Instructor's Solutions Manual for Thomas' Calculus, Eleventh Edition

to accompany

T ^HOMAS ’ C ^ALCULUS

E ^LEVENTH E ^DITION

B ASED ON THE O ^RIGINAL W ^{ORK BY}

George B. Thomas, Jr.

Massachusetts Institute of Technology

AS R ^{EVISED BY}

Maurice D. Weir

Naval Postgraduate School

J oel Hass

University of California, Davis

Frank R. Giordano

Naval Postgraduate School

I ^NSTRUCTOR ’ ^S

S ^OLUTIONS M ^ANUAL

P ^ART O ^NE

A RDIS • B ORZELLINO • B UCHANAN • M OGILL • N ELSON

Reproduced by Pearson Addison-Wesley from electronic files supplied by the authors.

Copyright © 2005 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 0-321-22653-4 1 2 3 4 5 6 BB 07 06 05 04

This Instructor's Solutions Manual contains the solutions to every exercise in the 11th Edition of THOMAS' CALCULUS by Maurice Weir, Joel Hass and Frank Giordano, 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).

Acknowledgments Solutions Writers

William Ardis, Collin County Community College-Preston Ridge Campus Joseph Borzellino, California Polytechnic State University

Linda Buchanana, Howard College Tim Mogill

Patricia Nelson, University of Wisconsin-La Crosse Accuracy Checkers

Karl Kattchee, University of Wisconsin-La Crosse Marie Vanisko, California State University, Stanislaus Tom Weigleitner, VISTA Information Technologies

Thanks to Rachel Reeve, Christine O'Brien, Sheila Spinney, Elka Block, and Joe Vetere for all their guidance and help at every step.

1.1 REAL NUMBERS AND THE REAL LINE

1. Executing long division, 0.1, 0.2, 0.3, 0.8, 0.9^"₉ œ ₉² œ ₉³ œ ₉⁸ œ ₉⁹ œ

2. Executing long division, 0.09, 0.18, 0.27, 0.81, 0.99₁₁^" œ ₁₁² œ ₁₁³ œ ₁₁⁹ œ ₁₁¹¹ œ 3. NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6.

a) NNT. 5 is a counter example.

b) NT. 2 < x < 6 Ê 2 2 < x2 < 6 Ê2 0 < x2 < 2.

c) NT. 2 < x < 6 Ê2/2 < x/2 < 6/2 Ê1 < x < 3.

d) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê1/6 < 1/x < 1/2.

e) NT. 2 < x < 6 Ê1/2 > 1/x > 1/6 Ê1/6 < 1/x < 1/2Ê6(1/6) < 6(1/x) < 6(1/2)Ê1 < 6/x < 3.

f) NT. 2 < x < 6 Ê x < 6Ê(x4) < 2 and 2 < x < 6 Êx > 2 Ê x < 2 Ê x + 4 < 2 Ê (x 4) < 2.

The pair of inequalities (x4) < 2 and (x 4) < 2 Ê | x4 | < 2.

g) NT. 2 < x < 6 Ê 2 > x > 6 Ê 6 < x < 2. But 2 < 2. So 6 < x < 2 < 2 or 6 < x < 2.

h) NT. 2 < x < 6 Ê 1(2) > 1(x) < 1(6) Ê 6 < x < 2

4. NT = necessarily true, NNT = Not necessarily true. Given: 1 < y 5 < 1.

a) NT. 1 < y 5 < 1 Ê 1 + 5 < y5 + 5 < 1 + 5 Ê4 < y < 6.

b) NNT. y = 5 is a counter example. (Actually, never true given that 4 y 6) c) NT. From a), 1 < y 5 < 1, Ê4 < y < 6Êy > 4.

d) NT. From a), 1 < y 5 < 1, Ê4 < y < 6Êy < 6.

e) NT. 1 < y 5 < 1 Ê 1 + 1 < y5 + 1 < 1 + 1 Ê0 < y4 < 2.

f) NT. 1 < y 5 < 1 Ê(1/2)( 1 + 5) < (1/2)(y 5 + 5) < (1/2)(1 + 5) Ê2 < y/2 < 3.

g) NT. From a), 4 < y < 6 Ê1/4 > 1/y > 1/6 Ê1/6 < 1/y < 1/4.

h) NT. 1 < y 5 < 1 Ê y 5 > 1 Ê y > 4 Ê y < 4 Ê y + 5 < 1 Ê (y 5) < 1.

Also, 1 < y 5 < 1 Ê y 5 < 1. The pair of inequalities (y 5) < 1 and (y5) < 1 Ê | y5 | < 1.

5. 2x 4 Ê x 2

6. 8 3x 5 3x 3 x 1 x

  Ê   Ê Ÿ ïïïïïïïïïñqqqqqqqqp1 7. 5x $ Ÿ ( 3x Ê 8xŸ10 Ê xŸ⁵₄

8. 3(2x)2(3x) Ê 6 3x 6 2x

0 5x 0 x x

Ê Ê ïïïïïïïïïðqqqqqqqqp0 9. 2x  ^"_# 7x⁷₆ Ê  ^"_# ⁷₆ 5x

x or x

Ê ^"₅ˆ¹⁰₆‰   ^"₃

10. ^{6 x}₄ ^{3x 4}₂ Ê 122x12x16

28 14x 2 x x

Ê Ê qqqqqqqqqðïïïïïïïïî2

11. (x⁴₅ 2)₃^"(x6) Ê 12(x2)5(x6) 12x 24 5x 30 7x 6 or x

Ê Ê ⁶₇

12. ^{x 5}₂ Ÿ^{12 3x}₄ Ê (4x20)Ÿ246x

44 10x x x

Ê Ÿ Ê ²²⁵ Ÿ qqqqqqqqqñïïïïïïïïî22/5 13. yœ3 or yœ 3

14. y œ3 7 or y œ Ê3 7 yœ10 or yœ 4

15. 2t œ5 4 or 2t & œ Ê4 2tœ 1 or 2tœ Ê œ 9 t ^"_# or tœ _#⁹ 16. 1 œt 1 or 1 œ Ê œ !t 1 t or t œ Ê œ2 t 0 or tœ2

17. 83sœ ⁹₂ or 83sœ Ê œ ⁹_# 3s ⁷_# or 3s œ ²⁵_# Ê œ s ⁷₆ or sœ²⁵₆ 18. ^s_# œ1 1 or ^s_# œ Ê1 1 _#^s œ2 or ^s_# œ ! Ê œ s 4 or sœ0

19. 2 x 2; solution interval ( 2 2) ß

20. 2 x 2; solution interval [ 2 2] x

2 2

Ÿ Ÿ ß qqqqñïïïïïïïïñqqqqp

21. Ÿ Ÿ3 t 1 3 Ê Ÿ Ÿ 2 t 4; solution interval [ 2 4] ß 22. 1 t 2 1 3Ê t 1;

solution interval ( 3 1) t

3 1

ß qqqqðïïïïïïïïðqqqqp

23. % 3y 7 4 Ê 33y11 Ê 1 y ¹¹₃ ; solution interval ˆ1ß¹¹₃‰

24. 1 2y " Ê 5 6 2y Ê 4 3 y 2;

solution interval ( 3 2) y

3 2

ß qqqqðïïïïïïïïðqqqqp

25. Ÿ Ÿ1 ^z₅ 1 1 Ê 0Ÿ Ÿ^z₅ 2 Ê 0Ÿ Ÿz 10;

solution interval [0 10] ß

26. Ÿ2 ^3z_# Ÿ1 2 Ê Ÿ1 ^3z_# Ÿ3 Ê Ÿ Ÿ²₃ z 2;

solution interval 2 z

2/3 2

ß ‘ qqqqñïïïïïïïïñqqqqp

2 3

27. ^"_# 3 ^"_x ^"_# Ê Ê_#^{7 "}_{x #}⁵ _#^{7 "}_{x #}⁵ x ; solution interval

Ê ²₇ ²₅ ˆ²₇ß²₅‰

28. 3 ²_x 4 3 Ê 1 ( ʲ_x 1 ^x_{# 7}^"

2 x x 2; solution interval 2 x

2/7 2

Ê Ê ²₇ ²₇ ˆ²₇ß ‰ qqqqðïïïïïïïïðqqqqp

29. 2s 4 or 2s  4 Ê   s 2 or sŸ 2;

solution intervals (_ß ß _2] [2 )

30. s  3 ^"_# or (s 3) ^"_# Ê   s _#⁵ or s   _#⁷

s or s ;

Ê   ⁵_# Ÿ ⁷_#

solution intervals s

7/2 5/2

‘ ‰

ˆ_ß ß _^{7# 5#} ïïïïïïñqqqqqqñïïïïïïî

31. 1 x 1 or ( " x)1 Ê x 0 or x2 x 0 or x 2; solution intervals ( ) (2 )

Ê _ß ! ß _

32. 23x5 or (2 3x)5 Ê 3x 3 or 3x7

x 1 or x ;

Ê ⁷₃

solution intervals ( 1) x

1 7/3

_ß ß _ ïïïïïïðqqqqqqðïïïïïïî ˆ⁷₃ ‰

33. ^r"_#  1 or ˆ^{r 1}_# ‰ 1 Ê   r 1 2 or r Ÿ 1 2

r 1 or r 3; solution intervals ( 3] [1 ) Ê   Ÿ _ß ß _ 34. ^3r₅ " ²₅ or ˆ^3r₅ " ‰ ²₅

or r or r 1

Ê ^3r₅ ⁷₅ Ê ^3r₅ ³₅ ⁷₃

solution intervals ( ) r

1 7/3

_ß " ß _ ïïïïïïðqqqqqqðïïïïïïî ˆ⁷₃ ‰

35. x^# # Ê k kx È2 2Ê È x È2 ;

solution interval Š È È2 2 ‹ x

È È

ß qqqqqqðïïïïïïðqqqqqqp # #

36. 4Ÿx ^# Ê 2Ÿk kx Ê x 2 or xŸ 2;

solution interval ( 2] [2 ) r

2 2

_ß ß _ ïïïïïïñqqqqqqñïïïïïïî 37. 4x^#9 Ê 2k kx 3 Ê 2 x 3 or 2 x 3

2 x 3 or 3 x 2;

Ê

solution intervals ( 3 2) (2 3) x

3 2 2 3

ß ß qqqqðïïïïðqqqqðïïïïðqqqp

38. x^"₉ ^{# "}₄ xÊ ^"₃ k k^"_# xÊ ^"₃ ^"_# or ^"₃ x ^"_#

x or x ;

Ê ^"₃ ^"_# ^"_# ^"₃

solution intervals x

1/2 1/3 1/3 1/2

ˆ ß ‰ ˆ ß ‰ qqqqðïïïïðqqqqðïïïïðqqqp

" " " "

# 3 3 #

39. (x1)^#4 Ê kx 1k 2 Ê 2 x 1 2

1 x 3; solution interval ( ) x

1 3

Ê "ß $ qqqqqqðïïïïïïïïðqqqqp

40. (x3)^# # Ê xk 3k È2

2 x 3 2 or 3 2 x 3 2 ;

Ê È È È È

solution interval 3 2 3 2 x

3 3

Š È È ‹

È È

ß qqqqqqðïïïïïïïïðqqqqp # #

41. x^# x 0 Ê x^#x + < ¹₄ ¹₄ Ê xˆ ¹₂‰² < ¹₄ ʹ x ¹₂ < ¹ ¹₂ Ê ¹₂ < x ¹₂ < ¹₂ Ê 0 < x < 1.

So the solution is the interval (0 1)ß

42. x^#  x 2 0 Ê x^#x + ¹₄   ⁹₄ Ê x¹ ¹₂ ¹   ³₂ Ê x ¹₂   ³₂ or ˆx ¹₂‰   ³₂ Ê x   2 or x Ÿ 1.

The solution interval is (_ß ß _1] [2 ) 43. True if a 0; False if a0.

44. kx œ Í 1k 1 x k (x 1)kœ Í  1 x 1 x 0 Í xŸ1 45. (1) ak œ bk (a b) or ak œ bk (a b);

both squared equal (ab)^# (2) abŸk kab œk k k ka b

(3) ak kœa or ak kœ a, so ak k^#œa ; likewise, b^# k k^# œb^#

(4) x^# Ÿy implies ^# Èx^#ŸÈy or x^# Ÿy for all nonnegative real numbers x and y. Let xœ ka b andk yœk k k ka b so that ak bk^#Ÿak k k ka bb^# Ê ak Ÿbk k k k ka b .

46. If a 0 and b 0, then ab 0 and abk kœabœk k k ka b .

If a0 and b0, then ab0 and abk kœabœ œ( a)( b) k k k ka b . If a 0 and b0, then abŸ0 and abk kœ (ab)œ(a)( b) œk k k ka b . If a0 and b 0, then abŸ0 and abk kœ (ab)œ ( a)(b)œk k k ka b . 47. Ÿ Ÿ3 x 3 and x Ê Ÿ^"_# ^"_# x 3.

48. Graph of xk k k k y Ÿ1 is the interior of “diamond-shaped" region.

49. Let be a real number > 0 and f(x) = 2x + 1. Suppose that | x 1 | < . Then | x 1 | < $ $ $ Ê 2| x 1 | < 2 $ Ê

| 2x | < 2 # $ Ê| (2x + 1) 3 | < 2 $ Ê| f(x) f(1) | < 2$

50. Let > 0 be any positive number and f(x) = 2x + 3. Suppose that | x 0 | < /2. Then 2| x 0 | < and% % %

| 2x + 3 3 | < . But f(x) = 2x + 3 and f(0) = 3. Thus | f(x) f(0) | < . % % 51. Consider: i) a > 0; ii) a < 0; iii) a = 0.

i) For a > 0, | a | œ a by definition. Now, a > 0 Ê a < 0. Let a = b. By definition, | b | œ b. Since b = a,

| a | œ ( a) œ a and | a | œ | a | œ a.

ii) For a < 0, | a | œ a. Now, a < 0Ê a > 0. Let a œ b. By definition, | b | œ b and thus | a| œ a. So again

| a | œ | a|.

iii) By definition | 0 | œ 0 and since 0 œ 0, | 0 | œ 0. Thus, by i), ii), and iii) | a | œ | a | for any real number.

52. i) Prove | x | > 0 Ê x > a or x < a for any positive number, a.

For x  0, | x | œ x. | x | > a Ê x > a.

For x < 0, | x | œ x. | x | > a Ê x > a Ê x < a.

ii) Prove x > a or x < a Ê | x | > 0 for any positive number, a.

a > 0 and x > a Ê | x | œ x. So x > a Ê | x | > a.

For a > 0, a < 0 and x < a Ê x < 0 Ê | x | œ x. So x < a Ê x > a Ê | x | > a.

53. a) 1 = 1 Ê | 1 | = 1 ʹ b

†

^"b¹ œ ^{l l}l lb ʹ ¹ ¹ ¹ b

†

^"b œ ^{l l}l lb Ê œ ʹ ¹^"b œ

b b b b

b b b b

¹ ¹ ¹ ¹ ¹ ¹

¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹

† †

"

b "

b) ^{l l}_{l lb}^a œ ¹ a

†

^"_b¹ œ ¹ ¹ ¹ ¹ a

†

^"_b œ ¹ ¹ a

†

_{l l}^"_b œ ^{l l}_{l l}^a_b

54. Prove Sn œk kan œk ka for any real number a and any positive integer n.n

a a a, so S is true. Now, assume that S a a is true form some positive integer .

k k^" œ k k^" œ _" kœ¸ ¸k œk kk 5

Since ak k^" œ k ka ^" and a¸ ¸^k œk ka , we have a^k ¸ ^k"¸œ ¸a^k†a^"¸œ¸ ¸a^k k ka^" œk k k ka ^k a ^"œ k ka ^k+". Thus, S_k" œ¸a^k"¸ œk ka ^k+" is also true. Thus by the Principle of Mathematical Induction, S_n œ l l œ l laⁿ a ⁿ is true for all n positive integers.

1.2 LINES, CIRCLES, AND PARABOLAS

1. ?xœ œ1 ( 3) 2, y? œ œ 2 2 4; dœÈ( x)? #( y)? # œÈ416œ2È5 2. ?xœ $ œ ( 1) 2, y? œ œ2 ( 2) 4; dœÈ( 2) #4#œ2È5

3. ?xœ 8.1 ( 3.2)œ 4.9, y? œ œ2 ( 2) 0; dœÈ( 4.9) ^#0^#œ4.9

4. ?xœ 0 È2œ È2, y? œ1.5 œ 4 2.5; dœÊŠÈ2‹^# ( 2.5)#œÈ8.25

5. Circle with center (!ß !) and radius 1. 6. Circle with center (!ß !) and radius È2.

7. Disk (i.e., circle together with its interior points) with center (!ß !) and radius È3.

8. The origin (a single point).

9. mœ^?_?^y_x œ ₂¹_{( 1)}² œ3 10. mœ^?_?^y_x œ ₂^{# "}_{( 2)} œ ₄³ perpendicular slopeœ ^"₃ perpendicular slopeœ₃⁴

11. mœ^?_?^y_x œ ³₁³₂ œ0 12. mœ^?_?^y_x œ _{# #}^# ₍⁰₎; no slope perpendicular slope does not exist perpendicular slopeœ0

13. (a) xœ 1 14. (a) xœÈ2 15. (a) xœ0 16. (a) xœ 1

(b) yœ⁴₃ (b) yœ 1.3 (b) yœ È2 (b) yœ0

17. P( 1 1), m ß œ Ê œ 1 y 1 1 xa ( 1) b Ê yœ x 18. P(2ß 3), mœ^"_# Ê œ y ( 3) ^"_#(x2) Ê yœ ^"_#x4

19. P(3 4), Q( 2 5) ß ß Ê mœ ^?_?^y_x œ ⁵₂⁴₃ œ Ê œ ₅^" y 4 ₅^"(x3) Ê yœ ₅^"x²³₅ 20. P( 8 0), Q( 1 3) ß ß Ê mœ^?_?^y_x œ ₁³_{( 8)}⁰ œ ₇³ Ê œ y 0 ₇³ax ( 8) b Ê yœ ₇³x²⁴₇ 21. mœ ⁵₄, bœ6 Ê yœ ⁵₄x6 22. mœ^"_#, bœ Ê3 yœ^"_#x3 23. mœ0, P( 12 ß 9) Ê yœ 9 24. No slope, Pˆ^"₃ß % ʉ xœ ^"₃ 25. aœ 1, bœ4 Ê (0 4) and (ß "ß0) are on the line Ê mœ ^?_?^y_x œ ⁰₁⁴₀ œ4 Ê yœ4x4 26. aœ2, bœ Ê6 (2 0) and (ß !ß 6) are on the line Ê mœ ^?_?^y_x œ ₀⁶₂⁰ œ3 Ê yœ3x6

27. P(5ß 1), L: 2x5yœ15 Ê mLœ Ê2 parallel line is y œ ( 1) 2(x5) Ê yœ 2x1

5 5 5

28. PŠÈ2 2 , L: ß ‹ È2x5yœÈ3ÊmLœ Êparallel line is y œ 2 Šx Š È2‹‹ Ê yœ x

È2 È2 È2

5 5 5 5

8

29. P(4 10), L: 6xß 3yœ5 Ê mLœ2 Ê m¼œ Ê" perpendicular line is y10œ "(x4) Ê yœ "x12

# # #

30. P(!ß1), L: 8x13yœ13 Ê mLœ 8 Ê m œ 13 Ê perpendicular line is yœ 13x1

13 ¼ 8 8

31. x-interceptœ4, y-interceptœ3 32. x-interceptœ 4, y-interceptœ 2

33. x-interceptœÈ3, y-interceptœ È2 34. x-interceptœ 2, y-interceptœ3

35. AxByœC _" Í yœ ^A_Bx^C_B^" and BxAyœC _# Í yœ ^B_Ax^C_A^#. Since ˆ^A_B‰ ˆ ‰_A^B œ 1 is the product of the slopes, the lines are perpendicular.

36. AxByœC " Í yœ Ax and AxByœC # Í yœ Ax . Since the lines have the same

B B B B

C_" C_#

slope ^A_B, they are parallel.

37. New positionœaxold?x yß old?ybœ # &ß ( 3 ( 6))œ $ß ( 3).

38. New positionœaxold?x yß old?ybœ(6 ß ( 6) 0 0)œ ß(0 0).

39. ?xœ5, y? œ6, B(3ß 3). Let Aœ ß(x y). Then x? œx_#x _" Ê 5œ Ê3 x xœ 2 and

y y y 6 3 y y 9. Therefore, A ( 9).

? œ _{# "} Ê œ Ê œ œ #ß

40. x? œ " " œ !, y? œ ! ! œ !

41. C(!ß2), aœ2 Ê x^# (y 2)^#œ4 42. C($ß0), aœ3 Ê (x3)^#y^#œ9

43. C( 1 5), a ß œÈ10 Ê(x1)^# (y 5)^# œ10

44. C("ß "), aœÈ2 Ê (x1)^# (y 1)^#œ2 xœ0 Ê (01)^# (y 1)^#œ2 Ê (y1)^#œ1

y 1 1 y 0 or y 2.

Ê œ „ Ê œ œ Similarly, yœ0 Ê xœ0 or xœ2

45. CŠÈ3ß 2 , a‹ œ2 Ê xŠ È3‹^# (y 2)^# œ4, xœ0 Ê Š0È3‹^# (y 2)^#œ4 Ê (y2)^# œ1

y 2 1 y 1 or y 3. Also, y 0 Ê œ „ Ê œ œ œ

x 3 (0 2) 4 x 3 0

Ê Š È ‹^{# #}œ Ê Š È ‹^#œ

x 3

Ê œ È

46. C 3ˆ ß^"_#‰, aœ5 Ê (x3)^#ˆy^"_#‰^# œ25, so

xœ0 Ê (03)^#ˆy^"_#‰^#œ25

y 16 y 4 y

Ê ˆ ^"_#‰^# œ Ê œ „ Ê^"_# œ _#⁹

or yœ ⁷_#. Also, yœ0 Ê (x3)^#ˆ0_#^{" #}‰ œ25

(x 3) x 3

Ê ^#œ ⁹⁹₄ Ê œ „ ^3È_#¹¹

x 3

Ê œ „^3È_#¹¹

47. x^#y^#4x4y % œ0 Ê x^# %B y^#4yœ 4

x 4x 4 y 4y 4 4

Ê ^{# #} œ

(x 2) (y 2) 4 C ( 2 2), a 2.

Ê ^{# #}œ Ê œ ß œ

48. x^#y^#8x4y16œ0 Ê x^#8xy^#4yœ 16

x 8x 16 y 4y 4 4

Ê ^{# #} œ Ê (x4)^# (y 2)^#œ4

C ( 2), a 2.

Ê œ %ß œ

49. x^#y^#3y œ4 0 Ê x^#y^#3yœ4

x y 3y

Ê ^{# #} œ⁹₄ ²⁵₄

x y C 0 ,

Ê ^#ˆ _#³‰^# œ²⁵₄ Ê œ ߈ _#³‰ aœ⁵_#.

50. x^#y^#4x œ⁹₄ 0

x 4x y

Ê ^{# #}œ⁹₄

x 4x 4 y

Ê ^{# #} œ²⁵₄ Ê (x2)^#y^#œ ²⁵₄

C (2 0), a . Ê œ ß œ⁵_#

51. x^#y^#4x4yœ0 Ê x^#4xy^#4yœ0

x 4x 4 y 4y 4 8

Ê ^{# #} œ (xÊ 2)^# (y 2)^#œ8

C(2 2), a 8.

Ê ß œÈ

52. x^#y^#2xœ3

x 2x 1 y 4

Ê ^{# #} œ Ê (x1)^#y^#œ4

C ( 1 0), a 2.

Ê œ ß œ

53. xœ _#^b_aœ ₂₍₁₎² œ1 Ê yœ(1)^#2(1) œ 3 4

V ( 4). If x 0 then y 3.

Ê œ "ß œ œ Also, yœ0 Ê x^#2x œ3 0

(x 3)(x 1) 0 x 3 or Ê œ Ê œ xœ 1. Axis of parabola is xœ1.

54. xœ _#^b_aœ ₂₍₁₎⁴ œ 2

Ê yœ ( 2)^# œ 4( 2) 3 1 V ( 2 1). If x 0 then y 3.

Ê œ ß œ œ Also, yœ0 Ê x^#4x œ3 0

(x 1)(x 3) 0 x 1 or Ê œ Ê œ xœ 3. Axis of parabola is xœ 2.

55. xœ _#^b_aœ _{2( 1)}⁴ œ2 Ê yœ (2)^#4(2)œ4

V (2 4). If x 0 then y 0.

Ê œ ß œ œ Also, yœ0 Ê x^# 4xœ0

x(x 4) 0 x 4 or x 0.

Ê œ Ê œ œ Axis of parabola is xœ2.

56. xœ _#^b_aœ _{2( 1)}⁴ œ2

Ê yœ (2)^#4(2) œ 5 1 V (2 1). If x 0 then y 5.

Ê œ ß œ œ Also, yœ0 Ê x^# 4x œ5 0

x 4x 5 0 x

Ê ^# œ Ê œ ^{4„ È}_# ⁴

no x intercepts. Axis of parabola is x 2.

Ê œ

57. xœ _#^b_aœ _{2( 1)}⁶ œ 3

Ê yœ ( 3)^# œ6( 3) 5 4 V ( 3 ). If x 0 then y 5.

Ê œ ß % œ œ Also, yœ0 Ê x^# 6x œ5 0

(x 5)(x 1) 0 x 5 or Ê œ Ê œ xœ 1. Axis of parabola is xœ 3.

58. xœ _#^b_aœ ₂₍₂₎¹ œ ₄^"

y 2 3

Ê œ ˆ ‰^"₄ ^# œ^"₄ ²³₈

V . If x 0 then y 3.

Ê œˆ^"ß ‰ œ œ

4 8 23

Also, yœ0 Ê 2x^# œx 3 0 x no x intercepts.

Ê œ^{1„ È}₄ ²³ Ê Axis of parabola is xœ^"₄.

59. xœ _#^b_aœ _2(1/2)¹ œ 1

Ê yœ^"_#( 1) ^# œ( 1) 4 ⁷₂

V . If x 0 then y 4.

Ê œ "߈ ⁷₂‰ œ œ Also, yœ0 Ê x^"_# ^# œx 4 0

x no x intercepts.

Ê œ ^{„ 1} ₁^{È 7} Ê Axis of parabola is xœ 1.

60. xœ _#^b_aœ _{2( 1/4)}² œ4

Ê yœ ^"₄(4)^#2(4) œ4 8

V (4 8) . If x 0 then y 4.

Ê œ ß œ œ Also, yœ0 xÊ ^{" #}₄ 2x œ4 0 Ê xœ ^„2_1/2^È8 œ „4 4È2.

Axis of parabola is xœ4.

61. The points that lie outside the circle with center (!ß0) and radius È7.

62. The points that lie inside the circle with center (!ß0) and radius È5.

63. The points that lie on or inside the circle with center ("ß0) and radius 2.

64. The points lying on or outside the circle with center (!ß2) and radius 2.

65. The points lying outside the circle with center (!ß0) and radius 1, but inside the circle with center (!ß0), and radius 2 (i.e., a washer).

66. The points on or inside the circle centered at (!ß !) with radius 2 and on or inside the circle centered at ( 2 0) with radius 2. ß

67. x^#y^#6y0 Ê x^# (y 3)^#9.

The interior points of the circle centered at (!ß 3) with radius 3, but above the line yœ 3.

68. x^#y^#4x2y4 Ê (x2)^# (y 1)^#9.

The points exterior to the circle centered at (2ß 1) with radius 3 and to the right of the line xœ2.

69. (x2)^# (y 1)^#6 70. (x4)^# (y 2)^#16

71. x^#y^#Ÿ2, x 1 72. x^#y^#4, (x1)^# (y 3)^#10

73. x^#y^#œ1 and yœ2x Ê 1œx^#4x^#œ5x^#

x and y or x and y .

Ê Š œ_È^" œ _È ‹ Š œ _È^" œ _È ‹

5 5 5 5

2 2

Thus, AŠ_È^"₅ß_Ȳ₅‹, BŠ_È^"₅ß _Ȳ₅‹ are the points of intersection.

74. x œy 1 and (x1)^#y^#œ1 Ê 1œ ( y)^#y^# œ2y^#

y and x or

Ê Š œ_È^"₂ œ " _È^"₂‹

y and x 1 . Thus,

Š œ _È^" œ _È^" ‹

2 2

AŠ" _È^" ß_È^" ‹ and B 1Š _È^" ß _È^" ‹

2 2 2 2

are intersection points.

75. y œx 1 and yœx ^# Ê x^# œx 1

x x 1 0 x .

Ê ^# œ Ê œ ^1„_#^È5 If xœ ¹_#^È5, then yœ œx 1 ³_#^È5. If xœ ¹_#^È5, then yœ œx 1 ³_#^È5. Thus, AŠ¹_#^È5ß³_#^È5‹ and BŠ¹_#^È5ß³_#^È5‹ are the intersection points.

76. yœ x and C œ (x 1) ^# Ê (x1)^# œx

x 3x 0 x . If

Ê ^# " œ Ê œ ^3„_#^È5 xœ³_#^È5, then yœ œx ^{È5 3}_# . If xœ³_#^È5, then yœ œ x ³_#^È5.

Thus, AŠ³_#^È5ß^{È5 3}_# ‹ and BŠ³_#^È5ß ³_#^È5‹ are the intersection points.

77. yœ2x^# œ 1 x ^# Ê 3x^#œ1

x and y or x and y .

Ê œ_È^"₃ œ ^"₃ œ _È^"₃ œ ^"₃

Thus, AŠ_È^"₃ß ^"₃‹ and BŠ_È^"₃ß ^"₃‹ are the intersection points.

78. yœ^x₄^# œ(x1) ^# Ê 0œ ^3x₄^# 2x1 Ê 0œ3x^#8x œ4 (3x2)(x2)

x 2 and y 1, or x and Ê œ œ^x₄^# œ œ ²₃ yœ^x₄^# œ₉^". Thus, A(2 1) and Bß ˆ²₃ß₉^"‰ are the intersection points.

79. x^#y^#œ œ1 (x1)^#y^# Ê x^# œ(x1)^#œx^#2x1

0 2x 1 x . Hence Ê œ Ê œ ^"_#

y^# œ " x^#œ ³₄ or yœ „^È_#³. Thus, AŠ^"_#ß^È_#³‹ and BŠ^"_#ß ^È_#³‹ are the intersection points.

80. x^#y^#œ œ1 x^# Êy y^#œy y(y 1) 0 y 0 or y 1.

Ê œ Ê œ œ

If yœ1, then x^#œ " y^# œ0 or xœ0.

If yœ0, then x^#œ 1 y^# œ1 or xœ „1.

Thus, A(0 1), B(ß "ß0), and C( 1 0) are the ß intersection points.

81. (a) A¸(69° 0 in), Bß ¸(68° .4 in) ß Ê mœ ^68° _.4 ^69°₀ ¸ 2.5°/in.

(b) A¸(68° .4 in), Bß ¸(10° 4 in) ß Ê mœ ^10° _{4 .4}^68° ¸ 16.1°/in.

(c) A¸(10° 4 in), Bß ¸(5° 4.6 in) ß Ê mœ ^5° _4.6 ^10°₄ ¸ 8.3°/in.

82. The time rate of heat transfer across a material, ^?_?^U_>, is directly proportional to the cross-sectional area, A, of the material, to the temperature gradient across the material, ^?_?^X_B (the slopes from the previous problem), and to a constant characteristic of the material. ^?_?^U_> œ-kA^?_?^X_B Êk =^???^Î . Note that ^?_?^U_> and ^?_?B^Xare of opposite sign because heat flow is toward lower

? U>

XB

A

temperature. So a small value of k corresponds to low heat flow through the material and thus the material is a good insulator.Since all three materials have the same cross section and the heat flow across each is the same (temperatures are

not changing), we may define another constant, K, characteristics of the material: K œ Þ?^" Using the values of ^X_B from

?X B

??

the prevous problem, fiberglass has the smallest K at 0.06 and thus is the best insulator. Likewise, the wallboard is the poorest insulator, with Kœ0.4.

83. pœkd1 and pœ10.94 at dœ100 Ê kœ ^10.94₁₀₀^" œ0.0994. Then pœ0.0994d1 is the diver's pressure equation so that dœ50 Ê pœ(0.0994)(50) œ1 5.97 atmospheres.

84. The line of incidence passes through (!ß1) and ("ß0) Ê The line of reflection passes through ("ß0) and (#ß ") m 1 y 0 1(x 1) y x 1 is the line of reflection.

Ê œ^{1 0}_#1 œ Ê œ Ê œ

85. Cœ ⁵₉(F32) and CœF Ê Fœ ⁵₉F¹⁶⁰₉ Ê F⁴₉ œ ¹⁶⁰₉ or Fœ 40° gives the same numerical reading.

86. mœ^37.1₁₀₀ œ_?¹⁴_x Ê x? œ _.371¹⁴ . Therefore, distance between first and last rows is É(14)^#ˆ_.371¹⁴‰^# ¸40.25 ft.

87. length ABœÈ(51)^# (5 2)^#œÈ16 œ9 5 length ACœÈ(41)^# # # œ( )^# È916œ5

length BCœÈ(45)# # ( 5)#œÈ149œÈ50œ5È2Á5

88. length ABœÊ(10)^#ŠÈ30‹^#œÈ1 œ3 2 length ACœÈ(20)# (0 0)#œÈ4 œ0 2 length BCœÊ(21)^#Š0È3‹^# œÈ1 œ3 2

89. Length ABœÈ( x)? ^#( y)? ^# œÈ1^#4^#œÈ17 and length BCœÈ( x)? ^#( y)? ^#œÈ4^#1^# œÈ17.

Also, slope ABœ⁴₁ and slope BCœ ₄^", so AB¼BC. Thus, the points are vertices of a square. The coordinate increments from the fourth vertex D(x y) to A must equal the increments from C to B ß Ê œ 2 x ?xœ4 and

1 y y x 2 and y 2. Thus D( 2) is the fourth vertex.

œ? œ " Ê œ œ #ß

90. Let Aœ ß(x 2) and Cœ ß(9 y) Ê Bœ ß(x y). Then 9 œx kAD and 2k œy kDC k Ê 2(9 x) 2(2y)œ56 and 9 œx 3(2y) Ê 2(3(2y))2(2y)œ56 Ê yœ Ê œ5 9 x 3(2 ( 5)) Ê xœ 12.

Therefore, Aœ ß( 12 2), Cœ ß (9 5), and Bœ ß ( 12 5).

91. Let A("ß "), B(#ß $), and C(2ß !) denote the points.

Since BC is vertical and has length BCk kœ3, let D (_" "ß4) be located vertically upward from A and D (_# "ß 2) be located vertically downward from A so that BCk kœkAD_"kœkAD_#kœ3. Denote the point D (x y). Since the slope of AB equals the slope of_$ ß CD we have _$ ^{y 3}_{x 2} œ Ê^"₃ 3y œ 9 x 2 or x3yœ11. Likewise, the slope of AC equals the slope of BD so that _$ ^y _x ⁰₂ œ₃² Ê 3yœ2x4 or 2x3yœ4.

Solving the system of equations x 3y we find x 5 and y 2 yielding the vertex D (5 ).

2x3yœ "" 4

œ œ œ _$ ß #

92. Let x, y , xa b Á ! and/or yÁ ! be a point on the coordinate plane. The slope, m, of the segment a!ß !b a to x, y is . A 90b ^y_x ^‰ rotation gives a segment with slope m^wœ œ _m^" _y^x. If this segment has length equal to the original segment, its endpoint will be ay, x or y, x , the first of these corresponds to a counter-clockwise rotation, the latter to a clockwiseb a b rotation.

(a) ("ß4); (b) (3ß 2); (c) (5 2); ß (d) (0 x);ß

(e) ( y 0); ß (f) ( y x); ß (g) (3ß 10)

93. 2xkyœ3 has slope ²_k and 4x œy 1 has slope 4. The lines are perpendicular when œ ²_k( 4) 1 or kœ 8 and parallel when œ ²_k 4 or kœ^"_#.

94. At the point of intersection, 2x4yœ6 and 2x3yœ 1. Subtracting these equations we find 7yœ7 or yœ1. Substitution into either equation gives xœ1 Ê (1 1) is the intersection point. The line through (1 1)ß ß and ("ß #) is vertical with equation xœ1.

95. Let M(a b) be the midpoint. Since the two trianglesß shown in the figure are congruent, the value a must lie midway between x and x , so a_{" #} œ ^x"_#^x#. Similarly, bœ^y"_#^y#.

96. (a) L has slope 1 so M is the line through P(2 1) with slope 1; or the line yß œ x 3. At the intersection point, Q, we have equal y-values, yœ œ x 2 x 3. Thus, 2xœ1 or xœ ^"_#. Hence Q has coordinates

. The distance from P to L the distance from P to Q .

ˆ^"_{# #}ß⁵‰ œ œÉˆ ‰_#^{3 #} ˆ _#³‰^#œÉ¹⁸₄ œ ^3È_#²

(b) L has slope ⁴₃ so M has slope and M has the equation 4y³₄ 3xœ12. We can rewrite the equations of the lines as L: x³₄yœ3 and M: B ⁴₃yœ4. Adding these we get ²⁵₁₂yœ7 so yœ ⁸⁴₂₅. Substitution into either equation gives xœ ⁴₃ˆ ‰⁸⁴₂₅ œ4 ¹²₂₅ so that Qˆ¹²₂₅ß⁸⁴₂₅‰ is the point of intersection. The distance from P to LœÉˆ4¹²₂₅‰^#ˆ6⁸⁴₂₅‰^#œ ²²₅.

(c) M is a horizontal line with equation yœb. The intersection point of L and M is Q("ßb). Thus, the distance from P to L is È(a1)^#0^#œ ka 1 .k

(d) If Bœ0 and AÁ0, then the distance from P to L is ¸^C_Ax as in (c). Similarly, if A_!¸ œ0 and BÁ0, the distance is ¸^C_By . If both A and B are_!¸ Á0 then L has slope ^A_B so M has slope . Thus,^B_A

L: AxByœC and M: Bx Ayœ Bx!Ay . Solving these equations simultaneously we find the!

point of intersection Q(x y) with xß œ ^{AC B Ay}_A^a#_B^!#^Bx!b and yœ ^{BC A Ay}_A^a#_B^!#^Bx!b. The distance from P to Q equals È( x)? ^#( y) , where ( x)? ^# ? ^#œŠ^x!a^A#B#b_A^{AC ABy}#_B# ^{!B x# !}‹^#

, and ( y) .

œ ^A#a^Ax_A^!_# ^By_B#^!#^Cb^# œ ^y!a^{A# B#}b_A^{BC A y}# _B# ^{# ! ABx!} œ ^B#a^Ax_A^!_# ^By_B#^!#^Cb^#

a b a b

#

? Š ‹#

Thus, È( x)? ^#( y)? ^#œÉ^aAx!_A#^By_B^!#^Cb# œ ^kAx_È!_A^By!_B^Ck.

# #

1.3 FUNCTIONS AND THEIR GRAPHS

1. domainœ _ß _( ); rangeœ ß _[1 ) 2. domainœ ß _[0 ); rangeœ _ß( 1]

3. domainœ !ß _( ); y in range Ê yœ _Èt^" , t0 Ê y^#œ ^"_t and y ! Ê y can be any positive real number Ê rangeœ !ß _( ).

4. domainœ ß _[0 ); y in range Ê yœ ₁^"_Èt, t0. If tœ0, then yœ1 and as t increases, y becomes a smaller and smaller positive real number Ê rangeœ ß(0 1].

5. 4z^# œ(2z)(2  z) 0 Í − ß œ z [ 2 2] domain. Largest value is g(0)œÈ4œ2 and smallest value is g( 2) œg(2)œÈ0œ0 Ê rangeœ ß[0 2].

6. domainœ ß( 2 2) from Exercise 5; smallest value is g(0)œ^"_# and as 0z increases to 2, g(z) gets larger and larger (also true as z0 decreases to 2) Ê rangeœ 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. yœÉˆ ‰^"_x " Ê "   ! Ê Ÿ^"_x x 1 and x !. So,

(a) No (x !Ñ; (b) No; division by undefined;! (c) No; if x  ", ^"_x " Ê " !^"_x ; (d) Ð!ß "Ó

10. yœÉ# ÈxÊ # Èx  ! ÊÈx  !and xÈ Ÿ #. xÈ   ! Ê   !x and xÈ Ÿ # Ê Ÿ %Þ x So, x! Ÿ Ÿ %. (a) No; (b) No; (c) Ò!ß %Ó

11. baseœx; (height)^#ˆ ‰_#^{x #}œx ^# Ê heightœ ^È_#³x; area is a(x)œ _#^"(base)(height)œ _#^"(x)Š^È_#³x‹œ^È₄³x ;^# perimeter is p(x)œ œx x x 3x.

12. sœside length Ê s^# œs^# d ^# Ê œ s ^d ; and area is aœs ^# Ê œ a ^"_#d^#

È2

13. Let Dœdiagonal of a face of the cube and j œthe length of an edge. Then j ^# D^#œd and (by Exercise 10)^# D^#œ j Ê j œ2 ^# 3 ^# d ^# Ê j œ _È^d₃. The surface area is 6j œ^{# 6d}₃^# œ2d and the volume is ^# j œ^$ Š ‹^d₃^{# $Î#}œ _3È^d$₃.

14. The coordinates of P are xˆ ßÈx so the slope of the line joining P to the origin is m‰ œ^È_x^x œ _È^"_x (x0). Thus, x, ˆ Èx‰œˆ_m^"#, _m^"‰.

15. The domain is a_ß _b. 16. The domain is a_ß _b.

17. The domain is a_ß _b. 18. The domain is Ð_ß !Ó.

19. The domain is a_ß ! !ß _b a b. 20. The domain is a_ß ! !ß _b a b.

21. Neither graph passes the vertical line test

(a) (b)

22. Neither graph passes the vertical line test

(a) (b)

x y 1

x y y 1 x

or or

x y y x

k k

Ú Þ Ú Þ

Û ß Û ß

Ü à Ü à

œ Í œ " Í œ œ " œ "

23. x 0 1 2 24. x 0 1 2

y 0 1 0 y 1 0 0

25. y 3 x, x 1 26. y

2x, 1 x

, x 0 x, 0 x

œ Ÿ œ

œ œ^"x Ÿ

27. (a) Line through a!ß !b and a"ß "b: yœx Line through a"ß "b and a#ß !b: yœ x 2

f(x) x, 0 x 1

x 2, 1 x 2

œ Ÿ Ÿ

œ Ÿ (b) f(x)

2, x

x

2 x

x œ

! Ÿ "

!ß " Ÿ # ß # Ÿ $

!ß $ Ÿ Ÿ % ÚÝ

Ý ÛÝ ÝÜ

28. (a) Line through a!ß2 and b a#ß !b: yœ x 2

Line through 2a ß "b and a&ß !b: mœ ^{! "}_{& #} œ ^"_$ œ ^"_$, so yœ ^"_$ax2b " œ ^"_$x^&_$

f(x) x , 0 x

x , x

œ # Ÿ #

# Ÿ &

œ " &

$ $

(b) Line through a"ß !b and a!ß $b: mœ_{! Ð"Ñ}^{$ !} œ $, so yœ $ $x Line through a!ß $b and a#ß "b: mœ ^{" $}_{# !} œ ^%_# œ #, so yœ # $x

f(x) x , x

x , x

œ $ $ " Ÿ !

# $ ! Ÿ # œ

29. (a) Line through a"ß "b and a!ß !b: yœ x Line through a!ß "b and a"ß "b: yœ "

Line through a"ß "b and a$ß !b: mœ ^!"_$" œ^"_# œ ^"_#, so yœ ^"_#ax " " œ b ^"_#x^$_# f(x)

x x

x

x x

œ

" Ÿ !

" ! Ÿ "

" $

Ú

ÛÜ ^"_# ^$_#

(b) Line through a#ß "b and a!ß !b: yœ^"_#x Line through a!ß #b and a"ß !b: yœ # #x Line through a"ß "b and a$ß "b: yœ "

f(x)

x x

x x

x œ

# Ÿ Ÿ !

# # ! Ÿ "

" " Ÿ $ Ú

ÛÜ

"

#

30. (a) Line through ˆ^T_#ß !‰ and Ta ß "b: mœ_{T Î#}^"!_{aT b} œ _T^#, so yœ _T^#ˆx^T_#‰ œ0 _T^#x "

f(x) , 0 x

x , x T

œ ! Ÿ Ÿ

" Ÿ J

T T

T #

# #

(b) f(x)

A, x

A x T

A T x

A x T

œ

! Ÿ

ß Ÿ

ß Ÿ

ß Ÿ Ÿ #

ÚÝ ÝÝ ÛÝ ÝÝ Ü

T T

T T

#

# $

$ #

#

31. (a) From the graph, ^x_# 1 ⁴_x Ê x− ß( 2 0) %ß _( ) (b) 1^x_# ⁴_x 1Ê ^x_# ⁴_x 0

x0: 1^x_# ⁴_x 0 Ê ^x# _2x^{2x 8} 0 Ê ^{(x 4)(x 2)}_#x 0 x 4 since x is positive;

Ê

x0: 1^x₂ ⁴_x 0 Ê ^x# _2x^{2x 8} 0 Ê ^{(x 4)(x 2)}_#x 0 x 2 since x is negative;

Ê

sign of (x4)(x2) ïïïïïðïïïïïðïïïïî 2

%

Solution interval: (#ß0) %ß _( )

32. (a) From the graph, _{x 1}³ _{x 1}² Ê x− _ß ß( 5) ( 1 1) (b) Casex 1: _{x 1}³ _{x 1}² Ê ^{3(x 1)}_{x 1} 2

3x 3 2x 2 x 5.

Ê Ê

Thus, x− _ß ( 5) solves the inequality.

1 x 1: 2

Case _{x 1}³ _{x 1}² Ê ^{3(x 1)}_{x 1}

3x 3 2x 2 x 5 which is true

Ê Ê

if x 1. Thus, x− ß( 1 1) solves the

inequality.

1 x: 3x 3 2x 2 x 5

Case _{x 1}³ _{x 1}² Ê Ê

which is never true if 1x, so no solution

here.

In conclusion, x− _ß ß( 5) ( 1 1).

33. (a) xÚ Û œ0 for x−[0 1) ß (b) xÜ Ý œ0 for x− ß( 1 0]

34. xÚ Û œ Ü Ýx only when x is an integer.

35. For any real number x, nŸ Ÿ "x n , where n is an integer. Now: nŸ Ÿ " Ê Ð "Ñ Ÿ Ÿ x n n x n. By definition: Ü Ý œ x n and xÚ Û œ Ê Ú Û œ n x n. So Ü Ý œ Ú Ûx x for all x− d.