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,SOLUTION MANUAL Calculus & Its Applications,
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,INSTRUCTOR’S SOLUTIONS
MANUAL


CALCULUS
AND ITS APPLICATIONS
THIRD EDITION

Marvin L. Bittinger
Indiana University Purdue University Indianapolis


David J. Ellenbogen
Community College of Vermont


Scott A. Surgent
Arizona State University


Gene F. Kramer
University of Cincinnati Blue Ash

,Product Manager: Tara Warrens
Content Producer: Anoop Chaturvedi




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, Table of Contents

Chapter R: Functions, Graphs, and Models ............................................................1
Chapter 1: Differentiation ...................................................................................101
Chapter 2: Exponential and Logarithmic Functions ...........................................249
Chapter 3: Applications of Differentiation .........................................................340
Chapter 4: Integration .........................................................................................550
Chapter 5: Applications of Integration ...............................................................691
Chapter 6: Functions of Several Variables .........................................................809
Chapter 7: Trigonometric Functions ...................................................................945
Chapter 8: Differential Equations .......................................................................994
Chapter 9: Sequences and Series ......................................................................1069
Chapter 10: Probability Distributions ...............................................................1161
Chapter 11: Systems and Matrices.....................................................................1218
Chapter 12: Combinatorics and Probability.......................................................1325




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iii

,
,Chapter 1
Differentiation
13. The notation lim is read “the limit, as x
x 2
Exercise Set 1.1
approaches 2 from the right.”

14. The notation lim is read “the limit, as x
1. The limit of the sequence is 2.8 and using the x 3
notation will give the expression x  2.8. approaches 3 from the left”.

2. The limit of the sequence is 0.3 and using the 15. The notation lim is read “the limit, as x
x 5
notation will give the expression x  0.3. approaches 5”.
3. The limit of the sequence is 3 and using the 16. The notation lim1 is read “the limit, as x
notation will give the expression x  3. x 2
1
approaches 2
”.
4. The limit of the sequence is 4.9 and using the
notation will give the expression x  4.9 . 17. The notation lim f  x  is read “the limit, as x
x4

2 approaches 4, of f  x  . ”
5. The limit of the sequence is and using the
3
2 18. The notation lim g  x  is read “the limit, as x
notation will give the expression x  . x 1
3 approaches 1, of g  x  . ”

4
6. The limit of the sequence is and using the 19. The notation lim F  x  is read “the limit, as x
3 x 5

4 approaches 5 from the left, of F  x  . ”
notation will give the expression x  .
3
20. The notation lim G  x  is read “the limit, as x
x4
7. The limit of the sequence is 5.4 and using the
notation will give the expression x  5.4. approaches 4 from the right, of G  x  . ”

8. The limit of the sequence is 0.3 and using the 21. a) As inputs x approach 1 from the left,
notation will give the expression x  0.3. outputs f  x  approach 3. Thus the limit
from the left is 3. That is,
9. The limit of the sequence is 1 and using the lim  f  x   3.
notation will give the expression x  1. x 1


10. The limit of the sequence is 0 and using the b) As inputs x approach 1 from the right,
outputs f  x  approach 3. That is,
notation will give the expression x  0.
lim f  x   3.
x 1
11. As x approaches 0, the value of x  5
approaches 5.
c) From parts (a) and (b) we find
12. We solve the equation: lim f  x   3.
x 1
2 x  6
x  3
Therefore, As x approaches 3, the value of x
approaches 6.




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,102 Chapter 1: Differentiation

22. a) As inputs x approach 3 from the left, outputs 26. We have lim F  x   4 and lim F  x   4.
f  x  approach 1. That is, lim f  x   1. x2 x2
x 3 Therefore, lim F  x   4.
x2

b) As inputs x approach 3 from the right,
27. As inputs x approach 2 from the left, outputs
outputs f  x  approach 2. Thus the limit
F  x  approach 4. Thus the limit from the left
from the right is 2. That is, lim f  x   2.
x 3 is 4. That is, lim  F  x   4.
x 2
c) From parts (a) and (b) we know that
As inputs x approach 2 from the right, outputs
lim f  x   1 and lim f  x   2. Since the
x 3 x 3 F  x  approach 2. Thus the limit from the right
limit from the left, 1, is not the same as the is 2. That is, lim  F  x   2.
limit from the right, 2, lim f  x  does not x 2
x 3
Since the limit from the left, 4, is not the same
exist. as the limit from the right, 2, we have
lim F  x  does not exist.
23. a) As inputs x approach 4 from the right, x 2
outputs g  x  approach 1 . Thus the limit
28. We have lim  F  x   0 and lim  F  x   0.
from the right is 1 . That is, x 5 x 5
lim g  x   1. Therefore, lim F  x   0.
x4 x 5


b) lim g  x   1 . 29. As inputs x approach 4 from the left, outputs
x  4
F  x  approach 2. Thus the limit from the left is
c) Since the limit from the left, 1 , is the same 2. That is, lim F  x   2.
x4
as the limit from the right, 1 , we have
As inputs x approach 4 from the right, outputs
lim g  x   1 .
x4 F  x  approach 2. Thus the limit from the right
is 2. That is, lim F  x   2.
24. a) As inputs x approach 2 from the left, x4
outputs g  x  approach 4. Thus the limit Since the limit from the left, 2, is the same as
the limit from the right, 2, we have
from the left is 4. That is, lim  g  x   4.
x 2 lim F  x   2.
x4

b) lim g  x   2. 30. We have lim F  x   0 and lim F  x   0.
x 2
x 6 x 6
Therefore, lim F  x   0.
c) lim g  x  does not exist. x 6
x 2
31. As inputs x approach 2 from the right, outputs
25. As inputs x approach 3 from the left, outputs F  x  approach 2. Thus the limit from the right
F  x  approach 5. Thus the limit from the left is
is 2. That is, lim  F  x   2.
5. That is, lim  F  x   5. x 2
x 3
As inputs x approach 3 from the right, outputs 32. As inputs x approach 2 from the left, outputs
F  x  approach 5. Thus the limit from the right F  x  approach 4. Thus the limit from the left is
is 5. That is, lim  F ( x)  5. 4. That is, lim  F  x   4.
x 3 x 2
Since the limit from the left, 5, is the same as
the limit from the right, 5, we have 33. lim F ( x)
x 
lim F  x   5 . As inputs x approaches negative infinity, F(x)
x 3
decreases without bound.
lim F ( x)  
x 




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, Exercise Set 1.1 103

34. lim F  x  42. We have lim G  x   0 and lim G  x   0 .
x  x 3 x 3
As inputs x approaches positive infinity, F  x  Therefore, lim G  x   0 .
x 3
decreases without bound.
lim F  x    43. lim G ( x)
x  x 

As inputs x approaches negative infinity, G  x 
35. As inputs x approach 2 from the left, outputs
G (x ) approach 1. Thus the limit from the left is decreases without bound.
lim G  x   
1. That is, lim  G  x   1 . x 
x 2
As inputs x approach 2 from the right, outputs 44. lim G  x 
G (x ) approach 1. Thus the limit from the right x 

is 1. That is, lim  G  x   1 . As inputs x approaches positive infinity, G  x 
x 2 increases without bound.
Since the limit from the left, 1, is the same as lim G  x   
the limit from the right, 1, we have x 
lim G  x   1 .
x 2
45. As inputs x approach 3 from the left, outputs
H  x  approach 0. Thus the limit from the left
36. We have lim G  x   3 and lim G  x   3 .
x 0 x 0 is 0. That is, lim  H  x   0 .
Therefore, lim G  x   3 . x 3
x 0 As inputs x approach 3 from the right, outputs
H  x  approach 0. Thus the limit from the right
37. As inputs x approach 1 from the left, outputs
G  x  approach 4. Thus the limit from the left is is 0. That is, lim  H  x   0 .
x 3

4. That is, lim G  x   4 . Since the limit from the left, 0, is the same as
x 1 the limit from the right, 0, we have
lim H  x   0 .
38. lim G  x   1 x 3
x 1

46. lim H  x   1 .
39. lim G  x   0 x 2
x 3
47. As inputs x approach 2 from the right, outputs
40. As inputs x approach 1 from the left, outputs H  x  approach 1. Thus the limit from the right
G  x  approach 4. Thus the limit from the left is is 1. That is,
4. That is, lim G  x   4 . lim  H  x   1 .
x 1 x 2
As inputs x approach 1 from the right, outputs
G  x  approach 1 . Thus the limit from the 48. We have lim  H  x   1 and lim  H  x   1 .
x 2 x 2
right is 1 . That is, lim G  x   1 . Therefore, lim H  x   1 .
x 1
x 2
Since the limit from the left, 4, is not the same
as the limit from the right, 1 , we have 49. As inputs x approach 1 from the left, outputs
lim G  x  does not exist. H  x  approach 4. Thus the limit from the left
x 1
is 4. That is,
41. As inputs x approach 3 from the right, outputs lim H  x   4 .
G  x  approach 0. Thus the limit from the right x 1

is 0. That is, lim G  x   0 . 50. lim H  x   2 .
x 3
x 1




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, 104 Chapter 1: Differentiation

51. lim H  x   1 . Since the limit from the left, 1, is the same as
x 3 the limit from the right, 1, we have
lim f  x   1 .
52. As inputs x approach 1 from the left, outputs x 1

H  x  approach 4. Thus the limit from the left
59. We have lim f  x   2 and lim f  x   2 .
is 4. That is, lim H  x   4 . x 0 x 0
x 1
The solution is continued on the next page. Therefore, lim f  x   2 .
x 0
As inputs x approach 1 from the right, outputs
H  x  approach 2 . Thus the limit from the right 60. As inputs x approach 3 from the left, outputs
is 2 . That is, lim H  x   2 . f  x  increase without bound. We say that the
x 1
Since the limit from the left, 4, is not the same limit from the left is infinity. That is,
as the limit from the right, 2 , we have lim  f  x    .
x 3
lim H  x  does not exist. As inputs x approach 3 from the right, outputs
x 1
f  x  decrease without bound. We say that limit
53. As inputs x approach 3 from the right, outputs from the right is negative infinity. That is,
H  x  approach 1. Thus the limit from the right lim  f  x   .
x 3
is 1. That is, Since the function values as x  3 from the left
lim H  x   1 . increase without bound, and the function values
x 3
as x  3 from the right decrease without
54. We have lim H  x   1 and lim H  x   1 . bound, the limit does not exist. We have,
x 3 x 3 lim f  x  does not exist.
Therefore, lim H  x   1 .
x 3
x 3
61. We have lim f  x    and lim f  x   .
lim H  x 
x 1 x 1
55.
x  Therefore, lim f  x  does not exist.
As inputs x approaches negative infinity, H  x 
x 1


decreases without bound. 62. As inputs x approach 3 from the left, outputs
lim H  x    f  x  approach 0. Thus the limit from the left is
x 
0. That is,
56. lim H ( x) lim f  x   0 .
x  x 3

As inputs x approaches positive infinity, H  x  As inputs x approach 3 from the right, outputs
inecreases without bound. f  x  approach 0.
lim H ( x)   Thus the limit from the right is 0. That is,
x 
lim f  x   0 .
x 3
57. We have lim f  x   1 and lim f  x   1 . Since the limit from the left, 0, is the same as
x2 x2 the limit from the right, 0, we have
Therefore, lim f  x   1 . lim f  x   0 .
x2 x 3


58. As inputs x approach 1 from the left, outputs 63. We have lim  f  x   0 and lim  f  x   0 .
f  x  approach 1. Thus the limit from the left is x 2 x 2
Therefore, lim f  x   0 .
1. That is, lim  f  x   1 . x 2
x 1
As inputs x approach 1 from the right, outputs
f  x  approach 1. Thus the limit from the right
is 1. That is, lim  f  x   1 .
x 1




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