Applied Calculus, Berresford - Downloadable Solutions Manual (Revised)

Contents
Diagnostic Test 1

Chapter 1 Functions 3

Chapter 2 Derivatives and Their Uses 41

Chapter 3 Further Applications of Derivatives 95

Chapter 4 Exponential and Logarithmic Functions 189

Chapter 5 Integration and Its Applications 222

Chapter 6 Integration Techniques 290

Chapter 7 Calculus of Several Variables 344

Chapter 8 Trigonometric Functions 415

Chapter 9 Differential Equations 454

Chapter 10 Sequences and Series 510

Chapter 11 Probability 552

Chapter Tests 595

Chapter Test Solutions 655

,DIAGNOSTIC TEST

Are you ready to study calculus?

Algebra is the language in which we express the ideas of calculus. Therefore, to un-
derstand calculus and express its ideas with precision, you need to know some algebra.

If you are comfortable with the algebra covered in the following problems, you are
ready to begin your study of calculus. If not, turn to the Algebra Appendix beginning
on page A.xxx and review the Complete Solutions to these problems, and continue
reading the other parts of the Appendix that cover anything that you do not know.

Problems Answers

1
False
1. True or False? 2
< 3
( 4, 5]
2. Express {x| 4 < x  5} in interval notation.
5
3. What is the slope of the line through the points (6, 7) and (9, 8)?
6
4. On the line y = 3x + 4, what value of y corresponds to x = 2?
a
5. Which sketch shows the graph of the line y = 2x 1?




✓p ◆ 2
x y2 True
6. True or False? =
y x
3
x=
7. Find the zeros of the function f (x) = 9x2 6x 1 1± 2
p

7 x2 + 5x
8. Expand and simplify x(8 x) (3x + 7).

x2 3x + 2 3, x 6= 0, x 6= 2} {x|x 6=
9. What is the domain of f (x) = ?
x3 + x2 6x
f (x + h) f (x) 5+h 2x
10. For f (x) = x2 5x, find the di↵erence quotient .
h




Diagnostic Test (in Front Matter)

,Exercises 1.1 3


Chapter 1: Functions
EXERCISES 1.1

1.  x 0  x  6 2.  x 3  x  5
–3 5


3.  x x  2 4.  x x  7
2 7

5. a. Since x = 3 and m = 5, then y, the 6. a. Since x = 5 and m = –2, then y, the
change in y, is change in y, is
y = 3 • m = 3 • 5 = 15 y = 5 • m = 5 • (–2) = –10

b. Since x = –2 and m = 5, then y, the b. Since x = –4 and m = –2, then y, the
change in y, is change in y, is
y = –2 • m = –2 • 5 = –10 y = –4 • m = –4 • (–2) = 8

7. For (2, 3) and (4, –1), the slope is 8. For (3, –1) and (5, 7), the slope is
1  3  4  2 7  (1) 7  1 8
42 2   4
53 2 2

9. For (–4, 0) and (2, 2), the slope is 10. For (–1, 4) and (5, 1), the slope is
20  2  2  1 1  4  3  3   1
2  ( 4) 2  4 6 3 5  ( 1) 5  1 6 2

11. For (0, –1) and (4, –1), the slope is
1  ( 1) 1  1 0
12.
2 2   
For 2, 1 and 5, 1 , the slope is
  0
40 4 4 1
 12
2  0  0 0
5  ( 2) 5 2 7

13. For (2, –1) and (2, 5), the slope is 14. For (6, –4) and (6, –3), the slope is
5  ( 1) 5  1 3  ( 4) 3  4
 undefined  undefined
22 0 66 0

15. Since y = 3x – 4 is in slope-intercept form, 16. Since y = 2x is in slope-intercept form, m = 2 and
m = 3 and the y-intercept is (0, –4). Using the the y-intercept is (0, 0). Using m = 2, we see that
slope m = 3, we see that the point 1 unit to the the point 1 to the right and 2 units up is also on
right and 3 units up is also on the line. the line.




 2016 Cengage Learning. All rights reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

, 4 Chapter 1: Functions


17. Since y =  12 x is in slope-intercept form, 18. Since y =  13 x + 2 is in slope-intercept form,
m=  1 and the y-intercept is (0, 0). Using
2
m =  13 and the y-intercept is (0, 2). Using
m=  1 , we see that the point 2 units to the
2
m =  13 , we see that the point 3 units to the right
right and 1 unit down is also on the line. and 1 unit down is also on the line.




19. The equation y = 4 is the equation of the hori- 20. The equation y = –3 is the equation of the hori-
zontal line through all points with y-coordinate zontal line through all points with y-coordinate
4. Thus, m = 0 and the y-intercept is (0, 4). –3. Thus, m = 0 and the y-intercept is (0, –3).




21. The equation x = 4 is the equation of the 22. The equation x = –3 is the equation of the vertical
vertical line through all points with x-coordinate line through all points with x-coordinate –3. Thus,
4. Thus, m is not defined and there is no y- m is not defined and there is no y-intercept.
intercept.




23. First, solve for y: 24. First, solve for y:
2 x  3 y  12 3x  2y  18
3 y  2 x  12 2y  3 x  18
y  2 x4
3 y   3 x 9
2
Therefore, m = 23 and the y-intercept is (0, –4).
Therefore, m =  32 and the y-intercept is (0, 9).




 2016 Cengage Learning. All rights reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.