lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS
MANUAL
Robert A. Adams
University of British Columbia
Calculus
Ninth Edition
Robert A. Adams
University of British Columbia
Christopher Essex
University of Western Ontario
dumperina
ISBN: 978-0-13-452876-2
Copyright © 2018 Pearson Canada Inc., Toronto, Ontario. All rights reserved. This work is protected by Canadian
copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning.
Dissemination or sale of any part of this work (including on the Internet) will destroy the integrity of the work and is
not permitted. The copyright holder grants permission to instructors who have adopted Calculus: A Complete Course,
Single-Variable Calculus, or Calculus of Several Variables by Adams and Essex, to post this material online only if the
use of the website is restricted by access codes to students in the instructor’s class that is using the textbook and
provided the reproduced material bears this copyright notice.
Downloaded by ted cage ()
, lOMoARcPSD|6566483
FOREWORD
These solutions are provided for the benefit of instructors using the textbooks:
Calculus: A Complete Course (9th Edition),
Single-Variable Calculus (9th Edition), and
Calculus of Several Variables (9th Edition)
by R. A. Adams and Chris Essex, published by Pearson Canada. For the most part, the solu-
tions are detailed, especially in exercises on core material and techniques. Occasionally some
details are omitted—for example, in exercises on applications of integration, the evaluation of
the integrals encountered is not always given with the same degree of detail as the evaluation of
integrals found in those exercises dealing specifically with techniques of integration.
Instructors may wish to make these solutions available to their students. However, students
should use such solutions with caution. It is always more beneficial for them to attempt exer-
cises and problems on their own, before they look at solutions done by others. If they examine
solutions as “study material” prior to attempting the exercises, they can lose much of the bene-
fit that follows from diligent attempts to develop their own analytical powers. When they have
tried unsuccessfully to solve a problem, then looking at a solution can give them a “hint” for
a second attempt. Separate Student Solutions Manuals for the books are available for students.
They contain the solutions to the even-numbered exercises only.
November, 2016.
R. A. Adams Chris Essex
Downloaded by ted cage ()
, lOMoARcPSD|6566483
CONTENTS
Solutions for Chapter P 1
Solutions for Chapter 1 23
Solutions for Chapter 2 39
Solutions for Chapter 3 81
Solutions for Chapter 4 108
Solutions for Chapter 5 177
Solutions for Chapter 6 213
Solutions for Chapter 7 267
Solutions for Chapter 8 316
Solutions for Chapter 9 351
Solutions for Chapter 10 392
Solutions for Chapter 11 420
Solutions for Chapter 12 448
Solutions for Chapter 13 491
Solutions for Chapter 14 538
Solutions for Chapter 15 579
Solutions for Chapter 16 610
Solutions for Chapter 17 637
Solutions for Chapter 18 644
Solutions for Chapter 18-cosv9 671
Solutions for Appendices 683
NOTE: “Solutions for Chapter 18-cosv9” is only needed by users of Calculus of Several Vari-
ables (9th Edition), which includes extra material in Sections 18.2 and 18.5 that is found in Cal-
culus: a Complete Course and in Single-Variable Calculus in Sections 7.9 and 3.7 respectively.
Solutions for Chapter 18-cosv9 contains only the solutions for the two Sections 18.2 and 18.9 in
the Several Variables book. All other Sections are in “Solutions for Chapter 18.”
It should also be noted that some of the material in Chapter 18 is beyond the scope of most
students in single-variable calculus courses as it requires the use of multivariable functions and
partial derivatives.
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, lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.1 (PAGE 10)
CHAPTER P. PRELIMINARIES 19. Given: 1=.2 " x/ < 3.
CASE I. If x < 2, then 1 < 3.2 " x/ D 6 " 3x, so 3x < 5
and x < 5=3. This case has solutions x < 5=3.
Section P.1 Real Numbers and the Real Line CASE II. If x > 2, then 1 > 3.2 " x/ D 6 " 3x, so 3x > 5
(page 10) and x > 5=3. This case has solutions x > 2.
Solution: ."1; 5=3/ [ .2; 1/.
2 20. Given: .x C 1/=x # 2.
1. D 0:22222222 ! ! ! D 0:2
9 CASE I. If x > 0, then x C 1 # 2x, so x $ 1.
1 CASE II. If x < 0, then x C 1 $ 2x, so x # 1. (not
2. D 0:09090909 ! ! ! D 0:09 possible)
11
Solution: .0; 1!.
3. If x D 0:121212 ! ! !, then 100x D 12:121212 ! ! ! D 12 C x.
Thus 99x D 12 and x D 12=99 D 4=33. 21. Given: x 2 " 2x $ 0. Then x.x " 2/ $ 0. This is only
possible if x # 0 and x $ 2. Solution: Œ0; 2!.
4. If x D 3:277777 ! ! !, then 10x " 32 D 0:77777 ! ! ! and
100x " 320 D 7 C .10x " 32/, or 90x D 295. Thus 22. Given 6x 2 " 5x $ "1, then .2x " 1/.3x " 1/ $ 0, so either
x D 295=90 D 59=18. x $ 1=2 and x # 1=3, or x $ 1=3 and x # 1=2. The latter
combination is not possible. The solution set is Œ1=3; 1=2!.
5. 1=7 D 0:142857142857 ! ! ! D 0:142857
23. Given x 3 > 4x, we have x.x 2 " 4/ > 0. This is possible
2=7 D 0:285714285714 ! ! ! D 0:285714 if x < 0 and x 2 < 4, or if x > 0 and x 2 > 4. The
3=7 D 0:428571428571 ! ! ! D 0:428571 possibilities are, therefore, "2 < x < 0 or 2 < x < 1.
Solution: ."2; 0/ [ .2; 1/.
4=7 D 0:571428571428 ! ! ! D 0:571428
note the same cyclic order of the repeating digits 24. Given x 2 "x $ 2, then x 2 "x "2 $ 0 so .x "2/.x C1/ $ 0.
5=7 D 0:714285714285 ! ! ! D 0:714285 This is possible if x $ 2 and x # "1 or if x # 2 and
x $ "1. The latter situation is not possible. The solution
6=7 D 0:857142857142 ! ! ! D 0:857142 set is Œ"1; 2!.
6. Two different decimal expansions can represent the same x 4
number. For instance, both 0:999999 ! ! ! D 0:9 and 25. Given: #1C .
2 x
1:000000 ! ! ! D 1:0 represent the number 1. CASE I. If x > 0, then x 2 # 2x C 8, so that
x 2 " 2x " 8 # 0, or .x " 4/.x C 2/ # 0. This is pos-
7. x # 0 and x $ 5 define the interval Œ0; 5!. sible for x > 0 only if x # 4.
8. x < 2 and x # "3 define the interval Œ"3; 2/. CASE II. If x < 0, then we must have .x " 4/.x C 2/ $ 0,
which is possible for x < 0 only if x # "2.
9. x > "5 or x < "6 defines the union ."1; "6/ [ ."5; 1/. Solution: Œ"2; 0/ [ Œ4; 1/.
10. x $ "1 defines the interval ."1; "1!. 3 2
26. Given: < .
x"1 xC1
11. x > "2 defines the interval ."2; 1/. CASE I. If x > 1 then .x " 1/.x C 1/ > 0, so that
3.x C 1/ < 2.x " 1/. Thus x < "5. There are no solutions
12. x < 4 or x # 2 defines the interval ."1; 1/, that is, the in this case.
whole real line. CASE II. If "1 < x < 1, then .x " 1/.x C 1/ < 0, so
13. If "2x > 4, then x < "2. Solution: ."1; "2/ 3.x C 1/ > 2.x " 1/. Thus x > "5. In this case all
numbers in ."1; 1/ are solutions.
14. If 3x C 5 $ 8, then 3x $ 8 " 5 " 3 and x $ 1. Solution: CASE III. If x < "1, then .x " 1/.x C 1/ > 0, so that
."1; 1! 3.x C 1/ < 2.x " 1/. Thus x < "5. All numbers x < "5
are solutions.
15. If 5x " 3 $ 7 " 3x, then 8x $ 10 and x $ 5=4. Solution: Solutions: ."1; "5/ [ ."1; 1/.
."1; 5=4!
27. If jxj D 3 then x D ˙3.
6"x 3x " 4
16. If # , then 6 " x # 6x " 8. Thus 14 # 7x 28. If jx " 3j D 7, then x " 3 D ˙7, so x D "4 or x D 10.
4 2
and x $ 2. Solution: ."1; 2! 29. If j2t C 5j D 4, then 2t C 5 D ˙4, so t D "9=2 or
17. If 3.2 " x/ < 2.3 C x/, then 0 < 5x and x > 0. Solution: t D "1=2.
.0; 1/ 30. Ifj1 " t j D 1, then 1 " t D ˙1, so t D 0 or t D 2.
2
18. If x < 9, then jxj < 3 and "3 < x < 3. Solution: 31. If j8 " 3sj D 9, then 8 " 3s D ˙9, so 3s D "1 or 17, and
."3; 3/ s D "1=3 or s D 17=3.
Copyright © 2018 Pearson Canada Inc. 1
Downloaded by ted cage ()
INSTRUCTOR’S SOLUTIONS
MANUAL
Robert A. Adams
University of British Columbia
Calculus
Ninth Edition
Robert A. Adams
University of British Columbia
Christopher Essex
University of Western Ontario
dumperina
ISBN: 978-0-13-452876-2
Copyright © 2018 Pearson Canada Inc., Toronto, Ontario. All rights reserved. This work is protected by Canadian
copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning.
Dissemination or sale of any part of this work (including on the Internet) will destroy the integrity of the work and is
not permitted. The copyright holder grants permission to instructors who have adopted Calculus: A Complete Course,
Single-Variable Calculus, or Calculus of Several Variables by Adams and Essex, to post this material online only if the
use of the website is restricted by access codes to students in the instructor’s class that is using the textbook and
provided the reproduced material bears this copyright notice.
Downloaded by ted cage ()
, lOMoARcPSD|6566483
FOREWORD
These solutions are provided for the benefit of instructors using the textbooks:
Calculus: A Complete Course (9th Edition),
Single-Variable Calculus (9th Edition), and
Calculus of Several Variables (9th Edition)
by R. A. Adams and Chris Essex, published by Pearson Canada. For the most part, the solu-
tions are detailed, especially in exercises on core material and techniques. Occasionally some
details are omitted—for example, in exercises on applications of integration, the evaluation of
the integrals encountered is not always given with the same degree of detail as the evaluation of
integrals found in those exercises dealing specifically with techniques of integration.
Instructors may wish to make these solutions available to their students. However, students
should use such solutions with caution. It is always more beneficial for them to attempt exer-
cises and problems on their own, before they look at solutions done by others. If they examine
solutions as “study material” prior to attempting the exercises, they can lose much of the bene-
fit that follows from diligent attempts to develop their own analytical powers. When they have
tried unsuccessfully to solve a problem, then looking at a solution can give them a “hint” for
a second attempt. Separate Student Solutions Manuals for the books are available for students.
They contain the solutions to the even-numbered exercises only.
November, 2016.
R. A. Adams Chris Essex
Downloaded by ted cage ()
, lOMoARcPSD|6566483
CONTENTS
Solutions for Chapter P 1
Solutions for Chapter 1 23
Solutions for Chapter 2 39
Solutions for Chapter 3 81
Solutions for Chapter 4 108
Solutions for Chapter 5 177
Solutions for Chapter 6 213
Solutions for Chapter 7 267
Solutions for Chapter 8 316
Solutions for Chapter 9 351
Solutions for Chapter 10 392
Solutions for Chapter 11 420
Solutions for Chapter 12 448
Solutions for Chapter 13 491
Solutions for Chapter 14 538
Solutions for Chapter 15 579
Solutions for Chapter 16 610
Solutions for Chapter 17 637
Solutions for Chapter 18 644
Solutions for Chapter 18-cosv9 671
Solutions for Appendices 683
NOTE: “Solutions for Chapter 18-cosv9” is only needed by users of Calculus of Several Vari-
ables (9th Edition), which includes extra material in Sections 18.2 and 18.5 that is found in Cal-
culus: a Complete Course and in Single-Variable Calculus in Sections 7.9 and 3.7 respectively.
Solutions for Chapter 18-cosv9 contains only the solutions for the two Sections 18.2 and 18.9 in
the Several Variables book. All other Sections are in “Solutions for Chapter 18.”
It should also be noted that some of the material in Chapter 18 is beyond the scope of most
students in single-variable calculus courses as it requires the use of multivariable functions and
partial derivatives.
Downloaded by ted cage ()
, lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.1 (PAGE 10)
CHAPTER P. PRELIMINARIES 19. Given: 1=.2 " x/ < 3.
CASE I. If x < 2, then 1 < 3.2 " x/ D 6 " 3x, so 3x < 5
and x < 5=3. This case has solutions x < 5=3.
Section P.1 Real Numbers and the Real Line CASE II. If x > 2, then 1 > 3.2 " x/ D 6 " 3x, so 3x > 5
(page 10) and x > 5=3. This case has solutions x > 2.
Solution: ."1; 5=3/ [ .2; 1/.
2 20. Given: .x C 1/=x # 2.
1. D 0:22222222 ! ! ! D 0:2
9 CASE I. If x > 0, then x C 1 # 2x, so x $ 1.
1 CASE II. If x < 0, then x C 1 $ 2x, so x # 1. (not
2. D 0:09090909 ! ! ! D 0:09 possible)
11
Solution: .0; 1!.
3. If x D 0:121212 ! ! !, then 100x D 12:121212 ! ! ! D 12 C x.
Thus 99x D 12 and x D 12=99 D 4=33. 21. Given: x 2 " 2x $ 0. Then x.x " 2/ $ 0. This is only
possible if x # 0 and x $ 2. Solution: Œ0; 2!.
4. If x D 3:277777 ! ! !, then 10x " 32 D 0:77777 ! ! ! and
100x " 320 D 7 C .10x " 32/, or 90x D 295. Thus 22. Given 6x 2 " 5x $ "1, then .2x " 1/.3x " 1/ $ 0, so either
x D 295=90 D 59=18. x $ 1=2 and x # 1=3, or x $ 1=3 and x # 1=2. The latter
combination is not possible. The solution set is Œ1=3; 1=2!.
5. 1=7 D 0:142857142857 ! ! ! D 0:142857
23. Given x 3 > 4x, we have x.x 2 " 4/ > 0. This is possible
2=7 D 0:285714285714 ! ! ! D 0:285714 if x < 0 and x 2 < 4, or if x > 0 and x 2 > 4. The
3=7 D 0:428571428571 ! ! ! D 0:428571 possibilities are, therefore, "2 < x < 0 or 2 < x < 1.
Solution: ."2; 0/ [ .2; 1/.
4=7 D 0:571428571428 ! ! ! D 0:571428
note the same cyclic order of the repeating digits 24. Given x 2 "x $ 2, then x 2 "x "2 $ 0 so .x "2/.x C1/ $ 0.
5=7 D 0:714285714285 ! ! ! D 0:714285 This is possible if x $ 2 and x # "1 or if x # 2 and
x $ "1. The latter situation is not possible. The solution
6=7 D 0:857142857142 ! ! ! D 0:857142 set is Œ"1; 2!.
6. Two different decimal expansions can represent the same x 4
number. For instance, both 0:999999 ! ! ! D 0:9 and 25. Given: #1C .
2 x
1:000000 ! ! ! D 1:0 represent the number 1. CASE I. If x > 0, then x 2 # 2x C 8, so that
x 2 " 2x " 8 # 0, or .x " 4/.x C 2/ # 0. This is pos-
7. x # 0 and x $ 5 define the interval Œ0; 5!. sible for x > 0 only if x # 4.
8. x < 2 and x # "3 define the interval Œ"3; 2/. CASE II. If x < 0, then we must have .x " 4/.x C 2/ $ 0,
which is possible for x < 0 only if x # "2.
9. x > "5 or x < "6 defines the union ."1; "6/ [ ."5; 1/. Solution: Œ"2; 0/ [ Œ4; 1/.
10. x $ "1 defines the interval ."1; "1!. 3 2
26. Given: < .
x"1 xC1
11. x > "2 defines the interval ."2; 1/. CASE I. If x > 1 then .x " 1/.x C 1/ > 0, so that
3.x C 1/ < 2.x " 1/. Thus x < "5. There are no solutions
12. x < 4 or x # 2 defines the interval ."1; 1/, that is, the in this case.
whole real line. CASE II. If "1 < x < 1, then .x " 1/.x C 1/ < 0, so
13. If "2x > 4, then x < "2. Solution: ."1; "2/ 3.x C 1/ > 2.x " 1/. Thus x > "5. In this case all
numbers in ."1; 1/ are solutions.
14. If 3x C 5 $ 8, then 3x $ 8 " 5 " 3 and x $ 1. Solution: CASE III. If x < "1, then .x " 1/.x C 1/ > 0, so that
."1; 1! 3.x C 1/ < 2.x " 1/. Thus x < "5. All numbers x < "5
are solutions.
15. If 5x " 3 $ 7 " 3x, then 8x $ 10 and x $ 5=4. Solution: Solutions: ."1; "5/ [ ."1; 1/.
."1; 5=4!
27. If jxj D 3 then x D ˙3.
6"x 3x " 4
16. If # , then 6 " x # 6x " 8. Thus 14 # 7x 28. If jx " 3j D 7, then x " 3 D ˙7, so x D "4 or x D 10.
4 2
and x $ 2. Solution: ."1; 2! 29. If j2t C 5j D 4, then 2t C 5 D ˙4, so t D "9=2 or
17. If 3.2 " x/ < 2.3 C x/, then 0 < 5x and x > 0. Solution: t D "1=2.
.0; 1/ 30. Ifj1 " t j D 1, then 1 " t D ˙1, so t D 0 or t D 2.
2
18. If x < 9, then jxj < 3 and "3 < x < 3. Solution: 31. If j8 " 3sj D 9, then 8 " 3s D ˙9, so 3s D "1 or 17, and
."3; 3/ s D "1=3 or s D 17=3.
Copyright © 2018 Pearson Canada Inc. 1
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