰ dy 0

"Î# œ 2880 'c4 10È16 y# dy 1140'c4 a16 y# b (2y) dy 0

0

œ 22,800 † (area of a quarter circle having radius 4) 23 (1140) ’a16 y# b œ 335,153.25 ft † lb

$Î# !

“

%

œ (22,800)(41) 48,640


382

Chapter 6 Applications of Definite Integrals

33. Intersection points: 3 x# œ 2x# Ê 3x# 3 œ 0 Ê 3(x 1)(x 1) œ 0 Ê x œ 1 or x œ 1. Symmetry suggests that x œ 0. The typical @/3-+6 strip has # # # center of mass: (µ x ßµ y ) œ Šxß 2x a3 x b ‹ œ Šxß x 3 ‹ , #

#

#

#

#

length: a3 x b 2x œ 3 a1 x b, width: dx, area: dA œ 3 a1 x# b dx, and mass: dm œ $ † dA œ 3$ a1 x# b dx Ê the moment about the x-axis is µ y dm œ œ

3 #

$ ax# 3b a1 x# b dx œ

3 #

&

2x$ 3

$ ’ x5

" x$ 3 “ "

œ 3$ ’x

3x“

" "

3 #

$ ax% 2x# 3b dx Ê Mx œ ' µ y dm œ

œ 3$ ˆ 5"

2 3

3$ 15

3‰ œ

œ 6$ ˆ1 "3 ‰ œ 4$ Ê y œ

Mx M

œ

(3 10 45) œ

32$ 5 †4 $

œ

8 5

32$ 5

3 #

$ 'c1 ax% 2x# 3b dx "

; M œ ' dm œ 3$ 'c1 a1 x# b dx "

. Therefore, the centroid is (xß y) œ ˆ!ß 85 ‰ .

34. Symmetry suggests that x œ 0. The typical @/3-+6 # strip has center of mass: (µ x ßµ y ) œ Šxß x# ‹ , length: x# , width: dx, area: dA œ x# dx, mass: dm œ $ † dA œ $ x# dx Ê the moment about the x-axis is µ y dm œ #$ x# † x# dx œ

$ #

x% dx Ê Mx œ ' µ y dm œ

$ #

'c22 x% dx œ 10$ cx& d ##

35. The typical @/3-+6 strip has: center of mass: (µ x ßµ y ) œ Œxß

#

4 x4 #

, length: 4

area: dA œ Š4 œ $ Š4

x# 4‹

x# 4 ‹dx,

width: dx,

mass: dm œ $ † dA

dx Ê the moment about the x-axis is #

Š4 x4 ‹

µ y dm œ $ †

x# 4,

#

Š4

x# 4‹

dx œ

$ #

Š16

moment about the y-axis is µ x dm œ $ Š4 œ

$ 2

’16x

% x& 5†16 “ !

$ #

œ

64

64 ‘ 5

œ

x% 16 ‹ x# 4‹

œ

16†$ †3 32†$

œ

3 2

and y œ

Mx M

œ

dx. Thus, Mx œ ' µ y dm œ

4

4

My M

x$ 4‹

† x dx œ $ Š4x

; My œ ' µ x dm œ $ '0 Š4x

128$ 5

œ $ (32 16) œ 16$ ; M œ ' dm œ $ '0 Š4 Ê xœ

dx; the

128†$ †3 5†32†$

x# 4‹

œ

dx œ $ ’4x

12 5

% x$ 12 “ !

x$ 4‹

dx œ $ ’2x#

œ $ ˆ16

64 ‰ 1#

œ

$ #

'04 Š16 16x ‹ dx %

% x% 16 “ !

32$ 3

‰ . Therefore, the centroid is (xß y) œ ˆ 3# ß 12 5 .

36. A typical 29+6 strip has: # center of mass: (µ x ßµ y ) œ Š y # 2y ß y‹ , length: 2y y# , width: dy, area: dA œ a2y y# b dy, mass: dm œ $ † dA œ $ a2y y# b dy; the moment about the x-axis is µ y dm œ $ † y † a2y y# b dy œ $ a2y# y$ b ; the moment # about the y-axis is µ x dm œ $ † ay 2yb † a2y y# b dy #

œ a4y y b dy Ê Mx œ ' µ y dm œ $ '0 a2y# y$ b dy $ #

#

œ $ ’ 23 y$ œ

$ #

yœ

ˆ 43†8 Mx M

2

%

œ

# y% 4 “!

32 ‰ 5

4†$ †3 3 † 4 †$

œ

œ $ ˆ 23 † 8 32$ 15

16 ‰ 4

œ $ ˆ 16 3

16 ‰ 4

œ

$ †16 12

œ

4$ 3

; My œ ' µ x dm œ

$ #

'02 a4y# y% b dy œ #$ ’ 34 y$ y5 “ #

$ ; M œ ' dm œ $ '0 a2y y# b dy œ $ ’y# y3 “ œ $ ˆ4 83 ‰ œ

2

# !

&

!

4$ 3

Ê xœ

œ 1. Therefore, the centroid is (xß y) œ ˆ 58 ß 1‰ .


My M

œ

$ †32†3 15†$ †4

œ

8 5

and

Chapter 6 Practice Exercises

383

37. A typical horizontal strip has: center of mass: (µ x ßµ y ) #

œ Š y #2y ß y‹ , length: 2y y# , width: dy, area: dA œ a2y y# b dy, mass: dm œ $ † dA œ (1 y) a2y y# b dy Ê the moment about the x-axis is µ y dm œ y(1 y) a2y y# b dy # œ a2y 2y$ y$ y% b dy œ a2y# y$ y% b dy; the moment about the y-axis is # µ x dm œ Š y 2y ‹ (1 y) a2y y# b dy œ " a4y# y% b (1 y) dy œ #

Ê Mx œ ' µ y dm œ '0 a2y# y$ y% b dy œ ’ 23 y$ 2

œ

16 60

œ

" #

" #

#

(20 15 24) œ $

Š 4†32 2%

2& 5

(11) œ

4 15

2' 6‹

2

44 40

œ

11 10

y$ 3

; My œ ' µ x dm œ '0

2

œ 4 ˆ 43 2

œ '0 a2y y# y$ b dy œ ’y# ‰ ˆ 38 ‰ œ œ ˆ 44 15

44 15

y% 4

4 5

œ ˆ4

8 3

16 ‰ 4

œ ˆ 16 3

16 4

32 ‰ 5

œ 16 ˆ "3

a4y# 4y$ y% y& b dy œ

86 ‰ œ 4 ˆ2 45 ‰ œ

# y% 4 “!

" #

# y& 5 “!

a4y# 4y$ y% y& b dy

œ

8 3

24 5

" #

" 4

25 ‰

’ 43 y$ y%

y& 5

; M œ ' dm œ '0 (1 y) a2y y# b dy

# y' 6 “!

2

Ê xœ

My M

‰ ˆ 83 ‰ œ œ ˆ 24 5

9 5

and y œ

Mx M

‰ . Therefore, the center of mass is (xß y) œ ˆ 95 ß 11 10 .

3 3 38. A typical vertical strip has: center of mass: (µ x ßµ y ) œ ˆxß 2x3$Î# ‰ , length: x$Î# , width: dx, area: dA œ x$Î# dx, µ 3 3 3 9$ mass: dm œ $ † dA œ $ † dx Ê the moment about the x-axis is y dm œ †$ dx œ dx; the moment about x$Î#

3 the y-axis is µ x dm œ x † $ x$Î# dx œ

(a) Mx œ $ '1

9

M œ $ '1

9

(b) Mx œ '1

9

" #

3 x$Î# x #

9$ #

ˆ x9$ ‰ dx œ

3$ x"Î#

#x$Î#

# *

20$ 9

’ x# “ œ "

3 ‰ ; My œ $ '1 x ˆ x$Î# dx œ 3$ 2x"Î# ‘ " œ 12$ ;

*

*

9 #

*

9

dx œ 6$ x"Î# ‘ " œ 4$ Ê x œ

ˆ x9$ ‰ dx œ

2x$

x$Î#

dx.

My M

œ

12$ 4$

œ 3 and y œ

Mx M

œ

ˆ 209$ ‰ 4$

œ

5 9

* 3 ‰ 3 ‰ x" ‘ * œ 4; My œ ' x# ˆ $Î# dx œ 2x$Î# ‘ " œ 52; M œ '1 x ˆ x$Î# dx " x 1

œ 6 x"Î# ‘ " œ 12 Ê x œ

9

My M

œ

13 3

and y œ

Mx M

9

œ

" 3

strip 39. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ 2 '0 (62.4)(2 y)(2y) dy œ 249.6'0 a2y y# b dy œ 249.6 ’y# b

2

2

# y$ 3 “!

œ (249.6) ˆ4 83 ‰ œ (249.6) ˆ 43 ‰ œ 332.8 lb strip 40. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ '0 75 ˆ 56 y‰ (2y 4) dy œ 75'0 ˆ 53 y 5Î6

b

5Î6

10 3

2y# 4y‰ dy

7 7 # 2 $ ‘ &Î' 50 ‰ 25 ‰ 125 ‰‘ #‰ ˆ 18 œ 75 '0 ˆ 10 dy œ 75 10 ˆ 67 ‰ ˆ 36 ˆ 32 ‰ ˆ 216 3 3 y 2y 3 y 6 y 3 y ! œ (75) 5Î6

œ (75) ˆ 25 9

175 216

250 ‰ 3†#16

‰ œ ˆ 9†75 #16 (25 † 216 175 † 9 250 † 3) œ

strip 41. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ 62.4'0 (9 y) Š2 † b

4

%

œ 62.4 6y$Î# 25 y&Î# ‘ ! œ (62.4) ˆ6 † 8

2 5

Èy 2 ‹

(75)(3075) 9†#16

¸ 118.63 lb.

dy œ 62.4'0 ˆ9y"Î# 3y$Î# ‰ dy 4

‰ (48 † 5 64) œ † 32‰ œ ˆ 62.4 5

(62.4)(176) 5

œ 2196.48 lb

strip 42. Place the origin at the bottom of the tank. Then F œ '0 W † Š depth ‹ † L(y) dy, h œ the height of the mercury column, h

strip depth œ h y, L(y) œ 1 Ê F œ '0 849(h y) " dy œ (849)'0 (h y) dy œ 849’hy h

œ

849 # 2 h .

Now solve

849 # 2 h

h

h

y# # “!

œ 849 Šh#

h# #‹

œ 40000 to get h ¸ 9.707 ft. The volume of the mercury is s2 h œ 12 † 9.707 œ 9.707 ft$ Þ


384


CHAPTER 6 ADDITIONAL AND ADVANCED EXERCISES 1. V œ 1 'a cf(x)d# dx œ b# ab Ê 1'a cf(t)d# dt œ x# ax for all x a Ê 1 [f(x)]# œ 2x a Ê f(x) œ É 2x1 a b

x

2. V œ 1 '0 [f(x)]# dx œ a# a Ê 1 '0 [f(t)]# dt œ x# x for all x a Ê 1[f(x)]# œ 2x 1 Ê f(x) œ É 2x1 1 a

x

3. s(x) œ Cx Ê '0 È1 [f w (t)]# dt œ Cx Ê È1 [f w (x)]# œ C Ê f w (x) œ ÈC# 1 for C 1 x

Ê f(x) œ '0 ÈC# 1 dt k. Then f(0) œ a Ê a œ 0 k Ê f(x) œ '0 ÈC# 1 dt a Ê f(x) œ xÈC# 1 a, x

x

where C 1. 4. (a) The graph of f(x) œ sin x traces out a path from (!ß !) to (!ß sin !) whose length is L œ '0 È1 cos# ) d). !

The line segment from (0ß 0) to (!ß sin !) has length È(! 0)# (sin ! 0)# œ È!# sin# !. Since the shortest distance between two points is the length of the straight line segment joining them, we have ! immediately that ' È1 cos# ) d) È!# sin# ! if 0 ! Ÿ 1 . #

0

(b) In general, if y œ f(x) is continuously differentiable and f(0) œ 0, then '0 È1 [f w (t)]# dt È!# f # (!) !

for ! 0. 5. We can find the centroid and then use Pappus' Theorem to calculate the volume. faxb œ x, gaxb œ x2 , faxb œ gaxb 1 Ê x œ x2 Ê x2 x œ 0 Ê x œ 0, x œ 1; $ œ 1; M œ '0 cx x2 ddx œ "# x2 13 x3 ‘0 œ ˆ "# 13 ‰ 0 œ 1

1 6

xœ

1 1 Î6

'01 xcx x2 ddx œ 6'01 cx2 x3 ddx œ 6 31 x3 41 x4 ‘10 œ 6ˆ 31 41 ‰ 0 œ 21

yœ

1 1 Î6

'01 12 ’x2 ax2 b2 “dx œ 3'01 cx2 x4 ddx œ 3 13 x3 15 x5 ‘10 œ 3ˆ 13 15 ‰ 0 œ 25 Ê The centroid is ˆ 12 , 25 ‰.

3 is the distance from ˆ 12 , 25 ‰ to the axis of rotation, y œ x. To calculate this distance we must find the point on y œ x that also lies on the line perpendicular to y œ x that passes through ˆ 12 , 25 ‰. The equation of this line is y 25 œ 1ˆx 12 ‰ Êxyœ

9 10 .

9 3 œ Éˆ 20

The point of intersection of the lines x y œ

1 ‰2 2

9 ˆ 20

2 ‰2 5

œ

1 . 10È2

9 and y œ x is ˆ 20 ,

9 10

Thus V œ 21Š 101È2 ‹ˆ 16 ‰ œ

9 ‰ 20 .

Thus,

1 . 30È2

6. Since the slice is made at an angle of 45‰ , the volume of the wedge is half the volume of the cylinder of radius height 1. Thus, V œ

" ˆ " ‰2 # ’1 # a1b

“œ

" #

and

1 8.

7. y œ 2Èx Ê ds œ É "x 1 dx Ê A œ '0 2Èx É "x 1 dx œ 3

4 3

(1 x)$Î# ‘ $ œ !

28 3

8. This surface is a triangle having a base of 21a and a height of 21ak. Therefore the surface area is " # # # (21a)(21ak) œ 21 a k. d# x dt#

9. F œ ma œ t# Ê

œaœ

t# m

Ê vœ

x œ 0 when t œ 0 Ê C" œ 0 Ê x œ W œ ' F dx œ '0

Ð12mhÑ"Î%

œ

(12mh)$Î# 18m

œ

F(t) †

12mh†È12mh 18m

œ

2h 3

dx dt

dx t$ dt œ 3m C; v œ 0 when t œ 0 Ê t% "Î% . 12m . Then x œ h Ê t œ (12mh)

dt œ '0

Ð12mhÑ"Î%

† 2È3mh œ

4h 3

t# †

$

t 3m

dt œ

" 3m

'

’ t6 “

Ð12mh)"Î%

0

Cœ0 Ê

dx dt

œ

t$ 3m

The work done is

" ‰ œ ˆ 18m (12mh)'Î%

È3mh


Ê xœ

t% 12m

C" ;

Chapter 6 Additional and Advanced Exercises 10. Converting to pounds and feet, 2 lb/in œ

†

2 lb 1 in

œ 24 lb/ft. Thus, F œ 24x Ê W œ '0

1Î2

12 in 1 ft

24x dx

"Î# " " ‰ œ c12x# d ! œ 3 ft † lb. Since W œ "# mv!# "# mv"# , where W œ 3 ft † lb, m œ ˆ 10 lb‰ ˆ 3# ft/sec # " œ 320 slugs, and v" œ 0 ft/sec, we have 3 œ ˆ #" ‰ ˆ 3#"0 v#! ‰ Ê v!# œ 3 † 640. For the projectile height, s œ 16t# v! t (since s œ 0 at t œ 0) Ê ds dt œ v œ 32t v! . At the top of the ball's path, v œ 0 Ê #

and the height is s œ 16 ˆ 3v#! ‰ v! ˆ 3v#! ‰ œ

v!# 64

3†640 64

œ

385

tœ

v! 3#

œ 30 ft.

11. From the symmetry of y œ 1 xn , n even, about the y-axis for 1 Ÿ x Ÿ 1, we have x œ 0. To find y œ MMx , we n use the vertical strips technique. The typical strip has center of mass: (µ x ßµ y ) œ ˆxß 1 2 x ‰ , length: 1 xn , width: dx, area: dA œ a1 xn b dx, mass: dm œ 1 † dA œ a1 xn b dx. The moment of the strip about the 1 1 n # n # " nb1 2n b 1 x-axis is µ y dm œ a1 x b dx Ê M œ ' a1 x b dx œ 2' " a1 2xn x2n b dx œ x 2x x ‘ #

œ1

2 n1

" #n 1

œ

x c1 # (n 1)(2n 1) 2(2n ") (n 1) (n 1)(#n 1)

œ

0 # 2n# 3n 1 4n 2 n 1 (n 1)(#n 1)

Also, M œ 'c1 dA œ 'c1 a1 xn b dx œ 2 '0 a1 xn b dx œ 2 x 1

yœ

Mx M

œ

1

#

†

2n (n 1)(2n 1)

1

(n 1) 2n

œ

xn b 1 ‘ " n1 !

n1

œ

2n# (n 1)(#n 1)

œ 2 ˆ1

" ‰ n1

#n 1 !

. œ

2n n1.

Therefore,

Ê ˆ!ß #n n 1 ‰ is the location of the centroid. As n Ä _, y Ä

n 2n 1

the limiting position of the centroid is ˆ!ß

" #

so

"‰ # .

12. Align the telephone pole along the x-axis as shown in the accompanying figure. The slope of the top length of pole is 9 ‰ ˆ 14.5 " 81 81 œ 8"1 † 40 † (14.5 9) œ 815.5 †40 40 11 ‰ y œ 891 8111†80 x œ 8"1 ˆ9 80 x is an

œ

11 81†80 .

Thus,

equation of the

line representing the top of the pole. Then, My œ 'a x † 1y# dx œ 1 '0 x 8"1 ˆ9 b

40

11 80

# x‰‘ dx

b 11 ‰# '040 x ˆ9 80 x dx; M œ 'a 1y# dx 40 40 11 ‰‘# ‰# dx. œ 1 '0 8"1 ˆ9 80 x dx œ 64"1 '0 ˆ9 11 80 x

œ

" 641

Thus, x œ

My M

¸

129,700 5623.3

¸ 23.06 (using a calculator to compute

the integrals). By symmetry about the x-axis, y œ 0 so the center of mass is about 23 ft from the top of the pole. 13. (a) Consider a single vertical strip with center of mass (µ x ßµ y ). If the plate lies to the right of the line, then µ µ b) $ dA Ê the plate's first moment the moment of this strip about the line x œ b is (x b) dm œ (x about x œ b is the integral ' (x b)$ dA œ ' $ x dA ' $ b dA œ My b$ A. (b) If the plate lies to the left of the line, the moment of a vertical strip about the line x œ b is ab µ x b dm œ ab µ x b $ dA Ê the plate's first moment about x œ b is ' (b x)$ dA œ ' b$ dA ' $ x dA œ b$ A My . 14. (a) By symmetry of the plate about the x-axis, y œ 0. A typical vertical strip has center of mass: (µ x ßµ y ) œ (xß 0), length: 4Èax, width: dx, area: 4Èax dx, mass: dm œ $ dA œ kx † 4Èax dx, for some a proportionality constant k. The moment of the strip about the y-axis is M œ ' µ x dm œ ' 4kx# Èax dx y

œ 4kÈa'0 x a

&Î#

dx œ

4kÈa 27

x

(Î# ‘ a

0

œ 4ka

"Î#

† a 2 7

(Î#

œ

œ 4kÈa'0 x$Î# dx œ 4kÈa 25 x&Î# ‘ 0 œ 4ka"Î# † 25 a&Î# œ a

Ê (xß y) œ

a

ˆ 5a ‰ 7 ß0

0

8ka% 7

. Also, M œ ' dm œ '0 4kxÈax dx

8ka$ 5

. Thus, x œ

a

My M

œ

8ka% 7

†

5 8ka$

œ

5 7

is the center of mass. y#

# # a (b) A typical horizontal strip has center of mass: (µ x ßµ y ) œ Œ 4a # ß y œ Š y 8a4a ß y‹ , length: a

width: dy, area: Ša œ 'c2a y kyk Ša 2a

a

y# 4a ‹

y# 4a ‹

dy, mass: dm œ $ dA œ kyk Ša

dy œ 'c2a y# Ša 0

y# 4a ‹

y# 4a ‹

dy '0 y# Ša 2a

dy. Thus, Mx œ ' µ y dm y# 4a ‹

dy


y# 4a

,

386

Chapter 6 Applications of Definite Integrals œ 'c2a Šay# 0

%

32a& 20a

œ 8a3

dy '0 Šay# 2a

y% 4a ‹ 8a% 3

y% 4a ‹

œ 0; My œ ' µ x dm œ 'c2a Š y 2a

32a& #0a

'c2a2a kyk ay# 4a# b Š 4a 4a y ‹ dy œ 32a" #

! y& #0a “ #a

dy œ ’ 3a y$ #

4a# 8a ‹

#

" 8a

œ

" 3 #a #

'c02a a16a% y y& b dy 32a" '02a a16a% y y& b dy œ 3#"a

œ

" 32a#

’8a% † 4a#

" 32a#

2a

œ

#

#

64a' 6 “

œ

" 16a#

#

’8a% y# 32a' 3 ‹

Š32a'

œ

#a

œ2†

#

16a% 4 ‹

#

Š2a † 4a

" #a

œ

! y' 6 “ #a

1 3#a#

’8a% y#

† 32 a32a' b œ

4 3

#a y' 6 “!

a% ;

'c2a2a kyk a4a# y# b dy

" 4a

%

" 4a

dy

" 16a#

'c02a a4a# y y$ b dy 4a" '02a a4a# y y$ b dy œ 4a" ’2a# y# y4 “ !

" 4a

yœ

c2a

’8a% † 4a#

M œ ' dm œ 'c2a kyk Š 4a 4ay ‹ dy œ

y# 4a ‹

kyk Ša

' kyk a16a% y% b dy

#

#

#a y& #0a “ !

2a

œ

64a' 6 “

’ 3a y$

%

%

$

a8a 4a b œ 2a . Therefore, x œ

" 4a

’2a# y#

œ ˆ 34 a% ‰ ˆ 2a"$ ‰ œ

My M

#a y% 4 “!

2a 3

and

œ 0 is the center of mass.

Mx M

15. (a) On [0ß a] a typical @/3-+6 strip has center of mass: (µ x ßµ y ) œ Šx,

È b# x # È a# x# ‹, #

length: Èb# x# Èa# x# , width: dx, area: dA œ ŠÈb# x# Èa# x# ‹ dx, mass: dm œ $ dA œ $ ŠÈb# x# Èa# x# ‹ dx. On [aß b] a typical @/3-+6 strip has center of mass: È # # (µ x ßµ y ) œ Šxß b # x ‹ , length: Èb# x# , width: dx, area: dA œ Èb# x# dx,

mass: dm œ $ dA œ $ Èb# x# dx. Thus, Mx œ ' µ y dm œ '0

a

" #

ŠÈb# x# Èa# x# ‹ $ ŠÈb# x# Èa# x# ‹ dx 'a

b

" #

Èb# x# $ Èb# x# dx

œ

$ #

'0a cab# x# b aa# x# bd dx #$ 'ab ab# x# b dx œ #$ '0a ab# a# b dx #$ 'ab ab# x# b dx

œ

$ #

cab# a# b xd ! #$ ’b# x

œ

$ #

aab# a$ b #$ Š 23 b$ ab#

a

b

x$ 3 “a

œ a$ 3‹

$ #

b$ 3‹

cab# a# b ad #$ ’Šb$

œ

$ b$ 3

$ a$ 3

œ $ Šb

$

a$ 3 ‹;

a$ 3 ‹“

Š b# a

My œ ' µ x dm

œ '0 x$ ŠÈb# x# Èa# x# ‹ dx 'a x$ Èb# x# dx a

b

œ $ '0 x ab# x# b a

œ

$ #

”

2 ab # x # b 3

$Î#

"Î#

dx $ '0 x aa# x# b a

a

$ 2 aa • #”

#

x# b 3

$Î#

# $Î#

#

œ ’ab a b

# $Î#

ab b

a

dx $ 'a x ab# x# b

$ 2 ab • #”

b

#

!

0

$ 3

"Î#

# $Î#

$ 3

“ ’0 aa b

x# b 3

"Î#

$Î#

• a

$ 3

“ ’0 ab# a# b #

#

$Î#

We calculate the mass geometrically: M œ $ A œ $ Š 14b ‹ $ Š 14a ‹ œ œ

$ ab $ a $ b 3

yœ (b) lim

œ

Mx M 4

b Ä a 31

†

4 $1 ab# a# b #

#

4 aa abb b 31(ab)

Ša

#

ab b ab

#

œ

4 31

$

$

a Š bb# a# ‹ œ

dx

b

4 (b a) aa# ab b# b 31 (b a)(b a)

“œ

$1 4

$ b$ 3

$ a$ 3

œ

$ ab $ a $ b 3

ab# a# b . Thus, x œ

œ Mx ; My M

œ

4 aa# ab b# b 31(a b)

2a 1

2a ‰ Ê (xß y) œ ˆ 2a 1 ß 1 is the limiting

; likewise

.

‹ œ ˆ 341 ‰ Š a

#

a# a# ‹ aa

#

œ ˆ 341 ‰ Š 3a 2a ‹ œ

position of the centroid as b Ä a. This is the centroid of a circle of radius a (and we note the two circles coincide when b œ a).


Chapter 6 Additional and Advanced Exercises 16. Since the area of the traingle is 36, the diagram may be labeled as shown at the right. The centroid of the triangle is ˆ 3a , 24 ‰ a . The shaded portion is 144 36 œ 108. Write ax, yb for the centroid of the remaining region. The centroid of the whole square is obviously a6, 6b. Think of the square as a sheet of uniform density, so that the centroid of the square is the average of the centroids of the two regions, weighted by area: 'œ

$'ˆ 3a ‰ "!)axb "%%

and ' œ

‰ $'ˆ 24 a "!)ayb "%%

which we solve to get x œ )

a *

and y œ

)a a " b . a

Set

x œ 7 in. (Given). It follows that a œ *, whence y œ œ

7 "*

'% *

in. The distances of the centroid ax, yb from the other sides are easily computed. (Note that if we set y œ 7 in.

above, we will find x œ 7 "* .) 17. The submerged triangular plate is depicted in the figure at the right. The hypotenuse of the triangle has slope 1 Ê y (2) œ (x 0) Ê x œ (y 2) is an equation of the hypotenuse. Using a typical horizontal strip, the fluid strip strip pressure is F œ ' (62.4) † Š depth ‹ † Š length ‹ dy

c2

c2

œ 'c6 (62.4)(y)[(y 2)] dy œ 62.4 'c6 ay# 2yb dy $

œ 62.4 ’ y3 y# “

#

‰‘ œ (62.4) ˆ 83 4‰ ˆ 216 3 36

'

‰ œ (62.4) ˆ 208 3 32 œ

(62.4)(112) 3

¸ 2329.6 lb

18. Consider a rectangular plate of length j and width w. The length is parallel with the surface of the fluid of weight density =. The force on one side of the plate is F œ ='cw (y)(j) dy œ =j ’ y# “ 0

#

! w

œ

=jw# #

. The

average force on one side of the plate is Fav œ œ

= w

#

’ y# “

! w

œ

=w #

. Therefore the force

= w

'c0w (y)dy

=jw# #

‰ œ ˆ =w # (jw) œ (the average pressure up and down) † (the area of the plate).


387

388


NOTES:


CHAPTER 7 TRANSCENDENTAL FUNCTIONS 7.1 INVERSE FUNCTIONS AND THEIR DERIVATIVES 1. Yes one-to-one, the graph passes the horizontal line test. 2. Not one-to-one, the graph fails the horizontal line test. 3. Not one-to-one since (for example) the horizontal line y œ 2 intersects the graph twice. 4. Not one-to-one, the graph fails the horizontal line test. 5. Yes one-to-one, the graph passes the horizontal line test 6. Yes one-to-one, the graph passes the horizontal line test 7. Not one-to-one since the horizontal line y œ 3 intersects the graph an infinite number of times. 8. Yes one-to-one, the graph passes the horizontal line test 9. Yes one-to-one, the graph passes the horizontal line test 10. Not one-to-one since (for example) the horizontal line y œ 1 intersects the graph twice. 11. Domain: 0 x Ÿ 1, Range: 0 Ÿ y

13. Domain: 1 Ÿ x Ÿ 1, Range: 1# Ÿ y Ÿ

12. Domain: x 1, Range: y 0

1 #

14. Domain: _ x _, Range: 1# y Ÿ


1 #

390

Chapter 7 Transcendental Functions

15. Domain: 0 Ÿ x Ÿ 6, Range: 0 Ÿ y Ÿ 3

16. Domain: 2 Ÿ x Ÿ 1, Range: 1 Ÿ y 3

17. The graph is symmetric about y œ x.

(b) y œ È1 x# Ê y# œ 1 x# Ê x# œ 1 y# Ê x œ È1 y# Ê y œ È1 x# œ f " (x) 18. The graph is symmetric about y œ x.

yœ

" x

Ê xœ

" y

Ê yœ

" x

œ f " (x)

19. Step 1: y œ x# 1 Ê x# œ y 1 Ê x œ Èy 1 Step 2: y œ Èx 1 œ f " (x) 20. Step 1: y œ x# Ê x œ Èy, since x Ÿ !. Step 2: y œ Èx œ f " (x) 21. Step 1: y œ x$ 1 Ê x$ œ y 1 Ê x œ (y 1)"Î$ Step 2: y œ $Èx 1 œ f " (x) 22. Step 1: y œ x# 2x 1 Ê y œ (x 1)# Ê Èy œ x 1, since x 1 Ê x œ 1 Èy Step 2: y œ 1 Èx œ f " (x) 23. Step 1: y œ (x 1)# Ê Èy œ x 1, since x 1 Ê x œ Èy 1 Step 2: y œ Èx 1 œ f " (x) 24. Step 1: y œ x#Î$ Ê x œ y$Î# Step 2: y œ x$Î# œ f " (x)


Section 7.1 Inverse Functions and Their Derivatives 25. Step 1: y œ x& Ê x œ y"Î& 5 Step 2: y œ È x œ f " (x); Domain and Range of f " : all reals; &

f af " (x)b œ ˆx"Î& ‰ œ x and f " (f(x)) œ ax& b

"Î&

œx

"Î%

œx

26. Step 1: y œ x% Ê x œ y"Î% Step 2: y œ %Èx œ f " (x); Domain of f " : x 0, Range of f " : y 0; %

f af " (x)b œ ˆx"Î% ‰ œ x and f " (f(x)) œ ax% b

27. Step 1: y œ x$ 1 Ê x$ œ y 1 Ê x œ (y 1)"Î$ Step 2: y œ $Èx 1 œ f " (x); Domain and Range of f " : all reals; $

f af " (x)b œ ˆ(x 1)"Î$ ‰ 1 œ (x 1) 1 œ x and f " (f(x)) œ aax$ 1b 1b 28. Step 1: y œ

" #

x

7 #

Ê "

" #

xœy

7 #

Step 2: y œ

" x#

Ê x# œ

" Èx

œ f " (x)

" y

Ê xœ

7 #

30. Step 1: y œ

" x$

" x"Î$ "

Step 2: y œ Domain of f

f af " axbb œ

" y

Ê xœ

Step 2: y

32. Step 1: y œ

œ

" Š "x ‹

œ x since x 0

" y"Î$

: x Á 0, Range of f " : y Á 0;

" $ ax "Î$ b

œ

" x "

œ x and f " afaxbb œ ˆ x"$ ‰

x3 x 2 Ê yax 2b 3 1 œ 2x axb; x1 œ f "

f af " axbb œ

" É x"#

3 " " œÉ (x); x œ f

31. Step 1: y œ Domain of f

œx

" Èy

Šx‹

Ê x$ œ

"Î$

œ x and f " (f(x)) œ 2 ˆ "# x 7# ‰ 7 œ (x 7) 7 œ x

Domain of f " : x 0, Range of f " : y 0; f af " (x)b œ "" # œ "" œ x and f " (f(x)) œ Š Èx ‹

œ ax$ b

Ê x œ 2y 7

Step 2: y œ 2x 7 œ f (x); Domain and Range of f " : all reals; f af " (x)b œ "# (2x 7) 7# œ ˆx 7# ‰ 29. Step 1: y œ

"Î$

"Î$

œ ˆ x" ‰

"

œx

œ x 3 Ê x y 2y œ x 3 Ê x y x œ 2y 3 Ê x œ

: x Á 1, Range of f " : y Á 2;

b3‰ ˆ 2x xc1 3 b3‰ ˆ 2x xc1 2 Èx Èx 3

œ

a2x 3b 3ax 1b a2x 3b 2ax 1b

œ

5x 5

œ x and f " afaxbb œ

3 2ˆ xx b c2‰ 3 3 ˆ xx b c2‰ 1

œ

2y 3 y1

2 ax 3 b 3 a x 2 b ax 3 b ax 2 b

œ

5x 5

œx

Ê yˆÈx 3‰ œ Èx Ê yÈx 3y œ Èx Ê yÈx Èx œ 3y Ê x œ Š y 3y 1‹

2

2

‰ œ f 1 a xb ; Step 2: y œ ˆ x 3x 1

Domain of f " : Ð_, 0Ó a1, _b, Range of f " : Ò0, 9Ñ a9, _b; f af " axbb œ f

"

2 Éˆ x 3x c1‰ 2

Éˆ x 3x c1‰ 3

afaxbb œ

; If x 1 or x Ÿ 0 Ê

Èx

3Š È x c 3 ‹

Èx

Š Èx c 3 ‹ 1

3x x1

0Ê

2 Éˆ x 3x c1‰ 2

Éˆ x 3x c1‰ 3

œ

3x xc1 3x xc1 3

œ

3x 3x 3ax 1b

œ

2

œ

9x ˆÈ x ˆÈ x 3 ‰‰ 2

œ

9x 9

œx


3x 3

œ x and

391

392


33. Step 1: y œ x2 2x, x Ÿ 1 Ê y 1 œ ax 1b2 , x Ÿ 1 Ê Èy 1 œ x 1, x Ÿ 1 Ê x œ 1 Èy 1 Step 2: y œ 1 Èx 1 œ f 1 axb; Domain of f " : Ò1, _Ñ, Range of f " : Ð_, 1Ó; 2

f af " axbb œ Š1 Èx 1‹ 2Š1 Èx 1‹ œ 1 2Èx 1 x 1 2 2Èx 1 œ x and f " afaxbb œ 1 Èax2 2xb 1, x Ÿ 1 œ 1 Éax 1b2 , x Ÿ 1 œ 1 lx 1l œ 1 a1 xb œ x 34. Step 1: y œ a2x3 1b 3 x Step 2: y œ É

5

1 2

1 Î5

Ê y5 œ 2x3 1 Ê y5 1 œ 2x3 Ê

y5 1 2

3 y œ x3 Ê x œ É

5

1 2

œ f 1 ax b ;

Domain of f " : a_, _b, Range of f " : a_, _b; 3 x f af " axbb œ Œ2ŠÉ

5

1 2 ‹

3

1 Î5

1

œ Š 2Š x

5

1 2 ‹

1‹

1 Î5

œ aax5 1b 1b

1 Î5

œ ax5 b

1 Î5

œ x and

5

1Î5 ’a2x3 1b “ 1

3

f " afaxbb œ Ê

2

3 a2x œÉ

35. (a) y œ 2x 3 Ê 2x œ y 3 Ê x œ y# 3# Ê f " (x) œ (c)

df ¸ dx xœ 1

36. (a) y œ

" 5

œ 2,

df c" dx ¹ xœ1

x7 Ê

" 5

œ

(c)

œ

" df c" 5 , dx ¹ xœ$%Î&

(c)

œ 4,

df c" dx ¹ xœ3

38. (a) y œ 2x# Ê x# œ Ê xœ (c)

df ¸ dx xœ&

" È2

" #

3

3 #

" #

"

(b)

(x) œ 5x 35

œ5

œ

(b) 5 4

x 4

" 4

(b)

y

Èy Ê f

3 2x œÉ 2 œ x

(b) x #

37. (a) y œ 5 4x Ê 4x œ 5 y Ê x œ 54 y4 Ê f " (x) œ df ¸ dx xœ1Î#

1 b 1 2

xœy7

Ê x œ 5y 35 Ê f df ¸ dx xœ 1

3

"

(x) œ

È x#

œ 4xk xœ5 œ 20,

df c" dx ¹ xœ&0

œ

" #È 2

x"Î# ¹

xœ50

œ

" #0


Section 7.1 Inverse Functions and Their Derivatives $

3 3 39. (a) f(g(x)) œ ˆÈ x‰ œ x, g(f(x)) œ È x3 œ x

w

#

w

(b)

w

(c) f (x) œ 3x Ê f (1) œ 3, f (1) œ 3; gw (x) œ 3" x#Î$ Ê gw (1) œ 3" , gw (1) œ

" 3

(d) The line y œ 0 is tangent to f(x) œ x$ at (!ß !); the line x œ 0 is tangent to g(x) œ $Èx at (0ß 0)

40. (a) h(k(x)) œ

" 4

ˆ(4x)"Î$ ‰$ œ x,

k(h(x)) œ Š4 † (c) hw (x) œ w

k (x) œ

x$ 4‹

"Î$

(b)

œx

3x# w w 4 Ê h (2) œ 3, h (2) 4 #Î$ Ê kw (2) œ "3 , 3 (4x)

œ 3; kw (2) œ

(d) The line y œ 0 is tangent to h(x) œ

x$ 4

" 3

at (!ß !);

the line x œ 0 is tangent to k(x) œ (4x)"Î$ at (!ß !) œ 3x# 6x Ê

41.

df dx

43.

df " dx ¹ x œ 4

df c" dx ¹ x œ f(3)

df " dx ¹ x œ f(2)

œ

45. (a) y œ mx Ê x œ

" m

œ

(b) The graph of y œ f 46. y œ mx b Ê x œ

y m

"

df dx

º

œ

xœ2

"

df dx

œ

º

xœ3

" ˆ 3" ‰

œ3

y Ê f " (x) œ "

" 9

œ

" m

œ 2x 4 Ê

42.

df dx

44.

dg " dx ¹x œ 0

b m

dg " dx ¹ x œ f(0)

œ

"

dg dx

º

œ

xœ0

"

df dx

º

œ

xœ5

œ

" 6

" 2

x

(x) is a line through the origin with slope

œ

df " dx ¹ x œ f(5)

Ê f " (x) œ

" m

x

b m;

" m.

the graph of f " (x) is a line with slope

" m

and y-intercept mb .

47. (a) y œ x 1 Ê x œ y 1 Ê f " (x) œ x 1 (b) y œ x b Ê x œ y b Ê f " (x) œ x b (c) Their graphs will be parallel to one another and lie on opposite sides of the line y œ x equidistant from that line.

48. (a) y œ x 1 Ê x œ y 1 Ê f " (x) œ 1 x; the lines intersect at a right angle (b) y œ x b Ê x œ y b Ê f " (x) œ b x; the lines intersect at a right angle (c) Such a function is its own inverse.


393

394


49. Let x" Á x# be two numbers in the domain of an increasing function f. Then, either x" x# or x" x# which implies f(x" ) f(x# ) or f(x" ) f(x# ), since f(x) is increasing. In either case, f(x" ) Á f(x# ) and f is one-to-one. Similar arguments hold if f is decreasing. 50. f(x) is increasing since x# x" Ê

" 3

x#

5 6

" 3

x" 56 ;

df dx

œ

" 3

51. f(x) is increasing since x# x" Ê 27x$# 27x"$ ; y œ 27x$ Ê x œ df dx

œ 81x# Ê

df " dx

œ

" ¸ 81x# 13 x"Î$

œ

" 9x#Î$

œ

" 9

df c" dx

Ê " 3

œ

df dx

œ 24x# Ê

dx

œ

" ¸ 24x# 12 Ð1 xÑ"Î$

œ

œ3

y"Î$ Ê f " (x) œ

" 3

x"Î$ ;

x#Î$

52. f(x) is decreasing since x# x" Ê 1 8x$# 1 8x"$ ; y œ 1 8x$ Ê x œ df c"

" ˆ "3 ‰

" 6(" x)#Î$

" #

(1 y)"Î$ Ê f " (x) œ

" #

(1 x)"Î$ ;

œ "6 (1 x)#Î$

53. f(x) is decreasing since x# x" Ê (1 x# )$ (1 x" )$ ; y œ (1 x)$ Ê x œ 1 y"Î$ Ê f " (x) œ 1 x"Î$ ; df dx

œ 3(1 x)# Ê

df c" dx

œ

" 3(1 x)# ¹ 1cx"Î$ &Î$

54. f(x) is increasing since x# x" Ê x# df dx

œ

5 3

x#Î$ Ê

df c" dx

œ

5 3

" ¹ x#Î$ x$Î&

œ

" 3x#Î$

œ "3 x#Î$

&Î$

x" ; y œ x&Î$ Ê x œ y$Î& Ê f " (x) œ x$Î& ; œ

3 5x#Î&

œ

3 5

x#Î&

55. The function g(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then f(x" ) Á f(x# ), so f(x" ) Á f(x# ) and therefore g(x" ) Á g(x# ). Therefore g(x) is one-to-one as well. 56. The function h(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then f(x" ) Á f(x# ), so " " f(x" ) Á f(x# ) , and therefore h(x" ) Á h(x# ). 57. The composite is one-to-one also. The reasoning: If x" Á x# then g(x" ) Á g(x# ) because g is one-to-one. Since g(x" ) Á g(x# ), we also have f(g(x" )) Á f(g(x# )) because f is one-to-one. Thus, f ‰ g is one-to-one because x" Á x# Ê f(g(x" )) Á f(g(x# )). 58. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers x" Á x# in the domain of g with g(x" ) œ g(x# ). For these numbers we would also have f(g(x" )) œ f(g(x# )), contradicting the assumption that f ‰ g is one-to-one. 59. (g ‰ f)(x) œ x Ê g(f(x)) œ x Ê gw (f(x))f w (x) œ 1 60. W(a) œ 'f(a) 1 ’af " (y)b a# “ dy œ 0 œ 'a 21x[f(a) f(x)] dx œ S(a); Ww (t) œ 1’af " (f(t))b a# “ f w (t) f(a)

a

#

#

œ 1 at# a# b f w (t); also S(t) œ 21f(t)'a x dx 21'a xf(x) dx œ c1f(t)t# 1f(t)a# d 21'a xf(x) dx Ê Sw (t) t

t

t

œ 1t# f w (t) 21tf(t) 1a# f w (t) 21tf(t) œ 1 at# a# b f w (t) Ê Ww (t) œ Sw (t). Therefore, W(t) œ S(t) for all t − [aß b]. 61-68. Example CAS commands: Maple: with( plots );#63 f := x -> sqrt(3*x-2); domain := 2/3 .. 4; x0 := 3; Df := D(f);

# (a)


Section 7.1 Inverse Functions and Their Derivatives

395

plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"], title="#61(a) (Section 7.1)" ); q1 := solve( y=f(x), x ); # (b) g := unapply( q1, y ); m1 := Df(x0); # (c) t1 := f(x0)+m1*(x-x0); y=t1; m2 := 1/Df(x0); # (d) t2 := g(f(x0)) + m2*(x-f(x0)); y=t2; domaing := map(f,domain); # (e) p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ): p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ): p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ): p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ): p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ): display( [p1,p2,p3,p4,p5], scaling=constrained, title="#63(e) (Section 7.1)" ); Mathematica: (assigned function and values for a, b, and x0 may vary) If a function requires the odd root of a negative number, begin by loading the RealOnly package that allows Mathematica to do this. See section 2.5 for details. (x 1) œ '0 tx e t dt œ lim ctx e t d b0 x '0 tx 1 e t dt œ lim ˆ beb 0x e! ‰ x>(x) œ x>(x) bÄ_

(c) >(n 1) œ n>(n) œ n!: n œ 0: >(0 1) œ >(1) œ 0!; n œ k: Assume >(k 1) œ k! n œ k 1: >(k 1 1) œ (k 1) >(k 1) œ (k 1)k! œ (k 1)! Thus, >(n 1) œ n>(n) œ n! for every positive integer n.

x

bÄ_

for some k 0; from part (b) induction hypothesis definition of factorial

x n n 32. (a) >(x) ¸ ˆ xe ‰ É 2x1 and n>(n) œ n! Ê n! ¸ n ˆ ne ‰ É 2n1 œ ˆ ne ‰ È2n1

(b)

533

n 10 20 30 40 50 60

ˆ ne ‰n È2n1 3598695.619 2.4227868 ‚ 10") 2.6451710 ‚ 10$# 8.1421726 ‚ 10%( 3.0363446 ‚ 10'% 8.3094383 ‚ 10)"

calculator 3628800 2.432902 ‚ 10") 2.652528 ‚ 10$# 8.1591528 ‚ 10%( 3.0414093 ‚ 10'% 8.3209871 ‚ 10)"


534

Chapter 8 Techniques of Integration

(c)

ˆ ne ‰n È2n1 3598695.619

n 10 ÐÑ

cos 3x

2e2x

ÐÑ

1 3 sin

4e2x

ÐÑ

19 cos 3x

33. e2x

Iœ

e2x 3

sin 3x

2e2x 9

13 9

Iœ

e2x 9

(3 sin 3x 2 cos 3x) Ê I œ

e2x 13

(3 sin 3x 2 cos 3x) C

sin 4x

3e3x

ÐÑ

4" cos 4x

9e3x

ÐÑ

" 16 sin 4x

3x

I œ e4 cos 4x 35.

calculator 3628800

3x

cos 3x 49 I Ê

ÐÑ

34. e3x

ˆ ne ‰n È2n1 e1Î12n 3628810.051

3e3x 16

sin 4x

I Ê

9 16

sin 3x

ÐÑ

sin x

3 cos 3x

ÐÑ

cos x

9 sin 3x

ÐÑ

sin x

Iœ

25 16

e3x 16

(3 sin 4x 4 cos 4x) Ê I œ

e3x 25

(3 sin 4x 4 cos 4x) C

I œ sin 3x cos x 3 cos 3x sin x 9I Ê 8I œ sin 3x cos x 3 cos 3x sin x Ê I œ sin 3x cos x83 cos 3x sin x C 36.

cos 5x

ÐÑ

sin 4x

sin 5x

ÐÑ

"4 cos 4x

25cos 5x

ÐÑ

" 16 sin 4

I œ "4 cos 5x cos 4x Ê Iœ

" 9

sin 5x sin 4x

9 I Ê 16 I œ "4 cos 5x cos 4x

5 16

sin 5x sin 4x

sin bx

aeax

ÐÑ

"b cos bx

a# eax

ÐÑ

b"# sin bx

ax

I œ eb cos bx Ê Iœ

25 16

(4 cos 5x cos 4x 5 sin 5x sin 4x) C ÐÑ

37. eax

5 16

eax a# b#

aeax b#

sin bx

a# b#

I Ê Ša

#

b# b# ‹ I

œ

eax b#

(a sin bx b cos bx)

(a sin bx b cos bx) C


Chapter 8 Additional and Advanced Exercises ÐÑ

38. eax

cos bx

aeax

ÐÑ

" b

a# eax

ÐÑ

b"# cos bx

Iœ

eax b

sin bx

Ê Iœ

" x

cos bx

a# b#

I Ê Ša

#

b# b# ‹ I

eax b#

œ

(a cos bx b sin bx)

(a cos bx b sin bx) C

eax a# b#

39. ln (ax)

aeax b#

sin bx

ÐÑ

1

ÐÑ

x

I œ x ln (ax) ' ˆ "x ‰ x dx œ x ln (ax) x C 40. ln (ax) " x

Iœ

" 3

ÐÑ

x#

ÐÑ

" 3

x$

x$ ln (ax) ' ˆ "x ‰ Š x3 ‹ dx œ $

41.

' 1 dxsin x œ '

42.

' 1 sin dxx cos x œ '

œ'

2 dz ‹ 1 z# 1 Š 2z # ‹ 1 z

Š

2 dz (1 z)#

" 3

x$ ln (ax) 9" x$ C

œ

2 1z

œ'

2 dz ‹ 1 z# 2z 1 z# 1 Š ‹ 1 z# 1 z#

Š

Cœ

2 1 tan ˆ #x ‰

2 dz 1 z# 2z 1 z#

C

œ'

œ ln k1 zk C

dz 1z

œ ln ¸tan ˆ x# ‰ 1¸ C 43.

'01Î2 1 dxsin x œ '01

44.

'11ÎÎ32

45.

) '01Î2 2 dcos '1 ) œ 0

œ

46.

dx 1 cos x

1 3È 3

'12Î12Î3

œ

" È2

1

Š

2 dz (1 z)#

œ '1ÎÈ3

2 dz ‹ 1 z# 1 z# 1Š ‹ 1 z#

Š

2 dz ‹ 1 z# 1 z# 2Š ‹ 1 z#

cos ) d) sin ) cos ) sin )

È$ z# “ 4 "

' sin t dt cos t œ ' œ

'11ÎÈ3

œ '0

1

œ '0

1

"

œ 1 2 z ‘ ! œ (1 2) œ 1

dz z#

" œ 1z ‘ "ÎÈ$ œ È3 1

2 dz 2 2z# 1 z#

œ '0

1

œ

2 dz z# 3

2 È3

’tan"

z È3 “

" !

œ

2 È3

tan"

" È3

È 31 9

œ ’ "# ln z

47.

œ

2 dz ‹ 1 z# 2z 1Š ‹ 1 z#

Š

È3

œ '1

È3

z# ‹ Š 2 dz# ‹ 1 z# 1z

Š1 Ô 2z Š1

z# ‹

× 2z # Š 1 z# ‹Ø Õ Š1 z# ‹

œ '1

œ Š "# ln È3 43 ‹ ˆ0 4" ‰ œ

2 dz ‹ 1 z# 2z 1 z# Š ‹ 1 z# 1 z#

Š

œ'

2 dz 2z 1 z#

œ'

2 a1 z# b dz 2z 2z$ 2z 2z$

ln 3 4

1 z# 2z

œ

" 4

œ

" È2

1 2 ln ¹ zz ¹C 1 È2

(ln 3 2) œ

" #

dz

" #

2 dz (z 1)# 2

È3

œ '1

Šln È3 1‹

È

tan ˆ t ‰ 1 È2

ln º tan ˆ #t ‰ 1 È2 º C #


535

536 48.

Chapter 8 Techniques of Integration

t dt ' 1cos cos 'Š t œ

œ'

49.

a1 z# b dz a1 z # b z #

1 z# ‹ Š 2 dz# ‹ 1 z# 1 z # 1 Š 1 z# ‹ 1z

œ'

dz z # a1 z # b

œ' '

' sec ) d) œ ' cosd) ) œ ' Š

2 dz ‹ 1 z# # Š 1 z# ‹ 1 z

2 a1 z# b dz a1 z # b # a 1 z # b a 1 z # b dz 1 z#

œ'

dz z#

œ'

2 dz 1 z#

œ'

œ ln k1 zk ln k1 zk C œ ln »

50.

' csc ) d) œ ' sind)) œ ' Š

2 dz ‹ 1 z# 2z Š ‹ 1 z#

1 tan Š )# ‹

2'

1 tan Š )# ‹ »

œ'

dz z

œ'

dz z# 1

2 dz (1 z)(1 z)

2 a1 z# b dz a1 z # b a 1 z # 1 z # b

œ "z 2 tan" z C œ cot ˆ #t ‰ t C œ'

dz 1z

'

dz 1z

C

œ ln kzk C œ ln ¸tan #) ¸ C


CHAPTER 9 FIRST-ORDER DIFFERENTIAL EQUATIONS 9.1 SOLUTIONS, SLOPE FIELDS AND EULER'S METHOD 1. y w œ x y Ê slope of 0 for the line y œ x. For x, y 0, y w œ x y Ê slope 0 in Quadrant I. For x, y 0, y w œ x y Ê slope 0 in Quadrant III. For kyk kxk, y 0, x 0, y w œ x y Ê slope 0 in Quadrant II above y œ x. For kyk kxk, y 0, x 0, y w œ x y Ê slope 0 in Quadrant II below y œ x. For kyk kxk, x 0, y 0, y w œ x y Ê slope 0 in Quadrant IV above y œ x. For kyk kxk, x 0, y 0, y w œ x y Ê slope 0 in Quadrant IV below y œ x. All of the conditions are seen in slope field (d). 2. y w œ y 1 Ê slope is constant for a given value of y, slope is 0 for y œ 1, slope is positive for y 1 and negative for y 1. These characteristics are evident in slope field (c).

3. y w œ xy Ê slope œ 1 on y œ x and 1 on y œ x. y w œ xy Ê slope œ 0 on the y-axis, excluding a0, 0b, and is undefined on the x-axis. Slopes are positive for x 0, y 0 and x 0, y 0 (Quadrants II and IV), otherwise negative. Field (a) is consistent with these conditions.

4. y w œ y2 x2 Ê slope is 0 for y œ x and for y œ x. For kyk kxk slope is positive and for kyk kxk slope is negative. Field (b) has these characteristics.


538

Chapter 9 First-Order Differential Equations

5.

6.

7. y œ 1 '1 at yatbbdt Ê x

8. y œ '1 1t dt Ê x

dy dx

dy dx

œ x yaxb; ya1b œ 1 '1 at yatbbdt œ 1; 1

œ x1 ; ya1b œ '1 1t dt œ 0; 1

9. y œ 2 '0 a1 yatbbsin t dt Ê x

dy dx

dy dx

dy dx

œ x y, ya1b œ 1

œ x1 , ya1b œ 0

œ a1 yaxbbsin x; ya0b œ 2 '0 a1 yatbbsin t dt œ 2; 0

dy dx

œ a1 ybsin x,

ya0b œ 2 10. y œ 1 '0 yatb dt Ê x

dy dx

œ yaxb; ya0b œ 1 '0 yatb dt œ 1; 0

1 ‰ # (.5)

11. y" œ y! Š1

y! x! ‹

dx œ 1 ˆ1

y# œ y " Š 1

y" x" ‹

dx œ 0.25 ˆ1

y$ œ y # Š 1

y# x# ‹

dx œ 0.3 ˆ1

dy dx

ˆ x" ‰ y œ 1 Ê P(x) œ

Ê yœ

" x

" x

' x † 1 dx œ x" Š x#

Ê y(3.5) œ

#

3.5 #

4 3.5

œ

4.25 7

, Q(x) œ 1 Ê

œ y, ya0b œ 1

œ 0.25,

0.25 ‰ (.5) 2.5

0.3 ‰ 3 (.5)

dy dx

œ 0.3,

œ 0.75;

' P(x) dx œ ' x" dx œ ln kxk œ ln x, x 0 Ê v(x) œ eln x œ x

C‹ ; x œ 2, y œ 1 Ê 1 œ 1

C 2

Ê C œ 4 Ê y œ

x #

4 x

¸ 0.6071

12. y" œ y! x! (1 y! ) dx œ 0 1(1 0)(.2) œ .2, y# œ y" x" (1 y" ) dx œ .2 1.2(1 .2)(.2) œ .392, y$ œ y# x# (1 y# ) dx œ .392 1.4(1 .392)(.2) œ .5622; dy 1 y

œ x dx Ê ln k1 yk œ

x# #

C; x œ 1, y œ 0 Ê ln 1 œ

" #

#

C Ê C œ #" Ê ln k1 yk œ x#

" #

#

Ê y œ 1 ea1x bÎ2 Ê y(1.6) ¸ .5416 13. y" œ y! (2x! y! 2y! ) dx œ 3 [2(0)(3) 2(3)](.2) œ 4.2, y# œ y" (2x" y" 2y" ) dx œ 4.2 [2(.2)(4.2) 2(4.2)](.2) œ 6.216, y$ œ y# (2x# y# 2y# ) dx œ 6.216 [2(.4)(6.216) 2(6.216)](.2) œ 9.6969; dy dx

œ 2y(x 1) Ê

dy y

œ 2(x 1) dx Ê ln kyk œ (x 1)# C; x œ 0, y œ 3 Ê ln 3 œ 1 C Ê C œ ln 3 1

Ê ln y œ (x 1)# ln 3 1 Ê y œ eÐx1Ñ ln 31 œ eln 3 ex 2x œ 3exÐx2Ñ Ê y(.6) ¸ 14.2765 #

#

14. y" œ y! y#! (1 2x! ) dx œ 1 1# [1 2(1)](.5) œ .5, y# œ y" y#" (1 2x" ) dx œ .5 (.5)# [1 2(.5)](.5) œ .5, y$ œ y# y## (1 2x# ) dx œ .5 (.5)# [1 2(0)](.5) œ .625; dy y#

œ (1 2x) dx Ê y" œ x x# C; x œ 1, y œ 1 Ê 1 œ 1 (1)# C Ê C œ 1 Ê

Ê yœ

" 1 x x#

Ê y(.5) œ

" 1 .5 (.5)#

œ4


" y

œ 1 x x#

Section 9.1 Solutins, Slope Fields and Euler's Method

539

#

15. y" œ y! 2x! ex! dx œ 2 2(0)(.1) œ 2, # # y# œ y" 2x" ex" dx œ 2 2(.1) eÞ1 (.1) œ 2.0202, # # y$ œ y# 2x# ex# dx œ 2.0202 2(.2) eÞ2 (.1) œ 2.0618, # # # # dy œ 2xex dx Ê y œ ex C; y(0) œ 2 Ê 2 œ 1 C Ê C œ 1 Ê y œ ex 1 Ê y(.3) œ eÞ3 1 ¸ 2.0942 16. y" œ y! ay! ex! b dx œ 2 a2 † e0 b (.5) œ 3, y2 œ y1 ay1 ex1 b dx œ 3 a3 † e0.5 b (.5) œ 5.47308, y3 œ y2 ay2 ex2 b dx œ 5.47308 a5.47308 † e1.0 b (.5) œ 12.9118, dy dx

œ y ex Ê

Ê y œ 2ee 17. y" y# y$ y% y& dy y

x

œ ex dx Ê lnlyl œ ex C; x œ 0, y œ 2 Ê ln 2 œ 1 C Ê C œ ln 2 1 Ê lnlyl œ ex ln 2 1

dy y

"

Ê ya1.5b œ 2ee

1.5

"

¸ 65.0292

œ 1 1(.2) œ 1.2, œ 1.2 (1.2)(.2) œ 1.44, œ 1.44 (1.44)(.2) œ 1.728, œ 1.728 (1.728)(.2) œ 2.0736, œ 2.0736 (2.0736)(.2) œ 2.48832; œ dx Ê ln y œ x C" Ê y œ Cex ; y(0) œ 1 Ê 1 œ Ce! Ê C œ 1 Ê y œ ex Ê y(1) œ e ¸ 2.7183

18. y" œ 2 ˆ 21 ‰ (.2) œ 2.4, ‰ y# œ 2.4 ˆ 2.4 1.2 (.2) œ 2.8, ‰ y$ œ 2.8 ˆ 2.8 1.4 (.2) œ 3.2, 3.2 y% œ 3.2 ˆ 1.6 ‰ (.2) œ 3.6, ‰ y& œ 3.6 ˆ 3.6 1.8 (.2) œ 4; dy y

œ

dx x

Ê ln y œ ln x C Ê y œ kx; y(1) œ 2 Ê 2 œ k Ê y œ 2x Ê y(2) œ 4 #

19. y" œ 1 ’ (È1) “ (.5) œ .5, 1 #

.5) y# œ .5 ’ (È “ (.5) œ .39794, 1.5 #

y$ œ .39794 ’ (.39794) “ (.5) œ .34195, È2 #

y% œ .34195 ’ (.34195) È2.5 “ (.5) œ .30497, y& œ .27812, y' œ .25745, y( œ .24088, y) œ .2272; dy " dx È y# œ Èx Ê y œ 2 x C; y(1) œ 1 Ê 1 œ 2 C Ê C œ 1 Ê y œ

" 1 #È x

Ê y(5) œ

" 1 #È 5

20. y" œ 1 a0 † sin 1b ˆ "3 ‰ œ 1, y# œ 1 ˆ "3 † sin 1‰ ˆ 3" ‰ œ 1.09350, y$ œ 1.09350 ˆ 23 † sin 1.09350‰ ˆ 3" ‰ œ 1.29089, y% œ 1.29089 ˆ 33 † sin 1.29089‰ ˆ 3" ‰ œ 1.61125, y& œ 1.61125 ˆ 43 † sin 1.61125‰ ˆ 3" ‰ œ 2.05533,

y' œ 2.05533 ˆ 53 † sin 2.05533‰ ˆ 3" ‰ œ 2.54694; y w œ x sin y Ê csc y dy œ x dx Ê lnlcsc y cot yl œ 12 x2 C Ê csc y cot y œ e 2 x C œ Ce 2 x 1 2

Ê

1 cos y sin y

1 2

2 2 2 œ Ce 2 x Ê cotˆ 2y ‰ œ Ce 2 x ; ya0b œ 1 Ê cotˆ 21 ‰ œ Ce0 œ C Ê cotˆ 2y ‰ œ cotˆ 21 ‰e 2 x 1

1

2 Ê y œ 2 cot1 Šcotˆ 12 ‰e 2 x ‹, ya2b œ 2 cot1 ˆcotˆ 12 ‰e2 ‰ œ 2.65591 1


1

¸ .2880

540

Chapter 9 First-Order Differential Equations

21. y œ 1 x a1 x0 y0 bex x0 Ê yax0 b œ 1 x0 a1 x0 y0 bex0 x0 œ 1 x0 a1 x0 y0 ba1b œ y0 dy dx

œ 1 a1 x0 y0 bex x0 Ê y œ 1 x a1 x0 y0 bex x0 œ

dy dx

xÊ

dy dx

œ xy

22. y w œ faxb, yax0 b œ y0 Ê y œ 'x fatbdt C, yax0 b œ 'x fatbdt C œ C Ê C œ y0 Ê y œ 'x fatbdt y0 x

x0

x

0

0

0

23-34. Example CAS commands: Maple: ode := diff( y(x), x ) = y(x); icA := [0, 1]; icB := [0, 2]; icC := [0,-1]; DEplot( ode, y(x), x=0..2, [icA,icB,icC], arrows=slim, linecolor=blue, title="#23 (Section 9.1)" ); Mathematica: To plot vector fields, you must begin by loading a graphics package.

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