SOLUTION MANUAL FOR Mathematical Applications 13th Edition Harshbarger

, SOLUTION MANUAL FOR Mathematical Applications 13th Edition
Harshbarger

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, TABLE OF CONTENTS
CHAPTER 0: ALGEBRAIC CONCEPTS……………………………………………………….2
CHAPTER 1: LINEAR EQUATIONS AND FUNCTIONS……………………………………23
CHAPTER 2: QUADRATIC AND OTHER SPECIAL FUNCTIONS………………………...55
CHAPTER 3: MATRICES…………………………………………………………………...…83
CHAPTER 4: INEQUALITIES AND LINEAR PROGRAMMING………………………….113
CHAPTER 5: EXPONENTIAL AND LOGARITHMIC FUNCTIONS………………………170
CHAPTER 6: MATHEMATICS OF FINANCE………………………………………………190
CHAPTER 7: INTRODUCTION TO PROBABILITY………………………………………..217
CHAPTER 8: FURTHER TOPICS IN PROBABILITY; STATISTICS………………………241
CHAPTER 9: DERIVATIVES…………………………………………………………………259
CHAPTER 10: APPLICATIONS OF DERIVATIVES………………………………………..298
CHAPTER 11: DERIVATIVES CONTINUED……………………………………………….330
CHAPTER 12: INDEFINITE INTEGRALS…………………………………………………...348
CHAPTER 13: DEFINITE INTEGRALS: TECHNIQUES OF INTEGRATION……………..369
CHAPTER 14: FUNCTIONS OF TWO OR MORE VARIABLES…………………………...402




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, Chapter 0: Algebraic Concepts
Exercises 0.1 __________________________________________________________________

1. 12  {1, 2, 3, 4,...} For problems 31 - 42, we have
U = {1, 2, 3, . . . , 9, 10}.
3. 6  {1, 2, 3, 4, 5}
31. A  {4, 6, 9, 10} since these are the only elements
5. a. {1, 2, 3, 4, 5, 6, 7} in U that are not elements of A.

33. B   {1, 2, 5, 6, 7, 9}
b. n( D )  7
A  B  {1, 2, 5, 7}
7. a. {x: x is a natural number greater than 2 and less
than 8} 35. A  B  1, 2, 3, 4, 5, 7, 8, 10
( A  B)  {6, 9}
b. n( D )  5
37. A  {4, 6, 9, 10}
9.   A since  is a subset of every set. A  B
B   {1, 2, 5, 6, 7, 9}
since every element of A is an element of B.
B  B since a set is always a subset of itself. A  B   {1, 2, 4, 5, 6, 7, 9, 10}

11. No. c  A but c  B. 39. B   {1, 2, 5, 6, 7, 9}
C   {1, 3, 5, 7, 9}
13. D  C since every element of D is an element of A  B  1, 2, 3, 5, 7, 8  1, 2, 5, 6, 7, 9
C.
 1, 2, 5, 7
15. A  B and B  A . (Also A  B .)  A  B    C   1, 2, 3, 5, 7, 9

17. Yes. A  B and B  A . Thus, A  B . 41. B   {1, 2, 5, 6, 7, 9}
A  B  1, 2, 3, 5, 7, 8  1, 2, 5, 6, 7, 9
19. No. D  E because 4  E and 4  D.
 1, 2, 5, 7
21. A and B are disjoint since they have no elements
in common. B and D are disjoint since they  A  B   C  3, 4, 6, 8, 9, 10  2, 4, 6, 8, 10
have no elements in common. C and D are  4, 6, 8, 10
disjoint.
For problems 43 - 46, we have
23. A  B  {4, 6} since 4 and 6 are elements of each U = {1, 2, 3, . . . , 8, 9}.
set.
43. A – B = {1, 3, 7, 9} – {3, 5, 8, 9} = {1, 7}
25. A  B =  since they have no common elements.
45. A – B = {2, 1, 5} – {1, 2, 3, 4, 5, 6} =  or 
27. A  B  {1, 2, 3, 4, 5}

29. A  B  {1, 2, 3, 4} or A  B  B.

47. a. A = {HSA, Mfg, Mgt, RT, S}
B = {TW, HAS, I, Mgt}
C = {Mfg, I, U, S}
b. No
c. B  is the set of industries with an expected employment does not increase from 2022 to 2032.
B  {Mfg, U, RT, S}
d. A = {TW, I, U}
A  B = {TW, I}
e. C  = {TW, HSA, Mgt, RT}

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, Chapter 0: Algebraic Concepts

B  C  = {TW, HSA, I, Mgt, RT}.
B  C  is the set of industries with an expected employment increase from 2022 to 2032 or the expected change
(positive or negative) in employment from 2022 to 2032 was greater than or equal to 500,000 jobs.


49. a. From the table, there are 100 white Republicans and 30 non-white Republicans who favor national
health care, for a total of 130.
b. From the table, there are 350 + 40 Republicans, and 250 + 200 Democrats who favor national health care,
for a total of 840.
c. From the table, there are 350 white Republicans, and 150 white Democrats and 20 non-whites who oppose
national health care, for a total of 520.

51. a. The key to solving this problem is to work from "the inside out". There
are 40 aides in E  F. This leaves 65 – 40 = 25 aides who speak English U
but do not speak French. Also we have 60 – 40 = 20 aides who speak E F
French but do not speak English. Thus there are 40 + 25 + 20 = 85 aides
40
who speak English or French. This means there are 15 aides who do not 25 20
speak English or French.
b. From the Venn diagram E  F has 40 aides. 15
c. From the Venn diagram E  F has 85 aides.
d. From the Venn diagram E  F  has 25 aides.

53. Since 12 students take M and E but not FA, and 15 take M and E, 3 take all three
classes. Since 9 students take M and FA and we have already counted 3, there are 6 U
taking M and FA which are not taking E. Since 4 students take E and FA and we
have already counted 3, there is only 1 taking E and FA but not taking M also. Since E F
20 students take E and we already have 16 enrolled in E, this leaves 4 taking only E. 40
25 20
Since 42 students take FA and we already have 10 enrolled in FA, this leaves 32 U
taking only FA. Since 38 students take M and we already have 21 enrolled in M, this Math 15
leaves 17 taking only M. 6 Fine Arts
a. In the union of the 3 courses we have 17 + 12 + 3 + 6 + 32 + 1 + 4 = 75 17 32
3
students enrolled. Thus, there are 100 – 75 = 25 students who are not 12 1
enrolled in any of these courses.
b. In M  E we have 17 + 12 + 3 + 6 + 1 + 4 = 43 enrolled. Economics
4 25
c. We have 17 + 32 + 4 = 53 students enrolled in exactly one of the courses.

55. a. and b.

U
A B
A  AB B
AB
A 
B O

O
Rh 




c. A : 34%; B  : 9% ; O  : 38% ; AB  : 3% ; O  : 7% ; A : 6% ; B  : 2% ; AB  : 1%




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, Chapter 0: Algebraic Concepts

Exercises 0.2 __________________________________________________________________

  1
1. a. Note that        , where  is 31. (2, 10); open interval
10  10 
1 33. 3  x  5
irrational and  is rational. The product
10
of a rational number other than 0 and an 35. x > 4
irrational number is an irrational number.
b. –9 is rational and an integer. 37. (, 4)  (3,  )  (3, 4)
9 3
c.   3 . This is a natural number, an 3 2 1 0 1 2
3 1 3 4 5
integer, and a rational number.
d. Division by zero is meaningless. 39. x > 4 and x  0 = (4, )

3. a. Commutative 2 1 0 1 2 3 4 5 6
b. Distributive
41. [0,  )  [1, 5]  [1,  )
5. a. Associative
b. Additive Identity 2 1 0 1 2 3 4 5 6

7. –6 < 0 43. (, 0)  (7,  )

9. –14 < –3 1 0 1 2 3 4 5 6 7

11  45. –0.000038585
11. 0.333    0.3333  
3 3 
47. 9122.387471
13. | 3 |  | 5 |  | 3  5 |
2500 2500
49.   3240.184509
[(1.16 )  1] 0.771561
15. 3  10  2  3  20  9  20  11
2 2



51. a. $300.00  $788.91  $1088.91
4  22 4  4 8
17.   4 b. Federal withholding
2 2 2 = 0.25(1088.91 – 54.45) = $258.62
c. Retirement: 0.05(1088.91) =
16  (4) 16  4 20 $54.45
19.   2
8  (2) 8  2 10 State tax = Retirement = $54.45
Local tax = 0.01(1088.91) = $10.89
Federal tax = $258.62 (from b.
above)
| 5  2 |  | 7 | | 3 |  | 7 | 3  7 4
21.    Social Security and Medicare tax
| 52| | 3| 3 3 d = 0.0765(1088.91) = $83.30
Total Withholding = $461.71
(3) 2  2  3  6 9  6  6 9 Take-home = 1088.91 – 461.71 =
23.   3 $627.20
4  22  3 443 3
53. a. Equation (1) is more accurate.
42  5  2  3 16  5  6 17 17 d Equation (1) gives
25.   
5  42 5  16 11 11 d y  5.7 10   260
 317 million
27. The entire line

29. (1, 3]; half-open interval


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, Chapter 0: Algebraic Concepts

b. For 2032, Equation (1) gives 55. a. 190,151  I  243, 725;
y  5.7 17   260 243, 726  I  609,350;
 356.9 million I  609,350
For 2032, Equation (2) gives b. T  $1160  0.12  $47,150  $11600 
y  0.3117   1117   240
2
 $5426 for I  $47,150
 337.4 million T  $5426  0.22  $100,525  $47,150 
 $17,168.50 for I  $100,525
c. 5426, 17,168.50
Equation (2) gives
y  0.3110   1110   240
2


 319 million

Exercises 0.3 __________________________________________________________________

1. a.  4    4  4  4  4   256
4

23. x 3  x 4  x3 4  x 7
b. 26  1  2  2  2  2  2  2  64
1
1 1 25. x 5  x3  x5  3  x 2 
3. a. 32   x2
32 9
2
3  3  3  9 x8
b.      1      27.  x8  4  x 4
2  2  2  4 x4

5. 1.2 y x 4  2.0736 y5
 y    y12
5  7
29.
y 7
7. 1.5 y x  5  0.1316872428
 
3
31. x 4  x 34  x12
9. 65  63  65 3  68
33. ( xy )2  x 2 y 2
8
10 1
11. 9
 1089  101 
10 10  2  24
4
16 16
35.  5    54  20
 
4
x  x5 x x
94  97 94 ( 7) 93
13.    93 ( 3)  90  1
93 93 93
x8
 
4
37. 2 x 2 y  2 4 x 8 y 4 
15. 3  
3 3
 33 3  39 16 y 4


2
2
3 9
2 
39. 8a 3b 2  2a b   16a
5 4  3 5 2  4
b
17.     
3 2 4  16a 2 b 2
16a 2
1 1 
19.  x 3  1  x 3  1  b2
x3 x3
y2
2 0
21. xy z  x 
1
1 
x   
41. 2 x 2  x 1 y 2   2
x2

x
2 x
x y
2
 2 2  2
xy
y2 y2

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, Chapter 0: Algebraic Concepts

3
 x   x3 1
3 55.    3   x 3
 x3  x 9 1 1 1 2
  2 8
43.  2   6
 9
 6
 9 6
y  y x y x y
57. P  1200, i  0.12, n  5

 a 2 b 1c 4 
3
 b2 
3
a c 6 4 3
a18 c12 S  P (1  i ) n
45.  4 3 0    6 4    2  
b6  1200(1  0.12)5
 a b c  a c   b 
 1200(1.12)5
2 x 2 1 1 1 1 1  $2114.81
47. a.  2   2  
(2 x ) 2
x 2
(2 x) 2
x 2
4x 2
2x 4 I  S – P  2114.81 –1200  $914.81
2
 2x  1 1 1 1 1 59. P  5000, i  0.115, n  6
b.     
 2 x 2 2x 2x 2 2
4x 2
4x 2
16 x 4 S  P (1  i ) n
2 x 2 1 1 1  5000(1  0.115)6
c. 2
 2 2
 2
 4
2x x 2x x  5000(1.115)6
2 x 2 1 1
  2x   2   $9607.70
2
d. 2
 2  4 x2  8
 2x x 2
x2 I  S – P  9607.70 – 5000  $4607.70

1 61. S = 15,000, n = 6, i = 0.115
49.  x1 n
x P  S 1  i 
6
 15, 000 1  0.115 
51.  2 x   2 x  8 x
3 3 3 3
6
 15, 000 1.115 

1 1 1 1 2  $7806.24
53.    x
(4 x 2 ) 4 x2 4



63. I  12.14 1.049 
t


Year 2021 2027 2032
a. t -value 11 17 22
b. Income
$20.5 $27.4 $34.8
(in trillions)

c. I  12.14 1.049 
20
 $31.60 trillion

120
65. y  t
1  5.25 1.066 
a. Year 2015 2020 2030
t  value 15 20 30
U.S. adults with diabetes
39.8 48.7 67.7
in millions
120
b. Year 2035: t  35 ; y  35
 76.9
1  5.25 1.066 
Increase between 2020 and 2035 is 50.0  76.9  26.9 million


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, Chapter 0: Algebraic Concepts

c. There are only a limited number of U.S. adults. To find the upper limit, which is 120 million, compute y for
large t-values:
Year 2100 2200 2300
t  value 100 200 300
Predicted number of
119.0 120.0 120.0
endangered species


67. H  24611.0514 
t


a. t  12 corresponds to 2022.
H  24611.0514 
12
b.  $4490.8 billion
H  24611.0514 
21
c.  $7050.8 billion

Exercises 0.4 __________________________________________________________________
2 6
 16  256 13. x 7/6  x 7
1. a. Since    we have
 3 9
1 1 1 1
256 16
  5.33 15.    x 5/ 4    5/ 4 
9 3 4 4 x 4
4 x5
b. 1.44  1.2
17. y1/ 4  y1/ 2  y (1/ 4)  (1/ 2)  y 3/ 4
 
3
c. 163/ 4  4
16  23  8

 
3
d. ( 16) 3/ 2  16 The square root of a 19. z 3/ 4  z 4  z (3/ 4)  (16/ 4)  z19/ 4
negative number is not real.
1
21. y 3/2  y 1  y ( 3/2) (2/2)  y 5/2 
2/3 2/3 2 y 5/2
 8   27   27   3  2 9
3. a.       3     
 27   8   8  2 4 1
 1   2  3
x3    
 
2
b. 64 2/3
 3
64 2
 4  16 23. 2
 x 3   3   x3  x

2/3 1 1 x3
c.  64   
 64   
2/3 2
3
64
y 5/ 2
25.  y ( 5/ 2)  ( 2/5)  y ( 25/10)  (4/10)
1 1 y 2/5
 
 4  2 16 1
 y 21/10  21/10
y
(6.12)   6.12 
4 4/9
5. 9
 2.237
( x 2/3 )3/ 4  x (2/3)(3/4)  x 2/4  x1/ 2
27.
7. m3  m3/2
1
29. ( x 1/2 )2  x1 
  x
4 1/ 4
9. m 2 n5  m 2 n5  m 2/ 4 n5/ 4  m1/ 2 n5/ 4

1 31. 64 x 4  8 x 2
11. 2x 2 2 x




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, Chapter 0: Algebraic Concepts

33. 128 x 4 y 5  64 x 4 y 4  2 y
2 3 2 3 6
49.   
 64  x 4  y 4  2 y  8 x 2 y 2 2 y 3 3 3 3 3

35. 3
40 x8 y 5  3 8 x 6 y 3  5 x 2 y 2 m2 x m m x mx
51.   
3
 3 8  x6  3 y3  3 5 x2 y 2 mx 2 x x x x

 2 x 2 y 3 5 x2 y 2 3 3 3
m2 x 3
m 3
m x2 mx 2
53.    
3 3 3 3 3
mx5 x4 x3  3 x x2 x x3
37. 12 x3 y  3 x 2 y  36 x5 y 2  36  x 5  y 2
3
mx 2
 6x2 y x 
x2

39. 63x5 y 3  28 x 2 y  9 x 4 y 2  7 xy  4 x 2  7 y 2 2 1 2
55.     x 2/3
2
 3x y 7 xy  2 x 7 y 3 x 3 2 3 x 2/3 3
3 2
 42 x y x
57. 3x x  3 x  x1/ 2  3 x3/ 2
12 x3 y12 4 x 2 y10 2 xy 5
41.   3 1/2 3
9 3 59. x  x
27 xy 2 2 2

1 1/2 1 1 1
4
61. x   1/2 
32a b 9 5
16b 4
b 2b 4 2 2 x 2 x
43.  4
8
  b
4
162a 17 81a 1 3a 2
17
63. a. R  8.5  I  1017/ 2  1017
2
45. ( A9 ) x  A9 x
b. I  109.0  1, 000, 000, 000
A9 x  A1
I 2011 109.0
9x  1 c.   102.1  125.9
I1989 106.9
1
x
9 5
 r 
65. a. S  1000 1  
 R  100 
x
47. 7
 R x /7
5
R x /7  R1  6.6 
b. S  1000 1    $1173.26
 100 
x
1
7
x7
100 13
67. a. P  0.924t13/100  0.924 t
b. Year t Population
2005 5 1.1390
2010 10 1.2464
2045 45 1.5156
2050 50 1.5365
Change from 2005 to 2010 : 0.1074 billion
Change from 2045 to 2050 : 0.0209 billion

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