INSTRUCTOR SOLUTIONS MANUAL - mttk.no

1 INSTRUCTOR S SOLUTIONS MANUAL to accompany ADAMS / ESSEX CALCULUS: A COMPLETE COURSE; CALCULUS: SINGLE VARIABLE; and CALCULUS: SEVERAL VARIABLES Eighth Edition Prepared by Robert A. Adams University of British Columbia Christopher Essex University of Windsor Ontario Toronto Copyright 2014 Pearson Canada Inc., Toronto, Ontario. Pearson Canada. 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.

2 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, Eighth Edition, by Adams/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. FOREWORDT hese SOLUTIONS are provided for the benefit of instructors using the textbooks:Calculus: A Complete Course (8th Edition),Single-Variable Calculus (8th Edition),andCalculus of Several Variables (8th Edition)by R.

3 A. Adams and Chris Essex, published by Pearson Education Canada. For the most part,the SOLUTIONS are detailed, especially in exercises on core material and techniques. Occasion-ally some details are omitted for example, in exercises on applications of integration, theevaluation of the integrals encountered is not always given with the same degree of detail asthe evaluation of integrals found in those exercises dealing specifically with techniques of may wish to make these SOLUTIONS available to their students. However, studentsshould use such SOLUTIONS with caution.

4 It is always more beneficial for them to attempt ex-ercises and problems on their own, before they look at SOLUTIONS done by others. If they ex-amine SOLUTIONS as study material prior to attempting the exercises, they can lose much ofthe benefit that follows from diligent attempts to develop their own analytical powers. Whenthey have tried unsuccessfully to solve a problem, then looking at a solution can give them a hint for a second attempt. SeparateStudent SOLUTIONS Manualsfor the books are availablefor students. They contain the SOLUTIONS to the even-numbered exercises , A.

5 2014 Pearson Canada for Chapter P1 SOLUTIONS for Chapter 123 SOLUTIONS for Chapter 240 SOLUTIONS for Chapter 382 SOLUTIONS for Chapter 4109 SOLUTIONS for Chapter 5179 SOLUTIONS for Chapter 6215 SOLUTIONS for Chapter 7270 SOLUTIONS for Chapter 8318 SOLUTIONS for Chapter 9353 SOLUTIONS for Chapter 10393 SOLUTIONS for Chapter 11421 SOLUTIONS for Chapter 12450 SOLUTIONS for Chapter 13494 SOLUTIONS for Chapter 14541 SOLUTIONS for Chapter 15583 SOLUTIONS for Chapter 16614 SOLUTIONS for Chapter 17641 SOLUTIONS for Chapter 18648 SOLUTIONS for Chapter 18 extended669 SOLUTIONS for Appendices699 NOTE: Chapter 18 extended is only needed by users ofCalculus of Several Variables (Eighth Edition)Copyright 2014 Pearson Canada S SOLUTIONS MANUALSECTION (PAGE 10)CHAPTER P.

6 PRELIMINARIESS ection Real Numbers and the Real Line(page 10) , then 99xD12 andxD12/99D4 , then 10x and100x 32/, or 90xD295. ThusxD295/90D59 the same cyclic order of the repeating digits5 different decimal expansions can represent the samenumber. For instance, both represent the number 0 andx 5 define the interval [0;5]. <2 andx 3 define the interval [ 3;2/. > 5 orx< 6 defines the union. ; 6/ . 5; /. 1 defines the interval. ; 1]. > 2 defines the interval.

7 2; /. <4 orx 2 defines the interval. ; /, that is, thewhole real 2x>4, thenx< 2. Solution:. ; 2 3xC5 8, then 3x 8 5 3 andx 1. Solution:. ;1] 5x 3 7 3x, then 8x 10 andx 5/4. Solution:. ;5/4] x4 3x 42, then 6 x 6x 8. Thus 14 7xandx 2. Solution:. ;2] x/ < , then 0<5xandx>0. Solution:.0; <9, then|x|<3 and 3<x<3. Solution:. 3;3 : 1/.2 x/ < I. Ifx<2, then 1< x/D6 3x, so 3x<5andx<5/3. This case has solutionsx<5 II. Ifx>2, then 1> x/D6 3x, so 3x>5andx>5/3. This case has solutionsx> :. ;5/3/ .2; /. :.xC1//x I. Ifx>0, thenxC1 2x, sox II.

8 Ifx<0, thenxC1 2x, sox 1. (notpossible)Solution:.0;1]. :x2 2x 0. 2/ 0. This is onlypossible ifx 0 andx 2. Solution: [0;2]. 6x2 5x 1, 1/.3x 1/ 0, soeitherx 1/2 andx 1/3, orx 1/3 andx 1 latter combination is not possible. The solution set is[1/3;1/2]. >4x, we 4/ >0. This is possibleifx<0 andx2<4, or ifx>0 andx2>4. Thepossibilities are, therefore, 2<x<0 or 2<x< .Solution:. 2;0/ .2; /. x 2, thenx2 x 2 0 2/.xC1/ is possible ifx 2 andx 1 or ifx 2 andx 1. The latter situation is not possible. The solutionset is [ 1;2]. :x2 I.

9 Ifx>0, thenx2 2xC8, so thatx2 2x 8 0, 4/.xC2/ 0. This ispossible forx>0 only ifx II. Ifx<0, then we must 4/.xC2/ 0,which is possible forx<0 only ifx : [ 2;0/ [4; /. :3x 1< I. Ifx>1 1/.xC1/ >0, so < 1/. Thusx< 5. There are no solutionsin this II. If 1<x<1, 1/.xC1/ <0, > 1/. Thusx> 5. In this case allnumbers in. 1;1/are III. Ifx< 1, 1/.xC1/ >0, so < 1/. Thusx< 5. All numbersx< 5are :. ; 5/ . 1;1/. |x|D3 thenxD |x 3|D7, thenx 3D 7, soxD 4 |2tC5| D4, then 2tC5D 4, sotD 9/2 ortD 1 |1 t|D1, then 1 tD 1, sotD0 2014 Pearson Canada (PAGE 10)ADAMS and ESSEX: CALCULUS |8 3s|D9, then 8 3sD 9, so 3sD 1 or 17, andsD 1/3 orsD17 s2 1 D1, thens2 1D 1, sosD0 |x|<2, thenxis in.]]

10 2;2/. |x| 2, thenxis in [ 2;2]. |s 1| 2, then 1 2 s 1C2, sosis in [ 1;3]. |tC2|<1, then 2 1<t< 2C1, sotis in. 3; 1/. |3x 7|<2, then 7 2<3x<7C2, soxis ;3/. |2xC5|<1, then 5 1<2x< 5C1, soxis in. 3; 2/. x2 1 1, then 1 1 x2 1C1, soxis in [0;4]. 2 x2 <12, thenx/2 lies between 2 .1/2 Thusxis ;5/. inequality|xC1|>|x 3|says that the distancefromxto 1 is greater than the distance fromxto 3, soxmust be to the right of the point half-way between 1and 3. Thusx> |x 3|<2|x| x2 3/2<4x2 3x2C6x 9>0 1/ >0. Thisinequality holds ifx< 3 orx> |a|Daif and only ifa 0.