COLLEGE ALGEBRA 5TH EDITION BY CYNTHIA Y YOUNG SOLUTIONS MANUAL by QuentinAxel

CHAPTER 1 Section 1.1 Solutions ----------------------------------------------------------------------------1.

5x = 35 1 1 ⋅ 5x = ⋅ 35 5 5

4 = −5 + y 5 + 4 = 5 + −5 + y

9=y

4p +5 = 9 4p = 4

5t + 11 = 18 5t = 7 t=7 5

n = 15

−50 = −5t 1 1 − ⋅ ( −50) = − ⋅ ( −5t ) 5 5 10 = t

3x − 5 = 7 3x = 12 x=4

18 = p

11.

p =1

13.

−3 + n = 12 3 + −3 + n = 3 + 12

1 6= p 3 1  3⋅6 = 3⋅ p 3 

1 n=3 5 1 5 ⋅ n = 5 ⋅3 5 n = 15 10.

24 = −3x 1 1 − ⋅ 24 = − ⋅ ( −3x ) 3 3 −8 = x

1 1 ⋅ 4t = ⋅ 32 4 4 t =8

x=7

4t = 32

14.

9m − 7 = 11

12.

2x + 4 = 5

9m = 18

2x = 1

m=2

x =1 2

7x + 4 = 21 + 24x 7x = 17 + 24x −17 x = 17 x = −1

15.

3x − 5 = 25 + 6 x 3x = 30 + 6 x −3x = 30 x = −10

Chapter 1 16.

17.

5x + 10 = 25 + 2 x 5x = 15 + 2 x 3x = 15 x=5

19.

21.

23.

25.

20 n − 30 = 20 − 5n 20 n = 50 − 5n 25n = 50

18.

14c + 15 = 43 + 7c 14c = 28 + 7c 7c = 28

n=2

20.

4( x − 3) = 2( x + 6) 4 x − 12 = 2 x + 12 2 x = 24 x = 12

c=4

5(2 y − 1) = 2(4 y − 3) 10 y − 5 = 8 y − 6 2 y = −1

y = − 12 22.

−3(4t − 5) = 5(6 − 2t ) −12t + 15 = 30 − 10t −15 = 2t − 15 2 = t

2(3n + 4) = −( n + 2) 6n + 8 = −n − 2 7n = −10 n = − 10 7

24.

2 ( x − 1) + 3 = x − 3 ( x + 1)

4 ( y + 6) − 8 = 2y − 4 ( y + 2)

2 x − 2 + 3 = x − 3x − 3

4 y + 24 − 8 = 2 y − 4 y − 8

2 x + 1 = −2 x − 3 4 x = −4

4 y + 16 = −2 y − 8

x = −1

y = −4

6 y = −24

26.

5 p + 6 ( p + 7) = 3( p + 2) 5 p + 6 p + 42 = 3 p + 6

3z + 15 − 5 = 4 z + 7 z − 14

11p + 42 = 3 p + 6

3z + 10 = 11z − 14 −8z = −24

8 p = −36 p=−

3 ( z + 5) − 5 = 4z + 7 ( z − 2 )

z =3

9 2

Section 1.1

27.

28. 7x − (2 x + 3) = x − 2 7x − 2x − 3 = x − 2

3x − (4 x + 2) = x − 5 3x − 4 x − 2 = x − 5

5x − 3 = x − 2

−x − 2 = x − 5

4x = 1 x = 14

3 = 2x 3 = x 2

29.

30. 2 − (4 x + 1) = 3 − (2 x − 1)

5 − (2 x − 3) = 7 − (3x + 5)

2 − 4x − 1 = 3 − 2x + 1

5 − 2 x + 3 = 7 − 3x − 5

1 − 4x = 4 − 2x

8 − 2 x = 2 − 3x x = −6

−3 = 2 x − 32 = x

31.

2a − 9 ( a + 6 ) = 6 ( a + 3 ) − 4 a −7a − 54 = 6a + 18 − 4a −7a − 54 = 2a + 18 −9a = 72 a = −8

32.

25 −  2 + 5 y − 3 ( y + 2 )  = −3 ( 2 y − 5 ) − 5 ( y − 1) − 3 y + 3 25 − [ 2 + 5 y − 3 y − 6 ] = −6 y + 15 − [5 y − 5 − 3 y + 3] 25 − 2 − 5 y + 3 y + 6 = −6 y + 15 − 5 y + 5 + 3 y − 3 29 − 2 y = −8 y + 17 6 y = −12 y = −2

Chapter 1 33.

32 −  4 + 6x − 5 ( x + 4 )  = 4 ( 3x + 4 ) − 6 ( 3x − 4 ) + 7 − 4 x  32 − [ 4 + 6x − 5x − 20 ] = 12x + 16 − [18x − 24 + 7 − 4x ] 32 − 4 − 6x + 5x + 20 = 12x + 16 − 18x + 24 − 7 + 4x 48 − x = −2x + 33 x = −15

34.

12 − 3 + 4 m − 6 ( 3m − 2 )  = −7 ( 2 m − 8 ) − 3 ( m − 2 ) + 3m − 5  12 − [3 + 4 m − 18m + 12 ] = −14 m + 56 − 3 [ m − 2 + 3m − 5] 12 − 3 − 4 m + 18m − 12 = −14 m + 56 − 3m + 6 − 9 m + 15 −3 + 14 m = −26 m + 77 40 m = 80 m=2

35.

20 − 4 c − 3 − 6 ( 2c + 3 )  = 5 ( 3c − 2 ) − 2 ( 7c − 8 ) − 4c + 7  20 − 4 [c − 3 − 12c − 18] = 15c − 10 − [14c − 16 − 4c + 7 ] 20 − 4c + 12 + 48c + 72 = 15c − 10 − 14c + 16 + 4c − 7 44c + 104 = 5c − 1 39c = −105 c=

36.

−35 13

46 − 7 − 8 y + 9 ( 6 y − 2 )  = −7 ( 4 y − 7 ) − 2 6 ( 2 y − 3 ) − 4 + 6 y  46 − [7 − 8 y + 54 y − 18] = −28 y + 49 − 2 [12 y − 18 − 4 + 6 y ] 46 − 7 + 8 y − 54 y + 18 = −28 y + 49 − 24 y + 36 + 8 − 12 y −46 y + 57 = −64 y + 93 18 y = 36 y=2

Section 1.1

37.

38.

1   1  24  z  = 24  z + 3   12   24  2 z = z + 72

1   1  60  m  = 60  m + 1 5   60  12 m = m + 60 11m = 60 m=

z = 72

60 11

39.

40.

x  2x  63   = 63  + 4 7  63  9x = 2 x + 252 7 x = 252

a  a  22   = 22  + 9   11   22  2a = a + 198 a = 198

x = 36

41.

42.

 3x   x 5 10  − x  = 10  −   5   10 2  6x − 10x = x − 25 −5x = −25

1  1   p 24  p  = 24  3 − 24  3   8 p = 72 − p 9 p = 72

x =5

p=8 43.

44.

 5y   2y 5  +  84  − 2 y  = 84   3   84 7  140 y − 168 y = 2 y + 60 −30 y = 60

5m    3m 4  72  2 m − +   = 72  8    72 3  144 m − 45 m = 3m + 96 96 m = 96

y = −2

m =1

Chapter 1 45.

46.

c  5 c 4  − 2c  = 4  −  4  4 2 c − 8c = 5 − 2c −5c = 5

p  5 8 p +  = 8  4  2 8 p + 2 p = 20 10 p = 20

c = −1

p=2 47.

48.

x −3 x − 4 x −6 − = 1− 3 2 6 x −3 x − 4  x −6 6⋅ − = 6 ⋅ 1 −  2  6   3  2( x − 3) − 3( x − 4) = 6 − ( x − 6) 2 x − 6 − 3x + 12 = 6 − x + 6 −x + 6 = −x + 12 6 = 12, which is false. Hence, no solution.

x − 5 x + 2 6x − 1 = − 3 5 15  x −5  x + 2 6x − 1 = 15 ⋅  − 15 ⋅ 1 −  3  15    5 15 − 5( x − 5) = 3( x + 2) − (6 x − 1) 15 − 5x + 25 = 3x + 6 − 6 x + 1 40 − 5x = −3x + 7 33 = 2 x 33 = x 2 1−

49.

50. 4   5  2y  − 5 = 2y   y   2y  8 − 10 y = 5 −10 y = −3

4   2  3x  + 10  = 3x   x   3x  12 + 30x = 2 30x = −10

y≠0

x≠0

x = − 13

3 y= 10

51.

52. 1    10  6x  7 −  = 6x   6x    3x  42 x − 1 = 20 42 x = 21 x=

5 7  6t   = 6t  2 +  3t   6t   7 = 12t + 10 −12t = 3

x≠0

1 2

−1 4

t≠0

Section 1.1

53.

54.

2   4  3a  − 4  = 3a   a   3a  6 − 12a = 4

4   5  x − 2 = x  x ≠ 0 x   2x  4 − 2x = 5 2 2x = 4 − 5 2 = 3 2

a≠0

−12a = −2 a=

x = 34

1 6

55.

56.

 x   2  + 5  = ( x − 2)  ( x − 2)   x≠2  x−2   x−2 x + 5( x − 2) = 2 x + 5x − 10 = 2 6x = 12 x=2 No solution since 2 was excluded from the solution set.

 n   n  + 2  = ( n − 5)  ( n − 5)   n≠5  n −5   n −5  n + 2( n − 5) = n 3n − 10 = n 2 n = 10 n=5 No solution since 5 was excluded from the solution set.

57.

58.

 2p   2  ( p − 1)   = ( p − 1)  3 +  p ≠1 p −1   p −1   2 p = 3( p − 1) + 2 2 p = 3p −3 + 2 2 p = 3 p −1 p =1 No solution since 1 was excluded from the solution set.

8   4t   (t + 2)   = (t + 2)  3 −  t ≠ −2 t+2  t+2 4t = 3(t + 2) − 8 4t = 3t + 6 − 8 t = −2 No solution since −2 was excluded from the solution set.

59.

 3x   2  − 4  = ( x + 2)  ( x + 2)   x ≠ −2 x+2   x+2 3x − 4( x + 2) = 2

−x − 8 = 2 x = −10

Chapter 1 60.

 5y   12  1 − 3  = (2 y − 1)  (2 y − 1)   y≠ 2  2y −1   2 y −1  5 y − 3(2 y − 1) = 12 5 y − 6 y + 3 = 12

− y + 3 = 12 y = −9

61. 1 1 −1 n ≠ −1,0 + = n n + 1 n ( n + 1)

62.

  1  1 1 x ( x − 1)  + x x 1 = − ( )    x ( x − 1)   x x −1   

LCD is n ( n + 1) . So,

First notice that x ≠ 0,1

( n + 1) + n = −1

( x − 1) + x = 1

n + 1 + n = −1 2 n = −2

2x = 2 x =1 But since we have already stipulated that x ≠ 1 , there is no solution.

n = −1 But since we have already stipulated that n ≠ −1, there is no solution.

63. 3 2 9 − = a a + 3 a ( a + 3)

64. 1 1 2 + = c − 2 c c (c − 2)

a ≠ 0, −3

c ≠ 0,2

LCD is c ( c − 2 ) . So,

LCD is a ( a + 3 ) . So,

c + (c − 2) = 2

3 ( a + 3 ) − 2a = 9

2c − 2 = 2 2c = 4 c=2 But since we have already stipulated

3a + 9 − 2a = 9 a=0 But since we have already stipulated that a ≠ 0, there is no solution.

that c ≠ 2, there is no solution.

Section 1.1

65. n −5 1 n −3 = − 6 ( n − 1) 9 4 ( n − 1)

n ≠1

LCD is 36 ( n − 1) . So,

( n − 5 )(36 )( n − 1) = 36 ( n − 1) − ( n − 3)(36 )( n − 1) 6 ( n − 1) 9 4 ( n − 1) 6 ( n − 5 ) = 4 ( n − 1) − 9 ( n − 3 )

So, the final solution is:

6 n − 30 = 4 n − 4 − 9n + 27 6 n − 30 = −5n + 23 11n = 53 53 n= 11

66. 5 3 6 + = m m − 2 m ( m − 2)

m ≠ 0,2

LCD is m ( m − 2 ) . So,

5 ( m − 2) + 3( m) = 6

5m − 10 + 3m = 6 8m − 10 = 6 8m = 16 m=2 Hence, no solution since we have already stipulated that m ≠ 2 .

67.

68. 2 1 x ≠ − 51 , 21 = 5x + 1 2 x − 1 2 ( 2 x − 1) = 1( 5x + 1)

3 2 n ≠ 14 , 5 2 = 4n − 1 2n − 5 3(2n − 5) = 2(4 n − 1) 6 n − 15 = 8n − 2 −13 = 2 n

4 x − 2 = 5x + 1 x = −3

n = − 13 2

Chapter 1 69.

70. t −1 3 t ≠1 = 1− t 2 3 (1 − t ) = 2 ( t − 1)

3 − 3t = 2t − 2 − 5t = −5 t =1 No solution since 1 was excluded from the solution set.

2−x 3 = x≠2 x−2 4 4(2 − x ) = 3( x − 2) 8 − 4 x = 3x − 6 14 = 7 x x=2 No solution since 2 is excluded from the solution set.

71.

72. 9 F = C + 32 5 9 F − 32 = C 5

P = 2 L + 2W P − 2 L = 2W W=

5 ( F − 32 ) = C 9 5 160 C= F− 9 9

73. Let x = number of texts sent. Solve: 20.25 = 3 ( 3 ) + 0.15x

P − 2L 2

74. Let x = number of miles she drove the car. Solve: 185 = 25(5) + 0.10 x 185 = 125 + 0.10 x 60 = 0.10 x 600 = x She drove the car 600 miles.

11.25 = 0.15x 11.25 x= = 75 0.15 Michael sent 75 texts.

75. Let x = number of systems sold. Solve: 72 ( 25 ) + 50 x = 2550

76. Let x = number of bottles sold. 20 x + 15.95 = 575.95 20 x = 560 x = 28 bottles Since there were 28 bottles and each of the four people ordered the same number of bottles, each person ordered 7 bottles.

1800 + 50 x = 2550 50 x = 750 x = 15

Dwayne sold 15 solar power systems. 68

Section 1.1

77. a. C ( x ) = 15,000 + 2,500x b. Solve for x: 15,000 + 2,500 x = 5,515,000

78. a. R( x ) = 5,000 + 0.75x b. Solve for x: 5,000 + 0.75x = 98,750

2,500 x = 5,500,000 x = 2,200 So, 2,200 days.

0.75x = 93,750 x = 125,000 So, 125,000 minutes.

79.

80.

Using a =

d with d = 600mg and c c = 125mg/5mL = 25mg/mL, we see that 600 mg = 24 mL . a= 25 mg / mL

d with d = 600mg and c c = 100mg/5mL = 20mg/mL, we see that 600 mg = 30 mL . a= 20 mg / mL

81.

82.

f =

Using a =

1 1 1 = + f d 0 di

λ≠0

LCD is d 0 di f . So, d 0 di = di f + d 0 f d 0 di − d 0 f = d i f d 0 ( di − f ) = di f d0 =

di f di − f

83. Should have subtracted 4x and added 7 to both sides. The correct answer is x = 5.

84. Forgot to distribute the negative sign through the parentheses.

85. Cannot cross multiply- must multiply by LCD first. The correct answer is p = 6 5 .

86. Should have eliminated x = 0, x = 1 from the domain first.

87. False x ≠ 0

88. False x ≠ 1, −2

Chapter 1 89. True

90. False x = 1 makes denominator = 0

91.

92. ax + b = c

a b x ⋅ −  = x ⋅c x x a − b = cx

a≠0

ax = c − b x=

c−b a

a−b c

93. b+c b −c = x ≠ ±a x+a x−a ( b + c )( x − a ) = ( b − c )( x + a ) bx − ba + cx − ca = bx + ba − cx − ca 2cx = 2ba

ba c

94. 1 1 2 + = y ≠ −a, a,1 y − a y + a y −1 LCD is ( y − a ) ( y + a )( y − 1) . So,

( y + a )( y − 1) + ( y − a )( y − 1) = 2 ( y − a )( y + a )

y 2 − y + ay − a + y 2 − y − ay + a = 2 ( y 2 + ay − ay − a 2 ) y 2 − y + ay −a + y 2 − y −ay +a = 2 y 2 +2ay −2ay − 2a 2

− 2 y = −2 a 2 − y = −a 2 y = a2 y ≥ 0

x ≠ 0, c ≠ 0

Section 1.1

95.

96.

1 x =1 x ≠ −1,0 1 1+ x 1 1 2  = 0 no solution 1− = 1+ x x x

1−

1 t =1

t ≠ 0,1 1 −1 t 1 1 −1 = t +  t = −1 t t

97.

a b x ≠ 0, − b c +1 1+ + c x a y= x + b + cx x ax y= b + x(c + 1) y=

y ( b + x(c + 1) ) = ax yb + xy(c + 1) − ax = 0 x [ y(c + 1) − a ] = − yb x= 98.

by a − y − cy

2 − a = 2 + 5 − 3a ( 2 ) 2 − a = 7 − 6a −5 = −5a a =1

Chapter 1 99. y1 = 3 ( x + 2 ) − 5x

100. y1 = −5 ( x − 1) − 7

y2 = 3x − 4

y2 = 10 − 9x

x =3

x=2 101. y1 = 2x + 6 y2 = 4 x − 2 x + 8 − 2

102. y1 = 10 − 20 x y2 = 10x − 30 x + 20 − 10

All real numbers All real numbers

Section 1.1

103. y1 =

x ( x − 1) x2

104. y2 = 1

y1 =

2x ( x + 3 ) x2

y2 = 2

No solution No solution 105. y1 = 0.035x + 0.029(8706 − x ) y2 = 285.03

106.

x = 5426

x = 7.95

y1 =

1 0.45 − 0.75x x

y2 =

1 9

Section 1.2 Solutions ---------------------------------------------------------------------------1. Let x = price without coupon 0.9x = 217.95 x = $242.17

2. Let x be the percentage of original price 51.80 x= = 0.7 = 70% 74 30% markdown

3. Let x = cost of pizza

4. Let x = take home pay Bills = 0.5x Investments = 0.2 x

Tom: 5.16 Chelsea:1 8 x Jeff:1 2 x 1 1 5.16 + x + x = x 8 2 41.28 + x + 4 x = 8x 3x = 41.28

Groceries = 560 Miscellaneous = 0.23x 0.5x + 0.2 x + 560 + 0.23x = x 0.93x + 560 = x 0.07 x = 560

x = $13.76

x = $8,000

5. Let x = original price 0.85x = 125,000 x = 147,058.82

6. Let x = price paid to publisher 1.25x = 79 79 x= = 63.2 1.25 Bookstore paid $63.20

Original price ≅ $147,058.82 Model price = $125,000 Savings = $22,058.82

Section 1.2

7. Let x = distance from Angela's home to the restaurant. Home → Train station =1 mile 3 1 On train → x In taxi → x 4 6 3 1 1+ x + x = x 4 6 LCD =12 12 + 9x + 2 x = 12 x 12 + 11x = 12 x x = 12 Angela travels 12 miles to the

8. Let x = distance from her house to VAB House → Park & Ride = 7 miles 5 Park & Ride → H.Q. = x 6 1 H.Q. → VAB = x 20 5 1 7+ x+ x = x 6 20 LCD = 60 420 + 50x + 3x = 60 x 420 = 7x x = 60

She travels 60 miles to the VAB.

restaurant.

9. x = hours awake 1 Class: x 3 1 Eating: x 5 Working out:

1 x 10

Studying: 3

10. Let x = calories for breakfast Dinner calories = 2x Lunch Calories = x + 100 Snack 1=100 Snack 2 =150 lunch snacks     + + 100 + 2 + 100 + 150 = 1550 x x x    breakfast

4 x + 350 = 1550 4 x = 1200 x = 300

Other things: 2.5 1 1 1 x + x + x + 3 + 2.5 = x 3 5 10 10 x + 6 x + 3x + 165 = 30 x 19x + 165 = 30 x

Breakfast: 300 Lunch: 400 Dinner: 600

11x = 165 x = 15 awake 9 hours of sleep

dinner

total

Chapter 1 11.

Fixed costs =15,000

12. Let x = number of sets of napkins

Variable costs =18.50x Total costs = 20,000

Fixed monthly costs =1329.50 Variable costs = 3.70 x

18.50 x + 15,000 = 20,000 18.50 x = 5000

Total budget =1870 1329.50 + 3.70x = 1870

Approximately 270 units can

3.70 x = 540.5 x = 146.08

be produced.

She can afford approximately 146

x = 270.27

sets of napkins. 13.

14. 2 1 x − 10 = x 3 4 5 x = 10 12

10 x = 2 x + 16 8x = 16 x=2

 12  x = 10   = 24 5 

15. Let the numbers be x, x + 2

16. Let the numbers be x, x + 1, x + 2

4 (x) = 2 + 3(x + 2)

x + ( x + 1) + ( x + 2 ) = 2  x + ( x + 1) 

4 x = 2 + 3x + 6 x =8

3x + 3 = 2 ( 2 x + 1) 3x + 3 = 4x + 2 x =1

The numbers are 8,10 .

The numbers are 1,2,3 .

Section 1.2

17. Let p = perimeter. First side =11

Second side =

18. w = width l = length = 2w + 1 p = 2l + 2w

1 p 5

20 = 2 ( 2w + 1) + 2w

1 Third side = p 4 1 1 11 + p + p = p 5 4 LCD = 20 220 + 4 p + 5 p = 20 p 220 = 11 p p = 20

20 = 4w + 2 + 2w 18 = 6w w =3 width = 3 feet length = 7 feet

The perimeter is 20 inches . 19.

20. w = width l = length = 2w + 40 p = 2l + 2w

w = width l = length = 3w + 2 p = 2l + 2w

260 = 2 ( 2w + 40 ) + 2w

28 = 2 ( 3w + 2 ) + 2w

260 = 4w + 80 + 2w 180 = 6w w = 30

28 = 6w + 4 + 2w 24 = 8w w =3

width = 30 yards length =100 yards

width = 3 inches length =11 inches

Chapter 1 21. r1 = radius of smaller circle r2 = radius of larger circle r2 = r1 + 3 Circumference of smaller circle = 2π r1 Circumference of larger circle = 2π r2 Ratio of circumferences =

2π r2 r2 2 = = 2π r1 r1 1

22. Perimeter of semicircle is half the circumference + twice the radius. 1 P = ( 2π r ) + 2r 2 P = (π + 2 ) r When the radius is increased by one, the perimeter doubles, so 2P = (π + 2 )( r + 1)

r2 = 2r1

2 (π + 2 ) r = (π + 2 ) ( r + 1)

2r1 = r1 + 3

2r = r + 1

r1 = 3

r =1

r1 = 3 feet, r2 = 6 feet

23.

24. x 4 = 225 3 3x = 900 x = 300

x 4 = 880 10 10 x = 3520 x = 352

The tree is 300 feet tall.

The oak tree is 352 feet tall.

25. Let x = length of alligator in feet. Solve: x 3.5 = 0.5 0.75 0.5x = 2.625 x = 5.25 The alligator is about 5.25 feet .

26. Let x = length of snake in inches. Fang 2 2.6 = = x Body 36 Solve: 2 x = 93.6 x = 46.8 The snake is about 3.9 feet = 46.8 inches .

Section 1.2

27. Let x = amount invested at 4%. 120,000 − x = amount invested at 7% Solve:

28. Let x = amount invested at 10%. 13,000 − x = amount invested at 14% Solve:

0.04x + 0.07 (120,000 − x ) = 7,800

0.10 x + 0.14 (13,000 − x ) = 1580

0.04x + 8400 − 0.07x = 7,800 − 0.03x = −600 x = 20,000

0.10 x + 1820 − 0.14 x = 1580 − 0.04x = −240 x = 6000

$20,000 at 4%, $100,000 at 7%

$6,000 at 10%, $7,000 at 14%

29. Let x = amount invested at 10%. 14,000 − x = amount invested at 2% 2 14,000 − x = amount invested at 40% 2 Interest earned = 16,610 − 14,000 = 2,610 Solve:  14,000 − x   14,000 − x  0.1x + 0.02   + 0.4   = 2610 2 2     0.1x + 140 − 0.01x + 2800 − 0.2x = 2610 − 0.11x = −330 x = 3,000 $3,000 at 10%, $5,500 at 2%, $5,500 at 40%

Chapter 1 30.

$2500 in money market ( rate = x ) $2500 in stock market ( rate = 3x ) Interest earned = $150 Solve: 2500 x + 2500 ( 3x ) = 150 2500 x + 7500 x = 150 10,000 x = 150 x = 0.015 Money market:1.5% Stock market: 4.5%

31.

Money for plants = 4200 − 2400 − 1500 = 300 Let x be the number of trees ( $32 each ) .

Let 33 − x be the number of shrubs ( $4 each ) . Solve: 32x + 4 ( 33 − x ) = 300 32x + 132 − 4 x = 300 28x = 168 x=6 6 trees, 27 shrubs

Section 1.2

32. Let x = lbs of turkey ( $6.32 per lb )

33. Let x = ml of 5% HCl

3.2 − x = lbs of cheese ( $4.27 per lb )

Solve: 100 − x = ml of 15% HCl

Solve:

0.05x + 0.15 (100 − x ) = 0.08 (100 )

6.32x + 4.27 ( 3.2 − x ) = 17.56

0.05x + 15 − 0.15x = 8 − 0.1x = −7

6.32x + 13.664 − 4.27x = 17.56 2.05x = 3.896 x = 1.9

x = 70 70 ml of 5% HCl

1.9 lbs of turkey,1.3 lbs of cheese

30 ml of 15% HCl

34. Let x = gallons of 100% alcohol Solve: 0.20(5) + 1.00 x = 0.50(5 + x ) 1 + x = 2.5 + 0.50 x 0.50 x = 1.5 x=

35. Let x = number of gallons to be drained. Solve: 0.40(5 − x ) + 1.00 x = 0.80(5) 2 − 0.40 x + x = 4 2 + 0.60 x = 4 0.60 x = 2

1.5 =3 0.5

x ≈ 3.3

3 gallons. About 3.3 gallons. 36.

Let x = $ of grant with 42.5% overhead 1,170,000 − x = $ of grant with 26% overhead Overhead generated = 0.39 (1,170,000 ) = 456,300 Solve: 0.425x + 0.26 (1,170,000 − x ) = 456,300 0.425x + 304,200 − 0.26 x = 456,300 0.165x = 152100 x = 921818..18 On campus ( 42.5% ) grant: $921,818

Off campus ( 26% ) grant: $248,182

Chapter 1 37. x = lbs of caramels ( $1.50/lb )

38. Let x = lbs of Jamaican Blue Mountain Coffee Solve: 12( x ) + 4.20(2 − x ) = 14.25 12 x + 8.40 − 4.20 x = 14.25

1.25 − x = lbs of gummy bears ( $2/lb ) Solve: 1.5x + 2 (1.25 − x ) = 2.50

7.80 x = 5.85

1.5x + 2.5 − 2 x = 2.50 − 0.5x = 0 x=0 No caramels,1.25lb of gummy bears

x = 0.75 She bought: 0.75 lbs Jamaican Blue Mountain 1.25 lbs of regular coffee beans.

39.

40.

distance = rate ⋅ time distance =100,000,000 miles

distance = 0.5 miles

rate = 670,616,629 mph

rate = 760 mph

time =

distance rate

time =

= 0.15 hours ≅ 9 minutes 41.

distance = rate ⋅ time

distance rate

= 0.0006579 hours ≅ 2.4 seconds

x + 0.047 x = 3.21 1.047 x = 3.21 x = 3.065 So, at the beginning of November, gas was $3.07 per gallon.

42. Let x = price in September. Then, since the price decreased by 40% by the end of Nov., the price for the TV would be 60% of the price in September. So, we solve the following equation for x: $299 = 0.60 x x ≈ $498

43. Let x = number of mL of distilled water (which has 0% salt). Solve for x: 0.03(100 mL ) + 0.00( x mL ) = 0.009(100 + x )mL 3mL + 0 mL = (0.9mL + 0.009x ) 2.1mL = 0.009x x ≈ 233mL

Section 1.2

44. Let x = number of mL of 20% solution. Solve for x: 0.00(100 mL ) + 0.20( x mL ) = 0.05(100 + x )mL 0.20x = (5 + 0.05x )mL 0.15x = 5mL x ≈ 33mL

45.

46.

boat speed: s = 16 mph

plane speed: s = 130 mph

upstream: r = s − c, t = 1 3 hours

upwind: r = s − w, t = 2 hours

downstream: r = s + c, t = 1 4 hours

downwind: r = s + w, t = 1.25 hours

rate ( r ) = boat speed ( s ) ± current speed ( c )

rate ( r ) = plane speed ( s ) ± wind speed ( w )

Distance is the same both ways ( rate ⋅ time ) Distance is the same both ways ( rate ⋅ time )

Solve:

Solve: 2 (130 − w ) = 1.25 (130 + w )

(16 − c )   = (16 + c )   1 3

1 4

260 − 2w = 162.5 + 1.25w 97.5 = 3.25w

4 (16 − c ) = 3 (16 + c ) 64 − 4c = 48 + 3c 7c = 16 c=

w = 30 mph

16 ≅ 2.3 mph 7

Chapter 1 47.

48.

rate of walker = rw

dist. of Southern route = d

rate of jogger = rw + 2

dist. of Northern route = d + 300

time of walker =1 hour

time of S route = 45 hours time of N route = 50 hours

2 3

time of jogger = hour

d d + 300 = 45 50 50d = 45 ( d + 300 )

rw (1) = ( rw + 2 )( 2 3 ) rw = 23 rw + 4 3 r = 34

1 3 w

50d = 45d + 13,500

rw = 4

5d = 13,500

walker: 4 mph jogger: 6 mph

d = 2700 S route = 2,700 miles N route = 3,000 miles

49. Let x = number of minutes it takes a rider to get to class Using Distance = Rate × Time, and the fact that since they use the same path, their distances are the same, we must solve the equation: 2(12 + x ) = 6( x ) 24 + 2x = 6x 24 = 4x x=6 So, it takes the bicyclist 6 minutes to get to class, and the walker 18 minutes.

50. Let x = number of minutes a car travels before catching the truck When it catches the truck, the car and truck will have traveled the same distance. So, we must solve: 70x = 50( x + 30) 70x = 50 x + 1500 x = 75 It takes the car 75 minutes to catch the truck.

Section 1.2

51. Let x = hours it takes Cynthia to paint house alone. Christopher can paint 1/15 house per hour. Cynthia can paint 1/x house per hour. Together they  1 1 paint  +  house per hour.  15 x 

1 1 1 + = 15 x 9 3x + 45 = 5x 2 x = 45 x = 22.5 Cynthia can paint the house alone in 22.5 hours.

52. Let x = number of hours it takes Morgan to complete the yard alone. Jay: 1/3 of the yard per hour Morgan: 1/x of the yard per hour Together: 1 hour for entire yard Solve: 1 1 + =1 3 x 1 1  3x ⋅  +  = 3x 3 x x + 3 = 3x x = 1.5 It takes 1.5 hours for Morgan to do the yard herself. 53. Tracey can do 1/4 of a delivery per hour, and Robin can do 1/6 of a delivery per hour. Together, they complete 1/4 + 1/6 = 1/(12/5) of the delivery in an hour. So, together, they complete the job in 2.4 hours. 54. 55. Joshua can do the job at a rate of 1/30 4 264 = job per minute. Amber can do it in 1/20 x1 5 job per minute. Together, they work at 4 x1 = 264 ( 5 ) (1 30 + 1 20 ) = 1 12 job per minute. They 4 x1 = 1320 will finish in 12 minutes x1 = 330 hertz

4 264 = 6 x2 4 x2 = 264 ( 6 ) 4 x2 = 1584

x2 = 396 hertz

Chapter 1 56. 10 220 = x1 12

10 220 = 15 x2

10 x1 = 220 (12 )

10 x2 = 220 (15 )

10 x1 = 2640

10 x2 = 3300

x1 = 264 hertz

x2 = 330 hertz

57. Let x = exam grade needed 86 + 80 + 84 + 90 Test average = = 85 4 To earn a "B": To earn an "A": 1 2 1 2 (85 ) + x = 80 (85 ) + x = 90 3 3 3 3 LCD = 3 LCD = 3 85 + 2x = 240 85 + 2x = 270 2 x = 155 2 x = 185

x = 77.5 58. Let x = exam grade needed 80 + 83 + 71 + 61 + 95 + x + x Avg = = 80 7 2 x + 390 = 80 7 2 x + 390 = 560

x = 92.5

59. Let x = # field goals 8 − x = # touchdowns 3x + 7 ( 8 − x ) = 48 3x + 56 − 7 x = 48

−4 x = −8 x=2

2 x = 170 x = 85

2 field goals, 6 touchdowns

60.

d +5 100 10 d d +5 = 100 12 100 10 12 10 d= (d + 5) 100 100 12d = 10d + 50 2d = 50 d = 25 DB: t =

100yds 100yds DB: 12secs 10secs Let d be the distance the TE runs. TE:

Then d + 5 is the distance the DB runs. Time spent running is the same for both. dist=rate ⋅ time, time=dist rate TE: t =

d 100 12

TE catches ball at 20 yard line and is tackled 25 yards later at the 45.

Section 1.2

61.

62.

( 42 )(5 ) = ( 60 )( x ) 210 = 60 x x = 3.5

(33 + 42 )( 4 ) = 60x ( 75 )( 4 ) = 60x x =5

Maria should sit 3.5 feet

Maria should sit 5 feet

from the center.

63. Let the board be 1 unit long. Let x = distance from Maria to fulcrum. 1 − x = distance from Max to fulcrum. 60x = 42 (1 − x ) 60x = 42 − 42x 102 x = 42 x ≅ 0.4 Fulcrum is 0.4 units from Maria and 0 .6 units from Max.

64. Let the board be 1 unit long. Let x = distance from Maria to fulcrum. 1 − x = distance from Max/Martin to fulcrum. 60x = ( 42 + 33 )(1 − x ) 60x = 75 − 75x 135x = 75 x = 0.55 Fulcrum is 0.56 units from Maria and 0.44 units from Max/Martin.

Chapter 1 65.

66. 1 1 1 = + f d 0 di

1 1 1 = + f d 0 di

f = 3, di = 5

f = 8, di = 2

1 1 1 = + 3 d0 5

1 1 1 = + 8 d0 2

LCD =15d 0

LCD = 8d 0

5d 0 = 15 + 3d 0

d 0 = 8 + 4d 0

2d 0 = 15

−3d 0 = 8

Object is d 0 = 7.5 cm from lens.

−8 ≈ −2.67 3 Object is 2.67 cm behind lens . d0 =

67.

68. 1 1 1 = + f d 0 di

1 1 1 = + f d 0 di

f = 2, di = 12 d 0

1 f = 8, di = d 0 2 1 1 1 = + 8 d0  1   d0  2  1 1 2 = + 8 d0 d0

1 1 1 = +1 2 d0 2 d0 1 2 = , d0 2 d0

Since 1

1 1 2 3 = + =  d0 = 6 2 d0 d0 d0

1 3 = 8 d0

Object distance = 6 cm

d 0 = 24 cm

69.

P = 2l + 2w P − 2l = 2w

P − 2l =w 2

70.

71.

P = 2l + 2w P − 2w = 2l

72. A = bh

C = 2π r

2 A = bh

C =r 2π

1 2

P − 2w =l 2

2A =h b

Section 1.2

73.

A = lw

74.

A =w l

75.

d = rt

V = lwh

76.

V =h lw

d =t r

V = π r2h V =h π r2

77. Let x = Janine’s average speed (in mph). Then, Tricia’s speed = (12 + x) mph. We must solve the equation: 2.5(12 + x ) + 2.5x = 320 30 + 2.5x + 2.5x = 320 5x = 290 x = 58 So, Janine’s average speed is 58 mph and Tricia’s average speed is 70 mph.

78. Let x = Rick’s average speed (in mph). Then, Mike’s speed = (8 + x) mph. We must solve the equation: 1.5(8 + x ) + 1.5x = 210 12 + 1.5x + 1.5x = 210 3x = 198 x = 66 So, Rick’s average speed is 66 mph, Mike’s average speed is 74 mph.

79. y = 11896.67 x + 132500

80. y = 11896.67 x + 132500

$191,983.35

$144,397

Chapter 1 81. Let x = number of times you play. Option A: y1 = 300 + 15x Option B: y2 = 150 + 42 x

Option B is better if you play about 5 times or less per month. . Option A is better if you play 6 times or more per month.

82. Let x = number of minutes used. Plan A: y1 = 30 + 0.1x Plan B: y2 = 50 + 0.03x

Plan A is better if you use about 285 minutes or less per month. Plan B is better if you use more than 285 minutes per month.

Section 1.3 Solutions -------------------------------------------------------------------------------1.

2. x − 5x + 6 = 0

v 2 + 7v + 6 = 0

( x − 3)( x − 2 ) = 0

(v + 6 )( v + 1) = 0

x − 3 = 0 or x − 2 = 0

v + 6 = 0 or v + 1 = 0

x = 3 or x = 2

v = −6 or v = −1

4. p − 8 p + 15 = 0

u 2 − 2u − 24 = 0

( p − 5)( p − 3) = 0

( u − 6)( u + 4) = 0

p = 5 or p = 3

u = 6 or u = −4

x = 12 − x

11x = 2 x 2 + 12

x 2 + x − 12 = 0

2 x 2 − 11x + 12 = 0

( x + 4 )( x − 3 ) = 0

( 2x − 3 )( x − 4 ) = 0

x + 4 = 0 or x − 3 = 0

2 x − 3 = 0 or x − 4 = 0

x = −4 or x = 3

x = 32 or x = 4

8. 3x 2 + 10 x − 8 = 0

16 x + 8x = −1 2

(3x − 2 )( x + 4 ) = 0

16 x 2 + 8x + 1 = 0

( 4x + 1)( 4x + 1) = 0

3x − 2 = 0 or x + 4 = 0

4x + 1 = 0

x = 23 or x = −4

x = −1 4 9.

10. 9y +1 = 6 y

4x = 4x 2 + 1

9 y2 − 6 y + 1 = 0 (3 y − 1)(3 y − 1) = 0

4x2 − 4x + 1 = 0 (2 x − 1)(2 x − 1) = 0

y = 13

x = 12

Chapter 1 11.

12. 8 y − 16 y = 0

3A2 = −12A

8y ( y − 2 ) = 0

3A2 + 12A = 0

8 y = 0 or y − 2 = 0

3A ( A + 4 ) = 0

y = 0 or y = 2

3A = 0 or A + 4 = 0 A = 0 or A = −4

13.

14. −4 x = 12 x

7 p 2 = 28 p

−4 x 2 − 12 x = 0

7 p2 − 28 p = 0

−4 x ( x + 3 ) = 0

7 p ( p − 4) = 0

x = −3, 0

p = 0, 4

15.

16. 9 p = 12 p − 4

4u 2 = 20u − 25

9 p 2 − 12 p + 4 = 0

4u 2 − 20u + 25 = 0

(3 p − 2 )(3 p − 2 ) = 0

(2u − 5)(2u − 5) = 0

3p − 2 = 0

u = 52

p = 23

17.

18. x −9 = 0

16v 2 − 25 = 0

( x + 3)( x − 3 ) = 0

(4v − 5)(4v + 5) = 0

x + 3 = 0 or x − 3 = 0

v = 5 4 or v = − 5 4

x = −3 or x = 3

Section 1.3

19.

20.

x ( x + 4 ) = 12

3t 2 − 48 = 0

3 ( t 2 − 16 ) = 0

x 2 + 4 x = 12

3 ( t + 4 )( t − 4 ) = 0

x 2 + 4 x − 12 = 0

( x + 6 )( x − 2 ) = 0

t + 4 = 0 or t − 4 = 0

x + 6 = 0 or x − 2 = 0

t = −4 or t = 4

x = −6 or x = 2

21.

22.

2 p − 50 = 0

5 y 2 − 45 = 0

2( p − 5)( p + 5) = 0

5( y − 3)( y + 3) = 0

p = −5 or p = 5

y = −3 or y = 3

2 ( p 2 − 25 ) = 0

5 ( y2 − 9 ) = 0

23.

24.

3x = 12

7v 2 = 28

3x 2 − 12 = 0

7v 2 − 28 = 0

3( x − 2)( x + 2) = 0

7(v − 2)(v + 2) = 0

x = −2 or x = 2

v = −2 or v = 2

3 ( x2 − 4 ) = 0

7 (v2 − 4 ) = 0

26.

25. x − 16 = 0

t 2 − 49 = 0

( x − 4 )( x + 4 ) = 0

(t − 7 )(t + 7 ) = 0

x = ±4

t = ±7

Chapter 1 28.

27. p −8 = 0

y 2 − 72 = 0

p2 = 8

y 2 = 72

p=± 8

y = ± 72

( 72 = 2 ⋅ 3 ) 3

p = ±2 2

y = ±6 2

29.

30. x +9 = 0

v 2 + 16 = 0

x 2 = −9

v 2 = −16

x = ±3i

v = ±4i

31.

32. 5 y − 20 = 0

3x 2 − 147 = 0

5( y 2 − 4) = 0

3( x 2 − 49) = 0

y2 = 4 y = ±2

x 2 = 49

x = ±7

33.

34.

( x − 3) = 36

( x − 1) = 25

x − 3 = ±6 x = 3±6

x − 1 = ±5 x = ±5 + 1

x = −3, 9

x = −4, 6

35.

36.

(3x + 5 ) = 16

( 4 y − 1) = 49

3x + 5 = ±4 3x = −5 ± 4

4 y − 1 = ±7 4y = 1± 7

−5 ± 4 3 1 x = −3, − 3

1± 7 4 3 y = − ,2 2

Section 1.3

38.

37.

( 2 x + 3 ) = −4

( 4x − 1) = −16

2 x + 3 = ±2i

4 x − 1 = ±4i 4 x = 1 ± 4i

2 x = −3 ± 2i x=

−3 ± 2i 2

x = 14 ± i

39.

40.

(5x − 2 ) = 27

(3x + 8) = 12

5x − 2 = ± 27

3x + 8 = ± 12

5x = 2 ± 3 3

3x = −8 ± 2 3

2 ±3 3 5

41.

−8 ± 2 3 3

42.

(1 − x ) = 9

(1 − x ) = −9

1 − x = ±3

1 − x = ±3i

−x = −1 ± 3 x = 1± 3

−x = −1 ± 3i

x = 1 ± 3i

x = −2, 4

44.

43. x + 6x

x 2 − 8x

1  2  ⋅6 = 3 = 9 2 

1  2  ⋅ 8  = 4 = 16 2 

x 2 + 6x + 9

x 2 − 8x + 16

45.

46. x − 12 x

x 2 + 20 x

1  2  ⋅12  = 6 = 36 2 

1  2  ⋅ 20  = 10 = 100 2 

x 2 − 12 x + 36

x 2 + 20x + 100

Chapter 1 47.

48.

x + 9x

x 2 − 3x

x 2 + 9x +

1   −3  9  ⋅ −3  =   = 4 2   2 

 1   9  81  ⋅9 =   = 4 2  2

81 4

x 2 − 3x +

49.

9 4

50.

x − 15x

x 2 + 11x

1   −15  225  ⋅ −15  =   = 4 2   2 

x 2 − 15x +

 1   11  121  ⋅11 =   = 4 2   2 

225 4

x 2 + 11x +

51.

121 4

52.

1 x2 − x 3

1 x2 − x 2 2

1 1 1 1  ⋅  =  =  2 2   4  16

1 1 1 1  ⋅  =  =  3 2   6  36

1 1 x2 − x + 2 16

1 1 x2 − x + 3 36 54.

53.

2 x2 + x 5 2

4 x2 + x 5 2

1 1 2 1  ⋅  =  =  2 5   5  25

4 1 4 2  ⋅  =  =  2 5   5  25

2 1 x2 + x + 5 25

4 4 x2 + x + 5 25

Section 1.3

56.

55. x − 2.4 x

x 2 + 1.6 x

1  2  ⋅ 2.4  = 1.2 = 1.44 2 

1  2  ⋅1.6  = 0.8 = 0.64 2 

x 2 − 2.4 + 1.44

x 2 + 1.6 x + 0.64

57.

58. x + 2x = 3

y2 + 8 y − 2 = 0

x 2 + 2x + 1 = 3 + 1

y 2 + 8 y + 16 = 2 + 16

( x + 1) = 4

( y + 4 ) = 18

x + 1 = ±2

y + 4 = ± 18

x = −1 ± 2

y = −4 ± 3 2

x = −3,1

59.

60. 2

t − 6t = −5

x 2 + 10 x = −21

t 2 − 6t + 9 = −5 + 9

x 2 + 10 x + 25 = −21 + 25

(t − 3) = 4

( x + 5) = 4

t − 3 = ±2

x + 5 = ±2

t = 1, 5

x = −7, − 3

62.

61. y − 4 y = −3

x 2 − 7x = −12

y 2 − 4 y + 4 = −3 + 4

7 7 x − 7x +   = −12 +   2 2

( y − 2) = 1 2

7 49 1  =  x −  = −12 + 2 4 4  7 1 x− =± 2 2 x = 3, 4

y − 2 = ±1 y = 1, 3

Chapter 1 63.

64. x − 5x = 9

x 2 + 7 x = −11

x 2 − 5x +

25 25 =9+ 4 4

x 2 + 7x +

49 49 = −11 + 4 4

5  61  x −  = 2 4 

7 5  x +  = 2 4 

5 61 =± 2 2 5 ± 61 x= 2

7 5 =± 2 2 −7 ± 5 x= 2

x−

65.

66. x + 3x = 4 9 9 x 2 + 3x + = 4 + 4 4

x 2 − 9x + 7 = 0

x 2 − 9x = −7 x 2 − 9x +

3  25  x +  = 2 4  3 5 x+ =± 2 2 x = −4,1

81 81 = −7 + 4 4

9  53  x +  = 2 4  9 53 =± 2 2 9 ± 53 x= 2 x+

Section 1.3

68.

67. 2 p + 8 p = −3

2 x 2 − 4 x = −3

2 ( p 2 + 4 p + 4 ) = −3 + 8

2 ( x 2 − 2 x + 1) = −3 + 2

2 ( p 2 + 4 p ) = −3

2 ( x 2 − 2 x ) = −3

2 ( p + 2) = 5

2 ( x − 1) = −1

5 2 5 p+2 = ± 2

( x − 1) =

−1 2 1 x −1 = ± i 2

( p + 2) =

−4 ± 10 2

x = 1±

69.

2 i 2

70. 2 x − 7 x = −3

3x 2 − 5x = 10

7   2  x 2 − x  = −3 2  

5   3  x 2 − x  = 10 3  

2 2  2 7 7  7 2  x − x +    = −3 + 2    2  4   4 

2 2  2 5 5  5 3  x − x +    = 10 + 3    3  6   6 

7   49  2  x −  = −3 + 2   4   16 

5   25  3  x −  = 10 + 3   6   36 

7 −3 49 25  + = x −  = 4 2 16 16  7 5 x− =± 4 4 1 x = ,3 2

5  10 25 145  = x −  = + 6 3 36 36  5 145 145 =± =± 6 36 6 (145 = 5 ⋅ 29 ) so the radical x−

can't be reduced. x=

5 ± 145 6

Chapter 1 71.

72. 2

t 2 2t 5 + + =0 3 3 6 5 t 2 + 2t = − 2 5 t 2 + 2t + 1 = − + 1 2 −3 2 (t + 1) = 2 3 t +1 = ± i 2

1 x − 2x = 2 4 1 x2 − 4x = 2 x2 − 4x + 4 =

1 +4 2

9 2 3 x−2 = ± 2

(x − 2) = 2

4 ±3 2 2

73.

t = −1 ±

6 i 2

74. t + 3t − 1 = 0

t 2 + 2t − 1 = 0

−3 ± 9 + 4 2

−3 ± 13 2

75.

1 −2 ± 4 + 4 8 = −1 ± 2 2

t = −1 ± 2

76. s + s +1 = 0

2 s 2 + 5s + 2 = 0

−1 ± 1 − 4 −1 ± −3 = 2 2

−1 ± i 3 2

−5 ± 25 − 16 −5 ± 9 = 4 4 −1 s = −2, 2

100

Section 1.3

77.

78. 3x − 3x − 4 = 0

4x 2 − 2x − 7 = 0

3 ± 9 + 48 1 57 = ± 6 2 6

3 ± 57 6

79.

2 ± 4 + 4 ⋅ 28 2 ± 116 = 8 8 2 (116 = 2 ⋅ 29 ) x=

2 ± 2 29 8

1 ± 29 4

80. 4 m2 + 7 m + 8 = 0

x − 2x + 17 = 0 2

2 ± 4 − 4 ⋅17 2 ± −64 = 2 2 2 ± 8i x= 2 x = 1 ± 4i

81.

−7 ± 49 − 4 ⋅ 32 −7 ± −79 = 8 8

7 79 m=− ± i 8 8

82. 5x + 7 x − 3 = 0

3x 2 + 5x + 11 = 0

−7 ± 49 + 60 10

−7 ± 109 10

5 107 x=− ± i 6 6

83. 1 2 2 1 x + x− =0 4 3 2 2 3x + 8x − 6 = 0

−5 ± 25 − 132 −5 ± −107 = 6 6

84. 1 2 2 1 x − x− =0 4 3 3 2 3x − 8x − 4 = 0

−8 ± 64 − 4(3)( −6) −8 ± 2 34 = 2(3) 2(3)

8 ± 64 − 4(3)( −4) 8 ± 4 7 = 2(3) 2(3)

−4 ± 34 3

4±2 7 3

101

Chapter 1 85. 2 ( −22 ) − 4 (1)(121) = 484 − 484 = 0

86. 2 ( −28) − 4 (1)(196 ) = 784 − 784 = 0

1 real solution ( repeated root )

87. 2 ( −30 ) − 4 ( 2 )( 68) = 900 − 544 = 356

88. 2 ( 27 ) − 4 ( −3)( 66 ) = 729 + 792 = 1521

2 real solutions ( distinct )

89.

( −7 ) − 4 ( 9 )( 8) = 49 − 288 = −239 2 complex solutions ( complex conjugate ) 2

90.

(5 ) − 4 ( −3)( −7 ) = 25 − 84 = −59 2 complex solutions ( complex conjugate ) 2

91.

92.

v 2 − 8v + 20 = 0

v − 8v − 20 = 0 2

(v − 10 )(v + 2 ) = 0

8 ± 64 − 80 8 ± −16 = 2 2 8 ± 4i v= 2 v = 4 ± 2i v=

v = −2,10

94.

93. 2

t + 5t − 6 = 0

t 2 + 5t + 6 = 0

(t + 6 )( t − 1) = 0

(t + 2 )(t + 3) = 0

t = −6,1

t = −2, − 3

95.

96.

( x + 3 ) = 16

( x + 3) = −16

x + 3 = ±4

x + 3 = ±4i

x = −7,1

x = −3 ± 4i

102

Section 1.3

97.

98.

( p − 2) = 4 p

( u + 5)2 = 16u

p2 − 4 p + 4 = 4 p

u 2 + 10u + 25 = 16u

p2 − 8 p + 4 = 0

u 2 − 6u + 25 = 0

6 ± 36 − 4(25) 6 ± 8i = 2 2 u = 3 ± 4i

8 ± 64 − 4(1)(4) 8 ± 4 3 = 2(1) 2

p = 4±2 3 99.

100. 8w + 2w + 21 = 0

8w 2 + 2w − 21 = 0

−2 ± 4 − 4 ⋅ 8 ⋅ 21 16 −2 ± −668 −2 ± 2i 167 = w= 16 16

( 4w + 7 )( 2w − 3) = 0

4w + 7 = 0 or 2w − 3 = 0 w=

−7 3 , 4 2

−1 ± i 167 8

101.

102. 3p − 9 p +1 = 0

3 p2 − 9 p − 1 = 0

9 ± 81 − 12 6

9 ± 81 + 12 6

9 ± 69 6

9 ± 93 6

103

Chapter 1 103.

104.

2 2 4 1 t − t− =0 3 3 5 LCD =15

1 2 2 2 x + x− =0 2 3 5 LCD = 30

10t 2 − 20t − 3 = 0

15x 2 + 20 x − 12 = 0

20 ± 400 + 120 20 20 ± 520 20 ± 2 130 t= = 20 20

−20 ± 400 + 4 ⋅15 ⋅12 30 −20 ± 1120 x= 30 −20 ± 4 70 x= 30

10 ± 130 10

x= 105.

−10 ± 2 70 15

106. x+

12 =7 x≠0 x x 2 + 12 = 7 x

x−

10 = −3 x ≠ 0 x x 2 − 10 = −3x

x 2 − 7 x + 12 = 0

x 2 + 3x − 10 = 0

( x − 3)( x − 4 ) = 0

( x + 5 )( x − 2 ) = 0

x − 3 = 0 or x − 4 = 0

x + 5 = 0 or x − 2 = 0

x = 3 or x = 4

x = −5 or x = 2

104

Section 1.3

107.

4 (x − 2) 3 −3 + = x ≠ 0,3 x −3 x x ( x − 3)

108. 5 3 =4+ y+4 y−2

y ≠ −4,2

LCD = ( y + 4 )( y − 2 )

LCD = x ( x − 3 )

5 ( y − 2 ) = 4 ( y + 4 )( y − 2 ) + 3 ( y + 4 )

4 x ( x − 2 ) + 3 ( x − 3 ) = −3 4 x 2 − 8x + 3x − 9 = −3

5 y − 10 = 4 ( y 2 + 2 y − 8 ) + 3 y + 12

4 x 2 − 5x − 6 = 0

5 y − 10 = 4 y 2 + 8 y − 32 + 3 y + 12

( 4x + 3 )( x − 2 ) = 0

−4 y2 − 6 y + 10 = 0

4 x + 3 = 0 or x − 2 = 0

2 y2 + 3 y − 5 = 0

x = −3 4 or x = 2

( 2 y + 5 )( y − 1) = 0 2 y + 5 = 0 or y − 1 = 0 y = − 52 or y = 1

109.

110. x − 0.1x − 0.12 = 0

y2 − 0.5 y + 0.06 = 0

( x − 0.4 )( x + 0.3) = 0

( y − 0.2 )( y − 0.3 ) = 0

x = −0.3, 0.4

y = 0.2, 0.3

111. P = −200t 2 + 1000t + 900 = 1700

112. P = −25t 2 + 175t + 15 = 265

−200t 2 + 1000t − 800 = 0

−25t 2 + 175t − 250 = 0

−200 ( t − 1)( t − 4 ) = 0

−25 ( t − 2 )( t − 5 ) = 0 t = 2 (January 2020) and t = 5 (April 2020)

−200 ( t 2 − 5t − 4 ) = 0

t = 1 (December 2019) and t = 4 (March 2020)

−25 ( t 2 − 7t + 10 ) = 0

105

Chapter 1 113. Solve P ( q ) = 0 : −100 + (0.2q − 3)q = 0 −100 + 0.2q 2 − 3q = 0 0.2q 2 − 3q − 100 = 0

q 2 − 15q − 500 = 0 q=

( −15 ) − 4(1)( −500) 2

15 ±

2(1)

15 ± 2,225 15 ± 47.17 = 2 2 = 31.085, −16.09 So, approximately 31,000 units must be sold to break even. =

114. Solve P ( q ) = 40 :

−100 + (0.2q − 3)q = 40 −100 + 0.2q 2 − 3q = 40 0.2q 2 − 3q − 140 = 0 q 2 − 15q − 700 = 0 ( q − 35)( q + 20) = 0

q = 35, −20 So, 35,000 units must be sold to achieve this profit level. 115. Solve P ( x ) = 460 : −5( x + 3)( x − 24) = 460

116. Solve P ( x ) = 630 : −5( x + 3)( x − 24) = 630

−5x 2 + 105x + 360 = 460

−5x 2 + 105x + 360 = 630

−5x 2 + 105x − 100 = 0

−5x 2 + 105x − 270 = 0

x 2 − 21x + 20 = 0

x 2 − 21x + 54 = 0

( x − 20)( x − 1) = 0

( x − 18)( x − 3) = 0

x = 1, 20 So, the smallest price increase that will produce a weekly profit of $460 is $1 per bottle.

x = 3, 18 So, the smallest price increase that will produce a weekly profit of $630 is $3 per bottle.

106

Section 1.3

117. Solve P (t ) = 160, 1 ≤ t ≤ 6 :

118. Solve P (t ) = 172, 1 ≤ t ≤ 6 :

−t + 13t + 130 = 160

−t 2 + 13t + 130 = 172

−t 2 + 13t − 30 = 0

−t 2 + 13t − 42 = 0

t 2 − 13t + 30 = 0

t 2 − 13t + 42 = 0

(t − 10)(t − 3) = 0

(t − 6)(t − 7) = 0

t = 3, 10 So, 160 people would have contracted the flu after 3 days.

t = 6, 7 So, 172 people would have contracted the flu after 6 days.

119. a. The width of useable space = (8.5 – 2(1)) inches = 6.5 inches The length of useable space = (11 – 2(1.25)) inches = 8.5 inches So, the amount of useable space is the area, namely (6.5 in)(8.5 in) = 55.25 in2.

120. a. The width of useable space = (8.5 – 2(1)) inches = 6.5 inches The length of useable space = (11 – 2(1)) inches = 9 inches So, the amount of useable space is the area, namely (6.5 in)(9 in) = 58.5 in2.

b. Let x = amount of margin reduction (in inches) Width of useable space = 8.5 – 2(1) + 2x = 6.5 + 2x Length of useable space = 11 – 2(1.25) + 2x = 8.5 + 2x So, the useable area is (6.5 + 2 x )(8.5 + 2 x ) = 55.25 + 30 x + 4 x 2

b. Let x = amount of margin reduction (in inches) Width of useable space = 8.5 – 2(1) + 2x = 6.5 + 2x Length of useable space = 11 – 2(1) + 2x = 9 + 2x So, the useable area is (6.5 + 2 x )(9 + 2 x ) = 58.5 + 31x + 4 x 2

c. 55.25 + 30x + 4x 2 − 55.25 = 4 x 2 + 30x This represents the increase in useable area of the paper.

c. 58.5 + 31x + 4 x 2 − 58.5 = 4x 2 + 31x This represents the increase in useable area of the paper.

d. Find x such that 10 ( 55.25 + 30x + 4x 2 ) = 11(55.25) .

d. Find x such that 15 ( 58.5 + 31x + 4x 2 ) = 16(58.5) .

Solving for x yields: 552.5 + 300 x + 40 x 2 = 607.75

Solving for x yields: 60 x 2 + 465x − 58.5 = 0

40 x 2 + 300 x − 55.25 = 0

12 x 2 + 93x − 11.7 = 0

8x 2 + 60x − 11.05 = 0

Continued onto next page.

107

Chapter 1

−60 ± 60 2 − 4(8)( −11.05) 2(8)

−93 ± 932 − 4(12)( −11.7) 2(12)

−60 ± 3,953.6 2.877 ≈ ≈ 0.2 16 16 So, about 0.2 inches.

−93 ± 9,210.6 2.9719 ≈ ≈ 0.1 24 24 So, about 0.1 inches.

121. Form a right triangle with legs of length x and 25in. and hypotenuse of length 32in. Then, by the Pythagorean Theorem, we solve: x 2 + 252 = 32 2

122. Form a right triangle with legs of length x and 20in. and hypotenuse of length 42in. Then, by the Pythagorean Theorem, we solve: x 2 + 20 2 = 42 2

x 2 = 399

x 2 = 1364

x = ± 399 ≈ ±20 So, the TV is approximately 20 inches high.

x = ± 1364 ≈ ±37 So, the TV is approximately 37 inches wide.

123.

124.

Let the numbers be x, x + 1.

Let the numbers be x, x + 2.

x + ( x + 1) = 35

x + ( x + 2 ) = 24

2 x = 34  x = 17

2 x = 22  x = 11

x ( x + 1) = 306

x ( x + 2 ) = 143

x 2 + x = 306

x 2 + 2 x = 143

x 2 + x − 306 = 0

x 2 + 2 x − 143 = 0

( x + 18)( x − 17 ) = 0

( x + 13)( x − 11) = 0

x = −18 ,17 So, the numbers are 17 and 18.

x = −13 ,11 So, the numbers are 11 and 13.

108

Section 1.3

125. Let l = length of the rectangle (in ft.) Then, the width w = l – 6 (in ft.) We must solve: 135 = lw 135 = l (l − 6)

l 2 − 6l − 135 = 0 (l − 15)(l + 9) = 0 l = 15, −9 So, the rectangle has: length 15ft. and width 9ft. 126.

Area = length ⋅ width = ( 2w + 2 )( w ) = 31.5 2w 2 + 2w − 31.5 = 0 −2 ± 4 + 4 ⋅ 2 ⋅ 31.5 4 −2 ± 256 −2 ± 16 w= = 4 4 Widths and lengths are always positive, so w=

14 7 7 = m, l = 2   + 2 = 9 m 4 2 2

127.

128.

s2 = A

1 Area = b ⋅ h = 60 2 h = 3b + 2 1 b ( 3b + 2 ) = 60 2 3 2 b + b = 60 2 3b2 + 2b − 120 = 0

( s + 3) = A + 69 2 ( s + 3) = s 2 + 69 2

s 2 + 6s + 9 = s 2 + 69 6 s − 60 = 0 s = 10 yards

(3b + 20 )( b − 6 ) = 0 b=

−20 , 6; h = 20 3

109

Chapter 1 130.

129. h = −16t + 100 Ground → h = 0

h = −16t 2 − 5t + 100 Ground → h = 0

−16t 2 + 100 = 0 100 t2 = 16 10 t = ± ( Time must be ≥ 0 ) 4 Impact with ground in 2.5 sec

16t 2 + 5t − 100 = 0

131.

−5 ± 25 + 4 ⋅16 ⋅100 32 −5 ± 6425 −5 ± 5 257 t= = 32 32 t ≅ 2.3 sec t=

132.

15 + 15 = r 2

d 2 = 902 + 902

r 2 = 450

d 2 = 16200

r = ± 450 = ±15 2

d = ± 16200 = ±90 2

r ≈ 21.2 feet

d ≅ 127 feet

133. volume = l ⋅ w ⋅ h v = ( x − 2 )( x − 2 )(1) 9 = ( x − 2)

134. volume = l ⋅ w ⋅ h = 12 l = 2w , h = 1 v = ( 2w − 2 )( w − 2 )(1) = 12

( 2w − 2 )( w − 2 ) = 12

x−2 = ±3 x = 2 ± 3 = −1,5 x =5

2w 2 − 6w + 4 = 12 2w 2 − 6w − 8 = 0 w 2 − 3w − 4 = 0

Original square was 5 ft × 5 ft

( w + 1)( w − 4 ) = 0 Original rectangle was 4 ft × 8 ft

110

Section 1.3

135. Let w = width of border Total area of garden + border = ( 8 + 2w )( 5 + 2w ) = 4w 2 + 26w + 40 Area of garden = 8 ⋅ 5 = 40

= 4w 2 + 26w Area of border = ( 4w 2 + 26w + 40 ) − 40   garden total

Volume of border = Area ⋅ depth ( depth = 4 in. =1 3 ft ) = ( 4w 2 + 26w ) (1 3 )

Volume = 27 ft 3 1 4w 2 + 26w ) = 27 ( 3 4w 2 + 26w = 81 4w 2 + 26w − 81 = 0 −26 ± 262 + 4 ⋅ 4 ⋅ 81 −26 ± 1972 = 2⋅4 8 w ≅ −8.8 , 2.3 w=

Width of border is 2.3 feet.

136.

π r2

π 62

= 18π 2 2 Volume of mulch = Area ⋅ depth =18π ⋅ d

Area of Rose Garden =

Volume of mulch = 54 ft 3 18π d = 54 54 d= ≅ 0.95 18π Mulch will be about 1 foot deep

111

Chapter 1 137.

Let x = days for Kimmie to complete job herself. x − 5 = days for Lindsey to complete job herself. 1 = % of job Kimmie can do per day. x 1 = % of job Lindsey can do per day. x −5 1 1 1 + = ( Together they can do it in 6 days.) x x −5 6 LCD = x ( x − 5 ) 6 x ≠ 0,5 6 ( x − 5 ) + 6x = x ( x − 5 ) 6x − 30 + 6x = x 2 − 5x x 2 − 17 x + 30 = 0

( x − 15 ) ( x − 2 ) = 0 x = 2 ,15 Kimmie alone:15 days Lindsey alone:10 days

138. 1 house per hour. 4 1 Ryan can clean house per hour. 6 1 1 10 5 = Together they can clean + = house per hour. 4 6 24 12 Jack can clean

They can clean the house in

12 = 2.4 hours. 5

112

Section 1.3

139.

140. Factored incorrectly

Forgot ±

t 2 − 5t − 6 = 0

( 2 y − 3) = 25 2

(t + 1)( t − 6 ) = 0

2 y − 3 = ±5 2y = 3 ± 5

t = −1, 6

y= 141.

3±5 = −1, 4 2

−a is imaginary for positive a

142. In completing the square we

9 9 3 a = − , so a = ± − = ± i 16 16 4

side. So, 2 ( x 2 − 2x + 1) = 3 + 2.

should add 2 ( not 1) to the right

The solutions are 1 ±

5 2

143. False x = −5 3 satisfies 1st equation but not 2 nd

144. True

145. True

146. True

147. If x = a is a repeated root for a quadratic equation, then ( x − a )2 = 0 . Simplifying yields:

148. If x = bi is a root for a quadratic equation, then so is x = −bi . ( x + bi )( x − bi ) = 0

x 2 − 2ax + a 2 = 0

Then, x 2 − b 2i 2 = 0 x 2 + b2 = 0

149.

( x − 2)( x − 5) = 0

150.

x 2 − 7 x + 10 = 0

x( x + 3) = 0 x 2 + 3x = 0

113

Chapter 1 151.

152.

A = P (1 + r )

1 2 gt 2 2s t2 = g s=

t=±

A 2 = (1 + r ) P A 1+ r = ± P

2s g

r = −1 ±

A P

154.

153. a +b =c 2

P = EI − RI 2

RI 2 − EI + P = 0

c = ± a2 + b2

This equation is quadratic in I . I=

155.

E ± E 2 − 4RP 2R

156. x − 4x = 0

3x − 6 x 2 = 0 3x(1 − 2x ) = 0

x 2 ( x − 2)( x + 2) = 0

x = 0, 1 2

x2 ( x2 − 4 ) = 0 x = 0, ± 2

157.

158. x + x − 4x − 4 = 0 3

x3 + 2x 2 − x − 2 = 0

( x + x ) − 4 ( x + 1) = 0

( x + 2x ) − ( x + 2 ) = 0

x 2 ( x + 1) − 4 ( x + 1) = 0

x2 ( x + 2 ) − ( x + 2 ) = 0

( x − 4 ) ( x + 1) = 0

( x − 1) ( x + 2 ) = 0

( x − 2)( x + 2)( x + 1) = 0

( x − 1)( x + 1)( x + 2) = 0

x = −1, ± 2

x = ±1, − 2

114

Section 1.3

159. −b + b 2 − 4ac x1 = 2a

x2 =

−b − b 2 − 4ac 2a

−b b 2 − 4ac b b 2 − 4ac + − − 2a 2a 2a 2a −2b −b = = 2a a

x1 + x2 =

160. −b − b 2 − 4ac x2 = 2a

−b + b 2 − 4ac x1 = 2a

( −b + b − 4ac )( −b − b − 4ac ) x ⋅x = 2

2a ⋅ 2a

b − ( b − 4ac ) 2

= 161.

(

)

(

)

4ac c = 4a 2 a

162.

x − 3 + 5  x − 3 − 5  = 0    ( x − 3 ) − 5  ( x − 3 ) + 5  = 0   

 x − ( 2 − i )   x − ( 2 + i )  = 0 ( x − 2 ) + i  ( x − 2 ) − i  = 0

(x − 2) − i2 = 0 2

( x − 3) − 5 = 0 2

x2 − 4x + 4 + 1 = 0

x − 6x + 9 − 5 = 0 2

x 2 − 4x + 5 = 0

x − 6x + 4 = 0 2

115

Chapter 1 163. Let x = speed in still air and y = time to make the trip with a tail wind. Using Distance = Rate × Time, we obtain the following two equations: With tail wind: ( x + 50) y = 600 (1) Against head wind: ( x − 50)( y + 1) = 600 ( 2 ) 600 Solve (1) for y: y = x + 50

Substitute this into (2) and solve for x:  600  ( x − 50)  + 1 = 600  x + 50   600 + x + 50  ( x − 50)   = 600  x + 50  ( x − 50)(650 + x ) = 600( x + 50) 650x − 32,500 − 50x + x 2 = 600 x + 30,000 x 2 − 62,500 = 0 ( x − 250)( x + 250) = 0 x = 250, − 250

So, the plane in still air travels at 250 mph.

164. Let c = Current rate. Down river: rate =10 + c in t hours

Substitute ( 3 ) into (1) to solve for c :

Up river: rate =10 − c in t + 1 hours.

(10 + c )(10 − c ) = 2c ( 24 )

Using Distance = Rate × Time,

100 − c 2 = 48c

we obtain:

−c 2 − 48c + 100 = 0

Distance down river:

c 2 + 48c − 100 = 0

d = (10 + c )( t ) = 24 (1) Distance up river: d = (10 − c )( t + 1) = 24 (2) Equating (1) and ( 2 ) yields:

(10 + c )(t ) = (10 − c )( t + 1)

10 − c   = 24  2c 

(10 + c ) 

−48 ± 482 + 400 c= 2 −48 ± 2704 −48 ± 52 = = 2 2 The rate must be postive, so c = 2 mph .

10t + ct = 10t + 10 − ct − c 2ct = 10 − c t=

10 − c ( 3) 2c

116

Section 1.3

165. 2 distinct real roots of ax 2 + bx + c = 0 are: x1 = If real roots are negatives of x1 , x2 , then x1* =

−b + b 2 − 4ac 2a

b − b2 − 4ac 2a

x2 =

x2* =

−b − b2 − 4ac 2a

b + b2 − 4ac 2a

Replace b with − b. So, ax 2 − bx + c = 0 .

166. ax 2 + bx + c = 0  x =

−b ± b2 − 4ac 2a

−b + b2 − 4ac −b − b2 − 4ac , x2 = 2a 2a 1 2a 1 −2a x1* = = x2* = = x1 −b + b2 − 4ac x2 b + b2 − 4ac

Label the roots as x1 =

Using these roots in a new quadratic equation,    2a −2a x− =0 x −    b + b2 − 4ac  −b + b2 − 4ac       2a 2a x+ x +   =0 b + b 2 − 4ac  b − b 2 − 4ac           2a 2a 2a 2a x2 + x   + x  +  =0 2 2 2 2  b + b − 4ac   b − b − 4ac   b − b − 4ac   b + b − 4ac  x + 2

(

)

(

2ax b + b 2 − 4ac + 2ax b − b 2 − 4ac

( b − b − 4ac )( b + b − 4ac ) 2

b a x2 + x + = 0 c c cx 2 + bx + a = 0

117

) + 4a = x + 4abx + 4a = 0 2

4ac

Chapter 1

167. Let x = speed of small jet (in mph). Then, the speed of the 757-jet = x+100 (mph) Form a right triangle depicting the relative position of the jets after two hours of flight. Using Distance = Rate × time, this triangle will have legs of length 2 x and 2( x + 100) , and hypotenuse of length 1000 miles. Using the Pythagorean Theorem then yields (2 x )2 + (2( x + 100))2 = 1000 2 4 x 2 + 4 x 2 + 800 x + 40,000 = 1,000,000 x 2 + 100x − 120,000 = 0 −100 ± 1002 + 4(120,000) −100 ± 700 = = −400 , 300 2 2 So, the speed of the small jet is 300mph and the speed of the 757-jet is 400mph. x=

168. Let x = speed of small boat (in mph). Then, the speed of the large boat = x+10 (mph). Form a right triangle depicting the relative position of the jets after three hours. Using Distance = Rate × time, this triangle will have legs of length 3x and 3( x + 10) , and hypotenuse of length 150 miles. Using the Pythagorean Theorem then yields (3x )2 + (3( x + 10))2 = 150 2 9x 2 + 9x 2 + 180 x + 900 = 22,500 x 2 + 10 x − 1200 = 0 ( x + 40)( x − 30) = 0 x = −40 , 30 So, the speed of the small boat is 30mph and the speed of the large boat is 40mph.

118

Section 1.3

169. x2 − x − 2 = 0

( x − 2 )( x + 1) = 0 x = −1, 2

170. x 2 − 2x + 2 = 0 2 ± 4 − 4 ⋅ 2 2 ± −4 = 2 2 x = 1± i

119

Chapter 1

171. (a) Consider x 2 − 2x − b = 0 . (1) For b = 8 , (1) factors as ( x − 4)( x + 2) = 0 , so that x = −2, 4 . Graphically, we let y1 = x 2 − 2 x, y2 = 8 and look for the intersection points of the graphs:

Note that they intersect at precisely the x-values obtained algebraically. So, yes, these values agree with the points of intersections.

(b) We do the same thing now for different values of b. b = −3 :

b = −1:

x − 2x + 3 = 0

x 2 − 2x + 1 = 0

2 ± 4 − 4(3) x= = 1± i 2 2 So, we don’t expect the graphs to intersect. Indeed, we have:

( x − 1)2 = 0 x =1 So, we expect the graphs to intersect once. Indeed, we have:

120

Section 1.3

b =0:

b =5: x − 2x = 0

x 2 − 2x − 5 = 0

x( x − 2) = 0 x = 0,2 So, we expect the graphs to intersect twice as in part (a). Indeed, we have:

2 ± 4 + 4(5) = 1± 6 2 So, we expect the graphs to intersect twice as in part (a). Indeed, we have: x=

172. (a) Consider x 2 + 2x − b = 0 . (1) For b = 8 , (1) factors as ( x + 4)( x − 2) = 0 , so that x = 2, −4 . Graphically, we let y1 = x 2 + 2 x, y2 = 8 and look for the intersection points of the graphs:

Note that they intersect at precisely the x-values obtained algebraically. So, yes, these values agree with the points of intersections.

121

Chapter 1

b) We do the same thing now for different values of b. b = −3 :

b = −1:

x + 2x + 3 = 0

x2 + 2x + 1 = 0

( x + 1)2 = 0

−2 ± 4 − 4(3) = −1 ± i 2 2 So, we don’t expect the graphs to intersect. Indeed, we have:

x = −1 So, we expect the graphs to intersect once. Indeed, we have:

b =0:

b =5:

x + 2x = 0

x 2 + 2x − 5 = 0

x( x + 2) = 0 x = 0, −2 So, we expect the graphs to intersect twice as in part (a). Indeed, we have:

−2 ± 4 + 4(5) = −1 ± 6 2 So, we expect the graphs to intersect twice as in part (a). Indeed, we have: x=

122

Section 1.4 Solutions ------------------------------------------------------------------------------1.

t −5 = 2

2t − 7 = 3

t −5 = 4

2t − 7 = 9 2t = 16

t=9

(4 p − 7)

4 p − 7 = 25

121 = 21 − p

4 p = 32

p = −100

t =8

11 = ( 21 − p )

p=8

u + 1 = −4

6. − 3 − 2u = 9

no solution

3 − 2u = −9

5x + 2 = 33 = 27

1 − x = −2 1 − x = −8

u + 1 = 16 u = 15

no solution

5x = 25

x=9

3 − 2u = 81

x =5

Check: 15 + 1 = 16 = 4

8. 3

5x + 2 = 3

2u = −78 u = −39 − 3 + 2 ⋅ 39 = −9

10. 1 3

1 3

( 4 y + 1) = −1

(5x − 1) = 4

4 y + 1 = −1 4 y = −2

5x − 1 = 64 5x = 65

1 2

x = 13

y=−

11. 12 + x = x

12. x = 56 − x

12 + x = x 2

x 2 = 56 − x

x 2 − x − 12 = 0

x 2 + x − 56 = 0

( x + 3)( x − 4 ) = 0

( x + 8)( x − 7 ) = 0

x = −3, 4

x = −8, 7

Check − 3 :

Check − 8:

12 − 3 = 9 ≠ −3

56 + 8 = 64 ≠ −8

Check 4 :

Check 7:

12 + 4 = 16 = 4

56 − 7 = 49 = 7

123

Chapter 1 14.

13. y=5 y

y 2 = 25 y

y 4 2

15. s =3 s−2

16. −2 s = 3 − s

s2 = 9 ( s − 2)

4s 2 = 3 − s

s 2 = 9s − 18

4s 2 + s − 3 = 0

y ( y − 25 ) = 0

y 16 2 y − 16 y = 0

y = 0,25

y ( y − 16 ) = 0

Check 0:

y = 0,16

s = 3,6

0=5 0 Check 25:

Check 0:

Check 3:

0 =0 4

3 = 3 3−2 = 3 1

y 2 − 25 y = 0

25 = 5 25

s 2 − 9s + 18 = 0

( s + 1)( 4s − 3) = 0

( s − 3 )( s − 6 ) = 0

3 4 Check − 1:

Check 16:

Check 6:

16 = 16 4

6 =3 6−2 =3 4

s = −1 ,

( −2 )( −1) = 3 + 1 2=2 Check 3 4 : 3 3 −2   = 3 − 4 4 −

17.

3 9 3 = ≠ 2 4 2

19.

18. 2x + 6 = x + 3

8 − 2x = 2x − 2

1 − 3x = x + 1

2x + 6 = ( x + 3)

8 − 2 x = 4 x − 8x + 4

1 − 3x = x 2 + 2x + 1

4x 2 − 6x − 4 = 0

x 2 + 5x = 0

2 x 2 − 3x − 2 = 0

x ( x + 5) = 0

x2 + 4x + 3 = 0

( x + 3)( x + 1) = 0 x = −3, − 1 Check − 3 : 2 ( −3 ) + 6 = −3 + 3 0 =0 Check − 1: 2 ( −1) + 6 = −1 + 3 4 =2

( 2x + 1)( x − 2 ) = 0

x = −5, 0

−1 x= , 2 2 −1 Check : 2

Check − 5 :

 −1   −1  8− 2  = 2  − 2  2   2 

1 =1

1 + 15 ≠ −4 Check 0 :

9 ≠ −3 Check 2: 8 − 4 = 2 (2) − 2 4 =2 124

Section 1.4

20.

21.

3x − 6 x − 1 = 3

2−x = x−2 2 − x = ( x − 2)

3x − 3 = 6 x − 1 x −1 = 2 x −1

2 − x = x2 − 4x + 4

(

x − 3x + 2 = 0 2

( x − 1)2 = 2 x − 1

( x − 2 )( x − 1) = 0

( x − 1)2 − 4( x − 1) = 0 ( x − 1)( x − 1 − 4) = 0 ( x − 1)( x − 5) = 0 x = 1 and 5

x = 1, 2

22.

23.

5x − 10 x + 2 = −10

3x − 6 x + 2 = 3

x − 2 x + 2 = −2

x −2 x + 2 =1

x+2 =2 x+2

(

( x + 2)2 = 2 x + 2

x −1 = 2 x + 2

)

(

( x − 1)2 = 2 x + 2

( x + 2)2 − 4( x + 2) = 0 ( x + 2)( x + 2 − 4) = 0 ( x + 2)( x − 2) = 0 x = −2 and 2

( x − 1)2 = 4( x + 2) x 2 − 2x + 1 = 4x + 8 x 2 − 6x − 7 = 0 ( x − 7)( x + 1) = 0 x = −1 , 7

125

)

Chapter 1 25.

24.

3 x + 4 − 2x = 9

2x − 4 x + 1 = 4

3 x + 4 = 2x + 9

x − 2 x +1 = 2

(3 x + 4 ) = (2x + 9) 2

x − 2 = 2 x +1

(

( x − 2)2 = 2 x + 1

)

9( x + 4) = 4x 2 + 36 x + 81 9x + 36 = 4x 2 + 36 x + 81

( x − 2)2 = 4( x + 1)

4x 2 + 27x + 45 = 0 (4x + 15)( x + 3) = 0

x2 − 4x + 4 = 4x + 4 x 2 − 8x = 0 x( x − 8) = 0

x = − 15 4 and − 3

x = 0, 8 27.

26. 2 x + 1 − 3x = −5

x2 − 4 = x − 1 x 2 − 4 = ( x − 1)

2 x + 1 = 3x − 5

( 2 x + 1) = (3x − 5) 2

x2 − 4 = x2 − 2x + 1 2x = 5

4( x + 1) = 9x 2 − 30 x + 25 4 x + 4 = 9x 2 − 30 x + 25

9x 2 − 34x + 21 = 0

5 2

(9x − 7)( x − 3) = 0 x = 79 , 3

29.

28.

30.

25 − x 2 = x + 1

x 2 − 2x − 5 = x + 1

25 − x 2 = ( x + 1)

x 2 − 2x − 5 = ( x + 1)

2 x 2 − 8x + 1 = x − 3 2

2 x 2 − 8x + 1 = ( x − 3 )

25 − x 2 = x 2 + 2 x + 1

x 2 − 2x − 5 = x 2 + 2x + 1

2 x 2 + 2 x − 24 = 0

−6 = 4 x

x 2 − 2x − 8 = 0

x 2 + x − 12 = 0

− 32 = x

( x − 4) ( x + 2 ) = 0

( x + 4 )( x − 3) = 0

No solution.

2 x 2 − 8x + 1 = x 2 − 6 x + 9

x = −2 , 4

x = −4 , 3

126

Section 1.4

31. 3x + 1 − 6 x − 5 = 1 3x + 1 = 6 x − 5 + 1

( 3x + 1) = ( 6x − 5 + 1) 2

3x + 1 = 6 x − 5 + 2 6 x − 5 + 1 3x + 1 = 6 x − 4 + 2 6 x − 5

(

( −3x + 5)2 = 2 6 x − 5

)

9x 2 − 30 x + 25 = 4(6 x − 5) 9x 2 − 30 x + 25 = 24 x − 20 9x 2 − 54 x + 45 = 0 (9x − 9)( x − 5) = 0 x = 1, 5

32. 2 − x + 6 − 5x = 6 2 − x = 6 − 6 − 5x

( 2 − x ) = ( 6 − 6 − 5x ) 2

2 − x = 36 − 12 6 − 5x + (6 − 5x ) 2 − x = 42 − 5x − 12 6 − 5x

−40 + 4 x = −12 6 − 5x −10 + x = −3 6 − 5x

( −10 + x ) = ( −3 6 − 5x ) 2

100 − 20 x + x 2 = 9(6 − 5x ) 100 − 20 x + x 2 = 54 − 45x x 2 + 25x + 46 = 0 ( x + 23)( x + 2) = 0 x = −23 , − 2

127

Chapter 1

33. x + 12 + 8 − x = 6 x + 12 = 6 − 8 − x

( x + 12 ) = (6 − 8 − x ) 2

x + 12 = 36 − 12 8 − x + (8 − x ) 2 x − 32 = −12 8 − x x − 16 = −6 8 − x

( x − 16 ) = ( −6 8 − x ) 2

x 2 − 32 x + 256 = 36(8 − x ) x 2 − 32 x + 256 = 288 − 36 x x 2 + 4 x − 32 = 0 ( x − 4)( x + 8) = 0 x = 4, −8

35.

34.

2x − 1 = 1 + x − 1

5 − x + 3x + 1 = 4

2x − 1 = 1 + 2 x − 1 + x − 1

5 − x = 4 − 3x + 1

( 5 − x ) = ( 4 − 3x + 1 ) 2

x −1 = 2 x −1

5 − x = 16 − 8 3x + 1 + (3x + 1) 5 − x = 17 + 3x − 8 3x + 1

( x − 5 )( x − 1) = 0

3 + x = 2 3x + 1

(

)

x 2 − 2x + 1 = 4x − 4 x 2 − 6x + 5 = 0

−12 − 4 x = −8 3x + 1 (3 + x )2 = 2 3x + 1

x 2 − 2 x + 1 = 4 ( x − 1)

x = 1,5

9 + 6 x + x 2 = 4(3x + 1) 9 + 6 x + x 2 = 12 x + 4 x 2 − 6x + 5 = 0 ( x − 5)( x − 1) = 0 x = 1,5

128

Section 1.4

36. 8 − x = 2 + 2x + 3

37. 3x − 5 = 7 − x + 2

38. x + 5 = 1+ x − 2

8 − x = 4 + 4 2 x + 3 + 2 x + 3 3x − 5 = 49 − 14 x + 2 + x + 2

x + 5 = 1+ 2 x − 2 + x − 2

− 3x + 1 = 4 2 x + 3

2 x − 56 = −14 x + 2

9x − 6 x + 1 = 16 ( 2 x + 3 )

x − 28 = −7 x + 2

6 = 2 x−2 9 = x−2

9x − 6 x + 1 = 32 x + 48

x 2 − 56 x + 784 = 49 ( x + 2 )

x = 11

9x 2 − 38x − 47 = 0

x 2 − 56 x + 784 = 49x + 98

2 2

( 9x − 47 )( x + 1) = 0 x = 479 , −1

x 2 − 105x + 686 = 0

( x − 98 )( x − 7 ) = 0 x = 7 , 98

40.

39.

41. Let u = x1 3

2+ x = x

2− x = x

2+ x = x

2− x = x

x = x−2

x =2−x

x = x2 − 4x + 4

x = 4 − 4x + x2

x 2 − 5x + 4 = 0

( x − 4 )( x − 1) = 0

( x − 1)( x − 4 ) = 0

x = 1, 4

42. Let u = x1 4

43. Let u = x 2

44. Let u = x 2

u 2 − 2u = 0

u 2 − 3u + 2 = 0

u 2 − 8u + 16 = 0

u (u − 2) = 0

( u − 1)( u − 2 ) = 0

u = 0,2

u = 1,2

(u − 4) = 0

u 2 + 2u = 0 u (u + 2) = 0 u = −2,0 x1 3 = 0 → x = 0 x1 3 = −2 → x = −8

u=4

=0→ x=0

x = 1 → x = ±1

x2 = 4

x1 4 = 2 → x = 16

x2 = 2 → x = ± 2

x = ±2

129

Chapter 1

45. Let u = x 2

46. Let u = x 4

47. Let u = 2 x + 1

2u 2 + 7u + 6 = 0

u 2 − 17u + 16 = 0

u 2 + 5u + 4 = 0

( 2u + 3)( u + 2 ) = 0

( u − 16 )( u − 1) = 0

( u + 4 )( u + 1) = 0 u = −1

x = −1

u = −3 2

u = −2

u =1

u = 16

x 2 = −3 2

x = −2

x =1

x = 16

x = ±i 3 2

x = ±i 2

x 2 = ±1

x 2 = ±4

u = −4 2 x + 1 = −4 2 x = −5

±i 6 x= 2

x = ±i 2

if x 2 = 1

if x 2 = 4

x = −5 2

x = ±1

x = ±2

if x = −1

if x 2 = −4

x = ±i

x = ±2i

2 x + 1 = −1 2 x = −2

48. Let u = x − 3

49. Let u = t − 1

50. Let u = 1 − y

u 2 + 6u + 8 = 0

4 u 2 − 9u + 2 = 0

2u 2 + 5u − 12 = 0 `

( u + 2 )( u + 4 ) = 0

( 4u − 1)( u − 2 ) = 0

( 2u − 3)( u + 4 ) = 0

u = −2

u = −4

u =1 4

u=2

u =3 2

u = −4

x − 3 = −2

x − 3 = −4

t −1 = 1 4

t −1 = 2

1− y = 3 2

1 − y = −4

x =1

x = −1

t=5 4

t =3

y = −1 2

y=5

51. Let u = x −4

52. Let x = u −1

53. Let u = y −1

u 2 − 17u + 16 = 0

2x 2 + 5x − 12 = 0

3u 2 + u − 4 = 0

( u − 16 )( u − 1) = 0

( 2x − 3 )( x + 4 ) = 0

( 3u + 4 )( u − 1) = 0

u =1

u = 16

x =3 2

x = −4

u = −4 3

u =1

x −4 = 1

x −4 = 16

u −1 = 3 2

u −1 = −4

y −1 = −4 3

y −1 = 1

x 2 = ±1

x2 = ±1 4

u=2 3

u = −1 4

y = −3 4

y =1

x = ±1, ± i

1 1 x = ± ,± i 2 2

130

Section 1.4

54. Let u = a −1

55. Let u = z1 5

5u 2 + 11u + 2 = 0

u 2 − 2u + 1 = 0

( 5u + 1)( u + 2 ) = 0

( u − 1) = 0 2

u = −1 5  a −1 = −1 5  u = −2  a = −2  −1

a = −5

u =1 

a = −1 2

z1 5 = 1 

z =1

56. Let u = x1 4 2u 2 + u − 1 = 0  (2u − 1)( u + 1) = 0 x 4 = 12 or 1

x 4 = −1  1

57. 5 ( x + 3) 3 = 32

 u = −1, 12

x = 161

58. ( x + 2) 3 = 16 4

x + 3 = 32 5 3

x + 2 = 16 4 3

(

x = −3 + 32 5 1

) = −3 + 2 = −3 + 8 = 5 3

( )

x = −2 + 16 4 1

= −2 + 23 = −2 + 8 = 6

59.

60. ( x + 1) 3 = 4

( x − 7) 3 = 81 4

x + 1 = ±4 2

x − 7 = ±81 4 x = 7 ± 27 x = −20 or x = 34

x = −1 ± 4 2 = −1 ± 8 3

x = −9 or x = 7

131

Chapter 1

61. Let u = t −1 3

62. u = t −1 3

6u 2 − u − 1 = 0

u2 − u − 6 = 0

( 3u + 1)( 2u − 1) = 0

( u − 3)( u + 2 ) = 0

u = −1 3

u =1 2

u =3

u = −2

t −1 3 = −1 3

t −1 3 = 1 2

t −1 3 = 3

t −1 3 = −2

t = ( −1 3 )

t = (1 2 )

t = 3−3

t = ( −2 )

t = 1 27

t = −1 8

−3

t = −27

−3

t =8

63.

−3

64. 3=

2 x ≠ −1 ( x + 1)

4 + 4 = 0 x ≠ −1 x +1

( x + 1) 2 3 ( x + 1) = 1 + 2 ( x + 1) 2 3 ( x + 1) − 2 ( x + 1) − 1 = 0

1 + 4 ( x + 1) + 4 ( x + 1) = 0

Let u = x + 1

3u 2 − 2u − 1 = 0

4u 2 + 4u + 1 = 0

(3u + 1)( u − 1) = 0

( 2u + 1) = 0

x + 1 = −1 3

u =1 x +1 = 1

x = −4 3

x=0

u = −1 3

( x + 1)

LCD = ( x + 1)

2 2

u = −1 2 x + 1 = −1 2 x = −3 2

132

Section 1.4

65.

66. 2

3 =2 2x + 1

1  1  − 12 = 0   +  2x − 1  2x − 1

( 2x + 1)

x ≠1 2

x ≠ −1 2

Let u =

1 2x − 1 2 u + u − 12 = 0

1 2x + 1 2 5u − 3u − 2 = 0

( u + 4 )( u − 3 ) = 0

(5u + 2 )( u − 1) = 0

−

Let u =

u = −4

u =3

u = −2 5

1 = −4 2x − 1 −4 ( 2 x − 1) = 1

1 =3 2x − 1 3 ( 2 x − 1) = 1

1 −2 = 2x + 1 5 −2 ( 2x + 1) = 5

−8x + 4 = 1

−4 x − 2 = 5

−8x = −3

6x − 3 = 1 6x = 4

x =3 8

x=2 3

x = −7 4

67. Let x = u 2 3

68. Let x = u 2 3

69. t4 − t2 − 6 = 0

x 2 − 5x + 4 = 0

x 2 + 5x + 4 = 0

Let u = t 2

( x − 4 )( x − 1) = 0

( x + 4 )( x + 1) = 0

u2 − u − 6 = 0

x=4

x =1

x = −4

x = −1

u = −2

u =3

u2 3 = 4

u2 3 = 1

u 2 3 = −4

u 2 3 = −1

t 2 = −2

t2 = 3

u = ±4 3 2

u = ±13 2

u = ( −4 )

u = ( −1)

t = ±i 2

t = 3,

u = ±8

u = ±1

12 u =  ( −4 )   

x=0

( u − 3)( u + 2 ) = 0

12 u = ( −1)   

u = [ 2i ]

u = ( ±i )

u = ±8i

u = ±i

1 =1 2x + 1 2x + 1 = 1 2x = 0

−4 x = 7

u =1

133

− 3

Chapter 1

70.

71. x3 − x 2 − 12x = 0

u = −2u − 1 2

u 4 = −2 u 2 − 1

x ( x 2 − x − 12 ) = 0

u 4 + 2u 2 + 1 = 0

x( x − 4)( x + 3) = 0

Let x = u 2 . Then, x 2 + 2 x + 1 = 0

x = 0, −3, 4

( x + 1) = 0  x = −1 2

So, u 2 = −1  u = i , −i u =i

72.

73. 2 y − 11y + 12 y = 0

4 p3 − 9 p = 0

y(2 y − 3)( y − 4) = 0

p(2 p − 3)(2 p + 3) = 0

y = 0, 4, 3 2

p = 0, ± 3 2

p ( 4 p2 − 9 ) = 0

y ( 2 y 2 − 11y + 12 ) = 0

74.

75. 25x = 4 x

u5 − 16u = 0

u ( u 4 − 16 ) = 0

25x3 − 4 x = 0

x ( 25x 2 − 4 ) = 0

u ( u 2 − 4 )( u 2 + 4 ) = 0

x(5x − 2)(5x + 2) = 0

u( u − 2)( u + 2)( u − 2i )( u + 2i ) = 0

x = 0, ± 5 2

u = 0, ± 2, ± 2i

77.

76. t − 81t = 0

x3 − 5x 2 − 9x + 45 = 0

t ( t 4 − 81) = 0

( x − 5x ) − ( 9x − 45 ) = 0 3

t ( t 2 − 9 )( t 2 + 9 ) = 0

x 2 ( x − 5) − 9( x − 5) = 0

t(t − 3)(t + 3)(t − 3i )(t + 3i ) = 0

( x − 9 ) ( x − 5) = 0

t = 0, ± 3, ± 3i

( x − 3)( x + 3)( x − 5) = 0

x = ±3, 5

134

Section 1.4

79.

78.

y( y − 5)3 − 14( y − 5)2 = 0

2 p − 3 p − 8 p + 12 = 0 3

( 2 p − 3 p ) − (8 p − 12 ) = 0 3

( y − 5)2 [ y( y − 5) − 14 ] = 0

( y − 5)2 ( y 2 − 5 y − 14 ) = 0

p 2 (2 p − 3) − 4(2 p − 3) = 0

( p − 4 ) (2 p − 3) = 0

( y − 5)2 ( y − 7)( y + 2) = 0

y = −2,5,7

( p − 2)( p + 2)(2 p − 3) = 0 p = ±2, 3 2

80.

81. v(v + 3) − 40(v + 3) = 0 3

x 4 − 2 x 4 − 3x 4 = 0 9

(v + 3)2 [v(v + 3) − 40 ] = 0

x 4  x 2 − 2 x − 3 = 0 1

(v + 3)2 ( v 2 + 3v − 40 ) = 0

x 4 ( x − 3)( x + 1) = 0 1

(v + 3)2 (v − 5)(v + 8) = 0

x = 0,3, −1

v = −8, −3,5

83.

82. u + u − 20u = 0

t 3 − 25t − 3 = 0

u 3 u 2 + u − 20  = 0

t − 3 t 2 − 25  = 0

u 3 ( u + 5)( u − 4) = 0

t − 3 (t − 5)(t + 5) = 0

t = ±5

u = −5,0, 4 −1

(Note: t 3 = 0 has no solution.)

84.

85. 4 x − 9x 9

− 15

y 2 − 5 y 2 + 6 y− 2 = 0

x − 5  4 x 2 − 9  = 0

y − 2  y 2 − 5 y + 6  = 0

x − 5 (2 x − 3)(2 x + 3) = 0

y − 2 ( y − 3)( y − 2) = 0

y = 2,3

x = ± 32 −1

(Note: x 5 = 0 has no solution.)

−1

(Note: y 2 = 0 has no solution.)

135

Chapter 1

87. Solve d (t ) = 3 . (Note: The rightside is 3, and not 3,000,000, because d(t) is measured in millions.) 3 t + 1 − 0.75t = 3

86. 4p −5p −6p 5

− 13

p− 3 ( 4 p2 − 5 p − 6 ) = 0 1

p − 3 (4 p + 3)( p − 2) = 0 1

3 t + 1 = 3 + 0.75t

p = − 3 4 ,2

(Note: p

− 13

(3 t + 1 ) = (3 + 0.75t ) 2

= 0 has no solution.)

9t + 9 = 9 + 4.5t + 0.5625t 2 0.5625t 2 − 4.5t = 0 t(0.5625t − 4.5) = 0 4.5 =8 t = 0, 0.5625 So, this occurs in January and September.

88. Solve d (t ) = 4 . (Note: The rightside is 4, and not 4,000,000, because d(t) is measured in millions.) 3 t + 1 − 0.75t = 4 3 t + 1 = 4 + 0.75t

(3 t + 1) = ( 4 + 0.75t ) 2

89.

wh = BSA for h, when w = 3,600 72 and BSA = 1.8. 72h = 1.8 3,600

Solve

72h = 1.8 60

9t + 9 = 16 + 6t + 0.5625t 2 0.5625t 2 − 3t + 7 = 0 No real solutions So, this never occurs. The demand for the product is never 4,000,000 units.

72 h = (1.8)(60) 72h = 1082 11,664 = 162 72 So, the height of such a female is 162 cm. h=

136

Section 1.4

90. Solve wh = BSA for w, 3,600 when h = 177 and BSA = 2.1. 177w = 2.1 3,600

91.

92.

C = 10 + a C =9

C = 5a + 1

9 = 10 + a 81 = 10 + a

20 = 5a + 1

a = 71 years old

5a = 399

C = 20 400 = 5a + 1 a=

177w = 2.1 60

399 = 79.8 5

79.8 years old

177w = (2.1)(60) 177w = 126 2 15,876 ≈ 90 177 So, the weight of such a male is about 90 kg. w=

93.

P = 5 t + 1 + 50 P = 85 2

94. S = 1200 + 10 2t = 1295

1200 + 10 2t = 1295

85 = 5 t 2 + 1 + 50

10 2t = 95

35 = 5 t 2 + 1

2t = 9.5 2t = 90.25 t = 45.125 ≈ 45 In the year 2035.

7 = t2 + 1 49 = t + 1 2

95. d d + , T =3 4 1100 d d 3= + 4 1100 LCD =1100

3300 = 275 d + d d + 275 d − 3300 = 0

t = 48

Let u = d

t = 48

u 2 + 275u − 3300 = 0

t = 4 3 ( t must be ≥ 0 ) t ≅ 7 months

−275 ± 2752 + 4 ⋅1 ⋅ 3300 u= 2 (1)

March

u = −286.5,11.5 d = 11.5 d = 132 feet

137

Chapter 1

96.

d =3  4

d = 12 

d = 144 feet

97.

98. L 9.8

1 = 2π 2

L 32

1 = 2π 2

L  1    =  2π  9.8 9.8 0.24824 m ≈ 2 = L 4π Convert to centimeters: 0.24824 m 100 cm ≈ 25 cm 1m

L  1    =  2π  32 32 0.81057 ft ≈ =L 4π 2 Convert to inches: 0.81057 ft 12 in ≈ 10 inches 1 ft

99.

100. 2

v2 c2

v 18 = 30 1 − 2 c

5 = 30 1 −

3 18 v2 = = 1− 2 5 30 c

1 5 v2 = = 1− 2 6 30 c 2

v2 3 1 = −   c2 5 16 v 2 = 25 c 2 16 v2 = c2 25 4 v= c 5 So, 80% of the speed of light.

101. t = 5 is extraneous; there is no solution.

v2 1 1 = −   c2 6 35 v 2 = 36 c 2 35 v2 = c2 36 35 c 6 So, about 98.6% of the speed of light.

102. x = −1 is extraneous. x=2

138

103. Forgot about the substitution u = x1 3 .

Section 1.4

104. x 2 = −1

105. True 3 Let u = ( 2 x − 1)

x = ± −i x = ±i ( not ± 1)

u 2 + 4u + 3 = 0 ( quadratic )

106. False Let u = t5

108. False

( x +2 + x) = ( x +5) 2

u5 + 2u + 1 = 0 ( not quadratic )

x+2+2 x x+2 + x = x+5

107. False 110.

109. Solve

x2 = x .

If x ≥ 0, then

Solve

x2 = −x .

x 2 = x , while if x < 0 , then If x ≥ 0, then

x 2 = − x . So, the solution set is [ 0, ∞ ) .

111.

x < 0 , then

x 2 = x , while if x 2 = − x . So, the

solution set is ( −∞,0 ] .

112. Let u = 3x + 2 x

Factor out x1 3

x1 3 ( 3x1 4 − x1 2 − 2 ) = 0

u= u u = 0,1

Let u = x1 4 . 3u − u 2 − 2 = 0 u 2 − 3u + 2 = 0

( u − 2 )( u − 1) = 0 3x 2 + 2 x = 0

3x 2 + 2 x = 1

u=2

u =1

x ( 3x + 2 ) = 0

3x 2 + 2 x − 1 = 0

x1 4 = 2

x1 4 = 1

(3x − 1)( x + 1) = 0

x = 16

x =1

x = 0, −2 3

x = −1,1 3

139

x 3 =0 1

x=0

Chapter 1

113. x + 6 + 11 + x = 5 3 + x

( x + 6 ) + 2 x + 6 11 + x + (11 + x ) = 25 ( 3 + x ) 2 x + 17 + 2 x + 6 11 + x = 75 + 25x 2 x + 6 11 + x = 58 + 23x 4 ( x + 6 )(11 + x ) = 529x 2 + 2668x + 3364

4 ( x 2 + 17 x + 66 ) = 529x 2 + 2668x + 3364 4 x 2 + 68x + 264 = 529x 2 + 2668x + 3364 525x 2 + 2600 x + 3100 = 0 21x 2 + 104 x + 124 = 0

( 21x + 62 )( x + 2 ) = 0 x=

−62 , x = −2 21

114.

(

)

2x x x 1 2 1 3  ( )  

(

) = 2 = 16

12 13

2x x ( x )

x  x ⋅ x1 2   x ⋅ x1 2 

=8 8 x≠0 x

8 x ⋅ x1 2 = x3 2 =   x   8 3  x =     x  

 8  64 =  = 2 x x

x3 = 64 x=4

140

Section 1.4

115. x −3 = 4 − x + 2 x − 3 = 16 − 8 x + 2 + x + 2 −21 = −8 x + 2 441 = 64 ( x + 2 ) = 64x + 128 313 = 64 x x=

313 ≅ 4.891 64

116. 2 x +1 = 1+ 3 − x 4 ( x + 1) = 1 + 2 3 − x + 3 − x 4x + 4 = 4 − x + 2 3 − x 5x = 2 3 − x 25x 2 = 4 ( 3 − x ) = 12 − 4 x 25x 2 + 4 x − 12 = 0 x= x≅

−4 ± 4 2 − 4 ( 25 )( −12 ) 2 ( 25 )

−4 ± 34.9 ≅ −0.778 , 0.62 50

117. −4 = x + 3 16 = x + 3 x = 13 ( Extraneous ) no solution

141

Chapter 1

118. x1 4 = −4 x1 2 + 21

4 x1 2 + x1 4 − 21 = 0 Let u = x1 4 to obtain 4u 2 + u − 21 = 0 u=

−1 ± 1 − 4 ( 4 )( −21) 2 (4)

−1 ± 18.4 ≅ −2.4, 2.2 8 x1 4 = 2.2 x1 4 = −2.4 u≅

x ≅ 22.2

no solution

119. 12

= −4 x

+ 21

+ 4x

− 21 = 0

x x

Let u = x

Graphically, let: 1 1 y1 = x 2 , y2 = −4x 4 + 21 .

to obtain

u 2 + 4u − 21 = 0 ( u + 7)( u − 3) = 0 u = −7,3 x1 4 = −7

x1 4 = 3

x = 81 Yes, the two solutions agree. no solution

142

Section 1.4

120. −1

−2

x = 3x − 10

Graphically, let: y1 = x −1 , y2 = 3x −2 − 10 .

3x −2 − x −1 − 10 = 0 Let u = x −1 to obtain 3u 2 − u − 10 = 0 (3u + 5)( u − 2) = 0 u = − 53 , 2 x −1 = − 5 3

x −1 = 2

x = − 35

x = 12

Yes, the two solutions agree.

121. x −2 = 3x −1 − 10

Graphically, let: y1 = x −2 , y2 = 3x −1 − 10 .

x −2 − 3x −1 + 10 = 0 Let u = x −1 to obtain u 2 − 3u + 10 = 0 3 ± 9 − 4(10)(1) 3 ± i 31 = 2 2 So, there are no real solutions. As such, we expect the graphs to not intersect. u=

Yes, the two solutions agree.

143

Section 1.5 Solutions ------------------------------------------------------------------------------1.

[3,∞ )

2. ( −∞, −2 )  [⎯⎯⎯⎯→

  ←⎯⎯⎯ ⎯) ... − 4 − 3 − 2 − 1 0 1 2 ...

... 0 1 2 3 4 5 6 ...

( −∞, −5]

( −7, ∞ )

  ←⎯⎯⎯ ⎯] ... − 7 − 6 − 5 − 4 − 3 − 2 − 1 0 ...

[ −2,3 )

... − 7 − 6 − 5 − 4 − 3 − 2 − 1 0 1...

[ −4, −1]

 [ ) 

 ]  [

... − 3 − 2 − 1 0 1 2 3 4 ...

... − 5 − 4 − 3 − 2 − 1 0 ...

( −3,5]

[0]

 (⎯⎯⎯⎯⎯⎯⎯→

( 0,6 )

  ( ]

 ( )

... − 3 − 2 − 1 0 1 2 3 4 5 ...

... − 2 − 1 0 1 2 3 4 5 6 7 8 ...

  •

10.

... − 3 − 2 − 1 0 1 2 3 ...

[ −7 ]

 • ... − 9 − 8 − 7 − 6

− 5 − 4 ...

12. ( −3,2]

11. [4,6]  [ ] 

 ( ] 

... 3 4 5 6 7 ...

... − 3 − 2 − 1 0 1 2 ...

13. [ −8, −6]

14. ( −∞,2)  ←⎯⎯ ⎯) ... 0 1 2 3 4 ...

 ] [ ... − 9 − 8 − 7 − 6 − 5 ...

15. no solution

16. no solution



... − 3 − 2 − 1 0 1 2 3 ...

144

Section 1.5

17. {x | 0 ≤ x < 2}

18.

{x | 0 < x ≤ 3}

19. {x | − 7 < x < −2 }

20. {x | − 3 ≤ x ≤ 2}

21. {x | x ≤ 6}

22. {x | x > 5}

23. {x | − ∞ < x < ∞ }

24. {x |4 ≤ x ≤ 4 }

25. −3 < x ≤ 7

( −3,7]

1 7 26. − ≤ x < 2 8

27. 3 ≤ x < 5

[3,5)

28. 4 < x ≤ 8 (4,8]

29. −2 ≤ x

[ −2, ∞ )

30. x < −3 ( −∞, −3 )

 1 7 − ,   2 8 

31. −∞ < x < 8 ( −∞,8)

32. −2 ≤ x < ∞ [ −2, ∞ )

33. ( −5,3 )

34. [ −5,7 )

 ( ) 

 [− − − − − − − − −)

... − 5 − 4 − 3 − 2 − 1 0 1 2 3 ...

. .. − 5 − 4 − 3 − 2 − 1 0 ... 7 8 9 ...

35. [ −6,5)

36. [ −6,1)

 [ ) 

 )  [

... − 7 − 6 − 5 ... 0 1 2 3 4 5 ...

... − 7 − 6 − 5 − 4 − 3 − 2 − 1 0 1 2 ...

37. [ −1,1]

38. ( −∞, −5 )   ←⎯⎯⎯ ⎯) ... − 7 − 6 − 5 − 3 − 2 − 1 0 ...

 [ ] ... − 2 − 1 0 1 2 3 ...

40. [ −3, ∞ )

39. [1,4)

 [ ) 

 [⎯⎯⎯⎯⎯⎯ →

... 0 1 2 3 4 ...

... − 3 − 2 − 1 0 1 2 3 4 ...

145

Chapter 1 41. [ −1,2 )

42. [ −2,5)

 [ ) 

 ) [

... − 3 − 2 − 1 0 1 2 3 4 5 ...

... − 2 − 1 0 1 2 3...

44. ( −∞, ∞ )

43. ( −∞, 4) ∪ (4, ∞ )

   



... 2 3 4 5 6 7 ...

... − 3 − 2 − 1 0 1 2 3 ...

45. ( −∞, −3] ∪ [3, ∞ )

46. ( −2,1]

]  [

  ( ]

... − 3 − 2 − 1 0 1 2 3 ...

... − 3 − 2 − 1 0 1 2 3 4 ...

47. ( −3,2 ]

48. ( −∞, ∞ )

  ( ]



... − 4 − 3 − 2 − 1 0 1 2 3 4 ...

... − 3 − 2 − 1 0 1 2 3 ...

49. no solution

50. no solution



... − 3 − 2 − 1 0 1 2 3 ...

51. ( −∞,2 ) ∪ [3,5 )

52. ( −∞, −5 ) ∪ [ 0,2 ]

53. ( −∞, −4] ∪ (2,5]

54. [ −12, −5) ∪ ( −2, ∞ )

55. [ −4, −2 ) ∪ ( 3,7 ]

56. ( −∞, −2 ) ∪ ( 2, ∞ )

57. ( −6, −3] ∪ [0, 4) 59.

x −3 < 7

58. ( −∞, −5] ∪ [ −1, ∞ ) 60.

x+4>9

61.

3x − 2 ≤ 4

x < 10

x >5

3x ≤ 6

( −∞,10 )

(5, ∞ )

x≤2

146

( −∞,2 ]

Section 1.5

62.

3x + 7 ≥ −8

63.

3x ≥ −15 x ≥ −5

[ −5, ∞ ) 65.

68.

−5 p ≥ 10 Divide by − 5 and flip sign p ≤ −2

( −∞, −2] 66.

3 − 2x ≤ 7 −2x ≤ 4 x ≥ −2

4 − 3x > −17

[ −2, ∞ )

( −∞,7 )

5 + 2 x > −15

69.

x > −10

( −10, ∞ )

( −∞, 4 )

2 x > −20

72.

2.7 x − 1.3 < 6.8 2.7 x < 8.1 x<3

( −∞,3)

−4u < 12 Divide by − 4 and flip sign u > −3

( −3, ∞ ) 67.

−3x > −21 x<7

−6 + 8x < 26 8x < 32 x<4

71. −1.8x + 2.5 > 3.4 −1.8x > 0.9 0.9 x< = −0.5 −1.8 ( −∞, −0.5 )

64.

−3x + 7 ≤ 22 −3x ≤ 15 x ≥ −5

[ −5, ∞ ) 70.

−4 x + 3 ≤ −21 −4 x ≤ −24 x≥6

[6, ∞ ) 73.

3 ( t + 1) > 2t 3t + 3 > 2t t +3 > 0 t > −3

( −3, ∞ )

147

Chapter 1

74.

76.

2 ( y + 5) ≤ 3 ( y − 4 )

75. 7 − 2 (1 − x ) > 5 + 3 ( x − 2 )

2 y + 10 ≤ 3 y − 12

7 − 2 + 2 x > 5 + 3x − 6

4 − 6 − 3x < 5

10 ≤ y − 12

5 + 2 x > 3x − 1 5 > x −1

−2 − 3x < 5 −3x < 7

x<6

x > −7 3

( −∞,6 )

( −7 3, ∞ )

22 ≤ y

[22, ∞ ) 77.

4 − 3(2 + x) < 5

78. y −3 y −2≤ 5 4 y  y −3  20 ⋅  − 2  ≤ 20 ⋅ 4  5  4( y − 3) − 2(20) ≤ 5 y

x+2 x −2 ≥ 3 2 LCD = 6

2 ( x + 2 ) − 2 ( 6 ) ≥ x (3) 2 x + 4 − 12 ≥ 3x

4 y − 12 − 40 ≤ 5 y

−8 ≥ x or x ≤ −8

−52 ≤ y

( −∞, −8] 79.

[ −52, ∞ )

80. 2 p +1 > −3 5  2 p +1  5⋅  > 5 ⋅ ( −3)  5  2 p + 1 > −15

t −5 ≤ −4 3 LCD = 3 t − 5 ≤ −4 ( 3 ) t − 5 ≤ −12

2 p > −16

t ≤ −7

p > −8

( −∞, −7]

( −8, ∞ )

148

Section 1.5

81.

82. Multiply by LCD = 6

s s −3 s 1 − > − 2 3 4 12 LCD = 12

4 y − 3 ( 5 − y ) < 10 y − 6 ( 2 + y ) 4 y − 15 + 3 y < 10 y − 12 − 6 y

83.

7 y − 15 < 4 y − 12

6 s − 4 ( s − 3 ) > 3s − 1

3 y − 15 < −12 3y < 3

2s + 12 > 3s − 1

6s − 4s + 12 > 3s − 1

y <1

s < 13

( −∞,1)

( −∞,13) 84.

−2 < x + 3 < 5 −5 < x < 2

−5 < x < 6

( −5,6 )

( −5,2 ) 85.

−8 ≤ 4 + 2 x < 8 −12 ≤ 2 x < 4 Divide by 2 −6 ≤ x < 2

1 < x + 6 < 12

86.

0<2+x ≤5

87.

−2 < x ≤ 3

( −2,3]

[ −6,2 ) 88.

[ −8, 4 ) 90.

89. 3 ≤ −2 − 5x ≤ 13 5 ≤ −5x ≤ 15 Divide by − 5 Flip the signs −1 ≥ x ≥ −3

[ −3, −1]

−3 < 1 − x ≤ 9 −4 < − x ≤ 8 Divide by − 1 Flip the signs −8 ≤ x < 4

1 A−3 < 7 2 1 6 < A < 10 2 Multiply by 2

1 0<2− y< 4 3 1 −2 < − y < 2 3 Multiply by − 3 Flip the signs

12 < A < 20

−6 < y < 6

(12,20 )

( −6,6 )

149

Chapter 1

91.

92. 1 1+ y 3 ≤ ≤ 2 3 4 Multiply by 3

2−z 1 ≤ 4 5 Multiply by 4

3 9 ≤ 1+ y ≤ 2 4 1 5 ≤ y≤ 2 4

4 5 6 −6 < − z ≤ − 5 Multiply by − 1 Flip the signs

−1 <

−4 < 2 − z ≤

1 5   2 , 4 

6>z≥

93. −0.7 ≤ 0.4x + 1.1 ≤ 1.3 −1.8 ≤ 0.4x ≤ 0.2 1.8 0.2 − ≤x≤ 0.4 0.4 −4.5 ≤ x ≤ 0.5

[ −4.5, 0.5]

6 5

6   5 ,6  94. 7.1 > 4.7 − 1.2x > 1.1 2.4 > −1.2 x > −3.6 −2 < x < 3 ( −2,3)

96. Low weight:

95. Low weight: 110 9 ) = 128  + 2 ( st 2 lbs

1 5 feet

High weight:

110 9 ) = 164  + 6 ( 6 lbs

1st 5 feet

128 ≤ w ≤ 164

114 ≤ w ≤ 150

1 5 feet

1st 5 feet

9 inches

105 9 ) = 114  + 1 ( 1 lbs

9 inches

105 9 ) = 150  + 5 ( 5 lbs 9 inches

97. Revenue = 100x ( x = # dresses )

98. If champ by age 2: Revenue = 30,000

Cost = 4000 + 20x Profit = Revenue − Cost

11,600 ≤ Cost ≤ 26,000

= 100 x − ( 4000 + 20 x ) > 0

Profit = revenue − cost

100 x − 4000 − 20 x > 0 80 x > 4000 x > 50

2000 + 400 ( 24 ) ≤ Cost ≤ 2000+1000 ( 24 )

30,000 − 26,000 ≤ profit ≤ 30,000 − 11,600 4,000 ≤ profit ≤ 18,400 $4,000 to $18,400

More than 50 dresses

150

Section 1.5

99. Solve: 5,000 + 1.75x ≥ 10,000 (Note: We changed from 10 to 10,000 on the right-side of the inequality because R(x) is measured in thousands of dollars.) 1.75x ≥ 5,000 x ≥ 2,857.14 So, must sell at least 285,700 units.

100. Solve: 5,000 + 1.75x ≥ 7,500 (Note: We changed from 7.5 to 7,500 on the right-side of the inequality because R(x) is measured in thousands of dollars.) 1.75x ≥ 2,500 x ≥ 1, 428.57 So, must sell at least 142,900 units.

101. Use the formula THR = ( HRmax − HRrest ) × I + HRrest

102. Use the formula THR = ( HRmax − HRrest ) × I + HRrest

with HRrest = 65, HRmax = 170 . Solve for I first when THR = 100 and then when THR = 140: 100 = (170 – 65)I + 65 35 = 105I I ≈ 0.33 So, about 33%. 140 = (170 – 65)I + 65 75 = 105I I ≈ 0.71 So, about 71%. So, can consider workouts between 33% and 71% intensity.

with HRrest = 75, HRmax = 175 . Solve for I first when THR = 110 and then when THR = 150: 110 = (175 – 75)I + 75 35 = 100I I ≈ 0.35 So, about 35%. 150 = (175 – 75)I + 75 75 = 100I I ≈ 0.75 So, about 75%. So, can consider workouts between 35% and 75% intensity.

103. Cable Charge: 30 + 3.99x (x = number of movies ordered) 53.94 ≤ 30 + 3.99x ≤ 73.89 23.94 ≤ 3.99x ≤ 43.89 6 ≤ x ≤ 11 Least movies: 6, Most movies: 11

104. Let x = number of ounces over 8 ounces Solve: 7.2 ≤ 6 + 0.3x ≤ 9 1.2 ≤ 0.3x ≤ 3

4 ≤ x ≤ 10 +8 oz. base The least number of ounces ordered was 12, and the most ordered was 18.

151

Chapter 1

105. Amount made: 400 + 0.18x ( x = number of views) 400 + 0.18x ≥ 535 0.18x ≥ 135 x ≥ 750 views

106. Amount of purchase: 1.5 + 1.25x (x = number of hot dogs) 1.5 + 1.25x ≥ 5 1.25x ≥ 3.5 x ≥ 2.8 At least three hot dogs.

107. Let x = grade on the 4th exam. 67 + 77 + 84 + x ≥ 80 4 67 + 77 + 84 + x ≥ 320

108. Let x = grade on the exam. 96 + 87 + 79 + 89 + x 80 ≤ ≤ 90 5 400 ≤ 351 + x ≤ 450 49 ≤ x ≤ 99 You would need to score between 49% and 99% on the final exam.

228 + x ≥ 320 x ≥ 92 109. Let x = invoice price. 27,999 27,999 <x< 1.30 1.15 $21,537.69 < invoice price < $24,346.96

110. Let x = invoice price. 42,599 42,599 <x< 1.30 1.15 $32,768.46 < invoice price < $37,042.61

111. 0.9 rT ≤ rR ≤ 1.1 rT

112. S ≥ 2 if N fluctuates by 10% N S > 2 or S > 2.2 N 1.1 N

113. 0.85L ≤ B ≤ 0.95L

114. 0.95ht ≤ hm ≤ 1.05ht

152

Section 1.5

115. Let x = number of times play. We want the smallest value of x for which 160 + 10 x ≤ 55x. Solving yields: 160 ≤ 45x 160 ≤x 3.56 ≈ 45 So, they would need to play 4 times in order to make the membership a better deal.

116. Let x = number of times play. We want the smallest value of x for which 125 + 10 x ≤ 40 x. Solving yields: 125 ≤ 30 x 125 ≤x 4.17 ≈ 30 So, they would need to play 5 times in order to make the membership a better deal.

117. Let T = amount of tax paid. Least amount of tax = $4543 Greatest amount of tax = 4543 + 0.22 ( 84,200 − 39, 475 )

118. Let T = amount of tax paid. Least amount of tax = $14382.50 Greatest amount of tax = 14382.50 + 0.24 (16,0725 − 84,200 )

= $14,382.50 So the range of taxes is: $4543 ≤ T ≤ $14,382.50

= $32,748.50 So the range of taxes is: $14,382.50 ≤ T ≤ $32,748.50

119. The correction should be [ −1, 4 ) , and the graph is not correct.

120. Performed union instead of intersection. ( 3, 4 )

121. You must reverse the sign.

122. x ≥ −2 corresponds to [ −2, ∞ )

123. True. In fact, the two inequalities are equivalent. 124. False. Need to switch the sign. 125. a and b

126. c and d

127. a and b

128. c and d

129. c

130. a and b

131. Mentally, realize that x ≤ −x holds only when the left-side is negative or zero. Hence, the solution set is ( −∞,0] .

132. Mentally, realize that x > −x holds only with the right-side is negative, which occurs when x > 0 . Hence, the solution set is (0, ∞ ) .

153

Chapter 1 133. Observe that ax + b < ax − c b < −c This is false because we are assuming that 0 < b < c , so that −c < b . Hence, the inequality has no solution.

134. Observe that −ax + b < −ax + c b<c This is true, by assumption. Hence, the solution set is all real numbers.

135. a)

b) 2.7x + 3.1 < 9.4x − 2.5 2.7x + 5.6 < 9.4x 5.6 < 6.7x x > 0.83582 (rounded)

c) yes

136. a)

b) −0.5x + 2.7 > 4.1x − 3.6 2.7 > 4.6 x − 3.6 6.3 > 4.6 x x < 1.36957 ( rounded )

c) yes

154

Section 1.5

137. a)

b) x − 3 < 2x − 1 < x + 4

−3 < x − 1 < 4 −2 < x < 5

( −2,5 ) c) yes

138. a) x − 2 < 3x + 4 ≤ 2x + 6 x − 2 < 3x + 4 and 3x + 4 ≤ 2x + 6 − 6 < 2x x≤2 and

b) Graphically, let y1 = x − 2, y2 = 3x + 4, y3 = 2x + 6

−3 < x

( −3,2] c) agree

155

Chapter 1

139. a)

b) x +3 < x +5 3<5 true for any x ∈ ( −∞, ∞ )

c) yes

140. a) 1 2

x − 3 > − 23 x + 1

7 6

x>4

b) Graphically, let y1 = 12 x − 3, y2 = − 23 x + 1

x > 4 ( 67 ) = 247

( 247 , ∞ ) c) agree

156

Section 1.6 Solutions -------------------------------------------------------------------------------1.

( x − 5 )( x + 2 ) ≥ 0

( x + 3 )( x − 1) < 0

CP's: x = −2, 5

CP's: x = −3,1

( −∞, −2 ] ∪ [5, ∞ )

( −3,1) 4.

3. u − 5u − 6 ≤ 0

u 2 − 6u − 40 > 0

( u − 6)( u + 1) ≤ 0

( u − 10)( u + 4) > 0

CP's: u = 6, −1 + − +  | |

CPs: − 4,10 + − +  | |

−1

−4

[ −1,6]

( −∞, −4) ∪ (10, ∞ )

6. p + 4p +3 < 0

p 2 − 2 p − 15 ≥ 0

( p + 3)( p + 1) < 0

( p − 5 )( p + 3 ) ≥ 0

CP's: p = −3, −1

CP's: p = −3,5

( −3, −1)

( −∞, −3] ∪ [5, ∞ )

8. 2t − t − 3 ≤ 0

3t 2 + 5t − 2 ≥ 0

( 2t − 3 )(t + 1) ≤ 0

(3t − 1)(t + 2 ) ≥ 0

CP's: t = −1,3 2

CP's: t = −2,1 3

[ −1,3 2]

( −∞, −2] ∪ [1 3, ∞ )

157

Chapter 1

10. 6v − 5v + 1 < 0

12t 2 − 37t − 10 < 0

(3v − 1)( 2v − 1) < 0

(3t − 10 )( 4t + 1) < 0

CP's: v = 1 3,1 2

CP's: t = −1 4,10 3

(1 3,1 2 )

( −1 4,10 3)

11.

12. 2 s − 5s − 3 ≥ 0

s 2 + 8s + 12 ≤ 0

( 2s + 1)( s − 3) ≥ 0

( s + 2 )( s + 6 ) ≤ 0

CP's: s = −1 2, 3

CP's: s = −6, −2

( −∞, −1 2] ∪ [3, ∞ )

[ −6, −2]

13.

14. y + 2 y − 4 ≥ 0 Note: Can't factor

y 2 + 3 y − 1 ≤ 0 Note: can't factor

To find CP's solve y 2 + 2 y − 4 = 0

To find CP's solve y 2 + 3 y − 1 = 0

−2 ± 2 2 − 4 (1)( −4 )

2 (1)

−2 ± 20 2 −2 ± 2 5 y= = −1 ± 5 2

−3 ± 32 − 4 (1)( −1) 2 (1)

−3 ± 13 2

 −3 − 13 −3 + 13  ,   2 2  

( −∞, −1 − 5  ∪ −1 + 5, ∞ )

158

Section 1.6

15.

16. x − 4x < 6

x 2 − 2x > 5

x2 − 4x − 6 < 0 CPs: Use quadratic formula:

x 2 − 2x − 5 > 0 CPs: Use quadratic formula:

4 ± 16 − 4(1)( −6) 4 ± 2 10 = 2 2 = 2 ± 10 + − +  | |

2 ± 4 − 4(1)( −5) 2 ± 2 6 = 2 2 = 1± 6 + − +  | |

2 − 10

2 + 10

1− 6

( 2 − 10,2 + 10 ) 17.

( −∞,1 − 6 ) ∪ (1 + 6, ∞ ) 18.

u − 3u ≥ 0

u 2 + 4u ≤ 0

u ( u − 3) ≥ 0

u (u + 4) ≤ 0

CP's: u = 0, 3

CP's: u = 0, −4

( −∞,0] ∪ [3, ∞ )

[ −4,0 ]

19.

20. x − 2x ≤ 0

x 2 + 3x ≥ 0

x (x − 2) ≤ 0

x ( x + 3) ≥ 0

CP's: x = 0,2

CP's: x = −3, 0

[ 0,2]

( −∞, −3] ∪ [ 0, ∞ )

159

Chapter 1

21.

22. x −9 > 0

x 2 − 16 ≥ 0

( x − 3)( x + 3 ) > 0

( x − 4 )( x + 4 ) ≥ 0

CP's: x = −3,3

CP's: − 4, 4

( −∞, −3) ∪ (3, ∞ )

( −∞, −4] ∪ [ 4, ∞ )

23.

24. 2

t − 81 < 0

t 2 − 49 ≤ 0

(t − 9 )( t + 9 ) < 0

(t − 7 )(t + 7 ) ≤ 0

CP's: t = −9,9 + − +  | | −9 9

CP's: t = −7,7 + − +  | | −7

[ −7,7]

( −9,9)

25.

26. z2 + 2 ≥ 0

z + 16 > 0 No critical points 2

( −∞, ∞ ) (consistent)

z + 16 > 0 for all z 2

( −∞, ∞ ) (consistent) 27.

28. y < −4 no real solution

y 2 ≤ −25 no real solution

 A real number squared     is always non-negative. 

160

Section 1.6

30.

29. −3 ≤ 0 x = 0 is CP x

3 ≤ 0 x = 0 is CP x

( 0,∞ )

( −∞,0 ) 32.

31. y > 0 CP's: y = −3, 0 y+3

y ≤ 0 CP's: y = 0,2 2−y

( −∞, −3 ) ∪ ( 0, ∞ )

( −∞,0 ] ∪ ( 2, ∞ )

33.

34. t +3 ≥0 t−4 CPs: − 3, 4 + − +  | | −3 4

2t − 5 <0 t−6 CPs: 5 2 ,6 + − +  | | 5 6 2

( −∞, −3] ∪ (4, ∞ )

( 5 2 ,6 )

35.

36. s +1 ≥0 ( 2 − s )( 2 + s )

s +5 ≤0 ( 2 − s )( 2 + s )

CP's: s = −2, − 1, 2

CP's: s = −5, − 2, 2

( −∞, −2 ) ∪ [ −1,2 )

[ −5, −2 ) ∪ ( 2, ∞ )

161

Chapter 1

37.

38. x −3 ≥0 x 2 − 25 x −3 ≥0 ( x − 5)( x + 5) CPs: 3, ± 5 − + − +   | | |

1− x ≤0 x2 − 9 1− x ≤0 ( x − 3)( x + 3) CPs: 1, ± 3 + − + −   | | |

−5

−3

( −5,3] ∪ (5, ∞ ) 39.

( −3,1] ∪ (3, ∞ ) 40.

2u + u < 3

u 2 − 3u ≥ 18

2u 2 + u − 3 < 0

u 2 − 3u − 18 ≥ 0

( 2u + 3)( u − 1) < 0

( u − 6 )( u + 3) ≥ 0

CP's: u = −3 2,1

CP's: u = −3, 6

( −3 2,1)

( −∞, −3] ∪ [6, ∞ )

42.

41.

−2t − t 2 −t ≥ 0 4 −t −2t − t 2 − t ( 4 − t ) ≥0 4 −t −2t − t 2 − 4t + t 2 ≥0 4 −t −6t ≥0 4 −t CP's: t = 0, 4

3t − 5t ≥ 0 t+2 3t 2 − 5t ( t + 2 ) ≥0 t+2 3t 2 − 5t 2 − 10t ≥0 t+2 −2t 2 − 10t ≥0 t+2 −2t ( t + 5 ) ≥ 0 CP's: t = −5, −2,0 t+2

( −∞,0 ] ∪ ( 4, ∞ )

( −∞, −5] ∪ ( −2,0] 162

Section 1.6

43.

3 p − 2 p (3 + p ) − <0 4 − p2 (2 − p)

44.

( p + 2) ≤ 0 −7 p − ( p − 10 )( p + 10 ) ( p + 10 ) −7 p − ( p + 2 )( p − 10 ) ≤0 ( p − 10 )( p + 10 )

p (3 − 2 p ) (3 + p ) < 0 − ( 2 − p )( 2 + p ) ( 2 − p )

p ( 3 − 2 p ) − ( 3 + p )( 2 + p ) <0 ( 2 − p )( 2 + p )

−7 p − p 2 + 8 p + 20 ≤0 ( p − 10 )( p + 10 ) − p 2 + p + 20 ≤0 ( p − 10 )( p + 10 )

3 p − 2 p2 − 6 − 5 p − p2 >0 ( 2 − p )( 2 + p ) −3 p 2 − 2 p − 6 <0 ( 2 − p )( 2 + p )

( − p + 5 )( p + 4 ) ≤ 0 ( p − 10 )( p + 10 )

3 p2 + 2 p + 6 >0 ( 2 − p )( 2 + p )

CP's: p = −10, −4,5,10

CP's: p = −2, 2

( −∞, −10 ) ∪ [ −4,5] ∪ (10, ∞ ) ( −2,2 ) 46.

45. 2

x <0 5 + x2

x2 ≤0 5 + x2

no real solution

x=0 •| ←⎯⎯⎯⎯⎯ → 0

48.

47.

 x2 + 2  − 2 <0 x +4

x + 10 >0 x 2 + 16 2

( −∞, ∞ ) (consistent)

| ←⎯⎯⎯⎯ → 0

163

Chapter 1

49.

(v − 3 )(v + 3) ≥ 0 v ≠ 3 (v − 3)

50.

(v − 1)

(v + 1) ≤ 0 v ≠ −1 (v + 1)

(v − 1) ≤ 0

v +3 ≥ 0 v ≥ −3

v ≤1

[ −3,3) ∪ (3, ∞ ) 51.

( −∞, −1) ∪ ( −1, −1] 52. 1 1 + ≤0 t−2 t+2 (t + 2) + (t − 2) ≤0 (t − 2)(t + 2) 2t ≤0 (t − 2)(t + 2) CPs: 0, ± 2 − + − +   | | |

2 1 + ≥0 t −3 t +3 2(t + 3) + (t − 3) ≥0 (t − 3)(t + 3) 3t + 3 ≥0 (t − 3)(t + 3) 3(t + 1) ≥0 (t − 3)(t + 3) CPs: − 1, ± 3 − + − +   | | | −3

−1

−2

( −∞, −2) ∪ [0,2)

( −3, −1] ∪ (3, ∞ )

164

Section 1.6

54.

53. 3 1 − ≤0 x+4 x−2 3( x − 2) − ( x + 4) ≤0 ( x + 4)( x − 2) 2 x − 10 ≤0 ( x + 4)( x − 2) 2( x − 5) ≤0 ( x + 4)( x − 2) CPs: − 4,2,5 − + − +   | | | −4

2 1 − ≥0 x − 5 x −1 2( x − 1) − ( x − 5) ≥0 ( x − 5)( x − 1) x+3 ≥0 ( x − 5)( x − 1) CPs: − 3,1,5 − + − +   | | | −3

[ −3,1) ∪ (5, ∞ )

( −∞, −4) ∪ (2,5]

56.

55. 1 1 p − 48 + − 2 >0 p + 4 p − 4 p − 16 2

( p − 4) + ( p + 4) − ( p 2 − 48 ) ( p + 4)( p − 4)

1 1 − −2≤ 0 p −3 p +3

( p + 3) − ( p − 3) − ( p2 − 9 )

( p + 3)( p − 3)

− ( p 2 − 2 p − 48 )

− p2 + 12 ≤0 ( p + 3)( p − 3)

>0 ( p + 4)( p − 4) −( p − 8)( p + 6) >0 ( p + 4)( p − 4) CPs: − 6, ± 4, 8 − + − + −  | | | | −6

−4

≤0

(

− p−2 3

)( p + 2 3 ) ≤ 0

( p + 3)( p − 3)

CPs: ± 3, ± 2 3 − + − + −  | | | |

−2 3

( −6, −4) ∪ (4,8)

−3

2 3

( −∞, −2 3  ∪ ( −3,3) ∪ 2 3, ∞ )

165

Chapter 1

57.

58. 1 1 3 − − 2 ≥0 p−2 p+2 p −4 ( p + 2) − ( p − 2) − 3 ≥0 ( p + 2)( p − 2) 1 ≥0 ( p + 2)( p − 2) CPs: ± 2 + − +  | | −2 2

2 1 1 − − 2 ≤0 2 p − 3 p +1 2 p − p − 3 2 1 1 − − ≤0 2 p − 3 p + 1 (2 p − 3)( p + 1) 2( p + 1) − (2 p − 3) − 1 ≤0 (2 p − 3)( p + 1) 4 ≤0 (2 p − 3)( p + 1) CPs: − 1, 3 2 + − +  | | 3 −1 2

( −∞, −2) ∪ (2, ∞ )

( −1, 3 2 ) 59.

60. −x + 130x − 3000 > 0

x 2 − 130 x + 3600 > 0

x 2 − 130x + 3000 < 0

( x − 40 )( x − 90 ) > 0

( x − 30 )( x − 100 ) < 0

CP's: x = 40,90

CP's: x = 30,100 Fewer than 40 or more than 90 orders Between 30 and 100 orders

166

Section 1.6

61.

62. Car is worth more than you owe:

Car is worth more than you owe:

t > 0 CP's: t = 0,3 t −3

 2 −t  − >0  4 −t 

(3, ∞ ) For years 3–5, the car is

(2, 4) For between 2 and 4 years,

worth more than you owe. You owe more than it's worth: t < 0 CP's: t = 0,3 t −3

the car is worth more than you owe. You owe more than it’s worth:  2 −t  − <0  4 −t 

( 0,2 ) ∪ ( 4, ∞ ) In the first 2 years and after the 4th year you owe more than the car is worth.

( 0,3) In the first 3 years you owe more than the car is worth.

63.

64. h = −16t + 1200t

h = −16t 2 + 600t

bullet is in the air if h > 0

−16t + 1200t > 0

−16t 2 + 600t > 0

−16t ( t − 75 ) > 0

−8t ( 2t − 75 ) > 0

CP's: t = 0,75

CP's: t = 0,75 2

( 0,75 )

( 0,75 2 )

Bullet is in the air for 75 sec

Bullet is in the air for 37.5 sec

167

Chapter 1

65.

66. Area = l ⋅ w

6.25t 2 − 25t + 325 ≤ 525

P = 2l + 2w = 100 100 − 2w l= 2  100 − 2w  A = l ⋅w =   (w ) 2  

6.25t 2 − 25t − 200 ≤ 0

50w − w 2 ≥ 600

CP's: t = −4, 8 + − + | | ←⎯⎯⎯⎯⎯⎯⎯ → −4 8

625t 2 − 2500t − 20000 ≤ 0 625 ( t 2 − 4t − 32 ) ≤ 0 625 ( t − 8 )( t + 4 ) ≤ 0

w 2 − 50w + 600 ≤ 0

[ −4,8]

( w − 20 )( w − 30 ) ≤ 0

t = 0 corresponds to Nov. 2014

CP's: w = 20, 30

t = 8 corresponds to July 2015 The price of $525 was reached in

[20,30]

July 2015 (t = 8).

20 ≤ width ≤ 30 20 ≤ length ≤ 30 Between 20 and 30 feet

67.

−5( x + 3)( x − 24) < 460

68.

−5( x + 3)( x − 24) > 550

−5x 2 + 105x + 360 < 460

−5x 2 + 105x + 360 > 550

−5x 2 + 105x − 100 < 0

−5x 2 + 105x − 190 > 0

x 2 − 21x + 20 > 0 ( x − 20)( x − 1) > 0 The solution set is ( −∞,1) ∪ (20, ∞ ) . So, a price increase less than $1 or greater than $20 per bottle.

x 2 − 21x + 38 < 0 ( x − 19)( x − 2) < 0 The solution set is (2,19) . So, a price increase between $2 and $19 per bottle.

168

Section 1.6

69. 400 ± 7 = 393, 407 1,360,000 1,360,000 ≤ price per acre ≤ 407 393 $3,341.52 ≤ price per acre ≤ $3460.56

70. 1000 ± 10 = 990,1010 1,000,000 1,000,000 ≤ price per acre ≤ 1010 990 $990.10 ≤ price per acre ≤ $1010.10

$3,342 to $3,461 per acre

71.

$990 to $1,010 per acre

72. Cannot divide by x.

Cannot take square root.

x − 3x > 0

u 2 − 25 < 0

x ( x − 3) > 0

( u − 5 )( u + 5 ) < 0 ( −5,5 )

( −∞,0 ) ∪ (3,∞ )

73. ( x − 2 )( x + 2 ) > 0 x ≠ 2 ( x + 2)

74. Can't cross-multiply x+4 1 + <0 3 x 3( x + 4) + x <0 3x 4 x + 12 <0 3x 4 ( x + 3) <0 3x x = 0, −3 are CP's

x−2 > 0 x>2 You must consider the x = −2 as a critical point.

( −3,0 ) 75. False ( −a, a )

76. False ( −∞, −a ] ∪ [ a, ∞ )

77. Assume that ax 2 + bx + c < 0 . If b2 − 4ac < 0 , then either there are infinitely many solutions or no real solution.

78. Assume that ax 2 + bx + c > 0 . If b2 − 4ac < 0 , then either there are infinitely many solutions or no real solution.

169

Chapter 1

79. x2 + a2 ≥ 0 True for all real values of x

80. x 2 − b2 < 0 x ≠ −b x+b ( x − b )( x + b ) < 0 x+b x−b < 0

( −∞, ∞ )

x<b

81.

( −∞, −b ) ∪ ( −b, b )

82. x +a ≥0 x2 + b2 2

a +b < 0 x2 a + bx 2 <0 x2 No real values for which

( −∞, ∞ )

this is true. No solution

83. y1 = 1.4 x 2 − 7.2 x + 5.3 y2 = −8.6 x + 3.7 Find when y1 > y2

( −∞, ∞ )

170

Section 1.6

84. y1 = 17x 2 + 50x − 19 y2 = 9x 2 + 2 Find when y1 < y2

( −6.65, 0.4 )

85. y1 = 11x 2 y2 = 8x + 16 Find when y1 < y2

( −0.8960,1.6233)

171

Chapter 1

86. y1 = 0.1x + 7.3 y2 = 0.3x 2 − 4.1 Find when y1 > y2

( −6,6.33 )

87. y1 = x y2 = x 2 − 3x y3 = 6 − 2x Find when y1 < y2 < y3 . ( −2,0)

88. y1 = x 2 + 3x − 5 y2 = − x 2 + 2 x + 10 Find when y1 ≥ y2 .

( −∞, −3] ∪ [2.5, ∞ )

172

Section 1.6

89. 2p 5− p y2 = 1 Find when y1 > y2 . y1 =

( 53 ,5 )

90. 3p 4− p y2 = 1 Find when y1 < y2 . y1 =

( −∞,1) ∪ (4, ∞ )

173

Section 1.7 Solutions -------------------------------------------------------------------------------1. x = −3 or x = 3

2. x = −2 or x = 2

3. No solution (absolute value is always non-negative)

4. No solution (absolute value is always non-negative)

5. t + 3 = −2

t +3 = 2

6. t − 3 = −2

t −3 = 2

t = −5

t = −1

t =1

t =5

7. p−7 = 3

p − 7 = −3

8. p+7 =3

p + 7 = −3

p = 10

p=4

p = −4

p = −10

9. 4 − y = −1

4 − y =1

10. 2 − y = −11

2 − y = 11

y=5

y=3

y = 13

y = −9

11.

13. 3 x = −9 x = −3

12.

4−y =2

7−y =5

4−y =2 y=2 4 − y = −2 y=6

7−y =5 y=2 7 − y = −5 y = 12

3x = 9

14. 5x = −50

5x = 50

x =3

x = −10

x = 10

174

Section 1.7

15. −7x = 21

16. −12 y = 144

−7x = 21

−7x = −21

−12 y = 144

−12 y = −144

x = −3

x =3

y = −12

y = 12

17. 2 x + 7 = −9

2x + 7 = 9

18. 2x − 5 = 7

2x − 5 = −7

2 x = −16

2x = 2

2x = 12

2 x = −2

x = −8

x =1

x=6

x = −1

19. 3t − 9 = 3 3t = 12

3t − 9 = −3 3t = 6

t=4

20. 4t + 2 = 2 4t = 0

4t + 2 = −2 4t = −4

t=2

t=0

t = −1

21. 7 − 2 x = −9 2 x = 16

7 − 2x = 9 2 x = −2

22. 6 − 3 y = 12 −3 y = 6

6 − 3 y = −12

x =8

x = −1

y = −2

y=6

23. 1 − 3y = 1 −3 y = 0

1 − 3 y = −1

24. 5−x = 2

−3 y = −2

− x = −3

5 − x = −2 − x = −7

y=0

y = 23

x =3

x=7

25. 3 − 2 x = −9

26. 7 + 4x = 5

No solution, because the absolute value of a number cannot be negative.

7 + 4x = 5 4x = −2 1 x=− 2

175

−3 y = −18

7 + 4 x = −5 4 x = −12 x = −3

Chapter 1

27. 4.7 − 2.1x = −3.3

4.7 − 2.1x = 3.3

2.1x = 8

2.1x = 1.4

28. 5.2 x + 3.7 = −2.4 5.2 x = −6.1

x = 80 21

x = 23

61 x = − 52

x = − 14

2 4 5 x− = 3 7 3 LCD = 21

30. 1 3 1 x+ =− 2 4 16 8x + 12 = −1 8x = −13

1 3 1 x+ = 2 4 16 8x + 12 = 1

x = −13 8

x = −11 8

x + 3 = −11 x = −14

29. 2 4 5 x− =− 3 7 3 LCD = 21 14 x − 12 = −35 14 x = −23

14 x − 12 = 35 14 x = 47

x = −23 14

x = 47 14

31. x −5 = 8

x − 5 = −8

32. x + 3 = 11

x −5 = 8

x = −3

x + 3 = 11

x = 13

x =8

33.

34.

5.2x + 3.7 = 2.4 5.2x = −1.3

8x = −11

3 x − 2 + 1 = 19

2 1− x − 4 = 2

3 x − 2 = 18

2 1− x = 6

x−2 =6

1− x = 3

x − 2 = 6 or x − 2 = −6

1 − x = 3 or 1 − x = −3

x = −4,8

35.

x = −2, 4

36. 5=7− 2−x

−1 = 3 − x − 3

−2 = − 2 − x

−4 = − x − 3

2 = 2−x

4 = x −3

2 − x = 2 or 2 − x = −2

x − 3 = 4 or x − 3 = −4

x = 0, 4

x = −1,7

176

Section 1.7

37. 2 p + 3 = 20 p + 3 = 10

p + 3 = −10

38. −3 p − 4 = −6

p − 4 = −2

p = −13

p−4 =2

p=2

p + 3 = 10

p−4 = 2

p=7

p=6

39.

40. 5 y − 2 − 10 = 4 y − 2 − 3

3 − y + 9 = 11 − 3 y + 9

y−2 =7

2 y+9 =8

y−2 =7

y − 2 = −7

y+9 = 4

y=9

y = −5

y+9 = 4

y + 9 = −4

y = −5

y = −13

41. 4 − x 2 = −1

4 − x =1

42. 7 − x 2 = −3

7 − x2 = 3

x2 = 5

x2 = 3

x 2 = 10

x2 = 4

x=± 5

x=± 3

x = ± 10

x = ±2

43. x 2 + 1 = −5

x +1 = 5

44. 4x2 − 9 = 0

x 2 = −6

x2 = 4

x2 =

4x 2 − 9 = 0

no solution

x = ±2

4x 2 = 9

9 4 3 x=± 2

45. 2 x 2 + 5 = 45

x = 20

46. x 2 − 1 = −5

x2 − 1 = 5

2 x 2 + 5 = 45

x = ±2 5

x 2 = −4

x2 = 6

x = ±2i

x=± 6

2 x 2 = 40

47. −7 < x < 7

48. −9 < y < 9

49. y ≤ −5 or y ≥ 5

50. x ≤ −2 or x ≥ 2

( −7,7 )

( −9,9 )

( −∞, −5] ∪ [5, ∞ )

( −∞, −2] ∪ [2, ∞ )

177

Chapter 1

51.

−7 < x + 3 < 7 −10 < x < 4

52.

−4 ≤ x + 2 ≤ 4 −6 ≤ x ≤ 2

[ −6,2]

( −10, 4 ) 53. x − 4 < −2 x<2

x−4 >2

54.

−3 < x − 1 < 3 −2 < x < 4

x>6 ( −∞,2 ) ∪ ( 6, ∞ )

55. −1 ≤ 4 − x ≤ 1 −5 ≤ − x ≤ −3 3≤x≤5

56. −3 < 1 − y < 3 −4 < − y < 2 4 > y > −2

[3,5]

( −2, 4 )

59.

( −2, 4 ) 57. ( −∞, ∞ )

58. no real solution

60. x − 7 > −2

x + 8 ≤ −7

all real numbers

no solution

62.

61. 2t + 3 < 5

3t − 5 > 1

−5 < 2t + 3 < 5 −8 < 2t < 2 −4 < t < 1

3t − 5 > 1 or 3t − 5 < −1 3t > 6 t>2

( −4,1)

63.

or 3t < 4 or t < 4 3

( −∞, 4 3 ) ∪ (2, ∞ )

64. 7 − 2y ≥ 3

6 − 5y ≤ 1

7 − 2 y ≥ 3 or 7 − 2 y ≤ −3

−2 y ≥ −4 or − 2 y ≤ −10 or y≤2 y≥5

−1 ≤ 6 − 5 y ≤ 1 −7 ≤ −5 y ≤ −5 7 ≥ y ≥1 5

( −∞,2] ∪ [5, ∞ )

[1, 7 5 ] 65. ( −∞, ∞ )

178

Section 1.7

67.

66. 4 − 3x ≤ −1 −3x ≤ −5 x≥5 3

4 − 3x ≥ 1

2 4x − 9 ≥ 3

or −3x ≥ −3 x ≤1

2 4x ≥ 12 4x ≥ 6

( −∞,1] ∪ [5 3, ∞ )

4 x ≥ 6 or 4 x ≤ −6 x ≥ 3 2 or x ≤ − 3 2 ( −∞, − 3 2 ] ∪ [ 3 2 , ∞ )

68.

69. 5 x −1 + 2 ≤ 7

2 x +1 − 3 ≤ 7

5 x −1 ≤ 5

2 x + 1 ≤ 10

x −1 ≤ 1

x +1 ≤ 5

−1 ≤ x − 1 ≤ 1 0≤x≤2

−5 ≤ x + 1 ≤ 5 −6 ≤ x ≤ 4

[0,2]

[ −6, 4]

70.

71. 3 x −1 − 5 > 4

3−2 x + 4 < 5

3 x −1 > 9

−2 x + 4 < 2

x −1 > 3

x + 4 > −1

x − 1 > 3 or x − 1 < −3

( −∞, ∞ )

x > 4 or x < −2

( −∞, −2 ) ∪ ( 4, ∞ ) 72.

73. 7 − 3 x + 2 ≥ −14

9 − 2x < 3

−3 x + 2 ≥ −21

− 2 x < −6

x+2 ≤7

2x > 6

−7 ≤ x + 2 ≤ 7 −9 ≤ x ≤ 5

2 x > 6 or 2x < −6 x > 3 or x < −3

[ −9,5]

( −∞, −3) ∪ (3, ∞ )

179

Chapter 1

74.

75. 4 − x +1 > 1

1 1 − < 1 − 2x < 2 2 3 1 − < −2 x < − 2 2 3 1 >x> 4 4

− x + 1 > −3 x +1 < 3 −3 < x + 1 < 3 −4 < x < 2

(1 4,3 4 )

( −4,2)

76.

77. 2 − 3x ≥ 2 2 − 3x ≤ −2

0 ≥ 3x

−1.8 < 2.6 x + 5.4 < 1.8 −7.2 < 2.6x < −3.6 −2.769 < x < −1.385

x≤0

( −2.769, −1.385 )

2 − 3x ≥ 2

4 ≤ 3x

4 3≤x

( −∞,0] ∪ [ 4 3, ∞ ) 78. 3.7 − 5.5x < −4.3 3.7 − 5.5x > 4.3 or −5.5x > 0.6 −5.5x < −8 x > 16 11

79. x2 −1 ≤ 8 x2 − 9 ≤ 0

x < − 556

( x − 3)( x + 3 ) ≤ 0

( −∞, − ) ∪ ( , ∞ ) 6 55

CP's: x = −3,3

16 11

[ −3,3]

−3 ≤ x ≤ 3

80. x 2 + 4 ≥ 29 x 2 − 25 ≥ 0

( x − 5 )( x + 5 ) ≥ 0

( −∞, −5] ∪ [5, ∞ )

CP's: − 5, 5

81. x − 2 < 7

82. x + 2 > 3

180

Section 1.7

83. x − 3 2 ≥ 1 2

84. x − 11 3 ≤ 5 3

85. x − a ≤ 2

86. x + 3 ≥ a

87. T − 83 ≤ 15

88. x − 97.8 ≤ 1.2

89. In order to win the hole, d < 4 . In order to have a tie, d = 4.

90. f − fC ≤ 15

91. (200 + 5x ) − (210 + 4.8x ) < 5 −10 + 0.2 x < 5 −5 < −10 + 0.2 x < 5 5 < 0.2 x < 15 25 < x < 75 So, when the number of units sold was between 25 and 75 units.

92. (200 + 5x ) − (210 + 4.8x ) < 3 −10 + 0.2 x < 3 −3 < −10 + 0.2 x < 3 7 < 0.2 x < 13 35 < x < 65 So, when the number of units sold is between 35 and 65 units.

93. The mistake was that x − 3 = −7 was not considered.

94. −7 < x − 3 < 7 is the appropriate inequality. The answer is (-4,10).

95. Did not reverse the inequality sign when divided by a negative.

96. Absolute value can never yield a negative number, so no solution.

97. True

98. True

99. False

100. x − 7 ≥ 0 x ≥ 7

181

Chapter 1

101.

−b < x − a < b a−b < x < a+b

102. a − x < −b a+b< x

( a − b, a + b ) 103. ( −∞, ∞ )

a−x >b

a−b > x

( −∞, a − b ) ∪ ( a + b, ∞ ) 104. No solution 106. No solution

105. x − a = −b

x−a = b

x = a−b

x =a+b

107. No solution

3x 2 − 7x + 2 > 8

108. 3x 2 − 7 x + 2 < − 8

3x 2 − 7x − 6 > 0

3x 2 − 7 x + 10 < 0

(3x + 2 )( x − 3) > 0

no solution

CP's: x = −2 3, 3

( −∞, −2 3 ) ∪ ( 3, ∞ ) 109. y1 = x − 7 y2 = x − 7

x≥7 Yes

182

Section 1.7

110. y1 = x + 1 y2 = x − 2 + 4 Do not coincide, yes.

111. y1 = 3x 2 − 7x + 2 y2 = 8

( −∞, − 23 ) ∪ (3, ∞ ) Yes

112.

y1 = 2.7x 2 − 7.9x + 5 y2 = 5.3x 2 − 9.2 x = −4.31, x = −0.38

( −∞, −4.31] ∪ [ −0.38, ∞ )

183

Chapter 1 113.

x x +1 y2 = 1 Find when y1 < y2 . y1 =

(− 12 , ∞)

114.

x x +1 y2 = 2 Find when y1 < y2 . y1 =

( −∞, −2) ∪ ( − 2 3 , ∞ )

184

Chapter 1 Review Solutions ----------------------------------------------------------------------1.

7x − 4 = 12

13d + 12 = 7d + 6

7x = 16

6d = −6

x = 16 7

d = −1

20 p + 14 = 6 − 5 p

4x − 28 − 4 = 4

25 p = −8

4x = 36

p = −8 25

x=9

3x + 21 − 2 = 4 x − 8

7c + 3c − 15 = 2c + 6 − 14 10c − 15 = 2c − 8 8c = 7

x = 27

c=7 8 7.

14 − [ −3 y + 12 + 9] = 8 y + 12 − 6 + 4

8. 6 − 4x + 2x − 14 − 52 = 6x − 12 + 6 [6x − 9 + 6 ]

14 + 3 y − 21 = 8 y + 10

−2x − 60 = 6x − 12 + 36 x − 18

−17 = 5 y

−2x − 60 = 42x − 30 −30 = 44x

y = −17 5

x = −2215 9.

10.

b≠0 12 − 3b = 6 + 4b 6 = 7b

3g + 9 g = 7 12 g = 7 g = 7 12

b=6 7

185

Chapter 1 11.

12.

LCD = 28

LCD = 6

4 (13x ) − 28x = 7 x − 2 ( 3 )

30b + b = 2b − 29

52 x − 28x = 7 x − 6

29b = −29

17 x = −6

b = −1

x = −6 17

13.

14. x ≠ −1,1

x≠0

LCD = ( x + 1)( x − 1)

LCD = x 1 − 4 x = 3 − 5x

4 ( x − 1) − 8 ( x + 1) = 3 ( x + 1)( x − 1)

−2 = −x 2=x

4 x − 4 − 8x − 8 = 3 ( x 2 − 1) −4 x − 12 = 3x 2 − 3

x=2

3x 2 + 4 x + 9 = 0 x= x=

−4 ± 4 2 − 4 ( 3 )( 9 ) 2 (3)

−4 ± −92 −4 ± 2i 23 = 6 6

2 23 x=− ± i 3 3

15.

16.

t ≠ −4,0 LCD = t ( t + 4 )

1 7 x≠− , 3 2 −2 ( 2 x − 7 ) = 3 ( 3x + 1)

2t − 7t − 28 = 6

−4 x + 14 = 9x + 3 13x = 11

2t − 7 ( t + 4 ) = 6 −5t = 34

x = 11 13

t = −34 5

186

Chapter 1 Review

17.

18.

x≠0

5 m ≠ − ,0 2 5 5 3− = 2+ m m 3m − 5 = 2 m + 5

LCD = 2 x 3 − 12 = 18x

−9 = 18x x=−

19.

1 2

m = 10

20.

7 x − 2 + 4 x = 3 [5 − 2 x ] + 12

LCD = 15

11x − 2 = 15 − 6 x + 12 17 x = 29

3x − ( x − 3 ) = −90

x = 29 17

2 x = −93

2 x + 3 = −90 x = −93 2

21.

22.

3x − 2 [3 y + 12 − 7 ] = y − 2 x + 6 x − 18

x+3 5x + 5 +2 = 1 + 2x 1 + 2x −5  x +3  1− 2y = 1− 2  =  1 + 2x  1 + 2x 5 ( x + 1) y+2 = 1 + 2 x = − ( x + 1) −5 1− 2y 1 + 2x y+2 =

3x − 6 y − 10 = y − 2 x + 6 x − 18

−x + 8 = 7 y x = 8 − 7y

187

Chapter 1 23.

Let x = total distance Drives:16 miles 3 Bus: x 4 1 Taxi: x 12 3 1 16 + x + x = x 4 12 LCD = 12 192 + 9x + x = 12 x 2 x = 192 x = 96 miles

24. B + L + D + 4S = 2000 D = 1.5B L = D − 100 = 1.5B − 100 1 1 S = L = (1.5B − 100 ) 4 4 B B B − 100  + 1.5    + 1.5 breakfast

lunch

dinner

1 + 4 ⋅ (1.5B − 100 ) = 2000 4  snacks

B + 1.5B − 100 + 1.5B + 1.5B − 100 = 2000 5.5B = 2200 B = 400 B = 400 L = 500 D = 600 S = 125

25.

x = number 1 1 12 + x = x 4 3 LCD = 12 144 + 3x = 4 x x = 144

26. x, x + 2, x + 4, x + 6 x + ( x + 2) + ( x + 4) + (x + 6) = 3 + 3( x + 6) 4 x + 12 = 3x + 21 x=9 9,11,13,15

188

Chapter 1 Review

27.

28.

P = 2l + 2w

x = Perimeter

l = 1 + 2w

1 1 10 + x + x = x 3 6 LCD = 18

P = 2 (1 + 2w ) + 2w P = 20

180 + 6 x + 3x = 18x

20 = 2 + 4w + 2w

9x = 180

6w = 18

x = 20 inches

w = 3 inches l = 7 inches

29. x = amount invested @ 20% 25000 − x = amount invested @ 8% Earned interest = 27600 − 25000 = 2600

30. $2500 in mutual funds $2500 in stock x = rate of mutual fund 4 x = rate of stock

0.2 x + 0.08 ( 25000 − x ) = 2600 0.2 x + 2000 − 0.08x = 2600

x ( 2500 ) + 4 x ( 2500 ) = 250

0.12 x = 600

2500 x + 10000 x = 250 12500 x = 250 x = 0.02

x = 5000 $5,000 @ 20%

Mutual Fund: 2% Stock: 8%

$20,000 @8%

31.

32.

x = ml of 5%

x = ounces of 8% 4 ounces of 20% Desired:12%

150 − x = ml of 10% 0.05x + 0.10 (150 − x ) = 0.08 (150 ) 0.05x + 15 − 0.10 x = 12

0.08x + 0.20 ( 4 ) = 0.12 ( x + 4 )

−0.05x = −3 x = 60 60 ml of 5%

Multiply by 100 8x + 80 = 12x + 48 32 = 4 x

90 ml of 10%

x = 8 8 ounces of 8%

189

Chapter 1 33.

34.

x = final exam grade 3x + 95 + 82 + 90 ≥ 90 6 3x + 267 ≥ 540

x = original price 0.80 x = 25000 x = 31250 $31,250

3x ≥ 273 At least 91

35.

36. b − 4b − 21 = 0

x 2 − 3x − 54 = 0

( b − 7 )( b + 3) = 0

( x − 9 )( x + 6 ) = 0

b = −3,7

x = −6,9

37.

38. x − 8x = 0 2

(3 y − 5 )( 2 y + 1) = 0

x ( x − 8) = 0

3y − 5 = 0 2 y + 1 = 0

x = 0,8

y = 5 3 or y = −1 2

39.

40. q = 169

c 2 = −36

q = ± 169

c = ± −36

q = ±13

c = ±6i

41.

42. d +7 = ± 4

2x − 4 = ± −64 2x − 4 = ±8i 2x = 4 ± 8i

d = −7 ± 2 d = −9, −5

x = 2 ± 4i

190

Chapter 1 Review

43.

44.

x − 4 x = 12

2 x 2 − 5x = 7

x 2 − 4 x + 4 = 12 + 4

5   2  x2 − x  = 7 2   5 25  25  2  x2 − x +  = 7 + 2 16  8 

( x − 2 ) = 16 2

x − 2 = ±4 x =2±4

5  81  2x −  = 4 8 

x = −2,6

5  81  x −  = 4  16  5 9 x− =± 4 4 5 9 x= ± 4 4 7 x = −1, 2

46.

45.

m2 − 8m = −15

x −x =8 1 1 x2 − x + = 8 + 4 4 2

m2 − 8m + 16 = −15 + 16

( m − 4) = 1 2

1  33  x −  = 2 4 

m − 4 = ±1 m = 4 ±1

1 33 x− =± 2 4 x=

m = 3,5

1 ± 33 2

191

Chapter 1 48.

47.

4 x 2 + 5x + 7 = 0 a = 4, b = 5, c = 7

3t − 4t − 7 = 0 a = 3, b = −4, c = −7 2

t= t=

− ( −4 ) ±

( −4 ) − 4 (3 )( −7 ) 2 (3)

4 ± 100 4 ± 10 = 6 6

t = −1,

−5 ± 52 − 4 ( 4 )( 7 )

7 3

2 ( 4)

−5 ± −87 8

−5 ± i 87 8

50.

49. 1 7 f − =0 3 6 LCD = 6

x 2 + 6x − 6 = 0 a = 1, b = 6, c = −6

48 f − 2 f − 7 = 0

8f 2 − 2

a = 48, b = −2, c = −7 f =

− ( −2 ) ±

f =

2 ± 1348 96

f =

2 ± 2 337 96

f =

1 ± 337 48

−6 ± 6 2 − 4 (1)( −6 ) 2 (1)

−6 ± 60 2 −6 ± 2 15 x= 2 x=

( −2 ) − 4 ( 48)( −7 ) 2 ( 48 ) 2

x = −3 ± 15

51.

52. a = 5, b = −3, c = −3 q= q=

− ( −3 ) ±

x − 7 = ± −12

( −3 ) − 4 (5 )( −3) 2 (5) 2

x = 7 ± 2i 3

3 ± 69 10

192

Chapter 1 Review

53.

54.

( 2x − 5 )( x + 1) = 0 x = −1,

g 2 + 3g − 3 = 0 a = 1, b = 3, c = −3

5 2

g= g=

55.

−3 ± 32 − 4 (1)( −3 ) 2 (1)

−3 ± 21 2

56. 7 x + 19x − 6 = 0

2b 2 + 1 = 7

( 7x − 2 )( x + 3) = 0

2b 2 = 6

x = −3,2 7

b2 = 3

b=± 3 57.

58. r2 =

S πh

r3 =

r=±

S πh

3V πh

r=3

3V πh

 negative radius is     non-physical 

59.

60.

vt = h + 16t v=

2π rh = A − 2π r 2

h + 16t 2 h = + 16t t t

193

A − 2π r 2 2π r

Chapter 1 62.

61.

−16t 2 + 500 = 0

1 A = bh 2 b = h +3 A = 2 1 2 = ( h + 3) h 2 4 = h 2 + 3h

16t 2 = 500 t2 =

500 16

t=±

h2 + 3h − 4 = 0

500 10 5 5 5 =± =± 16 4 2

Approximately 5.6 seconds

( h + 4 )( h − 1) = 0

h = −4,1 ( height must be positive ) h = 1 ft, b = 4 ft

63.

64. no solution

2x − 4 = 2 = 8 3

2 x = 12 x=6

65.

66. 2x − 7 = 3

x 2 = 7 x − 10

2 x = 7 + 243 = 250

x 2 − 7 x + 10 = 0

x = 125

( x − 2 )( x − 5 ) = 0

x = 2,5

68.

67.

( x − 4 ) = x + 5x + 6 2

2x − 7 = x + 3 x = 10

x 2 − 8x + 16 = x 2 + 5x + 6 13x = 10

 This answer would make   10  equal to a x=  the first  13    negative number  no solution

194

Chapter 1 Review

70.

69. x + 3 = 4 − 4 3x + 2 + 3x + 2

16 + 8 x − 3 + x − 3 = x − 5

−2x − 3 = −4 3x + 2

8 x − 3 = −18

2x + 3 = 4 3x + 2

no solution

( 2x + 3 ) = 16 ( 3x + 2 ) 2

4x 2 + 12 x + 9 = 48x + 32 4x 2 − 36 x − 23 = 0 x= x=

36 ± 36 2 − 4 ( 4 )( −23 ) 2 (4)

36 ± 1664 ≅ −0.6, 9.6 8

x ≅ −0.6 ( 9.6 doesn't check )

72.

71. x − 4 x + 4 = 49 − x 2

2x − 5 = 3 + x + 2

2 x − 4 x − 45 = 0 2

− ( −4 ) ±

2x − 5 = 9 + 6 x + 2 + x + 2

( −4 ) − 4 ( 2 )( −45 ) 2 (2)

x − 16 = 6 x + 2

( x − 16 ) = 36 ( x + 2 ) 2

4 ± 376 4 x ≅ −3.85 , 5.85

x 2 − 32 x + 256 = 36 x + 72

x ≅ 5.85

x 2 − 68x + 184 = 0 68 ± 682 − 4 (1)(184 )

2 68 ± 3888 x= 2 x ≅ 65.2

195

Chapter 1 73.

74. x = 3−x

15 + 2 x − 4 + x = 25

x + x −3 = 0

2 x − 4 + x = 10

2 2

−1 ± 1 − 4 (1)( −3 )

2 x − 4 = 10 − x

2 (1)

4 ( x − 4 ) = 100 − 20 x + x

−1 ± 13 ≅ −2.303, 1.3 2 x ≅ −2.303

4 x − 16 = 100 − 20 x + x

3x − 116 = −20 x 9x 2 − 696 x + 13456 = 400 x 9x 2 − 1096 x + 13456 = 0 x=

1096 ± 1096 2 − 4 ( 9 )(13456 )

18 1096 ± 846.6 ≅ 108,13.9 x≅ 18 x ≈ 13.9

75.

76.

(3x − 2 ) − 11( 3x − 2 ) + 28 = 0

u = x2

Let u = 3x − 2

u 2 − 6u + 9 = 0

u 2 − 11u + 28 = 0

( u − 3) = 0 2

( u − 4 )( u − 7 ) = 0

u =3

u = 4,7

x2 = 3

3x − 2 = 4

3x − 2 = 7

3x = 6  x = 2

3x = 9  x = 3

x=± 3

196

Chapter 1 Review

78.

77.

Let u = ( x − 4 )

x x ≠1 1− x u 2 + 2u − 15 = 0 u=

3u 2 − 11u − 20 = 0

(3u + 4 )( u − 5 ) = 0

( u + 5 )( u − 3) = 0

4 u = − ,5 3

u = −5,3 x 1− x −5 + 5 x = x

x 1− x 3 − 3x = x

4x = 5

4x = 3

−5 =

5 4

4 3 ( no solution )

(x − 4) = − 2

(x − 4) = 5 2

x−4 = ± 5 x = 4± 5

3 4

79.

80. −2

−1

y − 5y + 4 = 0

p −2 + 4 p −1 − 12 = 0

Let u = y −1

Let u = p −1

u 2 − 5u + 4 = 0

u 2 + 4u − 12 = 0

( u − 4)( u − 1) = 0

( u + 6)( u − 2) = 0

u = 4,1 So, we have:

u = −6,2 So, we have:

y −1 = 4  y = 1 4

p −1 = −6  p = − 1 6

y −1 = 1  y = 1

p −1 = 2  p = 1 2

197

Chapter 1 81.

82. 2x

Let u = x1 3

+ 3x − 5 = 0 13

Let u = x1 3

2u 2 − 3u − 5 = 0

2u 2 + 3u − 5 = 0

( 2u − 5 )( u + 1) = 0

( 2u + 5 )( u − 1) = 0

u = −1,

5 u = − ,1 2 x1 3 = −

 5 x = −   2 x=−

x1 3 = 1

5 2

x1 3 = −1

5 2 125 x= 8

x1 3 =

x =1

5 2

x = −1

125 8

83.

84. x

− 23

+ 3x

− 13

y− 2 − 2 y− 4 + 1 = 0

+2 = 0

Let u = x − 3 .

Let u = y − 4 .

u 2 + 3u + 2 = 0 ( u + 2)( u + 1) = 0 u = −2, −1

u 2 − 2u + 1 = 0

So, we have:

( u − 1)2 = 0 u =1 So, we have:

x − 3 = −2  x = ( −2)−3 = − 1 8

y− 4 = 1  y = 1

x − 3 = −1  x = ( −1)−3 = −1 1

198

Chapter 1 Review

85.

86.

Let u = x

Let u = x −1 2

u 2 + 5u − 36 = 0

u 2 − 4u + 3 = 0

( u + 9 )( u − 4 ) = 0

( u − 1)( u − 3) = 0

u = −9, 4

u = 1,3

−9 = x 2

4 = x2

1 = x −1 2

x = ±3i

x = ±2

x = 1−2 = 1

87.

3 = x −1 2 x = 3−2 =

1 9

88.

x + 4x − 32x = 0 3

9t3 − 25t = 0

x ( x 2 + 4x − 32 ) = 0

t ( 9t 2 − 25 ) = 0

x( x + 8)( x − 4) = 0

t(3t − 5)(3t + 5) = 0

x = 0, − 8, 4

t = 0, ± 5 3 90.

89.

4 x3 − 9x 2 + 4 x − 9 = 0

p − 3 p − 4 p + 12 = 0 3

( p − 3 p ) − 4( p − 3) = 0

( 4 x − 9x ) + ( 4 x − 9 ) = 0

p 2 ( p − 3) − 4( p − 3) = 0

x 2 ( 4x − 9 ) + ( 4x − 9 ) = 0

( p − 4 ) ( p − 3) = 0

( x + 1) ( 4x − 9 ) = 0

( p − 2)( p + 2)( p − 3) = 0

x = ±i , 9 4

p = ±2,3

91.

92. 2 ( t 2 − 9 ) − 20 ( t 2 − 9 ) = 0

p(2 p − 5) − 3(2 p − 5) = 0

(2 p − 5) [ p(2 p − 5) − 3] = 0

2 ( t 2 − 9 ) t 2 − 9 − 10  = 0  2

(2 p − 5) ( 2 p 2 − 5 p − 3 ) = 0

(

(2 p − 5)(2 p + 1)( p − 3) = 0

t 2 −19

)(

)

2(t − 3)(t + 3) t − 19 t + 19 = 0

p = − 12 , 5 2 , 3

t = ±3, ± 19

199

Chapter 1

94.

93. −1

9x 2 − 37x 2 + 4x − 2 = 0

y − 81y = 0 y−

x − 2 ( 9x 2 − 37 x + 4 ) = 0

81 =0 y

x − 2 (9x − 1)( x − 4) = 0 1

y − 81 =0 y ( y − 9)( y + 9) =0 y 2

x = 19 , 4

y = ±9

95. ( −∞, −4 ]

96. ( −1,7 ]

97. [ 2,6 ]

98. ( −1, ∞ )

99. x > −6

100. x ≤ 0

101. −3 ≤ x ≤ 7

102. −5 < x ≤ 2

103. x ≥ −4 [ −4, ∞ )

104. −4 ≤ x ≤ 4 [ −4, 4]

105.

107.

106.

 (⎯⎯→ ... 0 1 2 3 4 5 6 ...

[8,12]

( −∞,2 ]

 ←⎯⎯ ⎯] ... 0 1 2 3 4 ...

108. no solution 

 ]  [

... − 3 − 2 − 1 0 1 2 3 ...

... 7 8 9 10 11 12 ...

109.

( 4,∞ )

110.

3x < 5

6 x ≤ −2

x<5 3

x ≤ −1 3

( −∞,5 3)

( −∞, −1 3]

  ←⎯⎯⎯⎯ ⎯) 1 2 4 5 ... 0 3 3 1 3 3 ...

  ←⎯⎯⎯⎯ ⎯] 2 1 1 ... − 1 − 3 − 3 0 3 ...

200

Chapter 1 Review

111.

112.

x + 3 ≥ 18

2 x > −3

x ≥ 15

4<x≤9

x > −3 2

[15, ∞ )

( 4,9]

 (⎯⎯⎯⎯→

114.

 [⎯⎯⎯ →

 ( ]

... 13 14 15 16 ...

... 3 4 5 6 7 8 9 10...

− 2 − 32 − 1 − 21 0 ...

115.

−6 ≤ −4x − 7 ≤ 16 1 ≤ −4x ≤ 23 1 23 − ≥x≥− 4 4

8 ≤ 2 + 2x ≤ 9 6 ≤ 2x ≤ 7 3≤x≤7 2

[3,7 2]

 ]  [ 23

... − 6 − 4

LCD = 12 8 ≤ 2 (1 + x ) ≤ 9

 23 1   − 4 , − 4 

116.

6 < 2 + x ≤ 11

4x − 4 > 2x − 7

( −3 2, ∞ ) ...

113.

 [ ] 5 7 ... 2 3 2 4 ...

1 − 22 4 ... − 4 0 ...

117.

LCD = 18

72 + 65 + 69 + 70 + x ≥ 70 5 x + 276 ≥ 350

6 x + 2 ( x + 4 ) > 3x − 6 6 x + 2 x + 8 > 3x − 6

x ≥ 74

5x > −14 x > −14 5

So, the lowest score is 74.

( −14 5, ∞ )  (⎯⎯⎯⎯⎯⎯ → 13 14 12 ... − 3 − 5 − 5 − 5 ... 0 ...

201

Chapter 1

118.

119. Cost = 8500 + 50x

x 2 − 36 ≤ 0

Revenue = 300 x

( x − 6 )( x + 6 ) ≤ 0

Profit = 300 x − ( 8500 + 50 x ) > 0

CP's: x = −6,6

250 x > 8500 x > 34 So, greater than 34 suits.

[ −6,6] 121.

120. 6x − 7 x − 20 < 0

x 2 − 4x ≥ 0

( 3x + 4 )( 2x − 5 ) < 0

x ( x − 4) ≥ 0

4 5 CP's: x = − , 3 2

CP's: x = 0, x = 4

( −∞,0] ∪ [ 4, ∞ )

 4 5 − ,   3 2 122.

123. x + 9x + 14 ≤ 0

x 2 − 7x > 0

( x + 7 )( x + 2 ) ≤ 0

x (x − 7) > 0

CP's: − 7, −2

CP's: x = 0,7

[ −7, −2]

( −∞,0 ) ∪ ( 7, ∞ )

202

Chapter 1 Review

124.

125. x < −4

4 x 2 − 12 > 13x

no solution

4 x 2 − 13x − 12 > 0 (4x + 3)( x − 4) > 0

CPs: x = − 3 4 , 4 + − +  | | 4 − 34

( −∞, − 3 4 ) ∪ (4, ∞ )

127.

126. 3x ≤ x + 2 2

x <0 x≠3 x −3 CP's: x = 0,3

x 2 − 3x + 2 ≥ 0 ( x − 2)( x − 1) ≥ 0 CPs: x = 1,2 + − +  | | 1 2 ( −∞,1] ∪ [2, ∞ )

( 0,3) 129.

128. x −1 >0 x≠4 x−4 CP's:1,4

( −∞,1) ∪ ( 4, ∞ )

x 2 − 3x 18 ( 3 ) − ≥0 3 3 x 2 − 3x − 54 ≥0 3 ( x − 9 )( x + 6 ) ≥ 0 3 CP's: x = −6,9

( −∞, −6 ] ∪ [9, ∞ )

203

Chapter 1

130.

131.

( x − 7 )( x + 7 ) ≥ 0 x ≠ 7

3 1 − ≤0 x−2 x−4 3( x − 4) − ( x − 2) ≤0 ( x − 2)( x − 4) 2 x − 10 ≤0 ( x − 2)( x − 4) 2( x − 5) ≤0 ( x − 2)( x − 4) CPs: x = 2,4,5 − + − +   | | |

x −7 x+7 ≥ 0 x ≥ −7

[ −7,7 ) ∪ ( 7, ∞ )

( −∞,2) ∪ (4,5]

133. x2 + 9 ≥0 x −3 CP: 3 (since x 2 + 9 > 0, for all x ) − +   |

132. 4 2 ≤ x −1 x + 3 4 2 − ≤0 x −1 x + 3 4( x + 3) − 2( x − 1) ≤0 ( x − 1)( x + 3) 2( x + 7) ≤0 ( x − 1)( x + 3) CPs: x = −7, −3,1 − + − +   | | | −7

−3

(3, ∞ )

( −∞, −7] ∪ ( −3,1)

204

Chapter 1 Review

134.

135. 5x + 6 5x + 6  x− <0 x x x 2 − 5x − 6 <0 x ( x − 6)( x + 1) <0 x CPs: − 1,0,6 − + − +   | | |

x − 3 = −4

−1

no solution

136. 2 + x = −5 or 2 + x = 5 x = −7 or x = 3

( −∞, −1) ∪ (0,6)

137. 3x − 4 = −1.1

3x − 4 = 1.1

3x = 2.9

3x = 5.1

x ≈ 0.9667

x = 1.7

139.

138. x 2 − 6 = −3

x2 − 6 = 3

x2 = 3

x2 = 9

x=± 3

x = ±3

140. −4 < x < 4

( −4, 4 ) 141. x + 4 < −7 x < −11 ( −∞, −11) ∪ (3, ∞ )

−6 < x − 3 < 6 −3 < x < 9

(3,9 ) 142.

x+4>7 x >3

−4 ≤ −7 + y ≤ 4 3 ≤ y ≤ 11

[3,11]

205

Chapter 1

143. 2 x > 6

2x > 6

2 x < −6

x >3 ( −∞, −3) ∪ (3, ∞ )

x < −3

144. 4 + 2x 1 ≤− 3 7 LCD = 21 7 ( 4 + 2x ) ≤ −3 28 + 14x ≤ −3 14 x ≤ −31 x ≤ −31 14

4 + 2x 1 ≥ 3 7 7 ( 4 + 2x ) ≥ 3 28 + 14 x ≥ 3 14 x ≥ −25 x ≥ −25 14

( −∞, −31 14] ∪ [ −25 14, ∞ ) 145. ( −∞, ∞ )

146.

−4 ≤ 1 − 2x ≤ 4 −5 ≤ −2x ≤ 3 5 3 ≥x≥− 2 2

 3 5  − 2 , 2  147. T − 85 ≤ 10 or 75 ≤ T ≤ 95 149.

148. B − 0.08 ≤ 0.007

y1 = 0.031x + 0.017(4000 − x ) y2 = 103.14 x = 2,510

206

Chapter 1 Review

150. 1 0.2 − 0.16 x x 1 y2 = 4 x = 24.2 y1 =

151. (a) Consider x 2 + 4x − b = 0 . (1) For b = 5 , (1) factors as ( x − 1)( x + 5) = 0 , so that x = −5,1 .

Graphically, we let y1 = x 2 + 4 x, y2 = 5 and look for the intersection points of the graphs:

Note that they intersect at precisely the x-values obtained algebraically. So, yes, these values agree with the points of intersections.

(b) We do the same thing now for different values of b.

b = −5 :

b =0: x 2 + 4x = 0

x 2 + 4x + 5 = 0 x=

x( x + 4) = 0

−4 ± 16 − 4(5) −4 ± 2i = = −2 ± i 2 2

x = 0, −4

207

Chapter 1 So, we don’t expect the graphs to intersect. Indeed, we have:

So, we expect the graphs to intersect twice as in part (a). Indeed, we have:

b =7:

b = 12 : x 2 + 4 x − 12 = 0

x + 4x − 7 = 0 2

−4 ± 16 + 4(7) −4 ± 2 11 = = −2 ± 11 2 2

So, we expect the graphs to intersect twice as in part (a). Indeed, we have:

208

( x + 6)( x − 2) = 0 x = −6,2 So, we expect the graphs to intersect twice as in part (a). Indeed, we have:

Chapter 1 Review

152. (a) Consider x 2 − 4x − b = 0 . (1) For b = 5 , (1) factors as ( x + 1)( x − 5) = 0 , so that x = −1,5 .

Graphically, we let y1 = x 2 − 4 x, y2 = 5 and look for the intersection points of the graphs:

Note that they intersect at precisely the x-values obtained algebraically. So, yes, these values agree with the points of intersections.

(b) We do the same thing now for different values of b.

b = −5 :

b =0: x 2 − 4x = 0

x − 4x + 5 = 0 2

x( x − 4) = 0

4 ± 16 − 4(5) 4 ± 2i = 2±i = 2 2 So, we don’t expect the graphs to intersect. Indeed, we have: x=

x = 0, 4 So, we expect the graphs to intersect twice as in part (a). Indeed, we have:

209

Chapter 1 b =7:

b = 12 :

x − 4x − 7 = 0

x 2 − 4 x − 12 = 0

4 ± 16 + 4(7) 4 ± 2 11 = 2 ± 11 = 2 2 So, we expect the graphs to intersect twice as in part (a). Indeed, we have:

( x − 6)( x + 2) = 0

Graphically, let 1 1 y1 = 2x 4 , y2 = −x 2 + 6

153. 14

= −x

x = 6, −2 So, we expect the graphs to intersect twice as in part (a). Indeed, we have:

x1 2 + 2x1 4 − 6 = 0 Let u = x1 4 to obtain u 2 + 2u − 6 = 0 −2 ± 4 + 4(6) = −1 ± 7 2 x1 4 = −1 + 7 14 x = −1 − 7 4 x = −1 + 7 ≈ 7.34 no solution u=

(

)

210

Chapter 1 Review

Graphically, let 1 1 y1 = 2x − 2 , y2 = x − 4 + 6

154. 2 x −1 2 = x −1 4 + 6 2 x −1 2 − x −1 4 − 6 = 0 Let u = x −1 4 to obtain 2u 2 − u − 6 = 0 u=

−1 ± 1 + 4(2)(6) 1 ± 7 = = 2, − 32 2(2) 4 x −1 4 = 2

x −1 4 = − 32

x = 161

no solution

b) Graphically, let y1 = −0.61x + 7.62, y2 = 0.24x − 5.47

155. a) −0.61x + 7.62 > 0.24 x − 5.47 13.09 > 0.85x 15.4 > x ( −∞,15.4)

c) Agree

Graphically, let y1 = −0.5x + 7, y2 = 0.75x − 5

156. a) 1 3 − x +7 < x −5 2 4 −2 x + 28 < 3x − 20 48 < 5x 9.6 < x (9.6, ∞ )

c) Agree

211

Chapter 1

157. y1 = 0.2 x 2 − 2 y2 = 0.05x + 3.25 Find when y1 > y2 ( −∞, −5) ∪ (5.25, ∞ )

158. y1 = 12 x 2 − 7 x − 10 y2 = 2 x 2 + 2 x − 1 Find when y1 < y2

( − 35 , 32 )

159. 3p 7 −2p y2 = 1 y1 =

Find when y1 > y2

( 75 , 72 )

212

Chapter 1 Review

160. 7p 15 − 2 p y2 = 1 y1 =

Find when y1 < y2

( −∞, 53 ) ∪ ( 152 , ∞ )

161. y1 = 1.6 x 2 − 4.5 y2 = 3.2 Find when y1 < y2

( −2.19, −0.9 ) ∪ ( 0.9,2.19 )

162. y1 = 0.8x 2 − 5.4 x y2 = 4.5 Find when y1 > y2

( −∞, −0.75 ) ∪ ( 0.97,5.78 ) ∪ ( 7.5, ∞ )

213

Chapter 1 Practice Test Solutions ---------------------------------------------------------------1.

4 p − 7 = 6 p −1

−2 z + 2 + 3 = −3 z + 3 z − 3

−6 = 2 p

−2 z = −8

−3 = p

z=4

4. 3t = t − 28

8x 2 − 13x = 6

t 2 − 3t − 28 = 0 (t − 7)(t + 4) = 0

8x 2 − 13x − 6 = 0

t = −4,7

x = − 3 8 ,2

(8x + 3)( x − 2) = 0

6. 6 x − 13x − 8 = 0 2

3 5 − =0 x −1 x + 2 3( x + 2) − 5( x − 1) =0 ( x − 1)( x + 2) −2 x + 11 =0 ( x − 1)( x + 2)

(3x − 8)( 2x + 1) = 0 1 8 x=− , 2 3

x = 11 2

8. x 4 − 5x 2 − 36 = 0

5 30 +1− 2 =0 y −3 y −9

( x + 4 )( x − 9 ) = 0 2

5( y + 3) + ( y 2 − 9) − 30 =0 ( y − 3)( y + 3)

( x − 2i )( x + 2i )( x − 3)( x + 3) = 0 x = ±3, ± 2i

y 2 + 5 y − 24 =0 ( y − 3)( y + 3) ( y + 8) ( y − 3) ( y − 3) ( y + 3)

y = −8

214

Chapter 1 Practice Test

10. 2x + 1 + x = 7

Let u = x1 3

2x + 1 = 7 − x

2u 2 + 3u − 2 = 0

2 x + 1 = (7 − x )2

( 2u − 1)( u + 2 ) = 0

2 x + 1 = 49 − 14 x + x 2

u = −2,1 2

x 2 − 16 x + 48 = 0

So, x = −8, 18

( x − 12)( x − 4) = 0 x = 4 , 12

11.

12. x(3x − 5)3 − 2(3x − 5)2 = 0

3y − 2 = 9 − 6 3y + 1 + 3y + 1

(3x − 5)2 [ x(3x − 5) − 2 ] = 0

−12 = −6 3 y + 1 3y + 1 = 2

(3x − 5)2 ( 3x 2 − 5x − 2 ) = 0

3y + 1 = 4

(3x − 5)2 (3x + 1)( x − 2) = 0

3y = 3

x = 5 3 , − 13 , 2

y =1

14.

13.

F = 95 C + 32

x − 8x + 12 x = 0 7

x 3 ( x 2 − 8x + 12 ) = 0

F − 32 = 59 C

C = 59 ( F − 32 )

x 3 ( x − 6)( x − 2) = 0 1

x = 0,2,6

15.

P = 2 L + 2W P − 2W = 2 L L=

16. 7 − 5x > −18

−5x > −25 x<5

P − 2W 2

( −∞,5 )

215

Chapter 1

17.

18. 3x + 19 ≥ 5x − 15 17 ≥ x

−1 ≤ 3x + 5 < 26 −6 ≤ 3x < 21 −2 ≤ x < 7

( −∞,17]

[ −2,7)

34 ≥ 2x

19.

20. 2 x +8 1 < ≤ 5 4 2 8 < 5( x + 8) ≤ 10 −32 < 5x ≤ −30 −

3x ≥ 2 x 2 0 ≥ 2 x 2 − 3x 0 ≥ x(2 x − 3) CPs: x = 0, 3 2 + − +  | | 3 0 2

32 < x ≤ −6 5

 32   − , −6   5 

[ 0, 3 2 ] 22.

21. 3p − p − 4 ≥ 0

5 − 2x > 1

(3 p − 4 )( p + 1) ≥ 0

5 − 2 x > 1 or 5 − 2 x < −1

CP's: p = 34 , −1

−2 x > −4 or − 2 x < −6 x<2 or x > 3

( −∞, −1] ∪  34 , ∞ )

( −∞,2) ∪ (3, ∞ )

23.

24. x −3 ≤0 2x + 1 CPs: x = − 1 2 ,3 + − +  | | − 12 3

x+4 ≥0 x2 − 9 x+4 ≥0 ( x − 3)( x + 3) CPs: x = −4, ± 3 − + − +  | | | 3 −4 −3

( − 1 2 ,3]

[ −4, −3) ∪ (3, ∞ ) 216

Chapter 1 Practice Test

25. Let x = height of piling 1 Sand: x 4 Water:150 3 Air: x 5 1 3 x + 150 + x = x 4 5 LCD = 20 5x + 3000 + 12 x = 20 x 3x = 3000

26. Let x = purchase price 150,000 − x ≤ 15,000  10% of list price

135,000 ≤ Purchase Price ≤ 165,000 0.07 (135,000 ) ≤ commission ≤ 0.07 (165,000 ) $9, 450 ≤ commission ≤ $11,550

x = 1000 feet

27. Let x = number of ounces over 6 ounces. Total = 4 + 0.35x 5.75 ≤ 4 + 0.35x ≤ 7.15 1.75 ≤ 0.35x ≤ 3.15 5≤x≤9 +6 oz base 11 ≤ x ≤ 15

28. 9 16 Let x = height of the Instagram story. 9 2.8 = 16 x 9x = 44.8 x = 4.98 inches Story ratio:

Story: 2.8 in. × 4.98 in. 4 Post ratio: 5 Let x = height of the Instagram post. 4 2.8 = 5 x 4 x = 14 x = 3.5 inches Post: 2.8 in. × 3.5 in.

217

Chapter 1

29. 1 0.45 − x 0.75x 1 y2 = 9 x = 7.95 y1 =

30. y1 = 0.3 + 2.4 x 2 − 1.5 y2 = 6.3 Find when y1 ≤ y2 [ −1.768,1.768]

218

Chapter 1 Cumulative Review--------------------------------------------------------------------1.

5 ⋅ ( 7 − 3 ⋅ 4 + 2 ) = 5 ⋅ ( 7 − 12 + 2 )

( 4x b ) = 4 x b −3 4 −3

= 5 ⋅ ( −5 + 2 )

−3

9 −12

x9 = 64b12

= 5 ⋅ ( −3) = −15

(x y ) = x y = x (x y) x y y 2

−2 3

−6

−3

−6

−3

4. ( − x 4 + 2 x 3 ) + ( x 3 − 5x − 6 )

− ( 5x 4 + 4 x 3 − 6 x + 8 ) =

− x 4 + 2 x 3 + x 3 − 5x − 6 − 5x 4 − 4 x 3 + 6 x − 8 = −6 x 4 − x3 + x − 14

6. x ( x + 5)( x − 3) = x ( x + 2 x − 15)

3x3 − 3x 2 − 60x = 3x ( x 2 − x − 20 )

= x 4 + 2x3 − 15x 2

= 3x( x − 5)( x + 4)

  2a 3 + 2000 = 2  a 3 + 1000  =103  

− (x − 3 ) ( x + 1) 3 − x 5x − 15 ÷ = ⋅ x2 − 1 x + 1 ( x − 1) ( x + 1) 5 ( x − 3)

= 2( a + 10) ( a 2 − 10a + 100 )

= −

1 5( x − 1)

where x ≠ −1,1,3

10.

6x 5x 6 x( x + 2) − 5x( x − 2) − = x−2 x+2 x2 − 4 6 x 2 + 12 x − 5x 2 + 10 x = x2 − 4

x3 − x 2 − 30x = 0

x ( x 2 − x − 30 ) = 0 x( x − 6)( x + 5) = 0 x = −5,0,6

x 2 + 22 x = x2 − 4 where x ≠ −2,2

219

Chapter 1

12.

11. 2 7

45 6 + 3i 270 + 135i ⋅ = = 6 + 3i 6 − 3i 6 + 3i 45

x = x+9 1 8

16 x = 7 x + 504 9x = 504

x = 56

13.

14. 6 x 8x 7x − = 4− 5 3 15 18x − 40x = 60 − 7x −22x = 60 − 7x −15x = 60 x = −4

15. Tim rate: 1/9 job in one hour Chelsea and Tim combined rate: 1/5 job in one hour Let x = number of hours it takes Chelsea to complete job by herself Solve: 1 1 1 9 + x = 5

x −6 3 = , x≠6 6−x 2 x −6 3 − = x −6 2 3 −1 = 2 Since this statement is false, the equation has no solution.

16. y 2 + 36 = 0 y 2 = −36 y = ± −36 = ±6i

5x + 45 = 9x 45 = 4 x  11.25 = x It takes Chelsea 11.25 hours by herself.

220

Chapter 1 Cumulative Review

17.

18. x2 + x + 9 = 0

x + 12 x + 40 = 0 2

( x + 12x + 36 ) + 40 − 36 = 0

−1 ± 1 − 4(9) 2 −1 ± −35 = 2

( x + 6)2 + 4 = 0 ( x + 6)2 = −4

x + 6 = ± −4 = ±2i

x = −6 ± 2i

−1 ± i 35 2

20.

19. 4−x = x−4

3x −2 + 8x −1 + 4 = 0

4 − x = ( x − 4)2

Let u = x −1.

4 − x = x 2 − 8x + 16

3u 2 + 8u + 4 = 0

x 2 − 7 x + 12 = 0

(3u + 2)( u + 2) = 0

( x − 4)( x − 3) = 0

u = − 23 , −2 x −1 = − 23  x = − 32

x = 3, 4

x −1 = −2  x = − 12

21.

22. 0< 4−x ≤7 −4 < −x ≤ 3 4 > x ≥ −3 [ −3, 4)

4 x 2 < 9x − 11 4 x 2 − 9x + 11 < 0 9 ± 81 − 4(4)(11) 9 ± i 95 = 2(4) 8 No real solution x=

+ − + −  | | | 3 −3 −2

23. x+2 ≥0 9 − x2 x+2 ≥0 (3 − x )(3 + x ) CPs: x = −2, ±3

( −∞, −3) ∪ [ −2,3)

221

Chapter 1

24.

25. 4 − 5x 3 ≥ 7 14

1 2 7 x+ = 5 3 15

4 − 5x

3x + 10

3 7 14 2 4 − 5x ≥ 3

≥

7 15 15 3x + 10 = 7

2(4 − 5x ) ≥ 3 or 2(4 − 5x ) ≤ −3

3x + 10 = 7

8 − 10 x ≥ 3

8 − 10 x ≤ −3

3x = −3

3x = −17

−10 x ≥ −5

− 10 x ≤ −11

x = −1

x = − 173

x ≤ 12

11 x ≥ 10

3x + 10 = −7

11 ( −∞, 12 ] ∪ [ 10 , ∞)

26. 37 3 x − 27 = 0 8 8x 6 + 37 x3 − 216 = 0

x6 +

Graphically, let 37 y1 = x6 + x3 , y2 = 27 . 8

Let u = x3 . 8u 2 + 37u − 216 = 0

−37 ± 37 2 + 4(8)(216) 27 = −8, 2(8) 8

x3 = −8  x = −2 x3 =

27 3  x= 8 2

222

Chapter 1 Cumulative Review

27. 3x x−2 y2 = 1 Find when y1 < y2 y1 =

( −1, 12 )

223

CHAPTER 2 Section 2.1 Solutions -------------------------------------------------------------------------------1. (4,2)

2. ( −2,3)

3. ( −3,0)

4. ( −4, −2)

5. (0, −3)

6. (3, −1)

A is in Quadrant II. B is in Quadrant I. C is in Quadrant III. D is in Quadrant IV. E is on the y-axis. F is on the x-axis.

10.

The line being described is y = 2.

The line being described is x = −3.

224

Section 2.1

11.

12.

d = (1 − 5)2 + (3 − 3)2 = 16 = 4

d = ( −2 − ( −2 ))2 + (4 − ( −4 ))2 = 64 = 8

 1+ 5 3 + 3  M = ,  = (3,3) 2   2

 −2 + (−2) 4 + (−4)  M = ,  = ( −2,0) 2 2  

13.

14.

d = ( −1− 3) + (4 − 0) = 32 = 4 2

d = ( −3 −1)2 + ( −1 − 3)2 = 32 = 4 2

 −1 + 3 4 + 0  M = ,  = (1,2) 2   2

 −3 + 1 −1 + 3  M = ,  = ( −1,1) 2   2

15.

16. d = ( −10 − (−7)) + (8 − (−1)) 2

d = ( −2 − 7)2 + (12 −15)2

= 90 = 3 10

 −10 + (−7) 8 + (−1)  M = ,  2 2  

 −2 + 7 12 +15   5 27  M = , =  ,  2  2 2   2

 −17 7  =  ,   2 2

17.

18.

d = ( −3 − ( −7 )) + ( −1 − 2) 2

d = ( −4 − ( −9 ))2 + (5 − ( −7 ))2 = 169

= 25 = 5

= 13

1  −3 + (−7) −1+ 2   M = ,  =  −5,  2 2   2 

 −4 + (−9) 5 + (−7)   13  M = ,  =  − , − 1 2 2    2 

19.

20.

d = (0 − ( −4 ))2 + ( −7 − ( −5))2

d = ( −6 − ( −2 ))2 + ( −4 − ( −8))2

= 20 = 2 5

= 32 = 4 2  −6 + (−2) −4 + ( −8)  M = ,  2 2  

 0 + ( −4) −7 + ( −5)  M = ,  2 2  

= ( −4, − 6 )

= ( −2, − 6 )

225

Chapter 2

21.

22. 2

 1 7   1 10  d = − −  + −   2 2 3 3 

 1 9   7  2  d =  −  + −  −   5 5   3  3 

= 25 = 5

 1 7 1 10   − 2 + 2 3 + 3   3 11  M = , =  ,  2  2 6   2  

289 17 = 25 5

1 9 7 2 −   +  5 M =  5 5 , 3 3  =  1,  2   6  2  

23. 2

 2 1  1 1 d = − −  +− −  =  3 4  5 3 =

112 ⋅152 + 82 ⋅12 2 = 12 2 ⋅152

 11   8  −  +−   12   15 

4049 60

1 1  2 1 −3 + 4 −5 + 3   5 1  M = ,  = − ,  2 2   24 15    

24. 2

92 ⋅ 92 + 22 2 ⋅10 2 = 10 2 ⋅92

54,961 90

 7 1   1  7  d =  −  + −  −  =  5 2   9  3  =

 9   22    +   10   9 

 7 1 1  7  +−   + 9  3    19 10  5 2  M = , =  ,−  2 9  2   20    

226

Section 2.1

25.

26.

d = ( −1.5 − 2.1)2 + (3.2 − 4.7)2

d = ( −1.2 − 3.7)2 + ( −2.5 − 4.6)2

= ( −3.6)2 + (−1.5)2 = 3.9

= ( −4.9)2 + (−7.1)2 = 74.42

 −1.5 + 2.1 3.2 + 4.7  M = ,  2 2  

≅ 8.627  −1.2 + 3.7 −2.5 + 4.6  M= ,  2 2  

= (0.3, 3.95)

= (1.25, 1.05) 27.

28.

d = ( −14.2 −16.3) + (15.1− (−17.5)) 2

d = (1.1 − 3.3)2 + (2.2 − 4.4)2

= ( −30.5)2 + (32.6)2 = 1993.01

= 2(2.2)2 = 9.68 ≅ 3.111  1.1 + 3.3 2.2 + 4.4  M= ,  2 2  

≅ 44.64  −14.2 + 16.3 15.1+ (−17.5)  M= ,  2 2  

= (2.2, 3.3)

= (1.05, − 1.2)

29.

30.

( 3 − 3 ) + (5 2 − 2 ) = (0) + (4 2 ) = 4 2 2

 3+ 3 5 2+ 2 M =  ,  2 2   2 3 6 2  , =   = 2 2  

( 3,3 2 )

(

)) ( ( )) ( 4 5 ) + ( −2 3 ) = 92 = 2 23 (

3 5− − 5

+ −3 3 − − 3

 3 5 − 5 −3 3 − 3  M =  ,  2 2    2 5 −4 3  =  ,  = 5, −2 3 2   2

(

227

)

Chapter 2

31.

32. d= =

( ( )) (

3 − (−2)

(

)

1− − 2 1+ 2

) ( 2

3+2

)

(

)

 3 + ( −2)   2 

) (

)

( ) (

= 20 − 4 5 + 1 + 16 − 2 8 3 + 2 3 =

= 10 + 2 2 + 4 3  1+ − 2 M = ,  2 

(

2 5 −1 + 4 − 2 3

)

49 − 4 5 − 16 3

 2 5 +1 4 + 2 3  M =  ,  2 2    1+ 2 5  , 2 + 3  =   2 

 1 − 2 −2 + 3  =  ,  2  2 

33. d ( A, B) = ( −2 − 2)2 + (−2 − 5)2 = 65 d ( B,C ) = (2 − 5)2 + (5 − ( −1))2 = 45 d ( A,C ) = ( −2 − 5)2 + (−2 − (−1))2 = 50

The perimeter of the triangle rounded to two decimal places is 21.84 .

34. d ( D, E ) = (0 − 2)2 + (−3 − 3)2 = 40 d ( E ,C ) = (2 − 5)2 + (3 − (−1))2 = 5 d ( D,C ) = (0 − 5)2 + (−3 − (−1))2 = 29

The perimeter of the triangle rounded to two decimal places is 16.71.

228

Section 2.1

35. d ( A,D) = ( −2 − 0)2 + (−2 + 3)2 = 5 d ( D, E ) = (0 − 2)2 + (−3 − 3)2 = 40 d ( A, E ) = (2 − (−2))2 + (3 − (−2))2 = 41

The perimeter of the triangle rounded to two decimal places is 14.96.

36.

d ( B,C ) = ( 2 − 5)2 + (5 − (−1))2 = 45

d (C ,D) = (5 − 2 )2 + (−1− 3)2 = 5 d ( B, E ) = ( 2 − 2 )2 + (5 − 3)2 = 2 The perimeter of the triangle rounded to two decimal places is 13.71.

37. a = (3 − 3)2 + (5 − (−3))2 = 8 b = (0 − 3)2 + ( −3 − ( −3))2 = 3 c = (0 − 3)2 + (−3 − 5)2 = 82 + 32

Observe that c = a 2 + b 2 above, so it is a right triangle. It is not isosceles since all sides have different length.

229

Chapter 2

38. a = ( −2 − 0)2 + ( −2 − 2)2 = 20 b = (2 − 0)2 + ( −2 − 2)2 = 20 c = ( −2 − 2)2 + ( −2 − ( −2 ))2 = 4

Note that a 2 + b 2 ≠ c 2 , b 2 + c 2 ≠ a 2 , and a 2 + c 2 ≠ b 2 , so this is not a right triangle. It is isosceles since at least two sides have the same length.

39. a = (3 − ( −2 ))2 + ( −1 − ( −4 ))2 = 34 b = (1 − 3)2 + (1 − ( −1))2 = 8 c = (1 − ( −2))2 + (1 − ( −4))2 = 34

Note that a 2 + b 2 ≠ c 2 , b 2 + c 2 ≠ a 2 , and a 2 + c 2 ≠ b 2 , so this is not a right triangle. It is isosceles since at least two sides have the same length. 40. a = ( −3 − ( −3))2 + (3 − ( −3))2 = 6 b = ( −3 − 3)2 + (3 − 3)2 = 6 c = ( −3 − 3)2 + ( −3 − 3)2 = 6 2

Observe that c = a 2 + b 2 above, so it is a right triangle. Also, it is isosceles since at least two sides have the same length.

230

Section 2.1

42. Since d(Tampa, Orlando) = 80, a town midway between is 40 miles from Tampa. Also, since d (Tampa, Gainesville) = 100, a town midway between is 50 miles from Tampa. The third tower would be 64.03 miles from Tampa.

41.

The distance from Gainesville to Orlando is: (0 − 80)2 + (100 − 0)2 = 16, 400

43. Assume that Columbia is located at (0,0). Then, Atlanta is located at (−215,0) and Savannah is located at (0, −160) . So, d(Atlanta, Savannah) is:

(−215 − 0)2 + (0 − (−160))2 ≅ 268 miles

= 20 41 ≅ 128.06 miles 44.

45.

( −7 − 15 ) + ( −20 − 50 ) 2

 −9 + 6 3 + ( −4 )   3 1 ,  = − , −  2 2   2  2

( xm , ym ) = 

d = 5384 ≈ 73.4 yards 46.

−3 +1 2 + 4  ,  = ( −1, 3 ) 2   2

( xm , ym ) =  47.

231

Chapter 2

48. $3.00 $2.75 $2.50 $2.25 $2.00 $1.75 $1.50 0

49. Substituted the values incorrectly. It should have been

(9 − 2)2 + (10 − 7)2 .

So, d = 58. 50. The value of x1 is incorrect. The quantity (3 − ( −2))2 should be (3 + 2)2 . 51. Substituted the values incorrectly. It should have been  −3 + 7 4 + 9  13 ,   = ( 2, 2 ) . 2   2 52. The midpoint formula is incorrect. x +x y +y  It should be  1 2 , 1 2  . So, 2   2 M = ( −2, −3).

53. True. Follows directly from the distance formula. 54. True. Follows directly from the midpoint formula.

55. True. If ( x1 , y1 ) and ( x2 , y2 ) are in Quadrant I, then x1 , y1 , x2 , y2 are all positive. x +x y +y So, 1 2 and 1 2 are both positive, thereby placing the midpoint 2 2  x1 + x2 y1 + y2  ,  in Quadrant I.  2   2

232

Section 2.1

56. False. Observe that the midpoint of the segment joining (3,0) and (0, −3) is (0,0), which does not lie in any of the four quadrants.

57.

d = ( a − b )2 + ( b − a )2 = 2(a − b)2 =

2 a−b

a+b b+a M = ,  2   2 58. d = ( a − ( −a))2 + (b − ( −b))2

4(a 2 + b 2 ) = 2 a 2 + b 2

 a + ( −a ) b + ( −b)  M = ,  = ( 0,0) 2 2  

59. Since the midpoint is given by x +x y +y  M =  1 2 , 1 2  , the distance 2   2 from ( x1 , y1 ) to M is as follows: 2

x +x   y +y   d =  x1 − 1 2  +  y1 − 1 2  2   2   2

x +x   y +y   d =  x2 − 1 2  +  y2 − 1 2  2   2  

 2x − x − x   2 y − y − y  =  1 1 2  + 1 1 2  2 2     2

Similarly, the distance from ( x2 , y2 ) to M is as follows:

 2x − x − x   2 y − y − y  =  2 1 2  + 2 1 2  2 2     2 2

 x −x   y − y  =  2 1  + 2 1   2   2  1 2 2 = ( x1 − x2 ) + ( y1 − y2 ) 2

x −x   y −y  =  1 2  + 1 2   2   2  1 2 2 = ( x1 − x2 ) + ( y1 − y2 ) 2

60. Let: D1 be the diagonal connecting (0,0) to ( a + b, c) D2 be the diagonal connecting ( a, c) to ( b, 0)

a+b c  ,  , which coincides with the midpoint of D2 . The midpoint of D1 is   2 2 Hence, D1 and D2 intersect at their midpoints.

233

Chapter 2

61. Let P1 = ( a,b), P2 = (c, d ), d1 = distance from P1 to P2 , and d 2 = distance from P2 to P1. Observe that

d1 = ( a − c )2 + (b − d )2

d2 = (c − a )2 + (d − b)2 .

Since (c − a )2 = ( −( a − c ) ) = ( a − c )2 2

( d − b )2 = ( −( b − d ) ) = ( b − d )2 2

we see that d1 = d 2 , so that it does not matter what point is labeled “first” in the distance formula. 62. Let d1 = distance between ( −1, −1) and (0,0),

d2 = distance between (0,0) and (2,2)

d3 = distance between (2,2) and ( −1, −1) Using the distance formula yields d1 =

( −1 − 0 ) + ( −1 − 0 ) = 2

d2 =

(0 − 2) + (0 − 2) = 2 2

d3 =

( −1 − 2 ) + ( −1 − 2 ) = 3 2

Observe that d1 + d 2 = d3 . 63.

64.

x + x2 y1 + y2 z1 + z2  , ,  2  2  2  −1 + 1 2 + ( −4 ) 7 + ( −3 )  , ,   2 2  2 

( xm , ym , zm ) =  1

( 0, −1, 2 )

( x2 − x1 ) + ( y2 − y1 ) + ( z2 − z1 ) 2

d = 1 − ( −1)  + ( −4 ) − 2  + ( −3 ) − 7  2

( 2 ) + ( −6 ) + ( −10 ) 2

d = 4 + 36 + 100 = 140 = 2 35

234

Section 2.1

65.

66.

( 3.7 − (−2.3) ) + ( 6.2 − 4.1) 2

≅ 6.357

(5.2 − (−4.9) ) + (3.4 − (−3.2) ) 2

≅ 12.065

 3.7 − 2.3 6.2 + 4.1  , M =  2 2  

 5.2 − 4.9 3.4 − 3.2  M = ,  2 2  

= ( 0.7,5.15 )

= ( 0.15,0.1)

67.

68.

( 3.3 − 1.1) + ( 4.4 − 2.2 ) 2

≅ 12.241

= 2(2.2)2 ≅ 3.111

 3.3 + 1.1 4.4 + 2.2  , M=  = ( 2.2,3.3 ) 2 2  

( −1.3 − 2.3) + ( 7.2 − ( −4.5) )

 −1.3 + 2.3 7.2 − 4.5  M = ,  = ( 0.5,1.35 ) 2 2  

235

Section 2.2 Solutions -------------------------------------------------------------------------------1. a. 3(1) − 5 ≠ 2 No b. 3(−2) − 5 = −11 Yes

2. a. −2( −1) + 7 = 9 Yes b. −2(2) + 7 ≠ −4 No

3. a. b.

4. a. − 34 (8) + 1 ≠ 5 No b. − 34 ( −4) + 1 = 4 Yes

2 5

(5) − 4 = −2 Yes

2 5

( −5 ) − 4 ≠ 6 No

5. 2 a. − ( −2 + 1) − 2 = −3 Yes

6. a.

b. − ( 2 + 1) − 2 ≠ 8 No

( −3 ) + 2 ≠ −10 No 3 b. 21 ( 2 ) + 2 = 6 Yes

7. a. y = − 4 + 4 = 2 Yes

8. a. −1 + 3 − 4 = −2 Yes

b. y = − ( −4 ) + 4 ≠ 2 No

b. −2 + 3 − 4 ≠1 No

9. a. ( −1)2 − 2( −1) + 1 = 4 Yes b. (0)2 − 2(0) + 1 = 1 ≠ −1 No

10. a. ( −1)3 − 1 = −2 ≠ 0 No b. ( −2 )3 − 1 = −9 Yes

11. a. 7 + 2 = 3 Yes

12. a. 2 + 3 − 9 = 8 ≠ −4 No

1 2

b. 2 + 3 − ( −2) = 7 Yes

−6 + 2 = 2i ≠ 4 No

236

Section 2.2

13.

14.

−2 0 1

0 2 3

( −2,0) (0,2) (1,3)

15.

−1 0 2

−4 −1 5

( −1, −4 ) (0, −1) (2,5)

x −3 −2 −1 0 1

y = 1 − 2x − x2

( x, y) ( −3, −2) ( −2,1) ( −1,2) (0,1) (1, −2)

16.

1 2

y = x2 − x 2 0 − 14

1 2

0 2

−1 0

( x, y ) ( −1,2) (0,0) ( 12 , − 14 ) (1,0) (2,2)

237

−2 1 2 1 −2

Chapter 2

17.

18. x

1 2 5 10

y = x −1 0 1 2 3

( x, y )

(1,0) (2,1) (5,2) (10,3)

−2 −1 2 7

19.

20.

238

y = − x+2 0 −1 −2 −3

( x, y ) ( −2,0) ( −1, −1) (2, −2) (7, −3)

Section 2.2

21.

22.

23.

24.

25.

26.

239

Chapter 2

27. x-intercept: 2x − 0 = 6  x = 3 So, (3,0 ). y-intercept: 2(0) − y = 6  y = −6 So, (0, −6).

28. x-intercept: 4 x + 2(0) = 10  x = 52 So, ( 52 , 0). y-intercept: 4(0) + 2 y = 10  y = 5 So, (0,5).

29. x-intercept:

30. x-intercept: 3x + 0 = 12  x = 4 So, ( 4, 0 ) .

3x + 0 = 7  x =

7 7  So,  , 0  . 3 3 

y-intercept: 0 − 4 y = 12  y = −3 So, ( 0, − 3 ) .

y-intercept: 0 + y = 7  y = 7 So, ( 0, 7 ) .

31. x-intercepts: x 2 − 9 = 0  ( x − 3)( x + 3) = 0  x = ±3. So, ( ±3,0). y-intercept: (0)2 − 9 = y  y = −9 So, (0, −9).

32. x-intercepts: 4 x 2 − 1 = 0  (2 x − 1)(2 x + 1) = 0

33. x-intercept: x − 4 = 0  x = 4 So, (4,0). y-intercept: 4 = y. So, no y-intercept. 0−

34. x-intercept: 3 x − 8 = 0  x = 8 So, (8,0). y-intercept: 3 0 − 8 = y  y = −2 So, (0, −2).

 x = ± 12 . So, (± 12 ,0).

y-intercept: 4(0)2 − 1 = y  y = −1 So, (0, −1).

undefined

35. x-intercept: 1 = 0 has no solution. 2 x +4 So, no x-intercept. y-intercept: 1 = y  y = 14 So, (0, 14 ). 2 0 +4

36. x-intercepts: x 2 − x − 12 = 0  x 2 − x − 12 x = ( x − 4 )( x + 3) = 0  x = 4, −3 So, (4,0), (−3,0). y-intercept: 0 2 − 0 − 12 = y is undefined. So, 0 no y-intercept.

240

Section 2.2

37. x-intercepts: 4 x 2 + 0 = 16  x 2 − 4 = 0  ( x − 2 )( x + 2) = 0  x = ±2. So, ( ±2,0). y-intercept: 4(0)2 + y 2 = 16  y 2 − 16 = 0  ( y − 4 )( y + 4) = 0  y = ±4. So, (0, ±4).

38. x-intercepts: x2 − 0 = 9  x2 − 9 = 0  ( x − 3)( x + 3) = 0  x = ±3. So, ( ±3,0). y-intercept: 0 − y 2 = 9 which has no solution. So, no y-intercept.

39. d

40. e

41. a

42. c

43. b

44. d

45. ( −1, −3)

46. (2, 4)

47. (4, 7)

48. ( −12,12 )

49. ( −7,10)

50. (1,1)

51. (3,2), (−3,2), ( −3, −2)

52. (−1, −7), (1,7), (1, −7)

53. x-axis symmetry (Replace y by − y):

y-axis symmetry (Replace x by −x ):

x = ( − y )2 + 4

−x = y2 + 4

x = −( y 2 + 4) No, since not equivalent to the original.

x = y2 + 4 Yes, since equivalent to the original.

symmetry about origin (Replace y by −y and x by −x): − x = ( − y )2 + 4 = y 2 + 4 x = − ( y2 + 4 )

No, since not equivalent to the original. 54. x-axis symmetry (Replace y by −y):

y-axis symmetry (Replace x by −x ):

x = 2( − y ) + 3

−x = 2 y2 + 3

x = −(2 y 2 + 3) No, since not equivalent to the original.

x = 2 y2 + 3 Yes, since equivalent to the original.

symmetry about origin (Replace y by −y and x by −x): − x = 2( − y )2 + 3 = 2 y2 + 3 x = − ( 2 y2 + 3)

No, since not equivalent to the original. 241

Chapter 2

55. x-axis symmetry (Replace y by − y):

y-axis symmetry (Replace x by −x): y = ( −x )3 + ( −x )

−y = x + x 3

y = − ( x3 + x )

No, since not equivalent to the original. symmetry about origin (Replace y by −y and x by −x):

No, since not equivalent to the original.

− y = ( −x )3 + ( − x ) = − ( x3 + x )

y = x3 + x Yes, since equivalent to the original.

56. x-axis symmetry (Replace y by −y):

y-axis symmetry (Replace x by −x): y = ( −x )5 +1

− y = x +1 5

y = −( x5 + 1) No, since not equivalent to the original.

y = − x5 +1 No, since not equivalent to the original.

symmetry about origin (Replace y by −y and x by −x): − y = ( −x )5 +1 = −x5 +1

y = x5 − 1 No, since not equivalent to the original.

57. x-axis symmetry (Replace y by −y): x = − y = −1 y

y-axis symmetry (Replace x by −x): −x = y

x= y Yes, since equivalent to the original.

x=− y No, since not equivalent to the original.

symmetry about origin (Replace y by −y and x by −x ): −x = − y = −1 y = y

x=− y No, since not equivalent to the original.

242

Section 2.2

58. x-axis symmetry (Replace y by −y): x = − y − 2 = −1 y − 2

y-axis symmetry (Replace x by −x ): −x = y − 2

x = y −2 Yes, since equivalent to the original.

x = −( y − 2 ) No, since not equivalent to the original.

symmetry about origin (Replace y by −y and x by −x): −x = − y − 2 = −1 y − 2 = y − 2 x = − ( y − 2)

No, since not equivalent to the original. 59. x-axis symmetry (Replace y by −y):

y-axis symmetry (Replace x by −x):

x − ( − y ) = 100

( − x )2 − y 2 = 100

x 2 − y 2 = 100 Yes, since equivalent to the original.

x 2 − y2 = 100 Yes, since equivalent to the original.

symmetry about origin (Replace y by −y and x by −x): (− x )2 − (− y )2 = 100

x 2 − y 2 = 100 Yes, since equivalent to the original.

60. x-axis symmetry (Replace y by −y):

y-axis symmetry (Replace x by −x):

x 2 + 2( − y )2 = 30

( − x )2 + 2 y 2 = 30

x 2 + 2 y 2 = 30 Yes, since equivalent to the original.

symmetry about origin (Replace y by −y and x by −x): ( − x )2 + 2(− y )2 = 30

x 2 + 2 y 2 = 30 Yes, since equivalent to the original.

243

Chapter 2

61. x-axis symmetry (Replace y by −y):

y-axis symmetry (Replace x by −x):

−y = x 3

y = ( −x ) 3 2

y=x 3 Yes, since equivalent to the original. 2

y = −x 3 No, since not equivalent to the original. 2

symmetry about origin (Replace y by −y and x by −x): − y = ( −x ) 3 = x 3 2

y = −x 3 No, since not equivalent to the original. 2

62. x-axis symmetry (Replace y by −y):

y-axis symmetry (Replace x by −x ):

x = (− y) 3

−x = y 3 2

x = −y 3 No, since not equivalent to the original. 2

x=y3 Yes, since equivalent to the original. 2

symmetry about origin (Replace y by −y and x by −x): −x = ( − y ) 3 = y 3 2

x = −y 3 No, since not equivalent to the original. 2

63. x-axis symmetry (Replace y by −y):

y-axis symmetry (Replace x by −x):

x 2 + ( − y )3 = 1

( − x )2 + y 3 = 1

x 2 − y3 = 1 No, since not equivalent to the original.

x 2 + y3 = 1 Yes, since equivalent to the original.

symmetry about origin (Replace y by −y and x by −x): ( − x )2 + (− y )3 = 1

x 2 − y3 = 1 No, since not equivalent to the original.

244

Section 2.2

64. x-axis symmetry (Replace y by −y):

y-axis symmetry (Replace x by −x):

− y = 1 + x2

y = 1 + (− x )2

y = − 1 + x2 No, since not equivalent to the original.

y = 1 + x2 Yes, since equivalent to the original.

symmetry about origin (Replace y by −y and x by −x): − y = 1 + ( −x )2 = 1+ x 2 y = − 1+ x 2 No, since not equivalent to the original.

65. x-axis symmetry (Replace y by −y): 2 −y = x 2 y=− x No, since not equivalent to the original.

y-axis symmetry (Replace x by −x): 2 y= −x 2 y=− x No, since not equivalent to the original.

symmetry about origin (Replace y by −y and x by −x): 2 2 −y = =− −x x 2 y= x Yes, since equivalent to the original. 66. x-axis symmetry (Replace y by −y): x( − y ) = 1

y-axis symmetry (Replace x by −x): ( −x ) y = 1

−xy = 1 No, since not equivalent to the original.

symmetry about origin (Replace y by −y and x by −x): ( −x )( − y ) = 1 xy = 1 Yes, since equivalent to the original.

245

Chapter 2

67.

68.

69.

70.

71.

72.

246

Section 2.2

73.

74.

75.

76.

77.

78.

247

Chapter 2

79.

80.

81.

82.

83.

R (in km) 0.1 = 10−1 1 10,000 = 104

1 1   R, 2  2  R  R  (in Hz) 102 ( 10−1, 102 ) 1 ( 1, 1 ) 10−8 ( 104, 10−8)

Note: 1000 m = 1 km

248

Section 2.2

84.

λ (in m)

10−6 10−3 10−1

(in Hz) 3.0 × 1014 3.0 × 1011 3.0 × 109

3.0 ×108

From the plot of the points ( λ , f ) to the right, note that as wavelength decreases, frequency increases (and vice versa). 85. Consider the following graph:

86. Consider the following graph:

Either 2000 or 4000 units need to be sold to break even. The range of units that correspond to making a profit correspond to those x-values for which y > 0. This occurs for 2000 < x < 4000 .

249

Either 1000 or 3000 units need to be sold to break even. The range of units that correspond to making a profit correspond to those x-values for which y > 0. This occurs for 1000 < x < 3000 .

Chapter 2

88. a. The domain is the set of x such that 0.01x − 0.4 ≥ 0. Solving this inequality yields 0.01x ≥ 0.4

87. a. The domain is the set of x such that 0.01x − 0.01 ≥ 0. Solving this inequality yields 0.01x ≥ 0.01 x ≥ 1 or [1, ∞ ) This means that the demand model is defined when at least 1,000 units per day are demanded. b.

x ≥ 4 or [ 4, ∞ ) This means that the demand model is defined when at least 4,000 units per day are demanded. b. p (price per unit)

p (price per unit) 40 35

30 25 2

20 15 1

10 5

x (thousands of units demanded) 2

x (thousands of units demanded)

89. The equation is not linear – you need more than two points to plot the graph.

90. Note that −( −x )2 ≠ x 2 . Rather, −( −x )2 ≠ − x 2 . As such, the graph IS symmetric with respect to the y-axis.

91. To test for symmetry about the y-axis, one should replace x by −x, NOT y by − y. Doing so here yields the equation − x = y , which is not equivalent to the original equation x= y.

250

Section 2.2

92. To test for symmetry about the x-axis, replace y by − y, NOT x by −x. The conclusion should be that the graph is symmetric about the y-axis. The correct graph is given to the right:

93. False The correct conclusion would be that the point ( a, −b ) must also be on the graph. For instance, y 2 = x is symmetric about the x-axis and (1,1) is on the graph, but ( −1,1) is not.

94. True By definition of symmetry about the yaxis. 95. True 96. False. This is true only for linear equations.

97. x-axis symmetry (Replace y by − y): ax + b −y = cx3  ax 2 + b  y = −  3  cx  No, since not equivalent to the original. 2

y-axis symmetry (Replace x by − x): a( − x )2 + b ax 2 + b y= = c( − x )3 −cx3  ax 2 + b  y = −  3  cx  No, since not equivalent to the original.

symmetry about origin (Replace y by − y and x by − x): −y =

 ax 2 + b  a( − x )2 + b ax 2 + b ax 2 + b so that = − y = =   3 c( − x )3 −cx3 cx3  cx  Yes, since equivalent to the original.

251

Chapter 2

98. Consider y = ( x − a )2 − b 2 . x-intercepts: 0 = ( x − a)2 − b2  0 = ( x − a − b ) ( x − a + b )  x = a + b, a − b.

So, ( a + b, 0), (a − b,0).

y-intercept: y = (0 − a )2 − b 2 = a 2 − b2 . So, ( 0,a 2 − b2 ) . 100.

99.

Symmetric with respect to the y-axis.

Symmetric with respect to the origin.

101.

102.

Symmetric with respect to the x-axis, y-axis, and the origin.

252

Section 2.2

103.

104.

Symmetric with respect to the x-axis, y-axis, and the origin.

Symmetric with respect to the x-axis

253

Section 2.3 Solutions -------------------------------------------------------------------------------1.

3−6 = 3 1− 2

1− 9 = 4 2−4

−4 − 6 m= = 2 −1 − 4

5 − ( −3) = −2 −2 − 2

9 − ( −10) 19 = − −7 − 3 10

−3 − 6 9 = − 11− ( −2) 13

−1.7 − 5.2 6.9 m= = ≅ 2.379 0.2 − 3.1 2.9 9. m=

− − (− ) 3 4

1 4

2 3

− 56

1.7 − ( −2.3) 4.0 = = 1.25 −2.4 − ( −5.6) 3.2

10. 3 4 −7 16 m = 1 5 5 3 = − 55 = − 25 2 − (− 4 ) 4

= − 21 = −3

11. x-intercept: ( 0.5,0 )

y-intercept: ( 0, −1) slope: m =

−3 − 3 =2 −1 − 2

12. x-intercept: ( −2,0 )

y-intercept: ( 0,3 ) slope: m =

−3 − 3 3 = −4 − 0 2

13. x-intercept: (1,0 )

y-intercept: ( 0,1)

slope: m =

3 − ( −1) = −1 −2 − 2

rising

falling

14. x-intercept: ( 0,0 )

15. x-intercept: None y-intercept: ( 0,1) slope: m = 0 horizontal

16. x-intercept: ( −4,0 )

y-intercept: ( 0,0 ) slope: m =

3 − ( −2) = −1 −3 − 2

falling

254

y-intercept: None slope: Undefined vertical

Section 2.3

17. x-intercept: 0 = 2x − 3 So, ( 32 ,0 ) . 3 = x 2 y-intercept: y = 2(0) − 3 = −3. So, ( 0, −3) .

18. x-intercept: 0 = −3x + 2

So, ( 23 ,0 ) . =x y-intercept: y = −3(0) + 2 = 2. So, ( 0,2 ) . 2 3

19. x-intercept: 0 = − 12 x + 2

So, ( 4,0 ) . 4=x y-intercept: y = − 12 (0) + 2 = 2. So,

20. x-intercept: 0 = 13 x − 1 So, ( 3,0 ) . 3=x y-intercept: y = 13 (0) − 1 = −1 . So,

( 0,2 ) .

( 0, −1) .

255

Chapter 2

21. x-intercept: 4 = 2 x − 3(0) So, ( 2,0 ) . 2=x y-intercept: 4 = 2(0) − 3 y . So, ( 0,− 34 ) . − 34 = y

22. x-intercept: −1 = −x + 0 So, (1,0 ) . 1= x y-intercept: −1 = 0 + y . So, ( 0, −1) . −1 = y

23. x-intercept: −1 = 12 x + 12 (0)

24. x-intercept: 1 1 1 12 = 3 x − 4 (0)

−2 = x y-intercept: −1 = 12 ( 0) + 12 y −2 = y

So, ( −2,0 ) . . So, ( 0, −2 ) .

=x y-intercept: 1 1 1 12 = 3 (0) − 4 y 1 4

− 13 = y

256

So, ( 14 ,0 ) . . So, ( 0, − 13 ) .

Section 2.3

25. x-intercept: ( −1,0 ) y-intercept: None

26. x-intercept: None y-intercept: ( 0, −3 )

27. x-intercept: None y-intercept: ( 0,1.5 )

28. x-intercept: ( −7.5,0 ) y-intercept: None

257

Chapter 2

29. x-intercept: ( − 72 ,0 ) y-intercept: None

30. x-intercept: None y-intercept: ( 0, 53 )

31. y = 25 x − 2 m = 52 y-intercept: (0, −2)

32. y = 34 x − 3 m = 34 y-intercept: (0, −3)

33. y = − 13 x + 2 m = − 13 y-intercept: (0,2)

34. y = − 12 x + 4 m = − 12 y-intercept: (0, 4)

35. y = 4x − 3 m = 4 y-intercept: (0, −3)

36.

37. y = −2x + 4 m = −2 y-intercept: (0, 4)

38. y = 14 x − 12 m = 14 y-intercept: ( 0, − 12 )

39. y = 23 x − 2 m = 23 y-intercept: (0, −2)

40. y = −4x + 3 m = −4 y-intercept: (0, 3)

41. y = − 34 x + 6 m = − 34 y-intercept: (0,6)

42. y = − 58 x + 5 m = − 58 y-intercept: (0,5)

43. y = 2 x + 3

44. y = −2x + 1

258

y = x − 5 m = 1 y-intercept: (0, −5)

Section 2.3

45. y = − 13 x + 0 = − 13 x

46. y = 12 x − 3

47. y = 2

48. y = −1.5

49. x = 32

50. x = −3.5

51.

52.

−3 = 5(−1) + b 2=b So, the equation is y = 5x + 2 .

Find b:

2 = −3(−2) + b

53. Find b:

−4 = b So, the equation is y = −3x − 4 .

55. Find b:

−1 = 34 (1) + b − 74 = b

−1 = 2(1) + b −3 = b So, the equation is y = 2x − 3 .

Find b:

54. Find b:

−4 = −1(3) + b

−1 = b So, the equation is y = −x − 1.

3 = − 71 ( −5) + b 56. Find b: 16 7 =b

So, the equation is y = 34 x − 74 .

So, the equation is y = − 71 x + 167 .

57. Since m = 0, the line is horizontal.

58. Since m = 0, the line is horizontal.

So, the equation is y = 4 .

So, the equation is y = −3 .

59. Since m is undefined, the line is vertical. So, the equation is x = −1.

60. Since m is undefined, the line is vertical. So, the equation is x = 4 .

61. slope: −1 − 2 3 m= = −2 − 3 5 y-intercept: Use the point ( −2, −1) to find b:

62. slope: −3 − 1 4 m= = −4 − 5 9 y-intercept: Use the point ( −4, −3 ) to find b:

−1 = 35 ( −2) + b

−3 = 94 ( −4) + b

1 5

So, the equation is y = 53 x + 51 .

− 119 = b So, the equation is y = 94 x − 119 .

259

Chapter 2

63. slope: −1 − ( −6) m= = −5 −3 − ( −2) y-intercept: Use the point ( −3, −1) to find b:

64. slope: −8 − ( −2) 1 m= = −5 − 7 2 y-intercept: Use the point ( −5, −8 ) to find b:

−8 = 12 ( −5) + b

−1 = −5(−3) + b

− 112 = b

−16 = b So, the equation is y = −5x − 16 .

So, the equation is y = 12 x − 112 .

65. slope: −37 − ( −42) 1 = m= 20 − ( −10) 6 y-intercept: Use the point ( 20, −37 ) to find b:

66. slope: 12 − ( −12) m= =2 −8 − ( −20) y-intercept: Use the point ( −8,12 ) to find b:

−37 = 61 (20) + b

12 = 2( −8) + b

− 121 3 =b

28 = b So, the equation is y = 2x + 28 .

So, the equation is y = 61 x − 121 3 . 67. slope: m =

4 − ( −5) = −3 −1 − 2

68. slope: m =

3 − ( −3) 3 =− −2 − 2 2

y-intercept: Use the point ( −1,4 ) to find b:

y-intercept: Use the point ( −2,3 ) to find b:

4 = −3(−1) + b 1= b So, the equation is y = −3x + 1 .

3 = − 32 ( −2) + b 0=b So, the equation is y = − 32 x .

260

Section 2.3

69.

slope: m =

70. 3 4 1 2

− 3 = − 2 9 4 3 2

y-intercept: Use the point ( 12 , 34 ) to find b: 3 4

= 23 ( 21 ) + b

slope: m =

− 12 − 12 1 = − 23 − 73 3

y-intercept: Use the point ( − 23 , − 12 ) to find b:

− 12 = 13 ( − 23 ) + b

0=b

− 185 = b

So, the equation is y = 32 x .

So, the equation is y = 13 x − 158 .

71. Since m is undefined, the line is vertical. So, the equation is x = 3 .

72. Since m is undefined, the line is vertical. So, the equation is x = −5 .

7−7 = 0, the line is 3−9 horizontal. So, the equation is y = 7 .

−1 − ( −1) = 0 , the line is −2 − 3 horizontal. So, the equation is y = −1.

73. Since m =

74. Since m =

6−0 6 = . 0 − ( −5) 5 Since (0,6) is the y-intercept, b = 6.

76. Since m is undefined, the line is vertical. So, the equation is x = 0 .

75. The slope is m =

So, the equation is y = 65 x + 6 . 77. Since m is undefined, the line is vertical. So, the equation is x = −6 .

78. Since m is undefined, the line is vertical. So, the equation is x = −9 .

79. The slope is undefined. 80. Since m is undefined, the line is So, the equation of the line passing vertical. So, the equation is x = 13 . 3 2 2 1 through the points ( 5 , − 4 ) and ( 5 , 2 ) is x = 25 .

81. We identify the slope and yintercept, and then express the equation in slope-intercept form as y = x − 1.

82. We identify the slope and yintercept, and then express the equation in slope-intercept form as y = x + 1.

261

Chapter 2

83. We identify the slope and yintercept, and then express the equation in slope-intercept form as y = −2x + 3 .

84. We identify the slope and yintercept, and then express the equation in slope-intercept form as y = −4x + 2 .

85. We identify the slope and yintercept, and then express the equation in slope-intercept form as y = − 12 x + 1.

86. We identify the slope and yintercept, and then express the equation in slope-intercept form as y = − 52 x + 2 .

87. slope: Since parallel to y = 2x − 1, m = 2. y-intercept: Use the point ( −3,1) to find b:

88. slope: Since parallel to y = −x + 2, m = −1. y-intercept: Use the point (1,3 ) to find b:

1 = 2( −3) + b 7=b So, the equation is y = 2 x + 7 .

3 = −1(1) + b 4=b So, the equation is y = −x + 4 .

89. slope: Since perpendicular to 2x + 3 y = 12 (whose slope is − 23 ), m = 32 . y-intercept: Since (0,0) is assumed to be on the line and is the y-intercept, b = 0.

90. slope: Since perpendicular to x − y = 7 (whose slope is 1), m = −1. y-intercept: Since (0,6) is assumed to be on the line and is the y-intercept, b = 6. So, the equation is y = −x + 6 .

So, the equation is y = 32 x . 91. Since parallel to x-axis, the line is horizontal. So, its equation is of the form y = b. Since (3,5) is assumed to be on the line, the equation must be y =5.

92. Since parallel to y-axis, the line is vertical. So, its equation is of the form x = a. Since (3,5) is assumed to be on

the line, the equation must be x = 3 .

262

Section 2.3

93. Since perpendicular to y-axis, the line is horizontal. So, its equation is of the form y = b . Since ( −1,2) is assumed to be on the line, the equation must be y=2.

94. Since perpendicular to x-axis, the line is vertical. So, its equation is of the form x = a . Since ( −1,2) is assumed to be on the line, the equation must be x = −1.

95. slope: Since parallel to 12 x − 13 y = 5 (whose slope is 32 ), m = 32 . y-intercept: Use the point ( −2, −7 ) to find b:

96. slope: Since perpendicular to − 23 x + 23 y = −2 (whose slope is 94 ), m = − 94 . y-intercept: Use the point (1, 4 ) to find b: 4 = − 94 (1) + b

−7 = 32 ( −2) + b −4 = b So, the equation is y = 32 x − 4 . 97. Note that 8x + 10 y = −45 is equivalent to y = − 54 x − 29 . The slope is

− 54 . So, the slope of a line perpendicular to it is 54 . Using this as the slope, together with the point ( − 23 , 23 ) , we have the equation:

25 4

So, the equation is y = − 94 x + 254 . 98. Note that 6x + 14 y = 7 is equivalent to y = − 73 x + 12 . The slope is − 73 . So, the

slope of a line perpendicular to it is 73 . Using this as the slope, together with the point ( 65 ,3 ) , we have the equation: y − 3 = 73 ( x − 65 )

y − 23 = 54 ( x + 32 )

y − 3 = 73 x − 145

y − 23 = 54 x + 65 y= x+ 5 4

y = 73 x + 51

3 2

263

Chapter 2

99. Note that −15x + 35 y = 7 is equivalent to y = 73 x + 15 . The slope is

100. Note that 10x + 45 y = −9 is equivalent to y = − 29 x − 15 . The slope is

3 7 3 7

− 29 . So, the slope of a line parallel to it is − 29 . Using this as the slope, together

. So, the slope of a line parallel to it is . Using this as the slope, together

with the point ( 72 , 4 ) , we have the equation: y − 4 = 73 ( x − 72 )

with the point ( − 14 , − 139 ) , we have the equation: y + 139 = − 29 ( x + 14 )

y − 4 = 73 x − 23

y + 139 = − 29 x − 181

y = 73 x + 52

y = − 29 x − 32

101. Let h = number of hours a job lasts. The cost (in dollars) is given by C ( h ) = 1200 + 25h. So, for h = 32, C = 2000. This means that a 32-hour job will cost $2,000.

102. Let x = number of miles. The cost (in dollars) is given by C = 50 + 0.39x .

103. Let L = monthly loan payment and x = # times filled up gas tank. Then, 500 = L + 25x. So, if x = 5, then 500 = L + 125, so that L = 375. The monthly payment is $375. 104. Let B = monthly loan payment M = cost of filling gas tank once x = # times filled up gas tank

Then, the corresponding total monthly cost is given by C = B + Mx. We are given that (3,520) and (5,600) satisfy the equation. We compute M and B as follows: 520 − 600 = 40 y-intercept: Use (3,520) to find B: slope: M = 3−5 520 = B + 40(3) 400 = B Thus, the monthly loan payment is $400 and it costs $40 every time you fill up with gasoline.

264

Section 2.3

105. Let x = number of units sold.

106. Let x = number of units sold.

Operating Cost: C(x) = 1,300 + 3.50x Revenue: R(x) = 7.25x

Operating Cost: C(x) = 12,000 + 13.50x Revenue: R(x) = 27.25x

Find the value of x such that C(x) = R(x). 1,300 + 3.50x = 7.25x 1,300 = 3.75x 1,300 x= ≈ 347 units 3.75

Find the value of x such that C(x) = R(x). 12,000 + 13.50x = 27.25x 12,000 = 13.75x 12,000 x= ≈ 873 units 13.75

107. Assume that F and C are related by F = mC + b . We are given that (25,77) and (20,68) (where the points are listed in the form (C , F ) ) satisfy the equation. We compute m and b as follows: 77 − 68 9 slope: m = = y-intercept: Use (25,77) to find b: 25 − 20 5 77 = 59 (25) + b

32 = b So, the equation relating F and C is given by F = 95 C + 32 . The temperature for which F = C can be found by substituting F = C into the equation: C = 95 C + 32 − 54 C = 32 −40 = C Thus, −40 degrees Celsius = −40 degrees Fahrenheit. 108. Assume C = mx + b , where C is temperature (in degrees Celsius) and x is elevation (in feet). We are given that (0,15) and (500,14) satisfy the equation. So, We compute m and b as follows: 15 −14 1 slope: m = =− 0 − 500 500 y-intercept: Since (0,15) is assumed to be on the line and is the y-intercept, b = 15 .

1 x +15 . 500 The expected temperature at 2500 ft. is 10 degrees Celsius.

So, the equation is C = −

265

Chapter 2

109. Since (1900,67) and (2000,69) are assumed to be on the line, the slope of the 69 − 67 2 1 line is m = = = . So, the rate of change (in inches per year) is 2000 −1900 100 50 1 . 50 110. Since (1900,64) and (2000,67) are assumed to be on the line, the slope of the 67 − 64 line is m = = 0.03 . So, the rate of change (in inches per year) is 0.03 . 2000 −1900 111. Note: 6 pounds 4 ounces = 100 ounces 6 pounds 10 ounces = 106 ounces

Since (1900,100) and (2000,106) are assumed to be on the line, the slope of the line 106 −100 is m = = 0.06 . So, the rate of change (in ounces per year) is 0.06. 2000 −1900 To determine the expected weight in the year 2040, we need the equation of the line. So far, we know the form of the equation is y = 0.06x + b . Use (1900,100) to find b: 100 = 0.06(1900) + b −14 = b So, the equation is y = 0.06x − 14 . Therefore, in 2040, y = 0.06(2040) −14 = 108.4 oz. Since 108.4 oz. = 6.775 pounds, and 0.775 pounds = 12.4 oz., we expect a baby to weigh 6 pounds 12.4 oz in 2040. 112. Since (1906, 4.5) and (1957, 4) are assumed to be on the line, the slope of the 4.5 − 4 line is m = ≅ −0.01. So, the rate of change is −0.01. 1906 −1957 113. The y -intercept represents the flat monthly fee of $500 since x = 0 implies no sales made. 114. y -intercept: Represents the cost of the standard model with no customizations chosen. It is ( 0,35,000 ) .

266

Section 2.3

115. The rate of change is

33.8 − 35.1 = −0.26 in./yr. 5−0 Using the point ( 0,35.1) , and assuming that x = 0 corresponds to 2014, the equation that governs this trend is given by y − 35.1 = −0.26 ( x − 0 )  y = −0.26 x + 35.1.

Hence, in 2021, you can expect the average rainfall to be −0.26 ( 7 ) + 35.1 = 33.28 in. 116. The rate of change is

32 − 35 = −1.5°F/yr. 2−0 Using the point ( 0,35 ) , and assuming that x = 0 corresponds to Jan. 2017, the equation that governs this trend is given by y − 35 = −1.5 ( x − 0 )  y = −1.5x + 35.

Hence, in 2022, you can expect the average temperature to be −1.5 ( 5 ) + 35 = 27.5°F. . 117. The rate of change is

24 − 30 = −3 billion bags/yr. 2−0 Using the point ( 0,30 ) , and assuming that x = 0 corresponds to 2016, the equation that governs this trend is given by y − 30 = −3 ( x − 0 )  y = −3x + 30.

Hence, in 2021, you can expect −3 ( 5 ) + 30 = 15 billion plastic bags to be used in California.

267

Chapter 2

118. The rate of change is

6040 − 5686 = 177 dollars per year 2−0 Using the point ( 0,5686 ) , and remembering that x = 0 corresponds to 2016, the equation that governs this trend is given by y − 5686 = 177 ( x − 0 )  y = 177 x + 5686.

Hence, in 2022, you can expect the average credit card debt to be 177 ( 6 ) + 5686 = $6748 . 119. a. (1, 16(1) + 15.93) = (1, 31.93), (2, 16(2) + 19.18) = (2, 51.18), (5, 111.83) 31.93 − 0 b. m = = 31.93 This means that when you buy a single bottle, the cost 1− 0 per bottle of Hoison sauce is $31.93. 51.18 − 0 c. m = = 25.59 This means that when you buy 2 bottles of Hoison sauce, 2−0 the cost per bottle of Hoison sauce is $25.59. 111.83 − 0 d. m = = 22.37 This means that when you buy 5 bottles of Hoison 5−0 sauce, the cost per bottle of Hoison sauce is $22.37. 120. a. (1, 18.27), (2, 2(4) + 14.77) = (2, 22.77), (5, 35.93)

18.27 − 0 = 18.27 This means that when you buy a single bottle, the cost 1− 0 per bottle is $18.27.

b. m =

22.77 − 0 = 11.39 This means that when you buy 2 bottles the cost per 2−0 bottle of is $11.39.

c. m =

35.93 − 0 = 7.19 This means that when you buy 5 bottles the cost per bottle 5−0 of is $7.19.

d. m =

268

Section 2.3

121. The computations used to calculate the x- and y-intercepts should be reversed. So, the x-intercept is (3, 0) and the y-intercept is (0, − 2). 122. The denominator of the slope should be 4 − ( −2 ), resulting in m = − 13 . 123. The denominator and numerator in the slope computation should be switched, resulting in the slope being undefined. 124. These two are listed incorrectly: a. horizontal b. vertical 125. True. Since the equation mx + b = 0 has at most one solution.

126. False. Any vertical line x = a (a ≠ 0) does not have a y-intercept.

127. False. The lines are perpendicular.

128. True.

129. Any vertical line is perpendicular to a line with slope 0.

130. Any vertical line is parallel to a line with no slope.

131. Since B ≠ 0, Ax + By = C can be written as y = − BA x + CB . Since the desired line is parallel to this line, its slope m = − BA . Use the point (−B, A + 1) on the line to find the y-intercept: A + 1 = − BA ( −B ) + b A +1 = A + b b =1 So, the equation of the line in this case is y = − BA x + 1.

132. We know that the point ( B, A −1) is on the line. Since it is parallel to Ax + By = C, and the slope of this line is − BA (assuming B ≠ 0 ), the equation of the desired line is y − ( A − 1) = − BA ( x − B) y = ( A − 1) − BA x + A y = − BA x + (2A −1)

269

Chapter 2

133. We know that the point (−A, B − 1) is on the line. Since it is perpendicular to Ax + By = C, and the slope of this line is − BA (assuming B ≠ 0 ), the slope of the desired line is BA (assuming A ≠ 0). So, the equation of the desired line is y − ( B − 1) = BA ( x + A) y = ( B − 1) + BA x + B y = BA x + (2B −1)

134. Case 1: A ≠ 0 and B ≠ 0 In such case, Ax + By = C can be written as y = − BA x + CB . Since the desired line is perpendicular to this line, its slope m = BA . Use the point ( A, B + 1) on the line to find the y-intercept: B + 1 = BA ( A) + b

b =1 So, the equation of the line in this case is y = BA x + 1. Case 2: A = 0 and B ≠ 0 In such case, the line Ax + By = C can be written as y = CB , which is horizontal. Since the desired line is perpendicular to this one, it must be vertical. So, since ( A, B +1) is on the line, its equation is x = A, which is x = 0. Case 3: A ≠ 0 and B = 0 In such case, the line Ax + By = C can be written as x = CA , which is vertical. Since the desired line is perpendicular to this one, it must be horizontal. So, since ( A, B + 1) is on the line, its equation is y = B + 1, which is y = 1. 135. Let y1 = mx + b1 and y2 = mx + b2 , assuming that b1 ≠ b2 . At a point of intersection of these two lines, y1 = y2 . This is equivalent to mx + b1 = mx + b2 , which implies b1 = b2 , which contradicts our assumption. Hence, there are no points of intersection. 136. Solving m1x + b1 = m2 x + b2 yields

( m1 − m2 ) x = b2 − b1  x = This is the x-coordinate of the intersection point.

270

b2 − b1 . m1 − m2

Section 2.3

137. Slope of y1 = 17 x + 22 is 17. Slope of y2 = − 171 x − 13 is − 171 .

138. Slope of y1 = 0.35x + 2.7 is 0.35. Slope of y2 = 0.35x − 1.2 is 0.35. Since they have the same slope, parallel.

139. Slope of y1 = 0.25x + 3.3 is 0.25. Slope of y2 = −4 x + 2 is −4 .

140. Slope of y1 = 12 x + 5 is 12 . Slope of y2 = 2 x − 3 is 2. Since 12 (2) ≠ −1 and 12 ≠ 2 , they are neither parallel nor perpendicular.

Since 17 ( − 171 ) = −1 , perpendicular.

Since 0.25 ( −4 ) = −1 , they are perpendicular.

271

Chapter 2

141. Slope of y1 = 0.16 x + 2.7 is 0.16. Slope of y2 = 6.25x − 1.4 is 6.25. Since (0.16)(6.25) ≠ −1 and 0.16 ≠ 6.25, they are neither parallel nor perpendicular.

142. Slope of y1 = −3.75x + 8.2 is −3.75. Slope of y2 = 154 x + 65 is 154 . Since ( −3.75)( 154 ) = −1 they are perpendicular.

272

Section 2.4 Solutions -------------------------------------------------------------------------------1. 2 ( x − 1) + ( y − 2)2 = 9

2. 2 ( x − 3) + ( y − 4)2 = 25

3. 2 2 ( x − ( −3) ) + ( y − ( −4 ) ) = 102

4. 2 2 ( x − ( −1) ) + ( y − ( −2) ) = 42

( x + 3) + ( y + 4 ) = 100

( x + 1) + ( y + 2 ) = 16

5. 2 ( x − 5 ) + ( y − 7)2 = 81

6. 2 2 ( x − 2 ) + ( y − 8) = 36

7. 2 2 ( x − ( −11) ) + ( y −12 ) = 132

8. 2 2 ( x − 6 ) + ( y − ( −7) ) = 82

( x + 11) + ( y −12 ) = 169

( x − 6 ) + ( y + 7 ) = 64

9. 2 2 ( x − 0 ) + ( y − 0 ) = 22

10. 2 2 (x − 0) + ( y − 0) = 9

x2 + y2 = 4

x 2 + y2 = 9

11.

12. ( x − 3)2 + ( y − 0)2 = 2 2

( x − 0 )2 + ( y − 2)2 = 32 x 2 + ( y − 2)2 = 9

( x − 3)2 + y 2 = 4

13.

( x − 0) + ( y − 0) = ( 2 ) 2

14.

( x − ( −1) ) + ( y − 2 ) = ( 7 )

( x + 1) + ( y − 2 ) = 7

x2 + y2 = 2

16.

15.

( x − 5 ) + ( y − ( −3) ) = ( 2 3 ) 2

( x − ( −4 ) ) + ( y − ( −1) ) = (3 5 ) 2

( x − 5 ) + ( y + 3 ) = 12 2

( x + 4 ) + ( y + 1) = 45

273

Chapter 2

17.

18.

( x − ) + ( y − ( − 53 ) ) = ( ) 2 2 ( x − 32 ) + ( y + 53 ) = 161

( x − ( − 13 ) ) + ( y − ( − 72 ) ) = ( 52 ) 2 2 ( x + 13 ) + ( y + 72 ) = 245

19. 2 2 ( x − 1.3 ) + ( y − 2.7 ) = (3.2)2

20. 2 2 ( x − ( −3.1) ) + ( y − 4.2 ) = (5.5)2

( x − 1.3 ) + ( y − 2.7 ) = 10.24

( x + 3.1) + ( y − 4.2 ) = 30.25

2 2 3

1 2 4

21. The center is (1,3 ) and the radius is

22. The equation can be written as

( x − ( −1) ) + ( y − ( −3) ) = ( 11 ) . So, center is ( −1, −3 ) and radius is 2

25 = 5 .

11 .

23. The equation can be written as 2 2 2 ( x − 2 ) + ( y − ( −5) ) = ( 7 ) .

24. The equation can be written as 2 2 ( x − ( −3) ) + ( y − 7 ) = 92 .

25. The center is ( 4,9 ) and the radius

26. The equation can be written as

So, center is ( 2, −5 ) and radius is 7 .

So, center is ( −3,7 ) and radius is 9 .

( x − ( −1) ) + ( y − ( −2) ) = ( 8 ) . So, center is ( −1, −2 ) and radius is 8 = 2

20 = 2 5 .

2 2.

27. The center is ( 52 , 71 ) and the radius

4 9

28. The center is ( 12 , 13 ) and the radius is

= 23 .

9 25

= 35 .

29. The equation can be written as

30. The equation can be written as

( x − 1.5 ) + ( y − ( −2.7) ) = ( 1.69 ) . So, center is (1.5, −2.7 ) and radius is

( x − ( −3.1) ) + ( y − 7.4 ) = ( 56.25 ) . So, center is ( −3.1,7.4 ) and radius is

1.69 = 1.3 .

56.25 = 7.5 .

274

Section 2.4

31. The equation can be written as

32. The equation can be written as

( x − 0 ) + ( y − 0 ) = ( 50 ) . So, center is ( 0,0 ) and radius is

(x − 0) + ( y − 0) = ( 8 ) . So, center is ( 0,0 ) and radius is

50 = 5 2 .

8 =2 2.

33. Completing the square gives us: x 2 + y2 + 4x + 6 y − 3 = 0

34. Completing the square gives us: x 2 + y 2 + 2 x + 10y +17 = 0

x 2 + 4x + y2 + 6 y = 3

x 2 + 2x + y2 + 10y = −17

( x + 4x + 4 ) + ( y + 6 y + 9 ) = 3 + 4 + 9

( x + 2x + 1) + ( y +10y + 25 ) = −17 +1+ 25

( x + 2 ) + ( y + 3) = 16 2 2 ( x − ( −2) ) + ( y − ( −3) ) = 16 So, center is ( −2, −3 ) and radius

( x + 1) + ( y + 5 ) = 9 2 2 ( x − ( −1) ) + ( y − ( −5) ) = 9 So, center is ( −1, −5 ) and radius is

is 16 = 4 .

9 =3.

35. Completing the square gives us: x 2 + y 2 + 6x + 8 y − 75 = 0

36. Completing the square gives us: x 2 + y2 + 2x + 4 y − 9 = 0

x 2 + 6x + y2 + 8 y = 75

x 2 + 2x + y2 + 4 y = 9

( x + 6x + 9 ) + ( y + 8y +16 ) = 75 + 9 + 16 2

( x + 2x + 1) + ( y + 4y + 4 ) = 9 +1+ 4

( x + 3) + ( y + 4 ) = 100 2 2 ( x − ( −3) ) + ( y − (−4) ) = 100 So, center is ( −3, −4 ) and radius is 2

( x + 1) + ( y + 2 ) = 14 2 2 ( x − ( −1) ) + ( y − ( −2 ) ) = 14 So, center is ( −1, −2 ) and radius is 14 .

100 = 10 .

37. Completing the square gives us: x 2 + y 2 − 10 x − 14 y − 7 = 0 x 2 − 10 x

+ y 2 − 14 y = 7

( x − 10x + 25 ) + ( y − 14 y + 49 ) = 7 + 25 + 49 2

( x − 5 ) + ( y − 7 ) = 81 2

So, center is ( 5,7 ) and radius is 9.

275

Chapter 2

38. Completing the square gives us: x 2 + y 2 − 4 x − 16 y + 32 = 0 x 2 − 4x

+ y 2 − 16 y = −32

( x − 4x + 4 ) + ( y − 16 y + 64 ) = −32 + 4 + 64 2

( x − 2 ) + ( y − 8) = 36 2

So, center is ( 2,8 ) and radius is 6 .

39. Completing the square gives us: x 2 + y 2 − 2 y − 15 = 0

40. Completing the square gives us: x 2 + y2 + 2x − 8 = 0

x 2 + ( y2 − 2 y + 1) − 15 − 1 = 0

( x + 2x + 1) + y − 8 − 1 = 0 2

x 2 + ( y − 1)2 = 16 So, the center is (0,1) and radius is 4.

( x + 1)2 + y 2 = 9 So, the center is (−1, 0) and radius is 3.

41. Completing the square gives us: x 2 + y2 − 2x − 6 y + 1 = 0

42. Completing the square gives us: x 2 + y 2 − 8x − 6 y + 21 = 0

x 2 − 2x

+ y 2 − 6 y = −1

x 2 − 8x

+ y 2 − 6 y = −21

( x − 2x + 1) + ( y − 6 y + 9 ) = −1 + 1 + 9

( x − 8x + 16 ) + ( y − 6 y + 9 ) = −21 + 16 + 9

( x − 1) + ( y − 3 ) = 9 So, center is (1,3 ) and radius is 3.

( x − 4 ) + ( y − 3) = 4 So, center is ( 4,3 ) and radius is 2.

43. Completing the square gives us: x 2 + y 2 − 10x + 6 y + 22 = 0

x 2 − 10x

+ y 2 + 6y = −22

( x − 10x + 25 ) + ( y + 6y + 9 ) = −22 + 25 + 9 2

( x − 5 ) + ( y + 3) = 12 2 2 ( x − 5 ) + ( y − ( −3) ) = 12 2

So, center is ( 5, −3 ) and radius is 12 = 2 3 .

276

Section 2.4

44. Completing the square gives us: x 2 + y2 + 8x + 2 y − 28 = 0

x 2 + 8x

+ y2 + 2 y = 28

( x + 8x + 16 ) + ( y + 2y +1) = 28 +16 +1 2

( x + 4 ) + ( y + 1) = 45 2 2 ( x − ( −4) ) + ( y − ( −1) ) = 45 2

So, center is ( −4, −1) and radius is

45 = 3 5 .

45. Completing the square gives us: x2 + y2 − 6x − 4 y + 1 = 0 x 2 − 6x

+ y 2 − 4 y = −1

46. Completing the square gives us: x 2 + y 2 − 2 x − 10 y + 2 = 0 x 2 − 2x

+ y 2 − 10 y = −2

( x − 6x + 9 ) + ( y − 4 y + 4 ) = −1 + 9 + 4

( x − 2x + 1) + ( y − 10 y + 25 ) = −2 + 1 + 25

( x − 3 ) + ( y − 2 ) = 12 So, center is ( 3,2 ) and radius is

( x − 1) + ( y − 5 ) = 24 So, center is (1,5 ) and radius

is 24 = 2 6 .

47. Completing the square gives us: x 2 + y 2 − x + y + 14 = 0 x2 − x 2

1 4

( x − 12 ) + ( y + 12 ) = 14 2 2 ( x − 12 ) + ( y − ( − 12 ) ) = 14 So, center is ( 12 , − 12 ) and radius is 2

48. Completing the square gives us: x 2 + y 2 − 12 x − 32 y + 38 = 0

+ y 2 + y = − 14

(x − x + ) + ( y + y + ) = − + +

1 4

12 = 2 3 .

+ y 2 − 32 y = − 38

x 2 − 12 x

1 4

(x − x + ) + ( y − y + ) = − + + 2

1 2

1 16

3 2

3 8

9 16

( x − 14 ) + ( y − 34 ) = 14 So, center is ( 14 , 34 ) and radius is 2

= 12 .

277

1 16

9 16

1 4

= 12 .

Chapter 2

49. Completing the square gives us: x 2 + y 2 − 2.6 x − 5.4 y − 1.26 = 0 x 2 − 2.6 x

+ y 2 − 5.4 y = 1.26

( x − 2.6x + 1.69 ) + ( y − 5.4 y + 7.29 ) = 1.26 + 1.69 + 7.29 2

( x − 1.3) + ( y − 2.7 ) = 10.24 2

So, center is (1.3,2.7 ) and radius is 10.24 = 3.2 . 50. Completing the square gives us: x 2 + y 2 − 6.2 x − 8.4 y − 3 = 0 x 2 − 6.2 x

+ y 2 − 8.4 y = 3

( x − 6.2x + 9.61) + ( y − 8.4 y + 17.64 ) = 3 + 9.61 + 17.64 2

( x − 3.1) + ( y − 4.2 ) = 30.25 2

So, center is ( 3.1, 4.2 ) and radius is

30.25 = 5.5 .

51. The center is ( −1, −2 ) . The radius is the distance from ( −1, −2 ) to (1,0 ) , which is given by

( −1 − 1) + ( −2 − 0 ) = 8 . 2

52. The center is ( 4,9 ) .

The radius is the distance from ( 4,9 ) to

( 2,5 ) , which is given by

( 4 − 2 ) + ( 9 − 5 ) = 20 . 2

So, the equation is

( x − ( −1) ) + ( y − ( −2) ) = ( 8 ) 2

So, the equation is 2 2 ( x − 4 ) + ( y − 9 ) = 20.

( x + 1) + ( y + 2 ) = 8. 2

53. The center is ( −2,3 ) .

The radius is the distance from ( −2,3 )

to ( 3,7 ) , which is given by

54. The center is (1,1) .

The radius is the distance from (1,1) to

( −8, −5 ) , which is given by

( −2 − 3) + ( 3 − 7 ) = 41 . 2

(1 − ( −8) ) + (1 − ( −5) ) = 117 .

So, the equation is 2 2 ( x − ( −2) ) + ( y − 3 ) = 41

So, the equation is 2 2 ( x − 1) + ( y −1) = 117.

( x + 2 ) + ( y − 3 ) = 41. 2

278

Section 2.4

55. The center is ( −2, −5 ) .

56. The center is ( −3, −4 ) .

The radius is the distance from ( −2, −5 ) to (1, −9 ) , which is given by

( −2 − 1) + ( −5 − ( −9) ) = 5. 2

The radius is the distance from ( −3, −4 )

to ( −1, −8 ) , which is given by

So, the equation is 2 2 ( x − ( −2) ) + ( y − ( −5) ) = 25

( x + 2 ) + ( y + 5)2 = 25. 2

( −3 − ( −1) ) + ( −4 − ( −8) ) = 20 = 2 5. 2

So, the equation is 2 2 ( x − ( −3) ) + ( y − ( −4) ) = 20

( x + 3) + ( y + 4 ) = 20. 2

57. Assume that the phone tower is located at (0,0) . Then, the coordinates of your home are (33,95) . The distance from the phone tower is then 952 + 332 ≅ 100.568 miles, which exceeds the reception area. So, you cannot use the cell phone at home.

58. Assume that the phone tower is located at (0,0) . Then, the coordinates of your home are ( −87, −45) . The distance from the phone tower is then 87 2 + 452 ≅ 97.949 miles, which is within the reception area. So, you can use the cell phone at home.

59. Assume that the stake is located at (0,0) . In order to stay within the boundary of the yard, the leash must be no longer than 50 ft. The outer perimeter 2 2 for the dog would then be described by ( x − 0 ) + ( y − 0 ) = 502 , that is x 2 + y 2 = 2500 .

60. Now, the dog is restricted to one-fourth of the yard. Using the given diagram, the stake would be located at (25,25) . Also, the length of the leash would be reduced to 25 ft. The outer perimeter for the dog would then be described by 2 2 2 ( x − 25 ) + ( y − 25 ) = 25 . 625

61. The center of the inner circle is (0,0) . Since the diameter is 3000 ft., the

radius is 1500 ft. So, the equation of the inner circle is ( x − 0 ) + ( y − 0 ) = 15002 , 2

that is x 2 + y 2 = 2,250,000 .

279

Chapter 2

62. The center of the outer circle is (0,0) . Since the diameter is 6000 ft., the

radius is 3000 ft. So, the equation of the outer circle is ( x − 0 ) + ( y − 0 ) = 30002 , 2

that is x 2 + y 2 = 9 ×106. 63. Location of tower = center of reception area = (0,0). We also know the radius is 200 miles. Hence, the equation of the circle is

64. Assuming the fire station is located at the origin, we know that since the radius is 2 miles, the equation of the circle where fire has spread is

x 2 + y 2 = 40,000 .

x 2 + y2 = 4 .

65. The center of the Ferris wheel is 35 feet above the ground so it is located at (0, 35). Using r = 25 , we get the 2 equation x 2 + ( y − 35 ) = 625.

66. Inner circle: 2 2 ( x − 250 ) + ( y − 500 ) = 4002 Middle circle: 2 2 ( x − 250 ) + ( y − 500 ) = 8002 Outer circle: 2 2 ( x − 250 ) + ( y − 500 ) = 12002

67. a.

68. a.

( x − 3) + ( y − 3 ) = 3.52 2 2 Tower 2: ( x − 3 ) + ( y − 7 ) = 3.52 2 2 Tower 3: ( x − 7 ) + ( y − 3 ) = 3.52 2 2 Tower 4: ( x − 7 ) + ( y − 7 ) = 3.52

( x − 2.5 ) + ( y − 2.5 ) = 3.5 2 2 Tower 2: ( x − 2.5 ) + ( y − 7.5 ) = 3.52 2 2 Tower 3: ( x − 7.5 ) + ( y − 2.5 ) = 3.52 2 2 Tower 4: ( x − 7.5 ) + ( y − 7.5 ) = 3.52

Tower 1:

b. This placement of cell phone towers will provide cell phone coverage for the entire 10 mile by 10 mile square.

b. This placement of cell phone towers will not provide cell phone coverage for the entire 10 mile by 10 mile square.

Tower 1:

y 10

x 2

280

Section 2.4

69. The center should be (4, −3) .

70. The radius should be 2 .

71. The standard form of the equation of a circle requires that the right-side be non-negative. Since the radius would have to be −16 , which is not a real number, the result cannot be a circle.

72. The right-side, upon completing the square, should be 3 + 9 + 4 = 16 . So, the radius would be 4 rather than 2 3 .

73. True. Since the graph is comprised 74. True. Since the left-side is always of infinitely many points. non-negative. 76. True. The solution is (1, −3) .

75. True. Since the left-side is always non-negative.

77. Completing the square gives us: x 2 + y 2 + 10x − 6y + 34 = 0 x 2 + 10x

+ y 2 − 6y = −34

( x + 10x + 25 ) + ( y − 6y + 9 ) = −34 + 25 + 9 2

( x + 5 ) + ( y − 3) = 0 2

Thus, the graph consists of the single point (−5,3). 78. Completing the square gives us: x 2 + y2 − 4 x + 6 y + 49 = 0 x2 − 4x

+ y2 + 6 y = −49

( x − 4x + 4 ) + ( y + 6y + 9 ) = −49 + 4 + 9 2

( x − 2 ) + ( y + 3 ) = −36 2

Since no ordered pair of real numbers ( x, y ) can satisfy this equation, there is no graph.

281

Chapter 2

79. The diameter is equal to the distance between (5,2) and (1, −6), which is given

( 5 − 1) + ( 2 − (−6) ) = 80 = 4 5. So, the radius is 12 ( 4 5 ) = 2 5. 2

The center is the midpoint of the segment joining the points (5,2) and (1, −6) ,  5 + 1 2 + (−6)  which is given by  ,  = (3, −2). Therefore, the equation is then given 2   2 by:

( x − 3) + ( y − ( −2) ) = ( 2 5 ) 2

( x − 3) + ( y + 2 ) = 20 2

80. The diameter is equal to the distance between (3,0) and ( −1, −4), which is

given by

( 3 − ( −1) ) + ( 0 − ( −4) ) = 32 = 4 2. So, the radius is 12 ( 4 2 ) = 2 2. 2

The center is the midpoint of the segment joining the points (3,0) and (−1, −4),  3 + ( −1) 0 + ( −4)  which is given by  ,  = (1, −2). Therefore, the equation is then 2   2 given by:

( x − 1) + ( y + 2 ) = ( 2 2 ) 2

   8

81. Completing the square gives us: x 2 + y 2 + ax + by + c = 0 x 2 + ax

+ y 2 + by = −c

( x + ax + a4 ) + ( y + by + b4 ) = −c + a4 + b4 2

( x + a2 ) + ( y + b2 ) = −c + a42 + b42 2

If −c + a4 + b4 = 0, which is equivalent to 4c = a 2 + b 2 , then the graph will consist

(

)

of the single point − a2 , − 2b .

282

Section 2.4

82. Completing the square gives us: x 2 + y 2 + ax + by + c = 0 x 2 + ax

+ y 2 + by = −c

( x + ax + a4 ) + ( y + by + b4 ) = −c + a4 + b4 2

( x + a2 ) + ( y + 2b ) = −c + a42 + b42 2

If −c + a4 + b4 < 0, which is equivalent to 4c > a 2 + b 2 , then there is no graph. 83. Completing the square gives us: x 2 + y 2 − 2ax = 100 − a 2

( x − 2ax + a ) + y − a = 100 − a 2

( x − a ) + y2 = 100 2

So, the center is (a, 0) and radius is 10. 84. Completing the square gives us: x 2 + y 2 + 2by = 49 − b 2

x 2 + ( y 2 + 2by + b 2 ) − b 2 = 49 − b 2

x 2 + ( y + b )2 = 49 So, the center is (0, −b) and radius is 7.

85. The equation is given by 2 2 ( x − 2 ) + ( y + 5 ) = −20, so there is no graph.

86. The equation is given by 2 2 ( x − 1) + ( y + 3) = 0, so the graph consists of the single point (1, −3) .

87. Completing the square gives us: x 2 + y 2 + 10x − 6y + 34 = 0 x 2 + 10 x

+ y 2 − 6y = −34

( x + 10x + 25 ) + ( y − 6y + 9 ) = −34 + 25 + 9 2

( x + 5 ) + ( y − 3) = 0 2

Thus, the graph consists of the single point (−5,3).

283

Chapter 2

88. Completing the square gives us: x 2 + y 2 − 4 x + 6 y + 49 = 0 x2 − 4x

+ y 2 + 6 y = −49

( x − 4x + 4 ) + ( y + 6y + 9 ) = −49 + 4 + 9 2

( x − 2 ) + ( y + 3 ) = −36 2

Since no ordered pair of real numbers ( x, y) can satisfy this equation, there is no graph. 89. a. Completing the square yields ( x2 − 11x ) + ( y2 + 3y ) = 7.19

( x − 11x + 30.25 ) + ( y + 3y + 2.25 ) = 7.19 + 30.25 + 2.25 = 39.69 2

( x − 5.5 ) + ( y +1.5 ) = 39.69 (1) 2

Center (5.5, −1.5), Radius 6.3 b. Solving (1) for y yields:

( y + 1.5 ) = 39.69 − ( x − 5.5 ) 2

y + 1.5 = ± 39.69 − ( x − 5.5 )

y = −1.5 ± 39.69 − ( x − 5.5 )

d. The graphs in (a) and (c) are the same.

284

Section 2.4

90. a. Completing the square yields ( x2 + 1.2x ) + ( y2 − 3.2y ) = −2.11

( x + 1.2x + 0.36 ) + ( y − 3.2y + 2.56 ) = −2.11+ 0.36 + 2.56 = 0.81 2

( x + 0.6 ) + ( y −1.6 ) = 0.81 (1) 2

Center (−0.6, 1.6), Radius 0.9 b. Solving (1) for y yields:

( y − 1.6 ) = 0.81 − ( x + 0.6 ) 2

y − 1.6 = ± 0.81 − ( x + 0.6 ) y = 1.6 ± 0.81 − ( x + 0.6 )

d. The graphs in (a) and (c) are the same.

285

Section 2.5 Solutions --------------------------------------------------------------------------------1. Negative linear association because the data closely cluster around what is reasonably described as a line with negative slope. 2. There is no identifiable direction of association, and the data are certainly not linear and there is not even a recognizable nonlinear curve that describes a pattern to which the data conform. 3. Although the data seem to be comprised of two linear segments, the overall data set cannot be described as having a positive or negative direction of association. Moreover, the pattern of the data is not linear, per se; rather, it is nonlinear and conforms to an identifiable curve (an upside down V called the absolute value function). 4. Positive nonlinear association because the data are rising from left to right and cluster relatively closely around what is reasonably described as a nonlinear curve. 5. B because the association is positive, thereby eliminating choices A and C. And, since the data are closely clustered around a linear curve, the bigger of the two correlation coefficients, 0.80, is more appropriate. 6. A because the association is negative, thereby eliminating choices B and D. And, since the data are closely clustered around a linear curve, the correlation coefficient closer to –1 is more appropriate. 7. C because the association is negative, thereby eliminating choices B and D. And, the data are more loosely clustered around a linear curve than are those pictured in #6. So, the correlation coefficient is the negative choice closer to 0. 8. D because there is no real identifiable association, so that the correlation coefficient closest to 0 is the most appropriate.

286

Section 2.5

9. a. 20 15 10 5 0 -4

-2

-5

-10 -15

b. The data seem to be nearly perfectly aligned to a line with negative slope. So, it is reasonable to guess that the correlation coefficient is very close to −1. c. The equation of the best fit line is y = −3x + 5 with a correlation coefficient of r = −1. d. There is a perfect negative linear association between x and y. 10. a. 10 5 0 -10

-5

-5 -10 -15 -20

b. The data seem to be nearly perfectly aligned to a line with positive slope. So, it is reasonable to guess that the correlation coefficient is very close to 1. c. The equation of the best fit line is y = 2x with a correlation coefficient of r = 1. d. There is a perfect positive linear relationship between x and y.

287

Chapter 2

11. a. 5 0 -20

-10

-5 -10 -15 -20

b. The data tend to fall from left to right, so that the correlation coefficient should be negative. Also, the data do not seem to stray too far from a linear curve, so the r value should be reasonably close to –1, but not equal to it. A reasonable guess would be around –0.90. c. The equation of the best fit line is approximately y = −0.5844x − 3.801 with a correlation coefficient of about r = 0.9833. d. There is a strong (but not perfect) negative linear relationship between x and y. 12. a. 15 10 5 0 -1.5

-1

-0.5

-5

0.5

1.5

-10 -15 -20

b. The data tend to rise from left to right, so that the correlation coefficient should be positive. Also, the data do not seem to stray too far from a linear curve, so the r value should be reasonably close to 1, but not equal to it. A reasonable guess would be around 0.90. c. The equation of the best fit line is approximately y = 15.717x − 0.2945 with a correlation coefficient of about r = 0.9569. d. There is a strong (but not perfect) positive linear relationship between x and y.

288

Section 2.5

13. a. 6 4 2 0 -4

-2

-4 -6 -8

b. The data seem to rise from left to right, but it is difficult to be certain about this relationship since the data stray considerably away from an identifiable line. As such, it is reasonable to guess that r is a rather small value close to 0, say around 0.30. c. The equation of the best fit line is approximately y = 0.5x + 0.5 with a correlation coefficient of about 0.349. d. There is a very loose (bordering on unidentifiable) positive linear relationship between x and y. 14. a. 6 4 2 0 -10

-5

-2

-4 -6 -8

b. It is very difficult to discern if there is a rising or falling trend in the data as you move from left to right, and it does not appear to be reasonably described by a line. As such, it is reasonable to guess that r is a rather small value close to 0, say around 0.25 or less. c. The equation of the best fit line is approximately y = −0.2256x – 0.2352 with a correlation coefficient of about 0.297. d. The relationship between x and y is not really discernible.

289

Chapter 2

15. a.

-10

10 8 6 4 2 0 -5 -2 0 -4 -6 -8 -10

The equation of the best fit line is about y = 0.7553x – 0.4392 with a correlation coefficient of about r = 0.868. b. The values x = 0 and x = −6 are within the range of the data set, so that using the best fit line for predictive purposes is reasonable. This is not the case for the values x = 12 and x = −15. The predicted value of y when x = 0 is approximately −0.4392, and the predicted value of y when x = −6 is −4.971. c. Solve the equation 2 = 0.7553x – 0.4392 for x to obtain: 2.4392 = 0.7553x so that x = 3.229. So, using the best fit line, you would expect to get a y-value of 2 when x is approximately 3.229.

290

Section 2.5

16. a. 20 18 16 14 12 10 8 6 4 2 0 0

The equation of the best fit line is about y = 2.9832x – 13.508 with a correlation coefficient of about r = 0.909. b. None of the given x-values are within reasonable enough range of the data set to ensure that the best fit line should be used for predictive purposes for them. c. Solve the equation 2 = 2.9832x – 13.508 for x to obtain: 15.508 = 2.9832x so that x = 5.198. So, using the best fit line, you would expect to get a y-value of 2 when x is approximately 5.198.

291

Chapter 2

17. a. 20 15 10 5 0 -25

-20

-15

-10

-5

-5 0

-10 -15 -20

The equation of the best fit line is about y = −1.2631x – 11.979 with a correlation coefficient of about r = −0.980. b. The values x = −15, −6, and 0 are within the range of the data set, so that using the best fit line for predictive purposes is reasonable. This is not the case for the value x = 12. The predicted value of y when x = −15 is approximately 6.9675, the predicted value of y when x = −6 is about −4.4004, and the predicted value of y when x = 0 is −11.979. c. Solve the equation 2 = −1.2631x – 11.979 for x to obtain: 13.979 = −1.2631x so that x = −11.067. So, using the best fit line, you would expect to get a y-value of 2 when x is approximately −11.067.

292

Section 2.5

18. a. 20 15 10 5 0 -20

-10

-5

The equation of the best fit line is about y = −0.4077x + 2.9457 with a correlation coefficient of about r = −0.715. b. All of the given x-values are within reasonable range of the data set to be able to obtain a reasonable predicted y-value for each of them. The predicted value of y when x = −15 is about 9.0612, the predicted value of y when x = −6 is about 5.3919, the predicted value of y when x = 0 is 2.9457, and the predicted value of y when x = 12 is about −1.9467. c. Solve the equation 2 = −0.4077x + 2.9457 for x to obtain: −0.9457 = −0.4077x so that x = 2.3196. So, using the best fit line, you would expect to get a y-value of 2 when x is approximately 2.3196.

293

Chapter 2

19. a. The scatterplot for the entire data set is: 20 15 10 5 0 -4

-2

-5

-10 -15

The equation of the best fit line is y = −3x + 5 with a correlation coefficient of r = −1. b. The scatterplot for the data set obtained by removing the starred data point (5, −10) is:

-4

-2

16 14 12 10 8 6 4 2 0 -2 0 -4 -6

The equation of the best fit line of this modified data set is y = −3x + 5 with a correlation coefficient of r = −1. c. Removing the data point did not result in the slightest change in either the equation of the best fit line or the correlation coefficient. This is reasonable since the relationship between x and y in the original data set is perfectly linear, so that all of the points lie ON the same line. As such, removing one of them has no effect on the line itself.

294

Section 2.5

20. a. The scatterplot for the entire data set is:

-20

-10

4 2 0 -2 0 -4 -6 -8 -10 -12 -14 -16 -18

The equation of the best fit line is y = −0.5844x – 3.801 with a correlation coefficient of about r = −0.983. b. The scatterplot for the data set obtained by removing the starred data point (20, −16) is: 4 2 0 -20

-10

-2 0

-4 -6 -8 -10 -12 -14

The equation of the best fit line of this modified data set is y = −0.5597x − 3.7284 with a correlation coefficient of r = −0.971. c. Removing the data point did change both the best fit line and the correlation coefficient, but only very slightly.

295

Chapter 2

21. a. The scatterplot for the entire data set is: 20 15 10 5 0 -4

-2

-5

-10 -15 -20

The equation of the best fit line is y = −3.4776x + 4.6076 with a correlation coefficient of about r = −0.993. b. The scatterplot for the data set obtained by removing the starred data points (3, −4) and (6, −16) is: 20 15 10 5 0 -4

-2

-5

-10 -15 -20

The equation of the best fit line of this modified data set is y = −3.6534x + 4.2614 with a correlation coefficient of r = −0.995. c. Removing the data point did change both the best fit line and the correlation coefficient, but only very slightly.

296

Section 2.5

22. a. The scatterplot for the entire data set is: 70 60 50 40 30 20 10 0 -10

The equation of the best fit line is y = 3.0362x – 1.833 with a correlation coefficient of about r = 0.925. b. The scatterplot for the data set obtained by removing the starred data point (7, 25) is: 14 12 10 8 6 4 2 0 0

The equation of the best fit line of this modified data set is y = 2x with a correlation coefficient of r = 1. c. Removing the data point resulted in significant changes in both the best fit line and the correlation coefficient.

297

Chapter 2

23. a. 5 0 -20

-10

-5 -10 -15 -20 -25

b. The correlation coefficient is approximately r = −0.980. This is identical to the r-value from Problem 17. This makes sense because simply interchanging the x- and y-values does not change how the points cluster together in the xy-plane. c. The equation of the best fit line for the paired data (y, x) is x = −0.7607y – 9.4957. d. It is not reasonable to use the best fit line in (c) to find the predicted value of x when y = 23 because this value falls outside the range of the given data. However, it is okay to use the best fit line to find the predicted values of x when y = 2 or y = −16. Indeed, the predicted value of x when y = 2 is about −11.0171, and the predicted value of x when y = −16 is about 2.6755.

298

Section 2.5

24. a. 12 10 8 6 4 2 0 0

b. The correlation coefficient is approximately 0.909. This is identical to the rvalue from Problem 16. This makes sense because simply interchanging the x- and y-values does not change how the points cluster together in the xy-plane. c. The equation of the best fit line for the paired data (y, x) is x = 0.2773y + 5.0654. d. The only y-value for which it is reasonable to use the best fit line in (c) for predictive purposes is y = 2 because the other two values lie outside the range of values present in the data set. The predicted value of x when y = 2 is about x = 5.62.

299

Chapter 2

25. First, note that the scatterplot is given by 20 15 10 5 0 0

-5 -10

The paired data all lie identically on the vertical line x = 3. As such, you might think that the square of the correlation coefficient would be 1 and the best fit line is, in fact, x = 3. However, since there is absolutely no variation in the x-values for this data set, it turns out that in the formula for the correlation coefficient r=

the quantity

n  xy − (  x )(  y ) n x 2 − (  x ) ⋅ n y 2 − (  y ) 2

n x 2 − (  x ) turns out to be zero. (Check this on Excel for this 2

data set!) As such, there is no meaningful r-value for this data set. Also, the best fit line is definitely the vertical line x = 3, but the technology cannot provide it because its slope is undefined.

300

Section 2.5

26. First, note that the scatterplot is given by 0 -10

-5

-0.5 -1 -1.5 -2 -2.5

The paired data all lie identically on the horizontal line y = −2. As such, you might think that the square of the correlation coefficient would be 1 and the best fit line is, in fact, y = −2. However, since there is absolutely no variation in the y-values for this data set, it turns out that in the formula for the correlation coefficient r=

the quantity

n  xy − (  x )(  y ) n x 2 − (  x ) ⋅ n  y 2 − (  y ) 2

n y2 − (  y ) turns out to be zero. (Check this on Excel 2

for this data set!) As such, there is no meaningful r-value for this data set. However, this time, the technology is able to produce the equation of the best fit line (unlike in Problem 25) because the slope is defined – it is simply 0. 27. The y-intercept 1.257 is mistakenly interpreted as the slope. The correct interpretation is that for every unit increase in x, the y-value increases by about 5.175. 28. Since the slope of the best fit line will be negative, the correlation coefficient r should be negative as well. As such, r should equal −0.9913, not 0.9913.

301

Chapter 2

29. a. Here is a table listing all of the correlation coefficients between each of the events and the total points: r −0.567 0.842 0.227 0.587 −0.715 −0.448 0.173 0.446 0.542

Event 100 m Long Jump Shot Put High Jump 400 m 110 m Hurdle Discus Pole Vault Javelin

b. Long jump has the strongest correlation coefficient, r = 0.842. c. The equation of the best fit line between the two events in (b) is y = 939.98x + 1261.2. d. Evaluate the equation in (c) at x = 40 to get the total points are about 38,860. 30. Here is a table listing all of the correlation coefficients between each of the events and the total points: r −0.567 0.842 0.227 0.587 −0.715 −0.448 0.173 0.446 0.542

Event 100 m Long Jump Shot Put High Jump 400 m 110 m Hurdle Discus Pole Vault Javelin

a. The 400 m event has the second strongest relationship to the total points, r = −0.715. b. The equation of the best fit line between the two events in (a) is y = −232.03x + 19473.1. c. Since r = −0.715 indicates a reasonably strong relationship between 400 m and total points, we can make a reasonably accurate prediction. However, it will not be completely accurate. d. Evaluate the equation in (c) at x = 40 to get the total points are about 10,192.

302

Section 2.5

31. a. A scatterplot illustrating the relationship between % residents immunized and % residents with influenza is shown below.

b. The correlation coefficient between % residents immunized and % residents with influenza is r = −0.812. c. Based on the correlation coefficient (r = −0.812), we would believe that there is a strong relationship between % residents immunized and % residents with influenza. d. The equation of the best fit line that describes the relationship between % residents immunized and % residents with influenza is y = −0.27x + 32.20. e. Since r = −0.812 indicates a strong relationship between % residents immunized and % residents with influenza, we can make a reasonably accurate prediction. However it will not be completely accurate.

303

Chapter 2

32. a. The nursing home number of the outlier is 21. b. The scatterplot illustrating the relationship between % residents immunized and % residents with influenza WITH THE OUTLIER REMOVED is shown below.

c. The revised correlation coefficient between % residents immunized and % residents with influenza is r = −0.938. d. By removing the outlier, the correlation coefficient goes from −0.812 to −0.938 and as such, the strength of the relationship between % residents immunized and % residents with influenza is increased. e. The revised equation of the best fit line that describes the relationship between % residents immunized and % residents with influenza is y = −0.31x + 33.54.

304

Section 2.5

33. a. A scatterplot illustrating the relationship between average wait times and average rating of enjoyment is shown below.

b. The correlation coefficient between average wait times and average rating of enjoyment is r = 0.348. c. The correlation coefficient (r = 0.348) indicates a somewhat weak relationship between average wait times and average rating of enjoyment. d. The equation of the best fit line that describes the relationship between average wait times and average rating of enjoyment is y = 0.31x + 37.83. e. No, given the somewhat weak correlation coefficient between average wait times and average rating of enjoyment, one could not use the best fit line to produce accurate predictions.

305

Chapter 2

34. a. A scatterplot illustrating the relationship between average wait times and average rating of enjoyment for Park 1 is shown below.

b. The correlation coefficient between average wait times and average rating of enjoyment is r = 0.767. c. The correlation coefficient (r = 0.767) indicates a strong relationship between average wait times and average rating of enjoyment for Park 1. d. The equation of the best fit line that describes the relationship between average wait times and average rating of enjoyment is y = 0.71x + 22.91. e. Yes, given the strong correlation coefficient between average wait times and average rating of enjoyment for Park 1, one could use the best fit line to produce accurate predictions.

306

Section 2.5

35. a. A scatterplot illustrating the relationship between average wait times and average rating of enjoyment for Park 2 is shown below.

b. The correlation coefficient between average wait times and average rating of enjoyment is r = −0.064. c. The correlation coefficient (r = −0.064) indicates a weak relationship between average wait times and average rating of enjoyment for Park 2. d. The equation of the best fit line that describes the relationship between average wait times and average rating of enjoyment is y = −0.05x + 52.53. e. No, given the weak correlation coefficient between average wait times and average rating of enjoyment for Park 2, one could not use the best fit line to produce accurate predictions.

307

Chapter 2

36. Consider the combined scatterplot with best fit lines for Park 1 and Park 2 below.

As can be seen from the scatterplot and from reviewing the correlation coefficients, the relationship between average wait times and average rating of enjoyment for Park 1 in Florida (r = 0.767) is quite different from this relationship for Park 2 in California (r = −0.064). Specifically relationship between average wait times and average rating of enjoyment for Park 1 appears to be much stronger than the same relationship for Park 2.

308

Section 2.5

37. a. The scatterplot for this data set is given by 30 25 20 15 10 5 0 0

b. The equation of the best fit line is y = −1.9867x + 27.211 with a correlation coefficient of r = −0.671. This line does not seem to accurately describe the data because some of the points rise as you move left to right, while others fall as you move left to right; a line cannot capture both types of behavior simultaneously. Also, r being negative has no meaning here. c. The best fit is provided by QuadReg. The associated equation of the best fit quadratic curve, the correlation coefficient, and associated scatterplot are:

309

Chapter 2

38. a. The scatterplot for this data set is given by 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0

0.5

1.5

2.5

3.5

b. The equation of the best fit line is y = −0.3733x + 1.1647 with a correlation coefficient of r = −0.923. This line seems to provide a reasonably good fit for this data, although not perfect. c. The best fit is provided by ExpReg. The associated equation of the best fit exponential curve, the correlation coefficient, and associated scatterplot are:

310

Section 2.5

39. a. The scatterplot for this data set is given by

b. The equation of the best fit line is y = 0.3537x + 0.5593 with a correlation coefficient of r = 0.971. This line seems to provide a very good fit for this data, although not perfect. c. The best fit is provided by LnReg. The associated equation of the best fit logarithmic curve, the correlation coefficient, and associated scatterplot are:

311

Chapter 2

40. a. The scatterplot for this data set is given by 80 60 40 20 0 -20

-40 -60

b. The equation of the best fit line is y = 2.9237x – 5.2956 with a correlation coefficient of r = 0.206. This line seems to provide a terrible fit for the data. c. The best fit is provided by CubicReg. The associated equation of the best fit cubic polynomial curve, the correlation coefficient, and associated scatterplot are:

312

Chapter 2 Review Solutions ----------------------------------------------------------------------1 – 4.

( −4,2) is in Quadrant II. (4,7) is in Quadrant I. ( −1, −6 ) is in Quadrant III. (2, −1) is in Quadrant IV.

6. ( −2 − 4) + (0 − 3) = 2

(1 − 4)2 + (4 − 4)2 = 3

45 = 3 5

7. ( −4 − 2)2 + ( −6 − 7)2 =

205

( 14 − 13 ) + ( 121 − ( − 73 ) ) = 2

( − 121 ) + ( 1229 ) = 2

842 ≅ 2.418 12

9.  2+3 4 +8  5 ,   = ( 2 ,6 ) 2 2  

10.  −2 + 5 6 + 7  3 13 ,   = (2, 2 ) 2 2  

11.  2.3 + 5.4 3.4 + 7.2  ,  = ( 3.85,5.3 )  2 2  

12.  −a + a 2 + 4  ,  = (0,3)  2   2

13. The distance equals

14. The midpoint is  −5 + 10 −20 + 30  5 ,  = ( 2 ,5 ) .  2   2

( −5 − 10) + ( −20 − 30) = 2725 2

≅ 52.20

313

Chapter 2

15. x-intercepts: x 2 + 4(0)2 = 4  x = ±2 So, ( ±2,0 ) .

16. x-intercepts:

y-intercepts: 0 2 + 4 y 2 = 4  y = ±1 So, ( 0, ±1) .

17. x − 9 = 0  x = ±3 2

x-intercepts: So, ( ±3,0 ) .

1 ± 1 − 4(2) 2 1± i 7 = 2 Since these solutions are not real, there are no x-intercepts. y-intercept: y = 0 2 − 0 + 2  y = 2 So, ( 0,2 ) .

x2 − x + 2 = 0  x =

y-intercepts: 3i = 0 2 − 9 = y . Since this is not real, there is no y-intercept.

18. x-intercepts: x 2 − x −12 ( x − 4)( x + 3) = =0 x − 12 x − 12  x = −3, 4 So, ( −3,0 ) , ( 4,0 ) . 0 2 − 0 − 12 y-intercepts: y = = 1. 0 − 12 So, ( 0,1) .

19. x-axis symmetry (Replace y by −y ):

y-axis symmetry (Replace x by −x ):

x + (− y) = 4

x 2 + y3 = 4 Yes, since equivalent to the original.

x −y =4 No, since not equivalent to the original. 2

( − x )2 + y 3 = 4

symmetry about origin (Replace y by −y and x by −x ): ( − x )2 + (− y )3 = 4

x 2 − y3 = 4 No, since not equivalent to the original.

314

Chapter 2 Review y-axis symmetry (Replace x by −x):

20. x-axis symmetry (Replace y by −y ):

y = ( − x )2 − 2

−y = x − 2 2

y = −x 2 + 2 No, since not equivalent to the original.

y = x2 + 2 Yes, since equivalent to the original.

symmetry about origin (Replace y by −y and x by −x ): − y = ( − x )2 − 2 = x 2 − 2

y = −x 2 + 2 No, since not equivalent to the original.

21. x-axis symmetry (Replace y by − y ): x( − y ) = 4

−xy = 4 No, since not equivalent to the original.

y-axis symmetry (Replace x by −x ): ( −x ) y = 4 −xy = 4 No, since not equivalent to the original.

symmetry about origin (Replace y by −y and x by −x): ( −x )( − y ) = 4 xy = 4 Yes, since equivalent to the original. 22. x-axis symmetry (Replace y by − y ):

y-axis symmetry (Replace x by −x):

( − y )2 = 5 + x

y2 = 5 − x No, since not equivalent to the original.

y2 = 5 + x Yes, since equivalent to the original.

y 2 = 5 + (−x )

symmetry about origin (Replace y by −y and x by −x ): ( − y )2 = 5 + ( − x )

y2 = 5 − x No, since not equivalent to the original.

315

Chapter 2

23.

24.

25.

26.

316

Chapter 2 Review 27.

28.

29.

30.

31. y = −3x + 6 m = −3 y-intercept: (0,6) 32. y = − 34 x + 94 m = − 34 y-intercept: (0, 94 ) 33. y = − 32 x − 12

m = − 23 y-intercept: (0, − 21 )

34. y = − 83 x − 21 m = − 83 y-intercept: (0, − 21 )

317

Chapter 2

35.

x-intercept: 0 = 4x − 5 So, ( 54 ,0 ) . 5 x = 4 y-intercept: y = 4(0) − 5 = −5 . So, ( 0, −5 ) . slope: m = 4

36.

x-intercept: 0 = − 34 x − 3

So, ( −4,0 ) . −4 = x y-intercept: y = − 34 (0) − 3 = −3 . So,

( 0, −3) . slope: m = − 34

37.

x-intercept: x + 0 = 4 So, ( 4,0 ) .

y-intercept: 0 + y = 4 . So, ( 0, 4 ) . slope: x+y=4 So, m = −1 . y = −x + 4

318

Chapter 2 Review 38.

x-intercept: ( −4,0 ) y-intercept: None slope: Undefined

39.

x-intercept: None y-intercept: ( 0,2 ) slope: m = 0

40. x-intercept: − 12 x − 12 (0) = 3

x = −6 y-intercept: − 12 (0) − 12 y = 3 y = −6

So, ( −6,0 ) . So, ( 0, −6 ) .

slope: − 12 x − 12 y = 3 − 12 y = 12 x + 3 So, m = −1 . y = −x − 6

319

Chapter 2

41. y = 4 x − 3

42. y = 4

43. x = −3

44. y = − 23 x + 34 4 = −2(−3) + b

45. Find b:

−2 = b So, the equation is y = −2 x − 2 .

46. Find b:

16 = 34 (2) + b 29 2

So, the equation is y = 34 x + 292 .

47. Since m = 0 , the line is horizontal. So, the equation is y = 6 .

48. Since m is undefined, the line is vertical. So, the equation is x = 2 .

49.

50.

−2 − 3 5 slope: m = = −4 − 2 6 y-intercept: Use the point ( −4, −2 ) to find b: −2 = 56 ( −4) + b 4 3

slope: m =

4 −5 = −1 −1 − ( −2)

y-intercept: Use the point ( −1, 4 ) to find b: 4 = ( −1)(−1) + b

So, the equation is y = 56 x + 34 .

3=b So, the equation is y = − x + 3 .

51.

52.

slope: m =

− = −2 − − ( − 74 ) 1 2

5 2

3 4

y-intercept: Use the point ( − 34 , 21 ) to find b: 1 2

slope: m =

y-intercept: Use the point ( 3, −2 ) to find b: −2 = ( − 13 ) (3) + b

= −2 ( − 34 ) + b

−1 = b So, the equation is y = −2 x − 1 .

−2 − 2 1 =− 3 − ( −9) 3

−1 = b So, the equation is y = − 13 x − 1 .

320

Chapter 2 Review 53. slope: Since parallel to 2x − 3 y = 6 (whose slope is 23 ), m = 23 . y-intercept: Use the point ( −2, −1) to find b:

54. slope: Since perpendicular to 5x − 3 y = 0 (whose slope is 53 ), m = − 53 . y-intercept: Use the point ( 5,6 ) to find b:

−1 = 23 ( −2) + b 1 3

6 = − 53 (5) + b

9=b So, the equation is y = − 35 x + 9 .

So, the equation is y = 23 x + 13 .

55. slope: Since perpendicular to 23 x − 12 y = 12 (whose slope is 34 ), m = − 34 . y-intercept: Use the point ( − 34 , 52 ) to find b: 5 2

= − 34 ( − 34 ) + b

31 16

So, the equation is y = − 34 x + 1316 . 56. Case 1: B ≠ 0 Since the desired line is parallel to Ax + By = C (whose slope is − BA ), its slope is − BA . So, its equation is of the form y = − BA x + D . Use the fact that the point ( a + 2, b − 1) is on the line to see that b − 1 = − BA ( a + 2 ) + D , which is equivalent to D = b − 1 + BA ( a + 2) . Thus, the equation is y = − BA x + ( b − 1 + BA ( a + 2) ) .

Case 2: B = 0 The desired line is parallel to Ax = C , which is vertical. Since the line whose equation we seek passes through ( a + 2, b − 1) , its equation must be x = a + 2 .

321

Chapter 2

57. Since (1020,1324) and (950,1240) are assumed to be on the line, the slope of 1324 −1240 the line is m = = 1.2 . Use the point (1020,1324) to find y-intercept b: 1020 − 950 1324 = 1.2(1020) + b 100 = b Thus, the equation is y = 1.2 x + 100 , where x is the pretest score and y is the

posttest score. 58. Let C = 250 + 38t . If t = 1.5 , then C = 307 . So, it costs $307 for a job taking 1.5 hours.

59. 2 2 ( x − ( −2) ) + ( y − 3 ) = 62

( x + 2 ) + ( y − 3 ) = 36 2

( x − ( −6) ) + ( y − ( −8) ) = (3 6 ) 2

61. 2 2 ( x − 34 ) + ( y − 52 ) = 254 63. The center is ( −2, −3 ) and the

radius is

60.

81 = 9 .

( x + 6 ) + ( y + 8) = 54 2

62. 2 2 2 ( x − 1.2 ) + ( y − (−2.4) ) = (3.6 )

( x − 1.2 ) + ( y + 2.4 ) = 12.96 2

64. The center is ( 4, −2 ) and the radius is 32 = 4 2 .

322

Chapter 2 Review 66. Completing the square gives us: x2 + y2 + 4x − 2 y = 0

65. The center is ( − 34 , 21 ) and the radius is 16 36

x 2 + 4x

= 64 = 23 .

+ y2 − 2 y = 0

( x + 4x + 4 ) + ( y − 2 y +1) = 4 +1 2

( x + 2 ) + ( y − 1) = 5 2

( x − ( −2) ) + ( y −1) = ( 5 ) 2

So, center is ( −2,1) and radius is 67. Completing the square gives us: x 2 + y 2 − 4 x + 2 y + 11 = 0 x2 − 4x

68. Completing the square gives us: 3x 2 + 3 y 2 − 6 x − 7 = 0

+ y 2 + 2 y = −11

( x − 4x + 4 ) + ( y + 2 y + 1) = −11 + 4 + 1 2

3x 2 − 6 x

+ 3 y2 = 7

x2 − 2x

+ y 2 = 73

( x − 2x + 1) + ( y + 0 ) = +1 2

( x − 2 ) + ( y + 1) = −6 2

Since the left-side is always nonnegative, there is no graph. Hence, this is not a circle and so there is no center or radius in this case.

69. Completing the square gives us: 9x 2 + 9 y2 − 6x +12y − 76 = 0

x 2 − 23 x

+ y2 + 34 y = 769

(x − x + ) + ( y + y + ) = 2 3

1 9

7 3

( x − 1) + ( y − 0 ) = 103 So, center is (1,0 ) and radius is 103 . 2

4 3

4 9

76 9

( x − 13 ) + ( y + 23 ) = 9 2 2 ( x − 13 ) + ( y − ( − 23 ) ) = 9 2

So, center is ( 13 , − 23 ) and radius is 3.

323

+ 91 + 94

Chapter 2

70. Completing the square gives us: x 2 + y 2 + 3.2 x − 6.6 y − 2.4 = 0 x 2 + 3.2 x

+ y 2 − 6.6 y = 2.4

( x + 3.2x + 2.56 ) + ( y − 6.6 y + 10.89 ) = 2.4 + 2.56 + 10.89 2

( x + 1.6 ) + ( y − 3.3 ) = 15.85 2

So, center is ( −1.6,3.3 ) and radius is 15.85 ≅ 3.98 . 71. The center is ( 2,7 ) . The radius is the distance from ( 2,7 ) to ( 3,6 ) , which is

given by

( 2 − 3) + ( 7 − 6 ) = 2. So, the equation is ( x − 2 ) + ( y − 7 ) = 2 . 2

72. The center is the midpoint of the segment joining the points ( −2, −1) and  −2 + 5 −1+ 5  3 , (5,5), which is given by   = ( ,2 ) . The radius is equal to the 2  2  2 distance between the center and an endpoint of a diameter, say (−2, −1) . This is given by

( 32 − ( −2) ) + ( 2 − ( −1) ) = ( 72 ) + 32 = 12 85. 2

Therefore, the equation is

( x − 32 ) + ( y − 2 ) = 854 = 21.25 2

324

Chapter 2 Review 73. The triangle is depicted as follows:

Observe that 10 − ( −5) 3 = 10 − ( −10) 4 10 − ( −45) 11 mBC = =− 10 − 20 2 4 −5 − (−45) =− mAC = 3 −10 − 20 Since mAB ⋅ mAC = −1 , we know that AB is perpendicular to AC. Thus, the triangle is right. mAB =

Next, observe that d ( A, B) =

(10 − ( −10) ) + (10 − ( −5) ) = 25

d ( A,C ) =

( 20 − ( −10) ) + ( −45 − ( −5) ) = 50

d ( B,C ) =

(10 − ( −45) ) + (10 − 20 ) ≈ 55.9

So, it is not isosceles.

325

Chapter 2

74. The triangle is depicted as follows:

Observe that 8.4 − 2.1 = 0.75 4.2 − ( −4.2) 2.1− ( −10.5) mAC = = −1.2 −4.2 − 6.3 8.4 − ( −10.5) = −9 mBC = 4.2 − 6.3 No two sides are perpendicular, so the triangle is not right. mAB =

Next, observe that d ( A, B) =

(8.4 − 2.1) + ( 4.2 − ( −4.2) ) ≈ 10.5

d ( B,C ) =

( 8.4 − ( −10.5 ) + ( 4.2 − 6.3) ≈ 19.02

d ( A,C ) =

( 2.1 − ( −10.5) ) + ( −4.2 − 6.3) ≈ 16.4

So, it is not isosceles. Hence, the triangle is neither.

326

Chapter 2 Review 75.

76.

Symmetric with respect to x-axis, yaxis, and origin

Symmetric with respect to x-axis, y-axis, and origin

77. Slope of y1 = 0.875x + 1.5 is 0.875. Slope of y2 = − 78 x − 194 is − 78 . Since (0.875)( − 78 ) = −1, the lines are perpendicular.

78. Slope of y1 = −0.45x − 2.1 is −0.45. Slope of y2 = 56 − 209 x is − 209 . Since −0.45 = − 209 , the lines are parallel.

327

Chapter 2

79. Completing the square yields ( 9x2 − 6x ) + ( 9y2 +12y ) = 76 9 ( x 2 − 23 x ) + 9 ( y 2 + 34 y ) = 76

9 ( x 2 − 23 x + 19 ) + ( y 2 + 34 y + 94 ) = 76 + 1 + 4 = 81

( x − 13 ) + ( y + 23 ) = 9 2

(1)

Solving (1) for y yields:

( x − 13 ) + ( y + 23 ) = 9 2 2 ( y + 23 ) = 9 − ( x − 13 ) 2

y + 23 = ± 9 − ( x − 13 )

y = − 23 ± 9 − ( x − 13 )

The graph agrees with the graph in Exercise 69.

80. Completing the square yields ( x2 + 3.2x ) + ( y2 − 6.6y ) = 2.4

( x + 3.2x + 2.56 ) + ( y − 6.6y +10.89 ) = 2.4 + 2.56 +10.89 = 15.85 2

( x + 1.6 ) + ( y − 3.3) = 15.85 (1) 2

Solving (1) for y yields: 2 2 ( y − 3.3) = 15.85 − ( x +1.6 ) y − 3.3 = ± 15.85 − ( x +1.6 ) y = 3.3 ± 15.85 − ( x +1.6 )

The graph agrees with the graph in Exercise 70.

328

Chapter 2 Practice Test Solutions ---------------------------------------------------------------1.

d = ( −7 − 2)2 + ( −3 − ( −2 ))2 =

 −3 + 5 5 + (−1)  M = ,  = (1,2 ) 2   2

d = ( −2 − 3)2 + (4 − 6)2 =

 −2 + 3 4 + 6  1 , M =  = ( 2 ,5) 2 2   4. Suppose the coordinates of Chapel Hill are (0,0). Then, Durham is located at (8,10) and Raleigh is located at (28, −15) , as shown:

d (C ,R) = 282 + (−15)2 ≅ 31.76 d (C ,D) = 82 + 102 ≅ 12.81 d ( D,R) = (28 − 8)2 + (−15 −10)2 ≅ 32.02 So, the perimeter of the research triangle is approximately 77 miles .

5. Solve for y:

6. ( −3, −4)

5 = (3 − 6)2 + ( y − 5 )2 25 = 9 + ( y − 5)2 25 = 9 + y 2 −10 y + 25 0 = y 2 − 10 y + 9 0 = ( y − 9)( y −1) 1, 9 = y

329

Chapter 2

7. x-axis symmetry (Replace y by −y): x − (− y) = 5 2

x − y2 = 5 Yes, since equivalent to the original.

y-axis symmetry (Replace x by −x): −x − y2 = 5 No, since not equivalent to the original.

symmetry about origin (Replace y by −y and x by −x): − x − ( − y )2 = 5 −x − y2 = 5 No, since not equivalent to the original.

8. x-intercepts: 4 x 2 − 9(0)2 = 36  x 2 = 9  x = ±3 So, ( ±3,0 ) .

y-intercepts: 4(0)2 − 9y2 = 36, which has no real solution. So, no y-intercepts. 9.

10.

11. x-intercept: x − 3(0) = 6  x = 6. So, (6,0). y-intercept: 0 − 3y = 6  y = −2. So, (0, −2).

12.

13.

14. y = 4x + 3

2 3

x− y =2  − y =2− x 1 4

1 4

4x − 6 y = 12  − 6y = −4x +12

 y = 23 x − 2

2 3

 y = 83 x − 8

330

Chapter 2 Practice Test 15. 9−2 =1 4 − ( −3) y-intercept: Use the point ( −3,2 ) to

slope: m =

find b: 2 = 1( −3) + b 5=b So, the equation is y = x + 5 .

16. slope: Since parallel to y = 4x + 3 (whose slope is 4), m = 4 . y-intercept: Use the point (1,7 ) to find b: 7 = 4(1) + b 3=b So, the equation is y = 4 x + 3 .

331

Chapter 2

17. slope: Since perpendicular to 2 x − 4 y = 5 (whose slope is 12 ), m = −2 . y-intercept: Use the point (1,1) to find b: 1 = −2(1) + b 3=b So, the equation is y = −2x + 3 .

19. 4 − ( −2) slope: m = =2 1 − ( −2) y-intercept: Use the point (1, 4 ) to

find b: 4 = 2(1) + b 2=b So, the equation is y = 2 x + 2 .

18.

slope: m =

6−0 = −2 0 −3

y-intercept: Since (0,6) is on the line and is the y-intercept, b = 6. So, the equation is y = −2x + 6 .

20.

3 − ( −1) = −1 −2 − 2 y-intercept: Use the point ( −2,3 ) to find b: slope: m =

3 = ( −1)(−2) + b 1= b So, the equation is y = − x +1.

332

Chapter 2 Practice Test 22. Completing the square gives us: x 2 + y 2 −10x + 6 y + 22 = 0

21. 2 2 ( x − 6 ) + ( y − ( −7) ) = 82

x 2 − 10 x

( x − 6 ) + ( y + 7)2 = 64 2

+ y 2 + 6y = −22

( x − 10x + 25 ) + ( y + 6y + 9 ) = −22 + 25 + 9 2

( x − 5 ) + ( y + 3) = 12 2

( x − 5 ) + ( y − ( −3) ) = ( 12 ) So, center is ( 5, −3 ) and radius is 2

12 = 2 3 .

23. The center is ( 4,9 ) . The radius is the distance from ( 4,9 ) to ( 2,5 ) , which is

given by

( 4 − 2 ) + ( 9 − 5 ) = 20 . So, the equation is ( x − 4 ) + ( y − 9 ) = 20 . 2

24. The center is ( 0,0 ) and the radius is 93 million miles. Then, assuming that x

and y are measured in “millions of miles”, the equation is 2 2 ( x − 0 ) + ( y − 0 ) = 932 x 2 + y2 = 8649 .

333

Chapter 2

25. The triangle is depicted as follows:

Observe that 37.8 −16.8 = 2.5 0 − ( −8.4) 37.8 − 8.4 = −2.3 mBC = 0 −12.6 16.8 − 8.4 = −0.4 mAC = −8.4 − 12.6 Since mAB ⋅ mAC = −1 , AB is perpendicular to AC. So, the triangle is right. mAB =

Next, observe that d ( A, B) =

(37.8 − 16.8) + ( 0 + 8.4 ) 2

(37.8 − 8.4 ) + ( 0 − 12.6 )

= 511.56 d ( B,C ) =

= 1023.12 d ( A,C ) =

(16.8 − 8.4 ) + ( −8.4 − 12.6 ) 2

= 511.56 So, it is also isosceles. 26. Symmetric with respect to the origin, x-axis, and y-axis, as seen below.

334

Chapter 2 Cumulative Review -------------------------------------------------------------------1. 7−2 5 1 = = 7 + 3 10 2

(5x ) = 625x = 25x 3

25x

3. ( x − 4 ) ( x + 4) = [( x − 4 )(x + 4)] 2

=  x 2 − 16 

− 14

25x

− 14

4. 8x3 − 27 y3 = (2x)3 − (3y)3

= ( 2 x − 3 y ) ( 4x 2 + 6xy + 9y 2 )

= x 4 − 32 x 2 + 256 5. 1 x 1 x

− = + 1 5 1 5

5−x 5x 5+x 5x

x3 − 5x 2 − 4x + 20 = 0

5−x 5+x

x2 ( x − 5 ) − 4 ( x − 5) = 0

( x − 4 ) ( x − 5) = 0 2

( x − 2 )( x + 2)(x − 5) = 0 x = ±2,5 7. 8.

−36 ( 5 − 2i ) = 6i ( 5 − 2i ) = 30i − 12i 2 = 12 + 30i 15 − [5 + 3x − 4(2x − 6)] = 4(6x − 7) − [5(3x − 7) − 6x +10] 15 − [5 + 3x − 8x + 24] = 24x − 28 − [15x − 35 − 6x +10] 15 − (29 − 5x ) = 24 x − 28 − (9x − 25) 15 − 29 + 5x = 24 x − 28 − 9x + 25 −14 + 5x = 15x − 3 −11 = 10x 11 − 10 =x

335

Chapter 2

9. 5 3 5(4x − 1) − 3(4 x +1) 8x − 8 − =0  =0  =0 4x +1 4x −1 (4x − 1)(4x + 1) (4x − 1)(4x + 1) 8(x −1)  =0 (4x − 1)(4 x + 1)

So, x = 1. 10. Let x = amount invested in 5% CD. Solve: 0.05x + 0.08(17000 − x ) = 1075 0.05x + 1360 − 0.08x = 1075

11. 5 x 2 − 45 = 0

5 ( x2 − 9) = 0 5(x − 3)( x + 3) = 0

−0.03x = −285

x = ±3

x = 9500 $9500 was invested in CD and $7500 in

stock. 12.

13. Discriminant is equal to 4 – 4(5)(7) = −136 < 0. Hence, the equation has two complex conjugate solutions.

3x 2 + 6 x − 7 = 0

3 ( x 2 + 2x ) − 7 = 0

3 ( x 2 + 2 x + 1) − 7 − 3 = 0 3(x + 1)2 = 10 ( x + 1)2 = 103 x = −1 ±

14. r = ± p2 + q 2

10 3

15. x 2 + 3x − 10 = x − 2 x 2 + 3x − 10 = (x − 2)2 x 2 + 3x − 10 = x 2 − 4x + 4 7 x − 14 = 0 x=2

336

Chapter 2 Cumulative Review 16.

17. 6 < 14 x + 6 < 9

1 5 − + 4 = 0, x ≠ −2 2 (x + 2) x + 2

0 < 14 x < 3

1 − 5(x + 2) + 4( x + 2)2 =0 ( x + 2 )2

0 < x < 12 (0,12)

4 x + 11x + 7 =0 ( x + 2 )2 (4x + 7)( x + 1) =0 ( x + 2 )2 2

x = − 74 , − 1

18.

19.

−4 < 2 − x < 4 −6 < −x < 2 6 > x > −2

x − x ≥ 20 2

x 2 − x − 20 ≥ 0 ( x − 5 )( x + 4) ≥ 0 CPs: x = −4,5 + − + | | −4

( −2,6) 20.

So, the solution set is ( −∞, −4 ] ∪ [5, ∞ )

5 − 4x = 23 or 5 − 4 x = −23 −4x = 18

− 4 x = −28

x = − 92

x=7

21. x-axis symmetry: Substitute −y for y. This yields y = −4x , which is not equivalent to the original equation. Hence, not symmetric about x-axis. y-axis symmetry: Substitute −x for x. This yields y = −4x , which is not equivalent to the original equation. Hence, not symmetric about y-axis. origin symmetry: Substitute −y for y and −x for x. This yields y = 4x , which is equivalent to the original equation. Hence, symmetric about origin. 22. m = 54 y-intercept: To find b, use the fact that (5,1) is on the line: 1 = 54 (5) + b  b = −3

23. Since the line passes through (5,3) and is perpendicular to the x-axis, the line must be vertical and so, its equation is x = 5 .

So, equation of the line is y = 54 x − 3 .

337

Chapter 2

24. 5 3

− (− ) 2 3

7 3 7 − (− ) y-intercept: To find b, use the fact that ( 71 , 53 ) is on the line: m= 1

6 7

5 3

= 73 ( 71 ) + b  b = 34

25. The equation of the circle can be written as

( x − ( −5) ) + ( y − ( −3) ) = ( 30 ) 2

So, the center is (−5, −3) and the radius is

30.

So, equation of the line is y = 73 x + 34 . 26. d=

(5 − 7 ) + ( − 11 − 2 ) ≈ 5.8 2

 − 11 + 2 5 + 7  , M =   ≈ ( −0.7,3.8 ) 2 2  

27. Slope of y1 = 0.32 x + 1.5 is 0.32. Slope of y2 = − 165 x + 141 is − 165 . Since (0.32)( − 165 ) = −0.1, the lines are neither parallel nor perpendicular.

338

CHAPTER 3 Section 3.1 Solutions -------------------------------------------------------------------------------1. Function

2. Function

3. Not a function – 1pm and 4pm both map to two elements in the range.

4. Not a function – Pat maps to two elements of the range.

5. Function

6. Not a function – Both elements of the domain map to 2 elements of the range.

7. Not a function – 0 maps to both −3 and 3.

8. Not a function – 2 maps to both −2 and 2, and 5 maps to both −5 and 5.

9. Not a function – 4 maps to both −2 and 2, and 9 maps to both −3 and 3.

10. Function

11. Function

12. Function

13. Not a function – Since 1, −2 2 and 1,2 2 are both

14. Not a function – Since (1, −1) and (1,1) are both on the graph, it does not pass the vertical line test.

(

)

(

)

on the graph, it does not pass vertical line test. 15. Not a function – Since (1, −1) and (1,1) are both on the graph, it does not pass the vertical line test.

16. Function

17. Function

18. Function

19. Not a function – Since (0,5) and (0, −5) are both on the graph, it does not pass the vertical line test.

20. Not a function – Since (0, 4) and (0, −4) are both on the graph, it does not pass the vertical line test. 339

Chapter 3

21. Function

22. Function

23. Not a function – Since ( 0, −1) and ( 0, −3 ) are both on the

24. Function

graph, it does not pass the vertical line test. 25. a) 5

b) 1

c) −3

26. a) 1

b) −5

c) 0

27. a) 3

b) 2

c) 5

28. a) 0

b) 4

c) −5

29. a) −5

b) −5

c) −5

30. a) −2

b) −6

c) −4

31. a) 2

b) −8

c) −5

32. a) 2

b) 0

c) 3

33. 1

34. −1.5 and 3

35. –3 and 1

36. −7

37. For all x in the interval [ −4, 4]

38. For all x in the set [ −4,0) ∪ [4]

39. 6

40. −3

41. f ( −2) = 2( −2) − 3 = −7

42. G ( −3) = ( −3)2 + 2( −3) − 7 = −4

43. g (1) = 5 + 1 = 6

44. F ( −1) = 4 − ( −1)2 = 3

45. Using #41 and #43, we see that f ( −2) + g (1) = −7 + 6 = −1.

46. Using #42 and #44, we see that G ( −3) − F ( −1) = −4 − 3 = −7 .

47. Using #41 and #43, we see that 3 f ( −2) − 2 g (1) = 3( −7) − 2(6) = −33 .

48. Using #42 and #44, we see that 2F ( −1) − 2G ( −3) = 2(3) − 2( −4) = 14 .

49. Using #41 and #43, we see that f ( −2) 7 = − . g (1) 6

50. Using #42 and #44, we see that G ( −3) 4 = − . F ( −1) 3

340

Section 3.1

51.

52.

f (0) − f ( −2) ( 2(0) − 3 ) − ( −7) −3 + 7 2 = = = g (1) 6 6 3 G (0) − G ( −3) ( 0 + 2(0) − 7 ) − ( −4) −7 + 4 = = = −1 F ( −1) 3 3 2

53. f ( − x ) = 2( − x ) − 3 = −2 x − 3

54. − F (t ) = −(4 − t 2 ) = t 2 − 4

55. − g ( −t ) = −[5 + ( −t )] = −(5 − t ) = t − 5

56. G ( −x ) = ( −x )2 + 2( − x ) − 7 = x 2 − 2 x − 7

57.

f ( x + 1) − f ( x − 1) = [ 2( x + 1) − 3] − [ 2( x − 1) − 3] = [ 2 x + 2 − 3 ] − [ 2 x − 2 − 3] = [ 2 x − 1] − [ 2 x − 5] = 2x − 1 − 2x + 5 = 4

58. F (t + 1) − F (t − 1) =  4 − (t + 1)2  −  4 − (t − 1)2  =  4 − ( t 2 + 2t + 1)  −  4 − ( t 2 − 2t + 1)  =  4 − t 2 − 2t − 1 −  4 − t 2 + 2t − 1 = 4 − t 2 − 2t − 1 − 4 + t 2 − 2t + 1 = −4t

59.

g ( x + a ) − f ( x + a ) = [5 + ( x + a )] − [ 2( x + a ) − 3]

= [ 5 + x + a ] − [ 2 x + 2 a − 3] = 5 + x + a − 2 x − 2a + 3 = 8−x −a

341

Chapter 3

60. G ( x + b ) + F ( b ) = ( x + b )2 + 2( x + b ) − 7  +  4 − b 2 

= x 2 + 2bx + b 2 + 2x + 2b − 7 + 4 − b 2 = x 2 + 2bx + 2x + 2b − 3 61.

62.

f ( x + h ) − f ( x ) [ 2( x + h ) − 3] − [ 2x − 3] 2 x + 2 h − 3 − 2 x + 3 2h = = = = 2 h h h h

F (t + h ) − F (t )  4 − (t + h )  −  4 − t   4 − ( t + 2 ht + h )  −  4 − t  = = h h h 2 2 2  4 − t − 2 ht − h  −  4 − t  4 − t 2 − 2 ht − h 2 − 4 + t 2 = = h h 2 −2 ht − h − h(2t + h ) = = = − ( 2t + h ) h h 2

63.

g ( t + h ) − g (t ) [5 + (t + h ) ] − [5 + t ] 5 + t + h − 5 − t h = = = =1 h h h h

64. 2 2 G ( x + h ) − G ( x ) ( x + h ) + 2( x + h ) − 7  −  x + 2 x − 7  = h h 2 2 x + 2 hx + h + 2x + 2 h − 7 − x 2 − 2 x + 7 = h 2 2 hx + h + 2h h(2x + h + 2) = = = 2x + h + 2 h h

65. It follows directly from the computation in #57 with x = −2 that this equals 2.

66. It follows directly from the computation in #58 with t = −1 that this equals 2 − h .

342

Section 3.1

67. It follows directly from the computation in #59 with t = 1 that this equals 1 .

68. It follows directly from the computation in #60 with x = −3 that this equals h − 4 .

69. The domain is  . This is written using interval notation as ( −∞, ∞ ) .

70. The domain is  . This is written using interval notation as ( −∞, ∞ ) .

71. The domain is  . This is written using interval notation as ( −∞, ∞ ) .

72. The domain is  . This is written using interval notation as ( −∞, ∞ ) .

73. The domain is the set of all real numbers x such that x − 5 ≠ 0 , that is x ≠ 5. This is written using interval notation as ( −∞,5 ) ∪ ( 5, ∞ ) .

74. The domain is the set of all real numbers t such that t + 3 ≠ 0 , that is t ≠ −3. This is written using interval notation as ( −∞, −3 ) ∪ ( −3, ∞ ) .

75. The domain is the set of all real numbers x such that x 2 − 4 = ( x − 2)( x + 2) ≠ 0, that is x ≠ −2,2 . This is written using interval notation as ( −∞, −2 ) ∪ ( −2,2 ) ∪ ( 2, ∞ ) .

76. The domain is the set of all real numbers x such that x 2 − 1 = ( x − 1)( x + 1) ≠ 0, that is x ≠ −1,1 . This is written using interval notation as ( −∞, −1) ∪ ( −1,1) ∪ (1, ∞ ) .

77. Since x 2 + 1 ≠ 0 , for every real number x, the domain is  . This is written using interval notation as ( −∞, ∞ ) .

78. Since x 2 + 4 ≠ 0 , for every real number x, the domain is  . This is written using interval notation as ( −∞, ∞ ) .

79. The domain is the set of all real numbers x such that 7 − x ≥ 0 , that is 7 ≥ x. This is written using interval notation as ( −∞,7 ] .

80. The domain is the set of all real numbers t such that t − 7 ≥ 0, that is t ≥7. This is written using interval notation as [7, ∞ ) .

343

Chapter 3

81. The domain is the set of all real numbers x such that 2 x + 5 ≥ 0 , that is x ≥ − 52 . This is written using interval notation

82. The domain is the set of all real numbers x such that 5 − 2x ≥ 0 , that is 5 2 ≥ x. This is written using interval notation

as − 52 , ∞ ) .

as ( −∞, 52  .

83. The domain is the set of all real numbers t such that t 2 − 4 ≥ 0, which is equivalent to (t − 2)(t + 2) ≥ 0. CPs are −2, 2

84. The domain is the set of all real numbers x such that x 2 − 25 ≥ 0, which is equivalent to ( x − 5)( x + 5) ≥ 0. CPs are −5, 5

+ − +   | |

−2

−5

This is written using interval notation as ( −∞, −2 ] ∪ [ 2, ∞ ) .

This is written using interval notation as ( −∞, −5] ∪ [5, ∞ ) .

85. The domain is the set of all real numbers x such that x − 3 > 0 , that is x > 3. This is written using interval notation as ( 3, ∞ ) .

86. The domain is the set of all real numbers x such that 5 − x > 0, that is 5 > x. This is written using interval notation as ( −∞,5 ) .

87. Since 1 − 2x can be any real number, there is no restriction on x, so that the domain is ( −∞, ∞ ) .

88. Since 7 − 5x can be any real number, there is no restriction on x, so that the domain is ( −∞, ∞ ) .

89. The only restriction is that x + 4 ≠ 0 , so that x ≠ −4 . So, the domain is ( −∞, −4) ∪ ( −4, ∞ ) .

90. The only restriction is that x 2 − 9 = ( x − 3)( x + 3) ≠ 0 , so that x ≠ ±3 . So, the domain is ( −∞, −3) ∪ ( −3,3) ∪ (3, ∞ ) .

344

Section 3.1

91. The domain is the set of all real numbers x such that 3 − 2x > 0, that is 3 2 > x. This is written using interval notation

92. The domain is the set of all real numbers x such that 25 − x 2 > 0, which is equivalent to (5 − x )(5 + x ) > 0. CPs are −5, 5

as ( −∞, 32 ) .

− + −  | | −5

This is written using interval notation as ( −5,5 ) . 93. The domain is the set of all real numbers t such that t 2 − t − 6 > 0, which is equivalent to (t − 3)(t + 2) > 0. CPs are −2, 3

94. Since t 2 + 9 > 0 , for all real numbers t, there is no restriction. So, the domain is ( −∞, ∞ ) .

+ − +  | | −2

This is written using interval notation as ( −∞, −2 ) ∪ ( 3, ∞ ) . 95. The domain is the set of all real numbers x such that x 2 − 16 ≥ 0 , which is equivalent to ( x − 4)( x + 4) ≥ 0 . CPs are −4, 4

96. There is no restriction on x. So, the domain is ( −∞, ∞ ) .

+ − +  | | −4

This is written using interval notation as ( −∞, −4 ] ∪ [ 4, ∞ ) .

345

Chapter 3

97. The function can be written as x2 r( x ) = . So, the domain is the 3 − 2x set of real numbers x such that 3 − 2 x > 0, that is 32 > x. This is written

98. The function can be written as ( x − 1)2 p( x ) = 3 . So, the domain is the 2 x 9 − ( )5

using interval notation as ( −∞, 32 ) .

set of real numbers x such that x 2 − 9 = ( x − 3)( x + 3) ≠ 0, so that x ≠ ±3 . So, the domain is ( −∞, −3) ∪ ( −3,3) ∪ (3, ∞ ) .

99. The domain of any linear function is ( −∞, ∞ ) .

100. The domain of any quadratic function is ( −∞, ∞ ) .

101. Solve x 2 − 2 x − 5 = 3. x 2 − 2x − 8 = 0 ( x − 4)( x + 2) = 0

102. Solve 56 x − 34 = 23 . 10 x − 9 = 8 10 x = 17

x = −2, 4

103.

x = 17 10

2x( x − 5) − 12( x − 5) = 0

104. 3x( x + 3)2 − 6( x + 3)3 = 0

2( x − 5)2 [ x( x − 5) − 6 ] = 0

3( x + 3)2 [ x − 2( x + 3)] = 0

2( x − 5)2 ( x 2 − 5x − 6) = 0

3( x + 3)2 ( −x − 6 ) = 0

2( x − 5)2 ( x − 6)( x + 1) = 0

x = −3, −6

x = −1,5,6 105. Let x = number of people y = cost Then, y = 45x, x > 75 . The domain is ( 75,∞ ) .

106. Let x = number of miles driven and y = cost. Then, y = 15 + 0.25x, x ≥ 0 .

346

Section 3.1

107. Assume: 6am corresponds to x = 6 noon corresponds to x = 12 Then, the temperature at 6am is: T (6) = −0.7(6)2 + 16.8(6) − 10.8 = 64.8 F The temperature at noon is: T (12) = −0.7(12)2 + 16.8(12) − 10.8 = 90 F

108. h(4) = −16(4)2 + 128(4) = 256 ft

The domain is [ 0,∞ ) since we are starting at time t = 0 sec. 109. v(7) = 5 + 200 − 10(7) v(7) ≈ 16.402 thousand views v(15) = 5 + 200 − 10(15) v(7) ≈ 12.071 thousand views

110. Since

200 − 10 x is non-negative wherever it is defined, the lowest price must

200 − 10 x = 0. This happens when 200 − 10 x = 0 x = 20 So, the lowest views occur on day 20 and there were 5000 views at that time. Similarly, the highest views occur on the initial day of posting the video, day 0. This corresponds to v ( 0 ) = 5 + 200 − 10 ( 0 ) = 19.142 or 19,142 views. occur when

111. Start with a square piece of cardboard with dimensions 10 in. × 10 in. Then, cut out 4 square corners with dimensions x in. × x in., as shown in the diagram:

Upon bending all four corners up, a box of height x is formed. Notice that all four sides of the base of the resulting box have length 10 − 2x . The volume of the box, V ( x ) , is given by: V ( x ) = ( Length ) ⋅ ( Width ) ⋅ ( Height ) = (10 − 2 x )(10 − 2 x )( x ) = x(10 − 2 x )2 The domain is ( 0,5 ) . (For any other values of x, one cannot form a box.)

347

Chapter 3

112. The volume of a right circular cylindrical tank whose base radius is 10 ft and whose height is h is given by V ( h ) = π (10)2 h = 100π h . If the height is increased by 2 ft, the corresponding volume would be: V ( h + 2) = π (10)2 ( h + 2) = 100π h + 200π So, the volume increased by 200π cubic ft, which corresponds to 200π ⋅ 7.48gal ≅ 4700 gal . 113. T ( 4 ) = 9 tweets

114. The number of tweets increased by 2 from week 2 to week 3. T (3) − T ( 2 ) 8−6 2 tweets The number of tweets decreased by 3 from week 5 to week 6. T ( 6 ) − T (5 ) 5−8 −3 tweets

T ( 7 ) = 6 tweets

T ( 8 ) = 3 tweets

115. P (14) = − 14 (142 ) + 7(14) + 180 = 229 people 117. a. Let x = Length of the window (in inches) Then, width = x − 278 (inches) So, area of the window is A( x ) = x ( x − 278 ) = x 2 − 278 x. b. A(4.5) = 4.5 ( 4.5 −

27 8

) = 5.0625.

This means that the area of the window is approximately 5 square inches. c. A(8.5) = 8.5 ( 8.5 − 278 ) = 43.5625 . This would be the area of the window in the envelope. However, this is not possible for an envelope with dimensions 4 inches by 8 inches because the window would be larger than the entire envelope.

116. P (6) = − 14 (62 ) + 7(6) + 180 = 213 people 118. a. Let x = Length of the window (in inches) Then, width = x − 2.875 (inches) So, area of the window is A( x ) = x ( x − 2.875 ) = x 2 − 2.875x. b.

A(5.25) = 5.25 ( 5.25 − 2.875 )

. ≈ 12.46875 This means that the area of the window is approximately 12.5 square inches. c. A(10) = 10(10 − 2.875) = 71.25 . This would be the area of the window in the envelope. However, this is not possible for an envelope with dimensions 9.5 inches by 6 inches because the window would be larger than the entire envelope. 348

Section 3.1

119. Yes, because every input (year) corresponds to exactly one output (quarterback’s name).

120. (2010, Aaron Rodgers), (2012, Joe Flacco), (2014, Tom Brady), (2016, Tom Brady), (2018, Tom Brady), (2020, Tom Brady)

121. (1999, 3000), (2003, 4000), (2007, 5000), (2011, 6000), (2015, 7000).

122. Yes, for every input there corresponds a unique output.

123. a) F ( 50 ) = number of tons of

124. F (100 ) + g (100 ) + G (100 )

carbon emitted by cement in 1975 = 0 b) g ( 50 ) = number of tons of carbon emitted by solids in 1975 = 2000 c) H ( 50 ) = total number of tons of carbon emitted in 1975 = 5000

represents the total amount (in millions of metric tons) of carbon emitted in 2025 by cement, solids, and liquids.

125. Should apply the vertical line test to determine if the relationship describes a function. The given relationship IS a function in this case.

126. H (3) − H ( −1) ≠ H (3) + H (1) , in general. You cannot distribute −1 through in this manner.

127. f ( x + 1) ≠ f ( x ) + f (1) , in general. You cannot distribute the function f through the input at which you are evaluating it.

128. There are two mistakes. One, the computation 3 − t > 0 should be 3 − t ≥ 0 . And two, the statement directly preceding the computation should be, “What can 3 − t be?” The domain should be ( −∞,3] .

129. G ( −1 + h ) ≠ G ( −1) + G ( h ), in general. 130. f (1) = −1 means the point (1, −1) must satisfy the function. So, −1 = 1 − A − 1 0 = 1− A

131. False. Consider the function f ( x ) = 9 − x 2 on its domain [ −3,3] .

The vertical line test x = 4 doesn’t intersect the graph, but it still defines a function.

1= A

132. False. Consider the function f ( x ) = x 2 on its domain  .

349

Chapter 3

133. False. This simply means that a particular horizontal line intersects the graph twice. Consider f ( x ) = x 2 on  . In this case, f ( a ) = f ( −a ) , for all real numbers a.

134. True. The graph of the circle x 2 + y 2 = r 2 satisfies this relationship, but does not pass the vertical line test.

135.

136. f (1) = A(1) − 3(1) = −1 2

g (3) =

A − 3 = −1

1 is undefined only if b = 3 . b −3

A=2

137. C − ( −2) C + 2 = is undefined only if D = −2 . So, D − ( −2) D + 2 C − ( −1) C + 1 C + 1 F ( −1) = = = = −(C + 1) = 4 implies that C = −5 . D − ( −1) D + 1 −2 + 1 F ( −2) =

138. Many functions will work here. The easiest ones to construct are of the form b g(x) = . For such a function, certainly g (5) is undefined. In order for x −5 b b (1, −1) to be on the graph, it must be the case that −1 = = , so that b = 4. 1 − 5 −4 4 So, one function that works is g ( x ) = . x −5 139. The domain is the set of all real numbers x such that x 2 − a 2 = ( x − a )( x + a ) ≠ 0 , which is equivalent to x ≠ ± a . So, the domain is ( −∞, −a ) ∪ ( −a, a ) ∪ ( a, ∞ ) .

140. The domain is the set of all real numbers x such that x 2 − a 2 = ( x − a )( x + a ) ≥ 0 . CPs: x = ±a + − +  | | −a

So, the domain is ( −∞, −a ] ∪ [ a, ∞ ) .

350

Section 3.1

141.

The time of day when it is warmest is Noon ( x = 12) and the temperature is approximately 90 degrees. This model is only valid on the interval [6,18] since the values of T outside the

interval [6,18] are too small to be considered temperatures in Florida. 142.

The firecracker is airborne for 8 seconds after liftoff – this corresponds to the xintercept. The firecracker maintains a maximum height of 256 ft within 4 seconds of liftoff. The model only applies for 8 seconds since the firecracker will land after 8 seconds, and presumably will not travel through the ground afterwards.

351

Chapter 3

143.

144. Observe that S ( r + 3) = 4π ( r + 3)2  Radius increased by 3

Corresponding surface area

= 4π ( r 2 + 6r + 9 )

Least views 5000, most views 19,142. This agrees with Exercise 110.

= 4π r 2 + 4π ( 6r + 9 ) So, the surface area would increase by 4π ( 6r + 9 ) square mm.

145.

146.

The graph of y2 can be obtained by shifting the graph of y1 two units to the right.

The graph of y2 can be obtained by shifting the graph of y1 two units to the left.

352

Section 3.2 Solutions -------------------------------------------------------------------------------1. G ( −x ) = −x + 4 ≠ x + 4 = G ( x ) So, not even. −G ( −x ) = − ( −x + 4 ) = x − 4 ≠ G ( x )

2. h( − x ) = 3 − ( −x ) = 3 + x ≠ h( x ) So, not even. − h( − x ) = − ( 3 − ( − x ) ) = − ( 3 + x ) ≠ h( x )

So, not odd. Thus, neither. So, not odd. Thus, neither. 3. f ( − x ) = 3( −x )2 + 1 = 3x 2 + 1 = f ( x ) So, even. Thus, f cannot be odd.

4. F ( − x ) = ( − x )4 + 2( − x )2 = x 4 + 2x 2 = F ( x ) So, even. Thus, F cannot be odd.

5. g ( −t ) = 5( −t )3 − 3( −t ) = −5t3 + 3t

6. f ( − x ) = 3( −x )5 + 4( −x )3

So, not even. − g ( −t ) = − − ( 5t3 − 3t ) = g (t )

So, not even.

= − ( 5t3 − 3t ) ≠ g (t )

(

= − ( 3x 5 + 4 x 3 ) ≠ f ( x )

)

(

− f ( −x ) = − − ( 3x5 + 4x3 )

So, odd.

)

= 3x 5 + 4 x 3 = f ( x )

So, odd. 7. h( − x ) = ( − x )2 + 2( − x ) = x 2 − 2 x ≠ h( x ) So, not even. − h( −x ) = − ( x 2 − 2x ) = −x 2 + 2x ≠ h( x )

So, not odd. Thus, neither.

8. G ( − x ) = 2( − x )4 + 3( − x )3 = 2 x 4 − 3x3 ≠ G ( x ) So, not even. −G ( −x ) = − ( 2 x 4 − 3x3 ) ≠ G ( x )

So, not odd. Thus, neither.

353

Chapter 3

10. g ( − x ) = ( −x )−1 + ( − x )

h( − x ) = ( − x ) − ( − x ) 1

(

= − ( x −1 + x ) ≠ g ( x )

)

= − x 3 − x ≠ h( x )

So, not even.

((

− h( − x ) = − − x 3 − x 1

) ) = x − x = h( x ) 1

So, not even.

(

− g ( −x ) = − − ( x −1 + x )

)

So, odd.

= x −1 + x = g ( x ) So, odd. (Note: ( − x )−1 = −1x = − x1 = −( x )−1 )

11.

12.

f ( −x ) = −x + 5 = −1 x + 5

f ( − x ) = − x + ( − x )2

= x + 5 = f (x)

= −1 x + x 2 = f ( x ) So, even. Thus, f cannot be odd.

So, even. Thus, f cannot be odd. 13. f ( − x ) = − x = −1 x = f ( x )

14.

So, even. Thus, f cannot be odd.

15.

16.

f ( − x ) = ( − x ) = − x 3 = −1 x 3 = f ( x ) 3

G ( −t ) = ( −t ) − 3 = −(t + 3)

G ( −t ) = ( −t ) + 2 ≠ G (t )

= t + 3 ≠ G (t )

So, not even. −G ( −t ) = − ( −t ) + 2 ≠ G (t )

So, not even. −G ( −t ) = − t + 3 ≠ G (t )

So, not odd. Thus, neither.

So, not odd. Thus, neither. 17. G ( −t ) = −t − 3 = −(t + 3) ≠ G (t ) So, not even. −G ( −t ) = − −(t + 3) ≠ G (t ) 

18. f ( −x ) = 2 − ( −x ) = 2 + x ≠ f ( x ) So, not even. − f ( −x ) = − 2 + x ≠ f ( x ) So, not odd. Thus, neither.

Note: Cannot distribute −1 here

So, not odd. Thus, neither.

354

Section 3.2

19.

20. g ( − x ) = ( − x )2 + ( − x )

f ( − x ) = ( − x )2 + 2 = x 2 + 2 = f ( x ) So, even. Thus, f cannot be odd.

= x2 − x ≠ g( x)

So, not even. − g ( −x ) = − x 2 − x ≠ g ( x ) So, not odd. Thus, neither. 21.

22.

h( −x ) =

1 + 3 ≠ h( x ) −x So, not even.  1  1 −h( −x ) = −  + 3  = − 3 ≠ h( x )  −x  x So, not odd. Thus, neither.

1 − 2( −x ) −x 1  = −  − 2 x  ≠ h( x ) x  So, not even.  1  − h( − x ) = −  −  − 2 x     x 1 = − 2 x = h( x ) x So, odd.

23. Call the function h. h is not even since h(1) = 4 ≠ 0 = h( −1) . h is not odd since h(1) = 4 ≠ 0 = −h( −1) . Thus, neither.

24. Call the function h. h is not odd since h(2) = 3 ≠ −3 = −h( −2) . h is even since h( x ) = h( −x) for all x. (This can be visualized by reflecting the graph about the y-axis.)

h( −x ) =

25.

Increasing

( −∞,∞ ) [ −1, ∞ ) ( −1, ∞ )

Decreasing

( −3, −2)

Constant

( −∞, −3) ∪ ( −2, −1)

Domain Range

d) 0 e) −1 f) 2

355

Chapter 3

26.

Increasing

[ −4, ∞ ) ( −∞,3] (1,2 )

Decreasing

( −3,0) ∪ (2, ∞ )

Constant

( −4, −3 ) ∪ (0,1)

Domain Range

d) −1 e) 2 f) 1

27.

Increasing

[ −7,2] [ −5, 4] ( −4,0 )

Decreasing

( −7, −4) ∪ (0,2)

Constant

nowhere

Domain Range

d) 4 e) 1 f) −5

28.

Increasing

( −∞,∞ ) ( −∞,∞ ) ( −∞, −3 ) ∪ ( 3,∞ )

Decreasing

( −3,3)

Constant

nowhere

Domain Range

d) 0 e) approximately 3.5 f) approximately −3.5

29.

Increasing

( −∞,∞ ) ( −∞,∞ ) ( −∞, −3) ∪ ( 4,∞ )

Decreasing

nowhere

Constant

( −3, 4 )

Domain Range

d) 2 e) 2 f) 2

356

Section 3.2

30.

Range

( −∞,∞ ) ( −∞,∞ )

Increasing

nowhere

Decreasing

( −∞,∞ )

Constant

nowhere

Domain

d) 0 e) approximately 0.75 f) approximately −0.75

31.

Decreasing

( −∞,∞ ) [ −4, ∞ ) ( 0,∞ ) ( −∞,0 )

Constant

nowhere

Domain Range Increasing

d) −4 e) 0 f) 0

32. Domain Range Increasing Decreasing Constant

( −∞,∞ ) [0,∞ ) (3,∞ ) ( −∞, −3) ( −3,3)

d) 0 e) 0 f) 0

33.

Increasing

( −∞,0 ) ∪ ( 0,∞ ) ( −∞,0 ) ∪ ( 0,∞ ) ( −∞,0 ) ∪ ( 0,∞ )

Decreasing

nowhere

Constant

nowhere

Domain Range

d) undefined e) 3 f) −3

357

Chapter 3

34.

Decreasing

( −∞, 4 ) ∪ ( 4,∞ ) ( −∞,∞ ) ( −∞,0 ) ∪ ( 4,∞ ) ( 0, 4 )

Constant

nowhere

Domain Range Increasing

d) 4 e) 3 f) approximately 3.5

35.

( −∞,0 ) ∪ ( 0,∞ ) ( −∞,5 ) ∪ [7] ( −∞,0 ) (5,∞ ) ( 0,5 )

Domain Range Increasing Decreasing Constant

d) undefined e) 3 f) 7

36.

Decreasing

( −8,0 ) ∪ ( 0, 4 ] ( −4,3] ( −8, −5 ) ∪ ( 0, 4 ) ( −5,0 )

Constant

nowhere

Domain Range Increasing

d) undefined e) approximately 0 f) approximately 0

37.

( x + h )2 − ( x + h )  −  x 2 − x  x 2 + 2 hx + h2 − x − h − x 2 + x h ( 2x + h − 1) = = h h h = 2x + h − 1 38. ( x + h )2 + 2( x + h )  −  x 2 + 2 x  h

x 2 + 2 hx + h2 + 2 x + 2 h − x 2 − 2 x h ( 2 x + h + 2 ) = h h

= 2x + h + 2

358

Section 3.2

39. ( x + h )2 + 3( x + h )  −  x 2 + 3x  h

x 2 + 2 hx + h 2 + 3x + 3h − x 2 − 3x h ( 2 x + h + 3 ) = h h

= 2x + h + 3 40.  −( x + h )2 + 5( x + h )  −  −x 2 + 5x  −x 2 − 2 hx − h2 + 5x + 5h + x 2 − 5x = h h h ( −2x − h + 5 ) = = −2x − h + 5 h

41. ( x + h )2 − 3( x + h ) + 2  −  x 2 − 3x + 2  x 2 + 2hx + h 2 − 3x − 3h + 2 − x 2 + 3x − 2 = h h h ( 2x + h − 3) = = 2x + h − 3 h

42. ( x + h )2 − 2( x + h ) + 5  −  x 2 − 2 x + 5  x 2 + 2hx + h 2 − 2x − 2h + 5 − x 2 + 2x − 5 = h h h ( 2x + h − 2 ) = = 2x + h − 2 h

43.

 −3( x + h )2 + 5( x + h ) − 4  −  −3x 2 + 5x − 4  h 2 2 −3x − 6hx − 3h + 5x + 5h − 4 + 3x 2 − 5x + 4 = h h ( −6 x − 3h + 5 ) = −6 x − 3h + 5 = h

359

Chapter 3

44.

 −4( x + h )2 + 2( x + h ) − 3 −  −4 x 2 + 2 x − 3 h 2 −4x − 8hx − 4h 2 + 2x + 2h − 3 + 4x 2 − 2 x + 3 = h h ( −8x − 4 h + 2 ) = −8x − 4 h + 2 = h 45. 1 1 − 2 ( x + h ) 2x x x+h x−x−h 1 1 = − = − = h 2 h ( x + h ) 2 xh 2 xh ( x + h ) 2 xh ( x + h ) 2 xh ( x + h ) =

−h −1 = 2x h ( x + h ) 2x ( x + h )

46. 1 1 − (x + h) + 3 x + 3 = 1 1 x −3 x+ h+3 − = − h h ( x + h + 3 ) h ( x + 3 ) h ( x + h + 3 )( x + 3 ) h ( x + h + 3 )( x + 3 ) =

47. x + h +1− h

−1 x −3− x − h +3 −h = = h ( x + h + 3 )( x + 3 ) h ( x + h + 3 )( x + 3 ) ( x + h + 3 )( x + 3 )

48. x+h−2 − x−2 h

( x + 1) = x + h − x h

49. 33 − 13 27 − 1 = = 13 3 −1 2

50. 1 1 −2 3 −1 = 3 = − 13 3 −1 2

51. 3 −1 =1 3 −1

52. 2(3) − 2(1) 4 = = 2 3 −1 2

360

Section 3.2

53.

54.

(1 − 2(3)) − (1 − 2(1)) −5 − ( −1) = = −2 3 −1 2

( 9 − 3 ) − ( 9 − 1 ) = 0 − 8 = −4 2

3 −1

56.

55. 5 − 2(3) − 5 − 2(1) 3 −1

−1 − 3 = −1 2

32 − 1 − 3 12 − 1 3 8 = =1 3 −1 2

57. 2

 3  5  9   25  2 −  − 2 −  2   − 2   9 − 25  4  4  =  16   16  = 8 8 = −2 = −4 1 1 1 3  5 − −−  2 2 2 4  4

58. −0.19 + 1 − −0.36 + 1 0.81 − 0.64 0.9 − 0.8 0.1 10 = = = = −0.19 − ( −0.36 ) 0.17 0.17 0.17 17

59.

(

− 23 5

) − − ( 5 )  −8 (5 ) − ( −5) 3

23 5 − 3 5

35 3 25 35 3 25 =−3 ⋅3 =− = −7 3 25 5 5 25

60. 3.75 − 7 − 1.25 − 7 −3.25 − −5.75 3.25 − 5.75 = = = −1 3.75 − 1.25 2.5 2.5

61. 2 2  4  1  2  1  16 4 44   − −   −   5  2  3  2  25 − 9 225 22 = = = 4 2 2 2 15 − 5 3 15 15

62.

(3 2 ) + 6 − ( 2 ) + 6 2

3 2− 2

18 + 6 − 2 + 6 2 2 361

24 − 8 16 2 = ⋅ =4 2 2 2 2 2 2

Chapter 3

63.

5 4 5 − 3  − 5 − 3  5 − 5 − 5 − 4 0 −1 3 3 = = = −3 5 4 1 1 − 3 3 3 3 64. 3

3 1  27   1  27 1 26 25   − 25   25  −  − 25   5  5  =  125   125  = 5 5 = 5 = 13 3 1 2 2 2 − 5 5 5 5 5

65.

Increasing

( −∞,∞ ) ( −∞,∞ ) ( −∞,∞ )

Decreasing

nowhere

Constant

nowhere

Domain Range

66.

Decreasing

( −∞,∞ ) ( −2, ∞ ) (1,∞ ) ( −∞,1)

Constant

nowhere

Domain Range Increasing

362

Section 3.2

67.

Increasing

( −∞,∞ ) ( −∞,2 ] ( −∞,2 )

Decreasing

nowhere

Constant

( 2,∞ )

Domain Range

68. Range

( −∞,∞ ) {−1} ∪ (1,∞ )

Increasing

nowhere

Domain

Decreasing Constant

( −∞, −1) ( −1, ∞ )

Notes on the graph: There should be an open hole at ( −1,1), and a closed hole at ( −1, −1) . 69. Domain Range Increasing Decreasing Constant

( −∞,∞ ) [0,∞ ) ( 0,∞ ) ( −1,0 ) ( −∞, −1)

363

Chapter 3

70. Domain Range Increasing Decreasing Constant

( −∞,∞ ) [0,∞ ) ( 0,2 ) ( −∞,0 ) ( 2,∞ )

71.

Increasing

( −∞,∞ ) ( −∞,∞ ) ( −∞,∞ )

Decreasing Constant

nowhere nowhere

Domain Range

72.

Decreasing

( −∞,∞ ) [0,∞ ) ( 0,∞ ) ( −∞,0 )

Constant

nowhere

Domain Range Increasing

364

Section 3.2

73.

Decreasing

( −∞,∞ ) [1,∞ ) (1,∞ ) ( −∞,1)

Constant

nowhere

Domain Range Increasing

74.

Decreasing

( −∞,∞ ) ( −∞,∞ ) ( −∞, −1) ∪ ( 0,∞ ) ( −1,0 )

Constant

nowhere

Domain Range Increasing

75.

10 5-2*x 3*x-2

9 8

Decreasing

( −∞,2 ) ∪ ( −∞,2 ) (1,∞ ) ( 2,∞ ) ( −∞,2 )

Constant

nowhere

Domain Range Increasing

7 6 5 4 3 2

-3

365

-2

-1

Chapter 3

76.

10 3-0.5*x 4+1.5*x

Decreasing

( −∞, −2 ) ∪ ( −2,∞ ) (1,∞ ) ( −2, ∞ ) ( −∞, −2 )

Constant

nowhere

Domain Range Increasing

8 7 6 5 4 3 2 1

-3

77. Domain

Increasing

( −∞,∞ ) [ −1,3] ( −1,3)

Decreasing Constant

nowhere ( −∞, −1) ∪ ( 3,∞ )

Range

78.

Increasing

( −∞, −1) ∪ ( −1,3) ∪ ( 3,∞ ) [ −1,3] ( −1,3)

Decreasing Constant

nowhere ( −∞, −1) ∪ ( 3,∞ )

Domain Range

Notes on the graph: There should be open holes at ( −1, −1) and (3,3) .

366

-2

-1

Section 3.2

79.

Increasing

( −∞,∞ ) [1, 4] (1,2 )

Decreasing Constant

nowhere ( −∞,1) ∪ ( 2,∞ )

Domain Range

80.

Increasing

( −∞,1) ∪ (1,2 ) ∪ ( 2,∞ ) [1, 4] (1,2 )

Decreasing Constant

nowhere ( −∞,1) ∪ ( 2,∞ )

Domain Range

Notes on the graph: There should be open holes at (1,1) and (2, 4) . 81.

Decreasing

( −∞, −2 ) ∪ ( −2,∞ ) ( −∞,∞ ) ( −2,1) ( −∞, −2 ) ∪ (1,∞ )

Constant

nowhere

Domain Range Increasing

Notes on the graph: There should be open holes at ( −2,1), ( −2, −1), and (1,2) , and a closed hole at (1,0) .

367

Chapter 3

82.

Decreasing

( −∞,1) ∪ (1,∞ ) ( −∞,∞ ) ( −2,1) ( −∞, −2 ) ∪ (1,∞ )

Constant

nowhere

Domain Range Increasing

Notes on the graph: There should be open holes at ( −2, −1), (1,2), and (1,0), and a closed hole at ( −2,1) . 83.

Increasing

( −∞,∞ ) [0,∞ ) ( 0,∞ )

Decreasing

nowhere

Constant

( −∞,0 )

Domain Range

84.

Increasing

( −∞,1) ∪ (1,∞ ) [1,∞ ) (1,∞ )

Decreasing

nowhere

Constant

( −∞,1)

Domain Range

Notes on the graph: There should be an open hole at (1,1) .

368

Section 3.2

85.

Range

( −∞,∞ ) ( −∞,∞ )

Increasing

nowhere

Decreasing

( −∞,0 ) ∪ ( 0,∞ )

Constant

nowhere

Domain

Notes on the graph: There should be a closed hole at (0,0). 86.

Increasing

( −∞,∞ ) ( −∞,∞ ) ( −∞,0 ) ∪ ( 0,∞ )

Decreasing

nowhere

Constant

nowhere

Domain Range

Notes on the graph: There should be a closed hole at (0,0). 87.

Decreasing

( −∞,1) ∪ (1,∞ ) ( −∞, −1) ∪ ( −1,∞ ) ( −1,1) ( −∞, −1) ∪ (1,∞ )

Constant

nowhere

Domain Range Increasing

369

Chapter 3

88.

Decreasing

( −∞,1) ∪ (1,∞ ) ( −1,1) ∪ (1,∞ ) ( −1,1) ∪ (1,∞ ) ( −∞, −1)

Constant

nowhere

Domain Range Increasing

89.

Decreasing

( −∞,∞ ) ( −∞,2 ) ∪ [ 4,∞ ) ( −∞, −2 ) ∪ ( 0,2 ) ∪ ( 2,∞ ) ( −2,0 )

Constant

nowhere

Domain Range Increasing

Notes on the graph: There should be open holes at ( −2,2), (2,2) and closed holes at ( −2,1), (2, 4) . 90. Domain Range Increasing Decreasing Constant

( −∞, −1) ∪ ( −1,1) ∪ (1,∞ ) [1,∞ ) (1,∞ ) ( −∞, −1) ( −1,1)

Notes on the graph: There should be open holes at ( −1,1), (1,1) .

370

Section 3.2

91.

Increasing

( −∞,1) ∪ (1,∞ ) ( −∞,1) ∪ (1,∞ ) ( −∞,1) ∪ (1,∞ )

Decreasing

nowhere

Constant

nowhere

Domain Range

Notes on the graph: There should be an open hole at (1,1). 92.

Decreasing

( −∞,∞ ) ( −1, ∞ ) ( −1, ∞ ) ( −∞, −1)

Constant

nowhere

Domain Range Increasing

Notes on the graph: There should be an open hole at ( −1, −1) and a closed hole at ( −1,1) . 93. Profit is increasing from t = 10 to t = 12, which corresponds to Oct. to Dec. Profit is decreasing from t = 1 to t = 10, which corresponds to Jan to Oct. Profit never remains constant.

371

94. Profit is increasing from t = 1 to t = 8, which corresponds to Jan to Aug. Profit is decreasing from t = 8 to t = 12, which corresponds to Aug to Dec. Profit never remains constant.

Chapter 3

95. Let x = number of T-shirts ordered. The cost function is given by 10x, 0 ≤ x ≤ 50  C ( x ) = 9x, 50 < x ≤ 100 8x, x > 100 

96. Let x = number of new uniforms ordered. The cost function is given by 176.12 x, 0 ≤ x ≤ 50 C (x) =  159.73x, 50 < x ≤ 100

97. Let x = number of boats entered. The cost function is given by 250x, 0 ≤ x ≤ 10  0 ≤ x ≤ 10  250x,  2500 + 175 ⋅ ( x − 10) , x > 10 =  C ( x ) =     175x + 750, x > 10 Cost for first # of boats beyond first 10 10 boats  98.

 10 x 0 ≤ x < 6. C (x) =  x≥6 8.5x 99. Let x = number of people attending the reception. The cost function is given by 1000 0 ≤ x ≤ 100 + 35x,    Fee for reserving dining room  C (x) =  3500 + 25 ⋅ ( x − 100) , x > 100 1000 +   Cost for first # of guests beyond first 100  100 guests  Simplifying the terms above yields 1000 + 35x, 0 ≤ x ≤ 100 C (x) =  2000 + 25x, x > 100 100.

101.

0 < x ≤ 2. 19.95 C (x) =  19.95 + 10.05 ( x − 2 ) x > 2

x =1 139 C (x) =  139 + 125.1( x − 1) x > 1

102. x =1 111.2 C (x) =  111.2 + 100.08 ( x − 1) x > 1

372

Section 3.2

103. Let x = number of stained glass units sold. 100 +  700 + Total monthly cost is given by: C ( x ) =  Business Costs

Studio Rent

35x = 800 + 35x .   Cost of materials for x units

Revenue for x units sold is given by: R( x ) = 100x So, the total profit is given by: P ( x ) = R( x ) − C ( x ) = 100 x − ( 800 + 35x ) = 65x − 800 104. Let x = number of people who attend. Since it is assumed that each person eats 1 lb. of shrimp, it will cost 5x dollars for x people for the shrimp. So, the total cost is given by: C ( x ) = 30 + 5x The total revenue is given by: R( x ) = 10x So, the total profit is given by: P ( x ) = R( x ) − C ( x ) = 10x − (30 + 5x ) = 5x − 30 105. Observe that 0 < x ≤1 1.00, 1.00 + 0.20, 1< x ≤ 2  f (x) =  1.00 + 0.20 ( 2 ) , 2 < x ≤ 3   Using the greatest integer function, we have f ( x ) = 1.00 + 0.20  x , 0 ≤ x < 13.

106. Observe that 0 < x ≤1 2.19, 2.19 + 0.23, 1< x ≤ 2  f (x) =  2.19 + 0.23 ( 2 ) , 2 < x ≤ 3   Using the greatest integer function, we have f ( x ) = 2.19 + 0.23  x , 0 ≤ x < 13.

107.

108.  x   1+

t 

f ( t ) = 3 ( −1) , t ≥ 0

f ( x ) = ( −1) 100  , x ≥ 0

373

Chapter 3

109. a) 60, increasing f ( 2005 ) − f (1995 ) 2005 − 1995 7300 − 6700 10 60 b) −70 , decreasing f ( 2015 ) − f ( 2005 ) 2015 − 2005 6600 − 7300 10 −70 111.

110. a) 20, increasing f ( 2005 ) − f ( 2000 ) 2005 − 2000 7300 − 7200 5 20 b) −60 , decreasing f ( 2010 ) − f ( 2005 ) 2010 − 2005 7000 − 7300 5 −60

h(2) − h(1) ( −16(2) + 48(2) ) − ( −16(1) + 48(1) ) = = 0 ft/sec 2 −1 2 −1 2

112.

h(3) − h(1) ( −16(3) + 48(3) ) − ( −16(1) + 48(1) ) = 3 −1 3 −1 2

= −16 ft/sec

113. The first quarter starts at t = 1 and ends at t = 90. So, the average rate of change in d(t) during the first quarter is

(

) (

)

d (90) − d (1) 3 90 + 1 − 2.75(90) − 3 1 + 1 − 2.75(1) = ≈ 0.236 90 − 1 89 So, demand is increasing at an approximate rate of 236 units over the first quarter. 2

374

Section 3.2

114. The fourth quarter starts at t = 273 and ends at t = 365. So, the average rate of change in d(t) during the fourtht quarter is d (365) − d (273) = 365 − 273

(3 365 + 1 − 2.75(365)) − (3 273 + 1 − 2.75(273)) 2

92 ≈ 0.250 So, demand is increasing at an approximate rate of 250 units over the fourth quarter.

115. The domain is incorrect. It should be ( −∞, 0) ∪ (0, ∞ ). The range is also incorrect and should be (0, ∞ ). The graph is also incorrect. It should contain an open circle at (0, 0).

116. The open and closed holes at x = 1 should be switched, and then the range should be changed to [ −1, ∞ ) .

117. There should be no x in the function because the cost of the ticket is not multiplied by the person’s age.

118. The portion of C ( x ) for x > 10,000 should be: 0.02(10,000) + 0.04( x − 10,000)

119. True. This corresponds to y = mx + b with m = 1, b = 0 .

120. True. This corresponds to y = mx + b with m = 0 .

121. False. For instance, f ( x ) = x3 is always increasing.

122. True.

123. The individual pieces used to form f, namely ax , bx 2 , are continuous on  . So, the only x-value with which we need to be concerned regarding the continuity of f is x = 2 . For f to be continuous at 2, we need a(2) = b(2)2 , which is the same

as a = 2b . 124. Both x1 and − x1 are undefined at x = 0 . So, for every value of a, either a > 0 or a ≤ 0 . Hence, we would need to evaluate either x1 or − x1 at 0, which is not possible. So, this function cannot be continuous, for any value of a.

375

Chapter 3

125. The graph is odd.

126. The graph is even.

127. The graph is odd.

128. This is the graph of tan x (in #113).

129.

130.

Domain:  Range: The set of integers

376

Domain:  Range: The set of integers

Section 3.3 Solutions -------------------------------------------------------------------------------1. l Shift the graph of x 2 up 1 unit.

2. j Shift the graph of x 2 right 1 unit.

3. a Shift the graph of x 2 right 1 unit, then reflect over x-axis.

4. d Reflect the graph of x 2 over x-axis, then shift down 1 unit.

5. b Shift the graph of x 2 left 1 unit, then reflect over x-axis.

6. k Shift the graph of x 2 right 1 unit, reflect over x-axis, and then shift up 1 unit.

7. i Shift the graph of then shift up 1 unit.

8. h Reflect the graph of x over xaxis, then shift down 1 unit.

x right 1 unit,

9. c Shift the graph of x right 1 unit, then reflect over y-axis, and then shift down 1 unit.

10. e Reflect the graph of axis, then shift up 1 unit.

11. g Reflect the graph of x over yaxis, then reflect over x-axis, and then shift up 1 unit.

12. f Shift the graph of x right 1 unit, then reflect over y-axis, then reflect over x-axis, and then shift down 1 unit.

13. y = x + 3

14. y = x + 4

15. y = − x = x

16. y = − x

(since −x = −1 x = x ) 17. y = 3 x

18. y = 13 x

19. y = x3 − 4

20. y = ( x − 3)3

21. y = ( x + 1)3 + 3

22. y = − x3

23. y = (−x)3

24. y = x3

377

x over y-

Chapter 3

25.

26.

27.

28.

378

Section 3.3

29.

30.

31.

32.

33.

34.

379

Chapter 3

35.

36.

37.

38.

380

Section 3.3

39.

40.

41.

42.

43.

44.

381

Chapter 3

45.

46.

47.

48.

382

Section 3.3

49. Shift the graph of x 2 down 2 units.

50. Shift the graph of x 2 up 3 units.

51. Shift the graph of x 2 left 1 unit.

52. Shift the graph of x 2 right 2 units.

53. Shift the graph of x 2 right 3 units, and up 2 units.

54. Shift the graph of x 2 left 2 units, and up 1 unit.

383

Chapter 3

55. Shift the graph of x 2 right 1 unit, and then reflect over x-axis.

56. Shift the graph of x 2 left 2 units, and then reflect over x-axis.

57. Reflect the graph of x over y-

58. Reflect the graph of x over x-axis.

axis. (This yields the same graph as x since −x = −1 x = x .)

384

Section 3.3

59. Reflect the graph of x over x-

60. Since 1 − x + 2 = x − 1 + 2, shift the

axis, then shift left 2 units and down 1 unit.

graph of x right 1 unit, and up 2 units.

61. Vertically stretch the graph of x 2 by a factor of 2, then shift up 1 unit.

62. Vertically stretch the graph of x by

a factor of 2, then shift up 1 unit.

385

Chapter 3

63. Shift the graph of x right 2 units, 64. Since 2 − x = −( x − 2), reflect the then reflect over x-axis. graph of x over y-axis, then shift right 2 units.

65. Reflect the graph of x over xaxis, then shift left 2 units and down 1 unit.

66. Since 2 − x + 3 = −( x − 2) + 3,

reflect the graph of x over y-axis, then right 2 units and up 3 units.

386

Section 3.3

67. Shift the graph of 3 x right 1 unit, then up 2 units.

68. Shift the graph of 3 x left 2 units, then down 1 unit.

69.

70.

Shift the graph of

1 + 2 up two units. x

Shift the graph of

387

1 left two units. x+2

Chapter 3

71. Shift the graph of x1 left 3 units, then up 2 units.

72. Since 3−1x = − x1−3 , shift the graph of 1 x right 3 units, and then reflect over xaxis.

73. Shift the graph of x1 left 2 units, then reflect over x-axis, and then shift up 2 units.

74. Since 2 − − ( x1−1) = 2 + x1−1 , shift the

graph of x1 right 1 unit, then up 2 units.

388

Section 3.3

75. Reflect the graph of x over yaxis, then expand vertically by a factor of 5.

76. Reflect the graph of x over x-axis, then contract vertically by a factor of 5.

77. Completing the square yields f ( x ) = x 2 − 6 x + 11 = ( x 2 − 6 x + 9 ) + 11 − 9 = ( x − 3) + 2 2

So, shift the graph of x 2 right 3 units, then up 2 units.

389

Chapter 3

78. Completing the square yields f ( x ) = x 2 + 2x − 2 = ( x 2 + 2 x + 1) − 2 − 1 = ( x + 1) − 3 2

So, shift the graph of x 2 left 1 unit, then down 3 units.

79. Completing the square yields f ( x ) = − ( x 2 + 2x ) = − ( x 2 + 2 x + 1) + 1 = − ( x + 1) + 1 2

So, reflect the graph of x 2 over x-axis, then shift left 1 unit, then up 1 unit.

80. Completing the square yields f ( x ) = −x2 + 6x − 7

= − ( x2 − 6x ) − 7

= − ( x2 − 6x + 9 ) − 7 + 9 = − ( x − 3) + 2 2

So, reflect the graph of x 2 over x-axis, then shift right 3 units, then up 2 units.

390

Section 3.3

81. Completing the square yields f ( x ) = 2 x 2 − 8x + 3

= 2 ( x2 − 4x ) + 3

= 2 ( x2 − 4x + 4 ) + 3 − 8 = 2 ( x − 2) − 5 2

So, vertically stretch the graph of x 2 by a factor of 2, then shift right 2 units, then down 5 units.

82. Completing the square yields f ( x ) = 3x 2 − 6 x + 5

= 3 ( x 2 − 2x ) + 5

= 3 ( x 2 − 2 x + 1) + 5 − 3 = 3 ( x − 1) + 2 2

So, vertically stretch the graph of x 2 by a factor of 3, then shift right 1 unit, then up 2 units.

83.

85.

f ( x ) = 15x, g ( x ) = 15 ( x + 5 )

T ( x ) = 0.32 ( x − 12,200 )

84. The profit in a rainy year is given by P ( x − 10)(Cost of 1) , where x is the number of pallets sold. Since they are giving away 10 pallets in a rainy year, they don’t make a profit on the first 10. So, the profit would be P ( x − 10 ) . 86. The actual amount administered if the weight is overestimated by 3 ounces is A( x + 3) = x + 3 + 2.

391

Chapter 3

87. a. Use h = 162 to get 162w 9w BSA(w ) = = . 3,600 200 b. If she loses 3 kg, the new function is 9(w − 3) BSA(w − 3) = . 200

88. a. Use h = 180 to get 180w w BSA(w ) = = . 3,600 20 b. If he gains 5 kg, the new function is w +5 BSA(w + 5) = . 20

89. (b) shifted to the right 3 units

90. (c) is wrong – reflect over x-axis.

91. (b) should be deleted since 3 − x = x − 3 . The correct sequence of

92. (b) is wrong and (d) is misplaced. The correct sequence of steps would be: ( a ) → ( d ) → (∗) → (c ) , where (∗) = reflect over x-axis.

steps would be: ( a ) → (c )* → ( d ), where (c)* : Shift to the right 3 93. True. Since −x = −1 x = x .

94. False. y = − x is the reflection of

y = x over the y-axis. 95. True.

96. True.

97. The graph of y = f ( x − 3) + 2 is the graph of y = f ( x ) shifted right 3 units, then up 2 units. So, if the point (a, b ) is on the graph of y = f ( x ), then the point ( a + 3, b + 2) is on the graph of the translation y = f ( x − 3) + 2 . 98. The graph of y = − f ( − x ) + 1 is the graph of y = f ( x ) reflected over y-axis, then over x-axis, and then shifted up 1 unit. So, if the point ( a, b) is on the graph of y = f ( x ) , then the point ( −a, −b + 1) is on the graph of the translation y = − f ( −x ) + 1 .

392

Section 3.3

99. a.

Any part of the graph of y = f ( x ) that is below the x-axis is reflected above it for the graph of y = f ( x ) . 100. a.

 f ( x ), x ≥ 0 The relationship is described by: f ( x ) =   f ( −x ), x < 0

393

Chapter 3

101. a.

If a > 1 , then the graph is a horizontal compression. If 0 < a < 1, then the graph is a horizontal expansion. 102. a.

If 0 < a < 1, then the graph is a vertical compression. If a > 1 , then the graph is a vertical expansion.

394

Section 3.3

103. The graph of f is as follows:

104. The graph of g is as follows:

Each horizontal line in the graph of y = [[x]] is stretched by a factor of 2. Any portion of the graph that is below the x-axis is reflected above it. Also, there is a vertical shift up of one unit.

Any portion of the graph of y = [[x]] that is below the x-axis is reflected above it. Also, there is a vertical shift up of one unit and a vertical compression by a factor of 1 2 .

395

Section 3.4 Solutions -------------------------------------------------------------------------------1.

Domains: dom( f + g )   dom( f − g )  = ( −∞, ∞ ) dom( fg ) 

f ( x ) + g ( x ) = ( 2 x + 1) + (1 − x ) = x+2 f ( x ) − g ( x ) = ( 2 x + 1) − (1 − x )

f dom   = ( −∞,1) ∪ (1, ∞ ) g

= 2x + 1 − 1 + x = 3x f ( x ) ⋅ g ( x ) = ( 2 x + 1)(1 − x ) = 2x + 1 − 2x 2 − x = −2 x 2 + x + 1 f ( x ) 2x + 1 = g(x) 1 − x

Domains: dom( f + g )   dom( f − g )  = ( −∞, ∞ ) dom( fg ) 

f ( x ) + g ( x ) = ( 3x + 2 ) + ( 2x − 4 ) = 5x − 2 f ( x ) − g ( x ) = ( 3x + 2 ) − ( 2x − 4 )

f dom   = ( −∞,2 ) ∪ ( 2, ∞ ) g

= 3x + 2 − 2x + 4 = x+6 f ( x ) ⋅ g ( x ) = ( 3x + 2 ) ⋅ ( 2x − 4 ) = 6x 2 − 12x + 4x − 8 = 6x 2 − 8x − 8 f ( x ) 3x + 2 = g ( x ) 2x − 4

396

Section 3.4

Domains: dom( f + g )   dom( f − g )  = ( −∞, ∞ ) dom( fg ) 

f ( x ) + g ( x ) = ( 2x − x ) + ( x − 4 ) 2

= 3x 2 − x − 4

f ( x ) − g ( x ) = ( 2x 2 − x ) − ( x2 − 4 )

f dom   = ( −∞, −2 ) ∪ ( −2,2 ) ∪ ( 2, ∞ ) g

= 2x 2 − x − x 2 + 4 = x2 − x + 4

f ( x ) ⋅ g ( x ) = ( 2x 2 − x ) ⋅ ( x 2 − 4 ) = 2 x 4 − x 3 − 8x 2 + 4 x f ( x ) 2x 2 − x = 2 g( x) x −4

Domains: dom( f + g )   dom( f − g )  = ( −∞, ∞ ) dom( fg ) 

f ( x ) + g ( x ) = ( 3x + 2 ) + ( x − 25 ) 2

= x 2 + 3x − 23

f ( x ) − g ( x ) = ( 3x + 2 ) − ( x 2 − 25 )

f dom   = ( −∞, −5 ) ∪ ( −5,5 ) ∪ ( 5, ∞ ) g

= 3x + 2 − x 2 + 25 = −x 2 + 3x + 27

f ( x ) ⋅ g ( x ) = ( 3x + 2 ) ⋅ ( x 2 − 25 ) = 3x3 + 2 x 2 − 75x − 50 f ( x ) 3x + 2 = g ( x ) x 2 − 25

Domains: dom( f + g )  dom( f − g )   dom( fg )  = ( −∞,0 ) ∪ ( 0, ∞ ) f  dom     g  

5. 1+ x x 1 − x2 f ( x ) − g ( x ) = x1 − x = x 1 f (x) ⋅ g(x) = x ⋅ x = 1

f ( x ) + g ( x ) = x1 + x =

f ( x ) x1 1 = = 2 g(x) x x

397

Chapter 3

6. f ( x) + g( x) =

2x + 3 x − 4 + x − 4 3x + 2

( 2x + 3)(3x + 2 ) + ( x − 4 ) = ( x − 4 )(3x + 2 )

6 x 2 + 9x + 4 x + 6 + x 2 − 8x + 16 ( x − 4 )(3x + 2 )

7 x 2 + 5x + 22 ( x − 4 )( 3x + 2 )

f (x) − g(x) =

2x + 3 x − 4 − x − 4 3x + 2

( 2x + 3)(3x + 2 ) − ( x − 4 ) = ( x − 4 )( 3x + 2 )

6 x 2 + 9x + 4 x + 6 − x 2 + 8x − 16 ( x − 4 )( 3x + 2 )

5x 2 + 21x − 10 ( x − 4 )( 3x + 2 )

2x + 3 x − 4 ⋅ x − 4 3x + 2 2x + 3 = 3x + 2 2x + 3 f (x) x − 4 = x−4 g(x) 3x + 2 2 x + 3 3x + 2 = ⋅ x−4 x−4 ( 2x + 3 )(3x + 2 ) = 2 (x − 4)

f (x) ⋅ g(x) =

Domains: dom( f + g )  dom( f − g )   dom( fg )  = ( −∞, − 23 ) ∪ ( − 23 , 4 ) ∪ ( 4, ∞ ) f  dom     g  

398

Section 3.4

7. f ( x) + g( x) = x + 2 x = 3 x f ( x) − g( x) = x − 2 x = − x f ( x ) ⋅ g( x ) = x ⋅ 2 x = 2x

Domains: dom( f + g )   dom( f − g )  = [ 0, ∞ ) dom( fg )  f dom   = ( 0, ∞ ) g

1 f (x) x = = g( x) 2 x 2

8. f ( x ) + g( x ) = x − 1 + 2x2 f ( x ) − g( x ) = x − 1 − 2x2 f ( x ) ⋅ g( x ) = 2x2 x − 1 f (x) x −1 = g(x) 2x 2

9. f (x) + g(x) = 4 − x + x + 3 f (x) − g(x) = 4 − x − x + 3 f (x) ⋅ g(x) = 4 − x ⋅ x + 3 f (x) 4−x 4−x x +3 = = g(x) x +3 x +3

10.

f ( x ) + g ( x ) = 1 − 2x +

1 x

f ( x ) − g ( x ) = 1 − 2x − x1 f ( x ) ⋅ g ( x ) = 1 − 2x ⋅ x1 f (x) 1 − 2x = = x 1 − 2x 1 g( x) x

Domains: Must have both x − 1 ≥ 0 and 2 x 2 ≠ 0. So, dom( f + g )  dom( f − g )   dom( fg )  = [1, ∞ ) f  dom     g   Domains: Must have both 4 − x ≥ 0 and x + 3 ≥ 0 . So, dom( f + g )   dom( f − g )  = [ −3, 4 ] . dom( fg )  For the quotient, must have both 4 − x ≥ 0 and x + 3 > 0 . So, f dom   = ( −3, 4 ] . g Domains: Must have both 1 − 2x ≥ 0 and x ≠ 0 . So, dom( f + g )  dom( f − g )   dom( fg )  = ( −∞,0 ) ∪ ( 0, 12 ] f  dom     g  

399

Chapter 3

11.

( f  g )( x ) = 3 ( x + 2 ) − 1 = 3x + 6 − 1 = 3x + 5 ( g  f )( x ) = (3x − 1) + 2 = 3x + 1

Domains: dom ( f  g ) = ( −∞, ∞ ) , dom ( g  f ) = ( −∞, ∞ ) 12.

( f  g )( x ) = 2 (5 − x ) − 7 = 10 − 2x − 7 = 3 − 2x ( g  f )( x ) = 5 − ( 2x − 7 ) = 5 − 2x + 7 = 12 − 2x

Domains: dom ( f  g ) = ( −∞, ∞ ) , dom ( g  f ) = ( −∞, ∞ ) 13.

( f  g )( x ) = 2 ( x 2 − 3 ) + 1 = 2 x 2 − 6 + 1 = 2x 2 − 5 ( g  f )( x ) = ( 2x + 1) − 3 = 4 x 2 + 4 x + 1 − 3 = 4 x 2 + 4 x − 2 2

Domains: dom( f  g ) = ( −∞, ∞ ) = dom( g  f ) 14. ( f  g )( x ) = ( 2 − x ) − 1 = 4 − 4 x + x 2 − 1 = x 2 − 4 x + 3 2

( g  f )( x ) = 2 − ( x 2 − 1) = 2 − x 2 + 1 = − x 2 + 3

Domains: dom( f  g ) = ( −∞, ∞ ) = dom( g  f ) 15. 1 1 = ( x + 2) − 1 x + 1 1 1 + 2( x − 1) 1 + 2 x − 2 2 x − 1 ( g  f )( x ) = +2 = = = x −1 x −1 x −1 x −1 ( f  g )( x ) =

Domains: dom( f  g ) = ( −∞, −1) ∪ ( −1, ∞ ) ,

dom( g  f ) = ( −∞,1) ∪ (1, ∞ )

400

Section 3.4

16. 2 2 = (2 + x ) − 3 x − 1 2 2( x − 3) + 2 2 x − 6 + 2 2x − 4 ( g  f )( x ) = 2 + = = = x −3 x −3 x −3 x −3 ( f  g )( x ) =

Domains: dom( f  g ) = ( −∞,1) ∪ (1,3 ) ∪ ( 3, ∞ ) ,

dom( g  f ) = ( −∞,3 ) ∪ ( 3, ∞ )

17.

18. ( f  g )( x ) =

1 1 = x −1 x −1

( g  f )( x ) =

1 x −1

1 1− x −1 = x x 1 ( g  f )( x ) = x −1 Domains: dom( f  g ) = ( −∞,0 ) ∪ ( 0, ∞ ) ( f  g )( x ) =

Domains: dom( f  g ) = ( −∞,1) ∪ (1, ∞ )

dom( g  f ) = ( −∞,1) ∪ (1, ∞ )

dom( g  f ) = ( −∞, −1) ∪ ( −1,1) ∪ (1, ∞ )

19.

( f  g )( x ) = ( x + 5) − 1 = x + 4 ( g  f )( x ) = x − 1 + 5 Domains: dom( f  g ) : Must have x + 4 ≥ 0 . So, dom( f  g ) = [ −4, ∞ ) .

dom( g  f ) : Must have x − 1 ≥ 0 . So, dom( g  f ) = [1, ∞ ) .

20. ( f  g )( x ) = 2 − ( x 2 + 2 ) = 2 − x 2 − 2 = − x 2 ( g  f )( x ) =

(

2−x

) +2 =2−x+2 = 4−x 2

Domains: dom( f  g ) = [0,0] since − x 2 ≥ 0 only when x = 0. dom( g  f ) = ( −∞,2 ]

401

Chapter 3

21. 3

( f  g )( x ) = ( x − 4) 3  + 4 = x − 4 + 4 = x 1

3 ( g  f )( x ) = ( x3 + 4 ) − 4  =  x3  = x 1

Domains: dom( f  g ) = ( −∞, ∞ ) = dom( g  f ) 22.

(

)

(

( f  g )( x ) = 3 x 3 + 1 − 1 = 3 x 3 + 2 x 3 + 1 − 1 = 3 x 3 + 2 x 3 = 3 x 3 x 3 + 2 2

(

( g  f )( x ) = 3 x 2 − 1

)

+ 1 = ( x 2 − 1) + 1 2

Domains: dom( f  g ) = ( −∞, ∞ ) = dom( g  f ) 23.

24. ( f + g )(2) = f (2) + g (2)

( f + g )(10) = f (10) + g (10)

= 2 + 10  + 2 − 1

= 10 2 + 10  + 10 − 1

= 14 + 1 = 15

= 110 + 3 = 113

25.

26. ( f − g )(2) = f (2) − g (2)

( f − g )(5) = f (5) − g (5)

=  2 2 + 10  − 2 − 1

= 52 + 10  − 5 − 1

= 14 − 1 = 13

= 35 − 2 = 33

27.

28.

( f ⋅ g )(4) = f (4) ⋅ g (4)

( f ⋅ g )(5) = f (5) ⋅ g (5)

=  4 + 10  ⋅ 4 − 1

= 52 + 10  ⋅ 5 − 1

= 26 3

= 35(2) = 70

29.

30.

f f (10) 10 + 10 110 = = 3   (10) = g (10) 10 − 1 g

f f (2) 2 2 + 10 (2) = = = 14   g (2) 2 −1 g

402

Section 3.4

32.

31.

  f ( g (2)) = f   2 − 1  = 12 + 10 = 11  =1   

  f ( g (1)) = f   1 − 1  = 0 2 + 10 = 10  =0    34.

33.

  g ( f ( −3)) = g  ( −3)2 + 10  = 19 − 1    =19  

 2  g ( f (4)) = g  4 + 10  = 26 − 1 = 5  = 26   

= 3 2 35. 0 is not in the domain of g, so that g (0) is not defined. Hence, f ( g (0 )) is undefined.

36.

37. f ( g ( −3)) is not defined since g ( −3) is not defined.

38.

 2  g ( f (0)) = g  0 + 10  = 10 − 1 = 3  =10   

( ( 7 )) = g (( 7 ) + 10 ) 2

g f

= g (17) = 17 − 1 = 4 39.

( f  g )(4) = f ( g (4)) = f = f

(

4 −1

40. ( g  f )( −3) = g ( f ( −3)) = g ( ( −3)2 + 10 )

)

( 3 ) = ( 3 ) + 10 = 13 2

= g (19) = 19 − 1 = 3 2 42.

41.   f ( g (1)) = f  2(1) + 1 = 13     =3 

   1  2 f ( g (1)) = f   =1 + 1 = 2 − 2 1   =1    1  2  + 1 = = − 13 g ( f (2)) = g  2 − 2 5  =5 

g ( f (2)) = g ( 12 ) = 2 ( 12 ) + 1 = 2

403

Chapter 3

43.

44.

 2  f ( g (1)) = f  1 + 1 = 3 − 2 = 1  =2    2 g ( f (2)) = g  3 2 −   = 1 + 1 = 2   =1 

 2  f ( g (1)) = f  1 + 2  Since 3 is not in the  =3  domain of f, this is undefined. Likewise, g ( f (2)) is undefined since 2 is not in the domain of f. 46.

45.

  1 f ( g (1)) = f  2(1) − 3  = = 1   1   =1 

1   f ( g (1)) = f  1 + 3 = = 13 4 1 −  =4     1  g ( f (2)) = g   = 1+ 3 = 4 2 −1    =1 

g ( f (2)) = g ( 12 ) = 2 ( 12 ) − 3 = 2

47.

48.  2  + 5  = 6 −1 = f ( g (1)) = f  1  =6 

   1  3 1 f ( g (1)) = f  = − 2 − 3 = 3 − 72  −3   1  = −2  3  1 2 −3 = = − 14 g ( f (2)) = g      −1 − 3  =−1 

  2 − 1  = 12 + 5 = 6 g ( f (2)) = g    =1 

49. f ( g (1)) is undefined since g (1) is not defined.  1  g ( f (2)) = g  2  = g (1) , which is not  2 −3 defined. So, this is also undefined.

404

50.

f ( g (1)) = f ( 4 − 12 ) = f (3) =

3 = −3 2 −3 g ( f (2 )) is undefined since f (2) is not defined.

Section 3.4

51.

52.

f ( g (1)) = f (1 + 2(1) + 1) = f (4)

(

= 1 − ( −2)

)

g ( f (2)) = g (2 − 1) 3 = g (1) 1

( (

= (4 − 1) 3 = 3 3

(

)

(

f ( g (1)) = f (1 − 3) 3 = f ( −2) 3

)) 2

)

 2  = 1 −2 3  ,  <0  2

which is undefined . g ( f (2 )) is undefined since f (2) is not defined.

= 12 + 2(1) + 1 = 4

54.

53.

(3x + 2) − 2 3x = =x 3 3  x−2 g ( f ( x )) = 3  +2 = x−2+2 = x  3 

 x −1  f ( g ( x )) = 2   +1 = x −1 +1 = x  2  (2 x + 1) − 1 2 x = =x g ( f ( x )) = 2 2

f ( g ( x )) =

55. x =x ( x + 1) − 1 = x =  

f ( g ( x )) =

Since x ≥1

g ( f ( x )) =

56. f ( g ( x )) = 2 −

( x − 1 ) + 1 = ( x − 1) + 1 = x 2

( 2 − x ) = 2 − (2 − x ) 2

= 2−2+x = x g ( f ( x )) = 2 − ( 2 − x 2 ) = 2 − 2 + x 2 = x 2 = x

57.

58.

1 f ( g ( x )) = 1 = x x

3 f ( g ( x )) = 5 − ( 5 − x3 )  = 5 − 5 + x3 

1 g ( f ( x )) = 1 = x

=  x3  = x 3

g ( f ( x )) = 5 − (5 − x ) 3  = 5 − (5 − x ) 1

= 5 −5+ x = x

405

Chapter 3

59. 2

 x+9   x+9 f ( g ( x )) = 4   − 9 = 4  −9 = x  4   2 

( 4x − 9 ) + 9 2

g ( f ( x )) =

60.

4x 2 2x = = =x 2 2

61. 1 1 1 f ( g ( x )) = x +1 = x +1− x = 1 = x x −1 x x

 x3 + 1  3 3 f ( g ( x )) = 3 8   −1 = x = x  8  g ( f ( x )) =

1 + 1 1+ x −1 g ( f ( x )) = x −11 = x1−1 = x1−1 = x x

( 8 x − 1 ) + 1 = 8x − 1 + 1 = x 3

x −1

62.

f ( g ( x )) = g ( f ( x )) = 25 −

(

25 − x 2

) = 25 − ( 25 − x ) = x = x since x ≥ 0. 2

64. The most natural pairs are: f ( x ) = x1 g ( x ) = x 2 + 1

63. f ( x ) = 2 x + 5 x g ( x ) = 3x − 1 2

f ( x ) = x1+1 g ( x ) = x 2 65. f ( x ) = x2 67. f ( x ) =

g( x) = x − 3

3 x −2

66. f ( x ) = x

g(x) = x + 1

68. f ( x ) =

g( x ) = 1 − x2

x g(x) = x 3x + 2

69. F (C ( K )) = 59 ( K − 273.15 ) + 32 70. We need to calculate the composition function ( K  C )( F ) . Solve F = 59 C + 32 for C: C = 59 ( F − 32) Solve C = K − 273.15 for K: K = C + 273.15

So, ( K  C )( F ) = K (C ( F )) = K ( 59 ( F − 32) ) = 59 ( F − 32) + 273.15 = 59 F + 255.37 .

Thus, 32 F corresponds to 59 (32) + 255.37 = 273.15K , and 212 F corresponds to 59 (212) + 255.37 = 373.15K .

406

Section 3.4

71. Let x = number of linear feet of fence purchased. a. Let l = length of each side of the square pen 4l = x so that l = x4  Perimeter

A = area of the square pen = l 2 . 2 So, ( A  l )( x ) = A ( x4 ) = ( x4 ) . 2 b. A(100) = ( 100 4 ) = 625 ft 2

2 c. A(200) = ( 200 4 ) = 2500 ft 2

72. Let x = number of linear feet of fence purchased. a. Let l = length of a radius of the circular pen 2π l = x so that l = 2xπ  Circumference

A = area of the circular pen = π l 2 . 2 2 So, ( A  l )( x ) = A ( 2xπ ) = π ( 2xπ ) = 4xπ . 2500 b. A(100) = 100 4π = π 2

10,000 c. A(200) = 200 π 4π = 2

73. First, solve p = 3000 − 12 x for x: x = 2(3000 − p ) = 6000 − 2 p a. C ( x( p)) = C ( 6000 − 2 p ) = 2000 +10(6000 − 2 p) = 62,000 − 20 p b. R( x( p)) = 100(6000 − 2 p) = 600,000 − 200 p c. Profit P = R − C . So, P ( x( p )) = R( x( p )) − C ( x( p ))

= ( 600,000 − 200 p ) − ( 62,000 − 20 p ) = 538,000 − 180 p 74. First, solve p = 10,000 − 14 x for x: x = 4(10,000 − p ) = 40,000 − 4 p a. C ( x( p)) = C ( 40,000 − 4 p ) = 30,000 + 5(40,000 − 4 p) = 230,000 − 20 p b. R( x( p)) = 1000(40,000 − 4 p) = 40,000,000 − 4000 p c. Profit P = R − C . So, P ( x( p )) = R( x( p )) − C ( x( p ))

= ( 40,000,000 − 4000 p ) − ( 230,000 − 20 p ) = 39,770,000 − 3,980 p

407

Chapter 3

75. a. This gives the price of the shoes after a 25% discount followed by a $10 discount. ( f  g )( x ) = f ( g ( x ) ) = f ( 0.75x )

= 0.75x − 10 b. This gives the price of the shoes after a $10 discount followed by a 25% discount. ( g  f )( x ) = g ( f ( x ) )

= g ( x − 10 )

= 0.75 ( x − 10 )

= 0.75x − 7.5 c. Since 0.75x − 10 > 0.75x − 7.5 , typing in coupon f first would yield the bigger discount. 76. a. This gives the price of the shoes after a 10% discount followed by a 25% discount. ( g  h )( x ) = g ( h ( x ) )

= g ( 0.9x )

= 0.75 ( 0.9x ) = 0.675x b. This gives the price of the shoes after a 25% discount followed by a 10% discount. ( h  g )( x ) = h ( g ( x ) ) = h ( 0.75x )

= 0.9 ( 0.75x ) = 0.675x c. Since 0.675x = 0.675x it doesn’t matter what order he types in the codes. 77. a. A( r(t)) = π (10t − 0.2t 2 )

b. A( r(7)) = π (10(7) − 0.2(7)2 ) = 11,385 2

square miles

78. a. A( r(t)) = π ( 8t − 0.1t 2 )

b. A( r(5)) = π ( 8(5) − 0.1(5)2 ) = 4418 2

square miles

408

Section 3.4

79. 2

A(t ) = π 150 t  = 22,500π t ft 2

80. Volume of rectangular pool = length × width × height = 20 × 10 × h = 200 h . Let t = number of hours water has been pumped into the pool. Then, from the information we are given, volume = 50t . 50 t Thus, solving 50t = 200 h for h yields: h = 200 =t 4 Height of water as a function of time t

81. Let h = height of the fireworks above ground. Then, the distance between the family and the fireworks is given by

d ( h ) = ( h − 0)2 + (0 − 2)2 = h2 + 4 .

82. a. R ( n ) = 75n − 0.06 ( 75 ) n = 70.5n b. 70.5n − 1700 ≥ 0 70.5n ≥ 1700 n ≥ 24.11 Laura would need to rent her spare bedroom for 25 days or more in order to cover her mortgage payment that month. 83. Must exclude −2 from the domain.

84. Must also exclude −2 from the domain.

85. ( f  g )( x ) = f ( g ( x )) , not f ( x ) ⋅ g( x )

86. Domain is [3,∞ )

87. The mistake made was that ( f + g ) was multiplied by 2 when it ought to have been evaluated at 2.

409

Chapter 3

88. Didn’t distribute “ – ” to all parts of g ( x ) . Should have been:

f ( x ) − g ( x ) = ( x + 2) − ( x 2 − 4 ) = x + 2 − x2 + 4 = −x2 + x + 6

90. False. For example, consider the functions f ( x ) = x + 1, g ( x ) = 3 . Observe that f ( g (4)) = f (3) = 4

89. False. The domain of the sum, difference, or product of two functions is the intersection of their domains; the domain of the quotient is the set obtained by intersecting the two domains and then excluding all values where the denominator equals 0. 91. True 92. False 93.

g ( f (4)) = g (5) = 3

( g  f )( x ) =

1 1 = (x + a) − a x

Domain: x ≠ 0

94.

( g  f )( x ) =

1 1 1 = 2 = ( ax + bx + c ) − c ax + bx x(ax + b ) 2

Domain: x ≠ 0, − ba , c 95. ( g  f )( x ) =

96.

( x+a) −a = x+a−a = x 2

( g  f )( x ) =

1 b

1 = x ab 1 x ab

 1   a x  Domain: The domain depends on the values of a and b. For instance, if a = 1 and b = 3 , then domain 1 is ( −∞,0 ) ∪ ( 0, ∞ ) . If a = 2 and b = 1 ,

Domain: Must have x + a ≥ 0, so that x ≥ −a . So, domain is [ −a, ∞ ) .

then domain is ( 0,∞ ) .

410

Section 3.4

97. Notes on the graph: The dotted curve is the graph of y1 , while the thick, solid curve is the graph of y1 + y2 . Domain of y2 is [−7, 9].

98. y1 = 3 x + 5 1 3−x y1 y3 = y2 y2 =

Domain of y3 = ( −∞,3 )

The graph of y3 is as follows:

99.

y1 = x − 3x − 4 1 y2 = 2 x − 14 1 y3 = 2 ( y1) − 14 2

Domain of y3 = ( −∞, −3 ) ∪ ( −3, −1] ∪ [4,6) ∪ (6, ∞ )

411

Chapter 3

100. Notes on the graph: The dotted curve is the graph of y1 , while the thick, solid curve is the graph of y3 = y12 + 2 .

412

Section 3.5 Solutions -------------------------------------------------------------------------------1. Is a function. Not one-to-one April and October both map to 78 F .

2. Is a function. One-to-one

3. Is a function. One-to-one

4. Is a function. Not one-to-one Carrie and Michael both map to A, for instance.

5. Is a function. One-to-one

6. Is a function. One-to-one

7. Not a function since 4 maps to both 2 and −2 .

8. Is a function. Not one-to-one 0,1,2,3 all map to 1 in the range.

9. Is a function. Not one-to-one 0,2, −2 all map to 1 in the range, for instance.

10. Is a function. One-to-one

11. Is a function. Not one-to-one Doesn’t pass the horizontal line test. Both ( −1,1), (0,1) are on the graph.

12. Is a function. Not one-to-one Doesn’t pass the horizontal line test.

13. Is a function. One-to-one

14. Is a function. One-to-one

15. Is a function. Not one-to-one Doesn’t pass the horizontal line test.

16. Is a function. One-to-one

413

Chapter 3

17. Not one-to-one. Both (0,3), (6,3) lie on the graph.

19. f ( x1 ) = f ( x2 ) 

1 1 = x1 − 1 x2 − 1

 x2 − 1 = x1 − 1  x2 = x1

One-to-one

414

18. Not one-to-one. Both (0,5), (4,5) lie on the graph.

Section 3.5

20. f ( x1 ) = f ( x2 )  3 x1 = 3 x2 

( x ) =( x ) 3

 x1 = x2

One-to-one

21. f is not one-to-one since, for example, f ( −1) = f (1) = −3 .

22.

f ( x1 ) = f ( x2 ) 

x1 + 1 = x2 + 1

 x1 + 1 = x2 + 1  x1 = x2

One-to-one

415

Chapter 3

23.

f ( x1 ) = f ( x2 )  x13 − 1 = x23 − 1  x13 = x23  x1 = x2

One-to-one

24. f ( x1 ) = f ( x2 ) 

1 1 = x1 + 2 x2 + 2

 x2 + 2 = x1 + 2  x2 = x1

One-to-one

416

Section 3.5

25.

Given: f ( x ) = 2x + 1, f −1 ( x ) =

x −1 2

 x −1  f ( f −1 ( x ) ) = 2   +1 = x −1+1 = x  2  (2 x + 1) − 1 2 x = =x f −1 ( f ( x ) ) = 2 2

Notes on the graph: Thick, solid curve is the graph of f. Thin, solid curve is the graph of f −1 . Thick, dotted curve is the graph of y = x. 26. x−2 , f −1 ( x ) = 3x + 2 3 (3x + 2) − 2 3x f ( f −1 ( x ) ) = = =x 3 3  x−2 f −1 ( f ( x ) ) = 3  +2 = x−2+2 = x  3 

Given: f ( x ) =

Notes on the graph: Thick, solid curve is the graph of f. Thin, solid curve is the graph of f −1 . Thick, dotted curve is the graph of y = x.

417

Chapter 3

27.

Given: f ( x ) = −3x + 5; f −1 ( x ) =

x −5 −3

 x −5  f ( f −1 ( x ) ) = −3   +5 = x −5+5 = x  −3  −3x + 5 − 5 −3x f −1 ( f ( x ) ) = = =x −3 −3

Notes on the graph: Red curve is the graph of f . Blue curve is the graph of f −1. Green curve is the graph of y = x . 28.

1 − x −1 ; f ( x ) = 1 − 4x 4 1 − (1 − 4 x ) 4x f ( f −1 ( x ) ) = = =x 4 4  1− x  f −1 ( f ( x ) ) = 1 − 4  1 − (1 − x ) = x  4 

Given: f ( x ) =

Notes on the graph: Red curve is the graph of f . Blue curve is the graph of f −1. Green curve is the graph of y = x .

418

Section 3.5

29. Given: f ( x ) = x3 + 7; f −1 ( x ) = 3 x − 7

(

f ( f −1 ( x ) ) = 3 x − 7

) +7 = x −7+7 = x 3

f −1 ( f ( x ) ) = 3 x3 + 7 − 7 = 3 x3 = x

Notes on the graph: Red curve is the graph of f . Blue curve is the graph of f −1. Green curve is the graph of y = x . 30. 3 Given: f ( x ) = 3 x + 2; f −1 ( x ) = ( x − 2 ) f ( f −1 ( x ) ) = 3 ( x − 2 ) + 2 = x − 2 + 2 = x 3

(

f −1 ( f ( x ) ) = 3 x + 2 − 2

) =( x) = x 3

Notes on the graph: Red curve is the graph of f . Blue curve is the graph of f −1. Green curve is the graph of y = x .

419

Chapter 3

31. Given: f ( x ) = x − 1, x ≥ 1

f −1 ( x ) = x 2 + 1, x ≥ 0 f ( f −1 ( x ) ) =

x =x ( x + 1) − 1 = x =   2

Since x ≥ 0

f −1 ( f ( x ) ) =

( x −1) + 1 = x −1 + 1 = x 2

Notes on the graph: Thick, solid curve is the graph of f. Thin, solid curve is the graph of f −1 . Thick, dotted curve is the graph of y = x. 32. Given: f ( x ) = 2 − x 2 , x ≥ 0

f −1 ( x ) = 2 − x , x ≤ 2 f ( f −1 ( x ) ) = 2 −

( 2 − x ) = 2 − (2 − x ) = x 2

f −1 ( f ( x ) ) = 2 − ( 2 − x 2 ) = x 2 = x = x 

Since x ≥ 0

Notes on the graph: Thick, solid curve is the graph of f. Thin, solid curve is the graph of f −1 . Thick, dotted curve is the graph of y = x.

420

Section 3.5

33. Given: f ( x ) = x1 , f −1 ( x ) = x1 1 f ( f −1 ( x ) ) = 1 = x x

1 f −1 ( f ( x ) ) = 1 = x x

Notes on the graph: Thick, solid curve is the graph of f. Thin, solid curve is the graph of f −1 . Thick, dotted curve is the graph of y = x. 34. 1 Given: f ( x ) = (5 − x ) 3 , f −1 ( x ) = 5 − x3

f ( f −1 ( x ) ) = (5 − (5 − x3 )) 3 = (5 − 5 + x3 ) 3 1

= ( x3 ) 3 = x 1

f −1 ( f ( x ) ) = 5 − (5 − x ) 3  = 5 − (5 − x ) = x 1

Notes on the graph: Thick, solid curve is the graph of f. Thin, solid curve is the graph of f −1 . Thick, dotted curve is the graph of y = x.

421

Chapter 3

35. 1 1 , f −1 ( x ) = −3 2x + 6 2x 1 1 f ( f −1 ( x ) ) = = 1 −6+6  1  2 − 3 + 6 x  2x  1 = 1 =x

Given: f ( x ) =

1 1 −3 = −3 1  1  2 x+3  2 x + 6  = x +3−3 = x

f −1 ( f ( x ) ) =

Notes on the graph: Thick, solid curve is the graph of f. Thin, solid curve is the graph of f −1 . Thick, dotted curve is the graph of y = x.

36. 3 3 , f −1 ( x ) = 4 − x 4−x 3 3 f ( f −1 ( x ) ) = = 3 3  4−4−  4−4+ x x  x = 3⋅ = x 3 3 4−x f −1 ( f ( x ) ) = 4 − = 4− 3 ⋅ 3 3 4−x = 4−4+x = x

Given: f ( x ) =

Notes on the graph: Thick, solid curve is the graph of f. Thin, solid curve is the graph of f −1 . Thick, dotted curve is the graph of y = x.

422

Section 3.5

37. x +3 3 − 4x , f −1 ( x ) = x+4 x −1 3 − 4x 3 − 4 x + 3( x − 1) +3 x −1 f ( f −1 ( x ) ) = x − 1 = 3 − 4x 3 − 4 x + 4( x − 1) +4 x −1 x −1 3 − 4x + 3x − 3 −x x −1 = x −1 = −1 3 − 4 x + 4x − 4 x −1 x −1

Given: f ( x ) =

−x x − 1 ⋅ =x x − 1 −1

Notes on the graph: Thick, solid curve is the graph of f. Thin, solid curve is the graph of f −1 . Thick, dotted curve is the graph of y = x.

 x +3  4x + 12 3− 4  3− x+4 = x+4 f −1 ( f ( x ) ) = x + 3  x +3  −1   −1 x+4 x+4 −x 3x + 12 − 4x − 12 x+4 = x+4 = −1 x +3− x −4 x+4 x+4 −x x + 4 = ⋅ =x x + 4 −1

423

Chapter 3

38.

x −5 3x + 5 , f −1 ( x ) = 3−x x +1 3x + 5 3x + 5 − 5( x + 1) −5 x +1 f ( f −1 ( x ) ) = x + 1 = 3x + 5 3x + 3 − ( 3 x + 5 ) 3− x +1 x +1 −2 x 3x + 5 − 5x − 5 x +1 = x +1 = −2 3x + 3 − 3x − 5 x +1 x +1 −2 x x + 1 =x ⋅ = x + 1 −2

Given: f ( x ) =

 x −5  3x − 15 3 +5 +5 3 x −  f −1 ( f ( x ) ) =  = 3−x x −5  x −5  +1   +1 3−x 3−x  3x − 15 + 5(3 − x ) 3x − 15 + 15 − 5x 3−x 3−x = = x −5+3− x x −5 +3− x 3−x 3−x =

−2 x 3 − x =x ⋅ 3 − x −2

424

Notes on the graph: Thick, solid curve is the graph of f. Thin, solid curve is the graph of f −1 . Thick, dotted curve is the graph of y = x.

Section 3.5

40.

39.

Thick, solid curve is the graph of f −1 .

Notes on the graph: Thin, solid curve is the graph of f. Thick, dotted curve is the graph of y = x. Thick, solid curve is the graph of f −1 .

41.

42.

Notes on the graph: Thin, solid curve is the graph of f. Thick, dotted curve is the graph of y = x .

Notes on the graph: Thin, solid curve is the graph of f. Thick, dotted curve is the graph of y = x. Thick, solid curve is the graph of f −1 .

Notes on the graph: Thin, solid curve is the graph of f. Thick, dotted curve is the graph of y = x .

Thick, solid curve is the graph of f −1 .

425

Chapter 3

43.

44.

Notes on the graph: Thin, solid curve is the graph of f. Thick, dotted curve is the graph of y = x . Thick, solid curve is the graph of f −1 .

Notes on the graph: Thin, solid curve is the graph of f. Thick, dotted curve is the graph of y = x. Thick, solid curve is the graph of f −1 .

45.

46.

Notes on the graph: Thin, solid curve is the graph of f. Thick, dotted curve is the graph of y = x . Thick, solid curve is the graph of f −1 .

Notes on the graph: Thin, solid curve is the graph of f. Thick, dotted curve is the graph of y = x.

426

Thick, solid curve is the graph of f −1 .

Section 3.5

47. a. The point ( 2,5 ) lies on the graph of f . Thus, f ( 2 ) = 5. b. The point ( −3, 0 ) lies on the graph of f . Thus, x = −3.

c. Since the point ( −3, 0 ) lies on the graph of f , we know the point ( 0, −3 ) lies on

the graph of f −1. Thus, f −1 ( 0 ) = −3.

d. Since the point ( 0,3 ) lies on the graph of f , we know the point ( 3,0 ) lies on

the graph of f −1. Thus, x = 3. 48. a. The point ( 2, −3 ) lies on the graph of f . Thus, f ( 2 ) = −3. b. The point (1, 0 ) lies on the graph of f . Thus, x = 1.

c. Since the point ( 2, −3 ) lies on the graph of f , we know the point ( −3, 2 ) lies on

the graph of f −1. Thus, f −1 ( −3 ) = 2.

d. Since the point ( 0,1) lies on the graph of f , we know the point (1,0 ) lies on

the graph of f −1. Thus, x = 1. 49. a. The point ( −2, 4 ) lies on the graph of f . Thus, f ( −2 ) = −4. b. The point ( 2,0 ) lies on the graph of f . Thus, x = 2.

c. Since the point ( 2, 0 ) lies on the graph of f , we know the point ( 0, 2 ) lies on

the graph of f −1. Thus, f −1 ( 0 ) = 2.

d. Since the point ( −2, 4 ) lies on the graph of f , we know the point ( 4, −2 ) lies

on the graph of f −1. Thus, x = −2. 50. a. The point ( 5, 6 ) lies on the graph of f . Thus, f ( 5 ) = 6. b. The point (1, 4 ) lies on the graph of f . Thus, x = 1.

c. Since the point ( 2,5 ) lies on the graph of f , we know the point ( 5, 2 ) lies on

the graph of f −1. Thus, f −1 ( 5 ) = 2.

d. Since the point (1, 4 ) lies on the graph of f , we know the point ( 4,1) lies on

the graph of f −1. Thus, x = 4.

427

Chapter 3

51. Solve y = x − 1 for x: x = y +1 Thus, f −1 ( x ) = x + 1 .

Domains: dom ( f ) = rng ( f −1 ) = ( −∞, ∞ )

52. Solve y = 7 x for x: x = 71 y

Domains: dom ( f ) = rng ( f −1 ) = ( −∞, ∞ )

rng ( f ) = dom ( f −1 ) = ( −∞, ∞ )

Thus, f −1 ( x ) = 71 x . 53. Solve y = −3x + 2 for x: x = − 13 ( y − 2)

Domains: dom ( f ) = rng ( f −1 ) = ( −∞, ∞ ) rng ( f ) = dom ( f −1 ) = ( −∞, ∞ )

Thus, f −1 ( x ) = − 13 ( x − 2) = − 13 x + 23 . 54. Solve y = 2x + 3 for x: x = 12 ( y − 3)

Domains: dom ( f ) = rng ( f −1 ) = ( −∞, ∞ ) rng ( f ) = dom ( f −1 ) = ( −∞, ∞ )

Thus, f −1 ( x ) = 12 ( x − 3) .

Domains: dom ( f ) = rng ( f −1 ) = ( −∞, ∞ )

55. Solve y = x3 + 1 for x: x = 3 y −1

rng ( f ) = dom ( f −1 ) = ( −∞, ∞ )

Thus, f −1 ( x ) = 3 x − 1 .

Domains: dom ( f ) = rng ( f −1 ) = ( −∞, ∞ )

56. Solve y = x3 − 1 for x: x = 3 y +1

rng ( f ) = dom ( f −1 ) = ( −∞, ∞ )

Thus, f −1 ( x ) = 3 x + 1 .

Domains: dom ( f ) = rng ( f −1 ) = [3, ∞ )

57. Solve y = x − 3 for x: x = y2 + 3

Thus, f −1 ( x ) = x 2 + 3 .

rng ( f ) = dom ( f −1 ) = [ 0, ∞ )

58. Solve y = 3 − x for x: x = 3 − y2 Thus, f −1 ( x ) = 3 − x 2 .

Domains: dom ( f ) = rng ( f −1 ) = ( −∞,3] rng ( f ) = dom ( f −1 ) = [ 0, ∞ )

428

Section 3.5

59. Solve y = x 2 − 1 for x:

Domains: dom ( f ) = rng ( f −1 ) = [ 0, ∞ )

x = y +1

Thus, f −1 ( x ) = x + 1 .

rng ( f ) = dom ( f −1 ) = [ −1, ∞ )

60. Solve y = 2 x 2 + 1 for x:

Domains: dom ( f ) = rng ( f −1 ) = [ 0, ∞ )

2x 2 = y − 1 y −1 (since x ≥ 0) 2 x −1 . Thus, f −1 ( x ) = 2

rng ( f ) = dom ( f −1 ) = [1, ∞ )

61. Solve y = ( x + 2)2 − 3 for x:

Domains: dom ( f ) = rng ( f −1 ) = [ −2, ∞ )

x=+

y + 3 = ( x + 2)2

rng ( f ) = dom ( f −1 ) = [ −3, ∞ )

y + 3 = x + 2 (since x ≥ −2) −2 + y + 3 = x

Thus, f −1 ( x ) = −2 + x + 3 . 62. Solve y = ( x − 3)2 − 2 for x:

Domains: dom ( f ) = rng ( f −1 ) = [3, ∞ )

y + 2 = ( x − 3)2

rng ( f ) = dom ( f −1 ) = [ −2, ∞ )

y + 2 = x − 3 (since x ≥ 3) 3+ y+2 = x

Thus, f −1 ( x ) = 3 + x + 2 . 63. Solve y = x2 for x: xy = 2

Domains: dom ( f ) = rng ( f −1 ) = ( −∞,0 ) ∪ ( 0, ∞ ) rng ( f ) = dom ( f −1 ) = ( −∞,0 ) ∪ ( 0, ∞ )

x = 2y

Thus, f −1 ( x ) = x2 . 64. Solve y = − x3 for x: yx = −3

Domains: dom ( f ) = rng ( f −1 ) = ( −∞,0 ) ∪ ( 0, ∞ ) rng ( f ) = dom ( f −1 ) = ( −∞,0 ) ∪ ( 0, ∞ )

x = − 3y

Thus, f −1 ( x ) = − x3 .

429

Chapter 3

65. Solve y = 3−2x for x:

Domains: dom ( f ) = rng ( f −1 ) = ( −∞,3 ) ∪ ( 3, ∞ )

(3 − x ) y = 2

rng ( f ) = dom ( f −1 ) = ( −∞,0 ) ∪ ( 0, ∞ )

3 y − xy = 2 xy = 3 y − 2 x = 3 yy−2 Thus, f −1 ( x ) = 3 xx−2 = 3 − x2 . 66. Solve y = x7+ 2 for x:

Domains: dom ( f ) = rng ( f −1 ) = ( −∞, −2 ) ∪ ( −2, ∞ )

( x + 2) y = 7

rng ( f ) = dom ( f −1 ) = ( −∞,0 ) ∪ ( 0, ∞ )

2 y + xy = 7 xy = 7 − 2 y x = 7 −y2 y Thus, f −1 ( x ) = 7 −x2 x . 67. Solve y = 75x−+x1 for x: y(5 − x ) = 7 x + 1 5 y − xy = 7 x + 1 −7 x − xy = 1 − 5 y − x(7 + y ) = 1 − 5 y

Domains: dom ( f ) = rng ( f −1 ) = ( −∞,5 ) ∪ ( 5, ∞ )

rng ( f ) = dom ( f −1 ) = ( −∞, −7 ) ∪ ( −7, ∞ )

x = 57y+−y1

Thus, f −1 ( x ) = 5xx+−71 . 68. Solve y = 27x++x5 for x: y(7 + x ) = 2 x + 5 xy + 7 y = 2 x + 5 −2 x + xy = 5 − 7 y x( y − 2) = 5 − 7 y

Domains: dom ( f ) = rng ( f −1 ) = ( −∞, −7 ) ∪ ( −7, ∞ ) rng ( f ) = dom ( f −1 ) = ( −∞,2 ) ∪ ( 2, ∞ )

x = 5y−−72y

Thus, f −1 ( x ) = 5x−−72x .

430

Section 3.5

69. Not one-to-one

Calculate the inverse function piecewise: For x < 0 : Solve y = x1 for x:

70. One-to-one

y = x1 xy = 1 x = 1y

So, G −1 ( x ) = x1 on ( −∞,0 ) . For x ≥ 0 : Solve y = x for x: y= x x = y2 So, G −1 ( x ) = x 2 on ( 0,∞ ) . Thus, the inverse function is given by: 1  x , x < 0 G −1 ( x ) =  2 x , x ≥ 0

431

Chapter 3

71. One-to-one The portion of the graph for nonnegative x values is as follows. The graph for negative x-values is merely a reflection of this graph over the origin.

 x, x ≤ −1,  f −1 ( x ) =  3 x , − 1 < x < 1,  x, x ≥ 1  72. Not one-to-one

73. Solve y = 59 x + 32 for x: y − 32 = 95 x 5 9

( y − 32) = x

So, f −1 ( x ) = 59 ( x − 32) . The inverse function represents the conversion from degrees Fahrenheit to degrees Celsius.

74. Solve y = 12x for x :

y =x 12 So, f −1 ( x ) =

x . The inverse function represents the conversion from inches to feet. 12

432

Section 3.5

75. Let x = number of boats entered. The cost function is 250 x, 0 ≤ x ≤ 10   C ( x ) = 2500 + 175( x − 10), x > 10     = 175 x + 750 To calculate C −1 ( x ) , we calculate the inverse of each piece separately: y x For 0 ≤ x ≤ 10 : Solve y = 250x for x: x = 250 . So, C −1 ( x ) = 250 , for 0 ≤ x ≤ 2500 . −750 −750 For x > 10 : Solve y = 175x + 750 for x: x = y175 . So, C −1 ( x ) = x175 , for x > 2500 . Thus, the inverse function is given by: x , 0 ≤ x ≤ 2500  250 −1 C ( x ) =  x −750  175 , x > 2500

76. Let x = number of months after 2020. The value function is V ( x ) = 25000 x  − 1500  initial value

value lost

To calculate V −1 , solve y = 25000 − 1500x for x : x = So, V −1 ( x ) =

25000 − y . 1500

25000 − x , x ≥ 0. 1500

77. Let x = number of hours worked. Then the amount of the paycheck is given by E (x) = 30 x − 0.3 ( 30 x ) − 350   = 21x − 350   $30 per hour, overhead for x hours

30% of his fares

To calculate E −1 , solve y = 21x − 350 for x : x =

y + 350 . 21

x + 350 , x ≥ 0 . The inverse function tells him how many hours to 21 work when inputting the desired paycheck.

So, E −1 ( x ) =

433

Chapter 3

78. Let x = number of hours worked. Since the hourly rate for overtime pay is 1.5 (15 ) = 22.5 dollars per hour, we see

that the weekly earnings are described by the following function:  15x,  0 ≤ x ≤ 40  15x, 0 ≤ x ≤ 40 E ( x ) =  600 x − 40 ) = (  + 22.5   x > 40  22.5x − 300, x > 40 Pay for first  Amount of 40 hours  overtime pay To calculate E −1 , we calculate the inverse of each piece separately: x y For 0 ≤ x ≤ 40 : Solve y = 15x for x : x = . So E −1 ( x ) = , for 0 ≤ x ≤ 600 . 15 15 y + 300 y + 300 For x > 40 : Solve y = 22.5x − 300 for x : x = . So E −1 ( x ) = , for 22.5 22.5 x > 600. Thus, the inverse function is given by: x 0 ≤ x ≤ 600 15 , −1 E (x) =   x + 300 , x > 600  22.5 The inverse function tells you how many hours you need to work to attain a certain take-home pay. 79. Since we are only looking at 1 24-hour period, the domain is [0, 24]. For the range, note that T(t) is always increasing (being a translate of t3) and so, its minimum is T(0) and its maximum is T(24). So, the range is the approximate interval [97.5528, 101.70].

434

Section 3.5

80. Solve for t: T = 0.0003(t − 24)3 + 101.70 T − 101.70 = 0.0003(t − 24)3  T − 101.70  3   = (t − 24)  0.0003  1

 T − 101.70  3   = (t − 24)  0.0003  1

 T − 101.70  3 t=  + 24  0.0003  The right-side is the inverse of T. Changing the notation, we conclude that 1

 t − 101.70  T (t ) =   + 24  0.0003  −1

T −1 (1) = 3

t − 101.70 + 24 0.0003

T −1 (1) = 3

10000 ( t − 101.70 ) + 24 3

81. The domain of T −1 (t ) is the range of T(t) and is [97.5528, 101.70]. The range of T −1 (t ) is the domain of T(t) and is [0, 24]. 82. Solve for t when T = 99.5. Using the formula for the inverse from #72, we simply compute T −1 (99.5), which turns out to be approximately 4.57. So, around 5am. 83. Not a function since the graph does not pass the vertical line test.

84. To determine the points on the graph of the inverse of f, switch the order of the x and y in the ordered pairs rather than multiplying by −1 . So, in this case, since (3,3), (0, −4) are on the graph of f, the points (3,3), ( −4,0) are on the graph of f −1 .

85. Must restrict the domain to a portion on which f is one-to-one, say x ≥ 0 . Then, the calculation will be valid.

86. dom ( f −1 ) = rng ( f ) = [ 0, ∞ ) , not

435

[2,∞ ) .

Chapter 3

87. False. In fact, no even function can be one-to-one since the condition f ( x ) = f ( − x ) implies that the horizontal line test is violated.

88. False. The function f ( x ) = 0 is odd, but not one-to-one.

90. True. dom( f ) is inside ( −∞,0 ) and

91. ( b,0) since the x and y coordinates of all points on the graph of f are switched to get the corresponding points on the graph of f −1 .

rng ( f ) is inside ( 0,∞ ) . Since they are

switched for f −1 , dom( f −1 ) is inside ( 0,∞ )

and rng ( f −1 ) is inside ( −∞,0 ) . Thus, the graph of f −1 is in Quadrant IV.

89. False. Consider f ( x ) = x . Then, f −1 ( x ) = x also.

92. If ( a,0) is on the graph of f, then (0, a ) is on the graph of f −1 . This is its y-intercept.

93. The equation of the unit circle is x 2 + y 2 = 1 . The portion in Quadrant I is given by y = 1 − x 2 , 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 . To calculate the inverse of this function, solve for x:

y2 = 1 − x 2 , which gives us x = 1 − y 2 So, f −1 ( x ) = 1 − x 2 , 0 ≤ x ≤ 1 . The domain and range of both are [0,1] . 94. Let f ( x ) = xc , c ≠ 0 . To calculate the inverse of this function, solve for x: y = xc  x = cy  yx = c  y = xc .

Thus, f ( x ) = f −1 ( x ), x ≠ 0 . 95. As long as m ≠ 0 (that is, while the graph of f is not a horizontal line), it is one-to-one.

96. Assume m ≠ 0 . Then, solving y = mx + b for x yields: x = ym−b . So, the inverse of f ( x ) = mx + b is f −1 ( x ) = xm−b .

436

Section 3.5

97. Not one-to-one

98. One-to-one

99. Not one-to-one

100. One-to-one

437

Chapter 3

101. Not inverses. Had we restricted the domain of the parabola to [ 0, ∞ ) , then

they would have been.

103. Yes, it appears as though the given functions are inverses of each other.

438

102. Yes, it appears as though the given functions are inverses of each other.

104. No, they are not inverses.

Section 3.6 Solutions -------------------------------------------------------------------------------1. y = kx

2. s = kt

3. V = kx3

4. A = kx 2

5. z = km

6. h = k t

f =

k r2

10.

kw F= L

11. v = kgt

12. S = ktd

13.

14.

kT P

k PT

15. y = k x

16.

17. The general equation is d = kt . Using the fact that d = k (1) = r, we see that

18. The general equation is F = km . Using the fact that F = k (1) = a, we see that

k = r . So, d = rt .

k = a . So, F = ma .

19. The general equation is V = klw . Using the fact that V = k (2)(3) = 6h we

20. The general equation is A = kbh . Using the fact that A = k (5)(4) = 10 we see that

see that k = h . So, V = lwh .

k xz

k y= 3 t

439

1 1 . So, A = bh . 2 2

Chapter 3

21. The general equation is d = kr 2 . Using the fact that A = k (3)2 = 9π , we

22. The general equation is V = kr 3 . Using the fact that V = k (3)3 = 36π , we

see that k = π . So, A = π r 2 .

4 4 see that k = π . So, V = π r 3 . 3 3

23. The general equation is V = khr 2 . 4 Using the fact that V = k   (2)2 = 1, π 

24. The general equation is W = kRI 2 . Using the fact that W = k (100)(0.25)2 = 4 , we see that 25k 16 = 4 , so that k = . Hence, 4 25 16 W = RI 2 . 25

we see that k =

π 16

. So, V =

π 16

hr 2 .

25.

26.

The general equation is V =

k . Using P

k = 1000 , we see 400 that k = 400,000 . Hence, the fact that V =

400,000 . P

k . Using d2

k = 42 , we see that 16 2 10,752 k = 10,752 . Hence, I = . d2

the fact that I =

28.

27.

The general equation is F =

k . λL

Using the fact that k F= = 20π , we see that −6 (10 m )(105 m ) k = 2π . So, F =

The general equation is I =

2π . λL

The general equation is y =

k . Using xz

k = 32 , we see 4(0.05) that k = 4(0.05)(32) = 6.4 . Hence,

the fact that y =

440

6.4 . xz

Section 3.6

29.

30.

The general equation is t =

k . Using s

The general equation is W =

k . Using d2

k = 2.4 , we see that 8 19.2 k = 2.4(8) = 19.2 . Hence, t = . s

k = 180 , we see (0.2)2 that k = 180(0.2)2 = 7.2 . Hence,

31.

32.

k The general equation is R = 2 . Using I k = 0.4 , we see the fact that R = (3.5)2 that k = (3.5)2 (0.4) = 4.9 . Hence,

The general equation is y =

the fact that t =

4.9 . I2

the fact that W =

7.2 . d2

x z

k = 12 , we (0.2) 4 see that k = 12(0.2)(2) = 4.8 . Hence, Using the fact that y =

4.8 . x z

33.

34.

kL . A k (20) = 0.5 , we Using the fact that R = (0.4) see that 50 k = 0.5 , so that k = 0.01 . 0.01L . Hence, R = A

The general equation is F =

The general equation is R =

km . Using d

k (20) = 32 , we see that 8 12.8m k = 32 ( 208 ) = 12.8 . Hence, F = . d

the fact that F =

441

Chapter 3

35.

36.

km1m2 . d2 k (8)(16) = 20 , Using the fact that F = (0.4)2

k g . t2 k 16 Using the fact that w = = 20 , we (0.5)2

The general equation is F =

we see that k = Hence, F =

20(0.4)2 = 0.025 . (8)(16)

0.025m1m2 . d2

The general equation is w =

20(0.5)2 = 1.25 . Hence, see that k = 4

1.25 g . t2

37. S = 10.5W 38. Orange County: Assume that T = kP . We need to determine k. Using the data, we see that 2.60 = 40k , so that k = 0.065 . So, T = 0.065P . Seminole County: Assume that T = kP . We need to determine k. Using the data, we see that 0.84 = 12k , so that k = 0.07 . So, T = 0.07P . 39. Let S = speed of the object, and M = Mach number. We are given that S = kM . We also know that when S = 760 mph (at sea level), M = 1. As such, k = 760. Hence, for U.S. Navy Blue Angels, S = 760 (1.7 ) = 1292 mph .

40. Using the same model as in Exercise 39, we see that for the F-22A Raptor, S = 760 (1.9 ) = 1444 mph .

41. We are given that F = kH . Using the fact that F = 11 when H = 6.8, we see that 11 = 6.8k , so that k = 1.618. Hence, F = 1.618H .

42. Let S1 and S2 denote two abutting sections of a finger. It is known that S1 = kS2 . Using the fact that S2 = 5 when S1 = 8 , we see that k = 1.6 . So,

S1 = 1.6S2 .

442

Section 3.6

43. Assume Hooke’s law holds: F = kx. Using the fact that F = 30N when x = 10 cm, we see that k = 3 N . So, cm x, F = 3x . As such, 72N = 3 N cm so that x = 24 cm .

44. Using the same information as in Exercise 43, we see that the given information yields F= 3N 18 cm ) = 54N . cm (

45.

46.

(

y = money owed x = sales y = kx 85.6 = k (1712 ) k = 0.05 y = 0.05x

y = 0.05 (1800 )

)

(

)

k . We must first find k: x k 10,000 = 5.75 k = 57,500 So, the number of units when x = 6.50 is 57,500 d= = 8800 . 6.50 Here, d =

y = 90 47.

y = sweatshirts

48.

x = price y=

y = T-shirts x = price k y= x k 720 = 56 k = 40,320

k x

k 60 k = 81,000 1350 =

40,320 x 40,320 y= 18 y = 2240

81,000 x 81,000 y= 45 y = 1800

443

Chapter 3

k . Using the data for Earth, we obtain: D2 k 1400 w 2 = so that m (150,000 km )2

49. Use the formula I =

(

)

(

(150,000 km ) = 1400 w m2 (150,000,000 m ) = 3.15 ×1019 w m2 Hence, the intensity for Mars is given by: 3.15 ×1019 w 3.15 ×1019 w I= = ≈ 600 w 2 2 2 m ( 228,000 km ) ( 228,000,000 m ) k = 1400 w

50. Use the formula I =

1400 w

(

k = 1400 w

k . Using the data for Earth, we obtain: D2

(150,000 km )

so that

(150,000 km ) = (1400 w m ) (150,000,000 m ) = 3.15 ×10 w m ) 2

Hence, the intensity for Mercury is given by: 3.15 ×1019 w 3.15 ×1019 w I= = ≈ 9400 w 2 2 2 m (58,000 km ) (58,000,000 m ) 51. Use the formula I = kPt . Bank of America: 750 = k (25,000)(2) , so that k = 0.015, which corresponds to 1.5%. Navy Federal Credit Union: 1500 = k (25,000)(2) , so that k = 0.03, which corresponds to 3%.

52. Use the formula I = kPt . Observe that 1 3250 = k (130,000)   , so that 2 k = 0.05. So, 5% interest rate is needed to make $3250 in interest in 6 months.

53.

kT with T = 300K, P = 1 atm, and V = 4 ml to obtain V PV (1 atm ) (4) 4 4  275  11 k= or 0.92 atm . = = . Thus, P =  = T 300  4  12 300 300

Use the formula P =

444

Section 3.6

54.

kT with T = 300K, P = 1 atm, and V = 4 ml to obtain V PV (1 atm ) (4) 4 4  300  4 k= or 1.33 atm . = = . Thus, P =  = T 300  3  3 300 300

Use the formula P =

55. Should be y is inversely proportional to x.

56. y varies directly with the square of x ( x 2 ), NOT the square root ( x ).

57.

1 True. A = bh , so area is directly 2 proportional to both base and height.

58. False. Since d = rt , it follows that d r = . So, r is directly proportional to t d, but inversely proportional to t.

59. b

60. a

61. 7 11 Use the equation σ p21 = αC n2 k 6 L 6 with the following information (all converted to

meters):

2π ), and σ p21 = 7.1. −6 1.55 ×10 m Substituting this information into the equation yields α : Cn2 = 1.0 ×10 −13 , L = 2000m, λ = 1.55 ×10 −6 m (so that k = 7.1 = α (1.0 ×10

−13

6 )  1.552×π10−6  2000116

so that

α=

7.1

(1.0 ×10

−13

2π  6 )  1.55 ×10−6  2000116

Thus, the equation is given by σ p21 = 1.23C n2 k 6 L 6 . 7

445

≈ 1.23 .

Chapter 3

62. Use the equation σ s2p = αCn2 k 6 L 6 with the following information (all 7

converted to meters):

2π ), and σ s2p = 2.3. 1.55 ×10 −6 m Substituting this information into the equation yields α : Cn2 = 1.0 ×10 −13 , L = 2000m, λ = 1.55 ×10 −6 m (so that k = 2.3 = α (1.0 ×10

−13

6 2π   )  1.55 ×10−6  2000116

so that

α=

2.3

(1.0 ×10

−13

6 )  1.552×π10−6  2000116

≈ 0.399 .

Thus, the equation is given by σ s2p = 0.399Cn2 k 6 L 6 . 7

63. a. The least-squares regression line is y = −1.45x + 595.82 b. The variation constant is 968.09, and the equation of inverse variation is 968.08 y = 0.164 . x c. When the oil price is $49.82 in September of 2017, the predicted utilities stock index from the least-squares regression line is 524 and from the equation of inverse variation is 510. The least-squares regression line gives a closer approximation to the actual value of 530. 64. a. The least-squares regression line is y = −0.01x + 2.48

3.94 . x 0.192 c. When the oil price is $49.82 per barrel in September of 2017, the predicted 5year maturity rate from the least-squares regression line is 1.98 and from the equation of inverse variation is 1.86. The equation of inverse variation gives a closer approximation to the actual value of 1.80. b. The variation constant is 3.94, and the equation of direct variation is y =

65. a. The least-squares regression line is y = 8.55x + 18.42. b. The variation constant is 27.23, and the equation of direct variation is y = 27.23x 0.36 . c. When the maturity rate is 1.8% in September of 2017, the predicted new, privately owned housing units from the least-squares regression line is 34,067 and from the equation of direct variation is 33,647. The least-squares regression line gives a closer approximation to the actual value of 45,000. 446

Section 3.6

66. a. The least-squares regression line is y = 3.26x + 377.90. b. The variation constant is 161.50, and the equation of direct variation is y = 161.50 x 0.314 . c. There are 45,000 housing units in September of 2017. The predicted utilities stock index from the least-squares regression line is 525 and from the equation of direct variation is 534. The equation of direct variation gives a closer approximation to the actual value of 530. 67. a. y = −0.19x + 3.91 b. $1.63 c. $1.06

68.

4.15 x.26 b. $2.18 c. $2.05 a. y =

447

Chapter 3 Review Solutions ----------------------------------------------------------------------1. Yes

2. No, since 3 is assigned to two different values.

3. Yes

4. Yes

5. No, since both (0,6) and (0, −6) satisfy the equation, so that the graph fails the vertical line test.

6. No, since the graph fails the vertical line test.

7. Yes

8. Yes

9. No, since the graph fails the vertical line test.

10. Yes

11. (a) 2 (b) 4 (c) when x = −3, 4

12. (a) 2 (b) −2 (c) when x ≈ −2,3.2

13. (a) 0 (b) −2 (c) when x = −5, 2

14. (a) 7 (b) −3 (c) never

15. f (3) = 4(3) − 7 = 5

16. F (4) = 42 + 4(4) − 3 = 29

17. f ( −7) ⋅ g (3) = ( 4( −7) − 7 ) ⋅ 32 + 2(3) + 4

18. F (0) 3 = − g (0) 4

= −35 19 = −665

20. f (3 + h ) = 4(3 + h ) − 7 = 5 + 4 h

19.

f (2) − F (2) ( 4(2) − 7 ) − ( 2 + 4(2) − 3 ) = g (0) 4 1− 9 = −2 4 2

21.

f (3 + h ) − f (3) ( 4(3 + h ) − 7 ) − ( 4(3) − 7 ) 5 + 4 h − 5 = = = 4 h h h

448

Chapter 3 Review

22.

F (t + h ) − F (t ) ( (t + h ) + 4(t + h ) − 3 ) − ( t + 4t − 3 ) = h h 2 2 t + 2 ht + h + 4t + 4 h − 3 − t 2 − 4t + 3 = h 2 2 ht + h + 4 h h ( 2t + h + 4 ) = = = 2t + h + 4 h h 2

23. ( −∞, ∞ )

24. ( −∞, ∞ )

25. ( −∞, −4 ) ∪ ( −4, ∞ )

26. ( −∞, ∞ )

27. We need x − 4 ≥ 0 , so the domain is [ 4,∞ ) .

28. We need 2x − 6 > 0 , so the domain is ( 3,∞ ) .

29.

Solve 2 = f (5) =

D for D: 2 = D9 , so that D = 18 . 5 − 16 2

30. There are many such functions. The most natural one to construct has the form D f (x) = . Since ( 0, −4 ) is to lie on the graph of f, we substitute this ( x + 3)( x − 2) point into the equation for the function to find the corresponding value of D that D D = , so that D = 24 . Hence, one such will ensure this: −4 = (0 + 3)(0 − 2) −6 function is given by: 24 f (x) = . ( x + 3)( x − 2) 31. f ( −x ) = 2( −x ) − 7 = −(2 x + 7) ≠ f ( x ) So, not even. − f ( −x ) = − ( −(2 x + 7) ) = 2 x + 7 ≠ f ( x ) So, not odd. Thus, neither.

32. g ( − x ) = 7( − x )5 + 4( − x )3 − 2( − x )

= − ( 7 x5 + 4 x3 − 2 x ) ≠ g ( x )

So, not even.

(

− g ( −x ) = − − ( 7 x5 + 4 x3 − 2 x )

)

= 7 x5 + 4 x3 − 2 x = g ( x ) So, odd.

449

Chapter 3

33. h( − x ) = ( − x )3 − 7( −x ) = − ( x3 − 7x ) ≠ h( x )

So, not even. − h( − x ) = − − ( x3 − 7x ) = x3 − 7 x = h( x )

(

)

So, odd. 35. 1 1 f ( −x ) = ( −x ) 4 + ( −x ) = ( −x ) 4 − x ≠ f ( x ) So, not even. 1 1 − f ( −x ) = − x 4 − x = −x 4 + x ≠ f ( x )

(

)

So, not odd. Thus, neither.

34. f ( − x ) = ( − x )4 + 3( − x )2 = x 4 + 3x 2 = f ( x ) So, even. Hence, cannot be odd.

36. f ( − x ) = −x + 4 ≠ f ( x ) So, not even. − f ( −x ) = − −x + 4 ≠ f ( x ) So, not odd. Thus, neither.

37.

f ( −x ) =

1  1  + 3( − x ) = −  3 + 3x  ≠ f ( x ) So, not even. 3 ( −x ) x 

  1  1 − f ( −x ) = −  −  3 + 3x   = 3 + 3x = f ( x ) So, odd.  x  x 38. f ( −x ) =

1 + 3( −x )4 + −x = f ( x ) So, even. Hence, f cannot be odd. 2 ( −x )

39.

[ −4,7] [ −2, 4] (3,7 ) ( 0,3 ) ( −4,0 )

Domain Range Increasing Decreasing Constant 41.

40.

Increasing

nowhere

Decreasing

( −∞, −6 ) ∪ ( 6,∞ )

Constant

nowhere

42.

( 4 − 2 ) − ( 4 − 0 ) = −2 2

Range

( −∞, −6 ) ∪ ( 6,∞ ) ( −∞, −4 ) ∪ ( 6,∞ )

Domain

2(5) − 1 − 2(1) − 1 9 − 1 = = 2 5 −1 4

450

Chapter 3 Review

43. Domain: ( −∞, ∞ ) Range: ( 0,∞ )

44. Domain: ( −∞, ∞ ) Range: [ −3, ∞ )

Notes on the graph: There is an open hole at (0,0), and a closed hole at (0,2).

Notes on the graph: There are open holes at (0,4) and (1,5), and closed holes at (1,4) and (0, −3) .

45. Domain: ( −∞, ∞ )

46. Domain: ( −∞,0 ) ∪ ( 0, ∞ )

Notes on the graph: There is an open hole at (1,3), and a closed hole at (1, −1) .

Notes on the graph: There are open holes at (0,0) and (1,1), and a closed hole at (1, −2) .

Range: [ −1, ∞ )

Range: ( −∞, −2 ] ∪ ( 0, ∞ )

47. Let x = number of 30-minute periods. 25, 0 < x ≤ 2,  . Then, C ( x ) =  25 + 10.50( x − 2), x > 2

451

48. Let x = number of hours worked. Then, the weekly earnings are described by 30x, 0 ≤ x ≤ 40  1200 + 45( x − 40) , x > 40 E (x) =    Amount of overtime at time and a half. 

Chapter 3

49. Reflect the graph of x 2 over x-axis, then shift right 2 units and then up 4 units.

50. First, note that −x + 5 − 7 = − ( x − 5 ) − 7 = x − 5 − 7

So, shift the graph of x right 5 units and then down 7 units.

51. Shift the graph of 3 x right 3 units, and then up 2 units.

452

52. Shift the graph of x1 right 2 units, and then down 4 units.

Chapter 3 Review

53. Reflect the graph of x3 over x-axis, and then contract vertically by a factor of 2.

54. Expand the graph of x 2 vertically by a factor of 2, and then shift up 3 units.

55.

56.

453

Chapter 3

57.

58.

59. y = x + 3 Domain: [ −3, ∞ )

60. y = x − 4 Domain: [ 0,∞ )

61. y = x − 2 + 3 Domain: [ 2,∞ )

62. y = − x Domain: ( −∞,0 ]

63. y = 5 x − 6 Domain: [ 0,∞ )

64. y = 12 x + 3 Domain: [ 0,∞ )

65. y = ( x 2 + 4 x + 4 ) − 8 − 4 = ( x + 2 ) − 12 2

Domain:  or ( −∞, ∞ )

66. y = 2 ( x 2 + 3x ) − 5

= 2 ( x 2 + 3x + 94 ) − 5 − 29 = 2 ( x + 32 ) − 192 2

Domain:  or ( −∞, ∞ )

454

Chapter 3 Review

67. g ( x ) + h( x ) = ( −3x − 4 ) + ( x − 3 ) = −2x − 7

g ( x ) − h( x ) = ( −3x − 4 ) − ( x − 3 ) = −4x − 1

g ( x ) ⋅ h( x ) = ( −3x − 4 ) ⋅ ( x − 3 ) = −3x 2 + 5x + 12 g ( x ) −3x − 4 = h( x ) x −3

Domains: dom( g + h )   dom( g − h )  = ( −∞, ∞ ) dom( gh )  g dom   = ( −∞,3 ) ∪ ( 3, ∞ ) h

68. g ( x ) + h( x ) = ( 2 x + 3 ) + ( x 2 + 6 ) = x 2 + 2 x + 9

g ( x ) − h( x ) = ( 2 x + 3 ) − ( x 2 + 6 ) = −x 2 + 2 x − 3 g ( x ) ⋅ h( x ) = ( 2 x + 3 ) ⋅ ( x 2 + 6 )

= 2x3 + 3x 2 + 12x + 18 g ( x ) 2x + 3 = h( x ) x 2 + 6

Domains: dom( g + h )  dom( g − h )   dom( gh )  = ( −∞, ∞ )  g  dom    h  

69.

Domains: dom( g + h )  dom( g − h )   dom( gh )  = ( 0, ∞ )  g dom     h  

g ( x ) + h( x ) = x12 + x g ( x ) − h( x ) = x12 − x g ( x ) ⋅ h( x ) = x12 ⋅ x = 13 2 x

1 x2

g( x) 1 = = 5 h( x ) x x2

455

Chapter 3

70.

g ( x ) + h( x ) = =

( x + 3) + 2 ( 3x − 1) ( 2x − 4 )

7x + 1 2 ( x − 2)

g ( x ) − h( x ) =

g ( x ) ⋅ h( x ) =

3x − 1 x+3 + 2 ( x − 2) x − 2

3x − 1 x+3 − 2 ( x − 2) x − 2

( x + 3) − 2 (3x − 1) ( 2x − 4 )

−5x + 5 2 ( x − 2)

x + 3 3x − 1 ( x + 3 ) ⋅ ( 3x − 1) ⋅ = 2 2 (x − 2) x − 2 2 ( x − 2)

x+3 g(x) 2 ( x − 2) x +3 x−2 x +3 = = = ⋅ 3x − 1 h( x ) 2 ( x − 2 ) 3x − 1 2(3x − 1) x−2

Domains: dom( g + h )   dom( g − h )  = ( −∞,2 ) ∪ ( 2, ∞ ) dom( gh )  f dom   = ( −∞, 13 ) ∪ ( 13 ,2 ) ∪ ( 2, ∞ ) g

456

Chapter 3 Review

Domains: Must have both x − 4 ≥ 0 and 2 x + 1 ≥ 0 . So, dom( f + g )   dom( f − g )  = [ 4, ∞ ) . dom( fg )  For the quotient, must have both x − 4 ≥ 0 and 2x + 1 > 0 . So, f dom   = [ 4, ∞ ) . g

71. g ( x ) + h( x ) = x − 4 + 2 x + 1 g ( x ) − h( x ) = x − 4 − 2 x + 1 g ( x ) ⋅ h( x ) = x − 4 ⋅ 2 x + 1 g(x) x−4 = h( x ) 2x + 1

72. g ( x ) + h( x ) = ( x 2 − 4 ) + ( x + 2 ) = x 2 + x − 2 g ( x ) − h( x ) = ( x 2 − 4 ) − ( x + 2 ) = x 2 − x − 6 g ( x ) ⋅ h( x ) = ( x 2 − 4 ) ⋅ ( x + 2 )

Domains: dom( g + h )   dom( g − h )  = ( −∞, ∞ ) dom( gh ) 

g dom   = ( −∞, −2 ) ∪ ( −2, ∞ ) h

= x + 2x − 4x − 8 3

g(x ) x2 − 4 = = x − 2, x ≠ −2 h( x ) x + 2

73.

74.

( f  g )( x ) = 3 ( 2x + 1) − 4 = 6 x − 1

( f  g )( x ) = ( x + 3 ) + 2 ( x + 3 ) − 1 3

( g  f )( x ) = 2 ( 3x − 4 ) + 1 = 6 x − 7 Domains: dom( f  g ) = ( −∞, ∞ ) = dom( g  f )

= x3 + 9x 2 + 29x + 3

( g  f )( x ) = ( x3 + 2 x − 1) + 3 = x3 + 2x + 2

Domains: dom( f  g ) = ( −∞, ∞ ) = dom( g  f )

457

Chapter 3

75. 2

2 1 1 + 3(4 − x ) +3 4−x 4−x 2(4 − x ) 8 − 2 x = = 13 − 3x 13 − 3x x +3 1 1 = = ( g  f )( x ) = 2 4( x + 3) − 2 4 x + 10 4− x +3 x +3

( f  g )( x ) =

Domains: dom( f  g ) = ( −∞, 4 ) ∪ ( 4, 133 ) ∪ ( 133 , ∞ )

dom( g  f ) = ( −∞, −3 ) ∪ ( −3, − 52 ) ∪ ( − 52 , ∞ )

76.

( f  g )( x ) = 2

( x + 6 ) − 5 = 2( x + 6) − 5 2

= 2x + 7 ( g  f )( x ) =

2x 2 − 5 + 6

Domains: dom( f  g ) : Need both x + 6 ≥ 0 and 2 x + 7 ≥ 0 . Thus, dom( f  g ) = − 72 , ∞ ) . dom( g  f ) : Note 2x 2 − 5 + 6 ≥ 0 , for all values of x for which 2x 2 − 5 is defined. This is true when 2x 2 − 5 ≥ 0 . So, solving this inequality yields: 2x2 − 5 ≥ 0 x 2 − 52 ≥ 0

( x − )( x + ) ≥ 0 5 2

CPs: ±

5 2

+ | − | +  5 − 52 2

(

So, dom( g  f ) = −∞, −

5 2

) ∪ ( ,∞) . 5 2

458

Chapter 3 Review

77.

78.

( f  g )( x ) = x − 4 − 5 = x − 9 2

( f  g )( x ) =

= ( x − 3)( x + 3) ( g  f )( x ) =

( x −5) − 4 = x −9 2

( g  f )( x ) =

Domains: dom( f  g ) : Need ( x − 3)( x + 3) ≥ 0 . CPs: ±3

1 = x2 − 4 1 2 x −4 1 1 = 1 2 1 −4 x −4 x

( )

1 = 1− 4 x =

+ − | +  | −3 3

x 1 − 4x

Domains: dom( f  g ) : Need ( x − 2)( x + 2) > 0 .

So, dom( g  f ) = ( −∞, −3] ∪ [3, ∞ ) . dom( g  f ) : Need x − 5 ≥ 0 . Thus,

CPs: ±2

dom( g  f ) = [5, ∞ ) .

+ − | +  | −2 2

So, dom( g  f ) = ( −∞, −2 ) ∪ ( 2, ∞ ) . dom( g  f ) : Need 1 − 4 x ≠ 0 , so that x ≠ 14 . So,

dom( g  f ) = ( 0, 14 ) ∪ ( 14 , ∞ ) .

79. g (3) = 6(3) − 3 = 15 f ( g (3)) = f (15) = 4(15)2 − 3(15) + 2 = 857

80. g (3) = 32 + 5 = 14, but f ( g (3)) = f (14) is not defined.

f ( −1) = 4( −1)2 − 3( −1) + 2 = 9

f ( −1) = 4 − ( −1) = 5

g ( f ( −1)) = g (9) = 6(9) − 3 = 51

g ( f ( −1)) = g

g (3) = 5(3) + 2 = 17

f ( −1) =

82. g (3) = 32 − 1 = 8

81. f ( g (3)) = f (17) =

( 5 ) = ( 5 ) + 5 = 10

1 1 = 8 −1 7 1 1 f ( −1) = =− 2 −1 − 1

17 17 = 2(17) − 3 31

f ( g (3)) = f (8) =

−1 1 =− 2( −1) − 3 5

3  1  1 g ( f ( −1)) = g  −  =  −  − 1 = − 4  2  2

 1  1 g ( f ( −1)) = g  −  = 5  −  + 2 = 1  5  5

459

Chapter 3

84. f ( g (3)) is undefined since g (3) is not defined.   4 1 g ( f ( −1)) = g   = g ( −4) = 7 2  ( −1) − 2 

83.

(

f ( g (3)) = 3 3 − 4

) − ( 3 − 4 ) + 10 2

= ( −1)2 − ( −1) + 10 = 12

g ( f ( −1)) = g ( ( −1)2 + 1 + 10 ) = 3 12 − 4 = 2

85. Let f ( x ) = 3x 2 + 4 x + 7, g ( x ) = x − 2. Then, h( x ) = f ( g ( x )) .

x , g(x) = 3 x . 1− x Then, h( x ) = f ( g ( x )) .

87. Let f ( x ) =

1 , g( x ) = x2 + 7 . x Then, h( x ) = f ( g ( x )) .

88. Let f ( x ) = x , g ( x ) = 3x + 4 . Then, h( x ) = f ( g ( x )) .

89. The area of a circle with radius r(t ) is given by:

90. Since 42 = lw, l = 42w . So, the perimeter formula becomes: 84 + 2w 2 36 = 2l + 2w = 2 ( 42w ) + 2w = w so that 2w 2 − 36w + 84 = 0

(

A(t ) = π ( r(t ) ) = π 25 t + 2 2

)

= 625π (t + 2) in2

86. Let f ( x ) =

w 2 − 18w + 42 = 0

91. Yes

92. No, since Bill and Maria both received an A.

93. No, since both (2,3) and (3,3) lie on the graph of the function.

94. No, since −3 and 3 both map to 9.

95. Yes

96. Yes

97. Yes

98. No, since both (1,1) and ( −1,1) satisfy the equation.

460

Chapter 3 Review

99. One-to-one

100. Not one-to-one, since f ( −1) = f (1) = 1 , for instance.

101.

x−4 f ( f −1 ( x ) ) = 3  +4 = x−4+4 = x  3 

461

Chapter 3

102. f ( f −1 ( x ) ) =

1 1 = 1 + 7x  1 + 7x  −7 4 −7 x  4x  1 1 = = = x 1 1 +7−7 x x

103. f ( f −1 ( x ) ) =

( x − 4 ) + 4 = x = x, since 2

x ≥ 0.

104. 7 x + 2 + 2( x − 1) 7x + 2 +2 x −1 f ( f −1 ( x ) ) = x − 1 = 7x + 2 7x + 2 − 7( x − 1) −7 x −1 x −1

7x + 2 + 2x − 2 9x = = x 9 7x + 2 − 7x + 7

462

Chapter 3 Review

Domains: dom ( f ) = rng ( f −1 ) = ( −∞, ∞ )

105. Solve y = 2 x + 1 for x: x = 12 ( y − 1)

Thus, f −1 ( x ) = 12 ( x − 1) =

rng ( f ) = dom ( f −1 ) = ( −∞, ∞ )

x −1 . 2

Domains: dom ( f ) = rng ( f −1 ) = ( −∞, ∞ )

106. Solve y = x5 + 2 for x: x = 5 y−2

rng ( f ) = dom ( f −1 ) = ( −∞, ∞ )

Thus, f −1 ( x ) = 5 x − 2 .

Domains: dom ( f ) = rng ( f −1 ) = [ −4, ∞ )

107. Solve y = x + 4 for x: x = y2 − 4 −1

rng ( f ) = dom ( f −1 ) = [ 0, ∞ )

Thus, f ( x ) = x − 4 . 2

Domains: dom ( f ) = rng ( f −1 ) = [ −4, ∞ )

108. Solve y = ( x + 4)2 + 3 for x:

rng ( f ) = dom ( f −1 ) = [3, ∞ )

y −3 = x + 4

−4 + y − 3 = x Thus, f −1 ( x ) = −4 + x − 3 .

Domains: dom ( f ) = rng ( f −1 ) = ( −∞, −3 ) ∪ ( −3, ∞ )

109. Solve y = xx ++63 for x: ( x + 3) y = x + 6

rng ( f ) = dom ( f −1 ) = ( −∞,1) ∪ (1, ∞ )

xy + 3 y = x + 6 xy − x = 6 − 3 y x( y − 1) = 6 − 3 y x = 6y−−31y

Thus, f −1 ( x ) = 6x−−31x .

463

Chapter 3

Domains: dom ( f ) = rng ( f −1 ) = ( −∞, ∞ )

110. Solve y = 2 3 x − 5 − 8 for x:

rng ( f ) = dom ( f −1 ) = ( −∞, ∞ )

y +8 = 2 3 x −5

( 12 ( y + 8) ) = x − 5 3 5 + ( 12 ( y + 8) ) = x 3 Thus, f −1 ( x ) = 5 + ( 12 ( x + 8) ) . 3

111. Let x = total dollars worth of products sold. Then, S ( x ) = 98,000 + 0.11x.

Solving y = 98,000 + 0.11x for x yields: x =

1 ( y − 98,000 ) . 0.11

y − 98,000 . 0.11 This inverse function tells you the sales required to earn the desired income. Thus, S −1 ( x ) =

112. V ( s ) = 3s 2 , s ≥ 0 . Solving y = 3s 2 for s yields: s =

So, V −1 ( s ) =

1 3

s . This inverse function tells you the length s of a side of a base

required to get a desired volume. 113. The general equation is C = kr . Using the fact that C = k (1) = 2π , we see

that k = 2π . So, C = 2π r .

114. The general equation is V = klw . Using the fact that V = k (2)(6) = 12h, we see that k = h .

So, V = lwh .

115. The general equation is A = kr 2 . Using the fact that A = k52 = 25π , we see

that k = π . So, A = π r 2 .

116. The general equation is k . Using the fact that F= λL k F= = 20π −6 (10 m )(103 m )

we see that k = π 50 . So, F =

π . 50λ L

117. Assume that W = kH . We need to determine k . Using Cole’s data, we see that 357.75 = 27k , so that k = 13.25 . So, W = 13.25H . (Note that Dickson’s data also satisfies this equation.)

464

Chapter 3 Review

118.

3.50 = 0.07 T ( P ) = 0.07P 50 1.60 County B: = 0.08 T ( P ) = 0.08P 20 County A:

119. The graph of f is as follows:

Domain of f:

121. Domain Range Increasing Decreasing Constant

120. The graph of f is as follows:

( −∞, −1) ∪ (3, ∞ )

Domain of f: ( −∞, −3) ∪ ( −3,3) ∪ ( 3, ∞ )

( −∞,2 ) ∪ ( 2,∞ ) {−1,0,1} ∪ ( 2,∞ ) ( 2,∞ ) ( −∞, −1) ( −1,0 ) ∪ (0,1) ∪ (1,2)

Note on the graph: The vertical line at x = −1 should NOT be a part of the graph. The points (−1,2), (0,−1), (1,0), (2,1), and (2,3) on the graph should be open circles.” 465

Chapter 3

122.

Decreasing

( −2,2 ) ∪ ( 2,∞ ) [0,3) ∪ ( 4, ∞ ) ( −1,0 ) ∪ (1,2 ) ∪ ( 2,∞ ) ( −2, −1) ∪ ( 0,1)

Constant

nowhere

Domain Range Increasing

Note on the graph: This graph has open circles at (−2,3), (2,3), and (2,4) 123. The graphs of f and g are as follows:

The graph of f can be obtained by shifting the graph of g two units to the left. That is, f ( x ) = g ( x + 2) .

466

124. The graphs of f and g are as follows:

The graph of f can be obtained by shifting the graph of g one unit to the right and then reflecting it over the x-axis. That is, f ( x ) = − g ( x − 1) .

Chapter 3 Review

125. y1 = 2x + 3

y2 = 4 − x y1 y3 = y2 Domain of y3 = [ −1.5, 4 )

126. y1 = x 2 − 4 y2 = x 2 − 5 y3 = ( y1)2 − 5 Domain of y3 = ( −∞, −2 ] ∪ [ 2, ∞ )

467

Chapter 3

127. Yes, the function is one-to-one.

128. Yes, the functions are inverses, provided that g is equipped with the domain ( 0,∞ ) .

129. a) The line is approximately given by y = −39.02x + 1562. b) $1171.80 c) $1093.76

130. a) The line is approximately given by y = 1573.39x −0.1. b) $1222.78 c) $1200.69

468

Chapter 3 Practice Test Solutions----------------------------------------------------------------1. b (Not one-to-one since both (0,3) and ( −3,3) lie on the graph.)

2. a (Doesn’t pass the vertical line test.)

3. c

4. Observe that f (11) = 11 − 2 = 9 = 3 g ( −1) = ( −1)2 + 11 = 12 So, f (11) − 2 g ( −1) = 3 − 2(12) = −21 .

f x−2 5.   ( x ) = 2 Domain: [2, ∞ ) x + 11 g

g x 2 + 11 6.   ( x ) = Domain: (2, ∞ ) x−2 f

g ( f ( x )) =

( x − 2 ) + 11 = x − 2 + 11

( f + g )(6) = f (6) + g (6)

= 6 − 2 + ( 6 2 + 11) = 2 + 47 = 49

= x+9 Domain: [ 2,∞ ) 9.

( ( 7 )) = f (( 7 ) + 11) = f (18) 2

f g

= 18 − 2 = 4 11. f ( − x ) = 9( −x )3 + 5( −x ) − 3

= − 9x3 + 5x + 3 ≠ f ( x ) So, not even.

(

− f ( −x ) = − − 9x3 + 5x + 3

10. f ( − x ) = − x − ( − x )2 = x − x 2 = f ( x )

So, even. Therefore, f cannot be odd. 12. f ( − x ) = −2x = − x2 = − f ( x ) So, odd. Therefore, f cannot be even.

)

= 9 x 3 + 5x + 3 ≠ f ( x ) So, not odd. Thus, neither.

469

Chapter 3

13. f ( x ) = − x − 3 + 2

Reflect the graph of x over the x-axis, then shift right 3 units, and then up 2 units. Domain: [3,∞ ) Range: ( −∞,2 ]

14. f ( x ) = −2( x − 1)2 Reflect the graph of x 2 over the x-axis, then expand vertically by a factor of 2, and then shift right 1 unit. Domain: ( −∞, ∞ ) Range: ( −∞,0 ]

15.

 − x, x < −1  f ( x ) =  1, − 1 < x < 2  x2 , x ≥ 2  Domain: ( −∞, −1) ∪ ( −1, ∞ ) Range: [1,∞ )

Open holes at (−1,1) and (2,1); closed hole at (2,4)

16. (a) 1

(b) 1

(d) when x = −1.5,2

470

(e) when x = 3.5

Chapter 3 Practice Test

17. (a) −2 (b) 4 (c) −3 (d) when x = −3,2

18. (a) −3 (b) never (c) −1 (d) 1

19. (3( x + h )2 − 4( x + h ) + 1) − (3x2 − 4x + 1)

20.

3x 2 + 6xh + 3h 2 − 4x − 4h + 1 − 3x 2 + 4x − 1 h h ( 6x + 3h − 4 ) = = 6 x + 3h − 4 h =

(5 − 7( x + h ) ) − (5 − 7x ) = 5 − 7x − 7h − 5 + 7x = −7h = −7 h

21.

( 64 − 16(2) ) − ( 64 − 16(0) ) = 0 − 64 = −32 2

22. 10 − 1 − 2 − 1 3 − 1 1 = = 10 − 2 8 4

Domains: dom ( f ) = rng ( f −1 ) = [5, ∞ )

23. Solve y = x − 5 for x:

rng ( f ) = dom ( f −1 ) = [ 0, ∞ )

y2 = x − 5 y2 + 5 = x

Thus, f −1 ( x ) = x 2 + 5 . Domains: dom ( f ) = rng ( f −1 ) = [ 0, ∞ )

24. Solve y = x 2 + 5 for x:

y = x2 + 5

rng ( f ) = dom ( f −1 ) = [5, ∞ )

y − 5 = x, since x ≥ 0. −1

Thus, f ( x ) = x − 5 .

471

Chapter 3

Domains: dom ( f ) = rng ( f −1 ) = ( −∞,5 ) ∪ ( 5, ∞ )

25.

Solve y =

2x + 1 for x: 5−x (5 − x ) y = 2x + 1

rng ( f ) = dom ( f −1 ) = ( −∞, −2 ) ∪ ( −2, ∞ )

5 y − xy = 2 x + 1 5 y − 1 = x( y + 2) x=

Thus, f −1 ( x ) =

5y −1 y+2

5x − 1 . x+2

26. We compute the inverse of f piecewise: For x ≤ 0 : Solve y = −x for x: x = − y . So, f −1 ( x ) = − x on ( −∞,0 ] .

For x > 0 : Solve y = − x 2 ( ≤ 0 ) for x: x = − − y . So, f −1 ( x ) = − −x on ( 0,∞ ) . Thus, the inverse function is given by  − x, x ≥ 0 f −1 ( x ) =  − − x , x < 0 27. Can restrict to [ 0,∞ ) so that f will have an inverse. Also, one could restrict to

any interval of the form [ a, ∞ ) or ( −∞, −a ] , where a is a positive real number, to ensure f is one-to-one. 28. The point (5, −2) (switch x and y coordinates to get a point on the inverse.) 29. Let x = original price of a suit. Then, the sale price is x − 0.40 x = 0.60 x . Hence, the checkout price is S ( x ) = 0.60x − 0.30(0.60x ) = 0.42x .

30. Recall F = 59 C + 32 is the relationship between degrees Fahrenheit and degrees Celsius. If K = C + 273.15 , then C = K − 273.15 . Then, the composition function F (C ( K )) gives the relationship between Kelvin and degrees Fahrenheit. The composition is: F (C ( K )) = 59 ( K − 273.15 ) + 32 = 59 K − 459.67

472

Chapter 3 Practice Test

31. Consider f ( x ) = − 1 − x 2 , − 1 ≤ x ≤ 0 . (The graph of f is the quarter unit

circle in the third quadrant.) To find its inverse, solve y = − 1 − x 2 for x: y = − 1 − x2

( − y ) = 1 − x2 2

x 2 = 1 − y2 x = − 1 − y 2 since − 1 ≤ x ≤ 0

So, f −1 ( x ) = − 1 − x 2 (The graph looks identical to that of f.) 32. Solve r(t ) = 15 (At this point, the puddle just touches the sidelines.) 10 t = 15

33. 0 0 ≤ x < 6  f ( x ) =  5 6 ≤ x < 18  7 x ≥ 18 

t = 1.5 t = (1.5)2 = 2.25 So, after 2.25 hours, the puddle will reach the sidelines.

34. The general equation is y = kx 2 . Using the fact that y = k (5)2 = 8 , we see

36.

Range Increasing Decreasing

[ −4, −2 ) ∪ ( −2, 4] [0,5] ( −1,1) ∪ ( 3, 4 ) ( −2, −1) ∪ (1,3 ) ( −4, −2 )

Constant Note: This graph has an open hole at (−2,5).

473

km . P

k (2) = 20 , we 3 30m . see that k = 20 ( 32 ) = 30 . So, F = P Using the fact that F =

that k = 258 . So, y = 258 x 2

Domain

35. The general equation is F =

Chapter 3

37. Yes, the function is one-to-one.

474

Chapter 3 Cumulative Review -------------------------------------------------------------------1. 2 2 3+ 5 6+2 5 3+ 5 = ⋅ = = 4 2 3− 5 3− 5 3+ 5

2. 10 x 2 − 29x − 21 = (5x + 3)(2 x − 7) 3. x3 − 4 x x( x 2 − 4) x( x − 2)( x + 2) = = = x( x − 2) x+2 x+2 x+2 Domain: ( −∞, −2 ) ∪ ( −2, ∞ )

5. (8 − 9i )(8 + 9i ) = 64 + 81 = 145 + 0i

4. 1 6

x = − 51 x + 11

5x = −6x + 330 11x = 330 x = 30

7. Since 35.70 59.50 = 0.60 , the percent markdown is 40%.

6. 5 10 − 10 = , x≠0 x 3x 15 − 30x = 10 5 = 30 x x = 61

x(6 x + 1) = 12 6 x + x − 12 = 0 2

(3x − 4)(2 x + 3) = 0 x = 34 , − 23

475

Chapter 3

10. 2

x 1 −x = 2 5 2 5x − 10x = 2

x + 2 = −3 x + 2 = ( −3)3 = −27 x = −29

5 ( x 2 − 2x ) − 2 = 0

Check:

5 ( x 2 − 2x + 1) − 2 − 5 = 0

−29 + 2 = 3 −27 = −3 .

5( x − 1)2 = 7 ( x − 1)2 = 75 x = 1±

= 1± 5

7 5

11.

12. x − x − 12 = 0 4

Let u = x 2 .

−7 < 3 − 2 x ≤ 5 −10 < −2 x ≤ 2 5 > x ≥ −1 [ −1,5 )

u 2 − u − 12 = 0 ( u − 4)( u + 3) = 0 u = −3, 4 So, we have: x 2 = −3  x = ± i 3 x 2 = 4  x = ±2

13.

14.

x <0 x −5 CPs: x = 0,5

+ − +  | | 0

−1.3 ≤ 2.7 − 3.2x ≤ 1.3 −4 ≤ −3.2x ≤ −1.4 4 1.4 ≥x≥ 3.2 3.2  

=1.25

So, the solution set is ( 0,5 ) .

= 0.4375

[0.4375, 1.25]

476

Chapter 3 Cumulative Review

15. d=

(5.2 − ( −2.7) ) + ( 6.3 − ( −1.4) ) 2

= (7.9)2 + (7.7)2 ≈ 11.03  −2.7 + 5.2 −1.4 + 6.3  M = ,  = (1.25,2.45 ) 2 2  

16.

4.3 − ( −1.4) 5.7 = = 2.375 2.7 − 0.3 2.4

17. Since m = 0 , the line is horizontal. Hence, its equation is y = −3 . 18. x 2 + y 2 + 12 x − 18 y − 4 = 0

( x + 12x + 36 ) + ( y − 18 y + 81) = 4 + 36 + 81 2

( x + 6)2 + ( y − 9)2 = 121 So, the center is ( −6,9) and the radius is 11.

19. Since the center is ( −2, −1) , the general equation is ( x − ( −2))2 + ( y − ( −1))2 = r 2 . Now, use the fact that ( −4,3) lies on the circle to find r 2 . Indeed, observe that ( −4 − ( −2))2 + (3 − ( −1))2 = 20 = r 2 .

So, the equation is ( x + 2)2 + ( y + 1)2 = 20 . 20. We place the tower at the origin, and compute the distance between (0,0) and (85,23). Observe that

21. ( −∞,1) ∪ (1, ∞ )

852 + 232 ≈ 88.06 < 100 . So, you can use the cell phone here.

22. 5(4) − 5(2) 80 − 20 = = 30 4−2 2 2

23. g ( f ( −1)) = g ( 6 − ( −1) ) = 72 − 3 = 46

477

Chapter 3

24. y = x 2 + 3, x ≥ 0 To find the inverse, switch the x and y and solve for y: x = y2 + 3 x − 3 = y2 y = ± x − 3, x ≥ 3 Since the domain of the original function is x ≥ 0 , we use the positive root. So, f −1 ( x ) = x − 3 .

25. The general equation is r = k = 135 . So, r =

k k . Using the fact that r = = 45 , we see that 3 t

135 . t

26.

Decreasing

[ −1,1) ∪ (1,3] [0,1] ( −1,0 ) ∪ (1,2 ) ( 0,1) ∪ ( 2,3 )

Constant

nowhere

Domain Range Increasing

Note on the graph: There should be an open hole at x = 1 .

478

Chapter 3 Cumulative Review

27. Let f ( x ) = x 2 − 3x, g ( x ) = x 2 + x − 2 . Observe that 2 g ( h( x )) = ( h( x ) ) + ( h( x ) ) − 2 = x 2 − 3x

implies that h( x ) = x − 2 .

479

CHAPTER 4 Section 4.1 Solutions -------------------------------------------------------------------------------1. b Vertex ( −2, −5) and opens up

2. d Vertex (1,3) and opens up

3. a Vertex ( −3,2 ) and opens down

4. c Vertex (2,3) and opens down

5. Since the coefficient of x 2 is positive, the parabola opens up, so it must be b or d. Since the coefficient of x is positive, the x-coordinate of the vertex is negative. So, the graph is b. 6. As in #5, the parabola opens up, so it must be b or d. Since the coefficient of x is negative, the x-coordinate of the vertex is positive. So, the graph is d. 7. Since the coefficient of x 2 is negative, the parabola opens down, so it must be a or c. In comparison to #8, this one will grow more slowly in the negative ydirection. So, the graph is c. 8. Since the coefficient of x 2 is −2 , the parabola opens down and twice as quickly as the graph of y = − x 2 . So, the graph is a. 9.

10.

480

Section 4.1

11.

12.

13.

14.

15.

16.

481

Chapter 4

17.

18.

19.

20.

21.

22.

482

Section 4.1

23.

25.

24.

f ( x ) = ( x + 6x + 9 ) − 3 − 9

26.

= ( x + 3 ) − 12

= ( x + 4 ) − 14

27.

f ( x ) = − ( x +10x ) + 3

28.

= − ( x 2 +10 x + 25 ) + 3 + 25

= − ( x + 6 ) + 42

f ( x ) = 2 ( x + 4x ) − 2

30.

= 3 ( x − 32 ) + 174

= 2 ( x + 2 ) −10

f ( x ) = −4 ( x − 4x ) − 7

f ( x ) = 3 ( x 2 − 3x ) +11

= 3 ( x 2 − 3x + 94 ) +11− 274

= 2 ( x2 + 4x + 4 ) − 2 − 8

31.

f ( x ) = − ( x 2 +12x ) + 6

= − ( x 2 +12 x + 36 ) + 6 + 36

= − ( x + 5 ) + 28

29.

f ( x ) = ( x 2 + 8x +16 ) + 2 −16

32.

= −4 ( x 2 − 4 x + 4 ) − 7 + 16

f ( x ) = −5 ( x 2 − 20x ) − 36

= −5 ( x 2 − 20x +100 ) − 36 + 500

= −4 ( x − 2 ) + 9

= −5 ( x −10 ) + 464

483

Chapter 4

33.

( x − 8x ) + 3 = ( x − 8x + 1 6 ) + 3 − 8

f (x) =

1 2

34.

f ( x ) = − 13 ( x 2 −18x

)+4 = − ( x −18x + 81) + 4 + 27 1 3

= 12 ( x − 4 ) − 5

= − 13 ( x − 9 ) + 31

35.

36.

37.

38.

484

Section 4.1

39.

40.

41.

42.

43.

44.

485

Chapter 4

45. x=−

46. b −8 4 =− = 2a 2 (3) 3

x=−

b 6 3 =− = 2a 2 ( −5 ) 5 2

5 4 4 4 f   = 3  − 8  + 7 = 3 3 3 3

31 3 3 3 f   = −5   + 6   − 8 = − 5 5 5 5

4 5 Vertex =  ,  3 3

 3 31  Vertex =  , −  5 5 

47.

48.

−2 b 1 =− = x=− 2a 2 ( 33 ) 33

x=−

494  1   1   1  f   = 33   − 2   +15 = 33  33   33   33 

 1 494  Vertex:  ,   33 33 

b 4 2 =− =− 2a 2 (17 ) 17 2

 2  2  2 f  −  = 17  −  + 4  −  − 3  17   17   17  55 =− 17  2 55  Vertex:  − , −   17 17  50.

49.

b −7 =− =7 2a 1 2  2 1 2 39 f (7 ) = (7) − 7 (7) + 5 = − 2 2

x=−

2 b 3 =− 5 1 = 2a 2 (− 3 ) 5 2

13 23 103 3 f   = −   +  +4 = 25 35 55 5

 3 103  Vertex:  ,   5 25 

39   Vertex:  7, −  2   51.

52.

−0.3 b =− = −75 x=− 2a 2 ( −0.002 )

x=−

f ( −75 ) = −.002 ( −75 ) − 0.3 ( −75 ) + 1.7

f ( −25 ) = 0.05 ( −25 ) + 2.5 ( −25 ) − 1.5

= 12.95 Vertex: ( −75,12.95 )

= −32.75 Vertex: ( −25, −32.75 )

b 2.5 =− = −25 2a 2 ( 0.05 ) 2

486

Section 4.1

53. x=−

54. 3 7

x=−

b 15 =− = 2a  2  28 2−   5

− 23 b 7 =− =− 1 2a 2 (− 7 ) 3 2

2  15  3  15  829  15  f   = −   +  +2 = 5  28  7  28  392  28   15 829  Vertex:  ,   28 392 

1 7 2 7 1 8  7 f −  = − −  − − + = 7 3 3 3 9 9  3  7 8 Vertex:  − ,   3 9

55. Since the vertex is (−1, 4) , the

56. Since the vertex is (2, −3) , the

function has the form y = a( x + 1) + 4 . To find a, use the fact that the point (0,2) is on the graph: 2 = a( 0 + 1)2 + 4

function has the form y = a( x − 2 )2 − 3 . To find a, use the fact that the point (0,1) is on the graph: 1 = a( 0 − 2)2 − 3 1 = 4a − 3

−2 = a So, the function is y = −2( x + 1)2 + 4 .

57. Since the vertex is (2,5) , the function has the form y = a( x − 2 )2 + 5 . To find a, use the fact that the point (3,0) is on the graph: 0 = a(3 − 2)2 + 5 −5 = a So, the function is y = −5( x − 2)2 + 5 .

1= a So, the function is y = ( x − 2)2 − 3 . 58. Since the vertex is (1,3) , the function has the form y = a( x − 1)2 + 3 . To find a, use the fact that the point ( −2,0) is on the graph: 0 = a( −2 −1)2 + 3

0 = 9a + 3 − 13 = a So, the function is y = − 13 ( x − 1)2 + 3 .

487

Chapter 4

59. Since the vertex is (−1, −3) , the

60. Since the vertex is (0, −2) , the

function has the form y = a( x + 1) − 3 . To find a, use the fact that the point ( −4,2 ) is on the graph: 2 = a( −4 +1)2 − 3 2 = 9a − 3

function has the form y = a( x − 0 )2 − 2 . To find a, use the fact that the point (3,10) is on the graph: 10 = a(3 − 0)2 − 2 10 = 9a − 2

5 9

12 9

= 34 = a

So, the function is y = 59 ( x + 1)2 − 3 .

So, the function is y = 34 ( x − 0)2 − 2 .

61. Since the vertex is ( 12 , −43 ) , the

62. Since the vertex is ( − 65 , 23 ) , the

function has the form y = a( x − 12 )2 − 34 . To find a, use the fact that the point ( 34 , 0) is on the graph:

function has the form y = a( x + 56 )2 + 23 . To find a, use the fact that the point (0,0) is on the graph: 0 = a( 0 + 56 )2 + 23

0 = a( 34 − 12 )2 − 34

25 0 = 36 a + 32

0 = 161 a − 34 12 = a So, the function is y = 12( x − 12 )2 − 34 .

63. Since the vertex is (2.5, −3.5) , the function has the form y = a(x − 2.5)2 − 3.5 . To find a, use the fact that the point (4.5,1.5) is on the graph: 1.5 = a(4.5 − 2.5)2 − 3.5 1.5 = 4a − 3.5 5 4

So, the function is y = 54 ( x − 2.5)2 − 3.5 .

− 24 25 = a 5 2 2 So, the function is y = − 24 25 ( x + 6 ) + 3 .

64. Since the vertex is (1.8,2.7) , the function has the form y = a(x − 1.8)2 + 2.7 . To find a, use the fact that the point (−2.2, −2.1) is on the graph: −2.1 = a(−2.2 −1.8)2 + 2.7 −2.1 = 16a + 2.7 −0.3 = a So, the function is y = −0.3( x − 1.8)2 + 2.7 .

488

Section 4.1

65. Completing the square will enable you to identify the vertex of the parabola, which is precisely where the maximum occurs. P ( x ) = −0.0001( x 2 − 700,000x ) +12,500 = −0.0001( x 2 − 700,000 x + 350,000 2 ) +12,500 +12,250,000 = −0.0001( x − 350,000 ) +12,262,500 2

a. Maximum profit occurs when 350,000 units are sold. b. The maximum profit is P(350,000) = $12,262,500. 66. Completing the square will enable you to identify the vertex of the parabola, which is precisely where the minimum occurs. P ( x ) = 0.5x 2 − 20x + 1,600

= 0.5 ( x 2 − 40 x ) + 1,600

= 0.5 ( x 2 − 40 x + 400 ) + 1,600 − 200 = 0.5(x − 20)2 + 1, 400 a. Minimum profit occurs when x = 20, which corresponds to when 20,000 units are sold. b. The minimum profit is P(20) = 1,400 hundred dollars, which corresponds to $140,000. 67. Complete the square to identify the vertex. Since the coefficient of t 2 is negative, W(t) will be increasing to the left of the vertex and decreasing to the right of it. W (t ) = − 32 ( t 2 − 1390 t ) + 433 5

(

)

2 39 = − 23 t 2 − 1390 t + ( 39 + 433 20 ) 5 + 3 ( 20 ) 2

17,827 = − 23 ( t − 39 20 ) + 200 2

So, gaining weight through most of January 2020 and then losing weight during the second to eighteenth months, namely Feb 2020 to June 2021. 68. The maximum weight of the y-coordinate of the vertex and is 17,827 200 ≈ 89 kg .

489

Chapter 4

69. a. The maximum occurs at the vertex, which is (−5, 40) . So, the maximum height is 120 feet. b. If the height of the ball is assumed to be zero when the ball is kicked, and is zero when it lands, then we need to simply compute the x-intercepts of h and determine the distance between them. To this end, solve 2 8 0 = − 125 ( x + 5)2 + 40  40 ( 125  ± 25 = x + 5  x = −30,20 8 ) = ( x + 5)  = 625

So, the distance the ball covers is 50 yards. 70. a. The maximum occurs at the vertex, which is (30,50) . So, the maximum height is 150 feet. b. If the height of the ball is assumed to be zero when the ball is kicked, and is zero when it lands, then we need to simply compute the x-intercepts of h and determine the distance between them. To this end, solve 0 = − 405 ( x − 30)2 + 50  50 ( 450 ) = ( x − 30)2  ± 20 = x − 30  x = 10,50  = 400

So, the distance the ball covers is 40 yards. 71. Let x = length and y = width.

The total amount of fence is given by: 4 x + 3 y = 10,000 so that y =

10,000 − 4x 3

(1). The combined area of the two identical pens is 2xy . Substituting (1) in for y, we see that the area is described by the function:  10,000 − 4x  20,000 8 2 A( x ) = 2 x  =−3x + 3 x 3   Since this parabola opens downward (since the coefficient of x 2 is negative), the maximum occurs at the x-coordinate of the vertex, namely − 20,000 3 x = − 2ba = = 1250 . 2 ( − 83 )

10,000 − 4(1250) ≈ 1666.67 3 So, each of the two pens would have area ≅ 2,083,333 sq. ft.

The corresponding width of the pen (from (1)) is y =

490

Section 4.1

72. Let x = length of one of the four pastures, y = width of one of the four pastures. 30,000 − 8x The total amount of fence is given by: 8x + 5 y = 30,000 so that y = 5 (1). The combined area of the four identical pastures is 4xy . Substituting (1) in for y, we see that the area is described by the function:  30,000 − 8x  32 2 A( x ) = 4x   = − 5 x + 24,000 x 5   Since this parabola opens downward (since the coefficient of x 2 is negative), the maximum occurs at the x-coordinate of the vertex, namely −24,000 x = − 2ab = = 1875 2 ( − 325 )

30,000 − 8(1875) = 3000 5 So, each of the four pastures would have area 5,625,000 sq. ft.

The corresponding width of the pen (from (1)) is y =

b. Solve h(t ) = 0 .

73. a. Completing the square on h yields h(t ) = −16 ( t 2 − 2t ) + 100

−16(t − 1)2 + 116 = 0 (t − 1)2 = 116 16

= −16 ( t 2 − 2t + 1) + 100 + 16

t −1 = ±

= −16(t − 1)2 + 116 So, it takes 1 second to reach maximum height of 116 feet.

t = 1±

116 16 116 16

Since time must be positive, we conclude that the rock hits the water after about t = 1 + 116 16 ≅ 3.69 seconds (assuming the time started at t = 0 ).

74. Solve −16t 2 + 1200t = 0 :

−16t 2 + 1200t = 0 t( −16t + 1200) = 0 t = 0,75 So, the person has 75 seconds to get out of the way of the bullet.

491

Chapter 4

75.

A( x ) = −0.0003 ( x 2 − 31,000x ) − 46,075 = −0.0003 ( x −15,500 ) − 46,075 + 72,075 2

= −0.0003(x −15,500)2 + 26,000 Also, we need the x-intercepts to determine the horizontal distance. Observe −0.0003(x − 15,500)2 + 26,000 = 0 26,000 (x − 15,500)2 = ≈ 86,666,667 0.0003 x − 15,500 ≈ ±9309.49 x ≈ 15,500 ± 9309.49 = 6,309.49 and 24,809.49 So, the maximum altitude is 26,000 feet. over a horizontal distance of 8,944 feet. 76. a. Maximum height is the y-coordinate of the vertex of H ( x ) . Completing the square yields H (x) = −0.0128 ( x 2 − 78.125x ) = −0.0128 ( x 2 − 78.125x + 1525.87891) + 19.53125

= −0.0128(x − 39.06250)2 + 19.53125 So, the maximum height is 19.53125 feet. b. Need the x-intercepts. To find them, solve −0.0128x 2 + x = 0  x( −0.0128x + 1) = 0  x = 0,78.125 So, the ball travels 78.125 feet. 77. a. Maximum profit occurs at the x-coordinate of the vertex of P ( x ) . Completing the square yields P ( x ) = −0.4 ( x 2 + 200x ) + 20,000

= −0.4 ( x 2 + 200x +10,000 ) + 20,000 + 4,000 = −0.4(x −100)2 + 24,000

So, the profit is maximized when 100 boards are sold. b. The maximum profit is the y-coordinate of the vertex, namely $24,000.

492

Section 4.1

78. a. Since the vertex is (50,30) , the function has the form y = a( x − 50)2 + 30 . To find a, use the fact that the point (70,25) is on the graph: 25 = a(75 − 50)2 + 30

−5 = 400a so that − 0.0125 = a So, the function is y( x ) = −0.0125(x − 50)2 + 30 . b. Since y(90) = −0.0125(90 − 50)2 + 30 = 10 , you would expect 10 mpg. 79. First, completing the square yields P ( x ) = (100 − x)x −1000 − 20x

= −x 2 + 80x −1000 = − ( x 2 − 80 x ) −1000 = − ( x 2 − 80 x +1600 ) −1000 +1600

= −(x − 40)2 + 600 Now, solve −(x − 40)2 + 600 = 0:

−(x − 40)2 + 600 = 0 (x − 40)2 = 600 x − 40 = ± 600 so that x = 40 ± 600 ≈ 15.5, 64.5 So, 15 to 16 units to break even, or 64 or 65 units to break even.

80. Using #73, we see that the maximum profit occurs at the y-coordinate of the vertex, namely $600. 81. We are given that the vertex is ( h, k) = (0, 16) and that (9, 190.5) is on the graph. a. We need to find a such that the equation governing the situation is y = a(t − 0 )2 + 16 . (1) To do this, we use the fact that (9, 190.5) satisfies (1): 349 190.5 = a(9 − 0)2 +16  a = 162 349 2 Thus, the equation is y = 162 t + 16 , where y is measured in millions.

b. Note that the year 2016 corresponds to t = 20 . Observe that 349 y(20) = 162 (20)2 + 16 ≈ 878 million .

493

Chapter 4

82. a. We are given that the vertex is ( h, k) = (0, 49), that (16, 15.7) is on the graph (since 2014 corresponds to t = 16), and that the graph opens down (a < 0) since the peak occurs at the vertex. We need to find a such that the equation governing the situation is y = a(t − 0 )2 + 49 . (1) To do this, we use the fact that (16, 15.7) satisfies (1): 333 15.7 = a(16 − 0)2 + 49  a = − 2560 333 2 Thus, the equation is y = − 2560 t + 49 .

b. Note that the year 2016 corresponds to t = 18, so that 333 y(18) = − 2560 (18)2 + 49 ≈ 6.85. So, approximately 6.85%. 83. a. We are given that the vertex is ( h, k) = (225, 400), that (50, 93.75) is on the graph, and that the graph opens down (a < 0) since the peak occurs at the vertex. We need to find a such that the equation governing the situation is y = a(t − 225)2 + 400 . (1) To do this, we use the fact that (50, 93.75) satisfies (1): 93.75 = a(50 − 225)2 + 400  a = −0.01

Thus, the equation is y = −0.01(t − 225)2 + 400 . b. We must find the value(s) of t for which 0 = −0.01(t − 225)2 + 400 . To this end, 0 = −0.01(t − 225)2 + 400  (t − 225)2 = 40,000

 t − 225 = ± 40,000 = ±200  t = 225 ± 200 = 25, 425 So, it takes 425 minutes for the drug to be eliminated from the bloodstream.

494

Section 4.1

84. a. We know that the points (70, 20) and (50, 25) lie on this line. Hence, the 25 − 20 1 slope is m = 50 − 70 = − 4 . Using point-slope form, the price function is: p − 20 = − 14 ( x − 70) 

p(x) = − 14 x + 752

b. R( x ) = xp( x ) = − 14 x 2 + 752 x c. The maximum revenue occurs at the vertex of R( x ) . Completing the square yields − 14 x 2 + 752 x = − 14 ( x 2 −150x ) = − 14 ( x 2 −150 x + 5,625 ) + 5,625 4 = − 14 ( x − 75 ) + 5,625 4 2

The vertex is ( 75, 5,625 4 ) . So, he needs to wash 75 cars in order to maximize revenue. d. To maximize revenue, he should charge p( x ) = − 14 (75) + 752 = 754 = $18.75 . 85. The corrections that need to be made are vertex (−3, −1) and x-intercepts (−2, 0) (−4, 0). 86. b = −6, not 6

87. ƒ( x ) = −( x − 1)2 + 4. The negative must be factored out of the x2 and x terms.

88. Step 3 is wrong: f (9) = a(9 − 2)2 − 3 = 0

89. True. f ( x ) = a( x − h )2 + k , so that f (0) = ah 2 + k .

49a − 3 = 0 a = 493 So, f ( x ) =

3 49

(x − 2) − 3 . 2

91. False. The graph would not pass the vertical line test in such case, and hence wouldn’t define a function.

90. False. Consider f ( x ) = −1 − x 2 . 92. True. Consider f ( x ) = x 2 − 1 .

495

Chapter 4

93. Completing the square yields f ( x ) = ax 2 + bx + c

= a ( x 2 + ab x ) + c

(

)

= a x 2 + ab x + ( 2ba ) + c − a ( 2ba ) 2

= a ( x + 2ba ) + c − 4ba = a ( x + 2ab ) + 4ac4−a b 2

(

)

So, the vertex is − 2ba ,c − 4ba . Observe that f ( − 2ba ) = a ( − 2ba + 2ba ) + c − 4ba = c − 4ba .  2

94. Given that f ( x ) = a(x − h)2 + k , we have:

y-intercept: a(0 − h )2 + k = ah 2 + k , so that the y-intercept is ( 0, ah2 + k ) .

x-intercepts: Solve a( x − h )2 + k = 0 . a( x − h )2 + k = 0 ( x − h )2 = − ka x − h = ± − ka

(

)(

x = h ± − ka

)

So, the x-intercepts are h + − ka ,0 , h − − ka ,0 . 95. a. Let x = width of rectangular pasture, y = length of rectangular pasture. Then, the total amount of fence is described by 2x + 2 y = 1000 (so that y = 500 − x (1)).

The area of the pasture is xy . Substituting in (1) yields x(500 − x ) = − x 2 + 500 x . The maximum area occurs at the y-coordinate of the vertex (since the coefficient 2 2 of x 2 is negative); in this case this value is c − 4ba = 0 − 4500 ( −1) = 62,500. The maximum area of the rectangular pasture is 62,500 square feet. b. Let x = radius of circular pasture. 500 Then, the total amount of fence is described by 2π x = 1000 (so that x = 1000 2π = π (2)). 2 5002 The area of the pasture is π x 2 , which in this case must be (by (2)) π ( 500 π ) = π . The maximum area of the circular fence is approximately 79,577 square feet.

496

Section 4.1

96. I ( x ) = (90 + 5x)(600 − 10x) = −50x 2 + 2100x + 54,000 Completing the square then yields −50x 2 + 2100x + 54,000 = −50 ( x 2 − 42x ) + 54,000

= −50 ( x 2 − 42x + 441) + 54,000 + 22,050

= −50( x − 21)2 + 76,050 Since the coefficient of x 2 is negative, the maximum income must be $76,050, which occurs when there are 21 increases in room rate (i.e., at the vertex). Hence, the room rate that yields the maximum profit is 90 + 21(5) = $195 . 97. a. Vertex:

(−

5.7 2( −0.002 )

)

) , −23 − 4((5.7 −0.002 ) = (1425, 4038.25)

b. y-intercept: f (0) = −23 , so (0, −23) c. x-intercepts: Solve −0.002x 2 + 5.7x − 23 = 0 . Using part a., the standard form is −0.002(x − 1425)2 + 4038.25 = 0 . So, solving this equation yields: (x − 1425)2 = 2,019,125 x = 1425 ± 2,019,125

≅ 4.04, 2845.96 So, the x-intercepts are (4.04,0), (2845.96,0) . d. Axis of symmetry: x = 1425

497

Chapter 4

98. a. Since the vertex is (−0.5,1.7) , the

function has the form y = a( x + 0.5) + 1.7 . To find a, use the fact that the point (0, 4) is on the graph: 4 = a(0 + 0.5)2 + 1.7 4 = 0.25a + 1.7 2

9.2 = a So, the function is y = 9.2( x + 0.5)2 + 1.7 . c. Yes, they agree.

99. a. The equation obtained using this particular calculator feature is f ( x ) = −2x 2 + 12.8x + 4.32 . b. Completing the square yields f ( x ) = −2x 2 + 12.8x + 4.32

= −2 ( x 2 − 6.4x ) + 4.32

= −2 ( x 2 − 6.4x + 10.24 ) + 4.32 + 20.48 = −2(x − 3.2)2 + 24.8 So, the vertex is (3.2,24.8) . c. The graph is to the right. Yes, they agree with the given values.

498

Section 4.1

100. a. The equation obtained using this particular calculator feature is f ( x ) = 0.44x 2 + 2.92x − 12.10 . b. Completing the square yields f ( x ) = 0.44x 2 + 2.92x − 12.10

= 0.44 ( x 2 + 6.64 x ) − 12.10

= 0.44 ( x 2 + 6.64 x + 11.0224 ) − 12.10 − 4.85 = 0.44(x + 3.32)2 − 16.95 So, the vertex is (−3.32, −16.95) . c. The graph is to the right. Yes, they agree with the given values.

499

Section 4.2 Solutions -------------------------------------------------------------------------------1. Polynomial with degree 2

2. Polynomial with degree 5

3. Polynomial with degree 5

4. Polynomial with degree 9

5. Not a polynomial (due to the term 1 x 2)

6. Not a polynomial (due to the term 1 x 2)

7. Not a polynomial (due to the term 1 x 3)

8. Not a polynomial (due to the term 2 3x )

9. Not a polynomial (due to the terms 1 1 x , x2 )

10. Polynomial with degree 2

11. h linear function

12. g Parabola that opens down

13. b Parabola that opens up

14. f Note that −2x3 + 4 x 2 − 6 x = −2 x ( x 2 − 2 x + 3 ) .

Since x 2 − 2x + 3 is a parabola opening up with vertex (1,2), it has no real roots. So, this polynomial has only 1 xintercept at which it crosses. 15. e x3 − x 2 = x 2 ( x − 1) So, there are two x-intercepts: the graph is tangent at 0 and crosses at 1.

17. c − x 4 + 5x3 = − x3 ( x − 5) So, there are two x-intercepts (0, 5) and the graph crosses at each of them.

16. d Note that 2 x 4 − 18x 2 = 2x 2 ( x 2 − 9 )

= 2x 2 ( x − 3)(x + 3) There are three x-intercepts (0, 3, −3 ) and it crosses at each of them. 18. a x 5 − 5x 3 + 4 x = x ( x 4 − 5x 2 + 4 )

= x ( x 2 − 4 ) ( x 2 − 1)

= x( x − 1)( x + 1)( x − 2)( x + 2) So, there are five x-intercepts (0, 1, 2, −1, −2 ) and the graph crosses at each of them.

500

Section 4.2

19.

21.

23.

20.

22.

24.

501

Chapter 4

25.

26.

27. 3 (multiplicity 1) −4 (multiplicity 3)

28. −2 (multiplicity 3) 1 (multiplicity 2)

29. 0 (multiplicity 2) 7 (multiplicity 2) −4 (multiplicity 1)

30. 0 (multiplicity 3) −1 (multiplicity 4) 6 (multiplicity 1)

31. 0 (multiplicity 2) 1 (multiplicity 2) Note: x 2 + 4 = 0 has no real solutions.

32. 0 (multiplicity 2) −1 (multiplicity 1) 1 (multiplicity 1)

33.

x − 3x − 10 x = x ( x − 3x − 10 )

34. x 4 − 10 x3 + 21x 2 = x 2 ( x 2 − 10 x + 21)

35. 15x 4 + 7 x3 − 2 x 2 = x 2 (15x 2 + 7 x − 2 )

36. 6 x3 + 23x 2 + 20 x = x ( 6 x 2 + 23x + 20 )

= x ( x − 5) ( x + 2) So, the zeros are: 0 (multiplicity 1) 5 (multiplicity 1) –2 (multiplicity 1)

= x 2 ( 3x + 2 ) ( 5x − 1) So, the zeros are: 0 (multiplicity 2) − 23 (multiplicity 1) 51 (multiplicity 1)

= x2 ( x − 7 ) ( x − 3) So, the zeros are: 0 (multiplicity 2) 7 (multiplicity 1) 3 (multiplicity 1)

= x ( 2 x + 5 ) ( 3x + 4 ) So, the zeros are: 0 (multiplicity 1) − 52 (multiplicity 1) − 34 (multiplicity 1)

502

Section 4.2

37.

38.

8x + 6 x − 27x = x ( 8x + 6x − 27 ) 3

= x(2x − 3)(4x + 9)

2 x 4 + 5 x 3 − 3x 2 = x 2 ( 2 x 2 + 5 x − 3 )

= x 2 (2 x − 1)( x + 3)

So, the zeros are: 0 (multiplicity 1) 3 2 (multiplicity 1) − 94 (multiplicity 1)

So, the zeros are: 0 (multiplicity 2) 1 2 (multiplicity 1) −3 (multiplicity 1)

39.

40. 1.2 x 6 − 4.6 x 4 ≅ 1.2 x 4 ( x 2 − 3.83 )

−2.7 x − 8.1x = −2.7 x ( x + 3 ) 3

So, the zeros are: 0 (multiplicity 2) −3 (multiplicity 1) 41. 1 3

= x( x − 3.83)( x + 3.83) So, the zeros are approximately: 0 (multiplicity 1) 1.957 (multiplicity 1) −1.957 (multiplicity 1)

x 6 + 52 x 4 = 31 x 4 ( x 2 + 65 )    Always positive

So, the only zero is: 0 (multiplicity 4) 42. Note that 72 x5 − 34 x 4 + 21 x3 = 72 x3 ( x 2 − 218 x + 74 ) . To determine if the second term is

factorable, we complete the square:

(

)

21 x 2 − 218 x + 74 = x 2 − 218 x + ( 16 ) + 74 − ( 1621 ) 2

7 21 = ( x − 16 ) + 256   2

Always positive

Since ( x − is:

) + 2567 = 0 has no real solutions, we conclude that the only real zero

21 2 16

0 (multiplicity 3) 43. P ( x ) = x( x + 3)( x −1)( x − 2)

44. P ( x ) = x( x + 2)( x − 2)

45. P ( x ) = x(x + 5)(x + 3)(x − 2)(x − 6)

46. P ( x ) = x(x − 1)(x − 3)(x − 5)(x −10)

47. P ( x ) = (2x + 1)(3x − 2)(4x − 3)

48. P ( x ) = x(4x + 3)(3x +1)(2x − 1)

503

Chapter 4

49.

(

)(

P ( x ) = x − (1 − 2 ) x − (1 + 2 )

50.

)

(

)(

P ( x ) = x − (1 − 3) x − (1 + 3)

= x2 − 2x − 1

)

= x 2 − 2x − 2

51. P ( x ) = x 2 ( x + 2 )3

52. P ( x ) = ( x + 4)2 ( x − 5)3

53. P ( x ) = ( x + 3)2 ( x − 7)5

54. P ( x ) = x( x − 10)3

55. P ( x ) = x 2 (x + 1)(x + 3)2 (x − 3)2

56. P ( x ) = x(x − 1)2 (x + 5 )2 (x − 5 )2

57. f ( x ) = − ( x 2 + 6x + 9 ) = −(x + 3)2

58. f ( x ) = x 2 + 4x + 4 = ( x + 2)2 a. Zeros: −2 (multiplicity 2) b. Touches at −2 c. y-intercept: f (0) = 4, so (0, 4)

a. Zeros: −3 (multiplicity 2) b. Touches at −3 c. y-intercept: f (0) = −9, so (0, −9) d. End behavior: Behaves like y = − x 2 . Even degree and leading coefficient negative, so graph falls without bound to the left and right.

d. End behavior: Behaves like y = x 2 . Even degree and leading coefficient positive, so graph rises without bound to the left and right.

504

Section 4.2

59. f ( x ) = ( x − 2)3 a. Zeros: 2 (multiplicity 3) b. Crosses at 2 c. y-intercept: f (0) = −8, so (0, −8) d. End behavior: Behaves like y = x3 . Odd degree and leading coefficient positive, so graph falls without bound to the left and rises to the right.

60. f ( x ) = −( x + 3)3 a. Zeros: −3 (multiplicity 3) b. Crosses at −3 c. y-intercept: f (0) = −27, so (0, −27)

d. End behavior: Behaves like y = − x3 . Odd degree and leading coefficient negative, so graph falls without bound to the right and rises to the left.

61. f ( x ) = x3 − 9x = x( x − 3)( x + 3) a. Zeros: 0, 3, −3 (multiplicity 1) b. Crosses at each zero c. y-intercept: f (0) = 0, so (0,0) d. End behavior: Behaves like y = x3 . Odd degree and leading coefficient positive, so graph falls without bound to the left and rises to the right.

505

Chapter 4

62. f ( x ) = − x3 + 4 x 2 = −x 2 ( x − 4) a. Zeros: 0 (multiplicity 2), 4 (multiplicity 1) b. Crosses at 4, touches at 0 c. y-intercept: f (0) = 0, so (0,0)

d. End behavior: Behaves like y = − x3 . Odd degree and leading coefficient negative, so graph falls without bound to the right and rises to the left.

63. f ( x ) = − x3 + x 2 + 2 x = − x( x − 2)( x + 1) a. Zeros: 0, 2, −1 (multiplicity 1) b. Crosses at each zero c. y-intercept: f (0) = 0, so (0,0)

d. End behavior: Behaves like y = − x3 . Odd degree and leading coefficient negative, so graph falls without bound to the right and rises to the left.

64. f ( x ) = x3 − 6 x 2 + 9x = x( x − 3)2 a. Zeros: 0 (multiplicity 1) 3 (multiplicity 2) b. Crosses at 0, touches at 3 c. y-intercept: f (0) = 0, so (0,0) d. End behavior: Behaves like y = x3 . Odd degree and leading coefficient positive, so graph falls without bound to the left and rises to the right.

506

Section 4.2

65. f ( x ) = − x 4 − 3x3 = − x3 ( x + 3) a. Zeros: 0 (multiplicity 3) −3 (multiplicity 1) b. Crosses at both 0 and −3 c. y-intercept: f ( 0) = 0, so (0,0) d. End behavior: Behaves like y = − x 4 . Even degree and leading coefficient negative, so graph falls without bound to left and right.

66. f ( x ) = x5 − x3 = x3 (x 2 −1)

= x3 ( x − 1)(x +1) a. Zeros: 0 (multiplicity 3) 1 (multiplicity 1), −1 (multiplicity 1) b. Crosses at each of 0, 1, and −1 c. y-intercept: f ( 0) = 0, so (0,0)

d. End behavior: Behaves like y = x5 . Odd degree and leading coefficient positive, so graph falls without bound to the left and rises to the right. e.

67. f (x ) = 12x − 36x − 48x 6

= 12x 4 ( x 2 − 3x − 4 )

= 12x 4 ( x − 4)(x +1) a. Zeros: 0 (multiplicity 4), 4 (multiplicity 1), −1 (multiplicity 1) b. Touches at 0 and crosses at 4 and −1. c. y-intercept: f (0) = 0, so (0,0)

d. End behavior: Behaves like y = x6 . Even degree and leading coefficient positive, so graph rises without bound to the left and right.

507

Chapter 4

68.

e. f ( x ) = 7 x − 14x − 21x 5

= 7 x3 ( x 2 − 2 x − 3 )

= 7 x3 ( x − 3)(x +1) a. Zeros: 0 (multiplicity 3), 3 (multiplicity 1), −1 (multiplicity 1) b. Crosses at each of 0, 3, and −1 c. y-intercept: f (0) = 0, so (0,0)

d. End behavior: Behaves like y = x5 . Odd degree and leading coefficient positive, so graph falls without bound to the left and rises to the right. 69.

e. f ( x ) = 2 x − 6x − 8x 5

= 2 x3 ( x − 4)(x +1) a. Zeros: 0 (multiplicity 3), 4 (multiplicity 1), −1 (multiplicity 1) b. Crosses at each zero c. y-intercept: f (0) = 0, so (0,0)

d. End behavior: Behaves like y = x5 . Odd degree and leading coefficient positive, so graph falls without bound to the left and rises to the right.

70. f ( x ) = −5x 4 + 10x3 − 5x 2 = −5x 2 (x − 1)2 a. Zeros: 0 (multiplicity 2) 1 (multiplicity 2) b. Touches at each zero c. y-intercept: f (0) = 0, so (0,0) d. End behavior: Behaves like y = − x 4 . Even degree and leading coefficient negative, so graph falls without bound to left and right.

508

Section 4.2

71.

e. f ( x ) = x − x − 4x + 4 3

= ( x3 − x 2 ) − 4(x − 1)

= x 2 ( x − 1) − 4( x − 1) = ( x − 2 )( x + 2)(x − 1) a. Zeros: 1, 2, −2 (multiplicity 1) b. Crosses at each zero c. y-intercept: f (0) = 4, so (0, 4) d. End behavior: Behaves like y = x3 . Odd degree and leading coefficient positive, so graph falls without bound to the left and rises to the right. e.

72.

f ( x ) = x − x − x +1 3

= ( x3 − x 2 ) − ( x − 1) = x 2 ( x − 1) − 4( x − 1)

= ( x 2 − 1)( x −1) = ( x − 1)2 ( x +1) a. Zeros: −1 (multiplicity 1), 1 (multiplicity 2) b. Crosses at −1, touches at 1 c. y-intercept: f ( 0) = 1, so (0,1) d. End behavior: Behaves like y = x3 . Odd degree and leading coefficient positive, so graph falls without bound to the left and rises to the right.

509

Chapter 4

73. f ( x ) = −(x + 2)2 (x − 1)2 a. Zeros: −2 (multiplicity 2) 1 (multiplicity 2) b. Touches at both −2 and 1 c. y-intercept: f (0) = −4, so (0, −4)

d. End behavior: Behaves like y = − x 4 . Even degree and leading coefficient negative, so graph falls without bound to left and right.

74. f ( x ) = (x − 2)3 (x + 1)3 a. Zeros: −1 (multiplicity 3) 2 (multiplicity 3) b. Crosses at both −1and 2 c. y-intercept: f ( 0) = −8, so (0, −8)

d. End behavior: Behaves like y = x6 . Even degree and leading coefficient positive, so graph rises without bound to the left and right.

75. f ( x ) = x 2 (x − 2)3 (x + 3)2 a. Zeros: 0 (multiplicity 2) 2 (multiplicity 3) −3 (multiplicity 2) b. Touches at both 0 and −3 , and crosses at 2. c. y-intercept: f (0) = 0, so (0,0)

d. End behavior: Behaves like y = x7 . Odd degree and leading coefficient positive, so graph falls without bound to the left and rises to the right.

510

Section 4.2

76. f ( x ) = − x3 (x − 4)2 (x + 2)2 a. Zeros: 0 (multiplicity 3) 4 (multiplicity 2) −2 (multiplicity 2) b. Touches at 4 and −2 , and crosses at 0. c. y-intercept: f (0) = 0, so (0,0) d. Long-term behavior: Behaves like y = −x7 . Odd degree and leading coefficient negative, so graph falls without bound to the right and rises to the left.

77. a. zeros: −3 (multiplicity 1) −1 (multiplicity 2) 2 (multiplicity 1) b. degree of polynomial: even c. sign of leading coefficient: negative d. y-intercept: (0,6) e. f ( x ) = −(x + 1)2 (x − 2)(x + 3) .

78. a. zeros: −2 (multiplicity 1) 2 (multiplicity 2) 0 (multiplicity 1) b. degree of polynomial: even c. sign of leading coefficient: positive d. y-intercept: (0,0) e. f ( x ) = x(x + 2)(x − 2)2 .

79. a. zeros: 0 (multiplicity 2) −2 (multiplicity 2) 3 2 (multiplicity 1) b. degree of polynomial: odd c. sign of leading coefficient: positive d. y-intercept: (0,0) e. f ( x ) = x 2 (2x − 3)(x + 2)2 .

80. a. zeros: −3 (multiplicity 1) 0 (multiplicity 1) − 32 (multiplicity 1) 1 (multiplicity 2) b. degree of polynomial: odd c. sign of leading coefficient: negative d. y-intercept: (0,0) e. f ( x ) = − x(2x + 3)(x + 2)(x − 1)2 .

81. a. Revenue for the company is increasing when advertising costs are less than $400,000. Revenue for the company is decreasing when advertising costs are between $400,000 and $600,000. b. The zeros of the revenue function occur when $0 and $600,000 are spent on advertising. When either $0 or $600,000 is spent on advertising the company’s revenue is $0. 82. The company’s maximum revenue is $32,000,000 when $400,000 is spent on advertising.

511

Chapter 4

83. The velocity of air in the trachea is increasing when the radius of the trachea is between 0 and 0.45 cm and decreasing when the radius of the trachea is between 0.45 and 0.65 cm.

84. r ≅ 0.45 cm; v ≈ 5.265 meters per second. 85. From the data, the turning points occur between months 3 & 4, 4 & 5, 5 & 6, 6 & 7, and 7 & 8. The lowest degree polynomial that can represent her weight is a sixth-degree polynomial because there are five known turning points.

Note: There could be more turning points if there was more oscillation during these month-long periods, but the data doesn’t reveal such behavior.

512

Section 4.2

86. From the data, all we can deduce is that there must be a turning point between periods 2 and 3. The lowest degree polynomial modeling this behavior could be a second-degree polynomial.

87. From the data, there is a turning point at 2. Since it is a third degree polynomial, one would expect the price to go down.

88. From the data, we know that there is a turning point at 3. Since a third degree polynomial is used to model the stock, we expect there to be a turning point at 4, so that the stock should go up in the fifth period.

89. 4th degree The given graph has three turning points, and is assumed to be defined only on the interval [2012, 2020]. Since it seems to have 3 turning points, the lowest degree polynomial that would work is 4.

90. negative Judging from the fact that both ends point downward at the endpoints, the end behavior is best described by y = −x 2 n , for some positive integer n . So, the coefficient of the leading term must be negative.

91. If h is a zero of a polynomial, then ( x − h ) is a factor of it. So, in this case the function would be: P ( x ) = (x + 2)(x +1)(x − 3)(x − 4)

92. It should be similar to y = x 4 , which rises to the left and right without bound.

513

Chapter 4

93. ƒ( x ) = ( x − 1)2 (x + 2 )3 Yes, the zeros are −2 and 1. But you must remember that this is a fifthdegree polynomial. At the −2 zero the graph crosses, but it should be noted that at 1 it only touches at this value. The correct graph would be the following:

94. Should touch, not cross, at 1 and −1 since both have even multiplicity. The graph should look like:

95. False. A polynomial has general form f ( x ) = an x n + an −1x n −1 + ... + a1x + a0 .

So, f (0) = a0 .

96. True. Consider f ( x ) = x 2 + 1 97. True. 98. False. Consider f ( x ) = x 2 + 1 . Its range is [1,∞ ) .

99. A polynomial of degree n can have at most n zeros. 100. An nth degree polynomial can have at most (n – 1) turning points, and this would occur if it had n distinct real zeros, each with multiplicity 1.

514

Section 4.2

101. It touches at −1, so the multiplicity of this zero must be 2, 4, or 6. It crosses at 3, so the multiplicity of this zero must be 1, 3, or 5. Thus, the following polynomials would work: f ( x ) = (x + 1)2 (x − 3)5 , g(x) = (x + 1)4 (x − 3)3 , h( x ) = (x + 1)6 (x − 3)

102. A 5th degree polynomial satisfying these properties is of one of the following two forms. You can multiply each by a nonzero real constant to obtain others.

P1 ( x ) = ( x − 0)4 ( x − 4)

P2 ( x ) = ( x − 0)2 ( x − 4)3

103. Observe that x3 + (b − a)x 2 − abx = x ( x 2 + (b − a)x − ab ) = x( x − a )( x + b ) . So,

the zeros are 0, a, and −b .

515

Chapter 4

105. Note that x 4 + 2 x 2 + 1 = ( x 2 + 1) .    2

104. The function f ( x ) = x 2 (x − a)2 (x − b )2 has zeros of 0, a, and b, all of which touch the x-axis. The graph is as follows:

Always positive

So, there are no real zeros, and hence no x-intercepts. The graph is as follows:

106. First, note that 1.1x3 − 2.4x 2 + 5.2x = x (1.1x 2 − 2.4x + 5.2 ) .

Further, using the quadratic formula, the solutions of 1.1x 2 − 2.4x + 5.2 = 0 are: 2.4 ± −17.12 2.2

Hence, the only real zero (and hence, xintercept) of the above function is 0. The graph is as follows:

516

Section 4.2

107. From the following graph of both f ( x ) = −2 x5 − 5x 4 − 3x3 (solid curve) and y = −2x5 (dotted curve), we do conclude that they have the same end behavior.

108. From the following graph of both f ( x ) = x 4 − 6x 2 + 9 and y = x 4 , we do conclude that they have the same end behavior.

109. f ( x ) = x 4 − 15.9x3 +1.31x 2

+292.905x + 445.7025 The following are estimates: x-intercepts: (−2.25,0), (6.2,0), (14.2,0) zeros: −2.25 (multiplicity 2), 6.2 (multiplicity 1), 14.2 (multiplicity 1)

110. f (x ) = − x5 + 2.2x 4 + 18.49x3 − 29.878x 2

− 76.5x + 100.8 The following are estimates: x-intercepts: (−3.2,0), ( −2.5,0), (1.2,0), (2.5,0), (4.2,0) zeros: −3.2, −2.5, 1.2, 2.5, 4.2 (all have multiplicity 1)

517

Chapter 4

111. The graph is:

112. The graph is:

The coordinates of the relative extrema are (−2.56,−17.12), (−0.58,12.59), (1.27,−11.73).

The coordinates of the relative extrema are (−1.61,15.47), (−0.35,−9.95), (1.21,14.07), (2.36,−4.38).

518

Section 4.3 Solutions -------------------------------------------------------------------------------1.

2. 2x − 1 x + 3 2 x + 5x − 3

2 x + 11 x − 3 2 x + 5x − 3

−(2x 2 + 6 x )

−(2x 2 − 6 x )

− x −3 −( − x − 3 )

−(11x − 33)

So, Q( x ) = 2x − 1, r(x) = 0 .

So, Q( x ) = 2x + 11, r( x ) = 30 .

4. x−2

11x − 3

x −3 2 x − 5x + 6

2x + 1 x + 1 2 x + 3x + 1 2

−(2x 2 + 2x )

−( x 2 − 2 x ) − 3x + 6 −( −3x + 6)

x +1 −( x + 1)

So, Q( x ) = x − 3, r( x ) = 0 .

So, Q( x ) = 2x + 1, r( x ) = 0 .

6. 3x − 3 2 x − 2 3x − 9x − 5

x+5 x − 1 x + 4x − 3

−(3x 2 − 6 x )

−( x 2 − x )

− 3x − 5 −( −3x + 6)

5x − 3

−(5x − 5)

− 11

So, Q( x ) = 3x − 3, r( x ) = −11 .

So, Q( x ) = x + 5, r( x ) = 2 .

519

Chapter 4

8. 3x − 28 x + 5 3x − 13x − 10

3x + 2 x − 5 3x − 13x − 10

−(3x 2 + 15x )

−(3x 2 − 15x )

− 28x −10 −( −28x − 140)

2 x − 10

−(2x − 10)

130

So, Q( x ) = 3x − 28, r(x) = 130 .

So, Q( x ) = 3x + 2, r(x) = 0 .

10. x+4

x−4 2 x + 0x − 4

x−2

x+2 x + 0x − 9 2

−( x 2 − 2 x )

−( x 2 + 4 x ) − 4x − 4 −( −4x − 16)

2x − 9

−(2x − 4)

−5

So, Q( x ) = x − 4, r( x ) = 12 .

So, Q( x ) = x + 2, r( x ) = −5 .

11.

12. 3x + 5 3x − 5 9x + 0x − 25

5x − 5 x + 1 5x + 0x − 3

−(9x 2 − 15x )

−(5x 2 + 5x )

15x − 25

− 5x − 3 −( −5x − 5)

−(15x − 25) 0

So, Q( x ) = 3x + 5, r(x) = 0 .

So, Q( x ) = 5x − 5, r(x) = 2 .

520

Section 4.3

13.

14. 2x − 3 2x + 3 4 x + 0x − 9

4x 2 − 6x + 9 2 x + 3 8x3 + 0x 2 + 0x + 27

−(4x 2 + 6x )

−(8x3 + 12 x 2 )

− 6x − 9 −( −6x − 9)

−12x 2 + 0 x −( −12x 2 −18x )

18x + 27 −(18x + 27)

So, Q( x ) = 2x − 3, r(x) = 0 .

0 So, Q( x ) = 4x 2 − 6x + 9, r( x ) = 0 . 15.

16.

4x + 4x + 1 3 3x + 2 12 x + 20x 2 +11x + 2

6x2 + 7x + 2 2x + 1 12x3 + 20x 2 +11x + 2

−(12x3 + 8x 2 )

−(12x3 + 6 x 2 )

12x 2 +11x

14x 2 +11x

−(12x 2 + 8x )

−(14x 2 + 7 x )

3x + 2 −(3x + 2)

4x + 2 −(4x + 2)

So, Q( x ) = 4x 2 + 4x + 1, r( x ) = 0 .

So, Q( x ) = 6x 2 + 7x + 2, r( x ) = 0 .

521

Chapter 4

17.

18. 1 2

−2x3 − 34 x 2 − 92 x − 247

2 x + 1 4x3 + 0x 2 − 2x + 7

−3x + 2 6x 4 + 0x3 − 2x 2 + 0x + 5

2x − x − 2

−(6x 4 − 4 x3 )

−(4x3 + 2 x 2 ) − 2x 2 − 2x

4x3 − 2 x 2

−( −2x 2 − x )

−(4x3 − 83 x 2 )

− x+7

2 3

−( − x − 12 )

x 2 + 0x

−( 23 x 2 − 94 x )

15 2

So, Q( x ) = 2x − x − , r( x ) = 2

1 2

15 2

4 9

x+5

−( 94 x − 287 )

143 27

So, Q( x ) = −2x3 − 34 x 2 − 92 x − 274 , r( x ) = 143 27 .

20.

19.

4x − 10 x − 6 3 4 x − 12x 2 − x + 3 2

x − 12

x + 13

−(4x3 − 2x 2 )

12x 2 + 12x + 3 12 x3 + 16x 2 + 7x + 1 −(12x3 + 4 x 2 )

− 10x 2 − x

12x 2 + 7 x

−( −10x 2 + 5x )

−(12x 2 + 4x )

− 6x + 3 −( −6x + 3)

3x + 1 −(3x + 1)

So, Q( x ) = 4x 2 − 10x − 6, r(x) = 0 .

So, Q( x ) = 12x 2 + 12x + 3, r(x) = 0 .

522

Section 4.3

21.

−2x 2 − 3x − 9 x3 − 3x 2 + 0x + 1 − 2x5 + 3x 4 + 0x3 − 2x 2 + 0x + 0 −( −2x5 + 6 x 4 + 0x3 − 2x 2 ) − 3x 4 + 0x3 + 0x 2 + 0x −( − 3x 4 + 9x3 + 0x 2 − 3x) − 9x3 + 0x 2 + 3x + 0 −( − 9x3 + 27 x 2 + 0x − 9) − 27x 2 + 3x + 9 So, Q( x ) = −2x 2 − 3x − 9, r( x ) = − 27x 2 + 3x + 9 . 22. −3x 2 + 0 x + 73 3x 4 + 0x3 + 0x 2 − 2x +1 − 9x 6 + 0 x5 + 7 x 4 − 2 x3 + 0 x 2 + 0 x + 5 −( −9x6 + 0 x5 + 0x 4 + 6x3 − 3x 2 ) 7 x 4 − 8x3 + 3x 2 + 0x + 5 −(7x 4 + 0x3 + 0x 2 − 143 x + 73 ) − 8x3 + 3x 2 + 143 x + 38

So, Q( x ) = −3x 2 + 73 , r( x ) = −8x3 + 3x 2 + 143 x + 38 . 24.

23. x + 0x + 1 2 4 3 x + 0 x − 1 x + 0 x + 0x 2 + 0x −1

x 2 + 0x − 3 x 2 + 0 x + 3 x 4 + 0x3 + 0x 2 + 0x − 9

−( x 4 + 0 x 3 − x 2 )

−( x 4 + 0x3 + 3x 2 )

x 2 + 0x − 1

− 3x 2 + 0x − 9

−( x 2 + 0 x −1)

−( −3x 2 + 0x − 9)

So, Q( x ) = x 2 + 1, r( x ) = 0 . So, Q( x ) = x 2 − 3, r( x ) = 0 .

523

Chapter 4

25.

26.

6x 2 + x − 2 6x 4 + 7x3 + 0x 2 − 22x + 40

x2 + 0x − 1 4 x 2 + 0 x − 9 4x 4 + 0x3 −13x 2 + 0x + 9

−(6x 4 + x3 − 2x 2 )

−(4x 4 + 0 x3 − 9x 2 )

x +x+ 2

1 6

6 x3 + 2 x 2 − 22x

− 4x 2 + 0x + 9

−(6x3 + x 2 − 2x)

−( −4x 2 + 0 x + 9) 0

x 2 − 20 x + 40

So, Q( x ) = x 2 − 1, r( x ) = 0 .

−(x 2 + 61 x − 31 ) 121 − 121 6 x + 3

So, 121 Q( x ) = x 2 + x + 61 , r( x ) = − 121 6 x + 3 .

27.

−3x3 + 5.2 x 2 + 3.12x − 0.128 x − 0.6 − 3x 4 + 7x3 + 0x 2 − 2x + 1 −( −3x 4 + 1.8x3 ) 5.2x3 + 0 x 2

−(5.2x3 − 3.12 x 2 ) 3.12x 2 − 2 x

−(3.12x 2 − 1.872 x ) − 0.128x + 1 −( −0.128x + 0.0768) 0.9232

So, Q( x ) = −3x3 + 5.2x 2 + 3.12x − 0.128, r(x) = 0.9232 .

524

Section 4.3

28. 2x 4 + 1.8x3 − 2.38x 2 + 0.858x + 0.7722 x − 0.9 2 x5 + 0x 4 − 4x3 + 3x 2 + 0x + 5

−(2x5 − 1.8x 4 ) 1.8x 4 − 4 x3

−(1.8x 4 − 1.62 x3 ) − 2.38x3 + 3x 2 −( −2.38x3 + 2.142 x 2 ) 0.858x 2 + 0 x

−(0.858x 2 − 0.7722 x ) 0.7722x + 5

−(0.7722x − 0.69498) 5.69498

So, Q( x ) = 2x 4 + 1.8x3 − 2.38x 2 + 0.858x + 0.7722, r( x ) = 5.69498 . 29.

x 2 − 0.6x + 0.09 x 2 + 1.4x + 0.49 x 4 + 0.8x3 − 0.26x 2 − 0.168x + 0.0441 −(x 4 + 1.4x3 + 0.49x 2 ) − 0.6x3 − 0.75x 2 − 0.168x −( −0.6x3 − 0.84 x 2 − 0.294x) 0.09x 2 + 0.126 x + 0.0441 −(0.09x 2 + 0.126 x + 0.0441) 0 So, Q( x ) = x 2 − 0.6x + 0.09, r( x ) = 0 .

525

Chapter 4

30. x3 + 3.4 x 2 + 3.29x + 0.98 x 2 − 0.6 x + 0.09 x5 + 2.8x 4 + 1.34x3 − 0.688x 2 − 0.2919x + 0.0882

−(x5 − 0.6x 4 + 0.09x3 ) 3.4x 4 + 1.25x3 − 0.688x 2

−(3.4x 4 − 2.04 x3 + 0.306x 2 ) 3.29x3 − 0.994x 2 − 0.2919x

−(3.29x3 − 1.974x 2 + 0.2961x) 0.98x 2 − 0.588x + 0.0882

−(0.98x 2 − 0.588x + 0.0882) 0

So, Q( x ) = x3 + 3.4x 2 + 3.29x + 0.98, r(x) = 0 . 31.

32. −2

−5

7 −15 −10 15

−3

−6 −2 3

So, Q( x ) = 3x + 1, r(x) = 0 .

So, Q( x ) = 2x − 3, r( x ) = 0 .

33.

34. −1 7

−3 −7

8 18

− 10 15

4 4

9 19

So, Q( x ) = 7x − 10, r( x ) = 15 .

So, Q( x ) = 4x + 9, r( x ) = 19 .

35.

36. −2

−1 −2 3 4 −4 2 0 −6 4 −1

3 −2

1 1

1 4

So, Q( x ) = −x3 + 3x − 2, r( x ) = 0 .

3 0 −4 1 4 4 4

So, Q( x ) = x 2 + 4x + 4, r( x ) = 0 .

526

Section 4.3

37.

38. −1 1

0 0 0 − 1 1 −1

1 1

1 − 1 1 −1

−3 1

1 −3 9

So, Q( x ) = x3 − x 2 + x − 1, r( x ) = 2 .

So,

39.

40. −2

−2

4 −8 −8

1 −2

9 81

− 27 90

Q( x ) = x3 − 3x 2 + 9x − 27, r( x ) = 90 .

0 − 16

0 0 0 − 3 9 − 27

0 − 81

1 3 9

So, Q( x ) = x3 − 2 x 2 + 4 x − 8, r( x ) = 0 .

So, Q( x ) = x3 + 3x 2 + 9x + 27, r( x ) = 0 .

41.

42. −

1 2

2 − 5 −1 −1 2 −6

− 13

3 −8 0

3 −1 2

−1 3 −1 3 −9 3

So, Q( x ) = 2x 2 − 6 x + 2, r( x ) = 0 .

So, Q( x ) = 3x 2 − 9x + 3, r( x ) = 0 .

43.

44. 2 3

−3

−4

4 3

− 109

106 27

212 81

2 − 53

53 9

106 27

− 112 81

So, Q( x ) = 2 x3 − 53 x 2 + 539 x + 106 27 , r( x ) = − 112 81

3 4

−3

9 4

39 16

117 64

735 256

13 4

39 16

245 64

33 − 256

So, .

Q( x ) = 3x3 + 134 x 2 + 1396 x + 245 64 , 33 r( x ) = − 256

527

Chapter 4

45.

46. −1.5 2

9 −3

− 9 − 81 − 81 − 9 27 81

− 18 − 54

−0.8

5 − 5 10

So,

47. 1

1 0 0 −8

48. −1

1 1 1 −7 −7 − 4 − 4 −3

−2

−1 − 3

−1

1 −1

−3

−1

1 4

1 −7 −7 −4 −4

1 1

So, Q( x ) = 5x 2 − 5x + 10, r( x ) = 0 .

Q( x ) = 2 x + 6 x − 18x − 54, r( x ) = 0 . 3

5 −1 6 8 − 4 4 −8

1 3

So,

7 6

So,

Q( x ) = x + x + x − 7 x − 7 x − 4x − 4, 6

r ( x ) = −3

Q( x ) = x5 + 3x 4 − 3x3 + x 2 − x + 1, r( x ) = 6

49.

− 49

− 25

1225

− 44 5

− 220

− 245 5

−1225

5 − 44

− 44 5

− 245

− 245 5

So, Q( x ) = x5 + 5x 4 − 44 x3 − 44 5x 2 − 245x − 245 5, r( x ) = 0 . 50.

−4

−9

− 3

−3

−12 3

− 36

−1 − 3

−12 −12 3

So, Q( x ) = x5 + 3x 4 − x3 − 3x 2 −12x −12 3 , r(x) = 0 .

528

Section 4.3

51.

52. 2x − 7 3x − 1 6 x − 23x + 7

3x + 2 2x − 1 6x + x − 2

−(6 x 2 − 2 x )

−(6 x 2 − 3x )

− 21x + 7 −( −21x + 7)

4x − 2

−(4 x − 2)

So, Q( x ) = 2x − 7, r(x) = 0 .

So, Q( x ) = 3x + 2 r( x ) = 0 .

53.

54. 1 1 −1 − 9 9 1 0 −9 1

−9

−2

So, Q( x ) = x 2 − 9, r( x ) = 0 .

2 −6 −2 0

− 12 12

−6

So, Q( x ) = x 2 − 6, r( x ) = 0 .

55.

56. 2

−1

−5 1 0

2 4 16 36

−5

1 −5

1 0 4 1 2

8 18 36

So,

−1

− 10

25 − 120

585

24 − 117 575

So,

Q( x ) = x + 2 x + 8x + 18x + 36, 4

r( x ) = 71

Q( x ) = x3 − 5x 2 + 24 x − 117, r( x ) = 575 .

58.

57. x + 0x + 1 x + 0 x − 1 x + 0 x + 0 x + 0 x − 25 2

x x + 0x − 2 x + 0x + 0x − 8 2

−( x3 + 0x 2 − 2x)

−( x 4 + 0 x3 − x 2 )

2x − 8

x 2 + 0 x − 25

−( x 2 + 0 x − 1)

So, Q( x ) = x, r(x) = 2x − 8 .

− 24

So, Q( x ) = x 2 + 1, r( x ) = −24 .

529

Chapter 4

59.

60. 1 1 0 0 0 0 0 0 −1 1 1 1 1 1 1 1 1 1 1 1

3 1 0 0 0 0 0 − 27 3 9 27 81 243 729

1 1 1 0

1 3 9 27 81 243 702

So,

Q( x ) = x + x + x + x + x + x + 1, 6

r( x ) = 0

Q( x ) = x5 + 3x 4 + 9x3 + 27 x 2 + 81x + 243, r( x ) = 702

61. Area = length × width. So, solving for width, we see that width = Area ÷ length. So, we have: 3x 2 + 2 x +1 2x 2 + 0 x −1 6x 4 + 4x3 − x 2 − 2x −1 −(6x 4 + 0x3 − 3x 2 ) 4 x3 + 2 x 2 − 2x −(4x3 + 0 x 2 − 2x) 2x 2 + 0x −1 −(2x 2 + 0x −1) 0

Thus, the width (in terms of x) is 3x 2 + 2x + 1 feet . 62. Volume = (Area of base) × height. So, solving for height, we see that height = Volume ÷ (Area of base). So, we have: 3x + 1 4 3 2 5 4 3 2 6x + 4x − x − 2x −1 18x +18x + x − 7x − 5x −1

−(18x5 + 12x 4 − 3x3 − 6x 2 − 3x) 6x 4 + 4x3 − x 2 − 2x −1 −(6x 4 + 4x3 − x 2 − 2x −1) 0 So, the height (in terms of x) is 3x + 1 feet .

530

Section 4.3

63. Distance = Rate × Time. So, solving for Time, we have: Time = Distance ÷ Rate. So, we calculate ( x3 + 60 x 2 + x + 60 ) ÷ (x + 60) using synthetic division: −60

1 60 − 60 1

0 − 60 1

So, the time is x 2 + 1 hours . 64. Distance = Rate × Time. So, solving for Rate, we have: Rate = Distance ÷ Time. So, we calculate ( − x 2 − 5x + 50 ) ÷ (5 − x ) using long division: x + 10 − x + 5 − x − 5x + 50 2

−( − x 2 + 5 x ) − 10x + 50 −( −10x + 50) 0

So, the rate is x +10 yards per second . 65. In long division, you must subtract (not add) each term.

66. The zero of the divisor is used in synthetic division. So, 2 should replace −2 as the divisor.

67. Forgot the “0” placeholder.

68. Cannot use synthetic division with a quadratic divisor. Use long division instead.

69. True.

70. False. For instance, ( x3 − x2 + x −1) ÷ ( x − 1) = x2 +1.

71. False. Only use when the divisor has degree 1.

72. True.

531

Chapter 4

73. −b

2b − a

b 2 − 2ab

− ab 2

−b

− b 2 + ab

ab 2

1 b−a − ab 0 Since the remainder is 0 upon using synthetic division, YES, ( x + b ) is a factor of

x3 + (2b − a)x 2 + ( b2 − 2ab ) x − ab2 .

74.

−b 1

b2 − a 2

− a 2 b2

−b

− 2 b3 + a 2 b

2b 4 − a 2 b 2

1 − b 2b 2 − a 2 − 2b3 + a 2 b 2b 4 − 2a 2 b 2 Since the remainder is not 0 upon using synthetic division, NO, ( x + b ) is not a factor of the given polynomial, unless b = 0 , in which case the above simplifies to saying x − 0 is a factor of x 4 − a 2 x 2 .

75.

x + 2x + 1 x − 1 x + x2n − xn − 1 2n

−( x3 n − x 2 n ) 2x 2 n − x n

76. First, we rewrite the polynomial in a more familiar form using the substitution y = x n . Doing so yields x3 n + 5x 2 n + 8x n + 4 = y3 + 5y 2 + 8y + 4 . Now, apply synthetic division: −1 1 5 8 4 −1 − 4 − 4

−(2x 2 n − 2 x n ) x n −1 −( x n − 1) 0 So, Q( x ) = x 2 n + 2x n + 1, r( x ) = 0

Thus, y3 + 5 y 2 + 8y + 4 = ( y +1) ( y 2 + 4y + 4 ) = ( y + 1) ( y + 2 ) . Going back to the original polynomial, this says: 2

x3 n + 5x 2 n + 8x n + 4 = ( x n + 1)( x n + 2 ) . 2

532

Section 4.3

So, the graph is a line, as shown:

77. Long division gives us

2x − 1 x + 0 x + 5 2x − x +10x − 5 2

−(2x3 + 0 x 2 +10x) − x 2 + 0x − 5 −( −x 2 + 0 x − 5) 0

78. Synthetic division gives us 3 1 −3 4 − 12 3

1 0 4 0 So, the quotient is the quadratic function y = x 2 + 4 with a hole at 3, shown to the right:

79. Using synthetic division gives us: −2 1 2 0 − 1 − 2 1

−2

−1

So, the quotient is the cubic y = x3 − 1 with a hole at x = −2, as shown to the right:

533

Chapter 4

80. Long division yields:

x −3 x − 6x + 0x + 2x +1 x − 9x +18x + 2x − 5x − 3 4

−( x5 − 6 x 4 + 0x3 + 2x 2 + x) − 3x 4 + 18x3 + 0x 2 − 6x − 3 −( −3x 4 +18x3 + 0x 2 − 6x − 3) 0 Note that the graph of the original function is this line with a hole at approximately x = 0.8. So, the graph is the line y = x − 3 , as shown below:

81. Long division gives us −3x 2 + 8x − 5 2x + 3 − 6 x3 + 7x 2 +14x −15 −( −6x3 − 9x 2 ) 16x 2 +14 x −(16x 2 + 24x ) − 10x − 15 −( − 10x −15) 0 So, it is a quadratic function with a hole at x = − 32 .

534

Section 4.3

82. Long division gives us − x 3 + 9x 3x 2 + 4x − 2 − 3x5 − 4x 4 + 29x3 + 36x 2 −18x −( −3x5 − 4 x 4 + 2x3 ) 27x3 + 36x 2 − 18x −(27x3 + 36 x 2 −18x) 0

So, it is a third-degree polynomial with holes at

535

−2 ± 10 3

Section 4.4 Solutions -------------------------------------------------------------------------------1.

2. 1 3 −2 7 0 −8 3 1 8 8

3 So, f (1) = 0 .

−1 3 − 2 −3

7 0 −8 5 − 12 12

3 − 5 12 − 12 So, f ( −1) = 4 .

3. 1

−1

2 1 0 1

1 0

2 3 3

− 2 1 −1

2 3 3 4

2 −1 1 0

So, g ( −1) = 0 .

So, g(1) = 4 . 5.

6. −2

3 −2

−8

− 6 16 − 46

3 − 8 23 − 46 So, f ( −2) = 84 .

2 1 0

4 10 20 2 5 10

2 − 29

So, g ( 2) = 21 .

8. −7

2 − 29 −7

35 − 42

8 − 42

1 −5 6 0 Yes, −7 is a zero of P ( x ) .

1 4 − 21 0 Yes, 2 is a zero of P ( x ) .

10. −3 1 − 1 − 8 12 − 3 12 − 12

1 −4 4 Yes, −3 is a zero of P ( x ) .

1 1 −1 − 8 1 0

12 −8

1 0 −8 4 No, 1 is not a zero of P ( x ) .

536

Section 4.4

11.

12.

− 2

−5 −12 −3 12 0 −8

3 2

1 3

So, P ( x ) = ( x + 32 ) ( 2x − 8 )

−9

−3 −4 −12 0

= −3 ( x − 13 ) ( 3x + 4 )

= ( 2x + 3 ) ( x − 4 )

The zeros are −

−9

So, P ( x ) = ( x − 13 ) ( −9x − 12 )

= 2 ( x + 32 ) ( x − 4 ) 3 2

−9

= − ( 3x − 1) ( 3x + 4 )

The zeros are 13 and − 34 .

and 4 .

13.

14. 1 1 0 −13

3 1 3 − 10 − 24

−12

1 1 −12

1 6

So,

P ( x ) = (x − 1)(x + x − 12) . = ( x − 1)( x + 4)(x − 3) The zeros are 1, 3 and −4 .

P ( x ) = (x − 3)(x 2 + 6x + 8)

= ( x − 3)( x + 4)(x + 2 ) The zeros are 3, −2 and −4 .

15.

16.

1 2

− 13

3 − 14 7

2 1 −13

−6

−1

−12

3 − 15 12

So,

5 −4 0

So, P ( x ) = (x + 13 )(3x 2 −15x +12)

P ( x ) = (x − )(2x + 2x −12) 1 2

= 2(x − 12 )( x 2 + x − 6) = 2(x − 12 )( x + 3)(x − 2) = (2x − 1)( x + 3)(x − 2) The zeros are 12 , − 3 and 2 .

= 3(x + 13 )( x 2 − 5x + 4)

= 3(x + 13 )( x − 4)(x −1) = (3x + 1)( x − 4)(x −1) The zeros are 1, 4 and − 13 .

537

Chapter 4

17. Since −3 and 5 are both zeros, we know that (x + 3) and (x − 5) are factors

of P ( x ) and hence, must divide P ( x ) evenly. (Note: ( x + 3)( x − 5) = x 2 − 2x − 15 .) x2 + 4 x 2 − 2 x − 15 x 4 − 2x3 −11x 2 − 8x − 60 −( x 4 − 2x 3 −15x 2 ) 4x 2 − 8x − 60 − (4x 2 − 8x − 60) 0

So, P ( x ) = (x 2 + 4)(x 2 − 2x − 15) = (x 2 + 4)(x − 5)(x + 3) . The real zeros are

5 and − 3 . 18. Since −1 and 2 are both zeros, we know that (x +1) and (x − 2) are factors

of P ( x ) and hence, must divide P ( x ) evenly. (Note: ( x + 1)( x − 2) = x 2 − x − 2 .) x2 + 9 x 2 − x − 2 x 4 − x3 + 7 x 2 − 9x −18 −( x 4 − x3 − 2 x 2 ) 9x 2 − 9x −18 −( 9x 2 − 9x −18) 0

So, P ( x ) = ( x + 9 )(x − x − 2) = ( x + 9 )(x − 2)(x + 1) . The real zeros are 2 and −1. 2

538

Section 4.4

19. Since −3 and 1 are both zeros, we know that ( x + 3) and ( x − 1) are factors

of P ( x ) and hence, must divide P ( x ) evenly. (Note: ( x + 3)( x − 1) = x 2 + 2x − 3 .) x 2 − 2x + 2 x 2 + 2x − 3 x 4 + 0x3 − 5x 2 +10x − 6 −( x 4 + 2x3 − 3x 2 ) − 2 x3 − 2 x 2 + 10x −( −2x3 − 4 x 2 + 6x) 2x 2 + 4x − 6 −(2x 2 + 4 x − 6) 0

So, P(x) = (x 2 + 2x − 3)(x 2 − 2x + 2) = (x − 1)(x + 3)(x 2 − 2x + 2) . The real zeros are 1 and −3. 20. Since −2 and 4 are both zeros, we know that ( x + 2 ) and ( x − 4) are factors

of P ( x ) and hence, must divide P ( x ) evenly. (Note: ( x + 2 )( x − 4) = x 2 − 2x − 8 .) x2 − 2x + 5 x 2 − 2x − 8 x 4 − 4x3 + x 2 + 6x − 40 −( x 4 − 2 x3 − 8x 2 ) − 2 x3 + 9x 2 + 6x −( −2x3 + 4x 2 +16x) 5x 2 − 10 x − 40 −(5x 2 − 10 x − 40) 0

So, P(x) = ( x − 2x + 5)( x − 2x − 8) = ( x − 2x + 5)(x − 4)(x + 2) . The real zeros are 4 and − 2. 2

539

Chapter 4

21.

22.

−2 1

6 13 12 4 − 2 − 8 −10 − 4

1 1

4 1

−2 5

− 12 3

9 −9

−2 1

4 5 2 −2 −4 −2

1 1

5 1

3 6

−9 9

1 2

So,

So, P ( x ) = (x + 2) (x + 2x +1) 2

P ( x ) = (x − 1)2 (x 2 + 6x + 9)

= ( x + 2 )2 ( x +1)2 The zeros are −2 and − 1 , both with multiplicity 2.

= ( x − 1)2 ( x + 3)2 The zeros are −3 and 1, both with multiplicity 2.

23. Factors of 4: ±1, ± 2, ± 4 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 4

24. Factors of 4: ±1, ± 2, ± 4 Factors of −1 : ±1 Possible rational zeros: ±1, ± 2, ± 4

25. Factors of 12: ±1, ± 2, ± 3, ± 4, ± 6, ± 12 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 3, ± 4, ± 6, ± 12

26. Factors of 9: ±1, ± 3, ± 9 Factors of 1: ±1 Possible rational zeros: ±1, ± 3, ± 9

27. Factors of 8: ±1, ± 2, ± 4, ± 8 Factors of 2: ±1, ± 2 Possible rational zeros: ± 12 , ± 1, ± 2, ± 4, ± 8

28. Factors of −10 : ±1, ± 2, ± 5, ±10 Factors of 3: ±1, ± 3 Possible rational zeros: ±1, ± 2, ± 5, ± 10, ± 13 , ± 23 , ± 53 , ± 103

29. Factors of −20 : ±1, ± 2, ± 4, ± 5, ± 10, ± 20 Factors of 5: ±1, ± 5 Possible rational zeros: ±1, ± 2, ± 4, ± 5, ± 10, ± 20, ± 15 , ± 25 , ± 54

30. Factors of −21 : ±1, ± 3, ± 7, ± 21 Factors of 4: ±1, ± 2, ± 4 Possible rational zeros: ±1, ± 3, ± 7, ± 21, ± 12 , ± 14 , ± 23 ,

± 72 , ± 212 , ± 34 , ± 74 , ± 214

540

Section 4.4

31. Factors of 8: ±1, ± 2, ± 4, ± 8 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 4, ± 8 Testing: P (1) = P ( −1) = P ( 2) = P ( −4) = 0

32. Factors of 3: ±1, ± 3 Factors of 1: ±1 Possible rational zeros: ±1, ± 3 Testing: P (1) = P ( −1) = P( −3) = 0

33. Factors of −3 : ±1, ± 3 Factors of 2: ±1, ± 2 Possible rational zeros: ±1, ± 3, ± 12 , ± 32

34. Factors of −8 : ±1, ± 2, ± 4, ± 8 Factors of 3: ±1, ± 3 Possible rational zeros: ±1, ± 2, ± 4, ± 8, ± 13 , ± 23 , ± 34 , ± 83

Testing: P (1) = P (3) = P ( 1 2 ) = 0

Testing: P ( −2 ) = P (4) = P ( − 1 3 ) = 0

35. Number of sign variations for P ( x ) : 1 P ( −x ) = P ( x ) , so Number of sign variations for P (− x ) : 1 Since P ( x ) is degree 4, there are 4 zeros, the real ones of which are classified to the right: 36. Number of sign variations for P (x) : 0 P ( −x ) = P(x) , so Number of sign variations for P( −x ) : 0 Since P ( x ) is degree 4, there are 4 zeros, the real ones of which are classified to the right:

Positive Real Zeros 1

Negative Real Zeros 1

Positive Real Zeros 0

Negative Real Zeros 0

Positive Real Zeros 1

Negative Real Zeros 0

37. Number of sign variations for P ( x ) : 1 P ( −x ) = ( −x )5 − 1 = −x5 − 1 , so Number of sign variations for P (− x ) : 0 Since P(x) is degree 5, there are 5 zeros, the real ones of which are classified to the right:

541

Chapter 4

38. Number of sign variations for P ( x ) : 0 P ( − x ) = − x5 + 1 , so Number of sign variations for P (− x ) : 1 Since P ( x ) is degree 5, there are 5 zeros, the real ones of which are classified to the right:

Positive Real Zeros 0

Negative Real Zeros 1

39. Number of sign variations for P ( x ) : 2 P ( − x ) = − x5 + 3x3 + x + 2 , so Number of sign variations for P (− x ) : 1 Since P ( x ) is degree 5, there are 5 zeros, the real ones of which are classified to the right:

Positive Real Zeros 2 0

Negative Real Zeros 1 1

40. Number of sign variations for P ( x ) : 1 P ( −x ) = P ( x ) , so Number of sign variations for P (− x ) : 1 Since P ( x ) is degree 4, there are 4 zeros, the real ones of which are classified to the right:

Positive Real Zeros 1

Negative Real Zeros 1

Positive Real Zeros 1

Negative Real Zeros 1

41. Number of sign variations for P ( x ) : 1 P ( − x ) = −9x 7 − 2x5 + x3 + x , so Number of sign variations for P (− x ) : 1 Since P ( x ) is degree 7, there are 7 zeros. But, 0 is also a zero. So, we classify the remaining real zeros to the right:

542

Section 4.4

42. Number of sign variations for P ( x ) : 3 Positive Real Zeros 3 1

Negative Real Zeros 0 0

43. Number of sign variations for P ( x ) : 2 P ( −x ) = P(x) , so Number of sign variations for P( −x ) : 2 Since P ( x ) is degree 6, there are 6 zeros, the real ones of which are classified to the right:

Positive Real Zeros 2 0 2 0

Negative Real Zeros 2 2 0 0

44. Number of sign variations for P ( x ) : 1 P ( − x ) = −7x6 − 5x 4 − x 2 − 2x + 1 , so Number of sign variations for P (− x ) : 1 Since P ( x ) is degree 6, there are 6 zeros, the real ones of whicha are classified to the right:

Positive Real Zeros 1

Negative Real Zeros 1

Positive Real Zeros 4 2 0

Negative Real Zeros 0 0 0

P ( − x ) = −16x7 − 3x 4 − 2x − 1 , so Number of sign variations for P (− x ) : 0 Since P ( x ) is degree 7, there are 7 zeros, the real ones of which are classified to the right:

45. Number of sign variations for P ( x ) : 4 P ( − x ) = −3x 4 − 2x3 − 4x 2 − x − 11 , so Number of sign variations for P (− x ) : 0 Since P ( x ) is degree 4, there are 4 zeros, the real ones of which are classified to the right:

543

Chapter 4

46. Number of sign variations for P ( x ) : 2

Positive Real Zeros 2 2 0 0

P ( − x ) = 2x 4 + 3x3 + 7x 2 − 3x + 2 , so Number of sign variations for P (− x ) : 2 Since P ( x ) is degree 4, there are 4 zeros, the real ones of which classified to the right:

b. Factors of 6: ±1, ± 2, ± 3, ± 6 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 3, ± 6

47. a. Number of sign variations for P ( x ) : 0 P ( − x ) = − x3 + 6x 2 − 11x + 6 , so Number of sign variations for P (− x ) : 3 Since P ( x ) is degree 3, there are zeros, the real ones of which are classified as: Positive Real Zeros 0 0

c. Note that P ( −1) = P ( −2) = P ( −3) = 0 . So, the rational zeros are −1, − 2, − 3 . These are the only zeros since P has degree 3.

Negative Real Zeros 3 1

d. P ( x ) = (x + 1)(x + 2)(x + 3)

48. a. Number of sign variations for P ( x ) : 3 P ( − x ) = − x3 − 6x 2 − 11x − 6 , so Number of sign variations for P( −x ) : 0 Since P ( x ) is degree 3, there are 3 zeros, the real ones of which are classified as: Positive Real Zeros 3 1

Negative Real Zeros 0 0

Negative Real Zeros 2 0 2 0

b. Factors of −6 : ±1, ± 2, ± 3, ± 6 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 3, ± 6 c. Note that P (1) = P (2) = P(3) = 0 . So, the rational zeros are 1,2,3 . These are the only zeros since P has degree 3. d. P ( x ) = (x − 1)(x − 2)(x − 3)

544

Section 4.4

b. Factors of 7: ±1, ± 7 Factors of 1: ±1 Possible rational zeros: ±1, ± 7

49. a. Number of sign variations for P ( x ) : 2 P ( − x ) = − x3 − 7x 2 + x + 7 , so Number of sign variations for P( −x ) : 1 Since P ( x ) is degree 3, there are 3 zeros, the real ones of which are classified as:

Positive Real Zeros 2 0

c. Note that P (−1) = P (1) = P(7) = 0 . So, the rational zeros are −1, 1, 7 . These are the only zeros since P has degree 3.

Negative Real Zeros 1 1

d. P ( x ) = ( x + 1)( x − 1)( x − 7)

50. a. Number of sign variations for P ( x ) : 2 P ( − x ) = − x3 − 5x 2 + 4x + 20 , so Number of sign variations for P (− x ) : 1 Since P ( x ) is degree 3, there are 3 zeros, the real ones of which are classified as:

Positive Real Zeros 2 0

c. Note that P ( −2 ) = P (2) = P(5) = 0 . So, the rational zeros are −2, 2, 5 . These are the only zeros since P has degree 3.

Negative Real Zeros 1 1

d. P ( x ) = ( x + 2)( x − 2)( x − 5)

51. a. Number of sign variations for P ( x ) : 1 P ( − x ) = x − 6x + 3x + 10x, so Number of sign variations for P (− x ) : 2 Since P ( x ) is degree 4, there are 4 zeros, one of which is 0. We classify the remaining real zeros below: 4

Positive Real Zeros 1 1

Negative Real Zeros 2 0

b. Factors of 20: ±1, ± 2, ± 4, ± 5, ± 10, ± 20 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 4, ± 5, ± 10, ± 20

b. P ( x ) = x ( x3 + 6x 2 + 3x − 10 ) We list

the possible nonzero rational zeros below: Factors of −10 : ±1, ± 2, ± 5, ±10 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 5, ± 10 c. Note that P (0) = P (1) = P ( −2) = P ( −5) = 0 . So, the rational zeros are 0, 1, − 2, − 5 . These are the only zeros since P has degree 4. d. P ( x ) = x( x − 1)( x + 2)( x + 5)

545

Chapter 4

52. a. Number of sign variations for P ( x ) : 2 P ( − x ) = x 4 + x3 − 14x 2 − 24x, so Number of sign variations for P( −x ) : 1 Since P ( x ) is degree 4, there are 4 zeros, one of which is 0. We classify the remaining real zeros below:

Positive Real Zeros 2 0

Negative Real Zeros 1 1

53. a. Number of sign variations for P ( x ) : 4 P ( − x ) = x 4 + 7x3 + 27x 2 + 47x + 26 , so Number of sign variations for P (− x ) : 0 Since P ( x ) is degree 4, there are 4 zeros, the real ones of which are classified as: Positive Real Zeros 4 2 0

Negative Real Zeros 0 0 0

b. P ( x ) = x ( x3 − x 2 −14x + 24 ) We list

the possible nonzero rational zeros below: Factors of 24: ±1, ± 2, ± 3, ± 4, ± 6, ± 8, ±12, ± 24 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 3, ± 4, ± 6, ± 8, ±12, ± 24 c. Note that P (0) = P ( −4) = P (2) = P (3) = 0 . So, the rational zeros are 0, − 4, 2, 3 . These are the only zeros since P has degree 4. d. P ( x ) = x(x + 4)(x − 2)(x − 3) c. Note that P (1) = P (2) = 0 . After testing the others, it is found that the only rational zeros are 1, 2 . So, we at least know that ( x − 1)( x − 2) = x 2 − 3x + 2 divides P ( x ) evenly. To find the remaining zeros, we long divide: x 2 − 4x + 13 x 2 − 3x + 2 x 4 − 7x3 + 27x 2 − 47x + 26 −( x 4 − 3x3 + 2x 2 ) − 4 x3 + 25x 2 − 47x −( −4x3 +12x 2 − 8x )

b. Factors of 26: ±1, ± 2, ± 13, ± 26 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 13, ± 26

13x 2 − 39x + 26 −(13x 2 − 39x + 26) 0

Since x 2 − 4x + 13 is irreducible, the real zeros are 1 and 2. d. P ( x ) = (x − 1)(x − 2) ( x 2 − 4x +13 )

546

Section 4.4

54. a. Number of sign variations for P ( x ) : 3 P ( − x ) = x 4 + 5x3 + 5x 2 − 25x − 26 , so Number of sign variations for P( −x ) : 1 Since P ( x ) is degree 4, there are 4 zeros, the real ones of which are classified as:

Positive Real Zeros 3 1

Negative Real Zeros 1 1

c. Note that P (1) = P ( −2) = 0 . After testing the others, it is found that the only rational zeros are 1, − 2 . So, we at least know that ( x − 1)( x + 2) = x 2 + x − 2 divides P ( x ) evenly. To find the remaining zeros, we long divide: x 2 − 6 x + 13 x 2 + x − 2 x 4 − 5x3 + 5x 2 + 25x − 26 −( x 4 + x 3 − 2 x 2 ) − 6 x3 + 7 x 2 + 25x

b. Factors of −26 : ±1, ± 2, ±13, ± 26 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ±13, ± 26

−( −6x3 − 6 x 2 +12x) 13x 2 + 13x − 26 −(13x 2 + 13x − 26) 0

Since x − 6x + 13 is irreducible, the real zeros are 1 and −2. d. P ( x ) = (x − 1)(x + 2) ( x 2 − 6x +13 ) 2

55. a. Number of sign variations for P ( x ) : 2 P ( − x ) = −10x3 − 7x 2 + 4x + 1 , so Number of sign variations for P( −x ) : 1 Since P ( x ) is degree 3, there are 3 zeros, the real ones of which are classified as:

Positive Real Zeros 2 0

Negative Real Zeros 1 1

b. Factors of 1: ±1 Factors of 10: ±1, ± 2, ± 5, ± 10 Possible rational zeros: ±1, ± 12 , ± 15 , ± 101 c. Note that P (1) = P ( − 12 ) = P( 51 ) = 0 . So, the rational zeros are −1, − 12 , 51 . These are the only zeros since P has degree 3. d. P ( x ) = ( x − 1)(2x + 1)(5x − 1)

547

Chapter 4

56. a. Number of sign variations for P ( x ) : 3 P ( − x ) = −12x3 − 13x 2 − 2x − 1 , so Number of sign variations for P( −x ) : 0 Since P ( x ) is degree 3, there are 3 zeros, the real ones of which are classified as:

c. After testing, we conclude that the only rational zero is 1. To determine the other zeros, we use synthetic division: 1 12 −13 2 − 1 12 −1 1 12

Positive Real Zeros 3 1

Negative Real Zeros 0 0

−1

Since 12x − x + 1 is irreducible, there are no other rational zeros. 2

d. P ( x ) = (x − 1) (12x 2 − x +1)

b. Factors of −1: ±1 Factors of 12: ±1, ± 2, ± 3, ± 4, ± 6, ±12 Possible rational zeros: ±1, ± 12 , ± 13 , ± 14 , ± 61 , ± 112 57. a. Number of sign variations for P ( x ) : 1 P ( − x ) = −6x3 + 17x 2 − x − 10 , so Number of sign variations for P (− x ) : 2 Since P ( x ) is degree 3, there are 3 zeros, the real ones of which are classified as: Positive Real Zeros 1 1

Negative Real Zeros 2 0

b. Factors of −10 : ±1, ± 2, ± 5, ± 10 Factors of 6: ±1, ± 2, ± 3, ± 6 Possible rational zeros: ±1, ± 2, ± 5, ± 10, ± 21 , ± 31 , ± 61 , ± 23 ,

± 25 , ± 35 , ± 65 , ± 103 c. Note that P ( −1) = P ( − 52 ) = P ( 32 ) = 0 . So, the rational zeros are −1, − 52 , 32 . These are the only zeros since P has degree 3. d.

548

P ( x ) = (x + 1)(2x + 5)(3x − 2) = 6(x + 1) ( x + 52 ) ( x − 23 )

Section 4.4

58. a. Number of sign variations for P ( x ) : 1 P ( − x ) = −6x3 + x 2 + 5x − 2 , so Number of sign variations for P( −x ) : 2 Since P ( x ) is degree 3, there are 3 zeros, the real ones of which are classified as:

Positive Real Zeros 1 1

c. Note that P (1) = P ( − 12 ) = P( − 23 ) = 0 . So, the rational zeros are 1, − 12 , − 23 . These are the only zeros since P has degree 3. P ( x ) = (x − 1)(3x + 2)(2x +1) d. = 6(x − 1) ( x + 23 ) ( x + 12 )

Negative Real Zeros 2 0

59. a. Number of sign variations for P ( x ) : 4 P ( − x ) = x 4 + 2x3 + 5x 2 + 8x + 4 , so Number of sign variations for P( −x ) : 0 Since P ( x ) is degree 4, there are 4 zeros, the real ones of which are classified as:

Positive Real Zeros 4 2 0

Negative Real Zeros 0 0 0

b. Factors of 4 : ±1, ± 2, ± 4 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 4

b. Factors of −2 : ±1, ± 2 Factors of 6: ±1, ± 2, ± 3, ± 6 Possible rational zeros: ±1, ± 2, ± 12 , ± 13 , ± 61 , ± 23

c. Note that P(1) = 0 . After testing the others, it is found that the only rational zeros is 1 . Hence, by a, 1 must have multiplicity 2 or 4. So, we know that at least ( x − 1)2 = x 2 − 2 x + 1 divides P ( x ) evenly. To find the remaining zeros, we long divide: x2 + 4 x 2 − 2 x + 1 x 4 − 2x3 + 5x 2 − 8x + 4 −( x 4 − 2 x3 + x 2 ) 4 x 2 − 8x + 4 −(4x 2 − 8x + 4) 0

Since x + 4 is irreducible, the only real zero is 1 (multiplicity 2). 2

d. P ( x ) = (x − 1)2 (x 2 + 4)

549

Chapter 4

60. a. Number of sign variations for P ( x ) : 0 P ( − x ) = x 4 − 2x3 + 10x 2 − 18x + 9 , so Number of sign variations for P( −x ) : 4 Since P ( x ) is degree 4, there are 4 zeros, the real ones of which are classified as:

Positive Real Zeros 0 0 0

Negative Real Zeros 4 2 0

c. Note that P( −1) = 0 . After testing the others, it is found that the only rational zero is −1 . Hence, by a, −1 must have at least multiplicity 2. So, we know that ( x + 1)2 = x 2 + 2x + 1 divides P ( x ) evenly. To find the remaining zeros, we long divide: x2 + 9 x 2 + 2 x + 1 x 4 + 2 x3 +10x 2 +18x + 9 −( x 4 + 2 x3 + x 2 ) 9 x 2 + 1 8x + 9 −(9x 2 + 18x + 9)

b. Factors of 9: ±1, ± 3, ± 9 Factors of 1: ±1 Possible rational zeros: ±1, ± 3, ± 9

Since x 2 + 9 is irreducible, the only real zero is −1 (multiplicity 2). d. P ( x ) = (x + 1)2 (x 2 + 9)

61. a. Number of sign variations for P ( x ) : 1 P ( −x ) = P ( x ) , so Number of sign variations for P (− x ) : 1 Since P ( x ) is degree 6, there are 6 zeros, the real ones of which are classified as: Positive Real Zeros 1

Negative Real Zeros 1

b. Factors of −36 : ±1, ± 2, ± 3, ± 4, ± 6, ± 9, ±12, ±18, ± 36 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 3, ± 4, ± 6, ± 9, ±12, ±18, ± 36 c. Note that P ( −1) = P (1) = 0 . From a, there can be no other rational zeros for P.

We know that ( x + 1)( x − 1) = x 2 − 1 divides P ( x ) evenly. To find the remaining zeros, we long divide:

550

Section 4.4

x 4 + 13x 2 + 36 x 2 − 1 x6 + 0x5 +12x 4 + 0x3 + 23x 2 + 0x − 36 −( x6 + 0 x5 − x 4 ) 13x 4 + 0 x3 + 23x 2 −(13x 4 + 0 x3 −13x 2 ) 36x 2 + 0 x − 36 −(36x 2 + 0 x − 36) 0

Observe that x + 13x + 36 = (x + 9)(x + 4) , both of which are irreducible. So, the real zeros are: −1 and 1 d. P ( x ) = (x + 1)(x − 1)(x 2 + 9)(x 2 + 4) 4

62. a. Number of sign variations for P ( x ) : 2 P ( − x ) = x 4 + x3 − 16x 2 + 16 , so Number of sign variations for P(−x) : 2 Since P ( x ) is degree 4, there are 4 zeros, the real ones of which are classified as:

Positive Real Zeros 2 2 0 0

Negative Real Zeros 2 0 2 0

b. Factors of 16 : ±1, ± 2, ± 4, ± 8, ±16 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 4, ± 8, ±16 c. Note that P(1) = 0 . After testing, we conclude that the only rational zero is 1. The best we can do is estimate the remaining zeros graphically:

From the graph, we see that the other real zeros are approximately −3.35026, − 1.07838, and 4.42864. d. An approximate factorization of P ( x ) is: P ( x ) = (x − 1)(x + 3.35026)(x +1.07838)

551

(x − 4.2864)

Chapter 4

63. a. Number of sign variations for P ( x ) : 4 P ( − x ) = 4x 4 + 20x3 + 37x 2 + 24x + 5 , so Number of sign variations for P (− x ) : 0 Since P ( x ) is degree 4, there are 4 zeros, the real ones of which are classified as:

Positive Real Zeros 4 2 0

1 2

Negative Real Zeros 0 0 0

b. Factors of 5: ±1, ± 5 Factors of 4: ±1, ± 2, ± 4 Possible rational zeros: ±1, ± 5, ± 12 , ± 14 , ± 52 , ± 54 64. a. Number of sign variations for P ( x ) : 2 P ( − x ) = 4 x 4 + 8x3 + 7 x 2 − 30x + 50 , so Number of sign variations for P( −x ) : 2 Since P ( x ) is degree 4, there are 4 zeros, the real ones of which are classified as:

Positive Real Zeros 2 0 0 2

c. Note that P( 12 ) = 0 . After testing, we conclude that there the only rational zero is 12 , which has multiplicity 2 or 4. So, we know that at least (x − 12 )2 divides P ( x ) evenly:

1 2

− 20

− 24

−9

14 − 5

− 18

− 10

−8

− 16

So,

P ( x ) = (x − 12 )2 (4x 2 − 16x + 20) = 4(x − 12 )2 ( x 2 − 4x + 5)

c. After testing, we conclude that there are no rational zeros. The best we can do is graph the polynomial to locate any real zeros:

Negative Real Zeros 2 2 0 0

b. Factors of 50 : ±1, ± 2, ± 5, ± 10, ± 25, ± 50 Factors of 4: ±1, ± 2, ± 4 Possible rational zeros: ±1, ± 2, ± 5, ± 10, ± 25, ± 50,

± 12 , ± 52 , ± 225 , ± 14 , ± 54 , ± 254

As seen from the graph, there are no real zeros. d. An accurate factorization of P ( x ) is not possible since we don’t have values for the zeros.

552

Section 4.4

65.

66.

67.

68.

69. Observe that f (1) = 2, f (2) = −4 . So, by Intermediate Value Theorem, there must exist a zero in the interval (1,2). Graphically, we approximate this zero to be approximately 1.34, as seen below:

70. Observe that f (0) = 1, f (1) = −1. So, by Intermediate Value Theorem, there must exist a zero in the interval (0,1). Graphically, we approximate this zero to be approximately 0.74, as seen below:

553

Chapter 4

71. Observe that f (0) = −1, f (1) = 9 . So, by Intermediate Value Theorem, there must exist a zero in the interval (0,1). Graphically, we approximate this zero to be approximately 0.22, as seen below:

72. Observe that f ( −2 ) = 9, f ( −1) = −8 . So, by Intermediate Value Theorem, there must exist a zero in the interval ( −2, −1) . Graphically, we approximate this zero to be approximately –1.64, as seen below:

73. Observe that f ( −1) = 2, f (0) = −3 . So, by Intermediate Value Theorem, there must exist a zero in the interval (−1, 0). Graphically, we approximate this zero to be approximately −0.43, as seen below:

74. Observe that f ( −2 ) = 33, f ( −1) = −1 . So, by Intermediate Value Theorem, there must exist a zero in the interval (−2,−1). Graphically, we approximate this zero to be approximately −1.05, as seen below:

554

Section 4.4

75. a. P ( x ) = (46 − 3x 2 ) − (20 + 2x ) = −3x 2 − 2 x + 26, x ≥ 0 . b. Find all values of x such that: P ( x ) = −3x 2 − 2x + 26 = 0 : 2 ± ( −2)2 − 4(−3)(26) 2 ± 316 2 ± 17.78 = ≅ ≅ 2.63, 2(−3) −6 −6 So, approximately 263 subscribers will break even.

76. Solve P ( x ) = x3 − 5x 2 + 3x + 6 = 0, x ≥ 0 . We begin by applying the rational root theorem: Factors of 6: ±1, ± 2, ± 3, ± 6 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 3, ± 6 Note that P(2) = 0 , so that x − 2 must divide P ( x ) . Using synthetic division then yields: 77.

−3.30

−5

−6 −6

1 −3 −3

6 0

So, P ( x ) = ( x − 2)( x − 3x − 3) . Solving x 2 − 3x − 3 = 0 yields: 3 ± 9 + 12 3 ± 21 = ≅ −0.7912 ,3.79 x= 2 2 So, the company will break even if either 2 or approximately 3.79 units are sold. 2

P ( x ) = xp( x ) − C ( x ) = x(28 − 0.0002x ) − (20 x + 1,500)

= 28x − 0.0002x 2 − 20 x − 1,500 = −0.0002x 2 + 8x − 1,500 By Descartes Reule of Signs, there are either 0 or 2 positive real zeros. 78. Solve P(x) = 0. −0.0002 x 2 + 8x − 1,500 = 0 x 2 − 40,000 x + 7,500,000 = 0 (cleared the fractions) 40,000 ± 40,000 2 − 4(7,500,000) 2(1) ≈ 188 or 39,812 The break even points are 188 and 39,812 units. When fewer than 188 units or more than 34,812 units are produced and sold, profit is negative-money is lost. When the number of units being produced and sold is between 188 and 34,812 a profit is being made on the product.

555

Chapter 4

79. Solve C(t) = 0.

15.4 − 0.05t 2 = 0 1,540 − 5t 2 = 0 5(308 − t 2 ) = 0 t = ± 308 ≈ 17.55 hours So, it takes about 18 hours to eliminate the drug from the bloodstream. 80. Solve C(t) = 0.

60 − 0.75t 2 = 0 6,000 − 75t 2 = 0 75(80 − t 2 ) = 0 t = ± 80 ≈ 9 hours So, it takes about 9 hours to eliminate the drug from the bloodstream. 81. It is true that one can get 5 negative zeros here, but there may be just 1 or 3. Positive Real Zeros 0 0 0

Negative Real Zeros 5 3 1

82. Use 2, not −2 .

83. True

84. False. Consider f ( x ) = x 2 + 1 . There are no real zeros.

85. False. For instance, f ( x ) = ( x 2 + 1) ( x 2 + 2 ) cannot be factored over the reals. 86. False. This is one possibility, but if there are 2 or more such sign changes, then there could be fewer. 87. a

1 − ( a + b + c ) ( ab + ac + bc ) 1

− abc

− ab − ac

abc

−b−c

So, P ( x ) = ( x − a )( x − ( b + c )x + bc ) = ( x − a )( x − b )( x − c ) . So, the other zeros are b, c. 2

556

Section 4.4

88. a

−a +b−c a

b−c

− ( ab + bc − ac ) abc − abc ab − ac − bc

So, P ( x ) = ( x − a )( x 2 + ( b − c )x − bc ) = ( x − a )( x + b )( x − c ) . So, the other zeros are –b and c. 89.

Factors of −32: ±1, ± 2, ± 4, ± 8, ± 16, ± 32 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 4, ± 8, ± 16, ± 32 From the graph, we conclude that the only rational zero is 2.

90. Factors of −48: ±1, ± 2, ± 3, ± 4, ± 6, ± 8,

± 12, ± 16, ± 24, ± 48 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 3, ± 4, ± 6, ± 8, ±12, ± 16, ± 24, ± 48 From the graph, we conclude that 3 is the only real zero.

557

Chapter 4

91. Consider the graph to the right of P ( x ) = 12 x 4 + 25x3 + 56 x 2 − 7 x − 30 . a. From the graph, there are two real zeros, namely − 34 and 23 . As such, we know that both (3x − 2) and (4x + 3) are factors of P ( x ) and so, (3x − 2)(4 x + 3) = 12x 2 + x − 6 divides it evenly. Indeed, observe that x2 + 2x + 5 12 x 2 + x − 6 12 x 4 + 25x3 + 56 x 2 − 7 x − 30 −(12 x 4 + x3 − 6 x 2 ) 24 x3 + 62 x 2 − 7 x −(24 x3 + 2 x 2 −12 x ) 60 x 2 + 5x − 30 −(60 x 2 + 5x − 30) 0

b. Since x 2 + 2x + 5 is irreducible, P ( x ) factors as (3x − 2)(4 x + 3) ( x 2 + 2x + 5 ) .

558

Section 4.4

92. Consider the graph to the right of P ( x ) = −3x3 − x 2 − 7x − 49 . a. From the graph, there is one real zero, namely − 73 . As such, we know that (3x + 7) is a factor of P ( x ) and so, divides it evenly. Indeed, observe that −x 2 + 2x − 7 3x + 7 −3x3 − x 2 − 7 x − 49 −( −3x3 − 7 x 2 ) 6x 2 − 7x −(6 x 2 + 14 x ) − 21x − 49 −( −21x − 49) 0

b. Since −x + 2x − 7 is irreducible, we see that P ( x ) factors as 2

(3x + 7) ( −x 2 + 2x − 7 ) = −(3x + 7)

( x − 2x + 7 ) . 2

559

Section 4.5 Solutions -------------------------------------------------------------------------------1. P ( x ) = ( x + 2i )( x − 2i ) . Zeros are ±2i .

2. P ( x ) = ( x + 3i )( x − 3i ) . Zeros are ±3i .

3. Note that the zeros are x 2 − 2x + 2 = 0 

4. Note that the zeros are x2 − 4x + 5 = 0 

2 ± 4 − 4(2) 2 ± 2i = = 1± i 2 2 So, P ( x ) = (x − (1− i ))(x − (1+ i )) .

4 ± 16 − 4(5) 4 ± 2i = = 2±i 2 2 So, P ( x ) = (x − (2 − i ))(x − (2 + i )) .

5. Observe that P ( x ) = ( x 2 − 4 )( x 2 + 4 )

6. Observe that P( x ) = ( x2 − 9 ) ( x2 + 9 )

= (x − 2)( x + 2)(x − 2i )(x + 2i ) So, the zeros are ±2, ± 2i .

= (x − 3)( x + 3)(x − 3i )(x + 3i ) So, the zeros are ±3, ± 3i .

7. Observe that

P( x ) = ( x2 − 5 ) ( x2 + 5 )

(

)(

)

)(

)

= x − 5 x + 5 x − 5i x + 5i So, the zeros are ± 5, ± 5i . 8. Observe that

P( x ) = ( x2 − 3) ( x2 + 3)

(

)(

= x − 3 x + 3 x − 3i x + 3i So, the zeros are ± 3, ± 3i . 9. If i is a zero, then so is its conjugate –i. Since P ( x ) has degree 3, this is the only missing zero.

10. If −i is a zero, then so is its conjugate i. Since P ( x ) has degree 3, this is the only missing zero.

11. Since 2i and 3 − i are zeros, so are their conjugates −2i and 3 + i , respectively. Since P ( x ) has degree 4, these are the only missing zeros.

12. Since 3i and 2 + i are zeros, so are their conjugates −3i and 2 − i , respectively. Since P ( x ) has degree 4, these are the only missing zeros.

560

Section 4.5

13. Since 1− 3i and 2 + 5i are zeros, so are their conjugates 1+ 3i and 2 − 5i , respectively. Since P(x) has degree 6 and 2 is a zero with multiplicity 2, these are the only missing zeros.

14. Since 1− 5i and 2 + 3i are zeros, so are their conjugates 1 + 5i and 2 − 3i , respectively. Since P ( x ) has degree 6 and −2 is a zero with multiplicity 2, these are the only missing zeros.

15. Since −i and 1 − i are zeros, so are their conjugates i and 1+ i , respectively. Since 1− i has multiplicity 2, so does its conjugate. Since P ( x ) has degree 6, these are the only missing zeros.

16. Since 2i and 1+ i are zeros, so are their conjugates −2i and 1− i , respectively. Since 1+ i has multiplicity 2, so does its conjugate. Since P ( x ) has degree 6, these are the only missing zeros.

17. Let P(x) be the desired polynomial. Since 0 is a zero of P, x is a factor of P. Also, since 1± 2i is a conjugate pair, the following must divide into P(x) evenly: ( x − (1 − 2i ))( x − (1 + 2i )) = x 2 − 2x + 5 So, P(x) is given by x( x 2 − 2 x + 5) = x3 − 2x 2 + 5x.

18. Let P(x) be the desired polynomial. Since 0 is a zero of P, x is a factor of P. Also since 2 ± i is a conjugate pair, the following must divide into P(x) evenly: ( x − (2 − i ))( x − (2 + i )) = x 2 − 4 x + 5 So, P(x) is given by x( x 2 − 4x + 5) = x3 − 4x 2 + 5x.

19. Let P(x) be the desired polynomial. Since 1 is a zero of P, x – 1 is a factor of P. Also, since 1± 5i is a conjugate pair, the following must divide into P(x) evenly: ( x − (1 − 5i ))( x − (1 + 5i )) = x 2 − 2 x + 26 So, P(x) is given by ( x − 1)( x 2 − 2x + 26) = x3 − 3x 2 + 28x − 26. 20. Let P(x) be the desired polynomial. Since 2 is a zero of P, x – 2 is a factor of P. Also, since 4 ± i is a conjugate pair, the following must divide into P(x) evenly: ( x − ( 4 − i ))( x − (4 + i )) = x 2 − 8x + 17 So, P(x) is given by ( x − 2)( x 2 − 8x + 17) = x3 − 10x 2 + 33x − 34 .

561

Chapter 4

21. Let P(x) be the desired polynomial. Since 1± i is a conjugate pair, the following must divide into P(x) evenly: ( x − (1 − i ))( x − (1 + i )) = x 2 − 2x + 2 Also, since ±3i is a conjugate pair, the following must divide into P(x) evenly: ( x − 3i )( x + 3i ) = x 2 + 9 So, P(x) is given by ( x 2 + 9 ) ( x 2 − 2x + 2 ) = . x 4 − 2x3 + 11x 2 −18x +18

22. Let P(x) be the desired polynomial. Since ±i is a conjugate pair, the following must divide into P(x) evenly: ( x − i )( x + i ) = x 2 + 1 Also, since 1± 2i is a conjugate pair, the following must divide into P(x) evenly: ( x − (1 − 2i ))( x − (1 + 2i )) = x 2 − 2 x + 5 So, P(x) is given by ( x2 + 1)( x2 − 2x + 5 ) = . x 4 − 2x3 + 6x 2 − 2x + 5

23. Since 3i is a zero of P ( x ) , so is its

24. Since −i is a zero of P ( x ) , so is

conjugate −3i . As such, ( x − 3i ) ( x + 3i ) = x 2 + 9 divides P ( x ) evenly. Indeed, observe that x−4 x 2 + 0x + 9 x3 − 4x 2 + 9x − 36

its conjugate i . As such, ( x + i ) ( x − i ) = x 2 + 1 divides P ( x ) evenly. Indeed, observe that x +3 x 2 + 0x + 1 x3 + 3x 2 + x + 3

− ( x3 + 0 x 2 − 9x )

−( x3 + 0x 2 + x ) − 3x 2 + 0 x + 3

− 4x 2 + 0x − 36

− ( −4x 2 + 0 x − 36 )

−(3x 2 + 0x + 3) 0

So, P ( x ) = ( x + i )( x − i )( x + 3 )

So, P ( x ) = ( x − 3i )( x + 3i )( x − 4 ) So, the zeros are ±3i , 4.

So, the zeros are ±i, −3.

25. Since i is a zero of P ( x ) , so is its conjugate −i . As such, ( x + i ) ( x − i ) = x 2 + 1 divides P ( x ) evenly. Indeed, observe that

26. Since −2i is a zero of P ( x ) , so is its conjugate 2i . As such, ( x + 2i ) ( x − 2i ) = x2 + 4 divides P ( x ) evenly. Indeed, observe that

562

Section 4.5

3x − 2 x + 0x + 1 3x − 2 x + 3x − 2

5x + 4 x + 0x + 4 5x + 4 x + 20 x +16

−(3x3 + 0x 2 + 3x )

−(5x3 + 0x 2 + 20x )

− 2x 2 + 0x − 2

4 x 2 + 0 x + 16

−( −2x 2 + 0x − 2)

− (4x 2 + 0x + 16)

So, P ( x ) = ( x + i )( x − i )( 3x − 2 ) So, the zeros are ±i,

So, P ( x ) = ( x + 2i )( x − 2i )( 5x + 4 ) So, the zeros are ±2i , − 54

2 3

27. Since −2i is a zero of P ( x ) , so is its conjugate 2i. As such, (x − 2i )(x + 2i ) =

x 2 + 4 divides P(x) evenly. Indeed, observe that x 2 − 2 x − 15 x 2 + 0 x + 4 x 4 − 2 x3 − 11x 2 − 8x − 60

x2 − x − 2 x 2 + 0 x + 9 x 4 − x3 + 7 x 2 − 9x −18

−( x 4 + 0 x 3 + 4 x 2 )

−( x 4 + 0 x3 + 9x 2 )

− 2 x 3 − 15 x 2 − 8 x

− x3 − 2 x 2 − 9x

−( −2 x3 + 0 x 2 − 8x )

−( − x 3 + 0 x 2 − 9 x )

− 15x 2 − 60

− 2 x 2 − 18

−( −15x 2 − 60)

−( −2 x 2 −18)

So,

28. Since 3i is a zero of P ( x ) , so is its conjugate −3i. As such, ( x − 3i )(x + 3i ) = x 2 + 9 divides P ( x ) evenly. Indeed, observe that

P ( x ) = (x − 2i )(x + 2i ) ( x − 2x −15 )

So, P ( x ) = (x − 3i )(x + 3i ) ( x 2 − x − 2 )

= ( x − 2i )(x + 2i )(x − 5)(x + 3) So, the zeros are ±2i, − 3, 5 .

= ( x − 3i )(x + 3i )(x − 2)(x +1) So, the zeros are ±3i , − 1, 2 .

29. Since i is a zero of P(x) , so is its conjugate −i. As such, ( x − i )(x + i ) = x 2 + 1 divides P ( x ) evenly. Indeed, observe that

30. Since −2i is a zero of P ( x ) , so is its conjugate 2i. As such, ( x − 2i )(x + 2i ) = x 2 + 4 divides P ( x ) evenly. Indeed, observe that

563

Chapter 4

x 2 − 4x + 3 x 2 + 0x + 1 x 4 − 4x3 + 4x 2 − 4x + 3

x2 − x − 2 x 2 + 0 x + 4 x 4 − x3 + 2 x 2 − 4 x − 8

−( x 4 + 0 x 3 + x 2 )

So,

−( x 4 + 0 x 3 + 4 x 2 )

− 4x3 + 3x 2 − 4 x

− x3 − 2 x 2 − 4 x

−( −4 x3 + 0x 2 − 4x )

−( − x 3 + 0 x 2 − 4 x )

3x 2 + 3

− 2x 2 − 8

−(3x 2 + 3)

−( −2x 2 − 8)

P ( x ) = (x − i )(x + i ) ( x − 4x + 3 )

So, P ( x ) = (x − 2i )(x + 2i ) ( x 2 − x − 2 )

= ( x − i )(x + i )(x − 3)(x −1) So, the zeros are ±i, 1, 3 .

= ( x − 2i )(x + 2i )( x − 2 )(x + 1) So, the zeros are ±2i , − 1, 2 .

31. Since −3i is a zero of P ( x ) , so is its conjugate 3i. As such, ( x − 3i )(x + 3i ) = x 2 + 9 divides P ( x ) evenly. Indeed, observe that x 2 − 2x + 1 x 2 + 0 x + 9 x 4 − 2x3 +10 x 2 −18x + 9

32. Since 5i is a zero of P ( x ) , so is its conjugate −5i. As such, ( x − 5i )(x + 5i ) = x 2 + 25 divides P ( x ) evenly. Indeed, observe that

x 2 − 3x − 4 x 2 + 0 x + 25 x 4 − 3x3 + 21x 2 − 75x −100

−( x 4 + 0 x 3 + 9 x 2 )

−( x 4 + 0 x3 + 25x 2 )

− 2 x3 + x 2 − 18x

− 3x3 − 4 x 2 − 75x

−( −2 x3 + 0 x 2 −18x )

−( −3x3 + 0 x 2 − 75x )

x2 + 9

− 4 x 2 + 100

−( x 2 + 9 )

−( −4 x 2 +100)

So,

P ( x ) = ( x − 3i )( x + 3i ) ( x 2 − 2x + 1)

So, P ( x ) = (x − 5i )(x + 5i ) ( x 2 − 3x − 4 )

= ( x − 3i )( x + 3i )( x − 1)2 So, the zeros are ±3i and 1 (multiplicity 2) .

= (x − 5i )(x + 5i )(x − 4)(x +1) So, the zeros are ±5i , − 1, 4 .

564

Section 4.5

33. Since 1+ i is a zero of P ( x ) , so is its conjugate 1 – i. As such, (x − (1 + i ))(x − (1 − i )) = x 2 − 2x + 2 divides P ( x ) evenly. Indeed, observe that x2 + 2x − 7 x 2 − 2x + 2 x 4 + 0x3 − 9x 2 +18x −14

34. Since 1 − 2i is a zero of P ( x ) , so is its conjugate 1 + 2i. As such, ( x − (1 + 2i ))( x − (1 − 2i )) = x 2 − 2x + 5 divides P ( x ) evenly. Indeed, observe that x2 − 2x − 8 x 2 − 2x + 5 x 4 − 4x3 + x 2 + 6 x − 40

−( x 4 − 2 x3 + 2x 2 )

−( x 4 − 2x3 + 5x 2 )

2 x3 − 11x 2 + 18x

− 2 x3 − 4 x 2 + 6x

−(2x3 − 4 x 2 + 4x)

−( −2x3 + 4x 2 −10x)

− 7x 2 + 14x − 14

− 8x 2 + 16x − 40

−( −7x 2 +14 x −14)

−( −8x 2 +16x − 40)

0 Now, we find the roots of x + 2x − 7 : −2 ± 4 − 4(−7) x= = −1 ± 2 2 2 So, P ( x ) = ( x − (1+ i ))( x − (1− i )) ⋅

0 So,

P ( x ) = (x − (1+ 2i ))(x − (1− 2i )) ( x − 4 )( x + 2)

So, the zeros are 1 ± 2i, − 2, 4 .

( x − ( −1− 2 2 ))( x − ( −1+ 2 2 ))

So, the zeros are 1 ± i , − 1± 2 2 . 35. Since 3 − i is a zero of P(x) , so is its conjugate 3 + i. As such, (x − (3 + i ))(x − (3 − i )) = x 2 − 6x +10 divides P ( x ) evenly. Indeed, observe that x2 − 4 2 4 3 2 x − 6 x + 10 x − 6x + 6x + 24 x − 40 −( x 4 − 6x3 +10x 2 )

36. Since 2 + i is a zero of P ( x ) , so is its conjugate 2 − i. As such, (x − (2 + i ))(x − (2 − i )) = x 2 − 4x + 5 divides P ( x ) evenly. Indeed, observe that x 2 −1 2 4 3 2 x − 4x + 5 x − 4x + 4x + 4x − 5 −( x 4 − 4 x 3 + 5 x 2 )

− 4 x 2 + 24 x − 40

− x 2 + 4x − 5

−( −4 x 2 + 24 x − 40)

−( −x 2 + 4 x − 5)

565

Chapter 4

So, So, P ( x ) = (x − (3 + i ))(x − (3 − i ))(x − 2)(x + 2) P ( x ) = (x − (2 + i ))(x − (2 − i )) So, the zeros are 3 ± i, ± 2 . ( x − 1)( x + 1) So, the zeros are 2 ± i, ± 1. 37. Since 2 − i is a zero of P ( x ) , so is its conjugate 2 + i. As such, (x − (2 + i ))(x − (2 − i )) = x 2 − 4x + 5 divides P ( x ) evenly. Indeed, observe that x 2 − 5x + 4 x 2 − 4 x + 5 x 4 − 9x3 + 29x 2 − 41x + 20

38. Since 3 + i is a zero of P ( x ) , so is its conjugate 3 − i. As such, (x − (3 + i ))(x − (3 − i )) = x 2 − 6x +10 divides P ( x ) evenly. Indeed, observe that x2 − x − 2 x 2 − 6 x + 10 x 4 − 7 x3 +14 x 2 + 2 x − 20

−( x 4 − 4 x 3 + 5 x 2 )

−( x 4 − 6 x3 +10 x 2 )

− 5x3 + 24 x 2 − 41x

− x3 + 4x 2 + 2x

−( −5x3 + 20 x 2 − 25x )

−( − x3 + 6 x 2 −10 x )

4 x 2 − 16 x + 20

− 2 x 2 + 12 x − 2 0

−(4 x 2 − 16 x + 20)

−( −2 x 2 +12 x − 20)

So, P ( x ) = (x − (2 + i ))(x − (2 − i))(x −1)(x − 4) So, P ( x ) = (x − (3 + i ))(x − (3 − i )) So, the zeros are 2 ± i,1, 4 . ( x − 2 )( x + 1) So, the zeros are 3 ± i , − 1, 2. 39. Number of sign variations for P ( x ) : 2

P ( −x ) = −5x5 + 3x3 + 1 so the number of sign variations for P ( −x ) :1

Since P ( x ) is degree 5, there are 5 zeros. Finally, complex zeros come in pairs. POSITIVE REAL ZEROS 2

NEGATIVE REAL ZEROS 1

COMPLEX ZEROS 2

566

Section 4.5

40. Number of sign variations for P ( x ) : 2

P ( −x ) = 7 x5 + 12x 4 + 8x 2 − 5x −12 so the number of sign variations for P ( −x ) :1

Since P ( x ) is degree 5, there are 5 zeros. Finally, complex zeros come in pairs. POSITIVE REAL ZEROS 2

NEGATIVE REAL ZEROS 1

COMPLEX ZEROS 2

41. Number of sign variations for P ( x ) : 3

P ( −x ) = 14 x 4 − 3x3 − 2 x 2 − 5x − 1 so the number of sign variations for P ( −x ) :1

Since P ( x ) is degree 4, there are 4 zeros. Finally, complex zeros come in pairs. POSITIVE REAL ZEROS 3

NEGATIVE REAL ZEROS 1

COMPLEX ZEROS 0

42. Number of sign variations for P ( x ) : 2

P ( −x ) = 11x 4 − x 2 −12x + 3 so the number of sign variations for P ( − x ) : 2

Since P ( x ) is degree 4, there are 4 zeros. Finally, complex zeros come in pairs. POSITIVE REAL ZEROS 2

NEGATIVE REAL ZEROS 2

COMPLEX ZEROS 0

567

Chapter 4

43. x3 − x 2 + 9x − 9 = ( x3 − x 2 ) + 9 ( x − 1)

44. x3 − 2 x 2 + 4 x − 8 = ( x3 − 2x 2 ) + 4 ( x − 2 )

= x 2 ( x − 1) + 9 ( x − 1)

= x 2 ( x − 2) + 4 ( x − 2 )

= ( x + 3i )( x − 3i )( x − 1)

= ( x + 2i )( x − 2i )( x − 2)

= ( x 2 + 4 ) ( x − 2)

= ( x 2 + 9 ) ( x − 1)

45.

x − 5x + x − 5 = ( x − 5x ) + ( x − 5 )

46. x3 − 7 x 2 + x − 7 = ( x3 − 7 x 2 ) + ( x − 7 )

= x 2 ( x − 5) + ( x − 5 )

= x 2 ( x − 7) + ( x − 7 )

= ( x 2 + 1) ( x − 5)

= ( x 2 + 1) ( x − 7)

= ( x + i )( x − i )( x − 5)

= ( x + i )( x − i )( x − 7)

47. x3 + x 2 + 4 x + 4 = ( x3 + x 2 ) + 4 ( x + 1)

48. Consider P ( x ) = x3 + x 2 − 2 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2 . Note that 1 1 1 0 −2 1 2 2

= x 2 ( x + 1) + 4 ( x + 1) = ( x 2 + 4 ) ( x + 1)

= ( x + 2i )( x − 2i )( x + 1)

1 2

So, P ( x ) = (x − 1) ( x + 2x + 2 ) . Now, 2

we find the roots of x 2 + 2x + 2 : −2 ± 4 − 4(2) x= = −1 ± i 2 So, P ( x ) = (x − 1)(x − ( −1 + i ))(x − ( −1 − i )) .

568

Section 4.5

49. Consider P ( x ) = x3 − x 2 − 18 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 3, ± 6, ± 9, ± 18 . Note that 3 1 − 1 0 − 18 3 6 18 1

So, P ( x ) = (x − 3) ( x 2 + 2x + 6 ) . Now, we find the roots of x 2 + 2x + 6 : −2 ± 4 − 4(6) x= = −1 ± i 5 2 So, P ( x ) = (x − 3)(x − (−1+ i 5 ))

So,

− 20

−3

So, P ( x ) = (x + 1) ( x3 − 3x 2 + x − 3 )

= ( x + 1)( x − 3)(x + i )(x − i )

52. Consider P ( x ) = x 4 − x3 + 7x 2 − 9x − 18 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 3, ± 6, ± 9, ± 18 . Note that −1 1 − 1 7 − 9 − 18 − 1 2 − 9 18 1

−2

− 18

So, P ( x ) = (x + 1) ( x3 − 2x 2 + 9x −18 )

P ( x ) = (x + 3) ( x3 − 5x 2 + 4x − 20 ) = ( x + 3) ( x + 4 ) (x − 5)

= ( x + 1) ( x +1) (x − 3)

= ( x + 3)  x 2 (x − 5) + 4(x − 5) 

−3

51. Consider P ( x ) = x 4 − 2x3 − 11x 2 − 8x − 60 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 3, ± 4, ± 5, ± 6, ±10, ± 12, ±15, ± 20, ± 30, ± 60. Note that −3 1 − 2 −11 − 8 − 60 −3 15 − 12 60 −5

= ( x + 1)  x 2 (x − 3) + (x − 3) 

( x − ( −1− i 5 )).

50. Consider P ( x ) = x 4 − 2x3 − 2x 2 − 2x − 3 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 3 . Note that −1 1 − 2 − 2 − 2 − 3 −1 −1 3 3

= (x + 1)  x 2 (x − 2) + 9(x − 2) 

= ( x + 1) ( x 2 + 9 ) (x − 2)

= (x + 1)( x − 2)(x + 3i )(x − 3i )

= ( x + 3)( x − 5)( x + 2i )(x − 2i )

569

Chapter 4

53. Consider P ( x ) = x 4 − 4x3 − x 2 − 16x − 20 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 4, ± 5, ± 10, ± 20 Note that −1 1 − 4 −1 −16 − 20 −1 5 −4 20 1

So,

−5

− 20

1 −2

= ( x + 1)  x 2 (x − 5) + 4(x − 5) 

= ( x + 1) ( x 2 + 4 ) (x − 5)

= ( x − 1) ( x 2 + 9 ) (x − 2)

21 − 26

−6 2

21 −8

− 26 26

1 −4

2 1

= ( x − 1)( x − 2)(x + 3i )(x − 3i )

55. Consider P ( x ) = x 4 − 7x3 + 27x 2 − 47x + 26 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 13, ± 26 Note that 1 1 − 7 27 − 47 26

−6

= (x − 1)  x 2 (x − 2) + 9(x − 2) 

= ( x + 1)( x − 5)(x + 2i )(x − 2i )

− 18

So, P ( x ) = (x − 1) ( x3 − 2x 2 + 9x −18 )

P ( x ) = (x + 1) ( x − 5x + 4x − 20 ) 3

54. Consider P ( x ) = x 4 − 3x3 + 11x 2 − 27x + 18 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 3, ± 6, ± 9, ± 18 Note that 1 1 − 3 11 − 27 18 1 −2 9 − 18

56. Consider P ( x ) = x 4 − 5x3 + 5x 2 + 25x − 26 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 13, ± 26 Note that 1 1 −5 5 25 − 26

−2 1

−4

−4 −2

1 26 0 12 − 26

1 −6

So, P ( x ) = ( x − 1)( x − 2) ( x 2 − 4 x +13 ) .

So, P ( x ) = ( x − 1)( x + 2) ( x 2 − 6 x +13 ) .

Next, we find the roots of x 2 − 4x + 13 : 4 ± 16 − 4(13) = 2 ± 3i x= 2 So, P ( x ) = ( x − 1)( x − 2)( x − (2 − 3i ))

Next, we find the roots of x 2 − 6x + 13 : 6 ± 36 − 4(13) = 3 ± 2i x= 2 So, P ( x ) = ( x − 1)( x + 2)( x − (3 − 2i ))

( x − ( 2 + 3i ))

( x − (3 + 2i ))

570

Section 4.5

57. Consider P ( x ) = −x 4 − 3x3 + x 2 + 13x + 10 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 5, ± 10 Note that −1 1 3 − 1 −13 − 10

−1 − 2

2 2

−3 8

− 10 10

1 4

58. Consider P ( x ) = −x 4 − x3 + 12x 2 + 26x + 24 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24 Note that −3 1 1 − 12 − 26 − 24

−3

−2 4

−6 8

−8 8

So, P ( x ) = −( x +1)( x − 2) ( x + 4 x + 5 ) .

So, P ( x ) = −( x + 3)( x − 4) ( x 2 + 2 x + 2 ) .

Next, we find the roots of x 2 + 4x + 5 : −4 ± 16 − 4(5) = −2 ± i x= 2 So, P ( x ) = −( x +1)( x − 2)( x − (−2 − i ))

Next, we find the roots of x 2 + 2x + 2 : −2 ± 4 − 4(2) = −1 ± i x= 2 So, P ( x ) = −( x + 3)( x − 4)( x − (−1− i ))

( x − ( −2 + i ))

( x − ( −1+ i ))

59. Consider P(x) = x 4 − 2 x3 + 5x 2 − 8x + 4 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 4 . Note that 1 1 −2 5 −8 4 1 −1 4 − 4 1

So,

−1

−4

= ( x − 1)  x 2 ( x − 1) + 4( x − 1)

= ( x − 1) ( x + 4 ) ( x − 1)

So, P ( x ) = ( x + 1) ( x3 + x 2 + 9x + 9 )

P ( x ) = ( x − 1) ( x − x + 4 x − 4 ) 3

60. Consider P( x ) = x 4 + 2x3 + 10 x 2 + 18x + 9 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 3, ± 9 . Note that −1 1 2 10 18 9 −1 −1 − 9 − 9

= ( x + 1)  x 2 ( x +1) + 9( x +1)

= ( x + 1) ( x + 9 ) ( x + 1)

= ( x − 1)2 ( x + 2i )( x − 2i )

= ( x + 1)2 ( x + 3i )( x − 3i )

571

Chapter 4

61. Consider P ( x ) = x 6 + 12x 4 + 23x 2 − 36 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 4, ± 6, ± 9, ± 18, ± 36 Note that 1 1 0 12 0 23 0 − 36

13 13 36

−1 1 1 13 13 36 36 0 − 1 0 − 13 0 − 36 1 0 So,

62. Consider P ( x ) = x 6 − 2x5 + 9x 4 −16x3 + 24x 2

−32x + 16. By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 4, ± 8, ±16 Note that 1 1 − 2 9 − 16 24 − 32 16 1 −1 8 − 8 16 − 16 1 1

−1 1

8 0

−8 8

16 0

−16 0 16

P ( x ) = (x − 1)(x +1) ( x +13x + 36 ) 4

So,

= ( x − 1)( x +1) ( x 2 + 4 ) ( x 2 + 9 )

P ( x ) = (x − 1)2 ( x 4 + 8x 2 +16 ) = ( x − 1)2 ( x 2 + 4 )

= (x − 1)( x +1)(x − 2i )(x + 2i ) ( x − 3i )(x + 3i )

= ( x − 1)2 ( x − 2i )2 (x + 2i )2

63. Consider P ( x ) = 4x 4 − 20x3 + 37x 2 − 24x + 5 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 5, ± 12 , ± 52 , ± 14 , ± 54 Note that 1 4 − 20 37 − 24 5 2 1

−9

− 18

28 − 10

2 4

− 16

−8

4 4

− 21

− 42 2

So, P ( x ) = 4 ( x − 1 2 ) ( x − 4x + 5 ) . 2

64. Consider P ( x ) = 4x 4 − 44x3 + 145x 2 − 114x + 26 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 12 , ± 132 , ± 134 , ± 13, ± 26 Note that 1 4 − 44 145 − 114 26 2

14 − 5

124 − 52 − 20

− 40

So,

Next, we find the roots of x 2 − 4x + 5 : 4 ± 16 − 4(5) x= = 2±i 2

P ( x ) = ( x − 1 2 ) ( 4x − 40x + 104 ) 2

= 4 ( x − 1 2 ) ( x 2 − 10 x + 26 ) . 2

Next, we find the roots of x 2 − 10 x + 26 :

572

104 2

62 − 26

Section 4.5

So, 2 P ( x ) = 4 ( x − 1 2 ) (x − (2 − i ))(x − (2 + i ))

10 ± 100 − 4(26) = 5±i 2

= (2x −1)2 ( x − (2 − i ))(x − (2 + i ))

So, 2 P ( x ) = 4 ( x − 1 2 ) (x − (5 − i ))(x − (5 + i ))

65. Consider P ( x ) = 3x5 − 2x 4 + 9x3 − 6x 2 − 12x + 8 . By the Rational Zero Theorem, the only possible rational roots are ±1, ± 2, ± 4, ± 8, ± 13 , ± 23 , ± 34 , ± 83 Note that −1 3 − 2 9 − 6 −12 8

66. Consider P ( x ) = 2 x5 − 5x 4 + 4 x3 − 26 x 2 + 50 x − 25. By the Rational Zero Theorem, the only possible rational roots are ±1, ± 5, ± 25, ± 12 , ± 52 , ± 252 Note that 1 2 − 5 4 − 26 50 − 25

1 2

3 3

−3

− 14

20 − 8

−5

− 20

−2

12 − 8

−2

−8

−3

1 − 25

2 −1

2 2

So, P ( x ) = ( x − 1)( x + 1)( x − 23 )(3x 2 + 12)

1 − 25 25

25 0

0 − 25

−1

− 25

So, P ( x ) = ( x − 1)2 ( x − 52 )(2x 2 + 4 x +10)

= ( x − 1)( x + 1)(3x − 2)( x − 2i )( x + 2i )

= ( x − 1)2 (2 x − 5)( x 2 + 2 x + 5) Next, we find the roots of x 2 + 2x + 5 : −2 ± 4 − 4(5) x= = −1 ± 2i 2 So P ( x ) = ( x − 1)2 (2 x − 5) ( x + 1 − 2i )( x + 1 + 2i ).

573

Chapter 4

67. Yes—since there are no real zeros, this function is either always positive or always negative; the end behavior is similar to an even power function with a positive leading coefficient (rising in both ends). We expect profit to eventually rise (without bound!).

68. No. In such case, P(x) is always below the x-axis since the leading coefficient is negative. So, never have a positive profit.

69. No—since there is only one real zero (and it corresponds to producing a positive number of units), then that is the “break-even point”; the end behavior is similar to an odd power function with a negative leading coefficient (falling to the right). We expect profit to fall (without bound!) after the break-even point.

70. Yes. In such case, it crosses the x-axis and looks like y = x3 . So, profit is increasing.

71. Since the profit function is a third-degree polynomial we know that the function has three zeros and at most two turning points. Looking at the graph, we can see there is one real zero where t ≤ 0; There are no real zeros when t > 0; therefore the other two zeros must be complex conjugates. Therefore, the company always has a profit greater than approximately $5.1 million dollars and, in fact, the profit will increase towards infinity as t increases. 72. Since the profit function is a fourth-degree polynomial with a negative leading we know the function has four zeros and at most three turning points. The end behavior is towards negative infinity because of the negative leading coefficient of an even degree polynomial and there will be two real zeros; one where t ≤ 0 and one where t ≥ 6. The remaining two zeros are a complex conjugate pair. Therefore, the company will have profits of greater than approximately $5.1 million dollars during the first six months. Sometime later (t ≥ 6) the company’s profit will be zero. Then the company will start losing money and, in fact, the profit will decrease towards negative infinity as time increases. 73. Since the concentration function is a third degree polynomial, we know the function has three zeros and at most two turning points. Looking at the graph we can see there will be one real zero where t ≥ 8. The remaining zeros are a pair of complex conjugates. Therefore, the concentration of the drug in the blood stream will decrease to zero as the hours go by. Note that the concentration will not approach negative infinity since concentration is a non-negative quantity.

574

Section 4.5

74. Since the concentration function is a fourth degree polynomial, we know the function has four zeros. The negative leading coefficient indicates negative end behavior (opening down). Since the function opens down there is a real zero for t ≤ 0 and there will be a real zero for t ≥ 8. Note that the concentration will not approach negative infinity since concentration is a non-negative quantity. 75. Just because 1 is a zero does not mean −1 is a zero (holds only for complex conjugates). Should have used synthetic division with 1.

76. Possible rational roots include ± 12 from the Rational Zero theorem.

77. False. For example, consider P ( x ) = (x − 1)(x + 3) . Note that 1 is a zero of P, but −1 is not.

78. False. Complex zeros do not correspond to x-intercepts.

79. True. It has n complex zeros.

80. True.

81. No—if there is an imaginary zero, then its conjugate is also a zero. Therefore, all imaginary zeros can only correspond to an even-degree polynomial.

82. Yes. For example, P ( x ) = x 2 + 4 has zeros ±2i , both of which are imaginary.

83. Since bi is a zero of multiplicity 3, its conjugate −bi is also a zero of multiplicity 3. Hence, P ( x ) = ( x − bi )3 ( x + bi )3

84. Since a + bi is a zero of multiplicity 2, its conjugate a − bi is also a zero of multiplicity 2. Hence, 2 2 P ( x ) = ( x − (a + bi ) ) ( x − (a − bi ) )

= [( x − bi )( x + bi )]

= ( x − ( a + bi ) ) ( x − (a − bi ) ) 

= ( x2 + b2 )

=  x 2 − 2ax + ( a 2 + b 2 ) 

= x 6 + 3b 2 x 4 + 3b 4 x 2 + b6

575

Chapter 4

85. The possible combinations of zeros of P ( x ) are: Real Zeros 0 2 4

Complex Zeros 4 2 0

From the graph of P to the right, we note that all roots are complex.

86. The possible combinations of zeros of P ( x ) are: Real Zeros 0 2 4 6

Complex Zeros 6 4 2 0

From the graph of P to the right, we note that there are two real roots. 87. Consider the graph of P ( x ) to the right. Note that the only real zero is 3 5 , so that there must be four complex zeros. Observe that 0.6 − 5 3 − 25 15 − 20 12 −3 0 − 15 0 −12 −5

− 25

− 20

576

Section 4.5

So,

P ( x ) = (x − 0.6) ( −5x 4 − 25x 2 − 20 ) = −5(x − 0.6) ( x 4 + 5x 2 + 4 )

= −5(x − 0.6) ( x 2 + 4 ) ( x 2 +1) = −5(x − 0.6)(x − 2i )( x + 2i )( x − i )( x + i )

Zeroes at , ±i, ±2i 3 5

88. Consider the graph of P ( x ) to the right. Note that both 1.5 and −2.4 are real zeros, and −2.4 has multiplicity of 2 or 4 (since the graph touches the x-axis at this root). As such, we know that at least ( x − 1.5)( x + 2.4)2 divides P ( x ) evenly. Indeed, dividing P ( x ) by this yields the quotient x 2 − 1.2x + 0.40 , the roots of which are 0.6 ± 0.2i . Thus, P ( x ) = (x − 1.5)(x + 2.4)2 ⋅ (x − (0.6 + 0.2i ))(x − (0.6 − 0.2i ))

577

Section 4.6 Solutions -------------------------------------------------------------------------------1. Domain: ( −∞, −3 ) ∪ ( −3,∞ )

2. Domain: ( −∞, 4 ) ∪ ( 4,∞ )

3. Domain: ( −∞, − 13 ) ∪ ( − 13 , 1 2 ) ∪ ( 1 2 , ∞ )

4. Domain: ( −∞, 2 3 ) ∪ ( 2 3 , 7 ) ∪ ( 7, ∞ )

5. Note that x 2 + x − 12 = (x + 4)(x − 3) . Domain: ( −∞, −4 ) ∪ ( −4,3 ) ∪ ( 3,∞ )

6. Note that x 2 + 2 x − 3 = (x + 3)(x − 1) . Domain: ( −∞, −3 ) ∪ ( −3,1) ∪ (1,∞ )

7. Note that x 2 +16 is never zero. Domain: ( −∞,∞ )

8. Note that x 2 + 9 is never zero. Domain: ( −∞,∞ )

9. Note that 2 ( x 2 − x − 6 ) = 2(x − 3)(x + 2) .

10. Note that 2x 2 + 5x − 3 = ( 2x −1) ( x + 3 ) .

Domain: ( −∞, −2 ) ∪ ( −2,3 ) ∪ ( 3,∞ )

Domain: ( −∞, −3 ) ∪ ( −3, 12 ) ∪ ( 12 ,∞ )

11. Vertical Asymptote: x + 2 = 0 , so x = −2 is the VA. Horizontal Asymptote: Since the degree of the numerator is less than degree of the denominator, y = 0 is the HA.

12. Vertical Asymptote: 5 − x = 0 , so x = 5 is the VA. Horizontal Asymptote: Since the degree of the numerator is less than degree of the denominator, y = 0 is the HA.

13. Vertical Asymptote: x + 5 = 0 , so x = −5 is the VA. Horizontal Asymptote: Since the degree of the numerator is greater than degree of the denominator, there is no HA.

14. Vertical Asymptote: 2 x − 7 = 0 , so x = 72 is the VA. Horizontal Asymptote: Since the degree of the numerator is greater than degree of the denominator, there is no HA.

15. Vertical Asymptote: 5x 2 + 2 x − 3 = ( 5x − 3 ) ( x +1) = 0 , so

16. Vertical Asymptote: 2x 2 − 5x + 3 = ( 2x − 3 ) ( x −1) = 0 , so

x = 35 , x = −1 are the VAs. Horizontal Asymptote: Since the degree of the numerator is less than the degree of the denominator, y = 0 is the HA.

x = 32 , x = 1 are the VAs. Horizontal Asymptote: Since the degree of the numerator is less than the degree of the denominator, y = 0 is the HA.

578

Section 4.6

17. Vertical Asymptote: 6 x 2 − x − 15 = ( 3x − 5 ) ( 2x + 3 ) = 0, so

18. Vertical Asymptote: 2x 2 + 9x − 5 = ( x + 5 ) ( 2x −1) = 0, so

x = 53 , x = − 32 are the VAs. Horizontal Asymptote: Since the degree of the numerator is less than the degree of the denominator, y = 0 is the HA.

x = −5, x = 12 are the VAs. Horizontal Asymptote: Since the degree of the numerator is less than the degree of the denominator, y = 0 is the HA.

19. Vertical Asymptote: ( x + 1)( x − 1) = 0, so x = 1, x = −1 are

20. Vertical Asymptote: ( 2x + 1)( 3x − 2 ) = 0, so x = − 12 , x = 23 are

the VAs. Horizontal Asymptote: Since the degree of the numerator is equal to the degree of the denominator, y = 21 = 2 is the HA.

the VAs. Horizontal Asymptote: Since the degree of the numerator is equal to the degree of the denominator, y = 65 is the HA.

21. Vertical Asymptote: 6 x 2 + 5x − 4 = (2x − 1)(3x + 4) = 0 , so x = 12 , x = − 34 are the VAs. Horizontal Asymptote: Since the degree of the numerator is greater than degree of the denominator, there is no HA.

22. Vertical Asymptote: 3x 2 − 5x − 2 = (3x + 1)(x − 2) = 0 , so x = 2, x = − 13 are the VAs. Horizontal Asymptote: Since the degree of the numerator equals the degree of the denominator, y = 63 = 2 is the HA.

23. Vertical Asymptote: x 2 + 19 is never 0, so there is no VA. Horizontal Asymptote: Since the degree of the numerator equals the degree of the denominator, y = 13 is the HA.

24. Vertical Asymptote: (2x − 1) = 0 , so x = 12 is the VA. Horizontal Asymptote: Since the degree of the numerator is greater than degree of the denominator, there is no HA.

579

Chapter 4

25. Vertical Asymptote: (x − 0.5)(0.2 x + 0.3) = 0 , so x = 0.5, x = −1.5 are the VAs. Horizontal Asymptote: Since the degree of the numerator equals the degree of the denominator, y = 127 ≈ 1.71 is the HA.

26. Vertical Asymptote: x 2 − 0.25 = 0 , so x = 0.5, x = −0.5 are the VAs. Horizontal Asymptote: Since the degree of the numerator is greater than degree of the denominator, there is no HA.

27. To find the slant asymptote, we use synthetic division: −4 1 10 25 − 4 − 24

28. To find the slant asymptote, we use synthetic division: 3 1 9 20 3 36

1 6 1 So, the slant asymptote is y = x + 6 .

1 12 56 So, the slant asymptote is y = x + 12 .

29. To find the slant asymptote, we use synthetic division: 5 2 14 7 10 120

30. To find the slant asymptote, we use long division: 3x + 7 2 3 2 x − x − 30 3x + 4x − 6x +1

−(3x3 − 3x 2 − 90x)

2 24 127 So, the slant asymptote is y = 2x + 24 .

7x 2 + 84 x + 1 −(7x 2 − 7 x − 210) 91x + 211 So, the slant asymptote is y = 3x + 7 .

580

Section 4.6

31. To find the slant asymptote, we use long division:

4x + 112 2 x3 − x 2 + 3x −1 8x 4 + 7x3 + 0x 2 + 2x − 5 −(8x 4 − 4 x3 +12x 2 − 4x) 11x3 − 12 x 2 + 6x − 5 −(11x3 − 112 x 2 + 332 x − 112 ) − 132 x 2 − 212 x + 21 So, the slant asymptote is y = 4 x + 112 . 32. To find the slant asymptote, we use long division:

2x x + 0 x + 0x + 0 x + 0x −1 2x + 0x + 0x + 0x + 0x + 0x +1 5

−(2x6 + 0x5 + 0x 4 + 0x3 + 0x 2 − 2x ) 2x + 1 So, the slant asymptote is y = 2x . 33. b Vertical Asymptote: x = 4 Horizontal Asymptote: y = 0

34. d Vertical Asymptote: x = 4 Horizontal Asymptote: y = 3

35. a Vertical Asymptotes: x = 2, x = −2 Horizontal Asymptote: y = 3

36. f Vertical Asymptote: None Horizontal Asymptote: y = −3 Graph never goes above the x-axis

37. e Vertical Asymptotes: x = 2, x = −2 Horizontal Asymptote: y = −3

38. c Vertical Asymptote: x = −4 Horizontal Asymptote: None

581

Chapter 4

39. Vertical Asymptote: x = −1 Horizontal Asymptote: y = 0

40. Vertical Asymptote: x = 2 Horizontal Asymptote: y = 0

41. Vertical Asymptote: x = 1 Horizontal Asymptote: y = 2 Intercept: (0,0)

42. Vertical Asymptote: x = −2 Horizontal Asymptote: y = 4 Intercept: (0,0)

582

Section 4.6

43. Vertical Asymptote: x = 0 Horizontal Asymptote: y = 1 Intercept: (1,0)

44. Vertical Asymptote: x = 1 Horizontal Asymptote: y = 1 Intercepts: (−2,0), (0, −2)

45. Vertical Asymptotes: x = 0, x = −2 Horizontal Asymptote: y = 2 Intercepts: (3,0), (−1,0)

46. Vertical Asymptotes: x = 0, x = 3 Horizontal Asymptote: y = 3 Intercepts: (1,0), ( −1,0)

583

Chapter 4

47. Vertical Asymptote: x = −1 Intercept: (0,0) Slant Asymptote: y = x −1 x −1 2 x + 1 x + 0x + 0 −( x 2 + x ) −x+0 −( − x − 1) 1

48. Vertical Asymptote: x = −2 Intercepts: (3,0), ( −3,0) Slant Asymptote: y = x − 2 x−2 x + 2 x 2 + 0x − 9 −( x 2 + 2 x ) − 2x − 9 −( −2x − 4) −5

584

Section 4.6

49. Vertical Asymptotes: x = −2, x = 2 Intercepts: (0,0), ( − 12 ,0), (1,0) Slant Asymptote: y = 2x − 1 2x − 1 2 3 2 x + 0x − 4 2x − x − x + 0

−(2x3 + 0x 2 − 8x) − x2 + 7x + 0 −( − x 2 + 0 x + 4) 7x − 4

50. Vertical Asymptote: None Intercepts: (0,0), ( 13 ,0), ( −2,0) Slant Asymptote: y = 3x + 5 3x + 5 2 3 2 x + 0 x + 4 3x + 5x − 2x + 0

−(3x3 + 0 x 2 +12x) 5x 2 − 14 x + 0 −(5x 2 + 0x + 20) − 14x − 20

585

Chapter 4

51. Vertical Asymptote: x = 1, x = −1 Horizontal Asymptote: y = 1 Intercept: (0, −1)

52. Vertical Asymptote: None Horizontal Asymptote: y = −1 Intercepts: (1,0), ( −1,0), (0,1)

53. Vertical Asymptote: x = − 12 Horizontal Asymptote: y = 74 Intercept: (0,0)

54. Vertical Asymptote: x = − 13 Horizontal Asymptote: y = 274 Intercept: (0,0)

586

Section 4.6

55. Vertical Asymptotes: x = − 12 , x = 12 Horizontal Asymptote: y = 0 Intercepts: (0,1), ( 13 ,0), ( − 13 ,0)

56. Vertical Asymptotes: x = − 14 , x = 14 Horizontal Asymptote: y = 0 Intercepts: (0, −1), ( 51 ,0), ( − 51 ,0)

57. Vertical Asymptotes: x = 0 Slant Asymptote: y = 3x

58. Vertical Asymptotes: x = 0 Slant Asymptote: y = x Intercepts: (2,0), ( −2,0)

587

Chapter 4

59. Observe that ( x − 1)2 x −1 f (x) = = . ( x − 1)( x + 1) x +1

Open hole: (1,0) Vertical Asymptote: x = −1 Horizontal Asymptote: y = 1 Intercepts: (0, −1) and (1,0)

60. Observe that ( x + 1)2 x +1 f (x) = . = ( x − 1)( x + 1) x −1

Open hole: (−1,0) Vertical Asymptote: x = 1 Horizontal Asymptote: y = 1 Intercepts: (0, −1) Note: This graph has a hole at (−1,0).

61. Observe that ( x − 1)( x − 2)(x + 2) f (x) = ( x − 2 )( x 2 + 1) ( x − 1)( x + 2) . = x2 + 1

Open hole: (2,0) Vertical Asymptote: None Horizontal Asymptote: y = 1 Intercepts: (0, −2), (1,0), and ( −2,0)

588

Section 4.6

62. Observe that ( x − 1)( x − 3)(x + 3) f (x) = ( x − 3)( x 2 + 1) ( x − 1)( x + 3) . = x2 + 1

Open hole: (3,0) Vertical Asymptote: None Horizontal Asymptote: y = 1 Intercepts: (0, −3), (1,0), and ( −3,0)

63. Observe that 3x( x − 1) 3x − 3 f (x) = = . x( x − 2 )(x + 2) ( x − 2 )(x + 2)

Open hole: (0,0) Vertical Asymptote: x = 2, x = −2 Horizontal Asymptote: y = 0 Intercepts: (1,0)

64. Observe that −2x( x − 3) −2(x − 3) . f (x) = = x 2 +1 x ( x 2 +1)

Open hole: (0,0) Vertical Asymptote: None Horizontal Asymptote: y = 0 Intercepts: (3,0)

589

Chapter 4

65. a. x-intercept: (2,0) y-intercept: (0,0.5) b. horizontal asymptote: y = 0 vertical asymptotes: x = −1, x = 4 x−2 c. f ( x ) = ( x + 1)( x − 4)

66. a. x-intercepts: (−0.5,0), (3,0) y-intercept: (0,0.5) b. horizontal asymptote: y = 2 vertical asymptotes: x = −3, x = 2 (2x + 1)( x − 3) c. f ( x ) = ( x + 3)( x − 2)

67. a. x-intercept: (0,0) y-intercept: (0,0) b. horizontal asymptote: y = −3 vertical asymptotes: x = −4, x = 4

68. a. x-intercept: (0,0), (2,0) y-intercept: (0,0) b. horizontal asymptote: None vertical asymptote: x = −1 slant asymptote: y = x − 3 x( x − 2 ) c. f ( x ) = x +1

c. f ( x ) =

−3x 2 ( x + 4 )( x − 4 )

70. a. C ( 12 ) ≅ 0.0124

2 69. a. C(1) = 122(1) = 101 ≅ 0.0198 +100

b. Since 1 hour = 60 minutes, we have ) C(60) = 602(60 ≅ 0.0324 . 2 +100

b. C (1) ≅ 0.0243

c. Since 5 hours = 300 minutes, we have 2(300) C(300) = 300 ≅ 0. 2 +100

d. Even though the values temporarily get larger, eventually they decrease toward 0. So, the horizontal asymptote is y = 0 . So, after several days, C is approximately 0.

c. C ( 4 ) ≅ 0.0714

d. The horizontal asymptote is y = 0 . So, after several days, C is approximately 0. 71. a. N (0) = 52 wpm b. N(12) ≅ 107 wpm c. Since 3 years = 36 months, N(36) ≅ 120 wpm d. The horizontal asymptote is y = 130 . So, expect to type approximately 130 wpm as time goes on.

72. N (3) ≅ 78 , N (16) ≅ 267 The horizontal asymptote is y = 600 . So, expect to remember at most 600 names.

590

Section 4.6

73. The horizontal asymptote is y = 10. So most adult cats eventually eat about 10 ounces of food.

74. y (1000 ) =

1125 +3.75(1000 ) 1000

≈ $4.88. The horizontal asymptote is y = 3.75. So, the minimum price the sorority would have to pay according to this model is $3.75 per calendar.

75. Let w = width of the pen, l = length of the pen. Then, the area is given by wl = 500 , which is equivalent to l = 500 w (1). The amount of fence needed is given by the perimeter of the pen, namely 2w + 2l . 2 w 2 +1000 . Substituting in (1), we obtain the equivalent expression 2w + 2 ( 500 w )= w 76. Let w = width of the picture, l = length of the picture. Then, the area is given (1). by wl = 414, , which is equivalent to w = 414 l The width of the frame is represented by w + 2(4) and the legth of the frame is represented by l + 2(3.5). Then the area of the frame is given by A = (8 + w )(7 + l ). Substituting in (1), we obtain the equivalent expression (8l + 414)(7 + l ) A(l ) = ( 8 + 414 . l l ) (7 + l ) = 77. Solve for x:

−x3 + 10x 2 = 16 x − x 2 + 10x −16 = 0 ( x − 8)( x − 2) = 0  x = 2,8 So, must sell either 2,000 or 8,000 units to get this average profict. This yields an average value of $16 per unit. 78. Solve for x:

− x3 + 10 x 2 = 25 x −x 2 + 10x − 25 = 0 ( x − 5 )2 = 0  x = 5 So, must sell 5,000 units to get this average profict. This yields an average value of $25 per unit.

591

Chapter 4

22(14) + 24 ≈ 25.4 mcg/mL . 152 + 1 The times t for which this concentration is achieved are found by solving C (t ) = 25.4 , as follows: 22(t − 1) + 24 = 25.4 t2 + 1 22(t − 1) − 1.4 ( t 2 + 1) = 0 79. C(15) =

1.4t 2 − 22t + 23.4 = 0 22 ± 22 2 − 4(1.4)(23.4) ≈ 1, 15 2(1.4) So, there are two times, 1 hours and 15 hours, after taking the medication at which the concentration of the drug in the bloodstream is approximately 25.4 mcg/mL. The first time, approximately 1 hour, occurs as the concentration of the drug is increasing to a level high enough that the body will be able to maintain a concentration of approximately 25 mcg/mL throughout the day. The second time, approximately 15 hours, occurs many hours later in the day as the concentration of the medication in the bloodstream drops. t=

80. The times t for which this concentration is achieved are found by solving C (t ) = 25 , as follows: 22(t − 1) + 24 = 25 t2 + 1 22(t − 1) − ( t 2 + 1) = 0 t 2 − 22t + 23 = 0 22 ± 22 2 − 4(1)(23) t= ≈ 1.10 , 21 2(1) The concentration of the drug in the bloodstream is 25 mcg/mL approximately 21 hours after taking the medication. After 24 hours the concentration of the medication in the bloodstream has dropped to 24.9 mcg/mL. As the drug becomes inert during the 25th hour this concentration will drop to 0 mcg/mL. Thus it is important to take the next dose 24 hours after the previous does so that as the previous dose becomes inert the new dose has time to build up the concentration of the drug in the bloodstream. At the end of the 25th hour the previous dose will no longer be in the patient’s system but the new dose will provide a concentration of 24 mcg/mL.

592

Section 4.6

81. f (x) =

x −1 x −1 1 = = 2 x −1 ( x − 1) ( x + 1) x +1

82. x 2 + 1 = 0 has no real solution. So, there is no vertical asymptote.

with a hole at x = 1 . So, x = 1 is not a vertical asymptote. 83. In Step 2, the ratio of the leading coefficients should be −11 . So, the horizontal asymptote is y = −1.

84. “Degree of numerator = Degree of denominator – 1” is not the criterion for the existence of an oblique asymptote. In this case, there is a horizontal asymptote, namely y = 0 , but no oblique asymptote.

85. True. The only way to have a slant 86. False. Consider asymptote is for the degree of the 1 f ( x ) = . numerator to be greater than the degree ( x − 2 )( x + 2) of the denominator (by 1). In such The vertical asymptotes are case, there is no horizontal asymptote. x = −2, x = 2 . 87. False. This would require the denominator to equal 0, causing the function to be undefined.

88. True. Intersections with neither of these types of asymptotes creates a division by 0.

89. Vertical Asymptotes: x = c, x = −d Horizontal Asymptote: y = 1

90. There are no vertical asymptotes since x 2 + a 2 ≠ 0 . The horizontal asymptote is y = 3 . (Note: The actual values of a and b do not impact this result.)

91. Two such possibilities are: 4x 2 y= and ( x + 3)( x − 1)

92. f ( x ) =

4x5 ( x + 3)3 ( x − 1)2

593

x −3 is such a function. x 2 +1

Chapter 4

93.

94.

x−4 1 = has a hole ( x − 4 )( x + 2) x + 2 at x = 4 and a vertical asymptote at x = −2 .

2x + 1 1 = has a (2x + 1)(3x − 1) 3x − 1 hole at x = − 12 and a vertical asymptote

at x = 13 .

95.

96.

x − 2(3x + 1) −5x − 2 = x(3x + 1) x(3x + 1) Vertical asymptotes: x = 0, x = − 13 Horizontal asymptote: y = 0 Intercepts: ( − 25 ,0 )

− x + x 2 +1 x 2 − x + 1 = x( x 2 + 1) x( x 2 + 1) Vertical asymptotes: x = 0 Horizontal asymptote: y = 0

f (x) =

Intercepts: None since x 2 − x + 1 = 0 has no real solutions.

594

Section 4.6

97. a. Asymptotes for f: vertical asymptote: x = 3 horizontal asymptote: y = 0 Asymptotes for g: vertical asymptote: x = 3 horizontal asymptote: y = 2 Asymptotes for h: vertical asymptote: x = 3 horizontal asymptote: y = −3 b. The graphs of f and g are below. As x → ±∞, f (x) → 0 and g(x) → 2 .

c. The graphs of g and h are below. As

x → ±∞, g (x) → 2 and h(x) → −3 −3x + 10 2x − 5 . d. g ( x ) = , h( x ) = x −3 x −3 Yes, if the degree of the numerator is the same as the degree of the denominator, then the horizontal asymptote is the ratio of the leading coefficients for both g and h.

595

Chapter 4

98. a. Asymptotes for f: vertical asymptote: x = ±1 horizontal asymptote: y = 0 Asymptotes for g: vertical asymptote: x = ±1 horizontal asymptote: None slant asymptote: y = x Asymptotes for h: vertical asymptote: x = ±1 horizontal asymptote: None slant asymptote: y = x − 3 b. The graphs of f and g are below. As x → ±∞, f (x) → 0 and g(x) → x .

c. The graphs of g and h are below. As x → ±∞, g (x) → x and h(x) → x − 3

d. x3 + x x3 − 3x 2 + x + 3 = . , h( x ) x2 − 1 x2 − 1 Yes, if the degree of the numerator is the exactly one more than the degree of the denominator, then the quotient is the slant asymptote. g(x) =

596

Chapter 4 Review Solutions ----------------------------------------------------------------------1. b Opens down, vertex is ( −6,3)

2. c Opens up, vertex is (4,2)

3. a Opens up, vertex is ( − 12 , − 254 ) .

4. d Opens down, vertex is ( − 53 , 493 ) −3x 2 − 10 x + 8 = −3(x 2 + 103 x) + 8

x 2 + x − 6 = ( x 2 + x + 14 ) − 6 − 14

= −3(x 2 + 103 x + 259 ) + 8 + 253

= ( x + 12 )2 − 254

= −3(x + 53 )2 + 439

597

Chapter 4

10.

x − 3x − 10 = (x − 3x + ) −10 − 2

9 4

x 2 − 2 x − 24 = (x 2 − 2 x +1) − 24 − 1

9 4

= ( x − 32 )2 − 449

= ( x − 1)2 − 25

12. − 14 x 2 + 2x − 4 = − 14 (x 2 − 8x) − 4

11.

4 x + 8x − 7 = 4(x + 2x) − 7 2

= 4(x 2 + 2x +1) − 7 − 4

= − 14 ( x 2 − 8x +16) − 4 + 4

= 4(x + 1)2 −11

= − 14 ( x − 4 )2

13.

14.

15.

16.

598

Chapter 4 Review

17. Vertex is

( − ,c − ) = ( − 2

b 2a

b 4a

−5) ,12 − (4(13) 2

−5 2(13)

= ( 265 , 599 52 )

18. Vertex is

)

( − ,c − ) = ( −

b 2a

b2 4a

−0.12 2( −0.45)

( −0.12) ,3.6 − 4( −0.45) 2

451 = ( − 152 , 125 )

( − ,c − ) = ( −

)

b 2a

2 5

b2 4a

2 5

2( − 34 )

)

( 52 )2

,4 − 4( − ) 3 4

= ( 154 , 304 75 )

function has the form y = a( x + 2 )2 + 3 . To find a, use the fact that the point (1,4) is on the graph: 4 = a(1 + 2)2 + 3 4 = 9a + 3 =a

)

22. Since the vertex is (4,7) , the function has the form y = a( x − 4 )2 + 7 . To find a, use the fact that the point ( −3,1) is on the graph: 1 = a( −3 − 4)2 + 7

1 = 49a + 7 − 496 = a

So, the function is y = ( x + 2) + 3 . 1 9

,3 − (4(−4))

20. Vertex is

21. Since the vertex is (−2,3) , the

1 9

−4 2( 25 )

= (5, −7 )

19. Vertex is

( − ,c − ) = ( −

b2 4a

b 2a

So, the function is y = − 496 ( x − 4 )2 + 7 . 24. Since the vertex is ( − 52 , 74 ) , the function has the form y = a( x + 52 )2 + 74 . To find a, use the

23. Since the vertex is (2.7,3.4) , the function has the form y = a( x − 2.7)2 + 3.4 . To find a, use the fact that the point (3.2, 4.8) is on the graph: 4.8 = a(3.2 − 2.7)2 + 3.4

4.8 = 0.25a + 3.4 5.6 = a So, the function is y = 5.6( x − 2.7)2 + 3.4 .

599

fact that the point ( 12 , 53 ) is on the graph: 3 5 2 7 1 5 = a( 2 + 2 ) + 4 3 5

= 9a + 74

23 − 180 =a 23 ( x + 52 )2 + 74 . So, the function is y = − 180

Chapter 4

25. a. P ( x ) = R( x ) − C( x )

= ( −2x 2 + 12 x − 12 ) − ( 13 x + 2 ) = −2x 2 + 353 x − 14

b. Solve P ( x ) = 0 .

−2 x 2 + 353 x − 14 = 0

−2 ( x 2 − 356 x ) − 14 = 0

1225 −2 ( x 2 − 356 x + 1225 144 ) − 14 + 72 = 0 35 2 ) + 217 −2(x − 12 72 = 0 35 2 217 ( x − 12 ) = 144 35 x = 12 ±

217 144

= 35 ±12217 ≅ 4.1442433, 1.68909 c.

d. The range is approximately (1.6891, 4.144 ) , which corresponds to where the

graph is above the x-axis. 26. Area is A( x ) = (2x − 4)(x + 7) = 2x 2 + 10x − 28

600

Chapter 4 Review

27. Area is A( x ) = 12 (x + 2)(4 − x)

28. a. h(t ) = −12(t 2 − 203 t) 100 = −12(t 2 − 203 t + 100 9 ) + 12 ( 9 )

= − 21 ( x 2 − 2 x ) + 4 = − 12 ( x 2 − 2 x + 1) + 4 + 12 = − 12 ( x − 1)2 + 92 Note that A( x ) has a maximum at x = 1 (since its graph is a parabola that opens down). The corresponding dimensions are both base and height are 3 units.

= −12(t − 103 )2 + 400 3 So, the maximum height is approximately 133.33 units. b. Solve h(t) = 0 : t( −12t + 80) = 0 80 t = 0, 12 ≅ 6.67 So, after approximately 6.7 seconds, it will hit the ground.

29. Polynomial with degree 6

30. Polynomial with degree 5

31. Not a polynomial 1 (due to the term x 4 )

32. Polynomial with degree 3

33. d linear

34. b Parabola opens down

35. a 4th degree polynomial, looks like y = x 4 for x very large.

36. c 7th degree polynomial, looks like y = x7 for x very large.

37. Reflect the graph of x 7 over the xaxis

38. Shift the graph of x3 to the right 3 units.

601

Chapter 4

39. Shift the graph of x 4 down 2 units.

40. Shift the graph of x5 left 7 units, then reflect over the x-axis, and then move down 6 units and.

41. 6 (multiplicity 5) −4 (multiplicity 2)

42. 0 (multiplicity 1) 2 (multiplicity 3) −5 (multiplicity 1)

43. x5 − 13x3 + 36x = x(x 2 − 9)(x 2 − 4) = x( x + 3)(x − 3)(x + 2)(x − 2) So, the zeros are 0, − 2, 2, 3, − 3 , all with multiplicity 1.

44.

4.2x 4 − 2.6 x 2 ≅ 4.2x 2 ( x 2 − 0.619047 )

= 4.2x( x − 0.786795)(x + 0.786795) So, the zeros are approximately: 0 (multiplicity 2) 0.786795, −0.786795 (multiplicity 1) 45. x( x + 3)(x − 4)

46. (x − 2)( x − 4)(x − 6)(x + 8)

47. x( x + 52 )( x − 34 ) = x(5x + 2)(4x − 3)

48. ( x − ( 2 − 5 ))( x − ( 2 + 5 ))

602

Chapter 4 Review

50. ( x − 3)2 x3 (x + 1)2

49. ( x + 2 )4 ( x − 3)2 = x 4 − 2 x3 − 11x 2 + 12x + 36

51. f ( x ) = x 2 − 5x − 14 = (x − 7)(x + 2) a. Zeros: −2, 7 (both multiplicity 1) b. Crosses at both −2, 7 c. y-intercept: f (0) = −14, so (0, −14) d. Long-term behavior: Behaves like y = x2 . Even degree and leading coefficient positive, so graph rises without bound to the left and right.

52. f ( x ) = −(x − 5)5 a. Zeros: 5 (multiplicity 5) b. Crosses at 5 5 c. y-intercept: f (0) = − ( −5 ) = 3125,

so (0,3125) d. Long-term behavior: Behaves like y = − x5 . Odd degree and leading coefficient negative, so graph falls without bound to the right and rises to the left.

603

Chapter 4

53. f ( x ) = 6x 7 + 3x5 − x 2 + x − 4 a. Zeros: We first try to apply the Rational Root Test:

Need to actually graph the polynomial to determine the approximate zeros. From e., we see there is a zero at approximately (0.8748,0) with multiplicity 1. Factors of −4 : ±1, ± 2, ± 4 b. From the analysis in part a, we know Factors of 6: ±1, ± 2, ± 3, ± 6 the graph crosses at its only real zero. Possible rational zeros: c. y-intercept: f (0) = −4, so (0, −4) ±1, ± 2, ± 4, ± 12 , ± 13 , ± 23 , ± 34 , ± 61 d. Long-term behavior: Behaves like Unfortunately, it can be shown that none y = x7 . of these are zeros of f. To get a feel for the possible number of irrational and complex Odd degree and leading coefficient positive, so graph falls without bound root, we apply Descartes’ Rule of Signs: to the left and rises to the right. e. Number of sign variations for f ( x ) : 3 f ( − x ) = −6x7 − 3x5 − x 2 − x − 4 , so Number of sign variations for f (−x ) : 0 Since f ( x ) is degree 7, there are 7 zeros, classified as:

Positive Real Zeros

3 1

Negative Real Zeros 0 0

Imaginary Zeros

4 6

54. f ( x ) = − x 4 (3x + 6)3 (x − 7)3 a. Zeros: 0 (multiplicity 4) −2 (multiplicity 3) 7 (multiplicity 3) b. Touches at 0, and crosses at both −2, 7 c. y-intercept: f (0) = 0, so (0,0) d. Long-term behavior: Behaves like y = − x10 . Even degree and leading coefficient negative, so graph falls without bound to the right and left.

604

Chapter 4 Review

55. f ( x ) = (x − 1)(x − 3)(x − 7) a.

b. Zeros: 1, 3, 7 (all with multiplicity 1) c. Want intervals where the graph of f is above the x-axis. From a., this occurs on the intervals (1,3) and (7, ∞ ) .

So, between 1 and 3 hours, and more than 7 hours would be financially beneficial.

56. Consider the graph below. The peak seasons occur between months 2 & 5, and 7 & 9.

57. x−2

x+4 x + 2x − 6 2

−( x 2 − 2 x ) 4x − 6 −(4x − 8) 2

So, Q( x ) = x + 4, r( x ) = 2 .

605

Chapter 4

58.

59. x −1 2 x − 3 2 x − 5x − 1

2 x3 − 4 x 2 − 2x − 72

2x − 4

−(2x 2 − 3x )

4 x 4 − 16x3 + 12x 2 + x − 9

−(4x 4 − 8x3 )

− 2x −1 −( −2x + 3)

− 8x3 + 12 x 2 −( −8x3 + 16 x 2 )

−4

− 4x 2 + x

So, Q( x ) = x − 1, r( x ) = −4 .

−( −4x 2 + 8x ) − 7x − 9 −( −7x + 14) − 23 So, Q( x ) = 2x3 − 4x 2 − 2x − 72 , r( x ) = −23 .

60.

61. −2

−2 x + 2 x − 2 − 4x + 2x + 6x 2 − x + 2 2

2x + x − 4 2

4 5 −2 −8 −2 −4 −2 8

−( −4x 4 − 2 x3 + 8x 2 )

−4

So, Q( x ) = x + 2x + x − 4, r( x ) = 0 . 3

4 x3 − 2 x 2 − x

−(4x3 + 2 x 2 − 8x)

62.

− 4x2 + 7x + 2

−2

−( −4x 2 − 2 x + 8)

9x − 6

So, Q( x ) = −2x 2 + 2x − 2, r( x ) = 9x − 6 .

− 10

−2

1 −2

−6

So, Q( x ) = x − 2x − 6, r( x ) = 15 . 2

63. −8 1 0 −8

0 0 0 0 − 64 64 − 512 4096 − 32,768 262,144

1 − 8 64 − 512

4096

− 32,768 262,080

So, Q( x ) = x − 8x + 64x − 512x + 4096x − 32,768, r(x) = 262,080 . 5

606

Chapter 4 Review

64. 3 4

2 4 −2 0 2

3 2

33 8

51 32

153 128

3147 512

11 2

17 8

51 32

1049 128

5707 512

51 5707 So, Q( x ) = 2x 4 + 112 x3 + 178 x 2 + 32 x + 1049 128 , r ( x ) = 512 .

65.

66.

x+3 2 3 2 5x − 7x + 3 5x + 8x − 22x +1

3 1 2 −5 3 15

4 2 30 102

1 5

34 104

−(5x3 − 7 x 2 + 3x )

So, Q( x ) = x3 + 5x 2 + 10x + 34, r( x ) = 104 .

15x 2 − 25x + 1 −(15x 2 − 21x + 9) − 4x − 8 So, Q( x ) = x + 3, r( x ) = −4x − 8 . 67.

68. −1 1

−4 −1

2 5

−8 −7

1 −5

− 15

x −5 x + 0 x + 4 x − 5x + 4x − 20 2

−( x3 + 0x 2 + 4x)

So, Q( x ) = x 2 − 5x + 7, r( x ) = −15 .

− 5x 2 + 0 x − 20 −( −5x 2 + 0 x − 20) 0 So, Q( x ) = x − 5, r( x ) = 0 .

69. Area = length × width. So, solving for length, we see that length = Area ÷ width. So, in this case, 6x 4 − 8x3 − 10x 2 + 12x − 16 3x 4 − 4x3 − 5x 2 + 6x − 8 length = . = 2x − 4 x−2 We compute this quotient using synthetic division: 2 3 − 4 −5 6 −8 6 4 −2 8 3

−1

Thus, the length (in terms of x) is 3x3 + 2 x 2 − x + 4 feet . 607

Chapter 4

70. Let x = width (= length) of corner square. Then, the dimensions of the box formed are: width = 10 − 2 x length = 15 − 2x height = x Thus, the volume of the box is given by V ( x ) = x(10 − 2x)(15 − 2x) . 72.

71. −2

0 −7 1 −1 22 − 44 102 − 206

1 − 12

− 11 22 − 51 103

− 207

So, f ( −2 ) = −207 .

73. 1 6

1 0 − 7 1 −1 6 7 7 0 1

6 7 7 So, f (1) = 0 .

1 So, g(−1) = −2 .

0 −3 3 3

So, g(1) = 0 .

75. P( −3) = ( −3)3 − 5( −3)2 + 4( −3) + 2 = −82. So, it is not a zero.

2 0 −3 −1 −1 1 1

2 1

74. −1 1

1 1

−1 − 2

76. P ( −2 ) = P (2) = 0 Yes, they are zeros.

77. P(1) = 2(1)4 − 2(1) = 0 Yes, it is a zero.

78. P(4) = (4)4 − 2(4)3 − 8(4) = 96 No, it is not a zero.

79. P ( x ) = x ( x3 − 6x 2 + 32 )

80. Observe that since 3 is a zero, synthetic division yields: 3 1 − 7 0 36 3 − 12 − 36

Observe that since −2 is a zero, synthetic division yields: −2 1 − 6 0 32 − 2 16 − 32 1

− 8 16

P ( x ) = (x − 3)(x 2 − 4x − 12)

So, = x( x + 2)( x − 4 )2

So,

P ( x ) = x(x + 2)(x 2 − 8x + 16)

1 − 4 − 12

= ( x − 3)( x − 6)(x + 2)

608

Chapter 4 Review

81. P ( x ) = x 2 ( x3 − x 2 − 8x +12 )

We need to factor x3 − x 2 − 8x + 12 . To do so, we begin by applying the Rational Root Test:

82. x 4 − 32 x 2 − 144 = (x 2 − 36)(x 2 + 4) = ( x − 6 )( x + 6) ( x − 2i )(x + 2i )

Factors of 12: ±1, ± 2, ± 3, ± 4, ± 6, ± 12 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 3, ± 4, ± 6, ± 12 Observe that both −3 (multiplicity 1) and 2 (multiplicity 2) are zeros. So, P ( x ) = x 2 (x + 3)(x − 2)2 .

83. Number of sign variations for P ( x ) : 1 P ( − x ) = x 4 − 3x3 − 16 , so Number of sign variations for P ( −x ) : 1

84. Number of sign variations for P ( x ) : 1

85. Number of sign variations for P ( x ) : 5

P ( −x ) = −x5 − 6x3 + 4x − 2 , so Number of sign variations for P( −x ) : 2

Positive Real Zeros 1 1

Positive Real Zeros 1

Negative Real Zeros 1

P ( − x ) = − x9 + 2x 7 + x 4 + 3x3 − 2x − 1 , so Number of sign variations for P (−x ) : 2

Negative Real Zeros 2 0

Positive Real Zeros 5 5 3 3 1 1

609

Negative Real Zeros 2 0 2 0 2 0

Chapter 4

86. Number of sign variations for P ( x ) : 3

87. Factors of 6: ±1, ± 2, ± 3, ± 6 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 3, ± 6

P ( −x ) = −2x5 + 4x3 + 2x 2 − 7 , so Number of sign variations for P (−x ) : 2

Positive Real Zeros 3 1 3 1

Negative Real Zeros 2 2 0 0

88. Factors of −8 : ±1, ± 2, ± 4, ± 8 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 4, ± 8

89. Factors of 64: ±1, ± 2, ± 4, ± 8, ± 16, ± 32, ± 64 Factors of 2: ±1, ± 2 Possible rational zeros: ±1, ± 2, ± 4, ± 8, ± 16, ± 32, ± 64, ± 12

90. Factors of 2: ±1, ± 2 Factors of −4 : ±1, ± 2, ± 4 Possible rational zeros: ±1, ± 2, ± 12 , ± 14

91. Factors of 1: ±1 Factors of 2: ±1, ± 2 Possible rational zeros: ±1, ± 12

92. Factors of 3: ±1, ± 3 Factors of 12: ±1, ± 2, ± 3, ± 4, ± 6, ± 12 Possible rational zeros: ±1, ± 12 , ± 13 , ± 14 , ± 61 , ± 121 , ± 3, ± 32 , ± 34

The only rational zero is 12 .

The rational zeros are − 32 , 13 , 12 . 93. Factors of −16 : ±1, ± 2, ± 4, ± 8, ±16 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 4, ± 8, ±16

The rational zeros are 1, 2, 4, and −2 .

610

Chapter 4 Review

94. Factors of −2 : ±1, ± 2 Factors of 24: ±1, ± 2, ± 3, ± 4, ± 6, ± 8, ±12, ± 24 Possible rational zeros: ±1, ± 12 , ± 13 , ± 14 , ± 61 , ± 18 , ± 121 , ± 241 , ± 2, ± 23 There are no rational zeros. Indeed, see the graph to the right:

95. a. Number of sign variations for P ( x ) : 1 P ( −x ) = −x3 − 3x − 5 , so Number of sign variations for P (−x ) : 0 Since P ( x ) is degree 3, there are 3 zeros, the real ones of which are classified as:

Positive Real Zeros 1

Negative Real Zeros 0

b. Factors of −5 : ±1, ± 5 Factors of 1: ±1 Possible rational zeros: ±1, ± 5 c. −1 is a lower bound, 5 is an upper bound d. There are no rational zeros.

611

e. Not possible to accurately factor since we do not have the zeros. f.

Chapter 4

96. a. Number of sign variations for P ( x ) : 1 So, P ( x ) = (x − 2)(x 2 + 5x + 4) P ( −x ) = −x3 + 3x 2 + 6x − 8 , so = ( x − 2 )( x + 4)(x + 1) Number of sign variations for P ( −x ) : 2 Hence, the zeros are −4, 1, and 2. Positive Negative e. P ( x ) = (x − 2)(x + 4)(x + 1) Real Zeros Real Zeros f. 1 0

b. Factors of −8: ±1, ± 2, ± 4, ± 8 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 4, ± 8 c. −8 is a lower bound, 4 is upper bound (check by synthetic division) d. Observe that 2 1 3 −6 −8 2 10 8 1 5

97. a. Number of sign variations for P ( x ) : 3 So, P ( x ) = (x − 1)(x 2 − 8x + 12) P ( − x ) = − x3 − 9x 2 − 20x − 12 , so = ( x − 1)( x − 6)(x − 2) Number of sign variations for P ( −x ) : 0 Hence, the zeros are 1, 2, and 6. Positive Negative e. P ( x ) = (x − 1)(x − 6)(x − 2) Real Zeros Real Zeros f. 3 0

b. Factors of −12: ±1, ± 2, ± 3, ± 4, ± 6, ±12 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 3, ± 4, ± 6, ±12 c. −4 is a lower bound, 12 is upper bound (check by synthetic division) d. Observe that 1 1 − 9 20 − 12 1 − 8 12 1 − 8 12

612

Chapter 4 Review

98. a. Number of sign variations for P ( x ) : 2

So,

P ( − x ) = x 4 + x3 − 7x 2 − x + 6 , so Number of sign variations for P ( −x ) : 2

= (x − 1)( x + 2)(x − 3)(x + 1) Hence, the zeros are −2, −1, 1, and 3. e. P ( x ) = (x − 1)(x + 2)(x − 3)(x + 1) f.

Positive Real Zeros 2 2 0 0

Negative Real Zeros 2 0 2 0

b. Factors of 6: ±1, ± 2, ± 3, ± 6 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 3, ± 6 c. –3 is a lower bound, 6 is upper bound (check by synthetic division) d. Observe that 1 1 −1 − 7 1 6 1 0 −7 −6

−2 1 1

−7

−6

−2

−3

613

P ( x ) = (x − 1)(x + 2)(x 2 − 2x − 3)

Chapter 4

99. a. Number of sign variations for P ( x ) : 2 P ( −x ) = x 4 + 5x3 − 10x 2 − 20x + 24 , so Number of sign variations for P ( −x ) : 2

Positive Real Zeros 0 0 2 2

Negative Real Zeros 0 2 2 0

b. Factors of 24: ±1, ± 2, ± 3, ± 4, ± 6, ± 8, ±12, ± 24 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 3, ± 4, ± 6, ± 8, ±12, ± 24 c. −4 is a lower bound, 8 is an upper bound (check by synthetic division) d. Observe that 2 1 − 5 − 10 20 24 2 − 6 − 32 − 24

−1 1 1

−3

− 16

− 12

−1

−4

− 12

614

So, P ( x ) = (x − 2)(x + 1)(x 2 − 4x + 12) = ( x − 2 )( x + 1)(x + 2)(x − 6) Hence, the zeros are −2, −1, 1, and 6. e. P ( x ) = (x − 2)(x +1)(x + 2)(x − 6) f.

Chapter 4 Review

100. a. Number of sign variations for P ( x ) : 2 P ( − x ) = − x5 + 3x3 − 6x 2 − 8x , so Number of sign variations for P ( −x ) : 2 Since P ( x ) is degree 5, there are 5 zeros. Also note that 0 is a zero of P

Positive Real Zeros 2 2 0 0

2 1

0 −3 −6 8 2 4 2 −8

2 1

1 3

−4 4

1 3

So, P ( x ) = x(x − 1)(x − 2)(x 2 + 3x + 4) Now, solve x 2 + 3x + 4 = 0 using the quadratic formula: x = −3± 29−16 = −3±i2 7

Negative Real Zeros 2 0 2 0

So,

P ( x ) = x(x − 1)(x − 2)(x − ( −3+2i 7 )) ( x − ( −3−2i 7 ))

b. P ( x ) = x ( x 4 − 3x 2 − 6x + 8 )

101. x 2 + 25 = ( x − 5i )( x + 5i )

102. x 2 + 16 = ( x − 4i )(x + 4i )

103. Note that the zeros are x 2 − 2x + 5 = 0 

104. Note that the zeros are x2 + 4x + 5 = 0 

2 ± 4 − 4(5) 2 ± 4i = = 1 ± 2i 2 2 So, P ( x ) = (x − (1− 2i ))(x − (1+ 2i )) .

−4 ± 16 − 4(5) −4 ± 2i = = −2 ± i 2 2 So, P ( x ) = (x − (−2 − i ))(x − (−2 + i)) .

105. Since −2i and 3 + i are zeros, so are their conjugates 2i and 3 − i , respectively. Since P ( x ) has degree 4, these are the only missing zeros.

106. Since 3i and 2 − i are zeros, so are their conjugates −3i and 2 + i , respectively. Since P ( x ) has degree 4, these are the only missing zeros.

Factors of 8 : ±1, ± 2, ± 4, ± 8 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 4, ± 8 c. −2 is a lower bound. d. The rational zeros are 0, 1, 2. e. From d., we know that x( x − 1)(x − 2) divides P ( x ) evenly. We use synthetic division using 1 and 2 to determine the quotient P ( x ) ÷ x(x − 1)(x − 2) :

615

Chapter 4

107. Since i is a zero, then so is its conjugate –i. Also, since 2 – i is a zero of multiplicity 2, then its conjugate 2 + i is also a zero of multiplicity 2. These are all of the zeros.

108. Since 2i is a zero, then so is its conjugate –2i. Also, since 1 – i is a zero of multiplicity 2, then its conjugate 1 + i is also a zero of multiplicity 2. These are all of the zeros.

109. Since i is a zero of P ( x ) , so is its conjugate −i. As such, ( x − i )(x + i ) =

110. Since 2 − i is a zero of P ( x ) , so is its conjugate 2+i. As such, (x − (2 + i ))(x − (2 − i )) =

x 2 + 1 divides P ( x ) evenly. Indeed, observe that x 2 − 3x − 4 2 4 3 x + 0 x + 1 x − 3x − 3x 2 − 3x − 4

x 2 − 4x + 5 divides P ( x) evenly. Indeed, observe that x2 − 4 x 2 − 4 x + 5 x 4 − 4x3 + x 2 + 16x − 20

−( x 4 + 0 x 3 + x 2 )

−( x 4 − 4 x3 + 5x 2 )

− 3x3 − 4 x 2 − 3x

− 4 x 2 + 16 x − 20

−( −3x3 + 0 x 2 − 3x)

−( −4x 2 + 16x − 20)

− 4x 2 + 0x − 4

−( −4x 2 + 0x − 4)

So,

P ( x ) = (x − (2 + i ))(x − (2 − i )) ( x − 2 )( x + 2).

P ( x ) = (x − i )(x + i )(x 2 − 3x − 4) = ( x − i )(x + i )(x − 4)(x + 1)

616

Chapter 4 Review

111. Since −3i is a zero of P ( x ) , so is its conjugate 3i. As such, ( x − 3i )(x + 3i ) =

x 2 + 9 divides P ( x ) evenly. Indeed, observe that x2 − 2x + 2 x 2 + 0x + 9 x 4 − 2x3 + 11x 2 − 18x + 18

112. Since 1 + i is a zero of P ( x ) , so is its conjugate 1 − i. As such, (x − (1 + i ))(x − (1 − i )) =

x 2 − 2x + 2 divides P ( x) evenly. Indeed, observe that x2 + 2x − 3 x 2 − 2x + 2 x 4 + 0x3 − 5x 2 + 10x − 6

−( x 4 + 0x3 + 9x 2 )

−( x 4 − 2 x3 + 2x 2 )

− 2 x3 + 2 x 2 − 18x

2 x3 − 7 x 2 + 10x

−( −2x3 + 0x 2 − 18x )

−(2x3 − 4 x 2 + 4x)

2 x 2 + 0 x + 18

− 3x 2 + 6 x − 6

−(2x 2 + 0x + 18)

−( −3x 2 + 6 x − 6)

Next, we find the roots of x 2 − 2x + 2 : 2 ± 4 − 4(2) x= = 1± i 2 So, P ( x ) = (x − 3i )(x + 3i )

So, P ( x ) = (x − (1+ i ))(x − (1− i )) ( x − 1)( x + 3).

(x − (1 + i ))(x − (1 − i )). 113. P ( x ) = x 4 − 81 = ( x 2 − 9 )( x 2 + 9 )

114. P ( x ) = x3 − 6x 2 +12x = x ( x 2 − 6x +12 )

= (x − 3)( x + 3)(x − 3i )(x + 3i )

We need to find the roots of x 2 − 6x + 12 : 6 ± 36 − 4(12) x= = 3±i 3 2 So, P ( x ) = x(x − (3 + i 3))(x − (3 − i 3)).

115. x3 − x 2 + 4x − 4 = x 2 ( x − 1) + 4( x − 1) = ( x 2 + 4 ) ( x − 1) = (x − 2i )(x + 2i )(x − 1)

617

Chapter 4

116. P ( x ) = x 4 − 5x3 + 12x 2 − 2x − 20 Factors of −20: ±1, ± 2, ± 4, ± 5, ± 10, ± 20 Factors of 1: ±1 Possible rational zeros: ±1, ± 2, ± 4, ± 5, ± 10, ± 20 −1 1 − 5 12 − 2 − 20

−1 2

− 18

1 − 6 18 2 −8

− 20 20

1 − 4 10

So, P ( x ) = (x + 1)(x − 2) ( x 2 − 4x +10 ) . Next, we find the roots of x 2 − 4x + 10 : 4 ± 16 − 4(10) x= = 2±i 6 2 P ( x ) = (x + 1)(x − 2)(x − (2 + i 6 )) So, ( x − ( 2 − i 6 )). 119. Vertical Asymptote: x = −1 Horizontal Asymptote: none Slant Asymptote: y = 4 x − 4 To find the slant asymptote, we use long division: 4x − 4 2 x + 1 4 x + 0x + 0

117. Vertical Asymptote: x = −2 Horizontal Asymptote: Since the degree of the numerator equals the degree of the denominator, y = −1 is the HA.

118. Vertical Asymptote: x = 1 Horizontal Asymptote: Since the degree of the numerator is less than degree of the denominator, y = 0 is the HA.

120. Vertical Asymptote: None since x2 + 9 ≠ 0 Horizontal Asymptote: Since the degree of the numerator equals the degree of the denominator, y = 3 is the HA.

−(4x 2 + 4 x ) − 4x + 0 −( −4x − 4) 4

618

Chapter 4 Review

122. vertical asymptotes: x = −5 Horizontal asymptote: None Slant asymptote: y = −2 x +13 , as seen by the result of this synthetic division: −5 − 2 3 5 10 − 65

121. No vertical asymptotes, Horizontal asymptote: y = 2

− 2 13

123.

124.

125.

126.

619

− 60

Chapter 4

127. Note the hole at x = −7 .

128. Note the hole at x = − 12 .

129. The graph of f is to the right. The following are identified graphically: a. Vertex is (480, −1211) b. y-intercept: (0, −59) c. x-intercepts: (−12.14,0), (972.14,0) d. axis of symmetry is the line x = 480.

130. a. The general equation, using the b. 2 vertex, is f ( x ) = a(x − 2.4) − 3.1 . To find a, we use the fact that f (0) = 5.54 : 5.54 = a(0 − 2.4)2 − 3.1

8.64 = 5.76a a = 1.5 So, f ( x ) = 1.5(x − 2.4)2 − 3.1 . c. Yes, they agree.

620

Chapter 4 Review

131. The graph of f is as follows:

132. The graph of f is as follows:

x-intercepts: (−1,0), (0.4, 0), (2.8,0) zeros: −1, 0.4, 2.8, each with multiplicity 1 133. The graph is to the right. It is a linear function, as the following long division verifies: 5x − 4 2 3 2 3x − 7x + 2 15x − 47x + 38x − 8

−(15x3 − 35x3 +10x 2 ) − 12x 2 + 28x − 8 −( −12x 2 + 28x − 8) 0 Note: This graph has holes at x = and x = 2. 1 3

621

x-intercept: (−2.27,0) zeros: −2.27 (multiplicity 1)

Chapter 4

134. The graph is to the right: It is a quadratic function, as the following long division verifies: −4 x 2 + 2 x + 5 x − 3 −4 x3 + 14x 2 − x − 15 −( −4x3 + 12 x 2 )

2x 2 − x −(2x 2 − 6 x )

5 x − 15 − (5x − 15)

0 Note: This graph has a hole at x = 3.

135. The graph is to the right. a. The zeros are −2 (multiplicity 2), 3, and 4. b. P ( x ) = (x + 2)2 (x − 3)(x − 4)

622

Chapter 4 Review

136. The graph is to the right. a. The zeros are −1 (multiplicity 2) and 52 .

As such, we know that ( x + 1)2 (5x − 2) = 5x3 + 8x 2 + x − 2 divides P ( x ) evenly. So, −x 2 − 2x − 3 5x3 + 8x 2 + x − 2 −5x5 − 18x 4 − 32x3 − 24x 2 + x + 6 −( −5x5 − 8x 4 − x3 + 2x 2 ) − 10x 4 − 31x3 − 26x 2 + x −( −10x 4 − 16 x3 − 2x 2 + 4x) − 15x3 − 24 x 2 − 3x + 6 − ( −15x3 − 24 x 2 − 3x + 6)

So, P ( x ) = −(x + 1) (5x − 2 )(x + 2x + 3) . 2

623

Chapter 4

137. The graph is to the right. Note that the only real zero is 72 , and so we have: 7 2 1 − 2 − 91 2

2 So,

P ( x ) = ( x − 72 ) (2x 2 + 8x + 26) = (2x − 7)( x 2 + 4 x +13)

Need to find the roots of x 2 + 4x + 13 : −4 ± 16 − 4(13) x= = −2 ± 3i 2 138. The graph is to the right. Note that the real zeros are −5 and 32 ,

and so ( x + 5 )( x − 32 ) = x 2 + 72 x − 152 divides P ( x ) evenly. So, we have: −2 x 2 + 12 x − 20 x 2 + 72 x − 152 −2x 4 + 5x3 + 37x 2 − 160x + 150 −( −2x 4 − 7 x3 + 15x 2 )

12x3 + 22 x 2 − 160x −(12x3 + 42x 2 − 90x) − 20x 2 − 70 x + 150 − ( −20x 2 − 70 x + 150)

So, P ( x ) = −(2x − 3)(x + 5)(x − 6x + 10) . Next, we find the roots of x 2 − 6x + 10 : 6 ± 36 − 4(10) x= = 3±i . 2 P ( x ) = −(2x − 3)( x + 5) So, ( x − 3 − i )( x − 3 + i ). 2

624

Thus, P ( x ) = (2x − 7)(x + 2 − 3i )(x + 2 + 3i )

Chapter 4 Review

139. From the graph, we can see that there is a vertical asymptote at x = −1 and a horizontal asymptote at y = 2.

140. From the graph, we can see that there is a vertical asymptote at x = 2 and a horizontal asymptote at y = 4.

625

Chapter 4 Practice Test Solutions ---------------------------------------------------------------1.

2. y = − ( x 2 − 4x ) − 1 = −( x 2 − 4 x + 4) − 1 + 4 = −( x − 2 )2 + 3

y = − 21 ( x 2 − 6 x ) − 4 = − 12 ( x 2 − 6 x + 9) − 4 + 29

= − 12 ( x − 3)2 + 12 So, the vertex is (3, 12 ) .

4. Since the vertex is (−3, −1) , the function has the form y = a( x + 3)2 − 1 . To find a, use the fact that the point ( −4,1) is on the graph: 1 = a( −4 + 3)2 −1

1 = a −1 2=a So, the function is y = 2( x + 3) − 1 . 2

5. f ( x ) = x(x − 2)3 (x − 1)2 6. f ( x ) = x ( x3 + 6 x 2 − 7 ) a. Zeros: Certainly, 0 is a zero. To find the remaining three, we first try to apply the Rational Root Test: Factors of −7 : ±1, ± 7 Factors of 1: ±1 Possible rational zeros: ±1, ± 7 Observe that 1 is a zero. So, using synthetic division, we compute the quotient ( x3 + 6x2 − 7 ) ÷ ( x −1) : 1 1

6 1

1 7

626

0 −7 7 7 7

Chapter 4 Practice Test So, f ( x ) = x(x − 1)(x 2 + 7x + 7) . Now, we solve x 2 + 7x + 7 = 0 using the quadratic −7 ± 49 − 4(7)

= −7 ±2 21 formula: x = 2 So, there are four distinct x-intercepts. b. Crosses at all four zeros. c. y-intercept: f (0) = 0, so (0,0) d. Long-term behavior: Behaves like y = x 4 . Even degree and leading coefficient positive, so graph rises without bound to the left and right. e.

−2 x 2 − 2 x − 112 2 x 2 − 3x + 1 − 4x 4 + 2x3 − 7x 2 + 5x − 2

−( −4x 4 + 6 x3 − 2x 2 ) − 4 x3 − 5x 2 + 5x −( −4x3 + 6 x 2 − 2x) − 11x 2 + 7 x − 2 −( −11x 2 + 332 x − 112 ) − 192 x + 72 So, Q( x ) = −2x 2 − 2x − 112 , r( x ) = − 192 x + 72 .

627

Chapter 4

8. −2

9. 17

0 −4 0 2 − 10 − 34 68 − 128 256 − 516

17 − 34 64 − 128 258

− 526

So, Q( x ) = 17x 4 − 34x3 + 64x 2 − 128x + 258, r( x ) = −526 10. P( −1) = −1 − 2 + 5 − 7 + 3 + 2 = 0 . So, yes it is a zero.

3 1 1 − 13 3 12

−1 −3

12 − 12

1 4 −1 − 4 Yes, x − 3 is a factor of x 4 + x3 −13x 2 − x +12 .

11. 7

1 − 6 − 9 14 1

7 −14

−2

So, P ( x ) = (x − 7)(x 2 + x − 2) = ( x − 7 )( x + 2)(x − 1) 12. Since 3i is a zero, its conjugate −3i must also be a zero. So, we know that ( x − 3i )(x + 3i ) = x 2 + 9 divides P ( x ) evenly. This gives us the following, after long division: x 2 − 3x + 10 x 2 + 9 x 4 − 3x3 +19x 2 − 27x + 90 −( x 4 + 0 x3 + 9x 2 ) − 3x3 + 10 x 2 − 27x −( −3x3 + 0 x 2 − 27x) 10x 2 + 0 x + 90 −(10x 2 + 0 x + 90) 0

So, P( x ) = (x − 3i )(x + 3i )(x − 3x + 10) = (x − 3i )(x + 3i )(x − 5)(x + 2) . The zeros are ±3i , 5, −2 . 2

13. Yes, a complex zero cannot be an x-intercept.

628

Chapter 4 Practice Test 14. Number of sign variations for P ( x ) : 4 P ( − x ) = −3x5 + 2x 4 + 3x3 + 2x 2 + x + 1 , so Number of sign variations for P( −x ) : 1 Since P ( x ) is degree 5, there are 5 zeros that we classify as: Positive Real Zeros 4 2 0

Negative Real Zeros 1 1 1

Imaginary Zeros 0 2 4

15. Factors of 12: ±1, ± 2, ± 3, ± 4, ± 6, ± 12 Factors of 3: ±1, ± 3 Possible rational zeros: ±1, ± 2, ± 3, ± 4, ± 6, ± 12, ± 13 , ± 23 , ± 34 16. P ( x ) = − x3 + 4x = −x(x − 2)(x + 2) Zeros are 0,± 2 .

17.

P ( x ) = 2 x3 − 3x 2 + 8x −12 = x 2 (2 x − 3) + 4(2x − 3) = ( x 2 + 4 )(2x − 3) = ( x + 2i )( x − 2i )(2 x − 3) The only real zero is 3 2 , the other two are complex, namely ±2i .

629

Chapter 4

18. Factors of 9: ±1, ± 3, ± 9 Factors of 1: ±1 Possible rational zeros: ±1, ± 3, ± 9 Observe that 3 1 − 6 10 − 6 9

3 −9

−3 3

1 0

−3 3

19. The polynomial can have degree 3 since there are 2 turning points. See the graph below:

So, P ( x ) = (x − 3)2 (x 2 + 1) . So, the only real zero is 3 (multiplicity 2).

20. We need to solve x3 − 13x 2 + 47x − 35 = 0 . From the graph to the right, we observe that the zeros are 1, 5, and 7. So, you will break even at any of these values.

21. Since there are 2 turning points the polynomial must be at least degree 3.

630

Chapter 4 Practice Test 22. x-intercept: ( 92 ,0) y-intercept: (0,−3) Vertical Asymptote: x = −3 Horizontal Asymptote: y = 2 Slant Asymptote: None e.

23. Observe that x x g(x) = 2 = . x − 4 ( x − 2)(x + 2) x-intercept: (0,0) y-intercept: (0,0) Vertical Asymptote: x = ±2 Horizontal Asymptote: y = 0 Slant Asymptote: None e.

24. Observe that 3x 3 − 3 3( x3 − 1) . h( x ) = 2 = x − 4 ( x − 2)(x + 2) x-intercept: (1,0) y-intercept: (0, 34 ) Vertical Asymptote: x = ±2 Horizontal Asymptote: None Slant Asymptote: y = 3x , as seen by the following long division: 3x 2 3 2 x + 0 x − 4 3x + 0 x + 0 x − 3

631

Chapter 4

25. Observe that x −3 x −3 F (x) = 2 = . x − 2 x − 8 (x − 4)(x + 2) x-intercept: (3,0) y-intercept: (0, 38 ) Vertical Asymptote: x = −2, x = 4 Horizontal Asymptote: y = 0 Slant Asymptote: None

26.

27. a & b. The equation we obtain using the calculator is y = x 2 − 3x − 7.99 . We complete the square to put this into standard form: y = ( x − 1.5)2 − 10.24 c. x-intercepts: (−1.7,0) and (4.7,0) d. The graph is to the right. Yes, they agree.

632

Chapter 4 Practice Test 28. Observe that x(2 x − 3) + x 2 − 3x 3 x ( x − 2 ) f (x) = = x 2 − 3x x ( x − 3) So, there is an open hole at x = 0. Also, we have x-intercept: (2,0) y-intercept: none Vertical Asymptote: x = 3 Horizontal Asymptote: y = 3

The graph of f is to the right. Yes, they agree.

633

Chapters 1-4 Cumulative Test --------------------------------------------------------------------

(5x y ) = 125x y = 5 1. ( −10x y ) 100x y 4x y

−1

−2 3

−3

−6

2 2

= (2 x + 3)( y −1) 2x − 5 > 7

4 x − 36 x + 6x = ⋅ 2 x2 x + 9 x + 18 4(x − 3)( x + 3) x 2 (x + 6) ⋅ = x2 ( x + 6 )(x + 3) 2

4(x − 3) = 4 x −12

2x − 5 > 7

2x − 5 < −7

2x > 12

2 x < −2

x>6

x < −1

( −∞, −1) ∪ ( 6, ∞ )

5. Let x = number of minutes to do the job together. Then 1 1 1 = + x 75 60 300 = 4x + 5x 1 x = 300 9 = 33 3 min.

2xy − 2x + 3 y − 3 = 2 x( y −1) + 3( y − 1)

6. The discriminant is given by D = 32 − 4( −4)(15) = 249 > 0 So, there are two distinct real solutions.

7. 16 + x 2 = x + 2 16 + x 2 = ( x + 2)2 16 + x 2 = x 2 + 4x + 4 12 = 4x 3=x

8. x-axis: − y = −  x − 3 = − x + 3 ≠ y

No. y-axis:  − x − 3 = x − 3 = y Yes. Origin: −  −x − 3 = − x + 3 ≠ y No. So, only symmetric about y-axis.

9. Observe that x − 3 y = 8  y = 13 x − 83 So, slope is m = 13 . So, the equation of the line so far is y = 13 x + b . Now, use (4,1) to find b: 1 = 13 ( 4) + b  b = − 13 So, the equation is y = 13 x − 13 .

634

Chapters 1-4 Cumulative Test

10. The equation is equivalent to y = 3. So, there are no x-intercepts and the yintercept is (0,3). The graph is as follows:

11. Since the center is (0,6), the equation of the circle thus far is x 2 + ( y − 6)2 = r 2 . Use the fact that (1,5) is on the circle to now find the r2: 12 + (5 − 6)2 = r 2  r 2 = 2 . So, the equation is x 2 + ( y − 6)2 = 2 12. Need 6 x − 7 ≥ 0 , so that x ≥ 76 . So,

the domain is  76 ,∞ ) .

13. Observe that g ( − x ) = −x +10 ≠ g(x)

− g ( −x ) = − −x +10 ≠ g(x) So, neither. 15. Translate the graph of y = x to the right 1 unit and then up 3 units.

14. Shift the graph of y = x 2 left 1 unit, then reflect over the x-axis, and then move up 2 units.

16. f ( g ( x)) =

( x + 2 ) − 3 = x + 2 − 3 = x −1 2

Domain: [ −2, ∞ )

17. f ( −1) = 7 − 2( −1)2 = 5 g ( f ( −1)) = 2(5) − 10 = 0

18. To find the inverse, switch the x and y and solve for y: x = ( y − 4)2 + 2  x − 2 = ( y − 4 )2  y = 4 ± x − 2 Since we are assuming y ≥ 4 , use the positive root above. So,

f −1 ( x ) = 4 + x − 2, x ≥ 2 .

635

Chapter 4

19. Since the vertex is (−2,3), the equation so far is f ( x ) = a(x + 2)2 + 3 . Use (−1,4) to find a: 4 = a( −1 + 2)2 + 3 = a + 3  a = 1 So, f ( x ) = (x + 2)2 + 3 .

20. f ( x ) = −3.7x3 (x + 4) So, the zeros are 0 (multiplicity 3) and −4 (multiplicity 1)

21. Observe that

22. Observe that 3 2 3 −11 6

4 x + 4 x +1 −5x + 3 − 20 x − 8x 2 + 7x − 5 2

−( − 20x3 +12 x 2 )

So, Q( x ) = 2x + 9x + 16, r( x ) = 54 . 2

− 20x 2 + 7 x −( − 20x 2 +12 x ) − 5x − 5 −( −5x + 3) −8

So, Q( x ) = 4x 2 + 4x + 1, r( x ) = −8 23. P(x ) = 12 x3 + 29x 2 + 7x − 6 Factors of −6: ±1, ± 2, ± 3, ± 6 Factors of 12: ±1, ± 2, ± 3, ± 4, ± 6, ±12 Possible rational zeros: ±1, ± 2, ± 3, ± 4, ± 6, ± 12 , ± 13 ,

24. Since 5 is a zero, we know that x – 5 divides P ( x) evenly. So, 5 2 − 3 − 32 −15

± 14 , ± 61 , ± 121 , ± 32 , ± 23 , ± 34 Note that −2 12 29 7 −6 − 24 − 10 6

So,

−3

P ( x ) = (x − 5)(2x 2 + 7x + 3)

= ( x − 5 )(2 x + 1)(x + 3) So, the zeros are −3, − 12 , 5 .

So, P ( x ) = (x + 2)(12x 2 + 5x − 3) = ( x + 2 )(4x + 3)(3x − 1) So, the zeros are −2, − 34 , 31 .

25. Vertical asymptotes: x = ±2 Horizontal asymptote: y = 0

636

Chapters 1-4 Cumulative Test

26. Observe that x ( 2 x 2 − x − 1) x(2 x + 1) (x − 1) = f (x) = ( x − 1)( x + 1) ( x − 1) ( x + 1)

Open hole: (1, 32 ) x-intercepts: (0,0), ( − 12 , 0) y-intercept: (0,0) Vertical Asymptote: x = −1 Horizontal Asymptote: None Slant Asymptote: y = 2 x − 1 −1 2 1 0 −2 1 2 −1 1

27. Observe that 3(x + 1) x(2 x − 3) x-intercepts: (−1,0) y-intercept: none Vertical Asymptote: x = 0, x = 32 Horizontal Asymptote: y = 0 f (x) =

The graph of f is to the right. Yes, they agree. 28. Observe that 24x 2 − 6x − 18x 2 − 6x f (x) = (3x + 1)(4 x − 1) 6x( x − 2) = (3x + 1)(4x − 1) x-intercepts: (2,0), (0,0) y-intercept: (0,0) Vertical Asymptote: x = − 31 , x = 14 Horizontal Asymptote: y = 0.5

The graph of f is to the right. Yes, they agree.

637

CHAPTER 5 Section 5.1 Solutions -------------------------------------------------------------------------------1. 16 3.

1 52

2. 81 = 251

( ) =2 =4 2

5. 8 3 = 3 8 2

7. ( 19 )

− 32

(

( 9 ) = 3 = 27 3

= 641

6. 27 3 = 3 27

=9 2 =

1 43

8. ( 161 )

− 32

) =3 =9 2

= 16 2 = 3

( 16 ) = 4 = 64

9. −50 = −1

10. −60 = −1

11. 5.2780

12. 0.1895

13. 9.7385

14. 22.2740

15. 7.3891

16. 1.6487

17. 0.0432

18. 0.2431

19. f (3) = 33 = 27

20. h(1) = 10 2 = 100

21. g ( −1) = ( 161 ) = 16

22. f ( −2) = 3−2 = 312 = 91

−1

23. g ( − 12 ) = ( 161 ) 2 = 16 2 = 4

24. g ( − 32 ) = ( 161 ) 2 = 16 2 =

25. f ( e ) = 3e ≈ 19.81

26. g (π ) = ( 161 ) ≈ 0.0002

−1

−3

( 16 ) = 4 = 64 3

28. c y = 5 ( 5− x )

27. f y = 51 ( 5x )

Rising a factor of 5 slower than 5

Falling a factor of 5 quicker than 5− x , and y-intercept (0,5)

29. e y = − ( 5x )

30. d y = − ( 5− x )

The reflection of 5x over the x-axis

The reflection of 5− x over the x-axis.

638

Section 5.1

31. b y = − ( 5− x ) + 1

32. a y = 5x − 1 The graph of 5x shifted down 1 unit.

The reflection of 5− x over the x-axis and then shifted up 1 unit. 33. y-intercept: f (0) = 6 0 = 1 , so (0,1). HA: y = 0 Domain: ( −∞, ∞ ) Range: ( 0,∞ )

34. y-intercept: f (0) = 7 0 = 1 , so (0,1). HA: y = 0 Domain: ( −∞, ∞ ) Range: ( 0,∞ )

1  Other points:  −1,  , (1,6 )  6

1  Other points:  −1,  , (1,7 )  7

35. y-intercept: f (0) = 10 0 = 1, so (0,1). HA: y = 0

36. y-intercept: f (0) = 4 −0 = 1, so (0,1). HA: y = 0

Reflect graph of y = 10 x over y-axis. Domain: ( −∞, ∞ ) Range: ( 0, ∞ ) Other points: (1,0.1), (−1,10)

Reflect graph of y = 4 x over y-axis. Domain: ( −∞, ∞ ) Range: ( 0, ∞ )  1 Other points: 1,  , ( −1, 4 )  4

639

Chapter 5

37.

38.

1 y-intercept: f ( 0 ) =   = 1, so ( 0,1) . 2 HA: y = 0 Domain: ( −∞, ∞ )

3 y-intercept: f ( 0 ) =   = 1, so ( 0,1) . 4 HA: y = 0 Domain: ( −∞, ∞ )

 1 Other points: 1,  , ( −1, 2 )  2

4  3  Other points: 1,  ,  −1,  3  4 

39.

40.

Range: ( 0, ∞ )

−0

2 y-intercept: f ( 0 ) =   = 1, so ( 0,1) . 5 HA: y = 0

Range: ( 0, ∞ )

−0

1 y-intercept: f ( 0 ) =   = 1, so ( 0,1) . 3 HA: y = 0 x

2 Reflect graph of y =   over the 5 x-axis. Domain: ( −∞, ∞ )

1 Reflect graph of y =   over the 3 x-axis. Domain: ( −∞, ∞ )

2  5  Other points: 1,  ,  −1,  5  2 

 1 Other points: (1,3 ) ,  −1,   3

Range: ( 0, ∞ )

640

Section 5.1

41. y-intercept: f (0) = e 0 = 1, so (0,1). HA: y = 0

42. y-intercept: f (0) = −e 0 = −1, so (0,−1). HA: y = 0 Reflect graph of e x over the x-axis. Domain: ( −∞, ∞ ) Range: ( −∞,0 )

Reflect graph of y = e x over y-axis. Domain: ( −∞, ∞ ) Range: ( 0, ∞ )  1 Other points: 1,  , (−1, e)  e

1  Other points:  −1, −  (1, −e) e 

43. y-intercept: f (0) = 2 0 − 1 = 0, so (0,0). HA: y = −1

44. y-intercept: f (0) = 30 − 1 = 0, so (0,0). HA: y = −1

Shift graph of y = 2 x down 1 unit Domain: ( −∞, ∞ ) Range: ( −1, ∞ ) Other points: (2,3), (1,1)

Shift graph of y = 3x down 1 unit Domain: ( −∞, ∞ ) Range: ( −1, ∞ ) Other points: (2,8), (1,2)

641

Chapter 5

45. y-intercept: f (0) = 2 − e 0 = 1, so (0,1). HA: y = 2 Reflect graph of e x over the x-axis, then shift up 2 units. Domain: ( −∞, ∞ ) Range: ( −∞,2 )

46. y-intercept: f (0) = 1 + e −0 = 2, so (0,1). HA: y = 1 Reflect graph of e x over the y-axis, then shift up 1 unit. Domain: ( −∞, ∞ ) Range: (1, ∞ )

1 Other points: (1, 2−e), (−1, 2 − ) e

1 Other points: (−1, 1+e), (1, 1 + ) e

47. y-intercept: f (0) = e 0 +1 − 4 = e − 4, so (0, e − 4). HA: y = −4 Shift the graph of e x left 1 unit, then down 4 units. Domain: ( −∞, ∞ ) Range: ( −4, ∞ )

48. y-intercept: f (0) = 2 + e 0 −1 = 2 + 1e , so (0,2 + 1e ). HA: y = 2 Shift the graph of e x right 1 unit, then up 2 units. Domain: ( −∞, ∞ ) Range: ( 2, ∞ ) Other points: (1,3), (2, e + 2)

Other points: (−1,−3), (1, e 2 − 4)

642

Section 5.1

49. y-intercept: f (0) = 3 ⋅ e 0 = 3, so (0,3). HA: y = 0

50. y-intercept: f (0) = 2 ⋅ e −0 = 2, so (0,2). HA: y = 0

Expand the graph of y = e x horizon tally by a factor of 2, then expand vertically by a factor of 3. Domain: ( −∞, ∞ ) Range: ( 0, ∞ )

Reflect the graph of y = e x about the y-axis, then expand vertically by a factor of 2. Domain: ( −∞, ∞ ) Range: ( 0, ∞ )

Other points: (2,3e), (1, 3 e )

Other points: (0,2), ( −1,2e )

51. y-intercept: 0 −2 f (0) = 1 + ( 12 ) = 1 + 22 = 5 , so (0,5).

52. y-intercept: 0 +1 f (0) = 2 − ( 13 ) = 2 − 13 = 53 , so (0, 53 ).

HA: y = 1

Shift the graph of y = ( ) right 2 units, then up 1 unit. Domain: ( −∞, ∞ ) Range: (1, ∞ ) 1 x 2

Other points: (0,5), ( 2,2 )

HA: y = 2

Shift the graph of y = ( 13 ) left 1 unit, reflect over x-axis, then shift up 2 units. Domain: ( −∞, ∞ ) Range: ( −∞,2 ) x

Other points: (0, 53 ), ( −1,1)

643

Chapter 5

54. Use P ( t ) = P0 ( 2t / d ) .

53. Use P ( t ) = P0 ( 2t / d ) .

Here, P0 = 43,000, d = 6 ( 2010 − 2004 ) ,

Here, P0 = 7.1, d = 88, t = 48 (2090 − 2002).

t = 26 ( 2030 − 2004 )

( ) ≅ 10.4.

Thus, P (48) = 7.1 2

Thus, P ( 26 ) = 43,000 ( 2 26/6 ) ≅ 866,826

So, the expected population in 2050 is approximately 10.4 million.

So, the expected population in 2030 is approximately 899,826.

( )

55. Use P (t ) = P0 2 d .

56. Use P (t ) = P0 2 d .

Here, P0 = 1500, d = 5 (doubling time), t = 30.

Colin: P0 = 55,000, d = 15, t = 43 (65 − 22) .

( ) 30

( ) 43

Thus, P (30) = 1500 2 5 ≅ 96,000 . So,

Thus, P (43) = 55,000 2 15 ≅ 401,158 .

the cost per acre 30 years after initial investment is $96,000.

Cameron: P0 = 35,000, d = 10, t = 43 (65 − 22) .

( ) 43

Thus, P (43) = 35,000 2 10 ≅ 689, 441 . So, Cameron will have made more money by the time he retires. 57. Use A(t ) = A0 ( 12 ) h t

58. Use A(t ) = A0 ( 12 ) h t

Here, h = 119.77 days, A0 = 200 mg , t = 30 days

Here, h = 2.807 days, A0 = 300 mg , t = 7 days Thus, A(7) = 300 ( 12 ) 2.807 ≅ 53 . So, 53 mg remain after 1 week. 7

Thus, A(30) = 200 ( ) ≅ 168 . So, 168 mg remain after 30 days. 1 2

119.77

59. Use A(t ) = A0 ( 12 ) h

60. Use A(t ) = A0 ( 12 ) h

Here, h = 10 years, A0 = 8000, t = 14 years

Here, h = 1year, A0 = 1500, t = 4 years

Thus, A(14) = 8000 ( 12 ) 10 ≅ 3031 . So, the value after 14 years is $3031.

Thus, A(4) = 1500 ( 12 ) 1 ≅ 94 . So, the value after 4 years is $94.

644

Section 5.1

61. Use A(t ) = P (1 + nr )

62. Use A(t ) = P (1 + nr )

Here, P = 3200, r = 0.025, n = 4, t = 3 years . Thus, 4(3) A(4) = 3200 (1 + 0.025 ≅ 3448.42. 4 ) So, the amount in the account after 4 years is $3,448.42.

Here, P = 10,000, r = 0.035, n = 1, t = 5 years . Thus, 1(5) A(5) = 10,000 (1 + 0.035 ≅ 11,876.86 . 1 ) So, the amount in the account after 5 years is $11,876.86.

63. Use A(t ) = P (1 + nr )

64. Use A(t ) = P (1 + nr )

Here, A(18) = 32,000, r = 0.05, n = 365, t = 18. Thus, solving the above formula for P, 32,000 we have P = ≅ 13,011.03 . 365(18) (1 + 0.05 ) 365 So, the initial investment should be $13,011.03.

Here, A(15) = 80,000, r = 0.03, n = 52, t = 15 . Thus, solving the above formula for P, 80,000 we have P = ≅ 51,016.87 . 52(15) (1 + 0.03 ) 52 So, the initial investment should be $51,016.87.

65. Use A(t ) = Pe rt . Here, P = 3200, r = 0.02, t = 15 .

66. Use A(t ) = Pe rt . Here, P = 7000, r = 0.043, t = 10 .

Thus, A(15) = 3200e (0.02 )(15) ≅ 4319.55 . So, the amount in the account after 15 years is $4319.55.

Thus, A(10) = 7000e (0.043)(10) ≅ 10,760.80 . So, the amount in the account after 10 years is $10,760.80.

67. Use A(t ) = Pe rt . Here, A(20) = 38,000, r = 0.05, t = 20 . Thus, solving the above formula for P, 38,000 we have P = (0.05)(20 ) ≅ 13,979.42 . e So, the initial investment should be $13,979.42.

68. Use A(t ) = Pe rt . Here, A(18) = 80,000, r = 0.06, t = 18 . Thus, solving the above formula for P, 80,000 we have P = (0.06 )(18) ≅ 27,167.64 . e So, the initial investment should be $27,167.64.

69. Use C (t ) = C0 e −rt with r = 0.020,

70. Use C (t ) = C0 e −rt with r = 0.009,

C0 = 5 mg/L, t = 20 hours .

C0 = 4 mg/L, t = 4 hours .

C (20) = 5e −0.020(20 ) ≈ 3.4 mg/L .

C (4) = 4e −0.009( 4 ) ≈ 3.9 mg/L .

645

Chapter 5

71.

p (price per unit) 1.00 5.00 10.00 20.00 40.00

D(p)—approximate demand for product in units 1,955,000 1,020,500 452,810 89,147 3,455

60.00

134

80.00

90.00

72. D(91) ≈ 0.9. When the cost is $91 per unit, the demand for these units is less than a single unit (no demand at all). 73. The mistake is that 4 − 2 ≠ 4 2 . 1 1 1 Rather, 4 − 2 = 1 = . 42 2 1

74. The mistake is that 4 2 ≠ 3

Rather, 4 2 = 3

( 4) = 2 =8. 3

43 . 42

75. 2.5% needs to be converted to a decimal, 0.025.

76. Time is measured in years, so 6 months should be converted to 12 . Using t = 12 in the formula yields A = 5075.57 .

77. False. (0, −1) is the y-intercept.

78. True.

79. True. 3− x = (3−1 )x = ( 13 )

80. False. e is irrational, and hence cannot equal a finite decimal.

646

Section 5.1

81.

82.

Note on Graphs: Solid curve is y = e x and the dashed curve is y = ln x .

Note on Graphs: Solid curve is y = 3x and the dashed curve is y = log 3 x . 83.

84.

85. y-intercept: f (0) = be −0 +1 − a = be − a So, ( 0,be − a ) .

86. y-intercept: f (0) = a + b ⋅ e 0 +1 = a + be , so (0, a + be ) . HA: y = a

Horizontal asymptote: For x very large, be − x +1 ≈ 0 , so y = −a is the horizontal asymptote.

647

Chapter 5

87. The domain and range for x f ( x ) = b , where b > 1 , are:

Domain: ( −∞, ∞ ) Range: [1,∞ ) . Indeed, note that f ( x ) is defined piecewise, as follows: x  b , x ≥ 0 x b =  −x b , x < 0 Recall that the graph of b − x is the reflection of the graph of b x over the yaxis. 88.

89.

Note on Graphs: Solid curve is y = (1 + x1 )x and dashed curve (the horizontal asymptote) is y = e . 90. Since 2 < e < 3 , we have 2 x < e x < 3x , for all x, as is seen in the graphs to the right.

Note on Graphs: Solid curve is y = 2 x , dashed curve is y = e x , and dotted curve is y = 3x .

648

Section 5.1

91. The graphs are close on the interval ( −3,3 ) .

92. The graphs are close on the interval ( −3,3 ) .

Note on Graphs: Solid curve is y = e x

Note on Graphs: Solid curve is y = e − x

and thin curve is y = 1 + x + x2 + x6 + x24 .

and thin curve is y = 1 − x + x2 − x6 + x24 .

93. The graphs of f, g, and h are as follows:

94. The graphs of f, g, and h are as follows:

The horizontal asymptotes for f, g, and h are y = e, y = e 2 , y = e 4 . As x increases, f ( x ) → e, g ( x ) → e 2 , h( x ) → e 4 .

The horizontal asymptotes for f, g, and h are y = e, y = e −1 , y = e −2 . As x increases, f ( x ) → e, g ( x ) → e −1 , h( x ) → e −2 .

649

Chapter 5

95. a.

b. The best fit exponential curve is y = 228.34(0.9173)x with r 2 = 0.9628 . This best fit curve is shown below on the scatterplot. The fit is very good, as evidenced by the fact that the square of the correlation coefficient is very close to 1.

c. i. Compute the y-value when x = 6 to obtain about 136 degrees Fahrenheit. ii. The temperature of the soup the moment it was taken out of the microwave is the y-value at x = 0, namely about 228 degrees Fahrenheit. d. The shortcoming of this model for large values of x is that the curve approaches the x-axis, not 72 degrees. As such, it is no longer useful for describing the temperature beyond the x-value at which the temperature is 72 degrees.

650

Section 5.1

96. a.

b. The best fit exponential curve is y = 40.646(1.0746)x with r 2 = 0.976. This best fit curve is shown below on the scatterplot. The fit is very good, as evidenced by the fact that the square of the correlation coefficient is very close to 1.

c. Use the best fit exponential curve from (b) to answer the following: i. Compute the y-value when x = 10 to obtain about 84 degrees Fahrenheit. ii. The temperature of the soda the moment it was taken out of the refrigerator is the y-value at x = 0, namely about 41 degrees Fahrenheit. d. The shortcoming of this model for large values of x is that the curve continues to grow indefinitely in the y-direction rather than stopping once it hits 90 degrees. As such, it is no longer useful for describing the temperature beyond the x-value at which the temperature is 90 degrees.

651

Section 5.2 Solutions -------------------------------------------------------------------------------1. 53 = 125 1

2. 33 = 27 1

3. 814 = 3

4. 1212 = 11

5. 2 −5 = 321

6. 3−4 = 811

7. 10 −2 = 0.01

8. 10 −4 = 0.0001

9. 10 4 = 10,000

10. 103 = 1,000

11. ( 14 ) = 43 = 64

12. ( 61 ) = 62 = 36

13. ( 13 ) = 81

14. ( 71 ) = 343

15. 101 = 10

16. 71 = 7

17. e −1 = 1e

18. e1 = e

19. e 0 = 1

20. 100 = 1

21. e x = 5

22. e y = 4

23. x z = y

24. x y = z

25. log8 (512) = 3

26. log 2 (64) = 6

27. log5 251 = −2

28. log 2 321 = −5

29. log(0.00001) = −5

30. log(100,000) = 5

31. log 225 (15) = 12

32. log343 (7) = 13

8 33. log 2 5 ( 125 )=3

34. log 2 3 ( 278 ) = 3

35. log 127 (3) = − 13

36. log 11024 (4) = − 51

−3

−4

−2

−3

652

Section 5.2

37. ln 6 = x

38. ln 4 = −x

39. log y x = z

40. log y z = x

41. log 2 (1) = 0

42. log 5 (1) = 0

43.

44. log5 (3125) = x

log3 (729) = x

5x = 3125 x =5

3x = 729 x=6

45. log10 (107 ) = 7

46. log10 (10 −2 ) = −2

47.

48. log 17 (2401) = x

log 1 4 (4096) = x

( 14 ) = 4096

( 71 ) = 2401

4 − x = 4096

7 − x = 2401 x = −4

x = −6

50.

49. log 64 8 = x

log125 5 = x

64 x = 8

125x = 5 1 x= 3

1 2

51. undefined

52. undefined

53. undefined

54. undefined

55. 1.46

56. 3.37

57. 5.94

58. 2.58

59. undefined

60. undefined

61. −8.11

62. −3.52

653

Chapter 5

63. Must have x + 5 > 0 , so that the domain is ( −5, ∞ ) .

64. Must have 4 x − 1 > 0 , so that the domain is ( 14 ,∞ ) .

65. Must have 5 − 2 x > 0 , so that the domain is ( −∞, 52 ) .

66. Must have 5 − x > 0 , so that the domain is ( −∞,5 ) .

67. Must have 7 − 2 x > 0 , so that the domain is ( −∞, 72 ) .

68. Must have 3 − x > 0 , so that the domain is ( −∞,3 ) .

69. Must have x > 0 , so that the

70. Must have x + 1 > 0 , so that the

domain is ( −∞,0 ) ∪ ( 0, ∞ ) .

domain is ( −∞, −1) ∪ ( −1, ∞ ) .

71. Must have x 2 + 1 > 0 (which always occurs), so that the domain is .

72. Must have 1 − x 2 = (1 − x )(1 + x ) > 0 . CPs: −1, 1 − + −   | | −1

So, the domain is ( −1,1) . 73. b

74. e Reflect the graph of y = log 5 x over the y-axis.

75. c Reflect the graph of y = log 5 x over the x-axis and then the y-axis.

76. f Shift the graph of y = log 5 x left 3 units and down 1 unit.

77. d Since log 5 (1 − x ) − 2 = log 5 ( −( x − 1)) − 2 , Reflect the graph of y = log 5 x over the y-axis, then shift right 1 unit, and then shift down 5 units.

78. a Since − log 5 (3 − x ) + 2 = − log 5 ( −( x − 3)) + 2 , Reflect the graph of y = log 5 x over the y-axis, then shift right 3 units, then reflect over the x-axis, an then shift up 2 units.

654

Section 5.2

79. Shift the graph of y = log x right 1 unit. Domain: (1,∞ ) Range: ( −∞, ∞ )

80. Shift the graph of y = log x left 2 units. Domain: ( −2, ∞ ) Range: ( −∞, ∞ )

81. Shift the graph of y = ln x up 2 units. Domain: ( 0,∞ ) Range: ( −∞, ∞ )

82. Shift the graph of y = ln x down 1 unit. Domain: ( 0,∞ ) Range: ( −∞, ∞ )

655

Chapter 5

83. Shift the graph of y = log 3 x left 2 units, then down 1 unit. Domain: ( −2, ∞ ) Range: ( −∞, ∞ )

84. Shift the graph of y = log 3 x left 1 unit, then down 2 units. Domain: ( −1, ∞ ) Range: ( −∞, ∞ )

85. Reflect the graph of y = log x over the x-axis, then shift up 1 unit. Domain: ( 0,∞ ) Range: ( −∞, ∞ )

86. Reflect the graph of y = log x over the y-axis, then shift up 2 units. Domain: ( −∞,0 ) Range: ( −∞, ∞ )

656

Section 5.2

87. Shift the graph of y = ln x left 4 units. Domain: ( −4, ∞ ) Range: ( −∞, ∞ )

88. Since ln(4 − x ) = ln( −( x − 4)) , then shift the graph of y = ln x to the right 4 units, then reflect over the y-axis. Domain: ( −∞, 4 ) Range: ( −∞, ∞ )

89. Compress the graph of y = log x horizontally by a factor of 2. Domain: ( 0,∞ ) Range: ( −∞, ∞ )

90. Reflect the graph of y = ln x over the y-axis, then expand vertically by a factor of 2. Domain: ( −∞,0 ) Range: ( −∞, ∞ )

657

Chapter 5

91.

92.

 I  Use D = 10 log  . −12   1× 10  Here,  1× 10 −6  = 10 log(106 ) D = 10 log  −12   1×10 

 I  Use D = 10 log  . −12   1× 10  Here,  1× 101  = 10 log(1013 ) D = 10 log  −12   1× 10 

= 60 log(10) = 60 dB

= 130 log(10) = 130 dB

93.

 I  Use D = 10 log  . −12   1× 10  Here,  1× 10 −0.3  D = 10 log  = 10 log(1011.7 ) −12  1 10 ×   = 117 log(10) = 117 dB   

94.

 I  . Use D = 10 log  −12   1× 10  Here,  1× 10 −4.5  D = 10 log  = 10 log(107.5 ) −12   1× 10  = 75 log(10) = 75 dB

( )

95. Use M = 23 log 10E4.4 .

Here, M = 23 log

(

1.41×1017 10

4.4

96. Use M = 23 log 10E4.4 .

) = log (1.41×10 )

Here,

2 3

(

) = log (6.31×10 )

×10 M = 23 log 6.31 104.4

12.6

≈ 8.5

10.6

2 3

≈ 7.6

( )

97. Use M = 23 log 10E4.4 .

Here, M = 23 log

(

2×1014 10

4.4

98. Use M = 23 log 10E4.4 .

) = log ( 2 ×10 )

Here,

2 3

(

M = 23 log 810×10 4.4

9.6

≈ 6.6

) = log (8×10 ) 2 3

≈ 9.0

99. Use pH = − log10  H +  .

100. Use pH = − log10  H +  .

Here,

Here, −4

pH = − log10 (5.01×10 −11 )

pH = − log10 (5.01×10 ) ≈ 3.3

≈ 10.3

658

12.6

Section 5.2

101. Use pH = − log10  H +  .

102. Use pH = − log10  H +  .

Normal Rainwater: pH = − log10 (10 −5.6 ) = 5.6 Acid rain/tomato juice: pH = − log10 (10 −4 ) = 4

Here,

103. Use pH = − log10  H +  .

104. Use pH = − log10  H +  .

Here,

Here, pH = − log10 (10

−3.6

pH = − log10 (5.0 ×10 −13 ) ≈ 12.3

pH = − log10 (10 −4.2 )

)

= −( −3.6) log10 10 = 3.6

= −( −4.2) log10 10 = 4.2

105.

106.

 C  ln   500  Use t = −  . 0.0001216 Here,  100  1 ln  ln    500  5 =− t=−  0.0001216 0.0001216 ≅ 13,236 yr

 C  ln   500  Use t = −  . 0.0001216 Here,  40   1  ln  ln    500  12.5  =−  t=−  0.0001216 0.0001216 ≅ 20,771 yr

107.

108.

 3 ×10  dB = 10 log   ≈ −25 dB  1  So, the result is approximately a 25 dB loss. −3

 2 × 10 −4  dB = 10 log   ≈ −42 dB  3 

659

Chapter 5

109. a. Usage Super Low Frequency—Communication with Submarines Ultra Low Frequency—Communication within Mines Very Low Frequency—Avalanche Beacons Low Frequency—Navigation, AM Longwave Broadcasting Medium Frequency—AM Bradcasts, Amatuer Radio High Frequency—Shortwave broadcasts, Citizens Band Radio Very High Frequency—FM Radio, Television Ultra High Frequency—Television, Mobile Phones

660

Wavelength 10,000,000 m 1,000,000 m 100,000 m 10,000 m

Frequency 30 Hz 300 Hz 3000 Hz 30,000 Hz

1,000 m 100 m

300,000 Hz 3,000,000 Hz

10 m 0.050 m

30,000,000 Hz 6,000,000,000 Hz

Section 5.2

110. a. Color Wavelength Frequency Violet 400 nm 750 ×1012 Hz Cyan 470 nm 638 ×1012 Hz Green 480 nm 625 ×1012 Hz Yellow 580 nm 517 ×1012 Hz Orange 610 nm 491× 1012 Hz Red 630 nm 476 ×1012 Hz b.

111. log 2 4 = x is equivalent to 2 x = 4

112. log100 10 = x is equivalent to

(not x = 2 ).

100 x = 10 (not 10 x = 100 ).

113. The domain is the set of all real numbers such that x + 5 > 0 , which is written as ( −5, ∞ ) .

114. Must exclude x = 0 from the domain since ln 0 is not defined.

661

Chapter 5

115. False. The domain is all positive real numbers. 117. True.

116. False. There is no horizontal asymptote for y = ln x . There is, however, a vertical asymptote of x = 0 (i.e, the y-axis).

118. True. 119. Consider f ( x ) = − ln( x − a ) + b , where a, b are real numbers. Domain: Must have x − a > 0 , so that the domain is ( a, ∞ ) .

Range: The graph of f ( x ) is the graph of ln x shifted a units to the right, then reflected over the x-axis, and then shifted up b units. Through all of this movement, the range of y-values remains the same as that of ln x , namely ( −∞, ∞ ) . x-intercept: Solve − ln( x − a ) + b = 0 : − ln( x − a ) + b = 0 ln( x − a ) = b x − a = eb So, the x-intercept is ( a + e b ,0 ) .

x = a + eb

120. Consider f ( x ) = log( a − x ) − b , where a, b are real numbers. Domain: Must have a − x > 0 , so that the domain is ( −∞,a ) . Range: Note that f ( x ) = log( −( x − a )) − b . So, the graph of f ( x ) is the graph of log x shifted a units to the right, then reflected over the y-axis, and then shifted down b units. Through all of this movement, the range of y-values remains the same as that of ln x , namely ( −∞, ∞ ) . x-intercept: Solve log( a − x ) − b = 0 : log( a − x ) − b = 0 log( a − x ) = b

a − x = 10 b So, the x-intercept is ( a − 10 b ,0 ) .

x = a − 10 b

662

Section 5.2

121.

122.

123. The graphs are symmetric about the line y = x .

124. The graphs are symmetric about the line y = x .

Note on Graphs: Solid curve is y = e x and dashed curve is y = ln x .

Note on Graphs: Solid curve is y = 10 x and dashed curve is y = log x .

663

Chapter 5

125. The common characteristics are: x-intercept for both is (1,0). y-axis is the vertical asymptote for both. Range is (−∞,∞) for both. Domain is (0, ∞ ) for both.

Note on Graphs: Solid curve is y = log x and dashed curve is y = ln x 126. The function is defined everywhere except at 0.

127. The graphs of f, g, and h are below: We note that f and g have the same graph with domain ( 0,∞ ) .

128. The graphs of f, g, and h are below: We note that f and g have the same graph with domain ( 2,∞ ) .

664

Section 5.2

129. a.

b. A reasonable estimate for Vmax is about 156 μmol/min. c. Km is the value of [S] that results in the velocity being half of its maximum value, which by (b) is about 156. So, we need the value of [S] that corresponds to v = 78. From the graph, this is very difficult to ascertain because of the very small units. We can simply say that it occurs between 0.0001 and 0.0002. A more accurate estimate can be obtained if a best fit curve is known. d. i. The equation of the best fit logarithmic curve is v = 33.70 ln([S ]) + 395.80 with r 2 = 0.9984. It is shown on the scatterplot below.

ii. Using the equation, we must solve the following equation for [S]: 100 = 33.70 ln([S ]) + 395.80 −295.80 = 33.70 ln[S ] −8.77745 = ln[S ] e −8.77745 = S S ≈ 0.000154171

665

Chapter 5

130.

1 1 K 1 1 1 + , the slope of the line = m and x as is [S ] v v Vmax [S ] Vmax Km 1 . and the y-intercept is Vmax Vmax b. a. Thinking of y as

c. The equation of the best fit line is y = 0.0000005x + 0.0064 with r2 = 0.9998. It is shown on the below scatterplot. d. Using the information in (a), we solve the following equation:

= 0.0064 Vmax = 156.25 Vmax This value is very close to what we obtained in the previous problem. e. Using the information in (a) and (d), we solve the following equation: Km Km = = 0.0000005 Vmax 156.25 K m = 0.000078

666

Section 5.3 Solutions -------------------------------------------------------------------------------1.

2. log 9 1 = x

log 69 1 = x

log 1 ( 12 ) = x

9 =1

69 = 1

x=0

( 12 ) = 12

x =1

log3.3 3.3 = x

log10 10 = x

ln e3 = x

3.3x = 3.3 x =1

10 x = 108 x =8

e x = e3 x =3

7. log10 0.001 = x

log3 3 = x 7

10 x = 0.001 = 10 −3 x = −3

10.

log 2 8 = x 3

3x = 37 x=7

2x = 8 = 2 2 x = 32

11. 8log8 5 = 5

12. 2 log2 5 = 5

13. e ln( x +5) = x + 5

14. 10

log5 3 5 = x 1

5x = 3 5 = 5 3 x = 13 3

(

) = 3x 2 + 2 x + 1

log 3 x2 + 2 x +1

15. 53log5 2 = 5log5 2 = 23 = 8

16. 7 2 log7 5 = 7 log7 5 = 52 = 25

−2

17. 7 −2 log7 3 = 7 log7 3 = 3−2 = 19

1 18. e −2 ln10 = e ln10 = 10 −2 = 100

19.

20.

log b ( x y ) = log b ( x ) + log b ( y ) 3

= log b ( x ) + 2 log b ( y )

= 3 log b ( x ) + log b ( y )

21.

log b ( x y ) = log b ( x ) + log b ( y ) 3

= 3 log b ( x ) + 5 log b ( y )

log b ( xy 2 ) = log b ( x ) + log b ( y 2 )

22. log b ( x −3 y −5 ) = log b ( x −3 ) + log b ( y −5 )

= −3 log b ( x ) − 5 log b ( y )

667

Chapter 5

23. log b ( x3 y −2 ) = log b ( x3 ) + log b ( y −2 )

24.

= 3 log b ( x ) − 2 log b ( y )

25.

(

1 2

log b x y

1 3

2 = log b ( x ) + log b ( y ) 3

26.

) = log ( x ) + log ( y ) 1 3

1 2

)

(

log x 3 y2 = log b ( x ) + log b ( y 2 /3 )

log b

( r t ) = log ( r t ) 1 2

1 3

( )

= 12 log b ( x ) + 13 log b ( y )

( ) 1

= log b r 2 + log b t 3

= 12 log b ( r ) + 13 log b ( t ) 27. log b ( x 2 y3 z 2 ) = log b ( x 2 ) + log b ( y3 ) + log b ( z 2 ) = 2 log b ( x ) + 3 log b ( y ) + 2 log b ( z ) 28.

log b ( x 2 yz ) = log b ( x 2 ) + log b

( yz ) = 2 log ( x ) + 12 [log ( yz )] b

1 1 1 = 2 log b ( x ) + [log b ( y ) + log b ( z )] = 2 log b ( x ) + log b ( y ) + log b ( z ) 2 2 2 29.

30. r  1 1 log b  1  = log b r 3 − log b s 2  s2    = 13 log b ( r ) − 12 log b ( s ) 1 3

( )

 r4  log b  2  = log b ( r 4 ) − log b ( s 2 ) s  = 4 log b ( r ) − 2 log b ( s )

( )

31.

32.

 x  log b   = log b ( x ) − log b ( yz )  yz 

 xy  log b   = log b ( xy ) − log b ( z )  z  = log b ( x ) + log b ( y ) − log b ( z )

= log b ( x ) −  log b ( y ) + log b ( z )  = log b ( x ) − log b ( y ) − log b ( z )

668

Section 5.3

33.

(

)

log x 2 x + 5 = log ( x 2 ) + log

( x +5)

34.

log ( ( x − 3)( x + 2) ) = log( x − 3) + log( x + 2)

= log ( x 2 ) + log ( x + 5 ) 2 1

= 2 log x + 12 log ( x + 5 ) 35.

 x3 ( x − 2)2  3 2 ln   = ln ( x ( x − 2) ) − ln 2  x +5 

( x +5) 2

= ln ( x3 ) + ln ( ( x − 2)2 ) − ln ( x 2 + 5 )

1 2

= 3 ln( x ) + 2 ln( x − 2) − 12 ln ( x 2 + 5 ) 36.

 x +3 3 x −4  ln   = ln 4 x ( 1) +   = ln

( x + 3 x − 4 ) − ln (( x + 1) ) 4

( x + 3 ) + ln ( x − 4 ) − ln( x + 1) 3

= 12 ln( x + 3) + 13 ln( x − 4) − 4 ln( x + 1) 37.  x2 − 2x + 1   ( x − 1)2  = log  log    2  x −9   ( x − 3)( x + 3) 

= log ( ( x − 1)2 ) − log ( ( x − 3)( x + 3) )

= 2 log( x − 1) − [ log( x − 3) + log( x + 3)] = 2 log( x − 1) − log( x − 3) − log( x + 3)

38.

 x2 − x − 2   ( x − 2)( x + 1)  log  2  = log    ( x + 4)( x − 1)   x + 3x − 4  = log ( ( x − 2)( x + 1) ) − log ( ( x − 1)( x + 4) )

= log( x − 2) + log( x + 1) − [ log( x − 1) + log( x + 4)] = log( x − 2) + log( x + 1) − log( x − 1) − log( x + 4)

669

Chapter 5

39. 3 log b x + 5 log b y = log b ( x3 ) + log b y5

40.

41.

42.

2 log b u + 3 log b v = log b u 2 + log b v 3

= log b ( u 2v3 )

= log b ( x3 y5 )

5 log b u − 2 log b v = log b u − log b v 5

3 log b x − log b y = log b x3 − log b y

 u5  = log b  2  v 

 x3  = log b    y

43.  14 2  1 1/4 2 log b x + 2 log b y = log b x + log b y = log b  x y  4  

44.

 73  3 3 y −2 log b x + log b y = log b y − 2 log b x = log b y3/7 − log b x 2 = log b  2    7 7 x    45.

46. 1 2

1 2

log b x + log b y = log b x + log b y 2 3

(

= log b x 2 y 3

2 3

1 2

)

log b x − 23 log b y = log b x 2 − log b y 3  x2  = log b  2   y3    1

47. 2 log u − 3 log v − 2 log z = log u 2 −  log v3 + log z 2  = log u 2 − log ( v3 z 2 )  u2  = log  3 2  v z 

670

Section 5.3

48.

3 log u − log 2v − log z = log u3 − [ log 2v + log z ] = log u 3 − log ( 2vz )  u3  = log    2vz 

49. ln( x + 1) + ln( x − 1) − 2 ln( x 2 + 3) = ln( x + 1) + ln( x − 1) − ln( x 2 + 3)2

= ln [( x + 1)( x − 1)] − ln( x 2 + 3)2  x2 − 1  = ln  2 2   ( x + 3)  50. 1

ln( x − 1) + ln( x + 1) − 2 ln( x 2 − 1) = ln( x − 1) 2 + ln( x + 1) 2 − ln( x 2 − 1)2 = ln ( x − 1) 2 ( x + 1) 2  − ln( x 2 − 1)2   1 1  ( x − 1) 2 ( x + 1) 2  = ln    ( x 2 − 1)2    ( x − 1)( x + 1)  = ln    ( x 2 − 1)2   1

51. 1 2

ln( x + 3) − 13 ln( x + 2) − ln x = ln( x + 3) 2 −  ln( x + 2) 3 + ln x    1 1 = ln( x + 3) 2 −  ln x( x + 2) 3    1  ( x + 3) 2  = ln   x( x + 2) 13    1

(

671

)

Chapter 5

52. 1 3

ln( x 2 + 4) − 12 ln( x 2 − 3) − ln( x − 1) = ln( x 2 + 4) 3 −  ln( x 2 − 3) 2 + ln( x − 1)   1

(

)

1 1 = ln( x 2 + 4) 3 −  ln ( x − 1)( x 2 − 3) 2   

 ( x 2 + 4) 3  = ln  1   ( x − 1)( x 2 − 3) 2    1

53.

54. log 7 ≅ 1.2091 log 5

log 5 7 =

55.

log 4 19 =

log19 ≅ 2.1240 log 4

log 5 12 =

log 12 ≅ −0.4307 log 5

56.

log 1 5 = 2

log 5 ≅ −2.3219 log 12

57.

58. log 2.7 5.2 =

log 5.2 ≅ 1.6599 log 2.7

59.

log 7.2 2.5 =

log 2.5 ≅ 0.4642 log 7.2

logπ 2.7 =

log 2.7 ≅ 0.8677 log π

log 2 9 =

log 9 ≅ 6.3400 log 2

60. logπ 10 =

log10 ≅ 2.0115 log π

61.

62.

log 3 8 =

log 8 ≅ 3.7856 log 3

63.  I  Use D = 10 log  . −12   1×10  In this case, I = (1×10 −1 ) W m2 + (1×10 −6 ) W m2 = (1.00001× 10 −1 ) W m2 .     From music

From conversation

 1.00001×10  So, D = 10 log   ≅ 110 dB that you are exposed to. −12  1×10  −1

672

Section 5.3

 I  64. Use D = 10 log  . −12   1×10  In this case, I = (1×10 −10 ) W m2 + (1×10 −6 ) W m2 = (1.0001×10 −6 ) W m2 .       From whisperer

Normal

 1.0001× 10 −6  So, D = 10 log   ≅ 60 dB that you are exposed to. −12  1×10 

 E  65. Use M = 23 log  4.4  .  10  Here, the combined energy is (4.5 ×1012 ) + (7.8 ×108 ) joules. The corresponding  (4.5 ×1012 ) + (7.8 ×108 )  magnitude on the Richter scale is M = 23 log   ≅ 5.5 . 10 4.4  

 E  66. Use M = 23 log  4.4  .  10  Here, the combined energy is (5.2 ×1011 ) + (4.1×109 ) joules. The corresponding  (5.2 ×1011 ) + (4.1×109 )  magnitude on the Richter scale is M = 23 log   ≅ 4.9 . 10 4.4  

67. Cannot apply the quotient property directly. Observe that 3 log 5 − log 52 = 3 log 5 − 2 log 5 = log 5 . 68. Step 3 is wrong. Apply the product and quotient properties:  48  ln3 + ln16 − ln8 = ln(3 ⋅16) − ln8 = ln   = ln 6  8  69. Cannot apply the product and quotient properties to logarithms with different bases. So, you cannot reduce the given expression further without using the change of base formula. 70. Applied the power property incorrectly. Should be 2 log ( 53 ) = log ( 53 ) , 2

not ( log 53 ) . 2

673

Chapter 5

72.

71.

True. log10 = 1, so

ln e 1 True. log e = = ln10 ln10

1 = 1 , which log10

does equal ln e .

73. False. ln( xy )3 = 3 ln( xy ) = 3(ln x + ln y ) , which does not equal (ln x + ln y )3 , in general.

74.

True.

ln a log a ln10 ln a = ln b = log b ln10 ln b

75. Claim: log b ( M N ) = log b M − log b N

Proof: Let u = log b M , v = log b N . Then, b u = M , bv = N . Observe that log b ( M N ) = log b

( ) = log ( b ) = u − v = log M − log N u −v

bu bv

76. Claim: log b ( M p ) = p log b M

Proof: Let u = log b M . Then, bu = M . Observe that log b ( M p ) = log b ( b u ) = log b ( b u⋅ p ) = u ⋅ p = ( log b M ) ⋅ p = p log b M p

77. 6

3  x2   x2   x6  log b  3 −5  = log b  3 −5  = log b  9 −15   yz  y z  y z   

= log b ( x 6 ) − log b ( y9 z −15 )

= log b ( x 6 ) −  log b ( y9 ) + log b ( z −15 )  = 6 log b x − 9 log b y + 15 log b z 78. log b ( x1 ) = log b ( x −1 ) = − log b x

674

Section 5.3

79. Yes, they are the same graph.

80. No, they are not the same graph.

Note on Graphs: Solid curve is y = ln(2 + x ) and dashed curve is y = ln 2 + ln x 81. No, they are not the same graph.

82. Yes, they are the same graph.

Note on Graphs: Solid curve is x y = log log 2 and dashed curve is

y = log x − log 2 .

675

Chapter 5

83. No, they are not the same graph, even though the property is true.

Note on Graphs: The thin curve plus the dashed curve is y = ln( x 2 ) (domain all real numbers except 0) and the dashed curve only is y = 2 ln x (domain is (0, ∞ ) ). 85. Yes, the graphs of y = ln x and log x y= do coincide, and are as log e follows:

84. No, they are not the same graph.

Note on Graphs: Solid curve is y = (ln x )2 and dashed curve is y = 2 ln x .

86. Yes, the graphs of y = log x and ln x do coincide, and are as y= ln10 follows:

676

Section 5.4 Solutions -------------------------------------------------------------------------------1.

2. 3x = 81

5x = 125

3x = 3 4

5x = 53

x=4

x =3

1 49 x 7 = 7 −2 7x =

x = −2

4. 1 16 x 4 = 4 −2

6. 4

4x =

9−3 x = 81

(2 ) = 2 2 2x

9−3 x = 92 −3x = 2

2 4 x = 23 4x = 3 3 x= 4

x = −2

169 = 13 x

2 = 16 2x = 24

(13 ) = 13

x2 = 4 x = ±2

132 x = 13 2x = 1

2 x

( 23 ) = 278 −3 x +1 3 ( 23 ) = ( 23 ) = ( 23 ) x +1

x + 1 = −3 x = −4

x = 12 10.

11.

( ) = −2 x +1 2 ( 53 ) = ( 53 ) = ( 53 ) 3 x +1 5

e 2 x +3 = 1 = e 0 2x + 3 = 0

25 9

x = − 32

x + 1 = −2 x = −3

13.

12.

x 2 −1

= 1 = 10

2 3

8. x2

x=−

7 2 x −5 = 7 3 x − 4

2 x − 5 = 3x − 4

x2 − 1 = 0 x = ±1

x = −1

677

Chapter 5

14.

15. 2 x −3

2 x +12 = 27 x

53 x = ( 53 ) = 52 x −3

x 2 + 12 = 7x

125 = 5 x

x 2 − 7x + 12 = 0 ( x − 4)( x − 3) = 0

3x = 2 x − 3 x = −3

x = 3, 4

16.

17.

x 2 −3

9 x = 3x − 4 x

32 x = ( 32 ) = 3x − 4 x

x2 − 3 = 2x

x2 − 2x − 3 = 0 ( x − 3)( x + 1) = 0

2x = x 2 − 4x x 2 − 6x = 0 x( x − 6) = 0

x = −1,3

x = 0,6

18.

19. 16

x −1

e5 x −1 = e x +3

24 x−4 = ( 24 )

x −1

= 2x

5x − 1 = x 2 + 3

x 2 − 5x + 4 = 0 ( x − 4)( x − 1) = 0

4x − 4 = x2 x 2 − 4x + 4 = 0

x = 1, 4

( x − 2)2 = 0

x=2 20.

21.

x 2 −8 x

= 100

2x = 5

10 x −8 x = (102 ) = 102 x

log 2 2 x = log 2 5

x = log 2 5 ≈ 2.322

x 2 − 8x = 2 x x 2 − 10x = 0 x( x − 10) = 0 x = 0,10

678

Section 5.4

23.

22.

−x

2e 2 x = 14

= 15

log3 3− x = log3 15 − x = log3 15

e2x = 7 2 x = ln 7

x = − log3 15 ≈ −2.465

24.

ln 7 ≈ 0.973 2

25. −6 e

10 2 x −3 = 81

= −15

e3 x =

log (102 x −3 ) = log(81)

5 2

2 x − 3 = log(81)

5 3x = ln 2 5 ln x = 2 ≈ 0.305 3

2 x = 3 + log(81) ≅ 2.454 x = 3+ log(81) 2

27.

26. 23 x +1 = 21

3x +1 = 5 log3 (5) = x + 1

log 2 ( 23 x +1 ) = log 2 (21)

x = log3 (5) − 1 ≈ 0.465

3x + 1 = log 2 (21) 3x = −1 + log 2 (21) x = −1+ log3 2 (21) ≅ 1.131

28.

29. 2 x −1

27 = 23 x −1

= 35 log5 (35) = 2 x − 1 5

log 2 (27) = 3x − 1 x = 13 [ log 2 (27) + 1] ≈ 1.918

x = 12 [ log5 (35) + 1] ≈ 1.605

30.

31. 15 = 7

3− 2 x

3e x − 8 = 7

log 7 (15) = 3 − 2 x x=

3− log7 (15) 2

3e x = 15

≈ 0.804

ex = 5 x = ln5 ≈ 1.609

679

Chapter 5

32.

33. 5e + 12 = 27

9 − 2e 0.1x = 1

5e x = 15

2e 0.1x = 8

ex = 3

e 0.1x = 4 0.1x = ln 4 x = 10 ln 4 ≈ 13.863

x = ln3 ≈ 1.100

34.

21 − 4e

0.1x

35.

4e 0.1x = 16

2 ( 3x ) = 20

e =4 0.1x = ln 4 x = 10 ln 4 ≈ 13.863 0.1x

36.

2 ( 3x ) − 11 = 9

3x = 10 x = log3 (10) ≈ 2.096

37.

3 ( 2 ) + 8 = 35

e3 x + 4 = 22

ln ( e3 x + 4 ) = ln(22)

3 ( 2 x ) = 27

3x + 4 = ln(22)

2x = 9 x = log 2 (9) ≈ 3.170

3x = −4 + ln(22) x = ln(223 )− 4 ≅ −0.303

39.

38.

= 73

( ) = ln(73)

ln e x

x = ln(73) 2

x = ± ln(73) ≈ ±2.071

40. 3e

= 18

e2x = 6

ln ( e 2 x ) = ln(6) 2 x = ln(6) x = ln6 2 ≅ 0.896

680

4 (103 x ) = 20 103 x = 5

log (103 x ) = log 5 3x = log 5 x ≈ 0.233

Section 5.4

41. Note that e 2 x + 7e x − 3 = 0 is equivalent to ( e x ) + 7 ( e x ) − 3 = 0 . Let y = e x 2

and solve y2 + 7 y − 3 = 0 using the quadratic formula: −7 ± 7 2 − 4(1)( −3) −7 ± 61 = y= 2 2 So, substituting back in for y, the following two equations must be solved for x: −7 + 61 −7 − 61 ex = and e x = 2 2 −7 − 61 Since < 0 , the second equation has no real solution. Solving the first one 2  −7 + 61  yields x = ln   ≅ −0.904 . 2   42. Note that e 2 x − 4e x − 5 = 0 is equivalent to

( e ) − 4 ( e ) − 5 = ( e − 5 )( e + 1) = 0 . x 2

x So, e x − 5 = 0 or e 1 =0 . Solving the first equation yields e x = 5 , so that + No solution

x = ln5 ≈ 1.609 .

43.

(3 − 3 ) = 0 x

−x 2

3 x − 3− x = 0 3 x = 3− x x = −x x=0

44. (3x − 3− x )(3x + 3− x ) = 0

(3 ) − (3 ) = 0 x 2

−x 2

32 x − 3−2 x = 0 32 x = 3−2 x 2 x = −2 x x=0

681

45.

2 =1 e −5 2 = ex − 5 x

ex = 7 x = ln(7) ≈ 1.946

Chapter 5

46.

48.

47.

17 =2 e +4 17 = 2e x + 8

4 =8 3 − e3x 4 = 24 − 8e3 x

20 =4 6 − e2 x 20 = 24 − 4e 2 x

2e x = 9

4e 2 x = 4

8e3 x = 20

e x = 92

e2x = 1 2x = 0 x=0

e3 x = 52

x = ln ( 92 ) ≈ 1.504

49.

3x = ln ( 52 ) x ≈ 0.305

50. 4

log(2 x ) = 2

28 =4 10 x + 3 28 = 4 (10 x ) + 12

10 − 7 4 = 2 (102 x ) − 14 2x

51.

102 = 2 x 100 = 2 x 50 = x

10 x = 4

10 2 x = 9 2x = log10 (9)

x = log10 (4) ≈ 0.602

x ≈ 0.477

52.

54.

53.

log(5x ) = 3 5x = 10

log3 (2 x + 1) = 4

log 2 (3x − 1) = 3

2x + 1 = 3 2 x = 80 x = 40

3x − 1 = 23 3x = 9 x =3

x = 1000 5 = 200

55.

56. log 4 (5 − 2 x ) = −2

log 2 (4 x − 1) = −3 −3

2 = 4x − 1

4 −2 = 5 − 2 x

4 x = 18 + 1

79 2 x = 5 − 161 = 16 79 x = 32

x = 329

682

Section 5.4

57. ln x 2 − ln 9 = 0

58. log x 2 + log x = 3

59. log5 ( x − 4) + log5 x = 1

ln x 2 = ln 9

2 log x + log x = 3

x =9 x = ±3

3 log x = 3 log x = 1 x = 10

log5 ( x( x − 4) ) = 1

x( x − 4) = 5 x − 4x − 5 = 0 ( x − 5)( x + 1) = 0 2

x = 5, −1

60. log 2 ( x − 1) + log 2 ( x − 3) = 3

61. log( x − 3) + log( x + 2) = log(4 x ) log( x − 3)( x + 2) = log(4x ) ( x − 3)( x + 2) = 4 x

log 2 ( x − 1)( x − 3) = 3 ( x − 1)( x − 3) = x 2 − 4 x + 3 = 23

x2 − x − 6 = 4x

x 2 − 4x − 5 = 0 ( x − 5)( x + 1) = 0

x 2 − 5x − 6 = 0 ( x − 6)( x + 1) = 0

x = 5, −1

x = 6, −1

62. log 2 ( x + 1) + log 2 (4 − x ) = log 2 (6 x )

63. log(4 − x ) + log( x + 2) = log(3 − 2x ) log ( (4 − x )( x + 2) ) = log(3 − 2 x )

log 2 ( ( x + 1)(4 − x ) ) = log 2 (6 x )

(4 − x )( x + 2) = 3 − 2 x

( x + 1)(4 − x ) = 6 x

−x 2 + 2 x + 8 = 3 − 2 x

−x + 3x + 4 = 6x 2

x2 − 4x − 5 = 0 ( x − 5)( x + 1) = 0

x 2 + 3x − 4 = 0 ( x + 4)( x − 1) = 0

x = −1, 5

x = −4 , 1

683

Chapter 5

64.

65. log(3 − x ) + log( x + 3) = log(1 − 2x ) log ( (3 − x )( x + 3) ) = log(1 − 2 x ) (3 − x )( x + 3) = 1 − 2 x

−x 2 + 9 = 1 − 2x x2 − 2x − 8 = 0 ( x − 4)( x + 2) = 0

log 4 (4 x ) − log 4 ( x4 ) = 3

 4x  log 4  x  = 3  4  log 4 (16) = 3 Since the last line is a false statement, this equation has no solution.

x = −2, 4

66.

67.

log3 (7 − 2 x ) − log 3 ( x + 2) = 2

log(2x − 5) − log( x − 3) = 1  2x − 5  log   =1  x −3  2x − 5 = 10 x −3 2x − 5 = 10x − 30 8x = 25

 7 − 2x  log 3  =2  x+2  7 − 2x = 32 = 9 x+2 9x + 18 = 7 − 2 x 11x = −11 x = −1 68. log3 (10 − x ) − log 3 ( x + 2 ) = 1

 10 − x  log 3   =1  x+2  10 − x =3 x+2 10 − x = 3( x + 2) 10 − x = 3x + 6 4 = 4x 1= x

x = 258

69. ln x 2 = 5 x 2 = e5 x = ± e5 ≈ ±12.182

70.

log(3x ) = 2 102 = 3x 100 3

684

Section 5.4

72.

71.

ln(4x − 7) = 3

log(2 x + 5) = 2 2 x + 5 = 10

4x − 7 = e3

2x = 95 x = 47.5 73.

4x = 7 + e3 x = 7 +4e ≅ 6.771 3

74.

ln ( x + 1) = 4 2

log ( x 2 + 4 ) = 2 x 2 + 4 = 102 = 100

x2 + 1 = e 4 x2 = e 4 − 1

x 2 = 96

x = ± e 4 − 1 ≈ ±7.321

x = ± 96 ≈ ±9.798

75.

76. log(3x − 5) = −1

ln(2x + 3) = −2 2x + 3 = e

3x − 5 = 10 −1

−2

51 3x = 10

x = 12  −3 + e −2  ≈ −1.432

51 = 1.7 x = 30

77. log(2 − 3x ) + log ( 3 − 2x ) = 1.5

78. log 2 (3 − x ) + log 2 (1 − 2 x ) = 5 log 2 ( 3 − x )(1 − 2 x ) = 5

log ( (2 − 3x ) ( 3 − 2 x ) ) = 1.5

(3 − x )(1 − 2x ) = 25 = 32

(2 − 3x ) ( 3 − 2x ) = 101.5 ≅ 31.622

2 x 2 − 7 x + 3 = 32

6x 2 − 13x + 6 ≅ 31.622 6x 2 − 13x − 25.622 ≅ 0 Now, use the quadratic formula: 13 ± 783.93 x= , so that x ≈ 3.42 , −1.25 12

685

2 x 2 − 7 x − 29 = 0 x=

7 ± 49 − 4(2)( −29) ≈ −2.441, 5.941 2(2)

Chapter 5

79. ln x + ln( x − 2) = 4

80.

ln x( x − 2) = 4 x( x − 2) = e

ln(4 x ) + ln ( 2 + x ) = 2

ln ( (4 x ) ( 2 + x ) ) = 2

4x ( 2 + x ) = e 2

x2 − 2x − e 4 = 0

4 x 2 + 8x − e 2 = 0 Now, use the quadratic formula:

2 ± 4 − 4(1)( −e 4 ) 2 ≈ −6.456 , 8.456

−8 ± 64 − 4(4)( −e 2 ) x= 2(4) ≈ 0.6875, −2.6875

81. log 7 (1 − x ) − log 7 ( x + 2) = log 7 ( x )

 1− x  log 7   = log 7 ( x ) x+2 1− x =x x+2 1 − x = x( x + 2) x 2 + 3x − 1 = 0 There are no rational solutions since neither 1 nor −1 work. So, graph to find the real roots:

So, the solution is approximately 0.303.

82. log5 ( x + 1) − log5 ( x − 1) = log 5 x  x +1 log 5   = log 5 x  x −1  x +1 =x x −1 x + 1 = x2 − x x 2 − 2x − 1 = 0 2 ± 4 − 4(1)( −1) = 1± 2 2 ≈ −0.414 , 2.414

686

Section 5.4

83. ln x + 4 − ln x − 2 = ln x + 1 1 2

ln( x + 4) − 12 ln( x − 2) = 12 ln( x + 1) x+4 ln   = ln( x + 1)  x−2  x+4 = x +1 x−2 x + 4 = x2 − x − 2 x2 − 2x − 6 = 0 2 ± 4 − 4(1)( −6) = 1± 7 2 ≈ −1.646 , 3.646

84. log

( 1 − x ) − log ( x + 2 ) = log x  1− x  log   = log x  x+2   1− x  log   = log x x + 2   1− x =x x+2 1− x = x2 x+2 1 − x = x3 + 2 x 2 0 = x3 + 2 x 2 + x − 1

So, the solution is approximately 0.466.

687

Chapter 5

85. a. R(0) = 151 beats per minute. b. Solve for t: 151e −0.055t = 100 e −0.055t = 100 151 −0.055t = ln ( 100 151 )

1 ln ( 100 t = − 0.055 151 ) ≈ 7 min.

c. R(15) = 151e −0.055(15) ≈ 66 beats per minute. 86. a. Use V (t ) = V0 e k t .

Given that V0 = 45,000 , first find k using V(2) = 30,000: 30,000 = 45,000e k (2 ) 30,000 45,000

b. Solve for t: 45,000e −0.2027t = 20,000 e −0.2027t = 20,000 45,000 −0.2027t = ln ( 20,000 45,000 )

1 ln ( 20,000 t = − 0.2027 45,000 ) ≈ 4 years

k = 12 ln ( 30,000 45,000 ) ≈ −0.2027

So, V (t ) = 45,000e −0.2027t . 87. Use A(t ) = P (1 + nr ) .

88. Use A(t ) = P (1 + nr ) .

Here,

r = 0.035, n = 1 . In order to triple, if P is the initial investment, then we seek the time t such that A(t ) = 3P . So, we solve the following equation: 1( t ) 3P = P (1 + 0.035 1 )

r = 0.035, n = 12 . In order to triple, if P is the initial investment, then we seek the time t such that A(t ) = 3P . So, we solve the following equation: 12( t ) 3P = P (1 + 0.035 12 )

3 = (1.035)t log1.035 3 = t So, it takes approximately 31.9 years to triple.

3 = (1.0029)12t log1.0029 3 = 12t log1.0029 3 = t So, it takes approximately 31.62 years to triple. 1 12

688

Section 5.4

89. Use A(t ) = P (1 + nr ) . nt

Here, A = 20,000, r = 0.05, n = 4, P = 7500 . So, we solve the following equation: 4( t ) 20,000 = 7500 (1 + 0.05 4 ) 2.667 = (1.0125)4t log1.0125 2.667 = 4t log1.0125 2.667 = t So, it takes approximately 19.74 years. 1 4

 E  91. Use M = 23 log  4.4  .  10  Here, M = 7.4 . So, substituting in, we can solve for E:  E  7.4 = 23 log  4.4   10   E  11.1 = log  4.4   10  E 1011.1 = 4.4 10 11.1 4.4 10 ×10 ≈ 3.16 ×1015 = E 

90. Use A(t ) = Pe rt . Here, A = 15,000, r = 0.06, P = 9000 . Substituting into the above equation, we can solve for t: 15,000 = 9000e 0.06 t 1.667 = e 0.06 t ln1.667 = t So, it takes approximately 8.514 years. 1 0.06

 E  92. Use M = 23 log  4.4  .  10  Here, M = 8.3 . So, substituting in, we can solve for E:  E  8.3 = 23 log  4.4   10   E  12.45 = log  4.4   10  E 1012.45 = 4.4 10 12.45 4.4 10 ×10 ≈ 7.08 × 1016 = E 

=1015.5

So, it would generate 3.16 ×1015 joules of energy.

=1016.85

So, it would generate 7.08 ×1016 joules of energy.

689

Chapter 5

 I  93. Use D = 10 log  −12  .  10  Here, D = 120 . So, substituting in, we can solve for I:  I  120 = 10 log  −12   10   I  12 = log  −12   10  I 1012 = −12 10 1 = 1012 ×10 −12 = I So, the intensity is 1W m2 .

 I  94. Use D = 10 log  −12  .  10  Here, D = 100 . So, substituting in, we can solve for I:  I  100 = 10 log  −12   10   I  10 = log  −12   10  I 1010 = −12 10 −2 10 = 1010 ×10 −12 = I So, the intensity is 10 −2 W m2 .

95. Use A = A0 e −0.5 t . Here, A = 0.10A0 , where A0 is the initial amount of anesthesia. So, substituting into the above equation, we can solve for t: 0.10 A0 = A0 e −0.5 t

96. Use A = A0 e 0.03t . Here, A = 2 A0 , where A0 is the initial investment. So, substituting into the above equation, we can solve for t: 2A0 = A0 e 0.03t

0.10 = e

ln(0.10) = −0.5t − ln(0.10) = t So, it takes about 4.61 hours until 10% of the anesthesia remains in the bloodstream. 1 0.5

2 = e 0.03t ln 2 = 0.03t

−0.5 t

ln 2 = t So, it takes about 23.105 years until the initial investment doubles. 1 0.03

690

Section 5.4

97.

98.

200 . 1 + 24e −0.2t Here, N = 100 . So, substituting into the above equation, we can solve for t: 200 100 = 1 + 24e −0.2t 100 (1 + 24e −0.2t ) = 200

100,000 . 1 + 10e −2t Here, N = 50,000 . So, substituting into the above equation, we can solve for t: 100,000 50,000 = 1 + 10e −2t 50,000 (1 + 10e −2t ) = 100,000

Use N =

1 + 24e −0.2t = 2

50,000 + 500,000e −2t = 100,000

24e −0.2t = 1

e −2 t = 0.1 −2t = ln(0.1)

e −0.2t = 241 −0.2t = ln ( 241 )

1 ln ( 241 ) ≈ 15.89 t = − 0.2

So, it takes about 15.9 years. 99.

 E  Use M = 23 log  4.4  . For P waves:  10  Here, M = 6.2 . Substituting, we can solve

t = − 12 ln(0.1) ≈ 1.2 So, it takes 1.2 weeks until 50,000 Honda hybrids are on the road. For S waves: Here, M = 3.3 . So, substituting in, we can solve for E:

 E  3.3 = 23 log  4.4   10   E  4.95 = log  4.4   10  E 10 4.95 = 4.4 10 4.95 4.4 10 ×10 = E 

for E:

 E  6.2 = 23 log  4.4   10   E  9.3 = log  4.4   10  E 109.3 = 4.4 10 9.3 4.4 10 ×10 = E 

=109.35

=1013.7

(

13.7

So, the combined energy is 10

+ 109.35 ) joules. Hence, the reading on the Richter scale is:

 109.35 + 1013.7  M = 23 log   ≈ 6.2 10 4.4  

691

Chapter 5

100.

 I  Use D = 10 log  −12  .  10  Here, D = 100dB + 60dB = 160dB . So,     Concert

101. ln(4e x ) ≠ 4 x . Should first divide both sides by 4, then take the natural log: 4e x = 9 e x = 94

Converstaion

substituting in, we can solve for I:  I  160 = 10 log  −12   10   I  16 = log  −12   10  I 1016 = −12 10 16 −12 10 × 10 = I

ln( e x ) = ln( 94 ) x = ln( 94 )

102. Step 2 should be: 10 log(3 x ) = 10 log(3x ) = 1 3x = 10 x = 103

10 4 = I So, the intensity is 10 4 watts.

103. The correction that needs to be made is that x = −5 needs to be removed from the list of solutions. The domain of the logs cannot include a negative solution.

104. log x + log 2 ≠ log( x + 2) , in general. The computation should be: log(2 x ) = log 5 2x = 5 x = 52

105. True. 106. False. log( x 2 ) = 2 log( x ) ,whereas (log x )2 ≠ 2 log x , in general.

ln x 107. False. Since log x = ln10 ,

e log x = e ln10 = ( e ln x ) ln x

1 ln10

= x ln10 ≠ x .

108. True. Since the left side is always positive, and the right side is always negative.

692

Section 5.4

109. 3 2 1 1 − 2x  + 1) = 2 3 log b ( x ) + 2 log b ( x  = ( x −1)

2 log b ( x ) + 2 log b (1 − x ) = 4 log b ( x ) + log b (1 − x ) = 2

1 3

log b ( x ) + log b ( x − 1) 3

110.

2 ⋅ 12

log b ( x(1 − x ) ) = 2

log b x + log b ( x − 1) = 2 log b ( x( x − 1)) = 2 b 2 = x( x − 1) x2 − x − b2 = 0 Now, use the quadratic formula to find the solutions: x=

1 ± 1 − 4( −b 2 ) 2

1 + 1 + 4b 2 , 2

x(1 − x ) = b 2 x 2 − x + b2 = 0 Now, use the quadratic formula to find the solutions: x=

1 ± 1 − 4b 2 2

1 + 1 − 4b 2 1 − 1 − 4b 2 , 2 2 Note that we need to impose a restriction on b, namely we can only consider those values of b for which 1 − 4b2 ≥ 0 . This occurs for b ∈ [ − 12 , 12 ] .

1 − 1 + 4b 2 2   This is negative, for any value of b. So, it cannot be a solution.

Now, observe that for all such values of b, the integrand is positive and less than 1. Hence, both values of x above are positive, and thus are both solutions to the original equation. 112. Must have x 2 − a 2 = ( x − a )( x + a ) > 0 . CPs are a, − a .

111.

3000 y= 1 + 2e −0.2t y (1 + 2e −0.2t ) = 3000

+ − +   | | −a

y + 2 ye −0.2t = 3000

So, the identity is valid for any x in the set ( −∞, −a ) ∪ ( a, ∞ ) .

2 ye −0.2t = 3000 − y −y e −0.2t = 3000 2y

(

−y −0.2t = ln 3000 2y

(

)

−y t = −5 ln 3000 2y

)

693

Chapter 5

113. ex + e−x , for x ≥ 0, y ≥ 1 . (Need this restriction in order 2 for the function to be one-to-one, and hence have an inverse.) ex + e−x for x. Solve y = 2 2 y ± ( −2 y )2 − 4 2 y ± 4 y2 − 4 ex + e−x ex = = y= 2 2 2 2 2 y ± 2 y −1 x 2 = = y ± y2 − 1 2 e x + e1x ( e e)x +1 y= = Since dom( f ) = (0, ∞ ) = rng ( f −1 ) , we 2 2 x 2 use only y + y 2 − 1 here. Now to solve 2 y = ( e e)x +1 for x, take natural log of both sides: 2 ye x = ( e x )2 + 1 e x = y + y2 − 1 x 2 x (e ) − 2 y ( e ) + 1 = 0 x = ln y + y 2 − 1 Now, solve using the quadratic

Consider the function y =

)

(

formula:

Thus, inverse function is given by

)

(

f −1 ( x ) = ln x + x 2 − 1 , x ≥ 1 , x ≥ 1. 114. ex − e−x . (There is no need for a restriction on x and y 2 ex − e−x this time since the function is one-to-one on  .) Solve y = for x. 2 2 y ± ( −2 y )2 + 4 2 y ± 4 y2 + 4 ex − e−x ex = = y= 2 2 2 2 y ± 2 y2 +1 x 2 = = y ± y2 + 1 2 e x − e1x ( e e)x −1 y= = Since e x > 0 , we use only 2 2 x 2 y + y2 + 1 here. Now to solve for x, 2 y = ( e e)x −1 take natural log of both sides: 2 ye x = ( e x )2 − 1 e x = y + y2 + 1 x 2 x ( e ) − 2 y( e ) − 1 = 0 x = ln y + y 2 + 1 Now, solve using the quadratic formula: Thus, inverse function is given by

Consider the function y =

)

(

)

f −1 ( x ) = ln x + x 2 + 1 .

694

Section 5.4

115. Observe that ln(3x ) = ln( x 2 + 1) 3x = x 2 + 1 x 2 − 3x + 1 = 0 x = 3± 29− 4 = 3±2 5 These solutions agree with the graphical solution seen to the right.

Note on Graphs: Solid curve is y = ln(3x ) and the thin curve is y = ln( x 2 + 1) . 116. Observe that 2

10 x = 0.001x 10 x = (10 −3 ) 2

10 x = 10 −3 x x 2 = −3x x 2 + 3x = x( x + 3) = 0, so that x = 0, − 3 These solution agrees with the graphical solution seen to the right. Note on Graphs: Two graphs are provided since the range in values between the two solutions made it difficult to show both on the same graph. In both cases, the 2 solid curve is y = 10 x and the thin curve is y = 10 −3 x .

695

Chapter 5

117.

118.

Note on Graphs: Solid curve is y = 3x and the thin curve is y = 5x + 2 .

Note on Graphs: Solid curve is y = 2 log x and the thin curve is y = ln( x − 3) + 2 .

119. The graph of f ( x ) =

ex + e−x is 2

given below:

The domain is ( −∞, ∞ ) , and the graph is symmetric about the y-axis. There are no asymptotes.

120. The graph of f ( x ) =

ex + e−x is ex − e−x

given below:

The domain is ( −∞,0 ) ∪ ( 0, ∞ ) , and the graph is symmetric about the origin. The vertical asymptote is x = 0 , and there are two different horizontal asymptotes (one as x → ∞ and one as x → −∞ ). They are y = 1, y = −1 respectively. 696

Section 5.5 Solutions--------------------------------------------------------------------------------1. c (iv)

2. d (v)

3. a (iii)

4. b (ii)

5. f (i)

6. e (i)

7. Use N = N 0 e rt . Here, N 0 = 80, r = 0.0236. Determine N when t = 17 (determined by 2020 – 2003): N = 80e 0.0236(17 ) ≈ 119 million.

8. Use N = N 0 e rt . Here, N 0 = 13.7, r = 0.025. Determine N when t = 24 (determined by 2020 – 1996): N = 13.7e 0.025(24 ) ≈ 25 million.

9. Use N = N 0 e rt .

10. Use N = N 0 e − rt .

Here, N 0 = 103,800, r = 0.12. Determine t such that N = 260,000 :

Here, N 0 = 776,000, r = 0.015. Determine t such that N = 600,000: 600,000 = 776,000e −0.015 t

260,000 = 103,800e 0.12 t 260,000 103,800

= e 0.12 t

600,000 776,000

260,000 1 t = 0.12 ln ( 103,800 ) ≈ 7.7

= e −0.015 t

1 t = −0.015 ln ( 600,000 776,000 ) ≈ 17.148

So, the population would hit 260,000 sometime in the year 2010.

So, the population would hit 600,000 sometime in the year 2019.

11. Use N = N 0 e rt . Here, N 0 = 487.4, r = 0.165. Determine N when t = 13 (corresponds to the number of cell phone subscribers in 2020): N = 487.4e 0.165(13) ≈ 4163.4 There are approximately 4163.4 million subscribers in 2020.

12. Use N = N 0 e rt (t is measured in hours). Here, N 0 = 500, r = 0.20.

697

a. N (12) = 500e 0.20(12 ) ≈ 5,512 There are about 5,512 bacteria after 12 hours. b. N (24) = 500e 0.20( 24 ) ≈ 60,756 There are about 60,756 bacteria after 1 day.

Chapter 5

13. Use N = N 0 e rt .

Here, N 0 = 185,000, r = 0.30. Determine N when t = 12 : N = 185,000e 0.30(12 ) ≈ 6,770,673 The amount is approximately $6,770,673.

14. Use N = N 0 e rt (1). Assuming t = 0 corresponds to the year 2004, we have N (0) = N 0 = 230,000, (2) N (1) = 252,000 In order to determine N (12) , we need to first determine r. To this end, substitute (2) into (1) to obtain: 252,000 = 230,000e r  r = ln ( 252,000 230,000 )

(

12 ln 252,000

)

So, N (12) = 230,000e 230,000 ≈ 688,336 The amount is approximately $688,336. 15. Use N = N 0 e rt (t measured in months). Here, N 0 = 100 (million), r = 0.20. Determine N when t = 6 : N = 100e 0.20( 6 ) ≈ 332 million .

16. Use N = N 0 e rt (t measured in months). Here, N 0 = 50 (million), r = 0.12. Determine N when t = 3 : N = 50e 0.12(3) ≈ 72 million .

17. Use N = N 0 e rt (t measured in years). Here, N 0 = 1 (million), r = 0.025 . Determine N when t = 21: N = 1e 0.025( 21) ≈ 1.69 million.

18. Use N = N 0 e rt (t measured in years). Here, N 0 = 25 (million), r = 0.09 . Determine N when t = 21: N = 25e 0.09( 21) ≈ 165.48 million.

19. Use A = 100e −0.5 t . Observe that A(4) = 100e −0.5 ( 4 ) ≈ 13.53 ml

20. Use A = 100e −0.5 t . Observe that A(12) = 100e −0.5 (12) ≈ 0.25 ml

698

Section 5.5

21. a. Solve for k: 350 = 750 (1 − e − k (1) ) 350 750

22. a. Solve for k: k 2 525 = 850e ( ) 525 = e2k 850  525  ln   = 2k  850  1  525  k = ln   ≈ −0.2409 2  850 

= 1 − e−k

−k − ( 350 750 − 1) = e

ln (1 − 350 750 ) = − k

400 k = − ln ( 750 )

= − ln ( 158 ) ≈ 0.6286

P ( t ) = 850e −0.2409t

b. S (3) = 750 (1 − e −0.6286(3) ) ≈ 636. So,

about 636,000 MP3 players.

b. P (9) = 850e −0.2409(9) ≈ 97. So, about $97,000.

23. Use N = N 0 e rt (1). We know that N (5,730) = 12 N 0 (2). If N 0 = 5 (grams), find t such that N (t ) = 2 . To do so, we must first find r. To this end, substitute (2) into (1) to obtain: r (5,730 ) 1 1  5,730 ln ( 12 ) = r 2 =e

24. Use N = N 0 e rt (1). We know that N (1,600) = 12 N 0 (2). If N 0 = 5 (grams), find t such that N (t ) = 2 . To do so, we must first find r. To this end, substitute (2) into (1) to obtain: r (1,600 ) 1 1  1,600 ln ( 12 ) = r 2 =e

Now, solve for t:

2 = 5e

( )t

1 ln 1 5,730 2

2 = 5e 1,600

1 ln ( 25 ) = 5,730 ln ( 21 ) t

ln ( 52 )

1 5,730

ln ( 12 )

ln( 12 ) t

1 ln ( 25 ) = 1,600 ln ( 12 ) t

≈ 7575 years

ln ( 52 )

1 1,600

ln ( 21 )

25. Use N = N 0 e rt (1). We know that

Now, solve for t:

N (4.5 × 109 ) = 12 N 0 (2). Find t such that

0.98 N 0 = N 0 e 4.5×10

N (t ) = 0.98N 0 . To do so, we must first find r. To this end, substitute (2) into (1) to obtain:

1 2

= e r ( 4.5×10 ) 

1 4.5×109

ln ( 12 ) = r

699

≈ 2115 years

ln( 21 ) t

ln ( 0.98 ) 1 4.5×109

ln ( 12 )

≈ 131,158,556 years old

Chapter 5

26. Use N = N 0 e rt (1). We know that

1 2

= e12 r 

1 12

ln ( 12 ) = r

N (12) = 12 N 0 (2). Assuming that

Now, compute N (16) :

N 0 = 5, find N (16). To this end, substitute (2) into (1) to obtain:

N (16) = 5e 12

27. Use T = Ts + (T0 − Ts ) e − k t .

28. Use T = Ts + (T0 − Ts ) e − k t .

Here, T0 = 325, Ts = 72, and T (10) = 200 . Find T (30) . To do so, we must first find k. Observe 200 = 72 + ( 325 − 72 ) e − k (10 )

Here, T0 = 38, Ts = 75, and T (5) = 45 . Find T (20) . To do so, we must first find k. Observe 45 = 75 + ( 38 − 75 ) e − k (5)

128 = 253 e

k = − ln (

Now, T (30) = 72 + 253e

128 253

(

≈ 1.984 mg

−30 = −37 e −5 k

−10 k

1 10

( ) ⋅ 16

1 ln 1 2

30 k = − 15 ln ( 37 )

)

Now,

( )) ⋅ 30

1 ln 128 − − 10 253

T (20) = 75 − 37e



≈ 105 F .

(

( )) ⋅ 20

30 − − 15 ln 37

≈ 59 F .

29. Use T = Ts + (T0 − Ts ) e − k t (1).

30. Use T = Ts + (T0 − Ts ) e − k t (1).

Assume t = 0 corresponds to 7am. We know that T (0) = 85, T (1.5) = 82, Ts = 74. (2) Find t such that T (t ) = 98.6. We first use (2) in (1) to find k and T0 .

Assume t = 0 corresponds to 4am. We know T (0) = 90, T (1) = 86, Ts = 60. (2) Find t such that T (t ) = 98.6. We first use (2) in (1) to find k and T0 . 90 = 60 + (T0 − 60 ) e 0 = T0  T0 = 90

85 = 74 + (T0 − 74 ) e 0 = T0  T0 = 85 82 = 74 + ( 85 − 74 ) e

−1.5 k

 8 = 11e

1  k = − 1.5 ln ( 118 )

Now, solve for t: 98.6 = 74 + (85 − 74)e 24.6 = 11e

86 = 60 + ( 90 − 60 ) e − k  26 = 30e − k

−1.5 k

(

26  k = − ln ( 30 )

Now, solve for t: 98.6 = 60 + 30e

( ))t

1 ln 8 − − 1.5 11

38.6 = 30e

( )

1 ln 8 t 1.5 11

ln ( ) −3.8 ≈ 1 =t 8 1.5 ln ( 11 ) So, the victim died approximately 3.8 hours before 7am. So, by 8:30am, the victim has been dead for about 5.29 hours. 24.6 11

700

−1.76 ≈

( ( ))

26 t − − ln 30

( )

26 t ln 30

ln ( 38.6 30 )

=t 26 ln ( 30 ) So, the victim died approximately 1.76 hours before 4am. So, by 5am, the victim has been dead for about 2.76 hours.

Section 5.5

31. Use N = N 0 e − rt (1). We have N (0) = N 0 = 38,000

32. Use N = N 0 e − rt (1). We have N (0) = N 0 = 22,000

(2) N (1) = 32,000 In order to determine N (4) , we need to first determine r. To this end, substitute (2) into (1) to obtain: 32,000 = 38,000e − r  r = − ln ( 32,000 38,000 )

(2) N (2) = 14,000 In order to determine N (4) , we need to first determine r. To this end, substitute (2) into (1) to obtain: 14,000 = 22,000e − r (2 )  r = − 12 ln ( 14,000 22,000 )

Thus,

( (

− − ln 32,000 38,000

))

(

− − 1 ln 14,000

⋅4

N (4) = 38,000e ≈ 19,100 . The book value after 4 years is approximately $19,100.

)) ⋅ 4

N (4) = 22,000e 2 22,000 ≈ 8,900 . The book value after 4 years is approximately $8,900.

100,000 2,000,000 . 34. Use N = . −2t 1 + 10e 1 + 2e −4t 100,000 2,000,000 a. N (2) = a. N (2) = ≈ 84,520 ≈ 1,998,659 −2( 2 ) 1 + 10e 1 + 2e −4( 2) 100,000 2,000,000 b. N (30) = b. N (4) = ≈ 100,000 ≈ 2,000,000 −2(30 ) 1 + 10e 1 + 2e −4( 4 ) c. The highest number of new convertibles that will be sold is 100,000 since the smallest that 1 + 10e −2t can be is 1. 33. Use N =

35. Use N = N 0 e rt ( t measured in months) (1). Assuming that t = 0 (in months) corresponds to March 2020, we have N ( 0 ) = N 0 = 188,000 (2) N ( 2 ) = 1,600,000

In order to determine N ( 5 ) , we need to first determine r . To this end, substitute (2) into (1) to obtain: 1,600,000 = 188,000e ( )  r = r 2

Thus, N ( 5 ) = 188,000e

1  1,600,000  ln ⋅5 2  188,000 

1  1,600,000  ln 2  188,000 

≈ 39,724,934 cases.

701

Chapter 5

36. Use N = N o e rt (1). Assuming that t = 0 corresponds to 2004, we have N (0) = N 0 = 300,000

(2) N (3) = 630,000 In order to determine N (20), we first need to determine r. To this end, substitute (2) into (1) to obtain: 1  630,000  t 3 630,000 = 300,000e ( )  ln  3  300,000 

Thus, N (20) = 300,000e

1  630,000  ln ⋅20 3  300,000 

≈ 42,193,921 cases

10,000 = 5000 . 1 + 19e −1.56t 10,000 = 5000 (1 + 19e −1.56t )

38. Find t such that 1600 = 1200 . 1 + 0.6e −0.14t 1600 = 1200 (1 + 0.6e −0.14t )

37. Find t such that

2 = 1 + 19e −1.56t

400 = 720e −0.14t

1 = 19e −1.56t

400 1 t = − 0.14 ln ( 720 ) ≈ 4.20 years

e −1.56t = 191 1 t = − 1.56 ln ( 191 ) ≈ 1.89 years

39. Consider I ( r ) = e − r , whose graph is below. Note that the beam is brightest when r = 0 . 2

702

40. Since I (2) = e − 2 ≈ 0.0183 , about 1.83%. 2

Section 5.5

41.

42. 2 − ( x−752 ) 25

− ( x−802 )

Consider N ( x ) = 10e . a. The graph of N is as follows:

Consider N ( x ) = 10e . a. The graph of N is as follows: 16

b. Average grade is 80. 2

b. Average grade is 75.

c. N (60) = 10e

c. N (50) = 10e

) − (50−75 2 25

d. N (100) = 10e 43.

Use t = −

ln (1 −

−1

= 10e ≈ 4

)2 − (100−75 252

d. N (100) = 10e

44.

Use t = −

a. Here, P = 80,000, r = 0.09, n = 12, R = 750 . So, t=−

12 ln (1 +

0.09 12

)

) ≈ 18 years .

So, t = −

(

12 ln (1 +

0.09 12

)

≈2

Pr ln (1 − nR )

n ln (1 + nr )

a. Here, P = 20,000, r = 0.17, n = 12, R = 300 .

So, t = −

b. Here, P = 80,000, r = 0.09, n = 12, R = 1000 . 0.09) ln 1 − 80,000( 12(1000 )

≈2

)2 − (100−80 162

= 10e ≈ 4

). n ln (1 + nr )

(

−1

Pr nR

0.09) ln 1 − 80,000( 12(750 )

) − ( 60−80 2

(

0.17 ) ln 1 − 20,000( 12(300 )

12 ln (1 +

0.17 12

)

) ≈ 17.12 years .

b. Here, P = 20,000, r = 0.17, n = 12, R = 400 .

) ≈ 10 years .

So, t = −

703

(

0.17 ) ln 1 − 20,000( 12( 400 )

12 ln (1 +

0.17 12

)

) ≈ 7.30 years .

Chapter 5

45.

  i Use R = P  with − nt   1 − (1 + i )  R = 1,467, P = 200,000, i = 0.08 12 ≈ 0.006667 , and t is measured in years. a. Solve for t:   0.006667 1, 467 = 200,000    1 − (1 + 0.006667 )−12 t    −12 t 1, 467 − 1, 467(1.006667) = (0.006667)(200,000) 133.6 = 1, 467 (0.92336109)t 0.0910702 = (0.92336109)t ln(0.0910702) ≈ 30 years ln(0.92336109) b. Total amount paid = $1,467(30)(12) = $528,120. Since the original loan is $200,000, you’ve paid $328,120 in interest. t=

46.

  i Use R = P  with − nt   1 − (1 + i )  R = 2,934, P = 200,000, i = 0.08 12 ≈ 0.006667 , and t is measured in years. a. Solve for t:   0.006667 2,934 = 200,000    1 − (1 + 0.006667 )−12 t    −12 t 2,934 − 2,934(1.006667) = (0.006667)(200,000) 1600.6 = 2934 (0.92336109)t 0.545535106 = (0.92336109)t ln(0.545535106) ≈ 7.6 years ln(0.92336109) b. Total amount paid = $2,934(7.6)(12) = $267,580.80. Since the original loan is $200,000, you’ve paid $67,581 in interest. c. $328,120 − $67,581 = $260,539 is the amount of interest saved by doubling the payment. t=

704

Section 5.5

47. r = 0.07, not 7

48. r = 0.05, not 5

49. True

50. False. Should use logistic curve here.

51. False (Since there is a finite number of students at the school to which the lice can spread.)

52. True

53. Take a look at a couple of graphs for increasing values of c. For definiteness, let a = k = 1, and take c = 1 and c = 5, respectively. The graphs are:

54. Take a look at a couple of graphs for increasing values of k. For definiteness, let a = c = 1, and take k = 1 and k = 5, respectively. The graphs are:

As c increases, the model reaches the carrying capacity in less time.

As k increases, the model reaches the carrying capacity in less time.

705

Chapter 5

55. a. The graphs are below: For the same periodic payment, it will take Wing Shan fewer years to pay off the loan if she can afford to pay biweekly.

56. The graphs are below: a. For the same periodic payment, it will take Hong fewer years to pay off the loan if he can afford to pay biweekly.

b. 11.58 years c. 10.33 years d. 8.54 years, 7.69 years, respectively.

b. 5.12 years c. 4.50 years d. 3.35 years, 2.99 years, respectively.

706

Chapter 5 Review Solutions ----------------------------------------------------------------------1. 17,559.94

2. 1.58

3. 5.52

4. 1.24

5. 24.53

6. 23.14

7. 5.89

8. 0.01

9. 2 4 −( −2.2 ) = 26.2 ≅ 73.52

10. −21.3+ 4 = −25.3 ≅ −39.40

11. ( 52 )

12. ( 74 ) 5

1−6( 12 )

= ( 52 ) = 6.25 −2

5( 1 ) +1

= ( 74 ) = 16 49 2

13. b y-intercept (0, 14 )

14. a Note that y = −4( 12 )x and y-intercept (0, −4)

15. c y-intercept (0,11)

16. d y-intercept (0, −11)

17. y-intercept: (0, −1) HA: y = 0

18. y-intercept: (0,3) . HA: y = 4 Reflect the graph of 3x over the x-axis, then shift up 4 units.

Reflect the graph of ( 61 ) over the xx

axis.

707

Chapter 5

19. y-intercept: (0,2). HA: y = 1 1 Shift the graph of ( 100 ) up 1 unit. x

20. y-intercept: (0, −3). HA: y = −4 Shift the graph of 4 x down 4 units.

21. y-intercept: (0,1). HA: y = 0

22. y-intercept: (0, e −1 ). HA: y = 0

23. y-intercept: (0,3.2). HA: y = 0

24. y-intercept: (0,2 − e ). HA: y = 2

708

Chapter 5 Review

25. Use A(t ) = P (1 + nr )

26. Use A(t ) = P (1 + nr )

Here, P = 4500, r = 0.045, n = 2, t = 7 . Thus, 2(7 ) A(7) = 4500 (1 + 0.045 ≅ 6144.68. 2 ) So, the amount in the account after 7 years is $6144.68.

Here, A(8) = 25,000, r = 0.04, n = 4, t = 8 . Thus, solving the above formula for P, 25,000 we have P = ≅ 18,182.60 . 4(8) (1 + 0.044 ) So, the initial investment should be $18,182.60.

27. Use A(t ) = Pe rt . Here, P = 13, 450, r = 0.036, t = 15 . Thus, A(15) = 13, 450e (0.036 )(15) ≅ 23,080.29 . So, the amount in the account after 15 years is $23,080.29.

28. Use A(t ) = Pe rt . Here, A(10) = 15,000, r = 0.025, t = 10 . Thus, solving the above formula for P, 15,000 we have P = (0.025)(10 ) ≅ 11,682.01 . So, e the initial investment should be $11,682.01.

29. 43 = 64

30. 4 2 = 2

1 31. 10 −2 = 100

32. 16 2 = 4

33. log 6 216 = 3

34. log10 0.0001 = −4

4 35. log 2 ( 169 )=2

36. log512 8 = 13

38.

37. log 7 1 = x 7 =1 x

x=0

39.

log 4 256 = x 4 = 256 x=4 x

log 1 1296 = x 6

( 61 ) = 1296 = 64 x

x = −4

40. log10 1012 = 12 log10 10 = 12

41. 1.51

42. 3.47

43. −2.08

44. –0.90

45. Must have x + 2 > 0 , so that the domain is ( −2, ∞ ) .

709

Chapter 5

46. Must have 2 − x > 0 , so that the domain is ( −∞,2) .

47. Since x 2 + 3 > 0 , for all values of x, the domain is ( −∞, ∞ ) .

48. Must have 3 − x 2 = ( 3 − x )( 3 + x ) > 0 .

49. b

CPs: − 3, 3

− + −   | | −

(

)

So, the domain is − 3, 3 .

50. a Reflect the graph of log 7 x over the x-axis, then over the y-axis

51. d Shift the graph of log 7 x left 1 unit, then down 3 units. Also, VA is x = −1 .

52. c VA is x = 1

53. Shift the graph of log 4 x right 4 units, then up 2 units.

54. Shift the graph of log 4 x left 4 units, then down 3 units.

710

Chapter 5 Review

55. Reflect the graph of log 4 x over the x-axis, then shift down 6 units.

56. Reflect the graph of log 4 x over the y-axis, then over the x-axis, then expand vertically by a factor of 2, and then shift up 4 units.

57. Use pH = − log10  H +  .

58. Use pH = − log10  H +  .

Here, pH = − log10 (3.16 ×10 −7 )

Here, pH = − log10 (2.0 ×10 −3 )

= −  log(3.16) + log(10 −7 ) 

= −  log(2.0) + log(10 −3 ) 

= − [ log(3.16) − 7 ]

= − [ log(2.0) − 3]

≅ 6.5

≅ 2.7

59.

60.

 I  Use D = 10 log  . −12   1× 10  Here,  1× 10 −7  D = 10 log  = 10 log(105 ) −12  1 10 ×  

 I  Use D = 10 log  . −12   1× 10  Here,  1× 10 −4  D = 10 log  = 10 log(108 ) −12  1 10 ×  

= 50 log(10) = 50 dB

61. 1

= 80 log(10) = 80 dB

63. 6

62. log 2 16 = x 2 x = 16 = 4 x=2

711

Chapter 5 −3

64. e −3ln6 = e ln6

1 = 6 −3 = 216

65.

log c ( x a y b ) = log c ( x a ) + log c ( y b )

= a log c ( x ) + b log c ( y )

66. log 3 ( x 2 y −3 ) = log3 ( x 2 ) + log3 ( y −3 )

67.

68.

69.

 rs  log j  3  = log j ( rs ) − log j ( t3 ) t  = log j ( r ) + log j ( s ) − 3 log j ( t )

= 2 log 3 ( x ) − 3 log 3 ( y )

(

log x

)

x + 5 = log ( x ) + log

( x +5)

= log ( x ) + log ( x + 5 ) c

= c log x + 12 log ( x + 5 )

1 2

 a2  3 2 1 log  3 2  = log a 2 − log b 2 c 5  b2c5    3 2 1 = log a 2 −  log b 2 + log c 5    = 12 log ( a ) − 32 log ( b ) − 25 log ( c ) 1

( )

70. 1

 c 3d 3  3 1  c3d 3  log 7  6  = 3 log 7  6   e   e      1 = 13  log 7 c3d 3 − log 7 ( e 6 )    1 = 13  log 7 ( c3 ) + log 7 d 3 − log 7 ( e 6 )    = 13 3 log 7 ( c ) + 13 log 7 ( d ) − 6 log 7 ( e )  1

( )

72.

71. log 3 log 8 ≅ 0.5283

log 8 3 =

( )

73. 1 2

log log 5 ≅ −0.4307 log5 12 =

74. log1.4 log π ≅ 0.2939

logπ 1.4 =

712

log 2.5 log 3 ≅ 1.6681

log 3 2.5 =

Chapter 5 Review

75.

76. 2

1 4 = 256 x 4 = 4 −4 x = −4

3x = 81 = 34

x2 = 4 x = ±2 78.

77.

e x = e 4.8

e3x −4 = 1 = e 0 3x − 4 = 0 x=

x = 4.8

4 3

x = (4.8)2 = 23.04

80.

79.

( )

1 x+2 3

(3−1 )

x+2

= 81

100 x −3 = 10

= 34

(10 )

− x −2

2 2 x −3

= 10

So,

102( x −3) = 10

3 =3 −x − 2 = 4

2( x 2 − 3) = 1

x = −6

2x2 − 7 = 0

81.

x 2 = 72 x=±

7 2

82. e

2 x +3

− 3 = 10

22 x −1 + 3 = 17

e 2 x +3 = 13

2 2 x −1 = 14

log 2 ( 2 2 x −1 ) = log 2 (14)

2 x + 3 = ln13 x = −3+2ln13 ≈ −0.218

2 x − 1 = log 2 (14) x = 1+ log22 (14) ≅ 2.404

83. Note that e 2 x + 6e x + 5 = 0 is equivalent to

( e ) + 6 ( e ) + 5 = ( e + 5 )( e + 1) = 0 . x 2

x x Neither e + 5 =0 or e 1 =0 has a real solution. So, the original equation has  +

No solution

no solution.

713

Chapter 5

84.

85. 4e

0.1x

= 64

( 2 − 2 )( 2 + 2 ) = 0 (2 ) − (2 ) = 0 x

e 0.1x = 16

−x

x 2

0.1x = ln(16)

−x 2

2 2 x − 2 −2 x = 0

x ≈ 27.726

2 2 x = 2 −2 x 2 x = −2 x x=0

86.

87.

5 ( 2 ) = 25 x

10 2 = 3x 100 = 3x

2x = 5 x = log 2 5 ≅ 2.3219

88. log3 ( x + 2) = 4

log(3x ) = 2

100 3

89. log 4 ( x ) + log 4 2x = 8

log 4 ( 2 x 2 ) = 8

x + 2 = 34 x = 79

2 x 2 = 48

x 2 = 12 ( 48 ) x=±

1 2

(4 ) 4

x = 128 2 , −128 2 90.

91. log 6 ( x ) + log 6 (2 x − 1) = log 6 (3)

ln x 2 = 2.2

log 6 ( x(2 x − 1) ) = log 6 (3)

x 2 = e 2.2

x(2 x − 1) = 3

x = ± e 2.2 ≈ ±3.004

2x 2 − x − 3 = 0 (2 x − 3)( x + 1) = 0 x = 32 , −1

714

Chapter 5 Review

92.

93.

ln(3x − 4) = 7

log3 (2 − x ) − log3 ( x + 3) = log3 ( x )

3x − 4 = e

log3 ( 2x−+x3 ) = log 3 ( x )

3x = 4 + e 7 x=

4 + e7 3

2−x x +3

2 − x = x( x + 3)

≅ 366.88

x 2 + 4x − 2 = 0 x = −4 ± 216 +8 x = −2 + 6, −2 − 6 ≈ 0.449 94. 4 log( x + 1) − 2 log( x + 1) = 1 log( x + 1) = 0.5 x + 1 = 10 0.5 x = 10 0.5 − 1 ≈ 2.162

95. Use A(t ) = Pe rt . Here, A = 30,000, r = 0.05, t = 1 Substituting into the above equation, we can solve for P: 30,000 = Pe 0.05(1)

28,536.88 ≅ 30,000 =P e 0.05 So, the initial investment in 1 year CD should be approximately $28,536.88.

96. Use A(t ) = Pe rt . Here,

A = 4000, t = 2, P = 2800 (100 × 28shares) Substituting into the above equation, we can solve for r: 4000 = 2800e r (2 ) 10 7

2r = 4000 2800 = e

0.178 ≅ 12 ln( 107 ) = r So, about 17.8%.

715

Chapter 5

97. Use A(t ) = P (1 + nr )

Here, we know that r = 0.042, n = 4, A = 2P . We can substitute these in to find t. 2P = P (1 + 0.042 4 ) 2 = (1.0105 )

4( t )

98. Use A(t ) = Pe rt . Here, A = 22,500, r = 0.08, P = 9000 Substituting into the above equation, we can solve for t: 22,500 = 9000e 0.08t 22,500 9000

log1.0105 2 = 4t 16.6 ≅ 14 log1.0105 2 = t So, it takes approximately 16.6 years until the initial investment doubles. 99. Use N = N 0 e rt .

Here, N 0 = 3.08, r = 0.0175 . Determine N when t = 6 : N = 3.08e 0.0175⋅6 = 3.42 million The population in 2024 will be about 3.42 million.

= e 0.08t

1 11.453 ≅ 0.08 ln ( 22,500 9000 ) = t

So, it takes approximately 11.453 years.

100. Use N = N 0 e rt (1). We have N ( 0 ) = N 0 = 28.3

N ( 4 ) = 32.5

(2)

In order to determine N ( 28 ) , we need to first determine r. To this end, substitute (2) into (1) to obtain: 1  32.5  r 4 32.5 = 28.3e ( )  r = ln   4  28.3  Thus, N ( 28 ) = 28.3e

716

1  32.5  ln ⋅28 4  28.3 

≈ 74.55 million.

Chapter 5 Review

101. Use N = N 0 e rt (1). We have N (0) = N 0 = 1000

(2) N (3) = 2500 In order to determine N (6) , we need to first determine r. To this end, substitute (2) into (1) to obtain: 2500 2500 = 1000e r (3)  r = 13 ln ( 1000 )

Thus,

(

1 ln 2500 1000

N (6) = 1000e 3

)⋅ 6

102. Use N = N 0 e rt (1). We have N ( 0 ) = N 0 = 1,388,215 N (1) = 1, 418,041 (2)

In order to determine N ( 21) , we need to first determine r. To this end, substitute (2) into (1) to obtain: r1 1, 418,041 = 1,388,215e ( )   1, 418,041  r = ln    1,388,215 

≈ 6250 bacteria . Thus,

 1,418,041  ln ⋅21

N ( 21) = 1,388,215e  1,388,215 

≈ 2,169,351. 103. Use N = N 0 e − rt . We know that N (28) = 12 N 0 . Assuming that N 0 = 20 , determine t such that N = 5 . To do so, we first find r: − r (28) 1  r = − 281 ln ( 12 ) 2 N0 = N0e Now, solve: − ( − 1 ln( 1 ) )t 5 = 20e 28 2 1 4

104. Use N = N 0 e − rt . We know that N (25,000) = 12 N 0 . Assuming that N 0 = 100 , determine t such that N = 20 . To do so, we first find r: − r (25,000 ) 1 1  r = − 25,000 ln ( 12 ) 2 N0 = N0e Now, solve:

20 = 100e

( )

1 ln 1 t 2

= e 28

ln( ) 1 4

t = 1 ln( 1 ) ≈ 56 years 28

717

(

)

1 ln 1 t − − 25,000 (2)

ln( 51 )

( )

1 ln 1 25,000 2

≈ 58,048.2 years

Chapter 5

105. Use N = N 0 e − rt . Assuming that 2003 occurs at t = 0 , we know that N 0 = 5600, N (1) = 2420 . Find N (7) . To do so, we first find r: 2420 2420 = 5600e − r (1)  r = − ln ( 5600 )

Now, solve: N (7) = 5600e

(

)

2420 (7) − − ln( 5600 )

106. Use N = N 0 e − rt . We know that N 0 = 28,200, N (2) = 24,500 . Find N (6) . To do so, we first find r:

24,500 = 28,200e − r (2 )  r = − 12 ln ( 24,500 28,200 ) Now, solve: N (6) = 28,200e

≈ 16 fish

107. Use M = 1000 (1 − e −0.035 t ) .

(

))

− − 12 ln 24,500 (6 ) 28,200

≈ $18, 492.69 108. Use N = N 0 e rt .

Since t = 0 corresponds to 1998, we have M (12) = 1000 (1 − e −0.035 (12) ) ≈ 343 mice .

Here, N 0 = 50,000 r = 0.023. Determine N when t = 60 : N = 50,000e 0.023(60 ) ≈ 195,745 The population in 2030 is about 195,745.

109. The graph is as follows:

110. Let y1 = e − x + 2 , y2 = 3x + 1 . The graphs are as follows:

Using the calculator to compute the functional values for large values of x suggests that the HA is about y = 4.11 . (Note: The exact equation of the HA is y = e 2 .)

The coordinates of the point of intersection are about (0.785, 3.369).

718

Chapter 5 Review

111. Let 112. Let y1 = log 2.4 (3x − 1), y2 = log 0.8 ( x − 1) + 3.5. y1 = log 2.5 ( x − 1) + 2 , y2 = 3.5x −2 . The graphs are as follows: The graphs are as follows:

The coordinates of the point of intersection are about (2.376, 2.071).

113. The graphs agree on ( 0, ∞ ) , as seen below.

The intersection points occur at (28.09, 2.753) and (1.227, 0.379). 114. The graphs agree on (1, 3), as seen below.

719

Chapter 5

115. The graph is below. Domain: ( −∞, ∞ ) . Symmetric about the origin. Horizontal asymptotes: y = −1 (as x → −∞ ) , y = 1 (as x → ∞ )

116. The graph is below. Domain: ( −∞,0) ∪ (0, ∞ ) . Symmetric about the origin. Horizontal asymptote: y = 0 Vertical asymptote: x = 0

117. a. Using N = N 0 e rt with (0,4) and (18,2), we need to find r: 2 = 4e18r  r = 181 ln ( 12 ) ≈ −0.038508 So, the equation of dosage is given by N = 4e −0.038508t ≈ 4(0.9622)t . b. N = 4(0.9622)t c. Yes, they are the same.

118. a. Using N = N 0 e rt with (0,5600) and (1,2420), we need to find r: 2420 2420 = 5600e r  r = ln ( 5600 ) ≈ −0.8390 So, the equation of dosage is given by N = 5600e −0.8390 t ≈ 5600(0.4321428571)t . b. N (7) ≈ 15.76 , so you would expect 16 fish in 2010. c. Yes, they are the same.

720

Chapter 5 Practice Test---------------------------------------------------------------------------3

1. log10 x = x3 log10 = x3

log5 326 =

log 326 ≅ 3.60 log 5

log 13 81 = x

 e5 x  5x 4 ln   = ln( e ) − ln ( x( x + 1) ) 4 ( 1) + x x  

34 = ( 13 ) = 3− x x

= 5x − ( ln x + ln( x 4 + 1) )

−4 = x

= 5x − ln x − ln( x 4 + 1)

6. Note that e 2 x − 5e x + 6 = 0 is equivalent to

5. 2

e x −1 = 42

( e ) − 5 ( e ) + 6 = ( e − 3)( e − 2 ) = 0 . x 2

x 2 − 1 = ln 42

Thus,

x 2 = 1 + ln 42

e x − 3 = 0 or e x − 2 = 0

x = ± 1 + ln 42 ≈ ±2.177

ex = 3 x = ln3

ex = 2 x = ln 2

27e

0.2 x +1

32 x −1 = 15

= 300

2 x − 1 = log 3 15

e 0.2 x +1 = 300 27

x = 12 (1 + log 3 15 ) ≈ 1.732

0.2 x + 1 = ln( 300 27 ) 0.2x = −1 + ln( 300 27 ) x=

) −1+ ln( 300 27 0.2

≅ 7.04

3 ln( x − 4) = 6 ln( x − 4) = 2 x − 4 = e2 x = 4 + e 2 ≈ 11.389

721

Chapter 5

10. log(6 x + 5) − log(3) = log(2) − log( x )

log (

6 x +5 3

11.

) = log( )

6 x +5 3

ln(ln x ) = 1 ln x = e

2 x

x = e e ≈ 15.154

= x2

6 x 2 + 5x = 6 6 x 2 + 5x − 6 = 0 (3x − 2)(2 x + 3) = 0  x = 23 , − 32 12. log 2 (3x − 1) − log 2 ( x − 1) = log 2 ( x + 1)

13. log 6 x + log 6 ( x − 5) = 2

log 2 ( 3xx−−11 ) = log 2 ( x + 1)

log 6 ( x( x − 5) ) = 2

= x +1

3 x −1 x −1

x 2 − 5x = 36

3x − 1 = ( x + 1)( x − 1)

x 2 − 5x − 36 = 0

= x2 − 1

( x − 9)( x + 4) = 0

x 2 − 3x = 0  x( x − 3) = 0

x = −4 ,9

 x = 0, 3

14.

15.

ln( x + 2) − ln( x − 3) = 2

ln ( x( x + 3) ) = 1

 x+2 ln  =2  x −3  x+2 = e2 x −3 x + 2 = e 2 x − 3e 2

x 2 + 3x = e x 2 + 3x − e = 0 −3 ± 9 − 4( −e ) 2 ≈ 0.729, −3.729

( e − 1) x = 3e + 2 2

ln x + ln( x + 3) = 1

3e 2 + 2 ≈ 3.783 e2 − 1

722

Chapter 5 Practice Test

16.

17.

 2x + 3  log 2  =3  x −1  2x + 3 =8 x −1 2 x + 3 = 8x − 8 6x = 11

12 =6 1 + 2e x 12 = 6 + 12e x 6 = 12e x

x = ln ( 12 ) ≈ −0.693

x = 116 19. Must have x2x−1 > 0 and x 2 − 1 ≠ 0

18. ln x + ln( x − 3) = 2 ln x( x − 3) = 2 x − 3x = e 2

CPs −1,0,1 − + − +  | | |

−1

x − 3x − e = 0 2

So, the domain is ( −1,0 ) ∪ (1, ∞ ) .

3 ± 9 − 4(1)( −e ) ≈ 4.605, −1.605 2 2

20.

In order for the identity 10 log( 4 x −a ) = 4x − a to hold, we must have 4 x − a > 0  x > a4 . 21. y-intercept: f (0) = 3−0 + 1 = 2 . So, (0,2). x-intercept: None Domain: ( −∞, ∞ ) Range: (1, ∞ ) Horizontal Asymptote: y = 1 The graph is as follows:

723

22. 0 y-intercept: f (0) = ( 12 ) − 3 = −2 .

So, (0,−2). x-intercept: (−1.5850, 0) since x 0 = ( 12 ) − 3  x = log ( 1 ) 3 ≈ −1.5850 2

Domain: ( −∞, ∞ ) Range: ( −3, ∞ ) Horizontal Asymptote: y = −3 The graph is as follows:

Chapter 5

24. y-intercept: f (0) = log(1) + 2 = 2. So, (0,2). x-intercept: Must solve log(1 − x ) + 2 = 0. log(1 − x ) + 2 = 0  log(1 − x ) = −2

23. y-intercept: None x-intercept: Must solve ln(2 x − 3) + 1 = 0. ln(2 x − 3) = −1  2 x − 3 = 1e  x=

3 + 1e 2

1 − x = 10 −2

 3 + 1e  ,0  . So,   2  Domain: ( 32 , ∞ ) Range: ( −∞, ∞ ) Vertical Asymptote: x = 32 The graph is as follows:

25. Use A(t ) = P (1 + nr )

99  x = 100

99 ,0 ) . So, ( 100 Domain: ( −∞,1) Range: ( −∞, ∞ ) Vertical Asymptote: x = 1 The graph is as follows:

26. Use A(t ) = Pe rt . Here, P = 10,000, r = 0.05, t = 10 Thus, A(10) = 10,000e (0.05)(10 ) ≅ 16, 487.21 . So, after 10 years, to amount is approximately $16,487.

Here, P = 5000, r = 0.06, n = 4, t = 8 . Thus, A(8) = 5000 (1 + 0.06 4 )

4(8)

≅ 8051.62 .

So, the amount in the account after 8 years is $8051.62.

724

Chapter 5 Practice Test

28. Use N ( t ) = N 0 e rt .

27.

 I  Use D = 10 log  . −12   1× 10  Here,  1× 10 −3  D = 10 log  = 10 log(109 ) −12   1×10 

Here, N 0 = 800,000, r = 0.05 t = 20 ( 2020 − 2004 )

Thus, 0.05 20 N ( 20 ) = 800,000e ( )( ) ≈ 2,174,625 . So the population in 2024 is approximately 2,174,625.

= 90 log(10) = 90 dB

( )

For M = 6 :

29. Use M = 23 log 10E4.4 . Here,

For M = 5 :

( ) 9 = log ( )

6 = 23 log 10E4.4

( ) 7.5 = log ( ) 5 = 23 log 10E4.4

104.4

109 = 10E4.4

104.4

109 ×10 4.4 = E  

107.5 = 10E4.4

=1013.4

107.5 ×10 4.4 = E 

So, the energy here is 1013.4 ≈ 2.5 ×1013 joules.

=1011.9

So, the energy here is 1011.9 ≈ 7.9 ×1011 joules.

Thus, the range of energy is 7.9 × 1011 < E < 2.5 ×1013 joules. 30.

30 = 50e −0.0578t 3 5

= e −0.0578t

ln( 35 ) = −0.0578t 1 8.8 ≅ − 0.0578 ln( 53 ) = t

So, about 9 hours.

725

Chapter 5

31. Use N = N 0 e rt (1). We have N (0) = N 0 = 200

(2) N (2) = 500 In order to determine N (8) , we need to first determine r. To this end, substitute (2) into (1) to obtain: 500 = 200e r (2)  r = 12 ln ( 500 200 )

Thus, N (8) = 200e

( )⋅ 8

1 ln 500 2 200

2000 . 1 + 3e −0.4 t 1000(1 + 3e −0.4 t ) = 2000 1 + 3e

Now, solve: 40 = 100e

1 3

−0.4t = ln ( 13 )

1 t = − 0.4 ln ( 13 ) ≈ 3 days

(

)

1 ln 1 t − − 5730 (2)

ln( 40 )

t = 1 ln100( 1 ) ≈ 7574.65 years 5730

34. Use N ( t ) = N 0 e rt (1).

We have

3e −0.4 t = 1 −0.4 t

We know that N (5730) = 12 N 0 . Assuming that N 0 = 100 , determine t such that N = 40 . To do so, we first find r: − r (5730 ) 1 1  r = − 5730 ln ( 21 ) 2 N0 = N0e

≈ 7800 bacteria .

33. Solve 1000 = −0.4 t

32. Use N = N 0 e − rt .

N ( 0 ) = N 0 = 76 N ( 2 ) = 83

(2)

In order to determine N ( 22 ) , we need to first determine r. To this end, substitute (2) into (1) to obtain: 1  83  83 = 76e r ( 2 )  r = ln   2  76  Thus, N ( 22 ) = 76e

barrels.

726

1  83  ln ⋅22 2  76 

≈ 200.32 million

Chapter 5 Practice Test

35. The graph is below. Domain: ( −∞, ∞ ) . Symmetric about the origin. No asymptotes

36. Let y1 = 43− x , y2 = 2 x − 1 . The graphs are as follows:

The intersection occurs about when x = 2.22 .

727

Chapter 5 Cumulative Review -------------------------------------------------------------------1.

2. 1

8 5 1 2 x y 1 = x 3 x 2 y 5 y 5 = x 6 y2 − 12 − 8 5 x y 3

2 x −2

3−

2 x −2

2 +3 x −6

= 3 xx−−62−2 = x −2

3x − 4 3x − 8

5x − 4 x − 3 = 0 2

2 x + 13 = 2 + x + 3

(

4 ± 16 − 4(5)(−3) 2(5)

2 x + 13 = 2 + x + 3

x +6 = 4 x +3

(x + 6) = (4 x + 3 ) 2

x+4 ≤ −3 5 x + 4 ≤ −15 x ≤ −19

x 2 + 12 x + 36 = 16( x + 3) x 2 − 4 x − 12 = 0 ( x − 6 )( x + 2) = 0

x = −2, 6

( −∞, −19]

7. The line 4 x + 3 y = 6 is equivalent to y = − 34 x + 2 and so, has slope − 34 . A line perpendicular to it must have slope 3 4 . Since the desired line must pass through (7,6), we have y − 6 = 34 ( x − 7)  y = 34 x + 34

6. x −4 =9 2

x 2 − 4 = −9 or x 2 − 4 = 9 or x 2 = 13  x = ± 13

No solution

f ( x + h ) − f ( x ) ( 4( x + h ) − ( x + h ) ) − ( 4 x − x ) = h h 2 2 4 x + 4 h − ( x + 2 hx + h ) − 4 x + x 2 h(4 − 2 x − h ) = = = 4 − 2x − h h h 2

2 x + 13 = 4 + 4 x + 3 + ( x + 3)

2 ± 19 = 5

x 2 = −5

)

9. a. 1 b. 5 c. 1 d. undefined e. Domain: ( −2, ∞ ) Range: (0, ∞ ) f. Increasing on (4, ∞ ) Decreasing on (0,4) Constant on (−2,0)

728

Chapter 5 Cumulative Review

10. Shift the graph of y = x left 1 unit, and then reflect over y-axis.

12. The general equation is R = dk2 . To

11. Yes, f is one-to-one since f (x) = f ( y)  x − 4 = y − 4

find k, use R = 3.8, d = 0.02 . Observe k 3.8 = (0.02  k = 0.00152 . )2

 x−4 = y−4  x=y

. So, R = 0.00152 d2

13.

14. P ( x ) = C ( x + 5)2 ( x − 9)4 , where C is any constant.

f ( x ) = −4 x + 8x − 5 2

= −4 ( x 2 − 2 x ) − 5

= −4 ( x 2 − 2 x + 1) − 5 + 4 = −4( x − 1)2 − 1 So, the vertex is (1,−1). 15. Rewrite as: ( −x 4 − 4x3 + 3x2 + 7x − 20 ) ÷ ( x − ( −4))

Synthetic division then gives −4 − 1 −1

−4

− 20

− 12

−5

So, Q( x ) = −x + 3x − 5, r( x ) = 0 . 3

729

Chapter 5

16. Since 2 + i is a zero, so is its conjugate 2 − i . Hence, ( x − (2 + i ))( x − (2 − i )) = x 2 − 4 x + 5 divides P ( x ) evenly. Indeed, note that

x2 − 4x + 5

x 2 − 3x − 4 x 4 − 7 x3 + 13x 2 + x − 20

− ( x 4 − 4 x3 + 5x 2 ) − 3x 3 + 8x 2 + x

− ( −3x3 + 12 x 2 − 15x ) − 4 x 2 + 16 x − 20

− ( −4 x 2 + 16 x − 20 ) 0 So, P ( x ) = ( x − (2 + i ))( x − (2 − i ))( x − 4)( x + 1). 17. Vertical asymptote: x = 3 Horizontal asymptote: None Slant asymptote: y = x + 3 since x +3 2 x − 3 x + 0x + 7

18. The graph is as follows:

− ( x 2 + 0x − 9 ) 16

Vertical Asymptote: x = −1 Horizontal Asymptote: y = 3 19. ( 251 ) 2 = 25 2 = 53 = 125 −3

20. Use A = P (1 + nr ) . nt

We have P = 5400, r = 0.0275, n = 12, t = 4 . Thus, A ≈ $6,027.14 .

730

Chapter 5 Cumulative Review

21. log3 243 = log3 ( 35 ) = 5

22. 1 2

ln( x + 5) 2 −  ln( x + 1)2 + ln(3x ) = 1

23. 2 10 2 log( 4 x +9) = 10 log( 4 x +9) = (4 x + 9)2 = 121 4 x + 9 = ±11 4 x = −9 ± 11

ln( x + 5) 2 −  ln ( ( x + 1)2 (3x ) )  = 1

 ( x + 5) 2  ln   2  ( x + 1) (3x )  1

x = 14 ( −9 ± 11) = −5 , 0.5

24. 2

5x = 625 2

5x = 54

25. Use A = Pe rt . We know that P = 8500 and r = 0.04 . Determine t such that A = 12,000 . 12,000 = 8500e 0.04 t

x =4 x = ±2 2

26. Let y1 = e 3−2 x , y2 = 2 x −1 . The graphs are below:

ln( x + 5) − 2 ln( x + 1) − ln(3x ) =

1 ln ( 12,000 t = 0.04 8,500 ) ≈ 8.62 years

27. a. Using N = N 0 e rt with (0,6) and (28,3), we need to find r: 3 = 6e 28r  r = 281 ln ( 12 ) ≈ −0.247553

So, the equation of dosage is given by N = 6e −0.247553t ≈ 6(0.9755486421)t . b. N (32) ≈ 2.72 grams

The solution is approximately 1.21.

731

CHAPTER 6 Section 6.1 Solutions -------------------------------------------------------------------------------1.

5x − 21 = 14 x=7 Substitute x = 7 into (1) to find

y : y = 0. So the solution is ( 7,0 ) .

4 y + 15 = 3 y = −3 Substitute y = −3 into (1) to find x : x = −1. So the solution is ( −1, −3 ) .

(1) y = x − 7 Solve the system:  2 x + 3 y = 14 (2) Substitute (1) into (2) and solve for x. 2 x + 3 ( x − 7 ) = 14

(1) x = 2 y + 5 Solve the system:  3x − 2 y = 3 (2) Substitute (1) into (2) and solve for x. 3 ( 2 y + 5) − 2 y = 3

(1)  y = 3x − 5 Solve the system:  (2)  y = 7x + 3 Substitute (1) into (2) and solve for x. 3x − 5 = 7x + 3

(1) x = 6 y + 2 Solve the system:  (2) x = 4 y − 8 Substitute (1) into (2) and solve for y. 6y + 2 = 4y − 8

−4 x = 8 x = −2 Substitute x = −2 into (1) to find y :

2 y = −10 y = −5 Substitute y = −5 into (1) to find x : x = −28. So the solution is ( −28, −5 ) . Note: substituting y = −5

y = −11. So the solution is ( −2, −11) .

Note: substituting x = −2 into (2) would yield the same result.

into (2) would yield the same result.

 x − y = 1 (1) Solve the system:   x + y = 1 ( 2) Solve (1) for x: x = 1 + y (3) Substitute (3) into (2) and solve for y: (1 + y ) + y = 1

 x − y = 2 (1) Solve the system:   x + y = −2 ( 2 ) Solve (1) for x: x = 2 + y (3) Substitute (3) into (2) and solve for y: (2 + y ) + y = −2

1+ 2y = 1 y=0

2 y = −4 y = −2

732

Section 6.1

Substitute y = 0 into (1) to find x: x = 1 . So, the solution is (1,0 ) .

Substitute y = −2 into (1) to find x: x − ( −2) = 2, so x = 0. So, the solution is ( 0, −2 ) .

So, the solution is ( 8, −1) .

y=7 Substitute y = 7 into (1) to find x: x = −3. So, the solution is ( −3,7 ) .

10.

x + y = 7 (1) Solve the system:  x − y = 9 ( 2) Solve (1) for y: y = 7 − x (3) Substitute (3) into (2) and solve for x: x − (7 − x ) = 9 x−7+ x = 9 x=8 Substitute x = 8 into (1) to find y: y = −1.

2 x − y = 3 (1) Solve the system:   x − 3y = 4 (2) Solve (2) for x: x = 3 y + 4 (3) Substitute (3) into (1) and solve for y: 2(3 y + 4) − y = 3 5 y = −5 y = −1 Substitute y = −1 into (1) to find x: 2 x − ( −1) = 3 2x = 2 x = 1. So, the solution is (1, −1) .

 x − y = −10 (1) Solve the system:  (2) x + y = 4 Solve (1) for x: x = y − 10 (3) Substitute (3) into (2) and solve for y: ( y − 10) + y = 4 2 y = 14

4 x + 3 y = 3 (1) Solve the system:   2x + y = 1 ( 2 ) Solve (2) for y: y = 1 − 2x (3) Substitute (3) into (1) and solve for x: 4 x + 3(1 − 2 x ) = 3 4x + 3 − 6x = 3 −2x = 0 x=0 Substitute x = 0 into (3) to find that y =1

So, the solution is ( 0,1) .

733

Chapter 6

11.

12.

So, the solution is (1,2 ) .

So, the solution is ( −2,3 ) .

13.

14.

(1)  3x + y = 5 Solve the system:  2 x − 5 y = −8 ( 2 ) Solve (1) for y: y = 5 − 3x (3) Substitute (3) into (2) and solve for x: 2x − 5(5 − 3x ) = −8 2x − 25 + 15x = −8 17 x = 17 x =1 Substitute x = 1 into (3) to find y: y = 2

2u + 5v = 7 (1) Solve the system:  3u − v = 5 ( 2 ) Solve (2) for v: v = 3u − 5 (3) Substitute (3) into (1) and solve for u: 2u + 5(3u − 5) = 7 17u = 32 32 u = 17 32 into (2) to find v: Substitute u = 17 32 3( 17 ) − v = 5

 6 x − y = −15 (1) Solve the system:  2x − 4 y = −16 ( 2 ) Solve (1) for y: y = 6 x + 15 (3) Substitute (3) into (2) and solve for x: 2 x − 4(6 x + 15) = −16 2 x − 24 x − 60 = −16 −22 x = 44 x = −2 Substitute x = −2 into (3) to find y: y = 3

 m − 2 n = 4 (1) Solve the system:  3m + 2 n = 1 ( 2 ) Solve (1) for m: m = 2 n + 4 (3) Substitute (3) into (2) and solve for n: 3(2n + 4) + 2 n = 1 8n = −11 n = − 118 Substitute n = − 118 into (1) to find m: m − 2( − 118 ) = 4

11 v = 9617−85 = 17 .

m = 4 − 114 = 54 .

32 11 So, the solution is u = 17 , v = 17 .

So, the solution is m = 54 , n = − 118 .

15.

16.

 2 x + y = 7 (1) Solve the system:   −2 x − y = 5 ( 2 ) Solve (1) for y: y = 7 − 2 x (3) Substitute (3) into (2) and solve for x: −2 x − (7 − 2x ) = 5 −7 = 5 So, the system is inconsistent. Thus, there is no solution.

3x − y = 2 (1) Solve the system:  3x − y = 4 ( 2 ) Solve (1) for y: y = 3x − 2 (3) Substitute (3) into (2) and solve for x: 3x − (3x − 2) = 4 2=4 So, the system is inconsistent. Thus, there is no solution.

734

Section 6.1

17.

18.

 4r − s = 1 (1) Solve the system:  8r − 2 s = 2 ( 2 ) Solve (1) for s: s = 4r − 1 (3) Substitute (3) into (2) and solve for r: 8r − 2(4r − 1) = 2 2=2 So, the system is consistent. There are infinitely many solutions of the form (a, 4a – 1), where r is any real number.

−3 p + q = −4 (1) Solve the system:  6 p − 2q = 8 ( 2 ) Solve (1) for q: q = 3 p − 4 (3) Substitute (3) into (2) and solve for p: 6 p − 2(3 p − 4) = 8 8=8 So, the system is consistent. There are infinitely many solutions of the form (a, 3a – 4), where p is any real number.

19.

20.

(1)  5r − 3s = 15 Solve the system:   −10r + 6 s = −30 ( 2 ) Solve (1) for r: r = 51 (3s + 15) (3) Substitute (3) into (2) and solve for s: −10  51 (3s + 15) + 6 s = −30

−6 s − 30 + 6 s = −30 −30 = −30 So, the system is consistent. There are infinitely many solutions of the form ( a, 5 a3−15 ), where r is any real number.

−5 p − 3q = −1 (1) Solve the system:  10 p + 6q = 2 ( 2 ) Solve (1) for p: p = 51 ( −3q + 1) (3)

Substitute (3) into (2) and solve for q: 10  51 ( −3q + 1)  + 6q = 2

−6q + 2 + 6q = 2 2=2 So, the system is consistent. There are infinitely many solutions of the form ( a, −53a +1 ).

21.

22.

Solve (1) for x: x = 12 ( 3 y − 7 ) (3)

Solve (1) for x: x = 14 ( 5 y − 7 ) (3)

2 x − 3 y = −7 (1) Solve the system:  3x + 7 y = 24 ( 2 )

Substitute (3) into (2) and solve for y: 3 ⋅ 12 ( 3 y − 7 ) + 7 y = 24 9 2

4x − 5 y = −7 (1) Solve the system:  3x + 8 y = 30 ( 2 )

Substitute (3) into (2) and solve for y: 3 ⋅ 14 ( 5 y − 7 ) + 8 y = 30

y − 212 + 7 y = 24 23 2

15 4

y = 692  y = 3

Substitute this value back into (3) to obtain that x = 1.

y − 214 + 8 y = 30 47 4

y = 141 4  y =3

Substitute this value back into (3) to obtain that x = 2 .

735

Chapter 6

23.

 x − y = 0 (1) Solve the system:  2 3 − 3 x + 4 y = 2 ( 2 ) First, clear the fractions by multiplying both equations by 12 to obtain the equivalent system:  4 x − 3 y = 0 (3)  −8x + 9 y = 24 ( 4 ) Solve (3) for x: x = 34 y (5) Substitute (5) into (4) and solve for y: −8 ( 34 y ) + 9 y = 24 1 3

1 4

24.

 51 x + 32 y = 10 (1) Solve the system:  1 1 − 2 x − 6 y = −7 ( 2 ) First, clear the fractions by multiplying both equations by their respective LCDs to obtain the equivalent system:  3x + 10 y = 150 (3)   3x + y = 42 ( 4 ) Solve (4) for y: y = 42 − 3x (5) Substitute (5) into (3) and solve for x: 3x + 10 ( 42 − 3x ) = 150

−27 x + 420 = 150

3 y = 24 Substitute this back into (5) to obtain that x = 6 .

x = 10 Substitute this back into (5) to obtain that y = 12 .

25. Solve the system: (1)  7.2 x − 4.1y = 7  −3.5x + 16.5 y = 2.4 ( 2 )

26. Solve the system:  6.3x + 1.5 y = 10.5 (1)  −0.4 x + 2.2 y = −8.7 ( 2 )

Substitute (3) into (2) and solve for y: 1 −3.5 ( 7.2 ( 7 + 4.1y ) ) + 16.5 y = 2.4

Substitute (3) into (2) and solve for y: 1 −0.4 ( 6.3 (10.5 − 1.5 y ) ) + 2.2 y = −8.7

y = 0.4

y = −3.5

y=8

1 Solve (1) for x: x = 7.2 ( 7 + 4.1y ) (3)

Substitute this value back into (3) to obtain that x = 1.2 .

1 Solve (1) for x: x = 6.3 (10.5 − 1.5 y ) (3)

Substitute this value back into (3) to obtain that x = 2.5 .

736

Section 6.1

27.

 x − y = 2 (1) Solve the system:   x + y = 4 (2) Add (1) and (2) to eliminate y: 2x = 6 x = 3 (3) Substitute (3) into (1) and solve for y: 3− y = 2 y =1

So, the solution is ( 3,1) . 29.

28.

 x + y = 2 (1) Solve the system:   x − y = −2 ( 2 ) Add (1) and (2) to eliminate y: 2x = 0

x = 0 (3) Substitute (3) into (1) and solve for y: y=2 So, the solution is ( 0,2 ) .

30.

 x − y = −3 (1) Solve the system:   x + y = 7 (2) Add (1) and (2) to eliminate y: 2x = 4

 x − y = −10 (1) Solve the system:  ( 2) x+y=8 Add (1) and (2) to eliminate y: 2x = −2

x = 2 (3) Substitute (3) into (2) and solve for y: 2+ y =7 y=5

x = −1 ( 3 ) Substitute (3) into (2) and solve for y: −1 + y = 8 y=9

So, the solution is ( 2,5 ) .

So, the solution is ( −1,9 ) .

31.

32.

5x + 3 y = −3 (1) Solve the system:  3x − 3 y = −21 ( 2 ) Add (1) and (2) to eliminate y: 8x = −24 x = −3 ( 3 ) Substitute (3) into (1) and solve for y: 5( −3) + 3 y = −3 3 y = 12 y=4

So, the solution is ( −3, 4 ) .

−2 x + 3 y = 1 (1) Solve the system:   2x − y = 7 ( 2 ) Add (1) and (2) to eliminate x: 2y = 8 y = 4 (3) Substitute (3) into (2) and solve for x: 2x − 4 = 7 2 x = 11 x = 112

So, the solution is ( 112 , 4 ) .

737

Chapter 6

33.

34.

So, the solution is (1, − 72 ) .

y = 3 (3) Substitute (3) into (1) and solve for x: 3x + 2(3) = 6 3x = 0 x=0 So, the solution is ( 0,3 ) .

35.

36.

2 x − 7 y = 4 (1) Solve the system:  5 x + 7 y = 3 ( 2 ) Add (1) and (2) to eliminate y: 7x = 7 x = 1 (3) Substitute (3) into (2) and solve for y: 5(1) + 7 y = 3 7 y = −2

 3x + 2 y = 6 (1) Solve the system:  −3x + 6 y = 18 ( 2 ) Add (1) and (2) to eliminate x: 8 y = 24

y = − 72

 2 x + 5 y = 7 (1) Solve the system:  3x − 10 y = 5 ( 2 ) Multiply (1) by 2: 4 x + 10 y = 14 ( 3 ) Add (2) and (3) to eliminate y: 7 x = 19

 6 x − 2 y = 3 (1) Solve the system:  −3x + 2 y = −2 ( 2 ) Add (1) and (2) to eliminate y: 3x = 1 x = 13 ( 3 ) Substitute (3) into (1) and solve for y: 6( 13 ) − 2 y = 3

x = 197 (4) Substitute (4) into (1) and solve for y: 2( 197 ) + 5 y = 7 y=

2 − 2y = 3 y = − 12

11 35

11 So, the solution is ( 197 , 35 ).

So, the solution is ( 13 , − 12 ) .

37.

38.

(1)  2x + 5 y = 5 Solve the system:  −4 x − 10 y = −10 ( 2 ) Multiply (1) by 2: 4 x + 10 y = 10 ( 3 ) Add (2) and (3) to eliminate y: 0 = 0 (4) So, the system is consistent. Thus, there are infinitely many solutions of the form ( a, 5−52 a ), where x is any real number.

 11x + 3 y = 3 (1) Solve the system:  22 x + 6 y = 6 ( 2 ) Multiply (1) by −2: −22 x − 6 y = −6 ( 3 ) Add (2) and (3) to eliminate y: 0 = 0 (4) So, the system is consistent. Thus, there are infinitely many solutions of the form ( a, −113a +3 ), where x is any real number.

738

Section 6.1

39.

40.

y = 0 (5) Substitute (5) into (1) and solve for x: 3x − 2(0) = 12 x=4 So, the solution is ( 4,0 ) .

x = 3 (5) Substitute (5) into (1) and solve for x: 5(3) − 2 y = 7 So, the solution is ( 3, 4 ) .

41.

42.

3x − 2 y = 12 (1) Solve the system:   4 x + 3 y = 16 ( 2 ) Multiply (1) by 4: 12 x − 8 y = 48 ( 3 ) Multiply (2) by −3: −12 x − 9 y = −48 (4) Add (3) and (4) to eliminate x: −17 y = 0

 6 x − 3 y = −15 (1) Solve the system:  7 x + 2 y = −12 ( 2 ) Multiply (1) by 2: 12 x − 6 y = −30 (3) Multiply (2) by 3: 21x + 6 y = −36 (4) Add (3) and (4) to eliminate y: 33x = −66  x = −2 (5) Substitute (5) into (1) and solve for y: 6 ( −2 ) − 3 y = −15

−3 y = −3 y =1 So, the solution is ( −2,1) .

5x − 2 y = 7 (1) Solve the system:  3x + 5 y = 29 ( 2 ) Multiply (1) by 5: 25x − 10 y = 35 ( 3 ) Multiply (2) by 2: 6x + 10 y = 58 (4) Add (3) and (4) to eliminate y: 31x = 93

y=4

7 x − 4 y = −1 (1) Solve the system:   3x − 5 y = 16 ( 2 ) Multiply (1) by 5: 35x − 20 y = −5 ( 3 ) Multiply (2) by −4: −12x + 20 y = −64 (4) Add (3) and (4) to eliminate y: 23x = −69 x = −3 (5) Substitute (5) into (1) and solve for y: 7( −3) − 4( y ) = −1

y = −5 So, the solution is ( −3, −5 ) .

739

Chapter 6

43. Solve the system:  0.02 x + 0.05 y = 1.25 (1)  −0.06 x − 0.15 y = −3.75 ( 2 ) Multiply (1) by 3: 0.06 x + 0.15 y = 3.75 ( 3 ) Add (2) and (3) to eliminate y: 0 = 0 (4) So, the system is consistent. Thus, there are infinitely many solutions of the form ( a, 1.25 −50.02 a ), where x is any real number.

44. Solve the system:  −0.5x + 0.3 y = 0.8 (1)  −1.5x + 0.9 y = 2.4 ( 2 ) Multiply (1) by −3 : 1.5x − 0.9 y = −2.4 ( 3 ) Add (2) and (3) to eliminate y: 0 = 0 (4) So, the system is consistent. Thus, there are infinitely many solutions of the a + 0.8 form ( a, 0.50.3 ), where x is any real number.

45.

46.

 x + 12 y = 1 (1) Solve the system:  1 7  5 x + 2 y = 2 (2) Multiply (1) by −7 : − 73 x − 72 y = −7 ( 3 ) Add (2) and (3) to eliminate y: 32 − 15 x = −5

 12 x − 13 y = 0 (1) Solve the system:  3 3 1  2 x + 2 y = 4 (2) Multiply (1) by −3 : − 32 x + y = 0 ( 3 ) Add (2) and (3) to eliminate x: 3 3 2 y = 4

75 (4) x = 32 Substitute (4) into (1) and solve for y: 1 75 1 3 ( 32 ) + 2 y = 1

y = 12 (4) Substitute (4) into (1) and solve for x: 1 1 1 2 x − 3 (2) = 0

1 3

1 2

21 y = 96

1 2

x = 61

y = 167

x = 13

75 7 So, the solution is ( 32 , 16 ) .

So, the solution is ( 13 , 12 ) .

47. c Adding the equations yields 6x = 6, so that x = 1. Thus, 3(1) − y = 1, so that y = 2.

48. b Multiply first equation by 2 and add to the second. This yields y = 1, and upon substitution, x = 3.

49. d Subtracting the equations yields 0 = −4. Thus, the system is inconsistent and so, the lines should be parallel.

50. a Multiply first equation by 2 and add to the second. This yields 0 = 0. Thus, the system is consistent, and the lines should be the same.

740

Section 6.1

51.

Notes on the graph: Solid curve is y = −x Dashed curve is y = x So, the solution is (0,0).

52.

Notes on the graph: Solid curve is x − 3 y = 0 Dashed curve is x + 3 y = 0 So, the solution is (0,0).

53.

54.

Notes on the graph: Solid curve is 2x + y = −3 Dashed curve is x + y = −2 So, the solution is ( −1, −1).

Notes on the graph: Solid curve is x − 2 y = −1 Dashed curve is − x − y = −5 So, the solution is (3,2).

741

Chapter 6

55.

56.

Notes on the graph: Solid curve is 12 x − 23 y = 4 Dashed curve is 14 x − y = 6 So, the solution is (0, −6).

Notes on the graph: The curves 51 x − 52 y = 10 and 5 10 1 15 x − 6 y = 3 have the same graph pictured above. So, there are infinitely many solutions.

57.

58.

Notes on the graph: Solid curve is 1.6 x − y = 4.8 Dashed curve is −0.8x + 0.5 y = 1.5 So, there is no solution.

Notes on the graph: Solid curve is 1.1x − 2.2 y = 3.3 Dashed curve is −3.3x + 6.6 y = −6.6 So, there is no solution.

742

Section 6.1

59. Let x = number of Auspens in a kit, y = number of refill bottles in a kit. Must solve the system:  x + 40 y = 246 (1)  x=y (2)  Substituting (2) into (1) yields 41x = 246, x = 6. So, there are 6 Auspens per kit and 6 refill bottles per kit.

60. Let x = number of grams of 1% cream, y = number of grams of 10% cream. Must solve the system: 0.01x + 0.10 y = 0.03(454) (1)  (2) x + y = 454  Solve (2) for y: y = 454 – x. (3) Substitute (3) into (1): 0.01x + 0.10(454 − x ) = 0.03(454) 0.01x + 45.4 − 0.10x = 13.62 −0.09x = −31.78 x ≈ 353 Substitute this value of x back into (2) to see that y = 454 – 353 = 101. So, about 353 grams of 1% cream and 101 grams of 10% cream.

61. Let x = # of Montblanc pens y = # of Cross pens Must solve the system: x + y = 69 (1)   72 x + 10 y = 1000 ( 2 ) Solve (1) for y: y = 69 − x (3) Substitute (3) into (2): 72x + 10(69 − x ) = 1000  x = 5 Substitute this value of x into (3): y = 64 So, have 5 Montblanc pens and 64 Cross pens.

62. Let x = # of standard dumbbell sets y = # of deluxe dumbbells sets Must solve the system: x + y = 24 (1)   30 x + 45 y = 915 ( 2 ) Solve (1) for y: y = 24 − x (3) Substitute (3) into (2): 30x + 45(24 − x ) = 915  x = 11 Substitute this value of x into (3): y = 13 So, have 11 standard dumbbell sets and 13 deluxe dumbbell sets.

63. Let x = number of ml of 8% HCl. y = number of ml of 15% HCl. Must solve the system: 0.08x + 0.15 y = (0.12)(37) (1)  ( 2)  x + y = 37 Multiply (2) by −0.08 : −0.08x − 0.08 y = −2.96 ( 3 )

Add (1) and (3) to eliminate x: 0.07 y = 1.48  y ≅ 21.14 (4) Substitute (4) into (2) and solve for x: x ≅ 15.86 So, should use approximately 15.86 ml of 8% HCl and 21.14 ml of 15% HCl.

743

Chapter 6

64. Let x = # gallons of gas in 2.5% mix. y = # gallons of oil in 2.5% mix. Must solve the system:  0.025 ( x + y ) = y (1)  0.04 ( 350 − ( x + y ) ) = 10 − y ( 2 ) (1) is equivalent to: 0.975 y − 0.025x = 0 (3) (2) is equivalent to: 0.04x − 0.96 y = 4 (4)

Solve the equivalent system (3) & (4): Solve (3) for y: y = 0.025 0.975 x (5) Substitute (5) into (4) and simplify: x = 260 Substitute this value of x into (5) to get y = 203 . So, the 2.5% mixture contains 260 gallons of gas and 203 gallons of oil. The 4% mixture contains 80 gallons of gas and 103 gallons of oil. Thus, the 2.5% mixture contains 800 3 gallons and the 4% mixture contains 250 3 gallons.

65. Let x = total annual sales. Salary at Autocount: 15,000 + 0.10x Salary at Polk: 30,000 + 0.05x We need to find x such that 15,000 + 0.10x > 30,000 + 0.05x. Solving this yields: 0.05x > 15,000  x > 300,000 So, he needs to make at least $300,000 of sales in order to make more money at Autocount.

66. Let x = total made from homes sold. Salary for agents in resale: 0.06x Salary for agents in new homes: 15,000 + 0.015x We need to find x such that 0.06x > 15,000 + 0.015x. Solving this yields: 0.045x > 15,000  x > 333,333.33 So, an agent would need to sell at least $333,333.33 worth of homes to make more money in resale.

67. Let x = gallons used for highway miles y = gallons used for city miles Must solve the system: 26 x + 19 y = 349.5 (1)  x + y = 16 (2)  Multiply (2) by −19 : −19x − 19 y = −304 ( 3 )

Add (1) and (3) to eliminate y: 7x = 45.5  x = 6.5 (4) Substitute (4) into (2) and solve for y: y = 9.5 Thus, there are 169 highway miles and 180.5 city miles.

744

Section 6.1

68. Let x = number of tacos and y = number of burritos. Must solve the system: x + y = 405 (1)   (2) 3x + 8.5 y = 1501 Multiply (1) by −3 : −3x − 3 y = −1215 (3)

Add (2) and (3) to eliminate x : 5.5 y = 286  y = 52 (4) Substitute (4) into (1) and solve for x : x = 353 Thus, he sold 353 tacos and 52 burritos.

69. Let x = speed of the plane and y = wind speed.

Atlanta to Paris Paris to Atlanta

Rate (mph) x+y x−y

Time (hours) 8 10

Distance 4000 4000

So, using Distance = Rate × Time, we see that we must solve the system:  8( x + y ) = 4000 x + y = 500 (1) , which is equivalent to   10( x − y ) = 4000 x − y = 400 (2) Adding (1) and (2) yields: 2x = 900, so that x = 450. Substituting this into (1) then yields y = 50. So, the average ground speed of the plane is 450 mph, and the average wind speed is 50 mph. 70. Let x = speed of the plane and y = wind speed.

Trip Return Trip

Rate (mph) x+y x−y

Time (hours) 3 4

Distance 500 500

So, using Distance = Rate × Time, we see that we must solve the system: (1) x + y = 500  3( x + y ) = 500 3 , which is equivalent to   (2) 4( x − y ) = 500 x − y = 125 875 Adding (1) and (2) yields: 2x = 875 3 , so that x = 6 . Substituting this into (1) then yields y = 125 6 . So, the average ground speed of the plane is approximately145.83 mph, and the average wind speed is approximately 20.83 mph.

745

Chapter 6

71. Let x = amount invested in 10% stock y = amount invested in 14% stock We must solve the system: (1) 0.10 x + 0.14 y = 1260  x + y = 10,000 ( 2 )  Multiply (2) by −0.10: −0.10x − 0.10 y = −1000 ( 3 ) Add (1) and (3) to eliminate x: 0.04 y = 260

y = 6500 (4) Substitute (4) into (2) and solve for y: x = 3500 So, should invest $3500 in the 10% stock, and $6500 in the 14% stock.

72. Let x = amount invested in 5% stock y = amount invested in 7% stock We must solve the system: 0.05x + 0.07 y = 665 (1)  y = 2x (2)  Substitute (2) into (1) and solve for x: 0.05x + 0.07(2x ) = 665

0.19x = 665 x = 3500 (3) Substitute (3) into (2) to find y: y = 7000. So, should invest $3500 in the 5% stock, and $7000 in the 7% stock.

73. Let x = number of smartwatches sold. Cost equation: y = 35x + 735 (1) Revenue equation: y = 140x (2) We want the intersection of these two equations to find the break-even point. To this end, substitute (2) into (1) to find that x = 7. So, must sell 7 smartwatches to break even.

74. Let x = # of glasses of lemonade sold Cost equation: y = 0.10x + 15 (1) Revenue equation: y = 0.25x (2) We want the intersection of these two equations to find the break even point. To this end, substitute (2) into (1) to find that x = 100. So, must sell 100 glasses of lemonade to break even.

75. Every term in the first equation is not multiplied by −1 correctly. The equation should be −2x − y = 3, and the resulting solution should be x = 11, y = −25.

76. 0 = 12 is an inconsistent statement. Hence, there is no solution to the system.

77. Did not distribute −1 correctly. In Step 3, the calculation should be −( −3 y − 4) = 3 y + 4 and the resulting answer should be (2, −2).

78. Actually, the lines are coincident, so that there are infinitely many solutions to the system.

79. False. If the lines are coincident, then there are infinitely many solutions.

80. True. The intersection of such lines is guaranteed.

746

Section 6.1

81. False. The lines could be coincident.

82. False.

 Ax − By = 1  −Ax + By = −1 Adding the two equations results in 0 = 0, which means there are infinitely many solutions.

83. Substitute the pair (2, −3) into both Substitute (3) into (1) and solve for B: 2( −4) − 3B = −29 equations to get the system: −3B = −21 2 A − 3B = −29 (1)  B =7 2 A + 3B = 13 (2) A = −4, B = 7 . Hence, the solution is Add (1) and (2) to eliminate B: 4 A = −16

A = −4

(3)

84. The slopes are very close together, but are not equal. It may be difficult to distinguish on a graphing calculator unless just the right window is chosen. 85. Let x = # cups of pineapple juice for 2% drink, and y = # cups of pomegranate juice for 2% drink. Must solve the system: y  (1) x + y = 0.02  4− y  (100 − x )+( 4 − y ) = 0.04 (2) This system, after simplification, is equivalent to: (3) 0.02 x − 0.98 y = 0  0.04 x − 0.96 y = 0.16 (4) Substitute (3) into (4): y = 0.02 0.98 x (5) Substitute (5) into (4) and simplify: x = 7.84 Substitute this value of x back into (5): y = 0.16 So, she can make 8 cups of the 2% drink and 96 cups of the 4% drink.

86. Let x = # tablespoons of red for light purple solution, and y = # tablespoons of blue for light purple solution. Must solve the system: y  (1) x + y = 0.02  2− y  (30 − x )+(2 − y ) = 0.10 (2) This system, after simplification, is equivalent to: (3) 0.02 x − 0.98 y = 0  0.10 x − 0.90 y = 1.2 (4) Substitute (3) into (4): y = 0.02 0.98 x (5) Substitute (5) into (4) and simplify: x = 14.7 Substitute this value of x back into (5): y = 0.3 So, they can make 15 tablespoons of light purple and 17 tablespoons of deep purple.

747

Chapter 6

87.

88.

Notes on the graph: Solid curve is y = −1.25x + 17.5 Dashed curve is y = 2.3x − 14.1 The approximate solution is (8.9, 6.4).

Notes on the graph: Solid curve is y = 14.76 x + 19.43 Dashed curve is y = 2.76x + 5.22 The approximate solution is ( −1.2,2.0).

89.

90.

Notes on the graph: Solid curve is 23x + 15 y = 7 Dashed curve is 46x + 30 y = 14 Note that the graphs are the same line, so there are infinitely many solutions to this system. Infinitely many solutions: 7 − 23a a, 5

Notes on the graph: Solid curve is −3x + 7 y = 2 Dashed curve is 6x − 14 y = 3 Since the lines are parallel, there is no solution.

748

Section 6.1

91. The graph of the system of equations is as follows:

92. The graph of the system of equations is as follows:

The solution is approximately (2.426, −0.059).

The solution is approximately (−2.703, 4.139).

749

Chapter 6

Section 6.2 Solutions -------------------------------------------------------------------------------1. Solve the system:  x − y + z = 6 (1)  − x + y + z = 3 (2)  −x − y − z = 0 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (2): 2 z = 9  z = 92 (4)

Add (2) and (3): −2 x = 3  x = − 32 (5) Step 2: Solve for the remaining variable using (4) and (5). Substitute (4) and (5) into (1): − 32 − y + 92 = 6  y = −3

3. Solve the system: (1)  x+ y−z =2   −x − y − z = −3 (2) −x + y − z = 6 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (2): −2 z = −1  z = 12 (4) Add (2) and (3), and sub. in (4): −2 x − 2 z = 3  − 2 x − 2 ( 12 ) = 3  − 2x = 4  x = −2 (5) Step 2: Solve for the remaining variable using (4) and (5). Substitute (4) and (5) into (1): −2 + y − 12 = 2  y = 29

2. Solve the system: − x − y + z = −1 (1)  (2) −x + y − z = 3  x−y−z =5 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (2): −2x = 2  x = −1 (4) Add (2) and (3): −2 z = 8  z = −4 (5)

Step 2: Solve for the remaining variable using (4) and (5). Substitute (4) and (5) into (1): −( −1) − y + ( −4 ) = −1  y = −2 4. Solve the system:  x + y + z = −1 (1)  (2) −x + y − z = 3 −x − y + z = 8 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (3): 2 z = 7  z = 72 (4) Add (1) and (2): 2 y = 2  y = 1 (5) Step 2: Solve for the remaining variable using (4) and (5). Substitute (4) and (5) into (1): x + 1 + 72 = −1  x = − 112

750

Section 6.2

5. Solve the system: − x + y − z = −1 (1)  (2)  x−y−z =3  x+ y−z =9 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (2): −2 z = 2  z = −1 (4)

6. Solve the system:  x − y − z = 2 (1)  − x − y + z = 4 (2) − x + y − z = 6 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (2): −2 y = 6  y = −3 (4)

Add (2) and (3), and sub. in (4): 2x − 2 z = 12  2 x − 2 ( −1) = 12

Add (2) and (3): −2x = 10  x = −5 (5)

 2x = 10  x = 5 (5) Step 2: Solve for the remaining variable using (4) and (5). Substitute (4) and (5) into (1): −5 + y − ( −1) = −1  y = 3

Step 2: Solve for the remaining variable using (4) and (5). Substitute (4) and (5) into (1): −5 − ( −3) − z = 2  z = −4

7. Solve the system: (1)  x + y + z = −1  (2) −2 x + y + z = −10  x−y+z =5 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (3) to eliminate y: 2 x + 2 z = 4 (4) Add (2) and (3) to eliminate y: − x + 2 z = −5 (5) These steps yield the following system:  2x + 2 z = 4 (* )  − x + 2 z = −5

Step 2: Solve system (* ) from Step 1. Multiply (5) by –1: x − 2 z = 5 (6) Add (4) and (6) to eliminate z and solve for x: 3x = 9  x = 3 (7) Substitute (7) into (6) to find z: 3 − 2 z = −5  z = −1 (8) Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (7) and (8) into (1) to find y : 3 + y + ( −1) = −1  y = −3 Thus, the solution is: x = 3, y = −3, z = −1

751

Chapter 6

8. Solve the system: (1)  x + 2y + z = 6  (2)  −2 x − y + 3 z = 6  x−y+z =3 (3)  Step 1: Multiply (1) by −1: −x − 2 y − z = −6 (4) Add (3) and (4) to eliminate x and z : −3 y = −3  y = 1 (5) Substitute y = 1 into (1) and (2). These steps yield to following system:

x+z =4 (6)  (*)  (7) −2 x + 3z = 7 Step 2: Solve system (* ) from Step 1. Multiply (6) by 2: 2x + 2 z = 8 (8) Add (7) and (8) to eliminate x: 5z = 15  z = 3 (9) Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (5) and (9) into (1) to find x : x + 2(1) + 3 = 6  y = 1 Thus, the solution is: x = 1, y = 1, z = 3

9. Solve the system: (1)  x + y + 3z = 6  (2) 2 x − y + z = 5 2 x + y − z = −1 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (2) to eliminate y: 3x + 4 z = 11 (4) Add (2) and (3) to eliminate y: 4 x = 4  x = 1 (5)

Step 2: Since we inadvertently eliminated y and z, we can substitute (5) into (4) to find x: 3(1) + 4 z = 11  z = 2 (6) Step 3: Use the solutions found in Step 2 to find the value of the third variable in the original system. Substitute (5) and (6) into (1) to find x : 1 + y + 3 ( 2 ) = 6  y = −1 Thus, the solution is: x = 1, y = −1, z = 2

10. Solve the system: (1)  x + 3y − z = 8  (2)  − x + y + z = −4 x − 4y + z = 1 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (3) to eliminate x and z: 4 y = 4  y = 1 (4) Step 2: Since we inadvertently eliminated y and z, we can substitute (4) into (1) and (3) to create a new 2 × 2 system:

 x − z = 5 (5) (*)  x + z = 5 (6) Add (5) and (6) to eliminate z: 2x = 10  x = 5 (7) Step 3: Use the solutions found in Steps 1 and 2 to find the value of the third variable in the original system. Substitute (4) and (7) into (1) to find z : 5 + 3(1) − z = 8  z = 0 Thus, the solution is: x = 5, y = 1, z = 0

752

Section 6.2

11. Solve the system: 2 x − 3 y + 4 z = −3 (1)   − x + y + 2 z = 1 (2)  5x − 2 y − 3z = 7 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (2) by 3: −3x + 3 y + 6 z = 3 (4) Add (4) and (1) to eliminate y: −x + 10 z = 0 (5) Next, multiply (2) by 2: −2x + 2 y + 4 z = 2 (6) Add (6) and (3) to eliminate y: 3x + z = 9 (7) These steps yield the following system: −x + 10 z = 0 (5) (*)  3x + z = 9 (7) 12. Solve the system: (1)  x − 2y + z = 0  (2)  −2 x + y − z = −5 13x + 7 y + 5 z = 6 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (1) by 2: 2x − 4 y + 2 z = 0 (4) Add (4) and (2) to eliminate x: −3 y + z = −5 (5) Multiply (1) by −13 and then add to (3): 33 y − 8z = 6 (6) This yields the following 2 × 2 system:  − 3 y + z = −5 (5) (*)   33 y − 8z = 6 (6)

Step 2: Solve system (*) from Step 1. Multiply (5) by 3: −3x + 30 z = 0 (8) Add (8) and (7) to eliminate x and solve for z: 31z = 9 so that z = 319 (9) Substitute (9) into (7) to find x: −x + 10( 319 ) = 0 (10) x = 90 31 Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (9) and (10) into (2) to find y: 9 −( 90 31 ) + y + 2( 31 ) = 1 103 y − 72 31 = 1 so that y = 31 Thus, the solution is: 103 9 x = 90 31 , y = 31 , z = 31 .

Step 2: Solve system (*) from Step 1. Multiply (5) by 8 and then add to (6): 9 y = −34 y = − 349 (7) Substitute (7) into (5) to solve for z: −3( − 349 ) + z = −5 z = − 493 (8) Substitute (7) and (8) into (1) to find x: x − 2( − 349 ) + ( − 493 ) = 0 x = 799 (9)

Hence, the solution is: x = 799 , y = − 349 , z = − 493 .

753

Chapter 6

13. Solve the system:  −4 x + 3 y + 5 z = 2 (1)   2 x − 3 y − 2 z = −3 (2) −2 x + 4 y + 3z = 1 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (2) and (3) to eliminate x: y + z = −2 (4) Next, multiply (2) by 2: 4x − 6 y − 4 z = −6 (5) Add (5) and (1) to eliminate x: −3 y + z = −4 (6) These steps yield the following system:  y + z = −2 (4) (*)  −3 y + z = −4 (6) Step 2: Solve system (*) from Step 1.

Multiply (6) by −1 : 3 y − z = 4 (7) Add (4) and (7) to eliminate z and solve for y: 4 y = 2  y = 12 (8) Substitute (8) into (4) to find z: 1 2 + z = −2

14. Solve the system:  − x + 2 y + z = 5 (1)   2x − 2 y + 3z = 0 (2) −4 x + y − 2 z = 3 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (2) to eliminate y: x + 4 z = 5 (4) Next, multiply (3) by 2: −8x + 2 y − 4 z = 6 (5) Add (2) and (5) to eliminate y: −6x − z = 6 (6) These steps yield the following system:  x + 4 z = 5 (4) (*)   − 6 x − z = 6 (6) Step 2: Solve system (*) from Step 1. Multiply (6) by 4: −24x − 4 z = 24 (7)

Add (4) and (7) to eliminate z and solve for x: −23x = 29

z = − 52 (9) Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (8) and (9) into (2) to find y: 2 x − 3( 12 ) − 2( − 52 ) = −3

2 x = − 132 x = − 134 (10)

Thus, the solution is: x = − 134 , y = 21 , z = − 52 .

x = − 29 23 (8) Substitute (8) into (4) to find z: − 29 23 + 4 z = 5

4 z = 144 23 z = 36 (9) 23 Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (8) and (9) into (1) to find y: 29 36 23 + 2 y + 23 = 5

2 y = 50 23 y = 25 (10) 23 Thus, the solution is: 25 36 x = − 29 23 , y = 23 , z = 23 .

754

Section 6.2

15. Solve the system:  x − y + z = −1 (1)  y − z = −1 (2)  − x + y + z = 1 (3) 

16. Solve the system:  − y + z = 1 (1)  x − y + z = −1 (2)  x − y − z = −1 (3) 

Add (1) and (3) to eliminate both x and y: 2z = 0

Multiply (2) by −1 : −x + y − z = 1 ( 4) Add (3) and (4) to eliminate both x and y: −2 z = 0

z = 0 (4) Substitute (4) into (2) to find y: y − 0 = −1 (5) Substitute (4) and (5) into (1) to find x: x − ( −1) + 0 = −1 x = −2 (6) Thus, the solution is: x = −2, y = −1, z = 0 .

z = 0 (5) Substitute (5) into (1) to find y: −y + 0 = 1 y = −1 (6) Substitute (5) and (6) into (2) to find x: x − ( −1) + 0 = −1 x = −2 (6) Thus, the solution is: x = −2, y = −1, z = 0 .

17. Solve the system: 3x − 2 y − 3z = −1 (1)   x − y + z = −4 (2) 2 x + 3 y + 5 z = 14 (3) 

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (2) by 3: 3x − 3 y + 3z = −12 (4) Add (4) and (1) to eliminate z: 6x − 5 y = −13 (5) Next, multiply (2) by −5 : −5x + 5 y − 5z = 20 (6) Add (6) and (3) to eliminate z: −3x + 8 y = 34 (7) These steps yield the following system: 6 x − 5 y = −13 (5) (*)  −3x + 8 y = 34 (7)

755

Chapter 6

Step 2: Solve system (*) from Step 1. Multiply (7) by 2: −6x + 16 y = 68 (8) Add (5) and (8) to eliminate x and solve for y: 11y = 55 y = 5 (9) Substitute (9) into (5) to find x:

6x − 5(5) = −13 6x = 12

x = 2 (10) Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (9) and (10) into (2) to find z: 2 − 5 + z = −4 z = −1 (11) Thus, the solution is: x = 2, y = 5, z = −1. 18. Solve the system:  3x − y + z = 2 (1)  x − 2 y + 3z = 1 (2) 2 x + y − 3z = −1 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (3) to eliminate y: 5x − 2 z = 1 (4) Next, multiply (3) by 2: 4x + 2 y − 6 z = −2 (5) Add (2) and (5) to eliminate y: 5x − 3z = −1 (6) These steps yield the following system: 5x − 2 z = 1 (4) (*)  5x − 3z = −1 (6)

Step 2: Solve system (*) from Step 1. Multiply (4) by −1 : −5x + 2 z = −1 (7) Add (6) and (7) to eliminate x and solve for z: − z = −2 z = 2 (8) Substitute (8) into (4) to find x: 5x − 2(2) = 1 x = 1 (9) Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (8) and (9) into (2) to find y: 1 − 2 y + 3(2) = 1 y = 3 (10) Thus, the solution is: x = 1, y = 3, z = 2 . 756

Section 6.2

19. Solve the system:  −3x − y − z = 2 (1)   x + 2 y − 3z = 4 (2) 2 x − y + 4 z = 6 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (2) by 3: 3x + 6 y − 9z = 12 (4) Add (1) and (4) to eliminate x: 5 y − 10 z = 14 (5)

Next, multiply (2) by −2: −2x − 4 y + 6 z = −8 (6) Add (3) and (6) to eliminate x: −5 y + 10 z = −2 (7) These steps yield the following system:  5 y − 10 z = 14 (5) (*)  −5 y + 10 z = −2 (7) Step 2: Solve system (*) from Step 1. Adding (5) and (7) yields the false statement 0 = 12. Hence, this system has no solution.

20. Solve the system:  2x − 3 y + z = 1  x + 4 y − 2 z = 2  3x − y + 4 z = −3 

Step 2: Solve system (*) from Step 1. Multiply (5) by −2 : 22 y − 10 z = 6 (8) Add (7) and (8) to eliminate z and solve for y: 9 y = −3  y = − 13 (9) Substitute (9) into (5) to find x: −11( − 13 ) + 5 z = −3  z = − 34 (10) Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (9) and (10) into (1) to find x: 2 x − 3 ( − 13 ) − 34 = 1  x = 23 Thus, the solution is: x = 23 , y = − 13 , z = − 34 .

(1) (2) (3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (2) by −2: −2x − 8 y + 4 z = −4 (4) Add (1) and (4) to eliminate x: −11y + 5z = −3 (5) Next, multiply (2) by −3: −3x − 12 y + 6 z = −6 (6) Add (3) and (6) to eliminate x: −13 y + 10 z = −9 (7) These steps yield the following system: −11y + 5z = −3 (5) (*)  −13 y + 10 z = −9 (7)

757

Chapter 6

21. Solve the system: (1)  3x + 2 y + z = 4  −4 x − 3 y − z = −15 (2)  x − 2 y + 3z = 12 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (1) by −3: −9x − 6 y − 3z = −12 (4) Add (3) and (4) to eliminate z: −8x − 8 y = 0  x + y = 0 (5)

Next, multiply (2) by 3: −12x − 9 y − 3z = −45 (6) Add (3) and (6) to eliminate z: −11x − 11y = −33  x + y = 3 (7) These steps yield the following system:  x + y = 0 (5) (*)   x + y = 3 (7) Step 2: Solve system (*) from Step 1. Subtracting (5) and (7) yields the false statement 0 = 3. Hence, this system has no solution.

22. Solve the system:  3x − y + 4 z = 13  −4 x − 3 y − z = −15  x − 2 y + 3z = 12 

Next, multiply (3) by 4: 4x − 8 y + 12 z = 48 (6) Add (2) and (6) to eliminate z: −11y + 11z = 33  − y + z = 3 (7) These steps yield the following system:  y − z = − 235 (5) (*)   − y + z = 3 (7) Step 2: Solve system (*) from Step 1. Adding (5) and (7) yields the false statement 0 = − 85 . Hence, this system has no solution.

(1) (2) (3)

Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (3) by −3: −3x + 6 y − 9z = −36 (4) Add (1) and (4) to eliminate z: 5 y − 5 z = −23  y − z = − 235 (5)

758

Section 6.2

23. Solve the system:  −x + 2 y + z = −2 (1)   3x − 2 y + z = 4 (2) 2x − 4 y − 2 z = 4 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (1) by 3: −3x + 6 y + 3z = −6 (4) Add (4) and (2) to eliminate x: 4 y + 4 z = −2 (5) Next, multiply (1) by 2: −2x + 4 y + 2 z = −4 (6) Add (6) and (3) to eliminate x: 0 = 0 (7) 24. Solve the system: (1)  2x − y = 1   − x + z = −2 (2) −2 x + y = −1 (3) 

First, we solve the 2 × 2 system consisting of equations (1) and (3): Add (1) and (3): 0 = 0 So, there are infinitely many solutions to this system.

Hence, we know that the system has infinitely many solutions. Let z = a. Then, substituting this value into (5), we can find the value of y: 2 y = −2a − 1 y = −( a + 12 ) Now, substitute the values of z and y into (1) to find x: −x − 2( a + 12 ) + a = −2

−x − 2a − 1 + a = −2 x = 1− a Thus, the solution is x = 1 − a, y = −( a + 12 ), z = a , where a is any real number. To determine them, let y = a. Substitute this into (1) to find x: 2x − a = 1 x = 1+2a Now, substitute the values of x and y into (2) to find z: −( 1+2a ) + z = −2 z = 1+2a − 2 z = a 2−3 Thus, the solution is: x = 1+2a , y = a, z = a 2−3 .

759

Chapter 6

25. Solve the system:  x − y − z = 10 (1)  2x − 3 y + z = −11 (2)  − x + y + z = −10 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (3): 0 = 0 (4) Add (1) and (2): 3x − 4 y = −1 (5)

Hence, we know that the system has infinitely many solutions. 26. Solve the system: 2 x + y + z = −3 (1)   x + 2 y − z = 0 (2) x + y + 2 z = 5 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (2) to eliminate z: 3x + 3 y = −3 or x + y = −1 (4) Multiply (2) by 2: 2x + 4 y − 2 z = 0 (5) Add (5) and (3) to eliminate z: 3x + 5 y = 5 (6)

These steps yield the following system: x + y = −1 (4) (*)  3x + 5 y = 5 (6)

To determine them, let x = a. Substitute this into (5) to find y: 3a − 4 y = −1 Now, substitute the y = 14 (3a + 1) values of x and y into (1) to find z: a − 14 (3a + 1) − z = 10 z = 4 a −3a4−1− 40 z = a −441

Thus, the solution is: x = a, y = 14 (3a + 1), z = a −441 . Equivalently, x = 41 + 4a, y = 31 + 3a, z = a . Step 2: Solve system (*) from Step 1. Multiply (4) by −3 : −3x − 3 y = 3 (7) Add (6) and (7) to eliminate x and solve for y: 2y = 8 y = 4 (8) Substitute (8) into (4) to find x: x + 4 = −1 x = −5 (9) Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (8) and (9) into (2) to find z: −5 + 2(4) − z = 0 z = 3 (10) Thus, the solution is: x = −5, y = 4, z = 3 .

760

Section 6.2

27. Solve the system: (1)  3x1 + x2 − x3 = 1   x1 − x2 + x3 = −3 (2) 2 x + x + x = 0 (3) 2 3  1 Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (2) to eliminate x2 and x3 : 4 x1 = −2 x1 = − 12 (4) Add (2) and (3) to eliminate x2 :

3x1 + 2 x3 = −3 (5)

Step 2: Solve system (*) from Step 1. Substitute (4) into (5): 3( − 12 ) + 2 x3 = −3 2x3 = − 32 x3 = − 34 (6)

Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (4) and (6) into (1) to find x2 : 3( − 12 ) + x2 − ( − 34 ) = 1 x2 = 74 (7)

These steps yield the following system: x1 = − 12 (4)  (*)  3x1 + 2 x3 = −3 (5)

Thus, the solution is: x1 = − 12 , x2 = 74 , x3 = − 34 .

28. Solve the system: 2 x1 + x2 + x3 = −1 (1)   x1 + x2 − x3 = 5 (2)  3x − x − x = 1 (3) 3  1 2 Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (1) and (3) to eliminate x2 and x3 : 5x1 = 0

Step 2: Solve system (*) from Step 1. Substitute (4) into (5): 3(0) + 2 x2 = 4

x1 = 0 (4) Add (1) and (2) to eliminate x3 :

3x1 + 2 x2 = 4 (5)

These steps yield the following system: x1 = 0 (4)  (*)  3x1 + 2 x2 = 4 (5)

2 x2 = 4 x2 = 2 (6)

Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (4) and (6) into (2) to find x3 : 0 + 2 − x3 = 5 x3 = −3 (7) Thus, the solution is: x1 = 0, x2 = 2, x3 = −3 .

761

Chapter 6

29. Solve the system: 2 x + 5 y = 9 (1)   x + 2 y − z = 3 (2)  −3x − 4 y + 7 z = 1 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (2) by 7: 7x + 14 y − 7 z = 21 (4) Add (4) and (3) to eliminate z: 4x + 10 y = 22 (5)

Step 2: Solve system (*) from Step 1. Multiply (1) by −2 : −4x − 10 y = −18 (6) Add (5) and (6) to eliminate x and solve for y: 0 = 4 (7) Hence, we conclude from (7) that the system has no solution.

These steps yield the following system: 2 x + 5 y = 9 (1) (*)   4 x + 10 y = 22 (5) 30. Solve the system:  x − 2 y + 3z = 1 (1)  −2 x + 7 y − 9 z = 4 (2)  x + z = 9 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Solve (3) for z: z = 9 − x (4) Substitute (4) into (1) to eliminate z: x − 2 y + 3(9 − x ) = 1 −2x − 2 y = −26

Step 2: Solve system (*) from Step 1. Multiply (5) by −7 : −7x − 7 y = −91 (7) Add (6) and (7) to eliminate x and solve for y: 0 = −6 (8) Hence, we conclude from (8) that the system has no solution.

x + y = 13 (5) Substitute (4) into (2) to eliminate z: −2x + 7 y − 9(9 − x ) = 4 7 x + 7 y = 85 (6) These steps yield the following system: (5) x + y = 13 (*)  7x + 7 y = 85 (6)

762

Section 6.2

31. Solve the system: 2 x1 − x2 + x3 = 3 (1)   x1 − x2 + x3 = 2 (2)  −2x + 2 x − 2x = −4 (3) 1 2 3  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (1) by −1 : − 2 x1 + x2 − x3 = −3 (4) Add (2) and (4) to eliminate x2 and x3 : −x1 = −1 x1 = 1 (5) Substitute (5) into (2): 1 − x2 + x3 = 2

−x2 + x3 = 1 (6)

32. Solve the system: x1 − x2 − 2 x3 = 0 (1)   −2x1 + 5 x2 + 10x3 = −3 (2)  (3) 3x1 + x2 = 0  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Solve (3) for x2 : x2 = −3x1 (4) Substitute (4) into (1): x1 − ( −3x1 ) − 2 x3 = 0 4 x1 − 2x3 = 0 (5) Substitute (4) into (2): −2 x1 + 5( −3x1 ) + 10 x3 = −3

−17x1 + 10 x3 = −3 (6) These steps yield the following system: (5)  4 x1 − 2 x3 = 0 (*)   −17 x1 + 10 x3 = −3 (6)

Substitute (5) into (3): −2 + 2 x2 − 2 x3 = −4 x2 − x3 = −1 (7)

These steps yield the following system:  −x2 + x3 = 1 (6) (*)   x2 − x3 = −1 (7) Step 2: Solve system (*) from Step 1. Add (6) and (7): 0 = 0 (8) Hence, we conclude from (8) that there are infinitely many solutions. To determine them, let x3 = a . Substitute this value into (7) to see that x2 = a − 1 . Thus, the solution is: x1 = 1, x2 = −1 + a, x3 = a . Step 2: Solve system (*) from Step 1. Multiply (5) by 5: 20 x1 − 10 x3 = 0 (7) Add (6) and (7): 3x1 = −3 x1 = −1 (8) Substitute (8) into (5) to find x3 : 4( −1) − 2 x3 = 0

−2 = x3 (9) Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (8) into (4) to find x2 : x2 = −3( −1) = 3 Thus, the solution is: x1 = −1, x2 = −3, x3 = −2 .

763

Chapter 6

33. Solve the system: 2 x + y − z = 2 (1)   x − y − z = 6 (2) Since there are two equations and three unknowns, we know there are infinitely many solutions (as long as neither statement is inconsistent). To this end, let z = a . Add (1) and (2): 3x − 2 z = 8 (3) Substitute z = a into (3) to find x: x = 23 a + 83 (4) Finally, substitute z = a and (4) into (2) to find y: y = − 13 a − 103 Thus, the solution is x = 23 a + 83 , y = − 13 a − 103 , z = a .

34. Solve the system:  3x + y − z = 0 (1)  x + y + 7 z = 4 (2) Since there are two equations and three unknowns, we know there are infinitely many solutions (as long as neither statement is inconsistent). To this end, let x = a . Multiply (2) by −1 : −x − y − 7 z = − 4 (3) Add (1) and (3): 2x − 8z = −4 or x − 4 z = −2 (4) Substitute x = a into (4) to find z: a − 4 z = −2  z = 14 ( a + 2) (5) Finally, substitute x = a and (5) into (1) to find y: 3a + y − 14 ( a + 2) = 0  y = − 114 a + 12 Thus, the solution is x = a, y = 14 (1 − 11a ), z = 14 ( a + 2) .

35. Let x = number of basic widgets ($10 each) y = number of mid-price widgets ($12 each) z = number of top-of-the-line widgets ($15 each) Solve the system: x + y + z = 300 (1)   10x + 12 y + 15z = 3,700 ( 2 )  x=z (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. To this end, substitute (3) into both (1) and (2) to obtain the following system:

 y + 2 z = 300 ( 4 ) (*)  12 y + 25 z = 3,700 ( 5 ) Step 2: Solve system (*) from Step 1. Solve (4) for y: y = 300 − 2z. (6) Substitute (6) into (5) and solve for z: z = 100 (7) Substitute (7) into (6) to find y: y = 100 (8) Substitute (7) and (8) into (1) to find x: x = 100. So, 100 basic widgets, 100 mid-price widgets, and 100 top-of-the-line widgets are produced.

764

Section 6.2

36. Let x = number of basic widgets ($10 each) y = number of mid-price widgets ($12 each) z = number of top-of-the-line widgets ($15 each) Solve the system: x + y + z = 325 (1)   10x + 12 y + 15z = 3,825 ( 2 )  x = 2z (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. To this end, substitute (3) into both (1) and (2) to obtain the following system:

y + 3z = 325 ( 4 )  (*)  12 y + 35 z = 3,825 ( 5 ) Step 2: Solve system (*) from Step 1. Solve (4) for y: y = 325 − 3z. (6) Substitute (6) into (5) and solve for z: z = 75 (7) Substitute (7) into (6) to find y: y = 100 (8) Substitute (7) and (8) into (1) to find x: x = 150. So, 150 basic widgets, 100 mid-price widgets, and 75 top-of-the-line widgets are produced.

37. Let x = number of touchdowns y = number of extra points z = number of field goals Solve the system: (1)  x + y + z = 18  x = z+4 (2)  6 x + y + 3z = 66 (3) 

Subtract (4) from (5) and solve for z: 7z = 28 so that z = 4. Substitute this back into (4) to get y = 6. Finally, substitute z = 4 into (2) to get x =8 Thus, there were 8 touchdowns, 6 extra points, and 4 field goals.

Substitute (2) into both (3) into (1) and simplify to get a 2x2 system in y and z: y + 2 z = 14 (4 ) y + 9z = 42 (5)

765

Chapter 6

38. Let x = number of two-point shots y = number of three-point shots z = number of one-point free throws. We must solve the system: x = z + 7 (1)   x = 4 y + 4 (2)  2 x + 3 y + z = 155 (3)  Step 1: Solve (1) for z and (2) for y :  z = x − 7 (4)   x−4 (5)  y = 4

Substitute (4) and (5) into (3) to find x : x−4 2x + 3   + ( x − 7 ) = 155  4  8x + 3x − 12 + 4 x − 28 = 620 15x − 40 = 620  x = 44 (6) Step 2: Find the value of the remaining variables: Substitute (6) into (4) and (5): z = 44 − 7 = 37

39. Let x = # Sriracha chicken sand. y = # tuna sandwiches z = # roast beef sandwiches We must solve the system: (1) x + y + z = 14   (*) 350x + 430 y + 290 z = 4840 (2)  (3) 18x + 19 y + 5z = 190  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Solve (1) for x: x = 14 − y − z (4) Substitute (4) into (2) and simplify: 4 y − 3z = −3 (5) Substitute (4) into (3) and simplify: y − 13z = −62 (6) These steps yield the following system: 4 y − 3z = −3 (5) (*)   y − 13z = −62 (6)

Step 2: Solve system (*) from Step 1. Solve (6) for y: y = 13z − 62 (7) Substitute (7) into (5) and simplify: z = 5 (8) Substitute (8) into (7) and simplify: y = 3 (8) Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (8) and (9) into (1) to find x: x = 6 (10) Thus, there are: 6 Sriracha chicken sandwiches, 3 tuna sandwiches, 5 roast beef sandwiches.

44 − 4 = 10 4 So, there are 44 two-point shots, 10 three-point shots, and 37 one-point free throws. y=

766

Section 6.2

40. Let x = # Sriracha chicken sand. y = # tuna sandwiches z = # roast beef sandwiches We must solve the system: (1) x + y + z = 14   (*) 350x + 430 y + 290 z = 4380 (2)  (3) 18x + 19 y + 5z = 123  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Solve (1) for x: x = 14 − y − z (4) Substitute (4) into (2) and simplify: 4 y − 3z = −26 (5) Substitute (4) into (3) and simplify: y − 13z = −129 (6)

These steps yield the following system: 4 y − 3z = −26 (5) (*)   y − 13z = −129 (6) Step 2: Solve system (*) from Step 1. Solve (6) for y: y = 13z − 129 (7) Substitute (7) into (5) and simplify: z = 10 (8) Substitute (8) into (7) and simplify: y = 1 (8) Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (8) and (9) into (1) to find x: x = 3 (10) Thus, there are: 3 Sriracha chicken sandwiches, 1 tuna sandwiches, 10 roast beef sandwiches.

41. The system that must be solved is: 36 = 12 a(1)2 + v0 (1) + h0  2 1 40 = 2 a(2) + v0 (2) + h0  2 1 12 = 2 a(3) + v0 (3) + h0

Step 2: Solve the system in Step 1. Multiply (4) by −6 and add to (5): 2h0 = 0

which is equivalent to 72 = a + 2v0 + 2 h0 (1)  40 = 2a + 2v0 + h0 (2) 24 = 9a + 6v + 2 h (3) 0 0  Step 1: Obtain a 2 × 2 system: Multiply (1) by −2 , and add to (2): −2v0 − 3h0 = −104 (4) Multiply (1) by −9 , and add to (3): −12v0 − 16 h0 = −624 (5) These steps yield the following 2 × 2 system:  −2v0 − 3h0 = −104 (4)  −12v0 − 16h0 = −624 (5)

h0 = 0

(6)

Substitute (6) into (4): −2v0 − 3(0) = −104 −2v0 = −104 v0 = 52

(7)

Step 3: Find values of remaining variables. Substitute (6) and (7) into (1): 72 = a + 2(52) + 2(0) −32 = a Thus, the polynomial has the equation h = −16t 2 + 52t .

767

Chapter 6

42. The system that must be solved is: 84 = 12 a(1)2 + v0 (1) + h0  2 1 136 = 2 a(2) + v0 (2) + h0  2 1 156 = 2 a(3) + v0 (3) + h0 which is equivalent to 168 = a + 2v0 + 2 h0 (1)  136 = 2a + 2v0 + h0 (2) 312 = 9a + 6v + 2 h (3) 0 0  Step 1: Obtain a 2 × 2 system: Multiply (1) by −2 , and add to (2): −2v0 − 3h0 = −200 (4) Multiply (1) by −9 , and add to (3): −12v0 − 16 h0 = −1200 (5) These steps yield the following 2 × 2 system:  −2v0 − 3h0 = −200 (4)  −12v0 − 16 h0 = −1200 (5) Step 2: Solve the system in Step 1. Multiply (4) by −6 and add to (5): 2h0 = 0 h0 = 0

(6)

Substitute (6) into (4): −2v0 − 3(0) = −200 −2v0 = −200 v0 = 100

(7)

Step 3: Find values of remaining variables. Substitute (6) and (7) into (1): 168 = a + 2(100) + 2(0) −32 = a Thus, the polynomial has the equation h = 12 ( −32) t 2 + 100t . 

43. Since we are given that the points (20,30), (40,60), and (60, 40) , the system that must be solved is: 30 = a(20)2 + b(20) + c  2 60 = a(40) + b(40) + c 40 = a(60)2 + b(60) + c  which is equivalent to 30 = 400a + 20b + c (1)  60 = 1600a + 40b + c (2) 40 = 3600a + 60b + c (3)  Step 1: Obtain a 2 × 2 system: Multiply (1) by −4 , and add to (2): −40b − 3c = −60 (4) Multiply (1) by −9 , and add to (3): −120b − 8c = −230 (5) These steps yield the following 2 × 2 system:  −40b − 3c = −60 (4)  −120b − 8c = −230 (5) Step 2: Solve the system in Step 1. Multiply (4) by −3 and add to (5): c = −50 (6) Substitute (6) into (4): −40b − 3( −50) = −60 −40b = −210 b = 214 = 5.25 (7) Step 3: Find values of remaining variables. Substitute (6) and (7) into (1): 30 = 400a + 20(5.25) + ( −50) −25 = 400a 0.0625 = a Thus, the polynomial has the equation y = −0.0625x 2 + 5.25x − 50 .

−16

768

Section 6.2

44. Since we are given that the points (0, 24), (50,22), and (100,26), the system that must be solved is:  (0,24): a(0)2 + b(0) + c = 24  c = 24  2  (50,22): a(50) + b(50) + c = 22 (100,26): a(100)2 + b(100) + c = 26   2500a + 50b = −2 10,000a + 100b = 2 ↓ Elimination Method −5000a − 100b = 4 10,000a + 100b = 2

3 2500 ( 2500 ) + 50b = −2

3 + 50b = −2 50b = −5

1 10 3 130 x = 130: (130)2 − + 24 2500 10 = 31.28 ≅ 31 years old. b=−

3 1 x 2 − x + 24 2500 10

5,000a = 6 a=

3 2500

45. Let x = amount in money market y = amount in mutual fund z = amount in stock The system we must solve is: x = y + 6000 (1)   x + y + z = 20,000 (2)  0.03x + 0.07 y + 0.10 z = 1180 (3)  Step 1: Obtain a 2 × 2 system: Substitute (1) into both (2) and (3): ( y + 6000) + y + z = 20,000 0.03( y + 6000) + 0.07 y + 0.10 z = 1180 This yields the following system: 2 y + z = 14,000 (4)   (5) 0.10 y + 0.10 z = 1000

Step 2: Solve the system in Step 1. Multiply (4) by −0.10 and add to (5) to find y: −0.10 y = −400 y = 4000 (6) Substitute (6) into (1) to find x: x = 6000 + 4000 = 10,000 (7) Substitute (6) and (7) into (2) to find z: 10,000 + 4000 + z = 20,000 z = 6000 Thus, the following allocation of funds should be made: $10,000 in money market $4000 in mutual fund $6000 in stock

769

Chapter 6

46. Let x = amount in money market y = amount in mutual fund z = amount in stock The system we must solve is: x + y + z = 20,000 (1)   z = y + 6000 (2)  0.03x + 0.07 y + 0.10 z = 1680 (3)  Step 1: Obtain a 2 × 2 system: Substitute (2) into both (1) and (3): x + y + ( y + 6000) = 20,000 0.03x + 0.07 y + 0.10( y + 6000) = 1680 This yields the following system: x + 2 y = 14,000 (4)   (5) 0.03x + 0.17 y = 1080

47. Let x = # regular models skis y = # trick skis z = # slalom skis Solve the system: (1)  x + y + z = 110  2x + 3 y + 3z = 297 ( 2 )  x + 2 y + 5 z = 202 ( 3 )  Step 1: Obtain a 2 × 2 system: Solve (1) for x: x = 110 − y − z ( 4 ) Substitute (4) into (2), and simplify: y + z = 77 ( 5 ) Substitute (4) into (3), and simplify: y + 4 z = 92 ( 6 ) This yields the following system:  y + z = 77 ( 5 ) (*)   y + 4 z = 92 ( 6 )

Step 2: Solve the system in Step 1. Solve (4) for x: x = 14,000 − 2 y (6) Substitute (6) into (5) and solve for y: 0.03(14,000 − 2 y ) + 0.17 y = 1080 0.11y = 660 y = 6000 (7) Substitute (7) into (6) to find x: x = 14,000 − 2(6000) = 2000 (8) Finally, substitute (7) into (2) to find z: z = 6000 + 6000 = 12,000 Thus, the following allocation of funds should be made: $2,000 in money market $6,000 in mutual fund $12,000 in stock Step 2: Solve system (*) from Step 1. Subtract (6) – (5): 3z = 15  z = 5 (7) Substitute (7) into (5): y = 72 ( 8 ) Substitute (7) and (8) into (1): x + 72 + 5 = 110  x = 33 So, need to sell 33 regular model skis, 72 trick skis, and 5 slalom skis.

770

Section 6.2

48. Let x = # compact cars y = # intermediate cars z = # luxury models Solve the system: (1)  x + y + z = 500  200x + 300 y + 250 z = 128,750 ( 2 )  30x + 20 y + 45z = 15,625 (3)  Step 1: Obtain a 2 × 2 system: Solve (1) for x: x = 500 − y − z ( 4 ) Substitute (4) into (2), and simplify: 2 y + z = 575 ( 5 ) Substitute (4) into (3), and simplify: −2 y + 3z = 125 ( 6 ) This yields the following system:  2 y + z = 575 ( 5 ) (*)   − 2 y + 3z = 125 ( 6 ) 49. Let x = number of seats in pit Let y = number of seats in general admission Let z = number of lawn seats Solve the system: (1) x + y + z = 8500  (2) 3x = y 275x + 150 y + 100 z = 1,175,000 (3)  Step 1: Obtain a 2 × 2 system: Substitute (2) into (1) and simplify: 4x + z = 8500 Substitute (2) into (3) and simplify: 725x + 100 z = 1,175,000 This yields the following system: (4) 4x + z = 8500 (* )  725x + 100 z = 1175000 (5)

Step 2: Solve system (*) from Step 1. Add (6) + (5): 4 z = 700  z = 175 (7) Substitute (7) into (5): y = 200 ( 8 ) Substitute (7) and (8) into (1): x + 200 + 175 = 500  x = 125 So, 125 compact cars, 200 intermediate cars, and 175 luxury models.

Step 2: Solve system ( * ) from Step 1. Solve (4) for z : z = 8500 − 4x (6) Substitute (6) into (5): 725x + 100 ( 8500 − 4 x ) = 1,175,000

 x = 1000 (7) Substitute (7) into (6): z = 8500 − 4 (1000 )  z = 4500 Substitute (7) into (2): y = 3 (1000 ) = 3000 So, there are 1000 seats in the pit area, 3000 seats in general admission, and 4500 lawn seats.

771

Chapter 6

50. Let x = speed of IBM’s Blue Gene/L y = speed of IBM’s BGW z = speed of IBM’s ASC Purple Solve the system: (1)  x = y + 245  (2)  y = z + 22  x + y + z = 568 ( 3 )  Step 1: Obtain a 2 × 2 system: Substitute (1) into (3) : 2 y + z = 323 ( 4 ) Solve the system:  y = z + 22 ( 2 ) (*)   2 y + z = 323 ( 4 )

Step 2: Solve system (*) from Step 1. Substitute (2) into (4), and simplify: z = 93 ( 5 ) Substitute (5) into (2), and simplify: y = 115 ( 6 ) Substitute (5) and (6) into (3): x + 115 + 93 = 568  x = 360 So, the Blue Gene/L runs at 360 teraflops, the BGW at 115 teraflops, and the ASC Purple at 93 teraflops.

51. Equation (2) and Equation (3) must be added correctly – should be 2x − y + z = 2 . Also, should begin by eliminating one variable from Equation (1).

52. From the system  y + 3z = 5   y + 3z = 9 one should (upon subtracting the equations) conclude that 0 = −4 . So, the system is inconsistent, and hence, has no solution.

53. True

54. False. Consider the system x + y = 1  x + y = −1

772

Section 6.2

55. Substitute the given points into the equation to obtain the following system:  ( −2)2 + 4 2 + a( −2) + b(4) + c = 0  12 + 12 + a(1) + b(1) + c = 0  ( −2)2 + ( −2)2 + a( −2) + b( −2) + c = 0  which is equivalent to (after simplification) −2a + 4b + c = −20 (1)  a + b + c = −2 (2)   −2a − 2b + c = −8 (3) 

Multiply (2) by 2 and then, add to (1): 6b + 3c = −24 (5) Substitute (4) into (5) to find b: 6b + 3( −4) = −24 6b = −12 b = −2 (6) Finally, substitute (4) and (6) into (2) to find a: a + ( −2) + ( −4) = −2 a=4 Thus, the equation is x 2 + y2 + 4x − 2 y − 4 = 0 .

Multiply (2) by 2 and then, add to (3): 3c = −12  c = −4 (4)

56. Substitute the given points into the equation to obtain the following system:  (0)2 + 72 + a(0) + b(7) + c = 0  6 2 + 12 + a(6) + b(1) + c = 0  (5)2 + (4)2 + a(5) + b(4) + c = 0  which is equivalent to (after simplification) 7b + c = −49 (1)    6a + b + c = −37 (2) 5a + 4b + c = −41 (3)  Solve (1) for c: c = −7b − 49 (4) Substitute (4) into both (2) and (3): 6a + b + ( −7b − 49) = −37

which is equivalent to 6a − 6b = 12 (5)   5a − 3b = 8 (6) Multiply (6) by −2 and then, add to (5): −4a = −4 a = 1 (7) Substitute (7) into (6) to find b: 5(1) − 3b = 8 b = −1 (8) Finally, substitute (8) into (4) to find c: c = −7( −1) − 49 = −42 Thus, the equation is approximately x 2 + y 2 + x − y − 42 = 0 .

5a + 4b + ( −7b − 49) = −41

773

Chapter 6

57. We deduce from the diagram that the following points are on the curve: ( −2, 46), ( −1,51), (0, 44), (1,51), (2, 43) Substituting these points into the given equation gives rise to the system: 46 = a ( −2 )4 + b ( −2 )3 + c ( −2 )2 + d ( −2 ) + e  51 = a ( −1)4 + b ( −1)3 + c ( −1)2 + d ( −1) + e  4 3 2  44 = a ( 0 ) + b ( 0 ) + c ( 0 ) + d ( 0 ) + e  4 3 2 51 = a (1) + b (1) + c (1) + d (1) + e  4 3 2 43 = a ( 2 ) + b ( 2 ) + c ( 2 ) + d ( 2 ) + e Observe that the third equation above simplifies to e = 44 . Substitute this value into the remaining four equations to obtain the following system: 16a − 8b + 4c − 2d = 2 (1)  a − b + c − d = 7 (2)   a + b + c + d = 7 (3)  16a + 8b + 4c + 2d = −1 (4) Now, add (2) and (3): 2a + 2c = 14 (5) Add (1) and (4): 32a + 8c = 1 (6)  2a + 2c = 14 (5) Solve the system:  32a + 8c = 1 (6) Multiply (5) by −4 , and then add to (6): 24a = −55 a = − 55 (7) 24

774

Substitute (7) into (5) to find c: 2( − 55 24 ) + 2c = 14 223 c = 7 + 55 (8) 24 = 24

At this point, we have values for three of the five unknowns. We now use (7) and (8) to obtain a 2 × 2 system involving these two remaining unknowns: Substitute (7) and (8) into (1): 223 2 = 16( − 55 24 ) − 8b + 4( 24 ) − 2d 36 = −192b − 48d (9) Substitute (7) and (8) into (2): 223 7 = − 55 24 − b + 24 − d 0 = b + d (10) Solve the system: 36 = −192b − 48d (9)  (10)  0 = b+d Solve (10) for b: b = −d (11) Substitute (11) into (9) to find d: 36 = −192( −d ) − 48d

36 = 144d 1 4

= d (12)

Substitute (12) into (10) to find b: b = − 14 Hence, the equation of the polynomial is 4 223 2 1 3 1 y = − 55 24 x − 4 x + 24 x + 4 x + 44 .

Section 6.2

58. Let x = number of nickels y = number of dimes z = number of quarters The system that must be solved is: x + y + z = 30 (1)   0.05x + 0.10 y + 0.25 z = 4.60 (2)  z = x + 4 (3) 

Substitute (3) into both (1) and (2): x + y + ( x + 4) = 30 0.05x + 0.10 y + 0.25( x + 4) = 4.60 This simplifies to the following system: 2x + y = 26 (4)   0.30 x + 0.10 y = 3.60 (5) Multiply (4) by −0.10 , and then add to (5): 0.10x = 1.0 x = 10 (6) Substitute (6) into (4) to find y: 2(10) + y = 26 y = 6 (7) Finally, substitute (6) and (7) into (1) to find z: 10 + 6 + z = 30 z = 14 Thus, there are 10 nickels, 6 dimes, and 14 quarters.

59. Solve the system 2y + z = 3 (1)   4x − z = −3 (2)   (3) 7x − 3 y − 3z = 2  x − y − z = −2 (4) Step 1: Obtain a 3 × 3 system: Solve (2) for z: z = 4 x + 3 (5) Substitute (5) into each of (1), (3), and (4). Simplifying yields the following system:  2 y + 4 x = 0 (6)  −3 y − 5x = 11 (7)  − y − 3x = 1 (8)  Step 2: Solve the system in Step 1. Solve (6) for y: y = −2 x (9) Substitute (9) into both (7) and (8) to obtain the following two equations: −3( −2x ) − 5x = 11 so that x = 11 (10) −( −2x ) − 3x = 1 so that x = −1 (11) Observe that (10) and (11) yield different values of x. As such, there can be no solution to this system.

775

Chapter 6

60. Solve the system: (1) −2x − y + 2 z = 3  3x − 4 z = 2 (2)   2x + y = −1 (3)   −x + y − z = −8 (4) Solve (3) for y: y = −2x − 1 (5) Substitute (5) into (1): −2x − ( −2x − 1) + 2 z = 3 z = 1 (6) Substitute (5) into (4): −x + ( −2x − 1) − z = −8 −3x − z = −7 (7) 61. Solve the system  3x1 − 2 x2 + x3 + 2 x4 = −2 (1)  − x + 3x + 4 x + 3x = 4 (2)  1 2 3 4  (3) x1 + x2 + x3 + x4 = 0   5x1 + 3x2 + x3 + 2 x4 = −1 (4) Step 1: Obtain a 3 × 3 system: Add (2) and (3): 4 x2 + 5x3 + 4 x4 = 4 (5) Multiply (2) by 3, and then add to (1): 7x2 + 13x3 + 11x4 = 10 (6) Multiply (2) by 5, and then add to (4): 18x2 + 21x3 + 17 x4 = 19 (7) These steps yield the following system: (5)  4 x2 + 5x3 + 4 x4 = 4   7 x2 + 13x3 + 11x4 = 10 (6) 18x + 21x + 17 x = 19 (7) 3 4  2 Step 2: Solve the system in Step 1. Multiply (5) by −7 , (6) by 4, and then add them: 17 x3 + 16x4 = 12 (8) Multiply (6) by 18, (7) by −7 , and then add them: 87x3 + 79x4 = 47 (9)

Substitute (6) into (7) to find x: −3x − 1 = −7 x = 2 (8) Observe that substituting (6) and (8) into both (2) and (7) yields true statements. As such, it remains to find y. To do so, substitute (8) into (5): y = −2(2) − 1 = −5 All equations are satisfied by the solution x = 2, y = −5, z = 1 .

These steps yield the following system:  17 x3 + 16 x4 = 12 (8)  87x3 + 79x4 = 47 (9) Solve this system: Multiply (8) by −87 , (9) by 17, and then add : −49x4 = −245 x4 = 5 (10) Substitute (10) into (8): 17 x3 + 16(5) = 12 x3 = −4 (11) Now, substitute (10) and (11) into (5): 4 x2 + 5( −4) + 4(5) = 4 x2 = 1 (12) Finally, substitute (10) – (12) into (3): x1 + 1 − 4 + 5 = 0 . x1 = −2 Thus, the solution is: x1 = −2, x2 = 1, x3 = −4, x4 = 5 .

776

Section 6.2

62. Solve the system  5x1 + 3x2 + 8x3 + x4 = 1 (1) x + 2 x + 5x + 2 x = 3 (2)  1 2 3 4  4 x1 + x3 − 2 x4 = −3 (3)   x2 + x3 + x4 = 0 (4) Step 1: Obtain a 3 × 3 system: Multiply (2) by −5 , and then add to (1): −7 x2 − 17 x3 − 9x4 = −14 (5) Multiply (2) by −4 , and then add to (3): −8x2 − 19x3 − 10 x4 = −15 (6) These steps yield the following system: x2 + x3 + x4 = 0 (4)    −7 x2 − 17 x3 − 9x4 = −14 (5) −8x − 19x − 10 x = −15 (6) 2 3 4  Step 2: Solve the system in Step 1. Multiply (4) by 7, and then add to (5): −10 x3 − 2 x4 = −14 (7) Multiply (4) by 8, and then add to (6): −11x3 − 2 x4 = −15 (8)

These steps lead to the following system: −10 x3 − 2 x4 = −14 (7)  −11x3 − 2 x4 = −15 (8) Solve this system: Multiply (7) by −1 , and then add to (8): −x3 = −1

63. See the solution to #25.

64. See the solution to #26.

x3 = 1 (9) Substitute (9) into (7): −10 − 2 x4 = −14 x4 = 2 (10) Now, substitute (9) and (10) into (4): x2 + 1 + 2 = 0 x2 = −3 (11) Finally, substitute (9) and (10) into (3): 4 x1 + 1 − 2(2) = −3 x1 = 0

Thus, the solution is: x1 = 0, x2 = −3, x3 = 1, x4 = 2 .

65. Write the system in the form:  z = x − y − 10   z = −2x + 3 y − 11  z = x − y − 10  A graphical solution is given to the right: Notice that there are only two planes since the first and third equations of the system are the same. Hence, the line of intersection of these two planes constitutes the infinitely many solutions of the system. This is precisely what was found to be the case in Exercises 25 and 63.

777

Chapter 6

66. Write the system in the form:  z = −2 x − y − 3  z = x + 2 y  z = 1 −x − y + 5 ) 2(  A graphical solution is given to the right: In this case, there are three planes which intersect in the point (−5, 4, 3). It is difficult to see in the picture, but they all meet at this point. This is precisely what was found to be the case in Exercises 26 and 64. 67. If your calculator has 3dimensional graphing capabilities, then generate the graph of the system to obtain:

68. If your calculator has 3dimensional graphing capabilities, the generate the graph of the system to obtain:

It’s a bit difficult to tell from the graph, but the three planes intersect in a common point with coordinates ( − 807 , − 807 , 487 ) , which can be easily

The two planes intersect in a line, meaning there are infinitely many solutions to the system. This line is described by x = a, y = 30 a5−24 , z = 3a6−2 .

verified using your calculator.

778

Section 6.3 Solutions -------------------------------------------------------------------------------1. d x ( x 2 − 25 ) = x( x − 5)( x + 5) has three

2. c x ( x 2 + 25 ) has one linear factor and one

distinct linear factors

irreducible quadratic factor.

3. a x 2 ( x 2 + 25 ) has one repeated linear

factor and one irreducible quadratic factor.

4. f x 2 ( x − 5)( x + 5) has one repeated linear factor, and two other distinct linear factors.

5. b

6. e

x ( x + 25 ) has one linear factor and 2

x 2 ( x 2 + 25 ) has one repeated linear 2

one repeated irreducible quadratic factor.

factor and one repeated irreducible quadratic factor.

8. 9 9 = x − x − 20 ( x − 5 )( x + 4 )

8 8 = x − 3x − 10 ( x − 5 )( x + 2 )

A B + x −5 x + 4

2x + 5 2x + 5 = 2 3 2 x − 4x x (x − 4) =

10.

A B C + + 2 x−4 x x

A B + x −5 x + 2

x 2 + 2x − 1 x2 + 2x − 1 = 2 x 4 − 9x 2 x ( x − 3 )( x + 3 ) =

A B C D + 2+ + x x x −3 x +3

11. A B Cx + D 2 x3 − 3x 2 + 5 2 x 3 − 3x 2 + 5 = = + + 2 4 2 x − 81 ( x + 3)( x − 3)( x + 9) x + 3 x − 3 x + 9

12. 8x 2 + 12 x − 17 8x 2 + 12 x − 17 A B Cx + D = = + + 2 4 2 x − 16 ( x + 2)( x − 2)( x + 4) x + 2 x − 2 x + 4

779

Chapter 6

13.

14. 3x − x + 9 3

( x + 10 ) 2

= 2

Ax + B Cx + D + 2 x + 10 ( x 2 + 10 )2

5x3 + 2 x 2 + 4

( x + 13) 2

Ax + B Cx + D + 2 x + 13 ( x 2 + 13 )2

15. The partial fraction decomposition has the form: 1 A B = + (1) x ( x + 1) x x + 1 To find the coefficients, multiply both sides of (1) by x( x + 1) , and gather like terms: 1 = A( x + 1) + Bx 1 = ( A + B )x + A (2) Equate corresponding coefficients in (2) to obtain the following system: A + B = 0 (3) (*)  A = 1 (4)  Now, solve system (*) : Substitute (4) into (3) to see that B = −1. Thus, the partial fraction decomposition (1) becomes: 1 1 1 = − x ( x + 1) x x + 1

16. The partial fraction decomposition has the form: 1 A B (1) = + x ( x − 1) x x − 1 To find the coefficients, multiply both sides of (1) by x( x − 1) , and gather like terms: 1 = A( x − 1) + Bx 1 = ( A + B )x − A (2) Equate corresponding coefficients in (2) to obtain the following system: A + B = 0 (3) (*)   − A = 1 (4) Now, solve system (*) : From (4), we see that A = −1 . Substitute this value into (3) to see that B = 1. Thus, the partial fraction decomposition (1) becomes: 1 1 1 = − x ( x − 1) x − 1 x

17. Observe that simplifying the expression first yields x 1 . = x ( x − 1) x − 1 This IS the partial fraction decomposition.

18. Observe that simplifying the expression first yields x 1 . = x ( x + 1) x + 1 This IS the partial fraction decomposition.

780

Section 6.3

19. The partial fraction decomposition has the form: 9x − 11 A B (1) = + ( x − 3)( x + 5 ) x − 3 x + 5 To find the coefficients, multiply both sides of (1) by ( x − 3 ) ( x + 5) , and gather like terms: 9x − 11 = A( x + 5) + B ( x − 3 )

9x − 11 = ( A + B ) x + ( 5A − 3B ) (2) Equate corresponding coefficients in (2) to obtain the following system: (3)  A+ B = 9 (*)  5A − 3B = −11 (4) Now, solve system (*): Multiply (3) by −5 : −5A − 5B = −45 (5) Add (5) and (3) to solve for B: −8B = −56 B = 7 (6) Substitute (6) into (3) to find A: A=2 Thus, the partial fraction decomposition (1) becomes: 9x − 11 2 7 = + ( x − 3 )( x + 5 ) x − 3 x + 5

20. The partial fraction decomposition has the form: 8x − 13 A B (1) = + ( x − 2 )( x + 1) x − 2 x + 1 To find the coefficients, multiply both sides of (1) by ( x − 2 ) ( x + 1) , and gather like terms: 8x − 13 = A( x + 1) + B ( x − 2 ) 8x − 13 = ( A + B ) x + ( A − 2B ) (2)

Equate corresponding coefficients in (2) to obtain the following system: (3)  A+ B = 8 (*)  A − 2 B = −13 (4) Now, solve system (*): Multiply (3) by 2: 2A + 2 B = 16 (5) Add (5) and (4) to solve for A: 3A = 3 A = 1 (6) Substitute (6) into (3) to find B: 1+ B = 8 B =7 Thus, the partial fraction decomposition (1) becomes: 8x − 13 1 7 = + ( x − 2 )( x + 1) x − 2 x + 1

781

Chapter 6

21. The partial fraction decomposition has the form: 17 − x A B = + (1) ( x + 3)( x + 8) x + 3 x + 8 To find the coefficients, multiply both sides of (1) by ( x + 3)( x + 8) , and gather like terms: 17 − x = A( x + 8) + B ( x + 3) (2) 17 − x = (8A + 3B ) + ( A + B )x Equate corresponding coefficients in (2) to obtain the following system:  8A + 3B = 17 (3) (*)  (4) A + B = −1 22. The partial fraction decomposition has the form: 7 −6 x A B (1) = + ( 2x + 1)( 2x + 3 ) 2x + 1 2x + 3 To find the coefficients, multiply both sides of (1) by (2x + 1)(2 x + 3), and gather like terms: 7 − 6 x = A(2x + 3) + B (2x + 1) 7 − 6 x = (3A + B ) + (2 A + 2B )x (2) Equate corresponding coefficients in (2) to obtain the following system: (3) 3A + B = 7 (*)  (4)  2 A + 2 B = −6

Now, solve system (*) : Multiply (4) by –3: −3A − 3B = 3 (5) Add (5) and (3) to solve for A: 5A = 20 A = 4 (6) Substitute (6) into (4) to find B: B = −5 Thus, the partial fraction decomposition (1) becomes: 17 − x 4 5 = − ( x + 3)( x + 8) x + 3 x + 8

Now, solve system (*) : Multiply (3) by –2: −6A − 2B = −14 (5) Add (5) and (4) to solve for A: −4A = −20 A = 5 (6) Substitute (6) into (3) to find B: B = −8 Thus, the partial fraction decomposition (1) becomes: 7 −6 x 5 8 = − (2 x + 1)(2 x + 3) (2 x + 1) 2 x + 3

782

Section 6.3

23. The partial fraction decomposition has the form: 8x − 31 A B = + (1) ( x − 2)( x − 5) x − 2 x − 5 To find the coefficients, multiply both sides of (1) by ( x − 2 )( x − 5 ) , and gather like terms: 8x − 31 = A( x − 5) + B ( x − 2) (2) 8x − 31 = ( A + B )x + ( −5A − 2B ) Equate corresponding coefficients in (2) to obtain the following system: (3)  A+ B = 8 (*)  −5A − 2 B = −31 (4) 24. The partial fraction decomposition has the form: 14 x − 4 A B = + (1) (2 x + 5)(4 x − 3) 2 x +5 4x − 3 To find the coefficients, multiply both sides of (1) by (2 x + 5)(4 x − 3), and gather like terms: 14 x − 4 = A(4x − 3) + B (2x + 5) 14 x − 4 = (4 A + 2B )x + ( −3A + 5B ) (2) Equate corresponding coefficients in (2) to obtain the following system: (3) 4 A + 2 B = 14 (*)  (4)  −3A + 5B = −4

Now, solve system (*): Multiply (3) by 2: 2A + 2 B = 16 (5) Add (5) and (4) to solve for A: −3A = −15 A = 5 (6) Substitute (6) into (3) to find B: B =3 Thus, the partial fraction decomposition (1) becomes: 8x − 31 5 3 = + ( x − 2)( x − 5) x − 2 x − 5

Now, solve system (*): Multiply (3) by 3: 12 A + 6B = 42 (5) Multiply (4) by 4: −12A + 20 B = −16 (6) Add (5) and (6) to solve for B: 26 B = 26 B = 1 (7) Substitute (7) into (3) to find A: A=3 Thus, the partial fraction decomposition (1) becomes: 14 x − 4 3 1 = + ( 2x + 5 )( 4x − 3) 2x + 5 4x − 3

783

Chapter 6

25. The partial fraction decomposition has the form: 3x + 1 A B (1) = + 2 2 ( x − 1) x − 1 ( x − 1)

26. The partial fraction decomposition has the form: 9y − 2 A B (1) = + 2 2 ( y − 1) y − 1 ( y − 1)

To find the coefficients, multiply both 2 sides of (1) by ( x − 1) , and gather like

To find the coefficients, multiply both 2 sides of (1) by ( y − 1) , and gather like

terms:

3x + 1 = A( x − 1) + B

9 y − 2 = A( y − 1) + B

3x + 1 = Ax + ( B − A) (2) Equate corresponding coefficients in (2) to obtain the following system: A = 3 (3)  (*)  B − A = 1 (4) Substitute (3) into (4) to find B: B −3 =1  B = 4 Thus, the partial fraction decomposition (1) becomes: 3x + 1 3 4 = + 2 2 ( x − 1) x − 1 ( x − 1)

9 y − 2 = Ay + ( B − A) (2) Equate corresponding coefficients in (2) to obtain the following system: A=9 (3)  (*)  B − A = −2 (4) Substitute (3) into (4) to find B: B − 9 = −2  B = 7 Thus, the partial fraction decomposition (1) becomes: 9y − 2 9 7 = + 2 2 ( y − 1) y − 1 ( y − 1)

27. The partial fraction decomposition has the form: 4x − 3 A B (1) = + 2 2 ( x + 3) x + 3 ( x + 3)

Equate corresponding coefficients in (2) to obtain the following system: A=4 (3)  (*)  3A + B = −3 (4) Substitute (3) into (4) to find B: 3(4) + B = −3  B = −15 Thus, the partial fraction decomposition (1) becomes: 4x − 3 4 15 = − 2 2 ( x + 3) x + 3 ( x + 3)

To find the coefficients, multiply both 2 sides of (1) by ( x + 3 ) , gather like terms: 4 x − 3 = A( x + 3) + B 4 x − 3 = Ax + ( 3A + B ) (2)

784

Section 6.3

28. The partial fraction decomposition has the form: 3x + 1 A B (1) = + 2 2 (x + 2) x + 2 (x + 2) To find the coefficients, multiply both 2 sides of (1) by ( x + 2 ) , and gather like terms:

3x + 1 = A( x + 2) + B 3x + 1 = Ax + ( 2 A + B ) (2)

Equate corresponding coefficients in (2) to obtain the following system: A = 3 (3)  (*)  2 A + B = 1 (4) Substitute (3) into (4) to find B: 2(3) + B = 1  B = −5 Thus, the partial fraction decomposition 3x + 1 3 5 (1) becomes: = − 2 2 ( x + 2) x + 2 (x + 2)

29. The partial fraction decomposition has the form: 4 x 2 − 32 x + 72 A B C (1) = + + 2 2 ( x + 1)( x − 5 ) x + 1 x − 5 ( x − 5 )

To find the coefficients, multiply both sides of (1) by ( x + 1)( x − 5 ) , and gather 2

like terms: 4 x 2 − 32 x + 72 = A ( x − 5 ) + B ( x − 5 )( x + 1) + C ( x + 1) 2

= A ( x 2 − 10 x + 25 ) + B ( x 2 − 4 x − 5 ) + C ( x + 1) = ( A + B ) x 2 + ( −10 A − 4 B + C ) x + ( 25A − 5B + C ) (2)

Equate corresponding coefficients in (2) to obtain the following system: (3) A+ B = 4   (*) −10 A − 4 B + C = −32 (4)  25A − 5B + C = 72 (5)  Solve (3) for B: B = 4 − A (6) Substitute (6) into (4): −10A − 4(4 − A) + C = −32 which is equivalent to − 6A + C = −16 (7) Substitute (6) into (5): 25A − 5(4 − A) + C = 72 which is equivalent to 30A + C = 92 (8) Now, solve the 2 × 2 system: −6 A + C = −16 (7)   30 A + C = 92 (8) Multiply (7) by −1, and add to (8) to find A: 36A = 108 so that A = 3. Substitute this value for A into (7) to find C: −6(3) + C = −16 so that C = 2. Finally, substitute the value of A into (3) to find B: 3 + B = 4 so that B = 1.

785

Chapter 6

Thus, the partial fraction decomposition (1) becomes: 4 x 2 − 32 x + 72

( x + 1)( x − 5 )

3 1 2 + + x + 1 x − 5 ( x − 5 )2

30. The partial fraction decomposition has the form: 4x2 − 7x − 3 A B C (1) = + + 2 2 ( x + 2 )( x − 1) x + 2 x − 1 ( x − 1)

To find the coefficients, multiply both sides of (1) by ( x + 2 )( x − 1) , and gather 2

like terms: 4 x 2 − 7 x − 3 = A ( x − 1) + B ( x − 1)( x + 2 ) + C ( x + 2 ) 2

= A ( x 2 − 2 x + 1) + B ( x 2 + x − 2 ) + C ( x + 2 ) = ( A + B ) x 2 + ( −2 A + B + C ) x + ( A − 2 B + 2C ) (2)

Equate corresponding coefficients in (2) to obtain the following system: A+ B = 4 (3)   (*)  −2 A + B + C = −7 (4)  A − 2 B + 2C = −3 (5)  Solve (3) for B: B = 4 − A (6) Substitute (6) into (4): −2A + (4 − A) + C = −7 which is equivalent to − 3A + C = −11 (7) Substitute (6) into (5): A − 2(4 − A) + 2C = −3 which is equivalent to 3A + 2C = 5 (8) Now, solve the 2 × 2 system: −3A + C = −11 (7)  (8)  3A + 2C = 5 Add (7) and (8) to find C: 3C = −6 so that C = −2. Substitute this value for C into (7) to find A: −3A − 2 = −11 so that A = 3. Finally, substitute the value of A into (3) to find B: 3 + B = 4 so that B = 1. Thus, the partial fraction decomposition (1) becomes: 4x 2 − 7x − 3

( x + 2 )( x − 1)

3 1 2 + − x + 2 x − 1 ( x − 1)2

786

Section 6.3

31. The partial fraction decomposition has the form: 5x 2 + 28x − 6 A Bx + C (1) = + 2 2 ( x + 4 ) ( x + 3) x + 4 x + 3

To find the coefficients, multiply both sides of (1) by ( x + 4 ) ( x 2 + 3 ) , and gather like terms:

5x 2 + 28x − 6 = A ( x 2 + 3 ) + ( Bx + C )( x + 4 ) = Ax 2 + 3A + Bx 2 + Cx + 4 Bx + 4C = ( A + B ) x 2 + ( 4B + C ) x + ( 3A + 4C ) (2)

Equate corresponding coefficients in (2) to obtain the following system:  A + B = 5 (3)  (*)  4 B + C = 28 (4) 3A + 4C = −6 (5)  Solve (3) for B: B = 5 − A (6) Substitute (6) into (4): 4(5 − A) + C = 28 which is equivalent to − 4A + C = 8 (7) Now, solve the 2 × 2 system:  3A + 4C = −6 (5)  −4A + C = 8 (7) Multiply (7) by −4 and then add to (5) to find A: 19A = −38 so that A = −2. Substitute this value for A into (7) to find C: −4( −2) + C = 8 so that C = 0. Finally, substitute the value of A into (3) to find B: −2 + B = 5 so that B = 7. Thus, the partial fraction decomposition (1) becomes: 5x 2 + 28x − 6 −2 7x = + 2 2 ( x + 4 ) ( x + 3) x + 4 x + 3

32. The partial fraction decomposition has the form: x 2 + 5x + 4 A Bx + C (1) = + 2 2 (x − 2) (x + 2) x − 2 x + 2

To find the coefficients, multiply both sides of (1) by ( x − 2 ) ( x 2 + 2 ) , and gather like terms:

787

Chapter 6

x 2 + 5x + 4 = A ( x 2 + 2 ) + ( Bx + C )( x − 2 ) = Ax 2 + 2 A + Bx 2 + Cx − 2 Bx − 2C = ( A + B ) x 2 + (C − 2 B ) x + ( 2 A − 2C ) (2)

Equate corresponding coefficients in (2) to obtain the following system:  A + B = 1 (3)  (*)  −2 B + C = 5 (4)  2 A − 2C = 4 (5)  Solve (3) for B: B = 1 − A (6) Substitute (6) into (4): −2(1 − A) + C = 5 which is equivalent to 2A + C = 7 (7) Now, solve the 2 × 2 system: 2 A − 2C = 4 (5)   2 A + C = 7 (7) Multiply (7) by −1 and then add to (5) to find C: −3C = −3 so that C = 1. Substitute this value for C into (7) to find A: 2 A + 1 = 7 so that A = 3. Finally, substitute the value of A into (3) to find B: 3 + B = 1 so that B = −2. Thus, the partial fraction decomposition (1) becomes: x 2 + 5x + 4 3 −2 x + 1 = + 2 2 ( x − 2) (x + 2) x − 2 x + 2

33. The partial fraction decomposition has the form: −2 x 2 − 17 x + 11 A Bx + C = + 2 (1) 2 ( x − 7 ) (3x − 7x + 5 ) x − 7 3x − 7x + 5    Irreducible Quadratic Term

To find the coefficients, multiply both sides of (1) by ( x − 7 ) ( 3x 2 − 7 x + 5 ) , and gather like terms: −2 x 2 − 17 x + 11 = A ( 3x 2 − 7 x + 5 ) + ( Bx + C )( x − 7 ) = 3Ax 2 − 7 Ax + 5A + Bx 2 + Cx − 7 Bx − 7C = ( 3A + B ) x 2 + ( −7 A − 7 B + C ) x + ( 5A − 7C ) (2)

Equate corresponding coefficients in (2) to obtain the following system: 3A + B = −2 (3)   (*)  −7 A − 7 B + C = −17 (4)  5A − 7C = 11 (5)  788

Section 6.3

Solve (3) for B: B = −2 − 3A (6) Substitute (6) into (4): −7A − 7( −2 − 3A) + C = −17 which is equivalent to 14A + C = −31 (7) Now, solve the 2 × 2 system: 5A − 7C = 11 (5)  14 A + C = −31 (7) Multiply (7) by 7 and then add to (5) to find A: 103A = −206 so that A = −2. Substitute this value for A into (5) to find C: 5( −2) − 7C = 11 so that C = −3. Finally, substitute the value of A into (3) to find B: 3( −2) + B = −2 so that B = 4. Thus, the partial fraction decomposition (1) becomes: −2 x 2 − 17 x + 11 −2 4x − 3 = + 2 2 ( x − 7 ) (3x − 7x + 5 ) x − 7 3x − 7x + 5

34. The partial fraction decomposition has the form: 14 x 2 + 8x + 40 A Bx + C (1) = + 2 ( x + 5 ) ( 2 x − 3x + 5 ) x + 5 2 x 2 − 3x + 5   Irreducible Quadratic Term

To find the coefficients, multiply both sides of (1) by ( x + 5) ( 2x 2 − 3x + 5 ) , and gather like terms: 14 x 2 + 8x + 40 = A ( 2 x 2 − 3x + 5 ) + ( Bx + C )( x + 5 ) = 2 Ax 2 − 3Ax + 5A + Bx 2 + Cx + 5Bx + 5C = ( 2 A + B ) x 2 + ( −3A + 5B + C ) x + ( 5A + 5C ) (2) Equate corresponding coefficients in (2) to obtain the following system: 2 A + B = 14 (3)   (*)  −3A + 5B + C = 8 (4)  5A + 5C = 40 (5)  Solve (3) for B: B = 14 − 2A (6) Substitute (6) into (4): −3A + 5(14 − 2A) + C = 8 which is equivalent to − 13A + C = −62 (7) Now, solve the 2 × 2 system:  5A + 5C = 40 (5)  −13A + C = −62 (7) Multiply (7) by −5 and then add to (5) to find A: 70 A = 350 so that A = 5.

789

Chapter 6

Substitute this value for A into (5) to find C: 5(5) + 5C = 40 so that C = 3. Finally, substitute the value of A into (3) to find B: 2(5) + B = 14 so that B = 4. Thus, the partial fraction decomposition (1) becomes: 14 x 2 + 8x + 40 5 4x + 3 = + 2 2 ( x + 5 ) ( 2 x − 3x + 5 ) x + 5 2 x − 3x + 5

35. The partial fraction decomposition has the form: x3 Ax + B Cx + D (1) = 2 + 2 2 ( x2 + 9) x + 9 ( x2 + 9)

To find the coefficients, multiply both sides of (1) by ( x 2 + 9 ) , and gather like 2

terms:

x3 = ( Ax + B ) ( x 2 + 9 ) + (Cx + D ) = Ax3 + Bx 2 + 9Ax + 9B + Cx + D = Ax3 + Bx 2 + ( 9A + C ) x + ( 9B + D ) (2)

Equate corresponding coefficients in (2) to obtain the following system: A = 1 (3)   B = 0 (4)  (*)   9A + C = 0 (5) 9B + D = 0 (6) Substitute (3) into (5) to find C: 9(1) + C = 0 so that C = −9 Substitute (4) into (6) to find D: 9(0) + D = 0 so that D = 0 Thus, the partial fraction decomposition (1) becomes: x3

( x + 9) 2

9x x − x + 9 ( x 2 + 9 )2 2

36. The partial fraction decomposition has the form: x2 Ax + B Cx + D (1) = 2 + 2 2 ( x2 + 9 ) x + 9 ( x2 + 9 )

To find the coefficients, multiply both sides of (1) by ( x 2 + 9 ) , and gather like 2

terms:

790

Section 6.3

x 2 = ( Ax + B ) ( x 2 + 9 ) + (Cx + D ) = Ax3 + Bx 2 + 9Ax + 9B + Cx + D = Ax3 + Bx 2 + ( 9A + C ) x + ( 9B + D ) (2) Equate corresponding coefficients in (2) to obtain the following system: A = 0 (3)   B = 1 (4)  (*)   9A + C = 0 (5) 9B + D = 0 (6) Substitute (3) into (5) to find C: 9(0) + C = 0 so that C = 0 Substitute (4) into (6) to find D: 9(1) + D = 0 so that D = −9 Thus, the partial fraction decomposition (1) becomes:

( x + 9) 2

1 9 − x + 9 ( x 2 + 9 )2 2

37. The partial fraction decomposition has the form: Cx + D 2 x3 − 3x 2 + 7 x − 2 Ax + B (1) = 2 + 2 2 2 2 x 1 + ( x + 1) ( x + 1)

To find the coefficients, multiply both sides of (1) by ( x 2 + 1) , and gather like 2

terms:

2 x3 − 3x 2 + 7 x − 2 = ( Ax + B ) ( x 2 + 1) + (Cx + D ) = Ax3 + Bx 2 + Ax + B + Cx + D = Ax3 + Bx 2 + ( A + C ) x + ( B + D ) (2)

Equate corresponding coefficients in (2) to obtain the following system: A = 2 (3)   B = −3 (4)  (*)   A + C = 7 (5)  B + D = −2 (6) Substitute (3) into (5) to find C: 2 + C = 7 so that C = 5 Substitute (4) into (6) to find D: −3 + D = −2 so that D = 1

791

Chapter 6

Thus, the partial fraction decomposition (1) becomes: 2 x 3 − 3x 2 + 7 x − 2

( x + 1) 2

2x − 3 5x + 1 + 2 x + 1 ( x 2 + 1)2

38. The partial fraction decomposition has the form: Cx + D −x3 + 2 x 2 − 3x + 15 Ax + B (1) = 2 + 2 2 2 2 x 8 + x x 8 8 + + ( ) ( )

To find the coefficients, multiply both sides of (1) by ( x 2 + 8 ) , and gather like 2

terms:

− x3 + 2 x 2 − 3x + 15 = ( Ax + B ) ( x 2 + 8 ) + (Cx + D ) = Ax3 + Bx 2 + 8Ax + 8B + Cx + D

= Ax3 + Bx 2 + ( 8A + C ) x + ( 8B + D ) (2) Equate corresponding coefficients in (2) to obtain the following system: A = −1 (3)   B = 2 (4)  (*)   8A + C = −3 (5) 8B + D = 15 (6) Substitute (3) into (5) to find C: 8( −1) + C = −3 so that C = 5 Substitute (4) into (6) to find D: 8(2) + D = 15 so that D = −1 Thus, the partial fraction decomposition (1) becomes:

−x3 + 2x 2 − 3x + 15

( x + 8) 2

−x + 2 5x − 1 + 2 x + 8 ( x 2 + 8 )2

39. The partial fraction decomposition has the form: 3x + 1 3x + 1 3x + 1 A B Cx + D (1) = 2 = = + + 2 4 2 2 x − 1 ( x − 1)( x + 1) ( x − 1)( x + 1) ( x + 1) x − 1 x + 1 x + 1

To find the coefficients, multiply both sides of (1) by ( x − 1)( x + 1) ( x 2 + 1) , and gather like terms: 3x + 1 = A( x + 1) ( x 2 + 1) + B ( x − 1) ( x 2 + 1) + (Cx + D ) ( x − 1)( x + 1) = Ax3 + Ax 2 + Ax + A + Bx3 − Bx 2 + Bx − B + Cx3 + Dx 2 − Cx − D = ( A + B + C )x3 + ( A − B + D )x 2 + ( A + B − C )x + ( A − B − D ) (2)

792

Section 6.3

Equate corresponding coefficients in (2) to obtain the following system:  A + B + C = 0 (3)  A − B + D = 0 (4)  (*)   A + B − C = 3 (5)  A − B − D = 1 (6) To solve this system, first obtain a 3 × 3 system: Add (4) and (6): 2A − 2B = 1 (7) This enables us to consider the following 3 × 3 system: A + B + C = 0 (3)   A + B − C = 3 (5)  2 A − 2 B = 1 (7)  Now, to solve this system, obtain a 2 × 2 system: Add (3) and (5): 2A + 2 B = 3 (8) This enables us to consider the following 2 × 2 system: 2 A − 2B = 1 (7)  2 A + 2B = 3 (8) Add (7) and (8) to find A: 4A = 4 so that A = 1 Substitute this value of A into (7) to find B: 2(1) − 2B = 1 so that B = 12 . Now, substitute these values of A and B into (6) to find D: 1 − 12 − D = 1 so that D = − 12 Finally, substitute these values of A and B into (3) to find C: 1 + 12 + C = 0 so that C = − 23 . Thus, the partial fraction decomposition (1) becomes: 3x + 1 1 1 −3x − 1 = + + 4 x − 1 x − 1 2 ( x + 1) 2 ( x 2 + 1)

40. The partial fraction decomposition has the form: 2−x 2−x 2−x A B Cx + D (1) = 2 = = + + 2 4 2 2 x − 81 ( x − 9 )( x + 9 ) ( x − 3)( x + 3) ( x + 9 ) x − 3 x + 3 x + 9 To find the coefficients, multiply both sides of (1) by ( x − 3)( x + 3) ( x 2 + 9 ) , and gather like terms:

793

Chapter 6

2 − x = A( x + 3) ( x 2 + 9 ) + B ( x − 3) ( x 2 + 9 ) + (Cx + D ) ( x − 3)( x + 3) = Ax3 + 3Ax 2 + 9Ax + 27 A + Bx3 − 3Bx 2 + 9Bx − 27 B + Cx3 + Dx 2 − 9Cx − 9D = ( A + B + C )x3 + (3A − 3B + D )x 2 + (9A + 9B − 9C )x + (27A − 27B − 9D ) (2) Equate corresponding coefficients in (2) to obtain the following system: A + B + C = 0 (3)   3A − 3B + D = 0 (4)  (*)   9A + 9B − 9C = −1 (5) 27 A − 27 B − 9D = 2 (6) To solve this system, first obtain a 3 × 3 system: Multiply (4) by 9, and then add to (6): 54A − 54B = 2 (7) This enables us to consider the following 3 × 3 system:  A + B + C = 0 (3)  9A + 9B − 9C = −1 (5)  54A − 54 B = 2 (7) 

Now, to solve this system, obtain a 2 × 2 system: Multiply (3) by 9, and then add to (5): 18A + 18B = −1 (8) This enables us to consider the following 2 × 2 system: 54 A − 54 B = 2 (7)   18A + 18B = −1 (8) 1 Multiply (8) by 3, and then add to (7) to find A: 108A = −1 so that A = − 108 5 1 Substitute this value of A into (8) to find B: 18( − 108 ) + 18B= − 1 so that B = − 108 . Now, substitute these values of A and B into (3) to find C: 5 1 − 108 − 108 +C = 0 so that C = 181 Finally, substitute these values of A and B into (4) to find D: 5 1 3( − 108 ) − 3( − 108 ) + D = 0 so that D = − 19 . Thus, the partial fraction decomposition (1) becomes: 5 1 1 1 − 108 − 108 2−x 18 x − 9 = + + x 4 − 81 x − 3 x + 3 x 2 + 9

41. The partial fraction decomposition has the form: 5x 2 + 9x − 8 A Bx + C = + 2 (1) 2 x − 1 x + 2x − 1 ( x − 1) ( x + 2x − 1)  Irreducible Quadratic Term

794

Section 6.3

To find the coefficients, multiply both sides of (1) by ( x − 1) ( x 2 + 2 x − 1) , and gather like terms: 5x 2 + 9x − 8 = A ( x 2 + 2 x − 1) + ( Bx + C )( x − 1) = Ax 2 + 2 Ax − A + Bx 2 + Cx − Bx − C = ( A + B ) x 2 + ( 2 A − B + C ) x + ( −A − C ) (2)

Equate corresponding coefficients in (2) to obtain the following system: A + B = 5 (3)   (*) 2 A − B + C = 9 (4)  − A − C = −8 (5)  Solve (3) for B: B = 5 − A (6) Substitute (6) into (4):

3A + C = 14 (7)

Now, solve the 2 × 2 system:

− A − C = −8 (5)   3A + C = 14 (7) Add (5) and (7) to find A: 2A = 6 so that A = 3. Substitute this value for A into (7) to find C: 3(3) + C = 14 so that C = 5. Finally, substitute the value of A into (3) to find B: B = 2. Thus, the partial fraction decomposition (1) becomes: 5x 2 + 9x − 8 3 2x + 5 = + 2 2 ( x − 1) ( x + 2x − 1) x − 1 x + 2x − 1

42. The partial fraction decomposition has the form: 10 x 2 − 5x + 29 A Bx + C = + 2 (1) 2 x − 3 x + 4x + 5 ( x − 3) ( x + 4x + 5 )    Irreducible Quadratic Term

To find the coefficients, multiply both sides of (1) by ( x − 3 ) ( x 2 + 4x + 5 ) , and gather like terms: 10 x 2 − 5x + 29 = A ( x 2 + 4x + 5 ) + ( Bx + C )( x − 3 ) = Ax 2 + 4 Ax + 5A + Bx 2 + Cx − 3Bx − 3C = ( A + B ) x 2 + ( 4A − 3B + C ) x + ( 5A − 3C ) (2)

795

Chapter 6

Equate corresponding coefficients in (2) to obtain the following system: A + B = 10 (3)   (*) 4 A − 3B + C = −5 (4)  5A − 3C = 29 (5)  Solve (3) for B: B = 10 − A (6) Substitute (6) into (4):

7A + C = 25 (7)

Now, solve the 2 × 2 system:

5A − 3C = 29 (5)   7 A + C = 25 (7) Multiply (7) by 3: 21A + 3C = 75 ( 8 ) Add (5) and (8) to find A: 26 A = 104 so that A = 4. Substitute this value for A into (5) to find C: 5(4) − 3C = 29 so that C = −3. Finally, substitute the value of A into (6) to find B: B = 6. Thus, the partial fraction decomposition (1) becomes: 10 x 2 − 5x + 29 4 6x − 3 = + 2 2 ( x − 3 ) ( x + 4x + 5 ) x − 3 x + 4x + 5

43. The partial fraction decomposition has the form: 3x 3x A Bx + C = = + 2 (1) 3 2 x − 1 ( x − 1) ( x + x + 1) x − 1 x + x + 1

To find the coefficients, multiply both sides of (1) by ( x − 1) ( x 2 + x + 1) , and gather like terms:

3x = A ( x 2 + x + 1) + ( Bx + C ) ( x − 1) = Ax 2 + Ax + A + Bx 2 + Cx − Bx − C = ( A + B )x 2 + ( A − B + C ) x + ( A − C ) (2)

Equate corresponding coefficients in (2) to obtain the following system: A + B = 0 (3)   (*)  A − B + C = 3 (4)  A − C = 0 (5)  Solve (3) for B: B = −A (6) Solve (5) for C: C = A (7) Substitute (6) and (7) into (4): A − ( −A) + ( A) = 3 so that A = 1.

796

Section 6.3

Substitute this value of A into (6) to find B: B = −1 Finally, substitute this value of A into (7) to find C: C = 1 Thus, the partial fraction decomposition (1) becomes: 3x 1 1− x = + 2 3 x −1 x −1 x + x + 1

44. The partial fraction decomposition has the form: 5x + 2 5x + 2 A Bx + C = = + 2 (1) 3 2 x − 8 ( x − 2) ( x + 2 x + 4 ) x − 2 x + 2x + 4

To find the coefficients, multiply both sides of (1) by ( x − 2) ( x 2 + 2 x + 4 ) , and gather like terms: 5x + 2 = A ( x 2 + 2x + 4 ) + ( Bx + C ) ( x − 2) = Ax 2 + 2 Ax + 4 A + Bx 2 + Cx − 2 Bx − 2C

= ( A + B )x 2 + ( 2 A − 2 B + C ) x + ( 4 A − 2C ) (2) Equate corresponding coefficients in (2) to obtain the following system: A + B = 0 (3)   (*) 2 A − 2 B + C = 5 (4)  4 A − 2C = 2 (5)  Solve (3) for B: B = −A (6) Solve (5) for C: C = 12 (4A − 2) = 2 A − 1 (7) Substitute (6) and (7) into (4): 2A − 2( −A) + 2A − 1 = 5 so that A = 1. Substitute this value of A into (6) to find B: B = −1 Finally, substitute this value of A into (7) to find C: C = 2(1) − 1 = 1 Thus, the partial fraction decomposition (1) becomes: 5x + 2 1 −x + 1 = + 2 3 x − 8 x − 2 x + 2x + 4

45. The partial fraction decomposition has the form: f ( di + d 0 ) A B (1) = + di d 0 di d 0 To find the coefficients, multiply both sides of (1) by di d 0 , and gather like terms: f ( di + d 0 ) = Ad 0 + Bdi fdi + fd 0 = Ad 0 + Bdi

797

Chapter 6

Equate corresponding coefficients in (2) to obtain the following system:  A = f (3) (*)   B = f (4) Thus, the partial fraction decomposition (1) becomes f ( di + d 0 ) f f (5) = + di d 0 di d 0 f f 1 1 1 + = 1, or as + = . Hence, using (5) enables us to write the lens law as di d 0 d 0 di f

46. The partial fraction decomposition has the form: 1 A B = + (1) n( n + 1) n n + 1 To find the coefficients, multiply both sides of (1) by n( n + 1), and gather like terms: 1 = A( n + 1) + Bn = ( A + B )n + A (2) Equate corresponding coefficients in (2) to obtain the following system: A + B = 0 (3) (*)  A = 1 (4)  Substitute (4) into (3) to find B: 1 + B = 0 so that B = −1 Thus, the partial fraction decomposition (1) becomes 1 1 1 = − n( n + 1) n n + 1 Thus, we can use this to compute the following sum: 1 1 1 1 + + . . . + + = 1⋅ 2 2 ⋅ 3 998 ⋅ 999 999 ⋅1000 1 1   1 1   1 1   1 1  1 999 − − =  − + − + . . . + +  = 1− 1000 1000 1 2   2 3   998 999   999 1000 

A Bx + C + . Once x x2 + 1 1 2x + 3 this correction is made, the correct decomposition is + 2 . x x +1

47. The form of the decomposition is incorrect. It should be

48. Should long divide first since the degree of the numerator is larger than the degree of the denominator. 798

Section 6.3

49. False. The degree of the numerator must be less than or equal to the degree of the denominator in order to apply the partial fraction decomposition procedure. 50. True. 51. The first step in forming the partial fraction decomposition of x 2 + 4x − 8 is to factor the denominator. To do so, we begin by applying the x3 − x 2 − 4 x + 4 Rational Root Test: Factors of 4: ±1, ± 2, ± 4 Factors of 1: ±1 Possible Rational Zeros: ±1, ± 2, ± 4 Applying synthetic division to the zeros, one can see that 1 is a rational zero: 1 1 −1 − 4 4 1 0 −4 1

−4

So, x3 − x 2 − 4 x + 4 = ( x − 1)( x 2 − 4) = ( x − 1)( x − 2)( x + 2) . Now, the partial fraction decomposition has the form: x 2 + 4x − 8 x 2 + 4x − 8 A B C = = + + (1) 3 2 x − x − 4 x + 4 ( x − 1)( x − 2)( x + 2) x − 1 x − 2 x + 2 To find the coefficients, multiply both sides of (1) by ( x − 1)( x − 2)( x + 2) , and gather like terms: x 2 + 4 x − 8 = A( x − 2)( x + 2) + B ( x − 1)( x + 2) + C ( x − 1)( x − 2) = A( x 2 − 4) + B ( x 2 + x − 2) + C ( x 2 − 3x + 2) = ( A + B + C )x 2 + ( B − 3C )x + ( −4A − 2 B + 2C ) (2) Equate corresponding coefficients in (2) to obtain the following system: A + B + C = 1 (3)   (*)  B − 3C = 4 (4) −4 A − 2 B + 2C = −8 (5)  Now, solve system (*) : Multiply (3) by 4 and then, add to (5): 2B + 6C = −4 (6)

799

Chapter 6

This leads to the following 2 × 2 system: (4) B − 3C = 4  2B + 6C = −4 (6) Multiply (4) by −2 and then add to (6) to find C: 12C = −12 so that C = −1 Substitute this value of C into (4) to find B: B − 3( −1) = 4 so that B = 1. Finally, substitute these values of B and C into (3) to find A: A + 1 − 1 = 1 so that A = 1 . Thus, the partial fraction decomposition (1) becomes: x 2 + 4x − 8 1 1 1 = − + 3 2 x − x − 4x + 4 x − 1 x + 2 x − 2

52. Case 1: Assume c ≠ 0 . Then, the partial fraction decomposition has the form: ax + b ax + b A B (1) = = + 2 2 x −c ( x − c )( x + c ) x − c x + c To find the coefficients, multiply both sides of (1) by ( x − c )( x + c ) , and gather like terms: ax + b = A( x + c ) + B ( x − c ) = ( A + B ) x + ( Ac − Bc ) (2)

Equate corresponding coefficients in (2) to obtain the following system: A + B = a (3)  A+ B = a (*)  which is equivalent to (*)  b  Ac − Bc = b  A − B = c (4) Now, solve system (*) : Add (3) and (4) to find A: 2 A = a + bc so that A = 12 ( a + bc ) = ac2+c b . Substitute this value of A into (3) to find B: ac + b 2c + B = a

B = a − 12 ( a + bc ) = a2 − 2bc = ac2−c b

Thus, the partial fraction decomposition (1) in this case becomes: ac + b ac − b ax + b  ac + b ac − b  2c 2c = + = 21c  + 2 2 x −c x −c x +c x + c   x −c

Case 2: Assume c = 0 . Then,

ax + b ax + b a b = = + 2. x2 − c2 x2 x x

800

Section 6.3

53. The partial fraction decomposition has the form: 2x3 + x 2 − x − 1 2x3 + x 2 − x − 1 A B C D = = + 2+ 3+ (1) 4 3 3 x +x x ( x + 1) x x x x +1 To find the coefficients, multiply both sides of (1) by x3 ( x + 1) , and gather like terms: 2 x3 + x 2 − x − 1 = Ax 2 ( x + 1) + Bx ( x + 1) + C ( x + 1) + Dx3

= Ax3 + Ax 2 + Bx 2 + Bx + Cx + C + Dx3 = ( A + D ) x3 + ( A + B ) x 2 + ( B + C ) x + C (2) Equate corresponding coefficients in (2) to obtain the following system: A + D = 2 (3) A + B = 1 (4)  (*)   B + C = −1 (5)  C = −1 (6) Substitute (6) into (5) to find B: B − 1 = −1 so that B = 0 . Substitute this value of B into (4) to find A: A + 0 = 1 so that A = 1 . Substitute this value of A into (3) to find D: 1 + D = 2 so that D = 1 . Thus, the partial fraction decomposition (1) becomes: 2x3 + x 2 − x − 1 1 1 1 = + − 3 4 3 x +x x x +1 x

54. The partial fraction decomposition has the form: −x3 + 2x − 2 − x3 + 2 x − 2 A B C D E = = + 2+ 3+ 4+ (1) 5 4 4 x −x x ( x − 1) x x x x x −1 To find the coefficients, multiply both sides of (1) by x 4 ( x − 1) , and gather like terms: −x3 + 2 x − 2 = Ax3 ( x − 1) + Bx 2 ( x − 1) + Cx ( x − 1) + D ( x − 1) + Ex 4

= Ax 4 − Ax3 + Bx3 − Bx 2 + Cx 2 − Cx + Dx − D + Ex 4 = ( A + E ) x 4 + ( −A + B ) x3 + ( −B + C ) x 2 + ( −C + D ) x − D (2)

801

Chapter 6

Equate corresponding coefficients in (2) to obtain the following system:  A + E = 0 (3)  − A + B = −1 (4)  (*)  − B + C = 0 (5) −C + D = 2 (6)  −D = −2 (7)  From (7), we know that D = 2 . Substitute this value of D into (6) to find C: −C + 2 = 2 so that C = 0 . Substitute this value of C into (5) to find B: − B + 0 = 0 so that B = 0 . Substitute this value of B into (4) to find A: − A + 0 = −1 so that A = 1 . Substitute this value of A into (3) to find E: 1 + E = 0 so that E = −1 . Thus, the partial fraction decomposition (1) becomes: −x3 + 2 x − 2 1 2 1 = + 4− 5 4 x −x x x x −1

55. The partial fraction decomposition has the form: x5 + 2 Ax + B Cx + D Ex + F (1) = 2 + + 3 2 3 ( x2 + 1) x + 1 ( x2 + 1) ( x2 + 1)

To find the coefficients, multiply both sides of (1) by ( x 2 + 1) , and gather like 3

terms: x5 + 2 = ( Ax + B ) ( x 2 + 1) + (Cx + D ) ( x 2 + 1) + ( Ex + F )    2

x 4 + 2 x2 +1

= Ax + Bx + 2 Ax3 + 2 Bx 2 + Ax + B + Cx3 + Dx 2 + Cx + D + Ex + F 5

= Ax5 + Bx 4 + ( 2 A + C ) x3 + ( 2 B + D ) x 2 + ( A + C + E ) x + ( B + D + F ) (2)

Equate corresponding coefficients in (2) to obtain the following system: A = 1 (3)   B = 0 (4)   2 A + C = 0 (5) (*)   2B + D = 0 (6)  A + C + E = 0 (7)  B + D + F = 2 (8) Substitute (3) into (5) to find C: 2 + C = 0 so that C = −2. Substitute (4) into (6) to find D: 0 + D = 0 so that D = 0.

802

Section 6.3

Substitute the values of A and C into (7) to find E: 1 + ( −2) + E = 0 so that E = 1 . Substitute the values of B and D into (8) to find F: 0 + 0 + F = 2 so that F = 2 . Thus, the partial fraction decomposition (1) becomes:

x5 + 2

( x + 1)

x 2x x+2 − + 2 x + 1 ( x 2 + 1) ( x 2 + 1)3 2

56. The partial fraction decomposition has the form: x2 − 4 Ax + B Cx + D Ex + F (1) = 2 + + 3 2 3 ( x2 + 1) x + 1 ( x2 + 1) ( x2 + 1)

To find the coefficients, multiply both sides of (1) by ( x 2 + 1) , and gather like 3

terms: x 2 − 4 = ( Ax + B ) ( x 2 + 1) + (Cx + D ) ( x 2 + 1) + ( Ex + F )    2

x 4 + 2 x 2 +1

= Ax + Bx + 2 Ax3 + 2 Bx 2 + Ax + B + Cx3 + Dx 2 + Cx + D + Ex + F 5

= Ax5 + Bx 4 + ( 2 A + C ) x3 + ( 2 B + D ) x 2 + ( A + C + E ) x + ( B + D + F ) (2)

Equate corresponding coefficients in (2) to obtain the following system: A = 0 (3)   B = 0 (4)   2 A + C = 0 (5) (*)  (6)  2B + D = 1  A + C + E = 0 (7)  B + D + F = −4 (8) Substitute (3) into (5) to find C: 0 + C = 0 so that C = 0 . Substitute (4) into (6) to find D: 2(0) + D = 1 so that D = 1 . Substitute the values of A and C into (7) to find E: 0 + 0 + E = 0 so that E = 0 . Substitute the values of B and D into (8) to find F: 0 + 1 + F = −4 so that F = −5 . Thus, the partial fraction decomposition (1) becomes:

x2 − 4

( x + 1) ( x + 1) 2

803

−

( x + 1) 2

Chapter 6

57.

Notes on the graph: Solid curve: Graph of y1 , Dashed curve: Graph of y2 In this case, since the graphs coincide, we know y2 is, in fact, the partial fraction decomposition of y1 . 59.

Notes on the graph: Solid curve: Graph of y1 , Dashed curve: Graph of y2 In this case, since the graphs do not coincide, we know y2 is not the partial fraction decomposition of y1 .

58.

Notes on the graph: Solid curve: Graph of y1 , Dashed curve: Graph of y2 In this case, since the graphs coincide, we know y2 is, in fact, the partial fraction decomposition of y1 . 60.

Notes on the graph: Solid curve: Graph of y1 , Dashed curve: Graph of y2 In this case, since the graphs do not coincide, we know y2 is not the partial fraction decomposition of y1 .

804

Section 6.3

61. Since the graphs coincide, as seen below, y2 is the partial fraction decomposition of y1.

62. Since the graphs do not coincide, as seen below, y2 is not the partial fraction decomposition of y1.

805

Chapter 6

Section 6.4 Solutions -------------------------------------------------------------------------------1. d Above the line y = x , and do not include the line itself.

2. c Above the line y = x , and do include the line itself.

3. b Below the line y = x , and do not include the line itself.

4. a Below the line y = x , and do include the line itself.

The equation of dashed curve is y = x − 1 .

The equation of solid curve is y = −x + 1 .

The equation of the solid curve is y = −x .

The equation of the dashed curve is y = −x .

806

Section 6.4

10.

The equation of the solid curve is y = −3x + 2.

The equation of the dashed curve is y = 2 x + 3.

11.

12.

The equation of the solid curve is y = −2x + 1.

The equation of the dashed curve is y = 3x − 2.

13. Write the inequality as y < 14 (2 − 3x ).

14. Write the inequality as y > − 13 (2x + 6).

The equation of the dashed curve is y = 14 (2 − 3x ).

The equation of the dashed curve is y = − 13 (2x + 6).

807

Chapter 6

15.

16.

The equation of the dashed curve is y = − 53 x + 5.

The equation of the solid curve is y = 54 x − 4.

17. Write the inequality as y ≤ 2 x − 3 .

18. Write the inequality as y ≤ 2 x − 3 .

The equation of the solid curve is y = 2x − 3.

19.

20.

The equation of the solid curve is y = − 32 x + 3.

The equation of the solid curve is y = 52 x − 5.

808

Section 6.4

21.

22.

Notes on the graph: C1: y = x − 1 C2: y = x + 1

There is no common region, hence the system has no solution. Notes on the graph: C1: y = x + 1 C2: y = x − 1

23.

24.

There is no common region, hence the system has no solution. Notes on the graph: C1: y = 2x + 1 C2: y = 2x − 1

There is no common region, hence the system has no solution. Notes on the graph: C1: y = 2 x + 1 C2: y = 2 x − 1

809

Chapter 6

25.

26.

Notes on the graph: 2 4 C1: y = x − 3 3 1 5 C2: y = − x + 7 7

Notes on the graph: 3 C1: y = x − 3 4 2 7 C2: y = x − 5 5

27.

28.

Notes on the graph: 2 C1: y = x − 3 3 4 C2: y = − x + 2 3

Notes on the graph: 5 C1: y = − x − 4 2 C2: y = 6 x − 4

810

Section 6.4

29.

In this case, the common region is the line itself. Notes on the graph: C1 and C2: y = 2x

30. There is no solution in this case since there are no points that lie both strictly above and strictly below the given line. 31.

32.

811

Chapter 6

33.

34.

Notes on the graph: C1: y = x

35.

36.

Notes on the graph: C1: y = x C2: x = 0 C3: y = 4

Notes on the graph: C1: y = x

812

Section 6.4

37.

38.

Notes on the graph: C1: y = −x + 2 C2: y = 1 C3: x = 0

Notes on the graph: C1: y = −x + 4

39.

40.

Notes on the graph: C1: y = x + 1 C2: y = 3 C3: x = 0

Notes on the graph: C1: y = x − 2

813

Chapter 6

41.

42.

43.

44.

Notes on the graph: C1: y = x − 1 C2: y = − x + 3 C3: y = x + 2

Notes on the graph: C1: y = −x + 4 C2: y = x − 4 C3: y = −x − 4

814

Section 6.4

45.

Notes on the graph: C1: y = −x − 4 C2: y = x + 2 C3: y = −1 C4: y = 1

46.

Notes on the graph: C1: y = x + 2 C2: y = x − 2 C3: y = −x + 2 C4: y = −x − 2

47.

48.

Notes on the graph: C1: y = x + 3 C2: y = −x + 1

Notes on the graph: C1: y = −x + 2 C2: y = x − 3 815

Chapter 6

49. There is no solution since the regions do not overlap. 50.

51.

Notes on the graph: C1: y = x + 3 C2: y = −x + 3 52.

53. The shaded region is a triangle, seen below. Note that the base b has length 4, and the height h is 2. Hence, the area is: Area = 12 (4)(2) = 4 units 2

Notes on the graph: C1: y = x C2: y = 2

816

Section 6.4

54. Consider the following graph:

The shaded region is a right triangle.

55.

Area of R1: 12 (1)(5) = 25 units 2

The area is 12 (3)(3) = 92 units 2 .

Area of R2: 1(5) = 5 units 2 So, the area of the shaded region is 7.5 units2.

56.

Area of R1: 12 (1)(5) = 52 units 2 Area of R2: 1(5) = 5 units 2 So, the area of the shaded region is 7.5 units2.

817

Chapter 6

57. Let x = number of cases of water y = number of generators Certainly, x ≥ 0, y ≥ 0. Then, we also have the following restrictions: Cubic feet restriction: x + 20 y ≤ 2400 Weight restriction: 25x + 150 y ≤ 6000 So, we obtain the following system of inequalities: x ≥ 0, y ≥ 0  x + 20 y ≤ 2400 25x + 150 y ≤ 6000 

58. Let x = number of cases of plywood y = number of cases of tarps Certainly, x ≥ 0, y ≥ 0. Then, we also have the following restrictions: Cubic feet restriction: 60x + 10 y ≤ 1500 Weight restriction: 500x + 50 y ≤ 2000 So, we obtain the following system of inequalities: x ≥ 0, y ≥ 0  60 x + 10 y ≤ 1500 500 x + 50 y ≤ 2000 

Notes on the graph: C1: x + 20 y = 2400 C2: 25x + 150 y = 6000

Notes on the graph: C1: 60x + 10 y = 1500 C2: 500x + 50 y = 2000

818

Section 6.4

59. Let x = number of ounces of food A, y = number of ounces of food B. a.

275 ≤ 10x + 20 y, x ≥ 0, y ≥ 0 125 ≤ 15x + 10 y 200 ≤ 20x + 15 y b.

y 14 12 10 8 6 4 2

x 2

c. Two possible diet combinations are 2 ounces of food A and 14 ounces of food B or 10 ounces of food A and 10 ounces of food B. 60. Let x = number of ounces of food A, y = number of ounces of food B. a.

350 ≤ 15x + 25 y,

x ≥ 0, y ≥ 0

175 ≤ 25x + 10 y 225 ≤ 20x + 10 y b.

y 14 12 10 8 6 4 2

x 2

c. Two possible diet combinations are 4 ounces of food A and 15 ounces of food B or 6 ounces of food A and 12 ounces of food B. 819

Chapter 6

61. Let x = number of USB wireless mice, y = number of Bluetooth mice. a.

x ≥ 2 y, x + y ≥ 1000 b.

x ≥ 0, y ≥ 0

c. Two possible solutions would be for the manufacturer to produce 700 USB wireless mouse and 300 Bluetooth mouse or 800 USB wireless mouse and 300 Bluetooth mouse. 62. Let x = number of pieces of 0.5 mm pencil lead, y = number of pieces of 0.7 mm pencil lead. a.

x ≥ 1.5 y, x + y ≥ 10000 b.

x ≥ 0, y ≥ 0

c. Two possible solutions would be for the manufacturer to produce 7000 pieces of 0.5 mm pencil lead and 4000 pieces of 0.7 mm pencil lead or 8000 pieces of 0.5 mm pencil lead and 3000 pieces of 0.7 mm pencil lead. 820

Section 6.4

63. First, find the point of intersection of  P = 80 − 0.01x   P = 20 + 0.02 x Equating these and solving for x yields x = 2000. Then, substituting this into the first equation yields P = 60. Hence, the system for consumer surplus is:  P ≤ 80 − 0.01x   P ≥ 60 x≥0 

64. First, find the point of intersection of  P = 80 − 0.01x   P = 20 + 0.02 x Equating these and solving for x yields x = 2000. Then, substituting this into the first equation yields P = 60. Hence, the system for producer surplus is:  P ≥ 20 + 0.02 x   P ≤ 60 x≥0 

65. The graph of the system in Exercise 59 is as follows:

66. The graph of the system in Exercise 60 is as follows:

The consumer surplus is the area of the shaded region, which is 2 1 2 (20)(2000) = 20,000 units .

The producer surplus is the area of the shaded region, which is 2 1 2 (40)(2000) = 40,000 units .

67. The shading should be above the line.

68. The shading should not include the actual line (the line should be dashed).

69. True. The line cuts the plane into two half planes, and one must either shade above or below the line.

70. True. 71. False A dashed curve is used.

821

Chapter 6

72. False. The following linear system has no solution, as is seen graphically below: y > x +1  y < x −1

73. Given that a < b and c < d , the solution region is a shaded rectangle which includes the upper and left sides – shown below:

Notes on the graph: C1: y = x + 1 C2: y = x − 1

74. In the case when a > b and c > d , there is no solution since there is no common region.

75. For any value of b, the following system has a solution:  y ≤ ax + b   y ≥ −ax + b The solution region occurs in the first and fourth quadrants, and is shown graphically below:

If a = 0 , then the system becomes: y ≤ b  y ≥ b The solution to this system is the horizontal line y = b .

822

Section 6.4

76. For any value of b, the following system has a solution:  y ≤ ax + b   y ≥ −ax + b The solution region occurs in the second and third quadrants, and is shown graphically below:

77.

78.

The solid curve is the graph of y = 2x − 3 .

79.

80.

823

Section 6.5 Solutions -------------------------------------------------------------------------------1. Vertex

( −1, 4) (2, 4) ( −2, −1) (1, −1)

Objective Function z = 2x + 3 y z = 2( −1) + 3(4) = 10 z = 2(2) + 3(4) = 16 z = 2(−2) + 3( −1) = −7 z = 2(1) + 3( −1) = −1

2. Vertex

( −1, 4) (2, 4) ( −2, −1) (1, −1)

Objective Function z = 3x + 2 y z = 3( −1) + 2(4) = 5 z = 3(2) + 2(4) = 14 z = 3( −2) + 2(−1) = −8 z = 3(1) + 3(−1) = 0

So, the maximum value of z is 16, and the minimum value of z is −7.

So, the maximum value of z is 14, and the minimum value of z is −8.

3. Vertex

4. Vertex

( −1, 4 ) ( 2, 4 ) ( −2, −1) (1, −1)

Objective Function z = 4x − y z = 4 ( −1) − ( 4 ) = −8 z = 4 (2) − ( 4) = 4 z = 4 ( −2 ) − ( −1) = −7 z = 4 (1) − ( −1) = 5

( −1, 4 ) ( 2, 4 ) ( −2, −1) (1, −1)

Objective Function z = −x − 3 y z = − ( −1) − 3 ( 4 ) = −11 z = − ( 2 ) − 3 ( 4 ) = −14 z = − ( −2 ) − 3 ( −1) = 5 z = − (1) − 3 ( −1) = 2

So, the maximum value of z is 5, and the minimum value of z is –8.

So, the maximum value of z is 5, and the minimum value of z is –14.

5. Vertex

6. Vertex

( −1, 4) (2, 4) ( −2, −1) (1, −1)

Objective Function z = 1.5x + 4.5 y z = 1.5( −1) + 4.5(4) = 16.5 z = 1.5(2) + 4.5(4) = 21 z = 1.5( −2) + 4.5(−1) = −7.5 z = 1.5(1) + 4.5(−1) = −3

So, the maximum value of z is 21, and the minimum value of z is −7.5.

Objective Function z = 23 x + 35 y

( −1, 4)

26 z = 23 ( −1) + 53 (4) = 15

(2, 4)

56 z = 23 (2) + 35 (4) = 15

( −2, −1)

29 z = 23 ( −2) + 53 ( −1) = − 15

(1, −1)

z = 23 (1) + 53 ( −1) = 151

56 So, the maximum value of z is 15 , and 29 the minimum value of z is − 15 .

824

Section 6.5

7. The region in this case is:

Notes on the graph: C1: y = 4 + x C2: y = 0 C3: x = 0 Vertex

(0,0) (0, 4)

Objective Function z = 7x + 4 y z = 7(0) + 4(0) = 0 z = 7(0) + 4(4) = 16

So, the minimum value of z is 0.

8. The region in this case is: Vertex

(0, 4) (5,10)

Objective Function z = 3x + 5 y 20 65

Since the region is unbounded and the maximum must occur at a vertex, we conclude that the objective function does not have a maximum in this case.

825

Chapter 6 9. The region in this case is:

Notes on the graph: C1: y = 4 − x C2: y = −x C3: x = 0 Vertex

(0,0) (0, 4) Additional point (4,0)

Objective Function z = 4x + 3 y z = 4(0) + 3(0) = 0 z = 4(0) + 3(4) = 12 z = 4(4) + 3(0) = 16

Since the region is unbounded and the maximum must occur at a vertex, we conclude that the objective function does not have a maximum in this case. 10. The region in this case is:

Notes on the graph: C1: y = 10 − x C2: y = 0 C3: x = 0 Vertex

(0,0) (10,0) (0,10)

Objective Function z = 4x + 3 y z = 4(0) + 3(0) = 0 z = 4(10) + 3(0) = 40 z = 4(0) + 3(10) = 30

So, the minimum value of z is 0. 11. The region in this case is:

Notes on the graph: C1: y = x + 2 C2: y = −x + 6 C3: x = 4 Vertex

(2, 4) (0,0) (4,2) (0,2) (4, 0)

Objective Function z = 2.5x + 3.1y z = 2.5(2) + 3.1(4) = 17.4 z = 2.5(0) + 3.1(0) = 0 z = 2.5(4) + 3.1(2) = 16.2 z = 2.5(0) + 3.1(2) = 6.2 z = 2.5(4) + 3.1(0) = 10

So, the minimum value of z is 0.

826

Section 6.5

12. The region in this case is: Vertex

(1,5) (3,5) (2,4) (1,7) (3,7)

Objective Function z = 2.5x − 3.1y −13 −8 −7.4 −19.2 −14.2

The maximum value of z is −7.4.

13. The region in this case is:

Notes on the graph: C1: y = − x + 5 C2: y = − x + 7 C3: y = x + 5 C4: y = x + 3 We need all of the intersection points since they constitute the vertices: Intersection of y = x + 5 and y = −x + 7: x + 5 = −x + 7 2x = 2 x =1 So, the intersection point is (1,6). Intersection of y = x + 3 and y = − x + 7:

x + 3 = −x + 7 2x = 4 so that x = 2 So, the intersection point is (2,5). Intersection of y = x + 3 and y = −x + 5 : x + 3 = −x + 5 2 x = 2 so that x = 1 So, the intersection point is (1,4). Intersection of y = x + 5 and y = −x + 5 : x + 5 = −x + 5 2 x = 0 so that x = 0 So, the intersection point is (0,5). Now, compute the objective function at the vertices: Vertex

Objective Function z = 14 x + 25 y

(1,6)

z = 14 (1) + 25 (6) = 53 20

(2,5)

z = 14 (2) + 25 (5) = 52

(1,4)

z = 14 (1) + 25 (4) = 37 20

(0,5)

z = 14 (0) + 52 (5) = 2

So, the maximum value of z is 53 20 = 2.65.

827

Chapter 6 14. The region in this case is:

Now, compute the objective function at the vertices: Vertex

Objective Function z = 13 x − 52 y

(0,6)

− 125

(1,5)

− 53

(2,6)

26 − 15

(1,7)

37 − 15

37 So, the minimum value of z is − 15 .

15. Let x = number of Charley T-shirts y = number of Francis T-shirts

Profit from Charley T-shirts: Revenue – cost = 13x − 7 x = 6 x Profit from Francis T-shirts: Revenue – cost = 10 y − 5 y = 5 y So, to maximize profit, we must maximize the objective function z = 6x + 5 y. We have the following constraints: 7 x + 5 y ≤ 1000   x + y ≤ 180  x ≥ 0, y ≥ 0  The region in this case is:

Notes on the graph: C1: y = − 75 x + 200 C2: y = −x + 180 C3: x = 0 We need all of the intersection points since they constitute the vertices: Intersection of 7x + 5 y = 1000 and x + y = 180 : − x + 180 = − 75 x + 200 2 5

x = 20

x = 50 So, the intersection point is (50, 130). Now, compute the objective function at the vertices: Vertex

(0,180) ( 1000 7 , 0) (50, 130)

Objective Function z = 6x + 5 y z = 6(0) + 5(180) = 900 z = 6( 1000 7 ) + 5(0) ≅ 857.14 z = 6(50) + 5(130) = 950

So, to attain a maximum profit of $950, he should sell 130 Francis T-shirts, and 50 Charley T-shirts.

828

Section 6.5

16. Let x = # of “Got Plywood” Tshirts y = # of “Got Gas” T-shirts Profit from “Got Plywood” T-shirts: Revenue – cost = 13x − 8x = 5x Profit from “Got Gas” T-shirts: Revenue – cost = 10 y − 6 y = 4 y So, to maximize profit, we must maximize the objective function z = 5x + 4 y. We have the following constraints: 8x + 6 y ≤ 1400   x + y ≤ 200  x ≥ 0, y ≥ 0  The region in this case is:

Notes on the graph: 4 C1: y = 700 3 − 3 x C2: y = 200 − x C3: x = 0 We need all of the intersection points since they constitute the vertices: Intersection of 8x + 6 y = 1400 and x + y = 200 : 1400 −8 x = 200 − x 6

1400 − 8x = 1200 − 6x 100 = x So, the intersection point is (100,100). Now, compute the objective function at the vertices: Vertex

(175,0) (0, 200) (100,100)

Objective Function z = 5x + 4 y z = 5(175) + 4(0) = 875 z = 5(0) + 4(200) = 800 z = 5(100) + 4(100) = 900

Hence, in this case, in order to attain a maximum profit of $900, she should sell 100 of both types of T-shirts.

829

Chapter 6 17. Let x = # of desktops y = # of laptops

We must maximize the objective function z = 500x + 300 y subject to the following constraints: 5x + 3 y ≤ 90  700 x + 400 y ≤ 10,000   y ≥ 3x   x ≥ 0, y ≥ 0 The region in this case is:

We need all of the intersection points since they constitute the vertices: Intersection of 5x + 3 y = 90 and 700x + 400 y = 10,000 : 30 − 53 x = 25 − 74 x 360 − 20 x = 300 − 21x x = −60 So, the intersection point is ( −60,130). Not a vertex, however, since x must be non-negative (see the region). Intersection of C5 and C2: 3x = 25 − 74 x 19 4

x = 25 so that x = 100 19

300 So, the intersection point is ( 100 19 , 19 )

Now, compute the objective function at the vertices: Vertex

(0,25) (0,0) 300 ( 100 19 , 19 )

Notes on the graph: C1: y = 30 − 53 x C2: y = 25 − 74 x C3: x = 0 C4: y = 0 C5: y = 3x

Objective Function z = 500x + 300 y z = 500(0) + 300(25) = 7500 z = 500(0) + 300(0) = 0 140,000 300 z = 500( 100 19 ) + 300( 19 ) = 19

In order to attain a maximum profit of $7500, he must sell 25 laptops and 0 desktops.

830

Section 6.5

18. Let x = # of desktops y = # of laptops

We must maximize the objective function z = 500x + 300 y subject to the following constraints: 5x + 3 y ≤ 120  700 x + 400 y ≤ 30,000   y ≥ 3x   x ≥ 0, y ≥ 0 The region in this case is:

We need all of the intersection points since they constitute the vertices: Intersection of 5x + 3 y = 120 and 700x + 400 y = 30,000 : 75 − 74 x = 40 − 53 x 900 − 21x = 480 − 20x 420 = x So, the intersection point is (420, −660). Not a vertex, however, since y must be non-negative (see the region). Intersection of C4 and C2: 3x = 40 − 53 x 14 3

x = 40 so that x = 607

So, the intersection point is ( 607 , 180 7 ). Now, compute the objective function at the vertices: Vertex

(0,24) (0, 0)

( 607 , 1807 ) Notes on the graph: C1: y = 75 − 74 x C2: y = 40 − 53 x C3: x = 0 C4: y = 3x

Objective Function z = 500 x + 300 y z = 500(0) + 300(24) = 7200 z = 500(0) + 300(0) = 0 z = 500( 607 ) + 300( 180 7 ) = 12,000

They cannot make a fraction of a computer so they should sell 25 laptops and 8 desktops in order to obtain a maximum profit of $11,500.

831

Chapter 6 19. Let x = # first class cars y = # second class cars Let p denote the profit for each second class car. Then, the profit for each first class car is 2p. Hence, we seek to maximize the objective function z = py + 2 px = p( y + 2 x ). We have the following constraints: x + y = 30  2≤x≤4   y ≥ 8x  The feasible region in this case is simply the bolded segment in the below diagram:

We need all of the intersection points since they constitute the vertices: Intersection of x + y = 30 and y = 8x: −x + 30 = 8x 30 = 9x so that 103 = x

So, the intersection point is ( 103 , 803 ). Now, compute the objective function at the vertices: Vertex

(2, 16) (2, 28) ( 103 , 803 ) (3, 27)

Objective Function z = p(2 x + y ) z = p(2(2) + 16) = 20 p z = p(2(2) + 28) = 32 p z = p(2( 103 ) + 803 ) = 100 3 p z = p(2(3) + 27) = 33p

The maximum in this case would occur at ( 103 , 803 ). However, this is not tenable since one cannot have a fraction of a car. However, very near at the vertex (3, 27) the profit is very near to this one, as are the number of cars. Hence, to maximize profit, they should use: 3 first class cars and 27 second class cars.

Notes on the graph: C1: y = 8x C2: y = − x + 30 C3: x = 2

832

Section 6.5

20. Let x = # first class cars y = # second class cars

Let p denote the profit for each second class car. Then, the profit for each first class car is 2p. Hence, we seek to maximize the objective function z = py + 1.2 px = p( y + 1.2x ). We have the following constraints:  x + y = 30  1≤ x   y ≥ 10 x 

We need all of the intersection points since they constitute the vertices: Intersection of x + y = 30 and y = 10x: − x + 30 = 10x 30 = 11x =x 300 So, the intersection point is ( 30 11 , 11 ). Now, compute the objective function at the vertices: 30 11

Vertex

The feasible region in this case is simply the bolded segment in the below diagram:

Notes on the graph: C1: y = −x + 30 C2: y = 10 x C3: x = 1

(1, 10) (1, 29) 300 ( 30 11 , 11 )

Objective Function z = p(1.2x + y ) z = p(1.2(1) + 10) = 11.2 p z = p(1.2(1) + 29) = 30.2 p 300 z = p(1.2( 30 11 ) + 11 ) = 30.54 p

The maximum in this case would occur 300 at ( 30 11 , 11 ). However, this is not tenable since one cannot have a fraction of a car. However, very near at the point (2, 28), the profit would be 30.4p, which is larger than the profit at vertex (1, 29). Even though this is not a vertex, it is still in the region, and the restriction that the values of x and y requires one to consider pairs in the region close to the vertex at which the maximum would have occurred, and compare these values to the other vertices. In this case, to maximize profit, they should use: 2 first class cars and 28 second class cars.

833

Chapter 6 21. Let x = # of regular skis y = # of slalom skis

22. Let x = # dozen of crème-filled donuts y = # dozen of jelly-filled donuts

We must maximize the objective function z = 25x + 50 y subject to the following constraints:  x + y ≤ 400   x ≥ 200, y ≥ 80  x ≥ 0, y ≥ 0  The region in this case is:

We must maximize the objective function z = 1.20 x + 1.80 y subject to the following constraints:  10 ≤ x + y ≤ 30   3x + 2 y ≤ 120  x ≥ 0, y ≥ 0  The region in this case is:

Now, compute the objective function at the vertices:

Notes on the graph: C1: y = −x + 10 C2: y = −x + 30 C3: y = − 32 x + 60 Now, compute the objective function at the vertices:

Vertex

(200,80)

Objective Function z = 25x + 50 y z = 25(200) + 50(80)

(200,200)

= 9000 z = 25(200) + 50(200)

(320,80)

= 15,000 z = 25(320) + 50(80)

Vertex

(0,10) (0,30) (10,0) (30,0)

= 12,000 So, he should sell 200 of each type of ski.

Objective Function z = 1.20 x + 1.80 y 18 54 12 36

So, he should sell 0 crème-filled donuts and 30 jelly-filled donuts.

834

Section 6.5

23. Should compare the values of the objective function at the vertices rather than comparing the y-values of the vertices. 24. Since − x + y ≤ 0 is a constraint, you would need to shade on AND below that line. So, the region should be:

25. False. The region could be the entire plane (i.e., unconstrained). 26. True

Notes on the graph: C1: y = 2 − x C2: y = x 27. Assume that a > 2. Then, the region looks like:

Notes on the graph: C1: y = −ax − a C3: y = ax + a C2: y = −ax + a C4: y = ax − a Now, compute the objective function at the vertices: Vertex Objective Function z = 2x + y (0,a) z = 2(0) + a = a ( −1,0) z = 2( −1) + 0 = −2 (0, −a) z = 2(0) − a = −a z = 2(1) + 0 = 2 (1, 0) Since a > 2, the maximum in this case would occur at (0,a) and is a.

835

Chapter 6 28. Assume that a > b > 0. Then, the region looks like:

Intersection of y = a − x and y = (a − b) + x : (a − b) + x = a − x x = b2

So, the intersection point is ( b2 , a − b2 ) .

Intersection of y = ( a + b ) − x and y = a+x: a + x = (a + b) − x x = b2

So, the intersection point is ( b2 , a + b2 ) Now, evaluate the objective function at each of the vertices: Notes on the graph: C1: y = a + x C3: y = ( a − b ) + x C2: y = ( a + b ) − x C4: y = a − x

Vertex

We need all of the intersection points since they constitute the vertices: Intersection of y = ( a + b ) − x and y = (a − b) + x : (a − b) + x = (a + b) − x x=b So, the intersection point is (b, a).

( b2 ,a + b2 ) ( b2 ,a − 2b )

29. This is the same as #11 – the minimum occurs at (0, 0) and is 0.

(0,a) ( b, a)

Objective Function z = x + 2y z = 0 + 2( a ) = 2a z = b + 2(a) = b + 2a z = 2b + 2( a + 2b ) = 2a + 32b z = b2 + 2( a − b2 ) = 2a − b2

Since a > b > 0, the maximum in this case would occur at ( b2 , a + b2 ) and is 2a + 32b .

30. This is the same as #12 − the maximum at ƒ(2, 4) = −7.4

836

Section 6.5

31.

Notes on the graph: C1: y = x + 3.7 C2: y = −x + 11.2 C3: y = 4.5 Vertex

(0.8, 4.5) (6.7, 4.5) (3.75,7.45)

Objective Function z = 17 x + 14 y z = 17(0.8) + 14(4.5) = 76.6 z = 17(6.7) + 14(4.5) = 176.9 z = 17(3.75) + 14(7.45) = 168.05

The maximum occurs at (6.7, 4.5) and is 176.9. 32.

Notes on the graph: C1: y = −2.3x + 14.7 C2: y = 2.3x + 14.7 C3: y = 5.2 x + 3.7 C4: y = 2.3x + 1.5 Vertex

(2.9,8.1) (1.47,11.33)

Objective Function z = 1.2x + 1.5 y z = 1.2(2.9) + 1.5(8.1) = 15.63 z = 1.2(1.47) + 1.5(11.33)

= 18.76 (−0.76, −0.24) z = 1.2( −0.76) +1.5(−0.24) = −1.27

The minimum occurs at ( −0.76, −0.24) and is −1.27.

837

Chapter 6 33.

Vertex

(−2.2, 4.58) (0.444, 12.778) (3.75, −3.75)

Objective Function z = 4.5x + 1.8 y −1.656 24.9984 10.125

The maximum occurs at (0.444,12.778) and is approximately 25.

34.

Vertex

(−0.6, 3.12) (1.35, −1.17) (1.65, 7.17) (3.6, 2.88)

Objective Function z = 5.4x − 1.6 y −8.232 9.162 −2.562 14.832

The minimum occurs at (−0.6, 3.12) as is approximately −8.232.

838

Chapter 6 Review Solutions----------------------------------------------------------------------- r − s = 3 (1) 1. Solve the system:   r + s = 3 (2) Add (1) and (2): 2r = 6 so that r = 3 Substitute this value of r into (1) to find s: 3 − s = 3 so that s = 0. So, the solution is ( 3,0 ) .

 3x + 4 y = 2 (1) 2. Solve the system:   x − y = 6 ( 2) Multiply (2) by 4, and then add to (1) to eliminate y: 7 x = 26 x = 267 ( 3 ) Substitute (3) into (2) and solve for y: 26 7 −y =6 y = 267 − 6 = − 167

So, the solution is ( 267 , − 167 ) . 3. Solve the system:  − 4 x + 2 y = 3 (1)  4x − y = 5 ( 2 )  Add (1) and (2) to eliminate x: y = 8 ( 3) Substitute (3) into (2) and solve for x: 4x − 8 = 5 4 x = 13 x = 134

4. Solve the system:  0.25x − 0.5 y = 0.6 (1)   0.50 x + 0.25 y = 0.8 ( 2 ) Multiply (1) by 0.5 and then add to (2) to eliminate y: 0.625x = 1.1 x = 1.76 ( 3 ) Substitute (3) into (2) to find y: 0.5(1.76) + 0.25 y = 0.8 0.25 y = −0.08 y = −0.32

So, the solution is ( 134 ,8 ) .

So, the solution is (1.76, − 0.32 ) . x + y = 3 (1) 5. Solve the system:  x − y = 1 ( 2 ) Solve (1) for y: y = 3 − x (3) Substitute (3) into (2) and solve for x: x − (3 − x ) = 1 x −3+ x =1 2 x = 4 so that x = 2

 3x + y = 4 (1) 6. Solve the system:   2x + y = 1 ( 2 ) Solve (1) for y: y = 4 − 3x (3) Substitute (3) into (2) and solve for x: 2x + (4 − 3x ) = 1 4 − x =1 x =3

839

Chapter 6 Substitute this value of x into (3) to find y: y = 1. So, the solution is ( 2, 1) .

Substitute this value of x into (3) to find y: y = 4 − 3(3) = −5. So, the solution is ( 3, − 5 ) .

d = 138 Now, substitute this value of d into (3) to find c: c = 4 − 138 = 198

s = 167 Now, substitute this value of s into (3) to find r: r = 167 − 3 = − 75

So, the solution is ( 198 , 138 ) .

So, the solution is ( − 75 , 167 ) .

10.

Notes on the graph: Solid curve: y = − 12 x Dashed curve: y = 12 x + 2 So, the solution is ( −2,1).

Notes on the graph: Solid curve: 2x + 4 y = −2 Dashed curve: 4 x − 2 y = 3 So, the solution is (0.4, −0.7).

 4c − 4d = 3 (1) Solve the system:   c + d = 4 (2) Solve (2) for c: c = 4 − d (3) Substitute (3) into (1) and solve for d: 4(4 − d ) − 4d = 3 16 − 8d = 3 8d = 13

 5r + 2 s = 1 (1) Solve the system:   r − s = −3 ( 2 ) Solve (2) for r: r = s − 3 (3) Substitute (3) into (1) and solve for s: 5( s − 3) + 2 s = 1 7 s − 15 = 1 7 s = 16

840

Chapter 6 Review

11.

Notes on the graph: (Careful! The curves are very close together in a vicinity of the point of intersection.) Solid curve: 1.3x − 2.4 y = 1.6 Dashed curve: 0.7x − 1.2 y = 1.4

12.

Notes on the graph: Solid curve: 14 x − 34 y = 12 Dashed curve: 12 y + 14 x = 21 So, the solution is (20.4, −9.2).

So, the solution is (12,5.83). 13. Solve the system:  5x − 3 y = 21 (1)   − 2x + 7 y = −20 ( 2 ) To eliminate x, multiply (1) by 2: 10 x − 6 y = 42 (3) Multiply (2) by 5: −10 x + 35 y = −100 (4) Add (3) and (4): 29 y = −58  y = −2 Substitute this into (1) to find x: 5x = 15  x = 3 So, the solution is ( 3, −2 ) .

14. Solve the system:  6x − 2 y = −2 (1)   4x + 3 y = 16 ( 2 ) To eliminate y, multiply (1) by 3: 18x − 6 y = −6 (3) Multiply (2) by 2: 8x + 6 y = 32 (4) Add (3) and (4): 26x = 26  x = 1 Substitute this into (1) to find x: 6(1) − 2 y = −2  y = 4 So, the solution is (1, 4 ) .

841

Chapter 6 15. Solve the system:  10x − 7 y = −24 (1)  (2)  7x + 4 y = 1 To eliminate y, multiply (1) by 4: 40x − 28 y = −96 (3) Multiply (2) by 7: 49x + 28 y = 7 (4) Add (3) and (4): 89x = −89  x = −1 Substitute this into (1) to find y: −7 y = −14  y = 2 So, the solution is ( −1,2 ) .

16. Solve the system:  13 x − 29 y = 29 (1)  4 3 3  5 x + 4 y = − 4 (2) First, clear the fractions by multiplying (1) by 9 and (2) by 20: 3x − 2 y = 2 (3) 16 x + 15 y = −15 (4) To eliminate y, multiply (3) by 15: 45x − 30 y = 30 (5) Multiply (4) by 2: 32x + 30 y = −30 (6) Add (5) and (6): 77x = 0  x = 0 Substitute this into (3) to find y: −2 y = 2  y = −1 So, the solution is ( 0, −1) .

25 2 17. c The intersection point is ( 11 , 11 )

18. b Subtracting the two equations yields the false statement 0 = 4, so the lines must be parallel.

19. d Multiplying the first equation by 2 reveals that the two equations are equivalent. Hence, the graphs are the same line.

20. a The intersection point is (1, −1)

21. Let x = number of ml of 6% NaCl. y = number of ml of 18% NaCl. Must solve the system: 0.06 x + 0.18 y = (0.15)(42) (1)  x + y = 42 (2)  First, for convenience, simplify (1) to get the equivalent system: x + 3 y = 105 (3)   x + y = 42 (4)

22. Let x = gallons used for highway miles y = gallons used for city miles Must solve the system: 32 x + 18 y = 265 (1)  x + y = 12 ( 2 )  Solve (2) for y: y = 12 − x (3) Substitute (3) into (1) and solve for x: 32 x + 18(12 − x ) = 265 14 x + 216 = 265 14 x = 49 x = 3.5

842

Chapter 6 Review

Multiply (1) by −1, and then add to Now, substitute this value into (3) to find y: (2): y = 12 − 3.5 = 8.5 −2 y = −63 Thus, there are y = 632 = 31.5 (6) 32(3.5) = 112 highway miles Substitute (6) into (4) to find x: 18(8.5) = 153 city miles. x + 31.5 = 42 x = 10.5 So, should use approximately 10.5 ml of 6% NaCl and 31.5 ml of 18% NaCl. 23. Solve the system: (1)  x + y + z= 1   x − y − z = − 3 (2) −x + y + z = 3 (3)  Add (2) and (3): 0 = 0 (4) Hence, we know that the system has infinitely many solutions. Add (1) and (2): 2x = −2 so that x = −1 (5)

24. Solve the system:  x − 2 y + z = 3 (1)   2 x − y + z = − 4 (2)  3x − 3 y − 5 z = 2 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (1) by −1, and then add to (2) to eliminate z: x + y = −7 (4) Next, multiply (2) by 5 and add to (3) to eliminate z: 13x − 8 y = −18 (5) These steps yield the following system:  x + y = −7 (4) (*)  13x − 8 y = −18 (5)

Substitute this value of x into (1): −1 + y + z = 1 y + z = 2 (6) Let z = a. Substitute this into (6) to find z: y = −a + 2 Thus, the solutions are: x = −1, y = −a + 2, z = a .

Step 2: Solve system (*) from Step 1. Solve (4) for y: y = −x − 7 (6) Substitute (6) into (5) and solve for x: 13x + 8( x + 7) = −18 21x + 56 = −18 x ≅ −3.524 Substitute this value of x back into (6) to find y: y = −( −3.524) − 7 ≅ −3.476 Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute the above values of x and y into (1) to find z: −3.524 − 2( −3.476) + z = 3 z ≅ −0.429 Thus, the approximate solution is: x = −3.524, y = −3.476, z = −0.429 .

843

Chapter 6

25. Solve the system:  x + y + z = 7 (1)   x − y − z = 17 (2)  y + z = 5 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Subtract (1) - (2) and simplify: y + z = −5 (4) Solve the system:  y + z = 5 (3) (*)   y + z = −5 ( 4 ) Note that equating (3) and (4) yields the false statement 5 = −5. Hence, the system has no solution.

26. Solve the system: x + z = 3 (1)    − x + y − z = − 1 (2)  x + y + z = 5 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Add (2) and (3): 2 y = 4 so that y = 2 (4) Now, substitute (4) into (2) and consider (1) and (2) together. That is, consider the  x + z = 3 (5) 2 × 2 system:  −x − z = −3 (6) Add (5) and (6): 0 = 0 Hence, we know that the system has infinitely many solutions. To determine them, let x = a. Substitute this into (5) to find z: a + z = 3  z = 3 − a Thus, the solutions are: x = a, y = 2, z = 3 − a .

27. Since we are given that the points (16,2), (40,6), and (65, 4), the system that must be solved is: 2 = a(16)2 + b(16) + c  2 6 = a(40) + b(40) + c 4 = a(65)2 + b(65) + c  which is equivalent to 2 = 256a + 16b + c (1)  6 = 1600a + 40b + c (2) 4 = 4225a + 65b + c (3)  Step 1: Obtain a 2 × 2 system: Multiply (1) by −1, and add to (2): 1344a + 24b = 4 (4)

844

Chapter 6 Review

Multiply (1) by −1, and add to (3): 3969a + 49b = 2 (5) These steps yield the following 2 × 2 system: Step 2: Solve the system in Step 1. 1344a + 24b = 4 (4)  3969a + 49b = 2 (5) Multiply (4) by −49 and (5) by 24. Then, add them together: 29, 400a = −148 a ≅ −0.0050 (6) Substitute (6) into (4): 1344( −0.005) + 24b = 4 b ≅ 0.4486 (7) Step 3: Find values of remaining variables. Substitute (6) and (7) into (1): 256( −0.005) + 16(0.447) + c = 2 c ≅ −3.8884 Thus, the polynomial has the approximate equation y = −0.0050 x 2 + 0.4486 x − 3.8884 .

28. Let x = amount in IRA y = amount in mutual fund z = amount in stock The system we must solve is: x = y + 4000   x + y + z = 20,000  0.045x + 0.08 y + 0.12 z = 1525  We first simplify the third equation by multiplying 100 to get the following equivalent system: x = y + 4000 (1)   (2) x + y + z = 20,000  45x + 80 y + 120 z = 1,525,000 (3)  Step 1: Obtain a 2 × 2 system: Substitute (1) into both (2) and (3) to obtain the system 45( y + 4000) + 80 y + 120 z = 1,525,000 which is equivalent to  ( y + 4000) + y + z = 20,000 

845

Chapter 6

125 y + 120 z = 1,345,000 (4)  (5) 2 y + z = 16,000  Step 2: Solve the system in Step 1. Multiply (5) by −120 and add to (4) to find y: −115 y = −575,000 y = 5000 (6) Substitute (6) into (1) to find x: x = 5000 + 4000 = 9000 (7) Substitute (6) and (7) into (2) to find z: 5000 + 9000 + z = 20,000 z = 6000 Thus, the following approximate allocation of funds should be made: $9000 in IRA $5000 in mutual fund $6000 in stock

29. 4 A B C D = + + + 2 ( x − 1) ( x + 3)( x − 5) x − 1 ( x − 1) x + 3 x − 5 2

30. 7 A B C D = + + + 2 2 ( x − 9)(3x + 5) ( x + 4) x − 9 3x + 5 ( 3x + 5 ) x + 4

31. 12 A B C D = + + + 2 x(4 x + 5)(2x + 1) x 4x + 5 2x + 1 ( 2x + 1)2

32. 2 A B C D = + + + 2 ( x + 1)( x − 5)( x − 9) x + 1 x − 5 x − 9 ( x − 9 )2

33. 3 3 A B = = + x + x − 12 ( x − 3 )( x + 4 ) x − 3 x + 4 2

846

Chapter 6 Review

34. x 2 + 3x − 2 x 2 + 3x − 2 A B C = 2 = + 2 + 3 2 x + 6x x (x + 6) x x x+6

35. 3x3 + 4 x 2 + 56 x + 62

( x + 17 ) 2

Ax + B Cx + D + 2 x + 17 ( x 2 + 17 )2

36. x3 + 7 x 2 + 10

( x + 13) 2

Ax + B Cx + D + 2 x + 13 ( x 2 + 13 )2

37. The partial fraction decomposition has the form: 9x + 23 A B (1) = + ( x − 1) ( x + 7 ) x − 1 x + 7 To find the coefficients, multiply both sides of (1) by ( x − 1)( x + 7), and gather like terms: 9x + 23 = A( x + 7) + B ( x − 1)

38. The partial fraction decomposition has the form: 12x + 1 A B (1) = + (3x + 2) ( 2 x − 1) 3x + 2 2 x − 1 To find the coefficients, multiply both sides of (1) by (3x + 2) ( 2x − 1) , and

gather like terms: 12 x + 1 = A(2 x − 1) + B (3x + 2) = ( A + B )x + (7 A − B ) (2) = (2 A + 3B )x + ( − A + 2B ) (2) Equate corresponding coefficients in (2) Equate corresponding coefficients in (2) to obtain the following system: to obtain the following system:  A + B = 9 (3)  2 A + 3B = 12 (3) (*)  (*)  (4) − = 7 A B 23   −A + 2B = 1 (4) Now, solve system (*) : Now, solve system (*) : Add (3) and (4): 8A = 32  A = 4 Multiply (4) by 2: −2A + 4 B = 2 (5) Substitute this value of A into (3) to see Add (3) and (5): 7B = 14  B = 2 that B = 5. Substitute this value of B into (4) to see Thus, the partial fraction that A = 3. decomposition (1) becomes: Thus, the partial fraction 9x + 23 4 5 decomposition (1) becomes: = + ( x − 1) ( x + 7 ) x − 1 x + 7 12 x + 1 3 2 = + (3x + 2) ( 2 x − 1) 3x + 2 2 x − 1

847

Chapter 6

39. The partial fraction decomposition has the form: 13x 2 + 90x − 25 13x 2 + 90x − 25 A B C (1) = = + + 3 2x − 50 x 2 x( x − 5) ( x + 5 ) 2 x x − 5 x + 5

To find the coefficients, multiply both sides of (1) by 2 x( x − 5) ( x + 5 ) , and gather like terms: 13x 2 + 90 x − 25 = A( x 2 − 25) + B (2 x )( x + 5) + C (2 x )( x − 5) = ( A + 2 B + 2C )x 2 + (10B − 10C )x − 25A

(2)

Equate corresponding coefficients in (2) to obtain the following system: A + 2 B + 2C = 13 (3)  (*)  10B − 10C = 90 (4)  −25A = −25 (5)  Now, solve system (*) : Solve (5): A = 1 Substitute this value of A into (3): B + C = 6 (6) Note that (4) is equivalent to B − C = 9 (7). So, solve the system:  B + C = 6 (6 )   B − C = 9 (7) Add (6) and (7): 2B = 15  B = 152 Substitute this value of B into (6): C = − 32 Thus, the partial fraction decomposition (1) becomes: 13x 2 + 90 x − 25 1 15 3 = + − 3 2 x − 50 x 2x 2 ( x − 5 ) 2 ( x + 5 )

40. The partial fraction decomposition has the form: 5x 2 + x + 24 5x 2 + x + 24 A Bx + C (1) = = + 2 x3 + 8x x x +8 x ( x2 + 8)

To find the coefficients, multiply both sides of (1) by x ( x 2 + 8 ) , and gather like terms: 5x 2 + x + 24 = A( x 2 + 8) + x( Bx + C ) = ( A + B )x 2 + Cx + 8A (2)

848

Chapter 6 Review

Equate corresponding coefficients in (2) to obtain the following system: A + B = 5 (3)  (*)  C = 1 (4)  8A = 24 (5)  Solve (6): A = 3 Substitute this value of A into (4): 3 + B = 5  B = 2 Thus, the partial fraction decomposition (1) becomes: 5x 2 + x + 24 3 2 x + 1 = + 2 x 3 + 8x x x +8

41. The partial fraction decomposition has the form: 2 A B (1) = + x ( x + 1) x x + 1 To find the coefficients, multiply both sides of (1) by x( x + 1), and gather like terms: 2 = A( x + 1) + Bx 2 = ( A + B )x + A (2) Equate corresponding coefficients in (2) to obtain the following system:  A + B = 0 (3) (*)  A = 2 (4)  Now, solve system (*) : Substitute (4) into (3) to see that B = −2 . Thus, the partial fraction decomposition (1) becomes: 2 −2 2 = + x ( x + 1) x + 1 x

42. Observe that simplifying the expression first yields x 1 . = x ( x + 3) x + 3 This IS the partial fraction decomposition. 43. The partial fraction decomposition has the form: 5x − 17 A B = + (1) 2 2 (x + 2) x + 2 (x + 2)

849

Chapter 6 To find the coefficients, multiply both sides of (1) by ( x + 2 ) , and gather like 2

terms:

5x − 17 = A( x + 2) + B

5x − 17 = Ax + ( 2A + B ) (2) Equate corresponding coefficients in (2) to obtain the following system: A=5 (3)  (*)  2A + B = −17 (4) Substitute (3) into (4) to find B: 2(5) + B = −17 so that B = −27 Thus, the partial fraction decomposition (1) becomes: 5x − 17 5 27 = − 2 2 ( x + 2) x + 2 ( x + 2)

44. The partial fraction decomposition has the form: x3 Ax + B Cx + D (1) = 2 + 2 2 2 2 x 64 + x 64 x 64 + + ( ) ( )

To find the coefficients, multiply both sides of (1) by ( x 2 + 64 ) , and gather like 2

terms:

x3 = ( Ax + B ) ( x 2 + 64 ) + (Cx + D ) = Ax3 + Bx 2 + 64Ax + 64 B + Cx + D = Ax3 + Bx 2 + ( 64 A + C ) x + ( 64 B + D ) (2)

Equate corresponding coefficients in (2) to obtain the following system: A = 1 (3)   B = 0 (4)  (*)   64A + C = 0 (5) 64 B + D = 0 (6) Substitute (3) into (5) to find C: 64(1) + C = 0 so that C = −64 Substitute (4) into (6) to find D: 64(0) + D = 0 so that D = 0 Thus, the partial fraction decomposition (1) becomes: x3

( x + 64 ) 2

x 64x − 2 x + 64 ( x + 64 )2 2

850

Chapter 6 Review

45.

The solid curve is the graph of y = −2x + 3.

46.

The dashed curve is the graph of y = x − 4.

47. Shade above the dashed line y = 14 (5 − 2 x ).

48. Shade below the solid line y = 12 (4 − 5x ).

49. Shade above the solid line y = −3x + 2.

50. Shade below the dashed line y = x − 2.

851

Chapter 6

51. Shade below the solid line y = − 38 x + 2.

52. Shade above the solid line y = 29 x − 2

53.

54.

The common region is the line y = 3x. Notes on the graph: C1: y = x + 2 C2: y = x − 2 Since there is no region in common with both inequalities, the system has no solution.

852

Chapter 6 Review

55.

Notes on the graph: C1: y = x C2: x = −2

56.

Notes on the graph: C1: y = − 13 x + 2 C2: y = 2x − 8

57.

58.

Notes on the graph: C1: y = 34 x − 4 C2: y = 3 − 53 x

Notes on the graph: C1: y = −x − 4 C3: y = −2 C4: x = 8 C2: y = x − 3

853

Chapter 6

59. The region in this case is: Vertex

(0,0) (3,0) (0,3)

Objective Function z = 2x + y z = 2(0) + (0) = 0 z = 2(3) + (0) = 6 z = 2(0) + (3) = 3

So, the minimum value of z is 0.

60. The region in this case is: Vertex (0,0) (3,0) (3,3)

Objective Function z = 2x + 3y z = 2(0) + 3(0) = 0 z = 2(3) + 3(0) = 6 z = 2(3) + 3(3) = 15

So, the maximum value of z is 15.

854

Chapter 6 Review

61. The region in this case is: So, the intersection point is (4,4).

Vertex

(0,0) (0,8) (4, 4)

Objective Function z = 2.5x + 3.2 y z = 2.5(0) + 3.2(0) = 0 z = 2.5(0) + 3.2(8) = 25.6 z = 2.5(4) + 3.2(4) = 22.8

So, the maximum value of z is 25.6.

We need all of the intersection points since they constitute the vertices: Intersection of y = −x + 8 and y = x: −x + 8 = x 2x = 8 x=4

62. The region in this case is:

We need all of the intersection points since they constitute the vertices: Intersection of y = x + 10 and y = − x + 2: x + 10 = − x + 2 2 x = −8 x = −4 So, the intersection point is ( −4,6).

Vertex ( −4,6) (2,0) (2,12)

Objective Function z = 5x + 11y z = 5( −4) + 11(6) = 46 z = 5(2) + 11(0) = 10 z = 5(2) + 11(12) = 142

So, the minimum value of z is 10.

855

Chapter 6

63. The region in this case is: Vertex

( 3,0 )

Objective Function z = 3x − 5 y 9

(0,6)

−30

So, the minimum value of z is −30.

64. The region in this case is: Vertex

( 73 ,0 ) ( 157 , 74 ) (1,0)

Objective Function z = −2x + 7 y − 143 − 72 −2

So, the maximum value of z is − 72 .

856

Chapter 6 Review

65. Let x = number of ocean watercolor coaster sets y = number of geometric shape coaster sets Profit from ocean watercolor sets: Revenue – cost = 15x Profit from geometric shape sets: Revenue – cost = 8y So, to maximize profit, we must maximize the objective function z = 15x + 8 y. We have the following constraints: 4 x + 2 y ≤ 100   3x + 2 y ≤ 90  x ≥ 0, y ≥ 0  The region in this case is:

Notes on the graph: C1: y = 50 − 2 x C2: y = 45 − 32 x C3: y = 0 We need all of the intersection points since they constitute the vertices: Intersection of 4 x + 2 y = 100 and 3x + 2 y = 90 : 50 − 2 x = 45 − 32 x 5 = 12 x x = 10 So, the intersection point is (10, 30). Now, compute the objective function at the vertices:

Vertex (0,0) (0, 45) (10, 30) (25,0)

Objective Function z = 15x + 8 y z = 15(0) + 8(0) = 0 z = 15(0) + 8(45) = 360 z = 15(10) + 8(30) = 390 z = 15(25) + 8(0) = 375

So, to attain a maximum profit of $390, she should sell 10 ocean watercolor coaster sets and 30 geometric shape coaster sets.

857

Chapter 6

66. Let x = number of ocean watercolor coaster sets y = number of geometric shape coaster sets Profit from ocean watercolor sets: Revenue – cost = 15x Profit from geometric shape sets: Revenue – cost = 8y So, to maximize profit, we must maximize the objective function z = 15x + 8 y. We have the following constraints: 4 x + 2 y ≤ 300   3x + 2 y ≤ 90  x ≥ 0, y ≥ 0  The region in this case is:

Notes on the graph: C1: y = 150 − 2x C2: y = 45 − 32 x C3: y = 0 We need all of the intersection points since they constitute the vertices: Intersection of 4 x + 2 y = 300 and 3x + 2 y = 90 : 150 − 2 x = 45 − 32 x 105 = 12 x x = 210 So, the intersection point is (210, −270), which is not used to form the region since both x and y need to be nonnegative. Now, compute the objective function at the vertices:

Vertex (0, 45) (30, 0) (0,0)

Objective Function z = 15x + 8 y z = 15(0) + 8(45) = 360 z = 15(30) + 8(0) = 450 z = 15(0) + 8(0) = 0

So, to attain a maximum profit of $450, she should sell 30 ocean watercolor coaster sets and 0 geometric shape coaster sets.

858

Chapter 6 Review

67. The graph of this system of equation is as follows:

68. The graph of this system of equation is as follows:

The solution is (2, −3).

The solution is ( − 25 , 34 ) .

69. If your calculator has 3-dimensional graphing capabilities, graph the system to obtain the following:

70. If your calculator has 3-dimensional graphing capabilities, graph the system to obtain the following:

It is difficult to see from this graph, but zooming in will enable you to see that the solution is (3.6, 3, 0.8).

Although this may seem strange, observe that if you divide the first equation by 4 on both sides, you obtain the second equation. Hence, every point on this plane is a solution. These points are expressed as ( 0.75a − 1.5b + 3, a, b ) .

859

Chapter 6

71. Since the graphs coincide, y2 is the partial fraction decomposition of y1.

72. Since the graphs do not coincide, y2 is not the partial fraction decomposition of y1.

73. The region is as follows.

74. The region is as follows.

860

Chapter 6 Review

75. Vertex (−2.7, 2.6) (−7.5, −11.8) (1.8, 0.6)

Objective Function z = 6.2x + 1.5 y −12.84 −64.2 12.06

The maximum is 12.06 and occurs at (1.8, 0.6).

76.

Vertex (−2, −1.6) (−2, −11.2) (4, 17.6) (4, 8)

Objective Function z = 1.6 x − 2.8 y 1.28 28.16 −42.88 −16

The minimum is −42.88 and occurs at (4, 17.6).

861

Chapter 6 Practice Test----------------------------------------------------------------------------1. Solve the system:  x − 2 y = 1 (1)   − x + 3y = 2 ( 2) Add (1) and (2): y = 3 Substitute y = 3 into (1) to find x: x − 2(3) = 1 x = 7. So, the solution is ( 7,3 ) .

2. Solve the system:  3x + 5 y = −2 (1)   7x + 11y = −6 ( 2 ) Multiply (1) by 7, (2) by −3, and then add them together to eliminate x: 2y = 4

3. Solve the system: x − y = 2 (1)    − 2 x + 2 y = −4 ( 2 ) Multiply (1) by 2, and then add to (2): 0=0 So, the system is consistent. There are infinitely many solutions of the form x = a, y = a − 2 .

4. Solve the system:  3x − 2 y = 5 (1)   6x − 4 y = 0 ( 2 ) Multiply (1) by −2, and then add to (2): 0 = −10 Since this is a false statement, we conclude that the system has no solution.

y = 2 ( 3) Substitute (3) into (1) and solve for x: 3x + 5(2) = −2 3x = −12 so that x = −4 So, the solution is ( −4,2 ) .

862

Chapter 6 Practice Test

5. Solve the system:  x + y + z = − 1 (1)   2 x + y + z = 0 (2) − x + y + 2 z = 0 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (1) by −2, and then add to (2) to eliminate x: − y − z = 2 (4) Add (1) and (3) to eliminate x: 2 y + 3z = −1 (5) These steps yield system:  − y − z = 2 (4) (*)  2 y + 3z = −1 (5)

Step 2: Solve system (*) from Step 1. Multiply (4) by 2, and then add to (5): z = 3 (6) Substitute (6) into (4) to find y: − y − 3 = 2 so that y = −5 (7) Step 3: Use the solution of the system in Step 2 to find the value of the third variable in the original system. Substitute (6) and (7) into (1) to find x: x − 5 + 3 = −1 x =1 Thus, the solution is: x = 1, y = −5, z = 3 .

6. Solve the system:  6 x + 9 y + z = 5 (1)   2 x − 3 y + z = 3 (2) 10 x + 12 y + 2 z = 9 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (2) by −3, and then add to (1) to eliminate x: 18 y − 2 z = −4 (4) Multiply (2) by −5, and then add to (3) to eliminate x: 27 y − 3z = −6 (5) These steps lead to the system: 18 y − 2 z = −4 (4)  27 y − 3z = −6 (5)

which is equivalent to 9 y − z = −2 (6)  9 y − z = −2 Hence, we know that the system has infinitely many solutions. To determine them, let z = a. Substitute this into (6) to find y: y = 19 ( a − 2) Now, substitute the values of y and z into (2) to find x: 2x − 3( 19 ( a − 2)) + a = 3 2x = 13 ( a − 2) − a + 3 2x = − 23 a + 73 x = − 13 a + 76 Thus, the solution is: x = − 13 a + 76 , y = 19 ( a − 2), z = a .

863

Chapter 6

7. Solve the system: (1)  x − 2 y + 3z = 5   − 2 x + y + 4 z = 21 (2)  3x − 5 y + z = − 14 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Multiply (1) by 2, and then add to (2) to eliminate x: −3 y + 10 z = 31 (4) Multiply (1) by −3, and then add to (3) to eliminate x: y − 8z = −29 (5)

These steps lead to the system: −3 y + 10 z = 31 (4)   y − 8z = −29 (5) Step 2: Solve the system in Step 1. Multiply (5) by 3: 3 y − 24 z = −87 (6)

8. Solve the system:  x + z = 1 (1)   x + y = − 1 (2)  y + z = 0 (3)  Step 1: Obtain a system of 2 equations in 2 unknowns by eliminating the same variable in two pairs of equations. Solve (1) for x: x = 1 − z (4) Substitute (4) into (2) to eliminate x: y − z = −2 (5) These steps yield the following system:  y + z = 0 (3) (*)   y − z = −2 (5)

Now, add (3) and (5): 2 y = −2 y = −1 Substitute this value of y into (3) to find z: −1 + z = 0 z =1 Substitute this value of z into (1) to find x: x +1 = 1 x=0

Add (4) and (6): −14 z = −56  z = 4

Substitute this value of z into (6) to find y: y − 8(4) = −29  y = 3 Substitute the values of y and z into (1) to find x: x = −1.

Thus, the solution is: x = 0, y = −1, z = 1.

864

Chapter 6 Practice Test

Equate corresponding coefficients in (2) to obtain the following system: A + B = 2 (3) (*)  A = 5 (4)  Now, solve system (*) : Substitute (4) into (3) to see that B = −3.

9. The partial fraction decomposition has the form: 2x + 5 A B (1) = + x ( x + 1) x x + 1 To find the coefficients, multiply both sides of (1) by x( x + 1), and gather like terms: 2x + 5 = A( x + 1) + Bx

Thus, the partial fraction decomposition (1) becomes: 2x + 5 5 3 = − x ( x + 1) x x + 1

2x + 5 = ( A + B )x + A (2)

10. The partial fraction decomposition has the form: 3x − 13 A B (1) = + 2 ( x − 5) x − 5 ( x − 5)2 To find the coefficients, multiply both sides of (1) by ( x − 5)2 , and gather like terms: 3x − 13 = A( x − 5) + B = Ax + ( −5A + B ) (2) Equate corresponding coefficients in (2) to obtain the following system: A=3 (3)  (*)  −5A + B = −13 (4) Now, solve system (*) : Substitute (3) into (4) to see that B = 2. Thus, the partial fraction decomposition (1) becomes: 3x − 13 3 2 = + 2 ( x − 5) x − 5 ( x − 5)2

11. The partial fraction decomposition has the form: 7x + 5 A B (1) = + 2 ( x + 2) x + 2 ( x + 2)2 To find the coefficients, multiply both sides of (1) by ( x + 2)2 , and gather like terms: 7x + 5 = A( x + 2) + B = Ax + (2A + B ) (2) Equate corresponding coefficients in (2) to obtain the following system: A = 7 (3)  (*)  2 A + B = 5 (4) Now, solve system (*) : Substitute (3) into (4) to see that B = −9. Thus, the partial fraction decomposition (1) becomes: 7x + 5 7 9 = − 2 ( x + 2) x + 2 ( x + 2)2

865

Chapter 6 12. The partial fraction decomposition has the form: 1 1 A B = = + (1) 2 2 x + 5x − 3 ( 2 x − 1)( x + 3 ) 2 x − 1 x + 3

To find the coefficients, multiply both sides of (1) by (2x − 1)( x + 3), and gather like terms: 1 = A( x + 3) + B (2x − 1) = ( A + 2B )x + (3A − B ) (2) Equate corresponding coefficients in (2) to obtain the following system: A + 2 B = 0 (3) (*)   3A − B = 1 (4) Now, solve system (*) : Multiply (4) by 2, and then add to (3): 7 A = 2 so that A = 72 Substitute this value of A into (3) to find B: 2 7 + 2B = 0 B = − 71 Thus, the partial fraction decomposition (1) becomes: 2 1 1 7 7 = − 2 x 2 + 5x − 3 2 x − 1 x + 3

13. The partial fraction decomposition has the form: 5x − 3 A B C (1) = + + x ( x − 3 )( x + 3 ) x x − 3 x + 3

To find the coefficients, multiply both sides of (1) by x ( x − 3 )( x + 3 ) , and gather like terms: 5x − 3 = A ( x − 3 )( x + 3 ) + Bx ( x + 3 ) + Cx ( x − 3 ) = A ( x 2 − 9 ) + B ( x 2 + 3x ) + C ( x 2 − 3x )

= ( A + B + C ) x 2 + ( 3B − 3C ) x + ( −9A) (2) Equate corresponding coefficients in (2) to obtain the following system: A + B + C = 0 (3)  (*)  3B − 3C = 5 (4)  −9A = −3 (5) 

866

Chapter 6 Practice Test

Solve (5) for A: A = 13 (6) Substitute (6) into (1) to eliminate A: B + C = − 13 (7) Now, solve the 2 × 2 system: (4) 3B − 3C = 5  1  B + C = − 3 (7) Multiply (7) by 3, and then add to (4) to find B: 6 B = 4 so that B = 23 . Substitute this value for B into (7) to find C: 23 + C = − 13 so that C = −1. Finally, substitute the values of B and C into (1) to find A: A + 23 − 1 = 0 so that A = 13 . Thus, the partial fraction decomposition (1) becomes: 5x − 3 1 2 1 = + − x ( x − 3 )( x + 3 ) 3x 3 ( x − 3 ) x + 3 14. The partial fraction decomposition has the form: 5x − 3 A Bx + C (1) = + 2 2 x +9 x ( x + 9) x

To find the coefficients, multiply both sides of (1) by x ( x 2 + 9 ) , and gather like terms:

5x − 3 = A ( x 2 + 9 ) + ( Bx + C ) x = Ax 2 + 9A + Bx 2 + Cx = ( A + B ) x 2 + (C ) x + ( 9A) (2)

Equate corresponding coefficients in (2) to obtain the following system:  A + B = 0 (3)  (*)  C = 5 (4)  9A = −3 (5)  Solve (5) for A: A = − 13 (6) Substitute (6) into (3) to find B: − 13 + B = 0 so that B = 13 . Thus, the partial fraction decomposition (1) becomes: 1 − 13 5x − 3 3 x +5 = + 2 x2 + 9 x ( x + 9) x

867

Chapter 6 15.

16.

The dashed curve is the graph of y = 2x + 6.

The solid curve is the graph of y = 4x − 8.

17.

18.

Notes on the graph: C1: y = 4 − x C2: y = x − 2

Notes on the graph: C1: y = − 13 x + 2 C2: y = 2 x − 4

868

Chapter 6 Practice Test

19.

Notes on the graph: C1: y = 52 x − 4 C2: y = − 34 x + 3 21. The region in this case is:

20.

Notes on the graph: C1: y = x + 6 C2: y = −x + 4 We need all of the intersection points since they constitute the vertices: Intersection of y = −x + 3 and y = x + 1 : −x + 3 = x + 1 2x = 2 x =1 So, the intersection point is (1,2).

Vertex (0,3) (0,1) (1,2)

Notes on the graph: C1: y = −x + 3 C2: y = x + 1 C3: x = 0

Objective Function z = 5x + 7 y z = 5(0) + 7(3) = 21 z = 5(0) + 7(1) = 7 z = 5(1) + 7(2) = 19

So, the minimum value of z is 7.

869

Chapter 6 22. The region in this case is:

We need all of the intersection points since they constitute the vertices: Intersection of y = −x + 6 and y = 12 x + 2 : −x + 6 = 12 x + 2 x = 83

So, the intersection point is ( 83 , 103 ) . Vertex

(0,0) (0,2) (6,0) ( 83 , 103 )

Notes on the graph: C1: y = −x + 6 C2: y = 12 x + 2

Objective Function z = 3x + 6 y 0 12 18 28 MAX

So, the maximum value of z is 28. 23. Let x = amount in money market y = amount in aggressive stock z = amount in conservative stock The system we must solve is: y = z + 1000   x + y + z = 30,000  0.03x + 0.12 y + 0.06 z = 1890  First, simplify this system by multiplying the third equation by 100: y = z + 1000 (1)   x + y + z = 30,000 (2)  3x + 12 y + 6 z = 189,000 (3)  Step 1: Obtain a 2 × 2 system: Substitute (1) into both (2) and (3): x + ( z + 1000) + z = 30,000

3x + 12( z + 1000) + 6 z = 189,000

which are equivalent to (4)  x + 2 z = 29,000  3x + 18z = 177,000 (5) Step 2: Solve the system in Step 1. Multiply (4) by −3 and add to (5) to find z: 12 z = 90,000  z = 7500 (6) Substitute (6) into (1) to find y: y = 7500 + 1000 = 8500 (7) Substitute (6) and (7) into (2) to find x: x + 8500 + 7500 = 30,000 x = 14,000 Thus, the following allocation of funds should be made: $14,000 in money market $8,500 in aggressive stock $7,500 in conservative stock

870

Chapter 6 Practice Test

24. Let x = width of one of the small pens, y = length of one of the small pens. We must solve the following system:  4 xy = 245,000 (1) (combined area)  (2) (amount of fence needed) 8x + 5 y = 3150

= 61,250 (3) Solve (1) for y: y = 245,000 x 4x Substitute (3) into (2): 8x + 5( 61,250 x ) = 3150 8x 2 − 3150x + 306,250 =0 x 8x 2 − 3150x + 306,250 = 0 x = 3150 ± 16122,500 = 315016± 350 = 175, 218.75 Substitute each of these values into (3) to find the corresponding values of y: x = 175 : y = 61,250 175 = 350 x = 218.75 : y = 61,250 218.75 = 280 Hence, in the context of this problem, there are two solutions. Solution 1: The dimensions of each pen should be 175 × 350 ft. Solution 2: The dimensions of each pen should be 218.75 × 280 ft.

25. If your calculator has 3-dimensional graphing capabilities, use it to graph the system of equations to obtain:

It is difficult to see from the graph, but if you zoom in, you will see that the solution is (11,19,1).

871

Chapter 6

26.

Vertex (1, 12.7) (1, 0.1) (4.5, 1.5)

Objective Function z = x − 1.2 y −14.24 0.88 2.7

So, the minimum value is −14.24 and occurs at (1, 12.7)

872

Chapter 6 Cumulative Review--------------------------------------------------------------------2. (5 − 7i ) − (11 − 9i ) = −6 + 2i

1. 6 x − 11x + 5 = 30 x − 25 2

(6 x − 5) ( x − 1) 5 (6 x − 5)

x −1 5

Note that x ≠ 65 .

3. The LCD is x( x + 1), x ≠ 0, −1. Multiply both sides by the LCD and solve for x: 5( x + 1) − 5x = −5  5 = −5 Since this statement is false, the equation has no solution.

4. A = π r 2  πA = r 2  r = ±

Since r is a radius, we conclude that r=

2t 3t(t − 1) − ≥0  t −1 t −1 −t 2 + 3t t(3 − t ) ≥0 ≥0 t −1 t −1 CPs: t = 0,1,3 + − + − | | | 2

( − ( − )) + ( − − )

( 1920 ) + ( − 56 ) ≈ 1.263

3 4

1 5

2 3

1 2 6

−3+1 1−2  1 M =  4 5 , 6 3  = ( − 11 40 , − 4 ) 2 2  

The solution set is ( −∞,0 ] ∪ (1,3] .

6.2 − 5.0 = 0.5. To 5.6 − 3.2 find b, use the point (3.2, 5.0): 5.0 = 0.5(3.2) + b  b = 3.4 So, the line has equation y = 0.5x + 3.4. 7. The slope is m =

8. Need x 2 − 25 = ( x − 5)( x + 5) ≥ 0 . CPs: x = ±5 + − + | | −5

So, the domain is ( −∞, −5] ∪ [5, ∞ ) .

873

Chapter 6 9. f (4) − f (2) 54 − 52 = = − 58 4−2 4−2

10. Shift the graph of y = x1 right 2 units and then up 1 unit. Domain: ( −∞,2 ) ∪ ( 2, ∞ ) Vertical asymptote: x = 2 Horizontal asymptote: y = 1

11. f ( −1) = 3 −1 − 7 = −2 . So, g ( f ( −1)) = g ( −2) = 1.

12. To find the inverse, switch the x and y and solve for y: 5y + 2 x= y −3 xy − 3x = 5 y + 2 ( x − 5) y = 3x + 2 3x + 2 y= x −5 3 x + 2 So, f −1 ( x ) = . x −5

13. Since (0,7) is the vertex, we know that f ( x ) = a( x − 0)2 + 7. Use the point (2,−1) to find a: −1 = a(2 − 0)2 + 7  a = −2 Hence, f ( x ) = −2 x 2 + 7.

14. The only real zero is 0, with multiplicity 3.

874

Chapter 6 Cumulative Review

15. Factors of 10: ±1, ± 2, ± 5, ± 10 Factors of 2: ±1, ± 2 So, possible rational zeros: ±1, ± 2, ± 5, ± 10, ± 12 , ± 52 Observe that −1 2 7 − 18 − 13 10 23 − 10 −2 −5 2

− 23 10 18 − 10

−5

Thus, P ( x ) = ( x + 1)( x − 2) ( 2x 2 + 9x − 5 ) = ( x + 1)( x − 2)( x + 5)(2x − 1). So, the rational zeros are −5, −1, 12 ,2.

16. Factors of 5: ±1, ± 5 Factors of 4: ±1, ± 2, ± 4 So, possible rational zeros: ±1, ± 5, ± 12 , ± 52 , ± 14 , ± 54 Observe that − 1 2 4 − 4 13 18 5 −2 − 12

4 4

−8 −5

− 6 16

−2

− 10

−8

Thus, P ( x ) = ( x + 1 2 ) ( 4 x 2 − 8x + 20 ) = 4 ( x + 1 2 ) ( x 2 − 2 x + 5 ) . 2

2 ± 4 − 4(5) = 1 ± 2i. 2 2 So, P ( x ) = 4 ( x + 1 2 ) ( x − (1 + 2i ))( x − (1 − 2i )).

Next, we find the roots of x 2 − 2x + 5 : x =

875

Chapter 6

17. The graph is below. Domain: ( −∞, ∞ ) Range: ( −1, ∞ ) x-intercept: 0 = 5− x − 1  x = 0. So, (0,0). y-intercept: (0,0) Horizontal asymptote: y = −1

19. Shift the graph of y = ln x left 1 unit, and then down 3 units. The graph is as follows:

18. Use A = P (1 + nr ) . nt

Here, r = 0.052, t = 16, N (16) = 50,000, n = 365. Find P. 50,000 = P (1 + 0.052 365 )

365(16 )

50,000

(1 + 0.052 365 )

365(16 )

≈ 21,760.19

20. log 4.7 8.9 =

21.

ln8.9 ≈ 1.413 ln 4.7

5 (10 2 x ) = 37 10 2 x = 375 2 x = log ( 375 )

x = 12 log ( 375 ) ≈ 0.435

876

Chapter 6 Cumulative Review

22. Let x = cost of tall coffee, y = cost of donut Must solve the system: 4 x + 6 y = 14.10 (1)  5x + 4 y = 13.11 ( 2 ) Multiply (1) by −5: −20x − 30 y = −70.5 (3) Multiply (2) by 4: 20x + 16 y = 52.44 (4) Add (3) and (4): y = 1.29 Substitute this value of y into (1): x = 1.59 So, the cost of a tall coffee is $1.59 and the cost of a donut is $1.29.

23. Solve the system:  x + 6 y − z = −3   2x − 5 y + z = 9 x + 4 y + 2 z = 12 

(1) (2) (3)

To eliminate z, add (1) and (2): 3x + y = 6 (4) Multiply (1) by 2 and then add to (3): 3x + 16 y = 6 (5) Solve the system (4) & (5): Subtract (4) – (5): y = 0 Substitute this value of y into (4): x = 2 Finally, substitute these values of x and y into (1): z = 5.

24. By simply factoring the denominator, we find that the partial 1 fraction decomposition is: − ( x − 1)2

25. The region is as follows:

26. If your calculator has 3dimensional graphing capabilities, graph the system to obtain the following:

27. Yes. Since the two graphs overlap, we can see that y2 is the partial-fraction decomposition of y1.

It is difficult to see from this graph, but zooming in will enable you to see that the solution is approximately (4.8, 1, −2.8).

877

CHAPTER 7 Section 7.1 Solutions -------------------------------------------------------------------------------1. 2 × 3

2. 3 × 2

3. 1× 4

4. 4 × 1

5. 1× 1

6. 4 × 4

7.  3 −2 7     −4 6 −3

8. −1 1 2     1 −1 −4 

9.  2 −3 4 −3    −1 1 2 1   5 −2 −3 7 

10.  1 −2 1 0     −2 1 −1 −5   13 7 5 6 

11. 1 1 0 3     1 0 −1 2   0 1 1 5 

12. 1 −1 0 −4    0 1 1 3 

13. −4 3 5 2     2 −3 −2 −3   −2 4 3 1 

14. −1 2 1 5    2 −2 3 0   −4 1 −2 3 

15. −3x + 7 y = 2   x + 5y = 8

16. −x + 2 y + 4 z = 4   7x + 9 y + 3z = −3  4x + 6 y − 5z = 8 

878

Section 7.1

17. −x = 4    7x + 9 y + 3z = −3 4x + 6 y − 5z = 8 

18.  2x + 3 y − 4 z = 6   7x − y + 5z = 9

19. x = a  y = b

20.  3x + 5 z = 1   −4 y + 7 z = −3  2x − y = 8 

21. Not in row echelon form since Condition 3 is violated.

22. Not in row echelon form since Condition 3 is violated.

23. Reduced row echelon form.

24. In reduced form.

25. Not in row echelon form since Condition 4 is violated.

26. In reduced form.

27. Reduced row echelon form.

28. Not in row echelon form since Condition 2 is violated.

29. row echelon form, not reduced row echelon form.

30. In reduced form.

31.  1 −2 −3   0 7 5 

32. 1 2 5     2 −3 −4 

33. 1 2 4    1 3 −2 

34. −5 2 −4     1 −2 3 

35.  1 −2 −1 3     0 5 −1 0   3 −2 5 8 

36.  1 −2 1 3     0 1 −2 6   0 −6 2 4 

879

Chapter 7

37.  1 −2 5  0 1 1  0 −2 1   0 0 1 39. 1  0 0   0 41. 1  0 0   0

−1 2   −3 3  −2 5   −1 −6 

38. 1  0 0   0

0 5 −10 15   1 2 −3 4  0 72 −3 92   0 1 −1 −3

0 5 −10 −5   −3 −2  1 2 0 −7 6 3   0 8 −10 −9 

40. 1  0 0   0

0 1 0 0

0 0 1 0

0 1  0 −2  0 0  1 −3 

42. 1  0 0   0

0 −1 1 2 0 1 0 0

0 −3   0 −8  0 4  1 1 

0 1 0 0

4 2 1 0

0 27   0 −11 0 21   1 −3 

43. 1 2 4  2 3 2  R1 ↔ R2   ⎯⎯⎯→   2 3 2  1 2 4  2 3 2  R1 − 2 R2 →R2 ⎯⎯⎯⎯⎯ →   0 −1 −6  2 3 2  − R2 → R2 ⎯⎯⎯⎯ →   0 1 6 

44.  1 −1 3  −3 2 2  R1 ↔ R2   ⎯⎯⎯→    −3 2 2   1 −1 3  −3 2 2  R1 +3 R2 → R2 ⎯⎯⎯⎯⎯ →    0 −1 11 −3 2 2  − R2 → R2 ⎯⎯⎯⎯ →    0 1 −11

1 32 1  1 R →R 1 2 1 ⎯⎯⎯⎯ →   0 1 6  1 0 −8 R1 − 32 R2 → R1 ⎯⎯⎯⎯⎯ →   0 1 6 

1 − 23 − 23  − 13 R1 → R1 ⎯⎯⎯⎯ →   0 1 −11 1 0 −8  R1 + 23 R2 → R1 ⎯⎯⎯⎯⎯ →   0 1 −11

880

Section 7.1

45. 1 0 −4  − R1 + R2 →R2 1 0 −4  →   ⎯⎯⎯⎯⎯  1 3 1 −   0 3 3   1 0 −4  1R → R 2 2 ⎯3⎯⎯⎯ →  0 1 1 

46. −2 0 −6  − 12 R1 →R1 1 0 3 →   ⎯⎯⎯⎯  3 − 1 8   3 −1 8

1 0 3  −3 R1 + R2 → R2 ⎯⎯⎯⎯⎯ →  0 −1 −1 1 0 3 − R2 →R2 ⎯⎯⎯ ⎯ →  0 1 1

47.  1 −1 1 −1 1 −1 1 −1     R1 + R3 → R3  0 1 −1 −1 ⎯⎯⎯⎯→ 0 1 −1 −1  −1 1 1 1  0 0 2 0  1 −1 1 −1 R + R →R 1 0 0 −2    R12+ R23 →R12   ⎯⎯⎯⎯ → 0 1 −1 −1 ⎯⎯⎯⎯→ 0 1 0 −1 0 0 1 0  0 0 1 0  1 R →R 3 2 3

48.  0 −1 1 1  0    R2 − R3 → R3  1 −1 1 −1 ⎯⎯⎯⎯→  1  1 −1 −1 −1  0 1 − R2 → R2 1 R →R  3 2 3 ⎯⎯⎯⎯ → 0 0

−1 1 1   1 −1 1 −1  R1 ↔ R2   −1 1 −1 ⎯⎯⎯→  0 −1 1 −1  0 0 2 0  0 2 0  −1 1 −1 R + R + R →R 1 0 0 0   R12 + R23 →R3 2 1   1 −1 1  ⎯⎯⎯⎯⎯→ 0 1 0 1  0 0 1 0  0 1 0 

881

Chapter 7

49.

 3 −2 −3 −1 0 1 R1 −3 R2 →R1    R3 − 2 R2 → R3 → 1 −1 1 −1 1 −4  ⎯⎯⎯⎯⎯ 2 3 5 14  0 5 1 −1  R3 −5 R2 →R3 ⎯⎯⎯⎯⎯ → 0 1 0 0

−6 11   1 −1 1 −4   R1 ↔ R2   1 −4  ⎯⎯⎯→ 0 1 −6 11  0 5 3 22  3 22  1 −4  1 −1 1 −4   331 R3 →R3   −6 11  ⎯⎯⎯⎯ → 0 1 −6 11  0 0 1 −1 33 −33

1 −1 1 −4  1 0 0 2      R2 + 6 R3 → R2 R1 + R2 − R3 → R1 ⎯⎯⎯⎯⎯ → 0 1 0 5  ⎯⎯⎯⎯⎯ → 0 1 0 5  0 0 1 −1 0 0 1 −1 50.  3 −1 1 2  0 5    R1 −3 R2 → R1 →  1 −2 1 −2 3 1  ⎯⎯⎯⎯⎯ 2 1 −3 −1 2 1  1 −2  R1 ↔ R2 ⎯⎯⎯→  0 5  0 5  1 −2 1 R →R 2 5 2  − R3 →R3 → 0 1 ⎯⎯⎯⎯ 0 0

−8 −1  0 5 −8 −1  R3 −2 R2 →R3   3 1  ⎯⎯⎯⎯⎯ →  1 −2 3 1  0 5 −9 −3 −3 −1 1 −2 3 1  3 1  R3 −R2 →R3   −8 −1 ⎯⎯⎯⎯→ 0 5 −8 −1 0 0 −1 −2  −9 −3 3 1 1 −2 3 1  R2 + 58 R3 → R2   8 1 − 5 − 5  ⎯⎯⎯⎯⎯ → 0 1 0 3  0 0 1 2  1 2 

1 0 0 1    ⎯⎯⎯⎯⎯⎯ → 0 1 0 3  0 0 1 2  R1 + 2 R2 −3 R3 → R1

51. 1 R →R

2 1 −6 4  R1 −2 R2 →R2 2 1 −6 4  512 R12 →R12 1 12 −3 2  → →    ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯ 1 −2 2 −3 0 5 −10 10  0 1 −2 2  1 0 −2 1  R1 − 12 R2 → R1 ⎯⎯⎯⎯⎯ →  0 1 −2 2 

882

Section 7.1

52.

−3 −1 2 −1 R1 −3R2 →R1 0 5 −1 8  −51 RR12→→RR12 0 1 − 51 58  → →    ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯  −1 −2 1 −3  −1 −2 1 −3 1 2 −1 3 1 2 −1 3 R1 −2 R2 →R1 1 0 − 35 − 15  R1 ↔ R2 ⎯⎯⎯→ ⎯⎯⎯⎯⎯ →  8 1 8 1 0 1 − 5 5  0 1 − 5 5  53. −1 2 1 −2  −1 2 1 −2  −1 2 1 −2      R3 + 2 R1 →R3   R2 + 3 R1 →R2 → 0 4 4 −2  ⎯⎯⎯⎯⎯ →  0 4 4 −2   3 −2 1 4  ⎯⎯⎯⎯⎯  2 −4 −2 4   2 −4 −2 4   0 0 0 0  1 −2 −1 2  1 0 1 1  − R1 →R1 1 R →R     R + R → R 2 2 2 4 1 2 1 ⎯⎯⎯⎯ → 0 1 1 − 12  ⎯⎯⎯⎯⎯ → 0 1 1 − 12  0 0 0 0  0 0 0 0  54.  2 −1 0 1   0 −1 2 −3  0     R3 −2 R2 →R3  R1 + 2 R2 → R1 →  −1 0 1 −2  ⎯⎯⎯⎯⎯ →  −1  −1 0 1 −2  ⎯⎯⎯⎯⎯  −2 1 0 −1  −2 1 0 −1  0  0 −1 2 −3  −1 0   R1 ↔ R2  R1 + R3 → R3 ⎯⎯⎯⎯→  −1 0 1 −2  ⎯⎯⎯→  0 −1  0 0 0 0   0 0

 1 0 −1 2    ⎯⎯⎯⎯ → 0 1 −2 3  0 0 0 0  − R1 →R1 − R2 → R2

55. First, reduce the augmented matrix down to row echelon form:  2 3 1  R1 −2 R2 →R2  2 3 1  12 R1 →R1  1 32 12  → →    ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯  1 1 −2  0 1 5 0 1 5 

883

−1 2 −3   0 1 −2  1 −2 3  1 −2   2 −3  0 0 

Chapter 7

Now, from this we see that y = 5, and then substituting this value into the equation obtained from the first row yields: x + 32 (5) = 12

x = −7 Hence, the solution is x = −7, y = 5 . 56. First, reduce the augmented matrix down to row echelon form: 1 R →R

3 2 11 R1 −3R2 →R2 3 2 11  315 R12 →R12 1 23 113  → →    ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯ 1 112 0 5 25 − −     0 1 −5  Now, from this we see that y = −5 , and then substituting this value into the equation obtained from the first row yields: x + 23 ( −5) = 113 x=7

Hence, the solution is x = 7, y = −5 . 57. First, reduce the augmented matrix down to row echelon form: 3 4 1  R1 ↔ R2 1 −2 7  −3R1 + R2 →R2  1 −2 7  → 1 −2 7  ⎯⎯⎯→ 3 4 1  ⎯⎯⎯⎯⎯       0 10 −20  1 R →R  1 −2 7  2 10 2 ⎯⎯⎯⎯ →   0 1 −2  From this we see that y = −2, and then substituting this value into the equation obtained from the first row yields: x − 2 ( −2 ) = 7

x=3

Hence, the solution is x = 3, y = −2 .

58. First, reduce the augmented matrix down to row echelon form:  2 −1 3  R1 ↔ R2 1 −3 4  −2 R1 + R2 →R2 1 −3 4  51 R2 →R2 1 −3 4  → →  1 −3 4  ⎯⎯⎯→ 2 −1 3  ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯       0 5 −5   0 1 −1 From this we see that y = −1, and then substituting this value into the equation obtained from the first row yields: x − 3 ( −1) = 4

x =1

Hence, the solution is x = 1, y = −1.

884

Section 7.1

59. First, reduce the augmented matrix down to row echelon form: −1 2 3  2 R1 + R2 →R2 −1 2 3  →   ⎯⎯⎯⎯⎯  2 − 4 − 6    0 0 0 From the bottom row, we conclude that there are infinitely many solutions. To get the precise form of these solutions, let y = a. Then, substituting this value into the equation obtained from the first row yields: − x + 2a = 3 x = 2a − 3

Hence, the solution is x = 2a − 3, y = a . 60. First, reduce the augmented matrix down to row echelon form: 1 R →R 3 −1 −1 2 R1 −R2 →R2 3 −1 −1 −3 14 1R2 →1R2  1 − 13 − 13  →    ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯→  − − 6 2 2 0 4 4     0 1 1  Now, from this we see that y = 1, and then substituting this value into the equation obtained from the first row yields: x − 13 (1) = − 13

x=0 Hence, the solution is x = 0, y = 1. 61. First, reduce the augmented matrix down to row echelon form: 3 R1 → R1  23 13 98   2 1 38  R1 −2 R2 →R2 2 1 38  2 R2 → R2 ⎯⎯⎯⎯ → ⎯⎯⎯⎯⎯ → 1 1 3  1 3 1 1 2 2 4 4 2 0 0 − 3     

From the second row, we see that 0 = − 13 , which means that the system has no solution. 62. First, reduce the augmented matrix down to row echelon form: 10 R1 → R1  0.4 −0.5 2.08  4 −5 20.8 14 R1 →R1  1 −1.25 5.2  10 R2 → R2 ⎯⎯⎯⎯ →  →    ⎯⎯⎯⎯  − − 0.3 0.7 1.88 3 7 18.8 7 18.8      −3

1 −1.25 5.2  3.251 R2 →R2 1 −1.25 5.2  3 R1 + R2 → R2 ⎯⎯⎯⎯⎯ →  ⎯⎯⎯⎯→   1 10.58 0 3.25 34.4  0 Now, from this we see that y ≅ 10.58 , and then substituting this value into the equation obtained from the first row yields: x − 1.25(10.58) = 5.2 x ≅ 18.425 Hence, the solution is x = 18.425, y = 10.58 .

885

Chapter 7

63. First, rewrite the system as  x − y − z = 10  2 x − 3 y + z = −11  − x + y + z = −10  Now, reduce the augmented matrix down to row echelon form:  1 −1 −1 10  1 −1 −1 10   1 −1 −1 10  R1 + R3 → R3     − R2 →R2   R2 − 2 R1 → R2 → 0 −1 3 −31 ⎯⎯⎯⎯ → 0 1 −3 31  2 −3 1 −11 ⎯⎯⎯⎯⎯  −1 1 1 −10  0 0 0 0  0 0 0 0  From the last row, we conclude that the system has infinitely many solutions. To find them, let z = a. Then, the second row implies that y − 3a = 31 so that y = 31 + 3a. So, substituting these values of y and z into the first row yields x − (3a + 31) − a = 10, so that x = 4a + 41. Hence, the solutions are

x = 4a + 41, y = 31 + 3a, z = a . 64. First, rewrite the system as 2 x + y + z = −3  x + 2 y − z = 0 x + y + 2 z = 5  Now, reduce the augmented matrix down to row echelon form:  2 1 1 −3  2 1 1 −3  2 1 1 −3  R2 − R3 → R3     13 R2 →R2   R1 − 2 R2 → R2 → 0 −3 3 −3 ⎯⎯⎯⎯ → 0 −1 1 −1  1 2 −1 0  ⎯⎯⎯⎯⎯  1 1 2 5  0 1 −3 −5  0 1 −3 −5  1 1 12 12 −23   2 1 1 −3  −2 RR12→→RR12     − 12 R3 →R3 R2 + R3 → R3 ⎯⎯⎯⎯→ →  0 1 −1 1   0 −1 1 −1 ⎯⎯⎯⎯    0 0 −2 −6  0 0 1 3  From the last row, we conclude that z = 3. Then, the second row implies that y − 3 = 1 so that y = 4. So, substituting these values of y and z into the first row yields x + 12 (4) + 12 (3) = − 23 , so that x = −5. Hence, the solution is

x = −5, y = 4, z = 3 .

886

Section 7.1

65. Reduce the augmented matrix down to row echelon form: R →R 1 13 − 13 13  3 1 −1 1  314 R12 →R12  3 1 −1 1  R3 − 2 R2 → R3 1       3 R3 →R3 R1 −3 R2 →R2 → 0 4 −4 10  ⎯⎯⎯⎯ → 0 1 −1 52   1 −1 1 −3 ⎯⎯⎯⎯⎯ 0 1 − 13 2   2 1 1 0  0 3 −1 6    1

1 13 − 13 13  1 13 − 13 13     − 32 R3 → R3 R2 − R3 → R3 5 ⎯⎯⎯⎯→ → 0 1 −1 52  0 1 −1 2  ⎯⎯⎯⎯ 0 0 − 23 12  0 0 1 − 34      3 From the last row, we conclude that x3 = − 4 . Then, the second row implies that x2 − ( − 34 ) = 52 so that x2 = 74 . So, substituting these values of x2 and x3 into the first row yields x1 + 13 ( 74 ) − 13 ( − 34 ) = 13 , so that x1 = − 12 . Hence, the solution is x1 = − 12 , x2 = 74 , x3 = − 34 .

66. Reduce the augmented matrix down to row echelon form:  2 1 1 −1  2 1 1 −1  2 1 1 −1  R3 −3 R2 → R3     R3 − 4 R2 →R3   R1 − 2 R2 → R2 →  0 −1 3 −11 ⎯⎯⎯⎯⎯ →  0 −1 3 −11  1 1 −1 5  ⎯⎯⎯⎯⎯  3 −1 −1 1   0 −4 2 −14  0 0 −10 30  1 12 12 − 12    ⎯⎯⎯⎯→ 0 1 −3 11  0 0 1 −3    From the last row, we conclude that z = −3. Then, the second row implies that y − 3( −3) = 11 so that y = 2. So, substituting these values of y and z into the first row yields x + 12 (2) + 12 ( −3) = − 12 , so that x = 0. Hence, the solution is 1 R →R 1 2 1 − R2 → R2 1 R →R − 10 3 3

x = 0, y = 2, z = −3 .

887

Chapter 7

67. Reduce the augmented matrix down to row echelon form:  2 5 0 9 2 5 0 9  2 5 0 9  R3 +3 R2 →R3     R3 −2 R2 →R3   R1 − 2 R2 → R2 →  0 1 2 3  ⎯⎯⎯⎯⎯ → 0 1 2 3   1 2 −1 3  ⎯⎯⎯⎯⎯  −3 −4 7 1   0 2 4 10  0 0 0 4  1 52 0 92  1 R →R   1 2 1 ⎯⎯⎯⎯ → 0 1 2 3  0 0 0 4    From the last row, we have the false statement 0 = 4, so that we can conclude the system has no solution. 68. Reduce the augmented matrix down to row echelon form:  1 −2 3 1  1 −2 3 1  1 −2 3 1  R2 + 2 R3 → R2     71 R2 →R2   R1 − R3 → R3 → 0 7 −7 22  ⎯⎯⎯⎯ → 0 1 −1 227   −2 7 −9 4  ⎯⎯⎯⎯⎯  1 0 −2 2 −8 0 −2 2 −8 0 1 9  1 −2 3 1    R3 + 2 R2 →R3 ⎯⎯⎯⎯⎯ → 0 1 −1 227  0 0 0 − 127 

From the last row, we have the false statement 0 = − 127 , so that we can conclude the system has no solution. 69. Reduce the augmented matrix down to row echelon form: 1 − 12 12 32   2 −1 1 3  1 R →R 1 2 1     2 R2 + R3 → R3 → 0 1 −1 −1  1 −1 1 2  ⎯⎯⎯⎯⎯ 0 0 0 0   −2 2 −2 −4   From the last row, we conclude that the system has infinitely many solutions. To find them, let x3 = a. Then, the second row implies that x2 − a = −1 so that x2 = a − 1. So, substituting these values of x2 and x3 into the first row yields x1 − 12 ( a − 1) + 12 a = 32 , so that x1 = 1. Hence, the solutions are x1 = 1, x = a − 1, x3 = a .

888

Section 7.1

70. Reduce the augmented matrix down to row echelon form: 1 −1 −2 0   1 −1 −2 0   1 −1 −2 0  13 R2 →R2 R3 −3 R1 → R3       14 R3 →R3 R2 + 2 R1 → R2 →  0 3 6 −3 ⎯⎯⎯⎯ → 0 1 2 −1  −2 5 10 −3 ⎯⎯⎯⎯⎯ 3 0 1  3 1 0 0   0 4 6 0  0  2  1 −1 −2 0  1 −1 −2 0    2 R3 →R3   R2 − R3 →R3 ⎯⎯⎯⎯→ 0 1 2 −1 ⎯⎯⎯⎯ →  0 1 2 −1  1 0 0  0 0 1 −2  2 −1  From the last row, we conclude that x3 = −2. Then, the second row implies that x2 + 2( −2) = −1 so that x2 = 3. So, substituting these values of x2 and x3 into the first row yields x1 − 3 − 2( −2) = 0, so that x1 = −1. Hence, the solution is x1 = −1, x2 = 3, x3 = −2 . 71. Reduce the augmented matrix down to row echelon form: 1 R →R  2 1 −1 2  R1 −2 R2 →R2 2 1 −1 2  132 R12 →R12 1 12 − 12 1  → →   ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯ 10  1 1 −1 −1 6  0 3 1 −10  0 1 3 − 3  Since there are two rows and three unknowns, we know from the above calculation that there are infinitely many solutions to this system. (Note: The only other possibility in such case would be that there was no solution, which would occur if one of the rows yielded a false statement.) To find the solutions, let z = a. Then, from row 2, we observe that y + 13 a = − 103 so that y = − 13 ( a + 10). Then, substituting these values of y and z into the equation obtained from row 1, we see that x + 12 ( − 13 ( a + 10) ) − 12 a = 1

x − 23 a − 53 = 1 x = 23 a + 83

Hence, the solutions are x = 23 ( a + 4 ) , y = − 13 ( a + 10), z = a . 72. Reduce the augmented matrix down to row echelon form: 1 R →R

3 1 −1 0  R1 −3R2 →R2 3 1 −1 0  −3 12 1R2 →1R2 1 13 − 13 0  ⎯⎯⎯⎯⎯ → →      ⎯⎯⎯⎯ − − − 1 1 7 4 0 2 22 12      0 1 11 6 

889

Chapter 7

Since there are two rows and three unknowns, we know from the above calculation that there are infinitely many solutions to this system. (Note: The only other possibility in such case would be that there was no solution, which would occur if one of the rows yielded a false statement.) To find the solutions, let z = a. Then, from row 2, we observe that y + 11a = 6 so that y = −11a + 6. Then, substituting these values of y and z into the equation obtained from row 1, we see that x + 13 ( −11a + 6 ) − 13 a = 0 x − 113 a + 2 − 31 a = 0 x = 4a − 2 Hence, the solutions are x = 4a − 2, y = −11a + 6, z = a .

73. Reduce the augmented matrix down to row echelon form: 0 2 1 3   4 0 −1 −3   4 0 −1 −3     R3 −7 R4 →R3   0 2 1 3  R1 − 4 R4 →R4 0 2 1 3 R1 ↔ R2  4 0 −1 −3  ⎯⎯⎯⎯   → ⎯⎯⎯⎯⎯ →  7 −3 −3 2  7 −3 −3 2  0 4 4 16        1 −1 −1 −2  1 −1 −1 −2  0 4 3 5  1 0 − 14 − 34   4 0 −1 −3 14 R1 →R1     1 R →R R3 − 2 R2 →R3 0 2 1 3  212 R32 →R32 0 1 12 32  R4 − 2 R2 →R4   ⎯⎯⎯⎯⎯ → ⎯⎯⎯⎯ → 0 0 1 5   0 0 2 10      0 0 1 −1 0 0 1 −1  The last two rows require that z = 5 and z = −1 simultaneously. Since this is not possible, we conclude that the system has no solution. 74. Reduce the augmented matrix down to row echelon form: −2 −1 2 3  −2 −1 2 3  R4   RR12−+23RR44 →   → R2 0 3 −7 −22  R3 + 2 R4 → R3  3 0 −4 2  ⎯⎯⎯⎯⎯ →  2 1 0 −1  0 3 −2 −17       −1 1 −1 −8  0 −3 4 19  − 1 R →R

1 2 1  1 12 −1 − 32  −2 −1 2 3  1 R →R 2 3 2 1   − 5 R3 → R3   R2 − R3 →R3 0 3 −1 −22  1 R →R 0 1 − 13 − 223  R3 + R4 →R4 4   2 4 ⎯⎯⎯⎯⎯ → ⎯⎯⎯⎯→ 0 0 1 1   0 0 −5 −5      2  1   0 0 2  0 0 1

890

Section 7.1

Thus, Row 3now implies that z = 1 . Then, Row 2implies that y − 13 (1) = − 223 so that y = −7 . Then, Row 1implies that x + 12 ( −7) − (1) = − 23 so that x = 3 . Hence, the solution is x = 3, y = −7, z = 1 . 75. Reduce the augmented matrix down to row echelon form:  3 −2 1 2 −2   0 −5 −2 −1 −2  →R2   RR11 +−33RR32 →   R3 0 4 5 4 4 R4 −5 R3 →R4  −1 3 4 3 4  ⎯⎯⎯⎯⎯  → 1 1 1 1 0 1 1 1 1 0      5 3 1 2 −1  0 −2 −4 −3 −1 1 1 1 1 0  1 1 1 1 0  14 R2 →R2     − 15 R3 →R3 0 4 5 4 4  − 12 R4 →R4 0 1 54 1 1  R1 ↔ R3  ⎯⎯⎯⎯ → ⎯⎯⎯⎯→ 0 1 52 51 52  0 −5 −2 −1 −2      3 1 0 −2 −4 −3 −1 0 1 2 2 2  1 1 1 1 0  1 1 1 1 0      3 4 R2 − R3 → R2 0 0 17 0 1 52 51 52  20 5 5  R4 − R3 →R4 R2 ↔ R3   ⎯⎯⎯⎯→ ⎯⎯⎯⎯ → 3  4 0 1 52 51 52  0 0 17 20 5 5     8 13 8 13 1 1 0 1 5 10 10  0 1 5 10 10  1 1 1 1 0  1 1 1 1 0  20 R → R     3 17 3 0 1 52 51 52  R3 −R4 →R4  0 1 52 51 52  5 R →R 4  8 4 ⎯⎯⎯⎯ → ⎯⎯⎯⎯→ 12  12  0 0 1 16  0 0 1 16 17 17 17 17     13 35 175 1 0 0 1 16 16   0 0 0 272 272  1 1 1 1 0    272 R → R 0 1 25 51 52  4  35 4 ⎯⎯⎯⎯→ 12  0 0 1 16 17 17   0 0 0 1 5  Thus, Row 4 now implies that x4 = 5. 12 Then, Row 3 implies that x3 + 16 17 (5) = 17 so that x3 = −4. Then, Row 2 implies that x2 + 25 ( −4) + 15 (5) = 52 so that x2 = 1. Finally, Row 1 implies that x1 + 1 − 4 + 5 = 0 so that x1 = −2.

Hence, the solution is x1 = −2, x2 = 1, x3 = −4, x4 = 5 .

891

Chapter 7

76. Reduce the augmented matrix down to row echelon form: 5 3 8 1 1  5 3 8 1 1    R1 −5 R2 →R2   0 −7 −17 −9 −14  R3 − 4 R2 → R3  1 2 5 2 3  ⎯⎯⎯⎯⎯  →  4 0 1 −2 −3 0 −8 −19 −10 −15      1 1 0   0 1 1 1 0  0 1 5 3 8 1  5 3 8 1 1  1     R2 + 7 R4 → R2 0  0 0 −10 −2 −14  R4 ↔ R2 ↔R3 0 1 1 1 R3 +8 R4 → R3  ⎯⎯⎯⎯⎯ → ⎯⎯⎯⎯⎯ → 0 0 −10 −2 −14  0 0 −11 −2 −15      1 0  0 0 −11 −2 −15  0 1 1

1 35 58 1 35 85 51 51     0 1 1 1 0  R3 −R4 →R4 0 1 1  ⎯⎯⎯⎯→ ⎯⎯⎯⎯→ 0 0 1 0 0 1 51 75     15 2 0 0 1 11 11  0 0 0 1 53 58 51 51    0 1 1 1 0 55 R4 →R4  ⎯⎯⎯⎯ → 0 0 1 15 75    0 0 0 1 2  1 R →R 1 5 1 1 R →R − 10 3 3 1 R →R − 11 4 4

1 5

1 1 5 1 55

  0 7  5  2 55   1 5

Thus, Row 4 now implies that x4 = 2. Then, Row 3 implies that x3 + 51 (2) = 75 so that x3 = 1. Then, Row 2 implies that x2 + 1 + 2 = 0 so that x2 = −3. Finally, Row 1 implies that x1 + 35 ( −3) + 85 (1) + 51 (2) = 51 so that x1 = 0. 77. Reduce the matrix down to reduced row echelon form:  1 3 −5   1 3 −5  1 R →R 1 3 −5  R2 + 2 R1 → R2 2 5 2 → →   ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯   −2 −1 0   0 5 −10   0 1 −2  1 0 1  R1 −3 R2 → R1 ⎯⎯⎯⎯ ⎯ →   0 1 −2 

Hence, the solution is (1, −2 ) .

892

Section 7.1

78. Reduce the matrix down to reduced row echelon form: 5 −4 31  5 −4 31  5 −4 31  5 R →R R2 − 35 R1 →R2 2 47 2 → ⎯⎯⎯⎯ →   ⎯⎯⎯⎯⎯  47 188  3 7 −19   0 1 −4  0 5 − 5  5 0 15  1 0 3  1 R →R R1 + 4 R2 → R1 1 5 1 ⎯⎯⎯⎯⎯ → →  ⎯⎯⎯⎯  0 1 −4   0 1 −4 

Hence, the solution is ( 3, −4 ) . 79. Reduce the matrix down to reduced row echelon form:  2 1 −1 −2 R1 + R2 →R2 2 1 −1 − R2 + R1 →R1 2 0 −6  21 R1 →R1 1 0 −3 → → →   ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯  4 3 3  0 1 5  0 1 5  0 1 5  Hence, the solution is ( −3,5 ) . 80. Reduce the matrix down to reduced row echelon form:  4 −3 9  4 R2 →R2  4 −3 9  − 74 R1 + R2 →R2   4 −3 9   ⎯⎯⎯⎯ 29 → → 29 87   ⎯⎯⎯⎯⎯  0 −   7 2 −6   0 1 −3    4 4  4 0 0  14 R1 →R1  1 0 0  −3+ R1 → R1 ⎯⎯⎯⎯→  →  ⎯⎯⎯⎯   0 1 −3   0 1 −3

Hence, the solution is ( 0, −3 ) .

81. Reduce the matrix down to reduced row echelon form: 0 0 22  1 1 4  0 0 22  − 13 R2 → R2 3 R1 + R2 →R1 → →   ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯ 10   −3 −3 10   −3 −3 10  1 1 − 3  Since the first row translates to 0 = 22, which is false, the solution has no solution. 82. Reduce the matrix down to reduced row echelon form:  3 −4 12  0 0 0  0 0 0  − 61 R2 → R2 2 R1 + R2 →R1 → →   ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯  4  −6 8 −24   −6 8 −24  1 − 3 4  Since the first row holds for all values of x and y, there are infinitely many solutions to the system. Let x = a, so that from the second row, a − 34 y = 4  y = 34 a − 3. So, the solution is x = a, y = 34 a − 3 .

893

Chapter 7

83. Reduce the matrix down to reduced row echelon form:  1 −2 3 5  1 −2 3 5  R + R →R   −3R11 + R32 →R32   → 0 12 −13 −27   3 6 −4 −12  ⎯⎯⎯⎯⎯ 0 −6 9 21   −1 −4 6 16 

 1 −2 3 5   1 −2 3 5    51 R3 →R3   ⎯⎯⎯⎯⎯ → 0 12 −13 −27  ⎯⎯⎯⎯ → 0 12 −13 −27  0 0 0 0 5 15  1 3  1 −2 3 5  1 −2 3 5  1 R →R     13 R3 + R2 → R2 2 12 2 → 0 1 0 1 ⎯⎯⎯⎯⎯ → 0 12 0 12  ⎯⎯⎯⎯ 0 0 1 3  0 0 1 3  1 0 0 −2    R1 + 2 R2 −3 R3 → R1 ⎯⎯⎯⎯⎯⎯ → 0 1 0 1  0 0 1 3  R2 + 2 R3 → R3

Hence, the solution is ( −2,1,3 ) . 84. Reduce the matrix down to reduced row echelon form:  1 2 −1 6  R −2 R →R 1 2 −1 6  51 R2 →R2 1 2 −1 6    R32 −3R11→R32   12 R3 →R3   2 1 3 13 0 5 5 25 − − ⎯⎯⎯⎯⎯ → − − → 0 −1 1 −5      ⎯⎯⎯⎯  3 −2 3 −16   0 −8 6 −34  0 −4 3 −17 

1 2 −1 6  R + R →R 1 2 −1 6     − R2 3 →3R3 2  ⎯⎯⎯⎯⎯ → 0 −1 1 −5  ⎯⎯⎯⎯→ 0 −1 0 −2  0 0 1 −3  0 0 −1 3  1 0 0 −1 R1 + 2 R2 + R3 → R1   − R2 → R2 ⎯⎯⎯⎯⎯⎯ → 0 1 0 2  0 0 1 −3 R3 − 4 R2 →R3

Hence, the solution is ( −1,2, −3 ) .

894

Section 7.1

85. Reduce the matrix down to reduced row echelon form: 1 1 1 3  1 1 1 3  1 1 1 3    R1 −R2 →R2   R2 −R3 →R3   1 0 −1 1  ⎯⎯⎯⎯→ 0 1 2 2  ⎯⎯⎯⎯→ 0 1 2 2  0 1 −1 −4  0 1 −1 −4  0 0 3 6  1 1 1 3  1 1 1 3  1 R →R   R2 −2 R3 →R2   3 3 3 ⎯⎯⎯⎯ → 0 1 2 2  ⎯⎯⎯⎯⎯ → 0 1 0 −2  0 0 1 2  0 0 1 2 

1 0 0 3    R1 − R2 − R3 → R1 ⎯⎯⎯⎯⎯ →  0 1 0 −2  0 0 1 2  Hence, the solution is ( 3, −2,2 ) . 86. Reduce the matrix down to reduced row echelon form:  1 −2 4 2  1 −2 4 2  R −2 R →R 1 −2 4 2    4 R3 →R3   R32 −2 R11 →R32   → 2 −3 −2 −3 ⎯⎯⎯⎯⎯ → 0 1 −10 −7   2 −3 −2 −3  ⎯⎯⎯⎯  12 14 2 1 0 5 −4 −12  1 −2  4 −8

 1 −2 4 2  1 −2 4 2     461 R3 →R3  ⎯⎯⎯⎯⎯ → 0 1 −10 −7  ⎯⎯⎯⎯ → 0 1 −10 −7  1   0 0 46 23  1 2  0 0 R3 −5 R2 → R3

 1 −2 4 2   1 0 0 −4    R1 + 2 R2 − 4 R3 →R1   ⎯⎯⎯⎯⎯ → 0 1 0 −2  ⎯⎯⎯⎯⎯⎯ → 0 1 0 −2  0 0 1 12   0 0 1 12      R2 +10 R3 → R2

Hence, the solution is ( −4, −2, 12 ) . 87. Reduce the matrix down to reduced row echelon form:  1 2 1 3  R − 2 R →R 1 2 1 3  1 2 1 3   R32 −3R11→R32   R2 −R3 →R3   → 0 −5 1 1  ⎯⎯⎯⎯→  0 −5 1 1  2 −1 3 7  ⎯⎯⎯⎯⎯  3 1 4 5  0 −5 1 −4   0 0 0 5  The last row is equivalent to the statement 0 = 5, which is false. Hence, the system has no solution.

895

Chapter 7

88. Reduce the matrix down to reduced row echelon form:  1 2 1 3  R − 2 R →R  1 2 1 3 1 2 1 3    R32 −3R11→R32   R2 −R3 →R3   → 0 −5 1 1 ⎯⎯⎯⎯→ 0 −5 1 1   2 −1 3 7  ⎯⎯⎯⎯⎯  3 1 4 10  0 −5 1 1 0 0 0 0  1 0 75 175  1 2 1 3     R1 −2 R2 →R1  ⎯⎯⎯⎯ → 0 1 − 15 − 51  ⎯⎯⎯⎯⎯ → 0 1 − 51 − 51  0 0 0 0  0 0 0 0    Now, we must solve the dependent system:  x + 75 z = 175 5x + 7 z = 17 (1) which is equivalent to   1 1  5 y − z = −1 (2)  y−5 z = −5 Multiply (2) by 7 and add to (1) to obtain 5x + 35 y = 10, which is equivalent to x + 7 y = 2. Let x = a. Then, y = 2 −7 a and z = 17 −75 a . − 51 R2 →R2

89. Reduce the matrix down to reduced row echelon form: R1 − 14 R2 → R1 3 0 − 34 9  3 −1 1 8  3 −1 1 8  − 14 R2 → R2 R1 −3 R2 →R2 ⎯⎯⎯⎯⎯ → ⎯⎯⎯⎯⎯ →       7 0 1 − 4 1  1 1 −2 4  0 −4 7 −4 

1 0 − 14 3 1 R →R 1 3 1 ⎯⎯⎯⎯ →  7 0 1 − 4 1 Now, must solve the dependent system:  x − 14 z = 3 (1)  7  y − 4 z = 1 (2) Let z = a. Then, x = a4 + 3 and y = 74a + 1. 90. Reduce the matrix down to reduced row echelon form: 1 −2 3 10  1 −2 3 10  −3 0 1 9  3 R1 + R2 → R2 −3 R1 + R2 →R1 → →   ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯⎯   −3 0 1 9  0 −6 10 39   0 −6 10 39  Now, must solve the dependent system:  − 3x + z = 9 (1)  −6 y + 10 z = 39 ( 2 ) Multiply (1) by −10 and add to (2) to obtain 30x − 6 y = −51. Let x = a. Then, y = 30 a6+51 and z = 3a + 9.

896

Section 7.1

91. Reduce the matrix down to reduced row echelon form:  4 −2 5 20   4 −2 5 20  −28 0 −22 −144  R1 − 4 R2 → R2 −7 R1 + R2 →R1 → →   ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯⎯  −4  1 3 −2 6  0 −14 13 −4  0 −14 13

14 0 11 72  − 12 R1 → R1 ⎯⎯⎯⎯ →  0 −14 13 −4  Now, must solve the dependent system:  14 x + 11y = 72 (1)   − 14 y + 13z = −4 ( 2 ) Multiply (1) by 14 and (2) by 11 and add to obtain 196 x + 143z = 964. Let z = a. −11a and y = 1314a + 4 . Then, x = 7214 92. Reduce the matrix down to reduced row echelon form: 0 1 1 4  1 1 0 8  1 0 −1 4  R1 ↔ R2 R1 − R2 → R1   ⎯⎯⎯→   ⎯⎯⎯⎯→   1 1 0 8  0 1 1 4  0 1 1 4  Now, must solve the dependent system:  x − z = 4 (1)   y + z = 4 ( 2) Add (1) and (2) to obtain x + y = 4. Let x = a. Then, y = 4 − a and z = a − 4. 93. Reduce the matrix down to reduced row echelon form:  1 −1 −1 −1 1   1 −1 −1 −1 1  R2   RR22 −−22 RR13 →   → R3 1 2 3  R4 −3R3 → 0 3 3 4 1 R4 2 1  ⎯⎯⎯⎯⎯ →  1 −2 −2 −3 0  0 5 5 8 3       3 −4 1 5 −3  0 2 7 14 −3  1 −1 −1 −1 1   1 − 1 −1 − 1 1     R2 −3R3 →R2  0 3 3 4 1  R4 −2 R3 →R4  0 0 0 − 54 − 54  1 R →R 3  5 3 ⎯⎯⎯⎯ → ⎯⎯⎯⎯⎯ → 8 3  8 3  0 1 1 0 1 1 5 5 5 5     54 − 215   0 2 7 14 −3  0 0 5 5 1 −1 −1 −1 1  1 −1 −1 −1 1  1 R →R    3 8 3 8 3  R1 ↔ R3 5 3 0 1 1 0 1 1 − 54 R4 → R4 5 5  R3 ↔ R2 5 5    ⎯⎯⎯→ ⎯⎯⎯⎯→ 21  0 0 5 545 − 215  0 0 1 54 − 25 25     4 4 1  0 0 0 1 0 0 0 − 5 − 5  897

Chapter 7

1 −1 −1 −1 1  1 −1 −1 −1 1      8 3 0 1 1 0 1 0 0 2 R3 − 54 R → R3 R2 − R3 − 58 R4 → R2 5 5  25 4   ⎯⎯⎯⎯⎯ → ⎯⎯⎯⎯⎯⎯ → 0 0 1 0 −3  0 0 1 0 −3      0 0 0 1 1  0 0 0 1 1  1 0 0 0 1    0 1 0 0 2 R1 + R2 + R3 + R4 →R1  ⎯⎯⎯⎯⎯⎯→ 0 0 1 0 −3   0 0 0 1 1  Hence, the solution is w = 1, z = −3, y = 2, x = 1 94. Reduce the matrix down to reduced row echelon form:  1 −3 3 −2 4   1 −3 3 −2 4    R1 −R2 →R2   0 −5 4 −2 7  R1 − R3 →R3  1 2 −1 0 −3 ⎯⎯⎯⎯  → 1 0 3 2 3   0 −3 0 −4 1       0 1 1 5 6  0 1 1 5 6  1 −3 3 −2 4  1 −3    R2 + 5 R4 →R2 0 0 9 23 37  R2 −3R3 →R2 0 0 R3 + 3 R4 →R3  ⎯⎯⎯⎯⎯ → ⎯⎯⎯⎯⎯ → 0 0 3 11 19  0 0    0 1 1 5 6  0 1  1 −3  0 1 R2 ↔ R4 ⎯⎯⎯→  0 0   0 0  1 −3  0 1 R3 − 11 R →R3 3 4 ⎯⎯⎯⎯⎯ → 0 0  0 0

3 −2 4   0 −10 −20  3 11 19   1 5 6 

4  1 −3  − 101 R4 →R4  0 1 −20  13 R3 →R3 ⎯⎯⎯⎯→  0 0 31    0 −10 6  0 0

3 −2 4   1 5 6 1 113 313   0 1 2 

4 1 −3   6  R2 −R3 −5 R4 →R2 0 1 ⎯⎯ ⎯⎯⎯⎯ → 0 0 3   2  0 0

3 −2 4   0 0 −7  1 0 3  0 1 2 

3 1 3

−2 5 11

3 −2 1 5 1 0 0 1

898

Section 7.1

1 0 0  0 1 0 R1 +3 R2 −3 R3 + 2 R4 → R1 ⎯⎯⎯⎯⎯⎯⎯→  0 0 1  0 0 0 Hence, the solution is w = 2, x = −22, y = −7, z = 3 .

0 −22   0 −7  0 3   1 2 

95. Let x = number of touchdowns y = number of extra points z = number of field goals We must solve the system:  x + y + z = 15  x + y + z = 15   which is equivalent to  y − 2 z = 0 y = 2z  6 x + y + 3z = 51 6 x + y + 3z = 51  

Now, write down the augmented matrix, and reduce it down to row echelon form:  1 1 1 15  1 1 1 15  − R2 + R1 →R1 1 0 3 15    −6 R1 + R3 →   5 R2 + R3 →R3   → 0 1 −2 0   0 1 −2 0  ⎯⎯⎯⎯→ 0 1 −2 0  ⎯⎯⎯⎯⎯ 6 1 3 51 0 −5 −3 −39  0 0 −13 −39   1 0 3 15    ⎯⎯⎯⎯→  0 1 −2 0   0 0 1 3  Thus, Row 3 now implies that z = 3. Then, Row 2 implies that y − 2 ( 3 ) = 0 so that y = 6. 1 R →R − 13 3 3

Then, Row 1 implies that x + 6 + 3 = 15 so that x = 6. So, there were 6 touchdowns, 6 extra points, and 3 field goals. 96. Let x = number of two-point shots y = number of three-point shots z = number of one-point free throws We must solve the system: x = z + 10 x − z = 10     which is equivalent to  −2 y + z = −4 z = 2y − 4  2 x + 3 y + z = 224 2 x + 3 y + z = 224  

899

Chapter 7

Now, write down the augmented matrix, and reduce it down to row echelon form:  1 0 −1 10  1 0 −1 10    −2 R1 + R3 →R3   → 0 −2 1 −4   0 −2 1 −4  ⎯⎯⎯⎯⎯  2 3 1 224  0 3 3 204  1 0 −1 10  1 0 −1 10     R3 −R2 →R3  1 ⎯⎯⎯ ⎯→ 0 1 − 2 2  ⎯⎯⎯⎯→ 0 1 − 12 2  0 0 32 66  0 1 1 68    1 0 −1 10  2 R →R   3 3 3 ⎯⎯⎯ →  0 1 − 12 2  ⎯  0 0 1 44  − 12 R2 →R2 1 R →R 3 3 3

Thus, Row 3 now implies that z = 44. 1 Then, Row 2 implies that y − ( 44 ) = 2 so that y = 24. 2 Then, Row 1 implies that x − 44 = 10 so that x = 54. So, there were 54 two-point shots, 24 three-point shots, and 44 one-point free throws. 97. Let w = number of egg salad sandwiches x = number of club sandwiches y = number of turkey wraps z = number of roast beef sandwiches We must solve the system: w + x + y + z = 14  17w + 46 x + 45 y + 20 z = 526    18w + 19x + 5 y + 27 z = 168  36w + 20 x + 19 y + 34 z = 332 Now, write down the augmented matrix, and reduce it down to row echelon form:  1 1 1 1 14  1 1 1 1 14  36 R1 → R4   RR34 −−18   R1 → R3 0 29 28 3 288  R2 −17 R1 → R2 17 46 45 20 526  ⎯⎯⎯⎯⎯  → 18 19 5 27 168  0 1 −13 9 −84      36 20 19 34 332  0 −16 −17 −2 −172 

900

Section 7.1

1  R2 − 29 R3 →R2 0 R4 +16 R3 → R4  ⎯⎯⎯⎯⎯ → 0  0

14  1 1 1   0 405 −258 2724  R2 ↔R3 0 1 −13 ⎯⎯⎯→ 0 0 405 1 −13 9 −84    0 −225 142 −1516  0 0 −225 1 1 1 1 1 1 1 14     0 1 −13 9 −84  R4 + 225 R3 →R4 0 1 −13 1 R →R 3 405 3  ⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯→ 2724  0 0 1 0 0 1 − 258 405 405    0 0 0 0 0 −225 142 −1516  1 14  1 1 1   405 R → R 0 1 −13 9 −84  − 540 4 4  ⎯⎯⎯⎯→ 2724  0 0 1 − 258 405 405   1 2  0 0 0 Thus, Row 4 now implies that z = 2. 2724 Then, Row 3 implies that y − 258 405 (2) = 405 so that y = 8. Then, Row 2 implies that x − 13(8) + 9(2) = −84 so that x = 2. Then, Row 1 implies that w + 2 + 8 + 2 = 14 so that w = 2. So, there were 2 egg salad sandwiches, 2 club sandwiches, 8 turkey wraps, and 2 roast beef sandwiches 1

14   9 −84  −258 2724   142 −1516  1 14   −84  9 2724  − 258 405 405  540 − 405 − 1080 405   1

98. Let w = number of egg salad sandwiches x = number of club sandwiches y = number of turkey wraps z = number of roast beef sandwiches We must solve the system: w + x + y + z = 14  350w + 430x + 290 y + 430 z = 5180   17w + 46 x + 45 y + 20 z = 335   18w + 19x + 5 y + 27 z = 263 Now, write down the augmented matrix, and reduce it down to row echelon form:  1 1 1 1 14  1 1 1 1 14  350 R1 → R2   RR32 −−17   R1 →R3 0 80 −60 80 280  R4 −18 R1 →R4 350 430 290 430 5180  ⎯⎯⎯⎯⎯  →  17 46 45 20 335  0 29 28 3 97      5 27 263   18 19 0 1 −13 9 11 

901

Chapter 7

1 1 1 1 14  1   R3 −29 R2 →R3  0 R2 ↔ R4 R4 −80 R2 → R4  0 1 −13 9 11  ⎯⎯⎯⎯⎯ ⎯⎯⎯→ →  0 29 28 3 97  0    0 80 −60 80 280  0

14   1 −13 9 11  0 405 −258 −222   0 980 −640 −600  1 1 1 1 14  1 14  1 1 1     9 11  0 1 −13 9 11  R4 −980 R3 →R3  0 1 −13 1 R →R 3 405 3  ⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯→ 0 0 1   0 0 1 − 258 − 222 − 258 − 222 405 405 405 405    25,440  6360 0 0 980 −640 −600   0 0 0 − 405 − 405  1 14  1 1 1   405 R → R 0 1 −13 9 11  − 6360 4 4  ⎯⎯⎯⎯⎯ →  0 0 1 − 258  − 222 405 405   1 4   0 0 0 Thus, Row 4 now implies that z = 4. 222 Then, Row 3 implies that y − 258 405 (4) = − 405 so that y = 2. Then, Row 2 implies that x − 13(2) + 9(4) = 11 so that x = 1. Then, Row 1 implies that w + 1 + 2 + 4 = 14 so that w = 7. So, there were 7 egg salad sandwiches, 1 club sandwich, 2 turkey wraps, and 4 roast beef sandwiches. 1

99. From the given information, we must solve the following system: 34 = 12 a(1)2 + v0 (1) + h0  68 = a + 2v0 + 2 h0   2 1 36 = 2 a(2) + v0 (2) + h0 is equivalent to 72 = 4a + 4v0 + 2h0  12 = 9a + 6v + 2 h 2 1 0 0   6 = 2 a(3) + v0 (3) + h0 Now, write down the augmented matrix, and reduce it down to row echelon form:  1 2 2 68  R − 4 R →R 1 2 2 68  1 2 2 68    R32 −9 R11→R32   R3 −3R2 →R3   → 0 −4 −6 −200  ⎯⎯⎯⎯⎯ → 0 −4 −6 −200   4 4 2 72  ⎯⎯⎯⎯⎯  9 6 2 12  0 −12 −16 −600  0 0 2 0  Thus, Row 3 now implies that h0 = 0 ft. = initial height. Then, Row 2 implies that −4v0 − 6(0) = −200 so that v0 = 50 ft./sec. = initial velocity. Then, Row 1 implies that a + 2(50) + 2(0) = 68 so that a = −32 ft./sec 2 = acceleration. Thus, the equation of the curve is y = − 12 (32)t 2 + 50t + 0 = −16t 2 + 50t. 902

Section 7.1

100. From the given information, we must solve the following system: 54 = 12 a(1)2 + v0 (1) + h0 108 = a + 2v0 + 2 h0   2 1 66 = 2 a(2) + v0 (2) + h0 is equivalent to 132 = 4a + 4v0 + 2 h0   2 1  92 = 9a + 6v0 + 2 h0 46 = 2 a(3) + v0 (3) + h0 Now, write down the augmented matrix, and reduce it down to row echelon form:  1 2 2 108  R − 4 R →R 1 2 2 108  1 2 2 108    R32 −9 R11→R32   R3 −3R2 →R3   → 0 −4 −6 −300  ⎯⎯⎯⎯⎯ → 0 −4 −6 −300   4 4 2 132  ⎯⎯⎯⎯⎯  9 6 2 92  0 −12 −16 −880  0 0 2 20 

Thus, Row 3 now implies that 2 h0 = 20 so that h0 = 10 ft. = initial height. Then, Row 2 implies that −4v0 − 6(10) = −300 so that v0 = 60 ft./sec. = initial velocity. Then, Row 1 implies that a + 2(60) + 2(10) = 108 so that a = −32 ft./sec.2 = acceleration. Thus, the equation of the curve is y = − 12 (32)t 2 + 60t + 10 = −16t 2 + 60t + 10. 101. From the given information, we must solve the following system:  25 = a(16)2 + b(16) + c  25 = 256a + 16b + c   2 64 = a(40) + b(40) + c is equivalent to  64 = 1600a + 40b + c 40 = a(65)2 + b(65) + c 40 = 4225a + 65b + c   Now, write down the augmented matrix, and reduce it down to row echelon form: 25 1 1  1   256 16 1 25  16 256 256    2561 R1 →R1  → 1600 40 1 64  1600 40 1 64  ⎯⎯⎯⎯  4225 65 1 40   4225 65 1 40    25 1 1 1 1  − 601 R2 →R2 16 256 256 16 R → R    − 3 3185 3 → 0 ⎯⎯⎯⎯⎯→ 0 −60 − 1344 − 23616 256 256  ⎯⎯⎯⎯⎯ 0 0 − 3185  − 3969 − 95385 16 256 256    25 1 1 161  256 256   R3 − R2 →R3 1344 23616 ⎯⎯⎯⎯→ 0 1 15360 15360  63504 1344 1526160 23616  0 0 815360 − − 15360 815360 15360   R2 −1600 R1 → R1 R3 − 4225 R1 →R3

903

1 16

1 1

1 256 1344 15360 63504 815360

25 256 23616 15360 1526160 815360

    

Chapter 7 25 1  1 161   1 0.0625 0.0039 0.0977  256 256     1344 23616 1 0.0875 1.5375  ⎯⎯⎯⎯⎯⎯ →  0 1 15360 15360  ≅ 0 0 0 1 −34.76326   0 0 1 −34.76326   Thus, Row 3 now implies that c ≅ −34.76326. Then, Row 2 implies that b + 0.0875( −34.76326) = 1.5375 so that b ≅ 4.57928. Then, Row 1 implies that a + 0.0625(4.57928) + 0.0039( −34.76326) = 0.0977 so that a ≅ −0.052755. 1 R → R3 1344 − 63504 3 15360 815360

Thus, the approximate equation of the curve is y = −0.053x 2 + 4.58x − 34.76 . 102. From the given information, we must solve the following system:  18.4 = a(0)2 + b(0) + c  18.4 = 0a + 0b + c   2 20.3 = a(40) + b(40) + c is equivalent to  20.3 = 1600a + 40b + c 2  25.3 = 6724a + 82b + c   25.3 = a(82) + b(82) + c Now, write down the augmented matrix, and reduce it down to row echelon form: 0 1 18.4   0 6724 82 1 25.3 R3 ↔ R1 1600 40 1 20.3 ⎯⎯⎯→ 1600 40 1 20.3     6724 82 1 25.3  0 0 1 18.4  25.3 1  1 821 6724 6724  1 R →R , 1 R → R 1 1600 2 2 20.3  1 6724 1 ⎯⎯⎯⎯⎯⎯⎯ →  1 401 1600 1600   0 0 1 18.4  25.3 1 1 1  82 6724 6724  R2 − R1 → ⎯⎯⎯⎯ → 0 .012805 .0000476 .008925  0 0 1 18.4  25.3 1 1 821 6724 6724  1 R → R2 .012805 2 ⎯⎯ ⎯⎯⎯ → 0 1 .037195 .697  0 0 1 18.4  Thus, Row 3 now implies that c ≈ 18.4. Then, Row 2 implies that b + .037195 (18.4 ) = .697 so that b ≈ 0.0126.

25.3 1 Then, Row 1 implies that a + 821 ( 0.0126 ) + 6724 so that a ≈ 0.0008725. (18.4 ) = 6724 Thus, the approximate equation of the curve is y = 0.0008725x 2 + 0.0126x + 18.4. Finally, using this line to predict ages in future years, we see that in the year 2024 (which corresponds to x = 104), the average age would be y (104 ) ≈ 29.15 years.

904

Section 7.1

103. Let x = number of mL of 1.5% solution, y = number of mL of 30% solution. We must solve the system: x + y = 100   0.015x + 0.30 y = 0.05(100) Write down the augmented matrix, and reduce it down to row echelon form:  1 1 100  1000 R2 →R2  1 1 100    ⎯⎯⎯⎯→    0.015 0.30 5  15 300 5000 

15 300 5000  R1 −15 R2 →R2 15 300 5000  R1 ↔ R2 ⎯⎯⎯→ →   ⎯⎯⎯⎯⎯  1 1 100    0 285 3500  1 R →R 15 300 5000  151 R1 →R1 1 20 333.33 2 285 1 ⎯⎯⎯⎯ → →  ⎯⎯⎯⎯  0 1 12.28   0 1 12.28  1 0 87.73 R1 − 20 R2 → R1 ⎯⎯⎯⎯⎯ →  0 1 12.28  So, about 88mL of 1.5% solution and 12 mL of 30% solution. 104. Let x = number of grams of 1% cream, y = number of grams of 0.5% cream, z = number of grams of 0% cream. We must solve the system: x + y + z = 60   which is equivalent to 2z = y  0.01x + 0.005 y + 0 z = 0.007(60)  x + y + z = 60   y − 2z = 0  0.01x + 0.005 y + 0 z = 0.007(60)  Write down the augmented matrix, and reduce it down to row echelon form:  1 1 1 60  1 1 1 60    R3 −10 R1 →R3   → 0 1 −2 0   0 1 −2 0  ⎯⎯⎯⎯⎯ 10 5 0 420  0 −5 −10 −180 

1 1 1 60  1 1 1 60    R2 −R3 →R3   ⎯⎯⎯⎯ → 0 1 −2 0  ⎯⎯⎯⎯→ 0 1 −2 0  0 1 2 36  0 0 −4 −36  − 51 R3 → R3

905

Chapter 7

1 1 1 60  1 1 1 60      R2 + 2 R3 → R2 ⎯⎯⎯⎯→ 0 1 −2 0  ⎯⎯⎯⎯⎯ → 0 1 0 18  0 0 1 9  0 0 1 9  1 0 0 33   R1 − R2 − R3 → R1 ⎯⎯⎯⎯⎯⎯ → 0 1 0 18  0 0 1 9  − 14 R3 → R3

So, 33 grams of 1% cream, 18 grams of 0.5% cream, and 9 grams of 0% cream. 105. Let x = number of basic widgets ($12 each), y = number of mid-price widgets ($15 each), z = number of top-of-the-line widgets ($18 each). We must solve the system: x + y + z = 375 x + y + z = 375     12 x + 15 y + 18z = 5250 which is equivalent to 12 x + 15 y + 18z = 5250   x = 2y x − 2y = 0  

Write down the augmented matrix, and reduce it down to row echelon form:  1 1 1 375   1 1 1 375    R3 −12 R2 →R3   0  ⎯⎯⎯⎯⎯ →  1 −2 0 0   1 −2 0 12 15 18 5250   0 39 18 5250  0 3 1 375   1 −2 0 0    R1 ↔ R2   ⎯⎯⎯⎯→ 1 −2 0 0  ⎯⎯⎯→ 0 3 1 375  0 39 18 5250  0 39 18 5250   1 −2 0 0  1 −2 0 0  1 R →R     R3 − 29 R2 → R3 2 1 3 2 125  ⎯⎯⎯⎯⎯→ 0 1 13 125  ⎯⎯⎯⎯ → 0 1 3 0 39 18 5250  0 0 5 375   1 −2 0 0  1 −2 0 0  1 R →R 1 R →R     R − 3 3 2 3 2 5 3 ⎯⎯⎯⎯ →  0 1 13 125  ⎯⎯⎯⎯⎯ → 0 1 0 100   0 0 1 75  0 0 1 75  R1 − R2 → R1

 1 0 0 200    ⎯⎯⎯⎯⎯ →  0 1 0 100   0 0 1 75  So, 200 basic widgets, 100 mid-price widgets, and 75 top-of-the-line widgets will be produced. R1 + 2 R2 → R1

906

Section 7.1

106. Let x = number of basic widgets ($10 each), y = number of mid-price widgets ($12 each), z = number of top-of-the-line widgets ($15 each). We must solve the system: x + y + z = 350 x + y + z = 350     10 x + 12 y + 15 z = 4600 which is equivalent to 10 x + 12 y + 15 z = 4600   2x − z = 0 z = 2x   Write down the augmented matrix, and reduce it down to row echelon form:  1 1 1 350   1 1 1 350    R2 −10 R1 →R2   →  0 2 5 1100  10 12 15 4600  ⎯⎯⎯⎯⎯  2 0 −1  2 0 −1 0  0 

1 1 1 350  1   R2 + R3 →R3  ⎯⎯⎯⎯⎯ → 0 2 5 1100  ⎯⎯⎯⎯⎯ → 0 0 −2 −3 −700  0 1 1 1 1 1 1 350  1 R →R  12 R2 →R2   3 2 3 →  0 1 52 ⎯⎯⎯⎯ → 0 2 5 1100  ⎯⎯⎯⎯  0 0 1 0 0 1 200  R3 − 2 R1 → R3

1 1 350   2 5 1100  0 2 400  350   550  200 

1 1 1 350  1 0 0 100      R1 − R2 − R3 → R1 ⎯⎯⎯⎯⎯ → 0 1 0 50  ⎯⎯⎯⎯⎯⎯ → 0 1 0 50  0 0 1 200  0 0 1 200  So, 100 basic widgets, 50 mid-price widgets, and 200 top-of-the-line widgets will be produced. R2 − 52 R3 →R2

107. Let x = amount in money market (3%) y = amount in mutual fund (7%) z = amount in stock (10%) We must solve the system: x + y + z = 10,000 x + y + z = 10,000     x = y + 3000 which is equivalent to  x − y = 3000  0.03x + 0.07 y + 0.10 z = 540 3x + 7 y + 10 z = 54,000  

907

Chapter 7

Write down the augmented matrix, and reduce it down to row echelon form: 1 1 1 10000  R −3R →R  1 1 1 10000  1 3   R32 −R1 →   R2 1 1 0 3000 − ⎯⎯⎯⎯⎯ → 0 −2 −1 −7000    3 7 10 54000  0 4 7 24000  1 1 1 10000  1 1 1 10000  1 R →R 3 5 3   − 12 R2 →R2   ⎯⎯⎯⎯⎯ → 0 −2 −1 −7000  ⎯⎯⎯⎯ → 0 1 12 3500  0 0 5 10000  0 0 1 2000  1 1 1 10000  1 0 0 5500      R2 − 21 R3 → R2 R1 − R2 − R3 → R1 ⎯⎯⎯⎯⎯ → 0 1 0 2500  ⎯⎯⎯⎯⎯⎯ → 0 1 0 2500  0 0 1 2000  0 0 1 2000  So, they should invest $5500 in money market, $2500 in mutual fund, and $2000 in stock. R3 + 2 R2 → R3

108. Let x = amount in money market (3%) y = amount in mutual fund (7%) z = amount in stock (10%) We must solve the system: x + y + z = 10,000 x + y + z = 10,000     − y + z = 3000 z = y + 3000 which is equivalent to   0.03x + 0.07 y + 0.10 z = 840 3x + 7 y + 10 z = 84,000  

Write down the augmented matrix, and reduce it down to row echelon form:  1 1 1 10000   1 1 1 10000    R3 −3R1 →R3   →  0 −1 1 3000   0 −1 1 3000  ⎯⎯⎯⎯⎯  3 7 10 84000   0 4 7 54000  1 1 1 10000  − R2 →R2 1 1 1 10000    111 R3 →R3   ⎯⎯⎯⎯⎯ → 0 −1 1 3000  ⎯⎯⎯⎯ → 0 1 −1 −3000  0 0 11 66000  0 0 1 6000  Then, Row 3 implies that z = 6000. Then, Row 2 implies that y − 6000 = −3000 so that y = 3000. Then, Row 1 implies that x + 3000 + 6000 = 10000 so that x = 1000. So, they should invest $1000 in money market, $3000 in mutual fund, and $6000 in stock. 4 R2 + R3 → R3

908

Section 7.1

109. Let a = units of product x, b = units of product y, and c = units of product z. We employ matrix methods to solve the system:  20a + 25b + 150c = 2400  2a + 5b + 10c = 310   a + 0b + 12 c = 28   20 25 150 2400  1 R →R  4 5 30 480    R5 2 1−2 R31→R3   →  2 5 10 310   2 5 10 310  ⎯⎯⎯⎯⎯ 1  1 0  0 5 9 254  28  2  4 5 30 480   4 5 30 480      R + R → R R1 − 2 R2 →R2 2 3 3 ⎯⎯⎯⎯⎯ →  0 −5 10 −140  ⎯⎯⎯⎯→  0 −5 10 −140   0 5 9 254   0 0 19 114   4 5 30 480  1 R →R  1 54 152 120  1 4 1   R2 + 2 R3 →R2   ⎯⎯⎯⎯ →  0 1 −2 28  ⎯⎯⎯⎯⎯ →  0 1 0 40   0 0 1 6   0 0 1 6  1 0 0 25    R1 − 54 R2 − 15 R → R 1 2 3 ⎯⎯⎯⎯⎯⎯ → 0 1 0 40  0 0 1 6  Hence, there are 25 units of product x, 40 units of product y, and 6 units of product z. − 51 R2 → R2 1 R →R 3 19 3

110. Using the given three points generates the following system that must be solved in order to determine the coefficients of the quadratic function on which they lie:  a+b+c =5  4a − 2b + c = −10  c=4 

We solve using matrix methods: 1 1 1 5  1 1 1 5  1 1 1 5    R2 − 4 R1 →R2   R2 +3R3 →R2   → 0 −6 −3 −25  ⎯⎯⎯⎯⎯ → 0 −6 0 −13  4 −2 1 −10  ⎯⎯⎯⎯⎯  0 0 1 4  0 0 1 4  0 0 1 4 

909

Chapter 7

1 0 0 − 76  1 1 1 5     R1 −R2 −R3 →R1  ⎯⎯⎯⎯ → 0 1 0 136  ⎯⎯⎯⎯⎯ → 0 1 0 136  0 0 1 4  0 0 1 4    − 16 R2 →R2

Hence, the equation is y = − 76 x 2 + 136 x + 4 . 111. Let x = # of general admission tickets, y = # of reserved tickets, and z = # of end zone tickets. We solve the following system using matrix methods: y = x +5   z = x + 20  20 x + 40 y + 15 z = 2375 

 1 −1 0 −5  R1 −R2 →R2  1 −1 0 −5  1 −1 0 −5    15 R3 →R3   R3 − 4 R1 →R3   → 0 −1 1 15   1 0 −1 −20  ⎯⎯⎯⎯→  0 −1 1 15  ⎯⎯⎯⎯⎯  20 40 15 2375   4 8 3 475  0 12 3 495  1 −1 0 −5  1 −1 0 −5  R3 +12 R2 →R3 1 R →R     − R2 → R2 3 15 3 ⎯⎯⎯⎯⎯ → 0 1 −1 −15  ⎯⎯⎯⎯ → 0 1 −1 −15  0 0 15 675  0 0 1 45   1 0 0 25   1 −1 0 −5    R1 + R2 →R1   ⎯⎯⎯⎯→  0 1 0 30  ⎯⎯⎯⎯→ 0 1 0 30   0 0 1 45   0 0 1 45  Hence, there are 25 general admission tickets, 30 reserved tickets, and 45 end zone tickets. R2 + R3 → R2

112. Let x = # of minutes walking, y = # of minutes step-up exercise, and z = # of minutes weight training. a. # calories per minute for each exercise: cal walking: 1585min = 173 cal min cal = 92 cal min step-up: 1045min cal 137 weight training: 137 20 min = 20 cal min b. Solve the system using matrix methods: z = 2x   x + y + z = 60   17 x + 9 y + 137 z = 358 2 20 3

910

Section 7.1

2  1  173

0 −1 0  R −2 R →R  0 −2 −3 −120   601 R3 →2R3 1   1 1 60  ⎯⎯⎯⎯⎯ 1 1 60  → 1 9 137  340 270 411 21, 480  2 20 358  1 60  1 1  0 −2 −3 −120    R1 ↔R2   R3 −340 R2 → R3 1 60  ⎯⎯⎯→ 0 −2 −3 −120  ⎯⎯⎯⎯⎯→  1 1  0 −70 71 1080  0 −70 71 1080  1 1 1 60  1 1 1 60    1761 R3 →R3   ⎯⎯⎯⎯⎯ →  0 −2 −3 −120  ⎯⎯⎯⎯ → 0 −2 −3 −120   0 0 176 5280  0 0 1 30  R3 −35 R2 → R3

1 1 1 60   1 1 1 60    − 12 R2 →R2   ⎯⎯⎯⎯⎯ → 0 −2 0 −30  ⎯⎯⎯⎯ →  0 1 0 15  0 0 1 30   0 0 1 30   1 0 0 15    R1 − R2 − R3 →R1 ⎯⎯⎯⎯⎯→  0 1 0 15   0 0 1 30  Hence, 15 minutes walking, 15 minutes step-up exercise, and 30 minutes weight training. R2 + 3 R3 → R2

113. Since the points (4, 4), ( −3, −1), and (1, −3) are assumed to lie on the circle, they must each satisfy the equation. This generates the following system:  (4)2 + (4)2 + a(4) + b(4) + c = 0 4a + 4b + c = −32   2 2 ( −3) + ( −1) + a( −3) + b( −1) + c = 0 which is equivalent to  −3a − b + c = −10  (1)2 + ( −3)2 + a(1) + b( −3) + c = 0  a − 3b + c = −10   Write down the augmented matrix, and reduce it down to row echelon form: 1 1 1  4 4 1 −32  R − 4 R →R  4 4 1 −32  14 R1 →R1 −8 4 1 3 3 1    R2 +3R3 →R2   − 10 R2 →R2  →  0 −10 4 −40  ⎯⎯⎯⎯→ 0 1 − 52 4   −3 −1 1 −10  ⎯⎯⎯⎯⎯ 0 16 −3 8   1 −3 1 −10   0 16 −3 8    1 1 14 1 1 14 −8  −8  5   17 R3 →R3   R3 −16 R2 → R3 2 2 4  ⎯⎯⎯⎯⎯ → 0 1 − 5 4  ⎯⎯⎯⎯ → 0 1 − 5 0 0 175 −56  0 0 1 − 280  17    

Row 3 implies that c = − 280 17 .

911

Chapter 7 44 Then, Row 2 implies that b − 25 ( − 280 17 ) = 4 so that b = − 17 . 44 22 Then, Row 1 implies that a + ( − 17 ) + 14 ( − 280 17 ) = −8 so that a = − 17 . 44 y − 280 Hence, the equation of the circle is x 2 + y2 − 227 x − 17 17 = 0.

114. Since the points (0,7), (6,1), and (5, 4) are assumed to lie on the circle, they must each satisfy the equation. This generates the following system: (0)2 + (7)2 + a(0) + b(7) + c = 0 7b + c = −49    2 2  (6) + (1) + a(6) + b(1) + c = 0 which is equivalent to  6a + b + c = −37 (5)2 + (4)2 + a(5) + b(4) + c = 0 5a + 4b + c = −41   Write down the augmented matrix, and reduce it down to row echelon form:  1 61 61 − 376   0 7 1 −49  6 1 1 −37     R2 ↔ R1   61 R1 →R1  →  0 7 1 −49  6 1 1 −37  ⎯⎯⎯→ 0 7 1 −49  ⎯⎯⎯⎯ 5 4 1 −41 5 4 1 −41 5 4 1 −41   1  R3 −5 R1 →R3 ⎯⎯⎯⎯⎯ → 0 0 

1 6

7 19 6

1 − 376   71 R2 →R2  1 −49  ⎯⎯⎯⎯ → 0 61  1 0 −6 6  1 6

1 6

1 7

19 6

1 6

− 376   −7  − 616 

1 61 61 − 376  1 61 61 − 376    − 27 R3 →R3   R3 − 19 R → R3 6 2 ⎯⎯⎯⎯⎯ → 0 1 71 −7  ⎯⎯⎯⎯ → 0 1 71 −7  0 0 − 72 12  0 0 1 −42      Row 3 implies that c = −42 .

Then, Row 2 implies that b + 71 ( − 42) = −7 so that b = −1. Then, Row 1 implies that a + 61 ( −1) + 61 ( −42) = − 376 so that a = 1. Hence, the equation of the circle is x 2 + y 2 + x − y − 42 = 0. 115. Need to line up a single variable in a given column before forming the augmented matrix. Specifically, write the system as follows: −x + y + z = 2   x + y − 2 z = −3  x+ y + z =6 

912

Section 7.1

 −1 1 1 2  1 0 0 2      So, the correct matrix is  1 1 −2 −3 , and after reducing,  0 1 0 1  .  0 0 1 3   1 1 1 6  116. a. Need to subtract all entries of 2R1 from the respective entries of R2 . So, the

second row should be 0 −1 −1 0.

b. Same error as in part a. The third row should be 0 4 −1 − 12.

117. Row 3 implies that 1z = 0, which is valid. Also, Rows 2 and 1 imply y = 2, x = 1, respectively. So, the solution is x = 1, y = 2, z = 0. 118. Row 3 implies that the system is inconsistent since it requires 0z = 4. 119. False

120. False (See #47 for example.)

121. True

122. True

123. Assume year 1999 corresponds to 0, 2000 to 1, 2001 to 2, 2002 to 3, and 2003 to 4. Then, from the given data, we can infer that the following four points are on the graph of f ( x ) = ax 4 + bx3 + cx 2 + dx + e : (0, 44), (1,51), (2, 46), (3,51), (4, 44) Hence, they must all satisfy the equation. This leads to the following 5 × 5 system: 44 = a(0)4 + b(0)3 + c(0)2 + d (0) + e  4 3 2 51 = a(1) + b(1) + c(1) + d (1) + e  4 3 2 46 = a(2) + b(2) + c(2) + d (2) + e 51 = a(3)4 + b(3)3 + c(3)2 + d (3) + e  44 = a(4)4 + b(4)3 + c(4)2 + d (4) + e  which is equivalent to: 44 = e 51 = a + b + c + d + e  46 = 16a + 8b + 4c + 2d + e 51 = 81a + 27b + 9c + 3d + e  44 = 256a + 64b + 16c + 4d + e

913

Chapter 7

The corresponding augmented matrix is:  0 0 0 0 1 44    1 1 1 1 51   1  16 8 4 2 1 46     81 27 9 3 1 51   256 64 16 4 1 44    One can perform Gaussian elimination as in the previous problems, but this would be a fine example as to when using technology can be useful. Indeed, using a calculator with matrix capabilities can be used to verify the following choices for a, b, c, d, and e: 94 a = − 116 , b = 443 , c = − 223 6 , d = 3 , e = 44. Hence, the equation of the polynomial is 2 94 y = − 116 x 4 + 443 x3 − 223 6 x + 3 x + 44 .

124. Let x = number of nickels, y = number of dimes, z = number of quarters. We must solve the system:  x + y + z = 30  0.05x + 0.10 y + 0.25 z = 4.60 The corresponding augmented matrix is:  1 1 1 30     0.05 0.10 0.25 4.60  We solve this system by reducing the augmented matrix to row echelon form:  1 1 1 30  100 R2 →R2 1 1 1 30    ⎯⎯⎯⎯→    0.05 0.10 0.25 4.60  5 10 25 460 

1 1 1 30  15 R2 →R2 1 1 1 30  R2 −5 R1 → R2 ⎯⎯⎯⎯⎯ → →   ⎯⎯⎯⎯   0 5 20 310  0 1 4 62  Let z = t. Then, substituting this value into the equation obtained from the second row yields y = 62 − 4t, and then substituting these values of z and y into the equation obtained form the first row, we conclude that x = 3t − 32. Now, the restrictions on t in this problem are that (i) 0 ≤ 62 − 4t ≤ 30, (ii) 0 ≤ 3t − 32 ≤ 30, and (iii) t is an integer. Solving these inequalities yields 0 ≤ 62 − 4t ≤ 30 0 ≤ 3t − 32 ≤ 30 −62 ≤ − 4t ≤ −32 and 32 ≤ ≤ 62 3t t ≥8 15.5 ≥ ≤ 20.6 10.6 ≤ t

914

Section 7.1

Since both of these inequalities must hold simultaneously and t must be an integer, we conclude that the possible values of t are 11, 12, 13, 14, and 15. We verify that each of these is a solution below: t z =t x = 3t - 32 y = 62 − 4t 11 1 18 11 12 4 14 12 13 7 10 13 14 10 6 14 15 13 2 15 125.

126.

127. a.

128. a.

y = –0.24x2 + 0.93x + 6.09 b.

y = 0.36x2 – 0.92x – 17.6 b.

y = –0.24x2 + 0.93x + 6.09

y = 0.36x2 – 0.92x – 17.6

915

Section 7.2 Solutions -------------------------------------------------------------------------------1. Since corresponding entries of equal matrices must themselves be equal, we have x = −5, y = 1.

2. Since corresponding entries of equal matrices must themselves be equal, we have x = 10, y = 12

3.  y + 4 −1 7 −1 =  2 3x  2 9   Since corresponding entries of equal matrices must themselves be equal, we have y+4 =7 y =3 3x = 9  x = 3

4. 1   0 2 y  = 0 8    4 6  4 2 x −   Since corresponding entries of equal matrices must themselves be equal, we have 1 y = 8  y = 16 2 2 − x = 6  x = −4

5. Since corresponding entries of equal matrices must themselves be equal, we have  x + y = − 5 (1)   x − y = −1 ( 2 )  z =3  Solve the system: Add (1) and (2): 2x = −6  x = −3 Substitute this into (1) to obtain y = −2.

6. Since corresponding entries of equal matrices must themselves be equal, we have  x = 2 + y (1)  (2)  y=5 Solve the system: Substitute (2) into (1) to obtain x = 7.

7. Since corresponding entries of equal matrices must themselves be equal, we have  x − y = 3 (1)  x + 2 y = 12 ( 2 ) Solve the system: Multiply (1) by −1: −x + y = −3 ( 3 ) Add (2) and (3): 3 y = 9  y = 3 Substitute this into (1) to obtain x = 6.

8. Since corresponding entries of equal matrices must themselves be equal, we have  a 2 = 9 (1)   2b + 1 = 9 ( 2 )  2  b = 16 ( 3 )  2 a + 1 = −5 ( 4 )  Note that (2) implies b = 4, and this satisfies (3). Also, (4) implies a = −3, and this satisfies (1). 916

Section 7.2

9.  −1 3 0   0 2 1   −1 5 1  2 4 1  + 3 −2 4  = 5 2 5       

10.  2 −2  2 0    7 −1

11.  −2 4   2 −2     −1 3 

12.  −1 3 0   0 2 1   −1 1 −1   2 4 1  − 3 −2 4  =  −1 6 −3      

13. Not defined since B is 2 × 3 and C is 3 × 2

14. Not defined since A is 2 × 3 and D is 3× 2

15. This is not defined since the matrices have different orders.

16. This is not defined since the matrices have different orders.

17.  −2 6 0  4 8 2   

18.  8 −12  0 4    16 −8 

19.  0 −5   −10 5     −15 −5 

20.  0 −4 −2   −6 4 −8   

21.

22.

 −1 3 0   0 2 1   −2 6 0  0 6 3   −2 12 3 2 +3  = + =  2 4 1  3 −2 4   4 8 2  9 −6 12  13 2 14  0  3 −5 2   0 2 1   −1 3 0  0 4 2  3 − 9 2 −3 = +    =  3 −2 4  2 4 1  6 −4 8  −6 −12 −3 0 −16 5 

917

Chapter 7

23. Since A2 x 2 B2 x3 = ( AB )2 x3 , we can

24. Since B2 x 3C3 x 2 = ( BC )2 x 2 , we can

25. D3 x 3 E1x 2 is not defined, because D has 3 columns and E has 1 row.

26. E1x 2 E1x 2 is not defined, because E is not a square matrix.

27. Since D3 x 3C3 x 2 = ( DC )3 x 2 , we can say DC is defined and has order 3 × 2.

28. Since E1x 2 A2 x 2 = ( EA)1x 2 , we can say EA is defined and has order 1× 2.

29. Since E1x 2 B2 x3 = ( EB )1x3 , we can say EB is defined and has order 1× 3.

30. Since C3 x 2 B2 x 3 = (CB )3 x 3 , we can say CB is defined and has order 3 × 3.

31. Since DD is a square matrix, D3 x 3D3 x 3 = ( DD )3 x 3 . Thus, we can say DD is defined and has order 3 × 3. .

32. C3 x 2 D3 x 3 is not defined, because C has 2 columns and D has 3 rows.

33.  8 3 11 5   

34. [5]

35.  −3 21 6   −4 7 1     13 14 9 

36.  2 −2   20 21    13 14 

37.  0 2 0  3 0   2 −2   2 4 5  1 −1 = 20 21        2 1 3  2 5  13 14 

38.

say AB is defined and has order 2 × 3.

39. 3 6  −2 −2    17 24 

say BC is defined and has order 2 × 2.

0 2 0  First add A + E =  2 4 5  . Since you  2 1 3  cannot multiply a 3 × 2 matrix by a 3 × 3 matrix, this operation is not possible.

40.  −1 4 1   6 9 14     1 3 11

918

Section 7.2

41. Not possible you cannot multiply a 2 × 2 matrix by a 3 × 2 matrix.

42. Not possible since ED is 3 × 2 while C is 2 × 3. Matrices must have the same orders to be added.

43.

44. First multiply BA = [ −13 4 4 ] .

 2 0 −3 Then subtract B from the product: First multiply FB =  0 0 0  . Then [ −13 4 4] − [2 0 −3]  −2 0 3  = [ −15 4 7 ] . add the product to E:  2 0 −3  −1 0 1   1 0 −2  45. −4 [ 0 −15] = [ 0 60 ]  0 0 0  +  2 1 4 =  2 1 4  .        −2 0 3   −3 1 5   −5 1 8 

46.

47.

 −1 5   3 −15  −3 15 19  =  −45 −57   2 24   −6 −72 

0 2 0  [2 0 −3] 2 4 5  = [ −6 1 −9]  2 1 3 

48.  5 7 4   −5 35 10  9 21 10  +  15 0 5       0 42 14  =   24 21 15 

49.  2 0 −3  5 10 −5   0 0 0  +  0 15 5       −2 0 3  25 0 −10   7 10 −8  =  0 15 5   23 0 −7 

50.

51.  7 10   5 10  12 20  15 22  + 15 20  = 30 42       

52.

53.  −2 0 2   1   −4   4 2 8   0  =  −4        −6 2 10   −1  −16 

 −4 8 3  AA =  5 9 1   −5 10 −1  −2 0 2   −1 7 2     18 18 74   3 0 1   4 2 8  =  −12 2 16    −    6 2 10 

919

Chapter 7

54. 9 19 4   −5 35 10   4 54 14  8 6 −5  +  15 0 5  = 23 6 0       

57.

Yes response  0.70  A= =   No response   0.30 

55. Not possible since the inner dimensions are different. 56. 4 6 1 3 4 17    1 −2 −5 

Yes to "increase lung cancer"  0.89  B= =   Yes to "would shorten lives"  0.84 

 0.70  32.2  a. 46 A = 46  =  . This matrix tells us that out of 46 million people,  0.30  13.8  32.2 million said that they had tried to quit smoking, while 13.8 million said that they had not.  0.89   40.94  b. 46 B = 46  =  . This matrix tells us that out of 46 million people, 0.84  38.64  40.94 million believed that smoking would increase the chance of getting lung cancer, and that 38.64 million believed that smoking would shorten their lives.

58.

0.32   0.38  0.43  0.05  A=  (1991), B =  (2001), C =  ( 2011): C − B =       0.21 0.30  0.18  −0.12  This matrix tells us the percent increase from 2001 to 2011 of female graduate students in mathematics and computer science. In particular, there was a 5% increase of female graduate students in mathematics, and a 12% decrease of female graduate students in computer science.

 0.06  B –A =    0.09  This matrix tells us the percent increase from 1991 to 2001 of female graduate students in mathematics and computer science. In particular, there was a 6% increase of female graduate students in mathematics, 9% increase of female graduate students in computer science.

920

Section 7.2

59.

 0.595 0.630  120 M  153.30 M  , it follows that AB =  Since A =  and B =    . 130 M  122.42 M  0.472 0.506  This matrix tells us the number of registered voters and the number of actual voters. In particular, there are 153.30 million registered voters, and of those, 122.42 million actually voted.

60.  0.5 0.6  8 7 5   7.1 7  ⋅  0.3 0.1 =  . (The order is 2 × 2.) Here, BA =   7 6.8 6 8 8    0.2 0.3  Each entry of this matrix gives the respective applicant’s total score according to each of the two rubrics. Specifically, Applicant 1’s score according to Rubric 1 is 7.1, Applicant 1’s score according to Rubric 2 is 7, Applicant 2’s score according to Rubric 1 is 7, Applicant 2’s score according to Rubric 2 is 6.8.

61. A = [ 0.575 0.50 1.00 ] AB = [ 20,875.725]

63.

 7,523  B =  2,700  15,200 

 230 3 44 9   430 19 46 20   A=   290 5 45 19    330 5 47 24   460 6 88 18  860 38 92 40   2A =  580 10 90 38    660 10 94 48 

62. A = [85 75 100 ] AB = [51.25]

 0.25  B =  0.20   0.15 

This matrix represents the nutritional information corresponding to 2 sandwiches. 115 1.5 22 4.5  215 9.5 23 10   0.5A =  145 2.5 22.5 9.5    165 2.5 23.5 12  This matrix represents the nutritional information corresponding to one-half of a sandwich.

921

Chapter 7

64. BA = [6, 410 161 945 339] This corresponds to the nutritional information for food consumed during one week. 1 23 135 48.4 ] 7 BA = [ 915.7

65. 0.06 0  A = 0.02 0.1  0 0.3  0.228  AB =  0.081 0.015 

This corresponds to one day’s nutritional information (on average).

3.80  B=  0.05 

The entries in AB represent the total cost to run each type of automobile per mile.

66.

67.

2,736  12,000 AB =  972   180  Cost to run SUV: $2,736 Cost to run Hybrid: $972 Cost to run Electric Car: $180

2  10    N = 1  XN = 16   0   20  The nutritional content of the meal is 10 g of carbohydrates, 16 g of protein and 20 g of fat.

68.

69.

1  9   N = 0  XN = 15  2  16  The nutritional content of the meal is 9 g of carbohydrates, 15 g of protein and 16 g of fat.

 200   9.25    N =  25  XN = 13.25   0  15.75  Company 1 would charge $9.25, Company 2 would charge $13.25 and Company 3 would charge $15.75 respectively for 200 minutes of talking and 25 text messages. The better cell phone provider for this employee would be Company 1.

70. 125  N = 125  320 

59.25  XN = 71.35  59.10  Company 1 would charge $59.25, Company 2 would charge $71.35 and Company 3 would charge $59.10 for 125 minutes of talking, 125 text messages and 320 megabytes of data usage. The better cell phone provider for this employee would be Company 3.

922

Section 7.2

71. Not multiplying correctly. It should be: 3 2   −1 3 3( −1) + 2( −2) 3(3) + 2(5)   −7 19  1 4  ⋅  −2 5  = 1( −1) + 4( −2) 1(3) + 4(5)  =  −9 23         72. Not multiplying correctly. It should be: 3 2   −1 3 3( −1) + 2( −2) 3(3) + 2(5)   −7 19  1 4  ⋅  −2 5  = 1( −1) + 4( −2) 1(3) + 4(5)  =  −9 23         73. False. In general, a  b b  a b + a b a11b12 + a12 b22  a AB =  11 12  ⋅  11 12  =  11 11 12 21  a21 a22  b21 b22  a21b11 + a22 b21 a21b12 + a22 b22  74. False.  2 0 4  Let A =  1 4  and B =   . 1   −1 3  AB is defined, but BA is not. So, AB ≠ BA .

75. True 76. True 77.

 a2 + a a a11a12 + a12 a22  A2 = AA =  11 12 21  2 a21a11 + a22 a21 a22 + a21a12 

78. m = n since then the number of columns of A is equal to the number of rows of the matrix it is to be multiplied by, namely A. 79. Observe that 1 1 A= , 1 1 2 2  A2 = AA =  , 2 2   4 4  2 2 3 2 A =A A= = 2  4 4  2

22  2 1 1 2 = 2  1 1 = 2 A 2 2   

 2 n −1 2 n −1  1 1 n −1 A =  n −1 = 2 n −1   = 2 A, n ≥ 1 n −1  1 1 2 2     n

923

Chapter 7

80. Observe that 1 0  A=   0 1  1 0  A2 = AA =  =A 0 1   1 0  A =  = A, n ≥ 1 0 1  n

81. Am×n Bn× p is a matrix of order m × p . Since a product CD is defined if and only if the number of columns of C equals the number of rows of D, if C = D, then C has the same number of columns and rows. Applying this to the given product, we see that in order for

( A B ) to be defined, we need m = 2

m× n

n× p

82. AC is an m × m matrix, and CB is an n × n matrix. Since m ≠ n , AC ≠ CB (since they have different orders). 83.

 33 35   −96 −82   AB =   31 19    146 138 

86.

8 124 126   −2 114 148 −131 14   AA =   36 28 29 48     −6 11 189 87 

84. BA is not defined.

85. BB is not defined.

87.  5 −4 4   2 −15 −3    26 4 −8

924

88. 121 233   74  −503 −560 312    1072 −1006 −462 

Section 7.3 Solutions--------------------------------------------------------------------------------1.  −2 5   x  10   7 −2   y  =  −4      

2.  4 −8  x  10   3 5   y  = 15      

3.  1 −2   x   8   −3 1   y  = 6      

4. 7 −2   x   28  3 7   y  =  42      

5.  3 5 −1  x   2   1 0 2   y  = 17        −1 1 −1  z   4 

6.  1 −1 1   x  12   2 1 −3  y  =  6        −3 2 1   z  18 

7.  3 0 1   x  10   0 1 −2   y  =  4        1 2 0   z   6 

8.  1 1 −2 1   x  11  2 −1 3 0   y  17    =   −1 2 −3 4   z  12        0 1 4 6  w  19 

 8 −11 7 11 1 0  Since AB =  ⋅ =  = I , B must be the inverse of A.  −5 7   5 8   0 1 

10.

 7 −9   4 9   1 0  Since AB =  ⋅ =  = I , B must be the inverse of A.  −3 4   3 7   0 1 

11. 3 1   72 Since AB =  ⋅1 1 −2   7

 1 0  = = I , B must be the inverse of A. −  0 1  1 7

3 7

12. 2 3   51 Since AB =   ⋅ 1 1 −1  5

 1 0  = = I , B is the inverse of A. −  0 1  3 5

2 5

925

Chapter 7

13.

1 2   4 −2   −2 0  Since AB =  ⋅ =  ≠ I , B cannot be the inverse of A. 3 4   −3 1   0 −2 

14. 1 2  1 Since AB =   ⋅ 1 3 4   3

  53 1  =  13 5  ≠ I , B cannot be the inverse of A. 1 4  3 2 1 2

15. 1 −1 1  1 0 1  1 0 0  Since AB = 1 0 −1 ⋅ 1 −1 2  = 0 1 0  = I , B must be the inverse of A. 0 1 −1 1 −1 1  0 0 1 

16.  −1 0 −1  −1 1 −1 1 0 0  Since AB =  −1 1 −2  ⋅  −1 0 1  = 0 1 0  = I , B must be the inverse of A.  −1 1 −1  0 −1 1  0 0 1 

17. 2 0 1  0 2 1   2 4 4  Since AB = 0 3 1  ⋅ 0 3 0  =  2 9 2  ≠ I , B cannot be the inverse of A. 0 2 −1 2 0 2   −2 6 −2 

18. 1 0 0  1 0 0  1 0 0  Since AB = 0 2 0  ⋅ 0 12 0  =  0 1 0  = I , B must be the inverse of A. 0 0 3  0 0 13   0 0 1 

19.

0 −1 0 −1 A−1 = 11  = . 1 2  1 2 

20.

 1 −1  1 −1 A−1 = 11  = .  −2 3   −2 3 

21.  3 −2   − 394 ( 34 ) − 394 ( −2)   − 131 = 4 =  20 A−1 = −139  4 1  4 1  4  −5 3   − 39 ( −5) − 39 ( 3 )   39

926

8 39

−

4 117

 . 

Section 7.3

22.  2 A−1 = −11  3 1 2 − 3

−2   −2( 23 ) −2( −2)   − 34 = = 2 1  1 1  4   −2( − 3 ) −2( 4 )   3

4 . − 12 

23.

 1.7 −2.4   −0.1618 0.2284  1 A−1 = −10.51  −5.3 1.3  ≅  0.5043 −0.1237  .    

24.

 −3.2 −1.1  −1.391 −0.4783 1 . A−1 = 2.3  −4.6 −2.3 ≅  −2 −1    

25. Using Gaussian elimination, we obtain the following:  1 1 1 1 0 0 1 1 1 1 0 0    R2 + R3 →R3    1 −1 −1 0 1 0  ⎯⎯⎯⎯→ 1 −1 −1 0 1 0   −1 1 −1 0 0 1  0 0 −2 0 1 1   2 0 0 1 1 0  12 R1 →R1 1 0 0 12 12 1   − 2 R3 →R3  R1 + R2 →R1 ⎯⎯⎯⎯ →  1 −1 −1 0 1 0  ⎯⎯⎯⎯ → 1 −1 −1 0 1  0 0 −2 0 1 1  0 0 1 0 − 12 1 0 0  R1 − R2 → R2 ⎯⎯⎯⎯→  0 1 1  0 0 1 0  12 12  −1 1 1  So, A =  2 0 2 . 0 − 12 − 12 

1 2

− 0 − 12 1 2

1 2

0 1 0 0  R2 −R3 →R2  0  ⎯⎯⎯⎯→ 0 1 0 0 0 1 − 12 

927

1 2

0 − 12

0  0 − 12  0 1  2  − 12 

Chapter 7

26. Using Gaussian elimination, we obtain the following:  1 −1 1 1 0 0   1 −1 1 1 0 0    R1 −R2 →R2    1 1 1 0 1 0  ⎯⎯⎯⎯→  0 −2 0 1 −1 0   −1 2 −3 0 0 1   −1 2 −3 0 0 1  1 −1 1 1 0 0  1 −1 1 1 0 0    2 R3 + R2 →R3   R1 + R3 → R3 ⎯⎯⎯⎯→ 0 −2 0 1 −1 0  ⎯⎯⎯⎯⎯ → 0 −2 0 1 −1 0  0 1 −2 1 0 1  0 0 −4 3 −1 2  1 −1 1 1 0 0   1 0 1 12 12 0  − 12 R2 → R2   R1 + R2 →R1   − 14 R3 →R3 →  0 1 0 − 12 12 0  ⎯⎯⎯⎯ → 0 1 0 − 12 12 0  ⎯⎯⎯⎯ 3 1 1  0 0 1 − 34 14 − 12  0 0 1 − 4 4 − 2  1 0 0  R1 − R3 →R1 ⎯⎯⎯⎯→  0 1 0  0 0 1

5 4

− −

1 4 1 2 3 4

1 2 1 4

  0 − 12  1 2

 54 So, A−1 =  − 12  − 34

1 4 1 2 1 4

 0  . − 12  1 2

27. Using Gaussian elimination, we obtain the following: 1 0 1 1 0 0  1 0 1 1 0 0    R2 + R3 →R3    0 1 1 0 1 0  ⎯⎯⎯⎯→ 0 1 1 0 1 0   1 −1 0 0 0 1  1 0 1 0 1 1  1 0 1 1 0 0    R1 − R3 →R3 ⎯⎯⎯⎯→ 0 1 1 0 1 1  0 0 0 1 −1 −1 From the bottom row, we deduce that A is not invertible thus A−1 does not exist.

28. Using Gaussian elimination, we obtain the following: 1 2 −3 1 0 0  R −R →R 1 2 −3 1 0 0    R11 −R32 →R32   1 − 1 − 1 0 1 0   ⎯⎯⎯⎯→ 0 3 −2 1 −1 0  1 0 −4 0 0 1  0 2 1 1 0 −1 1 2 −3 1 2 −3 1 0 0  1 R →R    R3 − 2 R2 → R3 2 3 2 ⎯⎯⎯⎯ → 0 1 − 23 13 − 13 0  ⎯⎯⎯⎯⎯ → 0 1 − 23 0 0 73 0 2 1 1 0 −1  928

1 3

− 13

1 3

2 3

0  0 −1

Section 7.3

 1 2 −3  ⎯⎯⎯⎯ →  0 1 − 23  0 0 1 1 0  R1 − 2 R2 + 3 R3 → R1 ⎯⎯⎯⎯⎯⎯ → 0 1 0 0 3 R →R 3 7 3

1 3

−

1 7

2 7

4 7 3 7

0 1

1 7

1 3

0  1 2 −3 1 0  R2 + 32 R3 →R2  0  ⎯⎯⎯⎯⎯ →  0 1 0 73 − 71  0 0 1 71 72 − 73  8 − 75   74 78 − 75  7  So, A−1 =  73 − 71 − 72  − 71 − 72  . 2  71 72 − 73  − 73  7

0  − 72  − 73 

29. Using Gaussian elimination, we obtain the following:  2 4 1 1 0 0  1 R →R 1 2 12 12 0 0  R −R →R  R12−R33→R32   R2 2 1↔ R31  1 1 1 0 1 0 1 1 0 0 0 1 − ⎯⎯⎯⎯ →   ⎯⎯⎯⎯→   1 1 −1 0 1 0  1 1 0 0 0 1    1 2 12 12 0 0  1 2 12 12 0 0    R2 ↔ R3   R2 − 32 R3 →R2 3 1 → 0 0 1 0 −1 1  ⎯⎯⎯→ 0 1 2 2 −1 0  ⎯⎯⎯⎯⎯ 0 1 32 21 −1 0  0 0 1 0 −1 1      1 1 1 2 2 2 0 0 1 0 0 − 12 − 12 52     R1 − 2 R2 − 12 R3 → R1 3 1 1 1 → 0 1 0 12 − 23  2  0 1 0 2 2 − 2  ⎯⎯⎯⎯⎯⎯  0 0 1 0 −1 1  0 0 1 0 −1 1     − 12 − 12 52  1 − 23  . So, A−1 =  12 2  0 −1 1  30. Using Gaussian elimination, we obtain the following:  1 0 1 1 0 0  R −2 R →R 1 0 1 1 0 0 1 3   R32 −R1 →   R2 → 0 1 −2 −1 1 0  1 1 −1 0 1 0  ⎯⎯⎯⎯⎯ 2 1 −1 0 0 1  0 1 −3 −2 0 1  1 0 0  R −2 R →R 1 0 1 1 0 0  1 0 1   − R2 3 →R3 3 2   R3 − R2 → R3 ⎯⎯⎯⎯→ 0 1 −2 −1 1 1  ⎯⎯⎯⎯⎯ → 0 1 0 1 3 −2  0 0 −1 −1 −1 1  0 0 1 1 1 −1 1 0 0 0 −1 1    R1 − R3 → R1 ⎯⎯⎯⎯→ 0 1 0 1 3 −2  0 0 1 1 1 −1

929

Chapter 7

 0 −1 1  So, A−1 =  1 3 −2  .  1 1 −1

31. Using Gaussian elimination, we obtain the following:  1 1 −1 1 0 0  R −2 R →R 1 1 −1 1 0 0  1 3   R32 −R1 →   R2 1 − 1 1 0 1 0 ⎯⎯⎯⎯⎯ → 0 −2 2 −1 1 0    0 −3 1 −2 0 1  2 −1 −1 0 0 1  1 1 −1 1 0  1 1 −1 1 0 0 0 − 12 R2 →R2     − 13 R3 → R3 R R R − → 1 1 3 2 3 ⎯⎯⎯⎯ → 0 1 −1 12 − 12 0  ⎯⎯⎯⎯→ 0 1 −1 2 − 2 0  0 1 − 13 23 0 − 13  0 0 23 61 21 − 31       1 1 −1 1 0 0 1 1 −1 1 0 0  3 R →R   R2 + R3 →R2   3 1 1 2 3 ⎯⎯⎯⎯ → 0 1 −1 2 − 2 0  ⎯⎯⎯⎯→ 0 1 0 34 14 − 12  0 0 1 14 34 − 21  0 0 1 14 34 − 21  1 0 0  R1 − R2 + R3 → R1 ⎯⎯⎯⎯⎯→ 0 1 0 0 0 1  12 12 0  So, A−1 =  34 14 − 12  .  14 34 − 12 

1 2 3 4 1 4

1 2 1 4 3 4

0  − 12  − 21 

32. Using Gaussian elimination, we obtain the following: 1 −1 −1 1 0 0  3 −5 1 0 0 1    R3 ↔ R1   1 1 −3 0 1 0  ⎯⎯⎯→ 1 1 −3 0 1 0  3 −5 1 0 0 1  1 −1 −1 1 0 0  0 0 1 0 0 1 3 −5 1  3 −5 1 R1 −3 R2 → R2     R1 −3 R3 → R3 R2 − 4 R3 → R3 ⎯⎯⎯⎯⎯ → 0 −8 10 0 −3 1 ⎯⎯⎯⎯⎯ →  0 −8 10 0 −3 1   0 0 −6 12 −3 −3 0 −2 4 −3 0 1

930

Section 7.3

3 −5 1 0 0 1  3 −5 1 0 0 1     R2 + 54 R3 → R2 5 3 1 ⎯⎯⎯⎯ → 0 1 − 4 0 8 − 8  ⎯⎯⎯⎯⎯ → 0 1 0 − 52 1 12  0 0 0 0 1 −2 12 12  1 −2 12 12  − 16 R3 → R3 − 81 R2 →R2

3 0 0 − 212 92 3  1 0 0 − 72 32 1  1   3 R1 →R1   R1 +5 R2 − R3 → R1 ⎯⎯⎯⎯⎯ ⎯ → 0 1 0 − 52 1 21  ⎯⎯⎯⎯ → 0 1 0 − 52 1 21  0 0 1 −2 12 12  0 0 1 −2 12 12   − 72 32 1  So, A−1 =  − 52 1 12  .  −2 12 12 

33. First, write the system in the form AX = B: 3 −2   x   5  1 2   y  =  −1      −1

−1

3 −2   x  3 −2   5  1  2 2 . Since  , the solution is =  The solution is   =      8  −1 3  1 2   y  1 2   −1 x  1  2 2  5   1   y  = 8  −1 3   −1 =  −1 . So, x = 1, y = −1.       

34. First, write the system in the form AX = B: 5 −1  x   −4   2 −5   y  =  3       −1

−1

5 −1  x  5 −1  −4  1  −5 1  . Since  , the solution is =−  The solution is   =      23  −2 5   2 −5   y   2 −5   3  x  1  −5 1  −4   −1  y  = − 23  −2 5   3  =  −1 . So, x = −1, y = −1.       

35. First, write the system in the form AX = B: 1 −7   x  2  3 4   y  = 6       −1

−1

1 −7   x  1 −7  2  1  4 7 . , the solution is Since = The solution is   =  3 4     25  −3 1     y  3 4  6   x  1  4 7  2  2   y  = 25  −3 1  6  = 0  . So, x = 2, y = 0.        931

Chapter 7

36. First, write the system in the form AX = B: 1 3   x   3  5 2   y  =  −24       −1

−1

 x  1 3   3  1 3  1  2 −3 . Since  , the solution is =−  The solution is   =      13  −5 1  5 2   y  5 2   −24  x  1  2 −3  3   −6   y  = − 13  −5 1   −24  =  3  . So, x = −6, y = 3.       

37. First, write the system in the form AX = B:  2 −1  x  5   1 1   y  = 1       −1

−1

 2 −1  x  2 −1 5  1  1 1 . Since  , the solution is =  The solution is   =      3  −1 2  1 1   y  1 1  1   x  1  1 1  5   2   y  = 3  −1 2  1 =  −1 . So, x = 2 , y = −1 .       

38. First, write the system in the form AX = B:  2 −3  x  12  1 1   y  =  1       −1

−1

 2 −3   x  2 −3 12  1  1 3 . , the solution is = Since The solution is   =  1 1     5  −1 2     y  1 1   1   x  1  1 3  12   3   y  = 5  −1 2   1  =  −2  . So, x = 3, y = −2       

39. First, write the system in the form AX = B:  4 −9   x   −1  7 −3   y  =  5     2 −1

−1

 x   4 −9   −1  4 −9  1  −3 9  , the solution is = . Since The solution is   =       5 51  −7 4   7 −3  y  7 −3   2   x  1  −3 9   −1  12  1 1  y  = 51  −7 4   5  =  1  . So, x = 2 , y = 3 .      2  3

932

Section 7.3

40. First, write the system in the form AX = B:  7 −3   x   1   4 −5   y  =  − 7       5 −1

−1

 7 −3   x   7 −3   1  1  −5 3  , The solution is   =  = − Since .      7 23  −4 7   4 −5   y   4 −5   − 5  x  1  −5 3   1   52  3 2 the solution is   = −    − 7  =  3  . So, x = 5 , y = 5 . y 4 7 − 23      5 5

41. First, write the system in the form AX = B:  34 − 32   x  5  =   1 5    − 2 − 3   y  3  −1

−1

 34 − 32   x   3 − 23  5  12  − 35 The solution is   =  4 1 = − . Since  1    5 5 19  12  y   − 2 − 3  3  − 2 − 3  x  12  − 5 2  5   4  the solution is   = −  1 3 33    =   . So, x = 4, y = −3 . 19  2 4  3   −3  y

2 3 3 4

 , 

42. First, write the system in the form AX = B: 3  25 1  7  x  − 1 − 1    =  1  6 3   y  2 3  x   52 7  The solution is   =  1 1  y  − 2 − 3   x  15(14)  − 13 the solution is   =  17  12  y

−1

3  25 1  15(14)  − 31 7    1  . Since  − 1 − 1  = 17  12 6 3  2 − 73  1   −5  =   . So, x = −5, y = 7 . 2  1 7 5  6

− 73  , 2  5 

43. 0  12 12  −1 1 1  From #25, we know that A =  2 0 Further, in the current problem, we 2 . 0 − 12 − 12  1 x  x  0        −1 identify B =  −1 . Since A  y  = B is equivalent to  y  = A B = 0  , we  −1  z   z  1 

conclude that the solution of this system is x = 0, y = 0, z = 1.

933

Chapter 7

44.  54 14 12  From #26, we know that A−1 =  − 12 12 0  . Further, in the current problem,  − 34 14 − 12  0  x  x  1        −1 we identify B = 2  . Since A  y  = B is equivalent to  y  = A B = 1  , 1   z   z  0 

we conclude that the solution of this system is x = 1, y = 1, z = 0 .

45. From #27, we know that A−1 does not exist. 46.  74 78 − 57  From #28, we know that A−1 =  73 − 71 − 72  . Further, in the current problem,  71 72 − 73  1  x  x  4       −1 we identify B = 3  . Since A  y  = B is equivalent to  y  = A B =  0  , 0   z   z   1 

we conclude that the solution of this system is x = 4, y = 0, z = 1.

47.  − 12 − 12 52  1 From #29, we know that A−1 =  12 − 23  . Further, in the current problem, 2  0 −1 1   −5  x  x   −1       −1 we identify B =  7  . Since A y = B is equivalent to y = A B =  1  ,        0   z   z   −7 

we conclude that the solution of this system is x = −1, y = 1, z = −7 .

934

Section 7.3

48.  0 −1 1  From #30, we know that A =  1 3 −2  . Further, in the current problem, we  1 1 −1 3 x  x   −2        −1 identify B =  −3  . Since A  y  = B is equivalent to  y  = A B =  4  ,  −5   z   z   5  −1

we conclude that the solution of this system is x = −2, y = 4, z = 5 .

49.  12 12 0  From #31, we know that A−1 =  34 14 − 12  . Further, in the current problem, we  14 34 − 12  4 x  x  3        −1 identify B =  2  . Since A  y  = B is equivalent to  y  = A B =  5  ,  −3  z   z   4 

we conclude that the solution of this system is x = 3, y = 5, z = 4 .

50.  − 72 32 1  From #32, we know that A−1 =  − 52 1 12  . Further, in the current problem,  −2 12 12  0  x  x  7        −1 we identify B = 2 . Since A y = B is equivalent to y = A B =  4  ,          4   z   z   3 

we conclude that the solution of this system is x = 7, y = 4, z = 3 .

51.

1  −450,000,000,000  50  . = 9 ×109  −180,000,000,000  20  b. The price of a sweatshirt is $50 and the price of a t-shirt is $20.

a. A−1B = −

935

Chapter 7

52.

1,000,000,000  20  1 =  50,000,000 3,750,000,000  75  b. So, 20 hats were sold and 75 jackets were sold.

a. A−1B =

53. For Problems 53 – 58 we need to compute K −1 :  1 1 0 1 0 0  R + R →R 1 1 0 1 0 0    R13 −2 R2 1 →R2 3   R3 + 2 R2 →R3 → 0 1 1 1 1 0  ⎯⎯⎯⎯⎯ →  −1 0 1 0 1 0  ⎯⎯⎯⎯⎯  2 0 −1 0 0 1  0 −2 −1 −2 0 1  1 1 0 1 0 0  1 1 0 1 0 0    R2 −R3 →R2   R1 −R2 →R1 0 1 1 1 1 0  ⎯⎯⎯⎯→ 0 1 0 1 −1 −1 ⎯⎯⎯⎯→ 0 0 1 0 2 1  0 0 1 0 2 1  1 0 0 0 1 1  0 1 1      −1  0 1 0 1 −1 −1 . So, K = 1 −1 −1 .  0 0 1 0 2 1  0 2 1  0 1 1  As such, [55 10 −22 ] 1 −1 −1 = [10 1 23] , which corresponds to JAW. 0 2 1 

54. 0 1 1  Since K = 1 −1 −1 (see Problem #53), we see that 0 2 1  0 1 1  [31 8 −7] 1 −1 −1 = [8 9 16] , which corresponds to HIP. 0 2 1  −1

936

Section 7.3

55. 0 1 1  Since K = 1 −1 −1 (see Problem #53), we see that 0 2 1  0 1 1  [21 12 −2] 1 −1 −1 = [12 5 7] , which corresponds to LEG. 0 2 1  −1

56. 0 1 1  Since K = 1 −1 −1 (see Problem #53), we see that 0 2 1  0 1 1  [9 1 5] 1 −1 −1 = [1 18 13] , which corresponds to ARM. 0 2 1  −1

57. 0 1 1  Since K = 1 −1 −1 (see Problem #53), we see that 0 2 1  0 1 1  [ −10 5 20] 1 −1 −1 = [5 25 5] , which corresponds to EYE. 0 2 1  −1

58. 0 1 1  Since K = 1 −1 −1 (see Problem #53), we see that 0 2 1  0 1 1  [ 40 5 −17] 1 −1 −1 = [5 1 18] , which corresponds to EAR. 0 2 1  −1

937

Chapter 7

59. Using technology to compute, we see that −1  8 4 6  18  1 X =  6 10 5   21 = 1 10 4 8  22  1 So, the combination of 1 serving each of food A, B and C will create a meal of 18 g carbohydrates, 21 g of protein and 22 g of fat. 60. Using technology to compute, we see that −1  8 4 6  14   0  X =  6 10 5  25  =  2  10 4 8  16   1  The combination of 2 servings of food B and 1 serving of food C will create a meal of 14 g carbohydrates, 25 g of protein and 16 g of fat.

61. Using technology to compute, we see that −1  0.03 0.06 0.15   49.50  350  X =  0.04 0.05 0.18 52.00  =  400   0.05 0.07 0.13 58.50  100  The employee’s normal monthly usage is 350 minutes talking, 400 text messages and 100 megabytes of data usage.

62. Using technology to compute, we see that −1  0.03 0.06 0.15  82.50  350  X =  0.04 0.05 0.18  85.00  = 700   0.05 0.07 0.13  92.50   200  The employee’s normal monthly usage is 350 minutes talking, 700 text messages and 200 megabytes of data usage. 63. The third row of A is comprised of all zeros. Hence, det( A) = 0, which implies A is not invertible. Also, note that the identity matrix doesn’t occur on the left as it should once you have computed the inverse.

938

Section 7.3

64. This is not how the inverse is defined. Rather, 10 −5  1 10 −5   2 A−1 = 2(10 )1−5(3)   = 5  −3 2  =  − 3  −3 2     5

−1 . 2  5 

65. False. In general, −a12  a A−1 = a22 a11 −1 a21a12  22  , provided that a22 a11 − a21a12 ≠ 0  −a21 a11  0 0  66. False.   is a square matrix with determinant = 0, so that it has no 0 0  inverse.

67. By definition of inverse, the value of x for which A−1 does not exist would be the one for which 2x − 18 is 0, namely x = 9 , since we need to divide by the quantity 2(6) – 3(4) in order to form the inverse. 68. In this case, realize that if any of a, b, or c were 0, then the matrix A would have a row consisting of all zeros. In such case, the determinant of A would be 0, and hence, A would not be invertible. Assuming that none of them is zero, we obtain: 1 R →R  a 0 0 1 0 0  1ab R12 →R12 1 0 0 a1 0 0   a1 0 0  1   c R3 →R3   → 0 1 0 0 b1 0  . So, A−1 =  0 b1 0  .  0 b 0 0 1 0  ⎯⎯⎯⎯  0 0 c 0 0 1  0 0 1 0 0 1c   0 0 1c  69. Observe that a b   1  d −b   1   a b   d −b   ⋅ A ⋅ A−1 =  =     ⋅   c d   ad − bc  −c a   ad − bc   c d   −c a    ad − bc  0   ad bc 0 −   1 0  1 ad − bc = = =I =   ad − bc   ad − bc  0 1  ad − bc  0 0  ad − bc 

939

Chapter 7

70. Assume that a ≠ 0 . Applying Gaussian elimination to obtain the inverse reveals: b 1 0  a b 1 0  R2 − ac R1 →R2  a →    ⎯⎯⎯⎯⎯ bc c c d 0 1 0 d − a − a 1  b 1 1 0 a a ⎯⎯⎯⎯ →  bc c 0 d − a − a 1  1 0  1 ab a R →R a 2 da −bc 2 ⎯⎯⎯⎯ ⎯ → c a a  0 1 − a ( da −bc ) da −bc  1 0 1 + b c ( a ) − b a  a a a da − bc a da − bc R1 − ab R2 → R1  ⎯⎯⎯⎯⎯ → a 0 1  a c − a ( da −bc ) da − bc   Now, simplifying the result above yields:  1 0 1 + b c ( a ) − b a   1 0 da − bc + bc − b  a a a da − bc a da − bc a ( da − bc ) da − bc  =  c a a 0 1  a c 0 1 − da −bc  ) − ( − da bc   da bc da bc − − a    1 0 dad−bc − da b−bc  =  c a  0 1 − da −bc da −bc  1 R →R 1 a 1

(

 d Hence, A−1 =  da −cbc  − da −bc

)

(

)

(

)

− da b−bc   d −b  1 =  − da bc  −c a  . a   da − bc 

71. Since ad − bc = 0 the inverse does not exist. 72. In the attempt to calculate the inverse, applying the row operation  0 0 0 −2 1 0    R2 − 2R1 → R1 yields  2 4 −2 0 1 0  . As such, the 3 × 3 matrix to the left  0 1 3 0 0 1  of the vertical bar can never be transformed into the identity matrix, and so, the inverse doesn’t exist.

73. −  411 −1 A =  6008 57  751  −429  6008

115 6008

431 6008 −391 6008 28 751 145 6008

−1067 6008 731 6008 −85 751 1035 6008

    12  751 

103 751 −22 751 3 751

74.

1 0 −1 AA =  0  0

940

0 0 0 1 0 0  =I 0 1 0  0 0 1

Section 7.3

75.

 2.7 −3.1  x   9.82  The system in matrix form is given by    =  1.5 2.7   y   −1.62   x  2.7 −3.1 and its solution is   =    y  1.5 2.7 

−1

 9.82   −1.62  . Since  

−1

 2.7 −3.1 1  2.7 3.1 , we see that the solution is = 1.5 2.7  11.94  −1.5 2.7    x  1  2.7 3.1  9.82   1.8  =  y  11.94  −1.5 2.7   −1.62  =  −1.6  .        So, x = 1.8 , y = −1.6 .

76.

 3.7 −2.5   x   31.77  The system in matrix form is given by    =   −5.1 1.3   y   −39.07   x   3.7 −2.5  and its solution is   =    y   −5.1 1.3 

−1

 31.77   −39.07  .  

−1

 3.7 −2.5  1 1.3 2.5  , we see that the solution is Since  =−  7.94 5.1 3.7   −5.1 1.3  x  1 1.3 2.5   31.77   7.1   y  = − 7.94 5.1 3.7   −39.07  =  −2.2  .        So, x = 7.1, y = −2.2 .

77. The augmented matrix to enter into the calculator is:  5.1 7.3 1.2 12.51     2.3 −1.5 4.5 53.96   −8.1 5.4 −9.4 −130.35 

78. The augmented matrix to enter into the calculator is 12.4 −5.8 2.7 −60.92     −3.9 1.9 −0.6 18.73   6.4 −4.3 8.5 −62.79 

Solving then yields the solution of the system as ( 3.7, −2.4,9.3 ) .

Solving then yields the solution of the system as ( −1.6,5.5, −3.4 ) .

941

Section 7.4 Solutions -------------------------------------------------------------------------------1. (1)( 4 ) − ( 3 )( 2 ) = −2

2. (1)( −4 ) − ( −3 )( −2 ) = −10

3. ( 7 )( −2 ) − ( −5 )( 9 ) = 31

4. ( −3 )(15 ) − ( 7 )( −11) = 32

5. ( 0 )( −1) − ( 4 ) 7 = −28

6. ( 0 )( 0 ) − (1)( 0 ) = 0

7. ( −1.2 )(1.5 ) − ( −0.5 )( 2.4 ) = −0.6

8. ( −1.0 )( −2.8 ) − (1.5 )(1.4 ) = 0.7

9. ( 34 )( 98 ) − ( 2 ) ( 31 ) = 0

10. ( − 12 ) ( − 98 ) − ( 23 ) ( 14 ) = 185

11.

So, from Cramer’s Rule, we have: D −10 =5 x= x = −2 D D 12 y= y = = −6 D −2 Thus, the solution is x = 5, y = −6.

1 1 = (1)( −1) − (1)(1) = −2 1 −1

−1 1 = ( −1)( −1) − (11)(1) = −10 11 −1 1 −1 Dy = = (1)(11) − (1)( −1) = 12 1 11

Dx =

12. D=

1 1 = (1)( −1) − (1)(1) = −2 1 −1

−1 1 = ( −1)( −1) − ( −9 )(1) = 10 −9 −1 1 −1 = (1)( −9 ) − (1)( −1) = −8 Dy = 1 −9

Dx =

13. D=

3 2 = ( 3 )(1) − ( −2 )( 2 ) = 7 −2 1

−4 2 = ( −4 )(1) − ( 5 )( 2 ) = −14 5 1 3 −4 = ( 3 )( 5 ) − ( −2 )( −4 ) = 7 Dy = −2 5 Dx =

So, from Cramer’s Rule, we have: D 10 x= x = = −5 D −2 D −8 y= y = =4 D −2 Thus, the solution is x = −5, y = 4.

So, from Cramer’s Rule, we have: D −14 = −2 x= x = 7 D Dy 7 y= = =1 D 7 Thus, the solution is x = −2, y = 1.

942

Section 7.4

14. 5 3 = ( 5 )( −7 ) − ( 4 )( 3 ) = −47 4 −7

D= Dx =

1 3 = (1)( −7 ) − ( −18 )( 3 ) = 47 −18 −7

Dy =

5 1 = ( 5 )( −18 ) − ( 4 )(1) = −94 4 −18

15.

3 −2 5

= ( 3 )( 4 ) − ( 5 )( −2 ) = 22

Dx =

−1 −2 = ( −1)( 4 ) − ( −31)( −2 ) = −66 −31 4

Dy =

3 −1 = ( 3 )( −31) − ( 5 )( −1) = −88 5 −31

16.

1 −4

Dx = Dy = 17.

−7 −4 = ( −7 )( 8 ) − (19 )( −4 ) = 20 19 8 1 −7 3 19

7 −3 5

Dx = Dy =

= (1)( 8 ) − ( 3 )( −4 ) = 20

= ( 7 )( 2 ) − ( 5 )( −3 ) = 29

So, from Cramer’s Rule, we have: D −66 x= x = = −3 22 D Dy −88 y= = = −4 22 D Thus, the solution is x = − 3, y = −4.

So, from Cramer’s Rule, we have: D 20 x= x = =1 D 20 D 40 =2 y= y = D 20 Thus, the solution is x =1, y = 2.

= (1)(19 ) − ( 3 )( −7 ) = 40

−29 −3 0

So, from Cramer’s Rule, we have: D 47 x= x = = −1 D −47 Dy −94 = =2 y= D −47 Thus, the solution is x = −1, y = 2.

= ( −29 )( 2 ) − ( 0 )( −3 ) = −58

7 −29 = ( 7 )( 0 ) − ( 5 )( −29 ) = 145 5 0

943

So, from Cramer’s Rule, we have: D −58 x= x = = −2 D 29 D 145 y= y = =5 D 29 Thus, the solution is x = − 2, y = 5.

Chapter 7

18.

6 −2 = ( 6 )( 7 ) − ( 4 )( −2 ) = 50 4 7

Dx =

24 −2 = ( 24 )( 7 ) − ( 41)( −2 ) = 250 41 7

Dy =

6 24 = ( 6 )( 41) − ( 4 )( 24 ) = 150 4 41

19.

3 5 = ( 3 )(1) − ( −1)( 5 ) = 8 −1 1

Dx =

16 5 = (16 )(1) − ( 0 )( 5 ) = 16 0 1

Dy =

3 16 = ( 3 )( 0 ) − ( −1)(16 ) = 16 −1 0

20.

−2 −3 = ( −2 )( 7 ) − ( 4 )( −3 ) = −2 4 7

Dx =

15 −3 = (15 )( 7 ) − ( −33 )( −3 ) = 6 −33 7

Dy = 21. D=

−2

So, from Cramer’s Rule, we have: D 250 x= x = =5 D 50 Dy 150 y= = =3 D 50 Thus, the solution is x = 5, y = 3.

So, from Cramer’s Rule, we have: D 16 x= x = =2 D 8 Dy 16 y= = =2 D 8 Thus, the solution is x = 2, y = 2.

So, from Cramer’s Rule, we have: D 6 x= x = = −3 D −2 Dy 6 y= = = −3 D −2 Thus, the solution is x = − 3, y = −3.

15 = ( −2 )( −33 ) − ( 4 )(15 ) = 6 −33

22. 3 −5 = ( 3 )(10 ) − ( −6 )( −5 ) = 0 −6 10

Dx =

−5

−21 10

= ( 7 )(10 ) − ( −21)( −5 )

= −35 From this information alone, we can conclude that the system is inconsistent or dependent.

3 −5 = ( 3 )( −10 ) − ( 6 )( −5 ) = 0 6 −10

Dx =

7 −5 = ( 7 )( −10 ) − (14 )( −5 ) = 0 14 −10

3 7 = ( 3 )(14 ) − ( 6 )( 7 ) = 0 6 14 Hence, the system is inconsistent or dependent. Dy =

944

Section 7.4

23. D= Dx =

−3

−10 15

= ( 2 )(15 ) − ( −10 )( −3 ) = 0

4 −3 = ( 4 )(15 ) − ( −20 )( −3 ) = 0 −20 15

24.

−3 D= = ( 2 )( −15 ) − (10 )( −3 ) = 0 10 −15 2

Dx =

−3

20 −15

= ( 2 )( −20 ) − ( −10 )( 4 ) = 0 −10 −20 Hence, the system is inconsistent or dependent. Dy =

From this information alone, we can conclude that the system is inconsistent or dependent.

= ( 2 )( −15 ) − ( 20 )( −3 )

= 30 25. First, rewrite the system as:  6x + y = 2 .  12 x + y = 5 Now, compute the determinants from Cramer’s Rule: 6 1 = ( 6 )(1) − (12 )(1) = −6 D= 12 1 2 1 Dx = = ( 2 )(1) − ( 5 )(1) = −3 5 1 26. First, rewrite the system as: 12 x + 18 y = 9 .   4x + 3 y = 1 Now, compute the determinants from Cramer’s Rule: 12 18 D= = (12 )( 3 ) − ( 4 )(18 ) = −36 4 3 Dx =

9 18 = ( 9 )( 3 ) − (1)(18 ) = 9 1 3

= ( 6 )( 5 ) − (12 )( 2 ) = 6 12 5 So, from Cramer’s Rule, we have: D −3 1 = x= x = D −6 2 Dy 6 y= = = −1 D −6 Thus, the solution is x = 12 , y = −1. Dy =

12 9 = (12 )(1) − ( 4 )( 9 ) = −24 4 1 So, from Cramer’s Rule, we have: D 9 1 x= x = =− D −36 4 D −24 2 y= y = = D −36 3 Thus, the solution is x = − 14 , y = 23 . Dy =

945

Chapter 7

27. D=

0.3 −0.5 = ( 0.3 )(1) − ( 0.2 )( −0.5 ) 0.2 1

= 0.4 Dx =

−0.6 −0.5 2.4 1

= ( −0.6 )(1) − ( 2.4 )( −0.5 ) = −0.6

Dy =

= 0.84 So, from Cramer’s Rule, we have: D 0.6 x= x = = 1.5 D 0.4 D 0.84 y= y = = 2.1 D 0.4 Thus, the solution is x = 1.5, y = 2.1.

Dy =

28. D=

0.3 −0.6 = ( 0.3 )( 2.4 ) − ( 0.2 )( −0.6 ) 0.2 2.4

0.5 −0.4 10 3.6

0.5 −3.6 10 −14

= ( 0.5 )( −14 ) − (10 )( −3.6 ) = 29

So, from Cramer’s Rule, we have: D −18.56 x= x = ≅ −3.2 −3.6 −0.4 D 5.8 Dx = −14 3.6 D 29 y= y = ≅5 = ( −3.6 )( 3.6 ) − ( −14 )( −0.4 ) = −18.56 D 5.8 Thus, the solution is x ≈ −3.2, y ≈ 5. = ( 0.5 )( 3.6 ) − (10 )( −0.4 ) = 5.8

29. First, rewrite the system as:  17 x − y = −7 .  −15x − y = −7 Now, compute the determinants from Cramer’s Rule: 17 −1 D= = (17 )( −1) − ( −15 )( −1) −15 −1

= −32 Dx =

−7 −1 = ( −7 )( −1) − ( −7 )( −1) −7 −1

Dy =

−7

−15 −7

= (17 )( −7 ) − ( −15 )( −7 )

= −224 So, from Cramer’s Rule, we have: D 0 x= x = =0 D −32 D −224 y= y = =7 D −32 Thus, the solution is x = 0, y = 7.

946

Section 7.4

30. First, rewrite the system as: 9x + 2 y = −45 .   4x + 3 y = −20 Now, compute the determinants from Cramer’s Rule: 9 2 D= = ( 9 )( 3 ) − ( 4 )( 2 ) = 19 4 3 Dx =

9 −45

= ( 9 )( −20 ) − ( 4 )( −45 ) = 0 4 −20 So, from Cramer’s Rule, we have: D −95 x= x = = −5 D 19 D 0 y= y = =0 D 19 Thus, the solution is x = −5, y = 0. Dy =

−45 2 = ( −45 )( 3 ) − ( −20 )( 2 ) −20 3

= −95 31. First, rewrite the system as: 2 ( x1 ) − 3 1y = 2  .  1 5 ( x ) − 6 1y = 7 Now, compute the determinants from Cramer’s Rule: 2 −3 = ( 2 )( −6 ) − ( 5 )( −3 ) = 3 D= 5 −6

2 2 = ( 2 )( 7 ) − ( 5 )( 2 ) = 4 y 5 7 So, from Cramer’s Rule, we have: 1 Dx1 9 = = =3 x D 3 1 D1y 4 = = y D 3 Thus, the solution is x = 13 , y = 34 . D1 =

() ()

D1 = x

2 −3 = ( 2 )( −6 ) − ( 7 )( −3 ) = 9 7 −6

32. First, rewrite the system as: 2 ( x1 ) − 3 1y = −12  .  1 1 1 3 ( x ) + 2 y = 7 Now, compute the determinants from Cramer’s Rule: 2 −3 D= = ( 2 ) ( 12 ) − ( 3 )( −3 ) = 10 1 3 2

() ()

D1 = x

2 −12 = ( 2 )( 7 ) − ( 3 )( −12 ) = 50 3 7 So, from Cramer’s Rule, we have: 1 Dx1 15 3 = = = x D 10 2 1 D1y 50 = = =5 y D 10 Thus, the solution is x = 23 , y = 51 . D1 = y

−12 −3 = ( −12 ) ( 12 ) − ( 7 )( −3 ) = 15 1 7 2

947

Chapter 7

33. 3 1 0 0 −1 2 −1 2 0 2 0 −1 = 3 −1 +0 1 0 −4 0 −4 1 −4 1 0

= 3 ( 0 )( 0 ) − (1)( −1)  −1 ( 2 )( 0 ) − ( −4 )( −1)  + 0 ( 2 )(1) − ( −4 )( 0 )       =1

= −4

= 7 34. 1 1 0 2 −1 0 −1 0 2 0 2 −1 = − +0 0 5 0 −3 −3 5 0 −3 5

= ( 2 )( 5 ) − ( −3 )( −1)  − ( 0 )( 5 ) − ( 0 )( −1)  + 0 ( 0 )( −3 ) − ( 0 )( 2 )      =7

= 7 35.

2 1 −5 0 −1 3 −1 3 0 3 0 −1 = 2 − + ( −5) 0 7 4 7 4 0 4 0 7 = 2 ( 0 )( 7 ) − ( 0 )( −1)  − ( 3 )( 7 ) − ( 4 )( −1)  − 5 ( 3 )( 0 ) − ( 4 )( 0 )         =0

= 25

= −25 36. 2 1 −5 −7 0 3 0 3 −7 3 −7 0 = 2 − + ( −5) −6 0 4 0 4 −6 4 −6 0

= 2 ( −7 )( 0 ) − ( −6 )( 0 )  − ( 3 )( 0 ) − ( 4 )( 0 )  − 5 ( 3 )( −6 ) − ( 4 )( −7 )       =0

= −50

948

=10

Section 7.4

37. 1 1 −5 −7 −4 3 −4 3 −7 3 −7 −4 = − + ( −5) −6 9 4 9 4 −6 4 −6 9

= ( −7 )( 9 ) − ( −6 )( −4 )  − ( 3 )( 9 ) − ( 4 )( −4 )  − 5 ( 3 )( −6 ) − ( 4 )( −7 )        =−87

= 43

=10

= −180 38. −3 2 −5 8 2 1 2 1 8 1 8 2 =−3 −2 + ( −5) −6 9 4 9 4 −6 4 −6 9

= −3 ( 8 )( 9 ) − ( −6 )( 2 )  − 2 (1)( 9 ) − ( 4 )( 2 )  − 5 (1)( −6 ) − ( 4 )( 8 )          =84

=−38

= −64 39.

1 3 4 2 1 2 −1 −1 1 2 −1 1 = −3 +4 3 1 3 −2 −2 1 3 −2 1 = ( −1)(1) − ( −2 )(1)  − 3 ( 2 )(1) − ( 3 )(1)  + 4 ( 2 )( −2 ) − ( 3 )( −1)       =1

=−1

= 0 40. −7 2 5 7 7 3 4 4 3 8 8 7 3 4 = − 7 − 2 + 5 8 4 6 −1 6 −1 4 −1 4 6 = − 7 ( 3 )( 6 ) − ( 4 )( 4 )  − 2 ( 78 ) ( 6 ) − ( −1)( 4 )  + 5 ( 78 ) ( 4 ) − ( −1)( 3 )         =2

= 9.25

= 0

949

= 6.5

Chapter 7

41.

−3 1 5 0 6 2 6 2 0 −1 +5 2 0 6 = −3 7 −9 4 −9 4 7 4 7 −9 = −3 ( 0 )( −9 ) − ( 6 )( 7 )  − ( 2 )( −9 ) − ( 6 )( 4 )  + 5 ( 2 )( 7 ) − ( 4 )( 0 )         −42

−42

= 238 42. 1 −1 5 3 −3 6 = 1 4

−3 6 3 6 3 −3 −1 +5 9 0 4 0 4 9

= ( −3 )( 0 ) − ( 9 )( 6 )  + ( 3 )( 0 ) − ( 4 )( 6 )  + 5 ( 3 )( 9 ) − ( 4 )( −3 )         −54

−24

= 117 43. −2 1 −7 −2 14 4 14 4 −2 −1 + ( −7) 4 −2 14 = −2 1 8 0 8 0 1 0 1 8

= − 2 ( −2 )( 8 ) − (1)(14 )  −1 ( 4 )( 8 ) − ( 0 )(14 )  − 7 ( 4 )(1) − ( 0 )( −2 )         −30

= 0 44. 5 −2 −1 −9 −3 4 −3 4 −9 − ( −2) −1 4 −9 −3 = 5 8 −6 2 −6 2 8 2 8 −6

= 5 ( −9 )( −6 ) − ( 8 )( −3 )  + 2 ( 4 )( −6 ) − ( 2 )( −3 )  − ( 4 )( 8 ) − ( 2 )( −9 )      −18

= 304

950

Section 7.4

45. 3 4

−1

1 5

0 1 0 −12 −12 0 15 −12 = 34 5 − ( −1) +0 8 −2 0 −2 8 0 −2 = 34 ( 51 ) ( −2 ) − ( 0 )( −12 )  + 1 ( 0 )( −2 ) − ( 8 )( −12 )  + 0 ( 0 )( 0 ) − ( 8 ) ( 15 )       − 52

− 58

= 95.7

46. 0.2 5

−1

−1.4 2 5 2 5 −1.4 −1.4 2 = 0.2 −0 +3 −3 −1 −3 −1 0 0 −3 0 = 0.2 ( −1.4 )( −3 ) − ( 0 )( 2 )  − 0 ( 5 )( −3 ) − ( −1)( 2 )      −13

4.2

+3 ( 5 )( 0 ) − ( −1)( −1.4 )   −1.4

= −3.36 47.

1 1 −1 −1 1 1 1 1 −1 D = 1 −1 1 = 1 −1 + ( −1) 1 1 1 1 1 1 1 1 1 = 1 ( −1)(1) − (1)(1)  − 1 (1)(1) − (1)(1)  + ( −1) (1)(1) − (1)( −1)  = −4        =−2

0 1 −1 −1 1 4 1 4 −1 Dx = 4 −1 1 = 0 −1 + ( −1) 1 1 10 1 10 1 10 1 1 = 0 ( −1)(1) − (1)(1)  − 1 ( 4 )(1) − (10 )(1)  + ( −1) ( 4 )(1) − (10 )( −1)  = −8        =−2

=−6

951

=14

Chapter 7

Dy = 1 4 1 10

−1 1 =1 1

10 1

−0

1 1 1 1

+ ( −1)

1 10

= 1 ( 4 )(1) − (10 )(1)  − 0 (1)(1) − (1)(1)  + ( −1) (1)(10 ) − (1)( 4 )  = −12        = −6

−1 4 1 4 1 −1 Dz = 1 −1 4 = 1 −1 +0 1 10 1 10 1 1 1 1 10 = 1 ( −1)(10 ) − (1)( 4 )  −1 (1)(10 ) − (1)( 4 )  + 0 (1)(1) − (1)( −1)  = −20        =−14

So, by Cramer’s Rule,

x= 48.

Dx −8 = =2 D −4

Dy D

−12 =3 −4

Dz −20 = =5 D −4

−1 1 1 1 −1 1 −1 1 1 D = 1 1 −1 = ( −1) −1 +1 1 1 1 1 1 1 1 1 1 = ( −1) (1)(1) − (1)( −1)  − 1 (1)(1) − (1)( −1)  + 1 (1)(1) − (1)(1)  = −4        =2

−4 1 1 1 −1 0 −1 0 1 Dx = 0 1 −1 = −4 −1 +1 1 1 2 1 2 1 2 1 1 = −4 (1)(1) − (1)( −1)  −1 ( 0 )(1) − ( 2 )( −1)  + 1 ( 0 )(1) − ( 2 )(1)  = −12       =2

=−2

−1 −4 1 0 −1 1 −1 1 0 Dy = 1 0 −1 = ( −1) − ( − 4) +1 2 1 1 1 1 2 1 2 1 = ( −1) ( 0 )(1) − ( 2 )( −1)  + 4 (1)(1) − (1)( −1)  + 1 (1)( 2 ) − (1)( 0 )  = 8         =2

952

Section 7.4

−1 1 −4 Dz = 1

0 = ( −1)

1 0 1 2

−1

1 0

+ ( −4)

1 2

1 1 1 1

= ( −1) (1)( 2 ) − (1)( 0 )  − 1 (1)( 2 ) − (1)( 0 )  + ( −4) (1)(1) − (1)(1)  = −4        =2

So, by Cramer’s Rule, x=

Dx −12 = =3 D −4

8 = −2 −4

Dz −4 = =1 D −4

49. 3 8 2 5 3 −2 3 −2 5 D = −2 5 3 = 3 −8 +2 9 2 4 2 4 9 4 9 2 = 3 ( 5 )( 2 ) − ( 9 )( 3 )  − 8 ( −2 )( 2 ) − ( 4 )( 3 )  + 2 ( −2 )( 9 ) − ( 4 )( 5 )  = 1      =−17

=−16

=−38

28 8 2 5 3 34 3 34 5 −8 +2 Dx = 34 5 3 = 28 9 2 29 2 29 9 29 9 2 = 28 ( 5 )( 2 ) − ( 9 )( 3 )  − 8 ( 34 )( 2 ) − ( 29 )( 3 )  + 2 ( 34 )( 9 ) − ( 29 )( 5 )  = −2      =−17

=−19

=161

28 2

−2 3 −2 34 34 3 − 28 +2 Dy = −2 34 3 = 3 29 2 4 2 4 29 4 29 2 = 3 ( 34 )( 2 ) − ( 29 )( 3 )  − 28 ( −2 )( 2 ) − ( 4 )( 3 )  + 2 ( −2 )( 29 ) − ( 4 )( 34 )  = 3      = −19

=−16

=−194

3 8 28 5 34 −2 34 −2 5 −8 + 28 Dz = −2 5 34 = 3 9 29 4 29 4 9 4 9 29 = 3 ( 5 )( 29 ) − ( 9 ) ( 34 )  − 8 ( −2 )( 29 ) − ( 4 )( 34 )  + 28 ( −2 )( 9 ) − ( 4 )( 5 )  = 5      =−161

=−194

953

=−38

Chapter 7

So, by Cramer’s Rule,

x= 50.

Dx = −2 D

Dz =5 D

−1 5 1 6 1 6 5 1 =7 −2 + ( −1) −4 3 −5 3 −5 −4 −5 −4 3

7 D= 6

2 5

= 7 ( 5 )( 3 ) − ( −4 )(1)  − 2 ( 6 )( 3 ) − ( −5 )(1)  + ( −1) ( 6 )( −4 ) − ( −5 )( 5 )  = 86      =19

= 23

−1 2 −1 5 1 16 1 16 5 −2 + ( −1) Dx = 16 5 1 = ( −1) −4 3 −5 3 −5 −4 −5 −4 3 = ( −1) ( 5 )( 3 ) − ( −4 )(1)  − 2 (16 )( 3 ) − ( −5 )(1)  + ( −1) (16 )( −4 ) − ( −5 )( 5 )        =53

=19

=−39

= −86 −1 −1 16 1 6 1 6 16 − ( −1) + ( −1) Dy = 6 16 1 = 7 −5 3 −5 3 −5 −5 −5 −5 3 7

= 7 (16 )( 3 ) − ( −5 )(1)  − ( −1) ( 6 )( 3 ) − ( −5 )(1)  + ( −1) ( 6 )( −5 ) − ( −5 )(16 )        =53

= 23

=50

= 344 7 2 −1 5 16 6 16 6 5 −2 + ( −1) Dz = 6 5 16 = 7 −4 −5 −5 −5 −5 −4 −5 −4 −5 = 7 ( 5 )( −5 ) − ( −4 )(16 )  − 2 ( 6 )( −5 ) − ( −5 )(16 )  + ( −1) ( 6 )( −4 ) − ( −5 )( 5 )         = 39

=50

= 172 So, by Cramer’s Rule, x=

Dx −86 = = −1 D 86

Dy D

954

344 =4 86

Dz 172 = =2 D 86

Section 7.4

51. 3 0 5 4 3 0 3 0 4 −0 +5 D = 0 4 3 =3 2 0 2 −1 −1 0 2 −1 0 = 3 ( 4 )( 0 ) − ( −1)( 3 )  − 0 ( 0 )( 0 ) − ( 2 )( 3 )  + 5 ( 0 )( −1) − ( 2 )( 4 )  = −31        =3

11 Dx = −9 7

=−6

= −8

0 4

5 4 3 −9 3 −9 4 3 = 11 −0 +5 7 0 7 −1 −1 0 −1 0

= 11 ( 4 )( 0 ) − ( −1)( 3 )  − 0 ( −9 )( 0 ) − ( 7 )( 3 )  + 5 ( −9 )( −1) − ( 7 )( 4 )  = −62      =3

=−19

=−21

3 11 5 −9 3 0 3 0 −9 − 11 +5 Dy = 0 −9 3 = 3 7 0 2 0 2 7 2 7 0 = 3 ( −9 )( 0 ) − ( 7 )( 3 )  − 11 ( 0 )( 0 ) − ( 2 )( 3 )  + 5 ( 0 )( 7 ) − ( 2 )( −9 )  = 93       = − 21

=−6

=18

3 0 11 4 −9 0 −9 0 4 −0 + 11 Dz = 0 4 −9 = 3 2 7 2 −1 −1 7 2 −1 7 = 3 ( 4 )( 7 ) − ( −1)( −9 )  − 0 ( 0 ) ( 7 ) − ( 2 )( −9 )  + 11 ( 0 )( −1) − ( 2 )( 4 )  = −31     =19

=18

=−8

So, by Cramer’s Rule, x=

Dx −62 = =2 D −31

−2

D=4

1 =3

Dy D

93 = −3 −31

Dz −31 = =1 D −31

52.

6 −2

−2 0

−0

4 1 6 0

+ ( −2)

6 −2

= 3 ( 0 )( 0 ) − ( −2 )(1)  − 0 ( 4 )( 0 ) − ( 6 )(1)  + ( −2) ( 4 )( −2 ) − ( 6 )( 0 )  = 22       =2

=−6

955

=−8

Chapter 7

−2 0 1 24 1 24 0 Dx = 24 0 −0 + ( −2) 1 =7 −2 0 10 0 10 −2 10 −2 0 7

= 7 ( 0 )( 0 ) − ( −2 )(1)  − 0 ( 24 )( 0 ) − (10 )(1)  + ( −2) ( 24 )( −2 ) − (10 )( 0 )  = 110       =2

= −10

=−48

−2 24 1 4 1 4 24 Dy = 4 24 1 = 3 −7 + ( −2) 10 0 6 0 6 10 6 10 0 3

= 3 ( 24 )( 0 ) − (10 )(1)  − 7 ( 4 )( 0 ) − ( 6 ) (1)  + ( −2) ( 4 )(10 ) − ( 6 )( 24 )  = 220      =−6

=−10

Dz = 4

24 = 3

6 −2 10

−2 10

−0

4 24 6 10

=−104

6 −2

= 3 ( 0 )(10 ) − ( −2 )( 24 )  − 0 ( 4 )(10 ) − ( 6 )( 24 )  + 7 ( 4 )( −2 ) − ( 6 )( 0 )  = 88      = 48

=−104

So, by Cramer’s Rule, D 110 x= x = =5 D 22 53. 1

−1

D= 1

−1

1 =1

−2 −2

−1 1 −2 2

−1

Dy D

−2 2

=−8

220 = 10 22

+ ( −1)

Dz 88 = =4 D 22

−1

−2 −2

= 1 ( −1)( 2 ) − ( −2 )(1)  − 1 (1)( 2 ) − ( −2 )(1)  + ( −1) (1)( −2 ) − ( −2 )( −1)  = 0     =0

=−4

So, Cramer’s Rule implies that the system is either inconsistent or dependent. 54. 1

D = 1 −1 −2 −2

−1

−1 1 1 1 1 −1 1 =1 −1 + ( −1) −2 2 −2 2 −2 −2 2

=−4

So, Cramer’s Rule implies that the system is either inconsistent or dependent. 956

Section 7.4

55. 1

−1 1 1 1 1 −1 −1 +1 D = 1 −1 1 = 1 −1 −1 −1 1 1 −1 −1 1 −1 = 1 ( −1)( −1) − (1)(1)  − 1 (1)( −1) − ( −1)(1)  + 1 (1)(1) − ( −1)( −1)  = 0    =0

So, Cramer’s Rule implies that the system is either inconsistent or dependent. 56. 1

D= 1

−1 −1 = 1

−1

−1 −1 1

−1

= 1 ( −1)(1) − (1)( −1)  − 1 (1)(1) − ( −1)( −1)  + 1 (1)(1) − ( −1)( −1)  = 0    =0

So, Cramer’s Rule implies that the system is either inconsistent or dependent. 57. First, write the system in matrix form as:  1 2 3   x  11   −2 3 5   y  = 29        4 −1 8   z  19 

Now, apply Cramer’s rule to solve the system: 1 2 3 −2 5 −2 3 3 5 −2 +3 = 71 D = −2 3 5 = 1 −1 8 4 8 4 −1 4 −1 8 11 2 3 3 5 29 5 29 3 −2 +3 = − 213 Dx = 29 3 5 = 11 −1 8 19 8 19 −1 19 −1 8 1 11 3 −2 5 −2 29 29 5 −11 +3 = 71 Dy = −2 29 5 = 1 19 8 4 8 4 19 4 19 8 1 2 11 −2 29 −2 3 3 29 −2 + 11 = 284 Dz = −2 3 29 = 1 −1 19 4 19 4 −1 4 −1 19

957

Chapter 7

So, by Cramer’s rule: D D −213 71 x= x = = −3 y = y = =1 D D 71 71

Dz 284 = =4 D 71

58. First, write the system in matrix form as:  8 −2 5   x   36   3 1 −1  y  = 17        2 −6 4   z   −2  Now, apply Cramer’s rule to solve the system: 8 −2 5 1 −1 3 −1 3 1 D = 3 1 −1 = 8 +2 +5 = − 88 2 4 2 −6 −6 4 2 −6 4 36

−2

Dx = 17

−1 = 36

−2

8 36

1 −1 6

−1

−2

−2 −6

= −440

17 −1 3 −1 3 17 − 36 +5 = − 176 Dy = 3 17 91 = 8 2 4 2 −2 −2 4 2 −2 4 8 −2 36

1 17 3 17 3 1 +2 + 36 =0 Dz = 3 1 17 = 8 2 −2 2 −6 −6 −2 2 −6 −2 So, by Cramer’s rule: D D D x= x =5 y= y = 2 z = z =0 D D D

59. First, write the system in matrix form as:  1 −4 7   x   49   −3 2 −1  y  =  −17        5 8 −2   z   −24  Now, apply Cramer’s rule to solve the system: 1 −4 7 2 −1 −3 −1 −3 2 D = −3 2 −1 = 1 +4 +7 = −190 8 −2 5 −2 5 8 5 8 −2

958

Section 7.4

−4 7 −17 −1 −17 2 2 −1 +4 +7 = − 380 Dx = −17 2 −1 = 49 −24 −2 −24 8 8 −2 −24 8 −2 49

−17 −1 −3 −1 −3 −17 − 49 +7 = 570 Dy = −3 −17 −1 =1 −24 −2 5 −2 5 −24 5 −24 −2 −4 49 −3 −17 −3 2 2 −17 +4 + 49 = − 950 Dz = −3 2 −17 = 1 8 −24 5 −24 5 8 5 8 −24 So, by Cramer’s rule: Dy D D −950 −380 570 x= x = =2 y = = = −3 z= z = =5 D −190 D −190 D −190 1

60. First, write the system in matrix form as:  12 −2 7   x   25  1 1 −4   y  =  −2  4   −4 5 0   z   −56 

Now, apply Cramer’s rule to solve the system: 1 −2 7 2 1 1 −4 1 14 −4 1 1 4 D= 1 4 2 7 − = + + = 20 4 2 −4 0 5 0 −4 5 0 −4 5 25 −2 7 1 −2 −4 −4 −2 14 4 1 Dx = −2 4 25 2 7 − = + + = 80 4 −56 0 5 0 −56 5 0 −56 5 1 2

1 −4 1 −2 −2 −4 −2 −4 = 12 − 25 +7 = − 160 −56 0 −4 0 −4 −56 −4 −56 0

Dy = 1 1 2

−2

Dz = 1

1 4

−4

25 1 1 −2 1 14 −2 −2 = 12 4 +2 + 25 = 20 −4 −56 5 −56 −4 5 −56

959

Chapter 7

So, by Cramer’s rule: x=

Dx 80 = =4 D 20

Dy D

−160 = −8 20

Dz 20 = =1 D 20

61. Let x1 = 3, x2 = 5, x3 = 3, y1 = 2, y2 = 2, y3 = −4. Then, we have 3

5 1 5 2   2 1 Area = ± 5 2 1 = ± 12 3 −2 +1  3 1 3 −4  −4 1  3 −4 1 1 2

   = ± 3 ( 2 )(1) − ( −4 )(1)  − 2 ( 5 )(1) − ( 3 )(1)  + 1 ( 5 )( −4 ) − ( 3 )( 2 )           =6 =2 = − 26   1 = ± 2 ( −12 ) Hence, choosing the positive value above, we conclude that the area is 6 units2. Alternatively, we could compute the area by identifying the height and base of the triangle. Indeed, consider the following diagram: 1 2

Base = 2 Height = 6 Hence, Area = 2 1 1 2 (Base)(Height) = 2 (2)(6) = 6 units .

62. Let x1 = 2, x2 = 7, x3 = 7, y1 = 3, y2 = 3, y3 = 7. Then, we have 2 3 1 7 1 7 3  3 1 Area = ± 12 7 3 1 = ± 12  2 −3 +1  7 1 7 1 7 7  7 7 1    = ± 2 ( 3 )(1) − ( 7 )(1)  − 3 ( 7 )(1) − ( 7 )(1)  + 1 ( 7 )( 7 ) − ( 7 )( 3 )             =−4 =0 = 28   1 = ± 2 ( 20 ) 1 2

960

Section 7.4

Hence, choosing the positive value above, we conclude that the area is 10 units2. Alternatively, we could compute the area by identifying the height and base of the triangle. Indeed, consider the following diagram: Base = 5 Height = 4 Hence, we see that Area = 12 (Base)(Height) = 12 (5)(4) = 10 units2

63. Let x1 = 1, x2 = 3, x3 = −2, y1 = 2, y2 = 4, y3 = 5. Then, we have 2 1 3 1 3 4  4 1 Area = ± 4 1 = ± 12 1 −2 +1  5 1 −2 1 −2 5   −2 5 1 1 2

1 3

   = ± 1 ( 4 )(1) − ( 5 )(1)  − 2 ( 3 )(1) − ( −2 )(1)  + 1 ( 3 )( 5 ) − ( −2 )( 4 )           =−1 =5 = 23   1 = ± 2 (12 ) Hence, choosing the positive value above, we conclude that the area is 6 units2. 1 2

64. Let x1 = −1, x2 = 3, x3 = 2, y1 = −2, y2 = 4, y3 = 1. Then, we have −1 −2 1  4 1 3 1 3 4 Area = ± 3 4 1 = ± 12 ( −1) − ( −2) +1  1 1 2 1 2 1  2 1 1 1 2

   = ± ( −1) ( 4 )(1) − (1)(1)  − ( −2) ( 3 )(1) − ( 2 )(1)  + 1 ( 3 )(1) − ( 2 )( 4 )           =3 =1 =−5   1 = ± 2 ( −6 ) Hence, choosing the positive value above, we conclude that the area is 3 units2. 1 2

961

Chapter 7

65. Let x1 = 1, x2 = 2, y1 = 2, y2 = 4. The equation of the line through the two points ( x1 , y1 ) and ( x2 , y2 ) is:

x y 1 0 = 1 2 1 which simplifies to 2 0 =x

4 1

2 1 4 1

−y

1 1 2 1

1 2 2 4

= x ( 2 )(1) − ( 4 )(1)  − y (1)(1) − ( 2 )(1)  + 1 (1)( 4 ) − ( 2 )( 2 )  = −2x + y + 0        =−2

=−1

Hence, the equation of this line is y = 2x . 66. Let x1 = 0, x2 = 2, x3 = 1, y1 = 5, y2 = 0, y3 = 2. Observe that 0 5 1 0 1 2 1 2 0 2 0 1=0 −5 +1 2 1 1 1 1 2 1 2 1 = 0 ( 0 )(1) − ( 2 )(1)  − 5 ( 2 )(1) − (1)(1)  + 1 ( 2 )( 2 ) − (1)( 0 )         =−2

= −1 Since this determinant is not 0, we conclude that the points ( x1 , y1 ), ( x2 , y2 ), and ( x3 , y3 ) are not collinear.

67. The system we must solve is I1 − I 2 − I 3 = 0   1 −1 −1  I1   0         4 I1 + 0 I 2 + 2 I 3 = 16 , which is equivalent to  4 0 2   I 2  = 16  .  4 I + 4 I + 0 I = 24  4 4 0   I 3  24  2 3  1 The solution is −1  I1  1 −1 −1  0   I  =  4 0 2  16  .  2      I 3   4 4 0  24 

962

Section 7.4

We now compute the inverse:  −1 −1 1 0 0  R − 4 R →R 1 −1 −1 −1 0 0    R32 − 4 R11 →R32   R3 −2 R2 →R3 4 0 2 0 1 0 → 0 4 6 −4 1 0  ⎯⎯⎯⎯⎯ →   ⎯⎯⎯⎯⎯  4 4 0 0 0 1  0 8 4 −4 0 1   1 −1 −1 1 0 0  1 −1 −1 1 0 0    − 18 R3 →R3   R2 −6 R3 →R2 → 0 4 6 −4 1 0  ⎯⎯⎯⎯⎯ →  0 4 6 −4 1 0  ⎯⎯⎯⎯ 0 0 1 − 12 14 − 18   0 0 −8 4 −2 1   1 −1 − 1 1 0 0 1 −1 −1 1 0 0 1 R →R     R1 + R2 + R3 →R1 2 2 3 1 4 → 0 1 0 − 14 − 81 163  ⎯⎯⎯⎯⎯ →  0 4 0 −1 − 2 4  ⎯⎯⎯⎯ 1 1 1 1 1 1  0 0 1 − 2 4 − 8  0 0 1 − 2 4 − 8  1  1 0 0 14 8  1 1 0 1 0 − − 4 8   0 0 1 − 12 14 So, the solution is

   1 − 8 1 16 3 16

 I1   14    1 I 2  = − 4  I 3   − 12

1 8

− 81 1 4

  0   72     5  16  =  2  . − 18  24  1  1 16 3 16

So, I1 = 72 , I 2 = 52 , I 3 = 1. 68. The system we must solve is I1 − I 2 − I 3 = 0   1 −1 −1  I1  0         6 I1 + 0 I 2 + 3I 3 = 24 , which is equivalent to  2 0 1   I 2  = 8  . 6 I + 6 I + 0 I = 36  1 1 0   I 3  6  2 3  1 The solution is −1  I1  1 −1 −1 0   I  = 2 0 1  8  .  2      I 3  1 1 0  6  We now compute the inverse:  1 −1 −1 1 0 0  R −2 R →R  0 2 3 −2 1 0    R22 −2 R13 →R13   R1 + R3 →R3 2 0 1 0 1 0 →  2 0 1 0 1 0  ⎯⎯⎯⎯ →   ⎯⎯⎯⎯⎯  1 1 0 0 0 1  0 −2 1 0 1 −2 

963

Chapter 7

0  2  0 0  2  0

0  0 2 3 −2 1 0   14 R3 →R3   R1 −3R3 →R1 → 2 0 1 0 1 0  ⎯⎯⎯⎯→ 0 1 0 1 0  ⎯⎯⎯⎯ 0 0 1 − 12 12 − 12  0 4 −2 2 −2  2 3 −2 1

2 0 − 12 0

− 12 1

0 1 −

1 2

 0 1 0 − 14  1 2 0 0 2  0 0 1 − 12 So, the solution is:

− 14

0 1

1 2 1 2

 0  12 R1 →R1  → 2 0  ⎯⎯⎯⎯ 1 0 − 2 3 1 1 R →R 4  2 2 2  R2 ↔ R1 1  → 0 2  ⎯⎯⎯⎯ 0 − 12 

1 0 − 14

 I1   14 I  = − 1  2  4  I 3   − 12

 0   72   8  =  5     2 − 12  6  1 

3 2

1 4

− 14 1

0 1 − 12

1 2

1 4

1 0 − 14

− 14

0 1 − 12

1 2

0 1 0 0

  R2 −R3 →R2 0  ⎯⎯⎯⎯→ − 12  1 4  3  4  − 12  3 4

1 4 3 4

− 14 1 2

So, I1 = 72 , I 2 = 52 , I 3 = 1. 69. The second determinant should be subtracted; that is, it should be −3 2 . −1 1 −1

6 71. In Dx and Dy , the column    −3 should replace the column corresponding to the variable that is being solved for in each case. Precisely, 6 3 and Dx should be −3 −1

Dy should be

2 6 . −1 −3

70. The third determinant should be −3 0 3 . 1 4

72. x =

Dx D , not . Dx D

73. True

74. True

964

Section 7.4

75. False. Observe that 2 6 4 2 8 0 8 0 2 −6 +4 0 2 8 =2 0 10 4 10 4 0 4 0 10 = 2 ( 2 )(10 ) − ( 0 )( 8 )  − 6 ( 0 )(10 ) − ( 4 )( 8 )  + 4 ( 0 )( 0 ) − ( 4 )( 2 )         = 20

=−32

=−8

= 200 whereas 1 3 2 0 4 0 1  1 4 2 0 1 4 =2 1 −3 +2  0 5 2 5 2 0  2 0 5

  = 2  1 (1)( 5 ) − ( 0 )( 4 )  − 3 ( 0 )( 5 ) − ( 2 )( 4 )  + 2 ( 0 )( 0 ) − ( 2 )(1)             =5 =−8 = −2   = 2(25) = 50

76. True. (Note also that two rows are identical – the determinant will always be 0 in such cases.) 3 1 2 2 8 0 8 0 2 0 2 8 =3 −1 +2 1 2 3 2 3 1 3 1 2 = 3 ( 2 )( 2 ) − (1)( 8 )  − 1 ( 0 )( 2 ) − ( 3 )( 8 )  + 2 ( 0 )(1) − ( 3 )( 2 )  = 0          =−4

=−24

=−6

77. a 0 0 0 0 0 b b 0 0 b 0 =a −0 +0 0 c 0 c 0 0 0 0 c = a ( b )( c ) − ( 0 )( 0 )  − 0 ( 0 )( c ) − ( 0 )( 0 )  + 0 ( 0 )( 0 ) − ( 0 )( b )  = abc          = bc

965

Chapter 7

78. a1 b1 0 b2 0 0

c1 b c2 = a1 2 0 c3

c2 0 c2 0 b2 − b1 + c1 c3 0 c3 0 0

= a1 ( b2 )( c3 ) − ( 0 )( c2 )  − b1 ( 0 )( c3 ) − ( 0 )( c2 )  + c1 ( 0 )( 0 ) − ( 0 )( b2 )      = b2 c3

= a1b2c3

79. 1 -2 -1 3 0 1 2 4 1 2 4 0 2 4 0 1 4 0 1 2 = 1 3 2 4 − ( −2) 0 2 4 − 1 0 3 4 − 3 0 3 2 0 3 2 4 1 5 −4 1 −3 −4 1 −3 5 −3 5 −4 1 -3 5 -4 where 0 1 2 2 4 3 4 3 2 −1 +2 3 2 4 =0 −3 −4 −3 5 5 −4 −3 5 −4

= 0 ( 2 )( −4 ) − ( 5 )( 4 )  − 1 ( 3 )( −4 ) − ( −3 )( 4 )    =−28

+ 2 ( 3 )( 5 ) − ( −3 )( 2 )  = 42   = 21

4 1

2 4 0 4 0 2 0 2 4 =4 −1 +2 5 −4 1 −4 1 5 1 5 −4

= 4 ( 2 )( −4 ) − ( 5 )( 4 )  − 1 ( 0 )( −4 ) − (1)( 4 )     =−28

=−4

+2 ( 0 )( 5 ) − (1)( 2 )  = −112    =−2

966

(1)

Section 7.4

4 0 2 3 4 0 4 0 3 0 3 4 =4 −0 +2 1 −4 1 −3 −3 −4 1 −3 −4

= 4 ( 3 )( −4 ) − ( −3 )( 4 )  − 0 ( 0 )( −4 ) − (1)( 4 )       =0

=−4

+2 ( 0 )( −3 ) − (1)( 3 )  = −6   =−3

4 0 1 3 2 0 2 0 3 0 3 2 =4 −0 +1 1 5 1 −3 −3 5 1 −3 5

= 4 ( 3 )( 5 ) − ( −3 )( 2 )  − 0 ( 0 )( 5 ) − (1)( 2 )       = 21

=−2

+ 1 ( 0 )( −3 ) − (1)( 3 )  = 81   = −3

So, substituting the values of the determinants of the four individual 3 × 3 matrices, we conclude that the given determinant is 1(42) + 2( −112) − ( −6) − 3(81) = −419 3 2 = (3)( −4) − ( a )(2) = −12 − 2a. So, if a is chosen such a −4 that D = 0 , then there wouldn’t be a unique solution to the system. In this case, choose a = −6 .

80. Observe that D =

81.

a2 a3

b2 b3

a c2 = − b1 2 a3 c3

c2 a + b2 1 c3 a3

c1 a − b3 1 c3 a2

c1 c2

= − b1 ( a2 )( c3 ) − ( a3 )( c2 )  + b2 ( a1 )( c3 ) − ( a3 )( c1 )  −b3 ( a1 )( c2 ) − ( a2 )( c1 )  = −a2 b1c3 + a3b1c2 + a1b2c3 − a3b2c1 − a1b3c2 + a2 b3c1 The last line above is equal to the given right-side since addition and multiplication of real numbers is commutative.

967

Chapter 7

82.

a1 a2 a3

b1 b2 b3

c1 b c2 = a3 1 b2 c3

c1 a − b3 1 c2 a2

c1 a + c3 1 c2 a2

b1 b2

= a3 ( b1 )( c2 ) − ( b2 )( c1 )  − b3 ( a1 )( c2 ) − ( a2 )( c1 )  +c3 ( a1 )( b2 ) − ( a2 )( b1 )  = a3b1c2 − a3b2c1 − a1b3c2 + a2 b3c1 + a1b2c3 − a2 b1c3 = a1b2c3 + a2 b3c1 + a3b1c2 − a1b3c2 − a2 b1c3 − a3b2c1 The last line above is equal to the given right-side since addition and multiplication of real numbers is commutative. (Note also that this is equivalent to the answer found in Problem 81.) 83. −180

84. −64

85. −1019

86. −2287

87. First, write the system in matrix form as:  3.1 1.6 −4.8  x   −33.76  5.2 −3.4 0.5   y  =  −36.68       0.5 −6.4 11.4   z   25.96 

Now, apply Cramer’s rule to solve the system: 3.1 1.6 −4.8 −33.76 1.6 −4.8 D = 5.2 −3.4 0.5 , Dx = −36.68 −3.4 0.5 , 0.5 −6.4 11.4 25.96 −6.4 11.4 3.1 −33.76 −4.8 3.1 1.6 −33.76 Dy = 5.2 −36.68 0.5 , Dz = 5.2 −3.4 −36.68 0.5 25.96 11.4 0.5 −6.4 25.96 Using a calculator to compute the determinants, we obtain from Cramer’s rule that: Dy D D x = x = − 6.4 y = =1.5 z = z = 3.4 D D D

968

Section 7.4

88. First, write the system in matrix form as: 5.1   x   −89.2   −9.2 2.7  4.3 −6.9 −7.6   y  =  38.89        2.8 −3.9 −3.5   z   34.08 

Now, apply Cramer’s rule to solve the system: 5.1 5.1 −9.2 2.7 −89.2 2.7 D = 4.3 −6.9 −7.6 , Dx = 38.89 −6.9 −7.6 , 2.8 −3.9 −3.5 34.08 −3.9 −3.5 −9.2 −89.2 5.1 −9.2 2.7 −89.2 Dy = 4.3 38.89 −7.6 , Dz = 4.3 −6.9 38.89 2.8 34.08 −3.5 2.8 −3.9 34.08 Using a calculator to compute the determinants, we obtain from Cramer’s rule that: Dy D D x = x = 12.5 y = = −8.2 z = z = 9.4 D D D

969

Chapter 7 Review Solutions ----------------------------------------------------------------------1. 5 7 2    3 −4 −2 

2.  2.3 −4.5 6.8     −0.4 2.1 −9.1

3.  2 0 −1 3     0 1 −3 −2   1 0 4 −3 

4. −1 2 3 1     3 −2 4 −2   1 −1 −4 0 

5. Row echelon form. Not reduced row echelon form.

6. Row echelon form. Reduced row echelon form.

7. Not in row echelon form.

8. Not in row echelon form.

9.  1 −2 1     0 1 −1

10.  1 4 1    0 −10 1

11.  1 −4 3 −1     0 −2 3 −2   0 1 −4 8 

12.  −2 0 −4 −9 −2    3 −2  0 2 2  0 0 1 −2 4     0 −1 4 −5 7 

13. 1 3 0  R1 ↔ R2 3 4 1  R1 −3R2 →R2 3 4 1 →   ⎯⎯⎯→   ⎯⎯⎯⎯⎯  3 4 1  1 3 0  0 −5 1 1 R →R 1 3 1 − 51 R2 →R2

 1 34 13  R1 − 34 R2 →R1 1 0 35  ⎯⎯⎯⎯ → ⎯⎯⎯⎯⎯ → 1 1 − 0 1 5 0 1 − 5   

970

Chapter 7 Review

14.  1 2 −1 0   1 2 −1 0  1 2 −1 0        R3 − 4 R2 →R3 R2 + 2 R1 → R2 → 0 1 1 −1 ⎯⎯⎯⎯⎯ → 0 1 1 −1  0 1 1 −1 ⎯⎯⎯⎯⎯  −2 0 1 −2  0 4 −1 −2  0 0 −5 2   1 2 −1 0  R −R →R 1 0 0 54  2 3 2   R1 −2 R2 + R3 →R1   ⎯⎯⎯⎯ →  0 1 1 −1  ⎯⎯⎯⎯⎯⎯ → 0 1 0 − 35   0 0 1 − 52  0 0 1 − 52  − 51 R3 → R3

15.  4 1 −2 0   4 1 −2 0   4 1 −2 0  R1 − 4 R2 → R2       R1 + 2 R3 →R3 R3 −3 R2 → R2 →  0 1 2 0  ⎯⎯⎯⎯⎯ →  0 0 −6 24   1 0 −1 0  ⎯⎯⎯⎯⎯  −2 1 1 12   0 3 0 24   0 3 0 24  1 R →R

1 14 − 12 0   4 1 −2 0  134 R12 →R12  1 0 0 −4  1  R1 − 14 R2 + 12 R3 →R1    − 6 R3 →R3   R2 ↔ R3 ⎯⎯⎯→  0 3 0 24  ⎯⎯⎯⎯ → 0 1 0 8  ⎯⎯⎯⎯⎯⎯ → 0 1 0 8  0 0 1 −4   0 0 −6 24  0 0 1 −4   

16. 2 3 2 1  2 3 2 1  2 3 2 1      R2 + R3 →R3   R1 − 2 R3 → R3 → 0 −1 1 −2  ⎯⎯⎯⎯→ 0 1 −1 −2   0 −1 1 −2  ⎯⎯⎯⎯⎯  1 1 −1 6  0 1 4 −11 0 0 5 −13 1 R →R 1 2 1 1 32 1 12  R2 + R3 →R2 1 0 0 4  − R2 → R2 1 R →R   R1 − 23 R2 + R3 →R1   3 5 3 ⎯⎯⎯⎯ → 0 1 −1 2  ⎯⎯⎯⎯⎯⎯ → 0 1 0 − 53  0 0 1 − 135   0 0 1 − 135   

971

Chapter 7

17. First, reduce the augmented matrix down to row echelon form:  3 −2 2  − 12 R2 →R2 3 −2 2  R1 −3R2 →R2  3 −2 2  → ⎯⎯⎯⎯⎯ →   ⎯⎯⎯⎯ 7 1 − 2 4 1 1 2 − − 2    0 4 2  1 R →R 1 3 1 1 R →R 2 4 2

 1 − 23 23  ⎯⎯⎯⎯ → 7 0 1 8  Now, from this we see that y = 78 , and then substituting this value into the equation obtained from the first row yields: x − 23 ( 78 ) = 32 x = 54

Hence, the solution is x = 54 , y = 78 . 18. First, reduce the augmented matrix down to reduced row echelon form: 1 R →R  2 −7 22  R2 −2 R1 →R2 2 −7 22  −2 1711R2 →1 R2 1 − 72 11  →    ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯→  1 5 23 0 17 68 − −      0 1 −4   1 0 −3  R1 + 72 R2 → R1 ⎯⎯⎯⎯⎯ →   0 1 −4 

Hence, the solution is x = −3, y = −4 . 19. First, reduce the augmented matrix down to reduced row echelon form: 5 R2 → R2 9 1 − 15  − 211 R2 →R2 1 − 15 95  5 −1 9  R51 R1 −1 → R1 5 ⎯⎯⎯⎯⎯ →   ⎯⎯⎯⎯→     1 4 6  0 −21 −21 0 1 1 1 0 2  R1 + 15 R2 → R1 ⎯⎯⎯⎯⎯ →  0 1 1 

Hence, the solution is x = 2, y = 1. 20. First, reduce the augmented matrix down to reduced row echelon form: 1 R →R  8 7 10  318 R12→R12  1 78 54  R1 + R2 →R2 1 78 54  →  ⎯⎯⎯⎯→    ⎯⎯⎯⎯ 5 61 61  1 − 14 3  −3 5 42  0 24 4     1 78 54  R1 − 78 R2 →R1 1 0 −4  24 R → R 2 61 2 ⎯⎯⎯⎯ → →  ⎯⎯⎯⎯⎯  0 1 6 0 1 6    Hence, the solution is x = −4, y = 6 .

972

Chapter 7 Review

21. First, reduce the augmented matrix down to row echelon form:  1 −2 1 3   1 −2 1 3   1 −2 1 3  R3 −3 R1 → R3       R2 − R3 → R3 R2 − 2 R1 → R2 →  0 3 −1 −10  ⎯⎯⎯⎯→  2 −1 1 −4  ⎯⎯⎯⎯⎯ 0 3 −1 −10  3 −3 −5 2  0 3 −8 −7  0 0 7 −3  1 −2 1 3  1 R →R 3 7 3 1 R →R   2 3 2 ⎯⎯⎯⎯ →  0 1 − 13 − 103   0 0 1 − 73 

Now, from this we see that z = − 73 . Then, substituting this value into the equation obtained from the second row yields: y − 13 ( − 73 ) = − 103 so that y = − 73 21 Finally, substituting these values of y and z into the equation obtained from the first row subsequently yields: 3 74 x − 2( − 73 21 ) + ( − 7 ) = 3 so that x = − 21 . 73 3 Hence, the solution is x = − 74 21 , y = − 21 , z = − 7 .

22. We reduce the matrix down to reduced row echelon form. 3 −1 4 18  R −3R →R 0 −22 22 132  0 1 −1 −6    R12 −5 R33 →R12   − 221 R1 →R1   → 0 −33 29 170  ⎯⎯⎯⎯→ 0 −33 29 170  5 2 −1 −20  ⎯⎯⎯⎯⎯ 1 7 −6 −38  1 7 −6 −38 1 7 −6 −38

1 7 0 1 −1 −6  R1 ↔ R3 1 R →R    − R2 + 33 R1 → R2 2 4 2 ⎯⎯⎯⎯⎯ → 0 0 −4 −28 ⎯⎯⎯⎯ → 0 0  0 1 1 7 −6 −38  1 7 −6 −38 R +6 R −7 R →R  1   R12 ↔ R23 3 1  R2 + R3 → R3 ⎯⎯⎯⎯→  0 0 1 7  ⎯⎯⎯⎯⎯⎯ → 0  0 1 0 1   0 Hence, the solution is x = −3, y = 1, z = 7 .

973

−6 −38  1 7  −1 6  0 0 −3   1 0 1 0 1 7 

Chapter 7

23. We reduce the matrix down to reduced row echelon form: 1 −4 10 −61  1 −4 10 −61 R −3R →R 1 −4 10 −61  71 R2 →R2    R32 +5 R11 →R32   − 191 R3 →R3  → 0 7 −22 131  ⎯⎯⎯⎯→ 0 1 − 227 131 7   3 −5 8 −52  ⎯⎯⎯⎯⎯ 48 297  0 1 − 19 19   −5 1 −2 8  0 −19 48 −297  

1 −4 10 −61 1 −4 10 −61 19( 7 )     R2 + 227 R3 →R2 − R → R R2 − R3 → R3 3 3 131 82 ⎯⎯⎯⎯ → 0 1 − 227 → 0 1 − 227 131 → 7  ⎯⎯⎯⎯⎯ 7  ⎯⎯⎯⎯⎯ 82 410   0 0 1 −5  0 0 − 19(7) 19(7 )   1 −4 10 −61 1 0 0 1    R1 + 4 R2 −10 R3 →R1   0 1 0 3  ⎯⎯⎯⎯⎯⎯→ 0 1 0 3  0 0 1 −5  0 0 1 −5  Hence, the solution is x = 1, y = 3, z = −5 . 24. We reduce the matrix down to reduced row echelon form:  1 6 −3 − 172   4 −2 5 17  R − 4 R →R  0 −26 17 51  1 2 1     R1 ↔ R2  R3 + 2 R2 → R3 17  →  1 6 −3 − 172  ⎯⎯⎯→  0 −26 17 51   1 6 −3 − 2  ⎯⎯⎯⎯⎯  0 17 −5 −15   −2 5 1 2  0 17 −5 −15   

1 6 1 6 −3 − 172  −3 − 172    − 26159(17 ) R3 →R3  R2 − R3 → R3 51 51  ⎯⎯⎯⎯→ 0 1 − 17 − 26 → 1 − 17  ⎯⎯⎯⎯⎯ 26 26 − 26  ⎯⎯⎯⎯→  0 15   0 0 − 159 − 477  0 1 − 175 − 17 17(26 ) 17(26 )      1 6 −3 − 172  1 6 −3 − 172  1 0 0 21  17    R1 −6 R2 +3 R3 →R1   R2 + 26 R3 →R2 17 51  → 0 1 0 0  ⎯⎯⎯⎯⎯⎯ → 0 1 0 0   0 1 − 26 − 26  ⎯⎯⎯⎯⎯ 0 0 1 0 0 1 3  0 0 1 3  3       1 R →R − 26 2 2 1 R →R 3 17 3

Hence, the solution is x = 12 , y = 0, z = 3 .

974

Chapter 7 Review

25. First, reduce the augmented matrix down to row echelon form: 1 R →R 3 1 1 −4  R1 −3R2 →R2 3 1 1 −4  317 R12 →R12 1 13 13 − 34  → →    ⎯⎯⎯⎯⎯  ⎯⎯⎯⎯ 2 1 −2 1 −6   0 7 −2 14  0 1 − 7 2  Since there are two rows and three unknowns, we know from the above calculation that there are infinitely many solutions to this system. (Note: The only other possibility in such case would be that there was no solution, which would occur if one of the rows yielded a false statement.) To find the solutions, let z = a. Then, from row 2, we observe that y − 72 a = 2 so that y = 72 a + 2. Then, substituting these values of y and z into the equation obtained from row 1, we see that x + 13 ( 72 a + 2 ) + 31 a = − 34 x = − 73 a − 2

Hence, the solutions are x = − 73 a − 2, y = 72 a + 2, z = a . 26. First, reduce the augmented matrix down to row echelon form: 1 R →R 1 2 1 1 − 12 32 3  1 − 12 23 3  1 R →R  2 −1 3 6  R1 − R2 → R2 2 3 2 → 2   ⎯⎯⎯⎯→    ⎯⎯⎯⎯ 1 7 11 1 3 − 3 4  3 2 −1 12  0 − 6 6 −1 Since there are two rows and three unknowns, we know from the above calculation that there are infinitely many solutions to this system. (Note: The only other possibility in such case would be that there was no solution, which would occur if one of the rows yielded a false statement.) To find the solutions, let y = a. Then, from row 2, − 76 a + 116 z = −1  z = 117 a − 116 . Substituting these values into the equation obtained from row 1, we see that x − 12 a + 23 ( 117 a − 116 ) = 3 42 x + 10 22 a = 11 42 x = − 115 a + 11 42 , y = a, z = 117 a − 116 . Hence, the solutions are x = − 115 a + 11

27. From the given information, we must solve the following system: 2 = a(16)2 + b(16) + c 2 = 256a + 16b + c   2 6 = a(40) + b(40) + c is equivalent to 6 = 1600a + 40b + c   2 4 = 4225a + 65b + c 4 = a(65) + b(65) + c

975

Chapter 7

 256 16 1  2   Identify A = 1600 40 1 and B =  6  , and rewrite this system in matrix form:  4225 65 1  4  a  A  b  = B  c  a  The solution of this system is  b  = A−1B . In this case, rather than using Gaussian  c  elimination to compute the inverse, using technology would be beneficial. Indeed, in this case, the above simplifies to: a   −0.00503  b  = A−1B =  0.44857       c   −3.888 

Thus, the equation of the curve is y = −0.005x 2 + 0.45x − 3.89 . 28. Let x = amount invested in IRA account (4.5%) y = amount invested in mutual fund (8%) z = amount invested in stock (12%) Solve the system: x + y + z = 20,000 (1)   y = x + 3000 (2)  0.045x + 0.08 y + 0.12 z = 1877.50 (3) 

First, observe that substituting (2) into (1) and (3) to obtain the following 2 × 2 system: x + ( x + 3000) + z = 20,000  which is equivalent to  0.045x + 0.08( x + 3000) + 0.12 z = 1877.50 2 x + z = 17,000 (4)   0.125x + 0.12 z = 1637.50 (5)

976

Chapter 7 Review

 2 1 17,000  0.125 2 1 17,000  1 R →R 2 2   ⎯⎯⎯⎯→    0.125 0.12 1637.50  1 0.96 13,100  1 R →R

2 1 17,000  −2 0.9211 R2 1→R2 1 0.5 8500  ⎯⎯⎯⎯⎯ → →  ⎯⎯⎯⎯⎯   0 −0.92 −9200  0 1 10,000  Now, from this we see that z = 10,000. Then, substituting this value into the equation obtained from the second row yields: x + 0.5(10,000) = 8500 so that x = 3500 Finally, substituting this value of x into (2) yields: y = 3500 + 3000 = 6500. Therefore, he should make the following investments: $3500 in IRA, $6500 in mutual fund, and $10,000 in stock. R1 − 2 R2 → R2

29. Not defined since the matrices have different orders.

30. Not defined since the matrices have different orders.

31. 3 5 2  7 8 1   

32. 7 −1 9 8   

33.  4 −6  5 2  9 −4   0 2  + 9 7  =  9 9       

34. 6 0 9  1 5 −1 7 5 8  12 3 −3 + 3 7 2  = 15 10 −1      

35.

36.

37.

10 4  6 −9   4 13 18 14  − 0 3  = 18 11       3 15 −3 8 0 12   −5 15 −15  9 21 6  − 16 4 −4  =  −7 17 10        10 −15  10 4   0 −19  − = 0 5  18 14   −18 −9  

977

Chapter 7

38.

5 25 −5  8 0 12   −3 25 −17  15 35 10  − 16 4 −4  =  −1 31 14       

39.  −7 −11 −8  3 7 2   

40. 15 −8 15   29 −1 43  

41. 10 −13  18 −20   

42.  −17 −17   9 7  

43. 15 −8 15   2 0 3  17 −8 18  29 −1 43 +  4 1 −1 = 33 0 42       

44. 11 39 −1 30 94 5   

45. 10 9 20   22 −4 2   

46. Not defined

47.

1  1 0  6 4   −0.5 Since AB =  ⋅ = = I , B must be the inverse of A.  −1.5  0 1  4 2   1

48.

1 −2  1 2   −3 6  Since AB =  ⋅ =  ≠ I , B cannot be the inverse of A. 2 −4  2 −2   −6 12  49. 1 −2 6   − 71 Since AB = 2 3 −2  ⋅  72 0 −1 1   72 of A.

4 7

−

1 7

− 71

2  1 0 0  −2  = 0 1 0  = I , B must be the inverse −1 0 0 1 

978

Chapter 7 Review

50. 1 1   −2 −14 22  0 7 6 1    Since AB =  1 0 −4  ⋅  −2 −2 −2  =  −7 1 −23 ≠ I , B cannot be the 0 6   −4 −4 −4   −2 1 0   2 inverse of A.

51.  4 −2  1  4 −2   52 A−1 = 4(1)−(1−3)( 2 )   = 10  3 1  =  3 3 1     10

− 51  1  10 

6 −7  1 6 −7   38 A−1 = ( −2 )(6 )−1 ( −4 )(7)   = 16  4 −2  =  1  4 −2    4

− 167  − 18 

52.

53. 0 −1 1 0 −1 0 − 12  A−1 = ( 0 )( 0 )−1( −2)(1)   = 2 2 0  = 1 0  2 0     

54. −1

A =

1 (3)( 2 ) − ( −1)( −2)

2 1 1 2 1  12 2 3 = 4  2 3 =  1     2

1 4 3 4

  

55. Using Gaussian elimination, we obtain the following:  1 3 −2 1 0 0  1 3 −2 1 0 0    R2 −2 R1 →R2   → 0 −5 3 −2 1 0   2 1 −1 0 1 0  ⎯⎯⎯⎯⎯  0 1 −3 0 0 1  0 1 −3 0 0 1 

1  ⎯⎯⎯⎯⎯ → 0 0 1  R2 + 35 R3 → R2 ⎯⎯⎯⎯⎯ → 0  0 R2 + 5 R3 → R3

0 0 1 3 −2 − 15 R2 → R2 1 R →R   − 12 3 3 −2 1 0  ⎯⎯⎯⎯ → 0 1 − 53 0 0 1 −2 1 5 

−2 −5 3 0 −12 3

3 −2

1 2

1 6

− 14 − 121

2 5

− 15 − 121

0   0  − 125 

− 61

7 12

1 2

− 14 − 121

1 6

0  1 0 0  R1 −3 R2 + 2 R3 → R1 1  − 4  ⎯⎯⎯⎯⎯⎯ → 0 1 0 0 0 1 − 125 

979

1 6

− 121   − 14  − 125 

Chapter 7

 − 61 So, A−1 =  12  61

7 12

− 14 − 121

− 121  − 14  . − 125 

56. Using Gaussian elimination, we obtain the following:  0 1 0 1 0 0  − 1 R →R  4 1 2 0 1 0     R13↔3R2 3  → 0 1 0 1 0 0  4 1 2 0 1 0   ⎯⎯⎯⎯ 2 1 1   −3 −2 1 0 0 1  1 3 − 3 0 0 − 3  4 1 2 0 1 0 4 1 2 0 1 0   R3 + 53 R2 →R3   R1 − 4 R3 → R3 ⎯⎯⎯⎯⎯ →  0 1 0 1 0 0  ⎯⎯⎯⎯⎯ → 0 1 0 1 0 0   0 − 53 103 0 1 34   0 0 103 53 1 34      1 14 12 0 14 0  1 0 0 − 12 101 − 15  3 R →R 3 10 3 1 R →R 1 1   R1 − 4 R2 − 2 R3 →R1   1 4 1 ⎯⎯⎯⎯ → 0 1 0 1 0 0  ⎯⎯⎯⎯⎯⎯ → 0 1 0 1 0 0  2  0 0 1 12 103 25  0 0 1 12 103 5     − 12 101 − 15  So, A−1 =  1 0 0  . 2   12 103 5  57. Using Gaussian elimination, we obtain the following:  −1 1 0 1 0 0  R −2 R →R  −1 1 0 1 0 0    R12+ R3 →1 R3 2   − →  0 −1 2 −2 1 0  2 1 2 0 1 0   ⎯⎯⎯⎯⎯  1 2 4 0 0 1   0 3 4 1 0 1  1 −1 0 −1 0 0  R3 −5 R2 →R2 1 −1 0 −1 0 0  R3 + 3 R2 → R3   101 R3 →R3   − R1 →R1 ⎯⎯⎯⎯⎯ → 0 −1 2 −2 1 0  ⎯⎯⎯⎯⎯ → 0 5 0 5 −2 1  0 0 10 −5 3 1  0 0 1 − 12 103 101  1 −1 0 −1 0 0  1 0 0 0 − 52 51  1 R →R   R1 + R2 →R1   2 5 2 ⎯⎯⎯⎯ → 0 1 0 1 − 25 51  ⎯⎯⎯⎯ → 0 1 0 1 − 52 51  0 0 1 − 12 103 101  0 0 1 − 12 103 101  0 −1 So, A =  1  − 12

− 25 −

2 5

3 10

 .  1  10  1 5 1 5

980

Chapter 7 Review

58. Using Gaussian elimination, we obtain the following:  −4 4 3 1 0 0  R −3R →R  −4 4 3 1 0 0    R13 + 4 R22 →R32   →  0 12 11 1 4 0   1 2 2 0 1 0  ⎯⎯⎯⎯⎯  3 −1 6 0 0 1   0 −7 0 0 −3 1  − R →R  −4 4 3 1 0 0  − 714 R12 →R12 1 −1 − 34 − 14 0 0     121 R3 →R3  R2 ↔ R3 ⎯⎯⎯→ → 0 1 0 0 73 − 71   0 −7 0 0 −3 1  ⎯⎯⎯⎯ 11 1 1 0 1 12  0 12 11 1 4 0  0  12 3  1 −1 − 34 − 14 0 1 −1 − 34 − 14 0 0 12    R → R3 R3 − R2 → R3 3 3 11 3 ⎯⎯⎯⎯→ − 71  ⎯⎯⎯⎯ → 0 1 0 0 0 0 7 7 0 1 11 1 1  0 0 12 0 0 − 212 1 111 − 778 12 7    27 27  1 0 0 − 112 − 772  − 772   − 112 77 77   R1 + R2 + 34 R3 →R1 3 3 ⎯⎯⎯⎯⎯⎯ → 0 1 0 0 − 71  − 71  . So, A−1 =  0 7 7 12  12   0 0 1 111 − 778  111 − 778 77  77  1

59. The system in matrix form is: 3 −1  x  11 5 2   y  = 33 .      The solution is −1

 x  3 −1 11 1  2 1 11 5   y  = 5 2  33 = 11  −5 3 33 =  4  .            Hence, x = 5, y = 4 . 60. The system in matrix form is:  6 4   x  15   −3 −2   y  =  −1      The solution would be of the form −1

 x   6 4  15   y  =  −3 −2   −1 ,       4 6 but since    −3 −2 

−1

does not exist, there is no solution.

981

0  − 71  12  77 

Chapter 7

61. The system in matrix form is:  58 − 32   x   −3 3 5    =    4 6   y  16  The solution is: −1  x   58 − 32   −3 48  65 32   −3  8   y  =  3 5  16  = 49  − 3 5  16  = 12  .   4 6     4 8    Hence, x = 8, y = 12 . 62. The system in matrix form is:  1 1 −1  x   0   2 −1 3   y  = 18        3 −2 1   z  17  The solution is: −1  x  1 1 −1  0   y  = 2 −1 3  18  .        z  3 −2 1  17  We calculate the inverse below:  1 1 −1 1 0 0  R −2 R →R 1 1 −1 1 0 0    R32 −3R11→R32   2 − 1 3 0 1 0 → 0 −3 5 −2 1 0    ⎯⎯⎯⎯⎯  3 −2 1 0 0 1  0 −5 4 −3 0 1  1 1 −1 1 0 1 1 −1 1 0 − 13 R2 → R2    − 51 R3 → R3 R2 − R3 → R3 5 2 ⎯⎯⎯⎯ → 0 1 − 53 23 − 13 0  ⎯⎯⎯⎯→ 0 1 − 3 3 13 1 0 1 − 54 53 0 − 51  0 0 − 15 15   

0 − 13 − 31

1 1 −1 1 1 1 −1 1 0 0  5 R →R    R + 2 3 2 3 ⎯⎯⎯⎯→ 0 1 − 35 32 − 31 0  ⎯⎯⎯⎯⎯ →  0 1 0 137 5 3 1 1 0 0 1 − 13 13 − 13   0 0 1 − 13 1 2 1 0 0 135 3 13    R1 − R2 + R3 → R1 ⎯⎯⎯⎯⎯ → 0 1 0 137 − 392 − 135  5 0 0 1 − 131 − 133  13 − 15 R →R3 13 3

982

0  0 1 5 0 − 392 5 13

0   − 135  − 133 

Chapter 7 Review

Hence, the solution of the system is 1  x   135 3  y =  7 2    13 − 39 5  z   − 131 13

  0   112 13      97  −  18  =  − 13  −  17   3  2 13 5 13 3 13

97 Thus, the solution is x = 112 13 , y = − 13 , z = 3 .

63. We reduce down to reduced row echelon form: 3 −2 4 11  R −6 R →R 3 −2 4 11    R12−3R33→R32   R2 −9 R3 →R3 → 0 9 −44 −114  ⎯⎯⎯⎯⎯ → 6 3 −2 6  ⎯⎯⎯⎯⎯ 1 −1 7 20  0 1 −17 −49   3 −2 4  3 −2 4 11  19 R2 →R2 11  31 R1 →R1   1091 R3 →R3   R2 + 449 R3 →R2 →  0 1 − 449 −114  ⎯⎯⎯⎯⎯ →  0 9 −44 −114  ⎯⎯⎯⎯  0 0 109 327   0 0 109 327 

 1 − 23 34 113  1 0 0 1    R1 + 23 R2 − 34 R3 →R1   → 0 1 0 2   0 1 0 2  ⎯⎯⎯⎯⎯⎯ 0 0 1 3   0 0 1 3    Hence, the solution is x = 1, y = 2, z = 3 . 64. We reduce down to reduced row echelon form:  2 6 −4 11  R + 2 R →R 2 6 −4 11  12 R1 →R1 1 3 −2 112  1 2 2    − 71 R3 →R3   R3 + 4 R2 → R3 11  → 0 0 0 0  ⎯⎯⎯⎯ → 0 0 0 0   −1 −3 2 − 2  ⎯⎯⎯⎯⎯  4 5 6 20  0 −7 14 −2  0 1 −2 72  Let x = a. Then, substituting into the equation corresponding to row 3 yields a − 2 y = 72  y = 21 a − 71 . Then, substituting these two values into the equation corresponding to row 1 yields a + 3 ( 12 a − 71 ) − 2 z = 112

2 z = 52 a − 73 − 112 z = 54 a − 83 28 Thus, the solution is x = a, y = 12 a − 71 , z = 54 a − 83 28 .

983

Chapter 7

65. 2 4

= ( 2 )( 2 ) − ( 3 )( 4 ) = −8

3 2

67. 2.4 −2.3 = ( 2.4 )( −1.2 ) − ( 3.6 )( −2.3 ) 3.6 −1.2

66. −2 −4 = ( −2 )( 2 ) − ( −3 )( −4 ) = −16 −3 2 68. D=

− 14 3 4

4 = ( − 14 ) ( −4 ) − ( 34 ) ( 4 ) = −2 −4

= 5.4 69. D=

1 −1 1

Dx = Dy =

= (1)(1) − (1)( −1) = 2

2 −1 4

1 2 1 4

= ( 2 )(1) − ( 4 )( −1) = 6

= (1)( 4 ) − (1)( 2 ) = 2

70. First, write the system as  3 −1  x   −17   −1 5   y  =  43  .     

−1 = ( 3 )( 5 ) − ( −1)( −1) = 14 D= −1 5 3

Dx =

−17 −1 = ( −17 )( 5 ) − ( 43 )( −1) 43 5

= −42

So, from Cramer’s Rule, we have: D 6 x = x = =3 D 2 Dy 2 = =1 y= D 2 Thus, the solution is x = 3, y = 1 .

Dy =

−17

−1

= ( 3 )( 43 ) − ( −1)( −17 )

= 112 So, from Cramer’s Rule, we have: D −42 = −3 x= x = 14 D D 112 =8 y= y = D 14 Thus, the solution is x = −3, y = 8

984

Chapter 7 Review

71. Divide the first equation by 2. Then, we have: 1 2 = (1)( −2 ) − (1)( 2 ) = −4 D= 1 −2 Dx = Dy =

6 −2 1 6 1 6

= ( 6 )( −2 ) − ( 6 )( 2 ) = −24

= (1)( 6 ) − (1)( 6 ) = 0

72.

−1 1 = ( −1)( −6 ) − ( 2 )(1) = 4 2 −6 4 1 Dx = = ( 4 )( −6 ) − ( −5 )(1) = −19 −5 −6 −1 4 = ( −1)( −5 ) − ( 2 )( 4 ) = −3 Dy = 2 −5 D=

73. First, write the system as: −3x + 2 y = 40   2 x − y = 25 D=

Dx =

−3 2 = ( −3 )( −1) − ( 2 )( 2 ) = −1 2 −1 40

So, from Cramer’s Rule, we have: D −24 =6 x= x = −4 D D 0 =0 y= y = D −4 Thus, the solution is x = 6, y = 0 .

25 −1

= ( 40 )( −1) − ( 25 )( 2 )

So, from Cramer’s Rule, we have: D −19 x= x = 4 D D −3 y= y = 4 D Thus, the solution is x = − 194 , y = − 34 .

So, from Cramer’s Rule, we have: D −90 = 90 x= x = −1 D D −155 = 155 y= y = −1 D Thus, the solution is x = 90, y = 155 .

= −90 −3 40 Dy = = ( −3 )( 25 ) − ( 2 )( 40 ) 2 25 = −155

985

Chapter 7

74. First, write the system as  3 −4   x   20   −1 1   y  =  −6      

3 −4 = ( 3 )(1) − ( −1)( −4 ) = −1 −1 1

Dx =

20 −4 = ( 20 )(1) − ( −6 )( −4 ) = −4 −6 1

Dy =

3 20 = ( 3 )( −6 ) − ( −1)( 20 ) = 2 −1 −6

So, from Cramer’s Rule, we have: D −4 =4 x= x = D −1 D 2 = −2 y= y = D −1 Thus, the solution is x = 4, y = −2 .

75.

1 2 2 1 3 0 3 0 1 0 1 3 =1 −2 +2 2 0 2 −1 −1 0 2 −1 0 = 1 (1)( 0 ) − ( −1)( 3 )  − 2 ( 0 )( 0 ) − ( 2 )( 3 )  + 2 ( 0 )( −1) − ( 2 )(1)         =3

=−6

=−2

= 11 76. 0 −2

0 −3

7 =0

−1 −3

−3

−1 −3

− ( −2)

1 −3

0 −3 1 −1

= 0 ( −3 )( −3 ) − ( −1)( 7 )  − ( −2) ( 0 )( −3 ) − (1)( 7 )       =16

= −7

+ 1 ( 0 )( −1) − (1)( −3 )  = −11  =3

986

Chapter 7 Review

77. a 0 −a b

−b

b c −a c −a b c =a −0 + ( −b ) 0 −d 0 −d 0 0 0 −d

= a ( b )( −d ) − ( 0 )( c )  − 0 ( −a )( −d ) − ( 0 )( c )    =− bd

= ad

+( −b ) ( −a )( 0 ) − ( 0 )( b )  = −abd  =0

78.

−2 −4 6 0 3 2 3 2 0 − − + 2 0 3 = ( −2) ( 4) 6 −1 34 2 34 −1 2 −1 2 34 = ( −2) ( 0 ) ( 34 ) − ( 2 )( 3 )  − ( −4) ( 2 ) ( 34 ) − ( −1)( 3 )       =−6

= 92

+ 6 ( 2 )( 2 ) − ( −1)( 0 )  = 54   =4

79. 1

D = 2 −1 1 1

−2

−1 1 2 1 2 −1 1 =1 −1 + ( −2) 1 1 1 1 1 1 1

= 1 ( −1)(1) − (1)(1)  −1 ( 2 )(1) − (1)(1)  + ( −2) ( 2 )(1) − (1)( −1)  = −9       = −2

−2

Dx = 3

−1

1 = ( −2)

−1 1 1

−1

3 1 4 1

+ ( −2)

3 −1 4

= ( −2) ( −1)(1) − (1)(1)  − 1 ( 3 )(1) − ( 4 )(1)  + ( −2) ( 3 )(1) − ( 4 )( −1)  = −9         =−2

= −1

987

Chapter 7

1 −2 −2 3 1 2 1 2 3 − ( −2) + ( −2) Dy = 2 3 1 = 1 4 1 1 1 1 4 1 4 1 = 1 ( 3 )(1) − ( 4 )(1)  − ( −2) ( 2 )(1) − (1)(1)  + ( −2) ( 2 )( 4 ) − (1)( 3 )  = −9        =−1

1 1 −2 −1 3 2 3 2 −1 Dz = 2 −1 3 =1 −1 + ( −2) 1 4 1 4 1 1 1 1 4 = 1 ( −1)( 4 ) − (1)( 3 )  − 1 ( 2 )( 4 ) − (1)( 3 )  + ( −2) ( 2 )(1) − (1)( −1)  = −18        =−7

So, by Cramer’s Rule, x=

Dx −9 = =1 D −9

Dy D

−9 =1 −9

Dz −18 = =2 D −9

80.

−1 −1 1 2 −2 1 −2 1 2 D = 1 2 −2 = ( −1) − ( −1) +1 1 4 2 4 2 1 2 1 4 = ( −1) ( 2 )( 4 ) − (1)( −2 )  − ( −1) (1)( 4 ) − ( 2 )( −2 )  + 1 (1)(1) − ( 2 )( 2 )  = −5        =10

=−3

−1 1 2 −2 8 −2 8 2 Dx = 8 2 −2 = 3 − ( −1) +1 1 4 −4 4 −4 1 −4 1 4 3

= 3 ( 2 )( 4 ) − (1)( −2 )  − ( −1) ( 8 )( 4 ) − ( −4 )( −2 )  + 1 ( 8 )(1) − ( −4 )( 2 )  = 70      =10

= 24

=16

−1 3 1 8 −2 1 −2 1 8 Dy = 1 8 −2 = ( −1) −3 +1 2 4 2 −4 −4 4 2 −4 4 = ( −1) ( 8 )( 4 ) − ( −4 )( −2 )  − 3 (1)( 4 ) − ( 2 )( −2 )  + 1 (1)( −4 ) − ( 2 )( 8 )  = −68      = 24

988

= − 20

Chapter 7 Review

−1 −1

Dz = 1

8 = ( −1)

−4

1 −4

− ( −1)

2 −4

1 2 2 1

= ( −1) ( 2 )( −4 ) − (1)( 8 )  − ( −1) (1)( −4 ) − ( 2 )( 8 )  + 3 (1)(1) − ( 2 )( 2 )  = −13        =−16

=−20

So, by Cramer’s Rule, Dy D x = x = −14 y= = 13.6 D D

=−3

Dz = 2.6 D

81. 3 0

D=1 1

4 2 =3

0 1 −4

1 −4

−0

0 −4

1 1

0 1

= 3 (1)( −4 ) − (1)( 2 )  − 0 (1)( −4 ) − ( 0 )( 2 )  + 4 (1)(1) − ( 0 )(1)  = −14      = −6

−1 0

=−4

−3 2 −3 1 1 2 −0 +4 Dx = −3 1 2 = ( −1) −9 −4 −9 1 1 −4 −9 1 −4 = ( −1) (1)( −4 ) − (1)( 2 )  − 0 ( −3 )( −4 ) − ( −9 )( 2 )  + 4 ( −3 )(1) − ( −9 )(1)  = 30       = −6

3 −1

=30

−3 2 1 2 1 −3 − ( −1) +4 Dy = 1 −3 2 = 3 −9 −4 0 −4 0 −9 0 −9 −4 = 3 ( −3 )( −4 ) − ( −9 )( 2 )  − ( −1) (1)( −4 ) − ( 0 )( 2 )  + 4 (1)( −9 ) − ( 0 )( −3 )  = 50       =30

=−4

= −9

3 0 −1 1 −3 1 −3 1 1 −0 + ( −1) Dz = 1 1 −3 = 3 1 −9 0 −9 0 1 0 1 −9 = 3 (1)( −9 ) − (1)( −3 )  − 0 (1)( −9 ) − ( 0 )( −3 )  + ( −1) (1)(1) − ( 0 )(1)  = −19    =−6

=−9

989

Chapter 7

So, by Cramer’s Rule, x=

Dx 30 15 = =− D −14 7

Dy D

50 25 =− −14 7

Dz −19 19 = = D −14 14

82. 1

D = −1 −3 2

1 5 =1

−3

−1

−3

−1 −3 2

= 1 ( −3 )( −3 ) − (1)( 5 )  −1 ( −1)( −3 ) − ( 2 )( 5 )  + 1 ( −1)(1) − ( 2 )( −3 )  = 16    =4

= −7

−3 5 −2 5 −2 −3 −1 +1 Dx = −2 −3 5 = 0 −4 −3 −4 1 1 −3 −4 1 −3 = 0 ( −3 )( −3 ) − (1)( 5 )  − 1 ( −2 )( −3 ) − ( −4 )( 5 )  + 1 ( −2 )(1) − ( −4 )( −3 )  = −40        =4

=−14

= 26

−2 5 −1 5 −1 −2 −0 +1 Dy = −1 −2 5 =1 −4 −3 2 −3 2 −4 2 −4 −3 = 1 ( −2 )( −3 ) − ( −4 )( 5 )  − 0 ( −1)( −3 ) − ( 2 )( 5 )  + 1 ( −1)( −4 ) − ( 2 )( −2 )  = 34        = 26

=−7

1 1 0 −3 −2 −1 −2 −1 −3 −1 +0 Dz = −1 −3 −2 = 1 2 1 1 −4 2 −4 2 1 −4 = 1 ( −3 )( −4 ) − (1)( −2 )  − 1 ( −1)( −4 ) − ( 2 )( −2 )  + 0 ( −1)(1) − ( 2 )( −3 )  = 6        =14

So, by Cramer’s Rule, x=

Dx −40 5 = =− D 16 2

Dy D

34 17 = 16 8

990

Dz 6 3 = = D 16 8

Chapter 7 Review

83. Let x1 = 2, x2 = 4, x3 = −4, y1 = 4, y2 = 4, y3 = 3. Then, we have 2

4 1

4 1 4 4  4 1 4 1 = ± 12 2 −4 +1  3 1 −4 1 −4 3   −4 3 1

Area = ± 12 4

   = ± 2 ( 4 )(1) − ( 3 )(1)  − 4 ( 4 )(1) − ( −4 )(1)  + 1 ( 4 )( 3 ) − ( −4 )( 4 )           =1 =8 = 28   1 = ± 2 ( −2 ) Hence, choosing the positive value above, we conclude that the area is 1 unit2. 1 2

84. Let x1 = 0, x2 = 3, x3 = 1, y1 = −3, y2 = 0, y3 = 6. Observe that 0 −3 1 3

1=0

0 1 6 1

− ( −3)

3 1 1 1

3 0 1 6

= 0 ( 0 )(1) − ( 6 )(1)  − ( −3) ( 3 )(1) − (1)(1)  + 1 ( 3 )( 6 ) − (1)( 0 )  = 24        =−6

=18

Since this determinant is not 0, we conclude that the points ( x1 , y1 ), ( x2 , y2 ), and ( x3 , y3 ) are not collinear. 85. a.

y = 0.16x2 – 0.05x – 4.10 b.

86. a.

y = −0.29x2 – 2.57x – 4.86 b.

y = −0.29x2 – 2.57x – 4.86 y = 0.16x2 – 0.05x – 4.10

991

Chapter 7

87.  −238 206 50   −113 159 135     40 −30 0 

88.  −143 −41  −82 64   

89. The augmented matrix to enter into the calculator is  6.1 −14.2 75.495   .  −2.3 7.2 −36.495  Solving using the calculator then yields the solution of the system as (2.25, −4.35).

90. The augmented matrix to enter into the calculator is  7.2 3.2 −1.7 5.53     −1.3 4.1 2.8 −23.949   5.2 −1.8 6.2 48.596 

Solving using the calculator then yields the solution of the system as (4.15, −6.26, 2.54).

91. First, write the system in matrix form as:  4.5 −8.7   x   −72.33  −1.4 5.3   y  =  31.32       Now, apply Cramer’s rule to solve the system: 4.5 −8.7 −72.33 −8.7 4.5 −72.33 D= , Dx = , Dy = 31.32 5.3 −1.4 5.3 −1.4 31.32 Using a calculator to compute the determinants, we obtain from Cramer’s rule that: Dy D x = x = − 9.5 y = = 3.4 D D 92. First, write the system in matrix form as: 7.5   x   42.08  1.4 3.6  2.1 −5.7 −4.2   y  =  5.37       1.8 −2.8 −6.2   z   −9.86  Now, apply Cramer’s rule to solve the system: 1.4 3.6 7.5 42.08 3.6 7.5 D = 2.1 −5.7 −4.2 , Dx = 5.37 −5.7 −4.2 , −9.86 −2.8 −6.2 1.8 −2.8 −6.2

992

Chapter 7 Review

1.4 42.08

7.5

1.4

3.6

42.08

Dy = 2.1 5.37 −4.2 , Dz = 2.1 −5.7 5.37 1.8 −9.86 −6.2 1.8 −2.8 −9.86

Using a calculator to compute the determinants, we obtain from Cramer’s rule that: Dy D D x = x = 8.5 y = = − 1.2 z = z = 4.6 D D D

993

Chapter 7 Practice Test Solutions----------------------------------------------------------------1.

 1 −2 1  Augmented matrix is    −1 3 2 

 1 −2   x   1  Matrix equation is   =   −1 3   y  2 

 3 5 −2   3 5   x   −2  Augmented matrix is   Matrix equation is   =  7 11  y   −6  7 11 −6 

 6 9 1 5  6 9 1   x  5    Augmented matrix is  2 −3 1 3  Matrix equation is  2 −3 1   y  = 3  10 12 2 9  10 12 2   z  9  4.

 3 2 −10 2  Augmented matrix is   You cannot solve by finding the inverse  1 1 −1 5  since there are fewer equations than variables.

5. 5  1 3 5 1 3 1 3 5  R3 + 3 R1 →R3  2 7 −1 ⎯⎯⎯⎯⎯   R2 − 2 R1 →R2 →  0 1 −11 ⎯⎯⎯⎯⎯ → 0 1 −11    −3 −2 0   −3 −2 0  0 7 15 

6.  2 −1 1 3   0 −3 3 3  1 1 −1 0  R1 − 2 R2 → R1     R1 ↔R2   R3 −3 R2 → R3 → 1 1 −1 0  ⎯⎯⎯→ 0 −3 3 3   1 1 −1 0  ⎯⎯⎯⎯⎯  3 2 −2 1  0 −1 1 1  0 −1 1 1   1 1 −1 0   1 1 −1 0  1 0 0 1    − 31 R2 →R2   R1 −R2 →R1   ⎯⎯⎯⎯⎯ →  0 −3 3 3  ⎯⎯⎯⎯ →  0 1 −1 −1 ⎯⎯⎯⎯→ 0 1 −1 −1  0 0 0 0   0 0 0 0  0 0 0 0  R2 −3 R3 → R3

994

Chapter 7 Practice Test

7. Reduce the augmented matrix down to row echelon form: →

R R2 18 2  1 32 61 65   6 9 1 5 6 9 1 5  1 R →R 3 R1 −3 R2 → R2 27 3 1   R → R1     R3 −5 R2 → R3 6 1 →  0 18 −2 −4  ⎯⎯⎯⎯ →  0 1 − 19 − 29   2 −3 1 3  ⎯⎯⎯⎯⎯  0 1 − 19 − 29  10 12 2 9   0 27 −3 −6    3 5 7 1 1 1 0 3  1 2 6 6  6 3 R →R     − R R2 − R3 → R3 1 2 2 1 ⎯⎯⎯⎯→ 0 1 − 19 − 29  ⎯⎯⎯⎯⎯ → 0 1 − 19 − 29  0 0 0 0  0 0 0 0      From the last row, we conclude that the system has infinitely many solutions. To find them, let z = a. Then, the second row implies that y − 19 a = − 29 so that y = 19 a − 29 . So, substituting these values of y and z into the first row yields x + 32 ( 91 a − 92 ) + 61 a = 56 , so that x = − 13 a + 67 . Hence, the solutions 1

are x = − 13 a + 67 , y = 19 a − 29 , z = a . 8. Reduce the augmented matrix down to row echelon form: 2  − R2 →R2 1 23 − 103 23  3 2 −10 2  −313RR21+→R1R→1 R2 1 23 − 103 3 → →  ⎯⎯⎯⎯    ⎯⎯⎯⎯⎯ 1 1 −1 5  0 −1 −7 −13 0 1 7 13 From the last row, we conclude that the system has infinitely many solutions. To find them, let z = t . Then, the second row implies that y + 7t = 13 so that y = −7t + 13. So, substituting these values of y and z into the first row yields x + 23 ( −7t + 13) − 103 (t ) = 23 , so that x = 8t − 8. Hence, the solutions

are x = 8t − 8, y = −7t + 13, z = t . 9. 0 4 1 −2 5     −11 19   0 −1 3  ⋅  3 −5  =  −6 8     1 1   − 

10. Not possible since the matrices have different dimensions.

11.

4 3   5 −1  

−1

 −1 −3 1  −1 −3  191 = 4( −1)1−5(3)   = −19  −5 4  =  5  −5 4     19

995

 −  3 19

4 19

Chapter 7

12. Using Gaussian elimination, we obtain the following:  1 −3 2 1 0 0  R − 4 R →R 1 −3 2 1 0 0   R32 + R1 →1 R3 2   → 0 14 −8 −4 1 0   4 2 0 0 1 0  ⎯⎯⎯⎯⎯  −1 2 5 0 0 1  0 −1 7 1 0 1 

1 −3 2 1 −3 2 1 0 0 1 R →R    3 90 3 ⎯⎯⎯⎯⎯ → 0 14 −8 −4 1 0  ⎯⎯⎯⎯ → 0 14 −8 0 0 90 10 1 14  0 0 1 R2 +14 R3 → R3

1 −3 2  R2 + 8 R3 → R2 ⎯⎯ ⎯⎯⎯ → 0 14 0 0 0 1

− 289

98 90

1 9

1 90

1 −3 2 1  = 0 1 0 − 29 0 0 1 19 − 452   19 19 90 4  So, A−1 =  − 29 907 45  . 7   19 901 45 

0 7 90 1 90

 1 −3 2 0 1  R → R2 112  14 2 → 0 1 0 90  ⎯⎯⎯⎯ 14  0 0 1 90  

−4

1 9

1 90

0  0 14  90 

28 − 9(14)

98 90(14 )

1 9

1 90

1 0 0 0  R1 +3 R2 − 2 R3 → R1 4  → 0 1 0 45  ⎯⎯⎯⎯⎯⎯ 7  0 0 1 45 

− 29

19 90 7 90

1 9

1 90

1 9

0   112 90(14 )  7  45 

− 452  4  45  7  45 

13. No inverse – only square matrices can have inverses. 14. Write the matrix equation version of the system:  1 3   x   −1  −2 −5   y  =  4       Hence, the solution is: −1

x   1 3   −1   −5 −3   −1  −5 −3  −1  −7  1  y  =  −2 −5  ⋅  4  =  (1)( −5)−( −2 )(3)  2 1   ⋅  4  =  2 1  ⋅  4  =  2                   So, the solution is x = −7, y = 2 .

996

Chapter 7 Practice Test

15. The system in matrix form is:  3 −1 4   x  18   1 2 3   y  =  20  .       −4 6 −1  z   11  The solution is: −1  x   3 −1 4  18   y  =  1 2 3   20         z   −4 6 −1  11 

We calculate the inverse below:  3 −1 4 1 0 0   1 2 3 0 1 0  R −3 R → R   R1 ↔R2   R32 + 4 R11 →R23 →  1 2 3 0 1 0  ⎯⎯⎯→  3 −1 4 1 0 0  ⎯⎯⎯⎯⎯  −4 6 −1 0 0 1   −4 6 −1 0 0 1  1 2 3 0 1 0  1 2 3 0 1 0    R3 + 2 R2 →R3   − 17 R2 →R2 → 0 −7 −5 1 −3 0  ⎯⎯⎯⎯ → 0 −7 −5 1 −3 0  ⎯⎯⎯⎯⎯ 0 14 11 0 4 1  0 0 1 2 −2 1  1 0 1 2 3 0 1 2 3 0 1 0   R2 − 57 R3 →R2   R1 −2 R2 −3R3 →R1 3 13 5 1 11 → 0 1 0 − 7 − 57  ⎯⎯⎯⎯⎯⎯ → 7  0 1 7 − 7 7 0  ⎯⎯⎯⎯⎯  0 0 1 2 −2 1  0 0 1 2 −2 1   1 0 0 − 207 237 − 117    13 11 − 75  7 0 1 0 − 7  0 0 1 2 −2 1  Hence, the solution is  x   − 207    11  y =  − 7  z   2

− 117  18   −3 − 75  20  =  1  . −2 1  11   7  23 7 13 7

Thus, x = −3, y = 1, z = 7 . 16.

7 −5 2 −1

= (7)( −1) − (2)( −5) = 3

997

Chapter 7

17. 1 −2 −1 3 −5 4

−1

2 =1 0

−5 2 −1 0

− ( −2)

3 2 4 0

+ ( −1)

3 −5 4 −1

= 1 ( −5 )( 0 ) − ( −1)( 2 )  − ( −2) ( 3 )( 0 ) − ( 4 )( 2 )      =2

=−8

+ ( −1) ( 3 )( −1) − ( 4 )( −5 )  = −31  =17

18.

So, from Cramer’s Rule, we have: D x= x =7 D Dy =3 y= D Thus, the solution is x = 7, y = 3 .

−2 = (1)( 3 ) − ( −1)( −2 ) = 1 −1 3 1 −2 Dx = = (1)( 3 ) − ( 2 )( −2 ) = 7 2 3

Dy =

−1 2

= (1)( 2 ) − ( −1)(1) = 3

19.

3 5 −2 11 3 7 3 7 11 −5 −2 D = 7 11 3 = 3 −1 1 1 1 1 −1 1 −1 1 = 3 (11)(1) − ( −1)( 3 )  − 5 ( 7 )(1) − (1)( 3 )  − 2 ( 7 )( −1) − (1)(11)  = 58      =14

=−18

−6 5 −2 11 3 2 3 2 11 Dx = 2 11 3 = − 6 −5 −2 −1 1 4 1 4 −1 4 −1 1 = −6 (11)(1) − ( −1)( 3 )  − 5 ( 2 )(1) − ( 4 )( 3 )  − 2 ( 2 )( −1) − ( 4 )(11)  = 58      =14

=−10

998

=−46

Chapter 7 Practice Test

3 − 6 −2 Dy = 7

2 4

2 3 7 3 7 2 +6 −2 3 =3 4 1 1 1 1 4 1

= 3 ( 2 )(1) − ( 4 )( 3 )  + 6 ( 7 )(1) − (1)( 3 )  − 2 ( 7 )( 4 ) − (1)( 2 )  = −58          =−10

Dz = 7 11 1 −1

= 26

−6

11 2 7 2 7 11 −5 −6 2 =3 −1 4 1 4 1 −1 4

= 3 (11)( 4 ) − ( −1)( 2 )  − 5 ( 7 )( 4 ) − (1)( 2 )  − 6 ( 7 )( −1) − (1)(11)  = 116      = 46

=−18

= 26

So, by Cramer’s Rule, x=

Dx =1 D

Dy D

= −1

Dz =2 D

 0.4 0.6  4 7 3   = 5.4 4  . (The order is 2 × 2.) 20. Here, BA =  0.5 0.1 ⋅      6 5 4   0.1 0.3  5.3 5.3   Each entry of this matrix gives the respective applicant’s total score according to each of the two rubrics. Specifically, Applicant 1’s score according to Rubric 1 is 5.4, Applicant 1’s score according to Rubric 2 is 4, Applicant 2’s score according to Rubric 1 is 5.3, Applicant 2’s score according to Rubric 2 is 5.3.

21. Let x = amount invested in money market (2%) y = amount invested in conservative stock (4%) z = amount invested in aggressive stock (22%) Solve the system: y = x + 1000 (1)   (2) z = 2x  0.02 x + 0.04 y + 0.22 z = 1790 (3) 

First, observe that substituting (2) into (3) yields 0.46x + 0.04 y = 1790. Thus, we can consider this equation, together with (1) to get the following 2 × 2 system: x − y = −1000   0.46 x + 0.04 y = 1790 999

Chapter 7

Now, write this system in its equivalent matrix form: −1   x   −1000   1  0.46 0.04   y  =  1790       Solving then yields: −1

−1   −1000  x   1  y  = 0.46 0.04   1790          0.04 1   −1000  =  (1)(0.04 )−1(0.46 )( −1)     −0.46 1   1790    1  0.04 1   −1000   0.08 2   −1000  3500  =  0.5   =  =    −0.46 1   1790   −0.92 2   1790   4500  So, x = 3500, y = 4500. Substituting this value of x into (2) yields z = 2(3500) = 7000. Therefore, he should invest $3500 in money market, $4500 in conservative stock, and $7000 in aggressive stock.

22. a. The system we must solve is:  9a − 3b + c = 6   a + b + c = 12 25a + 5b + c = 7  b. The corresponding augmented matrix that should be entered into the calculator is  9 −3 1 6     1 1 1 12   25 5 1 7  Solving this system yields a = −0.34375, b = 0.8125, c = 11.53125. Hence, the desired quadratic function is y = −0.34375x 2 + 0.8125x + 11.53125. 23. The augmented matrix that should be entered into the calculator is  5.6 −2.7 87.28     −4.2 8.4 −106.26  Solving yields the solution (12.5, −6.4).

1000

Chapter 7 Practice Test

24. First, write the system in matrix form as:  1.5 2.6   x  18.34   −2.3 1.5   y  = 28.94       Now, apply Cramer’s rule to solve the system: 1.5 2.6 18.34 2.6 1.5 18.34 , Dx = , Dy = D= 28.94 1.5 −2.3 1.5 −2.3 28.94 Using a calculator to compute the determinants, we obtain from Cramer’s rule that: Dy D = 10.4 x = x = − 5.8 y = D D

1001

Chapter 7 Cumulative Review--------------------------------------------------------------------1.

2. Since the line is parallel to the y-axis, it is vertical. Hence, the equation is x = −2.

x − 6 x + 9 = 11 + 9 2

( x − 3)2 = 20 x − 3 = ± 20 = ±2 5 x =3±2 5

3. The radius is

4. The relation is NOT a function since if x = 6, for instance, then there are two corresponding outputs, namely ± 11.

( −3 − 1)2 + ( −1 − 2)2 = 5. So, the equation of the circle is ( x + 3)2 + ( y + 1)2 = 25.

(

)

(

)

So, both 6, 11 and 6, − 11 lie on the graph, preventing it from being a function.

5. g ( − x ) = 2 − ( − x )2 = 2 − x 2 = g ( x ) , so g is even. As such, since − g ( −x ) ≠ g ( x ) , g is not odd.

6. Translate to the right 4 units and vertically compress by a factor of 5.

7. f ( g ( x )) = ( x1 ) − 1 = x13 − 1.

8. To find the inverse, switch the x and y, and solve for y: y = 3 x −1

Domain: ( −∞,0 ) ∪ ( 0, ∞ ) .

x = 3 y −1 x +1 = 3 y y = ( x + 1)3 So, f −1 ( x ) = ( x + 1)3 . 9. Completing the square yields f ( x ) = 14 x 2 + 53 x − 256

= 14 ( x 2 + 125 x ) − 256

10.

(

p( x ) = C x + 7

) ( x − 0) ( x − 7 ) , 2

where C is a nonzero constant.

6 9 = 14 ( x + 125 x + 36 25 ) − 25 − 25 2

= 14 ( x + 65 ) − 35 2

So, the vertex is ( − 65 , − 35 ) .

1002

Chapter 7 Cumulative Review

11.

5x − 4 x + 0 x + 1 5x − 4 x + 0 x + 3 2

− ( 5 x 3 + 0 x 2 + 5x ) − 4 x 2 − 5x + 3

− ( −4 x 2 + 0 x − 4 ) − 5x + 7 Hence, Q( x ) = 5x − 4, r( x ) = −5x + 7. 12. Since 4i is a zero of P(x), so is −4i. Hence, ( x + 4i )( x − 4i ) = x 2 + 16 divides P(x) evenly. As such, we have x 2 + 2 x − 15 x 2 + 0 x + 16 x 4 + 2 x3 + x 2 + 32 x − 240

− ( x 4 + 0 x3 + 16 x 2 ) 2 x3 − 15x 2 + 32 x

− ( 2 x3 + 0x 2 + 32 x )

− 15x 2 − 240

− ( −15x 2 − 240 ) 0

Thus, P ( x ) = ( x + 16 )( x + 2x − 15 ) = ( x + 4i )( x − 4i )( x + 5)( x − 3). 2

13.

14. Use A = Pe rt . A = 5400e 0.025( 4 ) = $5,967.92

0.7x − 5x + 11 ( x − 3)( x + 2) Vertical asymptotes: x = 3, x = −2 Horizontal asymptote: y = 0.7 2

f (x) =

15. Must have x + 3 > 0 , so the domain is ( −3, ∞ ) .

16.

logπ 1 =

1003

ln1 =0 ln π

Chapter 7

17.

18. We must find t such that 3P = Pe 0.04 t

log 5 ( x + 2) + log5 (6 − x ) = log 5 (3x ) log5 ( x + 2)(6 − x ) = log5 (3x )

3 = e 0.04 t

( x + 2)(6 − x ) = 3x

ln3 = 0.04t

x − x − 12 = 0 ( x − 4)( x + 3) = 0

1 t = 0.04 ln3 ≈ 27.47 years

x = 4, −3

19. Write the system in matrix form as 2 3   x   6   1 1.5   y  =  −5.5  .      The solution would be −1

 x  2 3   6   y  = 1 1.5   −5.5  ,       2 3  but since    1 1.5 

−1

does not exist since

1 1.5

= 0 , the system has no solution.

20. Let x = price of a deluxe burger, y = price of an order of fries, z = price of a soda. We must solve the system: 6x + 5 y + 5 z = 24.34  8x + 4 y + 6 z = 28.42 3x + 2 y + 4 z = 13.51  To this end, observe that 6 5 5 24.34  1 R →R 6 5 5 24.34  6 5 5 24.34  2   R8 1 −2 2 R3 →   R1 −6 R2 →R2   R3 3 1 → 1 2 4 3.5525  ⎯⎯⎯⎯⎯ → 0 2 21 3.025  8 4 6 28.42  ⎯⎯⎯⎯⎯ 3 2 4 13.51  0 1 −3 −2.68  0 1 −3 −2.68

1  ⎯⎯⎯⎯ → 0 0  1 4 R →R  3 3 13 ⎯⎯⎯⎯ → 0 0  1 R →R 1 6 1 1 R →R 2 2 2

 1 65 4.05667   R2 −R3 →R3  1 14 1.5125  ⎯⎯⎯⎯→ 0 1 0 0 1 −3 −2.68  

4.05667   1 1.5125  4 13 4.1925  4 5 5 1 65 65 4.05667  4.05667  6 6 1  R2 − 4 R3 →R2   1 14 1.5125  ⎯⎯⎯⎯⎯ 1.19  → 0 1 0 0 0 1 0 1 1.29  1.29  

5 6

1004

5 6

Chapter 7 Cumulative Review

1 0 0 2.065    ⎯⎯⎯⎯⎯⎯ → 0 1 0 1.19  0 0 1 1.29  So, a deluxe burger costs $2.07, an order of fries costs $1.19, and a soda costs $1.29. R1 − 56 R2 − 56 R3 → R1

21. The region is as follows:

Vertex (0,0) (0,2) (2,0) 10 10 (7,7 )

z = 5x + 7 y 0 14 10 120 7 MAX

So, the maximum value of z is 120 7 and

occurs at ( 107 , 107 )

22. Observe that 11  17 R3 →R3 1 −2 3 11   1 −2 3 11  R − 4 R →R 1 −2 3    R32 −3R11→R32   131 R2 →R2  → 0 13 −13 −52  ⎯⎯⎯⎯ → 0 1 −1 −4   4 5 −1 −8 ⎯⎯⎯⎯⎯ 0 1 − 117 − 327   3 1 −2 1  0 7 −11 −32    1 −2 3 11  1 −2 3 11    74 R3 →R3   R2 − R3 → R3 ⎯⎯⎯⎯→ 0 1 −1 −4  ⎯⎯⎯⎯ → 0 1 −1 −4  4 4  0 0 0 0 1 1  7 7  

1 0 0 2   1 −2 3 11    R1 + 2 R2 −3R3 →R1   ⎯⎯⎯⎯→  0 1 0 −3 ⎯⎯⎯⎯⎯⎯ → 0 1 0 −3  0 0 1 1  0 0 1 1  So, the solution is x = 2, y = −3, z = 1. R2 + R3 → R2

78 −10 40  23.    26 0 14 

1005

Chapter 7

24. First, write the system as

 2 5   x   −1  −1 4   y  =  7  .     

The solution is given by −1

 x   2 5   −1  y  =  −1 4   7        1  4 −5   −1  −3 =  = 13  1 2   7   1 

So, the solution is x = −3, y = 1 25. First, write the system as  7 5  x   1   −1 4   y  =  −1 .     

−1 4

= ( 7 )( 4 ) − ( −1)( 5 ) = 33

Dx =

1 5 = (1)( 4 ) − ( −1)( 5 ) = 9 −1 4

Dy =

7 1 = ( 7 )( −1) − ( −1)(1) = −6 −1 −1

26. The graph of the given polynomial is as follows:

So, by Cramer’s Rule, we have D 3 x= x = D 11 Dy 2 =− y= 11 D

The x-intercepts are approximately (−6.53,0), (−0.84,0), and (2.37,0). The relative maximum occurs at approximately (−4.27, 51.55), and the relative minimum occurs at approximately (0.94, −19.03).

1006

Chapter 7 Cumulative Review

27. 7  131  35 12  −1 Using the calculator, we see that AB =  and ( AB ) =  − 19   −19 −14   262

1007

6 131 35 262

−

 .. 

CHAPTER 8 Section 8.1 Solutions -------------------------------------------------------------------------------1. Since A = 1, B = 1, C = −1, B 2 − 4 AC = 1 − 4 (1) ( −1) = 5 > 0. Hence,

2. Since A = 1, B = 1, C = 2, B 2 − 4 AC = 1 − 4 (1) ( 2 ) = −7 < 0. Hence,

this is the equation of a hyperbola.

this is the equation of an ellipse. Since A ≠ C , we know this ellipse is not a circle.

3. Since A = 2, B = 0, C = 2, B 2 − 4 AC = 0 − 4 ( 2 ) ( 2 ) = −16 < 0. Hence, this is the equation of an ellipse. Additionally, since A = C = 2, we know this ellipse is a circle.

4. Since A = 1, B = 0, C = 1, B 2 − 4 AC = 0 − 4 (1) (1) = −4 < 0. Hence, this is the equation of an ellipse. Additionally, since A = C = 1, we know this ellipse is a circle.

5. Since A = 2, B = 0, C = −1, B 2 − 4 AC = 0 − 4 ( 2 ) ( −1) = 16 > 0.

6. Since A = −1, B = 0, C = 2, B 2 − 4 AC = 0 − 4 ( −1) ( 2 ) = 8 > 0. Hence,

Hence, this is the equation of a hyperbola.

this is the equation of a hyperbola.

7. Since A = 5, B = 0, C = 20, B 2 − 4 AC = 0 − 4 ( 5 ) ( 25 ) = −500 < 0.

8. Since A = 4, B = 0, C = 8, B 2 − 4 AC = 0 − 4 ( 4 ) ( 8 ) = −64 < 0.

Hence, this is the equation of an ellipse. Since A ≠ C , we know this ellipse is not a circle.

9. Since A = 1, B = 0, C = 0, B 2 − 4 AC = 0 − 4 (1) ( 0 ) = 0. Hence, this is the equation of a parabola.

10. Since A = 0, B = 0, C = 1, B 2 − 4 AC = 0 − 4 ( 0 ) (1) = 0. Hence, this is the equation of a parabola.

11. Since A = 1, B = 0, C = 1, B 2 − 4 AC = 0 − 4 (1) (1) = −4 < 0. Hence,

12. Since A = 1, B = 0, C = 1, B 2 − 4 AC = 0 − 4 (1) (1) = −4 < 0. Hence,

this is the equation of an ellipse. Additionally, since A = C = 1, we know this ellipse is a circle.

1008

Section 8.2 Solutions -------------------------------------------------------------------------------1. c Opens to the right

2. a Opens to the left

3. d Opens down

4. b Opens up

5. c Vertex is (1,1) , opens to the right

6. d Vertex is (1, −1) , opens to the left

7. a Vertex is (−1, −1) , opens down

8. b Vertex is (1,1) , opens up

9. Vertex is (0,0) and focus is (0,3) . So, the parabola opens up. As such, the general form is x 2 = 4 py . In this case,

10. Vertex is (0,0) and focus is (2,0) . So, the parabola opens to the right. As such, the general form is y2 = 4 px . In this case, p = 2 , so the equation is

p = 3 , so the equation is x 2 = 12 y .

y 2 = 8x .

11. Vertex is (0,0) and focus is ( −5,0) . So, the parabola opens to the left. As such, the general form is y2 = 4 px . In this case, p = −5 , so the equation is

12. Vertex is (0,0) and focus is (0, −4) . So, the parabola opens down.

y 2 = −20x .

x 2 = −16 y .

13. Vertex is (3,5) and focus is (3,7) . So, the parabola opens up. As such, the general form is (x − 3)2 = 4 p( y − 5) . In this case, p = 2 , so the equation is

14. Vertex is (3,5) and focus is (7,5) . So, the parabola opens to the right. As such, the general form is ( y − 5 )2 = 4 p( x − 3) . In this case, p = 4 , so the equation is

( x − 3)2 = 8( y − 5) .

As such, the general form is x 2 = 4 py . In this case, p = −4 , so the equation is

( y − 5)2 = 16( x − 3) .

15. Vertex is (2, 4) and focus is (0, 4) . So, the parabola opens to the left. As such, the general form is ( y − 4 )2 = 4 p( x − 2) . In this case, p = −2 ,

so the equation is ( y − 4 )2 = −8( x − 2 ) .

16. Vertex is (2, 4) and focus is (2, −1) . So, the parabola opens down. As such, the general form is ( x − 2 )2 = 4 p( y − 4) . In this case, p = −5 , so the equation is (x − 2)2 = −20( y − 4) .

1009

Chapter 8

17. Focus is (2, 4) and the directrix is y = −2 . So, the parabola opens up. Since the distance between the focus and directrix is 6 and the vertex must occur halfway between them, we know p = 3 and the vertex is (2,1) . Hence, the equation is

18. Focus is (2, −2) and the directrix is y = 4 . So, the parabola opens down. Since the distance between the focus and directrix is 6 and the vertex must occur halfway between them, we know p = −3 and the vertex is (2,1) . Hence, the equation is

(x − 2)2 = 4(3)( y − 1) = 12( y − 1) .

(x − 2)2 = 4( −3)( y − 1) = −12( y − 1) .

19. Focus is (3, −1) and the directrix is x = 1 . So, the parabola opens to the right. Since the distance between the focus and directrix is 2 and the vertex must occur halfway between them, we know p = 1 and the vertex is (2, −1) . Hence, the equation is

20. Focus is ( −1,5) and the directrix is x = 5 . So, the parabola opens to the left. Since the distance between the focus and directrix is 6 and the vertex must occur halfway between them, we know p = −3 and the vertex is (2,5) . Hence, the equation is

( y + 1)2 = 4(1)( x − 2) = 4( x − 2) .

( y − 5)2 = 4( −3)( x − 2) = −12( x − 2) .

21. Since the vertex is ( −1,2) and the parabola opens to the right, the general form of its equation is ( y − 2 )2 = 4 p( x + 1) , for some p > 0 . The fact that (1,6) is on the graph implies: (6 − 2)2 = 4 p(2) 16 = 8p 2= p Hence, the equation is

22. Since the vertex is (2, −1) and the parabola opens up, the general form of its equation is ( x − 2 )2 = 4 p( y + 1) , for some p > 0 . The fact that (6,1) is on the graph implies: (6 − 2)2 = 4 p(2) 16 = 8p 2= p Hence, the equation is

( y − 2 )2 = 8( x + 1) .

( x − 2 )2 = 8( y + 1) .

1010

Section 8.2

23. Since the vertex is (2, −1) and the parabola opens down, the general form of its equation is ( x − 2 )2 = 4 p( y + 1) , for some p < 0 . The fact that (6, −3) is on the graph implies: (6 − 2)2 = 4 p( −3 + 1) 16 = −8p −2 = p Hence, the equation is ( x − 2 )2 = −8( y + 1) .

24. Since the vertex is ( −1,2 ) and the parabola opens to the left, the general form of its equation is ( y − 2 )2 = 4 p( x + 1) , for some p < 0 . The fact that ( −3,6 ) is on the graph implies: (6 − 2)2 = 4 p( −3 + 1) 16 = −8p −2 = p Hence, the equation is ( y − 2 )2 = −8( x + 1) .

25. Equation: x 2 = 8 y = 4(2)y (1) So, p = 2 and opens up. Vertex: (0,0) Focus: (0,2) Directrix: y = −2 Latus Rectum: Connects x-values corresponding to y = 2 . Substituting this into (1) yields: x 2 = 16 so that x = ±4 So, the length of the latus rectum is 8.

26. Equation: x 2 = −12 y = 4( −3)y (1) So, p = −3 and opens down. Vertex: (0,0) Focus: (0, −3) Directrix: y = 3 Latus Rectum: Connects x-values corresponding to y = −3 . Substituting this into (1) yields: x 2 = 36 so that x = ±6 So, the length of the latus rectum is 12.

1011

Chapter 8

27. Equation: y2 = −2x = 4(− 12 )x (1) So, p = − 12 and opens to the left. Vertex: (0,0) Focus: ( − 12 , 0) Directrix: x = 12 Latus Rectum: Connects y-values corresponding to x = − 12 . Substituting this into (1) yields: y2 = 1 so that y = ±1 So, the length of the latus rectum is 2. 28. Equation: y2 = 6 x = 4( 32 )x

(1)

So, p = and opens to the right. Vertex: (0,0) Focus: ( 32 , 0) Directrix: x = − 32 Latus Rectum: Connects y-values corresponding to x = 32 . Substituting this into (1) yields: y2 = 9 so that y = ±3 So, the length of the latus rectum is 6. 3 2

29. Equation: x 2 = 16 y = 4(4)y (1) So, p = 4 and opens up. Vertex: (0,0) Focus: (0, 4) Directrix: y = −4 Latus Rectum: Connects x-values corresponding to y = 4 . Substituting this into (1) yields: x 2 = 64 so that x = ±8 So, the length of the latus rectum is 16.

1012

Section 8.2

30. Equation: x 2 = −8 y = 4( −2)y (1) So, p = −2 and opens down. Vertex: (0,0) Focus: (0, −2) Directrix: y = 2 Latus Rectum: Connects x-values corresponding to y = −2 . Substituting this into (1) yields: x 2 = 16 so that x = ±4 So, the length of the latus rectum is 8. 31. Equation: y2 = 4x = 4(1)x (1) So, p = 1 and opens to the right. Vertex: (0,0) Focus: (1,0) Directrix: x = −1 Latus Rectum: Connects y-values corresponding to x = 1 . Substituting this into (1) yields: y 2 = 4 so that y = ±2 So, the length of the latus rectum is 4. 32. Equation: y2 = −16 x = 4( −4)x (1) So, p = −4 and opens to the left. Vertex: (0,0) Focus: ( −4,0) Directrix: x = 4 Latus Rectum: Connects y-values corresponding to x = −4 . Substituting this into (1) yields: y 2 = 64 so that y = ±8 So, the length of the latus rectum is 16.

1013

Chapter 8

33. Equation: ( y − 2 )2 = 4(1)( x + 3) (1) So, p = 1 and opens to the right. Vertex: (−3,2) Focus: ( −2,2) Directrix: x = −4 Latus Rectum: Connects y-values corresponding to x = −2 . Substituting this into (1) yields: ( y − 2 )2 = 4 so that y = 2 ± 2 = 4, 0 So, the length of the latus rectum is 4. 34. Equation: ( y + 2 )2 = 4( −1)( x − 1) (1) So, p = −1 and opens to the left. Vertex: (1, −2) Focus: (0, −2) Directrix: x = 2 Latus Rectum: Connects y-values corresponding to x = 0 . Substituting this into (1) yields: ( y + 2 )2 = 4 so that y = −2 ± 2 = −4, 0 So, the length of the latus rectum is 4. 35. Equation: ( x − 3)2 = 4( −2)( y + 1) (1) So, p = −2 and opens down. Vertex: (3, −1) Focus: (3, −3) Directrix: y = 1 Latus Rectum: Connects x-values corresponding to y = −3 . Substituting this into (1) yields: ( x − 3)2 = 16 so that x = 3 ± 4 = −1, 7 So, the length of the latus rectum is 8.

1014

Section 8.2

36. Equation: ( x + 3)2 = 4( −2)( y − 2 ) (1) So, p = −2 and opens down. Vertex: (−3,2) Focus: ( −3,0) Directrix: y = 4 Latus Rectum: Connects x-values corresponding to y = 0 . Substituting this into (1) yields: ( x + 3)2 = 16 so that x = −3 ± 4 = 1, − 7 So, the length of the latus rectum is 8. 37. Equation: ( x + 5 )2 = 4(− 12 )( y − 0 )

(1)

So, p = − 12 and opens down. Vertex: ( −5,0) Focus: ( −5, − 12 ) Directrix: y = 12 Latus Rectum: Connects x-values corresponding to y = − 12 . Substituting this into (1) yields: ( x + 5 )2 = 1 so that x = −5 ± 1 = −6, − 4 So, the length of the latus rectum is 2. 38. Equation: ( y − 0 )2 = 4( −4)( x + 1) (1) So, p = −4 and opens to the left. Vertex: ( −1,0) Focus: (−5,0) Directrix: x = 3 Latus Rectum: Connects y-values corresponding to x = −5 . Substituting this into (1) yields: y 2 = 64 so that y = ±8 So, the length of the latus rectum is 16.

1015

Chapter 8

39. Equation: Completing the square yields: y 2 − 4 y − 2x + 4 = 0

( y 2 − 4 y + 4) − 4 − 2x + 4 = 0 ( y − 2 )2 = 2x = 4( 12 )x (1) So, p = 12 and opens to the right. Vertex: (0,2) Focus: ( 12 , 2) Directrix: x = − 12 Latus Rectum: Connects y-values corresponding to x = 12 . Substituting this into (1) yields: ( y − 2 )2 = 1 so that y = 2 ± 1 = 1, 3 So, the length of the latus rectum is 2. 40. Equation: Completing the square yields: x 2 − 6 x + 2y + 9 = 0 ( x 2 − 6x + 9) − 9 + 2y + 9 = 0 ( x − 3)2 = −2 y = 4(− 12 ) y (1) So, p = − 12 and opens down. Vertex: (3,0) Focus: (3, − 12 ) Directrix: y = 12 Latus Rectum: Connects x-values corresponding to y = − 12 . Substituting this into (1) yields: ( x − 3)2 = 1 so that x = 3 ± 1 = 2, 4 So, the length of the latus rectum is 2.

1016

Section 8.2

41. Equation: Completing the square yields: y 2 + 2 y − 8x − 23 = 0 ( y 2 + 2 y + 1) − 1 − 8x − 23 = 0 ( y + 1)2 = 8x + 24 ( y + 1)2 = 8( x + 3) ( y + 1)2 = 4(2)( x + 3) (1) So, p = 2 and opens to the right. Vertex: ( −3, −1) Focus: (−1, −1) Directrix: x = −5 Latus Rectum: Connects y-values corresponding to x = −1 . Substituting this into (1) yields: ( y + 1)2 = 16 so that y = −1 ± 4 = 3, − 5 So, the length of the latus rectum is 8.

42. Equation: Completing the square yields: x 2 − 6 x − 4 y + 10 = 0 ( x 2 − 6 x + 9) − 9 − 4 y + 10 = 0 ( x − 3)2 = 4 y − 1 ( x − 3)2 = 4( y − 14 ) (1) So, p = 1 and opens up. Vertex: (3, 14 )

Focus: (3, 54 ) Directrix: y = − 34 Latus Rectum: Connects x-values corresponding to y = 54 . Substituting this into (1) yields: ( x − 3)2 = 4 so that x = 3 ± 2 = 1, 5 So, the length of the latus rectum is 4.

1017

Chapter 8

43. Equation: Completing the square yields: x2 − x + y − 1 = 0 ( x 2 − x + 14 ) − 14 + y − 1 = 0 ( x − 12 )2 = − y + 54 ( x − 12 )2 = −( y − 54 ) ( x − 12 )2 = 4( − 14 )( y − 54 ) (1) So, p = − 14 and opens down.

Vertex: ( 12 , 54 ) Focus: ( 12 ,1) Directrix: y = 32 Latus Rectum: Connects x-values corresponding to y = 1 . Substituting this into (1) yields: ( x − 21 )2 = 14 so that x = 12 ± 12 = 1,0 So, the length of the latus rectum is 1. 44. Equation: Completing the square yields: y2 + y − x + 1 = 0 ( y 2 + y + 14 ) − 14 − x + 1 = 0 ( y + 12 )2 = x − 34 ( y + 12 )2 = 4( 14 )( x − 34 ) (1) So, p = 14 and opens to the right.

Vertex: ( 34 , − 21 ) Focus: (1, − 12 ) Directrix: x = 12 Latus Rectum: Connects y-values corresponding to x = 1 . Substituting this into (1) yields: ( y + 12 )2 = 14 so that y = − 12 ± 12 = 0, −1 So, the length of the latus rectum is 1.

1018

Section 8.2

45. Assume the vertex is at (0,0). The distance from (0,0) to the opening of the dish is 2 feet. Identifying the opening as the latus rectum, we know that the focus will be at (0,2). So, the receiver should be placed 2 feet from the vertex. 46. Assume the vertex is at (0,0). The distance from (0,0) to the opening of the dish is 5 feet. Identifying the opening as the latus rectum, we know that the focus will be at (0,5). So, the receiver should be placed 5 feet from the vertex. 47. Assume the vertex is at the origin and that the parabola opens up. The general form of the equation, therefore, is x 2 = 4 py . Since the focus is at (0,2), p = 2 . Hence, the equation governing the shape of the lens is x 2 = 8 y , or equivalently, y = 18 x 2 . Since the latus rectum has length 5, the values of x are

restricted to the interval [ −2.5,2.5] for this model. Thus, the equation (with the

restricted domain) is:

y = 18 x 2 , for any x in [ −2.5,2.5] .

Alternatively, assuming the dish opens to the right, we obtain a parabola whose equation is the same as above, but with the roles of x and y switched, namely, x = 18 y 2 , for any y in [ −2.5,2.5] 48. Assume the vertex is at the origin and that the parabola opens up. Since the focus is at (0,3), p = 3 . So, the equation is x 2 = 4(3) y = 12 y , or equivalently, y = 121 x 2 . We need the x-values corresponding to y = 12 : x 2 = 12( 12 ) = 6 so that x = ± 6 . So, the lens is 2 6 cm wide, and is 0.5 cm from the vertex.

49. Assume the vertex is at (0,0) and that the parabola opens up. Since the rays are focused at (0, 40), we know that p = 40 . Hence, the equation is x 2 = 4(40) y = 160 y .

1019

50. Assume the vertex is at (0,0) and that the parabola opens up. Since sunlight is focused at (0, 25), we know that p = 25 . Hence, the equation is x 2 = 4(25) y = 100 y .

Chapter 8

51. From the diagram, we can assume that the vertex is at (0, 20) and that the parabola opens down. Hence, the general form of the equation is ( x − 0 )2 = 4 p( y − 20) (1) , for some p < 0 . Since the point (−40,0) is on the graph, we plug it into (1) to find p: ( −40 − 0)2 = 4 p(0 − 20) 1600 = −80 p −20 = p 2 So, the equation is ( x − 0) = 4( −20)( y − 20) which simplifies to x 2 = −80( y − 20) . Now, if the boat passes under the bridge 10 feet from the center, we need to compare the height of the bridge at x = 10 (or = −10) to the height of the boat (which is 17 feet). Indeed, from the equation, the height of the bridge at x = 10 is: (10)2 = −80( y − 20) 100 = −80 y + 1600 −1500 = −80 18.75 = y Since the mast is only 17 feet high, the boat will pass under the bridge without scraping the mast. 52. From the diagram, we can assume that the vertex is at (0, 25) and that the parabola opens down. Hence, the general form of the equation is ( x − 0 )2 = 4 p( y − 25) (1) , for some p < 0 . Since the point (10,0) is on the graph, we plug it into (1) to find p: (10 − 0)2 = 4 p(0 − 25) 100 = −100 p −1 = p 2 So, the equation is ( x − 0) = 4( −1)( y − 25) which simplifies to x 2 = −4( y − 25) . Now, if the RV straddles the center line, it will be able to pass under the bridge unhindered if the height of the bridge at x = 4 (and − 4) is bigger than 10 feet. Indeed, from the equation, the height of the bridge at x = 4 is: (4)2 = −4( y − 25) 16 = −4 y + 100 −84 = −4 y 21 = y Since the RV is only 10 feet high, it will make it under the bridge.

1020

Section 8.2

53. The focal length is 374.25 feet. Since the standard form of the equation is x 2 = 4 py , we see that p = 374.25 . So, the equation is x 2 = 1497 y . 54. From the given information, we infer that the vertex is at (0, 10) and opens up. So, the general form of the equation is ( y − 10) = 4 p( x − 0)2 (1) . Since the point ( −150,60) lies on the graph, substitute it into (1) to find p: (60 − 10) = 4 p( −150 − 0)2 50 = 4 p(22500)

p = 0.000555

Thus, the equation is ( y − 10) = 0.0022 x 2 . 55. Completing the square yields P ( x ) = −x 2 + 60 x − 500

= − ( x 2 − 60 x ) − 500

= − ( x 2 − 60 x + 900 ) − 500 + 900 = −( x − 30)2 + 400 So, the maximum profit of $400,000 is achieved when 3,000 units are produced. 56. Completing the square yields P ( x ) = −x 2 + 80 x − 1200

= − ( x 2 − 80 x ) − 1200

= − ( x 2 − 80 x + 1600 ) − 1200 + 1600 = −( x − 40)2 + 400 So, the maximum profit of $400,000 is achieved when 4,000 units are produced. 57. If the vertex is at the origin and the focus is at (3,0), then the parabola must open to the right. So, the general equation is y2 = 4 px , for some p > 0 .

58. If the vertex is at (3,2) and the focus is at (5,2), then the parabola must open to the right. So, the general equation is ( y − 2 )2 = 4 p( x − 3) , for some p > 0 .

59. True

60. False

1021

Chapter 8

61. False. It is the same distance that the 62. True. The endpoints of the latus focus is from the vertex, but on the rectum correspond to points on the opposite side. parabola determined by the segment parallel to the directrix which passes through the focus. 63. Derive the equation for x 2 = 4 py , where p > 0 . Let ( x, y) be any point on this parabola. Consult the diagram to the right. Using the distance formula, we find that

d1 = ( x − 0 )2 + ( y − p)2 d 2 = ( x − x )2 + ( y − (− p))2 = y + p Equate d1 and d 2 and simplify: ( x − 0 )2 + ( y − p)2 = y + p x 2 + ( y − p )2 = ( y + p)2 x 2 + y 2 − 2 py + p 2 = y 2 + 2 py + p 2 x 2 = 4 py,

as desired. 64. Derive the equation for y 2 = 4 px , where p > 0 . Let ( x, y) be any point on this parabola. Consult the diagram to the right. Using the distance formula, we find that

d1 = ( x − p )2 + ( y − 0)2 d 2 = ( x − (− p))2 + ( y − y)2 = x + p Equate d1 and d 2 and simplify: ( x − p )2 + y 2 = x + p ( x − p )2 + y2 = (x + p)2 x 2 − 2 px + p2 + y2 = x 2 + 2 px + p 2 y2 = 4 px,

as desired.

1022

Section 8.2

65.

66.

67. The vertex is (2.5, −3.5) . The parabola opens to the right since the square term is y 2 and p > 0 . The graph of ( y + 3.5)2 = 10( x − 2.5) is:

68. The vertex is ( −1.4, −1.7 ) . The parabola opens down since the square term is x 2 and p < 0 . The graph of (x + 1.4)2 = −5( y + 1.7) is:

69. The vertex is at (1.8, 1.5) and it should open to the left. Graphing this parabola on the calculator confirms this:

70. The vertex is at (−2.4, 3.2) and it should open up. Graphing this parabola on the calculator confirms this:

1023

Section 8.3 Solutions -------------------------------------------------------------------------------1. d x-axis is the major axis. The intercepts are (6,0), (−6,0), (0, 4), (0, −4) .

2. b y-axis is the major axis. The intercepts are (4,0), (−4,0), (0,6), (0, −6) .

3. a y-axis is the major axis. The intercepts are (2 2,0), (−2 2,0), (0,6 2 ), (0, −6 2 ) .

4. c x-axis is the major axis. The intercepts are ( 12 ,0), ( − 12 ,0), (0,1), (0, −1) .

Center: (0,0) Vertices: ( ±5,0 ) , ( 0, ±4 )

Center: (0,0) Vertices: ( ±7,0 ) , ( 0, ±3 )

Center: (0,0) Vertices: ( ±4,0 ) , ( 0, ±8 )

Center: (0,0) Vertices: ( ±5,0 ) , ( 0, ±12 )

1024

Section 8.3

Center: (0,0) Vertices: ( ±10,0 ) , ( 0, ±1)

11.

10. Write the equation in standard 2 2 form as x4 + y9 = 1 .

Center: (0,0) Vertices: ( ±2,0 ) , ( 0, ±3 ) 12. Write the equation in standard 2 2 form as x25 + y9 = 1 . 4

Center: (0,0) Vertices: ( ± 32 ,0 ) , ( 0, ± 19 )

100

Center: (0,0) Vertices: ( ± 52 ,0 ) , ( 0, ± 103 )

1025

Chapter 8

13. Write the equation in standard 2 2 form as x4 + 1y6 = 1 .

14.

Center: (0,0) Vertices: ( ±9,0 ) , ( 0, ±9 )

Center: (0,0) Vertices: ( ±2,0 ) , ( 0, ±4 ) 15. Write the equation in standard 2 2 form as x4 + y2 = 1 .

16. Write the equation in standard 2 2 form as x5 + y2 = 1 .

Center: (0,0) Vertices: ( ±2,0 ) , 0, ± 2

Center: (0,0) Vertices: ± 5,0 , 0, ± 2

(

)

(

1026

)(

)

Section 8.3

17. Since the foci are (−4,0) and (4,0) , we know that c = 4 and the center of the ellipse is (0,0) . Further, since the vertices are (−6,0) and (6,0) and they lie on the x-axis, the ellipse is horizontal and a = 6 . Hence, c 2 = a 2 − b2

18. Since the foci are (−1,0) and (1,0) , we know that c = 1 and the center of the ellipse is (0,0) . Further, since the vertices are (−3,0) and (3,0) they lie on the x-axis, the ellipse is horizontal and so, a = 3 . Hence, c 2 = a 2 − b2

16 = 36 − b2

1 = 9 − b2

b2 = 20 Therefore, the equation of the ellipse is

b2 = 8 Therefore, the equation of the ellipse is

x2 36

y + 20 =1 .

x2 9

+ y8 = 1 .

19. Since the foci are (0, −3) and (0,3) , we know that c = 3 and the center of the ellipse is (0,0) . Further, since the vertices are (0, −4) and (0, 4) and they lie on the y-axis, the ellipse is vertical and a = 4 . Hence, c 2 = a 2 − b2

20. Since the foci are (0, −1) and (0,1) , we know that c = 1 and the center of the ellipse is (0,0) . Further, since the vertices are (0, −2) and (0,2) and they lie on the y-axis, the ellipse is vertical and a = 2 . Hence, c 2 = a 2 − b2

9 = 16 − b 2

1 = 4 − b2

b2 = 7 Therefore, the equation of the ellipse is

b2 = 3 Therefore, the equation of the ellipse is

x2 7

+ 1y6 = 1 .

x2 3

21. Since the ellipse is centered at (0,0) , the length of the vertical major axis being 8 implies that a = 4 , and the length of the horizontal minor axis being 4 implies that b = 2 . Since the major axis is vertical, the equation of

the ellipse must be

x2 4

+ 1y6 = 1 .

+ y4 = 1 .

22. Since the ellipse is centered at (0,0) , the length of the horizontal major axis being 10 implies that a = 5 , and the length of the vertical minor axis being 2 implies that b = 1 . Since the major axis is horizontal, the equation of the ellipse

1027

x2 25

+ y2 = 1 .

Chapter 8

23. Since the vertices are (0,−7), (0,7) , we know that the ellipse is centered at (0,0) (since it is halfway between the vertices), the major axis is vertical, and a = 7 . Further, since the endpoints of the minor axis are ( −3,0), (3,0) , we know that b = 3 . Thus, the equation of

the ellipse is

x2 9

+ 4y9 = 1 .

24. Since the vertices are ( −9,0), (9,0) , we know that the ellipse is centered at (0,0) (since it is halfway between the vertices), the major axis is horizontal, and a = 9 . Further, since the endpoints of the minor axis are (0, −4), (0, 4) , we know that b = 4 . Thus, the equation of

the ellipse is

x2 81

y + 16 =1 .

25. c Center is (3, −2) and the major axis is vertical.

26. d Center is ( −3,2) and the major axis is vertical.

27. b Center is (3, −2) and the major axis is horizontal.

28. a Center is ( −3,2) and the major axis is horizontal.

29.

30.

Center: (1,2) Vertices: (−3,2), (5,2), (1,0), (1,4)

Center: (−1,−2) Vertices: (−7, −2), (5, −2), (−1,1), (−1,−5)

1028

Section 8.3

31. First, write the equation in

32. First, write the equation in standard

( x − 3) + ( y + 1) = 1 standard form: 2

(x − 2) + ( y + 2) = 1 form: 2

Center: ( 2, −2 )

Vertices: ( −5, −2 ) , ( 9, −2 ) , ( 2,1) , ( 2, −5 )

Center: ( 3, −1)

Vertices: (1, −1) , ( 5, −1) , ( 3,5 ) , ( 3, −7 ) 33. First, write the equation in 2 2 standard form: ( x +83 ) + ( y8−04 ) = 1

34. First, write the equation in standard 2 + 3 )2 form: ( x12 + ( y −34 ) = 1

Center: (−3,4) −2 2 − 3, 4 , 2 2 − 3, 4 , Vertices: −3, 4 + 4 5 , −3, 4 − 4 5

Center: (−3,4) −2 3 − 3, 4 , 2 3 − 3, 4 , Vertices: −3, 4 − 3 , −3, 4 + 3

( (

)( )(

)

( (

1029

)( )(

)

Chapter 8

35. First, write the equation in standard form by completing the square: x 2 + 4( y 2 − 6y) = −32

36. First, write the equation in standard form by completing the square: 25x 2 + 2( y 2 − 2y) = 48 25x 2 + 2( y2 − 2y + 1) = 48 + 2

x 2 + 4( y2 − 6y + 9) = −32 + 36

x 2 + 4( y − 3)2 = 4  x4 + ( y − 3)2 = 1

Center: (0,1) Vertices: ± 2,1 , (0, −4), (0,6)

(

Center: (0,3) Vertices: (−2,3), (2,3), (0,2), (0,4) 37. First, write the equation in standard form by completing the square: ( x 2 − 2 x) + 2( y 2 − 2y) = 5

)

38. First, write the equation in standard form by completing the square: 9(x 2 − 2x ) + 4y2 = 27

( x 2 − 2 x +1) + 2( y 2 − 2y +1) = 5 +1+ 2

9(x 2 − 2x + 1) + 4y2 = 27 + 9

( x − 1)2 + 2( y −1)2 = 8

9(x − 1)2 + 4 y2 = 36

( x −1)2 8

(

)

( x −1)2 4

+ ( y −41) = 1

Center: (1,1) Vertices: 1 ± 2 2,1 , (1,3), (1, −1)

25x 2 + 2( y − 1)2 = 50  x2 + ( y2−51) = 1

+ y9 = 1

Center: (1,0) Vertices: (−1,0), (3,0), (1,3), (1,−3)

1030

Section 8.3

39. First, write the equation in standard form by completing the square: 5(x 2 + 4x ) + ( y 2 + 6y) = 21

40. First, write the equation in standard form by completing the square: 9(x 2 + 4 x ) + ( y 2 + 2y) = −36

5(x 2 + 4x + 4) + ( y 2 + 6y + 9) = 21+ 20 + 9

9(x 2 + 4 x + 4) + ( y2 + 2y +1) = −36 + 36 + 1 9(x + 2)2 + ( y +1)2 = 1

5(x + 2)2 + ( y + 3)2 = 50 ( x+2 ) 10

1 9

+ 3) + ( y50 =1

Center: (−2,−3) Vertices: −2 ± 10, −3 , −2, −3 ± 5 2

(

( x+ 2 )2

)(

)

+ ( y + 1)2 = 1

Center: (−2,−1) Vertices: ( − 73 , −1) , ( − 53 , −1) , (−2,0), (−2, −2 )

41. Since the foci are ( −2,5), (6,5) and the vertices are (−3,5), (7,5) , we know: i) the ellipse is horizontal with center at ( −32+7 , 5) = (2,5) , ii) 2 − c = −2 so that c = 4 , iii) the length of the major axis is 7 − ( −3) = 10 . Hence, a = 5 . Now, since c 2 = a 2 − b2 , 16 = 25 − b 2 so that b 2 = 9 . Hence, the equation of the 2

−2 ) + ( y −95) = 1 . ellipse is ( x 25

42. Since the foci are (2, −2), (4, −2) and the vertices are (0, −2), (6, −2) , we know: i) the ellipse is horizontal with center at ( 0 +2 6 , −2) = (3, −2) , ii) 3 − c = 2 so that c = 1 , iii) the length of the major axis is 6 − 0 = 6 . Hence, a = 3 . Now, since c 2 = a 2 − b2 , 1 = 9 − b2 so that b2 = 8 . Hence, the equation of the ellipse 2

is ( x −93 ) + ( y +82 ) = 1 .

1031

Chapter 8

43. Since the foci are (4, −7), (4, −1) and the vertices are (4, −8), (4,0) , we know: i) the ellipse is vertical with center at (4, −82+ 0 ) = (4, −4) , ii) −4 − c = −7 so that c = 3 , iii) the length of the major axis is 0 − ( −8) = 8 . Hence, a = 4 . Now, since c 2 = a 2 − b 2 , 9 = 16 − b2 so that b 2 = 7 . Hence, the equation of the 2

ellipse is ( x −74 ) + ( y1+64 ) = 1 .

44. Since the foci are (2, −6), (2, −4) and the vertices are (2, −7), (2, −3) , we know: i) the ellipse is vertical with center at (2, −7 +2( −3) ) = (2, −5) , ii) −5 − c = −6 so that c = 1 , iii) the length of the major axis is −3 − ( −7) = 4 . Hence, a = 2 . Now, since c 2 = a 2 − b2 , 1 = 4 − b2 so that b2 = 3 . Hence, the equation of the ellipse 2

is ( x −32 ) + ( y +45) = 1 . 45. Since the length of the major axis is 8, we know that a = 4 ; since the length of the minor axis is 4, we know that b = 2 . As such, since the center is (3,2) and the 2

major axis is vertical, the equation of the ellipse must be ( x −43 ) + ( y1−62 ) = 1 . 46. Since the length of the major axis is 10, we know that a = 5 ; since the length of the minor axis is 2, we know that b = 1 . As such, since the center is (−4,3) and the 2

major axis is horizontal, the equation of the ellipse must be ( x+254 ) + ( y − 3)2 = 1 . 47. Since the vertices are (−1, −9), (−1,1) , we know that: i) the ellipse is vertical with center at ( −1, −92+1 ) = ( −1, −4) ,

ii) a = 1−(2−9) = 5 . Now, since the endpoints of the minor axis are (−4, −4), (2,−4) , we know that 2

b = 2 −(2−4 ) = 3 . Hence, the equation of the ellipse is ( x +91) + ( y +254 ) = 1 .

1032

Section 8.3

48. Since the vertices are (−2,3), (6,3) , we know that: i) the ellipse is horizontal with center at ( −22+6 , 3) = (2,3) ,

ii) a = 6 −(2−2 ) = 4 . Now, since the endpoints of the minor axis are (2,1), (2,5) , we know that 2

−2 ) b = 52−1 = 2 . Hence, the equation of the ellipse is ( x16 + ( y −43) = 1 .

49. Since the major axis has length 150 feet, we know that a = 75 ; since the minor axis has length 30 feet, we know that b = 15 . From the diagram we know that the y-axis is the major axis. So, since the ellipse is centered at the y2 x2 origin, its equation must be 225 + 5625 =1

50. Since both the major and minor axes have length 180 feet, we know that a = b = 90 . So, since the ellipse is centered at the origin, its equation must y2 x2 + 8100 = 1 , which simplifies to be 8100 x 2 + y2 = 8100 (a circle).

51. a. Since the major axis has length 150 yards, we know that a = 75 ; since the minor axis has length 40 yards, we know that b = 20 . So, since the ellipse is 2

y x + 400 =1. centered at the origin, its equation must be 5625 b. Since the football field is 120 yards long and it is completely surrounded by an elliptical track, we can compute the equation in part a. at x = 60 to find the corresponding y-values, and then subtract them to find the width at that particular position in the field: 400(2025) y2 = 400(1 − 3600 = 144 so that y = ±12 . 5625 ) = 5625 Hence, the width of the track at the end of the football field is 24 yards. Since the field itself is 30 yards wide, the track will NOT encompass the field. 2

52. Here we need to find the value of b so that when the equation of the track, 2 x2 + by2 = 1 , is evaluated at x = 60 , the difference in the corresponding namely 5625

y-values is at least 30. To find the smallest such value of b, we must solve: 602 302 5625 + b2 = 1 900 b2

2025 = 5625

) = 2500 b2 = 5625(900 2025

b = 50 Hence, the minor axis should be at least 100 yards wide in order to enclose the field.

1033

Chapter 8

53. Assume that the sun is at the origin. Then, the vertices of Pluto’s horizontal elliptical trajectory are: ( −4.447 ×108 ,0), (7.38 ×108 ,0) From this, we know that: i) The length of the major axis is 7.38 ×108 − ( −4.447 ×108 ) = 11.827 ×108 , and so the value of a is half of this, namely 5.9135 ×108 ;

ii) The center of the ellipse is

(

7.38×108 + ( −4.447×108 ) 2

)

,0 = (1.4665 × 108 ,0) ;

iii) The value of c is 1.4665×108 .

Now, since c 2 = a 2 − b2 , we have (1.4665 ×108 ) = ( 5.9135 ×108 ) − b 2 , so that 2

b2 = 3.2818 ×1017 . So, the equation of the ellipse must be

( x −(1.4665×10 )) 8

3.4969×10

( ) + 3.2818 =1 . ×1017 y −0

(Note: There is more than one correct way to set up this problem, and it begins with choosing the center of the ellipse. If, alternatively, you choose the center to be at (0,0), then the actual coordinates of the vertices and foci will change, and will result in the following slightly different equation for the trajectory: y2 x2 + 5,729,000,000 2 =1 5,914,000,0002 Both are equally correct!.) 54. Assume that the sun is at the origin. Then, the vertices of Earth’s horizontal elliptical trajectory are: ( −1.471×108 ,0), (1.526 ×108 ,0) From this, we know that: i) The length of the major axis is 1.526 ×108 − ( −1.471×108 ) = 2.997 ×108 , and so the value of a is half of this, namely 1.4985×108 ;

ii) The center of the ellipse is

(

1.526×108 + ( −1.471×108 ) 2

)

,0 = (0.0275 ×108 ,0) = (2.75×106 ,0) ;

iii) The value of c is 2.75×106 .

Now, since c 2 = a 2 − b2 , we have ( 2.75 ×106 ) = (1.4985 ×108 ) − b2 , so that 2

b = 2.245 ×10 . So, the equation of the ellipse must be 2

1034

( x −( 2.75×10 )) 6

2.246×10

( ) + 2.245 =1 . ×1016 y −0

Section 8.3

55. The length of the semi-major axis = a = 150 ×106 , so that a 2 = 2.25 ×1016 . Observe that the formula for eccentricity can be rewritten as follows: 2 e 2 = 1 − ab2 b2 a2

= 1 − e2

b 2 = a 2 (1 − e 2 ) b = a 1 − e2 Since e = 0.223 and (from above) a = 150,000,000, we find that b = 146,000,000. 2

y x =1 . Hence, the equation of the ellipse is 150,000,000 2 + 146,000,0002 2

56. The length of the semi-major axis = a = 350 ×106 , so that a 2 = 1.225 ×1017 . Observe that 0.634 = 1 − 1.225b×1017 2

×10 − b 0.40196 = 1.225 1.225×1017 17

4.9239 ×1016 = 1.225 ×1017 − b 2 7.326 ×1016 = b 2 2

Hence, the equation of the ellipse is 1.225x×1017 + 7.326y×1016 = 1 . 2

57. It approaches the graph of a straight line. Indeed, note that b2 b2 b2  1 − = 1  = 0  b =0. a2 a2 a2 Hence, a 2 x 2 + b 2 y 2 = a 2 b 2 becomes a 2 x 2 = 0 , and since a ≠ 0 , we conclude that x=0. 1 = 1−

58. The length of the semi-major axis is a = 17.8 AU, so that a 2 = 316.84 . Since 2 we also know that e ≈ 0.967 , we can use e 2 = 1 − ab2 to find b: b (0.967)2 = 1 − 316.84  b ≈ 20.56640 ≈ 4.53475 AU . 2

x2 y2 + = 1. So, the equation of the comet (using AU units) is 316.84 20.5664

1035

Chapter 8

59. a. Assuming the center of the trainer is located at (0,0), we must identify a 2 2 and b such that the ellipse traced by the pedals is of the form xa2 + by2 = 1.

The x-intercepts are ± a, and the distance between them (namely a – (−a) = 2a) should be 16. So, 2a = 16, so that a = 8. The minimum and maximum step heights occur at 2.5 and 12.5, respectively. So, the distance between them is 12.5 – 2.5 = 10. Hence, 2b = 10, so that b = 5. 2 y2 = 1. Thus, the equation of the ellipse is x64 + 25 b. p = π 2(64 + 25) ≈ 42 inches c. 1 step = 42 inches, and 1 mile = 5280 feet = 5280(12) inches = 63,360 inches. So, the number of steps for 1 mile is 63,360 42 ≈ 1,509 steps. 60. a. Assuming the center of the trainer is located at (0,0), we must identify a 2 2 and b such that the ellipse traced by the pedals is of the form xa2 + by2 = 1.

The x-intercepts are ± a, and the distance between them (namely a – (−a) = 2a) should be 18. So, 2a = 18, so that a = 9. The minimum and maximum step heights occur at 3.5 and 13.5, respectively. So, the distance between them is 13.5 – 3.5 = 10. Hence, 2b = 10, so that b = 5. 2 y2 = 1. Thus, the equation of the ellipse is x81 + 25 b. p = π 2(81 + 25) ≈ 46 inches c. 1 step = 46 inches, and 1 mile = 5280 feet = 5280(12) inches = 63,360 inches. So, the number of steps for 1 mile is 63,360 46 ≈ 1,377 steps.

a = ± 6 , b = ±2.

62. It should be c 2 = a 2 − b2 in place of c 2 = a 2 + b2 .

63. False. Since you could not deduce the coordinates of the endpoints of the minor axis from this information alone. Hence, you couldn’t determine the value of b.

64. True. The information provided enable you to determine the values of b and c. Then, using this information, the center and value of a can be found.

65. True. The equation would have 2 2 the general form xa2 + by2 = 1, so that

66. False. All circles are ellipses (with a = b ), but not all ellipses are circles.

61. It should be a 2 = 6, b 2 = 4, so that

substituting in −x for x and −y for y yields the exact same equation.

1036

Section 8.3

≅ 0.25 (from Problem 51) 67. Pluto: e = 1.4665×10 5.9135×108 8

2.75×10 Earth: e = 1.4985 ≅ 0.02 (from Problem 52) ×108 6

68. a. If e is close to 0, then c is close to 0. So, since a 2 − b 2 = c 2 , this would imply that a is close to b. So, the ellipse is nearly circular. b. If e is close to 1, then c is close to a. As such, using this fact, together with c 2 = a 2 − b 2 , we see that b 2 = a 2 − c 2 would be close to 0. So, the ellipse would be very narrow. c. If e = 12 , then ac = 12 , so that a = 2c . Then, since c 2 = a 2 − b2 ,

c 2 = 4c 2 − b2 b2 = 3c 2 so that b = 3 c 69. Note that from the graphs, we see that as c increases, the graph of the ellipse described by x 2 + cy 2 = 1 narrows in the y-direction and becomes more elongated.

70. Note that from the graphs, we see that as c increases, the graph of the ellipse described by cx 2 + y 2 = 1 narrows in the x-direction and becomes more elongated.

1037

Chapter 8

71. Note that from the graphs, we see that as c increases, the graph of the ellipse described by cx 2 + cy2 = 1 , which is equivalent to x 2 + y2 = 1c (a circle) shrinks towards the origin. (This makes sense since the radius is 1c , and as c gets larger, the value of 1c goes to 0.)

73. Note that from the graphs, we see that as c decreases, the major axis (along the x-axis) of the ellipse described by cx 2 + y 2 = 1 becomes longer.

72. This is not the graph of an ellipse. Rather, it is the graph of a hyperbola – see Section 8.4.

74. Note that from the graphs, we see that as c decreases, the major axis (along the y-axis) of the ellipse described by x 2 + cy 2 = 1 becomes longer.

1038

Section 8.4 Solutions -------------------------------------------------------------------------------1. b The transverse axis is the x-axis, so that the hyperbola will open left/right. The vertices are (−6,0), (6,0) .

2. a The transverse axis is the y-axis, so that the hyperbola will open up/down. The vertices are (0,−6), (0,6) .

3. d The transverse axis is the x-axis, so that the hyperbola will open left/right. The vertices are ( −2 2,0), (2 2,0) .

4. c The transverse axis is the y-axis, so that the hyperbola will open up/down. The vertices are (0, 12 ), (0, − 12 ) .

Notes on the Graph: Equations of the asymptotes are y = ± 54 x .

Notes on the Graph: Equations of the asymptotes are y = ± 73 x .

Notes on the Graph: Equations of the asymptotes are y = ± 12 x .

Notes on the Graph: Equations of the asymptotes are y = ± 125 x .

1039

Chapter 8

10.

Notes on the Graph: Equations of the asymptotes are y = ± 101 x .

Notes on the Graph: First, write the equation in standard 2 2 form as y4 − x9 = 1 . The equations of the asymptotes are y = ± 23 x .

11.

12.

Notes on the Graph: First, write the equation in standard 2 2 form as y9 − x1 = 1 . The equations of

Notes on the Graph: First, write the equation in standard 2 2 form as x25 − y9 = 1 . The equations of the

the asymptotes are y = ±

asymptotes are y = ± 253 x .

27 2

1040

100

Section 8.4

13.

14.

Notes on the Graph: First, write the equation in standard 2 2 form as x4 − 1y6 = 1 . The equations of the asymptotes are y = ±2x .

Notes on the Graph: First, write the equation in standard 2 2 form as y81 − x81 = 1 . The equations of the asymptotes are y = ± x .

15.

16.

Notes on the Graph: First, write the equation in standard 2 2 form as y4 − x2 = 1 . The equations of the asymptotes are y = ± 22 x .

Notes on the Graph: First, write the equation in standard 2 2 form as x5 − y2 = 1 . The equations of the asymptotes are y = ±

1041

2 5

x = ± 510 x .

Chapter 8

17. Since the foci are (−6,0), (6,0) and the vertices are (−4,0), (4,0) , we know that: i) the hyperbola opens right/left with center at (0,0) , ii) a = 4, c = 6 , Now, since c 2 = a 2 + b2 , 36 = 16 + b 2 so that b 2 = 20 . Hence, the equation of the 2

y x hyperbola is 16 − 20 =1 . 2

18. Since the foci are ( −3,0), (3,0 ) and the vertices are ( −1,0), (1,0) , we know that: i) the hyperbola opens right/left with center at (0,0) , ii) a = 1, c = 3 , Now, since c 2 = a 2 + b2 , 9 = 1 + b 2 so that b 2 = 8 . Hence, the equation of the 2

hyperbola is x 2 − y8 = 1 . 19. Since the foci are (0, −4), (0, 4) and the vertices are (0, −3), (0,3) , we know that: i) the hyperbola opens up/down with center at (0,0) , ii) a = 3, c = 4 , Now, since c 2 = a 2 + b2 , 16 = 9 + b 2 so that b 2 = 7 . Hence, the equation of the

hyperbola is

y2 9

− x7 = 1 . 2

20. Since the foci are (0, −2), (0,2) and the vertices are (0, −1), (0,1) , we know that: i) the hyperbola opens up/down with center at (0,0) , ii) a = 1, c = 2 , Now, since c 2 = a 2 + b2 , 4 = 1 + b2 so that b2 = 3 . Hence, the equation of the

hyperbola is y 2 − x3 = 1 . 2

21. Since the center is (0,0) and the transverse axis is the x-axis, the general form 2

of the equation of this hyperbola is xa2 − by2 = 1 with asymptotes y = ± ab x . In this 2

case, ba = 1 so that a = b . Thus, the equation simplifies to xa2 − ay2 = 1 , or 2

equivalently x 2 − y 2 = a 2 .

1042

Section 8.4

22. Since the center is (0,0) and the transverse axis is the y-axis, the general form 2

of the equation of this hyperbola is ay2 − xb2 = 1 with asymptotes y = ± ab x . In this 2

case, ab = 1 so that a = b . Thus, the equation simplifies to ay2 − xa2 = 1 , or 2

equivalently y 2 − x 2 = a 2 . 23. Since the center is (0,0) and the transverse axis is the y-axis, the general form 2

of the equation of this hyperbola is ay2 − xb2 = 1 with asymptotes y = ± ab x . In this 2

case, ab = 2 so that a = 2b . Thus, the equation simplifies to (2yb )2 − xb2 = 1 , or equivalently

y2 4

− x 2 = b2 .

24. Since the center is (0,0) and the transverse axis is the x-axis, the general form 2

of the equation of this hyperbola is xa2 − by2 = 1 with asymptotes y = ± ab x . In this 2

case, ba = 2 so that 2a = b . Thus, the equation simplifies to xa2 − (2ya )2 = 1 , or 2

equivalently x 2 − y4 = a 2 . 25. c The transverse axis is parallel to the x-axis, so that the hyperbola will open left/right. The vertices are (1, −2 ), (5, −2) and the center is (3, −2) .

26. d The transverse axis is parallel to the x-axis, so that the hyperbola will open left/right. The vertices are (−5,2), ( −1,2) and the center is ( −3,2) .

27. b The transverse axis is parallel to the y-axis, so that the hyperbola will open up/down. The vertices are ( −2,8), ( −2, −2 ) and the center is ( −2,3) .

28. a The transverse axis is parallel to the y-axis, so that the hyperbola will open up/down. The vertices are (2, −8), (2,2) and the center is (2, −3) .

1043

Chapter 8

29.

30.

Notes on the Graph: The equations of the asymptotes are y = ± 12 ( x − 1) + 2 .

Notes on the Graph: The equations of the asymptotes are y = ±2( x + 2) −1 .

31.

32.

Notes on the Graph: The equations of the asymptotes are y = ±3 ( x − 3 ) − 1 .

Notes on the Graph: The equations of the asymptotes are 7 y = ± ( x + 2) + 2 . 3

1044

Section 8.4

33.

Notes on the Graph: The equations of the asymptotes are y = ± 101 ( x − 4) − 3 = ± 1010 (x − 4) − 3 . 35.

34.

Notes on the Graph: The equations of the asymptotes are y = ± 12 ( x + 3) + 4 .

Notes on the Graph: First, write the equation in standard form by completing the square: x2 − 4x − 4 y2 = 0

( x − 4x + 4 ) − 4 y = 0 + 4 2

( x − 2 )2 − 4 y 2 = 4 ( x − 2 )2 4

− y2 = 1 The equations of the asymptotes are y = ± 12 ( x − 2 ) .

1045

Chapter 8

36.

Notes on the Graph: First, write the equation in standard form by completing the square: −9x 2 + y2 + 2 y = 8 −9x 2 + ( y 2 + 2 y +1) = 8 + 1 −9x 2 + ( y + 1) = 9 2

−x 2 + ( 9 ) = 1 y +1

( y +1)2

− x2 = 1 The equations of the asymptotes are y = ±3x − 1 . 9

37.

Notes on the Graph: First, write the equation in standard form by completing the square: −9 ( x 2 + 2x ) + 4 ( y2 − 2y ) = 41

−9 ( x 2 + 2 x +1) + 4 ( y2 − 2y +1) = 41− 9 + 4 −9(x + 1)2 + 4( y −1)2 = 36 2

( y −1)2 9

− ( x +41) + ( y −91) = 1 − ( x +41) = 1 The equations of the asymptotes are y = ± 32 ( x + 1) + 1 .

1046

Section 8.4

38.

Notes on the Graph: First, write the equation in standard form by completing the square: 25 ( x 2 − 2 x ) − 4 ( y 2 + 2 y ) = 79

25 ( x 2 − 2 x + 1) − 4 ( y2 + 2 y + 1) = 79 + 25 − 4 25( x − 1)2 − 4( y + 1)2 = 100 ( x −1)2 4

− ( y2+51) = 1 The equations of the asymptotes are y = ± 52 ( x − 1) − 1 .

39.

Notes on the Graph: First, write the equation in standard form by completing the square: ( x 2 − 6x ) − 4 ( y2 + 4 y ) = 8

( x − 6x + 9 ) − 4 ( y + 4 y + 4 ) = 8 + 9 − 16 2

( x − 3)2 − 4( y + 2)2 = 1 2

( x − 3)2 − ( y +12 ) = 1 4

The equations of the asymptotes are y = ± 12 ( x − 3) − 2 . 40.

Notes on the Graph: First, write the equation in standard form by completing the square: −4 ( x 2 + 4x ) + ( y 2 − 2y ) = 19 −4 ( x 2 + 4x + 4 ) + ( y2 − 2y +1) = 19 −16 + 1 −4(x + 2)2 + ( y −1)2 = 4 2

−( x + 2 )2 + ( y−41) = 1 ( y−1)2 4

− ( x + 2 )2 = 1 The equations of the asymptotes are y = ±2( x + 2) +1 .

1047

Chapter 8

41. Since the vertices are (−2,5), (6,5) , we know that the center is (2,5) and the transverse axis is parallel to the x-axis. Hence, 2 − a = −2 so that a = 4 . Also, since the foci are (−3,5), (7,5) , we know that 2 + c = −3 so that c = −5 . Now, to find b, we substitute the values of c and a obtained above into c 2 = a 2 + b2 to see that 2

−2 ) − ( y −95) = 1 . 25 = 16 + b2 so that b 2 = 9 . Hence, the equation of the hyperbola is ( x16

42. Since the vertices are (1, −2), (3, −2) , we know that the center is (2, −2) and the transverse axis is parallel to the x-axis. Hence, 2 − a = 1 so that a = 1 . Also, since the foci are (0, −2), (4, −2) , we know that 2 + c = 0 so that c = −2 . Now, to find b, we substitute the values of c and a obtained above into c 2 = a 2 + b2 to see that 4 = 1 + b2 so that b 2 = 3 . Hence, the equation of the hyperbola is 2

( x − 2 )2 − ( y+32 ) = 1 .

43. Since the vertices are (4, −7), (4, −1) , we know that the center is (4, −4) and the transverse axis is parallel to the y-axis. Hence, −4 − a = −7 so that a = 3 . Also, since the foci are (4, −8), (4,0) , we know that −4 − c = −8 so that c = 4 . Now, to find b, we substitute the values of c and a obtained above into c 2 = a 2 + b 2 to see that 16 = 9 + b2 so that b2 = 7 . Hence, the equation of the hyperbola is ( y + 4 )2 9

− ( x −74 ) = 1 .

44. Since the vertices are (2, −6), (2, −4) , we know that the center is (2, −5) and the transverse axis is parallel to the y-axis. Hence, −5 − a = −6 so that a = 1 . Also, since the foci are (2, −7), (2, −3) , we know that −5 − c = −7 so that c = 2 . Now, to find b, we substitute the values of c and a obtained above into c 2 = a 2 + b2 to see that 4 = 1 + b2 so that b 2 = 3 . Hence, the equation of the hyperbola is 2

( y + 5 )2 − ( x−32 ) = 1 .

45. Assume that the stations coincide with the foci and are located at ( −75,0), (75,0) . The difference in distance between the ship and each of the two stations must remain constantly 2a, where (a,0) is the vertex. Assume that the radio signal speed is 186,000 mi sec and the time difference is 0.0005 sec . Then, using distance = rate × time, we obtain: 2a = (186,000)(0.0005) = 93 so that a = 46.5 So, the ship will come ashore between the two stations 28.5 miles from one and 121.5 miles from the other. 1048

Section 8.4

46. Assume that the stations coincide with the foci and are located at (−150,0), (150,0) . The difference in distance between the ship and each of the two stations must remain constantly 2a, where (a,0) is the vertex. Assume that the radio signal speed is 186,000 mi sec and the time difference is 0.0007 sec . Then, using distance = rate × time, we obtain: 2a = (186,000)(0.0007) = 130.2 so that a = 65.1 So, the ship will come ashore between the two stations 84.9 miles from one and 215.1 miles from the other. 47. Here, we want a = 45 . So, 2a = 90 . Using the same radio speed, observe that 90 = 186,000(t) , so that t = 0.000484 sec .

48. Here, we want a = 100 . So, 2a = 200 . Using the same radio speed, observe that 200 = 186,000(t) , so that t = 0.00107 sec .

49. Assume that the vertices are (0,a), (0, −a ) . Then, 2a = 2 so that a = 1 , and the center is at (0,0). Further, the transverse axis is parallel to the y-axis. Since the foci are (0,1.5), (0, −1.5) , we know that c = 1.5 . Thus, to find b, we substitute these values of a and c into c 2 = a 2 + b 2 to see that (1.5)2 = 12 + b2 , and so b 2 = 1.25 . Hence, the x = 1 , which is equivalent to y 2 − 54 x 2 = 1 . equation of the hyperbola is y 2 − 1.25 2

50. Consider the following diagram:

We must find the value of y such that if we were to place the explosive at the point (0, y) , it would be heard at Alpha and Bravo at the same times as such an explosion would be heard if placed at (1200,0). To do so, we see that Time heard at Bravo: distance from (0, y) to B tBII = r

3000 2 + y 2 r Time heard at Alpha: Since distance = rate × time, we have: distance from (0, y) to A y Time heard at Bravo: tAII = = r r distance from (1200,0) to B 2700 I tB = = I II Hence, we solve tA = tA : r r y 300 Time heard at Alpha: r = r so that y = 300 . distance from (1200,0) to A 300 Thus, the explosive should be placed at tAI = = r r (0,300). =

1049

Chapter 8

51. Find the x values when y = 150 (which correspond to the top of the cooling tower): x2 150 2 − =1 8,100 16,900 x2 22,500 39, 400 = 1+ = 8,100 16,900 16,900  39, 400  x 2 = 8,100   ≈ 18,884.02367  16,900  x = ± 18,884.02367 ≈ ±137.4 So, the diameter is 137.4 – (−137.4) = 275 feet.

52. Find the x values when y = −300 (which correspond to the bottom of the cooling tower): x2 (−300)2 − =1 8,100 16,900 x2 90,000 106,900 = 1+ = 8,100 16,900 16,900  106, 900  x 2 = 8,100   ≈ 51,236.09467  16,900  x = ± 51,236.09467 ≈ ±226.35 So, the diameter is 226.35 – (−226.35) = 453 feet.

53. The transverse axis should be vertical. The points are (3,0), ( −3,0) and the vertices are (0,2), (0, −2) .

54. Here, a = 1, b = 2 . So, the vertices are (1,0), ( −1,0) and the points are (0,2), (0, −2) .

55. False. You won’t be able to find the value of b in such case without more information.

56. True. Since the foci and vertices yield the values of a and c, you can find b using c 2 = a 2 + b2 .

57. True. Since the general forms of 2 2 the equations are xa2 − by2 = 1 and

58. False.

y2 a2

− xb2 = 1 , substituting −x for x and 2

−y for y doesn’t change the equation – hence, the symmetry follows. 1050

Section 8.4

59. Assume the center of the hyperbola is (0,0). Assume that the two asymptotes are perpendicular. We separate our discussion into two cases, depending on which axis is the transverse axis. Case 1: Here, the equations of the asymptotes are y = ± ab x . Hence, in order for them to be perpendicular, the products of their slopes must be −1 . So, we have 2 ( ab ) ( − ab ) = −1 , which simplifies to − ab2 = −1 so that a 2 = b2 . Hence, in such case, the 2

equation is ay2 − xa2 = 1 , which is equivalent to y 2 − x 2 = a 2 . 2

Case 2: Here, the equations of the asymptotes are y = ± ab x . Hence, in order for them to be perpendicular, the products of their slopes must be −1 . So, we have 2 ( ba ) ( − ab ) = −1 , which simplifies to − ab2 = −1 so that a 2 = b2 . Hence, in such case, the 2

equation is xa2 − ay2 = 1 , which is equivalent to x 2 − y 2 = a 2 . 2

60. Since the vertices are (3, −2), ( −1, −2) , and the center is the midpoint of the segment connecting them, the center must be (1, −2) . Moreover, since the form of the vertices is ( h ± a, k ) ( which in this case is (1 ± a, −2) ) we see that 1 + a = 3 so that a = 2 . It remains to find b. At this point, we need to use the fact that we are given that the equations of the asymptotes are y = 2 x − 4, y = −2 x . Since the transverse axis is parallel to the x-axis, we know that the slopes of the asymptotes for such a hyperbola are ± ab . Thus, ± ab = ±2 . Since a = 2 , this implies 2

that b = ±4 . Hence, the equation of the hyperbola must be ( x −41) − ( y1+62 ) = 1 . 61. Observe from the graphs that as c increases, the graphs of the hyperbolas described by x 2 − cy 2 = 1 become more and more squeezed down towards the x-axis.

1051

Chapter 8

62. Observe from the graphs that as c increases, the vertices of the graphs of the hyperbolas described by cx 2 − y 2 = 1 get closer to the origin, thereby causing the hyperbolas to rise more and more steeply.

63. Note from the graphs that as c decreases, the vertices of the hyperbola whose equation is given by cx 2 − y 2 = 1 are located at ( ± 1c ,0 ) , and are moving away from the origin:

64. Note from the graphs that as c decreases, the vertices of the hyperbola whose equation is given by x 2 − cy 2 = 1 remain at ( ±1,0 ) , but the graphs open outward more narrowly:

1052

Section 8.5 Solutions -------------------------------------------------------------------------------1.  x 2 − y = −2 (1) Solve the system  − x + y = 4 (2) Add (1) and (2) to get an equation in terms of x: x2 − x − 2 = 0 ( x − 2 )( x + 1) = 0 x = 2, − 1 Now, substitute each value of x back into (2) to find the corresponding values of y: x = 2 : − 2 + y = 4 so that y = 6

x = −1: − (−1) + y = 4 so that y = 3 So, the solutions are (2,6) and (−1,3) . 2.  x 2 + y = 2 (1) Solve the system  2 x + y = −1 (2) Multiply (2) by −1 , and then add to (1) to get an equation in terms of x: x 2 − 2x − 3 = 0 ( x − 3)( x + 1) = 0 x = 3, − 1 Now, substitute each value of x back into (2) to find the corresponding values of y: x = 3 : 2(3) + y = −1 so that y = −7 x = −1: 2(−1) + y = −1 so that y = 1

So, the solutions are (3, −7) and ( −1,1) . 3.  x 2 + y = 1 (1) Solve the system  2 x + y = 2 (2) Multiply (2) by −1 , and then add to (1) to get an equation in terms of x: x 2 − 2x + 1 = 0 ( x − 1)2 = 0 x =1

1053

Chapter 8

Now, substitute this value of x back into (1) to find the corresponding value of y: x = 1: 12 + y = 1 so that y = 0 So, the solution is (1,0) . 4.  x 2 − y = 2 (1) Solve the system  −2 x + y = −3 (2) Add (1) and (2) to get an equation in terms of x: x 2 − 2 x = −1 x 2 − 2x + 1 = 0 ( x − 1)2 = 0 x =1 Now, substitute this value of x back into (1) to find the corresponding value of y: x = 1: 12 − y = 2 so that y = −1

So, the solution is (1, −1) . 5.  x 2 + y = −5 (1) Solve the system  − x + y = 3 (2) Multiply (2) by −1 and then add to (1) to get an equation in terms of x: x 2 + x = −8

x2 + x + 8 = 0 −1± 1−4(8)

x= 2 Since the values of x are not real numbers, there is no solution to this system.

6. x 2 − y = −7 (1) Solve the system   x + y = −2 (2) Add (1) and (2) to get an equation in terms of x: x 2 + x = −9 x2 + x + 9 = 0 −1± 1−4(9)

x= 2 Since the values of x are not real numbers, there is no solution to this system.

1054

Section 8.5

x 2 + y2 = 1 (1) Solve the system  2  x − y = −1 (2) Multiply (2) by −1 , and then add to (1) to get an equation in terms of y: y2 + y = 2 y2 + y − 2 = 0 ( y + 2 )( y − 1) = 0 y = −2, 1 Now, substitute each value of y back into (2) to find the corresponding values of x: y = −2 : x 2 − (−2) = −1 so that x 2 = −3 (No real solutions)

y = 1:

x 2 − 1 = −1 x 2 = 0 so that x = 0

So, the solution is (0,1) . 8.

x 2 + y2 = 1 (1) Solve the system  2  x + y = −1 (2) Multiply (2) by −1 , and then add to (1) to get an equation in terms of y: y2 − y = 2 y2 − y − 2 = 0 ( y − 2 )( y + 1) = 0 y = 2, − 1 Now, substitute each value of y back into (2) to find the corresponding values of x: y = 2 : x 2 + 2 = −1 so that x 2 = −3 (No real solutions)

y = −1: x 2 − 1 = −1 x 2 = 0 so that x = 0 So, the solution is (0, −1) .

1055

Chapter 8

9.  x 2 + y2 = 3 (1) Solve the system  2 4 x + y = 0 (2) Multiply (1) by −4 , and then add to (2) to get an equation in terms of y: −4y 2 + y = −12 4 y 2 − y − 12 = 0 1± 1− 4( 4 )( −12) 2( 4 )

= 1± 8193

≅ 1.862, − 1.612 Now, substitute each value of y back into (2) to find the corresponding values of x: y = 1.862 : 4x 2 + 1.862 = 0 so that x 2 = −0.466 (No real solutions)

y = −1.612 : 4 x 2 −1.612 = 0 x 2 = 0.403 so that x ≅ ±0.635 So, the solutions are (0.63, −1.61) and ( −0.63, −1.61) .

10.  x 2 + y2 = 6 (1) Solve the system  2 −7x + y = 0 (2) Multiply (1) by 7, and then add to (2) to get an equation in terms of y: 7y 2 + y = 42 7 y 2 + y − 42 = 0 y=

−1± 1− 4(7 )( −42 ) 2(7)

≅ −1±34.307 14

≅ 2.379, − 2.52 Now, substitute each value of y back into (2) to find the corresponding values of x: y = 2.379 : − 7 x 2 + 2.379 = 0

x 2 = 0.340 so that x = ±0.583 y = −2.52 : − 7x 2 − 2.52 = 0 x 2 = −0.36 (No real solutions) So, the solutions are (0.583,2.379) and (−0.583,2.379) .

1056

Section 8.5

11.  x 2 + y2 = −6 (1) Solve the system  2 −2x + y = 7 (2) Multiply (1) by 2, and then add to (2) to get an equation in terms of y: 2 y2 + y = −5 2 y2 + y + 5 = 0 y=

−1± 1−4(2 )(5) 2(2)

Since the values of y are not real numbers, there is no solution to this system.

12.  x 2 + y2 = 5 (1) Solve the system  2 3x + y = 9 (2) Multiply (1) by −3 , and then add to (2) to get an equation in terms of y: −3 y 2 + y = −6 3 y2 − y − 6 = 0 y=

1± 1−4(3)( −6 ) 2(3)

= 1± 6 73

≅ 1.591, − 1.257 Now, substitute each value of y back into (2) to find the corresponding values of x: y = 1.591: 3x 2 + 1.591 = 9

x 2 = 2.470 x ≅ ±1.572 y = −1.257 : 3x 2 −1.257 = 9 x 2 = 3.419 x ≅ ±1.849 So, the solutions are (1.572,1.591), (−1.572,1.591), (1.849, − 1.257), ( −1.849, − 1.257) .

1057

Chapter 8

13.

 x + y = 2 (1) Solve the system  2 2 x + y = 2 (2) Solve (1) for y: y = 2 − x (3) Substitute (3) into (2) to get an equation in terms of x: x 2 + 4 − 4x + x 2 = 2

2x2 − 4x + 2 = 0 2(x − 1)2 = 0 x =1 Now, substitute this value of x back into (3) to find the corresponding value of y: y = 2 − (1) = 1

So, the solution is (1,1) .

14.

 x − y = −2 (1) Solve the system  2 2 (2) x + y = 2 Solve (1) for x: x = y − 2 (3) Substitute (3) into (2) to get an equation in terms of y: ( y − 2 )2 + y 2 = 2

y2 − 4 y + 4 + y2 = 2 2 y2 − 4 y + 2 = 0 2( y − 1)2 = 0 y =1 Now, substitute this value of y back into (3) to find the corresponding value of x: x = 1 − (2) = −1

So, the solution is ( −1,1) .

1058

Section 8.5

15. 2 x 2 + y 2 = 64 (1) Solve the system  2 2 2 x − y = 36 (2) Solve (1) for y 2 : y2 = 64 − 2 x 2 (3) Substitute (3) into (2) to get an equation in terms of x : 2x 2 − ( 64 − 2 x 2 ) = 36

4 x 2 − 64 = 36 4x 2 = 100 x 2 = 25 x = ±5 Now, substitute each of these values of x back into (3) to find the corresponding values of y :

x = 5 : y2 = 64 − 2 ( 5 ) = 14  y = ± 14 2

x = −5 : y 2 = 64 − 2 ( −5 ) = 14  y = ± 14 So, the solutions are 2

(5, 14 ) ,(5, − 14 ) ,( −5, 14 ) ,( −5, − 14 ) . 16.

x 2 + 2 y2 = 3 (1) Solve the system  2  2 y − x = 1 (2) Solve (1) for x 2 : x 2 = 3 − 2 y 2 (3) Substitute (3) into (2) to get an equation in terms of y :

2 y2 − (3 − 2 y2 ) = 1

4 y2 − 3 = 1 4 y2 = 4 y2 = 1 y = ±1 Now, substitute each of these values of x back into (3) to find the corresponding values of x :

y = 1: x 2 = 3 − 2 (1) = 1  x = ±1 2

y = −1: x 2 = 3 − 2 ( −1) = 1  x = ±1 2

So, the solutions are ( −1,1) , (1,1) , ( −1, −1)(1, −1) .

1059

Chapter 8

17.

xy = 4 (1)  Solve the system  2 2 x + y = 10 (2) Solve (1) for y: y = x4 (3) Substitute (3) into (2) to get an equation in terms of x: x 2 + ( x4 ) = 10 2

= 10 x 2 + 16 x2 x 4 +16 = 10 x2 x 4 +16 10x 2 − 2 =0 x2 x 4 2 x − 10 x + 16 =0 x2 ( x2 − 8) ( x2 − 2 ) = 0 x2 x 2 − 8 = 0 or x 2 − 2 = 0

Hence, x = ±2 2 or x = ± 2 . Now, substitute each of these values of x back into (3) to find the corresponding value of y: x = 2 2 : y = 2 4 2 = 22 x = −2 2 : y = −24 2 = − 22 x = 2 : y = 42 = 42 x = − 2 : y = − 4 2 = − 42

So, the solutions are (after rationalizing)

( 2 2, 2 ) , ( −2 2, − 2 ) , ( 2, 2 2 ) , and ( − 2, − 2 2 ) .

1060

Section 8.5

18.

xy = −3 (1)  Solve the system  2 2 x + y = 12 (2) Solve (1) for y: y = − x3 (3) Substitute (3) into (2) to get an equation in terms of x: x 2 + ( − x3 ) = 12 2

x 2 + x92 = 12 x4 + 9 = 12 x2 x 4 + 9 12 x 2 − 2 = 0 (4) x2 x 4 2 x − 12 x + 9 =0 x2

Let u = x 2 (5). Use (5) to rewrite (4) as the equivalent equation: u 2 − 12u + 9 = 0 Solve this equation using the quadratic formula: 12 ± 144 − 4(9) 12 ± 108 u= = 2 2 12 ± 6 3 = = 6±3 3 2 Now, substituting these back into (5) yields the following two equations for x: x2 = 6 + 3 3 x2 = 6 − 3 3 and x ≅ ±3.346 x ≅ ±0.896 Finally, substitute each of these values of x back into (3) to find the corresponding value of y: 3 ≅ 0.897 x = −3.346: y = − −3.346 3 ≅ −0.897 x = 3.346: y = − 3.346 3 ≅ 3.348 x = −0.896: y = − −0.896 3 ≅ 3.348 x = 0.896: y = − 0.896

So, the solutions are ( −3.346,0.897 ) , (3.346, −0.897 ) , ( −0.896,3.348) , and ( 0.896, −3.348) .

1061

Chapter 8

19.  y = x 2 − 3 (1) Solve the system   y = −4 x + 9 (2) Substitute (1) into (2) to get an equation in terms of x: x 2 − 3 = −4 x + 9 x 2 + 4 x − 12 = 0 ( x + 6 )( x − 2) = 0 x = −6, 2 Now, substitute this value of y back into (1) to find the corresponding value of y: x = −6 : y = (−6)2 − 3 = 33 x =2:

y = (2)2 − 3 = 1

So, the solutions are ( 2,1) and ( −6,33 ) .

20.  y = − x 2 + 5 (1) Solve the system  (2)  y = 3x − 4 Substitute (1) into (2) to get an equation in terms of x: −x 2 + 5 = 3x − 4 x 2 + 3x − 9 = 0 −3 ± 9 − 4(−9) −3 ± 3 5 = ≅ 1.854, − 4.854 2 2 Now, substitute these values of x back into (2) to find the corresponding value of y: x = 1.854 : y = 3(1.854) − 4 = 1.562 x=

x = −4.854 : y = 3(−4.854) − 4 = −18.562 So, the solutions are (1.854,1.562 ) and ( −4.854, −18.562 ) .

1062

Section 8.5

21. x 2 + xy − y 2 = 5 (1) Solve the system  x − y = −1 (2)  Solve (2) for x: x = y − 1 (3) Substitute (3) into (1) to get an equation in terms of y: ( y − 1)2 + ( y − 1)y − y 2 = 5 y2 − 2 y + 1 + y2 − y − y2 = 5 y2 − 3 y − 4 = 0 ( y − 4 )( y + 1) = 0 y = 4, − 1 Now, substitute these values of y back into (3) to find the corresponding values of x: y = 4 : x = 4 −1 = 3

y = −1: x = −1−1 = −2 So, the solutions are (3, 4) and (−2, −1) . 22. x 2 + xy + y 2 = 13 (1) Solve the system  x + y = −1 (2)  Solve (2) for x: x = − y −1 (3) Substitute (3) into (1) to get an equation in terms of y: ( − y − 1)2 + ( − y − 1)y + y 2 = 13 y 2 + 2 y + 1 − y 2 − y + y 2 = 13 y2 + y − 12 = 0 ( y + 4 )( y − 3) = 0 y = −4, 3 Now, substitute these values of y back into (3) to find the corresponding values of x: y = −4 : x = −(−4) −1 = 3

y = 3:

x = −3 −1 = −4

So, the solutions are (3, −4) and ( −4,3) .

1063

Chapter 8

23.

2 x − y = 3 (1)  Solve the system  2 2 x + y − 2 x + 6 y = −9 (2) Solve (1) for y: y = 2x − 3 (3) Substitute (3) into (2) to get an equation in terms of x: x 2 + (2 x − 3)2 − 2x + 6(2x − 3) = −9

x 2 + 4 x 2 − 12x + 9 − 2x +12x −18 + 9 = 0 5x 2 − 2 x = 0 x(5x − 2) = 0 x = 0, 52 Now, substitute these values of x back into (3) to find the corresponding values of y: x = 0 : y = 2(0) − 3 = −3 x = 52 : y = 2( 25 ) − 3 = − 115

So, the solutions are (0, −3) and ( 52 , − 115 ) .

24.

x 2 + y2 − 2x − 4 y = 0 (1) Solve the system  −2x + y = −3 (2)  Solve (2) for y: y = 2x − 3 (3) Substitute (3) into (1) to get an equation in terms of x: x 2 + (2 x − 3)2 − 2x − 4(2x − 3) = 0

x 2 + 4 x 2 − 12x + 9 − 2x − 8x +12 = 0 5x 2 − 22x + 21 = 0 ( x − 3)(5x − 7) = 0 x = 3, 75 Now, substitute these values of x back into (3) to find the corresponding values of y: x = 3 : y = 2(3) − 3 = 3 x = 75 : y = 2( 75 ) − 3 = − 51

So, the solutions are (3,3) and ( 75 , − 15 ) .

1064

Section 8.5

25. 4 x 2 + 12 xy + 9y 2 = 25 (1) Solve the system  −2 x + y = 1 (2)  Solve (2) for y: y = 2x + 1 (3) Substitute (3) into (1) to get an equation in terms of x: 4 x 2 + 12 x(2x + 1) + 9(2x +1)2 = 25 4 x 2 + 24 x 2 + 12x + 9(4x 2 + 4x +1) − 25 = 0   36 x2 + 36 x + 9

64x 2 + 48x − 16 = 0 4 x 2 + 3x −1 = 0 ( x + 1)(4 x − 1) = 0 x = −1, 14 Now, substitute these values of x back into (3) to find the corresponding values of y: x = −1: y = 2( −1) + 1 = −1 x = 14 : y = 2( 14 ) + 1 = 23

So, the solutions are ( −1, −1) and ( 14 , 32 ) .

26. −4 xy + 4 y 2 = 8 (1) Solve the system  3x + y = 2 (2)  Solve (2) for y: y = 2 − 3x (3) Substitute (3) into (1) to get an equation in terms of x: −4x( 2 − 3x ) + 4(2 − 3x)2 = 8 −8x + 12 x 2 + 4(4 −12x + 9x 2 ) − 8 = 0  16− 48x +36 x2

48x 2 − 56 x + 8 = 0 6 x 2 − 7 x +1 = 0 ( x − 1)(6 x − 1) = 0 x = 1, 61 Now, substitute these values of x back into (3) to find the corresponding values of y: x = 1: y = 2 − 3(1) = −1 x = 61 : y = 2 − 3( 61 ) = 23

So, the solutions are (1, −1) and ( 61 , 23 ) .

1065

Chapter 8

27. x3 − y3 = 63 (1) Solve the system   x − y = 3 (2) Solve (2) for x: x = y + 3 (3) Substitute (3) into (1) to get an equation in terms of x: ( y + 3)3 − y3 = 63    ( y +3 )( y2 + 6 y+9)

( y3 + 6 y 2 + 9y + 3y 2 +18y + 27) − y3 − 63 = 0 9 y 2 + 27 y − 36 = 0 y2 + 3 y − 4 = 0 ( y + 4 )( y − 1) = 0 y = −4, 1 Now, substitute these values of y back into (3) to find the corresponding values of x: y = −4 : x = −4 + 3 = −1

y = 1:

x = 1+ 3 = 4

So, the solutions are (4,1) and (−1, −4) .

28.

x3 + y3 = −26 (1) Solve the system   x + y = −2 (2) Solve (2) for y: y = −x − 2 (3) Substitute (3) into (1) to get an equation in terms of x: x3 + ( − x − 2)3 

= −26

[ − ( x + 2 )] =( −1) ( x + 2 ) =−( x+2 ) 3

x3 − ( x + 2)3 = −26 x3 − ( x + 2)(x 2 + 4x + 4) = −26 x3 − ( x3 + 4x 2 + 4x + 2x 2 + 8x + 8) + 26 = 0 −6x 2 − 12 x +18 = 0 x 2 + 2x − 3 = 0 ( x + 3)( x − 1) = 0 x = −3, 1 Now, substitute these values of x back into (3) to find the corresponding values of y:

1066

Section 8.5

x = −3 : y = −(−3) − 2 = 1 x = 1:

y = −(1) − 2 = −3

So, the solutions are (−3,1) and (1, −3) .

29.  4x 2 − 3xy = −5 (1) Solve the system  2 − x + 3xy = 8 (2) Add (1) and (2) to get an equation in terms of x: 3x 2 = 3 x2 = 1 x = −1, 1 Now, substitute these values of x back into (2) to find the corresponding values of y: x = 1: − (1)2 + 3(1)( y ) = 8 3y = 9 so that y = 3 x = −1:

− ( −1) + 3( −1)( y ) = 8 − 3y = 9 so that y = −3 2

So, the solutions are (1,3) and (−1, −3) .

30. 2 x 2 + 5xy = 2 (1) Solve the system  2  x − xy = 1 (2) Multiply (2) by 5, and then add to (1) to get an equation in terms of x: 7x 2 = 7 x2 = 1 x = −1, 1 Now, substitute these values of x back into (2) to find the corresponding values of y: (1)2 − (1)( y ) = 1 x = 1: y=0 x = −1:

(−1) − (−1)( y ) = 1 2

y=0

So, the solutions are (1,0) and (−1,0) .

1067

Chapter 8

31.

32.

log x (2y) = 3 (1) Solve the system   log x ( y) = 2 (2) This system is equivalent to: 3  x = 2 y (3)  2 x = y (4) Substitute (4) into (3) to get an equation in terms of x: x3 = 2( x 2 )

 log x ( y ) = 1 (1) Solve the system  1 log x (2 y ) = 2 (2) This system is equivalent to:  x = y (3)  12 x = 2 y (4) Substitute (3) into (4) to get an equation in terms of x: 1 x 2 = 2 x (Square both sides)

x3 − 2 x 2 = 0

x = 4x 2

x2 ( x − 2) = 0

4x 2 − x = 0

x = 0, 2 Recall that 0 is not an allowable base for a logarithm. So, although x = 0 would give rise to a corresponding y (namely y = 0 ) such that the pair would solve system (3) – (4), the equations (1) – (2) would not be defined for such x and y values. So, we only substitute the value x = 2 back into (4) to find the corresponding value of y that will yield a solution to the original system: x = 2 : y = 2(2) = 4

So, the solution is (2, 4) .

x(4x − 1) = 0 x = 0 , 14 Recall that 0 is not an allowable base for a logarithm. So, although x = 0 would give rise to a corresponding y (namely y = 0 ) such that the pair would solve system (3) – (4), the equations (1) – (2) would not be defined for such x and y values. So, we only substitute the value x = 14 back into (3) to find the corresponding value of y that will yield a solution to the original system: x = 14 : y = 14

So, the solution is ( 14 , 14 ) .

1068

Section 8.5

33.

1 1  x3 + y2 = 17 (1)  Solve the system   1 − 1 = −1 (2)  x3 y 2 Add (1) and (2) to get an equation in terms of x: 2 = 16 x3 2 = 16x3

x3 = 18 so that the only real solution is x = 12 Now, substitute this value of x back into (2) to find the corresponding value of y: 1 1 − 2 = −1 3 ( 12 ) y 8−

1 = −1 y2 1 9 = 2 so that y2 = 19 . Hence, y = ± 13 . y

So, the solutions are ( 12 , 13 ) and ( 21 , − 31 ) .

34.

3 5 2  x 2 + y2 = 6 (1)  Solve the system   4 − 9 = 0 (2)  x 2 y2 Multiply (1) by 3, and then add to (2) to get an equation in terms of x: 10 5 = x2 2 5x 2 = 20

x 2 = 4 so that x = ±2. Now, substitute these values of x back into (2) to find the corresponding value of y: 4 9 x =2: − 2 =0 2 (2) y

1069

Chapter 8

9 so that y2 = 9. Hence, y = ±3. y2 4 9 − 2 =0 x = −2 : 2 ( −2 ) y 9 1 = 2 so that y2 = 9. Hence, y = ±3. y 1=

So, the solutions are (2,3), (2, −3), ( −2,3), and (−2, −3) . 35.

36.

2 x + 4 y = −2 (1) Solve the system  2 4 6x + 3 y = −1 (2) Note that the left side of (1) (and (2)) is always ≥ 0 , while the right side is negative. Hence, there are no real values of x and y that can satisfy either equation, not to mention both simultaneously. Hence, there is no solution of this system.

x 2 + y2 = −2 (1) Solve the system  2 2 x + y = −1 (2) Note that the left side of (1) (and (2)) is always ≥ 0 , while the right side is negative. Hence, there are no real values of x and y that can satisfy either equation, not to mention both simultaneously. Hence, there is no solution of this system.

37.

38.

Notes on the graph: Solid curve: y = x 2 − 6x + 11 Dashed curve: y = −x + 7 The solutions of the system are (1,6) and (4,3).

Note that in this case the graph of the equation ( x − 2 )2 + ( y − 1)2 = 0 consists of the single point (2,1), which does not satisfy the second equation (the graph of which is the solid curve). Hence, this system has no solution.

1070

Section 8.5

39. The graphs of the two equations are as follows:

The two solutions are (−1, −1) and (1, 2). 40. The graphs of the two equations are as follows:

Since the graphs do not intersect, there is no solution.

1071

Chapter 8

41. Let x and y be two numbers. The system that must be solved is:  x + y = 10 (1)  2 2 x − y = 40 (2) Solve (1) for y: y = 10 − x (3) Substitute (3) into (2) to get an equation in x: x 2 − (10 − x )2 = 40

42. Let x and y be two numbers. The system that must be solved is:  x − y = 3 (1)  2 2  x − y = 51 (2) Solve (1) for x: x = y + 3 (3) Substitute (3) into (2) to get an equation in y: ( y + 3)2 − y 2 = 51 y2 + 6 y + 9 − y 2 = 51 6 y = 42

x 2 − 100 + 20x − x 2 = 40 20x = 140 x=7 Substitute this value of x into (3) to find the corresponding value of y: y = 10 − 7 = 3 So, the two numbers are 3 and 7.

y=7 Substitute this value of y into (3) to find the corresponding value of x: x = 7 + 3 = 10 So, the two numbers are 10 and 7.

43. Let x and y be two numbers. The system that must be solved is: 1  xy = 1 − 1 (1) x y  xy = 72 (2)  First, simplify (1): 1 1 xy xy = 1 1 = y − x = y−x x − y xy 1 = y − x (3)

 xy = 72 So, we now solve the following simplified system:  y − x = 1 To do so, solve (3) for y: y = x + 1 (4) Substitute (4) into (2) to get an equation in x: x( x + 1) = 72 x 2 + x − 72 = 0 ( x + 9)( x − 8) = 0 x = −9, 8

1072

(2) (3)

Section 8.5

Substitute these values of x into (4) to find the corresponding values of y: x = −9 : y = (−9) +1 = −8 x =8: y = (8) +1 = 9 So, there are two pairs of numbers that will work, namely the pair 8 and 9, and the pair −8 and − 9 . 44. Let x and y be two numbers. x + y = 9 (1)  The system that must be solved is:  x − y  xy = 80 (2)  First, simplify (1): x + y = 9(x − y) = 9x − 9y 8x − 10 y = 0 (3) xy = 80 (2)  So, we now solve the following simplified system:  8x − 10 y = 0 (3) 4 To do so, solve (3) for y: y = 5 x (4) Substitute (4) into (2) to get an equation in x: x( 54 x ) = 80 4 5

x 2 = 80 x 2 = 54 (80) = 100

x = ±10 Substitute these values of x into (4) to find the corresponding values of y: x = −10 : y = 54 ( −10) = −8 x = 10 : y = 54 (10) = 8 So, there are two pairs of numbers that will work, namely the pair 8 and 10, and the pair −10 and − 8 .

45. Let x = width of rectangular pen (in cm) y = length of rectangular pen. The system that must be solved is: 2x + 2 y = 36 (1) (perimeter)  xy = 80 (2) (area)  Solve (1) for y: y = 18 − x (3)

1073

Chapter 8

Substitute (3) into (2) to get an equation in x: x(18 − x ) = 80 18x − x 2 = 80 x 2 − 18x + 80 = 0 (x − 10)( x − 8) = 0 so that x = 10, 8 Substitute these values of x into (3) to find the corresponding values of y: x = 10 : y = 18 −10 = 8

x =8: y = 18 − 8 = 10 Hence, in terms of the context of the problem, both x values result in the same solution, namely that the dimensions of the rectangular pen should be 8 cm ×10 cm . 46. Assume that circles C1 and C2 are centered at the same point, say the origin. Let x = radius of circle C1 y = radius of circle C2 The system that must be solved is: 2π x + 2π y = 16π (1) (perimeter)  2 2  π x + π y = 34π (2) (area) First, simplify equations (1) and (2) to get the following equivalent system:  x + y = 8 (3)  2 2 x + y = 34 (4) Solve (3) for y: y = 8 − x (5) Substitute (3) into (4) to get an equation in x: x 2 + (8 − x )2 = 34

x 2 + 64 − 16 x + x 2 = 34 2 x 2 − 16 x + 30 = 0 x 2 − 8x + 15 = 0 ( x − 5)( x − 3) = 0 x = 5, 3 Substitute these values of x into (5) to find the corresponding values of y: x = 5: y = 8−5 = 3 x = 3: y = 8−3 = 5 Hence, in terms of the context of the problem, both x values result in the same solution, namely that the radii of the two concentric circles should be 5 units and 3 units. 1074

Section 8.5

47. Let x = width of one of the two congruent rectangular pens (in ft) y = length of one of the two congruent rectangular pens. The system that must be solved is: (1) (total amount of fence needed) 3x + 4 y = 2200   x ⋅ 2 y = 200,000 (2) (combined area of the two pens) Solve (1) for y: y = 14 (2200 − 3x ) (3) Substitute (3) into (2) to get an equation in x: x ⋅ 2 [ 14 (2200 − 3x ) ] = 200,000

2200x − 3x 2 = 400,000 3x 2 − 2200x + 400,000 = 0 x=

2200 ±

4,840,000 − 4,800,000 2(3)

= 2200 ± 6 40,000

= 22006± 200 = 400, 1000 3 Substitute these values of x into (3) to find the corresponding values of y: x = 400 : 14 (2200 − 3(400)) = 250 1000 1 x = 1000 3 : 4 (2200 − 3( 3 )) = 300 Hence, in terms of the context of the problem, there are two distinct solutions. Solution 1: Each of the two congruent rectangular pens should be 400 ft × 250 ft , so that the two combined have dimensions 400 ft × 500 ft .

Solution 2: Each of the two congruent rectangular pens should be 1000 ft × 300 ft , 3 so that the two combined have dimensions 1000 ft × 600 ft . 3 48. Let x = width of rectangular pen (in ft) y = length of rectangular pen. The system that must be solved is: xy = 11,250 (1)  (2)  y = 2x Substitute (2) into (1) to get an equation in x: x(2x ) = 11,250 x 2 = 5625 x = 75, −75

1075

Chapter 8

Note that we omit x = −75 as a possibility since width must be nonnegative. So, we substitute only the value x = 75 into (2) to find the corresponding value of y: y = 2(75) = 150 Hence, in terms of the context of the problem, the dimensions of the rectangular pen should be 75 ft × 150 ft . 49. Let r = speed of the professor (in m min ) t = number of minutes it took James to complete the 400m race

We have the following information: Distance (in m) Rate (in m min ) r 400 Professor

Time (in min) 5 t+  1 + 3     Head start

= t + 83

Finished 1 min 40 sec after Jerermy

t 5r 400 James Since Distance = Rate × Time, the following system must be solved: 8  400 = r ( t + 3 ) (1)  (2)  400 = 5rt 80 Solve (2) for t: t = r (3) Substitute (3) into (1) to get an equation in x: 400 = r ( 80r + 83 ) 400 = r

(

240 + 8x 3r

)

1200 = 240 + 8r 960 = 8r so that 120 = r Hence, the professor’s average speed was 120 m min = 2 m sec , while James’ average

speed was 600 m min = 10 m sec . (Note: Though the value of t was not needed to answer the question posed in the problem, one could substitute this value of r into (3) to find that the corresponding value of t is 23 , meaning that it took James 40 sec to complete the race).

1076

Section 8.5

50. Let x = your speed (in m min ) y = number of minutes it took James to complete the 800m race We have the following information: Distance (in m) Rate (in m min ) Time (in min) 1 x 800 You t+  1 + = t + 34 3    Head start

Finished 20 sec after Jerermy

t 800 2x James Since Distance = Rate × Time, the following system must be solved: 800 = x ( t + 34 ) (1)  (2) 800 = 2xt Solve (2) for t: t = 400 (3) x Substitute (3) into (1) to get an equation in x: 4 800 = x ( 400 x + 3) 800 = x

(

1200 + 4x 3x

)

2400 = 1200 + 4x 1200 = 4x so that 300 = x Hence, your average speed was 300 m min , while James’ average speed was 600 m min . (Note: Though the value of t was not needed to answer the question posed in the problem, one could substitute this value of x into (3) to find that the corresponding value of y is 34 , meaning that it took James 1 min 20 sec to complete the race).

51. In general, y 2 − y ≠ 0 . Must solve this system using substitution.

52. Once you have the x-values, you should substitute them into the second equation to obtain: x = −1: y = −2 x = 1: y = 2 So, the solutions are (1,2) and ( −1, −2) .

53. False. There are at most two intersection points. Solving a system of the form  y = ax 2 + bx + c   y = dx + e

amounts to solving the quadratic equation ax 2 + bx + c = dx + e . Such an equation can have at most two real solutions. So, the graphs can intersect in at most two points.

1077

Chapter 8

54. True. There are at most three intersection points. Solving a system of the form  y = ax3 + bx 2 + cx + d   y = ex + f

amounts to solving the cubic equation ax3 + bx 2 + cx + d = ex + f . Such an equation can have at most three real solutions. So, the graphs can intersect in at most three points. 55. False. For example, the system  x − y =1  2 2 x + y = 5 cannot be solved using elimination since there is no common term in both equations.

56. False. For example, consider the system  y 4 − 5xy3 + x3 = −3  7 3  y + 5 y + x − 3x = 4

57. Let y = an x n + an −1x n −1 + ⋅ ⋅ ⋅ + a1x + a0 be a given nth degree polynomial. Then, consider a system of the form: y = an x n + an −1x n−1 + ⋅ ⋅ ⋅ + a1x + a0 (1)   2 2 2 (2) ( x − h ) + ( y − k) = r Substitute (1) into (2) to obtain: ( x − h )2 + (an x n + an −1x n −1 + ⋅ ⋅ ⋅ + a1x + a0 − k)2 = r 2 (3)    Has degree 2 n

The degree of the left side of (3) is 2n , so there can be at most 2n real solutions. Consequently, there are at most 2n intersection points of the graphs of the equations of (1) and (2). 58. Let y = an x n + an −1x n −1 + ⋅ ⋅ ⋅ + a1x + a0 be a given nth degree polynomial. Then, consider a system of the form:  y = an x n + an −1x n −1 + ⋅ ⋅ ⋅ + a1x + a0 (1)  (2)  y = mx + b Substitute (2) into (1) to obtain: mx + b = an x n + an −1x n−1 + ⋅ ⋅ ⋅ + a1x + a0 0 = an x n + an −1x n −1 + ⋅ ⋅ ⋅ + ( a1 − m )x + ( a0 − b ) (3) The degree of the right side of (3) is n, so there can be at most n real solutions. Consequently, there are at most n intersection points of the graphs of the equations of (1) and (2).

1078

Section 8.5

59.

60.

y = x +1 Consider  . y = 1  Any system in which the linear equation is the tangent line to the parabola at its vertex will have only one solution.

x 2 + y2 = 1 Consider  . y = −x 2 −1  Any system in which the quadratic equation is the tangent to the circle at a point on the circle will have only one solution.

61.

62.

Notes on the graph: Solid curve: y = e x Dashed curve: y = ln x Since the two graphs never intersect, the system has no solution.

Notes on the graph: Solid curve: y = 10 x Dashed curve: y = log x Since the two graphs never intersect, the system has no solution.

63.

Notes on the graph: Solid curve: 2x3 + 4 y 2 = 3 Dashed curve: xy3 = 7 The approximate solution of the system is ( −1.57, −1.64) .

1079

Chapter 8

64. Here is the graph of the first curve. Notice the scale positions it within the boundary of the second curve - you can hardly see that curve since the scale is so small. Graph of 3x 4 − 2 xy + 5y 2 = 19 :

Graph of the pair of equations (the dashed curve is the graph of x 4 y = 5 ).

Since the two graphs never intersect, the system has no solution. 65. The graph of the nonlinear system is given by:

66. The graph of the nonlinear system is given by:

The points of intersection are approximately (1.067, 4.119), (1.986, 0.638), and (–1.017, –4.757).

The points of intersection are approximately (1.035, 3.966), (1.899, –2.385), (–1.011, 4.664), and (–1.759, –2.165).

1080

Section 8.6 Solutions -------------------------------------------------------------------------------1. b

2. i

3. j

4. g

5. h

6. f

7. c

8. l

9. d

10. a

11. k

12. e

13.

14.

The equation of the solid curve is y = x2 − 2 .

The equation of the solid curve is y = −x 2 + 3 .

15.

16.

The equation of the dashed curve is x 2 + y2 = 4 .

The equation of the dashed curve is x 2 + y 2 = 16 .

1081

Chapter 8

17. Begin by writing the inequality in standard form (so the radius and center are easily identified): ( x 2 − 2x + ) + ( y2 + 4 y + ) ≥ − 4

( x − 2x +1 ) + ( y + 4y + 4 ) ≥ − 4 +1+ 4 2

18. Begin by writing the inequality in standard form (so the radius and center are easily identified): ( x 2 + 2x + ) + ( y2 − 2 y + ) ≤ 2

( x + 2x +1 ) + ( y − 2y +1 ) ≤ 2 +1+ 1 2

( x − 1) + ( y + 2 ) ≥ 1 2

( x + 1) + ( y − 1) ≤ 4

The equation of the solid curve is 2 2 ( x − 1) + ( y + 2 ) = 1 .

The equation of the solid curve is 2 2 ( x + 1) + ( y − 1) = 4 .

19.

20.

The equation of the solid curve is x 2 y2 + =1. 4 3

1082

Section 8.6

21. Upon completing the square, the inequality becomes 9(x − 1)2 + 16( y + 3)2 > 144

22.

23.

24.

25.

26. Upon completing the square, the inequality becomes 2 ( x + 4 )2 − ( y −252 ) ≥ 1 36

1083

Chapter 8

27.

28.

The equation of the solid curve is y = e .

The equation of the solid curve is y = ln x .

29.

30.

The equation of the dashed curve is y = −x 4 .

The equation of the dashed curve is y = − x3 . 31.

32.

1084

Section 8.6

33.

34.

35.

36.

37. First write the inequalities as: y > x2 − 1

38. First write the inequalities as: y2 < 1 − x

y < 1 − x2

y2 < 1 + x

Notes on the graph: C1: y = x 2 − 1 C2: y = −x 2 + 1

Notes on the graph: C1: x = − y 2 + 1 C2: x = y2 − 1 1085

Chapter 8

40.

39. First write the inequalities as: y ≥ x2 x ≥ y2

Notes on the graph: C1: y = x 2 C2: x = y 2 41.

42. Since the line y = 6 is tangent to the circle at (0,6), and neither boundary is included, there are no points (x, y) that satisfy both inequalities simultaneously. That is, there are no points strictly within the circle and strictly above the given line. Hence, the system has no solution.

Notes on the graph: C1: x 2 + y 2 = 36 C2: 2x + y = 3

1086

Section 8.6

43.

44.

45.

46.

47.

48.

1087

Chapter 8

49.

50.

Notes on the graph: C1: y = exp( x ) C2: y = ln( x ) C3: x = 0 51. The shaded region is a semi-circle of radius 3, seen below. Hence, the area is:

(

Area= 21 π ( 3 )

) = π units 9 2

Notes on the graph: C1: y = 10 x C2: y = log( x ) C3: x = 0 52. The shaded region is a quarter circle of radius 5 in the second quadrant, seen below. Hence, the area

( ( 5 ) ) = π units

is: Area= 14 π

53. There is no common region here – it is empty, as is seen in the graph below:

5 4

54. The curves should be dashed, and there is no solution – the wrong areas are shaded.

55. False. Shade below the parabola, and include the parabola itself in the region; the inequality is y ≤ x 2

1088

Section 8.6

56. False. The following system has no solution, as is seen graphically below: y > x +1   y < x −1

57. Assume that a and b are both positive. Must have a 2 ≤ b 2 , which is equivalent to: a 2 − b 2 = ( a − b )(a + b) ≤ 0 This requires that a − b ≤ 0 and a+b≥0 .   Automatically holds since a and b are assumed to be positive.

This implies that a ≤ b and a ≥ −b , and so −b ≤ a ≤ b . Since a is assumed to be positive, this simplifies to 0≤a≤b.

Notes on the graph: C1: y = x +1 C2: y = x −1

58. No, since x 2 + y 2 ≥ 0 , for all values of x and y, there can never be a point ( x, y ) that satisfies the equation if x and y are assumed to be real numbers. (If x and y can be imaginary, then there would be solutions.)

1089

Chapter 8

59. Begin by writing the inequality in standard form (so the radius and center are easily identified): ( x 2 − 2x + ) + ( y2 + 4 y + ) ≥ − 4

( x − 2x +1 ) + ( y + 4y + 4 ) ≥ − 4 +1+ 4 2

60. Begin by writing the inequality in standard form (so the radius and center are easily identified): ( x 2 + 2x + ) + ( y2 − 2 y + ) ≤ 2

( x + 2x + 1 ) + ( y − 2y +1 ) ≤ 2 +1+1 2

( x − 1) + ( y + 2 ) ≥ 1 2

( x + 1) + ( y − 1) ≤ 4

The solid curve is the graph of 2 2 ( x + 1) + ( y − 1) = 4 .

The solid curve is the graph of 2 2 ( x − 1) + ( y + 2 ) = 1. 61.

62.

The solid curve is the graph of y = e x .

The solid curve is the graph of y = ln x .

1090

Section 8.6

63.

64.

Notes on the graph: C1: y = exp( x ) C2: y = ln( x ) C3: x = 0

Notes on the graph: C1: y = 10 x C2: y = log( x ) C3: x = 0

65. From the calculator, the dark shaded region is the one desired:

66. From the calculator, the dark shaded region is the one desired:

1091

Chapter 8 Review Solutions ----------------------------------------------------------------------1. False. The focus is always in the region “inside” the parabola.

2. False. It opens to the right (in the direction of positive x).

3. True.

4. False. The center is (−1,3).

5. Vertex is (0,0) and focus is (3,0) . So, the parabola opens to the right. As such, the general form is y 2 = 4 px . In this case, p = 3 , so the equation is

6. Vertex is (0,0) and focus is (0,2) . So, the parabola opens up. As such, the general form is x 2 = 4 py . In this

y 2 = 12x .

7. Vertex is (0,0) and the directrix is x = 5 . So, the parabola opens to the left. Since the distance between the vertex and the directrix is 5, and the focus must be equidistant from the vertex, we know p = −5 . Hence, the equation is

case, p = 2 , so the equation is x 2 = 8 y .

8. Vertex is (0,0) and the directrix is y = 4 . So, the parabola opens down. Since the distance between the vertex and the directrix is 4 and the focus must be equidistant from the vertex, we know p = −4 . Hence, the equation is ( x − 0 )2 = 4(−4)( y − 0) = −16 y 

( y − 0 )2 = 4(−5)( x − 0) = −20 x 

y = −20 x

x 2 = −16 y

9. Vertex is (2,3) and focus is (2,5) . So, the parabola opens up. As such, the general form is ( x − 2 )2 = 4 p( y − 3) . In this case, p = 2 , so the equation is

10. Vertex is (−1, −2) and focus is (1, −2) . So, the parabola opens to the right. As such, the general form is ( y + 2 )2 = 4 p( x + 1) . In this case, p = 2 ,

( x − 2 )2 = 8( y − 3) .

so the equation is ( y + 2 )2 = 8( x + 1) .

11. Focus is (1,5) and the directrix is y = 7 . So, the parabola opens down. Since the distance between the focus and directrix is 2 and the vertex must occur halfway between them, we know p = −1 and the vertex is (1,6) . Hence, the equation is

12. Focus is (2,2) and the directrix is x = 0 . So, the parabola opens to the right. Since the distance between the focus and directrix is 2 and the vertex must occur halfway between them, we know p = 1 and the vertex is (1,2) . Hence, the equation is

(x − 1)2 = 4( −1)( y − 6) = −4( y − 6) .

( y − 2)2 = 4(1)( x − 1) = 4( x − 1) .

1092

Chapter 8 Review

13. Equation: x 2 = −12 y = 4( −3)y (1) So, p = −3 and opens down. Vertex: (0,0) Focus: (0, −3) Directrix: y = 3 Latus Rectum: Connects x-values corresponding to y = −3 . Substituting this into (1) yields: x 2 = 36 so that x = ±6 So, the length of the latus rectum is 12. 14. Equation: x 2 = 8 y = 4(2)y (1) So, p = 2 and opens up. Vertex: (0,0) Focus: (0,2) Directrix: y = −2 Latus Rectum: Connects x-values corresponding to y = 2 . Substituting this into (1) yields: x 2 = 16 so that x = ±4 So, the length of the latus rectum is 8. 15. Equation: y2 = x = 4( 14 )x

(1)

So, p = and opens to the right. Vertex: (0,0) Focus: ( 14 , 0) Directrix: x = − 14 Latus Rectum: Connects y-values corresponding to x = 14 . Substituting this into (1) yields: y2 = 14 so that y = ± 12 So, the length of the latus rectum is 1. 1 4

1093

Chapter 8

16. Equation: y2 = −6x = 4(− 32 )x

(1)

So, p = − and opens to the left. Vertex: (0,0) Focus: ( − 32 , 0) Directrix: x = 32 Latus Rectum: Connects y-values corresponding to x = − 32 . Substituting this into (1) yields: y 2 = 9 so that y = ±3 So, the length of the latus rectum is 6. 3 2

17. Equation: ( y + 2 )2 = 4(1)( x − 2 ) (1) So, p = 1 and opens to the right. Vertex: (2, −2) Focus: (3, −2) Directrix: x = 1 Latus Rectum: Connects y-values corresponding to x = 3 . Substituting this into (1) yields: ( y + 2 )2 = 4 so that y = −2 ± 2 = −4, 0 So, the length of the latus rectum is 4. 18. Equation: ( y − 2 )2 = 4( −1)( x + 1) (1) So, p = −1 and opens to the left. Vertex: (−1,2) Focus: ( −2,2) Directrix: x = 0 Latus Rectum: Connects y-values corresponding to x = −2 . Substituting this into (1) yields: ( y − 2 )2 = 4 so that y = 2 ± 2 = 4, 0 So, the length of the latus rectum is 4.

1094

Chapter 8 Review

19. Equation: ( x + 3)2 = 4( −2)( y − 1) (1) So, p = −2 and opens down. Vertex: ( −3,1) Focus: (−3, −1) Directrix: y = 3 Latus Rectum: Connects x-values corresponding to y = −1. Substituting this into (1) yields: ( x + 3)2 = 16 so that x = −3 ± 4 = 1, − 7 So, the length of the latus rectum is 8. 20. Equation: ( x − 3)2 = 4( −2)( y + 2) (1) So, p = −2 and opens down. Vertex: (3, −2) Focus: (3, −4) Directrix: y = 0 Latus Rectum: Connects x-values corresponding to y = −4 . Substituting this into (1) yields: ( x − 3)2 = 16 so that x = 3 ± 4 = −1, 7 So, the length of the latus rectum is 8. 21. Equation: Completing the square yields: x 2 + 5x + 2 y + 25 = 0 ( x 2 + 5x + 254 ) = −2y − 25 + 254 ( x + 52 )2 = −2 y − 745 ( x + 52 )2 = −2( y + 785 ) ( x + 52 )2 = 4( − 12 )( y + 758 ) (1) So, p = − 12 and opens down.

Vertex: ( − 52 , − 758 ) Focus: ( − 52 , − 798 ) Directrix: y = − 718 Latus Rectum: Connects x-values corresponding to y = − 798 . Substituting this into (1) yields: ( x + 52 )2 = 1 so that x = − 52 ±1 = − 32 , − 72 So, the length of the latus rectum is 2. 1095

Chapter 8

22. Equation: Completing the square yields: y2 + 2 y − 16x +1 = 0

( y 2 + 2 y +1) −1 = 16x ( y + 1)2 = 16 x ( y + 1)2 = 4(4)x (1) So, p = 4 and opens to the right. Vertex: (0, −1) Focus: (4, −1) Directrix: x = −4 Latus Rectum: Connects y-values corresponding to x = 4 . Substituting this into (1) yields: ( y + 1)2 = 64 so that y = −1 ± 8 = 7, − 9 So, the length of the latus rectum is 16. 23. Assume the vertex is at (0, −2) . Then, the general equation of this parabola is x 2 = 4 p( y + 2) (1). Since ( −5,0) is on the graph, we can substitute it into (1) to see: 25 = 4 p(2) = 8 p so that p = 258 So, the focus is at (0, 98 ) , and the receiver should be placed 258 = 3.125 feet from the vertex. 24. Assume that the parabola has vertex at (0,40) and opens down. Then, the form of the equation is ( x − 0 )2 = 4 p( y − 40) (1). Since the point (15,0) is on the graph, we can substitute it into (1) to find p: 225 = −160 p so that p = −1.406 . So, (1) becomes x 2 = −5.625( y − 40) . We need to determine the y-value when x = 4 . If it is larger than 14, then the RV will pass under the bridge without a problem. Observe that 16 = −5.625( y − 40) = −5.625 y + 225 . 37.16 = y Hence, the RV will pass under the bridge without a problem.

1096

Chapter 8 Review

25.

26.

27. Write the equation in standard y2 form as x 2 + 25 =1.

28. Write the equation in standard form 2 x2 as 16 + y8 = 1 .

1097

Chapter 8

29. Since the foci are (−3,0) and (3,0) , we know that c = 3 and the center of the ellipse is (0,0) . Further, since the vertices are (−5,0) and (5,0) and they lie on the x-axis, the ellipse is horizontal and a = 5 . Hence, c 2 = a 2 − b2

30. Since the foci are (0,−2) and (0,2) , we know that c = 2 and the center of the ellipse is (0,0) . Further, since the vertices are (0, −3) and (0,3) and they lie on the y-axis, the ellipse is vertical and a = 3 . Hence, c 2 = a 2 − b2

9 = 25 − b2

4 = 9 − b2

b2 = 16 Therefore, the equation of the ellipse is

b2 = 5 Therefore, the equation of the ellipse is

x2 25

y + 16 =1 .

x2 5

31. Since the ellipse is centered at (0,0) , the length of the horizontal major axis being 16 implies that a = 8 , and the length of the horizontal minor axis being 6 implies that b = 3 . Since the major axis is vertical, the equation

of the ellipse is 33.

x2 9

+ 6y4 = 1 .

+ y9 = 1 .

32. Since the ellipse is centered at (0,0) , the length of the horizontal major axis being 30 implies that a = 15 , and the length of the vertical minor axis being 20 implies that b = 10 . Since the major axis is horizontal, the equation of the ellipse 2

y x + 100 =1 . is 225 2

34. Write the equation in standard form 2 2 as ( x +63 ) + ( y1−204) = 1 .

1098

Chapter 8 Review

35. First, write the equation in standard form by completing the square: 4(x 2 − 4 x ) + 12( y 2 + 6y) = −123

36. First, write the equation in standard form by completing the square: 4(x 2 − 2 x ) + 9( y 2 − 8y) = −147 4(x 2 − 2x + 1) + 9( y2 − 8y + 16) = −147

4(x 2 − 4 x + 4) + 12( y2 + 6y + 9) = −123

+16 + 108 4(x − 2) + 12( y + 3) = 1 2

( x − 2 )2 1 4

+4 + 144 4(x − 1)2 + 9( y − 4)2 = 1 ( x −1)2 1 4

+ ( y +13) = 1

+ ( y −14 ) = 1 9

37. Since the foci are (−1,3), (7,3) and the vertices are (−2,3), (8,3) , we know: i) the ellipse is horizontal with center at ( −22+8 , 3) = (3,3) , ii) 3 − c = −1 so that c = 4 , iii) the length of the major axis is 8 − ( −2) = 10 . Hence, a = 5 . Now, since c 2 = a 2 − b2 , 16 = 25 − b 2 so that b 2 = 9 . Hence, the equation of the 2

−3 ) + ( y −93) = 1 . ellipse is ( x25

38. Since the foci are (1, −3), (1, −1) and the vertices are (1, −4), (1,0) , we know: i) the ellipse is vertical with center at (1, −42+ 0 ) = (1, −2) , ii) −2 − c = −3 so that c = 1 , iii) the length of the major axis is 0 − ( −4) = 4 . Hence, a = 2 . Now, since c 2 = a 2 − b2 , 1 = 4 − b2 so that b2 = 3 . Hence, the equation of the ellipse 2

is ( x −31) + ( y +42 ) = 1 .

1099

Chapter 8

39. Assume that the sun is at the origin. Then, the vertices of Jupiter’s horizontal elliptical trajectory are: ( −7.409 ×108 ,0), (8.157 ×108 ,0) From this, we know that: i) The length of the major axis is 8.157 ×108 − ( −7.409 ×108 ) = 1.5566 ×109 , and so the value of a is half of this, namely 7.783 ×108 ;

ii) The center of the ellipse is

(

−7.409×108 + (8.157×108 ) 2

)

,0 = (3.74 ×107 ,0) ;

iii) The value of c is 3.74 ×107 .

Now, since c 2 = a 2 − b2 , we have ( 3.74 ×107 ) = ( 7.783 ×108 ) − b2 , so that 2

b2 = 6.044 × 1017 . Hence, the equation of the ellipse must be

( x −(3.74×10 )) 7

6.058×10

( ) + 6.044 =1 . ×1017 y −0

(Note: There is more than one correct way to set up this problem, and it begins with choosing the center of the ellipse. If, alternatively, you choose the center to be at (0,0), then the actual coordinates of the vertices and foci will change, and will result in the following slightly different equation for the trajectory: y2 x2 + 777,400,000 2 =1 778,300,0002 Both are equally correct!.) 40. Assume that the sun is at the origin. Then, the vertices of Mars’ horizontal elliptical trajectory are: ( −2.07 × 108 ,0), (2.49 ×108 ,0) From this, we know that: i) The length of the major axis is 2.49 ×108 − ( −2.077 ×108 ) = 4.56 ×108 , and so the value of a is half of this, namely 2.28 ×108 ;

ii) The center of the ellipse is

(

2.49×108 + ( −2.07×108 ) 2

)

,0 = (2.1× 107 ,0) ;

iii) The value of c is 2.1×107 .

Now, since c 2 = a 2 − b2 , we have ( 2.1× 107 ) = ( 2.28 ×108 ) − b 2 , so that 2

b2 = 5.154 ×1016 . Hence, the equation of the ellipse must be

( x −(2.1×10 )) 7

5.1984×10

( ) + 5.154 =1 . ×1016 y −0

1100

Chapter 8 Review

41.

42.

Notes on the Graph: The equations of the asymptotes are y = ± 83 x .

Notes on the Graph: The equations of the asymptotes are y = ± 79 x .

43.

44.

Notes on the Graph: First, write the equation in standard 2 form as x25 − y 2 = 1 . The equations of the asymptotes are y = ± 51 x .

Notes on the Graph: First, write the equation in standard 2 2 form as y8 − 1x6 = 1 . The equations of the asymptotes are y = ± 22 x .

1101

Chapter 8

45. Since the foci are ( −5,0), (5,0) and the vertices are ( −3,0), (3,0) , we know that: i) the hyperbola opens right/left with center at (0,0) , ii) a = 3, c = 5 , Now, since c 2 = a 2 + b2 , 25 = 9 + b2 so that b2 = 16 . Hence, the equation of the

hyperbola is

x2 9

− 1y6 = 1 .

46. Since the foci are (0, −3), (0,3) and the vertices are (0, −1), (0,1) , we know that: i) the hyperbola opens up/down with center at (0,0) , ii) a = 1, c = 3 , Now, since c 2 = a 2 + b2 , 9 = 1 + b 2 so that b 2 = 8 . Hence, the equation of the

hyperbola is y 2 − x8 = 1 . 2

47. Since the center is (0,0) and the transverse axis is the y-axis, the general form 2

of the equation of this hyperbola is ay2 − xb2 = 1 with asymptotes y = ± ab x . In this 2

case, ab = 3 so that a = 3b . Thus, the equation simplifies to (3yb )2 − xb2 = 1 , or 2

y2 9

equivalently y9 − x 2 = b 2 . If we took b = 1, then this simplifies to

− x2 = 1 .

48. Since the center is (0,0) and the transverse axis is the y-axis, the general form 2

of the equation of this hyperbola is ay2 − xb2 = 1 with asymptotes y = ± ab x . In this 2

case, ab = 12 so that 2a = b . Thus, the equation simplifies to ay2 − (2xa )2 = 1 , or equivalently y 2 − x4 = a 2 . 2

1102

Chapter 8 Review

49.

50.

Notes on the Graph: The equations of the asymptotes are y = ±2( x − 2) +1 .

Notes on the Graph: The equations of the asymptotes are y = ± 12 ( x + 3) + 4 .

51.

Notes on the Graph: First, write the equation in standard form by completing the square: 8 ( x 2 − 4 x ) − 10 ( y 2 + 6y ) = 138 8 ( x 2 − 4 x + 4 ) − 10 ( y 2 + 6y + 9 ) = 138 + 32 − 90 8(x − 2)2 − 10( y + 3)2 = 80 ( x − 2 )2 10

− ( y +83) = 1 The equations of the asymptotes are y = ± 25 ( x − 2) − 3 . 52.

Notes on the Graph: First, write the equation in standard form by completing the square: 2 ( x 2 + 6x ) − 8 ( y 2 − 2y ) = −6 2 ( x 2 + 6 x + 9 ) − 8 ( y2 − 2y + 1) = −6 + 18 − 8 2(x + 3)2 − 8( y − 1)2 = 4 ( x + 3 )2 2

− ( y −11) = 1 2

The equations of the asymptotes are y = ± 12 ( x + 3) + 1 .

1103

Chapter 8

53. Since the vertices are (0,3), (8,3) , we know that the center is (4,3) and the transverse axis is parallel to the x-axis. Hence, 4 − a = 0 so that a = 4 . Also, since the foci are (−1,3), (9,3) , we know that 4 + c = −1 so that c = −5 . Now, to find b, we substitute the values of c and a obtained above into c 2 = a 2 + b2 to see that 25 = 16 + b2 so that b2 = 9 . Hence, the equation of the hyperbola is ( x − 4 )2 16

− ( y −93) = 1 .

54. Since the vertices are (4, −2), (4,0) , we know that the center is (4, −1) and the transverse axis is parallel to the y-axis. Hence, −1 − a = −2 so that a = 1 . Also, since the foci are (4, −3), (4,1) , we know that −1 − c = −3 so that c = 2 . Now, to find b, we substitute the values of c and a obtained above into c 2 = a 2 + b2 to see that 4 = 1 + b2 so that b2 = 3 . Hence, the equation of the hyperbola is ( y −1)2 1

− ( x −34 ) = 1 .

55. Assume that the stations coincide with the foci and are located at (−110,0), (110,0) . The difference in distance between the ship and each of the two stations must remain constantly 2a, where (a,0) is the vertex. Assume that the radio signal speed is 186,000 mi sec and the time difference is 0.00048 sec . Then, using distance = rate × time, we obtain: 2a = (186,000)(0.00048) = 89.28 so that a = 44.64 So, the ship will come ashore between the two stations 65.36 miles from one and 154.64 miles from the other. 56. Assume that the stations coincide with the foci and are located at (−200,0), (200,0) . The difference in distance between the ship and each of the two stations must remain constantly 2a, where ( a, 0) is the vertex. Assume that the radio signal speed is 186,000 mi sec and the time difference is 0.0008 sec . Then, using distance = rate × time, we obtain: 2a = (186,000)(0.0008) = 148.8 so that a = 74.4 So, the ship will come ashore between the two stations 125.6 miles from one and 274.4 miles from the other.

1104

Chapter 8 Review

57. x 2 + y = −3 (1) Solve the system   x − y = 5 (2) Add (1) and (2) to get an equation in terms of x: x2 + x − 2 = 0 ( x + 2 )( x − 1) = 0 x = −2, 1 Now, substitute each value of x back into (2) to find the corresponding values of y: x = −2 : − 2 − y = 5 so that y = −7

x = 1:

1 − y = 5 so that y = −4

So, the solutions are (−2, −7) and (1, −4 ) . 58.

 x 2 + y = 2 (1) Solve the system  2 2 x + y = 4 (2) Solve (1) for y: y = 2 − x 2 (3) Substitute (3) into (2) to get an equation in terms of x: x2 + 4 − 4x2 + x 4 = 4 − 3x 2 + x 4 = 0 − x 2 (3 − x 2 ) = 0

−x 2

( 3 − x)( 3 + x) = 0

x = 0, ± 3 Now, substitute these values of x back into (3) to find the corresponding values of y: x = 0 : y = 2 − (0)2 = 2

(

x = ± 3: y =2− ± 3

) = 2 − 3 = −1 2

So, the solutions are (0,2), (− 3, −1), and ( 3, −1) .

1105

Chapter 8

59.

 x 2 + y2 = 5 (1) Solve the system  2 2 x − y = 0 (2) Solve (2) for y: y = 2 x 2 (3) Substitute (3) into (1) to get an equation in x: x 2 + (2 x 2 )2 = 5 x 2 + 4x 4 = 5

4x 4 + x2 − 5 = 0

( 4x + 5 ) ( x − 1) = 0 2

x = ±1 Now, substitute each value of x back into (2) to find the corresponding values of y: x = 1: y = 2(1)2 = 2 x = −1: y = 2(−1)2 = 2

So, the solutions are (1,2) and (−1,2) . 60.

61.

 x + y = 16 (1) Solve the system  2 2 6x + y = 16 (2) Multiply (1) by −1, and then add to (2): 5x 2 = 0 x = 0 (3) Substitute (3) into (1) to find the corresponding values of y: 0 2 + y 2 = 16 y = ±4 2

 x + y = 3 (1) Solve the system  2 2 x + y = 4 (2) Solve (1) for y: y = 3 − x (3) Substitute (3) into (2) to get an equation in terms of y: x 2 + (3 − x )2 = 4 x 2 + 9 − 6x + x 2 = 4

So, the solutions are (0, 4) and (0, −4) .

1106

2x 2 − 6x + 5 = 0 x=

6 ± 36 − 40 2(2)

which are not real numbers Hence, the system has no solution.

Chapter 8 Review

62.

xy = 4 (1)  Solve the system  2 2 x + y = 16 (2) Solve (1) for y: y = x4 (3) Substitute (3) into (2) to get an equation in terms of x: 2 x 2 + ( x4 ) = 16

x 2 + 16 = 16 x2 x 4 +16 = 16 x2 x 4 +16 16 x 2 − 2 =0 x2 x 4 2 x − 16x + 16 =0 x2 x 4 − 16 x 2 + 16 = 0 (4) Let u = x 2 (5) . Then, (4) can be written in the equivalent form u 2 − 16u + 16 = 0 . Solve this equation using the quadratic formula:

16 ± 256 − 4(16 ) 2

16 ± 192 2

= 16 ±28 3

= 8± 4 3 Substituting both of these values into (5) now leads to the following two equations in x: x2 = 8 + 4 3 x2 = 8 − 4 3 and x ≅ ±3.864 x ≅ ±1.035 Now, substitute each of these values of x back into (3) to find the corresponding value of y: 4 ≅ 1.035 x = 3.864: y = 3.864 4 ≅ −1.035 x = −3.864: y = −3.864 4 ≅ 3.865 x = 1.035: y = 1.035 4 ≅ −3.865 x = −1.035: y = −1.035 So, the approximate solutions are (3.864,1.035 ) , ( −3.864, −1.035 ) ,

(1.035,3.865 ) , and ( −1.035, −3.865 )

63. x 2 + xy + y 2 = −12 (1) Solve the system  x−y=2 (2)  Solve (2) for x: x = y + 2 (3) Substitute (3) into (1) to get an equation in terms of y: ( y + 2 )2 + ( y + 2)y + y 2 = −12 y 2 + 4 y + 4 + y 2 + 2y + y 2 = −12 3y 2 + 6 y + 16 = 0 so that y = Hence, the system has no solution.

−6 ± 36 − 4(3)(16 ) 2

1107

which are not real numbers

Chapter 8

64.  x − y 2 = −9 (1) Solve the system  (2) 3x + y = 3 2 Solve (1) for x: x = y − 9 (3) Substitute (3) into (1) to get an equation in terms of y: 3( y 2 − 9) + y = 3 3 y 2 + y − 30 = 0 (3y + 10)( y − 3) = 0 so that y = − 103 , 3 Now, substitute these values of y back into (3) to find the corresponding values of x: 2 y = − 103 : x = ( − 103 ) − 9 = 199 y = 3 : x = (3) − 9 = 0 2

So, the solutions are (0,3) and ( − 103 , 199 ) . 65. x3 − y3 = −19 (1) Solve the system   x − y = −1 (2) Solve (2) for x: x = y −1 (3) Substitute (3) into (1) to get an equation in terms of x: ( y − 1)3 − y3 = −19  ( y −1)( y2 − 2 y+1)

( y3 − 2 y 2 + y − y 2 + 2y −1) − y3 +19 = 0 −3 y 2 + 3 y +18 = 0 y2 − y − 6 = 0 ( y + 2)( y − 3) = 0 so that y = −2, 3 Now, substitute these values of y back into (3) to find the corresponding values of x: y = −2 : x = −2 − 1 = −3, y = 3 : x = 3 − 1 = 2

So, the solutions are (2,3) and (−3, −2) .

1108

Chapter 8 Review

66.

2 x 2 + 4xy = 9 (1) Solve the system  2  x − 2xy = 0 (2) Multiply (2) by 2 and then add to (1) to get an equation in terms of x: 4x 2 = 9

x 2 = 94 x = ± 32 Now, substitute these values of x back into (2) to find the corresponding values of y: ( 32 )2 − 2( 23 )( y ) = 0 x = 32 : 3 y = 94 so that y = 34 x = − 32 :

(− 32 )2 − 2(− 32 )( y ) = 0 3 y = − 94 so that y = − 34

So, the solutions are ( 32 , 34 ) and ( − 32 , − 34 ) . 67.

1 2  x 2 + y2 = 15 (1)  Solve the system   1 − 1 = −3 (2)  x 2 y2 Add (1) and (2) to get an equation in terms of x: 3 = 12 x2 12x 2 = 3 x 2 = 14 so that x = ± 12 .

1109

Chapter 8

Now, substitute these values of x back into (2) to find the corresponding value of y: 1 1 1 1 x = 12 : 1 2 − 2 = −3 x = − 12 : − 2 = −3 1 2 (2) y y (− 2 ) 4−

1 = −3 y2 1 7= 2 y

4−

1 = −3 y2 1 7= 2 y

7 y2 = 1

y2 = 71

y=±

1 7

So, the solutions are ( 12 , 17 ), ( − 12 , 17 ), ( 12 , − 17 ), and ( − 12 , − 17 ) . 68.

x 2 + y2 = 2 (1) Solve the system  2 2 x + y = 4 (2) Add (1) and (2) to obtain the false statement 0 = 6. Hence, there are no real values of x and y that can satisfy both equations simultaneously. Hence, there is no solution of this system. 69.

70.

1110

Chapter 8 Review

71.

72.

73.

74.

75.

76.

1111

Chapter 8

77.

78.

79.

80.

81. The vertex is at (0.6, −1.2), and it should open down. The graph below confirms this:

82. The vertex is at (2.8, 0.2) and it should open to the right. The graph below confirms this:

1112

Chapter 8 Review

83. From the graphs, we see that as c increases the minor axis (along the x-axis) of the ellipse whose equation is given by 4(cx)2 + y 2 = 1 decreases.

84. From the graphs, we see that as c increases the minor axis (along the y-axis) of the ellipse whose equation is given by x 2 + 4(cy)2 = 1 decreases.

85. From the graphs, we see that as c increases, the vertices of the hyperbola whose equation is given by 4(cx)2 − y 2 = 1 are at ( ± 21c ,0 ) are moving towards the origin:

86. From the graphs, we see that as c increases, the vertices of the hyperbola whose equation is given by x 2 − 4(cy)2 = 1 remain at ( ±1,0 ) , but the graphs open more narrowly:

87. The graph of this nonlinear system is as follows:

88. The graph of this nonlinear system is as follows:

The points of intersection are approximately (0.635, 2.480), (–0.635, 2.480), (–1.245, 0.645), and (1.245, 0.645).

The points of intersection are approximately (0.876, 1.458) and (1.350, –1.781).

1113

Chapter 8

89. The dark region below is the one desired:

90. The dark region below is the one desired:

1114

Chapter 8 Practice Test Solutions----------------------------------------------------------------1. c Parabola opens to the right

2. b Parabola opens up

3. d Ellipse is horizontal

4. e Hyperbola whose transverse axis is the x-axis

5. f Ellipse is vertical

6. a Hyperbola whose transverse axis is y-axis.

7. Vertex is (0,0) and focus is (−4,0) . So, the parabola opens to the left. As such, the general form is y 2 = 4 px . In this case, p = −4 , so the equation is

8. Vertex is (0,0) and the directrix is y = 2 . So, the parabola opens down. Since the distance between the vertex and directrix is 2, we know p = −2 . Hence,

y 2 = −16x .

the equation is x 2 = 4( −2) y = −8 y .

9. Vertex is ( −1,5) and focus is ( −1,2 ) . So, the parabola opens down. As such, the general form is ( x + 1)2 = 4 p( y − 5) . In this case, p = −3 , so the equation is

10. Vertex is (2, −3) and the directrix is x = 0 . So, the parabola opens to the right. Since the distance between the vertex and directrix is 2, we know p = 2 . Hence, the equation is

(x + 1)2 = −12( y − 5) .

( y + 3)2 = 4(2)( x − 2) = 8( x − 2) .

11. Since the foci are (0, −3), (0,3) and the vertices are (0, −4), (0, 4) , we know: i) the ellipse is vertical with center at (0,0) , ii) c = 3 , iii) the length of the major axis is 4 − ( −4) = 8 . Hence, a = 4 . Now, since c 2 = a 2 − b2 , 9 = 16 − b2 so that b 2 = 7 . Hence, the equation of the

ellipse is

x2 7

+ 1y6 = 1 .

12. Since the foci are ( −1,0), (1,0) and the vertices are ( −3,0), (3,0) , we know: i) the ellipse is horizontal with center at (0,0) , ii) c = 1 , iii) the length of the major axis is 3 − ( −3) = 6 . Hence, a = 3 . Now, since c 2 = a 2 − b2 , 1 = 9 − b2 so that b2 = 8 . Hence, the equation of the ellipse

x2 9

+ y8 = 1 .

1115

Chapter 8

13. Since the foci are (2, −4), (2, 4) and the vertices are (2, −6), (2,6) , we know: i) the ellipse is vertical with center at (2, −62+6 ) = (2,0) , ii) c = 4 , iii) the length of the major axis is 6 − ( −6) = 12 . Hence, a = 6 . Now, since c 2 = a 2 − b2 , 16 = 36 − b2 so that b 2 = 20 . Hence, the equation of the 2

−2 ) y ellipse is ( x 20 + 36 =1 .

14. Since the foci are ( −6, −3), ( −5, −3) and the vertices are ( −7, −3), ( −4, −3) , we know: i) the ellipse is horizontal with center at ( −7 +2 −4 , −3) = ( − 112 , −3) , ii) c = 12 , iii) the length of the major axis is −4 − ( −7) = 3 . Hence, a = 32 .

Now, since c 2 = a 2 − b2 , ( 21 )2 = ( 23 )2 − b2 so that b2 = 2 . Hence, the equation of the ellipse is

( x+ 11 )2 2 2.25

+ ( y+23 ) = 1 .

15. Since the vertices are ( −1,0), (1,0) , and the center is the midpoint of the segment connecting them, the center must be (0,0). Moreover, since the form of the vertices is ( h ± a, k ) ( which in this case is (0 ± a,0) ) we see that a = 1 . It remains to find b. At this point, we need to use the fact that we are given that the equations of the asymptotes are y = ±2x . Since the transverse axis is parallel to the x-axis, we know that the slopes of the asymptotes for such a hyperbola are ± ab . Thus, ± b1 = ±2 , which implies that b = ±2 . Hence, the equation of the 2

hyperbola must be x 2 − y4 = 1 . 16. Since the vertices are (0, −1), (0,1) , and the center is the midpoint of the segment connecting them, the center must be (0,0). Moreover, since the form of the vertices is ( h ± a, k ) ( which in this case is (0 ± a,0) ) we see that . a = 1 . It remains to find b. At this point, we need to use the fact that we are given that the equations of the asymptotes are y = ± 13 x . Since the transverse axis is parallel to the y-axis, we know that the slopes of the asymptotes for such a hyperbola are ± ab . Thus, ± ab = ± 13 . Since a = 1 , this implies that b = ±3 . Hence, the equation of

the hyperbola must be y 2 − x9 = 1 . 2

1116

Chapter 8 Practice Test

17. Since the vertices are (2, −4), (2, 4) , we know that the center is (2,0) and the transverse axis is parallel to the y-axis. Hence, 0 − a = −4 so that a = 4 . Also, since the foci are (2, −6), (2,6) , we know that 0 − c = −6 so that c = 6 . Now, to find b, we substitute the values of c and a obtained above into c 2 = a 2 + b2 to see that 36 = 16 + b2 so that b2 = 20 . Hence, the equation of the hyperbola is y2 16

−2 ) − ( x20 =1 .

18. Since the vertices are (−6, −3), (−5,−3) , we know that the center is ( −5.5, −3) and the transverse axis is parallel to the x-axis. Hence, −5.5 − a = −6 so that a = 0.5 . Also, since the foci are ( −7, −3), ( −4, −3) , we know that −5.5 − c = −7 so that c = 1.5 . Now, to find b, we substitute the values of c and a obtained above into c 2 = a 2 + b2 to see that (1.5)2 = (0.5)2 + b2 so that b2 = 2 . 2

+5.5) − ( y +23) = 1 . Hence, the equation of the hyperbola is ( x0.25

19.

Notes on the Graph: First, write the equation in standard form by completing the square: 9 ( x 2 + 2 x ) − 4 ( y 2 − 4 y ) = 43

9 ( x 2 + 2 x + 1) − 4 ( y 2 − 4 y + 4 ) = 43 + 9 − 16 9(x + 1)2 − 4( y − 2)2 = 36 ( x +1)2 4

− ( y −92 ) = 1 The equations of the asymptotes are y = ± 32 ( x + 1) + 2 . 20.

First, write the equation in standard form by completing the square: 4(x 2 − 2x ) + ( y2 +10y) = −28 4(x 2 − 2 x + 1) + ( y2 +10y + 25) = −28 + 4 + 25 4(x − 1)2 + ( y + 5)2 = 1 ( x−1)2 1 4

1117

+ ( y + 5 )2 = 1

Chapter 8

Equation: Completing the square yields: y 2 + 4 y = 16x − 20

21.

( y 2 + 4 y + 4) = 16x − 20 + 4 ( y + 2 )2 = 16x − 16 ( y + 2)2 = 16( x − 1) ( y + 2 )2 = 4(4)( x − 1) (1) So, p = 4 and opens to the right and its vertex is (1, −2) .

23. Assume that the vertex is at the origin and that the parabola opens up. We are given that p = 1.5 . Hence, the equation is x 2 = 4(1.5) y = 6y, − 2 ≤ x ≤ 2 .

22.

24. Assume that the sun is at the origin. Then, the vertices of Uranus’s horizontal elliptical trajectory are: ( −2.739 × 109 ,0), (3.003 ×109 ,0) From this, we know that: i) The length of the major axis is 3.003 ×109 − ( −2.739 ×109 ) = 5.742 ×109 , and so the value of a is half of this, namely 2.871×109 ;

ii) The center of the ellipse is

(

3.003×109 + ( −2.739×109 ) 2

)

,0 = (1.32 × 108 ,0) ;

iii) The value of c is 1.32 × 10 . 8

Now, since c 2 = a 2 − b2 , we have (1.32 × 108 ) = ( 2.871×109 ) − b 2 , so that 2

b2 = 8.225 ×1018 . Hence, the equation of the ellipse must be

( x−(1.32×10 )) 8

8.24×10

2 2

+ 8.225y×1018 = 1 . 1118

Chapter 8 Practice Test

25.

 x + y = 4 (1) Solve the system:  2  −x + y = −4 ( 2 ) Solve (2) for y: y = x 2 − 4 (3) Substitute (3) into (1) and solve for x: x2 + x2 − 4 = 4 2

2x2 = 8 x = ±2 Substitute these back into (3) to find y, and conclude that the solutions of the system are ( ±2,0 ) .

26. Solve the system: (1)  x + 3y = 7  y = x2 − 1 ( 2)  Substitute (2) into (1) and solve for x: x + 3 ( x 2 − 1) = 7 3x 2 + x − 10 = 0 (3x − 5)( x + 2) = 0 x = 53 , − 2 Substitute these back into (2) to find y, and conclude that the solutions of the system are ( 53 , 169 ) and ( −2,3 ) .

27.

28.

29.

30.

1119

Chapter 8

31. The graph using the calculator is as follows:

32. The graph of this nonlinear system is as follows:

The points of intersection are approximately (2.457, 3.001) and (−2.457, −3.001).

1120

Chapter 8 Cumulative Review -------------------------------------------------------------------1.

−5x − 3 ≤ 12 −5x ≤ 15 x ≥ −3 So, the solution set is [ −3, ∞ ) .

(2 − x) = −16 2 − x = ±4i 2

x = 2 ± 4i

3. 3

m = 41 3

− ( − 14 )

− ( − 61 )

= 2

4. x 2 − 3.2 x + y 2 + 4.4y = 0.44

( x − 3.2x + 2.56 ) + ( y + 4.4y + 4.84 ) = 0.44 + 2.56 + 4.84 2

(x − 1.6)2 + ( y + 2.2)2 = 7.84

Center: (1.6, −2.2) Radius: 2.8 f ( x + h ) − f (x) ( 5 − ( x + h) ) − ( 5 − x ) −2hx − h 2 5. = = = −2x − h h h h 2

6. a. −1 b. 7 c. 0 d. 16 e. Domain: (−∞,∞) Range: (−∞,∞) f. increasing on (0,5) decreasing on ( −∞, 0) ∪ (5,∞) constant – never Note: This graph has open holes at (0,0) and (5,9). 7. This is undefined since f (−1) is undefined.

1121

Chapter 8

8. The general form of the function is f ( x ) = a(x − h)2 + k . Since we know the vertex is (1.5, 2.5), this becomes f ( x ) = a(x − 1.5)2 + 2.5 . To find a, use the fact f (−0.5) = −0.5 :

9. 0 (multiplicity 2), −2 (multiplicity 3) 10. Factors of −2: ±1, ± 2 Factors of 6: ±1, ± 2, ± 3, ± 6 Possible rational roots: ±1, ± 12 , ± 13 , ± 61 , ± 2, ± 23

−0.5 = a(−0.5 −1.5)2 + 2.5 a = −0.75 Thus, f ( x ) = −0.75(x − 1.5)2 + 2.5 . 11. Vertical asymptote: x = 2 Horizontal asymptote: none Slant asymptote: y = x + 2 , as seen from the synthetic divison below: 2 1 0 3 2 4 1 2 7

12. y-intercept: ( 0, 15 )

Domain: ( −∞, ∞ ) Range: (0,∞) Horizontal asymptote: y = 0

13. Use A = Pe rt . 85,000 = Pe 0.055(15) 85,000 P = 0.055(15) ≈ $37,250 e

1122

Chapter 8 Cumulative Review

14. Reflect the graph of y = log x over the x-axis, and then shift it up 2 units, to obtain:

15. −3 10 −3log10 = 10 log10 = 0.001 16. ln 6 − 3x − ln x + 2 = ln x  6 − 3x  ln   = ln x 2 x +   1

 6 − 3x    =x  x+2  6 − 3x = x2 x+2 6 − 3x = x3 + 2x 2 2

x3 + 2 x 2 + 3x − 6 = 0 Possible rational zeros: ±1, ± 2, ± 3,± 6 Note that 1 1 2 3 −6 1 3 1 3

17. Let x = cost of a soda and y = cost of a soft pretzel. Solve the system:  3x + 2 y = 6.77 (1)   5x + 4 y = 12.25 ( 2 ) Multiply (1) by −5 and (2) by 3, and add: − 15x − 10 y = −33.85 + 15x + 12 y = 36.75 2 y = 2.90 y = 1.45 Substitute this value of y into (1) to solve for x: 3x + 2(1.45) = 6.77  x = 1.29 So, a soda costs $1.29 and a soft pretzel costs $1.45.

6 0

Hence, x3 + 2 x 2 + 3x − 6 = (x −1) ( x 2 + 3x + 6 ) =0 so that the only solution is x = 1.

1123

Chapter 8

18. The partial fraction decomposition is of the form 3x + 5 A Bx + C = + 2 2 ( x − 3) ( x + 5 ) x − 3 x + 5 Multiply by the LCD to obtain 3x + 5 = A ( x 2 + 5 ) + (Bx +C )(x − 3) = (A + B)x 2 + (C − 3B)x + (5A − 3C ) . We must solve the following system:  A + B = 0 (1)   C − 3B = 3 (2)  5A − 3C = 5 ( 3 )  Solve (1) for B: B = −A ( 4 ) Substitute (4) into (2): C + 3A = 3 ( 5 ) Solve the 2 × 2 system:  5A − 3C = 5 ( 3 )   3A + C = 3 ( 5 ) Multiply (5) by 3 and add to (3) to see that A = 1 . Substitute this back into (5) to obtain C = 0 . Finally, substitute the value of A into (1) to see that B = −1 . Hence, the partial fraction decomposition is −x 3x + 5 1 = + 2 2 ( x − 3) ( x + 5 ) x − 3 x + 5

19.

20. 1 −2 1 7  1 −2 1 7  R2 +3 R1 → R2 →    ⎯⎯ ⎯ ⎯⎯  −3 1 2 −11 0 −5 5 10  1 −2 1 7  − 15 R2 → R2 ⎯⎯ ⎯⎯⎯ →   0 1 −1 −2 

Let z = a . Then, x = a + 3 and by substituting these two values into the equation corresponding to the first row of the last matrix, we see that y = a − 2 .

1124

Chapter 8 Cumulative Review

21.

16 −4 12  9 12 −21 2B − 3A =   − 18 0 −2  0 3 15  7 −16 33  =  18 −3 −17 

22. Write the system in matrix form as  25 40   x   −12  75 −105   y  =  69       Observe that 25 40 D= = −5,625 75 −105

Dx =

−12 40 = −1,500 69 −105

Dy =

25 −12 = 2,625 75 69

So, 2625 7 4 x = 1500 5625 = 15 , y = − 5625 = − 15 .

23. Since the vertices are (6,3) and (6,−7), the center is ( 6, 3−27 ) = (6, −2) . Using the foci, we see that −2 + c = 2 and − 2 − c = −6 , so that c = 4 . Also, we have −2 + a = 3 and − 2 − a = −7 , so that a = 5 . As such, c 2 = a 2 − b 2  b = 3 . Thus, the equation of the ellipse is 2 ( x − 6 )2 + ( y +252 ) = 1 . 9

24. Since the vertices are (5,−2) and (5,0), the center of the hyperbola is (5,−1). Also, we have a = 1. Thus, the equation so far is ( x − 5 )2 2 ( y + 1) − = 1. b2 To find b, we use the fact that c = 2, so that c 2 = a2 + b2  b = 3 Thus, the desired equation is ( x − 5 )2 ( y + 1)2 − =1 3

25. Solve the system:  x + y = 6 (1)  2 2  x + y = 20 (2)

Solve (1) for y: y = 6 − x ( 3 )

1125

Chapter 8

Substitute (3) into (2) and solve for x: x 2 + (6 − x )2 = 20 2 x 2 − 12 x + 36 = 20 x 2 − 6x + 8 = 0 ( x − 4 )( x − 2) = 0 x = 2, 4 Substitute each value into (1) to find the corresponding y-value. This yields two solutions, namely (2,4) and (4,2).

26. The dashed region is the solution to this nonlinear system:

27. The graph from the calculator is as follows:

28. The dark region is the one desired:

1126

CHAPTER 9 Section 9.1 Solutions--------------------------------------------------------------------------------1. 1,2,3, 4

2. 1, 4,9,16

3. 1,3,5,7

4. x, x 2 , x3 , x 4

5. 1 2 3 4 , , , 2 3 4 5

8. 2,

3 4 5 2, , , 2 3 4

4 8 16 , , , 2!  3!  4!  =

4 2⋅1

n! n! 1 = = , ( n + 1)! ( n + 1) n ! n +1

8 16 = 3⋅2⋅1 4⋅3⋅2⋅1

so the terms are

4 2 which is equivalent to 2, 2, , 3 3

9. − x 2 , x3 , − x 4 , x5

10. 1, − 4,9, − 16

11. −1 1 −1 1 , , , 2 ⋅3 3⋅ 4 4 ⋅5 5 ⋅6

12. 1 4 9 0, , , 9 16 25

which is equivalent to 13. 9

1 1   =  2  512 15. 4 −7 =

1 16,384

−1 1 −1 1 , , , 6 12 20 30

14. 15

(16 )

1 4

15 256

16. 21/12 = 12 2

1127

1 1 1 1 , , , 2 3 4 5

Chapter 9 17.

18.

( −1) 19! = −1 = −1 21⋅ 20 420 ( 21) !

( −1) (12 )(15 ) = 180

19.

20.

1 99 = 2 10 100

1   101  10,201  1+  =  =  100   100  10,000

1−

21. log10 23 = 23 log10 = 23

22. e ln 49 = 49

23. an = 2 n, n ≥ 1

24. an = 3n, n ≥ 1

25. an = n3 , n ≥ 1

26. an = 2 n − 1, n ≥ 1

27.

28.

an =

1 , n ≥1 ( n +1) n

1 an =   , n ≥ 1 2

29.

30. n

 2  ( −1) 2 , n ≥1 an = ( −1)   = 3n 3

an =

31. an = ( −1)

32.

n+1

, n ≥1

3n −1 , n ≥1 2n

an = ( −1)

n +1

n , n ≥1 n+2

33. 9! 9 ⋅ 8 ⋅ 7! = = 72 7! 7!

34. 4! 4! 1 = = 6! 6 ⋅ 5 ⋅ 4! 30

35. 29! 29 ⋅ 28 ⋅ 27! = = 812 27! 27!

36. 32! 32 ⋅ 31 ⋅ 30! = = 992 30! 30!

37. 75! 75! 1 = = 77! 77 ⋅ 76 ⋅ 75! 5852

38. 100! 100! 1 = = 103! 103⋅102 ⋅101⋅ 100! 1,061,106

1128

Section 9.1

39. 97! 97 ⋅ 96 ⋅ 95 ⋅ 94 ⋅ 93! = = 83,156,160 93! 93!

40. 101! 101⋅100 ⋅ 99 ⋅ 98! = = 999,900 98! 98!

41.

42. ( n + 2 ) ! = ( n + 2 )( n + 1) n ! = n + 2 n + 1 ( )( ) n! n!

( n −1) ! ( n −1) ! = 1 = ( n +1) ! ( n + 1) n ( n − 1) ! ( n +1) n 43.

44.

( 2n + 3 ) ! = ( 2n + 3) ( 2n + 2 ) ( 2n + 1) ! ( 2n + 1) ! ( 2n + 1) ! = ( 2n + 3) ( 2n + 2 )

( 2n + 2 ) ! = ( 2n + 2 ) ( 2n + 1)( 2n ) ( 2n − 1) ! ( 2n − 1) ! ( 2n − 1) ! = ( 2n + 2 ) ( 2 n +1)( 2n )

45. 7,10,13,16

46. 2,3, 4,5

47. 60,56,52, 48

48. 1, −1, −3, −5

49. 1,2,6,24

50. 2,6,24,120

51.

100,

52. 25 3

5 20 5 20, 2 , 2 , 92 , 2 3  4 

100 50 , , 2!  3! 4!   =50

50 25 = 25 = 25 = 3⋅4⋅3⋅2 72 3⋅2 3

which is equivalent to 100, 50,

25 25 , . 3 72

5 144

which is equivalent to 20, 5,

53. 1,2,2, 4 (Note that a3 = a2 ⋅ a1 , a4 = a3 ⋅ a2 )

54.

55.

56. 1, − 1, 2 ( −1) +1, 3 ( −1) + 2 ( −1)     

1, − 1, − (1) + ( −1)  , ( −1) + ( −2 )        2

=−2

which is equivalent to 1, − 1, − 2,5

1, 2,

5 5 , 9 144

a a 1 , 4 (Note that a3 = 1 , a4 = 2 ) a2 a3 2

=−1

=−5

which is equivalent to 1, − 1, − 1, − 5 (Note that a3 = 2a2 + 1a1 , a4 = 3a3 + 2a2 .)

1129

Chapter 9 57. 2 ⋅ 5 = 10

58. 7 ⋅ 5 = 35

59. 02 + 12 + 22 + 32 + 4 2 = 30

60.

61.

62.

1 1 1 25 1+ + + = 2 3 4 12

 (2n − 1) =1+ 3 + 5 + 7 + 9 +11 = 36

 ( n + 1) =2 + 3 + 4 + 5 + 6 + 7 = 27

n=1

63.

1n = 1,for all n. So, 1n = 1 = 5 (1) = 5.

64. 20 + 21 + 22 + 23 + 2 4 = 1+ 2 + 4 + 8 +16 = 31

65. 1 − x + x 2 − x3

66. − x + x 2 − x3 + x 4

n =0

67. 0

2 2 2 2 2 2 + + + + + 0! 1!  2!  3!  4! 5!  =

4 2⋅1

8 3⋅2

16 4⋅3⋅2

32 5⋅4⋅3⋅2

4 2 4 + + 3 3 15 4 109 =7+ = 15 15

68. 1 1 1 1 1 1 − + − + − 0! 1! 2! 3! 4! 5! 1 1 1 1 11 =1−1+ − + − = 2 6 24 120 30

= 1+ 2 + 2 +

69.

70. 2

1 (1)x (2)x 2 (3)x3 (4)x 4 − − − − 0! 1! 2! 3! 4! 3 4 x x = 1 − x − x2 − − 2 6

1 x x x x + + + + 0! 1! 2! 3! 4! x 2 x3 x 4 = 1+ x + + + 2 6 24

71.

72.

( 0.1) = 2 = 2.2 = 20 2 0

1 − 0.1

0.9

 1   5 10 5⋅   = = 5.5 = 509 1 0.9 1− 10

1130

Section 9.1

73. Not possible. (The result is infinite.)

74. Not possible. (The result is infinite.)

75.

76.

 ( −1) n= 0

∞

1 2n

 ( −1)

n= 0

77. ∞

2 78.

n−1

n

n=1

79.

80.

( n + 1) ! = n( n + 1)   n =1 ( n − 1) ! n =1 6

∞ 2 n −1 2n =   n =1 ( n −1) ! n =0 n ! ∞

(Remember that 0! = 1! = 1.)

81. ∞

 ( −1) n =1

82. n −1

n −1

∞

xn  n =1 ( n −1) ! 6

x n x =  ( −1) n! ( n −1) ! n=0

83.

84. 72

 0.06  A72 = 20,000 1 +  ≈ $28,640.89. 12   So, she has approximately $28,640.89 in her account after 6 years (or 72 months).

 0.05  A12 = 7,000 1 +  ≈ $8,125.28 4   So, she has approximately $8,125.28 in her account after 3 years (or 36 months).

85. Let n = number of years experience. Then, the salary per hour is given by

86. Here, an = 480,000 + 75,000 ( n −1) , n ≥ 1.

an = 20 + 2n, n ≥ 0. So, a20 = 20 + 2(20) = 60. Thus, a paralegal with 20 years experience would make $60 per hour.

So, an = salary for a 3-year career. n =1

1131

Chapter 9 88. Let n = # 20 minute periods. Then, the number of E.coli cells is

87. Let n = number of years on the job. Then, the salary is given recursively by a0 = 30,000 an =

described by: a0 = 2

an −1 + 0.03an −1 = 1.03an −1      

an = 2 ⋅ an −1 = 2 n+1 , n ≥ 1.

raise

previous year

After 12 hours, 36 20-minute periods have passed and so, a36 E.coli cells exist, which equals 237 ≈ 1.374 × 1011. After 48 hours, 144 20-minute periods have passed and a144 E.coli cells exist, which equals 2145 ≈ 4.46 × 10 43.

89. Let n = number of years. Then, the number of T-cells in body is given by: a1 = 1000

90. a3 = 3.8 + 1.6 ( 3 ) = 8.6 billion in sales

a4 = 3.8 + 1.6 ( 4 ) = 10.2 billion in sales 1 4 1 an = ( 8.6 + 10.2 ) = 9.4  2 n=3 2 represents the average sales for the years 2020 and 2021.

So,

an+1 = 1000 − 75n, n ≥ 1. We must find n such that an+1 ≤ 200. To do so, note that the above formula can be expressed explicitly as an+1 = 1000 − 75n. So, we must solve 1000 − 75n = 200 : 75n = 800 n ≈ 10.7 As such, after approximately 10.7 years, the person would have full blown AIDS.

91. Observe that A1 = 100,000 (1.001 − 1) = 100

92. Observe that A1 = 50,000 (1.001 − 1) = 50

A2 = 100,000 (1.001) − 1 = 200.10

A2 = 50,000 (1.001) − 1 = 100.05

( ) A = 100,000 ( (1.001) − 1) ≈ 300.30 2

( ) A = 50,000 ( (1.001) − 1) ≈ 150.15 2

1132

Section 9.1

( ) A = 100,000 ( (1.001) − 1) ≈ 3663.72

( ) A = 50,000 ( (1.001) − 1) ≈ 2457.27

A4 = 100,000 (1.001) − 1 ≈ 400.60

A4 = 50,000 (1.001) − 1 ≈ 200.30

93.

94. At x = 2 1 3 5

1 1+ x 2 1+ x + x2 3

≈ 7.0

1+ x + x2 + x3! + x4! 2

= 239,704.7886 So, approximately $239,705.

≈ 6.33

1+ x + x2 + x3! 2

A6 = 195,000(1.035)6

Approx. of e 2 ≈ 7.38906 95.

96.  (1.06)n −1  FV = 5000    0.06 

At x = 1.1 0.1

( x −1)

( x − 1) − ( x −21)

0.0953

( x − 1) − ( x −21) + ( x −31) 2

( x − 1) − ( x −21) + ( x −31) − ( x −41) 2

( x − 1) − ( x −21) + ( x −31) 2

−( 4) + ( 5) x −1

n 1 2 3 4 5

0.0950

x −1

0.09531 0.095310 Approx of ln(1.1)

FV 5,000 10,300 15,918 21,873.08 28,185.4648

97. The mistake is that 6! ≠ 3!2!, but rather 6! = 6 ⋅ 5 ⋅ 4 ⋅3⋅ 2 ⋅1. 98. The mistake is that ( 2 n − 2 ) ! ≠ ( 2 n − 2 )( 2n − 4 )( 2n − 6 ) ⋅  ⋅ (2) The terms should be consecutive and decrease by 1 (not 2). Therefore, it should be

( 2n − 2 ) ! = ( 2n − 2 ) ( 2n − 3)( 2n − 4 ) ⋅  ⋅ (1).

The same error is made when computing ( 2n + 2 ) !.

1133

Chapter 9 100. Same error as 93.

99.

 1, n = 1,3,5, = −1, n = 2, 4,6, So, the terms should all be the opposite sign.

( −1)

n+1

103. False. Let ak = 1 = bk and n = 5. Then, observe that n

k =1

 ak bk = 1 = 1⋅5 = 5

101. True 102. True 104. False. Let a = 3, b = 2. Then, observe that

( a !) ( b !) = (3!)( 2!) = ( 3⋅ 2 ⋅1)( 2 ⋅1) = 12 whereas

whereas  n   n   5   5  ⋅ a b   k    k  =  1 ⋅  1 = 25.  k =1   k =1   k =1   k =1  105.

( ab ) ! = ( 2 ⋅ 3) ! = 6! = 720.

106. a1 = C

a1 = C

a2 = (C ) + D

a2 = D (C )

a3 = (C + D ) + D = C + 2D

a3 = D ( DC ) = D 2C

a4 = D ( D 2C ) = D3C

a4 = (C + 2D ) + D = C + 3D

107.

(1 + 5 ) − (1− 5 ) = 2 5 = 1 F = 1

2 5

(1 + 5 ) − (1− 5 ) 2

1 2

2 5

F2 =

108. n

1 + 2 5 + 5 −1 + 2 5 − 5 = =1 4 5

an = an −1 , n ≥ 2

7 7 =7 2 1

7 =7 4 1

7 =7 8 1

7 = 7 16 1

In general, an = 7 2 n−1 , n ≥ 2.

1134

Section 9.1

109. Using a calculator yields the following table of values: n

 1 1+   n 2.704813 ≈ 2.705 2.716923 ≈ 2.717 2.718146 ≈ 2.718 2.718268 2.718280 2.718281693 ↓ e

100 1000 10000 100000 1000000 10000000 ↓ ∞ 110. n

an +1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

1 1 2 3 5 8 13 20 33 53 86 139 225 364 589 953 1542 2498 4037

1 2 3 5 8 13 20 33 53 86 139 228 364 589 953 1542 2495 4037

111. The calculator gives 109 15 , which agrees with Exercise 61.

an +1 an 1 2 1.5 1.66 1.6 1.625 1.53846 1.65 1.60606 1.62264 1.616279 1.618705 1.61777 1.618131 1.617996 1.618048 1.618028 1.618036

11 112. The calculator gives 30 , which agrees with Exercise 62.

1135

Section 9.2 Solutions--------------------------------------------------------------------------------1. Yes, d =3.

2. Yes, d = −3.

3. No.

4. No.

5. Yes, d = − 0.03.

6. Yes, d = 0.5.

7. Yes, d = 2/3.

8. Yes, d =1 3.

9. Yes, d = 1/4.

10. No

11. No.

12. No.

13. First four terms: 3,1, − 1, − 3 So, yes and d = − 2.

14. First four terms: −7, − 4, − 1,2 So, yes and d =3.

15. First four terms: 1, 4,9,16 So, no.

16. First four terms are: 1, 2!4 , 3!9 , 164!    4 =2 2

9 =3 3 2 2

16 = 2 4 3 2 3

3 2 which is equivalent to 1,2, , . So, no. 2 3 17. First four terms: 2,7,12,17 So, yes and d = 5.

18. First four terms: 1, − 3, − 7, − 11 So, yes and d = − 4.

19. First four terms: 0,10,20,30 So, yes and d = 10.

20. First four terms: 4,12,20,28 So, yes and d =8.

21. First four terms: −1,2, − 3, 4 So, no.

22. First four terms: 2, − 4,6, − 8 So, no.

7 5 23. First four terms: − , −3, − , −2 2 2 1 So, yes and d = 2 25. an = 11 + ( n −1) 5 =5n + 6

24. First four terms:

So, yes and d = −

5 1 1 , , − , −1 4 2 4

3 4

26. an = 5 + ( n −1)11=11n − 6

1136

Section 9.2

27. an = − 4 + ( n −1) (2) = − 6 + 2n 29. an = 0 + ( n −1)

2 2 2 = n− 3 3 3

28. an = 2 + ( n − 1) ( −4 ) = − 4 n + 6

1  −3  −3 30. an = − 1 + ( n − 1)   = n − 4  4  4

31. an = 0 + ( n − 1) e = en − e

32. an = 1.1 + ( n −1) ( −0.3 ) = − 0.3n +1.4

33. Here, a1 = 7, d =13.

34. Here, a1 = 7, d = − 6.

So, an = 7 + ( n −1)13 = 13n − 6.

So, an = 7 + ( n −1) ( −6 ) = −6n +13

Thus, a10 = 130 − 6 = 124.

Thus, a19 = −6 (19 ) +13 = −101.

35. Here, a1 = 9, d = − 7.

36. Here, a1 =13, d = 6.

So, an = 9 + ( n −1) ( −7 ) = − 7n + 16

So, an = 13 + ( n −1) ( 6 ) = 6n + 7

Thus, a100 = − 700 + 16 = − 684 .

Thus, a90 = 6 ( 90 ) + 7 =547.

37. Here, a1 = 13 , d = 14 . So, an = 13 + ( n − 1) ( 14 ) . Thus, a21 = 13 + (20)( 14 ) = 163 . 38. Here, a1 = 15 , d = 13 . So, an = 51 + ( n −1) ( 31 ) . Thus, a33 = 51 + (32)( 31 ) = 163 15 . 39. a5 = 44

Find d :a17 = a5 + 12d

a17 = 152

152 = 44 + 12d

Find a1:a5 = a1 + 4 ( 9 ) 44 = a1 + 36

a21 = −55

= 9n − 1

8 = a1

9=d

40. a9 = −19

So, an = 8 + ( n −1) 9

Find d :a21 = a9 + 12d Find a1:a9 = a1 + 8 ( −3 ) So, an = 5 + ( n − 1) ( −3 ) −55 = −19 +12d −36 = 12d −3 = d

−19 = a1 − 24 5 = a1

= −3n + 8

41. a7 = −1 Find d : a17 = a7 + 10d Find a1:a7 = a1 + 6 ( −4 ) So, an = 23 + ( n −1) ( −4 ) a17 = −41

−41 = −1+10d −40 = 10d −4 = d

−1 = a1 − 24 23 = a1

1137

= −4 n + 27

Chapter 9 Find d : a21 = a8 + 13d Find a1 : a8 = a1 + 7 ( 5 ) So, an = 12 + ( n −1) ( 5 )

42. a8 = 47 a21 =112

112 = 47 + 13d 65 =13d d =5

47 = a1 + 35

= 5n + 7

12 = a1

2 2 Find d : a22 = a4 + 18d Find a1 : a4 = a1 + 3   So, an =1 + ( n − 1) 3 3 2 1 15 = 3 + 18d a22 =15 3 = a1 + 2 = n+ 3 3 12 =18d 1 = a1 2 3 =d

43. a4 = 3

 1  −1  44. a11 = − 3 Find d : a31 = a1 + 20d Find a1 : a11 = a1 + 10  −  So, an = 2 + ( n −1)    2  2  1 5 a31 = − 13 −13 = − 3 + 20d −3 = a1 − 5 =− n+ 2 2 −10 = 20d 2 = a1 1 d =− 2 45.

46.

k =1

 2k = 2  k = 2 [2 + 46] =552

k =0

↑ k =1

 5k = 5 k = 2 [5 + 100] =1050 Note that the 1st term when k = 0 is 0 .

47.

48.

n =1

 ( −2n + 5 ) = − 2  n +  5 n =1

 (3n − 10 ) = −10 +  (3n − 10 ) n=0

n =1

17 = −10 + [ −7 + 41] 2 = 279

30 = [3 − 55] = −930 + 150 = − 780 2 49.

50.

  0.5 j =   0.5 j −  0.5 j   j =3 j =1   j =1 3 14 = [ 0.5 + 7 ] − = 0.5 [102 ] =51 2 2 14

33  1

33 

 4 = 2  4 + 4  =140.25 j =1

1138

Section 9.2

51. 2 + 7 + 12 +  + 62 . We need to find n such that an = 62 . To this end, observe    arithmetic

that a1 = 2 and d = 5 . Thus, since the sequence is arithmetic, an = 2 + ( n −1) 5 =5n − 3 . So, we need to solve an = 5n − 3 = 62 :

5n = 65, so that n = 13 . Therefore, this sum equals

13 [2 + 62] = 416 . 2

52. 1 − 3 − 7 − ... − 75 . We need to find n such that an = −75 . To this end, observe  arithmetic

that a1 = 1 and d = − 4 . Thus, since the sequence is arithmetic, an = 1 + ( n − 1)( −4 ) = − 4n + 5 . So, we need to solve an = −4 n + 5 = − 75 :

−4n = − 80, so that n = 20 . Therefore, this sum equals

20 [1− 75] = −740 . 2

53. 4 + 7 + 10 +  + 151 . We need to find n such that an = 151 . To this end,   arithmetic

observe that a1 = 4 and d = 3 . Thus, since the sequence is arithmetic, an = 4 + ( n − 1) 3 =3n + 1 . So, we need to solve an = 3n + 1 =151 :

3n =150, so that n = 50 . Therefore, this sum equals =

50 ( 4 + 151) = 3875 . 2

54. 2 + 0 − 2 −  − 56 . We need to find n such that an = − 56 . To this end, observe  arithmetic

that a1 = 2 and d = − 2 . Thus, since the sequence is arithmetic, an = 2 + ( n − 1) ( − 2 ) = − 2n + 4 .

So, we need to solve an = −2 n + 4 = − 56 : −2n = − 60, so that n = 30 . 30 Therefore, this sum equals ( 2 − 56 ) = − 810 . 2

1139

Chapter 9 55. 1 1 1 13 − − − − . We need to find n such that an = − 132 . To this end, observe 6 6 2  2  arithmetic

that a1 = 61 and d = − 13 . Thus, since the sequence is arithmetic, 1 1 1 1 1 13  −1  an = + ( n −1)   = − n + . So, we need to solve an = − n + = − : 6 3 2 3 2 2  3  −1 n = − 7, so that n = 21 . 3 21  1 13  21  1−39  133 − =  . Therefore, this sum equals =−  2 6 2  2  6  2 56. 11 7 17 14 + + + + . We need to find n such that an = 143 . To this end, observe 12 2  3 61 arithmetic

7 11 3 1 11 and d = − = = . Thus, since the sequence is arithmetic, that a1 = 12 6 12 12 4 11 2 1 1 an = + ( n −1)   = n + . 12 3 4 4 1 2 14 1 12 n = = 4, so that n = 16 . So, we need to solve an = n + = : 4 3 3 4 3 16  11 14   111− 56  134 . Therefore, this sum equals =  +  = 8  = 2 12 3   12  3 57. Here, d = 4 and a1 = 1 . Hence, an = 1 + ( n − 1)(4) = 4 n − 3 . So,

58. Here, d = 3 and a1 = 2 . Hence, an = 2 + ( n − 1)(3) = 3n − 1 . So,

Sn = 2n ( a1 + an ) = 2n (1 + 4 n − 3 ) = n(2 n − 1). Sn = 2n ( a1 + an ) = 2n ( 2 + 3n − 1) = 2n (3n +1) .

As such, S18 = 18(35) = 630.

As such, S21 = 212 (64) = 672.

59. Here, d = − 12 and a1 = 1 . Hence,

60. Here, d = − 32 and a1 = 3 . Hence,

Sn = 2n ( a1 + an ) = 2n (1− 21 n + 32 )

Sn = 2n ( a1 + an ) = 2n ( 3 − 32 n + 92 )

an = 1 + ( n −1) ( − 12 ) = − 12 n + 23 . So,

= 2n ( − 21 n + 52 ) .

As such, S43 = 434 ( 5 − 43 ) = −

817 . 2

an = 3 + ( n −1) ( − 32 ) = − 32 n + 92 . So,

= 2n ( − 23 n + 152 ) .

As such, S37 = 374 (15 − 3(37) ) = −888.

1140

Section 9.2

61. Here, d = 10 and a1 = −9 . Hence, an = −9 + ( n − 1)(10) = 10 n − 19 . So,

62. Here, d = 10 and a1 = −2 . Hence, an = −2 + ( n − 1)(10) = 10 n − 12 . So,

= n(5n − 14). As such, S18 = 18(5(18) − 14) = 1368.

= 2n (10 n − 14). As such, S21 = 21(5(21) − 7) = 2058.

Colin a1 = 28,000 d =1500 So an = 28,000 + ( n −1) (1500 ) = 26,500 + 1500n Thus, a10 = 41,500. After 10 years, Colin will have accumulated 102 (28,000 + 41,500) = $347,500.

Camden a1 = 25,000 d = 2000 So, an = 25,000 + ( n −1) 2000 = 23,000 + 2000n So, a10 = 43,000. After 10 years, Camden will have accumulated 102 (25,000 + 43,000) = $340,000.

Jasmine a1 = 80,000 d = 2000. So, an = 80,000 + ( n −1) ( 2000 ) = 78,000 + 2000n Thus, a15 =108,000. After 15 years, Jasmine will have accumulated 152 (80,000 + 108,000) = $1,410,000.

Megan a1 = 90,000 d =5000 So, an = 90,000 + ( n −1) 5000 = 85,000 + 5000n Thus, a15 = 160,000. After 15 years, Megan will have accumulated 152 (90,000 + 160,000) = $1,875,000.

Sn = 2n ( a1 + an ) = 2n ( −9 + 10 n − 19 )

Sn = 2n ( a1 + an ) = 2n ( −2 + 10 n −12 )

63.

64.

65. We are given that a1 = 22 . We seek  ai . Observe that d = 1 . So, i =1

an = 22 + ( n −1) (1) = 21 + n . Thus, a25 = 21 + 25 = 46 . As such, using the formula

Sn =

25 n 25 ( a1 + an ) , we conclude that  ai = = [ 22 + 46] = 850 . 2 2 i =1

1141

Chapter 9 20

66. We are given that a1 = 1 . We seek  ai . Observe that d = 1 . So, i =1

an = 1 + ( n − 1) (1) = n . Thus, a20 = 20 . As such, using the formula Sn = 20

we conclude that  ai = = i =1

n ( a1 + an ) , 2

20 [1 + 20] = 210 . So, there are 210 tulips in each delta. 2

67. We are given that a = 1 and n = 66. We need to find a66 and d . Note that 66

a = 49, 401 = 2 (1+ a ) . 66

i =1

49,368 = a66 . 33 Now, since an = a1 + ( n −1) d , we can substitute n = 66, a1 = 1 in to find d :

Hence, 49, 401 = 33 (1 + a66 ) , so that 1496 =

1496 = 1 + ( 66 −1) d , so that 1495 = 65d and hence, d = 23. So, there were 1496 glasses on the bottom row and each row had 23 fewer glasses than the one below.

68. From the information given in the problem, we infer that a1 = 1 , a25 = 25 , and d = 1 . 25 Since the sequence is arithmetic, the sum must be [1 + 25] = 325 logs in the pile . 2 69. We are given that a1 = 16, d = 32, n = 10 . Hence, an = 16 + ( n −1) ( 32 ) = −16 + 32 n .

So, a10 = −16 + 320 = 304 . Now, observe that 10 10 ai = [16 + 304]  2 i =1 = 1600 feet in 10 seconds

70. We are given that ai = 4.9 , d = 9.8 , n = 10 . Hence, an = 4.9 + ( n −1) ( 9.8 ) = −4.9 + 9.8n

So, a10 = 93.1 . Now, observe that 10

 a = 2 [ 4.9 + 93.1] i

i =1

1142

= 490 meters in 10 seconds

Section 9.2

71. The sum is 20 + 19 + 18 + . . . + 1. This is arithmetic with a1 = 20, d = −1 . So, an = 20 + ( n − 1)( −1) = 21 − n . Thus, Sn = 2n (20 + 21 − n ) = 2n (41 − n ) . So, S20 = 10(21) = 210 . There are 210 oranges in the display.

72. The sum is 45000 + 46500 + … + an . This is arithmetic with a1 = 45000, d = 1500 . So, an = 45000 + ( n − 1)(1500) = 1500 n + 43500 a. a35 = 96,000 -- salary in year 35 b. Total earnings in 35 years = S35

= 352 ( 45000 + 1500(35) − 43500 ) = 2, 467,500

73. a1 + a2 + a3 + 26 + 27 + 28 + ... + an Here, d = 1 , so that a1 = 23. a. There are 23 seats in the first row. b. an = 23 + ( n − 1)(1) = n + 22. So, Sn = 2n (23 + n + 22) , and hence, S30 = 1125 seats in the theater.

74. Here d = 2e and a1 = 1e . Hence, an = 1e + ( n − 1)( 2e ) = − 1e + 2e n . So,

Sn = 122 ( 1e + 23e ) = 144 e . Since there are 12

terms all told, the total is S12 = 144 e .

75. an = a1 + ( n −1) d , not a1 + nd

76. Here, d = −2 since consecutive terms decrease by 2

77. There are 11 terms, not 10. So, 11 n = 11 , and thus, S11 = (1 + 21) = 121. 2

78. Sn =

79. False. In a series you are adding terms, while in a sequence you are not.

80. False. For instance,

n n ( an + a1 ) , not Sn = ( an − a1 ) 2 2

∞

 ( 2i + 1) doesn’t exist. All FINITE i =1

arithmetic series can be computed, however. 81. True, since d must be constant. If a sequence alternates, the difference between consecutive terms would need to change sign.

82. False. If the terms are decreasing, d would be negative.

1143

Chapter 9 83. n +1

a + ( a + b ) +  + ( a + nb ) =  ( a + ( k − 1)b ) = k =1

k =−29

( n + 1)( 2a + nb ) n +1 ( a + a + nb ) = 2 2

84. First, observe that  ln e k =  k = −29 − 28 −  −1+ 0 +1+  + 30 .

Now, we simplify as follows: = − 29 − 28 −  − 2 − 1 0 30 + 29 + 28 +  + 2 + 1 30

So, the sum is 30. 85. As n gets larger, n12 goes to zero.

(

)

So, v = R k12 − n12 gets closer to kR2 , which in this case is 27,419.5.

86. Here, a1 = −4, d = 6 . So, an = −4 + ( n − 1)(6) = 6 n − 10 . Thus, Sn = 2n ( −4 + 6 n − 10) = n(3n − 7) . Solve n(3n − 7) = 570 : 3n 2 − 7 n − 570 = 0 7 ± 49 + 4(3)(570) 6 = 15 or − 383

100

87. Compute  i − you should get

88. Compute  2i − you should get

5050.

2550.

i =1

89. Compute  ( 2i − 1) − you should

i =1

90. Same answer.

i =1

get 2500. 91. 18,850

92. 12,320

1144

Section 9.3

Section 9.3 Solutions--------------------------------------------------------------------------------1. Yes, r = 3

2. Yes, r = 2

No,

9 16 ≠ for instance 4 9

1 1 No, 9 ≠ 16 for instance 1 1 4 9

Yes, r =

1 2

Yes, r = −

1 2

7. Yes, r = 1.7

8. Yes, r = 2.2

9. 6,18,54,162, 486

10. 17,34,68,136,272

11. 1, − 4,16, − 64,256

12. −3,6, − 12,24, − 48

13. 10,000, 10,600, 11,236,

14. 10000, 8000, 6400, 5120, 4096

11,910.16, 12,624.77 15. 2 1 1 1 1 , , , , 3 3 6 12 24

16. 1 1 1 1 1 ,− , ,− , 10 50 250 1250 6250

17. an = 5 ( 2 )

18. an = 12 ( 3 )

n−1

19. an = 1( −3 )

23.

an =

20. an = −4 ( −2 )

n−1

21. an = 1000 (1.07 )

n−1

16  1  −  3  4

n−1

22. an = 1000 ( 0.5 ) 24.

n−1

an =

1145

1 n−1 (5) 200

n−1

Chapter 9 25. Since a1 = −2, r = −2 , we have

an = −2 ( −2 ) In particular, 7 −1 a7 = −2 ( −2 ) = −128.

n−1

26. Since a1 = 1, r = −5 , we have

an = ( −5 )

n−1

In particular, 10−1 a10 = ( −5 ) = −1,953,125.

27.

28.

1 Since a1 = , r = 2 , we have 3 1 n−1 an = ( 2 ) 3 1 13−1 In particular, a13 = ( 2 ) = 4096 3 . 3

1 Since a1 = 100, r = , we have 5 n −1 1 an = 100   5 In particular,

29.

30.

Since a1 = 1000, r =

1 , we have 20

 1  an = 1000    20  In particular,

n −1

≈ 6.10 ×10 −16.

31.

9 −1

= 2.56 ×10 −4.

4 Since a1 = 1000, r = − , we have 5  4 an = 1000  −   5 In particular,

15 −1

 1  a15 = 1000    20 

1 a9 = 100   5

(1 − r ) .

 4 a8 = 1000  −   5

8 −1

= −209.7152.

32.

Use the formula Sn = a1

1− r 1 Since a1 = , n = 13, r = 2, the given 3 1  1− 213  8191 sum is S13 =  . = 3  1− 2  3

n −1

(1 − r ) . n

Use the formula Sn = a1

1− r Since a1 = 1, n = 11, r = 13 , the given sum

  1 11   1−    3 is S11 = 1    ≈ 1.5. 1    1− 3   

1146

Section 9.3

33.

34.

(1 − r ) . Use the formula S = a n

1− r Since a1 = 2, n = 10, r = 3 , the given sum is  1 − 310  S10 = 2   = 59,048 .  1− 3 

1− r Since a1 = 1, r = 4, n = 10 , the given sum

35.

36.

(1 − r ) . Use the formula S = a

 1 − 410  is S10 = 1  = 349,525 .  1− 4 

(1 − r ) . Use the formula S = a

1− r Since a1 = 2, r = 0.1, n = 11 , the given sum is  1 − ( 0.1)11  S11 = 2   = 2.2 .  1 − 0.1    (Note: The sum starts with n = 0 , so there are 11 terms.)

1− r Since a1 = 3, r = 0.2, n = 12 , the given sum is  1 − ( 0.2 )12  S12 = 3   ≈ 3.75  1 − 0.2    (Note: The sum starts with n = 0 , so there are 12 terms.)

37.

38.

(1 − r ) . Use the formula S = a n

1− r Since a1 = 2, r = 3, n = 8 , the given sum is  1 − 38  S8 = 2   = 6560 .  1− 3 

2 Since a1 = , r = 5, n = 9 , the given sum 3 2  1 − 59  is S9 =   ≈ 325,520.6667 . 3  1− 5 

39.

40.

(1 − r ) . Use the formula S = a n

 1 − 214  is S14 = 1  = 16,383 .  1− 2 

1− r

(1 − r ) . Use the formula S = a n

1− r Since a1 = 1, r = 2, n = 14 , the given sum n

1− r

1 Since a1 = 1, r = , n = 14 , the given sum 2   1 14   1−    2 is S14 = 1    ≈ 2.0 . 1    1−  2  

1147

Chapter 9 41.

42.

Use the formula S∞ =

a1 . 1− r

Use the formula S∞ =

1 , the given sum is 2 1 S∞ = = 2. 1 1− 2

a1 . 1− r

Since a1 = 1, r =

1 1 Since a1 = , r = , the given sum is 3 3 1 1 S∞ = 3 = . 1 2 1− 3

43.

44.

Use the formula S∞ =

a1 . 1− r

Use the formula S∞ =

a1 . 1− r

1 1 Since a1 = − , r = − , the given sum is 3 3 1 − 3 =−1 . S∞ = 4  1 1−  −   3

1 Since a1 = 1, r = − , the given sum is 2 1 2 = . S∞ =  1 3 1−  −   2

45. Not possible - diverges.

46. Not possible - diverges.

47.

48.

a Use the formula S∞ = 1 . 1− r 1 Since a1 = −9, r = , the given sum is 3 −9 27 S∞ = =− . 1 2 1− 3

Use the formula S∞ =

a1 . 1− r Since a1 = 10,000, r = 0.05 , the given 10,000 ≈ 10,526 . sum is S∞ = 1 − 0.05 49. Use the formula S∞ =

a1 . 1− r

1 Since a1 = −8, r = − , the given sum is 2 −8 16 =− . S∞ = 3  1 1−  −   2

a1 . 1− r Since a1 = 200, r = 0.04 , the given sum is 200 625 = . S∞ = 1 − 0.04 3

50. Use the formula S∞ =

1148

Section 9.3

52. The sum can be written as ∞ 3 ( 101 ) 1 1 n 3 , which equals = . ( )  10 1 − 101 3 n =1

51.

The sum is

0.4 2 = . 1 − 0.4 3

54. Diverges – infinite sum

53. 0

The sum is

(0.99) = 100. 1 − 0.99

55. Observe that a1 = 34,000, r = 1.025 . We need to find a12 . Observe that the nth term is given

56. Observe that a1 = 22,000, r = 1.15 .

a12 = 34000 (1.025 )

a10 = 22,000 (1.15 )

≈ 44,610.95 So, the salary after 12 years is approximately $44,610.95.

≈ 77,393.28 So, the salary after 10 years is approximately $77,393.28.

57. Since a1 = 2000, r = 0.5 , we see that

58. BMW a1 = 35,000

We need to find a10 . Observe that the nth term is given by n −1 n 1 − by an = a1r n −1 = 34,000 (1.025 ) . Hence, an = 22,000 (1.15 ) . Hence,

an =

2000 ( 0.5 )  n

Value of laptop after n years

In particular, we have a4 =

2000 ( 0.5 ) = 125  4

Worth when graduating from college.

a7 =

2000 ( 0.5 ) = 16  7

Worth when graduating from graduate school.

Honda a1 = 25,000

r = 0.8

r = 0.9

an = 35,000 ( 0.8 )

an = 25,000 ( 0.9 )

n −1

a10 = 35,000 ( 0.8 )

a10 = 25,000 ( 0.9 )

n −1

= 9685.51 = 4697.62 So, the Honda will be worth (much) more in 10 years.

59. Since a1 = 100, r = 0.7 , we see that

60. Since a1 = 200, r = 0.65 , we see that

an = 100 ( 0.7 ) .

an = 200 ( 0.65 ) .

We need to find a5 . We see that

We need to find a8 . We see that

a5 = 100 ( 0.7 ) ≈ 17 feet.

a8 = 200 ( 0.65 ) ≈ 6.37 feet.

1149

Chapter 9 62. Since a1 = 20000, r = 1.05 , we see

61. Since 37,800 a1 = 36,000, r = = 1.05 , 36,000

we see that an = 36,000 (1.05 )

n −1

that an = 20,000 (1.05 )

n −1

So, in particular, we have 51 a52 = 20,000 (1.05 ) ≈ 240,815 . So, there will be about 240,815 hits per week one year from now.

So, in particular, 10 a11 = 36,000 (1.05 ) ≈ 58,640 . As such, in the year 2010, there will be approximately 58,640 students. 63. Since a1 = 1000, r = 0.90 , we know

64. Since a1 = 0.01, r = 2 , we see that

that an = 1000 ( 0.9 )

an = 0.01( 2 ) . He would be paid $10,737,418.24 to work on January 31. Hence, the total paid out in January is given by: 31  1 − 231  n −1 0.01( 2 ) = 0.01   n =1  1− 2 

n−1

First, find the value of n0 such that an0 ≤ 1. Since 1000(0.9)n −1 < 1  (0.9)n −1 < 0.001

 n > ln(0.001) ln(0.9) ≈ 67

≅ $21, 474,836.47

We see that after 67 days, he will pay less than $1 per day. Next, the total amount of money he paid in January is given by: 31 1 − 0.931 ) ( n −1 ≈ 9618 1000 ( 0.9 ) = 1000  1 − 0.9 n =1 65. nt

 r Use the formula A = P  1 +  .  n Note that in the current problem, r = 0.05, n = 12, so that the formula becomes: 12t

12t  0.05  A = P 1 +  = P (1.0042 ) 12   n Let t = , where n is the number of 12 months of investment. Also, let n An = P (1.0042 ) .

First deposit of $100 gains interest for 36 36 months: A36 = $100 (1.0042 ) Second deposit of $100 gains interest 35 for 35 months: A35 = $100 (1.0042 )  Last deposit of $100 gains interest for 1 1 month: A1 = $100 (1.0042 )

Now, sum the amount accrued for 36 deposits: 36

A1 +  + A36 = 100 (1.0042 ) n =1

1150

n −1

Section 9.3

Note that here a1 = 100, r = 1.0042 . So,

So, they saved $3,877.64 in 3 years.

 1 − (1.0042 )36  So, S36 = 100    1 − 1.0042    ≈ 3877.64  Last deposit of $50 gains interest for 1 1 week: A1 = 50 (1.00077 ) Now, sum the amount accrued for 52 weeks:

66. nt

 r Use the formula A = P  1 +  . Note  n that in the current problem, r = 0.04, n = 52, so the formula becomes:

A1 +  + A52 =  50 (1.00077 )

52t

52t  0.04  A = P 1 +  = P (1.00077 ) 52   n , where n is number of weeks Let t = 52 of investment. Also, let n An = P (1.00077 ) .

n −1

n =1

Note that here a1 = 50, r = 1.00077 . So,

 1 − (1.00077 )52  So, S52 = 50    1 − 1.00077    ≈ 2651.71 First deposit of $50 gains interest for 52 So, he saved $2,651.71 in one year. 52 weeks: A52 = 50 (1.00077 )

Second deposit of $50 gains interest for 51 51 weeks: A51 = 50 (1.00077 ) 67. nt

 r Use the formula A = P  1 +  . Note that in the current problem,  n  0.06  r = 0.06, n = 52, so that the formula becomes: A = P  1 +  52   n , where n is the number of weeks of investment. Let t = 52 n Also, let An = P (1.0012 ) .

1151

52t

= P (1.0012 )

52t

Chapter 9 26 week investment: First deposit of $500 gains interest for 26 26 weeks: A26 = 500 (1.0012 ) Second deposit of $500 gains interest for 25 25 weeks: A25 = 500 (1.0012 )

52 week investment: First deposit of $500 gains interest for 52 52 weeks: A52 = 500 (1.0012 ) Second deposit of $500 gains interest 51 for 51 weeks: A51 = 500 (1.0012 )

 Last deposit of $500 gains interest for 1 1 week: A1 = 500 (1.0012 )

Now, the amount accrued for 26 weeks:

Now, the amount accrued for 52 weeks:

A1 +  + A26 =  500 (1.0012 )

n −1

n =1

Note that here a1 = 500, r = 1.0012 . So,

A1 +  + A52 =  500 (1.0012 )

n −1

n =1

 1 − (1.0012 )26  S26 = 500   ≈ 13,196.88 .  1 − 1.0012   

Note that here a1 = 500, r = 1.0012 . So,  1 − (1.0012 )52  S52 = 500   ≈ 26,811.75 .  1 − 1.0012   

68.

 Last deposit of $300 gains interest for 1 1 month: A1 = $300 (1.0042 )

 r Use the formula A = P  1 +  .  n Note that in the current problem, r = 0.05, n = 12, so that the formula becomes:

Now, sum the amount accrued for 36 deposits: 60

A1 +  + A60 =  300 (1.0042 )

12t

12t  0.05  A = P 1 +  = P (1.0042 ) 12   n Let t = , where n is the number of 12 months of investment. Also, let n An = P (1.0042 ) .

n −1

n =1

First deposit of $300 gains interest for 60 60 months: A36 = $300 (1.0042 ) Second deposit of $300 gains interest for 59 59 months: A35 = $300 (1.0042 )

1152

Note that here a1 = 300, r = 1.0042 . So,  1 − (1.0042 )60  So, S60 = 300    1 − 1.0042    ≈ 20, 422.66 So, she saved $20,422.66 in 5 years.

Section 9.3

69. Here, a1 = 320,000 , and 1+ r = 1+ 0.038 = 1.038 . n−1 So, an = 320,000 (1.038 ) , n ≥ 1.

70. Here, a1 = 9 and r = 13 . Each time the ball bounces, it travels up and down by the same distance. So, the total distance traveled is: ∞ 18 ( 13 ) n 9 +  2 ⋅ 9 ( 13 ) = 9 + = 18 ft. 1 − 13 n =1

Then, a15 = 539, 402. Thus, after 15 years, it is worth $539, 402.

72. The finite sum is 1 − 230 1⋅ = 230 − 1 = 1,073,741,823 . 1− 2 This makes Option 1 the better deal because it is $1,063,741,823 more than Option 2.

71.

The sum is

1 2

1 − 12

= 1.

74. Should be a1 = 2 , so that you have

73.

Should be r = −

1 3

 2⋅2

k −1

k =1

75. Should use r = −3 all the way through the calculation. Also, a1 = −12(not 4).

76. Formula for S∞ only applies if r < 1.

77. False. 1 1 1 1, − , , − is geometric with a1 = 1, 2 4 8 1 r=− 2

78. False. Finite geometric series can always be computed, but the sum of an infinite geometric series only

79. True. We could have either of the following: 1 1 1 1, , , (Here r = .) 2 4 2 1 1 1 1 1, − , , − , (Here r = − .) 2 4 8 2

80. False. Common ratio must be < 1 in

exists if r < 1.

absolute value in order for an infinite geometric series to exist.

1153

Chapter 9 82. Since a1 = 1, r = 2, n = 21 , and we can simplify the given series as:

81. ∞

a + a ⋅ b + + a ⋅ b n +  =  ab n −1

This sum exists for b < 1. In such case, the sum is S∞ =

 log102 =  2k

n=1

k =0

 1− 2  we have S21 = 1  = 2,097,151 .  1− 2  21

a . (Here, a1 = a,r = b, 1− b

a1 ) 1− r

83. This decimal can be represented as ∞ 47 47 100 1 n 47 = = . ( )  100 1 1 − 100 99 n =1

84.

2 is the sum of a geometric 1− x series, then a1 = 2 and r = x. Hence, If

∞

the series is  2x n −1 . n =1

a. First five terms: 2 + 2 x + 2x 2 + 2x3 + 2x 4 b. This series converges for x < 1 . 85. Since a1 = 1, r = −2, n = 50 , we have 50

 1 ( −2 ) k =1

k −1

 1 − ( −2 )50  = 1   1 − ( −2 )    = −375,299,968,947,541 ∞

1 since as 1− x n=0 n → ∞ , the graph of yn = 1 + …+ x n gets 1 . closer to the graph of 1− x Notes on the graph: Solid curve: y1 = 1 + x + x 2 + x3 + x 4 1 , assuming |x| < 1. Dashed curve: y2 = 1− x 87. We expect that  x n =

1154

86. Yes, you should get: 1 1 3 = . a1 = 1, r = and S∞ = 1 2 3 1− 3

Section 9.3

88. The plot of the two functions is:

89. The plot of the two functions is:

1 assuming The series will sum to 1+ x |x| < 1.

The series will sum to assuming that |x| < ½

1155

1 , 1 − 2x

Chapter 9 Section 9.4 Solutions –------------------------------------------------------------------------------1. Claim: n 2 ≤ n3 , for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . 12 ≤ 13 is true since 12 = 13 = 1 . Step 2: Assume the statement is true for n = k : k 2 ≤ k 3 Show the statement is true for n = k +1 : ( k + 1)2 ≤ ( k + 1)3 ( k + 1)2 = k 2 + 2k + 1

≤ k 3 + 2 k + 1 (by assumption) ≤ k 3 + 3k + 1 (since 2k ≤ 3k, for k > 0) ≤ k 3 + 3k 2 + 3k + 1 (since 3k 2 > 0) = ( k + 1)3 This completes the proof. 2. Claim: If 0 < x < 1, then 0 < x n < 1, for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . The statement If 0 < x < 1, then 0 < x < 1 tautologically true. Step 2: Assume the statement is true for n = k : If 0 < x < 1, then 0 < x k < 1 Show the statement is true for n = k + 1 : If 0 < x < 1, then 0 < x k +1 < 1 Starting from 0 < x k < 1, multiply through the inequality by x (which is k fine since x > 0 . This yields the inequality 0 < x ⋅ x < 1 ⋅ x , which  = x k +1

k +1

simplifies to 0 < x < x (1). Since from Step 1 we know that x < 1 , we can use this in (1) to further conclude that 0 < x k +1 < x < 1, so that 0 < x k +1 < 1, as desired. This completes the proof. 3. Claim: 2n ≤ 2 n , for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . 2(1) ≤ 21 is true since both terms equal 2. Step 2: Assume the statement is true for n = k : 2k ≤ 2 k Show the statement is true for n = k + 1 : 2(k + 1) ≤ 2 k +1

2(k + 1) = 2 k + 2 ≤ 2 k + 21 (by assumption) ≤ 2 k + 2 k (since 2 ≤ 2 k , for k > 0) = 2(2 k ) = 2 k +1 This completes the proof.

1156

Section 9.4

4. Claim: 5n < 5n +1 , for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . 51 < 51+1 = 52 = 25 is clearly true. Step 2: Assume the statement is true for n = k : 5k < 5k +1 Show the statement is true for n = k + 1 : 5k +1 < 5k + 2 5k +1 = 5k ⋅ 5 < 5k +1 ⋅ 5 (by assumption) = 5k + 2 This completes the proof. 5. Claim: n ! > 2 n , for all n ≥ 4 . Proof. Step 1: Show the statement is true for n = 4 . 4! > 2 4 since 4! = 4 ⋅ 3 ⋅ 2 ⋅1 = 24 and 2 4 = 16 . Step 2: Assume the statement is true for n = k : k ! > 2 k Show the statement is true for n = k +1 : ( k + 1)! > 2 k +1 ( k + 1)! = ( k + 1) ⋅ k !

> (k + 1) ⋅ 2 k (by assumption) > 2 ⋅ 2k

(since k + 1 > 2 for k ≥ 4)

= 2 k +1 This completes the proof. 6. Claim: For any c ≥ 1 , (1 + c)n ≥ nc, for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . (1 + c)1 ≥ 1 ⋅ c is clearly true. Step 2: Assume the statement is true for n = k : (1 + c )k ≥ k c Show the statement is true for n = k + 1 : (1 + c )k +1 ≥ ( k + 1)c ck + c ≤ (1 + c)k + c ≤ (1 + c )k + (1+ c ) ≤ (1 + c )k + (1+ c )k = 2(1 + c)k ≤ (1 + c )(1+ c )k (since c ≥ 1) = (1 + c)k +1

This completes the proof.

1157

Chapter 9 7. Claim: n( n +1)(n − 1) is divisible by 3, for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . (1)(1+ 1)(1 − 1) = 1(2)(0) = 0 , which is clearly divisible by 3. Step 2: Assume the statement is true for n = k : k ( k +1)(k − 1) is divisible by 3 Show the statement is true for n = k + 1 : (k + 1)( k + 2)( k ) is divisible by 3 First, note that since by assumption k ( k +1)(k − 1) is divisible by 3 , we know that there exists an integer m such that k ( k + 1)(k − 1) = 3m (1) . Now, observe that (k + 1)( k + 2)(k) = (k 2 + 3k + 2)(k) = k 3 + 3k 2 + 2k (At this point, write 2k = 3k − k and group the terms as shown.)

= ( k 3 − k ) + (3k 2 + 3k) = 3m + 3( k 2 + k)

(by (1) )

= 3(m + k 2 + k) Now, choose p = m + k 2 + k to see that you have expressed ( k + 1)( k + 2)( k) as 3p , thereby showing ( k + 1)( k + 2)(k) is divisible by 3. This completes the proof. 8. Claim: n3 − n is divisible by 3, for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . 13 − 1 = 0 , which is clearly divisible by 3. Step 2: Assume the statement is true for n = k : k 3 − k is divisible by 3 Show the statement is true for n = k + 1 : (k + 1)3 − ( k + 1) is divisible by 3 First, note that since by assumption k 3 − k is divisible by 3 , we know that there exists an integer m such that k 3 − k = 3m (1) . Now, observe that ( k + 1)3 − ( k + 1) = ( k 3 + 3k 2 + 3k +1) − ( k +1) = k 3 + 3k 2 + 2k (At this point, write 2k = 3k − k and group the terms as shown.) = ( k 3 − k ) + (3k 2 + 3k) = 3m + 3( k 2 + k)

(by (1) )

= 3(m + k + k) 2

Now, choose p = m + k 2 + k to see that you have expressed ( k + 1)3 − ( k + 1) as 3p , thereby showing ( k + 1)3 − ( k + 1) is divisible by 3. This completes the proof. 1158

Section 9.4

9. Claim: n2 + 3n is divisible by 2, for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . 12 + 3(1) = 4 = 2( 2) , which is clearly divisible by 2. Step 2: Assume the statement is true for n = k : k 2 + 3k is divisible by 2 Show the statement is true for n = k +1 : (k + 1)2 + 3( k + 1) is divisible by 2 First, note that since by assumption k 2 + 3k is divisible by 2 , we know that there exists an integer m such that k 2 + 3k = 2m (1) . Now, observe that ( k + 1)2 + 3( k + 1) = ( k 2 + 2k +1) + 3 ( k +1) = k 2 + 5k + 4

(At this point, write 5k = 3k + 2 k and group the terms as shown.) = ( k 2 + 3k ) + (2k + 4) = 2 m + 2( k + 2) (by (1) ) = 2(m + k + 2) Now, choose p = m + k + 2 to see that you have expressed ( k + 1)2 + 3( k + 1) as 2 p , thereby showing ( k + 1)2 + 3( k + 1) is divisible by 2. This completes the proof.

10. Claim: n( n +1)(n + 2) is divisible by 6, for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . (1)(1+ 1)(1 + 2) = 1(2)(3) = 6 , which is clearly divisible by 6. Step 2: Assume the statement is true for n = k : k ( k +1)(k + 2) is divisible by 6 Show the statement is true for n = k +1 : (k + 1)( k + 2)(k + 3) is divisible by 6 First, note that since by assumption k ( k +1)(k + 2) is divisible by 6 , we know that there exists an integer m such that k ( k + 1)(k + 2) = 6m (1) . Now, observe that (k + 1)( k + 2)(k + 3) = (k +1)( k 2 + 5k + 6)

= k 3 + 5k 2 + 6 k + k 2 + 5k + 6 = k 3 + 6 k 2 + 11k + 6 (At this point, write 6k 2 = 3k 2 + 3k 2 and 11k = 2 k + 9k and group the terms as shown.)

1159

Chapter 9 = ( k 3 + 3k 2 + 2k) + (3k 2 + 9k + 6) = 6 m + 3( k 2 + 3k + 2)

(by (1) )

= 6 m + 3 ( k + 2)(k +1)    For any k, one of these consecutive integers must be even. So, the product must be divisible by 2.

= 6m + 3(2 s ), for some integer s = 6(m + s) Now, choose p = m + s to see that you’ve expressed ( k + 1)( k + 2)(k + 3) as 6 p , thereby showing ( k + 1)( k + 2)(k + 3) is divisible by 6. This completes the proof. 11. Claim: 2 + 4 + ...+ 2n = n(n +1), for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . 2 = 1(2 ) , which is clearly true. Step 2: Assume the statement is true for n = k : 2 + 4 + ...+ 2k = k (k +1) Show the statement is true for n = k +1 : 2 + 4 + ...+ 2(k +1) = ( k +1)(k + 2) Observe that 2 + 4 + ... + 2k + 2( k +1) = ( 2 + 4 + ... + 2k ) + 2(k +1) = k ( k + 1) + 2( k + 1) (by assumption) = ( k + 2 )( k +1) This completes the proof.

12. Claim: 1 + 3 + ... + (2n − 1) = n2 , for all n ≥ 1 Proof. Step 1: Show the statement is true for n = 1 . 1 = 12 , which is clearly true. Step 2: Assume the statement is true for n = k : 1 + 3 + ... + (2k − 1) = k 2 Show the statement is true for n = k + 1 : 1 + 3 + ... + (2(k + 1) − 1) = ( k + 1)2 1 + 3 + ... + (2k − 1) + (2( k + 1) − 1) = (1 + 3 + ... + (2k − 1) ) + (2k + 1)  2 k +1

= k 2 + (2 k + 1) (by assumption) = (k + 1)2 (factor) This completes the proof.

1160

Section 9.4

13. Claim: 1+ 3 + ... + 3n = 3 2 −1 , for all n ≥ 1 Proof. Step 1: Show the statement is true for n = 1 . 2 1 + 31 = 3 2−1 = 4 is clearly true. n+1

Step 2: Assume the statement is true for n = k : 1+ 3 + ... + 3k = 3 2 −1 k+1

Show the statement is true for n = k + 1 : 1+ 3 + ... + 3k +1 = 3 2 −1 k+2

1+ 3 + ... + 3k + 3k +1 = (1+ 3 + ... + 3k ) + 3k +1

= 3 2 −1 + 3k +1 (by assumption) k+1

k +1

−1+2(3k +1 ) 2

k+1

= 3(3 2 )−1 = 3 2 −1 k+2

This completes the proof.

14. Claim: 2 + 4 + ... + 2 n = 2 n +1 − 2, for all n ≥ 1 Proof. Step 1: Show the statement is true for n = 1 . 2 = 21+1 − 2 is clearly true. Step 2: Assume the statement is true for n = k : 2 + 4 + ... + 2 k = 2 k +1 − 2 Show the statement is true for n = k + 1 : 2 + 4 + ... + 2 k +1 = 2 k + 2 − 2 2 + 4 + ... + 2 k + 2 k +1 = ( 2 + 4 + ... + 2 k ) + 2 k +1 = ( 2 k +1 − 2 ) + 2 k +1 (by assumption) = 2(2 k +1 ) − 2 = 2k +2 − 2

This completes the proof. n +1) 15. Claim: 12 + ... + n2 = n ( n +1)(2 , for all n ≥ 1 6 Proof. Step 1: Show the statement is true for n = 1 . +1) 12 = 1(1+1)(2(1) = 66 = 1 is clearly true. 6 k +1) Step 2: Assume the statement is true for n = k : 12 + ... + k 2 = k ( k +1)(2 6

Show the statement is true for n = k +1 : 1 + ... + (k +1) = 2

1161

2 k+3   ( k +1)( k + 2 )(2( k +1)+1) 6

Chapter 9 12 + ... + k 2 + ( k + 1)2 = (12 + ... + k 2 ) + ( k + 1)2 +1) = k ( k +1)(2k + ( k + 1)2 (by assumption) 6

= k ( k +1)(2k +61)+6( k +1)

= 2 k +3 k +k 6+6k +12k +6 3

= 2 k +9 k6 +13k +6 3

= ( k +1)( k +62 )(2k +3) This completes the proof. 2

16. Claim: 13 + ... + n3 = n ( n4+1) , for all n ≥ 1 Proof. Step 1: Show the statement is true for n = 1 . 2

13 = 1 (1+1) = 1 is clearly true. 4 2

Step 2: Assume the statement is true for n = k : 13 + ... + k 3 = k ( k4+1)

Show the statement is true for n = k + 1 : 13 + ... + ( k + 1)3 = ( k +1) 4( k + 2 ) 13 + ... + k 3 + ( k +1)3 = (13 + ... + k 3 ) + ( k +1)3 2

= k ( k4+1) + ( k + 1)3 (by assumption) 3

= k ( k +1) 4+ 4( k +1) =

(

( k +1)2 k 2 + 4( k +1)

)

4 2

= ( k +1) 4( k + 2)

This completes the proof. n 17. Claim: 11⋅2 + 21⋅3 + ... + n ( n1+1) = n+1 , for all n ≥ 1

Proof. Step 1: Show the statement is true for n = 1 . 1 1 1 1⋅2 = 1+1 = 2 , which is clearly true. Step 2: Assume the statement is true for n = k : 11⋅2 + 21⋅3 + ... + k ( k1+1) = kk+1

Show the statement is true for n = k + 1 : 11⋅2 + 21⋅3 + ... + ( k +1)(1 k + 2) = kk++12

1162

Section 9.4

1 1⋅2

(

)

1 + 21⋅3 + ... + k ( k1+1) + ( k +1)(1 k +2 ) = 1⋅21 + 2⋅3 + ... + k ( k1+1) + ( k +1)(1 k +2)

= kk+1 + ( k +1)(1 k +2 ) (by assumption) = ( kk (+k1)(+ 2k)++21) = ( kk+1+)(2kk ++12 ) 2

= ( k (+k1)(+1k)+ 2 ) = ( kk++12 ) This completes the proof. 18.

1 2⋅3

+ ... + ( n +1)(1 n + 2 ) = 2( nn+ 2 ) , for all n ≥ 1

Proof. Step 1: Show the statement is true for n = 1 . 1 1 1 , which is clearly true. 2⋅3 = 2(1+ 2 ) = 2(3)

Step 2: Assume the statement is true for n = k :

1 2⋅3

+ ... + ( k +1)(1 k +2) = 2( kk+ 2 )

Show the statement is true for n = k +1 :

1 2⋅3

1 + ... + ( k + 2 )(1 k +3) = 2(k+ k +3)

1 2⋅3

(

)

1 + ... + ( k +1)(1 k +2 ) + ( k +2 )(1 k +3) = 2⋅3 + ... + ( k +1)(1 k +2 ) + ( k +2 )(1 k +3)

= 2( kk+ 2) + ( k + 2 )(1 k +3) (by assumption) +3 )+ 2 = 2(kk(+k2)( k + 3) 3k +2 = 2(kk ++2)( k + 3) 2

( k + 2 ) ( k +1)

= 2 ( k +2 ) ( k +3 ) = 2(kk+1+3)

This completes the proof. 19. Claim: (1⋅ 2) + (2 ⋅3) + ... + n( n +1) = n ( n +13)( n + 2 ) , for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . +2 ) (1⋅ 2) = 1(1+1)(1 = 1(2)(3) = 2 , which is clearly true. 3 3 Step 2: Assume the statement is true for n = k : (1⋅ 2) + (2 ⋅3) + ... + k ( k +1) = k ( k +13)( k + 2 ) Show the statement is true for n = k + 1 : (1⋅ 2) + (2 ⋅3) + ... + (k + 1)( k + 2) = ( k +1)( k +3 2 )( k +3)

1163

Chapter 9 (1⋅ 2) + (2 ⋅3) + ... + k ( k +1) + ( k + 1)( k + 2) = ( (1⋅ 2) + (2 ⋅3) + ... + k ( k +1) ) +( k + 1)( k + 2) = k ( k +13)( k + 2 ) + ( k +1)( k + 2) (by assumption) =

k ( k +1)( k + 2 )+3( k +1)( k +2 ) 3

= ( k +1)( k +3 2 )( k +3) (factor out ( k + 1)( k + 2))

This completes the proof. +7 ) 20. Claim: (1⋅ 3) + (2 ⋅ 4) + ... + n( n + 2) = n ( n +1)(2n , for all n ≥ 1 . 6 Proof. Step 1: Show the statement is true for n = 1 . + 7) (1⋅ 3) = 1(1+1)(2(1) = 1(26)(9) = 3 , which is clearly true. 6 Step 2: Assume the statement is true for n = k : +7 ) (1⋅ 3) + (2 ⋅ 4) + ... + k ( k + 2) = k ( k +1)(2k 6 Show the statement is true for n = k +1 :

(1⋅ 3) + (2 ⋅ 4) + ... + ( k +1)( k + 3) =

2 k+9   ( k +1)( k + 2)(2( k +1)+7 ) 6

(1⋅ 3) + (2 ⋅ 4) + ... + k ( k + 2) + (k +1)(k + 3) = ( (1⋅ 3) + (2 ⋅ 4) + ... + k ( k + 2) ) +( k + 1)( k + 3) +7 ) = k ( k +1)(2k + (k +1)( k + 3) 6

(by assumption) =

k ( k +1)(2k + 7 )+6( k +1)( k +3) 6

( k +1)[ k (2k + 7 )+6( k +3)] 6

( k +1) 2 k +13k +18   6 2

= ( k +1)( k +62 )(2k +9)

This completes the proof. 21. Claim: 1+ x + ... + x n−1 = 1−x 1−x , for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . 1 1 = 1−x 1−x , which is clearly true. n

1164

(factor out (k + 1))

Section 9.4

Step 2: Assume the statement is true for n = k : 1+ x + ... + x k −1 = 1−x 1−x

k +1

Show the statement is true for n = k + 1 : 1+ x + ... + x k = 1−x 1−x 1+ ... + x k −1 + x k = (1+ ... + x k −1 ) + x k k = 1−x 1−x + x k

x (1−x ) = 1− x +1−x k

x −x x = 1− x +1−x k +1

= 1−x 1−x This completes the proof. 22. Claim:

1 2

+ 14 +... + 21n = 1 − 21n , for all n ≥ 1 .

Proof. Step 1: Show the statement is true for n = 1 . 1 1 2 = 1 − 21 , which is clearly true.

Step 2: Assume the statement is true for n = k :

1 2

+ 14 +... + 21k = 1 − 21k

Show the statement is true for n = k +1 :

1 2

+ 14 +... + 2k1+1 = 1− 2k1+1

1 2

( = (1 − ) +

)

+ 14 +... + 21k + 2k1+1 = 21 + 14 +... + 21k + 2k1+1 1 2k

1 2 k +1

k +1

= 2 2k−+12 +1 k +1

= 22k +−1 1 = 1 − 2k1+1 This completes the proof. 23. Claim: a1 + ( a1 + d ) + ... + (a1 + (n −1)d ) = 2n [ 2a1 + (n −1)d ] , for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . a1 = 12 [ 2a1 + (1−1)d ] = 12 [ 2a1 ] , which is clearly true. Step 2: Assume the statement is true for n = k : a1 + ( a1 + d ) + ... + (a1 + (k −1)d ) = k2 [ 2a1 + (k −1)d ]

1165

Chapter 9 Show the statement is true for n = k +1 : a1 + ( a1 + d ) + ... + (a1 + ((k +1) −1)d ) =

k +1 2

k       2a1 + (k +1−1)d   

Observe that a1 + ( a1 + d ) + ... + (a1 + (k −1)d ) + (a1 + kd ) = ( a1 + ( a1 + d ) + ... + (a1 + (k −1)d ) ) + (a1 + kd ) = k2 [ 2a1 + ( k −1)d ] + (a1 + kd )

= ka1 + k ( k2−1) d + a1 + kd = ( k + 1)a1 + kd( k2−1 +1) = ( k + 1)a1 + kd( k2+1 ) = ( k2+1) 2a1 + kd ( k2+1 ) = ( k2+1) [ 2a1 + kd ] This completes the proof.

( )

24. Claim: a1 + ra1 + ... + r n−1a1 = 11−r−r a1 , for all n ≥ 1 . n

Proof. Step 1: Show the statement is true for n = 1 . 1 a1 = 11−−rr a1 , which is clearly true. 

( ) =1

( ) Show the statement is true for n = k +1 : a + ra + ... + r a = ( )a

Step 2: Assume the statement is true for n = k : a1 + ra1 + ... + r k −1a1 = 11−r−r a1 k

1−r k +1 1−r

Observe that a1 + ra1 +...+ r k −1a1 + r k a1 = ( a1 + ra1 +...+ r k −1a1 ) + r k a1

( )

= 11−r−r a1 + r k a1 k

a 1− a1r k + r k a1 −r k+1a1

= a1 (1−r )1+−rr a1 (1−r ) = 1

1− r

(

k +1

)

= 1−r1−r a1

This completes the proof. 25. Label the disks 1, 2, and 3 (smallest = 1 and largest = 3), and label the posts A, B, and C. The following are the moves on would take to solve the problem in the fewest number of step. (Note: The manner in which the disks are stacked (from top to bottom) on each peg form the contents of each cell; a blank cell means that no disk is on that peg in that particular move.)

1166

Section 9.4

Initial placement

Move 1 Move 2 Move 3

Post A 1 2 3 2 3 3 3

Post B

1 1

Move 4 Move 5 Move 6

Post C

1 1

3 2 3 1 2 3

Move 7

2 1 2 1 2 2

So, the puzzle can be solved in as few as 7 steps. An argument as to why this is actually the fewest number of steps is beyond the scope of the text. (Note: Alternatively, we could have initially placed disk 1 on post C, and proceeded in a similar manner.) 26. Label the disks 1, 2, 3, and 4 (smallest = 1 and largest = 4), and label the posts A, B, and C. The following are the moves on would take to solve the problem in the fewest number of step. (Note: The manner in which the disks are stacked (from top to bottom) on each peg form the contents of each cell; a blank cell means that no disk is on that peg in that particular move.) (This solution continues onto the next page.)

Initial placement

Move 1

Move 2

Post A 1 2 3 4 2 3 4 3 4

Post B

Post C

1167

Chapter 9 Move 3 Move 4 Move 5 Move 6 Move 7

3 4 4

1 4 1 4 4

3 2 3 1 2 3 1 2 3

Move 8

2 3 3

Move 9 Move 10

Move 11

Move 13

1 2 1 2 2

Move 14

Move 12

Move 15

1 2 1 2 2

1 4 1 4 4 3 4 3 4 2 3 4 1 2 3 4

So, the puzzle can be solved in as few as 15 steps. An argument as to why this is actually the fewest number of steps is beyond the scope of the text. (Note: Alternatively, we could have initially placed disk 1 on post C, and proceeded in a similar manner.) 27. Using the strategy of Problem 25 (and 26), this puzzle can be solved in as few as 31 steps. Have fun trying it!! There are many classical references that discuss this problem, as well as several internet sites.

1168

Section 9.4

28. What follows is not a formal proof, but rather an intuitive discussion that hints at what the proof ought to be. To establish a more formal argument, one needs to know how to work with recurrence relations, a topic not addressed in the text. But, see if you can follow the spirit of the argument as presented below.

First, note that the general strategy used to solve the problem with n disks is to: - Move the topmost n − 1 disks from the left peg to one of the other two, - Move the bottommost disk from left to right, - Move the n −1 disks from its current location to the right. And, this process is replicated inductively until you are down to one disks. As such, since this amounts to two steps applied in succession for each of the n disks, this would imply that the number of steps would be 2 n . However, 1 is subtracted since you start off with the problem with an initial configuration. So, truly, the problem can be solved in as few as 2 n − 1 moves. 29. If n = 2, then the number of wires needed is 2(22−1) = 1 . Assume the formula holds for k cities. Then, if k cities are to be connected directly to each other, the number of wires needed is k ( k2−1) . Now, if one more city is added, then you must have k additional wires to connect the telephone to this 2 additional city. So, the total wires in such case is k + k ( k2−1) = 2 k + 2k − k = k ( k2+1) . Thus, the statement holds for k+1 cities. Hence, we have proven the statement by induction. 30. For n = 3, the formula becomes (3−2)(180)=180. The formula is true when n =3 (a triangle), since it is known that the sum of the interior angles of a triangle is 180 degrees. Assume the formula is true for k sided regular polygon. Then, the sum of the interior angles is (k−2)(180). Looking at a few particular examples (for particular values of k) we observe that you can divide a k-sided polygon into (k−2) triangles, and each time we increase the number of sides by one, one additional triangle is formed. As such, the interior angle sum increases by 1. Hence, adding one side to a k-side polygon then has interior angle sum 180 + (k−2)(180) = 180(k−1), thereby showing the statement holds for k+1. Hence, we have proven the statement by induction. 31. False. You first need to show that S1 is true.

32. False. You must show that if Sk is true, then Sk +1 is true, for all values of k, not just the first pair of consecutive values of k.

1169

Chapter 9 n

1)(3 n +3 n −1) 33. Claim:  k 4 = n ( n +1)(2 n +30 , for all n ≥ 1 . 2

k =1

Proof. Step 1: Show the statement is true for n = 1 . 2 +3(1) −1) 14 = 1(1+1)(2(1)+1)(3(1) = 1 , which is clearly true. 30 p

Step 2: Assume the statement is true for n = p :  k 4 = p ( p +1)(2 p +301)(3 p +3 p−1) 2

k =1

p+1

Show the statement is true for n = p +1 :  k 4 = ( p +1)( p + 2 )(2( p+1)+301)(3( p+1) +3( p+1)−1) 2

k =1

Observe that p +1

 k =  k + ( p + 1) 4

k =1

p ( p +1)(2 p +1)(3 p2 +3 p−1) 30

+ ( p + 1)4

= p ( p +1)(2 p +1)(3 p30+3 p−1)+30( p+1) =

(by assumption)

( p +1)  p (2 p +1)(3 p2 + 3 p−1) +30( p+1)3    30 ( p +1) 6 p 4 + 39 p3 + 91 p2 + 89 p+30 

  (1) = 30 Next, note that multiplying out these terms yields ( p +1)( p + 2 )(2( p +1) +1)(3( p +1)2 + 3( p +1) −1) 30

= =

( p +3 p + 2 )( 2 p +3)(3 p +6 p +3+3 p +3−1) 2

( 2 p +9 p +13 p+6 )(3 p +9 p+5) 3

30 6 p5 + 45 p 4 +130 p3 +180 p2 +119 p+30 30

= (2) Comparing (1) and (2), we see they are, in fact, equal. This is precisely what we needed to show to establish the claim. This completes the proof. n

n + 2 n −1) 34. Claim:  k 5 = n ( n +1) (2 , for all n ≥ 1 . 12 2

k =1

Proof. Step 1: Show the statement is true for n = 1 . 2 2 2 + 2(1) −1) 15 = 1 (1+1) (2(1) = 1 , which is clearly true. 12 p

p + 2 p−1) Step 2: Assume the statement is true for n = p :  k 5 = p ( p +1) (2 12 k =1

1170

Section 9.4 = 2 p2 +6 p+3

p +1

Show the statement is true for n = p +1 :  k 5 = k =1

 ( p +1)2 (( p +1) +1)2 (2( p+1)2 + 2( p+1) −1) 12

Observe that p +1

k =1

 k 5 =  k 5 + ( p + 1)5 =

p2 ( p +1)2 (2 p2 + 2 p−1) 12 2

+ ( p + 1)5

(by assumption)

= p ( p +1) (2 p +122 p−1)+12( p+1) =

( p +1)2  p2 (2 p2 + 2 p−1) +12( p+1)3    12 ( p +1)  2 p 4 +14 p3 + 35 p2 + 36 p+12 

  (1) = 12 Next, note that multiplying out these terms yields 2 2 2 ( p +1)2 (( p +1)+1)2 (2( p+1)2 +2( p+1)−1) = ( p+1) ( p+2)12(2 p +6 p+3) 12 2

p +6 p+3) = ( p +1) ( p + 4 p+4)(2 12 2

= ( p +1) (2 p +14 p12+35 p +36 p+12 ) (2) Comparing (1) and (2), we see they are, in fact, equal. This is precisely what we needed to show to establish the claim. This completes the proof.

35. Claim: (1 + 11 ) ⋅ (1 + 12 ) ⋅ ... ⋅ (1 + 1n ) = n + 1, for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . (1 + 11 ) = 1 + 1 = 2 , which is clearly true.

Step 2: Assume the statement is true for n = k : (1 + 11 ) ⋅ (1 + 21 ) ⋅ ... ⋅ (1 + k1 ) = k + 1

Show the statement is true for n = k + 1 : (1 + 11 ) ⋅ (1 + 21 ) ⋅ ... ⋅ (1+ k1+1 ) = (k + 1) + 1   = k +2

Observe that (1 + 11 ) ⋅ (1 + 21 ) ⋅ ...⋅ (1 + k1 ) ⋅ (1 + k1+1 ) = (1 + 11 ) ⋅ (1 + 21 ) ⋅ ... ⋅ (1 + k1 ) ⋅ (1 + k1+1 ) = (k + 1) ⋅ (1 + k1+1 ) (by assumption) = ( k + 1) = k +2 This completes the proof.

1171

(

k +1+1 k +1

)

Chapter 9 36. Claim: x + y is a factor of x 2n − y 2 n , for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . x 2 − y 2 = ( x + y )(x − y ) , so that x + y is a factor. Step 2: Assume the statement is true for n = k : x + y is a factor of x 2 k − y 2 k Show the statement is true for n = k +1 : x + y is a factor of x 2( k +1) − y 2( k +1) Observe that 2( k +1) x − y 2( k +1) = x 2k x 2 − y 2k y 2 = x 2k x 2 − y 2 k x 2 + y 2 k x 2 − y 2 k y 2 (Add zero.) = x 2 ( x 2 k − y 2 k ) + y 2k ( x 2 − y 2 )

= x 2 ( x 2 k − y2 k ) + y 2k ( x + y )(x − y) = x 2 ( x + y) p(x, y) + y 2 k (x + y)(x − y) (by assumption) (1) where p( x, y) is some polynomial in x and y. Hence, since ( x + y ) can be factored out of both terms in (1), we conclude that, in fact, ( x + y ) is

indeed a factor of x 2( k +1) − y 2( k +1) . This completes the proof. 37. Claim: ln ( c1 ⋅ c2 ⋅ ... ⋅cn ) = ln(c1 ) +... + ln(cn ), for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . ln ( c1 ) = ln(c1 ) is clearly true. Step 2: Assume the statement is true for n = k : ln ( c1 ⋅ c2 ⋅ ... ⋅ck ) = ln(c1 ) +... + ln(ck ) Show the statement is true for n = k +1 : ln ( c1 ⋅ c2 ⋅ ... ⋅ck +1 ) = ln(c1 ) +... + ln(ck +1 )

Observe that     ln  c1 ⋅ c2 ⋅ ... ⋅ ck ⋅ck +1  = ln(c1 ⋅...⋅ck ) + ln(ck +1 )     Treat as a single   quantity  = ln(c1 ) + ... + ln(ck ) + ln(ck +1 ) (by assumption) This completes the proof. 255 38. 2,706,800. Yes. 39. 256 . Yes.

1172

Section 9.5

Section 9.5 Solutions -------------------------------------------------------------------------------1. 7 7! 7! 7 ⋅ 6 ⋅ 5 ⋅ 4! = = = 35  =  3  ( 7 − 3 ) !3! 4!3! 4! ( 3 ⋅ 2 ⋅1) 2. 8 8! 8! 8⋅ 7 ⋅ 6! = = = 28  =  2  ( 8 − 2 ) !2! 6!2! 6! ( 2 ⋅1) 3. 10  10! 10! 10 ⋅ 9 ⋅ 8! = = = 45  =  8  (10 − 8 ) !8! 2!8! 8! ( 2 ⋅1) 4.  23  23! 23! 23⋅ 22 ⋅ 21! = = = 253  = 21! ( 2 ⋅1)  21  ( 23 − 21) !21! 2!21! 5. 17  17! 17! 1 = =1 =  =  0  (17 − 0 ) !0! 17! 0! 0! 6. 100  100! 100! 1 = =1 =  =  0  (100 − 0 ) !0! 100! 0! 0! 7.  99  99! 99! 1 = =1 =  =  99  ( 99 − 99 ) !99! 0! 99! 0! 8.  52  52! 52! 1 = =1 =  =  52  ( 52 − 52 ) !52! 0! 52! 0! 9.  48  48! 48! 48⋅ 47 ⋅ 46 ⋅ 45! = = = 17,296  = 45! ( 3 ⋅ 2 ⋅1)  45  ( 48 − 45 ) !45! 3!45!

1173

Chapter 9 10.  29  29! 29! 29 ⋅ 28 ⋅ 27 ⋅ 26! = = = 3654  = 26! ( 3 ⋅ 2 ⋅1)  26  ( 29 − 26 ) !26! 3!26! 11. 4 4 ( x + 2 )4 =    x 4 −k 2 k k =0  k  4 4 4 4 4 =   x 4 2 0 +   x3 21 +   x 2 22 +   x1 23 +   x 0 2 4 0 1 2 3 4 4! 4! 4! 4! 4! x 4 20 + 3!1! x3 21 + 2!2! x 2 2 2 + 1!3! x1 23 + 0!4! x0 24 = 4!0!      =1

= x 4 + 8x3 + 24x 2 + 32x +16 12. 5 5 ( x + 3)5 =    x5−k 3k k =0  k  5 5 5 5 5 5 =   x5 30 +   x 4 31 +   x 3 32 +   x 2 33 +   x1 34 +   x 0 35 0 1 2 3 4 5 5! 5! 5! 5! 5! 5! x5 30 + 4!1! x 4 31 + 3!2! x3 32 + 2!3! x 2 33 + 1!4! x1 34 + 0!5! x 0 35 = 5!0!       =1

=10

= x5 + 15x 4 + 90x3 + 270x 2 + 405x + 243 13. 5 5 ( y − 3)5 =    y5−k ( −3)k k =0  k  5 5 5 5 5 =   y5 ( −3)0 +   y 4 ( −3)1 +   y3 ( −3)2 +   y 2 ( −3)3 +   y1 ( −3)4 0 1 2 3 4 5 +   y 0 ( −3)5 5 5! 5! 5! 5! 5! y5 ( −3)0 + 4!1! y 4 ( −3)1 + 3!2! y3 ( −3)2 + 2!5!3! y 2 ( −3)3 + 1!4! = 5!0! y1 ( −3)4 + 0!5 y 0 ( −3)5      ! =1

=10

= y − 15 y + 90y − 270y + 405y − 243 5

1174

Section 9.5

14. 4 4 ( y − 4)4 =    y 4 −k (−4)k k =0  k  4 4 4 4 4 =   y 4 ( −4 )0 +   y3 (−4)1 +   y 2 ( −4)2 +   y1 ( −4)3 +   y 0 ( −4)4 0 1 2 3 4 4! 4! 4! 4! 4! y 2 ( −4)2 + 4!0! y1 (−4)3 + 4!0! = 4!0! y 4 ( −4 )0 + 4!0! y3 (−4)1 + 4!0! y 0 ( −4)4      =1

= y − 16 y + 96y − 256 y + 256 4

15. 5  5  5−k  1  1   x −  =  k  x  −  2  k =0     2 5

0 1 2 3  5   1  5  1   5   1  5  1  =   x5  −  +   x 4  −  +   x3  −  +   x 2  −   0  2  1  2   2  2  3  2  4 5  5  1  1  5 0  1  + x −  + x −   4   2  5  2  1

5! 4  1  5! 3  1  5! 5  1  5! 2  1  = x −  + x −  + x −  + x −  5!0!  2   4!1!  2   3!2!  2   2!3!  2   =5

=10

5! 1  1  5! 0  1  x −  + x −  1! 0!5!  2  4!  2    =5

5 5 5 5 1 = x 5 − x 4 + x3 − x 2 + x − 2 2 4 16 32

1175

=10

Chapter 9 16. 4  4  4− k  1  1  + = x    k x   3  k =0   3  4

0 1 2 3 4 4 4  1  4 3  1  4 2  1  4 1  1  4 0  1  =  x   + x   + x   + x   + x   0  3  1  3  2  3  3  3  4  3  0

4! 4  1  4! 3  1  4! 2  1  4! 1  1  4! 0  1  x   + x   + = x   + x   + x   4!0!  3   3!1!  3   2!2!  3  1 !3!  3   0!4!  3   =1

4 2 4 1 = x 4 + x3 + x 2 + x + 3 27 81 3

17. 5 5 ( x + y )5 =    x5−k y k k =0  k  5 5 5 5 5 5 =   x5 y 0 +   x 4 y1 +   x3 y 2 +   x 2 y3 +   x1 y 4 +   x 0 y5 0 1 2 3 4 5 5! 5! 5! 5! 5! = 55!!0! x5 y 0 + 4!1! x 4 y1 + 3!2! x 3 y 2 + 2!3! x 2 y3 + 1!4! x1 y 4 + 0!5! x 0 y5       =1

=10

= x + 5x y + 10x y +10x y + 5xy + y 5

18. 6 6 ( x − y )6 =    x 6 −k ( − y)k k =0  k  6 6 6 6 =   x 6 ( − y )0 +   x5 ( − y )1 +   x 4 ( − y )2 +   x3 ( − y )3 0 1 2 3

6 6 6 +   x 2 ( − y )4 +   x1 ( − y )5 +   x 0 ( − y )6 4 5 6 6! 6! 6! 6! = 6!0! x3 ( − y )3 x 6 ( − y )0 + 5!1! x5 ( − y )1 + 4!2! x 4 ( − y )2 + 3!3!     =1

6! 2!4!



=15

x ( − y) + 2

=15

6! 1!5!

 =6

=20

6! x ( − y ) + 0!6! x 0 ( − y )6  1

= x − 6x y + 15x y − 20x y + 15x y − 6xy5 + y6 6

1176

Section 9.5

19. 3 3 ( x + 3 y )3 =    x3−k (3y)k k =0  k  3 3 3 3 =   x3 (3 y )0 +   x 2 (3 y )1 +   x1 (3 y )2 +   x 0 (3 y )3 0 1 2 3 3! 3! 3! 3! = 3!0! x3 (3 y )0 + 2!1! x 2 (3 y )1 + 1!2! x1 (3y)2 + 0!3! x 0 (3 y )3     =1

= x + 9x y + 27xy + 27y 3

20. 3 3 (2 x − y )3 =    (2 x )3−k (− y )k k =0  k  3 3 3 3 =   (2x )3 ( − y )0 +   (2 x )2 (− y )1 +   (2 x )1 (− y )2 +   (2 x )0 ( − y )3 0 1 2 3 3 0 2 1 1 2 3! 3! 3! 3! (2 x ) ( − y ) + 2!1! (2 x ) (− y ) + 1!2! (2 x ) (− y ) + 0!3! (2 x )0 ( − y )3 = 3!0!     =1

= 8x3 − 12 x 2 y + 6 xy2 − y3

21. 3 3 (5x − 2)3 =    (5x )3−k (−2)k k =0  k  3 3 3 3 =   (5x )3 (−2)0 +   (5x )2 ( −2)1 +   (5x )1 (−2)2 +   (5x )0 ( −2)3 0 1 2 3 3! 3! 3! 3! = 3!0! (5x )3 ( −2)0 + 2!1! (5x )2 (−2)1 + 1!2! (5x )1 (−2)2 + 0!3! (5x )0 (−2)3     =1

= 125x − 150x + 60x − 8 3

1177

Chapter 9

22. 3 3 ( a − 7b )3 =    (a)3−k (−7b)k k =0  k  3 3 3 3 =   ( a )3 ( −7b)0 +   (a )2 (−7b )1 +   ( a )1 ( −7b)2 +   (a)0 ( −7b )3 0 1 2 3 3! 3! 3! = 3!0! ( a )3 ( −7b )0 + 2!1! (a)2 (−7b)1 + 1!2! ( a )1 ( −7b)2 + 0!3!3! (a)0 (−7b)3     =1

= a − 21a b +147ab − 343b 3

23. 4  4  4−k ( x1 + 5 y )4 =    ( x1 ) (5 y )k , x ≠ 0 k =0  k  4 4 4 3 4 2 4 1 =   ( x1 ) (5 y )0 +   ( x1 ) (5 y )1 +   ( x1 ) (5 y )2 +   ( x1 ) (5 y )3 0 1 2 3 4 0 +   ( x1 ) (5 y )4 4 4! 4! 1 4! 4! 4! 1 1 1 1 (5 y )0 + 3!1! (5 y )1 + 2!2! (5 y )2 + 1!3! (5 y )3 + 0!4! = 4!0! (5 y )4 (x) (x) (x) (x) (x) 4

= x14 + 20x3y + 150x2y + 500xy + 625 y 4

24. 4 4 4−k (2 x + 3y )4 =    ( 2x ) ( 3y )k , y ≠ 0 k =0  k  4 4 4 4 4 3 2 1 =   ( 2 x ) ( 3y )0 +   ( 2x ) ( 3y )1 +   ( 2x ) ( 3y )2 +   ( 2 x ) ( 3y )3 0 1 2 3

4 0 +   ( 2 x ) ( 3y )4 4 4! 4! 4! 4! 2 x ( 3 )0 + 3!4!1! ( 2x ) ( 3y )1 + 2!2! 2 x ( 3 )2 + 1!3! 2 x ( 3 )3 + 0!4! 2 x ( 3 )4 = 4!0! ( ) y  ( ) y ( ) y ( ) y 4

= 16x + 4

96x3 y

216 x2 y2

216x y3

81 y4

1178

Section 9.5

25. 4 4 −k 4 ( x 2 + y 2 )4 =    ( x 2 ) ( y 2 )k k =0  k  4 3 2 1 4 4 4 4 =   ( x 2 ) ( y 2 )0 +   ( x 2 ) ( y 2 )1 +   ( x 2 ) ( y2 )2 +   ( x 2 ) ( y 2 )3 0 1 2 3 0 4 +   ( x 2 ) ( y 2 )4 4 4! 4! 4! 4! x 2 ( y2 )0 + 3!1! x 2 ( y 2 )1 + 2!2! x 2 ( y 2 )2 + 1!3! x 2 ( y 2 )3 = 4!0! ( ) ( ) ( ) ( ) 4

4! + 0!4! x ( y 2 )4 ( ) 2 0

= x + 4 x y 2 + 6 x 4 y 4 + 4 x 2 y6 + y8 8

26. 3 3 ( r 3 − s 3 )3 =    (r 3 )3−k (−s 3 )k k =0  k  3 3 3 3 =   ( r 3 )3 ( − s 3 )0 +   ( r 3 )2 (−s 3 )1 +   ( r 3 )1 (−s3 )2 +   ( r3 )0 ( −s 3 )3 0 1 2 3 3! 3! 3! 3! = 3!0! (r 3 )3 ( − s 3 )0 + 2!1! (r 3 )2 ( −s 3 )1 + 1!2! (r 3 )1 ( −s 3 )2 + 0!3! (r 3 )0 ( −s 3 )3     =1

= r 9 − 3r 6 s 3 + 3r 3 s 6 − s 9

27. 5 5 ( ax + by )5 =    (ax)5−k (by)k k =0  k  5 5 5 5 =   (ax)5 ( by )0 +   (ax)4 ( by )1 +   (ax)3 ( by )2 +   (ax)2 (by )3 0 1 2 3

5 5 +   ( ax )1 (by)4 +   ( ax )0 (by)5 4 5

1179

Chapter 9 5! 5! 5! 5! = 5!0! ( ax )5 ( by )0 + 4!1! ( ax )4 ( by )1 + 3!2! ( ax )3 ( by )2 + 2!3! ( ax )2 ( by )3     =1

=10

5! ( ax ) ( by ) + 1!5!4! ( ax ) ( by ) + 0!5!   1

= a5 x5 + 5a 4 bx 4 y + 10a3b 2 x3 y 2 + 10a 2 b3 x 2 y3 + 5ab 4 xy 4 + b5 y5

28. 5 5 ( ax − by )5 =    ( ax )5−k ( −by )k k =0  k  5 5 5 5 =   ( ax )5 ( −by )0 +   ( ax )4 ( −by )1 +   ( ax )3 ( −by )2 +   ( ax )2 ( −by )3 0 1 2 3 5 5 +   ( ax )1 ( −by )4 +   ( ax )0 ( −by )5 4 5 5! 5! 5! 5! ( ax )5 ( −by )0 + 4!1! ( ax )4 ( −by )1 + 3!2! ( ax )3 ( −by )2 + 2!3! ( ax )2 ( −by )3 = 5!0!     =1

=10

5! 5! ( ax ) ( −by ) + 0!5 ( ax ) ( −by ) + 1!4!  ! 1

= a5 x5 − 5a 4 bx 4 y + 10a3b2 x3 y 2 − 10a 2 b3 x 2 y3 + 5ab 4 xy 4 − b5 y5

29. 6 6−k 6 ( x + 2 )6 =    x 2k , x ≥ 0 k =0  k  6 5 6 6 6 =   x 2 0 +   x 21 +   0 1 2

( )

6 +  4 6! = 6!0!  =1

( x ) 2 +  3  ( x ) 2 4

( x ) 2 +  5  ( x ) 2 +  6  ( x ) 2 2

( x) 2 + ( x) 2 + ( x) 2 + ( x) 2 6

6! 5 !1!

6! + 2!4!  =15

6! 4!2!

6! 3!3!

=15

( x) 2 + ( x) 2 + ( x) 2 2

6! 1!5!

6! 0!6!

1180

=20

= x + 12 x 2 + 60x +160x 2 + 240x +192x 2 + 64 3

Section 9.5

30. 4 k 4 (3 + y )4 =    34 − k y , y ≥ 0 k =0  k  0 1 4 4 4 =   34 y +   33 y +   32 1 2 0

( )

4! 34 = 4!0!  =1

( )

( y ) +  3  3 ( y ) +  4  3 ( y ) 4

( y) + 3 ( y) + 3 ( y) + 3 ( y) + 3 ( y) 0

4! 3!1!

4! 2!2!

4! 1!3!

= 81 + 108y 2 + 54 y + 12y 2 + y 3

4! 0!4!

31. 4  4  3 4− k 1 4 k 3 1 ( a 4 + b 4 )4 =    a 4 b , a,b ≥ 0 k =0  k  4 3 4 1 0 4 3 3 1 1 4 3 2 1 2 =   a 4 b 4 +  a 4 b 4 +  a 4 b 4 0 1 2

( ) ( )

( )( )

4 3 1 1 3 4 3 0 1 4 +  a 4 b 4 +  a 4 b 4 3 4

( )( )

( ) (b ) +  (a ) (b ) +  (a ) (b )

( ) (b ) +  (a ) (b )

4! a4 = 4!0!  3

( )( )

4! 3!1!

+ 14!!3! a 4  3

4! 2!2!

4! 0! 4!

= a 3 + 4a 4 b 4 + 6a 2 b 2 + 4a 4 b 4 + b 9

32. 3  3  2 3−k 13 k 2 1 ( x 3 + y 3 )3 =    x 3 y k =0  k   3  2 3 1 0  3  2 2 1 3 1  3  2 3 1 13 2  3  2 3 0 13 3 y y y =  x 3 y 3 +  x 3 +  x +  x 0 1 2 3

( ) ( )

( )( ) 3

3! 2!1!

3! 1!2! =3

= x + 3x 3 y 3 + 3x 3 y 3 + y 2

( )( )

( ) ( y ) +  (x ) ( y ) +  (x ) ( y ) +  (x ) ( y )

3! x3 = 3!0!  =1

( )( )

1181

3! 0!3! =1

Chapter 9

33. 4 k  4  1 4−k 1 2 y , x, y ≥ 0 ( x 4 + 2 y )4 =    x 4 k =0  k  0 1 2 4 1 4 4 1 3 4 1 2 =  x 4 2 y +  x 4 2 y +  x 4 2 y 1 2 0

( ) (

( )(

)

( )(

)

( )(

)

4 3 4 1 1 4 1 0 +  x 4 2 y +  x 4 2 y 3 4

( )(

)

( ) (2 y ) +  (x ) (2 y ) +  (x ) (2 y )

( ) (2 y ) +  (x ) (2 y )

4! x4 = 4!0!  1

)

4! 3!1!

4! 2!2!

4! 0!4!

= x + 8x 4 y 2 + 24x 2 y + 32x 4 y 2 +16y 1

4! x4 + 1!3!  3

34. 8 8− k k 8 1 1 ( x − 3 y 4 )8 =    x −3y 4 , x, y ≥ 0 k =0  k  8 7 0 1 8 8 8 1 1 =   x −3 y 4 +   x −3 y 4 +   0 1 2

( ) (

)

( )(

8 +  3 8 +  6

( )(

)

( x ) ( −3 y ) 6

( x ) ( −3y ) +  4  ( x ) ( −3y ) +  5  ( x ) ( −3y ) 5

8 8 ( x ) ( −3y ) +  7  ( x ) ( −3y ) +  8  ( x ) ( −3y ) = ( x ) ( −3 y ) + ( x ) ( −3 y ) + ( x ) ( −3 y )    2

8! 8!0!

8! 7!1!

8! 6!2!

= 28

8! + 5!3! 

( x ) ( −3 y ) +  ( x ) ( −3 y ) +  ( x ) ( −3 y )

8! + 2!6! 

( x ) ( −3y ) +  ( x ) ( −3y ) +  ( x ) ( −3y )

= 56

= 28

8! 4!4!

8! 3!5!

= 70

8! 0!8!

= x − 24x 2 y 4 + 252x y 2 −1512x 2 y 4 + 5670x 2 y 7

−13,608x 2 y 4 + 20, 412xy 2 − 17, 496 x 2 y 4 + 6561y 2 3

=56

8! 1!7!

1182

Section 9.5

35.

( x + 2 ) = x5 + 5x 4 ( 2 ) +10x3 ( 2 ) +10x2 ( 2 ) + 5x ( 2 ) + ( 2 ) 5

= x5 + 10x 4 + 40x3 + 80x 2 + 80x + 32

36. 6 2 3 4 5 6 ( y − 3) = y6 + 6y5 ( −3) +15y4 ( −3) + 20y3 ( −3) +15y2 ( −3) + 6y ( −3) + ( −3) = y6 − 18 y5 +135y 4 − 540y3 +1215y2 −1458y + 729

37. ( r − s )4 = r 4 + 4r 3 ( −s) + 6r 2 ( − s)2 + 4r( − s)3 + ( − s)4

= r 4 − 4r 3 s + 6r 2 s 2 − 4rs 3 + s 4

38.

( x + y ) = ( x ) + 7 ( x ) ( y ) + 21( x ) ( y ) + 35 ( x ) ( y ) + 35 ( x ) ( y ) + 21( x ) ( y ) + 7 ( x ) ( y ) + ( y ) 2

2 7

2 6

2 2

2 1

2 5

2 1

2 6

2 2

2 4

2 3

2 7

= x14 + 7x12 y2 + 21x10 y 4 + 35x8 y6 + 35x6 y8 + 21x 4 y10 + 7x 2 y12 + y14

39.

( ax + by ) = (ax)6 + 6(ax)5 (by) +15(ax)4 (by)2 + 20(ax)3 (by)3 6

+ 15(ax)2 (by)4 + 6(ax)1 (by)5 + (by)6 = a 6 x 6 + 6a 5 bx5 y + 15a 4 b 2 x 4 y 2 + 20a3b3 x3 y3 +15a 2 b 4 x 2 y 4 + 6ab5 xy5 + b6 y6

40. (x + 3 y )4 = x 4 + 4x3 (3y)1 + 6x 2 (3y)2 + 4x1 (3y)3 + (3y)4 = x 4 + 12x3 y + 54x 2 y 2 +108xy3 + 81y 4

41. 10  10! 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6! 6 (16)x 6 = This term is   x10−4 (2)4 = x = 3360 x6 4 4!6! (4 ⋅ 3 ⋅ 2 ⋅ 1) 6!   So, the desired coefficient is 3360.

1183

2 4

Chapter 9

42. 9 9! 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5! 5 (81) y5 = This term is   y9− 4 (3)4 = y = 10,206y5 4 5!4! (4 ⋅ 3 ⋅ 2 ⋅1) 5!   So, the desired coefficient is 10,206.

43. 8 8! 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4! 4 (81) y 4 = This term is   y8− 4 ( −3)4 = y = 5670y 4 4 4!4! (4 ⋅ 3 ⋅ 2 ⋅1) 4!   So, the desired coefficient is 5670.

44. 12  12! 12 ⋅11⋅10 ⋅ 9 ⋅ 8 ⋅ 7! ( −1)x5 = This term is   x12−7 ( −1)7 = ( −1)x5 = −792 x5 7!5! (5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1) 7! 7 So, the desired coefficient is −792 .

45. 7 7! 7 ⋅ 6 ⋅ 5 ⋅ 4! (8 ⋅ 81)x3 y 4 = This term is   (2x )7 − 4 (3 y )4 = (648)x3 y 4 = 22,680x3 y 4 4!3! (3 ⋅ 2 ⋅1) 4! 4 So, the desired coefficient is 22,680.

46. 9 9! 9 ⋅ 8 ⋅ 7! 9 −7 7 ( 9)( − 78,125 )x 2 y7 = (−703,125)x 2 y7   (3x ) ( −5 y ) = 7!2! This term is  7  (2 ⋅1) 7! = ( − 25,312,500)x 2 y7 So, the desired coefficient is −25,312,500 .

47. 8 8! 8 4 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4! 8 4 x y = 70x8 y 4 x y = This term is   ( x 2 )8−4 ( y )4 = 4 4!4! ⋅ ⋅ ⋅ 1) 4! (4 3 2   So, the desired coefficient is 70.

48. 10  10! 6 8 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6! 6 8 rs = This term is   r10 − 4 ( − s 2 )4 = r s = 210r 6 s 8 4 4!6! ⋅ 3 ⋅ 2 ⋅ 1) 6! (4   So, the desired coefficient is 210.

1184

Section 9.5

49.  40  40! 40 ⋅ 39 ⋅ 38 ⋅ 37 ⋅ 36 ⋅ 35 ⋅ 34! Since   = = = 3,838,380 , there are 3,838,380 6! 34!  6  6!34! such 6-number lottery numbers.

50.  60  60! 60 ⋅ 59 ⋅ 58 ⋅ 57 ⋅ 56 ⋅ 55 ⋅ 54! Since   = = = 50,063,860 , there are 50,063,860 6! 54!  6  6!54! such 6-number lottery numbers.

51.  52  52! 52 ⋅ 51 ⋅ 50 ⋅ 49 ⋅ 48 ⋅ 47! Since   = = = 2,598,960 , there are 2,598,960 5! 47!  5  5!47! possible 5-card poker hands.

52. 108  108! 108 ⋅107 ⋅ ... ⋅ 98 ⋅ 97! Since  = = 344,985,116,800,000 , there are = 11! 97!  11  97!11! 344,985,116,800,000 different 11-card Canasta hands.

53.  7  7!  7  7!   ≠ ,  =  5  5!  5  5!2! 54. Should replace powers of y by powers of 2y . More precisely, the expansion should be (x + 2 y )4 = x 4 + 4x3 (2y) + 6x 2 (2y)2 + 4x(2y)3 + (2y)4 = x 4 + 8x3 y + 24x 2 y 2 + 32xy3 +16y 4

55. False, there are 11 terms.

56. True

57. True

0 58. False. This is only defined for n = 0 , and the result in such case is   = 1 . 0

1185

Chapter 9

59. n  n  Claim:   =   , 0 ≤ k ≤ n, for any n ≥ 1 . k n−k n  n  n! n! n! Proof. Observe that   = = = =  − ( n − k )) !  n − k   k  k !( n − k)! ( n − k ) !k ! ( n − k ) ! (n  This is k written in a different form.

60. Observe that 2 n = (1 + 1)n n n n n =  1n −010 +  1n −111 + . . . +  1n − k1k + . . . +  101n 0 1 k n n n n =  +   + ... +   0 1 n since all powers of 1 equal 1.

61. The binomial expansion of (1 − x)3 is 1 − 3x + 3x 2 − x3 . Notes on the Graph: Heavy Solid Curve: y1 = 1 − 3x + 3x 2 − x3 Heavy Dashed Curve: y2 = −1 + 3x − 3x 2 + x3 Heavy Dotted Curve: y3 = (1 − x )3 The graphs of y1 and y3 are the same, so you only see two graphs.

62. The binomial expansion of ( x + 3)4 is x 4 + 12 x3 + 54x 2 +108x + 81. Notes on the Graph: Heavy Solid Curve: y3 = x 4 +12x3 + 54x 2 +108x + 81 Heavy Dashed Curve: y2 = x 4 + 4 x3 + 6 x 2 + 4x +1 Heavy Dotted Curve: y1 = ( x + 3)4 The graphs of y1 and y3 are the same, so you only see two graphs. 1186

Section 9.5

63. We see from the graph that as each term is added, the graphs of the respective functions get closer to the graph of y4 = (1 − x )3 when 1 < x < 2 . However, when x > 1 , this is no longer true. Notes on the graph: Heavy solid curve: y4 Heavy dashed curve: y1 Heavy dotted curve: y2 Thin dashed curve: y3

64. We see from the graph that as each term is added, the graphs of the respective functions get closer to the graph of y4 = (1 − x1 )3 when 1 < x < 2 . However, when 0 < x < 1 , this is no longer true. Notes on the graph: Heavy solid curve: y4 Heavy dashed curve: y1 Heavy dotted curve: y2 Thin dashed curve: y3

65. As each new termp is added, the corresponding graph of the curve is a better approximation to the graph of 3 y = (1 + x1 ) , for 1 < x < 2 . The series of functions does not get closer to this graph if 0 < x < 1 .

1187

66. As each new term is added, the corresponding graph of the curve is a better approximation to the graph of y = e x , for −1 < x < 1 . The series of functions does not get closer to this graph if 1 < x < 2 .

Section 9.6 Solutions--------------------------------------------------------------------------------2.

1. 6

P4 =

6! 6! 6 ⋅ 5 ⋅ 4 ⋅3⋅ 2! = = 2! ( 6 − 4 ) ! 2!

P3 =

7! 7! 7 ⋅ 6 ⋅ 5 ⋅ 4! = = = 210 4! ( 7 − 3) ! 4!

= 360

4. 9

P5 =

9! 9! 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4! = = 4! ( 9 − 5 ) ! 4!

P4 =

= 3024

= 15,120

8! 8! = = 8! = 40,320 8 P8 = (8 − 8) ! 0!

8. P3 =

13! 13! 13⋅12 ⋅11⋅ 10! = = 10! (13 − 3) ! 10!

9! 9! 9 ⋅ 8 ⋅ 7 ⋅6 ⋅ 5! = = 5! ( 9 − 4 ) ! 5!

P6 =

P3 =

= 1716

6! 6! = = 720 ( 6 − 6 ) ! 0!

20! 20! 20 ⋅19 ⋅18⋅ 17! = = 17! ( 20 − 3) ! 17!

= 6840

9. 10

C5 =

10! 10! 10 ⋅ 9 ⋅ 8⋅7 ⋅6 ⋅ 5! = = = 252 (10 − 5 ) !5! 5!5! 5! (5 ⋅ 4 ⋅3⋅ 2 ⋅1)

10. 9

C4 =

9! 9! 9 ⋅ 8 ⋅ 7 ⋅6 ⋅ 5! = = = 126 ( 9 − 4 ) !4! 5!4! 5! ( 4 ⋅ 3 ⋅ 2 ⋅1)

11. 50

C6 =

50! 50 ⋅ 49 ⋅ 48⋅ 47 ⋅ 46 ⋅ 45 ⋅ 44! = = 15,890,700 44! ( 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1) ( 50 − 6 ) !6!

1188

Section 9.6

12. 50

C10 = =

50! 50! = (50 −10 ) !10! 40!10! 50 ⋅ 49 ⋅ 48⋅ 47 ⋅ 46 ⋅ 45 ⋅ 44 ⋅ 43⋅ 42 ⋅ 41⋅ 40! = 10,272,278,170 40! (101)

14. 8C8 = 1

13. 7

C7 =

7! 7! = =1 ( 7 − 7 ) !7! 0!7!

15. 30

16.

C4 =

30! 30 ⋅ 29 ⋅ 28 ⋅ 27 ⋅ 26! = 26! ( 4 1) (30 − 4 ) !4!

C5 =

13! 13⋅12 ⋅11⋅10 ⋅ 9 ⋅ 8! = 8! ( 51) (13 − 5 ) !5!

= 1287

= 27, 405

17. 45

C8 =

45! 45 ⋅ 44 ⋅ 43⋅ 42 ⋅ 41⋅ 40 ⋅ 39 ⋅38⋅ 37! = = 215,553,195 37! ( 8 ⋅ 7 1) ( 45 − 8 ) !8!

18. 30

C4 =

30! 30! 30 ⋅ 29 ⋅ 28⋅ 27 ⋅ 26! = = = 27, 405 26! ( 41) (30 − 4 ) !4! 26!4!

19. 4 ⋅ 3 ⋅ 2 = 24 different system configurations

20. 4 ⋅ 5 ⋅3 ⋅ 2 = 120 different houses to choose from

21. 2 ⋅ 3 ⋅

22. 4 ⋅ 2 ⋅ 3 ⋅ 5

2

= 12 different types

of invitations

= 120

main starch vegetable appetizer course

color writing envelopes

different dinner combinations

23. Each slot can have 0,1,,9 (10 24. Each slot can have a,, z (26 choices) Can allow repetition of digits choices).Can allow repetition of letters and order is important. So, there are and order is important. So, there are 4 10 ⋅10 ⋅10 ⋅10 = 10 = 10,000 different pin 26 4 = 456,976 different passwords. numbers

1189

Chapter 9

25.

15! 15 ⋅14 ⋅13 ⋅12 ⋅ 11! = = 32,760 different leadership teams 11! 11!

26.  3 ⋅  2 ⋅  4 ⋅  1 = 24 president

vice secretary treasurer president

possible election outcomes

27. Each of 20 questions has 4 answer choices. So, there are 4 20 ≈ 1.1× 1012 possible ways to answer the questions on the exam.

28. Each of 25 questions has 5 answer choices. So, there are 525 ( = 2,980,232,239 ×108 ) different ways to complete the exam. 29. The number of 5-digit zip codes where any of 10 digits can be used in each slot is 105 =100,000. The number of 5-digit zip codes where 0 cannot be used in 1st and last slot: 1st—9 digits 2nd—10 digits 3rd —10 digits 4th —10 digits 5th —9 digits 2 3 9 ×10 = 81,000 such zip codes. 30. 1st slot—24 (eliminate O and I) 2nd slot—24 3rd slot—24 4th slot—8 (eliminate 0 and 1) 5th slot—8 6th slot—8 243 × 83 = 7,077,888 license plate numbers

31. The number of 3-digit area codes where any of 10 digits can be used in the second and third slot is but 0 and 1 cannot be used in the first slot is: 1st: 8 digits 2nd: 10 digits 3rd: 10 digits

8 × 102 = 800 such area codes

 31 31! 31⋅ 30 ⋅ 29 ⋅ 28! 32.   = = = 4, 495 different ice cream combinations 3! 28!  3  3! ( 31− 3 ) ! (Note: Here, order is not important.)

33. 30! ≈ 2.65 ×1032

34. 4! = 24 different seating arrangements

1190

Section 9.6

⋅

35. 40 1st #

⋅

2nd #

= 59,280

3rd #

(1 less digit (2 less digits to choose from) to choose from)

36. 50 ⋅ 49 ⋅ 48 = 117,600 different safe combinations 37.

(Note: Here, order is important.)

1000! = 1000 ⋅999 ⋅ 998 = (1000 − 3) ! 997,002,000 possible winning scenarios 1000

So, there are 59,280 different locker combinations.

38. 100 P3 = 100 ⋅ 99 ⋅ 98 = 970,200 different possible placings of 1st, 2nd, and 3rd. (Note: Here, order is important.)

P3 =

39.  53  53!  =  6  6! ( 53 − 6 ) ! 53⋅ 52 ⋅ 51⋅ 50 ⋅ 49 ⋅ 48⋅ 47! 6! 47! = 22,957, 480 different 6number combinations =

40. 41. 53   53! 53⋅ 52 ⋅ 51⋅ 50 ⋅ 49 ⋅ 48! =  = 5! 48!  5  5! ( 53 − 5 ) ! = 2,869,685 different 5-number combinations (Note: Here, order is not important.)

 52  52!  =  5  5! ( 52 − 5 ) ! 52 ⋅ 51⋅ 50 ⋅ 49 ⋅ 48⋅ 47! ( 51) 47! = 2,598,960 different 5-card =

hands (Note: Here, order is not important.)

42.  52  52!  =  7  7! ( 52 − 7 ) ! 52 ⋅51⋅50 ⋅ 49 ⋅ 48⋅ 47 ⋅ 46 ⋅ 45! ( 71) 45! = 133,784,560 different 7-card =

 52  52! 52 ⋅ 51 ⋅ 50! 43.   = = 50! 2  2  ( 52 − 2 ) !2! = 1326 different blackjack hands

hands

1191

Chapter 9

44. 4 aces—must be one of the cards 4 kings 4 queens 16 possible choices for 2nd card 4 jacks 4 tens So, 4 ⋅ 16 = 64 different hands

45.

 64  64! ≈ 4.9 × 1014   = 16 16!48!  # possible scenarios for Sweet Sixteen

1st card 2 nd card

yielding 21.

46.

47. 16  ⋅ 16  = 256

 64  64! 64 ⋅ 63 ⋅ 62 ⋅61⋅ 60! =   = 4 60!4! 60! ( 4 1) 

1st slot 2 nd slot AFC NFC

possible scenarios for Super Bowl

# possible scenarios for Final Four

= 635,376

48. 6 ⋅ 6 = 36 possible scenarios at this

49.

stage

votes themselves off) = 56 = 15,625 voting combinations for Survivor when down to 6 contestants

50. 12! different orderings of 12 contestants If the contestants must alternate male/female, then the number of ways to do this are:

51.

6 

1 6 choices for first male

6 

5 

2 3 4 6 choices for 5 choices for first female second male. And so on...

5

5  5

(Nobody

player 6

 6   6   6!   6 ⋅ 5 ⋅ 4     ⋅  =   =  3   3   3!3!   3!  2

= 20 2 = 400

 1 1   11

5

player 1 player 2 player 3

AFC NFC

52. 6! ⋅ 6! = 720 ⋅ 720 = 518, 400

So, the number of possibilities is: 2 6 2 ⋅ 52 ⋅12 = ( 6!) = 7202 = 518, 400

1192

53. The combination formula n C r should be used instead.

Section 9.6

54. Order is important here (since president, vice president, secretary, and treasurer are all different positions with different possibilities). So, should use permutation formula instead.

55. True

56. False. Suppose you have 6 objects, 4 of type A and 2 of type B. Then, the number of distinguishable permutations 6! is = 15 . 4!2! However, the number of combinations 6! = 20 . taken 3 at a time is 3!3!

57. False. Since there will be repetition when permuting the letters in ABBA that are not distinguishable while no such repetition occurs in permuting ABCD.

58. True 59. Observe that Cr ( n − r −1) ! ( r +1) ! = ( n − r −1) ! ( r +1) r ! = r +1 n! = n! ( n − r )!r ! ( n − r ) ( n − r −1) ! r ! n − r n C r +1 n

60. Observe that n! ( n − r ) ! = n ! ⋅ ( n − r + 1) ! = ( n − r + 1) ( n − r )! = n − r + 1 n Pr = ( ) n! n! (n − r )! (n − r )! n Pr −1 ( n − ( r − 1) ) !

61. C ( n, r ) ⋅ r ! =

62. C ( n, r ) =

n! ⋅ r! (n − r)! ⋅ r !

n! (n − r)!

n! n! = = C ( n, n − r ) . (n − r)! ⋅ r ! ( n − ( n − r ))! ⋅( n − r )!

63. Use a calculator to do so, if you have one.

64. Use a calculator to do so, if you have one.

65. a. 5,040 b. 5,040 c. Yes. d. This is true because n Pr = r ! n Cr .

66. a. 95,040 b. 95,040 c. Yes. d. This is true because n Pr = r ! n Cr .

1193

Section 9.7 Solutions--------------------------------------------------------------------------------1. Sum the two dice. So, the possible rolls constitute the sample space: S = {2,312} 2. S = {HHH , HHT , HTH , HTT ,THH ,THT ,TTH ,TTT } H H T H T H T H H T T T H T 3. S = {BBBB, BBBG , BBGB , BBGG , BGBB, BGBG , BGGB, BGGG ,GBBB,GBBG , GBGB, GBGG , GGBB,GGBG ,GGGB,GGGG}

B G

G G

B G

B G B G B G B G B G B G B G B G

4. S = {H1, H 2,, H 6,T 1, T 2,, T 6}

1194

Section 9.7

5. S = {RR, RB, RW , BB, BR, BW ,WR,WB} (Can’t have WW since there is only one white ball in the container.) R R B W B B R W R W B

6. {FF , FS , FJ , SF , SS , SJ , JF , JS , JJ } F F S J F S S J F J S J Note: For problems 7-10, the sample space is given in Problem 2. (Note that there are 8 distinct elements all with the same probability of occurring.) 7.

8. 1 P (all heads) = P ( HHH ) = 8

P (TTH or THT or HTT ) =

3 8

10. P (at least two heads) =

P (at least one head) = 1 − P (TTT ) =

P (exactly one head) =

P ( HHT or HTH or THH or HHH )

7 8

1 2

Note: For Problems 11-16, there are 36 different possible outcomes of throwing 2 dice, assuming that the two dice are distinguishable from each other. The sample space is given in Example 4 on page 841 of the text.

1195

Chapter 9

12. Note that the sum of the two dice is odd for the following rolls: (1,2 ) , (1, 4 ) , (1.6 )

11. 2 1 = P ( sum = 3 ) =   36 18 (1,2 ) or ( 2,1)

13. P ( sum is even ) = 1 − P ( sum is odd ) = 1−

1 1 = 2 2

( 2,1) , ( 2,3) , ( 2.5 ) ( 3,2 ) , ( 3, 4 ) , ( 3.6 ) ( 4,1) , ( 4,3) , ( 4,5 ) (5,2 ) , (5, 4 ) , (5,6 ) ( 6,1) , ( 6,3) , ( 6,5 )

So, P (sum is odd) =

18 1 = . 36 2

14. Note that the sum of the two dice is prime if the sum is 2, 3, 5, 7, or 11. Consulting the sample space in text (page 759), we see this can occur 15 distinct 15 5 = . ways. So, P ( sum is prime ) = 36 12 15. Note that the sum of two dice can be greater than 7 only when the sum is 8, 9, 10, 11, or 12. Consulting sample space on page 759 of the text, we see that this can occur in 15 distinct ways. So, 5 P ( sum > 7 ) = 15 36 = 12

16. Note that the sum of two dice can be less than 7 only when the sum is 2, 3, 4, 5, or 6. Consulting sample space, we see that this can occur in 15 distinct ways. So, 5 P ( sum < 7 ) = 15 36 = 12

Note: For Problems 17-20, the sample space is S = set of 52 distinct cards . 17. There are 12 face cards (4 kings, 4 queens, 4 jacks). So, P(drawing a 40 = 10 nonface card) = 52 13 .

18. There are 26 black cards. So, P(drawing a black card) = 12 .

19. There are 4 twos, 4 fours, 4 sixes, 4 eights. So, P(drawing a 2, 4, 6, or 8) 4 = 16 52 = 13 .

20. There are 4 threes, 4 fives, 4 sevens, 4 nines, 4 aces. So, P (drawing a 3, 5, 7, 9, or ace) 20 = 52 = 135 .

1196

Section 9.7

21. P ( not E1 ) = 1 − P ( E1 ) = 1 −

22. 1 3 = 4 4

P ( not E2 ) = 1 − P ( E2 ) = 1 −

23. If E1 , E2 are mutually exclusive, then 3 P ( E1 ∪ E2 ) = P ( E1 ) + P ( E2 ) = 4

1 1 = 2 2

24. Since E1 , E2 are not mutually exclusive, we see that P ( E1 ∪ E 2 ) = P ( E1 ) + P ( E2 ) −P ( E1 ∩ E2 )

1 1 1 5 + − = 4 2 8 8

25. If E1 , E2 are mutually exclusive, then

26. Since E1 , E2 are independent,

P ( E1 and E2 ) = 0 .

P ( E1 and E2 ) = P ( E1 ) ⋅ P ( E2 ) =

1 1 1 ⋅ = 4 2 8

27. a.  52  52! 52 ⋅ 51 ⋅ 50 ⋅ 49 ⋅ 48! =  = 4! 48!  4  4!48!

28. a.  52  52! 52 ⋅ 51 ⋅ 50! =  = 50! ( 2 ⋅1)  2  50!2!

= 270,725 b. Number of ways to get 4 13  spades:   = 715 4 So, P(getting 4 spades) 715 = ≅ 0.0026 270,725 c. Since there are 13 ways of getting 4 of a kind, we see that 13 P ( 4 of a kind ) = ≈ 0.00005 270,725 So, there is about a 0.005% chance of getting four of a kind.

= 1326 different 2-card hands b. Number of ways of getting ace ace card  + f  

1197

4 possibilities

12 possibilities

is 4 ⋅ 12 = 48   card 1

card 2

48 ≈ 0.037 1326 So, there is about a 3.7% chance of getting 21.

So, P ( getting 21) =

Chapter 9

29. Since drawing a 7 and drawing an 8 are mutually exclusive events, P (drawing 7 or 8) = P(draw 7) + P(draw 8) 4 4 8 2 = + = = ≈ 0.154 52 52 52 13 So, there is about a 15.4% chance of drawing a 7 or an 8.

30. Since drawing a red 7 or a black 8 are mutually exclusive events, P (drawing red 7 or black 8) = P (draw red 7) + P(draw black 8) = 2 2 4 1 + = = ≈ 0.077 52 52 52 13 So, there is about a 7.7% chance of drawing a red 7 or a black 8.

31. Since these two events are independent, we see that 4 4 16 4 P (7 on 1st draw AND 8 on 2nd draw) = ⋅ = = ≈ 0.006 . 52 51 2652 663 So, there is about a 0.6% chance of getting a 7 on the first draw and an 8 on the second draw. 32. Since these two events are independent, we see that 2 2 4 1 P (red 7 on 1st draw AND black 8 on 2nd draw) = ⋅ = = ≈ 0.002 . 52 51 2652 663 So, there is about a 0.2% chance of getting a red 7 on the first draw and a black 8 on the second draw. 33. Sample space S has 25 elements, all elements are presumed equally likely. 1 = 0.03125 . So, P (5 daughters) = 32 Thus, there is a 3.1% chance of getting 5 daughters.

34. Sample space S has 2 4 elements, all elements are presumed equally likely. 1 = 0.0625 . So, P (4 sons) = 16 Thus, there is a 6.25% chance of getting 4 sons.

35. Sample space has 25 elements, all elements presumed equally likely. So, P ( ≥ 1 boy ) = 1 − P ( 0 boys )

36. Sample space has 26 elements, all elements presumed equally likely. So, P ( ≥ 1 girl ) = 1− P ( 0 girls )

= 1 − P ( 5 daughters )

1 31 = = 0.96875 32 32 So, there is a 96.9% chance of having at least one boy.

= 1−

1198

1 63 = = 0.984375 64 64 So, there is about 98.4% chance of having at least one girl. = 1−

Section 9.7

37. Assume independence of spins. Then, P(red on 1st spin AND 2nd AND 3rd AND 4th) =P(red on 1st) ⋅ P(red on 2nd)  P(red 4

 18  on 4th) =   ≈ 0.0503 . So, about a  38  5.03% chance.

38. Again, assume independence of the spins. Then, P(green on 1st spin AND green on 2nd 2

 2  spin) =   ≈ 0.0028 . So, about a  38  0.28% chance. 8 10 So, probability of 10 infected is, by

39. P(not defective) = 9/10. Since you are drawing the 8 DVD players from different batches, the 8 choices are independent. So, the probability that 8 none of the 8 are defective is ( 109 ) ≈ 0.43 . So, about 43% chance.

 8 independence,   ≈ 0.1073 .  10  So, about 10.7% chance.

41. The events are independent since they are chosen at random. So, P(1st number even AND 2nd number even) = P (1st number even ) ⋅

42. The events are independent since chosen at random. P(1st number odd AND 2nd number odd) = P (1st number odd ) ⋅

P ( 2nd number even ) =

1 1 1 ⋅ = 2 2 4

8 8 64 ⋅ = 15 15 225 So, about 28.4% chance.

P ( 2nd number odd ) =

So, 25% chance.

43. Number of 2 card hands is given by:  52  52! 52 ⋅ 51 ⋅ 50! = = 1326  = 2 ⋅ 50!  2  2!52! Number of possible blackjack hands is 4 ⋅ 16 = 64   ace

40. P(non defective) =

face or ten

64 ≈ 0.0483 1326 So, about a 4.8% chance.

So, P(blackjack) =

1199

44. Assume independence of hands dealt and assume all cards are gathered and reshuffled. Then, probability of getting blackjack on 1st hand AND blackjack on 2nd hand is given by ( 0.0483 ) ( 0.0483) ≈ 0.0023 since the two events are independent. So, about a 0.23% chance.

Chapter 9

45. Assume independence of games from week to week. Then, the probability of 16

1 going 16 − 0 is   ≈ 0.00001526 . So, 2 about a 0.001526% chance.

46. Assume independence of games from week to week. Then, the probability of winning at least 1 game = 1 − P ( lose all ) 16

1 = 1−   2 ≈ 1 − 0.0000153 = 0.9999847 So, it is practically a certainty.

47. a. An outcome is of the form: (gene from mother, gene from father) So, the sample space is given by S = { (Brown, Blue), (Brown, Brown), (Blue, Brown), (Blue, Blue) } b. ¼ since outcomes are equally likely c. ¾

48. a. An outcome is of the form: (gene from mother, gene from father) So, the sample space is given by S = { (Brown, Brown), (Blue, Brown) } b. 0 since (Blue, Blue) is not a possible outcome. c. 1 since Brown is dominant.

49.

50. a. 4  48  4! 48! ⋅   ⋅   = 2 3  2! ⋅ 2! 3! ⋅ 45!       

 52  52! 52 ⋅ 51⋅ 50 ⋅ 49 ⋅ 48 =  = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1  5  47! ⋅ 5! = 2,598,960

Ways to choose 3 aces

Ways to choose 3 non-aces

= 103,776 b. Using #49 for total number of 5card hands, the probability of a hand 103,776 ≈ 0.04 . About 4% in (a) is 2,598,960 chance.

51. Using #49 for the total number of 5card hands, we obtain  13    13! 1287 8!⋅5! 5 = = 2,598,960 2,598,960 2,598,960

1200

52. Using #49 for the total number of 5-card hands, we obtain 4 4    4! 4!  2   3  = 2!2! ⋅ 3!1! = 0.00001 2,598,960 2,598,960

Section 9.7

53. The events aren’t mutually exclusive. So, 4 13 1 16 4 probability = + − = = 52 52 52 52 13  2of spades

54. The tacit assumption made here is that the order in which they were born is important. This is not the 3 case. It should be . 8

55. True. Since all events are mutually exclusive, 1 = P ( E1 ∪ E2 ∪ E3 ) =

56. False

57. False

58. False. The probabilities are, in fact, equal since both scenarios can 1 only occur with probability . 32

59. The probability that Person 1 has 1 . Now, birthday on Day A is simply 365 to compute the probability that Persons 1 and 2 have the same birthday, we note, first and foremost, that these events are independent since the people are being chosen at random. Since there are 365 days in a year, we can see that the probability of two people having the same birthday is 1 1 365 ( 365 ⋅ 365 ) = 3651 = 0.0027 .

60. The probability of more than two people having the same birthday is equal to 1 − P (no 2 people having same bday) =

61. Sample space S = {31,32,33,34,35,36, 41, 4243, 44, 45, 46}

62. Let x = P (4) . Then, P (3) = 2 x . As such, since the only events possible are getting 3 or getting 4, we know that P (4) + P (3) = 1  x + 2 x = 1 

P ( E1 ) + P ( E2 ) + P ( E3 ) .

So, P ( Sum is 2,5,or 6 )

= P ( Sum is 2 ) + P ( Sum is 5 ) + P ( Sum is 6 )

= 0 + 122 + 122

1 1 1 − 365 ⋅ 364 ≅ 0.999924733 So, from a probabilistic point of view, in a group of 30 people, it is highly likely that at least two of them will have the same birthday.

3x = 1  x = 13

= 13 ≈ 0.333

So, P(rolling 3) = 23 ≈ 0.667 .

63-64. Do so if you have access to a computer.

1201

Chapter 9

65. Use n = 10 and k = 2 in the formula provided to obtain the approximate probability of 0.2907.

1202

66. Use n = 8 and each of k = 2, k = 1, and k = 0 in the formula provided (and add the results) to obtain the approximate probability of 0.8652.

Chapter 9 Review Solutions ----------------------------------------------------------------------1. 1,8,27,64

2. 1! 2! 3! 4! , , , , 1 2 3 4 which is equivalent to 1,1,2,6 (Note that n ! n ( n − 1) ! an = = = ( n − 1) ! .) n n

3. 5,8,11,14

4. ( −1) x3 , ( −1) x 4 , ( −1) x5 , ( −1) x6 1

which is equivalent to − x3 , x 4 , − x5 , x6 .

5. 5

64 6561

32 2 as =   = ≈ 0.13  3  243

a8 =

8. a10 = 1 +

( −1) (15 − 1) ! = − (14 ) ! 15 (16 ) ! 15 (15 +1) ! 15

a15 =

=−

9. an = ( −1)

1 14! =− 3600 15 ⋅16 ⋅15 ⋅ 14!

n +1

10.

3n, n ≥ 1

 n, n odd  an =  1  n , n even

11. an = ( −1) , n ≥ 1

12. an = 10 n −1 , n ≥ 1

13.

1 11 = 10 10

8! 8⋅ 7 ⋅ 6! = = 56 6! 6!

14.

1203

20! 20! 1 = = 23! 23⋅ 22 ⋅ 21⋅ 20! 10,626

Chapter 9

15.

16.

n ( n −1) ! n ( n −1) ! 1 = = ( n +1) ! ( n + 1) n ( n − 1) ! n +1

(n − 2)! =

17. 5,3,1, −1

18. 2 2 2 1,2 ⋅1, 4 ⋅1,16 ⋅1 =1, 4,16,256 

19. 2 2 1,2, ( 2 ) ⋅ (1) , 4 ⋅2    =32

( n − 2)! 1 = n ( n − 1) ( n − 2 ) ! n ( n −1)

=16

256

20. 1,2,

which is equivalent to 1,2, 4,32 .

(2)  1   2

  4 

1 4

=32

1 which is equivalent to 1,2, ,32 . 4

21.

22.

 3 = 3 (5 ) = 15

 n = 1 + 4 + 9 + 16 =

n =1

144 + 36 + 16 + 9 144

205 144

24.

23. 6

 (3n + 1) = 2 ( 4 +19 ) = 69

2 k +1

 k ! = 2 + 1 + 2! + 3! + 4! + 5!

n =1

k =0

8 4 8 218 =2+4+4+ + + = 3 3 15 15

26.

25. 7

( −1)

2

 2n

n−1

n=1

28.

27. ∞

∞

x ( Note: 2! = 2, 3! = 6, 4! = 24 )  n= 0 n !

 ( −1) n= 0

1204

x n +1 n!

Chapter 9 Review 60

 0.04  29. A60 = 30,000  1 +  = $36,639.90 12   So, the amount in the account after 5 years is $36,639.90. 30. an = 450,000 + 30,000( n − 1), n ≥ 1 we have  3( 4) 

31. Yes, it is arithmetic with d = −2 .

a = 450,000 ( 4 ) + 30,000  2  = 1,980,000

32. No

So, the total salary for 4 years is $1,980,000. Also, the salary in the 4th year is given by a4 = 450,000 + 30,000(3) = 540,000

33. Yes, it is arithmetic with 1 d= . 2

34. Yes, it is arithmetic with d = −1 .

35.

n=1





an =

( n + 1) ! = ( n + 1) n ! = n + 1

n! n! Yes, it is arithmetic with d = 1 .

36. an = 5n − 5 Yes, it is arithmetic with d = 5 .

37. an = a1 + ( n −1) d

= −4 + ( n − 1) ( 5 ) = 5n − 9

38. an = 5 + ( n − 1) ( 6 ) = 6 n −1

39. 2 5  2 an = 1 + ( n − 1)  −  = − n + 3 3  3

40. an = 0.001 + ( n −1) ( 0.01) = 0.01 n − 0.009

1205

Chapter 9

41. We are given that a5 = ai + ( 5 − 1) d = 13 ,

42. We are given that a7 = −14 = ai + ( 7 −1) d ,

In order to find an , we must first find d. To do so, subtract the above formulae to eliminate a1 : a1 + 4d = 13

In order to find an , we must first find d. To do so, subtract the above formulae to eliminate a1 : a1 + 6d = −14

− a1 + 16d = 37

− a1 + 9d = −23

a17 = ai + (17 −1) d = 37 .

a10 = −23 = ai + (10 −1) d .

− 12d = −24

− 3d = 9 Hence, d = −3 . So, an = a1 + ( n −1) ( −3 ) .

Hence, d = 2 . So, an = a1 + ( n −1) ( 2 ) . To find a1 , observe that a5 − a1 = 2 ( 4 ) = 8

To find a1 , observe that a7 − a1 = ( −3 ) ( 6 ) = −18

So, 13 − a1 = 8 so that a1 = 5 . Thus, an = 5 + ( n −1) ( 2 ) = 2 n + 3, n ≥ 1

So, −14 − a1 = −18 , so that a1 = 4 . Thus, an = 4 + ( n − 1) ( −3 ) = −3n + 7, n ≥ 1

43. We are given that a8 = 52 = a1 + ( 8 −1) d ,

44. We are given that a11 = −30 = a1 + (11−1) d ,

In order to find an , we must first find d. To do so, subtract the above formulae to eliminate a1 : a1 + 7d = 52

In order to find an , we must first find d. To do so, subtract the above formulae to eliminate a1 : a1 + 10d = −30

− a1 + 20d = 130

− a1 + 20d = −80

a21 = 130 = a1 + ( 21−1) d .

a21 = −80 = a1 + ( 21−1) d .

−13d = −78 Hence, d = 6 . So, an = a1 + ( n −1) ( 6 ) .

−10d = 50 Hence, d = −5 . So, an = a1 + ( n −1)( −5 ) .

To find a1 ,observe that a8 − a1 = 6 ( 7 ) = 42

To find a1 ,observe that a11 − a1 = −5 (10 ) = −50

So, 52 − a1 = 42 , so that a1 = 10 . Thus, an = 10 + ( n −1) 6 = 6 n + 4, n ≥ 1 .

So, −30 − a1 = −50 , so that a1 = 20 . Thus, an = 20 + ( n −1) ( −5 ) = −5n + 25, n ≥ 1 . 1206

Chapter 9 Review 20

45.  3k = k =1

20 (3 + 60 ) = 630 2

46.  ( n + 5 ) = n=1

47. The sequence is arithmetic with n = 12 , a1 = 2 , and a12 = 68 . So, the given sum is 12 ( 2 + 68) = 420 . 2

15 ( 6 + 20 ) = 195 2

48. The sequence is arithmetic with 1 −31 n = 17 , a1 = , and a17 = . So, the 4 4 given sum is 17  1 31  17  −30   = −63.75 .  − =  2 4 4  2  4 

49. Bob an = 45,000 + ( n −1) 2000 15 

Tania an = 38,000 + 4000 ( n −1)



15 

  a = 2  45,000 + 73,000   = 885,000



  a = 2 38,000 + 94,000   = 990,000

n=1 =a15 =a15     So, Bob earns $885,000 in 15 years, while Tania earns $990,000 in 15 years. n=1

50. The sequence 16, 48,80, , is arithmetic with a1 = 16 and d = 32 . So, the nth term is given by an = 16 + ( n −1) 32 = 32n −16 . So, a5 = 144 . Further, 5

 a = 2 [16 + 144] = 400  . So, she would have fallen 400 feet in 5 seconds. n

n=1

51. Yes, it is geometric with r = −2 . 53. Yes, it is geometric with r =

52. No 54. Yes, it is geometric with r = 10 .

1 . 2

55. 3,6,12,24, 48

56. 10,

57. 100, − 400, 1600, − 6400, 25,600

10 10 10 10 , , , 4 16 64 256

58. −60, 30, − 15,

1207

15 15 ,− 2 4

Chapter 9

59. an = a1r n −1 = 7 ⋅ 2 n −1 , n ≥ 1

60.

1 an = 12   3 61. an = 1( −2 )

n−1

n −1

, n ≥1

62.

, n ≥1

an =

63. an = 2 ( 2 )

n−1

32  −1    5  4 

n−1

, n ≥1

64.

,n ≥1

So, a25 = 2 ( 2 ) = 33,554, 432

1 n−1 (2) , n ≥ 1 2 1 9 a10 = ( 2 ) = 256 2

65.

66.

an =

n−1

 −1  an = 100   , n ≥ 1  5  So, in particular,

 −1  an = 1000   , n ≥ 1  2  So, in particular,

 −1  a12 = 100   = −2.048 × 10 −6  5 

1000  −1  a11 = 1000   = . 1024  2 

67.

68.

 1  n−1 1 1 − 3  19,682 = 4920.50   3 =  = 2  1− 3  4 n=1  2  9

  1 11  n−1 1−    11 1  2 1  1  = 1     = 2  1 −   1    2148  n=1  2  1−  2   4094 = 2048

69. 8

 5 (3) n=1

n−1

1 − 38  = 5  = 16, 400  1− 3 

1208

Chapter 9 Review

70. First, note that the given sum is not in the standard form since the exponent on the common ratio is n rather than n − 1 . So, you must first simplify the expression to put it into standard form, and then apply the known formula, as follows: 7 2 n 7 10 n −1 (5) =  (5)  n =1 3 n =1 3 =

71. 1 2 1− 3

10 1 − 57  195,310 = 3  1 − 5  3

72. 2  −1  1   5   = 25 = 1 1 6 30 1+ 5 5

73. an = 48,000 (1.02 )

n−1

74. an = 15000 ( 0.80 )

, n ≥1

So, in particular, 13−1 a12 = 48000 (1.02 ) = 60,875.61. So, the salary after 12 years is $60,875.61.

n−1

So, in particular, 3−1 a3 = 15,000 ( 0.80 ) = 9600 . So, the boat will be worth $9600 after 3 years.

75. Claim: 3n ≤ 3n , for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . 3(1) ≤ 31 is true since both terms equal 3. Step 2: Assume the statement is true for n = k : 3k ≤ 3k Show the statement is true for n = k +1 : 3( k + 1) ≤ 3k +1 3( k + 1) = 3k + 3 ≤ 3k + 31 (by assumption) ≤ 3k + 3k (since 3 ≤ 3k , for k ≥ 1) = 2(3k ) < 3(3k ) = 3k +1

This completes the proof.

1209

,n ≥1

Chapter 9

76. Claim: 4 n < 4 n +1 , for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . 41 < 41+1 = 42 = 16 is clearly true. Step 2: Assume the statement is true for n = k : 4 k < 4 k +1 Show the statement is true for n = k + 1 : 4 k +1 < 4 k + 2 4 k +1 = 4 k ⋅ 4 < 4 k +1 ⋅ 4 (by assumption) = 4k +2 This completes the proof.

77. Claim: 2 + 7 + ... + (5n − 3) = 2n (5n − 1), for all n ≥ 1 . Proof. Step 1: Show the statement is true for n = 1 . 2 = 12 (5(1) − 1) = 12 (4) is clearly true. Step 2: Assume the statement is true for n = k : 2 + 7 + ... + (5k − 3) = k2 (5k − 1) Show the statement is true for n = k + 1 : 2 + 7 + ... + (5( k + 1) − 3) = k2+1 (5( k + 1) − 1) Observe that 2 + 7 + ... + (5k − 3) + (5( k +1) − 3) = ( 2 + 7 + ... + (5k − 3) ) + (5( k +1) − 3) = k2 (5k − 1) + (5( k +1) − 3) (by assumption) = k (5 k −1)+22(5 k + 2 ) = 5 k +29 k + 4 = 2

( k +1)(5 k + 4 ) 2

= k2+1 (5( k +1) −1)

This completes the proof.

78. Claim: 2 n > ( n + 1)2 , for all n ≥ 3 . Proof. Step 1: Show the statement is true for n = 3 . 18 = 2(3)2 > (3 + 1)2 = 16 is clearly true. Step 2: Assume the statement is true for n = k : 2 k 2 > ( k + 1)2 Show the statement is true for n = k +1 : 2( k + 1)2 > ( k + 1 + 1)2

1210

Chapter 9 Review

2( k + 1)2 = 2 k 2 + 4 k +1 > ( k + 1)2 + 4 k +1 = k 2 + 2k + 1 + 4k + 1 = k 2 + 4 k + 4 + (2 k − 2)    >1 since k ≥3

> (k + 2)

This completes the proof.

79. 11 11! 11! 11⋅10 ⋅ 9 ⋅ 8! = = = 165  = 8! ( 3 ⋅ 2 ⋅1)  8  (11− 8 ) !8! 3!8!

80. 10  10! 10! 1 = = =1  =  0  (10 − 0 ) !0! 10! 0! 0!

81.

82.  47  47! 47! =  =  45  ( 47 − 45 ) !45! 2!45!

 22  22! 22! 1 = = =1  =  22  ( 22 − 22 ) !22! 22! 0! 0!

47 ⋅ 46 ⋅ 45! = 1081 45! ( 2 ⋅1)

83. 4 4 ( x − 5 )4 =    x 4 −k (−5)k k =0  k  4 4 4 4 4 =   x 4 ( −5)0 +   x3 (−5)1 +   x 2 (−5)2 +   x1 (−5)3 +   x 0 ( −5)4 0 1 2 3 4 4! 4! 4! 4! 4! = 4!0! x 4 ( −5)0 + 3!1! x3 (−5)1 + 2!2! x 2 ( −5)2 + 1!3! x1 (−5)3 + 0!4! x 0 ( −5 )4     

= x − 20x +150 x − 500x + 625 4

1211

Chapter 9

84. 5 5 ( x + y )5 =    x5−k y k k =0  k  5 5 5 5 5 5 =   x5 y 0 +   x 4 y1 +   x3 y 2 +   x 2 y3 +   x1 y 4 +   x 0 y5 0 1 2 3 4 5 5! 5! 5! 5! 5! x 4 y1 + 3!2! x 3 y 2 + 2!3! x 2 y3 + 1!4! x1 y 4 + 0!5! x 0 y5 = 55!!0! x5 y 0 + 4!1!      

=10

= x5 + 5x 4 y + 10 x3 y 2 +10 x 2 y3 + 5xy 4 + y5

85. 3 3 (2 x − 5 )3 =    (2 x )3−k (−5)k k =0  k  3 3 3 3 =   (2 x )3 (−5)0 +   (2 x )2 ( −5)1 +   (2 x )1 (−5)2 +   (2 x )0 ( −5)3 0 1 2 3 3! 3! 3! = 3!0! (2 x )3 ( −5)0 + 2!1! (2 x )2 (−5)1 + 13!!2! (2 x )1 ( −5)2 + 0!3! (2x )0 (−5)3    

= 8x − 60 x + 150 x −125 3

86. 4 4 −k 4 ( x 2 + y3 ) 4 =    ( x 2 ) ( y 3 ) k k =0  k  4 3 2 1 4 4 4 4 =   ( x 2 ) ( y3 )0 +   ( x 2 ) ( y3 )1 +   ( x 2 ) ( y3 )2 +   ( x 2 ) ( y3 )3 0 1 2 3 0 4 +   ( x 2 ) ( y3 )4 4 4! 4! 4! 4! x 2 ( y3 )0 + 3!1! x 2 ( y3 )1 + 2!2! x 2 ( y3 )2 + 1!3! x 2 ( y3 )3 = 4!0! ( ) ( ) ( ) ( ) 4

4! x 2 ( y3 )4 + 0!4! ( ) 0

= x + 4x y + 6x y + 4x y + y 8

1212

Chapter 9 Review

87.

5 ( x + 1) =    ( x )5−k (1)k , x ≥ 0 k =0  k  5 5 5 5 =   ( x )5 (1)0 +   ( x )4 (1)1 +   ( x )3 (1)2 +   ( x )2 (1)3 0 1 2 3 5 5 +   ( x )1 (1)4 +   ( x )0 (1)5 4 5 5

5! 5! 5! = 55!!0 ! ( x )5 (1)0 + 4!1! ( x )4 (1)1 + 3!2! ( x )3 (1)2 + 2!3! ( x )2 (1)3    

=10

5! 5! ( x ) (1) + 0!5! ( x ) (1) + 1!4!   1

= x 2 + 5x 2 + 10 x 2 + 10 x + 5x 2 + 1 5

88. 6  6  2 6 −k 13 k 2 1 y ( x 3 + y 3 )6 =    x 3 k =0  k   6  2 6 13 0  6  2 3 5 13 1  6  2 3 4 13 2 +  x =  x 3 +  x y y y 0 1 2

( ) ( )

( )( )

 6  2 3 1 3  6  2 2 13 4  6  2 3 1 13 5  6  2 3 0 13 6 +  x y y y +  x 3 y 3 +  x 3 +  x 3 4 5 6

( )( )

( ) ( y ) +  (x ) ( y ) +  (x ) ( y ) +  (x ) ( y ) 6

6! = 6!0! x3  2

( )( )

6! 5!1!

6! 4!2!

=15

6! 1!5!

6! 3!3!

= 20

( ) ( y ) +  (x ) ( y ) +  (x ) ( y )

+ 2!6!4! x 3  =15

6! 0! 6!

= x 4 + 6 x 3 y 3 + 15x 3 y 3 + 20x 2 y +15x 3 y 3 + 6x 3 y 3 + y 2 10

89. ( r − s )5 = r 5 s 0 + 5r 4 ( −s ) + 10r 3 ( − s )2 + 10r 2 ( −s )3 + 5r( −s )4 + ( −s )5 r 0

= r 5 − 5r 4 s + 10r 3 s 2 − 10r 2 s3 + 5rs 4 − s 5

90. ( ax + by )4 = ( ax )4 ( by )0 + 4( ax )3 ( by )1 + 6( ax )2 ( by )2 + 4( ax )1 ( by )3 + ( ax )0 ( by )4 = a 4 x 4 + 4a 3bx3 y + 6a 2 b 2 x 2 y 2 + 4ab3 xy3 + b 4 y 4

1213

Chapter 9

8 8! 8 ⋅ 7 ⋅ 6! (4)x 6 = 91. This term is   x8−2 ( −2 )2 = (4)x6 = 112 x 6 2!6! (2 ⋅1) 6! 2 So, the desired coefficient is 112. 7 7! 7 ⋅ 6 ⋅ 5 ⋅ 4! ( 27) y 4 = (27) y 4 = 945 y 4 92. This term is   33 y7 −3 = 3 3!4! (3 ⋅ 2 ⋅ 1) 4!   So, the desired coefficient is 945. 93. This term is 6 6! 6 ⋅ 5 ⋅ 4! 6−4 4 (4 ⋅ 625)x 2 y 4 = (2500)x 2 y 4 = 37,500x 2 y 4   (2 x ) (5 y ) = 4 4!2! (2 ⋅ 1) 4!   So, the desired coefficient is 37,500.

8 8! 8 4 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4! 8 4 94. This term is   ( r 2 )8− 4 ( − s )4 = rs = r s = 70r 8 s 4 4!4! (4 ⋅ 3 ⋅ 2 ⋅1) 4! 4 So, the desired coefficient is 70.  53  53! 53 ⋅ 52 ⋅ 51 ⋅ 50 ⋅ 49 ⋅ 48 ⋅ 47! 95. Since   = = = 22,957, 480 , there are 6! 47!  6  6!47! 22,957,480 different 6-number combinations. 108  108! 108 ⋅107 ⋅ ... ⋅ 96 ⋅ 95! 96. Since  = = 20,592,957,740,000,000 , there = 13! 95!  13  95!13! are 20,592,957,740,000,000 different 13-card Canasta hands. 98.

97. 7

P4 =

7! 7! 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3! = = = 840 3! ( 7 − 4 ) ! 3!

99. 12

P9 =

9! 9! = = 9! = 362,880 ( 9 − 9 ) ! 0!

100.

P5 =

12! 12! 12 ⋅11 ⋅10 ⋅ 9 ⋅ 8 ⋅ 7! = = 7! (12 − 5 ) ! 7!

= 95,040

1214

P1 =

10! 10! 10 ⋅ 9! = = = 10 9! (10 −1) ! 9!

Chapter 9 Review

101. 12

12! 12! 12 ⋅11⋅10 ⋅9 ⋅8⋅ 7! = = = 792 (12 − 7 ) !7! 7!5! 7! ( 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1)

C7 =

102. 40

C5 =

40! 10! 40 ⋅ 39 ⋅ 38⋅37 ⋅36 ⋅ 35! = = = 658,008 35! ( 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1) ( 40 − 5 ) !5! 35!5!

103. 9

C9 =

9! 9! = =1 ( 9 − 9 ) !9! 0!9!

104. 53

C6 =

53! 53⋅ 52 ⋅ 51⋅ 50 ⋅ 49 ⋅ 48⋅ 47! = = 22,957, 480 6!47! 6! 47!

105. There are  3 ⋅  2 ⋅  5 = 30 different cars to choose from. Models

Interior

Exterior

106. Since each slot can have one of 26 letters, and order is important, there are 266 = 30,895,776 different passwords. 107. Since all 4 jobs are different with different responsibilities, we know order is important in how we fill the slots. As such, there are 10! 10! 10 ⋅ 9 ⋅ 8⋅7 ⋅ 6! = = = 5040 different leadership teams. 10 P4 = 6! (10 − 4 ) ! 6! 108. Since each of the 6 slots can be filled with one of 24 letters and 8 digits, and order is important, there are 326 = 1,073,741,824 different license plate numbers. 109. There are  5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120 different seating arrangements. Seat 1

Seat 2

Seat 3

Seat 4

Seat 5

Since there are 8 home games in a season, it would take 120 8 = 15 years to go through all possible arrangements.

110. 60

P3 =

60! 60! 60 ⋅ 59 ⋅ 58⋅ 57! = = = 205,320 57! ( 60 − 3) ! 57! 1215

Chapter 9

111. Since the prizes are different, we know order is important. So, there are 100! 100! 100 ⋅ 99 ⋅ 98⋅97 ⋅ 96! = = = 94,109, 400 different ways to 100 P4 = 96! (100 − 4 ) ! 96! distribute the prizes.

117  117! 117 ⋅116 ⋅ 115! 112. Since  = = 6786 , there are 6786 different = 2! 115!  2  2!115! matchups.  52  52! 52 ⋅ 51 ⋅ 50 ⋅ 49 ⋅ 48 ⋅ 47 ⋅ 46! 113. Since   = = = 20,358,520 , there are 6! 46!  6  6!46! 20,358,520 different 6-card hands. 114. Since  8 ⋅  32 = 256 , there are 256 different 2-card combinations (using Ace

Face or 10

two decks) that lead to blackjack.

115. The sample space has 2 4 = 16 elements and there is only one way to get all heads. So, P ( HHHH ) = 161 = 0.0625 . So, there is a 6.25% chance of getting all heads. 116. Since the sum of two dice is odd for the following rolls, we know that 18 1 P (sum is odd) = = . 36 2 Odd sum rolls: (1,2 ) , (1, 4 ) , (1,6 ) , ( 2,1) , ( 2,3 ) , ( 2,5 ) , ( 3,2 ) , ( 3, 4 ) , ( 3,6 ) ,

( 4,1) , ( 4,3) , ( 4,5 ) , (5,2 ) , (5, 4 ) , (5,6 ) , ( 6,1) , ( 6,3) , ( 6,5 )

117. (The sample space for this problem can be found in Example 4 of Section 9.7 of the text.) P (not getting 7) = 1 − P (getting 7) = 1 − 366 = 30 36 = 0.833 So, there is about an 83.3% chance of not getting 7. 1 118. P (diamond) = 13 52 = 4

119. P (not E1 ) = 1 − P ( E1 ) = 1 − 13 = 23 ≈ 66.7%

120. Since E1 and E 2 are mutually exclusive, P ( E1 ∪ E2 ) = P ( E1 ) + P ( E2 ) = 13 + 12 = 65

1216

Chapter 9 Review

121. Since E1 and E 2 are not mutually exclusive, P ( E1 ∪ E2 ) = P ( E1 ) + P ( E2 ) − P ( E1 ∩ E 2 ) = 13 + 12 − 14 = 127 ≈ 58.3%

122. Since E1 and E 2 are independent, P ( E1 ∩ E2 ) = P ( E1 ) ⋅ P ( E 2 ) = 13 ⋅ 12 = 61 . 123. Since these two events are mutually exclusive, we see that P(drawing ace or drawing 2) = P(drawing ace) + P(drawing 2) = 524 + 524 = 528 = 132 ≈ 15.4% . 124. Since these two events are independent, we see that P(drawing ace on the first draw and drawing 2 on the second draw) = P(drawing ace on the first draw)  P(drawing 2 on the second draw) = 16 4 4 52 ⋅ 51 = 2652 . 125. The sample space in this problem has 25 = 32 elements. So, we see that P(having at least one girl) = 1 − P (having no girls) = 1 − P(having all boys) 31 = 1 − 321 = 32 = 96.88% 1 126. ( 12 ) ⋅ ( 21 ) = 4096   11

Win

129.

127.

5369 3600

128. The sum is infinite.

Lose

34,875 14

130. 6556

131. The graphs of the two curves are given by:

132. Yes, and the sum is approximately 7.4215 since n

1 π e = ≈ 7.4215    = e π −e n=0  π  1− ∞

The series will sum to

1 . 1 + 2x

1217

Chapter 9

133. The sum is 99,900. Yes, it agrees with Exercise 77.

134. The graphs are given by

Yes, it confirms the proof in Exercise 78.

135. As each new term is added, the graphs become better approximations of the graph of y = (1 + 2 x )4 for −0.1 < x < 0.1 . The series does not get closer to this graph for 0.1 < x < 1 .

136. As each new term is added, the graphs become better approximations of the graph of y = (1 − 2 x )4 for −0.1 < x < 0.1 . The series does not get closer to this graph for 0.1 < x < 1 .

137. a. 11,440 b. 11,440 c. Yes d. This is true because nr!Pr = n Cr .

138. a. 20,358,520 b. 20,358,520 c. Yes d. This is true because n Pr r! = n C r .

139. Using the formula with n = 10 and k = 9 (and then cubing the result since the events are independent) yields the approximate probability of 0.0722.

140. Using the formula with n = 8 and k = 9 (and then squaring the result since the events are independent) yields the approximate probability of 0.7795.

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Chapter 9 Practice Test----------------------------------------------------------------------------1. x n−1

2. geometric with r = x

1 − xn 3. Sn = 1− x

4.  x n

∞

n=0

5. We must have x < 1 in order for the sum in Problem 4 to exist. In such case, 1 . the sum is 1−x

∞

 ( 31 ) = n

n=1

50 2

1 3

= 1

1− 3

1 2

1 − ( 14 )

 3⋅( ) = 3⋅( ) ⋅ 1 − 1 n 4

1 4

n=1

(3 + 2(50) + 1) = 25 (104 ) = 2600

1 4

= 1 − ( 14 ) ≈ 1 10

9. 100

100

 (5n − 3) = 2 ( 2 + 497 ) = 24,950 n=1

2 10. Claim: 2 + 4 + ...+ 2n = n + n , for all n ≥ 1 . = n ( n +1)

Proof. Step 1: Show the statement is true for n = 1 . 2 = 1 + 1 = 12 + 1 = 2 , which is clearly true. Step 2: Assume the statement is true for n = k : 2 + 4 + ...+ 2 k = k ( k +1) Show the statement is true for n = k + 1 : 2 + 4 + ...+ 2(k +1) = ( k +1)( k + 2) Observe that 2 + 4 + ... + 2 k + 2( k +1) = ( 2 + 4 + ... + 2k ) + 2( k +1)

= k ( k + 1) + 2( k + 1) (by assumption) = ( k + 2 )( k +1) This completes the proof.

11. 7! 7⋅ 6 ⋅ 5⋅4⋅3⋅ 2! = = 2520 2! 2!

1219

Chapter 9

12.

5 5  Note that (2 x + y )5 = (2 x )5 +   (2 x )4 y +   (2 x )3 y 2 + ... 1  2 5  5! 8x3 y 2 = 80 x3 y 2 . Hence,   (2 x )3 y 2 = 2!3! 2 14. k k! k! = =1  =  k  ( k − k ) ! k ! 0! k !

13.  15  15! 15 ⋅14 ⋅13 ⋅ 12! = = 455  = 3! 12! 12  12!3! 15. 14

P3 =

14! 14! 14 ⋅13 ⋅12 ⋅ 11! = = = 2184 11! (14 − 3) ! 11!

16. 200

C3 =

200! 200! 200 ⋅199 ⋅198⋅ 197! = = = 1,313, 400 197! ( 3 ⋅ 2 ⋅1) ( 200 − 3) !3! 197!3!

17. 5 5− k 5 k ( x 2 + x1 )5 =    ( x 2 ) ( x1 ) , x ≠ 0 k =0  k  5 4 3 5 5 5 0 1 2 =   ( x 2 ) ( x1 ) +   ( x 2 ) ( x1 ) +   ( x 2 ) ( x1 ) 0 1 2 2 1 0 5 5 5 3 4 5 +   ( x 2 ) ( x1 ) +   ( x 2 ) ( x1 ) +   ( x 2 ) ( x1 ) 3 4 5 2 5! 5! 5! 1 1 = 5!0! + 4!1! x2 ( x2 ) ( x1 ) +  3!2! ( x ) ( x ) ( ) (x)  5

=10

2 5! 5! 5! 1 + 2!3! + 1!4! x ( x2 ) ( x1 ) +  0!5! ( x ) ( x ) ( ) ( )  2 2

1 3 x

=10

= x + 5x + 10x +10 x + 10

5 x2

1220

1 x5

Chapter 9 Practice Test

18. 4 4 4 4 (3x − 2 )4 = (3x )4 (2)0 +   (3x )3 (−2) +   (3x)2 (−2)2 +   (3x)(−2)3 +   (3x )0 ( −2)4 1  2 3  4 4! 4! 4! 54 x3 + 36x 2 − 24 +16 = 81x 4 − 3!1! 2!2! 1!3! = 81x 4 − 216x3 + 216x 2 − 96x +16

19. Intuitively, there are more permutations than combinations since order is taken into account when determining the number of permutations, while order does not increase the number of combinations. More precisely, observe that n! 1  n!  1 =  nCr =  = n Pr . r !( n − r)! r !  (n − r)!  r ! Since r! ≥ 1, it follows that r1! ≤ 1 , so that n Pr ≥ n Cr . 20. Since order is important here, the number of ways of choosing 3 horses out of 15 is 15! 15! 15 ⋅14 ⋅13⋅ 12! = = = 2730 . 15 P3 = 12! (15 − 3) ! 12! So, the probability of NOT winning is 1 − 2730 15! ≈ 1 .

21. P(red) = 18 38 ≅ 0.47 . So, about 47% chance. 22. Since all spins are mutually independent, we see that P(red on 1st AND red on 2nd AND red on 3rd AND red on 4th AND red on 5th) = P(red on 1st)  P(red on 2nd )  P(red on 3rd)  P(red on 4th)  P(red on 5th) = 5 ( 1838 ) ≅ 0.0238 So, about 2.38% chance. 23. Since spins are independent, the previous history has no effect on the probability of a particular spin. So, this probability is again 18 38 ≅ 0.47 . 24. The desired probability is as follows: 3 3 1 ⋅ ⋅ 84 ⋅ 10 9 7     Black on first draw

Blue on second draw

Red on third draw

1221

Red on fourth draw

⋅

6  Green on fifth draw

1 = 420

Chapter 9

25. Since these two events are not mutually exclusive (since there is an ace of diamonds), the probability is: P (ace or diamond) = P (ace) + P(diamond) − P(ace AND diamond) 16 1 4 = 524 + 13 52 − 52 = 52 = 13 ≈ 0.308

26. a. 8! = 40,320 b.

1 8!

≈ 0.00002

27. Expanding this using a calculator reveals that the constant term is 184,756.

28. Using a calculator to compute this sum yields 73,375 12 .

29. Since only the lockers with an odd number of multiples will be left open, we see that the open lockers are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196. Then, an = n 2 , n ≥ 1.

30. Neither: not arithmetic because there is no common difference., not 3 5 geometric because ≠ . 2 3

1222

Chapter 9 Cumulative Review -------------------------------------------------------------------2. Let w = width of rectangle. Then, l = 2w − 5 . Given that the perimeter is 38 inches, we have: 2w + 2(2w − 5) = 38 2w + 4w − 10 = 38 6w = 48 w =8 Thus, l = 11 and so, the dimensions are 8 inches by 11 inches.

1. 10x + 1 5 + 4x 10x + 1 5 + 4x + = − 3x − 2 2 − 3x 3x − 2 3 x − 2 10x + 1 − ( 5 + 4x ) = 3x − 2 6x − 4 = 3x − 2 2 ( 3x − 2 ) =2 = 3x − 2

x −5 > 3

5 ± ( −5) − 4(2)(11) 5 ± −63 = 2(2) 4 2

x= =

x − 5 > 3 or x − 5 < −3 x > 8 or x < 2 So, the solution set is ( −∞,2 ) ∪ ( 8,∞ ) .

5 ± 3i 7 4

5. Since the slope is undefined, the line is vertical and hence, is described by an equation of the form x = a . Since the point (−8,0) is on the line, the equation is x = −8 .

6. x-intercept: 3x − 5(0) = 15  x = 5. So, (5,0). y-intercept: 3(0) − 5( y ) = 15  y = −3. So, (0, −3). Slope m = 35

7. f ( x + h) − f (x) h ( x + h )2 − 3( x + h ) + 2 − ( x 2 − 3x + 2 ) = h 2 2 x + 2hx + h − 3x − 3h + 2 − x 2 + 3x − 2 = h h ( 2x + h − 3) = = 2x + h − 3 h

8. f ( g (x)) = 4 −

1 1 2 x +1

= 4 − (2x + 1)

= −2x + 3 provided that x ≠ − 12 .

Domain: ( −∞, − 12 ) ∪ ( − 12 , ∞ )

1223

Chapter 9

10. f ( x ) = −0.04x + 1.2x − 3 2

= −0.04 ( x 2 − 30 x ) − 3

P ( x ) = ( x 2 + 9 )( x 2 −1) = (x + 3i )(x − 3i )(x −1)(x +1)

= −0.04 ( x 2 − 30 x + 225 ) − 3 + 9

= −0.04(x − 15)2 + 6 So, the vertex is (15,6).

11. Vertical asymptote: x = 3 Horizontal asymptote: y = −5

12. Use A = P (1 + nr ) . nt

Here, A = 65,000, r = 0.047, n = 52, t = 17 . So, we have 65,000 = P (1 + 0.047 52 )

52(17 )

29,246.21 ≈

65,000

(1+ 0.047 52 )

52 (17 )

So, about $29,246.21. ln6 13. log 2 6 = ln2 ≈ 2.585

14. ln(5x − 6) = 2 5x − 6 = e 2 x=

15. Solve the system:  8x − 5 y = 15 (1)  8  y = 5 x + 10 ( 2 ) Substitute (2) into (1): 8x − 5 ( 58 x + 10 ) = 15  0 = 65 Since this results in a false statement, the system has no solution.

6 + e2 ≈ 2.678 5

16. Using matrix methods, we have  2 −1 1 1  R1 − 2 R2 →R2 →   ⎯⎯ ⎯⎯⎯ 1 − 1 4 3    2 −1 1 1  12 R1 →R1 1 − 12 12 12  →    ⎯⎯ ⎯⎯  0 1 −7 −5  0 1 −7 −5  Let z = a. Then, y = 7a − 5 (using 2nd row). Hence, substituting these into the equation corresponding to the first row, we see that: x − 12 (7a − 5) + 12 a = 21  x = 3a − 2

1224

Chapter 9 Cumulative Review

17. The region is as follows: Vertex (1,4) (1,2) (3,2)

z = 4x + 5y 24 MAX 14 22

So, the maximum is z(1,4) = 24.

Note on the graph: C1: y = 5 – x

18.  1 5 −2 3  R −3R →R 1 5 −2 3    R32 −2 R11 →R32   − → 0 −14 8 −12  3 1 2 3   ⎯⎯ ⎯⎯⎯  2 −4 4 10  0 −14 8 4  1 5 −2 3    R2 − R3 →R3 ⎯⎯⎯⎯→ 0 −14 8 −12  0 0 0 −16  The last row suggests the system has no solution. 19. 9 0  11 2 −1 99 18 −9  1 2   9 1 4  = 29 4 7      

1225

Chapter 9

20. 2 5 −1 4 0 1 0 1 4 1 4 0 =2 −5 −1 1 3 −2 3 −2 1 −2 1 3 = 2(12) − 5(3) −1(9) = 0

22.

21. Since (3,5) is the vertex and the directrix is x = 7, the parabola opens to the left and has general equation −4 p( x − 3) = ( y − 5)2 To find p, note the distance from vertex to directrix is 4 units, so p = 4. Hence, the equation is −16(x − 3) = ( y − 5)2 1 x = − ( y − 5 )2 + 3 16 23. 2 0 21 2 2 23 + + + = 1 + 1 + 23 + 13 = 3 1! 2! 3! 4!

24. The sequence can be described by an = 5 ⋅ 3n −1 , n ≥ 1. So, it is geometric. 25. 210 = 1024

26. Using a calculator to compute this sum yields 20,090.

27. Using a calculator to expand this quantity then reveals that the constant term is −3,432.

1226