SOLUTIONS MANUAL For Algebra and Trigonometry, Real Mathematics, Real People, 7Edition, Ron Larson by StudyGuide

Algebra and Trigonometry, Real Mathematics, Real People, 7e Ron Larson (Solutions Manual All Chapters, 100% Original Verified, A+ Grade) CHAPTER P Prerequisites Section P.1

Real Numbers ......................................................................................... 2

Section P.2

Exponents and Radicals ......................................................................... 7

Section P.3

Polynomials and Factoring...................................................................13

Section P.4

Rational Expressions ............................................................................23

Section P.5

The Cartesian Plane .............................................................................. 33

Section P.6

Representing Data Graphically ............................................................ 41

Chapter P Review ........................................................................................................43 Chapter P Test ..............................................................................................................50

C H A P T E R Prerequisites

Section P.1 Real Numbers 16. {2.3030030003, 0.7575, − 4.63,

1. rational

10, − 2, 0.3, 8} (a) Natural number: 8 (b) Whole number: 8 (c) Integers: −2, 8 (d) Rational numbers: 0.7575, − 4.63, − 2, 0.3, 8

2. Irrational 3. prime 4. variables, constants 5. terms

(e) Irrational numbers: 2.3030030003,

6. Yes. 5 − 2 = 2 − 5  3 = −3 = 3

17.

7. (c) Commutative Property of Addition: a + b = b + a 8. (d) Associative Property of Multiplication: ( ab ) c = a ( bc )

18.

2 3

(a) Natural numbers: 5, 1 (b) Whole numbers: 5, 0, 1 (c) Integers: −9, 5, 0, 1, − 4, − 1 (d) Rational numbers: −9, − 27 , 5, 23 , 0, 1, − 4, − 1 (e) Irrational number: 14.

{ 5, − 7, − , 0, 3.12, , − 2, − 8, 3} 7 3

(a) (b) (c) (d)

5 4

Natural number: 3 Whole numbers: 0, 3 Integers: −7, 0, − 2, − 8, 3 Rational numbers: −7, − 73 , 0, 3.12, 54 , − 2, − 8, 3

(e) Irrational number:

15. {2.01, 0.666, − 13, 0.010110111, 1, − 10, 20} (a) Natural numbers: 1, 20 (b) Whole numbers: 1, 20 (c) Integers: −13, 1, − 10, 20 (d) Rational numbers: 2.01, 0.666, − 13, 1, − 10, 20 (e) Irrational number: 0.010110111

(a) Natural numbers:

6 3

(b) Whole numbers:

6 3

(since it equals 2), 3 ,3

{25, − 17, − , 9, 3.12, π , 6, − 4, 18} 12 5

1 2

(a) Natural numbers: 25,

9 = 3, 6, 18

(b) Whole numbers: 25,

9 = 3, 6, 18

{−9, − , 5, , 2, 0, 1, − 4, − 1} 7 2

}

2 , − 7.5, − 2, 3, − 3

(e) Irrational numbers: −π , 12 2

12. (a) Multiplicative Identity Property: a ⋅ 1 = a 13.

1 2

(d) Rational numbers: − 13 , 63 , − 7.5, − 2, 3, − 3

10. (b) Distributive Property: a ( b + c ) = ab + ac Associative Property of Addition: ( a + b) + c = a + (b + c)

6 3

1 3

9. (e) Additive Inverse Property: a + ( − a ) = 0

11. (f)

{−π , − , ,

(e) Irrational number: 12 π 19.

5 16

= 0.3125

20.

17 4

= 4.25

21.

41 333

= 0.123

22.

3 7

= 0.428571

23. − 100 = −9.09 11 24. − 218 = −6.60 33 4 64 32 25. 6.4 = 6 = = 10 10 5

26. − 7.5 = − 7

5 75 15 = − = − 10 10 2

27. −12.3 = −12

3 123 = − 10 10

Section P.1

(b)

29. −1 < 2.5

(b)

−3

−2

−1

3 >− 7 2 2

(b)

7 2

34. − < − 8 7

(b) −3

−2

−1

3 4

− −

x 0

−1

44. (a) The inequality − 9 < x ≤ − 6 is the set of all real

35. − 34 < − 85

numbers greater than − 9 and then less than or equal to − 6.

5 8

−1

(b)

−1 2

−9 −8 −7 −6 −5 −4 −3 −2 −1

> 23 2 5 3 6

45.

x < 0; ( −∞, 0 )

46.

y ≥ 0;  0, ∞ )

37. (a) The inequality x ≤ 5 is the set of all real numbers less than or equal to 5. x 0

−3

−1

43. (a) The inequality −1 ≤ x < 0 is the set of all negative real numbers greater than or equal to −1.

3 7

−8

(b)

−2

3 2

−4 −3 −2 −1

−2

−1

42. (a) The inequality 0 ≤ x ≤ 5 is the set of all real numbers greater than or equal to zero and less than or equal to 5.

− 3.5

(b)

32. −3.5 < 1

5 6

41. (a) The inequality −2 < x < 2 is the set of all real numbers greater than −2 and less than 2.

−5 −4 −3 −2 −1

36.

31. − 4 < 2

−

30. −6 < −2.5

33.

40. (a) The inequality x ≥ 4 is the set of all real numbers greater than or equal to 4.

87 187 28. 1.87 = 1 = 100 100

−4

Real Numbers

−2

−1

47. z ≥ 10, 10, ∞ ) 48.

y ≤ 25, ( −∞, 25

49. 9 ≤ t ≤ 24;  9, 24  50. −1 ≤ k < 3;  −1, 3 ) 51. 0 < m ≤ 5 or (0, 5] 52. 2.5% ≤ r ≤ 5%;  0.025, 0.05 53. −3 ≤ x < 8 or  −3, 8 ) 54. −4 < x ≤ 4 or ( −4, 4  55. −a, a + 4 56.

( −c + 2, c + 1)

Chapter P

Prerequisites

57. The interval ( −6, ∞ ) consists of all real numbers greater than −6. 58. The interval ( −∞, 4  consists of all real numbers less than or equal to 4. 59. The interval ( −∞, 2  consists of all real numbers less

63. −3 − −3 = −3 − 3 = −6 64.

−1 − −2 = (1) − ( 2 ) = −1

65.

−5 −5 = = −1 −5 5

66. − 3 − 3 = − 3(3) = − 9

than or equal to 2. 60. The interval (1, ∞ ) consists of all real numbers greater than 1. 61.

−10 = − ( −10 ) = 10

62.

0 =0

x +1

67. (a) If x > −1  x + 1 > 0, then

x +1 x +1

(b) If x < −1  x + 1 < 0, then

x +1 x − 2

68. (a) If x > 2  x − 2 > 0, then (b) If x < 2  x − 2 > 0, then 69.

x − 2 x − 2 x − 2

y − 4 x = −3 − 4 ( 2 ) = −3 − 8 = −11 = 11

70.

x − 2 y = −2 − 2 −1 = 2 − 2 (1) = 0

71.

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

= =

x +1 = 1. x +1 −( x + 1) x +1

= −1.

x − 2 = 1. x − 2 − ( x − 2) x − 2

= −1. 77. − −1 < − ( −1) since −1 < 1. 78. − ( −2 ) > − 2 since − ( −2 ) = 2. 79. d (126, 75 ) = 75 − 126 = 51 80. d ( −126, − 75 ) = −126 − ( −75 )

= −126 + 75

12 + 2 14 7 = = = 4 4 2

72.

3 x − 2y 2x + y

= =

3 −2 − 2 ( −4 ) 2 ( −2 ) + ( −4 )

3(2) + 8 14 7 = = ( −4 ) + ( −4 ) −8 4

= −51 = 51

(

)

( ) = 142 = 7 = 7

81. d − 52 , 92 = 92 − − 52 82. d ( 14 , 114 ) = 14 − 114

= − 104 = 104 = 25

73.

−3 > − −3 since 3 > −3.

83. d ( 165 , 112 = 112 − 165 = 128 75 ) 75 75

74.

−4 = 4 since −4 = 4 and 4 = 4.

84. d − 15 , 7 = 73 − − 15 8 3 8

75. −5 = − 5 since − 5 = −5. 76. − −6 < −6 since −6 = 6 and

(

)

= 101 ( ) = 5624+ 45 = 101 24 24

85. d ( x, 5 ) = x − 5 and d ( x, 5 ) ≤ 3

Thus, x − 5 ≤ 3.

− −6 = − ( 6 ) = −6.

Section P.1

Real Numbers

88. d ( y, a ) = y − a and d ( y, a ) ≤ 3, So, y − a ≤ 3.

86. d ( x, − 10 ) = x − ( −10 ) = x + 10 , and d ( x, − 10 ) ≥ 6. So, x + 10 ≥ 6.

87. d ( y, 0 ) = y − 0 = y and d ( y, 0 ) ≥ 6

Thus, y ≥ 6.

Receipts

Expenditures

Receipts − Expenditures

89. 1992 $1091.2 billion $1381.5 billion

$290.3 billion

90. 1996 $1453.1 billion $1560.5 billion

$107.4 billion

91. 2000 $2025.2 billion $1789.0 billion

$236.2 billion

92. 2004 $1880.1 billion $2292.8 billion

$412.7 billion

93. 2008 $2524.0 billion $2982.5 billion

$458.5 billion

94. 2012 $2450.2 billion $3537.1 billion

$1086.9 billion

95. Budgeted Expense, b

Actual Expense, a

a−b

0.05b

$113,356

$656

$5635

$112,700

The actual expense difference is greater than $500 (but is less than 5% of the budget) so it does not pass the “budget variance test.” 96. Budgeted Expense, b

Actual Expense, a

a−b

0.05b

$9772

$372

0.05 ( $9400 ) = $470

$9400

Because the difference between the actual expenses and the budget is less than $500 and less than 5% of the budgeted amount, there is compliance with the “budget variance test.” 97. Budgeted Expense, b

Actual Expense, a

a−b

0.05b

$37,335

$265

$1880

$37,600

Because the difference between the actual expenses and the budget is less than $500 and less than 5% of the budgeted amount, there is compliance with the “budget variance test.” 98. Budgeted Expense, b

Actual Expense, a

a−b

0.05b

$25,263

$537

0.05( 25,800) = $1290

$25,800

The actual expense difference is greater than $500 (but is less than 5% of the budget) so it does not meet the “budget variance test.” 99. 7 x + 4 Terms: 7x, 4 Coefficient of 7 x : 7 100. 2 x − 9 Terms: 2 x, − 9 Coefficient of 2 x : 2 101.

103. 4 x 3 +

Terms: 4 x 3 ,

3 x 2 , − 8 x, − 11

Coefficient of 3 x 2 : 3 Coefficient of −8 x : − 8 102. 7 5 x 2 + 3

Terms: 7 5 x 2 , 3 Coefficient of 7 5 x 2 : 7 5

x , −5 2

Coefficient of 4 x 3 : 4 x 1 Coefficient of : 2 2

3 x 2 − 8 x − 11 Terms:

x −5 2

104. 3 x 4 +

2 x3 5

Terms: 3 x 4 ,

2 3 x 5

Coefficient of 3 x 4 : 3 2 2 Coefficient of x 3 : 5 5

Chapter P

Prerequisites

105. 2 x − 5

(a) 2 ( 4 ) − 5 = 8 − 5 = 3

(b) 2 ( − ) − 5 = −1 − 5 = −6 1 2

109. 2 ( x + 3 ) = 2 x + 6

Distributive Property 110. ( z − 2 ) + 0 = z − 2

106. 4 − 3x

(a) 4 − 3 ( 2 ) = 4 − 6 = −2

4 − 3 ( − 65 ) = 4 + 156 = 4 + 25 = 132

(b)

107. x 2 − 4

(a) ( 2 ) − 4 = 4 − 4 = 0 2

(b) ( −2 ) − 4 = 4 − 4 = 0 2

x2 108. x+4

(1)

(a)

Additive Identity Property 111. x + 9 = 9 + x Commutative Property of Addition 112.

(h + 6)

( h + 6 ) = 1, h ≠ −6

Multiplicative Inverse Property 113. − y + ( y + 10) = ( − y + y ) + 10 = 10

Associative Property of Addition 2

12 + 4

1 1 = 1+ 4 5

( −4 ) = 16 , undefined 2

(b)

114.

1 7

−4 + 4 0 Division by zero is undefined.

( 7 ⋅ 12 ) = ( 17 ⋅ 7 )12 Associative Property of Multiplication = 1 ⋅ 12

Multiplicative Inverse Property

= 12

Multiplicative Identity Property

115.

3 16

+ 165 = 168 = 21

126. 60° − 23° = 37° change

116.

6 7

− 57 = 71

127. False. The number 0 is nonnegative but positive.

117.

5 8

− 125 + 61 = 15 − 10 + 244 = 249 = 83 24 24

128. False. If a > 0 and b < 0, then ab < 0.

118.

10 11

60 59 + 336 − 13 = 66 + 12 − 13 = 66 66 66 66

129. False. For example, 3 > 2, but 13 < 12 .

119.

x 4 x x 2 x 3x x + = + = = 6 12 6 6 6 2

2 x x 4 x 5x 9 x + = + = 120. 5 2 10 10 10 12 1 12 8 96 ÷ = ⋅ = 121. x 8 x 1 x 122.

11 3 11 4 44 ÷ = ⋅ = x 4 x 3 3x

123. ( 25 ÷ 4 ) − ( 4 ⋅ 83 ) = ( 25 ⋅ 14 ) − 128 = 101 − 23 15 = 101 − 10 = − 14 = − 57 10

130. (a)

n 5 n

1 5

0.5 10

0.01 500

0.0001 50,000

0.000001 5,000,000

(b) As n approaches 0, 5 n approaches infinity ( ∞ ) . That is, 5 n increases without bound. 131. (a) − A is negative, − A < 0, because A > 0. (b) −C is positive, −C > 0 because C < 0. (c) B − A is negative, B − A < 0, because B < 0 and − A < 0. (d) A − C is positive, A − C > 0 because −C > 0 and A > 0.

124. ( 35 ÷ 3 ) − ( 6 ⋅ 84 ) = ( 35 ⋅ 13 ) − ( 3 ) = 15 − 3 = 15 − 155 = − 145

125. d ( 57, 236 ) = 236 − 57 = 179 miles

Section P.2 132. (a) Matches graph (ii).

Exponents and Radicals

133. When u and v have the same sign, u + v = u + v . For example if

(b) Matches graph (i). A range of prices can only include zero and positive numbers with at most two decimal places. So, a range of prices can be represented by whole numbers and some noninteger positive fractions. A range of lengths can only include positive numbers. So, a range of lengths can be represented by positive real numbers.

u = 2 and v = 1, then 2 + 1 = 2 + 1 , or if

u = −2 and v = −1, then −2 + ( −1) = −2 + −1 . If u and v have different signs, then u + v < u + v . For example if u = 2 and v = −1, 2 + ( −1) < 2 + −1 . Finally, u + v >| u + v , no matter the signs of u and v. 134. a ≤ 0; If the original value of a is negative, then a

results in a positive number. Because a is negative, the expression a = a states that a is equal to a negative number, which can never happen. So, if a is originally negative, a must equal −a, which is a positive value.

Section P. 2 Exponents and Radicals 1. exponent, base

14. (a) 24( − 2)

2. square root

4. index, radicand

(b)

7. The conjugate of 2 + 3 5 is 2 − 3 5. 8. An expression involving radicals is in simplest form when the following conditions are satisfied:

All possible factors have been removed from the radical. All fractions have radical-free denominators. The index of the radical is reduced.

16. (a)

17. (a)

9. No, −10.767 × 10 is not written in scientific notation. It

should be −1.0767 × 10 4.

(

)

10. 64 is both a perfect square 8 = 64 and a perfect cube

( 4 = 64 ). 3

11. (a) 3 ⋅ 33 = 34 = 81

32 1 1 = 2 = 34 3 9

53 12. (a) 2 = 51 = 5 5 (b) 42 ⋅ 42 = 44 = 256

(− 32 ) = (− 9)3 = − 729 3

(23 ⋅ 32 ) = (8 ⋅ 9)2 = (72)2 = 5184 2

(b)

24 3 = − − 32 4

52 25  5 (b)   = 2 = 8 64 8

(b)

13. (a)

6. power, index

(b)

(− 2)

15. (a) ( 4 ⋅ 3) = 123 = 1728

5. rationalizing

2. 3.

(b) − 7 0 = −1

3. principal nth root

−5

18. (a)

1 1 3 2 5 + = + = 2 3 6 6 6

2 −1 + 3−1 =

(3 ) = 3 = 27 −1

−3

() ( 4 )( 3 )

4 4 19 4 ⋅ 3− 2 4 12 48 16 = = 19 = ⋅ = = −2 −1 1 1 2 ⋅3 9 1 9 3 12

(b) 3−1 + 2 − 2 =

1 1 4 + 3 7 + = = 3 4 12 12

19. When x = −3,

2 x 3 = 2 ( −3 ) = 2 ( −27 ) = −54. 3

20. When x = 2, − 3 x 4 = −3 ( 2 ) = −3 (16 ) = −48. 4

21. When x = 4, 5( − x) = 5( − 4) = 5(1) = 5. 0

(43 ) = 40 = 1 0

6 = 64 = 1296 6−3

Chapter P

Prerequisites

22. When x = 7,

 1  7 23. When x = 2, 7 x −2 = 7 2 −2 = 7  2  = . 2  4

( )

6 x 0 − ( 6 x ) = 6 ( 7 ) − ( 6 ⋅ 7 ) = 6 (1) − 1 = 5. 0

24. When x = − 5, 20 x − 2 + x −1 = 20( − 5)

( ) (b) x ( 3 x ) = 3 x

−2

+ ( − 5)

x3 x2 = x5

25. (a)

(

) (b) −5 x ( 4 x ) = −20 x 3

27. (a)

( 3 x ) = 32 x 2 = 9 x 2

(b)

(4 x ) = 1, x ≠ 0 3

4 5 1024 = ≈ 1.405 93 729

37.

43 − 1 = 34 ( 64 − 1) = ( 81)( 63 ) = 5103 3 −4

38.

32 − 2 9 − 2 7 = = ≈ 0.538 4 2 − 3 16 − 3 13

39. 973.50 = 9.735 × 10 2

40. 28,022.2 = 2.80222 × 10 4

( ) = 6z (16z ) = 96z (b) ( 3 x ) ( 2 x ) = ( 27 x )( 4 x ) = 108 x 4

6z2 2z5 5

41. 10,252.484 = 1.0252484 × 10 4 29

7x2 7 = 7 x 2 − 3 = 7 x −1 = x3 x

29. (a)

12 ( x + y )

(b)

9( x + y)

44. −5,222,145 = −5.222145 × 106 45. 0.0002485 = 2.485 × 10 −4

2 4 ( x + y) , x + y ≠ 0 3

3x 2 y 4 y2 (b) = = , x ≠ 0, y ≠ 0 2 2 2 15 x y 5 15( xy ) 31. (a)

−1 2  x 2 y −2 −1  = x 2 y −2 = x , x ≠ 0 2   y

(b)

 a  b  b b b  −2    = 2 ⋅ 3 = 5 , b ≠ 0 a a a  b  a 

(

)

−2

(

) (5x2 z 6 ) 3

−3

( −4 ) ( 52 ) = ( −64 )( 25) 3

= −1600 34.

(8 )(10 ) ≈ 0.244

35.

36 729 = ≈ 2.125 73 343

48. −0.000125005 = −1.25005 × 10 −4 49. 57,300,000 = 5.73 × 107 square miles 50. 9,460,000,000,000 = 9.46 × 1012 kilometers 51. 0.0000899 = 8.99 × 10 −5 gram per cm3

4 3  64  81  5184 32. (a)     =  3  4  = y y y y y7       

(b) 5 x 2 z 6

46. 0.0000025 = 2.5 × 10 −6 47. −0.0000025 = −2.5 × 10 −6

r5 1 = r9 r4 3x 2 y 4

30. (a)

42. 525,252,118 = 5.25252118 × 108 43. −1110.25 = −1.11025 × 10 3

3 −1 4 = ( x + y) 3

−4

36.

28. (a)

33.

1 4 1 3  1  = 20  + = − = . − 25 5 5 5 5  

4 z 4 −2 z 3 = −8z 7

26. (a)

−1

= 1, x ≠ 0, z ≠ 0

52. 0.000003281 foot = 3.281 × 10−6 foot 53. 1.08 × 10 4 = 10,800 54. −4.816 × 108 = −481,600,000 55. − 7.65 × 10 − 7 = − 0.000000765 56. 5.098 × 10−10 = 0.0000000005098 57. 5.14 × 102 = 514

58. 1.5 × 107 = 15,000,000 degrees Celsius 59. 9.0 × 10 −5 = 0.00009 meter

Section P.2 60. 1.6022 × 10 −19 = 0.00000000000000000016022 coulomb

77. − 6

61.

( 2.0 × 10 )( 3.0 × 10 ) = 6.0 × 10 = 60,000

78.

62.

(1.4 × 10 )( 5.2 × 10 ) = (1.4 )( 5.2 ) × 10

−3

−2

7.0 × 10 5 7.0 = × 108 = 1.75 × 108 = 175,000,000 4.0 × 10 −3 4.0 −3

64.

3.0 × 10 3.0 = × 10 −5 = 0.5 × 10 −5 = 5.0 × 10 −6 6.0 × 102 6.0 = 0.000005

( ) = 5 × 10 = 50,000

66.

25 × 108 = 52 × 10 4

65.

( )

8 × 1015 = 3 23 × 10 5 = 2 × 10

= 200,000

( 9.3 × 10 ) ( 6.1 × 10 ) ≈ 4.907 × 10 3

67. (a)

−4

( 2.414 × 10 ) ≈ 1.479 (b) (1.68 × 10 ) 5

0.11   750  1 +  365  

(b)

69. (a)

(b) 70. (a)

(b)

− 243 −3 1 = = − 9 9 3

79.

452 ≈ 12.651

80.

−273 = ( −27 )

800

≈ 954.448

81.

( 6.1)

82.

( 3.4 )

83.

84.

(1.2 ) 75 + 3 8 ≈ 14.499

85.

−5 + 33 ≈ 0.149 5

86.

−2.9

2.5

90 − ( 4.13 ) 17 ≈ − 281.088 −2

− 68 + 4 ≈ − 0.817 0.1

87.

3.14

+ 3 5 ≈ 2.709

88.

10 − π 2 ≈ −8.605 2.5

89.

( 2.8 ) + 1.01 × 106 ≈ 1,010,000.128 −2

( 20 ) = 20

92.

(2.65 × 10− 4 )

≈ 0.064

( − 3 x) 4 = 3 x 12 ⋅ 3 = 36 = 6

93.

9.9 × 106 ≈ 56.093

− 49 is not possible. Not a real number.

73. − 3 −64 = − 3 ( −4 ) = − ( −4 ) = 4

94.

≈ 9.390

72.

40 x5

= 3

45 =

95. (a)

75.

≈ 21.316

91.

121 = 11 = 11

≈ 0.005

4.5 × 109 ≈ 67,082.039

71.

≈ −7.225

90. 2.12 × 10 −2 + 15 ≈ 3.894

74.

67,000,000 + 93,000,000 0.0052 ≈ 30,769,230,769.2 ≈ 3.077 × 1010

(7.3 × 104 ) 4

1 1 = − 729 3

68. (a)

= 7.28 × 10 3 = 7280

63.

Exponents and Radicals

40 x5 = 3 8 x3 = 2 x 5x2

9⋅5 = 3 5 13

(b)

96. (a)

32 a 2  23 ⋅ 2 2 a 2  =  2 b2  b 

= 23

4a2 b2

125 = 5 − 625 is not possible. Not a real number.

76. − 7 −128 = −( − 2) = 2

(b)

54 = 3 33 ⋅ 2 = 3 3 2

( ) = 4 xy 2 x

32 x 3 y 4 = 2 4 ⋅ 2 ⋅ x ⋅ x 2 ⋅ y 2

Chapter P

97. (a)

Prerequisites

16 x5 = 3 8 ⋅ 2 ⋅ x3 ⋅ x 2 = 2 x3

So, 5 = 32 + 4 2 ⋅ 107.

98. (a)

3x 4 y 2 = x 4 3 y 2

(b)

160 x8 z 4 = 5 32 ⋅ 5 ⋅ x5 ⋅ x3 ⋅ z 4 = 2 x 5 5 x3 z 4

108.

99. (a) 2 50 + 12 8 = 2 25 ⋅ 2 + 12 4 ⋅ 2

(

)

(

= 2 5 2 + 12 2 2

32 + 4 2 = 9 + 16 = 25 = 5

106.

25 ⋅ 3 ⋅ x 2 5x 3 = y4 y2

75 x 2 y − 4 =

(b)

2x2

)

109.

= 10 2 + 24 2

1 3 8 3

= 3

= 34 2

(b) 10 32 − 6 18 = 10 16 ⋅ 2 − 6 9 ⋅ 2

(

) ( )

= 10 4 2 − 6 3 2

= 40 2 − 18 2

100. (a) 5 x − 3 x = 2 x

110.

(b) −2 9 y + 10 y = −2 3 ⋅ y + 10 y

3 5+ 6

= −6 y + 10 y = 4 y

= 7 ( 4 ) 5 x − 2 ( 5) 5 x = 28 5 x − 10 5 x

102. (a) 5 10 x − 90 x = 5 10 x − 3 ⋅ 10 ⋅ x

= 2 10 x 2 = 2 x 10

(

(b) 8 3 27 x − 12 3 64 x = 8 33 ⋅ x

1 2

= 24 x

− 2x

= 22 x

= 22 3 x

3 3 = 11 11

104.

5 + 3 ≈ 3.968 and 5+3 = 8 ≈ 2.828

105.

5 + 3 > 5 + 3.

32 + 2 2 = 9 + 4 = 13 ≈ 3.606

( 14 + 2 ) 10 14 + 2 2

5+ 6

5− 6

⋅

( 5 − 6)

5− 6

5−6

( 6 − 5)

3 3 3 1 ⋅ = = 3 3 3 3 3

113.

5+ 3 = 3 =

5+ 3 5− 3 . 3 5− 3 5−3

( 5 − 3)

(

103.

Thus,

14 − 4

112.

) − (4 ⋅ x) 13

( 14 + 2 )

14 + 2

12 2 3 3⋅ 3 3 = = 3= = 2 2 1⋅ 3 3

= 5 10 x 2 − 3 10 x 2

14 + 2

⋅

111.

= 18 5 x 2

83 4 = 43 4 2

(b) 7 80 x − 2 125 x = 7 16 ⋅ 5 x − 2 25 ⋅ 5 x

101. (a) 3 x + 1 + 10 x + 1 = 13 x + 1

14 − 2

= 22 2

3 3

= =

⋅3

14 − 2

3 3

⋅

114.

2 5− 3

)

7 −3 = 4

7 −3 7 +3 ⋅ 4 7 +3 7−9 −1 = = 4 7 +3 2 7 +3

(

) (

)

−1 2 7 +6

Thus, 5 > 32 + 2 2 .

Section P.2 4

115. (a)

(b)

32 = 32 4 = 31 2 = 3

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

= 3 ( x + 1) 6

116. (a)

(b)

x =x

(3x ) = 3x 4

117.

= ( x + 1)

128.

= x

(

− 144

1 Answer 32

120.

614.125 Given

( 614.125)

121.

−243 Answer

( −243)

122.

−216 Given

( −216 )

123.

813 Given

813 4 Answer

124.

16 5 Answer

16 5 4 Given

2 x

x 4 3 y2 3

( xy )

( )

23 2 x 2 12

2 x

1 132. − ( 125 )

Answer

5 −1 2 ⋅ 5 x 5 2 x = 5−1 x = , x > 0 53 2 x 3 2 5

1 = 323 5

( 32 ) 5

4 =  9 −1 3

−4 3

= ( −27 )

= − (125 )

35 = (351 2 )

133. 134.

23 2 x 3 21 2 x 4

(

2x = (2x )

(2)

1 8

= 3 −27 = 3 ( −3 ) = −3 3

) = − ( 5) = −625 4

= 351 4 = 4 35

) = (2x) = 2x 12

(

243 ( x + 1) = 243 ( x + 1) 

(

= 243 ( x + 1)

2 x

(

Answer

41 2 2 = 91 2 3

= − 1251 3

Given

135.

−1 2

 1  131.  −   27 

(1 32)1 5 Given

= 23 2 −1 2 x 3 − 4 = 21 x −1 =

126.

9 130.   4

) Given

(5x )

Answer

119.

(2x )

5−1 2 ⋅ 5 x 5 2

Rational Exponent Form

118. − 144 Answer

125.

x −3 ⋅ x1 2 x1 2 ⋅ x1 1 2 +1− 3 2 − 3 = 3 2 3 = x( ) ( ) −1 32 x ⋅x x ⋅x 1 = x −3 = 3 , x > 0 x

129. 32 − 3 5 =

64 Given

Radical Form 3

127.

Exponents and Radicals

)

= 4 243 ( x + 1) = 4 3 ⋅ 34 ( x + 1)

x 4 3 y2 3 43 − 13 23 − 13 = x ( ) ( ) y( ) ( ) x1 3 y1 3

= 3 4 3 ( x + 1)

= xy1 3 , x ≠ 0, y ≠ 0

136.

13 128a 7b = (128a 7b)   

= (128a 7b)

= 6 128a 7b = 6 64 ⋅ 2 ⋅ a 6 ⋅ a ⋅ b = 2a 6 2ab

Chapter P

137.

Brazil:

Prerequisites 142. False. For example, let a = 5 and b = 4.

2.19 × 1012 = 1.09 × 104 2.01 × 108

(a + b)2 = (5 + 4)2 = 92 = 81, whereas

Canada:

1.83 × 10 = 5.29 × 104 3.46 × 107

Germany:

3.59 × 1012 = 4.43 × 104 8.11 × 107

(5)2 + ( 4)2 = 25 + 16 = 41. 143. True.

1 1 + x y

India:

1.76 × 10 = 1.44 × 103 1.22 × 109

x −1 + y −1 =

Iran:

4.12 × 1011 = 5.16 × 103 7.99 × 107

y x + xy xy

Ireland:

2.21 × 1011 = 4.62 × 104 4.78 × 106

y + x xy

Mexico:

1.33 × 1012 = 1.12 × 104 1.19 × 108

x + y xy

0.274( 2.51 × 108 ) ≈ 6.8774 × 107 tons

Paper:

138.

0.089( 2.51 × 108 ) ≈ 2.2339 × 107 tons

Metals:

0.046( 2.51 × 108 ) ≈ 1.1546 × 107 tons

Glass:

0.127( 2.51 × 108 ) ≈ 3.1877 × 107 tons

Plastics:

Yard waste: 0.135( 2.51 × 10 ) ≈ 3.3885 × 10 tons 8

0.329(2.51 × 108 ) ≈ 8.2579 × 107 tons

Other: 139. For h = 7,

t = 0.03 12 5 2 − (12 − 7 )  = 0.03 125 2 − 55 2    ≈ 13.288 seconds. x k +1 x k +1 = 1 = xk. x x

( a ) = ( 2 ) = 8 = 64, whereas a( ) = 2( ) = 2 = 512. k

side of package B is about 2(6.3) = 12.6 inches, and 8 < 12.6. So, the length x of a side of package A is less than twice the length of a side of package B. 145. For a ≠ 0, 1 =

a a1 = = a 1 −1 = a 0 . a a1

146. Consider x 2 = n, x a positive integer.

141. False. For example, let a = 2, n = 3 and k = 2. Then n

about 6.3 inches (6.33 ≈ 250). Twice the length of a

Thus, a 0 = 1.

140. True. For x ≠ 0,

144. The length of a side of package A is about 8 inches (83 = 512), and the length of a side of package B is

Unit digit of x Unit digit of n = x 2 1 1 2 4 3 9 4 6 5 5 6 6 7 9 8 4 9 1 0 0 Therefore, the possible digits are 0, 1, 4, 5, 6, and 9 thus 5233 is not an integer because its unit digit is 3.

147. No. Rationalizing the denominator produces a number equivalent to the original fraction; squaring does not. 2

 5  25 5 3 5 3 ≠ ⋅ =   = 3 3 3 3  3

Section P.3

Polynomials and Factoring

Section P.3 Polynomials and Factoring 1. n, an

19. −8 + x 7 = x 7 − 8 Standard form Degree: 7 Leading coefficient: 1

2. monomial 3. First, Outer, Inner, Last 4. prime 5. A polynomial is completely factorial when each of its factors are prime. 6. Four guidelines for factoring polynomials are as follows: (1) Factor out any common factors using the Distributive Property. (2) Factor according to one of the special polynomial forms. (3) Factor as ax 2 + bx + c = ( mx + r )( nx + s ) .

(4) Factor by grouping. 7. 7 is a polynomial of degree zero. Matches (d).

20. 23 − x 3 = − x 3 + 23 Standard form Degree: 3 Leading coefficient: −1 21. 1 − x + 6 x 4 − 2 x 5 = −2 x 5 + 6 x 4 − x + 1 Standard form Degree: 5 Leading coefficient: −2 22. − x 6 + 5 − 4 x 5 + x 3 = − x 6 − 4 x 5 + x 3 + 5 Standard

form Degree: 6 Leading coefficient: − 1 23. This is a polynomial: − 8 y 2 + 2 y.

1 is not a polynomial. x2

8. −3 x 5 + 2 x 3 + x is a trinomial of degree 5. Matches (e).

24. 5 x 4 − 2 x 2 +

9. −4 x 3 + 1 is a binomial with leading coefficient −4. Matches (b).

25.

x 2 − x 4 is not a polynomial.

26.

x2 + 2 x − 3 1 2 1 1 = x + x − is a polynomial. 6 6 3 2

27.

(4 x + 1) + ( − x + 9) = 3 x + 10

28.

( t − 3) + (6t − 4t ) = 7t − 4t − 3

29.

(8 x + 5) − ( 6 x − 12 ) = 8 x + 5 − 6 x + 12 = 2 x + 17

30.

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

10. 6x is a monomial of positive degree. Matches (a). 11.

3 4

x 4 + x 2 + 14 is a trinomial with leading coefficient 34 .

Matches (f). 12.

x + 2 x − 4 x + 1 is a third-degree polynomial with leading coefficient 1. Matches (c).

13. −2 x 3 + 4 x is one possible answer.

14. 8 x 5 + 14 is one possible answer.

= − x 2 + 3x − 5

15. −15 x 4 + 2 x is one possible answer.

31.

( 2 x − 9 x − 20 ) + ( −2 x + 10 x ) = x − 20

16. 2 x 3 + 4 x + 2 is one possible answer.

32.

( y − 6 y + 3) + ( 5y − 2 y + y − 10 )

17. 3 x + 4 x 2 + 2 = 4 x 2 + 3 x + 2 Standard form Degree: 2 Leading coefficient: 4 18.

x 2 − 4 − 3 x 4 = −3 x 4 + x 2 − 4 Standard form Degree: 4 Leading coefficient: −3

= 6 y3 − 2 y2 − 5 y − 7 33.

(15x − 6 ) − ( −8.1x − 14.7 x − 17) 2

= 15 x 2 − 6 + 8.1x 3 + 14.7 x 2 + 17 = 8.1x 3 + 29.7 x 2 + 11

34. (13.6w4 − 14 w − 17.4) − (16.9 w4 − 9.2 w + 13) = 13.6 w4 − 14 w − 17.4 − 16.9 w4 + 9.2w − 13 = 3.3w4 − 4.8w − 30.4 35. 5 z( z − 8) = 5 z 2 − 40 z 36.

( 16 x + 1)(2 x ) = 13 x + 2 x 2

3   3  37.  5 − y  ( − 4 y ) = ( 5 )( − 4 y ) −  y  ( − 4 y ) 2   2  2 = 6 y − 20 y

Chapter P

(

Prerequisites

)

38. − 7 x 4 − x3 = − 28 x + 7 x 4

56.

( 3x + 2y) = ( 3x) + 3( 3x) ( 2y) + 3( 3x)( 2y) + ( 2y) 3

)

( )

39. 3 x x 2 − 2 x + 1 = 3 x x 2 + 3 x ( −2 x ) + 3 x (1) 3

(

( x − 3)( x + 3) = ( x ) − 3 =

58.

(1.5 y + 0.6)(1.5 y − 0.6) = 2.25 y 2 − 0.36

)

( ) − y ( 2 y ) − y ( −3)

40. − y 4 y + 2 y − 3 = − y 4 y

= − 4 y4 − 2 y3 + 3 y2 41.

( x + 3)( x + 4 ) = x 2 + 4 x + 3 x + 12 FOIL = x 2 + 7 x + 12

42.

1 4

−

61.

= 28 x 2 − 29 x + 6

− +

(3a − 3)

47.

( 2 − 5 x ) = 4 − 2 ( 2 )( 5 x ) + 25 x

= 9a − 18a + 9

3x3

− 15 x 2

− 3x 4

−

− 12 x 2 3

−

= 25 x 2 + 80 xy + 64 y 2

( x − 9 )( x + 9 ) = x 2 − 92 = x 2 − 81

50.

( 5 x + 6 )( 5 x − 6 ) = ( 5 x ) − 62 = 25 x 2 − 36

51.

( x + 2 y )( x − 2 y ) = x 2 − ( 2 y ) = x 2 − 4 y2

52.

( 2r − 5)( 2r + 5) = ( 2r ) − 5 = 4r − 25

53.

( x + 1) = x 3 + 3 x 2 (1) + 3 x (12 ) + 13

+ −

3x + x +

2 4

+ 12 x + − 2x

−

4 x2 3x2

2x4

x + 6 x3

4 x2

2x4

+ 5x3

5x2

( y − 4 ) = y 3 − 3 y 2 ( 4 ) + 3 y ( 4 ) − 43 3

= y 3 − 12 y 2 + 48 y − 64 55.

( 2 x − y ) = ( 2 x ) − 3 ( 2 x ) y + 3 ( 2 x ) y 2 − y3 3

= 8 x 3 − 12 x 2 y + 6 xy 2 − y3

+ 10 x + 8

63. ( x + z ) + 5 ( x + z ) − 5 = ( x + z ) − 52 = x 2 + 2 xz + z 2 − 25 2

= ( x − 3y) − z2 = x2 − 2 x (3 y ) + (3 y ) − z 2 2

= x 2 − 6 xy + 9 y 2 − z 2 65. ( x − 3 ) + y  = ( x − 3 ) + 2 y ( x − 3 ) + y 2 = x 2 − 6 x + 9 + 2 xy − 6 y + y 2 2

= x 2 + 2 xy + y 2 − 6 x − 6 y + 9

= x3 + 3x2 + 3x + 1

54.

− 19 x − 5

2 x2

+ x − 5 − 20 x

x − 5 4x + 1

64. ( x − 3 y ) + z  ( x − 3 y ) − z 

49.

+ +

( 5 x + 8 y ) = 25 x 2 + 2 ( 5 x )(8 y ) + 64 y 2

Answer: −3 x − x − 12 x − 19 x − 5

= 4 − 20 x + 25 x 2 = 25 x 2 − 20 x + 4 48.

x2 4 x2

62.

− 3x 4

46.

3x − 4x

( 7 x − 2 )( 4 x − 3) = 28 x 2 − 21x − 8 x + 6 FOIL

x2 − 9

= 3.24 y 2 − 18 y + 25

= 6 x − 7x − 5

(4 y + 7)2 = 16 y 2 + 56 y + 49

1 16

(1.8 y − 5) = (1.8 y ) + 2 (1.8 y )( −5) + ( −5)

( x − 5)( x + 10 ) = x 2 + 10 x − 5 x − 50 FOIL

45.

44.

25 2 5  x + 15 x + 9 59.  x + 3 = 4 2  60.

( 3 x − 5)( 2 x + 1) = 6 x 2 + 3 x − 10 x − 5 FOIL

1 4

= x 2 + 5 x − 50 43.

57.

= 3x − 6 x + 3x 2

= 27x3 + 54x2 y + 36xy2 + 8y3

= 7 x 4 − 28 x

(

66. ( x + 1) − y  = ( x + 1) + 2 ( x + 1)( − y ) + ( − y ) = x 2 + 2 x + 1 − 2 xy − 2 y + y 2 2

= x 2 − 2 xy + y 2 + 2 x − 2 y + 1 67. 5 x − 40 = 5 ( x − 8 ) 68. 4 y + 20 = 4 ( y + 5 )

Section P.3

(

69. 2 x 3 − 6 x = 2 x x 2 − 3

)

Polynomials and Factoring

71. 3 x ( x − 5 ) + 8 ( x − 5 ) = ( 3 x + 8 )( x − 5 ) 72. 5( x + 1) − x( x + 1) = ( x + 1)(5 − x)

70. 3z 4 − 6 z 2 + 9 z = 3z ( z 3 − 2 z + 3)

= −( x + 1)( x − 5) 73.

( 5 x − 4 ) + ( 5 x − 4 ) = ( 5 x − 4 ) ( 5 x − 4 ) + 1 = ( 5 x − 4 )( 5 x − 3 ) 2

( ) = ( x + 3)( − 3x 2 − 7 x)

74. 2 x( x + 3) − 3x( x + 3) = ( x + 3) 2 x − 3 x( x + 3) = ( x + 3) 2 x − 3 x 2 − 9 x 2

= − x( x + 3)(3x + 7) 75.

x 2 − 36 = ( x + 6)( x − 6)

76.

x − 81 = ( x + 9 )( x − 9 )

87. 4 x 2 − 12 x + 9 = ( 2 x ) − 2 ( 2 x )( 3 ) + 32 2

= ( 2 x − 3)

(

)

77. 48 y 2 − 27 = 3 16 y 2 − 9 = 3 ( 4 y ) − 32 2

= 3 ( 4 y + 3 )( 4 y − 3 )

(

78. 50 − 98 z 2 = 2 25 − 49 z 2

(

= 2 52 − ( 7 z )

88. 25z 2 − 10 z + 1 = ( 5z ) − 2 ( 5z )(1) + 1 2

)

= ( 5z − 1)

89. 4 x 2 − 43 x + 19 = ( 2 x ) − 2 ( 2 x ) ( 13 ) + ( 13 ) 2

)

= ( 2 x − 13 )

)

= 2 ( 5 + 7 z )( 5 − 7 z )

90.

= −2 ( 7z + 5 )( 7 z − 5 )

9 y2 −

2 3 1 1 1 y+ = (3y ) − 2 (3y )   +   2 16 4 4

79. 4 x − 19 = ( 2 x ) − ( 13 ) = ( 2 x + 13 )( 2 x − 13 )

1  =  3y −  4 

y − 49 = ( y ) − 7 = ( y + 7 )( y − 7 )

(12 y − 1) =

80. 81.

25 36

5 6

( x − 1) − 4 = ( x − 1) + 2  ( x − 1) − 2  = ( x + 1)( x − 3 ) 2

82. 25 − ( z + 5 ) = 52 − ( z + 5 ) 2

(

91. x3 − 8 = ( x) − ( 2) 3

= ( x − 2)( x 2 + 2 x + 4)

)(

= 5 − ( z + 5) 5 + ( z + 5) = ( 5 − z − 5 )( 5 + z + 5 )

92.

)

84.

x 2 + 10 x + 25 = x 2 + 2 ( 5 )( x ) + 52 = ( x + 5 )

85.

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

1 2

)

86.

x − x + = x − 2( ) x + ( ) = ( x −

2 3

4 3

4 9

1 2

2 3

= ( z + 1)( z 2 − z + 1)

x 2 − 4 x + 4 = x 2 − 2 ( 2 ) x + 22 = ( x − 2 )

93. z 3 + 1 = ( z ) + (1)

83.

1 4

y 3 − 125 = ( y ) − (5)

= ( y − 5)( y 2 + 5 y + 25)

= − z ( z + 10 )

94. x3 + 64 = ( x) + ( 4) 3

= ( x + 4)( x 2 − 4 x + 16)

)

95.

x3 +

1 3 1 = ( x) +   27  3

1  1 1  =  x +  x 2 − x +  3 3 9  

Chapter P

96. w3 −

Prerequisites

27 1 3 1 = w3 − = ( w) −   216 8  2

97. 125v3 − 1 = (5v) − (1) 3

= (5v − 1)( 25v 2 + 5v + 1)

1  1 1  =  w −  w2 + x +  2  2 4 

98. 343a 3 + 8 = (7 a ) + ( 2) 3

= (7 a + 2)( 49a 2 − 14a + 4) 99.

( y + 1) − x3 = ( y + 1) − ( x) 3

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

= ( y − x + 1)( y 2 + 2 y + 1 + xy + x + x 2 )

= ( − x + y + 1)( x 2 + y 2 + xy + x + 2 y + 1) 100. ( 2 x − z ) + 125 y 3 = ( 2 x − z ) + (5 y ) 3

2 = ( 2 x − z ) + 5 y ( 2 x − z ) − 5 y ( 2 x − z ) + 25 y 2   

= ( 2 x + 5 y − z )( 4 x 2 − 4 xz + z 2 − 10 xy + 5 yz + 25 y 2 ) = ( 2 x + 5 y − z )( 4 x 2 + 25 y 2 + z 2 − 10 xy − 4 xz + 5 yz ) 101. x 2 + x − 2 = ( x + 2 )( x − 1)

113.

1 8

102. x 2 + 6 x + 8 = ( x + 4 )( x + 2 ) 103. s − 5s + 6 = ( s − 3 )( s − 2 )

(

x 2 − 961 x − 161 = 18 x 2 − 121 x − 12

)

( x − )( x + )

1 8

1 96

3 4

2 3

( 4 x − 3)( 3 x + 2 )

114.

104. t 2 − t − 6 = t 2 + 2t − 3t − 6 = ( t + 2 )( t − 3 ) 105. 20 − y − y 2 = ( 5 + y )( 4 − y )

or − ( y + 5 )( y − 4 )

or ( 19 x − 2 )( 19 x + 4 )

(

= ( x + 5) x 2 − 5

(

118. x 3 − x 2 + 3 x − 3 = x 2 ( x − 1) + 3 ( x − 1)

110. 8 x 2 − 45 x − 18 = ( x − 6 )( 8 x + 3 )

(

)

119. x 2 + x − 20 = x 2 + 5 x − 4 x − 20

= x( x + 5) − 4( x + 5)

or ( 2 − 5u )( u + 3 )

(

112. −6 x + 23 x + 4 = − 6 x − 23 x − 4

)

= x 2 + 3 ( x − 1)

= − ( 5u − 2 )( u + 3 )

)

= x 2 + 1 ( x − 5)

109. 5 x 2 + 26 x + 5 = ( 5 x + 1)( x + 5 )

111. −5u − 13u + 6 = − 5u + 13u − 6

)

117. x 3 − 5 x 2 + x − 5 = x 2 ( x − 5 ) + ( x − 5 )

108. 2 x 2 − x − 21 = ( 2 x − 7 )( x + 3 )

)

116. x 3 + 5 x 2 − 5 x − 25 = x 2 ( x + 5 ) − 5 ( x + 5 )

107. 3 x 2 + 13 x − 10 = ( 3 x − 2 )( x + 5 )

)

= − ( x − 4 )( 6 x + 1) or ( 4 − x )( 6 x + 1)

= 811 ( x − 18 )( x + 36 )

= ( x − 1) x 2 + 2

= ( 8 − z )( 3 + z )

x 2 + 29 x − 8 = 811  x 2 + 18 x − 648 

115. x 3 − x 2 + 2 x − 2 = x 2 ( x − 1) + 2 ( x − 1)

106. 24 + 5z − z 2 = 24 + 8z − 3z − z 2

1 81

= ( x + 5)( x − 4) 120. b 2 − 11b + 18 = b 2 − 9b − 2b + 18

= b(b − 9) − 2(b − 9) = (b − 9)(b − 2)

Section P.3 121. 6 x 2 + x − 2

a = 6, c = −2, ac = −12 = 4 ( −3) , and

Polynomials and Factoring

136. 5 − x + 5 x 2 − x 3 = 1( 5 − x ) + x 2 ( 5 − x )

(

= (5 − x ) 1 + x2

4 − 3 = 1 = b. Thus, 6 x 2 + x − 2 = 6 x 2 + 4 x − 3 x − 2

= −u 2 ( u + 2 ) + 3 ( u + 2 )

(

= ( 2 x − 1)( 3 x + 2 ) .

6 + 4 = 10 = b. Thus, 3 x 2 + 10 x + 8 = 3 x 2 + 4 x + 6 x + 8

(

138. x 4 − 4 x 3 + x 2 − 4 x = x 3 ( x − 4 ) + x ( x − 4 )

(

)

(

)

(

)

= x 2 − 4 ( 2 x + 1) = ( x + 2 )( x − 2 )( 2 x + 1) 140. 3 x 3 + x 2 − 27 x − 9 = x 2 ( 3 x + 1) − 9 ( 3 x + 1)

(

)

= x 2 − 9 ( 3 x + 1) = ( x + 3 )( x − 3 )( 3 x + 1)

126. x 3 − 9 x 2 = x 2 ( x − 9 ) 127. x − 2 x + 1 = ( x − 1)

139. 2 x 3 + x 2 − 8 x − 4 = x 2 ( 2 x + 1) − 4 ( 2 x + 1)

125. y3 − y = y y 2 − 1 = y ( y + 1)( y − 1)

(

= x x + 1 ( x − 4)

)

(

)

= x3 + x ( x − 4 )

123. 10 x 2 − 40 = 10 x 2 − 4 = 10 ( x + 2 )( x − 2 ) 124. 7z 2 − 63 = 7 z 2 − 9 = 7 ( z + 3)( z − 3)

)

= 3 − u2 ( u + 2 )

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

)

137. 3u − 2u 2 + 6 − u3 = −u3 − 2u2 + 3u + 6

= 2 x (3x + 2 ) − ( 3x + 2 )

122. a = 3, c = 8, ac = 24 = 6 ( 4 ) , and

141.

( x + 1) − 4 x = ( x + 1) + 2 x  ( x + 1) − 2 x  = ( x + 2 x + 1)( x − 2 x + 1) 2

128. 9 x 2 − 6 x + 1 = ( 3 x − 1)

= ( x + 1) ( x − 1) 2

129. 1 − 4 x + 4 x 2 = (1 − 2 x ) = ( 2 x − 1) 2

(

130. 16 − 6 x − x 2 = − x 2 + 6 x − 16

142.

)

= ( x + 8 )( 2 − x )

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

131. 2 x 2 + 6 x − 2 x 3 = − 2 x 3 + 2 x 2 + 6 x = − 2 x( x 2 − x − 3)

(

( x + 8) − 36 x = ( x + 8) − ( 6 x ) = ( x + 8 ) − 6 x  ( x + 8 ) + 6 x     = ( x − 6 x + 8 )( x + 6 x + 8 ) 2

= − ( x + 8 )( x − 2 )

132. 7 y 2 + 15 y − 2 y 3 = − y 2 y 2 − 7 y − 15

)

= − y ( 2 y + 3 )( y − 5 )

(

)

(

143. 3t 3 + 24 = 3 t 3 + 8 = 3 ( t + 2 ) t 2 − 2t + 4

(

)

(

)

144. 4 x 3 − 32 = 4 x 3 − 8 = 4 ( x − 2 ) x 2 + 2 x + 4

(

)

145. 4 x ( 2 x − 1) + 2 ( 2 x − 1) = 2 ( 2 x − 1) 2 x + ( 2 x − 1) 2

= 2 ( 2 x − 1)( 4x − 1)

133. 9 x 2 + 10 x + 1 = ( 9 x + 1)( x + 1)

)

146. 5 ( 3 − 4 x ) − 8 ( 3 − 4 x )( 5 x − 1) 2

134. 13 x + 6 + 5 x 2 = 5 x 2 + 13 x + 6

= ( 3 − 4 x ) 5 ( 3 − 4 x ) − 8 ( 5 x − 1) 

= 5 x 2 + 10 x + 3 x + 6

= ( 3 − 4 x ) 15 − 20 x − 40 x + 8 

= ( 5 x + 3 )( x + 2 ) 135. 3 x 3 + x 2 + 15 x + 5 = x 2 ( 3 x + 1) + 5 ( 3 x + 1)

(

= ( 3 x + 1) x 2 + 5

= ( 3 − 4 x )( 23 − 60 x )

)

Chapter P

Prerequisites

147. 2 ( x + 1)( x − 3 ) − 3 ( x + 1) ( x − 3 ) 2

148. 7 ( 3 x + 2 ) (1 − x ) + ( 3 x + 2 )(1 − x ) 2

= ( x + 1)( x − 3 ) 2 ( x − 3 ) − 3 ( x + 1) 

= ( 3 x + 2 )(1 − x ) ( 21x + 14 + 1 − x ) 2

= ( x + 1)( x − 3 )( − x − 9 )

= ( 3 x + 2 )(1 − x ) ( 20 x + 15 ) 2

= − ( x + 1)( x − 3 )( x + 9 )

)

= ( 3 x + 2 )(1 − x ) 7 ( 3 x + 2 ) + (1 − x ) 

= ( x + 1)( x − 3 ) 2 x − 6 − 3 x − 3

(

= 5 ( 3 x + 2 )(1 − x ) ( 4 x + 3 ) 2

(

149. (a) 1000 1 + r 2 = 1000 1 + 2r + r 2

)

= 1000r 2 + 2000r + 1000

(b)

1 12 %

2 12 %

1020.10

1030.23

1040.40

1050.63

1060.90

1000 (1 + r )

155. x 2 + 3 x + 2 = ( x + 2 )( x + 1)

= 2 (13 − x )( 2 )( 9 − x )( x )

= 4 x ( −1)( x − 13 )( −1)( x − 9 )

= 4 x ( x − 13 )( x − 9 )

When x = 1: V = 4 (1)( −12 )( −8 ) = 384 cubic inches. When x = 2: V = 4 ( 2 )( −11)( −7 ) = 616 cubic inches. When x = 3: V = 4 ( 3 )( −10 )( −6 ) = 720 cubic inches.

(

151. (a) T = R + B = 1.1x + 0.0475 x 2 − 0.001x + 0.23

)

= 0.0475 x 2 + 1.099 x + 0.23

(b)

T feet

75.95

120.19

204.36

(c) As the speed x increases, the total stopping distance increases. 152. (a) Estimates will vary. Actual safe loads for x = 12:

)

S 6 = 0.06 (12 ) − 2.42 (12 ) + 38.71 2

= 335.2561( using a calculator )

(

S8 = 0.08 (12 ) − 3.30 (12 ) + 51.93 2

= 568.8225 ( using a calculator )

)

1 x

x 1

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

x mi hr

(

x 1

1 x

x 1

157. 3 x 2 + 7 x + 2 = ( 3 x + 1)( x + 2 ) x

x 1

Difference in safe loads = 568.8225 − 335.2561 = 233.6 pounds (b) The difference in safe loads decreases in magnitude as the span increases.

x 1 x

1 x

x 1 1

153. a 2 − b 2 = ( a + b )( a − b )

Matches model (a). 154. ab + a + b + 1 = ( a + 1)( b + 1)

Matches model (b).

Section P.3 158. 2 x 2 + 7 x + 3 = ( 2 x + 1)( x + 3 ) x

1 x

(

159. A = π ( r + 2 ) − π r 2 = π ( r + 2 ) − r 2    2 2 = π r + 4r + 4 − r  = π ( 4r + 4 ) = 4π ( r + 1) 2

5 8

= ( 2 x − 5) ( 5x − 4) 15( 2 x − 5) + 8 ( 5x − 4)  3

= ( 2 x − 5) ( 5x − 4) 30 x − 75 + 40 x − 32 3

= ( 2 x − 5) ( 5x − 4) ( 70 x − 107) 3

( ) = ( x + 6x + 9 − ) = ( x + 6 x + 9 − 16 ) = ( x + 6 x − 7) 5 8

= 85 x 2 + 6 x + 9 − 10 2

163. ( 2 x − 5) ( 3)( 5x − 4) ( 5) + ( 5x − 4) ( 4)( 2 x − 5) ( 2)

160. Area = 12 ( x + 3 ) ( 54 ) ( x + 3 ) − 12 ( 5 )( 4 ) 5 8

)

) ( 2 x ) + ( x + 1) ( 3x ) = 3 x ( x + 1) 2 x + ( x + 1)    = 3 x ( x + 1) ( 3 x + 1)

162. x 3 ( 3 ) x 2 + 1

(

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

1 1

( )

= 4 x 3 ( 2 x + 1) 2 x 2 + ( 2 x + 1)  3

1 x

x 1

161. x 4 ( 4 )( 2 x + 1) ( 2 x ) + ( 2 x + 1) 4 x 3

Polynomials and Factoring

164.

80 5

( x − 5) ( 2)( 4x + 3)( 4) + ( 4x + 3) (3) ( x − 5) ( x ) = ( x − 5) ( 4x + 3) 8( x − 5) + 3x ( 4x + 3)    = ( x − 5) ( 4x + 3) (12x + 17x − 40) 3

= 85 ( x + 7 )( x − 1)

165.

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

166.

3(5 x − 1) − (3 x + 1)(3)(5 x − 1) (5)

( 2 x + 3)

(2 x + 3)4 4( 2 x + 3)[− 2 x + 4] = 4 ( 2 x + 3) − 8( 2 x + 3)( x − 2) = 4 ( 2 x + 3) − 8( x − 2) = ( 2 x + 3)3 3(5 x − 1) (5 x − 1) − (5)(3 x + 1) 2

(5 x − 1)6

4( 2 x + 3) ( 2 x + 3) − ( 4 x − 1)

(5 x − 1)6

3(5 x − 1) [5 x − 1 − 15 x − 5] 2

(5 x − 1)6

3(5 x − 1) ( −10 x − 6) 2

(5 x − 1)6

− 6(5 x − 1) (5 x + 3) 2

(5 x − 1) 6(5 x + 3) = − (5 x − 1)4

Chapter P

Prerequisites

167. For x 2 + bx − 15 = ( x + m )( x + n ) to be

factorable, b must equal m + n where mn = −15. Factors of − 15

Sum of factors

(15)( −1)

15 + ( −1) = 14

( −15)(1)

−15 + 1 = −14

( 3)( −5)

3 + ( −5 ) = −2

( −3)( 5)

−3 + 5 = 2

The possible b-value are 14, − 14, − 2, or 2. 168. For x 2 + bx − 12 to be factorable, b must equal m + n where mn = −12.

Factors of − 12

Sum of factors

(1)( −12 )

1 − 12 = −11

( −1)(12 )

−1 + 12 = 11

( 2 )( −6 )

169. For x 2 + bx + 50 = ( x + m )( x + n ) to be

factorable, b must equal m + n where mn = 50. Factors of 50

Sum of factors

( 50 )(1)

( −50 )( −1)

−51

( 25)( 2 )

( −25 )( −2 )

−27

(10 )( 5 )

( −10 )( −5 )

−15

The possible b-values are 51, − 51, 27, − 27, 15, or − 15. 170. For x 2 + bx + 24 to be factorable, b must be equal to m + n where mn = 24.

Factors of 24

Sum of factors

2 − 6 = −4

( 24 )(1)

( −2 )( 6 )

−2 + 6 = 4

( −24 )( −1)

−25

( 3)( −4 )

3 − 4 = −1

(12 )( 2 )

( −3)( 4 )

−3 + 4 = 1

( −12 )( −2 )

−14

(8 )( 3)

( −8 )( −3 )

−11

( 6 )( 4 )

( −6 )( −4 )

−10

The possible b-values are 11, − 11, 4, − 4, 1 or − 1.

The possible b-values are 25, − 25, 14, − 14, 11, − 11, 10 or − 10.

Section P.3

Polynomials and Factoring

171. For x2 + x + c to be factorable, the factors of c must add up to 1.

Possible c-values

Factors of c that add up to 1

−2

(2)(−1) = − 2 and 2 + (−1) = 1

−6

(3)( − 2) = − 6 and 3 + (− 2) = 1

− 20

(5)(− 4) = − 20 and 5 + (− 4) = 1

These are a few possible c-values. There are many correct answers. If c = − 2: x 2 + x − 2 = ( x + 2)( x − 1) If c = − 6: x 2 + x − 6 = ( x + 3)( x − 2) If c = − 20: x 2 + x − 20 = ( x + 5)( x − 4) 172. For x 2 − 9 x + c to be factorable, the factors of c must add up to − 9.

Possible c-values

Factors of c that add up to − 9

( −1)(−8) = 8 and −1 + (− 8) = − 9

(− 2)(− 7 ) = 14 and − 2 + (− 7) = − 9

( − 3)(− 6) = 18 and − 3 + ( − 6) = − 9

These are a few possible c-values. There are many correct answers. If c = 8: x 2 − 9 x + 8 = ( x − 1)( x − 8) If c = 14: x 2 − 9 x + 14 = ( x − 2)( x − 7) If c = 18: x 2 − 9 x + 18 = ( x − 3)( x − 6) 173. For 2 x 2 + 5 x + c to be factorable, the factors of 2c must add up to 5.

Possible c-values

Factors of 2c that add up to 5

(1)( 4 ) = 4 and 1 + 4 = 5

( 2 )( 3) = 6 and 2 + 3 = 5

−3

−6

( 6 )( −1) = −6 and 6 + ( −1) = 5

−7

−14

( 7 )( −2 ) = −14 and 7 + ( −2 ) = 5

−12

−24

(8 )( −3) = −24 and 8 + ( −3) = 5

These are a few possible c-values. There are many correct answers. If c = 2 : 2x 2 + 5 x + 2 = ( 2 x + 1)( x + 2 ) If c = 3 : 2 x 2 + 5 x + 3 = ( 2 x + 3 )( x + 1) If c = −3 : 2 x 2 + 5 x − 3 = ( 2 x − 1)( x + 3 ) If c = −7 : 2 x 2 + 5 x − 7 = ( 2 x + 7 )( x − 1) If c = −12 : 2 x 2 + 5 x − 12 = ( 2 x − 3 )( x + 4 )

Chapter P

Prerequisites

174. For 3 x 2 − 10 x + c to be factorable, of 3c must add up to −10.

Possible c-values

Factors of 3c that must add up to − 10

( −9 )( −1) = 9 and − 9 − 1 = −10

( −3)( −7 ) = 21 and − 3 − 7 = −10

( −4 )( −6 ) = 24 and − 4 − 6 = −10

−8

−24

( 2 )( −12 ) = −24 and − 12 + 2 = −10

Other c-values are possible. The above values yield the following factorizations. There are many correct answers. If c = 3 : 3 x 2 − 10 x + 3 = ( 3 x − 1)( x − 3 ) If c = 7 : 3 x 2 − 10 x + 7 = ( 3 x − 7 )( x − 1) If c = 8 : 3 x 2 − 10 x + 8 = ( 3 x − 4 )( x − 2 ) If c = −8 : 3 x 2 − 10 x − 8 = ( 3 x + 2 )( x − 4 ) 175. V = π R 2 h − π r 2 h

(a) V = π h  R − r  = π h ( R − r )( R + r ) 2

R+r . The thickness 2 of the shell is R − r. Therefore,  R+r  V = π h ( R + r )( R − r ) = 2π  ( R − r )h  2 

(b) The average radius is

= 2π (average radius)(thickness)h.

182. (a) The box could have been created by cutting squares of length x from the corners of the piece of cardboard. The original dimensions of the cardboard are 52 inches × 42 inches.

(b) The degree is 3 because the volume is length × width × height, and each dimension contains an x. (c) x(52 − 2 x )( 42 − 2 x ) = 4 x( 26 − x )( 21 − x ) The possible values of x are 0 < x < 21.

176. kQx − kx 2 = kx ( Q − x )

183. If two polynomials have degree m and n, then their product is degree m + n.

177. False. The product of the two binomials is not always a second-degree polynomial. For instance,

184. ( x + y ) ≠ x 2 + y 2 because you cannot just distribute the

( x + 2 )( x − 3) = x − x − 6 is a fourth-degree 2

polynomial. 178. False. The product of the two binomials is not always a trinomial. For example, ( x + 2 )( x − 2 ) = x 2 − 4. 179. False. For example, ( x 2 − 3x + 1) + (− x 2 + x − 2) = − 2 x − 1, which is a

first-degree polynomial. 180. False. The sum of a third-degree polynomial and a fourth-degree polynomial will always be a fourth-degree polynomial. 181. False. (3 x − 6)( x + 1) = 3( x − 2)( x + 1)

squares. You have to use the FOIL Method.

( x + y ) = x 2 + 2 xy + y2 ≠ x 2 + y2 2

185. To cube a binomial difference, cube the first term. Next, subtract 3 times the square of the first term times the second term. Next, add 3 times the first term times the square of the second term. Finally, subtract the cube of the second term.

( x − y ) = x 3 − 3 x 2 y + 3 xy2 − y3 3

186. A polynomial is in factored form when each of its factors is prime (it cannot be factored any further using integer coefficients).

(

)

187. 9 x 2 − 9 x − 54 = 9 x 2 − x − 6 = 9 ( x + 2 )( x − 3)

The error in the problem in the book was that 3 was factored out of the first binomial but not out of the second binomial. ( 3x + 6)( 3x − 9) = 3( x + 2)( 3)( x − 3) = 9( x + 2)( x − 3)

Section P.4 188. Answers will vary. Sample answer: x 2 − 3

Rational Expressions

190. x 3n + y 3n = ( x n + y n )( x 2 n − x n y n + y 2 n )

189. x 2 n − y 2 n = ( x n + y n )( x n − y n )

Section P.4 Rational Expressions 1. domain

15. The domain of the expression x2 − 2x − 3 ( x − 3)( x + 1) = is the set of all real 2 9x − 1 (3x + 1)(3x − 1)

2. rational expression 3. complex fractions 4. lesser 5. A rational expression is in simplest form when its numerator and denominator have no common factors aside from ±1. 6. Values that make the denominator equal to zero are excluded from the domain of a rational expression. 2

7. The domain of the polynomial x + 7 x − 3 is the set of all real numbers. 8. The domain of the polynomial 6 x 2 − x − 10 is the set of all real numbers. 9. The domain of the polynomial 5 x 2 + 1, x > 0, is the set of all positive real numbers. 10. The domain of the polynomial 9 x − 4, x ≤ 0, is the set of all negative real numbers. x is the set of all real x + 2 numbers except x = − 2, which would result in division

11. The domain of the expression

by zero, which is undefined. 1− x is the set of all real 4 − x numbers except x = 4, which would result in division by zero, which is undefined.

12. The domain of the expression

13. The domain of the expression x( x + 3) x 2 + 3x is the set of all real = 2 2 x + 14 x + 49 ( x + 7) numbers except x = − 7, which would result in division by zero, which is undefined. 14. The domain of the expression ( x + 4)( x − 2) = x + 4 , x ≠ 2, is the x2 + 2x − 8 = x2 − 4 ( x + 2)( x − 2) x + 2

1 numbers except x = ± , which would result in division 3 by zero, which is undefined.

16. The domain of the expression

( x + 3) is the set of all real x2 + 6 x + 9 = 2 2 x − 10 x + 25 ( x − 5) 2

numbers except x = 5, which would result in division by zero, which is undefined. 17. The domain of the radical expression x + 10 is the set of all real numbers greater than or equal to −10, because the square root of a negative number is not a real number. 18. The domain of the radical expression x − 7 is the set of all real numbers greater than or equal to 7 because the square root of a negative number is not a real number. 19. Because 12 − 3 x ≥ 0  x ≤ 4, the domain of the radical expression 12 − 3x is the set of all real numbers less than or equal to 4. 20. Because 6 − 4 x ≥ 0  x ≤

3 , the domain of the 2

6 − 4x is the set of all real 3 numbers less than or equal to . 2 radical expression

21. Because x + 1 > 0  x > −1, the domain of the radical expression

1 is the set of all real numbers x +1

greater than −1. 22. Because x − 5 > 0  x > 5, the domain of the radical 1 expression is the set of all real numbers greater x −5 than 5.

set of all real numbers except x = ± 2, which would result in division by zero, which is undefined.

Chapter P

Prerequisites

23.

5(3x ) 5 (3x ) 5 = = , x≠0 2 x ( 2 x )( 3 x ) 6 x2

35.

x−5 x−5 1 = =− , x≠5 10 − 2 x −2 ( x − 5 ) 2

36.

12 − 4 x −4 ( x − 3 ) = = −4, x ≠ 3 x −3 x −3

37.

y2 − 16 ( y + 4 )( y − 4 ) = = y − 4, y ≠ −4 y+4 y+4

38.

x 2 − 25 ( x + 5 )( x − 5 ) = = − ( x + 5) , x ≠ 5 5− x −1 ( x − 5 )

39.

x 3 + 5 x 2 + 6 x x ( x + 2 )( x + 3 ) = x2 − 4 ( x + 2 )( x − 2 )

Missing factor: 3x

( ) ( )( )

( )

24.

2 x 2 x 2 = = , x≠0 2 2 2 3x 3x 4 3x x Missing factor: x 2

25.

3 3 ( x + 1) = , x ≠1 4 4 ( x + 1)

Missing factor: ( x + 1) 26.

2 2 ( x − 3) = , x≠3 5 5 ( x − 3)

Missing factor: x − 3 27.

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

28.

4 ( x − 1)( x + 2 ) 4 ( x + 2)

40.

x ( x + 3) x−2

, x ≠ −2

x 2 + 8 x − 20 ( x + 10 )( x − 2 ) = x 2 + 11x + 10 ( x + 10 )( x + 1) =

, x ≠ −2

x−2 , x ≠ −10 x +1

Missing factor: x + 2

41.

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

y 2 − 7 y + 12 ( y − 3 )( y − 4 ) y − 4 = = , y≠3 y 2 + 3 y − 18 ( y + 6 )( y − 3 ) y + 6

42.

− ( x + 10 ) −10 − x = x 2 + 11x + 10 ( x + 10 )( x + 1)

( x + 3)( x − 1) , x ≠ 1 2 2 ( x − 1)

=−

Missing factor: x − 1

1 , x ≠ −10 x +1

2 − x + 2 x2 − x3 ( 2 − x ) + x ( 2 − x ) = x−2 − (2 − x ) 2

5x (3x ) 3x 15 x = = , x≠0 10 x 5x (2) 2

43.

29.

30.

6 y2 ( 3) 18 y 2 3 = = ,y≠0 60 y 5 6 y 2 10 y 3 10 y3

31.

3 xy 3 xy 3y = 2 = 2 x y + x x ( y + 1) x( y + 1)

(

44.

x2 − 9 x2 − 9 = 2 2 x + x − 9x − 9 x − 9 ( x + 1)

(

33.

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

−4 y ( 2 y − 1) 5 ( 2 y − 1)

=−

4y 1 , y≠ 5 2

9 x 2 + 9 x 9 x ( x + 1) 9 x = = , x ≠ −1 2x + 2 2 ( x + 1) 2

)

1 = , x ≠ ±3 x +1 45.

)

= − 1 + x2 , x ≠ 2

( )

y 2 x2 2 x2 y 2 x2 = = , y≠0 xy − y y ( x − 1) x − 1

( 2 − x ) (1 + x 2 ) −(2 − x)

(

)

32.

34.

46.

(

)

( z − 2 ) z + 2z + 4 z3 − 8 = = z−2 2 z + 2z + 4 z2 + 2z + 4 2

y ( y − 3 )( y + 1) y3 − 2 y2 − 3 y = y3 + 1 y ( + 1) y 2 − y + 1

(

y ( y − 3) y2 − y + 1

)

, y ≠ −1

Section P.4 47.

x2 + 2 x − 3 x −1 x+3

−4

−3

−2

−1

Undef.

−1

The expressions are equivalent except at x = 1. In fact, 48.

x −3 x2 − x − 6

1 2

1 3

1 4

Undef.

1 6

1 7

1 8

1 x+2

1 2

1 3

1 4

1 5

1 6

1 7

1 8

Rational Expressions

( x + 3)( x − 1) = x + 3, x ≠ 1. x2 + 2x − 3 = x −1 x −1 55.

56.

3( x + y ) 4

x + y 3( x + y ) 2 3 = ⋅ = , x ≠ −y 2 4 x+y 2

2x − y 3y − 6x 2x − y y2 − 6 y + 5 ÷ 2 = ⋅ y −1 y − 6y + 5 y −1 − 6x + 3y =

The expressions are equivalent except at x = 3. In fact,

x −3 x −3 1 = = , x ≠ 3. x − x − 6 ( x − 3)( x + 2 ) x + 2

2 x − y ( y − 5)( y − 1) ⋅ − 3( 2 x − y ) y −1

= −

π r2

πr2

y −5 , y ≠ 1, 5, 2 x 3

49.

x −1 5 1 ⋅ = , x ≠1 x − 1 25 ( x − 2 ) 5 ( x − 2 )

57.

50.

x ( x − 3) x + 13 x ( x − 3) x + 13 ⋅ = 3 ⋅ 3 5 5 x (3 − x ) x ( x − 3)( −1)

( x + 5)  x+5 58. Area of shaded portion:   = 2 4  

51.

52.

4r 2

( x + 5) 4 = ( x + 5) 4 Ratio: 2 ( x + 3)( x + 5) ( 2 x + 3) 2

x+5 , x ≠ −5 4 ( 2 x + 3)

4 ( y − 4 ) 2 ( y + 3) 4 y − 16 4 − y ÷ = ⋅ 5 y + 15 2 y + 6 5 ( y + 3 ) − ( y − 4 )

60.

2 x − 1 1 − x 2 x − 1 − 1 + x 3x − 2 − = = x+3 x+3 x+3 x+3

8 8 = − , y ≠ −3, 4 −5 5

61.

6 ( x + 3 ) − x ( 2 x + 1) 6 x − = 2x + 1 x + 3 ( 2 x + 1)( x + 3)

( t − 3 )( t + 2 )( t + 3) t2 − t − 6 t + 3 ⋅ = t 2 + 6t + 9 t 2 − 4 ( t + 3 )2 ( t + 2 )( t − 2 )

(

)

( y − 2) y + 2y + 4 4y 4y y3 − 8 ⋅ 2 = ⋅ 3 2y y − 5y + 6 2y3 ( y − 2)( y − 3) =

(

Area of total figure: ( 2 x + 3)( x + 5)

5 x 5+ x x +5 + = = x −1 x −1 x −1 x −1

t −3 = , t ≠ −2 t + 3 ( )( t − 2 )

54.

59.

53.

( 2r )

x + 13 x + 13 , x≠3 =− 5x2 −5 x 2

r r2 r r2 − 1 ÷ 2 = − ⋅ 1− r r −1 r −1 r2 (r + 1)(r − 1) r = − ⋅ r −1 r2 r +1 , r ≠ −1, 1 = − r

6 x + 18 − 2 x 2 − x ( 2 x + 1)( x + 3)

−2 x 2 + 5 x + 18 ( 2 x + 1)( x + 3)

=−

2 x 2 − 5 x − 18 ( 2 x + 1)( x + 3)

), y≠2

2 y2 + 2y + 4 y2 ( y − 3)

Chapter P

Prerequisites

62.

3 ( 3 x + 4 ) + 5 x ( x − 1) 3 5x + = x − 1 3x + 4 ( x − 1)( 3 x + 4 )

67.

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

5 x 2 + 4 x + 12 = ( x − 1)( 3 x + 4 ) 63.

3 5 3 5 2 + = − =− x−2 2− x x−2 x−2 x−2

64.

5 ( −1) 2x 5 2x − = − x − 5 5 − x x − 5 ( −1)( 5 − x )

68. −

2x − 2 + 2x + 2 + 1 ( x + 1)( x − 1) 4x + 1

( x + 1)( x − 1)

( (

) )

− x2 + 1 1 2 1 2x 1 + 2 − 3 = + − x x + 1 x + x x x2 + 1 x x2 + 1 x x2 + 1 =

(

)

−x − 1 + 2x − 1

(

)

x x2 + 1

=−

x − 2x + 2

(

)

x x2 + 1

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

−x2 − 3 ( x +1)( x − 2)( x − 3)

x2 + 3 ( x +1)( x − 2)( x − 3)

x − 6x − 6 + 5x2 x 2 ( x + 1)

5x2 − 5x − 6 x 2 ( x + 1)

2 10 2 10 + = + x2 − x − 2 x2 + 2x − 8 ( x − 2)( x +1) ( x + 4)( x − 2)

69.

1 6 5 1 6 5 − 2 + = − 2 + x2 + x x x +1 x( x + 1) x x +1 x − 6( x + 1) + 5 x 2 x 2 ( x + 1)

2 ( x + 4)

( x − 2 )( x + 1)( x + 4 ) 10 ( x + 1) + ( x − 2 )( x + 1)( x + 4 )

2 x + 8 + 10 x + 10 ( x − 2 )( x + 1)( x + 4 )

12 x + 18 ( x − 2 )( x + 1)( x + 4 )

(

1 1 x x − = − x − x − 2 x2 − 5x + 6 ( x − 2)( x +1) ( x − 2)( x − 3)

70.

=−

66.

2 ( x − 1) 2 ( x + 1) 1 + + ( x + 1)( x − 1) ( x + 1)( x − 1) ( x + 1)( x − 1)

2x −5 2x + 5 = − = x−5 x−5 x−5

65.

6 ( 2 x + 3) ( x − 2 )( x + 1)( x + 4 )

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

3 x( x + 3) − x( x) − 2( x + 3)( x − 3) x( x − 3)( x + 3)

3 x 2 + 9 x − x 2 − 2 x 2 + 18 x( x − 3)( x + 3)

9 x + 18 x( x − 3)( x + 3)

)

Section P.4

71.

x  x 2  2 − 1  2 − 2   =  ( x − 2)  x − 2   1    =

76.

x−2 1 1 ⋅ = , x≠2 2 x−2 2

 x−4

72.

 x−4    1  =  x 2 16   x 2 − 16  −      4x 4x   4x 

( x − 4 ) =  1 

 x 4 4 − x  

x−4 4x ⋅ 2 1 x − 16 x−4 4x = ⋅ 1 ( x + 4 )( x − 4 ) =

73.

74.

4x , x ≠ 0, 4 x+4

 x2    2 3 2  ( x + 1)  ( x + 1)  = x ⋅  x  ( x + 1)2 x   3  ( x + 1)    x ( x + 1) = 1 = x 2 + x, x ≠ 0, − 1

= =

x − ( x + h)

hx 2 ( x + h )

t2 − 1

 t 2 − (t 2 − 1)     t 2 − 1  t 2 − 1   =  t2 1 t2 − 1 t2 1

(

hx ( x + h ) 2

)

−h ( 2 x + h ) 2

2x + h x2 ( x + h)

, h≠0 2

) x x− 5

x 5 − 5 x −3 = x −3 x8 − 5 =

1 t2

t2 − 1

(

80.

(

) x x− 2

x 5 − 2 x −2 = x −2 x 7 − 2 =

⋅

t2 − 1 1

(

79.

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

81. x2 x2 + 1

−5

−4

−5

=−

( x + 1) 2

82. 2x ( x − 5) − 4x2 ( x − 5) = 2x ( x − 5) ( x − 5 − 2x) −3

x 2 − x 2 + 2 xh + h2

=−

−

hx 2 ( x + h )

 1 1  1 1  − 2  − 2 2 2  ( x + h ) x   ( x + h) x  x2 ( x + h )2  = ⋅ 2 h h x2 ( x + h)

 1   1   x−   x−  2 x  2 x 2 x  = ⋅ 2 x x x 2x − 1 = , x>0 2x

x +1 , x≠0 x −1

x   x+h  x + h +1 − x +1   h  ( x + h )( x + 1) x ( x + h + 1)  −    ( x + h + 1)( x + 1) ( x + h + 1)( x + 1)    = h 1  ( x + h )( x + 1) x ( x + h + 1)  1 = − ⋅  ( x + h + 1)( x + 1) ( x + h + 1)( x + 1)  h   2 2  x + x + hx + h − x − xh − x  1 = ⋅   h ( x + h + 1)( x + 1)   1 1 h , h≠0 = ⋅ = ( x + h + 1)( x + 1) h ( x + h + 1)( x + 1)

  78. 

 x2 − 1    2 x  x  = x −1 ⋅ 2 2 x  ( x − 1)  ( x − 1)    x    ( x + 1)( x − 1) = 2 ( x − 1) =

75.

77.

Rational Expressions

−4

= 83. 2x2 ( x −1) − 5( x −1) 12

−1 2

2x( −x − 5)

( x − 5)

= ( x −1) =

−1 2

−2x( x + 5)

( x − 5)

( 2x ( x −1) − 5) 2

2x3 − 2x2 − 5

( x −1)

Chapter P

84. 4 x 3 ( 2 x − 1)

Prerequisites − 2 x ( 2 x − 1)

−1 2

= 2 x ( 2 x − 1)

−1 2

= 2 x ( 2 x − 1)

−1 2

(2 x ( 2 x − 1) − 1) 2

( 2 x ( 4 x − 4 x + 1) − 1) 2

(8 x − 8 x + 2 x − 1) 2 x ( 8 x − 8 x + 2 x − 1) = = 2 x ( 2 x − 1)

−1 2

( 2 x − 1)

85.

86.

2 x 3 2 − x −1 2 x = x2

−1 2

( 2 x − 1) = 2 x − 1 2

x 2 ( x −1 2 ) − 3x1 2 ( x 2 ) x

x5 2

x3 2 − 3x5 2 x4

x3 2 (1 − 3 x) x4 3x − 1 = − 52 x =

87.

(

)

− x2 x2 + 1

−1 2

(

)

+ 2 x x2 + 1

(

)

( x + 1) =

−3 2

x x2 + 1 2

−3 2

(

)

− x x2 + 1 + 2   x3

3 − x 3 − x + 2    = −x − x + 2 32 x2 x2 x2 + 1

(

( x − 1) ( x + x + 2 )

)

=−

88.

(

)

(

x 3 4 x −1 2 − 3 x 2 83 x − 3 2 x6

(

) = 4x − 8x 52

)

x2 x2 + 1 12

(

4 x1 2 x 2 − 2 x6

) = 4 ( x − 2) 2

x11 2

( x + 5) ( ) ( 4 x + 3) ( 4 ) − ( 4 x + 3) ( 2 x ) = 2 ( 4 x + 3) ( x + 5) − x ( 4 x + 3) 89. ( x + 5) ( x + 5) 2 ( −3 x − 3 x + 5 ) = ( x + 5) 4 x + 3 2 ( 3 x + 3 x − 5) =− ( x + 5) 4 x + 3 2

−1 2

1 2

−1 2

( 2 x + 1) 3 ( x − 5) − ( x − 5) ( 12 ) ( 2 x + 1) ( 2 ) = ( x − 5) ( 2 x + 1) 90. 12

−1 2

2x + 1

3 ( 2 x + 1) − ( x − 5 )    2x + 1

−1 2

( x − 5) ( 6 x + 3 − x + 5) 32 ( 2 x + 1) 2 ( x − 5) ( 5 x + 8 ) = 32 ( 2 x + 1) 2

Section P.4

91.

x + 4 − 4

= = =

92.

z−3 − z = 3 =

93.

= =

97. Probability =

)

(

x + 4 +

1 x + 4 +

x x

−3

( z − 3 + z ) 3( z − 3 + z ) x+2 − 2 x+2 + 2 ⋅ x x+2 + 2

−1 z−3 + z

95.

1− x −1 = x

( x + 2) − 2

( x +2 + 2)

(

x 1

( x + 5 + 5) 1

(

−x

, x≠0

= = =

)

1− x +1

1 ,x ≠ 0 1− x +1

4+ x − 2 = x

( x + 5 + 5)

x+5 + 5

( 1 − x + 1)

= −

( x + 5) − 5

(1 − x) − (1)

, x≠0

96.

1− x +1 1− x +1

)

x+5 − 5 x+5 + 5 ⋅ x x+5 + 5 x

1− x −1 ⋅ x

x+2 + 2

x+5 − 5 = x =

( x+4+

x + 4 + x + 4 +

⋅

( x + 4) − ( x)

( z − 3) − z

94.

z −3 − z z −3 + z ⋅ 3 z −3 + z

x+2 − 2 = x =

x + 4 − 4

Rational Expressions

4+ x − 2 ⋅ x

4+ x + 2 4+ x + 2

(4 + x) − (4) x

( 4 + x + 2)

(

)

4+ x + 2

1 ,x ≠ 0 4+ x + 2

x ( x 2) Area shaded rectangle x 2 2 x = = ⋅ = Area large rectangle x ( 2 x + 1) 2 x + 1 2 2 ( 2 x + 1)

Chapter P

Prerequisites

Shaded area 1 4 ⋅ ( x + 2 )( x + x + 4 ) (area of trapezoid) 2 x 98. Probability = = 1 4 Total area ( x + 4 ) ( x + 2 ) + x ( x + 2 ) 2   ( area of triangle )

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

4 ⋅ 2 ( x + 2)

4 ( x + 4 )( x + 2 ) 1 + x    8( x + 2)

⋅

( x + 4 )( x + 2 ) 1 + x  4



( x + 4 )( x + 2 ) 1 + x  4

8( x + 2)



8( x + 2)

( x + 4 )( x + 2 )( x + 4 ) ( x + 4 )2

In Exercises 99 and 100, use the formula

 24 ( NM − P )    N   r= . NM   P +  12   99. (a)

M = $525

(

N = ( 4 )(12 ) = 48 P = $20,000

(

)

 24 ( 48 )( 475 ) − 20,000    48     r=  48 )( 475 )  (  20,000 +  12   r ≈ 0.0639  6.39%  24 ( NM − P )    24 ( NM − P ) N    = N r= 12 P + NM NM    P + 12  12   = r=

24 ( NM − P ) N

100. (a) N = ( 5 )(12 ) = 60

P = $28,000

M = $475

(b)



1 

⋅

288 ( NM − P ) 12 = 12 P + NM N (12 P + NM )

288 ( 48 ⋅ 475 − 20,000 )

48 (12 ⋅ 20,000 + 48 ⋅ 475 )

≈ 0.0639  6.39%

)

 24 ( 60 )( 525 ) − 28,000    60     r=  60 )( 525 )  (  28,000 +  12   r ≈ 0.0457  4.57%  24 ( NM − P )    24 ( NM − P ) N    = N (b) r = 12 P + NM NM    P + 12  12   = r=

24 ( NM − P ) N

⋅

288 ( NM − P ) 12 = 12 P + NM N (12 P + NM )

288 ( 60 ⋅ 525 − 28,000 )

60 (12 ⋅ 28,000 + 60 ⋅ 525 )

101. Copy rate =

≈ 0.0457  4.57%

50 pages 1 minute

(a) The time required to copy one page =

1 minute 50

 1  (b) The time required to copy x pages = x    50  x = minutes 50 (c) The time required to copy  1  12 120 pages = 120   = minutes or 2.4 minutes  50  5

Section P.4

102. (a) RT =

(b) RT = =

103. (a)

Rational Expressions

1 1 1 1 + + R1 R2 R3 1 R2 R3 + R1 R3 + R1 R2 R1 R2 R3 R1R2 R3 R2 R3 + R1 R3 + R1R2

( 6 )( 4 )(12 ) ( 4 )(12 ) + ( 6 )(12 ) + ( 6 )( 4 ) 288 = 2 ohms 144

Year

2005

2006

2007

2008

Births (in millions)

4.152

4.266

4.341

4.265

Population (in millions)

295.9

298.5

301.2

303.8

Year

2009

2010

2011

2012

Births (in millions)

4.125

4.027

3.971

3.939

Population (in millions)

306.5

309.1

311.7

314.4

(b) The models are close to the actual data. (c) The ratio of the number of births B to the number of people P is given by B = P =

(d)

0.06815t 2 − 0.9865t + 3.948 0.06815t 2 − 0.9865t + 3.948 1 0.01753t 2 − 0.2530t + 1 = ⋅ 2.64t + 282.7 0.01753t 2 − 0.2530t + 1 2.64t + 282.7 0.06815t 2 − 0.9865t + 3.948

(0.01753t 2 − 0.2530t + 1)(2.64t + 282.7)

Year

2005

2006

2007

2008

Ratio

0.0140

0.0143

0.0144

0.0140

Year

2009

2010

2011

2012

Ratio

0.0135

0.0130

0.0127

0.0125

The ratio has remained fairly constant over time.

Chapter P

Prerequisites

104. (a) t

0 75

2 55.9

12 41.7

6 45

4 48.3

14 41.3

16 41.1

18 40.9

8 43.3 20 40.7

10 42.3

 3   n ( n + 1)( 2 n + 1)  3 110. 9     − n    n 6   n  =

22 40.6

( x + h) 107.

− 2

1 x2

h 3 x + 3 xh + h 2

)

= = = =

108.

109.

1 1 − 2 ( x + h) 2x h

113.

x+h − x = h =

x2 − ( x + h) hx 2 ( x + h )

(

x 2 − x 2 + 2 xh + h 2 hx 2 ( x + h )

)

−2 x − h x ( x + h) 2

115.

h (2x ) 2 ( x + h)

−1 , h≠0 2x ( x + h)

(

h x+h + x

)

5x3 5x3 = 3 2 x + 4 2 x3 + 2

(

)

There are no common factors, so this expression is in reduced form. In this case, factors of terms were incorrectly cancelled.

4  n ( n + 1)( 2n + 1)   4  2 ( n + 1)( 2n + 1) +8   + 2n   =   n 6 3 n  =

ax − b is undefined for values of a, b, and x such b − ax that b = ax.

2x − 2 ( x + h)

( x+h + x)

114. No.

, h≠0

−2h h4 x ( x + h )

x+h + x However, the expression still has a radical in the denominator, so the expression is not in simplest form.

x+h − x x+h + x ⋅ h x+h + x x+h−x

−2 xh − h 2 hx 2 ( x + h )

112. False. The domain of the left-hand side is all x ≠ 1, unlike the domain of the right-hand side, which is all real numbers x.

h = 3 x 2 + 3 xh + h2 , h ≠ 0 1

) ( x − 1) is all

x ≠ 1, unlike the domain of the right-hand side.

(

)

(

( x + h ) − x 3 = x 3 + 3 x 2h + 3 xh2 + h3 − x 3 106. =

)

111. False. For n odd, the domain of x 2 n − 1

h = 2 x + h, h ≠ 0

−3

9 2 n 2 + 3n + 1 − 6

(

( x + h ) − x 2 = x 2 + 2 xh + h2 − x 2 105. h h (2x + h)

(

2 18n 2 + 27n + 3 = 2 3 2 = 6 n + 9n + 1 , n ≠ 0 2

 4t 2 + 16t + 75  (b) T = 10  2  appears to be approaching 40.  t + 4t + 10 

9 ( n + 1)( 2 n + 1)

)

2 2n2 + 3n + 1 + 24

3 4n2 + 6n + 26 = , n≠0 3

116. The negative sign in front of the second fraction was not distributed through the numerator before the fractions were added. 117. Answers will vary. For example, let x = y = 1 : 1 1 1 1 = ≠ + =2 1+1 2 1 1 118. Answers will vary. Sample answer: When t = 0, the percent is 100%. After one week, the percent drops by one-half and then starts to increase as the time increases, slowly approaching 100% again.

Section P.5

119.

(

The Cartesian Plane

)

−4 2 2 x −2 − x −4 x 2 x − 1 = 1 1 2 x2 − 1 = x4

2 x −2 − x −4 x 4 2 x 2 − 1 ⋅ 4 = 1 x x4 Answers will vary.

Section P.5 The Cartesian Plane 1. Cartesian

15.

2. Distance Formula

3. Midpoint Formula

4 2

4. ( x − h ) + ( y − k ) = r 2 , center, radius 2

(3, 8)

(− 2, − 2.5)

5. The x-axis is the horizontal real number line. Matches (c). 6. The y-axis is the vertical real number line. Matches (f).

−8 − 6 −4 −2

(0.5, − 1)

−4

(5, − 6)

−6 −8 y

16. 3

7. The origin is the point of intersection of the vertical and horizontal axes. Matches (a).

(− 52 , 2) 2

( 32 , 1)

8. The quadrants are four regions of the coordinate plane. Matches (d).

−3

−2

−1

x −1 −2

9. An x-coordinate is the directed distance from the y-axis. Matches (e).

−3

2 1, − 1 2

(

)

(3, − 3)

17.

( −5, 4 )

18.

( 2, − 3)

12. A : ( 23 , − 4 ) ; B : ( 0, − 2 ) ; C : ( −3, 25 ) ; D : ( −6, 0 )

19.

( 0, − 6 )

13.

20.

( −11, 0 )

21.

x > 0  The point lies in Quadrant I or in Quadrant IV. y < 0  The point lies in Quadrant III or in Quadrant IV.

10. A y-coordinate is the directed distance from the x-axis. Matches (b). 11. A : ( 2, 6 ) , B : ( −6, − 2 ) , C : ( 4, − 4 ) , D : ( −3, 2 )

(0, 5) 4

(− 4, 2) 2 −4

2 −2

(− 3, − 6)

x > 0 and y < 0  ( x, y ) lies in Quadrant IV.

−2

(1, − 4)

−4

22. If x < 0 and y < 0 then ( x, y ) is in Quadrant III.

−6

23. y

14.

5 4 3 2 1 (0, 0)

(−4, 0) −5 −4 −3 −2 −1

(−5, −5)

−2 −3 −4 −5

x = −4  x is negative  The point lies in Quadrant II or in Quadrant III. y > 0  The point lies in Quadrant I or Quadrant II.

x = −4 and y > 0  ( x, y ) lies in Quadrant II. x

1 2 3 4 5

(4, −2)

24. If x > 2 and y = 3 then ( x, 3) is in Quadrant I. 25.

y < −5  y is negative  The point lies in either Quadrant III or Quadrant IV.

Chapter P

Prerequisites

26. If x > 4 then ( x, y ) is in Quadrant I or IV.

32.

27. If − y > 0, then y < 0. x < 0  The point lies in Quadrant II or in Quadrant III. y < 0  The point lies in Quadrant III or in Quadrant IV. x < 0 and y < 0  ( x, y ) lies in Quadrant III. 28. If ( − x, y ) is in Quadrant IV, then ( x, y ) must be in

33.

Quadrant III.

40 30 20 10

both negative. Hence, ( x, y ) lies in either Quadrant I or

34.

Quadrant III.

− 20 − 30

Month (1 ↔ January)

( 6, − 3 ) , ( 6, 5 ) ( 6 − 6 ) + ( 5 − ( −3) ) = 02 + 82 = 64 = 8 2

( −11, 4 ) , ( −1, 4 ) d=

30. If xy < 0, then x and y have opposite signs. This happens in Quadrants II and IV.

0 − 10

29. If xy > 0, then either x and y are both positive, or

( −1 − ( −11) ) + ( 4 − 4 ) = 10 + 0 2

= 100 = 10

31.

35.

180,000

( − 2, 6), (3, − 6)

150,000

120,000 90,000

( 3 − ( −2 ) ) + ( −6 − 6 ) = 5 + ( −12 ) 2

= 25 + 144 = 169 = 13

60,000 30,000

36.

2002 2004 2006 2008 2010 2012

Sales (in millions of dollars)

Recorded low temperature (°F)

(8, 5), (0, 20) d=

Year

( 0 − 8 ) + ( 20 − 5) = 82 + 152 2

= 64 + 225 = 289 = 17 37. ( 2, 6), ( − 5, 5)

(− 5 − 2) + (5 − 6) 2

d =

(− 7) + (−1) 2

50 = 5 2

38. ( − 3, − 7), (1, −15) 2

1 − ( − 3) + −15 − ( − 7)

d =

39.

( , ) , ( 2, − 1) d = ( − 2 ) + ( + 1) 1 2

40.

9 4

277 36

277 6

49 16

457 144

80 = 4 5

(9.5, − 2.6), (− 3.9, 8.2)

43. (a)

( − + 1) + (3 − ) 1 9

5 4

(1, 1) , ( 4, 5) d=

( 4 − 1) + ( 5 − 1) 2

= 32 + 4 2 = 25 = 5

457 12

( 4, 5) , ( 4, 1)

≈ 1.78

d = 1 − 5 = −4 = 4

( −4.2, 3.1) , ( −12.5, 4.8) d=

( 9.5 + 3.9 ) + ( −2.6 − 8.2 )

= 296.2 ≈ 17.21

≈ 2.77

2 3

= 179.56 + 116.64

+ 499 =

(4) + (− 8)

( − 23 , 3) , ( − 1, 54 ) d=

41.

4 3

42.

4 3

1 2

( −4.2 + 12.5) + ( 3.1 − 4.8 )

= 68.89 + 2.89 = 71.78 ≈ 8.47

( 4, 1) , (1, 1)

d = 1 − 4 = −3 = 3 (b)

32 + 42 = 9 + 16 = 25 = 52

Section P.5 44. (a)

(1, 0 ) , (13, 5) d=

d1 =

d2 =

= 169 = 13

(13, 5) , (13, 0 )

d3 =

(1, 0 ) , (13, 0 ) 5 + 12 = 25 + 144 = 169 = 13

( 9 − ( −1) ) + ( 4 − 1)

( 9, 4 ) , ( 9, 1) ( 9, 1) , ( −1, 1) 10 2 + 32 = 100 + 9 = 109 =

( 109 )

(1, 5) , ( 5, − 2 ) (1 − 5) + ( 5 − ( −2 ) ) 2

( −4 ) + ( 7 )

4 + 7 = 16 + 49 = 65 =

( 65 )

( 4 − 2 ) + ( 0 − 1) = 4 + 1 = 5

d2 =

( 4 + 1) + ( 0 + 5) = 25 + 25 = 50

d3 =

( 2 + 1) + (1 + 5) = 9 + 36 = 45

( 5 ) + ( 45 ) = ( 50 ) 2

( 3 + 2 ) + ( 2 − 4 ) = 25 + 4 = 29

d3 =

(1 + 2 ) + ( −3 − 4 ) = 9 + 49 = 58

( 4 − 2 ) + ( 9 − 3) 2

= 4 + 36 = 40 = 2 10 d2 =

( −2 − 4 ) + ( 7 − 9 ) 2

( −2 − 2 ) + ( 7 − 3) 2

51. Find the distances between pairs of points.

d1 =

d2 =

= 16 + 16 = 32 = 4 2 Because d1 = d2 , the triangle is isosceles.

47. Find the distances between pairs of points.

(1 − 3) + ( −3 − 2 ) = 4 + 25 = 29

d3 =

d = 1 − 5 = −4 = 4

= 36 + 4 = 40 = 2 10 2

(1, 5) , (1, − 2 ) d = 5 − ( −2 ) = 5 + 2 = 7 = 7 (1, − 2 ) , ( 5, − 2 )

d1 =

= 16 + 49 = 65

(b)

50. Find the distances between the pairs of points.

d = −1 − 9 = −10 = 10

Because d1 = d2 , the triangle is isosceles.

d = 1 − 4 = −3 = 3

( 5 − ( −1) ) + (1 − 3 )

49. Find the distances between pairs of points. 2

= 109

Because d12 + d2 2 = d3 2 , the triangle is a right triangle.

( −1, 1) , ( 9, 4 )

( 20 ) + ( 20 ) = ( 40 )

= 10 2 + 32

46. (a)

( 5 − 3) + (1 − 5)

(b)

= 36 + 4 = 40

d = 1 − 13 = −12 = 12

= 4 + 16 = 20

d = 5−0 = 5 =5

45. (a)

( 3 − ( −1) ) + ( 5 − 3)

= 16 + 4 = 20

= 12 2 + 52

(b)

48. Find the distances between pairs of points.

(13 − 1) + ( 5 − 0 )

The Cartesian Plane

Because d12 + d3 2 = d2 2 , the triangle is a right triangle.

d1 =

( 0 − 2 ) + ( 9 − 5) = 4 + 16 = 20 = 2 5

d2 =

( −2 − 0 ) + ( 0 − 9 ) = 4 + 81 = 85

d3 =

( 0 − ( −2 ) ) + ( −4 − 0 ) = 4 + 16 = 20 = 2 5

d4 =

( 0 − 2 ) + ( −4 − 5) = 4 + 81 = 85

Opposite sides have equal lengths of 2 5 and 85, so the figure is a parallelogram. 52. Find the distances between pairs of points

d1 =

( 0 − 3) + (1 − 7 ) = 9 + 36 = 45 = 3 5

d2 =

( 3 − 4 ) + ( 7 − 4 ) = 1 + 9 = 10

d3 =

( 4 − 1) + ( 4 + 2 ) = 9 + 36 = 45 = 3 5

d4 =

( 0 − 1) + (1 + 2 ) = 1 + 9 = 10

Opposite sides have equal lengths of 3 5 and figure is a parallelogram.

10. The

Chapter P

Prerequisites

53. First show that the diagonals are equal in length.

d1 = 0 − ( −3) + ( 8 − 1) = 9 + 49 = 58 2

( 2 − ( −5)) + ( 3 − 6 ) = 49 + 9 = 58 2

d2 =

Now use the Pythagorean Theorem to verify that at least one angle is 90° (and hence, they are all right angles). d3 =

(

)

d4 =

( −3 − ( −5) ) + (1 − 6 ) = 4 + 25 = 29

Thus, d3 + d 4 = d1 .

(0, 90)

75 50 25

(300, 25) (0, 0)

d1 =

( 300 − 0 ) + ( 25 − 0 ) 2

d1 =

( 3 − 2 ) + (1 − 4 ) = 1 + 9 = 10

d2 =

( 4 − 1) + ( 3 − 2 ) = 9 + 1 = 10

Now use the Pythagorean Theorem to verify that at least one angle is 90°. d3 =

( 4 − 2) + (3 − 4) = 4 + 1 = 5

d4 =

( 2 − 1) + ( 4 − 2 ) = 1 + 4 = 5

( 300 − 0 ) + ( 25 − 90 ) 2

(120 − 0 ) + (150 − 0 ) 2

(0, − 360) represent the location of Austin, and (− 514, 0) represent the location of Albuquerque.

(− 514 − 0) + 0 − (− 360) 2

d = =

(− 514) + 360

393,796 ≈ 627.53 mi

59. (a) 8

(2, 4)

(1, 2)

(4, 3)

(0, 0) −2

(3, 1) 1

(8, 6)

1 3

(b)

( 45 − 10 ) + ( 40 − 15) = 352 + 252 2

x 2

−2

−1

55. d =

58. Let (0, 0) represent the location of Oklahoma City,

−1

= 36,900 ≈ 192.1 km

Thus, d3 + d 4 = d1 .

and (120, 150 ) represent the destination, Rome.

54. First show that the diagonals are equal in length.

300

57. Let ( 0, 0 ) represent the point of departure, Naples,

−2

250

= 94,225 ≈ 307.0 feet

(−3, 1) −2

200

Distance from ( 300, 25 ) to home plate:

(2, 3)

150

Distance (in feet)

d2 =

d2 d4

100

Distance from ( 300, 25 ) to third base:

(0, 8)

x 50

= 90,625 ≈ 301.0 feet

−4

100

0 − ( −5 ) + ( 8 − 6 ) = 25 + 4 = 29

−6

125

(−5, 6)

56. Distance (in feet)

8+0 6+0 8 6 ,   =  ,  = ( 4, 3 ) 2  2 2  2

= 1850 = 5 74 ≈ 43 yards

Section P.5 y

60. (a)

The Cartesian Plane y

64. (a)

(2, 8)

(1, 12)

8 6

10 8

6 4

− 10 − 8

−6

−2

(9, 0) −2

(− 7, − 4)

 1 + 9 12 + 0   10 12  , (b)   =  ,  = (5, 6) 2  2 2  2

−4

 −7 + 2 −4 + 8   −5 4   5  ,   =  ,  = − , 2 2   2 2  2   2

65. (a)

61. (a)

(b)

y 5 2

(5, 4)

(− 52 , 43 )

3 2

(− 1, 2)

−1

(b)

( 12, 1)

1 2

x 1

−5

−2

−1

 −1 + 5 2 + 4   4 6  ,   =  ,  = ( 2, 3 ) 2  2 2  2

62. (a)

x 2

−2

(b)

−3

1 −1 − 2

 − 25 + 12 ,   2

4 3

+ 1   −4 2 7 3  , =  2   2 2  7  =  −1,  6 

(2, 10) 10

1 2

66. (a)

−

3 6

−2

−1

(10, 2)

(− 13 , − 13 )

x 2

(b) 63. (a)

 2 + 10 10 + 2   12 12  ,  = ,  = ( 6, 6 ) 2   2 2   2

(b)

10 8 6

−4 −6

(b)

x 4

−3

 ( −1 3 ) − (1 6 ) ( −1 3 ) − (1 2 )  ,     2 2   5   −1 2 −5 6   1 , =  = − , −  2   4 12   2

67. (a)

−8 −6 −4 −2

(− 16 , − 12 )

(− 4, 10)

−2

(6.2, 5.4)

(4, − 5)

(− 3.7, 1.8)

4 2

 4 − 4 −5 + 10   0 5   5  ,   =  ,  =  0,  2  2 2  2  2

−4

−2

(b)

 6.2 − 3.7 5.4 + 1.8   2.5 7.2  , ,  =  2 2 2     2 = (1.25, 3.6 )

Chapter P

Prerequisites

68. (a)

78. r =

(− 16.8, 12.3)

1 2

 0+6 0+8 Center:  ,  = ( 3, 4 ) 2   2

15 10

( x − 3) + ( y − 4 ) = 25 2

(5.6, 4.9) − 20 − 15 − 10

−5

(b)

 −16.8 + 5.6 12.3 + 4.9  ,   2 2    −11.2 17.2  = ,  = ( −5.6, 8.6 ) 2   2

69. Calculate the midpoint.

 2009 + 2013 924 + 1423  ,   = ( 2011, 1182.5) 2 2  

The revenue for Texas Roadhouse was $1182.5 million in 2011. 70. Calculate the midpoint.

 2009 + 2013 1106 + 1439  ,   = ( 2011, 1272.5) 2 2  

71.

( x − 0) + ( y − 0) = 5 2

79. Because the circle is tangent to the x-axis, the radius is 1.

( x + 2 ) + ( y − 1) = 1 2

80. Because the circle is tangent to the y-axis, the radius is 3.

( x − 3) + ( y + 2 ) = 9 2

81. The center is the midpoint of one of the diagonals of the square.  7 + ( −1) −2 + ( −10 )  , Center:   = ( 3, − 6 )   2 2   The radius is one half the length of a side of the square. 1 Radius: 7 − ( −1) = 4 2

(

)

Circle: ( x − 3 ) + ( y + 6 ) = 16 2

The revenue for Papa John’s Intl. was $1272.5 million in 2011. 2

( 6 − 0 ) + (8 − 0 ) = 2 100 = 5

2 x

−6 −4 −2

(− 1, − 2)

(7, − 2)

x 2 + y 2 = 25

72.

( x − 0 ) + ( y − 0 ) = 62 2

(− 1, − 10)

x + y = 36

73.

( x − 2 ) + ( y + 1) = 42 2 2 ( x − 2 ) + ( y + 1) = 16

74.

( x + 5) + ( y − 3) = 22 2 2 ( x + 5) + ( y − 3) = 4

75.

( x + 1) + ( y − 2 ) = r 2 2 2 ( 0 + 1) + ( 0 − 2 ) = r 2  r 2 = 5 2 2 ( x + 1) + ( y − 2 ) = 5

(7, − 10)

− 12

82. The center is the midpoint of one of the diagonals of the square.  8 + ( −12 ) 10 + ( −10 )  , Center:   = ( −2, 0 )   2 2   The radius is one half the length of a side of the square. 1 Radius: 8 − ( −12 ) = 10 2

(

)

Circle: ( x + 2 ) + y 2 = 100 2

y 16

76. r =

( 3 − ( −1)) + ( −2 − 1) = 16 + 9 = 5 2

( x − 3) + ( y + 2 ) = 52 = 25 2

 −4 + 4 −1 + 1  77. Center:  ,  = ( 0, 0 ) 2   2

(− 12, 10)

(8, 10)

12 8 4

−8 −4 −4

x 4

−8

(− 12, − 10) − 12

(8, − 10)

( 4 − 0 ) + (1 − 0 ) = 17 2

x + y 2 = 17

Section P.5 83. From the graph, you can estimate the center to be ( 2, − 1) and the radius to be 4.

1 2 Center:  , −   2 3

84. From the graph, you can estimate the center to be ( −3, 1)

Radius:

and the radius to be 5.

( x + 3) + ( y − 1) = 25 2

4 3

90.

x 2

( x + 4.5) + ( y − 0.5) 2

= 2.25 y

Center: ( − 4.5, 0.5)

2 −2

Radius: 1.5

−2

3 2

−4

−6

−7 −6

−4

−2 −1 −1 −3 −4

Center: (0, 0)

Radius: 8

91. The x-coordinates are increased by 2, and the y-coordinates are increased by 5. Original vertex Shifted vertex ( −1, − 1) ( −1 + 2, − 1 + 5) = (1, 4 )

6 4 2 −10

−6 −4 −2

2 4 6

−4 −6

( −2, − 4 ) ( 2, − 3)

−10

y 6

Radius: 3

( −3, 6 ) ( −1, 3) ( −3, 0 )

4 2 −2

x 2

−2 −4

= 4

Center: ( −1, 3) Radius: 2

( −3 + 6, 6 − 3) = ( 3, 3) ( −1 + 6, 3 − 3) = ( 5, 0 ) ( −3 + 6, 0 − 3) = ( 3, − 3)

93. The x-coordinates are decreased by 1, and the y-coordinates are increased by 3. Original vertex Shifted vertex ( 0, 2 ) ( 0 − 1, 2 + 3) = ( −1, 5)

−6

( x + 1) + ( y − 3)

( −2 + 2, − 4 + 5) = ( 0, 1) ( 2 + 2, − 3 + 5 ) = ( 4, 2 )

92. The x-coordinates are increased by 6, and the y-coordinates are decreased by 3. Original vertex Shifted vertex ( −5, 3) ( −5 + 6, 3 − 3) = (1, 0 )

( x − 6) + y 2 = 9 Center: (6, 0)

88.

x 1

−2

86. x 2 + y 2 = 64

87.

−3

−6

1 −1

Radius: 5

−2

85. x 2 + y 2 = 25

Center: (0, 0)

1 2 16   89.  x −  +  y +  = 3 3 9    

( x − 2 ) + ( y + 1) = 16 2

The Cartesian Plane

( −3, 5) ( −5, 2 ) ( −2, − 1)

y 6

4 3 2

−5 − 4 −3 − 2 − 1 −1 −2

x 1

( −3 − 1, 5 + 3) = ( −4, 8 ) ( −5 − 1, 2 + 3) = ( −6, 5) ( −2 − 1, − 1 + 3) = ( −3, 2 )

94. The x-coordinates are decreased by 3, and the y-coordinates are decreased by 2. Original vertex Shifted vertex (1, − 1) (1 − 3, − 1 − 2 ) = ( −2, − 3)

( 3, 2 ) (1, − 2 )

( 3 − 3, 2 − 2 ) = ( 0, 0 ) (1 − 3, − 2 − 2 ) = ( −2, − 4 )

Chapter P

Prerequisites

95. (a) The point (65, 83) represents a final exam score of

83, given an entrance score of 65. (b) No. There are many variables that will affect the final exam score. 96. In the last two years, six performers have been elected to the Rock and Rock Hall of Fame. If this pattern continues, then there will be six performers elected in 2015. 97. True. The side joining ( −8, 4 ) and ( 2, 11) has

( −8 − 2 ) + ( 4 − 11) = 149. The side joining ( 2, 11) and ( −5, 1) has length 2

length

( 2 + 5) + (11 − 1) = 149. 2

98. False. The polygon could be a rhombus. For example, consider the points ( 4, 0 ) , ( 0, 6 ) , ( −4, 0 ) , and ( 0, − 6 ) . 99. The y-coordinate of a point on the x-axis is 0. The x-coordinate of a point on the y-axis is 0. x1 + x2 y + y2 we have: and ym = 1 2 2 2 xm = x1 + x2 2 ym = y1 + y2

100. Since xm =

2 xm − x1 = x2

2 ym − y1 = y2

So, ( x2 , y2 ) = ( 2 xm − x1 , 2 ym − y 1 ) . (a)

(b)

( x2 , y2 ) = ( 2 xm − x1, 2 ym − y1 ) = ( 2 ( 4 ) − 1, 2 ( −1) − ( −2 ) ) = ( 7, 0 ) ( x2 , y2 ) = ( 2 xm − x1, 2 ym − y1 ) = ( 2 ( 2 ) − ( −5) , 2 ( 4 ) − 11) = ( 9, − 3)

 x + x2 y1 + y2  101. Midpoint of segment:  1 ,  2   2 Midpoint between y +y  x +x ( x1, y1 ) and  1 2 2 , 1 2 2  :  

x1 + x2 y + y   2x + x + x 2y1 + y1 + y2   y1 + 1 2   1 1 2  x1 + 2  2  = 2 2 , ,   2 2 2 2          3x1 + x2 3y1 + y2  = ,  4   4 Midpoint between  x1 + x2 y1 + y2  ,   and ( x2 , y2 ) : 2   2 y1 + y2  x1 + x2   x + x + 2x2 y1 + y2 +2y2  + y2   1 2  2 + x2  2 2 2 , ,   =  2 2 2 2          x +3x y +3y  = 1 2 , 1 2  4   4 (a)

 3x1 + x2 3y1 + y2   3 (1) + 4 3 ( −2 ) − 1  , ,   =  4   4 4  4  7 7 = , −  4 4 

 x1 + x2 y1 + y2   1 + 4 −2 −1  , ,   =  2   2 2   2 3 5 = , −  2 2  x1 + 3x2 y1 + 3y2   1 + 3( 4) −2 + 3( −1)  , ,    =  4   4 4  4 

5  13 = , −  4 4  3x + x 3y + y   3( −2) + 0 3( −3) + 0  (b)  1 2 , 1 2  =  ,   4   4 4  4 

9  3 = − , −  4  2

 x1 + x2 y1 + y2   −2 + 0 −3 + 0  , ,   =  2   2 2   2

3  =  −1, −  2   x1 + 3x2 y1 + 3y2   −2 + 0 −3 + 0  , ,   =  4   4 4   4 3  1 = − , −  4  2

Section P.6

(a) x0 < 0, − y0 > 0

0+a+b 0+c a+b c , ,   = 2 2   2 2 

So, ( x0 , − y0 ) is in Quadrant III. Matches (ii).

(c)

−2 x0 > 0, y0 > 0

Midpoint between ( a, 0 ) and ( b, c ) :

So, ( −2 x0 , y0 ) is in Quadrant I. Matches (iii).

a+b 0+c a+b c , ,   = 2   2 2  2 Therefore, the diagonals of the parallelogram intersect at their midpoints.

x0 < 0, 12 y0 > 0 So, ( x0 , 12 y0 ) is in Quadrant II. Matches (iv).

(d)

103. Midpoint between ( 0, 0 ) and ( a + b, c ) :

102. x0 < 0, y0 > 0

(b)

Representing Data Graphically

− x0 > 0, − y0 < 0 So, ( − x0 , − y0 ) is in Quadrant IV. Matches (i).

Section P.6 Representing Data Graphically 12. Interval

1. Line plots

Tally

|||| |||| |||| |||| |||| |||| |||| | [0, 1000) [1000, 2000) | | | | | | | | [2000, 3000) | | | [3000, 4000) [4000, 5000) [5000, 6000) | [6000, 7000) |

2. Line graphs 3. Line plot matches (c). 4. Bar graph matches (d). 5. Histogram matches (a). 6. Line graph matches (b).

Number of states (including the District of Columbia)

7. (a) The price $3.52 occurred with the greatest frequency. (b) The range of prices is $3.98 − $3.42 = $0.56.

1000 2000 3000 4000 5000 6000 7000

8. (a) The weight of 900 pounds occurred with the greatest frequency (9). (b) The weights range from 600 to 1300 pounds. The range of weights is 1300 − 600 = 700 pounds.

40 35 30 25 20 15 10 5

13.

Quiz Scores

The score of 15 occurred with the greatest frequency.

98 100

Exam Scores

Month

Answers will vary. Sample answer: The amount of precipitation decreases at a fairly constant rate from January to July, and then it starts to increase at a fairly constant rate until December.

The scores of 81 and 85 occurred with the greatest frequency. 11. Interval Tally 20

Number of states

[7, 10) | | [10, 13) | | | | | | | | | | | | [13, 16) | | | | | | | | | | | | | | | [16, 19) | | | | | | | | | [19, 22) | | | |

7 6 5 4 3 2 1 Ja Fe nua br ry u M ary ar c A h pr M il a Ju y ne Ju Se Aug ly pt us em t O b N cto er ov b D em er ec be em r be r

10.

Precipitation (in inches)

Students enrolled in public schools

16 12 8 4 7

Percent of individuals living below the poverty level

Chapter P

Prerequisites 19. The price increased at a fairly constant rate from 2005 to 2008.

120 100

20. The price decreased dramatically from 2008 to 2009.

80 60

21. In 2008, the price was about $3.50 and dropped to about $3.05 in 2010, which is a decrease of $3.50 − $3.05 $0.45 = ≈ 0.129 = 12.9%. $3.50 $3.50

2013

2011

2009

2007

2005

2003

2001

1999

20 1997

Revenue (in billions of dollars)

14.

Year

Answers will vary. Sample answer: As time progresses from 1997 to 2013, the revenue of Costco Wholesale increases at a fairly constant rate. 15.

22. In 2009, the price was about $2.60 and rose to about $3.90 in 2013, which is an increase of $3.90 − $ 2.60 $1.30 = = 0.50 = 50%. $ 2.60 $ 2.60

Year

2008

2009

2010

Differences in tuition charges (in dollars)

16,681

17,058

16,693

23. In December, the price paid for one dozen Grade A large eggs was approximately $2.03.

Year

2011

2012

2013

24. The highest price was $2.03 and the lowest price was $1.83. The difference is $2.03 − $1.83 = $0.20.

Differences in tuition charges (in dollars)

17,110

17,291

18,044

25. Because $2.03 − $1.92 = $0.11 is the greatest difference

between months, the greatest rate of increase occurred from November to December.

Private

598

−126

742

2011–2012

2012–2013

Public

483

340

Private

664

1093

2012 2011 2010 2009 2008 2007 2006 2005

35 30 25 20 15 10 5

Answers will vary. Sample answer: From 2000 to 2012, the percent of wives who earned more income than their husbands increases at a fairly constant rate.

18. 2012 1990

Philadelphia, PA Houston, TX

28. Trade deficit (in billions of dollars)

14,000

12,000

10,000

8,000

6,000

4,000

Men Women

College enrollment (in thousands)

900 800 700 600 500 400 300 200 100

Chicago, IL 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013

City

27.

Year

2,000

Year

17.

26. Answers will vary. Sample answer: The highest prices seemed to occur in the winter months, while the lowest prices seemed to occur in the summer months. According to the data, a price of about $2.10 per dozen seems reasonable. If the trend continues, the price should be within $0.10 of the actual price in February 2014.

2012

325

2010

239

2008

221

2006

Public

2004

2010-2011

2002

2009–2010

2000

2008–2009

Percent of wives who earned more than their husbands (United States)

16.

Los Angeles, CA

Year

New York, NY 2

Population (in millions)

Answers will vary. Sample answer: From 2004 to 2008, the trade deficit increased at a fairly constant rate, then dropped significantly in 2009, and then increased at a fairly constant rate until 2013, when it dropped slightly.

Chapter P Review 29.

32. A histogram has a portion of the real number line as its horizontal axis, and the bars are not separated by spaces. A bar graph can be either horizontal or vertical. The labels are not necessarily numbers, and the bars are usually separated by spaces.

4.5

1999

30.

2014

Answers will vary. Sample answer: A histogram is best because the data are percents within a year that do not relate to increasing or decreasing behavior. Female athletes Male athletes

5000

33. A line plot and a histogram both use a portion of the real number line to order numbers. A line plot is especially useful for ordering small sets of data, and recording the frequency of each value; while a histogram is more useful to organize large sets of data and then grouping the data into intervals and plotting the frequency of the data in each interval. 34. The second graph is misleading because the vertical scale is too small which makes small changes look large. Answers will vary. 35. Answers will vary. Line plots are useful for ordering small sets of data. Histograms or bar graphs can be used to organize larger sets. Line graphs are used to show trends over time.

4000 3000 2000

2012

2010

2008

2006

1000 2002

Number of athletes (in thousands)

6000

2004

31.

Year

Answers will vary. Sample answer: A double bar graph is best because there are two different sets of data within the same time interval that do not deal primarily with increasing or decreasing behavior.

Chapter P Review 1.

{11, − 14, − , , 6 , 0.4} 8 9

5 2

(a) Natural number: 11 (b) Whole number: 11 (c) Integers: 11, − 14

(a)

1 3

= 0.3

(b)

9 25

= 0.36 9 25

1 12

(d) Rational numbers: 11, − 14, − 89 , 25 , 0.4

(e) Irrational number: 2.

{ 15, − 22, − , 0, 5.2, } 3 7

1 3

5 12

1 2

7 12

5. (a) The inequality x ≥ − 6 is the set of all real numbers

greater than or equal to − 6.

(a) Natural numbers: none (b) Whole number: 0 (c) Integers: −22, 0

(b)

(d) Rational numbers: −22, − 103 , 0, 5.2, 73

(e) Irrational number: 3.

1 4

0.3 = 13 < 259 = 0.36

10 3

1 6

(a)

5 6

= 0.83

(b)

7 8

= 0.875

5 6

< 87

17 24

3 4

x −7 −6 −5 −4 −3 −2 −1

6. (a) The inequality x < 1 is the set of all real numbers less than 1.

(b)

x −2

−1

5 6

7 8

11 12

23 24

Chapter P

Prerequisites

7. (a) The inequality − 4 ≤ x ≤ 0 is the set of all real

numbers greater than or equal to − 4 and less than or equal to 0. (b)

x −6 −5 −4 −3 −2 −1

19. 0 + ( a − 5) = a − 5

Additive Identity Property 20.

8. (a) The inequality 7 ≤ x < 10 is the set of all real numbers greater than or equal to 7and less than 10. x 5

10 11 12

10. d ( −123, − 9 ) = −9 − ( −123 ) = 114 = 114

11.

Commutative Property of Addition

(b)

( t + 1) + 3 = 3 + ( t + 1)

x −7 ≥6

21.

2 y + 4 ⋅ = 1, y ≠ − 4 y + 4 2

Multiplicative Inverse Property 22. 1 ⋅ (3 x + 4) = 3 x + 4

Multiplicative Identity Property 23. (a)

(b)

x = 3 : 9 ( 3) − 2 = 27 − 2 = 25

( a b )(3ab ) = 3a b 2

−2

2 +1 4 − 2

40(b − 3)

75(b − 3)

8 15(b − 3)

(b)

3−4 m−1n−3 92 n 3 81 1 = 4 = = −2 −3 9 mn 3 mmn3 81m 2 m 2 −1

14.

x − 11x + 24 (a) x = −2 :

( −2 ) − 11( −2 ) + 24 = 4 + 22 + 24 = 50 x = 2 : 22 − 11( 2 ) + 24 = 4 − 22 + 24 = 6 2

(b) 15.

− 2x + 3 x (a) x = 0: not defined (b) x = 6:

16.

4x x −1

− 2(6) + 3 −12 + 3 −9 3 = = = − 6 6 6 2

4 ( −1) −4 = =2 ( −1) − 1 −2

(a)

x = −1:

(b)

x = 1 : not defined

17. 2 x + (3 x − 10) = ( 2 x + 3 x ) − 10

Associative Property of Addition 18. 4(t + 2) = 4 ⋅ t + 4 ⋅ 2

Distributive Property

36u 0v −3 36u ( )v −3−1 3u 2 = = 3u 2v −4 = 4 −2 12u v 12 v

26. (a) ( a 4b −3c 0 ) a 2 =

= 3a 3b 2

0 − −2

25. (a)

13. 9 x − 2

x = −1: 9 ( −1) − 2 = −9 − 2 = −11

12. d ( y, − 30 ) = y − ( −30 ) = y + 30 and

(a)

(4 y 2 ) = 43 y 6 = 64 y 4 , y ≠ 0 24. (a) (b)

d ( y, − 30 ) < 5, so y + 30 < 5.

( −2z ) = ( −2 ) z3 = −8z3

(a b c ) 4 −3 0

−1

(b)

b3 a2

−1

 y −2   x 2   1   x 2 y 2     −2  =  2     x   y   xy   1   xy 2  x 2 y 2  =    1  1  = x 3 y 4 , x ≠ 0, y ≠ 0

27. 2,585,000,000 = 2.585 × 10 9 28. −3,250,000 = −3.25 × 106 29. − 0.000000125 = −1.25 × 10 − 7 30. 0.00000008064 = 8.064 × 10−8 31. 1.28 × 105 = 128,000 32. −4.002 × 102 = −400.2 33. 1.80 × 10−5 = 0.0000180 34. −4.02 × 10−2 = −0.0402

Chapter P Review

35.

( 78 ) = ( 78 ) = 78 = 78 (5 x )

36. 5

38.

= ( xy )

= 6 xy

x −1

20 2 5 5 5 5 = = ⋅ = 4 4 2 5 2 5

54.

2 − 11 = 3 =

5 ⋅ a ⋅ a = 5a a

55. 64 5 2 =

( 2 − 11)

57.

( 2 + 11 ) 3 ( 2 + 11 ) −3 2 + 11

1 1 1 1 = = 2 = 2 23 1 3 64 4 16 64

(

)

(− 3x−1 6 )(− 2 x1 2 ) = 6 x −1 6 x1 2 = 6 x −1 6 + 3 6 = 6x2 6

2 x3 3 2 x3 x 3 21 3 x = = 2= 3 27 3 3 3 48 − 27 = 3 ⋅ 4 2 − 3 ⋅ 32

= 6 x1 3 , x ≠ 0

58.

( x − 1) ( x − 1) 13

−1 4

= 4 3 −3 3 = 3

59. 15 x − 2 x + 3 x + 5 − x

47. 8 3 x − 5 3 x = 3 3 x

= − 4 x 4 + x 3 + x 2 − x − 10

Leading coefficient: − 4 61.

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

= 3x2 + 7x − 1

= 2x 2x + 2x

50. 3 14 x 2 − 56 x 2 = 3 x 14 − 2 x 14

62.

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

63.

( 2x − 5x + 10x − 7) + ( 4x − 7x − 2)

= x 14 1 3− 5

1 3+ 5 ⋅ 3− 5 3+ 5

3+ 5 3+ 5 = 9−5 4

Standard form

Degree: 4

= −72 y

= ( 2 x + 1) 2 x

Standard form

60. − 4 x 4 + x 2 − 10 − x + x 3

48. −11 36 y − 6 y = −11( 6 ) y − 6 y

8 x 3 + 2 x = 2 ⋅ 22 ⋅ x ⋅ x 2 + 2 x

, x ≠1

= −2 x 5 − x 4 + 3 x 3 + 15 x 2 + 5 Degree: 5 Leading coefficient: −2

= 3⋅4⋅ 2 + 4⋅7⋅ 2

51.

1 3 −1 4 1 12

46. 3 32 + 4 98 = 3 2 5 + 4 2 ⋅ 72

49.

= ( x − 1) = ( x − 1)

= 40 2

−9

( 64 ) = 8 = 32,768

56. 64 −2 3 =

2 − 11 2 + 11 ⋅ 3 2 + 11

3 ⋅ 52 ⋅ x 2 5 x = 2 y4 y

x +1 x −1

x +1

53.

125 3 53 5 = 3 = 216 6 6

x +1

⋅

x −1

64 x 6 = 5 ( 2 x )( 2 x ) = 2 x 5 2 x

75 x 2 = y4

43.

45.

52.

81 9⋅9 9 3 = = = 144 12 ⋅ 12 12 4

41.

44.

12 xy = ( xy )   

25a = 5

= 5 x

39.

42.

8 ⋅ 5 4 = 5 23 ⋅ 5 22 = 23 5 ⋅ 22 5 = 25 5 = 2

37.

40.

= 2 x3 − 5x2 + 4 x2 + 10 x − 7 x − 7 − 2 = 2 x 3 − x 2 + 3x − 9

64.

(6 x − 4 x − x + 3 − 20 x ) − (16 + 9 x − 11x ) 4

= 6 x 4 − 4 x 3 − x + 3 − 20 x 2 − 16 − 9 x 4 + 11x 2 = −3 x 4 − 4 x 3 − 9 x 2 − x − 13

Chapter P

(

Prerequisites 84. 9 x 2 − 251 = ( 3 x − 15 )( 3 x + 15 )

)

65. −2 a a 2 + a − 3 = −2 a 3 − 2 a 2 + 6 a 66.

( y − 4 y )( y ) = y − 4 y

67.

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

85.

87.

( z + 1)( 5z − 6 ) = 5z − 6 z + 5z − 6 2

( x + 8)( x − 8) = x2 − 82 = x2 − 64

70.

( 7 x − 4)

(

= ( 4 x − 3) 16 x 2 + 12 x + 9

= (7 x) − 2(7 x)( 4) + ( 4) 2

( x − 4 ) = x 3 − 3 x 2 ( 4 ) + 3 x ( 4 ) − 43 3

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

89.

x 2 − 6 x − 27 = ( x − 9)( x + 3)

90.

x 2 − 9 x + 14 = ( x − 2 )( x − 7)

73. ( m − 7 ) + n  ( m − 7 ) − n  = ( m − 7 ) − n 2 = m 2 − 14 m + 49 − n 2

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

(

= ( x − 4 ) x2 − 3

94.

(

a = 2, c = −15, ac = −30 = ( −6 ) 5 and

Distributive Property

(

= ( x − 6 )( x − 1)( x + 1)

95. 2 x 2 − x − 15

( x + 3)( x + 5) = x ( x + 5) + 3( x + 5)

)

−6 + 5 = −1 = b

)

So, 2 x 2 − x − 15 = 2 x 2 − 6 x + 5 x − 15

= 2 x ( x − 3) + 5 ( x − 3)

= 2500r 2 + 5000r + 2500

= ( 2 x + 5 )( x − 3) .

77. 7 x + 35 = 7 ( x + 5)

96. 6 x + x − 12

78. 4b − 12 = 4 ( b − 3)

a = 6, c = −12, ac = −72 = ( −8 ) 9 and

(

79. 2 x 3 + 18 x 2 − 4 x = 2 x x 2 + 9 x − 2

−8 + 9 = 1 = b

)

(

80. −6 x 4 − 3 x 3 + 12 x = −3 x 2 x 3 + x 2 − 4 81.

)

= ( x − 6) x2 − 1

(

)

x3 − 6 x2 − x + 6 = x2 ( x − 6 ) − ( x − 6 )

74. ( x − y ) − 4  ( x − y ) + 4  = ( x − y ) + 4 2 = x 2 − 2 xy + y 2 − 16

76. 2500 1 + r 2 = 2500 1 + 2r + r 2

)

2 92. 3x + 14 x + 8 = ( 3x + 2 )( x + 4 )

93.

= 8 x 3 + 12 x 2 + 6 x + 1

75.

)

91. 2 x 2 + 21x + 10 = ( 2 x + 1)( x + 10 )

= x − 12 x + 48 x − 64

72.

= 49 x 2 − 56 x + 16

71.

(

x 3 + 216 = x 3 + 6 3 = ( x + 6 ) x 2 − 6 x + 36

88. 64 x 3 − 27 = ( 4 x ) − 33

= 5z 2 − z − 6 69.

86. 4 x 2 − 4 x + 1 = ( 2 x − 1)( 2 x − 1) = ( 2 x − 1)

= x 2 + 13 x + 36 68.

x 2 + 6 x + 9 = ( x + 3 )( x + 3) = ( x + 3 )

Thus, 6 x 2 + x − 12 = 6 x 2 − 8 x + 9 x − 12

= 2 x (3x − 4 ) + 3 (3x − 4 )

)

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

82. 8( 2 − y ) − ( 2 − y ) = ( 2 − y ) 8 − ( 2 − y ) 2

= ( 2 − y )(8 − 2 + y )

= ( 2 x + 3)( 3 x − 4 ) . 97. Domain: all x 98. Domain: x < 0 99. Domain: all x ≠ 23

= ( 2 − y )(6 + y ) = − ( y − 2)( y + 6)

83.

x2 − 169 = ( x + 13)( x − 13)

Chapter P Review 100. Domain: all x ≥ −12

106.

101.

4 x2 4x2 x = = 2 , x≠0 4 x + 28 x 4 x x 2 + 7 x +7

102.

6 xy 6 xy 6y = = , x≠0 xy + 2 x x ( y + 2 ) y + 2

103.

(

2x −1 x2 − 1 2 x − 1 ( x + 1)( x − 1) ⋅ 2 = ⋅ x + 1 2x − 7x + 3 x + 1 ( 2 x − 1)( x − 3 )

)

107.

x2 ( 5x − 6 ) 2x + 3

x 2 − x − 30 ( x − 6 )( x + 5 ) x − 6 = = , x ≠ −5 x 2 − 25 ( x + 5)( x − 5) x − 5

x 2 − 9 x + 18 ( x − 6 )( x − 3 ) x − 3 , x≠6 = = 104. 8 x − 48 8( x − 6) 8

108.

(

)(

4x − 6

( x − 1)

)

x ( 5x − 6 ) 2 x + 3 5x = ⋅ 2x + 3 2x + 3 5x x (5x − 6) 3 = , x ≠ 0, − 5 2

2 x 2 − 3x 4 x − 6 x2 + 2 x − 3 = ⋅ 2 x + 2 x − 3 ( x − 1)2 2 x2 − 3x

( x − 2 )( x + 2 ) ⋅ 1 = ( x − 2 )( x + 2 ) x 2

x −1 1 , x ≠ , −1 x −3 2 2

( x − 2 )( x + 2 ) ⋅ x 2 + 2 x2 − 4 x2 + 2 105. 4 ⋅ = 2 2 2 x − 2x − 8 x x2 x − 4 x2 + 2

2 ( 2 x − 3) ( x + 3)( x − 1) ⋅ 2 x ( 2 x − 3) ( x − 1) 2 ( x + 3) x ( x − 1)

, x ≠ −3,

3 2

1 , x ≠ ±2 x2

( x − 1) ( x + 2 ) + ( x − 1) + ( x + 2 ) 1 1 + = 109. x − 1 + x + 2 x −1 ( x + 2 )( x − 1) 2

( x − 2 x + 1) ( x + 2 ) + 2 x + 1 2

( x + 2 )( x − 1)

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

( x + 2 )( x − 1)

110. 2 x + = =

111.

x − x+3

( x + 2 )( x − 1)

3 1 − 2 ( x − 4) 2 ( x + 2)

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

(

2 ( x − 4 )( x + 2 )

)

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

4 x 3 − 8 x 2 − 32 x + 2 x + 10 2 ( x − 4 )( x + 2 )

4 x 3 − 8 x 2 − 30 x + 10 2 ( x − 4 )( x + 2 )

2 x 3 − 4 x 2 − 15 x + 5 ( x − 4 )( x + 2 )

(

)

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

(

)

(

)

(

)

Chapter P

Prerequisites

112.

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

(

)

113.

(

x + x +1+ x −1− x + x

( x − 1) ( x 2 + x + 1)

1 1 − −1 y−x 1 x y = ⋅ = ,x≠y xy ( x − y )( x + y ) xy ( x + y ) x 2 − y2

)

( 2 x + 3) − ( 2 x − 3) 1 1 − ( 2 x − 3)( 2 x + 3) 114. 2 x − 3 2 x + 3 = 1 1 ( 2 x + 3) − 2 x − 2x 2x + 3 2 x ( 2 x + 3) 6

(

)

115. x3 2 x 2 + 1

−4

+ x( 2 x 2 + 1)

−3

= =

x(3 x 2 + 1)

(2 x2 + 1)

5 x1 2 − 2 x5 2 x3

−4

(− 52 , 10) 8

x1 2 (5 − 2 x 2 )

4 2

−6

(8, − 3)

−2

x 2

−2

120.

−2 −4

−4

Quadrant II

x 4

4x 3 , x≠− , 0 2x − 3 2

119.

2 2

−2

2 x ( 2 x + 3) 6 ⋅ − + x x 2 3 2 3 3 ( )( )

(3x2 + 1)

5 − 2x2 = x5 2

117.

−4

= x 2 (5 x − 3 2 ) − 2 x 3 ( x −1 2 )

3 2 x ( 2 x + 3)

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

116.

( 2 x − 3)( 2 x + 3)

5 4

−6

−8

− 10

1 −1 −1

Quadrant IV

(6.5, 0.5) x 1

−2

118.

−9

−6

−3 x

−3

Quadrant I

−3

−6

(− 4, − 9)

−9

Quadrant III

125. ( 5.6, 0 ) , ( 0, 4.2 )

1,400

(a)

1,300 1,200

1,100

1,000

900

(0, 4.2)

2013

2012

2011

2010

2009

3 2008

121.

Revenue (in millions of US dollars)

Chapter P Review

−1 −1

122. The revenues increased from 2008 to 2013.

(b) d =

123. ( −3, 8 ) , (1, 5)

(a)

(− 3, 8)

(5.6, 0)

Year

( 0 − 5.6 ) + ( 4.2 − 0 ) 2

( −5.6 ) + ( 4.2 ) 2

= 31.36 + 17.64

= 49 = 7 (1, 5)

 5.6 + 0 0 + 4.2  (c) Midpoint:  ,  = ( 2.8, 2.1) 2   2

4 2

−4

(b) d =

126. ( 3.8, 2.6 ) , ( −1.2, − 9.4 )

−2

(1 − ( −3)) + ( 5 − 8) 2

(a) 2

= 42 + ( −3) = 16 + 9 2

−8 −6 −4 −2 −2

124. ( −12, 5) , ( 4, − 7)

−12

(4, − 7)

−6 −9 − 12

( 4 − ( −12 )) + ( −7 − 5) 2

= 16 + ( −12 ) 2

( −5) + ( −12 ) 2

 3.8 + ( −1.2 ) 2.6 + ( −9.4 )  , (c) Midpoint:     2 2    2.6 −6.8  , =  = (1.3, − 3.4 ) 2   2

−9 −6

( −1.2 − 3.8 ) + ( −9.4 − 2.6 )

= 25 + 144 = 169 = 13

(−1.2, −9.4)

(b) d =

(b)

−6

13   −3 + 1 8 + 5   (c) Midpoint:  ,  =  −1,  2   2   2

(− 12, 5)

x 4

−4

= 25 = 5

(a)

(3.8, 2.6)

= 256 + 144 = 400 = 20  −12 + 4 5 + ( −7 )  (c) Midpoint:  ,  = ( −4, − 1)   2 2  

127. Radius:

( 3 − ( −5) ) + ( −1 − 1) = 64 + 4 = 68 2

Circle: ( x − 3) + ( y + 1) = 68 2

 −4 + 10 6 − 2  128. Center:  ,  = ( 3, 2 ) 2   2 Radius: 2 2 2 1 1 (10 + 4) + ( −2 − 6) = 2 142 + ( −8) = 65 2

Circle: ( x − 3) + ( y − 2 ) = 65 2

Prerequisites

( 4 − 2, 8 − 3) = ( 2, 5) ( 6 − 2, 8 − 3) = ( 4, 5) ( 4 − 2, 3 − 3) = ( 2, 0 ) ( 6 − 2, 3 − 3) = ( 4, 0 )

130. The x-coordinates are increased by 4, and the y-coordinates are increased by 5. Original vertices Shifted vertices

( 0, 1) ( 3, 3) ( 0, 5) ( −3, 3)

100

( 0 + 4, 1 + 5) = ( 4, 6 ) ( 3 + 4, 3 + 5) = ( 7, 8) ( 0 + 4, 5 + 5) = ( 4, 10 ) ( −3 + 4, 3 + 5) = (1, 8)

Max Min

60 40 20

Jan Feb Mar Apr May Jun

Month

134.

0.6 0.5 0.4 0.3 0.2 0.1 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013

( 4, 8) ( 6, 8) ( 4, 3) ( 6, 3)

133. Temperature (°F)

129. The x-coordinates are decreased by 2, and the y-coordinates are decreased by 3. Original vertices Shifted vertices

Retail price (in dollars)

Chapter P

Year

131.

60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100

Running Shoe Prices

The price increased from 2004 to 2008, dropped slightly in 2009 and 2010, then stayed constant from 2011 to 2013.

The price of $100 occurs with the greatest frequency (4). 0, 4 ) 4, 8 )

Tally |||| |||| | ||||

8, 12 ) |||| 12, 16 ) |||| 16, 20 ) || 20, 24 )

Number of players

132. Interval

10 8 6 4 2

24, 28 ) 28, 32 ) |

8 12 16 20 24 28 32

Average number of points per game

Chapter P Test 1. − 103 ≈ −3.3 and − −4 = −4, hence − 103 > − −4 . 2. d = −16 − 38 = −54 = 54 3. 5 ⋅ (1 − x ) ⋅ 2 = 5 ⋅ 2 ⋅ (1 − x )

−3

 32  23 8  2  4. (a)   =  2  = 6 = 3 729 3   2 

(b)

5 ⋅ 125 = 5 ⋅ 5 5 = 25

(c)

5.4 × 108 5.4 = × 105 = 1.8 × 105 3 × 103 3

(d)

(3 × 10 ) = 3 × (10 )

Commutative Property of Multiplication

= 27 × 1012 = 2.7 × 1013

Chapter P Test

( ) = 3z 4 ⋅ z = 12z

3z 2 2 z 3

5. (a)

(b)

(

15. 2 x 4 − 3 x 3 − 2 x 2 = x 2 2 x 2 − 3 x − 2

(u − 2) (u − 2) = (u − 2) = −4

−3

−1

(u − 2)

= x ( 2 x + 1)( x − 2 ) 7

16.

x y  3 3x   = −2 2 = 2 3 x y y   −2

(c)

x3 + 2 x2 − 4 x − 8 = x2 ( x + 2 ) − 4 ( x + 2 )

(

= ( x + 2) x2 − 4

= ( x + 2) ( x − 2) 2

9 z 8 z − 3 2 z = 9 z ⋅ 2 2 z − 3z 2 z = 15z 2 z

(b)

(

17. 8 x 3 − 64 = 8 x 3 − 8

−5 16 y + 10 y = −5 ⋅ 4 y + 10 y

(

= 8( x − 2) x2 + 2 x + 4

16 3 23 ⋅ 2 2 3 2 = = v5 v3v2 v v2

= −2 x 5 − x 4 + 3 x 3 + 3 Standard form

x except x = ± 4, because division by zero is

Degree: 5 Leading coefficient: −2

undefined.

( x + 3) − 3x + (8 − x ) = x + 3 − 3x − 8 + x 2

= 2 x2 − 3x − 5

( 2 x − 5 ) ( 4 x 2 + 6 ) = 8 x 3 + 12 x − 20 x 2 − 30

(b) The domain of the radical expression 7 − x is all real numbers x such that x ≤ 7, because the square root of a negative number is not a real number. 19. (a)

= 8 x 3 − 20 x 2 + 12 x − 30

10.

11.

8x 24 8x 24 + = − x −3 3− x x −3 x −3 8 x − 24 8 ( x − 3 ) = = x −3 x −3 = 8, x ≠ 3

16 3

= 3

8⋅2

16 1 2 2 3 ⋅ 2 21 3 2 2 3

= 4 ⋅ 22 3 = 4 3 4

(b)

6 1− 3

= =

2  4 2  − ÷ 2  x x +1  x −1 =

)

18. (a) The domain of the rational expression x + 3 x + 3 is all real numbers = x 2 − 16 x + 4)( x − 4) (

7. 3 − 2 x 5 + 3 x 3 − x 4

)

3 = 8  x 3 − ( 2 )   

= −10 y (c)

)

= ( x + 2 )( x + 2 )( x − 2 )

6. (a)

)

−7

6 1+ 3 ⋅ 1− 3 1+ 3

(

) = −3 1 + 3 ( ) 1− 3

6 1+ 3

x+2 + 2 ( x + 2) − 2

x+2 + 2 x

2 x −1 x −1 , x ≠ ±1 ⋅ = x 4 2x

20. Shaded region = ( area big triangle )

2 ( x + 1) − 2 x x ( x + 1)

(c)

( x − 1)( x + 1) ⋅

12.

( x + 5 )( x − 5 ) = x − ( 5 ) = x − 5

13.

( x − 2 ) = x 3 − 3 x 2 ( 2 ) + 3 x ( 2 2 ) − 23

1 x+2 − 2

= x 3 − 6 x 2 + 12 x − 8

x+2 + 2 x+2 + 2

− ( area small triangle )

⋅

1 2

( 3 x ) ( 3 x ) − 12 ( 2 x ) ( 23 3 x )

= 12 3 3 x 2 − 23 3 x 2 = ( 32 − 23 ) 3 x 2 = 65 3 x 2

14. ( x + y ) − z  ( x + y ) + z  = ( x + y ) − z 2 = x 2 + 2 xy + y 2 − z 2 2

Chapter P

21.

Prerequisites

y 6

(− 2, 5)

5 3 2 1

−2 −1

(6, 0) x 1

−2

 −2 + 6 5 + 0   5  , Midpoint:   =  2,  2   2  2

( −2 − 6 ) + ( 5 − 0 ) 2

= 64 + 25 = 89 ≈ 9.43 22. (a) The endpoints of a diameter ( − 3, 4) and (1, − 8)

shifted five units to the left are ( − 3 − 5, 4) = ( − 8, 4) and (1 − 5, − 8) = ( − 4, − 8). (b) Use the midpoint of the diameter to find the center.

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

( h, k ) = 

Use the distance from the center to an endpoint of a diameter to find the radius. 2

− 6 − ( − 8) + ( − 2 − 4)

d =

(2) + (− 6) 2

So, the equation of the circle is ( x + 6) + ( y + 2) = 40. 2

2012

2008

2004

2000

1996

1992

72 68 64 60 56 52 48 44 40 36 1988

Number of votes (in millions)

23.

Year

C H A P T E R 1 Functions and Their Graphs Section 1.1

Graphs of Equations ...................................................................... 54

Section 1.2

Lines in the Plane .......................................................................... 61

Section 1.3

Functions ....................................................................................... 73

Section 1.4

Graphs of Functions ...................................................................... 80

Section 1.5

Shifting, Reflecting, and Stretching Graphs .................................. 90

Section 1.6

Combinations of Functions............................................................ 97

Section 1.7

Inverse Functions ........................................................................ 107

Chapter 1 Review .............................................................................................. 120 Chapter 1 Test ................................................................................................... 133

C H A P T E R 1 Functions and Their Graphs Section 1.1 Graphs of Equations 1.

solution point

graph

(1.2, 3.2 ) : 3.2 = 4 − 1.2 − 2 ?

3.2 = 4 − −0.8

Three common approaches that can be used to solve problems mathematically are algebraic, graphical, and numerical.

3.2 = 4 − 0.8 ?

3.2 = 3.2 Yes, the point is on the graph

Steps sketching the graph of an equation by point-plotting are: 1.

If possible, rewrite the equation so that one of the variables is isolated on one side of the equation. Make a table of values showing several solution points. Plot these points on a rectangular coordinate system.

Connect the points with a smooth curve or line.

x+4

(a)

( 0, 2 ): 2 = 0 + 4

(b)

y = x −1 + 2 (a)

1 = 1+ 2 1 ≠ 3 No, the point is not on the graph. (b)

(b)

(3.2, 4.2) : 4.2 = (3.2) − 1 + 2 4.2 = 2.2 + 2 4.2 = 4.2 Yes, the point is on the graph.

2= 4 2=2 Yes, the point is on the graph.

(2, 1) : 1 = (2) − 1 + 2

9. 2 x − y − 3 = 0

(12, 4 ) : 4 = 12 + 4

(a)

(1, 2 ) : 2 (1) − 2 − 3 = 0 −3 ≠ 0

4 = 16

No, the point is not on the graph.

4=4 Yes, the point is on the graph. 6.

(b)

0=0 Yes, the point is on the graph.

y = 5− x

(a)

(1, 2 ) : 2 =

5 − (1)

2 = 4 2=2

10.

x 2 + y 2 = 20

(a)

( 5, 0 ) : 0 =

5 − ( 5)

0 =

0=0 Yes, the point is on the graph. 7.

y = 4− x−2

(a)

2 ?

( 3, − 2 ) : 32 + ( −2 ) = 20 9 + 4 = 20 13 ≠ 20 No, the point is not on the graph.

Yes, the point is on the graph. (b)

(1, − 1) : 2 (1) − ( −1) − 3 = 0

(b)

( −4, 2 ) : ( −4 ) + 22 = 20 2

16 + 4 = 20 20 = 20 Yes, the point is on the graph.

(1, 5) : 5 = 4 − 1 − 2 5 ≠ 4 −1 No, the point is not on the graph.

Section 1.1 11.

( 52 , 34 ) : 34 = ( 52 ) − 3( 52 ) + 2 ?

? 25 3 = − 15 + 2 4 4 2 3 4

−3

− 52

−1

(

Solution point ( − 3, − 1) − 52 , 0

3 4

Yes, the point is on the graph. (b)

(− 2, 8) : 8 = (− 2) − 3(− 2) + 2

10 8

8 ≠ 12 No, the point is not on the graph.

2 −6

y = 13 x3 − 2 x2 (a)

( 2, − ) : ( 2 ) − 2 ( 2 ) =? − 16 3

1 3

16 3

−2

8 3

−8 = − 24 ?

− 3 =− − 163 = −

16 3 16 3 16 3

−1

Solution point

( −27 ) − 2 ( 9 ) =? 9 ?

−9 − 18 = 9 −27 ≠ 9 No, the point is not on the graph.

−3 −2 −1

2 3 4 5 6 7

−2 −3

13. 3 x − 2 y = 2  y = 23 x − 1

16. 6 x − 2 y = −2 x 2  y = x 2 + 3 x

−2

2 3

−4

−1

1 2

( −2, − 4 ) ( 0, − 1) ( 23 , 0 ) (1, 12 ) ( 2, 2 )

Solution point

( 3, 3)

7 6 5 4

2 ?

( −3, 9 ) : 13 ( −3) − 2 ( −3) = 9 3

( −1, 3) ( 0, 0 ) (1, − 1) ( 2, 0 )

Yes, the point is on the graph.

1 3

−4

15. 2 x + y = x 2  y = x 2 − 2 x

⋅ 8 − 2 ⋅ 4 = − 163 8 3

(b)

) ( 0, 5) (1, 7) ( 2, 9)

8 = 4+6+ 2

12.

14. − 4 x + 2 y = 10  y = 2 x + 5

y = x 2 − 3x + 2

(a)

Graphs of Equations

−4

−3

−2

Solution point

( −4, 4 ) ( −3, 0 ) ( −2, − 2 ) ( 0, 0 ) (1, 4 ) y

5 5 4 3 2 1 − 5 − 4 −3 −2 − 1

4 3 2 1 x 1 2 3 4 5

−5 −4

−2 −1

x 1

−3

−4 −5

17. y = 2 x has one intercept ( 0, 0 ) .

Matches graph (b). 18.

y = 4 − x 2 has intercepts ( 0, 4 ) , ( 2, 0 ) , and ( −2, 0 ) .

Matches graph (d).

Chapter 1

Functions and Their Graphs

19.

y = 9 − x 2 has intercepts ( 0, 3 ) , ( − 3, 0 ) , and ( 3, 0 ) .

26. y = 1 − x

Matches graph (c). 20.

y = x − 3 has intercepts ( 0, − 3 ) , ( 3, 0 ) , and ( −3, 0 ) .

5 4

Matches graph (a). 21.

(

y = x 3 − 3 x has intercepts −

( 3, 0).

)

3, 0 , (0, 0), and

−4

−3

−2

−1

Matches graph (e). 22.

(

x = 5 − y 2 has intercepts (5, 0), 0,

(0, − 5 ).

)

27.

5 , and

x −1

y = x−2 y

Matches graph (f ).

4 3

23.

y=2−x

2 1

y −1

x 1

−1

28.

y=4− x

1 −3

−2

−1

y x 1

−1

−2

3 2

24.

y = x −3

−4 − 3 −2 −1 −1

−2 −3

3 2 1 x

−5 −4 −3 − 2 −1

29.

x = y2 − 1

2 3 4 5 y

−2 3 2

−6 −7

25.

−2

y = x −3

−2

−3

30.

x = y2 + 4

4 1

3 x 1

−1

2 1 −1 −1

x 1

−2 −3 −4

Section 1.1 31.

y = 5 − 32 x

Intercepts:

37.

(

y = x x+3

Intercepts: ( 0, 0 ) , ( −3, 0 )

) (0, 5)

10 , 0 , 3

− 10

32. y =

Graphs of Equations

2 x −1 3

38.

y = (6 − x ) x

Intercepts: ( 0, 0 ) , ( 6, 0 )

3  Intercepts: ( 0, − 1) ,  , 0  2 

10 − 10 − 10

33.

39.

y = 3 x −8

Intercepts: ( 8, 0 ) , ( 0, − 2 )

y = x + 2 −3

Intercepts: ( − 5, 0), (1, 0), (0, −1)

10 − 10 − 10

34.

40.

y = 3 x +1

Intercepts: ( −1, 0 ) , ( 0, 1)

y = − x −3 +1

Intercepts: ( 2, 0), ( 4, 0), (0, − 2)

10 − 10 − 10

35.

41.

2x x −1

y = x2 − 4 x + 3

Intercepts: ( 3, 0 ) , (1, 0 ) , ( 0, 3 )

Intercept: ( 0, 0 )

10 − 10 − 10

10 36. y = 2 x + 2 Intercept: ( 0, 5 )

42.

Intercepts: ( 2, 0 ) , ( −4, 0 ) , ( 0, − 4 ) 10

− 10

x2 + 2 x − 8 2

− 10

Chapter 1

Functions and Their Graphs

43.

y = x2 ( x − 4) + 4x

49.

= x3 − 4 x2 + 4 x

(

Graphing these equations with a graphing utility shows that the graphs are identical. The Associative Property of Multiplication is illustrated.

50.

y =1− x

51.

− 10

y = 5− x 4

− 10

−3

y = −10 x + 50 (a) (b) Xmin = −10 Xmax = 10

52.

y = x 2 ( x − 3)

−6

Ymax = 100 Yscl = 25

(a)

y = x + 2 −1

(b)

Range Window 53.

Xmin = −5 Xmax = 1

(b)

−4

( − 1, y ) = ( − 1, − 4 ) ( x, 6 ) ≈ ( 3.49, 6 )

( −0.5, y ) ≈ ( −0.5, 2.47 ) ( x, − 2) ≈ ( − 1.58, − 2) , ( 0.40, − 2) , (1.37, − 2) 6

Ymax = 1 Yscl = 1

(

y = x5 − 5x

(a)

Xscl = 1 Ymin = −3

y1 = 14 x 2 − 8

( 3, y ) ≈ ( 3, 1.41) ( x, 3) = ( −4, 3) 4

Xscl = 2 Ymin = −50

47.

−2

Range Window

46.

Graphing these equations with a graphing utility shows that their graphs are identical. The Multiplicative Inverse Property is illustrated.

Intercepts: ( 0, 1) , (1, 0 )

45.

) x 1+ 3

(

y1 = x 2 + 3 ⋅ y2 = 1

− 10

44.

)

y2 = 2 x − 1

Intercepts: ( 0, 0 ) , ( 2, 0 )

− 10

(

y1 = 15 10 x 2 − 1   

−9

)

−6

y 2 = 14 x 2 − 2 Graphing these equations with a graphing utility shows that the graphs are identical. The Distributive Property is illustrated. 48.

y1 = 12 x + ( x + 1) y2 = 23 x + 1

54.

y = x2 − 6 x + 5 (a) (b)

( 2, y ) ≈ ( 2, 3 ) ( x, 1.5) ≈ ( 0.65, 1.5) , (1.42, 1.5 ) ( 4.58, 1.5) , ( 5.35, 1.5) 8

Graphing these equations with a graphing utility shows that their graphs are identical. The Associative Property of Addition is illustrated. −3

Section 1.1 55.

x 2 + y 2 = 16

59.

y 2 = 16 − x 2

2 ?

(1, 3) : (1 − 1) + ( 3 − 2 ) = 25 2

1 ≠ 25

Use y1 = 16 − x 2 and y2 = − 16 − x 2 .

(b)

−9

2 ?

( −2, 6 ) : ( −2 − 1) + ( 6 − 2 ) = 25 2

( −3) + ( 4 ) =? 25 2

25 = 25 Yes

−6

56.

x 2 + y 2 = 36 2

( x − 1) + ( y − 2 ) = 25 (a)

y = ± 16 − x 2

Graphs of Equations

y = 36 − x

(c)

( 5, − 1) : ( 5 − 1) + ( −1 − 2 ) =? 25 2

( 4 ) + ( −3) =? 25 2

y = ± 36 − x 2

25 = 25 Yes

Use y1 = 36 − x 2 and y2 = − 36 − x 2 . 8

(d) − 12

( 0, 2 + 2 6 ) : ( 0 −1) + ( 2 + 2 6 − 2) =? 25 ( −1) + ( 2 6 ) =? 25 2

25 = 25

−8

Yes

57.

60.

( x − 1) + ( y − 2 ) = 49 2 2 ( y − 2 ) = 49 − ( x − 1) 2

y − 2 = ± 49 − ( x − 1)

( x + 2 ) + ( y − 3 ) = 25 2

(a) 2

y = 2 ± 49 − ( x − 1)

( −2, 3 ) : ( −2 + 2 ) + ( 3 − 3 ) = 0 ≠ 25 No 2

(b) ( 0, 0 ) : ( 0 + 2 ) + ( 0 − 3) = 4 + 9 2

= 13 ≠ 25 No

Use y1 = 2 + 49 − ( x − 1) and

(c)

(1, − 1) : (1 + 2) + ( −1 − 3) = 9 + 16 = 25 Yes

y2 = 2 − 49 − ( x − 1) .

(d)

( −1, 3 − 2 6 ) : ( −1 + 2 ) + (3 − 2 6 − 3)

= 1 + 24 = 25 Yes

− 15

61. (a)

500,000

− 10

58.

( x − 3 ) + ( y − 1) = 25 2 2 ( y − 1) = 25 − ( x − 3) 2

y − 1 = ± 25 − ( x − 3 )

Use y1 = 1 + 25 − ( x − 3 )

y2 = 1 − 25 − ( x − 3 ) . 2

and

Algebraically, y = 500,000 − 47,000t = 500,000 − 47,000(5.8)

= 227,400. (c) Using the zoom and trace features, when y = 156,000, t ≈ 7.3. Algebraically,

(b) Using the value feature, when t = 5.8, y = 227,400.

y = 1 ± 25 − ( x − 3 )

y = 500,000 − 47,000t 156,000 = 500,000 − 47,000t

−6

7.3 ≈ t.

−5

Chapter 1

62. (a)

Functions and Their Graphs

9000

(b) When t = 0, y =

The y-intercept is 63.6, which represents the life expectancy in 1940.

(b) Using the zoom and trace features, when y = 5545.25, t ≈ 3.93.

63.6 + 0.97t 1 + 0.01t 63.6 + 0.97t 70.1 = 1 + 0.01t 70.1 + 0.701t = 63.6 + 0.97t

y =

y = 8250 − 689t

Algebraically,

5545.25 = 8250 − 689t 3.93 ≈ t.

6.5 = 0.269t

Algebraically, y = 8250 − 689t

24.2 ≈ t

= 8250 − 689(5.5)

So, in the year 1964, the life expectancy was 70.1.

= 4460.50. 63. (a)

Year New houses (in thousands) Year New houses (in thousands)

2006

2007

2008

2009

410.5

290.9

198.1

132.1

2010

2011

2012

2013

93.0

80.7

95.3

136.7

The model fits the data well. (b)

63.6 + 0.97(0) = 63.6. 1 + 0.01(0)

(d) Graphically, when t = 38, y = 72.8. Algebraically, 63.6 + 0.97t p y = 1 + 0.01t 63.6 + 0.97(38) = 1 + 0.01(38) = 72.8.

So, in the year 1978, the life expectancy was 72.8. 65. False. y = x 2 − 1 has two x-intercepts, (1, 0 ) and

( −1, 0 ) . Also, y = x 2 + 1 has no x-intercepts.

420

66. False. The line y = 0 has an infinite number of x-intercepts. 6

The model fits the data well. (c) In 2015, t = 15. y = 13.42(15) − 294.1(15) + 1692 2

= 30  $300,000

67. Option 1: w1 = 3000 + 0.07 x

Option 2: w2 = 3400 + 0.05 x (x is amount of sales) w1 = w2 3000 + 0.07 x = 3400 + 0.05 x 0.02 x = 400 x = 20,000

In 2017, t = 17. y = 13.42(17) − 294.1(17) + 1692 2

= 570.68  $570,680

Yes, the answers seem reasonable. Answers will vary.

If sales equal $20,000, the options are equivalent. For sales less than $20,000, choose option 2. For sales greater than $20,000, choose option 1.

(d) Using the zoom and trace features, there were 100,000 new houses during the years 2009 and 2012.

y 8000

64. (a)

100

6000

y = 3400 + 0.05x

4000 2000 0

The model fits the data well.

y = 3000 + 0.07x x 20,000

40,000

Section 1.2 68. (a) Xmin = −9 Xmax = 9

Lines in the Plane

(− 4, 1): 1 = (− 4) − 1 − 4

(c)

Xscl = 1 Ymin = −6

1 = 25 − 4 1 ≠ 21

Ymax = 6 Yscl = 1

No, the point is not on the graph. ?

( 2, − 3): − 3 = ( 2) − 1 − 4

(b) x-intercepts: ( −1, 0), (3, 0)

−3 = 1 − 4

y-intercept: (0, − 3)

−3 = −3 Yes, the point is on the graph. 69.

( 9 x − 4 ) + ( 2 x 2 − x + 15) = 2 x 2 + 8 x + 11

70.

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

= −3 x 4 + 8 x 2 − 5

Section 1.2 Lines in the Plane (a) iii

slope

parallel

They are perpendicular to each other.

Since x = 3 is a vertical line, all horizontal lines are perpendicular and have slope m = 0.

(b) i

(d) ii

(e) iv

11.

m=2 8

1 ( x − 8) is in point-slope 4 form, the point (8, −1) lies on the line.

Since the line y − ( −1) =

m=1

m = −3

m=0

(2, 3)

x 2

12.

m=4

m = −2

m=1

2 (a) m = . Since the slope is positive, the line rises. 3 Matches L2 .

(a) m = 0. The line is horizontal. Matches L2 . 3 (b) m = − . Because the slope is negative, the line 4 falls. Matches L1. (c) m = 1. Because the slope is positive, the line rises. Matches L3 .

Slope =

(−4, 1) −6

−2 −2 −4

13. Slope =

0 − ( −10) 10 5 = =− −4 − 0 −4 2 4

−12

(−4, 0)

(0, −10)

rise 3 = run 2

10. The line appears to go through (0, 8) and (2, 0). 8−0 = −4 Slope = 0−2

m is undefined.

(b) m is undefined. The line is vertical. Matches L3 . (c) m = −2. The line falls. Matches L1. 8.

−12

14. Slope =

−4 − 4 = −4 4−2 6

(2, 4) −6

(4, − 4) −6

Chapter 1

15. Slope =

Functions and Their Graphs

4−1 3 = ; slope is undefined. − 6 − ( − 6) 0

26. m = − 3, ( − 3, 6) y − 6 = − 3( x + 3) y − 6 = −3x − 9

y = −3x − 3

(−6, 4)

y −10

(−6, −1)

(−3, 6)

−2

16. Slope =

12 − 9 3 = 6−4 2

−6

−4

−2

−4

−6

(6, 12) (4, 9)

−5

16 −1

17. Since m = 0, y does not change. Three additional points are (0, 1), (3, 1), and (−1, 1). 18. Since m = 0, y does not change. Three additional points are (0, − 2), (1, − 2), and (4, − 2). 19. Since m is undefined, x does not change and the line is vertical. Three additional points are (1, 1), (1, 2), and (1, 3).

1 27. m = − , (2, − 3) 2 1 y − ( −3) = − ( x − 2) 2 1 y + 3 = − x +1 2 1 y =− x−2 2 y 1 −2

20. Because m is undefined, x does not change. Three additional points are (−4, 0), (−4, 3), and (−4, 5).

1 , y increases 1 for every increase of 2 units 2 in x. Three additional points are (9, −1), (11, 0), and (13, 1).

23. Since m =

1 24. Since m = − , y decreases 1 for every increase of 3 3 units in x. Three additional points are ( 2, − 7), (5, − 8), and (8, − 9).

x 1

−1

(2, − 3)

−3

21. Since m = −2, y decreases 2 for every unit increase in x. Three additional points are (1, − 11), (2, − 13), and (3, − 15). 22. Since m = 4, y increases 4 for every unit increase in x. Three additional points are ( − 4, 8), ( − 3, 12), and (− 2, 16).

−1

−4 −5

28. m =

3 , ( −2, − 5) 4 3 y + 5 = ( x + 2) 4 3 3 y −5 = x + 4 2 3 7 y = x − 4 2 y x

−2

2 −2

25. m = 3, (0, − 2) y + 2 = 3( x − 0) y = 3x − 2  3x − y − 2 = 0

(−2, −5)

y 2 1 −2

−1

x 1

−1 −2

(0, −2)

Section 1.2 29. m is undefined, (6, − 1) x=6

m =

y 6

1.7 − 1.5 0.2 1 = = 13 − 7 6 30 1 ( x − 7) 30 1 7 y − 1.5 = x − 30 30 1 19 y = x + 30 15 y − 1.5 =

4 2 2

(6, −1)

−2

−4 −6

When x = 19: y =

30. m is undefined, (−10, 4) x = 10 vertical line

m = 8

(− 10, 4)

−8

555,000 − 348,000 207,000 = = 23,000 13 − 4 9

y − 348,000 = 23,000( x − 4)

4 x

−4

1 19 = $1.9 million (19) + 30 15

34. Begin by letting x = 4 correspond to 2004. Then using the points ( 4, 348,000) and (13, 555,000), you have

−12

33. Begin by letting x = 7 correspond to 2007. Then using the points (7, 1.5) and (13, 1.7), you have

vertical line

−4 −2

Lines in the Plane

y − 348,000 = 23,000 x − 92,000

y = 23,000 x + 256,000

−4

When x = 19:

−8

y = 23,000(19) + 256,000 = $693,000 y

 1 3 31. m = 0,  − ,   2 2

( −3

−1

x 1

−1

2 x −3 3

2 3 y-intercept: (0, − 3)

Slope:

horizontal increase of 3 units.

4 2

−10 −12 −14 −16

y =

The line passes through (0, − 3) and rises 2 units for each

y − (−8.5) = 0( x − 2.3) y + 8.5 = 0 y = − 8.5 horizontal line

(2.3, −8.5)

−2

32. m = 0, (2.3, − 8.5)

−4 −6

(

− 1, 3 2 2 2 1

3 y− =0 2 3 horizontal line y= 2

2 4 6 8 10

− 3y = − 2x + 9

3 1  y − = 0 x +  2 2 

−8 −6 −4 −2

35. 2 x − 3 y = 9

36. 3 x + 4 y = 1 4 y = −3 x + 1 1 −3 y= x+ 4 4 Slope: −

3 4

 1 y-intercept:  0,   4  1 The line passes through  0,  and falls 3 units for each  4 horizontal increase of 4 units.

Chapter 1

Functions and Their Graphs

37. 2 x − 5 y + 10 = 0 −5 y = −2 x − 10 2 y= x+2 5 Slope:

2 5

43. 5 x − y + 3 = 0 y = 5x + 3

(a) Slope: m = 5 y-intercept: (0, 3) y

(b) 5

y-intercept: (0, 2)

4 3

The line passes through (0, 2) and rises 2 units for each horizontal increase of 5 units. 38. 4 x − 3 y − 9 = 0 −3 y = − 4 x + 9 4 y = x −3 3 Slope:

4 3

y-intercept: (0, − 3) The line passes through (0, − 3) and rises 4 units for each horizontal increase of 3 units. 39.

x = −6 Slope is undefined; no y-intercept.

−4

(b) 5 4

(0, 3) 2 1 x

−1

45. 5 x − 2 = 0 2 x= 5

(a) Slope: undefined No y-intercept y

(b)

2  The line is horizontal and passes through  0, −  . 3  42. 2 x − 5 = 0 2x = 5 5 x= 2

2 3 y-intercept: (0, 3)

The line is horizontal and passes through (0, 12).

2  y-intercept:  0, −  3 

−1

(a) Slope: m = −

40. y = 12 Slope: 0 y-intercept: (0, 12)

Slope: 0

−2

44. 2 x + 3 y − 9 = 0 3 y = −2 x + 9 2 y =− x+3 3

The line is vertical and passes through ( − 6, 0).

41. 3 y + 2 = 0 3 y = −2 2 y=− 3

−3

(0, 3)

2 1 x

−1

−1 −2

Slope is undefined; no y-intercept.

5  The line is vertical and passes through  , 0  . 2 

Section 1.2 46. 3 x + 7 = 0

49. The slope is

7 x=− 3

(a) Slope: undefined

y = 2x − 5

2 1

−1

3 5 − 1 2 2 50. The slope is = =− . 4 − (−1) 5 2 1 y − ( −1) = − ( x − 4) 2 1 y +1= − x + 2 2 1 y = − x +1 2

−1 −

−3

x 1

−1 −2 −3

47. 3 y + 5 = 0

51. (5, − 1), (−5, 5) 5+1 y +1 = ( x − 5) −5 − 5 3 y = − ( x − 5) − 1 5 3 y=− x+2 5

5 y=− 3

(a) Slope: m = 0 5  y-intercept:  0, −  3  y

(b)

−3 − ( −7) 4 = = 2. 1 − ( −1) 2 y − (−3) = 2( x − 1) y + 3 = 2x − 2

No y-intercept (b)

Lines in the Plane

−2

−1

−2

−1

(0, − 53 )

−2

52. (4, 3), (−4, − 4) −4 − 3 y−3= ( x − 4) −4 − 4 7 y − 3 = ( x − 4) 8 7 1 y= x− 8 2

−3

48. −11 − 4 y = 0 − 4 y = 11 11 4 (a) Slope: m = 0 y = −

11   y-intercept:  0, −  4  (b)

−6

y −4

2 1

−3

−2

−1

x 1

−1 −2 −3 −4

)0, − 114)

53. (−8, 1), (−8, 7) Since both points have an x-coordinate of –8, the slope is undefined and the line is vertical. x +8 = 0 4

− 10

−4

Chapter 1

54.

(−1, 6), (5, 6)

Functions and Their Graphs

6−6 y −6 = ( x − (−1)) 5 − ( −1)

y − 6 = 0( x + 1) y −6 = 0 y = 6 7

3  9 9  1 57.  − , −  ,  , −  5  10 5   10

9 3 − + 3 1   y+ = 5 5 x +  9 1  5 10  + 10 10 3 6 1  y+ = − x+  5 5 10  y=−

−6

6 18 x− 5 25 6

−1

 1 1 5 55.  2,  ,  ,   2 2 4 5 1 − 1 4 2 y− = ( x − 2) 2 1 −2 2 1 1 y = − ( x − 2) + 2 2 1 3 y=− x+ 2 2 3

−2

−9

−6

3 3  4 7 58.  ,  ,  − ,  4 2  3 4 7 3 − 3 3  y − = 4 2 x −  2 −4 − 3  4 3 4 3 3  3 y− = − x−  2 25  4 y−

−1

56.

3 3 9 =− x+ 2 25 100 3 159 y=− x+ 25 100

(1, 1) ,  6, − 3  2





2 −1 y −1 = 3 ( x − 1) 6 −1 1 y − 1 = − ( x − 1) 3 1 1 y −1 = − x + 3 3 1 4 y=− x+ 3 3

−

−3

3 −1

59. (1, 0.6), (−2, − 0.6) −0.6 − 0.6 ( x − 1) −2 − 1 y = 0.4( x − 1) + 0.6 y = 0.4 x + 0.2

y − 0.6 =

5 2

−6

−3

−3 −2

Section 1.2 60. (−8, 0.6), (2, − 2.4)

Lines in the Plane

63. L1 : (0, − 1), (5, 9)

−2.4 − 0.6 ( x + 8) 2 − (−8) y − 0.6 = − 0.3( x + 8) y = − 0.3 x − 1.8 y − 0.6 =

9 +1 =2 5−0 L2 : (0, 3), (4, 1)

m1 =

1− 3 1 1 =− =− 4−0 2 m1

m2 =

L1 and L2 are perpendicular. −6

64. L1 : (−2, − 1), (1, 5)

L2 : (1, 3), (5, − 5)

61.

5 − ( −1) 6 = =2 1 − ( −2) 3

m1 =

−5

−1

−5 − 3 −8 = = −2 5 −1 4

m2 =

The lines are neither parallel nor perpendicular. −5

65. L1 : (3, 6), (−6, 0) 4

0−6 2 = −6 − 3 3

m1 =

−5

 7 L2: (0, − 1),  5,   3

7 +1 2 m2 = 3 = = m1 5−0 3

−3 5

−5

L1 and L2 are parallel. 66. L1 : (4, 8), (−4, 2)

−5

The first graph does not show both intercepts. The third graph is best because it shows both intercepts and gives the most accurate view of the slope by using a square setting. 62.

2−8 −6 3 = = − 4 − 4 −8 4 1  L2 : (3, − 5),  −1,   3

m1 =

m2:

(1/ 3) − (−5) 16 / 3 4 = =− −1 − 3 −4 3

The lines are perpendicular. −5

y = 2x −

− 10

(a) Parallel slope: m = 2

y − 1 = 2( x − 2) y = 2x − 3

− 80 10

−5

3 2

Slope: m = 2

−5

67. 4 x − 2 y = 3

(b) Perpendicular slope: m = −

13 −2

The second graph does not give a good view of the intercepts. The third graph is best because it gives the most accurate view of the slope by using a square setting.

1 2

1 y − 1 = − ( x − 2) 2 1 y=− x+2 2

Chapter 1

Functions and Their Graphs

68.

x+y=7 y = −x + 7 Slope: m = −1

71. 6 x + 5 y = 9 5 y = − 6x + 9 6 9 y = − x + 5 5 6 Slope: m = − 5

(a) Parallel slope: m = −1 y − 2 = −1( x + 3) y = −x − 1 (b) Perpendicular slope: m = 1 y − 2 = 1( x + 3) y= x+5

(a) Parallel slope: m = −

6 ( x + 3.9) 5 6 y + 1.4 = − x − 4.68 5 6 y = − x − 6.08 5 y + 1.4 = −

69. 3 x + 4 y = 7 3 7 x+ 4 4 3 Slope: m = − 4 y=−

(a) Parallel slope: m = −

(b) Perpendicular slope: m =

3 4

3 3 x+ 4 8 4 (b) Perpendicular slope: m = 3

5 ( x + 3.9) 6 5 y + 1.4 = x + 3.25 6 5 y = x + 1.85 6

y=−

7 4 2 = x+  8 3 3 y=

72. 5 x + 4 y = 1 y=−

4 127 x+ 3 72

3 2

(b) Perpendicular slope: m = −

5 4

5 y − 2.4 = − ( x + 1.2) 4 y = −1.25 x + 0.9

(b) Perpendicular slope: m = 0.8

y − 2.4 = 0.8( x + 1.2) y = 0.8 x + 3.36

3 8 x− 2 5

2 2 y +1 = −  x −  3 5 2 11 y=− x− 3 15

5 = −1.25 4

(a) Parallel slope: m = −

3 2 y +1 =  x −  2 5 y=

5 1 x+ 4 4

Slope: m = −

70. 3 x − 2 y = 6 3 y = x −6 2 3 Slope: m = 2

(a) Parallel slope: m =

5 6

y + 1.4 =

7 3 2 y− = − x+  8 4 3

y−

6 5

73. 2 3

x − 4 = 0 vertical line Slope is undefined.

74.

(a)

x − 3 = 0 passes through (3, − 2) and is vertical.

(b)

y = −2 passes through (3, − 2) and is horizontal.

y−2 =0 y = 2 horizontal line Slope: m = 0 (a)

y = −1 passes through (3, −1) and is horizontal.

(b)

x − 3 = 0 passes through (3, −1) and is vertical.

Section 1.2 75.

y+2=0 y = −2 horizontal line

76.

2 2 x and y = x + 2 are parallel. Both are 3 3 3 perpendicular to y = − x. 2

y = 1 passes through (− 5, 1) and is horizontal. x + 5 = 0 passes through (− 5, 1) and is vertical.

x + 5 = 0 vertical line

x + 2 = 0 passes through (−2, 4) and is vertical. y = 4 passes through (−2, 4) and is horizontal.

77. The slope is 2 and (−1, −1) lies on the line. Hence,

1 1 83. The lines y = − x and y = − x + 3 are parallel. Both 2 2 are perpendicular to y = 2 x − 4. y = −1x + 3 2

y = 2x − 4

− 15

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

− 10

y = −1x 2

y = −2 x − 1. 1 79. The slope of the given line is 2. Then y2 has slope − . 2 Hence,

84. The lines y = x − 8 and y = x + 1 are parallel. Both are perpendicular to y = − x + 3. 10

y=x−8

y = −x + 3 − 15

1 y − 2 = − ( x − (−2)) 2 1 y − 2 = − ( x + 2) 2 1 y = − x + 1. 2

y=x+1 − 10

85.

1 80. The slope of the given line is 3. Then y2 has slope − . 3 Hence,

1 y − 5 = − ( x − (−3)) 3 1 y − 5 = − ( x + 3) 3 1 y = − x + 4. 3

rise 3 x = = 1 run 4 (32) 2 3 x = 4 16 4 x = 48 x = 12 The maximum height in the attic is 12 feet. rise run −12 −2000 = 100 x −12 x = ( −2000)(100)

86. Slope =

81. The lines y = − 4 x and y =

− 15

y = −3x 2

78. The slope is −2 and (−1, 1) lies on the line. Hence,

y = 2 x −6

y − ( −1) = 2( x − ( −1)) y + 1 = 2( x + 1) y = 2 x + 1.

y = −4x

y = 2x + 2

−9

Slope is undefined. (a) (b)

82. The lines y =

Slope: m = 0 (a) (b)

Lines in the Plane

y = 4x

1 x are perpendicular. 4

2 x = 16,666 ft ≈ 3.16 miles 3

y = 1x 4

−10

Chapter 1

87. (a)

Functions and Their Graphs

Years

Slope

2005–2006

24.088 − 23.104 = 0.984

2006–2007

28.857 − 24.088 = 4.769

2007–2008

31.944 − 28.857 = 3.087

2008–2009

30.990 − 31.944 = − 0.954

2009–2010

35.123 − 30.990 = 4.133

2010–2011

46.554 − 35.123 = 11.431

2011–2012

48.017 − 46.554 = 1.463

The greatest increase was $11.431 billion from 2010 to 2011. The greatest decrease was $954 million from 2008 to 2009.

y − 19.7 = 8.65 x − 60.55 y = 8.65 x − 40.85

(d) There was an average increase in profit of approximately $8.65 million per year from 2007 to 2013. (e) When x = 17, y = 8.65(17) − 40.85 = $106.2 million. Answers will vary. For Exercises 89–92, t = 15 corresponds to 2015. 89.

(b) Using the points (5, 23.104) and (12, 48.017), the 48.017 − 23.104 = 3.559. 12 − 5 Then y − 23.104 = 3.559( x − 5)

V − 2540 = 125t − 1875 V = 125t + 665

slope is m =

90.

y − 23.104 = 3.559 x − 17.795

(15, 156), m = 5.50 V − 156 = 5.50(t − 15) V − 156 = 5.5t − 82.5 V = 5.5t + 73.5

y = 3.559 x + 5.309

(15, 2540), m = 125 V − 2540 = 125(t − 15)

91.

(d) When x = 17: y = 3.559(17 ) + 5.309 y = $ 65.812 billion

(15, 20,400), m = − 2000 V − 20,400 = −2000(t − 15) V − 20,400 = −2000t + 30,000

Answers will vary.

V = −2000t + 50,400

88. (a)

92.

Sales

(15, 245,000), m = − 5600 V − 245,000 = −5600(t − 15) V − 245,000 = −5600t + 84,000

V = −5600t + 329,000

30 15 x 7

93. (a)

9 10 11 12 13

Year (7 ↔ 2007)

(b)

Years

Slope

2007–2008

24.4 − 19.7 = 4.7

2008–2009

30.7 − 24.4 = 6.3

2009–2010

38.4 − 30.7 = 7.7

2010–2011

50.4 − 38.4 = 12.0

2011–2012

57.3 − 50.4 = 6.9

2012–2013

71.6 − 57.3 = 14.3

The greatest increase was $14.3 million from 2012 to 2013. The least increase was $4.7 million from 2007 to 2008.

(b)

(0, 25,000), (10, 2000) 2000 − 25,000 V − 25,000 = (t − 0) 10 − 0 V − 25,000 = −2300t V = −2300t + 25,000 25,000

t 0 1 2 3 4 V 25,000 22,700 20,400 18,100 15,800 t V

6 11,200

7 8900

8 6600

9 4300

(c)

t = 0: V = −2300(0) + 25,000 = 25,000 t = 1: V = −2300(1) + 25,000 = 22,700 etc.

5 13,500

10 2000

Section 1.2 94. (a) Using the points (0, 32) and (100, 212), you have

(b)

R = 80t

(c)

(d)

9 90 = C + 32 5 9 58 = C 5 32.2 ≈ C

49.25t = 36,500 t ≈ 741.1 h

96. (a)

−17.8° 0°

47 − 50 ( p − 580) 625 − 580 −1 ( p − 580) x − 50 = 15 1 266 x =− p+ 15 3

(b)

9 (−10) + 32 5 F = −18 + 32 F=

1500

If p = 655, x = 45 units. Algebraically, x = −

1 266 (655) + = 45. 15 3

9 F = (177) + 32 5 F = 318.6 + 32 F = 350.6 10° 50°

100

9 68 = C + 32 5 9 36 = C 5 20 = C

−10° 14°

(580, 50), (625, 47) x − 50 =

F = 14

C F

P = 0: 49.25t − 36,500 = 0

9 (10) + 32 5 F = 18 + 32 F = 50

F = 90°:

C = 177°:

P = R −C P = 80t − (36,500 + 30.75t )

9 0 = C + 32 5 9 −32 = C 5 −17.8 ≈ C F=

F = 68°:

R = tp (t hours at $p per hour )

P = 49.25t − 36,500

C = 10°:

C = −10°:

C = 36,500 + 11.25t + 19.50t

R = t (80)

9 F = C + 32 5

F = 0°:

C = 36,500 + 30.75t

212 − 32 180 9 = = 100 − 0 100 5 9 F − 32 = (C − 0) 5 9 F = C + 32. 5 m=

(b)

95. (a)

Lines in the Plane

20° 68°

1 266 (595) + = 49. 15 3

97. (a) Using the points (1994, 73,500) and ( 2013, 98,097),

the slope is 98,097 − 73,500 24,597 = ≈ 1295. 2013 − 1994 19 The average annual increase in enrollment was about 1295 students per year. m =

32.2° 90°

177° 350.6°

(b) 1996: 73,500 + 2(1,295) = 76,090 students

2006: 73,500 + 12(1,295) = 89,040 students 2011: 73,500 + 17(1,295) = 95,515 students

(c) Using m = 1295 and letting x = 4 correspond to 1994, y − 73,500 = 1295( x − 4) y − 73,500 = 1295 x − 5180 y = 1295 x + 68,320

The slope is 1295 and it determines the average increase in enrollment per year from 1994 to 2013.

Chapter 1

Functions and Their Graphs

98. Answers will vary. Sample answer: Slope is the rate of change over an interval; average rate of change is the slope of the line passing through the first and last points of a plot.

103. −1

99. False. The slopes are different:

−2

4−2 2 = −1 + 8 7 7+4 11 =− −7 − 0 7

x y + =1 4 −2 3 −8 2 − x + 4y = 3 3 −2 x + 12 y = −8

100. False.

The equation of the line joining (10, − 3) and (2, − 9) is −9 + 3 ( x − 10) y+3= 2 − 10 3 y + 3 = ( x − 10) 4 3 21 y= x− . 4 2

For x = −12, y =

a and b are the x- and y-intercepts. 104.

−3 −1

3 21 ( −12) − 4 2

x y + =1 1 5 2 1 5 5x + y = 2 2 10 x + y = 5

= −19.5 −37 ≠ 2 = −18.5 101.

a and b are the x- and y-intercepts.

−3

105. −5

x y + = 1 7 −3

106.

− 3 x + 7 y + 21 = 0

a and b are the x- and y-intercepts. 102.

x y + =1 2 9 9 x + 2 y − 18 = 0 x y + =1 a b x y + =1 −5 − 4 4 x + 5 y + 20 = 0

107. −8

4 −2

x y + =1 −6 2

x  y = 2 1 +  6  x y= +2 3

a and b are the x- and y-intercepts.

108.

x y + =1 −1/ 6 −2 / 3 3 −6 x − y = 1 2 12 x + 3 y + 2 = 0 x y + =1 a b x y + =1 3/4 4/5 4 3 3 x+ y= 5 4 5 16 x + 15 y − 12 = 0

Section 1.3

Functions

109. The slope is positive and the y-intercept is positive. Matches (a).

110. The slope is negative and the y-intercept is negative. Matches (b).

(d) The y-intercept is 600 and the slope is m = −100,

111. Both lines have positive slope, but their y-intercepts differ in sign. Matches (c).

which represents the decrease in the value of the computer each year. Matches graph (iv). 117. Yes. x + 20

112. The lines intersect in the first quadrant at a point ( x, y ) where x < y. Matches (a).

118. Yes. 3 x − 10 x 2 + 1 = −10 x 2 + 3 x + 1

113. No. The line y = 2 does not have an x-intercept.

119. No. The term x −1 =

114. No. x = 1 cannot be written in slope-intercept form because the slope is undefined.

polynomial.

115. Yes. Once a parallel line is established to the given line, there are an infinite number of distances away from that line, and thus an infinite number of parallel lines. 116. (a) The slope is m = −10. This represents the decrease

in the amount of the loan each week. Matches graph (ii). (b) The y-intercept is 13.5 and the slope is m = 2, which represents the increase in hourly wage per unit produced. Matches graph (iii).

1 causes the expression to not be a x

120. Yes. 2 x 2 − 2 x 4 − x3 + 2 = −2 x 4 − x3 + 2 x 2 + 2 121. No. This expression is not defined for x = ± 3. 122. No. 123. x 2 − 6 x − 27 = ( x − 9)( x + 3) 124. x 2 + 11x + 28 = ( x + 4)( x + 7) 125. 2 x 2 + 11x − 40 = (2 x − 5)( x + 8) 126. 3 x 2 − 16 x + 5 = (3 x − 1)( x − 5) 127. Answers will vary.

Section 1.3 Functions 1.

domain, range, function

independent, dependent

No. The input element x = 3 cannot be assigned to more than exactly one output element.

To find g( x + 1) for g( x ) = 3 x − 2, substitute x with the quantity x + 1. g( x + 1) = 3( x + 1) − 2

= 3x + 3 − 2 = 3x + 1

No. The National Football Conference, an element in the domain, is assigned to three elements in the range, the Giants, the Saints, and the Seahawks; The American Football Conference, an element in the domain, is also assigned to three elements in the range, the Patriots, the Ravens, and the Steelers.

10. Yes. Each element, or state, in the domain is assigned to exactly one element, or electoral votes, in the range. 11. Yes, the table represents y as a function of x. Each domain value is matched with only one range value. 12. No, the table does not represent a function. The input values of 0 and 1 are each matched with two different output values.

No. The domain of the function f ( x ) = 1 + x is [ −1, ∞) which does not include x = −2.

The domain of a piece-wise function must be explicitly described, so that it can determine which equation is used to evaluate the function.

Yes. Each domain value is matched with only one range value.

14. Yes, the graph represents a function. Each input value is matched with one output value.

No. The domain value of −1 is matched with two output values.

15. (a) Each element of A is matched with exactly one element of B, so it does represent a function. (b) The element 1 in A is matched with two elements, −2 and 1 of B, so it does not represent a function. (c) Each element of A is matched with exactly one element of B, so it does represent a function.

13. No, the graph does not represent a function. The input values 1, 2, and 3 are each matched with two outputs.

Chapter 1

Functions and Their Graphs

16. (a) The element c in A is matched with two elements, 2 and 3 of B, so it is not a function. (b) Each element of A is matched with exactly one element of B, so it does represent a function. (c) This is not a function from A to B (it represents a function from B to A instead). 17. Both are functions. For each year there is exactly one and only one average price of a name brand prescription and average price of a generic prescription. 18. Since b(t ) represents the average price of a name brand prescription, b( 2009) ≈ $151. Since g (t ) represents the average price of a generic prescription, g ( 2006) ≈ $31. 19. x 2 + y 2 = 4  y = ± 4 − x 2

Thus, y is not a function of x. For instance, the values y = 2 and y = −2 both correspond to x = 0. 20. x = y 2 + 1 y = ± x −1

31. f (t ) = 3t + 1

(a) (b) (c)

f (2) = 3(2) + 1 = 7 f (−4) = 3(−4) + 1 = −11 f (t + 2) = 3(t + 2) + 1 = 3t + 7

32. g( y) = 7 − 3 y

(a) (b) (c)

g(0) = 7 − 3(0) = 7 7 7 g   = 7 − 3  = 0 3 3 g ( s + 5) = 7 − 3( s + 5) = 7 − 3s − 15 = − 3s − 8

33. h(t ) = t 2 − 2t

(a)

h ( 2 ) = 2 2 − 2( 2 ) = 0

(b)

h(1.5) = (1.5) − 2(1.5) = − 0.75

(c)

h( x − 4) = ( x − 4) − 2( x − 4)

= x 2 − 8 x + 16 − 2 x + 8

This is not a function of x. For example, the values y = 2 and y = −2 both correspond to x = 5. 21. y = x 2 − 1

This is a function of x. 22. y = x + 5

This is a function of x. 1 23. 2 x + 3 y = 4  y = (4 − 2 x ) 3 Thus, y is a function of x. 24. x = − y + 5  y = − x + 5

This is a function of x. 25. y 2 = x 2 − 1  y = ± x 2 − 1

Thus, y is not a function of x. For instance, the values y = 3 and y = − 3 both correspond to x = 2. 26. x + y = 3  y = ± 3 − x 2

Thus, y is not a function of x. 27. y = 4 − x

This is a function of x. 28.

y = 3 − 2 x  y = 3 − 2 x or y = − (3 − 2 x) Thus, y is not a function of x.

= x 2 − 10 x + 24 34. V (r ) =

4 3 πr 3 4 π (3)3 = 36π 3

(a)

V (3) =

(b)

4 27 9π 3 4 3 V = π  = ⋅ π = 3 8 2 2 3 2

(c)

V (2r ) =

4 32π r 3 π (2r )3 = 3 3

35. f ( y ) = 3 − y

(a)

f (4) = 3 − 4 = 1

(b)

f (0.25) = 3 − 0.25 = 2.5

(c)

f (4 x 2 ) = 3 − 4 x 2 = 3 − 2 x

36. f ( x ) = x + 8 + 2

(a)

f (−4) = −4 + 8 + 2 = 4

(b)

f (8) = 8 + 8 + 2 = 6

(c)

f ( x − 8) = x − 8 + 8 + 2 = x + 2

37. q( x ) =

(a)

29. x = −7 does not represent y as a function of x. All values of y correspond to x = −7.

(b)

30. y = 8 is a function of x, a constant function.

(c)

1 x2 − 9

1 1 1 = = undefined (−3)2 − 9 9 − 9 0 1 1 1 q(2) = = =− (2)2 − 9 4 − 9 5 1 1 1 q( y + 3) = = 2 = 2 2 ( y + 3) − 9 y + 6 y + 9 − 9 y + 6 y q(−3) =

Section 1.3

38. q(t ) =

2t 2 + 3 t2

44.

2(2)2 + 3 8 + 3 11 = = (2)2 4 4

(a)

q(2) =

(b)

2(0)2 + 3 q(0) = Division by zero is undefined. (0)2

(c)

q( − x ) =

39. f ( x ) =

(a)

2(− x )2 + 3 2 x 2 + 3 = ( − x )2 x2

45.

x x

41.

f (9) =

(b)

f (−9) =

(c)

f (t ) =

42.

x ≤ 0 x >0

(a)

f (−2) = ( −2)2 − 4 = 4 − 4 = 0

(b)

f (0) = 0 2 − 4 = −4

(c)

f (1) = 1 − 2(12 ) = 1 − 2 = −1

x + 2, x < 0  f ( x ) = 4, 0 ≤ x < 2 x 2 + 1, x ≥ 2 

(b) f (0) = 4

−9

= −1

 1, t > 0 = t −1, t < 0 t

46.

(a)

f (5) = 5 + 4 = 9

(b)

f ( −5) = −5 + 4 = 9

(c)

f (t ) = t + 4

2 x + 1, x < 0 f ( x) =  2x + 2, x ≥ 0 f ( −1) = 2( −1) + 1 = −1 f (0) = 2(0) + 2 = 2 f (2) = 2(2) + 2 = 6

2 x + 5, x ≤ 0 f ( x) =   2 − x, x > 0

(a) f ( − 2) = 2( − 2) + 5 = 1

(b) f (0) = 2(0) + 5 = 5

5 − 2 x, x < 0  f ( x ) = 5, 0 ≤ x <1 4 x + 1, x ≥ 1 

(a) f ( − 4) = 5 − 2( − 4) = 13

f ( x) = x + 4

(a) (b) (c)

(a) f ( − 2) = ( − 2) + 2 = 0

f (0) is undefined.

40.

2  x − 4, f ( x) =  2 1 − 2 x ,

Functions

(b) f (0) = 5 (c) f (1) = 4(1) + 1 = 5 47.

f ( x) = ( x − 1)

{(− 2, 9), (−1, 4), (0, 1), (1, 0), (2, 1)} 48.

f ( x) = x2 − 3

{(−2, 1), (−1, − 2), (0, − 3), (1, − 2), (2, 1)} 49.

f ( x) = x + 2

{(−2, 4), (−1, 3), (0, 2), (1, 3), (2, 4)} 50.

f ( x) = x + 1

{(−2, 1), (−1, 0), (0, 1), (1, 2), (2, 3)}

2  x + 2, x ≤ 1 f ( x) =  2  2x + 2, x > 1

(a) (b)

f ( −2) = ( −2)2 + 2 = 6 f (1) = (1)2 + 2 = 3

(c)

f (2) = 2(2)2 + 2 = 10

Chapter 1

51. h(t ) =

Functions and Their Graphs

1 t +3 2

55.

1 1 1 −5 + 3 = −2 = (2) = 1 2 2 2 1 1 1 1 h( −4) = −4 + 3 = −1 = (1) = 2 2 2 2 1 1 h(−3) = −3 + 3 = 0 = 0 2 2

h(−5) =

9x − 4 = 0 5 9x − 4 = 0 f ( x) =

9x = 4 x =

56.

f ( x) =

1 1 1 1 −2 + 3 = 1 = (1) = 2 2 2 2 1 1 1 h(−1) = −1 + 3 = 2 = (2) = 1 2 2 2

h(−2) =

4 9

2x − 3 =0 7 2x − 3 = 0

2x = 3 3 x= 2

−5

−4

−3

−2

−1

h( t )

1 2

57.

f ( x) = 5x 2 + 2 x − 1 Since f ( x ) is a polynomial, the domain is all real numbers x.

58. g( x ) = 1 − 2 x 2 52.

f (s) =

f (0) = =

f (1)

s−2

Because g ( x ) is a polynomial, the domain is all real numbers x.

s−2 0−2 0−2 1− 2

2 = −1 −2

1 = −1 −1

59. h(t ) =

Domain: all real numbers except t = 0

1− 2 3 1 −2 3 2 2 = = −1 f = 2 3 −2 −1 2 2 5 1 −2 5 2 = 2 =1 f = 1 2 5 −2 2 2 4−2 2 = =1 f (4) = 4−2 2

53.

Domain: all real numbers except y = −5 61.

62.

f (s )

−1

x=−

1 5

f ( x) = 3 x − 4 Domain: all real numbers x

f ( x) = 5x + 1 = 0 5 x = −1

3y y+5

y+5≠0 y ≠ −5

x=5 54.

60. s( y) =

3 2

f ( x ) = 15 − 3 x = 0 3 x = 15

4 t

f ( x) = 4 x2 + 3x x 2 + 3 x = x ( x + 3) ≥ 0

5 2

Domain: x ≤ −3 or x ≥ 0 63. g( x ) =

1 3 − x x+2

Domain: all real numbers except x = 0, x = −2 64.

10 x2 − 2 x x2 − 2 x ≠ 0 x( x − 2) ≠ 0 h( x ) =

Domain: all real numbers except x = 0, x = 2

Section 1.3 y+2

65. g( y ) =

71.

y − 10

C 2π 2

y > 10

C2  C  A=π  = 4π  2π 

Domain: all y > 10

x+6 66. f ( x) = 6+ x x + 6 ≥ 0 for numerator and x ≠ −6 for denominator.

67.

A = π r 2 , C = 2π r

y − 10 > 0

Functions

72.

1 bh, in an equilateral triangle b = s and: 2 2

Domain: all x > −6

s s 2 = h2 +   2

f ( x) =

s h = s2 −   2

16 − x 2 6

−9

4s 2 s 2 3s − = 4 4 2

1 3s s⋅ = 2 2

3s 2 4

−6

Domain: [− 4, 4] Range: [0, 4] 68.

f ( x) = x2 + 1 b=s

−9

73. (a) From the table, the maximum volume seems to be 1024 cm3, corresponding to x = 4.

(b)

−3

s 2

1200

Domain: all real numbers Range: 1 ≤ y 0

69. g( x ) = 2 x + 3

Yes, V is a function of x.

V = length × width × height = (24 − 2 x )(24 − 2 x ) x

= x(24 − 2 x )2 = 4 x(12 − x )2

−2

Domain: 0 < x < 12

Domain: ( −∞, ∞ )

(d)

1200

Range: [0, ∞ ) 70.

g ( x) = 3x − 5 0

The function is a good fit. Answers will vary. −4

8 −1

Domain: all real numbers Range: y ≥ 0

Chapter 1

74.

Functions and Their Graphs

1 1 (base)(height) = xy. 2 2

Since (0, y ), (2, 1), and ( x, 0) all lie on the same line, the slopes between any pair of points are equal. 1− y 1− 0 = 2−0 2− x 2 1− y = 2−x y =1−

(b)

2 x = 2− x x−2

The domain is x > 2, since A > 0. A = l ⋅ w = (2 x ) y = 2 xy

(c)

f (5) = 11.575, and represents the revenue in May: $11,575.

(d)

f (11) = 4.63, and represents the revenue in November: $4630.

(e) The values obtained from the model are close approximations to the actual data. 79. (a) The independent variable is t and represents the year. The dependent variable is n and represents the numbers of miles traveled. (b) t 0 1 2 3 4 5

But y = 36 − x 2 , so A = 2 x 36 − x 2 , 0 < x < 6. 76. (a)

V = (length)(width)(height) = yx 2 But, y + 4 x = 108, or y = 108 − 4 x. Thus, V = (108 − 4 x ) x 2 . Since y = 108 − 4 x > 0

n(t)

3.95

3.96

3.98

3.99

4.00

4.02

n(t)

4.03

4.04

4.05

4.07

4.08

4.09

4 x < 108 x < 27.

80. (a)

Domain: 0 < x < 27 (b)

7 ≤ x ≤ 12  −1.97 x + 26.3, f ( x) =  2 0.505 x − 1.47 x + 6.3, 1 ≤ x ≤ 6

Answers will vary.

1 1  x  x2 Therefore, A = xy = x  . = 2 2  x − 2  2x − 4

75.

78. (a) The independent variable is x and represents the month. The dependent variable is y and represents the monthly revenue.

F ( y) = 149.76 10 y 5 / 2

12,000

F ( y) 26, 474 149,760 847,170 2,334,527 4,792,320

(Answers will vary.) 0

(c) The highest point on the graph occurs at x = 18. The dimensions that maximize the volume are 18 × 18 × 36 inches. 77. (a)

Total cost = Variable costs + Fixed costs C = 68.75 x + 248,000

(b) Revenue = Selling price × Units sold

R = 99.99 x (c) Since P = R − C P = 99.99 x − (68.75 x + 248,000) P = 31.24 x − 248,000.

F increases very rapidly as y increases. (b)

5,000,000

Xmin = 0 Xmax = 50 Xscl = 10 Ymin = 0 Ymax = 5,000,000 Yscl = 500,000

(c) From the table, y ≈ 22 ft (slightly above 20). You could obtain a better approximation by completing the table for values of y between 20 and 30. (d) By graphing F ( y) together with the horizontal line y2 = 1,000,000, you obtain y ≈ 21.37 feet. 81. Yes. If x = 30, y = − 0.01(30) + 3(30) + 6 2

y = 6 feet Since the child trying to catch the throw is holding the glove at a height of 5 feet, the ball will fly over the glove.

Section 1.3

82. (a)

f ( 2013) − f ( 2005)

89.

≈ $525 million/year

2013 − 2005

S(t)

217.3

136.9

237.4

518.8

981.1

S(t)

1624.2

2448.2

3453.1

4638.9

The model approximates the data well. 83.

f ( x) = 2 x

90.

f ( x) = x + 3 Domain: [ −3, ∞) or x ≥ −3 Range: [0, ∞ ) or y ≥ 0

91. No. f is not the independent variable. Because the value of f depends on the value of x, x is the independent variable and f is the dependent variable. 92. (a) The height h is a function of t because for each value of t there is exactly one corresponding value of h for

0 ≤ t ≤ 2.6. (b) The height after 0.5 second is about 20 feet. The height after 1.25 seconds is about 28 feet. (c) From the graph, the domain is 0 ≤ t ≤ 2.6. (d) The time t is not a function of h because some values of h correspond to more than one value of t.

84. g( x ) = 3 x − 1

g( x + h) = 3( x + h) − 1 = 3 x + 3h − 1 g( x + h) − g( x) = (3 x + 3h − 1) − (3 x − 1) = 3h

93. 12 −

g( x + h) − g( x ) 3h = = 3, h ≠ 0 h h

94.

f ( x ) = x 2 − x + 1, f (2) = 3

f (2 + h) − f (2) (2 + h)2 − (2 + h) + 1 − 3 = h h 4 + 4h + h 2 − 2 − h + 1 − 3 = h 2 h + 3h = = h + 3, h ≠ 0 h 86.

f ( x) = x + 2

Range: [2, ∞ ) or y ≥ 2

f ( x + c ) − f ( x ) 2( x + c ) − 2 x = c c 2c = = 2, c ≠ 0 c

85.

= =

f ( x + h) = ( x + h)3 + ( x + h) = x 3 + 3 x 2 h + 3 xh 2 + h3 + x + h

95.

= 3 x 2 h + 3 xh 2 + h3 + h

( x + 5)( x − 4)( x − 1)

2 x ( x − 4)

( x + 5)( x − 1)( x − 4)

3x − 3 + 2 x − 8 x

( x + 5)( x − 4)( x − 1) 2x2 − 5x − 3 ( x + 5)( x − 4)( x − 1)

2x3 + 11x2 − 6 x x + 10 x(2 x2 + 11x − 6)( x + 10) ⋅ 2 = 5x 2 x + 5x − 3 5x(2 x − 1)( x + 3)

(2 x − 1)( x + 6)( x + 10) 5(2 x − 1)( x + 3) ( x + 6)( x + 10) 1 = , x ≠ 0, 5( x + 3) 2

f ( x + h) − f ( x ) h(3 x 2 + 3 xh + h 2 + 1) = = 3 x 2 + 3 xh + h 2 + 1, h ≠ 0 h h

88. True. The first number in each ordered pair corresponds to exactly one second number.

3( x − 1)

= h(3 x 2 + 3 xh + h 2 + 1)

87. False. The range of f ( x ) is ( −1, ∞ ).

4 12( x + 2) − 4 12 x + 20 = = x+2 x+2 x+2

3 2x + 2 x 2 + x − 20 x + 4x − 5 3 2x = + ( x + 5)( x − 4) ( x + 5)( x − 1) =

f ( x) = x3 + x

f ( x + h) − f ( x ) = ( x 3 + 3 x 2 h + 3 xh2 + h3 + x + h) − ( x 3 + x )

Domain: [0, ∞ ) or x ≥ 0

This represents the increase in sales per year from 2005 to 2013. (b)

Functions

96.

x+7 x −7 x + 7 2( x − 9) x + 7 ÷ = . = , x≠9 2( x − 9) 2( x − 9) 2( x − 9) x − 7 x −7

Chapter 1

Functions and Their Graphs

Section 1.4 Graphs of Functions 1.

decreasing

even

Domain: 1 ≤ x ≤ 4 or 1, 4 

No. If a vertical line intersects the graph more than once, then it does not represent y as a function of x.

12.

If f (2) ≥ f (2) for all x in (0, 3), then (2, f (2)) is a relative maximum of f.

Since f ( x ) =  x  = n, where n is an integer and n ≤ x, the input value of x needs to be greater than or equal to 5 but less than 6 in order to produce an output value of 5. So the interval [5, 6) would yield a function value of 5.

−6

Domain: ( −∞, ∞) Range: [−1, ∞) 13. f ( x ) =

x + 2 3

−3

Domain: all real numbers, ( −∞, ∞ )

3 −1

x+2 ≥ 0

f (0) = 1

x ≥ −2

Domain: all real numbers, ( −∞, ∞ )

Domain: [− 2, ∞)

Range: all real numbers, ( −∞, ∞ )

Range: [0, ∞ )

f (0) = 2

−4

Range: ( −∞, 1]

f ( x) = x 2 − 1

14. h(t ) = 4 − t 2

Domain:  −4, 4 

4 − t2 ≥ 0  t2 ≤ 4

Range: 0, 4 

f (0) = 4

10. Domain: all real numbers, ( −∞, ∞ ) −3

Range: [ −3, ∞)

3 −1

f (0) = −3

Domain: [ −2, 2]

11. f ( x ) = − 2 x 2 + 3

Range: [0, 2]

−6

15.

f ( x) = x + 3 7

−4

Domain: ( −∞, ∞) Range: ( −∞, 3]

−9

3 −1

Domain: ( −∞, ∞ ) Range: [0, ∞ )

Section 1.4

16.

f ( x) = −

1 x−5 4

19.

20. −6

x − y2 = 1  y = ± x − 1

21. 0.25 x 2 + y 2 = 25

1 2 x + y2 = 1 4

17. (a) Domain: ( −∞, ∞ )

(b) Range: [ −2, ∞) (c)

A vertical line intersects the graph more than once, so y is not a function of x. Graph the circle as

f ( x ) = 0 at x = −1 and x = 3.

(d) The values of x = −1 and x = 3 are the x-intercepts of the graph of f. (e)

f (0) = −1

(f) The value of y = −1 is the y-intercept of the graph of f. (g) The value of f at x = 1 is f (1) = −2. The coordinates of the point are (1, − 2). (h) The value of f at x = −1 is f ( −1) = 0. The coordinates of the point are (−1, 0). (i)

y1 = 22.

24.

25.

(e)

f (0) = 4

(f)

The value of y = 4 is the y-intercept of the graph of f.

(g) The value of f at x = 1 is f (1) = 3.

f ( x) = x2 − 4 x

f ( x) = x3 − 3x 2 + 2 f is increasing on ( −∞, 0) and (2, ∞).

f ( x ) = 0 at x = −4 and x = 2.

(d) The values of x = −4 and x = 2 are the x-intercepts of the graph of f.

x 2 = 2 xy − 1

The graph is decreasing on ( −∞, 2) and increasing on (2, ∞).

(b) Range: ( −∞, 4] (c)

1 1 4 − x2 and y2 = − 25 − x2 . 2 2

A vertical line intersects the graph just once, so y is a x2 + 1 . function of x. Solve for y and graph y1 = 2x 3 23. f ( x ) = x 2 f is increasing on ( −∞, ∞ ).

The coordinates of the point are ( −3, f ( −3)) or ( −3, 2).

18. (a) Domain: ( −∞, ∞ )

1 2 x 2

y is not a function of x. The vertical line x = 2 intersects the graph twice. Graph y1 = x − 1 and y2 = − x − 1.

Domain: ( −∞, ∞ ) Range: (−∞, 0]

A vertical line intersects the graph just once, so y is a 1 function of x. Graph y1 = x 2 . 2

−4

Graphs of Functions

f is decreasing on (0, 2).

26.

f ( x) = x2 − 1

The graph is decreasing on ( −∞, − 1) and increasing on (1, ∞ ). 27.

f ( x) = 3

(a)

The coordinates of the point are (1, 3). (h) The value of f at x = −1 is f ( −1) = 3.

−6

The coordinates of the point are (−1, 3). (i)

The coordinates of the point are ( −3, f ( −3)) or ( −3, 1).

6 −2

(b)

f is constant on ( −∞, ∞ ).

Chapter 1

28.

f ( x) = x

Functions and Their Graphs 33.

(a)

−6

f ( x) = x + 1 + x − 1 6

6 −6

6 −2

−4

(b) Increasing on (1, ∞ ), constant on ( −1, 1), decreasing on ( −∞, − 1)

(b) Increasing on ( −∞, ∞ ) 29.

f ( x) = x 2 / 3 (a)

34.

f ( x) = − x + 4 − x + 1

(a)

1 −10

−6

6 −2

−9

(b) Increasing on (0, ∞ )

(b) Increasing on ( −∞, − 4), constant on ( −4, − 1), decreasing on ( −1, ∞ )

Decreasing on ( −∞, 0) 30.

f ( x) = − x3 / 4 (a)

35.

f ( x) = x2 − 6 x

1 −1

−6

−5

−10

(b) Decreasing on (0, ∞ ) 31.

Relative minimum: (3, − 9)

f ( x) = x x + 3

(a)

36.

f ( x) = 3 x 2 − 2 x − 5 6

−9 −9 −3

−6

(b) Increasing on (−2, ∞)

Relative minimum: (0.33, − 5.33)

Decreasing on ( −3, − 2) 32.

37.

y = − 2 x3 − x 2 + 14 x

f ( x) = x 3 − x

(a)

4 −6 −6

6 −20 −4

(b) Increasing on ( −∞, 2) Decreasing on ( 2, 3)

Relative minimum: ( −1.70, −16.86) Relative maximum: (1.37, 12.16)

Section 1.4 38.

43.

y = x 3 − 6 x 2 + 15

Graphs of Functions

f ( x) = x3 − 3x 4

−4

−6

−4

− 18

Relative minimum: (1, − 2)

Relative minimum: (4, − 17)

Relative maximum: ( −1, 2)

Relative maximum: (0, 15) 44.

39. h( x ) = ( x − 1) x

f ( x) = − x3 + 3x 2 6

−1

−5

7 −2

−1

Relative minimum: (0, 0)

Relative minimum: (0.33, − 0.38) (0, 0) is not a relative maximum because it occurs at the endpoint of the domain [0, ∞).

Relative maximum: (2, 4) 45.

f ( x) = 3x2 − 6 x + 1 5

40. g( x ) = x 4 − x 4 −5

−3

Relative minimum: (1, − 2)

−2

Relative maximum: (2.67, 3.08) 41.

46.

f ( x) = 8 x − 4 x 2 5

f ( x) = x2 − 4 x − 5 2 −6

−5

−3

Relative maximum: (1, 4)

− 10

Relative minimum: (2, − 9) 42.

47.

f ( x ) =  x  + 2 y

f ( x ) = 3 x 2 − 12 x

6 5

3 − 10

3 2

−5 −4 − 13

Relative minimum: (2, − 12)

−1

−2 −3

Chapter 1

Functions and Their Graphs

48.

f ( x ) =  x  − 3

 1  1 53. s( x ) = 2  x −  x    4   4

y 2

1 −3 − 2 − 1

−2

−9

−3 −4

−6

Domain: ( −∞, ∞ )

−7

49.

Range: [0, 2)

f ( x) =  x − 1 − 2

Sawtooth pattern

 1  1 54. g( x ) = 2  x −  x    4   4

4 3 2 1 −3 −2 −1

−2 −3

−9

−4

50.

f ( x) =  x + 2 + 1

Domain: ( −∞, ∞ ) Range: [0, 2)

y 6

Sawtooth pattern

5 4 3

55.

1 −5

−2 −1

2 x + 3, x < 0 f ( x) =  3 − x, x ≥ 0

−2

−3

51.

f ( x) = 2 x

−3

−1 −1

−2

−3

−4

4 2 −6

−4

−2

56.

 x + 6, x ≤ − 4 f ( x) =  3x − 4, x > − 4 y

4 2

52.

f ( x ) =  4 x 

− 12

−4 −2

4 6

−4

y 2 1 −3

−2

−1

− 16

Section 1.4

57.

 x + 4, f ( x) =   4 − x ,

x<0

61.

x≥0

 2 x + 1, f ( x) =  2  x − 2,

Graphs of Functions

x ≤ −1 x > −1

3 −4 −3 −2 −1

1 −4 −3 − 2 −1 −1

x −3 −4

−2

−5

−3

58.

1 − ( x − 1)2, f ( x) =   x − 2 ,

x≤2 x>2

3 + x, 62. h( x ) =  2  x + 1,

x<0 x≥0

7 6 5 4 3

4 3 2

1 −1

−5

59.

1 2 3 4 5

−2 −3

63.

 x + 3,  f ( x ) = 3, 2x − 1, 

−2 −1

f ( x ) = 5 is even. 8

x≤0 0<x≤2 −9

x>2

−4

64.

f ( x ) = −9 is even.

1 −9

1 −4

−2 − 1 −1

−2 −11

−3

65. 60.

x + 5, x ≤ − 3  g ( x) = 5, −3 < x < 1 5 x − 4, x ≥ 1 

f ( x ) = 3 x − 2 is neither even nor odd. 6

−9

y −6

7 6

66.

4 3 2 1 −4 −3 −2 −1 −2 −3

f ( x) = 4 − 5 x is neither even nor odd. 14

1 2 3 4

−6

6 −2

Chapter 1

Functions and Their Graphs

67. h( x) = x 2 + 6 is even. 11

−9

 5  74.  − , − 7   3  5  (a) If f is even, another point is  , − 7  . 3  5  (b) If f is odd, another point is  , 7  . 3 

9 −1

68.

f ( x ) = − x 2 − 8 is even. 2 −18

75.

(− 2, − 9) (a) If f is even, another point is ( 2, − 9). (b) If f is odd, another point is ( 2, 9).

76. (5, − 1) −22

69.

f ( x ) = 1 − x is neither even nor odd. 3

(a) If f is even, another point is ( −5, − 1). (b) If f is odd, another point is ( −5, 1). 77. ( x, − y ) (a) If f is even, another point is ( − x , − y ).

−4

(b) If f is odd, another point is ( − x, y ) . 78. (2 a, 2 c )

−1

70. g(t ) = t − 1 is neither even nor odd. 3

(a) If f is even, another point is ( −2 a , 2c ). (b) If f is odd, another point is ( −2 a, − 2c). 79. (a)

−3

= t 2 − 2t − 3 ≠ f (t ) ≠ − f (t ) f is neither even nor odd.

−2

71.

f ( x ) = x + 2 is neither even nor odd.

f ( −t ) = (−t )2 + 2(−t ) − 3

(b)

3 6

−6

−5

1 −6

−1

72.

f ( x ) = − x − 5 is neither even nor odd. 6

−3

The graph is neither symmetric with respect to the origin nor with respect to the y-axis. So, f is neither even nor odd. (c) Tables will vary. f is neither even nor odd. 80. (a)

= x6 − 2 x2 + 3 = f ( x) f is even.

−6

3  73.  , 4  2 

f (− x ) = (− x )6 − 2(− x )2 + 3

(b)

 3  (a) If f is even, another point is  − , 4 .  2   3  (b) If f is odd, another point is  − , − 4 . 2  

−3 −5

The graph is symmetric with respect to the y-axis. So, f is even. (c) Tables will vary. f is even.

Section 1.4 81. (a)

g(− x ) = (− x )3 − 5(− x )

84. (a)

= − x3 + 5x

f (− x ) = (− x ) (− x ) + 5

= −x −x + 5 ≠ f ( x) ≠ − f ( x) f is neither even nor odd.

= −( x 3 − 5 x ) = − g( x ) g is odd.

(b)

Graphs of Functions

(b)

6 −7 −4

4 −5 −6

The graph is symmetric with respect to the origin. So, g is odd. (c) Tables will vary. g is odd. 82. (a)

h( − x) = ( − x) − 4( − x) 5

The graph is neither symmetric with respect to the origin nor with respect to the y-axis. So, f is neither even nor odd. (c) Tables will vary. f is neither even nor odd. 85. (a)

(

= − x5 + 4 x3

= 4 3 −s

= − ( x5 − 4 x3 )

= 4s 2 3

h is odd.

)

= g (s) g is even.

= − h( x )

(b)

g (− s) = 4(− s) 2 3

(b)

−4

−2

2 −2

− 12

The graph is symmetric with respect to the origin. So, h is odd. (c) Tables will vary. h is odd. 83. (a)

The graph is symmetric with respect to the y-axis. So, g is even. (c) Tables will vary. g is even. 86. (a)

f ( − s ) = 4( − s )

= − 4s 3 5

f ( − x ) = ( − x ) 1 − ( − x )2

= − f (s)

= − x 1 − x2 = − f ( x) f is odd.

(b)

f is odd. (b)

1 −1 −2

2 − 10 −1

The graph is symmetric with respect to the origin. So, f is odd. (c) Tables will vary. f is odd.

Chapter 1

Functions and Their Graphs

87.

f ( x) = 4 − x ≥ 0

90.

f ( x) = x2 − 4 x y

y 5

−2

2 1

−3

−1

x −4

f ( x) ≥ 0

f ( x) ≥ 0 4−x≥0

x − 4x ≥ 0 x( x − 4) ≥ 0 (−∞, 0], [4, ∞)

4≥x ( −∞, 4]

88.

−1

f ( x) = 4 x + 8

91. (a)

C2 is the appropriate model.

The cost of the first hour is $1.00 and the cost increases $0.50 when the next hour begins, and so on.

10 8

(b)

2 −6

−4

−2

x 0

f ( x) ≥ 0

(

)

C2 7 + 16 = C2 (7.167)

4x + 8 ≥ 0

= 1.00 − 0.50 − (7.167 − 1)

4x ≥ −8

= 4.50

x ≥ −2

The cost for parking for 7 hours and 10 minutes is $4.50.

[− 2, ∞) f ( x) = x 2 − 9 ≥ 0

92. C = 23.20 + 2 x, x > 0

2 −6

−4

−2

50 2

−4

− 10

f ( x) ≥ 0 x2 − 9 ≥ 0 x2 ≥ 9 x ≥ 3 or x ≤ −3 (−∞, − 3], [3, ∞)

Cost of overnight delivery (in dollars)

89.

45 40 35 30 25 20 15 1 2 3 4 5 6 7 8 9 10

Package weight (in pounds)

93. h = top – bottom = (− x 2 + 4 x − 1) − 2

= − x 2 + 4 x − 3, 1 ≤ x ≤ 3 94. h = top – bottom = 3 − (4 x − x 2 ) = 3 − 4 x + x2 , 0 ≤ x ≤ 1

Section 1.4 95. (a)

108. The graph of the greatest integer function is a series of steps with a closed circle at the left end and an open circle at the right end. The graph of a line with a slope of zero is one continuous horizontal line with no steps.

650

10 450

109. f is an even function.

(a) g ( x) = − f ( x) is even because

(b) The number of cooperative homes and condos was decreasing from 2010 to early 2011 and increasing from early 2011 to 2013.

g ( − x) = − f ( − x) = − f ( x) = g ( x). (b) g ( x) = f ( − x) is even because

Graphs of Functions

g ( − x) = f ( − ( − x )) = f ( x) = f ( − x) = g ( x ).

Interval Intake Pipe Drain Pipe 1 Drain Pipe 2

(d) g ( x) = − f ( x + 3) is neither even nor odd because

[0, 5]

Open

Closed

[5, 10]

Open

Closed

g ( − x) = − f ( − x + 3) = − f ( − ( x − 3))

[10, 20]

Closed

= − f ( x − 3) ≠ g ( x ) nor − g ( x ).

[20, 30]

Closed

Open

[30, 40]

Open

[40, 45]

Open

Closed

Open

[45, 50]

Open

[50, 60]

Open

Closed

110. (a)

(b) Using the graph from part (a), the domain is ( −∞, ∞ ) and the range is ( −∞, ∞ ). (c) Using the graph from part (a), you can see that the graph is increasing on −1 < x < 1, and the graph is

97. False. The domain of f ( x ) = x 2 is the set of all real numbers.

decreasing on x < −1 and x > 1. (d) Using the graph from part (a), there is a relative minimum at ( −1, − 2) and a relative maximum at (1, 2).

98. False. The domain must be symmetric about the y-axis. 99. c 111.

100. d

1000

f ( x ) = a2 n +1 x 2 n +1 + a2 n −1 x 2 n −1 +  + a3 x 3 + a1 x

101. b

f ( − x ) = a2 n +1 ( − x )2 n +1 + a2 n −1 ( − x )2 n −1 +  + a3 ( − x )3 + a1 ( − x )

102. e

Therefore, f ( x ) is odd.

= − a2 n +1 x 2 n +1 − a2 n −1 x 2 n −1 −  − a3 x 3 − a1 x = − f ( x )

103. a

112.

f ( x ) = a2 n x 2 n + a2 n − 2 x 2 n − 2 +  + a2 x 2 + a0 f ( − x ) = a 2 n ( − x ) 2 n + a2 n − 2 ( − x ) 2 n − 2 +  + a 2 ( − x ) 2 + a 0

104. f 105. No. Each y-value corresponds to two distinct x-values when y > 0. 106. No. Each y-value corresponds to two distinct x-values when − 5 < y < 5.

 −1,  107. Yes, f ( x ) = 0, 1,  

−1 ≤ x < 0 0 ≤x< 1 1 ≤ x<2

= a2 n x 2 n + a 2 n − 2 x 2 n − 2 +  + a2 x 2 + a 0 = f ( x )

f ( − x ) = f ( x ); thus, f ( x ) is even.

113. − 2 x 2 + 11x + 3

Terms: − 2 x 2 , 11x, 3 Coefficients: − 2, 11 114. 10 + 3x

Terms: 3 x , 10 Coefficient: 3

115.

Chapter 1

Functions and Their Graphs 118. f ( x ) = x x − 3

x − 5x2 + x3 3 x Terms: , − 5 x 2 , x 3 3 1 Coefficients: , − 5, 1 3

116. 7 x 4 +

(a) f (3) = 3 3 − 3 = 0 (b) f (12) = 12 12 − 3 = 12 9 = 12(3) = 36 (c) f (6) = 6 6 − 3 = 6 3

2 x2 − x

Terms: 7 x 4 ,

119. f ( x ) = x 2 − 2 x + 9 f (3 + h) = (3 + h)2 − 2(3 + h) + 9 = 9 + 6h + h 2 − 6 − 2h + 9

2 x2 , − x

= h 2 + 4h + 12

2, −1

Coefficients: 7,

f (3) = 32 − 2(3) + 9 = 12 f (3 + h) − f (3) (h 2 + 4h + 12) − 12 h(h + 4) = = h h h = h + 4, h ≠ 0

117. f ( x ) = − x 2 − x + 3

(a) f (4) = −(4)2 − 4 + 3 = −17 (b) f (−5) = −(−5)2 − (−5) + 3 = 12 (c) f ( x − 2) = −( x − 2)2 − ( x − 2) + 3

120. f ( x ) = 5 + 6 x − x 2

f (6 + h ) = 5 + 6(6 + h ) − (6 + h ) 2

= −( x 2 − 4 x + 4) − x + 2 + 3

= 5 + 36 + 6 h − (36 + 12 h + h 2 )

= − x2 + 3x + 1

= −h2 − 6h + 5 f (6) = 5 + 6(6) − 6 2 = 5 f ( h + 6) − f (6) ( − h 2 − 6 h + 5) − 5 h ( − h − 6) = = h h h = − h − 6, h ≠ 0

Section 1.5 Shifting, Reflecting, and Stretching Graphs 1. 2.

Horizontal shifts, vertical shifts, and reflections are rigid transformations. (a) (b) (c) (d)

h(x)

− f ( x ), f ( − x )

c > 1, 0 < c < 1 y

−6

f (x)

g(x)

h(x)

−2

g(x)

f (x)

−2

−6

−4

−2

4 2

f(x)

g(x) −4

−2

−4

h(x)

−4

g(x) f (x)

ii iv iii i

−6

−2

−8

−6

g(x) 2

−4 −6 −8

h(x)

−8

f (x)

Section 1.5 10.

Shifting, Reflecting, and Stretching Graphs y

16.

4 3

f (x)

g(x)

−6

−4

−1 −1

f (x)

−2

y 4 3

h(x)

−2

h(x)

17.

−4

−6

−3

f (x)

h(x)

−2

h(x)

−2

11.

g(x)

−2

h(x)

2 1

f(x)

f(x) 3

g(x)

−4

18.

12. 6

g(x)

−4

f (x)

−2

h(x)

−4

− 5 f (x)

f(x) −6

y 2

h(x)

g(x) −4

g(x)

−6

19. (a)

13. 8

f (x)

−6

g(x)

−2

f (x)

(1, 2) (0, 1) 1

(b)

g(x)

(3, 3)

−4

14.

(4, 4)

−4

y = f ( x) + 2

h(x)

y = − f ( x) y

4 2

−6

−4

h(x) −1 −2

(3, − 1) (4, − 2)

−3

h(x)

g(x)

−2 −1

(1, 0) 1

−6

15.

(0, 1)

−2 −3

f(x)

−4

Chapter 1 (c)

Functions and Their Graphs 20. (a)

y = f ( x − 2)

y = f (x) –1

(− 2, 3)

4 3

(0, 2)

(6, 2)

(5, 1)

(3, 0) 1

−1

−2

−1 −1

(2, − 1)

−2

(3, − 2)

−2

(d)

y = f ( x + 3)

(b)

y = f ( x + 2) y

(1, 2)

−1

−1 −4

−3

−2

−1

(− 1, 4)

(1, 0)

−3 −2 −1

−2

(3, 2)

(1, 3)

2 1

(0, − 2)

(2, 0)

−3

−1

y = f (− x) y

(− 3, 1) −3

(d)

y 2

(− 1, 0)

−2 −1 −1

−2

−1

−4 −5

g(0) = f ( 12 (0) ) = f (0) = −1

−3

−2

(g) Let g( x ) = f ( x ) . Then from the graph,

(5, 1) (3, 0)

(2, − 3)

(e)

(2, 4)

4 3

g(6) = f ( 12 (6) ) = f (3) = 1

(0, 3)

g (8) = f ( 12 (8) ) = f (4) = 2.

(− 1, 0) −3

−1

(− 3, − 1)

(0, − 4)

g(2) = f ( 12 (2) ) = f (1) = 0

−2

(0, − 1)

1 2

(4, − 1)

(− 4, 2)

−4

(1, − 1)

(4, 4)

−1

(− 1, 0)

y = 2 f ( x)

−5

−2

(f )

(− 2, 3)

(0, 1)

(− 3, − 1)

(e)

(− 4, 4)

(− 2, 0) −3

(1, − 1)

−1

4 3

(8, 2)

2 1 −1 −2

(2, 0) 2

(0, − 1)

(6, 1) 3

−3 −4

Section 1.5 y

(f) (− 2, 2)

y = x+2

(0, 32 )

−2

(1, 0)

−1

33. Reflection in the x-axis and a vertical shift one unit upward of y = x : y = 1 − x

(

3, − 1 2

−1

)

34. Reflection in the x-axis and a vertical shift one unit upward of y = x 3 : y = 1 − x 3

−2

(g) Let g ( x ) = f (2 x ). Then from the graph,

y = − x = x , therefore y = − x is identical to y = x .

g (0) = f (2(0)) = f (0) = 3

( ) g ( ) = f ( 2 ( ) ) = f (3) = −1. g ( 12 ) = f 2 ( 12 ) = f (1) = 0 3 2

37. y = (− x )2 is a reflection in the y-axis. In fact, y = (− x )2 = x 2 , therefore y = (− x )2 is identical to y = x2 .

3 2

38. y = − x 3 is a reflection of f ( x ) = x 3 in the x-axis. However, since y = − x 3 = ( − x )3 , either a reflection in the x-axis or a reflection in the y-axis produces the same graph.

4 3

(0, 3)

2 1

−3

−2

( 12 , 0)

−1 −1

1 1 is a reflection of f ( x ) = in the y-axis. −x x 1 1 = − , either a reflection in the However, since y = −x x y-axis or a reflection in the x-axis produces the same graph.

39. y = 2 3 , −1 2

(

)

21. The graph of f ( x ) = x 2 should have been shifted one unit to the left instead of one unit to the right. 22. The graph of f ( x ) = x 2 should have been shifted one unit to the right instead of one unit downward. 23. y = x + 2 is f ( x ) = two units.

35. y = − x is f ( x ) reflected in the x-axis. 36. y = − x is a reflection in the y-axis. In fact

g ( −1) = f (2( −1)) = f ( −2) = 4

(−1, 4)

32. Horizontal shift two units to the left of y = x :

3 2

Shifting, Reflecting, and Stretching Graphs

x shifted vertically upward

1 1 − 5 is f ( x ) = shifted vertically five units x x downward.

24. y =

1 1 is a reflection of f ( x ) = in the x-axis. x x 1 1 , either a reflection in the However, since y = − = x −x x-axis or a reflection in the y-axis produces the same graph.

40. y = −

41. y = 4 x is a vertical stretch of f ( x ) = x .

25. y = ( x − 4)3 is f ( x ) = x 3 shifted horizontally four units to the right.

42. p( x ) =

1 2 x −1 x is a vertical shrink of f ( x ) = . 2 4

26. y = x + 5 is f ( x ) = x shifted horizontally five units to

43. g( x ) =

1 3 x is a vertical shrink of f ( x) = x 3. 4

the left. 27. y = x 2 − 2 is f ( x ) = x 2 shifted vertically two units downward. 28. y = x − 2 is f ( x ) = to the right.

x shifted horizontally two units

45. f ( x ) = 4 x is a horizontal shrink of f ( x ) = x .

However, since f ( x ) = 4 x = 2 x , it also can be described as a vertical stretch of f ( x ) = x .

29. Horizontal shift three units to left of y = x : y = x + 3 (or vertical shift three units upward) 30. Horizontal shift two units to the left of y =

44. y = 2 x is a vertical stretch of f ( x ) = x .

1 1 :y= x x+2

46. y =

1 1 x = x is a vertical shrink of f ( x ) = x . 2 2

31. Vertical shift three units downward of y = x 2:

y = x2 − 3

Chapter 1

Functions and Their Graphs 51. (a)

47. f ( x ) = x 3 − 3 x 2

(b) g( x ) = 2 − ( x + 5)2 is obtained from f by a horizontal shift to the left five units, a reflection in the x-axis, and a vertical shift upward two units. y (c)

g −5

f ( x) = x 2

3 2 1

−4

g( x ) = f ( x + 2) = ( x + 2)3 − 3( x + 2)2 is a horizontal shift two units to the left. 1 1 h( x ) = f ( x ) = ( x 3 − 3 x 2 ) is a vertical shrink. 2 2 48. f ( x ) = x 3 − 3 x 2 + 2 f

−2 −3 −4 −5 −6 −7

(b) g ( x) = ( x − 10) + 5 is obtained from f by a 2

horizontal shift 10 units to the right and a vertical shift 5 units upward.

−4

g( x ) = f ( x − 1) = ( x − 1) − 3( x − 1) + 2 is a horizontal shift one unit to the right.

18 16 14 12 10 8 6 4 2

49. f ( x ) = x 3 − 3 x 2 4

−6

(c)

h( x ) = f (3 x ) = (3 x )3 − 3(3 x )2 + 2 is a horizontal shrink.

−2 −1

52. (a) f ( x) = x 2

−4

−5 −4

(d) g ( x ) = 2 − f ( x + 5)

−9 −8 −7

−2

2 4 6 8 10 12 14 16 18

(d) g ( x) = f ( x − 10) + 5 −4

1 1 f ( x ) = − ( x 3 − 3 x 2 ) is a reflection in the 3 3 x-axis and a vertical shrink. g( x ) = −

h( x ) = f (− x ) = (− x )3 − 3(− x )2 is a reflection in the y-axis.

53. (a) (b)

f ( x) = x 2 g( x ) = 3 + 2( x − 4)2 is obtained from f by a horizontal shift four units to the right, a vertical stretch of 2, and a vertical shift upward three units.

(c)

y 7 6 5

50. f ( x ) = x 3 − 3 x 2 + 2

2 1

−6

−1

(d)

−4

g ( x ) = 3 + 2 f ( x − 4)

g( x ) = − f ( x ) = −( x − 3 x + 2) is a reflection in the x-axis. h( x ) = f (2 x ) = (2 x )3 − 3(2 x )2 + 2 is a horizontal shrink.

Section 1.5 f ( x) = x 2 2 1 (b) g( x ) = − ( x + 2 ) − 2 is obtained from f by 4 a horizontal shift two units to the left, a vertical 1 shrink of , a reflection in the x-axis, and a 4 vertical shift two units downward. y (c)

54. (a)

Shifting, Reflecting, and Stretching Graphs 57. (a)

(b)

f ( x) = x 3 g( x ) = ( x − 1)3 + 2 is obtained from f by a horizontal shift one unit to the right and a vertical shift upward two units.

(c)

y 5 4 3

3 2 1

2 1

−7 −6 −5 −4 −3 −2 −1

1 2 3

−3 −4 −5 −6 −7

(d) 55. (a)

−3 −2 −1

(d) 58. (a)

1 g( x ) = − f ( x + 2) − 2 4

(b)

g ( x ) = f ( x − 1) + 2

f ( x) = x 3 g( x ) = −( x + 3)3 − 10 is obtained from f by a horizontal shift 3 units to the left, a reflection in the x-axis, and a vertical shift 10 units downward.

f ( x) = x3

(b) g ( x) = 13 ( x − 2) is obtained from f by a 3

horizontal shift two units to the right followed by a vertical shrink of 13.

− 12 − 10 − 8 − 6 − 4 − 2

−4

(c)

−6

−8

2 1 −1

−1

−2 −3

(d) g ( x) = 13 f ( x − 2) 56. (a)

(d)

g( x ) = − f ( x + 3) − 10

59. (a)

f ( x) =

1 x

(b)

g( x ) =

1 − 9 is obtained from f by a x +8

horizontal shift eight units to the left and a vertical shift nine units downward.

f ( x) = x3

(b) g ( x) = 12 ( x + 1) is obtained from f by a 3

(c)

y 2

horizontal shift one unit to the left and a vertical shrink of 12.

−14

−1

−6

5 4 3 2 1 −5 −4 − 3

−6 −4 −2 −2 −4

(c)

−10

−8 −10 −12 −14 1 2 3 4 5

−2 −3 −4 −5

(d)

g ( x ) = f ( x + 8) − 9

(d) g( x) = 12 f ( x + 1)

Chapter 1

60. (a)

(b)

Functions and Their Graphs

1 x

f ( x) =

63. (a)

1 g( x ) = + 4 is obtained from f by a x−7

(b)

1 x + 3 − 1 is obtained from f by a 2

g( x ) = −

horizontal shift three units to the left, a vertical shrink, a reflection in the x-axis, and a vertical shift one unit downward.

horizontal shift seven units to the right and a vertical shift four units upward. (c)

f ( x) = x

(c)

5 4 3 2 1

8 6 2 −2 −2

8 10 12 14

−5 −4

−2 −1 −3 −4 −5

−4 −6

(d) 61. (a)

(b)

1 2 3 4 5

g ( x ) = f ( x − 7) + 4

(d)

g ( x) = −

64. (a)

f ( x) =

f ( x) = x g( x ) = −2 x − 1 − 4 is obtained from f by a

horizontal shift one unit to the right, a vertical stretch of 2, a reflection in the x-axis, and a vertical shift downward four units. (c)

(b)

−8 −6 −4 −2 −2

g ( x) = − 3 x + 1 − 6 is obtained from f by a

horizontal shift one unit to the left, a reflection in the x-axis, a vertical stretch of 3, and a vertical shift six units downward.

1 f ( x + 3) − 1 2

− 6 − 4 −2 −2

−4

8 10

−4

−6

− 10

(d) 62. (a)

(b)

− 12

− 14

− 14 − 16

g( x ) = −2 f ( x − 1) − 4

(d)

f ( x) = x

g( x ) =

1 x − 2 − 3 is obtained from f by a 2

horizontal shift two units to the right, a vertical shrink, and a vertical shift three units downward. (c)

65. (a)

(b)

150

1 2 3 4 5

−3 −4 −5

(d)

N (t ) is a horizontal shift of 26.17 units to the right, a vertical shift of 126.5 units upward, a reflection in the t-axis (horizontal axis), and a vertical shrink of 0.03.

5 4 3 2 1

−1

g ( x) = − 3 f ( x + 1) − 6

1 g( x ) = − f ( x − 2) − 3 2

(c)

G ( t ) = N ( t + 9) 2

= − 0.03(t + 9) − 26.17 + 126.5 = − 0.03(t − 17.17) + 126.5 2

To make a horizontal shift 9 years backward (9 units left), add 9 to t.

Section 1.6 66.

(a) D(t ) is a horizontal shift of 3.7 units to the left and a vertical stretch of 1.9. (b)

Combinations of Functions

75. The vertex is approximately (2, − 4) and the graph opens upward. Matches (c). 76. The graph of f is y = x 3 shifted to the left approximately four units, reflected in the x-axis, and shifted upward approximately two units. Matches (b) and (d).

77. Answers will vary. 0

t + 3.7 =

78. Since y = f ( x + 2) − 1 is a horizontal shift of two units to the left and a vertical shift one unit downward, the point (0, 1) will shift to ( −2, 0), (1, 2) will shift to ( −1, 1), and (2, 3) will shift to (0, 2). 79. (a) Since 0 < a < 1, g( x ) = ax 2 will be a vertical

shrink of f ( x ) = x 2 . (b) Since a > 1, g ( x ) = ax 2 will be a vertical stretch of f ( x ) = x 2 . 80. (a)

on the intervals ( − ∞, − 2) and (1, ∞)

In the year 2015, the depreciation of assets will be approximately $6 million. (d) G (t ) = D (t + 2) = 1.9

(b) Increasing on the intervals ( − ∞, − 2) and (1, ∞), and decreasing on the interval ( − 2, 1)

(t + 2) + 3.7

(c)

= 1.9 t + 5.7

10 + 2 =3 2+2

81. Slope L1 :

68. True. y = x + 6 and y = − x + 6 are identical because a reflection in the y-axis of y = x + 6 will be identical to itself. Additionally it is an even function. 69. y = f (− x ) is a reflection in the y-axis, so the x-intercepts are x = −2 and x = 3.

Increasing on the intervals ( − ∞, −1) and ( 2, ∞), and decreasing on the interval ( −1, 2)

To make a horizontal shift 2 years backward (2 units left), add 2 to t. 67. False. y = f (− x ) is a reflection in the y-axis.

Increasing on the interval ( − 2, 1) and decreasing

Slope L2 :

9−3 3 = 3 +1 2

Neither parallel nor perpendicular 82. Slope L1 :

3 − (−7) 10 = =2 4 − (−1) 5

70. y = 2 f ( x ) is a vertical stretch, so the x-intercepts are the same: x = 2, − 3.

Slope L2 :

−7 − 5 −12 = =4 −2 − 1 −3

71. y = f ( x ) + 2 is a vertical shift, so you cannot determine the x-intercepts.

Neither parallel nor perpendicular

72. y = f ( x − 3) is a horizontal shift 3 units to the right, so the x-intercepts are x = 5 and x = 0.

83. Domain: All x ≠ 9 84. f ( x ) =

73. The vertex is approximately at (2, 1) and the graph opens upward. Matches (c). 74. The domain is [0, − ∞) and (0, − 4) is approximately on the graph, and f ( x ) < 0. Matches (a) and (b).

x−5 x−7

Domain: x ≥ 5 and x ≠ 7 85. Domain: 100 − x 2 ≥ 0  x 2 ≤ 100  −10 ≤ x ≤ 10 86. f ( x ) = 3 16 − x 2

Domain: all real numbers

Section 1.6 Combinations of Functions 1.

addition, subtraction, multiplication, division

g( x )

composition

inner, outer

Chapter 1

Since ( fg )( x ) = 2 x( x 2 + 1) and f ( x ) = x 2 + 1, g ( x ) = 2 x, and ( fg )( x ) = ( gf )( x ) = (2 x ) f ( x ).

Functions and Their Graphs 12.

f ( x ) = 2 x − 5, g ( x ) = 1 − x

(a) (b)

Since ( f  g )( x ) = f ( g( x )) and ( f  g )( x ) = f ( x + 1), 2

g ( x ) must equal x + 1.

(c)

7. 3

= −2 x 2 + 7 x − 5

(d)

1 −1

= 3x − 6 ( fg )( x ) = (2 x − 5)(1 − x ) = 2 x − 2 x2 − 5 + 5x

−2

( f + g )( x ) = 2 x − 5 + 1 − x = x − 4 ( f − g )( x ) = 2 x − 5 − (1 − x ) = 2x − 5 −1 + x

−1

Domain: 1 − x ≠ 0

x ≠1

−2 y

13.

4 3 2

f ( x) = 3x 2 , g ( x) = 6 − 5 x

(a)

1 −4 −3 −2 − 1 −1

 f 2x − 5   ( x) = 1− x g

( f + g )( x) = f ( x) + g ( x) = 3 x 2 + (6 − 5 x ) = 3x2 − 5 x + 6

−2

(b) ( f − g )( x) = f ( x) − g ( x)

−3 −4

= 3 x 2 − (6 − 5 x ) = 3x 2 + 5 x − 6

(c)

7 6 5

( fg )( x) = f ( x) ⋅ g ( x) = 3 x 2 (6 − 5 x ) = 18 x 2 − 15 x 3 = −15 x 3 + 18 x 2

2 1 −3 −2 −1

f ( x) f 3x 2 (d)  ( x) = = g ( x) 6 − 5x g

10.

Domain: all x ≠

6 5

14.

1 −4 −3

−1 −1

f ( x) = 2 x + 5, g ( x) = x 2 − 9

(a)

= x2 + 2x − 4

−2 −3 −4

11.

f ( x ) = x + 3, g ( x ) = x − 3

(a) (b)

( f + g )( x ) = f ( x ) + g( x ) = ( x + 3) + ( x − 3) = 2 x ( f − g )( x ) = f ( x ) − g ( x ) = ( x + 3) − ( x − 3) = 6

(c)

( fg )( x ) = f ( x )g( x ) = ( x + 3)( x − 3) = x 2 − 9

(d)

 f f ( x) x + 3 =   ( x) = g g ( x) x − 3  

( f + g )( x) = (2 x + 5) + ( x2 − 9)

(

)

2 (b) ( f − g )( x) = ( 2 x + 5) − x − 9

= − x2 + 2 x + 14 (c)

( fg )( x) = (2 x + 5)( x2 − 9) = 2 x3 + 5 x 2 − 18 x − 45

f 2x + 5 (d)  ( x) = 2 x −9 g Domain: all x ≠ ± 3

Domain: all x ≠ 3

Section 1.6 15.

f ( x ) = x 2 + 5, g( x ) = 1 − x (a)

( f + g )( x ) = x 2 + 5 + 1 − x

(b)

( f − g)( x ) = x 2 + 5 − 1 − x

(c)

( fg )( x ) = ( x 2 + 5) 1 − x

18. f ( x) =

f x2 + 5   ( x) = 1− x g Domain: x < 1

(d)

16.

f ( x) = x2 − 4, g( x) =

x2 x +1

x x2 + 1 x2 2 (b) ( f − g)( x) = x − 4 − 2 x +1 (a)

( f + g)( x) = x 2 − 4 +

( x − 4 )  x x+ 1  = x x x+ 1− 4 2

(c)

( fg )( x ) =

(d)

f x 2   ( x) = x − 4 ÷ 2 g x +1  

( f + g )( x) =

x 1 x4 + x + 1 + 3 = 3 x +1 x x ( x + 1)

(b)

( f − g )( x) =

x 1 x4 − x − 1 − 3 = 3 x +1 x x ( x + 1)

(c)

( fg )( x) =

(d)

f x 1 ÷ 3  ( x ) = 1 g x x +  

Domain: x2 − 4 ≥ 0 and x ≠ 0

f ( x) =

1 1 , g( x ) = 2 x x

1 1 x +1 + = 2 x x2 x 1 1 x −1 (b) ( f − g)( x ) = − 2 = 2 x x x 1 1 1 (c) ( fg)( x ) = ⋅ 2 = 3 x x x 1  f  (d)   ( x ) = x = x, x ≠ 0 1 g x2 Domain: x ≠ 0 (a)

x x3 ⋅ x +1 1

x4 x +1

19. ( f + g )(3) = f (3) + g (3)

= (32 − 1) + (3 − 2) = 8 +1 = 9 20. ( f − g )( −2) = f ( −2) − g( −2) = ((−2)2 − 1) − (−2 − 2) = 3 − (−4) = 7

x ≥ 2 or x ≤ −2 17.

x 1 1 ⋅ = 2 x + 1 x3 x ( x + 1)

Domain: x ≠ 0, x ≠ −1

( x + 1) x − 4 x2

x 1 , g ( x) = 3 x +1 x

(a)

Combinations of Functions

21. ( f − g )(0) = f (0) − g (0) = (0 − 1) − (0 − 2)

( f + g)( x ) =

=1 22. ( f + g )(1) = f (1) + g (1)

= (12 − 1) + (1 − 2) = 0 + (−1) = −1 23. ( fg )( − 6) = f ( − 6) g ( − 6) =

((− 6) − 1)((− 6) − 2) 2

= (35)( − 8) = − 280

24. ( fg )( 4) = f ( 4) g ( 4) =

((4) − 1)(4 − 2) 2

= (15)( 2) = 30

100

Chapter 1

Functions and Their Graphs 33.

 f  f ( −5) 25.   ( −5) = g g ( −5)  

h g f

−4

( −5)2 − 1 = −5 − 2 24 = −7 24 =− 7

−3

34.

 f  f (0) 26.   (0) = g(0) g 0 −1 = 0−2 1 = 2

−3

35.

−5

27. ( f − g )(t + 1) = f (t + 1) − g (t + 1)

((t + 1) − 1) − ((t + 1) − 2) 2

−2

36.

= (t + 2t + 1 − 1) − (t − 1)

−6

= t2 + t + 1

f −3

28. ( f + g )(t − 3) = f (t − 3) + g (t − 3)

((t − 3) − 1) + (t − 3 − 2) 2

37.

f ( x ) = 3 x, g ( x ) = −

= (t 2 − 6t + 9 − 1) + (t − 5)

x3 x3 , ( f + g)( x ) = 3 x − 10 10

= t 2 − 5t + 3

− 14

29. ( fg )( −5t ) = f ( −5t ) g ( −5t )

f+g

= ((−5t )2 − 1)(−5t − 2)

− 10

= (25t 2 − 1)(−5t − 2)

For 0 ≤ x ≤ 2, f ( x ) contributes more to the magnitude.

= −125t 3 − 50t 2 + 5t + 2

For x > 6, g( x ) contributes more to the magnitude.

30. ( fg )(3t 2 ) = f (3t 2 ) g(3t 2 )

= ((3t 2 )2 − 1)(3t 2 − 2) = (9t 4 − 1)(3t 2 − 2) = 27t 6 − 18t 4 − 3t 2 + 2

f (t − 4) f 31.  (t − 4) = g g (t − 4)  

(t − 4) − 1 (t − 4) − 2

38.

x , g( x ) = x , 2 x ( f + g)( x ) = + x 2

f ( x) =

f+g

t 2 − 8t + 15 = t −6  f f (t + 2) 32.   (t + 2) = g g (t + 2)  

−4

g 14

−2

g ( x ) contributes more to the magnitude of the sum for 0 ≤ x ≤ 2. f ( x ) contributes more to the magnitude of the

sum for x > 6 .

(t + 2)2 − 1 = (t + 2) − 2 =

t 2 + 4t + 3 , t≠0 t

Section 1.6 39.

f ( x ) = 3 x + 2, g( x ) = − x + 5,

43.

( f + g )( x ) = 3 x + 2 − x + 5

44.

−6

f ( x ) contributes more to the magnitude in both intervals.

40.

f ( x ) = 3 x + 5, g( x ) = 5 − x

( ) (b) ( g  f ) ( x) = g ( f ( x ) ) = g(3x + 5) = 5 − (3x + 5) = −3x

f+g f

1 f ( x ) = x 2 − , g( x ) = −3 x 2 − 1, 2 1 3  ( f + g)( x ) =  x 2 −  + (−3 x 2 − 1) = −2 x 2 − 2 2 

f 6

f+g

f ( x ) = x 3 , g( x ) = (a) (b) (c)

45. (a)

−6

41. f ( x) = 2 x , g ( x) = x + 4 2

( f  g )( x) = f ( g ( x)) = f ( x + 4) = 2( x + 4)

( g  f )( x) = g ( f ( x)) = g (2 x2 )

= x =x ( g  f )( x ) = g( f ( x ))

( x −1) = ( x −1) +1 =g

4x2 − 7

7 ∪ x ≥ 2

7 2

Domain of f : x + 3 ≥ 0  x ≥ − 3 x 2  

Domain:

x +3 2

x + 3 ≥ 0  x ≥ −6 2

47. (a) Domain of f: all real numbers (b) Domain of g: all x ≥ 0

( f  g )( x ) = f ( g ( x ) ) = f ( x ) =

( x ) + 1 = x + 1, x ≥ 0 2

Domain: x ≥ 0 48. (a) Domain of f : x ≥ 0 (b) Domain of g: all real numbers

(c)

( f  g )( x ) = f ( g( x ))

= 3 ( x 3 + 1) − 1

(c)

Domain of f: x − 7 ≥ 0 or x ≥ 7

( f  g )( x) = f   =

( f  g )( x ) = f ( g ( x ) ) = f ( x 4 ) = ( x 4 )

1/ 4

Domain: all real numbers

= f ( x 3 + 1)

(b)

( f  g ) (0) is not defined.

(c)

f ( x ) = x − 1, g( x ) = x + 1 (a)

) ( )

Domain of g : all real numbers

= 2 x2 + 4

42.

(

(b)

= (2 x2 ) + 4

    1 ( g  f )( x ) = g f ( x ) = g x 3 = x3

( f  g )( x) = f ( g ( x)) = f (4 x 2 ) =

(c)

( f  g )(0) = 2(0 + 4) = 32

(c)

= 2 x 2 + 16 x + 32

(c)

Domain of g: all real numbers

46. (a)

( f  g )( x ) = f ( g ( x ) ) = f  x  =  x  = x 3

Domain: all real x ≤ −

g( x ) contributes more to the magnitude on both intervals.

(b)

1 x

(b)

−4

(a)

101

(a) ( f  g ) ( x) = f g ( x ) = f (5 − x) = 3(5 − x) + 5 = 20 − 3x

−9

Combinations of Functions

49. (a)

Domain of f : all x ≠ 0

(b)

Domain of g: all x ≠ 3

(c)

( f  g )( x) = f 

1   =  x + 3

1 = x +3 1 x +3

Domain: all x ≠ − 3

= ( x − 1) + 1 = x ( f  g )(0) = 0

102

Chapter 1

Functions and Their Graphs

50. (a) Domain of f: all x ≠ 0 (b) Domain of g: all x ≠ 0  1  (c) ( f  g )( x ) = f   = 2 x, x ≠ 0  2x 

56. (a)

= 3 ( x 3 − 1) + 1 = 3 x 3 = x Domain: all real numbers

Domain: all x ≠ 0

( g  f )( x ) = g ( f ( x ) ) = g ( 3 x + 1 )

51. (a) Domain of f: all real numbers (b) Domain of g: all real numbers

(c)

( f  g )( x ) = f ( g ( x ) ) = f ( x 3 − 1)

=  3 x + 1 −1   = ( x + 1) − 1 = x

( f  g )( x ) = f ( g ( x ) ) = f ( 3 − x ) = (3 − x ) − 4 = − x − 1 = x + 1

(b)

Domain: all real numbers 52. (a)

f°g=g°f −6

Domain of f : all x ≠ 0

(b)

Domain of g: all real numbers

(c)

( f  g )( x) = f ( g ( x)) = f ( x − 5) =

−4

2 x −5

They are equal.

Domain: all x ≠ 5

57. (a)

( f  g )( x ) = f ( g ( x ) ) = f ( 3 x + 9 ) 1 = (3 x + 9) − 3 = x 3

53. (a) Domain of f: all real numbers

(b) Domain of g: all x ≠ ±2 (c)

Domain: all real numbers 1 ( g  f )( x ) = g f ( x ) = g  3 x − 3   

1 1 ( f  g )( x ) = f g ( x ) = f  x 2 − 4  = x 2 − 4 + 2  

(

)

(

1  = 3 x − 3  + 9 = x 3 

Domain: x ≠ ±2 54. (a) Domain of f: all x ≠ ±1 (b) Domain of g: all real numbers 3 (c) ( f  g )( x ) = f ( x + 1) = ( x + 1)2 − 1 3 3 = 2 = x + 2 x x ( x + 2) Domain: all x ≠ 0, − 2 55. (a)

(b)

− 13

They are equal. 58. (a)

Domain: all real numbers

( f  g )( x ) = ( g  f )( x ) =

x = x1 / 4

Domain: all x ≥ 0 2

(b)

= x + 4, x ≥ −4

(b)

f°g=g°f −6

( f  g )( x ) = f ( g ( x ) ) = f ( x 2 ) = x 2 + 4 ( g  f )( x ) = g ( f ( x ) ) = g ( x + 4 ) = ( x + 4 )

)

f°g=g°f

8 −1

−6

−1

f°g

g°f 0

They are equal. 6

They are not equal.

Section 1.6

59. (a)

( f  g )( x ) = f ( g ( x ) ) = f ( x 6 ) = ( x 6 )

= x4

Domain: all real numbers

( g  f )( x ) = g ( f ( x ) ) = g ( x 2 3 ) = ( x 2 3 ) = x 4 6

(b)

1 1 [(4 x + 1) − 1] = [4 x ] = x 4 4 1  1  ( g  f )( x ) = g  ( x − 1)  = 4  ( x − 1) + 1 = ( x − 1) + 1 = x 4  4  (b)They are equal because x = x.

f°g=g°f −1

They are equal. 60. (a)

( f  g )( x ) = f ( g ( x ) ) = f ( − x 2 + 1) = − x 2 + 1 Domain: all real numbers

( g  f )( x ) = g ( f ( x ) ) = g ( x ) = − x + 1

103

62. (a) ( f  g)( x ) = f (4 x + 1) =

Combinations of Functions

f ( g( x ))

g( f ( x ))

–1

= 1 − x2

(b)

( f  g)( x ) = f ( g( x)) = f ( x 2 − 5) = ( x 2 − 5) + 6 = x 2 + 1 ( g  f )( x ) = g( f ( x )) = g

f°g −6

( x +6) =( x +6) −5 2

= ( x + 6) − 5 = x + 1, x ≥ −6

g°f

x 2 + 1 ≠ x + 1.

(b) They are not equal because −4

(c)

They are not equal. 61. (a)

63. (a)

( f  g )( x) = f ( g ( x)) = f ( 15 ( x − 4))

= 5( 15 ( x − 4)) + 4

f ( g( x ))

g( f ( x ))

–2

–1

= x − 4 + 4 = x

64. (a)

( g  f )( x) = g ( f ( x)) = g ( 5 x + 4) = 15 ((5 x + 4) − 4) = 15 (5 x)

( x + 10 ) =  x + 10  − 4 3

( g  f )( x ) = g( x 3 − 4) = 3 ( x 3 − 4) + 10 = 3 x 3 + 6 (b) They are not equal because x + 6 ≠ 3 x 3 + 6.

= x

(c)

f ( g( x ))

g( f ( x ))

–2

g ( x)

− 54

− 35 − 52

− 15

( f  g )( x)

0 1

f ( x)

4 9 14 19

( g  f )( x) 0

2 2

3 3

= ( x + 10) − 4 = x + 6

(b) They are equal because x = x. (c)

( f  g)( x ) = f

−2

104

Chapter 1

65. (a)

Functions and Their Graphs

( f  g )( x ) = f ( g( x )) = f (2 x 3 ) = 2 x 3

( )

( g  f )( x ) = g( f ( x )) = g x = 2 x

(b) They are equal because 2 x 3 = 2 x . (c)

f ( g( x ))

g( f ( x ))

–1

f ( x ) = 3 x − 4 and g( x ) = x 2 or

f ( x ) = 3 − x and g( x ) = 4 − x 2 or

f ( x ) = 9 x and g( x ) = ( x 2 − 4)3 .

74. Let g( x ) = 9 − x and f ( x) =

( f  g )( x ) = f ( g( x )) = f ( − x ) =

6 6 = 3( − x ) − 5 −3 x − 5

−6  6   6  ( g  f )( x ) = g   = − =  3x − 5   3x − 5  3x − 5 6 −6 . ≠ (b) They are not equal because −3 x − 5 3 x − 5

67. (a)

(b)

f ( g( x ))

g( f ( x ))

−

6 5

3 − 4

−

6 11

–6

−

3 7

−

3 2

( f + g )(3) = f (3) + g(3) = 2 + 1 = 3

 f f (2) 0 = =0   (2) = g g (2) 2  

68. (a) (b)

( f − g )(1) = f (1) − g(1) = 2 − 3 = −1 ( fg )(4) = f (4) ⋅ g(4) = 4 ⋅ 0 = 0

69. (a)

( f  g )(3) = f ( g (3)) = f (1) = 2

(b)

( g  f )(2) = g ( f (2)) = g (0) = 4

70. (a) (b)

73. Let f ( x ) = 3 x and g( x ) = x 2 − 4, then ( f  g )( x ) = h( x ). This answer is not unique. Other possibilities are

66. (a)

(c)

72. Let g( x ) = 1 − x and f ( x ) = x 3 , then ( f  g )( x ) = h( x ). This answer is not unique. Another possibility is f ( x ) = ( x + 1)3 and g( x ) = − x.

( f  g )(1) = f ( g(1)) = f (3) = 2 ( g  f )(3) = g( f (3)) = g(2) = 2

then ( f  g ) ( x ) = h( x ). This answer is not unique. Another possibility is f ( x ) = 9 + x and g( x ) = − x.

1 and g( x ) = x + 2, x then ( f  g )( x ) = h( x ). This is not a unique solution. Other possibilities are

75. Let f ( x ) =

1 1 and g ( x) = x or f ( x) = and x+2 x +1 g ( x ) = x + 1. f ( x) =

4 , then ( f  g )( x ) = h( x ). x2 This answer is not unique. Another possibility is 4 f ( x ) = and g( x ) = (5 x + 2)2 . x 77. Let f(x) = x2 + 2x and g(x) = x + 4, then ( f  g )( x ) = h( x ). 76. Let g( x ) = 5 x + 2 and f ( x ) =

This answer is not unique. Another possibility is f(x) = x and g(x) = (x + 4)2 + 2(x + 4). 78. Let g( x ) = x + 3 and f ( x ) = x 3 2 + 4 x1 2, then ( f  g )( x ) = h( x ). This answer is not unique.

Another possibility is f ( x ) = ( x + 1)3 2 + 4( x + 1)1 2 and g( x ) = x + 2. 79. (a)

(b)

T ( x ) = R( x ) + B( x ) =

3 1 x + x2 4 15

300

T B R 0

(c)

B( x ) contributes more to T ( x ) at higher speeds.

71. Let f ( x ) = x 2 and g( x ) = 2 x + 1, then ( f  g )( x ) = h( x ). This is not a unique solution.

Another possibility is f ( x ) = ( x + 1)2 and g( x ) = 2 x.

Section 1.6 80. (a)

Year

2002

2003

2004

2005

222.92

238.52

252.88

266

1115.36 1222.46 1325.84 1425.5

144.12

151.82

159.28

2006

2007

2008

2009

277.88

288.52

297.92

306.08

1521.44 1613.66

1702.16 1786.94

173.48

186.72

Year 2010

2011

2012

313

318.68

323.12

1868 1945.34 2018.96

199

204.78

105

3000

yT y2 y1

166.5

Year

180.22

(b)

Combinations of Functions

y3 13

yT represents the total out-of-pocket payments, insurance premiums, and other types of payments in billions of dollars spent on health consumption expenditures in the United States and Puerto Rico for each year t.

192.98

81. (a)

(b) (c)

210.32

r( x) =

x 2

A(r ) = π r 2 ( A  r )( x ) = A(r ( x )) 2

The models are a good fit for the data. The variation of the model from the actual data is small in comparison to the sizes of the numbers.

1 x x = A   = π   = π x2 4 2 2 A  r represents the area of the circular base of the x tank with radius . 2 82. ( A  r )(t ) = A(r (t )) = A(0.6t )

= π (0.6t )2 = 0.36π t 2 ( A  r )(t ) gives the area of the circle as a function of time.

83. (a)

(

)

Since T = S1 + S2 , T = 973 + 1.3t 2 + (349 + 72.4t ) 2

T = 1.3t + 72.4t + 1322 (b)

3000

T S2 S1 9

84. (a)

(

)

Since P = R − C , P = (341 + 3.2t ) − 254 − 9t + 1.1t 2 − (341 + 3.2t ) 2

P = −1.1t + 12.2t + 87

(b)

450

R C P 9

106

Chapter 1

85. (a)

Functions and Their Graphs

(

( N  T )(t ) = N (T (t )) = N ( 2t + 1)

)

90. True. ( f  g )( x ) = f g ( x ) is only defined when g ( x )

is in the domain of f.

= 10( 2t + 1) − 20( 2t + 1) + 600 2

= 40t 2 + 590

N  T represents the number of bacteria after t hours outside the refrigerator.

( )

(b) ( N  T )(12) = 10 252 − 20( 25) + 600 = 6350 There are 6350 bacteria in a refrigerated food product after 12 hours outside the refrigerator. (c) 40t 2 + 590 = 1200 40t 2 = 610 t 2 = 15.25 t ≈ ± 3.9

N = 1200 when t ≈ 3.9 hours. 86. (a)

Area = π r 2 , r (t ) = 5.25 t. Hence

91. (a) If f ( x ) = x 2 and g( x ) = 2

12 1  1  f ( g( x )) =  = = = h( x ).  2 ( x − 2)2  x − 2  ( x − 2) 1 (b) If f ( x ) = and g( x ) = x 2 , then x −2 1 f ( g( x )) = 2 ≠ h( x ). x −2 1 (c) If f ( x ) = and g( x ) = ( x − 2)2 , then x 1 f ( g( x )) = = h( x ). ( x − 2)2 92. Let f ( x ) and g( x ) be odd functions, and define h( x ) = f ( x ) g( x ). Then h(− x ) = f (− x )g( − x )

= [ − f ( x )][ − g( x )] since f and g are both odd = f ( x )g( x ) = h( x ).

( A  r )(t ) = π 5.25 t  = 27.5625π t, t ≥ 0   (b) ( A  r )(36) = 27.5625π (36) = 992.25π ≈ 3117 square meters A = 6250 = 27.5625π t  t ≈ 72.2 hours (c) 87. First, write the distance each plane is from point P. The plane that is 200 miles from point P is traveling at 450 miles per hour. Its distance is 200 − 450t. Similarly, the other plane is 150 − 450t from point P.

So, the distance between the planes s(t ) can be found using the distance formula (or the Pythagorean Theorem): s(t ) = (200 − 450t )2 + (150 − 450t )2 s(t ) = 50 162t 2 − 126t + 25

88. (a)

R ( p ) = p − 2000

(b)

S ( p ) = 0.91 p

(c)

( R  S )( p ) = 0.91 p − 2000 This is the cost if the discount is taken before the rebate.

( S  R )( p ) = 0.91( p − 2000) This is the cost if the rebate is taken before the discount. (d)

( R  S )(24,795) = $20,563.45 ( S  R)(24,795) = $20,743.45 The discount first yields a lower cost because the discount is applied to the full amount and then the rebate is taken.

1 , then x −2

Thus, h is even. Let f ( x ) and g( x ) be even functions, and define h( x ) = f ( x ) g( x ). Then h(− x ) = f ( − x ) g( − x )

= f ( x ) g( x ) since f and g are both even = h( x ). Thus, h is even. 93. The product of an odd function and an even function is odd. Let f be odd and g even. Then ( fg )(− x ) = f (− x )g(− x ) = − f ( x ) g( x ) = −( fg )( x ). Thus, fg is odd. 94. Let A, B, and C be the three siblings, in decreasing age. 1 Then A = 2 B and B = C + 6. 2 1  (a) A = 2 B = 2  C + 6  = C + 12 2 

(b) If A = 16, then B = 8 and C = 4.

1 95. From Exercise 94, A = 2 B and B = C + 6. 2 (a)

2( B − 6) = C and B =

1 A. Hence, 2

1  C = 2  A − 6  = A − 12. 2  (b) If C = 2, then B = 7 and A = 14.

89. False. g( x ) = x − 3

Section 1.7 96. (a)

Matches L2 because each S-value is half the p-value. Matches L1 because each S-value is 5 less than the p-value.

(b)

Matches L4 because ( g  f )( p ) represents

(c)

Matches L3 because ( f  g )( p ) represents

107

97. Three points on the graph of y = − x 2 + x − 5 are (0, − 5), (1, − 5), and (2, − 7).

1 3 x − 4 x2 + 1 5 are (0, 1), (1, − 2.8), and (−1, − 3.2).

98. Three points on the graph of y =

subtracting $5 after applying a 50% discount. (d)

Inverse Functions

99. Three points on the graph of x 2 + y 2 = 49 are

(7, 0), (− 7, 0), and (0, 7).

subtracting $5 from the price and then taking a 50% discount.

x x2 − 5 1 1   are (0, 0), 1, −  , and  −1,  . 4 4  

100. Three points on the graph of y =

Section 1.7 Inverse Functions 1. 2.

inverse, f −1

y=x

one-to-one

If a function is one-to-one, no two x-values in the domain can correspond to the same y-value in the range. Therefore, a horizontal line can intersect the graph at most once.

No. If both the points (1, 4) and (2, 4) lie on the graph of a function, then the function is not one-to-one; it would not pass the Horizontal Line Test.

f ( x) = 6 x 1 f −1 ( x ) = x 6

f(f

−1

(

( x) = x − 3 ( x)) = f ( x − 3) = ( x − 3) + 3 = x

−1

f −1 ( f ( x)) = f −1 ( x + 3) = ( x + 3) − 3 = x

x −1 2 f −1 ( x) = 2 x + 1 f ( x) =

11.

2x + 1 − 1 2x = = x 2 2  x − 1  x − 1 f −1 ( f ( x)) = f −1   = 2 +1  2   2  f ( f −1 ( x)) = f ( 2 x + 1) =

= x −1+1 = x

f ( x) = 4( x − 1)

12.

1  1  f f ( x) = f  x  = 6 x  = x 6  6  1 f −1 f ( x ) = f −1 ( 6 x ) = ( 6 x ) = x 6 1 f ( x) = x 3

(

range, domain

f ( x) = x + 3

10.

)

−1

( x) = 14 x + 1

(

)

( )=x f ( f ( x)) = f −1 ( 4( x − 1)) = 14 ( 4( x − 1)) + 1 =

(

1 (3x ) = x 3 1  1  f ( x ) = f −1  x  = 3  x  = x 3   3 

)

f −1

(

)

f ( x) = x + 11 f −1 ( x) = x − 11

4 14 x

−1

= x −1+1 = x

f −1 ( x ) = 3 x

f f −1 ( x ) = f ( 3 x ) =

)

f ( f −1 ( x)) = f ( 14 x + 1) = 4 ( 14 x + 1) − 1

13.

f ( x) = 3 x

f −1 ( x ) = x 3

(

) ( ) f ( f ( x )) = f ( x ) = ( x ) = x f f −1 ( x ) = f x 3 = 3 x 3 = x −1

−1

f ( f −1 ( x)) = f ( x − 11) = ( x − 11) + 11 = x f −1 ( f ( x)) = f −1 ( x + 11) = ( x + 11) − 11 = x

108

Chapter 1

14.

f ( x ) = x7

Functions and Their Graphs

21.

f −1 ( x ) = 7 x

(

−1

)

) ( x −4)

(

g f ( x) = g

15. The inverse is a line through ( −1, 0 ) . Matches

graph (c).

( x −4) +4 = x 2

16. The inverse is a line through (0, 6) and (6, 0). Matches graph (b).

g f

17. The inverse is half a parabola starting at (1, 0). Matches graph (a). 18. The inverse is a reflection in y = x of a third-degree equation through (0, 0). Matches graph (d). 19.

( ) = ( x + 4) − 4 = x

f g ( x ) = f x2 + 4 , x ≥ 0

( ( x )) = f ( x ) = ( x ) = x f ( f ( x )) = f ( x ) = x = x f f

−1

f ( x ) = x − 4; g ( x ) = x 2 + 4, x ≥ 0

Reflections in the line y = x 22.

f ( x ) = x3 , g ( x ) = 3 x 3

(

−6

)

f g( x) = f

( 9− x) = 9−( 9− x)

= 9 − (9 − x) = x

) (

(

)

(

)

g f ( x ) = g 9 − x2 = 9 − 9 − x2 = x2 = x

f ( x ) = 9 − x 2, x ≥ 0

g( x) = 9 − x , x ≤ 9

( ) ( x)=( x) = x g ( f ( x )) = g ( x ) = x = x f g( x) = f

f −4

Reflections in the line y = x 20.

1 1 , g( x) = x x x 1 1 1 f g( x) = f   = = 1 ÷ = 1⋅ = x 1 x x 1 x

f ( x) =

(

)

1 x 1 1 g f ( x) = g  = = 1 ÷ = 1⋅ = x 1 x x 1 x

(

)

Reflections in the line y = x 23.

f ( x ) = 1 − x3 , g ( x ) = 3 1 − x

( ) ( 1 − x ) = 1 − ( 1 − x ) = 1 − (1 − x ) = x g ( f ( x ) ) = g (1 − x ) = 1 − (1 − x ) = x = x f g( x) = f

f=g −6

f 6 −6

g −4

The graphs are the same.

−4

Reflections in the line y = x

Section 1.7

24. f ( x ) =

Inverse Functions

109

1 1− x , x ≥ 0; g ( x ) = , 0 < x ≤1 1+ x x

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

(

)

 1  1+ x 1 x 1−  − x x +1 1 + x  1 + x 1 + x 1 + x  1   = = = ⋅ =x g( f ( x )) = g  = 1 1  1  1+ x 1 1+ x    1+ x 1+ x 1+ x  4

g f 0

Reflections in the line y = x 25. (a)

 2x + 6  f ( g( x )) = f  −  7  

26. (a)

7  2x + 6  = − − −3= x 2 7   7  g( f ( x )) = g  − x − 3   2  =−

(b)

2 ( − 72 x − 3 ) + 6 7

(b)

g −9

−6

−9

(c)

−6

(c)

(4 x + 9) − 9 =x 4  x−9  x−9 g( f ( x )) = g   = 4 +9= x 4    4  f ( g( x )) = f (4 x + 9) =

X −9 4 Y2 = 4 X + 9

Y1 =

−3

5 9

Y3 = Y1 (Y2 )

7 Y1 = − X − 3 2 2X + 6 Y2 = − 7 Y3 = Y1 (Y2 )

Y4 = Y2 (Y1 ) 27. (a)

f ( g( x )) = f

( x − 5) = ( x −5) + 5 = x 3

g( f ( x )) = g( x 3 + 5) = 3 ( x 3 + 5) − 5 = x

Y4 = Y2 (Y1 )

(b)

−4 −2

2 4

f g −7

−4

(c)

Y1 = X 3 + 5 Y2 = 3 X − 5

Y3 = Y1 (Y2 ) Y4 = Y2 (Y1 )

X −2

−1 0 1 4

–2 –1 0 1 4

110

Chapter 1

Functions and Their Graphs

( 2x ) = x 28. (a) f ( g( x )) = f ( 2 x ) = 2 3

30. (a)

 x 4 + 10  f ( g ( x)) = f   3    x 4 + 10  = 4 3  − 10 = x, x ≥ 0 3  

 x3   x3  g( f ( x )) = g   = 3 2   = x  2   2  (b)

(

g ( f ( x)) = g 4 3x − 10

( 3x − 10 ) + 10 = 3x − 10 + 10

g −6

3 10 = x, x ≥ 3

−4

(c)

Y1 =

X3 2

Y2 = 3 2 X

Y3 = Y1 (Y2 ) Y4 = Y2 (Y1 ) 29. (a)

−2

0 2 4 6

(b)

x 4 + 10 ,x ≥ 0 3 Y3 = Y1 (Y2 )

Y2 =

Y4 = Y2 (Y1 )

= 8 + ( − x − 8 )2 = 8 + ( x − 8) = x, x ≥ 8 24

−6

Y1 = − x − 8 Y2 = 8 + x 2 , x = 0

Y3 = Y1 (Y2 ) Y4 = Y2 (Y1 )

14 −1

g( f ( x )) = g(− x − 8 )

(c)

−1

= − x 2 = −( − x ) = x, x ≤ 0

= − (8 + x 2 ) − 8

−9

f ( g( x )) = f (8 + x 2 )

(b)

)

8 9

15 20

31. (a)

10 3

2 4

5 5

11 7

20 20 10

x f ( g( x )) = f   2 x = 2  = x 2 g( f ( x )) = g(2 x ) 2x = =x 2

(b)

f g −6

−4

(c)

Y1 = 2 x

x Y2 = 2 Y3 = Y1 (Y2 ) Y4 = Y2 (Y1 )

X –4 −2 0 2 4

–4 −2 0 2 4

Section 1.7

32. (a)

 x − 5  x − 5 f ( g ( x )) = f  −  = − 3 − +5 3  3    = x −5+5 = x

34. (a)

 2x + 3  5x  x −1  + 3   x = = − 1 = x, x ≠ 1 5  2x + 3   x −1  − 2 x −1  

3x = = x 3

 x+3 g( f ( x )) = g    x−2

g −6

 x+3 5x 2 +3 x − 2   x = = − 2 = x, x ≠ 2 5  x+3  x − 2  −1 x 2 −  

f −4

(b)

x −5 3 Y3 = Y1 (Y2 )

Y2 = −

(c)

20 14

5 −7 − 25

20 14

5 −7 − 25 4

 5x + 1  f ( g( x )) = f  −   x −1 

1 3

38. Yes. No two x-values in the domain correspond to the same y-value in the range. 39. Not a function 40. It is the graph of a function, but not one-to-one. 41. It is the graph of a one-to-one function. 42. It is the graph of a one-to-one function.

43. It is the graph of a one-to-one function.

44. Not a function

− 10

x −1 x+5 5x + 1 Y2 = − x −1 Y3 = Y1 (Y2 )

Y4 −2 −1

37. No. Both x-values, –3 and 0, in the domain correspond to the y-value 6 in the range.

Y3 −2 −1

36. No. Both elements, 1 2 hour and 1 hour, in the domain correspond to the same element, $40, in the range.

 x −1  6x 5 +1 x + 5   x =− = + 5 = x, x ≠ −5 6  x −1   x + 5  −1 x +5  

− 12

X −2 −1

35. Yes. No two elements, number of cans, in the domain correspond to the same element, price, in the range.

 x −1  g( f ( x )) = g    x+5

x+3 x−2 2x + 3 Y2 = x −1 Y3 = Y1 (Y2 )

Y1 =

Y4 = Y2 (Y1 )

 5x + 1  6x  − x −1  −1   x = = − 1 = x, x ≠ 1 6  5x + 1  − x −1  + 5 x −1  

(c)

−6

− 5 −3 0

(b)

f − 10

Y4 −5 −3 0

33. (a)

Y4 = Y2 (Y1 ) X

111

 2x + 3  f ( g( x )) = f    x −1 

(− 3x + 5) − 5 g ( f ( x)) = g ( − 3x − 5) = −

(b)

Inverse Functions

1 x 2 f is one-to-one because a horizontal line will intersect the graph at most once.

45. f ( x ) = 3 −

Y1 =

Y4 = Y2 (Y1 )

−2 −1 0 1 2

−4

8 −2

112

Chapter 1

Functions and Their Graphs

46.

f ( x) = 2 x1 3 + 5

51.

f ( x ) = 10 f is not one-to-one because the horizontal line y = 10 intersects the graph at every point on the graph.

f does pass the Horizontal Line Test, so f is one-to-one. 10

−9

9 − 12

−2

47. h( x ) =

12 −2

x2 x2 + 1

52.

f ( x) = −0.65 f is not one-to-one because it does not pass the Horizontal Line Test.

h is not one-to-one because some horizontal lines intersect the graph twice.

−3 −3

−2

−1

4−x 6 x2 g does not pass the Horizontal Line Test, so g is not one-to-one.

48. g( x ) =

53. g( x ) = ( x + 5)3

g is one-to-one because a horizontal line will intersect the graph at most once. 4

− 10 −6

6 −4

−2

54.

49. h( x ) = 16 − x 2

f ( x) = x 5 − 7 f is one-to-one because it passes the Horizontal Line Test.

h is not one-to-one because some horizontal lines intersect the graph twice.

6 − 15

−6

− 10

−2

50.

f ( x ) = −2 x 16 − x

55. 2

f is not one-to-one because it does not pass the Horizontal Line Test.

h( x) = x − x − 4 h is not one-to-one because some horizontal lines intersect the graph more than once. 8

16 − 12 − 10

10 −8 − 16

Section 1.7

56.

f ( x) = −

x −9

61.

x+7

f is not one-to-one because it does not pass the Horizontal Line Test. 1 −6

−3

57. 3

62.

The graph of the function passes the Horizontal Line Test and does have an inverse function. 58.

y 4 3 2

63.

1 −1 −1

Inverse Functions

3x + 4 6 5 3x + 4 −9 y= 5 3y + 4 x= −6 5 5x = 3y + 4 5x − 4 = 3y 5x − 4 =y 3 5x − 4 f −1 ( x ) = 3 f is one-to-one and has an inverse function.

113

f ( x) =

1 f ( x) = 3 x + 5 −8 y = 3x + 5 x = 3y + 5 x − 5 = 3y −7 x−5 =y 3 x−5 f −1 ( x ) = 3 f is one-to-one and has an inverse function.

1 is not one-to-one. x2 For example, f (1) = f (−1) = 1. f ( x) =

−2 −3

The graph of the function does not pass the Horizontal Line Test and does not have an inverse function. 59.

f ( x) = x 4 f is not one-to-one. For example, f (2) = f (−2) = 16.

−4

4 −1

64. h( x ) =

4 is not one-to-one. x2

For example, h(1) = h(−1) = 4.

−4

4 −1 −6

60. g( x ) = x 2 − x 4 g is not one-to-one. For example, g(1) = g(−1) = 0. 1 −4

6 −1

65.

f ( x ) = ( x + 3)2 , x ≥ −3, y ≥ 0 y = ( x + 3) , x ≥ −3, y ≥ 0 x = ( y + 3)2 , y ≥ −3, x ≥ 0 x = y + 3, y ≥ −3, x ≥ 0 y = x − 3, x ≥ 0, y ≥ −3

−5

−3

f is one-to-one and has an inverse function.

Chapter 1

114

Functions and Their Graphs

66. q( x ) = ( x − 5)2

70.

y = ( x − 5)2 , x ≤ 5

f ( x) =

x = ( y − 5)2 , y ≤ 5

− x = y − 5, y ≤ 5

f ( x) =

−2 −1

y=− x +5 q is one-to-one and has an inverse function. 67.

x2 , x ≥ 0 is one-to-one. x +1 2

3 f ( x) = 2 x + 3  x ≥ − , y ≥ 0 2 3 y = 2 x + 3, x ≥ − , y ≥ 0 2 3 x = 2 y + 3, y ≥ − , x ≥ 0 2 3 x 2 = 2 y + 3, x ≥ 0, y ≥ − 2 x2 − 3 3 y= , x ≥ 0, y ≥ − 2 2

x2 , x ≥ 0, 0 ≤ y < 1 x +1 2

y =

x2 x +1

x =

y2 y +1

x( y 2 + 1) = y 2 xy 2 + x = y 2 xy 2 − y 2 = − x y 2 ( x − 1) = − x y2 =

f is one-to-one and has an inverse function. 6

x 1− x

y =

x , 0 ≤ x < 1, y ≥ 0 1− x

f −1 ( x ) =

x , 0 ≤ x < 1, y ≥ 0 1− x

f does have an inverse. −3

−2

f ( x ) = x − 2  x ≥ 2, y ≥ 0

68.

y = x − 2, x ≥ 2, y ≥ 0 71.

x = y − 2, y ≥ 2, x ≥ 0 x = y − 2, x ≥ 0, y ≥ 2

x = 4y − 9

x + 2 = y, x ≥ 0, y ≥ 2 f is one-to-one and has an inverse function.

− 12

x +9 = y 4 x +9 f −1 ( x) = 4

− 10

Reflections in the line y = x

−1

f ( x ) = x − 2 , x ≤ 2, y ≥ 0

72.

y = x−2 x = y − 2 , y ≤ 2, x ≥ 0

f −1

x + 9 = 4y

69.

f ( x) = 4 x − 9 y = 4x − 9

−3

−9

x = −( y − 2) since y − 2 ≤ 0. x = −y + 2 y = − x + 2, x ≥ 0, y ≤ 2 f is one-to-one and has an inverse function.

−6

f ( x) = 3 x y = 3x −6 x = 3y x =y 3 x f −1 ( x ) = 3 Reflections in the line y = x

f −1 6

−4

Section 1.7 f ( x) = x 5

73.

y = x5 −6

x = y5

79.

f −1 6

y=5 x

−4

f −1 ( x ) = 5 x Reflections in the line y = x

y = x +1

x = y3 + 1 3

−6

f −1 ( x) = 3

3 −4

x −1 = y −1

y = 3

f −1

80.

f ( x) = x , x ≤ 0, y ≥ 0 4

−

−9

f −1

x = y

f ( x) = − 4 x , x ≥ 0, y ≤ 0 Reflections in the line y = x −1

76.

f ( x ) = x 2, x ≥ 0 y=x

x=y

−6

x 2 = 16 − y 2

x 6

f −1 f

x 6

36 y

x +2= y

Domain of f : x ≥ 2 −1

Domain of f : x ≥ 0

f ( x ) = 16 − x 2 , − 4 ≤ x ≤ 0 x = 16 − y 2 , − 4 ≤ y ≤ 0

f −1 ( x ) = x + 2

y = 16 − x 2

4 ,x ≠ 0 x

x = y−2 4

f ( x) = 4 − x , 0 ≤ x ≤ 2 The graphs are the same. 78.

4 x

x = ( y − 2)2

f = f −1

y = 4 − x2 −1

−4

y = ( x − 2)2

x = 4 − y2

y2 = 4 − x 2

Reflections in the line y = x

x 2 = 4 − y2

f −1

81. If you let f ( x ) = ( x − 2)2, x ≥ 2, then f has an inverse function. [Note: You could also let x ≤ 2.]

f ( x) = 4 − x 2 , 0 ≤ x ≤ 2 y= 4−x

f −1

−6

36 , x>0 x2 36 f −1 ( x ) = 2 , x > 0 x

Reflections in the line y = x 77.

f f −1

x =y

x= x2 =

f −1 ( x ) = x

f ( x) = y=

y = x x = y4

Reflections in the line y = x

f ( x) = x − 1 Reflections in the line y = x 75.

115

4 x

x −1 = y

4 ,x ≠ 0 x3 4 y = 3 x 4 x = 3 y

f ( x) =

y3 =

f ( x) = x 3 + 1

74.

Inverse Functions

Range of f : y ≥ 0 Range of f −1 : y ≥ 2

f −6

f −1 −4

y 2 = 16 − x 2

y = − 16 − x 2

f −1 ( x ) = − 16 − x 2 , 0 ≤ x ≤ 4 Reflections in the line y = x

116

Chapter 1

Functions and Their Graphs

82. If you let f ( x ) = x 4 + 1, x ≥ 0, then f has an inverse

function. [Note: You could also let x ≤ 0. ]

86. If you let f ( x ) = ( x − 4)2 , x ≥ 4, then f has an inverse function. [Note: You could also let x ≤ 4.]

y = x4 + 1

y = ( x − 4)2

x = y4 + 1

x = ( y − 4)2 x = y−4

x − 1 = y4 4

y= x +4

x −1 = y

f −1 ( x ) = x + 4

f −1 ( x) = 4 x − 1

Domain of f : x ≥ 0

Range of f : y ≥ 1

Domain of f −1 : x ≥ 1 Range of f −1 : y ≥ 0 83. If you let f ( x ) = x + 2 , x ≥ −2, then f has an inverse

function. [Note: You could also let x ≤ −2.]

Domain of f : x ≥ 4 Domain of f

−1

Range of f : y ≥ 0 Range of f −1 : y ≥ 4

:x≥0

87. If you let f ( x) = − 2 x 2 − 5, x ≥ 0, then f has an

inverse function. [Note: You could also let x ≤ 0. ] y = − 2x2 − 5

y = x+2

x = − 2 y2 − 5

x= y+2 x −2 = y

x + 5 = − 2 y2

f −1 ( x ) = x − 2 Domain of f : x ≥ –2 −1

Domain of f : x ≥ 0

−

x +5 = y2 2

−

x +5 = y 2 f −1 ( x) =

Range of f : y ≥ 0 Range of f −1: y ≥ –2

−

x +5 2

Domain of f : x ≥ 0

Range of f : y ≤ − 5

−1

84. If you let f ( x ) = x − 2 , x ≥ 2, then f has an inverse

function. [Note: You could also let x ≤ 2.]

y = x −2 x = y−2 x+2= y

Domain of f : x ≤ − 5 Range of f −1 : y ≥ 0 1 2 x + 1, x ≥ 0, then f has an inverse 2 function. [Note: You could also let x ≤ 0. ]

88. If you let f ( x) =

1 2 x +1 2 1 x = y2 + 1 2 1 x − 1 = y2 2 2( x − 1) = y 2 y =

f −1 ( x ) = x + 2 Domain of f : x ≥ –2 –1

Domain of f : x ≥ 0

Range of f : y ≥ 0 Range of f –1: y ≥ –2

85. If you let f ( x ) = ( x + 3)2 , x ≥ −3 then f has an inverse function. [Note: You could also let x ≤ −3. ]

y = ( x + 3)

x = ( y + 3)

2( x − 1) = y f −1 ( x) =

2( x − 1)

f −1 ( x) =

2x − 2

x = y+3 y = x −3

Domain of f : x ≥ 0

f −1 ( x) = x − 3

Range of f : y ≥ 1

−1

Domain of f : x ≥ 1 Range of f −1 : y ≥ 0

Domain of f : x ≥ −3

Range of f : y ≥ 0

Domain of f −1 : x ≥ 0

Range of f −1 : y ≥ −3

Section 1.7

function. [Note: You could also let x ≤ 4.]

101. f ( x ) = x 3 + x + 1

y = x − 4 +1

(a) and (b)

y = x − 3 because x ≥ 4 x = y −3 y = x +3

f f −1 −6

f −1 ( x ) = x + 3

−3

Domain of f : x ≥ 4

Range of f : y ≥ 1

Domain of f −1 : x ≥ 1

Range of f −1 : y ≥ 4

90. If you let f ( x ) = − x − 1 − 2, x ≥ 1, then f has an inverse

102. f ( x) = x 4 − x 2

function. [Note: You could also let x ≤ 1.]

(a) and (b)

y = − x − 1 − 2 = −( x − 1) − 2 because x ≥ 1

y = −x − 1 x = –y – 1 x + 1 = −y

−3

f −1 ( x ) = − x − 1 Range of f : y ≤ −2

Domain of f : x ≤ −2 Range of f −1 : y ≥ 1

−2 −1 1

−4 −2 2

f −1

−1

f ( x)

f −2

Domain of f : x ≥ 1

f −1 ( x ) −2 −1 1 3

x −4 −2 2

103. f ( x) =

(a) and (b)

1 −4 −3

3x 2 x2 + 1

−1

−6

f −1

−2 −4

−3

92.

x 4

f ( x) −3

−2

−1 −2

0 6

x −3 −2

0 6

The graph of the inverse relation is not an inverse function since it does not satisfy the Vertical Line Test.

f −1 ( x ) 4 3 −1 −2

6 5

104. f ( x ) =

4 3 2

−3 −2 −1

4x 2

x + 15

(a) and (b) 4

−2

−3

93. f

−1

(− 4) = 4 because f (4) = − 4.

94. g −1 (0) = −2 because g(−2) = 0. 95. ( f  g)(2) = f (3) = −2 96. g( f (−4)) = g(4) = 6 97.

117

100. ( f −1  g −1 )(6) = f −1 ( g −1 (6)) = f −1 (4) = −4

89. If you let f ( x ) = x − 4 + 1, x ≥ 4, then f has an inverse

91.

Inverse Functions

f −1 ( g(0)) = f −1 (2) = 0

98. ( g −1  f )(3) = g −1 (−2) = −3 99. ( g  f −1 )(2) = g(0) = 2

−6

f −1 −4

1 x − 3, f −1 ( x ) = 8( x + 3), 8

g( x) = x 3 , g −1 ( x ) = 3 x .

( )

105. ( f −1  g −1 )(1) = f −1 ( g −1 (1)) = f −1 3 1

(

)

= 8 3 1 + 3 = 8(1 + 3) = 32

118

Chapter 1

Functions and Their Graphs

106. ( g −1  f −1 )(−3) = g −1 ( f −1 (−3)) = g −1 (8(−3 + 3)) = g −1 (0) = 3 0 = 0

107.

113. ( f  g)( x) = f (g( x)) = f (2 x − 5) = (2 x − 5) + 4 = 2 x − 1. Now find the inverse function of ( f  g)( x) = 2 x − 1.

y = 2x − 1 x = 2y − 1 x + 1 = 2y x +1 y= 2 x + 1 ( f  g)−1 ( x ) = 2

( f −1  f −1 )(− 6) = f −1( f −1(− 6)) = f −1 (8[− 6 + 3]) = f −1 ( − 24) = 8( − 24 + 3) = −168

108.

( g −1  g −1)(4) = g −1( g −1(4))

( )

= g −1 3 4 = 109. f

−1

3 3

4 =

Note that ( f  g )−1 ( x ) = ( g −1  f −1 )( x ); see Exercise 111.

( x) = 8( x + 3) and g ( x) = −1

( f −1  g −1 )( x) = f −1( g −1( x))

( )

= f −1 3 x

(

)

= 8 3 x + 3 = 8

x + 24

110. g −1 ( x) = 3 x and f −1 ( x) = 8( x + 3).

( g −1  f −1)( x) = g −1( f −1( x)) = g

−1

(8( x + 3))

= 3 8( x + 3)

114. (g  f )( x) = g( f ( x)) = g( x + 4) = 2( x + 4) − 5 = 2 x + 8 − 5 = 2 x + 3.

Now find the inverse function of ( g  f )( x) = 2 x + 3.

y = 2x + 3 x = 2y + 3 x − 3 = 2y x −3 =y 2 x −3 ( g  f )−1 ( x ) = 2 Note that ( g  f )−1 = f −1  g −1. 115. (a) Yes, f is one-to-one. For each European shoe size, there is exactly one U.S. shoe size.

= 23 x + 3

(b)

f (11) = 44

In Exercises 111–114, f ( x ) = x + 4, f −1 ( x ) = x − 4, x+5 . g( x ) = 2 x − 5, g −1 ( x ) = 2

(c)

f −1 ( 43) = 10 because f (10) = 43.

(d)

f ( f −1 ( 41)) = f (8) = 41

(e)

f −1( f (12)) = f −1( 45) = 12

111. ( g −1  f −1 )( x ) = g −1 ( f −1 ( x ))

= g −1 ( x − 4) ( x − 4) + 5 = 2 x +1 = 2

116. If two functions are inverse functions of each other, then

g ( g −1 ( x)) = g −1 ( g ( x)) = x, so g −1( g (6)) = 6.

112. ( f −1  g −1 )( x ) = f −1 ( g −1 ( x ))

 x+5 = f −1    2  x+5 = −4 2 x +5−8 = 2 x −3 = 2

Section 1.7 117. (a) CALL ME LATER corresponds to numerical values: 3 1 12 12 0 13 5 0 12 1 20 5 18. Using f to encode,

f (3) = 19 f (1) = 9 f (12) = 64 f (12) = 64 f (0) = 4 f (13) = 69 f (5) = 29 f (0) = 4 f (12) = 64 f (1) = 9 f (20) = 104 f (5) = 29 f (18) = 94

Inverse Functions

119

119. False. f ( x ) = x 2 is even, but f −1 does not exist. 120. True. If (0, b) is the y-intercept of f, then (b, 0) is the

x-intercept of f −1. 121. Yes. The inverse would give the time it took to complete n miles. 122. This situation could not be represented by a one-to-one function. Since the population was the greatest in 2012, it would have increased sometime from 2005 to 2012, reached a maximum in 2012, and would have decreased sometime from 2012 to 2015. Therefore, the graph would not pass the Horizontal Line Test, and not be one-to-one. 123. No. The function oscillates. 124. This situation could not be represented by a one-to-one function because height remains constant after a certain age.

x−4 . (b) For f ( x ) = 5 x + 4, f ( x ) = 5

125. The graph of f −1 is a reflection of the graph of f in the line y = x.

Using f −1 to decode, f −1 (119) = 23

126. If the domain of f is [0, 9] and the range is [ −3, 3], and

−1

f (44) = 8 −1

f (9) = 1 −1

f (104) = 20 f −1 (4) = 0 f −1 (104) = 20 −1

f (49) = 9

f −1 (69) = 13 f −1 (29) = 5 Converting from numerical values to letters, the message is WHAT TIME. 118. (a) y = 12 + 0.55 x x − 12 = 0.55 y x − 12 = y 0.55 x − 12 , where y −1 is the number of So, y −1 = 0.55 units produced while x is the hourly wage. 40

127. (a) The function f will have an inverse function because no two temperatures in degrees Celsius will correspond to the same temperature in degrees Fahrenheit. (b) f −1 (50) would represent the temperature in degrees Celsius for a temperature of 50° Fahrenheit. 128. Yes. The function would pass the Horizontal Line Test and therefore have an inverse function. 129. The constant function f ( x) = c, whose graph is a horizontal line, would never have an inverse function.

x = 12 + 0.55 y

(b)

since the graphs of f and f −1 can be described as if the point (a, b) lies on the graph of f, then the point (b, a) lies on the graph of f –1, then the domain of f –1 is [ −3, 3] and the range is [0, 9] .

130. (a) No, the graphs are not reflections of each other in the line y = x. (b) Yes, the graphs are reflections of each other in the line y = x. (c) Yes, the graphs are reflections of each other in the line y = x. (d) Yes, the graphs are reflections of each other in the line y = x. 131. We will show that ( f  g )−1 ( x ) = ( g −1  f −1 )( x ) for all x in their domains. Let y = ( f  g)−1 ( x )  ( f  g )( y) = x then

f ( g( y)) = x  f −1 ( x ) = g( y). Hence, ( g −1  f −1 )( x ) = g −1 ( f −1 ( x )) = g −1 ( g( y)) = y = ( f  g)−1 ( x ). Thus, g −1  f −1 = ( f  g )−1.

120

Chapter 1

Functions and Their Graphs 137. x = 5. No, it does not pass the Vertical Line Test.

132. If f is one-to-one, then f −1 exists. If f is odd, then

f (− x) = − f ( x). Consider f ( x ) = y ↔ f −1 ( y) = x. Then

138. y − 7 = − 3

f −1 (− y) = f −1 (− f ( x )) = f −1 ( f ( − x )) = − x = − f −1 ( y).

y = 4

Thus, f −1 is odd.

Yes, y is a function of x.

133.

27 x = 9 x, x ≠ 0 3x2

139. x 2 + y = 5

5 x 2 y( y + 5) 5 x 2 y 2 + 25 x 2 y = = 5 xy, x( y + 5) ≠ 0 134. xy + 5 x x( y + 5)

y = − x2 + 5 Yes, y is a function of x. 140. x − y 2 = 0

x 2 − 36 ( x − 6)( x + 6) x + 6 = = = − x − 6, x ≠ 6 135. 6−x −( x − 6) −1

y2 = x y=± x No, y is not a function of x.

x 2 + 3 x − 40 ( x − 5)( x + 8) x + 8 = = , x≠5 136. 2 x − 3 x − 10 ( x − 5)( x + 2) x + 2

Chapter 1 Review 1. y = − 12 x + 2

−2

1 2

Solution point

( −2, 3) ( 0, 2 ) ( 2, 1) ( 3, 12 ) ( 4, 0 )

y 5 4 3 1

−3 −2 −1

1 2 3 4

−2 −3 −4 −5

2. y = x 2 − 6 x

−1

−9

−8

Solution point

( −1, 7 ) ( 0, 0 ) ( 3, − 9 ) ( 4, − 8) ( 6, 0 )

2 −6 −4 − 2

8 10 12

−4 −6 −8 −10

Chapter 1 Review 3.

y = 14 x4 − 9 x2

121

y = x2 x + 4

10 − 15

−9

− 90

Intercepts: (0, 0), ( ± 6, 0) 4.

9 −1

Intercept: ( 0, 0 )

y = 12 x3 − 8x

y = 17,500 − 1400 t , 0 ≤ t ≤ 6

(a) −9

X min = 0 X max = 6 Xscl = 1 Y min = 9000 Y max = 18,000 Yscl = 1000

−18

Intercepts: (0, 0), ( ± 4, 0) 5.

y = x 9 − x2

(b) 18,000

−9

0 9,000

−6

Intercepts: ( 0, 0 ) , ( ±3, 0 )

8. (a)

4000

(b)

(c)

Year

2004

2005

2006

2007

2008

2009

2010

2011

2012

2013

Revenue

261.6

295.9

377.3

505.7

681.3

903.9

1173.6

1490.4

1854.2

2265.2

When R = 500, t ≈ 6.96. So, 2007 is the first year that the revenues exceed $5 billion.

10.

(−3, 2)

−4

−2

(8, 2)

−4 x

(− 3, − 1)

2−2 0 = =0 8 − (−3) 11

−2

(3, − 1)

−3 −4

−4

−2 −1

−5

−1 − ( −1) 0 = =0 −3 − 3 −6

122

Chapter 1

Functions and Their Graphs y

11.

15.

(− 5, 9)

(− 4.5, 6)

8 6

(2.1, 3)

2 − 10 − 8

(− 5, − 1)

−6

−2

−6

−2

m =

10 m = = ; m is undefined. −5 − ( −5) 0

y 2

(8, 2)

−1

−2

8.1 = 3 2.7

−3 −4

(8, − 1) (− 2.7, − 6.3)

−3 −4

−5 −6 −7

−5

2 − ( −1)

m =

3 = ; m is undefined. 0

8−8 y

13.

m =

17. (a)

6 4 2

( ) 3 ,1 2

(5, 52)

m =

(− 34 , 56 ) 1 −2 −3

3 y − 5 = − ( x + 3) 2

2 y − 10 = −3x − 9 3x + 2 y − 1 = 0

−1

(Other answers possible.) 18. (a)

−1

1 ( x − 2) 4 4y + 4 = x − 2 y +1 =

(2 + 4, −1 + 1) = (6, 0) (6 + 4, 0 + 1) = (10, 1) (10 + 4, 1 + 1) = (14, 2)

−2

0 − ( − 2.7)

(b) Three additional points:

( 5 2) − 1 = 3 2 = 3 5 − ( 3 2) 72 7

14.

1.8 − ( − 6.3)

− x + 4y + 6 = 0

−2

−3

1 −2

(0, 1.8)

−6 −5 −4 −3 −2

1 1

3−6 −3 30 5 = = − = − 2.1 − ( − 4.5) 6.6 66 11

16.

−4

9 − ( −1)

12.

−4

( 12 , − 52 )

5  5 5 15 10 − −  + 6  2 6 6 m = = = 3 3 1 3 2 5 − − − − − 4 2 4 4 4 10 4 8 = − ⋅ = − 3 5 3

(b) Three additional points:

(−3 + 2, 5 − 3) = (−1, 2) (−1 + 2, 2 − 3) = (1, − 1) (1 + 2, − 1 − 3) = (3, − 4) (Other answers possible.)

Chapter 1 Review

19. (a)

3 ( x − 0) 2 3 y +5 = − x 2 3 y = − x −5 2 3 x + 2 y + 10 = 0 y +5 = −

(b) Three additional points:

(0 + 2, − 5 − 3) = (2, − 8) (2 + 2, − 8 − 3) = (4, −11) (4 + 2, −11 − 3) = (6, −14) (Other answers possible.)

4 ( x − 0) 5 4 y −1 = x 5 y −1 =

20. (a)

4 − x + y −1 = 0 5 4x − 5 y + 5 = 0

123

24. (0, 0), (0, 10) m=

10 − 0 10 = 0−0 0

The slope is undefined and the line is vertical.

x=0 2

−3

−2

5  5  25.  , −1,  , 3 6  6  m =

3 − ( −1) 4 = 5 5 0 − 6 6

The slope is undefined and the line is vertical. x =

5 6 6

(b) Three additional points:

(0 − 5, 1 − 4) = ( − 5, − 3) (0 + 5, 1 + 4) = (5, 5) (5 + 5, 5 + 4) = (10, 9) (Other answers possible.) 21. (a)

y − 6 = 0( x + 2) = 0 y = 6 (horizontal line) y −6 = 0

(b) Three additional points:

(0, 6), (1, 6), (− 1, 6) (Other answers possible.)

−6 −2

 4  4 26.  7, ,  9,   3  3 4 4 − 3 = 0 = 0 m = 3 9 − 7 2

The line is horizontal. y −

y − 8 = 0( x + 8) = 0 y = 8 (horizontal line) y −8 = 0 (b) Three additional points: (0, 8), (1, 8), (2, 8)

22. (a)

(Other answers possible.) 23. (2, − 1), (4, − 1)

4 = 0( x − 7) 3 4 y = 3 2

−3

−2

−1 − ( −1) 0 m= = =0 4−2 2

The line is horizontal.

y − (−1) = 0( x − 2) y = −1 1 −3

−3

124

Chapter 1

27.

( −1, 0 ) , ( 6, 2 )

Functions and Their Graphs

(b) Perpendicular slope: m = −

2−0 2 = 6 − (−1) 7 2 y − 0 = ( x + 1) 7 2 2 y= x+ 7 7

4 5

4 y − ( −2) = − ( x − 3) 5

5y + 10 = −4 x + 12 4 x + 5y − 2 = 0 y=−

4 2 x+ 5 5

−6

−4

28.

−6

(1, 6 ) , ( 4, 2 ) 2 − 6 −4 = 4 −1 3 4 y − 6 = − ( x − 1) 3 4 22 y=− x+ 3 3

29.

2 5 2 x + and m = − 3 3 3 2 (a) Parallel slope: m = − 3 2 y − 3 = − ( x + 8) 3 3 y − 9 = −2 x − 16

32. 2 x + 3 y = 5  y = − 4

−1

−4

2x + 3 y + 7 = 0

( 3, − 1) , ( −3, 2 ) 2 − ( −1) 3 1 m= = =− −3 − 3 −6 2 1 y − ( −1) = − ( x − 3) 2 1 1 y=− x+ 2 2

2  5   30.  − , 1 ,  −4,  9  2   2 7 −1 − 14 9 9 m= = =  5  − 3 27 −4 −  −  2  2 y −1 = y=

−6

5 23 x− 4 4

3 ( x + 8) 2 2 y − 6 = 3 x + 24

3x − 2 y + 30 = 0 y=

3 x + 15 2 30

−8

5 5 x − 2 and m = 4 4 5 (a) Parallel slope: m = 4 5 y − ( −2) = ( x − 3) 4

3 2

y−3=

14 62 x+ 27 27

4 y + 8 = 5x − 15 0 = 5x − 4 y − 23

(b) Perpendicular slope: m =

−4

14   5   x −  −  27   2 

31. 5 x − 4 y = 8  y =

2 7 y =− x− 3 3

−3

− 40

−10

33. (a) Not a function. 20 is assigned two different values. (b) Function 34. (a) Function (b) Not a function. w is assigned two different values and u is unassigned. 35. 16 x 2 − y 2 = 0  y = ±4 x No, y is not a function of x. Some x-values correspond to two y-values. For example, x = 1 corresponds to

y = 4 and y = −4. 36.

x 3 + y2 = 64  y = ± 64 − x 3 No, y is not a function of x. Some x-values correspond to two y-values. For example, x = 0 corresponds to y = 8 and y = −8.

Chapter 1 Review 37.

y = 2x − 3

48.

(a)

This is a function of x. 38.

y = −2 x + 10

y = x2 + 4 This is a function of x.

41.

f ( x ) = x2 + 1

(b)

f (1) = 12 + 1 = 2

(b)

f ( −3) = ( −3) + 1 = 10

(c)

f b = b

( ) ( ) +1 = b +1

(d)

f ( x − 1) = ( x − 1) + 1 = x 2 − 2 x + 2

2017: 55.75 million; Answers will vary.

f ( x ) = 2 x2 + 3x − 1

49.

f ( x + h ) = 2 ( x + h ) + 3( x + h ) − 1 2

= 2 x 2 + 4 xh + 2 h 2 + 3 x + 3h − 1

(a)

g ( −1) =

(b)

g ( 3 ) = 32 + 1 = 9 + 1 = 10

(c)

g (3x ) =

(d)

g ( x + 2) =

f ( x + h) − f ( x ) = h

( −1) + 1 = 1 + 1 = 2 2

( 2 x + 4 xh + 2h + 3x + 3h − 1) − ( 2 x + 3x − 1) 2

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

43. The domain of f ( x ) =

x −1 is all real numbers x ≠ −2. x+2

44. The domain of f ( x ) =

x2 is the set of all real numbers. 2 x +1

= 4 x + 2h + 3, h ≠ 0 f ( x ) = x2 − 3x + 5

50.

f ( x + h ) = ( x + h ) − 3( x + h ) + 5 2

= x 2 + 2 xh + h 2 − 3 x − 3h + 5

f ( x + h) − f ( x ) h

f ( x ) = 25 − x 2

( 5 + x )( 5 − x ) ≥ 0

(

x 2 + 2 xh + h 2 − 3 x − 3h + 5 − x 2 − 3 x + 5 h 2 xh + h2 − 3h h h ( 2 x + h − 3)

= 2 x + h − 3, h ≠ 0

The domain is  −5, 5 .

x 2 − 16 ≥ 0

= =

25 − x 2 ≥ 0

f ( x ) = x 2 − 16

h 4 xh + 2h 2 + 3h = h

( 3x ) + 1 = 9 x2 + 1 = x2 + 4x + 5

46.

2015: 62.75 million; 2016: 59.6 million;

42. g ( x ) = x 2 + 1

45.

2013: 66.95 million; 2014: 65.2 million;

(a)

n(t ) 66.95 68 68.35 68

y = 1− x This is a function of x.

40.

n(t ) 64 63.8 64.9 65.2

This is a function of x. 39.

125

51.

−6

x ≥ 16 The domain is (−∞, − 4] ∪ [4, ∞).

47.

(a) C ( x) = 17,500 + 5.25 x

−4

Domain: all real numbers x Range: y ≤ 3

(b) P( x) = R( x) − C ( x)

P = 8.43x − (5.25 x − 17,500) P = 3.18 x − 17,500

)

126

Chapter 1

52.

Functions and Their Graphs 58.

−8 −9

9 −1

−5

Domain: all real numbers x Range: 5, ∞ ) 53.

Domain: all real numbers x Range: −3, ∞ ) 59. y − 4 x = x 2 A vertical line intersects the graph just once, so y is a function of x. Solve for y and graph y1 = x 2 + 4 x.

−5

10 −1

Domain:  −3, ∞ )

60. 3 x + y 2 − 2 = 0 A vertical line intersects the graph more than once, so y is not a function of x. Solve for y and graph y1 = −3 x + 2

Range:  4, ∞ ) 54.

and y2 = − −3 x + 2. 61.

f ( x ) = x3 − 3x (a)

−9

−5

Domain: 5, ∞ )

−6

Range: ( −∞, 2  55.

(b) Increasing on ( −∞, − 1) and (1, ∞ ) Decreasing on ( −1, 1)

62. −9

f ( x ) = x2 − 9 (a)

−4

Domain: 36 − x 2 ≥ 0  x 2 ≤ 36  −6 ≤ x ≤ 6 Range: 0 ≤ y ≤ 6 56.

−9

−4

(b) Increasing on ( 3, ∞ ) Decreasing on ( −∞, − 3)

−9

63. −4

f ( x) = x x − 6 (a)

Domain: ( −∞, − 3 , 3, ∞ ) Range:  0, ∞ ) 0

57.

(b) Increasing on ( 6, ∞ ) −3

12 −1

Domain: all real numbers x Range: [2, ∞)

Chapter 1 Review

64.

x +8

f ( x) =

127

 1 x + 3, x < 0 68. f ( x) =  2  4 − x 2 , x ≥ 0

(a)

y 5 4

−16

−4

2 1 −5 −4 −3 −2 −1

(b) Increasing on ( −8, ∞ )

(

f ( x ) = x2 − 4

)

3 4 5

−2 −3 −4 −5

Decreasing on ( −∞, − 8 ) 65.

69. f ( x) =  x − 3

y 3 2 1

−18

−4 −3 −2 −1

−4

−6

4 − 12

5 4 3 2

− 12

−6 −5 −4

−1

Relative maximum: ( 0, − 1)

1 2 3 4

−2 −3 −4 −5

Relative minimum: ( 2.67, − 10.48 )

71. f ( − x) = ( − x) + 6 2

3 x + 5, x < 0 67. f ( x) =   x − 4, x ≥ 0

= x2 + 6 = f ( x)

y 5

f is even. 14

2 1

−2 −3 −4 −5

70. f ( x) =  x + 2

f ( x ) = x3 − 4 x2 − 1

−1

−3

Relative maximum: ( 0, 16 )

−5 −4 −3

−2

Relative minima: ( −2, 0 ) and ( 2, 0 )

66.

1 2

4 5

−6

6 −2

The graph is symmetric with respect to the y-axis. So, f is even.

128

Chapter 1

Functions and Their Graphs

72.

f ( −x ) = ( −x ) − ( −x ) − 1

75. f ( − x ) = 3( − x )5 2 ≠ f ( x ) and f (− x) ≠ − f ( x)

= x2 + x − 1

f is neither even nor odd. (Note that the domain of f is x ≥ 0.)

≠ f ( x)

and f ( − x ) ≠ − f ( x )

f is neither even nor odd. 5

−3

6 −1

−5

The graph is neither symmetric with respect to the origin nor with respect to the y-axis. So, f is neither even nor odd.

−3

The graph is neither symmetric with respect to the origin nor with respect to the y-axis. So, f is neither even nor odd. 73.

(

f (−x) = (−x) − 8 2

(

= x2 − 8

)

76.

f (− x ) = 3( − x) 2 5 = 3 x 2 5 = f ( x ) f is even. 7

2 −6

= f ( x)

−1

f is even.

The graph is symmetric with respect to the y-axis. So, f is even.

77. f ( − x) = 2( − x) 4 − ( − x) −6

= − 2x 4 − x2

(

−8

= − 2x 4 − x2

The graph is symmetric with respect to the y-axis. So, f is even.

= − f ( x)

)

f is odd.

74. f ( − x) = 2( − x) − ( − x) 3

= − 2 x3 + x = − (2 x3 − x)

−3

= − f ( x) −4

and f ( − x) ≠ − f ( x)

The graph is symmetric with respect to the origin. So, f is odd.

f is odd. 2

78. −3

−2

The graph is symmetric with respect to the origin. So, f is odd.

f ( − x ) = ( − x ) ( − x )2 − 1 = − x x2 − 1

−6

= − f ( x) −4 f is odd. The graph is symmetric with respect to the origin. So, f is odd.

Chapter 1 Review 79. Horizontal shift three units to the right of 1 1 f ( x) = : y = x x−3 80. Horizontal shift two units to the right, followed by a vertical shift one unit upward of f ( x ) = x 2 : y = ( x − 2)2 + 1 81. Vertical shift three units upward of f ( x ) = x : y = x + 3

87. h( x ) =

(a)

1 −6 x

f ( x) =

1 x

(b) The graph of h is a vertical shift six units downward of f. y (c) 1 −5 −4 −3 −2 −1

82. Horizontal shift three units to the right, followed by a reflection in the x-axis of f ( x ) = x : y = − x − 3

1 2 3 4 5

−2 −3 −4 −5

83.

129

3 1 − 8− 7

− 4− 3− 2− 1 −1

3 4

−2

(d)

h( x) = f ( x) − 6

88. h( x ) = −

−3 −4

y = f (− x) is a reflection in the y-axis. 84.

1 + 3 x

(a) f ( x ) =

1 x

(b) The graph of h is a reflection in the x-axis and a vertical shift three units upward of f.

y 8 7 6 5 4 3 2 1

(c)

y 8 7

1 2 3 4

− 4 −3

6 7 8

4 2 1

−3 −4

y = − f ( x) is a reflection in the x-axis.

−5 −4 −3 −2 −1

1 2 3 4 5

−2

85.

(d) h( x) = − f ( x) + 3

8 7 6 5

89. h( x ) = ( x − 2)3 + 5

3 2 1 −4

− 2 −1 −2 −3 −4

1 2 3 4 5 6

y = f ( x) + 2 is a vertical shift two units upward.

(a) f ( x ) = x 3 (b) The graph of h is a horizontal shift of f two units to the right, followed by a vertical shift five units upward of f. y (c) 7 6 5 4 3 2 1

86. 3 2

1 2 3 4 5 6

− 3 −2

8 9

−2 −3 −4 −5 −6 −7 −8 −9

−2 −1

1 2 3 4 5 6 7 8

−2 −3

(d)

h( x ) = ( x − 2) 3 + 5 = f ( x − 2) + 5

y = f ( x − 1) is a horizontal shift one unit to the right.

130

Chapter 1

Functions and Their Graphs

90. h( x) = ( − x) − 8

93. h( x) = 3x + 9

(a) f ( x) = x 2

(a) f ( x) = x

(b) The graph of h is a reflection in the y-axis, followed by a vertical shift eight units downward of f.

(b) The graph of h is a vertical shift nine units upward, and a horizontal shrink of f.

(c)

4 2 −8 −6 − 4

−2

8 6 4 2 −5 −4 −3 −2 −1

−10

94.

x −6

(a) f ( x) =

y 2 −2 −2

h( x) = 2 x + 8 − 1 h( x) = 2( x + 4) − 1

(b) The graph of h is a reflection in the x-axis and a vertical shift six units downward of f. (c)

(d) h( x) = f (3 x) + 9

(d) h( x) = f ( − x) − 8 91. h( x) = −

1 2 3 4 5

8 10 12 14

(a) f ( x) = x (b) h is a horizontal shift four units to the left, followed by a vertical shift one unit downward, and a horizontal shrink of f. (c)

y 10 8

−4 −6 −8

− 10 − 12

−12 −10 −8 −6

− 14

−2 −2

(d) h( x) = f ( 2( x + 4)) − 1

(a) f ( x ) = x (b) The graph of h is a horizontal shift one unit to the right and a vertical shift four units upward of f. y 7 6

95. ( f − g)(4) = f (4) − g(4)

= 3 − 2(4) − 4 = −5 − 2 = −7 96. ( f + h)(5) = f (5) + h(5) = −7 + 77 = 70

5 4 3 2 1

(d)

−6

92. h( x ) = x − 1 + 4

−1 −1

−4

(d) h( x) = − f ( x) − 6

(c)

h( x) = f ( x − 1) + 4

97. ( f + g)(25) = f (25) + g(25) = −47 + 5 = −42 98. (g − h)(1) = g(1) − h(1) = 1 − 5 = −4 99. ( fh)(1) = f (1)h(1) = (3 − 2(1))(3(1)2 + 2)

= (1)(5) = 5 g(1) 1 g = 100.   (1) = h h (1) 5  

Chapter 1 Review 101. (h  g)(5) = h(g(5))

131

106. f ( x) = 3 x , g ( x) = ( x + 2) 2

( 5) = 3 ( 5 ) + 2 = 17

( f  g )( x ) = f ( ( x + 2 ) ) = 3 ( x + 2 ) = h( x )

3 , g ( x) = x + 2 x

107. f ( x) =

102. (g  f )(−3) = g( f (−3))

( f  g )( x ) = f ( x + 2 ) = x + 2 = h( x)

= g ( 9) = 9 =3

6 , g ( x) = 3x + 1 x3

108. f ( x) =

103. f ( x) = x 2, g ( x) = x + 3

( f  g )( x ) = f ( x + 3)

( f  g )( x ) = f ( 3 x + 1) = (3 x + 1)3 = h( x)

= ( x + 3)2 = h( x )

109.

3600

y1 + y 2

104. f ( x) = x , g ( x) = 1 − 2 x

( f  g )( x ) = f (1 − 2 x ) = (1 − 2 x ) = h( x) 3

105. f ( x) = x , g ( x) = 4 x + 2

8 1400

( f  g )( x ) = f ( 4 x + 2 ) = 4 x + 2 = h( x) 110. For 2017, t = 17. 3 2 3 2 y1 + y2 = − 5.824(17) + 182.05(17) − 1832.8(17) + 7515 + 3.398(17) − 103.63(17) + 1106.5(17) − 2543     = 3,369.342 thousand students or 3,369,342 students

111. f ( x) = 6 x 1 f −1 ( x ) = x 6

x−4 5

114. f ( x ) =

f −1 ( x ) = 5 x + 4

1  1  f ( f −1 ( x)) = f  x  = 6  x  = x 6  6  1 f −1 ( f ( x )) = f −1 (6 x ) = (6 x ) = x 6

 x−4 f −1 ( f ( x)) = f −1    5   x−4 = 5 +4= x−4+4= x  5 

112. f ( x) = x + 5

f −1 ( x ) = x − 5

f ( f −1 ( x )) = f ( x − 5 ) = ( x − 5) + 5 = x −1

f ( f ( x )) = f 113. f ( x ) =

−1

5x + 4 − 4 5x = =x 5 5

f ( f −1 ( x )) = f (5 x + 4) =

( x + 5) = ( x + 5) − 5 = x

1 x+3 2

(a)

f ( x ) = 2( x − 3) = 2 x − 6

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

)

(

) 3 − (34− 4 x) = 44x = x 5

(b)

1  1  = 2 x + 3 − 3 = 2 x  = x 2  2 

(

g f ( x) =

−1

f ( f −1 ( x )) = f ( 2( x − 3) )

3− x 4 3− x  f g( x) = 3 − 4  = 3 − (3 − x) = x  4 

115. f ( x) = 3 − 4 x, g ( x) =

−6

−3

(c)

Y1 = 3 – 4X Y =

3− X

4 Y3 = Y1 (Y2) Y4 = Y2 (Y1)

−2 −1 0

1 2

132

Chapter 1

Functions and Their Graphs

116. f ( x ) = x + 1, g( x ) = x 2 − 1, x ≥ 0

(a)

121.

f ( g( x)) = ( x 2 − 1) + 1 = x 2 = x g ( f ( x )) =

( x +1) −1 = x +1−1 = x 2

(b)

g f

−6

1 x−5 2 1 x = y−5 2 1 x+5= y 2 y = 2( x + 5) y=

f −1 ( x ) = 2 x + 10 −4

7x + 3 8 1 y = (7 x + 3) 8 1 x = (7 y + 3) 8 8x = 7 y + 3

f ( x) =

122.

(c)

x+1

Y2 = X 2 – 1, X ≥ 0

0 1 2 3 4

Y1 =

Y3 = Y1 (Y2) Y4 = Y2 (Y1) 117.

8x − 3 = 7 y 1 f −1 ( x ) = (8 x − 3) 7

−9

f ( x) = − 5 x3 − 3

123. −6

y = − 5 x3 − 3

1 x − 3 passes the Horizontal Line Test, and hence 2 is one-to-one and has an inverse function.

x = − 5 y3 − 3

f ( x) =

118.

x + 3 = − 5 y3 x +3 = y3 −5

f −1 ( x) = 3 −6

6 −2

y = 5x3 + 2

124. 2

f ( x ) = ( x − 1) does not pass the Horizontal Line Test, so f is not one-to-one and does not have an inverse function. 119.

x = 5 y3 + 2 x − 2 = 5 y3 x−2 = y3 5

− 11

f −1 ( x) = 3

1 −1

h(t ) = (t + 5)

does not pass the Horizontal Line Test,

−2

x−2 5

f ( x ) = x + 10

125.

y = x + 10, x ≥ −10, y ≥ 0

and hence is not one-to-one and does not have an inverse function. 120.

x +3 x +3 = −3 −5 5

y + 10, y ≥ −10, x ≥ 0

x = y + 10 2

x − 10 = y 10

f −1 ( x ) = x 2 − 10, x ≥ 0

−4

5 passes the Horizontal Line Test, and x −4 hence is one-to-one and has an inverse function. g ( x) =

Chapter 1 Test 126. f ( x ) = 4 6 − x , x ≤ 6, y ≥ 0

y=4 6−x

x = 4 6 − y , y ≤ 6, x ≥ 0 x 2 = 16(6 − y) = 96 − 16 y 16 y = 96 − x 2 96 − x 2 16 96 − x 2 −1 f ( x) = , x≥0 16 y=

127.

133

1 2 x , x≤0 9 1 y = 5 − x2 9 1 2 x = 5− y 9 1 2 x−5= − y 9

128. f ( x) = 5 −

−9( x − 5) = y 2

f −1 ( x) = − −9( x − 5) = − 3 5 − x The negative square root is chosen as f −1 since the

1 2 x + 1, x ≥ 0 4 1 y = x2 + 1 4 1 x = y2 + 1 4 1 x − 1 = y2 4 f ( x) =

domain of f is ( − ∞ , 0 .

4( x − 1) = y 2

f −1 ( x) = 4( x − 1) = 2 x − 1 The positive square root is chosen as f −1 since the domain of f is [0, ∞).

Chapter 1 Test 1. y = 2 x − 1

3. y = 2 x2 − 4 x y

y 6

6 5 4 3 2

1 −4 −3 −2 −1

−2 −1

−2

−3

(

) ( 12 , 0)

Intercepts: (0, 1), − 12 , 0 ,

Intercepts: (0, 0), ( 2, 0) 4. y = x3 − x

2. y = − 23 x + 3

y 5

2 1

−4 −3 − 2

1 −3 −2 −1 −1

−2

Intercepts: (0, 3),

( 92 , 0)

−1

−2 −3 −4

Intercepts: ( −1, 0), (0, 0), (1, 0)

Chapter 1

134

Functions and Their Graphs

5. y = − x2 + 9

10.

y 10

f ( x ) = x + 2 − 15 (a)

f (−8) = −8 + 2 − 15 = 6 − 15 = −9

(b)

f (14) = 14 + 2 − 15 = 16 − 15 = 1

(c)

f (t − 6) = t − 6 + 2 − 15 = t − 4 − 15

11. 3 − x ≥ 0  domain is all x ≤ 3.

2 −6

−4

−2

12. Total cost = Variable costs + Fixed costs

C = 24.60 x + 25,000

Intercepts: ( − 3, 0), (3, 0), (0, 9) 6. y =

Revenue = Price per unit × Number of units R = 101.50 x

x−2

Profit = Revenue − Cost

P = 101.50 x − ( 24.60 x + 25,000)

6 5

P = 76.90 x − 25,000

4 3 2

13. f (− x ) = 2(− x )3 − 3( − x )

1 −1

= −2 x 3 + 3 x = − f ( x ) Odd

−2 −3

14. f ( − x ) = 3(− x )4 + 5( − x )2

Intercept: ( 2, 0)

= 3x 4 + 5x 2 = f ( x) Even

7. 5x + 2 y = 3 2 y = −5 x + 3 5 3 y=− x+ 2 2 5 Slope = − 2

15. h ( x ) =

)

(0, 2). 16. g(t ) = t + 2 − t − 2

5 y − 4 = − ( x − 0) 2 5 y=− x+4 2 5x + 2 y − 8 = 0 (b) Perpendicular line slope :

By graphing g, you see that the graph is increasing on (−2, 2), and constant on (−∞, − 2) and (2, ∞). 17.

2 5

2 ( x − 0) 5 2 y= x+4 5 2 x − 5 y + 20 = 0

4 − (−1) 5 = = −1 −3 − 2 −5 y + 1 = −1( x − 2) y = −x + 1

Slope =

No. For some x there corresponds more than one value of 1 y. For instance, if x = 1, y = ± . 3

− 15

−7

y−4=

(

By graphing h, you see that the graph is increasing on (−2, 0) and (2, ∞) and decreasing on (−∞, − 2) and

5 (a) Parallel line slope : − 2

1 4 1 x − 2 x2 = x2 x2 − 8 4 4

Relative minimum: (−3.33, − 6.52) Relative maximum: (0, 12) 18.

−9

−4

Relative minimum: (1.34, 1.10) Relative maximum: ( −1.34, 6.90)

Chapter 1 Test 19. (a) f ( x ) = x 3 (b) The graph of g is a horizontal shift five units to the right, a vertical stretch of 2, a reflection in the x-axis, and a vertical shift three units upward of f.

23. f ( x ) = x 3 + 8 Yes, f is one-to-one and has an inverse function. y = x3 + 8 x = y3 + 8

(c)

135

x − 8 = y3

8 3

x −8 = y

f −1 ( x ) = 3 x − 8

4 2

−2

−4

20. g ( x ) =

−7 x − 14

−7 ( x + 2 )

(a) f ( x) =

24. f ( x ) = x 2 + 6 No, f is not one-to-one, and does not have an inverse function.

3x x 8 Yes, f is one-to-one and has an inverse function. 3 y = x 3/ 2 , x ≥ 0, y ≥ 0 8 3 x = y 3/ 2 , y ≥ 0, x ≥ 0 8 8 x = y 3/ 2 3

25. f ( x ) =

(b) The graph of g is a reflection in the y-axis, a horizontal shrink, and a horizontal shift two units to the left. y (c) 6 5 4

2/ 3

=y 2/3

8  f −1 ( x ) =  x  , x ≥ 0 3 

3 2 1 − 7 − 6 − 5 − 4 −3 − 2 − 1

8   x 3 

−2

21. (a)

f ( x) = x

(b) g is obtained from f by a vertical stretch of 4 followed by a vertical shift seven units downward. y (c) 2 1

−5 −4 −3 −2

1 2 3 4 5

−7 −8

22. (a)

( f − g )( x ) = x 2 − 2 − x Domain: x ≤ 2

(b)

(c)

f  x2  ( x) = 2−x g Domain: x < 2

( f  g )( x ) = f ( 2 − x ) = 2 − x Domain: x ≤ 2

(d)

( g  f )( x ) = g ( x 2 ) = 2 − x 2

C H A P T E R 2 Solving Equations and Inequalities Section 2.1

Linear Equations and Problem Solving .............................................138

Section 2.2

Solving Equations Graphically ..........................................................147

Section 2.3

Complex Numbers..............................................................................158

Section 2.4

Solving Quadratic Equations Algebraically ......................................162

Section 2.5

Solving Other Types of Equations Algebraically .............................175

Section 2.6

Solving Inequalities Algebraically and Graphically .........................189

Section 2.7

Linear Models and Scatter Plots ........................................................201

Chapter 2 Review .......................................................................................................206 Chapter 2 Test ............................................................................................................222 Chapters P–2 Cumulative Test ................................................................................224

C H A P T E R 2 Solving Equations and Inequalities Section 2.1 Linear Equations and Problem Solving 1.

equation

ax + b = 0

extraneous

formulas

The equation x + 1 = 3 is a conditional equation.

x 1 + 1 = of fractions, multiply 2 4 both sides of the equation by the least common denominator of all the fractions, which is 4. To clear the equation

(b)

5 4 − =3 2x x 5 4 ? (a) − =3 2 ( −1 2 ) ( −1 2 )

3=3 x = − is a solution. 1 2

(b)

(c)

(d)

(c)

5 4? − =3 2 ( 4) 4 3 − ≠3 8 x = 4 is not a solution. 5 4 − is undefined. 2 (0) 0

x = 0 is not a solution. 5 4 ? − =3 2 (1 4 ) 1 4 −6 ≠ 3

138

x 6 x 19 + = 2 7 14 (a) x = − 2

1 4

is not a solution.

(d)

− 2 6 ( −2 ) ? 19 + = 2 7 14 −14 − 24 ? 19 = 14 14 −38 ? 19 = 14 14 x = −2 is not a solution. x =1

1 6 (1) ? 19 + = 2 7 14 19 19 = 14 14 x = 1 is a solution. 1 x= 2

1 2 6 (1 2 ) ? 19 + = 2 7 14 7 2 + 6 ? 19 = 14 14 19 19 ≠ 28 14 1 x = is not a solution. 2 x=7 7 6 ( 7 ) 19 + = 2 7 14 7 19 +6 = 2 14 19 ? 19 = 2 14 x = 7 is not a solution.

Section 2.1

(b)

(c)

(d)

10.

−3 + 4 ? +3=4 6 19 ≠4 6 x = −3 is not a solution.

0+4 ? +3=4 6 10 ≠4 3 x = 0 is not a solution.

139

(d) x = 16

x+4 +3= 4 6 (a)

Linear Equations and Problem Solving

16 − 8 ? = −1 3 3

8 ? = −1 3 2 ≠ −1 3

x = 16 is not a solution. 11. 2 ( x − 1) = 2 x − 2 is an identity by the Distributive

Property. It is true for all real values of x.

21 + 4 ? +3= 4 6 23 ≠4 6 x = 21 is not a solution.

12. −5 ( x − 1) = −5 ( x + 1) is a contradiction. There are no

32 + 4 ? +3= 4 6 4=4 x = 32 is a solution.

13.

x −8 = −1 3

14.

real values of x for which it is true. −5 x + 5 = −5 x − 5

5 ≠ −5

are no real values of x for which it is true. x 2 − 2 x − 15 = x 2 − 2 x − 14

−15 ≠ −14

for all real values of x. 15. ( x + 6) = ( x + 8)( x + 2) is a conditional. There are

−4 ≠ −1 3

real values of x for which the equation is not true (for example, x = 0 ).

x = 4 is not a solution. (b) x = 0

16. ( x + 1)( x − 5) = ( x + 3)( x − 1) is a conditional. There

are real values of x for which the equation is not true (for example, x = 0 ).

0 −8 ? = −1 3 3

−8 ? = −1 3 −2 ≠ −1 3

x = 0 is not a solution. (c) x = −19 3

4−8 ? = −1 3 3

x 2 − 8 x + 5 = ( x − 4 ) − 11 is an identity since

( x − 4 ) − 11 = x 2 − 8 x + 16 − 11 = x 2 − 8 x + 5. It is true

(a) x = 4 3

( x + 3 )( x − 5 ) = x 2 − 2 ( x + 7 ) is a contradiction. There

1 4x = is conditional. There are real values of x x +1 x +1 for which the equation is not true (for example, x = 0) .

17. 3 +

18.

5 3 + = 24 is conditional. There are real values of x for x x which the equation is not true (for example, x = 1) .

−19 − 8 ? = −1 3 3

− 27 ? = −1 3 −3 = −1 3

x = −19 is a solution.

140

Chapter 2

Solving Equations and Inequalities

3x 4 x − =4 8 3 9 x − 32 x =4 24 −23x = 96 96 x=− 23 3x 4 x − and y2 = 4 in the same Method 2: Graph y1 = 8 3 viewing window. These lines intersect at 96 x ≈ −4.1739 ≈ − . 23 3z z 20. Method 1: − =6 8 10 3 1  z −  = 6  8 10   22  z  = 6  80  6 ( 80 ) 240 = ≈ 21.8182 z= 22 11 3x x − and y2 = 6 in the same Method 2: Graph y1 = 8 10

19. Method 1:

viewing window. The lines intersect at 240 x ≈ 21.8182 ≈ . 11 21. Method 1:

2x 4 + 5x = 5 3 2 x + 25 x 4 = 5 3 20 27 x = 3 20 20 x= = 3 ( 27 ) 81

2x 4 + 5x and y2 = in the same Method 2: Graph y1 = 5 3 viewing window. These lines intersect at 20 x ≈ 0.2469 ≈ . 81 4y 16 − 2y = 3 5 4 y − 6 y 16 = 3 5 48 −2 y = 5 24 y=− 5 4x 16 − 2 x and y2 = in the same Method 2: Graph y1 = 3 5 viewing window. These lines intersect at 24 x = −4.8 = − . 5

23.

3x − 5 = 2 x + 7 3x − 2 x = 7 + 5 x = 12

24.

5x + 3 = 6 − 2 x 5x + 2 x = 6 − 3 7x = 3 3 x= 7

25. 3 ( y − 5 ) = 3 + 5 y

3 y − 15 = 3 + 5 y −18 = 2 y y = −9 26. 4( z − 3) + 3 z = 1 + 8 z 4 z − 12 + 3 z = 1 + 8 z 7 z − 12 = 1 + 8 z − z = 13 z = −13

27.

x x − =3 5 2 2 x − 5x =3 10 −3 x = 30 x = −10

28.

x 3x + = −5 4 2 x  3x +  = 4( − 5) 4 2  4 3 x + 2 x = − 20 5 x = − 20 x = −4

29.

5x − 4 2 = 5x + 4 3 3( 5x − 4 ) = 2 ( 5x + 4 ) 15 x − 12 = 10 x + 8

22. Method 1:

5 x = 20 x=4 30.

10 x + 3 1 = 5x + 6 2 20 x + 6 = 5 x + 6 15 x = 0 x=0

Section 2.1

31.

32.

2 3z = 4z ( z − 4) + 5 10 2 8 3z z − + = 4z 5 5 10 7z 8 − = 4z 10 5 8 33 − = z 5 10 16 − = z 33

3x 1 + ( x − 2 ) = 10 2 4 3x 1 ( 4 )  2  + ( 4 ) 4 ( x − 2 ) = ( 4 )10  

Linear Equations and Problem Solving

36.

(

1 3 4 + = x − 2 x + 3 x2 + x − 6 1 3 4 x2 + x − 6 + x2 + x − 6 = x2 + x − 6 2 x −2 x +3 x + x −6 ( x + 3) + 3( x − 2) = 4

)

37.

38.

−1 = x

35.

1 1 10 + = x − 3 x + 3 x2 − 9 ( x + 3) + ( x − 3) = 10 x2 − 9 x2 − 9 2 x = 10 x=5

7 4

3=2+

5 3

2 z+2

2 z+2 z+2=2 1=

z=0 2

39.

( x − 4 )( x − 2 ) 

1 2 + x−4 x−2



( x − 4 )( x − 2 )  x − 4 x − 2  =  x − 4 + x − 2  ( x − 4 )( x − 2 ) )( )     ( 2 = 1( x − 2 ) + 2 ( x − 4 )

x − 11 x −9 = + 2 x x x − 11 x − 9 + 2x = x x x − 11 3x − 9 = x x x − 11 = 3x − 9 − 2 = 2x

)

1 2 + =0 x x−5 1( x − 5 ) + 2 x = 0

49 = 98 y 1 =y 2

34.

(

3x − 5 = 0 3x = 5

17 + y 32 + y + = 100 y y 17 + y 32 + y ( y ) y + ( y ) y = 100 ( y ) 17 + y + 32 + y = 100 y 49 + 2 y = 100 y

)

7 x − 2 = 40 7 x = 42

33.

(

x + 3 + 3x − 6 = 4 4x − 3 = 4 4x = 7

6 x + ( x − 2 ) = 40

x =6

141

2 = x − 2 + 2x − 8 12 = 3 x 4=x In the original equation, x = 4 yields a denominator of 0. So, x = 4 is an extraneous solution, and the original equation has no solution.

40.

2 5 1 + = x ( x − 2) x x − 2  2 5  1  x ( x − 2)  x ( x − 2) + = 2 x x − ) x   x − 2   ( 2 + 5( x − 2) = x 2 + 5 x − 10 = x 4x = 8 x=2 In the original equation, x = 2 yields a denominator of 0. So, x = 2 is an extraneous solution, and the original equation has no solution.

142

Chapter 2

Solving Equations and Inequalities

41.

3 4 1 + = x ( x − 3) x x − 3 3 + 4 ( x − 3) = x 3 + 4 x − 12 = x 3x = 9

48.

x=3 In the original equation, x = 3 yields a denominator of 0. So, x = 3 is an extraneous solution, and the original equation has no solution. 3 ( x + 5) 6 2 − = x x + 3 x ( x + 3)

42.

x ( x + 3)

3 ( x + 5) 6 2 − x ( x + 3) = x ( x + 3) x x+3 x ( x + 3) 6 ( x + 3) − 2 x = 3 ( x + 5) 6 x + 18 − 2 x = 3 x + 15 4 x + 18 = 3 x + 15

x = −3 In the original equation, x = −3 yields a denominator of 0. Thus, x = −3 is an extraneous solution, and the original equation has no solution. 43.

2 47. V = π r h V h= 2 πr

1 bh 2 2 A = bh A=

1 V = π r 2h 3 3V = π r 2 h 3V h= 2 πr

49. Female: y = 0.386 x − 19.20

For y = 43,

43 = 0.386 x − 19.20 62.2 = 0.386 x 161.14 ≈ x. The height of the female is about 161.14 centimeters. 50. Male: y = 0.442 x − 29.37 ?

48 = 0.442(175) − 29.37 48 ≈ 47.98 Yes, the estimated height of a male with a 48-centimeter thigh bone is about 175 centimeters. 51. (a)

2A =h b 44.

1 (a + b)h 2 2 A = ah + bh A=

2 A − ah = bh

46.

(b)

l = 1.5w P = 2l + 2 w = 2 (1.5w ) + 2 w = 5w

2 A − ah =b h

45.

(c)

r  A = P 1 +  n  A P= nt r  1 + n   

r  P = A 1 +  n 

− nt

25 = 5w  w = 5 m and l = (1.5 )( 5 ) = 7.5 m

Dimensions: 7.5 m × 5 m 52. (a) h

(b)

P = 2 h + 2 w = 2 ( 23 w ) + 2 w = 43 w + 2 w = 103 w

A = P + Prt A − P = Prt A−P r= Pt

h = 23 w

(c)

P = 3 = 103 w  w = 109 = 0.9 m and h = 23 ( 0.9 ) = 0.6 m

Dimensions: 0.6 m × 0.9 m

Section 2.1

53. (a) Test Average =

test 1 + test 2 + test 3 + test 4 4

test 1 + test 2 + test 3 + test 4 (b) Test Average = 4 93 + 91 + 84 + x 90 = 4 268 + x 90 = 4 360 = 268 + x 92 = x You must earn at least 92 points on the fourth test to earn an A in the course. 54. Sales: Monday, $150 Tuesday, $125

Wednesday, $75 Thursday, $180 Friday, $x 150 + 125 + 75 + 180 + x Average: = 150 5 530 + x = 150 5 530 + x = 750 x = $220 55. Model: distance = ( rate )( time )

The salesperson drove 50 km in a half hour, therefore the 50 km km = 100 . rate is 1 2 hr hr Since the salesperson continues at the same rate to travel a total distance of 250 km, the time distance 250 km = = 2.5 hours. required is rate 100 km hr 56. Model: distance = ( rate )( time )

Total distance = ( rate #1)( time #1) + ( rate #2 )( time #2 )  mi   mi  336 =  58  t +  52  ( 6 − t )  hr   hr  336 = 58t + 312 − 52t 24 = 6t 4=t The salesperson traveled for 4 hours at 58 mph and then 2 hours at 52 mph.

Linear Equations and Problem Solving

143

57. Model: ( Distance ) = ( rate )( time1 + time 2 )

Labels: Distance = 2 ⋅ 200 = 400 miles, rate = r, distance 200 time1 = hours, = rate1 55

time 2 =

distance 200 = hours rate 2 40

 200 200  + Equation: 400 = r   40   55  1600 2200  3800 400 = r  + r = 440  440  440 43.6 ≈ r The average speed for the round trip was approximately 46.3 miles per hour. 58. Rate =

Distance 50 kilometers = = 100 kilometers/hour 1 Time hours 2 Total distance 300 kilometers = Rate 100 kilometers/hour = 3 hours

Total time =

59. Let x = height of the pine tree. x 36 = 20 24 24 x = 720

x = 30 feet The pine tree is approximately 30 feet tall. 60. (a)

6 ft 30 ft

(b) Model:

5 ft

( height of pole )

( height of pole's shadow ) =

( height of person )

height of person's shadow

Labels: height of pole = h, height of pole's shadow = 30 + 5 = 35 feet, height of person = 6 feet, height of person's shadow = 5 feet h 6 Equation: = 35 5 6 h = ⋅ 35 = 42 5 The pole is 42 feet tall.

144

Chapter 2

61.

I = Prt

Solving Equations and Inequalities 66. Let x = amount invested in 8 × 10 frames, and y = amount invested in 5 × 7 frames. x + y = 4500  y = 4500 − x

200 = 8000r ( 2) 200 = 16,000r r =

0.25 x + 0.22 y = 0.24 ( 4500 ) = 1080

200 16,000

0.25 x + 0.22 ( 4500 − x ) = 1080

r = 0.0125, or 1.25%

0.03 x = 90 x = 3000 So, $3000 is invested in 8 × 10 frames and $4500 − $3000 = $1500 is invested in 5 × 7 frames.

62. Let x = amount in 4 12 % fund.

Then 12,000 − x = amount in 5% fund. 560 = 0.045 x + 0.05 (12,000 − x ) 560 = 0.045 + 600 − 0.05 x 0.005 x = 40 x = 8000 You must invest $8000 in the 4 12 % fund and 12,000 − 8000 = $4000 in the 5% fund. 63. Model: Total pounds at $5.25 = pounds at $2.50 + pounds at $8.00 Labels: x = pounds of peanuts at $2.50

67.

68. Let x = length of side of square I, and y = length of side of square II. 4 x = 20  x = 5

4 y = 32  y = 8 Hence, square III has side of length 5 + 8 =13. Area = 132 = 169 square inches

100 − x = pounds of walnuts at $8.00

Equation: 100(5.25) = x( 2.50) + (100 − x)(8.00) 525 = 2.5 x + 800 − 8 x

1 bh 2 2 A 2 (182.25 ) h= = = 27 ft b 13.5

69. (a) h

275 = 5.5 x l

50 = x

The mixture contains 50 pounds at $2.50 and 100 − 50 = 50 pounds at $8.00. 64. Initially, the forester has 2 33

64 33

(b)

9 2

gallons of oil.

64 33 + x 40 = 2 33 1 which gives x = 16 33 gallon. 65. Model: Total profit = profit on notebooks + profit on tablet

l = 3w, h = (1 12 ) w

V = lwh = ( 3w )( w ) ( 32 w ) = 2304

gallons of gas and

64 2 + =2 33 33 64 33 = 32 2 33 Suppose she adds x gallons of gas.

w3 = 2304

w3 = 512 w = 8 inches Dimensions: 24 × 8 × 12 inches 70.

V =

4 3 πr 3

4 3 π r = 6255 3 18,765 r3 = 4π r = 3

18,765 ≈ 11.43 cm 4π

Labels: x = amount invested in notebook computers 40,000 − x = amount invested in tablet computers

Equation: (0.24)( 40,000) = 0.20 x + 0.25( 40,000 − x ) 9600 = 0.2 x + 10,000 − 0.25 x − 400 = − 0.05 x 8000 = x So, $8000 is invested in notebook computers and $40,000 − $8000 = $32,000, invested in tablet computers. © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Section 2.1 71. Solve the temperature for C. 9 C + 32 5 9 F − 32 = C 5 F =

5 ( F − 32) = C 9

5 (73 − 32)  C ≈ 22.8° 9 5 F = 74°: C = (74 − 32)  C ≈ 23.3° 9 5 F = 76°: C = (76 − 32)  C ≈ 24.4° 9 5 F = 78°: C = (78 − 32)  C ≈ 25.6° 9 5 F = 82°: C = (82 − 32)  C ≈ 27.8° 9 5 F = 79°: C = (79 − 32)  C ≈ 26.1° 9 5 F = 74°: C = (74 − 32)  C ≈ 23.3° 9 F = 73°: C =

Linear Equations and Problem Solving

73. Let x = the wind speed, then the rate to the city = 600 + x, the rate from the city = 600 − x, the distance to the city = 1500 kilometers, the distance traveled so far in the return trip = 1500 − 300 = 1200 kilometers. Distance Time = Rate 1500 1200 = 600 + x 600 − x 1500 ( 600 − x ) = 1200 ( 600 + x ) 900,000 − 1500 x = 720,000 + 1200 x 180,000 = 2700 x 200 =x 3 200 km h Wind speed: 3

74. Let h = height of the building in feet.

Temperature (°C)

145

4 ft

80 ft

1 3 2 ft

Not drawn to scale

24 22

4 feet h feet = 80 feet 3.5 feet h 4 = 80 3.5 3.5h = 320

10:00 A.M. 11:00 A.M. 12:00 P.M. 1:00 P.M. 2:00 P.M. 3:00 P.M. 4:00 P.M. 5:00 P.M. 6:00 P.M.

Time

h ≈ 91.4 feet

72. Solve the temperature for C.

9 C + 32 5 9 F − 32 = C 5 F =

5 ( F − 32) = C 9 5 C = (74.3 − 32) 9 5 C = ( 42.3) 9 C = 23.5°

75.

W1 x = W2 ( L − x )

50 x = 75 (10 − x ) 50 x = 750 − 75 x 125 x = 750 x = 6 feet from the 50-pound child 76.

W1 x = W2 ( L − x ) W1 = 200 pounds W2 = 550 pounds L = 5 feet 200x = 550 ( 5 − x ) 200 x = 2750 − 550 x 750 x = 2750 x = 113 feet

146

Chapter 2

Solving Equations and Inequalities

77. False. x ( 3 − x ) = 10 is a quadratic equation.

87. y = 5 − 4 x y

− x 2 + 3 x = 10 or x 2 − 3 x + 10 = 0 5 4 3 2 1

78. False.

Volume of cube = ( 9.5) = 857.375 cubic inches 3

Volume of sphere = 43 π ( 5.9 ) ≈ 860.29 cubic inches 3

−4 −3 −2 −1

One answer is a = 2, c = 1 and b = 3 ( 2 x + 3 = x ) and

88.

80. You need a ( 0 ) + b = c or b = c. So, a can be any real

3x − 5 3 5 3 1 +2= x− +2= x− 2 2 2 2 2 y

number except 0. One answer is a = 1 and b = c = 4 : x + 4 = 4.

5 4 3 2 1

1 81. You need a   + b = c or a + 4b = 4c. One answer is 4 1 a = 2 and b = 12 . So, 2 + 4   = 4 = 4c  c = 1. 2 1 The equation is 2 x + = 1. 2

−5 −4 −3 −2 −1

89. y = ( x − 3) − 7 2

a = −1 and b = 2. So, −1( − 2.5 ) + 2 = 4.5 = c.

The equation is − x + 2 = 4.5.

83. In the original equation, x = 1 yields a denominator of zero. So, x = 1 is an extraneous solution and therefore cannot be a solution to the equation.

−4 −2 −2

−4 −6 −8

84. (a)

Height of building Height of pole = Length of building's shadow Length of pole's shadow (b)

90. y = 4 −

1 2 x 3

x 4 = 30 3 6

( x − 3)( x − 1)

y 5

3 2 1

3 4 + x − 3 x −1

To clear this equation of fractions, find the least common denominator (LCD) of all terms in the equation and multiply every term by this LCD. It is possible to introduce an extraneous solution because you are multiplying by a variable. To determine whether a solution is extraneous, substitute the answer for the variable in the original equation or graph the original equation. 86.

1 2 3 4 5

−3 −4 −5

82. You need a ( −2.5 ) + b = c. One answer is

85.

−2 −3 −4

79. You need a ( −3 ) + b = c ( −3 ) or b = ( c − a )( −3 ) = 3 ( a − c ) .

another is a = 6, c = 1, and b = 15 ( 6 x + 15 = x ) .

1 2 3 4 5 6

−5

−2 −1

1 2 3

10 + 2c = 20 − 2c 4c = 10 c = 25

−2 −3 −4 −5

91.

y = − 12 x + 4 − 1 y 3 2 1

5 x + 2c = 12 + 4 x − 2c, x = 2 5 ( 2 ) + 2c = 12 + 4 ( 2 ) − 2c

−9 −8 −7 −6 −5

−3 −2 −1

−2 −4 −5 −6 −7

Section 2.2 92.

Solving Equations Graphically

147

y = x − 2 + 10 y 18 16 12 10 8 6 4 2

− 10 − 8 − 6 − 4 − 2

2 4 6 8 10

Section 2.2 Solving Equations Graphically 1. x-intercept, y-intercept

11.

2. zero

0 = x x + 2  x = 0, − 2  ( 0, 0 ) , ( −2, 0 ) x -intercepts

3. The x-intercepts of y = f ( x ) are ( −1, 0 ) and (1, 0 ) .

Let x = 0 :

y = 0 0 + 2 = 0  ( 0, 0 ) y-intercept

4. The y-intercept of y = g ( x ) is ( 0, − 1) . 5. The zeros of the function f are x = −1 and x = 1.

12. y = − 12 x x + 3 + 1

Let y = 0 :

6. The solutions of the equation f ( x ) = g ( x ) are

x=− 7.

1 2

0 = − 12 x x + 3 + 1  12 x x + 3 + 1  x x + 3 = 2

and x = 1.

 x 2 ( x + 3) = 4  x 3 + 3 x 2 − 4 = 0

(

y = x−5

 x = 1  (1, 0 )

13. y =

Let y = 0 :

Let y = 0:

Let x = 0 :

0 =

y=− 9.

4x − 8 x

0 = − 34 x − 3  34 x = −3  x = −4  ( −4, 0 ) x -intercept

3 4

4x − 8 x 0 = 4x − 8

( 0 ) − 3 = −3  ( 0, − 3) y-intercept

8 = 4x

y = x2 + 2 x + 2

2 = x  ( 2, 0) x -intercept

Let y = 0 : x + 2 x + 2 = 0  no x-intercepts Let x = 0 : 2

Let x = 0:

y = ( 0 ) + 2 ( 0 ) + 2 = 2  ( 0, 2 ) y-intercept 2

y =

10. y = 4 − x Let y = 0 :

0 = 4 − x2  x = 2, − 2  ( 2, 0) , ( −2, 0) x-intercepts Let x = 0 :

y = 4 − 02 = 4  ( 0, 4 ) y-intercept

( x = −2 is impossible ) Let x = 0  y = 1  ( 0, 1) y-intercept

Let x = 0 : y = 0 − 5  y = −5  ( 0, − 5 ) y-intercept

y = − 34 x − 3

)

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

Let y = 0 : 0 = x − 5  ( 5, 0 ) x -intercept

y= x x+2 Let y = 0 :

14.

4(0) − 8 0

is impossible. No y-intercepts

3x − 1 4x Let y = 0 : y=

0 = 3 x − 1  x = 13  ( 13 , 0 ) x -intercept

Let x = 0 : 0 = −1 is impossible. No y-intercepts

148

Chapter 2

Solving Equations and Inequalities

15.

xy − 2 y − x + 1 = 0 Let y = 0 :

y = 20 − ( 3 x − 10 )

19.

0 = − x + 1 = 0  x = 1  (1, 0 ) x -intercept

Let x = 0 :

−2 y + 1 = 0  y = 12  ( 0, 12 ) y -intercept

− 35

16. xy − x + 4 y = 3

− 10

x-intercept: 0 = 20 − ( 3 x − 10 )

Let y = 0: x(0) − x + 4(0) = 3

0 = 20 − 3 x + 10 0 = 30 − 3 x 3 x = 30

−x = 3 x = − 3  ( − 3, 0) x-intercept

x = 10  (10, 0 )

Let x = 0:

y-intercept: y = 20 − 3 ( 0 ) − 10 

( 0) y − ( 0) + 4 y = 3

y = 30  ( 0, 30 )

4y = 3 y =

3  3   0,  y -intercept 4  4

20. y = 10 + 2 ( x − 2 ) 12

17. y = 3( x − 2) − 5 4 − 12

− 12

12 −4

0 = 10 + 2 ( x − 2 )

x-intercept:

0 = 10 + 2 x − 4

− 16

0 = 6 + 2x −2x = 6

x-intercept: 3( x − 2) − 5 = 0 3x − 6 − 5 = 0

x = −3  ( −3, 0 )

3 x = 11 x =

y-intercept: y = 10 + 2 ( 0 − 2 )

11  11    , 0 3 3 

y-intercept: y = 3(0 − 2) − 5

y = 6  ( 0, 6 )

4 (3 − x ) = 0

y = −11  (0, −11)

18.

f ( x ) = 4 (3 − x )

21.

y = −6 − 5

3− x = 0 x =3

y = 4 ( x + 3) − 2

−12 − 14

10 −2

x-intercept: 0 = 4 ( x + 3 ) − 2 0 = 4 x + 10 4 x = −10

x = − 25  ( − 25 , 0 )

y-intercept: y = 4 ( 0 + 3) − 2

y = 10  ( 0, 10 )

12 −3

22.

f ( x ) = 3 ( x − 5) + 9 3 x − 15 + 9 = 0 3x = 6 x=2

−4

Section 2.2 f ( x ) = x3 − 6 x2 + 5x

23.

x3 − 6 x 2 + 5x = 0

(

)

x x2 − 6 x + 5 = 0 x ( x − 5 )( x − 1) = 0

Solving Equations Graphically

149

27. 2.7 x − 0.4 x = 1.2 2.3 x = 1.2 1.2 12 x= = ≈ 0.522 2.3 23 f ( x ) = 2.7 x − 0.4 x − 1.2 = 0 2.3 x − 1.2 = 0

x = 0, 5, 1

x ≈ 0.522

28. 3.6 x − 8.2 = 0.5 x

−2

3.1x = 8.2 x =

− 15

f ( x ) = 3.6 x − 8.2 − 0.5 x = 0

f ( x ) = x 3 − 9 x 2 + 18 x

24.

8.2 82 = 3.1 31 3.1x − 8.2 = 0

x 3 − 9 x 2 + 18 x = 0 x ( x − 3 )( x − 6 ) = 0

x =

x = 0, 3, 6

82 31

29. 12( x + 2) = 15( x − 4) − 3

12 x + 24 = 15 x − 60 − 3 −4

− 3 x = − 87

x = 29 f ( x ) = 12( x + 2) − 15( x − 4) + 3 = 0

− 12

f ( x) =

25.

− 3 x + 87 = 0

x +1 x − 2 − +1 2 7

x +1 x − 2 − +1= 0 2 7 7( x + 1) − 2( x − 2) + 14 = 0

x = 29

30. 1200 = 300 + 2 ( x − 500 )

900 = 2 x − 1000 1900 = 2 x x = 950 f ( x ) = 300 + 2 ( x − 500 ) − 1200 = 0

7 x + 7 − 2 x + 4 + 14 = 0 5 x + 25 = 0 5 x = − 25

300 + 2 x − 1000 − 1200 = 0 2 x − 1900 = 0 x = 950

x = −5 9

− 15

31.

−7

f ( x) = x − 3 −

26.

10 =0 x x 2 − 3 x − 10 = 0 x −3−

( x − 5)( x + 2 ) = 0 x = 5, − 2 4

−6

10 x

3x 1 + ( x + 2 ) = 10 2 4 6x x 1 + = 10 − 4 4 2 7 x 19 = 4 2 38 x= 7 3x 1 f ( x) = + ( x + 2 ) − 10 = 0 2 4 3 1 1 x + x + − 10 = 0 2 4 2 7 19 x− =0 4 2 7 x − 38 = 0 38 x= 7

−4

150

32.

Chapter 2

Solving Equations and Inequalities

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

36.

12 x − 36 = 25 x − 125 13 x = 89

7 17 x= 6 2 51 x= ≈ 7.286 7 2 1 f ( x ) = x + ( x − 5) − 6 = 0 3 2 4 x + 3 ( x − 5 ) − 36 = 0

89 ≈ 6.846 13 x−3 x−5 − =0 f ( x) = 25 12 12 ( x − 3) − 25 ( x − 5 ) = 0 x=

12 x − 36 − 25 x + 125 = 0 −13 x + 89 = 0 13 x − 89 = 0 x ≈ 6.846

4 x + 3 x − 15 − 36 = 0 7 x − 51 = 0 x ≈ 7.286 33.

0.60 x + 0.40 (100 − x ) = 1.2

37.

0.60 x + 40 − 0.40 x = 1.2 0.20 x = −38.8

0.60 x + 40 − 0.40 x − 1.2 = 0 0.20 x + 38.8 = 0 x = −194 0.75 x + 0.2(80 − x) = 3.9 0.75 x + 16 − 0.2 x − 3.9 = 0

38.

0.55 x + 12.1 = 0 55 x = −1210

−x = 9

f ( x) = 0.75 x + 0.2(80 − x) − 3.9 = 0

x = −9 x −5 x−3 − −1 = 0 f ( x) = 10 5 x − 5 − 2 ( x − 3) − 10 = 0

0.55 x + 12.1 = 0 55 x + 1210 = 0 x = − 22

x − 5 − 2 x + 6 − 10 = 0 −x − 9 = 0 x+9=0 x = −9

x − 3 3x − 5 = 35. 3 2 2 ( x − 3) = 3 ( 3 x − 5)

9 ≈ 1.286 7 x − 3 3x − 5 − =0 f ( x) = 3 2 2 ( x − 3) − 3 ( 3 x − 5 ) = 0 x=

2 x − 6 − 9 x + 15 = 0 −7 x + 9 = 0 7x − 9 = 0 x ≈ 1.286

x −5 x −3 − =1 10 5 ( x − 5) − 2 ( x − 3) = 10 x − 5 − 2 x + 6 = 10

x = − 22

2 x − 6 = 9 x − 15 9 = 7x

x−5 x + = 10 4 2 ( x − 5) + 2 x = 40 3 x = 45 x = 15 x−5 x + − 10 = 0 f ( x) = 4 2 x − 5 + 2 x − 40 = 0 3 x − 45 = 0 x = 15

x = −194 f ( x ) = 0.60 x + 0.40 (100 − x ) − 1.2 = 0

34.

x −3 x −5 = 25 12 12 ( x − 3 ) = 25 ( x − 5 )

39.

( x + 2) = x2 − 6 x + 1 2

x2 + 4 x + 4 = x2 − 6 x + 1 10 x = −3 x = − 103 = −0.3 f ( x ) = ( x + 2) − x2 + 6x − 1 = 0 2

x2 + 4x + 4 − x2 + 6x − 1 = 0 10 x + 3 = 0 x = −0.3

Section 2.2 40.

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

Solving Equations Graphically

151

46. 4 x 3 + 12 x 2 − 26 x − 24 = 0 x ≈ −4.206, − 0.735, 1.941

x2 + 2 x + 1 + 2 x − 4 = x2 − x − 2 5x = 1

x = = 0.2 1 5

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

(

−6

)

x + 2x + 1 + 2x − 4 − x − x − 2 = 0 2

− 40

x + 4x − 3 − x + x + 2 = 0 5x − 1 = 0

x5 − 3x3 + 3 = 0

x = 0.2

41.

x5 = 3x3 − 3

47.

x ≈ −1.861

x3 + x + 4 = 0 x ≈ −1.379

−3

3 −2

−6 −1

x5 = 3 + 2 x3

48.

42. 2 x + x + 4 = 0

x5 − 3 − 2 x3 = 0 x ≈ 1.638

x ≈ −1.128

8 −6 −12

−8

43.

1 4

−6

( x − 10 x + 17 ) = 0 2

2 =3 x+2

49.

x ≈ 2.172, 7.828

2 −3= 0 x+2

x ≈ −1.333 −3

2 −8

−3

(

)

44. − 12 x 2 − 6 x + 6 = 0 x ≈ 1.268, 4.732

−6

1 = −2 x −3

50.

−5

1 +2=0 x −3

x = 2.5

−7

45. 2 x 3 − x 2 − 18 x + 9 = 0 x = −3.0, 0.5. 3.0 30

−4

8 −2

−5

− 20

152

Chapter 2

Solving Equations and Inequalities

5 3 =1+ x x+2

51.

4x + 1 + 2 = 8

56.

4x + 1 − 6 = 0

5 3 −1− 3 =0 x x+2 x ≈ −3.162, 3.162

x = −1.75, 1.25 4

−8

−6

−6 − 16

52.

x−2 =3

57.

3 3 +1 = x x+2 3 3 +1− =0 x x −1

x −2 −3= 0 x = 11 5

x ≈ −1.303, 2.303

−1

2 −5 −4

x−4 =8

58.

x − 4 −8 = 0

−4

x = 68

− x + 1 = −3

53.

− x +1 + 3 = 0

x = − 4, 2

100

− 10

−6

− x − 2 = −6

54.

− 10

59. 2 −

x +5 =1

1−

x +5 = 0 x = −4

− x − 2 + 6 = 0

x = − 4, 8 8

−6

−7

55.

−2

−4

3x − 2 − 1 = 4

60. 8 −

x +9 = 6

2−

x +9 = 0 x = −5

3x − 2 − 5 = 0

x ≈ −1.0, 2.333

−10

2 −6

−3

−6

Section 2.2 61. (a)

−1

3.2 x − 5.8

−9

−5.8

−2.6

0.6

3.8

7.0

Solving Equations Graphically

153

Because of the sign change, 1 < x < 2. (b)

1.5

1.6

1.7

3.2 x − 5.8 −1 −0.68 −0.36 1.8

1.9

2.0

3.2 x − 5.8 −0.04 0.28 0.6 Because of the sign change, 1.8 < x < 1.9. To improve accuracy, evaluate the expression for values in this interval and determine where the sign changes. Let y1 = 3.2 x − 5.8. The graph of y1 crosses the x-axis at x = 1.8125. 2 −5

−8

62. 0.3( x − 1.8) − 1 = 0

x 0.3( x − 1.8) − 1

− 0.94

− 0.64

− 0.34

− 0.04

0.26

0.56

0.86

Because of the sign change, 5 < x < 6. x

5.0

5.1

5.2

5.3

5.4

0.3( x − 1.8) − 1

− 0.04

− 0.01

0.02

0.05

0.08

Because of the sign change, 5.1 < x < 5.2. To improve accuracy, evaluate the expression in this interval, and determine where the sign changes. Let y1 = 0.3( x − 1.8) − 1. The graph of y1 crosses the x-axis at x ≈ 5.13. 4

−5

−6

63. y = 6 − x

65.

y = 3x − 2 6 − x = 3x − 2 8 = 4x x = 2  y = 6−2 = 4 ( x, y ) = ( 2, 4)

64. y = 2 x − 3

y = 9− x 2x − 3 = 9 − x 3x = 12 x = 4  y = 9−4 = 5 ( x, y ) = ( 4, 5)

2x + y = 6  y = 6 − 2x −x + y = 0  y = x 6 − 2x = x 6 = 3x x =2 y = x =2

( x, y ) = ( 2, 2 )

66.

x − y = −4  x = y − 4 x + 2 y = 5  x = −2 y + 5 y − 4 = −2 y + 5 3y = 9 y = 3  x = 3 − 4 = −1

( x, y ) = ( −1, 3)

154

Chapter 2

Solving Equations and Inequalities

67.

x − y = 10  y = x − 10 x + 2y = 4  y = − x + 2 1 2

72.

x − 10 = − 12 x + 2 2 x − 20 = − x + 4 3 x = 24

1 x +1 3 5 11 5 x − 2 y = 11  y = x − 2 2 ( x, y ) = (3, 2) x − 3y = −3  y =

x = 8  y = 8 − 10 = −2

( x, y ) = (8, − 2 )

(3, 2)

−9

y = 1x + 1

x − 4 y = 1  y = 14 x − 14 4 x − 4 = 14 x − 14

−8

73.

y=x y = 2 x − x2

16 x − 16 = x − 1

( x, y ) = ( 0, 0 ) , (1, 1)

15 x = 15 x = 1  y = 4x − 4 = 0

( x, y ) = (1, 0 ) −4

69.

y = x2 − x + 1

74.

y = 2x − x2

y = 4 − x2 y = 2x −1

( x, y ) = (1.449, 1.898) , ( −3.449, − 7.899 )

y = ( −1) − ( −1) + 1 = 3 2

( x, y ) = ( −1, 3)

(1.449, 1.898)

y = − x 2 + 3x + 1 y = − x2 − 2x − 4

5 x = −5 x = −1 y = − ( −1) + 3 ( −1) + 1 = −3 2

( x, y ) = ( −1, − 3)

75.

y = 4 − x2

x3 − y = 3  y = x3 − 3 2x + y = 5  y = 5 − 2x

( x, y ) = (1.670, 1.660 ) 4

( x, y ) = ( 4,1)

y = 5 − 2x

−6

y = 9 − 2x −4

(4, 1)

−4

y = 2x − 1

(− 3.449, − 7.899) − 9

y = 9 − 2x y = x −3

−4

− 11

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

y=x −3

x2 − x + 1 = x2 + 2x + 4 −3 = 3x x = −1

71.

(1, 1)

(0, 0)

y = x2 + 2x + 4

70.

y = 52 x − 11 2

68. 4 x − y = 4  y = 4 x − 4

y = x3 − 3

y=x− 3

Section 2.2 76.

y = 2 x2

80. (a)

y = x4 − 2 x2

−4

(− 2, 8)

(b) Domain: 0 ≤ x ≤ 55 55

y = 2x 2

(2, 8)

0 4

(0, 0) −4

y = x 4 − 2x

1 + 0.73205 1.73205 = 1 − 0.73205 0.26795 ≈ 6.464079 ≈ 6.46 1 + 0.73205 1.73205 (b) = 1 − 0.73205 0.26795 1.73 ≈ 0.27 ≈ 6.407407 ≈ 6.41 Yes, the more rounding performed, the less accurate the result.

77. (a)

81. (a) Divide into two regions. Then find the area of each region and add.

Total area = Area 1 + Area 2 A( x) = 4 ⋅ x + 4 ⋅ 2x = 4x + 8x = 12 x

(b)

1 + 0.86603 1.86603 = 1 − 0.86603 0.13397 ≈ 13.92871538 ≈ 13.93 1 + 0.86603 1.86603 (b) = 1 − 0.86603 0.13397 1.87 ≈ 0.13 ≈ 14.38461538 ≈ 14.38 Yes, the more rounding performed, the less accurate the result.

78. (a)

x ( 280 − x ) + 63 54 (b) Domain: 0 ≤ x ≤ 280

79. (a)

240

A = 12 bh

A ( x ) = 12 ( x ) ( 23 x + 1) = 13 x 2 + 21 x

(b) 0

Area 2

Area 1

155

A = x + 0.33 ( 55 − x )

( x, y ) = ( 0, 0 ) , ( 2, 8 ) , ( −2, 8 ) 12

Solving Equations Graphically

240

280

and x = 164.5 miles.

(b)

T = I + S = x + 10,000 − 12 x = 10,000 + 12 x If S = 6600 = 10,000 − 12 x  12 x = 3400

 x = $6800 (c) If T = 13,800 = 10,000 + 12 x  3800 = 12 x

 x = $7600 (d) If T = 12,500 = 10,000 + 12 x then

x = 5000. Thus, S = 10,000 − 12 x = $7500.

156

Chapter 2

Solving Equations and Inequalities

84. y = 16.9t + 574, 0 ≤ t ≤ 12

86. (a) VTOTAL = VRECTANGULAR SIDEWALL + VTRIANGULAR SIDEWALL = l ⋅ w ⋅ ( pool width ) + 12 ⋅ b ⋅ h ⋅ ( pool width )

(a) Let t = 0 and find y. y = 16.9(0) + 574 = 574 The median weekly earnings of full-time workers was $574 in 2000. (b) The slope m is 16.9. The median weekly earnings of full-time workers increases by $16.90 every year. (c) Answers will vary. (d) Answers will vary. Sample answer: Algebraically: Let y = 800 and solve for t.

= ( 4 )( 40 )( 20 ) + ( 12 ) ( 40 )( 5 )( 20 ) = 5200 cubic feet

gallons   Number of gallons = 5200 ft 3  7.48  ft 3   = 38,896 gallons (c) The base of the pool passes through the points ( 0, 0 ) and ( 40, 5).

5−0 1 = 40 − 0 8 1 y − 0 = ( x − 0) 8 1 y= x 8

85. M = 43.4t + 5355, 1 ≤ t ≤ 13 W = 28.4t + 5398, 1 ≤ t ≤ 13 6000

)

Graphically: Using the zoom and trace features, find t when y = 800.

(a)

(

(b)

(d) For 0 ≤ d ≤ 5, by similar triangle,

d b =  b = 8 d. 5 40

1 1 bd ( 20 ) = ( 8d ) d ( 20 ) = 80d 2 , 0 ≤ d ≤ 5. 2 2 For 5 < d ≤ 9, 1 V = ( 5 )( 40 )( 20 ) + ( d − 5 )( 40 )( 20 ) 2 = 2000 + 800 d − 4000 = 800 d − 2000, 5 < d ≤ 9. So, V =

1 5000

The point of intersection is approximately ( 2.9, 5479.4). So, in 2002, both states had the same population. (b) 43.4t + 5355 = 28.4t + 5398

15t = 43

(0, 5)

t ≈ 2.9

(0, 0) d

The point of intersection is approximately ( 2.9, 5479.4). So, in 2002, both states had the same

(0, 9) (0, 5)

population.

(e)

(d) Find t = 16. M = 43.4(16) + 5355 = 6049.4 W = 28.4(16) + 5398 = 5852.4

5000 4000 3000 2000 1000 1 2 3 4 5 6 7 8 9

The population of Maryland will be 6,049,400 and the population of Wisconsin will be 5,852,400. Answers will vary.

(40, 5)

(0, 0)

Volume (in cubic feet)

(c) The slopes of the linear models represent the change in population per year. Since the slope of the model for Maryland is greater than that of Wisconsin, the population of Maryland is growing faster.

(40, 5)

Depth (in feet)

(f )

720

2000

3600

5200

(g) V = 4800 : 800d − 2000 = 4800 800 d = 6800 d = 8.5 feet

Section 2.2 87. True.

(b)

Solving Equations Graphically

157

f ( x ) = x2 − 1 Algebraically: x = −1 and x = 1 are zeros.

88. True. A line must intersect at least one axis.

f ( −1) = ( −1) − 1 = 0 2

x 99 = 89. x − 1 100 100 x = 99 x − 99

f (1) = (1) − 1 = 0 2

Numerically:

x = −99 The approximate answer −99.1 is not a good answer, even though the substitution yields a small error.

−2

−1

y = x2 − 1

−1

90. (a) From the table, f ( x ) = 0 for x = 3.

So, the zeros are −1 and 1. Graphically:

(b) From the table, g ( x ) = 0 for x = −2. (c) From the table, g ( x ) = − f ( x ) for x = 1. In this case,

f ( x ) = −6 and g ( x ) = 6. (d) From the table, f ( x ) = −6 g ( x ) . In this case,

−6

f ( x ) = −12 and g ( x ) = 2, for x = −1. 91. (a)

−2

(c)

y = 2x + 2 Algebraically: Let y = 0 : 2 x + 2 = 0

y = 2 x + 2, y = x 2 − 1 Algebraically:

2 x = −1 x = −1

x −3= 0 x=3

( −1, 0 ) x-intercept Let x = 0 : y = 2 ( 0 ) + 2

( 3, 8 )

( 0, 2 ) y-intercept

x = −1  y = 2 ( −1) + 2 = 0

Numerically:

−2 −1

y = 2 x + 2 −2

x +1 = 0 x = −1

x = 3  y = 2 ( 3) + 2 = 8

y=2

2 x + 2 = x2 − 1 x2 − 2 x − 3 = 0 ( x − 3)( x + 1) = 0

( −1, 0 )

Numerically:

−1

y = 2x + 2

y = x2 − 1

−1

The x-intercept is ( −1, 0 ) , The y-intercept is ( 0, 2 ) . Graphically: 4

The points of intersection are ( −1, 0 ) and ( 3, 8) . −4

Graphically: 10

−2

y1 (3, 8)

−9

(−1, 0)

9 −2

158

92.

93.

Chapter 2 12 5 3

⋅

3 3

Solving Equations and Inequalities

12 3 4 3 = 5 ( 3) 5

10 = 14 − 2

10 ⋅ 14 − 2

94.

( 14 + 2) ( 14 ) − (2) 10( 14 + 2) = =

14 + 2 14 + 2

14 − 4

95.

( 14 + 2)

= 2+

⋅

8 − 11

8 + 11 8 − 11

(

3 8 − 11 64 − 11

(

) = 3 (8 − 11 ) 53

)

3 10 + 1 14 3 10 + 1 14 3 10 + 1 = = 90 − 1 89 3 10 − 1 3 10 + 1 14

⋅

(

)

96. ( x − 6)(3 x − 5) = 3 x 2 − 5 x − 18 x + 30

= 3 x 2 − 23 x + 30 97.

( 3x + 13)( 4 x − 7) = 12 x2 + 31x − 91

98.

( 2 x − 9)( 2 x + 9) = 4 x2 − 81

99.

( 4 x + 1) = ( 4 x + 1)( 4 x + 1) = 16 x 2 + 8 x + 1 2

Section 2.3 Complex Numbers 1. (a) ii (b) iii (c) i 2.

11. 4 +

−9 = 4 +

= 4 + 3i

−1, − 1

3. complex, a + bi

12. 7 −

− 25 = 7 − = 7 − 5i

4. To multiply two complex numbers, ( a + bi )( c + di ) , the

FOIL Method can be used; ( a + bi )( c + di ) = ac + adi + bci + bdi 2

= ( ac − bd ) + ( ad + bc ) i.

13. 12 = 12 + 0i

= 12 14. − 3 = − 3 + 0i = −3

5. The additive inverse of 2 − 4i is − 2 + 4i. 6. The complex conjugate of 2 − 4i is 2 + 4i. 7. a + bi = −9 + 4i a = −9 b=4

15. − 8i − i 2 = − 8i − ( −1)

= 1 − 8i 16. 2i 2 − 6i = 2( −1) − 6i = − 2 − 6i

8. a + bi = 12 + 5i a = 12 b=5

17.

( −16 ) + 5 = ( 16i) + 5 2

= ( 4i ) + 5 2

9. ( a − 1) + ( b + 3 ) i = 5 + 8i

= 16i 2 + 5

a −1 = 5  a = 6 b+3=8b =5

= 16( −1) + 5 = −11

10. ( a + 6 ) + 2bi = 6 − 5i 2b = −5 b=− a+6=6 a=0

25i

5 2

18. − i −

( − 23 ) = − i − ( 23 i) 2

= − i − 23i 2 = − i − 23( −1) = 23 − i

Section 2.3 19.

−0.09 = 0.09i = 0.3i

20.

−0.0004 = 0.02i

21.

159

 3 5   5 11   3 5   5 11  25.  + i  +  + i  =  +  +  +  i 2 2  3 3  2 3 2 3  9 + 10 15 + 22 i = + 6 6 19 37 = + i 6 6

( 4 + i ) − ( 7 − 2i ) = ( 4 − 7 ) + (1 + 2 ) i = −3 + 3i

22.

Complex Numbers

(11 − 2i ) − ( −3 + 6i ) = (11 + 3) + ( −2 − 6 ) i

23. 13i − (14 − 7i ) = 13i − 14 + 7i = −14 + 20i

3 7  5 1  3 5 7 1 26.  + i  −  − i  =  −  +  +  i 4 5  6 6  4 6 5 6 1 47 =− + i 12 30

24. 22 + ( −5 + 8i ) − 9i = ( 22 − 5 ) + ( 8 − 9 ) i

27.

= 14 − 8i

= 17 − i

(1.6 + 3.2i ) + ( −5.8 + 4.3i ) = −4.2 + 7.5i

28. −( −3.7 − 12.8i ) − ( 6.1 − 16.3i ) = 3.7 + 12.8i − 6.1 + 16.3i

= ( 3.7 − 6.1) + (12.8 + 16.3) i = −2.4 + 29.1i

(

29. 5 +

) (

− 27 − −12 +

) (

− 48 = 5 + 3 3 i − −12 + 4 3 i

)

= 5 + 3 3 i + 12 − 4 3 i

(

)

= 17 + 3 3 − 4 3 i = 17 − 30.

( 7 + −18 ) + (3 + −32 ) = ( 7 + 3 2i) + (3 + 4 2i) = ( 7 + 3) + ( 3 2 + 4 2 ) i = 10 + 7 2i

31.

32.

( 6i )( 2i ) = 12i = ( 2 3 ) ( −1) = −2 3

−6 ⋅ −2 =

−5 ⋅ −10 =

( 5i )( 10i )

= 50 i 2 = 5 2 ( −1) = −5 2 33.

( −10 ) = ( 10i ) = 10i = −10

34.

( −75 ) = ( 75i ) = 75i = −75

38.

= 12 − 22i − 6 = 6 − 22i

39. 4i ( 8 + 5i ) = 32i + 20i 2

= 32i + 20 ( −1) = −20 + 32i

40. −3i ( 6 − i ) = −18i − 3 = −3 − 18i 41.

( 14 + 10i )( 14 − 10i ) = 14 − 10i

42.

( 3 + 15i )( 3 − 15i ) = ( 3 )( 3 ) − 3 15i + 3 15i − ( 15i )( 15i ) = 3 − 15i 2 = 3 + 15 = 18

35. 4 ( 3 + 5i ) = 4 ( 3 ) + 4 ( 5 ) i 43.

( 6 + 7i ) = ( 6 ) + 2 ( 6 )( 7i ) + ( 7i ) 2

=3+i+2 = 5+i

= ( 36 − 49 ) + 84i

= −30 + 18i

(1 + i )( 3 − 2i ) = 3 − 2i + 3i − 2i 2

= 36 + 84i + 49i 2

36. −6 ( 5 − 3i ) = ( −6 )( 5 ) + ( −6 )( −3i )

37.

= 14 + 10 = 24

= 12 + 20i

( 6 − 2i )( 2 − 3i ) = 12 − 18i − 4i + 6i 2

= −13 + 84i 44.

( 5 − 4i ) = ( 5) − 2 ( 5)( 4i ) + ( 4i ) 2

= 25 − 40i + 16i 2 = ( 25 − 16 ) − 40i = 9 − 40i

160

Chapter 2

45.

( 4 + 5i ) − ( 4 − 5i ) = ( 4 + 5i ) + ( 4 − 5i )  ( 4 + 5i ) − ( 4 − 5i )  = 8 (10i ) = 80i

46.

Solving Equations and Inequalities

(1 − 2i ) − (1 + 2i ) = 1 − 4i + 4i 2 − (1 + 4i + 4i 2 ) 2

55.

6 6 −i −6i −6i = ⋅ = = = −6i i i −i −i 2 1

56.

−5 i −5i 5 ⋅ = = i 2i i −2 2

57.

2 2 4 + 5i 8 + 10i 8 10 = ⋅ = = + i 4 − 5i 4 − 5i 4 + 5i 16 + 25 41 41

58.

3 1 + i 3 + 3i 3 + 3i 3 3 ⋅ = = = + i 1 − i 1 + i 1 − i2 2 2 2

59.

3−i 3−i 3−i = ⋅ 3+i 3+i 3−i

= 1 − 4i + 4i 2 − 1 − 4i − 4i 2 = −8i 47. 6 + 2i is the complex conjugate of 6 − 2i.

(6 − 2i)(6 + 2i) = 36 − 4i 2 = 36 + 4 = 40

9 − 6i + i 2 9 − i2 9 − 6i − 1 = 9+1 8 − 6i = 10 4 3 = − i 5 5

48. 3 − 5i is the complex conjugate of 3 + 5i.

(3 + 5i)(3 − 5i) = 9 − 25i 2 = 9 + 25 = 34 49. −1 −

7 i is the complex conjugate of −1 +

( −1 +

)(

7 i.

)

7 i = 1 − 7i 2

7 i −1 −

=1+ 7 = 8

50. − 4 +

3 i is the complex conjugate of − 4 −

(− 4 −

)(

3 i.

60.

)

3 i = 16 − 3i 2

3i −4 +

= 16 + 3 = 19 − 29 =

51. −

61.

29 i

29 i is the complex conjugate of

(− 29 i)( 29 i) = − 29i

8 − 7i 1 + 2i 8 + 16i − 7i − 14i 2 ⋅ = 1 − 2i 1 + 2i 1 − 4i 2 22 + 9i 22 9 = = + i 5 5 5 i

( 4 − 5i )

29 i.

−10 =

i −9 + 40i ⋅ −9 − 40i −9 + 40i −40 − 9i = 81 + 40 2 40 9 i =− − 1681 1681

10 i

− 10 i is the complex conjugate of

(− 10 i)( 10 i) = −10i

10 i.

62.

(9 +

)(

6 i.

)

6 i = 81 − 6i 2 = 81 + 6 = 87

54. − 8 −

(− 8 +

15 i is the complex conjugate of − 8 +

)(

15 i − 8 −

( 2 + 3i )

5i −5 − 12i ⋅ −5 + 12i −5 − 12i

6 i is the complex conjugate of 9 −

6i 9 −

−25i + 60 25 + 144 60 25 = − i 169 169

= 10

53. 9 +

i 16 − 25 − 40i

= 29

52.

)

15 i = 64 − 15i 2 = 64 + 15 = 79

63. 15 i.

2 (1 − i ) − 3 (1 + i ) 2 3 − = 1+ i 1− i (1 + i )(1 − i ) 2 − 2i − 3 − 3i 1+1 −1 − 5i 1 5 = =− − i 2 2 2

Section 2.3

64.

2i ( 2 − i ) 5( 2 + i ) 2i 5 + = + 2 + i 2 − i ( 2 + i )( 2 − i ) ( 2 + i )( 2 − i ) 4i − 2i 2 + 10 + 5i = 4 − i2 12 + 9i 12 9 = = + i 5 5 5

65.

i 2i 3i + 8i 2 + 6i − 4i 2 + = 3 − 2i 3 + 8i ( 3 − 2i )( 3 + 8i )

73. (a)

74. (a)

67. −6i + i = −6i i + i

114

16 3

28 2

z1 = 5 + 2i

1 1 1 1 1 = + = + z z1 z2 5 + 2i 3 − 4i =

( 3 − 4 i ) + ( 5 + 2i ) ( 5 + 2i )( 3 − 4i )

8 − 2i 23 − 14i 23 − 14i  8 + 2i  z=   8 − 2i  8 + 2i  =

212 − 66i ≈ 3.118 − 0.971i 68 z1 = 16i + 9 =

(b)

z2 = 20 − 10i

1 1 1 1 1 = + = + z z1 z2 9 + 16i 20 − 10i

( 20 − 10i ) + ( 9 + 16i ) ( 9 + 16i )( 20 − 10i )

29 + 6i 340 + 230i 340 + 230i  29 − 6i  11,240 + 4630i z=  = 29 + 6i  29 − 6i  877 =

= 6i − 1 = −1 + 6i

68. 4i 2 − 2i 3 = −4 + 2i 69.

i 45 = i 4

= −6 ( −1) i + ( −1)

≈ 12.816 + 5.279i

( −75 ) = ( 5 3i ) = 5 ( 3 ) i = 125 ( 3 3 ) ( −i ) 3

( ) = (1) = 1

i 20 = i 4

z 2 = 3 − 4i

−4 + 9i 9 + 18i + 16 −4 + 9i 25 − 18i = ⋅ 25 + 18i 25 − 18i −100 + 72i + 225i + 162 = 252 + 182 62 + 297i = 949 62 297 = + i 949 949

161

( ) i = (1) i = i (c) i = ( i ) i = (1) i = i = −i (d) i = ( i ) i = (1) i = i = −1 (b)

1+ i 3 1 + i −i 3 4+i − = ⋅ − ⋅ 66. 4−i i i −i 4 − i 4 + i −i + 1 12 + 3i = − 1 16 + 1 5 20 = − i 17 17

Complex Numbers

= −375 3i

75. False. A real number a + 0i = a is equal to its conjugate. 76. False. i 44 + i150 − i 74 − i109 + i 61 = 1 − 1 − i + i = 1

70.

( −2 ) = ( 2i ) = 8i = 8i i = −8

77. False. For example, (1 + 2i ) + (1 − 2i ) = 2, which is not an

71.

1 1 i i i = ⋅ = = =i i3 i3 i i 4 1

78. True. Let z1 = a1 + b1i and z2 = a2 + b2i. Then

72.

4 2

1 1 8i 8i 1 = 3 = ⋅ = = i 3 2 ( 2i ) 8i −8i 8i −64i 8

imaginary number.

z1z2 = ( a1 + b1i )( a2 + b2i ) = ( a1a2 − b1b2 ) + ( a1b2 + b1a2 ) i = ( a1a2 − b1b2 ) − ( a1b2 + b1a2 ) i = ( a1 − b1i )( a2 − b2i ) = a1 + b1i a2 + b2i = z1 z2 .

162

Chapter 2

Solving Equations and Inequalities (b) Product of the binomials: ( x + 5)( 2 x − 1) = 2 x 2 − x + 10 x − 5

79. True. Let z1 = a1 + b1i and z2 = a2 + b2i. Then

z1 + z2 = ( a1 + b1i ) + ( a2 + b2i )

= 2 x2 + 9x − 5 Product of the complex numbers: (1 + 5i )( 2 − i ) = 2 − i + 10i − 5i 2

= ( a1 + a2 ) + ( b1 + b2 ) i = ( a1 + a2 ) − ( b1 + b2 ) i

= ( a1 − b1i ) + ( a2 − b2i )

= 7 + 9i The product of the binomials results in a seconddegree trinomial. The product of the complex numbers is a complex number because of the property i 2 = 1. (c) Answers will vary.

= a1 + b1i + a2 + b2i = z1 + z2 . 80. (i) 2 = 2 + 0i is the point ( 2, 0) that matches D.

(ii) 2i = 0 + 2i is the point (0, 2) that matches C. (iii) − 2 + i is the point ( − 2, 1) that matches A.

( 4 x − 5)( 4 x + 5) = 16 x 2 − 20 x + 20 x − 25

83.

= 16 x 2 − 25

(iv) 1 − 2i is the point (1, − 2) that matches B. 81. The error is not simplifying before multiplying. The correct method is −6 −6 = 6 i 6 i = 6i 2 = −6. 82. Given the binomials: x + 5 and 2 x − 1 and the complex numbers: 1 + 5i and 2 − i

(a) Sum of the binomials: ( x + 5) + ( 2 x − 1) = 3x + 4

( x + 2 ) = x 3 + 3 x 2 2 + 3 x ( 2 ) + 23 3

84.

= x 3 + 6 x 2 + 12 x + 8

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

1 2

85.

1 2

= 3 x 2 + 232 x − 2

( 2 x − 5) = 4 x 2 − 20 x + 25 2

86.

Sum of the complex numbers:

(1 + 5i ) + ( 2 − i ) = 3 + 4i

Answers will vary. Sample answer: The coefficient of the x-terms of the binomials are the same as the real part of the complex numbers. The constant terms of the binomials are the same as the coefficients of the imaginary part of the complex numbers.

Section 2.4 Solving Quadratic Equations Algebraically 1.

quadratic equation

discriminant

Four methods to solve a quadratic equation are: factoring, extracting square roots, completing the square, and using the Qualitative Formula.

The height of an object that is falling is given by the equation s = −16t 2 + v0 t + s0 , where s is the height, vo is the initial velocity, and so is the initial height.

2 x2 = 3 − 5x

Standard form; 2 x 2 + 5 x − 3 = 0 6.

x 2 = 25 x + 26

Standard form: x 2 − 25 x − 26 = 0

1 5

( 3x − 10 ) = 12 x 2

3 x 2 − 10 = 60 x Standard form: 3 x 2 − 60 x − 10 = 0 x ( x + 2) = 3x2 + 1

x2 + 2 x = 3x2 + 1 2

−2 x + 2 x − 1 = 0

( −1) ( −2 x 2 + 2 x − 1) = −1( 0 ) Standard form: 2 x 2 − 2 x + 1 = 0 9. 15 x 2 + 5 x = 0

5 x(3 x + 1) = 0 5x = 0  x = 0 3x + 1 = 0  x = −

1 3

Section 2.4 10. 9 x 2 − 21x = 0

Solving Quadratic Equations Algebraically 18.

3 x (3 x − 7 ) = 0

− x 2 − 11x = 30 − x 2 − 11x − 30 = 0

3x = 0  x = 0

x 2 + 11x + 30 = 0

7 3

( x + 5)( x + 6) = 0

3x − 7 = 0  x =

x + 5 = 0  x = −5 x + 6 = 0  x = −6

11. x 2 − 10 x + 21 = 0

( x − 7)( x − 3) = 0 x−7 = 0  x = 7

19.

x−3 = 0  x = 3 12.

x − 10 x + 9 = 0

( x − 9 )( x − 1) = 0

20.

x−9=0 x =9 x −1 = 0  x = 1

( x + a ) − b2 = 0  ( x + a ) + b  ( x + a ) − b  = 0    2

x + a + b = 0  x = −a − b x + a − b = 0  x = −a + b

x 2 + 2 ax + a 2 = 0

( x + a) = 0 2

x+a=0

13. x 2 − 8 x + 16 = 0

( x − 4)

x = −a

= 0

21.

x − 4 = 0

22. 2

4 x + 12 x + 9 = 0

23. 3 x 2 = 81

2x + 3 = 0 2 x = −3

x 2 = 27

x=−

x = ±

3 2

3x2 = 8 − 2 x 3x 2 + 2 x − 8 = 0

(3x − 4)( x + 2) = 0 4 3x − 4 = 0  x = 3 x + 2 = 0  x = −2 2 x 2 = 19 x + 33

16.

x 2 = 144 x = ± 144 = ±12

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

15.

x 2 = 49 x = ± 49 = ±7

x = 4

14.

x = ± 3 3 ≈ ± 5.20

24. 9 x 2 = 36 x2 = 4 x = ±

x = ±2

25. x 2 + 12 = 112

2 x 2 − 19 x − 33 = 0

x 2 = 100

( 2 x + 3)( x − 11) = 0

x = ± 100

2x + 3 = 0  x = −

3 2

x − 11 = 0  x = 11

x = ±10

26. x 2 − 3 = 78

− x 2 − 11x = 28

x 2 = 81

− x 2 − 11x − 28 = 0

x = ±

17.

163

x + 11x + 28 = 0

x = ±9

( x + 4)( x + 7) = 0 x + 4 = 0  x = −4 x + 7 = 0  x = −7

164

Chapter 2

27.

( x − 12 ) = 16

Solving Equations and Inequalities

35.

x 2 − 6 x = −2

x − 12 = ± 16 = ±4 x = 12 ± 4 x = 8, 16 28.

x2 − 6 x + 2 = 0 x 2 − 6 x + 32 = −2 + 32

( x − 3) = 7 2

x −3= ± 7

( x − 5) = 25 2

x = 3± 7

x − 5 = ±5 x =5±5

36.

x 2 + 8 x + 14 = 0

x = 0, 10 29.

x 2 + 8 x = −14 2

x + 8 x + 4 2 = −14 + 16

( 3 x − 1) + 6 = 0 2 ( 3 x − 1) = −6 2

( x + 4) = 2 2

x+4=± 2

3 x − 1 = ± −6 = ± 6 i x= 30.

1 6 ± i ≈ 0.33 ± 0.82i 3 3

x = −4 ± 2 37.

x 2 − 4 x + 13 = 0 x 2 − 4 x + 4 = −13 + 4

( 2 x + 3) + 25 = 0 2 ( 2 x + 3) = −25 2

( x − 2 ) = −9 2

x − 2 = ±3i x = 2 ± 3i

2 x + 3 = ± −25 = ±5i x = − 23 ± 25 i = −1.50 ± 2.50i

31.

x=2

32.

38.

( x − 7 ) = ( x + 3) x − 7 = ± ( x + 3) x − 7 = + ( x + 3)  impossible x − 7 = − ( x + 3)  2 x = 4 2

x 2 − 6 x + 9 = −34 + 9

( x − 3) = −25 2

x − 3 = ±5i x = 3 ± 5i 39.

( x + 5) = ( x + 4 ) ( x + 5) = ± ( x + 4 ) 2

x 2 − 6 x + 34 = 0

x 2 + 8 x + 32 = 0 x 2 + 8 x + ( 4) = − 32 + 16 2

( x + 4)

x + 5 = x + 4, not possible

x = − 4 ± 4i

x = − 29 = −4.50

x 2 + 4 x + 4 = 32 + 4

40.

x 2 + 18 x + 117 = 0 x 2 + 18 x + 81 = −117 + 81

( x + 9 ) = −36 2

( x + 2 ) = 36 2

x + 2 = ±6 x = −2 ± 6 x = −8, 4

−16

x + 4 = ± 4i

2 x = −9

x 2 + 4 x = 32

= −16

x + 4 = ±

x + 5 = − ( x + 4)

33.

x + 9 = ±6i x = −9 ± 6i 41. −6 + 2 x − x 2 = 0

( x − 2 x + 1) = −6 + 1 2

34.

x2 − 2 x − 3 = 0 x2 − 2 x + 1 = 3 + 1

( x − 1) = 4 2

x − 1 = ±2 x =1± 2

( x − 1) = −5 2

x − 1 = ± −5 = ± 5i x = 1 ± 5i

x = − 1, 3

Section 2.4 42. − x 2 + 6 x − 16 = 0

Solving Quadratic Equations Algebraically 9 x 2 − 12 x = 14 4 14 x2 − x = 3 9

46.

x 2 − 6 x + 16 = 0 x 2 − 6 x + (3) = −16 + 9 2

( x − 3)

x2 −

= −7

x −3 = ±

−7

x −3 = ±

4  2  14 4 x+  = + 3 9 9 3 2

2  18  x−  = 3 9  2

x = 3±

2  x−  =2 3  2 x− =± 2 3 2 x= ± 2 3

43. 9 x 2 − 18 x + 3 = 0 1 x2 − 2 x + = 0 3

1 3 1 2 2 2 x − 2x + 1 = − + 1 3 2 2 ( x − 1) = 3 x2 − 2 x = −

47.

y = ( x + 3) − 4 2

− 12

(b) The x-intercepts are ( −1, 0 ) and ( −5, 0 ) . 2

4 = ( x + 3)

± 4 = x+3 −3 ± 2 = x

5 = 0 4 5 2 x 2 − 4 x + ( 2) = +4 4 21 2 ( x − 2) = 4 x2 − 4x −

x = −1 or x = −5

The x-intercepts are ( −1, 0 ) and ( −5, 0 ) . 48.

x−2 = ±

21 4

x−2 = ±

21 2 21 2

2 x + 5x − 8 = 0 5 x2 + x − 4 = 0 2 5 25 25 x2 + x + =4+ 2 16 16 2

5 89  x+  = 4  16  x+

0 = ( x + 3) − 4

(c)

6 3

44. 4 x 2 − 16 x − 5 = 0

45.

−6

2 x =1± 3

x = 2±

(a)

2 x −1 = ± 3

x =1±

165

y = 1 − ( x − 2) (a)

3 −7

−9

(b) The x-intercepts are (1, 0 ) and ( 3, 0 ) . (c)

0 = 1 − ( x − 2)

( x − 2) = 1 2

x−2 = ± 1 x = 2 ± 1 = 3, 1

The x-intercepts are ( 3, 0 ) and (1, 0 ) .

5 89 =± 4 4 x=−

5 89 ± 4 4

166

Chapter 2

Solving Equations and Inequalities

49.

y = −4 x 2 + 4 x + 3 (a)

52.

y = 2 x2 − x − 1 = 0 7

−9

−7

−3

−6

(b) The x-intercepts are ( −0.5, 0 ) and (1.5, 0 ) . 0 = −4 x 2 + 4 x + 3 4x − 4x = 3

(c)

(

Two real solutions 53.

y = 47 x 2 − 8 x + 28 = 0 9

)

4 x2 − x = 3 x 2 − x = 34 x 2 − x + ( 12 ) = 34 + ( 12 ) 2

−2

(x − ) =1 1 2

13 −1

One real solution

x − 12 = ± 1

54.

x = 12 ± 1

y = 13 x 2 − 5 x + 25 = 0

x = 32 or x = − 12

The x-intercepts are ( 32 , 0 ) and ( − 12 , 0 ) . 50.

y = x2 + 3x − 4 (a)

No real solutions

− 11

55.

y = −0.2 x 2 + 1.2 x − 8 = 0 4

−9

−8

(b) The x-intercepts are (1, 0 ) and ( −1, 0 ) . (c)

0 = x2 + 3x − 4

− 12

0 = ( x + 4 )( x − 1)

No real solution

0 = x + 4  x = −4 0 = x −1 x = 1

56.

y = 9 + 2.4 x − 8.3 x 2 = 0

The x-intercepts are ( −4, 0 ) and (1, 0 ) . 51.

y = x2 − 4 x + 4 = 0 −9

9 −2

Two real solution −6

9 −2

One real solution

57. x 2 − 9 x + 19 = 0 x = = = =

−b ±

b 2 − 4ac 2a

− ( − 9) ± 9± 9± 2

(− 9) − 4(1)(19) 2(1) 2

81 − 76 2 5

Section 2.4 58.

x 2 − 10 x + 22 = 0

x= =

Solving Quadratic Equations Algebraically 62.

−b ± b2 − 4ac 2a − ( −10 ) ±

8x = 4 − x2 x2 + 8x − 4 = 0

( −10 ) − 4 (1)( 22 ) 2 (1) 2

10 ± 100 − 88 2 10 ± 2 3 = =5± 3 2

x = =

59.

−3 ± 9 − 4 ( 8 )

b 2 − 4ac 2a

− (8) ±

(8) − 4(1)( − 4) 2(1)

−8 ±

64 + 16 2

− 8 ± 80 2

−8 ± 4 5 2

63. 20 x 2 − 20 x + 5 = 0 4 x2 − 4x + 1 = 0 x =

−5 ± 25 − 4 (16 )

2 −5 ± −39 = 2 5 39 =− ± i 2 2

61.

b 2 − 4ac 2a

− ( − 4) ± 4±

( − 4)2 − 4( 4)(1) 2( 4)

16 − 16 8

4±0 8 1 = 2

x2 + 4x − 8 = 0

−b ±

4x = 8 − x2

x =

= −4 ± 2 5

x 2 + 5 x + 16 = 0 x=

−b ±

x + 3x + 8 = 0 2 −3 ± −23 = 2 3 23i =− ± 2 2

60.

167

−b ± − ( 4) ± −4 ±

b 2 − 4ac 2a

( 4) − 4(1)( − 8) 2(1)

64. 9 x 2 − 18 x + 9 = 0

x2 − 2 x + 1 = 0

16 + 32 2

− 4 ± 48 2

−4 ± 4 3 2

x= =

−b ± b 2 − 4ac 2a − ( −2 ) ±

( −2 ) − 4 (1)(1) 2 (1) 2

2± 0 2 =1 =

= −2 ± 2 3

168

Chapter 2

Solving Equations and Inequalities

65. 16 x 2 + 24 x + 9 = 0

x = = =

70.

(

− ( 24) ± − 24 ±

(24) − 4(16)(9) 2(16) 2

576 − 576 32

− 24 ± 0 32 3 = − 4

71.

= =

72.

b − 4ac 2a

− (30) ±

(30) − 4(9)(25) 2(9) 2

73.

− b ± b3 − 4ac 2a 2 ± 4 − 4 (13 4 )

2 2 ± −9 = 2 3 =1± i 2

74.

2 ( 4)

−16 ± −16 8 −16 ± 4i = 8 1 = −2 ± i 2 =

19 4 19 2 + 4 x + 4x + 4 = − 4 3 2 ( x + 2) = − 4 3 x + 2 = ± − 4 x2 + 4 x = −

x + 2 = ±

68. 9 x 2 − 6 x + 37 = 0

3 i 2

x = −2 ±

−b ± b − 4 ac 2a 6 ± 36 − 1332 = 18 1 = ± 2i 3

13 4

13 =0 4

67. 4 x 2 + 16 x + 17 = 0

x2 − 2x = −

− 30 ± 0 18 5 = − 3

−16 ± 16 2 − 4 ( 4 )(17 )

x2 − 2 x +

900 − 900 18

−b ± b 2 − 4 ac 2a

( x − 1) = −1 x − 1 = ± −1 = ±i x =1± i

x+3=9 x =6 x + 3 = −9  x = −12

− 30 ±

( x + 3) = 81 x + 3 = ±9

66. 9 x 2 + 30 x + 25 = 0

−b ±

x ( x + 3) = 0 x=0 x + 3 = 0  x = −3

x =

)

11 x 2 + 3 x = 0

b 2 − 4ac 2a

−b ±

11x 2 + 33 x = 0

69.

x − 3x − 4 = 0

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

75.

3 i 2

5 x 2 = 3x + 1 5 x 2 − 3x − 1 = 0 x = =

−b ±

b 2 − 4ac 2a

− ( − 3) ±

(− 3) − 4(5)(−1) 2(5) 2

9 + 20 10 3 ± 29 = 10 =

3±

Section 2.4 4 x2 = 7 x + 3

76.

Solving Quadratic Equations Algebraically 81.

4x − 7x − 3 = 0 x = = = =

− ( − 7) ±

( − 7) − 4( 4)( − 3) 2( 4)

(

)

2 x 2 + x − 30 = 2 x 2 + 2 x − 60 = 0

(other answers possible)

82.

( x + 2 )( x − 1) = 0

7±

49 + 48 8

x2 + x − 2 = 0 Multiply by 2 to find a second equation.

7±

2 x2 + x − 2 = 2 x2 + 2 x − 4 = 0

(

−b ± b 2 − 4 ac x= 2a

83.

−7 ± 7 2 − 4 ( 2 )( −3) 2 ( 2)

−7 ± 73 4 7 73 =− ± 4 4 =

10 x 2 − 11x + 3 = 0

( 5 x − 3)( 2 x − 1) = 0 or

2x −1 = 0 2x = 1 1 x= 2

( 2 x − 3) = 0 2

(other answers possible) 85.

(

2 80. 16 x − 24 x + 9 = 0

( 4 x − 3) = 0 2

3 x= 4

x 2 − 75 = 0 1 Multiply by to find a second equation. 5 1 2 1 x − 75 = x 2 − 15 = 0 5 5 (other answers possible)

2x = 3 3 x= 2

4x = 3

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

2x − 3 = 0

4x − 3 = 0

)

2 4  84.  x +   x −  = 0 3 3  2 8 x2 − x − = 0 3 9 Multiply by 9 to find a second equation. 2 8  9 x2 − x −  = 9x2 − 6 x − 8 = 0 3 9 

79. −4 x 2 + 12 x − 9 = 0 4 x 2 − 12 x + 9 = 0

7 6  x+ x− =0 3 7   ( 3 x + 7 )( 7 x − 6 ) = 0 21 21x 2 + 31x − 42 = 0 1 Multiply by to find a second equation. 7 1 31 21x 2 + 31x − 42 = 3 x 2 + x − 6 = 0 7 7 (other answers possible)

(

78. −10 x 2 + 11x − 3 = 0

5x − 3 = 0 5x = 3 3 x= 5

)

(other answers possible)

77. 2 x 2 + 7 x − 3 = 0

( x + 6 )( x − 5) = 0 x 2 + x − 30 = 0 Multiply by 2 to find a second equation.

b 2 − 4ac 2a

−b ±

169

86.

)

( x − 2 5 )( x + 2 5 ) = 0 x − (2 5 ) = 0 2

x 2 − 20 = 0 1 Multiply by to find a second equation. 2 1 2 1 x − 20 = x 2 − 10 = 0 2 2 (other answers possible)

(

)

170 87.

Chapter 2

Solving Equations and Inequalities

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

S = x 2 + 4 xh

92.

561 = x 2 + 4 x( 4) 561 = x 2 + 16 x

561 + 64 = x 2 + 16 x + 64

x − 2 x + 1 − 12 = 0

625 = ( x + 8)

x − 2 x − 11 = 0 Multiply by 5 to find a second equation.

(

)

5 x 2 − 2 x − 11 = 5 x 2 − 10 x − 55 = 0

(other answers possible) 88.

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

625 = x + 8

− 8 ± 25 = x Because x > 0, x = 17. The dimensions are 17 feet by 17 feet by 4 feet. 93. (a)

4 x + 3 y = 100 ( amount of fence ) 1 (100 − 4 x ) 3 Domain: 0 < x < 25 y=

x 2 − 4 x + 4 − 45 = 0 x 2 − 4 x − 41 = 0

Area = A ( x ) = ( 2 x ) y = 2 x

Multiply by 2 to find a second equation.

(

)

2 x 2 − 4 x − 41 = 2 x 2 − 8 x − 82 = 0

8 x ( 25 − x ) 3 8 200 = − x2 + x 3 3 y Area 92 368 ≈ 123 3 3

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

(b)

( x − 2) + 1 = 0 2

x2 − 4 x + 5 = 0 Multiply by −1 to find a second equation.

(

)

− x2 − 4 x + 5 = − x2 + 4 x − 5 = 0

(other answers possible) 90.  x − ( 3 + 4i )   x − ( 3 − 4i )  = 0

 ( x − 3 ) − 4 i   ( x − 3 ) + 4i  = 0   

( x − 3) + 16 = 0

(

4 6

224

76 3

304

68 3

52 3

44 3

1088 3

≈ 363

400 416 1232 3

≈ 411

Approximate dimensions for maximum area: x = 12, y = 523 , or 24 m × 523 m

x 2 − 6 x + 25 = 0 Multiply by −1 to find a second equation.

1 (100 − 4 x ) 3

(other answers possible)

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

(c)

450

)

− x 2 − 6 x + 25 = − x 2 + 6 x − 25 = 0

(other answers possible)

Approximate dimensions for maximum area: x = 12.5, y = 16.67, or 25 m × 503 m

91. (a) w

(d) The graphs of y1 = 83 x ( 25 − x ) and y2 = 350 intersect at x = 7.5 and x = 17.5. The dimensions are therefore 15 m × 23 13 m and 35 m × 10 m.

w + 14

w ( w + 14 ) = 1632

(b) 2

w + 14 w − 1632 = 0

(c)

( w + 48)( w − 34 ) = 0 w = 34, length = w + 14 = 48 Width: 34 feet, length: 48 feet

8 3

(e)

x ( 25 − x ) = 350

−8 x 2 + 200 x = 1050 2

4 x − 100 x + 525 = 0

( 2 x − 35)( 2 x − 15) = 0 x = 352 or x = 152

Section 2.4

Solving Quadratic Equations Algebraically

171

V = Length ⋅ Width ⋅ Height = 576 cm 3

94.

( x )( x )( 4 ) = 576 4 x 2 = 576 x 2 = 144 x = 12 cm

Because x = 12 cm, the original piece of material is 12 + 4 + 4 = 20 cm by 20 cm. 95. (a) s = −16t 2 + v0t + s0

= −16t 2 + (0)t + 2080 = −16t 2 + 2080 (b)

2080

2016

1824

1504

1056

480

− 224

(c) From the table, the object reaches the ground between 10 and 12 seconds, (10, 12). 0 = −16t 2 + 2080 16t 2 = 2080 t 2 = 130 t = ± 130 t = ±11.40

The object reaches the ground after 11.40 seconds. s = −16t 2 + v0t 2 + s0

96. (a)

97. (a)

v0 = 45, s0 = 5.5

s = −16t 2 + v0t + s0 0 = −16t 2 + 8000

s = −16t + 45t + 5.5

16t 2 = 8000

(b)

s ( 12 ) = −16 ( 12 ) + 45 ( 12 ) + 5.5 = 24 feet

t 2 = 500

(c)

−16t 2 + 45t + 5.5 = 6

t = 10 5 ≈ 22.36 seconds.

16t − 45t + 0.5 = 0 Using the Quadratic Formula, t ≈ 2.801 seconds.

(b) Distance =

( 600 miles hour ) (10 5 seconds ) ( 3600 seconds hour )

(

)

= 61 10 5 miles ≈ 3.73 miles ≈ 19,677.4 feet

The curves y = −16t 2 + 45t + 5.5 and y = 6 intersect at t ≈ 2.801.

98. (a)

s ( t ) = −16t 2 + 1100

(b) The pellets hit the ground when 16t 2 = 1100  t ≈ 8.29 seconds. 95 95 miles per hour = miles per second. 3600 The plane travels 95 ⋅ 8.29 ≈ 0.22 mile, or 1162 feet. 3600

172

Chapter 2

Solving Equations and Inequalities

99. (a) S = − 0.143t 2 + 3.73t + 32.5, 5 ≤ t ≤ 13

Find t when S = 50. 50 = − 0.143t 2 + 3.73t + 32.5 0 = − 0.143t 2 + 3.73t − 17.5 t = t = t =

− (3.73) ± − 3.73 ±

(3.73) − 4( − 0.143)( −17.5) 2( − 0.143) 2

13.9129 − 10.01 − 0.286

− 3.73 ± 3.9029 − 0.286

t ≈ 6.13 and t ≈ 19.95 (not in the domain of the model)

In 2006, the average salary was $50,000. (b)

47.46

49.73

51.60

53.19

54.49

55.50

56.23

56.67

56.82

(c)

(d)

55.5 = − 0.143t 2 + 3.73t + 32.5 0 = − 0.143t 2 + 3.73t − 23 t = t = t =

− (3.73) ± − 3.73 ±

(3.73)2 − 4(− 0.143)(− 23) 2( − 0.143) 13.9129 − 13.156 − 0.286

− 3.73 ± 0.7569 − 0.286

t = 10

and t ≈ 16.1 (not in the domain of the model)

In 2010, the average salary was $55,500. (e) Answers will vary. Sample answer: For some years, the model may be used to predict the average salaries for years beyond 2013. However, the model eventually would yield values that begin to decrease and eventually become negative.

Section 2.4

Solving Quadratic Equations Algebraically

173

100. (a) D = 0.051t 2 + 0.20t + 5.0, 5 ≤ t ≤ 14

Find t when D = 10. 10 = 0.051t 2 + 0.20t + 5.0 0 = 0.051t 2 + 0.20t − 5.0 t = t =

− (0.20) ± − 0.20 ±

(0.20)2 − 4(0.051)(− 5.0) 2(0.051)

0.04 + 1.02 0.102

− 0.20 ± 1.06 0.102 t ≈ 8.13, and t ≈ −12.05 (not in the domain of the model) t =

In 2008, total public debt reached $10 trillion. (b)

(c)

7.28

8.04

8.90

9.86

10.93

12.10

13.37

14.74

16.22

17.80

(d) 20 = 0.051t 2 + 0.20t + 5.0 0 = 0.051t 2 + 0.20t − 15.0 t = t =

− (0.20) ± − 0.20 ±

(0.20)2 − 4(0.051)(−15.0) 2(0.051)

0.04 + 3.06 0.102

− 0.20 ± 3.10 0.102 t ≈ 15.3, and t ≈ −19.2 (not in the domain of the model) t =

In 2015, the total public debt will reach $20 trillion. (e) Answers will vary. Sample answer: For some years, the model may be used to predict the total public debt for years beyond 2014. However, the model eventually would yield values that are very high.

174

Chapter 2

Solving Equations and Inequalities

101. C = 0.45 x 2 − 1.73 x + 52.65, 10 ≤ x ≤ 25

(a)

300

(b) When C = 150, x = 16.797° C. (c) If the temperature is increased from 10° C to 20° C, the oxygen consumption increases by a factor of approximately 2.5, from 80.35 to 198.05. 102. F = − 0.091s 2 + 1.639 s + 2.20, 5 ≤ s ≤ 65

(a)

(b) Use the maximum feature of a graphing utility to find the greatest fuel efficiency. So, the car should travel at a speed of 42.91 miles per hour for the greatest fuel efficiency of 37.36 miles per gallon. (c) When the average speed of the car is increased from 20 miles per hour to 30 miles per hour, the fuel efficiency will increase from 27.34 miles per gallon to 34.18 miles per gallon, which is a factor of approximately 1.25. 105. False. The solutions are complex numbers.

103. (a) x 2 + 152 = l 2

x 2 + 225 = l 2 2

106. False. You can only draw a conclusion about the factors if their product is 0, not 8.

x 2 + 225 = 5625

107. False. The solutions are either both imaginary or both real.

(b) x + 225 = (75) 2

x = 5400 x = ±

108. (a)

x ( ax + b ) = 0

5400

x = ± 30 6

x=0

x ≈ ± 73.5

Because x > 0, the distance is about 73.5 feet. 104. Let u be the speed of the eastbound plane. Then

u + 50 = speed of northbound plane. 2

ax + b = 0  ax = −b b x=− a 2 (b) ax − ax = 0 ax ( x − 1) = 0

u ( 3 )  + ( u + 50 ) 3 = 2440 2     9u + 9 ( u + 50 ) = 2440 2

ax = 0 x=0

x −1 = 0  x = 1

18u + 900u + 22,500 = 2440 2 18u 2 + 900u − 5,931,100 = 0 u=

−900 ±

ax 2 + bx = 0

( 900 ) − 4 (18 )( −5,931,100 ) 2 (18 ) 2

−900 ± 427,849,200 36 ≈ 549.57 =

109. Add the two solutions and the radicals cancel. −b + b 2 − 4 ac −b − b 2 − 4 ac + 2a 2a b =− a

u + 50 = 599.57 So, the eastband plane is traveling at about 550 mph and the northband plane is traveling at about 600 mph.

Section 2.5 110. Multiply the two solutions and the radicals disappear.  −b + b2 − 4 ac  −b − b 2 − 4 ac    P =    2a 2a   

Solving Other Types of Equations Algebraically

175

112. (a) Because the equation 0 = ( x − 1) + 2 has no real 2

solutions, its graph has no x-intercepts. So, the equation matches graph (ii).

( −b ) − ( b2 − 4ac ) 2

(b) Because the equation 0 = ( x + 1) − 2 has two real

4a 2 4 ac c = 2 = 4a a

solutions, its graph has two x-intercepts. So, the equation matches graph (i).

111. (a) 3 x + 5 x − 11 = 0 : Quadratic Formula; The equation does not factor and it is not easily solved using completing the square, nor is it the type to extract square roots. (b) x 2 + 10 x = 3 : Complete the square; The equation is easily solved by completing the square since the leading coefficient is 1, and the linear term is even. (c) x 2 − 16 + 64 = 0 : Factoring; The equation is easily factored, since it is a perfect square trinomial. (d) x 2 − 15 = 0 : Extracting square roots; The equation is

of the type x 2 − d = 0, and is easily solved by extracting square roots.

(

)

(

113. x 5 − 27 x 2 = x 2 x 3 − 27 = x 2 ( x − 3 ) x 2 + 3 x + 9

(

114. x 3 − 5 x 2 − 14 x = x x 2 − 5 x − 14

)

= x ( x − 7 )( x + 2 ) 115. x 3 + 5 x 2 − 2 x − 10 = x 2 ( x + 5 ) − 2 ( x + 5 )

(

)

= x 2 − 2 ( x + 5)

(

= x+ 2

)( x − 2 ) ( x + 5)

116. 5 ( x + 5 ) x1 3 + 4 x 4 3 = x1 3 5 ( x + 5 ) + 4 x 

= x1 3 ( 9 x + 25 ) 117. Answers will vary. (Make a Decision)

Section 2.5 Solving Other Types of Equations Algebraically 1. polynomial 2.

7. 7 x3 + 63x = 0 7 x ( x 2 + 9) = 0

x ( x − 3)

7x = 0  x = 0

3. To eliminate or remove the radical from the equation x + 2 = x, square each side of the equation to produce

x + 9 = 0  x2 = − 9 2

x = ±

the equation x + 2 = x 2 .

x = ± 3i

4. The equation x 4 − 2 x + 4 = 0 is not of quadratic type. 5.

4 x 4 − 16 x 2 = 0

(

)

4 x2 x2 − 4 = 0 4 x 2 ( x − 2 )( x + 2 ) = 0 x = 0, ± 2 6.

8 x 4 − 18 x 2 = 0

(

)

−9

x 3 + 512 = 0

( x + 8)( x 2 − 8 x + 64) = 0 x + 8 = 0  x = −8 2

x − 8 x + 64 = 0 x =

2 x2 4 x2 − 9 = 0 2 x 2 ( 2 x + 3 )( 2 x − 3 ) = 0 x = 0, ± 23

= = =

b 2 − 4ac 2a

−b ±

− ( − 8) ±

(− 8) − 4(1)(64) 2(1) 2

8±

64 − 256 2

8±

−192 2

8 ± 8 3i = 2 = 4 ± 4 3i

176

Chapter 2

Solving Equations and Inequalities

9. 5 x 3 + 30 x 2 + 45 x = 0

(

( 36t − 7)( t + 1) = 0 (6t + 7 )(6t − 7 ) ( t + 1) = 0

5x x + 6 x + 9 = 0

5 x ( x + 3) = 0 2

x + 3 = 0  x = −3

6t + 7 = 0  t = −

9 x 4 − 24 x 3 + 16 x 2 = 0

6t − 7 = 0  t =

(

)

x (3x − 4 ) = 0 2

x =0 

x=0

3x − 4 = 0 

x = 43

4 x 4 − 65 x 2 + 16 = 0

16.

( 4 x − 1)( x − 16 ) = 0 2

2 x + 1 = 0  x = − 12

x3 − 5x2 − x + 5 = 0

2 x − 1 = 0  x = 12

x 2 ( x − 5) − ( x − 5) = 0

x + 4 = 0  x = −4 x−4=0 x =4

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

17. 3 x − 10 = 0

x−5=0 x =5

3 x = 10 9 x = 100

x + 1 = 0  x = −1 x −1 = 0  x = 1 4

x = 100 9

x + 2 x − 8 x − 16 = 0 x ( x + 2) − 8( x + 2) = 0 3

18. 3 x − 6 = 0

3 x =6

( x − 8) ( x + 2 ) = 0 ( x − 2)( x + 2x + 4)( x + 2) = 0 3

x =2

−2 ± 4 − 16 = −1 ± 3i 2 x−2=0 x = 2 x+2=0 x =−2

( x ) = (2) 2

x2 + 2x + 4 = 0  x =

( x − 3)( x − 1) = 0 2

x − 10 − 4 = 0

19.

x − 10 = 4 x − 10 = 16 x = 26

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

x 4 − 5 x 2 − 36 = 0

( x − 9)( x + 4 ) = 0 ( x + 3)( x − 3) ( x + 4 ) = 0 2

x+3=0 x−3= 0 x2 + 4 = 0 

x=−3 x=3 x 2 = − 4 = ±2i

2x + 5 + 3 = 0

20.

2 x + 5 = −3 No solution

x+ 3 =0 x=− 3 x− 3 =0 x= 3

x=4

x4 − 4 x2 + 3 = 0

13.

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

x3 + 5 = 5x2 + x

11.

12.

7 6

7 6 t 2 + 1 = 0  t = ±i

x 2 9 x 2 − 24 x + 16 = 0

14.

5x = 0  x = 0

10.

36t 4 + 29t 2 − 7 = 0

15.

)

21.

6x + 9 = 0 3

6x = − 9

( 6 x ) = (− 9) 3

6 x = − 729 x = −

243 2

Section 2.5 22. 2 3 x − 10 = 0

28.

x = 10

x + 5 = 2x + 3

x = 5

x + 5 = 4 x 2 + 12 x + 9

4 x 2 + 11x + 4 = 0

−11 ± 121 − 64 −11 ± 57 = 8 8 − 11 − 57 = (extraneous) and 8 − 11 ± 57 8

x = 125 23.

2x + 1 + 8 = 0 3

2 x + 1 = −8 2 x + 1 = −512 2 x = −513 x = − 513 2 = −256.5

24.

29.

x + 1 = 3x + 1 x + 1 = 3x + 1 0 = 2x x=0

30.

x + 5 = 2x − 5 x + 5 = 2x − 5 10 = x x = 10

4x − 3 + 2 = 0

( 4 x − 3)

= −2

4 x − 3 = −8 4 x = −5 x = − 45 25.

5 x − 26 + 4 = x 5 x − 26 = x − 4

31.

5 x − 26 = x 2 − 8 x + 16 x 2 − 13 x + 42 = 0

x−7 =0 x =7

x − 5 = 8 x − 31

32. 6 x − 7 x − 3 = 0 6x − 3 = 7 x

x 2 − 10 x + 25 = 8 x − 31

( 6 x − 3) = ( 7 x ) 2

x − 18 x + 56 = 0

( x − 4 )( x − 14 ) = 0

36 x 2 − 36 x + 9 = 49 x 36 x 2 − 85 x + 9 = 0

x − 4 = 0  x = 4, extraneous

( 9 x − 1)( 4 x − 9 ) = 0

x − 14 = 0  x = 14

27.

( 2 x − 1)( x + 5) = 0 ( x = −5 is not possible.) Note: You can see graphically that there is only one solution.

x−6=0 x =6

x − 8 x − 31 = 5

2x + 9 x − 5 = 0

x = 12  x = 14

( x − 6 )( x − 7 ) = 0

26.

177

x + 5 − 2x = 3

( x ) = (5) 3

Solving Other Types of Equations Algebraically

9 x − 1 = 0  x = 19 , extraneous

x + 1 − 3x = 1

4 x − 9 = 0  x = 94

x + 1 = 3x + 1 x + 1 = 9 x2 + 6x + 1 2

0 = 9 x + 5x 0 = x ( 9 x + 5) x=0 9 x + 5 = 0  x = − 95 , extraneous

33.

x − x −5 =1 x =1+ x − 5

( x ) = (1 + x − 5 ) 2

x =1+ 2 x − 5 + x − 5 4=2 x−5 2 = x−5 4 = x−5 9=x

178

Chapter 2

Solving Equations and Inequalities

34. x + x − 20 = 10

( x + 6)

39.

x = 10 − x − 20

( x + 6)  = (1)2 3   x +6 =1

x = 100 − 20 x − 20 + x − 20

x = −5

−80 = −20 x − 20

40.

4 = x − 20 16 = x − 20

( x − 1)

x −1 = 0

3 x −5 = −

x − 1 = 82 3 x −1 = 4 x=5

36 = x

35. 3 x − 5 +

( x − 9)

= 25

( x − 9 )   

= ( 25 )

41.

x −1

(3 x − 5 ) = (− x − 1)

x − 9 = ( 5)

x − 9 = ( −5 )

9 x − 45 = x − 1

x − 9 = 125 x = 134

x − 9 = −125 x = −116

x =

11 , extraneous 2

( x − 7)

 ( x − 7 )2 3   

= ( 9)

42.

No solution

4 x − 3 = 6 x − 17 + 3 16 ( x − 3 ) = ( 6 x − 17 ) + 9 + 6 6 x − 17 10 x − 40 = 6 6 x − 17 5 x − 20 = 3 6 x − 17

43.

x − 7 = 27

x − 7 = −27

x = 34

x = −20

x2 − 5x + 6 = 0

( x − 3)( x − 2 ) = 0

( x − 7 )( 25 x − 79 ) = 0

x−3= 0 x =3

x=7 ( x = 79 is extraneous.) 25

x−2=0 x =2

44.

3 x1 3 + 2 x 2 3 = 5

( x − x − 22 ) = 16 43

x 2 − x − 22 = ±163 4

2 x 2 3 + 3 x1 3 − 5 = 0

( 2 x + 5)( x − 1) = 0 13

x 2 − x − 22 = ±8 x − x − 30 = 0  x = −5, 6 2

 x = − 125 8  x =1

9t 2 3 + 24t 1 3 + 16 = 0

(3t + 4 )( 3t + 4 ) = 0 13

x 2 − 5 x − 2 = ( −2 ) = −8

25 x 2 − 254 x + 553 = 0

38.

x − 7 = ( −3 )

( x − 5x − 2) = −2

25 x 2 − 200 x + 400 = 9 ( 6 x − 17 )

x1 3 = − 25 x1 3 = 1

x − 7 = (3)

36. 4 x − 3 − 6 x − 17 = 13

9( x − 5) = x − 1 8 x = 44

37.

32 23

( x ) = (10 − x − 20 ) 2

3t 1 3 + 4 = 0 3t 1 3 = −4 4 64 t1 3 = −  t = − 3 27

x 2 − x − 14 = 0  x = 45. 3 x ( x − 1)

+ 2 ( x − 1)

1 ± 57 2 32

( x − 1) 3 x + 2 ( x − 1) = 0 12 ( x − 1) ( 5 x − 2 ) = 0 12 ( x − 1) = 0  x − 1 = 0  x = 1 12

5 x − 2 = 0  x = 52 , extraneous

Section 2.5 46.

4 x 2 ( x − 1)

2 x 2 x ( x − 1) 

+ 6 x ( x − 1)

Solving Other Types of Equations Algebraically

51.

43 + 3 ( x − 1)  = 0 

2 x ( x − 1) 2 x + 3 ( x − 1)  = 0 13

2 x ( x − 1)

( 5 x − 3) = 0

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

1 1 − =3 x x +1 1 1 = x ( x + 1)( 3 ) x ( x + 1) − x ( x + 1) x x +1 x + 1 − x = 3 x ( x + 1) 1 = 3x2 + 3x 0 = 3x2 + 3x − 1

2x = 0  x = 0 x −1 = 0  x = 1 5 x − 3 = 0  x = 35 47.

a = 3, b = 3, c = −1 x=

52.

−3 ±

2x − x − 6 = 0

x2 + 2 x − 3 = 0

( 2 x + 3)( x − 2 ) = 0

( x − 1)( x + 3) = 0 3 2

x−2 =0 x =2

x = 1, − 3 53.

4 5 x − = x 3 6 x 4 5 (6 x ) x − (6 x ) 3 = (6 x ) 6 24 − 10 x = x 2

1 3 1 1 + 5t = 0  t = − 5 1 + 3t = 0  t = −

( x + 12 )( x − 2 ) = 0 x + 12 = 0  x = −12 x−2=0  x =2

20 − x =x x 20 − x = x 2

( 3 x + 1)( 2 x − 1) = 0

0 = ( x + 5 )( x − 4 ) x + 5 = 0  x = −5 x−4=0 x =4 4x + 1 =

1 1 − =0 x x2 6 x2 − x − 1 = 0

6−

54.

1 1 x=− , 3 2

0 = x 2 + x − 20

50.

1 8 + + 15 = 0 t2 t 1 + 8t + 15t 2 = 0

(1 + 3t )(1 + 5t ) = 0

x 2 + 10 x − 24 = 0

49.

4 x + 8 − 3x − 3 = x2 + 3x + 2

48.

( 3) − 4 ( 3)( −1) −3 ± 21 = 2 ( 3) 6

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

2 x2 = 6 + x

2x + 3 = 0  x = −

179

3 x

3 ( x ) 4 x + ( x )1 = ( x ) x 4 x2 + x = 3 4 x2 + x − 3 = 0

( 4 x − 3)( x + 1) = 0 3 4 x + 1 = 0  x = −1

55.

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

x −x−4=0 x=

1 ± 1 − 4 ( −4 )

2 1 17 x= ± 2 2

4x − 3 = 0  x =

180

56.

Chapter 2

Solving Equations and Inequalities

1 x + =3 x −4 x +2 1 x ( x + 2)( x − 2) x2 − 4 + ( x + 2)( x − 2) x + 2 = 3( x + 2)( x − 2) x + x − 2 = 3x2 − 12 2

3x2 − 2x − 10 = 0

 t   t  58. 8   − 2 −3= 0 − t 1    t −1  t Let = x. Then: t −1 8x2 − 2 x − 3 = 0

( 2 x + 1)( 4 x − 3) = 0

a = 3, b = −2, c = −10 x= =

− ( −2 ) ±

1 3 x=− , x= 2 4

( −2 ) − 4 ( 3)( −10 ) 2 ( 3) 2

t 1 1 = −  2 t = −t + 1  t = t −1 2 3 t 3 =  4t = 3t − 3  t = −3 t −1 4

2 ± 124 2 ± 2 31 1 ± 31 = = 6 6 3 2

 s   s  57. 6   + 5 −6 = 0  s +1  s +1

59.

2 x − 5 = 11 2 x − 5 = 11 or 2 x − 5 = −11

Let u = s ( s + 1) .

6u2 + 5u − 6 = 0

( 3u − 2 )( 2u + 3) = 0

2 s = s +1 3 3 s =− 2 s +1 61.

2x = − 6

x = 8

x = −3

60. 3 x + 2 = 7

2 3u − 2 = 0  u = 3 2u + 3 = 0  u = −

2 x = 16

3x + 2 = 7

or 3 x + 2 = − 7

3x = 5

3x = − 9

5 x = 3

x = −3

3 2

 s=2  s=−

3 5

x = x 2 + x − 24 x = x 2 + x − 24

− x = x 2 + x − 24

0 = x 2 − 24

0 = x 2 + 2 x − 24

0 = ( x + 6)( x − 4)

x = 24 x = ±

x = ±2 6

x + 6 = 0  x = −6 x − 4 = 0  x = 4, extraneous

x = − 2 6, extraneous x = 2 6 62.

x 2 + 6 x = 3x + 18 x 2 + 6 x = 3x + 18

x 2 + 6 x = − (3 x + 18)

x 2 + 3 x − 18 = 0

( x + 6)( x − 3) = 0 x + 6 = 0  x = −6 x−3 = 0  x = 3

x 2 + 6 x = − 3 x − 18 2

x + 9 x + 18 = 0

( x + 3)( x + 6) = 0 x + 3 = 0  x = −3 x + 6 = 0  x = −6

Section 2.5 63.

181

x + 1 = x2 − 5 x +1 = x2 − 5

− ( x + 1) = x2 − 5 x + x−4=0

x2 − x − 6 = 0

( x − 3)( x + 2) = 0

x =3

−1 ± 1 − 4 ( −4)

1 x=− − 2 1 x=− + 2

x = −2, is extraneous

64.

Solving Other Types of Equations Algebraically

2 17 2 17 , extraneous 2

x − 15 = x 2 − 15 x x − 15 = x 2 − 15 x

0 = x 2 − 16 x + 15

x − 5 = − x 2 + 15 x 0 = x 2 − 14 x − 5

0 = ( x − 15)( x + 1) x − 15 = 0  x = 15 x + 1 = 0  x = −1

x = x =

− ( −14) ±

( −14) − 4(1)(− 5) 2

2 14 ±

216 2

x = 7 ± 3 6, extraneous 65.

y = x3 − 2 x2 − 3x

66.

(a)

y = 2 x 4 − 15 x 3 + 18 x 2 (a)

40 −3

−9

−7

(b) x-intercepts: ( −1, 0 ) , ( 0, 0 ) , ( 3, 0 ) (c)

0 = x3 − 2 x2 − 3x

− 200

(b) x-intercepts: ( 0, 0 ) , ( 32 , 0 ) , ( 6, 0 ) (c)

0 = 2 x 4 − 15 x 3 + 18 x 2

(

0 = x ( x + 1)( x − 3)

= x 2 2 x 2 − 15 x + 18

x=0 x + 1 = 0  x = −1 x −3= 0 x =3

= x ( 2 x − 3 )( x − 6 )

(d) The x-intercepts are the same as the solutions.

)

0 = x2  x = 0 0 = 2 x − 3  x = 32 0 = x −6 x =6

x-intercepts: ( 0, 0 ) , ( 32 , 0 ) , ( 6, 0 ) (d) The x-intercepts are the same as the solutions.

182

Chapter 2

67.

y = x 4 − 10 x 2 + 9 (a)

Solving Equations and Inequalities 69.

y = 11x − 30 − x (a)

−5

−1

−5

− 20

(b) x-intercepts: ( ±1, 0 ) , ( ±3, 0 ) (c)

0 = x 4 − 10 x 2 + 9

(

)(

0 = x −1 x − 9

(b) x-intercepts: ( 5, 0 ) , ( 6, 0 )

)

x = 11x − 30 x 2 = 11x − 30 x 2 − 11x + 30 = 0

x + 1 = 0  x = −1

( x − 5)( x − 6 ) = 0

x −1 = 0  x = 1

x−5=0 x =5

x + 3 = 0  x = −3

x −6 = 0 x =6

x −3= 0 x = 3 (d) The x-intercepts are the same as the solutions.

68.

0 = 11x − 30 − x

(c)

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

(d) The x-intercepts and the solutions are the same.

y = x − 29 x + 100 (a)

70.

y = 2 x − 15 − 4 x

150

(a) −10

−4

−150

(b) x-intercepts: ( ±2, 0 ) , ( ±5, 0 ) (c)

0 = x 4 − 29 x 2 + 100

(

)(

= x − 4 x − 25

−5

(b) (c)

)

= ( x + 2 )( x − 2 )( x + 5 )( x − 5 ) 0 = x + 2  x = −2 0 = x−2 x =2 0 = x + 5  x = −5 0 = x−5 x =5

x-intercepts: ( −2, 0 ) , ( 2, 0 ) , ( −5, 0 ) , ( 5, 0 ) (d) The x-intercepts are the same as the solutions.

x-intercept: ( 23 , 0 ) 0 = 2 x − 15 − 4 x 15 − 4 x = 2 x 15 − 4 x = 4 x 2 0 = 4 x 2 + 4 x − 15 0 = ( 2 x + 5 )( 2 x − 3 )

0 = 2 x + 5  x = − 25 0 = 2 x − 3  x = 32

x = − 25 is extraneous. The x-intercept is ( 23 , 0 ) .

(d) The x-intercept and the solution are the same.

Section 2.5 71.

y = 3x − 3 x − 4 (a)

Solving Other Types of Equations Algebraically

73.

1 4 − −1 x x −1

(a)

−6

183

−4

−6

−24

(b) x-intercept: ( 3.09164, 0 ) (c)

(b) x-intercept: ( −1, 0 )

3 x − 3 x − 4 = 0. Let y = x .

(c)

3y 2 − 3 y − 4 = 0 y= y=

− ( −3 ) ±

( −3) − 4 ( 3)( −4 ) 2 ( 3) 2

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

3 ± 57 6

Because y = x =

1 4 − −1 x x −1 0 = ( x − 1) − 4 x − x ( x − 1) 0=

3 ± 57 , then 6

(3 + 57 ) ≈ 3.09164 x=

0 = x2 + 2 x + 1 x + 1 = 0  x = −1 (d) The x-intercept and the solution are the same.

(3 − 57 ) ≈ 0.57503 extraneous x= 2

74. y = x − 5 +

(a)

6 (d) The x-intercept and the solution are the same. 72.

−4

0.5

(b) x-intercepts: ( − 2, 0), ( 4, 0)

−3

(c)

− 0.5

0 = 7 x + 36 − 5 x + 16 − 2 7 x + 36 = 2 + 5 x + 16

( 7 x + 36 ) = ( 2 + 5x + 16 ) 2

7 = 0 x +3 ( x − 5)( x + 3) + 7 = 0 x −5+

x 2 − 2 x − 15 + 7 = 0

(b) x-intercepts: ( 0, 0 ) , ( 4, 0 ) (c)

−6

y = 7 x + 36 − 5 x + 16 − 2 (a)

7 x + 3

7 x + 36 = 4 + 4 5 x + 16 + 5 x + 16 7 x + 36 = 5 x + 20 + 4 5 x + 16

x2 − 2x − 8 = 0

( x − 4)( x + 2) = 0 x − 4 = 0  x = 4 x + 2 = 0  x = −2 x-intercepts: ( − 2, 0), ( 4, 0) (d) The x-intercepts and the solutions are the same.

2 x + 16 = 4 5 x + 16 x + 8 = 2 5 x + 16 x + 16 x + 64 = 4 ( 5 x + 16 ) 2

x 2 + 16 x + 64 = 20 x + 64 x2 − 4 x = 0 x ( x − 4) = 0 x=0 x−4=0 x =4 (d) The x-intercepts and the solutions are the same.

184

Chapter 2

75.

y=x+

Solving Equations and Inequalities

9 −5 x +1

(a)

77.

y = x +1 − 2 (a)

8 −10

−10

−4 −18

(b)

x -intercept: ( 2, 0 )

(c)

0=x+

(b) x-intercepts: (1, 0 ) , ( −3, 0 ) (c)

9 −5 x +1

2 = x +1 − 2 x + 1 = 2 or − ( x + 1) = 2

9 − 5 ( x + 1) x +1 0 = x2 + x + 9 − 5x − 5 0 = x ( x + 1) + ( x + 1)

x =1

− x −1 = 2 −x=3 x = −3 (d) The x-intercepts and the solutions are the same.

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

78.

x-intercept: ( 2, 0 )

(d) The x-intercept and the solution are the same. 76. y = 2 x +

(a)

0 = x +1 − 2

y = x −2 −3 7

(a)

8 − 2 x −5

−7

−5

−4

(b) x-intercepts: ( 5, 0 ) , ( −1, 0 ) (c) −7

0 = x −2 −3 3= x −2 First equation: x − 2 = 3  x = 5

(b) x-intercept: (3, 0) 8 − 2 = 0 x −5 2 x( x − 5) + 8 − 2( x − 5) = 0 2x +

2 x 2 − 10 x + 8 − 2 x + 10 = 0

Second equation: − ( x − 2 ) = 3 − ( x − 2) = 3

− x + 2 = 3  x = −1 (d) The x-intercepts and the solutions are the same.

2 x 2 − 12 x + 18 = 0 x2 − 6x + 9 = 0

( x − 3)2 = 0 x = 3

x-intercept: (3, 0) (c) The x-intercept and the solution are the same.

Section 2.5

79. Let x = the number of students. Then

Solving Other Types of Equations Algebraically

185

1700 1700 is the original cost per student. So, − 7.50 is the reduced cost per student x x

after 6 more students join the trip. Verbal model: Cost per student × Number of students = Total cost  1700  − 7.50 ( x + 6) = 1700   x  10,200 − 7.5 x − 45 = 1700 1700 + x 10,200 − 7.5 x − 45 + = 0 x 7.5 x 2 + 45 x − 10,200 = 0 x 2 + 6 x − 1360 = 0 x = x =

− ( 6) ±

(6) − 4(1)(−1360) 2(1)

−6 ±

36 + 5440 2

−6 ±

5476 2 − 6 − 74 x = , extraneous 2 − 6 + 74 = 34 students in the original group x = 2 x =

80. Let x = monthly rent for the apartment. The original rent for each student is x 3. By adding a fourth student, the rent is x 4. So, x x = + 75 3 4 4 x = 3 x + 900

x = 900. The monthly rent is $900.

r  8055.19 = 30001 +  2  r  1 +  2  1+

r  81. A = P1 +  , n = 12, t = 10 n  r   11,752.45 = 75001 +  12  

r  82. A = P1 +  , n = 2, t = 20 n 

(12)(10)

( 2)( 20)

= 2.685063 r = 1.025000 2 r = 0.025000 2 r ≈ 0.050, or 5.0%

120

r   1 +  12   1+

= 1.566993

r = 1.00375 12 r = 0.00375 12 r ≈ 0.045, or 4.5%

186

Chapter 2

Solving Equations and Inequalities

83. T = 75.82 − 2.11x + 43.51 x , 5 ≤ x ≤ 40 Find x when T = 212°F.

212 = 75.82 − 2.11x + 43.51 x 2.11x − 43.51 x + 136.18 = 0 Let y = x . 2.11y 2 − 43.51y + 136.18 = 0

y= =

− ( −43.51) ±

( −43.51) − 4 ( 2.11)(136.18) 2 ( 2.11) 2

43.51 ± 743.7609 4.22

y= x =

43.51 ± 743.7609 4.22 2

 43.51 ± 743.7609  x =   4.22   x ≈ 14.806 pounds per square inch

( x ≈ 281.332 is not in the given domain, 5 ≤ x ≤ 40.) 84.

0.2 x + 1 = C = 2.5 0.2 x + 1 = 6.25 0.2 x = 5.25 x = 26.25,or 26,250 passengers

85. The hypotenuse of the right triangle is 9 16 x 2 + 9 = . 16 4 Total cost = Cost of powerline over land + Cost of powerline under water

x 2 + ( 34 ) = x 2 + 2

(a)

24 dollars feet ⋅ 5280 ⋅ ( 8 − x ) miles foot mile +

30 dollars feet 16 x 2 + 9 ⋅ 5280 ⋅ miles foot mile 4

C = 1,013,760 − 126,720 x + 39,600 16 x 2 + 9

(b) Find C when x = 3 (miles). C = 1,013,760 − 126,720 ( 3 ) + 39,600 16 ( 3 ) + 9 2

= $1,123,424.95

(c) Find x, when C = 1,098,662.40. 1,013,760 − 126,720 x + 39,600 16 x 2 + 9 = 1,098,662.40 39,600 16 x 2 + 9 − 126,720 x − 84,902.4 = 0 39,600 16 x 2 + 9 = 126,720 + 84,902.4 16 x 2 + 9 = 3.2 x + 2.144 16 x 2 + 9 = 10.24 x 2 + 13.7216 x + 4.596736 5.76 x 2 − 13.7216 x + 4.403264 = 0 Use the Quadratic Formula. x ≈ 1.997  2 miles x ≈ 0.385  0.382 mile

Section 2.5 (d)

Solving Other Types of Equations Algebraically

187

1,600,000

(e) Using the graph from part (d), if x ≈ 1 mile, the cost is minimized, C ≈ $1,085,340. 86. d = 1002 + h2

(a)

300

240

If d = 200, h ≈ 173. (b) h

160

165

170

175

180

185

188.68

192.94

197.23

201.56

205.91

210.3

When d = 200, 170 < h < 175. (c)

1002 + h2 = 200 1002 + h2 = 2002 h2 = 2002 − 1002

h ≈ 173.205 (d) Solving graphically or numerically yields an approximate solution. An exact solution is obtained algebraically. 87. (a)

Year

2008

2009

2010

2011

2012

Crimes (in millions)

11.17

10.73

10.41

10.23

10.20

(b) According to the table, in 2008, the number of crimes committed fell below 11 million. (c) C =

1.49145t 2 − 35.034t + 309.6, 8 ≤ t ≤ 12

Find t when C = 11. 11 =

1.49145t 2 − 35.034t + 309.6

(11)2 = ( 1.49145t 2 − 35.034t + 309.6 )

121 = 1.49145t 2 − 35.034t + 309.6 0 = 1.49145t 2 − 35.034t + 188.6 t = t =

− ( − 35.034) ± 35.034 ±

( − 35.034)2 − 4(1.49145)(188.6) 2(1.49145)

1227.381156 − 1125.14988 2.9829

35.034 ± 102.231276 2.9829 t ≈ 8.36, 15.13

t =

Because t ≈ 15.13 is not in the domain of the model, t ≈ 8.36. So, in 2008, the number of crimes committed fell below 11 million. © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

188

Chapter 2 (d)

Solving Equations and Inequalities

In 2008, the number of crimes fell below 11 million. 88. (a) and (b) 4000

0 2000

The model is a good fit for the data. (c) Find t when S = 4. S = 110.386t −

16.4988 + 2167.7, 1 ≤ t ≤ 12 t

16.4988 + 2167.7 t 16.4988 − 1832.3 0 = 110.386t − t 0 = 110.386t 2 − 1832.3t − 16.4988

4000 = 110.386t −

t =

−( −1832.3) ±

(−1832.3) − 4(110.386)(−16.4988) 2(110.386) 2

3,357,323.29 + 7284.9461472 220.772 1832.3 ± 3,364,608.23615 t = 220.772 t ≈ 16.61, − 0.01 t =

1832.3 ±

Since t ≈ − 0.01 is not in the domain of the model, t ≈ 16.61. In 2016, the average MLB player salary will exceed $4 million. 89. False. An equation can have any number of extraneous solutions. For example, see Example 7. 90. False. Consider x = 0  x = 0. 91. (a) The distance between (1, 2 ) and ( x, − 10 ) is 13.

(b) The distance between ( −8, 0 ) and ( x, 5) is 13.

( x − 1) + ( −10 − 2 ) = 13 2 2 ( x − 1) + ( −12 ) = 132

( x + 8 ) + ( 5 − 0 ) = 13 2 ( x + 8 ) + 52 = 132

x 2 − 2 x + 1 + 144 = 169

x 2 + 16 x + 64 + 25 = 169

x 2 − 2 x − 24 = 0

x 2 + 16 x − 80 = 0

( x + 4 )( x − 6 ) = 0

( x + 20 )( x − 4 ) = 0

x + 4 = 0  x = −4

x + 20 = 0  x = −20 x−4=0  x =4

x −6 = 0 x =6

Section 2.6

Solving Inequalities Algebraically and Graphically

x + x − a = b, x = 20

92.

20 + 20 − a = b One solution is a = 19 and b = 21. Another solution is a = b = 20. 93. Dividing each side of the equation by x loses the solution x = 0. The correct method is to first factor:

189

94. (a) The expression represents the volume of water in the tank.

(b) Solve the equation for x and substitute that value into the expression x 3 to find the volume of the cube. 95.

8 3 16 9 25 + = + = 3x 2 x 6 x 6 x 6 x

x 3 − 25 x = 0

(

)

x x 2 − 25 = 0 x=0 x 2 = 25 x = ±5 96.

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

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

97.

x−4

( x − 2 )( x + 2 )( x − 1)

2 2 2 2  − 3 −  = −3+ z+2  z  z+2 z = =

98. 25 y 2 ÷

2z − 3( z + 2 ) z + 2 ( z + 2 )

y = 125   , y ≠ 0 x

z ( z + 2)

2 z − 3z 2 − 6 z + 2 z + 4 z ( z + 2)

−3z 2 − 2 z + 4 = z ( z + 2)

 5  xy = 25 y 2   5  xy 

99. x 2 − 22 x + 121 = 0

( x − 11) = 0 2

x = 11 100. x ( x − 20 ) + 3 ( x − 20 ) = 0

( x + 3)( x − 20 ) = 0 x = −3, 20

Section 2.6 Solving Inequalities Algebraically and Graphically 1. double

8. x ≤ 2

2. −a ≤ x ≤ a

Matches (a).

3. x ≤ −a, x ≥ a

9. − 2 < x < 2

4. zeros, undefined values 5. The inequalities x − 4 < 5 and x > 9 are not equivalent. The first simplifies to x − 4 < 5  x < 9, which is not equivalent to the second, x > 9. 6. The Transitive Property of Inequalities is as follows: a < b and b < c  a < c. 7. x < 2 Matches (d).

Matches (f). 10. − 2 < x ≤ 2

Matches (b). 11. − 2 ≤ x < 2

Matches (e). 12. − 2 ≤ x ≤ 2

Matches (c).

190

Chapter 2

Solving Equations and Inequalities

13. 5 x − 12 > 0 (a) x = 3

15. −1 <

(a)

5 ( 3 ) − 12 > 0

? 3 − ( −1) ?

−1 <

≤1 2 −1 < 2 ≤ 1 No, x = −1 is not a solution.

3>0 Yes, x = 3 is a solution. (b) x = −3 ?

5 ( −3 ) − 12 > 0

(b)

−1 <

≤1 2 3− 5 −1 < ≤1 2 3− 5 Note: ≈ 0.382 2

5 ( 25 ) − 12 > 0 >0

Yes, x = 25 is a solution. (d)

Yes, x = 5 is a solution. (c) x = 1

3 2

? 3 −1 ?

5 ( 23 ) − 12 > 0

−1 <

≤1 2 −1 < 1 ≤ 1 Yes, x = 1 is a solution.

− 92 > 0 No, x = 32 is not a solution.

(d)

14. − 5 < 2 x − 1 ≤ 1

3−5 ? ≤1 2 −1 < − 1 ≤ 1 No, x = 5 is not a solution. ?

− 5 < 2( 2) − 1 ≤ 1 − 5 < 3 ≤1

16.

No, x = 2 is not a solution.

x − 10 ≥ 3 (a)

(b) x = − 2

3 ≥3

−5 < − 5 ≤ 1 No, x = − 2 is not a solution.

(b)

Yes, x = 13 is a solution. x = −1 ?

−1 − 10 ≥ 3

x = 0

−11 ≥ 3

− 5 < 2(0) − 1 ≤ 1 − 5 < −1 ≤ 1

Yes, x = 0 is a solution. (d) x = −

x = 13 13 − 10 ≥ 3

− 5 < 2( − 2) − 1 ≤ 1

x=5 −1 <

(a) x = 2

(c)

x= 5 ? 3− 5 ?

−27 > 0 No, x = −3 is not a solution. (c) x = 25

1 2

3− x ≤1 2 x = −1

1 2

? ?  1 − 5 < 2 −  − 1 ≤ 1  2 −5 < − 2 ≤ 1

1 Yes, x = − is a solution. 2

(c)

Yes, x = −1 is a solution. x = 14 ?

14 − 10 ≥ 3 4 ≥3

Yes, x = 14 is a solution. (d) x = 8 ?

8 − 10 ≥ 3 −2 ≥ 3

No, x = 8 is not a solution.

Section 2.6 17. 6 x > 42

Solving Inequalities Algebraically and Graphically 25.

−1 ≤ x ≤ 3

x 1

1 ≤ 2x + 3 ≤ 9 − 2 ≤ 2x ≤ 6

x > 7 0

191

x −2

18. −10 x ≤ 40

26.

x ≥ −4 −4

−3

−2

−1

13 3

−2

−1

≥ x > − 83

13 3

−3 −2 −1

27.

−8 ≤ 1 − 3 ( x − 2 ) < 13 −8 ≤ −3 x + 7 < 13 −15 ≤ −3 x < 6 5 ≥ x > −2  −2 < x ≤ 5

1 2

x −2 −1

x 1

28.

21. 2(1 − x) < 3 x + 7

0 ≤ 2 − 3 ( x + 1) < 20 0 ≤ −3 x − 1 < 20 1 ≤ −3 x < 21

2 − 2 x < 3x + 7 − 5x < 5

− 13 ≥ x > −7  −7 < x ≤ − 13

x > −1

−1 3

x −2

−1

−8 −7 −6 −5 −4 −3 −2 −1

22. 2 x + 7 < 3 ( x − 4 )

29.

2 x + 7 < 3 x − 12 19 < x  x > 19

23.

3 4

1 ≤ 14 x 4≤x

−

9 15 <x< 2 2

−

9 2

−6 −4 −2

x≥4 2

24.

2x − 3 <4 3 −12 < 2 x − 3 < 12 −4 <

x −6 ≤ x −7

x 3

3 + 27 x > x − 2 21 + 2 x > 7 x − 14 35 > 5 x 7>x x<7

x 0

−9 < 2 x < 15

x 18

−8 ≤ 1 − 3 x + 6 < 13

x ≥ 12

20. 3 x + 1 ≥ 2 + x 2x ≥ 1

−1

−8

x −3

− 83 < x ≤ 133

19. 4 x + 7 < 3 + 2 x 2 x < −4 x < −2 −4

−8 ≤ −3 x + 5 < 13 −13 ≤ −3 x < 8

x −5

−1

30.

15 2 0

x+3 <5 2 0 ≤ x + 3 < 10 −3 ≤ x < 7 0≤

x −4 −3 −2 −1 0 1 2 3 4 5 6 7

x 5

192

Chapter 2

Solving Equations and Inequalities

31. 5 − 2 x ≥ 1

y≤0

(b)

2x − 3 ≤ 0

−2 x ≥ −4 x≤2

2x ≤ 3 x ≤ 23

36. 9

−9

−5

−6

−2

32. 20 < 6 x − 1

Using the graph, (a) y ≤ 5 for x ≤ 6, and (b) y ≥ 0 for

x > 27

x ≥ − 32 .

Algebraically: y≤5

(a)

−9

2 3

x +1≤ 5 2 3

−6

33. 3 ( x + 1) < x + 7

x≤6 y≥0

(b)

3x + 3 < x + 7

x≤4

2 3

2x < 4

x +1≥ 0 2 3

x<2

x ≥ −1 x ≥ − 32

37.

−9

−2

−6 −2

34. 4 ( x − 3) > 8 − x

Using the graph, (a) −1 ≤ y ≤ 3 for 53 ≤ x ≤ 3, and (b)

x>4 6

−9

−6

35. −4

y ≤ 0 for x ≥ 83 . Algebraically: (a) −1 ≤ y ≤ 3 −1 ≤ − 3 x + 8 ≤ 3 − 9 ≤ −3 x ≤ − 5 3 ≥ x ≥ 53  53 ≤ x ≤ 3 (b) y≤0 −3 x + 8 ≤ 0 8 ≤ 3x 8 ≤ x  x ≥ 83 3

−5

Using the graph, (a) y ≥ 1 for x ≥ 2 and (b) y ≤ 0 for x ≤ 23 . Algebraically: (a)

y ≥1 2x − 3 ≥ 1 2x ≥ 4 x≥2

Section 2.6 38.

Solving Inequalities Algebraically and Graphically 44.

193

x + 14 + 3 ≥ 17 x + 14 ≥ 14

−6

x + 14 ≤ −14 or x + 14 ≥ 14 x ≤ −28 or x≥0

6 −2

Using the graph, (a) 0 ≤ y ≤ 3 for −2 ≤ x ≤ 4 and (b) y ≥ 0 for x ≤ 4. Algebraically: (a) 0≤ y≤3

− 35 − 28 − 21 − 14 − 7

1 − x < 12 − 12 < 1 − x < 12

−2 ≤ − 12 x ≤ 1

− 32 < − x < − 12

4 ≥ x ≥ −2  −2 ≤ x ≤ 4 y≥0 − 12 x + 2 ≥ 0

3 2

> x > 12

1 2

< x < 23 1 2

2 ≥ 12 x 4≥xx≤4

−1

x 2

4 − 5x < 3 −3 < 4 − 5 x < 3 −7 < −5 x < −1

−3 −2 −1

x ≤1 2

7 5

> x > 51

1 5

< x < 75

1 5

x ≤2 x −4 −3 −2 −1

7 5

−2 ≤ x ≤ 2

41.

3 2

46. 3 4 − 5 x < 9

5 x > 10 5 x < −10 or 5 x > 10 x < −2 or x > 2

40.

45. 10 1 − x < 5

0 ≤ − 12 x + 2 ≤ 3

(b)

39.

47.

x 2

y = x −3 7

x−7 ≤6 −6 ≤ x − 7 ≤ 6 1 ≤ x ≤ 13 −3

1 0

42.

13 2

10 12 14

x − 20 > 4 x − 20 > 4 or x − 20 < − 4 x > 24 or x < 16

Graphically, (a) y ≤ 2 for 1 ≤ x ≤ 5 and (b) y ≥ 4 for x ≤ −1 or x ≥ 7. Algebraically: (a) y≤2 x−3 ≤ 2 −2 ≤ x − 3 ≤ 2

x 14

43.

(b)

x −3 ≥5 2 x − 3 ≥ 10 or x − 3 ≤ −10 x ≥ 13 or x ≤ −7 −7

13 0

1≤ x ≤ 5 y≥4 x −3 ≥ 4

x − 3 ≤ −4 or x ≤ −1 or

x − 3 ≥ 10

−9 −6 −3

9 −1

x −3≥ 4 x≥7

9 12 15

194

Chapter 2

48.

y = 12 x + 1

Solving Equations and Inequalities 55. All real numbers more than 3 units from −1 x +1 > 3

56. All real numbers at least 5 units from 3 x −3 ≥ 5 − 14

57.

−4

( x − 5)( x + 1) > 0

Graphically, (a) y ≤ 4 for −10 ≤ x ≤ 6 and (b) y ≥ 1 for x ≤ −4 or x ≥ 0. Algebraically: (a) y≤4 1 2

Testing the intervals ( −∞, − 1) , ( −1, 5) and ( 5, ∞ ) , we have x 2 − 4 x − 5 > 0 on ( −∞, − 1) and ( 5, ∞ ) .

Similarly, x 2 − 4 x − 5 < 0 on ( −1, 5 ) . 58.

−5 ≤ 12 x ≤ 3

x + 1 ≤ −1 or 1 x ≤ −2 or 2 x ≤ −4 or

x2 − 3x − 4 > 0

( x − 4 )( x + 1) > 0

−10 ≤ x ≤ 6 y ≥1 1 + ≥1 x 1 2 1 2

Key numbers: −1, 5

x +1 ≤ 4 −4 ≤ 12 x ≤ 4

(b)

x2 − 4 x − 5 > 0

Key numbers: −1, 4 Testing the intervals ( −∞, − 1) , ( −1, 4 ) , and ( 4, ∞ ) , we

1 2

x +1≥1 1 x≥0 2 x≥0

49. The midpoint of the interval  −3, 3 is 0. The interval

represents all real numbers x no more than three units from 0. x−0 ≤3 x ≤3 50. The midpoint of the interval ( − 1, 1) is 0. The two

intervals represent all real numbers x more than one unit from 0. x − 0 >1

x >1 51. The midpoint of the interval ( − 5, 3) is −1. The two

have x 2 − 3 x − 4 > 0 on ( −∞, − 1) and ( 4, ∞ ) . Similarly, x 2 − 3 x − 4 < 0 on ( −1, 4 ) .

59. 2 x 2 − 4 x − 3 > 0

4 ± 16 + 24 10 =1± 4 2 Testing the intervals  10   10 10  , 1+  −∞, 1 −  ,  1 −  , and 2 2 2    

Key numbers: x =

  10 , ∞  , you have 2 x 2 − 4 x − 3 > 0 on  1 +  2    10 10  , 1+  1 −  . Similarly, 2 x 2 − 4 x − 3 < 0 2 2     10   10 , ∞ . on  −∞, 1 −  ∪  1 +   2 2    

intervals represent all real numbers x at least four units from −1. x − ( −1) ≥ 4 x +1 ≥ 4 52. The midpoint of the interval ( 4, 10) is 7. The interval

represents all real numbers x less than three units from 7. x − 7 < 3

53. All real numbers less than 10 units from 6 x − 6 < 10 54. All real numbers no more than 8 units from −5

x+5 ≤8

Section 2.6

60.

Solving Inequalities Algebraically and Graphically

− 2x2 + x + 5 = 0

x2 − 6 x − 7 < 0

2x − x − 5 = 0 Key numbers: x =

1 ± 1 − 4 ( 2 )( −5 ) 2 (2)

( x + 1)( x − 7 ) < 0

1 ± 41 4

Key numbers: x = −1, x = 7

Test intervals:  1 41   1 41 1 41  , +  −∞, −  ,  −  , and 4 4 4 4 4 4     1  41 , ∞  +  4 4  

Test intervals: ( −∞, − 1)  ( x + 1)( x − 7 ) > 0

( −1, 7 )  ( x + 1)( x − 7 ) < 0 ( 7, ∞ )  ( x + 1)( x − 7 ) > 0 Test: Is ( x + 1)( x − 7 ) < 0? Solution set: ( −1, 7 )

 1 41  2 x 2 − x − 5 > 0 on  −∞, −  and  4 4  

 41 , ∞  , and 2 x 2 − x − 5 < 0 on  4  41 1 41  , + . 4 4 4 

−1 −2

7 0

x 8

( x + 2 ) < 25 2

65.

x 2 + 4 x + 4 < 25 x 2 + 4 x − 21 < 0

( x + 7)( x − 3) < 0

61. There are no key numbers because − x 2 + 6 x − 10 ≠ 0. The

only test interval is ( −∞, ∞ ) . − x + 6 x − 10 < 0 for all x. 2

Key numbers: x = −7, x = 3 Test intervals: ( −∞, − 7 ) , ( −7, 3) , ( 3, ∞ )

62. There are no key numbers because 3x 2 + 8 x + 6 ≠ 0.

The only test interval is ( − ∞, ∞ ). 3x 2 + 8 x + 6 > 0 for

Test: Is ( x + 7 )( x − 3) < 0?

all real numbers x. 63.

x 2 − 6 x + 9 < 16

64.

1  + 4 1  − 4

195

Solution set: ( −7, 3 )

x + 4x + 4 ≥ 9

−8 −6 −4 −2

( x + 5)( x − 1) ≥ 0 Key numbers: x = −5, x = 1

66.

Test: Is ( x + 5)( x − 1) ≥ 0? Solution set: ( −∞, − 5 ∪ 1, ∞ ) x 0

x 4

( x − 3) ≥ 1 2

x − 3 ≥ 1 or x≥4

Test intervals: ( −∞, − 5) , ( −5, 1) , (1, ∞ )

−6 −5 −4 −3 −2 −1

−7

x2 + 4 x − 5 ≥ 0

x − 3 ≤ −1 x≤2 x

67.

x3 − 4 x 2 ≥ 0 x 2 ( x − 4) ≥ 0 Key numbers: x = 0, x = 4 Test intervals: ( − ∞, 0), (0, 4), ( 4, ∞ ) Test: Is x 2 ( x − 4) ≥ 0? Solution set: x = 0, x = 4 and

(4, ∞)  x = 0 ∪ [4, ∞) x −2 −1

196

Chapter 2

Solving Equations and Inequalities

68.

x5 − 3x 4 ≤ 0 x 4 ( x − 3) ≤ 0

2 x 3 + 13 x 2 − 8 x − 52 ≥ 0 x 2 ( 2 x + 13 ) − 4 ( 2 x + 13 ) ≥ 0

Key numbers: x = 0, x = 3

( 2 x + 13) ( x 2 − 4 ) ≥ 0 ( 2 x + 13)( x + 2 )( x − 2 ) ≥ 0

Test intervals: ( − ∞, 0), (0, 3), (3, ∞ ) Test: Is x 4 ( x − 3) ≤ 0?

Key numbers: x = − 132 , x = −2, x = 2

Solution set: ( − ∞, 0), x = 0, (0, 3), and

Test intervals: ( −∞, − 132 ) , ( − 132 , −2 ) , ( −2, 2 ) , ( 2, ∞ )

x = 3  ( − ∞, 3]

Test: Is ( 2 x + 13)( x + 2 )( x − 2 ) ≥ 0?

x 1

Solution set:  − 132 , − 2  , 2, ∞ )

2 x3 + 5x 2 − 6 x − 9 > 0

69.

− 13 2

( x + 1)( x + 3)( 2 x − 3) > 0

(

)(

Test intervals: ( −∞, 3), ( − 3, −1), −1, 32 , 32 , ∞

)

No solution 74. 4 x 2 + 12 x + 9 ≤ 0

x 2

3 x 2 − 11x + 16 = 0. In fact, 3 x 2 − 11x + 16 > 0 for all x.

Solution set: ( −3, − 1) , ( , ∞ )

Since b2 − 4 ac = −71 < 0, there are no real solutions of

3 2

73. 3 x 2 − 11x + 16 ≤ 0

Test: Is ( x + 1)( x + 3)( 2 x − 3) > 0?

3 2

−8 −6 −4 −2

Key numbers: x = −3, x = −1, x = 32

−4 −3 −2 −1

2 x 3 + 13 x 2 − 8 x − 46 ≥ 6

72.

( 2 x + 3) ≤ 0 2

2x + 3 = 0

2 x 3 + 3 x 2 − 11x − 6 < 0

70.

x = − 23

( x − 2 )( x + 3)( 2 x + 1) < 0

−

Key numbers: x = −3, x = − 12 , x = 2

(

)(

)

−3

Test intervals: ( −∞, − 3), − 3, − 12 , − 12 , 2 , ( 2, ∞ ) Test: Is ( x − 2)( x + 3)( 2 x + 1) < 0?

x 2 ( x − 3) − ( x − 3) > 0

1 1   Solution set:  −∞, ,  , ∞  2  2  

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

1 2

Key numbers: x = 3, x = −1, x = 1 Test intervals: ( −∞, − 1) , ( −1, 1) , (1, 3) , ( 3, ∞ ) Test: Is ( x − 3 )( x + 1)( x − 1) > 0? Solution set: ( −1, 1) , ( 3, ∞ ) x 1

1 2

x − 3x 2 − x + 3 > 0

> 0

Test: Is ( 2 x − 1) > 0?

−1

1 1   Test intervals:  −∞, ,  , ∞  2  2  

x − 3 x − x > −3

71.

Key number: x =

x 0

( 2 x − 1)

−1 −6 −5 −4 −3 −2 −1

x −1

75. 4 x 2 − 4 x + 1 > 0

Solution set: ( − ∞, − 3) , ( − 12 , 2 ) 2

−2

3 2

−3 −2 −1

76.

x 1

x2 + 3x + 8 > 0

Because x 2 + 3 x + 8 ≠ 0, there are no real key numbers. x 2 + 3 x + 8 is always positive.

Solution set: all real numbers x x

−4 −3 −2 −1

Section 2.6 77. (a)

f ( x ) = g ( x ) when x = 1.

(b)

f ( x ) ≥ g ( x ) when x ≥ 1.

(c)

f ( x ) > g ( x ) when x > 1.

78. (a)

f ( x ) = g ( x ) when x = −1 or x = 3.

(b)

f ( x ) ≥ g ( x ) when x ≤ −1 or x ≥ 3.

(c)

f ( x ) > g ( x ) when x < −1 or x > 3.

79.

Solving Inequalities Algebraically and Graphically

y ≥ 36

(b)

x 3 − x 2 − 16 x + 16 ≥ 36 x 3 − x 2 − 16 x − 20 ≥ 0

( x + 2 )( x − 5)( x + 2 ) ≥ 0 y ≥ 36 when x = −2, 5 ≤ x < ∞. 81.

y = −x2 + 2x + 3 6

1 −x>0 x 1 − x2 >0 x Key numbers: x = 0, x = ±1 Test intervals: ( −∞, − 1) , ( −1, 0 ) , ( 0, 1) , (1, ∞ )

−5

Test: Is

Using the graph, (a) when x ≤ −1 or x ≥ 3 and (b) y ≥ 3 when 0 ≤ x ≤ 2. Algebraically: (a) − x 2 + 2 x + 3 ≤ 0

1 − x2 ≥ 0? x

Solution set: ( −∞, − 1) ∪ ( 0, 1)

−2

x −2

82.

x2 − 2x − 3 ≥ 0

( x − 3)( x + 1) ≥ 0 Key numbers: x = −1, x = 3

−1

you obtain x ≤ −1 or x ≥ 3.

x2 − 2x ≤ 0 x ( x − 2) ≤ 0

Key numbers: x = 0, x = 2

1 −4<0 x 1 − 4x <0 x

1 4

Testing the intervals ( −∞, 0 ) , ( 0, 2 ) and ( 2, ∞ ) , −1

you obtain 0 ≤ x ≤ 2. y = x 3 − x 2 − 16 x + 16

83.

− 10

− 24

Using the graph, (a) y ≤ 0 when −∞ < x ≤ −4 and 1 ≤ x ≤ 4, and (b) y ≥ 36 when x = −2 and 5 ≤ x < ∞. Algebraically: (a) y ≤ 0

x+6 −2≤0 x +1 x + 6 − 2 ( x + 1) ≤0 x +1 4− x ≤0 x +1 Key numbers: x = −1, x = 4 Test intervals: ( −∞, − 1) , ( −1, 4 ) , ( 4, ∞ ) 4− x ≤ 0? x +1 Solution set: ( −∞, − 1) ∪ 4, ∞ )

Test: Is

x 3 − x 2 − 16 x + 16 ≤ 0 x ( x − 1) − 16 ( x − 1) ≤ 0 2

1 4

 1 1  Test intervals: ( −∞, 0 ) ,  0,  ,  , ∞  4 4     1 − 4x Test: Is < 0? x 1  Solution set: ( −∞, 0 ) ∪  , ∞  4 

−x2 + 2 x + 3 ≥ 3 − x2 + 2 x ≥ 0

80.

Key numbers: x = 0, x =

Testing the intervals ( −∞, − 1) , ( −1, 3) , and ( 3, ∞ ) , (b)

197

−2 −1

( x − 1) ( x 2 − 16 ) ≤ 0

y ≤ 0 when −∞ < x ≤ −4, 1 ≤ x ≤ 4.

198

84.

Chapter 2

Solving Equations and Inequalities

x + 12 −3≥ 0 x+2 x + 12 − 3 ( x + 2 ) ≥0 x+2 6 − 2x ≥0 x+2

86.

5x x2 + 4 4

−6

Key numbers: x = −2, x = 3

−4

Using the graph, y ≥ 1 when 1 ≤ x ≤ 4 and y ≤ 0 when −∞ < x ≤ 0. Algebraically: (a) y ≥1 5x ≥1 x2 + 4

Test intervals: ( −∞, − 2 ) , ( −2, 3) , ( 3, ∞ ) Test: Is

6 − 2x ≥ 0? x+2

Solution set: ( −2, 3 x

−2

85.

−1

(

5x − x2 + 4

) ≥0

x2 + 4 ( x − 4 )( x − 1)

3x x−2

≤0 x2 + 4 Key numbers: x = 1, x = 4

Test intervals: ( −∞, 1) , (1, 4 ) , ( 4, ∞ )

−6

Test: Is

( x − 4)( x −1) ≤ 0? x2 + 4

−4

Solution set: 1, 4 

Using the graph, (a) y ≤ 0 when 0 ≤ x < 2 and (b) y ≥ 6 when 2 < x ≤ 4. Algebraically: y≤0 (a)

(b)

3x ≤0 x −2 Key numbers: x = 0, x = 2

Test: Is

Test intervals: ( −∞, 0 ) , ( 0, 2 ) , ( 2, ∞ ) Test: Is

87.

y≥6

(b)

3x ≥6 x−2 3x − 6 ( x − 2) ≥0 x−2 −3 x + 12 ≥0 x−2 Key numbers: x = 2, x = 4

Test intervals: ( −∞, 2 ) , ( 2, 4 ) , ( 4, ∞ ) Test: Is

− 3 x + 12 ≥ 0? x−2

Solution set: ( 2, 4 

5x ≤ 0? x2 + 4

Solution set: ( −∞, 0 

3x ≤ 0? x−2

Solution set: 0, 2 )

y≤0 5x ≤0 x2 + 4 Key number: x = 0 Test intervals: ( −∞, 0 ) , ( 0, ∞ )

x−5 x−5≥0 x≥5

Domain: 5, ∞ ) 88.

6 x + 15

6 x + 15 ≥ 0 6 x ≥ −15 x ≥ −

5 2

 5  Domain: − , ∞   2 

Section 2.6

P (t ) = 450 at about t = 5. So, in 2005, the

− x 2 + x + 12 ≥ 0 x 2 − x − 12 ≤ 0

( x − 4)( x + 3) ≤ 0

population was 450,000. (b) Find the interval of t when P(t ) < 450. Using the graph, P (t ) < 450 for 0 < t < 5. So, from 2003 to

Key numbers: x = − 3, x = 4

2005, the population was less than 450,000.

Test intervals: ( −∞, − 3) > 0

Find the interval of t when P(t ) > 450. Using the

(− 3, 4) < 0 (4, ∞) > 0 Domain: [− 3, 4]

graph, P (t ) > 450 for 5 < t < 12. So, from 2005 to 2012, the population was greater than 450,000. 94. (a) Find t when P(t ) = 400. Using the graph, P (t ) = 400 at about t = 9. So, in 2009, the

2 x2 − 8

90.

2 x2 − 8 ≥ 0 2( x − 2)( x + 2) ≥ 0

population was 400,000. (b) Find the interval of t when P(t ) < 400. Using the graph, P (t ) < 400 for 9 < t < 12. So, from 2009 to

Key numbers: x = − 2, x = 2

2012, the population was less than 400,000.

Test intervals: ( −∞, − 2) > 0

Find the interval of t when P(t ) > 400. Using the

(− 2, 2) < 0 (2, ∞) > 0 Domain: ( − ∞, − 2] ∪ [2, ∞) 91.

graph, P (t ) > 400 for 3 < t < 9. So, from 2003 to 2009, the population was greater than 400,000. 95. (a)

s = −16t 2 + v0 t + s0 s = −16t 2 + 160t

3 x 2 − 20 x − 7

s = 16t (10 − t )

3 x 2 − 20 x − 7 ≥ 0

(3x + 1)( x − 7) ≥ 0

s = 0 when t = 10 seconds.

(b)

s = −16t 2 + 160t > 384

1 Key numbers: x = − , x = 7 3

16t 2 − 160t + 384 < 0

1  Test intervals:  − ∞, −  > 0 3 

Key numbers: t = 6, 4 s > 384 when 4 < t < 6.

 1   − , 7 < 0  3 

16 ( t − 6 )( t − 4 ) < 0

96. (a)

1  Domain:  −∞, −  ∪ [7, ∞) 3  4

2x2 + 4x + 3

2x2 + 4x + 3 ≥ 0

There are no key numbers.

s = −16t 2 + v0 t + s0 = −16t 2 + 128t

(7, ∞) > 0

92.

199

93. (a) Find t when P(t ) = 450. Using the graph,

− x 2 + x + 12

89.

Solving Inequalities Algebraically and Graphically

= 16t ( 8 − t )

(b)

t = 8 seconds −16 t 2 + 128t < 128 16 t 2 − 128t + 128 > 0 Key numbers: t ≈ 1.17, 6.83

( 0, 1.17) , ( 6.83, 8)

Test interval: ( − ∞, ∞) > 0 Domain: ( − ∞, ∞)

200

Chapter 2

Solving Equations and Inequalities

97. D = 0.0743t 2 + 0.628t + 42.61, 0 ≤ t ≤ 22

(a)

98. (a) and (b) 350

100

(c) For y ≥ 200, x ≥ 181.5 pounds (d) The model is not accurate. The data are not linear. Other factors include muscle strength, height, physical condition, etc.

(b) The number of doctorate degrees was between 50 and 60 thousand for about 6 ≤ t ≤ 11, or (1996, 2001). (c) 50 ≤ 0.0743t 2 + 0.628t + 42.61 ≤ 60 Find the key numbers.

300

In Exercises 99–102, use the following equations.

Bed Bath & Beyond: B = 86.5t + 342, 0 ≤ t ≤ 13

Williams-Sonoma: W = − 2.92t 2 + 52.0t + 381, 0 ≤ t ≤ 13

50 = 0.0743t + 0.628t + 42.61 0 = 0.0743t + 0.628t − 7.39 and

0.0743t 2 + 0.628t + 42.61 = 60 0.0743t 2 + 0.628t − 17.39 = 0 Key numbers: t ≈ 6.6, t ≈ −15.1, t ≈ − 20.1 and

t ≈ 11.6

99.

B(t ) ≥ 900 86.5t + 342 ≥ 900 86.5t ≥ 558 t ≥ 6.5

Note: t ≈ −15.1 and t = − 20.1 are not in the

domain. Solution set: 6.6 < t < 11.6, which corresponds

During the year 2006, there were at least 900 Bed Bath & Beyond stores.

to (1996, 2001) W (t ) ≤ 600

100. 2

− 2.92t + 52.0t + 381 ≤ 600 − 2.92t 2 + 52.0t − 219 ≤ 0

Key numbers: t ≈ 6.83, t ≈ 10.97

After testing the intervals, t ≤ 6.83 and t ≥ 10.97. From 2000 to 2006 and 2011 to 2013, there were at most 600 Williams-Sonoma stores. 101.

B (t ) = W ( t ) 86.5t + 342 = − 2.92t 2 + 52.0t + 381 2.92t 2 + 34.5t − 39 = 0 t ≈ 1.04, t ≈ −12.85 (not in the domain of the models) In 2001, there were about the same number of Bed Bath & Beyond stores as Williams-Sonoma stores.

102.

B ( t ) ≥ W (t ) 86.5t + 342 ≥ − 2.92t 2 + 52.0t + 381 2.92t 2 + 34.5t − 39 ≥ 0 t ≥ 1.04 Note: t ≈ −12.85 is not in the domain.

During the year 2001, the number of Bed Bath & Beyond stores exceeded the number of WilliamsSonoma stores.

105. When 200 ≤ v ≤ 400, 1.2 < t < 2.4. 106. For t < 3, 0 < v < 500. 107. False. If −10 ≤ x ≤ 8, then 10 ≥ − x and − x ≤ −8. 108. True. 23 x 2 + 3 x + 6 ≥ 0 for all x. 109. False. Cube roots have no restrictions on the domain.

103. When t = 2: v ≈ 333 vibrations per second. 104. For v = 600, t ≈ 3.6 mm. © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Section 2.7 110. (iv) a < b (ii) 2 a < 2b (iii) 2a < a + b < 2b a+b <b (i) a < 2 111. (a) When solving this inequality, you do not need to reverse the inequality symbol. So, in the solution, the symbol will be ≤ . Matches (iv). (b) When solving this inequality, you need to reverse the inequality symbol. So, in the solution, the symbol will be ≥ . Matches (ii). (c) When solving this inequality, you will use a double inequality that results in a bounded interval. Matches (iii). (d) When solving this inequality, you will use two separate inequalities. The solution will contain two unbounded intervals. Matches (i).

113.

201

y = 12 x x = 12 y x =y 12 x f −1 ( x ) = 12 y = 5x + 8 x = 5y + 8 x − 8 = 5y

114.

1 ( x − 8) = y 5 1 f −1 ( x ) = ( x − 8 ) 5

y = x3 + 7

115.

x = y3 + 7

112. (a) The polynomial equals zero at x = a and at x = b.

x − 7 = y3

(b) f ( x) > 0 on the intervals ( − ∞ , a ) and (b, ∞), and f ( x) < 0 on the interval ( a, b). When x < a or x > b , the factors have the same sign so the product is positive. When a < x < b , the factors have opposite signs so the product is negative.

Linear Models and Scatter Plots

x−7 = y

−1

( x) = 3 x − 7 y = 3 y −7

116.

x = 3 y −7 x3 = y − 7

x3 + 7 = y

f −1 ( x ) = x3 + 7 117. Answers will vary. (Make a Decision)

Section 2.7 Linear Models and Scatter Plots positive

6. (a)

regression or linear regression

negative

No. The closer the correlation coefficient r is to 1, the better the fit.

(a)

Score on second quiz

16 14 12 10 8 6 4 2

x 2 4 6 8 10 12 14 16

Monthly sales (in thousands of dollars)

Score on first quiz 60

(b) No. Quiz scores are dependent on several variables, such as study time, class attendance, etc.

50 40

7. Negative correlation

30 20

8. No correlation

10 x 1

Years of experience

9. No correlation 10. Positive correlation

(b) Yes, the data appears somewhat linear. The more experience x corresponds to higher sales y.

202

Chapter 2

Solving Equations and Inequalities

11. (a) and (b)

15. (a) and (b) y

−4 −3 −2

−2

−3

−1

−4

the line is y = x. 12. (a) and (b) y

16. (a) and (b)

4 3

6 5

−4 −3 −2 −1 −1

−2

−3

−1

the line is y = − 2 x. 13. (a) and (b) 17.

y 6

x 1

4 3 −4

1 x

−3 −2 −1 −1

(b)

−2

x Linear equation Given data

Using the points (0, 2) and ( − 2, 1), an equation of

(c)

the line is y =

5 −1

1 x + 2. 2

14. (a) and (b)

–1 1.14 1

0 1.6 2

2 2.52 3

4 3.44 3

0 2.8 2

1 1.5 1

2 0.2 1

The model fits the data well. 18.

–3 0.22 0

y = −1.3 x + 2.8 (a)

7 6 5 −7

4 3

−2

(b)

1 −1 −1

(c)

Using the points ( 2, 3) and (3, 2), an equation of

x Linear equation Given data

the line is y = − x + 5.

The model fits the data fairly well.

–2 5.4 6

–1 4.1 4

Section 2.7 19. (a) Elongation (in centimeters)

Linear Models and Scatter Plots

203

20. (a) and (c)

400

6 5 4 3 2 0

1 F 20

100

The model fits the data well.

Force (in kilograms)

(b)

(b) d = 0.07F − 0.3 (c) d = 0.066 F or F = 15.13d + 0.096 (d) If F = 55, d = 0.066(55) ≈ 3.63 centimeters.

S = 15.28t + 145.8

(d) For 2020, t = 20 and S = 15.28(20) + 145.8 = 451.4 million subscribers.

Answers will vary. 21. (a) and (c) 400

The model fits the data well. (b) V = 22.8t + 12.8 (d) For 2015, t = 15 and V = 22.8(15) + 12.8 = $354.8 million. For 2020, t = 20 and V = 22.8( 20) + 12.8 = $468.8 million. Answers will vary. (e) 22.8; The slope represents the average annual increase in salaries (in millions of dollars). 22. (a) and (c) 80

The model fits the data well. (b)

S = 1.61t + 39.6

(d) For 2020, t = 20 and S = 1.61( 20) + 39.6 = $71.8 thousand. For 2022, t = 22 and S = 16.1( 22) + 39.6 = $75.82 thousand. Answers will vary. 23. (a) and (c) 9000

0 8400

(b) P = 37.3t + 8416 (d)

Year Actual Model

2008 8711 8714.4

2009 8756 8751.7

2010 8792 8789

2011 8821 8826.3

2012 8868 8863.6

2013 8899 8900.9

204

Chapter 2

Solving Equations and Inequalities

(e) For 2050, t = 50 and P = 37.3(50) + 8416 = 10,280.8 hundred people or 10,280,800 people. Answers will vary. 24. (a) and (c) 600

0 540

(b) P = 5.9t + 506 (d)

Year Actual Model

2009 560 559.1

2010 564 565

2011 568 570.9

2012 577 576.8

2013 583 582.7

The model fits the data well. (e) For 2050, t = 50 and P = 5.9(50) + 506 = 800.5 thousand people or 800,500 people. Answers will vary. 25. (a) y = 45.70 x + 108.0

(b)

27. (a)

T = − 0.015t + 4.79 r ≈ − 0.866

250

(b) The negative slope means that the winning times are generally decreasing over time. 1 150

(c)

y = 45.70(1.5) + 108.0 = 176.55 or $176,550. 26. (a)

1800

5 1440

The first three points and the last four points are approximately linear. (b) T1 = 97t + 908 T2 = 12.7t + 1624 6 ≤ t ≤ 8 97t + 908, (c) T =  12.7 t 1624, 8 + < t ≤ 12  1800

5.2

0 3.6

(d)

Year

1956

1960

1964

1968

1972

Actual Model

4.91 4.70

4.84 4.64

4.72 4.58

4.53 4.52

4.32 4.46

Year

1976

1980

1984

1988

1992

Actual Model

4.16 4.40

4.15 4.34

4.12 4.28

4.06 4.22

4.12 4.16

Year

1996

2000

2004

2008

2012

Actual

4.12

4.10

4.09

4.05

4.02

Model

4.10

4.04

3.98

3.92

3.86

The model does not fit the data well. (e) The closer r is to 1, the better the model fits the data. (f ) No; The winning times have leveled off in recent years, but the model values continue to decrease to unrealistic times.

5 1440

(d) Answers will vary.

Section 2.7 28. (a)

205

(iii) y = 0.68 x + 2.7 r = 0.62

l = 0.34d + 77.9 r ≈ 0.993

The slope is steeper than the slope of the line in (ii). The line is not a good fit for the data, so the correlation coefficient will not be close to 1.

(b) Yes, r is close to 1. (c)

Linear Models and Scatter Plots

120

100

The data fits the model well. (d) For d = 112, l = 0.34(112) + 77.9 = 115.98 or

(b) Model (i) is the best fit for its data because its r-value is −0.986 , and therefore r is closest to 1.

about 116 centimeters. 29. True. To have positive correlation, the y-values tend to increase as x increases.

33.

30. False. The closer the correlation coefficient is to –1 or 1, the better a line fits the data.

f ( x) = 2 x 2 − 3x + 5

(a)

f (−1) = 2 + 3 + 5 = 10

(b)

f (w + 2) = 2( w + 2)2 − 3( w + 2) + 5 = 2 w 2 + 5w + 7

31. Answers will vary. 32. (a) (i) y = −0.62 x + 10.0 r = −0.986 The data are decreasing, so the slope and correlation coefficient are negative.

34.

g ( x) = 5 x 2 − 6 x + 1

(a)

g(−2) = 5(4) − 6(−2) + 1 = 33

(b)

g( z − 2) = 5( z − 2)2 − 6( z − 2) + 1 = 5 z 2 − 26 z + 33

(ii) y = 0.41x + 2.7 r = 0.973 The slope is less steep than the slope of the line in (iii). The line is a better fit for the data than the line in (iii), so the correlation coefficient will be greater. 12

35. 6 x + 1 = −9 x − 8 15 x = −9 9 3 x=− =− 15 5 36. 3( x − 3) = 7 x + 2 −11 = 4x 11 x=− 4 37.

8 x 2 − 10 x − 3 = 0

(4 x + 1)(2 x − 3) = 0 1 3 x=− , 4 2

38. 10 x 2 − 23 x − 5 = 0 (2 x − 5)(5 x + 1) = 0

5 1 , − 2 5

206

Chapter 2

Solving Equations and Inequalities

Chapter 2 Review 1.

3 =5 x−4

x =

(a)

11 2

3 ? =5 −4 2

6 + 11

3 ? 6 + 3 =5

2 ? 6 ( 3) + 1 = 3+3 3 38 ? 19 = 6 3 Yes, x = 3 is a solution. (c) x = 0 6+

6+ 2 ≠ 5 is not a solution. No, x = 11 2 (b)

x=0

3 ? =5 0−4 3 ? 6− = 5 4 5.25 ≠ 5 No, x = 0 is not a solution. (c) x = −2 3 ? 6+ =5 −2 − 4 1? 6− =5 2 5.5 ≠ 5 No, x = −2 is not a solution. (d) x = 1 3 ? 6+ =5 1− 4

2 6x + 1 = x+3 3 (a) x = −3 2 6+ is undefined. −3 + 3 No, x = −3 is not a solution. (b) x = 3 6+

2 ? 6 (0) + 1 = 0+3 3 20 ? 1 = 3 3 No, x = 0 is not a solution. 2 (d) x = − 3 2 ? 6 ( −2 3 ) + 1 6+ = 3 ( − 2 3) + 3 6+

6 ? = −1 7 No, x = − 23 is not a solution. 6+

6 −1 = 5 5=5 Yes, x = 1 is a solution.

x x + = 7 18 10 5 x + 9 x = 630 14 x = 630 x = 45

x x + = 9 2 7 7 x + 2 x = 126 9 x = 126 x = 14

5 13 = x − 2 2x − 3 10 x − 15 = 13 x − 26 11 = 3 x x=

11 3

Chapter 2 Review

10 12 = x + 1 3x − 2 10 ( 3 x − 2 ) = 12 ( x + 1)

12.

30 x − 20 = 12 x + 12 18 x = 32 32 16 = x= 18 9

5 1 2 + = 2 x − 5 x + 5 x − 25 5 ( x + 5) + ( x − 5) 2 = 2 x 2 − 25 x − 25 6 x + 20 = 2 6 x = −18 x = −3

13. September’s profit + October’s profit = 689, 000 Let x = September’s profit. Then x + 0.12 x = October’s profit.

2 = 10 x −1 2 = −4 x −1 2 = −4 ( x − 1)

7. 14 +

x + ( x + 0.12 x ) = 689,000

2.12 x = 689,000 x = 325,000 x + 0.12 x = 364,000 September: $325,000; October: $364,000

2 = −4 x + 4 4x = 2 1 x= 2

14. Let x = the number of quarts of pure antifreeze.

30% of (10 − x ) + 100% of x = 50% of 10

2 8. 10 + =4 x −1 10 ( x − 1) + 2 = 4 ( x − 1)

0.30 (10 − x ) + 1.00 x = 0.50 (10 ) 3 − 0.30 x + 1.00 x = 5

10 x − 10 + 2 = 4 x − 4 6x = 4 4 2 x= = 6 3

207

0.70 x = 2 2 0.70 20 6 = = 2 liters 7 7

4 5 = 6+ x x 6x − 4 = 6x + 5 6−

15. (a)

−4 ≠ 5 h

No solutions 10. 2 −

1 4 = 2 + 2 x2 x

2 x2 − 1 = 2 x2 + 4

−1 ≠ 4

(b)

No solutions 11.

9x 4 − =3 3x − 1 3x + 1 9 x ( 3 x + 1) − 4 ( 3 x − 1) = 3 ( 3 x − 1)( 3 x + 1)

(

)

27 x 2 + 9 x − 12 x + 4 = 3 9 x 2 − 1 27 x 2 − 3 x + 4 = 27 x 2 − 3 −3 x = −7 x=

2m 75 cm

h 2 8 = = 8 34 3 64 1 h= = 21 meters 3 3

16. 12,000 ( 0.025 ) + 10,000 x = 870

10,000 x = 570 x = 0.057, or 5.7%

7 3

208

17.

Chapter 2

Solving Equations and Inequalities 23. 5 ( x − 2 ) − 1 = 0

9 C + 32 5 9 F − 32 = C 5 F =

2 −4

5 ( F − 32) = C 9

Let F = 28.3.

−6

Solution: x = 2.2

5 ( 28.3 − 32) = C 9 5 ( − 3.7) = C 9 − 2.06 ≈ C

24. 12 − 5 ( x − 7) = 0 10

The average temperature is about − 2.06° C. 18. Basketball: 2π r = 30  r =

−5

− 10

Solution: x = 9.4

4 4  15  V = π r 3 = π   ≈ 455.95 in.3 3 3 π  Baseball: 2π r = 9.25  r =

3x 2 + 3 = 5 x

25.

3x 2 − 5 x + 3 = 0

9.25 2π

4  9.25  3 π  ≈ 13.4 in. 3  2π 

19. − x + y = 3

−1

Let x = 0 : y = 3, y-intercept: ( 0, 3)

No real solution

Let y = 0 : x = −3, x-intercept: ( −3, 0 ) 20.

Let x = 0 : − 5 y = 20  y = −4, y -intercept: ( 0, − 4)

1 3 x − 2x + 4 = 0 3

Let y = 0 : x − 5 ( 0 ) = 20  x = 20, x-intercept: ( 20, 0 ) 21.

1 3 x + 4 = 2x 3

26.

x − 5 y = 20

y = x 2 − 9 x + 8 = ( x − 8)( x − 1) Let x = 0 : y = 0 2 − 9 ( 0 ) + 8  y = 8,

−4

y -intercept: ( 0, 8 )

4 −2

Let y = 0 : 0 = x 2 − 9 x + 8  0 = ( x − 8 )( x − 1)

Solution: x ≈ − 3.135

 x = 1, 8, x -intercepts: (1, 0 ) , ( 8, 0 )

0.3x 4 = 5 x + 2

27.

22. y = 25 − x 2

Let x = 0 : y = 25 − ( 0 )  y = 25, 2

y -intercept: ( 0, 25 )

0.3x 4 − 5 x − 2 = 0 1 −2

Let y = 0 : 0 = 25 − x 2 = ( 5 − x )( 5 + x )

 x = ±5, x -intercepts: ( 5, 0 ) , ( −5, 0 )

−9

Solutions: x ≈ − 0.402, 2.676

Chapter 2 Review

0.8x 4 = 0.5 x − 5

28.

209

y = −x + 7

32.

y = 2 x3 − x + 9

0.8 x 4 − 0.5 x + 5 = 0

2x − x + 9 = −x + 7

2 x3 + 2 = 0 x3 + 1 = 0

−3

( x + 1) ( x 2 − x + 1) = 0  x = −1  y = − ( −1) + 7 = 8 ( x, y ) = ( −1, 8)

No real solution 29.

3 x + 5 y = −7 − x − 2 y = 3  x = −2 y − 3

33.

3−

3 ( −2 y − 3) + 5 y = 7

34.

− 50 + 8 = 8 + 5 2 i

− y − 9 = −7 y = −2 x = −2 ( −2 ) − 3 = 1

( x, y ) = (1, − 2) 30.

−16 = 3 − 4i

35. − i + 4i 2 = − i + 4( −1) = −4 − i

36. 7i − 9i 2 = 7i − 9( −1)

x−y=3

= 9 + 7i

2 x + y = 12  y = −2 x + 12

x − ( −2 x + 12 ) = 3

37. ( 2 + 13i ) + (6 − 5i ) = ( 2 + 6) + (13 − 5) i = 8 + 8i

x + 2 x − 12 = 3 3 x − 12 = 3 3 x = 15 x=5 5− y =3 y =2

1 3  1 3  1 1  3 3 38.  + i  −  − i  =  −  +  + i 2 4  2 4  2 2  4 4   = 0+

( x, y ) = ( 5, 2 )

31. 4 x 2 + 2 y = 14 2x + y = 3  y = − 2x + 3

4 x 2 + 2( − 2 x + 3) = 14 4 x 2 − 4 x + 6 = 14 4x2 − 4x − 8 = 0

40.

(1 + 6i )( 5 − 2i ) = 5 − 2i + 30i + 12 = 17 + 28i

41.

( −16 + 3)( −25 − 2 ) = ( 4i + 3)( 5i − 2 )

x − x−2 = 0

= −20 − 8i + 15i − 6

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

= −26 + 7i

42.

y = − 2( −1) + 3 = 5

( 5 − −4 )( 5 + −4 ) = ( 5 − 2i )( 5 + 2i ) = 25 + 4

( x, y ) = (2, −1) x + 1 = 0  x = −1

3 i 2

39. 5i (13 − 8i ) = 65i − 40i 2 = 40 + 65i

x−2 = 0  x = 2

2 3 i 4

= 29

43.

( x, y ) = (−1, 5)

−9 + 3 + −36 = 3i + 3 + 6i = 3 + 9i

44. 7 − −81 + −49 = 7 − 9i + 7i

= 7 − 2i 45.

(10 − 8i )( 2 − 3i ) = 20 − 30i − 16i + 24i2 = −4 − 46i

210

Chapter 2

Solving Equations and Inequalities

46. i ( 6 + i )( 3 − 2i ) = i (18 + 3i − 12i + 2 )

57.

( x − 8)( x + 1) = 0

= i ( 20 − 9i ) = 9 + 20i

47.

x2 − 7 x − 8 = 0 x −8 = 0  x = 8

( 3 + 7i ) + ( 3 − 7i ) = ( 9 + 42i − 49 ) + ( 9 − 42i − 49 ) 2

x + 1 = 0  x = −1

= −80

48.

( 4 − i ) − ( 4 + i ) = (16 − 8i − 1) − (16 + 8i − 1) 2

58.

( x + 6)( x − 3) = 0

= −16i

49.

50.

51.

52.

6 + i 6 + i −i −6i − i 2 = ⋅ = i i −i −i 2 −6i + 1 = = 1 − 6i 1 4 −4 −i 4i 4 = ⋅ = = i −3i 3i −i 3 3 3 + 2i 5 − i 15 + 10i − 3i + 2 ⋅ = 5+i 5−i 25 + 1 17 7 i = + 26 26 1 − 7i 2 − 3i 2 − 21 − 17i ⋅ = 2 + 3i 2 − 3i 4+9 19 17 =− − i 13 13

x 2 + 3 x − 18 = 0 x + 6 = 0  x = −6 x −3 = 0  x = 3

59.

( x + 4 ) = 18 2

x + 4 = ± 18 x = −4 ± 3 2

60.

( x + 1) = 24 2

x + 1 = ± 24 = ±2 6 x = −1 ± 2 6

61. (3 x − 1) + 4 = 0 2

(3x − 1)

3 x = 1 ± 2i x =

49 x = 9 49 9

x = ± 7 3

(5 x − 3)

= −16

5x − 3 = ±

−16

5 x − 3 = ± 4i

8x = 2x2

5 x = 3 ± 4i x =

0 = 2 x( x − 4) 2x = 0  x = 0 x −4 = 0  x = 4 6 x = 3x

1 2 ± i 3 3

62. (5 x − 3) + 16 = 0

0 = 2x2 − 8x

55.

−4

3 x − 1 = ± 2i

54.

= −4

3x − 1 = ±

53. 9 x 2 = 49

x = ±

2 2

0 = 3x − 6 x 0 = 3x ( x − 2 ) 3x = 0  x = 0 x −2 =0 x =2

3 4 ± i 5 5

63. x 2 + 4 x − 9 = 0

x2 + 4x + 4 = 9 + 4

( x + 2)

= 13

x + 2 = ± 13 x = −2 ±

56. 16 x 2 = 25 25 x 2 = 16

x=±

25 16

= ± 54

Chapter 2 Review 64. x 2 − 6 x − 5 = 0

71.

2x2 + 9x − 5 = 0

( 2 x − 1)( x + 5) = 0

x2 − 6x = 5

1 2 x + 5 = 0  x = −5

x2 − 6 x + 9 = 5 + 9

( x − 3)

2x −1 = 0  x =

= 14

x − 3 = ± 14 x = 3± 65.

72.

4 x2 + x − 5 = 0

( 4 x − 5)( x − 1) = 0

x − 12 x + 30 = 0

4x + 5 = 0  x = −

x − 12 x = −30

( x − 6) = 6 2

73. − x 2 − x + 15 = 0 x 2 + x − 15 = 0

x −6 = ± 6 x =6± 6

x2 + 6x − 3 = 0

− b ± b 2 − 4ac x= 2a =

− b ± b 2 − 4ac 2a −1 ± 1 − 4 ( −15 )

2 −1 ± 61 = 2

−6 ± 62 − 4 (1)( −3) 2 (1)

74. − x 2 − 3 x + 2 = 0

−6 ± 48 = 2 = −3 ± 2 3

x 2 + 3x − 2 = 0 x =

x 2 − 3x = 28

67.

x − 3x − 28 = 0

( x − 7)( x + 4) = 0

x−7 = 0  x = 7 x + 4 = 0  x = −4

x 2 + 3x = 40

68.

75.

x 2 + 3x − 40 = 0

x + 8 = 0  x = −8 x 2 − 10 x = 9 2

x − 10 x + 25 = 9 + 25 2

= 34

x−5 = ± x = 5± 70.

76.

34 34

x2 + 8x = 7

( x + 4)

x = −4 ±

−3 ±

9+8

2 − 3 ± 17 2

x2 + 6x −1 = 0

− b ± b 2 − 4ac 2a −6 ±

( 6 ) − 4 ( −1) 2

2 −6 ± 40 = 2 = −3 ± 10

= 23

x + 4 = ±

(3) − 4(1)(− 2) 2(1)

x 2 + 4 x + 10 = 0

x 2 + 8 x + 16 = 7 + 16 2

− (3) ±

x−5 = 0  x = 5

( x − 5)

b 2 − 4ac 2a

−b ±

− b ± b 2 − 4ac 2a −4 ± 16 − 40 = 2 −4 ± −24 = 2 = −2 ± 6i

( x − 5)( x + 8) = 0

69.

5 4

x −1 = 0  x = 1

x 2 − 12 x + 36 = −30 + 36

66.

211

212

Chapter 2

Solving Equations and Inequalities 79. (a)

77. 2 x 2 − 6 x + 21 = 0 − b ± b 2 − 4ac 2a 6 ± 36 − 168 = 4 6 ± −132 = 4 3 33 = ± i 2 2

2000

0 1000

(b) In 2008, the average cost per day reached $1800. (c) 1800 = − 0.54t 2 + 82.6t + 1136

0 = − 0.54t 2 + 82.6t − 664 t =

78. 2 x 2 − 8 x + 11 = 0

x= =

− b ± b 2 − 4ac 2a

− (82.6) ±

(82.6) − 4(− 0.54)(− 664) 2( − 0.54) 2

t ≈ 8.51, 144.45 (not in the domain of the model) In 2008, the average cost per day reached $1800.

( −8) − 4 ( 2)(11) 2 ( 2) 2

8±

(d) Answers will vary.

8 ± −24 4 8 ± 2 6i = 4 6 = 2± i 2 =

80. (a) 10,000

(b) In 2010, the revenues reached $4 billion dollars. (c) 4000 = 86.727t 2 − 839.83t + 2967.9 0 = 86.727t 2 − 839.83t − 1032.1 t =

− ( − 839.83) ±

(− 839.83) − 4(86.727)(−1032.1) 2(86.727) 2

t ≈ −1.10 (not in the domain of the model), 10.79 In 2010, the revenue reached $4 billion. (d) 10,000 = 86.727t 2 − 839.83t + 2967.9 0 = 86.727t 2 − 839.83t − 7032.1 t =

− ( − 839.83) ±

(− 839.83) − 4(86.727)(− 7032.1) 2(86.727) 2

t ≈ − 5.38 (not in the domain of the model), 15.07 In 2015, the revenue will reach $10 billion.

Chapter 2 Review

81.

3 x 3 − 26 x 2 + 16 x = 0

(

88.

)

x 3 x 2 − 26 x + 16 = 0

2 3 x −8 = 0  x =8

3x − 2 = 0  x =

89.

36 x 3 − x = 0

)

6x + 1 = 0  x = −

90.

1 6

x − 2 = 64 x = 66

5 x 4 − 12 x 3 = 0 x 3 ( 5 x − 12 ) = 0

91. 2 x − 5 x + 3 = 0

2x + 3 = 5 x

x 3 = 0 or 5x − 12 = 0 x=0 x = 125

(2 x + 3)2 = (5

9 4 x −1 = 0  x =1

4x − 9 = 0  x =

( x − 4 )( x + 3) = 0 2

x2 − 4 = 0 x2 = 4

92. or

x2 + 3 = 0 x 2 = −3

x = ± 4 = ±2

87.

x = ± 3i

( x − 5)( x + 1) = 0

3x − 2 = ( 4 − x )

3 x − 2 = 16 − 8 x + x 2 0 = 18 − 11x + x 2

0 = x − 9  x = 9, extraneous

3x − 2 = 4 − x

0 = ( x − 9 )( x − 2 )

x4 − 4 x2 − 5 = 0

x −5=0

(4 x − 9)( x − 1) = 0

x 4 − x 2 − 12 = 0

)

4 x 2 − 13x + 9 = 0

x2 = 0  x = 0 4 x − 6 = 0  x = 32

4 x 2 + 12 x + 9 = 25 x

4 x3 − 6 x2 = 0

x2 ( 4 x − 6) = 0

86.

x −2 −8 = 0 x−2 =8

1 6x − 1 = 0  x = 6

x+4=9 x=5

x=0

85.

x = ± −2 = ± 2i

x+4 =3

( x + 4 ) = ( 3)

x ( 6 x + 1)( 6 x − 1) = 0

84.

or x 2 + 2 = 0 x 2 = −2

x = ± −4 = ±2i

x 36 x 2 − 1 = 0

83.

x2 + 4 = 0 x 2 = −4

x=0

(

3 x 4 + 18 x 2 + 24 = 0 x4 + 6 x2 + 8 = 0

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

x ( 3 x − 2 )( x − 8 ) = 0

82.

213

0 = x−2 x =2

x +1 = 0

x2 = 5

x 2 = −1

x=± 5

x = ± −1 = ±i

2 x 4 − 22 x 2 + 56 = 0

( x − 4 )( 2 x − 14 ) = 0 2

x2 − 4 = 0 x2 = 4 x = ± 4 = ±2

or 2 x 2 − 14 = 0 2 x 2 = 14 x2 = 7 x=± 7

214

Chapter 2

Solving Equations and Inequalities

93.

2x + 3 + x − 2 = 2

97.

( 2 x + 3 ) = (2 − x − 2 ) 2

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

( x + 1) = ( −4 x − 2 ) x 2 + 2 x + 1 = 16 ( x − 2 ) 2

( x + 4 ) + 5x ( x + 4 ) = 0 12 ( x + 4 ) 1 + 5 x ( x + 4 ) = 0 12 ( x + 4 ) ( 5x 2 + 20 x + 1) = 0 12 ( x + 4 ) = 0 or 5x 2 + 20 x + 1 = 0 12

x = −4

x 2 − 14 x + 33 = 0

( x − 3)( x − 11) = 0

94.

98.

(

8x2 x2 − 4

−20 ± 2 95 10 95 5

) + ( x − 4) = 0 13

25 x = 36 + 12 x − 1 + x − 1

( x − 2 ) ( x + 2 ) ( 3 x − 2 )( 3 x + 2 ) = 0 13

24 x − 35 = 12 x − 1

x−2 =0 x =2

576 x − 1680 x + 1225 = 144 ( x − 1)

x + 2 = 0  x = −2

576 x 2 − 1824 x + 1369 = 0

3 x − 2 = 0  x = 23

x= =

− ( −1824 ) ±

( −1824 ) − 4 ( 576 )(1369 ) 2 ( 576 ) 2

3 x + 2 = 0  x = − 23

99.

1824 ± 172,800 1824 ± 240 3 = 1152 1152

38 + 5 3 24 38 − 5 3 x= , extraneous 24 x=

( x − 1) − 25 = 0 23 ( x − 1) = 25 2 ( x − 1) = 253

x + 4 = 0  x = −4 x −1 = 0  x = 1

100.

( x + 2)

3x 1 5 = − x 2 2 2 3x = − 5 x 3x2 + 5x − 2 = 0

( x + 2 )( 3 x − 1) = 0

x = 126 or x = −124 34

x 3 1 + = 8 8 2x 4 x+3= x x2 + 3x − 4 = 0

( x + 4 )( x − 1) = 0

x − 1 = ± 253 x = 1 ± 125

96.

( x − 4 ) 8 x + x − 4  = 0 ( x − 4) (9x − 4) = 0

5 x − x −1 = 6 5 x = 6 + x −1

95.

−20 ± 400 − 20 10

x = −2 ±

x = 3, extraneous or x = 11, extraneous

No solution (You can verify that the graph of y = 2 x + 3 + x − 2 − 2 lies above the x-axis.)

x + 2 = 0  x = −2

= 27

3x − 1 = 0  x =

x + 2 = 274 3 x + 2 = 81 x = 79

101.

1 3

5 3 =1+ x x + 2 5( x + 2) = x( x + 2) + 3( x) 5 x + 10 = x 2 + 2 x + 3x 5 x + 10 = x 2 + 5 x + 5 10 = x 2 ± 10 = x

Chapter 2 Review

102.

6 8 + = 3 x x +5 6( x + 5) + 8( x) = 3( x)( x + 5)

104.

6 x + 30 + 8 x = 3x 2 + 15 x

2x 3 − =1 x2 − 1 x + 1 2x 3 − =1 ( x − 1)( x + 1) x + 1

0 = 3x 2 + x − 30

2 x − 3( x − 1) = ( x − 1)( x + 1)

0 = (3 x + 10)( x − 3)

2 x − 3x + 3 = x 2 − 1

3 x + 10 = 0  x = −

x2 + x − 4 = 0

10 3

x =

x −3 = 0  x = 3

2 16 3+ = 2 x x

103.

215

3x 2 + 2 x = 16

3x + 2 x − 16 = 0

(3x + 8)( x − 2) = 0

105.

8 3x + 8 = 0  x = − 3 x − 2 = 0  x = 2

−b ± − (1) ±

b 2 − 4ac 2a

(1) − 4(1)(− 4) 2(1) 2

−1 ± 17 2

x − 5 = 10 x − 5 = −10 or x − 5 = 10 x = −5 x = 15

106.

2x + 3 = 7 2 x + 3 = 7 or 2 x + 3 = −7

107.

2x = 4

2 x = −10

x=2

x = −5

x2 − 3 = 2 x

x 2 − 3 = −2 x

x2 − 2 x − 3 = 0

x2 + 2 x − 3 = 0

( x − 3)( x + 1) = 0

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

x = 3 or x = −1

x = −3 or x = 1

The only solutions to the original equation are x = 3 or x = 1, ( x = −3 and x = −1 are extraneous.) 108.

x2 − 6 = x x2 − 6 = x or 2 x − x−6=0 ( x − 3)( x + 2 ) = 0 x −3= 0 x =3 x + 2 = 0  x = −2, extraneous

x2 − 6 = − x x2 + x − 6 = 0 ( x + 3)( x − 2 ) = 0 x−2=0 x =2 x + 3 = 0  x = −3, extraneous

216

Chapter 2

Solving Equations and Inequalities

109.

Let x = number of investors. 240,000 240,000 = + 20,000 x x+2  240,000   240,000  x ( x + 2)  + 20,000  x ( x + 2 ) = x    x+2 

(

)

(

)

240,000 ( x + 2 ) = 240,000 x + 20,000 x ( x + 2 ) 240,000 x + 480,000 = 240,000 x + 20,000 x 2 + 40,000 x 2

20,000 x + 40,000 x − 480,000 = 0 x 2 + 2 x − 24 = 0

( x + 6 )( x − 4 ) = 0 x+6=0

x−4=0

x = −6 ( x = −6 is not in the original domain.) There are 4 investors currently in the group. 110.

x=4

Let x = number of students. 1700 1700 = + 7.50 x x+6  1700   1700  + 7.50  x ( x + 6 ) x ( x + 6)  =  x   x+6 

(

)

(

)

1700 ( x + 6 ) = 1700 x + 7.50 x ( x + 6 )

1700 x + 10,200 = 1700 x + 7.5 x 2 + 45 x 7.5 x 2 + 45 x − 10,200 = 0 75 x 2 + 450 x − 102,000 = 0 3 x 2 + 18 x − 4080 = 0 x 2 + 6 x − 1360 = 0

( x + 40 )( x − 34 ) = 0 x + 40 = 0

x − 34 = 0

x = −40

x = 34

( x = −40 is not in the original domain.) There are 34 students currently in the group. 111.

Let x = average speed originally from Portland to Seattle. 145 145 12 = + x x + 40 60 1  145   145 5 x ( x + 40 )  +  5 x ( x + 40 ) = 40 5 x x +   

(

)

725 x + 29000 = 725 x + x 2 + 40 x x 2 + 40 x − 29,000 = 0 Using the Quadratic Formula: x=

−40 ±

( 40 ) − 4 (1)( −29,000 ) 2 (1) 2

−40 ± 117,600 ≈ 151.5 mi h 2 So, on the return trip from Seattle to Portland, the average speed is x + 40 = 191.5 mph. =

Chapter 2 Review 112.

Let x = average speed on the first trip. 56 56 1 = + x x +8 6 1 56 ( x + 8 ) = 56 x + x ( x + 8 ) 6 1 448 = x ( x + 8 ) 6 x 2 + 8 x − 2688 = 0

113.

r  A = P 1 +  n 

217

nt 12 ( 6 )

r   1196.95 = 1000  1 +  12   72

r   1.19695 =  1 +   12  1 72 r 1+ = (1.19695 ) 12 r = 0.03, 3%

( x − 48)( x + 56 ) = 0 x = 48 So, the average speed on the return trip is 48 + 8 = 56 mph.

114.

r  A = P 1 +  n 

r  2465.43 = 1500  1 +  4 

4 (10 )

r  1.64362 =  1 +  4  r 1 + = 1.643621 40 4 r = 0.05, 5% 115. (a)

Year

2000

2001

2002

2003

2004

Enrollment (in millions)

3.09

3.17

3.25

3.33

3.40

Year

2005

2006

2007

2008

2009

Enrollment (in millions)

3.48

3.55

3.62

3.69

3.76

Year

2010

2011

2012

Enrollment (in millions)

3.83

3.89

3.96

(b)

( 0.51049t + 9.5287 ) = (3.5) 2

0.51049t + 9.5287 = 12.25 0.51049t = 2.7213 t ≈ 5.33

In 2005, enrollment reached 3.5 million.

218

Chapter 2

Solving Equations and Inequalities

(e) Find t when S = 6. 0.51049t + 9.5287 = 6

( 0.51049t + 9.5287 ) = (6) 2

0.51049t + 9.5287 = 36 0.51049t = 26.4713 t ≈ 51.85

In 2051, enrollment will reach 6 million. Answers will vary. (f) Answers will vary. 116. (a)

Year

2003

2004

2005

2006

2007

Population (in millions)

8.07

8.03

8.02

8.01

8.03

Year

2008

2009

2010

2011

2012

Population (in millions)

8.06

8.11

8.18

8.26

8.36

(b)

(c) In 2010, the population of New York reached 8.2 million. (d) Find t when P = 8.2. 0.13296t 2 − 1.4650t + 68.243 = 8.2

( 0.13296t − 1.4650t + 68.243) = (8.2) 2

0.13296t 2 − 1.4650t + 68.243 = 67.24 0.13296t 2 − 1.4650t + 1.003 = 0 t =

− ( −1.4650) ±

(−1.4650)2 − 4(0.13296)(1.003) 2(0.13296)

t ≈ 0.73 (not in the domain model), 10.28

In 2010, the population of New York reached 8.2 million. (e) Find t when P = 8.5. 0.13296t 2 − 1.4650t + 68.243 = 8.5

( 0.13296t − 1.4650t + 68.243) = (8.5) 2

0.13296t 2 − 1.4650t + 68.243 = 72.25 0.13296t 2 − 1.4650t − 4.007 = 0 t =

− ( −1.4650) ±

( −1.4650) − 4(0.13296)( −4.007) 2(0.13296) 2

t ≈ − 2.27 (not in the domain model), 13.29

In 2013, the population will reach 8.5 million. Answers will vary. (f) Answers will vary.

Chapter 2 Review

219

123. x ≤ 4

117. 8 x − 3 < 6 x + 15 2 x < 18 x<9

−4 ≤ x ≤ 4

 −4, 4 

( −∞, 9)

x x

−4

−2

− 1 0 1 2 3 4 5 6 7 8 9 10

124. x − 2 < 1

118. 9 x − 8 ≤ 7 x + 16 2 x ≤ 24 x ≤ 12 ( −∞, 12

−1 < x − 2 < 1 1< x < 3

(1, 3)

x 9

119. 12 ( 3 − x ) > 13 ( 2 − 3 x )

125. x − 32 > 32

3(3 − x ) > 2 ( 2 − 3x )

x − 32 < − 32 x<0

9 − 3x > 4 − 6 x 3 x > −5

x − 32 > 32 x>3

or or

( −∞, 0 ) ∪ ( 3, ∞ )

x > − 35

( − 35 , ∞ ) −

−3 −2 −1

126. x − 3 ≥ 4

5 3 x

−2

−1

x − 3 ≥ 4 or x≥7

120. 4 ( 5 − 2 x ) ≥ 12 ( 8 − x )

( −∞, − 1 ∪ 7, ∞ )

8(5 − 2x ) ≥ 8 − x

40 − 16 x ≥ 8 − x 32 ≥ 15 x

−2 −1 0 1 2 3 4 5 6 7 8 9

127. 4 3 − 2 x ≤ 16

32 x ≤ 15

( −∞,

32 15

3 − 2x ≤ 4



−4 ≤ 3 − 2 x ≤ 4 −7 ≤ −2 x ≤ 1

32 15 x 0

x−3≤ 4 x ≤ −1

≥ x ≥ − 12

7 2

− 12 ≤ x ≤ 27

121. −2 < − x + 7 ≤ 10 −9 < − x ≤ 3

 − 12 , 72 

9 > x ≥ −3 −3 ≤ x < 9

−1

 −3, 9 )

7 2

x −1

x −6

−3

128. x + 9 + 7 > 19

122. −6 ≤ 3 − 2 ( x − 5 ) < 14

x + 9 > 12

−6 ≤ 13 − 2 x < 14 −19 ≤ −2 x < 1

x + 9 > 12 or x>3

≥ x > − 12

( −∞, − 21) , ( 3, ∞ )

19 2

− 12 < x ≤ 192

 1 19  − ,   2 2 −1 2

x + 9 < −12 x < −21

− 24 − 18 − 12

−6

19 2

x − 1 0 1 2 3 4 5 6 7 8 9 10

220

Chapter 2

Solving Equations and Inequalities

x2 − 2x > 3

129.

x ( x − 4 )( x + 4 ) ≥ 0

x − 2x − 3 > 0

( x − 3)( x + 1) > 0

Key numbers: x = 0, x = 4, x = −4

Key numbers: x = −1, x = 3

Test intervals: ( −∞, − 4 ) , ( −4, 0 ) , ( 0, 4 ) , ( 4, ∞ )

Test intervals: ( −∞, − 1) , ( −1, 3) , ( 3, ∞ )

Test: Is x ( x − 4 )( x + 4 ) ≥ 0?

Test: Is ( x − 3)( x + 1) > 0?

Solution set  −4, 0  ,  4, ∞ )

Solution set: ( −∞, −1) ∪ ( 3, ∞ )

−4

−3 −2 −1

130.

x 3 − 16 x ≥ 0

133.

Test intervals: ( −∞, 0 ) , ( 0, 53 ) , ( 35 , ∞ ) Test: Is 4 x 2 ( 3 x − 5) < 0?

Test intervals: ( −∞, − 3) , ( −3, 9 ) , ( 9, ∞ )

Solution set: ( −∞, 0 ) , ( 0, 35 )

Test: Is ( x − 9 )( x + 3) < 0?

5 3

Solution set: ( −3, 9 )

x −2

x 6

135.

4 x − 23 x ≤ 6

131.

Key numbers: x = 0, x = 53

Key numbers: x = −3, x = 9

4 x 2 ( 3 x − 5) < 0

( x − 9 )( x + 3) < 0

134. 12 x 3 − 20 x 2 < 0

x 2 − 6 x − 27 < 0

−9 −6 −3

−2

4 x − 23 x − 6 ≤ 0

( x − 6 )( 4 x + 1) ≤ 0 Key numbers: x = 6, x = −

−1

1 + 3 > 0 x 1 + 3x > 0 x 1 Key numbers: x = − , x = 0 3

1 4

Test intervals: ( −∞, − 14 ) , ( − 14 , 6 ) , ( 6, ∞ )

1  1   Intervals:  − ∞, − ,  − , 0 , (0, ∞) 3  3  

Solution set:  − 14 , 6 

Test: Is

Test: Is ( x − 6 )( 4 x + 1) ≤ 0?

1  Solution set:  − ∞, −  ∪ (0, ∞) 3 

−1 4

x −1

1 + 3x > 0? x

−1 3 x

6x2 + 5x ≥ 4

132.

−2

−1

6 x2 + 5x − 4 ≥ 0 4 1 Key numbers: x = − , x = 3 2

4  4 1 1   Test intervals:  − ∞, − ,  − , ,  , ∞  3  3 2  2   Test: Is ( 3x + 4 )( 2 x − 1) ≥ 0? 4  1  Solution set:  −∞, −  ∪  , ∞  3  2  −4

−1

9 − x ≥ 0 x 9 − x2 ≥ 0 x (3 − x)(3 + x) ≥ 0 x

Key numbers: x = − 3, x = 0, x = 3 Test intervals: ( − ∞, − 3), ( − 3, 0), (0, 3), (3, ∞ )

1 2

−2

9 ≥ x x

136.

(3x + 4)(2 x − 1) ≥ 0

Test: Is 4 x

( 3 − x )( 3 + x ) ≥ 0?

x Solution set: ( − ∞, − 3] ∪ (0, 3] x

−4 −3 −2 −1

Chapter 2 Review

(0.1)(3.69) = 0.369

3x + 8 −4≤0 x −3 3 x + 8 − 4 ( x − 3) ≤0 ( x − 3)

141.

20 − x ≤0 x −3 x − 20 ≥0 x −3 Key numbers: x = 3, x = 20

142. h − 50 ≤ 30

≈ $0.37 per gallon You may be overcharged $0.37 × 15 gallons ≈ $5.55.

Test intervals: ( −∞, 3) , ( 3, 20 ) , ( 20, ∞ )

−30 ≤ h − 50 ≤ 30 20 ≤ h ≤ 80 Minimum 20, maximum 80 143. (a) Grade-point average

x − 20 ≥ 0? x−3 Solution set: ( −∞, 3) ∪ 20, ∞ ) Test: Is

3 x

138.

Exam score

(b) Yes, the relationship is approximately linear. Higher entrance exam scores, x, are associated with higher grade-point averages, y.

−5

16 − x

40 30 20 10 x

Distance bent (in centimeters)

(b) Yes. Answers will vary.

16 − x 2 ≥ 0

(4 − x)(4 + x) ≥ 0 Key numbers: x = − 4, 4 Test intervals: ( − ∞, − 4), ( − 4, 4), ( 4, ∞ )

(4 − x)(4 + x) ≥ 0 on [− 4, 4]

140.

5 10 15 20 25 30 35

145. (a) 139.

144. (a)

− x−2 < 0? x+5 Solution set: (-∞, − 5) ∪( −2, ∞) 0

1 x

Test: Is

−2

65 70 75 80 85 90 95

Test intervals: ( −∞, − 5 ) , ( −5, − 2 ) , ( −2, ∞ )

−4

x +8 −2<0 x+5 x + 8 − 2 ( x + 5) <0 x+5 −x − 2 <0 x+5 Key numbers: x = −5, x = −2

−6

Time to failure (in hours)

Speed (in meters per second)

137.

221

40 35 30 25 20 15 10 5 t 1

Time (in seconds)

Domain: [− 4, 4]

s ≈ 10t − 0.4 Approximations will vary. (c) s = 9.7t + 0.4; r ≈ 0.999

(d)

x 2 − 5 x − 14

(b)

s ( 2.5) = 9.7 ( 2.5) + 0.4 ≈ 24.7 m/sec

x 2 − 5 x − 14 ≥ 0

( x + 2)( x − 7) ≥ 0 Key numbers: x = − 2, 7 Test intervals: ( − ∞, − 2), ( − 2, 7), (7, ∞ )

( x + 2)( x − 7) ≥ 0 on (− ∞, − 2] ∪ [7, ∞) Domain: ( − ∞, − 2] ∪ [7, ∞ ) © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

222

Chapter 2

Solving Equations and Inequalities 151. They are the same. A point ( a, 0 ) is an x-intercept if it is a

146. (a) y = − 0.0089 x + 4.086

(b)

4.3

5 3.5

(c) Answers will vary. Sample answer: Yes, the model fits the data well. (d) Answers will vary. Sample answer: No, eventually the model will yield results including negative times that would be physically impossible.

solution point of the equation. In other words, a is a zero of the function. b 152. ax + b = 0. x = − . Then: a (a) If ab > 0, then x < 0. (b) If ab < 0, then x > 0.

( − 8)( − 8 ) .

In fact, − 8 − 8 = 8i 8i = − 8. −4 ≠ 4i. In fact,

154. The error is

147. False. A function can have only one y-intercept. (Vertical Line Test) 148. False. (1 + 2i ) + (1 − 2i ) = 2, a real number.

−8 −8 ≠

153. The error is

−i

( −4 − 1) = −i ( 2i − 1) = 2 + i.

( ) =1 =1 (b) i = i ( i ) = i (1) = i (c) i = i ( i ) = ( −1)(1) = −1 (d) i = i ( i ) = −i (1) = −i

155. (a)

i 40 = i 4 25

149. False. The slope can be positive, negative, or 0. 150. An identity is an equation that is true for every real number in the domain of the variable. A conditional equation is true for just some (or even none) of the real numbers in the domain.

Chapter 2 Test 1.

12 27 −7=− +6 x x 39 = 13 x 39 = 13 x 3= x x=3

8 + 5i 8 + 5i − i = ⋅ i i −i = =

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

(

12 x + 8 − 27 x + 18 x = −3 9 x − 4

)

− 8i − 5i 2 − i2 − 8i − 5( −1) − ( −1)

= 5 − 8i

( 2i − 1) ⋅ 2 − 3i = 6 − 2 + 4i + 3i 4+9 ( 3i + 2 ) 2 − 3i

−27 x 2 + 30 x + 8 = −27 x 2 + 12

30 x = 4 x=

9. f ( x ) = 3 x 2 − 6 = 0

2 15

x ≈ ±1.414

3. ( −8 − 3i ) + ( −1 − 15i ) = −9 − 18i

(

) (

2 −6

)

4 7 + i 13 13

4. 10 + −20 − 4 − −14 = 6 + 2 5i + 14 i

(

)

= 6 + 2 5 + 14 i

−6

5. ( 2 + i )( 6 − i ) = 12 + 6i − 2i + 1 = 13 + 4i 6. ( 4 + 3i ) − ( 5 + i ) = (16 + 24i − 9) − ( 25 + 10i − 1) 2

= −17 + 14i

Chapter 2 Test 10. f ( x ) = 8 x 2 − 2 = 0

16. 5 x 2 + 7 x + 6 = 0

x = ±0.5 1 3

11.

−7 ±

f ( x ) = x3 − 4 x2 + 5x = 0 x=0 6

(7) − 4(5)(6) 2(5) 2

49 − 120 10

− 7 ± − 71 10

− 7 ± 71i 10

3 x3 − 4 x 2 − 12 x + 16 = 0

17. −4

x ( 3x − 4 ) − 4 ( 3x − 4 ) = 0 2

( x − 4 ) ( 3x − 4 ) = 0

−2

12.

− ( 7) ±

−3

b 2 − 4ac 2a

−b ±

x =

−3

f ( x ) = x − x3 = 0

x 2 − 4 = 0  x 2 = 4 = ±2

x = 0, ± 1

3x − 4 = 0  x =

−6

223

18.

4 3

x + 22 − 3 x = 6 22 − 3 x = 6 − x 22 − 3 x = ( 6 − x )

−4

22 − 3 x = 36 − 12 x + x 2

13. x 2 − 15 x + 56 = 0

x 2 − 9 x + 14 = 0

( x − 7)( x − 8) = 0

( x − 2 )( x − 7 ) = 0

x −7 = 0  x = 7

x−2=0 x=2 x − 7 = 0  x = 7, extraneous

x −8 = 0  x = 8

14.

x2 + 12 x − 2 = 0 x=

− b ± b 2 − 4ac 2a

19.

−12 ± 12 − 4 ( −2 )

2 −12 + 152 = 2 = −6 ± 38 2

15. 4 x − 81 = 0 2

4 x = 81 81 x2 = 4 9 x=± 2

( x + 6 ) = 16 23

x 2 + 6 = 163 2 = 64

x 2 = 58 x = ± 58 ≈ ±7.616 20.

8 x − 1 = 21 8 x − 1 = 21 or − ( 8 x − 1) = 21 8 x = 22

−8 x = 20

x = − 52

11 4

21. 6 x − 1 > 3 x − 10 3x > − 9 x>−3

( − 3, ∞ ) x −4 −3 −2 −1

224

Chapter 2

Solving Equations and Inequalities

22. 2 x − 8 < 10

24.

x −8 < 5 −5 < x − 8 < 5 3< x

< 13 x

3 4 5 6 7 8 9 10 11 12 13

23.

6 x2 + 5x + 1 ≥ 0

( 3 x + 1)( 2 x + 1) ≥ 0

8 − 5x ≤ −2 2 + 3x 8 − 5x + 2 ≤ 0 2 + 3x 8 − 5 x + 2( 2 + 3x) ≤ 0 2 + 3x x + 12 ≤ 0 2 + 3x Key numbers: x = −12, x = −

Key numbers: x = − , x = − 1 3

1 2

2 3

1  1 1  1   Test intervals:  −∞, −  ,  − , −  ,  − , ∞  2 2 3 3      

2  2   Test intervals: ( − ∞, −12),  −12, − ,  − , ∞  3  3  

Test: Is ( 3 x + 1)( 2 x + 1) ≥ 0?

Test: Is

Solution set: ( −∞, − 12 ) , ( − 13 , ∞ )

2  Solution set:  − ∞, −  3 

−1 −1 2

x + 12 ≤ 0? 2 + 3x

2 3

x −2

−1

− 12 − 10 − 8 − 6 − 4 − 2

25.

−16t 2 + 224t > 350 −16t 2 + 224t − 350 > 0

Key numbers: t =

− ( 224) ±

(224)2 − 4(−16)(− 350) 2( −16)

t ≈ 1.8, t ≈ 12.2 Test intervals: ( − ∞, 1.8), (1.8, 12.2), (12.2, ∞ ) Test: Is −16t 2 + 224t − 350 > 0? −16t 2 + 224t − 350 > 0 on (1.8, 12.2)

The projectile exceeds 350 feet over the interval (1.8 seconds, 12.2 seconds). 26.

C = 2.485t + 30.36 85 = 2.485t + 30.36 54.64 = 2.485t t ≈ 21.99 In 2021, the average monthly cost of table will be $85.

Chapters P–2 Cumulative Test 1.

14 x 2 y −3 7 x3 , x≠0 = −1 2 32 x y 16 y5

2. 8 60 − 2 135 − 15 = 16 15 − 6 15 − 15 = 9 15

( x − 2 ) ( x 2 + x − 3) = x 3 + x 2 − 3 x − 2 x 2 − 2 x + 6 = x3 − x 2 − 5x + 6

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

28 x 4 y 3 = 2 x 2 y 7 y

x −1

( x + 3)( x + 1)

4. 4 x − 2 x + 5 ( 2 − x )  = 4 x −  −3 x + 10 

= −10 + 7 x = 7 x − 10

Chapters P–2 Cumulative Test 7. 36 − ( x − 4) = 6 − ( x − 4) 6 + ( x − 4) 2

14.

225

y= 4−x

= [6 − x + 4][6 + x − 4]

= ( − x + 10)( x + 2)

5 4 3

= − ( x − 10)( x + 2)

(

)

x − 5x − 6 x = − x 6 x + 5x − 1

= x ( x + 1)(1 − 6 x )

(

)

15. (a)

= 2 (3 − 2 x ) 9 + 6 x + 4 x2

)

y = −x + 3 x+y=3

(b) Three additional points: ( −1, 4 ) , ( 0, 3 ) , (1, 2 )

 1  = − , − 2  2 

(Answers not unique.) 2

 5  7  2 Distance =  −  −   + ( −8 − 4 )  2  2 

16. (a)

= 36 + 144 = 180

(b) Three additional points: ( 0, 0 ) , (1, − 2 ) , ( 2, − 4 ) (Answers not unique.)

 1  11. Center:  − , 8  = ( h, k )  2  Radius: r = 4

17. (a) Vertical line: x = − 37 or x + 37 = 0

(b) Three additional points: ( − 37 , 0 ) , ( − 37 , 1) , ( − 37 , 2 )

2 1   x +  + ( y + 8 ) = 16 2 

(Answers not unique.)

x − 3 y + 12 = 0 −3 y = − x − 12 x y= +4 3

18. 6 x − y = 4 has slope m = 6. (a) Parallel line has slope m = 6.

y − 3 = 6 ( x − 2) y = 6x − 9

1 (b) Perpendicular line has slope m = − . 6

8 7 6 5

y − 3 = − 61 ( x − 2 ) y = − 61 x + 103

3 2 1

−5 −4 −3 −2 −1

19.

1 2 3 4 5

−2

y = x2 − 9 y 2 1

− 6− 5− 4

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

= 6 5 ≈ 13.42

13.

8−4 = −1 −5 − ( −1)

Slope =

y − 8 = −1( x + 5 )

( ( −7 2 ) + ( 5 2 ) , 4 + ( −8) ) 10. Midpoint =

12.

1 2 3 4 5

−2 −3 −4 −5

= − x ( 6 x − 1)( x + 1)

9. 54 − 16 x 3 = 2 27 − 8 x 3

−5 −4 −3 −2 −1

−1 −2 −3 −4 −5 −6

x 1 2

4 5 6

f ( x) =

x x−2 5 5 = 5−2 3

(a)

f ( 5) =

(b)

f ( 2 ) is undefined.

(c)

f ( 5 + 4s ) =

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

− 10

226

Chapter 2

20.

3 x − 8, x < 0 f ( x) =  2  x + 4, x ≥ 0

21.

Solving Equations and Inequalities 26. No, for some x-values, correspond two values of y.

(a)

f ( −8 ) = 3 ( −8 ) − 8 = −32

(b)

f ( 0 ) = 02 + 4 = 4

(c)

f ( 4 ) = 4 + 4 = 20

27.

− 25

− 15

Decreasing on ( −∞, 5) , increasing on ( 5, ∞ )

( −∞, ∞ )

22. 5 + 7t ≥ 0 7t ≥ −5

28. (a) r ( x) =

1 2

(b) h( x) =

t ≥ − 75

−  , ∞)

x =

1 f ( x) is a vertical shrink of f. 2

x + 2 = f ( x) + 2 a vertical shift two

units upward of f.

5 7

23. 9 − s 2 ≥ 0

x + 2 = − f ( x + 2) is a horizontal shift

two units to the left, followed by a reflection in the x-axis of f.

9 ≥ s2  −3, 3 24.

29. ( f + g )( x) = ( x 2 + x) + (3x − 2) = x 2 + 4 x − 2

( − ∞, − 53 ) ∪ ( − 52 , ∞ )

25. g ( − x ) = 3 ( − x ) − ( − x )

= −3 x + x 3 = − g ( x )

Odd function. 30. ( g − f )( x) = ( x 2 + x)(3x − 2) − ( x 2 + x) = − x 2 + 2 x − 2

(

)

31. ( g  f )( x) = g ( f ( x)) = 3 x 2 + x − 2 = 3x 2 + 3x − 2 32. ( fg )( x) = ( x 2 + x)(3x − 2) = 3x3 − 2 x 2 + 3 x 2 − 2 x = 3x3 + x 2 − 2 x 33.

f ( x) = − 5 x + 4 has an inverse function.

36. 4 x 3 − 12 x 2 + 8 x = 0 4 x ( x 2 − 3 x + 2) = 0

y = −5x + 4 x = −5 y + 4

4 x( x − 2)( x − 1) = 0

x − 4 = −5 y

4x = 0  x = 0

x −4 y = −5

x − 2 = 0  x = 2

1 4 f ( x) = − x + 5 5

x −1 = 0  x =1

−1

34. f ( x) = ( x − 1) is not one-to-one, so f −1( x) does 2

not exist. 35.

37.

5 10 = x x −3 5( x − 3) = 10 x 5 x − 15 = 10 x

f ( x) = 3 x + 2 has an inverse function. y = 3 x + 2

− 5 x = 15 x = −3

x = 3 y + 2 x −2 = 3 y y = ( x − 2)

f −1 ( x) = ( x − 2)

Chapters P–2 Cumulative Test 38.

3x + 4 − 2 = 0

42.

3x + 4 = 2

39.

or 3 x + 4 = 2

3x = − 6

3x = − 2

x = −2

x = −

x>−

2 3

( x + 1) = (− x + 9) 2

x 2 + 1 = x 2 − 18 x + 81 18 x = 80 40 x = 9

x<−

3 2

2 ( x − 2)

≤0 x +1 Key numbers: x = −1, 2 Test intervals: ( −∞, − 1) , ( −1, 2 ) , ( 2, ∞ ) 2 ( x − 2)

≤ 0? x +1 Solution set: ( − 1, 2

44. V =

4 3 3 V πr  r = 3 3 4π 3 ( 370.7 ) 4π ≈ 4.456 inches =3

( − ∞, 1207 

45. (a) Let x and y be the lengths of the sides. 2 x + 2 y = 546  y = 273 − x

120 7 x

A = xy = x ( 273 − x )

2 x 2 + x ≥ 15

41.

43.

Test: Is

x −x −6≤ +6 5 2 x x + ≤ 12 2 5 7x ≤ 12 10 120 x≤ 7

1 4

x −4 −3 −2 −1

x2 + 1 = − x + 9

40.

or 7 + 8 x < −5 8x < −12

( − ∞, − 32 ) ∪ ( − 14 , ∞ )

x2 + 1 + x − 9 = 0

7 + 8x > 5 7 + 8x > 5 8 x > −2

3x + 4 = − 2

227

(b)

25,000

2 x + x − 15 ≥ 0

( 2 x − 5)( x + 3) ≥ 0 0

5 2

Key numbers: x = −3,

5 5   Test intervals: ( −∞, − 3) ,  −3,  ,  , ∞  2 2     Test: Is ( 2 x − 5)( x + 3) ≥ 0?

273

Domain: 0 < x < 273 (c) If A = 15,000, then x = 76.23 or 196.77. Dimensions in feet: 76.23 × 196.77 or 196.77 × 76.23

5  Solution set: ( −∞, − 3 ∪  , ∞  2  5 2 x −4 −3 −2 −1

Chapter 2

Solving Equations and Inequalities

6000 5000 4000 3000 2000 1000 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013

46. (a)

Profit (in millions of dollars)

228

Year

McDonald’s profits appear to be increasing at a fairly constant rate. (b) P = 403.05t + 722.69; r ≈ 0.9811 (c)

7000

(d) 2015: P(15) = 403.05(15) + 722.69 = $6768.44 million

2018: P(18) = 403.05(18) + 722.69 = $7977.59 million (e) Answers will vary. Sample answer: Yes, if the net profits continue to follow the model, it can be used to predict future years.

C H A P T E R 3 Polynomial and Rational Functions Section 3.1

Quadratic Functions............................................................................230

Section 3.2

Polynomial Functions of Higher Degree ...........................................240

Section 3.3

Real Zeros of Polynomial Functions .................................................252

Section 3.4

The Fundamental Theorem of Algebra .............................................267

Section 3.5

Rational Functions and Asymptotes ..................................................275

Section 3.6

Graphs of Rational Functions ............................................................282

Section 3.7

Quadratic Models ...............................................................................297

Chapter 3 Review .......................................................................................................302 Chapter 3 Test ............................................................................................................317

C H A P T E R 3 Polynomial and Rational Functions Section 3.1 Quadratic Functions 1.

nonnegative integer, real

quadratic, parabola

Yes, f ( x ) = ( x - 2)2 + 3 is in the form

11. 4 3 2

f ( x ) = a( x - h) + k. The vertex is (2, 3).

f ( x ) = ( x - 2)2 opens upward and has vertex (2, 0). Matches graph (c).

f ( x ) = 3 - x 2 opens downward and has vertex (0, 3). Matches graph (d).

−6 −5 −4 − 3 −2 −1 −1

No, the graph of f ( x ) = -3 x 2 + 5 x + 2 is a parabola opening downward, therefore there is a relative maximum at its vertex.

1 1

−2

The graph of y = ( x + 3)2 is a horizontal shift three units to the left of y = x 2 . y

12. 1 −6 −5 −4 −3 −2 −1 −1

f ( x ) = x 2 + 3 opens upward and has vertex (0, 3). Matches graph (b).

−2 −3 −4

f ( x ) = -( x - 4)2 opens downward and has vertex (4, 0). Matches graph (a).

The graph of y = -( x + 3)2 - 1 is a reflection in the x-axis, a horizontal shift three units to the left, and a vertical shift one unit downward of y = x 2 .

2 1 −4 −3 −2

13.

−2

−3

−4

−5

−6

3 2

The graph of y = - x 2 is a reflection of y = x 2 in the x-axis.

1 − 5 −4 −3 −2 −1 −1

10.

−2

The graph of y = ( x + 1)2 is a horizontal shift one unit to

4 3

the left of y = x 2 .

2 1 −4 −3 −2 −1

14. 4 3

−2 −3

1 2

The graph of y = x - 1 is a vertical shift downward one unit of y = x 2 .

−4 −3 −2

−1

−2 −3 −4

The graph of y = - x 2 + 2 is a reflection in the x-axis and a vertical shift two units upward of y = x 2 .

230

Section 3.1 15.

Quadratic Functions

231

23. h( x) = x 2 − 2 x + 1

= ( x − 1)

5 4

A parabola opening upward with vertex (1, 0)

3 2 1 −2 − 1 −1

24. g ( x) = x 2 + 16 x + 64

= ( x + 8)

−2

A parabola opening upward with vertex ( − 8, 0)

The graph of y = ( x - 3)2 is a horizontal shift three units to the right of y = x 2 .

25.

f ( x) = x 2 - x +

16. 2

5 4 æ 1ö 5 1 = çç x 2 - x + ÷÷ + çè 4 ÷ø 4 4

= ( x 2 - x) +

1 −2 −1 −1

5 4

−2 −3

æ 1ö = çç x - ÷÷÷ + 1 èç 2ø

−4 −5 −6 2

The graph of y = -( x - 3) + 1 is a horizontal shift three units to the right, a reflection in the x-axis, and a vertical shift one unit upward of y = x 2 . 17. f ( x) = 20 − x 2

= − x 2 + 20 A parabola opening downward with vertex (0, 20)

æ1 ö A parabola opening upward with vertex çç , 1÷÷ çè 2 ÷ø

26. f ( x ) = x 2 + 3 x +

1 4

1 4 æ 2 ö 1 9 9 = çç x + 3 x + ÷÷÷ + èç 4ø 4 4 = ( x 2 + 3x) +

æ 3ö = çç x + ÷÷ - 2 çè 2 ÷ø

18. f ( x) = x + 8 2

A parabola opening upward with vertex (0, 8) 19. f ( x) =

1 2 x −5 2

A parabola opening upward with vertex (0, − 5) 20. f ( x) = − 6 −

1 2 x 4

1 = − x2 − 6 4 A parabola opening downward with vertex (0, − 6) 21. f ( x) = ( x + 3) − 4 2

A parabola opening upward with vertex ( − 3, − 4)

æ 3 ö A parabola opening upward with vertex çç- , - 2÷÷ çè 2 ÷ø

27. f ( x ) = -x 2 + 2 x + 5 = -( x 2 - 2 x ) + 5 = -( x 2 - 2 x + 1) + 5 + 1

= -( x - 1)2 + 6 A parabola opening downward with vertex (1, 6)

28. f ( x ) = - x 2 - 4 x + 1 = -( x 2 + 4 x ) + 1 = -( x 2 + 4 x + 4) + 1 + 4 = -( x + 2)2 + 5 A parabola opening downward with vertex (- 2, 5)

22. f ( x) = ( x − 7) + 2 2

A parabola opening upward with vertex (7, 2)

232

Chapter 3

Polynomial and Rational Functions

29.

h( x ) = 4 x 2 - 4 x + 21

32.

= 4( x - x ) + 21 æ æ1ö 1ö = 4 çç x 2 - x + ÷÷÷ + 21 - 4 çç ÷÷÷ èç èç 4 ø 4ø

= ( x 2 + 10 x + 25) + 14 − 25 = ( x + 5)2 − 11

æ 1ö = 4 çç x - ÷÷ + 20 çè 2 ÷ø

x 2 + 10 x + 14 = 0 x=

æ1 ö A parabola opening upward with vertex çç , 20÷÷÷ èç 2 ø

30.

f ( x) = 2 x2 - x + 1 æ 1 ö = 2 çç x 2 - x ÷÷÷ + 1 èç 2 ø

= −5 ± 11 A parabola opening upward with vertex (- 5, - 11) and

x-intercepts (−5 ± 11, 0) 33.

= −( x 2 − 2 x + 1) + 1 + 15

= ( x + 8 x ) + 11 2

= ( x + 8 x + 16) + 11 - 16

= −( x − 1) + 16 2

= ( x + 4)2 - 5

−( x 2 − 2 x − 15) = 0

x 2 + 8 x + 11 = 0

−( x + 3)( x − 5) = 0

-8  82 - 4(1)(11) 2(1)

x + 3 = 0  x = −3 x −5 = 0  x = 5

x-intercepts (−4 ± 5, 0).

f ( x) = −( x 2 − 2 x − 15)

= −( x 2 − 2 x + 1 − 1) + 15

31. g( x ) = x 2 + 8 x + 11

-8  64 - 44 2 -8  20 = 2 -8  2 5 = 2 = -4  5 A parabola opening upward with vertex (- 4, - 5) and

−10 ± 100 − 56 2

−10 ± 44 2 −10 ± 2 11 = 2

æ 1ö 7 = 2 çç x - ÷÷÷ + èç 4ø 8

−10 ± 10 2 − 4(1)(14) 2(1)

æ æ1ö 1 1ö = 2 çç x 2 - x + ÷÷ + 1 - 2 çç ÷÷ çè çè 16 ÷ø 2 16 ÷ø

æ1 7ö A parabola opening upward with vertex çç , ÷÷÷ èç 4 8 ø

f ( x ) = x 2 + 10 x + 14 = ( x 2 + 10 x ) + 14

A parabola opening downward with vertex (1, 16) and x-intercepts ( − 3, 0) and (5, 0) 34.

f ( x) = −( x 2 + 3x − 4) 9 9  = − x 2 + 3x + −  + 4 4 4  9 9  = − x 2 + 3x +  + + 4 4 4  9  25  = − x 2 + 3x +  + 4 4  2

3 25  = − x +  + 2 4  −( x 2 + 3x − 4) = 0 −( x + 4)( x − 1) = 0 x + 4 = 0  x = −4 x −1 = 0  x =1

 3 25  A parabola opening downward with vertex  − ,   2 4 and x-intercepts ( − 4, 0) and (1, 0)

Section 3.1 35.

= −2( x 2 − 8 x ) − 31 = −2( x 2 − 8 x + 16) − 31 + 32 = −2( x − 4) 2 + 1

4 = 4a 1 = a.

2( −2)

Thus, f ( x ) = ( x + 2) 2 − 1.

−16 ± 256 − 248 −4 −16 ± 8 −4 −16 ± 2 2

= 4±

−4 1

 

1 2

 

2, 0 

36. f ( x ) = -4 x 2 + 24 x - 41

= -4( x 2 - 6 x) - 41 = -4( x 2 - 6 x + 9) - 41 + 36 = -4( x - 3) 2 - 5 -4 x 2 + 24 x - 41 = 0 2

39. ( − 2, 5) is the vertex, so f ( x) = a( x + 2) 2 + 5. Since the graph passes through the point (0, 9),

9 = a(0 + 2) 2 + 5 4 = 4a 1 = a. Thus, f ( x ) = ( x + 2)2 + 5.

2 2 A parabola opening downward with vertex (4, 1) and x-intercepts  4 ±

3 = a (0 + 2) 2 − 1

−16 ± 16 − 4( −2)( −31)

f ( x ) = a( x + 2)2 − 1 Since the graph passes through (0, 3),

3 = 4a − 1

− 2 x 2 + 16 x − 31 = 0

233

38. ( −2, − 1) is the vertex.

f ( x ) = −2 x 2 + 16 x − 31

Quadratic Functions

-24  24 - 4(-4)(-41) 2(-4)

-24  576 - 656 -8 -24  -80 = -8 =

No real solution A parabola opening downward with vertex (3, - 5) and no x-intercepts 37. (- 1, 4) is the vertex. f ( x ) = a( x + 1)2 + 4

Since the graph passes through the point (1, 0),

0 = a(1 + 1) 2 + 4

40. (4, 1) is the vertex, so f ( x) = a ( x − 4) 2 + 1. Since the graph passes through the point (6, − 7), −7 = a (6 − 4) 2 + 1 −7 = 4 a + 1 −8 = 4 a −2 = a.

Thus, f ( x ) = −2( x − 4)2 + 1. 41. (1, − 2) is the vertex, so f ( x) = a ( x − 1) 2 − 2. Since the graph passes through the point ( −1, 14), 14 = a ( −1 − 1) 2 − 2 14 = 4a − 2 16 = 4a 4 = a.

Thus, f ( x ) = 4( x − 1)2 − 2. 42. ( −4, − 1) is the vertex, so f ( x) = a ( x + 4) 2 − 1. Since the graph passes through the point ( −2, 4), 4 = a (−2 + 4) 2 − 1 5 = 4a 5 a= . 4

5 Thus, f ( x ) = ( x + 4)2 − 1. 4

0 = 4a + 4 −1 = a . Thus, f ( x ) = −( x + 1)2 + 4. Note that ( −3, 0) is on the parabola.

234

Chapter 3

Polynomial and Rational Functions 2

1 1   43.  , 1  is the vertex, so f ( x) = a  x −  + 1. 2 2   

21   Since the graph passes through the point  −2, −  , 5  2

21 1  = a  −2 −  + 1 5 2  21 25 − = a +1 5 4 26 25 − = a 5 4 104 − = a. 125 −

0 = x2 − 6x + 9 0 = ( x − 3)2 x=3 49. y = x 2 − 4 x 3

−4

−5

x-intercepts: (0, 0), (4, 0)

Thus, f ( x ) = −

48. y = x 2 − 6 x + 9 x-intercept: (3, 0)

104  1  x −  + 1. 125  2

0 = x2 − 4 x 2

1  1   44.  − , − 1  is the vertex, so f ( x) = a  x +  − 1. 4 4   

17   Since the graph passes through the point  0, −  ,  16 

0 = x ( x − 4) x = 0 or x = 4

50. y = −2 x 2 + 10 x 15

17 1  = a  0 +  −1 16 4  17 1 − = a −1 16 16 1 1 − = a 16 16 a = −1. −

−9

15 −1

x-intercepts: (0, 0), (5, 0)

1  Thus, f ( x ) = −  x +  − 1. 4 

45. y = x 2 − 4 x − 5 x-intercepts: (5, 0), ( − 1, 0)

0 = −2 x 2 + 10 x 0 = x( −2 x + 10) x = 0, x = 5 51. y = 2 x 2 − 7 x − 30 5 − 20

0 = x2 − 4 x − 5 0 = ( x − 5)( x + 1) x = 5 or x = −1

46. y = 2 x 2 + 5 x − 3

1  x-intercepts:  , 0  , ( −3, 0 ) 2  0 = 2 x 2 + 5x − 3 0 = (2 x − 1)( x + 3) 1 x = , −3 2

− 40

 5  x-intercepts:  − , 0  , (6, 0)  2  0 = 2 x 2 − 7 x − 30 0 = (2 x + 5)( x − 6) 5 x = − or x = 6 2

47. y = x 2 + 8 x + 16 x-intercept: ( −4, 0)

0 = x 2 + 8 x + 16 0 = ( x + 4)2 x = −4

Section 3.1 56.

52. y = 4 x 2 + 25 x − 21 10 −8

235

f ( x ) = x( x − 10) = x 2 − 10 x, opens upward. g( x ) = − x ( x − 10) = − x 2 + 10 x, opens downward.

Note: f ( x ) = ax ( x − 10) has x-intercepts (0, 0) and (10, 0) for all real numbers a ≠ 0.

57.

− 70

x-intercepts: ( −7, 0), (0.75, 0)

  1  f ( x ) = [ x − ( −3)]  x −  −   (2), opens upward   2  1  = ( x + 3)  x +  (2) 2  = ( x + 3)(2 x + 1)

0 = 4 x 2 + 25 x − 21 = ( x + 7)(4 x − 3) 3 x = −7, 4 1 2 53. y = − ( x − 6 x − 7) 2

= 2 x2 + 7x + 3

g( x ) = −(2 x 2 + 7 x + 3), opens downward

= −2 x 2 − 7 x − 3 Note: f ( x ) = a ( x + 3)(2 x + 1) has x-intercepts ( −3, 0)

 1  and  − , 0  for all real numbers a ≠ 0.  2 

−6

58.

−3

x-intercepts: ( −1, 0), (7, 0) 1 0 = − ( x 2 − 6 x − 7) 2 0 = x2 − 6 x − 7 0 = ( x + 1)( x − 7) x = −1, 7 54. y =

Quadratic Functions

7 2 ( x + 12 x − 45) 10 10

− 18

− 60

x-intercepts: (3, 0), ( − 15, 0) 7 0 = ( x 2 + 12 x − 45) 10 0 = x 2 + 12 x − 45 = ( x − 3)( x + 15) x = 3, − 15 55. f ( x ) = [ x − ( −1)]( x − 3), opens upward = ( x + 1)( x − 3)

= x2 − 2 x − 3 g ( x ) = −[ x − ( −1)]( x − 3), opens downward

  5  f ( x ) = 2  x −  −   ( x − 2)   2  5  = 2  x +  ( x − 2) 2  = 2 x 2 + x − 10, opens upward g ( x ) = − f ( x ), opens downward g( x ) = −2 x 2 − x + 10

5   5  Note: f ( x ) = a  x +  ( x − 2) has x-intercepts  − , 0  2    2  and (2, 0) for all real numbers a ≠ 0. 59. Let x = the first number and y = the second number. Then the sum is x + y = 110  y = 110 − x. The product is P ( x ) = xy = x (110 − x ) = 110 x − x 2. P ( x ) = − x 2 + 110 x

= −( x 2 − 110 x + 3025 − 3025) = −[( x − 55)2 − 3025] = −( x − 55)2 + 3025 The maximum value of the product occurs at the vertex of P ( x ) and is 3025. This happens when x = y = 55. 60. Let x = first number and y = second number. Then x + y = 66 or y = 66 − x. The product P is given by P ( x ) = xy. P ( x ) = x (66 − x ) = 66 x − x 2

= −( x + 1)( x − 3)

= −( x 2 − 66 x )

= −( x 2 − 2 x − 3)

= −( x 2 − 66 x + 332 ) + 332

= −x + 2x + 3 Note: f ( x ) = a ( x + 1)( x − 3) has x-intercepts ( −1, 0) and (3, 0) for all real numbers a ≠ 0.

= −( x − 33)2 + 1089 The maximum P occurs at the vertex where x = 33, so y = 66 − 33 = 33. Therefore, the two numbers are 33 and 33.

236

Chapter 3

Polynomial and Rational Functions

61. Let x be the first number and y be the second number. Then x + 2 y = 24  x = 24 − 2 y. The product is

(e) The area is maximum when x = 50 and 200 − 2(50) 100 y= = .

P = xy = (24 − 2 y ) y = 24 y − 2 y 2. Completing the square,

2000

P = −2 y 2 + 24 y = −2( y 2 − 12 y + 36) + 72 = −2( y − 6)2 + 72.

The maximum value of the product P occurs at the vertex of the parabola and equals 72. This happens when y = 6 and x = 24 − 2(6) = 12.

64. (a)

62. Let x = first number and y = second number. Then

1 x + 3 y = 42, y = (42 − x ). The product is 3 1 1 P( x ) = xy = x (42 − x ) = 14 x − x 2. 3 3 1 2 P ( x ) = − x + 14 x 3 1 2 = − ( x − 42 x ) 3 1 2 = − ( x − 42 x + 441) + 147 3 1 = − ( x − 21)2 + 147. 3 The maximum value of the product is 147, and occurs 1 when x = 21 and y = (42 − 21) = 7. 3 63. (a)

100

1 4 x + 3 y = 200  y = (200 − 4 x ) 3 1 8x  A = 2 xy = 2 x (200 − 4 x ) = (50 − x ) 3 3

(b) x

Area

1 [ 200 − 4(2)] 3

2 xy = 256

1 [ 200 − 4(4)] 3

2 xy ≈ 491

1 [ 200 − 4(6)] 3

2 xy = 704

1 [ 200 − 4(8)] 3

2 xy = 896

1 [ 200 − 4(10)] 3

2 xy ≈ 1067

1 [ 200 − 4(12)] 3

2 xy = 1216

Area

1 [ 200 − 4(20)] 3

2 xy = 1600

1 [ 200 − 4(22)] 3

2 xy ≈ 1643

1 [ 200 − 4(24)] 3

2 xy = 1664

1 [ 200 − 4(26)] 3

2 xy = 1664

1 [ 200 − 4(28)] 3

2 xy ≈ 1643

1 [ 200 − 4(30)] 3

2 xy = 1600

1 (b) Radius of semicircular ends of track: r = y 2 Distance around two semicircular parts of track: 1  d = 2π r = 2π  y  = π y 2  (c) Distance traveled around track in one lap: d = π y + 2 x = 200 π y = 200 − 2 x y=

200 − 2 x

 200 − 2x  (d) Area of rectangular region: A = xy = x   π  

Maximum area when x = 25, y = 33

1 3

Section 3.1

(c)

8 x(50 − x) 3

66.

2000

Maximum when x = 25, y = 33

1 3

8 x(50 − x ) 3 8 = − ( x 2 − 50 x ) 3 8 2 = − ( x − 50 x + 625 − 625) 3 8 = − [( x − 25)2 − 625] 3 8 5000 = − ( x − 25)2 + 3 3

(d)

5000 3 square feet. This happens when x = 25 feet and (200 − 4(25)) 100 = y= feet. The dimensions are 3 3 1 2 x = 50 feet by 33 feet. 3 (e) The result are the same. The maximum area occurs at the vertex and is

65. (a)

120

250

3 (b) When x = 0, y = feet. 2 (c) The vertex occurs at −b −9 / 5 3645 = = ≈ 113.9. x= 2a 2(−16 / 2025) 32 The maximum height is 2

−16  3645  9  3645  3   +  + 2025  32  5  32  2 ≈ 104.0 feet. (d) Using a graphing utility, the zero of y occurs at x ≈ 228.6, or 228.6 feet from the punter. y=

237

4 24 y = − x2 + x + 12 9 9 The maximum height of the dive occurs at the vertex, 24 −b =− 9 = 3. x= 2a  −4  2   9 

Quadratic Functions

4 24 (3) + 12 = 16. The height at x = 3 is − (3)2 + 9 9 The maximum height of the dive is 16 feet. 67. (a)

A = lw 100 − 2x x−6

A = (100 − 2 x )( x − 6) A = −2 x 2 + 112 x − 600 (b)

Y1 = −2 x 2 + 112 x − 600

950

960

966

968

966

960

The area is maximum when x = 28 inches. 68. R( p) = −10 p 2 + 1580 p

(a) When p = $50, R (50) = $54,000. When p = $70, R (70) = $61,600. When p = $90, R (90) = $61,200. (b) The maximum R occurs at the vertex, −b p= 2a −1580 = $79 p= 2( −10) (c) When p = $79, R (79) = $62, 410.

238

Chapter 3

69. (a)

Polynomial and Rational Functions 75. The parabola opens downward and the vertex is ( −2, − 4) . Matches (c) and (d).

4500

76. The parabola opens upward and the vertex is (1, 3) . Matches (a).

(b) Using the graph, during 1963 the maximum average annual consumption of cigarettes appears to have occurred and was 4110 cigarettes per person. (c) Yes, the warning had an effect because the maximum consumption occurred in 1963 and consumption decreased from then on. (d) In 2010, C (60) = 1057 cigarettes per person.

1057 ≈ 3 cigarettes per day. 365 70. (a) According to the model, t = t =

78. The graph of f ( x ) = x 2 − z would be a vertical shift

z units downward of g( x ) = x 2. 79. The graph of f ( x ) = z( x − 3)2 would be a vertical stretch ( z > 1) and horizontal shift three units to the right of

to the right of g( x ) = x 2.

−b or 2a

80. The graph of f ( x ) = zx 2 + 4 would be a vertical stretch ( z > 1) and vertical shift four units upward of g( x ) = x 2.

The graph of f ( x ) = zx 2 + 4 would be a vertical shrink (0 < z < 1) and vertical shift four units upward of

P (3) ≈ 82,430,000

So, the population in 2003 was about 82,430,000 people. (b) When t = 75, P(75) = 6,106,000. No, the population of Germany would not be expected to decrease this much. 71. True.

g( x ) = x 2. 2

b   b2   81. For a < 0, f ( x ) = a  x +  is a maximum  + c − 2a   4a  

when x =

c−

−12 x 2 − 1 = 0

−b . In this case, the maximum value is 2a

b2 . Hence, 4a

25 = −75 −

12 x 2 = −1, impossible

72. True. The graph of f ( x) = a( x − 5) , where a ≠ 0, is a 2

parabola with vertex (5, 0). Because the vertex is the x-intercept, the function has exactly one x-intercept. 73. False, The standard form of the function

b2 4( −1)

−100 = 300 − b2 400 = b2 b = ±20. 2

f ( x) = 3x2 + 6x + 7 is f ( x) = 3( x + 1) + 4 with 2

vertex ( − 1, 4). The standard form of the function

f ( x) = 3x + 6x − 1 is f ( x) = 3( x + 1) − 4 with 2

vertex ( − 1, − 4). So, the functions do not have the same vertex.

−b −10 10 5 =− =− =− . 2a 2(−4) 8 4 −b −30 −30 −5 = = = . For g( x ), 2a 2(12) 24 4 5 In both cases, x = − is the axis of symmetry. 4

74. True. For f ( x ),

units to the right of g( x ) = x 2 .

g( x ) = x 2. The graph of f ( x ) = z( x − 3)2 would be a vertical shrink (0 < z < 1) and horizontal shift three units

− 95.9 ≈ 3 or 2003. 2( −14.82)

77. The graph of f ( x ) = ( x − z )2 would be a horizontal shift z

b   b2   82. For a < 0, f ( x ) = a  x +  is a maximum  + c − 2a   4a  

when x =

c−

−b . In this case, the maximum value is 2a

b2 . Hence, 4a

48 = −16 −

b2 4(−1)

−192 = 64 − b2 b2 = 256 b = ±16.

Section 3.1 2

b   b2   83. For a > 0, f ( x ) = a  x +  is a minimum  + c − 2a   4a  

when x =

c−

87.

−b . In this case, the minimum value is 2a

b2 . Hence, 4a

Quadratic Functions

239

y = ax 2 + bx − 4 (1, 0) on graph: 0 = a + b − 4 (4, 0) on graph: 0 = 16 a + 4 b − 4 From the first equation, b = 4 − a. Thus, 0 = 16 a + 4(4 − a ) − 4 = 12 a + 12  a = −1 and

hence b = 5, and y = − x 2 + 5 x − 4.

b2 4 40 = 104 − b2

88. (a)

10 = 26 −

Since the graph of the function

R(t ) = at 2 + bt + c is a parabola that opens upward, the value of a is positive. (b) Because a > 0, the function has a minimum, which represents the year the company had the least b revenue. So, an expression is t = − . 2a

b2 = 64 b = ±8. 2

b   b2   84. For a > 0, f ( x ) = a  x +  is a minimum  + c − 2a   4a  

(c) Because the axis of symmetry intersects the graph at the vertex, and the minimum is at the vertex, use t = 4 and t = 14 to find the minimum. The 4 + 14 18 = = 9. minimum is t = 2 2

−b . In this case, the minimum value is when x = 2a b2 c − . Hence, 4a b2 4 −200 = −100 − b2

So, the company had the least revenue in 2009.

−50 = −25 −

(d) Assuming the model is still valid today, the revenues will be increasing.

b 2 = 100 b = ±10. 85. Let x = first number and y = second number. Then x + y = s or y = s − x. The product is given by P = xy or P = x ( s − x ). P = x(s − x )

P = sx − x 2 The maximum P occurs at the vertex when x =

−b . 2a

−s s = 2(−1) 2 s s s When x = , y = s − = . 2 2 2 x=

So, the numbers x and y are both

s . 2

86. If f ( x ) = ax 2 + bx + c has two real zeros, then by the − b ± b 2 − 4 ac Quadratic Formula they are x = . 2a

The average of the zeros of f is −b − b 2 − 4 ac −b + b 2 − 4 ac −2 b + b 2a 2a = 2a = − . 2 2 2a

This is the x-coordinate of the vertex of the graph.

89. Model (a) is preferable. a > 0 means the parabola opens upward and profits are increasing for t to the right of the vertex

t≥−

b . (2 a)

90. x + y = 8  y = 8 − x 2 − x +8− x =6 3 5 − x +8 =6 3 5 − x = −2 3 x = 1.2 y = 8 − 1.2 = 6.8

The point of intersection is (1.2, 6.8). 91.

1 x +1 4 12 x − 40 = x + 4

y = 3 x − 10 =

11x = 44 x=4 y = 3(4) − 10

y = 12 − 10 = 2 The graphs intersect at (4, 2).

240

Chapter 3

Polynomial and Rational Functions

92.

y = x + 3 = 9 − x2

y = x 3 + 2 x − 1 = −2 x + 15

93.

x 3 + 4 x − 16 = 0

x + x−6=0 ( x + 3)( x − 2) = 0 x = −3, x = 2 y = −3+3=0

y=2+3=5 Thus, ( −3, 0) and (2, 5) are the points of intersection.

(

)

( x − 2) x 2 + 2 x + 8 = 0 x=2 y = −2(2) + 15 = −4 + 15 = 11 The graphs intersect at (2, 11). 94. Answers will vary. (Make a Decision)

Section 3.2 Polynomial Functions of Higher Degree 1.

continuous

n, n − 1

(a) solution (b) ( x − a ) (c) ( a, 0)

touches, crosses

No. If f is an even-degree fourth-degree polynomial function, its left and right end behavior is either that it rises left and right or falls left and right.

No. Assuming f is an odd-degree polynomial function, if its leading coefficient is negative, it must rise to the left and fall to the right.

1 4 4  14. f ( x ) = − x 3 + x 2 − has y-intercept  0, −  . 3 3 3  Matches graph (d). 15. f ( x ) = x 4 + 2 x 3 has intercepts (0, 0) and ( − 2, 0). Matches graph (g). 16. f ( x ) =

1 5 9 x − 2 x 3 + x has intercepts 5 5

(0, 0), (1, 0), ( −1, 0), (3, 0), and ( −3, 0). Matches (b).

17. The graph of g ( x) = ( x − 3) is a horizontal shift 3

three units to the right of f ( x) = x3. y 4

Because f is a polynomial, it is a continuous on

 x1 , x2  and f ( x1 ) < 0 and f ( x2 ) > 0. Then f ( x ) = 0 for some value of x in  x1, x2 . 8.

The real zero in  x3 , x4  is of even multiplicity, since the graph touches the x-axis but does not cross the x-axis.

f ( x ) = −2 x + 3 is a line with y-intercept ( 0, 3) . Matches

graph (f).

2 1 −3 −2

−2 −3 −4

18. The graph of g( x) = x3 − 3 is a horizontal shift

three units downward of f ( x) = x3. y

10. f ( x ) = x 2 − 4 x is a parabola with intercepts ( 0, 0 ) and

(4, 0) and opens upward. Matches graph (h).

11. f ( x ) = −2 x 2 − 5 x is a parabola with x-intercepts (0, 0)

1 − 4 −3 − 2

12. f ( x ) = 2 x 3 − 3 x + 1 has intercepts (0, 1), (1, 0),  1 1   1 1  3, 0  and  − + 3, 0  . − − 2 2 2 2     Matches graph (a).

−2

 5  and  − , 0  and opens downward. Matches graph (c).  2 

−5

19. The graph of g( x) = − x3 + 4 is a reflection in the x-axis

and a vertical shift four units upward of f ( x) = x3. y 6

1 13. f ( x ) = − x 4 + 3 x 2 has intercepts (0, 0) and 4

(

5 3

)

±2 3, 0 . Matches graph (e).

2 1 − 4 −3 − 2

−2

Section 3.2 20. The graph of g ( x) = ( x − 2) − 3 is a horizontal shift 3

Polynomial Functions of Higher Degree

25.

two units to the right and a vertical shift three units

−8

downward of f ( x) = x3.

241

g f

y −20

Yes, because both graphs have the same leading coefficient.

1 −3 −2

26.

−2

−3 −4

−9

−5

−6 −4

21. The graph of g ( x) = −( x − 3) is a horizontal shift 3

Yes, because both graphs have the same leading coefficient.

three units to the right and a reflection in the x-axis of f ( x) = x3.

27.

y −9

3 2

−6

1 −3 −2

No, because the graphs have different leading coefficients.

−2

28.

−3

−4

22. The graph of g ( x) = ( x + 4) + 1 is a horizontal shift 3

−5

four units to the left and a vertical shift one unit upward of f ( x) = x3. 4

29. f ( x ) = 2 x 4 − 3 x + 1

3 2

Degree: 4

1 − 4 −3 − 2

−2 −4

30. h( x ) = 1 − x 6

f g

− 12

Degree: 6 12

−8

Yes, because both graphs have the same leading coefficient. 24.

Leading coefficient: 2 The degree is even and the leading coefficient is positive. The graph rises to the left and right.

−3

23.

− 20

No, because the graphs have different leading coefficients.

−6

Leading coefficient: –1 The degree is even and the leading coefficient is negative. The graph falls to the left and right. 31. g( x ) = 5 −

− 12

7 x − 3x2 2

Degree: 2 12

−8

Yes, because both graphs have the same leading coefficient.

Leading coefficient: –3 The degree is even and the leading coefficient is negative. The graph falls to the left and right.

242

Chapter 3

32. f ( x ) =

Polynomial and Rational Functions

1 3 x + 5x 3

37. (a)

f ( x ) = 3 x 2 − 12 x + 3

(

Leading coefficient:

1 3

The degree is odd and the leading coefficient is positive. The graph falls to the left and rises to the right. 33. f ( x ) =

4 ± 16 − 4 =2± 3 2

(b)

2 −7

6 x5 − 2 x 4 + 4 x 2 − 5 x 3

Degree: 5 Leading coefficient:

− 10

(c)

6 =2 3

3 x 7 − 2 x5 + 5 x3 + 6 x 2 34. f ( x ) = 4

38. (a)

(

2 ± 4 − 4( −1) =1± 2 2

(b)

1 −8

−11

(c)

)

Leading coefficient: −

x ≈ −0.414, 2.414; the answers are approximately

the same. 39. (a)

Degree: 2

g (t ) =

2 3

The degree is even and the leading coefficient is negative. The graph falls to the left and right. 36. f (s ) = −

)

3 4

The degree is odd and the leading coefficient is positive. The graph falls to the left and rises to the right.

2 2 t − 5t + 3 3

(

g( x ) = 5 x 2 − 2 x − 1 = 0 x=

Degree: 7

35. h(t ) = −

x ≈ 3.732, 0.268; the answers are approximately the

same.

The degree is odd and the leading coefficient is positive. The graph falls to the left and rises to the right.

Leading coefficient:

)

= 3 x2 − 4 x + 1 = 0

Degree: 3

1 ( t + 1) (t − 1)(t 2 + 1) = 0 2

t = ±1

(b)

7 3 s + 5s 2 − 7s + 1 8

(

1 4 1 t − 2 2

)

−6

Degree: 3

7 Leading coefficient: − 8 The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right.

−2

t = ±1; the answers are the same.

1 3 2 x ( x − 9) 4

x = 0, ± 3

(b)

− 18

− 12

(c)

x = 0, ± 3; the answers are the same.

Section 3.2 f ( x) = x 5 + x3 − 6 x

41. (a)

Polynomial Functions of Higher Degree

45. (a)

( ) = x ( x + 3 ) ( x − 2) = 0

f ( x ) = x 3 − 4 x 2 − 25 x + 100

= x 2 ( x − 4) − 25( x − 4)

= x x 4 + x2 − 6 2

= ( x 2 − 25)( x − 4) = ( x − 5)( x + 5)( x − 4) = 0

x = 0, ± 2

(b)

x = ±5, 4

(b)

−6

130

6 −6

6 − 10

−4

(c)

x = 0, 1.414, − 1.414; the answers are approximately

the same.

x = 4, 5, − 5; the answers are the same.

0 = 4 x3 + 4 x2 − 7x + 2

g ( t ) = t − 6 t + 9t

42. (a)

243

(

= t t − 6t + 9

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

)

= t (t − 3) = 0 t = 0, ± 3

(b)

1 2

x = −2, (b)

−9

9 −2

−6

(c)

x = 0, ± 1.732; the answers are approximately the

(c)

same. 47. (a)

f ( x ) = 2 x 4 − 2 x 2 − 40

43. (a)

y = 4 x 3 − 20 x 2 + 25 x 0 = 4 x 3 − 20 x 2 + 25 x

( ) = 2 ( x + 4 ) ( x + 5 )( x − 5 ) = 0 = 2 x 4 − x 2 − 20

0 = x (2 x − 5)2

x=± 5

1 x = −2, ; the answers are the same. 2

(b)

5 2

x = 0,

5 − 10

−2

6 −4

− 45

(c)

x = 0,

x = 2.236, − 2.236; the answers are approximately

the same.

48. (a)

5 ; the answers are the same. 2

y = x5 − 5x3 + 4 x

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

= x( x 4 − 5 x 2 + 4)

= 5( x 2 + 1)( x 2 + 2) > 0 No real zeros

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

44. (a)

(b)

x = 0, ± 1, ± 2

(b) −6

−9

−5

−6

(c)

x = 0, ± 1, ± 2; the answers are the same.

244

Chapter 3

Polynomial and Rational Functions 59. f ( x ) = 2 x 4 − 6 x 2 + 1

49. f ( x ) = x 2 − 25 = ( x + 5)( x − 5) x = ±5 (multiplicity 1)

−6

50. f ( x ) = 49 − x 2

= (7 − x )(7 + x )

x = ±7 (multiplicity 1) 51.

h( t ) = t 2 − 6 t + 9 = (t − 3)2 t = 3 (multiplicity 2) 2

52. f ( x ) = x + 10 x + 25

−4

Zeros: x ≈ ± 0.421, ± 1.680 Relative maximum: (0, 1) Relative minima: (1.225, − 3.5), ( −1.225, − 3.5)

3 60. f ( x ) = − x 4 − x 3 + 2 x 2 + 5 8

= ( x + 5)2 x = −5 (multiplicity 2)

53. f ( x ) = x 2 + x − 2

−6

= ( x + 2)( x − 1) x = −2, 1 (multiplicity 1)

54. f ( x ) = 2 x 2 − 14 x + 24 = 2( x 2 − 7 x + 12) = 2( x − 3)( x − 4) x = 3, 4 (multiplicity 1)

4 − 10

Zeros: −4.142, 1.934 Relative maxima: (0.915, 5.646), ( −2.915, 19.688) Relative minimum: (0, 5) 61. f ( x ) = x 5 + 3 x 3 − x + 6 11

55. f (t ) = t 3 − 4t 2 + 4t = t (t − 2)2 t = 0 (multiplicity 1), 2 (multiplicity 2) 4

56. f ( x ) = x − x − 20 x

= x 2 ( x 2 − x − 20)

= x 2 ( x + 4)( x − 5) x = −4 (multiplicity 1), 5 (multiplicity 1),

0 (multiplicity 2)

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

57. f ( x ) =

−5 ± 25 − 4( −3) 5 37 =− ± 2 2 2 ≈ 0.5414, − 5.5414 (multiplicity 1)

5 2 8 4 x + x− 3 3 3 1 = (5 x 2 + 8 x − 4) 3 1 = (5 x − 2)( x + 2) 3 2 x = , − 2 (multiplicity 1) 5

58. f ( x ) =

−9

9 −1

Zero: x ≈ −1.178 Relative maximum: ( −0.324, 6.218) Relative minimum: (0.324, 5.782) 62. f ( x ) = −3 x 3 − 4 x 2 + x − 3 1 −6

−7

Zero: −1.819 Relative maximum: (0.111, − 2.942) Relative minimum: ( −1, − 5) 63. f ( x ) = −2 x 4 + 5 x 2 − x − 1 4

−6

−4

Zeros: −1.618, − 0.366, 0.618, 1.366 Relative minimum: (0.101, − 1.050) Relative maxima: ( −1.165, 3.267), (1.064, 1.033)

Section 3.2 64.

f ( x) = 3x 5 − 2 x2 − x + 1

71.

(

)

( 2)

= x2 − 2x + 1 − 2 = x2 − 2x −1

Zeros: −0.737, 0.548, 0.839 Relative minimum: (0.712, − 0.177) Relative maximum: ( −0.238, 1.122) 65. f ( x) = ( x − 0)( x − 7) = x2 − 7x

Note: f ( x) = a ( x 2 − 2 x − 1) has zeros 1 + 2 and 1 − 2 for all nonzero real numbers a.

72.

Note: f ( x ) = a( x − 0)( x − 7) = ax( x − 7) has zeros

4 + 3 and 4 − 3 for all nonzero real numbers a.

73.

68. f ( x) = ( x − 0)( x − 1)( x − 6) = x3 − 7 x2 + 6x

(

)

Note: f ( x) = a ( x − 2)[( x − 2) 2 − 5] has zeros 2, 2 + 5,

all nonzero real numbers a.

and 2 − 5 for all nonzero real numbers a.

f ( x) = ( x − 4)( x + 3)( x − 3)( x − 0) 74.

= x − 4 x − 9 x + 36 x

(

)

(

)

f ( x) = ( x − 3)  x − 2 + 7   x − 2 − 7     = ( x − 3) ( x − 2 ) − 7  ( x − 2 ) + 7     2 = ( x − 3)(( x − 2) − 7)

Note: f ( x ) = a( x 4 − 4 x 3 − 9 x 2 + 36 x ) has zeros 4, − 3, 3, and 0 for all nonzero real numbers a.

= ( x − 3)( x 2 − 4 x − 3) = x3 − 7 x 2 + 9 x + 9

f ( x ) = ( x − ( −2))( x − ( −1))( x − 0)( x − 1)( x − 2)

Note: f ( x) = a( x − 3)( x 2 − 4 x − 3) has zeros 3, 2 ± 7 for all nonzero real numbers a.

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

= x( x 4 − 5 x 2 + 4)

)

= x3 − 6 x 2 + 7 x + 2

Note: f ( x) = ax( x − 1)( x − 6) has zeros 0, 1, and 6 for

(

f ( x ) = ( x − 2 )  x − 2 + 5   x − 2 − 5     = ( x − 2) ( x − 2 ) − 5  ( x − 2 ) + 5     = ( x − 2)[( x − 2) 2 − 5]

for all nonzero real numbers a.

)

Note: f ( x) = a ( x 2 − 8 x + 13) has zeros

nonzero real numbers a.

= (x − 4)( x 2 − 9)x

(

= x 2 − 8 x + 13

Note: f ( x) = a( x + 2)( x − 5) has zeros −2 and 5 for all

Note: f ( x ) = ax( x + 2)( x + 4) has zeros 0, −2, and −4

)

= x 2 − 8 x + 16 − 3

66. f ( x) = ( x + 2)( x − 5) = x2 − 3x − 10

67. f ( x) = ( x − 0)( x + 2)( x + 4) = x3 + 6x2 + 8x

(

f ( x) =  x − 4 + 3   x − 4 − 3     =  ( x − 4 ) − 3  ( x − 4 ) + 3     = ( x − 4) 2 − 3

0 and 7 for all nonzero real numbers a.

)

f ( x ) =  x − 1 + 2   x − 1 − 2    

= ( x − 1) 2 −

−2

70.

(

245

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

−3

69.

Polynomial Functions of Higher Degree

75.

f ( x ) = ( x + 2)2 ( x + 1) = x 3 + 5 x 2 + 8 x + 4

= x − 5x + 4x Note: f ( x ) = ax ( x + 2)( x + 1)( x − 1)( x − 2) has zeros −2, − 1, 0, 1, 2, for all nonzero real numbers a.

Note: f ( x ) = a( x + 2)2 ( x + 1) has zeros −2, − 2, and −1 for all nonzero real numbers a. 76.

f ( x ) = ( x − 3)( x − 2)3

= x 4 − 9 x 3 + 30 x 2 − 44 x + 24 Note: f ( x ) = a( x − 3)( x − 2)3 has zeros 3, 2, 2, 2 for all nonzero real numbers a.

246

Chapter 3

Polynomial and Rational Functions

77.

f ( x ) = ( x + 4)2 ( x − 3)2

84.

y = − x 4 + x 3 + 3x 2 − 6x

= x 4 + 2 x 3 − 23 x 2 − 24 x + 144 Note: f ( x ) = a( x + 4)2 ( x − 3)2 has zeros −4, − 4, 3, 3 for all nonzero real numbers a. 78.

4 2 − 18

− 14

− 10

−6

f ( x ) = ( x − 5)3 ( x − 0)2

= x 5 − 15 x 4 + 75 x 3 − 125 x 2 Note: f ( x ) = a( x − 5)3 x 2 has zeros 5, 5, 5, 0, 0 for all nonzero real numbers a. 79.

f ( x ) = −( x + 1)2 ( x + 2)

= − x3 − 4 x 2 − 5x − 2 Note: f ( x ) = a( x + 1)2 ( x + 2)2 , a < 0, has zeros −1, − 1, − 2, rises to the left, and falls to the right. 80.

85. (a)

The degree of f is odd and the leading coefficient is 1. The graph falls to the left and rises to the right. (b) f ( x ) = x 3 − 9 x = x( x 2 − 9) = x( x − 3)( x + 3) Zeros: 0, 3, − 3 (c) and (d) y

f ( x ) = −( x − 1)2 ( x − 4)2 4 2

= − x 4 + 10 x 3 − 33 x 2 + 40 x − 16

Note: f ( x ) = a( x − 1)2 ( x − 4)2, a < 0, has zeros 1, 1, 4, 4, falls to the left, and falls to the right. y

81. 3 2 1 −3

−1

1 2 3 4 5 6 7

−2

−5 −6 −7

−8 −6 −4 −2 −4 −6 −8

2 4 6 8 10

86. (a) The degree of g is even and the leading coefficient is 1. The graph rises to the left and rises to the right. (b) g ( x ) = x 4 − 4 x 2 = x 2 ( x 2 − 4) = x 2 ( x − 2)( x + 2) Zeros: 0, 2, − 2

y = −x 3 + 3x − 2

82.

−4 −3

2 −14 −12 −10 −8 −6 −4

−1

−4

87. (a) The degree of f is odd and the leading coefficient is 1. The graph falls to the left and rises to the right.

5 4 3

−3 −2

y = x 4 + x 3 − 3x 2 − x + 2

83.

(b)

1 2 3 4 5 6 7

f ( x ) = x 3 − 3 x 2 = x 2 ( x − 3)

Zeros: 0, 3 (c) and (d) y 5 4 3 2 1

y = x 5 − 5x 2 − x + 2

−4 −3 −2 −1

1 2

4 5 6

Section 3.2 88. (a) The degree of f is odd and the leading coefficient is 3. The graph falls to the left and rises to the right.

(b) f ( x ) = 3 x 3 − 24 x 2 = 3 x 2 ( x − 8)

Polynomial Functions of Higher Degree

247

91. (a) The degree of f is odd and the leading coefficient is 1. The graph falls to the left and rises to the right. (b) f ( x ) = x 3 + 3 x 2 − 9 x − 27 = x 2 ( x + 3) − 9( x + 3) = ( x 2 − 9)( x + 3)

Zeros: 0, 8 (c) and (d)

= ( x − 3)( x + 3)

Zeros: 3, − 3

50 25 2 3 4 5 6 7

8 4 − 20

−175 −200 −225

89.

(a) The degree of f is even and the leading coefficient is −1. The graph falls to the left and falls to the right. (b) f ( x ) = − x 4 + 9 x 2 − 20 = −( x 2 − 4)( x 2 − 5) Zeros: ±2, ± 5 (c) and (d)

− 12

= ( x 3 + 8)( x 2 − 4)

2 −8

92. (a) The degree of h is odd and the leading coefficient is 1. The graph falls to the left and rises to the right. (b) h( x ) = x 5 − 4 x 3 + 8 x 2 − 32 = x 3 ( x 2 − 4) + 8( x 2 − 4)

− 12

8 12 16 20

−4

4 6 8 10 12

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

Zeros: −2, 2 (c) and (d) y

8 4

90. (a) The degree of f is even and the leading coefficient is −1. The graph falls to the left and falls to the right. (b) f ( x ) = − x 6 + 7 x 3 + 8 = −( x 3 + 1)( x 3 − 8)

Zeros: −1, 2 (c) and (d) 20 16 12

3 4 5

−8

− 32

−5 −4 − 3 −2

−5 − 4 −3

3 4 5

93. (a) The degree of g is even and the leading coefficient is 1 − . The graph falls to the left and falls to the right. 4 1 1 (b) g(t ) = − (t 4 − 8t 2 + 16) = − (t 2 − 4)2 4 4 Zeros: −2, − 2, 2, 2

−5 −4 −3

3 4 5

−4 −5 −6 −7 −8 −9

248

Chapter 3

Polynomial and Rational Functions

94.

(a) The degree of g is even and the leading coefficient is 1 . The graph rises to the right and rises to the left. 10 1 (b) g( x ) = ( x + 1)2 ( x − 3)2 10 Zeros: −1, 3

f (x ) = − 2x 3 − 6x 2 + 3

(a) −5

The function has three zeros. They are in the intervals ( − 3, − 2), ( − 1, 0), and (0, 1). (b) Zeros: −2.810, − 0.832, 0.642

9 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1

1 2 3 4 5

–2.83 –0.277

–0.86 –0.166

0.62

0.217

–2.82 –0.137

–0.85 –0.0107

0.63

0.119

–0.84 –0.048

0.64

0.018

–2.80 –0.136

–0.83

0.010

0.65

–0.084

–2.79 –0.269

–0.82

0.068

0.66

–0.189

–2.81

f (x) = x3 − 3x 2 +3

(a)

−6

95.

96.

≈0

97. g( x ) = 3 x 4 + 4x 3 − 3 −5

(a)

−3

−6

The function has three zeros. They are in the intervals ( − 1, 0), (1, 2) and (2, 3). (b) Zeros: −0.879, 1.347, 2.532 x

–0.9

–0.159

1.3

−5

( − 2, − 1) and (0, 1).

(b) Zeros: −1.585, 0.779 0.127

2.5

–0.125

1.31 0.09979

2.51

–0.88 –0.0047

1.32 0.07277

2.52 –0.0482

–0.87

1.33 0.04594

2.53 –0.0084

–0.86 0.14514

1.34

0.0193

2.54 0.03226

–0.85 0.21838

1.35 –0.0071

2.55 0.07388

–0.84

1.36 –0.0333

2.56 0.11642

0.2905

The function has two zeros. They are in the intervals

–0.89 –0.0813

0.0708

–0.087

–1.6

0.2768

0.75 –0.3633

–1.59 0.09515

0.76 –0.2432

–1.58 –0.0812

0.77 –0.1193

–1.57 –0.2524

0.78 0.00866

–1.56 –0.4184

0.79 0.14066

–1.55 –0.5795

0.80

–1.54 –0.7356

0.81 0.41717

0.2768

Section 3.2 98. h( x ) = x 4 − 10 x 2 + 2

(a)

Polynomial Functions of Higher Degree

249

100. f (x )= x 3 − 4x 2 − 2x +10

4 − 10

(a)

−4

− 24 − 20

The function has four zeros. They are in the intervals (0, 1), (3, 4), ( − 1, 0), and ( − 4, − 3).

(b) Notice that h is even. Hence, the zeros come in symmetric pairs. Zeros: ±0.452, ± 3.130. Because the function is even, we only need to verify the positive zeros.

99.

The function has three zeros. They are in the intervals (−2, − 1), (1, 2), and (3, 4). (b) Zeros: −1.537, 1.693, 3.843 x

− 1.56

−0.4108

1.66

0.2319

0.42

0.26712

3.09

−2.315

−1.55

−0.2339

1.67

0.16186

0.43

0.18519

3.10

−1.748

− 1.54

−0.0587

1.68

0.09203

0.44

0.10148

3.11

−1.171

−1.53

0.11482

1.69

0.02241

0.45

0.01601

3.12

−0.5855

− 1.52

0.28659

1.70

−0.047

0.46

−0.0712

3.13

0.01025

−1.51

0.45665

1.71

−0.1162

0.47

−0.1602

3.14

0.61571

− 1.50

0.625

1.72

−0.1852

0.48

−0.2509

3.15

1.231

3.82

−0.2666

3.83

−0.1537

3.84

−0.0393

3.85

0.07663

The function has two zeros. They are in the intervals

3.86

0.19406

( −1, 0) and (3, 4).

3.87

0.313

3.88

0.43347

f ( x) = x 4 − 3x3 − 4 x − 3

(a)

−4

− 25

(b) Zeros: −0.578, 3.418 x

−0.61

0.2594

3.39

−1.366

− 0.60

0.1776

3.40

−0.8784

− 0.59

0.09731

3.41

−0.3828

− 0.58

0.0185

3.42

0.12071

− 0.57

−0.0589

3.43

0.63205

− 0.56

−0.1348

3.44

1.1513

−0.55

−0.2094

3.45

1.6786

101. f ( x ) = x 2 ( x + 6) 35

−12

8 −5

No symmetry Two x-intercepts

250

Chapter 3

Polynomial and Rational Functions

102. h( x) = x3 ( x − 3)

107. g( x ) =

1 ( x + 1)2 ( x − 3)(2 x − 9) 5 14

−6

− 14

−1 −6

No symmetry Two x-intercepts

No symmetry Three x-intercepts

1 103. g(t ) = − (t − 4)2 (t + 4)2 2 10 − 10

1 108. h( x ) = ( x + 2)2 (3 x − 5)2 5

−6

− 150

Symmetric with respect to the y-axis Two x-intercepts

1 104. g( x) = ( x + 1)2 ( x − 3)3 8

6 −3

No symmetry Two x-intercepts 109. (a)

2 −6

Volume = length × width × height

Because the box is made from a square, length = width. Thus: Volume = ( length ) × height = ( 36 − 2x ) x 2

(b) Domain: 0 < 36 − 2x < 36 −36 < − 2x < 0 18 > x > 0

−6

No symmetry Two x-intercepts

(c)

105. f ( x ) = x 3 − 4 x = x ( x + 2)( x − 2) 6

−9

−6

Symmetric with respect to the origin Three x-intercepts 106. f ( x ) = x 4 − 2 x 2

Height, x

Length and Width

Volume, V

36 − 2(1)

1 36 − 2(1) = 1156

36 − 2(2)

2 36 − 2(2) = 2048

36 − 2(3)

3 36 − 2(3) = 2700

36 − 2(4)

4 36 − 2(4) = 3136

36 − 2(5)

5 36 − 2(5) = 3380

36 − 2(6)

6 36 − 2(6) = 3456

36 − 2(7)

7 36 − 2(7) = 3388

−3

−2

Symmetric with respect to the y-axis Three x-intercepts

Maximum volume is in the interval 5 < x < 7.

Section 3.2 (d)

Polynomial Functions of Higher Degree

251

(c)

3500

350

5 3300

7 0

x = 6 when V ( x ) is maximum.

110. (a)

0 The model fits the data well. 3500

V ( x ) = length × width × height

= ( 24 − 2 x )( 24 − 4 x ) x = 8 x (12 − x )( 6 − x )

(b) Domain: 0 < x < 6 (c)

Answers will vary. Sample answer: You could use the model to estimate production in 2015 because the result is somewhat reasonable, but you would not use the model to estimate the 2020 production because the result is unreasonably high.

V 720 600 480 360

114. True. f ( x ) = x 6 has only one zero, 0.

240 120 1

Maximum occurs at x = 2.54. 111. The point of diminishing returns (where the graph changes from curving upward to curving downward) occurs when x = 200 The point is (200, 160) which corresponds to spending $2,000,000 on advertising to obtain a revenue of $160 million. 112.

115. The graph will always cross the x-axis. 116. False. The left-hand and the right-hand behaviors of the graph are the same. 117. True. The degree is odd and the leading coefficient is −1. 118. False. The graph touches at x = 1, but does not cross the x-axis there. 119. False. The graph crosses the x-axis at x = −3 and x = 0.

120. False. The graph rises to the left, and rises to the right. 121. 0

Point of diminishing returns: (15.2, 27.3 ) 15.2 years 113. (a)

will have another x-intercept at (3, 0) of odd multiplicity (crossing the x-axis). 15

200

y3 −5

The model fits the data well.

−6

The graph of y3 will fall to the left and rise to the right. It

(b)

122. (a) Degree: 3 Leading coefficient: Positive (b) Degree: 2 Leading coefficient: Positive (c) Degree: 4 Leading coefficient: Positive (d) Degree: 5 Leading coefficient: Positive 123. ( f + g )( −4 ) = f ( −4 ) + g ( −4 ) = −59 + 128 = 69

252

Chapter 3

Polynomial and Rational Functions

124. ( g − f )( 3 ) = g ( 3 ) − f ( 3 ) = 8 ( 3 )

− 14 ( 3 ) − 3

130. 2 x 2 − x ≥ 1

= 72 − 39 = 33

−1 2 x

 4  4  4  8•16  125. ( f  g) −  = f −  g −  = (−11)    7  7  7  49  =−

−2

)

127. ( f  g )( −1) = f g ( −1) = f ( 8 ) = 109 128. ( g  f )( 0 ) = g f ( 0 ) = g ( −3 ) = 8 ( −3 ) = 72 2

131.

− 39 − 26 − 13 x

−6 −4

−2

5x − 2 ≤ 4 x − 7 7

129. 3 ( x − 5 ) < 4 x − 7 − 10 − 8

2x + 1 ≥ 0 and x − 1 ≥ 0 or 2x +1 ≤ 0 and x − 1 ≤ 0 1 1      x ≥ − 2 and x ≥ 1 or  x ≤ − 2 and x ≤ 1     1 or x≥ 1 x ≤ − 2

f −24 4 126.   ( −1.5 ) = = = − − 1.5 18 3 g g ( )  

)

( 2 x + 1)( x − 1) ≥ 0

f ( −1.5 )

(

2 x2 − x − 1 ≥ 0

1408 ≈ − 28.7347 49

(

−1

x 13

5x − 2 − 4 ≤ 0 x − 7 5x − 2 − 4 ( x − 7) ≤ 0 x − 7 x + 26 ≤ 0 x − 7

3 x − 15 < 4 x − 7 −8 < x

 x + 26 ≥ 0 and x − 7 < 0  or  x + 26 ≤ 0 and x − 7 > 0  or  x ≤ −26 and x > 7   x ≥ −26 and x < 7  −26 ≤ x < 7

impossible

132. x + 8 − 1 ≥ 15 x

− 32 − 24 − 16 − 8

x + 8 ≥ 16 x + 8 ≥ 16 or x + 8 ≤ − 16 x ≥ 8

or x ≤ − 24

Section 3.3 Real Zeros of Polynomial Functions 1.

f ( x ) is the dividend, d ( x ) is the divisor, q( x ) is the quotient, and r ( x ) is the remainder.

According to the Remainder Theorem, if you divide f ( x ) by x − 4 and the remainder is 7, then f (4) = 7.

improper

8. To check whether ( x − 2) is a factor of

constant term, leading coefficient

f ( x) = x3 + 6 x 2 − 5x + 3, the synthetic division format

Descartes’s Rule, Signs

should appear as 2 1 6 − 5 3 and if it is a factor, the

upper, lower

remainder should be 0.

According to Descartes’s Rule of Signs, given that f (− x ) has 5 variations in sign, there are either 5, 3, or 1 negative real zeros.

Section 3.3 x + 2 9. x + 3 x 2 + 5 x + 6

Real Zeros of Polynomial Functions

x 2 − x − 20 13. x − 3 x − 4 x 2 − 17 x + 6 3

x 2 + 3x

x3 − 3x2

2x + 6

− x 2 − 17 x

2x + 6

− x2 + 3x

− 20x + 6 − 20 x + 60

x + 5x + 6 = x + 2, x ≠ −3 x + 3

− 54 3

x − 4 x − 17 x + 6 54 = x 2 − x − 20 − x −3 x −3

So, x2 + 5x + 6 = ( x + 3)( x + 2).

10.

5x + 3 x − 4 5 x 2 − 17 x − 12

x2 − 3x + 1 14. 4 x + 5 4 x − 7 x 2 − 11x + 5 3

5 x 2 − 20 x

−4 x 3 + 5 x

3 x − 12 3 x − 12

− 12x 2 − 11x −12x 2 − 15 x

4x + 5 − 4x + 5

5 x 2 − 17 x − 12 = 5 x + 3, x ≠ 4 x−4

So, 5x2 − 17 x − 12 = ( x − 4)(5x + 3).

4 x − 7 x − 11x + 5 5 = x 2 − 3 x + 1, x ≠ − 4x + 5 4

x 2 + 3 x − 18 11. x + 2 x + 5 x − 12 x − 36 3

15.

x3 + 2 x 2

7 x 2 − 14 x + 28 x + 2 7x + 0 x2 + 0 x + 3 3

7 x 3 + 14 x 2

3 x 2 − 12 x

− 14 x 2

3x 2 + 6 x

−14 x 2 − 28 x

−18 x − 36

28 x + 3 28x +56

− 18 x − 36 0 3

− 53 3

7x + 3 53 = 7 x 2 − 14 x + 28 − x+2 x+2

x + 5 x − 12 x − 36 = x 2 + 3 x − 18, x ≠ − 2 x + 2

(

)

So, x3 + 5x 2 − 12 x − 36 = ( x + 2) x 2 + 3x − 18 . x 2 − 25 12. 2 x − 3 2 x − 3 x − 50 x + 75 3

253

1 2 16. 2 x + 1 8 x 4 + 0 x3 + 0 x 2 + 0 x − 5 4 x3 − 2 x 2 + x −

8 x 4 + 4 x3

2 x3 − 3x2

− 4 x3

− 50 x + 75

−4 x 3 − 2 x 2

− 50 x + 75

2 x2

0 2 x 3 − 3 x 2 − 50 x + 75 3 = x 2 − 25, x ≠ 2x − 3 2

(

)

So, 2 x3 − 3x 2 − 50 x + 75 = ( 2 x − 3) x 2 − 25

= ( 2 x − 3)( x + 5)( x − 5).

2x2 + x −x − 5 1 −x − 2 9 − 2 9 8x4 − 5 1 3 2 2 = 4x − 2x + x − − 2x + 1 2 2x + 1

254

Chapter 3

Polynomial and Rational Functions

5x − 1 17. 2 x 2 + 1 10 x 3 − 2 x 2 + 5 x − 1 10 x 3

22. ( x − 1) = x − 3x + 3x − 1

+ 5x − 2 x2

−1

x4 − 3x3 + 3x2 − x

−1

3x3 − 3x2 + x

− 2x

3x3 − 9x2 + 9 x − 3

10 x3 − 2 x 2 + 5 x − 1 = 5x − 1 2 x2 + 1

18.

6 x2 + 8 x + 3 x4

x2 + 2 x + 4 2 4 3 x − 2 x + 3 x + 0 x + 3x2 + 0 x + 1 2x

+ 0x 2

4x − 6x + 1 2

4 x − 8 x + 12 2 x − 11 2

x + 3x + 1 2 x − 11 = x2 + 2 x + 4 + 2 x2 − 2 x + 3 x − 2x + 3

x 19. x 2 + 1 x 3 + 0 x 2 + 0 x − 9

x −9 x+9 =x− 2 x2 + 1 x +1

20.

x2 x − 1 x + 0 x + 0 x + 0 x2 + 0 x + 7 3

− x2 x2

2 x3 − 4 x2 + 2 x

7 − 1 26 18 75 222

= 2x −

17 x − 5

( x − 1)

248

26. −6 2

14 − 12

− 20 − 12

7 192

− 32

199

27. 2 9

− 18 18

− 17 x + 5

( x − 1)

2 x 3 + 14 x 2 − 20 x + 7 199 = 2 x 2 + 2 x − 32 + x+6 x+6

2x 21. x 2 − 2 x + 1 2 x 3 − 4 x 2 − 15 x + 5

6 x 3 + 7 x 2 − x + 26 248 = 6 x 2 + 25 x + 74 + x −3 x −3

x5 + 7 x2 + 7 = x2 + 3 3 x −1 x −1

2 x 3 − 4 x 2 − 15 x + 5

3 −2

6 25 74

24. −3 5 18 7 − 6 − 15 − 9 6

25. 3 6

− x−9

( x − 1)

5 x 3 + 18 x 2 + 7 x − 6 = 5 x 2 + 3 x − 2, x ≠ −3 x+3

+ x

6 x2 − 8x + 3

3 x 3 − 17 x 2 + 15 x − 25 = 3 x 2 − 2 x + 5, x ≠ 5 x−5

= x +3+

−2

2 x3 − 4 x2 + 6 x

( x − 1)

23. 5 3 − 17 15 − 25 15 − 10 25

x 4 − 2 x3 + 3x2

x +3

x3 − 3x2 + 3x − 1 x4

− 16 32 0 − 32 − 16

9 x 3 − 18 x 2 − 16 x + 32 = 9 x 2 − 16, x ≠ 2 x−2

28. −2 5

0 − 10

6 8 20 − 52

5 − 10 26 − 44 5x3 + 6 x + 8 44 = 5 x 2 − 10 x + 26 − x+2 x+2

Section 3.3 29. −8 1

0 0 512 − 8 64 − 512

34.

x+3 x + 2x − 3 + 2 = x+3 x2 + 2 x − 1 = x+3 = y1

1 9 81 0 x 3 − 729 = x 2 + 9 x + 81, x ≠ 9 x−9 − 23 −7

31. − 1 4 16 −2 2

3 3 2

−4

− 18

9 2

3 4

9 8

− 12

35.

y2 = x 2 − 8 +

39 x2 + 5

( x − 8 )( x + 5) + 39 2

x2 + 5 x − 8 x + 5 x 2 − 40 + 39 = x2 + 5 4 2 x − 3x − 1 = x2 + 5 = y1

1 2

−9

−6

36.

y2 = x 2 − =

(

1 x2 + 1

)

x2 x2 + 1 − 1

x2 + 1 x4 + x2 − 1 = x2 + 1 = y1

− 15

3 49 4 8 3x3 − 4 x2 + 5 1 3 49 = 3x 2 + x + + 3 − 12 x 2 4 8 x− 2 4 33. y2 = x − 2 + x+2 ( x − 2 )( x + 2 ) + 4 = x+2 x2 − 4 + 4 = x+2 x2 = x+2 = y1 3

− 15 15

4 14 − 30 0 3 2 4 x + 16 x − 23 x − 15 1 = 4 x 2 + 14 x − 30, x ≠ − 1 2 x+ 2

32.

2 x+3 ( x − 1)( x + 3) + 2 2

− 729 729

0 0 9 81

255

y2 = x − 1 +

1 − 8 64 0 x 3 + 512 2 = x − 8 x + 64, x ≠ −8 x +8

30. 9 1

Real Zeros of Polynomial Functions

− 10 −6

6 −2

37.

f ( x ) = x 3 − x 2 − 14 x + 11, k = 4

−1

− 14

12 − 8

1 3 −2 3 f ( x ) = ( x − 4)( x 2 + 3 x − 2) + 3 f (4) = (0)(26) + 3 = 3

256

Chapter 3

38.

f ( x ) = 15 x 4 + 10 x 3 − 6 x 2 + 14, k = −

−

Polynomial and Rational Functions

2 15 10 − 6 3 − 10 0 0 −6

(d)

2 3

14 8 4 − 3 34 4 3

− 14

2 2+3 2

1 3+ 2

2 (b) 1 2

(c)

−8

3 2

40. − 5 1

− 5 5−2 5 1 2− 5

0 4

3 4

2 5

3 2

0 3 6 18

0 −1 0 3 63 189 564 1692

−6

− 12

−4

4−4 3

10 − 2 3

−2−4 3

−2−2 3

f (1 − 3) = 0

−5 1

− 35

−5 2

−7 4 − 6 − 26

−2

− 14

−7

− 10 4

− 215

− 211

8−4 2

46.

(b)

(a) 1 2

0 2

−7 2

3 −5

−5

−2

−2 2

0 −4 2 −4

−7 8 1

−7

0 1 1

1 2 13 − 2

4 − 12 −2 4 − 16 −8 4

= f (1) 3 −2 1

= f (−2)

(c)

5 4

4 (d) −10 4

3 13 − 4 1 1 − = f  4 2

= h( −5)

f ( x ) = 4 x 4 − 16 x 3 + 7 x 2 + 20 (a) 1 4 − 16 7 0

= g( −1)

= h(−2)

−7

1 − 10

= g(3)

= h(3)

−5

f ( x) = 2 x3 − 7 x + 3

− 13

−2−4 2

f (2 + 2 ) = 0

1 2

−2

−6−3 2

f ( x ) = [ x − (2 + 2 )][ −3 x + (2 − 3 2 ) x + 8 − 4 2 ]

(c)

4 – 39

−8

(b)

−6

2−3 2

−3

−4

5 −5

−5

−3

3 4

1 − 3 − 13 − 22 = h(2) −2 1 − 5 −7 4

(d)

= g(1)

−1 0 5 −4

(c)

= g(2)

175

188 564 1695

1 21

f ( x ) = ( x − 1 + 3)[4 x − (2 + 4 3) x − (2 + 2 3)]

43.

0 3 0 −2 2 −5

(b)

42. 2 + 2

45. h( x ) = x 3 − 5 x 2 − 7 x + 4 (a) 3 1 − 5 −7

f ( x ) = ( x + 5)( x 2 + (2 − 5) x − 2 5) + 6 f (− 5) = 6 41. 1 − 3 4

−1 5

2 −2

−2 5

22 0 5

(d)

−5 −4

0 3 86 172

0 3 2 2

−1 2

−1 44

4 11

) ( x + (3 + 2 ) x + 3 2 ) − 8 f ( 2 ) = 0 ( 4 + 6 2 ) − 8 = −8 (

f ( x) = x − 2

3 2 5 = f (2)

44. g( x ) = 2 x 6 + 3 x 4 − x 2 + 3 (a) 2 2 0 3 0

−2

−7 8 1

0 4 4

2 34  f ( x ) =  x +  (15 x 3 − 6 x + 4) + 3 3   2  34 f −  =  3 3 39.

2 2

− 12

−5

15 = f (1)

− 110

220

− 24

− 110

− 16

240 = f ( −2)

135

675

135

695

− 16

= f (5) 20

− 40

560 − 5670

56,700

− 56

567 − 5670

56,720

= f ( −10)

Section 3.3 47. 4 1 0 −13 −12

1 4

51.

(

3 1

)

0 − 31 −6

36 − 30

1 −6

(

)

52. 2 −

= ( x + 6)( x − 5)( x − 1)

2 3

3 ) x + 2 3 ) (  3 )( x + 2 x + 3x + 2 3 ) 3 )  x( x + 2) + 3 ( x + 2) 3 )( x + 3 )( x + 2)

3 x 2 + 2 +  2

−1 2−

3 2 −17 12 63 49. − 2 − 3 30 − 63

−13 − 3 7 −3 5

2 1−

5 −6 − 3 5

5 1 1−

5 −6 − 3 5

2+ 1

6+3 5

(

x − x − 13x − 3 =  x − 2 −  3

2 x 3 − 17 x 2 + 12 x + 63

(

3  =  x + ( 2 x 2 − 20 x + 42) 2 

= x − 2+

= ( 2 x + 3)( x 2 − 10 x + 21)

53. (a)

−2 2

= ( 2 x + 3)( x − 3)( x − 7)

1 −4

−5 6

) ( 5 )( x − 2 −

)

5  x − 2 + 

5  ( x + 3) 

)

5 ( x + 3)

2 −2

2 −3 1 0 (b) 2 x − 3 x + 1 = (2 x − 1)( x − 1) Remaining factors: (2 x − 1), ( x − 1) (c) f ( x ) = ( x + 2)(2 x − 1)( x − 1) 1 (d) Real zeros: −2, , 1 2

3 Zeros: − , 3, 7 2

1 60 − 89 41 − 6 3 20 − 23 6 60 − 69

5 1

Zeros: − 6, 1, 5

50.

Zeros: ± 3, − 2

x3 − 31x + 30 = ( x + 6) x 2 − 6 x + 5

2 − 20 42

( = (x − = (x − = (x −

Zeros: −3, −1, 4 48. − 6 1

3 3+2 3

x3 + 2 x 2 − 3x − 6 = x −

= ( x − 4)( x + 1)( x + 3)

257

−3 −6

2 2+

x − 13x − 12 = ( x − 4) x + 4 x + 3 3

Real Zeros of Polynomial Functions

1  60 x 3 − 89 x 2 + 41x − 6 =  x − (60 x 2 − 69 x + 18) 3 

−8

= (3x − 1)( 20 x − 23 x + 6) 2

−3

= (3x − 1)(5 x − 2)( 4 x − 3)

1 2 3 Zeros: , , 3 5 4

54. (a)

−3 3 3

2 − 19 6 21 − 6 −9 −7

3 x 2 − 7 x + 2 = (3 x − 1)( x − 2) Remaining factors: (3 x − 1), ( x − 2) (c) f ( x ) = ( x + 3)(3 x − 1)( x − 2) 1 (d) Real zeros: −3, , 2 3

(b)

−6

−20

258

Chapter 3

55. (a)

5 1

Polynomial and Rational Functions

−4 5

1 −4 1

− 15 5

58 − 50

− 40 40

1 − 10

1 −4

58. (a)

(b)

− 10 8 12 − 8

5 1 2 − 1 − 10 1 0 −5 2 2 0 − 10 0 2 x 2 − 10 = 2( x − 5 )( x + 5)

Remaining factors: ( x − 5), ( x + 5) f ( x ) = (2 x − 1)( x + 5)( x − 5 ) 1 (d) Real zeros: , ± 5 2

(c)

1 −3 2 0 2 (b) x − 3 x + 2 = ( x − 2)( x − 1) Remaining factors: ( x − 2), ( x − 1) (c) f ( x ) = ( x − 5)( x + 4)( x − 2)( x − 1) (d) Real zeros: 5, − 4, 2, 1

−6

20 −7

7 −10

59. −200

56. (a)

−2 8 8

4 8

− 14

− 71

− 10

− 16

22 − 24

− 30

− 11

− 30 32

− 11 8

p = factor of −3 q = factor of 1

Possible rational zeros: ±1, ± 3

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

12 − 12

8 2 0 −3 2 (b) 8 x + 2 x − 3 = (4 x + 3)(2 x − 1) Remaining factors: (4 x + 3), (2 x − 1) (c) f ( x ) = ( x + 2)( x − 4)(4 x + 3)(2 x − 1) 3 1 (d) Real zeros: −2, 4, − , 4 2

f ( x) = x3 + 3x2 − x − 3

Rational zeros: ±1, − 3 60.

f ( x ) = x 3 − 4 x 2 − 4 x + 16

p = factor of 16 q = factor of 1 Possible rational zeros: ±1, ± 2, ± 4, ± 8, ± 16 f ( x ) = x 2 ( x − 4) − 4( x − 4) = ( x − 4)( x 2 − 4)

50 −6

Rational zeros: 4, ± 2 61. −400

57. (a)

1 6 − 2

−9

− 14

−3

− 19

6 38 − 28 0 2 (b) 6 x + 38 x − 28 = (3 x − 2)(2 x + 14) Remaining factors: (3 x − 2), ( x + 7) (c) f ( x ) = (2 x + 1)(3 x − 2)( x + 7) 1 2 (d) Real zeros: − , , − 7 2 3 320

−9

f ( x ) = 2 x 4 − 17 x 3 + 35 x 2 + 9 x − 45

p = factor of −45 q = factor of 2 Possible rational zeros: ±1, ± 3, ± 5, ± 9, ± 15, ± 45, 1 3 5 9 15 45 ± , ± , ± , ± , ± , ± 2 2 2 2 2 2 Using synthetic division, −1, 3, and 5 are zeros. f ( x ) = ( x + 1)( x − 3)( x − 5)(2 x − 3) 3 Rational zeros: −1, 3, 5, 2

3 −20

Section 3.3 62.

f ( x ) = 4 x 5 − 8 x 4 − 5 x 3 + 10 x 2 + x − 2

68. (a)

f ( x ) = −3 x 3 + 20 x 2 − 36 x + 16

f (− x ) = 3 x 3 + 20 x 2 + 36 x + 16 0 sign changes  No negative real zeros (b) Possible rational zeros: 1 2 4 8 16 ± , ± , ± , ± , ± , ± 1, ± 2, ± 4, ± 8, ± 16 3 3 3 3 3

1 1 , ± 2 4 Using synthetic division, −1, 1, and 2 are zeros. f ( x ) = ( x + 1)( x − 1)( x − 2)(2 x − 1)(2 x + 1) 1 Rational zeros: ±1, ± , 2 2

Possible rational zeros: ±2, ± 1, ±

(c)

f ( x) = 2 x 4 − x3 + 6 x2 − x + 5

−6

4 variations in sign  4, 2, or 0 positive real zeros

f ( x) = 3x 4 + 5x3 − 6 x2 + 8 x − 3

3 sign changes  3 or 1 positive real zeros f (− x ) = 3 x 4 − 5 x 3 − 6 x 2 − 8 x − 3 1 sign change  1 negative real zero

65. g( x ) = 4 x 3 − 5 x + 8

2 variations in sign  2 or 0 positive real zeros g ( − x ) = −4 x 3 + 5 x + 8 1 variation in sign  1 negative real zero

(d) Zeros: 69.

f ( x ) has variations in sign  3 or 1 positive real zeros. f (− x ) = −2 x 4 − 13 x 3 − 21x 2 − 2 x + 8 has 1 variation in sign  1 negative real zero. 1 (b) Possible rational zeros: ± , ± 1, ± 2, ± 4, ± 8 2

(a)

g(− x ) = −2 x − 4 x − 5 No sign change  no negative real zeros

(a)

1 (d) Real zeros: − , 1, 2, 4 2

70. (a)

f ( x) = x3 + x 2 − 4 x − 4

f (− x ) = − x + x + 4 x − 4 has 2 variations in sign  2 or 0 negative real zeros. (b) Possible rational zeros: ±1, ± 2, ± 4 (c) 1 −6

f ( x ) = 4 x 4 − 17 x 2 + 4

2 sign changes  0 or 2 positive real zeros

f ( x ) has 1 variation in sign  1 positive real zero. 3

−8

1 sign change  1 positive real zero

67.

2 , 2, 4 3

f ( x ) = −2 x 4 + 13 x 3 − 21x 2 + 2 x + 8

66. g( x ) = 2 x 3 − 4 x 2 − 5 3

−4

f (− x ) = 2 x 4 + x 3 + 6 x 2 + x + 5 0 variations in sign  0 negative real zeros

64.

259

3 sign changes  3 or 1 positive real zeros

p = factor of −2 q = factor of 4

63.

Real Zeros of Polynomial Functions

f (− x ) = 4 x 4 − 17 x 2 + 4 2 sign changes  0 or 2 negative real zeros 1 1 (b) Possible rational zeros: ± , ± , ± 1, ± 2, ± 4 4 2 9

−7

(d) Real zeros: −2, − 1, 2

− 15

(d) Zeros: ± 2, ±

1 2

260

Chapter 3

Polynomial and Rational Functions

71.

f ( x ) = 32 x 3 − 52 x 2 + 17 x + 3 f ( x ) has 2 variations in sign  2 or 0 positive real zeros. f ( − x ) = −32 x 3 − 52 x 2 − 17 x + 3 has 1 variation in sign  1 negative real zero. (b) Possible rational zeros: 1 1 1 1 1 3 3 3 3 3 ± , ± , ± , ± , ± , ± 1, ± , ± , ± , ± , ± , ±3 32 16 8 4 2 32 16 8 4 2

(a)

4 2 − 3 − 12 8 8 20 32 2

−4

−3 2

−3 −6

− 12 8 27 − 45

−9

3 1 ,− 4 8

f ( x ) has 1 variation in sign  1 positive real zero. f (− x ) = x 4 + x 3 − 29 x 2 + x − 30 has 3 variations in signs  3 or 1 negative real zeros. (b) Possible rational zeros: ±1, ± 2, ± 3, ± 5, ± 6, ± 10, ± 15, ± 30

(a)

200

(d) Real zeros: −5, 6 73. f ( x ) = x 4 − 4 x 3 + 15

0 0

0 15 0 0

16 − 16 525 2705

21 105

541 2689

5 is an upper bound. −3 1 1

−4

− 16

− 3 21

− 63

141

− 7 21

− 47

125

Real zeros: − 2, 2

3 2 0 0 6 18

−8 54

3 138

2 6 18

141

3 is an upper bound. 0

−8

− 128

544

2 −8

− 136

547

−4 2

0 15

4 is an upper bound. 0 −1 1 − 4 0 −1 5 − 5

−5 5 −5

−4 0 25 105

76. f ( x ) = 2 x 4 − 8 x + 3

−350

5 1

−3 is a lower bound.

−8

4 1 −4 0 4 0

−3 is a lower bound.

75. f ( x ) = x 4 − 4 x 3 + 16 x − 16

f ( x ) = x 4 − x 3 − 29 x 2 − x − 30

(c)

− 37

Real zeros: − 2.152, 0.611, 3.041

−2

(d) Real zeros: 1,

8 40

4 is an upper bound.

(c)

72.

74. f ( x ) = 2 x 3 − 3 x 2 − 12 x + 8

−1 is a lower bound. Real zeros: 1.937, 3.705

− 4 is lower bound.

Real zeros: 0.380, 1.435

77. P ( x ) = x 4 −

25 2 x +9 4

1 = (4x 4 − 25x 2 + 36) 4 1 = (4x 2 − 9) (x 2 − 4) 4 1 = (2 x + 3)(2 x − 3)( x + 2)( x − 2) 4

The rational zeros are ±

3 and ± 2. 2

Section 3.3 1 78. f ( x ) = (2 x 3 − 3 x 2 − 23 x + 12) 2

Possible rational zeros: 1 3 ± 1, ± 2, ± 3, ± 4, ± 6, ± 12, ± , ± 2 2

−3

− 23

− 12

5 −3 0 1 2 f ( x ) = ( x − 4)(2 x + 5 x − 3) 2 1 = ( x − 4)(2 x − 1)( x + 3) 2

Real Zeros of Polynomial Functions

81. f ( x ) = x 3 − 1

=(x − 1)(x 2 + x + 1) Rational zeros: 1 ( x = 1) Irrational zeros: 0 Matches (d). 82. f ( x ) = x 3 − 2

1 ,4 2

79. f ( x ) = x 3 −

1 2 1 x −x+ 4 4

1 = (4 x 3 − x 2 − 4 x + 1) 4 1 2 = [ x (4 x − 1) − (4 x − 1)] 4 1 = (4 x − 1)( x 2 − 1) 4 1 = (4x − 1)(x + 1)(x − 1) 4 1 The rational zeros are and ± 1. 4 1 80. f ( z ) = (6 z 3 + 11z 2 − 3z − 2) 6 1 1 2 1 Possible rational zeros: ±1, ± 2, ± , ± , ± , ± 2 3 3 6

−2 6

−3

−2

− 12

6 −1

−1

1 f ( x ) = ( z + 2)(6 z 2 − z − 1) 6 1 = ( z + 2)(3z + 1)(2 z − 1) 6 1 1 Rational zeros: − 2, − , 3 2

(

=(x − 3 2 ) x 2 + 3 2 x + 3 4

Rational zeros: − 3,

261

)

Rational zeros: 0

(

Irrational zeros: 1 x = 3 2

)

Matches (a). 83. f ( x ) = x 3 − x = x ( x + 1)( x − 1)

Rational zeros: 3 ( x = 0, ± 1) Irrational zeros: 0 Matches (b). 84. f ( x ) = x 3 − 2 x = x ( x 2 − 2)

(

=x x+ 2

)( x − 2 )

Rational zeros: 1 ( x = 0)

(

Irrational zeros: 2 x = ± 2

)

Matches (c). 85. y = 2 x 4 − 9 x 3 + 5 x 2 + 3 x − 1

Using the graph and synthetic division, − −

1 2 2

−9 −1

5 5

3 −5

−1 1

− 10 10

−2

1 is a zero. 2

1  y =  x +  (2 x 3 − 10 x 2 + 10 x − 2) 2  x = 1 is a zero of the cubic, so y = (2 x + 1)( x − 1)( x 2 − 4 x + 1).

For the quadratic term, use the Quadratic Formula.

4 ± 16 − 4 =2± 3 2

1 The real zeros are − , 1, 2 ± 3. 2

262

Chapter 3

Polynomial and Rational Functions

86. y = x 4 − 5 x 3 − 7 x 2 + 13 x − 2

89. 5 x 4 + 9 x3 − 19 x 2 − 3x = 0 x(5 x3 + 9 x 2 − 19 x − 3) = 0

Using the graph and synthetic division, 1 and –2 are zeros.

x = 0 is a real zero. 1 3 Possible rational zeros: ±1, ± 3, ± , ± 5 5 −3 5 9 −19 − 3

y = ( x − 1)( x + 2)( x 2 − 6 x + 1)

For the quadratic term, use the Quadratic Formula.

6 ± 36 − 4 =3±2 2 2

−6

−1

x = −3 is a real zero.

The real zeros are 1, − 2, 3 ± 2 2.

Use the Quadratic Formula. 5 x 2 − 6 x − 1 = 0

87. y = −2 x 4 + 17 x 3 − 3 x 2 − 25 x − 3

x =

3 are Using the graph and synthetic division, −1 and 2 zeros.

y = −( x + 1)(2 x − 3)( x 2 − 8 x − 1)

−15

Real zeros: x = 0, −3, 90.

3±

8 ± 64 + 4 = 4 ± 17 2 3 The real zeros are −1, , 4 ± 17. 2

x = 0 is a real zero.

Using the graph and synthetic division, 2 and –1 are zeros.

Possible rational zeros: ±1, ± 2, ± 4, ± 8, ± 4 4

− 11 16

− 22 20

8 −8

−2

y = −( x − 2)( x + 1)( x 2 − 4 x − 2)

x = 4 is a real zero.

For the quadratic term, use the Quadratic Formula.

Use the Quadratic Formula. 4x2 + 5x − 2 = 0

4 ± 16 + 8 =2± 6 2

4 x 4 − 11x 3 − 22 x 2 + 8 x = 0 x(4 x 3 − 11x 2 − 22 x + 8) = 0

88. y = − x 4 + 5 x 3 − 10 x − 4

For the quadratic term, use the Quadratic Formula.

3±

The real zeros are 2, − 1, 2 ± 6.

1 1 ,± 4 2

−5 ± 57 8

Real zeros: x = 0, 4,

−5 ± 57 8

91. z 4 − z 3 − 2z − 4 = 0

Possible rational zeros: ±1, ± 2, ± 4 −1 1

−1 0 −1 2

1 −2 2 1 1

−2 −2

−4 4

−4

−2

2 −4

z = −1 and z = 2 are real zeros.

z 4 − z 3 − 2 z − 4 = ( z + 1)( z − 2)( z 2 + 2) = 0 The only real zeros are –1 and 2. You can verify this by graphing the function f ( z ) = z 4 − z 3 − 2 z − 4.

Section 3.3 92. 4 x 3 + 7 x 2 − 11x − 18 = 0

94.

Possible rational zeros: ±1, ± 2, ± 3, ± 6, ± 9, ± 18, 1 1 3 3 9 9 ± , ± , ± , ± , ± , ± 4 2 4 2 4 2 −2 4

7 −8

− 11 2

− 18 18

4 −1

−9

Real Zeros of Polynomial Functions

x5 − x 4 − 3x3 + 5x2 − 2 x = 0 x( x 4 − x 3 − 3 x 2 + 5 x − 2) = 0 x = 0 is a real zero. −1 − 3 5 −2 1 0 −3 2

1 1 1 1

x = −2 is a real zero.

x= Real zeros: x = −2,

1 ± 145 8

93. 2 y 4 + 7 y 3 − 26 y 2 + 23 y − 6 = 0

Possible rational zeros: 1 3 ± , ± 1, ± , ± 2, ± 3, ± 6 2 2

1 2 2 2

− 26 4

7 1

23 − 6 − 11 6

8 − 22

− 22

− 12

2 10

− 12

−6 2 2

− 12

−2

−2 2

Real zeros: y = −6, 1,

1 2

x = 1 and x = −2 are real zeros.

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

Real zeros: −2, 0, 1 95. 4 x 4 − 55 x 2 − 45 x + 36 = 0

Possible rational zeros: 1 3 3 9 9 ± , ± , ± 1, ± , ± 2, ± , ± 3, ± 4, ± , ± 6, ± 9, ± 18 2 4 2 4 2 4 4

0 − 55 16 64

4 16

−3 4

16 − 12

−3

4 2

−3 3

1 4 2

−

( y + 6)( y − 1)2 (2 y − 1) = 0

0 −3 2 −2 4 −2

1 −2

1 y = , y = 1, y = −6, and y = 1 are real zeros. 2

−3

−2

Use the Quadratic Formula. 4 x 2 − x − 9 = 0

263

3 2

− 45 36

36 − 36

−9

9 −9 − 12 9

6 −6

1 3 x = 4, x = −3, x = , and x = − are real zeros. 2 2 ( x − 4)( x + 3)(2 x − 1)(2 x + 3) = 0

Real zeros: x = 4, − 3,

1 3 ,− 2 2

264 96.

Chapter 3

Polynomial and Rational Functions

6 x 4 + 33x3 − 69 x + 30 = 0

x 5 + 5 x 4 − 5 x 3 − 15 x 2 − 6 x = 0

3( 2 x 4 + 11x3 − 23x + 10) = 0

x( x 4 + 5 x 3 − 5 x 2 − 15 x − 6) = 0

1 5 Possible rational zeros: ±1, ± 2, ± 5, ±10, ± , ± 2 2

x = 0 is a real zero.

−2 2 2

0 − 23

− 4 −14

28 −10

7 −14

−5 2

7 −14 −10

−3

−1 1

5 −1

−5

− 15

−6

−4

−9

−6

x = −1 is a real zero.

−9

−6

2 1

15 − 5 1

Possible rational zeros: ±1, ± 2, ± 3, ± 6

x = −2 is a real zero.

0 1

x = −5 is a real zero. 3( 2 x 4 + 11x3 − 23 x + 10) = 0

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

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

1 Real zeros: x = −2, −5, , 1 2 97.

98.

x = 2 is a real zero.

Use the Quadratic Formula.

x2 + 6 x + 3 = 0

x = −3 ± 6

Real zeros: x = 0, − 1, 2, − 3 ± 6 99. 8x5 + 6 x4 − 37 x3 − 36 x 2 + 29 x + 30 = 0

Possible rational zeroes:

8 x 4 + 28 x 3 + 9 x 2 − 9 x = 0 x(8 x 3 + 28 x 2 + 9 x − 9) = 0 x = 0 is a real zero.

Possible rational zeros: 1 3 9 1 3 9 ±1, ± 3, ± 9, ± , ± , ± , ± , ± , ± , 8 8 8 2 2 2 1 3 9 ± , ± , ± 4 4 4 −3 8

28 − 24

9 − 12

−9 9

−3

1 1 1 3 3 ±1, ± 3, ± 5, ±10, ±15, ± 30, ± , ± , ± , ± , ± , 2 4 8 2 4 3 5 5 5 15 15 15 ± ,± , ± ,± ,± ,± ,± 8 2 4 8 2 4 8 6 − 37 − 36

1 8

8 14 − 23 − 59 − 30 2 8 14 − 23 − 59 − 30 16

8 30

x = 2 is a real zero. −1 8

Real zeros: x = 0, − 3,

x = 1 is a real zero.

Use the Quadratic Formula. 8x2 + 4 x − 3 = 0

− 8 − 22 −15 8

−1± 7 4

14 − 23 − 59 − 30

x = −3 is a real zero.

x = −1 is a real zero. −1 ± 7 4

8 x5 + 6 x 4 − 37 x3 − 36 x 2 + 29 x + 30 = 0

( x − 1)( x − 2)( x + 1)(8x2 + 22 x + 15) = 0 ( x − 1)( x − 2)( x + 1)(4 x + 5)(2 x + 3) = 0 x = −

5 3 and x = − are real zeros. 4 2

5 3 Real zeros: x = 1, 2, −1, − , − . 4 2

Section 3.3 100. 4 x 5 + 8 x 4 − 15 x 3 − 23 x 2 + 11x + 15 = 0 Possible rational zeros: 1 1 3 3 5 ±1, ± 3, ± 5, ± 15, ± , ± , ± , ± , ± , 2 4 2 4 2 5 15 15 ± , ± , ± 4 2 4 1 4 4

8 − 15 4 12

− 23 −3

11 − 26

15 − 15

−3

− 26

− 15

−1 4

12 −4

−3 −8

− 26 11

− 15 15

− 11

− 15

−1 4

− 11

− 15

−4

− 15

4 4 6

− 15 15

4 10

3 2

5 − 2

(a)

x = 0, 3, 4, ±1.414

(b)

3 1 1

−7

− 24

− 12

−6

−4

−2

x = 3 is a zero.

4 1 1

(c)

−4 4

−2 0

8 −8

−2

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

104. g( x ) = 6 x 4 − 11x 3 − 51x 2 + 99 x − 27

(a) (b)

x = ±3.0, 1.5, 0.333 3 6 − 11 − 51

99 − 90

− 27 27

6 7 x = 3 is a zero.

− 30

3 5 x = 1, x = −1, x = −1, x = , and x = are real zeros. 2 2 3 5 Real zeros: x = 1, − 1, − 1, , − 2 2

− 30

− 18

−9

− 11

x = −3 is a zero.

(c)

101. h(t ) = t 3 − 2t 2 − 7t + 2 105. (a)

(a) Zeros: −2, 3.732, 0.268

−2 −2

−7 8

2 −2

−4

t = −2 is a zero.

g( x ) = ( x − 3)( x + 3)(6 x 2 − 11x + 3) = ( x − 3)( x + 3)(3 x − 1)(2 x − 3) 110

(b) The model fits the data well. (c) S = − 0.0223 x 3 + 0.825 x 2 − 3.58 x + 12.6 30 − 0.0223 − 0.0223

(a) Zeros 6, 5.236, 0.764 − 12 6

40 − 36

−6

0.825 − 3.58 12.6 − 0.669

102. f (s ) = s 3 − 12 s 2 + 40 s − 24

(b) 6 1

x = 4 is a zero.

h( x ) = x ( x − 3)( x − 4)( x 2 − 2)

−3 6

(b) −2 1

265

103. h( x ) = x 5 − 7 x 4 + 10 x 3 + 14 x 2 − 24 x

10 − 10

Real Zeros of Polynomial Functions

− 24 24 s = 6 is a zero. 0

0.156

4.68

1.1 45.6

In 2020 ( x = 30), the model predicts approximately 45.6 subscriptions per 100 people. This is not a reasonable prediction because you would not expect the number of subscriptions to decrease.

266

Chapter 3

Polynomial and Rational Functions

106. (a) 20,000

108. y = −5.05 x 3 + 3857 x − 38,411.25, 13 ≤ x ≤ 18 2700

(a) 0

(b) In 1960 (t = 0), there were approximately 3,167,000 employees. − 4.4 318.5 16,655

(b) The second air-fuel ratio of 16.89 can be obtained by finding the second point where the curves y and y1 = 2400 intersect.

6.37 333.1 19,852

50 − 0.088 10.77 − 0.088

14.6

3197

−5.05 x 3 + 3857 x − 40,811.25 = 0.

In 2010 (t = 50), the model predicts approximately 19,852,000 employees.

By synthetic division:

107. (a) Combined length and width: 2

Volume = l ⋅ w ⋅ h = x y = 4 x 2 (30 − x ) 18,000

Dimension with maximum volume: 20 × 20 × 40 13,500 = 4 x (30 − x ) 3

4 x − 120 x + 13,500 = 0

y = a( x − 1)2 ( x + 2), a < 0 Since f (0) = −4, a = −2.

1 −30 0 3375 15 −225 −3375 1 −15 −225

y = −2( x − 1)2 ( x + 2)

112. The zeros are 1, –1, and –2. The graph rises to the right.

( x − 15)( x 2 − 15 x − 225) = 0

Using the Quadratic Formula, x = 15 or The value of is negative.

4 is a zero of f. 7

111. The zeros are 1, 1, and –2. The graph falls to the right.

x 3 − 30 x 2 + 3375 = 0 15

1 110. False, f   ≈ −7.896 ≠ 0. 7

(c)

2720.75

75.75 − (−75.75)2 − 4( −5.05)(2720.75) ≈ 16.89 2( −5.05)

109. False, −

0 3857 −40,811.25 −75.75 −1136.25 40,811.25

The positive zero of the quadratic −5.05 x 2 − 75.75 x + 2720.75 can be found using the Quadratic Formula.

= x 2 (120 − 4 x )

−5.05

−5.05 −75.75

4 x + y = 120  y = 120 − 4 x

(b)

15 ± 15 5 . 2

15 − 15 5 is not possible because it 2

y = a( x − 1)( x + 1)( x + 2), a > 0 Since f (0) = −4, a = 2. y = 2( x − 1)( x + 1)( x + 2)

113. f ( x ) = −( x + 1)( x − 1)( x + 2)( x − 2) 114. Use synthetic division. 3

−k 2k −12 3 9 − 3k 27 − 3k

1 3 − k 9 − k 15 − 3 k Since the remainder 15 − 3k should be 0, k = 5.

Section 3.4

115. (a)

(b) (c)

x2 − 1 = x + 1, x ≠ 1 x −1 x3 − 1 = x 2 + x + 1, x ≠ 1 x −1 x4 − 1 = x 3 + x 2 + x + 1, x ≠ 1 x −1

The Fundamental Theorem of Algebra

117. 4 x 2 − 17 = 0 4 x 2 = 17 x2 =

In general, n

x − 1 n −1 = x + x n − 2 + . . . + x + 1, x ≠ 1. x −1

116. (a) f has 1 negative real zero. f ( − x ) has only 1 sign change. (b) Because f has 4 sign changes, f has either 4 or 2 positive real zeros. The graph either turns upward and rises to the right or it turns upward and crosses the x-axis, then turns downward and crosses the x-axis and then turns back upward and rises to the right. (c) No. No factor of 3 divided by a factor of 2 is equal 1 to − . 3 3 (d) Use synthetic division. If r = 0, then x − is a 2 factor of f. Otherwise, if each number in the last row 3 is either positive or 0, then x = is an upper bound. 2

267

17 4

x = ±

17 4

x = ±

17 2

118. 25 x 2 − 1 = 0 25 x 2 = 1 x2 =

1 25 1 25

x = ± x = ±

1 5

119. 3x 2 − 11x − 20 = 0

(3x + 4)( x − 5) = 0 3x + 4 = 0

or x − 5 = 0 4 3

x = 5

−4 ±

42 − 4(6)( − 3)

x = − 120. 6 x 2 + 4 x − 3 = 0 x =

2(6)

x =

− 4 ± 88 12

x =

− 4 ± 2 22 12

x =

− 2 ± 22 6

Section 3.4 The Fundamental Theorem of Algebra 1.

Fundamental Theorem, Algebra

irreducible, reals

The Linear Factorization Theorem states that a polynomial function f of degree n, n > 0 , has exactly n linear factors f ( x ) = a( x − c1 )( x − c2 ) ( x − cn ).

Since complex zeros occur in conjugate pairs, if a fourthdegree polynomial function has zeros –1, 3, and 2i, then – 2i is also a zero.

5. f ( x) = − 2 x4 + 32 has exactly 4 zeros. Matches (c).

6. f ( x) = x5 − x3 has exactly 5 zeros. Matches (d). 7. f ( x) = x3 + 3x 2 + 2 x has exactly 3 zeros. Matches (b). 8. f ( x ) = x − 14 has exactly 1 zero. Matches (a). 9.

f ( x) = x 2 + 5 x2 + 5 = 0 x2 = − 5 x = ±

−5

x = ±

268

Chapter 3

Polynomial and Rational Functions

10.

f ( x) = x3 + 9 x

16.

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

x3 + 9x = 0 x ( x 2 + 9) = 0 x=0

= ( x + 2)( x − 2)( x 2 + 1)

x2 + 9 = 0

Zeros: ±2, ±i

x 2 = −9

The only real zeros are x = −2, 2. This corresponds to the x-intercepts of (−2, 0) and (2, 0) on the graph.

x = ± −9 x = ±3i

17. h( x ) = x 2 − 4 x + 1

11. f ( x) = 3x4 − 48

h has no rational zeros. By the Quadratic Formula, the zeros are

3x4 − 48 = 0 3( x4 − 16) = 0

4 ± 16 − 4 = 2 ± 3. 2 h( x ) =  x − 2 + 3   x − 2 − 3     x=

( x − 4)( x + 4) = 0 2

x2 − 4 = 0

x2 + 4 = 0

x2 = 4

x2 = − 4

x = ±2

x = ±

(

−4

)

18. g( x ) = x 2 + 10 x + 23 g has no rational zeros. By the Quadratic Formula, the zeros

−10 ± 8 = −5 ± 2. 2 g( x ) =  x − (−5 + 2 )  x − (−5 − 2 )   

are x =

12. f ( x) = 2x5 − 2x

2 x5 − 2 x = 0 2 x( x 4 − 1) = 0

= ( x + 5 + 2 )( x + 5 − 2 )

2 x( x 2 − 1)( x2 + 1) = 0

19.

2x = 0

x2 − 1 = 0

x2 + 1 = 0

x = 0

x =1

x = −1

x = ±1

x = ±

f ( x ) = x 2 − 12 x + 26 f has no rational zeros. By the Quadratic Formula, the zeros are

12 ± ( −12)2 − 4(26) = 6 ± 10. 2 f ( x ) =  x − (6 + 10 )  x − (6 − 10 )   

−1

x = ±i 13. f ( x) = x3 − 4 x 2 + x − 4 = x 2 ( x − 4) + 1( x − 4)

= ( x − 4)( x 2 + 1) Zeros: 4, ± i The only real zero of f(x) is x = 4. This corresponds to the x-intercept of (4, 0) on the graph.

= ( x − 6 − 10 )( x − 6 + 10 )

20.

−6 ± 62 − 4(−2) = −3 ± 11. 2 f ( x ) =  x − (−3 + 11)  x − (−3 − 11)   

f ( x ) = x 3 − 4 x 2 − 4 x + 16

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

The zeros are x = 2, −2, and 4. This corresponds to the xintercepts of (−2, 0) , (2, 0), and (4, 0) on the graph.

= ( x + 3 − 11)( x + 3 + 11)

21.

f ( x ) = x 2 + 25 Zeros: ±5i f ( x ) = ( x + 5i )( x − 5i )

22.

f ( x ) = x 2 + 36 Zeros: ±6i f ( x ) = ( x + 6i )( x − 6i )

f ( x ) = x 4 + 4 x 2 + 4 = ( x 2 + 2)2

Zeros: ± 2i, ± 2i f(x) has no real zeros and the graph of f(x) has no x-intercepts.

f ( x) = x2 + 6 x − 2 f has no rational zeros. By the Quadratic Formula, the zeros are

= x 2 ( x − 4) − 4( x − 4)

15.

(

= ( x − 2 − 3)( x − 2 + 3 )

x = ± 2i

14.

)

Section 3.4

23.

29.

f ( x ) = 16 x 4 − 81 = (4 x 2 − 9)(4 x 2 + 9) = (2 x − 3)(2 x + 3)(2 x + 3i )(2 x − 3i ) 3 3 Zeros: ± , ± i 2 2

24.

1 ± 1 − 4(56) 2

1 ± −223 2

1 223 ± i 2 2

0 48

(

The zeros are 30.

0 75

(

27.

The zeros are 31.

f ( x ) = x 4 + 10 x 2 + 9 = ( x 2 + 1)( x 2 + 9) = ( x + i )( x − i )( x + 3i )( x − 3i )

28.

)

= (3 x − 2)( x 2 + 25) = (3 x − 2)( x + 5i )( x − 5i )

4 ± 16 + 12 x= =2± 7 2

)( x − 2 − 7 )

2  f ( x ) =  x −  3 x 3 + 75 3 

26. h( x ) = x 2 − 4 x − 3

(

5 , 4i, − 4i. 3

f ( x ) = 3 x 3 − 2 x 2 + 75 x − 50 2 is a zero: Using synthetic division, 3 2 3 −2 75 −50 3 2 0 50 3

1 223 ± i 2 2  1 223i  1 223i  f (z) =  z − + z− −    2 2  2 2   

h( x ) = x − 2 + 7

)

= (3 x − 5)( x 2 + 16) = (3 x − 5)( x + 4i )( x − 4i )

Zeros:

Zeros: 2 ± 7

5  f ( x ) =  x −  3 x 2 + 48 3  

f ( z ) = z 2 − z + 56 z=

2 , 5i, − 5i. 3

f (t ) = t 3 − 3t 2 − 15t + 125 Possible rational zeros: ±1, ± 5, ± 25, ± 125 −5

1 −3 −15 125 −5 40 −125 1 −8

The zeros of f(x) are x = ±i and x = ±3i.

By the Quadratic Formula, the zeros of t 2 − 8t + 25 are

f ( x ) = x 4 + 29 x 2 + 100

269

f ( x ) = 3 x 3 − 5 x 2 + 48 x − 80 5 is a zero: Using synthetic division, 3 5 3 −5 48 −80 3 5 0 80 3

f ( y) = 81y 4 − 625 = (9 y 2 + 25)(9 y 2 − 25) = (3 y + 5i )(3 y − 5i )(3 y + 5)(3 y − 5) 5 5 Zeros: ± , ± i 3 3

25.

The Fundamental Theorem of Algebra

8 ± 64 − 100 = 4 ± 3i. 2

= ( x + 25)( x + 4) Zeros: x = ±2i, ± 5i

The zeros of f(t) are t = −5 and t = 4 ± 3i.

f ( x ) = ( x + 2i )( x − 2i )( x + 5i )( x − 5i )

f (t ) = t − (−5) t − (4 + 3i) t − (4 − 3i) = ( t + 5 )( t − 4 − 3i )( t − 4 + 3i )

270

Chapter 3

32.

f ( x ) = x 3 + 11x 2 + 39 x + 29 −1

Polynomial and Rational Functions 36. h( x ) = x 4 + 6 x 3 + 10 x 2 + 6 x + 9 9 −3 1 6 10 6

1 11 39 29 −1 −10 −29 1 10

−3 −9 −3 −9 −3

−10 ± 16i = −5 ± 2i 2

Zeros: x = −1,

33.

6 3 2 1 Possible rational zeros: ±6, ± , ± 3, ± , ± 2, ± , ± 1, ± 5 5 5 5 1 5 −9 28 6 − 5 − 1 2 −6

By the Quadratic Formula,

= ( x − 7 − 3)( x − 7 + 3)

2 ± 4 − 4(6) = 1 ± 5i 2 1 Zeros: − , 1 ± 5i 5

(

)

(

)(

)

2 ± 4 − 16 = 1 ± 3i. 2 2 Zeros: − , 1 ± 3i 3

(

35. g( x ) = x 4 − 4 x 3 + 8 x 2 − 16 x + 16

Possible rational zeros: ±1, ± 2, ± 4, ± 8, ± 16 2 1 −4 8 −16 16 2 −4 8 −16

(

12 ± (−12)2 − 4(34) = 6 ± 2. 2

The zeros are 6 + 2 and 6 − 2. (b) f ( x) =  x − 6 + 2   x − 6 − 2    

f ( s) = (3s + 2)( s − 1 + 3i )( s − 1 − 3i )

f ( x ) = x 2 − 12 x + 34

By the Quadratic Formula,

Factoring the quadratic,

)

−4

38. (a)

= (3s + 2)(s 2 − 2 s + 4)

)

f ( s ) = 3s 3 − 4 s 2 + 8 s + 8

(

= ( 5 x + 1) x − 1 − 5i x − 1 + 5i

−8

)

1  f ( x) = 5  x +   x − 1 + 5i   x − 1 − 5i    5  

14 ± (−14)2 − 4(46) = 7 ± 3. 2

f ( x ) =  x − (7 + 3)  x − (7 − 3)   

(b)

are those of x 2 − 2 x + 6 :

1 −2

0 −3

The zeros are 7 + 3 and 7 − 3.

By the Quadratic Formula, the zeros of 5 x − 10 x + 30

−3

f ( x ) = x 2 − 14 x + 46.

37. (a)

0 2

34.

f ( x ) = 5 x − 9 x + 28 x + 6

(

h( x ) = ( x + 3 ) ( x + i )( x − i )

5 −10 30

1 0 1 0 Zeros: x = −3, − 3, ± i

f ( x ) = ( x + 1)( x + 5 + 2i )( x + 5 − 2i ) 3

g( x ) = ( x − 2 )( x − 2 ) x 2 + 4

) ( ) = ( x − 6 − 2 )( x − 6 + 2 ) (c) x-intercepts: ( 6 + 2,0 ) ( 6 − 2,0 ) 40

−2

13 − 10

)

= ( x − 2 ) ( x + 2i )( x − 2i ) 2

The zeros of g are 2, 2, and ±2i .

Section 3.4 39. (a)

f ( x ) = 2 x 3 − 3 x 2 + 8 x − 12

(

= ( 2 x − 3) x 2 + 4

The Fundamental Theorem of Algebra

42. (a)

)

(

3  (c) x-intercept:  , 0  2 

−4

(b)

f ( x) = ( x + 2 )( x + 4 + i )( x + 4 − i )

−50

f ( x ) = 2 x 3 − 5 x 2 + 18 x − 45 = (2 x − 5)( x 2 + 9) 5 and ±3i . The zeros are 2

(b)

−8 ± 82 − 4(17) 2

−8 ± −4 = −4 + i 2 The zeros are −2, − 4 + i, and − 4 − i.

40. (a)

− 20

f ( x ) = ( 2 x − 5 )( x + 3i )( x − 3i )

5  (c) x-intercept:  , 0  2 

− 10

43. (a)

f ( x ) = x 4 + 25 x 2 + 144

(

)

The zeros are ±3i, ± 4i.

−4

(b)

(

)(

f ( x ) = x 2 + 9 x 2 + 16

)

= ( x + 3i )( x − 3i )( x + 4i )( x − 4i ) (c) No x-intercepts

−80

200

f ( x ) = x 3 − 11x + 150

(

= ( x + 6 ) x 2 − 6 x + 25

)

Use the Quadratic Formula to find the zeros of x 2 − 6 x + 25.

−5

44. (a)

f ( x ) = x 4 − 8 x 3 + 17 x 2 − 8 x + 16

( )( = ( x + 1) ( x − 4 )

6 ± ( −6) − 4(25) = 3 ± 4i. 2 The zeros are – 6, 3 + 4i, and 3 − 4i.

= x 2 + 1 x 2 − 8 x + 16

f ( x ) = ( x + 6 )( x − 3 + 4i )( x − 3 − 4i )

5 −10

(b)

)(

= x 2 + 9 x 2 + 16

41. (a)

)

Use the Quadratic Formula to find the zeros of x 2 + 8 x + 17.

The zeros are (b)

f ( x ) = x 3 + 10 x 2 + 33 x + 34 = ( x + 2 ) x 2 + 8 x + 17

3 and ±2i. 2 f ( x ) = (2 x − 3)( x + 2i )( x − 2i )

271

)

The zeros are i, − i, 4, and 4. (b)

(

)

f ( x) = x 2 + 1 ( x − 4 )

300

40 − 10

− 300

−4

8 −5

272

Chapter 3

Polynomial and Rational Functions

45. f ( x) = ( x − 5)( x − i )( x + i )

51. (a) f ( x ) = a ( x + 1)( −2 )( x − i )( x + i )

= ( x − 5)( x + 1)

(

f (1) = 8 = a(2)( −1)(2)  a = −2

= x3 − 5 x 2 + x − 5

f ( x ) = −2 ( x + 1)( x − 2 )( x − i )( x + i )

Note: f ( x) = a( x3 − 5 x 2 + x − 5), where a is any

(

Note: f ( x ) = a( x 3 − 3 x 2 + 16 x − 48) , where a is any nonzero real number, has zeros 3, ± 4i.

f (0) = − 6 = a( −1)( 4)(3)  a = f ( x) =

47. Because 2i is a zero, so is −2i.

f ( x) = ( x − 1) ( x − 2i )( x + 2i ) 2

= ( x 2 − 2 x + 1)( x 2 + 4) = x 4 − 2 x3 + 5 x 2 − 8 x + 4 Note: f ( x) = a( x 4 − 2 x3 + 5 x 2 − 8 x + 4), where a is

(

53. (a) f ( x) = a( x + 1) x − 2 −

(

f ( x) = 3( x + 1) x − 2 −

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

)(

5i x − 2 +

)(

5i x − 2 +

(

)

Note: f ( x) = a( x 4 + 6 x3 + 10 x 2 + 6 x + 9), where a

54. (a)

is any nonzero number, has zeros − 3, − 3, ± i. 49. Because 1 + 2i is a zero, so is 1 − 2i.

(

)(

f ( x ) = ( x − 0 )( x + 5) x − 1 − 2i x − 1 + 2i

( )( ) = ( x + 5 x )( x − 2 x + 3 )

= 3x − 9 x + 15 x + 27

= x 4 + 6 x3 + 10 x 2 + 6 x + 9

(

)(

f ( x) = a ( x + 2 ) x − 2 − 2 2i x − 2 + 2 2i

( ) = a ( x + 2 ) ( x − 4 x + 12 )

)

= a ( x + 2) x2 − 4 x + 4 + 8 2

)

f ( −1) = −34 = a(1)(17)  a = −2

(

)(

f ( x ) = −2 ( x + 2 ) x − 2 − 2 2 i x − 2 + 2 2i

= x2 + 5x x2 − 2 x + 1 + 2

(b)

(

f ( x ) = −2 ( x + 2 ) x − 4 x + 12 3

)

= −2 x + 4 x − 8 x − 48

= x 4 + 3 x 3 − 7 x 2 − 15 x 2

Note: f ( x ) = a( x + 3 x − 7 x + 15 x ), where a is any

nonzero real number, has zeros 0, − 5, 1 ± 2 i. 50. Because 1 + 2i is a zero, so is 1 ± 2 i.

(

)(

f ( x ) = ( x − 0 )( x − 4 ) x − 1 − 2i x − 1 + 2i

)(

)

(b) f ( x) = 3( x + 1) x 2 − 4 x + 9

= ( x 2 + 6 x + 9)( x 2 + 1)

= x − 4x x − 2x + 1 + 2 3

f ( 2) = 45 = a(3)( 4 − 8 + 9)  a = 3

f ( x) = ( x + 3) ( x − i)( x + i )

)(

3i x +

= a( x + 1)( x 2 − 4 x + 9)

48. Because i is a zero, so is −i.

(

1 2

= a( x + 1)( x 2 − 4 x + 4 + 5)

any nonzero number, has zeros 1, 1, ± 2i.

(

)

1 2 ( x + 3x − 4)( x2 + 3) 2 1 3 1 9 = x 4 + x3 − x 2 + x + 6 2 2 2 2

= ( x − 1) ( x + 4)

1 ( x − 1)( x + 4) x − 2

(b) f ( x) =

)(

3i x +

= a( x − 1)( x + 4)( x 2 + 3)

= x 3 − 3 x 2 + 16 x − 48

52. (a) f ( x) = a( x − 1)( x + 4) x −

= ( x − 3)( x 2 + 16)

)

= −2 x + 2 x + 2 x + 2 x + 4

f ( x ) = ( x − 3)( x − 4i )( x + 4i )

)(

(b) f ( x ) = −2 x 2 − x − 2 x 2 + 1

nonzero real number, has zeros 5, ± i. 46.

)

= a ( x + 1)( x − 2 ) x 2 + 1

)

55.

f ( x) = x 4 − 6 x 2 − 7 (a)

(

)(

)

f ( x) = x2 − 7 x 2 + 1

( )( )( ) (c) f ( x ) = ( x − 7 )( x + 7 ) ( x + i )( x − i ) (b)

f ( x) = x − 7 x + 7 x 2 + 1

= x − 6 x + 11x − 12 x Note: f ( x ) = a( x 4 − 6 x 3 + 11x 2 − 12 x ), where a is any

nonzero real number, has zeros 0, 4, 1 ± 2 i.

Section 3.4 56.

f ( x ) = x 4 + 6 x 2 − 27 (a) (b) (c)

57.

60.

( )( ) f ( x ) = ( x + 9 ) ( x + 3 )( x − 3 ) f ( x) = x2 + 9 x2 − 3

(

)(

f ( x ) = ( x + 3i )( x − 3i ) x + 3 x − 3

(

)(

f ( x) = x2 − 6 x2 − 2 x + 3

)

( )( x − 6 ) ( x − 2 x + 3) (c) f ( x ) = ( x + 6 )( x − 6 )( x − 1 − 2i )( x − 1 + 2 i ) (b)

58.

f ( x ) = x 4 − 3 x 3 − x 2 − 12 x − 20

)

f ( x) = x2 + 4 x2 − 3x − 5

(b)

 3 + 29  3 − 29  f ( x) = x2 + 4  x − x−     2 2   

)(

(a)

(

−5i

15i

2 3 + 10i 15i −10i −15i 2

3 3 The zero of 2 x + 3 is x = − . The zeros of f are x = − 2 2 and x = ±5i. Alternate solution Since x = ±5i are zeros of f ( x),

( x + 5i)( x − 5i) = x 2 + 25 is a factor of f ( x ). By long division we have: 2x + 3 x 2 + 0 x + 25

2 x 3 + 3 x 2 + 50 x + 75 2 x 3 + 0 x 2 + 50 x 3x 2 + 0 x + 75 3x 2 + 0 x + 75 0

(

)

Thus, f ( x ) = x 2 + 25 ( 2 x + 3 ) and the zeros of

3 f are x = ±5i and x = − . 2

1 1 0 The zeros of f are 3i, − 3i, and −1 .

61. g( x ) = x 3 − 7 x 2 − x + 87. Since 5 + 2i is a zero, so is 5 − 2i. 1 −7 −1 87 5 + 2i 5 + 2i −14 + 6i −87 1 −2 + 2i −15 + 6i

1 −2 + 2i −15 + 6i 5 − 2i 15 − 6i

5 − 2i

 3 + 29  3 − 29  f ( x ) = ( x + 2i )( x − 2i )  x − x−     2 2   

2 3 + 10i

1 9 9 3i −9 + 3i −9 1 1 + 3i 3i −3i −3i

−3i

)

f ( x ) = 2 x 3 + 3 x 2 + 50 x + 75 Since is 5i a zero, so is −5i . 2 3 50 75 5i 10i −50 + 15i −75

1 1 + 3i

f ( x) = x + 6

(

)

273

f ( x) = x3 + x 2 + 9 x + 9 Since is 3i is a zero, so is −3i .

f ( x ) = x 4 − 2 x 3 − 3 x 2 + 12 x − 18 (a)

The Fundamental Theorem of Algebra

The zero of x + 3 is x = −3. The zeros of f are −3, 5 ± 2i. . 62.

g ( x) = 4 x3 + 23x3 + 34 x − 10 Since −3 + i is a zero, so is −3 − i. −3 + i

−3 − i

4 4 4

34 −10

− 12 + 4i −37 − i

11 + 4i 11 + 4i

−3−i −3 − i

− 12 − 4i

3+i

−1

0 1 The zero of 4 x − 1 is x = . The zeros of g(x) are 4 1 x = −3 ± i and x = . 4

Alternate solution Since −3 ± i are zeros of g(x), [ x − ( −3 + i )][ x − ( −3 − i )] = [( x − 3) − i ][( x + 3) + i ] = ( x + 3)2 − i 2 = x 2 + 6 x + 10 is a factor of g(x). By long division we have: 4x −1 x 2 + 6 x + 10

4 x 3 + 23 x 2 + 34 x − 10 4 x 3 + 24 x 2 + 40 x − x 2 − 6 x − 10 − x 2 − 6 x − 10 0

Thus, g( x ) = ( x 2 + 6 x + 10)(4 x − 1) and the zeros of g are 1 x = −3 ± i and x = . 4

274

Chapter 3

Polynomial and Rational Functions

63. h( x ) = 3 x 3 − 4 x 2 + 8 x + 8 Since 1 − 3i is a zero, so

66.

is 1 + 3i. −4

1 − 3i

8 8 − 3 − 3 3i −10 − 2 3i

1 + 3i

3 −1 − 3 3i −2 − 2 3i 3 −1 − 3 3i −2 − 2 3i 3 + 3 3i

2 + 2 3i

−2

f ( x ) = 25 x 3 − 55 x 2 − 54 x − 18 Since

1 −2 + 2i (−2 + 2i) = is a zero, so is 5 5

−55

− 75

The zero of 25 x − 75 is x = 3. The zeros of f are

x = 3,

−2 ± 2i . 5

f ( x ) = x 4 + 3 x 3 − 5 x 2 − 21x + 22 (a) The root feature yields the real roots 1 and 2, and the complex roots −3 ± 1.414i. (b) By synthetic division: 1 1 3 −5 −21 22 1 4 −1 −22 1 4 −1 −22 2

68.

1 6

The complex roots of x 2 + 6 x + 11 are

−6 ± 62 − 4(11) x= = −3 ± 2 i. 2

8 −14 18 −9 6 −6 9 − 8 12

f ( x ) = 25 x 3 − 55 x 2 − 54 x − 18 (a) The root feature yields the real root 3 and the complex roots −0.4 ± 0.2828i. (b) By synthetic division: 25 −55 −54 −18 75 60 18 25

The complex roots of 25 x 2 + 20 x + 6 are

1 4 −1 −22 2 12 22

8 ± 64 − 4(8)(12) 1 5 = ± i. 2(8) 2 2

65.

The complex roots of 8 x 2 − 8 x + 12 are

25 −65 + 5 2i −30 − 15 2i −10 − 5 2i 30 + 15 2i 25

0 −2 − 2i 5

−2 ± 4 − 4(1)(10) = −1 ± 3i. 2

3 4

−54 −18 18 24 − 15 2i

−10 − 5 2i

4 14 20 −2 −4 −20

67. h( x ) = 8 x 3 − 14 x 2 + 18 x − 9 (a) The root feature yields the real root 0.75, and the complex roots 0.5 ± 1.118i. (b) By synthetic division:

−2 − 2 i . 5 −2 + 2i 5

The complex roots of x 2 + 2 x + 10 are

2 The zero of 3 x + 2 is x = − . The zeros of h are 3 2 x = − , 1 ± 3i. 3 64.

f ( x ) = x 3 + 4 x 2 + 14 x + 20 (a) The root feature yields the real root – 2 and the complex roots −1 ± 3i. (b) By synthetic division:

−20 ± 400 − 4(25)6 −2 ± 2i = . 2(25) 5

69. To determine if the football reaches a height of 50 feet, set h(t ) = 50 and solve for t.

−16t 2 + 48t = 50 −16t 2 + 48t − 50 = 0 Using the Quadratic Formula: t=

−(48) ± 482 − 4( −16)(−50) 2(−16)

−(48) ± −896 −32

Because the discriminant is negative, the solutions are not real, therefore the football does not reach a height of 50 feet.

Section 3.5 70. First find the revenue quation, R = xP . R = x (140 − 0.001x )

R = 140 x − 0.001x Then P = R − C

75.

81  7  , −  2 4  

P = −0.001x 2 + 100 x − 150,000. Next, set P = 3,000,000 and solve for x.

76. f ( x ) = − x 2 + x + 6 2

−0.001x 2 + 100 x − 3,150,000 = 0 Using the Quadratic Formula

1  25  = − x −  + 2 4  A parabola opening downward with vertex

−(100) ± (100)2 − 4(−0.001)(−3,150,000) 2(−0.001)

−100 ± −260 . −0.002 Because the discriminant is negative, the solutions are not real. Therefore, there is no price P that will yield a profit of $3 million. x=

 1 25   ,  2 4 

77.

f ( x) = 6 x 2 + 5 x − 6 2

5  169  = 6 x +  − 12 24   A parabola opening upward with vertex

71. False, a third-degree polynomial must have at least one real zero. 72. False. Because complex zeros occur in conjugate pairs, [ x + (4 − 3i )] is also a factor of f.

f ( x) = x 2 − 7 x − 8 2

P = (140 x − 0.001x 2 ) − (40 x + 150,000)

275

7  81  =x−  − 2 4  A parabola opening upward with vertex

−0.001x 2 + 100 x − 150,000 = 3,000,000

Rational Functions and Asymptotes

169   5 − , −  24   12

78.

73. Answers will vary.

f ( x ) = 4 x 2 + 2 x − 12 2

1  49  = 4 x +  − 4 4  A parabola opening upward with vertex 49   1 − , −  4 4  

74. (a) A vertical shift k units upward, where 0 < k < 4

(b) A vertical shift 4 units upward (c) A vertical shift k units downward, where k > 0 (d) A vertical shift k units upward, where k > 4

Section 3.5 Rational Functions and Asymptotes 1.

rational functions

vertical asymptote

To determine the vertical asymptote(s) of the graph of 9 y= , find the real zeros of the denominator of the x −3 equation. (Assuming no common factors in the numerator and denominator)

No, the x-axis, y = 0, is the horizontal asymptote of the 2x graph of y = 2 because the numerator’s degree is 3x − 5 less than the denominator’s degree

1 x −1 (a) Domain: all x ≠ 1 (b) x f ( x)

f ( x)

0.5

−2

1.5

0.9

−10

1.1

f ( x) =

0.99

−100

1.01

100

0.999

−1000

1.001

1000

f ( x)

0.25

−5

−0.16

0.1

−10

−0.09

100

0.01

−100

−0.0099

1000

0.001

−1000

−0.00099

276

Chapter 3

f ( x) =

Polynomial and Rational Functions

5x x −1

(a) Domain: all x ≠ 1 (b)

f ( x) =

3 x −1

(a) Domain: all x ≠ 1 (b)

f ( x)

0.5

−5

1.5

0.5

1.5

0.9

−45

1.1

0.9

1.1

0.99

−495

1.01

505

0.99

300

1.01

300

0.999

−4995

1.001

5005

0.999

3000

1.001

3000

f ( x)

6.25

−5

4.16

0.75

−5

0.5

5.55

−10

4.54

0.33

−10

0.27

100

5.05

−100

4.950495

100

0.03

−100

0.0297

1000

5.005

−1000

4.995

1000

0.003

−1000

0.003

f ( x) =

3x x −1

f ( x) =

3x2 x2 − 1

(a) Domain: all x ≠ 1 (b)

(a) Domain: all x ≠ 1 (b) x

f ( x)

0.5

1.5

0.5

−1

1.5

5.4

0.9

1.1

0.9

−12.79

1.1

17.29

0.99

297

1.01

303

0.99

−147.8

1.01

152.3

0.999

2997

1.001

3003

0.999

−1498

1.001

1502.3

f ( x)

3.75

−5

−2.5

3.125

−5

3.125

3.33

−10

−2.727

3.03

−10

3.03

100

3.03

−100

−2.970

100

3.0003

−100

3.0003

1000

−1000

−2.997

1000

−1000

3.003

(c) f approaches −∞ from the left of 1, and ∞ from the right of 1. f approaches ∞ from the left of –1, and −∞ from the right of –1.

Section 3.5

10.

f ( x) =

4x x2 − 1

16.

(a) Domain: all x ≠ ±1 (b) x

f ( x)

0.5

−2.66

1.5

4.8

0.9

−18.95

1.1

20.95

0.99

−199

1.01

201

0.999

−1999

1.001

2001

f ( x)

−0.833

−5

−0.833

0.40

−10

−0.40

100

0.04

−100

−0.04

1000

0.004

−1000

−0.004

17.

18.

12.

2 x+2 Vertical asymptote: x = −2 Horizontal asymptote: y = 0 Matches graph (a). f ( x) =

1 x −3 Vertical asymptote: x = 3 Horizontal asymptote: y = 0 Matches graph (d).

1− x 14. f ( x ) = x Vertical asymptote: x = 0 Horizontal asymptote: y = −1 Matches graph (e).

15.

x+2 x+4 Vertical asymptote: x = −4 Horizontal asymptote: y = −1 Matches graph (f). f ( x) = −

1 x2 Vertical asymptote: x = 0 Horizontal asymptote: y = 0 or x-axis f ( x) =

f ( x) =

19.

f ( x) =

x−2 x−4 Vertical asymptote: x = 4 Horizontal asymptote: y = 1 Matches graph (b). f ( x) =

( x − 2)

3 x − 4x + 4 2

2x2 2 x2 = x + x − 6 ( x + 3 )( x − 2 ) 2

Vertical asymptotes: x = −3, x = 2 Horizontal asymptote: y = 2 20.

f ( x) =

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

Vertical asymptotes: x = −2, x = 2 Horizontal asymptote: y = 1 21.

f ( x) =

x (2 + x) 2x − x

x ( x + 2)

− x ( x − 2)

=−

x+2 , x≠0 x−2

Vertical asymptote: x = 2 Horizontal asymptote: y = −1 Hole at x = 0

f ( x) =

4x + 1 13. f ( x ) = x Vertical asymptote: x = 0 Horizontal asymptote: y = 4 Matches graph (c).

277

Vertical asymptote: x = 2 Horizontal asymptote: y = 0 or x-axis

(c) f approaches −0.833 ∞ from the left of 1, and ∞ from the right of 1. f approaches −∞ from the left of –1, and ∞ from the right of –1. 11.

Rational Functions and Asymptotes

22.

f ( x) =

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

Vertical asymptote: x = 32 Horizontal asymptote: y = 12 Hole at x = −1 23. f ( x) =

( x + 4)( x − 4) x 2 − 16 = 2 x + 8x x( x + 8)

Vertical asymptotes: x = 0, and x = − 8 Horizontal asymptote: y = 1 No hole 24.

f ( x) =

3 − 14 x − 5 x 2 − ( x + 3 ) ( 5 x − 1) − ( 5 x − 1) = = 2x + 1 3 + 7x + 2 x2 ( x + 3) ( 2 x + 1)

Vertical asymptote: x = − 12 Horizontal asymptote: y = − 52 Hole at x = −3

278

Chapter 3

25. f ( x) =

Polynomial and Rational Functions

5x2 − 2 x − 6 x2 + 4

30. f ( x ) =

(a) Domain: all real numbers

g( x ) = x + 3 (a) Domain of f: all real x except x = 3 Domain of g: all real x (b) Vertical asymptote: f has none. g has none. Hole: f has a hole at x = 3; g has none. (c) x 3 0 1 2 4 f ( x) 3 5 Undef. 7 4 7 g( x ) 3 5 6 4

(b) Continuous (c) Vertical asymptote: none Horizontal asymptote: y = 5 26.

27.

3x2 + 1 x2 + x + 9 (a) Domain: all real numbers. The denominator has no real zeros. [Try the Quadratic Formula on the denominator.] (b) Continuous (c) Vertical asymptote: none Horizontal asymptote: y = 3 f ( x) =

( x + 4 )( x − 1) x2 + 3x − 4 = − x 3 + 27 − ( x − 3) x 2 + 3 x + 9

f ( x) =

(

−9

(

−4

Graphing utilities are limited in their resolution and therefore may not show a hole in a graph.

31.

)

f ( x) =

g( x ) = x + 4 (a) Domain of f: all real x except x = 4 Domain of g: all real x (b) Vertical asymptote: f has none. g has none. Hole: f has a hole at x = 4; g has none. (c) x

5 5

6 6

3 7 7

Undef. 8

5 9 9

7 11 11

−5

(d)

(d) 6 10 10

( x + 1) ( x − 1) x − 1 x2 − 1 = = , x ≠ −1 2 x − 2 x − 3 ( x − 3) ( x + 1) x − 3

x −1 x −3 (a) Domain of f: all real x except x = 3 and x = −1 Domain of g: all real x except x = 3 (b) Vertical asymptote: f has a vertical asymptote at x = 3. g has a vertical asymptote at x = 3. Hole: f has a hole at x = −1; g has none. (c) x 0 3 −2 −1 1 2 4 3 1 f ( x) Undef. 0 −1 Undef. 3 5 3 3 1 1 g( x ) 0 −1 Undef. 3 5 2 3

x 2 − 16 ( x + 4) ( x − 4) = = x + 4, x ≠ 4 x−4 x−4

f ( x) g( x )

f ( x) =

g( x ) =

(a) Domain: all real numbers x except x = −2 (b) Not continuous at x = −2 (c) Vertical asymptote: x = −2 Horizontal asymptote: y = 43 29.

6 9 9

)

4 x3 − x2 + 3 4 x3 − x2 + 3 = 3 3 x + 24 3( x + 2) x2 − 2 x + 4

f ( x) =

5 8 8

(d)

(a) Domain: all real numbers x except x = 3 (b) Not continuous at x = 3 (c) Vertical asymptote: x = 3 Horizontal asymptote: y = 0 or x-axis 28.

x 2 − 9 ( x + 3) ( x − 3) = = x + 3, x ≠ 3 x −3 x −3

(e) Graphing utilities are limited in their resolution and therefore may not show a hole in a graph. −9

−4

(e) Graphing utilities are limited in their resolution and therefore may not show a hole in a graph.

Section 3.5

f ( x) =

32.

( x + 2) ( x − 2) x + 2 x2 − 4 = = , x≠2 x − 3 x + 2 ( x − 2) ( x − 1) x − 1

37. f ( x ) =

x+2 x −1 (a) Domain of f: all real x except x = 1 and x = 2 Domain of g: all real x except x = 1 (b) Vertical asymptote: f has a vertical asymptote at x = 1. g has a vertical asymptote at x = 1. Hole: f has a hole at x = 2; g has none. (c)

g( x ) =

f ( x) g( x )

−3 1 4 1 4

−2

0 0

−2

Undef.

−2

Undef.

−1

1 − 2 1 − 2

−5

x2 − 9 x2 + 5 ( x − 3)( x + 3) x2 + 5

38. h( x) =

x3 + 8 x 2 − 11

The zero of h corresponds to the zero of the numerator and is x = − 2. 3 5 2 5 2

6 x −3 x +3 = x −3

39. g ( x) = 1 +

The zero of g corresponds to the zero of the numerator and is x = − 3. 7

−12 12 = 3+ 2 x +2 x +2 3x 2 + 18 = 2 x + 2

40. f ( x) = 3 − −4

(e) Graphing utilities are limited in their resolution and therefore may not show a hole in a graph. 33.

1 x As x → ±∞, f ( x ) → 4. As x → ∞, f ( x ) → 4 but is less than 4. As x → −∞, f ( x ) → 4 but is greater than 4.

36.

2x − 1 x −3 As x → ±∞, f ( x ) → 2 . As x → ∞, f ( x ) → 2 but is greater than 2. As x → −∞, f ( x ) → 2 but is less than 2.

41. h( x ) = =

2x − 1 x2 + 1 As x → ±∞, f ( x ) → 0. As x → ∞, f ( x ) → 0 but is greater than 0. As x → −∞, f ( x ) → 0 but is less than 0.

x 2 − x − 20 x2 + 7 ( x − 5)( x + 4) x2 + 7

The zeros of h correspond to the zeros of the numerator and are x = − 4, 5. 42. g ( x) =

( x − 6)( x − 2) x 2 − 8 x + 12 = 2 x + 4 x2 + 4

The zeros of g correspond to the zeros of the numerator and are x = 2, 6.

f ( x) =

There are no real zeros.

f ( x) = 4 −

1 34. f ( x) = 2 + x −3 As x → ±∞, f ( x ) → 2. As x → ∞, f ( x ) → 2 but is greater than 2. As x → −∞, f ( x ) → 2 but is less than 2. 35.

279

The zeros of f correspond to the zeros of the numerator and are x = ± 3.

(d)

Rational Functions and Asymptotes

43. f ( x) =

x 2 + 4 x − 21 ( x + 7)( x − 3) x+7 ,x ≠ 3 = = 1 3 x2 − 4 x + 3 x x − − ( )( ) x −1

The zero of f corresponds to the zero of the numerator and is x = − 7. 44. h( x) =

2 x 2 + 11x + 5 ( 2 x + 1)( x + 5) = 3 x 2 + 13x − 10 (3x − 2)( x + 5) 2x + 1 , x ≠ −5 3x − 2

The zero of h corresponds to the zero of the numerator 1 and is x = − . 2

280

Chapter 3

45. N =

(a)

Polynomial and Rational Functions

20(5 + 3t ) , 0≤t 1 + 0.04t

48. S =

N (5) ≈ 333 deer N (10) = 500 deer N (25) = 800 deer (c) The herd is limited by the horizontal asymptote: 60 N= = 1500 deer 0.04

(b)

46. (a) Use data 1  1   1   1    16, 3  ,  32, 4.7  ,  44, 9.8  ,  50, 19.7  ,         1    60, 39.4  .  

1 = −0.007 x + 0.445 y y= (b)

(a)

1200

150n ,n ≥ 0 n + 100 2

(b) S (5) =

S ( 20) =

= 6000 tablets

150(10)

= 7500 tablets

(10)2 + 100 150( 20)

( 20)2 + 100

= 6000 tablets

25,000 p , 0 ≤ p < 100 100 − p 300,000

100 0

16 3.0

32 4.5

60 40

(b) C (15) =

(Answers will vary.) (c) No, the function is negative for x = 70.

C (50) =

47. (a)

150(5)

(5)2 + 100

S (10) =

(a)

1 0.445 − 0.007 x

M 200

400

600

7.3

50 10.5

800 1000

t 0.472 0.596 0.710 0.817 0.916 M 1200 1400 1600 1800 2000

C (90) =

25,000(15)

100 − (15) 25,000(50)

100 − (50) 25,000(90)

100 − (90)

38 M + 16,965 and t = 1.056. 10( M + 500)

If you do this, you obtain M ≈ 1306 grams.

= $25,000 = $225,000

(c) No. The function is undefined at p = 100. 50. False. A rational function can have at most n vertical asymptotes, where n is the degree of the denominator.

t 1.009 1.096 1.178 1.255 1.328 The greater the mass, the more time required per oscillation. The model is a good fit to the actual data. (b) You can find M corresponding to t = 1.056 by finding the point of intersection of

= $4411.76

51. False. For example, f ( x ) =

1 has no vertical x +1 2

asymptote.

x −1 x3 − 8 x−2 (b) f ( x ) = ( x + 1)2

52. (a) f ( x ) =

2( x + 3)( x − 3) 2 x 2 − 18 = 2 ( x + 2)( x − 1) x + x −2

−2( x + 2)( x − 3) −2 x 2 + 2 x + 12 = ( x + 1)( x − 2) x2 − x − 2

(e) f ( x) =

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

3x3 − 6 x 2 − 3x − 6 x3 − 9 x

Section 3.5 53. No. If x = c is also a zero of the denominator of f, then f is undefined at x = c, and the graph of f may have a hole of vertical asymptote at x = c. 54. Yes. If the graph of f, a rational function, has a vertical asymptote at x = 4 , then f may have a common factor of x − 4. For example, ( x − 4)( x + 1) f ( x) = . ( x − 4)( x − 4)

Rational Functions and Asymptotes

4−0 ( x − 0) −9 − 0 −9 y = 4 x 4x + 9y = 0

60. y − 0 =

x+9

61. 2

x−4

x + 5x + 6 x2 − 4x 9x + 6 9 x − 36 42

55. −5

x2 + 5x + 6 42 = x+9+ x−4 x−4

−10

3x3 − 5x2 + 4 x − 5 3x3 and y2 = 2 are 2 2x − 6x + 7 2x approximately the same graph as x → ∞ and x → −∞. Therefore as x → ±∞, the graph of a rational function

The graphs of y1 =

an x n . bm x m

1 − 10 15

62. 3

3 − 21 −7 −6

1 2

x − 10 x + 15 −6 = x−7+ x −3 x −3

an x n +  a1 x + a0 appears to be very close to the bm x m +  b1 x + b0

graph of y =

63.

2 x2 − 9 x2 + 5

2 x 4 + 0 x 3 + x 2 + 0 x − 11 2 x4 + 10 x 2

56. (a) There is a vertical asymptote at x = 1, so D (1) is

−9 x 2 − −9 x 2 −

undefined. (b) There is a horizontal asymptote at y = 2. So, the degrees of N ( x ) and D( x ) are equal. of the leading coefficient of N ( x ) to the leading coefficient of D( x) is 2, and not 1. So, statements (a) and (c) are false.

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

57. y − 2 =

1+ 5 ( x + 6) −6 − 4 −10 y + 10 = 6 x + 36 3 x + 5 y + 13 = 0

58. y − 1 =

10 − 7 ( x − 2) = 3( x − 2) 3−2 y = 3x + 1 3x − y + 1 = 0

11 45 34

2 x 4 + x 2 − 11 34 = 2x2 − 9 + 2 x2 + 5 x +5

59. y − 7 =

281

64.

2 x 4 − 3x3 + 6 x 2 − 9 x +

27 2

2x + 3

4 x5 + 0 x 4 + 3 x3 + 0 x 2 + 0 x − 10 4 x5 + 6 x 4 −6 x 4 + 3 x3 −6 x 4 − 9 x3 12 x3 12 x3 + 18 x 2 −18 x 2 −18 x 2 − 27 x 27 x − 10 81 27 x + 2 101 − 2 4 x 5 + 3 x 3 − 10 27 101 4 3 2 = 2 x − 3x + 6 x − 9x + − 2x + 3 2 4x + 6

282

Chapter 3

Polynomial and Rational Functions

Section 3.6 Graphs of Rational Functions 1.

slant, asymptote

vertical

y 1 −6 −5 −4 −3

Yes. Because the numerator’s degree is exactly 1 greater than that of the denominator, the graph of f has a slant asymptote.

−1 −1

−6 −7

−1 − 4 is a reflection in the x + 2 y-axis, a horizontal shift two units to the left and a vertical 1 shift four units downward of the graph of f ( x) = . x

The graph of g ( x) =

3 1 1

g( x ) =

2 +1 x

−2

−1 + 2 is a reflection in the x-axis x and a vertical shift two units upward of the graph of 1 f ( x) = . x The graph of g ( x) =

−3

−4 −3 −2 −1 −1

−2

The slant asymptote’s equation is that of the quotient, y = x − 1.

f −4

The graph of g is a vertical shift one unit upward of the graph of f. 10. g( x ) =

−6

2 x −1

−4 −2 −2

8 10 12

f g

−6

−4 −6

−8

−4

1 is a horizontal shift six units The graph of g ( x) = x−6 1 to the right of the graph of f ( x) = . x

The graph of g is a horizontal shift one unit to the right of the graph of f. 11. g( x ) = −

2 x

3 −6

1 −1

−2 −3 −4

g −4

The graph of g is a reflection in the x-axis of the graph of f.

−5

1 − 1 is a horizontal shift x −3 three units to the right and a vertical shift one unit 1 downward of the graph of f ( x) = . x The graph of g ( x) =

12. g( x ) =

1 x+2 4

g −6

f 6

−4

The graph of g is a horizontal shift two units to the left, and vertical shrink of the graph of f. .

Section 3.6 13. g ( x) =

3 −2 x2

283

1 x+2

17. f ( x ) =

 1 y-intercept:  0,   2 Vertical asymptote: x = −2 Horizontal asymptote: y = 0

−6

Graphs of Rational Functions

−4

3 − 2 is a vertical shift two units x2 3 downward of the graph of f ( x) = 2 . x

The graph of g ( x) =

(0, 12 )

−3

−1 −1

3 14. g ( x) = − 2 2  x 

−2

−4

−3

−1

−

1 2

−1

1 2

1 3

f ( x) =

1 x −6

−6

g −4

3 The graph of g ( x) = − 2 2  is a reflection in the x  x-axis and vertical stretch of the graph of f ( x) = 15. g ( x) =

18.

3 . x2

( x − 2)

1  y-intercept:  0, −  6  Vertical asymptote: x = 6 Horizontal asymptote: y = 0 y

g f

6 4

g f

−6

−1

−2

The graph of g ( x) =

( x − 2)

is a horizontal shift

3 two units to the right of the graph of f ( x) = 2 . x 16. g ( x) =

1 3    4  x2 

−2

(0, − 16 (

−4

−6

−1

−

1 1 1 1 − − − 7 6 4 2

1 2

1 4

−6

6 −2

The graph of g ( x) = graph of f ( x) =

1 3    is a vertical shrink of the 4  x2 

3 . x2

284

Chapter 3

Polynomial and Rational Functions

5 + 2x 2x + 5 = 1+ x x +1  5  x-intercept:  − , 0   2 

19. C ( x ) =

21.

f (t ) =

1 − 2t 2t − 1 =− t t

1  t-intercept:  , 0  2  Vertical asymptote: t = 0 Horizontal asymptote: y = −2

y-intercept: (0, 5) Vertical asymptote: x = −1 Horizontal asymptote: y = 2

( 12 , 0)

−2

−1

(0, 5)

(− 52 , 0)

−3 −6

−4

− 4 −3 − 2

− 2 −1

1 2

C(x)

1 2

7 2

−

5 −3 2

−1 −

20. P ( x ) =

−1

1 − 3x 3x − 1 = 1− x x −1

22. g( x ) =

1  x-intercept:  , 0  3  y-intercept: (0, 1)

3 2

1 2x + 5 +2= x+2 x+2

 5 y-intercept:  0,   2

Vertical asymptote: x = 1 Horizontal asymptote: y = 3 y

 5  x-intercept:  − , 0   2  Vertical asymptote: x = −2 Horizontal asymptote: y = 2

5 4

(− 52 , 0(

( 13 , 0)

(0, 1) −2

−1

(0, 52 (

2 1

−4 −3

−1 −1

−2

−1

5 −1 2

5 2

9 4

−4 −

3 2

Section 3.6

23.

f ( x) =

x2 x −4

Intercept: (−1, 0) Vertical asymptotes: x = 0 and x = 4 Horizontal asymptote: y = 0

Intercept: (0, 0) Vertical asymptotes: x = 2, x = −2 Horizontal asymptote: y = 1 y-axis symmetry

8 6

(0, 0)

−6

−4

285

4( x + 1) x ( x − 4)

25. g( x ) =

Graphs of Rational Functions

−2

(− 1, 0) −6 −4 −2

8 10

−4 −6

−4

−8

−6

− 4 −1

−1

4 3

−

1 3

4 3

24. g( x ) =

−

1 3

−2 −1

−

1 3

−

8 3

−3 −

16 2 4 3 5

7 3

26. h( x ) =

x x2 − 9

2 x 2 ( x − 3)

Vertical asymptotes: x = 0, x = 3 Horizontal asymptote: y = 0

Intercepts: (0, 0) Vertical asymptotes: x = ±3 Horizontal asymptote: y = 0 Origin symmetry

4 3 2

y 5 4 3 2 1 −2 −1 −2 −3 −4 −5

−5

−

1 −3

(0, 0) 1

4 5

−4 −2

5 4 − 16 7

2 5

−

2 5

4 7

5 16

−2

−

1 Undef. 10

1 2

Undef.

1 8

−1 −

286

Chapter 3

27.

f ( x) =

Polynomial and Rational Functions

3x 3x = x 2 − x − 2 ( x + 1)( x − 2)

29.

Intercept: (0, 0) Vertical asymptotes: x = −1, 2 Horizontal asymptote: y = 0

f ( x) =

x2 + 3x x ( x + 3) x = = , x 2 + x − 6 ( x − 2)( x + 3) x − 2

x ≠ −3 Intercept: (0, 0) Vertical asymptote: x = 2 (There is a hole at x = −3. ) Horizontal asymptote: y = 1

y 4

(0, 0) −2

4 2 −8 −6 −4 −2 −2

−4

28.

−3

−

9 10

−

3 2

9 4

6 5

−6 −8

2x 2x f ( x) = 2 = x + x − 2 ( x + 2)( x − 1)

Intercept: (0, 0) Vertical asymptotes: x = −2, 1 Horizontal asymptote: y = 0

−2 −1

1 2

−1

Undef.

30. g( x ) =

5( x + 4) 5( x + 4) 5 , = = x 2 + x − 12 ( x + 4)( x − 3) x − 3

x ≠ −4 Vertical asymptote: x = 3 Horizontal asymptote: y = 0

y 4 3

2 1 −3

1 3

−1 −1

6 4

(0, 0)

−2

−3 −2

8 10

−4

−3

−1

1 2

4 3 − 5 2

−

−4

−

4 5

3 5

−6 −8

Hole at x = −4

x y

−4

3 1 5 5 − Undef. − Undef. 3 2

4 5

Section 3.6

31.

f ( x) =

x 2 − 1 ( x + 1)( x − 1) = = x − 1, x +1 x +1

x ≠ −1 The graph is a line, with a hole at x = −1.

35. g ( x) =

Graphs of Rational Functions

287

3x − 4 −x 2

−6

8 6

−6

Domain: x ≠ 0 or ( −∞, 0) ∪ (0, ∞ )

4 2 −8 −6 −4 −2

Vertical asymptote: x = 0 Horizontal asymptote: y = − 3

−4 −6 −8

32.

f ( x) =

36. h( x) =

x 2 − 16 = x + 4, x ≠ 4 x−4 y

2x − 1 x +5 6

− 12

12 10

−6

Domain: x ≠ − 5 or ( − ∞, − 5) ∪ ( − 5, ∞ )

6 4 2 −8 −6

−2 −2

Vertical asymptote: x = − 5 Horizontal asymptote: y = 2

−4

37. g ( x) =

Hole at x = 4 33.

f ( x) =

5 x2 + 1 7

2+x x+2 =− 1− x x −1 4

−6

6 −1

−5

Vertical asymptote: None

−4

Domain: x ≠ 1 or (−∞, 1) ∪ (1, ∞) Vertical asymptote: x = 1 Horizontal asymptote: y = −1 34.

f ( x) =

Domain: all real numbers or ( − ∞, ∞ )

3− x x −3 = 2− x x−2

Horizontal asymptote: y = 0 38. g( x ) = −

x ( x − 2)2

1 −4

−4

Domain: x ≠ 2 or ( −∞, 2) ∪ ( 2, ∞ ) Vertical asymptote: x = 2 Horizontal asymptote: y = 1

−7

Vertical asymptote: x = 2 Horizontal asymptote: y = 0 Domain: all real numbers except x = 2 or (−∞, 2) ∪ (2, ∞)

288

Chapter 3

39.

f ( x) =

Polynomial and Rational Functions

x +1 x +1 = x 2 − x − 6 ( x − 3)( x + 2)

43. h( x ) =

−9

− 12

There are two horizontal asymptotes, y = ±6. 44.

f ( x) =

x+4 x2 + x − 6

−x 9 + x2 4

−6

−8

Vertical asymptotes: x = 3, x = −2 Horizontal asymptote: y = 0 Domain: all real numbers except x = 3, − 2 f ( x) =

x2 + 1 8

−6

40.

−4

There are two horizontal asymptotes, y = ±1. −4

Domain: all real numbers except x = − 3 and x = 2 or (−∞, − 3) ∪ ( − 3, 2) ∪ (2, ∞)

45. g( x ) =

4| x−2| x +1 16

Vertical asymptotes: x = −3, x = 2 Horizontal asymptote: y = 0 − 24

41.

f ( x) =

20 x 1 19 x − 1 − = x 2 + 1 x x ( x 2 + 1)

− 16

There are two horizontal asymptotes, y = ±4 and one vertical asymptote, x = −1.

− 15

46.

f ( x) =

− 10

Domain: all real numbers except x = 0, or

(−∞, 0) ∪ (0, ∞ )

− 30

Vertical asymptote: x = 0 Horizontal asymptote: y = 0 42.

−8 | 3 + x | 8 | 3 + x | = 2−x x−2

− 25

There are two horizontal asymptotes, y = −8 and y = 8, and one vertical asymptote, x = 2.

1  30  1 f ( x) = 5  − = − − − x x x x + 2) 4 2 ( 4)(  

47.

f ( x) =

4( x − 1)2 x2 − 4 x + 5 12

− 12

−8

Domain: all real numbers except x = −2 and x = 4 Vertical asymptotes: x = −2 , x = 4 Horizontal asymptote: y = 0

− 10

14 −4

The graph crosses its horizontal asymptote, y = 4.

Section 3.6 3x4 − 5x + 3 x4 + 1

48. g( x ) =

52.

−9

Graphs of Rational Functions

289

x3 x =x+ 2 x −1 x −1

f ( x) =

Intercept: (0, 0) Vertical asymptotes: x = ±1 Slant asymptote: y = x Origin symmetry

9 −3

The graph crosses its horizontal asymptote, y = 3. 49.

2 x2 + 1 1 = 2x + x x

f ( x) =

2 −6

Vertical asymptote: x = 0 Slant asymptote: y = 2 x Origin symmetry

−4

−2

53. g( x ) =

6 4 2 −6

−4

x3 1 4x = x+ 2 2 x2 − 8 2 2x − 8

Intercept: (0, 0) Vertical asymptotes: x = ±2 1 Slant asymptote: y = x 2 Origin symmetry

y = 2x

−2

y=x (0, 0)

−6

y 2

1− x x Intercepts: (1, 0), (−1, 0)

50. g( x ) =

6 4

(0, 0)

Vertical asymptote: x = 0 Slant asymptote: y = − x

−8 −6 −4

y= 1x 2

8 6 4 2

(1, 0)

−8 −6 −4 −2 −2

(− 1, 0)

54. 6

y = −x

−8

1 x2 = x +1+ x −1 x −1

51. h( x ) =

x3 4x =x− 2 2 x +4 x +4

Intercept: (0, 0) Vertical asymptotes: None Slant asymptote: y = x Origin symmetry

−4 −6

f ( x) =

Intercept: (0, 0) Vertical asymptote: x = 1 Slant asymptote: y = x + 1 y

y 6 4 2

(0, 0) −6

−4

−2

−4

y = x −6

6 4

y=x+1

(0, 0) −4

−2

−4

290

Chapter 3

Polynomial and Rational Functions

55.

x 3− x3 + 2 x2 + 4 x 2 f ( x) = = +1+ 2 2 x2 + 1 2 2x + 1

59.

Intercepts: (−2.594, 0), (0, 4) x Slant asymptote: y = + 1 2 y

x2 = 1 x = ±1

7 6 5 4

(0, 4)

60. −5

1 −x x x-intercepts: ( ±1, 0) 1 0= −x x 1 x= x y=

−2 −1

1 2 3 4 5 y = 1x + 1 2

−2 −3

2 x2 − 5x + 5 3 56. f ( x ) = 2x − 1 + x−2 x−2 5  y-intercept:  0, −  2  

Vertical asymptote: x = 2 Slant asymptote: y = 2 x − 1

2 x x-intercepts: (1, 0), (2, 0) 2 0 = x −3+ x y = x −3+

0 = x2 − 3x + 2 0 = ( x − 1)( x − 2) x = 1, 2

61.

Domain: all real numbers except x = −1 Vertical asymptote: x = −1 Slant asymptote: y = 2 x − 1

y 15 12

9 6

y = 2x − 1

3 −9 −6 −3

(0, − 52)

9 12 15

2x x −3 x-intercept: (0, 0) 2x 0= x −3 0 = 2x 0=x y=

− 12

− 10

−9

x +1 57. y = x −3 x-intercept: ( −1, 0) x +1 0= x −3 0 = x +1 −1 = x

58.

2 x2 + x 1 = 2x − 1 + x +1 x +1

62.

x2 + 5x + 8 2 = x+2+ x+3 x+3

Domain: all x ≠ −3 Vertical asymptote: x = −3 Slant asymptote: y = x + 2 8

− 14

−8

Section 3.6

63.

1 + 3x2 − x3 1 1 = 2 + 3 − x = −x + 3 + 2 x2 x x

68.

Domain: all real numbers except 0 or ( −∞, 0) ∪ (0, ∞) Vertical asymptote: x = 0 Slant asymptote: y = − x + 3

1 2 3 Horizontal asymptote: y = 2 No slant asymptotes

 4 Hole at x = 2,  2,   5

69. y=

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

Domain: all real numbers except –4 or ( −∞, − 4) ∪ ( −4, ∞ ) x-intercepts: ( −4.61, 0), (2.61, 0)  3 y-intercept:  0,   2 Vertical asymptote: x = −4 1 Slant asymptote: y = − x + 1 2

65.

f ( x) =

x 2 − 5 x + 4 ( x − 4)( x − 1) = ( x − 2)( x + 2) x2 − 4

Vertical asymptotes: x = 2, x = −2 Horizontal asymptote: y = 1 No slant asymptotes, no holes x 2 − 2 x − 8 ( x − 4)( x + 2) = 66. f ( x ) = x2 − 9 ( x − 3)( x + 3) Vertical asymptotes: x = 3, x = −3

2 x3 − x2 − 2 x + 1 x2 + 3x + 2

( x − 1)( x + 1)(2 x − 1) ( x + 1)( x + 2)

( x − 1)(2 x − 1) , x ≠ −1 x+2

f ( x) =

2 x2 − 3x + 1 15 = 2x − 7 + . x+2 x+2

Vertical asymptote: x = −2 No horizontal asymptote Slant asymptote: y = 2 x − 7 Hole at x = −1, ( −1, 6)

−6

f ( x) =

Long division gives

− 16

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

Vertical asymptote: x = −

−4

64.

291

x≠2

−12

f ( x) =

Graphs of Rational Functions

70.

2 x3 + x2 − 8x − 4 x2 − 3x + 2 ( x − 2)( x + 2)(2 x + 1) = ( x − 2)( x − 1) ( x + 2)(2 x + 1) = , x≠2 x −1

f ( x) =

Long division gives f ( x ) = 2 x + 7 +

9 . x −1

Vertical asymptote: x = 1 No horizontal asymptote Slant asymptote: y = 2 x + 7 Hole at x = 2, (2, 20)

Horizontal asymptote: y = 1 No slant asymptotes, no holes 67.

f ( x) =

2 x 2 − 5 x + 2 (2 x − 1)( x − 2) 2 x − 1 , = = 2 x 2 − x − 6 (2 x + 3)( x − 2) 2 x + 3

x≠2 3 2 Horizontal asymptote: y = 1 No slant asymptotes  3 Hole at x = 2,  2,   7

Vertical asymptote: x = −

292

Chapter 3

71.

Polynomial and Rational Functions

1 4 + x+5 x

74. y =

6 1 − x +3 x + 4 6

6 − 15 − 11

− 14

−6

x-intercept: ( −4, 0) 1 4 0= + x+5 x 4 1 − = x x+5 −4( x + 5) = x

x-intercept: ( − 4.2, 0)

6 1 − = 0 x +3 x + 4 6 1 = x +3 x + 4 6 x + 24 = x + 3 5 x = − 21

−4 x − 20 = x −5 x = 20 x = −4

72. y =

21 5 x = − 4.2 x = −

1 5 − x−2 x

2 x +1

75. y = x −

6 − 12

12 −8

−8

x-intercept: ( 2.5, 0)

−6

1 5 − = 0 x − 2 x 1 5 = x − 2 x x = 5 x − 10 10 = 4 x

x −

2 = 0 x +1 x =

2 x +1

x2 + x = 2 x2 + x − 2 = 0

5 = x 2 2.5 = x 73. y =

x-intercepts: ( − 2, 0) and (1, 0)

( x + 2)( x − 1) = 0 x = − 2, x = 1

2 3 − x + 2 x −1 6

−10

−6

x-intercept: ( −8, 0 ) 2 3 = x + 2 x −1 2 x − 2 = 3x + 6 −8 = x

76. y = 2 x −

Section 3.6

Graphs of Rational Functions

8 x

78. y = 2 x − 1 +

1 x−2

−8

−6

x-intercepts: ( − 2, 0) and ( 2, 0)

2x −

−3

8 = 0 x 8 2x = x

2 x2 = 8

3  x-intercepts: (1, 0 ) ,  , 0  2   1 2x − 1 + =0 x−2 1 = 1 − 2x x−2 1 = −2 x 2 + 5 x − 2

x = 4

2 x − 5x + 3 = 0

x = ±2

( x − 1)( 2 x − 3) = 0

1 x +1

x = 1,

77. y = x + 2 −

79. y = x + 1 +

−10

−6

x-intercepts: ( −2.618, 0 ) , ( −0.382, 0 )

x2 + 3x + 2 = 1 x2 + 3x + 1 = 0

−3 ± 9 − 4 2 3 5 =− ± 2 2 ≈ −2.618, − 0.382

2 x −1

−10

1 x+2

3 2

x+2=

293

−4

No x-intercepts 2 x +1+ =0 x −1 2 = −x − 1 x −1 2 = −x2 + 1 2

x +1= 0 No real zeros

80. y = x + 2 +

2 x+2 6

− 11

−6

No x-intercepts x+2+

2 =0 x+2 2 = −x − 2 x+2

2 = −x2 − 4 x − 4 x2 + 4 x + 6 = 0

Because b2 − 4 ac = 16 − 24 < 0, there are no real zeros. © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 3

294

81. y = x + 3 −

Polynomial and Rational Functions 83. (a) 0.25 ( 50 ) + 0.75 ( x ) = C ( 50 + x )

2 2x − 1

12.5 + 0.75 x =C 50 + x 50 + 3 x =C 200 + 4 x 3 x + 50 C= 4 ( x + 50 )

−5

7 −1

x-intercepts: ( 0.766, 0 ) , ( −3.266, 0 ) x +3−

2 =0 2x − 1 x+3=

(b) Domain: x ≥ 0 and x ≤ 1000 − 50 = 950 Thus, 0 ≤ x ≤ 950. 1 (c)

2 2x −1 0

2 x2 + 5x − 3 = 2 2 x2 + 5x − 5 = 0

−5 ± 25 − 4 ( 2 )( −5 ) 4

As the tank fills, the rate that the concentration is increasing slows down. It approaches the horizontal 3 asymptote C = = 0.75. When the tank is full 4

( x = 950 ) , the concentration is C = 0.725.

−5 ± 65 4 ≈ 0.766, − 3.266

82. y = x − 1 −

950

84. (a) Area = xy = 500

500 x (b) Domain: x > 0 y (c)

2 2x − 3

120 −5

100

80 60

−3

x-intercepts: ( 0.219, 0 ) , ( 2.281, 0 )

2 x −1− =0 2x − 3

x −1 =

2 2x − 3

2 x2 − 5x + 3 = 2

5 ± 25 − 8 4 5 ± 17 ≈ 0.219, 2.281 4

For x = 30, y =

500 2 = 16 meters. 30 3

85. (a) A = xy and

( x − 2)( y − 4) = 30

2 x2 − 5x + 1 = 0 x=

30 x−2 30 4 x + 22 y=4+ = x−2 x−2 2 x ( 2 x + 11) .  4 x + 22  Thus, A = xy = x  = x−2  x−2  (b) Domain: Since the margins on the left and right are each 1 inch, x > 2, or ( 2, ∞ ). y−4=

(c)

100

The area is minimum when x ≈ 5.87 in. and y ≈ 11.75 in.

Section 3.6 86. (a) The line passes through the points ( a, 0 ) and

( 3, 2 ) and has a slope of 2−0 2 = . 3−a 3−a 2 y−0= ( x − a) 3−a 2 ( x − a ) −2 ( a − x ) y= by the point-slope form = 3−a −1( a − 3)

2 ( a − x) a−3

b=a

h = y when x = 0, so h =

295

C 0.2 x 2 + 10 x + 5 = , x>0 x x

0.5

20.1

15.2

12.9

12.3

C 30 25 20

, 0≤ x≤a

(b) The area of a triangle is A =

88. C =

Graphs of Rational Functions

15 10

1 bh. 2

2 ( a − 0) a −3

5 5

2a . a−3

1  2a  a2 A = a = 2  a−3 a−3 a2 9 = a+3+ a−3 a−3 Vertical asymptote: a = 3

The minimum average cost occurs when x = 5. 3t 2 + t 89. C = 3 , 0≤t t + 50 (a) The horizontal asymptote is the t-axis, or C = 0. This indicates that the chemical eventually dissipates.

(b)

0 0

Slant asymptote: A = a + 3 A is a minimum when a = 6 and A = 12. x   200 87. C = 100  2 + , 1 ≤ x x + 30   x

The maximum occurs when t ≈ 4.5. (c) Graph C together with y = 0.345. The graphs intersect at t ≈ 2.65 and t ≈ 8.32 . C < 0.345 when 0 ≤ t ≤ 2.65 hours and when t > 8.32 hours. 90. (a) Rate × Time = Distance or

Distance = Time Rate

100 100 200 + = =4 x y 50

300

25 25 + =1 x y 0

300

25 y + 25 x = xy 25 x = xy − 25 y

The minimum occurs when x = 40.4 ≈ 40.

25 x = y ( x − 25) 25 x x − 25 (b) Vertical asymptote: x = 25 Horizontal asymptote: y = 25 (c)

150 87.5 66.7 56.3 50

45.8 42.9

The results in the table are unexpected. You would expect the average speed for the round trip to be the average of the average speeds for the two parts of the trip. © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

296

Chapter 3 (d)

Polynomial and Rational Functions

150

x2 + x − 2 x −1 ( x + 2)( x − 1) = = x + 2, x ≠ 1 x −1 Since g(x) is not reduced ( x − 1) is a factor of both the numerator and the denominator, x = 1 is not a horizontal asymptote.

96. g( x ) =

(e) No, it is not possible to average 20 miles per hour in one direction and still average 50 miles per hour on the round trip. At 20 miles per hour you would use more time in one direction than is required for the round trip at an average speed of 50 miles per hour.

−7

91. (a) A = − 0.14t + 3.1 −4

There is a hole at x = 1. 97. Horizontal asymptotes: 0

(b) A =

1 0.035t + 0.253

(c) Answers will vary. 92. (a) Domain: t ≥ 0; the model is valid only after the elk have been released. (b) At t = 0, P = 10. 2.7 ≈ 27 is the 0.1 limit. So, 27 elk is the expected population.

If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote. If the degree of the numerator is less than the degree of the denominator, then there is a horizontal asymptote at y = 0. If the degree of the numerator is equal to the degree of the denominator, then there is a horizontal asymptote at the line given by the ratio of the leading coefficients. Vertical asymptotes: Set the denominator equal to zero and solve. Slant asymptotes: If there is no horizontal asymptote and the degree of the numerator is exactly one greater than the degree of the denominator, then divide the numerator by the denominator. The slant asymptote is the result, not including the remainder. a has a slant asymptote x+2 y = x + 1 and a vertical asymptote x = −2. a 0 = 2 +1+ 2+2 a 0 =3+ 4 a = −3 4 a = −12 12 x 2 + 3 x − 10 Hence, y = x + 1 − = . x+2 x+2

98. (a) y = x + 1 + 93. False. The graph of a rational function is continuous when the polynomial in the denominator has no real zeros. x crosses its horizontal asymptote x3 + 1 y = 0 at x = 0.

94. False. f ( x ) =

95. h( x ) =

6 − 2 x 2(3 − x ) = = 2, x ≠ 3 3− x 3− x

Since h(x) is not reduced and (3 − x ) is a factor of both the numerator and the denominator, x = 3 is not a horizontal asymptote. There is a hole in the graph at x = 3. 4

−6

−4

a has slant asymptote y = x − 2 and x+4 vertical asymptote at x = −4. We determine a so that y has a zero at x = 3 : a a 0 =3−2+ = 1 +  a = −7 3+4 7 −7 x 2 + 2 x − 15 Hence, y = x − 2 + = . x+4 x+4

(b) y = x − 2 +

Section 3.7 −3

104.

512 x 8 99.   =   = 3 x x 8     100.

101.

( ) 4 x2

−2

(4x ) 2

= 2

−5

Domain: −11 ≤ x ≤ 11 105.

5 − 20

x −2 ⋅ x 1 2 x ⋅ x1 2 x3 2 102. −1 5 2 = 2 5 2 = 9 2 x ⋅x x ⋅x x

103.

Range: 0 ≤ y ≤ 11

37 6 = 36 6 = 3 31 6

− 11

1 x3

Domain: all x Range: y ≤ 0

106.

− 15 −6

297

− 15

1 16 x 4

Quadratic Models

6 −1 − 10

Domain: all x Range: y ≥ 6

Domain: all x Range: y ≤ 9 107. Answers will vary.

Section 3.7 Quadratic Models 1.

quadratic

r 2 = 0.9688, the closer the value of r 2 is to 1, the better fit of the model.

Quadratic

Linear

Neither

Quadratic

(a)

(b) The linear model is better. (c) y = 0.14 x + 2.2, linear (d) 4

(e)

2.1

2.4

2.5

2.8

2.9

Model

2.2

2.4

2.5

2.6

2.8

3.0

3.2

3.4

3.5

3.6

Model

2.9

3.0

3.2

3.4

3.5

3.6

Chapter 3

298

10. (a)

Polynomial and Rational Functions 12. (a)

−3

0 5200

(b) The quadratic model is better. (c) y = 0.006 x 2 − 0.23 x + 10.5, quadratic 15 (d)

−3

(b) The quadratic model is better. (c) y = −17.79 x 2 + 354.8 x + 6163 (d) 8000

0 5200

8000

(e)

−2

−1

10.7

10.4

10.3

10.1

6140

6815

7335

7710

7915

7590

Model

10.7

10.4

10.3

10.1

Model

6163

6801

7298

7651

7863

7932

9.9

9.6

9.4

9.2

9.0

7975

7700

7325

6820

6125

5325

Model

9.9

9.7

9.5

9.3

9.2

9.0

Model

7859

7643

7286

6785

6143

5358

13. (a)

11. (a)

5000

− 1000

(b) The quadratic model is better. (c)

y = 5.50 x 2 − 274.6 x + 2822

(d)

5000

(b) The linear model is better. (c) y = 2.48 x + 1.1 (d) 30

− 1000

(e)

− 450

Actual, y

4.0

6.5

8.8

10.6

13.9

− 59.5

− 470

Model, y

3.6

6.1

8.5

11.0

13.5

− 520

− 55

625

1630

Actual, y

15.0

17.5

20.1

24.0

27.1

− 605.5

− 466

− 51.5

638

1602.5

Model, y

16.0

18.5

20.9

23.4

25.9

Actual, y

2845

4350

Model, y

2842

4356.5

Actual, y

2795

1590

650

− 30

Model, y

2822

1586.5

626

Actual, y

− 615

Model, y

Section 3.7 14. (a)

16. (a)

− 11

299

−6

− 10

(b) Quadratic model is better. (c) y = −0.283 x 2 + 0.25 x + 35.6 (d) 40

(b) Linear model is better. (c) y = −0.89 x + 5.3 12 (d)

− 11

− 10

−6

(e)

15. (a)

Quadratic Models

(e)

−6

−4

−2

Actual, y

10.7

9.0

7.0

5.4

3.5

Actual, y

34.3

33.8

32.6

30.1

27.8

22.5

Model, y

10.6

8.9

7.1

5.3

3.5

Model, y

35.0

33.8

32.1

29.8

26.9

23.5

Actual, y

1.7

−0.1

−1.8

−3.6

−5.3

Actual, y

19.1

14,8

9.4

3.7

−1.6

Model, y

1.7

0.0

−1.8

−3.6

−5.4

Model, y

19.5

14.9

9.8

4.1

−2.2

650

17. (a)

0 350

55 0

(b) Quadratic is better. (c) y = 0.14 x 2 − 9.9 x + 591 (d) 650

(b) A quadratic model is P = 0.1601t 2 − 2.206t + 7.65. The coefficient of determination is r 2 ≈ 0.9185. (c)

0 350

(e) 0

Actual, y

587

551

512

478

436

430

Model, y

591

545

506

474

449

431

Actual, y

424

420

423

429

444

Model, y

420

416

419

429

446

(d) According to the graph, the minimum occurs at (6.889, 0.0509). So, June is the month in which the normal precipitation is the least. (e) According to the table, July is the month in which the normal precipitation is the least. The answers from (d) and (e) are not the same.

300

Chapter 3

18. (a)

Polynomial and Rational Functions Use the value t ≈ 16.03, which corresponds to the year 2016. So, in the year 2016, the percent of the U.S. population who use the Internet will fall below 60%.

240

0 200

This is not a good model for predicting future years. In fact, by 2022, the model gives negative values for the percentage.

(b) A quadratic model for the data is S = − 0.457t 2 + 15.03t + 119.5. (c)

20. (a)

240

0 200

(d) To determine when sales will be less than $200 billion, let S = 200 and solve for t.

250

(b)

250

− 0.457t 2 + 15.03t + 119.5 = 200 0.457t 2 − 15.03t + 80.5 = 0

Use the Quadratic Formula to find the value(s) of t. t =

15.03 ±

15.03 − 4(0.457)(80.5)

t ≈ 6.7 or t ≈ 26.2

Use the value t ≈ 26.2, and round up to t ≈ 27, which corresponds to the year 2027. So, in the year 2027, the annual sales will be less than $200 billion. This is not a good model for predicting future years. In fact, by 2040 the model gives negative values for sales. 19. (a)

The model fits the data well. The y-values of the model are close to the actual y-values.

2(0.457)

(e) The cubic model is a better fit because its coefficient of determination is closer to 1 and its y-values are closest to the actual y-values. (f)

The model fits the data fairly well. The y-values of the model are fairly close to the actual y-values.

(b) A quadratic model for the data is P = − 0.805t 2 + 19.10t − 39.3. (c)

250

Year

2013

2014

2015

145

143

174

Cubic model

153

161

174

Quadratic model

142

138

134

Answers will vary.

(d) To determine when the percent P is less than 60%, let P = 60 and solve for t.

− 0.805t 2 + 19.10t − 39.3 = 60 − 0.805t 2 + 19.10t − 99.3 = 0 t =

− (19.10) ±

19.10 2 − 4( − 0.805)( − 99.3) 2( − 0.805)

t ≈ 7.69 or t ≈ 16.03

Section 3.7 21. (a)

(b) A linear model for the data is T = 1.23t + 102.4. The coefficient of determination is r 2 ≈ 0.9468.

23. True. A quadratic model with a positive leading coefficient opens upward. So, its vertex is at its lowest point. 24. False. A quadratic model could be a better fit for data that are positively correlated. 25. The model is above all data points.

118

0 100

(d) A quadratic model for the data is T = − 0.076t 2 + 2.14t + 100.8. The coefficient of determination is r 2 ≈ 0.9870. (e)

118

26. Because the data appears more quadratic than linear, a quadratic model would be the better fit. Therefore the S 2 -value of 0.9995 must be that of the quadratic model. 27. (a)

( f  g )( x ) = f ( x 2 + 3) = 2( x 2 + 3) − 1 = 2 x 2 + 5

(b) ( g  f )( x ) = g(2 x − 1) = (2 x − 1)2 + 3 = 4 x 2 − 4 x + 4 28. (a)

( f  g)( x ) = f (2 x 2 − 1) = 5(2 x 2 − 1) + 8 = 10 x 2 + 3

(b) (g  f )( x) = g(5x + 8) = 2(5x + 8)2 − 1 = 50 x2 + 160 x + 127 29. (a)

( f  g)( x ) = f ( 3 x + 1 ) = x + 1 − 1 = x

(b)

( g  f )( x ) = g( x 3 − 1) = 3 x 3 − 1 + 1 = x

(f) The quadratic model is a better fit. Its coefficient of determination is closer to a value of 1 and its y-values are closest to the actual y-values.

30. (a)

( f  g)( x ) = f ( x 3 − 5) = 3 x 3 − 5 + 5 = x

(g) To determine when the number of households with televisions T in homes will reach 120 million, let T = 120 and solve for t.

31. f is one-to-one. y = 2x + 5

0 100

Linear model:

1.23t + 102.4 = 120 1.23t = 17.6 t ≈ 14.3 Because t ≈ 14.3 corresponds to the year 2014, in 2014 the number of households with televisions will reach 120 million. Quadratic model:

− 0.076t 2 + 2.14t + 100.8 = 120 − 0.076t 2 + 2.14t − 19.2 = 0 Use the Quadratic Formula. t =

301

22. True. A quadratic model with a negative leading coefficient opens downward. So, its vertex is at its highest point.

118

0 100

(c)

Quadratic Models

− 2.14 ±

2.142 − 4( − 0.076)( −19.2) 2( − 0.076)

No real solutions

(b)

( g  f )( x ) = g( 3 x + 5) =  3 x + 5  − 5 = x  

x = 2y + 5 2y = x − 5 ( x − 5) x−5  f −1 ( x ) = y= 2 2 32. f is one-to-one. x−4 y= 5 y−4 x= 5

5 x + 4 = y  f −1 ( x ) = 5 x + 4 33. f is one-to-one on [0, ∞).

y = x 2 + 5, x ≥ 0 x = y 2 + 5, y ≥ 0 y2 = x − 5 y = x − 5  f −1 ( x ) = x − 5, x ≥ 5

So, it is not possible to determine when the number of households with televisions will reach 120 million.

302

Chapter 3

Polynomial and Rational Functions

34. f is one-to-one. y = 2 x 2 − 3, x ≥ 0

36. For −2 + 4i, the complex conjugate is −2 − 4i. (−2 + 4i )(−2 − 4i ) = 4 − 16i 2

x = 2 y − 3, y ≥ 0 ( x + 3) y2 = 2 x+3 y=  f −1 ( x ) = 2

= 4 + 16 = 20

x+3 2 2x + 6 , x ≥ −3 2

37. For −5i, the complex conjugate is 5i.

(−5i )(5i) = −25i 2 = 25 38. For 8i, the complex conjugate is −8i.

(8i)(−8i ) = −64i 2 = 64

35. For 1 − 3i, the complex conjugate is 1 + 3i. (1 − 3i )(1 + 3i ) = 1 − 9i 2 =1+ 9 = 10

Chapter 3 Review y

4. 1

6 5

− 5 −4 − 3

−1 −1

−2

−3

−4 −4 −3 − 2 −1 −1

−5 −6 −7

−2

The graph of y = x + 1 is a vertical shift one unit

The graph of y = − ( x + 1) is a horizontal shift one

upward of y = x 2 .

unit to the left and a reflection in the x-axis of y = x 2 .

1 −7 −6 −5 −4 −3 −2 −1 −1

4 3

−2

−3

1 −4 −3 − 2 −1 −1

−4

−5 −6

−2

−7

The graph of y = − x 2 + 5 is a reflection in the x-axis

The graph of y = − ( x + 5) − 1 is a reflection in the

and a vertical shift five units upward of y = x 2 .

x-axis, a horizontal shift five units to the left, and a vertical shift one unit downward of y = x 2 .

12 10

2 1

6 4

− 2 −1 −1

2 −4 −2 −2

to the right of y = x .

−2

The graph of y = ( x − 6) is a horizontal shift six units 2

The graph of y = − ( x − 2) + 2 is a reflection in the 2

x-axis, a horizontal shift two units to the right, and a vertical shift two units upward of y = x 2 .

Chapter 3 Review 7.

The graph of f ( x ) = ( x + 3 / 2 ) + 1 is a parabola opening 2

303

Use Quadratic Formula.

 3  upward with vertex  − , 1  , and no x-intercepts.  2 

−5 ± 41 2

 −5 + 41   −5 − 41  , 0 ,  , 0     2 2    

6 5

4 12

10 8

1 −4

−3

−2

−1

4 2 2

The graph of f ( x ) = ( x − 4) − 4 is a parabola opening upward with vertex (4, − 4). x-intercepts: ( x − 4)2 − 4 = 0

( x − 4)2 = 4 x − 4 = ±2 x = 6, 2 (6, 0), (2, 0)

y 14

− 10 − 8 − 6

−2

−4

10.

f ( x ) = 3 x 2 − 12 x + 11 = 3( x 2 − 4 x ) + 11 = 3( x 2 − 4 x + 4 − 4) + 11 = 3( x − 2)2 − 1 The graph of f is a parabola opening upward with vertex (2, − 1).

12 10 8

x-intercepts: 3( x − 2)2 − 1 = 0

6 4

2 −6 − 4 − 2

8 10 12

3 x 2 − 12 x + 11 = 0

−4

1 2 ( x + 5 x − 4) 3 1 25 25  4 =  x2 + 5x + − − 3 4 4  3

Use Quadratic Formula.

f ( x) =

1 5  41 x+  − 3 2  12

The graph of f is a parabola opening upward with 41   5 vertex  − , −  . 12   2

6± 3 3

6+ 3  6− 3  , 0 ,     3   3    y 14 12 10 8 6

1 5  41 = 0. x-intercepts:  x +  − 3 2  12

4 2 − 6 −4 − 2

8 10

1 2 x + 5x − 4 = 0 3 x2 + 5x − 4 = 0

(

)

304

Chapter 3

Polynomial and Rational Functions

11. ( − 3, 4) is the vertex, so f ( x) = a( x + 3) + 4. 2

(c)

Since the graph passes through the point (0, − 5), − 5 = a(0 + 3) + 4 2

− 9 = a(3)

− 9 = 9a −1 = a.

The graph of A is a parabola opening downward, with vertex (4, 8). Therefore, when x = 4, the maximum area is 8 square units. Yes, graphically and algebraically the same dimensions result.

Thus, f ( x) = − ( x + 3) + 4. 2

12. ( 2, −1) is the vertex, so f ( x) = a( x − 2) − 1. 2

Since the graph passes through the point (5, 2),

14. 6 x + 4 y = 1500 Total amount of fencing Area enclosed A = 3 xy 1 Because y = (1500 − 6 x ), 4 1 A = 3 x   (1500 − 6 x ) 4

2 = a ( 5 − 2) − 1 2

3 = a(3)

3 = 9a 1 = a. 3 Thus, f ( x) =

9 = − x 2 + 1125 x. 2 −b −1125 The vertex is at x = = = 125.  9 2a 2 −   2

1 ( x − 2)2 − 1. 3

13. (a) A = xy

If x + 2 y − 8 = 0, then y =

8− x . 2

8− x  A = x   2  1 A = 4 x − x2 , 0 < x < 8 2

(b)

1 2 x + 4x 2 1 = − ( x 2 − 8 x) 2 1 = − ( x 2 − 8 x + 16 − 16) 2 1 = − ( x − 4)2 + 8 2

A=−

1 (1500 − 6(125)) = 187.5, and the 4 dimensions are 375 feet by 187.5 feet.

Thus x = 125 feet, y =

15. 3 2

1 − 4 −3 − 2

8 0

Using the graph, when x = 4, the area is a maximum. 8−4 When x = 4, y = = 2. 2

−4 −5

The graph of g ( x) = x3 − 2 is a vertical shift two units downward of f ( x) = x3.

Chapter 3 Review y

16.

20.

305

f ( x ) = − x 4 + 2 x 3 ; g( x ) = − x 4

4 −12

2 −6 −4 − 2

8 10

−12

The graph of g ( x) = ( x − 5) is a horizontal shift 3

five units to the right of f ( x) = x3. y

17. 3 2

22. g ( x) = − 3x3 + 4 x2 − 1

1 −2

21. f ( x) = x2 − x − 6

The degree is even and the leading coefficient is positive. The graph rises to the left and right.

− 6 −5 − 4

The graphs have the same end behavior. Both functions are of the same degree and have negative leading coefficients.

The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right.

−2 −3 −4

23. (a)

The graph of g ( x) = − ( x + 4) is a reflection in the 3

)

= x ( x − 2 )( x + 1) = 0 2

x-axis and a horizontal shift four units to the left of

f ( x) = x3.

(

x 4 − x3 − 2 x2 = x2 x2 − x − 2

Zeros: x = −1, 0, 0, 2 (b)

18.

−6

10 8

−4

4 2 −8 − 6 − 4 −2

24. (a)

(

)

−2 x 3 − x 2 + x = − x 2 x 2 + x − 1

= − x ( 2 x − 1)( x + 1) = 0 The graph of g ( x) = − ( x − 2) + 6 is a reflection in

Zeros: x = −1, 0, 12

the x-axis, a horizontal shift two units to the right, and a

(b)

vertical shift six units upward of f ( x) = x3. 19.

f ( x) =

−6

1 3 1 x − 2 x + 1; g( x ) = x 3 2 2

−4

− 18

25. (a)

g − 12

The graphs have the same end behavior. Both functions are of the same degree and have positive leading coefficients.

) (

(

)(

)

t 3 − 3t = t t 2 − 3 = t t + 3 t − 3 = 0 Zeros: t = 0, ± 3

(b)

−6

−4

306

Chapter 3

Polynomial and Rational Functions −10 ± 100 − 112 2

26. (a) For the quadratic, x =

= −5 ± 3i.

33. (a) The degree of f is even and the leading coefficient is 1. The graph rises to the left and rises to the right. (b) f ( x ) = x 4 − 2 x 3 − 12 x 2 + 18 x + 27

= ( x 4 − 12 x 2 + 27) − (2 x 3 − 18 x )

Zeros: −8, − 5 ± 3i

= ( x 2 − 9)( x 2 − 3) − 2 x( x 2 − 9)

− ( x + 6 ) − 8 = x 3 + 18 x 2 + 108 x + 224 3

(

= ( x 2 − 9)( x 2 − 3 − 2 x )

)

= ( x + 8 ) x + 10 x + 28 = 0

(b)

= ( x + 3)( x − 3)( x 2 − 2 x − 3) = ( x + 3)( x − 3)( x − 3)( x + 1) Zeros: −3, 3, 3, − 1 (c) and (d)

2 − 16

− 14

27. (a)

−4

Zeros: x = 0, − 3, − 3 (b)

−2

−40

34. (a) The degree of f is odd and the leading coefficient is −3 . The graph rises to the left and falls to the right.

−5

(b)

t 4 − 4t 2 = 0

f ( x ) = −3 x 3 − 2 x 2 + 27 x + 18 = − ( x − 3)( x + 3)( 3 x + 2 )

Zeros: x = ±3, −

)

t2 t2 − 4 = 0

2 3

t (t + 2)(t − 2) = 0 Zeros: t = 0, 0, ± 2

(b)

−30

−8

(

−20

28. (a)

40 30

−7

−4 −5

35.

30. f ( x) = ( x + 1)( x − 0)( x − 3)( x − 5)

f ( x) = x3 + 2 x 2 − x − 1 (a) f ( −3) < 0, f ( −2) > 0  zero in ( −3, − 2) f ( −1) > 0, f (0) < 0  zero in ( −1, 0) f (0) < 0, f (1) > 0  zero in (0, 1)

= x 4 − 7 x3 + 7 x 2 + 15 x

(

)(

f ( x ) = ( x − 3) x − 2 + 3 x − 2 − 3 3

− 30

= x3 − 12 x − 16

31.

1 − 20

−2

)

−6

= x − 7 x + 13 x − 3 32.

(

f ( x) = ( x + 7) x − 4 + 6

)( x − 4 − 6 )

−4

(b) Zeros: −2.247, − 0.555, 0.802

= x 3 − x 2 − 46 x + 70

Chapter 3 Review 36.

f ( x) = x 4 − 6 x 2 − 4 (a) f ( −3) > 0, f ( −2) < 0  zero in ( −3, − 2) f (2) < 0, f (3) > 0  zero in (2, 3)

4 8 x+ 3 9

40.

3x − 2 4 x2 + 0 x + 7

4 x2 −

−5

8 x 3 8 x+7 3 8 16 x− 3 9

−14

(b) Zeros: ±2.570 37.

307

79 9

f ( x ) = 0.24 x − 2.6 x − 1.4 (a) f ( −3) < 0, f ( −2) > 0  zero in ( −3, − 2) f ( −1) > 0, f (0) < 0  zero in ( −1, 0) f (3) < 0, f (4) > 0  zero in (3, 4)

79 4 x2 + 7 4 8 4 8 79 = x+ + 9 = x+ + 3x − 2 3 9 3x − 2 3 9 27 x − 18

x2 − 2

41.

x2 − 1 x 4 − 3x 2 + 2

−6

x4 − x2

−2 x 2 + 2 −2 x 2 + 2

−5

(b) Zeros: −2.979, − 0.554, 3.533 7 38. f ( x ) = 2 x + x 3 − 2 2 (a) f ( −2) > 0, f ( −1) < 0  zero in ( −2, − 1) f (0) < 0, f (1) > 0  zero in (0, 1)

Thus,

x − 3x + 2 = x 2 − 2, ( x ≠ ±1). x2 − 1

3x2 + 4

42.

x2 − 1 3x 4 − 3x2 − 1

−5

3x 4 − 3x2 4 x2 − 1

4 x2 − 4 3 −4

Thus,

(b) Zeros: −1.897, 0.738 39.

8x + 5 3 x − 2 24 x 2 − x − 8

5x + 2

43. 2

x − 3 x + 1 5 x − 13 x − x + 2 5 x 3 − 15 x 2 + 5 x

24 x 2 − 16 x

2 x2 − 6 x + 2 2 x2 − 6 x + 2 0

15x − 8 15 x − 10 2 24 x 2 − x − 8 2 Thus, = 8x + 5 + . 3x − 2 3x − 2

3x4 + x2 − 1 3 = 3x2 + 4 + 2 . x2 − 1 x −1

5 x 3 − 13 x 2 − x + 2 = 5 x + 2, x2 − 3x + 1 1    x ≠ (3 ± 5 )  . 2  

Thus,

Chapter 3

308

Polynomial and Rational Functions

44.

3x 2 + 5x + 8

51. (a)

−3

(b)

−2

2 x + 0 x − 1 6 x 4 + 10 x 3 + 13 x 2 − 5 x + 2

10 x 3 + 0 x 2 − 5 x 16 x 2 − 0 + 2

2 20 3

Thus, 46. 1

(a)

53.

48. 5

0.25 x − 4 x 3 1 3 9 2 36 = x − x + 9 x − 18 + . x+2 4 2 x+2

54.

49.

2 3

(a)

19.5

Thus, 50.

1 2

Thus,

9 4

2 x3 + 2 x2 − x + 2 1 94 = 2 x2 + 3x + + . x − (1/ 2) 2 x − (1/ 2)

4 2 − 10 8 − 10 2 4 2 − 10

20 −20

− 10 2

1 −4 −7 22 24 −2 12 −10 −24 12

1 −6 5 12 3 −9 −12 0

x 2 − 3 x − 4 = ( x − 4)( x + 1) Remaining factors: ( x − 4), ( x + 1) (c) f ( x ) = ( x + 2)( x − 3)( x − 4)( x + 1) (d) Zeros: −2, 3, 4, − 1

(b)

2 2 −1 2 1 23 14 1 2

1 −3 −4 ( x − 3) is a factor.

0 0

6 x 4 − 4 x 3 − 27 x 2 + 18 x 2 = 6 x 3 − 27 x, x ≠ . x − (2 / 3) 3

2 3

−2

6 −4 −27 18 0 4 0 −18 0 0 −27

( 2)

1 −6 5 ( x + 2) is a factor.

0.1x + 0.3 x 2 − 0.5 19.5 = 0.1x 2 + 0.8 x + 4 + . x−5 x−5

4−5 2

−8

f ( x ) = x 4 − 4 x 3 − 7 x 2 + 22 x + 24

Thus,

2 2

x 2 + 8 x + 7 = ( x + 1)( x + 7) Remaining factors: ( x + 1), ( x + 7) (c) f ( x ) = ( x − 4)( x + 1)( x + 7) (d) Zeros: 4, − 1, − 7

0.1 0.3 0 −0.5 0.5 4 20 0.1 0.8 4

(b)

Thus,

1 8 7 0 ( x − 4) is a factor.

9 −18 36

− 29

1 4

−5

0 0 −4 0 1 − 2 9 −18 36

0.25

−8

f ( x ) = x 3 + 4 x 2 − 25 x − 28 (a) 4 1 4 −25 −28

2 x 3 + 6 x 2 − 14 x + 9 3 Thus, = 2 x2 + 8x − 6 + . x −1 x −1

47. −2

2 2 2 −5 4−5 2

−6

2 8

(b)

8 −6

2 −5

−8 52 −208 832 −3296

3 x − 10 x + 12 x − 22 58 = 3 x 2 + 2 x + 20 + . x−4 x−4

−4

2 −13 52 −208 824 −3276 = g(−4)

2 6 −14

8 −40 100 −156 = f ( −2)

52. g(t ) = 2t 5 − 5t 4 − 8t + 20

3 −10 12 −22 12 8 80 3

1 10 −24 20 44 −2 −16 80 −200 1

Thus, 6 x 4 + 10 x 3 + 13 x 2 − 5 x + 2 10 = 3x2 + 5x + 8 + 2 . 2 x2 − 1 2x − 1 45. 4

7 −45 155 −421 = f ( −3)

6 x 4 + 0 x3 − 3x2 10 x 3 + 16 x 2 − 5 x

16 x 2 + 0 − 8 10

1 10 −24 20 44 −3 −21 135 −465

55.

f ( x ) = 4 x 3 − 11x 2 + 10 x − 3 Possible rational zeros: ±1, ± 3, ± 23 , ± 43 , ± 12 , ± 41 Zeros: 1, 1,

3 4

Chapter 3 Review 56.

f ( x ) = 10 x 3 + 21x 2 − x − 6 Possible rational zeros: ±1, ± 2, ± 3, ± 6, ± 15 , ± 101 , ± 12 , ± 103 , ± 52 ,

61. f ( x) = x3 − 4 x 2 + x + 6

Possible rational zeros: ±1, ± 2, ± 3, ± 6 Use synthetic division to check x = 2.

± 53 , ± 65 , ± 32 Zeros: −2,

1 2

, −

2 1 −4

3 5

Rewrite in polynomial form and factor. x2 − 2x − 3 = 0

( x − 3)( x + 1) = 0

4 −3 4 − 3 4 1 5

4 1 5 2 All entries positive; x = 1 is upper bound.

− 14

x +1 = 0

x = 3

x = −1

62. f ( x) = x3 + x 2 − 28 x − 10

Possible rational zeros: ±1, ± 2, ± 5, ±10 Use synthetic division to check x = 5.

4 −3 4 −3 −1 1 − 54

5 1 1 − 28 −10

Alternating signs; x = − 14 is lower bound.

1 6

x = 5 is a zero.

3 Real zero: x = 4

Rewrite in polynomial form and use the Quadratic Formula. x2 + 6 x + 2 = 0

2 −5 −14 8 16 88 592

x =

2 11 74 600 All positive  x = 8 is upper bound. −4 2 −5 −14 8 − 8 52 −152 2 −13 38 −144 Alternating signs  x = −4 is lower bound. 1 Real zeros: x = −2, , 4 2

x −3 = 0

Real zeros: x = −1, 2, 3

4 −4 5 − 174

60. 8

x = 2 is a real zero.

f (− x ) = −2 x − 3 x − 2 x − 1 has no variations in sign  0 negative real zeros. 59. 1

1 − 2 −3

f ( x ) = 2 x 5 − 3 x 2 + 2 x − 1 has three variations in sign  1 or 3 positive real zeros. 5

2 −4 −6

57. g( x ) = 5 x 3 − 6 x + 9 has two variation in sign  0 or 2 positive real zeros. g(− x ) = −5 x 3 + 6 x + 9 has one variation in sign  1 negative real zero. 58.

309

x =

− 6 ± 28 2

x =

−6 ± 2 7 = −3 ± 2

Zeros: x = 5, − 3 ± 63.

(6) − 4(1)(2) 2(1)

−6 ±

f ( x ) = 6 x 4 − 25 x 3 + 14 x 2 + 27 x − 18

Possible rational zeros: ±1, ± 2, ± 3, ± 6, ± 9, ± 18, ± 12 , ± 32 , ± 92 , ± 13 ,

± 23 , ± 16 Use synthetic division to check x = −1 and x = 3. −1

6 −25 14 27 −18 −6 31 −45 18 6 −31 45 −18

6 −31 45 −18 18 −39 18 6 −13

6 x 4 − 25 x 3 + 14 x 2 + 27 x − 18 = ( x + 1)( x − 3)(6 x 2 − 13 x + 6) = ( x + 1)( x − 3)(3 x − 2)(2 x − 3)

Zeros: x = −1, 3,

2 3 , 3 2

310

Chapter 3

Polynomial and Rational Functions

64.

f ( x ) = 5 x 4 + 126 x 2 + 25 2

72.

f ( x ) = 2 x 3 − 5 x 2 + 9 x + 40

(

= ( 2 x + 5) x 2 − 5 x + 8

= (5 x + 1)( x + 25) No real zeros 65.

x + 6x + 9 = 0

5 5 7 Zeros: − , ± i 2 2 2  5 7  5 7  f ( x ) = ( 2 x + 5)  x − + i  x − − i   2 2  2 2  

( x + 3) = 0 2

x+3=0 x = −3 66.

x 2 − 10 x + 25 = 0

73.

( x − 5) = 0 2

(

f ( x ) = 2 x 4 − 5 x 3 + 10 x − 12 2 −5 0 10 −12 4 −2 −4 12

x−5=0 x=5 67.

2 −1 −2

x 3 + 16 x = 0

)

x = 0 x 2 + 16 = 0

−

x = −16

2 −4

)

(

x = 0 x + 144 = 0

(

= ( x − 2 )( 2 x + 3 ) x 2 − 2 x + 2

x 2 = −144 x = ±12i

Zeros: x = 0, − 3 3

(

Quadratic Formula: x =

x = 4 is a zero. Applying the Quadratic Formula on

Zeros: −1,

x − 3 x + 6,

Zeros: 4,

3 15 ± i. 2 2

3 15 3 15 + i, − i 2 2 2 2

 3 + 15i  3 − 15i  h ( x ) = ( x − 4) x − x−     2 2   

= 1 ± i.

= ( x + 1)( 3 x − 1) x 2 − 2 x + 4

74. g ( x ) = 3 x 4 − 4 x 3 + 7 x 2 + 10 x − 4

1 −7 18 −24 4 −12 24

3 ± 9 − 4 (6)

2 ± 4 − 4 (2)

f ( x ) = ( x − 2 )( 2 x + 3)( x − 1 + i )( x − 1 − i )

Zeros: x = − 2, 8, 8, 8

)

3 Zeros: 2, − , 1 ± i 2

70. f ( x) = ( x − 8) ( x + 2)

71. h( x ) = x 3 − 7 x 2 + 18 x − 24

)

By the Quadratic Formula, applied to x 2 – 2 x + 2,

69. f ( x) = − 4 x( x + 3)

3 is a zero. 2 3  f ( x ) = ( x − 2) x +  2 x2 − 4 x + 4 2 

x x 2 + 144 = 0

−3

x=−

68. x 3 + 144 x = 0

2 −1 −2 6 −3 6 −6

3 2

x = ±4i

x = 2 is a zero.

x x 2 + 16 = 0

(

5 ± 25 − 32 5 ± 7 i = 2 2

Quadratic Formula: x =

)

1 , 1 ± 3i 3

)

2 ± 4 − 16 = 1 ± 3i 2

(

)(

g ( x ) = ( x + 1)( 3 x − 1) x − 1 + 3i x − 1 − 3i

75.

f ( x ) = x 5 + x 4 + 5 x3 + 5x 2

(

= x2 x3 + x 2 + 5x + 5

)

= x 2  x 2 ( x + 1) + 5 ( x + 1) 

(

= x 2 ( x + 1) x 2 + 5

(

)

)(

= x ( x + 1) x + 5i x − 5i 2

)

Zeros: 0, 0, − 1, ± 5i

Chapter 3 Review 76.

f ( x ) = x5 − 5x3 + 4 x

80. (a)

f ( x ) = 2 x 3 − 9 x 2 + 22 x − 30

(

= ( 2 x − 5) x 2 − 2 x + 6

( ) = x ( x − 4 )( x − 1)

= x x4 − 5x2 + 4 2

f ( x ) = x ( x − 2 )( x + 2 )( x − 1)( x + 1) Zeros: 0, ± 1, ± 2

(a)

(b)

(

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

)

2±

( −2) − 4 ( 2 ) 2

2 Zeros: 2, 1 + i, 1 − i (b)

81.

= 1 ± i.

(

= ( x + 3 ) x 2 − 8 x + 17

( −8 ) − 4 (17 ) 2

2 = 4±i Zeros: −3, 4 + i, 4 − i (b)

)

8 ± −4 2

)(

)

(

)(

x 4 + 34 x 2 + 225 = x 2 + 9 x 2 + 25

)

Zeros: ±3i, ± 5i (b)

( x + 3i )( x − 3i )( x + 5i )( x − 5i )

( )( ) = ( x + 1) ( x + 5) = ( x + i )( x − i )( x + 5) 2

Zeros: ±i, − 5, − 5 (c) x-intercept: ( −5, 0 ) 83. Since 3i is a zero, so is −3i. f ( x) = ( x − 5)( x − 3i)( x + 3i)

f ( x ) = −3 x 3 − 19 x 2 − 4 x + 12

(

= − ( x + 1) 3 x 2 + 16 x − 12

= ( x − 5)( x 2 + 9)

)

= x3 − 5 x 2 + 9 x − 45

−19 3

−4 16

12 −12

84. Since −i is a zero, so is i.

− 3 − 16

= ( x + 6)( x 2 + 1)

3 x + 16 x − 12 = 0

( 3x − 2 )( x + 6 ) = 0 2 3 x + 6 = 0  x = −6 2 Zeros: −1, , −6 3 3x − 2 = 0  x =

(b)

(

f ( x ) = ( 2 x − 5 ) x − 1 − 5i x − 1 + 5i

= x 2 + 1 x 2 + 10 x + 25

f ( x ) = ( x + 3)( x − 4 − i )( x − 4 + i )

−1 − 3

5 , 1 + 5i , 1 − 5 i 2

82. (a), (b) f ( x ) = x 4 + 10 x 3 + 26 x 2 + 10 x + 25

= 1 ± 5i

f ( x ) = x 3 − 5 x 2 − 7 x + 51

8±

f ( x ) = x 4 + 34 x 2 + 225 (a)

f ( x ) = ( x − 2 )( x − 1 − i )( x − 1 + i )

( −2 ) − 4 ( 6 )

)

5  (c) x-intercept:  , 0 2 

By the Quadratic Formula for x 2 − 2 x + 2,

2±

Zeros:

77. f ( x ) = x 3 − 4 x 2 + 6 x − 4

311

f ( x ) = − ( x + 1)( 3 x − 2 )( x + 6 )

2  (c) x-intercepts: ( −1, 0 ) , ( −6, 0 ) ,  , 0  3 

f ( x) = ( x + 6)( x − i )( x + i ) = x3 + 6 x 2 + x + 6 85. Since − 2 − 4i is a zero, so is − 2 + 4i. f ( x) = ( x + 2)( x + 2 − 4i)( x + 2 + 4i) 2 = ( x + 2) ( x + 2) + 16  

= ( x + 2)  x 2 + 4 x + 20 = x3 + 6 x 2 + 28 x + 40

312

Chapter 3

86. Since − 5 +

Polynomial and Rational Functions

2i is a zero, so is − 5 −

2i.

(

f ( x ) = ( x − 1) x + 5 −

)(

2i x + 5 +

92.

)

2 = ( x − 1) ( x + 5) + 2  

= ( x − 1)  x 2 + 10 x + 27 = x 3 + 9 x 2 + 17 x − 27

87.

93.

f ( x ) = x 4 − 2 x 3 + 8 x 2 − 18 x − 9 (a) f ( x ) = ( x 2 + 9)( x 2 − 2 x − 1) (b) For the quadratic

x 2 − 2 x − 1, x =

2 ± (−2)2 − 4(−1) = 1 ± 2. 2

)( x − 1 − 2 ) (c) f ( x ) = ( x + 3i )( x − 3i ) ( x − 1 + 2 )( x − 1 − 2 ) (

)(

94.

f ( x) = x 2 + 9 x − 1 + 2

88.

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

1 ± ( −1)2 − 4(−4) 1 17 = ± 2 2 2  1 17  1 17  2 f ( x) =  x − − x− +   x − 3x + 4   2 2  2 2  

(b) x =

(

2 2 = x 2 − 3 x − 18 ( x − 6)( x + 3) (a) Domain: all x ≠ 6, − 3 (b) Not continuous (c) Horizontal asymptote: y = 0 Vertical asymptotes: x = 6, x = −3 f ( x) =

f ( x) =

2 x2 + 3 x2 + x + 3

The denominator x 2 + x + 3 has no real zeros. (a) Domain: all x (b) Continuous (c) Horizontal asymptote: y = 2 Vertical asymptote: none

f ( x ) = x 4 − 4 x 3 + 3 x 2 + 8 x − 16 (a)

4x x −8 (a) Domain: all x ≠ 8 (b) Not continuous (c) Horizontal asymptote: y = 4 Vertical asymptote: x = 8 f ( x) =

95.

)

3 ± ( −3)2 − 4(4) 3 7 = ± i 2 2 2  1 17  1 17  3 7  f ( x) =  x − − x− + x− + i      2 2  2 2  2 2    3 7  i  x − −  2 2  

96.

7+ x 7−x (a) Domain: all x ≠ 7 (b) Not continuous (c) Horizontal asymptote: y = −1 Vertical asymptote: x = 7 f ( x) =

6x 6x = x 2 − 1 ( x + 1)( x − 1) (a) Domain: all x ≠ ±1 (b) Not continuous (c) Horizontal asymptote: y = 0 Vertical asymptotes: x = ±1 f ( x) =

89. Zeros: −2i, 2i ( x + 2i )( x − 2i ) = x 2 + 4 is a factor.

97.

f ( x ) = ( x 2 + 4)( x + 3) Zeros: ±2i, − 3

Vertical asymptotes: x = ±

)

f ( x ) = x 2 − 4 x + 9 ( 2 x + 1) 1 Zeros: x = − , 2 ± 5i 2

91.

2−x x+3 (a) Domain: all x ≠ −3 (b) Not continuous (c) Horizontal asymptote: y = −1 Vertical asymptote: x = −3 f ( x) =

3 6 =± 2 2

(b) Not continuous (c) Horizontal asymptote: y = 2

( x − 2 − 5i )( x − 2 − 5i ) = ( x − 2 ) + 5 = x − 4 x + 9 is a 2

(

4 x2 2 x2 − 3

(a) Domain: all x ≠ ±

90. Zeros: 2 + 5i, 2 − 5i factor.

f ( x) =

98.

3 6 =± 2 2

3 x 2 − 11x − 4 x2 + 2 (a) Domain: all x (b) Continuous (c) Horizontal asymptote: y = 3 No vertical asymptote f ( x) =

Chapter 3 Review

99.

2 x − 10 2( x − 5) 2 = = , x≠5 x 2 − 2 x − 15 ( x − 5)( x + 3) x + 3 (a) Domain: all x ≠ 5, − 3 (b) Not continuous (c) Vertical asymptote: x = −3 (There is a hole at x = 5. ) Horizontal asymptote: y = 0 f ( x) =

100. f ( x) =

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

(a) Domain: all x ≠ − 6, 0 (b) Not continuous (c) Vertical asymptotes: x = − 6, x = 0 Horizontal asymptote: y = 0 101. f ( x) = = =

3x 2 − 15 x3 − 5 x 2 − 24 x 3( x 2 − 5)

x( x 2 − 5 x − 24) 3( x 2 − 5)

x( x − 8)( x + 3)

(a) Domain: all x ≠ − 3, 8 (b) Not continuous (c) Vertical asymptotes: x = − 3, x = 0, x = 8 Horizontal asymptote: y = 0 102. f ( x) =

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

(a) Domain: all x ≠ 0, 4 (b) Not continuous (c) Vertical asymptotes: x = 0, x = 4 Horizontal asymptote: y = 0 103. C =

528 p , 0 ≤ p < 100 100 − p 528(25) = $176 million. 100 − 25 528(50) When p = 50, C = = $528 million. 100 − 50 528(75) When p = 75, C = = $1584 million. 100 − 75

(a) When p = 25, C =

(b)

313

1.568 x − 0.001 , x>0 6.360 x + 1 The moth will be satiated at the horizontal asymptote, 1.568 y= ≈ 0.247 mg. 6.360

104. y =

x2 − 5x + 4 x2 − 1 ( x − 4)( x − 1) = ( x − 1)( x + 1) x−4 = , x ≠1 x +1 Vertical asymptote: x = −1 Horizontal asymptote: y = 1 No slant asymptotes Hole at x = 1

105. f ( x ) =

2 x2 − 7x + 3 2 x2 − 3x − 9 ( x − 3)(2 x − 1) = ( x − 3)(2 x + 3) 2x − 1 = , x≠3 2x + 3

106. f ( x ) =

3 2 Horizontal asymptote: y = 1 No slant asymptotes Hole at x = 3

Vertical asymptote: x = −

3x2 + 5x − 2 x +1 (3 x − 1)( x + 2) = x +1 Vertical asymptote: x = −1 Horizontal asymptote: none Long division gives:

107. f ( x ) =

3x + 2 x + 1 3x2 + 5x − 2 3x 2 + 3x 2x − 2 2x + 2 −4 Slant asymptote: y = 3 x + 2

5000

100

Answers will vary. (c) No. As p → 100, C approaches infinity.

314

Chapter 3

Polynomial and Rational Functions −1 x + 2 x-intercept: none

2 x2 + 5x + 3 x−2 (2 x + 3)( x + 1) = x−2 Vertical asymptote: x = 2 Horizontal asymptote: none Long division gives:

110. f ( x ) =

108. f ( x ) =

1  y-intercept:  0, −  2  Vertical asymptotes: x = − 2 Horizontal asymptote: y = 0

2x + 9 x − 2 2 x2 + 5x + 3 2 x2 − 4 x 9x + 3 9 x − 18 21

−6

−4

−3

−1

1 4

1 2

−1

−

1 2

−

y 4

Slant asymptote: y = 2 x + 9 109. f ( x ) =

3 2

1 +3 x

1 − 6 − 5 − 4 −3

−1

 1  x-intercept:  − , 0   3 

2 0, − 1 2

(

−2

)

−3 −4

y-intercept: none Vertical asymptote: x = 0 Horizontal asymptote: y = 3 x

−2

−1

5 2

7 2

2x −1 x−5  1 1  Intercepts:  0,  ,  , 0   5 2  Vertical asymptote: x = 5 Horizontal asymptote: y = 2

111. f ( x ) =

−4

−1

1 2

1 5

−

1 4

6 5 4

(− 13 , 0)

−4 −3 − 2 − 1 −2

10 8 6 4 2

−6 −4 −2

( 12 , 0) −− 46

(0, 15) 4 6 8 10 12 14

−8 − 10

1 4

Chapter 3 Review x −3 x−2 x-intercept: (3, 0)

4 ( x − 1) 2 y-intercept: (0, 4) Vertical asymptote: x = 1 Horizontal asymptote: y = 0

112. f ( x ) =

114. f ( x ) =

 3 y-intercept:  0,   2 Vertical asymptote: x = 2 Horizontal asymptote: y = 1

−1

4 3

3 2

1 2

2 3

−2

−1

4 9

8 7 6 5 4 3

y 6 5 4 3 2

− 3 − 2 −1 −2 −3 −4 −5 −6

(0, 32) −4 −3 −2 −1

3 4 5 6 7 8

315

2 3 4 5 6

−2

(3, 0)

2 x2 x2 − 4 Intercept: (0, 0) y-axis symmetry Vertical asymptotes: x = ±2 Horizontal asymptote: y = 2

115. f ( x ) =

2 ( x + 1)2 Intercept: (0, 2) Horizontal asymptote: y = 0 Vertical asymptote: x = −1

113. f ( x ) =

−4

−3

−2

2 9

1 2

2 9

−6

−4

−1

9 4

8 3

−

2 3

−

2 3

8 3

9 4

y 6

(0, 0) 4 3 2 1

−6 −5 −4 −3 −2 −1

−6

−4

(0, 2)

1 2 3 4

316

Chapter 3

Polynomial and Rational Functions

5x x2 + 1 Intercept: (0, 0) Symmetry: origin Horizontal asymptote: y = 0 No vertical asymptotes

2 x2 + 7 x + 3 2 = 2x + 5 − x +1 x +1  1  Intercepts: (0, 3), (−3, 0),  − , 0   2 

116. f ( x ) =

−3

3 − 2

−2

−1

−2

5 − 2

118. f ( x ) =

Vertical asymptote: x = −1 Slant asymptote: y = 2 x + 5 0

5 2

3 2

−5

−4

−3

−2

−

9 2

−

7 3

y 5 4 3 2 1

−4 −3 −2 −1

y 7 6 5

(0, 0) 1 2 3 4 5

(0, 3)

−3 −4 −5

−7 −6 −5 −4

−2

1 2 3

y = 2x + 5

x2 − x + 1 7 = x+2+ x −3 x −3 1  Intercept:  0, −  3 

117. f ( x ) =

119. (a)

Vertical asymptote: x = 3 Slant asymptote: y = x + 2

−4

−3

−

1 3

−3

10.5

y 16 12 8

(0, − 13)

−4

y=x+2 4

800

20(4 + 3(5)) = 304 thousand fish 1 + 0.05(5) 20(4 + 3(10)) 10 years: N (10) = = 453.3 thousand fish 1 + 0.05(10) 20(4 + 3(25)) 25 years: N (25) = = 702.2 thousand fish 1 + 0.05(25) (b) The maximum number of fish is N = 1,200,000. The graph of N has a horizontal asymptote at N = 1200 or 1,200,000 fish.

5 years: N (5) =

−8

Chapter 3 Test

317

121. Quadratic model

120. (a) 2 in.

30 in.2

122. Neither 123. (a) 6700

2 in.

0 6150

(b)

( x − 4)( y − 4) = 30  y = 4 +

30 x−4

(b) 6700

30   Area = A = xy = x  4 + x − 4    4 x − 16 + 30  = x  x−4   2 x (2 x + 7) = x−4 (c) Domain: x > 4 (d) 200

0 6150

9.48 by 9.48

(d) The cubic model is a better fit because it more closely follows the pattern of the data. (e) Answers will vary. Sample answer: For 2015, let t = 15. S (15) = −1.088(15) + 26.61(15) − 150.9(15) + 6459 3

≈ 6511 So, there will be about 6511 commercial FM radio stations in 2015. 124. False. A fourth degree polynomial with real coefficients can have at most four zeros. Since –8i and 4i are zeros, so are 8i and −4i. 125. False. For example, (1 + i ) + (1 − i ) = 2 is a real number.

Chapter 3 Test 1. y = x 2 + 5 x + 6 25 25 − + 6 4 4 25  1  =  x2 + 5x +  − 4 4  = x2 + 5x +

5 1  = x +  − 2 4  

When x = 0, y = 6. When y = 0,

x2 + 5x + 6 = 0

( x + 3)( x + 2) = 0 x = − 3, − 2. Intercepts: (0, 6), ( − 3, 0), ( − 2, 0)

 5 1 Vertex:  − , −   2 4

318

Chapter 3

Polynomial and Rational Functions

Let y = a( x − h)2 + k. The vertex (3, − 6) implies that

2 0 −5 0 −3 4 8 6 12

y = a( x − 3) − 6. For (0, 3) you obtain

3 = a(0 − 3)2 − 6 = 9a − 6  a = 1. 2

Thus, y = ( x − 3) − 6 = x − 6 x + 3. 3. f ( x) = 4 x3 − 12 x 2 + 9 x

= x( 2 x − 3)

3 , multiplicity 2 2

−2

9 x−2

0 −6 5 −1 −6 12 −12 14

3 −6 f ( −2) = 13

Zeros: x = 0, multiplicity 1 x =

2 x + 4 x + 3x + 6 +

= x( 4 x 2 − 12 x + 9)

− 7 13

Possible rational zeros: ±1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24, ±

1 2 4. (a) y = x + 3x + 5 20

1 3 , ± , 2 2

5 − 10

The function has a maximum at x=

− ( 3) −b = = 30. 2a  1  2 −   20 

1 2 So, y = − ( 30 ) + 3( 30 ) + 5 20 1 =− ( 900 ) + 90 + 5 20 = 50.

− 35

Rational zeros: −2,

3 2

1 2 10. Possible rational zeros: ±1, ± 2, ± , ± 3 3 5

(b) The constant term 5 represents the height at which the ball was thrown. Yes, changing this value would act as a vertical shift upward or downward.

−9

−7

f ( x) = − x3 + 7 x + 6

Rational zeros: ±1, −

11.

f ( x ) = x 3 − 7 x 2 + 11x + 19

= ( x + 1)( x 2 − 8 x + 19)

9 6

For the quadratic,

3 −12 −9 −6 −3

9 12

−6

2x 6. x + 2 2 x + 0 x + 5 x − 3 2

2 x3

2 3

+ 4x x −3

2 x3 + x 2 + 5 x − 3 x − 3 = 2x + 2 x2 + 2 x + 2

8 ± 64 − 4(19) = 4 ± 3 i. 2 Zeros: −1, 4 ± 3 i x=

f ( x ) = ( x + 1)( x − 4 + 3 i ) ( x − 4 − 3 i ) 12. Since 2 + i is a zero, so is 2 − i.

f ( x) = ( x − 0)( x − 2)  x − ( 2 − i ) 2 − ( 2 + i ) = x( x − 2)( x − 2 + i )( x − 2 − i ) = ( x − 2)( x 2 − 4 x + 4 − i 2 ) = x( x − 2)( x 2 − 4 x + 5)

= x( x3 − 6 x 2 + 13x − 10) = x 4 − 6 x3 + 13x 2 − 10 x

Chapter 3 Test 13. Since 1 −

3i is a zero, so is 1 +

3i.

(

3i ( x − 2)( x − 2) 

) ( 3i )( x − 1 −

f ( x) =  x − 1 − 

)

3i  x − 1 − 

(

= x −1+

)

17.

2 x2 + 9 5x2 + 2

f ( x) =

Horizontal asymptote: y =

3i ( x − 4 x − i ) + 4 2

319

2 5

y-axis symmetry  9 Intercept:  0,   2

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

= ( x 2 − 2 x + 4)( x 2 − 4 x + 4)

= x 4 − 6 x3 + 16 x 2 − 24 x + 16

(0, 92)

14. Since 1 + i is a zero, so is 1 − i.

f ( x) = ( x − 0)  x − (1 + i )  x − (1 − i)

2 1

= x( x − 1 − i)( x − 1 + i ) = x( x 2 − 2 x + 1 − i 2 ) = x( x 2 − 2 x + 2)

−3

18.

(a)

−2

−1

800

= x3 − 2 x 2 + 2 x 15. h( x) =

4 −1 x2

(b) A = −11.313t 2 + 241.07t − 585.7 (c)

(− 2, 0)

3 2 1

(2, 0)

800

x 6

−2 −3 −4

The model fits the data well.

Vertical asymptote: x = 0 Intercepts: (2, 0), ( −2, 0) Symmetry: y-axis Horizontal asymptote: y = −1 x2 + 2 3 = x +1+ x −1 x −1 Vertical asymptote: x = 1 Intercept: (0, − 2) Slant asymptote: y = x + 1

16. g( x ) =

(d) For 2015, let t = 15.

A(15) = −11.313(15) + 241.07(15) − 585.7 2

= $484.9 billion For 2020, let t = 20.

A( 20) = −11.313( 20) + 241.07( 20) − 585.7 2

= − $289.5 billion (e) No, the model is not useful for predicting future years beyond 2103. In fact, the model yields negative values by 2019.

10 8 6 4

y=x+1

−8 −6 −4

−4

(0, − 2)

−6

C H A P T E R 4 Exponential and Logarithmic Functions Section 4.1

Exponential Functions and Their Graphs ..........................................322

Section 4.2

Logarithmic Functions and Their Graphs .........................................334

Section 4.3

Properties of Logarithms ....................................................................343

Section 4.4

Solving Exponential and Logarithmic Equations .............................353

Section 4.5

Exponential and Logarithmic Models ...............................................369

Section 4.6

Nonlinear Models ...............................................................................379

Chapter 4 Review .......................................................................................................384 Chapter 4 Test ............................................................................................................395 Chapters 3–4 Cumulative Test .................................................................................397

C H A P T E R 4 Exponential and Logarithmic Functions Section 4.1 Exponential Functions and Their Graphs 1.

transcendental

natural exponential, natural

The graph of f ( x + 1) is a horizontal shift one unit to

10. f ( x) = 10 x −2 1 100

the left of f ( x ) = 5 . 4.

The formula A = Pert gives the balance A of an account earning interest compounded continuously.

(3.4)6.8 ≈ 4112.033 1/ 3

−1 1 10

100

Asymptote: y = 0 Intercept: (0, 1) Increasing y 5

≈ 1.063

1.2

5−π ≈ 0.006

8.6 −3( − 2 ) = 8.63 2 ≈ 9220.217

g( x ) = 5

3 2 1

−3

x y

−2

−1

1 25

1 5

Asymptote: y = 0

−2

−1

11. f ( x) = 5− x = ( 15 )

x y

Intercept: (0, 1)

−2

−1

1 5

1 25

Intercept: (0, 1)

Decreasing

4 3

−1

Asymptote: y = 0

Increasing

−2

−1

3 1

−2

322

−1

Section 4.1 12. h ( x ) = 10 − x

Exponential Functions and Their Graphs

15. g ( x ) =

−2

−1

100

1 1 10

2 1 100

Asymptote: y = 0 Intercept: (0, 1) Decreasing

( 54 )

−x

( 54 )

323

−2

−1

25 16

5 4

4 5

16 25

Asymptote: y = 0 Intercept: (0, 1)

Decreasing

7 6 5 4

1 −3

−2

13. h( x) =

−1

−4 −3 −2 −1 −1

()

5 x 4

−2

−1

16 25

4 5

5 4

25 16

16. f ( x ) = ( 32 )

Asymptote: y = 0 Intercept: (0, 1)

−x

9 4

3 2

2 3

4 9

y 9 8 7 6 5 4 3 2

4 3 2

−5 − 4 −3 −2 −1

−2

−1

4 9

2 3

3 2

9 4

Asymptote: y = 0 Intercept: (0, 1) Increasing

1 2 3 4 5

17. f ( x ) = 2 x − 2 rises to the right. Asymptote: y = 0 Intercept: ( 0, 14 ) Matches graph (d). 18. f ( x ) = 2 − x is positive and decreasing. Matches graph (a).

19. f ( x ) = 2 x − 4 rises to the right. Asymptote: y = −4 Intercept: (0, − 3) Matches graph (c).

9 8 7 6 5 4 3 2

−5 −4 −3 −2 −1

−1

x y

−2

14. g ( x) = ( )

= ( 23 )

3 2

Asymptote: y = 0 Intercept: (0, 1) Decreasing

Increasing

−4 −3 −2 −1 −1

1 2 3 4 5

324

Chapter 4

Exponential and Logarithmic Functions

20. f ( x ) = 2 x + 1 is increasing and has (0, 2) intercept. Matches graph (b). 21. The graph of g( x ) = 3 x − 5 is a horizontal shift five units

to the right of f ( x ) = 3 x. y

x+4

3 25. The graph of g( x ) = −   is a reflection in the 5 x-axis and a horizontal shift four units to the left of x

3 f ( x) =   . 5 y

5 4 3 2

−8

−6

−2

1 −2 −1

to the left of f ( x) = 10 x. y

−6

26. The graph of g( x ) = −0.3 x + 5 is a reflection in the x-axis and a vertical shift five units upward of f ( x ) = 0.3 x.

−4

22. The graph of g ( x) = 10 x + 4 is a horizontal shift four units

−2

5 4

3 2

− 6 −5 − 4 − 3 −2 − 1 −1

23. The graph of g ( x) = 6 + 5 is a vertical shift five units x

upward of f ( x) = 6 x.

−1

units right and a vertical shift three units downward of f ( x) = 0.4 x.

−2

27. The graph of g ( x) = 0.4 x − 2 − 3 is a horizontal shift two

−3

4 3

1 −4 −3 −2 −1 −1

1 −3 − 2 −1 −1

−4

x+2

2 − 3 is a horizontal shift two 28. The graph of g ( x) =   3 units to the left and a vertical shift three units downward

5 3

2 1 1

−2 −4

−3

five units upward of f ( x ) = −2 x.

−3

−2

24. The graph of g( x ) = 5 − 2 x = − 2 x + 5 is a vertical shift

−5 −4 −3 −2

 2 of f ( x) =   .  3 y

f −5

− 3 −2 −1 −1 −2 −3

−4

Section 4.1 −x

1 29. The graph of g ( x) =   + 2 is a reflection in the  4 y-axis and a vertical shift two units upward of x

1 f ( x) =   .  4

33. e 9.2 ≈ 9897.129 34. e − ( −3 / 4) = e3 / 4 ≈ 2.117

36. −5.5e −200 = 7.611 × 10 −87 ≈ 0 f

7 6

37.

5 4 3

f ( x) = ( 25 )

−2

−1

0.16

0.4

2.5

6.25

f ( x)

1 −4 −3 −2 −1 −1

9 8 7 6 5 4 3 2 1

− ( x + 4)

1 30. The graph of g( x ) =   is a reflection in the 2 y-axis and a horizontal shift four units to the left of x

1 f ( x) =   . 2

−5 − 4 −3 −2 −1 y

325

35. 50e 4(0.02) ≈ 54.164

Exponential Functions and Their Graphs

Asymptote: y = 0

7 6 5

38.

4 3 2

5 f ( x) =   2

1 − 6 −5 − 4 − 3 − 2 − 1 −1

1 2 3 4 5

f ( x)

−x

−2

−1

6.25

2.5

0.4

0.16

31.

(a)

9 8 7 6 5 4 3

100

(b) e0.5 ≈ 1.6487

−5 −4 −3 −2 −1

1 2 3 4 5

Asymptote: y = 0 39.

f ( x) = 6 x

x 24

32. (a)

f ( x)

−2

−1

0.03

0.17

(b)

9 8 7 6 5 4 3 2 1

100

e 3 ≈ 20.086

−5 − 4 − 3 −2 − 1

1 2 3 4 5

326

Chapter 4

40.

f ( x ) = 2 x −1

Exponential and Logarithmic Functions

−1

f ( x)

y = 3x − 2 + 1

43.

0.25

0.5

x y

3 4

−5 −4 − 3 −2 −1

1 2 3 4 5

−4 − 3 −2 −1

−3 0.33

1 2 3 4 5 6

−2

0 9

−1 3

x y

1 27

−2 −1.75

−5 −4 −3 −2

1 2 3 4

1 14

2 62

1 2 3 4 5

Asymptote: y = −2

Asymptote: y = 0

f ( x ) = e− x

45. − x2

x −2 0.06

0 2

−2 −3 −4 −5

3 2 1 −6 −5 − 4 −3 −2 − 1

−1 −1

5 4 3 2 1

9 8 7

0 1

−1 0.5

1 0.5

2 0.06

f ( x)

−2

−1

7.39

2.72

0.37

0.14

9 8 7 6 5 4 3

9 8 7 6 5 4 3 2

−5 −4 − 3 −2 − 1

4 10

x y

y = 4 x +1 − 2

44.

f ( x) = 3x + 2

y=2

Asymptote: y = 1

Asymptote: y = 0

42.

9 8 7 6 5 4 3 2

9 8 7 6 5 4 3 2 1

x f ( x)

1 1.33

41.

0 1.11

−1 1.04

1 1 2 3 4 5

Asymptote: y = 0

−5 − 4 − 3 −2 − 1

1 2 3 4 5

Asymptote: y = 0

Section 4.1 46. s(t ) = 3e −0.2 t

49.

−1

s (t )

3.66

2.46

2.01

1.65

1.35

Exponential Functions and Their Graphs f ( x) = 2 + e x −5

x f ( x)

1 2 3 4 5 6

6 4.72

9.39

t −1

Asymptote: y = 0 f ( x ) = 3e

5 3

4 2.37

9 8 7 6 5 4 3

2 1

47.

3 2.14 y

9 8 7 6 5 4

−4 −3 −2 −1

1 2 3 4 5 6 7 8 9

Asymptote: y = 2

x+4

50. g ( x) = e x + 1 + 2

−6

−5

−3

f ( x)

−4

−2

0.41

1.10

8.15

22.17

−3

−2

−1

g ( x)

2.14

2.37

4.72

9.39

9 8 7 6 5 4 3 2 1

y 7 6 5 3

−9 −8 −7 −6 −5 −4 −3 −2 −1

Asymptote: y = 0

− 6 −5 − 4 − 3 −2 −1 −1

Asymptote: y = 2

48. f ( x) = 2e x − 3

51. s(t ) = 2e − 0.12t

f ( x)

0.27

0.74

5.44

14.78

−2

−1

s (t )

2.54

2.26

1.77

1.57

y 7

6 5

4 3

1 −2 −1 −1

327

Asymptote: y = 0

x 1 −4 −3 −2 −1 −1

Asymptote: y = 0

Chapter 4

328

Exponential and Logarithmic Functions

52. g ( x) = e0.5 x − 1

55.

−1

g ( x)

− 0.39

0.65

1.72

3.48

f ( x) =

−6 2 − e 0.2 x

(a)

− 15

y − 10

(b)

5 4 3 2 1 −4

−1

−20

−10

3.4

f ( x)

−3.03

−3.22

−6

−34

−230

x f ( x)

−2

Asymptote: y = −1 f ( x) =

53.

3.46 −2617

3.47 3516

10 1.11

20 0.11

−1 5.1

−0.1 3.2

−0.01 3

0.289

2715

7.7

6.3

6.1

5 8.4

4 26.6

Horizontal asymptotes: y = −3, y = 0 Vertical asymptote: x ≈ 3.47

8 1 + e −0.5 x

(a)

56.

f ( x) =

6 2 − e0.2 / x

(a) −9

9 −1

(b)

−9

9 −2

−30

−20

−10

(b)

f ( x)

≈0

0.05

7.95

≈8

x f ( x)

Horizontal asymptotes: y = 0, y = 8 54. g( x ) =

8 1 + e−0.5 / x

(a)

f ( x)

0.2 ln 2 undef.

−10 5.9

Horizontal asymptote: y = 6 −9

Vertical asymptote: x =

9 −2

(b)

−15 5.9

x f ( x)

57.

−15 3.93

−2 3.5

−1 3.0

−0.2 0.61

2000

−0.1 0.05 0

x f ( x)

0 undef.

0.01 8

Horizontal asymptotes: y = 4

0.2 7.4

1 5.0

0.2 ≈ 0.2885 ln 2

100 0

5 4.2

Intersection: (86.350, 1500) 58.

15,000

Vertical asymptote: x = 0 0

600 0

Intersection: (482.831, 12,500)

Section 4.1 59.

Exponential Functions and Their Graphs

329

64. f ( x) = x3e− x + 2

(a) − 12

0 − 0.5 −3

Intersection: ( − 8.419, 0.2) 60.

12 −1

(b) Increasing on ( − ∞, 3)

Decreasing on (3, ∞ ) (d) Relative minimum: (3, 9.933)

−5

65. P = 2500, r = 2% = 0.02, t = 10

−1

Intersection: ( − 2.114, 0.9)

Compounded n times per year: nt

61.

f ( x) = x e

(a)

Compounded continuously: A = Pert = 2500e(0.02)(10)

−3

9 −1

(b) Decreasing on (−∞, 0), (2, ∞) Increasing on (0, 2) (c) Relative maximum: (2, 4e −2 ) ≈ (2, 0.541) Relative minimum: (0, 0) 62.

10 n

r 0.02    A = P  1 +  = 2500  1 +  n n   

2 −x

f ( x ) = 2 x 2 e x +1

$3047.49

$3050.48

$3051.99

$3053.00

365

Continuous

$3053.49

$3053.51

66. P = 2500, r = 6% = 0.06, t = 10 10

(a)

Compounded n times per year: nt

− 13

5 −2

(b) Increasing on (−∞, − 2) and (0, ∞) Decreasing on (−2, 0) (c) Relative maximum: (−2, 2.943) Relative minimum: (0, 0) 63. f ( x) = x 3e x 6

(a)

−10

10 n

r 0.06    A = P  1 +  = 2500  1 +  n n    

Compounded continuously: A = Pert = 2500e(0.06)(10)

$4477.12

$4515.28

$4535.05

$4548.49

365

Continuous

$4555.07

$4555.30

2 −2

(b) Decreasing on ( − ∞, − 3) Increasing on ( − 3, ∞ ) (d) Relative minimum: ( − 3, −1.344)

Chapter 4

330

Exponential and Logarithmic Functions

67. P = 2500, r = 4% = 0.04, t = 20

68. P = 2500, r = 3% = 0.03, t = 40

Compounded n times per year: nt

Compounded n times per year:

r 0.04    A = P  1 +  = 2500  1 +  n n    

20 n

r 0.03    A = P  1 +  = 2500  1 +  n n    

40 n

Compounded continuously: A = Pe rt = 2500 e(0.03)(40)

Compounded continuously: A = Pe rt = 2500 e(0.04)(20)

$5477.81

$5520.10

$5541.79

$5556.46

$8155.09

$8226.66

$8263.21

$8287.87

365

Continuous

365

Continuous

$5563.61

$5563.85

$8299.88

$8300.29

69. P = 12,000, r = 4% = 0.04, A = pert = 12,000e0.04 t

$12,489.73 $17,901.90 $26,706.49 $39,841.40 $59,436.39 $88,668.67

70. P = 12,000, r = 6% = 0.06, compounded continuously: A = pert = 12,000e0.06 t

$12,742.04

$21,865.43

$39,841.40

$75,595.77

$132,278.12

$241,026.44

71. P = 12,000, r = 3.5% = 0.035, A = pert = 12,000e0.035t

$12,427.44

$17,028.81

$24,165.03

$34,291.81

$48,662.40

$69,055.23

72. P = 12,000, r = 2.5% = 0.025, compounded continuously: A = pert = 12,000e0.025t

73.

$12,303.78

$15,408.31

$19,784.66

$25,404.00

$32,619.38

$41,884.12

48   0.12  − 1  1 +  12   A = 25     0.12   12  

 1.0148 − 1  = 25    0.01  = $1530.57

74.

60   0.09  −1  1 +  12   A = 100     0.09   12   = $7542.41

Section 4.1

76.

y1 = 500 (1 + 0.07 )

y3 = 500e

(b)

5000

 1  5700 = 10 (1) = 10 grams. (a) When t = 0, Q = 10   2 (b) When t = 2000,

Q = 10 ( 12 )

0.07 x

2000

 1  5700 79. Q = 10   2

0.07   y2 = 500  1 +  4 

y3 has the highest return.

After 20 years:

y2 − y1 = 2003.20 − 1934.84 = $68.36

2000/5700

≈ 7.84 grams.

331

(d) Never. The graph has a horizontal asymptote at Q = 0.

24   0.03  − 1  1 +   12   A = 75    0.03   12   = $1852.71

77. (a)

(c)

Mass of 14C (in grams)

75.

72   0.06  −1  1 +  12   A = 200     0.06   12   = $17,281.77

Exponential Functions and Their Graphs

12 10 8 6 4 2 4000

8000

Time (in years)

80. (a)

y3 − y2 = 2027.60 − 2003.20 = $24.40 y3 − y1 = 2027.60 − 1934.84 = $92.76 0

1 78. Q = 25   2

t 1600

Using the graph, the cost of an oil change in 10 years is $39.79. (b)

(a) When t = 0, 0

 1  1600 Q = 25   = 25(1) = 25 grams. 2

(b) When t = 1000, 1000

 1  1600 Q = 25   ≈ 16.21 grams. 2

Using a table, the cost of an oil change in 10 years is C (10) ≈ $39.79. (c) Algebraically, the cost of an oil change in 10 years is C (10) = 26.88(1.04)

≈ $39.79.

332

Chapter 4

81. (a)

(b)

Exponential and Logarithmic Functions

41.35

41.62

41.89

42.16

42.44

42.71

42.99

43.27

43.56

43.84

44.13

44.41

44.70

44.99

45.29

45.58

45.88

46.18

46.48

46.78

47.09

47.40

47.71

48.02

48.33

48.64

48.96

49.28

49.60

49.93

50.25

50.58

50.91

51.24

51.58

51.91

52.25

52.59

52.93

53.28

53.63

Because t ≈ 52.3, the population of California will exceed 51 million in the year 2053.

Section 4.1 82. (a)

Exponential Functions and Their Graphs y

87.

35,000

333

y = 4x

y = 3x

5 4 0

3 2

(b)

V (t )

25,072.00

20,057.60

16,046.08

12,836.86

−4 − 3 −2 −1 −1

V (t )

10,269.49

8215.59

6572.47

5258.98

V (t )

4206.38

3365.11

(c) Because the graph of V has a horizontal asymptote at V = 0 as t approaches positive infinity, the sedan will never have no value. 83. True. f ( x ) = 1x is not an exponential function. 84. False. e is an irrational number. 85. The graph decreases for all x and has a positive y-intercept. Matches (d). 86.

a b c

−6

−4

−2

d e f

−2

(a)

y = 2 − x  reflection in y-axis of y = 2 x

(b)

y = e − x  reflection in y-axis of y = e x

(c)

y = 10 − x  reflection in y-axis of y = 10 x

(d)

y = 10 x  increases most rapidly of all a x

(e)

y = e x  increases between y = 10 x and y = 2 x

(f)

y = 2 x  increases least rapidly of all a

89. eπ ≈ 23.14, π e ≈ 22.46

90. 210 = 1024, 10 2 = 100

y2 = x 2

210 > 10 2

y3 = x 3 y4 = x

91. 5−3 = 0.008, 3−5 ≈ 0.0041 5 −3 > 3 −5

y5 = x y5 y2

1  1 92. 41\ 2 = 2,   =  2 16

y4 9

y3 −4

(a)

eπ > π e

y1 = e x

−9

The solution set to 3 x < 4 x is x > 0. 88.

y1 = e x increases at the fastest rate.

(b) For any positive integer n, e x > x n for x sufficiently large. That is, e x grows faster than x n . (c) A quantity is growing exponentially if its growth rate is of the form y = cerx . This is a faster rate than any polynomial growth rate.

 1 41\ 2 >    2

93. f has an inverse because f is one-to-one. y = 5x − 7 x = 5y − 7 x + 7 = 5y f −1 ( x ) =

1 ( x + 7) 5

334

Chapter 4

Exponential and Logarithmic Functions

94. f is one-to-one, so it has an inverse. 2 5 f ( x) = − x + 3 2 2 5 y=− x+ 3 2 2 5 x=− y+ 3 2 5 2 x− =− y 2 3 3 5 − x−  = y 2 2 3 15 f −1 ( x ) = − x + 2 4

95. f has an inverse because f is one-to-one. y = 3 x+8 x = 3 y+8 x3 = y + 8 x3 − 8 = y f −1 ( x ) = x 3 − 8

96. f is not one-to-one, so it does not have an inverse function. 97. Answers will vary.

Section 4.2 Logarithmic Functions and Their Graphs 1. logarithmic function

21. g a = 4  log g 4 = a

2. common, natural 3. a

log a x

22. nt = 10  log n 10 = t

23. log2 16 = log2 24 = 4

4. loge x = ln x 5. If log a b = c, then a c = b.

1 1 24. log16   = − 2 4

6. If ln x = ln7, then x = 7.

because 16 −1 / 2 =

7. log4 64 = 3  43 = 64 8. log3 81 = 4  3 = 81 4

9. log

1 7 49

= −2  7 = −2

11. log32 4 =  32 2 5

12. log16 8 = 34  16

= log10 (10−3 ) = −3 26.

g (100,000) = log10 (100,000)

= log10 (105 )

=4 =8

13. log2 2 = 12  21 2 = 2 14. log5 3 25 =

 1   1  25. g   = log10  1000  1000    

1 49

1 1 = −3  10−3 = 1000 10. log10 1000

2  52 3 = 3 25 3

1 1 = . 161 / 2 4

=5 27. log10 345 ≈ 2.538

4 28. log10   ≈ −0.097 5 29. 6 log10 14.8 ≈ 7.022

15. 53 = 125  log5 125 = 3

30. 1.9 log10 (4.3) ≈ 1.204

16. 82 = 64  log8 64 = 2

31. log 7 x = log 7 9

1 = 3  log81 3 = 4

32. log5 5 = log 5 x

1/ 4

17. 81 18. 9

3/2

3 = 27  log 9 27 = 2

19. 6−2 = 361  log6 361 = −2

x=9

5 = x 33. log 4 42 = x

2=x

20. 10−3 = 0.001  log10 0.001 = −3

Section 4.2 34. log3 3−5 = x

Logarithmic Functions and Their Graphs 44.

f ( x ) = 4 x and g( x ) = log 4 x are inverse functions of

each other.

−5 = x

35. log8 x = log8 10 −1 x = 10 −1 =

6 5 4 3 2 1

1 10

36. log 4 43 = log 4 x

f g

−3 −2 − 1

43 = x

1 2 3 4 5 6 7

37. log4 43 x = (3 x )log 4 4 = 3 x

45.

y = log10 ( x + 2)

Domain: x + 2 > 0  x > −2 Vertical asymptote: x = −2 log10 ( x + 2) = 0

38. 6log6 36 x = 36 x 39. 3log2 ( 12 ) = 3log2 (2 −1 )

x + 2 =1 x = −1

= 3(−1) = −3 log4 16 = 14 log4 42 = 14 (2) = 21

40.

1 4

41.

f ( x ) = 6 x and g ( x ) = log 6 x are inverse functions of

x-intercept: (−1, 0) y

each other.

5 4 3 2 1

f −4 −3

1 2 3 4 5 6 7

−2 −3

42.

46.

f ( x ) = 5 x and g( x ) = log5 x are inverse functions of

each other.

−1

1 2 3 4 5 6

y = log10 ( x − 1)

Domain: x − 1 > 0  x > 1 Vertical asymptote: x = 1 log10 ( x − 1) = 0

x −1 = 1 x=2

7 6 5 4 3 2 1

x-intercept: (2, 0)

−3 −2 −1

1 2 3 4 5 6 7

5 4 3 2 1

−2 −3

43.

f ( x ) = 15x and g( x ) = log15 x are inverse functions of

each other. y 8 7 6 5 4 3 2 1 −2 −1

−2 −3 −4 −5

−3 −2 −1

−2 −3 −4

64 = x

7 6 5 4 3 2 1

335

−1

2 3 4 5 6 7 8 9

−2 −3 −4 −5

g 1 2 3 4 5 6 7 8

−2

336 47.

Chapter 4

Exponential and Logarithmic Functions

y = 1 + log10 x

50.

Domain: x > 0 Vertical asymptote: x = 0 1 + log10 x = 0

y = 2 + log10 ( x + 1)

Domain: x + 1 > 0  x > −1 Vertical asymptote: x = −1 2 + log10 ( x + 1) = 0

log10 x = −1

log10 ( x + 1) = −2

1 x = 10 = 10 −1

1 x + 1 = 10−2 = 100 99 x = − 100

1  x-intercept:  , 0   10 

99 , 0) x-intercept: ( − 100

5 4 3 2 1 −2 −1

5 4 3 x

1 2 3 4 5 6 7 8

−2 −3 −4 −5

48.

−3 −2

y = 2 − log10 x

51.

Domain: x > 0 Vertical asymptote: x = 0 2 − log10 x = 0

Matches graph ( b ) .

x = 10 2 = 100

52.

x-intercept: (100, 0)

f ( x ) = − log3 x Asymptote: x = 0 Point on graph: (1, 0 )

y 7 6 5 4 3 2 1 −1 −2 −3

f ( x ) = log3 x + 2 Asymptote: x = 0 Point on graph: (1, 2 )

log10 x = 2

49.

1 2 3 4 5 6 7

Matches graph ( c ) . 53. 30

120

f ( x ) = − log3 ( x + 2 ) Asymptote: x = −2 Point on graph: ( −1, 0 )

Matches graph ( d ) .

y = 1 + log10 ( x − 2)

54.

Domain: x − 2 > 0  x > 2 Vertical asymptote: x = 2 1 + log10 ( x − 2) = 0

f ( x ) = log3 (1 − x ) Asymptote: x = 1 Domain: 1 − x > 0  x < 1 Point on graph: ( 0, 0 )

log10 ( x − 2) = −1

Matches graph ( a ) .

x − 2 = 10−1 1 10 21 x= 10

55. The graph of g ( x ) = − log10 x is a reflection in the

x−2=

y-axis of the graph of f ( x ) = log10 x. y

56. The graph of g ( x ) = log10 ( x + 7 ) is a horizontal shift 7

5 4 3 2 1

 21  x-intercept:  , 0   10  −1

−2 −3 −4 −5

units to the left of the graph of f ( x ) = log10 x. 1

3 4 5 6 7 8 9

57. The graph of g ( x ) = 4 − log2 x is a reflection in the

x-axis followed by a vertical shift four units upward of the graph of f ( x ) = log2 x.

Section 4.2 58. The graph of g ( x ) = − 3 + log 2 x is a vertical shift three

units downward of the graph of f ( x ) = log2 x. 59. The graph of g ( x ) = −2 + log8 ( x − 3) is a horizontal

shift three units to the right and a vertical shift two units downward of the graph of f ( x ) = log8 x. 60. The graph of g ( x ) = 4 + log8 ( x − 1) is a horizontal shift

one unit to the right and a vertical shift four units upward of the graph of f ( x ) = log8 x. ln 6 = 1.7917

61.

e1.7917 ≈ 6 ln 4 = 1.3862

62.

e1.3862 = 4

63. ln e = 1  e1 = e 64. ln e3 = 3  e3 = e3 65. ln e =

1  e1 / 2 = e 2

Logarithmic Functions and Their Graphs 81. ln e 2 = 2 (Inverse Property) 82. −ln e = −1 (Inverse Property) 83. e ln 1.8 = 1.8 (Inverse Property) 84. 7 ln e 0 = 7 ln 1 = 7(0) = 0 85. e ln 1 = e(0) = 0 86. e ln 22 = 22 (Inverse Property) 87. ln eln e = 1 ( ln e = 1)

1 88. ln  4  = ln e−4 = −4 (Inverse Property) e  89.

f ( x ) = ln ( x − 1) Domain: x > 1 The domain is (1, ∞ ) . Vertical asymptote: x − 1 = 0  x = 1

1 1 = −2  e−2 = 2 e2 e

x-intercept:

67. ln 9 = 2.1972  e 2.1972 = 9

ln ( x − 1) = 0

66. ln

1 68. ln e =  e1 / 3 = 3 e 3

x − 1 = e0 x −1 = 1

69. e3 = 20.0855  ln 20.0855 = 3 70. e 0 = 1  ln 1 = 0 71. e1.3 = 3.6692  ln 3.6692 = 1.3 72. e 2.5 = 12.1842  ln 12.1824 = 2.5 73.

x=2 The x-intercept is ( 2, 0 ) .

y = ln ( x − 1)  e y + 1 = x

1.02

1.14

8.39

−4

−2

e = 1.3956  ln 1.3956 = 13

74.

1 = e −4 = 0.0183  ln 0.0183 = −4 e4

75.

e3 = 4.4816  e3 / 2 = 4.4816  ln 4.4816 = 23

76. e3 / 4 = 2.1170  ln 2.1170 = 34 77. ln 11 ≈ 2.398

337

5 4 3 2 1 −2 −1

2 3 4 5 6 7 8

−2 −3 −4 −5

78. ln 18.31 ≈ 2.907

1 79. − ln   ≈ 0.693 2 80. 3 ln

0.65 ≈ − 0.646

Chapter 4

338

Exponential and Logarithmic Functions

90. h ( x ) = ln( x + 1)

92.

f ( x ) = ln ( 3 − x )

Domain: x + 1 > 0  x > −1 The domain is ( −1, ∞ ) .

Domain: 3 − x > 0  x < 3

Vertical asymptote: x + 1 = 0  x = −1

Vertical asymptote: 3 − x = 0  x = 3

x-intercept:

The domain is ( −∞, 3) .

ln ( 3 − x ) = 0

ln ( x + 1) = 0

e0 = 3 − x

e0 = x + 1 1= x +1

1=3− x 2=x

0 = x. The x-intercept is ( 0, 0 ) .

The x-intercept is ( 2, 0 ) .

y = ln( x + 1)  e y − 1 = x

y = ln ( 3 − x )  x = 3 − e y

−0.39

1.72

6.39

19.09

2.95

2.86

2.63

0.28

− 12

−3

−2

−1

5 4 3 2 1 −3 −2

5 4 3 2 x

1 2 3 4 5 6 7

−6 −5 −4 −3 −2 −1

1 2

−2 −3 −4 −5

91. g ( x ) = ln ( − x )

93. g ( x ) = ln ( x + 8) is a horizontal shift eight units to the

left of f ( x) = ln x.

Domain: − x > 0  x < 0 The domain is ( −∞, 0 ) .

94. g ( x ) = ln ( x − 4 ) is a horizontal shift four units to the

Vertical asymptote: − x = 0  x = 0 x-intercept: 0 = ln ( − x )

right of f ( x ) = ln x. 95. g ( x ) = ln x − 5 is a vertical shift five units downward of

e0 = − x −1 = x

f ( x ) = ln x. 96. g ( x ) = ln x + 4 is a vertical shift four units upward of

The x-intercept is ( −1, 0 ) .

f ( x ) = ln x.

y = ln ( − x )  e y = − x

−0.14

−0.37

−1

−2.72

−7.39

−2

−1

right and a vertical shift two units upward of f ( x ) = ln x. 98. g ( x ) = ln ( x + 2 ) − 5 is a horizontal shift two units to

the left and a vertical shift five units downward of f ( x ) = ln x.

5 4 3 2 1 −8 −7 −6 −5 −4 −3 −2 −1

97. g ( x ) = ln ( x − 1) + 2 is a horizontal shift one unit to the

1 2

−2 −3 −4 −5

Section 4.2 x x − ln 2 4

99. f ( x ) =

(a) −6

−1

−4

−1

(b) Domain: ( 0, ∞ )

(b)

100. g ( x ) = 6 x ln x 5

(a)

x+2 > 0; Critical numbers: 1, − 2 x −1 Test intervals: ( −∞, − 2 ) , ( −2, 1) , (1, ∞ ) Testing these three intervals, we see that the domain is ( −∞, − 2 ) ∪ (1, ∞ ) .

(d) Relative minimum: ( 2, 1.693)

−6

(d) There are no relative maximum or minimum values. 6

−3

 2x  104. f ( x ) = ln    x+2 4

(a)

Domain: (0, ∞ ) (c) Decreasing on (0, 0.368)

−6

Increasing on (0.368, ∞ ) (d) Relative minimum: (0.368, − 2.207) 101. h( x) =

(a)

339

 x+2 103. f ( x ) = ln    x −1 

(a)

Logarithmic Functions and Their Graphs

−4

(b)

14ln x x

2x > 0; Critical numbers: 0, − 2 x+2 Test intervals: ( −∞, − 2 ) , ( −2, 0 ) , ( 0, ∞ )

Testing these three intervals, we see that the domain is ( −∞, − 2) ∪ ( 0, ∞) .

( 0, ∞ ) .

(d) There are no relative maximum or minimum values.

−1

(b) Domain: (0, ∞ ) (c) Increasing on (0, 2.718) Decreasing on ( 2.718, ∞ )

 x2  105. f ( x ) = ln    10 

(d) Relative minimum: ( 2.718, 5.150) 102. f ( x ) =

x ln x

(a)

(a) −6

−6

x2 > 0  x ≠ 0; Domain: all x ≠ 0 10 (c) The graph is increasing on ( 0, ∞ ) and decreasing

(b)

−2

−4

(b) Domain: ( 0, 1) , (1, ∞ )

on ( −∞, 0 ) . (d) There are no relative maximum or minimum values.

(c) Increasing on ( 2.72, ∞ ) Decreasing on ( 0, 1) , (1, 2.72 ) (d) Relative minimum: ( 2.72, 2.72 ) © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

340

Chapter 4

Exponential and Logarithmic Functions 109. f ( t ) = 80 − 17log10 ( t + 1) , 0 ≤ t ≤ 12

 x  106. f ( x ) = ln  2   x +1

(a)

−1

(a)

f ( 0 ) = 80 − 17log10 ( 0 + 1) = 80

(b)

f ( 2 ) = 80 − 17log10 ( 2 + 1) ≈ 71.89

(c)

f (11) = 80 − 17log10 (11 + 1) ≈ 61.65 100

−4

(b) Domain: x > 0 (c) The graph is increasing on ( 0, 1) and decreasing

on (1, ∞ ) . (d) Relative maximum: (1, − 0.693)

110. (a)

360

107. f ( x ) = ln x

(a)

0 140

110

The model is a good fit. −1

(b) T > 300°F when p > 67.3 pounds per square inch.

−1

(b)

ln x ≥ 0  x ≥ 1; Domain: x ≥ 1

[The graph of T and y = 300 intersect at p = 67.3. ]

(c)

T ( 74 ) = 306.48°F

(a)

−1

11 −1

(b) Domain: x > 0 (c) The graph is decreasing on ( 0, 1) and increasing on (1, ∞ ) . (d) Relative minimum: (1, 0 ) 111. t =

ln K 0.035

(a)

19.80

39.61

51.79

59.41

65.79

71.00

It takes about 19.80 years for the principal to double. 100

(b)

−5

20 − 10

Section 4.2  I  112. β = 10 log10  −12   10  (a)

 1  I 1: β = 10log10  −12  = 10log10 (1012 ) = 10(12)  10  = 120 decibles

 x  113. t = 16.625ln   , x > 750  x − 750 

(b)

116. True. log 3 27 = log3 33 = 3 117. 5 = log b 32 b 5 = 32 = 2 5 b=2 118.

119.

Interest for 30-year loan is 323,179.20 − 150,000 = $173,179.20.

Interest for 10-year loan is 199,108.80 − 150,000 = $49,108.80. 30

1 2 = log b    81  1 b2 = 81 1 b2 =   9 1 b = 9

(897.72 )( 30 )(12 ) = 323,179.20 (1659.24 )(10 )(12 ) = 199,108.80

4 = log b 81 b 4 = 81 = 34 b=3

 897.72  16.625ln   ≈ 30 years  897.72 − 750   1659.24  16.625ln   ≈ 10 years  1659.24 − 750 

114. (a)

341

115. False. You would reflect y = 6 x in the line y = x.

 10−2  (b) I = 10−2 : β = 10log10  −2  = 10log10 (1010 ) = 10(10)  10  = 100 decibles (c) No, this is a logarithmic scale.

(a)

Logarithmic Functions and Their Graphs

120.

1 3 = log b    64  1 b3 = 64 1 b3 =    4 1 b = 4

121. The vertical asymptote is to the right of the y-axis, and the graph increases. Matches (b). 100

1500

Using the graph, the required rate of ventilation in a room with 300 cubic feet of air space is approximately 17.66 ft 3 / min. (b) To determine the rate of ventilation per child in a room of 30 students with a system that moves 450 ft 3 per minute, divide the volume per minute by the number of students.

122. The vertical asymptote is to the left of the y-axis. Matches (b). 123. f ( x ) = log a x is the inverse function of

g ( x ) = a x , where a > 0, a ≠ 1.

450 ft 3 / min = 15 ft 3 / min/ student 30 students (c) Using the graph from part (a), the minimum required air space per child is approximately 382 ft 3 . (d) Yes. The air space per child is 384 ft 3 , which exceeds the minimum requirement of 382 ft 3 .

342

Chapter 4

124. (a)

Exponential and Logarithmic Functions

126. (a) The graph of m is exponential, so m = f ( x ). The

graph of n is logarithmic, so n = g ( x ).

f 0

(b) Because f ( x ) and g ( x) are inverses, and

f ( a ) = b, g (b) = a.

−2

(b) Using the graph from part (a), g ( x ) = x

approaches infinity. (c)

(a)

f ( x)

f g 0

−2

than g ( x ) = x 3 .

128.

102

104

0.322 0.230 0.046

106

0.00092 0.0000138

f ( t ) = 75 − 6 ln ( t + 1)

60 = 75 − 6 ln ( t + 1)

15 5 = 6 2 t = e5/ 2 − 1 ≈ 11.18 months

ln ( t + 1) =

g 20

−2

Or, you could graph f ( t ) and y = 60 together in the

When n = 4, f ( x ) = ln x is increasing at a greater rate

same viewing window, and determine their point of intersection.

than g ( x ) = x1 / 4 .

129. x 2 + 2 x − 3 = ( x + 3)( x − 1)

130. x 2 + 4 x − 5 = ( x + 5)( x − 1)

g 0

(b) As x increases without bound, f ( x ) approaches 0 .

When n = 3, f ( x ) = ln x is increasing at a greater rate

ln x x

127. f ( x ) =

increasing at a greater rate than f ( x ) = ln x as x

131. 12 x 2 + 5 x − 3 = ( 4 x + 3)( 3 x − 1)

−2

When n = 5, f ( x ) = ln x is increasing at a greater rate

132. 16 x 2 − 54 x − 7 = (8 x + 1)( 2 x − 7)

than g ( x ) = x . 1/ 5

As n increases, n = 2, 3, 4,  , the rate at which y = x n increases decreases. 125. (a)

135. 2 x 3 + x 2 − 45 x = x 2 x 2 + x − 45

y3 −4

= x ( 2 x − 9 )( x + 5 )

(

1 1 1 2 3 4 ( x −1) + ( x −1) − ( x −1) +. 2 3 4

)

= x ( 3 x + 4 )( x − 3 )

137. ( f + g)( 2) = f ( 2) + g ( 2) = [3( 2) + 2] + [23 − 1] = 8 + 7 = 15

y = ln x −4

)

136. 3 x 3 − 5 x 2 − 12 x = x 3 x 2 − 5 x − 12

−4

(b) Pattern is ( x −1) −

134. 36 x 2 − 49 = ( 6 x − 7 )( 6 x + 7 )

(

y = ln x

133. 16 x 2 − 25 = ( 4 x + 5)( 4 x − 5)

138. ( f − g )( −1) = ( −1) − ( −2 ) = 1

−4

As you use more terms, the graph better approximates the graph of ln x on the interval (0, 2).

Section 4.3 139. The graphs of y1 = 5 x − 7 and y2 = 7 + 5 x do not

Properties of Logarithms

343

141. The graphs of y1 = 3 x − 2 and y2 = 9 intersect when

intersect because they are parallel. So, no real solutions.

x ≈ 27.667 or

83 . 3

−9

−6

−5

40 −2

140. The graphs of y1 = −2 x + 3 and y2 = 8 x intersect at

x = 0.3.

142. The graphs of y1 = x − 11 and y2 = x + 2 do not

intersect. No real solution. 6

−6

−10

−2

50 −10

Section 4.3 Properties of Logarithms 1.

change-of-base n

log au , ln u + ln v

Using the change of base formula: ln 24 log3 24 = ln 3

x Using the quotient property: ln x − ln 2 = ln   . 2

(a)

log 5 x =

log10 x log10 5

(b)

log5 x =

ln x ln 5

(a)

log3 x =

log10 x log10 3

(b)

log3 x =

ln x ln 3

(a)

log1 6 x =

log10 x log10 x = log10 1 6 −log10 6

(b)

log1 6 x =

ln x ln x = ln 1 6 −ln 6

(a)

log1 4 x =

log10 x −log10 x = log10 (1 4) log10 4

(b)

log1 4 x =

ln x − ln x = ln (1 4) ln 4

(a)

 3 log a   =  10 

 3 log10    10  log10 a

(b)

 3 loga   =  10 

 3 ln    10  ln a

10. (a)

(b)

 4 log10    5  4 log a   =  5 log10 a  4 ln    5  4 log a   =  5 ln a

11. (a)

log 2.6 x =

log10 x log10 2.6

(b)

log2.6 x =

ln x ln 2.6

12. (a)

log 7.1 x =

log10 x log10 7.1

(b)

log 7.1 x =

ln x ln 7.1

13. log3 7 =

ln 7 ≈ 1.771 ln 3

14. log9 4 =

ln 4 ≈ 0.631 ln 9

15. log1 2 16 =

ln 16 =−4 ln (1 2 )

344

Chapter 4

16. log1 8 4 =

Exponential and Logarithmic Functions

ln 64 ln (1 8 )

29.

−5

ln 0.9 ≈ −0.059 ln 6

18. log 4 0.045 =

ln 0.045 ≈ −2.237 ln 4

19. log15 1460 =

ln 1460 ≈ 2.691 ln15

ln( x + 1) ln 3

ln 82 = = −2 − ln 8 17. log6 0.9 =

f ( x ) = log 3 ( x + 1) =

−4

30.

f ( x ) = log 2 ( x − 1) =

ln( x − 1) ln 2

−1

ln175 20. log 20 175 = ≈ 1.724 ln 20

21. ln 20 = ln(4 ⋅ 5) = ln 4 + ln 5

−4

31.

f ( x ) = log1 2 ( x − 2) =

22. ln 500 = ln(53 ⋅ 4)

= ln 5 − ln 4 = 2 ln 5 − ln 4 24. ln

−1

25 = ln 25 − ln 4 4 2

ln( x − 2) ln( x − 2) = ln (1 2 ) − ln 2

= ln 53 + ln 4 = 3ln 5 + ln 4 23. ln

−4

32.

f ( x) = log1 2 ( x + 2) =

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

5 = ln 5 − ln 2 2 = ln 5 − ln 41 / 2

−4

= ln 5 − ln 4 1 2

−4

25. log b 8 = log b 23 = 3 log b 2

33.

f ( x ) = log1 4 ( x 2 ) =

≈ 3(0.3562) ≈ 1.0686

ln x 2 ln x 2 = ln (1 4 ) − ln 4

26. log b 30 = log b (2 ⋅ 3 ⋅ 5)

= logb 2 + logb 3 + logb 5

−6

≈ 0.3562 + 0.5646 + 0.8271 ≈ 1.7479 −4

27. log b 16 = logb 16 − log b 25 25 = log b 2 4 − log b 52 = 4 log b 2 − 2 logb 5 = 4(0.3562) − 2(0.8271) = −0.2294

28. log b 3 = 12 log b 3

≈ 12 (0.5646) ≈ 0.2823

Section 4.3

34.

f ( x ) = log3 x =

Properties of Logarithms

345

39. log 2 (4 2 ⋅ 34 ) = log 2 4 2 + log 2 34

ln x ln 3

= 2 log2 4 + 4 log 2 3 = 2 log2 22 + 4 log2 3

= 4 log 2 2 + 4 log 2 3 −2

= 4 + 4 log2 3

40. log 3 (9 2 ⋅ 2 4 ) = log 3 9 2 + log 3 2 4

−4

35.

 x  ln ( x 2) ln ( x 2 ) f ( x) = log1 5   = = ln 5 ln 5 2 5

( ) + log 2

= log3 32

= log3 34 + 4 log3 2 = 4 + 4 log3 2 = 4(1 + log3 2)

−2

41. ln(5e6 ) = ln 5 + ln e6 = ln 5 + 6 = 6 + ln 5

42. ln(8e3 ) = ln8 + ln e3 = ln8 + 3

−3

x 36. f ( x) = log1 3   3   = log1 3 x − log1 3 3 = log1 3 ( x) − log1 3 ( 13 )

6 43. ln 2 = ln 6 − ln e2 e = ln 6 − 2 ln e = ln 6 − 2 −1

= log1 3 ( x) + 1 =1+

ln x ln ( 13 )

=1+

ln x ln1 − ln 3

1 45. log 5 250 = log 5 1 − log5 250 = 0 − log5 (125 ⋅ 2)

= − log 5 (53 ⋅ 2) = −[log 5 53 + log 5 2] = −[3log 5 5 + log 5 2] = −3 − log 5 2

ln x =1+ − ln3

46. − ln 24 = − ln(23 ⋅ 3)

= − ln 23 − ln3 = −3ln 2 − ln3 = −(3ln 2 + ln 3)

ln x =1− ln3 3

−2

 e5  44. ln   = ln e5 − ln 7 7 = 5 − ln 7

47. log1010 x = log1010 + log10 x = 1 + log10 x 48. log10100 x = log10100 + log10 x = 2 + log10 x

−3

37. log 4 8 = log 4 2 3 = 3 log 4 2

= 3log 4 41 2 = 3 ( 12 ) log 4 4 = 23 38. log 9 243 = log 9 (81 ⋅ 3)

= log9 81 + log9 3

t 49. log10   = log10 t − log10 8 8 7 50. log10   = log10 7 − log10 z z

51. log8 x 4 = 4 log8 x

= log9 92 + log9 91/2

52. log6 z −3 = −3log6 z

1 5 = 2 2

53. ln z = ln z1 / 2 =

= 2+

1 ln z 2

346

Chapter 4

Exponential and Logarithmic Functions

54. ln 3 t = ln t 1 / 3 =

 ( x + 1)( x − 1)   x2 − 1  61. ln  3  = ln     x x3    

1 ln t 3

= ln ( x + 1) + ln ( x − 1) − ln x 3

55. ln xyz = ln x + ln y + ln z 56. ln

= ln ( x + 1) + ln ( x − 1) − 3 ln x, x > 1

xy = ln x + ln y − ln z z

 x  = ln x − ln x 2 + 1 62. ln   x 2 + 1   

57. log 6 ab3c − 2 = log 6 a + log 6 b3 + log 6 c − 2

= log 6 a + 3log 6 b − 2log 6 c

(

58. log 4 xy 6 z 4 = log 4 x + log 4 y 6 + log 4 z 4

= ln x −

= log 4 x + 6 log 4 y + 4 log 4 z 59. ln 3

1/ 2

 x2  x2 = ln  3  3 y y 

= 4log b x +

64. logb

1  x2  ln   2  y3 

(

)

y − log b z 5

= log b x 4 + log b

y − log b z 5

1 log b y − 5log b z 2

x y4 = log b x y 4 − logb z 4 z4 = log b x1 / 2 + log b y 4 − log b z 4 =

1 ln x 2 − ln y3 2 1 = ( 2 ln x − 3ln y ) 2 3 = ln x − ln y 2 =

(

1 (ln x 4 − ln y3 ) 3 1 = ( 4ln x − 3ln y ) 3 4 = ln x − ln y 3

1/ 2

1 ln x 2 + 1 2

 x4 y  = log b x 4 63. log b   z 5   

x4 1 x4 ln = y3 3 y3 =

60. ln

)

= ln x − ln x 2 + 1

)

1 log b x + 4 log b y − 4 logb z 2

65. y1 = ln  x 2 ( x − 4), y2 = 2ln x + ln ( x − 4)

(a)

−2

22 −2

(b)

Error

3.22

4.28

4.99

5.55

6.00

6.40

Error

3.22

4.28

4.99

5.55

6.00

6.40

y1 = ln  x 2 ( x − 4) = ln x 2 + ln ( x − 4) = 2ln x + ln ( x − 4) = y2 .

Section 4.3

Properties of Logarithms

347

66. y1 = ln 9 x3 , y2 = ln 9 + 3ln x

(a)

−1

−5

(b)

Error

2.20

4.28

5.49

6.36

7.03

7.57

Error

2.20

4.28

5.49

6.36

7.03

7.57

 x4  y1 =   , y2 = 4 ln x − ln ( x − 2 )  x−2 10

(a)

(b)

x y1

0 Error

Error

4.39

4 4.85

5.34

4.39

4.85

5.34

5.78

6.17

6.53

5.78

6.17

6.53

= 4 ln x − ln ( x − 2 ) = y2 .

Chapter 4

348

Exponential and Logarithmic Functions

 x  1 68. y1 = ln  , y2 = ln x − ln ( x + 3) x + 3 2  

(a)

1 −2

−5

(b)

Error

−1.39

−1.26

−1.24

−1.25

−1.28

−1.30

Error

−1.39

−1.26

−1.24

−1.25

−1.28

−1.30

(c) The graphs and table suggest that y1 = y2 . In fact,  x  y1 = ln    x + 3 = ln x1 2 − ln ( x + 3) 1 ln x − ln ( x + 3) 2 = y2 . =

69. ln x + ln 4 = ln 4 x

79. ln ( x − 2) + ln 2 − 3ln y = ln ( x − 2) + ln 2 − ln y 3 = ln 2( x − 2) − ln y 3 

70. ln y + ln z = ln yz 71. log 4 z − log 4 y = log 4

z y

72. log 5 8 − log 5 t = log 5

8 t

75.

(

)

= ln

(

ln x 2 + 4 + ln x = ln x 2 + 4

(

= ln x 3 y 2 − ln z 4

5/2 5 log 7 ( z − 4 ) = log 7 ( z − 4 ) 2

1 2

2( x − 2)

80. 3ln x + 2 ln y − 4 ln z = ln x 3 + ln y 2 − ln z 4

73. 4log 3 ( x + 2 ) = log 3 ( x + 2 ) 74.

= ln

= ln x x 2 + 4

)

x 3 y2 z4

81. ln x − 2[ln( x + 2) + ln( x − 2)] = ln x − 2 ln[( x + 2)( x − 2)]

= ln x − 2 ln( x 2 − 4)

) + ln x 1/ 2

76. 2 ln x + ln( x + 1) = ln x 2 + ln( x + 1)

= ln( x 2 ( x + 1)) = ln( x 3 + x 2 ) 77. ln x − 3ln( x + 1) = ln x − ln( x + 1)3 x = ln ( x + 1)3

= ln x − ln( x 2 − 4)2 x = ln 2 ( x − 4)2 2 82. 4 ln z + ln(z + 5) − 2ln(z − 5) = 4 ln z( x + 5) − ln(z − 5) 4

= ln  z(z + 5) − ln(z − 5)2 = ln

z 4 (z + 5)4 (z − 5)2

78. ln x − 2 ln( x + 2) = ln x − ln( x + 2)2 x = ln ( x + 2)2

Section 4.3

83.

Properties of Logarithms

349

1 1 2 ln( x + 3) + ln x( x 2 − 1) = ln( x + 3)2 + ln x − ln( x 2 − 1) 3 3 2 1 = lnx( x + 3)  − ln x2 −1       3

(

1 x( x + 3) = ln 2 x −1 3 = ln

x( x + 3)

)

x2 −1

x   − ln ( x − 1)  84. 2  ln x − ln ( x + 1) − ln ( x − 1)  = 2  ln x 1 +  

86.

(

x   = 2  ln 2   x − 1  x  = ln  2   x − 1 85.

(

−12

(b)

)

  64  y2 = ln  2   2  x + 1  (a)

)

y1 = 2  ln8 − ln x 2 + 1   

(

(a)

  x = 2  ln   ( x + 1)( x − 1) 

)

2 y1 = 2  ln 6 + ln x 2 + 1  , y2 = ln 36 x 2 + 1     

−1 4.97

3.58

1 4.97

2 6.80

4.97

3.58

4.97

6.80 5 10.10 10.10

8.19

4 9.25

8.19

9.25

−9

y1 = 2  ln 6 + ln( x 2 + 1) 

= 2 ln 6 + 2 ln( x 2 + 1) = ln 6 2 + ln( x 2 + 1)2

−6

(b)

−8

−4.1899

−4 −1.5075

−2 0.9400

4.1589

−4.1899

−1.5075

0.9400

4.1589

x y1

2 0.9400

4 −1.5075

8 −4.1899

0.9400

−1.5075

−4.1899

= ln 36( x 2 + 1)2 

= y2 . 87.

(

(a)

−2

(

)

y1 = 2  ln8 − ln x + 1    8 64 = 2 ln 2 = ln = y2 . 2 x +1 x2 + 1 2

(

)

y1 = ln x + 12 ln ( x + 1) , y2 = ln x x + 1 , x > 0

−3

(b)

ERROR 0.34657 1.2425 2.5053 3.5015

y1 = ln x + 12 ln ( x + 1) = ln x + ln ( x + 1)

= ln  x x + 1  = y2 .  

350

Chapter 4

Exponential and Logarithmic Functions

88.

 x  y1 = 12 ln x − ln ( x + 2) , y2 = ln    x + 2 

100. 3ln e5 = 15ln e = 15  1  1 1 12 101. ln   = ln(1) − ln e = 0 − ln e = − 2 2  e

(a) −1

102. ln 5 e3 = ln e3 5 = −5

(b)

2 103. y1 = ln x

y2 = 2 ln x

0 Error

1 −1.10

2 −1.04

−1.06

Error

−1.10

−1.04

−1.06

3 3 ln e = 5 5

(a)

−9

4 −1.10

−1.14

−1.18

−1.10

−1.14

−1.18

−6

(The domain of y2 is x > 0.)

(b)

y1 = ln x − ln ( x + 2 ) 1 2

= ln x1 / 2 − ln ( x + 2 )  x  = ln    x + 2 

−8

−4

4.1589

2.7726

1.3863

2.7726

Error

1.3863

2.7726

(c) The graphs and table suggest that y1 = y2 for x > 0. The functions are not equivalent because the domains are different.

= y2 . 89. log 3 9 = 2 log 3 3 = 2

2 104. y1 = 2(ln2 + ln x), y2 = ln4 x 8

(a) 90. log 6 61 = 1 −9

3.4 2 91. log4 16 = 3.4log4 (4 ) = 6.8log4 4 = 6.8

−4

1 = log5 5−3 = −3log5 5 = −3(1) = −3 92. log5 125

93. log 2 ( −4) is undefined. −4 is not in the domain of log 2 x. 94. log 4 (−16) is undefined because −16 is not in the domain

of log 4 x.

= log5125 = log5 53 = 3 95. log 5 375 − log5 3 = log 5 375 3 12 52 96. log4 2 + log4 32 = log4 4 + log4 4

= 12 log 4 4 + 25 log 4 4

(b)

x y1

−1 Error

0 Error

1 1.39

2 2.77

1.39

Error

1.39

2.77

3.58

4 4.16

4.61

3.58

4.16

4.61

= 12 (1) + 25 (1)

= 2 ln ( 2 x )

= ln ( 2 x )

97. ln e 3 − ln e 7 = 3 − 7 = −4 98. ln e 6 − 2ln e 7 = 6ln e − 14ln e = 6 − 14 = − 8 99. 2 ln e 4 = 2(4)ln e = 8

= ln 4 x 2 = y2 for x ≠ 0. The domain of y1 is all real x such that x > 0, however the domain of y2 is all real x such that x ≠ 0.

Section 4.3 Properties of Logarithms

 I  107. β = 10 log10  −12   10 

2 105. y1 = ln( x − 2) + ln( x + 2), y2 = ln( x − 4)

(a)

(a) −9

β = 10  log10 I − log10 10 −12  = 10  log10 I − (−12)

= 10  log10 I + 12 

−4

(b)

351

= 120 + 10 log10 I

−4

−3

Error

1.61

2.48

1.61

Error

1.61

2.48

3.04

3.47

2.48

3.04

3.47

(b)

10 −4 10 −6 10 −8 10 −10 10 −12 10 −14

−20

108. To find an equation that relates y and x, take the natural log of each of the x- and y-values in the table.

X = ln x

Y = ln y

ln1 = 0

−2.631

ln 2 ≈ 0.6931

−2.120

x > −2.

ln3 ≈ 1.0986

−1.911

Using the properties of logarithms,

ln 4 ≈ 1.3863

−1.595

y1 = ln( x − 2) + ln( x − 2)

ln 5 ≈ 1.6094

−1.435

ln 6 ≈ 1.7918

−1.259

= ln[( x − 2)( x + 2)] = ln( x 2 − 4) = y2 For x > − 2.

Using the linear regression feature of a graphing utility, find a linear model for the data.

The domain of y1 is all real x such that x > −2, however the domain of y2 is all real x such that x < −2 or x > 2.

Y = 0.757 X + ( − 2.656)

So, ln y = 0.757ln x − 2.656. 80

109. (a)

4 2 2 106. y1 = 14 ln[ x ( x + 1)], y2 = ln x + 14 ln( x + 1) 6

(a) −9

(b) T − 21 = 54.4(0.964)t −6

T = 21 + 54.4(0.964)t

(b)

−10

3.4564

ERROR ERROR ERROR 0.17329 2.4240 3.4564

−1

0.17329 ERROR 0.17329 2.4240 3.4564 0

The data (t, T − 21) fits the model (c) No, the expressions are not equivalent. The domain of y1 is all x ≠ 0, whereas the domain of y2 is x > 0.

T − 21 = 54.4(0.964)t . The model T = 21 + 54.4(0.964)t fits the original data.

352

Chapter 4 (c)

Exponential and Logarithmic Functions

ln(T − 21) = −0.0372t + 3.9971, linear model T − 21 = e −0.0372 t + 3.9971 T = 21 + 54.4e

−0.0372 t

= 21 + 54.4(0.964)t

116. False. For example, let x = e .

Then f ( x) = ln e = 12 ln e = 12 > 0, but

e < e.

117. The error is an improper use of the Quotient Property of logarithms.

x2 x2 + 4

= ln x 2 − ln x 2 + 4 = 2 ln x − ln( x 2 + 4)1 / 2

(d)

0.07

x 2 ln x g( x ) = ln 2 h( x ) = ln x − ln 2

118. f ( x ) = ln

1 = 0.00121t + 0.01615, linear model T − 21

1 0.00121t + 0.01615 1 T = 21 + 0.00121t + 0.01615

T − 21 =

f ( x ) = h( x ) by the Quotient Property. 4

g −2

f=h 10

−4

119. The natural logarithms of 14 team integers between 1 and 20 can be approximated using ln2, ln3, and ln 5.

= 2 ln x − 12 ln( x 2 + 4)

110. If y = ab x , then ln y = ln(ab x ) = ln a + x ln b, which is 1 1 linear. If y = , then = cx + d. y cx + d

ln1 = 0, ln 2 ≈ 0.6931, ln3 ≈ 1.0986, ln 5 ≈ 1.6094 ln 2 ≈ 0.6931 ln 3 ≈ 1.0986 ln 4 = ln 2 + ln 2 ≈ 0.6931 + 0.6931 = 1.3862 ln 5 ≈ 1.6094 ln 6 = ln 2 + ln 3 ≈ 0.6931 + 1.0986 = 1.7917 ln8 = ln 23 = 3ln 2 ≈ 3(0.6931) = 2.0793

111. True. By the Product Property: ln(uv) = ln u + ln v

ln 9 = ln32 = 2 ln3 ≈ 2(1.0986) = 2.1972 ln10 = ln 5 + ln 2 ≈ 1.6094 + 0.6931 = 2.3025

112. False. For example, let x = 2 and a = 1.

ln12 = ln 2 2 + ln 3 = 2 ln 2 + ln3 ≈ 2(0.6931) + 1.0986 = 2.4848 ln15 = ln 5 + ln 3 ≈ 1.6094 + 1.0986 = 2.7080

Then f ( x − a) = ln(2 − 1) = 0, but f ( x) − f (a) = ln2 − ln1 = ln2. 113. False.

ln x ≠ 12 ln x 12

In fact, ln x

= 12 ln x.

114. False. For example, let n = 2 and x = e.

Then [ f ( x )]n = [ln e]2 = 1, but nf ( x ) = 2 ln e = 2. 115. True. By the definition of the natural logarithmic function, for x > 0, y = ln x if and only if x = e y .

So, if y < 0, then x is a number between 0 and 1, or 0 < x < 1.

ln16 = ln 2 4 = 4 ln 2 ≈ 4(0.6931) = 2.7724 ln18 = ln 32 + ln 2 = 2 ln 3 + ln 2 ≈ 2(1.0986) + 0.6931 = 2.8903 ln 20 = ln 5 + ln 22 = ln 5 + 2 ln 2 ≈ 1.6094 + 2(0.6931) = 2.9956 120. y = ln x has an x-intercept at (0, 1) and does not

increase as much as y = ln x 2 . So, y = ln x matches graph B. y = ln x 2 has an x-intercept at (0, 1) and increases faster

than y = ln x. So, y = ln x 2 matches graph D.

y = ln 2 x has an x-intercept at (0, 0.5), so it matches graph C.

y = ln 2 has a constant function, so it matches graph A.

Section 4.4 Solving Exponential and Logarithmic Equations 121. No. The domains are not the same.

353

(8 x3 y 2 ) 3 (64 x3 y 4 ) 4 23 ⋅ x 3 y 2 ) ( = 3 ( 26 ⋅ x 3 y 4 ) 4

The domain of y1 = ln[ x( x − 2)] is (−∞, 0), (2, ∞). This

123. (64 x3 y 4 )

−3

(8x3 y 2 ) = 4

can be found by solving the quadratic inequality, x( x − 2) > 0. The domain of y2 = ln x + ln( x − 2) is (2, ∞). This can be found by the intersection of the intervals (0, ∞) and (2, ∞), the domains of each term respectively. z

a 122. Let y = log a x and z = loga b x, then a y = x =   and b z

1 y−z   =a b 1 = a( y − z) z b 1 y−z y  1  log a x log a   = . = − 1  1 + log a   = b z z    b  log a b x

124. xy( x −1 + y −1 )−1 =

212 ⋅ x12 y 8 2 ⋅ x9 ⋅ y12

x3 2 ⋅ y4

x3 ,x ≠ 0 64 y 4

xy x −1 + y −1

xy 1 1 + x y xy = y+x xy =

125.

( xy)2 , x ≠ 0, y ≠ 0 x+y

24 xy −2 24 xx3 3x 4 = = , x≠0 16 x −3 y 16 yy 2 2 y3 −3

 2 x2   3y  126.   = 2   2x   3y 

(3 y )3 (2 x 2 )3

27 y 3 , y≠0 8 x6

Section 4.4 Solving Exponential and Logarithmic Equations 1.

(a) (b) (c) (d)

x=y x=y x x

extraneous

ln e 7 = 7 because of the Inverse Property.

Yes. 5x = 125

5x = 53 x=3 5.

Inverse Property log 4 x = 2  4 2 = x 16 = x

4 2 x − 7 = 64

(a)

x=5

4 ( ) = 43 = 64 Yes, x = 5 is a solution. (b) x = 2 2 5 −7

4()

2 2 −7

= 4 −3 = 641 ≠ 64

No, x = 2 is not a solution.

To solve 3 + ln x = 10, isolate the logarithmic term by subtracting 3 from each side of the equation.

354

Chapter 4

23 x+1 = 128

Exponential and Logarithmic Functions 5  12. log 6  x  = 2 3 

x = −1

(a)

2 ( ) = 2 −2 = 14 3 −1 +1

(b)

No, x = −1 is not a solution. x=2 2 ( ) = 2 7 = 128 Yes, x = 2 is a solution. 3 2 +1

3e x + 2 = 75

5  5  ln  x  ln  ⋅ 20.2882  3 3 =   ≈ 1.965 ≠ 2; (a) x ≈ 20.2882;  ln 6 ln 6 No, x ≈ 20.2882 is not a solution. 108  5 108  (b) x = ; log 6  ⋅  = log 6 ( 36 ) = 2; 5 3 5 

3e( −2 + e ) + 2 = 3ee ≠ 75 No, x = −2 + e25 is not a solution. (b) x = −2 + ln 25 3e (

(c)

−2 + ln 25 ) + 2

3e1.2189 + 2 = 3e3.2189 ≈ 75 Yes, x ≈ 1.2189 is an approximate solution. 10. 4e

(a)

= 60 x = 1 + ln15 4e(

1 + ln 15 ) −1

13. ln ( x − 1) = 3.8

(a)

4e3.7081−1 = 4e2.7081 ≈ 60 Yes, x ≈ 3.7081 is an approximate solution. (c) x = ln16 4eln16 −1 ≈ 23.5 ≠ 60 No, x = ln16 is not a solution. 11. log 4 ( 3 x ) = 3 43 = 3 x 64 ≈ 21.333 3

(a) Yes, x ≈ 21.3560 is an approximate solution. (b) No, x = −4 is not a solution. 64 is a solution. (c) Yes, x = 3

x = 1 + e3.8

(

)

ln 1 + e3.8 − 1 = ln e3.8 = 3.8

Yes, x = 1 + e3.8 is a solution. (b) x ≈ 45.7012

ln ( 45.7012 − 1) = ln ( 44.7012 ) ≈ 3.8

= 4 e ln 15 = 4 (15 ) = 60

Yes, x = 1 + ln15 is a solution. (b) x ≈ 3.7081

(c)

= 3e ln 25 = 3 ( 25 ) = 75

Yes, x = −2 + ln 25 is a solution. x ≈ 1.2189

x −1

108 is a solution. 5 5  ln  (21.6)  3  = 2; Yes, x = 7.2 x = 21.6;  ln 6 is a solution.

Yes, x =

x = −2 + e25

(a)

(c)

Yes, x ≈ 45.7012 is an approximate solution. x = 1 + ln 3.8

ln (1 + ln3.8 − 1) = ln ( ln3.8 ) ≈ 0.289

No, x = 1 + ln 3.8 is not a solution. 14. ln ( 2 + x ) = 2.5

(a)

(

)

x = e2.5 − 2; ln 2 + e2.5 − 2 = ln e2.5 = 2.5; Yes, 2.5

x = e − 2 is a solution. 4073 4073 4073   (b) x ≈ is ; ln  2 +  ≈ 2.5; Yes, x ≈ 400 400 400   an approximate solution. 1 1 1  (c) x = ; ln  2 +  ≈ 0.9163 ≠ 2.5; No, x = is not 2 2 2   a solution.

Section 4.4 Solving Exponential and Logarithmic Equations 15. f ( x) = 64 x , g ( x) = 8

18.

355

f ( x ) = 2 − x +1 − 3 g ( x ) = 13

f g

g −9

−1

Point of intersection: ( 12 , 8)

−12

Algebraically: 64 x = 8

Point of intersection: ( −3, 13)

= 2

Algebraically: 2− x +1 − 3 = 13 2 − x +1 = 16 = 2 4 −x + 1 = 4 x = −3

6x = 3 x = 16.

1 2

f ( x ) = 27 x , g ( x ) = 9

19. f ( x ) = 4 log3 x

g ( x ) = 20

−9

−2 − 50

2  Point of intersection:  , 9  3 

275 −5

Point of intersection: ( 243, 20 )

Algebraically: 27x = 9 27 x = 27 2 3

Algebraically: 4 log 3 x = 20 log3 x = 5

x = 23

17.

f ( x ) = 5 x − 2 − 15

x = 35 = 243 20.

g ( x ) = 10

f ( x ) = 3log5 x = 3 ⋅ g( x) = 6

g f

−4

ln x ln5

−2

− 20

Point of intersection: ( 4, 10 ) Algebraically: 5

x −2

− 15 = 10 5x − 2 = 25 = 52 x−2 =2 x=4

−8

Point of intersection: ( 25, 6 ) Algebraically: 3log 5 x = 6 log 5 x = 2

x = 52 = 25

356

Chapter 4

21.

f ( x ) = ln e x +1

Exponential and Logarithmic Functions x

1 28.   = 32 2

g( x) = 2x + 5

1 1   =  2 2 x = −5

−9

−5

29.

−6

Point of intersection: ( −4, − 3)

81 2   = 16 3 −x

3 3   =  2   2 −x = 4

Algebraically: ln e x +1 = 2 x + 5 x + 1 = 2x + 5

x = −4

−4 = x 22.

f ( x ) = ln e x − 2 = x − 2 g ( x ) = 3x + 2

64 3 30.   = 27 4

−8

Point of intersection: ( −2, − 4 ) Algebraically: x − 2 = 3 x + 2 −4 = 2 x

31.

32.

33.

24. 3x = 243

e x = 66

( )

6 10 x = 216

log10 10 x = log10 36

1 625 1 5x = 4 = 5−4 5 x = −4

x = log10 36 ≈ 1.5563

25. 5 x =

( )

34. 5 8 x = 325

8 x = 65 x = log8 65

1 49

7 x = 7 −2 x = −2

35.

ln65 ≈ 2.0075 ln8

2 x+3 = 256 2 x ⋅ 2 3 = 256

1 27.   = 64 8

e x = 14

10 x = 36

3 x = 35 x=5

−3

ln e x = ln 66 x = ln 66

4 x = 42 x=2

−x

ln e x = ln14 x = ln14

x = −2

26. 7 x =

3 3  4 = 4     x = −3

23. 4 x = 16

3 4  4 = 3    

−9

−x = 2 x = −2

2 x = 32 x=5

Alternate solution: 2 x + 3 = 28

x+3=8 x=5

Section 4.4 Solving Exponential and Logarithmic Equations 36.

4 x − 1 = 256

47. ln e x = x 2 ln e = x 2

4 x − 1 = 4 −4

48. ln e2 x −1 = 2 x − 1

x −1 = 4 x = 5

49. e

37. ln x − ln 5 = 0 ln x = ln 5

(

ln x 2 −3

) = x2 − 3

50. eln x = x 2

x=5

51. −1 + ln e2 x = −1 + 2 x = 2 x − 1

38. ln x − ln 2 = 0 ln x = ln 2

52. −4 + e ln x = −4 + x 4

= x4 − 4

x=2 2

53. 5 − eln( x +1) = 5 − x 2 + 1

39. ln x = −9

eln x = e −9

= 4 − x2 1 e9

x = e −9 =

( x + 2) = 3 − ( x2 + 2) 54. 3 − ln e

= 1 − x2

40. ln x = −14 e ln x = e −14 1 x = e −14 = 14 e

55.

83 x = 360 ln83 x = ln 360 3 x ln8 = ln 360 ln 360 3x = ln8 1 ln 360 x= 3 ln8 x ≈ 0.944

56.

65 x = 3000 ln 6 5 x = ln 3000 (5 x ) ln 6 = ln 3000 ln 3000 5x = ln 6 ln 3000 x= ≈ 0.894 5ln 6

41. log x 625 = 4 x 4 = 625 x 4 = 54 x=5

42. log x 81 = 2

x2 = 81 x =9 43. log11 x = −1

x = 11−1 1 x= 11 44. log10 x = −

357

57. 5− t 2 = 0.20 =

1 4

x = 10−1 4 =

1 4 10

≈ 1.778 45. ln(2 x − 1) = 5 2 x − 1 = e5

1 + e5 ≈ 74.707 2

1 5

t 1 − ln 5 = ln   2 5 t − ln 5 = − ln 5 2 t =1 2 t =2

46. ln(3 x + 5) = 8

e8 = 3 x + 5 1 x = (e8 − 5) 3 ≈ 991.986

358 58.

Chapter 4

Exponential and Logarithmic Functions

4−2t = 0.0625

65. 5(23 − x ) − 13 = 100

5(2 3 − x ) = 113 113 23 − x = 5  113  ln 2 3 − x = ln    5 

ln 4−2t = ln 0.0625

(− 2t )ln 4 = ln 0.0625 − 2t =

ln 0.0625 ln 4

t = −

ln 0.0625 =1 2ln 4

3− x =

59. 250e0.02 x = 10,000

x =3−

e0.02 x = 40 ln 40 0.02 x ≈ 184.444

66. 6(8−2 − x ) + 15 = 2601

6(8−2 − x ) = 2586 8 −2 − x = 431 (−2 − x ) ln8 = ln 431

60. 100e0.005 x = 125,000

e0.005 x = 1250

ln 431 ln8 ln 431 x = −2 − ln8 x ≈ −4.917

−2 − x =

0.005 x = ln1250 ln1250 0.005 x ≈ 1426.180 x=

61. 500e− x = 300 3 e− x = 5 3 − x = ln 5

x = − ln

ln 2 ln (113 5 )

ln 2 x ≈ −1.498

0.02 x = ln 40 x=

ln (113 5 )

12 t

67.

0.10   1 +  12  

=2 12 t

 12.1    =2  12   12.1  (12t ) ln   = ln 2  12 

3 5 = ln ≈ 0.511 5 3

62. 1000e−4 x = 75 3 e −4 x = 40

3 ln e −4 x = ln 40 3 −4 x = ln 40 1 3 x = − ln ≈ 0.648 4 40 63. 7 − 2e x = 1 − 2e x = − 6 ex = 3 x = ln 3 ≈ 1.099 x

64. −14 + 3e = 11

3e x = 25 25 ex = 3 25 x ln e = ln 3 25 ≈ 2.120 x = ln 3

1 ln 2 12 ln (12.1 12 )

t ≈ 6.960 3t

68.

0.878    16 +  = 30 26  

0.878   3t ln  16 +  = ln30 26   ln 30 0.878   3ln  16 + 26   3.4012 ≈ ≈ 0.409 8.3241

 (1 + 0.005) x  69. 5000   = 250,000  0.005 

5000(1.005) x = 1250 1.005x = 0.25 x ln(1.005) = ln 0.25 ln 0.25 x= ln(1.005) x ≈ −277.951

Section 4.4 Solving Exponential and Logarithmic Equations  (1 + 0.01) x  70. 250   = 150,000  0.01 

77.

250(1.01) x = 1500 1.01x = 6 x ln1.01 = ln 6 ln 6 ln1.01 x ≈ 180.070 x=

71.

e 2 x − 4e x − 5 = 0 (e x − 5)(e x + 1) = 0

78.

e x = 5 or e x = −1 x = ln 5 ≈ 1.609 (e x = −1 is impossible.)

e2 x − e x + 6 = 0

72.

(e x + 2)(e x − 3) = 0 e x = − 2 or e x = 3 x = ln ( − 2) has no real solutions. x = ln 3 ≈ 1.099

79.

e x = e x −2 x = x2 − 2

73.

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

e2 x = e x − 8 2 x = x2 − 8

74.

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

75. e x − 3 x = e x − 2 x2 − 3x = x − 2

x2 − 4 x + 2 = 0 4 ± 16 − 8 2 x =2± 2 x ≈ 3.414, 0.586 x=

76.

e− x = e x − 2 x − x2 = x2 − 2 x

359

400 = 350 1 + e− x 400 8 = 1 + e− x = 350 7 1 e− x = 7 1 − x = ln   = − ln 7 7 x = ln 7 ≈ 1.946 525 = 275 1 + e− x 525 1 + e− x = 275 525 250 10 e− x = −1 = = 275 275 11 10 − x = ln 11 10 11 x = − ln = ln ≈ 0.095 11 10 40 = 200 1 − 5e−0.01 x 40 1 1 − 5e −0.01 x = = 200 5 4 5e −0.01 x = 5 4 −0.01 x e = 25  4  −0.01x = ln    25  x=

ln ( 4 25)

−0.01 x ≈ 183.258

80.

50 = 1000 1 − 2e −0.001x 50 = 0.05 1 − 2e −0.001 x = 1000 2e −0.001 x = 0.95 e −0.001 x = 0.475 −0.001x = ln 0.475 ln 0.475 −0.001 x ≈ 744.440 x=

2 x2 − 2 x = 0 2 x ( x − 1) = 0 x = 0, 1

360

Chapter 4

Exponential and Logarithmic Functions 83. 20(100 − e x / 2 ) = 500

81. e3 x = 12

(a)

0.6 0.7 0.8 0 .9 1.0 e3 x 6.05 8.17 11.02 14.88 20.09

(a)

(

) 1756 1598 1338 908 200

Using the table, a solution to the equation must lie within (8, 9).

(b) −3

20 100 − e x 2

Using the table, a solution to the equation must lie in (0.8, 0.9). (b)

2200

3 −2

(c)

−2

e3 x = 12 ln e3 x = ln12 3 x = ln12

(c)

(

)

20 100 − e x / 2 = 500

100 − e

x = ln12 ≈ 0.828 1 3

x 2

= 25

−e x 2 = −75

82. 4e5 x = 24

(a)

12 − 200

e x 2 = 75

0.1

0.2

0.3

0.4

0.5

4e5 x

6.59

10.87

17.93

29.56

48.73

ln e x 2 = ln 75 x 2 = ln 75 x = 2 ln 75 ≈ 8.635

Using the table, a solution to the equation must lie within (0.3, 0.4). (b)

−3

(c)

4e5 x = 24 e5 x = 6 ln e5 x = ln 6 5 x = ln 6 x = 15 ln 6 x ≈ 0.358

84. 11(77 − e x − 4 ) = 264

(a)

11(77 − e x − 4 )

817.10

765.72

626.06

246.42

−785.55

Using the table, a solution to the equation must lie within (7, 8).

77 − e x − 4 = 24 − e x − 4 = − 53 e x − 4 = 53 ln e x − 4 = ln 53

900

(b)

x − 4 = ln 53 x = 4 + ln 53 −1

x ≈ 7.970

Section 4.4 Solving Exponential and Logarithmic Equations 0.065   85.  1 +  365  

365 t

92.

361

h(t ) = e −0.125t − 8 10

1.000178365t = 4 The zero of y = 1.000178365t − 4 is t ≈ 21.330.

− 24

− 10

2.471   86.  4 −  = 21 40  

Zero at t ≈ −16.636

3.938225 = 21

93. ln x = −3

The zero of y = 3.9382259 t − 21 is t ≈ 0.247. 87.

94. ln x = −4 e ln x = e −4

7000 = 2 5 + e3 x The zero of y =

88.

7000 − 2 is x ≈ 2.720. 5 + e3 x

x = e −4 = e1 / 4 ≈ 0.018

95. ln 4 x = 2.1

4 x = e2.1

119 =7 6x e − 14 The zero of y =

x = e−3 ≈ 0.050

119 − 7 is x ≈ 0.572. e6 x − 14

96. ln2 x = 1.5

2 x = e1.5

89. g( x ) = 6e1− x − 25

6 −6

1 2.1 e ≈ 2.042 4

1 1.5 e ≈ 2.241 2

97. log 5 (3x + 2) = log5 ( − x)

3x + 2 = − x 4x = − 2

− 30

Zero at x ≈ −0.427 90.

x = −

f ( x ) = 3e3 x 2 − 962

1 2

98. log9 ( 4 + x) = log9 ( 2 x)

1000

4 + x = 2x − 10

4 = x

99. −2 + 2 ln 3 x = 17 2 ln 3 x = 19

− 1100

Zero at x ≈ 3.847 91. h(t ) = e0.125 t − 8 8

− 20

−4

19 2 3 x = e19 2 1 x = e19 2 3 x ≈ 4453.242

ln 3 x =

100. 3 + 2 ln x = 10 2 ln x = 7

7 = 3.5 2 x = e3.5 ≈ 33.115

ln x =

362

Chapter 4

Exponential and Logarithmic Functions

101. 7 log 4 (0.6 x ) = 12 log 4 (0.6 x ) =

108. ln ( x − 4) = 5 2

12 7

412 7 = 0.6 x =

( x − 4)2 = e5 x − 4 = ± e5 2

3 x 5

5 x = 412 7 3 ≈ 17.945

102. 4 log10 ( x − 6) = 11

x = 4 ± e5 2 x ≈ 16.182 or x ≈ − 8.182

1 2  x  1 log 4  =  x −1  2

109. log 4 x − log 4 ( x − 1) =

11 4 x − 6 = 1011 / 4

log10 ( x − 6) =

4 log4 ( x x −1) = 41 2 x =2 x −1 x = 2( x − 1) x = 2x − 2

x = 6 + 1011 4 ≈ 568.341 103. log10 ( z 2 + 19) = 2 z 2 + 19 = 102

2=x

z 2 + 19 = 100

110. log3 x + log3 ( x − 8) = 2

log3[ x( x − 8)] = 2

z = 81

32 = x( x − 8) = x 2 − 8 x

z = ±9

x2 − 8x − 9 = 0

104. log12 x 2 = 6

( x − 9 )( x + 1) = 0

x 2 = 126

x=9

x 2 = 2,985,984

( x = −1 is extraneous.)

x = ±1728

111.

105. ln x + 2 = 1

x + 2 = e1 x + 2 = e2 x = e2 − 2 ≈ 5.389 106. ln x − 8 = 5 1 ln( x − 8) = 5 2 ln( x − 8) = 10 10

x −8 = e

x = 8 + e10 ≈ 22,034.466

107. ln( x + 1)2 = 2 2

e ln( x +1) = e 2 ( x + 1)2 = e 2

x + 1 = e or x + 1 = −e x = e − 1 ≈ 1.718 or x = −e − 1 ≈ −3.718

ln ( x + 5) = ln ( x − 1) + ln ( x + 1) ln ( x + 5) = ln ( x + 1)( x − 1) ln ( x + 5) = ln ( x 2 − 1) x + 5 = x2 − 1 x2 − x − 6 = 0

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

(extraneous solution )

So, x = 3 is the only solution. 112. ln ( x + 1) − ln ( x − 2) = ln x, x > 2

 x +1  ln   = ln x  x−2 x +1 =x x−2 x + 1 = x ( x − 2) = x2 − 2 x x2 − 3x − 1 = 0 x=

3 ± 9 + 4 3 ± 13 = 2 2

Taking the positive solution, x =

3 + 13 ≈ 3.303. 2

Section 4.4

(

)

= 10 2

1+ x

x 7 5 6 8 4 3ln 5x 8.99 9.66 10.20 10.67 11.07

(a)

1+ x 8x

363

116. 3 ln 5 x = 10

113. log10 8 x − log10 1 + x = 2 log10

Solving Exponential and Logarithmic Functions

Using the table, a solution must fall within the

interval ( 5, 6 ) .

8 x = 100 + 100 x

(b)

8 x − 100 x − 100 = 0 2 x − 25 x − 25 = 0

x= =

25 ± 252 − 4 ( 2 )( −25 )

−4

(c)

25 ± 5 33 4

Choosing the positive value, we have x ≈ 180.384.

(

x ≈ 13.431 and

3 ln 5 x = 10 10 ln 5 x = 3 e ln 5 x = e10 / 3

5 x = e10 / 3 1 x = e10 / 3 ≈ 5.606 5

)

114. log10 4 x − log10 12 + x = 2, x > 0 log10

4x 12 + x 4x 12 + x

117. 6log3 ( 0.5 x ) = 11

= 10 2 = 100

(a)

x − 25 x − 300 = 0, Quadratic in x=

25 ±

( −25 ) − 4 ( −300 ) 2

6log3 ( 0.5x ) 9.79 10.22 10.63 11.00 11.36

4 x = 1200 + 100 x 2

Using the table, a solution must fall within the interval (14, 15) .

25 ± 1825 = 2

Taking the positive root and squaring, x ≈ 1146.5.

(b)

115. ln 2 x = 2.4 0

x 3 5 6 2 4 ln 2x 1.39 1.79 2.08 2.30 2.48

(a)

Using the table, a solution must fall within the interval (5, 6). (b)

(c)

6 log3 ( 0.5 x ) = 11 log3 ( 0.5 x ) = 3

log3 ( 0.5 x )

11 6

= 311 / 6

0.5 x = 311 / 6 x = 2(311 / 6 ) ≈ 14.988

−2

10 −2

(c)

ln 2 x = 2.4

e ln 2 x = e2.4 2 x = e2.4 x=

1 2.4 e ≈ 5.512 2

364

Chapter 4

Exponential and Logarithmic Functions

118. 5log10 ( x − 2 ) = 11

125. y1 = 7

(a)

150

155

160

165

y2 = 2 x −1 − 5

170

5log10 ( x − 2 ) 10.85 10.92 10.99 11.06 11.13

Intersection: ( 4.585, 7)

Using the table, a solution must fall within the interval (160, 165).

y2 − 12

(b)

14 −6

126. y1 = 4 0

(c)

y2 = 3x +1 − 2

200

5log10 ( x − 2 ) = 11 log10 ( x − 2 ) = 10

log10 ( x − 2 )

Intersection: ( 0.631, 4 )

11 5

= 1011 / 5 11 / 5

x − 2 = 10

−9

x = 2 + 1011 / 5 ≈ 160.489

−4

119. log10 x = x 3 − 3

127. y1 = 80 3

Graphing y = log10 x − x + 3, you obtain two zeros, x ≈ 1.469 and x ≈ 0.001.

120. log10 x = ( x − 3)

y2 = 4e −0.2 x

Intersection: ( −14.979, 80)

2 100

Graphing y = log10 x = ( x − 3) , you obtain two zeros, 2

x ≈ 2.2386 and x ≈ 3.758.

y2 − 30

30 − 10

121. log10 x + e0.5 x = 6

Graphing y = log10 x + e0.5 x − 6, you obtain one zero,

128. y1 = 500 y2 = 1500e − x 2

x ≈ 3.398.

Intersection: (2.197, 500)

122. e x log10 x = 7

800

Graphing y = e x log10 x − 7 you obtain one zero,

x ≈ 2.764.

123. ln ( x + 2) − 3

y2 x−2

+ 10 = 5

−2 − 200

Graphing y = ln ( x + 2) − 3x − 2 + 5 you obtain two zeros, x ≈ −1.993 and x ≈ 3.738. 2

124. ln x − e = 3 − ln x

Graphing y = 2ln x 2 − e x − 3 you obtain three zeros,

129. y1 = 3.25

y2 = 12 ln( x + 2) Intersection: (663.142, 3.25) 4

x ≈ − 0.546, x ≈ 0.846, and x ≈ 1.569.

y1 y2 0

700

Section 4.4 130. y1 = 1.05

 1  x  3ln   − 1 = 0 x  

1 3ln   − 1 = 0 (since x > 0) x 1 1 ln   = x 3 1 = e1 3 x x = e −1 3 ≈ 0.717

y2 12

Intersection: (10.166, 1.05) 131. 2 x 2 e2 x + 2 xe2 x = 0 (2 x 2 + 2 x )e2 x = 0

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

(since e 2x ≠ 0)

x = 0, − 1 132. − x 2e− x + 2 xe− x = 0 (− x 2 + 2 x )e − x = 0

(since e − x ≠ 0)

− x2 + 2 x = 0 − x( x − 2) = 0

x = 0, 2 −x

133. − xe + e = 0 ( − x + 1)e − x = 0

134. e

−2 x

(since e

−x

≠ 0)

− 2 xe = 0 (1 − 2 x )e −2 x = 0

 y ln   = bx  a

or x =

1 (ln y − ln a ) b

y a = e − bx

(since e −2 x ≠ 0)

x = 12

ln ( y a ) = ln e − bx ln ( y a ) = −bx 1 − ln ( y a ) = x b

135. 2 x ln x + x = 0 x(2 ln x + 1) = 0

2 ln x + 1 = 0

y = ebx a  y ln   = ln ebx  a

140. Exponential decay y = ae− bx

−2 x

1 − 2x = 0

139. Exponential growth y = aebx

1  y =x ln b  a 

−x

−x + 1 = 0 x =1

365

1 138. 3 x ln   − x = 0 x

y2 = ln x − 2 = 12 ln( x − 2)

Solving Exponential and Logarithmic Functions

(since x > 0)

ln x = − 12

1 1 or x = − (ln y − ln a ) = (ln a − ln y) b b

x = e −1 2 ≈ 0.607 136.

1 − ln x =0 x2 1 − ln x = 0 ( x > 0)

ln x = 1 x = e ≈ 2.718 137.

1 + ln x =0 2 1 + ln x = 0 ln x = −1 x = e −1 =

1 ≈ 0.368 e

366

Chapter 4

Exponential and Logarithmic Functions

141. Gaussian model

y = ae

145. (a)

2000 = 1000e0.025t 2 = e0.025t ln 2 = 0.025t ln 2 t= ≈ 27.73 years 0.025

(b)

3000 = 1000e0.025t 3 = e0.025t ln3 = 0.025t ln3 t= ≈ 43.94 years 0.025

146. (a)

2000 = 1000e0.0375t 2 = e0.0375t ln 2 = 0.0375t ln 2 t= ≈ 18.48 years 0.0375

(b)

3000 = 1000e0.0375t 3 = e0.0375t ln3 = 0.0375t ln 3 t= ≈ 29.30 years 0.0375

− ( x − b )2 c

2 y = e−( x − b) c a 2  y ln   = ln e − ( x − b ) c  a

( x − b)2  y ln   = −  a c  y c ln   = −( x − b)2  a c(ln y − ln a ) = −( x − b)2 −c(ln y − ln a ) = ( x − b)2 c(ln a − ln y) = ( x − b)2 ± c(ln a − ln y) = x − b b ± c(ln a − ln y) = x 142. Logarithmic model y = a + b ln x y − a = b ln x y−a = ln x b y−a

e b = e ln x y−a

e b =x

143. (a) 2000 = 1000e 0.07 t 2 = e0.07 t

ln 2 = 0.07t

(b)

ln 2 t= ≈ 9.90 years 0.07 3000 = 1000e 0.07 t 3 = e0.07 t ln 3 = 0.07t ln 3 t= ≈ 15.69 years 0.07

144. (a) 2000 = 1000e0.06 t

2 = e0.06 t ln 2 = 0.06t t=

147. p =

1 1+ e

− (t − 93) 22.5

(a) 50%  p = 0.50 0.50 =

(

0.50 1 + e

− (t − 93) 22.5

1+e

1 1+e

− (t − 93) 22.5

)=1

− (t − 93) 22.5

= 2

e − (t − 93) 22.5 = 1

(

)

ln e − (t − 93) 22.5 = ln1 − (t − 93) 22.5 = 0 − (t − 93) = 0 t − 93 = 0 t = 93 months

ln 2 ≈ 11.55 years 0.06

(b) 3000 = 1000e0.06 t

3 = e0.06 t ln 3 = 0.06t ln 3 t= ≈ 18.31 years 0.06

Section 4.4 (b) 80%  p = 0.80

)

1 1+e

− (t − 93) 22.5

0.80 1 + e − (t − 93) 22.5 = 1 1+e

− (t − 93) 22.5

= 54

e − (t − 93) 22.5 = 14

(

367

149. y = 11,912 − 2340.1ln t

0.80 =

(

Solving Exponential and Logarithmic Functions

)

ln e − (t − 93) 22.5 = ln 14 − (t − 93) 22.5 = ln 14 − (t − 93) =

22.5ln 14

t − 93 = − 22.5ln 14 t = 93 − 22.5ln 14

To find when there were 6300 commercial banks in the United States, let y = 6300 and solve for t. 6300 = 11,912 − 2340.1ln t − 5612 = − 2340.1ln t 2.398 = ln t e 2.398 = eln t e 2.398 = t 11.0 ≈ t

There were about 6300 commercial banks in 2011. 150. V = 6.7e − 48.1 t , t > 0 10

(a)

t ≈ 124.2

So, round up to about 125 months. 0

148. (a)

110

(b) From the graph we see horizontal asymptotes at y = 0 and y = 100. These represent the lower and upper percent bounds. 100 50 = (c) Males: 1 + e −0.6114( x − 69.71) 1 + e−0.6114( x − 69.71) = 2 e−0.6114( x − 69.71) = 1 −0.6114( x − 69.71) = ln1 −0.6114( x − 69.71) = 0 x = 69.71 inches

Females:

50 =

100 1 + e −0.66607( x − 64.51)

1 + e −0.66607( x − 64.51) = 2 e −0.66607( x − 64.51) = 1 −0.66607( x − 64.51) = ln1 −0.66607( x − 64.51) = 0 x = 64.51 inches

1500

(b) As t → ∞, V → 6.7. Horizontal asymptote: y = 6.7 The yield will approach 6.7 million cubic feet per acre. (c) 1.3 = 6.7e− 48.1 t 1.3 = e − 48.1 t 6.7 13 −48.1 = ln t 67 −48.1 ≈ 29.3 years t= ln (13 67 )

( )

151. T = 20 1 + 7 2 − h   

(a)

175

(b) We see a horizontal asymptote at y = 20. This represents the room temperature. 100 = 20[1 + 7(2− h )] (c)

( ) 4 = 7(2 ) 5 = 1 + 7 2− h −h

4 = 2− h 7 4 ln   = ln 2− h 7 4 ln   = − h ln 2 7 ln ( 4 7 ) =h − ln 2 h ≈ 0.81 hour

368

Chapter 4

Exponential and Logarithmic Functions

152. N = −152,656 + 111,959.9ln t

(a) To find when 150,537 seniors intending to major in engineering would take the SAT exam, let

157. f ( x) = log a x, g ( x) = a x , a > 1.

(a)

a = 1.2 20

N = 150,537 and solve for t.

150,537 = −152,656 + 111,959.9ln t 303,193 = 111,959.9ln t

2.708 = ln t e

2.708

= e

2.708

= t

− 10

The curves intersect twice: (1.258, 1.258) and (14.767, 14.767)

ln t

(b) If f ( x ) = log a x = a x = g( x ) intersect exactly once,

15.0 ≈ t So, 150,537 seniors intending to major in engineering would take the SAT exam in 2015.

then x = log a x = a x  ax1 / x . The graphs of y = x1 x and y = a intersect once for

a = e1 e ≈ 1.445. Then

(b) 180,000

( ) = x  e = x  = e.

log a x = x  e1 e

x e

For a = e1 e , the curves intersect once at (e, e). 8 60,000

The point (15, 150,537) lies on the graph, so the year is 2015. (c) Answers will vary. x

153. False. The equation e = 0 has no solutions. 154. False. A logarithmic equation can have any number of extraneous solutions. For example ln(2 x − 1) + ln( x + 2) = ln( x 2 − x − 5) has two extraneous solutions, x = −1 and x = −3. 155. The error is that both sides of the equation should be divided by 2, before taking the natural log. x

2e = 10 ex = 5 x

ln e = ln 5 x = ln 5 156. The domain of the function y = log 3 x + log 3 ( x − 8) − 2 is (8, ∞).

(c) For 1 < a < e1 e , the curves intersect twice. For a > e1 e , the curves do not intersect. 158. Yes. The doubling time is given by 2 P = Pert 2 = ert ln 2 = rt t=

The time to quadruple is given by 4 P = Pert 4 = ert ln 4 = rt t=

ln 4 ln 22 2 ln 2  ln 2  = = = 2  r r r  r 

which is twice as long. 159. f ( x ) = 3 x 3 − 4

So, x = −1 is an extraneous solution and x = 9 is an actual solution of the equation.

ln 2 . r

2 1 −4 −3 −2 −1 −1

−2 −3

Section 4.5 160. f ( x) = x − 2 − 8

Exponential and Logarithmic Models

369

x − 9, x ≤ 1 162. f ( x) =  2  x + 1, x > 1

12 9

4 −12 −12 −8

12 16

−8

− 12 − 9 − 6 − 3 −3

−12

−6

−16

−9

9 12

− 15

2 x + 1, x < 0 161. f ( x) =  2 x ≥ 0 − x , y 2 1 −4 −3 −2

−2 −3 −4 −5 −6

Section 4.5 Exponential and Logarithmic Models 1.

(a) Exponential growth model: y = aebx , b > 0 (b) Exponential decay model: y = ae − bx , b > 0 a (c) Logistic growth model: y = 1 + be− rx − ( x − b )2 c

(d) Gaussian model: y = ae (e) Natural logarithmic model: y = a + b ln x (f) Common logarithmic modal: y = a + b log10 x

y = 6e − x 4 This is an exponential decay model. Matches graph (e).

y = 6 + log10 ( x + 2) This is a logarithmic model, and contains ( − 1, 6). Matches graph (b).

normally

sigmoidal

A Gaussian model has a graph that is bell-shaped.

Gaussian model

The model y = 120e −0.25 x would represent an exponential decay model because the exponent's coefficient is negative.

Matches graph (a).

a 1 + be−rx would have two distinct horizontal asymptotes, y = a as x → ∞, and y = 0 as x → −∞.

The graph of a logistic growth model, y =

y = 2e x 4 This is an exponential growth model.

10. y = 3e − ( x − 2) / 5

11. y = ln ( x + 1)

This is a logarithmic model. Matches graph (d). 12. y =

4 1 + e −2 x

Logistics model Matches graph (f).

Matches graph (c).

370

Chapter 4

Exponential and Logarithmic Functions

13. Since, A = 10,000e0.04t , the time to double is given by

20,000 = 10,000e0.04t 2 = e0.04t ln 2 = 0.04t t =

ln 2 ≈ 17.3 years. 0.04

Amount after 10 years:

17. Since A = 5000ert and A = 5665.74 when t = 10, we have the following. 5665.74 = 5000e10 r 5665.74 = e10 r 5000  5665.74  ln   = 10r  5000 

A = 10,000e0.04(10) ≈ $14,918.25

≈ 0.0125 = 1.25%

14. Since, A = 2000e0.02t , the time to double is given by

4000 = 2000e0.02t 2 = e0.02t ln 2 = 0.02t t =

ln 2 ≈ 34.7 years. 0.02

The time to double is given by 10,000 = 5000e0.0125t 2 = e0.0125t ln 2 = 0.0125t t=

Amount after 10 years:

A = 2000e0.02(10) ≈ $2442.81 15. Since A = 7500ert and A = 15,000 when t = 21, we have the following. 15,000 = 7500e21r 2 = e21r ln 2 = 21r r=

16. Since A = 1000ert and A = 2000 when t = 12, we have the following. 2000 = 1000e12 r 2 = e12 r ln 2 = 12r ln 2 r= ≈ 0.058 = 5.8% 12

Amount after 10 years: A = 1000 e0.058(10) ≈ $1786.04

ln 2 ≈ 55.5 years. 0.0125

18. Since A = 300ert and A = 385.21 when t = 10, we have the following. 385.21 = 300e10 r 385.21 10 r =e 300  385.21  ln   = 10r  300 

ln2 ≈ 0.033 = 3.3% 21

Amount after 10 years: A = 7500e0.033(10)  $10,432.26

1  5665.74  ln   10  5000 

1  385.21  ln   10  300 

≈ 0.025 = 2.5%

The time to double is given by 600 = 300e0.025t 2 = e0.025t ln 2 = 0.025t t=

ln 2 ≈ 27.7 years. 0.025

19. Since A = Pe0.045t and A = 100,000 when t = 10, we have the following. 100,000 = Pe 0.045(10) 100,000 = P ≈ $63,762.82 e 0.45

The time to double is given by 127,525.64 = 63,762.82e 0.045t 2 = e0.045t ln 2 = 0.045t t=

ln 2 ≈ 15.4 years. 0.045

Section 4.5 20. Since A = Pe0.035t and A = 2500 when t = 10, we have

Exponential and Logarithmic Models

23.

371

the following: 2500 = Pe0.035(10) 2500 = P ≈ $1761.72 e 0.35

Because the graph of A = e0.07t increases at a faster rate than A = 1 + 0.075t , continuous compounding results in faster growth.

The time to double is given by 3523.44 = 1761.72e0.035t 2 = e0.035t

24.

ln 2 = 0.035t t =

ln 2 ≈ 19.8 years. 0.035 0

3P = Pert

21.

365t

 0.055  increases at a Because the graph of A = 1 + 365   faster rate than A = 1 + 0.06t , daily compounding results in faster growth.

3 = ert ln 3 = rt ln 3 =t r 2% 4% 6% 8% 10% 12% r ln 3 t= 54.93 27.47 18.31 13.73 10.99 9.16 r

25.

1 = e1600 k 2 ln (1 2 ) k= 1600

So, y = Ce kt

0.14

= 10e ( ) ≈ 6.48 grams.

[ln 1 2 1600]1000

3P = P (1 + r )t

22.

3 = (1 + r )t ln 3 = ln(1 + r )

26. t

ln 3 = t ln(1 + r ) ln 3 =t ln(1 + r )

2% 4% 6% 8% 10% 12% ln 3 t= 55.48 28.01 18.85 14.27 11.53 9.69 ln(1 + r )

ln 1 2 57301000 = 3e  ( ) = 2.66 grams.

27.

0.14

1 C = Ce k (5730) 2 1 = e5730 k 2 1 ln 2 k = 5730

So, y = Ce kt

1 C = Ce k (1600) 2

1 C = Ce k (32,800) 2 1 = e32,800 k 2 ln (1 2) k = 24,100 ln 1 2 32,8001000 So, 1.5 = Ce  ( )

1.5 ≈ C (0.97909) C ≈ 1.53 grams.

372

Chapter 4

28.

1 C = Cek (24,100) 2

Exponential and Logarithmic Functions 33. (0, 4)  a = 4

(5, 1)  1 = 4eb (5)  b =

1 = e 24,100 k 2 ln (1 2 ) k= 24,100

=−

 ln (1 2 ) 24,100 1000

 0.4 = Ce  0.4 ≈ C (0.97165)

C ≈ 0.41 g 29.

34. (0, 5)  a = 5 1 2

= 5e3b

1 10

= e 3b

ln 101 = 3b kt

≈ 5.31 grams.

1 C = Ce k (87.74) 2 1 = e87.74 k 2 ln (1 2) k = 87.74 y = Ce kt

So,

So, y = 4e −0.2773 x .

ln 101 = ln e3b

ln 1 2 432.21000 = 26.4e  ( )

30.

1 ln 4 ≈ −0.2773 5

(3, 12 )  12 = 5eb(3)

1 C = Ce k (432.2) 2 1 = e 432.2 k 2 ln (1 2) k = 432.2 So, y = 26.4e

1 1 ln   5 4

ln 1 2 87.741000 0.1 = Ce  ( )

0.1 ≈ C (0.003707) C ≈ 269.73 grams.

ln 101 = b 3 b ≈ − 0.7673

So, y = 5e − 0.7673 x . 35. (a) According to the model, P = 90e0.013t , the population was increasing because the coefficient of the exponent is positive.

(b) P (6) = 90e0.013(6) ≈ 97.301, or 97,301 people P (9) = 90e 0.013(9) ≈ 101.171, or 101,171 people P (12) = 90e 0.013(12) ≈ 105.194, or 105,194 people

P = 90e0.013t 116 = 90e0.013t

31. y = ae

1 = ae b (0)  1 = a 10 = e b (3) ln 10 = 3b ln 10 = b  b ≈ 0.7675 3 So, y = e0.768 x .

32. (0, 1)  a = 1

(4, 5)  5 = eb(4)

116 = e0.013t 90  116  ln   = 0.013t  90   116  ln    90  = t 0.013 19.5 ≈ t In 2019, the population is expected to be about 116,000.

ln 5 = ln e 4b ln 5 = 4b ln 5 = b 4 b ≈ 0.4024 So, y = e0.4024 x .

Section 4.5

Exponential and Logarithmic Models

373

36. (a) Australia: y = 22.3 when t = 13  22.3 = ae13b

Hungary: y = 9.9 when t = 13 

9.9 = ae13b

y = 25.1 when t = 25  25.1 = ae 25b

y = 9.6 when t = 25 

9.6 = ae 25b

25.1 = a e 25b

9.6 = a e 25b

22.3 = ae13b

9.9 = ae13b

 25.1  22.3 =  25b e13b e 

 9.6  9.9 =  25b e13b e 

22.3 = 25.1e−12b

9.9 = 9.6 e−12b 9.9 = e −12b 9.6  9.9  ln   = −12b  9.6 

22.3 = e −12b 25.1  22.3  ln   = −12b  25.1 

− 0.0026 ≈ b

0.0099 ≈ b

So, a =

25.1 25.1 = 25(0.0099) ≈ 19.6. e 25b e

So, a =

9.6 9.6 = 25(− 0.0026) ≈ 10.2. e 25b e

An exponential growth model for Australia is y = 19.6e0.0099t .

An exponential decay model for Hungary is y = 10.2e − 0.0026t .

So, in 2040, the population will be about 19.6e0.0099(40) = 29.1 million.

So, in 2040, the population will be about 10.2e − 0.0026(40) = 9.2 million.

Canada: y = 34.6 when t = 13  34.6 = ae13b

Philippines: y = 105.7 when t = 13  105.7 = ae13b

y = 37.6 when t = 25  37.6 = ae 25b 37.6 = a e 25b  37.6  34.6 =  25b e13b e 

34.6 = 37.6 e −12b 34.6 = e −12b 37.6  34.6  ln   = −12b  37.6  0.0069 ≈ b So, a =

37.6 37.6 = 25(0.0069) ≈ 31.6. e 25b e

y = 128.9 when t = 25  128.9 = ae 25b 128.9 = a e 25b 105.7 = ae13b  128.9  105.7 =  25b e13b  e  105.7 = 128.9 e −12b 105.7 = e −12b 128.9  105.7  ln   = −12b  128.9  0.0165 ≈ b

128.9 128.9 = 25(0.0165) ≈ 85.3. e 25b e

An exponential growth model for Canada is y = 31.6e0.0069t .

So, a =

So, in 2040, the population will be about 31.6e0.0069(40) = 41.6 million.

An exponential growth model for Philippines is y = 85.3e0.0165t . So, in 2040, the population will be about 85.3e0.0165(40) = 165.0 million.

374

Chapter 4

Exponential and Logarithmic Functions

Turkey: y = 80.7 when t = 13  80.7 = ae13b

38. (a)

In 2008, t = 8 and P = 7.

y = 90.5 when t = 25  90.5 = ae 25b

P = 5.4e kt

90.5 = a e 25b

7 = 5.4e k (8) 7 = e8 k 5.4  7  ln   = 8k  5.4 

80.7 = ae13b  90.5  80.7 =  25b e13b e  80.7 = 90.5 e −12b

1  7  ln   = k 8  5.4 

 80.7  ln   = −12b  90.5  0.0096 ≈ b

So, a =

0.0324 ≈ k

(b) For 2018, let t = 18 and find P.

90.5 90.5 = 25(0.0096) = 71.2. e 25b e

P = 5.4e0.0324t P = 5.4e0.0324(18)

An exponential growth model for Turkey is y = 71.2e0.0096t . So, in 2040, the population will be about 71.2e0.0096(40) = 104.5 million.

P ≈ 9.676 The population in 2018 will be about 9676. 39.

(b) The coefficient of the exponent, b, determines rate of growth. As the magnitude of b increases, so does the rate of growth.

y = Cekt 1 C = Ce5700 k 2 1 = e5700 k 2 ln ( 12 ) = 5700 k

(c) The constant b determines whether the equation is a growth or a decay model. If b < 0, then the model is a decay (decreasing) model, and if b > 0, then model is a growth (increasing) model.

ln ( 12 ) 5700

37. (a) In 2011, t = 11 and P = 412.6.

The ancient charcoal has 15% as much radioactive carbon as it did originally.

P = 339.2e kt 412.6 = 339.2e k (11)

0.15C = Ce

412.6 = e11k 339.2  412.6  ln   = 11k  339.2 

0.15 = e

(

( ln (1 / 2 ) 5700 ) t )

ln (1/ 2 ) 5700 t

(

)

ln (0.15) = ln (1/ 2 ) 5700 t ln ( 0.15 )

1  412.6  ln   = k 11  339.2 

( ln (1 2 ) 5700 )

0.0178 ≈ k

15,600.7 ≈ t

(b) For 2018, let t = 18 and find P.

or about 15,601 years ago

P = 339.2e

0.0178t

P = 339.2e

0.0178(18)

P ≈ 467.309 The population in 2018 will be about 467,309.

y = Ce kt

40.

1 C = Ce(1600)k 2 1 ln   = 1600k 2 ln (1 2) 1600

= k  ln(1 2)  100  

When t = 100, y = Ce  1600

≈ 0.958C , or 95.8%.

Section 4.5 41. (a) Use the points (0, 39,780) and ( 2, 25,459),

Exponential and Logarithmic Models

42. (a) Use the points (0, 1200) and ( 2, 650),

where t is the time since the purchase and V is the value.

39,780 − 25,459 = − 7160.5 0−2

m =

375

m =

1200 − 650 = − 275 0− 2

The V-intercept is (0, 39,780), so the linear model is

The V-intercept is (0, 1200), so the linear model is

V = − 7160.5t + 39,780.

V = − 275t + 1200.

(b) Using the points (0, 39,780) and ( 2, 25,459), find a.

(b) Using the points (0, 1200) and ( 2, 650), find a.

V = ae kt

39,780 = ae k (0)

1200 = ae k (0)

39,780 = a

1200 = a So, V = 1200 ekt .

So, V = 39,780 e .

Find k using the second point.

Find k using the second point. 25,459 = 39,780 e

650 = 1200 e k (2)

k ( 2)

650 = e2k 1200  650  ln   = 2k  1200 

25,459 = e2k 39,780  25,459  ln   = 2k  39,780 

1  650  ln   = k 2  1200 

1  25,459  ln   = k 2  39,780 

− 0.3066 ≈ k

− 0.2231 ≈ k

An exponential model is V = 1200 e − 0.3066t .

An exponential model is V = 39,780 e− 0.2231t . (c)

(c)

40,000

(d) The exponential model has a greater depreciation rate the first year. (e) Using the graph, the value of the sedan with the linear model is greater for 0 < t < 2, and the value of the sedan with the exponential model is greater for t ≥ 2.

1300

(d) The exponential model has a greater depreciation rate the first year. (e) Using the graph, the value of the laptop with the linear model is greater for 0 < t ≤ 2, and the value of the sedan with the exponential model is greater for t > 2. 43. y = 0.0266e (

− x −100 )

(a)

450

, 70 ≤ x ≤ 115

0.05

115

(b) Maximum point is x = 100, the average IQ score.

376

Chapter 4

Exponential and Logarithmic Functions

44. S = 10(1 − e kx )

46.

x = 5 (in hundreds)

y= (a)

S = 2.5 (in thousands)

1000

2.5 = 10(1 − e k (5) )

(a)

0.25 = 1 − e5 k e5 k = 0.75 5k = ln 0.75 k ≈ −0.0575

(

S = 10 1 − e −0.0575 x

p(t ) =

)

1000 1 + 9e −0.1656 t

1000 ≈ 203 animals 1 + 9e −0.1656(5) 1000 500 = 1 + 9e−0.1656 t p(5) =

(a) (b)

1 + 9e−0.1656 t = 2 9e −0.1656 t = 1 e −0.1656 t = t=

1 9 −ln (1 9 ) 0.1656

(b) For t = 19, y ≈ 662. For t = 30, y ≈ 663. 663 (c) As t → ∞, y → = 663, limiting value. 1 (d) Answers will vary.

(b) When x = 7, S = 10(1 − e −0.0575(7) ) ≈ 3.314 which corresponds to 3314 units. 45.

663 , 0 ≤ t ≤ 18 1 + 72e −0.547 t

 I  47. R = log10   = log10 I  I = 10 R  I0  (a)

I = 107 = 10,000,000

(b)

I = 108.3 ≈ 1,999,526,231

(c)

I = 106.3 ≈ 1,995,262

 I  48. R = log10   = log10 I  I0  (a)

R = log10 39,811,000 ≈ 7.6

(b)

R = log1012,589,000 ≈ 7.1

(c)

R = log10 251,200 ≈ 5.4

≈ 13 months

1300

100

49. β (I ) = 10 log10 ( I I 0 ) , where I 0 = 10 −12 watt per square meter.

 10 −10  = 10 log10 10 2 = 20 decibels −12   10 

(a)

β (10 −10 ) = 10 ⋅ log10 

(b)

β (10 −5 ) = 10 ⋅ log10 

(c)

β (10 0 ) = 10 ⋅ log10 

 10 −5  = 10 log10 10 7 = 70 decibels −12   10 

 10 0  = 10 log10 1012 = 120 decibels −12  10  

Section 4.5

 I   I  50. β = 10 log10   = 10 log10  −12  I  10   0  10−4  β (10 ) = 10 log10  −12  = 10 log10 (108 )∂  10  −4

(a)

= 80 decibels

(b)

β (10 −3 ) = 90 decibels

(c)

β (10 −2 ) = 100 decibels

pH − 1 = − log10 [H + ]

−(pH − 1) = log10 [H + ] 10 − (pH −1) = [H + ] 10 − pH +1 = [H + ] 10 − pH ⋅ 10 = [H + ]

    0.03t   57. (a) u = 230,000  − 1 12t   1 1 −       1 + 0.03 12   

 I    I0 

10 β 10 =

377

The hydrogen ion concentration is increased by a factor of 10.

β = 10 log10 

51.

56.

Exponential and Logarithmic Models

I I0

100,000

I = I 0 10 β 10 % decrease =

I 0 108.8 − I 0 107.2 × 100 I 0 108.8

= 97.5%

 I    I0 

(b) From the graph, when u = 230,000, t ≈ 53.08 years. Yes, a mortgage of approximately 58. (a) P = 200,000, r = 0.0425, M = 983.88

I 10 β 10 = I0

12t

Pr  r   u = M − M − 1 +  12 12   

I = I 0 10 β 10 I 10 % decrease = 0

94 years will result in about $460,000 in interest.

β = 10 log10 

52.

9.3

− I 0 10 I 0 10 9.3

= 983.88 − (983.88 − 708.33)(1 + 0.003542)

12t

8.0

× 100 ≈ 95%

= 983.88 − 275.55(1.003542)

12t

v = (983.88 − 708.33)(1.003542)

12 t

−5

53. pH = − log10 [H ] = − log10 [2.3 × 10 ] ≈ 4.64 54.

5.8 = − log10  H + 

−5.8

= 275.55(1.003542)

12 t

1000

= [H ]

[H ] ≈ 1.58 × 10 −6 moles per liter

u 0

55.

pH = − log10 [H + ] +

−pH = log10 [H ] 10 − pH = [H + ]

Hydrogen ion concentration of grape 10−3.5 = −8.0 Hydrogen ion concentration of baking soda 10

(b) In earlier years, the majority of the monthly payment goes toward interest. The interest and principle are equal when t ≈ 13.7 ≈ 14 years. (c) P = 200,000, r = 0.0425, M = 1238.47

u = 1238.47 − (1238.47 − 708.33)(1.003542)

12 t

v = (1238.47 − 708.33)(1.003542)

12 t

= 104.5 ≈ 31,623 times

1300

u 0

u = v when t ≈ 3.7 years. Answers will vary.

378

Chapter 4

Exponential and Logarithmic Functions 66. A logistic growth model or exponential decay model will have a horizontal asymptote of its graph represent the limiting value of a population.

 T − 70  59. t = −10 ln    98.6 − 70 

60.

At 9:00 A.M. we have t = −10 ln ( 85.7 − 70 ) ( 98.6 − 70 )  ≈ 6 hours.

67. An exponential growth model will have a graph which shows a steadily increasing rate of growth.

Thus, we can conclude that the person died 6 hours before 9 A.M., or 3:00 A.M.

68. A logarithmic model will have only a vertical asymptote occur for its graph.

 T − 40  t = −5.05 ln   ( t = 0 is 11 A.M.)  0 − 40 

 T − 40  7 = −5.05 ln    0 − 40  −7  T − 40  = ln   5.05  −40   T − 40  −7 5.05  =e  −40  T = 40 − 40e −7 5.05 ≈ 29.998 ≈ 30 < 32

Hence, the steaks do not thaw out in time. 61. False. The domain could be all real numbers. 62. False. For example, y =

3 does not have an 1 + 2e −0.5 x

1 69. 4 x − 3 y − 9 = 0  y = (4 x − 9) 3 4 Slope: 3 Matches (a). 9  Intercepts: (0, − 3),  , 0  4  2 70. 2 x + 5 y − 10 = 0  y = − x + 2 5 2 Slope: − 5 Matches (b) Intercepts: (5, 0), (0, 2) 71.

x-intercept.

 100  Intercepts: (0, 25),  , 0  9 

63. The graph of a Gaussian model will not have an x-intercept because no x-value will yield a negative y-value. 64. (a) Logarithmic model; The only asymptote of the graph is a vertical asymptote; The Richter scale is one example. (b) Logistic growth model; The graph increases then decreases from left to right; The spread of a virus is one example. (c) Exponential decay model; A horizontal asymptote of the graph represents the limiting value of the population; Carbon dating is one example. (d) Exponential growth model; The model shows a steady increasing rate of growth; Population growth is one example. (e) Quadratic model; The minimum value of the function occurs at the average value of the independent variable; The average value of a population is one example. (f) Gaussian model; The maximum value of the function occurs at the average value of the independent variable; The average SAT score for a class is one example.

y = 25 − 2.25 x Slope: −2.25 Matches (d).

72.

73.

x y + = 1  y = −2 x + 4 2 4 Slope: –2 Matches (c). Intercepts: (2, 0), (0, 4). f ( x) = 2 x 3 − 3x 2 + x − 1 The graph falls to the left and rises to the right.

74.

f ( x ) = −4 x 4 − x 2 + 5 The graph falls to the left and falls to the right.

75. g( x ) = −1.6 x 5 + 4 x 2 − 2

The graph rises to the left and falls to the right. 76. g( x ) = 7 x 6 + 9.1x 5 − 3.2 x 4 + 25 x 3

The graph rises to the left and rises to the right.

65. A Gaussian model will have the maximum value occur at the average value of the independent variable.

77.

2 −8 3 − 9 8 0 12 2 0 3

78.

−5

2 x3 − 8x2 + 3x − 9 3 = 2 x2 + 3 + x−4 x−4

Section 4.6

Nonlinear Models

0 0 −3 −5 25 −125

1 640

1 −5 25 −128

641

379

x4 − 3x + 1 641 = x 3 − 5 x 2 + 25 x − 128 + x+5 x+5

79. Answers will vary.

Section 4.6 Nonlinear Models 1.

y = ax b

ln b x y = ae( )

16.

A scatter plot can be created to provide an idea of what type of model will best fit a set of data.

An exponential model with r 2 ≈ 0.967 would be a better fit for a set of data as compared to a power model with r 2 ≈ 0.901.

Logarithmic model

Linear model

Quadratic model

Exponential model

Exponential model 17.

Linear model 18.

10. Logistic model

11. Quadratic model

Logarithmic model

12. Linear model 13.

19.

y = 4.752(1.2607) x Coefficient of determination: 0.96773 14

Logarithmic model 14.

20.

y = 3.964(1.4084) x Coefficient of determination: 0.99495 140

Linear model 15.

Exponential model

Chapter 4

380

Exponential and Logarithmic Functions

21. y = 7.707(0.8070)

27.

Coefficient of determination: 0.9448

y = 1.985 x 0.760 Coefficient of determination: 0.99686 14

22. y = 82.640(0.9530)

28.

Coefficient of determination: 0.9314

y = 3.397 x1.650 Coefficient of determination: 0.99788 150

120

0 −4

23.

y = 2.083 + 1.257 ln x Coefficient of determination: 0.98672

29.

24.

y = 9.027 + 2.537 ln x Coefficient of determination: 0.96884

30.

y = 525.428 x −0.226 Coefficient of determination: 0.99549 500

y = 16.103 x −3.174 Coefficient of determination: 0.88161

25. y = − 4.730 ln x + 10.353

Coefficient of determination: 0.9661

0 300

31. (a)

Exponential model: S = 2952.790(1.1203) ; t

r 2 = 0.9821 Power model: S = 877.314t1.0326 ; r 2 = 0.9782

(b) Exponential model: 14,000

26.

y = 20.076 − 5.027 ln x Coefficient of determination: 0.99977 15

Power model: 14,000

(c) The exponential model with r 2 ≈ 0.9821 is a better fit, as compared to the power model with r 2 ≈ 0.9782. © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Section 4.6

Nonlinear Models

381

32. (a) Linear model: B = 15.51t + 149.0; r 2 ≈ 0.9898 Exponential model: B = 179.203(1.0539) ; r 2 ≈ 0.9895 t

Logarithmic model: B = 143.44 ln t − 24.2; r 2 ≈ 0.9807 (b) The linear model fits the data best with a coefficient of determination closest to 1, r 2 = 0.9898. (c) B = 15.51t + 149.0 = 15.51(17) + 149.0 = 412.67 The number of beds in 2017 will be 412,670. Answers will vary. 33. (a) Linear model: P = 7.57t + 352.7; r 2 ≈ 0.9786 600

(b) Power model: P = 248.55t 0.2351; r 2 ≈ 0.9438 600

600

(d) Logarithmic model: P = 111.6 ln t + 168.6; r 2 ≈ 0.9289 600

(e) The exponential model from part (c) is the best fit because the coefficient of determination, r 2 ≈ 0.9866, is closet to 1. (f) Linear Model: Year

2014

2015

2016

2017

2018

Population (in thousands) 534.4 542.0 549.5 557.1 564.7

Power Model: Year

2014

2015

2016

2017

2018

Population (in thousands) 524.7 529.7 534.7 539.4 544.1

Exponential Model: Year

2014

2015

2016

2017

2018

Population (in thousands) 536.6 545.2 553.9 562.8 571.8

382

Chapter 4

Exponential and Logarithmic Functions

Logarithmic Model: Year

2014

2015

2016

2017

2018

Population (in thousands) 523.5 528.0 532.4 536.6 540.7

(g) and (h) Answers will vary. h = 0 is not in the domain of the logarithmic function. (b) h = 0.863 − 6.447ln p (c) 35

34. (a)

0 0 −5

(d) Answers will vary.

(d) For p = 0.75, h ≈ 2.71 km.

37. (a) Logistic model: P =

(e) For h = 13, p ≈ 0.15 atmosphere. 35. (a) Exponential model: S = 3.723(1.1117)

1.2

88.2676 1 + 901.9894e−0.2560 x

100

(b)

(b) S = 3.723e ln(1.1117) = 3.723e 0.1059 t

The model fits the data well.

S = 3.723e 0.1059(18) ≈ 25.05

The annual sales in 2018 will be about $25.05 billion. Answers will vary. 36. (a) Linear model: T = −1.24t + 73.0

38. (a) Logistic model: y =

68.3744 1 + 4.9648e−0.3934 x

Power model: y = 13.299 x0.6550 (b) Logistic model:

30 0

The data do not appear to be linear.

Power model:

(b) Quadratic model: T = 0.034 t 2 − 2.26 t + 77.3

The data appear to be quadratic. A quadratic model might not be a good model for predicting the temperature of the liquid when t = 60 because the temperature of the water should decrease, and not increase as the model would predict.

The logistic model fits the data better because the data are closest to the model. (c) Let x = 17, and find y using the logistic model.

y =

68.3744 1 + 4.9648e− 0.3934(17)

≈ 69.95

The length of a 17-year old yellowtail snapper is about 69.95 centimeters. 39. True

Section 4.6

383

46. 1.2 x + 3.5y = 10.5

40. False. Rewrite b as b = e In b . Then, y = ab x = ae(In b ) x = aecx .

35 y = −12 x + 105 y = − 12 x + 105 35 35

41. Answers will vary.

= − 12 x+3 35

42. (i) The scatterplot appears to be an exponential model, so the equation is y = 1.05e0.02288 x , r 2 = 0.968.

Slope: − 12 35 y-intercept: (0, 3)

(ii) The scatterplot appears to be logarithmic model, so 2

the equation is y = 1.06 ln x + 0.96, r = 0.9991. 43.

Nonlinear Models

2 x + 5 y = 10 5y = −2 x + 10

8 6

y = − 25 x + 2

2 −4 −2 −2

Slope: − 25

−4

y-intercept: (0, 2)

−6

47.

f ( x ) = a( x − h )2 + k

4 3

f ( x ) = a( x + 1)2 + 2 2

1 = a(0 + 1) + 2 1= a +2

1 −2 −1 −1

Point: (0, 1)

−1 = a f ( x ) = −( x + 1)2 + 2

−2 −3

44. 3x − 2 y = 9

48.

y= x− 3 2

Slope:

9 2

3 = a(0 − 2)2 − 1 Point: 3 = 4a − 1 4 = 4a 1= a

2 1 1

f ( x ) = a( x − h )2 + k f ( x ) = a( x − 2)2 − 1 Vertex: (2, − 1)

3 2

y-intercept: ( 0, − 29 )

−2 −1 −1

Vertex: (−1, 2)

(0, 3)

f ( x ) = ( x − 2)2 − 1

49.

−2

f ( x ) = a( x − h )2 + k

−3

f ( x ) = a( x − 3)2 + 2 Vertex: (3, 2)

−4

0 = a(4 − 3)2 + 2 Point: (4, 0) 0=a+2 −2 = a f ( x ) = −2( x − 3)2 + 2

45. 0.4 x − 2.5 y = 12.5 25 y = 4 x − 125 4 y = 25 x −5

50.

4 = 0.16 Slope: 25

y-intercept: (0, − 5)

f ( x ) = a( x − 2)2 − 2 Vertex: (2, − 2)

1 −4 −3 −2 −1 −1 −2 −3 −4 −6

f ( x ) = a( x − h )2 + k

0 = a(0 − 2)2 − 2 Point: (0, 0) 0 = 4a − 2 2 = 4a 1 2

f ( x) = 12 ( x − 2)2 − 2

−7

384

Chapter 4

Exponential and Logarithmic Functions

Chapter 4 Review 1.

(1.45)

2π

10.

≈ 10.3254

Intercept: ( 0, 1)

2. 7 11 ≈ 635.1453 3. 60 4.

2( − 0.5)

= 60

−1

f ( x ) = 0.3 x Horizontal asymptote: x-axis or y = 0 Decreasing on: ( −∞, ∞)

≈ 0.0167

−3 3 2 25 ( ) ≈ 5.12 × 10 −7 ≈ 0

5. f ( x) = 4 x − 3

−2

−1

100 9

10 3

3 10

9 100

 1 Intercept: (0, 4− 3 ) =  0,   64 

Horizontal asymptote: x-axis

Increasing on: ( −∞, ∞)

Matches graph (c).

6. f ( x ) = 4 − x

−3

Intercept: ( 0, 1)

−2

−1

11. f ( x) = 8− x

Horizontal asymptote: x-axis Decreasing on: ( −∞, ∞ )

Intercept: (0, 1) Horizontal asymptote: x-axis or y = 0

Matches graph (d).

Increasing on: ( −∞, ∞)

7. f ( x ) = −4 x Intercept: ( 0, − 1)

−2

Horizontal asymptote: x-axis Decreasing on: ( −∞, ∞ )

−1

1 8

1 64

Matches graph (b).

7 6

8. f ( x ) = 4 x + 1

Intercept: ( 0, 2 ) Horizontal asymptote: y = 1 Increasing on: ( −∞, ∞ )

Matches graph (a).

−4 −3 −2 −1 −1

9. f ( x) = 8x

12. g( x ) = 0.3− x

Intercept: (0, 1) Horizontal asymptote: x-axis or y = 0 Increasing on: ( −∞, ∞)

Intercept: ( 0, 1) Horizontal asymptote: x-axis or y = 0 Increasing on: ( −∞, ∞ )

−2

−1

−2

−1

1 64

1 8

0.09

0 .3

y 7

1 3

1 9

5 4

3 2

−4 −3 −2 −1 −1

2 1

−2

−1

Chapter 4 Review 13. h ( x ) = e x −1

385

16. f ( x) = − 2e− x −2

h( x)

0.37

2.72

7.39

20.09

f ( x ) −14.78

−1

− 5.44

−2

− 0.74 − 0.27

2 1

9 8 7 6 5 4 3 2 1

−5 −4 −3 −2 −1

2 3 4 5

−2

−5 −4 − 3 −2 − 1

−8

1 2 3 4 5

Horizontal asymptote: y = 0

Horizontal asymptote: y = 0 17.

14. f ( x ) = e x + 2 x

−4

−3

−2

−1

f ( x)

0.14

0.37

2.72

7.39

f ( x ) = 4 e −0.5 x

−1

f ( x)

6.59

2.43

1.47

0.89

0.54

y y

9 8 7

9 8 7 6

4 3 2 1

2 1 −7 −6 −5 −4 − 3 −2 − 1

1 2 3

−2 − 1

Horizontal asymptote: y = 0

18. f ( x) = 2 + e x − 3

15. h( x) = − e x + 3 x

−2

−1

h( x )

2.86

2.63

0.28 − 4.39

x f ( x)

2.05

2.14

2.37

3.00

4.72

8 7 6 5 4 3

5 4 3 2 1 −5 −4 −3 −2 −1

1 2 3 4 5 6 7 8

1 2 3 4 5

−2 −3 −4 −5

Horizontal asymptote: y = 0

1 −2 −1

1 2 3 4 5 6 7 8

−2

Horizontal asymptote: y = 2

386

Chapter 4

19.

A = Pert = 10,000e0.08 t

Exponential and Logarithmic Functions

$10,832.87 $22,255.41 $49,530.32 $110,231.76 $245,325.30 $545,981.50

20. r = 3% = 0.03, A = 10,000e0.03t

$10,304.55 $13,498.54 $18,221.19 $24,596.63 $33,201.17 $44,816.89

()

21. V (t ) = 30,795 54 (a)

28.

32,000

35 = 243 log3 243 = 5

e3.2 ≈ 24.532

29.

ln 24.532 ≈ 3.2 0

( ) = $19,708.80.

(b) After 2 years, V ( 2) = 30,795 54

1.6094

≈ 5 −3

31.

1   =8 2

log1 2 8 = −3

t / 14

(a) Initial quantity: Q ( 0 ) = 50 ( 12 )

0 / 14

(b) After 10 years: Q (10 ) = 50 ( 12 )

= 50 grams

10 / 14

(c)

ln 5 ≈ 1.6094

30. 2

≈ 30.5 grams

−2

32.

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

33. log6 216 = log6 63 0

23. log 5 625 = 4 54 = 625 24. log 9 81 = 2 2

9 = 81

25. log64 2 =

= 3log6 6

1 6

641 6 = 2

34. log 7 1 = 0 1 35. log 4   = log 4 4 −1 4 = − log 4 4

( )

= −1 36. log10 0.00001 = log10 10−5 = −5

 1  26. log10   = −2  100  1 10 −2 = 100

27.

4 3 = 64 log 4 64 = 3

Chapter 4 Review

387

40. f ( x) = log10 ( x + 2) − 3

37. g ( x) = − log10 x + 5 Domain: (0, ∞)

Domain: ( − 2, ∞)

Vertical asymptote: x = 0 x-intercept: (10,000, 0)

Vertical asymptote: x = − 2 x-intercept: (998, 0)

9 8 7 6 5 4 3 2 1

1 −3 −2 −1

1 2 3 4 5 6 7

−2

−1

1 2 3 4 5 6 7 8 9

−4 −5 −6 −7 −8 −9

38. g ( x) = log10 ( x − 3)

41. ln(21.5) ≈ 3.068

Domain: (3, ∞) Vertical asymptote: x = 3

42. ln 0.46 ≈ −0.777 43. ln 6 ≈ 0.896

x-intercept: ( 4, 0)

44. ln ( 65 ) ≈ −0.182

y 5 4 3 2 1

f ( x ) = ln x + 3

45.

−1

1 2

4 5 6 7 8 9

Domain: (0, ∞) Vertical asymptote: x = 0 x-intercept: (0.05, 0)

−2 −3 −4 −5

39. f ( x) = log10 ( x − 1) + 6 −1

Domain: (1, ∞) Vertical asymptote: x = 1 x-intercept: (1 + 10−6 , 0) = (1.000001, 0) y 9 8 7 6 5 4 3 2 1 −1

8 −1

f ( x) = ln( x − 3)

46.

Domain: (3, ∞) Vertical asymptote: x = 3 x-intercept: (4, 0) 3

2 3 4 5 6 7 8 9

x −3

388

Chapter 4

Exponential and Logarithmic Functions

1 ln x 2

47. h( x ) =

52. log1 2 10 =

Domain: x > 0 Vertical asymptote: x = 0 x-intercept: (1, 0)

log1 2 10 =

53. log14 15.6 =

−1

log14 15.6 =

−3

54. log3 0.28 =

48. f ( x) = 4 ln x

log3 0.28 =

Domain: (0, ∞) Vertical asymptote: x = 0 x-intercept: (1, 0)

log10 10 1 = ≈ − 3.322 log10 1 2 log10 1 2 ln 10 ≈ − 3.322 ln (1 2)

log10 15.6 ≈ 1.041 log10 14 ln 15.6 ≈ 1.041 ln 14 log10 0.28 ≈ −1.159 log10 3

ln 0.28 ≈ −1.159 ln3 ln ( x − 2)

55. f ( x ) = log 2 ( x − 2) =

ln 2

−1

−4

−2

56.

18,000 49. t = 50 log10 18,000 − h (a)

0 ≤ h < 18,000

(b)

100

ln x ln3

f ( x ) = 2 − log3 x = 2 − 7

−1

11 −1

20,000

57. f ( x) = − log1 2 ( x − 2)

=−

Vertical asymptote: h = 18,000 (c) The plane climbs at a faster rate as it approaches its absolute ceiling. 18,000 ≈ 5.46 minutes. (d) If h = 4000, t = 50 log10 18,000 − 4000 x   50. t = 12.542 ln   , x > 1000  x − 1000 

−1

log10 9 ≈ 1.585 log10 4

ln 9 log4 9 = ≈ 1.585 ln 4

−4

58.

f ( x) = log1 3 ( x − 1) + 1

(a) For x = 1254.68, t ≈ 20 years. (b) For x = 1254.68, t = 20, the total amount paid is (1254.68)(20)(12) = $301,123.20. The interest is 301,123.20 − 150,000.00 = $151,123.20. 51. log 4 9 =

ln( x − 2) ln( x − 2) ln( x − 2) = = − ln(1 2) ln(1 2) ln 2

ln( x − 1) ln( x − 1) +1 = − +1 ln(1 3) ln 3

−1

−3

Chapter 4 Review 59. logb 9 = logb 32

70. ln

= 2 logb 3 = 2(0.5646) = 1.1292

389

x 1 = ln x1 2 − ln 4 = ln x − ln 4 4 2 1 = ln x − 2 ln 2 2

 x+3 71. ln   = ln( x + 3) − ln( xy)  xy  = ln( x + 3) − ln x − ln y

4 60. log b   = log b 2 2 − log b 32 9

= 2 logb 2 − 2 logb 3 = 2(0.3562) − 2(0.5646)

72. ln

= −0.4168

xy 5

= ln x + ln y 5 − ln z1 2

1 = ln x + 5ln y − ln z 2

61. logb 5 = logb 51 2 1 log b 5 2 1 = (0.8271) 2 = 0.41355 =

73. log2 9 + log2 x = log2 (9 x ) 74. log6 y − 2 log6 z = log6 y − log6 z 2

= log6

62. logb 50 = logb [2 ⋅ 5 ]

= logb 2 + 2logb 5

75.

= 0.3562 + 2(0.8271)

y z2

1 ln(2 x − 1) − 2 ln( x + 1) = ln 2 x − 1 − ln( x + 1)2 2

= 2.0104

= ln

63. ln(5e −2 ) = ln 5 + ln e −2 = ln 5 − 2 ln e

2x − 1 ( x + 1)2

76. 5 ln( x − 2) − ln( x + 2) + 3ln( x) = ln( x − 2)5 − ln( x + 2) + ln( x)3

= ln 5 − 2

= ln

64. ln e 5 = ln e 5 2 5 5 = ln e = 2 2

77. ln 3 +

65. log10 200 = log10 (2 ⋅ 100)

( x − 2)5 x3 ( x + 2)

 3(4 − x 2 )1 3  1 3 3 4 − x2 ln(4 − x 2 ) − ln x = ln   = ln 3 x x  

78. 3  ln x − 2 ln( x 2 + 1)  + 2 ln 5 = ln x 3 − ln( x 2 + 1)6 + ln 52 25 x 3 = ln 2 ( x + 1)6

= log10 2 + log10 10 2 = log10 2 + 2 66. log10 0.002 = log10 (2 ⋅ 10−3 )

79. s = 25 −

= log10 2 + log10 10 −3 = log10 2 − 3

(a)

13ln ( h 12 ) ln 3

67. log5 5 x 2 = log5 5 + log5 x 2 = 1 + 2 log5 x 68. log4 16 xy2 = log4 16 + log4 x + log4 y2

= 2 + log 4 x + 2 log 4 y (b) 69. log10

5 y x2

= log10 5 y − log10 x

= log10 5 + log10 y − log10 x 2 = log10 5 +

1 log10 y − 2 log10 x 2

h s

4 38

6 33.2

8 29.8

10 27.2

12 25

14 23.2

390

Chapter 4

Exponential and Logarithmic Functions

80. To find when the average score decreases to 68, set f (t ) = 68, and solve for t.

90. log5 ( 2 x + 27) = 2

52 = 2 x + 27

68 = 85 − 17log10 (t + 1)

25 = 2 x + 27

−17 = −17 log10 (t + 1)

− 2 = 2x

1 = log10 (t + 1)

−1 = x

101 = t + 1 9=t 9 months

91. ln x = 4

81. 10 x = 10,000

92. ln x = −3

10 x = 10 4 x=4

x = e 4 ≈ 54.598 x = e −3 ≈ 0.0498

93. ln ( x − 1) = 2 e2 = x − 1

82. 7 x = 343

x = 1 + e2

7x = 73 x =3

94. ln ( 2 x + 1) = −4

1 1 = = 6−3  x = −3 83. 6 x = 216 63 84.

1 1296 1 6x − 2 = 4 6

e −4 = 2 x + 1 e −4 − 1 2 1 − e4 = 2e 4

6x − 2 =

6 x − 2 = 6− 4 x − 2 = −4

95. 3e −5 x = 132 e−5 x = 44

−5 x = ln 44 ln 44 x=− ≈ −0.757 5

x = −2

1 16 2 x +1 = 2 −4

85. 2 x +1 =

x + 1 = −4 x = −5 86. 4 x / 2 = 64 4 x / 2 = 43 x =3 2 x=6

96.

14 e 3 x + 2 = 560

e3 x + 2 = 40 ln e3 x + 2 = ln 40 3 x + 2 = ln 40 x=

( ln 40 ) − 2 ≈ 0.563 3

97. 2 x + 13 = 35 2 x = 22

87. log8 x = 4 84 = x 4096 = x 88. log x 729 = 6 x 6 = 729 x = 6 729 = 3

x ln 2 = ln 22 ln 22 x= ≈ 4.459 ln 2 98. 6 x − 28 = −8 6 x = 20

x ln 6 = ln 20 ln 20 x= ≈ 1.672 ln 6

89. log 2 ( x − 1) = 3 23 = x − 1 x=9

Chapter 4 Review

( )

99. −4 5 x = −68

106. ln 5 x = 4.5

e 4.5 = 5 x

5 x = 17 x ln 5 = ln17 ln17 x= ≈ 1.760 ln 5

( )

100. 2 12 x = 190

1 4.5 e =x 5 18.003 ≈ x 107. ln x − ln 5 = 2 x ln   = 2 5 x e2 = 5 5e2 = x 36.945 ≈ x

12 x = 95 x ln12 = ln 95 ln 95 x= ≈ 1.833 ln12 101. 2 e x − 3 − 1 = 4

2e x − 3 = 5 e x −3 =

108. ln x − ln 3 = 4 x ln   = 4 3 x e4 = 3 3e 4 = x 163.794 ≈ x

5 2

5 x − 3 = ln   2 5 x = 3 + ln   ≈ 3.916 2

1 2 1 ex / 2 = 2 x 1 = ln   = − ln 2 2 2 x = −2 ln 2 = − ln 4 ≈ −1.386

102. −e x / 2 + 1 =

103. e 2 x − 7e x + 10 = 0

( e − 5)( e − 2 ) = 0 x

e x = 5  x = ln 5 ≈ 1.609 e x = 2  x = ln 2 ≈ 0.693

104. e 2 x − 6 e x + 8 = 0

( e − 4 )( e − 2 ) = 0 x

e x = 4 or e x = 2 x = ln 4 or x = ln 2 x ≈ 1.386 x ≈ 0.693

391

109. ln x + 1 = 2 1 ln ( x + 1) = 2 2 ln ( x + 1) = 4

x + 1 = e4 x = e4 − 1 ≈ 53.598 110.

ln x + 40 = 3 ln ( x + 40 )

1/ 2

1 ln ( x + 40 ) = 3 2 ln ( x + 40 ) = 6 e6 = x + 40 e6 − 40 = x 363.429 ≈ x

105. ln 3 x = 6.4

e6.4 = 3 x 1 6.4 e =x 3 200.615 ≈ x

392

Chapter 4

Exponential and Logarithmic Functions

log 4 ( x + 5) = log 4 (13 − x ) − log 4 ( x − 3)

111.

117. x ln x + x = 0 x ( ln x + 1) = 0

 13 − x  log 4 ( x + 5) = log 4    x −3 13 − x x + 5 = x −3 ( x + 5)( x − 3) = 13 − x

ln x = −1 x = e −1 =

x 2 + 2 x − 15 = 13 − x

118.

x 2 + 3 x − 28 = 0

( x + 7)( x − 4) = 0 x + 7 = 0  x = − 7 (extraneous) x − 4 = 0  x = 4

112. log 5 ( x + 2 ) − log 5 ( x ) = log 5 ( x + 5 )

 x+2 log5   = log5 ( x + 5)  x  x+2 = x+5 x x + 2 = x2 + 5x x2 + 4 x − 2 = 0 x=

−4 ± 16 − 4 ( −2 )

2 x = −2 + 6 ≈ 0.449

= −2 ± 6

x = 1 − 10 −1 = 0.9

114. log10 ( − x − 4 ) = 2

− x − 4 = 102 = 100 − x = 104 x = −104

116. 2 xe + e

x = e− 1 ≈ 0.368 119. To find the time for a $7550 deposit to double earning 6.9% interest compounded continuously, set A = 15,100

and solve for t using the formula A = pert .

15,100 = 7550 e0.069t 2 = e0.069t ln 2 = 0.069t ln 2 =t 0.069 10.05 ≈ t

4   500 = 50001 −  4 + e − 0.0005 x   4 0.1 = 1 − 4 + e − 0.0005 x 4 − 0.9 = − 4 + e − 0.0005 x 4 + e − 0.0005 x = 4.4444 e − 0.0005 x = 0.4444 − 0.0005 x = ln 0.4444

( since e ≠ 0 ) x

x = −1 2x

ln x = − 1

(a) Find demand x for a price p = $500.

10 −1 = 1 − x

( since x > 0)

4   120. p = 50001 − − 0.0005 x  4 e +  

113. log10 (1 − x ) = −1

x +1 = 0

1 + ln x =0 x2 1 + ln x = 0

1 ≈ 0.368 e

About 10.05 years

(Other zero is extraneous.)

115. xe x + e x = 0 ( x + 1) e x = 0

( since x > 0 )

ln x + 1 = 0

x =

ln 0.4444 − 0.0005

x ≈ 1622 televisions

( 2 x + 1) e2 x = 0

( since e ≠ 0 ) 2x

2x + 1 = 0 x=−

1 2

Chapter 4 Review (b) Find demand x for a price p = $350. 4   350 = 50001 − − 0.0005 x  + e 4   4 0.07 = 1 − 4 + e −0.0005 x 4 − 0.93 = − 4 + e − 0.0005 x 4 + e − 0.0005 x = 4.3011 e − 0.0005 x = 0.3011 0.0005 x = ln 0.3011 x =

ln 0.3011 − 0.0005

x ≈ 2401 televisions

121. y = 3e − 2 x 3

393

127. P = 8.06e kt

Use t = 1 and P = 8.21 to find k.

8.21 = 8.06ek (1) 8.21 = ek 8.06  8.21  ln   = k  8.06   8.21  ln   = k  8.06  0.0184 ≈ k So, P = 8.06e 0.0184t . For 2020, let t = 20, and find P. P = 8.06e 0.0184(20) ≈ 11.645427

Decreasing exponential Matches graph (e).

In the year 2020, the population will be about 11,645,427. 128. (a)

122. y = 4e2 x 3

0.1

Intercept: ( 0, 4 ) Increasing Matches graph (b). 123. y = ln ( x + 3 )

Logarithmic function shifted to left Matches graph (f). 124. y = 7 − log ( x + 3 )

Vertical asymptote: x = −3 Decreasing Matches graph (d). 125. y = 2e − ( x + 4 ) 3

150

y = 0.0499 (

− x − 74 ) 128 2

(b) The average score corresponds to the maximum, 74. 129. N =

62 1 + 5.4e − 0.24t

(a) To find the number of weeks t to read 45 words per minute, let N = 45 and solve for t. 45 =

Gaussian model Matches graph (a). 126. y =

6 1 + 2 e −2 x

Logistic model Matches graph (c).

62 1 + 5.4e − 0.24t

1 + 5.4e − 0.24t = 1.3778 5.4e −0.24t = 0.3778 e − 0.24t = 0.0700 − 0.24t = ln 0.0700 t =

ln 0.0700 − 0.24

t ≈ 11.08 It took 11.08 weeks.

394

Chapter 4

Exponential and Logarithmic Functions

(b) Let N = 61 and find t.

61 =

62 1 + 5.4e − 0.24t

1 + 5.4e − 0.24t = 1.0164

(d) Find t when N = 3697 using the linear model.

= 0.0164

N = 40.4t + 2728

0.0164 5.4  0.0164  − 0.24t = ln    5.4 

3679 = 40.4t + + 2728

5.4e

− 0.24 t

e − 0.24t =

951 = 40.4t 951 = t 40.4 23.5 ≈ t

 0.0164  ln   5.4  t =  − 0.24

In the 2023–2024 school year, about 3,697,000 girls will participate.

t ≈ 24.15 It took 24.15 weeks.

134. (a) h =

 I  130. R = log10   = log10 I  I = 10 R  I0 

(b)

(a) (b)

R = 7.1  I = 107.1 ≈ 12,589,254

(c)

R = 3.8  I = 103.8 ≈ 6310

131. Logistic model 132. Logarithmic model

R = 9.2  I = 109.2 ≈ 1,584,893,192

21.253 1 + 24.284−0.131 x

(c) The model fits the data well. (d) As x → ∞, h → 21.3 m. So, the graph of h has a horizontal asymptote of h = 21.253 to the right.

ex e True (by properties of exponents).

135. e x −1 = e x ⋅ e−1 =

133. (a) Linear model: N = 40.4t + 2728; r 2 ≈ 0.9836

Exponential model: N = 2737.170(1.0135) ; r 2 ≈ 0.9832 t

136. ln x + ln y = ln ( xy ) (by the Product Property)

Power model: N = 2688.8t 0.0645 ; r 2 ≈ 0.8748

137. False. The domain of f ( x ) = ln ( x ) is x > 0.

(b) Linear model:

x 138. True. ln   = ln x − ln y (by the Quotient Property) y

3300

139. Since 1 < 2 < 2, then 21 < 2 2 < 22  2 < 2 2 < 4. 0 2700

140. (a)

Exponential model:

y = ex

3300

−4

y1 −1

(b) Pattern  0 2700

i =0

x i!

y4 = 1 + x +

Power model: 3300

x2 x3 x 4 + + 2! 3! 4!

The graph of y4 closely approximates y = e x near ( 0, 1) .

0 2700

Chapter 4 Test

395

Chapter 4 Test 1.

f ( x ) = 10 − x

−2

−1

f ( x)

100

0.1

0.01

log7 7−0.89 = −0.89 log7 7 = −0.89

4.6 ln e2 = 4.6(2) ln e = 9.2

5 − log101000 = 5 − log10103 = 5−3= 2

9 8

2 1 −5 −4 −3 −2 − 1

(

1 2 3 4 5

y 1

Intercept: (0, 1)

−1 −1

f ( x ) = −6 x − 2

f ( x)

−0.03

−1

−6

−36

−4 −5 −6 −7

8. f ( x) = ln( x + 4)

2 1 1

Domain: ( − 4, ∞)

3 4 5 6

Vertical asymptote: x = − 4

−2 −3 −4 −5 −6 −7 −8

x-intercept: ( − 3, 0) y 5 4 3 2

Horizontal asymptote: y = 0 1   Intercept:  0, −  36  

−5 −4

−2 −1

−2

−1

f ( x) 0.9817 0.8647 0 −6.3891 −53.5982 y

9. f ( x) = 1 + ln( x − 6)

Domain: (6, ∞) Vertical asymptote: x = 6

−5 −4 − 3 − 2 −1

1 2 3 4 5

−2 −3 −4 −5

f ( x ) = 1 − e2 x

−3

−2

−4 −3 −2 −1

)

x-intercept: 10 −6 , 0 ≈ (0, 0)

Horizontal asymptote: y = 0

f ( x ) = − log10 x − 6

Domain: x > 0 Vertical asymptote: x = 0

1 2 3 4 5

−2 −3 −4 −5 −6 −7 −8

Horizontal asymptote: y = 1 Intercept: (0, 0)

x-intercept: (6.37, 0) y 8 6 4 2 −2 −2

8 10 12 14

−4 −6 −8

396

Chapter 4

10. log7 44 =

ln 44 ≈ 1.945 ln 7

11. log 2 5 1.3 =

12. log12 64 =

Exponential and Logarithmic Functions

23.

ln 1.3 ≈ − 0.286 ln ( 2 5)

1025 =5 8 + e4 x 1025 = 40 + 5e 4 x 985 = 5e 4 x e4 x = 197 4 x = ln(197)

ln 64 ≈ 1.674 ln 12

13. log2 3a 4 = log2 3 + log2 a 4 = log2 3 + 4log2 a 14. ln

15. ln

24. − xe − x + e − x = 0

5 x = ln 5 + ln x − ln 6 6 1 = ln 5 + ln x − ln 6 2

e − x (1 − x ) = 0 x =1 25. log10 x − log10 (8 − 5 x ) = 2

x x +1 = ln x + ln x + 1 − ln(2) − ln e 4 2e 4 1 = ln x + ln ( x + 1) − ln2 − 4 2

 x  log10  =2  8 − 5x  x 8 − 5x 800 − 500 x = x 800 = 501x 800 x= ≈ 1.597 501 10 2 =

16. log313 + log3 y = log3 (13 y)  x4  x 17. 4 ln x − 4 ln y = ln x 4 − ln y 4 = ln  4  = ln   y y   x (2 x − 3)  18. ln x − ln( x + 2) + ln(2 x − 3) = ln    x+2 

26. 2 x ln x − x = 0 2 ln x = 1, ( x ≠ 0)

1 2 x = e1 2 ≈ 1.649

ln x =

19. 3 x = 81 = 34 x=4 20.

52 x = 2500

2 x ln 5 = ln 2500 1 ln 2500 ≈ 2.431 x= 2 ln 5

1 ln(197) ≈ 1.321 4

27.

1 2

= 1ek (22) (half-life is 22 years)

ln 12 = 22k k = 221 ln 12 = − 221 ln 2 ≈ −0.03151 A = e −0.03151(15) ≈ 0.6234 or 62.34% remains

21. log 7 x = 6 76 = x x = 117, 649

22. log10 ( x − 4) = 5

10 5 = x − 4 x = 105 + 4 = 100,004

Chapters 3–4 Cumulative Test

397

28. (a) Logarithmic model: R = 81.2970 ln t − 144.646

Exponential model: R = 2.006(1.3266)

Power model: R = 0.099t 2.5639 (b) Logarithmic model: 90

Exponential model: 90

Power model: 90

≈ 430.914 In 2019, the revenue will be about $430,914,000,000.

Chapters 3–4 Cumulative Test 2. f ( x ) = x 2 − 6 x + 5

1. f ( x ) = − ( x − 2 ) + 5 2

f ( x ) = ( x − 3) − 4 2

5 4

6 5 4 3 2 1

3 2 1 −2 −1

−2 −1

2 3 4

6 7 8

−2 −3 −4

398

Chapter 4

Exponential and Logarithmic Functions

f ( x ) = x + 2 x 2 − 9 x − 18

f ( x) = x2 ( x + 2) − 9 ( x + 2)

(

f ( x) = ( x + 2) x − 9 2

f ( x ) = 3x 4 + 10 x3 − 27 x 2 − 10 x (a) Since f ( x ) has one variation in signs, f has 1 positive

)

real zero.

f ( x ) = ( x + 2 )( x − 3 )( x + 3 )

f ( − x ) = 3 ( − x ) + 10 ( − x ) − 27 ( − x ) − 10 ( − x ) 4

= 3x 4 − 10 x 3 − 27 x 2 + 10 x

Since f ( − x ) has two variations in signs, f has

5 −15 −10 −5

5 10 15 20

2 negative real zeros. (b) Possible rational zeros: ± 12 , ± 23 , ± 1, ± 53 , ± 2, ± 10 , ± 5, ± 10 3 (c)

200

−8

f ( x ) = − x3 + 10 x2 − 27 x + 18

(a) Since f ( x ) has three variations in sign, f has 3 or 1

− 350

positive real zeros.

(d) f ( x ) = 3 x 4 + 10 x 3 − 27 x 2 − 10 x

f ( − x ) = − ( − x ) + 10 ( − x ) − 27 ( − x ) + 18 3

(

= x 3 x 3 + 10 x 2 − 27 x − 10

= x3 + 10 x 2 + 27 x + 18 Since f ( − x ) has no variation in sign, f has no

negative real zeros. (b) Possible rational zeros: ± 1, ± 2, ± 3, ± 6, ± 9, ± 18 (c)

−27

−10

(

f ( x ) = x ( x − 2 ) 3 x 2 + 16 x + 5

)

= x ( x − 2 )( 3 x + 1)( x + 5 ) −8

The real zeros of f are x = − 5, x = − 13 , x = 0,

and x = 2.

−5

(d)

−1

−27

−1

−18

−1

(

x3 + 2 x2 + 4 x + 8 = ( x + 2) x2 + 4

x + 2 = 0  x = −2 x2 + 4 = 0 x 2 = −4

f ( x ) = − x 3 + 10 x 2 − 27 x + 18

x = ± −4  ±2i

(

) = − ( x − 1) ( x − 9 x + 18 ) = ( x − 1) − x 2 + 9 x − 18

Zeros: −2, ± 2i

= − ( x − 1)( x − 6 )( x − 3)

)

g ( x ) = x3 + 4 x 2 − 11

The real zeros of f are x = 1, x = 3, and x = 6.

4 −6

− 12

x ≈ 1.424

Chapters 3–4 Cumulative Test

4x + 2 x + 3 4 x 2 + 14 x − 9

5x 5x = x 2 + x − 6 ( x + 3)( x − 2)

13. f ( x ) =

4 x 2 + 12 x

399

Vertical asymptote: x = −3, 2 Horizontal asymptote: y = 0

2x − 9 2x + 6

− 15

6 4

4 x 2 + 14 x − 9 15 = 4x + 2 − x+3 x+3

2 −5

−20

288

268

2 3

−4

−2

−6

x2 − 3x + 8 6 = x −1+ x−2 x−2

14. f ( x ) =

2 x − 5 x + 6 x − 20 268 = 2 x 2 + 7 x + 48 + x −6 x −6

Vertical asymptote: x = 2 Slant asymptote: y = x − 1

10. f ( x) = ( x + 6)( x + 1)( x − 1)

= ( x + 6)( x 2 − 1)

12 8

= x + 6x − x − 6

11. Since 1+ 3i is a zero, so is 1− 3i.

(

)

(

)

− 12 − 8

(

−4

)

f ( x ) = ( x − 0 )( x − 2 )  x − 1 + 3i   x − 1 − 3i     = x ( x − 2) x − 2x + 4

(

= x x3 − 4 x2 + 8 x − 8 4

)

15. g ( x ) = y

1 +2 x−5

-2

-4

The graph of g ( x ) is a horizontal shift 5 units to the

8 6 4

−4

Vertical asymptote: x = 3 Horizontal asymptote: y = 2

−2

2x 12. f ( x) = x −3

−2

− 12

= x − 4x + 8x − 8x

−4

−8

= x ( x − 2 )  x 2 − 2 x + 1 − 3i 2   

(

−4

−2

right and a vertical shift 2 units upward of the graph 1 of f ( x ) = . x 1 +2=0 x −5 1 =−2 x −5 1 = − 2 ( x − 5)

1 = − 2 x + 10 − 9 = − 2x 9 =x 2 9 Zero: x = 2

400

Chapter 4

Exponential and Logarithmic Functions

16. Answers will vary. Sample answer: y =

4x2 x +1 2

21.

f ( x) =

ln x x

(a) Domain: all real numbers x > 0

17. The graph of g ( x) = 3x +1 − 5 is a horizontal shift one

graph of f ( x) = 3x.

0 =

18. The graph of g ( x) = − log10 ( x + 3) is a reflection in

f ( x) = 8− x +1 + 2

(1, 0)

(a) Domain: all real numbers x

y-intercept: y =

(b) x-intercept:

0 = 8− x +1 + 2 − 2 = 23(− x +1)

−(0) +1 +2 y-intercept: y = 8

Vertical asymptote: x = 0 22.

f ( x) = log 6( 2 x − 1) + 1

= 8+2

(a) Domain: all real numbers x > 12

= 10

(b) x-intercept:

(0, 10) (c) Horizontal asymptote: y = 2 20.

ln 0  undefined 0

No y-intercept (c) Horizontal asymptote: y = 0

− 2 = 8− x +1 No real solution, so no x-intercept

1 (ln x) x

1 = 0  undefined x ln x = 0  x = 1

the x-axis and a horizontal shift 3 units to the left of f ( x) = log10 x. 19.

ln x x

(b) x-intercept: 0 =

unit to the left and a vertical shift 5 units downward of the

0 = log 6( 2 x − 1) + 1 −1 = log 6( 2 x − 1) 6−1 = 6log6 (2 x −1)

f ( x) = 2e0.01x

1 6

= 2x − 1

(a) Domain: all real numbers x

7 6

= 2x

0 = 2e0.01x

7 12

= x

(b) x-intercept:

0 = e

0.01x

(127 , 0)

ln 0 = ln e0.01x

y-intercept: y = log 6 2(0) − 1 + 1

No real solution, so no x-intercept

y = log 6 ( −1) + 1

0.01 0 y-intercept: y = 2e ( )

No real solution, so no y-intercept

= 2e 0 = 2 (0, 2) (c) Horizontal asymptote: y = 0

( )

1 , log 1 23. Because 2 − 4 = 16 2 16 = − 4.

24. ln e10 = 10 25. ln

1 = ln e− 3 = − 3 e3

26. log516 =

ln 16 ≈ 1.723 ln 5

27. log9 6.8 =

ln 6.8 ≈ 0.872 ln 9

Chapters 3–4 Cumulative Test

ln 8.61 ≈ − 7.484 ln ( 3 4 )

28. log3 4 ( 8.61) =

37. ln(2 x − 5) − ln x = 1 ln

 3  ln ( 3 2 ) 29. log 7 8   = ≈ − 3.036  2  ln ( 7 8 )  x2 − 4  2 30. ln  2  = ln[( x − 2)( x + 2)] − ln( x + 1) x 1 +  

2x − 5 =1 x 2x − 5 e= x ex = 2 x − 5

x(e − 2) = −5 x=

= ln( x − 2) + ln( x + 2) − ln( x 2 + 1)  x +1 31. 2 ln x − ln( x − 1) + ln( x + 1) = ln x 2    x −1 

32.

−5 <0 e−2

 −5  No solution because ln   does not exist. e−2

450,000 + 5 x x C (10,000 ) ≈ $50

8e 2 x = 56

38. C ( x ) =

e =7 ln e 2 x = ln 7 2 x = ln 7

C (100,000 ) ≈ $9.50

x = 12 ln 7 ≈ 0.973

C (1,000,000 ) ≈ $5.45 Limiting average cost is $5.00.

33. 6 x − 4 − 29 = 12

6 x − 4 = 41

39. (a)

ln 6 x − 4 = ln 41 ln 41 ln 6 ln 41 x = 4+ ≈ 6.073 ln 6

x−4=

(b) 40. (a)

34. log 2 x + log 2 5 = 7

log 2 ( 5 x ) = 7 5 x = 27 5 x = 128 x=

128 = 25.6 5

35. 250e 0.05 x = 500,000

e0.05 x = 2000 0.05 x = ln 2000 x = 20 ln 2000 x ≈ 152.018 36. 2 x 2 e 2 x − 2 xe 2 x = 0 2

401

(2 x − 2 x )e

2x2 − 2 x = 0 2 x ( x − 1) = 0

(b)

 0.06  A = 200 1 +  12  

A = 200e

0.06 ( 30 )

30(12 )

≈ $1204.52

≈ $1209.93

1200 1 + 3e − t 5 1200 1200 p (0) = = = 300 1 + 3e0 1 + 3 p (t ) =

p ( 5) =

(c)

1200 1200 = ≈ 570.44 −5 5 1 + 3e 1 + 3e−1 1200 1 + 3e − t 5 = 1200

800 =

(

800 1 + 3e − t 5 1 + 3e

)

−t 5

= 1.5

−t 5

= 0.5 1 e−t 5 = 6

t 1 − = ln   5 6

1 t = −5ln   ≈ 9.0 years 6

x = 0, 1

402

Chapter 4

41. (a)

(b)

Exponential and Logarithmic Functions

Let x and y be the lengths of the sides. 2 x + 2 y = 546  y = 273 − x A = xy = x(273 − x ) 25,000

273

Domain: 0 < x < 273 (c)

If A = 15,000, then x = 76.23 or 196.77. Dimensions in feet: 76.23 × 196.77 or 196.77 × 76.23

42. (a) Quadratic model: S = −1.16963t 2 + 40.2326t + 147.853; r 2 ≈ 0.9905

Exponential model: S = 229.067(1.0652) ; r 2 ≈ 0.9486 t

Power model: S = 160.681t 0.4294 ; r 2 ≈ 0.9921 (b) Quadratic model: 600

Exponential model: 600

Power model: 600

(c) The power model is the best fit because its coefficient of determination, r 2 ≈ 0.9921, is closest to 1. (d) Using the power model, let t = 20 and find S. S = 160.681( 20)

0.4294

≈ 581.603 In 2020, the annual sales will be $581,603,000,000. Answers will vary.

C H A P T E R 5 Trigonometric Functions Section 5.1

Angles and Their Measure .................................................................404

Section 5.2

Right Triangle Trigonometry .............................................................413

Section 5.3

Trigonometric Functions of Any Angle ............................................422

Section 5.4

Graphs of Sine and Cosine Functions................................................440

Section 5.5

Graphs of Other Trigonometric Functions ........................................451

Section 5.6

Inverse Trigonometric Functions .......................................................462

Section 5.7

Applications and Models....................................................................473

Chapter 5 Review .......................................................................................................483 Chapter 5 Test ............................................................................................................500

C H A P T E R 5 Trigonometric Functions Section 5.1 Angles and Their Measure 15. (a) Since − 90° < −150° < −180°, −150° lies in

Trigonometry

angle

standard position

coterminal

radian

angular

One-half revolution of a circle is equal to 180° or π radians.

16. (a) Since 0° < 87.9° < 90°, 87.9° lies in Quadrant I.

(b) Since − 360° < − 8.5° < − 270°, − 8.5° lies in Quadrant IV.

The sum of two positive complementary angles is 90° or

π 2

Quadrant III. (b) Since 270° < 282° < 360°, 282° lies in Quadrant IV.

radians.

The angles 315° and −225° are not coterminal. See figure.

17. (a) Since 90° < 132°50′ < 180°, 132°50′ lies in

Quadrant II. (b) Since −360° < − 336° 30′ < − 270°, − 336° 30′ lies in Quadrant I. 18. (a) Since −270° < − 245.25° < − 180°, − 245.25° lies in Quadrant II. (b) Since 0° < 12.35° < 90°, 12.35° lies in

Quadrant I.

19. (a)

45° y

3158 x

22258

45° x

10. The angle

2π π is obtuse because it is greater than and 3 2

less than π .

(b)

90° y

11.

The angle shown is approximately 210°.

90° x

12.

The angle shown is approximately −45°. 13. (a) Since 0° < 55° < 90°, 55° lies in Quadrant I. (b) Since 180° < 215° < 270°, 215° lies in Quadrant III. 14. (a) Since 90° < 121° < 180°, 121° lies in Quadrant II. (b) Since 180° < 181° < 270°, 181° lies in Quadrant III.

404

Section 5.1 20. (a)

60°

(b)

Angles and Their Measure

405

−120°

60° x

− 120°

(b) 180°

23. (a)

405° y

180°

405°

21. (a)

− 30°

(b)

− 780° y

−780°

− 30°

(b) 150°

24. (a) y

−450° y

150°

− 450°

22. (a)

270°

(b) y

600° y

600°

270° x

406

Chapter 5

Trigonometric Functions

25. (a) Coterminal angles for 52° : 52° + 360° = 412° 52° − 360° = −308°

28. (a) Coterminal angles for − 445° :

− 445° + 360° = − 85° − 445° + 720° = 275°

(b) Coterminal angles for −36° : −36° + 360° = 324° −36° − 360° = −396°

(b) Coterminal angles for 230° : 230° + 360° = 590° 230° − 360° = −130°

26. (a) Coterminal angles for 114° : 114° + 360° = 474° 114° − 360° = −246° (b) Coterminal angles for −390° : −390° + 720° = 330° −390° + 360° = −30°

  45  29. 64° 45′ = 64° +   = 64.75°  60 

27. (a) Coterminal angles for 300° :

30. −124° 30′ = −124.5°    18   30  31. 85° 18′ 30′′ = 85° +   +  ≈ 85.308°   60   3600 

300° + 360° = 660°

32. −408° 16′ 25′′ ≈ −408.274°

300° − 360° = − 60°

  36  33. −125° 36′′ = −125° −  = −125.01°  3600 

(b) Coterminal angles for −740° : −740° + 1080° = 340° −740° + 720° = −20°

34. 330° 25′′ ≈ 330.007°

35. 51° 22′ 30′′ − 38° 17′ 15′′ = (51° − 38°) + ( 22′ − 17′) + (30′′ − 15′′) = 13° 5′ 15′′

36. 120° 45′ 29′′ − 12° 36′ 3′′ = (120° − 12°) + ( 45′ − 36′) + ( 29′′ − 3′′) = 108° 9′ 26′′

37. 48° 18′ − 25° 16′ 59′′ = ( 48° − 25°) + (18′ − 16′) + (0′′ − 59′′) = ( 48° − 25°) + (17′ − 16′) + (60′′ − 59′′) = 23° 1′ 1′′

38. 36° 8′ 43′′ − 81° 17′′ = (36° − 81°) + (8′ − 0′) + ( 43′′ − 17′′) = − 45° 8′ 26′′

39. 280.6° = 280° + 0.6(60)′ = 280° 36′ 40. −115.8° = −115° 48′ 41. −345.12° = −345° 7′ 12′′ 42. 490.75° = 490° 45′

 180°  43. −0.355 = −0.355    π 

≈ −20.34° = −20° 20′ 24′′

46. Complement: Not possible; 129° is greater than 90°. Supplement: 180° − 129° = 51° 47. Complement: Not possible; 167° is greater than 90°. Supplement: 180° − 167° = 13° 48. Complement: 90° − 87° = 3° Supplement: 180° − 87° = 93° 49.

 180°  44. 0.7865 = 0.7865    π  ≈ 45.0631° = 45° + (0.0631)(60′) = 45° + 3′ + 0.786(60′′) = 45° 3′ 47′′ 45. Complement: 90° − 24° = 66° Supplement: 180° − 24° = 156°

The angle shown is approximately 2 radians. 50.

The angle shown is approximately − 4 radians.

Section 5.1

51. (a)

Since 0 <

π 2

π 6

lies in Quadrant I.

58. (a)

Angles and Their Measure

3π 4 y

5π 3π 5π (b) Since π < lies in Quadrant III. < , 4 2 4

3π 4

5π 5π 52. (a) Since lies in Quadrant II. < < π, 2 6 6 5π 3π 5π (b) Since −2π < − lies in Quadrant I. <− , − 3 2 3 3π 7π 7π lies in Quadrant IV. < < 2π , 2 4 4 5π 11π 11π (b) Since lies in Quadrant II. < < 3π , 2 4 4

53. (a) Since

(b)

4π 3 y

5π 5π lies in Quadrant IV. <− < 0, 2 12 2 3π 13π 13π (b) Since − < − lies in Quadrant II. < −π , 2 9 9

54. (a) Since −

55. (a) Since −

π 2

56. (a) Since π < 3.5 <

57. (a)

π 2

4π 3 x

< −1 < 0, − 1 lies in Quadrant IV.

(b) Since −π < −2 < −

(b) Since

π 2

, − 2 lies in Quadrant III.

59. (a)

−

7π 4

3π , 3.5 lies in Quadrant III. 2

< 2.25 < π , 2.25 lies in Quadrant II. x

3π 2

− y

(b)

3π 2 x

(b)

−

407

−

7π 4

5π 2 y

5π − 2

2 y

π − 2

408

Chapter 5

60. (a)

Trigonometric Functions

11π 6

62. (a) 2 y y

11π 6

x x

(b) (b)

−

− 3π

2π 3

− 3π x

2π − 3

63. (a) 61. (a)

5π

 π  π 30° = 30°  =  180°  6

 π  5π (b) 150° = 150°  =  180°  6

5π

64. (a) x

 π  7π 315° = 315°  =  180°  4

 π  2π (b) 120° = 120°  =  180°  3

π  π  65. (a) 18° = 18°  = 10  180°  4π  π  (b) −240° = −240°  =− 3  180° 

(b) −4 y

−4

66. (a)

11π  π  − 330° = − 330°  =− 6  180° 

 π  4π (b) 144° = 144°  =  180°  5 67. (a)

(b)

68. (a)

(b)

3π 3π  180°  = = 270° 2 2  π  −

7π 7π  180°  =−   = −210° 6 6  π 

 180°  −4π = −4π   = −720°  π   180°  3π = 3π   = 540°  π 

Section 5.1

7π 7π  180°  =   = 420° 3 3  π 

69. (a)

(b)

−

13π 13π  180°  =−   = −39° 60 60  π 

70. (a)

−

15π 15π  180°  =−   = −450° 6 6  π 

28π 28π  180°  =   = 336° 15 15  π 

(b)

 π  71. 115° = 115°   ≈ 2.007 radians  180°   π  72. 83.7° = 83.7°   ≈ 1.461 radians  180°   π  73. −216.35° = −216.35°   ≈ −3.776 radians  180°   π  74. −46.52° = −46.52°   ≈ −0.812 radian  180°   π  75. −0.78° = −0.78°   ≈ −0.014 radian  180°   π  76. 395° = 395°   ≈ 6.894 radians  180°  77.

78.

π 7

83. (a) Coterminal angles for

 180°  79. 6.5π = 6.5π   = 1170°  π   180°  80. −4.2π = −4.2π   = −756°  π   180°  81. −2 = −2   ≈ −114.592°  π   180°   102.6  ° 82. −0.57 = −0.57   = −  ≈ −32.659°  π   π 

409

13π 6 π 11π − 2π = − 6 6 6

+ 2π =

2π : 3

(b) Coterminal angles for 2π 8π + 2π = 3 3 2π 4π − 2π = − 3 3

7π : 6

84. (a) Coterminal angles for 7π 19π + 2π = 6 6 7π 5π − 2π = − 6 6

5π : 4

(b) Coterminal angles for 5π 13π + 2π = 4 4 5π 3π − 2π = − 4 4

85. (a) Conterminal angles for

9π : 4

9π π − 2π = 4 4 9π 7π − 4π = − 4 4

π  180°    ≈ 25.714° 7 π 

5π 5π  180°   900 ° =  =  ≈ 81.818° 11 11  π   11 

Angles and Their Measure

(b) Coterminal angles for −

2π : 15

2π 28π + 2π = 15 15 2π 32π − − 2π = − 15 15 −

86. (a) Conterminal angles for −

7π : 8

7π 9π + 2π = 8 8 7π 23π − − 2π = − 8 8

−

(b) Coterminal angles for

π 12

25π 12 π 23π − 2π = − 12 12

+ 2π =

410

Chapter 5

87. Complement:

Trigonometric Functions

π 2

−

99. s = rθ , θ in radians

6 2π Supplement: π − = 3 3

 2π  s = 27   = 18π meters ≈ 56.55 meters  3 

88. Complement: Not possible;

Supplement: π −

90. Complement:

π 2

Supplement: π −

100. r = 12 centimeters, θ =

2π π is greater than . 3 2

101. r =

3π π = 4 4

89. Complement: Not possible;

Supplement: π −

3π π is greater than . 2 4

2π π = 3 3 −

π 6

 3π  s = rθ = 12   = 9π centimeters ≈ 28.27 cm  4 

102. r =

π 3

5π 6

3π π 91. Complement: Not possible; is greater than . 2 2 Supplement: Not possible;

3π is greater than π . 2

3π 4

θ s

= =

π 2

feet ≈ 22.92 feet

3 9 = meters ≈ 0.72 meter 4π 3 4π

103. r =

s 82 328 miles ≈ 34.80 miles = = θ 135° (π 180° ) 3π

104. r =

s 8 48 inches ≈ 1.39 inches = = θ 330° (π 180° ) 11π

105. The angle between Omaha and Dallas:

θ = 41° 15′ 50′′ − 32° 47′ 39′′

92. Complement: Not possible;

12π π is greater than . 2 5

= 8° 28′ 11′′ ≈ 0.1478 radian s = θ r = (0.1478)(4000) ≈ 591.2 miles

Supplement: Not possible;

12π is greater than π . 5

106. The angle between Seattle and San Francisco:

93.

s = rθ 8 = 15θ 8 θ= radian 15

94. θ =

= 9° 49′ 42′′ ≈ 0.1715 radian s = θ r = (0.1715)(4000) ≈ 686.1 miles 107. θ =

s 10 5 = = radian r 22 11

95. s = rθ 35 = 14.5θ

θ=

θ = 47° 37′ 18′′ − 37° 47′ 36′′

70 ≈ 2.414 radians 29

96. r = 80 kilometers, s = 160 kilometers s 160 θ= = = 2 radians r 80 97. s = rθ , θ in radians

 π  s = 14 (180 )   = 14π ≈ 43.982 inches  180  98. r = 9 feet, θ = 60° =

π 3

π  s = rθ = 9   = 3π feet 3

s 450 = ≈ 0.07056 radian ≈ 4.04° r 6378

≈ 4° 2′ 33.02′′ 108. θ =

s 2.5 25 5 = = = radian ≈ 23.87° 6 60 12 r

109. θ =

s 24 = = 4.8 rad ≈ 275.02° 5 r

1 revolutions 2 = 360° + 180° = 540° = 2π + π = 3π radians

110. (a) Single axel: 1

1 revolutions 2 = 360° + 360° + 180° = 900° = 2π + 2π + π = 5π radians

(b) Double axel: 2

1 revolutions 2 = 360° + 360° + 360° + 180° = 1260° = 2π + 2π + 2π + π = 7π radians

Section 5.1

(6378 + 1250)2π s rθ = = t t 110 = 435.71 km / min

111. Linear speed =

s = rθ 1 2π s = (2π ) = feet 3 3

Therefore, the chain moves 2π 3 feet, as does the

Angular speed = ( 2π )( 4800) = 9600π rad/min

smaller rear sprocket. Thus, the angle θ of the smaller sprocket is ( r = 2 inches = 2 12 feet )

7.25 = 3.625 in. 2

s rθ θ = = r = r (angular speed) t t t = 3.625(9600π ) = 109,327.4 in./min

θ=

Speed =

 14  14π s = θ r = ( 4π )   = feet  12  3

Angular speed = 2π (1050 rev/min )

Speed =

= 2100π rad/min

θ s rθ = = r = r (angular speed ) t t t = 9.75( 2100π )

(b) Since the arc length of the tire is (14π ) 3 feet and the cyclist is pedaling at a rate of one revolution per second, we have:

≈ 64,324.1 in./min

feet  1 mile  14π Distance =   ( n revolutions)  3 revolutions  5280 feet  7π n miles. = 7920

Revolutions 480 = = 28,000 rev/h 114. (a) Hour (1/ 60 )

Angular speed = 2π (28, 800) = 57, 600π rad / h 25 2 5 Radius of wheel = = miles (12 in. ft)( 5280 ft mi) 25,344

Distance = Rate ⋅ Time

(c)

14π  1 mile  = feet per second (t seconds)  3  5280 feet  7π t miles = 7920

θ s rθ = = r = r ( angular speed ) t t t 5 125π = ⋅ 57,600π = ≈ 35.70 miles h 25,344 11

Speed =

The functions are both linear.

(b) Let x = spin balance machine rate.  x  70 = r (angular speed) = r  2π .  (1/60)   =

115. (a)

5 120π x  x ≈ 941.18 rev min 25,344

Revolutions/min 10,000 500 = = = rev/sec seconds/min 60 3  500  1000 π rad/sec Angular speed = 2π   = 3  3 

(b) Radius of disc =

12 cm 1m ⋅ = 0.06 m 2 100 cm

s rθ = = r (angular speed ) t t  1000  = 0.06 π  ≈ 62.83 m/sec  3 

Speed =

s (14π ) 3 14π = = feet per second t 1 second 3

14π feet 3600 seconds 1 mile × × 3 seconds 1 hour 5280 feet ≈ 10 miles per hour

19.5 in. = 9.75 in. 2

Speed =

s (2π ) 3 feet = = 4π and the arc length of the r 2 12 feet

tire in feet is:

113. (a) Revolutions = 1050 rev/min

(b) Radius of motorcycle wheel =

411

116. (a) Arc length of larger sprocket in feet:

112. (a) Revolutions = 4800 rev/min

(b) Radius of saw blade =

Angles and Their Measure

  180  117. False, 1 radian =  ≈ 57.3°, so one radian is much  π 

larger than one degree. 118. False, −1260° is coterminal with 180°, and therefore lies on the negative x-axis. 119. True:

2π π π 8π + 3π + π + + = = π = 180° 3 4 12 12

120. Let A be the area of a circular sector of radius r and central angle θ . Then A

πr2 121. A =

θ 2π

 A=

1 2 r θ. 2

1 2 1 π 50 r θ = (10)2 ⋅ = π square meters 2 2 3 3

412

Chapter 5

Trigonometric Functions

12 . 15 1 1  12  Hence, A = r 2θ = 152   = 90 ft 2 . 2 2  15 

122. Because s = rθ , θ =

128. The graph of g is a vertical shift four units downward of f ( x ) = x 3 . y 2 1

1 123. A = r 2θ , s = rθ 2

−4 −3 −2

−2 −3

1 (a) θ = 0.8  A = r 2 (0.8) = 0.4r 2 Domain: r > 0 2 s = rθ = r (0.8) Domain: r > 0 8

−6

129. The graph of g is a reflection in the x-axis and a vertical shift two units upward of f ( x ) = x 3 .

y 0

The area function changes more rapidly for r > 1 because it is quadratic and the arc length function is linear.

3 1 −4 −3 −2

1 102 θ = 50θ Domain: 0 < θ < 2π 2 s = rθ = 10θ Domain: 0 < θ < 2π

( )

(b) r = 10  A =

−2 −3

130. The graph of g is a horizontal shift three units to the left and a reflection in the x-axis of f ( x ) = x 3 .

320

s 0

4 2π

3 2

124. Angles B and C are conterminal with angle A because the initial and terminal sides are the same.

1 −5 −4 −3

125. Answers will vary.

−2

126. Answers will vary.

−4

−3

127. The graph of g is a horizontal shift one unit to the right of f ( x ) = x 3 .

131. The graph of g is a horizontal shift one unit to the left and a vertical shift three units downward of f ( x ) = x 3 .

1 −4 −3 −2

−4 −3 −2

−2 −4 −5

Section 5.2

Right Triangle Trigonometry

413

132. The graph of g is a horizontal shift five units to the right and a vertical shift one unit upward of y = x 3 . y 4 3 2 1 −2

−2 −3 −4

Section 5.2 Right Triangle Trigonometry 1.

(a) (b) (c) (d) (e) (f)

iii vi ii v i iv

8. 13

b 2

b = 13 − 52 = 169 − 25 = 12

hypotenuse, opposite, adjacent

elevation, depression 13

sin θ =

opp 5 = hyp 13

cosθ =

adj 12 = hyp 13

tan θ =

opp 5 = adj 12

cscθ =

hyp 13 = opp 5

secθ =

hyp 13 = adj 12

cot θ =

adj 12 = opp 5

Figure for Exercises 4–6 4.

The side opposite θ has length 5.

The side adjacent to θ has length 12.

The hypotenuse has length 13.

θ 2

adj = 41 − 9 = 1600 = 40 opp 9 = sin θ = hyp 41 adj 40 = cosθ = hyp 41 opp 9 = adj 40 adj 40 = cot θ = opp 9 hyp 41 = sec θ = adj 40 hyp 41 = csc θ = opp 9 tan θ =

hyp = 82 + 152 = 17 opp 8 sin θ = = hyp 17 adj 15 cosθ = = hyp 17 opp 8 = tan θ = adj 15 hyp 17 = cscθ = opp 8 hyp 17 = sec θ = adj 15 adj 15 = cot θ = opp 8

414

Chapter 5

Trigonometric Functions

10.

12. c

θ 2

C = 18 + 12 = 468 = 6 13 3 13 sin θ = = 13 6 13

adj =

152 − 62 =

sin θ =

opp 6 2 = = hyp 15 5

cos θ =

adj 3 21 = = hyp 15

21 5

tan θ =

opp 6 = = adj 3 21

2 2 21 = 21 21

cscθ =

hyp 5 = opp 2

secθ =

hyp = adj

5 5 21 = 21 21

cot θ =

adj = opp

21 2

cosθ =

6 13 18 3 tan θ = = 12 2 13 csc θ = 3

2 13 13

13 sec θ = 2 2 cot θ = 3 11. 10

θ 8

2.5 θ 2

opp = 2.52 − 2 2 = 1.5

opp 6 3 = = hyp 10 5 adj 8 4 cosθ = = = hyp 10 5 opp 6 3 tan θ = = = adj 8 4 hyp 10 5 csc θ = = = opp 6 3 hyp 10 5 sec θ = = = adj 8 4 adj 8 4 cot θ = = = opp 6 3

opp 1.5 3 = = hyp 2.5 5 adj 2 4 cosθ = = = hyp 2.5 5 opp 1.5 3 tan θ = = = adj 2 4 hyp 2.5 5 csc θ = = = opp 1.5 3 hyp 2.5 5 sec θ = = = adj 2 4 adj 2 4 cot θ = = = opp 1.5 3 sin θ =

189 = 3 21

opp = 10 2 − 8 2 = 6

sin θ =

θ 12

adj =

102 − 42 =

sin θ =

opp 4 2 = = hyp 10 5

cos θ =

adj 2 21 = = hyp 10

21 5

tan θ =

opp 4 = = adj 2 21

2 2 2 = 21 21

cscθ =

hyp 10 5 = = opp 4 2

secθ =

hyp 10 5 21 = = adj 21 2 21

cot θ =

adj 2 21 = = opp 4

84 = 2 21

21 2

The function values are the same since the triangles are similar and the corresponding sides are proportional.

Section 5.2 13.

415

15. 6

Given: sin θ =

5 opp = 6 hyp

Given: sec θ = 4 =

52 + ( adj) = 6 2 2

opp = 15

adj 11 = cosθ = hyp 6

sin θ =

opp 15 = hyp 4

tan θ =

opp 5 5 11 = = adj 11 11

cosθ =

adj 1 = hyp 4

cot θ =

adj 11 = opp 5

tan θ =

opp = 15 adj

cot θ =

adj 1 15 = = opp 15 15

csc θ =

hyp 4 4 15 = = opp 15 15

hyp 6 6 11 = = adj 11 11 hyp 6 = csc θ = opp 5

sec θ =

4 hyp = 1 adj

( opp ) + 12 = 42

adj = 11

14.

Right Triangle Trigonometry

16.

Given: cot θ = 5 =

5 adj = 1 opp

hyp = 52 + 12 = 26 opp 1 26 sin θ = = = hyp 26 26 adj 5 5 26 = = hyp 26 26 opp 1 tan θ = = adj 5

2 10

θ 3

Given: cosθ =

cosθ =

csc θ =

hyp 26 = = 26 opp 1

sec θ =

hyp 26 = adj 5

3 adj = 7 hyp

opp = 72 − 32 = 40 = 2 10 sin θ =

opp 2 10 = hyp 7

tan θ =

opp 2 10 = adj 3

csc θ =

hyp 7 7 10 = = opp 2 10 20

sec θ =

hyp 7 = adj 3

cot θ =

adj 3 3 10 = = opp 2 10 20

416

Chapter 5

Trigonometric Functions

17.

19. 13

Given: cot θ =

3 adj = 2 opp

32 + 2 2 = hyp 2

Given: tan θ = 3 =

3 opp = 1 adj

3 + 1 = ( hyp ) 2

13 = hyp 2 13 = hyp

sin θ =

hyp = 10

adj 3 3 13 = = hyp 13 13 opp 2 = tan θ = adj 3

opp 3 10 = hyp 10

sin θ =

cosθ =

adj 10 = hyp 10 hyp sec θ = = 10 adj

cosθ =

18.

cot θ =

adj 1 = opp 3

csc θ =

hyp 10 = opp 3

opp 2 2 13 = = hyp 13 13

sec θ =

hyp 13 = adj 3

csc θ =

hyp 13 = opp 2

20. 8

Given: sin θ =

273

Given: csc θ =

17 hyp = 4 adj

adj = 82 − 32 = 55 adj 55 cosθ = = hyp 8 opp 3 3 55 tan θ = = = adj 55 55 1 8 csc θ = = sinθ 3 1 8 8 55 sec θ = = = cosθ 55 55

adj = 172 − 42 = 273 opp 4 sin θ = = hyp 17 cosθ =

adj 273 = hyp 17

tan θ =

opp = adj

sec θ =

1 17 17 273 = = cosθ 273 273

cot θ =

1 = tanθ

4 273

273 4

3 opp = 8 hyp

4 273 273

cot θ =

Function

1 55 = tan θ 3

θ (deg) θ (rad)

21. sin

30°

22. cos

60°

23. tan

60°

24. sec

45°

3 3

π 4

Function Value

1 2 1 2 3 2

Section 5.2

θ (deg) θ (rad)

Function 25. csc

45° 60°

27. cos

30°

28. sin

45°

29. cot

45°

30. tan

30°

31. (a)

sin 10° ≈ 0.1736

(b)

cos 80° ≈ 0.1736

32. (a)

3 3

3 2

2 2

3 3

(b)

34. (a)

(b)

sec 42° 12′ = sec 42.2° =

(

)

30 ° sin 48 + 607 + 3600

(

)

10 ° = sec 56° 10′′ = sec 56 + 3600

(b)

36. (a)

(b)

tan

π 8

1 cosθ

42. cot θ =

1 tan θ

43. tan θ =

sin θ cosθ

44. cot θ =

cosθ sin θ

49. tan ( 90° − θ ) = cot θ

≈ 1.3430

tan (π 16 )

50. cot ( 90° − θ ) = tanθ 51. sec ( 90° − θ ) = cscθ

(

)

10 ° cos 56 + 3600

35. Make sure that your calculator is in radian mode.

41. sec θ =

48. cos ( 90° − θ ) = sinθ

cos 8° 50′ 2′′ ≈ 0.9881

cot

1 sin θ

47. sin ( 90° − θ ) = cosθ

1 ≈ 1.3499 cos 42.2°

≈ 1.7884

(a)

40. csc θ =

46. 1 + tan 2 θ = sec2 θ

tan 18.5° ≈ 0.3346

csc 48° 7′ 30" =

1 cot θ

52. csc ( 90° − θ ) = sec θ 3 1 , cos60° = 2 2 sin 60° tan 60° = = 3 cos60°

53. sin 60° =

(a)

≈ 5.0273 (b)

sin30° = cos60° =

1 2

(c)

cos30° = sin60° =

3 2

(d)

cot 60° =

≈ 0.4142

sec (1.54 ) =

1 ≈ 32.4765 cos (1.54 )

cos (1.25 ) ≈ 0.3153

37. sin θ =

1 csc θ

38. cos θ =

1 sec θ

417

45. sin 2 θ + cos2 θ = 1

1 ≈ 0.3346 (b) cot 71.5° = tan 71.5° 33. (a)

39. tan θ =

26. cot

Function Value

Right Triangle Trigonometry

cos60° 1 3 = = sin 60° 3 3

418

Chapter 5

Trigonometric Functions 58. tan β = 3 ( β lies in Quadrant I or III.)

1 3 , tan30° = 2 3 1 csc30° = =2 sin 30°

54. sin30° =

(a) (b) (c)

(d)

3 cot 60° = tan ( 90° − 60°) = tan30° = 3 (1 2) = 3 = 3 sin 30° cos30° = = tan 30° 2 2 3 3 3

(

cot 30° =

cot β =

(b)

sec 2 β = 1 + tan 2 β cos β =

)

1 3 3 3 = = = 3 tan 30° 3 3

= =

3 2 4 1 1 sin θ = = csc θ 3

55. cscθ = 3, secθ =

(a) (b)

cosθ =

1 2 2 = secθ 3

(c)

tan θ =

sin θ 13 2 = = cosθ 4 2 2 3

(d)

sec ( 90° − θ ) = cscθ = 3

(

)

1 1 + tan 2 β 1 1+9 1 10 10 10

tan ( 90° − β ) = cot β =

(d)

csc β = 1 + cot 2 β = 1 +

1 1 6 = = tan θ 2 6 12

cot θ =

(c)

cot 90  − θ = tan θ = 2 6

(d)

1 2 6 sin θ = tan θ cosθ = 2 6   = 5 5

)

( )

60.

csc 2θ sin 2 θ = 1

61. csc θ tan θ =

1 sin θ 1 ⋅ = = sec θ sin θ cosθ cosθ

62. cot θ sin θ =

cosθ sin θ = cosθ sin θ

63.

57. cot α = 4

(1 + cosθ )(1 − cosθ ) = 1 − cos2 θ

(

(b) 1 + cot α = csc α 2

= sin θ 2

64.

1 + ( 4) = csc 2 α 2

( cscθ + cot θ )( cscθ − cot θ ) = csc2 θ − cot 2 θ =1

17 = csc α 2

17 = cscα

65.

secθ − cosθ secθ cosθ = − secθ secθ secθ

=1−

1 1 +   = sec 2 α  4 17 = sec 2 α 16 17 = sec α 4 (d) tan (90° − α ) = cot α = 4

)

= sin 2 θ + cos2 θ − cos2 θ

1 1 = (a) tan α = cot α 4

1 10 = 9 3

 1  2  2  sin θ = 1  sin θ  1 =1

(b)

1 3

(c)

 1  59. tan θ cot θ = tan θ   =1  tan θ 

56. sec θ = 5, tan θ = 2 6 1 1 (a) cosθ = = sec θ 5

(

1 1 = tan β 3

(a)

cos θ

(1 cosθ )

= 1 − cos 2 θ = sin 2 θ

66.

tan θ + cot θ tan θ cot θ = + tan θ tan θ tan θ cot θ =1+ (1 cot θ )

= 1 + cot 2 θ = csc 2 θ

Section 5.2

67. (a)

(b)

68. (a)

(b)

sin θ =

1 π  θ = 30° = 2 6

csc θ = 2  θ = 30° =

cosθ =

69. (a)

secθ = 2  θ = 60° =

(b)

cot θ = 1  θ = 45° =

70. (a)

(b)

π 6

π 4

π 78. (a)

(b)

1 π  θ = 30° = 2 6

cos θ =

2 π  θ = 45° = 2 4

72. (a)

cot θ =

3 3 3 π = 3  θ = 60° = tan θ = 3 3

secθ = 2 1 2

y 3  y = 105tan 30° = 105 ⋅ = 35 3 105 3 105 105 105 210 r = = = = 70 3 cos30° = r cos30° 3 2 3

x 3 15 3 74. cos30° =  x = 15cos30° = 15 ⋅ = 15 2 2 y  1  15  y = 15sin 30° = 15   = sin 30° = 15 2 2 x 1  x = 16 cos60° = 16   = 8 16 2

sin 60° =

 3 y  y = 16sin60° = 16  =8 3  2  16  

76. cot 60° =

x 1 38 3  x = 38cot 60° = 38 ⋅ = 38 3 3

sin60° =

1500 ft

1500 3000 1 sin θ = 2 θ = 30° sin θ =

80.

2 π  θ = 45° = 2 4

73. tan 30° =

75. cos60° =

50 2 25 2 = ft sec rate down the zip line 6 3 50 25 = ft sec vertical rate 6 3

3000 ft

(b)

cosθ =

50 = 1  θ = 45° 50

L2 = 502 + 502 = 2 ⋅ 502  L = 50 2 feet

3  θ = 60° =

(b)

(c)

tan θ =

79.

tan θ =

71. (a)

sin 35.4° =

≈ 173.8 feet per minute

2 3 π  θ = 60° = cscθ = 3 3 sin θ =

419

x 896.5 x = 896.5sin 35.4° ≈ 519.3 feet (b) 1693.5 − 519.3 = 1174.2 feet above sea level 896.5 (c) minutes to reach top 300 519.3 Vertical rate = (896.5 300 )

77. (a)

2 π  θ = 45° = 2 4

tan θ = 1  θ = 45° =

Right Triangle Trigonometry

opp adj w tan 58° = 100 w = 100 tan 58° ≈ 160 feet tan θ =

81. h 3.5° 13

9° c not drawn to scale

cot 9° =

c h

13 + c h 13 Subtracting, = cot 3.5° − cot 9° h cot 3.5° =

13 cot 3.5° − cot 9° 13 ≈ ≈ 1.295 ≈ 1.3 miles. 16.3499 − 6.3138

38 38 38 76 3 r = = = sin60° 3 r 3 2

420

Chapter 5

Trigonometric Functions

82.

(x1, y1)

84. (a)

30°

85°

y sin 30° = 1 56 1 y1 = ( sin30° )( 56 ) =   ( 56 ) = 28 2 x1 cos30° = 56  3 x1 = cos30° ( 56 ) =  56 = 28 3  2  ( )  

( x1, y1 ) = ( 28,

3, 28

)

sin85° =

(c)

h = 20 sin 85° ≈ 19.9 meters

(d) As the breeze becomes stronger and the angle the balloon makes with the ground decreases, the side of the triangle labeled h will decrease in height. (e)

(x2, y2)

Angle, θ

80°

70°

60°

50°

Height

19.7

18.8

17.3

15.3

Angle, θ

40°

30°

20°

10°

Height

12.9

10.0

6.8

3.5

(f ) As the angle the balloon makes with the ground approaches 0°, the height h of the balloon approaches 0 meters.

60°

sin 60° =

h 20

(b)

y2 56

 3 y2 = sin60° ( 56 ) =  56 = 28 3  2  ( )   x cos60° = 2 56 1 x2 = ( cos60° )( 56 ) =   ( 56 ) = 28 2

( x2 , y2 ) = ( 28, 28 3 ) 83. (a)

85. True.

sin 60° csc 60° = sin 60°

1 =1 sin 60°

86. True.

sec30° =

h 6 16

θ 5

6 h (b) tan θ = and tan θ = 5 21 6 h Thus, = . 5 21 6 ( 21) = 25.2 feet (c) h = 5

2 3 = csc60° 3

87. False.

sin 45° + cos 45° =

2 2 + = 2 ≠1 2 2

88. True. cot 2 10° − csc 2 10° = −1 cot 2 10° − (1 + cot 2 10°) = −1 −1 = −1

Section 5.2 89. False.

95.

sin 60° = sin 30°

3 2 12

Right Triangle Trigonometry

f ( x ) = −e3 x x

−1

f ( x ) − 0.05

3 ≠ sin 2°

421

− 20.09 − 403.43

−1

90. False.

tan (5°)  = tan 2 5°   2

−4 −3 −2 −1 −1

tan 25° = ( tan 5°)( tan 5°)

−3

0.4663 ≠ 0.0077

−5

cos θ

−7

20°

40°

60°

Horizontal asymptote: y = 0

80°

0.9397 0.7660 0.5000 0.1736

96.

f ( x ) = 2 + ex x

f ( x)

cosθ = sin ( 90° − θ ) θ and 90° − θ are complementary angles.

(b) side y

0° 0

20° 0.3420

40° 0.6428

60° 0.8660

80° 0.9848

cos θ

0.9397

0.7660

0.5000

0.1736

tan θ

0.3640

0.8391

1.7321

5.6713

sin θ

(b) Sine and tangent are increasing; cosine is decreasing. sin θ (c) In each case, tan θ = . cosθ 94.

f ( x)

2.05

22.09

405.43

4 3 1 −4 −3 −2 −1 −1

Horizontal asymptote: y = 2 97.

f ( x ) = −4 + e3 x x

f ( x)

−1

−3.95

−3

16.09 399.43

y 3

f ( x ) = e3x x

y y (c) Because sin θ = and cos(90° − θ ) = , r r sin θ = cos (90° − θ ).

−1

(a) side y

93. (a)

−6

sin ( 90° − θ ) 1 0.9397 0.7660 0.5000 0.1736

92.

−4

91. Yes, with the Pythagorean Theorem. 0°

−2

−1

0.05

20.09 403.43

1 −4 −3 −2 −1 −1

−2 −3

−5

7 6

Horizontal asymptote: y = −4

5 4 3 2 1 −4 −3 −2 −1 −1

Horizontal asymptote: y = 0

422

Chapter 5

Trigonometric Functions

98.

f ( x ) = log3 x =

ln x ln 3

100. f ( x ) = − log3 x = − 4

−2

−4

Domain: ( 0, ∞ )

Vertical asymptote: x = 0

x-intercept: (1, 0 ) 99.

ln x ln 3

f ( x ) = log3 x + 1 =

ln x +1 ln 3

x-intercept: (1, 0 )

101. f ( x ) = log3 ( x − 4 ) =

−2

ln ( x − 4 ) ln 3

−2

−4

Domain: ( 0, ∞ )

Domain: ( 4, ∞ )

Vertical asymptote: x = 0 1  x-intercept:  , 0  3 

Vertical asymptote: x = 4 x-intercept: ( 5, 0 )

Section 5.3 Trigonometric Functions of Any Angle 1.

y r

csc θ

y x

r x

cos θ

cot θ

odd

even

Reference angle

13.

( x, y ) = ( 4, 3) r = 16 + 9 = 5 y 3 sin θ = = r 5 x 4 cos θ = = r 5 y 3 tan θ = = x 4 r 5 cscθ = = y 3 r 5 secθ = = x 4 x 4 cot θ = = y 3

10. cos θ is positive in Quadrants I and IV because x > 0. 11. sin θ = 0 at the quadrant angles θ = 0 and π . 12. cos 170° ≠ cos 10° because cos θ < 0 in Quadrant II and cos θ > 0 in Quadrant I.

Section 5.3 14.

( x, y ) = ( −8, − 15)

17.

r = 64 + 225 = 17

( x, y ) = ( − 3, − 1) 1 y =− 2 r

sin θ =

x − 3 = 2 r 3 y tan θ = = 3 x r csc θ = = −2 y cosθ =

r −2 3 = 3 x x cot θ = = 3 y

sec θ =

x = 12, y = −5

r = 122 + ( −5 ) = 13 2

18.

5 y −5 = =− 13 r 13 x 12 cosθ = = r 13 5 y −5 =− tan θ = = 12 x 12 13 r 13 =− csc θ = = 5 y −5

x = −1, y = 1

( −1) + 1 = 2

sin θ =

1 2 y = = 2 r 2

2 x −1 = =− 2 r 2 1 y = −1 tan θ = = x −1

cosθ =

csc θ =

r 2 = = 2 y 1

r 2 = =− 2 x −1 x −1 = −1 cot θ = = y 1

sec θ =

( 5 ) + (−2) = 2

sin θ =

y −2 2 = = − r 3 3

cosθ =

x = r

9 = 3

5 3

y −2 2 5 = = − x 5 5 r 3 3 = = − cscθ = y −2 2

r 13 = x 12 12 x 12 =− cot θ = = 5 y −5

5, y = − 2

tan θ =

sec θ =

x =

r =

sin θ =

16.

423

r = 3 +1 = 2

15 y sin θ = = − 17 r 8 x cosθ = = − 17 r y 15 tan θ = = x 8 17 r csc θ = = − 15 y 17 r secθ = = − 8 x x 8 cot θ = = y 15 15.

Trigonometric Functions of Any Angle

19.

secθ =

r = x

cot θ =

x 5 5 = = − y −2 2

3 3 5 = 5 5

( x, y ) = ( 7, 24 ) r = 49 + 576 = 25 y 24 = r 25 x 7 cosθ = = r 25 y 24 tan θ = = x 7 r 25 csc θ = = y 24 r 25 sec θ = = x 7 x 7 cot θ = = y 24 sin θ =

424

Chapter 5

20.

x = 8, y = 15 2

Trigonometric Functions 23.

r = 8 + 15 = 17 y 15 sin θ = = r 17 x 8 cosθ = = r 17 y 15 tan θ = = x 8 r 17 csc θ = = y 15 r 17 sec θ = = x 8 x 8 cot θ = = y 15 21.

( x, y ) = ( −

21, 2

)

(− 21) + 2 = 2

r =

sin θ =

2 y = 5 r

cos θ =

21 x = − 5 r

25 = 5

2 2 21 y = − = − 21 x 21 r 5 cscθ = = y 2 tan θ =

( x, y ) = ( 5, − 12) r = 52 + ( −12 ) = 25 + 144 = 169 = 13

secθ =

r 5 5 21 = − = − x 21 21

cot θ =

x 21 = − y 2

y 12 =− r 13 x 5 cosθ = = r 13 y 12 tan θ = = − x 5 r 13 csc θ = = − y 12

24.

sin θ =

y 10 5 = = r 26 13 x −24 −12 cosθ = = = r 26 13 y 10 5 tan θ = = =− x −24 12 r 13 cscθ = = y 5

sin θ =

r 13 =− x 12 x 12 cot θ = = − y 5

secθ =

) = 2

16 = 4

7 y = − 4 r x 3 cos θ = = − r 4

( −24 ) + (10 ) = 26 2

sin θ =

x = −24, y = 10

(− 3)2 + (−

r =

r 13 sec θ = = x 5 x 5 cot θ = = − y 12

22.

x = − 3, y = −

25.

tan θ =

y − 7 = = x −3

cscθ =

r 4 4 7 = − = − 7 y 7

secθ =

r 4 = − x 3

cot θ =

x 3 7 −3 = = y 7 − 7

7 3

( x, y ) = ( −10, − 8 ) r=

( −10 ) + (−8)2 = 164 = 2 41

sin θ =

−8 − 4 41 y = = 41 r 2 41

−10 5 41 x = =− r 2 41 41 y −8 4 tan θ = = = x −10 5 r 41 cscθ = = − y 4

cosθ =

r 41 =− x 5 x 5 cot θ = = y 4

secθ =

Section 5.3 26.

x = 3, y = −9

r = 32 + ( −9 ) = 90 = 3 10 2

sin θ =

−9 3 10 y = =− 10 r 3 10

x 3 10 = = r 3 10 10 y −9 = −3 tan θ = = x 3

cosθ =

csc θ =

r 10 =− y 3

r = 10 x x 1 cot θ = = − y 3

sec θ =

27. sin θ < 0  θ lies in Quadrant III or Quadrant IV. cos θ < 0  θ lies in Quadrant II or Quadrant III. sin θ < 0 and cos θ < 0  θ lies in Quadrant III. 28. sec θ > 0 and cotθ < 0 r x > 0 and < 0 x y θ lies in Quadrant IV. 29. cot θ > 0  θ lies in Quadrant I or Quadrant III. cos θ > 0  θ lies in Quadrant I or Quadrant IV. cot θ > 0 and cos θ > 0  θ lies in Quadrant I. 30. tan θ > 0 and csc θ < 0 y r > 0 and < 0 x y lies in Quadrant III. θ

y 3 31. sin θ = =  x 2 = 25 − 9 = 16 r 5 θ in Quadrant II  x = −4 y 3 sin θ = = r 5 4 x cosθ = = − 5 r 3 y tan θ = = − 4 x r 5 csc θ = = y 3 5 r sec θ = = − 4 x 4 x cot θ = = − y 3

Trigonometric Functions of Any Angle

425

x −4 =  y =3 r 5 θ in Quadrant III  y = −3 y 3 sin θ = = − r 5 x 4 cosθ = = − r 5 y 3 tan θ = = x 4 5 csc θ = − 3 5 sec θ = − 4 4 cot θ = 3

32. cosθ =

33. sin θ < 0  y < 0 y −15 tan θ = =  r = 17 x 8 y 15 sin θ = = − 17 r x 8 cos θ = = r 17 17 r cscθ = = − 15 y r 17 secθ = = x 8 8 x cot θ = = − 15 y 34. csc θ =

r 4 =  x = ± 15 y 1

cot θ < 0  x = − 15 y 1 sin θ = = r 4 15 x cosθ = = − 4 r 15 y tan θ = = − 15 x csc θ = 4 sec θ = −

4 15 15

cot θ = − 15

426

Chapter 5

35. sec θ =

Trigonometric Functions 39. To find a point on the terminal side of θ , use any point

r 2 =  y2 = 4 − 1 = 3 x −1

on the line y = − x that lies in Quadrant II. ( −1, 1) is

sin θ ≥ 0  y = 3

one such point. x = −1, y = 1, r = 2

y 3 sin θ = = r 2 x 1 cosθ = = − r 2 y tan θ = = − 3 x r 2 3 cscθ = = y 3 r secθ = = −2 x x 3 cot θ = = − y 3 36. sin θ = 0 and

≤θ ≤

2 cosθ = cos π = −1 sin θ tan θ = =0 cosθ cscθ is undefined. secθ = −1 cot θ is undefined.

sin θ =

y 1 2 = = r 2 2

x 1 2 =− =− r 2 2 y tan θ = = −1 x r csc θ = = 2 y r sec θ = = − 2 x x cot θ = = −1 y cosθ =

3π θ =π 2

40. To find a point on the terminal side of θ , use any point

on the line y = 13 x that lies in Quadrant I. ( 3, 1) is one such point. x = 3, y = 1, r =

37. cot θ is undefined  θ = nπ . π 3π ≤θ ≤  θ = π , y = 0, x = − r 2 2 y sin θ = = 0 r x −r cos θ = = = −1 r r y 0 tan θ = = = 0 x x r cscθ = is undefined. y r secθ = = −1 x cot θ is undefined. 38. tan θ is undefined and π ≤ θ ≤ 2π  θ =

32 + 12 =

sin θ =

y = r

1 10 = 10 10

cosθ =

x = r

3 3 10 = 10 10

tan θ =

1 y = x 3

cscθ =

r = y

10 = 1

r 10 = x 3 x 3 cot θ = = = 3 y 1

secθ =

3π . 2

sin θ = −1 cosθ = 0 tan θ is undefined. csc θ = −1 sec θ is undefined. cot θ = 0

Section 5.3 41. To find a point on the terminal side of θ , use any point

46.

on the line y = 2 x that lies in Quadrant III. ( −1, − 2 ) is one such point. x = −1, y = −2, r = 5 2 2 5 y sin θ = = − =− 5 r 5 1 5 x =− =− 5 r 5 y −2 tan θ = = =2 x −1

cosθ =

csc θ =

5 5 r = =− 2 y −2

5 r sec θ = = =− 5 x −1 x −1 1 cot θ = = = y −2 2

42. 4 x + 3 y = 0  y = −

4 x 3

4    x, − x  Quadrant IV, x > 0 3   r = x2 +

Trigonometric Functions of Any Angle

( x, y ) = (1, 0 ) csc 0 =

47.

3π r 1 = = = −1 2 y −1

( x, y) = (1, 0) sec 0 =

49.

r 1 =  undefined y 0

( x, y ) = ( 0, − 1) csc

48.

r 1 = =1 x 1

( x, y) = ( −1, 0) cot π =

50. csc

51.

427

π 2

1 x = −  undefined y 0

1 sin

1 =1 1

θ = 120 ° θ ′ = 180 ° − 120 ° = 60 ° y

16 2 5 x = x 9 3

sin θ =

4 y ( −4 3 ) x = =− 5 r ( 5 3) x

cosθ =

3 x x = = r ( 5 3) x 5

4 y ( − 4 3) x = =− x x 3 5 3 x ( ) =−5 r csc θ = = y ( − 4 3) x 4 tan θ =

r ( 5 3) x 5 = = x x 3 x x 3 =− cot θ = = y ( −4 3 ) x 4

θ ′ = 60°

θ = 120° x

52. θ = 225 ° θ ′ = 225 ° − 180 ° = 45 ° y

sec θ =

43.

r 1 = = −1 x −1

( x, y ) = ( 0, 1) cot

45.

θ ′ = 45°

( x, y) = ( −1, 0) sec π =

44.

θ = 225°

π 2

x 0 = =0 y 1

( x, y) = ( 0, − 1) sec

53. θ = 150 ° θ ′ = 180 ° − 150 ° = 30 °

3π r 1 = =  undefined 2 x 0

θ ′ = 30°

θ = 150° x

428

Chapter 5

Trigonometric Functions

54. θ = 315 ° θ ′ = 360 ° − 315 ° = 45 °

3π 4

58. θ =

θ ′= π −

3π π = 4 4 y

θ = 315° x

θ ′ = 45°

θ′=

θ=

π 4

3π 4 x

55. θ = − 45° is coterminal with 315 °. θ ′ = 360 ° − 315 ° = 45 ° y

5π 7π is coterminal with . 6 6 7π π θ ′= −π = 6 6

59. θ = −

θ = − 45° x

θ ′ = 45°

θ′=

56. θ = − 330 ° θ ′ = − 330 ° + 360 ° = 30 °

π 6

5π θ=− 6

θ = − 330°

θ ′ = 30°

7π 6 7π π θ′ = −π = 6 6

60. θ = x

57. θ =

5π 3

θ=

θ ′ = 2π −

5π π = 3 3

θ′=

7π 6 x

π 6

θ=

5π 3

61. θ = θ′=

π 3

11π 6

θ ′ = 2π −

11π π = 6 6 y

θ=

11π 6

θ′=

π 6

Section 5.3

62. θ = −

θ ′=

π 5π is coterminal with . 3 3

Trigonometric Functions of Any Angle

429

66. θ = − 165 ° θ ′ = 180 ° − 165 ° = 15 °

3 y

θ ′ = 15°

π θ′= 3 θ=−

θ = −165° x

5π 3

11π 9π . is coterminal with 5 5 9π π = θ ′ = 2π − 5 5

67. θ = − 63. θ = 208 ° θ ′ = 208 ° − 180 ° = 28 °

θ = 208° θ ′ = 28°

θ=− x

11π 5

θ′=

64. θ = 322 ° θ ′ = 360 ° − 322 ° = 38 °

68. θ =

θ′ =

π 5

10π 4π . is coterminal with 3 3 4π π −π = 3 3 y

θ = 322° x

θ ′ = 38°

θ=

10π 3 x

π θ′= 3

65. θ = − 292 ° θ ′ = 360 ° − 292 ° = 68 ° y

69. θ = 1.8 lies in Quadrant II. θ ′ = π − 1.8 ≈ 1.342

θ = − 292°

θ ′ = 68° x

θ ′ ≈ 1.342

θ = 1.8 x

430

Chapter 5

Trigonometric Functions

70. θ = − 6.5 lies in Quadrant IV and is coterminal to

about 6.066 radians. θ ′ = 2π − 6.066 ≈ 0.217 y

θ ′ ≈ 0.217

71. θ = 225°, θ ′ = 360° − 225° = 45°, Quadrant III

2 2 2 cos 225° = − cos 45° = − 2 tan 225° = tan 45° = 1

sin 225° = − sin 45° = −

72. θ = 300°, θ ′ = 360° − 300° = 60°, Quadrant IV

3 2

1 2 tan 300° = − tan 60° = − 3

cos 300° = cos 60° =

73. θ = − 750 ° is coterminal with 330 °, Quadrant IV. θ ′ = 360 ° − 330 ° = 30 ° 1 sin ( −750°) = − sin 30° = − 2 3 cos ( −750°) = cos 30° = 2 3 tan ( −750°) = − tan 30° = − 3 74. θ = − 495 ° is coterminal with 225°, Quadrant III. θ ′ = 225 ° − 180 ° = 45 ° 2 2 2 cos ( −495°) = − cos 45° = − 2 tan ( −495°) = tan 45° = 1

sin ( −495° ) = − sin 45° = −

5π , Quadrant IV 3 5π π = θ ′ = 2π − 3 3 π 5π 3 = −sin =− sin 3 3 2 π 1 5π = cos = cos 3 3 2 π 5π = − tan = − 3 tan 3 3

75. θ =

3π π 2 = sin = 4 4 2 π 3π 2 = − cos = − cos 4 4 2 π 3π = − tan = −1 tan 4 4

sin

θ = − 6.5

sin 300° = − sin 60° = −

3π , Quadrant II 4 3π π θ ′= π − = 4 4

76. θ =

5π 7π is coterminal with , Quadrant III. 6 6 7π π θ′ = −π = 6 6 π 1  5π  sin  −  = − sin = − 6 2  6 

77. θ = −

π 3  5π  cos  −  = − cos = − 6 2  6  π  5π  tan  −  = tan = 6 6  

3 3

4π 2π is coterminal with , Quadrant II. 3 3 2π π θ ′= π − = 3 3

78. θ = −

π 3  4π  sin  − =  = sin 3 3 2   π 1  4π  cos  −  = − cos = − 3 3 2   π  4π  tan  −  = − tan = − 3 3  3  11π 3π is coterminal with , Quadrant II. 4 4 3π π = θ ′= π − 4 4 11π 2 π = sin = sin 4 4 2 11π 2 π = − cos = − cos 4 4 2 11π π = − tan = −1 tan 4 4

79. θ =

Section 5.3

10π 4π is coterminal with , Quadrant III. 3 3 4π π θ ′= −π = 3 3

80. θ =

84.

cot θ = −3 1 + ( −3) = csc 2 θ 2

10 = csc 2 θ csc θ > 0 in Quadrant II.

10 = csc θ 1 csc θ = sin θ sin θ =

17π 7π , Quadrant III. is coterminal with 6 6 7π π −π = θ ′= 6 6 1  17π  π  sin  −  = −sin   = − 2  6  6

81. θ = −

85.

π 3  20π  sin  − =−  = −sin 3 2  3  π 1  20π  cos  −  = − cos = − 3 2  3 

cscθ = −2 cot 2 θ = csc 2 θ − 1 cot 2 θ = ( −2 ) − 1 2

cot 2 θ = 3 cot θ < 0 in Quadrant IV.

cot θ = − 3

3  17π  π  tan  −  = tan   =  6  6 3 −20π 4π is coterminal with , Quadrant III. 82. θ = 3 3 4π π θ ′= −π = 3 3

1 1 10 = = csc θ 10 10

1 + cot 2 θ = csc 2 θ

3  17π  π  cos  −  = − cos   = − 2  6  6

86. cosθ =

5 8

1 1  sec θ = sec θ cosθ 1 8 = sec θ = 58 5

cosθ =

87.

secθ = −

9 4

1 + tan 2 θ = sec 2 θ

π  20π  tan  −  = tan = 3 3 3   sin θ = −

431

1 + cot 2 θ = csc 2 θ

10π π 3 sin = −sin =− 3 3 2 10π 1 π cos = − cos = − 3 3 2 10π π tan = tan = 3 3 3

83.

Trigonometric Functions of Any Angle

tan 2 θ = sec 2 θ − 1 2

 9 tan 2 θ =  −  − 1  4 65 2 tan θ = 16

3 5

sin 2 θ + cos2 θ = 1 cos2 θ = 1 − sin 2 θ

tan θ > 0 in Quadrant III. 2

 3 cos2 θ = 1 −  −   5 9 cos2 θ = 1 − 25 16 cos2 θ = 25 cos θ > 0 in Quadrant IV. 4 cosθ = 5

tan θ =

88.

65 4

tan θ = −

5 4  cot θ = − 4 5

1 + cot 2 θ = csc 2 θ 1+

16 = csc 2 θ 25

csc θ = ±

41 5

Quadrant IV  csc θ = −

41 5

432

Chapter 5

89. sin θ =

Trigonometric Functions

2 and cosθ > 0  θ is in Quadrant I. 5

cosθ = − 1 − sin 2 θ = 1 − tan θ =

4 21 = 25 5

sin θ 25 2 2 21 = = = cos θ 21 21 5 21

1 5 = sin θ 2 1 5 5 21 = = secθ = cos θ 21 21 cscθ =

cot θ =

1 21 = tan θ 2

90. cosθ = −

3 and sin θ < 0  θ is in Quadrant III. 7

sin θ = − 1 − cos 2 θ = − 1 −

9 − 40 −2 10 = = 49 7 7

3 and tan θ < 0  θ is in Quadrant IV. 2 1 2 =− sin θ = csc θ 3 4 5 cosθ = 1 − sin 2 θ = 1 − = 9 3

93. cscθ = −

sec θ =

1 3 3 5 = = cosθ 5 5

tan θ =

−2 3 −2 sin θ 2 5 = = =− cosθ 5 5 3 5

cot θ =

1 5 =− tan θ 2

4 and cot θ > 0  θ is in Quadrant III. 3 1 3 cosθ = =− sec θ 4

94. secθ = −

9 7 =− 16 4

tan θ =

sin θ −2 10 7 2 10 = = cosθ −3 7 3

sin θ = − 1 − cos2 θ = − 1 −

cot θ =

1 3 3 10 = = tan θ 2 10 20

csc θ =

1 4 4 7 =− =− sin θ 7 7

1 7 10 −7 = =− sin θ 2 10 20 1 7 secθ = =− cosθ 3

tan θ =

sin θ − 7 4 7 = = cosθ 3 −3 4

cot θ =

1 3 3 7 = = tan θ 7 7

cscθ =

91. tan θ = − 4 and cos θ < 0  θ is in Quadrant II.

sec θ = − 1 + tan 2 θ = − 1 + 16 = − 17 cosθ =

1 −1 17 = =− sec θ 17 17

 17  4 17 = sin θ = tan θ cosθ = ( −4 )  −  17  17  

1 17 17 = = sin θ 4 17 4 1 1 =− cot θ = tan θ 4

csc θ =

95. sin105 ° ≈ 0.9659 96. sec 235° =

1 ≈ −1.7434 cos235°

97. cos ( −110°) ≈ −0.3420 98. sin( − 220°) ≈ 0.6428 99. tan

2π ≈ 0.8391 9

100. tan

11π ≈ 0.8391 9

92. cot θ = 5 and sin θ > 0  θ is in Quadrant I. 1 1 = tan θ = cot θ 5 1 26 secθ = 1 + tan 2 θ = 1 + = 25 5

1  8π  101. csc  − =  9  sin  − 8π   9    ≈ −2.9238

1 5 5 26 = = secθ 26 26

 −15π  102. cos  ≈ − 0.9749  14 

cosθ =

26  1   5 26  sin θ = tan θ cosθ =     =  5 26 26    cscθ =

1 26 = = 26 sin θ 26

Section 5.3 103. f (t ) = cos t

Trigonometric Functions of Any Angle

108. (a)

(a) f (3) = cos 3 ≈ − 0.9900 (b) f (π ) = cos π = −1 104. f (t ) = sin t

(a) f ( 2) = sin 2 ≈ 0.9093

(b)

(b) f (0) = sin 0 = 0 105. f (t ) = tan t

(a) f (0.3) = tan 0.3 ≈ 0.3093

109. (a)

 π  π (b) f  −  = tan  −  = −1  4  4

106. f (t ) = cos t

(a) f (0.1) = cos 0.1 ≈ 0.9950 1  2π   2π  (b) f  −  = cos −  = − 2  3   3 

1 π  reference angle is 30° or and θ 6 2 is in Quadrant I or Quadrant II.

107. (a) sin θ =

(b)

433

π 2  reference angle is 45° or 4 2 and θ is in Quadrant I or IV. Values in degrees: 45°, 315° π 7π Values in radians: , 4 4 cosθ =

π 2  reference angle is 45° or and 2 4 θ is in Quadrant II or III. Values in degrees: 135°, 225° 3π 5π Values in radians: , 4 4 cosθ = −

π 2 3  reference angle is 60° or and 3 3 θ is in Quadrant I or Quadrant II. Values in degrees: 60°, 120° π 2π Values in radians: , 3 3 csc θ =

cot θ = − 1  reference angle is 45° or

π 4

and θ

is in Quadrant II or Quadrant IV. Values in degrees: 135°, 315° 3π 7π , Values in radians: 4 4

Values in degrees: 30°, 150° π 5π Values in radians: , 6 6 1 π (b) sin θ = −  reference angle is 30° or and θ 6 2 is in Quadrant III or Quadrant IV. Values in degrees: 210°, 330° 7π 11π , Values in radians: 6 6 110. (a)

π −1 3 = −  reference angle is 30° or , and θ is in Quadrant II or Quadrant IV. 6 3 3 Values in degrees: 150°, 330° cot θ = −

3  tan θ =

5π 11π , 6 6 π 1 (b) cscθ = 2  sin θ =  reference angle is or 30 °, and θ is in Quadrant I or Quadrant II. 2 6 Values in degrees: 30°, 150° π 5π Values in radians: , 6 6 Values in radians:

434

Chapter 5

111. (a)

(b)

Trigonometric Functions

π 2 3 3  cos θ = −  reference angle is or 30 °, and θ is in Quadrant II or Quadrant III. 3 2 6 Values in degrees: 150°, 210° 5π 7π , Values in radians: 6 6 secθ = −

π 3  reference angle is or 30 °, and θ is in Quadrant II or Quadrant IV. 6 3 Values in degrees: 150°, 330° tan θ = −

Values in radians: 112. (a)

5π 11π , 6 6

tan θ = 1  reference angle is 45° or

π 4

, and θ is in Quadrant II or Quadrant III.

Values in degrees: 45°, 225° Values in radians: (b)

(b)

1 = 2 Values in degrees: 45°, 315° secθ =

2  cosθ =

Values in radians: 113. (a)

π 5π π 2  reference angle is 45° or , and θ is in Quadrant I or Quadrant IV. 4 2

π 7π 4

−1

121. f (30°) = sin 30° =

− 0.4

1 2

g (30°) = cos 30° =

114. (a) 0.7 (b) − 0.8

3 2 1 3 1+ 3 + = 2 2 2

115. (a) 0.25 and 2.89 (b) 1.82 and 4.46

(a)

sin 30° + cos30° =

116. (a) 3.99 and 5.14 (b) 0.72 and 5.56

(b)

cos30° − sin 30° =

(c)

 3 3  cos30°  =  2  = 4  

(d)

3  1  3  sin 30° cos30° =     =  2 2 4   

(e)

sin ( 2 ⋅ 30° ) = sin 60° =

3 2

(f)

cos ( −30° ) = cos30° =

3 2

117. (a)

(b)

118. (a)

(b)

119. (a)

(b)

120. (a)

(b)

sin ( − t ) = − sin t = − csc( − t ) = −

1 3

1 1 = − = −3 sin t 13

3 4 1 1 4 = − = − sec( − t ) = cos t 34 3 cos( − t ) = cos t = −

cos t = cos( − t ) = − sec( − t ) =

1 5

1 1 = − = −5 cos ( − t ) 15

sin t = − sin ( − t ) = − csc t = −

3 −1 2

3 8

1 1 8 = − = − sin ( − t ) 38 3

Section 5.3

3 2 1 g (60°) = cos 60° = 2

122. f (60°) = sin 60° =

(a)

sin 60° + cos 60° =

3 1 3 +1 + = 2 2 2

(b)

cos60° − sin 60° =

1 3 1− 3 − = 2 2 2

(c)

Trigonometric Functions of Any Angle

(d)

 − 2  2  1 =− sin 315° cos315° =   2   2  2   

(e)

sin ( 2 ⋅ 315°) = sin 630° = sin 270° = −1

(f)

cos ( −315° ) = cos 45° =

2 2

124. f ( 225°) = sin 225° = − sin 45° = −

2 2

2 1 1 cos60° =   = 2 4  

g ( 225°) = cos 225° = − cos 45° = −

2 2

(d)

 3  1  3 sin 60° cos60° =  =  2   2  4  

(a)

sin 225° + cos 225° = −

(e)

sin ( 2 ⋅ 60° ) = sin120° =

(b)

(f)

1 cos ( −60° ) = cos60° = 2

cos225° − sin 225° =

(c)

− 2  1  cos225°  =  2  = 2  

(d)

 − 2  − 2  1 = sin 225° cos225° =   2   2  2   

(e)

sin ( 2 ⋅ 225°) = sin 450° = sin 90° = 1

(f)

cos ( −225° ) = − cos 45° = −

3 2

435

2 2 − 2 2 =− 2 − 2 − 2  − =0  2  2   2

2 123. f (315°) = sin 315° = − sin 45° = − 2 2 2

g (315°) = cos 315° = cos 45° =

2 2 + =0 2 2

(a)

sin 315° + cos315° = −

(b)

2 − 2  − cos315° − sin 315° = = 2 2  2 

(c)

 2 2 1 =  cos315°  =   2  2  

2 2

125. f ( −150°) = sin ( −150°) = − sin 30° = −

1 2

g ( −150°) = cos( −150°) = − cos 30° = −

3 2

(a)

sin ( −150°) + cos( −150°) = −

1  3 −1 − + −  = 2  2  2

(b)

cos( −150°) − sin ( −150°) = −

3  1 1− 3 − −  = 2 2  2

(c)

 2 3 3 cos( −150°) =  −  = 2 4  

(d)

3  1  sin ( −150°) cos( −150°) =  −  ⋅  −  =  2   2 

3 4

(e)

sin 2( −150°) = sin ( − 300°) = sin 60° =

3 2

(f)

cos −  ( −150°) = cos 150° = − cos 30° = −

3 2

436

Chapter 5

Trigonometric Functions

3 2 1 g ( − 300°) = cos( − 300°) = cos 60° = 2

126. f ( − 300°) = sin ( − 300°) = sin 60° =

(a)

sin ( − 300°) + cos( − 300°) =

3 1 1+ 3 + = 2 2 2

(b)

cos( − 300°) − sin ( − 300°) =

1  3 1 − 3 −  = 2  2  2

(c)

2 1 1 cos( − 300°) =   = 4  2

(d)

 3  1  sin ( − 300°) cos( − 300°) =     =  2  2 

(e)

sin 2( − 300°) = sin ( − 600°) = sin 60° =

(f)

cos − ( − 300°) = cos 300° = cos 60° =

3 4 3 2

1 2

7π 7π π 1 127. f   = sin   = − sin = − 6 2  6   6 

π 3  7π   7π  = − g   = cos  = − cos 6 2  6   6  (a)

sin

7π 7π 1 3 1+ 3 + cos =− − =− 6 6 2 2 2

(b)

cos

7π 7π − 3  1  1 − 3 − sin = −−  = 6 6 2 2  2

(c)

2 7π   − 3  3  = cos  =  6   2  4 

(d)

sin

(e)

7π π 3  7π  sin  2 ⋅ = sin = sin = 6  3 3 2 

(f)

π 3  −7π  cos   = − cos 6 = − 2  6 

7π 7π  1   3 3 cos =  −  − = 6 6  2   2  4

Section 5.3 5π 5π π 1 128. f   = sin   = sin = 6 2  6   6 

Trigonometric Functions of Any Angle

437

5π 5π π 3 130. f  −  = sin  −  = sin = 3 2  3   3  π 1  5π   5π  g  −  = cos −  = cos = 3 2  3   3 

π 3  5π   5π  = − g   = cos  = − cos 6 2  6   6  (a)

sin

5π 5π 1 − 3 1 − 3 + cos = + = 6 6 2 2 2

(a)

 5π   5π  sin  −  + cos −  =  3   3 

(b)

cos

5π 5π − 3 1 1+ 3 − sin = − =− 6 6 2 2 2

(b)

1  3 1 − 3  5π   5π  cos −  − sin  −  = −  = 2  2  2  3   3 

(c)

2 5π   − 3  3  cos 6  =  2  = 4    

(c)

  5π  1 1 cos − 3  =  2  = 4     

(d)

sin

5π 5π  1   − 3  3 =   cos =− 6 6  2   2  4

(d)

 5π   5π   3   1  sin  −  cos −  =    =  3   3   2   2 

(e)

5π π 3  5π  sin  2 ⋅  = sin 3 = − sin 3 = − 2 6  

(e)

  5π   π  10π  = sin  2 −   = sin  −  = sin 3  3    3 

(f)

π 3  −5π  cos   = − cos 6 = − 2 6  

(f)

  5π   π 1  5π  cos −  −   = cos  = cos = 3 2  3    3 

sin

3 −1 2

(b)

cos

(c)

4π   1  1  cos 3  =  − 2  = 4    

(d)

sin

(e)

8π 2π 3  4π  sin  2 ⋅ = sin = sin = 3  3 3 2 

(f)

4π 4π  − 3   1  3 = cos  −  = 3 3  2   2  4

π 1  − 4π  cos   = − cos 3 = − 2 3  

3 4

3 2

g ( − 270°) = cos( − 270°) = cos 90° = 0

4π 4π − 3 1 1+ 3 + cos = − =− 3 3 2 2 2 4π 4π 1 − 3 − sin = − − = 3 3 2  2 

131. f ( − 270°) = sin( − 270°) = sin 90° = 1

4π 4π π 3 129. f   = sin   = − sin = − 3 2  3   3  π 1  4π   4π  g   = cos  = − cos = − 3 2  3   3 

(a)

3 1 1+ 3 + = 2 2 2

(a)

sin ( − 270°) + cos( − 270°) = 1 + 0 = 1

(b)

cos( − 270°) − sin( − 270°) = 0 − (1) = −1

(c)

cos( − 270°) = 0 2 = 0

(d)

sin( − 270°) cos( − 270°) = (1)(0) = 0

(e)

sin 2( − 270°) = sin( −540°) = sin 180° = 0

(f)

cos−( − 270°) = cos 270° = 0

132. f (180°) = sin180° = sin 0° = 0 g (180°) = cos180° = −1

(a) (b)

sin180° + cos180° = 0 − 1 = −1

(c)

cos180° = ( −1) = 1

(d)

sin180° cos180° = 0 ( −1) = 0

(e) (f)

cos180 ° − sin180 ° = − 1 − 0 = − 1 2

sin ( 2 ⋅180°) = sin360° = sin 0° = 0 cos ( −180°) = cos 180° = −1

438

Chapter 5

Trigonometric Functions

7π 3π  7π  = sin = −1 133. f   = sin 2 2  2  7π 7π  7π  = cos = 0 g   = cos 2 2  2  (a) (b)

7π 7π + cos = −1 + 0 = −1 2 2 7π 7π − sin = 0 − ( −1) = 1 cos 2 2

 πt π  136. S = 2.7 + 0.142t + 2.2 sin  −   6 2

(a) January 2014 ( t = 1) :  π S = 2.7 + 0.142 + 2.2sin  −   3 ≈ 0.9367  94 units

sin

(b) February 2015 ( t = 14) :

 11π  S = 2.7 + 0.142 (14 ) + 2.2 sin    6  ≈ 3.588  359 units

7π   2 cos 2  = 0 = 0   7π 7π = ( −1)( 0 ) = 0 sin cos 2 2  7π  = sin 7π = sin π = 0 sin  2 ⋅ 2  

π  − 7π  cos = 0  = cos 2  2 

π  S = 2.7 + 0.142 ( 5 ) + 2.2 sin   3 ≈ 5.315  532 units

(d) June 2015 ( t = 18) :

 5π  S = 2.7 + 0.142 (18 ) + 2.2 sin    2 

5π π  5π  = sin =1 134. f   = sin 2 2  2 

π 5π  5π  = cos = 0 g   = cos 2 2  2  (a) (b)

5π 5π + cos =1+ 0 =1 2 2 5π 5π − sin = 0 − 1 = −1 cos 2 2

sin

5π   2 cos 2  = 0 = 0   5π 5π = (1)( 0 ) = 0 cos 2 2  5π  sin  2 ⋅  = sin 5π = 0 2   sin

π  −5π  cos   = cos 2 = 0  2 

≈ 7.456  746 units

137. sin θ =

(a) θ = 30 ° 6 6 d= = = 12 miles sin 30° (1 2 ) (b) θ = 90 ° 6 6 d= = = 6 miles sin 90° 1 (c) θ = 120 ° 6 d= ≈ 6.9 miles sin120° 138. y(t ) = 14 cos 6t

(a) y (0) = 14 cos(6 ⋅ 0) = 14 cos 0

7π  πt 135. T = 76.4 + 16cos −  6  6 (a) January (t = 1): T (1) = 76.4 + 1 cos( − π ) = 60.4°F (b) July (t = 7): T (7) = 76.4 + 16 cos(0) = 92.4°F π  (c) October (t = 10): T (10) = 76.4 + 16 cos  2 = 76.4°F

6 6 d= d sin θ

= 14 (1) = 0.25 foot

(b)

( ) = 14 cos(6 ⋅ 14 ) = 14 cos( 32 )

y 14

≈ 0.02 foot

()

(

)

1 cos 3 4

≈ − 0.25 foot 139. True. Since 0 < cos θ < 1 for 0 < θ < π 2, or Quadrant I, sin θ then sin θ < = tan θ . cosθ

Section 5.3 140. True. Since y < x for 0° < θ < 45°, then y x = sin θ < cosθ = . r r

Trigonometric Functions of Any Angle

439

142. True. If 90° < θ < 180°, cos θ < 0, and since

cos2 θ = 1 − sin 2 θ , then cosθ = − 1 − sin2 θ .

141. False. If 90° < θ < 180°, sinθ > 0. 143. (a)

0°

20°

40°

60°

80°

sin θ

0.3420

0.6428

0.8660

0.9848

sin (180° − θ )

0.3420

0.6428

0.8660

0.9848

(b) It appears that sinθ = sin (180° − θ ) . 144. (a)

0°

20° 0.9397

40° 0.7660

−1

− 0.9397

− 0.7660

60° 0.5

80° 0.1736

− 0.5

− 0.1736

cos θ

cos (180° − θ )

θ cos θ

cos (180° − θ )

(b) cosθ = − cos (180° − θ ) 145. (a)

147. h(t ) is an odd function. 148. As θ increases from 0° to 90 °, x decreases from 12 centimeters to 0 centimeters, and y increases from 0 centimeters to 12 centimeters. Therefore, sinθ = y 12 increases from 0 to 1, and cosθ = x 12 decreases from 1 to 0. Thus, tan θ = y x begins at 0 and increases without bound. When θ = 90°, the tangent is undefined. 149. If your classmate used a calculator and found π tan   = 0.0274224385, then the calculator was in  2

0.3

0.6

 3π  −θ  cos   2 

− 0.296

− 0.565

− sin θ

− 0.296

− 0.565

0.9

1.2

1.5

 3π  cos  −θ   2 

− 0.783

− 0.932

− 0.997

3 x = 21 x=7

− sin θ

− 0.783

− 0.932

− 0.997

152. 44 − 9 x = 61

the wrong mode, degrees instead of radians. 150. Answers will vary. 151.

9 x = −17 17 x = − ≈ −1.889 9

 3π  − θ  = − sin θ (b) cos   2  146.

3 x − 7 = 14

0.3

tan t

0.309

0.684

 π tan  t +  2 

Undef.

− 3.233

− 1.462

tan ( t + π )

0.309

0.684

tan ( t + 2π )

0.309

0.684

153.

x2 − 2x − 5 = 0 2 ± 4 + 20 =1± 6 2 x ≈ −1.449, 3.449 x=

154. 2 x 2 + x − 4 = 0 −1 ± 1 + 4(4)(2) −1 ± 33 = 4 4 x ≈ −1.686, 1.186

0.9

1.2

1.5

tan t

1.260

2.572

14.101

 π tan  t +  2 

− 0.794

− 0.389

− 0.071

tan ( t + π )

1.260

2.572

14.101

tan ( t + 2π )

1.260

2.572

14.101

The period of the tangent function is π because tan(t + π n) = tan t , n = 0, 1, 2, 3, 

440

Chapter 5

Trigonometric Functions

3 x+2 = x −1 9

155.

156.

27 = ( x − 1)( x + 2) 2

x + x − 29 = 0 −1 ± 1 + 4(29) −1 ± 117 = 2 2 x ≈ −5.908, 4.908 x=

5 x+4 = 2x x 10 x = x 2 + 4 x

x2 − 6x = 0 x ( x − 6) = 0 x = 6 ( x = 0 extraneous)

5.4 Graphs of Sine and Cosine Functions 1.

amplitude

one cycle

2π b

phase shift

The period of y = sin x is 2π .

2π The period of y = cos bx is . b

The constant d of y = sin x + d is a vertical shift of d units.

The amplitude of y = − 4.5sin x is − 4.5 = 4.5.

f ( x ) = sin x

(a) x-intercepts: ( −2π , 0), ( −π , 0), (0, 0), (π , 0), (2π , 0) (b) y- intercept: (0, 0) 3π   π π   3π   (c) Increasing on:  −2π , − , − , , , 2π  2   2 2   2   3 π π π 3 π     Decreasing on:  − , −  ,  , 2   2 2   2  3π   π  (d) Relative maxima:  − , 1 ,  , 1   2  2   π   3π  Relative minima:  − , − 1  ,  , − 1   2   2 

10.

12.

2π 2π = b 3 Amplitude: a = 2 Period:

13.

14.

2π =π 2 Amplitude: 3 = 3 Period:

x 3

2π 2π = = 6π b (1 3)

Amplitude: a = −3 = 3 15.

2 y = sin π x 3 2π Period: =2

Amplitude:

16.

2 2 = 3 3

3 πx y = cos 2 2 2π 2π Period: = =4 b π ( 2) Amplitude: a =

17.

3 2

y = −2 sin x

2π = 2π 1 Amplitude: −2 = 2

Period:

(d) Relative maxima: ( −2π , 1) , ( 0, 1) , ( 2π , 1)

y = 3sin 2 x

2π = 4π 12

y = −3sin Period:

Decreasing on: ( −2π , − π ), (0, π )

11.

x 2

Amplitude: 5 = 5

Relative minima: ( −π , − 1) , (π , − 1)

y = 5cos Period:

f ( x ) = cos x

 3π   π  π   3π  (a) x-intercepts:  − , 0  ,  − , 0  ,  , 0  ,  , 0  2 2 2 2       (b) y-intercept: ( 0, 1)

y = 2 cos3 x

18.

2x 5 2π 2π Period: = = 5π 25 b y = − cos

Amplitude: a = −1 = 1

Section 5.4

19.

1 4x y = cos 4 3 2π 3π Period: = 43 2 Amplitude:

20.

1 1 = 4 4

5 x y = cos 2 4 2π 2π Period: = = 8π b 14

Graphs of Sine and Cosine Functions

32. The graph of g is a vertical shift two units upward of the graph of f. 33.

f ( x ) = sin x Period: 2 π Amplitude: 1 g ( x ) = −4 sin x Period: 2 π Amplitude: −4 = 4 y

5 Amplitude: a = 2 21.

f ( x ) = sin x

The graph of g is a horizontal shift of f, π units to the left. 23.

24.

−2 −3 −4

34. f ( x ) = sin x x g( x ) = sin 3 0

g ( x ) = − cos 2 x The graph of g is a reflection in the x-axis of the graph of f.

sin x

f ( x ) = sin 3 x, g ( x ) = sin ( −3 x )

sin

x 3

The graph of g is a reflection of f about the y-axis (or, about the x-axis). 25.

π 2

f ( x ) = cos 2 x

− 3π 2

The graph of g is a horizontal shift to the right π units of the graph of f (a phase shift). f ( x ) = cos x, g ( x ) = cos ( x + π )

g ( x ) = sin ( x − π )

22.

3π 2

2π

−1

1 2

3 2

f ( x) = cos 2 x

g f

g ( x) = 3 + cos 2 x The graph of g is a vertical shift three units upward of the graph of f. 26.

27.

f ( x ) = cos 4 x g ( x ) = −2 + cos 4 x The graph of g is a vertical shift of two units downward of the graph of f.

f ( x) = sin x, g ( x) = 5sin ( − x)

The graph of g is a reflection in the y-axis and has five times the amplitude of f. 28.

1 f ( x ) = sin x, g( x ) = − sin x 2 The amplitude of the graph of g is one-half that of f. g is a reflection of f in the x-axis.

6π

31. The graph of g is a horizontal shift π units to the right of the graph of f.

−2

35. f ( x ) = cos π x

Period: 2 Amplitude: 1 g ( x ) = 1 + cos π x is the graph of f

shifted vertically one unit upward. y 3 2

29. The graph of g has twice the amplitude as the graph of f. 30. The period of g is one-half the period of f.

441

−2

−1

1 −1

442

Chapter 5

Trigonometric Functions

36. f ( x) = 4sin x

Period: 2π Amplitude: 4 g ( x ) = 4sin x − 1 is the graph of f

π  39. f ( x ) = sin x, g( x ) = cos  x −  2  Period: 2π Amplitude: 1 2

shifted vertically one unit downward. y

f=g −2π

2π

−2

2 − 3π 2

−π 2

π 2

3π 2

π  Conjecture: sin x = cos  x −  2  3π   40. f ( x ) = sin x, g ( x ) = cos  x +  2  

g −6

1 x 37. f ( x) = − sin 2 2 Period: 4π 1 Amplitude: 2

sin x

2 1

3π   cos  x +  2  

3π 2 −1

−1

2π 0 0

x g ( x) = 2sin 4

f=g −2π

2π

Period: 8π Amplitude: 2

−2

3π   Conjecture: sin x = cos  x +  2  

y 4 3

41. f ( x ) = cos x

1 −6 π

2π

6π

π  π  g( x ) = − sin  x −  = sin  − x = cos x  2  2

−3 −4

f=g −2π

2π

38. f ( x) = 2cos 2 x −2

Period: π

π  Conjecture: cos x = − sin  x −  2 

Amplitude: 2 g ( x) = − cos 4 x

42. f ( x ) = cos x, g ( x ) = − cos ( x − π )

Period: π2 Amplitude: 1 y 2

2 0

−1

3π 2 0

−1

cos x − cos ( x − π ) g

2π

1 1

f=g −2π

2π

−2 −2

Conjecture: cos x = − cos ( x − π )

Section 5.4 43. y = 4 sin x

Graphs of Sine and Cosine Functions 3 cos x 4

46. y =

Period: 2π

Amplitude: 4 Key points: π 3π ( 0, 0 ) ,  , 4  , (π , 0 ) ,  , − 4  , ( 2π , 0 ) 2   2 

3 4

Amplitude:

Key points: 3   3π 3  3 π      0, 4  ,  2 , 0  ,  π , − 4  ,  4 , 0  ,  2π , 4           

4 3

2 1 − 3π 2

443

−π 2

π 2

3π 2

−2π

2π

−1

44. y = 5sin x

−2

Period: 2π Amplitude: 5 Key points: π 3π ( 0, 0 ) ,  , 5  , (π , 0 ) ,  , − 5  , ( 2π , 0 ) 2    2 

47. y = cos

x 2

Period: 4π Amplitude: 1 Key points: (0, 1), (π , 0), (2π , − 1), (3π , 0), (4π , 1)

y 8

6 3

4 2 − 3π 2

45. y =

−π 2

π 2

3π 2

−6

−1

−8

−2

2π

4π

−3

1 cos x 4

48.

Period: 2π Amplitude:

1 4

y = sin

x 4

Period: 8π Amplitude: 1

Key points: 1   3π 1  1 π      0, 4  ,  2 , 0  ,  π , − 4  ,  2 , 0  ,  2π , 4           

Key points: (0, 0), ( 2π , 1), ( 4π , 0), (6π , −1), (8π , 0) y 2

1 1 −6 π

0.5

π 2 −0.5

−

π 2

2π

− 2π

2π

6π

x −2

−1

444

Chapter 5

Trigonometric Functions

π π  49. y = sin  x −  ; a = 1, b = 1, c = 4 4 

π  52. y = 3cos  x +  2 

Period: 2π

Amplitude: 1

Amplitude: 3

Shift: Set x −

π 4

= 0 and x −

π 4

= 2π

9π 4

Key points: π   3π   5π   7π   9π  , 0,  , −1  ,  , 0  , 0  ,  , 1 ,  4   4   4   4   4 

Horizontal shift of f ( x ) = 3cos x,

π 2

units to the left

Key points: π 3π ( 0, 0 ) ,  , − 3  , (π , 0 ) ,  , 3  , ( 2π , 0 ) 2   2  y

4 3

y 3

−π 2 −2

1 −π

3π 2

π 2

−3 −4

−2

π  Note: 3cos  x +  = −3sin x 2 

−3

50. y = sin ( x − π )

53. y = 1 − sin

Period: 2π

2π x 3

Period: 3

Amplitude: 1 Horizontal shift of f ( x) = sin x, π units to the right π   3π  Key points: (0, 0),  , −1 , (π , 0),  , 1 , (2π , 0) 2    2  y

Amplitude: 1 Vertical shift one unit upward and a reflection in the 2π x . x-axis of f ( x ) = sin 3 3  3  9  Key points: (0, 1),  , 0  ,  , 1 ,  , 2  , ( 3, 1) 4  2  4 

−π 2 −1

π 2

3π 2

3 1 −4 −3 −2 −1 −1

−2

−2 −3

51. y = −8 cos( x + π )

−4

Period: 2π Amplitude: 8 Key points:  π  , 0  , ( 0, 8 ) ,  , 0  , (π , − 8 )  2  2 

( −π , − 8 ) ,  −

10 8

2 −2 π

−π

−2 −4 −6 −8 − 10

2π

Section 5.4 54. y = 2cos x − 3

Graphs of Sine and Cosine Functions

Period: 2π

 x π 55. y = cos  −   2 4

Amplitude: 2

Period: 4π

Vertical shift downward three units of the graph of f ( x) = 2cos x.

Amplitude:

Key points: π   3π  (0, −1),  , − 3  ,(π , − 5),  , − 3  , (2π , −1) 2   2 

Shift:

−π

−1

2π

x π − = 2π 2 4

and

9π 2

Key points: 2   7π  π 2   3π   5π   9π 2  , 0,  ,   , ,  , 0,  , − ,  2 3 2 2 3 2          2 3

1 −2π

2 3

x π − =0 2 4

445

y 2

−4 −5

−6

−4 π

4π

−1 −2

56.

y = − 2 cos( 4π x + 1)

1 2

Period:

Amplitude: 2 Shift: 4π x + 1 = 0 x = −

and 4π x + 1 = 2π 1 4π

and

x =

2π − 1 1 1 = − 4π 2 4π

1  1 1  3 1  1 1  1  1  , 0 ,  − , 2 ,  − , 0 ,  − , − 2 Key points:  − , − 2 ,  − π π π π π 4 8 4 4 4 8 4 2 4           y 2 1

1 4

−4 −1

57. y = −2sin

2π x 3

58. y = −10 cos

πx 6

Amplitude: 2

Amplitude: 10

2π Period: =3 2π / 3

Period:

−6

−4

2π = 12 π /6

− 18

− 12

446

Chapter 5

59. y = −4 + 5cos

Trigonometric Functions 63. y = −2sin(4 x + π )

πt 12

Amplitude: 2

Amplitude: 5

Period:

2π Period: = 24 π /12

π 2 4

−π − 30

−4 − 20

π 2 64. y = −4sin  x −  3 3

2π x 60. y = 2 − 2sin 3

Amplitude: 4

Amplitude: 2

Period: 3π

2π Period: =3 2π / 3

− 12

−8 −6

π  65. y = cos  2π x −  + 1 2 

−2

2 x π 61. y = − cos  −  3 2 4

Amplitude: 1 Period: 1

2 Amplitude: 3

Period:

2π = 4π 1/ 2

−3

− 4π

3 −1

5π

πx π  66. y = 3cos  + −2 2  2 Amplitude: 3

−2

Period: 4

5 62. y = cos(6 x + π ) 2

2 6

−6

5 Amplitude: 2

Period:

2π π = 6 3

−6

67. y = 5sin(π − 2 x) + 10

Amplitude: 5 −

π 2

Period: π 20

−6

−π

π −4

Section 5.4 68. y = 5cos(π − 2 x) + 6

Graphs of Sine and Cosine Functions

447

73. f ( x) = a cos x + d

Amplitude: 5

Amplitude:

Period: π

−2 − ( −4) =1 2

Reflected in the x-axis:

a = −1 −4 = −1 cos 0 + d −π

69. y =

d = −3 y = −3 − cos x

1 sin120π t 100

Amplitude: Period:

74. y = a cos x + d

1 100

Amplitude:

1 2

Period: 2π

1 60

Reflected in x-axis, a = − 0.02

1 2

d = −4 − π 180

π 180

− 0.02

1 y = −4 − cos x 2

75. f ( x) = a sin(bx − c)

1 70. y = cos(50π t ) 100

Amplitude: Period:

Amplitude: a = 3

1 100

Since the graph is reflected about the x-axis, we have a = −3. 2π Period: =π b =2 b Phase shift: c = 0

1 25 0.04

Thus, f ( x) = −3sin2 x. − 0.06

0.06

76. − 0.04

71. f ( x) = a cos x + d 1 Amplitude: [8 − 0] = 4 2

Since f ( x ) is the graph of g( x) = 4cos x reflected about the x-axis and shifted vertically four units upward, we have a = −4 and d = 4. Thus, f ( x ) = −4 cos x + 4 = 4 − 4 cos x. 72. f ( x) = a cos x + d Amplitude:

1 [7 − (−5)] = 6 2

Since f ( x ) is the graph of g ( x) = 6 cos x reflected about the x-axis and shifted vertically one unit upward, we have a = −6 and d = 1. Thus, f ( x) = −6 cos x + 1.

y = a sin(bx − c) Amplitude: 2  a = 2 Period: 4π 2π 1 = 4π  b = 2 b Phase shift: c = 0 x y = 2sin   2

77. f ( x) = a sin(bx − c) Amplitude: a = 1 Period: 2π  b = 1 Phase shift: bx − c = 0 when x =

π 4

π π  (1)   − c = 0  c = 4 4 π  Thus, f ( x ) = sin  x −  . 4 

448

Chapter 5

Trigonometric Functions

78. y = a sin(bx − c)

82. v = 0.85sin

Amplitude: 2  a = 2

(a)

πt 3

Period: 4 2π π =4b= 2 b

4π

−2

c π Phase shift: = − 1  c = − b 2

(b)

πx π y = 2sin  +   2 2

(c)

Time for one cycle = one period =

60 = 10 cycles per min 6 (d) The period would decrease. Cycles per min =

79. y1 = sin x y2 = −

83. S = 74.50 + 43.75cos

1 2

In the interval [ −2π , 2π ], sin x = − x=−

2π = 6sec π /3

1 when 2

(a)

5π π 7π 11π , − , , . 6 6 6 6

y1 −2π

2π

120

πt

(b) Maximum sales: December (t = 12) Minimum sales: June (t = 6)

 2π t  + 10.9  84. C = 30.3 + 21.6sin  365  

−2

80.

y1 = cos x y2 =

(a)

1 2

y1 = y2 when x = −

5π π π 5π ,− , , 3 3 3 3

Period:

2π 2π = = 365 days b (2π / 365)

This is to be expected: 365 days = 1year. (b) The constant 30.3 gallons is the average daily fuel consumption. 60 (c)

y2 − 2π

2π

y1 0

−2

81. P = 100 − 20 cos

Consumption exceeds 40 gallons/day when 124 ≤ x ≤ 252. (Graph C together with y = 40. ) (Beginning of May through part of September)

8π t 3

130

85. h = 25sin 0

1.5

Period:

365

2π 3 = (8π / 3) 4

1 heartbeat 4  heatbearts/second = 80 heartbeats/minute (3/4) 3

(a)

π 15

(t − 75) + 30

135

(b) Minimum: 30 − 25 = 5 feet Maximum: 30 + 25 = 55 feet

Section 5.4

Graphs of Sine and Cosine Functions

449

86. (a) Yes, y is a function of t because for each value of t there corresponds one and only one value of y. (b) The period is approximately 2(0.375 − 0.125) = 0.5 second. The amplitude is approximately 1 (2.35 − 1.65) = 0.35 centimeter. 2 (c) One model is y = 0.35 sin 4π t + 2. (d)

0.9

Answers will vary. 87.

(a) and (c) y 1.2

0.6 0.4 0.2 − 10

10 20 30 40 50 60

The model fits the data well. (b) y = 0.5sin (0.210 x + 2.830) + 0.5 (d) Period =

2π ≈ 30 days 0.21

(e) June 21, 2017 There are 366 days in the year 2016 plus 172 days of the year 2017, so t = 538. So, y = 0.5 sin 0.210(538) + 2.830 + 0.5 ≈ 0.708 or 70.8%. 88.

(a) C (t ) = 57.1 + 27.3 sin (0.489 x − 1.903) (b)

(c)

The model fits the data well

(d) Quillayute: 57.5 °F Chicago: 57.1°F The constant term gives the average daily high temperatures. (e) Quillayute: period = Chicago: period: =

2π ≈ 11.1 months 0.566

2π ≈ 12.8 months 0.489

The period of each model is approximately 1 year. (f) Chicago has a greater temperature variability. The amplitude determines the variability.

450

Chapter 5

Trigonometric Functions

89. True. The period of f ( x) = sin x is 2π , the period of g( x) = sin( x + 2π ) is also 2π , and g is a horizontal shift of 2π to the left which is equivalent to a horizontal shift of 2π to the right. 90. True. The period is

(c)

98. (a)

1 that of y = cos x. 2

and h( x ) = [ f ( x )]2 , we make the conjecture that h( x ) is even. (b) In Exercise 97, g( x ) = sin x is odd and we saw that

y = sin ( x + π 2 ) about the x-axis.

h( x ) = sin 2 x is even. Therefore, for g ( x ) odd and

b=2

b=3 x

2π

b=1

h( x ) = [ g( x )]2 , we make the conjecture that h( x ) is even. (c) Form part (c) of 97, the product of an even function and an odd function is odd. So, we make the conjecture that h ( x ) is odd. 99. (a)

−2

The value of b affects the period of the graph. b = 2 → 2 cycles b = 3 → 3 cycles Answers will vary. Sample answer: Given y = d + a sin (bx − c), the variable a will determine the amplitude of the graph, b will be used to 2π determine the period using the formula: period = . b The variables c and d will determine any horizontal shift and vertical shift respectively. 95. The amplitude is 4 and the period 2π . Since (0, − 4) is on the graph, matches (a). 96. The period is 4π and the amplitude is 1. Since (0, 1) and (π , 0) are on the graph, matches (c).

1.0

1.0 0.1

0.001

0.01

Undef.

1.0

0.9983 0.8415

1.1

0.8

sin x approaches 1. x sin x (c) As x approaches 0, approaches 1. x As x → 0, f ( x ) =

100. (a)

x −1 1 − cos x −0.4597 x x 1 − cos x x

(b)

−0.1

−0.01

−0.05

−0.005 −0.0005

0.001

0.01

0.1

Undef. 0.0005 0.005

0.05

0.4597

−0.001

0.5

−1

−2

−0.01 −0.001

0.8415 0.9983

−1

−0.1

−1

(b)

h( x ) = cos2 x is even.

−π

(b)

x sin x x x sin x x

b = 12 → 12 cycle

h(x) = cos2 x

In Exercise 97, f ( x ) = cos x is even and we saw that h( x ) = cos2 x is even. Therefore, for f ( x ) even

y = − cos x is a reflection of the graph of

97. (a)

−2

92. True. Since sin ( x + π 2 ) = cos x, the graph of

93.

h(x) = sin x cos x

−π

2π 20π = . 3 /10 3

91. False. The amplitude is

94.

h( x ) = sin x cos x is odd.

h( x ) = sin x is even. 2

h(x) = sin2 x

−0.5

1 − cos x approaches 0. x 1 − cos x (c) As x approaches 0, approaches 0. x As x → 0, f ( x ) =

−π

−2

Section 5.5 101. (a)

Graphs of Other Trigonometric Functions

− 2π

2π

x7 7!

Next term for cosine approximation: −

x6 6!

−2

The polynomial function is a good approximation of the sine function when x is close to 0. (b)

451

− 2π

2π

2 −2 − 2π

2π

−2

− 2π

2π

The polynomial function is a good approximation of the cosine function when x is close to 0.

−2

The accuracy increased. 102. (a) For c ≠ 0, the value of c shifts the graph of f ( x ) = sin x c units to the left when c < 0 and c units to the right when c > 0.

π   π 7π  (b) Because the graph of y = − cos x +  has one cycle of the graph on the interval − , , the graph has the 4    4 4  π   π   3π   7π  key points  − , −1,  , 0 ,  , 1, and  , −1. So, the graph is equivalent to the green graph.  4  4   4   4  103.

104.

(−1, 4)

(2, 5)

2 1

4 3

−4 −3 −2 −1 −1

2 1 −4 −3 − 2 −1 −1

Slope =

4 3

(0, 1) 1

−2 3

(3, −2)

−3 −4

5 −1 =2 2−0

−2 − 4 −6 3 = =− 3+1 4 2

 180°  105. 8.5 = 8.5   ≈ 487.014°  π   180°  106. −0.48 = −0.48   ≈ −27.502°  π  107. Answers will vary. (Make a Decision)

Section 5.5 Graphs of Other Trigonometric Functions 1.

vertical

reciprocal

The tangent and cotangent functions have a period of π and a range of all real numbers.

The factor e2 x is the damping factor of the function f ( x ) = e2 x sin 2 x.

452

Chapter 5

f ( x ) = tan x

Trigonometric Functions

(a) x-intercepts: (−2π , 0), (−π , 0), (0, 0), (π , 0), (2π , 0) (b) y-intercept: (0, 0) (c) Increasing on:

y = tan

x 5

Period: 5π Two consecutive asymptotes: x = −

3π   3π π  π π   −2π , − ,  − , − ,  − , , 2   2 2  2 2   π 3π   3π   ,  ,  , 2π  2 2   2  Never decreasing

−

5π 4

−1

(d) No relative extrema (e) Vertical asymptotes: x = − 6.

3π π π 3π , − , , 2 2 2 2

3 2 1

f ( x ) = cot x

 3π   π   π   3π  (a) x-intercepts: − , 0, − , 0,  , 0,  , 0  2   2  2   2  (b) No y-intercepts (c) Decreasing on: ( −2π , − π ), (−π , 0), (0, π ), (π , 2π ) Never increasing (d) No relative extrema (e) Vertical asymptotes: x = −2π , − π , 0, π , 2π 7.

− 8π

10.

 π   3π  Relative mixima:  − , − 1 ,  , − 1  2   2  (e) Vertical asymptotes: x = ±π , ± 2π , 0

6π

Two consecutive asymptotes: x = −

−

−1

and

π 6

y 3 2 1 −π 2

f ( x ) = csc x

 3π   π  (d) Relative minima:  − , 1 ,  , 1  2  2 

2π

Period:

f ( x ) = sec x

(a) No x-intercepts (b) No y-intercepts (c) Increasing on: π  π 3π   3π      − 2 , − π  ,  −π , − 2  ,  2 , π  ,  π , 2          Decreasing on: 3π   π   π   3π    −2π , − 2  ,  − 2 , 0  ,  0, 2  ,  2 , 2π         

− 2π

y = tan3x

(a) No x-intercepts (b) y-intercept: (0, 1) (c) Increasing on: 3π   3π    π  π   −2π , − 2  ,  − 2 , − π  ,  0, 2  ,  2 , π          Decreasing on π   π   3π   3π    −π , − 2  ,  − 2 , 0  ,  π , 2  ,  2 , 2π          (d) Relative minima: ( −2π , 1), (0, 1), (π , 1) Relative minima: ( −π , − 1), (π , − 1) 3π π π 3π (e) Vertical asymptotes: x = − , − , , 2 2 2 2 8.

5π 5π and x = 2 2

11.

−π 6

π 6

π 2

y = −2 tan 2 x

Period:

π 2

Two consecutive asymptotes: 2 x = − 2x =

π 2

x=

−

π 2

x=−

π 4

0 0

π 8

−2

Section 5.5

12. y = − 4 tan

Graphs of Other Trigonometric Functions 14.

x 3

Two consecutive asymptotes:

−

y = 3cot π x

π =1 π Two consecutive asymptotes: x = 0, x = 1 Period:

Period: 3π

453

3π 4

−4

x π 3π = ±  x = ± 3 2 2

1 4

1 2

3 4

−3

2 1 −3 − 4π

2π

−2

−1

4π

−4 −8

13.

15.

1 x cot 2 2

Period:

π 12

1 y = − sec x 2

Period: 2π Two consecutive asymptotes: π π x=− , x= 2 2 x

= 2π

x =0 x=0 2 x = π  x = 2π 2

2 1 2

− 0.707

0 − 0.5

π 4

− 0.707

1 −π

3π 2

−

1 2

16.

4 2

−6

−4

−π

−

Two consecutive asymptotes:

1 sec x 4

Period: 2π Two consecutive asymptotes: π π x=− , x= 2 2

−

0.354

0.25

0.354

3 2 1 − 2π

−π

2π

454

Chapter 5

17.

y = 3 csc

Period:

Trigonometric Functions 19.

x 2

Period: 2 Shift graph of sec π x down three units. Two consecutive asymptotes:

2π = 4π 12

1 1 x=− , x= 2 2

Two consecutive asymptotes: x =0 x=0 2 x = π  x = 2π 2 x y

4.243 3

− 0.125

0.25

−1.586 −2 −1.586

−2

−1

3π 2

−3

4.243

−5

y = sec π x − 3

−4

10 8 6 4 2

20.

− 3π − 2π − π

2π 3π 4π

y = − csc

2π π = 4 2

Period:

π π Two consecutive asymptotes: x = − , x = 8 8 π π − x 0 16 16

−8 − 10

18.

y = 2 sec 4x + 2

x 3

4.828

2π Period: = 6π (1 3)

Two consecutive asymptotes: x = 0, x = 3π

3 2 1

2π

4π

−1.155

1.155

−π −1

π 4

−1

π 2

−2

21.

− 4π

−π 2

y = 2 tan

πx 4

π Period: =4 π 4 π

4π

Two consecutive asymptotes:

−2

πx

−3

π 2

πx 4

=−

π 2

, x = −2

 x=2 y

−1

−2

6 4 2

−6

−2

Section 5.5

22.

y =

1 tan π x 2

Graphs of Other Trigonometric Functions

25.

Period: 1 1 1 Two consecutive asymptotes: x = − , x = 2 2 1 4 1 − 2

−

x y

1 sec ( 2 x − π ) 2 2π Period: =π 2 y=

Two consecutive asymptotes: x = ±

1 4 1 2

3 2 1

−1

455

−

− 0.707

− 0.5

− 0.707

x y

−2

−3

3 2 1

23.

y = − csc (π − x )

−π 2

Period: 2π

π 2

Two consecutive asymptotes:

π − x = 0  x = π π − x = 2π  x = − π x y

−

26.

2π =π 2 Two consecutive asymptotes:

3π 4

−

−1

y = csc ( 2π − x )

Period:

2x − π = 0  x =

2x − π = π  x = π

6 4

x − 2π

−π

2π

−2

3π 8

− 1.414

−1

− 1.414

−4 y

−6

24.

5 4 3

y = − sec ( x + π ) Period: 2π Two consecutive asymptotes: x = ±

−

1.414

π 2

−π

−

π 2

−1 −2 −3 −4 −5

π 2

1.414

8 6

− 2π

−2

2π

−4 −6 −8

456

27.

Chapter 5

Trigonometric Functions

π  y = 2 cot  x +  2  Period: π Two consecutive asymptotes: π π x + = 0  x = − 2 2 π π x + = π  x = 2 2 π

−

29.

y = 2 csc3 x =

Period:

2 sin ( 3 x )

2π 3 10

−π

−10

−2 −π

y −10

30. −π

−2

y = − csc 4 x y =

−4

−1 sin 4 x

Period:

−6

π 2 3

1 cot ( x + π ) 4 Period: π

28. y =

−

π 2

Two consecutive asymptotes: −3

x +π =π  x = 0 x + π = 2π  x = π

x y

4 1 4

3π 4 1 4

−

π − 2 −2 −3 −4 −5

31.

π 2

3π 2

π 2

−3

5 4 3 2 1 3π − 2

π 2

y = −2sec 4 x −2 = cos 4x

Period:

π 2 4

−π 2

π 2

−4

−π 2

π 2

−4

Section 5.5

32.

1 1 sec π x = 4 4co s π x Period: 2 y=

Graphs of Other Trigonometric Functions

457

35. tan x = 1 The solutions appear to be:

x=−

7π 3π π 5π , − , , 4 4 4 4

(or in decimal form: −5.498, − 2.356, 0.785, 3.927) 2

−2

36. cot x = −1 The solutions appear to be:

−2

x = −

(or in decimal form: − 3.927, − 0.785, 2.356, 5.498)

37. sec x = 2

−2

33.

The solutions appear to be:

1 πx π  y = sec  +  3  2 2 1 = πx π  3cos  +   2 2

π 5π x = ± ,± 3 3 (or in decimal form: − 5.236, −1.047, 1.047, 5.236) 38. csc x = 2

Period: 4

The solutions appear to be: 7π 5π π 3π x=− , − , , 4 4 4 4 (or in decimal form: −5.498, − 3.927, 0.785, 2.356)

−6

5π π 3π 7π ,− , , 4 4 4 4

39. cot x = − 3 −2

The solutions appear to be:

x=−

−6

7π π 5π 11π , − , , 6 6 6 6

(or in decimal form: −3.665, − 0.524, 2.618, 5.760)

−2

40. sec x = − 2 34.

1 1 csc ( 2 x − π ) = 2 2sin ( 2 x − π )

The solutions appear to be: 5π 3π 3π 5π x=− , − , , 4 4 4 4 (or in decimal form: −3.927, − 2.356, 2.356, 3.927)

Period: π 3

−π

41. The graph of f ( x ) = sec x has y-axis symmetry. Thus, the function is even. 42.

−3 3

−π

f ( x ) = tan x tan( − x ) = − tan x Thus the function is odd and the graph of y = tan x is symmetric with the origin.

43. The function 1 sin 2 x has origin symmetry. Thus, the function is odd. f ( x ) = csc 2 x =

−3

458

Chapter 5

44.

f ( x ) = cot 2 x

Trigonometric Functions

48.

f (− x) = cot(−2 x) =

π  f ( x ) = tan  x −  2  π  f (− x ) = tan  − x −  2    π  = tan  −  x +   2      π  sin  −  x +   2    =   π  cos  −  x +   2    π  sin  x +  2  =− π  cos  x +  2  π   = − tan  x +  2  π  = − tan  x −  2 

−2π

Equivalent cos x cot x = sin x 49.

−3

3 −1

It appears that y1 = y2 .

1 + cot 2 x = csc2 x 50.

y1 = sec 2 x − 1, y2 = tan 2 x 3

− 3π 2

f ( x ) = sec ( x + π )

3π 2 −1

It appears that y1 = y2 .

= sec ( − ( x − π ) ) 1 = cos ( − ( x − π ) )

1 + tan 2 x = sec 2 x tan 2 x = sec 2 x − 1

51.

1 = cos ( x − π )

f ( x ) = x cos x As x → 0, f ( x ) → 0.

Odd function  3π  f =0  2 

= sec ( x − π ) = sec ( x + π )

Matches graph (d).

Thus, the function is even. y1 = sin x sec x, y2 = tan x 4

−2π

y1 = 1 + cot 2 x, y2 = csc 2 x 3

f (− x ) = sec ( − x + π )

47.

2π

−4

Thus, the function is odd. 46.

cos x 1 and y2 = cot x = sin x tan x 4

cos(−2 x) cos(2 x) =− = − f ( x) sin(−2 x) sin(2 x)

Thus, the function is odd. 45.

y1 =

52.

f ( x ) = x sin x As x → 0, f ( x ) → 0, and f ( x ) ≥ 0 for all x.

2π

Matches graph (a). 53. g( x ) = x sin x

−4

It appears that y1 = y2 . 1 sin x = = tan x sin x sec x = sin x cos x cos x

As x → 0, g ( x ) → 0.

Odd function g ( 2π ) = 0 Matches graph (b).

Section 5.5 54. g ( x ) = x cos x

Graphs of Other Trigonometric Functions

58.

y = 2− x

2 2

cos x

Even function As x → 0, g( x) → 0.

Matches graph (c). 55.

−5

−2

Since 2

−3

59.

y = ±e

touches y = e cos x at x = nπ . −x

−x

y = e cos x has x -intercepts at cos x = 0 or x = f ( x) = e

−2 x

π 2

+ nπ .

sin x 2

−3

60.

−2

Since the damping factor is e −2 x , −e

−2 x

≤e

y = ±e

−2 x

cos x ≤ e

−2 x

touches y = e

sin x at x =

+ nπ .

2 y = e−2 x sin x has x -intercepts at sin x = 0 or x = nπ . 57. h ( x ) = 2 − x

2 4

y = ±2

− x2 2

sin x

2 2

cos x ≤ 2− x

touches y = 2

2 2

− x2 2

cos x at x = nπ .

cos x has x-intercepts at x =

π 2

+ nπ .

f ( x ) = tan x

π+

(a)

As x →

(b)

As x →

(c)

As x → −

(d)

As x → −

π−

, f ( x ) → −∞. , f ( x ) → ∞.

π+ 2

π− 2

, f ( x ) → −∞. , f ( x ) → ∞.

f ( x ) = sec x

π+

(a)

As x →

(b)

As x →

(c)

As x → −

(d)

As x → −

. −2 x

is the damping factor,

≤ 2− x

y = 2

−e − x ≤ e − x cos x ≤ e − x.

− x2 2

−2

Since the damping factor is e − x

56.

f ( x) = e − x cos x

−x

459

π− 2

, f ( x ) → −∞. , f ( x ) → ∞.

π+ 2

π− 2

, f ( x ) → ∞. , f ( x ) → −∞.

61. −5

−2

Since 2 − 2− x

− x2 4

2 4

y = ±2

is the damping factor,

≤ 2− x − x2 4

y = 2− x

2 4

sin x ≤ 2− x

touches y = 2

2 4

− x2 4

sin x at x =

sin x has x-intercepts at x = nπ .

π 2

62. + nπ .

f ( x ) = csc x (a)

As x → 0 + , f ( x ) → ∞.

(b)

As x → 0 − , f ( x ) → −∞.

(c)

As x → π + , f ( x ) → −∞.

(d)

As x → π − , f ( x ) → ∞.

f ( x ) = cot x (a)

As x → 0 + , f ( x ) → ∞.

(b)

As x → 0 − , f ( x ) → −∞.

(c)

As x → π + , f ( x ) → ∞.

(d)

As x → π − , f ( x ) → −∞.

63. tan x = d=

7 d 7 = 7 cot x tan x

− 36

460

Chapter 5

Trigonometric Functions

27 d 27 = 27 sec x d= cos x

64. cos x =

68.

 − 7π   5π  + π  = tan  −  is True. Because tan   2   2  1 x  undefined, the graph of y = − tan  + π  has a 8 2  vertical asymptote at x = − 7π .

69. True. For x → −∞, 2 x → 0. −

π 2

65. (a)

70. True. Because 1 y = csc x = sin x for a given value of x, the y-coordinate of csc x is the reciprocal of the y-coordinate of sin x.

π 2

0.6

4π

71. True. y = sec x =

−0.6

(b) The displacement function is not periodic, but damped. It approaches 0 as t increases.

If the reciprocal of y = sin x is translated π 2 units to the left, then 1 1 = = sec x. y= π  cos x  sin  x +   2

1700 66. (a) = 850 rev/min 2 (b) The direction of the saw is reversed.  π  π  (c) L =  + φ  + cot φ  , 0 < φ < 2 2   

(d)

φ L

0.3 306.2

0.6 217.9

0.9 195.9

1.2 189.6

1.5 188.5

1 cos x

72.

f ( x ) = 2sin x, g( x ) =

(a)

1 csc x 2

(e) Straight line lengths change faster. 500 (f)

f g π

0 −1

π 2

67. (a) Yes. For each t there corresponds one and only one value of y. (b) One way to determine the frequency is to note that the time between the first and second maximum points is t = 0.7622 − 0 = 0.7622. Thus, the frequency is approximately (0.7622) −1 = 1.3 oscillations per second. (c)

y = 12(0.221)t cos(8.2t ) To obtain this model, first fit an exponential model y = abt to the data points (0, 12), (0.7622, 3.76), and

(1.5476, 1.16). This yields y = 12(0.2210)t . Using 2π ≈ 8.2 for the cosine term, you obtain the 0.7622 model above. (d) (e)

ln 0.221 ≈ −1.51  y = 12e −1.5t cos(8.2t ) 14

f ( x ) > g( x ) for

(c)

As x → π , 2sin x → 0 and

<x<

5π 6

(b)

1 csc x → ∞, since g( x ) 2

is the reciprocal of f ( x). 73.

f ( x ) = tan

πx

1 πx , g( x ) = sec 2 2 2

(a)

g −1

f −3

1  (b) The interval in which f < g is  −1,  . 3  1  (c) The interval in which 2 f < 2 g is  −1,  , which 3  is the same interval as part (b).

−8

Section 5.5 74. (a)

77.

Graphs of Other Trigonometric Functions

y=x+

y = tan x

(b)

y3 =

4

π 

−3

−2

78. (a) Period is

41

y4 = y3 +

π  9

 sin 9π x  

−1

−0.1

tan x 1.5574 1.0033 x x

tan x x

Undef.

2 x 3 16 x 5 are + 3! 5!

 π π similar on the interval  − ,  .  2 2

(b)

−2

The graphs of y1 = tan x and y2 = x +

−3

75. (a)

1 1 1  sin π x + sin 3π x + sin 5π x + sin 7π x  3 5 7  2

(c)

2x 3 16x 5 + 3! 5!

−3

461

π 2

 π π and graph is increasing on  − ,  .  4 4

Matches (i). (b) Period is

−0.01 −0.001 1.0

1.0

0.001 0.01 1.0

1.0

π 2

π  and  , 1  is on the graph. 8 

Matches (ii).

0.1

79. Distributive Property 1

1.0033 1.5574

80. Multiplicative Inverse Property 81. Additive Identity Property 82. Associative Property of Addition

83. Not one-to-one 84. Not one-to-one −1.1

1.1

As x → 0, f ( x ) =

85. One-to-one 14 , y≥0 3 14 x = 3 y − 14, y ≥ , x ≥ 0 3 x 2 = 3 y − 14 1 y = ( x 2 + 14 ) 3

tan x → 1. x

y = 3x − 14, x ≥

tan 3 x 3x

−1

−0.1

−0.01

−0.0475 1.0311 1.0003

−0.001 1.0 0.1

1 f −1 ( x) = ( x 2 + 14), x ≥ 0 3

0.001

0.01

tan 3 x 3x

Undef.

1.0

1.0003 1.0311 −0.0475

86. One-to-one f ( x) = 4 x − 5 y = ( x − 5) , x ≥ 5 14

(b)

x = ( y − 5)

x4 = y − 5 −1

tan 3 x → 1. As x → 0, f ( x ) = 3x

x + 5 = y f −1 ( x) = x 4 + 5, x ≥ 0

462

Chapter 5

Trigonometric Functions

87.

y = x2 + 3x − 4 Domain: (−∞, ∞) x-intercepts: x 2 + 3x − 4 = 0

89.

Domain: (−∞, ∞) x-intercept: none y-intercept: y = 30 +1 + 2 = 3 + 2 = 5

( x + 4 )( x − 1) = 0

(0, 5)

x + 4 = 0  x = −4 x −1 = 0  x = 1 x = −4 x = 1 (−4, 0), (1, 0)

Horizontal asymptote: y = 2 90.

y-intercept: y = 0 2 + 3 ( 0 ) − 4 = −4

f ( x) =

x−7 x−7 = x + 4 x + 4 ( x + 2)2 2

Domain: all real x ≠ −2

(0, –4) No asymptotes 88.

f x = 3 x +1 + 2

x-intercept: x − 7 = 0

x=7

y = ln x 4

(7, 0)

Domain: all real x ≠ 0

y-intercept: y =

x-intercept: x = 1 x = ±1

0−7 −7 = 2 (0 + 2) 4

(1, 0), (−1, 0)

7   0, − 4   

y-intercept: none

Vertical asymptote: x = −2

Vertical asymptote: x = 0

Horizontal asymptote: y = 0

Section 5.6 Inverse Trigonometric Functions 1.

y = sin −1 x, − 1 ≤ x ≤ 1

y = arccos x, 0 ≤ y ≤ π

−1

The inverse sine function can be denoted as sin x or arcsin x. 1 , arccos x = cos−1 x. cos x

No. arccos x ≠

(a)

y = arcsin

(b)

y = arcsin0  sin y = 0 for −

(a)

y = arccos

y=0

(b)

1 1 π  cos y = for 0 ≤ y ≤ π  y = 2 2 3

10. (a)

≤y≤

(a)

y = arcsin ( −1)  sin y = −1 for

(b)

≤ y ≤

π 2

 y = −

π 2

π 4

< y <

π 4

 y =

π 4

y = arccos ( −1)  cos y = −1 for 0 ≤ y ≤ π  y = π

1 1 π π π  sin y = for − ≤ y ≤  y = 2 2 2 2 6

y = arccos0  cos y = 0 for 0 ≤ y ≤ π  y =

(b)

9. (a)

(b)

−

y = arctan1  tan y = 1 for −

(a)

3  tan y = 3 π π π − < y <  y = 4 4 6 y = arctan

3 for 3

3 3 π π is  sin y = for − ≤ y ≤ 2 2 2 2 not possible. y = arcsin

y = cos −1 ( − 2)  cos y = − 2 for 0 ≤ y ≤ π is not possible.

(b)

 2 2 y = sin −1 − for   sin y = − 2  2  π π π − ≤ y ≤  y = − 2 2 4

y = arccos 1  cos y = 1 for 0 ≤ y ≤ π  y = 0

Section 5.6 11. (a)

(

−

(b)

12. (a)

)

3  tan y = −

y = arctan −

< y <

 y = −

π 3

 3 y = arccos    cos y =  2  π 0 ≤ y ≤ π  y = 6

(

3 for 2

−1.0

)

(b)

14.

0 0.2014 0.4115 0.6435 0.9273 1.5708

0.8

1 −1 −2

(c)

5 π π for − ≤ y ≤ 2 2 2

−1

−2

The graphs are the same. (d) Intercept: ( 0, 0 ) ; symmetric to the origin

−0.8

−0.6

−0.4

−0.2

3.1416

2.4981

2.2143 1.9823

1.7722

0.2

0.4

0.6

0.8

1.5708 1.3694 1.1593 0.9273 0.6435

1.0 0

−10

−8

−6

−4

−2

−1.4711 −1.4464 −1.4056 −1.3258 −1.1071

0 1.1071 1.3258 1.4056 1.4464 1.4711

(b) 2 1

4 − 10

−5

−2

(c)

0.6

−2

0.4

−1

2 for 2

16. (a)

(b)

0.2

−1

−0.2

3 for

 3 3 y = tan −1 −  = tan y = − 3 for 3   π π π − < y <  y = − 4 4 6

y = arccos x (a) x

−0.4

2 y = arcsin  sin y = 2 π π π − ≤ y ≤  y = 2 2 4

5 13. (a) y = sin −1 = sin y = 2 is not possible.

−0.6

−1.5708 −0.9273 −0.6435 −0.4115 −0.2014

0 ≤ y ≤ π is not possible.

(b)

−0.8

463

(b)

3  cos y = −

y = arccos −

15. (a)

3 for

Inverse Trigonometric Functions

−1

(c)

−10

−2

−1

The graphs are the same.  π (d) Intercepts:  0,  , (1, 0 ) ; no symmetry 2 

The graphs are the same. (d) Horizontal asymptotes: y = ±

π 2

464

Chapter 5

17.

y = arctan x

Trigonometric Functions

π 

  π x = − 3  y = −  tan  −  = − 3  3   3 

33. g ( x ) = arcsin ( x + 3) is a horizontal shift three units to the left of f ( x ) = arcsin x. y

3   π 3 y=− x=−  tan  −  = −  6 3   6 3 

π    x = 1  tan = 1  4 4   y = arccos x

18.

3π 4 π 2 π 4

−4

−2

−1

x = −1  y = π ( cos π = −1) 1 2π  2π 1 x=− y= =−   cos 2 3  3 2

3  π 3 y= x=  cos =  6 2  6 2 

−

y 3π 4 π 2 π 4

21. tan −1 0.75 ≈ 0.64

−1

22. arctan 15 ≈ 1.50

−

3π 4

f ( x ) = arcsin x. y

25. arccos( − 0.9) ≈ 2.69

3π 4 π 2

26. arcsin ( −1.3) is not possible. 27. arctan ( −6 ) ≈ −1.41

−3

−2

28. tan −1 5.9 ≈ 1.40

−1

−

36.

3π 4

g ( x ) = − arcsin x is a reflection in the x-axis of f ( x ) = arcsin x. y 3π 4 π 2

7 31. arctan   ≈ 1.29 2  95  32. tan −1  −  ≈ −1.50  7 

35. g ( x ) = arcsin ( − x ) is a reflection in the y-axis of

24. sin −1 0.56 ≈ 0.59

5 30. arccos   is not possible.  3

3π 4

the right of f ( x ) = arcsin x.

20. cos −1 0.28 ≈ 1.29

 3 29. sin −1  −  ≈ − 0.85  4

34. g ( x ) = arcsin ( x − 4 ) is a horizontal shift four units to

19. arcsin 0.45 ≈ 0.47

23. cos −1 3.5 is not possible.

−3

−2

−1

−

3π 4

Section 5.6

37. g ( x ) = arccos ( x + π ) is a horizontal shift π units to

41.

Inverse Trigonometric Functions

g ( x ) = arctan x + 1 is a vertical shift one unit upward

of f ( x) = arctan x.

the left of f ( x ) = arccos x.

3π 4 π 2

5π 4

−2

−1

π 4 −5

38.

−4

−3

−2 −1

−

465

π 2 3π − 4 −

π 4

g ( x ) = arccos x − 5 is a vertical shift five units

42. g ( x ) = arctan ( x − z ) is a horizontal shift two units to the right of f ( x ) = arctan x.

downward of f ( x ) = arccos x.

3π 4 π 2 π 4

1 −5 −4 −3 −2 −1

1 2 3 4 5

−2

−1

−π 2 3π − 4

−4 −5 −6

(

)

39. g ( x ) = arccos ( − x − 2 ) = arccos − ( x + 2 ) is a reflection in the y-axis and a horizontal shift two units to the left of f ( x ) = arccos x.

43.

g ( x) = − arctan x − 3 is a reflection in the x-axis and a

vertical shift three units downward of f ( x) = arctan x. y

π 2

y 5π 4

−3

−2

−1

1 −

−3

−2

−1

−π 4

π 2

−π

π 4 −4

40. g ( x ) = arccos ( − x ) − 3 is a reflection in the y-axis and a vertical shift three units downward of f ( x ) = arccos x.

44. g ( x ) = arctan ( − x ) + 4 is a reflection in the y-axis and a vertical shift four units upward of f ( x ) = arctan x. y

2π

π π π 2 −2

−1

−3

−2

−π 2

−π −π

466

Chapter 5

Trigonometric Functions

x 8

45. tan θ =

50. Let y be the third side. Then

θ = arctan

y2 = 32 − x2 = 9 − x2  y = 9 − x2 .

x 8

sin θ =

tan θ =

46.

4 x

cosθ =

9 − x2 9 − x2  θ = arccos 3 3 x x  θ = arctan 9 − x2 9 − x2

51. Let y be the hypotenuse. Then y 2 = ( x + 1) + 2 2 = x 2 + 2 x + 5  y = x 2 + 2 x + 5. 2

4 θ = arccos x

sin θ = cosθ =

x +1 2

x + 2x + 5 2

 θ = arcsin

x +1 2

x + 2x + 5 2

 θ = arccos x2 + 2x + 5 x2 + 2 x + 5 x +1 x +1 tan θ =  θ = arctan 2 2

θ 4

52. Let y be the hypotenuse. Then

x+2 5

47. sin θ =

cosθ =

x x  θ = arcsin 3 3

y 2 = ( x + 2 ) + 32 = x 2 + 4 x + 13  y = x 2 + 4 x + 13. 2

 x+2   5 

θ = arcsin 

sin θ =

x+2

x + 4 x + 13 3

 θ = arcsin

x+2 2

x + 4 x + 13 3

 θ = arc cos x 2 + 4 x + 13 x 2 + 4 x + 13 x+2 x+2 tan θ =  θ = arctan 3 3

cosθ = 5

x+2 2

53. cos(arccos 0.3) = 0.3

54. tan ( arctan 25) = 25

x +1 10

48. tan θ =

 x +1   10 

θ = arctan 

55. Because the domain of the arcsine function is −1 ≤ x ≤ 1, sin arcsin (1.7) is not possible.

(

)

56. sin arcsin ( −0.1) = −0.1 x+1

θ 10

49. Let y be the third side. Then

y2 = 22 − x2 = 4 − x2  y = 4 − x2 x x sin θ =  θ = arcsin 2 2

cosθ = tan θ =

4 − x2 4 − x2  θ = arccos 2 2 x x  θ = arctan 4 − x2 4 − x2

57. tan tan −1 ( − 0.5) = − 0.5 58. Because the domain of the arccosine function is −1 ≤ x ≤ 1, cos (cos −1 ( − 2)) is not possible.  11π  3 π  59. arctan  tan  = −  = arctan  − 6 3 6      4π  3 π  60. arcsin  sin  = −  = arcsin  − 3  2 3   

5π  π  −1 61. sin −1  sin  = sin 1 = 2 2  

Section 5.6 3π  π  −1 62. cos−1  cos  = cos 0 = 2  2  5π  π  −1 63. sin −1  tan  = sin 1 = 4 2  

Inverse Trigonometric Functions

467

 12  68. Let u = arctan  −  ,  5  12 π tan u = − , − < u < 0. 5 2 y

3π   −1 64. cos−1  tan  = cos ( −1) = π 4   4 65. Let y = arctan . Then 3 4 π tan y = , 0 < y < , and 3 2 4 sin y = . 5

5 −12

 hyp 13  12   csc arctan  −   = csc u = =− 5 opp 12     2 2 π 69. Let y = arccos  −  . Then, cos y = − , < y < π and 3 2  3

5 . 3

sin y =

66. Let y = arcsin sin y =

24 . Then 25

24 7 , and cos y = . 25 25

−2 25

 3 70. Let y = arctan  −  . Then,  5

3 π tan y = − , − < y < 0 and sec y = 5 2

4 67. Let u = arcsin . 5 4 sin u = 5 4  sec  arcsin  = sec u 5 

34 . 5

y 34

−3

hyp 5 = adj 3

u 3

468

Chapter 5

Trigonometric Functions 75. Let y = arccos ( x + 2 ) , cos y = x + 2.

 3 71. Let u = arcsin  −  ,  4 3 π sin u = − , − < u < 0. 4 2

Opposite side:

1 − ( x + 2)

sin y =

1 − ( x + 2)

= − x2 − 4 x − 3

1 7

1 − (x + 2)2

−3

x+2

 −3 3 7  3  tan  arcsin  −   = tan u = =− 4 7 7    

76. Let u = arcsin ( x − 1) , sin u = x − 1 =

5 72. Let u = arctan , 8 5 π tan u = , 0 < u < . 8 2 5 8  cot  arctan  = cot u = 8 5  

x −1 . 1

x −1

u 2x − x 2

sec arcsin ( x − 1)  = sec u =

hyp = adj

1 2 x − x2

x x 77. Let y = arccos . Then cos y = , and 5 5 89 u

25 − x 2 . x

tan y =

5 x

25 − x 2

y x

73. Let y = arctan x. Then, 1 tan y = x and cot y = . x x2 + 1

4 4 78. Let u = arctan , tan u = . x x

x 2 + 16

y 1

x 74. Let u = arctan x, tan u = x = . 1

x2 + 1

4 adj x  = cot  arctan  = cot u = x opp 4  x x . Then tan y = 79. Let y = arctan and 7 7

csc y =

7 + x2 . x

u 1

opp sin ( arctan x ) = sin u = = hyp

7 + x2

x x2 + 1

y 7

Section 5.6 x−h x−h , sin u = . r r

80. Let u = arcsin

84. If arccos

r 2 − (x − h) 2

x−h  cos  arcsin  = cos u = r  

r2 − ( x − h)

arccos

14 14 81. Let y = arctan . Then tan y = and x x

85.

  14 . Thus, y = arcsin  .  2  196 + x  196 + x 

4x − x 2

x−2

x−2 4 x − x2 = arctan , since 2 < x < 4 2 x−2

y = 2 arccos x Domain: −1 ≤ x ≤ 1 Range: 0 ≤ y ≤ 2π

Vertical stretch of f ( x ) = arccos x

196 + x 2

2π

−2

36 − x 2 82. If arcsin = u, then sin u = 6

36 − x 2 . 6

86.

x 2 Domain: −2 ≤ x ≤ 2 y = arcsin

Range: − 6

arcsin

π 2

≤y≤

π 2

Horizontal stretch of f ( x ) = arcsin x

36 − x 2

π 2

−3

x −π 2

36 − x 2 x = arccos 6 6 3

83. Let y = arccos cos y =

469

x−2 x−2 . = u, 2 < x < 4 then cos u = 2 2

x−h

sin y =

Inverse Trigonometric Functions

x 2 − 2 x + 10

3 2

x − 2 x + 10

y = arcsin

( x − 1) + 9 2

x −1

( x − 1) + 9 2

x2 2 2x 1 10

f ( x ) = arcsin ( x − 2) Domain: 1 ≤ x ≤ 3 π π Range. − ≤ y ≤ 2 2

. Then,

x −1

and sin y =

87.

Horizontal shift two units to the right of f ( x ) = arcsin x π 2

. Thus,

= arcsin

x −1 x 2 − 2 x + 10

−π 2

x − 1

y 3

470

Chapter 5

Trigonometric Functions

88. g ( t ) = arccos ( t + 2 )

92.

f ( t ) = 4 cos π t + 3 sin π t

4  = 4 2 + 32 sin  π t + arctan  3 

Domain: −3 ≤ t ≤ −1 Range: 0 ≤ y ≤ π This is the graph of y = arccos t shifted two units to the left.

4  = 5sin  π t + arctan  3 

π 6

−9 −4

−6

89.

f ( x ) = arctan

The two forms are equivalent.

x 4

Domain: all real numbers Range: −

< y <

Horizontal stretch of y = arctan x π 2

−

90.

< y <

95. As x → ∞, arctan x →

π 2

−3

99. (a) 5.5 m

17 m

(b) tan θ = 3

h  12.5 tan (0.574) 12.5 ≈ 8.08 meters

100. β = arctan

f ( t ) = 3cos 2 t + 3sin 2t

3  = 32 + 32 sin  2t + arctan  3  = 3 2 sin ( 2t + arctan1)

π  = 3 2 sin  2t +  4  6

2π

5.5  θ ≈ 0.574 radian or 32.9° 8.5

π 2

−2π

π 2

91.

96. As x → −1+ , arcsin x → −

Horizontal shrink of y = arctan x

−

98. As x → −∞, arctan x → −

Domain: all real numbers

97. As x → −1+ , arccos x → π .

π 2

f ( x ) = arctan 3 x

Range: −

94. As x → 1− , arccos x → 0.

−8

93. As x → 1− , arcsin x →

(a)

3x , x>0 x2 + 4

1.5

− 0.5

(b) β is maximum when x = 2 feet. (c) The graph has a horizontal asymptote at β = 0. As the camera moves further from the picture, the angle subtended by the camera approaches 0.

−6

The two forms are equivalent.

Section 5.6

101. (a) tan θ =

Inverse Trigonometric Functions

108. y = arcsec x if and only if sec y = x where

6 x

x ≤ −1 ∪ x ≥ 1 and 0 ≤ y < π 2 or π 2 < y ≤ π . The

6 θ = arctan   x (b) When x = 10,

domain of y = arcsec x is ( −∞, − 1 ∪ 1, ∞ ) and the range is  0, π 2 ) ∪ (π 2, π  .

 6   ≈ 0.54 radian, ( ≈ 31° ) .  10 

θ = arctan 

When x = 3, 6

θ = arctan   ≈ 1.11 radians, ( ≈ 63° ) . 3

π 2

 

x 20

102. (a) tan θ =

−2

θ = arctan

x 20

−1

 π   π Range:  − , 0  ∪  0,  2  2   y

π 2

5π π because the range of = 1  arctan1 = 4 4

 π π arctan x is  − ,   2 2 sin x cos x

−1

−π 2

Thus,

(b) arcsin x = arccos x when x = (c) arcsin x > arccos x when

2 . 2

2 < x ≤ 1. 2

107. y = arc cot x if and only if cot y = x, − ∞ < x < ∞ and 0 < y < π. y

1 1 = tan y and y = arctan   . x x

π + arctan (1 x ) , x < 0  Hence, graph y = π 2, x = 0. arctan (1 x ) , x>0  (b) y = arcsec x if and only if π or x = sec y, x ≤ −1 or x ≥ 1, and 0 ≤ y < 2 π < y < π. 2 1 1 Thus, = cos y and y = arccos   . Hence graph x x 1 y = arccos , x ∈ ( −∞, − 1 ∪ 1, ∞ ) . x (c) y = arccsc x if and only if x = csc y, x ≤ −1 or

π 2

−1

−2

110. (a) y = arccot x if and only if x = cot y, − ∞ < x < ∞ and 0 < y < π .

106. (a) arcsin x < arccos x when −1 ≤ x <

−2

Domain: ( −∞, − 1 ∪ 1, ∞ )

1 π 103. False. arcsin = 2 6

105. False. tan x =

109. y = arccsc x if and only if csc y = x.

(b) When x = 5, 5 θ = arctan ≈ 14.0°, ( 0.24 rad ) . 20 When x = 12, 12 θ = arctan ≈ 31.0°, ( 0.54 rad ) . 20

104. False. tan

471

x ≥ 1, and − 1

Thus,

π 2

≤ y < 0 or 0 < y ≤

π 2

1 1 = sin y and y = arcsin   . Hence, graph x x

1 y = arcsin   , x ∈ ( −∞, − 1 ∪ 1, ∞ ) . x

472

Chapter 5

Trigonometric Functions

111. y = arcsec 2  sec y = 2 and π π π 0≤ y< ∪ < y≤π  y = 2 2 4

(c)

112. y = arcsec 1  sec y = 1 and π π 0≤ y< ∪ < y≤π  y =0 2 2

(d)

(

a = 0, b = 3 Area = arctan 3 − arctan 0 ≈ 1.25 − 0 = 1.25 = 1.25 a = −1, b = 3

Area = arctan3 − arctan ( −1)  π ≈ 1.25 −  −  ≈ 2.03  4

)

113. y = arccot − 3  cot y = − 3 and 0< y<π  y =

5π 6

119.

114. y = arccsc 2  csc y = 2 and π π π − ≤ y<0∪0< y≤  y = 2 2 6

120.

115. Let y = arcsin ( − x ) . Then,

121.

sin y = − x − sin y = x

122.

sin ( − y ) = x

4 4 2 2 3

tan y = − x, −

π 2

<y<

5 5 2 10

(

)

< −y <

π 2

csc θ =

− y = arctan x y = − arctan x

sec θ =

cot θ =

2 Let α = arcsin x  sin α = x. Let β = arccos x  cos β = x. Hence, sin a = cos β  a and β are complementary angles  a + β =

 arcsin x + arccos x =

2 118. Area = arctan b − arctan a (a) a = 0, b = 1 Area = arctan1 − arctan 0 π π = − 0 = ≈ 0.785 4 4 (b) a = −1, b = 1

Area = arctan1 − arctan ( −1) π  −π  π = −  = ≈ 1.571 4  4  2

π 2

2 2 5

5 2 4

11 6 5 11

5 11 11

6 11 11

6 5

6 11

2 5

π 2

2 2 + 12 = 5

2 5 5

5 5

5 2 sec θ = 5 csc θ =

11 5

Adjacent side:

cosθ =

6 2 − 52 = 11

124. tan θ = 2, 0 < θ <

sin θ =

5 2 2

5 6

arctan tan ( − y ) = arctan x

117. arcsin x + arccos x =

5 5

tan θ =

2 3 3

2 3 3 = 6 3

cosθ =

− tan y = x tan ( − y ) = x, −

Adjacent side:

y = arctan ( − x )

116.

3 3

123. sin θ =

Therefore, arcsin ( − x ) = − arcsin x.

2 2

2 3

− y = arcsin x y = − arcsin x.

θ 1

1 cot θ = 2

Section 5.7

125. cosθ =

3 4

126. sec θ = 3, 0 < θ <

Opposite side:

16 − 9 =

sin θ =

7 4

tan θ =

7 3

cot θ =

3 3 7 = 7 7

secθ =

Adjacent side:

4 4 7 = 7 7

473

π 2

9 −1 = 8 = 2 2

2 2 sin θ = 3 1 cosθ = 3 tan θ = 2 2

4 3

cscθ =

Applications and Models

2 2

3 2 = csc θ = 4 2 2 cot θ =

1 2 2

2 4

Section 5.7 Applications and Models 1.

harmonic motion

bearing

No. N 20° E means a direction first of due north, then 20° east of north.

The amplitude of the simple harmonic motion,

d = 3sin 5.

B = 90° − 7.4° = 82.6°

π 2

is 3.

a a 20.5 b= = ≈ 157.84 b tan A tan 7.4° a a 20.5 sin A =  c = = ≈ 159.17 c sin A sin 7.4° tan A =

9. Given: a = 6, b = 12

Given: A = 30°, b = 10

c 2 = a 2 + b2  c = 36 + 144 ≈ 13.42

B = 90° − 30° = 60°

tan A =

a  a = b tan A = 10 tan30° ≈ 5.77 b b b 10 cos A =  c = = ≈ 11.55 c cos A cos30° tan A =

8. Given: A = 7.4°, a = 20.5

Given: B = 60°, c = 15

a 6 1 1 = =  A = arctan ≈ 26.57° b 12 2 2 B = 90° − 26.57° = 63.43°

10. Given: a = 25, b = 37

a 2 + b2 

c =

tan A =

a  25   A = arctan  ≈ 34.05° b  37 

tan B =

b  37   B = arctan  ≈ 55.95° a  25 

A = 90° − 60° = 30° b 15 3 ≈ 12.99 sin B =  b = c sin B = 15sin 60° = c 2 a 15 = 7.50 cos B =  a = c cos B = 15cos 60° = c 2

Given: B = 71°, b = 14 b b 14 a= = ≈ 4.82 a tan B tan 71° b b 14 = ≈ 14.81 sin B =  c = c sin B sin 71° A = 90° − 71° = 19° tan B =

1994 ≈ 44.65

11. Given: a = 16, c = 54

b =

c2 − a2 

sin A =

2660 ≈ 51.58

a 16  16  =  A = arcsin   ≈ 17.24° 54 c  54 

B = 90° − 17.24° = 72.76°

Chapter 5

474

Trigonometric Functions

12. Given: b = 1.32, c = 18.9

a = c 2 − b2 = 355.4676 ≈ 18.85 cos A =

b 1.32  1.32  =  A = arccos   ≈ 86.00° c 18.9  18.9 

sin B =

b  1.32   B = arcsin   ≈ 4.00° c  18.9 

1 b tan θ 2 1 23 45 °  + h = (3.1) tan  62 +  ≈ 2.96 cm 2 60 3600   h =

h 20 h = 20sin 75°

19. sin 75° =

13. A = 12°15′, c = 430.5

B = 90° − 12°15′ = 77°45′ a 430.5 a = 430.5sin12°15′ ≈ 91.34 b cos12°15′ = 430.5 b = 430.5cos12°15′ ≈ 420.70

≈ 19.32 ft

sin12°15′ =

h b2

18. tan θ =

20 h

c = 430.5

12°15′ b

75°

14. Given: B = 65°12′, a = 145.5

20.

A = 90° − 65°12′ = 24°48′ a a 145.5 cos B =  c = = ≈ 346.88 c cos B cos ( 65°12′ ) tan B =

15. tan θ =

75 w w cos16° = 75 75 w= cos16° w ≈ 78.02 feet

h b2

h 16. tan θ = b2 1 b tan θ 2 1 h = (13) tan 48.5° ≈ 7.35 ft 2

cos16° =

b  b = a tan B = 145.5 tan ( 65°12′ ) ≈ 314.89 a

1 b tan θ 2 1 h = ( 8 ) tan 52° ≈ 5.12 in. 2

w 16°

21.

sin 66° =

h 120

120 sin 66° = h 109.63 feet ≈ h

h =

17. tan θ =

h b 2

1 b tan θ 2 1 38 °  h = (16) tan  71 +  ≈ 24.10 m 2 60  

120

h =

66°

So, the cadet on the rope is about 109.63 + 4 = 113.63 feet above the ground.

Section 5.7

Applications and Models

475

26. (a)

22.

28 ft

49° 40′ 70°

50 ft

37°

(b) Let x be the height of the church and y be the height of the church and steeple. x y tan 37° = and tan 49°40' = 50 50 x = 50 tan 37° and y = 50 tan 49°40'

28 x 28 x = tan 70°

tan 70° =

h = y − x = 50( tan 49°40' − tan 37°)

≈ 10.19 ft

x 23. sin 31.5° = 4000 x = 4000 sin 31.5° ≈ 2089.99 feet

h ≈ 21.21 feet

(c)

So, the width of the cottage is 2(10.19) = 20.38 feet.

350  d ≈ 3071.91 ft d 350 tan 4° =  D ≈ 5005.23ft D Distance between ships: D − d ≈ 1933.3 ft

27. tan 6.5° =

6.5°

31.5°

24. sin 9.5° =

4°

350

4000

x 4

28.

28° 10 km

55° d x

9.5°

a  a = 100 tan 28° 100

a+s 100 a + s = 100 tan 39.75°

tan 39.75° =

s = 100 tan 39.75° − a = 100 tan 39.75° − 100 tan 28° ≈ 30 feet

39.75°

28° D

d  d ≈ 7 kilometers 10 D cot 28° =  D ≈ 18.8 kilometers 10 cot 55° =

Distance between towns: D − d = 18.8 − 7 = 11.8 kilometers a  x = a cot 57° x a tan16° = x + ( 55 6 )

29. tan 57° =

tan16° =

55°

x = 4 sin 9.5° ≈ 0.66 mile

25. tan 28° =

cot16° =

a a cot 57° + ( 55 6 ) a cot 57° + ( 55 6 ) a

55 6  a ≈ 3.23 miles ≈ 17,054 feet

a cot16° − a cot 57° = 100

28° P1

P2 a

57°

16°

H x

550 60

476

Chapter 5

Trigonometric Functions

h h , tan10° = x x − 18 h h x= ,x= + 18 tan 2.5 tan10° h h h + 18 tan10° = + 18 = tan 2.5° tan10° tan10° h tan10° = h tan 2.5° + 18 ( tan10° )( tan 2.5° )

30. tan 2.5° =

18 ( tan10° )( tan 2.5° ) tan10° − tan 2.5°

≈ 1.04 miles ≈ 5515 feet

2.5°

34.

275(s)

sin18° = s=

h 275 sin18°

10,000 ≈ 117.7 seconds. 275 sin18° 16,000 ≈ 188.3 seconds. If h = 16,000, s = 275 sin18°

θ = 90° − 42° = 48° h 217 217sin 48° = h 161.26 feet ≈ h sin 48° =

h 10° x − 18 not drawn to scale

31. sin α =

h 275(s)

If h = 10,000, s =

35. (a)

x not drawn to scale

18°

4000  α ≈ 14.03° 16,500

217

42°

θ ≈ 90° − α ≈ 75.97°

(b) 16,500

4000

θ d

32. tan θ =

250 2.5 ( 5280 )

(c)



250   ≈ 1.09° 2.5(5280)  

θ = arctan  2.5 miles θ

250 feet

not drawn to scale

33. Since the airplane speed is ft  sec  ft   275 sec  60 min  = 16,500 min , after one minute its    distance traveled is 16,500 feet. 16500

(d)

tan θ =

h d

h tan θ

d (10°) ≈ 914.55, d ( 20°) ≈ 443.060, d (30°) ≈ 279.31, d ( 40°) ≈ 192.18, d (50°) ≈ 135.31

(e) Even though the angle measure θ increases in equal increments in the table in part ( a ) , the distance d does not increase in equal increments.

Answers will vary.

18°

a 16,500 a = 16,500sin18°

sin18° =

≈ 5099 ft

Section 5.7 36. (a)

38. (a) 1

h 30 30 sin 45° = h 21.21 feet ≈ h

sin 45° =

17 3 ft

1 2 tan θ = 1 17 3  1  12  θ = tan −1  2   17 1   3 θ ≈ 35.80° 12

(c)

(b)

   21.21  θ = tan −1    d 

21.21 tan 25° d ≈ 45.48 feet 21.21 d 21.21 d= tan 30° d ≈ 36.74 feet 36.74 feet ≤ d ≤ 45.48 feet Let θ = 30°  tan30° =

20 2 6 5 14

l 2 = ( h + 14 ) + 100 2 2

l = h 2 + 28h + 10,196 100 h

39. 22

 100    l 

θ = arccos 

If θ = 35°, then cos35° =

100 . l

(a)

100

tan α =

100 cos35° l ≈ 122.08 feet

28 15

 28   15 

α = tan −1 

α ≈ 61.82°

When l = 122.08 feet, using part (a) :

tan β =

122.082 = ( h + 14 ) + 100 2 2

4903.5264 = ( h + 14 )

α 15

70.025 = h + 14 56.03 feet ≈ h

21.21 d

100

(c)

Let θ = 25°  tan 25° =

h d

h

(c)

cos θ =

tan θ =

θ = tan −1   d

37. (a)

(b)

458

(b)

477

12 2 ft

Applications and Models

28 100

 28   100 

β = tan −1 

β ≈ 15.64° (b) 28 42° x

28 x 28 x= tan 42° x ≈ 31.10 feet

tan 42° =

478

Chapter 5

Trigonometric Functions l

40.

45 3 =  θ ≈ 56.31° 30 2

43. tan θ =

200

Bearing: N 56.31°W

150 A

B 45

Port 2

l =

(a)

200 + 150

62,500 = 250 feet

150 tan A = 200  150  A = arctan   ≈ 36.87°  200  200 tan B = 150  200  B = arctan   ≈ 53.13°  150  (b) d = r ⋅ t d t = r  1 mile  (250 feet )   5280 feet  t =   1 hour (35 miles/hour )  3600 seconds  

Ship

44. tan θ =

85  θ ≈ 27.98° 160

Bearing: S 27.98° W N

Plane

160

≈ 4.87 seconds

Airport

41. 90° − 29° = 61°

( 20 )( 6 ) = 120 nautical miles a  a = 120sin 61° 120 ≈ 104.95 nautical miles south b cos 61° =  b = 120cos61° 120 ≈ 58.18 nautical miles west

sin 61° =

45. θ = 32°, φ = 68° Note: ABC forms a right triangle.

(a)

α = 90° − 32° = 58° Bearing from A to C: N 58° E N

d W

b 61°

42. c = 600 (1.5 ) = 900

y = 900sin38° ≈ 554.1 miles north x = 900cos38° ≈ 709.2 miles east

θ A

(b)

120

ϕ C γ β α 50 β

β = θ = 32° γ = 90° − φ = 22° C = β + γ = 54° tan C =

d d  tan 54° =  d ≈ 68.82 m 50 50

c 52° 38° x

y E

Section 5.7

46.

d  x = d cot14° x d d d = tan 34° =  30 − x 30 − d cot14° y tan14° =

b r b = cos30°r

51. cos30° =

30 − d cot14° d d cot 34° = 30 − d cot14°

3 5 3 x −  m1 = 2 2 2 L2 : x + y = 1  y = − x + 1  m2 = −1

tan α =

−1 − ( 3 2 )

1 + ( −1)( 3 2 )

−5 2 =5 −1 2

α = arctan 5 ≈ 78.7°

tan α =

α = arctan

35 = 17.5 2 a sin15° = c a = c sin15° = 17.5sin15°  4.53

Distance = 2 a ≈ 9.06 centimeters c

15°

1 5

m2 − m1 1 + m2 m1

52. c =

48. L1 = 2 x + y = 8  m1 = −2 L2 = x − 5 y = −4  m2 =

b 30° r

 3r  = 2 = 3r  2   

30 ≈ 5.46 kilometers d= cot 34° + cot14°

479

r 2

3r b= 2 y = 2b

cot 34° =

47. L1 : 3 x − 2 y = 5  y =

Applications and Models

10 a 10 a = cos35°

53. cos35° =

m2 − m1 1 + m2 m1

1 − ( −2) = arctan 5 1 = arctan ( 3 23 ) ≈ 74.7° 1 + 5 ( −2)

a ≈ 12.2

b 10 b = 10 ⋅ tan 35°

tan 35° =

49. The diagonal of the base has a length of

a 2 + a 2 = 2 a. Now, we have: a 1 tan θ = = 2a 2 1 ≈ 35.3° θ = arctan 2

b ≈ 7.0

54.

1 a = 92 + 62 2 1 a ≈ 10.8 2 a ≈ 21.6 Let θ be a base angle of the isosceles triangle. tan θ =

6 9 6  

θ = arctan   9

a 2 = 2 a θ = arctan 2  54.7°

θ ≈ 33.7°

50. tan θ =

b 10.8 b = 10.8 ⋅ tan 33.7°

tan 33.7° =

b ≈ 7.2 a

c =

(10.8) + (7.2) 2

c ≈ 13.0

480

Chapter 5

Trigonometric Functions

55. d = 0 when t = 0, a = 8, period = 2

Use d = a sin ω t since d = 0 when t = 0. Period:

2π

=2ω =π

Thus, d = 8sin π t. 56. Displacement at t = 0 is 0  d = a sin ω t.

2π

=6ω =

Thus, d = 3sin

πt 3

(a)

Maximum displacement: a =

(c)

π 3

Use d = a cos ω t since d = 3 when t = 0. 2π 4 = 1.5  ω = π Period: ω 3 4 Thus, d = 3cos π t. 3 58. Displacement at t = 0 is − 2  d = a cos ω t

t= 61. d = −

(a)

= 10  ω =

Thus, d = − 2cos

πt 5

π 5

59. d = 4cos8π t (a) Maximum displacement = amplitude = 4 ω 8π (b) Frequency: = 2π 2π = 4 cycles per unit of time

(c)

When t = 5, d = 4cos ( 8π ( 5 ) ) = 4.

(d) Last positive value of t for which d = 0 : 4cos8π t = 0 cos8π t = 0

8π t =

62. d =

(a)

1 1 = 2 20π 40 ⋅

1 16

ω 140π = 2π 2π = 70 cycles per unit of time

1 sin140π ( 0 )  = 0. 16  Least positive value of t for which d = 0 : 1 − sin140π t = 0 16 sin140π t = 0 140π t = π π 1 = t= 140π 140 1 sin 792π t 64

Maximum displacement: a =

of time. (c)

When t = 5, d = −

(b) Frequency:

1 1 = t= ⋅ 2 8π 16

Maximum displacement: amplitude =

π π

1 sin140π t 16

(b) Frequency:

Amplitude: a = 2 2π

2π

20π t =

57. d = 3 when t = 0, a = 3, period = 1.5

Period:

1 1 = 2 2

20π = 10 cycles per unit of the time 2π 1 1 When t = 5, d = cos  20π ( 5 )  = . 2 2 Least positive value of t for which d = 0 : 1 cos 20π t = 0 2 cos 20π t = 0

(b) Frequency:

(d)

Amplitude: a = 3

Period:

1 60. d = cos20π t 2

ω 2π

When t = 5, d =

1 1 = 64 64

792π = 396 cycles per unit 2π

1 sin 792π ( 5 )  = 0. 64

(d) Least positive value of t for which d = 0 : 1 sin 792π t = 0 64 sin 792π t = 0 792π t = π t=

π 792π

1 792

Section 5.7 63. d = a sin ω t 2π Period = =

67.

(a)

Applications and Models

L1 + L2

0.1

8 sin 0.1

6 cos 0.1

86.16

0.2

8 sin 0.2

6 cos0.2

46.39

0.3

8 sin 0.3

6 cos 0.3

33.35

0.4

8 sin 0.4

6 cos 0.4

27.06

L1 + L2

0.5

8 sin 0.5

6 cos0.5

23.52

0.6

8 sin 0.6

6 cos0.6

21.44

0.7

8 sin 0.7

6 cos0.7

20.26

0.8

8 sin 0.8

6 cos0.8

19.76

1 frequency

2π

1 = ω 264 ω = 2π (264) = 528π

64. At t = 0, buoy is at its high point  d = a cos ω t.

Distance from high to low = 2 a = 3.5 7 4 Returns to high point every 10 seconds: 2π π Period = = 10  ω = 5 ω 7 πt d = cos 4 5 a =

65. y =

(b)

1 cos16t , t > 0 4

(a)

0.5

−0.5

(b) (c)

Period:

2π π = seconds 16 8

1 cos 16t = 0 when 4 π π 16t =  t = seconds. 2 32

66. (a)

481

8 ft

θ 0

6 ft

(b) The period is 12 months. This makes sense because there are 12 months in a year.

sin θ =

8 L1

8 sin θ 8 6 L1 + L2 = + sin θ cosθ

L1 =

cosθ =

6 L2

L2 =

6 cosθ

(d)

1.5

Graphically, the minimum length is approximately 19.75 feet. The answer from part (b) and the graphical estimate are approximately equal.

482

Chapter 5

Trigonometric Functions

68. (a) and (b)

72. False. Aviation uses bearings measured clockwise from north. So S 25° W would translate.

Base 1

Base 2

Altitude

Area

8 + 16 cos10°

8 sin 10°

22.1

8 + 16 cos20°

8 sin 20°

42.5

8 + 16 cos30°

8 sin 30°

59.7

25°

8 + 16 cos 40°

8 sin 40°

72.7

8 + 16 cos50°

8 sin 50°

80.5

8 + 16 cos60°

8 sin 60°

83.1

73. Given:

8 + 16 cos70°

8 sin 70°

80.7

0°

W 205°

205°

Maximum ≈ 83.1 square feet (c)

A = 12 ( b1 + b2 ) h

32° E

= 12 [8 + 8 + 16 cosθ ]8sin θ = 64(1 + cosθ )sin θ (d) Maximum area is approximately 83.1 square feet for θ = 60°.

32°

100

E 58° 0

69. (a) and (b)

S 58° E

Average sales (in millions of dollars)

8 7 6 5 4 3 2 1

58°

32°

10 12

N 58° W

Month (1 ↔ January)

74. (a) The amplitude is 4 centimeters.

s = 3.15sin ( 0.524t + 1.5 ( 86 ) ) + 4 The model fits the data well. 2π (c) Period = ≈ 11.99 or 12 months 0.524 The period of 12 months is measurable because outer wear sales would be seasonal. (d) The amplitude is 3.15 and gives the maximum displacement or greatest and least average sales from 4 + 3.15 = $7.15 billion to 4 − 3.15 = $0.85 billion. 70. True Simple harmonic motion is given by a sin ω t or a cos ω t , when a is a real number not another function of t.

(b) The period is 3 seconds. (c) Because the maximum displacement from the point of equilibrium is the amplitude, the equation is of the form d = a cosω t. 75.

y − 2 = 4 ( x + 1)

4x − y + 6 = 0 76.

1 1 y−0= − x−  2 3 1 2 y = −x + 3 3x + 6 y − 1 = 0

71. False, Simple harmonic motion is given by a sin ω t or a cos ω t tangent a, function.

Chapter 5 Review

77.

79. Domain: ( −∞, ∞ )

6−2 4 =− −2 − 3 5 4 y − 2 = − ( x − 3) 5 5 y − 10 = −4 x + 12 4 x + 5 y − 22 = 0 Slope =

80. Domain: ( −∞, ∞ ) 81. Domain: ( −∞, ∞ )

1 3+ 2 3 1 4 m= = =− 3 −1 2 − 1 4 −3 4

78.

483

2 4 1 = − x−  3 3 4 3 y + 2 = −4 x + 1 y+

82. Domain: ( −∞, 7] 83. Domain: ( 6, ∞ ) 84. Domain: ( − ∞, − 3) ∪ ( − 3, 3) ∪ ( 3, ∞ )

4x + 3y + 1 = 0

Chapter 5 Review 1.

40°

260°

(a)

5. (a)

− 405° x

45°

(b) Quadrant IV (c) −405° + 720° = 315° −405° + 360° = −45° (b) Quadrant I (c) 45° + 360° = 405° 45° − 360° = −315° 4.

6. (a)

(a)

− 135°

210° x

(b) Quadrant III (c) −135° + 360° = 225° −135° − 360° = −495° (b) Quadrant III (c) 210° + 360° = 570° 210° − 360° = −150°

16 45  °  7. 135° 16′ 45′′ = 135 + +  ≈ 135.279° 60 3600   40  °  8. −234° 40′′ = −  234 +  ≈ −234.011° 3600  

34 19 °  + 9. − 6° 34′ 19′′ = −  6 +  ≈ − 6.572° 60 3600   24 9 °  + 10. 242° 24′ 9′′ =  242 +  ≈ 242.403° 60 3600   11. 135.29° = 135° + (0.29)(60)′ = 135° 17′ 24′′

484

Chapter 5

Trigonometric Functions

12. 25.8° = 25° 48′

21. (a)

13. −85.36° = −[85 + 0.36(60′)] = −85° 21′ 36′′ 14. −327.93° = −327° 55′ 48′′

15. Complement of 5°: 90° − 5° = 85°

−

5π 6

Supplement of 5°: 180° − 5° = 175°

16. Complement of 84°: 90° − 94° = 6° Supplement of 84°: 180° − 84° = 96°

(b) Quadrant III 5π 7π (c) − + 2π = 6 6 5π 17π − − 2π = − 6 6

17. Complement: Not possible, 253° is greater than 90°.

Supplement: Not possible; 253° is greater than 180°. 18. Complement: 180° Not possible; 180° is greater than 90°. Supplement of 180°: 180° − 108° = 72°

22. (a)

− y

19. (a)

9π 4 x

4π 3 x

(b) Quadrant IV 9π 7π + 4π = (c) − 4 4 9π π − + 2π = − 4 4

(b) Quadrant III 4π 10π (c) + 2π = 3 3 4π 2π − 2π = − 3 3

23. − 90° ⋅ 24. 225° ⋅

20. (a) 11π 6

25. 478° ⋅

π rad 180°

26. −151° ⋅ 27.

(b) Quadrant IV 11π 23π (c) + 2π = 6 6 11π π − 2π = − 6 6

π rad 180°

= − =

π 2

rad ≈ −1.571 rad

5π rad ≈ 3.927 rad 4

≈ 8.343 rad ≈ − 2.635 rad

5π 5π  180°  = ≈ 128.571° 7 7  π 

28. −

3π 3π  180°  =−  = −108° 5 5  π 

 180°  29. −3.5 = −3.5   ≈ −200.535°  π   180°  30. 1.55 = 1.55   ≈ 88.808°  π 

Chapter 5 Review

31. Complement:

−

Supplement: π −

π 8

π 8 =

3π 8

40. (a)

28 miles per hour =

7π 8

−

Number of revolutions per minute:

2464 7392 = ≈ 336.1 rev/min (7 / 3)π 7π 7π π is greater than 10 2

33. Complement: Not possible, Supplement: π − 34.

7π 3π = 10 10

Complement: Not possible; Supplement: Not possible;

35.

36.

8π π is greater than . 5 2

8π is greater than π . 5

7392 14,724 ≈ 2112 rad / min (2π ) = 7π 7

s 6ft = = 1.2 rad r 5ft

s 31 in. = = 2.583 rad r 12 in.

92 − 42 = 81 − 16 = 65.

opp 65 = hyp 9 adj 4 = cosθ = hyp 9 sin θ =

tan θ =

s = rθ

θ =

(b)

41. The opposite side is

s = rθ

θ =

28(5280) = 2464 ft/min 60

 1 7 Circumference of wheel: C = π  2  = π  3 3

5π 12 π 11π = Supplement: π − 12 12

32. Complement:

485

opp 65 = adj 4

1 9 9 65 = = sin θ 65 65 1 9 = secθ = cosθ 4

cscθ =

cot θ =

1 4 4 65 = = tan θ 65 65

37. s = rθ

s = 20(138°) s=

π 180°

46π ≈ 48.171 m 3

38. s = rθ = (15)

π 3

= 5π ≈ 15.71 cm θ

39. In one revolution, the arc length traveled is s = 2π r = 2π (6) = 12π cm. The time required for one 1 revolution is t = minutes. 500

Linear speed =

s 12π = = 6000π cm/min t 1 500

486

Chapter 5

Trigonometric Functions

42. The hypotenuse is

sin θ =

15 3 41

122 + 152 = 369 = 3 41.

5 41 41

cosθ =

4 41 = cosθ = 41 3 41 15 5 = tan θ = 12 4 41 41 = csc θ = 5 5 41 41

sec θ =

4 41 4 cot θ = 5

44. The adjacent side is 102 − 22 = 96 = 4 6. 2 1 = sin θ = 10 5

tan θ =

4 6 2 6 = 10 5 2

4 6 csc θ = 5 secθ =

6 12

5 6 12

2 6 12 =2 6 cot θ = 6

41 4

θ 96 = 4 6

3 41

45. Given: sin θ =

7 10

102 − 7 2 =

Adjacent side = cosθ =

51 10

tan θ =

7 7 51 = 51 51

opp 10 5 5 61 = = = hyp 2 61 61 61

tan θ =

7 7 51 = 51 51

adj 12 6 6 61 = = = hyp 2 61 61 61 opp 10 5 = = tan θ = adj 12 6

cot θ =

51 7

θ 12

43. The hypotenuse is 122 + 102 = 244 = 2 61.

sin θ = cosθ =

csc θ =

1 61 = sin θ 5

1 61 = cosθ 6 1 6 = cot θ = tan θ 5

10 10 51 = 51 51 10 cscθ = 7

secθ =

θ 51 2

θ 12

Chapter 5 Review

51. (a)

cos

5 3

(b)

sec

5 2 2

52. (a)

46. Given: cosθ = sin θ = tan θ = cot θ = sec θ = csc θ =

5 3 2 3 5

2 5 5

1 17 4

cosθ =

(b) 3

3 5 5

≈ 0.7071 ≈ 1.4142

53. cosθ secθ = cos θ ⋅

1 =1 cos θ

cot θ + tan θ tan θ =1+ = 1 + tan 2 θ = sec 2 θ cot θ cot θ

54.

17 17

55. tan 62° =

4 17 17

w 125

x = 125tan 62° ≈ 235 feet

cot θ = 4

56. (a)

17 secθ = 4 csc θ = 17

152 ft

30°

9 48. Given: cot θ = 40

(b)

40 41 9 cosθ = 41 40 tan θ = 9 41 sec θ = 9 41 csc θ = 40

sin θ =

(c)

h  h = 152sin 30° 152 1 h = 152   = 76 feet 2

sin 30° =

57. x = 12, y = 16, r = 144 + 256 = 400 = 20 y 4 = r 5 x 3 cosθ = = r 5 y 4 tan θ = = x 3 r 5 csc θ = = y 4 sin θ =

r 5 = x 3 x 3 cot θ = = y 4

 3π  tan   ≈ 0.5095  20  1  3π  cot  =   20  tan(3π / 20) ≈ 1.9626

1 4

47. Given: tan θ =

sin θ =

2 3

487

sec θ =

θ 9

49. (a) (b)

cos84° ≈ 0.1045 sin 6° ≈ 0.1045

50. (a)

csc(52° 12′ ) =

(b)

sec(66°7′ ) =

1 1 = ≈ 1.2656 sin(52° 12′ ) sin 52.2°

1 1 = ≈ 2.470 ′ cos(66° 7 ) cos(66.1167°)

488

Chapter 5

Trigonometric Functions

58. x = 2, y = 10, r = 4 + 100 = 104 = 2 26

sin θ =

10 5 26 y = = 26 r 2 26

2 26 x = cosθ = = 26 r 2 26 y 10 =5 tan θ = = x 2 26 r csc θ = = 5 y r sec θ = = 26 x x 1 cot θ = = y 5

sin θ =

y = r

481 24

cosθ =

x = r

481 24

tan θ =

y 5 8 15 = = x 2 3 16

csc θ =

r 481 = y 15

15 481 481

16 481 481

r 481 = x 16 x 16 cot θ = = y 15

sec θ =

59. x = −7, y = 2, r = 49 + 4 = 53 sin θ =

2 5 481 2 5 61. x = and y = , r =   +   = 3 8 24 3 8

2 2 53 y = = 53 r 53

62. x = −

7 7 53 x =− =− 53 r 53 2 y tan θ = = − 7 x

sin θ =

y −2 3 −1 26 = = =− r 2 26 3 26 26

53 2 53 sec θ = − 7 7 cot θ = − 2

cosθ =

x −10 3 −5 5 26 = = =− r 2 26 3 26 26

y −2 3 1 = = x −10 3 5 r csc θ = = − 26 y

cosθ =

csc θ =

60. x = 3, y = − 7, r =

sin θ =

tan θ =

32 + ( − 7) =

7 7 58 y = − = − 58 r 58

3 3 58 = 58 58 7 y = − tan θ = 3 x

x = cosθ = r

58 r = − cscθ = 7 y 58 r = 3 x 3 x = − cot θ = 7 y

secθ =

10 2 2 26  10   2  , y = − , r = −  + −  = 3 3 3 3 3    

r 26 =− x 5 x cot θ = = 5 y

sec θ =

6 63. secθ = , tan θ < 0  θ is in Quadrant IV. 5

r = 6, x = 5, y = − 36 − 25 = − 11

y 11 =− r 6 x 5 cosθ = = r 6

sin θ =

tan θ =

y 11 =− x 5

csc θ = − sec θ =

6 11 11

6 5

cotθ = −

5 11 11

Chapter 5 Review 3 64. sin θ = , cosθ < 0  θ is in Quadrant II. 8

68. θ = 210°

Reference angle: θ ' = 210° − 180° = 30°

y = 3, r = 8, x = − 55 sin θ =

y 3 = r 8

cosθ =

55 x =− r 8

489

θ = 210° x

θ ′ = 30°

y 3 3 55 =− =− x 55 55 8 csc θ = 3 tan θ =

8 55 sec θ = − =− 55 55

5π π −π = 4 4

x 2 = −  y = ± 21, sin θ > 0  y = 21 r 5

sin θ =

21 y = 5 r

cosθ =

2 x =− r 5

tanθ =

−2 y 21 x 2 21 =− =− cot θ = = x y 2 21 21

66. tanθ =

5π 4

Reference angle: θ ' =

55 cot θ = − 3

65. cos θ =

69. θ =

5 5 21 r = = 21 y 21 5 5 r =− sec θ = = x −2 2 csc θ =

12 12 y = − = −  r = 13, sinθ > 0  y = 12, x = −5 5 x x

y 12 sin θ = = r 13 x 5 cosθ = = − 13 r r 13 csc θ = = y 12 r 13 13 sec θ = = =− x −5 5 x 5 cot θ = = − y 12

θ=

θ′=

70. θ =

5π 4 x

π 4

9π π is coterminal with . 4 4

Reference angle: θ ' =

π 4

θ′= θ=

π 4

9π 4

71. θ = − 264° is coterminal with 96°.

67. θ = 330°

Reference angle: θ ' = 180° − 96° = 84°

Reference angle: θ ' = 360° − 330° = 30°

θ ′ = 84° x

θ = − 264°

θ = 330° θ ′ = 30°

490

Chapter 5

Trigonometric Functions

72. θ = − 635° is coterminal with 85°.

Reference angle: θ ' = 85° y

76. 315° is in Quadrant IV with reference angle 45°.

sin 315° = − sin 45° = −

2 2

2 2 tan 315° = − tan 45° = −1

cos315° = cos 45° = θ ′ = 85° x

θ = − 635°

77. − 225° is coterminal with 135° in Quadrant II with

reference angle 45°. sin ( − 225°) = sin ( 45°) =

73. θ = −

6π 4π is coterminal with . 5 5

θ' = π −

cos ( − 225°) = − cos ( 45°) = − tan ( − 225°) =

Reference angle: 4π π = 5 5

2 2

−

2 2

2 2 = −1 2 2

78. −315° is coterminal with 45° in Quadrant I.

2 2 2 cos( −315°) = 2 tan(−315°) = 1

sin(−315°) = θ′=

π 5

6π θ=− 5

79. sin(4π ) = sin(0) = 0

17π π 74. θ = − is coterminal with . 3 3

Reference angle: θ ' =

3  7π  π  sin   = sin  3  = 2  3     7π  π  1 cos   = cos  3  = 2  3   

π 3 x

θ =−

17π 3

75. 240° is in Quadrant III with reference angle 60° .

3 2 1 cos240° = − cos60° = − 2 − 3 2 tan 240° = = 3 −1 2

sin 240° = − sin60° = −

tan ( 4π ) = 0 80. 7π / 3 is coterminal with π / 3 in Quadrant I.

θ′=

cos ( 4π ) = 1

 7π  π  tan   = tan   = 3  3  3

81. −9π / 4 is coterminal with 7π / 4 in Quadrant IV with reference angle π / 4 .

2  9π  π  sin  −  = − sin   = − 2  4  4 2  9π  π  cos  −  = cos   =  4  4 2  9π  − 2 / 2 = −1 tan  −  = 2 /2  4 

Chapter 5 Review 82. sin ( − 3π ) = sin π = 0

90.

cos ( − 3π ) = cos π = −1 tan ( − 3π ) = tan π = 0

3 f ( x ) = sin x 5 3 Amplitude: 5 y

83. tan 33° ≈ 0.6494 84. csc105° = 85. sec

12π = 5

1 ≈ 1.0353 sin105°

1 2

1 ≈ 3.2361  12π  cos    5 

 π 86. sin  −  ≈ −0.3420  9

− 3π 2

−π 2

91. Period:

2π

92. Period:

y 8 6

2π = 4π 1 ( 2)

Amplitude:

4 2

π 2

3π 2

2π =π 2 Amplitude: 3.4

94. Period:

88. f ( x ) = 7 cos x

2π =4 π ( 2)

Amplitude: 4

Amplitude: 7 y

95.

2 −π

3 2

93. Period:

−8

−2π

π Amplitude: 5

Amplitude: 6

−π 2

3π 2

π 2

−1

87. f ( x ) = 6 sin x

− 3π 2

491

−2

2π

y = 3cos2π x Amplitude: 3 2π Period: =1 2π

−4 −6 −8

1 89. f ( x ) = cos x 5 1 Amplitude: 5

−1

1 2

−2 −3

y 1 1 2

−2π

−π 2 −1

π 2

2π

−1

492

Chapter 5

Trigonometric Functions

96.

y = −2sin π x 2π Period: =2 π Amplitude: −2 = 2

5 x 99. f ( x ) = − cos 2 4 5 2

Amplitude:

Reflected in x-axis Period:

y 3

2π = 8π 14 y 5 4 3 2 1

1 x

1 −1 −2

− 4π

−3

4π

−3 −4 −5

 2x  97. f ( x) = 5sin  −   5 

1 πx 100. f ( x ) = − sin 2 4

Amplitude: 5 Period:

2π = 5π 25

1 2

Amplitude: Period: 8

y 6

y 1

2 − 6π

8π

4π

−2

6π

−6

−4

−2

−1 2

−6

−1

 x 98. f ( x ) = 8cos  −   4 Period:

2π

(1 4 )

5 101. f ( x ) = sin( x − π ) 2

= 8π Amplitude:

Amplitude: 8

5 2

Period: 2π

Reflected in y-axis

Shift:

x −π = 0 x =π

x − π = 2π x = 3π

and y

4π −4 −6 −8

8π

4 3 1 −1

−2 −3 −4

Chapter 5 Review 102. f ( x) = 3cos( x + π )

493

π  105. f ( x ) = − 3sin  2 x −  − 1 4 

Period: 2π

  π  = − 3sin 2  x −  − 1 8   

Amplitude: 3 This is the graph of y = 3cos x shifted to the left π units. y

Amplitude: 3 2π = π 2

Period:

Reflection in x-axis

3 2 1 −π

Horizontal shift

unit right

Vertical shift downward one unit −3

−4

103. f ( x ) = 2 − cos

πx

−π

Amplitude: 1

−π 2

−3 −4 −5

106. f ( x) = 4 − 2cos(4 x + π )

Reflection in x-axis Vertical shift upward two units

Amplitude: 2

Period:

−4 − 3 − 2 −1 −1

−2

−3 −4

1 104. f ( x ) = sin π x − 3 2

−π 2

−π 4

−2

π 4

π 2

Reflection in x-axis

1 2

Horizontal shift

Period: 2

π 4

unit to the left

Vertical shift four units upward

Vertical shift downward three units y

−2

2π =4 Period: (π 2 )

Amplitude:

π 2

−1

107. f ( x) = a cos(bx − c)

Amplitude: 3 Period: π

−1

a = 3, b = 2, c = 0

−2

f ( x ) = 3cos ( 2 x )

−3 −4

Chapter 5

494

Trigonometric Functions

108. f ( x) = a cos(bx − c)

Amplitude:

113. f ( x ) = − tan

1 2

Two consecutive asymptotes:

a = 12 , b = π , c = 0

x = −2, x = 2

1 cos π x 2

f ( x) =

Reflected in x-axis y

109. f ( x) = a cos (bx − c)

3 Amplitude: 4 Period: 2π Shift:

4 2

−6

4 3 π a = − , b = 1, c = 4 4 3 π  f ( x) = − cos  x −  4 4 

−4

−2

−6

114. f ( x) = 4 tan π x

π =1 π

Two consecutive asymptotes:

Amplitude: 4 Period: π

1 1 x=− , x= 2 2

a = − 4, b = 2, c =

111. S = 48.4 − 6.1cos

−2 −4

Period:

110. f ( x) = a cos(bx − c)

Shift:

π =4 (π 4 )

Period:

Period: 2

πx

4 2

πt 6

Maximum sales: t = 6 (June) Minimum sales: t = 12 1 tan 2 x − 1 4

115. f ( x ) =

(December) 56

Period:

Two consecutive asymptotes: 0

112. S = 56.25 + 9.50sin

2x = −

πt 6

2x =

Maximum sales: t = 3

 x =

π 4

Minimum sale: t = 9

3 2

(September)

 x = −

(March)

−π

Chapter 5 Review

116. f ( x ) = 2 + 2 tan

Period:

x 3

1 π  119. f ( x ) = cot  x −  2 2  Period: π

= 3π

495

Two consecutive asymptotes:

Two consecutive asymptotes: 3π 3π x=− , x= 2 2

x−

π 2

=0 x=

2 3π x− =π  x = 2 2

6 5 4 3

4 2

− 3π − 2π

3π 4 π

x − 2π

−π

−1

−4

2π

−6

117. f ( x ) = 3cot

Period:

x 2

π 4  120. f ( x ) = 4cot  x +  = π 4   tan  x +  4 

= 2π

Two consecutive asymptotes: x =0 x=0 2 x = π  x = 2π 2

Period:

=π

Two consecutive asymptotes: x+

π 4

=0 x=−

4 3π x+ =π  x = 4 4

y 6

4 2

− 4π

4 2π

− 2π

4π

− 2π

1 πx 1 118. f ( x ) = cot = 2 2 2 tan π x 2 Period:

2π

121. f ( x ) = 4 sec x

Two consecutive asymptotes:

x = 0, x = 2

π 2

, x=

3π 2

y 2

−1

Period: 2π

π =2 (π 2 )

Two consecutive asymptotes:

−2

x −2

2π

3π

−4 −6

496

Chapter 5

122. f ( x) = −

Trigonometric Functions

1 1 csc x = − 2 2sin x

Period: 2π

π  125. f ( x) = sec  x +  4  Two consecutive asymptotes:

Two consecutive asymptotes: x = π , x = 2π y

x = − x =

−π

2π

3π

3π 4

π 4

Secant function shifted

to left

−1 3

−2

123. f ( x ) =

1 csc 2 x 4

Period:

2π =π 2

− 2π

−1

2π

−2 −3

Two consecutive asymptotes: 2x = 0  x = 0 2x = π  x =

−π

π 2

1 1 126. f ( x) = csc(2 x − π ) = 2 2sin(2 x − π ) Period:

2π =π 2

Two consecutive asymptotes:

2x − π = π  x = π

2 1

2x − π = 0  x =

π 2

−2 −3

1 1 124. f ( x ) = sec 2π x = 2 2 cos2π x

Period:

−π

−π 2

2π =1 2π

π 2

−1

Two consecutive asymptotes:

1 2π x =  x = 2 4 3π 3 2π x = x= 2 4

127. f ( x) =

1 1 π x  π x  sec  , g ( x) = cos   4 4  2   2  1

−6

−1

Answers will vary. −1

−1

Chapter 5 Review

π π   128. f ( x) = csc  x + , g ( x) = sin  x +  4 4  

134. f ( x) =

497

1  π 1 π  sec  2 x − , g ( x) = cos  2 x −  3  4 3 4 

− 2π

2π

−π

−2

−1

Answers will vary.

129. f ( x) = 4csc ( 2 x − π ), g ( x) = 4sin ( 2 x − π )

135. f ( x ) = e sin2 x x

−π

−4

−8

− 20

Answers will vary.

Since the damping factor is e − x , − e − x ≤ e − x sin 2 x ≤ e − x .

130. f ( x) = − 2sec ( 4 x + π ), g ( x) = − 2cos ( 4 x + π )

y = ±e − x touches y = e − x sin 2 x at x =

−π

π 4

π 2

f ( x ) = e− x sin2 x has x-intercepts at sin 2 x = 0 or x = nπ . 2

136. f ( x ) = e cos x x

−5

Answers will vary. 131. f ( x ) = 2sec ( x − π ) , g ( x ) = 2cos ( x − π )

−4

−5 −4π

4π

y = ±e x touches y = e x cos x at x = nπ .

−5

f ( x ) = e x cos x has x- intercepts at cos x = 0

Answers will vary. 132. f ( x) = 2csc ( x − π ), g ( x) = 2sin ( x − π ) 5

− 4π

Since the damping factor is e x , − e x ≤ e x cos x ≤ e x .

or x =

π 2

+ nπ .

137. f ( x ) = 2 x cos x 14

4π

−10

−5

Answers will vary.

π π   133. f ( x) = − csc  3 x − , g ( x) = − sin  3 x −  2 2   2

− 14

Since the damping factor is 2 x, −2 x ≤ 2 x cos x ≤ 2 x.

y = ±2 x touches y = 2 x cos x at x = nπ . − 4π 3

4π 3

−2

Answers will vary.

y = 2 x cos x has x- intercepts at cos x = 0 or x =

π 2

+ nπ .

498

Chapter 5

Trigonometric Functions 153. Let y = arcsin( x). Then

138. f ( x) = x sin π x 6

sin y = x = −9

sec y =

1 − x2

−6

Since the damping factor is x,

1 − x2 1 − x2

− x ≤ x sin π x ≤ x. y = ± x touches y = x sin π x at x =

1 + n. 2

x and 1

y = x sin π x has x- intercepts at sin π x = 0 or x = n.

y 1 − x2

1 π  π  1 139. (a) Because sin  −  = − , arcsin  −  = − . 6 2 2 6    

154. Let u = arcsin 10x. Then sin u = 10 x and

 3 3 5π  5π  , arccos  − (b) Because cos   = −  2  = 6 . 2  6   

csc ( arcsin10 x ) = csc u =

 3 3 π  π , arcsin  − 140. (a) Because sin  −  = −  2  = − 3 . 2  3  

1 2π  2π   1 . (b) Because cos   = − , arccos  −  = 2 3  3   2 141. (a) Because cos

π 2

= 0, cos −1 0 =

π 2

u 1 − 100x 2

155. Let y = arccos

142. (a) Because cos 0 = 1, cos −1 1 = 0.

(

3, tan −1 −

)

3 = − . 3

146. cos

( −0.94) ≈ −1.22

−1

( −0.12) ≈ 1.69

147. arctan 21 ≈ 1.52 148. arctan ( −12 ) ≈ −1.49

4 − x2

144. arcsin 0.63 ≈ 0.68 145. sin

x2 x2 . Then cos y = and 2 4−x 4 − x2

(4 − x ) − ( x ) sin y =

143. arccos ( 0.42 ) ≈ 1.14

−1

10x

 π π (b) Because tan  −  = −1, tan −1 ( −1) = − . 4  4

 π (b) Because tan  −  = −  3

hyp 1 = . opp 10 x

16 − 2 x 2 4−x

2 4 − 2 x2 . 4 − x2

4 − x2

(4 − x 2 ) 2 − (x 2) 2

y x2

149. tan −1 (0.81) ≈ 0.68 150. tan −1 6.4 ≈ 1.42 151. sin θ =

x  x   θ = arcsin   16  16 

152. tan θ =

x +1  x +1  θ = arctan   20  20 

Chapter 5 Review x x 156. Let u = arccos . Then cos u = and 2 2

x  tan  arccos  = tan u 2  =

161. tan14° =

y  y = 37,000 tan14° ≈ 9225.1 feet 37,000

x+y  x + y = 37,000 tan 58° ≈ 59,212 feet 37,000 x = 59,212.4 − 9225.1 ≈ 49,987.2 feet The towns are approximately 50,000 feet apart or 9.47 miles. tan 58° =

4 − x2 . x

32° 2

76°

58° 37,000

4 − x2

14° y

x u

157. tan θ = 70 45

d1   d1 ≈ 483  d1 + d 2 = 1217  650  d2  d 2 ≈ 734  cos 25° =  810

162. sin 48° =

θ = tan −1 ( 70 45 ) θ ≈ 57.26° 158.

499

h 2.5 2.5 tan 25° = h

d3   d3 ≈ 435  650  d3 − d4 ≈ 93 d4 sin 25° =  d4 ≈ 342   810

tan 25° =

cos 48° =

1.166 ≈ h The shuttle is about 1.166 miles off the ground. 159. sin(1°10′) =

a 3.5

 7°  a = 3.5sin (1°10′ ) = 3.5sin   ≈ 0.0713 km or 7 meters 6 3.5

93  θ ≈ 4.4° 1217 D  D ≈ 1217sec 4.4° ≈ 1221 sec 4.4° ≈ 1217 tanθ ≈

The distance is about 1221 miles and the bearing is N 85.6° E.

1° 10′ not drawn to scale

 12  θ = arctan   ≈ 0.1194 or 6.84°  100  h sin(θ )= 4 h = 4sin ( 0.1194 ) = 0.48 mile or 2534 feet

(answer depends on angle θ accuracy) h

48°

12 160. tan θ = 100

25° 48° 65° 810 650 D θ d1 d2

d4 C

163. Use cosine model with amplitude 3 feet.

Period: 15 seconds  2π  d = 3cos  t  15 

164. Use a cosine model with amplitude

Period: 2 seconds  Period: 2 =

7 = 3.5 inches. 2

2π b=π b

y = 3.5cosπ t

Chapter 5

500

Trigonometric Functions

165. False. y = sinθ is a function, but it is not one-to-one.

171. (a)

166. True. y = cos x is not one-to-one on the interval − π 2 ≤ x ≤ π 2. The graph would fail the Horizontal Line Test.

−π

The polynomial function is a good approximation for the arctangent function when x is close to 0. x9 (b) Next term: 9 x3 x5 x7 x9 arctan x ≈ x − + − + 3 5 7 9 The accuracy of the approximation increases as more terms are added.

168. False. The sine of a non-acute angle θ is equal to the sine of the reference angle θ ′ after the appropriate + or − sign is applied.

(a)

0.672s 2 3000

1.28°

5.12° 11.40° 19.72° 29.25° 38.88°

−10

167. False. The sine or cosine functions are used to model simple harmonic motion.

169. tan θ =

60 −π

(b) θ increases at an increasing rate. The function is not linear. 170.

−10

−

3π 2

−6

The polynomial function is a good approximation for the secant function when x is close to 0.

Chapter 5 Test y

1. (a)

2. angular speed =

5π 3

θ time

100 kilometers 100,000 meters 250   m sec hour 3600 seconds 9 x

s r 250 m = 9 sec 0.625 m 4 = 44 rad sec 9

Since s = θ r , θ =

(b)

(c)

5π 11π + 2π = 3 3 π 5π − 2π = − 3 3 5π 180° ⋅ = 300° 3 π

x = −1, y = 4, r = 1+16 = 17

sin θ =

4 17

cosθ = −

4 17 17 =−

17 17

tan θ = −4 17 4 sec θ = − 17 csc θ =

cot θ = −

1 4

Chapter 5 Test

tan θ =

7 2

cosθ =

49 53 +1 = 4 2

Period:

2 53 53

2π = 3π 23

Amplitude: 2 Reflection in the x-axis

7 53 sin θ = tan θ cosθ = 53

y 3

53 = csc θ = 7 7 53 2 cot θ = 7

−2 π − π

53 2

sec θ =

2x 3

10. g ( x ) = − 2 sin

tan 2 θ + 1 = sec 2 θ  secθ =

501

−1

2π

−2 −3

θ = 255°

11.

f ( x) =

θ ′ = 255° − 180° = 75° Period:

1 tan 4 x 2

π 4

Two consecutive asymptotes: x=−

θ = 255° x

π 8

,x=

π 8

θ ′ = 75° 2 1

sec θ =

1 < 0  Quadrants II or III cosθ

π 2

π 4

tanθ > 0  Quadrants I or III Hence, Quadrant III. 7.

cosθ = −

3 2

θ = 150°, 210° 8.

1 1 = 1.030  sin θ = and θ in sin θ 1.030 Quadrant 1 or II.

csc θ =

Using a calculator, θ = 1.33, 1.81 radians. 9.

3 cosθ = − , sin θ > 0, Quadrant II 5 4 sin θ = 5 5 sec = − 3 3 cot θ = − 4 4 tan θ = − 3 5 csc θ = 4

1 1 sec ( x − π ) − 4 is y = sec x shifted π units to 2 2 the right, and 4 units downward.

12. f ( x ) =

Period: 2π Two consecutive asymptotes: x −π = −

π 2

x=

2 3π x −π =  x = 2 2

y 1 −π

−1

−2 −3 −4

−7

502

Chapter 5

Trigonometric Functions

13. f ( x ) = 2 cos( −2 x ) + 3

19. Let u = arccos

Amplitude: 2

2 2  cos u = . 3 3

2 5  Then tan  arccos  = tan u = . 3 2 

Reflection in the y-axis Shifted vertically upward 3 units y

 x 20. f ( x) = 2arcsin   2

6 5

−3

1 −π 2

−π

π 2

−1

−π

π  14. f ( x ) = 2 cot  x −  2  Period: π

21.

f ( x) = 2arccos x 2π

Two consecutive asymptotes: x−

π 2

=0 x=

;

−π 2

2 π 3π x− =π  x = 2 2

π 2

3π 2

−2

−3

22. f ( x) = arctan

−4

π  15. f ( x) = 2 csc  x +  2  π 2

−10

x+ x+

16.

− 2π

π 2

=0x=− =π  x =

to the left −π

Two consecutive asymptotes:

x 2

π 2

Period: π Shifted

−2

23.

−4

−π 2

2π

80 mi

17.

64 mi

6 0

tanθ =

−2

−4

Period is 2.

2π 1 b= b 2

Horizontal shift

π 4

x π y = 2 cos  +  2 4

π units to the left, so c = − . 4

80 96

80 ≈ 51.34° 64 Bearing: S 51.34° W

Not periodic

θ = arctan

18. Amplitude: a = 2

Period: 4π =

Airport −6

Plane

24.

At t = 0, the ball starts at its lowest point, so d = acosω t. Distance from low point to high point: 2 a = 12  a = 6

Returns to lowest point every 2 seconds: Period =

2π

= 2  ω = π

So, the equation is d = − 6 cos π t.

C H A P T E R 6 Analytic Trigonometry Section 6.1

Using Fundamental Identities ............................................................504

Section 6.2

Verifying Trigonometric Identities ....................................................514

Section 6.3

Solving Trigonometric Equations ......................................................524

Section 6.4

Sum and Difference Formulas ...........................................................535

Section 6.5

Multiple-Angle and Product-to-Sum Formulas ................................548

Chapter 6 Review .......................................................................................................567 Chapter 6 Test ............................................................................................................582

C H A P T E R 6 Analytic Trigonometry Section 6.1 Using Fundamental Identities 1.

(a) (b)

1 , matches (iii). csc u 1 cos u = , matches (i). sec u 1 tan u = , matches (ii). cot u

sin u =

9. cot θ = −1, sin θ = −

(a)

sin 2 u = 1 − cos2 u, matches (ii).

(b)

sec 2 u = 1 + tan 2 u, matches (iii).

(c)

csc 2 u = 1 + cot 2 u, matches (i).

2 , θ is in Quadrant IV. 2

tan θ =

1 = −1 cot θ

cos θ =

− 2 2 sin θ = = −1 tan θ

secθ =

1 = cosθ

cscθ =

1 2 = − = − sin θ 2

2 = 2

2 2

sin u

secu

− tanu

sin x =

1 1 = − csc x 2

1 3 sin x = , cos x = , x is in Quadrant I. 2 2

cos x =

−1 2 sin x 3 = = − tan x 2 3 3

sec x =

1 2 2 3 = − = − cos x 3 3

cot x =

1 = tan x

tan x =

10. tan x =

12 3 2

1 3

3 3

cot x = 3 csc x = 2 sec x =

8. cosθ =

2 3

2 3 3

1 , sin θ = 2

3 , θ is in Quadrant I. 2

sin θ = cosθ

3 2 = 12

cot θ =

1 = tan θ

1 = 3

secθ =

1 = 2 cosθ 1 = sin θ

3 = 3

−25 7  x is in Quadrant III. , sec x = 24 24 24 cot x = 7 24 cos x = − 25 7 sin x = − 1 − cos2 x = − 25 1 25 csc x = =− sin x 7

11. tan x =

tan θ =

cscθ =

3 , csc x = − 2, x is in Quadrant III. 3

3 = 1 3 3

2 2 3 = 3 3

12. cot φ = −5, sin φ =

26 , φ is in Quadrant II. 26

cot φ = cot φ ⋅ sinφ = −

5 26 26

1 1 =− cot φ 5 1 26 csc φ = = = 26 sin φ 26 tan φ =

sec φ =

504

1 26 −26 = =− cosφ 5 26 5

Section 6.1 17 8 , sin φ = , φ is in Quadrant II. 15 17 15 cosφ = − 17 17 csc φ = 8 8 17 8 tan φ = =− 15 −15 17 15 cot φ = − 8

13. sec φ = −

4 π  3 14. cos  − x  = , cos x = , x is in Quadrant I. 5 2  5

cot x =

1 1 5 = =− sec θ − 5 5 1 1 = cot θ = tan θ 2

cosθ =

sin θ = − 1 − cos2 θ = − 1− cscθ =

1 4 −2 2 5 =− = =− 5 5 5 5

1 5 =− sin θ 2

cot θ =

1 1 = secθ 4 1 15 = − 15 − 15

15 1 sin θ = tan θ cosθ = − 15   = − 4  4

(

2 2 5  x is in Quadrant II. , tan x = − 3 5 4 5 =− 9 3

1 5 =− tan x 2

1 3 5 =− cos x 5 1 3 csc x = = sin x 2

16. csc x = 5, cos x > 0, x is in Quadrant I. 1 1 = sin x = csc x 5 cos x =

2 6 5

tan x =

sin x 1 5 6 = ⋅ = cos x 5 2 6 12

1 5 5 6 = = cos x 2 6 12 1 =2 6 cot x = tan x

secθ = − tan 2 θ + 1 = − 5

tan 2 θ = sec 2 θ − 1 = 16 − 1 = 8  tan θ = − 15

sec x =

17. tan θ = 2, sin θ < 0  θ is in Quadrant III.

cosθ =

2 2 15. sin ( − x ) = − sin x = −  sin x = 3 3

cos x = − 1 − sin 2 x = − 1 −

505

18. secθ = 4, tan θ < 0, θ is in Quadrant IV.

3 4 sin x = 1 −   = 5 5   sin x 3 5 3 tan x = = ⋅ = cos x 5 4 4 1 5 csc x = = sin x 3 1 5 sec x = = cos x 4 1 4 cot x = = tan x 3

sin x =

Using Fundamental Identities

cscθ =

)

1 4 4 15 = − = − sin θ 15 15

19. cscθ is undefined and cosθ < 0  θ = π . sin θ = 0

cosθ = −1 tan θ = 0 cot θ is undefined. secθ = −1 20. tanθ is undefined, sin θ > 0.

θ=

tan θ =

sin θ is undefined  cosθ = 0. cosθ

sin θ = 1 − 02 = 1 1 =1 sin θ 1 is undefined. sec θ = cosθ cosθ 0 cot θ = = =0 sin θ 1 csc θ =

21. sec x cos x =

1 ⋅ cos x = 1 cos x

Matches (d). 22. tan x csc x =

sin x 1 1 = = sec x cos x sin x cos x

Matches (a).

506

Chapter 6

Analytic Trigonometry

23. cot 2 x − csc 2 x = cot 2 x − (1 + cot 2 x ) = −1

35. sin φ (csc φ − sin φ ) = sin φ csc φ − sin 2 φ

Matches (b). 24.

= sin φ ⋅

(1 − cos x ) csc x = sin x  sin1 x  = sin x 2

= 1 − sin 2 φ

= cos2 φ

Matches (f).

36. cos x(1 + tan 2 x) = cos x(sec 2 x)

sin ( − x )

− sin x 25. = = − tan x cos ( − x ) cos x

 1  = cos x 2   cos x  1 = cos x

Matches (e). 26.

sin (π 2 ) − x  cos x = = cot x cos (π 2 ) − x  sin x

= sec x

Matches (c).  1  27. sin x sec x = sin x   = tan x  cos x  Matches (b). 28. cos2 x ( sec 2 x − 1) = cos2 x tan 2 x = sin 2 x

37.

csc x 1 sin x 1 = ⋅ = = sec x cot x sin x cos x cos x

38.

sec θ 1 = sin θ = tan θ csc θ cosθ

39. sec α

Matches (c).

sin α 1 = ( sin α ) cot α tan α cos α

29. sec 4 x − tan 4 x = ( sec 2 x + tan 2 x )( sec 2 x − tan 2 x )

(

)

= sec 2 x + tan 2 x (1) 2

= sec x + tan x

Matches (f). 30. cot x sec x =

cos x 1 1 ⋅ = sin x cos x sin x

= csc x Matches (a).

sec 2 x − 1 tan 2 x sin 2 x 1 = = ⋅ 2 = sec2 x 31. 2 2 2 sin x sin x cos x sin x Matches (e). 32.

cos2 (π 2 ) − x  cos x

sin 2 x sin x sin x = cos x cos x = tan x sin x

40.

cos x sin x = cos x sin x

 sin β  34. cos β tan β = cos β   = sin β  cos β 

1 cos α  ( sin α )   =1 cos α  sin α 

1 + tan 2 θ sec 2 θ = 2 sec θ sec 2 θ =1

1 π  = cot x 41. sin  − x  csc x = cos x ⋅ sin x 2  π  42. cot  − x  cos x = tan x cos x 2  sin x = ⋅ cos x = sin x cos x 43.

cos2 y 1 − sin 2 y = 1 − sin y 1 − sin y

Matches (d). 33. cot x sin x =

1 − sin 2 φ sin φ

(1 + sin y )(1 − sin y ) 1 − sin y

= 1 + sin y 44.

1 1 = = sin 2 x cot 2 x + 1 csc 2 x

45. cot 2 x − cot 2 x cos2 x = cot 2 x (1 − cos2 x )

cos2 x 2 sin x = cos2 x sin 2 x

46. sec 2 x tan 2 x + sec 2 x = sec 2 x ( tan 2 x + 1)

(

)

= sec 2 x sec 2 x = sec 4 x

Section 6.1

47.

cos2 x − 4 ( cos x + 2 )( cos x − 2 ) = = cos x + 2 cos x − 2 cos x − 2

48.

csc 2 x − 1 ( csc x − 1)( csc x + 1) = = csc x + 1 csc x − 1 csc x − 1

60.

Using Fundamental Identities

sec x − 1 − ( sec x + 1) 1 1 − = sec x + 1 sec x − 1 ( sec x + 1)( sec x − 1)

sec x − 1 − sec x − 1 sec 2 x − 1 −2 = tan 2 x  1  = −2  2   tan x  = −2cot 2 x =

( ) = ( sec x ) = sec x

49. tan 4 x + 2 tan 2 x + 1 = tan 2 x + 1

( ) = ( cos x ) = cos x

50. 1 − 2sin 2 x + sin 4 x = 1 − sin 2 x 2

61. tan x −

51. sin 4 x − cos 4 x = ( sin 2 x + cos2 x )( sin 2 x − cos2 x )

62. tan x +

= (1) ( sin 2 x − cos2 x ) = sin 2 x − cos2 x

(

)(

4 4 2 2 2 2 52. sec x − tan x = sec x + tan x sec x − tan x

)

sec 2 x tan 2 x − sec2 x −1 = = = − cot x tan x tan x tan x cos x sin x cos x = + 1 + sin x cos x 1 + sin x sin x + sin 2 x + cos2 x = cos x (1 + sin x )

= sec x + tan x

53. csc3 x − csc 2 x − csc x + 1 = csc 2 x ( csc x − 1) − ( csc x − 1)

= cot 2 x ( csc x − 1)

63.

54. sec3 x − sec 2 x − sec x + 1 = sec 2 x ( sec x − 1) − ( sec x − 1)

= (sec 2 x − 1)(sec x − 1) = tan 2 x(sec x − 1) 55.

( sin x + cos x ) = sin x + 2sin x cos x + cos x 2

(

)

= sin 2 x + cos2 x + 2sin x cos x = 1 + 2sin x cos x

sin x + 1 cos x (1 + sin x )

1 cos x = sec x

= ( csc 2 x − 1) ( csc x − 1)

cos x 1 + sin x cos2 x + 1 + 2sin x + sin 2 x + = 1 + sin x cos x cos x (1 + sin x ) =

2 + 2sin x cos x (1 + sin x )

2 = 2sec x cos x

tan x 1 + sec x tan x + (1 + sec x ) + = 1 + sec x tan x (1 + sec x ) tan x 2

64.

tan 2 x + 1 + 2sec x + sec 2 x (1 + sec x ) tan x

= sec 2 x + 2tan x

sec 2 x + 2sec x + sec 2 x (1 + sec x ) tan x

57.

( csc x + 1)( csc x − 1) = csc2 x − 1 = cot 2 x

2sec 2 x + 2sec x (1 + sec x ) tan x

58.

( 5 − 5sin x )( 5 + 5sin x ) = 25 − 25sin 2 x

56. (1 + tan x) = 1 + 2tan x + tan 2 x 2

= 1 + tan 2 x + 2tan x

(

= 25 1 − sin 2 x

)

= 25cos2 x

59.

1 1 1 − cos x + 1 + cos x + = 1 + cos x 1 − cos x (1 + cos x )(1 − cos x )

2 1 − cos2 x 2 = sin 2 x = 2 csc 2 x =

507

2sec x ( sec x + 1)

(1 + sec x ) tan x

2sec x = tan x  1  cos x  = 2    cos x  sin x  2 = sin x = 2 csc x

Chapter 6

508

65.

66.

67.

Analytic Trigonometry 72. y1 = cos x + sin x tan x, y2 = sec x

sin x  cos x  = sin x   tan x  sin x  = cos x

sin 2 y 1 − cos2 y = 1 − cos y 1 − cos y

(1 + cos y )(1 − cos y ) 0

1.4

73. y1 =

0.6

0.8

1.0

1.2

1.4

y1 = y2 0

1.5

Conjecture: y1 = y2 74. y1 = sec 4 x − sec 2 x, y2 = tan 2 x + tan 4 x

0.2

0.4

0.6

0.8

1.0

1.2

1.4

y1 0.0428 0.2107 0.6871 2.1841 8.3087 50.3869 1163.6143 y2 0.0428 0.2107 0.6871 2.1841 8.3087 50.3869 1163.6143

1.0

1.2

1.4

1200

y1 = y 2

y2 0.1987 0.3894 0.5646 0.7174 0.8415 0.9320 0.9854 0

y1 = y2

0.4

y1 0.1987 0.3894 0.5646 0.7174 0.8415 0.9320 0.9854

1.0

0.2

y2 1.2230 1.5085 1.8958 2.4650 3.4082 5.3319 11.6814

π  71. y1 = cos  − x  , y2 = sin x 2  

0.8

cos x 1 + sin x , y2 = 1 − sin x cos x

y1 1.2230 1.5085 1.8958 2.4650 3.4082 5.3319 11.6814

5 tan x − sec x 5 ( tan x − sec x ) 70. ⋅ = tan x + sec x tan x − sec x tan 2 x − sec 2 x 5 ( tan x − sec x ) = −1 = 5 ( sec x − tan x )

0.6

1.5

Conjecture: y1 = y2

3 sec x + tan x 3 ( sec x + tan x ) ⋅ = sec x − tan x sec x + tan x sec 2 x − tan 2 x 3 ( sec x + tan x ) = 1 = 3 ( sec x + tan x )

0.4

1.2

1 − cos y

= tan 4 x ( csc x − 1)

0.2

1.0

y1 = y2

0.8

tan 2 x csc x − 1 tan x ( csc x − 1) ⋅ = csc x + 1 csc x − 1 csc 2 x − 1 2 tan x ( csc x − 1) = cot 2 x 2 = tan x ( csc x − 1) tan 2 x

69.

0.6

y2 1.0203 1.0857 1.2116 1.4353 1.8508 2.7597 5.8835

= 1 + cos y 68.

0.4

y1 1.0203 1.0857 1.2116 1.4353 1.8508 2.7597 5.8835

tan y = tan y sin 2 y + cos2 y

0.2

1.5

Conjecture: y1 = y2 75. y1 = cos x cot x + sin x = csc x

1.5

Conjecture: y1 = y2 −2 π

2π

−4

Section 6.1 76. sin x ( cot x + tan x ) = sec x

Using Fundamental Identities

y1 and y 2 =

509

1 = sec θ cosθ

4 − 2π

2π

y1 = y2

−2π

2π

−4 −4

 1  1 − cos x  77. y1 =  sin x  cos x 

y1 and y 2 =

= tan x

1 = cot θ tan θ

4 4

−2π − 2π

2π

−4 −4

1  1 + sin θ cosθ  + 78. y1 =  2  cosθ 1 + sin θ 

1  1 + sin θ cosθ  + = sec θ . 2  cosθ 1 + sin θ 

79. x = 5 sin θ , 0 < θ <

y1 and y2 = sin θ

π 2

25 − x = 25 − (5sin θ )2 2

−2π

= 25 − 25sin 2 θ 2π

y1 and y2 = cosθ 4

80. x = 2 cosθ , 0 < θ < y2

−2π

(

= 25 1 − sin 2 θ

2π

−4

y1 and y2 = tan θ

2π

−4

1 = csc θ sin θ

4 − x2 =

4 − ( 4cos 2 θ )

4(1 − cos 2 θ )

4sin 2 θ

81. x = 3 tan θ , 0 < θ <

−2π

= 2sin θ

y1 and y 2 =

)

= 25cos2 θ = 5 cosθ

−4

−2π

It appears that

2π

π 2

x2 + 9 =

(3tan θ )2 + 9

9 tan 2 θ + 9

9 ( tan 2 θ + 1)

9sec 2 θ

= 3secθ

−4

510

Chapter 6

Analytic Trigonometry

82. x = 10 tan θ , 0 < θ <

89. x = 3tan θ , 0 < θ <

(10 tan θ ) + 100 2

x 2 + 100 =

(

(9 + x 2 ) = 3

)

= 100 tan 2 θ + 1 = 100sec θ = 10 sec θ 2

83. x = 7sin θ , 0 < θ <

49 − x 2 =

49 − (7sin θ )

49 − 49sin 2 θ

49 1 − sin 2 θ

49cos 2 θ

(

64 − 64cos 2 θ

64 1 − cos 2 θ

64sin 2 θ

9 (sec 2 θ )  

π 2 ( 4secθ )2 − 16  

(16sec2 θ + 16)

16(sec 2 θ + 1)  

16 ( tan 2 θ )  

= ( 4 tan θ )

)

= 64 tan 3 θ

Let y1 = sin x and y 2 = 1 − cos2 x , 0 ≤ x < 2π .

π 2

4 x + 9 = 9 tan θ + 9 2

9 + (1 + tan 2 θ )  

91. sin θ = 1 − cos2 θ

= 8sin θ

85. 2 x = 3tan θ , 0 < θ <

( x 2 − 16) =

(

(9 + 9 tan 2 θ )

90. x = 4secθ , 0 < θ <

64 − (8cosθ )

64 − x 2 =

9 + (3tan θ )2   

= 27sec3 θ

= 7 cosθ

84. x = 8cosθ , 0 < θ <

= (3secθ )

)

y1 = y2 for 0 ≤ x ≤ π , so we have sin θ = 1 − cos2 θ for 0 ≤ θ ≤ π .

= 9sec θ = 3sec θ 2

86. 3 x = 2 tan θ , 0 < θ <

π 2

2π

9 x + 4 = 4 tan θ + 4 2

= 4sec 2 θ = 2sec θ 87. x = 2 sinθ , 0 < θ <

π 2

2 − x 2 = 2 − 2sin 2 θ = 2cos2 θ = 2 cosθ 88. x = 5 cosθ , 0 < θ <

−2

92. cosθ = − 1 − sin 2 θ

Let y1 = cosθ and y2 = − 1 − sin 2 θ . y1 = y2 for

≤θ ≤

3π . 2

5 − x 2 = 5 − 5cos2 θ

y1 2π

= 5sin θ = 5 sin θ 2

−2

Section 6.1 93. sec θ = 1 + tan 2 θ

Let y1 =

Using Fundamental Identities

3π y1 = y2 for 0 ≤ x < and < x < 2π , so we 2 2

= ln cos x (1 + sin x )

(

)

(1 + tan t ) 2

= ln

have sec θ = 1 + tan θ for π 3π < θ < 2π . 0 ≤ θ < and 2 2

= ln

tan t 1 tan 2 t + tan t tan t

= ln cot t + tan t y2 y1

101. csc 2 θ − cot 2 θ = 1

2π

(a) csc 2 132° − cot 2 132° ≈ 1.8107 − 0.8107 = 1 2π 2π (b) csc 2 − cot 2 ≈ 1.6360 − 0.6360 = 1 7 7

−4

94. tan θ = sec 2 θ − 1

102. tan 2 θ + 1 = sec2 θ

1 − 1, 0 ≤ θ < 2π . cos2 θ

Let y1 = tan θ and y2 = y1 = y2 for 0 ≤ θ <

π 2

, π ≤θ <

3π . 2

y2 0

(

100. ln cot t + ln 1 + tan 2 t = ln  cot t 1 + tan 2 t 

1 + sin x sec x

99. ln (1 + sin x ) − ln sec x = ln

1 and y 2 = 1 + tan 2 x , 0 ≤ x < 2π . cos x

511

2π

( tan 346°) + 1 ≈ 1.0622 2





( sec346°) =  cos346°  ≈ 1.0622 2

  (b) θ = 3.1 (tan3.1)2 + 1 ≈ 1.00173 2

 1  (sec3.1)2 =   ≈ 1.00173  cos3.1 

−10

cosθ = ln cot θ sin θ

95. ln cosθ − ln sin θ = ln

96. ln cot θ + ln sin θ = ln cot θ ⋅ sin θ = ln

cosθ ⋅ sin θ sin θ

= ln cosθ

97. ln sec x + ln sin x = ln sec x ⋅ sin x = ln

(a) θ = 346°

1 ⋅ sin x cos x

= ln tan x

tan x 98. ln tan x − ln sin x = ln sin x sin x cos x = ln sin x

= ln

1 cos x

π  103. cos  − θ  = sin θ 2   (a) θ = 80° cos ( 90° − 80° ) = sin80° (b) θ = 0.8

0.9848 = 0.9848

π  cos  − 0.8  = sin 0.8 2  0.7174 = 0.7174 104. sin ( − θ ) = − sin θ

(a) θ = 250° sin ( − 250°) ≈ 0.9397 − (sin 250°) ≈ 0.9397

(b) θ =

π 4

 π sin  −  = − 0.7071  4  π −  sin  = − 0.7071 4 

= ln sec x

512

Chapter 6

Analytic Trigonometry

105. μW cosθ = W sinθ W sin θ μ= = tan θ W cosθ

111. As x →

1 sin x − sin x cos x cos x = sin x (sec 2 x − 1)

107. sec x tan x − sin x =

113. As x →

π− 2

1 → 1. cos x

, tan x → ∞ and cot x → 0.

114. As x → π + ,

= sin x ⋅ tan 2 x

sin x → 0 and csc x =

108. True for all θ ≠ nπ .

 1  sin θ ⋅ csc θ = sin θ   =1  sin θ 

1 → −∞. sin x

115. sin θ cosθ = ± 1 − sin 2 θ

109. False.

110. sin θ =

, sin x → 1 and csc x → 1.

cos x → 1 and sec x =

= tan x − cot x

cos0sec

112. As x → 0 + ,

106. sec x − csc x = (1 + tan x ) − (1 + cot x ) 2

π−

π 4

sin θ sin θ =± cosθ 1 − sin 2 θ 1 csc θ = sin θ 1 sec θ = ± 1 − sin 2 θ tan θ =

≠1 a

and cosθ =

a 2 + b2

 sin 2 θ + cos 2 θ =   2

a 2 + b2 = 2 =1 a + b2  sin θ  1 + tan 2 θ = 1 +    cosθ 

sin 2 θ cos 2 θ

cos 2 θ + sin 2 θ cos 2 θ

1 cos 2 θ

= sec 2 θ  cosθ  1 + cot 2 θ = 1 +    sin θ 

  2 2 a +b  b

cot θ = ±

1 − sin 2 θ sin θ

The sign + or − depends on the choice of θ . 116. cosθ sin θ = ± 1 − cos2 θ sin θ 1 − cos2 θ =± cosθ cosθ 1 1 csc θ = =± sin θ 1 − cos2 θ tan θ =

=1+

   + 2 2 a +b   a

a2 b2 + 2 2 a +b a + b2

=1+

b a 2 + b2

1 cosθ 1 cosθ cot θ = =± tan θ 1 − cos2θ

sec θ =

The sign + or − depends on the choice of θ .

cos 2 θ sin 2 θ

sin 2 θ + cos 2 θ sin 2 θ

1 sin 2 θ

= csc 2 θ Answers will vary.

Section 6.1  1  sin θ   1 +  secθ (1 + tan θ ) cosθ  cosθ   117. = secθ + cscθ  1   1    +   cos θ   sin θ 

121.

Using Fundamental Identities

x x2 x x2 + = + x − 25 x − 5 ( x + 5)( x − 5) x − 5 2

 1  cos θ + sin θ     cosθ  cos θ  =  sin θ + cos θ sin θ cosθ cos θ + sin θ cos 2 θ = sin θ + cosθ sin θ cos θ

122.

cos θ + sin θ sin θ cos θ ⋅ cos 2 θ sin θ + cos θ

sin θ cos θ

sin θ + cos θ sin 2 θ 1 sin θ cosθ

Period:

2π

x3 + 5x2 + x ( x + 5)( x − 5)

x( x 2 + 5 x + 1) ( x + 5)( x − 5)

2 x 2 + 8 x − 7 x 2 + 28 ( x + 2)( x − 2)( x + 4)

−5 x 2 + 8 x + 28 ( x + 2)( x − 2)( x + 4)

= 2

Amplitude: 1 y 1

−1

−1 −2

124. f ( x ) = −2 tan

sin θ cosθ + cos θ sin θ 2

Period:

πx 2

π =2 π 2 y 4

1 x x − 8 + x( x + 5) + = 119. x + 5 x −8 ( x + 5)( x − 8)

6x 3 6x 3 120. − = + x−4 4−x x−4 x−4 6x + 3 = x−4

−3

sin θ + cosθ sin θ cosθ = ⋅ sin 2 θ 1

x2 + 6 x − 8 ( x + 5)( x − 8)

x + x 2 ( x + 5) ( x + 5)( x − 5)

2x 7 2 x( x + 4) − 7( x + 2)( x − 2) − = x2 − 4 x + 4 ( x + 2)( x − 2)( x + 4)

 1  sin θ + cosθ     sin θ  sin θ  =  sin 2 θ + cos 2 θ sin θ cosθ

123. f ( x) = − sin π x − 1

 1  cosθ   1 +  cscθ (1 + cot θ ) sin θ  sin θ  118. =  tan θ + cot θ  sin θ   cosθ    +   cosθ   sin θ 

513

−3

−1

−2 −3 −4

514

Chapter 6

Analytic Trigonometry 126. f ( x) = cos( x − π ) + 3

1 π  125. f ( x ) = cot  x +  2 4 

Period: = 2π Amplitude: 1

Period: π

y 5

4 3

−π

−1

x 1

−2

− 2π

−π

−1

2π

Section 6.2 Verifying Trigonometric Identities 1. cot u

13.

2. sin u

csc 2 x 1 sin x 1 = ⋅ = cot x sin 2 x cos x sin x ⋅ cos x = csc x ⋅ sec x

3. tan u 4. cos u

14.

5. cos2 u 6. cot u

sin 2 t sin 2 t = = cos2 t 2 tan t sin 2 t cos2 t

15. cos2 β − sin 2 β = (1 − sin 2 β ) − sin 2 β = 1 − 2 sin 2 β

7. −sin u 8. − csc u

16.

cot 2 β + csc 2 β = (csc 2 β − 1) + csc 2 β = 2csc 2 β − 1

9. No. Algebraic techniques must be used to produce a valid proof to verify a trigonometric identity.

17. tan 2 θ + 6 = (tan 2 θ + 1) + 5

10. No. Conditional equations are equations that are true for only some of the values in its domain.

= sec 2 θ + 5

 1  11. sin t csc t = sin t   =1  sin t 

12. sec y cos y =

18.

3 + sin 2 z = 3 + (1 − cos 2 z ) = 4 − cos 2 z

1 cos y = 1 cos y

19. (1 + sin x )(1 − sin x ) = 1 − sin 2 x = cos2 x 20. tan 2 y(csc 2 y − 1) = tan 2 y cot 2 y = 1

21. y1 =

secθ − 1 secθ − 1 = 1 1 − cosθ 1− secθ

secθ − 1 = secθ − 1 secθ = secθ − 1 ⋅

secθ secθ − 1

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.0203

1.0857

1.2116

1.4353

1.8508

2.7597

5.8835

1.0203

1.0857

1.2116

1.4353

1.8508

2.7597

5.8835

= secθ = y2

y1 = y2

1.5

Section 6.2

Verifying Trigonometric Identities

515

1 −1 csc x − 1 sin x = 22. y1 = 1 − sin x 1 − sin x 1 − sin x 1 = sin x 1 − sin x 1 = sin x = csc x = y2

0.2

0.4

0.6

0.8

1.0

1.2

1.4

y1 5.0335 2.5679 1.7710 1.3940 1.1884 1.0729 1.0148 y2 5.0335 2.5679 1.7710 1.3940 1.1884 1.0729 1.0148 10

y1 = y2 0

1.5

23. y1 = (1 + cot 2 x) cos 2 x = (csc 2 x) cos 2 x

 1  =  2  cos 2 x  sin x  =

cos 2 x sin 2 x

= cot 2 x = y2

0.2

0.4

0.6

0.8

1.0

1.2

1.4

24.3360

5.5943

2.1366

0.9433

0.4123

0.1511

0.0297

24.3360

5.5943

2.1366

0.9433

0.4123

0.1511

0.0297

y1 = y 2

1.5

Chapter 6

516

Analytic Trigonometry

24. y1 = (1 + tan 2 x) sin 2 x = (sec 2 x) sin 2 x

 1  2 sin x =  2   cos x  =

sin 2 x cos 2 x

= tan 2 x = y2

0.2

0.4

0.6

0.8

1.0

1.2

1.4

0.0411

0.1788

0.4680

1.0602

2.4255

6.6160

33.6155

0.0411

0.1788

0.4680

1.0602

2.4255

6.6160

33.6155

0.8

1.0

1.2

1.4

y1 = y 2

1.5

1 − sin x sin x 1 − sin 2 x = sin x cos2 x = sin x cos x = cos x ⋅ sin x = cos x ⋅ cot x

25. csc x − sin x =

0.2

4.8348

2.1785 1.2064 0.6767 0.3469 0.1409 0.0293

4.8348

2.1785 1.2064 0.6767 0.3469 0.1409 0.0293

0.4

0.6

y1 = y 2

1.5

Section 6.2

26.

Verifying Trigonometric Identities

517

1 − cos x cos x 1 − cos2 x = cos x sin 2 x = cos x  sin x  = sin x    cos x 

y1 = sec x − cos x =

= sin x tan x = y2

0.2

0.4

0.6

0.8

1.0

1.2

1.4

y1 0.0403 0.1646 0.3863 0.7386 1.3105 2.3973 5.7135 y2 0.0403 0.1646 0.3863 0.7386 1.3105 2.3973 5.7135 6

y1 = y2

1.5

27. sin x + cos x cot x = sin x + cos x

cos x sin x

sin 2 x + cos2 x sin x 1 = sin x = csc x =

0.2

0.4

0.6

0.8

1.0

1.2

1.4

y1 5.0335 2.5679 1.7710 1.3940 1.1884 1.0729 1.0148 y2 5.0335 2.5679 1.7710 1.3940 1.1884 1.0729 1.0148 5

y1 = y2

1.5

518

Chapter 6

Analytic Trigonometry

28.

y1 = cos x + sin x tan x

31. The error is in line 1: cot ( − x ) = −cot x.

sin 2 x = cos x + cos x cos2 x + sin 2 x = cos x 1 = cos x = sec x = y2

0.2

0.4

32. There are two errors in line 1:

sec ( −θ ) = sec θ and sin ( −θ ) = −sin θ .

( ) = ( tan x )

33. sec 4 x − 2 sec 2 x + 1 = sec 2 x − 1 2

0.6

0.8

1.0

1.2

y2 1.0203 1.0857 1.2116 1.4353 1.8508 2.7597 5.8835

34.

y1 = y2

1.5

1 1 cot x + tan x + = tan x cot x tan x ⋅ cot x

29.

sin x cos x − tan x − cot x cos x sin x = tan x + cot x sin x + cos x cos x sin x sin 2 x − cos2 x x cos x = sin sin 2 x + cos2 x sin x cos x sin 2 x − cos2 x sin x cos x = ⋅ sin x cos x sin 2 x + cos2 x sin 2 x − cos2 x = 1 = sin 2 x − cos2 x = 1 − cos2 x − cos2 x

= cot x + tan x x

0.2

0.4

0.6

0.8

1.0

1.2

y2 5.1359 2.7880 2.1458 2.0009 2.1995 2.9609 5.9704 6

y1 = y 2

30.

y1 =

1 1 − sin x csc x

= csc x − sin x = y2

0.2

0.4

0.6

π  35. cot  − x  csc x = tan x csc x 2  sin x 1 = ⋅ cos x sin x 1 = = sec x cos x

π  cos  − x  2   = sin x 36. cos x π  sin  − x  2  = tan x

1.5

= 1 − 2cos2 x

1.4

y1 5.1359 2.7880 2.1458 2.0009 2.1995 2.9609 5.9704

37.

0.8

1.0

1.2

1.4

csc(− x ) 1 sin(− x ) = sec(− x ) 1 cos( − x ) cos(− x ) sin( − x ) cos x = −sin x = −cot x =

y1 4.8348 2.1785 1.2064 0.6767 0.3469 0.1409 0.0293 y2 4.8348 2.1785 1.2064 0.6767 0.3469 0.1409 0.0293 10

y1 = y2

= tan 4 x

1.4

y1 1.0203 1.0857 1.2116 1.4353 1.8508 2.7597 5.8835

38.

(1 + sin y ) 1 + sin (− y) = (1 + sin y )(1 − sin y ) = 1 − sin 2 y

1.5

= cos2 y

Section 6.2

(

39. sin1 2 x cos x − sin 5 2 x cos x = sin1 2 x cos x 1 − sin 2 x

)

44.

= sin1 2 x cos x ⋅ cos2 x = cos3 x sin x

40. sec 6 x(sec x tan x ) − sec 4 x(sec x tan x ) = sec 4 x(sec x tan x )(sec 2 x − 1) = sec 4 x (sec x tan x )tan 2 x = sec 5 x tan 3 x

41.

cos x − cos y sin x − sin y + sin x + sin y cos x + cos y =

cos2 x − cos2 y + sin 2 x − sin 2 y (sin x + sin y)(cos x + cos y) 1−1 = (sin x + sin y)(cos x + cos y) =0

43.

tan x + cot y 1 1 = + = tan y + cot x tan x cot y cot y tan x cos θ cos θ 1 + sin θ = ⋅ 1 − sin θ 1 − sin θ 1 + sin θ cos θ (1 + sin θ ) = 1 − sin 2 θ cos θ (1 + sin θ ) = cos2 θ 1 + sin θ = cos θ 1 sin θ = + cos θ cos θ = sec θ + tan θ

(

cos θ 1 − sin 2 θ

sin θ cos θ 1 − sin 2 θ = sin θ cos θ cos 2 θ = sin θ cos θ cos θ = sin θ = cot θ

(cos x + cos y)(cos x − cos y) + (sin x + sin y)(sin x − sin y) (sin x + sin y)(cos x + cos y)

519

(sec θ − tan θ )(csc θ + 1) = sec θ csc θ + sec θ − tan θ csc θ − tan θ 1 1 1 sin θ 1 sin θ = + − − cos θ sin θ cos θ cos θ sin θ cos θ 1 1 1 sin θ = + − − sin θ cos θ cos θ cos θ cos θ 1 sin θ = − sin θ cos θ cos θ cos θ − sin 2 θ cos θ = sin θ cos 2 θ =

42.

Verifying Trigonometric Identities

)

π  45. sin 2  − x  + sin 2 x = cos2 x + sin 2 x = 1 2  π  46. sec 2 y − cot 2  − y  = sec 2 y − tan 2 y = 1 2  π  47. sin x csc  − x  = sin x sec x 2 

 1  = sin x    cos x  = tan x π  48. sec 2  − x  − 1 = csc 2 x − 1 = cot 2 x 2 

49. 2 sec 2 x − 2 sec 2 x sin 2 x − sin 2 x − cos2 x = 2 sec 2 x(1 − sin 2 x ) − (sin 2 x + cos2 x )

(

)

= 2sec 2 x cos2 x − 1 1 = 2⋅ ⋅ cos2 x − 1 cos2 x = 2 −1 = 1 50. csc x(csc x − sin x) +

sin x − cos x + cot x sin x

= csc 2 x − csc x sin x + 1 −

cos x + cot x sin x

= csc 2 x − 1 + 1 − cot x + cot x = csc 2 x

51.

cot x tan x 1 = = sec x cos x cos x

520

Chapter 6

Analytic Trigonometry

52.

1 + csc θ 1  cos θ  − cot θ = cos θ  1 + − sec θ sin θ  sin θ 

57.

cos θ (sin θ + 1 − 1) sin θ = cos θ =

53. secθ cot θ =

1 cosθ ⋅ cosθ sin θ

58.

1 = sin θ = cscθ

54. sin θ csc θ − sin 2 θ = sin θ

1 − sin 2 θ sin θ

= 1 − sin 2 θ = cos θ

55.

cos θ + (1 + sin θ ) 1 + sin θ cosθ + = cos θ 1 + sin θ (1 + sin θ ) cosθ

cot α csc α + 1 cot α (csc α + 1) ⋅ = csc α − 1 csc α + 1 csc 2 α − 1 cot α (csc α + 1) = cot 2 α csc α + 1 = cot α

1 + sin α = 1 − sin α

59.

sin β 1 + cos β sin β (1 + cos β ) ⋅ = 1 − cos β 1 + cos β 1 − cos2 β sin β (1 + cos β ) = sin 2 β 1 + cos β = sin β

cos 2 θ + 1 + 2sin θ + sin 2 θ = (1 + sin θ ) cosθ

sin 2 θ + cos 2 θ + 1 + 2sin θ = (1 + sin θ ) cosθ

2 + 2sin θ

(1 + sin θ ) cosθ 2(1 + sin θ ) = (1 + sin θ ) cos θ

1 − cos β = 1 + cos β

60.

2 = cosθ

= 2secθ

tan θ 1 + sec θ tan 2 θ + (1 + sec θ )2 56. + = 1 + sec θ tan θ (1 + sec θ )tan θ =

tan 2 θ + 1 + 2 sec θ + sec 2 θ (1 + sec θ ) tan θ

sec 2 θ − 1 + 1 + 2 sec θ + sec 2 θ = (1 + sec θ )tan θ 2 sec 2 θ + 2 sec θ (1 + sec θ ) tan θ 2 sec θ (sec θ + 1) = (1 + sec θ )tan θ 2 sec θ = tan θ  1  cos θ  = 2    cos θ  sin θ  2 = sin θ = 2 csc θ =

1 + sin α 1 + sin α ⋅ 1 − sin α 1 + sin α

(1 + sin α )2 1 − sin 2 α

(1 + sin α )2 cos 2 α

1 + sin α cos α 1 − cos β 1 − cos β ⋅ 1 + cos β 1 − cos β

(1 − cos β )

1 − cos 2 β sin 2 β

1 − cos β sin β

61. It appears that y1 = 1. Analytically,

1 1 tan x + 1 + cot x + 1 + = cot x + 1 tan x + 1 (cot x + 1)(tan x + 1) tan x + cot x + 2 = cot x tan x + cot x + tan x + 1 tan x + cot x + 2 tan x + cot x + 2 = 1. =

−3

−2

Section 6.2 62. The function appears to be y = cos x. Analytically, cos x sin x ⋅ cos x + 1 − tan x sin x − cos x cos x sin x cos x = + 1 − ( sin x cos x ) sin x − cos x

Verifying Trigonometric Identities

64. The function appears to be y = csc t. Analytically, cot 2 t csc t 1 + cot 2 t = csc t csc 2 t = = csc t. csc t

y = sin t +

cos2 x sin x cos x − cos x − sin x cos x − sin x cos x (cos x − sin x ) = = cos x. cos x − sin x =

521

−2

65. ln cot θ = ln

−2

63. It appears that y1 = sin x. Analytically,

= ln

1 cos2 x 1 − cos2 x sin 2 x − = = = sin x. sin x sin x sin x sin x

cos θ sin θ

= ln cos θ − ln sin θ

66. ln sec θ = ln

1 cos θ

= ln cos θ

−2

−1

= − ln cos θ 67. sin 2 25° + sin 2 65° = cos 2 (90° − 25°) + sin 2 65°

= cos 2 65° + sin 2 65° = 1 68. cos 2 18° + cos 2 72° = sin 2 (90° − 18°) + cos 2 72°

= sin 2 72° + cos 2 72° = 1 69. cos 2 20 ° + cos 2 52 ° + cos 2 38 ° + cos 2 70 ° = cos 2 20 ° + cos 2 52 ° + sin 2 (90 ° − 38 °) + sin 2 (90 ° − 70 °)

= cos2 20° + cos2 52° + sin 2 52° + sin 2 20° = (cos2 20° + sin 2 20°) + (cos2 52° + sin 2 52°) =1+1 = 2 70. sin 218° + sin 2 40° + sin 2 50° + sin 2 72°

= (sin 2 18° + sin 2 72°) + (sin 2 40° + sin 2 50°)

= (sin 2 18° + cos2 18°) + (sin 2 40° + cos2 40°) =1+1 = 2 71. cot 6 x = cot 4 x ⋅ cot 2 x = cot 4 x(csc 2 x − 1) = cot 4 x csc 2 x − cot 4 x

72. sec 4 x tan 2 x = sec 2 x (1 + tan 2 x )tan 2 x = (tan 2 x + tan 4 x ) sec 2 x

73. (sin 2 x − sin 4 x) cos x = sin 2 x(1 − sin 2 x)cos x = sin 2 x ⋅ cos 2 x ⋅ cos x = cos3 x sin 2 x

522

Chapter 6

Analytic Trigonometry

74. 1 − 2 cos 2 x + 2 cos 4 x = 1 − 2 cos 2 x + cos 4 x  + cos 4 x

(

= 1 − cos2 x

) + cos x 2

= sin 4 x + cos4 x

75. Let θ = sin −1 x  sin θ = x =

x . 1

From the diagram,

(

)

tan sin −1 x = tan θ =

x 1 − x2

79. (a)

h sin(90° − θ ) h cosθ = = h cotθ sinθ sinθ

(b)

15°

30°

45°

60°

75°

90°

18.66

8.66

2.89

1.34

80. cos x − csc x cot x = cos x −

θ 1 − x2

76. Let θ = sin −1 x  sin θ = x =

x . 1

 sin 2 x − 1  = cos x   2  sin x   1 − sin 2 x  = −cos x   2  sin x   cos2 x  = −cos x  2   sin x  = −cos x cot 2 x

From the diagram, cos(sin −1 x ) = cos θ = 1 − x 2 . 1

θ 1 − x2

77. Let θ = sin −1

x −1 x −1  sin θ = . 4 4

81. False. Just because the equation is true for one value of θ , you cannot conclude that the equation is an identity. For example, sin 2

From the diagram, x −1  x −1  tan  sin −1 .  = tanθ = 4   16 − ( x − 1)2

π 4

+ cos2

π 4

= 1 ≠ 1 + tan 2

π 4

( )

82. False. For example, sin 12 ≠ sin 2 (1) . sin x sin x 1 − cos x = ⋅ 1 + cos x 1 + cos x 1 − cos x sin x (1 − cos x ) = 1 − cos2 x sin x (1 − cos x ) = sin 2 x 1 − cos x = sin x (b) Not true for x = 0 because (1 − cos x ) sin x is not

83. (a) 4

x −1

θ 16 − (x − 1)2

78. Let θ = cos−1

x +1 x +1  cos θ = . 2 2

From the diagram, x +1  tan  cos −1  = tan θ = 2  

defined for x = 0. 4 − ( x + 1)2 . x +1

4 − (x + 1)2

θ x+1

Section 6.2

tan x 1 − cos x cos x sin x cos x cos x = 1 − cos2 x sin x = sin 2 x 1 1 cos x = ⋅ sin x 1 cos x sec x = tan x (b) Not true for x = π because sec x tan x is not defined for x = π .

84. (a)

85.

tan x = sec x − cos x

Verifying Trigonometric Identities

90. The correct identity is sin θ = 1 − cos 2 θ . 3π . 2  3π  Then sin   = −1 ≠ 1 − 0 = 1.  2 

Let θ =

91. The correct identity is 1 − cos2 θ = sin 2 θ .

Let x =

3π . 4

 2  3π  Then 1 − cos   = 1 −  −   4   2  =1+

a 2 − u 2 = a 2 − a 2 sin 2θ

86.

a 2 − u2 = a 2 − a 2cos2θ

Let x =

=1+

a 2 + u2 = a 2 + a 2 tan 2θ = a 2 sec 2θ = a secθ

π  = sin 2nπ +  6 

= a 2 (sec 2θ − 1) = a 2 tan 2 θ = a tanθ

Because the angles are all coterminal to the angle

sec2 x − 1 = tan x .

7π . Then 4

 7π  sec 2   − 1 =  4 

2 2

 (12n + 1)π   12n π  π +  93. sin   = sin  6 6 6     

u2 − a 2 = a 2 sec 2θ − a 2

Let x =

2 2

 3π  ≠ sin   =  4 

= a 2 (1 + tan 2θ )

89. The correct identity is

3π . 4

 2  3π  Then 1 − cos   = 1 −  −  2  4   

= a 2 sin 2θ = a sinθ

88.

2 2

92. The correct identity is 1 + tan 2 x = sec2 x.

= a 2 (1 − cos2θ )

87.

2 2

 3π  ≠ sin   =  4 

= a 2 (1 − sin 2θ ) = a 2 cos2 θ = a cosθ

523

π 6

, and

 (12n + 1)π  1 1 π  sin   = , sin   = , for all integers n. 6 2 6 2    

( 2) − 1 2

2 −1

=1  7π  ≠ tan   = −1.  4 

524

Chapter 6

94. sin θ =

Analytic Trigonometry

a b and cos θ = c c

96.

 1    −1 2 sec θ − 1  cos θ  = 2 sec 2 θ  θ     cosθ 

f ( x ) = −2 x − 3

−2

−0.03125

−0.125

−0.5

−2 − 8

y 2

c   −1 b =   2 c   b

−2 −4 −6 −8

(c 2 − b 2 ) b 2

Horizontal asymptote: y = 0 as x → −∞

c2 b2

97. f ( x) = 2− x − 1

c − b2 = c2 2

By the Pythagorean theorem, a = c − b .

−2

c2 b2 − 1 = c2 b2 =

−3

−2

−0.75

−0.9375

a c2

a =   c

5 4 3

= sin θ 2

2 1

95.

f ( x) = 2 x + 3

−4 − 3 −2 − 1

−4

−2

3.0625

3.25

−2

Horizontal asymptote: y = −1 as x → −∞ 98. f ( x) = 2 x + 1 + 5

y 8 6

−4

−2

5.125

5.5

−4

−2

12 10 8

Horizontal asymptote: y = 3 as x → −∞

4 2 −4 − 3 −2 − 1 −2

Horizontal asymptote: y = 5 as x → ∞

Section 6.3 Solving Trigonometric Equations 1. 2.

general quadratic

No. The solutions to cos x = 0 are x =

π 2

+ nπ .

Section 6.3 4.

No. To solve sec x sin 2 x = sec x, first subtract sec x from both sides to set the equation equal to 0.

5. tan x −

3 = 0

(a) x =

π 3

: tan

π 3

−

3 =

4π 4π : tan − 3 3

(b) x = 6.

3 −

3 =

3 = 0

3 −

2sin 2 x − sin x − 1 = 0 (a)

(b)

(a)

π 2 7π

π  π  : 2sin 2   − sin   − 1 = 2 − 1 − 1 = 0 2 2 7 π    7π  1  1 2

: 2 sin   − sin  6  − 1 = 2  4  −  − 2  − 1 = 0  6       

: csc 4

π 6

(b)

3 tan 2 2 x − 1 = 0 2

  2π    1  2 π : 3  tan   −1 = 0   − 1 = 3tan 6 − 1 = 3  12 12   3  

(a)

(b)

  10π   5π 2 5π x= : 3  tan    − 1 = 3tan 6 − 1  12 12  

11.

8. 2cos2 4 x − 1 = 0

π  π  : 2cos 2  4 ⋅  − 1 = 2cos 2   − 1 16   4 2

 2 = 2   − 1  2   2 = 2  − 1  4

sin x = 0

x = 180° 13.

2 2 x = 135°, 225°

cos x = −

3 2

14. cos x =

x = 30°, 330° 15. tan x = 1

x = 45°, 225°

1 = 2  − 1 = 0  2 (b) x =

0=0 5π 5 π 5π x= : csc 4 − 4csc 2 =0 6 6 6 (2)4 − 4(2)2 = 0 16 − 4(4) = 0 0=0

12. cos x = −1

 1  = 3 −  − 1 = 0  3

x = 0°, 180°

16 − 4(4) = 0

5π  5π  : sec  −2 =2−2 =0 3  3 

(a) x =

1 π  x = : sec   − 2 = −2 3 cos (π 3) 3 x=

− 4 csc 2

( 2 ) − 4(2)2 = 0

=2−2 =0 (b)

525

10. csc 4 x − 4csc2 x = 0

3 = 0

sec x − 2 = 0 (a)

Solving Trigonometric Equations

3π 3π   2  3π  : 2cos 2  4 ⋅  − 1 = 2cos   − 1 16 16    4  2

 2 = 2  −  − 1  2   2 = 2  − 1  4 1 = 2  − 1 = 0  2

16. tan x = − 3

x = 120°, 300° 3 2 5π 7π x= , 6 6

17. cos x = −

18. sin x = − x=

1 2

7π 11π , 6 6

526

Chapter 6

Analytic Trigonometry 29. 2sin x + 1 = 0

19. cot x = −1 3π 7π , x= 4 4

2sin x = −1 sin x = −

3 2

20. sin x =

7π + 2 nπ 6 11π + 2 nπ x= 6 x=

2π 3

3 3

21. tan x = −

30.

5π 11π , x= 6 6

π 4

7π 4

π 4

π 6

π 3

31.

3 csc x − 2 = 0 3 csc x = 2 csc x =

2 3

sin x =

3 2

3 3

1 2

5π 3

27. tan x = −1 3π 7π , x= 4 4 2 28. csc x = − 2  sin x = − 2 5π 7π , x= 4 4

= 2nπ 3 2π or x = = 2nπ 3 x =

7π 6

26. sec x = 2  cos x = x=

2 2

7π 4

25. cot x = 3  tan x = x=

1 2

7π 11π , 6 6

24. sec x = 2  cos x = x=

2 5π x= + 2 nπ 4 7π x= + 2 nπ 4

23. csc x = −2  sin x = − x=

2 sin x + 1 = 0 sin x = −

2 2

22. cos x =

1 2

32. cot x + 1 = 0 cot x = −1 3π + nπ x= 4

33. sec2 x − 4 = 0 4 sec 2 x = 3 sec x = ±

2 3

cos x = ±

3 2

π 6

+ nπ or x =

5π + nπ 6

34. 3cot 2 x − 1 = 0 1 cot 2 x = 3

cot x = ±

3 3

+ nπ 3 2π x= + nπ 3 x=

Section 6.3 35. sin x(sin x + 1) = 0

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

3π + 2nπ 2

tan x = 0

36. cos x(cos x − 1) = 0 cos x = 0 x=

π 2

3tan 2 x − 1 = 0

x = 0, π

tan 2 x =

or cos x = 1 + nπ or

3 3 π 5π 7π 11π , , x = , 6 6 6 6

1 2

sin x cos x − 3cos x = 0 cos (sin x − 3) = 0

38. 3sin x + 1 = sin x

cos x = 0 or

2sin x = − 1

x =

1 2

44.

π 3π , 2 2

csc 2 x = 2

sec x csc x = 2 csc x

or sec x − 2 = 0 csc x = 0 sec x = 2 No solution

csc x = ± 2 1

2 3π 5π 7π , , x= , 4 4 4 4

40. tan x − 1 = 0 tan 2 x = 1 tan x = ±1 π 3π 5π 7π x= , , , 4 4 4 4

46.

sec 2 x − tan 2 x = sec x − tan x 1 = sec x − tan x

cos x (cos x − 1) = 0

π 2

3π 2

sec x + tan x = 1 (sec x + tan x )(sec x − tan x ) = sec x − tan x

cos3 x − cos x = 0

, 3 3

Since 2sin 2 x + 1 > 0, there are no solutions.

cos3 x = cos x

cos x = 0

π 5π

45. 2sin x + csc x = 0 1 2 sin x + =0 sin x 2 sin 2 x + 1 = 0

41.

sin x = 3

sec x csc x − 2csc x = 0 csc x(sec x − 2) = 0

39. csc2 x − 2 = 0

sin x = ±

sin x − 3 = 0

No solution

7π 11π , 6 6

x =

sin x cos x = 3cos x

43.

2π 4π , 3 3

sin x = −

1 3

tan x = ±

2cos x = −1

x =

1 3

tan x = ±

x = 2 nπ

37. cos x + 1 = − cos x

cos x = −

527

3tan 3 x = tan x

42.

sin x = 0 or sin x = −1 x = nπ or x =

Solving Trigonometric Equations

cos2 x − 1 = 0 cos x = ±1 x = 0, π

Hence, sec x + tan x = sec x − tan x  tan x = 0.

sec x = 1, tan x = 0  x = 0

Chapter 6

528

Analytic Trigonometry

sec 2 x + tan 2 x − 1 = 0

47.

51. 4sin 2 x = 2cos x + 1

sec 2 x + (sec 2 x − 1) − 1 = 0

4sin 2 x − 2cos x − 1 = 0

2sec 2 x − 2 = 0

y = 4sin 2 x − 2 cos x − 1

2(sec x − 1) = 0 2

x ≈ 0.8614, 5.4218

sec 2 x = 1

1 =1 cos 2 x

2π

cos x = 1 cos x = ±1 cos x = 1 or cos x = −1 x = π

x = 0

−4

csc2 x − 3csc x − 4 = 0

sin 2 x + cos x + 1 = 0

48.

csc2 x − 3csc x = 4

52.

(1 − cos2 x ) + cos x + 1 = 0

1 3 − −4 sin 2 x sin x

x ≈ 0.2527, 2.8889, 4.7124

cos2 x − cos x − 2 = 0 (cos x − 2)(cos x + 1) = 0

cos x = 2, Impossible cos x + 1 = 0  x = π

2π

2sin x + 3sin x = − 1 −10

49. 2sin 2 x + 3sin x + 1 = 0

53.

y = 2sin 2 x + 3sin x + 1 x ≈ 3.6652, 4.7124, 5.7596

csc x + cot x = 1 csc x + cot x − 1 = 0 y=

1 cos x + −1 sin x sin x

π  x ≈ 1.5708   2 2π

−1

2 2 50. 2sec x + tan x = 3 2

2π

2sec x + tan x − 3 = 0 −8

2 y= + tan 2 x − 3 cos2 x x ≈ 0.5236, 2.6180, 3.6652, 5.7596 7

54.

4sin x = cos x − 2 4sin x − cos x + 2 = 0

y = 4sin x − cos x + 2 x ≈ 3.8930, 6.0217 7

2π

0 −1

2π

−3

Section 6.3

55.

60. (a)

cos x cot x =3 1 − sin x cos x cot x −3= 0 1 − sin x cos x y= −3 (1 − sin x )tan x x ≈ 0.5236, 2.6180

−1

(b) cos2 x = e− x + x − 1 (c) Point of intersection: ( 0.9510, 0.3374 ) x =0 4

x π x 3π = + 2 nπ or = + 2 nπ 4 2 4 2

−2

x = 2π + 8nπ or x = 6π + 8nπ

1 + sin x = 2 cos 2 x

Combining, x = 2π + 4nπ .

1 + sin x −2 = 0 cos 2 x

62. sin

1 + sin x − 2 cos 2 x

y =

2π

61. cos

56.

x ≈ 0.5236, 2.6180

x =0 2 x = nπ 2 x = 2 nπ

63. tan 4 x = 1

4x = 2π

x = −4

57. (a)

529

Solving Trigonometric Equations

+ nπ

π 16

nπ 4

64. tan 2 x = −1

3π + nπ 4 nπ 3π x = + 8 2

2x = 2π

−2

(b) sin 2 x = x 2 − 2 x (c) Points of intersection: (0, 0), (1.7757, − 0.3984) 58. (a)

65. sec 4 x = 2

4x =

5π + 2nπ 3 5π nπ x= + 12 2

+ 2nπ or 4 x = +

nπ 2

66. sin 2 x = −

3 2

π 12

2π

−2

(b) cos x = x + x 2 (c) Point of intersection: (0.5500, 0.8525) 59. (a)

2π

−2

(b) sin x = e − 4 x (c) Points of intersection: ( 0.3194, 0.0986 ) , ( 2.2680, 0.5878 )

4π 5π + 2nπ or 2 x = + 2nπ 3 3 2π 5π x= x= + nπ + nπ 3 6

2x =

67. cos

x 2 = 2 2

x 7π x π = + 2nπ or = + 2nπ 2 4 2 4 π 7π x = + 4nπ x= + 4nπ 2 2

Chapter 6

530

68. csc

x = 4

Analytic Trigonometry 75.

x sin = 4

1 = 2

Graph y1 = x tan x − 1 and estimate the zeros.

2 2

x ≈ 0.8603, 3.4256

x π = + 2nπ or 4 4 x = π + 8nπ

69. y = sin

πx 2

x tan x − 1 = 0

x 3π = + 2nπ 4 4 x = 3π + 8nπ

2π

−8

From the graph in the textbook we see that the curve has x-intercepts at x = −1 and at x = 3.

76. 2 x sin x − 2 = 0 Graph y1 = 2 x sin x − 2 and estimate the zeros. x ≈ 1.1142, 2.7726

y = sinπ x + cosπ x

70.

From the graph in the textbook, we see that the curve has x-intercepts at x = −0.25, 0.75,1.75, and 2.75. πx  y = tan 2  −3  6 

71.

−12

From the graph in the textbook, we see that the curve has x-intercepts at x = ±2. πx  y = sec 4  −4  8 

72.

2π

77. sec 2 x + 0.5 tan x − 1 = 0

Graph y1 =

1 + 0.5tan x − 1 and estimate the zeros. (cos x )2

From the graph in the textbook, we see that the curve has x-intercepts at x = −2, 2. 2π

73. 2cos x − sin x = 0

−2

Graph y1 = 2cos x − sin x and estimate the zeros.

x ≈ 1.1071, 4.2487

78. csc2 x + 0.5cot x − 5 = 0

2π

x = 0, x ≈ 2.6779, 3.1416, 5.8195

1  1  − 5 and estimate the zeros. Graph y1 =   +  sin x  2 tan x 5

−4

2π

74. 2sin x + cos x

Graph y1 = 2sin x + cos x and estimate the zeros.

−5

x ≈ 0.5153, 2.7259, 3.6569, 5.8675

x ≈ 2.6779, 5.8195

79. 6sin 2 x − 7sin x + 2 = 0

Graph y1 = 6sin 2 x − 7sin x + 2 and estimate 2π

the zeros. 16

−4

2π

0 −2

Section 6.3 80. 2 tan 2 x + 7 tan x − 15 = 0

85.

Solving Trigonometric Equations

531

2sin 2 x + 5cos x = 4 2sin 2 x + 5cos x − 4 = 0

Graph y1 = 2 tan 2 x + 7 tan x − 15 and estimate

2 (1 − cos 2 x) + 5cos x − 4 = 0

the zeros. 24

2 − 2cos 2 x + 5cos x − 4 = 0 − 2cos 2 x + 5cos x − 2 = 0

2π

2cos 2 x − 5cos x + 2 = 0

(2cos x − 1)(cos x − 2) = 0

−24

x ≈ 0.9828, 1.7682, 4.1244, 4.9098 81.

2cos x − 1 = 0 2cos x = 1

tan 2 x + tan x − 12 = 0 tan x + 4 = 0

tan x = 3

x =

tan x = − 4 x = arctan ( − 4) + nπ

x = arctan 3 + nπ 82.

86.

3π x = + nπ 4

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

sec 2 x − 6 tan x + 4 = 0

sin x =

tan 2 x − 6 tan x + 5 = 0

( tan x − 1)( tan x − 5) = 0

84.

π 4

x =

tan x − 5 = 0

tan x = 1

tan x = 5 + nπ

x = arctan5 + nπ

( tan x − 1)( tan x + 2) = 0 or

tan x = 1 x =

π 4

tan x + 2 = 0 tan x = − 2

+ nπ

x = arctan ( − 2) + nπ

1 2

π 6

No solution + 2nπ ,

5π + 2nπ 6

x ≈ −1.154, 0.534

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

sin x = 3

 π π 87. 3 tan 2 x + 5 tan x − 4 = 0,  − ,   2 2

88.

sec2 x + tan x − 3 = 0

tan x − 1 = 0

sin x − 3 = 0

2sin x = 1

(1 + tan 2 x) − 6 tan x + 4 = 0

x =

2cos 2 x + 7sin x = 5

2sin 2 x − 7 cos x + 3 = 0

x = arctan 2 + nπ

5π + 2nπ 3

− 2sin 2 x + 7sin x − 3 = 0

tan x = 2

tan x − 1 = 0

+ 2nπ ,

2 − 2sin 2 x + 7sin x − 5 = 0

tan x − 2 = 0

tan x = −1

83.

2 (1 − sin 2 x) + 7sin x − 5 = 0

( tan x + 1)( tan x − 2) = 0 or

No solution

2cos 2 x + 7sin x − 5 = 0

tan 2 x − tan x − 2 = 0

tan x + 1 = 0

cos x = 2

1 cos x = 2

( tan x − 3)( tan x + 4) = 0 tan x − 3 = 0

cos x − 2 = 0

y = cos2 x − 2 cos x − 1 = 0, [0, π ] x ≈ 1.998

 π π 89. 4 cos2 x − 2 sin x + 1 = 0,  − ,   2 2 x ≈ 1.110

90.

y = 2 sec 2 x + tan x − 6 = 0,

 π π − 2 , 2   

x ≈ −1.035, 0.870

532

Chapter 6

91.

f ( x) = sin2 x (a)

Analytic Trigonometry 94. f ( x ) = cos2 x − sin x

2π

−2

Maxima: (3.6652,1.25), (5.7596, 1.25)

Maxima: (0.7854, 1), (3.9270,1) Minima: ( 2.3562, − 1) , ( 5.4978, − 1) (b)

(a)

(b)

cos x ( 2sin x + 1) = 0

2cos2 x = 0 cos2 x = 0 2x =

Minima: (1.5708, − 1), (4.7124, 1) −2sin x cos x − cos x = 0

π 2

cos x = 0  x = + nπ

π 2

+ nπ

1 7π 11π sin x = −  x = + 2 nπ , + 2nπ 2 6 6

nπ x= + 4 2

The zeros are 1.5708, 3.6652, 5.7596, and 4.7124.

The zeros are 0.7854, 2.3562, 3.9270, and 5.4978. 92. f ( x) = cos 2x

95. f ( x) = sin x + cos x 2

(a)

2π

−2

Maximum: (0.7854, 1.4142)

−2

Minimum: (3.9270, − 1.4142)

Maxima: (0, 1), (3.1416, 1), (6.2832, 1) Minima: (1.5708, − 1), (4.7124, − 1) (b)

(b)

−2sin 2 x = 0

cos x − sin x = 0 cos x = sin x

sin x cos x tan x = 1 1=

sin 2 x = 0 2 x = nπ x=

nπ 2

5π + 2nπ 4

(a)

2π

−4

−2

Maxima: (0.5236, 1.5), (2.6180, 1.5)

Maxima: (1.0472, 1.25), (5.2360, 1.25) Minima: (0, 1), (3.1416, − 1), (6.2832, 1) (b) 2sin x cos x − sin x = 0 sin x(2 cos x − 1) = 0

sin x = 0  x = nπ cos x =

+ 2nπ ,

96. f ( x) = 2sin x + cos2 x

93. f ( x ) = sin 2 x + cos x

The zeros are 0.7854 and 3.9270.

The zeros are 0, 1.5708, 3.1416, 4.7124, and 6.2832.

(a)

π 1 5π  x = + 2nπ , + 2 nπ 2 3 3

The zeros are 0, 1.0472, 3.1416, 5.2360, and 6.2832.

Minima: (1.5708, 1), (4.7124, − 3.0) (b)

2cos x − 4sin x cos x = 0

2cos x(1 − 2sin x) = 0 π 3π cos x = 0  x = + 2 nπ , + 2 nπ 2

2 1 π 5π 1 − 2sin x = 0  sin x =  x = + 2nπ , + 2nπ 2 6 6 The zeros are 0.5236, 2.618, 4.712, and 1.571.

Section 6.3

97. f ( x ) = tan

Solving Trigonometric Equations

533

98. Graph y = cos x and y = x on the same set of axes. Their point of intersection gives the value of c such that

πx 4

f (c) = c  cos c = c.

tan0 = 0, but 0 is not positive. By graphing πx y = tan

− x, you see that the smallest positive fixed 4 point is x = 1.

y = cos x

2 (0.739, 0.739)

y=x 3

−3

−2

c ≈ 0.739 99. S = 74.50 − 43.75cos

πt 6

S 36.6 52.6 74.5 96.4 112.4 118.3 112.4 96.4 74.5 52.6 36.6 30.8 S > 100 for t = 5, 6, 7 (May, June, July) 100. U = 58.3 + 32.5 cos

πt

101. f ( x ) = cos

Let U = 75 and solve for t. 75 = 58.3 + 32.5cos 16.7 = 32.5cos

πt 6

16.7 πt = cos 32.5 6 16.7 π t   arccos  =  32.5  6 6

 16.7  arccos  =t  32.5  1.96 ≈ t

6  16.7   and 2π − arccos   = t π  32.5   10.03 ≈ t Therefore over the interval (0, 1.96) and (10.03, 12),

U ≥ 75, which approximately translates to the months November, December, and January.

1 x

(a) The domain of f ( x) is all real numbers except 0. (b) The graph has y-axis symmetry and a horizontal asymptote at y = 1. (c) As x → 0, f ( x) oscillates between − 1 and 1. (d) There are an infinite number of solutions in the interval [ −1, 1].

1 π π + 2nπ 2 x= = + nπ = 2 π (2n + 1) x 2 (e) The greatest solution appears to occur at x ≈ 0.6366. 102. f ( x ) =

sin x x

(a) Domain: all real numbers except x = 0. (b) The graph has y-axis symmetry. Horizontal asymptote: y = 0 (c) As x → 0, f ( x) → 1. (d)

sin x x = 0 has four solutions in the interval

[−8, 8].

( sin x )   = 0

1 x sin x = 0 x = −2π , − π , π , 2π

Chapter 6

534

Analytic Trigonometry

103.

1 (cos8t − 3sin8t ) 12

1 (cos8t − 3sin8t ) = 0 12 cos8t = 3sin8t 1 = tan8t 3 8t = 0.32175 + nπ nπ t = 0.04 + 8 In the interval 0 ≤ t ≤ 1, t ≈ 0.04, 0.43, and 0.83 second. 104. y1 = 1.56e

−0.22 t

107. A = 2 x cos x, 0 ≤ x ≤

(a)

π 2

−2

The maximum area of A ≈ 1.12 occurs when x ≈ 0.86. (b) A ≥ 1 for 0.6 < x < 1.1 108. (a)

cos 4.9t intersects y2 = −1 at t ≈ 1.96

(and other points). The displacement does not exceed one foot from equilibrium after t = 1.96 seconds.

(b) T = 4.45sin (0.519t − 2.492) + 84.4

(c)

−2

105.

1 2 v0 sin 2θ 32 1 2 310 = (100) sin 2θ 32 sin 2θ = 0.992 r =

2θ ≈ 1.444

(d) The constant term, 84.4, gives the average daily high temperature of 84.4° F.

2θ ≈ π − 1.287 ≈ 1.698

θ ≈ 0.849 ≈ 48.64°

(e) Graph the line y = 86 to show that the values on the interval [5, 10], corresponding to the months of May, June, July, August, September, and October, have a normal daily high temperature above 86° F.

106. f ( x) = 3sin(0.6 x − 2)

(a) Zero: sin(0.6 x − 2) = 0 0.6 x − 2 = 0

0.6 x = 2 2 10 x= = 0.6 3 g( x ) = −0.45 x 2 + 5.52 x − 13.70

109. y1 = 2sin x y2 = 3 x + 1

From the graph we see that there is only one point of intersection. 110. y1 = 2sin x 1 x +1 2 By inspecting the graphs of y1 and y2 , it appears they y2 =

g −5

For 3.5 ≤ x ≤ 6 the approximation appears to be good. Answers will vary. (c)

The model fits the data well.

θ ≈ 0.722 ≈ 41.37°

(b)

intersect at three points. 111. False. The equation has no solution because −1 ≤ sin x ≤ 1.

−0.45 x 2 + 5.52 x − 13.70 = 0 −5.52 ± (5.52)2 − 4(−0.45)(−13.70) 2(−0.45) x ≈ 3.46, 8.81 x=

The zero of g on [0, 6] is 3.46. The zero is close to the zero 103 ≈ 3.33 of f. © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Section 6.4

Sum and Difference Formulas

112. True. 2sin t − 1 = 0

113. Answers will vary.

1 2

114. Answers will vary.

sin t = t =

π 5π 6

or two solutions in [0, 2π )

2sin 4t − 1 = 0 sin 4t = 4t =

 π  115. 124° = 124°   ≈ 2.164 radians  180°   π  116. 486° = 486°   ≈ 8.482 radians  180° 

1 2

π 5π 13π 17π 25π 29π 37π 41π

, , , , , , 6 6 6 6 6 6 6 π 5π 13π 17π 25π 29π 37π 41π t = , , , , , , , 24 24 24 24 24 24 24 24 or eight solutions in [0, 2π ) 6

535

 π  117. −0.41° = −0.41°   ≈ −0.007 radian  180°   π  118. −210.55° = −210.55°   ≈ −3.675 radians  180° 

119. Answers will vary. (Make a Decision.)

Section 6.4 Sum and Difference Formulas 10. (a) sin(405° + 120°)

1. sin u cos v − cos u sin v

= sin 405° cos120° + cos 405° sin120°

2. cos u cos v − sin u sin v

tan u + tan v 3. 1 − tan u tan v

2  1 2 3 6− 2  = −  + 2  2  2  2  4

(b) sin 405° + sin120° =

4. sin u cos v + cos u cos v

tan u − tan v 1 + tan u tan v

Sample answer:

2+ 3 2

 2π 5π   9π  11. (a) sin  + = sin    = −1 6   3  6 

5. cos u cos v + sin u sin v 6.

2 3 + = 2 2

(b) sin

sin195° = sin (135° + 60°) = sin135° cos60° + cos135° sin 60°

2π 5π 3 1 3 +1 + sin = + = 3 6 2 2 2

π π π π π π  12. (a) cos  +  = cos cos − sin sin 4 3 4 3 4 3 2 1 2 3 =    − 2  2  2  2 

8. Sample answer:

π π π π π cos   = cos  −  = cos cos  12   3 4 3 4 π

+ sin sin 3 4

2− 6 4

(b) cos

π 4

+ cos

π 3

2 1 + = 2 2

2 +1 2

1 2 1 3 (b) cos(240°) − cos0° = − − 1 = − 2 2

9. (a) cos(240° − 0°) = cos(240°) = −

536

Chapter 6

Analytic Trigonometry

13. sin105° = sin(60° + 45°) = sin 60° cos 45° + sin 45° cos 60°

15. sin195° = sin(225° − 30°) = sin 225° cos30° − sin 30° cos 225° = − sin 45° cos30° + sin 30° cos 45°

3 2 2 1 = ⋅ + ⋅ 2 2 2 2 2 = 3 +1 4

(

=−

)

1 2 3 1 = ⋅ − ⋅ 2 2 2 2 2 = 1− 3 4

=−

)

3 +1

tan 225° − tan 30° 1 + tan 225° tan 30° tan 45° − tan 30° = 1 + tan 45° tan 30°

3 +1 1+ 3 ⋅ 1− 3 1+ 3

4+2 3 = −2 − 3 −2

14. 165° = 135° + 30°

( 3 3) = 3 − 3 ⋅ 3 − 3 1 + ( 3 3) 3 + 3 3 − 3

12 − 6 3 = 2− 3 6

1−

16. 255° = (300° − 45°)

sin165° = sin(135° + 30°) = sin135° cos30° + sin 30° cos135° = sin 45° cos30° − sin 30° cos 45° 2 3 1 2 2 ⋅ − ⋅ = 3 −1 2 2 2 2 4 cos165° = cos(135° + 30°) = cos135° cos30° − sin135° sin 30° = − cos 45° cos30° − sin 45° cos30°

(

)

sin 255 ° = sin(300° − 45°) = sin300° cos 45° − cos300° cos 45°  3  2 1 2  − 6 − 2 = − − =  2  2 2  2  4    

cos255° = cos(300°− 45°) = cos300° cos 45° + sin300° sin 45° 1 2   3 2 − 6+ 2 =  =  + −  2  2   2  2 4

2 3 2 1 =− ⋅ − ⋅ 2 2 2 2

(

)

2 3 +1 4 tan165° = tan(135° + tan 30°) =−

tan135° + tan 30° 1 − tan135° tan 30° − tan 45° + tan 30° = 1 + tan 45° tan 30° =

( 3 / 3) = −2 + 3 1 + ( 3 / 3)

−1 +

( 3 + 1)

tan 60° + tan 45° 1 − tan 60° tan 45° 1− 3

2 3 2 1 2 ⋅ − ⋅ =− 2 2 2 2 4

tan195° = tan(225° − 30°)

tan105° = tan(60° + 45°)

)

= cos225° cos30° + sin 225° sin 30° = − cos 45° cos30° − sin 45° cos30°

= cos 60° cos 45° − sin 60° sin 45°

(

cos195° = cos(225° − 30°)

cos105° = cos(60° + 45°)

(

2 3 1 2 2 1− 3 ⋅ + ⋅ = 2 2 2 2 4

tan 255° = tan(300° − 45°)

= =

tan 300° − tan 45° 1 + tan 300° tan 45° − 3 −1

(

)

1 + − 3 (1)

1+ 3 3 −1

1+ 3 3 +1 ⋅ 3 −1 3 +1

4+2 3 = 2+ 3 2

Section 6.4

17. −

11π π 7π = − 12 4 6

18. −

7π   11π  π sin  −  = sin  −  6   12  4 π π 7π 7π = sin cos − cos sin 4 6 4 6 = =

6 + 4

2 − 4

2 3  − + 2  2  −

6 − 4

6 + 4

3 2 6− 2  1 2 = −  + = 2 2 2 2 4  

π 7π tan − tan 7π   11π  π 4 6 tan  −  = tan  −  = 6  1 + tan π tan 7π  12  4 4 6  3 1 −   3   =  3 1 + (1)   3 

3 2  1 2 6+ 2 −−  = 2 2  2 2 4

 19π   2π 9π  cos  − −  = cos   4   12   3 2π 9π 2π 9π cos sin = cos + sin 3 4 3 4

2 1 −  2  2

= −

19π 2π 9π = − 12 3 4

7π   11π  π cos −  = cos −  6   12  4 π π 7π 7π = cos cos + sin sin 4 6 4 6 =

537

 19π   2π 9π  sin  − −  = sin   4   12   3 2π 9π 2π 9π cos sin = sin − cos 3 4 3 4

2 3 2 1 − − −  2  2  2  2 −

Sum and Difference Formulas

2π 9π − tan tan  19π   2π 9π  3 4 − tan  −  = tan  = 4  1 + tan 2π tan 9π  12   3 3 4 = =

− 3 −1

(

)

1 + − 3 (1) 3 +1 3 −1

=2+ 3

 3 1 −    3  =  3 1 +    3  =

3− 3+

3 3

3− 3+

3 3− ⋅ 3 3−

9−6 3 +3 9−3

12 − 6 3 = 2− 6

3 3

538

Chapter 6

Analytic Trigonometry

19. 285° = 225° + 60°

21. −105° = 30° − 135°

sin ( −105°) = sin (30° − 135°)

sin 285° = sin (225° + 60°) = sin225° cos60° + cos225° sin60°

= sin (30°) cos (135°) − cos (30°) sin (135°)

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

2   3  2  1  =   − −  −      2  2   2  2 

= −

6 + 4

2 6 − = − 4 4

cos285° = cos( 225° + 60°)

2 + 4

= −

cos ( −105°) = cos (30° − 135°) = cos (30°) cos (135°) + sin (30°) sin (135°)

 2  1   2  3  =  −   −  −     2  2   2  2 

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

6 − 4

2 6 + = 4 4

tan 285° = tan ( 225° + 60°)

−

6 + 4

2 − 4

( 3)

tan (30°) − tan (135°)

1 + tan (30°) tan (135°) 3 3 − ( −1)

1 − (1)

1+ 1−

3 3

1+ 1−

3 1+ ⋅ 3 1+

3+ 3−

1+ 2 3 +3 1−3

12 + 6 3 = 2 + 6

4+ 2 3 = −2 − −2

3 3

)

3 3+1

3 3+ = 3 3−

1−

( 3 3)

3 3+ ⋅ 3 3+

3 3

= sin ( −135° ) cos30° − cos ( −135° ) sin 30°

(

)

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

( 3 − 1)

cos15° = cos ( 45° − 30°) = cos 45° cos30° + sin 45° sin 30° 2

( 3 + 1) 4

( 3 +1) tan 45° − tan 30° 1 + tan 45° tan 30°

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

(

3 3 ( −1)

sin ( −165° ) = sin ( −135° − 30° )

sin15° = sin ( 45° − 30° ) = sin 45° cos30° − cos 45° sin30°

tan15° = tan ( 45° − 30° ) =

22. −165° = 135° − 30°

20. 15° = 45° − 30°

 2  3   2   1  = + =  2   2   2   2      

tan ( −105°) = tan (30° − 135°)

tan 225° + tan 60° = 1 − tan 225° tan 60°

2 − 4

= cos225° cos60° − sin225° sin60°

= −

2 = 4

−

 2 3  2 1 2− 6 = − − − =  2  2  2  2 4     cos( −165°) = cos(−135° − 30°) = cos(−135°)cos30° + sin(−135°)sin 30°  2 3  2 1 − 6 − 2 = − + − =  2  2  2  2 4     tan ( −165° ) = tan ( −135° − 30° )

= = =

tan ( −135° ) − tan 30°

1 + tan ( −135° ) tan 30° 1− 3 3 1+1

( 3 3)

3− 3 3+ 3

3− 3 3− 3 ⋅ 3+ 3 3− 3

12 − 6 3 = =2− 3 6

12 − 6 3 =2− 3 6

Section 6.4

23.

Sum and Difference Formulas

539

13π 3π π = + 12 4 3 sin

13π 3π π π 3π  3π π  cos + sin cos = sin  +  = sin 12 3 4 3 3 4  4

2 1 3 2 2− 6 ⋅ +  − = 2 2 2  2  4 13π 3π π 3π π  3π π  = cos  +  = cos cos cos − sin sin 12 3 4 3 4 3  4 =

 2   1   2  3  6+ 2 = − =−  −2   2   2   2  4      13π  3π π  tan ( 3π 4 ) + tan (π 3 ) = tan  + = tan 12 3  1 − tan ( 3π 4 ) tan (π 3 )  4

24.

( −1) + 3 = 1 − ( −1) 3

3 −1 3 +1

=2− 3

5π π π = + 12 4 6 π π π π  5π  π π  sin   = sin  +  = sin cos + sin cos 4 6 6 4  12  4 6

2 3 1 2 6+ 2 + = 2 2 2 2 4

π π π π  5π  π π  cos   = cos  +  = cos cos − sin sin 4 6 4 6  12  4 6 =

2 3 2 1 6− 2 − = 2 2 2 2 4

6+ 2  5π  sin ( 5π 12 ) = =2+ 3 tan  = 12 cos 5 12 π 6− 2 ( )  

25. −

7π π 3π = − 12 6 4

π 3π 3π π  7π   π 3π  − sin sin  − cos  = sin  −  = sin cos 4  6 4 4 6  12  6     1 2 2 3 2+ 6 = − −  =− 2  2  2  2  4 π 3π π 3π  7π   π 3π  + sin sin cos  −  = cos  −  = cos cos 4  6 4 6 4  12  6 =

3 2 1 2 = − + ⋅ 2  2  2 2

2− 6 4

 7π   π 3π  tan (π 6 ) − tan ( 3π 4 ) tan  −  = tan  − = 4  1 + tan (π 6 ) tan ( 3π 4 )  12  6 =

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

540

Chapter 6

Analytic Trigonometry

26. −

23π 5π π = − − 12 3 4

π  23π   5π  5π  π   5π   π  −  = sin  −  cos   − cos −  sin   sin  −  = sin  − 4  12   3  3  4  3  4  3  2   1  2  =    −      2  2   2  2  6 − 4

π  23π   5π  5π   π   5π   π  −  = cos −  cos   + sin  −  sin   cos −  = cos − 4  12   3  3  4  3  4  1  2   3  2  =   +    2  2   2   2 

2 + 4

 5π  π  tan  −  − tan   π  23π   5π 3   4 tan  − −  =  = tan  − π 5 12 3 4   π      1 + tan  −  tan    3  4 =

3 −1

( 3)(1)

3 −1 = 1+ 3

3 −1 ⋅ 3 +1

3 −1 3 +1 3 −1 3− 2 3 +1 = 3−1 3 −1

4− 2 3 = 2− 2

27. sin 60° cos10° − cos 60° sin10° = sin (60° − 10°)

= sin (50°) 28. sin110° cos80° + cos110° sin80° = sin (110° + 80° ) = sin (190° ) 29.

tan 325° − tan116° = tan ( 325° − 116° ) 1 + tan 325° tan116°

= tan 209° 30.

tan154° − tan 49° = tan (154° − 49° ) = tan105° 1 + tan154° tan 49°

31. cos

32. sin

π 9

cos

π 7

− sin

π 9

sin

π 7

π π   16π  = cos  +  = cos   9 7  63 

4π π 4π π  4π π   41π  cos + cos sin = sin  +  = sin   9 8 9 8 9 8    72 

33. sin3.5x cos1.2 y + cos3.5 x sin1.2 y = sin(3.5x + 1.2 y) 34. cos0.96 y cos 0.42z + sin0.96 y sin0.42 z = cos(0.96 y − 0.42z)

Section 6.4

35. sin

π 12

cos

π 4

+ cos

π 12

sin

π π = sin  +  4  12 4 

= cos ( 90° )

π  = sin   3

36. cos

cos

5π π − tan 6 6 = tan  5π − π  39.   5π π 6  6 1 + tan tan 6 6

3 2

tan

3π 3π π  π 3π  − sin sin = cos  +   16 16  16 16 16

 2π  = tan    3 

π = cos    4 =

=− 3

2 2

40.

37. sin120° cos60° − cos120° sin 60° = sin (120° − 60° )

tan 25° + tan110° = tan ( 25° + 110° ) 1 − tan 25° tan110° = tan135° = −1

= sin ( 60° ) =

41.

541

38. cos120° cos30° + sin120° sin30° = cos (120° − 30° )

Sum and Difference Formulas

3 2

y1 = cos ( x + π ) cos ( x − π ) = ( cos x ⋅ cos π − sin x ⋅ sin π )  cos x cos π + sin x sin π  =  − cos x   − cos x  = cos2 x = y2

0.2

0.4

0.6

0.8

1.0

1.2

1.4

y1 0.9605 0.8484 0.6812 0.4854 0.2919 0.1313 0.0289

y2 0.9605 0.8484 0.6812 0.4854 0.2919 0.1313 0.0289 2

y1 = y2 0

42.

1.5

y1 = sin ( x + π ) sin ( x − π ) = sin x cos π + sin π cos π  sin x cos π − sin π cos x  =  − sin x   − sin x  = sin 2 x = y2

0.2

0.4

0.6

0.8

1.0

1.2

1.4

0.0395

0.1516

0.3188

0.5146

0.7081

0.8687

0.9711

0.0395

0.1516

0.3188

0.5146

0.7081

0.8687

0.9711

5 and u is in Quadrant II. 13

cos v = −

3 and v is in Quadrant II. 5

y1 = y2 0

1.5

For Exercises 43 – 46; sin u =

13 −12

u x

v −3

542

Chapter 6

Analytic Trigonometry

43. sin ( u + v ) = sin u cos v + cos u sin v

tan u + tan v 1 − tan u tan v 7 3 + 24 4 =  7  3  1 −     24  4 

47. tan ( u + v ) =

 5  − 3   − 12  4  =   +    13  5   13  5  63 − 3 48 = − =− 13 65 65

25 25 32 4 24 = = ⋅ = 7 24 25 3 1− 32

tan u + tan v 1 − tan u tan v  −5  4  12  +  − 3     =   5  4  1 −  −  −   12  3  7 7 − − 4 = = 4 5 4 1− 9 9 7 9 63 =− ⋅ =− 4 4 16

44. tan ( u + v ) =

48. cos ( u + v ) = cos u cos v − sin u sin v  − 24  − 4   − 7  − 3  =  −    25  5   25  5  96 21 3 = − = 125 125 5

49. sin ( v − u ) = sin v cos u − cos v sin u  − 3  − 24   − 4  − 7  =  −    5  25   5  25  72 28 44 = − = 125 125 125

45. cos(u − v) = cos u cos v + sin u sin v  12  3   5  4  =  −  −  +     13  5   13  5  36 4 56 = + = 65 13 65

50. cos ( u − v ) = cos u cos v + sin u sin v  − 24  − 4   − 7  − 3  =  +    25  5   25  5  96 21 117 = + = 125 125 125

46. sin ( u − v ) = sin u cos v − cos u sin v  5  − 3   −12  4  =   −    13  5   13  5  − 3 48 33 = + = 13 65 65

51. sin ( arcsin x + arccos x ) = sin ( arcsin x ) cos ( arccos x )

+ sin ( arccos x ) cos ( arcsin x )

For Exercises 47 – 50 :

= x ⋅ x + 1 − x2 ⋅ 1 − x2

7 and u is in Quadrant III. sin u = − 25

= x2 + 1 − x2 =1

4 cos v = − and v is in Quadrant III. 5 y

u −24 −7

v −4

x −3

1 − x2

α 1 − x2

θ = arcsin x

α = arccos x

Section 6.4 52. Let:

v = arcsin x u = arccos x and sin v = x cos u = x

55. sin −1 ( −1) = −

+ sin ( arccos x ) sin ( arcsin x )

(

57. sin −1 1 =

4 x2 + 1

and cos−1 0 =

= 1.

)

59. Let θ = cos −1

( x) −

= cos π = −1

1 4 x2 + 1

(

1 − x2

2 x2 − 1 − x2

(

= sin u cos v − cos u sin v 2x

2 π π  cos sin −1 1 + cos−1 0 = cos  +  2 2

sin ( arctan 2 x − arccos x ) = sin ( u − v )

because sin

)

58. sin −1 1 =

v = arccos x cos v = x

tan u = 2 x

(

1 − x2

u = arctan 2 x and

53. Let:

cos−1 1 = 0 because cos 0 = 1. π  sin sin −1 1 + cos−1 1 = sin  + 0  = 1 2  

v x

)

sin cos−1 ( −1) + π = sin (π + π ) = sin 2π = 0

= 2 x 1 − x2

56. cos −1 ( −1) = π

= x 1 − x2 + 1 − x2 x

1 − x2

543

2 π  π π  −1 sin  + sin ( −1)  = sin  −  = sin 0 = 0 2  2 2

cos ( arccos x − arcsin x ) = cos ( arccos x ) cos ( arcsin x )

Sum and Difference Formulas

4 x2 + 1

)

Let φ = sin −1

4 4  cosθ = . 5 5

3 3  sin φ = . 5 5

4 3  sin  cos −1 + sin −1  = sin (θ + φ ) 5 5  = sin θ cos φ + cos θ sinφ 24  3  4   4  3  =    +    = 25  5  5   5  5 

4x 2 + 1

1 − x2

5 v

54. Let:

+ sin ( arcsin x ) sin ( arctan 2 x )

1 4 x2 + 1

Let φ = sin −1 135  sin φ = 135 .

(

)

sin cos−1 35 − sin −1 135 = sin (θ − φ ) = sin θ cosφ − cosθ sin φ

= ( 45 )( 12 − ( 53 )( 135 ) = 33 13 ) 65

4 x2 + 1

2 x2 + 1 − x2 4 x2 + 1

4x 2 + 1

θ 1

60. Let θ = cos −1 35  cosθ = 35 .

cos ( arcsin x − arctan 2 x ) = cos ( arcsin x ) cos ( arctan 2 x )

ϕ 4

u = arcsin x and v = arctan 2 x sin u = x tan v = 2 x

= 1 − x2

1 − x2

v 1

544

Chapter 6

Analytic Trigonometry

π π π  61. sin  + x  = sin cos x + sin x cos 2 2 2 

π π   cos  x +  − cos  x −  = 1 6 6  

70.

= (1) cos x + 0 = cos x

π π  π π   cos x cos − sin x sin  −  cos x cos + sin x sin  = 1 6 6  6 6 

62. sin ( 3π − x ) = sin 3π cos x − cos3π sin x

−2sin x sin

=1 6 1 −2sin x   = 1 2

= ( 0 ) cos x − ( −1) sin x = sin x 63. tan ( x + π ) − tan (π − x ) = =

tan x + tan π tan π − tan x − 1 − tan x ⋅ tan π 1 + tan π tan x

sin x = −1 3π x= 2

tan x  tan x  −−  = 2 tan x 1 1  

tan x + tan π + 2 ( sin x cos π + cos x sin π ) = 0 1 − tan x tan π tan x + 0 + 2 sin x ( −1) + cos x ( 0 )  = 0 1 − tan x(0) tan x − 2sin x = 0 1 sin x = 2sin x cos x sin x = 2sin x cos x

65. sin ( x + y ) + sin ( x − y ) = sin x cos y + sin y cos x

66. cos ( x + y ) + cos ( x − y )

tan ( x + π ) + 2sin ( x + π ) = 0

71.

1 − tan θ π  tan (π 4 ) − tan θ 64. tan  − θ  = = 4 1 tan π 4 tan θ 1 + + tan θ ( )  

+ sin x cos y − sin y cos x = 2 sin x cos y

= cos x cos y − sin x sin y + cos x cos y + sin x sin y = 2 cos x cos y

sin x (1 − 2 cos x ) = 0

67. cos ( x + y ) cos ( x − y ) = cos x cos y − sin x sin y 

sin x = 0

cos x cos y + sin x sin y 

(

)

or cos x =

x = 0, π

)

= cos2 x cos2 y − sin 2 x sin 2 y = cos2 x 1 − sin 2 y − sin 2 x sin 2 y

(

= cos2 x − sin 2 y cos2 x + sin 2 x = cos2 x − sin 2 y

= sin x cos y + cos x sin y  sin x cos y − cos x sin y 

(

)

= sin 2 x cos2 y − cos2 x sin 2 y = sin 2 x 1 − sin 2 y − cos2 x sin 2 y 2

(

)

= sin x − sin y sin x + cos x = sin x − sin y

π π   sin  x +  + sin  x −  = 1 3 3  

69. sin x cos

π 3

+ cos x sin

π 3

+ sin x cos

π 3

− cos x sin

π 2

5π 3

(

)

2 1 − sin 2 x − 3sin x = 0 2

2sin x + 3sin x − 2 = 0

=1 3 2sin x ( 0.5 ) = 1 x=

π π  2 sin x cos + cos x sin  + 3tan ( − x ) = 0 2 2  sin x 2 cos x − 3 =0 cos x 2cos2 x − 3sin x = 0

sin x = 1

1 2

π  2sin  x +  + 3tan (π − x ) = 0 2 

72.

68. sin ( x + y ) sin ( x − y )

5π

( 2sin x − 1)( sin x + 2 )  sin x = 2  x = 6 , 6 π π   73. Graph y1 = cos  x +  + cos  x −  and y2 = 1, and 4 4   find the point(s) of intersection. 2

y2 2π

y1 −2

x ≈ 0.7854, 5.4978

Section 6.4

π 3π    74. Graph y1 = sin  x +  − cos  x +  and approximate 2 2    the zeros.

78.

Sum and Difference Formulas

545

1 1 y = sin 2t + cos2t 3 4

(a)

2 10

2π

−1

−2

(b)

x ≈ 0.7854, 3.9270

π  75. Graph y1 = sin  x +  + cos2 x and approximate the 2  zeros. 3

2π

1 1 a= , b= , B=2 3 4

C = arctan

3 b = arctan ≈ 0.6435 4 a

5 1 1 y ≈   +   sin ( 2t + 0.6435 ) = sin ( 2t + 0.6435 ) 3 4 12     5 (c) Amplitude: 12 1 B 2 1 (d) Frequency: = = = period 2π 2π π

−1

π 2

, π,

3π 2

79. False. cos ( u ± v ) = cos u cos v + sin u sin v

π  76. Graph y1 = cos  x −  − sin 2 x and approximate the 2  zeros.

11π  11π 11π  80. True. sin  x − = sin x cos − cos x sin 2  2 2  = 0 − cos x ( −1) = cos x

2π

81. (a) Domains of f and g are all real numbers, h ≠ 0. (b)

−3

x = 0, 77.

π 2

0.01

,π

x  t x t y1 + y2 = A cos2π  −  + A cos2π  +  T λ T λ    

0.02

0.05

0.1

0.2

0.5

f (h) 0.4957 0.4913 0.4781 0.4559 0.4104 0.2674

g(h) 0.4957 0.4913 0.4781 0.4559 0.4104 0.2674 (c)

0.6

  2π t   2π x   2π t   2π x   cos  = A cos  + sin     sin    λ   T   λ    T    2π t   2π x   2π t   2π x   + A cos   cos  λ  − sin  T  sin  λ   T           2π t   2π x  = 2 A cos   cos  λ   T   

0.2

0.6

(d) As h → 0, f → 12 and g → 12 . In fact, f = g. 82. (a)

From the graph, sin u ≈ 0.7, sin v ≈ 0.7, and sin (u + v) ≈ 0. So, sin (u + v ) ≠ sin u + sin v.

(b)

From the graph , sin u ≈ 0.7, sin v ≈ 0.7, and sin (u − v) ≈ −1. So, sin (u − v ) ≠ sin u − sin v.

546

Chapter 6

Analytic Trigonometry

83. cos ( nπ + θ ) = cos nπ cos θ − sin nπ sin θ

= ( −1) ( cosθ ) − ( 0 )( sin θ ) n

= ( −1) ( cosθ ) , where n is an integer. n

84. sin ( nπ + θ ) = sin nπ cosθ + sin θ cos nπ

= ( 0 )( cosθ ) + ( sin θ )( −1)

= ( −1) ( sin θ ) , where n is an integer. n

85. C = arctan b  tan C = b  sin C = a

cos C =

a2 + b2

88. sin2θ − cos2θ ; a = 1, b = −1, B = 2 π b (a) C = arctan = arctan ( −1) = − a 4 sin 2θ − cos2θ = a 2 + b 2 sin ( Bθ + C )

π  = 2 sin  2θ −  4  (b) Because b > 0 in the formula, we write the given expression as − ( − sin 2θ + cos2θ ) ; a = −1, b = 1, B = 2, π a C = arctan   = arctan ( −1) = − . 4 b

Hence,

a2 + b2

− ( − sin 2θ + cos2θ ) = − a 2 + b 2 cos ( Bθ − C )

a + b sin ( Bθ + C ) 2

 a = a + b  sin Bθ ⋅ +  2 a + b2  = a sin Bθ + b cos Bθ 2

86. C = arctan

a  sin C = b

a 2

a +b

 ⋅ cos Bθ   2 2 a +b 

π  = − 2 cos  2θ +  . 4 

, cos C =

89. 12sin3θ + 5cos3θ ; a = 12, b = 5, B = 3 b 5 (a) C = arctan = arctan ≈ 0.3948 a 12

12sin3θ + 5cos3θ = a 2 + b2 sin ( Bθ + C )

a + b2

≈ 13sin ( 3θ + 0.3948 )

a 2 + b2 cos ( Bθ − C )   b a = a 2 + b2  cos Bθ ⋅ + sin Bθ ⋅   2 2 2 2  a +b a +b   = b cos Bθ + a sin Bθ

a 12 (b) C = arctan = arctan ≈ 1.1760 b 5

12sin3θ + 5cos3θ = a 2 + b2 cos ( Bθ − C ) ≈ 13cos ( 3θ − 1.1760 )

= a sin Bθ + b cos Bθ 87. sinθ + cosθ ; a = 1, b = 1, B = 1 (a)

C = arctan

π b = arctan1 = a 4

sin θ + cosθ = a 2 + b 2 sin ( Bθ + C )

π  = 2 sin  θ +  4 

π a (b) C = arctan = arctan1 = b 4 sin θ + cosθ = a + b cos ( Bθ − C ) 2

π  = 2 cos  θ −  4 

90. 3sin2θ + 4cos2θ ; a = 3, b = 4, B = 2 (a)

C = arctan

b 4 = arctan ≈ 0.9273 a 3

3sin 2θ + 4cos2θ = a 2 + b2 sin ( Bθ + C ) ≈ 5sin ( 2θ + 0.9273)

(b) C = arctan

a 3 = arctan ≈ 0.6435 b 4

3sin 2θ + 4cos2θ = a 2 + b2 cos ( Bθ − C ) ≈ 5cos ( 2θ − 0.6435)

91. C = arctan

b π = a=0 a 2

a 2 + b2 = 2  b = 2

B =1

π  2sin  θ +  = ( 0 )( sin θ ) + ( 2 )( cosθ ) = 2 cosθ 2 

Section 6.4

92. C = −

a a = arctan    = −1 4 b b  a = −1, b = 1

a 2 + b2 = 2 Hence, B = 1 and   π  π 5  5cos  θ +  = 2 cos θ −  −   4 2    4  5 = − sin θ + cos θ  2 5 5 =− sin θ + cosθ 2 2 5 2 5 2 =− sin θ + cosθ . 2 2

y 1 = m 1x + b 1

 m2 − m1    1 + m1m2 

θ = arctan  m2 = 1 m1 =

1 3

1   1 − 3   = arctan 2 − θ = arctan  1 + 1    3 

(

)

3 = 15°

95. From the figure, it appears that u + v = w. Assume that u, v, and w are all in Quadrant I. From the figure: s 1 = 3s 3 s 1 tan v = = 2s 2 s tan w = = 1 s tan u + tan v tan ( u + v ) = 1 − tan u tan v (1 3) + (1 2 ) = 1 − (1 3)(1 2 ) tan u =

θ δ β

547

94. For m2 > m1 > 0, the angle θ between the lines is:

93.

Sum and Difference Formulas

y 2 = m 2x + b2

m1 = tan α and m2 = tan β

β + δ = 90°  δ = 90° − β α + θ + δ = 90°  α + θ + (90° − β ) = 90°  θ = β − α

56 1 − (1 6 )

=1 = tan w

So, θ = arctan m2 − arctan m1. For y = x and y = m2 =

3 x you have m1 = 1 and

θ = arctan 3 − arctan 1 = 60° − 45° = 15° 96. cos(u + v + w) = cos ( u + v ) + w

= cos ( u + v ) cos w − sin ( u + v ) sin w

= ( cos u cos v − sin u sin v ) cos w − ( sin u cos v + cos u sin v ) sin w = cos u cos v cos w − sin u sin v cos w − sin u cos v sin w − cos u sin v sin w

97. x = 0 : y = −

1 ( 0 − 10 ) + 14 = 5 + 14 = 19 2

y-intercept: ( 0, 19 ) y = 0: 0 = −

1 1 ( x − 10 ) + 14 = − x + 19  x = 38 2 2

x-intercept: ( 38, 0 )

98. y = 0 : x 2 − 3 x − 40 = ( x − 8 )( x + 5 ) = 0 x-intercepts: ( 8, 0 ) , ( −5, 0 )

x = 0  y = −40 y-intercept: ( 0, − 40 )

99. x = 0 : 2 ( 0 ) − 9 − 5 = 9 − 5 = 4 y-intercept: ( 0, 4 ) y = 0: 2 x − 9 = 5  x = 7, 2 x-intercepts: ( 2, 0 ) , ( 7, 0 )

100. y = 0 : 2 x x + 7 = 0  x = 0, − 7 x-intercepts: ( 0, 0 ) , ( −7, 0 )

x=0y=0 y-intercept: ( 0, 0 )

548

Chapter 6

Analytic Trigonometry

Section 6.5 Multiple-Angle and Product-to-Sum Formulas 1.

1 + cos2u 2

1  cos(u − v ) + cos(u + v)  2

u+v u−v −2sin   sin    2   2 

1 − cos u 2

tan

u 2

(f)

(g)

(h)

tan 2 u

(a)

sin2u = 2sin u cos u; matches (ii).

(b)

cos2u = 1 − 2sin 2 u; matches (i). 2 tan u tan 2u = ; matches (iii). 1 − tan 2 u

(c)

1 cos2θ 25 = 7 1 csc 2θ = sin 2θ 25 = 24 1 cot 2θ = tan 2θ 1 − tan 2 θ = 2 tan θ

sec 2θ =

1 (a) sin u cos v = sin(u + v ) + sin(u − v )  ; 2 matches (ii). 1 (b) cos u sin v = sin(u + v ) − sin(u − v)  ; 2 matches (iii). 1 (c) cos u cos v = cos(u − v) + cos(u + v)  ; 2 matches (i).

3 7 1−  4  16 = = 3 3 2  2 4 =

10.

7 2 7 ⋅ = 16 3 24

(a) (b) (c)

12 13 5 cosθ = 13 12 tan θ = 5

sin θ =

(d) cos2θ = cos2 θ − sin 2 θ

θ 4

(a) (b) (c) (d)

3 5 4 cosθ = 5 3 tan θ = 4 sin 2θ = 2 sin θ cos θ sin θ =

(e)

 3  4  = 2     5  5  24 = 25

(e)

cos2θ = cos2 θ − sin 2 θ 2

4 3 =  −   5 5 7 = 25

 5   12  =  −   13   13  119 =− 169 2 tan θ tan 2θ = 1 − tan 2 θ  12  2  5 =  2  12  1−    5 24 = 5 119 − 25 120  24  25  =   − =−  119  25  119 

Section 6.5

(f)

1 tan 2θ 1 − tan 2 θ = 2 tan θ

cot 2θ =

 12  1−    5  =  12  2   5 

(h)

119 = 25 24 5 119 5 =− ⋅ 25 24 119 =− 120 1 sec 2θ = cos2θ 116 =− 119 1 csc 2θ = sin 2θ =

14. sin 2 x sin x = cos x x ≈ 0.7854, 1.5708, 2.3562, 3.9270, 4.7124, 5.4978 Analytically: 2sin x cos x sin x − cos x = 0 cos x (2 sin 2 x − 1) = 0

2 2 π 3π 5π 7π x= , , , 4 4 4 4

sin x = ±

15. cos 2 x − cos x = 0 x ≈ 0, 2.094, 4.189, (6.283 not in interval) Analytically: cos2 x − cos x = 0

11. sin 2 x − sin x = 0 Solutions: 0, 1.047, 3.142, 5.236 Analytically: sin 2 x − sin x = 0 2sin x cos x − sin x = 0 sin x (2 cos x − 1) = 0 sin x = 0 or 2 cos x − 1 = 0

or 2sin x − 1 = 0 1 π 3π sin 2 x = x= , 2 2 2

cos x = 0

169 120

x = 0, π 5π x = 0, ,π , 3 3

549

13. 4 sin x cos x = 1 x ≈ 0.2618, 1.3090, 3.4034, 4.4506 Analytically: 4sin x cos x = 1 2sin(2 x ) = 1 1 sin(2 x ) = 2 π 5π 13π 17π 2x = , , , 6 6 6 6 π 5π 13π 17π x= , , , 12 12 12 12

−

(g)

Multiple-Angle and Product-to-Sum Formulas

cos x = x=

2cos2 x − 1 − cos x = 0 (2cos x + 1)(cos x − 1) = 0

1 2

π 3

5π 3

12. sin 2 x + cos x = 0 Solutions: 1.5708, 3.6652, 4.7124, 5.7596 Analytically: sin 2 x + cos x = 0 2sin x cos x + cos x = 0 cos x(2sin x + 1) = 0 1 sin x = − cos x = 0 2 π 3π 7π 11π x= , x= , 2 2 6 6

1 cos x = − , cos x = 1 2 2π 4π x= , , 0, (2π not in interval) 3 3

16. cos 2 x + sin x = 0 x ≈ 1.5708, 3.6652, 5.7596 Algebraically: cos2 x + sin x = 0 1 − 2sin 2 x + sin x = 0 2 sin 2 x − sin x − 1 = 0 (2sin x + 1)(sin x − 1) = 0 1 7π 11π sin x = −  x = , 2 6 6 sin x = 1

x=

π 2

550

Chapter 6

Analytic Trigonometry

17. sin 4 x = −2sin 2 x

19. tan 2 x − cot x = 0 x ≈ 0.5236, 1.5708, 2.6180, 3.6652, 4.7124, 5.7596 Analytically: 2 tan x = cot x 1 − tan 2 x 2 tan x = cot x(1 − tan 2 x )

x ≈ 0, 1.571, 3.142, 4.712 sin 4 x = −2sin 2 x sin 4 x + 2sin 2 x = 0 2sin 2 x cos2 x + 2sin 2 x = 0 2sin 2 x (cos2 x + 1) = 0 2sin 2 x = 0 or cos2x + 1 = 0 sin 2x = 0

cos2x = −1

2 x = nπ

2 x = π + 2 nπ

n x= π 2

x = 0,

18.

2 tan x = cot x − cot x tan 2 x

π 2

x= , π,

3π 2

2 tan x = cot x − tan x 3tan x = cot x 3tan x − cot x = 0

+ nπ

1 =0 tan x 3tan 2 x − 1 =0 tan x

3tan x −

π 3π 2

( sin2 x + cos2 x ) = 1 2

x ≈ 0.0, 0.7854, 1.5708, 2.3562, 3.1416, 3.9270, 4.7124, 5.4978

Analytically: sin 2 2 x + 2sin 2 x cos2 x + cos2 2 x = 1 2sin 2 x cos2 x = 0 sin 4 x = 0 4 x = nπ nπ x= 4

or 3tan 2 x − 1 = 0 1 π 3π x= , tan 2 x = 2 2 3

cot x = 0

3 3 π 5π 7π 11π x= , , , 6 6 6 6 π π 5π 7π 3π 11π x= , , , , , 6 2 6 6 2 6 tan x = ±

3π , , 2 4 5π 3π 7π π, , , 4 2 4

x = 0,

1 (3tan 2 x − 1) = 0 tan x cot x(3tan 2 x − 1) = 0

20. tan 2 x − 2 cos x = 0 x ≈ 0.5236, 1.5708, 2.6180, 4.7124 Algebraically: tan 2 x = 2cos x 2 tan x = 2cos x 1 − tan 2 x  sin x sin 2 x  = cos x  1 −  cos x cos2 x   π 3π    Note: cos x = 0  x = ,  2 2   sin x = cos2 x − sin 2 x sin x = 1 − 2sin 2 x 2sin 2 x + sin x − 1 = 0 (2sin x − 1)(sin x + 1) = 0 1 π 5π sin x =  x = , 2 6 6 3π sin x = −1  x = 2 π cos x = 0  x = 2

Section 6.5 3 3π 4 21. sin u = − , < u < 2π  cos u = 5 2 5 y

Multiple-Angle and Product-to-Sum Formulas

551

1 3π 1  sin u = − and 23. tan u = , π < u < 2 2 5 2 cos u = − 5 y

u −3

−2 −1

24  3 4 sin 2u = 2sin u cos u = 2 ⋅  −  ⋅ = − 25  5 5 16 9 7 − = cos 2u = cos 2 u − sin 2 u = 25 25 25 2 ( − 3 4) 2 tan u 24 tan 2u = = =− 2 1 − tan u 1 − ( 9 16 ) 7

22. cos u = −

2 π and <u<π 2 3 y

 1  2  4 sin 2u = 2sin u cos u = 2  −  − = 5  5 5  2

 2   1  3 cos2u = cos2 u − sin 2 u =  −  −−  = 5 5 5     2 (1 2 ) 2 tan u 4 tan 2u = = = 2 1 − tan u 1 − (1 4 ) 3 3π < u < 2π 2

24. cot u = −6, y

−2

x u

−1

not drawn to scale

sin 2u = 2sin u cos u

 5  2  = 2 −  3   3    4 5 9 cos2u = cos2 u − sin 2 u =−

2  2  5  = −  −   3   3 

 1  6  12 sin 2u = 2sin u cos u = 2  −  =− 37 37  37   36 1 35 cos2u = cos2 u − sin 2 u = − = 37 37 37 2 ( −1 6 ) 2 tan u 12 −2 6 tan 2u = = = =− 1 − tan 2 u 1 − ( −1 6 )2 35 36 35

1 9 2 tan u tan 2u = 1 − tan 2 u  5 2 −  2   =  2  5 1− −  2    =−

− 5 5 1− 4 − 5 = 1 − 4

552

Chapter 6

25. sec u = −2,

Analytic Trigonometry

27. 6sin x cos x = 3( 2sin x cos x) = 3sin 2x

<u<π

28. 14sin x cos x = 7( 2sin x cos x) = 7sin 2x

29. cos 2 x −

u x

−1

1 1 1 = ( 2 cos 2 x − 1) = cos2 x 2 2 2

(

)

30. 10sin 2 x − 5 = 5 2sin 2 x − 1

= − 5(1 − 2sin x) = − 5cos2 x 2

sin 2u = 2sin u cos u

31. sin 2 x =

 3  1  = 2 −  2   2   

1 − cos2 x 1 cos2 x = − 2 2 2 2

3 =− 2

−2π

2π

cos2u = cos u − sin u 2  1  3  = −  −   2   2 

−2

32.

1 3 1 = − =− 4 4 2 2 tan u tan 2u = 1 − tan 2 u  3 2 −  1   =  2  3 1− −  1   

f ( x ) = cos2 x = 2

−2π

2π

−2

33. f ( x) = cos3 x = (cos x)(cos 2 x)  1 + cos 2 x  = (cos x)  2   cos x + cos x cos 2 x = 2

−2 3 1− 3

= 3

26. csc u = 3, 0 < u <

− 2π

3 u 2 2

sin u =

2π

−2

34.

1 2 2 1 , cos u = , tan u = = 3 3 2 2

4 2  1  2 2  sin 2u = 2sin u cos u = 2   = 9  3  3  cos 2u = cos 2 u − sin 2 u = tan 2u =

1 + cos2 x 1 cos2 x = + 2 2 2

(

8 1 7 − = 9 9 9

f ( x ) = sin 3 x  1 − cos2 x  = sin x   2  

2 4

−2π

2π

−2

)

2 2 4 2 tan u 4 2 = = 2 1 − tan u 1 − (1 8) 7

Section 6.5

(

)(

35. cos 4 x = cos2 x cos2 x

)

1 + 2cos 2 x + cos 2 2 x 4 1 + 2cos 2 x + (1 + cos 4 x ) 2

4 2 + 4cos 2 x + 1 + cos 4 x = 8 3 + 4cos 2 x + cos 4 x = 8 1 = ( 3 + 4cos 2 x + cos 4 x ) 8

36.

x  1 + cos2 x  (sin x )( cos x ) =  1 − cos2   2 2   2

(

37. sin 8 x = sin 2 x

 1 + cos2 x   1 + cos2 x  =     2 2

Multiple-Angle and Product-to-Sum Formulas

1 − cos 2 2 x 4 1  1 + cos 4 x  = 1 −  4 2  1 = ( 2 − 1 − cos 4 x ) 8 1 = (1 − cos 4 x ) 8 =

 1 − cos2 x  =   2

)

553

1 cos 4 2 x − 4 cos3 2 x + 6 cos2 2 x − 4 cos2 x + 1 16 2 1  = cos2 2 x − 4 cos2 x cos2 2 x + 6 cos2 2 x − 4 cos2 x + 1   16  2  1 + cos 4 x   1 + cos 4 x   − 4 cos2 x        2 2 1   =  16   1 + cos 4 x   +6   − 4 cos2 x + 1      2 1  1  1 + 2 cos 4 x + cos2 4 x − 2 cos2 x − 2 cos2 x cos 4 x  = 4  16   +3 + 3cos 4 x − 4 cos2 x + 1  =

(

)

(

)

(

)

1 1 1  1  + cos 4 x + (1 + cos8 x ) − 6 cos2 x  4 2 8  16   −2 cos2 x cos 4 x + 3cos 4 x + 4 

1  35 7  1  + cos 4 x + cos8 x − 6 cos2 x  8 2 8  16   −2 cos2 x cos 4 x 

1 35 + 28cos 4 x + cos8 x − 48cos2 x    128  −16 cos2 x cos 4 x 

38. cos8 x 4

4  1 + cos2 x  = (cos 2 x) =   2   1 = (1 + 4cos 2 x + 6cos2 2 x + 4cos3 2 x + cos4 2 x) 16 2 1  1 + cos 4 x  2 2 1 + 4cos2 x + 6 =  + 4cos2 x(cos 2 x) + (cos 2 x)   16  2   

1  1 + cos 4 x   1 + cos4 x   1 + 4cos2 x + 3 ⋅ (1 + cos 4 x) + 4cos2 x +   16  2 2     

1 1  1 + 4cos2 x + 3 + 3cos 4 x + 2cos 2 x + 2cos2 x cos4 x + (1 + 2cos4 x + cos 2 4 x)  16  4 

1 1  1 + cos8 x   1 + 4cos2 x + 3 + 3cos 4 x + 2cos 2 x + 2cos2 x cos4 x + 1 + 2cos4 x +    16  4 2   

1 1 1  1 + 4cos2 x + 3 + 3cos 4 x + 2cos 2 x + 2cos2 x cos4 x + 1 + 2cos4 x + (1 + cos8 x)  16  4 2 

1 1 1 1  1 + 4cos2 x + 3 + 3cos 4 x + 2cos 2 x + 2cos2 x cos4 x + 1 + 2cos4 x + + cos8 x  16  4 2 2 

1 3 1 1  1 + 4cos2 x + 3 + 3cos 4 x + 2cos 2 x + 2cos2 x cos4 x + + cos 4 x + cos8 x  16  8 2 8 

1  35 7 1  + 6cos2 x + cos 4 x + 2cos 2 x cos 4 x + cos8 x 16  8 2 8  1 = [35 + 48cos 2 x + 28cos 4 x + 16cos 2 x cos 4 x + cos8 x] 128 =

554

Chapter 6

Analytic Trigonometry

39. sin2 x cos4 x = sin 2 x cos2 x cos2 x  1 − cos2 x  1 + cos2 x  1 + cos2 x  =    2 2 2     1 (1 − cos2 x )(1 + cos2 x )(1 + cos2 x ) 8 1 = 1 − cos2 2 x (1 + cos2 x ) 8 1 = 1 + cos2 x − cos2 2 x − cos3 2 x 8 1  1 + cos 4 x   1 + cos 4 x   = 1 + cos2 x −   − cos2 x   8 2 2     1 = 2 + 2 cos2 x − 1 − cos 4 x − cos2 x − cos2 x cos 4 x  16  1  1 1  = 1 + cos2 x − cos 4 x −  cos2 x + cos6 x   16  2 2  1 = ( 2 + 2 cos2 x − 2 cos 4 x − cos2 x − cos6 x ) 32 1 = ( 2 + cos2 x − 2 cos 4 x − cos6 x ) 32

43. cos2

(

)

(

)

40. sin 4 x cos2 x = sin 2 x sin 2 x cos2 x  1 − cos2 x  1 − cos2 x  1 + cos2 x  =    2 2 2    

44. sin 2

(

x 1 − cos x = 2 2 1 1 = − cos x 2 2 1 = (1 − cos x ) 2

 1 + cos 2 x  1 − cos2 x  45. cos 2 x tan 2 x =    2   1 + cos2 x  1 − cos 2 x 2 1 = (1 − cos 2 x) 2 =

 1 − cos2 x  1 − cos2 x  46. sin 2 x tan 2 x =    2   1 + cos2 x 

1 (1 − cos2 x ) 1 − cos2 2 x 8 1 = 1 − cos2 x − cos2 2 x − cos3 2 x 8 1  1 + cos 4 x   1 + cos 4 x   = 1 − cos2 x −   + cos2 x   8 2 2     1 = (2 − 2 cos2 x − 1 − cos 4 x + cos2 x + cos2 x cos 4 x ) 16 1 1  1  =  1 − cos2 x − cos 4 x + cos2 x + cos6 x  2 16  2  =

x 1 + cos x = 2 2 1 1 = + cos x 2 2 1 = (1 + cos x ) 2

)

 1 + cos 4 x  1 − 2 cos2 x +   2   = 2 (1 + cos2 x )

)

1 ( 2 − 2 cos2 x − 2cos 4 x + cos2 x + cos6 x ) 32 1 = ( 2 − cos2 x − 2 cos 4 x + cos6 x ) 32 =

1 − cos 4 x 2 1 1 = − cos4 x 2 2 1 = (1 − cos 4 x ) 2

41. sin 2 2 x =

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

42. cos2 2 x =

1 − 2 cos2 x + cos2 2 x 2 (1 + cos2 x )

3 1 − 2 cos2 x + cos 4 x 2 2 = 2 (1 + cos2 x ) =

47. sin 4

3 − 4 cos2 x + cos 4 x 4 (1 + cos2 x )

x  2 x  2 x  =  sin  sin  2  2  2

 1 − cos x  1 − cos x  =   2 2    1 2 = 1 − 2 cos x + cos x 4 1 1 + cos2 x  =  1 − 2cos x +  4 2 

(

)

1 ( 2 − 4cos x + 1 + cos2 x ) 8 1 = ( 3 − 4 cos x + cos2 x ) 8 =

Section 6.5

48. cos 4

x  2 x  2 x  =  cos  cos  2  2  2

sin

(b)

cos

(c)

tan

(d)

cot

(e)

sec

1 + (15 17 ) 1 + cosθ 16 4 4 17 cos = = = = = 2 2 2 17 17 17

(f)

csc

1 − (15 17 ) 1 − cosθ 17 sin = = = 2 2 2 17 θ sin θ 8 17 8 1 = = = tan = 2 1 + cosθ 1 + (15 17 ) 12 4

(g)

2sin

(

)

1 ( 2 + 4 cos x + 1 + cos2 x ) 8 1 = ( 3 + 4 cos x + cos2 x ) 8 =

(b) (c) (d)

sec

cos (θ 2 )

(1 + cosθ ) 2

(h)

csc

sin (θ 2 )

1 + (15 17 )

(1 − cosθ ) 2

1 1 − (15 17 ) 2

1 − ( 7 25 ) 1 − cosθ 18 3 = = = 2 2 50 5

1 + ( 7 25 ) 4 1 + cosθ = = 2 2 5

2 2 2 2 2

sin (θ 2 )

cos (θ 2 )

1 4 = tan (θ 2 ) 3

1 5 = cos (θ 2 ) 4

1 5 = sin (θ 2 ) 3

θ 24 cos = sin θ = 2 2 25 θ θ θ 3 6 2 cos tan = 2sin = 2   = 2 2 2 5 5 θ

(f) (g)

 1   4 17  8 θ θ 2sin cos = 2   = , ( = sin θ ) = 2 2  17   17  17

(h)

2 cos

θ 2

= 2 sin

θ 2

2 17 17

2 2+ 3 2 2 2+ 3 2

1 + cos150°  150  ° cos75° = cos   = 2  2 

8 17 15 cosθ = 17

3 2

1− 2

sin θ =

3 2

1 − cos150°  150  ° 51. sin 75° = sin   = 2  2 

1 + cosθ 1 + (15 17 ) cot = = =4 2 sin θ 8 17

tan

3 4

1 17

555

= 17

17 = 4

(e)

50. (a)

 1 + cos x  1 + cos x  =   2 2    1 = 1 + 2 cos x + cos2 x 4 1 1 + cos2 x  =  1 + 2 cos x +  4 2 

49. (a)

Multiple-Angle and Product-to-Sum Formulas

2− 3 2 = 2

2− 3 2

 150  ° 1 − cos150° tan 75° = tan   = sin150°  2 

3 2

556

Chapter 6

Analytic Trigonometry

(

1− 3 2 1 − cos330° 1  = 52. sin165° = sin  ⋅ 330° = 2  2 2 1 = 2− 3 2

(

)

 135 ° 54. sin 67°30′ = sin   =  2 

(

 225 ° 53. sin112°30′ = sin   =  2  1+

)

2 2+

1 − cos 225° 2

2 2

 225 ° cos112°30′ = cos  = −  2  = −

= − =

2+ 2

1 + cos 225° 2

2 2− 2 2 2 2− 2

2 2

= −

2 2

2+ 2 − 2

2 2 + 2 −2

= −1 −

1 + cos135° 2

2 2− 2 2 2 2− 2

2 2 2 2

= =

π 8

2 2

2 2 +2 2

=1+

55. sin

2 2 2 2

2 2

1−

 225 ° tan112°30′ = tan    2  1 − cos 225° = sin 225° 2 2 2 − 2

2+ 2

 135 ° tan 67°30′ = tan    2  1 − cos135° = sin135°

2 2

1−

 135 ° cos 67°30′ = cos  =  2 

2 2 2

2 2

)

1+ 3 2 1 + cos330° 1  cos165° = cos  ⋅ 330°  = − = 2 2 2  1 2+ 3 =− 2 sin 330° −1 2 −1 1  tan165° = tan  ⋅ 330°  = = = 2+ 3 2  1 + cos330° 1 + 3 2 = 3 −2

1 − cos135° 2

 1  π  1 − cos π 4 1 = sin     = = 2− 2 2 4 2 2    

1 + cos (π 4 ) 1  1  π  = cos     = = 2+ 2 2 4 2 2     sin (π 4 )  1  π  2 2 π = = 2 −1 tan = tan     = 8  2  4   1 + cos (π 4 ) 1 + 2 2

cos

(

)

Section 6.5

56. sin

(

1− 3 2  1  π  1 − cos(π 6) = sin     = = 12 2 2  2  6  1 = 2− 3 2

)

Multiple-Angle and Product-to-Sum Formulas

58. sin

1 + cos (π 6 ) 1  1  π  π cos = cos     = = 2+ 3 12 2 2  2  6  sin (π 6 )  1  π  12 π tan = tan     = = =2− 3 12 2 6 1 + cos 6 π ( ) 1+ 3 2   

(

1+ 3 2 1 − cos ( 7 π 6 ) 7π   1 7π  57. sin  = sin  ⋅ = =   2 2  12  2 6  =

)

cos

2− 3 2

2+ 2

(

)

1 3 −2

= −

= −2 − 3

(

1+ −

2 2

)

( 2 2)

1−

2 1−

tan

1 + cos (5π 4) 2

5π  1 5π  = cos ⋅  = − 8 2 4  = −

sin ( 7π 6 ) − (1 2 )  7π   1 7π  tan  = =  = tan  ⋅ =  12   2 6  1 + cos ( 7π 6 ) 1 − 3 2

2 4

(

2 2 2

1− 3 2 1 + cos ( 7 π 6 )  7π   1 7π  cos  =−  = cos  ⋅ =− 2 2  12  2 6 

)

2 2

( 2 2)

)

2+ 3 2

=−

(

1− −

)

1 − cos (5π 4)

5π  1 5π  = sin  ⋅  = 8 2 4  =

557

= −

2 2

= −

2− 2 4

= −

2− 2

sin (5π 4) 5π  1 5π  = tan  ⋅  = 8  2 4  1 + cos (5π 4) =

−

(

1+ −

−

2 2 = 2− 2 2 2 2

2 2

)

− 2 2 − 2

2+ − 2 ⋅ 2 − 2 2+

−2 2 − 2 2

= −

2 2

2 −1

558

Chapter 6

59. cos u =

Analytic Trigonometry

7 π ,0 < u < 25 2

u cos  = 2

= 25

= 1 − cos u 2

u sin   = 2

1−

7 25

2 25 + 7 25 2 32 25 2

 u  1 − cos u tan   = sin u 2 7 1− 25 = 24 25 18 = 25 24 25 18 = 24 3 = 4

18 25 2

3 2 5 2 3 = 5 =

60. sin u =

7 25

4 2 5 2 4 = 5

25 − 7 7 2

1 + cos u 2

5 π 12 < u < π  cos u = − , 13 2 13

1 − cos u 1 + (12 13) 5 26 u = = sin   = 2 2 26 2 1 + cos u 1 − (12 13) 26 u = = cos   = 2 2 26 2 sin u 5 13 5 u = = =5 tan   =  2  1 + cos u 1 − (12 13) 1 y

13 u 12

Section 6.5

61. cot u = 3 and π < u <

Multiple-Angle and Product-to-Sum Formulas

559

3π 2

u −3 −1

1 − cos u u sin   = 2 2 3 10

1+ =

1 + cos u u cos   = − 2 2 =−

2 10 + 3 10 2

62. tan u = −

= −

1 26 = − 26 26 1 + cos u = 2 25 = 26

( 10 + 3)

−5 sin u = 13 12 cos u = 13 1 − cos u = 2

=−

1 10 − 3 10 2 5 1 =− 50 − 15 10 10

u sin   = −  2

10 + 3 10 = 1 − 10

=−

5 3π , < u < 2π 12 2

10 − 3 2 10

10 − 3 10 =− 20

1 10 + 3 10 1 = 50 + 15 10 2 5 10

u cos  =  2

10 − 3 10 2

=−

10 + 3 10 = 20 =

=−

10 + 3 2 10

3 10

1−

 u  1 − cos u tan   = sin u 2 3 1+ 10 = 1 − 10

1 − (12 13) = − 2

1 + (12 13) = 2

x −5

13 − 12 2(13)

13 + 12 2(13)

5 5 26 = 26 26

1 − (12 13)  u  1 − cos u = = tan   = − 5 13 2 sin u  

13 − 12 1 13 = − 5 5 − 13

560

Chapter 6

Analytic Trigonometry

5 3π 63. csc u = − , π < u < 3 2 3 4 sin u = − , cos u = − 5 5

67. −

68. −

1 − cos ( x − 1) 2 sin

−3

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

sin

 x −1 = − sin    2 

x + cos x = 0 2 x x sin + cos2   = 0 2 2

69.

−4

sin 4 x = − tan 4 x cos 4 x

=−

1 + ( 4 5) 1 − cos u 3 3 10 u = = = sin   = 2 2 10 10 2 1 − ( 4 5) 1 + cos u −1 10 u cos   = − =− = =− 2 2 10 10 2 1 4 5 + ( ) = −3  u  1 − cos u tan   = = sin u −3 5 2

(1 − cos8 x ) 2 (1 + cos8 x ) 2

1 − cos8 x =− 1 + cos8 x

x x +1 = 0 sin − 1 = 0 2 2 x −1 x sin = sin = 1 2 2 2 x 7π 11π x π = , = 2 6 6 2 2 7π 11π x= , x =π 3 3 Only x = π is in the interval [0, 2π ). 2sin

64. sec u =

7 π ,0 < u < 2 2 y

3 5

70.

cos u =

2 3 5 , sin u = 7 7

sin

u = 2

1 − cos u = 2

1 − ( 2 7) = 2

5 70 = 14 14

cos

u = 2

1 + cos u = 2

1 + ( 2 7) = 2

9 = 14

sin

65.

66.

)

1 − cos6 x = sin 3 x 2

x + cos x − 1 2

x + cos x − 1 = 0 2 ±

3 3 14 = 14 14

5 5 5 5 5 u 1 − cos u 1 − ( 2 7) 7 tan = = = = = = 2 sin u 15 3 3 5 3 5 3 57 7

(

h ( x ) = sin

1 − cos x = 1 − cos x 2 1 − cos x = 1 − 2 cos x + cos2 x 2 1 − cos x = 2 − 4 cos x + 2 cos2 x

2 cos2 x − 3cos x + 1 = 0

( 2 cos x − 1)( cos x − 1) = 0 2 cos x − 1 = 0

π 5π 3

cos x − 1 = 0 cos x = 1

1 cos x = 2 ,

x=0

1 + cos4 x 4x = cos = cos2 x 2 2

Section 6.5

71.

x − sin x = 0 2

cos

1 + cos x = sin x 2 1 + cos x = sin 2 x 2 1 + cos x = 2sin 2 x

1 + cos x = 2 − 2 cos2 x 2 cos2 x + cos x − 1 = 0

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

cos x = x=

π 5π 3

g ( x ) = tan

72. tan

cos x + 1 = 0 cos x = −1 x =π

x − sin x 2

x − sin x = 0 2 1 − cos x = sin x sin x 1 − cos x = sin 2 x 1 − cos x = 1 − cos2 x

cos2 x − cos x = 0 cos x ( cos x − 1) = 0

or cos x − 1 = 0 cos x = 1 π 3π x= , 2 2 x=0

cos x = 0

73. 10cos 75° cos15° = 5 cos ( 75° − 15° ) + cos ( 75° + 15° )  = 5 ( cos 60° + cos90° )

74. 6sin 45° cos15° = 3 sin ( 45° + 15° ) + sin ( 45° − 15° )  = 3 ( sin 60° + sin 30° )

75. 6sin

76.

4cos

π 3

cos

sin

π 3

1  π π   π π  = 6 ⋅ sin  +  + sin  −   2  3 3  3 3  2π 2π   = 3  sin + sin 0  = 3sin 3 3  

 1   π 5π  5π  π 5π    = 4  sin  +  − sin  3 − 6    6 2 3 6        

Multiple-Angle and Product-to-Sum Formulas

561

5 cos(θ − 3θ ) − cos(θ + 3θ ) 2 5 = cos( − 2θ ) − cos( 4θ ) 2 5 = (cos 2θ − cos 4θ ) 2

77. 5sin θ sin 3θ =

1  78. 3sin 2α sin 3α = 3  cos ( 2α − 3α ) − cos ( 2α + 3α )   2  3 3 = cos ( −α ) − cos ( 5α )  = ( cos α − cos5α ) 2 2

79. sin ( x + y ) cos ( x − y ) 1 sin ( x + y ) + ( x − y ) + sin ( x + y ) − ( x − y )   2 1 = ( sin 2 x + sin 2 y ) 2

(

)

(

)

80. sin ( x + y ) sin ( x − y ) 1 cos ( x + y ) − ( x − y ) − cos ( x + y ) + ( x − y )   2 1 = ( cos2 y − cos2 x ) 2 =

(

)

(

)

 5θ + θ   5θ − θ  81. sin 5θ − sin θ = 2 cos   sin    2   2  = 2 cos 3θ sin 2θ

 3θ + θ   3θ − θ  82. sin 3θ − sin θ = 2cos  sin    2   2  = 2cos 2θ sin θ  6x + 2x   6x − 2x  83. cos6 x + cos2 x = 2 cos   cos   2 2     = 2 cos 4 x cos 2 x

 x + 5x   x − 5x  84. sin x + sin 5 x = 2sin   cos    2   2  = 2 sin 3 x cos ( −2 x ) = 2 sin 3 x cos2 x

85. sin (α + β ) − sin (α − β )  α + β +α − β   α + β −α + β  = 2cos   sin   2 2    

= 2cosα sin β

  7π   π  = 2 sin   − sin  −      2 6  π 7π  = 2  sin + sin   6 2

562

Chapter 6

Analytic Trigonometry

 φ + α + φ − α   φ + α − (φ − α )  86. cos(φ + α ) − cos(φ − α ) = − 2sin   sin  2 2     = − 2sin φ sin α

π π   87. cos  θ +  − cos  θ −  2 2    θ + (π 2 ) + θ − (π 2 )   θ + ( π 2 ) − θ + ( π 2 )  = −2sin   sin       2 2     = −2sin θ sin

π 2

= −2sin θ

 5π 3π   5π 3π  + −  5π 3π 4  sin  4 4  92. sin − sin = 2 cos  4 4 4 2 2        

= 2 cos π sin

 75° + 15°   75° − 15°  89. sin 75° + sin15° = 2sin  cos  2 2     = 2sin ( 45°)cos(30°)

94.

 2  3  = 2     2  2  =

cos 2 x − cos 6 x = 0 −2 sin 4 x sin ( −2 x ) = 0

2 sin 4 x sin 2 x = 0 sin 4 x = 0 or sin 2 x = 0

6 2

4 x = nπ

 3 = (0)  = 0  2 

95.

π   3π π  3π  4 + 4  4 − 4 3π π − cos = − 2sin  91. cos sin   4 4 2 2         π  π   2  = − 2sin  sin    2   2   

2 x = nπ

nπ 2 π π 3π 5π 3π 7π x = 0, , , , π, , , 4 2 4 4 2 4 x=

120° + 60°   120° − 60°  90. cos120° + cos60° = 2cos cos  2 2     = 2cos(90°)cos(30°)

nπ 4

cos2 x −1 = 0 sin 3 x − sin x cos2 x =1 sin 3 x − sin x cos2 x =1 2 cos2 x sin x 2sin x = 1 sin x = x=

π  π  = − 2sin  sin   2 4  2 = − 2(1)  = −  2 

 2 = 2 ( −1)  =− 2  2   

sin 4 x cos2 x = 0 sin 4 x = 0 or cos2 x = 0 4 x = nπ π 2 x = + nπ 2 nπ x= π nπ 4 x= + 4 2 π π 3π 5π 3π 7π x = 0, , , , π, , , 4 2 4 4 2 4

 x + ( π 2 ) + x − (π 2 )   x + (π 2 ) − x + (π 2 )  = 2sin   cos       2 2    

93. sin 6 x + sin 2 x = 0  6x + 2x   6x − 2x  2sin   cos  =0 2    2 

π π   88. sin  x +  + sin  x −  2 2  

= 2sin x cos

1 2

π 5π 6

Section 6.5 sin 2 3 x − sin 2 x = 0

96.

( sin 3 x + sin x )( sin 3x − sin x ) = 0 ( 2sin 2 x cos x )( 2 cos2 x sin x ) = 0 sin 2 x = 0  x = 0,

, π,

3π or 2

2 3π cos x = 0  x = , or 2 2 π 3π 5π 7π cos2 x = 0  x = , , , or 4 4 4 4 sin x = 0  x = 0, π

103. cos2 2α − sin 2 2α = cos 2 ( 2α )  = cos 4α

(

105. ( sin x + cos x ) = sin 2 x + 2sin x cos x + cos2 x 2

(

= 1 + sin 2 x

α 3

cos

α 3

Figure for 97–100 2

107.

99.

π  π  sin α cos β = cos  − α  sin  − β  2  2  5 4 4    =    =  13  5  13  12  3  36 100. cos α sin β =    =  13  5  65

π  π  cos α sin β = sin  − α  cos  − β  2 2      12  3  36 =    =  13  5  65

1 2α sin 2 3

sin y 2

2sin y cos y + sin y − 2sin 3 y sin y sin y( 2cos 2 y + 1 − 2sin 2 y )

 5  4  4 sin α cos β =    =  13  5  13

(2sin y cos y)cos y + (1 − 2sin 2 y)sin y

25 144  5  cos2 α = 1 − sin 2 α = 1 −   = 1 − = 169 169  13 

1  α α   α α  sin  +  + sin  −   2  3 3  3 3 

sin 2 y cos y + cos2 y sin y sin y

2  12  144 98. cos2 α = ( cos α ) =   =  13  169

sin( 2 y + y ) sin3 y = sin y sin y =

144 25  12  sin α = 1 − cos α = 1 −   = 1 − = 169 169  13  2

)

= sin 2 x + cos2 x + 2sin x cos x

25  5  97. sin 2 α =   =  13  169

)

= ( cos2 x )(1) = cos2 x

106. sin

)(

104. cos4 x − sin 4 x = cos2 x − sin 2 x cos2 x + sin 2 x

5 β 4 α

563

Multiple-Angle and Product-to-Sum Formulas

sin y 2

= 1 − 2sin y + 2cos2 y 108.

cos3β cos ( 2 β + β ) = cos β cos β cos2 β cos β − sin 2 β sin β = cos β

(1 − 2sin β ) cos β − 2sin β cos β sin β 2

(

)

cos β

= 1 − 2sin β − 2sin 2 β 2

= 1 − 4sin β 2

1 sin 2θ 1 = 2 sin θ cosθ 1 1 = ⋅ sin θ 2 cosθ csc θ = 2 cosθ

101. csc 2θ =

102. tan

u 1 − cos u 1 cos u = = − = csc u − cot u 2 sin u sin u sin u

564

Chapter 6

109. sec 2θ =

Analytic Trigonometry

Case 2:

1 1 = 2 cos2θ cos θ − sin 2 θ

1 cos θ

 x + y  x − y 2 cos  sin  2   2  sin x − sin y = cos x + cosy  x + y  x − y 2 cos  cos   2   2 

(

1 − sin 2 θ cos2 θ sec θ 1 − tan 2 θ sec 2 θ

)

= =

(

x−y sin    2  = x−y cos    2 

)

1 − sec 2 θ − 1 sec θ 2 − sec 2 θ 2

110. sec

x−y = tan    2 

u 1 = 2 cos ( u 2 )

sin x + sin y 114. = cos x − cos y

=±

2 1 + cos u

=±

2sin u sin u (1 + cos u )

=±

2sin u sin u + sin u cos u

=±

(2sin u) (cos u) (sin u) (cos u) + (sin u cos u) (cos u)

=±

2 tan u tan u + sin u

x−y cos    2  =− x−y sin    2 

= 2 x 1 − x2

= cos2 β cos β − sin 2 β sin β

(

)

= cos β − sin β cos β − 2sin β cos β sin β 2

x−y = − cot    2 

115. Let θ = arcsin x. sin(2arcsin x ) = 2sin(arcsin x )cos(arcsin x )

111. cos3β = cos ( 2 β + β ) 2

116. Let θ = arccos x. cos(2arccos x ) = cos2 (arccos x ) − sin 2 (arccos x )

= x 2 − (1 − x 2 )

= cos β − sin β cos β − 2sin 2 β cos β 3

= 2 x2 − 1

= cos β − 3sin β cos β 3

x+y x−y 2sin   cos  2  2     x+y x−y −2sin   sin    2   2 

112. sin 4β = 2sin2β cos2β

( (

)

= 2  2 sin β cos β cos β − sin β    = 2  2 sin β cos β 1 − sin 2 β − sin 2 β   

(

= 4 sin β cos β 1 − 2 sin 2 β

)

θ 1 − x2

Figure for 115 and 117

113. Case 1: sin x + sin y = cos x + cos y

 x + y  x − y 2sin  cos   2   2   x + y  x − y 2 cos  cos   2   2 

x+y sin    2  = x+y cos    2 

1 − x2

Figure for 116 and 118

117. Let θ = arcsin x. cos(2 arcsin x ) = 1 − 2 sin 2 (arcsin x ) = 1 − 2 x2

118. Let θ = arccos x. sin(2arccos x ) = 2sin(arccos x )cos(arccos x )

= 2 1 − x 2 ( x) = 2 x 1 − x 2

x+y = tan    2 

Section 6.5

Multiple-Angle and Product-to-Sum Formulas

565

122. (a) f ( x) = cos2 x + sin x = 0 1 − 2sin 2 x + sin x = 0 1 + x2

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

θ 1

sin x = −

Figure for 119 and 120

sin x = 1  x =

119. Let θ = arctan x. sin(2arctan x ) = 2sin(arctan x )cos(arctan x ) x 1 =2 ⋅ 2 1+ x 1 + x2 2x = 1 + x2

(b)

π 2

−2 sin 2 x + cos x = 0

−4sin x cos x + cos x = 0 cos x(1 − 4sin x ) = 0 x=

π 2

3π 1 1 , arcsin   , π − arcsin   2 4 4

x ≈ 1.5708, 4.7124, 0.2527, 2.8889

120. Let θ = arctan x. cos(2arctan x ) = 1 − 2sin 2 (arctan x )   x = 1 − 2  1 + x 2    2 x2 =1− 1 + x2 1 − x2 = 1 + x2

1 7π 11π , x= 2 6 6

1 2 v0 sin 2θ 32 Let r = 130 feet and v0 = 75 feet per second, and find θ . 1 130 = (75)2 sin 2θ 32 4160 = 5625sin 2θ 4160 = sin 2θ 5625  4160  sin −1   = 2θ  5625 

123. r =

121. (a) f ( x) = sin2 x − sin x = 0 2sin x cos x − sin x = 0

sin x(2 cos x − 1) = 0 sin x = 0  x = 0, π , 2π 1 π 5π cos x =  x = , 2 3 3 (b) 2 cos 2 x − cos x = 0 2(2 cos2 x − 1) − cos x = 0

47.69° ≈ 2θ 23.85° = θ

(0° < θ < 90°)

θ θ  b 2 124. (a) sin   =  b = 20sin 2  2  10

θ θ  h cos   =  h = 10 cos 2  2  10 1 1 θ  θ A = bh =  20sin  10 cos  2 2 2  2

4 cos2 x − cos x − 2 = 0

1 ± 1 + 32 1 ± 33 = 8 8  1 ± 33  x = arccos   8     1 ± 33  x = 2π − arccos   8    x ≈ 0.5678, 2.2057, 4.0775, 5.7154

cos x =

= 100sin

cos 2 2 θ θ  (b) A = 50  2sin cos  = 50 sin θ 2 2 

The area is maximum when θ =

π 2

, A = 50.

125.

x θ 1 − cosθ  θ = 2r sin 2 , x = 4r sin  = 4r 2 2 2  2

= 2r (1 − cosθ )

566

Chapter 6

Analytic Trigonometry

θ 1 = 2 M θ 1 θ π (a) sin = = 1  =  θ = π = 180° 2 1 2 2 θ 1 2 (b) sin = = 2 4.5 9 θ 2 = arcsin   ≈ 0.2241 2 9 θ ≈ 0.4482 ≈ 25.7° (c) M = 1  Speed 760 mph Speed = 4.5  Speed = 3420 mph M = 4.5  760 θ 1 − cosθ 1 (d) sin = = 2 2 M 1 − cosθ 1 = 2 2 M 2 1 − cosθ = 2 M 2 cosθ = 1 − 2 M

126. sin

127. False. If x = π , sin

x π = sin = 1, whereas 2 2

130. (a) Using the graph, sin 2u ≈ 1 and 2sin u cos u ≈ 2(0.7)(0.7) ≈ 1.

Because 1 = 1, sin 2u = 2sin u cos u. (b) Using the graph,

cos2u ≈ 0 and cos2 u − sin 2 u ≈ (0.7) − (0.7) = 0. 2

Because 0 = 0, cos 2u = cos 2 u − sin 2 u. 131. (a)

(− 1, 4) 4 3

(5, 2)

2 1 −2 −1 −1

−2 −3 −4

(5 + 1)2 + (2 − 4)2 = 40 = 2 10

(b) Distance:

 −1 + 5 4 + 2  (c) Midpoint:  ,  = (2, 3) 2   2 y

132. (a) 10

(6, 10)

1 − cos π − = −1. 2

6 4 2

128. True. sin(π ) = 0 and y = 4 − 8sin π = 4, a maximum. 2

129. (a) f ( x ) = sin 4 x + cos4 x. From Example 5 and Exercise 23, 1 1 f ( x) = (3 − 4cos 2 x + cos 4 x) + (3 + 4cos 2 x + cos 4 x) 8 8 3 1 = + cos 4 x 4 4 (b) Sample answer: f ( x ) = sin 2 x ⋅ sin 2 x + cos2 x ⋅ cos2 x 2

−8 −6 −4 −2 −2

(− 4, − 3) − 4 −6

(b) Distance:

(6 + 4)2 + (10 + 3)2 = 269

 6 − 4 10 − 3   7  (c) Midpoint:  ,  =  1,  2   2  2

 1 − cos2 x   1 + cos2 x  =  +  2 2     1 1 = [2 + 2 cos2 2 x ] = (1 + cos2 2 x ) 4 2

(c)

f ( x ) = sin 4 x + 2sin 2 x cos2 x + cos4 x − 2sin 2 x cos2 x

= (sin 2 x + cos2 x )2 − 2sin 2 x cos2 x = 1 − 2sin 2 x cos2 x 1 (d) f ( x ) = 1 − sin 2 2 x 2 (e) Answer will vary.

Chapter 6 Review y

133. (a) 3

135. (a) Complement: 90° − 55° = 35° Supplement: 180° − 55° = 125° (b) Complement: None Supplement: 180° − 162° = 18°

( 43 , 52(

(− 1, 12 ( 1 −2

−1

136. (a) Complement: None Supplement: 180° − 109° = 71° (b) Complement: 90° − 78° = 12° Supplement: 180° − 78° = 102°

−1 −2

(b) Distance: 2

4  5 1  − ( −1)  +  −  3   2 2

49 + 4 4

85 9

85 3

 ( −1) + ( 4 3) (1 2) + (5 2)   1 3  , (c) Midpoint:   =  ,  2 2   6 2 y

134. (a)

−3

−2

( ( 1

( −2

138. (a) Complement:

π 2

−

− 0.95 ≈ 0.6208

Supplement: π − 0.95 ≈ 2.1916

π   140. s = r θ  s = (15)110° ⋅  ≈ 28.7979 in. 180°  

−3

(b) Distance:

8π 4π = 18 9 π 17π Supplement: π − = 18 18 π 9π π (b) Complement: − = 2 20 20 9π 11π Supplement: π − = 20 20 2

π   139. s = r θ  s = ( 21) 35° ⋅  ≈ 12.8282 cm 180°  

−1

(

Supplement: π − 2.76 ≈ 0.3816 1 2 , 3 3

−1

− 1, − 3 2

137. (a) Complement:

(b) Complement: None

2 1

567

1  2 3  + 1 +  +  = 3  3 2

233 6

Chapter 6 Review 1 = sec x cos x 1 2. = csc x sin x 1.

1 = cos x sec x

1 = sin x csc x

1 − cos2 x = sin x

csc2 x − 1 = cot x

π  7. cot  − x  = tan x 2  

π  8. sec  − x  = csc x 2  9. sec(− x) = sec x 10. tan(− x ) = − tan x

4 3 11. sin x = , cos x = , Quadrant I 5 5 sin x 4 tan x = = cos x 3 3 cot x = 4 5 sec x = 3 5 csc x = 4

568

Chapter 6

Analytic Trigonometry 1 2 π  13. sin  − x  = cos x = , = 2 2 2 

2 13 12. tan θ = , sec θ = , Quadrant I 3 3

cosθ =

3 13 13

sin x = −

2  3 13  2 13 sin θ = tan θ cosθ =  = 3  13  13 csc θ = cot θ =

3 2

1 2

=−

2 , Quadrant IV 2

tan x = −1 cot x = −1 sec x = 2 csc x = − 2

13 2

2 2 π  , Quadrant II 14. csc − θ  = − 3, sin θ = 3 2  π  csc − θ  = secθ = − 3 2  cscθ =

1 3 3 2 = = sin θ 4 2 2

cosθ =

1 1 = − secθ 3

tan θ = − cot θ =

15.

16.

17.

sec2 θ − 1 = −

sin 2 θ + cos2 θ 1 = sin θ sin θ = csc θ

secθ

− tan θ = secθ sin θ cos θ = 1 cosθ −

sin θ cosθ = − ⋅ cos θ 1 = − sin θ

18.

sec 2 ( −θ ) csc θ 2

8 = −2 2

1 1 2 = − = − tan θ 4 2 2

1 1 = = cos2 x tan 2 x + 1 sec 2 x

tan ( −θ )

( − 3) − 1 = −

sec 2 θ csc 2 θ  1    cos2 θ  =  1   sin 2 θ    2 1    sin θ  =    2  cos θ   1 

19. tan 2 θ (csc 2 θ − 1) = tan 2 θ (cot 2 θ )  1  = tan 2 θ  2  = 1  tan θ 

20. csc 2 x(1 − cos2 x ) = csc 2 x(sin 2 x ) = 1 π  21. tan  − x  sec x = cot x sec x 2   cos x 1 1 = = = csc x . sin x cos x sin x

22.

23.

24.

sin(− x )cot x = π  sin  − x  2 

( − sin x ) 

cos x    sin x  = −1 cos

sec 2 x − 1 (sec x − 1)(sec x + 1) = = sec x + 1 sec x − 1 sec x − 1 sin 2 α − cos 2 α (sin α + cos α )(sin α − cos α ) = sin 2 α − sin α cos α sin α (sin α − cos α ) sin α + cos α = = 1 + cot α sin α

= tan 2 θ

Chapter 6 Review

25. sin −1 2 x cos x =

34. cos 4 x csc2 x = (cos2 x) csc2 x 2

cos x sin1 2 x

= (1 − sin 2 x) csc2 x 2

sin x cos x ⋅ sin x sin x

 1  = (1 − 2sin 2 x + sin 4 x) 2   sin x  1 = − 2 + sin 2 x sin 2 x

cos x = sin x = cot x sin x sin x 1 1 cos x − . sin 2 x sin x sin x 1 − cos x = sin 2 x

= csc2 x + sin 2 x − 2

26. csc 2 x − csc x cot x =

35.

27. cos x(tan 2 x + 1) = cos x sec 2 x 1 = sec 2 x = sec x sec x 2

29. sin 3 θ + sin θ cos2 θ = sin θ (sin 2 θ + cos2 θ ) = sin θ

36.

1 − cos x =

(1 − cos x )

37.

csc( − x ) csc x cos x =− =− = − cot x sec( − x ) sec x sin x

38.

1 + sec( − x ) 1 + sec x = sin(− x ) + tan(− x ) − sin x − tan x

30. cot 2 x − cos2 x =

= cos2 x cot 2 x

1 + sec x − sin x (1 + sec x ) 1 =− = − csc x sin x =

π  31. csc 2  − x  − 1 = sec 2 x − 1 = tan 2 x 2  π  32. tan  − x  sec x = cot x sec x 2   cos x 1 . = sin x cos x 1 = = csc x sin x sin 4 x cos 2 x = (sin 2 x) cos 2 x

39. 2 sin x − 1 = 0 1 sin x = 2

= (1 − cos 2 x + cos 4 x) cos 2 x = cos 2 x − 2cos 4 x + cos 6 x

+ 2 nπ 6 5π x= + 2 nπ 6 x=

= (1 − cos 2 x ) cos 2 x

1 + cos x 1 + cos x

sin x sin 2 x = 1 + cos x 1 + cos x

cos x − cos2 x sin 2 = cos2 x[csc 2 x − 1]

(1 − sin θ ) 1 − sin θ (1 − sin θ )2 = = cos2 θ cosθ cosθ

Note: We can drop the absolute value on 1 − sin θ since it is always nonnegative.

= cot x tan 2 x 1 = tan 2 x = tan x tan x

1 − sin θ 1 − sin θ 1 − sin θ (1 − sin θ )2 = . = 1 + sin θ 1 + sin θ 1 − sin θ 1 − sin 2 θ

28. sec x cot x − cot x = cot x(sec x − 1)

33.

569

40. tan x + 1 = 0 tan x = −1 3π x= + nπ 4 41.

sin x = 3 − sin x 2sin x = 3 sin x =

3 2

+ 2 nπ 3 2π x= + 2 nπ 3 x=

570

Chapter 6

Analytic Trigonometry

42. 4 cos x = 1 + 2 cos x 2 cos x = 1

2cos 2 x − cos x = 0 cos x( 2cos x − 1) = 0

1 2

cos x = x=

47.

cos x = 0 or

x =

+ 2 nπ

3 5π x= + 2 nπ 3

2cos x − 1 = 0

+ nπ

cos x =

1 3

cot x =

3 3

x =

44.

π 3

cos x =

48.

sin x(sin x + 1) = 0 sin x = 0

49.

1 2

sin x = ±

sin x − tan x = 0 sin x sin x − =0 cos x sin x cos x − sin x = 0

sin x = 0

or cos x − 1 = 0

x = nπ

50.

cos x = 1 x = 2 nπ

csc x − 2 cot x = 0

1 (1 − 2 cos x ) = 0 sin x 1 cos x = 2

46. 4 tan 2 x − 1 = tan 2 x 3tan 2 x = 1 1 tan 2 x = 3 1 tan x = ± 3

+ 2 nπ 3 5π + 2 nπ x= 3 x=

3 2

+ nπ 3 2π x= + nπ 3 x=

3π + 2nπ 2

sin x (cos x − 1) = 0

45. 3csc 2 x = 4 4 csc 2 x = 3 3 sin 2 x = 4

or sin x = −1

x = nπ or x =

+ 2 nπ 3 5π x= + 2 nπ 3 x=

sin 2 x + sin x = 0

+ nπ

1 sec x − 1 = 0 2 sec x = 2

+ 2nπ 3 5π x = + 2nπ 3 x =

43. 3 3cot x = 3 cot x =

1 2

51.

2cos2 x − cos x − 1 = 0 (2cos x + 1)(cos x − 1) = 0 2 cos x + 1 = 0 or cos x − 1 = 0 1 cos x = − cos x = 1 2 2π 4π x= x=0 , 3 3

+ nπ 6 5π x= + nπ 6 x=

Chapter 6 Review 52.

53.

59. 2sin 2 x + 1 = 0

2sin2 x − 3sin x + 1 = 0 (2sin x − 1)(sin x − 1) = 0 1 sin x = or sin x = 1 2 π 5π π or x = x= , 6 6 2 cos2 x + sin x = 1 1 − sin 2 + sin x = 1 sin x(sin x − 1) = 0 sin x = 0

sin 2 x = −

7π 11π + 2nπ or 2 x = + 2nπ 6 6 7π 11π + nπ + nπ x = x = 12 12

60. 2cos 4 x + 3 = 0

− 3 2 5π 7π 4x = + 2 nπ or 4 x = + 2 nπ 6 6 5π nπ 7π nπ or x = x= + + 24 2 24 2

cos 4 x =

sin x = 1 x=

π 2

sin 2 x + 2cos x = 2 (1 − cos2 x ) + 2 cos x − 2 = 0 cos2 x − 2 cos x + 1 = 0

61.

(cos x − 1)2 = 0 cos x = 1 x=0

2sin 2 3 x − 1 = 0 1 sin 2 3 x = 2 2 2 π nπ 3x = + 4 2 π nπ x= + 12 6

sin 3 x = ±

55. 2sin 2 x = 2

sin 2 x =

2 2

3π 9π 11π , , 4 4 4 π 3π 9π 11π , , x= , 8 8 8 8

2x =

56.

62. 4cos2 2 x − 3 = 0 3 cos2 2 x = 4

3 tan3 x = 0 tan 3 x = 0 3 x = kπ x = 0, x =1 2 1 x sin = 2 2 π x = 2 6

cos2 x = ±

2x =

π 3

2π 4π 5π , π, , 3 3 3

57. 2sin

x =

58. cos

π 3

x = 0 3 x π = 3 2 3π x = 2

1 2

2x =

x = 0, π 54.

571

π 6

π 12

3 2

5π + nπ 6 5π nπ x= + 12 2

+ nπ or 2 x = +

nπ or 2

63. tan 2 x − 2 tan x = 0 tan x( tan x − 2) = 0

x 5π or = 2 6 5π x = 3

tan x = 0 or tan x − 2 x = nπ

= 0

tan x = 2 x = arctan 2 + nπ

64.

3cos2 x + 5cos x = 0 cos x(3cos x + 5) = 0 cos x = 0  x = cos x = −

π 2

3π 2

5 (impossible) 3

572 65.

Chapter 6

Analytic Trigonometry

3cos 2 x − sin x − 1 = 0 3(1 − sin x) − sin x − 1 = 0 2

sin

69.

2+ 6  1   2   2  3  =  −   −  +    = 2 4  2   2   2   

3 − 3sin 2 x − sin x − 1 = 0 3sin 2 x + sin x − 2 = 0

(3sin x − 2)(sin x + 1) = 0

cos

3sin x − 2 = 0

2 sin x = or sin x = −1 3 3π x = + 2nπ 2 2 x = arcsin   + 2nπ 3  2 or x = − arcsin  + ( 2n + 1)π  3 66.

sec 2 x + 6 tan x + 4 = 0 (1 + tan 2 x ) + 6 tan x + 4 = 0 tan 2 x + 6 tan x + 5 = 0 (tan x + 1)(tan x + 5) = 0 tan x = −1

or tan x = −5 x = arctan(−5) + π ≈ 1.7682

3π 7π x= , 4 4

x = arctan( −5) + 2π ≈ 4.9098

67. sin 285° = sin(315° − 30°) = sin 315° cos30° − cos315° sin 30°

 2  3   2   1  6+ 2 = − − =−  2   2   2   2  4      cos285° = cos(315° − 30°) = cos315° cos30° + sin 315° sin 30°  2  3   2  1  6− 2 = + − =  2   2   2   2  4      tan 285° = −

6+ 2 6− 2

= −2 − 3

68. sin 345° = sin(300° + 45°) = sin 300° cos45° + cos300° sin 45°

3 2 1 2 2− 6 + . = 2 2 2 2 4 cos345° = cos(300° + 45°) = cos300° cos 45° − sin 300° sin 45°

31π 11π 3π 3π 11π  11π 3π  = sin  +  = sin cos + sin cos 12 4 6 4 4 6  6

31π 11π 3π 11π 3π  11π 3π  = cos  + − sin cos sin  = cos 12 6 4 6 4 6 4    3  2   1  2  2− 6 =    − 2  −  − 2   2  = 2 4     

tan

31π sin(31π / 12) 2+ 6 = = = −2 − 3 12 cos(31π / 12) 2− 2

π  17π   7π 70. sin  −   = sin  3  12   4 7π π 7π π = sin cos − cos sin 4 3 4 3  2  1   2  3  =  −   −      2  2   2  2 

= cos

tan 345° =

sin 345° = cos345°

π 7π 7π π cos + sin sin 4 3 4 3

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

2 6 − = 4 4

2 − 4

π  17π   7π −  tan   = tan  3  12   4 7π π tan − tan 4 3 = 7π π 1 + tan tan 4 3 =

( −1) − ( 3 ) 1 + ( −1)( 3 )

−1 − 3 1− 3

=−

1 2 3 2 2+ 6 = . + = 2 2 2 2 4 2− 6 = 3 −2 2+ 6

2 + 4

2 6 − = − 4 4 π  17π   7π −  cos  = cos 3  12   4 = −

= −

1+ 1−

= −

1+ 2 3 +3 −2

3 1+ ⋅ 3 1+

3 3

4+2 3 2

= 2+

71. sin 45° cos 60° − cos 45° sin 60° = sin ( 45° − 60°) = sin ( −15°)

Chapter 6 Review 72. cos 45° cos120° − sin 45° sin120° = cos(45° + 120°) = cos(165°) 73.

tan 25° + tan 50° = tan(25° + 50°) = tan 75° 1 − tan 25° tan 50°

74.

tan 63° − tan112° = tan(63° − 112°) = tan( −49°) 1 + tan63° tan112°

7 4 and cos v = − , and u is in 5 25 Quadrant I and v is in Quadrant III.

For Exercises 75–80, sin u =

79. cos ( u + v ) = cos u cos v − sin u sin v

 3  7   4  24  =   −  −   −   5  25   5  25  21 96 =− + 125 125 75 3 = = 125 5 80. cos(u − v) = cos u cos v + sin u sin v

 3  7   4  24  =   −  +   −   5  25   5  25  21 96 117 =− − =− 125 125 125

y −7

− 24

75. sin(u + v) = sin u cos v + cos u sin v

 4  7   3  24  =   −  +   −   5  25   5  25  28 72 100 4 =− − =− =− 125 125 125 5 tan u + tan v 1 − tan u tan v  4   24   +  3 7 =      4  24  1 −     3  7  100 4 = 21 = − −25 3 7

76. tan(u + v) =

tan u − tan v 77. tan(u − v) = 1 + tan u tan v  4   24   −  3 7 =      4  24  1 +     3  7  44 − 44 = 21 = − 39 117 7

573

π π π  81. cos  x +  = cos x cos − sin x sin 2 2 2   = (cos x )(0) − (sin x )(1) = − sin x 3π  3π 3π  82. sin  x − = sin x cos − sin cos x  2  2 2  = (sin x)(0) − (−1)(cos x ) = cos x

83. cos(5π − x ) = cos 5π cos x − sin 5π sin x = ( −1)cos x + (0)sin x = − cos x

84. sin(π − x) = sin π cos x − sin x cos π = (0)(cos x ) − (sin x )( −1)

= sin x 85. cos3 x = cos(2 x + x) = cos2 x cos x − sin 2 x sin x = (cos2 x − sin 2 x )cos x − 2sin x cos x sin x = cos3 x − 3sin 2 x cos x = cos3 x − 3cos x(1 − cos2 x ) = cos3 x − 3cos x + 3cos3 x = 4 cos3 x − 3cos x

86.

sin(α + β ) sin α cos β + cosα sin β = cos α cos β cos α cos β sin α cos β cosα sin β = + cosα cos β cosα cos β = tan α + tan β

78. sin(u − v) = sin u cos v − cos u sin v

 4  7   3  24  =   −  −  −  −   5  25   5  25  28 72 44 =− + = 125 125 125

574

Chapter 6

Analytic Trigonometry 5 π 89. sin u = , 0 < u < 7 2

π π   87. sin  x +  − sin  x −  = 2 2 2   2 cos x sin

π 2

= 2 (Sum-to-Product)

cos x =

2 2

π 4

7π 4

π π   88. cos  x +  − cos  x −  = 1 4 4    −2sin x sin

π 2

5 u

2 6

sin 2u = 2sin u cos u

= 1 (Sum-to-Product)

sin x = −

2 2

5π 7π , x= 4 4

 5  2 6  = 2    7   7  20 6 49 cos2u = cos2 u − sin 2 u =

 2 6   5 2 = −  7   7    24 25 1 = − =− 49 49 49 2 tan u tan 2u = 1 − tan 2 u  5  2  2 6 =  2  5  1−   2 6  5 6 = 25 1− 24 5 = 6 1 − 24 5  24  = −  6 1  120 =− 6 =

−120 6 6

= −20 6

Chapter 6 Review

90.

4 3π < u < 2π cos u = , 5 2 −3 −3 sin u = , tan u = 5 4 24  −3  4  sin 2u = 2sin u cos u = 2    = − 5 5 25    16 9 7 cos2u = cos2 u − sin 2 u = − = 25 25 25 sin 2u 24 =− tan 2u = cos2u 7

575

2 π 91. tan u = − , < u < π 9 2 y

−9

4 85 85 + 1 =  sec u = − 81 81 9 −9 85 2 85 cos u = , sin u = (tan u )(cos u ) = 85 85 sin 2u = 2sin u cos u sec 2 u = tan 2 u + 1 =

−3

 2 85  −9 85  36 = 2    = −  85  85  85   4  77 cos 2u = 1 − 2sin 2 u = 1 − 2   =  85  85 sin 2u 36 =− tan 2u = cos 2u 77 92.

cos u = −

2 3π ,π < u < 2 5

sin 2 u = 1 − cos 2 u = 1 −

4 1 1 5 =  sin u = − = − 5 5 5 5

4  1  2  sin 2u = 2sin u cos u = 2 −  −  = 5 5  5  4 1 3 cos 2u = cos 2 u − sin 2 u = − = 5 5 5 sin 2u 4 tan 2u = = cos 2u 3 93. 6sin x cos x = 3[2sin x cos x ] = 3sin x 2 x

−2

97.

= 4sin x cos x (2 cos2 x − 1) = 8cos3 x sin x − 4 cos x sin x

r = 321 v0 2 sin 2θ

sin 2θ = 0.5 2θ = 30° or 2θ = 180° − 30° = 150° θ = 15° θ = 75°

95. 1 − 4sin 2 x cos2 x = 1 − (2sin x cos x )2

96. sin 4 x = 2 sin 2 x cos 2 x = 2[2sin x cos x (cos2 x − sin 2 x )]

100 = 321 (80)2 sin 2θ

94. 4sin x cos x + 2 = 2(2sin x cos x ) + 2 = 2 sin 2 x + 2

= 1 − sin 2 2 x = cos2 2 x

98.

r = 321 v0 2 sin 2θ

77 = 321 (50)2 sin 2θ sin 2θ = 0.9856 2θ = 80.2649° or 99.7351° θ = 40.13° or 49.87° 99. tan 2 4 x =

1 − cos8 x 1 + cos8 x

576

Chapter 6

Analytic Trigonometry 3

 1 − cos2 x  1 2 3 100. sin 6 x =   = 8 (1 − 3cos2 x + 3cos 2 x − cos 2 x ) 2   1  1 + cos 4 x   1 + cos 4 x   = 1 − 3cos2 x + 3   − cos2 x   8 2 2     1 3 3 1 1  =  1 − 3cos2 x + + cos 4 x − cos2 x − cos2 x cos 4 x  8 2 2 2 2  1 1  =  5 − 7cos2 x + 3cos 4 x − [cos2 x + cos6 x ]  16  2  1 = (10 − 15cos2 x + 6 cos 4 x − cos6 x ) 32 2

 1 + cos 4 x  101. cos4 2 x =   2   1 = (1 + 2 cos 4 x + cos2 4 x ) 4 1 1 + cos8 x  =  1 + 2 cos 4 x +  4 2  1 = (2 + 4cos 4 x + 1 + cos8 x ) 8 1 = (3 + 4cos 4 x + cos8 x ) 8 2

 1 − cos 4 x  102. sin 4 2 x =   2   1 = (1 − 2cos 4 x + cos 2 4 x) 4 1 1 + cos8 x  = 1 − 2cos 4 x +  4 2  1 3 1  =  − 2cos 4 x + cos8 x  8 2 2  1 = ( 3 + cos8 x − 4cos 4 x ) 8 2

 1 + cos2 x   1 − cos2 x  103. cos4 x sin 4 x =     2 2     2 (1 + 2 cos2 x + cos 2 x )(1 − 2 cos2 x + cos2 2 x ) = 16 1 + cos 4 x  1 + cos 4 x    1 + 2 cos2 x +  1 − 2 cos2 x +  2 2    = 16 (3 + 4 cos2 x + cos 4 x )(3 − 4 cos2 x + cos 4 x ) = 64 1 ( 3 + cos 4 x ) + 4 cos2 x  ( 3 + cos 4 x ) − 4 cos2 x  = 64  1 = [(3 + cos 4 x )2 − 16 cos2 2 x ] 64 1 = [9 + 6 cos 4 x + cos2 4 x − 16 cos2 2 x ] 64 1  1 + cos8 x 1 + cos 4 x  9 + 6 cos 4 x + = − 16 .  64  2 2  1 3 1  + cos8 x − 2 cos 4 x  64  2 2  1 = (cos8 x − 4 cos 4 x + 3) 128 =

Chapter 6 Review  1 − cos 4 x  1 − cos 4 x  104. sin 2 2 x tan 2 2 x =   1 + cos 4 x  2    1 − 2 cos 4 x + cos2 4 x = 2(1 + cos 4 x )

 1 + cos8 x  1 − 2 cos 4 x +   2   = 2(1 + cos 4 x ) 3 − 4 cos 4 x + cos8 x = 4(1 + cos 4 x ) 1 − cos30°  30  ° 105. sin15° = sin   = 2  2 

1−

3 2

157°30′ = 157.5° 1  sin (157.5°) = sin  ⋅ 315°  2   1 − cos315° 2

1−

3 2

2+ 3 2 sin 30°  30  ° tan15° = tan   =  2  1 + cos30° 1 2 = 3 1+ 2 1 2 = 2+ 3 2 1 = 2+ 3 =

2 2 2

2−

2 2

1  cos(157.5°) = cos ⋅ 315°  2  1 + cos315° 2

= −

2− 3 = 2 ° 30 1 cos30° +   cos15° = cos   = 2  2  =

106.

577

= −

2 2 2

= −

2 2

1  tan (157.5°) = tan  ⋅ 315°  2  sin 315° = 1 + cos315° = =

− 2 2 1+ 2 2 −

(2 +

2 2

)

2 2

− 2 2− ⋅ 2+ 2 2−

−2 2 + 2 4−2

−2 2 + 2 2

=1−

2 2

=2− 3

578

Chapter 6

Analytic Trigonometry

1 − cos(7π / 4)  7π   1 7π  107. sin   = sin  ⋅ = 2  8  2 4  =

1−

109. sin u =

( 2 / 2) = 2 − 2 2

1 + cos(7π / 4)  7π   1 7π  cos   = cos  ⋅ =− 2  8  2 4  1+

=−

( 2 / 2) = − 2 + 2

(

u sin   =  2

− 2 = =1− 2 2+ 2 2/2

)

1 − cos(11π / 16)  11π   1 11π  108. sin   = sin  ⋅ = 2  12  2 6  1−

u cos  =  2

( 3 / 2) = 2 − 3 2

1 + cos(11π / 6)  11π   1 11π  cos   = cos  ⋅ =− 2  12  2 6  =−

1 − cos u 2 1 − (5 13) 2

8 13 = 2

4 = 13

2 2 13 = 13 13

18 13 = 2

9 = 26

3 3 13 = 13 13

1 + cos u 2 1 + (5 13) 2

8 1 − (5 13) 2  u  1 − cos u 13 tan   = = = = 12 sin u 35 3  2 13

( 3 / 2) = − 2 + 3

2 2 sin(11π / 6)  11π   1 11π  tan   = tan  ⋅ =  12   2 6  1 + cos(11π / 6) −1 / 2 −1 = = = −2 + 3 2+ 3 1+ 3 / 2

(

2 2 sin(7π / 4)  7π   1 7π  tan   = tan  ⋅ =  8   2 4  1 + cos(7π / 4) − 2/2

12 π 5  cos u = ,0 < u < 13 2 13

)

3 3π 4 110. cos u = − , π < u <  sin u = − 5 2 5 y −3 −4

u x

u sin   =  2 = u cos  =  2

1 − cos u 2 1 − ( − 3 5) = 2

8 5 = 2

4 = 5

2 2 5 = 5 5

1 + cos u 2

2 1 5 5 = − = − = − 2 5 5 8 1 − ( − 3 5)  u  1 − cos u tan   = = = 5 = −2 4 sin u ( − 4 5)  2 − 5 1 + ( − 3 5) = − 2

Chapter 6 Review

45 π < u < π , 2 2

111. tan u = −

45 3 5 2 = and cos u = − 7 7 7

 sin u =

u −2

9 7 = 2

1 − ( − 2 7) = 2

3 5

1 − cos u 2

u sin   =  2

579

9 = 14

3 3 14 = 14 14

1 + cos u 2

u cos  =  2

5 7 = 2

1 + ( − 2 7) = 2

5 70 = 14 14 9 1 − ( − 2 7) 45 3 5  u  1 − cos u = = 7 = = tan   = sin u 5 3 5 21 5  2 3 5 5 5 =

(

112.

sec u = 6,

)

3π 1 35 < u < 2π  cos u = and sin u = − 2 6 6

u sin   =  2 = u cos  =  2 = −

− 35

1 − cos u 2 1 − (1 6) 2

5 6 = 2

5 5 = = 12 2 3

15 6

1 + cos u 2 1 + (1 6)

= −

7 6 = − 2

1 − ( 2 6)  u  1 − cos u tan   = = = sin u  2 − 35 6

(

)

7 7 21 = − = − 12 6 2 3 5 6

35 − 6

5 −

= −

5 35 35 = − 35 7

580

Chapter 6

Analytic Trigonometry

113. −

1 + cos8 x = − cos 4 x 2

115.

sin10 x = tan 5 x 1 + cos10 x

114.

1 − cos6 x  6x  = sin   = sin3 x 2  2 

116.

1 − cos12 x  12 x  = tan   = tan 6 x sin12 x  2 

117. Volume V of the trough will be area A of the isosceles triangle times the length l of the trough. 1 V = A ⋅ l, A = bh 2 θ θ h  h = 0.5cos cos = 2 0.5 2 θ b/2 b θ  = 0.5sin sin = 2 0.5 2 2

A = 0.5sin 0.5cos = (0.5)2 sin cos = 0.25sin cos m 2 2 2 2 2 2 2

V = (0.25)(4)sin cos = sin cos m 3 2 2 2 2

4m b 0.5 m θ

h 0.5 m

not drawn to scale

118. Volume V of the trough will be area A of the isosceles triangle times the length l of the trough. 1 V = A ⋅ l , A = bh 2 h θ θ cos =  h = 0.5cos 2 0.5 2 θ b/2 b θ sin =  = 0.5sin 2 0.5 2 2

A = 0.5sin 0.5cos = (0.5)2 sin cos = 0.25sin cos m 2 2 2 2 2 2 2

V = (0.25)(4)sin cos = sin cos m3 2 2 2 2 θ θ 1 θ θ 1 V = sin cos =  2sin cos  = sin θ m3 2 2 2 2 2 2 Volume is maximum when θ = π /2.

4m b 0.5 m θ

h 0.5 m

not drawn to scale

Chapter 6 Review

119. 6sin

π 4

cos

π 4

1  π π   π π  = 6  sin  +  + sin  +    4 4  2  4 4   π  = 3  sin + sin 0  = 3 2   1 [cos(15° − 45°) − cos(15° + 45°)] 2 = 2  cos ( −30° ) − cos ( 60° ) 

128. y =

581

1 1  10 sin  8t − arctan  2 3 

2π

120. 4sin15° sin 45° = 4 ⋅

= 2 [ cos(30°) − cos(60°)]

1 [cos(5α − 4α ) − cos(5α + 4α )] 2 1 = ( cos α − cos9α ) 2

121. sin 5α sin 4α =

1 122. cos6θ sin 8θ = sin(6θ + 8θ ) − sin ( 6θ − 8θ )  2 1 = [ sin14θ + sin(2θ )] 2 1 = ( sin14θ − sin 2θ ) 2  9θ  θ  123. cos5θ + cos 4θ = 2 cos   cos   2   2  5θ   θ  124. sin 3θ + sin 2θ = 2 cos   cos    2  2

π π π   125. sin  x +  − sin  x −  = 2 cos x sin x = 2 cos x 4 4 4  

π π   126. cos  x +  − cos  x −  6 6   π π   x+ 6 +x− 6  = −2sin   2     π π  x+ 6 −x+ 6  π sin   = −2 sin x sin 2 6     127. y = 1.5sin8t − 0.5cos8t 3 1  1/2  a = , b = − , B = 8, C = arctan  −  2 2  3/2  2

 3 1  1  y =   +   sin  8t + arctan  −   2 2  3   y=

 1  1  10 sin  8t + arctan  −   2  3  

−2

129. The amplitude is

130. Frequency =

131. False, if

π 2

10 . 2

1 4 = period π

<θ <π 

π 4

θ 2

π 2

θ  which is in Quadrant I  cos   > 0. 2

132. False. sin( x + y) = sin x cos y + cos x sin y 133. True. 4sin(− x )cos(− x ) = 4( − sin x )(cos x ) = −4sin x cos x = −2(2 sin x cos x ) = −2sin 2 x 1  134. 4sin 45° cos15° = 4  [sin(45° + 15°) + sin(45° − 15°)]  2  = 2[sin 60° + sin 30°]

 3 1  3 +1 = 2 +  = 2  = 1 + 3  2 2    2  135. Answers will vary. See page 350. 136. No. cosθ = ± 1 − sin 2 θ 137. No. A trigonometric equation with an infinite number of solutions does not have to be an identity. For example, the 1 π general solution of sin x = is x = + 2 nπ and 6 2 5π x= + 2 nπ , which produces an infinite number 6 of solutions. 138. No. The equation a sin x − b = 0 has no solution when b b a < b . If a < b , then sin x = and > 1. But the a a

range of sin x is [ −1, 1], therefore sin x  1.

582

Chapter 6

Analytic Trigonometry

π  139. y1 = sec 2  − x  = csc 2 x 2 

140. y1 =

y2 = cot 2 t

cos3 x cos x

y2 = (2sin x )2

csc 2 x = cot 2 x + 1 Let y3 = y2 + 1 = cot 2 x + 1 = y1.

From the graphs, y3 = − y2 + 1 = y1.

Chapter 6 Test 1.

3 tan θ = , cos θ < 0  θ in Quadrant III 2 9 13 13 + 1 =  sec θ = − 4 4 2 2 2 13 =− cosθ = − 13 13

6. Conjecture: y1 = y2 Algebraically,

sec 2 θ = tan 2 θ + 1 =

y1 = sin x + cos x cot x = sin x +

3  2 13  3 13 sin θ = tan θ cosθ =  −  = −  2  13  13 csc θ = − cot θ =

13 3 13

2 3 1 . 2 sin β = 1 sin 2 β

sec 4 x − tan 4 x [(sec2 x) + (tan2 x)][sec2 x − tan2 x] = sec2 x + tan2 x sec2 x + tan2 x = sec2 x − tan 2 x = 1

cosθ sin θ cos2 θ + sin 2 θ 1 + = = = csc θ secθ sin θ cosθ sin θ cosθ sin θ cosθ

5. Since tan 2 θ = sec 2 θ − 1 for all θ , then tan θ = − sec 2 θ − 1 in Quadrants II and IV.

Thus,

π 2

sin 2 x + cos2 x sin x

1 = csc x. sin x 4

−2π

2. csc 2 β (1 − cos2 β ) =

13 3

=−

< θ ≤ π and

cos2 x sin x

2π

−4

7. sin θ .sec θ = sin θ = sin θ

1 = tan θ cosθ

8. sec 2 x tan 2 x + sec 2 x = sec 2 x(tan 2 x + 1) = sec 4 x 1 1 cosα + sin α + csc α + sec α sin α cosα . 9. = = sin α cosα sin α + cosα sin α + cosα (sin α + cosα )

1 cos2 α + sin 2 α = sin α cosα sin α cosα cosα sin α = + = cot α + tan α sin α cos α =

3π < θ < 2π . 2

3π  π  10. sin  x −  + sin  + x  2   2  π 3π 3π   π   =  sin x cos − cos x sin  +  sin cos x − cos sin x  2 2   2 2   = sin x ⋅ 0 − cos x( −1) + 1 ⋅ cos x − 0 ⋅ sin x = cos x + cos x = 2cos x

11. sin ( nπ + θ ) = sin nπ cosθ + cos nπ sin θ

= 0 + ( −1)n sin θ

12. (sin x + cos x )2 = sin 2 x + cos2 x + 2sin x cos x = 1 + sin 2 x

= ( −1) sin θ n

Chapter 6 Test

13. cos 255° = cos( 225° + 30°)

21.

= cos225°cos30° − sin225°sin30°

−

14. sin 4 x tan 2 x =

6 + 4

csc x − 2 = 0  csc x = 2  sin x =

22. 5sin x − x = 0

sin x cos2 x

Let y1 = 5sin x − x and graph the equation. Use the zero or root features to approximate the x-intercepts.

1 (10 − 15cos2 x + 6 cos 4 x − cos6 x ) . 32 (1 + cos2 x ) / 2

1  10 − 15cos2 x + 6 cos 4 x − cos6 x    16  1 + cos2 x 

− 2π

 3θ + θ   3θ − θ  17. cos3θ − cos θ = − 2sin  sin    2   2  = − 2sin 2θ sin θ

tan 2 x + tan x = 0 tan x(tan x + 1) = 0 tan x = 0  x = 0, π 3π 7π , 4 4

sin 2α − cos α = 0 2sin α cosα − cosα = 0 cos α (2sin α − 1) = 0

cosα = 0  α =

π 2

−4

x ≈ − 2.596, x = 0, x ≈ 2.596 23.

= 2[sin 6θ − sin(−2θ )] = 2(sin 6θ + sin 2θ )

19.

2π

x ≈ − 2.596

1 16. 4 cos2θ sin 4θ = 4  [sin(2θ + 4θ ) − sin(2θ − 4θ )] 4

tan x + 1 = 0  tan x = −1  x =

x ≈ 2.596

x=0

sin 4θ 4θ = tan = tan 2θ 15. 1 + cos 4θ 2

18.

csc2 x − csc x − 2 = 0 (csc x − 2)(csc x + 1) = 0 1 π 5π x= , 2 6 6 3π csc x + 1 = 0  csc x = −1  sin x = −1  x = 2

 2  3   2  1  =  −   −  −     2  2   2  2  =

583

3π 2

1 π 5π 2sin α − 1 = 0  sin α =  α = , 2 6 6

1 4 . = 5 5 5 2 tan u 2(2) 4 tan 2u = = =− 2 2 1 − tan u 1 − 2 3

sin 2u = 2sin u cos u = 2

3  2  cos2u = 1 − 2sin 2 u = 1 − 2   =− 5  5 24.

sin[(θ / 2) + (α / 2)] sin(θ / 2)

3 sin (θ 2 ) + 30° = 2 sin (θ 2 ) 3sin

3sin

2 2

θ  θ  = 2 sin cos30° + cos sin 30° 2  2  = 3 sin

θ 2

+ cos

θ 2

(3 − 3 ) sin 2 = cos 2 tan

θ 2

1 3− 3

 1  = arctan   ≈ 38.26° 2 3− 3  θ ≈ 76.52°

20. 4cos2 x − 3 = 0 3 cos2 x = 4 cos x = ±

π 6

3 2

5π 7π 11π , , 6 6 6

C H A P T E R 7 Additional Topics in Trigonometry Section 7.1

Law of Sines .......................................................................................586

Section 7.2

Law of Cosines ...................................................................................592

Section 7.3

Vectors in the Plane ............................................................................600

Section 7.4

Vectors and Dot Products...................................................................615

Section 7.5

Trigonometric Form of a Complex Number .....................................622

Chapter 7 Review .......................................................................................................646 Chapter 7 Test ............................................................................................................660 Chapters 5–7 Cumulative Test ................................................................................663

C H A P T E R 7 Additional Topics in Trigonometry Section 7.1 Law of Sines 1. oblique 2.

b sin B

1 1 1 bc sin A; ab sin C ; ac sin B 2 2 2

11. Given: A = 80° 15′, B = 25° 30′, b = 2.8 km C = 180° − 80° 15′ − 25° 30′ = 74° 15′

b 2.8 (sin A) = (sin80° 15′) ≈ 6.41km sin B sin 25° 30′ b 2.8 c= (sin C ) = (sin 74° 15′) ≈ 6.26 km sin B sin 25° 30′ a=

4. Two angles and one side determine a unique triangle, that is AAS and ASA. 5. The two cases AAS (two angles and one side) or ASA (angle side angle) and SSA (two sides and an angle opposite) can be solved using the Law of Sines. 6. Yes, the longest side of an oblique triangle is always opposite the largest angle of the triangle. 7. Given: A = 25°, B = 60°, a = 12 in.

C = 180° − 25° − 60° = 95° a 12 (sin B) = (sin60°) ≈ 24.59 in. sin A sin 25° a 12 c= (sin C ) = (sin 95°) ≈ 28.29 in. sin A sin 25°

12. Given: A = 88° 35′, B = 22° 45′, b = 50.2 yd C = 180° − 88° 35′ − 22° 45′ = 68° 40′

b 50.2 (sin A) = (sin88° 35′) ≈ 129.77 yd sin B sin 22° 45′ b 50.2 c= (sin C ) = (sin 68° 40′) ≈ 120.92 yd sin B sin 22° 45′

13. Given: A = 36°, a = 8, b = 5 b sin A 5 sin(36°) = ≈ 0.3674  B ≈ 21.6° 8 a C = 180° − A − B ≈ 180° − 36° − 21.6° = 122.4°

sin B =

8. Given: A = 35°, B = 55°, a = 18 mm

C = 180° − 35° − 55° = 90° a 18 (sin B) = (sin 55°) ≈ 25.71 mm sin A sin 35° a 18 c= (sin C ) = (sin 90°) ≈ 31.38 mm sin A sin 35°

9. Given: A = 45°, B = 15°, c = 20 cm C = 180° − 45° − 15° = 120° a =

c 20 (sin A) = (sin 45°) ≈ 16.33 cm sin C sin 120°

b =

c 20 (sin B) = (sin 15°) ≈ 5.98 cm sin C sin 120°

10. Given: A = 20°, B = 30°, c = 30 ft C = 180° − 20° − 30° = 130°

586

a =

c 30 (sin A) = (sin 20°) ≈ 13.39 ft sin C sin 130°

b =

c 30 (sin B) = (sin 30°) ≈ 19.58 ft sin C sin 130°

8 a (sin C ) = sin(122.4°) ≈ 11.49 sin A sin(36°)

14. Given: A = 76°, a = 34, b = 21 b sin A 21 sin 76° = ≈ 0.5993  B ≈ 36.8° a 34 C = 180° − 76° − 36.8° ≈ 67.2°

sin B =

a 34 sin C = sin 67.2° ≈ 32.30 sin A sin 76°

15. Given: A = 35°, B = 40°, c = 10 C = 180° − 35° − 40° = 105° a =

10 ⋅ sin 35° c ≈ 5.94 (sin A) = sin C sin 105°

b =

10 ⋅ sin 40° c ≈ 6.65 (sin B ) = sin C sin 105°

16. Given: A = 120°, B = 45°, c = 16 C = 180° − 120° − 45° = 15° a =

16 ⋅ sin 120° c ≈ 53.54 (sin A) = sin C sin 15°

b =

16 ⋅ sin 45° c ≈ 43.71 (sin B) = sin C sin 15°

Section 7.1 17. Given: A = 110°, a = 125, b = 100 b sin A 100 sin110° = ≈ 0.75175  B ≈ 48.74° sin B = a 125 C = 180° − A − B ≈ 21.26° 125 sin 21.26° a (sin C ) = c= ≈ 48.23 sin A sin110° 18. Given: A = 145°, a = 14, b = 4 b ⋅ sin A 4 ⋅ sin 145° = sin B = a 14 ≈ 0.16388  B ≈ 9.43° C = 180° − A − B ≈ 180° − 145° − 9.43° = 25.57° c =

a 14 (sin C ) = (sin 25.57°) ≈ 10.53 sin A sin 145°

19. Given: A = 102.4°, C = 16.7°, a = 21.6 B = 180° − A − C = 180° − 102.4° − 16.7° = 60.9°

a 21.6 (sin B) = (sin 60.9°) ≈ 19.32 sin A sin102.4° a 21.6 c= (sin C ) = (sin16.7°) ≈ 6.36 sin A sin102.4°

20. Given: A = 24.3°, C = 54.6°, c = 2.68

c 2.68 sin 24.3° ≈ 1.35 (sin A) = sin C sin 54.6° c 2.68 sin101.1° ≈ 3.23 b= (sin B) = sin C sin 54.6°

5 8

A = 180° − B − C = 180° − 28° − 104° = 48° 29 a b= (sin B) = 8 sin 28° ≈ 2.29 sin A sin 48° 29 a c= (sin C ) = 8 sin104° ≈ 4.73 sin A sin 48°

23. Given: A = 110°15′, a = 48, b = 16 b sin A 16 sin110° 15′ = ≈ 0.31273  B ≈ 18° 13′ a 48 C = 180° − A − B ≈ 180° − 110° 15′ − 18° 13′ = 51° 32′ sin B =

C = 180° − 55° − 42° = 83°

b =

3 sin 55° c ≈ 0.62 (sin A) = 4 sin C sin 83° 3 sin 42° 4

48 a (sin C ) = (sin 51° 32′) ≈ 40.06 sin A sin110° 15′

24. Given: B = 2°45′, b = 6.2, c = 5.8 c sin B 5.8 sin2° 45′ sin C = = ≈ 0.04488  C ≈ 2.57°(2° 34′) 6.2 b A = 180° − B − C ≈ 174.68° (174° 41′) 6.2 sin174° 41′ b (sin A) ≈ a= ≈ 11.97 sin B sin 2° 45′ 25. Given: A = 76°, a = 18, b = 20 b sin A 20 sin 76° = ≈ 1.078 sin B = a 18

No solution 26. Given: A = 110°, a = 125, b = 200 A obtuse and a < b  No solution 27. Given: A = 120°, a = 25, b = 25

28. Given: A = 60°, a = 9, c = 10

c ⋅ sin A 10 ⋅ sin 60° = ≈ 0.9623 a 9 C ≈ 74.21° or 105.79°

sin C =

Case 1

C ≈ 74.21° B ≈ 180° − 60° − 74.21° = 45.79° b =

c ≈ 0.51 (sin B) = sin C sin 83°

a 9 ⋅ sin 45.79° ≈ 7.45 (sin B) = sin A sin 60° C 74.21°

7.45

3 22. Given: A = 55°, B = 42°, c = 4

a =

587

A is obtuse and a = b  No solution

B = 180° − A − C = 101.1°

21. Given: B = 28°, C = 104°, a = 3

Law of Sines

60°

45.79°

Case 2 C ≈ 105.79° B ≈ 180° − 60° − 105.79° = 14.21° b =

a 9 ⋅ sin 14.21° ≈ 2.55 (sin B) = sin A sin 60° C

2.55 105.79° 60° A

10 9

14.21°

Chapter 7

588

Additional Topics in Trigonometry

29. Given: A = 58°, a = 11.4, b = 12.8 b sin A 12.8 sin 58° = ≈ 0.9522  B ≈ 72.21° or 107.79° sin B = a 11.4 Case 1

32. Given: A = 60° and a = 10 10 sin 60°

(a) One solution: If b ≤ a = 10 or if b = (right triangle)

B ≈ 72.21° C = 180° − 58° − 72.21° = 49.79° c=

a 11.4 (sin C ) = (sin 49.79°) ≈ 10.27 sin A sin 58°

a = 10

a = 10 = h

or 60°

12.8

60°

11.4

58°

(b) Two solutions: If a = 10 < b <

72.21°

10 sin 60°

Case 2 B ≈ 107.79° C ≈ 180° − 58° − 107.79° = 14.21°

a 11.4 c= (sin C ) = (sin14.21°) ≈ 3.30 sin A sin 58° C A

12.8

60°

11.4

58°

107.79°

10 sin 60°

30. Given: A = 58°, a = 4.5, b = 12.8

a = 10

a < h = b sin 58° 4.5 < 10.86 A

No solution

33. Given: A = 10° and a = 10.8

31. Given: A = 36° and a = 5

(a) One solution: If b ≤ a = 5 or if b =

5 sin 36°

(a) One solution: If b ≤ a = 10.8 or b =

a=5 or

36°

(b) Two solutions: If a = 5 < b <

36°

5 sin 36°

a=5=h

5 a=

10°

5 sin 36°

b A

10.8 sin10°

a = 10.8

b A

a = 10.8

10°

a = 10.8

(b) Two solutions: If a = 10.8 < b <

a=5

36°

a = 10.8

10°

A b

10.8 sin10°

(right triangle)

(right triangle) b

60°

10.8 sin10° a = 10.8

10°

a=5

36°

Section 7.1 34. Given: A = 88° and a = 315.6 315.6 (a) One solution: If b ≤ a = 315.6 or if b = sin 88° (right triangle)

Law of Sines

589

39. Area = 12 ac sin B = 12 (103)(58) sin 75° 15′ ≈ 2888.6 square units

40. Area = 12 ab sin C = 12 (16)(20) sin85° 45′ ≈ 159.6 square units

a = 315.6

b h

a = 315.6 = h

41. C = 180° − 94° − 30° = 56°

h =

88° A

(b) Two solutions: If a = 315.6 < b <

a = 315.6

40 (sin 30°) ≈ 24.12 meters sin 56°

88°

315.6 sin88°

42. Given: A = 74° − 28° = 46°, B = 180° − 41° − 74° = 65°, c = 100

C = 180° − 46° − 65° = 69° c 100 a= (sin A) = (sin 46°) ≈ 77 meters sin C sin 69° A

100

46°

65°

88° 69°

A a = 315.6

315.6 (c) No solution: If b > sin 88°

a sin B 500 sin(46°) = ≈ 0.4995 720 b A ≈ 29.97°

43. sin A =

∠ACD = 90° − 29.97° ≈ 60° a = 315.6

b h

Bearing: S 60° W or (240° in plane navigation) C 500 44° 46°

720

88° A

35. Area = 12 ab sin C = 12 (6)(10) sin(110°) ≈ 28.2 square units

36. Area = 12 ac sin B

44. Angle CAB = 70° Angle B = 20° + 14° = 34°

(a) 20°

= (92)(30) sin(130°) 1 2

≈ 1057.1 square units

70°

34° 16

37. Area = 12 bc sin A = 12 (8)(10) sin (150°) = 20 square units

38. Area = 12 ab sin C = 12 ( 4)(6) sin (120°)

14°

(b)

16 h = sin 70° sin 34°

(c)

16 sin 34° ≈ 9.52 meters sin 70°

≈ 10.39 square units

590

Chapter 7

Additional Topics in Trigonometry

45. (a)

48. d A

(b)

B 2 mi

Not drawn to scale

Third angle in triangle = α

1 A = 20°, B = 90° + 63° = 153°, c = 10   = 2.5 4 C = 180° − 20° − 153° = 7° c 2.5 sin153° b= (sin B) = ≈ 9.31 sin C sin 7° d ≈ b sin A ≈ 9.31 sin 20° ≈ 3.2 miles A

θ + α + (180° − φ ) = 180°  α = φ − θ d 2 = sin θ sin α 2 sin θ 2 sin θ d= = sin α sin(φ − θ )

b C

49. (a)

(b) 3000 ft

5.45 ≈ 0.0934 58.36 α ≈ 5.36°

sin α =

d 58.36 d sin θ =  sin β = sin β sin θ 58.36  d sin θ    58.36 

β = sin −1 

5.45

3000 sin 1 2 (180° − 40° ) 

(b)

(c)

 π  s ≈ 40°   4385.71 ≈ 3061.80 feet  180° 

sin 40°

≈ 4385.71 feet

47. ∠ACD = 65° ∠ADC = 180° − 65° − 15° = 100°

∠CDB = 180° − 100° = 80° ∠B = 180° − 80° − 70° = 30° b 30 a= (sin A) = (sin15°) ≈ 15.53 km sin B sin30° b 30 c= (sin C ) = (sin135°) ≈ 42.43 km sin B sin 30° (Colt Station) C 80°

65°

65° 70° D

A (Pine Knob)

15°

B a

46. (a)

40°

2.5

58.36 α

58.36  58.36  d = sin β   = sin(84.64° − θ ) θ sin sin θ   (d)

10°

20°

30°

40°

50°

60°

324.1

154.2

95.2

63.8

43.3

28.1

a B (Fire)

Section 7.1

Law of Sines

591

sin α sin β = 9 18 sin α = 0.5 sin β α = arcsin(0.5 sin β )

50. (a)

(b) 0

−1

Domain: 0 < β < π Range: 0 < α ≤ π 6

γ = π − α − β = π − β − arcsin(0.5 sin β )

(c)

c 18 = sin γ sin β 18 sin γ c= sin β 18 sin[π − β − arcsin(0.5 sin β )] = sin β (d)

Domain: 0 < β < π Range: 9 < c < 27 (e)

0.4

0.8

1.2

1.6

2.0

2.4

2.8

0.1960

0.3669

0.4848

0.5234

0.4720

0.3445

0.1683

25.95

23.07

19.19

15.33

12.29

10.31

9.27

As β → 0, c → 27. As β → π , c → 9. 51. False. If just the three angles are known, the triangle cannot be solved. 52. True. No angle could be 90°. 53. False. The cases that give two angles and a side do have unique solutions, they are AAS and ASA.

54. Yes, the Law of Sines can be used to solve a right triangle if you are given at least one side and one angle, or two sides. Answers will vary. 55. Answers will vary. A = 36°, a = 5 (a) b = 4 one solution (b) b = 7 two solutions [h = b sin A < a < b] (c) b = 10 no solution [a < h = b sin A]

592

Chapter 7

Additional Topics in Trigonometry

56. Distance from (0, 0) to (4, 3) : 2

(4 − 0) + (3 − 0) = 5 A is acute. (a) (b) (c)

a ≥ 5, a = 3 3<a<5 a<3

57. tan θ =

sin θ − 12 13 12 = =− cosθ 5 13 5

1 13 = cos θ 5 1 5 cot θ = =− tan θ 12 1 13 cscθ = =− sin θ 12

1 59. 6 sin8θ cos3θ = 6   [sin(8θ + 3θ ) + sin(8θ − 3θ )] 2 = 3(sin11θ + sin 5θ )

1 60. 2 cos2θ cos5θ = 2   [cos(2θ − 5θ ) + cos(2θ + 5θ )] 2 = cos3θ + cos7θ 61.

1 π 5π 1 1   π 5π  5π  π =   sin  + cos sin  − sin  −  3 6 3 3 2   6 3  3  6

secθ =

58. cot θ =

1 15 = − tan θ 8

sin θ =

1 8 = csc θ 17

62.

1   11π   3π  sin   − sin  −  6   6   2 

1   11π   3π  sin   + sin   6   6   2 

5 3π 5π 5 1   3π 5π   3π 5π   sin sin = ⋅ cos  − +  − cos   2 4 6 2 2  4 6  6   4

cos θ = cot θ ⋅ sin θ

5  π   19π   cos  −  − cos   4   12   12  

5  π   19π   cos   − cos   4   12   12  

15  15  8  =  −   = − 17  8  17  1 17 sec θ = = − cos θ 15

Section 7.2 Law of Cosines 1.

c 2 = a 2 + b 2 − 2 ab cos C

1 − bh, 2

No. AAS, two angles and a side opposite, would use the Law of Sines.

Given: a = 12, b = 16, c = 18

s ( s − a )( s − b)( s − c )

No. ASA, two angles and the included side, would use the Law of Sines.

Yes. SAS, two sides and the included angle, would use the Law of Cosines.

Yes. SSS, three sides, would use the Law of Cosines.

b2 + c2 − a2 162 + 182 − 122 = 2bc 2(16)(18) ≈ 0.75694  A ≈ 40.80° b sin A ≈ 0.8712  B ≈ 60.61° sin B = a C ≈ 180° − 60.61° − 40.80° = 78.59°

cos A =

8. Given: a = 8, b = 18, c = 12 b 2 + c 2 − a 2 182 + 12 2 − 82 = ≈ 0.9352  A ≈ 20.74° 2bc 2(18)(12) c sin A ≈ 0.5312  C ≈ 32.09° sin C = a B ≈ 180° − 32.09° − 20.74° = 127.17°

cos A =

9. Given: a = 8.5, b = 9.2, c = 10.8 b 2 + c 2 − a 2 9.2 2 + 10.82 − 8.52 = ≈ 0.6493  A ≈ 49.51° 2bc 2(9.2)(10.8) b sin A 9.2 sin 49.51° ≈ ≈ 0.82315  B ≈ 55.40° sin B = a 8.5 C ≈ 180° − 55.40° − 49.51° = 75.09°

cos A =

Section 7.2

Law of Cosines

593

10. Given: a = 4.2, b = 5.4, c = 2.1 a 2 + b 2 − c 2 4.2 2 + 5.4 2 − 2.12 = ≈ 0.9345  C ≈ 20.85° 2 ab 2(4.2)(5.4) a sin C 4.2sin 20.8° ≈ ≈ 0.7102  A ≈ 45.38° sin A = c 2.1 B ≈ 180° − 20.85° − 45.38° = 113.77°

cos C =

11. Given: a = 10, c = 15, B = 20° b 2 = a 2 + c 2 − 2 ac cos B = 100 + 225 − 2(10)(15)cos20° ≈ 43.0922  b ≈ 6.56 mm cos A =

b 2 + c 2 − a 2 43.0922 + 225 − 100 ≈ 2bc 2(6.56)(15) ≈ 0.8541  A ≈ 31.40°

C ≈ 180° − 20° − 31.40° = 128.60°

12. Given: a = 10.4, c = 12.5, B = 50° 30′ = 50.5°

b2 = a2 + c2 − 2ac cos B = 10.42 + 12.52 − 2(10.4)(12.5)cos50.5° ≈ 99.0297  b ≈ 9.95 ft b2 + c2 − a2 99.0297 + 12.52 − 10.42 ≈ 2bc 2(9.95)(12.5) ≈ 0.5914  A ≈ 53.75° = 53° 45′ C ≈ 180° − 50.5° − 53.75° = 75.75° = 75° 45′ cos A =

13. Given: a = 11, b = 15, c = 21 cos A =

225 + 441 − 121 b2 + c2 − a 2 = ≈ 0.8651  A ≈ 30.11° 2bc 2(15)( 21)

sin B =

15 sin 30.11° b sin A ≈ ≈ 0.6841  B ≈ 43.16° 11 a

C ≈ 180° − 30.11° − 43.16° = 106.73°

14. Given: a = 9, b = 3, c = 11

b2 + c 2 − a 2 32 + 112 − 92 = ≈ 0.7424  A ≈ 42.1° cos A = 2bc 2(3)(11) a 2 + b2 − c 2 92 + 32 − 112 = ≈ −0.574  C ≈ 125.0° cos C = 2 ab 2(9)(3) B = 180° − A − C ≈ 12.9° 15. Given: A = 50°, b = 15, c = 30 a 2 = b 2 + c 2 − 2bc cos A = 225 + 900 − 2(15)(30)cos50° ≈ 546.49  a ≈ 23.38 cos B =

a 2 + c 2 − b 2 546.49 + 900 − 225 ≈ 2 ac 2(23.4)(30) ≈ 0.8708  B ≈ 29.4°

C = 180° − A − B ≈ 180° − 50° − 29.5° = 100.6°

16. Given: C = 108°, a = 10, b = 7 c 2 = a 2 + b 2 − 2 ab cos C = 10 2 + 72 − 2(10)(7)cos108° ≈ 192.2624  c ≈ 13.9 sin C sin108° sin B = b= (7) ≈ 0.4789  B ≈ 28.7° c 13.9 A = 180° − 108° − 28.7° = 43.3°

17. Given: A = 120°, b = 6, c = 7 a 2 = b 2 + c 2 − 2bc cos A = 36 + 49 − 2(6)(7) cos 120° = 127  a ≈ 11.27

cos B =

127 + 49 − 36 a 2 + c2 − b2 ≈ 2ac 2(11.27)(7) ≈ 0.8873  B ≈ 27.46°

C = 180° − A − B ≈ 180° − 120° − 27.46° = 32.54°

594

Chapter 7

Additional Topics in Trigonometry

18. Given A = 48°, b = 3, c = 14

a 2 = b 2 + c 2 − 2bc cos A

c 2 = a 2 + b 2 − 2ab cos C

= 9 + 196 − 2(3)(14) cos 48°

≈ 148.793  a ≈ 12.20

cos B =

a 2 + c2 − b2 148.84 + 196 − 9 ≈ 2ac 2(12.2)(14)

≈ 0.9831  B ≈ 10.54°

19. Given: a = 75.4, b = 48, c = 48

sin B =

 4 7  4  7  =   +   − 2   cos 43° 9 9  9  9  ≈ 0.2968  c ≈ 0.54 2

b 2 + c 2 − a 2 482 + 482 − 75.42 = 2bc 2(48)(48) ≈ −0.2338  A ≈ 103.5°

b sin A 48 sin(103.5°) ≈ 75.4 a ≈ 0.6190  B ≈ 38.2°

cos A =

3 3 ,b = 8 4

c 2 = a 2 + b 2 − 2ab cos C 2

 3  3  3  3  =   +   − 2   cos 101° 8 8      8  4  ≈ 0.8105  c ≈ 0.90 2

20. Given: a = 1.42, b = 0.75, c = 1.25 2

24. Given: C = 101°, a =

C = B ≈ 38.2° (Because of roundoff error, A + B + C ≠ 180°.)

 4 7 2   + 0.54 −   a +c −b 9  9 ≈ cos B =  4 2ac 2 (0.54) 9 ≈ − 0.2413  B ≈ 103.96° C = 180° − A − B ≈ 180° − 43° − 103.96° = 33.04° 2

C = 180° − A − B ≈ 180° − 48° − 10.54° = 121.46°

cos A =

4 7 ,b = 9 9

23. Given: C = 43°, a =

b + c − a (0.75) + (1.25) − (1.42) = 2bc 2(0.75)(1.25) = 0.05792  A ≈ 86.7°

a2 + c2 − b2 (1.42)2 + (1.25)2 − (0.75)2 = 2ac 2(1.42)(1.25) ≈ 0.8497  B ≈ 31.8° C = 180° − 86.7°− 31.8° ≈ 61.5° cos B =

21. Given: B = 8°15′ = 8.25°, a = 26, c = 18 b 2 = a 2 + c 2 − 2ac cos B = 262 + 182 − 2(26)(18) cos(8.25°) ≈ 73.6863  b ≈ 8.58 c sin B 18 sin(8.25°) sin C = ≈ 8.58 b ≈ 0.3  C ≈ 17.51° ≈ 17° 31′ A = 180° − B − C ≈ 180° − 8.25° − 17.51° = 154.24° ≈ 154° 14′

22. Given: B = 10° 35′ ≈ 10.583°, a = 40, c = 30

a +c −b 2ac

cos B =

 3  3 2   + 0.90 −   8  4 ≈    3 2 (0.90) 8 ≈ 0.575  B ≈ 54.90°

C = 180° − A − B ≈ 180° − 101° − 54.90° = 24.1°

25. d 2 = 42 + 82 − 2(4)(8)cos30° ≈ 24.57  d ≈ 4.96 2φ = 360° − 2θ  φ = 150° c 2 = 42 + 82 − 2(4)(8)cos150° ≈ 135.43 c ≈ 11.64 8 c d

30°

26. c 2 = 252 + 352 − 2(25)(35) cos120° = 2725  c ≈ 52.2 2θ = 360° − 2(120°) = 120°  θ = 60° d 2 = 252 + 352 − 2(25)(35)cos60° = 975  d ≈ 31.22 35

b2 = a 2 + c 2 − 2 ac cos B ≈ 140.8268  b ≈ 11.87 a sin B sin A = ≈ 0.6189  A ≈ 141.75° ≈ 141° 45′ b C = 180° − A − B = 27.67° or 27° 40′

120°

25 d

Section 7.2 102 + 14 2 − 20 2 2(10)(14) φ ≈ 111.8° 2θ = 360° − 2(111.80°)

595

252 + 17.52 − 252 2(25)(17.5) α ≈ 69.513° β ≈ 180° − α ≈ 110.487°

30. cosα =

27. cos φ =

a 2 = 17.52 + 252 − 2(17.5)(25) cos110.487°

θ = 68.2° 2

Law of Cosines

a ≈ 35.18

d = 10 + 14 − 2(10)(14) cos68.2° d ≈ 13.86

z = 180° − 2α ≈ 40.974

252 + 35.182 − 17.52 2(25)(35.18) μ ≈ 27.772° θ = μ + z ≈ 68.7°

cos μ =

ϕ 10

ω = 180° − μ − β ≈ 41.741° φ = ω + α ≈ 111.3°

40 2 + 602 − 802 1 ≈ −  θ ≈ 104.5° 2(40)(60) 4 2φ ≈ 360° − 2(104.5°) = 151°  φ = 75.5°

28. cosθ =

17.5

c2 ≈ 402 + 602 − 2(40)(60) cos75.5° = 4000 c ≈ 63.25

a 25

60 c

17.5

μ z

31.

20 m A 40° 15 m

29. cos α =

15 + 12.5 − 10 = 0.75  α ≈ 41.41° 2(15)(12.5)

Given: b = 15 m, c = 20 m, A = 40° Given two sides and included angle, use the Law of Cosines.

152 + 102 − 12.52 cos β = = 0.5625  β ≈ 55.77° 2(15)(10) δ = 180° − 41.41° − 55.77° ≈ 82.82°

a 2 = b 2 + c 2 − 2bc cos A = 225 + 400 − 2(15)( 20) cos 40°

μ = 180° − δ ≈ 97.18° b2 = 12.52 + 10 2 − 2(12.5)(10) cos(97.18°) ≈ 287.50 b ≈ 16.96 10 sin ω = sin μ ≈ 0.585  ω ≈ 35.8° 16.96 12.5 sin ∈ = sin μ ≈ 0.731  ∈ ≈ 47° 16.99 θ = α + ω ≈ 77.2°

φ = β + ∈ ≈ 102.8° β

10 15

12.5 α

12.5

μ μ

b sin A 15 sin 40° ≈ a 12.86 ≈ 0.7498  B ≈ 48.57° C ≈ 180° − 48.57° − 40° ≈ 91.43°

sin B =

32.

11 cm C

≈ 165.3733  a ≈ 12.86 m

97° 43°

Given: A = 11 cm, B = 97°, C = 43° Given two angles and a side, use the Law of Sines. A = 180° − 97° − 43° = 40° c =

a sin C 11 sin 43° = ≈ 11.67 cm sin A sin 40°

b =

a sin B 11 sin 97° = ≈ 16.99 cm sin A sin 40°

596

Chapter 7

Additional Topics in Trigonometry

33. Given: a = 8, c = 5, B = 40° Given two sides and included angle, use the Law of Cosines. b2 = a2 + c2 − 2ac cos B = 64 + 25 − 2(8)(5) cos40° ≈ 27.7164  b ≈ 5.26 b2 + c2 − a2 (5.26)2 + 25 − 64 ≈ 2bc 2(5.26)(5) ≈ −0.2154  A ≈ 102.44° C ≈ 180° − 102.44° − 40° ≈ 37.56°

39. Given: a = 12, b = 24, c = 18 a+b+c s= = 27 2

Area = s(s − a)(s − b)(s − c) = 27(15)(3)(9)

cos A =

34. Given: a = 10, b = 12, C = 70° Given two sides and included angle, use the Law of Cosines.

≈ 104.57 square inches 40. Given: a = 25, b = 35, c = 32 a+b+c s= = 46 2

Area = s(s − a)(s − b)(s − c) = 46(21)(11)(14)

c2 = a2 + b2 − 2ab cos C = 100 + 144 − 2(10)(12) cos70°

≈ 385.70 square meters

≈ 161.9152  c ≈ 12.72 bsin C 12sin70° ≈ ≈ 0.8865  B ≈ 62.44° c 12.72 A ≈ 180° − 62.44° − 70° ≈ 47.56°

sin B =

35. Given: A = 24°, a = 4, b = 18 Given two sides and an angle opposite one of them, use the Law of Sines. h = b sin A = 18sin 24° ≈ 7.32 Because a < h, no triangle is formed. 36. Given: a = 11, b = 13, c = 7 Given three sides, use the Law of Cosines. a 2 + c 2 − b 2 121 + 49 − 169 = ≈ 0.0065  B ≈ 89.63° cos B = 2 ac 2(11)(7) a sin B 11sin89.63° ≈ ≈ 0.8461  A ≈ 57.79° sin A = b 13 C ≈ 180° − 57.79° − 89.63° ≈ 32.58° 37. Given: A = 42°, B = 35°, c = 1.2 Given two angles and a side, use the Law of Sines. C = 180° − 42° − 35° = 103°

c sin A 1.2sin 42° = ≈ 0.82 sin C sin103° c sin B 1.2sin 35° = ≈ 0.71 b= sin C sin103° a=

41. Given: a = 5, b = 8, c = 10 a + b + c 23 s= = = 11.5 2 2

Area = s(s − a)(s − b)(s − c) = 11.5(6.5)(3.5)(1.5) ≈ 19.81 square units 42. Given: a = 12, b = 17, c = 8 s =

Area = =

A = 180° − B − C ≈ 180° − 49.21° − 95° = 35.79° a =

25 sin 35.79° c sin A = ≈ 14.68 sin C sin 95°

 37  13  3  31        2  2  2  2 

30,303 30,303 = 16 4 ≈ 43.52 square units 43. Given: a = 1.24, b = 2.45, c = 1.25 s =

a + b + c = 2.47 2

Area =

s( s − a)( s − b)( s − c)

2.47(1.23)(0.02)(1.22)

Given two sides and an angle opposite one of them, use the Law of Sines. b sin C 19 sin 95° = 25 c ≈ 0.7571  B ≈ 49.21°

s( s − a)( s − b)( s − c)

38. Given: C = 95°, b = 19, c = 25

sin B =

a + b + c 12 + 17 + 8 37 = = 2 2 2

≈ 0.27 square units 44. Given: a = 2.4, b = 2.75, c = 2.25 s =

a + b + c = 3.7 2

Area = =

s( s − a)( s − b)( s − c) 3.7(1.3)(0.95)(1.45)

≈ 2.57 square units

Section 7.2

45. Given: a = 1, b = s =

1 3 ,c = 2 4

216 miles

135 3 15 = 4096 64 ≈ 0.18 square unit 3 5 3 ,b = ,c = 5 8 8

a + b + c 4 = 2 5

s( s − a )( s − b)( s − c)

Area = =

 4  1  7  17        5  5  40  40 

476 40,000

≈ 0.11 square unit 47. Angle at B = 180° − 80° = 100°

C 165 miles 17.2° B

72.8° 59.7°

 9  1  5  3        8  8  8  8 

46. Given: a =

E S

s =

s( s − a )( s − b)( s − c)

597

49.

a + b + c 9 = 2 8

Area =

Law of Cosines

368 miles 13.1°

a = 165, b = 216, c = 368 1652 + 3682 − 2162 ≈ 0.9551 2(165)(368) B ≈ 17.2° 2162 + 3682 − 1652 cos A = ≈ 0.9741 2(216)(368) A ≈ 13.1°

cos B =

(a) Bearing of Minneapolis (C) from Phoenix (A) N (90° − 17.2° − 13.1°) E N 59.7° E (b) Bearing of Albany (B) from Phoenix (A) N (90° − 17.2°) E N 72.8° E 50. C = 180° − 53° − 67° = 60° c 2 = a 2 + b2 − 2 ab cos C

= 36 2 + 482 − 2(36)(48)(05) = 1872 c ≈ 43.3 mi

b2 = 2402 + 3802 − 2(240)(380)cos100° ≈ 233,673.4  b ≈ 483.4 meters

80°

36 mi

380

240 b

2 2 + 32 − (4.5)2 48. cosθ = ≈ −0.60417 2(2)(3) θ ≈ 127.2°

60°

53° E

67° 48 mi S

51. The largest angle is across from the largest side. 6502 + 5752 − 7252 cos C = 2(650)(575) c ≈ 72.3° C

575

650

725

598

Chapter 7

Additional Topics in Trigonometry

52.

RS = 82 + 10 2 = 164 = 2 41 ≈ 12.8 feet

1 1 16 2 + 10 2 = 356 = 89 ≈ 9.4 feet 2 2 10 tan P = 16 5 P = arctan ≈ 32.0° 8 PQ =

QS = 82 + 9.42 − 2(8)(9.4) cos32° ≈ 24.81 ≈ 5.0 feet 53. The angles at the base of the tower are 96° and 84°. The longer guy wire g1 is given by:

g12 = 752 + 1002 − 2(75)(100) cos96° ≈ 17,192.9  g1 ≈ 131.1 feet The shorter guy wire g2 is given by: g2 2 = 752 + 100 2 − 2(75)(100) cos84° ≈ 14,057.1  g2 ≈ 118.6 feet

54.

A = 180° − 40° − 20° = 120° (sin 20°) x= (10) sin120° ≈ 3.95 feet C

57. (a)

72 = 1.52 + x 2 − 2(1.5)( x ) cosθ 49 = 2.25 + x 2 − 3 x cosθ

x 2 − 3 x cosθ = 46.75

(b)

 3 cosθ   3 cosθ  x 2 − 3 x cosθ +   = 46.75 +    2   2 

50° x 40°

3 cosθ  187 9 cos2 θ  x − 2  = 4 + 4  

A 10

3 cosθ 187 + 9 cos2 θ =± 2 4 Choosing the positive values of x, we have 1 x = 3 cosθ + 9 cos2 θ + 187 . 2

20°

x−

70° B

55. s =

a + b + c 140 + 150 + 160 = = 225 2 2

Area = =

)

(

(c)

s( s − a)( s − b)( s − c) 225(85)(75)(65)

≈ 9655.79 square units 56. The height is h = 70 sin 70° ≈ 65.778. Area = base × height = (100)(65.778) ≈ 6577.8 square meters

2π

(d) Note that x = 8.5 when θ = 0 and θ = 2π , and x = 5.5 when θ = π . Thus, the distance is 2(8.5 − 5.5) = 2(3) = 6 inches.

Section 7.2 58. (a)

d2 = 102 + 72 − 2(10)(7) cosθ  d = 149 − 140 cosθ

 10 2 + 72 − d 2   149 − d 2  (b) θ = arccos    = arccos   2(10)(7)   140  360° − θ (360° − θ )π (c) s = (2π r ) = 360° 45° (d)

62. (a)

Law of Cosines

Because all three sides and no angles are given, use the Law of Cosines.

(b) Because two angles and a side are given, use the Law of Sines. 63. Given: a = 12, b = 30, A = 20° a 2 = b 2 + c 2 − 2bc cos A 12 2 = 30 2 + c 2 − 2(30)(c) cos20°

d (inches)

c 2 − (60 cos20°)c + 756 = 0

θ (degrees)

60.9°

69.5°

88.0°

Solving this quadratic equation, c ≈ 21.97, 34.41. For c = 21.97,

s (inches)

20.88

20.28

18.99

d (inches)

θ (degrees)

98.2°

109.6°

122.9°

139.8°

s (inches)

18.28

17.48

16.55

15.37

The other angles are determined by the Law of Sines.

61. (a)

a+b+c a+b+c , not . 2 3

a2 + c2 − b2 122 + 34.412 − 302 ≈ ≈ 0.5183  B ≈ 58.8° 2ac 2(12)(34.41) C ≈ 180° − 58.8° − 20° = 101.2°. cos B =

Using the Law of Sines, b sin A 30 sin 20° sin B = = a 12 ≈ 0.8551  B ≈ 58.8° or 121.2°. c=

1  2bc + b + c − a  bc   2  2bc  2

1 ( b + c ) 2 − a 2   4 1 = ( b + c ) + a  ( b + c ) − a  4 b+c+a b+c−a = ⋅ 2 2 a + b + c −a + b + c = ⋅ 2 2 =

(b)

For c = 34.41,

For B = 58.8°, C = 180° − 58.8° − 20° = 101.2° and

1 1  b2 + c 2 − a 2  bc(1 + cos A) = bc 1 +  2 2  2bc  2

a2 + c2 − b2 122 + 21.972 − 302 ≈ ≈ −0.5184  B ≈ 121.2° 2ac 2(12)(21.97) C ≈ 180°−121.2°− 20° = 38.8°. cos B =

59. True. The third side is found by the Law of Cosines.

60. False. s =

599

a sinC ≈ 34.42. sin A

For B = 121.2°, C = 180° − 121.2° − 20° = 38.8° and a sin C c= ≈ 21.98. sin A This gives the same result as using the Law of Cosines. An advantage of using the Law of Cosines is that it is easier to choose the correct value to avoid the ambiguous case. Its disadvantage is that there are more computations. The opposite is true for the Law of Sines.

1 1  b2 + c 2 − a 2  bc(1 − cos A) = bc 1 −  2 2  2bc  1  2bc − b2 − c 2 + a 2  = bc   2  2bc  1 2 2 = a − (b − c)   4 1 = ( a − b + c )( a + b − c )  4  a − b + c  a + b − c  =   2 2   

600

Chapter 7

Additional Topics in Trigonometry

1 + cos C C  64. (a) Since 0 < C < 180°, cos   = . 2 2 Hence,

(

1 + a 2 + b2 − c2 C  cos   = 2 2

1 − cos C C  (b) Since 0 < C < 180°, sin   = . 2 2 Hence,

) 2ab = 2ab + a + b − c . 2

4 ab

On the other hand, 1 1  s(s − c) = (a + b + c)  (a + b + c) − c  2 2  1 1 (a + b + c) (a + b − c) 2 2 1 = (( a + b)2 − c 2 ) 4 1 2 = ( a + b 2 + 2 ab − c 2 ). 4

1 − (a 2 + b 2 − c 2 ) (2 ab) C  sin   = = 2 2

On the other hand, 1 1  (s − a)(s − b) =  (a + b + c) − a   (a + b + c) − b  2 2    1 1 (b + c − a ) ( a + c − b ) 2 2 1 = [c − ( a − b)][ c + (a − b)] 4 1 2 = [ c − ( a − b)2 ] 4 1 2 = (c − a 2 − b 2 + 2 ab). 4

Thus,

s(s − c) = ab

a 2 + b 2 + 2 ab − c 2 and we have 4 ab

C  verified that cos   = 2

s(s − c) . ab

2 ab − a 2 − b 2 + c 2 . 4 ab

Thus,

(s − a)( s − b) = ab

c 2 − a 2 − b 2 + 2 ab C  = sin   . 4 ab 2

 π π 65. Because sin  −  = −1, arcsin( −1) = − . 2 2    3 3 5π  5π  66. Because cos  = − , arccos − .  = 2 6  6   2  π  67. Because tan   = 6

 3 3 π , tan −1   = . 3 6  3 

π  68. Because tan   = 3, tan −1 3

( 3 ) = π3 .

Section 7.3 Vectors in the Plane 1. directed line segment 2. initial, terminal

11. u = 6 − 2, 5 − 4 = 4, 1 = v 12. u = −3 − 0, − 4 − 4 = −3, − 8

3. magnitude

v = 0 − 3, − 5 − 3 = −3, − 8

4. vector

u=v

5. standard position 6. multiplication, addition 7. resultant 8. linear combination, horizontal, vertical 9. Two directed line segments that have the same magnitude and direction are equivalent. 10. A unit vector has a magnitude of 1.

13. Initial point: (0, 0) Terminal point: (1, 3) v = 1 − 0, 3 − 0 = 1, 3 v = (1)2 + (3)2 = 10 ≈ 3.16

14. Initial point: (0, 0) Terminal point: (4, − 2) v = 4 − 0, − 2 − 0 = 4, − 2 v = 4 2 + (−2)2 = 20 = 2 5 ≈ 4.47

Section 7.3

601

23. Initial point: ( − 23 , − 1)

15. Initial point: (2, 2) Terminal point: (−1, 4)

Terminal point: ( 12 , 45 )

v = −1 − 2, 4 − 2 = −3, 2 2

Vectors in the Plane

v = (−3) + 2 = 13 ≈ 3.61

1 2

− ( − 23 ) , 45 − ( −1) = 2

7 9 v =   +  = 6 5

16. Initial point: (−1, − 1) Terminal point: (3, 5)

7 6

, 95

4141 ≈ 2.1450 30

24. Initial point: ( 25 , − 2 )

v = 3 − (−1), 5 − (−1) = 4, 6

 2 Terminal point:  1,   5

v = 4 2 + 6 2 = 52 = 2 13 ≈ 7.21

v = 1 − 25 , 25 − ( −2 ) = − 23 , 125

17. Initial point: (3, − 2) Terminal point: (3, 3)

3 89  3   12  ≈ 2.8302 v = −  +   = 2 5 10    

v = 3 − 3, 3 − ( −2) = 0, 5 v =5

25. −v

18. Initial point: (−4, − 1) Terminal point: (3, − 1)

v = 3 − (−4), − 1 − (−1) = 7, 0

v = 72 + 0 2 = 7

19. Initial point: (−3, − 5) Terminal point: (5, 1)

−v

v = 5 − (−3), 1 − (−5) = 8, 6

26. 3u

v = (8)2 + (6)2 = 64 + 36 = 100 = 10

20. Initial point: ( − 2, 7)

Terminal point: (5, −17)

v = 5 − ( − 2), −17 − 7 = 7, − 24 v =

(7)2 + (− 24)2 =

49 + 576 =

625 = 25

21. Initial point: (0.6, 3)

27. u + v y

Terminal point: ( − 3, − 0.6)

v = − 3 − 0.6, − 0.6 − 3 = − 3.6, − 3.6 v =

(− 3.6) + (− 3.6)

25.92 ≈ 5.09

u+v

12.96 + 12.96

22. Initial point: ( − 4.5, − 2)

Terminal point: ( 2, 4.5) v = 2 − ( − 4.5), 4.5 − ( − 2) = 6.5, 6.5 v = =

(6.5)2 + (6.5)2 =

42.25 + 42.25

84.25 ≈ 9.19

602

Chapter 7

Additional Topics in Trigonometry

28. u − v

33. u + 2 v

2v x

u + 2v x

u−v

−v u

29. 2 v − u

1 v 2

34. u +

2v 2v − u 1 v 2

−u

30. v +

u + 1v

1 u 2

35. 2 v −

1 u 2 y

v + 1u 2

2v − 1 u

1 u 2

−1u 2

31. 2u y

36. 3v + 2u y

3v x

3v + 2u

2u 2u

32. − 3 v y

v x

u −3v

Section 7.3 37. u = 4, 2 , v = 8, 0

603

38. u = 5, 3 , v = −4, 0

(a)

(a) u + v = 12, 2

u + v = 5, 3 + −4, 0 = 1, 3 y

y 5

u+v

2 1

−2 −1

10 12

−4

(b)

−2

−1

u − v = 5, 3 − −4, 0 = 9, 3 y

y 5

u−v

−v

v −6

(b) u − v = − 4, 2

u+v

−2

−8 −6 −4 −2 −1

1 −2

−3

(c)

2 u − 3v = 2 5, 3 − 3 −4, 0 = 22, 6 y

2u − 3v

3 2

− 25 − 20 − 15 − 10 −5 −1

−v 2

−1

− 3v

u−v

2 x

−2

1 (d) v + 4u = 4, 0 + 16, 8 = 20, 8 2

−4

(d)

−3v 4

−1

1 v + 4u = 1 2 2

12 16 20 24

−4, 0 + 4 5, 3 = 18, 12

y 1 v + 4u 2

6 4

1 v + 4u 2

−8

2 −4 −2

Vectors in the Plane

1 v 2

12 16 20 24

1 v 2

12 16 20

604

Chapter 7

Additional Topics in Trigonometry

39. u = −6, − 8 , v = 2, 4

(a)

40. u = 0, − 5 , v = − 4, 10

u + v = −4, − 4

(a) u + v = 0, − 5 + − 4, 10 = − 4, 5

−6

2 −4

6 4

u+v

−4

−6 −5 − 4 −3 −2

−6

(b)

u+v

−8

−6

u − v = −8, − 12

(b) u − v = 0, − 5 − − 4, 10 = 4, −15

2 −10 −8 −6 −4 −2

−2 −1 −2 −4

−v u

−6

u−v

−v

−8

− 10

−10

− 12

−12

− 14

u−v

− 16

(c)

2 u − 3v = −12, − 16 − 6, 12 = −18, − 28

5 −24 −20 −16 −12 −8 −4

−8

−3v

1 v + 4u = 1 2 2

2u − 3v

− 48

2, 4 + 4 −6, − 8 = −23, − 30 y

1 v 2

(d)

1 1 v + 4u = − 4, 10 + 4 0, − 5 = − 2, −15 2 2

y 10

1 v 2

− 10

− 15 − 20

− 30

− 25

− 3v

− 40

−30

1 v + 4u 2

10 12

− 32

−25

− 25 − 20 − 15 − 10 − 5

− 24

−20

(d)

− 16

−15

2u − 3v

−4

−3 1 v + 4u 2

−1

− 10 − 15

Section 7.3 41. u = i + j, v = 2i − 3 j (a) u + v = 3i − 2 j

1 −1

−1

−2

u+v

(b)

y 1

5 4

−1

−v

−1

(c)

u−v

8 −2 4

2u −2

(d)

−2

(d)

2u − 3v

1 1 15 7 v + 4u = −i + j + 4 2i − j = i− j 2 2 2 2 y

1 1 v 2

3 2

1 v + 4u 2

−3

−3v

−6

−2

−4

1 v + 4u = 1 (2i − 3j) + (4i + 4 j) = 5i + 5 j 2 2 2

−1 −1

−10 −8 −6 −4 −2

2u − 3v = (4i − 2 j) − (−3i + 3 j) = 7i − 5 j

−3v

−v

−3

2u − 3v = (2i + 2 j) − (6i − 9 j) = −4i + 11j 2u − 3v

−2

(c)

u − v = 3i − 2 j

−1

−2

u − v = −i + 4 j

−2

2 −1

−4

−3

u+v

−1

−3

u−v

−2

(b)

605

42. u = 2i − j, v = −i + j (a) u + v = i

−2

Vectors in the Plane

1 1 v 2

−2

−1 x

−2 −3 −4

1 v + 4u 2

43. w = u + v 44. v = w − u 45. u = w − v 46. 2 v = 2( w − u ) = 2w − 2u

606

Chapter 7

47.

u = 6, 0

Additional Topics in Trigonometry

55. w = 2 j

u = 12 (2 j) = j

u =6 1 6, 0 = 1, 0 6

Unit vector: 48.

u = 0, − 2

56.

w = −3i w =3 u = 13 ( −3i) = − i

u =2

  1  57. 10 u  = 10  u     

Unit vector: 1 0, − 2 = 0, − 1 2 49.

10 − 3, 4 25 10 = − 3, 4 5 = 2 − 3, 4 =

v = −1, 1 v = 2 1 v v

Unit vector:

50.

= −

1 , 2

1 2

= −

2 , 2

2 2

  1  58. 3 u  = 10 u     

(− 2) + 22 = 2

8 = 2 2

1 2 2 , v = − v 2 2 2 2 = −

1 1 , 2 2

= −

2 2 , 2 2

v = −24, − 7 = (−24)2 + (−7)2 = 25 , − 257 Unit vector: 251 −24, − 7 = − 24 25

52.

v = − 9, 12 v =

(− 9) + 122 = 2

 −12, − 5  2 2  (−12) + (− 5)  1

3 −12, − 5 169

Unit vector:

51.

= − 6, 8

v = − 2, 3 v =

 − 3, 4  2  ( − 3) + 42  1

3 −12, − 5 13 36 15 = − ,− 13 13 =

 1    1 u  = 8 5, 6  59. 8   52 + 6 2   u      8 5, 6 = 61 =

40 61 48 61 , 61 61

40 61 48 61 i+ j 61 61

225 = 15

Unit vector: 151 − 9, 102 = − 35 , 45 53. u =

1 1 1 4 3 v= (4 i − 3 j) = (4i − 3 j) = i − j v 5 5 5 16 + 9

54. w = i − 2 j 1 1 1 (i − 2 j) = (i − 2 j) u= w= w 5 12 + (−2)2 =

5 2 5 i− j 5 5

Section 7.3

= 32 (2i − j) = 3i − 23 j

  1 = 3 4, − 4   42 + ( −4)2     1  = 3 4, − 4  4 2 

3 2

607

67. v = 32 u

 1  60. v = 3  u  u   

Vectors in the Plane

, −

= 3, − 23 y

−1

3 u 2

3 2 −3 2 , 2 2

3 2 3 2 i− j 2 2

−2

3 w 4 3 = (i + 2 j) 4 3 3 = i + j = 4 2

68. v =  1    1 61. 7  3, 4  u  = 7  32 + 4 2   u      7 = 3, 4 5 21 28 = , 5 5

21 28 j = i+ 5 5

 1  62. v = 10  u  u     1  = 10  2, − 3   4+9  = =

20 13 20 13

, −

i−

30 13

63. v = 4 − ( −3 ) , 5 − 1 = 7, 4 = 7i + 4 j

3 w 4

−1

69. v = u + 2w = (2i − j) + 2(i + 2 j)

= 4i + 3 j = 4, 3 y

u + 2w

3 2 1 3

64. 65.

3 − 0, 6 − ( −2 ) = 3, 8 = 3i + 8 j

0 − ( −6 ) , 1 − 4 = 6, − 3 = 6 i − 3 j

66. v = 2 − ( −1) , 3 − ( −5 ) = 3, 8 = 3i + 8 j

−1

20 13 30 13 = i− j 13 13

3 3 , 4 2

−1

70. v = − u + w = −(2 i − j) + ( i + 2 j) = − i + 3 j = −1, 3 y

−u + w

−u −2

−1

608

Chapter 7

Additional Topics in Trigonometry

78. v = − 7 i − 6 j, Quadrant III

71. v = 12 (3u + w ) = 12 3 ( 2i − j) + ( i + 2 j)  = 12 ( 6i − 3 j + i + 2 j) = 72 i − 12 j =

7 2

, − 12

v =

( − 7 ) + ( − 6)

tan θ =

−6 6 =  θ ≈ 220.60° −7 7

49 + 36 =

79. v = 3 cos0°, 3 sin 0°

= 3, 0

1 w 2

4 3

−1

1 4 (3u + w) 2

−1

3 u 2

−2

θ = 0° (3, 0)

−4 −3 −2 −1

−2 −3 −4

72. v = 2 u − w

80. v = cos 45°, sin 45°

= 2 ( 2 i − j) − (i + 2 j) = 2(i − 3 j)

= 2i − 6 j = 2, − 6

2 , 2

2 2 y

1 −4 − 3 −2 −1

(

−2

−3

−1

−4 −6

)

θ = 45°

2u −2w

2 2 , 2 2

−2

2(u − w)

−7

73.

81. v =

v = 5 ( cos60°i + sin 60° j) v = 5, θ = 60°

74.

v = 8 ( cos135°i + sin135° j )

v = 12i + 15 j, Quadrant I

(

7 7 3 − , 4 4

v = 122 + 152 = 369 = 3 41 tan θ =

)

y 4 3 2

15  θ ≈ 51.3° 12

−4 − 3 −2 −1 −1

76. v = 8i − 3 j, Quadrant IV

θ = 120° 1

−2 −3

v =

82 + ( − 3) =

tan θ =

−3  θ ≈ 339.44° 8

7  1  7 3  ⋅  − ,   2  2  2  2 

7 7 3 = − , 4 4

v = 8, θ = 135°

75.

7 7 cos 120°, sin 120° 2 2

64 + 9 =

−4

77. v = −2i + 5 j

v = (−2)2 + 52 = 29 tan θ = − 25 Since v lies in Quadrant II, θ ≈ 111.8°.

Section 7.3 5 5 cos 135°, sin 135° 2 2

82. v =

5 2  5 2  − ,   2  2  2  2 

= −

−

5 2 5 2 , 4 4

)

3 2

10 3 10 i+ j 5 5

10 3 10 , 5 5 y

θ = 135°

(

4 3

−4

−3

−2

−1

609

  1 85. v = 2   (i + 3 j) 2 2   3 +1  2 = (i + 3 j) 10

(

Vectors in the Plane

−1

2 1

)

θ ≈ 71.57°

−4 −3 −2 −1

−2

10 3 10 , 5 5

−2

−3

−3 −4

83. v = 3 2 cos150°, 3 2 sin150° = −

  1 86. v = 3  (3i + 4 j)  32 + 4 2    3 = (3i + 4 j) 5 9 12 9 12 = i+ j= , 5 5 5 5

3 6 3 2 , 2 2 y

(

−

3 6 3 2 , 2 2

) 2

θ = 150° −4 −3 −2 −1

−2

−3

−4

1 −4 −3 −2 −1

84. v = 4 3 cos 90°, sin90° = 0, 4 3

−4

87.

u = 5 cos0°, 5 sin 0° = 5, 0 v = 5 cos90°, 5 sin 90° = 0, 5

u + v = 5, 0 + 0, 5 = 5, 5

−4 −3 −2 −1 −1

−3

(0, 4

θ ≈ 53.13° 1

−2

( 95 , 125 )

θ = 90° 1

88.

u = 4 cos 60°, 4 sin 60° = 2, 2 3 v = 4 cos90°, 4 sin 90° = 0, 4 u + v = 2 + 0, 2 3 + 4 = 2, 2 3 + 4

89.

u = 20 cos 45°, 20 sin 45° = 10 2, 10 2 v = 50 cos 160°, 50 sin 160° ≈ − 46.98, 17.10 u + v = − 32.84, 31.24

90.

u = 35 cos25°, 35 sin 25° = 31.72, 14.79 v = 50 cos120°, 50 sin120° = −25, 25 3 u + v ≈ 6.72, 58.09

610

Chapter 7

91.

v=i+ j

Additional Topics in Trigonometry

v = 400 cos25°, sin 25°

94.

w = 2(i − j) u = v − w = −i + 3 j

u = 300 cos135°, sin135° u + v = 150.39, 381.18

v = 2

u + v = 409.77

w =2 2

 381.18   = 68.47°  150.39 

θ = arctan 

v − w = 10 2

cos α =

v + w − v−w

2 v w

2 + 8 − 10 2 2 ⋅2 2

=0 600

α = 90°

400

409.77

200

68.47°

v u

α −1

−400

−1

400

Resultant Force: u + v = ( 45 + 60 cosθ ) i + 60 sinθ j u + v = (45 + 60 cosθ )2 + (60 sinθ )2 = 90

v = 3i + j w = 2i − j u = v − w = i + 2j 2

cos α =

200

95. Force One: u = 45i Force Two: v = 60 cosθ i + 60 sinθ j

−2

92.

−200

v + w − v−w 2 v w

10 + 5 − 5 2 10 5

α = 45° 3 2

−3 −2 −1

−1

α v

u 3

2 2

96. Force One: u = 3000 i Force Two: v = 1000 cosθ i + 1000 sinθ j Resultant Force: u + v = (3000 + 1000 cosθ )i + 1000 sinθ j

2025 + 5400 cosθ + 3600 = 8100 5400 cosθ = 2475 2475 cosθ = ≈ 0.4583 5400 θ ≈ 62.7°

u + v = (3000 + 1000 cosθ )2 + (1000 sinθ )2 = 3750

9,000,000 + 6,000,000 cosθ + 1,000,000 = 14,062,500 6,000,000 cosθ = 4,062,500

−2 −3

93.

4,062,500 ≈ 0.6771 6,000,000 θ ≈ 47.4°

cosθ =

u = 400 cos25°i + 400 sin 25° j v = 300 cos70°i + 300 sin 70° j u + v ≈ 465.13i + 450.96 j 2

u + v ≈ (465.13) + (450.96) ≈ 647.85  450.96   ≈ 44.11°  465.13 

α = arctan  y

500

647.85

400

200

−100 −100

Vertical component of velocity: 70 sin 40° ≈ 45.0 ft sec 98. Vertical component of velocity: 1200 sin 6° ≈ 125.43 ft/sec

Horizontal component of velocity: 1200 cos 6° ≈ 1193.43 ft/sec

300

100

97. Horizontal component of velocity: 70 cos 40 ° ≈ 53.62 ft sec

44.11° 100 200 300 400 500

Section 7.3

99. Left cable: u = u (cos155.7°i + sin 155.7° j)

Tow line 2: v = u (cos(−θ )i + sin(−θ ) j) Resultant:

u cos155.7° + v cos 44.5° = 0

u + v = 6000i =  u cosθ + u cos(−θ ) i 

u sin155.7° + v sin 44.5° = 20,240

6000 = 2 u cosθ  u ≈ 3000 sec θ

 −0.9114 u + 0.7133 v = 0

T = u = 3000 sec θ

0.4115 u + 0.7009 v = 20,240

Domain: 0° ≤ θ < 90°

Solving this system for u and v yields: (b)

Tension of left cable: u = 15,484 pounds Tension of right cable: v = 19,786 pounds  100. Cable AC : u = 10i − 24 j The vector lies in Quadrant IV and its reference angle is  12  arctan   .  5   12  12    u = u cos  arctan  i − sin  arctan  j 5  5       Cable BC : v = −20 i − 24 j The vector lies in Quadrant III and its reference angle is 6 arctan   . 5

611

101. (a) Tow line 1: u = u (cosθ i + sin θ j)

Right cable: v = v (cos44.5°i + sin 44.5° j)

u + v = 20,240 j 

Vectors in the Plane

10°

3046.3 3192.5 3464.1 3916.2

(c)

7500

20°

30°

40°

50°

60°

4667.2 6000.0

(d) The tension increases because the component in the direction of the motion of the barge decreases. 102. (a) u = u (cos(90° − θ )i + sin(90° − θ ) j) v = v (cos(90° + θ )i + sin(90° + θ ) j)

 6 6    v = v  − cos  arctan  i − sin  arctan  j 5 5    

u + v = 100 j  u cos(90° − θ ) + v cos(90° + θ ) = 0

Resultant: u + v = −5000 j

100 = u sin ( 90° − θ ) + u sin ( 90° + θ )

 u = v , and

12  6   u cos  arctan  − v cos  arctan  = 0 5 5    12  6   − u sin  arctan  − v sin  arctan  = −5000 5  5  

Solving this system of equations yields: TAC = u ≈ 3611.1 pounds TBC = v ≈ 2169.5 pounds

= 2 u cosθ 50 = 50sec θ . cosθ Domain: 0° ≤ θ < 90°

Hence, u = T =

(b)

10°

20°

30°

40°

50°

60°

50.8

53.2

57.7

65.3

77.8

100

(c)

120

(d) The vertical component of the vectors decreases as θ increases.

612

Chapter 7

Additional Topics in Trigonometry

103. (a) u = 220i, v = 150 cos30°i + 150 sin30° j

(

)

u + v = 220 + 75 3 i + 75 j u+v = tanθ =

(b)

( 220 + 75 3 ) + 75 ≈ 357.85 newtons 2

75 220+75 3

 θ ≈ 12.1°

u + v = 220i + (150 cosθ i + 150 sinθ j) M = u + v = 2202 + 1502 (cos2 θ + sin 2 θ ) + 2(220)(150) cosθ

= 70,900 + 66,000 cosθ = 10 709 + 660 cosθ  15 sin θ  α = arctan    22 + 15 cosθ  (c)

0°

30°

60°

90° 120° 150° 180°

M 370.0 357.9 322.3 266.3 194.7 117.2 70.0

(d)

0°

12.1° 23.8° 34.3° 41.9° 39.8° 0°

500

180

(e) For increasing θ , the two vectors tend to work against each other, resulting in a decrease in the magnitude of the resultant.

(

104. (a) T = T cos ( 90° + θ ) i + sin ( 90° + θ ) j

)

u + w + T = 0  u + T cos ( 90° + θ ) = 0 u − T sin θ = 0

− 1 + T sin ( 90° + θ ) = 0 − 1 + T cosθ = 0  T = secθ , for 0° ≤ θ < 90°

and u = T sin θ = sec θ sin θ = tan θ for 0° ≤ θ < 90°. (b)

T ⏐⏐u⏐⏐ 0

Section 7.3

Vectors in the Plane

613

105. Airspeed: v = 860(cos302°i + sin302° j) Groundspeed: u = 800(cos310°i + sin310° j) w+v=u w = u − v = 800(cos310°i + sin 310° j) − 860(cos302°i + sin302° j) ≈ 58.50 i + 116.49 j w = 58.502 + 116.492 ≈ 130.35 km hr

 116.49   ≈ 63.3°  58.50 

θ = arctan 

Direction: N 26.7° E y

106. (a)

112. F1 = 10, 0 , F2 = 5 cosθ , sin θ

580 mph

(a) F1 + F2 = 10 + 5 cosθ , 5 sin θ

E S

28°

F1 + F2 = (10 + 5 cosθ )2 + (5 sinθ )2

45° 60 mph

= 100 + 100 cosθ + 25 cos2 θ + 25 sin2 θ x

= 5 4 + 4 cosθ + cos2 θ + sin2 θ = 5 4 + 4 cosθ + 1

(b)

w = 60 cos 45°, sin 45° = 30 2, 30 2

(c)

v = 580 cos118°, sin118° = −272.3, 512.1

(d)

w + v = −229.9, 554.5 w + v ≈ 600.3 mph

(e)

554.5  θ ≈ 112.5° −229.9 Direction: N 22.5° W (or 337.5° using airplane navigation) tan θ =

107. True by definition. 108. True. u =

v v

109. True. In fact, a = b = 0. 110. True. u = 1 = a 2 + b 2 111. (a) The angle between them is 0°. (b) The angle between them is 180°. (c) No. At most it can be equal to the sum when the angle between them is 0°.

= 5 5 + 4 cosθ (b)

2π

(c) Range: [5, 15] Maximum is 15 when θ = 0. Minimum is 5 when θ = π . (d) The magnitude of the resultant is never 0 because the magnitudes of F1 and F2 are not the same. 113. Answers will vary. Possible answers: (a) To add two vectors geometrically, first position them (without changing their lengths or directions) so that the initial point of the second vector v coincides with the terminal point of the first vector u. Then, the sum u + v is the vector formed by joining the initial point of the first vector u with the terminal point of the second vector v. (b) The product of a vector v and a scalar k is the vector that is k times as long as v. If k is positive, then kv has the same direction as v, and if k is negative, then kv has the opposite direction as v.

614

Chapter 7

Additional Topics in Trigonometry

114. (a) True. a and d have the same magnitude, are parallel, and are pointing in opposite directions. (b) True. c and s have the same magnitude, are parallel, and are pointing in the same direction. (c) True. By definition of vector addition. (d) False. v − w = − s (e) True. a = −d, w = −d, a + w = −d + ( −d) = −2d (f ) True. a = −d , a + d = −d + d = 0 (g) False. u − v = 2u and −2(b + t) = −2( −2u) = 4u (h) True. a = w , b = t , t − w = b − a 115. Let v = (cosθ )i + (sinθ ) j. v = cos2 θ + sin 2 θ = 1 = 1 Therefore, v is a unit vector for any value of θ .

117.

116. The following program is written for a TI-82 or TI-83 graphing calculator. The program sketches two vectors u = ai + bj and v = ci + dj in standard position, and then sketches the vector difference u − v using the parallelogram law. PROGRAM: SUBVECT :Input “ENTER A”, A :Input “ENTER B”, B :Input “ENTER C”, C :Input “ENTER D”, D :Line (0, 0, A, B) :Line (0, 0, C, D) :Pause :A−C→E :B−D → F :Line (A, B, C, D) :Line (A, B, E, F) :Line (0, 0, E, F) :Pause :ClrDraw :Stop

u = 5 − 1, 2 − 6 = 4, − 4 v = 9 − 4, 4 − 5 = 5, − 1 u − v = −1, − 3 v − u = 1, 3

118.

u = 80 − 10, 80 − 60 = 70, 20 v = −20 − (−100), 70 − 0 = 80, 70 u − v = 70 − 80, 20 − 70 = −10, − 50 v − u = 80 − 70, 70 − 20 = 10, 50

119. (a) The muzzle velocity of a bullet is a vector. Sample answer: It will have both direction and speed. (b) The price of a company’s stock is a scalar. Sample answer: It can be represented as a single value. (c) The air temperature of a room is a scalar. Sample answer: It can be represented by a single value. (d) The weight of an automobile is a scalar. Sample answer: It is represented by a single value. (e) The volume of a fish tank is a scalar. Sample answer: It is represented by a single value. (f) The current of a river is a vector. Sample answer: It will have both direction and speed.

 6x4  12 x 4 y 5 y 2 120.  −2  14 x −1 y 5 = x  7y 

(

)

123.

( 5ab )( a b )( 2a b ) = ( 5ab ) a1 21b 2

−3 0

−2

= 12 x 3 y 7 , x ≠ 0, y ≠ 0 121.



−2





−1



( 5s t )  503st  = 103st , s ≠ 0 5 −5

(

122. (18 x ) ( 4 xy ) 3 x 0

−1

)

16 x 2 y 2 (3) = x

(

)(

)

(

)(

)

( )

5 , b≠0 4a 2

124. 2.1 × 10 9 3.4 × 10 −4 = 7.14 × 105 125. 6.5 × 10− 6 3.8 × 104 = 24.7 × 10− 2 = 2.47 × 10− 1

= 48 xy2 , x ≠ 0

Section 7.4

7 csc x = −14

+ nπ 2 cos x = −1  x = π + 2 nπ

(

)

sin x = 0 or

sin x =

csc x = − 2 sin x = −

127. sin x 2 sin x + 2 = 0 − 2 2

5π + 2 nπ 4 7π x= + 2 nπ 4

x =

(10 cot θ )2 + 100

100 cot 2θ + 100

100(cot 2θ + 1)

100 csc2θ

1 2

π 3

7π 11π + 2nπ , + 2nπ 6 6

x 2 + 100 =

128. 3 sec x + 4 = 10 sec x = 2

1 2

130. Let x = 10 cot θ .

x = nπ or x =

cos x =

615

129. 7 csc x + 18 = 4

126. cos x(cos x + 1) = 0 cos x = 0  x =

Vectors and Dot Products

= 10 csc θ + 2 nπ ,

5π + 2nπ 3

131. Let x = 7 sec θ .

x 2 − 49 =

(7sec θ )2 − 49

49 sec2θ − 49

49(sec2θ − 1)

49 tan 2θ

= 7 tan θ

Section 7.4 Vectors and Dot Products 1. Yes. Let u = u1 , u2 and v = v1 , v2 . u ⋅ v = u1v1 + u2 v2 = v1u1 + v2u2 = v ⋅ u 2. The dot product of two orthogonal vectors is u ⋅ v = 0.

9. u ⋅ v = 5, 1 ⋅ 3, − 1 = 5 ( 3) + 1( −1) = 14 10. u ⋅ v = 2, 6 ⋅ − 3, 7 = ( 2)( − 3) + (6)(7) = 36 11.

u ⋅ u = 2(2) + 2(2) = 8, scalar

3. The dot product of two vectors is a scalar. 4.

u⋅v u v

12. v ⋅ w = −3, 4 ⋅ 1, − 4 = (−3)(1) + 4( −4) = −19, scalar

 u⋅v  5.  2  v  v    6.

proj F PQ

u = 2, 2

13.

u = 2, 2 , v = −3, 4 u ⋅ 2 v = 2 u ⋅ v = 4( −3) + 4(4) = 4, scalar

  PQ , F ⋅ PQ

7. u ⋅ v = 6, 3 ⋅ 2, − 4 = 6(2) + 3(−4) = 0

14. 4 u ⋅ v = 4 2, 2 ⋅ −3, 4

= 4( −6 + 8) = 8, scalar

8. u ⋅ v = −4, 1 ⋅ 5, − 4 = ( −4)(5) + 1( −4) = −24

616

Chapter 7

Additional Topics in Trigonometry

( ) 2, 2 = ( 3 ( −3 ) + ( −12 )( 4 ) ) 2, 2

22.

= −57 2, 2

23. u = −1, 0 , v = 0, 2

15. (3w ⋅ v )u = 3 1, − 4 ⋅ −3, 4

= −114, − 114 , vector

16.

( u ⋅ 2v ) w = ( 2, 2 ⋅ 2 −3, 4 ) 1, − 4 = ( 2 ( −6 ) + 2 ( 8 ) ) 1, − 4

u = 9i u = u ⋅ u = 9(9) + 0 = 81 = 9

cosθ =

24. u = 4, 4 , v = −2, 0 cosθ =

= 4 1, − 4 = 4, − 16 , vector

= 17.

u = −5, 12 u = u ⋅ u = (−5)2 + 122 = 13

18.

u = − 8, 15 u =

19.

(− 8) + (15)

64 + 225 =

( 4 2 ) (2)

2 2 θ = 135°

289 = 17

u⋅v −10 + 12 2 1 = = = u v 290 2 290 2 10 29

(



θ = arccos 

u = 20i + 25 j

cos θ =

u = 6i − 10 j

)(

)

1   ≈ 86.63° 290 

26. u = 7 i − 2 j, v = − 8i + 6 j

u = u ⋅ u = 6(6) + (−10)( −10) = 136 = 2 34 21.

4( −2) + 4(0)

=−

cos θ =

u = u ⋅ u = (20)2 + (25)2 = 1025 = 5 41 20.

u⋅v u v

25. u = 2i + 6 j, v = − 5i + 2 j

u⋅u

u⋅v 0 = = 0  θ = 90° u v (1)(2)

u⋅v − 56 − 12 − 68 34 = = = u v 10 53 5 53 53 (10)

(

)



34   ≈ 159.08° 5 53  

θ = arccos −

u = −4 j u = u ⋅ u = (−4)( −4) = 4

27. u = 2 i, v = −3 j u⋅v 0 cosθ = = = 0  θ = 90° u v (2)(3) 28. u = 4 j, v = − 9i cos θ =

u⋅ v 0 = = 0 u v (9)(4)

θ = 90° 29.

π  π 1 3  u =  cos  i +  sin  j = i + j 3 3 2 2     3π   3π  2 2  v =  cos i+ j  i +  sin j= 4   4  2 2  u = v =1 cosθ =

u⋅v 2   3  2  − 2 + 6  1  = u ⋅ v =   − +   =  u v 4  2   2   2   2  − 2+ 6  5π  = 75° =  4 12  

θ = arccos 

Section 7.4

Vectors and Dot Products

617

2 2 π  π  30. u = cos   i + sin   j = i+ j 2 2 4 4

1 3  2π   2π  v = cos   i + sin   j = − i + j 3 3 2 2     − u⋅v = u v

cosθ =

( 2 4) + ( 6 4) = 6 − 2  θ = 75° = 5π (1)(1)

31. u = 2i − 4 j, v = 3i − 5 j

cos θ =

13 170  θ ≈ 4.40° 170

(− 7)(− 8) + (− 4)(2) u⋅v = u v 65 68

(

)(

65 2 17

−1

5 4 3 2 1

−3 −4

−5

− 9 − 8 −7 −6 −5 −4 −3

−6

u⋅v 48 + 10 = u v 40 89 29 890  θ ≈ 13.57° 890

2 −2

6 −2

−4

−2 −3 −4 −5

32. u = 6i − 2 j, v = 8i − 5 j

≈ 0.72199

−2

cosθ =

)

θ ≈ 43.78°

1 −1

12 34. u = − 7 i − 4 j, v = − 8i + 2 j

u⋅v 6 + 20 = u v 20 34

cosθ =

35. P = (1, 2), Q = (3, 4), R = (2, 5)    PQ = 2, 2 , PR = 1, 3 , QR = −1, 1   8 2 PQ ⋅ PR cosα =   =  α = arccos ≈ 26.57° 5 PQ PR 2 2 10   PQ ⋅ QR cos β =   = 0  β = 90° PQ QR

(

)( )

γ ≈ 180° − 26.6° − 90° = 63.43°

−6 −8 −10

33. u = −6i − 3 j, v = −8i + 4 j u⋅v −6(−8) + ( −3)(4) cosθ = = u v 45 80 36 = = 0.6 60 θ ≈ 53.13°

36. P = (−3, − 4), Q = (1, 7), R = (8, 2)   PQ = 4, 11 , QR = 7, − 5 ,   PR = 11, 6 , QP = −4, − 11   110 PQ ⋅ PR cosα = =  α ≈ 41.41° PQ PR 137 157   27 QR ⋅ QP cos β =   =  β ≈ 74.45° QR QP 74 137

(

)(

(

)(

)

γ ≈ 180° − 41.41° − 74.45° = 64.14°

8 6

4 2

−10 −8 −6 −4

−2

−4 −6 −8

618

Chapter 7

Additional Topics in Trigonometry

37. P = (−3, 0), Q = (2, 2), R = (0, 6)    PQ = 5, 2 , QR = −2, 4 , PR = 3, 6 ,  QP = −5, − 2   27 PQ ⋅ PR cosα =   =  α ≈ 41.63° PQ PR 29 45   2 QR ⋅ QP  β ≈ 85.24° cos β =   = QR QP 20 29

(

)(

(

)(

43. u ⋅ v = 10, − 6 ⋅ 9, 15 = (10)(9) + ( −6)(15) = 90 − 90 =0 u and v are orthogonal.

)

44. u ⋅ v = 12, 4 ⋅

1  1 = (12)   + (4)  −  4  3 4 5 =3− = − ≠ 0 3 3 u and v are not orthogonal.

φ = 180° − 41.6° − 85.2° ≈ 53.13° 38. P = (−3, 5), Q = (−1, 9), R = (7, 9)   PQ = 2, 4 , QR = 8, 0 ,   PR = 10, 4 , QP = −2, − 4   36 PQ ⋅ PR cos α =   =  α ≈ 41.63° PQ PR 20 116   −16 QR ⋅ QP  β ≈ 116.57° cos β =   = QR QP 8 20

(

)(

(

γ ≈ 180° − 41.6° − 116.6° = 21.80° 39. u ⋅ v = u

v cosθ

45. u ⋅ v = 0, 1 ⋅ 1, − 1

)

= (0)(1) + (1)(−1) = 0 − 1 = −1 ≠ 0 u and v are not orthogonal.

46. u = 2 i − 2 j, v = − i − j u ⋅ v = 0  u and v are orthogonal. 47. u = 10, 20 and v = −18, 9

3π = (9)(36) cos 4  2 = 324  −  2   

u ≠ kv  Not parallel u ⋅ v = 10, 20 ⋅ −18, 9

= (10)( −18) + ( 20)(9) = −180 + 180 = 0

= −162 2 40. u ⋅ v = u

u and v are orthogonal.

v cosθ

= (4)(12)cos

1 1 , − 4 3

1 = 48   = 24 3 2

48. u = 15, 9 and v = −5, − 3

u = kv  15, 9 = (−3) −5, − 3  k = −3 u and v are parallel.

41.

u = 4, v = 10, θ =

u⋅v = u

2π 3

3 7 49. u = − i + i and v = 12 i − 14 j 5 10 3 7 1  1  u = kv  − i + j =  −  (12 i − 14 j)  k = − 5 10 20 20  

v cosθ

= (4)(10)cos

2π 3

 1 = 40  −   2 = −20

42.

u and v are parallel. 50.

u = 100, v = 250, θ =

u⋅v = u

3 2 = 12,500 3 = 25,000 ⋅

u⋅v = −

v cosθ

= (100)(250) cos

9 3 i + 3j and v = − 5i + j 10 2 u ≠ kv  Not parallel u = −

π 6

9 3 , 3 ⋅ − 5, 10 2

 9  3 =  − ( − 5) + (3)   10   2 9 9 = + = 9 ≠ 0  not orthogonal. 2 2

Section 7.4 51. u ⋅ v = 2, − k ⋅ 3, 2 = 6 − 2k = 0  k = 3 52. u ⋅ v = 8, 4 ⋅ 2, − k = 16 − 4k = 0  k = 4

55. u ⋅ v = −3k, 2 ⋅ −6, 0 = 18k = 0  k = 0 56. u ⋅ v = 4, − 4k ⋅ 0, 3 = −12k = 0  k = 0

u ⋅ v v w1 = projvu =   v2   

18 + 8 26 6, 4 = 3, 2 = u v= 36 + 16 52

62. Because u and v are parallel, the projection of u onto v is u. 63. projv u = 0 since they are perpendicular. projv u =

u⋅v v

v = 0, since u ⋅ v = 0.

65. u = 2, 6 For v to be orthogonal to u, u ⋅ v must equal 0.

14 84 14  14  =  v = 6, 1 = , 37 37 37 37   84 14 , 37 37

14 = 6, 1 37 = −

10 60 , 37 37

84 14 10 60 + − , u = w1 + w 2 = , 37 37 37 37 58. u = 4, 2 , v = 1, − 2  u⋅v  w1 = projv u =  2  v = 0 1, − 2 = 0, 0  v    w 2 = u − w1 = 4, 2 − 0, 0 = 4, 2 u = 4, 2 + 0, 0 59. u = 0, 3 , v = 2, 15  u⋅v  45 2, 15 w1 = projv u =  2  v =  v  229   45 2, 15 w 2 = u − w1 = 0, 3 − 229 90 12 6 , = − = −15, 2 229 229 229 u = w1 + w 2 =

u⋅v

projv u =

64. Because u and v are orthogonal, the projection of u onto v is 0.

57. u = 2, 2 , v = 6, 1

w 2 = u − w1 = 2, 2 −

619

61. projv u = u since they are parallel.

53. u ⋅ v = 1, 4 ⋅ 7k , − 5 = 7k − 20 = 0  k = 20 7 54. u ⋅ v = −3k, 5 ⋅ 2, − 4 = −6k − 20 = 0  k = − 103

Vectors and Dot Products

45 6 2, 15 + −15, 2 229 229

60. u = −5, − 1 , v = −1, 1

Two possibilities: 6, − 2 and −6, 2 66. u = −7, 5 For v to be orthogonal to u, u ⋅ v must equal 0. Two possibilities: 5, 7 , −5, − 7 1 3 i− j 2 4 For v to be orthogonal to u, u ⋅ v must equal 0.

67. u =

Two possibilities:

3 1 3 1 i + j and − i − j 4 2 4 2

5 68. u = − i − 3 j 2 For v to be orthogonal to u, u ⋅ v must equal 0.

Two possibilities: v = 3i −

v 69. W = proj PQ

5 5 j and v = −3i + j 2 2

  PQ , where PQ = 4, 7 and v = 1, 4

     32   v ⋅ PQ   proj v = PQ   2  PQ =  65  4, 7    PQ      32 65  65 = 32 W = proj v PQ =  PQ  65   

( )

 u⋅v  4 w1 = projv u =  2  v = −1, 1 = 2 −1, 1  v  2   w 2 = u − w1 = −5, − 1 − 2 −1, 1

= −3, − 3 = 3 −1, − 1 u = −3, − 3 + −2, 2

620

Chapter 7

Additional Topics in Trigonometry

70. P = (1, 3), Q = (−3, 5), v = −2i + 3 j  work = proj v PQ , where PQ  PQ = −4, 2 and v = −2, 3 .

71. u = 1225, 2445 and v = 12.00, 10.25

u ⋅ v = (1225)(12.00) + (2445)(10.25) = $39,761.25 u ⋅ v is the total amount paid to both levels of employees at the temp agency. (b) To increase wages by 2%, multiply v by 1.02 or 1.02v = 12.24, 10.46 . (a)

     14   v ⋅ PQ   = proj v PQ   2  PQ =  20  −4, 2    PQ      14 20  work = proj 20 = 14 v PQ =  PQ  20   

(

)

72. u = 3240, 2450 , v = 3.25, 3.50 (a) u ⋅ v = 3240(3.25) + 2450(3.50) = $19,105

This gives the total revenue earned. 1 (b) To increase prices by 2 percent, multiply v by 1.025 or 1.025 v = 1.025 3.25, 3.50 = 3.33, 3.59 . 2

73. (a)

 F = −30,000 j, Gravitational force v = cos(d°), sin(d°)

 F⋅v  w1 = projvF =  2  v = ( F ⋅ v ) v = −30,000 sin(d°) cos d°, sin d°  v   

= −30,000 sin d ° cos d °, − 30,000 sin 2 d ° Force needed: 30,000 sin(d °) (b)

(c)

0°

1°

2°

3°

4°

5°

Force

523.6

1047.0

1570.1

2092.7

2614.7

6°

7°

8°

9°

10°

Force

3135.9

3656.1

4157.2

4693.0

5209.4

w2 = F − w1 = −30,000 j + 2614.7(cos(5°)i + sin(5°) j) ≈ 2604.75i − 29,772.11j

w2 ≈ 29,885.8 pounds 74. Force due to gravity: F = −5400 j v = cos10°i + sin10° j

w1 = projv F =

F⋅v v

v = ( F ⋅ v )v

= 0, − 5400 ⋅ cos10°, sin10° v = −937.70 v Magnitude: 937.70 1b, to keep the vehicle from rolling down the hill Force perpendicular to the hill: w 2 = F − w1 = −5400 j + 937.7[cos10°i + sin10° j] ≈ 923.45i − 5237.17 j

w 2 ≈ (923.45)2 + (−5237.17)2 ≈ 5318 lb

Section 7.4 75. (a)

Vectors and Dot Products

621

86. Let u and v be two adjacent sides of the rhombus, u = v . The diagonals are u + v and u − v.

F = 250 cos30°, sin30°  PQ = d 1, 0

(u + v) ⋅ (u − v) = u ⋅ u + v ⋅ u − u ⋅ v − v ⋅ v

 3 w = F ⋅ PQ = 250 d 2 = 125 3 d

= u − v =0 Hence, the diagonals are perpendicular.

(b)

u+v

Work

100

v u−v

5412.66 10,825.32 21,650.64

87. Use the Law of Cosines on the triangle:

76.

u−v = u + v −2 u

35 pounds 22°

v cosθ

= u + v − 2u ⋅ v

200 ft

W = (cos22°)(35)(200) ≈ 6490.29 ft-lb u−v

77. Work = (cos 25°)(50)(15) ≈ 679.7 foot-pounds θ

78. Work = (cos20°)(25)(12) ≈ 281.9 foot-pounds

79. True. u ⋅ v = 0

88. u ⋅ (cv + dw) = c(u ⋅ v) + d(u ⋅ w) = c0 + d 0 = 0

80. False. Work is a scalar, not a vector.

89. From trigonometry, x = cosθ and y = sinθ . Thus, u = x, y = cosθ i + sinθ j.

81. u ⋅ v = 0  they are orthogonal (unit vectors).

82. The dot product is a scalar, not a vector. 5, 8 ⋅ −2, 7 = (5)(−2) + (8)(7)

= −10 + 56 = 46

83. If u is a unit vector, then u ⋅ u = 1. The angle between u u⋅u and if and itself is 0. So, since cosθ = u u

θ = 0, cos = 1  84. (a) (b) (c)

u⋅u = 1. u u

u ⋅ v = 0  u and v are orthogonal and θ = u ⋅ v > 0  cosθ > 0  0 ≤ θ < u ⋅ v < 0  cosθ < 0 

π 2

θ x

(b)

projv u = u  u and v are parallel. projv u = 0  u and v are orthogonal.

90. From trigonometry, π  π  x = cos  − θ  and y = sin  − θ  . 2  2  .

π  π  Thus, u = x, y = cos  − θ  i + sin  − θ  j. 2 2    

<θ ≤π

j u

85. (a)

(x, y)

θ x

622

Chapter 7

Additional Topics in Trigonometry

91. g( x ) = f ( x − 4) is a horizontal shift of f four units to the right.

98. (7 − 4i)(7 + 4i) = 49 + 16 = 65

92. g( x) = − f ( x) is a reflection of f with respect to the x-axis.

99.

93. g( x ) = f ( x) + 6 is a vertical shift of f six units upward. 94. g( x ) = f (2 x ) is a horizontal shrink of f. 95. 3i(4 − 5i) = 12i + 15 = 15 + 12i 96. −2i(1 + 6i) = −2i + 12 = 12 − 2i 97. (1 + 3i )(1 − 3i ) = 1 − (3i )2 = 1 + 9 = 10

100.

3 2 3 1− i 2 2 + 3i + = ⋅ + ⋅ 1 + i 2 − 3i 1 + i 1 − i 2 − 3i 2 + 3i 3 − 3i 4 + 6i = + 2 13 39 − 39i + 8 + 12i = 26 47 27 = − i 26 26

6 4+i 3 1 − i 24 + 6i 3 − 3i ⋅ − ⋅ = − 4 − i 4 + i 1+ i 1− i 17 2 48 + 12i − 51 + 51i = 34 3 63 =− + i 34 34

Section 7.5 Trigonometric Form of a Complex Number 1. absolute value

8. 10 − 3i = 10 2 + ( −3)2

2. DeMoivre’s

= 109

3. nth root

Imaginary axis

4. The trigonometric form of the complex number z = a + bi is given by z = r (cosθ + i sinθ ), where

10 8 6 4 2

a = r cosθ , b = r sinθ , r = a 2 + b2 , and tanθ = b a . 5. In the trigonometric form z = r cosθ + i sin θ , r is the distance from the origin to the point (a, b). 6. In the trigonometric form z = r cosθ + i sin θ , θ is the angle (measured counterclockwise) from the positive real axis to the line segment.

−8 −6 −4 −2

Real axis

8 10

−4 −6 −8 −10

10 − 3i

9. −5 − 12i = 52 + 12 2

= 169 = 13 Imaginary axis

7. 9 + 7i = (9)2 + (7)2

= 81 + 49 = 130

−10 −8

−6

−4

−2

Real axis

Imaginary axis 8

−6

9 + 7i

−8

−10

−5 − 12i

2 −2

−2 −4

−12

Real axis

10. − 4 + 6i =

(− 4) + (6)

52 = 2 13

Imaginary axis

− 4 + 6i

7 6 5 4 3 2

−5 −4 −3 −2 −1

Real axis

Section 7.5

Trigonometric Form of a Complex Number

623

16. z = − i

11. 9i = 9

r = 02 + (−1)2 = 1

Imaginary axis

tan θ undefined  θ =

z = cos

6 4 2

−6 −4 −2

−2

Real axis

3π 2

3π 3π + i sin 2 2

17. z = −4 r = (−4)2 + (0)2 = 16 = 4 0 = 0 θ =π −4 z = 4(cos π + i sin π )

tan θ =

12.

−2i = 2 Imaginary axis

18. z = 5

5 4 3 2 1

r = (5)2 + (0)2 = 25 = 5

−4 −3 −2 −1

1 2 3 4 5

−2i

−2 −3 −4 −5

13.

Real axis

0 = 0 θ = 0 5 z = 5(cos0 + i sin 0)

tan θ =

19. z = −3 − 3i r = (−3)2 + (−3)2 = 9 + 9 = 18 = 3 2 −3 5π =1θ = −3 4 5π 5π   z = 3 2  cos + i sin  4 4  

tan θ =

−5 = 5 Imaginary axis 3

20. z = 3 + 3i

2 1

−5 −5

−4

−3

−2

−1

Real axis

−2 −3

14.

8 =8

r = 32 + 32 = 3 2 3 π =1θ = 3 4 π π  z = 3 2  cos + i sin  4 4 

tan θ =

21. z = 3 − i

Imaginary axis

−2

tan θ =

−2

( 3 ) + ( −1) = 2

8 2

Real axis

−4 −6

15. z = 2i r = (0) 2 + (2) 2 = 4 = 2 2 , undefined  θ = π 2 0 z = 2 ( cos 0 + i sin 0 ) tan θ =

−1 3

θ =

11π 6

11π 11π   + i sin z = 2  cos  6 6   22. z = −1 + 3i

( −1) + ( 3 ) = 4 = 2 2

3 2π = − 3 θ = −1 3 2π 2π   + i sin z = 2  cos  3 3  

tan θ =

624

Chapter 7

Additional Topics in Trigonometry 26. z = −10i

23. z = −8i 2

r = 0 + (−8) = 64 = 8 8 3π tan θ = − , undefined  θ = 0 2 3π 3π   + i sin z = 8  cos  2 2   Imaginary axis

1 2 3 4 5

−2 −3 −4 −5 −6 −7 −8

(0)2 + (−10)2 =

100 = 10

−10 3π , undefined  θ = 0 2 3π 3π   + i sin z = 10 cos  2 2  

tan θ =

Imaginary axis

2 1 −4 −3 −2 −1

r =

Real axis

2 −8 −6 − 4 −2

Real axis

−4 −6 −8 −10

−8i

−10i

−12 −14

24. z = 4i r=4

27. z = 7 − 7i

π 4 tan θ = , undefined  θ = 0 2 π π  z = 4  cos + i sin  2 2  Imaginary axis

5 4 3 2 1 −4 −3 −2 −1

r =

7 2 + ( − 7)

98 = 7 2

7 7π = −1  θ = 7 4 7π 7π   + i sin z = 7 2  cos  4 4  

tan θ = −

Imaginary axis

1 2 3 4 5

Real axis

−2 −3 −4 −5

−2

Real axis

−4 −6

7 − 7i

−8

25.

z = 5i

− 10 2

r = (0) + (5) = 25 = 5 5 π , undefined  θ = 0 2 π π  z = 5  cos + i sin  2 2 

tan θ =

Imaginary axis

π 2 =1θ = 2 4 π π  z = 2 2  cos + i sin  4 4 

tan θ =

5 4 3 2 1

3 2 1

−1

r = 4+4 =2 2

Imaginary axis

−3 −2 −1

28. z = 2 + 2i

Real axis

−4 −3 −2 −1

2 + 2i

1 2 3 4 5

Real axis

−2 −3 −4 −5

Section 7.5

32.

29. z = 3 + i

( 3) +1 = 4 = 2 2

tan θ =

Trigonometric Form of a Complex Number

π 3 θ = 3 6

π π  z = 2  cos + i sin  6 6 

z = 5 − 5i

r = (5) 2 + (− 5) 2 = 25 + 25 = 50 = 5 2 −5 = − 1  θ = 7π 4 5 7π 7π   + i sin z = 5 2  cos 4 4  

tan θ =

Imaginary axis 2 1

Imaginary axis

−2 −1

Real axis

−2

−3

3+i

−4

4 − 4i

−5

−1

−6

Real axis

−1

(

33. z = −2 1 + 3i

30. z = 1 − 3i

r = 1+ 3 = 2 3 5π  θ = 300° = 1 3 5π 5π   z = 2  cos + i sin 3 3  

tan θ = −

)

( −2 ) + ( −2 3 ) = 16 = 4 2

3 4π = 3 θ = 1 3 4π 4π   z = 4  cos + i sin  3 3  

tan θ =

Imaginary axis

−4

5 4 3 2 1 −4 −3 −2 −1

−2

1−

−3

−2(1 + 3i)

r = (1)2 + (1)2 = 2 1 =1θ = π 4 1 π π  z = 2  cos + i sin  4 4 

tan θ =

Imaginary axis

34. z = −

5 2

r =

5  2

−4

( 3 + i) 2

 5  3  +  (1)   2 

100 4

25 = 5

tan θ =

1 = 3

3 7π θ = 3 6

7π 7π   + i sin z = 5 cos  6 6  

Imaginary axis

1+i

Real axis

2 1

−1 −2

Real axis

−1

Real axis

31. z = 1 + i

−1

−3

−2

1 2 3 4 5

−2 −3 −4 −5

−2

625

−5 −4 −3 −2

−1

Real axis

−2 5 2

− ( 3 + i)

−3 −4

626

Chapter 7

Additional Topics in Trigonometry

35. z = −7 + 4i

38. z = 6 r =6

r = 49 + 16 = 65 4  θ ≈ 2.62 radians or 150.26° −7 z = 65 ( cos 2.62 + i sin 2.62 )

tan θ =

θ =0 z = 6(cos0 + i sin 0) Imaginary axis

Imaginary axis

−7 + 4i

10 8 6 4 2

4 2

−8

−6

−4

Real axis

−2

−8 −6 −4 −2

2 4 6 8 10

Real axis

−4 −6 −8 −10

−2 −4

39. z = 3 + 3i

36. z = 5 − 3i

r =

52 + ( − 3)

tan θ = −

3  θ ≈ − 30.9° or 329.0° ≈ 5.743 rad 5

34 (cos 5.743 + i sin 5.743)

z ≈ Imaginary axis

r = 9 + 3 = 12 = 2 3 3 π  θ = or 30° 3 6 π π  z = 2 3  cos + i sin  6 6 

tan θ =

Imaginary axis

5 4 3 2 1

−1

Real axis

−2

3 + 3i

−4 −3 −2 −1

−3

−2 −3 −4 −5

5 − 3i

−4

Real axis

1 2 3 4 5

37. z = 3 40. z = − 2 2 + i

r = 32 + 0 2 = 3 0 tan θ = = 0  θ = 0° 3 z = 3(cos0° + i sin 0°)

r =

−4 −3 −2 −1 −2 −3 −4 −5

tan θ =

Imaginary axis

5 4 3 2 1

(− 2 2 ) + (1) = 2

9 = 3

1 2 = −  θ ≈ 160.529 ≈ 2.802 rad 4 −2 2

z ≈ 3(cos 2.802 + i sin 2.802) Imaginary axis

3 1 2 3 4 5

Real axis

−2 2 + i

−3

−2

−1

Real axis

−1 −2

Section 7.5

Trigonometric Form of a Complex Number 44. z = −3 + i

41. z = −1 − 5i 2

r = 1 + 5 = 26 −5 tan θ = = 5  θ ≈ 258.69° ≈ 4.515 rad −1 z = 26 ( cos 4.515 + i sin 4.515 )

r = 9 + 1 = 10 1 tanθ =  θ ≈ 161.6° ≈ 2.820 rad −3 z ≈ 3.16(cos161.6° + i sin161.6°)

Imaginary axis

2 1

−5 −4 −3 −2 −1

Real axis

5 4 3 2 1

−3 + i

−4 −3 −2 −1

−6

42. z = 1 + 3i tan θ = 13 = 3  θ ≈ 71.6° ≈ 1.249 rad z ≈ 10 ( cos71.6° + i sin 71.6° ) Imaginary axis

r = 18 + 49 = 67 −7 tan θ = ≈ −1.6499  θ ≈ 301.22° ≈ 5.257 rad 3 2 z = 67 ( cos301.22° + i sin 301.22° ) Imaginary axis

1 + 3i

2 1

−4 −3 −2 −1 2

43. z = 5 − 2i 25 + 4 =

tan θ = − z ≈

29 ≈ 5.385

2  θ ≈ 338.199° ≈ 5.903 rad 5

29 (cos 5.903 + i sin 5.903)

3 2 − 7i

46. z = −8 − 5 3i r = 64 + 75 = 139 tan θ =

−5 3 ≈ 227.3° ≈ 3.967 rad −8

Imaginary axis

5 4 3 2 1

−2 −3 −4 −5

Real axis

z ≈ 11.79 ( cos227.3° + i sin 227.3°)

Imaginary axis

−4 −3 −2 −1

1 2 3 4 5

−2 −3 −4 −5 −6 −7 −8

Real axis

−1

r =

Real axis

45. z = 3 2 − 7i

r = 12 + 32 = 10

1 2 3 4 5

−2 −3 −4 −5

− 1 − 5i

−1

627

4 5

5 − 2i

10 8 6 4 2

Real axis

−8 −6 −4

−8 − 5 3i

2 4 6 8 10

Real axis

−4 −6 −8

628

Chapter 7

Additional Topics in Trigonometry

1 π π 3   47. 6  cos + i sin  = 6  + i  = 3 + 3 3i   3 3 2 2    

 1 3  i 51. 2(cos120° + i sin120°) = 2  − +  2  2  

Imaginary axis

8 7 6 5 4 3 2 1

= −1 + 3i Imaginary axis

3 + 3 3i

−4 −3 −2 −1

Real axis

1 2 3 4 5

−4

−1 +

−3

−1

−2

−1

Real axis

−2

 5π 5π  3 1  + i sin +i  48. 8  cos  = 8  − 6 6  2    2 = −4 3 + 4i Imaginary axis

−3

 2  2  + i 52. 5(cos135° + i sin135°) = 5  −   2    2  

−4 3 + 4i

=−

4 3 2 1

−7 −6 −5 −4 −3 −2

−1

Imaginary axis

Real axis

−

−2

5 2 5 2 + i 2 2

4 3

−3

3π 3π  15 2 15 2  49. 3.75  cos + i sin + i =− 4 4  8 8 

Imaginary axis

−

15 2 15 2 + i 8 8

5 2 5 2 + i 2 2

−3

−4

53.

−2

Real axis

−1

5 5 3 1  cos( − 30°) + i sin( − 30°) =  − i  2 2 2 2 

−4

−3

−2

Real axis

−1

Imaginary axis

5 3 5 − i 4 4

−1 2

π π  50. 1.5  cos + i sin  = 1.5(0 + i) = 1.5i 2 2 

−1

Imaginary axis

5 4 3 2 1 −4 −3 −2 −1 −2 −3 −4 −5

1 2 3 4 5

Real axis

−1

5 3 5 − i 4 4

−2

1.5i

Real axis

54.

1 1 2 2  cos( − 45°) + i sin ( − 45°) =  − i  4 4 2 2  =

2 2 − i 8 8

Imaginary axis

−1

Real axis

2 2 − i 8 8

Section 7.5 3π 3π   + i sin  = 2 3 (0 − i ) = − 2 3i 12  cos 2 2  

55.

Imaginary axis

−2

−1

Real axis

−1

Trigonometric Form of a Complex Number

629

62. 2(cos 73° + i sin 73°) ≈ 0.5847 + 1.9126i   π π    3π 3π  63. 3 cos + i sin  9 cos + i sin  2 2 2 2        π 3π  3π  π = (3)(9) cos +  + i sin  +  2 2 2 2      = 27(cos 2π + i sin 2π )

−2 −3

= 27(cos 0 + i sin 0)

−2 3 i

−4

48 (cos 0 + i sin 0) = 4 3 (1 + 0i )

56.

= 4 3 Imaginary axis

 π π    7π 7π  64. 4 cos + i sin  5 cos + i sin  12    12 12    12  π 7π  7π  π = ( 4)(5) cos + +  + i sin   12  12   12   12 2π 2π   = 20 cos + i sin  3 3  

4 3 2 1 −1

4 3 1

Real axis

−2

2 6π 6π    9π 9π  + i sin  9 cos + i sin  65.   cos 3 7 7 14 14       9π  9π   2    6π  6π =  (9) cos + +  + i sin   14  14   3   7  7

−3 −4

57. 3 cos (18° 45′ ) + i sin (18° 45′ )  ≈ 2.8408 + 0.9643i Imaginary axis

  3π   3π  = 6 cos   + i sin    2    2  3 π π    π π  66.   cos + i sin   6  cos + i sin   2 6 6 4 4      

2.8408 + 0.9643i

1 1

Real axis

−1 −2

58. 6  cos ( 230°30′ ) + i sin ( 230°30′ )  ≈ −3.8165 − 4.6297i Imaginary axis

 3 π π π π  ( 6 ) cos  6 + 4  + i sin  6 + 4  2     

5π 5π   = 9  cos + i sin  12 12   67.  35 ( cos140° + i sin140° )   23 ( cos60° + i sin 60° )  = ( 35 )( 23 ) cos (140° + 60° ) + i sin (140° + 60° )  = 109 ( cos200° + i sin 200° )

−5

−4

−3

−2

−1

Real axis

1 4  68.  ( cos115° + i sin115° )   ( cos300° + i sin300° )  2 5   

−2 −3

14   cos (115° + 300° ) + i sin (115° + 300° )  2  5 

2 2 ( cos 415° + i sin 415°) = 5 ( cos55° + i sin 55°) 3

−4

−3.8165 − 4.6297i −5

7π 7π   59. 5 cos + i sin  ≈ − 3.8302 + 3.2139i 9 9  

3π 3π   + i sin 60. 12  cos  ≈ −3.71 + 11.41i 5 5  

69. (cos 290° + i sin 290°)(cos 200° + i sin 200°)

= cos( 290° + 200°) + i sin ( 290° + 200°) = cos 490° + i sin 490° = cos 130° + i sin 130°

61. 9 ( cos58° + i sin 58° ) ≈ 4.7693 + 7.6324i

630

Chapter 7

Additional Topics in Trigonometry

70. (cos5° + i sin5°)(cos20° + i sin20°) = cos(5° + 20°) + i sin(5° + 20°)

76.

= cos25° + i sin 25° 71.

72.

5 ( cos75° + i sin 75° )

9  cos ( 20° − 75° ) + i sin ( 20° − 75° )   5 9 =  cos ( −55° ) + i sin ( −55° )  5 9 = ( cos305° + i sin 305° ) 5 =

cos50° + i sin 50° = cos(50° − 20°) + i sin(50° − 20°) cos20° + i sin 20° = cos30° + i sin30° 2 ( cos120° + i sin120° ) 4 ( cos 40° + i sin 40° )

1  cos (120° − 40° ) + i sin (120° − 40° )   2 1 = (cos80° + i sin80°) 2

77. (a)

73.

9 ( cos20° + i sin 20° )

π π  1 + i = 2  cos + i sin  4 4  (b) (2 − 2i )(1 + i ) 7π 7π   π π  = 2 2  cos + i sin  2  cos + i sin  4 4   4 4  = 4(cos2π + i sin 2π ) = 4

5(cos 2π + i sin 2π ) 4(cos π + i sin π )

5 cos( 2π − π ) + i sin ( 2π − π ) 4 5 = (cos π + i sin π ) 4 =

7π 7π + i sin 4 4 = cos 3π + i sin 3π cos π + i sin π 4 4

cos

74.

75.

( = 6 ( cos ( −48° ) + i sin ( −48° ) )

= 6 cos ( 54° − 102° ) + i sin ( 54° − 102° )

7π 7π   + i sin 3 − 3i = 3 2  cos  4 4  

7π 7π  7π 7π    = 3 2  cos + i sin + i sin  2  cos  4 4  4 4    7π 7π   = 6  cos + i sin  = −6i 2 2  

)

= 6 ( cos312° + i sin312° )

79. (a)

(2 − 2i)(1 + i) = 2 + 2 = 4

7π 7π   + i sin 1 − i = 2  cos  4 4   (b) (3 − 3i )(1 − i )

18 ( cos54° + i sin 54° )

3 ( cos102° + i sin102° )

7π 7π   + i sin 2 − 2i = 2 2  cos  4 4  

(c)

( 3 − 3i )(1 − i ) = 3 − 3 − 6i = −6i

π π  2 + 2i = 2 2  cos + i sin  4 4    7π   7π   + i sin  − 1 − i = 2 cos  −    4    4 

(b)

(2 + 2i )(1 − i )  π π      7π    7π    =  2 2  cos + i sin    2  cos  + i sin     4 4       4   4     = 4 ( cos 0 + i sin 0 ) = 4

(c)

(2 + 2i )(1 − i ) = 2 − 2i + 2i − 2i 2 = 2 + 2 = 4

Section 7.5

Trigonometric Form of a Complex Number

631

π π  80. (a) 3 + 3i = 3 2  cos + i sin  4 4  π π  2  cos + i sin  4 4 

1+i =

 π π    π π   (b) (3 + 3i )(1 + i ) = 3 2  cos + i sin   2  cos + i sin  4 4   4 4      π π π  π = 6 cos  +  + i sin  +  4 4 4 4     

π π  = 6 cos + i sin  2 2  = 6i (c) (3 + 3i)(1 + i ) = 3 + 6i + 3i 2

= 0 + 6i = 6i 81. (a) −2i = 2 cos ( −90° ) + i sin ( −90° ) 

1 + i = 2 ( cos45° + i sin 45° ) (b) −2i(1 + i) = 2[cos( −90°) + i sin(−90°)]  2 (cos 45° + i sin 45°)    = 2 2 cos ( −45° ) + i sin ( −45° )   1 1  =2 2 − i  = 2 − 2i 2   2

π π  3i = 3  cos + i sin  2 2 

82. (a)

83. (a)

1 + 2i = 3 ( cos 0.955 + i sin 0.955 )

(b)

(

3i 1 + 2i

)

11π 11π   + i sin 3 − i = 2  cos  6 6   (b)

  π π  = 3  cos + i sin    3 ( cos 0.955 + i sin 0.955 )   2 2    

(

−2i

( 3 − i)

3π 3π   11π 11π   = 2  cos + i sin + i sin  2  cos  2 2 6 6     20π 20π   = 4  cos + i sin  6 6  

 π  π  = 3 3  cos  + 0.955  + i sin  + 0.955   2 2      

(c)

3π 3π   −2i = 2  cos + i sin  2 2  

≈ −3 2 + 3i

 1 3  i = 4 − −  2 2  

)

3i 1 + 2 i = 3i + 3 2 i 2 = −3 2 + 3i

= −2 − 2 3i (c)

−2 i

( 3 − i ) = −2 3 i − 2

632

Chapter 7

84. (a)

(b)

Additional Topics in Trigonometry

3π 3π + i sin 2 2 π π  1 + 3i = 2  cos + i sin  3 3  −i = cos

(

−i 1 + 3i

)

87. (a)

(b)

π π 3π 3π    =  cos + i sin 2 cos + i sin  2 2   3 3  11π 11π   = 2  cos + i sin 6 6  

(

)

( −i ) 1 + 3 i = − i + 3

85. (a)

2 = 2 ( cos0 + i sin 0 )

5π 5π   + i sin 1 − 3i = 2  cos  3 3   5 ( cos0.927 + i sin 0.927 ) 2 ( cos5π 3 + i sin 5π 3 ) =

5  5π  5π    cos  0.927 −  + i sin  0.927 −  2  3  3  

5 cos1.974 + i sin1.974  2  3  3 3 ≈ − 3+ + 1 i   4   4 

 3 1 = 2 −i  = 3 −i  2 2   (c)

3 + 4i = 5 ( cos0.927 + i sin 0.927 )

(c)

3 + 4i 1 + 3i 3 + 3 3i + 4i + 4 3i ⋅ = 1 − 3i 2 1 − 3i 1 + 3i =

7π 7π   + i sin 1 − i = 2  cos  4 4   (b)

7π 7π   2(1 − i ) = 2 2  cos + i sin  4 4    2 2  =2 2 − i  2 2   = 2 (1 − i ) = 2 − 2i

2 (1 − i ) = 2 − 2i

4 3 3 3   =  − 3  + 1 + i  4 4    

88. (a)

(b)

−4 = 4(cos π + i sin π )

(c)

π π  2 + 2i = 2 2  cos + i sin  4 4  π π  1 + 3i = 2  cos + i sin  3 3  2 + 2i

1 + 3i

2 2 π π   π π   cos  −  + i sin  −   2  4 3    4 3 

  π   π  = 2  cos  −  + i sin  −    12   12   

π π  1 + i = 2  cos + i sin  4 4  5π 5π   (b) −4 (1 + i ) = 4 2  cos + i sin  4 4    2 2  = 4 2 − − i  2 2   = −4 − 4 i −4(1 + i) = −4 − 4i

(3 − 4 3 ) + ( 4 + 3 3 ) i

(c)

1+ 3 1− 3 i + 2 2

89. (a)

(b)

(c)

(

)

2 + 2i 1 − 3i 2 + 2 3 + 2 − 2 3 i ⋅ = 4 1 + 3i 1 − 3i 1+ 3 1− 3 + i 2 2

5 = 5 ( cos0 + i sin 0 )

π π  2 + 2i = 2 2  cos + i sin  4 4  5 5(cos0 + i sin 0) = π π 2 + 2i  2 2  cos + i sin  4 4  =

5   π  π   cos  −  + i sin  −   4 2 2    4 

5  2 2  5 5 i = − i −  2  4 4 2 2 2

5 2 − 2i 5(2 − 2i ) 5 5 ⋅ = = − i 2 + 2i 2 − 2i 4+4 4 4

Section 7.5 90. (a)

2 = 2(cos0 + i sin0)

2  11π   11π   =  cos  −  + i sin  −  3 −i 2  6   6  2

3 1 + i 2 2

(c)

91. (a)

2 3 −i

3 +i

⋅

3 +i

( 3 + i) = 3 + 1 i 4

2 2 1 (1 + i ) 2 (1 + i ) = 2 ( 2i ) = i 2  2  2 z3 = z2 z = i  1 + i ) = ( ( i − 1) 2  2  z2 =

Imaginary axis

z3 =

3π 3π   −1 + i = 2  cos + i sin  4 4   (b)

(c)

92. (a)

4i 4   π 3π   π 3π   =  cos  −  + i sin  −  2 4 4  −1 + i 2   2 4   π  π  =  cos  −  + i sin  −   2  4  4  4  2 2  i  = 2 − 2i = −   2 2 2 

4i −1 − i −4i + 4 ⋅ = = 2 − 2i 2 −1 + i −1 − i

π π  2i = 2  cos + i sin  2 2 

2i 1 − 3i

2   π 5π   π 5π    cos  −  + i sin  −  2  2 3  3  2

 7π   7π  − 3 1 = cos  − + i  + i sin  − = 6 2 2    6  (c)

2i − 2 3 3 1 ⋅ = =− + i 4 2 2 1 − 3i 1 + 3i 2i

1 + 3i

2 (−1 + i) z 2 = i 2 2 z= (1 + i) 2 z 4 = −1 Real −2

−1

axis

−1 −2

The absolute value of each power is 1.

(

)

1 1 + 3i 2 1 1 1 1 + 3 i = −1 + 3 i z 2 = 1 + 3i 2 2 2 1 1 1 + 3i z 3 = z 2 z = −1 + 3 i 2 2 Imaginary = −1 axis

94. z =

(

) (

(

) (

z 4 = z3z

(

1 = (−1) 1 + 3i 2 1 = −1 − 3 i 2

5π 5π   1 − 3i = 2  cos + i sin  3 3   (b)

2 2 1 ( i − 1) 2 (1 + i ) = 2 ( −2 ) = −1 2

z 4 = z3 z =

π π  4i = 4  cos + i sin  2 2 

633

2 (1 + i ) 2

93. z =

11π 11π   3 − i = 2  cos + i sin  6 6   (b)

Trigonometric Form of a Complex Number

(

)

z3 =

)

2 (−1 + i) z 2 = i 2 2 z= (1 + i) 2 z 4 = −1 Real −2

−1

axis

−1 −2

The absolute value of each is 1. 95. Let z = x + iy. z = 3  3 =

x2 + y 2  9 = x2 + y 2

Circle of radius 3 Imaginary axis 4 2 1 −4

−2 −1

Real axis

−2 −4

634

Chapter 7

Additional Topics in Trigonometry

96. Let z = x + iy.

100. Let z = x + iy.

z = 5  5 = x + y  25 = x + y 2

Circle of radius 5

z = 8  8 = x 2 + y 2  x 2 + y 2 = 64 Circle of radius 8

Imaginary axis

6 4

−6

−2

Real axis

−4 −2

97. Let z = x + iy.

z = 4  4 = x + y  x + y = 16 2

Circle of radius 4 Imaginary axis 6

101. θ =

6 Let z = x + iy such that: π y 1 1 y y= tan =  = x 6 x x 3 3 Line Imaginary axis

−2

Real axis

−6

−2

−4

−6

Real axis

θ=π

−6

Real axis

98. Let z = x + iy.

z = 6  6 = x 2 + y 2  x 2 + y 2 = 36 102. θ =

Circle of radius 6 Imaginary axis 9

3 −9

−3

Real axis

4 Let z = x + iy such that: π y y tan =  1 =  y = x 4 x x Line Imaginary axis

−9

θ =π

99. Let z = x + iy.

Real axis

z = 7  7 = x 2 + y 2  x 2 + y 2 = 49 Circle of radius 7 Imaginary axis 8 6 4 2 −8

−4 −2

Real axis

−4 −6 −8

Section 7.5

103. θ =

3 Let z = x + iy such that: tan

y = 3 x y = 3 x y = 3x

Line

θ=π

3π 4 Let z = x + iy such that: 3π y tan = 4 x y = −1 x y = −x Line

106. θ =

θ = 3π 4

Real axis

2 Let z = x + iy such that:

y but in this case the tangent is undefined. = x 2 Therefore it is the vertical line, x = 0. tan

635

Imaginary axis

104. θ =

Trigonometric Form of a Complex Number

Imaginary axis

θ=π

2 Real axis

  π π  107. (1 + i )3 =  2  cos + i sin   4 4    3 3π 3π  = 2  cos + i sin  4 4  

( )

 2 2  = 2 2 − + i  2  2   = −2 + 2i  π π   108. (2 + 2i )6 = 2 2  cos + i sin   4 4   

6 6π 6π  = 2 2  cos + i sin  4 4   3π 3π   = 512  cos + i sin  2 2  

(

2π 3 Let z = x + iy such that: 2π y y tan =  − 3 =  y = − 3x 3 x x Line

105. θ =

)

= −512i 6

  3π 3π   109. (−1 + i )6 =  2  cos + i sin  4 4    6 18π 18π  = 2  cos + i sin  4 4  

( )

Imaginary axis

θ = 2π 3

Real axis

9π 9π   = 8  cos + i sin  2 2     π π   = 8 cos  4π +  + i sin  4π +   2 2     

π π  = 8  cos + i sin  2 2  = 8(0 + i ) = 8i

636

Chapter 7

Additional Topics in Trigonometry

 7π 7π  8  + i sin 110. (3 − 3i ) = 3 2  cos  4 4   

( ) = (3 2 ) cos (0) + i sin (0) 8

= 3 2 cos (14π ) + i sin (14π )

117. 2 ( cos π + i sin π )  = 28 ( cos8π + i sin8π ) = 256(1) = 256

= 104,976(1 + 0i ) = 104,976

(

111. 2

3 −i

)

  π π  116.  2  cos + i sin   = 212 (cos6π + i sin 6π ) 2 2    = 212 = 4096

118. (cos0 + i sin 0)20 = cos0 + i sin 0 =1

  11π 11π  = 22 cos + i sin  6 6   

  55π 55π  = 225  cos + i sin  6 6      7π 7π  = 225  cos + i sin  6 6     3 1  = 64 − − i  2 2  

119.  4 ( cos10° + i sin10° )  = 46 ( cos60° + i sin 60° ) 1 3  = 4096  + i 2 2   = 2048 + 2048 3i 4

120. 6(cos 15° + i sin 15°) = 64 (cos 60° + i sin 60°)

1 3  = 64  + i  2 2  

= − 32 3 − 32i

(

112. 4 1 − 3i

)

  5π 5π   = 4 2  cos + i sin  3 3    = 4 23 ( cos5π + i sin 5π )  = 32(−1) = −32 3

113. 5(cos 140° + i sin 140°) = 53 (cos 420° + i sin 420°)

= 125(cos 60° + i sin 60°) 1 3  = 125 + i  2 2   =

125 125 3 + i 2 2

114. 3 ( cos150° + i sin150° )  = 34 (cos600° + i sin600°)

= 81(cos240° + i sin 240°) = 81( − cos60° − i sin60°) =−

81 81 3 i − 2 2 10

π 2

+ i sin

π 2

= 648 + 648 3i 2

  π π  π π  121. 3  cos + i sin   = 32  cos + i sin  8 8 4 4      2 2 = 9 +i   2 2   9 9 = 2+ 2i 2 2 5

  π π  π π  122. 2  cos + i sin   = 32  cos + i sin  10 10 2 2     = 32i

(

)

123. (3 − 2i )5 =  13 cos ( 5.6952 ) + i sin ( 5.6952 )    = 169 13 cos ( 28.476 ) + i sin ( 28.476 )  ≈ −597 − 122i 124.

5π 5π   + i sin 115.  cos  4 4   25π 25π = cos + i sin 2 2 π π   = cos  12π +  + i sin  12π +  2 2   = cos

1 3  = 1296 + i  2 2  

( 5 − 4i ) =  21 ( cos ( 5.2221) + i sin ( 5.2221)) 4

= 441 cos ( 20.8884 ) + i sin ( 20.8884 )  ≈ −199 + 393.55i 4

125.  2 ( cos1.25 + i sin1.25 )  = 2 4 (cos5 + i sin 5) ≈ 4.5386 − 15.3428i 5

126.  4 ( cos2.8 + i sin 2.8 )  = 45 (cos14 + i sin14) ≈ 140.02 + 1014.38i

Section 7.5 6

4π 4π   1   127.  − 1 + 3i  =  cos + i sin 2 3 3     = cos8π + i sin8π =1

(

)

128. 2 −1 4 (1 − i ) is a fourth root of − 2 if −2 =  2 −1 4 (1 − i )  . 4

(

2 −1 4 (1 − i )  = 2 −1 4  

) (1 − i ) 4

= 2 −1 (1 − i )

= 12 (1 − i ) (1 − i ) 2

= 12 (−2i )(−2i )

( )

= 12 4i 2

= 12 (−4) = −2

π π  129. 2i = 2  cos + i sin  2 2  Square roots: π π  2  cos + i sin  = 1 + i 4 4  5π 5π   2  cos + i sin  = −1 − i 4 4  

π π  130. 5i = 5  cos + i sin  2 2  Square roots: π π 10 10  + 5  cos + i sin  = i 4 4 2 2   5π 5π  10 10  + i sin − 5  cos i =− 4 4 2 2  

3π 3π   + i sin 131. −3i = 3  cos  2 2   Square roots: 3π 3π  6 6  + i sin + 3  cos i =− 4 4  2 2  7π 7π  6 6  + i sin − 3  cos i = 4 4  2 2 

Trigonometric Form of a Complex Number

637

3π 3π   + i sin 132. −6i = 6  cos  2 2   Square roots: 3π 3π   + i sin 6  cos  = − 3 + 3i 4 4   7π 7π   + i sin 6  cos  = 3 − 3i 4 4  

7π 7π   + i sin 133. 2 − 2i = 2 2  cos  4 4   Square roots: 7π 7π   + i sin 81 4  cos  ≈ −1.554 + 0.644i 8 8   15π 15π   + i sin 81 4  cos  ≈ 1.554 − 0.644i 8 8  

π π  134. 2 + 2i = 2 2  cos + i sin  4 4  Square roots: π π  81 4  cos + i sin  ≈ 1.554 + 0.644i 8 8  9π 9π   81 4  cos + i sin  ≈ −1.554 − 0.644i 8 8  

π π  135. 1 + 3i = 2  cos + i sin  3 3  Square roots: π π 6 2  + 2  cos + i sin  = i 6 6 2 2   7π 7π  6 2  + i sin − 2  cos i =− 6 6 2 2  

5π 5π   + i sin 136. 1 − 3i = 2  cos  3 3   Square roots:  5π 5π  3 1   + i sin + i 2  cos  = 2  − 6 6 2 2     6 2 + i 2 2 11π 11π  6 2  + i sin − 2  cos i = 6 6  2 2  =−

638

Chapter 7

Additional Topics in Trigonometry

137. (a) Square roots of 5(cos120° + i sin120°) :

  120° + 360°k   120° + 360°k   5 cos   + i sin    , k = 0, 1 2 2      (b)

k =0:

5(cos60° + i sin60°)

k = 1:

5(cos240° + i sin 240°)

Imaginary axis 3

1 −3

−1

Real axis

−3

(c)

5 15 5 15 + i, − − i 2 2 2 2

138. (a) Square roots of 16 ( cos60° + i sin60° ) :   60° + 360°k   60° + 360°k   16 cos   + i sin    , k = 0, 1 2 2      k = 0 : 4(cos30° + i sin 30°) k = 1 : 4(cos210° + i sin 210°)

(b)

Imaginary axis 6

2 −6

−2

Real axis

−6

(c)

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

2π 2π   139. (a) Fourth roots of 81 cos + i sin : 3 3   4

Imaginary axis

(b)

  ( 2π 3) + 2kπ   ( 2π 3) + 2kπ  81 cos   + i sin   4 4     

k = 0, 1, 2, 3

−4

π π  k = 0: 3 cos + i sin  6 6  2π 2π   k = 1: 3 cos + i sin  3 3  

−2

−1

Real axis

−2 −4

(c)

3 3 3 + i 2 2

7π 7π   k = 2: 3 cos + i sin  6 6  

−

5π 5π   k = 3: 3 cos + i sin  3 3  

3 3 3 + i 2 2

−

3 3 3 − i 2 2

1 3 3 − i 2 2

Section 7.5

5π 5π   + i sin 140. (a) Fifth roots of 32  cos : 6 6  

Trigonometric Form of a Complex Number

π π  142. (a) Fourth roots of 625i = 625  cos + i sin  : 2 2 

  ( 5π 6 ) + 2 kπ   ( 5π 6 ) + 2 kπ   32 cos   + i sin       5 5      k = 0, 1, 2, 3, 4

  (π 2 ) + 2 kπ   (π 2 ) + 2 kπ   625  cos   + i sin       4 4      k = 0, 1, 2, 3

π π  k = 0 : 2  cos + i sin  6 6  17π 17π   k = 1 : 2  cos + i sin  30 30  

π π  k = 0 : 5  cos + i sin  8 8  5π 5π   + i sin k = 1 : 5  cos  8 8  

29π 29π   k = 2 : 2  cos + i sin  30 30   41π 41π   k = 3 : 2  cos + i sin  30 30  

9π 9π   + i sin k = 2 : 5  cos  8 8   13π 13π   + i sin k = 3 : 5  cos  8 8  

53π 53π   k = 4 : 2  cos + i sin  30 30  

(b)

Imaginary axis

(b)

Imaginary axis 3

−6

−3

−0.8135 −1.827i, 1.486 −1.338i 3π 3π   + i sin 141. (a) Cube roots of −25i = 25  cos : 2 2  

  ( 4π 3) + 2kπ   ( 4π 3) + 2kπ   125  cos   + i sin    ,  3 3       k = 0, 1, 2 4π 4π   k = 0 : 5  cos + i sin 9 9   10π 10π   k = 1: 5  cos + i sin 9 9   16π 16π   k = 2 : 5  cos + i sin 9 9  

k = 2 : 3 25 cos11π 6 + i sin11π 6 

Imaginary axis

(b)

1 −1

)

k = 1 : 3 25 cos7π 6 + i sin 7π 6 

−2

Real axis

4.619 + 1.913i, − 1.913 + 4.619i, −4.619 − 1.913i, 1.913 − 4.619i

(

k = 0 : 3 25 cos π 2 + i sin π 2 

−4

143. (a) Cube roots of 125 4π 4π   1 + 3i = 125  cos − + i sin : 2 3 3  

  3π   3π  + 2 kπ  + 2 kπ      3 25 cos  2  + i sin  2  3 3             k = 0, 1, 2

Imaginary axis

−6

(c)

(b)

−2

−4

Real axis

−3

Real axis

−6

−2

2.9240i, − 2.5323 − 1.4620i, 2.5323 − 1.4620i

−2

Real axis

−4 −6

−4

(c)

639

(c)

0.8682 + 4.9240i, − 4.6985 − 1.7101i, 3.8302 − 3.2139i

640

Chapter 7

Additional Topics in Trigonometry

144. (a) Cube roots of

3π 3π   −4 2 (1 − i ) = 8  cos + i sin : 4 4  

146. (a) Fourth roots of i = cos

k = 0 : cos

19π 19π   + i sin k = 2 : 2  cos  12 12  

(b)

+ i sin

Real axis

−2

π π  145. (a) Fourth roots of 256i = 256 cos + i sin  : 2 2    (π 2) + 2kπ   (π 2) + 2kπ  256 cos   + i sin   4 4     

(c)

0.9239 + 0.3827i, − 0.3827 + 0.9239i, −0.9239 − 0.3827i, 0.3827 − 0.9239i

147. (a) Fifth roots of 1 = cos0 + i sin 0 : 2 kπ 2 kπ cos , k = 0, 1, 2, 3, 4 + i sin 5 5 k = 0 : cos 0 + i sin 0

2π 2π + i sin 5 5 4π 4π k = 2 : cos + i sin 5 5 6π 6π k = 3: cos + i sin 5 5 8π 8π k = 4 : cos + i sin 5 5

k = 0, 1, 2, 3

k = 1: cos

π π  k = 0: 4 cos + i sin  8 8  5π 5π   k = 1: 4 cos + i sin  8 8   9π 9π   k = 2: 4 cos + i sin  8 8   13π 13π   k = 3: 4 cos + i sin  8 8  

(b)

Imaginary axis 2

Imaginary axis 5

−2

−1 −2 −3

Real axis

−2

−3

−5

Imaginary axis

(b)

8 8 5π 5π k = 1 : cos + i sin 8 8 9π 9π k = 2 : cos + i sin 8 8 13π 13π k = 3 : cos + i sin 8 8

π π  k = 0 : 2  cos + i sin  4 4  11π 11π   + i sin k = 1 : 2  cos  12 12  

  (π 2 ) + 2 kπ   (π 2 ) + 2 kπ   1  cos   + i sin       4 4      k = 0, 1, 2, 3

  ( 3π 4 ) + 2 kπ   ( 3π 4 ) + 2 kπ   8  cos   + i sin       3 3      k = 0, 1, 2

Imaginary axis

+ i sin

(b)

1 2 3

−5

Real axis

−2

−1.5307 + 3.6955i − 3.6955 + 1.5307i 1.5307 − 3.6955i © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Section 7.5 148. (a) Cube roots of 1000 = 1000(cos0 + i sin0) :

Trigonometric Form of a Complex Number 150. (a) Fourth roots of −4 = 4(cos180° + i sin180°) :

2 kπ 2 kπ   1000  cos + i sin  3 3   k = 0, 1, 2 k = 0 : 10(cos0 + i sin 0)

  180° + 360k   180° + 360k  4 cos   + i sin   , 4 4      k = 0, 1, 2, 3

2π 2π   k = 1 : 10  cos + i sin  3 3   4π 4π   k = 2 : 10  cos + i sin  3 3  

(b)

Imaginary axis

k = 0:

2(cos45° + i sin45°)

k = 1:

2(cos135° + i sin135°)

k = 2:

2(cos225° + i sin225°)

k = 3:

2(cos315° + i sin315°)

(b)

Imaginary axis 2

8 6 4

−8 −6 −4 −2

2 4 6 8

−2

Real axis

−6 −8

149. (a) Cube root of −125 = 125(cos180° + i sin180°)

3π 3π   128(−1 + i ) = 128 2  cos + i sin : 4 4  

  180° + 360 k   180° + 360 k   3 125 cos   + i sin   , 3 3      k = 0, 1, 2

(128 2 ) cos  3π 4 5+ 2kπ  + i sin  3π 4 5+ 2kπ  15

k = 0 : 5(cos60° + i sin 60°)

k = 0, 1, 2, 3, 4

k = 1 : 5(cos180° + i sin180°) k = 2 : 5(cos300° + i sin 300°)

3π 3π   + i sin k = 0 : 2 2  cos  20 20   11π 11π   k = 1 : 2 2  cos + i sin  20 20  

Imaginary axis 4 3 2 1

−4 −3 −2 −1

2 3 4

19π 19π   k = 2 : 2 2  cos + i sin  20 20   27π 27π   k = 3 : 2 2  cos + i sin  20 20  

Real axis

−2 −3 −4

(c)

Real axis

−2

(b)

641

5 5 3 5 5 3 + i, − 5, − i 2 2 2 2

7π  7π  + i sin k = 4 : 2 2  cos  4 4   (b)

Imaginary axis

−2 −1

Real axis

−2

(c)

2.52 + 1.28i, − 0.44 + 2.79i, −2.79 + 0.44i, − 1.28 − 2.52i, 2 − 2i

642

Chapter 7

Additional Topics in Trigonometry

152. (a) Fifth roots of 7π 7π   + i sin 4(1 − i ) = 4 2  cos : 4 4  

(4 2 ) cos 7π 4 5+ 2kπ  + i sin  7π 4 5+ 2kπ  15

(

k = 0, 1, 2, 3, 4

Note: 4 2

k = 0:

7π 7π   2  cos + i sin  20 20  

k = 1:

3π 3π   2  cos + i sin  4 4  

k = 2:

23π 23π   2  cos + i sin  20 20  

k = 3:

31π 31π   2  cos + i sin  20 20  

k = 4:

39π 39π   2  cos + i sin  20 20  

(b)

)

( 32 )

(32)1 5 =

Imaginary axis

−2 −1

Real axis

−2

153. x 4 − i = 0 x4 = i

The solutions are the fourth roots of i = cos

π 2

+ i sin

π 2

  ( π 2 ) + 2 kπ   ( π 2 ) + 2 kπ   1 cos   + i sin    ,  4 4       k = 0, 1, 2, 3 4

k = 0 : cos

+ i sin

Imaginary axis

8 8 5π 5π k = 1 : cos + i sin 8 8 9π 9π k = 2 : cos + i sin 8 8 13π 13π k = 3 : cos + i sin 8 8

−2

Real axis

−2

Section 7.5 154. x 3 + 1 = 0

Trigonometric Form of a Complex Number 156. x 3 − 27 = 0

x = −1 The solutions are the cube roots of −1 = 1(cos π + i sin π ).

  π + 2 kπ   π + 2 kπ   1 cos   + i sin   3 3      k = 0, 1, 2

x 3 = 27 The solutions are the cube roots of 27 = 27 [ cos0 + i sin 0 ] :

  0 + 2 kπ   0 + 2 kπ   27  cos   + i sin   3  3      k = 0, 1, 2

π π π 1 3  π 1 cos + i sin  = cos + i sin = + i 3 3 3 3 2 2  k = 1: cosπ + i sin π = −1 k =0:

k = 0 : 3 27 [ cos 0 + i sin 0] = 3 ( cos 0 + i sin 0 ) = 3

k = 2 : cos

643

5π 5π 1 3 + i sin = − i 3 3 2 2

Imaginary axis 2

−2

2π 2π  2π 2π    + i sin = 3  cos + i sin k = 1: 3 27 cos  3 3  3 3    3 3 3 =− + i 2 2 4π 4π  4π 4π    + i sin k = 2 : 3 27 cos  = 3  cos 3 + i sin 3  3 3     3 3 3 i =− − 2 2

Imaginary axis

Real axis

4 2

−2 −4

155. x 5 + 243 = 0 x = −243

  π + 2 kπ   π + 2 kπ   243 cos   + i sin    , k = 0, 1, 2, 3, 4 5 5     

π π  k = 0 : 3  cos + i sin  5 5  3π 3π   + i sin k = 1: 3  cos  5 5   5π 5π   + i sin k = 2 : 3  cos  = 3(cos π + i sin π ) 5 5   7π 7π   + i sin k = 3 : 3  cos  5 5   9π 9π   + i sin k = 4 : 3  cos  5 5   Imaginary axis

Real axis

157. x 4 + 16i = 0 x 4 = −16i

The solutions are the fourth roots of 3π 3π   −16i = 16 cos + i sin  : 2 2     ( 3π 2 ) + 2 kπ   ( 3π 2 ) + 2 kπ   4 16 cos   + i sin    , 4 4       k = 0, 1, 2, 3   3π   3π   k = 0: 2 cos   + i sin  8       8    7π   7π   k = 1 : 2 cos   + i sin  8   8        11π   11π   k = 2: 2 cos   + i sin  8       8    15π   15π   k = 3: 2 cos   + i sin  8       8 

5 4 1 −2 −1

−4

The solutions are the fifth roots of −243 = 243[cos π + i sin π ] :

−5 −4

−2

−2 −1

4 5

Imaginary axis

Real axis

−4 −5

1 −3

−1

Real axis

−3

644

Chapter 7

Additional Topics in Trigonometry 160. x 4 + (1 + i ) = 0

158. x 6 − 64i = 0 6

x = 64i The solutions are the sixth roots of π π  64i = 64  cos + i sin  : 2 2 

x 4 = −1 − i

The solutions are the fourth roots of −1 − i = 2 ( cos225° + i sin 225° ) :

  (π 2) + 2 kπ   (π 2 ) + 2 kπ   64  cos  + i sin    ,  6 6       k = 0, 1, 2, 3, 4, 5

  225° + 360°k   225° + 360°k   2 cos   + i sin   4 4      k = 0, 1, 2, 3

π π   k = 0 : 2  cos + i sin  ≈ 1.932 + 0.5176i 12 12   5π 5π   k = 1 : 2  cos + i sin  0.5176 + 1.932i 12 12   3π 5π   k = 2 : 2  cos + i sin  −1.414 + 1.414i 4 4   13π 13π   k = 3 : 2  cos + i sin ≈ −1.932 − 0.5176i 12 12   17π 17π   k = 4 : 2  cos + i sin ≈ −0.5176 − 1.932i 12 4   7π 7π   k = 5 : 2  cos + i sin ≈ 1.414 − 1.414i 4 4  

k = 0 : 8 2(cos56.25° + i sin 56.25°) ≈ 0.6059 + 0.9067i

k = 1 : 8 2(cos146.25° + i sin146.25°) ≈ −0.9067 + 0.6059i k = 2 : 8 2(cos236.25° + i sin 236.25°) ≈ −0.6059 − 0.9067i k = 3 : 8 2 ( cos326.25° + i sin 326.25° ) ≈ 0.9067 − 0.6059i Imaginary axis 2

−2

Real axis

−2

Imaginary axis

161. E = I ⋅ Z

−3

Real axis

= (10 + 2i )(4 + 3i ) = (40 + 6) + (30 + 8)i = 34 + 38i 162. E = I ⋅ Z = (12 + 2i )(3 + 5i )

−3

159. x 3 − (1 − i ) = 0 x3 = 1 − i The solutions are the cube roots of

1 − i = 2 ( cos315° + i sin315° ) :   315° + 360°k   315° + 360°   2 cos   + i sin   , 3 3      k = 0, 1, 2 3

k = 0 : 6 2 ( cos105° + i sin105° ) k = 1: 6 2 ( cos225° + i sin 225° ) k = 2 : 6 2 ( cos345° + i sin 345° )

= (36 − 10) + (60 + 6)i = 26 + 66i E I 5 + 5i 2 − 4i = . 2 + 4i 2 − 4 i (10 + 20) + (10 − 20)i = 4 + 16 30 − 10i = 20 3 1 = − i 2 2

163. Z =

Imaginary axis 2

−2

Real axis

−2

Section 7.5

Trigonometric Form of a Complex Number 172. z = r (cosθ + i sinθ )

E I 4 + 5i 10 − 2i . = 10 + 2i 10 − 2i (40 + 10) + (50 − 8)i = 100 + 4 50 + 42i = 104 25 21 = + i 52 52 E 165. I = Z 12 + 24i 12 − 20i . = 12 + 20i 12 − 20i (144 + 480) + (288 − 240)i = 144 + 400 624 + 48i = 544 39 3 = + i 34 34

164. Z =

z = r (cosθ − i sin θ ) = r (cos( −θ ) + i sin( −θ ))

173. (a) zz = [r (cosθ + i sin θ )][r (cos( −θ ) + i sin( −θ ))] = r 2 [cos(θ − θ ) + i sin(θ − θ )] = r 2 [cos0 + i sin 0] = r2

r (cosθ + i sin θ ) r [cos( −θ ) + i sin (−θ )] z r = [cos(θ − (−θ )) + i sin(θ − (−θ ))] r = cos2θ + i sin 2θ z

(b)

174.

z = r (cosθ + i sin θ ) − z = −r (cosθ + i sin θ ) = r (− cosθ − i sin θ ) = r (cos(θ + π ) + i sin(θ + π ))

175. Let a = 0 and b = π in Euler’s Formula: e a + bi = e a (cos b + i sin b)

E 166. I = Z 15 + 12i 25 − 24i . = 25 + 24i 25 − 24i (375 + 288) + (300 − 360)i = 625 + 576 663 − 60i = 1201 663 60 i = − 1201 1201

e 0 +π i = e0 (cos π + i sin π ) eπ i = −1 πi

e +1 = 0 Imaginary axis

176.

3  1  1 167. True.  1 − 3i  =  − i  = −1 2  2 2 

168. False.

)

( 3 + i ) = 2 + 2 3i ≠ 8i 2

169. True. z1z2 = r1r2 cos (θ1 + θ2 ) + i sin (θ1 + θ2 )  = 0 if and

only if r1 = 0 and or r2 = 0. 170. False. They are equally spaced around the circle centered at the origin with radius n r.

r ( cosθ1 + i sin θ1 ) cosθ 2 − i sin θ 2 z . 171. 1 = 1 z2 r2 (cosθ 2 + i sin θ 2 ) cosθ 2 − i sin θ 2 =

r1 [cosθ1 cosθ 2 + sin θ1 sin θ 2 r2 (cos θ 2 + sin 2 θ 2 ) 2

+ i(sin θ1 cosθ 2 − sin θ 2 cosθ1 )] r = 1 [cos (θ1 − θ 2 ) + i sin(θ1 − θ 2 )] r2

z0 30°

1 −1

(

645

Real axis

−1

(a) Because one of the fourth roots is shown, there are three roots not shown. (b) The other three roots also lie on the circle, with arguments of θ = 120°, 210°, and 300°. π  177. d = 16 cos  t  4  Maximum displacement: 16

Lowest possible t-value: 178. d =

π 4

π 2

t =2

1 5π sin t 7 4

1 7 5π 4 Lowest possible t -value: t = π  t = 4 5 Maximum displacement:

646

Chapter 7

Additional Topics in Trigonometry

1 179. d = cos(12π t ) 8 Maximum displacement:

180. d =

1 8

Lowest possible t-value: 12π t =

1 sin 60π t 12

Maximum displacement:

π 2

t =

1 24

1 12

Lowest positive t-value: 60π t = π  t =

1 60

Chapter 7 Review 1. Given: A = 32°, B = 50°, a = 16 C = 180° − 32° − 50° = 98° a sin B 16sin 50° b= = ≈ 23.13 sin A sin 32° a sin C 16sin 98° c= = ≈ 29.90 sin A sin 32°

2. Given: A = 38°, B = 58°, a = 12 C = 180° − 38° − 58° = 84° a sin B 12sin 58° b= = ≈ 16.53 sin A sin 38° a sin C 12sin84° c= = ≈ 19.38 sin A sin 38°

3. Given: B = 16°, C = 98°, a = 8.4

A = 180° − 16° − 98° = 66° 8.4 sin 16° a sin B b = = ≈ 2.53 sin A sin 66° c =

8.4 sin 98° a sin C = ≈ 9.11 sin A sin 66°

4. Given: B = 95°, C = 45°, a = 104.8

A = 180° − 95° − 45° = 40° b =

a sin B 104.8 sin 95° = ≈ 162.42 sin A sin 40°

c =

a sin C 104.8 sin 45° = ≈ 115.29 sin A sin 40°

5. Given: A = 60° 15′ = 60.25°, B = 45° 30′ = 45.5°, b = 4.8 C = 180° − 60.25° − 45.5° = 74.25° = 74° 15′

b sin A 4.8sin 60.25° = ≈ 5.84 sin B sin 45.5° b sin C 4.8sin 74.25° c= = ≈ 6.48 sin B sin 45.5° a=

6. Given:

A = 82° 45′ = 82.75°, B = 28° 45′ = 28.75°, b = 40.2 C = 180° − 82.75° − 28.75° = 68.5° = 68° 30′ b sin A 40.2sin82.75° = ≈ 82.91 a= sin B sin 28.75° b sin C 40.2sin 68.5° = ≈ 77.76 c= sin B sin 28.75°

7. Given: A = 75°, a = 51.2, b = 33.7

b sin A 33.7 sin 75° = ≈ 0.63578  B ≈ 39.48° a 51.2 C = 180° − A − B = 180° − 75° − 39.48° = 65.52°

sin B =

c =

a sin C 51.2 sin 65.52° = ≈ 48.24 sin A sin 75°

8. Given: A = 15°, a = 5, b = 10 b sin A 10sin15° = ≈ 0.5176  B ≈ 31.2° or 148.8° sin B = a 5 Case 1 : B ≈ 31.2° C ≈ 180° − 15° − 31.2° = 133.8° a sin C 5 sin133.8° ≈ ≈ 13.94 c= sin A sin15° Case 2 : B ≈ 148.8°

C ≈ 180° − 15° − 148.8° = 16.2° a sin C 5 sin16.2° ≈ ≈ 5.4 c= sin A sin15° © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 7 Review

647

9. Given: B = 25°, a = 6.2, b = 4 Two Solutions a sin B 6.2 sin 25° = ≈ 0.6551  A1 ≈ 40.92° b 4 C1 ≈ 180° − 25° − 40.92° = 114.08°

sin A =

c1 =

b sin C 4 sin 114.08° = ≈ 8.64 sin B sin 25°

a sin B 6.2 sin 25° = ≈ 0.6551  A1 ≈ 180° − 40.92° = 139.08° b 4 C2 ≈ 180° − 25° − 139.08° = 15.92°

sin A =

c2 =

b sin C 4 sin 15.92° = ≈ 2.60 sin B sin 25°

10. Given: B = 150°, a = 10, b = 3 a sin B 10sin150° = = 1.6667 > 1 sin A = b 3 No solution

h 75 h = 75sin 28° ≈ 35.21 feet

15. sin 28° =

x 75 x = 75cos28° ≈ 66.22 feet

cos28° =

11. Given: A = 33°, b = 7, c = 10

1 Area = bc sin A 2 1 = (7)(10)sin33° 2 ≈ 19.06 square units

H x H = x tan 45° ≈ 66.22 feet

tan 45° =

Height of tree: H − h ≈ 31 feet

12. Given: B = 80°, a = 4, c = 8

1 Area = ac sin B 2 1 = (4)(8)(0.9848) 2 ≈ 15.76 square units 13. Given: C = 122°, b = 18, a = 29

1 Area = ab sin C 2 1 = (29)(18)sin122° 2 ≈ 221.34 square units 14. Given: C = 100°, a = 120, b = 74

1 ab sin C 2 1 = (120)(74)sin100° 2 ≈ 4372.5 square units

H 75

45° 28°

16.

a 400 = sin 75° sin 37.5°

400sin 75° ≈ 634.7 feet sin 37.5° w sin 67.5° = a w = 634.7 sin 67.5° ≈ 586.4 feet a=

Tree 37°30′

Area =

15°

22°30′ 67°30′

75°

400

648

Chapter 7

Additional Topics in Trigonometry

17. Given: a = 9, b = 12, c = 20 2

a +b −c 2ab 81 + 144 − 400 = 2(9)(12) ≈ −0.8102  C ≈ 144.1°

cos C =

18. Given: a = 7, b = 15, c = 19

a 2 + b2 − c 2 ≈ −0.4143  C ≈ 114.5° 2 ab a 2 + c 2 − b2 ≈ 0.6955  B ≈ 45.9° cos B = 2 ac A = 180° − B − C = 19.6° cos C =

a sin C c 9sin(144.1°) = 20 ≈ 0.264  A ≈ 15.3 B = 180° − 144.1° − 15.3° = 20.6° sin A =

19. Given: a = 6.5, b = 10.2, c = 16 cos A =

b 2 + c 2 − a 2 10.22 + 162 − 6.52 = ≈ 0.97  A ≈ 13.19° 2bc 2(10.2)(16)

a 2 + c 2 − b2 6.52 + 162 − 10.22 = ≈ 0.93  B ≈ 20.98° 2 ac 2(6.5)(16) C = 180° − A − B ≈ 145.83° cos B =

20. Given: a = 6.2, b = 6.4, c = 2.1 cos A =

b 2 + c 2 − a 2 6.4 2 + 2.12 − 6.2 2 = ≈ 0.26  A ≈ 75.06° 2bc 2(6.4)(2.1)

a 2 + c 2 − b2 6.2 2 + 2.12 − 6.4 2 = ≈ 0.07  B ≈ 85.84° 2 ac 2(6.2)(2.1) C = 180° − A − B ≈ 19.10° cos B =

21. Given: C = 65°, a = 25, b = 12 c 2 = a 2 + b2 − 2 ab cos C = 252 + 12 2 − 2(25)(12)cos65° ≈ 515.4290  c ≈ 22.70 a sin C 25sin 65° = ≈ 0.998  A ≈ 86.38° sin A = c 22.70 B = 180° − A − C ≈ 28.62°

22. Given: B = 48°, a = 18, c = 12 b2 = a 2 + c 2 − 2 ac cos B = 182 + 122 − 2(18)(12)cos 48° ≈ 178.9356  b ≈ 13.38

c sin B 12sin 48° = ≈ 0.666663  C ≈ 41.81° 13.38 b A = 180° − B − C ≈ 90.19°

sin C =

23. a 2 = 52 + 82 − 2(5)(8)cos152° ≈ 159.6  a ≈ 12.63 ft

b2 = 52 + 82 − 2(5)(8)cos28° ≈ 18.36  b ≈ 4.285 ft a

8 5

28°

= 300 2 + 4252 − 2(300)(425)cos(180 ° − 65°) ≈ 378,392.66 b ≈ 615.1 meters

25. Given: a = 3, b = 6, c = 8 a +b+c 3+6+8 s= = = 8.5 2 2 Area = s(s − a )(s − b)(s − c) = 8.5(8.5 − 3)(8.5 − 6)(8.5 − 8) = 58.4375 ≈ 7.64 square units 26. Given: a = 15, b = 8, c = 10 a + b + c 15 + 8 + 10 = = 16.5 2 2 Area = s(s − a )(s − b)(s − c ) = 16.5(1.5)(8.5)(6.5) s=

(Answers may vary.)

24. b 2 = a 2 + c 2 − 2 ac cos B

≈ 36.98 square units

27. Given: a = 64.8, b = 49.2, c = 24.1 a + b + c 64.8 + 49.2 + 24.1 = = 69.05 s= 2 2 Area = s( s − a )(s − b)(s − c) = 69.05(4.25)(19.85)(44.95) ≈ 511.7 square units

Chapter 7 Review 28. Given: a = 8.55, b = 5.14, c = 12.73 a + b + c 8.55 + 5.14 + 12.73 s= = = 13.21 2 2 Area = s(s − a)(s − b)(s − c) = 13.21(4.66)(8.07)(0.48) ≈ 15.4 square units

33. Initial point: ( 0, 10 ) Terminal point: ( 7, 3 ) v = 7 − 0, 3 − 10 = 7, − 7 v = 72 + (−7)2 = 98 = 7 2 34. Initial point: (1, 5 ) Terminal point: (15, 9 )

29. Vector u: Initial point: ( − 2, 1)

Terminal point: ( 4, 6)

v = 15 − 1, 9 − 5 = 14, 4

u = 4 − ( − 2), 6 − 1 = 6, 5

v = 142 + 42 = 212 = 2 53

6 2 + 52 =

u =

35. (a)

Vector v: Initial point: (0, − 2)

u + v = −1, − 3 + −3, 6 = −4, 3 y

Terminal point: (6, 3)

v = 6 − 0, 3 − ( − 2) = 6, 5 6 2 + 52 =

v =

5 4

u+v

3 2

u = v

−7 −6 −5 −4 −3

−1

30. Vector u: Initial point: ( − 3, 2) u = −1 − ( − 3), − 4 − 2 = 2, − 6

2 + ( − 6)

u =

(b)

40 = 2 10

2 −6

−4

−2

v = 3 − 1, − 2 − 4 = 2, − 6 2

u = v

−8

40 = 2 10

−v

−6

u−v

−10

(c)

3u = −3, − 9

31. Initial point: ( −5, 4) Terminal point: (2, − 1)

y 2

v = 2 − (−5), − 1 − 4 = 7, − 5

−8

−6

−4

v = 72 + (−5)2 = 74

−2

−6

32. Initial point: ( 0, 1)  7 Terminal point:  6,   2 7 5 v = 6 − 0, − 1 = 6, 2 2

u − v = 2, − 9

Terminal point: (3, − 2) 2 2 + ( − 6)

Vector v: Initial point: (1, 4)

v =

Terminal point: ( −1, − 4)

649

−8 −10

(d)

2 v + 5u = −6, 12 + −5, − 15 = −11, − 3 y

169 13 5 = v = 62 +   = 4 2 2

−20 −16

2v + 5u 5u −16

650

Chapter 7

36. (a)

Additional Topics in Trigonometry

u + v = 4, 5 + 0, − 1 = 4, 4

37. (a)

u + v = −5, 2 + 4, 4 = −1, 6 y

y 8

u+v

1 −2 −1 v

−1

−6

−4

−2

u − v = 4, 6

(b)

u − v = −9, − 2

u− v

6 5

2 1

− 10

−v −2 −1 −1

u− v

−2

−6

(c)

3u = −15, 6

u 3

6 −15 −12 −9

(d)

−3

−6

−3

−6 −9

2 v + 5u = 0, − 2 + 20, 5 = 20, 23

(d)

2 v + 5u = 8, 8 + −25, 10 = −17, 13 y

y 25

24 20

2v + 5u 20

2v + 5u

16 12

8 4

− v −4

−2

−3

−4

−2

(b)

−2

−25 −20 − 15 −10 −5 −5

−10

Chapter 7 Review

38. (a)

u + v = 1, − 8 + 3, − 2 = 4, − 10

39. (a)

u + v = (2 i − j) + (5i + 3 j) = 7 i + 2 j y

y 5

2 −4

−2

v 4

−2

u+v

2 1

−4 −6

−1 −1

−8

(b)

− 5 −4 −3 −2

u −2

u−v

−3

−5 −6

− 10

−4

4 −4

−1 −1

−8

−4 −5

(d)

2 v + 5u = (10i + 6j ) + (10i − 5j ) = 20 i + j

(d) 2v + 5u = 6, − 4 + 5, − 40 = 11, − 44

12 8

8 2

10 12

− 24

−4 −8

4 −4

− 16

− 48

−3

− 16

− 24

−2

− 12

− 20

3u = 6 i − 3 j y

− 40

−4

−6 −8

− 32

1 x

−v

−2 −8

−4

− 12 − 8

−v 2

(c)

u − v = −3 i − 4 j

u −v

−3

u − v = −2, − 6

−6

−2

u+v

− 10

(b)

651

2v + 5u 8

− 12

2v + 5u

652

Chapter 7

Additional Topics in Trigonometry

40. (a) u + v = ( − 7i − 3 j) + ( 4i − j) = − 3i − 4 j

41.

1 1 10 10 v = (10i + 3 j) = i + j = ,1 3 3 3 3 y

6 4

2 −10 −8

6 4

u + v −4 −8

−2 −2

−10

(b) u − v = ( − 7i − 3 j) − ( 4i − j) = −11i − 2 j y

42.

−v 1 2

−1

u−v u

10 12

v = 10i + 3j 1 3 3 v = 5i + j = 5, 2 2 2

−6

−4

−14 −12 −10

1 v 3

−6

x 5

−2

−3

−4

2 1

−5

−2 −1

1 v 2 2

8 10 12 14

−2

−3

2 x

−21 −18 −15 −12 −9 −6

43. w = 4u + 5v = 4(6i − 5 j) + 5(10i − 3j)

−2

= 74i − 5 j = 74, − 5

−4

−6

50 40 30 20 10

−8

−10 −12

(d) 2 v + 5u = 2( 4i − j) + 5( − 7i − 3 j)

−10 −20 −30 −40 −50

= (8i − 2 j) + ( − 35i − 15 j) = − 27i − 17 j

20 30 40

4u + 5v

44. 3v − 2u = 3 (10i + 3j) − 2 ( 6i − 5 j) = 18i + 19 j

y − 40 − 32 −24 −16 − 8

= 18, 19 y

−9

− 12

5u 2v + 5u

3v − 2u

− 15

−2u

− 18

−12 −6 −6

3v 6

12 18 24 30

−12 −18

45.

u =6

Unit vector:

1 0, − 6 = 0, − 1 6

Chapter 7 Review

46.

v =

56. v = 8i − j

( −12 ) + ( −5 ) = 169 = 13 2

v = 82 + ( −1) = 65 2

12 5 Unit vector: − , − 13 13

82 + ( −15)

47. v =

tan θ =

64 + 225 =

289 = 17

1 8 15 Unit vector: 8, −15 = ,− 17 17 17 =

48.

8 15 i − i 17 17

w =7 1 ( −7 i ) = − i 7

49. Initial point: ( − 8, 3) Terminal point: (1, − 5) u = 1 − ( − 8), − 5 − 3 = 9, − 8 = 9i − 8 j

50. Initial point: ( − 2, 7) Terminal point: (5, − 9)

−1 , θ in Quadrant IV  θ ≈ 352.9° 8

57. Force One: u = 85i Force Two: v = 50 cos15°i + 50 sin15° j Resultant Force: u + v = ( 85 + cos15° ) i + ( 50sin15° ) j

(85 + 50 cos15°) + ( 50sin15°) 2

u+v =

Unit vector:

653

= 852 + 8500 cos15° + 502 = 133.92 Ib 50sin15°  θ ≈ 5.5° from the 85-pound force. tan θ = 85 + 50cos15°  3 1  58. Rope One: u = u ( cos30°i − sin 30° j) = u  i − j  2 2   Rope Two:  3 1  v = u ( − cos30°i − sin 30° j ) = u  − i − j  2 2   Resultant: u + v = − u j = −180 j

u = 5 − ( − 2), − 9 − 7 = 7, −16 = 7i − 16 j

51. v = 7(cos 120°i + sin 120° j)

v = 7

u = 180

Therefore, the tension on each rope is u = 180 pounds. 59. By symmetry, the magnitudes of the tensions are equal. T = T ( cos120°i + sin120° j)

θ = 120° 52. v = 3 ( cos150°i + sin150° j)

1 100 200 T sin120° = (200)  T = = ≈ 115.5 lb 2 3 2 3

v = 3, θ = 150°

53. v = 5i + 4 j v = 52 + 4 2 = 41 tan θ =

4  θ ≈ 38.7° 5

54. v = −3i − 3j v =

( −3) + ( −3) = 3 2

tan θ =

−3 = 1  θ ≈ 225° −3

55. v = −4i + 7 j

⏐⏐T⏐⏐

60°

120°

200 lb

60. Airplane velocity: u = 430 cos315°, sin 315°

Wind velocity: w = 35 cos60°, sin 60° u + w ≈ 321.5559, − 273.7450 u + w ≈ 422.3 mph

v =

( −4 ) + 7 = 65

tan θ =

7 , θ in Quadrant II  θ ≈ 119.7° −4

⏐⏐T⏐⏐

−273.7450 321.5559 ≈ −0.8513  θ ≈ −40.4°

tan ( u + w ) =

The bearing from the north is 90° + 40.4° = 130.4°.

654

Chapter 7

Additional Topics in Trigonometry

61. u ⋅ v = − 6, 7 ⋅ − 3, − 5 = ( − 6)( − 3) + (7)( − 5) = −17

u⋅v = 0  θ = 90° u v

73. cosθ =

62. u ⋅ v = −7, 12 ⋅ −4, − 14 = ( −7)( −4) + (12)( −14)

= 28 − 168 = −140

4 3

63. u ⋅ v = 6, − 1 ⋅ 2, 5 = 6(2) + ( −1)(5) = 7

64. u ⋅ v = ( 8i − 7 j ) ⋅ ( 3i − 4 j) = 24 + 28 = 52

−4 − 3 −2 −1 −1 −2

65. u ⋅ u = (6)(6) + ( − 4)( − 4) = 52, scalar

22 + 12 =

66. v − 3 =

−3 −4

5 − 3, scalar

u⋅v u v

74. cosθ =

67. (u ⋅ v ) v = (6)( 2) + ( − 4)(1) 2, 1

−18 − 2

= 8 2, 1

40 10 = −1  θ ≈ 180°

= 16, 8 , vector

68. (u ⋅ v ) u = (6)( 2) + ( − 4)(1) 6, − 4

8 6

= 8 6, − 4

−8 − 6 −4

u⋅v = u v

(

−8 24

)( ) 3

70. u = 3,

3 , v = 4, 3 3

cos θ =

(3)(4) + u⋅v = u v 12

71.

v = cos

cosθ =

1 2

, −

8 6

2 −4 − 2 −2

8 10 12

−6 −8

))

76. cosθ = =

u⋅v u v 42.93 − 11.2

35.93 81.61 ≈ 0.5860 θ ≈ 54.1° y

2 2 , 2 2

1 3 v = cos300°i + sin300° j = , − 2 2 cosθ =

−4

11π 12

u = cos 45°i + sin 45° j =

720 − 17.36 83.2 109.61

)) ( (

u⋅v = u v

≈ 0.5722  θ ≈ 55.098°

− 3 2 2 − 1 2 2 u⋅v = u v (1)(1)

θ ≈ 165° or

72.

(

−8

75. cos θ =

5π 5π 3 1 i + sin j= − , 6 6 2 2

(

−6

 θ ≈ 160.5°

21  θ ≈ 22.41° ≈ 0.391 radian 2 149 7π 7π i + sin j= 4 4

−4

( 3)(3 3) ( )( 43)

u = cos

v −2

69. u = 2 2, − 4 , v = − 2, 1 cosθ =

= 48, − 32 , vector

2 4− 6 4 u⋅v =  θ = 105° (1)(1) u v

8 6

4 2

−10 −8 − 6 −4

−2

−4 −6 −8

Chapter 7 Review

77. u = 12, − 8 , v = 6, 9 u ⋅ v = (12)(6) + ( − 8)(9)

= 0  u and v are orthogonal.

78. u ⋅ v = 8, − 4 ⋅ 5, 10 = 40 − 40 = 0  orthogonal 79. u = 8, 5 , v = −2, 4 u ⋅ v = 8(−2) + (5)(4) = 4 ≠ 0  u and v are not orthogonal. u ≠ kv  u and v are not parallel. Neither 3 3 80. − v = − 20, − 68 = −15, 51 = u  parallel. 4 4

81. 1, − k ⋅ 1, 2 = 1 − 2 k = 0  k = 12 82.

− 2, 4 ⋅ 6, − k = −12 − 4k = 0  k = − 3

83.

k , − 1 ⋅ 2, − 2 = 2 k + 2 = 0  k = −1

84.

k , − 2 ⋅ 1, 4 = k − 8 = 0  k = 8

88. u = −3, 5 , v = −5, 2  u⋅v  25 −5, 2 projv u =  2  v =  v  29   125 50 38 95 = u − projv u = −3, 5 − − , , 29 29 29 29 u= −

125 50 38 95 + , , 29 29 29 29

89. 48 inches = 4 feet

Work = 18,000 ( 4 ) = 72,000 ft ⋅ lb 90. Force = 500 sin12° ≈ 104 lbs 91. z = 5 + 3i Imaginary axis 5 4

85. u = −4, 3 , v = −8, − 2  u⋅v  13  26  projv u =  2  v =   −8, − 2 = − 4, 1  v  17  68    u − projv u = −4, 3 −

−52 −13 , 17 17

5 + 3i

3 2 1 −2 −1 −1

52 13 16 64 , − + − , − 17 17 17 17

86. u = 5, 6 , v = 10, 0  u⋅v   50  projv u =  2  v =   10, 0 = 5, 0  v   100    u − projv u = 5, 6 − 5, 0 = 0, 6

u = 5, 0 + 0, 6

5 5 9 9 + − , u= − , 2 2 2 2

Real axis

−3

z =

( 5) + ( 3) = 34 2

92. z = −10 − 4i Imaginary axis 4 2 −10 −8 −6 −4

Real axis

−2 −4

− 10 − 4i

−6 −8

z =

( −10 ) + ( −4 ) = 116 = 2 29 2

93. z = − 7i Imaginary axis

87. u = 2, 7 , v = 1, − 1

 u⋅v  −5 5 5 projvu =  2  v = 1, − 1 = − ,  v  2 2 2   5 5 9 9 = , u − projvu = 2, 7 − − , 2 2 2 2

−2

16 64 = − , 17 17 u= −

655

1 − 4 −3 −2 −1

Real axis

−2 −3 −4 −5 −6 −7

z =

−7i

02 + ( − 7)

= 7

656

Chapter 7

Additional Topics in Trigonometry

94. z = 6i 102.

Imaginary axis

3 1 cos( 230°− 95°) + i sin( 230°− 95°)  = ( cos135°+ i sin135°) 9 3

7 6

5 4 3 2 1 −4 −3 −2 −1

Real axis

02 + 62 = 6

z = 95.

z = 4i

θ =

π π  104. (a) 4 + 4i = 4 2  cos + i sin  4 4  5π 5π   −1 − i = 2  cos + i sin 4 4  

0 + 42 = 4

z =

π 2

π π  z = 4 cos + i sin  2 2  96.

π π  5π 5π   (b) 4 2  cos + i sin  2  cos + i sin 4 4 4 4     3π 3π   = 8  cos + i sin  = −8i 2 2   (c) (4 + 4i )( −1 − i ) = −4 − 4i − 4i + 4 = −8i

z = −9 z =

( − 9) + 0 2 = 9 2

θ = π z = 9(cos π + i sin π )

97.

z = − 3 −i z = 3 +1 = 2

7π 6 7π 7π   z = 2  cos + i sin 6 6  

θ=

98.

z = − 3 3 + 3i z =

θ =

(− 3 3 ) + 3 = 6 2

5π 6

5π 5π   + i sin z = 6 cos  6 6  

5 π π   π π  3π   3π 99.  cos + i sin 4cos + i sin  =10cos + i sin  2 2 2 4 4 4 4           2π π  5π   2π π   5π 100. 6 cos +  + i sin +  = 6 cos + i sin  6 6  3 6     3 6 101.

7π 7π   103. (a) 2 − 2i = 2 2  cos + i sin 4 4   π π  3 + 3i = 3 2  cos + i sin  4 4  7π 7π  π π   (b) 2 2  cos + i sin 3 2  cos + i sin  4 4  4 4   = 12 ( cos2π + i sin2π ) = 12

20 ( cos320° + i sin 320° ) 5 ( cos80° + i sin80° )

7π 7π   105. (a) 3 − 3i = 3 2  cos + i sin 4 4   π π  2 + 2i = 2 2  cos + i sin  4 4  7 π 7 π   + i sin 3 2  cos  3π 3π  4 4  3  (b) =  cos + i sin  π π   2 2 2  2 2  cos + i sin  4 4  3 3 = ( −i ) = − i 2 2 3 − 3i ( 2 − 2i ) 6 − 12i − 6 12i 3 (c) ⋅ = =− =− i 2 + 2i ( 2 − 2i ) 8 8 2

5π 5π   106. (a) −1 − i = 2  cos + i sin  4 4   5π 5π   −2 − 2i = 2 2  cos + i sin  4 4   −1 − i 2 1 = ( cos0 + i sin 0 ) = 2 − 2i 2 2 2 −1 − i −1 − i 1 = = (c) − 2 − 2i 2(−1 − i ) 2

(b) −

= 4 [ cos240° + i sin 240°]

Chapter 7 Review

657

  π π  4π   4π 107.  5  cos + i sin   = 54  + i sin 12 12 12 12      π π  = 625  cos + i sin  3 3  1 3  = 625  + i  2 2    =

625 625 3 + i 2 2

  4π 4π   4π  5  4π 108.  2  cos + i sin + i sin  = 2   15 15 3 3       1 3  = 32  − − i  2 2    = −16 − 16 3i

109. (2 + 3i )6 ≈  13 ( cos56.3° + i sin 56.3° )    = 133 ( cos337.9° + i sin 337.9° )

110. (1 − i )8 =  2 ( cos315° + i sin 315° ) 

= 16 ( cos2520° + i sin 2520° ) = 16 ( cos0° + i sin 0° ) = 16 5π 5π   111. − 3 + i = 2  cos + i sin 6 6   Square roots: 5π 5π   + i sin ≈ 0.3660 + 1.3660i 2  cos 12 12   17π 17π   + i sin ≈ − 0.3660 − 1.3660i 2  cos 12 12  

112. 6

≈ 133 ( 0.9263 − 0.3769i ) ≈ 2035 − 828i

11π 11π   + i sin 3 − i = 2  cos 6 6   Square roots: 11π 11π   + i sin ≈ −1.3660 + 0.3660i 2  cos 12 12   23π 23π   + i sin ≈ 1.3660 − 0.3660i 2  cos 12 12  

3π 3π   113. − 4i = 4 cos + i sin  2 2  

Square roots:

 3π 3π  2 2  4  cos + i sin +i  = −  = 2 − 4 4  2    2  2 7π 7π  2  4  cos + i sin −i  =  = 2 4 4  2    2

2 +

2 −

2i 2i

3π 3π   114. − 9i = 9 cos + i sin  2 2  

Square roots:

 3π 3π  2 2 3 2 3 2  9  cos + i sin +i + i  = −  = 3 − 4 4  2  2 2   2  2 7π 7π  2 3 2 3 2  9  cos + i sin −i − i  =  = 3 4 4  2  2 2   2

658

Chapter 7

Additional Topics in Trigonometry

3π   3π 115. (a) Sixth roots of − 729i = 729 cos + i sin  : 2 2    3 π 2 2 π 3 π 2 + k ( ) ( ) + 2kπ  , 6 729  cos + i sin  6 6   k = 0, 1, 2, 3, 4, 5

π π  116. (a) Fourth roots of 256i = 256  cos + i sin  : 2 2   π 2 + 2 k π π 2 + 2 ( ) ( ) kπ  4 256  cos + i sin  4 4   k = 0, 1, 2, 3

π π  k = 0 : 3  cos + i sin  4 4  7π 7π   + i sin k = 1 : 3  cos 12 12  

π π  k = 0 : 4  cos + i sin  8 8  5π 5π   k = 1: 4  cos + i sin 8 8  

11π 11π   + i sin k = 2 : 3  cos 12 12   5π 5π   + i sin k = 3 : 3  cos 4 4  

9π 9π   k = 2 : 4  cos + i sin 8 8   13π 13π   k = 3 : 4  cos + i sin 8 8  

19π 19π   + i sin k = 4 : 3  cos 12 12   23π 23π   + i sin k = 5 : 3  cos 12 12  

(b)

Imaginary axis

5 3 1

(b)

Imaginary axis

−3

1 2 3

Real axis

−2 −3

5 4

−5 1 −4

−2

4 5

Real axis

(c)

−2 −4 −5

(c)

3.696 + 1.531i, − 1.531 + 3.696i, − 3.696 − 1.531i, 1.531 − 3.696i

117. (a) Cube roots of 8 = 8(cos0 + i sin 0) :

  2π k  2π k  8  cos   + i sin 3  , k = 0, 1, 2 3     k = 0 : 2 ( cos 0 + i sin 0 )

3 2 3 2 + i, − 0.7765 + 2.898i, − 2.898 + 0.7765i, 2 2 3 2 3 2 − − i, 0.7765 − 2.898i, 2.898 − 0.7765i 2 2 i, 0.7765 − 2.898i, 2.898 − 0.07765i

2π 2π   k = 1 : 2  cos + i sin 3 3   4π 4π   + i sin k = 2 : 2  cos  3 3  

(b)

Imaginary axis

1 −3

−1

Real axis

−3

Chapter 7 Review 118. (a) Fifth roots of −1024 = 1024(cosπ + i sin π ) :

119. x 4 + 256 = 0 x 4 = −256 = 256(cos π + i sin π )

π + 2π k π + 2π k   1024  cos + i sin , 5 5   k = 1, 2, 3, 4 5

  π + 2π k   π + 2π k   −256 = 4 cos   + i sin   4 4      k = 0,1, 2, 3 4

π π  k = 0 : 4  cos + i sin  5 5  3π 3π   k = 1: 4  cos + i sin  5 5  

π π 4 2 4 2  k = 0 : 4  cos + i sin  = i + 4 4 2 2  = 2 2 + 2 2i

5π 5π   k = 2 : 4  cos + i sin  = 4(cos π + i sin π ) 5 5   7π 7π   + i sin k = 3: 4  cos  5 5   9π 9π   + i sin k = 4 : 4  cos  5 5   (b)

3π 3π  4 2 4 2  k = 1: 4  cos i + i sin  = − + 4 4 2 2   = −2 2 + 2 2i 5π 5π  4 2 4 2  k = 2 : 4  cos i + i sin − =− 4 4 2 2   = −2 2 − 2 2i

Imaginary axis

7π 7π  4 2 4 2  k = 3: 4  cos i + i sin − = 4 4  2 2 

1 −3 −2 − 1

659

2 3

= 2 2 − 2 2i

Real axis

Imaginary axis 5

−5

(c)

3 2

3.236 ± 2.351i, − 1.236 ± 3.804i, − 4 −5

−3 −2

2 3

Real axis

−2 −3 −5

π π  120. x 5 = 32i = 32  cos + i sin  2 2   π π + 2 2 k    π 2 + 2π k   5 32  cos   + i sin   , 5 5      k = 0, 1, 2, 3, 4

π π   k = 0 : 2  cos + i sin  10 10   π π  k = 1: 2  cos + i sin  = 2i 2 2  9π 9π   + i sin k = 2 : 2  cos 10 10   13π 13π   + i sin k = 3 : 2  cos 10 10   17π 17π   + i sin k = 4 : 2  cos 10 10  

Imaginary axis

5 4 3 1 − 4 −3

3 4 5

Real axis

−3 −4 −5

660

Chapter 7

Additional Topics in Trigonometry

122. x 4 = −81 = −81(cos π + i sin π )

121. x 3 + 8i = 0 3

x = −8i 3π 3π   −8i = 8  cos + i sin  2 2    ( 3π 2 + 2π k ) + i sin ( 3π 2 + 2π k )  , k = 0, 1, 2 −8i = 3 8 cos  3 3   π π  k = 0 : 2  cos + i sin  = 2i 2 2  3

7π 7π   k = 1 : 2  cos + i sin = − 3 −i 6 6   11π 11π   k = 2 : 2  cos + i sin = 3 −i 6 6  

π + 2π k π + 2π k )   81  cos + i sin  , k = 0, 1, 2, 3 4 4   π π  k = 0 : 3  cos + i sin  4 4  4

3π 3π   k = 1 : 3  cos + i sin  4 4   5π 5π   + i sin k = 2 : 3  cos 4 4   7π 7π   + i sin k = 3 : 3  cos 4 4   Imaginary axis

5 4

Imaginary axis

−4

−2

4 5

Real axis

−2 1

−3

−1

−4 −5

Real axis

123. True. sin 90° is defined in the Law of Sines.

−3

124. False. There may be no solution, one solution, or two solutions.

Chapter 7 Test 1. A = 36°, B = 98°, c = 16 C = 180° − 36° − 98° = 46°

c sin A ≈ 13.07 sin C a b= sin B ≈ 22.03 sin A a=

2. a = 2, b = 4, c = 5 a 2 + b2 − c 2 2 ab 4 + 16 − 25 = = −0.3125  C ≈ 108.21° 2(2)(4)

cos C =

a 2 + c 2 − b2 2 ac 4 + 25 − 16 = = 0.65  B ≈ 49.46° 2(2)(5)

cos B =

3. A = 35°, b = 8, c = 12 a 2 = b 2 + c 2 − 2bc ⋅ cos A = 64 + 144 − 2(8)(12)cos35° ≈ 50.7228 a ≈ 7.12 sin A sin B = b ≈ 0.6443  B ≈ 40.11° a C = 180° − A − B ≈ 104.89°

4. B = 24°, C = 68°, b = 12.2 A = 180° − B − C = 180° − 24° − 68° = 88° a =

12.2 sin 88° b sin A = ≈ 29.98 sin B sin 24°

c =

b sin C 12.2 sin 68° = ≈ 27.81 sin B sin 24°

b2 + c2 − a 2 2bc 16 + 25 − 4 = = 0.925  A ≈ 22.33° 2(4)(5)

cos A =

Chapter 7 Test

5. A = 25°, b = 28, a = 18 sin A (sin 25°)28 b= ≈ 0.6574 a 18 B1 ≈ 41.10° or B2 ≈ 180° − 41.1° ≈ 138.90°

sin B =

7. Law of Cosines: a 2 = b 2 + c 2 − 2bc cosθ = 480 2 + 5652 − 2(480)(565)cos80° = 455,438.2  a ≈ 674.9 ft

Case 1: B1 = 41.10° C = 180° − A − B1 ≈ 113.90°

a + b + c 55 + 85 + 100 = = 120 2 2 A = s( s − a )(s − b)(s − c )

8. s =

a sin C ≈ 38.94 sin A Case 2 : B2 = 138.90° c=

= 120(65)(35)(20) ≈ 2336.7 square meters

C = 180° − A − B2 ≈ 16.10° c=

a sin C ≈ 11.81 sin A

661

9. w = 4 − ( −8), 1 − ( −12) = 12, 13 w = 12 2 + 132 = 313 ≈ 17.69

6. No triangle possible (5.2 ≤ 10.1) 10. u = − 7, − 4 , v = 4, 6 (a) 2 v + u = 2 4, 6 + − 7, − 4 = 8, 12 + − 7, − 4 = 1, 8 (b) u − 3v = − 7, − 4 − 3 4, 6

= − 7, − 4 − 12, 18 = −19, − 22 (c) 5u − 4 v = 5 − 7, − 4 − 4 4, 6 = − 35, − 20 − 16, 24 = − 51, − 44 11. u = −5, 2 , v = −1, − 10 (a)

(b)

u − 3 v = ( 2 i + 3 j ) − 3 ( − i − 2 j ) = 5i + 9 j

= −2, − 20 + −5, − 2

(c)

5u − 4 v = 5 ( 2i + 3 j) − 4 ( −i − 2 j) = 14i + 23 j

u − 3v = −5, 2 − 3 −1, − 10 = −5, 2 − −3, − 30 = −2, 32

(c)

5u − 4 v = 5 −5, 2 − 4 −1, − 10 = −25, 10 − −4, − 40 = −21, 50

12. (a) (b) (c)

2v + u = 2 ( −i − 2 j) + ( 2i + 3 j) = − j

2v + u = 2 −1, − 10 + −5, 2 = −7, − 18

(b)

13. (a)

2 v + u = 2 ( 6 i + 9 j ) + ( i − j) = 13i + 17 j u − 3v = ( i − j) ) − 3 ( 6 i + 9 j ) = −17i − 28 j

5u − 4 v = 5 ( i − j) − 4 ( 6i + 9 j) = −21i − 41j

14.

v = 24i − 7 j u = v =

v v

(24) + (− 7) 2

1 (24i − 7 j) 25 24 7 u = i − j = 25 25

476 + 49 =

625 = 25

u =

15. 12

3, −5 3, −5

= =

12 34

24 7 ,− 25 25

3, −5 =

36 34

−60 34

18 34 −30 34 , 17 17

662

Chapter 7

Additional Topics in Trigonometry

16. 250 ( cos 45°i + sin 45° j ) , first force

130 ( cos ( −60°) i + sin ( −60°) j) , second force

  2  2  3   1   + 130    i + 250  Resultant: 250   + 130  −   j  2    2        2   2  

(

) (

250

)

45°

= 125 2 + 65 i+ 125 2 − 65 3 j Magnitude:

u = −9, 4 , v= 1, 2

  5π 5π   40π 40π  8 + i sin 23. 3  cos   = 3  cos 6 + i sin 6  6 6        1 3  i = 38  − +  2 2   

u ⋅ v = −9, 4 ⋅ 1, 2

= ( −9 )(1) + ( 4 )( 2 ) = −9 + 8 = −1

18. cos θ =

130

 125 2 − 65 3  Direction: θ = arctan   θ ≈ 14.9°  125 2 + 65   

17.

− 60°

(125 2 + 65) + (125 2 − 65 3 ) ≈ 250.15 2

= −3280.5 + 3280.5 3i 6561 6561 3i =− + 2 2

−8 u⋅v =  θ ≈ 105.9° u v 53 ( 4 )

19. Find the dot products. u ⋅ v = 6, −10 ⋅ − 5, − 3 = (6)( − 5) + ( −10)( − 3)

24.



−37 185 37 , −5, − 1 = = w1 26 26 26

w 2 = u − w1 = 6, 7 −

21.

185 37 −29 145 , = , 26 26 26 26



7π 7π   + i sin  4 4  

42π 42π   = 5832  cos + i sin = 5832i 4 4  

= − 30 + 30 = 0 Yes. Because u ⋅ v = 0, u and v are orthogonal.

20. projv u =



( 3 − 3i ) = 3 2  cos

1 3  π π  25. 128 1 + 3i = 256  +  2 2 i  = 256  cos 3 + i sin 3     + 2π k (π 3) + 2π k  π 3 + i sin Fourth roots: 4 256  cos  4 4  

(

)

u = w1 + w 2

k = 0, 1, 2, 3

z =2 2

π π   Four roots are: 4  cos + i sin  ≈ 3.8637 + 1.0353i 12 12  

3π 3π   z = 2 2  cos + i sin 4 4  

22. 100(cos240° + i sin 240°) = −50 − 50 3i

7π 7π   4  cos + i sin ≈ −1.0353 + 3.8637i 12 12   13π 13π   4  cos + i sin ≈ −3.8637 − 1.0353i 12 12   19π 19π   4  cos + i sin  ≈ 1.0353 − 3.8637i 12 12  

Chapters 5–7 Cumulative Test

663

26. x 4 = 625i

π π  Fourth roots of 625i = 625  cos + i sin  2 2    (π / 2 ) + 2π k   (π / 2 ) + 2π k   625  cos   + i sin     4 4      k = 0, 1, 2, 3 4

π π  Four roots are: 5  cos + i sin  8 8  5π 5π   + i sin 5  cos  8 8   9π 9π   + i sin 5  cos  8 8   13π 13π   + i sin 5  cos  8 8   Imaginary axis 8 6

−8 −6

Real axis

−4 −6 −8

Chapters 5–7 Cumulative Test 1. θ = −150° (a)

 180°  2. 2.55 rad = 2.55   ≈ 146.1°  π 

441 841  21  = 3. sec 2 θ = 1 + tan 2 θ = 1 +  −  = 1 + 400 400  20 

841 29 = (since θ lies in Quadrant IV) 400 20 20 cos θ = 29 sec θ =

θ = − 150°

−150° + 360° = 210° π 5 (c) ( −150° ) = − π radians 180° 6 (d) θ ′ = 30° 1 (e) sin θ = csc θ = −2 2

(b)

cosθ = − tan θ =

3 2

3 3

sec θ = − cot θ =

3 3

=−

4. f ( x ) = 3 − 2 sin π x y

3 2

2 3 3

−3

= 3

−2

−1

664

Chapter 7

Additional Topics in Trigonometry

8. cos (arccos 3.6) is not possible. The domain of arccosine

1 π  tan  x −  2 2 

5. f ( x) =

is [−1, 1].

Period: π Vertical asymptotes at x − x −

π 2

= − =

π 2

 x = 0

 x  π

3 9. Let θ = arctan . 4 3 tan θ = 4

3 3  sin  arctan  = 4 5  −π

−π 2

π 2

−1

−2

θ 4

6. f ( x ) = sec( x + π ) − 2

10. Let θ = arccos 2x.

Period: 2π Vertical shift downward two units

1 − 4x2

3 2

− 2π

−1

2π

sin (arccos 2 x ) = sin θ =

−2

11. 7.

)

(

1 − 4 x2

sin θ − 1 cosθ sin 2 θ − 2sin θ + 1 − cos2 θ − = cosθ sin θ − 1 cosθ ( sin θ − 1)

Amplitude: 3 Cosine curve reflected about the x-axis Period: 2  h ( x ) = −3cos (π x ) Answer: a = −3, b = π , c = 0

(

1 − 4x2 = 1

= =

sin 2 θ − 2sin θ + sin 2 θ cosθ ( sin θ − 1) 2sin θ ( sin θ − 1) cosθ ( sin θ − 1)

= 2 tan θ

)

12. cot 2 α sec 2 α − 1 = cot 2 α tan 2 α = 1 13. sin ( x + y ) sin ( x − y ) = [sin x cos y + cos x sin y ][sin x cos y − sin y cos x ]

= sin 2 x cos 2 y − sin 2 y cos 2 x

= sin 2 x (1 − sin 2 y ) − sin 2 y (1 − sin 2 x ) = sin 2 x − sin 2 x sin 2 y − sin 2 y + sin 2 y sin 2 x = sin 2 x − sin 2 y 1 2 ( 2 sin x cos x ) 4 1 2 = ( sin 2 x ) 4 1 1 − cos 4 x = ⋅ 4 2 1 = (1 − cos 4 x ) 8

14. sin 2 x cos2 x =

15. sin 2 x + 2sin x + 1 = 0

( sin x + 1) = 0 2

sin x + 1 = 0 sin x = −1 x=

3π + 2 nπ 2

Chapters 5–7 Cumulative Test

16. 3tan θ − cot θ = 0 1 =0 3tan θ − tan θ 3tan 2 θ − 1 = 0 tan θ = ±

2 2 22. sec θ = tan θ + 1 =

cosθ = −

1 3

θ =

π 6

+ nπ ,

5π + nπ 6

sin

17. Graph y = cos2 x − 5cos x − 1 on [ 0, 2π ) .

Roots are x ≈ 1.7646, 4.5186.

18.

2π

=±

3 ( Quadrant III ) 5

1 − cosθ 2 2 5 =± =± 2 5 5

in Quadrant II: sin

θ 2

2 5 5

sin x cos2 x = sin x cos x cos x 1 1 = ( 2sin x cos x ) = sin 2 x 2 2

24. tan x (1 − sin 2 x ) =

− 10

Zeros: x ≈ 1.047, 5.236 1 + sin x cos x Algebraically: + =4 cos x 1 + sin x

1  cos ( 3θ − θ ) − cos ( 3θ + θ )  2 1 = ( cos 2θ − cos 4θ ) 2

25. sin 3θ sin θ =

1 + 2sin x + sin 2 x + cos2 x =4 cos x (1 + sin x ) 2 + 2sin x =4 cos x (1 + sin x )

1 ( sin ( 3 x + 2 x ) + sin ( 3 x − 2 x ) ) 2 1 = ( sin 5 x + sin x ) 2

26. sin 3 x cos 2 x =

2 =4 cos x 1 cos x = 2 x=

19.

25 5  secθ = −  9 3

 8x + 4 x   8x − 4 x  23. cos8 x + cos 4 x = 2 cos   cos  2  2     = 2 cos6 x cos 2 x

2 0

665

π 5π ,

3 3

27.

sin A b = 0.2569  B ≈ 14.9° a C = 180° − 46° − 14.9° = 119.1°

28. sin B = 2π

2 cos3 x 2 cos3 x 1 = = = csc x sin 4 x − sin 2 x 2 cos3 x ⋅ sin x sin x

c= −12

Zeros: x ≈ 0.785, 3.927 Algebraically: tan3 x − tan2 x + 3tan x − 3 = 0

tan2 x ( tan x − 1) + 3( tan x − 1) = 0

( tan x + 3) ( tan x − 1) = 0 2

π 5π tan x = 1  x = , 4 4 12 5 12  cos u =  tan u = 13 13 5 3 4 4 cos v =  sin v =  tan v = 5 5 3 (12 5 ) − ( 4 3 ) = 16 tan u − tan v tan ( u − v ) = = 1 + tan u tan v 1 + (12 5 )( 4 3 ) 63

20. sin u =

21. tan ( 2θ ) =

29.

a ( sin C ) ≈ 17.0 sin A

A = 32°, b = 8, c = 10 a 2 = b 2 + c 2 − 2bc cos A = 64 + 100 − 2(8)(10)cos32° = 28.3123  a  5.32 b sin A sin B = ≈ 0.7967  B ≈ 52.82° a C = 180° − B − A = 95.18°

30. B = 180° − 24° − 101° = 55° a sin B ≈ 20.14 b= sin A a sin C ≈ 24.13 c= sin A

2 (1 2 ) 2 tan θ 4 = = 1 − tan 2 θ 1 − (1 4 ) 3

666

Chapter 7

Additional Topics in Trigonometry

b2 + c2 − a2 = 0.8982  A ≈ 26.1° 2bc a 2 + c 2 − b2 cos B = = 0.8355  B ≈ 33.3° 2bc C = 180° − 26.1° − 33.3° = 120.6°

31. cos A =

1 1 32. A = bh = 19 ⋅ 14sin82° 2 2 ≈ 131.7 sq. in.

3π 3π 3π   + i sin : 3 2  cos 4 4 4  

 3 1  40. 6 3  − + i = −9 + 3 3i  2 2   

41.  4 ( cos30° + i sin30° )  6 ( cos120° + i sin120° )  = 24 ( cos ( 30° + 120° ) + i sin ( 30° + 120° ) ) = 24 ( cos150° + i sin150° )

 3 1 = 24  − + i  = −12 3 + 12i  2  2  

39. z = 3 2, θ =

82°

42. 2 + i = 5 ( cosθ + i sin θ ) , where tan θ =

33. s =

 θ ≈ 0.4636

30 + 41 + 45 = 58 2

Area = =

θ θ  Square roots: z1 = 51 4  cos + i sin  2 2 

s( s − a )( s − b)( s − c)

  θ + 2π   θ + 2π   z2 = 51 4  cos   + i sin  2   2      z1 ≈ 1.4533 + 0.3436i

58( 28)(17)(13) ≈ 599.09 square meters

34. u = 3,5 = 3i + 5 j

z2 ≈ −1.4553 − 0.3436i

35. v = 4i − 2 j v =

42 + ( − 2)

Unit vector: u =

= 1

2 5

20 = 2 5

(4i − 2 j)

2 1 ,− 5 5

2 5 5 ,− 5 5

2 5 5 i − j 5 5

36. u = 3i + 4 j, v = 6i − 9 j u ⋅ v = (3)(6) + ( 4)( − 9) = −18

1, 2k . 2, − 1 = 0

8 − 10 1 5 1, 5 = − , − = w1 26 13 13

w 2 = u − w1 = 8, − 2 − − 105 21 ,− 13 13 u = w1 + w 2 =

 0 + 2π k   0 + 2π k  cos  , k = 0, 1, 2  + i sin  3 3     k = 0 : cos0 + i sin 0 = 1 k = 1: cos

2π 2π 1 3 + i sin =− + i 3 3 2 2

k = 2 : cos

4π 4π 1 3 + i sin =− − i 3 3 2 2

44. x 4 + 1296 = 0 x 4 = − 625

2 − 2k = 0 k =1

38. projvu =

43. 1 = 1( cos 0 + i sin 0 )

Four fourth roots of −1296 = 1296(cos π + i sin π ) are:

u⋅v = 0

37.

1 2

1 5 ,− 13 13

π + 2π k π + 2π k   + i sin 1296  cos ; k = 0, 1, 2, 3 4 4  

π π  6 cos + i sin  4 4  3π 3π   + i sin 6 cos  4 4   5π 5π   + i sin 6 cos  4 4   7π 7π   + i sin 6 cos  4 4  

Chapters 5–7 Cumulative Test

45. tan18° =

h 200

tan16°45′ =

k 200

Hence, f = h − k = 200 tan18° − 200 tan16°45′' ≈ 4.8 ≈ 5 feet.

48. cos A =

602 + 1252 − 1002 = 0.615  A ≈ 52.05° 2 ( 60 )(125)

cos B =

1002 + 1252 − 602 = 0.881  B ≈ 28.24 2 (100 )(125 )

667

Angle between vectors = A + B ≈ 80.3° 60

100

16° 45′ 200

18°

125

A 60

Not drawn to scale

46. Given the maximum displacement is 7 inches when  2π  t = 0, use d = a cos t  with a period of 8 seconds.  b  a = 7 and

2π π = 8  b = 4 b

So, d = 7 cos

47. 30°

v1 = 500(cos 30°, sin 30°) = airplane

135°

v2 = 50(cos 135°, sin 130°) = wind

45°

v = v1 + v 2 = 500 cos30°, sin30° + 50 cos135°, sin135° v ≈ 397.7, 285.4 v ≈

( 397.7 ) + ( 285.4 ) ≈ 489.45 km/hr 2

θr

 285.4   ≈ 35.66°  θ = 90° − θr = 54.34°  397.7 

θr = tan −1 

The direction of the airplane is 54.34° at an airspeed relative to the ground of 489.45 km/hr.

C H A P T E R 8 Linear Systems and Matrices Section 8.1

Solving Systems of Equations ...........................................................670

Section 8.2

Systems of Linear Equations in Two Variables ................................682

Section 8.3

Multivariable Linear Systems ............................................................694

Section 8.4

Matrices and Systems of Equations ...................................................712

Section 8.5

Operations with Matrices ...................................................................726

Section 8.6

The Inverse of a Square Matrix .........................................................737

Section 8.7

The Determinant of a Square Matrix .................................................745

Section 8.8

Applications of Matrices and Determinants ......................................754

Chapter 8 Review .......................................................................................................761 Chapter 8 Test ............................................................................................................784

C H A P T E R 8 Linear Systems and Matrices Section 8.1 Solving Systems of Equations 1.

system, equations

solution

substitution

solution

break-even point

If the graphs of the equations of a system do not intersect, then the system has no solution.

(a)

(b)

− ( −2 ) − ( −9 ) = 11

7≠3 11 = 11 No, ( −2, − 9 ) is not a solution. (c)

4 ( 0 ) − ( −3 ) = 1

15 ≠ 3

− 92 ≠ 11

No, ( − 32 , 6 ) is not a solution.

3≠1 −3 ≠ −6

(d)

No, ( 0, − 3) is not a solution.

3=3 11 = 11 Yes, ( − 74 , − 374 ) is a solution.

6 ( −1) + ( −5 ) = 6

(c)

4 (−

6(−

(a)

) − ( −3 ) = 1

0 ≠ −2e −2 −6 ≠ 2

) + ( 3 ) =− 6

No, ( −2, 0 ) is not a solution.

−9 ≠ 1 −6 = −6

No, ( − , 3 ) is not a solution. 3 2

(d)

4(−

1 2

6(−

1 2

(b)

) − ( −3 ) = 1

−2 = −2 2=2 Yes, ( 0, − 2 ) is a solution.

) + ( −3 ) = 6 1=1

Yes, ( − , − 3 ) is a solution. 1 2

(a)

(c)

−3 ≠ −2 3≠2 No, ( 0, − 3) is not a solution.

−2 − ( −13 ) = 11

3=3 11 = 11 Yes, ( 2, − 13 ) is a solution.

−3 =− 2e0 3 ( 0 ) − ( −3 ) = 2

4 ( 2 ) + ( −13 ) = 3 2

−2 =− 2e0 3 ( 0 ) − ( −2 ) = 2

−6 = −6

0 =− 2e−2 3 ( −2 ) − 0 = 2

3 2

4 ( − 1) − ( − 5 ) = 1

1=1 −11 ≠ −6 No, ( −1, − 5 ) is not a solution.

4 ( − 74 ) + ( − 374 ) = 3 − ( − 74 ) − ( − 374 ) = 11

3 2

4 ( − 23 ) + ( 6 ) = 3 − ( − 23 ) − ( 6 ) =11

6 ( 0 ) + ( −3 ) =− 6

(b)

4 ( −2 ) + ( −9 ) = 3

(d)

−5 =− 2e−1 ?

3 ( −1) − ( −5 ) = 2

−5 ≠ −2e −1 2=2 No, ( −1, − 5 ) is not a solution. 670

Section 8.1

10. (a)

− log10 (100 ) + 3 = 1 1 9

(100 ) + 1 = 289 1=1 109 9

≠

28 9

No, (100, 1) is not a solution. (b)

− log10 10 + 3 = 2 1 9

(10 ) + 2 = 289 2=2 = 289

28 9

Yes, (10, 2 ) is a solution. (c)

− log10 1 + 3 = 3 1 9

(1) + 3 =

28 9

3=3 28 9

28 9

Yes, (1, 3 ) is a solution. (d)

Solving Systems of Equations

13.  x − y = −4 Equation 1  2  x − y = −2 Equation 2 Solve for y in Equation 1: y = x + 4

Substitute for y in Equation 2: x 2 − ( x + 4 ) = −2 Solve for x: x 2 − x − 2 = 0  ( x + 1)( x − 2 ) = 0  x = −1, 2 Back-substitute x = −1 : y = −1 + 4 = 3 Back-substitute x = 2 : y = 2 + 4 = 6

Answer: ( −1, 3 ) , ( 2, 6 ) 14. −2 x + y = −5 Equation 1  2 2  x + y = 25 Equation 2

Solve for y in Equation 1: y = 2 x − 5 Substitute for y in Equation 2: x 2 + ( 2 x − 5) = 25 2

Solve for x: x 2 + 4 x 2 − 20 x + 25 = 25 5 x 2 − 20 x = 0

5x ( x − 4 ) = 0

− log10 1 + 3 = 1 1 9

(1) + 1 = 289 3≠1 10 9

≠ 289

No, (1, 1) is not a solution. 11.  2 x + y = 6 Equation 1  − x + y = 0 Equation 2 Solve for y in Equation 1: y = 6 − 2 x

Substitute for y in Equation 2: − x + ( 6 − 2 x ) = 0 Solve for x: − 3 x + 6 = 0  x = 2

x = 0, 4 Back-substitute x = 0 : y = −5 Back-substitute x = 4 : y = 3 Answer: ( 0, − 5 ) , ( 4, 3 ) 15.  3 x + y = 2 Equation 1  3  x − 2 + y = 0 Equation 2 Solve for y in Equation 1: y = 2 − 3 x

Substitute for y in Equation 2: x 3 − 2 + ( 2 − 3 x ) = 0

(

12.  x − y = −4 Equation 1   x + 2 y = 5 Equation 2

Solve for x in Equation 1: x = y − 4 Substitute for x in Equation 2: ( y − 4 ) + 2 y = 5 Solve for y: 3 y − 4 = 5  y = 3 Back-substitute y = 3 : x = 3 − 4 = −1

Answer: ( −1, 3)

)

Solve for x: x 3 − 3 x = 0  x x 2 − 3 = 0  x = 0, ± 3 Back-substitute: x = 0 : y = 2 x = 3 : y = 2−3 3

Back-substitute x = 2 : y = 6 − 2 ( 2 ) = 2

Answer: ( 2, 2 )

671

x =− 3 : y =2+3 3

Answer: ( 0, 2 ) ,

( 3, 2 − 3 3 ), ( − 3, 2 + 3 3 )

16.  x + y = 0 Equation 1  3  x − 5 x − y = 0 Equation 2

Solve for y in Equation 1: y = − x Substitute for y in Equation 2: x 3 − 5 x − ( − x ) = 0

(

)

Solve for x: x 3 − 4 x = 0  x x 2 − 4 = 0  x = 0, ± 2 Back-substitute x = 0 : y = −0 = 0 Back-substitute x = 2 : y = −2 Back-substitute x = −2 : y = − ( −2 ) = 2 Answer: ( 0, 0 ) , ( 2, − 2 ) , ( −2, 2 )

672

Chapter 8

Linear Systems and Matrices

17. − 72 x − y = −18 Equation 1  2 3 Equation 2 8 x − 2 y = 0

Solve for x in Equation 1: − 27 x = y − 18  x = − 27 y + 367 Substitute for x in Equation 2: 8 ( − 27 y + 367 ) − 2 y3 = 0 2

Solve for x:

(

)

−2 y 3 + 8 494 y 2 − 144 y + 3649 = 0 49 3

49 y − 16 y + 576 y − 5184 = 0

( y − 4 ) ( 49 y2 + 180 y + 1296 ) = 0 Hence, y = 4 and x = − 27 ( 4 ) + 367 = 4. Answer: ( 4, 4 ) 18.  y = x 3 − 3 x 2 + 4 Equation 1  Equation 2  y = −2 x + 4 Substitute for y in Equation 1: −2 x + 4 = x 3 − 3 x 2 + 4 Solve for x: 0 = x3 − 3x2 + 2 x

(

0 = x x − 3x + 2

Solve for x: 4 x + ( 2 x + 2 ) − 5 = 0  6 x − 3 = 0  x = 12

Back-substitute x = 12 : y = 2 x + 2 = 2 ( 12 ) + 2 = 3

Answer: ( 12 , 3 )

22. 6 x − 3 y − 4 = 0 Equation 1   x + 2 y − 4 = 0 Equation 2 Solve for x in Equation 2: x = 4 − 2 y

Substitute for x in Equation 1: 6 ( 4 − 2 y ) − 3 y − 4 = 0 Solve for y: 24 − 12 y − 3 y − 4 = 0  −15 y = −20  y = 43

)

0 = x ( x − 2 )( x − 1)  x = 0, 1, 2

Back-substitute x = 0 : y = −2 ( 0 ) + 4 = 4 Back-substitute x = 1 : y = −2 (1) + 4 = 2 Back-substitute x = 2 : y = −2 ( 2 ) + 4 = 0 Answer: ( 0, 4 ) , (1, 2 ) , ( 2, 0 ) 19.  x + y = 0  4 x + 3 y = 10

21. 2 x − y + 2 = 0 Equation 1   4 x + y − 5 = 0 Equation 2 Solve for y in Equation 1: y = 2 x + 2 Substitute for y in Equation 2: 4x + (2x + 2) − 5 = 0

Equation 1 Equation 2

Back-substitute y = 43 : x = 4 − 2 y = 4 − 2 ( 43 ) = 43 Answer: ( 43 , 43 )

23. 1.5 x + 0.8 y = 2.3  15 x + 8 y = 23  0.3 x − 0.2 y = 0.1  3 x − 2 y = 1

Solve for y in Equation 2: −2 y = 1 − 3 x 3x − 1 2  3x − 1  Substitute for y in Equation 1: 15 x + 8   = 23  2  15 x + 12 x − 4 = 23 y=

27 x = 27

Solve for y in Equation 1: y = − x Substitute for y in Equation 2 and solve for x: 4 x + 3( − x) = 10 4 x − 3 x = 10

Then, y =

3x − 1 = 2

3 (1) − 1 2

x =1

= 1.

Answer: (1, 1)

x = 10 Back-substitute in Equation 1: y = − x y = −(10) y = −10

24. − 0.5 x + 4 y = 7.8   0.2 x − 1.6 y = − 3.6

Solve for x in Equation 1: x = 1 − 2 y Substitute for x in Equation 2: 5 (1 − 2 y ) − 4 y = −23

Equation 2

Multiply both equations by 10: − 5 x + 40 y = 78

Answer: (10, −10) 20.  x + 2 y = 1 Equation 1  5 x − 4 y = −23 Equation 2

Equation 1

2 x − 16 y = − 36

Solve for x in Equation 2: x = 8 y − 18 Substitute for x in Equation 1: − 5(8 y − 18) + 40 y = 78 Solve for y: 90 − 40 y + 40 y = 78 90 ≠ 78

No solution

Solve for y: −14 y = −28  y = 2 Back-substitute y = 2 : x = 1 − 2 y = 1 − 4 = −3 Answer: ( −3, 2 )

Section 8.1

29.  x + y = 18,000  400 0.04 x + 0.02 y =

25.  15 x + 12 y = 8 Equation 1   x + y = 20 Equation 2 Solve for x in Equation 2: x = 20 − y

Substitute for x in Equation 1: 15 ( 20 − y ) + 12 y = 8

Equation 2

Solve for y in Equation 1: y = 18,000 − x Substitute for y in Equation 2: 0.04 x + 0.02(18,000 − x) = 400

Back-substitute y = 403 : x = 20 − y

Solve for x: 0.04 x + 360 − 0.02 x = 400

= 20 − 403 = 203

0.02 x = 40 x = 2000

Back-substitute: y = 18,000 − 2000 = 16,000

26.  12 x + 34 y = 10 Equation 1 3  4 x − y = 4 Equation 2 Solve for y in Equation 2: y = 34 x − 4

Answer: ( 2000, 16,000) $2000 at 4% and $16,000 at 2%

Substitute for y in Equation 1: 12 x + 43 ( 43 x − 4 ) = 10 Solve for x: 1 x + 169 x − 3 = 10  17 x = 13  x = 208 2 16 17 Back-substitute x = Answer: (

208 17

88 17

208 17

: y=

3 4

)

( )−4= 208 17

Equation 1 Equation 2

Substitute for y in Equation 2: 0.06 x + 0.03(18,000 − x) = 840 Solve for x: 0.06 x + 540 − 0.03x = 840 0.03 x = 300 x = 10,000

Solve for y in Equation 1: y = 5 + 35 x

Substitute for y in Equation 2: −5 x + 3 ( 5 + 53 x ) = 6 Solve for x: − 5 x + 15 + 5 x = 6

Back-substitute: y = 18,000 − 10,000 = 8000 Answer: (10,000, 8000) $10,000 at 6% and $8000 at 3%

15 ≠ 6 Inconsistent

No solution + y = 2  − 12 y = 4 3 x 

30.  x + y = 18,000  840 0.06 x + 0.03 y =

Solve for y in Equation 1: y = 18,000 − x 88 17

27. − 35 x + y = 5 Equation 1  −5 x + 3 y = 6 Equation 2

2 − 3 x

673

Equation 1

Solve for y: 4 + 103 y = 8  y = 403

Answer: ( 203 , 403 )

28.

Solving Systems of Equations

31.  x + y = 18,000  0.056 x 0.068 y = 1182 + 

Equation 1 Equation 2

Solve for y in Equation 1: y = 18,000 − x Substitute for y in Equation 2: 0.056 x + 0.068(18,000 − x) = 1182

Solve for y in Equation 1: y = 2 + 23 x

(

)

Substitute for y in Equation 2: 3 x − 12 2 + 23 x = 4

Solve for x: 0.056 x + 1224 − 0.068 x = 1182 −0.012 x = − 42

3 x − 1 − 13 x = 4 8x = 5 3

x = 15 8

( )

Substitute for x in Equation 1: y = 2 + 23 15 8

x = 3500

Back-substitute: y = 18,000 − 3500 = 14,500 Answer: (3500, 14,500) $3500 at 5.6% and $14,500 at 6.8%

y = 2 + 54 y = 13 4

Answer:

(158 , 134 )

674

Chapter 8

Linear Systems and Matrices

32.  x + y = 18,000 Equation 1  + = 0.0275 x 0.0425 y 684 Equation 2  Solve for y in Equation 1: y = 18,000 − x Substitute for y in Equation 2: 0.0275 x + 0.0425 (18,000 − x ) = 684

36. 2 x 2 + y = 3 Equation 1   x + y = 4 Equation 2 Solve for y in Equation 2: y = − x + 4

Substitute for y in Equation 1: 2 x 2 + (− x + 4) = 3 2x2 − x + 1 = 0

0.0275 x + 765 − 0.0425 x = 684

Solve for x:

No real solution

− 0.015 x = −81

x = 5400 Back-substitute: y = 18,000 − 5400 = 12,600 Answer: (5400, 12,600) $5400 at 2.75% and $12,600 at 4.25%

37.  x 3 − y = 0 Equation 1   x − y = 0 Equation 2 Solve for y in Equation 2: y = x

Substitute for y in Equation 1: x 3 − x = 0 Solve for x: x ( x − 1)( x + 1) = 0  x = 0, 1, − 1

33.  x 2 − 2 x + y = 8 Equation 1  x − y = −2 Equation 2 

Back-substitute: x = 0  y = 0 x =1 y =1

Solve for y in Equation 2: y = x + 2 Substitute for y in Equation 1: x 2 − 2 x + ( x + 2) = 8

x = −1  y = −1 Answer: ( 0, 0 ) , (1, 1), ( − 1, − 1)

x − x −6 = 0 ( x − 3)( x + 2) = 0

38.

x = 3, − 2

y = − x = x3 + 3x2 + 2 x

x =3 y =5

x3 + 3x2 + 3x = 0

x = −2  y = 0 Answer: (3, 5), (−2, 0)

x x2 + 3x + 3 = 0

(

)

x=0 y=0

34. 2 x 2 − 2 x − y = 14 Equation 1  2 x − y = −2 Equation 2  Solve for y in Equation 2: y = 2 x + 2 Substitute for y in Equation 1: 2 x 2 − 2 x − ( 2 x + 2 ) = 14

2 x 2 − 4 x − 16 = 0

Answer: (0, 0) 39. −2 x + y = 7   x + 3y = 0 y

− 2x + y = 7

x + 3y = 0

x − 2x − 8 = 0

3 2

(− 3, 1) 1

( x − 4 )( x + 2 ) = 0 x = 4  y = 10 x = −2  y = −2

x = 4, − 2

−6 −5

−3 −2 −1 −1 −2 −3

Point of intersection: ( − 3, 1)

Answer: ( 4, 10 ) , ( −2, − 2 ) 35. 2 x 2 − y = 1 Equation 1   x − y = 2 Equation 2 Solve for y in Equation 2: y = x − 2

Substitute for y in Equation 1: 2 x − ( x − 2 ) = 1

40.  x + y = 8  4 x + 4 y = 0 y

2x2 − x + 1 = 0

No real solution

x+y=8

8 6 4 2 −6 −4 −2 −2

−4

4x + 4y = 0

−6

No solution

Section 8.1

Solving Systems of Equations

675

= 3   −x − y = 3 45. − x − y  2  2 2 2 x + y − 4 x − 21 = 0  ( x − 2) + y = 25

41.  x − 2 y = − 3  5 x + 6 y = 17 y

x − 2y = −3

x 2 − 4x − 21 + y 2 = 0

8 6

(− 3, 0)

(1, 2)

−3 −2 −1 −1

−2

−6 −4

−x − y = 1

2 2 46.  y 2 − x 2 + 9 = 0  x − y = 9    3 1 − 12 x + y = 32 − 2 x + y = 2 

x2 − y2 = 9 8

1 2

− 1x + y = 3 2

6 4

− 5x + 2y = − 2

(5, 4)

(− 3, 0) −8

−2

x − 2y = 6

−4

Points of intersection: ( − 3, 0), ( 2, − 5)

(− 1, − 3.5)

−8

42. −5 x + 2 y = − 2   x − 2y = 6

(2, − 5)

Point of intersection: (1, 2)

−5 −4 −3 −2 −1

−6

5x + 6y = 17

−3

−4

−5

−6

Point of intersection: ( −1, − 3.5) 43.  x 2 + y = −1  − x + 2 y = 5

−8

Points of intersection: ( − 3, 0), (5, 4) 47. 7 x + 8 y = 24  y1 = − 87 x + 3  1  x − 8 y = 8  y2 = 8 x − 1

6 4

− x + 2y = 5 −4 −2

−4

x2 + y = − 1

−4

Point of intersection: ( 4, − 0.5)

−6 −8 − 10

48.  x − y = 0  y1 = x  5 5 x − 2 y = 6  y2 = 2 x − 3

No real solution 44. x 2 − y = 3   x − y = 1

y 3 0

2 1 −4 −3 −2

(− 1, − 2)

−1

(2, 1) 1

−4

Point of intersection: ( 2, 2 )

x2 − y = 3

x−y=1

Points of intersection:

( −1, − 2), ( 2, 1)

676

Chapter 8

Linear Systems and Matrices

49.

x − y 2 = −1  y 2 = x + 1  y1 = x + 1

53.  y = ex   x − y + 1 = 0  y = x + 1

y2 = − x + 1

x − y = 5  y3 = x − 5

6 −3 −5

13 −1

Point of intersection: ( 0, 1)

−6

Points of intersection: ( 8, 3) , ( 3, − 2 ) 50.

54.  y = −4e − x   y + 3 x + 8 = 0  y = −3 x − 8

x − y 2 = −2  y1 = x + 2 , y2 = − x + 2

x − 2 y = 6  y3 = 12 ( x − 6 )

2 −9

−6

18 − 10

Point of intersection: ( −0.490, − 6.530 )

−8

Points of intersection: ( 2, − 2 ) , (14, 4 ) 51.

55.  x + 2 y = 8  y1 = 4 − x 2  y = 2 + ln x  y2 = 2 + ln x 

x 2 + y 2 = 8  y1 = 8 − x 2 , y2 = − 8 − x 2 y = x 2  y3 = x 2

−6

−1 −1

−4

Points of intersection: (1.540, 2.372 ) , ( −1.540, 2.372 )

x + y = 25  y1 = 25 − x 2

52.

y2 = − 25 − x 2

Point of intersection: ( 2.318, 2.841) 56.  y = −2 + ln ( x − 1)  1 3 y + 2 x = 9  y = 3 ( 9 − 2 x ) 6

( x − 8 ) + y2 = 41  y3 = 41 − ( x − 8 )2 2

y4 = − 41 − ( x − 8 )

−6

−8

−6

−8

Points of intersection: ( 3, 4 ) , ( 3, − 4 )

Point of intersection: ( 5.309, − 0.539 ) 57.  y = x + 4   y = 2 x + 1 9

−3 −3

Point of intersection: ( 2.25, 5.5 )

Section 8.1 58.  x − y = 3  y = x − 3   x − y = 1  y = x − 1

Solving Systems of Equations

677

62.  x + y = 4 Equation 1  2  x + y = 2 Equation 2 Solve for y in Equation 1: y = 4 − x

Substitute for y in Equation 2: x 2 + ( 4 − x ) = 2 −2

Solve for x: x 2 − x + 2 = 0 No real solutions because the discriminant in the Quadratic Formula is negative. Inconsistent. No solution

−4

Point of intersection: ( 4, 1)

63. 3 x − 7 y + 6 = 0 Equation 1  x 2 − y 2 = 4 Equation 2 

59.  x 2 + y 2 = 169  y1 = 169 − x 2 and   y2 = − 169 − x 2  2 1 2  x − 8 y = 104  y3 = 8 x − 13

Solve for y in Equation 1: y =

 3x + 6  Substitute for y in Equation 2: x 2 −   =4  7 

 9 x 2 + 36 x + 36  Solve for x: x 2 −  =4 49  

−27

3x + 6 7

(

−18

)

49 x 2 − 9 x 2 + 36 x + 36 = 196

Points of intersection: ( 0, − 13) , ( ±12, 5)

40 x − 36 x − 232 = 0

60.  x 2 + y 2 = 4  y1 = 4 − x 2   y2 = − 4 − x 2  2 2 2 x − y = 2  y3 = 2 x − 2

10 x 2 − 9 x − 58 = 0  x =

9 ± 81 + 40 ( 58 ) 20

x=

29 , −2 10

29 3 x + 6 3 ( 29 10 ) + 6 21 : y= = = 10 7 7 10 3x + 6 Back-substitute x = −2 : y = =0 7

Back-substitute x =

−6

29 21 , 10 Answer: ( 10 ) , ( −2, 0 )

−4

Points of intersection: ( 0, − 2 ) ,

(

1 2

) (

7, 32 , − 12 7, 23

or ( 0, − 2 ) , ( ±1.323, 1.500 )

)

64.  x 2 + y 2 = 25 Equation 1  2 x + y = 10 Equation 2 Solve for y in Equation 2: y = 10 − 2 x

61.  y = 2 x Equation 1  2 = + y x 1 Equation 2 

Substitute for y in Equation 1: x 2 + (10 − 2 x ) = 25 2

Solve for x: x 2 + 100 − 40 x + 4 x 2 = 25  x 2 − 8 x + 15 = 0

Substitute for y in Equation 2: 2 x = x 2 + 1 Solve for x:

 ( x − 5 )( x − 3 ) = 0  x = 3, 5

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

Back-substitute x = 3 : y = 10 − 2(3) = 4 Back-substitute x = 5 : y = 10 − 2(5) = 0

Back-substitute x = 1 in Equation 1: y = 2 x = 2 Answer: (1, 2 )

Answer: ( 3, 4 ) , ( 5, 0 ) 65.

x 2 + y2 = 1 x+y=4

Graphing y1 = 1 − x 2 , y2 = − 1 − x 2 and y3 = 4 − x, you see that there are no points of intersection. No solution

678

Chapter 8

66.

x 2 + y2 = 4

Linear Systems and Matrices

x−y=5

Graphing y1 = 4 − x 2 , y2 = − 4 − x 2 and y3 = x − 5, you see that there are no points of intersection. No solution 67.  y = 2 x + 1   y = x + 2

71.  y = x 3 − 2 x 2 + 1 Equation 1  2 Equation 2  y = 1 − x Substitute for y in Equation 2: x3 − 2 x2 + 1 = 1 − x2 Solve for x: x 3 − x 2 = 0

x 2 ( x − 1) = 0  x = 0, 1

Back-substitute: x = 0  y = 1 x =1 y = 0 Answer: ( 0, 1) , (1, 0 )

−5

72.  y = x 3 − 2 x 2 + x − 1 Equation 1  2 Equation 2  y = − x + 3 x − 1

7 −2

Point of intersection: ( 14 , 32 ) or ( 0.25, 1.5 )

(

68.  y = 2 x − 1   y = x + 1

0 = x x2 − x − 2

Back-substitute x = 0 in Equation 2:

4 x2 − 4 x + 1 = x + 1

y = −02 + 3 ( 0 ) − 1 = −1

4 x − 5x = 0

Back-substitute x = 2 in Equation 2:

x(4 x − 5) = 0

y = −22 + 3 ( 2 ) − 1 = 1

x = 0 extraneous x = 45  y = 23

Answer: ( ,

3 2

)

0 = x ( x − 2 )( x + 1)  x = 0, 2, − 1

2x − 1 = x + 1

5 4

Substitute for y in Equation 1: − x2 + 3x − 1 = x3 − 2 x2 + x − 1 Solve for x: 0 = x 3 − x 2 − 2 x

Back-substitute x = −1 in Equation 2:

y = − ( −1) + 3 ( −1) − 1 = −5 2

) or (1.25, 1.5)

69.  y − e − x = 1  y = e − x + 1   y − ln x = 3  y = ln x + 3

Answer: ( 0, − 1) , ( 2, 1) , ( −1, − 5 ) 73.  xy − 1 = 0  5 2 x y + 1 = 0 − − 

Equation 1 Equation 2

Solve for y in Equation 1: xy = 1 y = −3

6 −1

Point of intersection: ( 0.287, 1.751) 70. Graph y = 4 − 2 ln x and y = e

1 Substitute for y in Equation 2: − 5 x − 2  + 1 = 0  x Solve for x: − 5 x 2 − 2 + x = 0 5x2 − x + 2 = 0

x =

−3

1 x

x =

−( −1) ± 1±

− 39 10

(−1) − 4(5)(2) 2(5) 2

 No real solution

Point of intersection: (1.262, 3.534 )

Section 8.1 74.  xy − 2 = 0 Equation 1  x y + 4 = 0 Equation 2 − 3 2 

Solve for y in Equation 1: y =

679

77. C = 5.5 x + 10,000, R = 3.29 x

R =C 3.29x = 5.5 x + 10,000

2 x

3.29x − 10,000 = 5.5 x

Substitute for y in Equation 2: 2 3 x − 2   + 4 = 0 (x cannot be 0.) x

10.8241x2 − 65,800x + 100,000,000 = 30.25x 10.8241x2 − 65,830.25x + 100,000,000 = 0 15,000 x ≈ 3133 units

3x 2 + 4 x − 4 = 0

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

5,000

In order for the revenue to break even with the cost, 3133 units must be sold, R = $10,308.

2  Answer:  , 3  , ( −2, − 1) 3 

75.

Solving Systems of Equations

78. C = 7.8 x + 18,500, R = 12.84 x R =C

C = 8650 x + 250,000, R = 9950 x R=C 9950 x = 8650 x + 250,000

12.84 x = 7.8 x + 18,500 12.84 x − 7.8 x − 18,500 = 0 Quadratic in

1300 x = 250,000 x ≈ 192 units

25,000

3,500,000

C C

R 0

400

x ≈ 1464 units, R ≈ $18,798

R ≈ $1,910,400

76.

5,000

79. 2 l + 2 w = 30  l + w = 15

C = 2.65 x + 350,000, R = 4.15 x R=C 4.15 x = 2.65 x + 350,000 1.50 x = 350,000

l = w+3

( w + 3) + w = 15 2 w = 12 w=6

l =w+3=9

x ≈ 233,333 units, R = $968,333

Dimensions: 6 meters × 9 meters

2,500,000

80. 2l + 2 w = 280  l + w = 140

w = l − 20  l + ( l − 20 ) = 140

C 0

2l = 160

500,000

l = 80 w = l − 20 = 80 − 20 = 60 Dimensions: 60 × 80 centimeters

81. N = 360 − 24 x Animated film N = 24 + 18 x Horror film (a) Week x

Animated

336 312 288 264 240 216 192 168 144 120

Horror

42 60

96 114 132 150 168 186 204 222 240

(b) For x = 8, N = 168.

680

Chapter 8 (c)

Linear Systems and Matrices

360 − 24 x = 24 + 18 x

336 = 42 x x =8

N = 24 + 18 ( 8 ) = 168 (d) The answers are the same. (e) During week 8, the same number (168 ) were rented. 82. (a) Pellet stove: ys = 19.15 x + 3650

(b) Electric furnace: y f = 33.25 x + 2780 (c)

7000

85. 2l + 2 w = 40  l + w = 20  w = 20 − l lw = 96  l ( 20 − l ) = 96

20l − l 2 = 96 0 = l 2 − 20l + 96 0 = ( l − 8 )( l − 12 )

yf ys

l = 8 or l = 12 0

100

l = 12, w = 8 If the length is supposed to be greater than the width, we have l = 12 miles and w = 8 miles.

(d) Substitute ys for y f : 19.15 x + 3650 = 33.25 x + 2780 870 = 14.1x x ≈ 61.70 million Btu of heat

(e) The pellet stove will cost more to use for heat until 61.70 million Btu of heat is used. After that, the electric furnace will have a higher cost. 83. (a) The total cost will be the sum of the variable cost plus the fixed cost (initial cost). C = 9.45 x + 16,000 The total revenue is the selling price times the number of units sold. R = 55.95 x (b) 30,000

86.

A = 12 bh 1 = 12 a 2 a2 = 2 a= 2 The dimensions are b = h = 2 inches and hypotenuse = 2 inches.

R C a 0

700

87. (a)

The break-even point is the point of intersection of the cost function and the revenue function. 55.95 x = 9.45 x + 16,000 46.5 x = 16,000 x ≈ 344 units 84. I = 0.01x + 33,000  I = 0.025 x + 30,000

(b)

Equation 1  x + y = 20,000  + = 0.055 x 0.075 y 1300 Equation 2  22,000

First company Second company

0.01x + 33,000 = 0.025 x + 30,000 3000 = 0.015 x 200,000 = x For sales greater than $200,000, the annual salary of $30,000 is the better offer.

25,000

0.055 x + 1500 − 0.075 x = 1300 −0.02 x = −200 x = 10,000 Back-substitute: y = 20,000 − 10,000 = 10,000 To earn $1300 in interest, $10,000 should be invested in each fund, earning 5.5% and 7.5%.

Section 8.1 (e)

88. V = ( D − 4 ) , 5 ≤ D ≤ 40 2

V = 0.79 D 2 − 2 D − 4, 5 ≤ D ≤ 40 (a) 750 Doyle

Solving Systems of Equations

681

At one point during 2016, the populations of Colorado and Minnesota were equal.

90. (a) E = 77.982 x 2 − 1280.6 x + 8202.9 P = 988 x − 6421

Scribner

(b)

7000

P 5

(b) The two graphs intersect at D = 24.72. Algebraically:

( D − 4 ) = 0.79D − 2 D − 4 2

D − 8 D + 16 = 0.79 D 2 − 2 D − 4

0.21D 2 − 6 D + 20 = 0 D ≈ 24.72, 3.9 Since 5 ≤ D ≤ 40, the scales agree when

(d) E = 77.982 x 2 − 1280.6 x + 8202.9   P = 988 x − 6421 Set E = P :

Year

Colorado

Minnesota

2013

5265

5417

2014

5339

5450

2015

5413

5483

x =

2016

5487

5516

x ≈ 9.64  10

2017

5561

5549

2018

5635

5582

(e) The graphical results and the algebraic results are the same.

2019

5709

5615

91. False. You could solve for x first.

2020

5783

5648

92. False. There could be four points of intersection.

6000

M C 20

0 4000

The point of intersection is approximately (16.71, 5539.34). (d)

77.982 x 2 − 1280.6 x + 8202.9 = 988 x − 6421

(b) In 2017, the population of Colorado, 5,561,000, is greater than the population of Minnesota, 5,549,000. (c)

(c) The point of intersection is about (9.64, 3103.32), so the first year when the revenues of Priceline.com will be greater is 2010.

D ≈ 24.72 inches.

89. (a)

 C = 74.0t + 4303  M = 33.0t + 4988

Set C = M : 74.0t + 4303 = 33.0t + 4988

77.982 x 2 − 2268.6 x + 14,623.9 = 0 Use the Quadratic Formula.

685 ≈ 16.71 41

C = 74.0(16.71) + 4303 = 5539.34

(16.71, 5539.34)

(− 2268.6) − 4(77.982)(14,623.9) 2(77.982) 2

For example, x 2 + y 2 = 4 and y = x 2 − 3. 93. The system has no solution if you arrive at a false statement, such as 4 = 8, or you have a quadratic equation with a negative discriminant, which would yield imaginary roots. 94. (a) The line y = 2 x intersects the parabola y = x 2 at two points, ( 0, 0 ) and ( 2, 4 ) .

(b) The line y = 0 intersects y = x 2 at (0, 0) only. (c) The line y = x − 2 does not intersect y = x 2 . (Other answers possible.) 95. Answers will vary. For example,

(a)

3 x + y = 3  3 x + y = 5

(b)

3 x + y = 4  2 x + y = 2

(c)

6 x + 3 y = 9  2 x + y = 3

41.0t = 685 t =

−( − 2268.6) ±

682

Chapter 8

Linear Systems and Matrices

96. Answers will vary. For example, y = x − 3  2 y = x − 4 97. (a)

300

y = x2

3  102.  , 0  , ( 4, 6 ) 5  6 − 0 6 30 m= = = 4 − 35 175 17

30 ( x − 4) 17 30 120 y−6 = x− 17 17 30 18 y= x− 17 17

y = 2x

y−6 = y = x4

−5

−3

5 − 30

−2

y = 4x

(b) Based on the graphs in part (a), it appears that for b > 1, there are three points of intersection for the x

graphs of y = b and y = x when b is an even number. 98. (a) The point of intersection is approximately (2500, 150,000). This is the break-even point where cost equals revenue.

(b) (i) For values of 0 ≤ x < 2500 the overall cost C is greater than the revenue R.

103. f ( x ) =

Domain: all x ≠ 6 Vertical asymptote: x = 6 Horizontal asymptote: y = 0 2x − 7 3x + 2 Domain: all x ≠ − 23

104. f ( x ) =

(ii) For values of x > 2500, the revenue R is greater than the cost C. 99.

( 3, 4 ) , (10, 6 ) 6−4 2 m= = 10 − 3 7 2 y − 4 = ( x − 3) 7 2 6 y−4= x− 7 7 2 22 y= x+ 7 7

Vertical asymptote: x = − 23 Horizontal asymptote: y = 23

x2 + 2 x 2 − 16 Domain: all x ≠ ±4 Vertical asymptotes: x = ±4 Horizontal asymptote: y = 1

105. f ( x ) =

106. f ( x ) = 3 −

3−3 =0 10 − 6 The line is horizontal.

x +1 x2 + 1 Domain: all real numbers x Horizontal asymptote: y = 0

107. f ( x ) =

y=3

101. ( 4, −2 ) , ( 4, 5 )

2 3x2 − 2 = x2 x2

Domain: all x ≠ 0 Vertical asymptote: x = 0 Horizontal asymptote: y = 3

100. ( 6, 3 ) , (10, 3 ) m=

5 x −6

x−4 x 2 + 16 Domain: all real numbers x Horizontal asymptote: y = 0

108. f ( x ) =

x=4

A vertical line cannot be written slope-intercept form.

Section 8.2 Systems of Linear Equations in Two Variables 1.

method, elimination

equivalent

A system of linear equations with no solution is inconsistent.

A system of linear equations with at least one solution is consistent.

A system of linear equations where the lines are coincident or identical is consistent.

Section 8.2 6.

A system of linear equations in which the lines do not intersect or are parallel has no solution and is inconsistent.  2 x + y = 5 Equation 1   x − y = 1 Equation 2 Add to eliminate y: 3 x = 6  x = 2 Substitute x = 2 in Equation 2: 2 − y = 1  y = 1

Systems of Linear Equations in Two Variables 10.  2 x − y = 3 Equation 1   4 x + 3 y = 21 Equation 2

Multiply Equation 1 by 3: 6 x − 3 y = 9 Add this to Equation 2 to eliminate y: 10 x = 30  x = 3 Substitute x = 3 in Equation 1: 2(3) − y = 3  y = 3 Answer: ( 3, 3 )

Answer: ( 2, 1) −8

−4

−5

−3

2x + y = 5

 x + 3 y = 1 Equation 1   − x + 2 y = 4 Equation 2 Add to eliminate x: 5 y = 5  y = 1 Substitute y = 1 in Equation 1: x + 3(1) = 1  x = −2

Answer: ( −2, 1) x + 3y = 1

−x + 2y = 4

−8

−2x + 2y = 5

x−y=2

−6

−4

12.  3 x − 2 y = 5 Equation 1   −6 x + 4 y = −10 Equation 2

 x + y = 0 Equation 1  3 x + 2 y = 1 Equation 2

Multiply Equation 1 by −2: −2 x − 2 y = 0 Add this to Equation 2 to eliminate y: x = 1 Substitute x = 1 in Equation 1: 1 + y = 0  y = −1 Answer: (1, − 1) x+y=0

4x + 3y = 21

11.  x − y = 2 Equation 1   −2 x + 2 y = 5 Equation 2 Multiply Equation 1 by 2: 2 x − 2 y = 4 Add this to Equation 2: 0 = 9 There are no solutions.

−3

2x − y = 3

x−y=1

683

Multiply Equation 1 by 2 and add to Equation 2: 0+0=0

There are infinitely many solutions. All points on the line 3 x − 2 y = 5. 4

−6x + 4y = −10

−5

4 −4

−6

−4

3x + 2y = 1

3x − 2y = 5

13.  x + 2 y = 3 Equation 1   x − 2 y = 1 Equation 2 Add to eliminate y: 2x = 4 x=2 Substitute x = 2 into Equation 1: 2 + 2y = 3

2y = 1 y = 12 Answer: ( 2, 12 )

684

Chapter 8

Linear Systems and Matrices

14.  3 x − 5 y = 2 Equation 1   2 x + 5 y = 13 Equation 2

Add to eliminate y: 5 x = 15 x=3 Substitute x = 3 into Equation 1: 3(3) − 5 y = 2 9 − 5y = 2 −5 y = −7 y = 75

15.  2 x + 3 y = 18 Equation 1   5 x − y = 11 Equation 2 Multiply Equation 2 by 3: 15 x − 3 y = 33 Add this to Equation 1 to eliminate y: 17 x = 51  x = 3 Substitute x = 3 in Equation 1: 6 + 3 y = 18  y = 4

Answer: ( 3, 4 ) 16.  x + 7 y = 12 Equation 1  3 x − 5 y = 10 Equation 2

Multiply Equation 1 by −3: −3 x − 21y = −36 Add this to Equation 2 to eliminate x: −26 y = −26  y = 1 Substitute y = 1 in Equation 1: x + 7 = 12  x = 5

Answer: (5, 1) Equation 1 Equation 2

Multiply Equation 1 by 3: 9 r + 6 s = −18 Multiply Equation 2 by −1: − 2r − 6 s = − 3 Add to eliminate s: 7 r = − 21 r = −3

Substitute r = − 3 in Equation 1: 3( − 3) + 2 s = − 6 3 s = 2

16r + 50 s = 55 18s = 15 s=

5 6

Substitute s = 65 in Equation 1:

Answer: ( 3, 75 )

17. 3r + 2 s = − 6  2r + 6 s = 3

18.  8r + 16 s = 20 Equation 1  16r + 50 s = 55 Equation 2 Multiply Equation 1 by − 2 : − 16 r − 32 s = −40 Add to Equation 2 to eliminate r: −16r − 32 s = −40

5 8r + 16   = 20 6 40 8r + = 20 3 20 8r = 3 5 r= 6

Answer: ( 65 , 65 ) 19. 5u + 6v = 24 Equation 1  3u + 5v = 18 Equation 2

Multiply Equation 1 by 3 and Equation 2 by ( −5 ) : 15u + 18v = 72 −15u − 25v = −90

Add to eliminate u: −7v = −18  v =

18 7

Substitute v = 187 in Equation 2: 3u + 5 ( 187 ) = 18  u = 127

Answer: ( 127 , 187 )

20.  3u + 11v = 4 Equation 1   −2u − 5v = 9 Equation 2 Multiply Equation 1 by 2 and Equation 2 by 3:  6u + 22 v = 8   −6u − 15v = 27

Adding, 7v = 35  v = 5. Then, 3u + 11( 5 ) = 4  u = −17. Answer : ( −17, 5)

3  Answer:  − 3,  2 

Section 8.2 21. − 6 x + 5 y = −15   4 x + 12 y = 10

Equation 1 Equation 2

Systems of Linear Equations in Two Variables 1 3 x + 4 y = 1 25.  1 2 x − 3 y = 0

Equation 1 Equation 2

Multiply Equation 1 by 2: −12 x + 10 y = − 30

Multiply Equation 1 by 4: 12 x + y = 4

Multiply Equation 2 by 3: 12 x + 36 y = 30

Multiply Equation 2 by 3: 6 x − y = 0

Add to eliminate x:

Add to eliminate y: 18 x = 4

46 y = 0

685

x = 92

y = 0

Substitute y = 0 in Equation 1:

Substitute x = 92 into Equation 1:

− 6 x + 5(0) = −15

3 92 + 14 y = 1

()

6 x = −15

2 + 1y 3 4

1y 4

= 13

5 x = 2

y = 43

5  Answer:  , 0  2  22. 9 x + 3 y = 18  2 x − 7 y = −19

Answer: Equation 1 Equation 2

Multiply Equation 1 by 2: 18 x + 6 y = 36

26.  12 x − 2 y = − 52  − x + 4 y = 5

Equation 2

0 = 0

Add to eliminate x:

There are an infinite number of solutions. All points on the line − x + 4 y = 5.

69 y = 207

Substitute y = 3 into Equation 1:

Equation 1

Multiply Equation 1 by 2 and add to Equation 2:

Multiply Equation 2 by − 9: −18 x + 63 y = 171

y = 3

( 92 , 34 )

27. 2 x − 5 y = 0

x − y =3 y = x −3 2 x − 5( x − 3) = 0

9 x + 3(3) = 18 9x = 9

− 3 x = −15

x =1

x = 5, y = 2 Matches (b). One solution; consistent

Answer: (1, 3) 23. 1.8 x + 1.2 y = 4 Equation 1   9 x + 6 y = 3 Equation 2

Multiply Equation 1 by ( −5 ) : −9 x − 6 y = −20 Add this to Equation 2: 0 = −17 Inconsistent; no solution 24.  3.1x − 2.9 y = −10.2 Equation 1  Equation 2  31x − 12y = 34 Multiply Equation 1 by − 10 : −31x + 29 y = 102 Add this to Equation 2: 17 y = 136  y = 8. Substituting this value into Equation 2: 31x − 12 ( 8 ) = 34  31x = 130  x = 130 31

, 8) Answer: ( 130 31

28. −7 x + 6 y = −4 14 x − 12 y = 8 Lines coincide. Matches (a). Infinite number of solutions; consistent 29. 2 x − 5 y = 0  −2 y = 4  2 x − 3 y = −4  y = −2, x = −5

One solution; consistent Matches (c). 30.

7 x − 6 y = −6 −7 x + 6 y = −4 Parallel lines Matches (d). Inconsistent; no solution

Chapter 8

686

Linear Systems and Matrices

31. 5 x + 3 y = 6 Equation 1  3 x − y = 5 Equation 2

Multiply Equation 2 by 3 and add to eliminate y:

36.  14 x + 16 y = 1 Equation 1   −3 x − 2 y = 0 Equation 2 Multiply Equation 1 by 12 and add to Equation 2:

5x + 3y = 6

0 = 12

9 x − 3 y = 15

Inconsistent: no solution

14 x = 21 x = 23

Substitute x = 32 into Equation 1: 5 ( 32 ) + 3 y = 6 + 3y = 6

15 2

x + y = 3  x − y = 4

3 y = − 32 y = − 12

Answer: ( , − 12 ) 3 2

32.  x + 5 y = 10 Equation 1  3 x − 10 y = −5 Equation 2

Multiply Equation 1 by 2 and add to eliminate y: 2 x + 10 y = 20 3 x − 10 y = −5 5 x = 15 x=3 Substitute x = 3 into Equation 1:

3 + 5 y = 10 5y = 7 y = 75

Answer: ( 3, 75 )

0=8

Inconsistent; no solution Equation 1 Equation 2

Multiply Equation 1 by −6 and add to Equation 2: 0=0

There are an infinite number of solutions. All points on the line 4 x + y = 4. 3 1  x + y = 8 Equation 1 35.  49 3  4 x + 3 y = 8 Equation 2 Multiply Equation 1 by −3 : − 94 x − 3 y = − 83

Add this to Equation 2: 0 = 0 There are an infinite number of solutions. The solutions consist of all ( x, y ) satisfying 3 4

Add to eliminate y: 2 x = 7  x =

7 2

7 into Equation 2: 2 7 1 −y=4 y=− 2 2

Substitute x =

Answer: ( 27 , − 12 )

y −1 2x + 5 38.  + = −1 3  2  2x − y = 12 

Equation 1 Equation 2

Multiply Equation 1 by 6 and Equation 2 by 2: 6 x + 2 y = −19  4 x − 2 y = 24

Add to eliminate y: 10 x = 5

33.  52 x − 32 y = 4 Equation 1 1 3  5 x − 4 y = −2 Equation 2 Multiply Equation 2 by −2 and add to Equation 1:

34.  23 x + 16 y = 23   4x + y = 4

 x + 2 y −1 37.  + = 1 Equation 1 4  4  x − y = 4 Equation 2  Multiply Equation 1 by 4:

x = Substitute x =

1 2

1 into Equation 2: 2

1 2  − y = 12 2 y = −11 1  Answer:  , −11 2  39.  −5 x + 6 y = −3   20 x − 24 y = 12

Equation 1 Equation 2

Multiply Equation 2 by 14 and add: −5 x + 6 y = −3 5x − 6 y = 3 0=0 There are an infinite number of solutions. All points on the line −5 x + 6 y = −3.

x + y = 18 , or 6 x + 8 y = 1.

Section 8.2 40.  7 x + 8 y = 16  −14 x − 16 y = − 4

Equation 1 Equation 2

Multiply Equation 2 by 12 : − 7 x − 8 y = − 2 7 x + 8 y = 16

Add to eliminate y:

687

1 1 and Y = . y x

 X + 3Y = 2 Equation 1   4 X − Y = −5 Equation 2

0 ≠ 14

Equation 1

Hence, X = 2 − 3Y = 2 − 3 (1) = −1.

Equation 2

Inconsistent: no solution

Subtract to eliminate X: 13Y = 13  Y = 1

Multiply Equation 1 by 4 and Equation 2 by −10:  10 x − 12 y = 6  −10 x + 12 y = 36

Add to eliminate y: 0 ≠ 42

Inconsistent: no solution 42. 6.3 x + 7.2 y = 5.4 Equation 1  5.6 x + 6.4 y = 4.8 Equation 2

7 x + 8 y = 6 ( Divide by 0.9 )  7 x + 8 y = 6 ( Divide by 0.8 ) There are an infinite number of solutions. All points on the line 7 x + 8 y = 6. 43. 0.2 x − 0.5 y = −27.8 Equation 1  0.3 x + 0.4 y = 68.7 Equation 2

Multiply Equation 1 by 40 and Equation 2 by 50 :  8 x − 20 y = −1112  15 x + 20 y = 3435

Adding the equations eliminates y: 23 x = 2323  x = 101 Substitute x = 101 into Equation 1:

8 (101) − 20 y = −1112  y = 96

Answer: (101, 96 ) 44. 0.2 x + 0.6 y = −1 Equation 1   x − 0.5 y = 2 Equation 2

Multiply Equation 1 by 10 and Equation 2 by 2 : 2 x + 6 y = −10  2x − y = 4 Subtract to eliminate x: 7 y = −14  y = −2

45. Let X =

Multiply Equation 1 by 4:  4 X + 12Y = 8   4 X − Y = −5

−7x − 8y = − 2

41. 2.5 x − 3 y = 1.5   x − 1.2 y = − 3.6

Systems of Linear Equations in Two Variables

Hence, x = 2 + 0.5 ( y ) = 2 + 0.5 ( −2 ) = 1.

1 1 = −1, y = = 1 X Y

Answer: ( −1, 1) 46. Let X =

1 1 and Y = . y x

2 X − Y = 5 Equation 1  6 X + Y = 11 Equation 2 Adding the equations, 8 X = 16  X = 2 Hence, Y = 11 − 6 X = 11 − 12 = −1. 1 1 1 x= = , y = = −1 X 2 Y 1  Answer:  , − 1 2 

 y = − 53 x − 52 47. 3x + 5 y = − 2    4 x − y = 5  y = 4 x − 5

The system is consistent. There is one solution, (1, −1). 6

−9

−6

 y = 13 x 48. − x + 3 y = 0    1 14  3 x − 9 y = 14  y = 3 x − 3

The lines are parallel, so the system is inconsistent. There is no solution. 6

−9

−6

Answer: (1, − 2 )

Chapter 8

688

Linear Systems and Matrices

8 49.   6 x + 3 y = − 8   y = − 2 x − 3   8 1 4 − x − 2 y = 3  y = − 2 x − 3

53.  6 y = 42  y = 7  6 x − y = 16  y = 6 x − 16 9

The system is consistent. The solution set consists of all points on the line 6 x + 3 y = − 8. 4

−6

−4

 y = − 12 x − 2 50. − 14 x − 12 y = 1     5 x + y = 1  y = − 5 x + 1 The system is consistent. There is one solution,

Answer: ( 236 , 7 ) ≈ ( 3.833, 7 ) 54.  4 y = −8  y = −2  7 x − 2 y = 25  y = ( 7 x − 25 ) 2 4

−7

( 23 , − 73 )

−8

Answer: ( 3, − 2 )

−6

−4

55.  32 x − 15 y = 8  y = 5 ( 23 x − 8 )  1 −2 x + 3 y = 3  y = 3 ( 3 + 2 x )

15 1 51. 3.2 x − 16 y = 7.5  y = 5 x − 32    9 1  x − 5 y = −9  y = 5 x + 5

The lines are parallel, so the system is inconsistent.

−3

There is no solution.

15 −2

Answer: ( 6, 5 )

−6

−4

 y = 3 x − 9 52.  −6 x + 4 y = − 9 2 4    3 9 4.5 x − 3 y = 6.75  y = 2 x − 4

3 5 2 3 18  3 56.  x − y = −9  y =  x + 9  = x + 2 54 5 4  10  + 28 x ( ) x 14  = +  − x + 6 y = 28  y = 6 6 3 10

The system is consistent. The solution set consists of all points on the line − 6 x + 4 y = − 9, or y = 32 x − 94 . 4

−4

14 −2

Answer: ( 8, 6 ) −6

−4

Section 8.2

57.

1 1 1 1 x+y=− y=− − x 3 3 3 3 1 5x − 3y = 7  y = ( 5x − 7) 3 1 −2

Systems of Linear Equations in Two Variables

689

61. 3x − 5 y = 7 Equation 1   2 x + y = 9 Equation 2

Multiply Equation 2 by 5: 10 x + 5 y = 45 Add this to Equation 1: 13 x = 52  x = 4 Back-substitute x = 4 into Equation 2: 2 ( 4) + y = 9  y = 1

Answer: ( 4, 1)

−3

Answer: (1, − 0.667 ) 58. 5 x − y = −4  y = 5 x + 4

2 x + 35 y = 25  y = 35 ( −2 x + 25 ) 3

62. − x + 3 y = 17 Equation 1  4 x + 3 y = 7 Equation 2 Subtract Equation 2 from Equation 1 to eliminate y: −5 x = 10  x = −2 Substitute x = −2 in Equation 1:

− ( −2 ) + 3 y = 17  y = 5

Answer: ( −2, 5 ) −4

2 −1

Answer: ( −0.4, 2 ) 59. 0.5 x + 2.2 y = 9  y = 1 2.2 ( 9 − 0.5 x )   6 x + 0.4 y = −22  y = 1 0.4 ( −22 − 6 x ) 10

63.  y = 2 x − 5 Equation 1   y = 5 x − 11 Equation 2 Set Equation 1 equal to Equation 2: 2 x − 5 = 5 x − 11

−3 x = −6 x=2 Substitute x = 2 into Equation 1: y = 2 ( 2 ) − 5 = −1

Answer: ( 2, − 1) − 12

6 −2

Answer: ( −4, 5 ) 60. 2.4 x + 3.8 y = −17.6  y = ( −2.4 x − 17.6 ) 3.8   4 x − 0.2 y = − 3.2  y = ( 4 x + 3.2 ) 0.2 1 −6

Equation 1 64. 7 x + 3 y = 16  y = x + 2 Equation 2  Substitute Equation 2 into Equation 1: 7 x + 3 ( x + 2 ) = 16 7 x + 3 x + 6 = 16 10 x = 10 x =1 Substitute x = 1 into Equation 2: y =1+ 2 = 3

Answer: (1, 3 ) −7

Answer: ( −1, − 4 )

65.  x − 5 y = 21  6 x + 5 y = 21

Adding the equations, 7 x = 42  x = 6. Back-substituting, x − 5 y = 6 − 5 y = 21  −5 y = 15  y = −3 Answer: ( 6, − 3)

690

Chapter 8

Linear Systems and Matrices

66.  y = −2 x − 17 Equation 1  Equation 2  y = 2 − 3x Set Equation 1 equal to Equation 2: −2 x − 17 = 2 − 3 x

x = 19 Substitute x = 19 into Equation 1: y = −2 (19 ) − 17 = −55

(

the solution. One possible system:  4 x + 1 y = 5 3 2   4 x − 2 y = − 27

73.

Answer: (19, − 55 )

Equation 1 67. −5 x + 9 y = 13  y = x − 4 Equation 2  Substitute Equation 2 into Equation 1: −5 x + 9 ( x − 4 ) = 13

68.  4 x − 3 y = 6 Equation 1  −5 x + 7 y = −1 Equation 2 Multiply Equation 1 by 5 and Equation 2 by 4. 20 x − 15 y = 30

−20 x + 28 y = −4 Adding, 13 y = 26  y = 2.

Then, 4 x − 3( 2 ) = 6  x = 3.

Answer: ( 3, 2 )

69. There are infinitely many systems that have the solution (5, 0). One possible system:  1 x + y = 1 5  3 x − 2 y = 15

70. There are infinitely many systems that have ( − 6, 1) as the

solution. One possible system:  4 x + 2 y = − 22  2 9 − 3 x + 5 y =

Demand = Supply 500 − 0.4 x = 380 + 0.1x

−0.5 x = −120 x = 240 units p = 500 − 0.4 ( 240 ) = $404

Answer : ( 240, 404 ) 74.

−5 x + 9 x − 36 = 13 4 x = 49 49 x= 4 49 Substitute x = into Equation 2: 4 49 33 y= −4= 4 4  49 33  Answer:  ,  4   4

)

72. There are infinitely many systems that have − 34 , 12 as

Supply = Demand 25 + 0.1x = 100 − 0.05 x

0.15 x = 75 x = 500 p = 75

Answer : ( 500, 75) 75.

Demand = Supply 140 − 0.00002 x = 80 + 0.00001x

60 = 0.00003 x x = 2,000,000 units p = $100.00

Answer : ( 2,000,000, 100 ) 76.

Supply = Demand 225 + 0.0005 x = 400 − 0.0002 x

0.0007 x = 175 x = 250,000 p = 350

Answer: ( 250,000, 350 ) 77. Let x = the ground speed and y = the wind speed.

3.6 ( x − y ) = 1800 Equation 1 x − y = 500   3 ( x + y ) = 1800 Equation 2 x + y = 600 2x = 1100 = 550 x Substituting x = 550 in Equation 2: 550 + y = 600 y = 50 Answer : x = 550 mph, y = 50 mph

71. There are infinitely many systems that have ( 2.5, − 4) as

the solution. One possible system:

 2 x + 1 y = 4 4  − 4 x − 3 y = 2

Section 8.2

Systems of Linear Equations in Two Variables

691

78. Let x be the boat speed and y be the current speed.

 1( x − y ) = 20 2  3 ( x + y ) = 20

Equation 1: upstream against the current Equation 2: downstream with the current

Multiply Equation 2 by 32 . x − y = 20  x + y = 30

Equation 1 Equation 2

Add to solve for x: 2 x = 50 x = 25

Answer: ( 25, 5) The speed of the motorboat is 25 mph and the speed of the current is 5 mph. 79. (a)  A + C = 1175  15 A + 12C = 16,275

Equation 1 Equation 2

(b) Multiply Equation 1 by −15: −15 A − 15C = −17,625   15 A + 12C = 16,275

80. (a) Let x = amount at 25% solution and y = amount at 50% solution.

 x+ y = 30 Total liters  0.25 x + 0.5 y = 0.4 ( 30 ) = 12 40% acid solution (b) 30

( Equation 1) ( Equation 2 )

Add to eliminate A: − 3C = −1350 0

C = 450

So, A = 1175 − 450 = 725.

There were 725 adult tickets and 450 child tickets sold.

(d) Solve Equation 1 for y and substitute into Equation 2: y = 30 − x

Answers will vary. (c)

0.25 x + 0.5 ( 30 − x ) = 12

600

0.25 x + 15 − 0.5 x = 12 −0.25 x = −3

600 300

900

The point of intersection is (725, 450), so A = 725 and C = 450.

x = 12 liters of 25% acid solution Substitute x = 12 into Equation 1: y = 30 − 12 = 18 liters of 50% acid solution 81. Let M = number of oranges and let R = number of grapefruits.

M+ R = 16 Equation 1   0.95 M 1.05 R 15.90 Equation 2 + =  Solving for R in Equation 1: R = 16 − M . Substituting into Equation 2: 0.95M + 1.05 (16 − M ) = 15.9 0.95M + 16.8 − 1.05M = 15.9 0.9 = 0.1M M =9 Hence, R = 16 − 9 = 7. 9 oranges and 7 grapefruits

692

Chapter 8

Linear Systems and Matrices

82. Let x = number of cheeseburgers and y = number of fries. 2 x + y = 830 Equation 1  3x + 2 y = 1360 Equation 2 Multiply Equation 1 by −2 and add to eliminate y

− 4 x − 2 y = −1660   3x + 2 y = 1360 − x = − 300 x = 300 Substitute into Equation 1: 2 ( 300 ) + y = 830

Let y = number of pairs of $89.95 shoes. x + y = 250   79.50 x + 89.95 y = 20,711

Solve for y in the first equation and substitute into the second equation: 79.50 x + 89.95( 250 − x) = 20,711 −10.45 x + 22,487.5 = 20,711 −10.45 x = −1776.5 x = 170

y = 230 A cheeseburger contains 300 calories and an order of fries contains 230 calories. Answer: ( 300, 230 )

83. (a) S − 149.9t = 415.5  S − 183.1t = 117.3

84. Let x = number of pairs of $79.50 shoes.

Equation 1 Equation 2

Multiply Equation 2 by −1 and add to eliminate S:  S − 149.9t = 415.5  − S + 183.1t = −117.3

33.2t = 298.2 t ≈ 8.98 S − 149.9(8.98) = 415.5 S ≈ 1761.60 Answer: (8.98, 1761.60) Answers will vary. (b) In 2018, both retailers had sales of about $1761.60 million. (c) The coefficient of the t-term is the annual increase in sales for each retailer. (d) If the coefficient of the t-term was the same for each retailer, then each company would have the same annual increase in sales. The lines would then be parallel, so the graph would not have any points of intersection and the retailers would never have the same amount of sales.

y = 250 − x = 80

Yes. The shoe store sold 170 pairs of $79.50 shoes and 80 pairs of $89.95 shoes. 85.  5b + 10a = 20.2  −10b − 20a = −40.4  10b + 30a = 50.1 10b + 30a = 50.1 

10 a = 9.7 a = 0.97 b = 2.1 Least squares regression line: y = 0.97 x + 2.1 86.  5b + 10a = 11.7  −10b − 20a = −23.4  10b + 30a = 25.6  10b + 30a = 25.6

10a = 2.2 a = 0.22 5b + 10 ( 0.22 ) = 11.7 b = 1.9 Least squares regression line: y = 0.22 x + 1.9 87.  6b + 15a = 23.6  − 30b − 75a = −118  15b + 55a = 48.8  30b + 110a = 97.6 35a = − 20.4

a ≈ − 0.58 15b + 55( − 0.58) ≈

48.8

15b ≈

80.7

b ≈

5.39

Least squares regression line: y = − 0.58 x + 5.39 88.  7b + 21a = 13.1  − 21b − 63a = − 39.3  21b + 91a = 2.8  21b + 91a = − 2.8 28a = − 42.1

a ≈ −1.504 7b + 21( −1.504) ≈

13.1

7b ≈ 44.68 b ≈

6.38

Least squares regression line: y = −1.504 x + 6.38

Section 8.2

4b + 7.0a = 174  28b + 49a = 1218  −28b − 54a = −1288 7b + 13.5a = 322  Adding, −5a = −70  a = 14, b = 19. Thus, y = 14 x + 19 (b) Using a graphing utility, you obtain y = 14 x + 19. (c) 60

89. (a)

(d)

If x = 1.6, (160 pounds acre ) ,

Systems of Linear Equations in Two Variables

96. (a) The lines appear parallel with positive slope and one with a positive y-intercept and the other with a negative y-intercept.

(b) No, based only on the graph shown, you cannot assume the lines are parallel, and have no point of intersection. Therefore, you cannot conclude that the system is inconsistent. 97. u sin x + v cos x = 0 u cos x − v sin x = sec x Multiply the first equation by sin x, the second by cos x , and add the equations:

u sin 2 x + u cos2 x = sec x ⋅ cos x

y = 14 (1.6 ) + 19 = 41.4 bushels per acre. 90. (a)

105 3.00b + 3.70a =  3.70 b 4.69 a 123.9 + = 

Solving this system, you obtain a = −44.21 and b = 89.53 (b) y = −44.21x + 89.53 y = −44.21x + 89.53 (c) 50

693

u =1 Hence, v cos x = −u sin x = − sin x v = − tan x.

u cos2 x + v sin 2 x = 0

98.

u ( −2sin 2 x ) + v ( 2 cos2 x ) = csc 2 x Multiply the first equation by 2 sin 2x, the second by cos 2x, and add the equations: 2v sin 2 2 x + 2v cos 2 2 x = csc 2 x ⋅ cos 2 x 2v = cot 2 x v = 12 cot 2 x

(d) For x = 1.75, y ≈ 12 candy bars 91. True. A consistent linear system has either one solution or an infinite number of solutions.

Hence, u cos2 x + 12 cot 2 x ⋅ sin 2 x = 0 u = − 12 .

99. −11 − 6 x ≥ 33 −6 x ≥ 44 x ≤ − 446 = − 223

92. True. If the lines were not parallel, there would be one solution. 93. False. At times, only a reasonable approximation is possible graphically. 94. No, it is not possible for a consistent system of linear equations to have exactly two solutions. Either the lines will intersect once or they will coincide and then the system would have infinite solutions.

− 22 3

− 10

−8

−7

−6

−5

100. −6 ≤ 3 x − 10 < 6 4 ≤ 3x < 16 4 3

≤ x

< 163

4 3

16 3

x 1

95.  x + 3 y = 9 Equation 1  2 x + 6 y = k Equation 2 (a) For the system to have infinitely many solutions, the line must coincide, therefore k = 18. (Equation 2 must be 2 times Equation 1.) (b) For the system to have no solution, the lines must be parallel (but not coincide), therefore k ≠ 18.

−9

101. x − 8 < 10

−10 < x − 8 < 10 −2 < x < 18 x − 2 0 2 4 6 8 10 12 14 16 18

102. x + 10 ≥ −3 is true for all x,

since x + 10 ≥ 0. x

−4 −3 −2 −1

694

Chapter 8

Linear Systems and Matrices 106. ln x − 5ln ( x + 3 ) = ln x − ln ( x + 3 )

103. 2 x 2 + 3 x − 35 < 0

( 2 x − 7 )( x + 5) < 0

= ln

Critical numbers: 72 , − 5 Testing the three intervals, −5 < x < 27 . 7 2

( x + 3)

107. log9 12 − log9 x = log9

108.

−6 −5 −4 −3 −2 −1 0 1 2 3 4

12 x

1 1 1 log6 3 + log6 x = log6 ( 3 x ) 4 4 4 = log6 ( 3x )

104. 3 x 2 + 12 x < 0

3x ( x + 4 ) > 0

Critical numbers: 0, − 4 Checking the three intervals, you obtain x < −4 and x > 0.

109. 2 ln x − ln ( x + 2 ) = ln x 2 − ln ( x + 2 )  x2  = ln    x+2

−6 −5 −4 −3 −2 −1

110.

105. ln x + ln 6 = ln 6 x

12 1 ln x 2 + 4 − ln x = ln x 2 + 4 − ln x 2  x2 + 4   = ln    x  

(

)

(

)

111. Answers will vary. (Make a Decision)

Section 8.3 Multivariable Linear Systems 1. row-echelon

2. ordered triple

2 (8 )

3. Gaussian

− 3 ( 3 ) =− 14 No ?

5 ( −1) + 2(3) = 8

4. nonsquare

No, ( 0, − 1, 3 ) is not a solution.

5. three-dimensional

6. partial fraction decomposition

(d) (1) − ( 0 ) + ( 4 ) = 1 2 (1)

7. A consistent system with exactly one solution is independent. 8. A consistent system with infinitely many solutions is dependent. ?

9. (a) 3 ( 3 ) − ( 5 ) + ( −3 ) =1

Yes

− 3 ( 4 ) = −14 Yes ?

Yes

No, (1, 0, 4 ) is not a solution. 10. (a)

3 (1) + 4 ( 5 ) − 6 = 17 Yes ?

− 3 ( −3 ) =− 14 No

5 (1) − 5 + 2 ( 6 ) = − 2 No

2 (1) − 3(5) + 7 ( 6 ) = − 21 No

5 ( 5 ) + 2 ( −3 ) = 8

No, (1, 5, 6 ) is not a solution.

No, ( 3, 5, − 3) is not a solution. ?

(b) 3 ( −1) − ( 0 ) + ( 4 ) =1 2 ( −1)

5( 0) + 2 ( 4) = 8

2 ( 3)

Yes

− 3 ( 4 ) = − 14 Yes ?

5 ( 0 ) + 2(4)=8

(b)

3 ( −2 ) + 4 ( −4 ) − 2 = 17

5 ( −2 ) − ( −4 ) + 2 ( 2 ) =− 2 Yes ?

Yes

Yes, ( −1, 0, 4 ) is solution.

2 ( −2 ) − 3 ( −4 ) + 7(2) =− 21 No

No, ( −2, − 4, 2 ) is not a solution.

Section 8.3

(c)

3 (1) + 4 ( 3 ) − ( −2 ) = 17 Yes

(b)

Multivariable Linear Systems ?

−4 ( − 332 ) −( −10 ) − 8 (10 ) =− 6 No

5 (1) − 3 + 2 ( −2 ) =− 2 Yes

− 10

3 ( 0 ) + 4 ( 7 ) − ( 0 ) = 17 No

(c)

5 ( 0 ) − 7 + 2 ( 0 ) =− 2 No ?

4 ( 0 ) + (1) − (1) = 0

(d)

Yes

=−

9 4

No No

+ 4 =0 ?

Yes Yes

 1 3  5? 4 −  +   −  − =0 Yes  2 4  4  1 3  5? 7 Yes −8  − − 6   +  −  =− 4  2 4  4 9 Yes 4

=−

13. 2 x − y + 5z = 16 Equation 1  y + 2 z = 2 Equation 2   z = 2 Equation 3 

Back-substitute z = 2 into Equation 2: y + 2(2) = 2 y = −2

Yes, ( − 12 , 34 , − 54 ) is a solution.

Back-substitute z = 2 and y = −2 into Equation 1: 2 x −( −2 ) + 5 ( 2 ) = 16 2x = 4 x=2

Answer: ( 2, − 2, 2 ) 14. 4 x − 3 y − 2 z = 21  6y − 5z = −8   z = −2 

Equation 1 Equation 2 Equation 3

 1 1  3? 4  −  +   −  −  = 0 No  2 6  4  1 1  3? 7 −8  −  −6   + −  =− No 4  2 6  4

Back-substitute z = −2 into Equation 2:

 1 1 3 −  −    2 6

y = −3 Back-substitute y = −3 and z = −2 into Equation 1:

=−

9 No 4

No, ( − 12 , 16 , − 34 ) is not a solution. 12. (a)

Yes, ( − , − 4, 4 ) is a solution.

No, ( − 23 , 45 , − 45 ) is not a solution.

(d)

11 2

 3 5  5? 4 −  +   −  − =0  2 4  4  3 5  5? 7 −8  −  − 6   +  −  =− 4  2 4  4

 1 3 3 −  −    2 4

4 ( − 112 ) − 7 ( −4 )

No, ( 0, 1, 1) is not a solution.

(c)

= 0 Yes

−4 ( − 112 ) −( −4 ) − 8 ( 4 ) =− 6 Yes −4

7 −8 ( 0 ) − 6 (1) + (1) =− No 4 ? 9 =− 3 ( 0 ) − (1) No 4

 3 5 3 −  −    2 4

+ 12

No, ( 18 , − 12 , 21 ) is not a solution.

(b)

4 ( 81 ) − 7 ( − 12 )

No, ( 0, 7, 0 ) is not a solution. 11. (a)

−4 ( 81 ) −( − 21 ) − 8 ( 21 ) =− 6 No − 12

2 ( 0 ) − 3 ( 7 ) + 7 ( 0 ) =− 21 Yes

= 0 Yes

No, ( − 332 , − 10, 10 ) is not a solution.

Yes, (1, 3, − 2 ) is a solution. (d)

+ 10

4 ( − 332 ) − 7 ( −10 )

2 (1) − 3 ( 3 ) + 7 ( −2 ) =− 21 Yes

695

−4 ( −2 ) −( −2 ) − 8 ( 2 ) = − 6 Yes −2

4 ( −2 ) − 7 ( −2 )

6 y − 5 ( −2 ) = −8

6y + 10 = −8 6 y = − 18

4 x − 3 ( −3 ) − 2(−2) = 21 4 x + 9 + 4 = 21 4x = 8

= 0 Yes ?

= 6 Yes

Answer: ( 2, − 3, − 2 )

x=2

Yes, ( −2, − 2, 2 ) is solution.

696

Chapter 8

Linear Systems and Matrices

15. 2 x − y − 3z = 10 Equation 1  y + z = 12 Equation 2   z = 2 Equation 3 

Back-substitute z = 2 into Equation 2: y + 2 = 12 y = 10 Back-substitute y = 10 and z = 2 into Equation 1: 2 x + 10 − 3 ( 2 ) = 10

Answer: ( 3, 10, 2 )

2x = 6 x =3

16.  x − y + 2 z = 22 Equation 1   3 y − 8 z = −9 Equation 2  z = −3 Equation 3 

Back-substitute z = −3 into Equation 2:

3 y = 8 ( −3 ) − 9 = −33  y = − 11

Back-substitute y = −11 and z = −3 into Equation 1: x = −11 − 2 ( −3 ) + 22 = 17 Answer: (17, − 11, − 3 ) 17. 4 x − 2 y + z = 8 Equation 1  − y + z = 4 Equation 2   z = 11 Equation 3 

Back-substitute z = 11 into Equation 2: − y + 11 = 4 − y = −7 y=7 Back-substitute y = 7 and z = 11 into Equation 1: 4 x − 2 ( 7 ) + 11 = 8

4 x − 14 + 11 = 8 4 x = 11 x = 114

Answer:

( 114 , 7, 11)

19.  x − 2 y + 3z = 5   − x + 3 y − 5z = 4 2 x − 3z = 0 

Equation 1 Equation 2 Equation 3

Add Equation 1 to Equation 2. y − 2 z = 9 New Equation 2 This is the first step in putting the system in row-echelon from.  x − 2 y + 3z = 5  y − 2z = 9  2 x − 3z = 0  20.  x − 2 y + 3 z = 5 Equation 1  − x + 3 y − 5 z = 4 Equation 2  − 3 z = 0 Equation 3 2 x Add −2 times Equation 1 to Equation 3. 4 y − 9 z = − 10 New Equation 3 This is a step in putting the system in row-echelon form.  x − 2 y + 3z = 5   − x + 3 y − 5z = 4  4 y − 9 z = 70  21.  x + y + z = 6  2x − y + z = 3 3 x −z=0 

Equation 1 Equation 2 Equation 3

x + y + z = 6   − 3y − z = − 9   − 3 y − 4 z = −18 x + y + z =6   − 3y − z = − 9  − 3z = −9 

( −2 ) Eq. 1+Eq. 2 ( −3) Eq. 1+ Eq. 3 ( −1) Eq. 2+ Eq. 3

−3z = −9  z = 3 −3 y − 3 = −9  y = 2 x + 2 +3 = 6  x =1 Answer: (1, 2, 3)

18. 5 x − 8 z = 22 Equation 1   3 y − 5 z = 10 Equation 2  z = −4 Equation 3  Back-substitute z = −4 in Equation 2:

3 y − 5 ( −4 ) = 10  y = − 103

Back-substitute z = −4 in Equation 1: 5 x − 8 ( −4 ) = 22  x = − 2

Answer: ( −2, − 103 , − 4 )

Section 8.3 22. x + y + z = 5  x − 2 y + 4 z = −1  3 y + 4 z = −1 

Equation 1 Equation 2 Equation 3

x + y + z = 5  3 y − 3z = 6   3 y + 4 z = −1 

Eq. 1 + ( −1) Eq. 2

x + y + z = 5  3 y − 3z = 6   − 7z = 7 

Eq. 2 + ( −1) Eq. 3

 x + y + z = 5  y − z = 2    z = −1 

( 12 ) Eq. 2 (− 17 ) Eq. 3

z = −1 y − ( −1) = 2  y = 1 x + (1) + ( −1) = 5  x = 5 Answer: (5, 1, −1) 23. 2 x + 2z = 2  x y 5 3 + =4   3y − 4z = 4 

Equation 1 Equation 2 Equation 3

 x+ z = 1 ( 12 ) Eq. 1   5x + 3y = 4  3 y 4 z − = 4   x + z= 1  3 y − 5z = −1 ( −5 ) Eq. 1 + Eq. 2   3y − 4z = 4      

+ z= 1 3 y − 5z = −1 z= 5

( −1) Eq. 2 + Eq. 3

3y − 5 ( 5) = − 1  y = 8

x + 5 =1  x = − 4

Multivariable Linear Systems

24. 2 x + 4 y + z = 2  −2 y − 3 z = − 8   x − z = −1 

697

Equation 1 Equation 2 Equation 3

2 x + 4 y + z = 2  −2 y − 3z = − 8   4 y + 3z = 4 

Eq. 1 + ( − 2) Eq. 3

z = 2 2 x + 4 y +  − 2 y − 3 z = − 8   −3 z = −12 

(2) Eq. 2 + Eq. 3

2 x + 4 y + z = 2   −2 y − 3 z = − 8   z = 4 

(− 13 ) Eq. 3

z = 4 − 2 y − 3( 4) = − 8  y = − 2 2 x + 4( − 2) + ( 4) = 2  x = 3 Answer: (3, − 2, 4) 25.  4 x + y − 3z = 11 Equation 1  2 x − 3 y + 2 z = 9 Equation 2  x + y + z = −3 Equation 3   x + y + z = −3 Interchange Equations  2 x − 3 y + 2 z = 9 1 and 3.  4 x + y − 3z = 11   x + y + z = −3  = 15  − 5y   − 3 y − 7 z = 23

( −2 ) Eq. 1 + Eq. 2 ( −4 ) Eq. 1 + Eq. 3

y = −3  −3 ( −3 ) − 7 z = 23

 −7z = 14  z = −2

x + ( −3 ) + ( −2 ) = −3  x = 2 Answer: ( 2, − 3, − 2 )

Answer: ( −4, 8, 5 )

698

Chapter 8

Linear Systems and Matrices

26. 5 x − 3 y + 2 z = 3 Equations 1  2 x + 4y − z = 7 Equations 2  x − 11y + 4 z = 3 Equations 3 

29. 3 x + 3 y + 5z = 1 Equation 1  3 x + 5 y + 9z = 0 Equation 2 5 x + 9 y + 17 z = 0 Equation 3 

 x − 11y + 4 z = 3 Interchange rows.  5 x − 3 y + 2 z = 3 2 x + 4 y − z = 7   x − 11y + 4z = 3   52 y − 18 z = −12 −5Eq. 1 + Eq. 2  26y − 9 z = 1 −2Eq. 1 + Eq. 3   x − 11y + 4z = 3   52 y − 18 z = −12  0 = 7 − 12 Eq. 2 + Eq. 3 

6 x + 6 y + 10 z = 2 2 Eq. 1  3 x + 5 y + 9z = 0 5 x + 9 y + 17z = 0 

Inconsistent; no solution.

 x − 3 y − 7z = 2   84 y + 180z = −36 6 Eq. 2  84 y + 182 z = −35 3.5 Eq. 3 

27. 3 x − 2 y + 4 z = 1 Equation 1   x + y − 2z = 3 Equation 2 2 x − 3 y + 6 z = 8 Equation 3   x + y − 2z = 3 Interchange  3 x − 2 y + 4 z = 1 Equations 1 and 2. 2 x − 3 y + 6 z = 8   x + y − 2z = 3   − 5 y + 10 z = −8 −3 Eq. 1 + Eq. 2  − 5 y + 10 z = 2 −2 Eq. 1 + Eq. 3   x + y − 2z = 3   − 5 y + 10 z = −8  0 = 10 − Eq. 2 + Eq. 3  Inconsistent; no solution.

28.  2 x + 4y + z = −4 Equation 1  2 x − 4 y + 6 z = 13 Equation 2  4 x − 2y + z = 6 Equation 3  2 x + 4y + z = −4   − 8 y + 5z = 17 −Eq. 1 + Eq. 2  − 10 y − z = 14 −2Eq. 1 + Eq. 3  2 x + 4y + z = −4   −40 y + 25z = 85 5 Eq. 2  −40 y − 4 z = 56 4 Eq. 3   2 x + 4y + z = − 4   − 40 y + 25z = 85  − 29 z = −29 − Eq. 2 + Eq. 3  −29z = −29  z = 1

 x − 3 y − 7z = 2 − Eq. 3 + Eq. 1  3 x + 5 y + 9z = 0 5 x + 9 y + 17z = 0   x − 3 y − 7z = 2   14 y + 30z = − 6 −3 Eq. 1 + Eq. 2  24 y + 52 z = −10 −5 Eq. 1 + Eq. 3 

 x − 3 y − 7z = 2   84 y + 180z = −36  2 z = 1 −Eq. 2+ Eq. 3  2 z = 1  z = 12

84 y + 180 ( 12 ) = −36  y = − 32 x − 3 ( − 32 ) − 7 ( 12 ) = 2  x = 1

Answer: (1, − 23 , 12 )

30. 2 x + y + 3z = 1 Equation 1  2 x + 6 y + 8z = 3 Equation 2 6 x + 8 y + 18 z = 5 Equation 3   2 x + y + 3z = 1  5 y + 5z = 2 − Eq. 1 + Eq. 2   5 y + 9z = 2 −3Eq. 1 + Eq. 3   2 x + y + 3z = 1  5 y + 5z = 2   4z = 0 − Eq. 2 + Eq. 3  4z = 0  z = 0 5 y + 5 ( 0 ) = 2  y = 25

2 x + 25 + 3 ( 0 ) = 1  x = 103

Answer: ( 103 , 25 , 0 )

−40 y + 25 (1) = 85  y = − 23

2 x + 4 ( − 32 ) + 1 = −4  x = 12

Answer: ( 12 , − 23 , 1)

Section 8.3 31. 3 x − 3 y + 6z = 6 Equation 1   x + 2 y − z = 5 Equation 2 5 x − 8 y + 13z = 7 Equation 3   x − y + 2z = 2    x + 2y − z = 5  5 x − 8 y + 13z = 7   x − y + 2z = 2  3y − 3z = 3    − 3 y + 3z = −3  x − y + 2z = 2  y− z= 1   0= 0  + z= 3 x  y− z= 1  Let z = a, then: y = a +1

( ) Eq. 1 1 3

( −1) Eq. 1 + Eq. 2 ( −5) Eq. 1 + Eq. 3

x = −a + 3 Answer: ( − a + 3, a + 1, a ) 32.  − x + 3 y + z = 4 Equation 1  4 x − 2 y − 5z = −7 Equation 2  2x + 4 y − 3z = 12 Equation 3  − x + 3 y + z = 4  10y − z = 9 4 Eq. 1 + Eq. 2   10y − z = 20 2 Eq. 1 + Eq. 3   − x + 3y + z = 4  10y − z = 9   0 = 11 − Eq. 2 + Eq. 3  No solution; inconsistent

33.  x − 2 y + 3z = 4 Equation 1   3x − y + 2z = 0 Equation 2  x + 3 y − 4 z = −2 Equation 3   x − 2 y + 3z = 4  5 y − 7 z = −12 −3Eq. 1 + Eq. 2   5 y − 7 z = − 6 −1Eq. 1 + Eq. 3   x − 2 y + 3z = 4  5 y − 7 z = −12   0= 6 − Eq. 2 + Eq. 3  No solution; inconsistent

Multivariable Linear Systems

699

34.  x + 4 z = 13 Equation 1   4 x − 2 y + z = 7 Equation 2  2 x − 2 y − 7 z = −19 Equation 3  + 4 z = 13 x   − 2 y − 15z = −45 −4Eq. 1 + Eq. 2  − 2 y − 15z = −45 −2Eq. 1 + Eq. 3  + 4 z = 13 x   − 2 y − 15z = −45  0 = 0 − Eq. 2 + Eq. 3  z=a

y = − 152 a + 452 x = − 4 a + 13 Answer: ( −4 a + 13, − 152 a + 452 , a ) 35.  x + 2 y + z = 1 Equation 1   x − 2 y + 3z = −3 Equation 2 2 x + y + z = −1 Equation 3  x + 2y + z = 1    − 4 y + 2 z = −4   − 3 y − z = −3

( −1) Eq. 1 + Eq. 2 ( −2 ) Eq. 1 + Eq. 3

x + 2y + z = 1   y − 12 z = 1 ( − 14 ) Eq. 2    3 y + z = 3 ( −1) Eq. 3 x + 2y + z = 1  y − 12 z = 1   5 z = 0 ( −3 ) Eq. 2 + Eq. 3  2 z=0 y =1− 0 =1 x + 2 y + z = 1  x = 1 − 2 = −1

Answer: ( −1, 1, 0 ) 36.  3 x − 2 y − 6 z = −4 Equation 1   −3x + 2 y + 6 z = 1 Equation 2  x − y − 5z = −3 Equation 3   x − y − 5z = −3 Interchange the equations.   3x − 2 y − 6 z = −4 −3x + 2 y + 6 z = 1   x − y − 5z = −3  y + 9 z = 5 −3 Eq. 1 + Eq. 2   − y − 9 z = −8 3 Eq. 1 + Eq. 3   x − y − 5z = −3  y + 9z = 5   0 = −3 Eq. 2 + Eq. 3 

No solution; inconsistent

700

Chapter 8

Linear Systems and Matrices

37.  x + 4 z = 1 Equation 1   x + y + 10 z = 10 Equation 2 2 x − y + 2z = −5 Equation 3 

40.  2x − 3 y + z = −2 Equation 1   −4 x + 9 y = 7 Equation 2  2 x − 3 y + z = −2  3 y + 2 z = 3 2 Eq. 1 + Eq. 2 

 x + 4z = 1  y + 6 z = 9 − Eq. 1 + Eq. 2   − y − 6 z = −7 −2Eq. 1 + Eq. 3 

+ 3z = 1  2x   3y + 2 z = 3 Let z = a, then:

 x + 4z = 1  y + 6z = 9   0 = 2 Eq. 2 + Eq. 3 

2 y = − a +1 3 3 1 x=− a+ 2 2 1 2  3  Answer:  − a + , − a + 1, a  2 3  2 

No solution; inconsistent 38.  x − 2 y + z = 2 Equation 1   2 x + 2 y − 3z = −4 Equation 2  5x + z = 1 Equation 3 

41. 12 x + 5 y + z = 0   23 x + 4 y − z = 0 23x + 4 y − z = 0  12 x + 5 y + z = 0

 x − 2y + z = 2  6y − 5z = −8 ( −2 ) Eq. 1 + Eq. 2    10y − 4 z = −9 ( −5) Eq. 1 + Eq. 3  x − 2y + z = 2  6y − 5z = −8   13 z = 133 ( − 35 ) Eq. 2 + Eq. 3 3 

x − 2 y + z = 2  x = 2 + 2 ( − 12 ) − 1 = 0

Answer: ( 0, − 12 , 1)

39.  x − 2 y + 5z = 2 Equation 1  − z = 0 Equation 2 4 x  x − 2 y + 5z = 2  8y − 21z = −8 −4 Eq. 1 + Eq. 2   x − 2 y + 5z = 2  y − 218 z = −1 18 Eq. 2  − 14 z = 0 2 Eq. 2 + Eq. 1  x  y − 218 z = −1  Let z = a, then y = 218 a − 1 and x = 14 a

Answer: ( 14 a, 218 a − 1, a )

Equation 1 Equation 2 Interchange the equations.

 x + 6y + 3z = 0 2 Eq. 2 − Eq. 1   − 67 y − 35z = 0 −12 Eq. 1 + Eq. 2

To avoid fractions, let z = 67a, then: −67 y − 35 ( 67a ) = 0

z =1 6 y − 5z = −8  6 y = −8 + 5 = −3  y = −

Eq.2 + Eq.1

1 2

y = −35a x + 6 ( −35a ) + 3 ( 67a ) = 0 x = 9a

Answer: ( 9a, − 35a, 67a )

42. 10 x − 3 y + 2 z = 0 Equation 1   19 x − 5 y − z = 0 Equation 2  x − y + 5z = 0 2 Eq. 1 − Eq. 2  19 x − 5 y − z = 0  x − y + 5z = 0   14y − 96 z = 0 −19 Eq. 1 + Eq. 2 Infinite number of solutions. Let z = 7a. Then 96 z 96 ( 7a ) = = 48a 14 14 x = y − 5z = 48a − 5 ( 7a ) = 13a

48  13  Answer: (13a, 48a, 7a ) or  a, a, a  7  7 

43. There are an infinite number of linear systems that have ( 3, − 4, 2 ) as their solution.

One possible system is: 3 + ( −4 ) + 2 = 1

 x+ y+ z= 1  2 ( 3 ) + ( −4 ) + 2 = 4   2 x + y + z = 4  x + y − 3 z = −7 3+ ( −4 ) − 3 ( 2 ) = −7 

Section 8.3 44. There are an infinite number of linear systems that have ( −5, − 2, 1) as their solution.

Multivariable Linear Systems

701

46. There are an infinite number of linear systems that have ( − 32 , 4, − 7 ) as their solution.

One possible system is: 1( −5 ) + 1( −2 ) + 1 = −6

Once possible system is: 2 ( − 32 ) + 4 − ( −7 ) = 8  2x + y − z = 8  3 4 ( − 2 ) + 2 ( 4 ) + ( −7 ) = −5   4 x + 2 y + z = −5  −2 x + 5 y − 3 z = 44 −2 ( − 32 ) + 5 ( 4 ) − 3 ( −7 ) = 44 

 x + y + z = −6  2 ( −2 ) + 1 = −3   2 y + z = −3  2z = 2 2 (1) = 2 

45. There are an infinite number of linear systems that have ( −6, − 12 , − 74 ) as their solution.

One possible system is: −6 + ( − 12 ) + 2( − 74 ) = −10

 x + y + 2z = −10  − ( −6) + 12( − ) + 8( − ) = −14  −x + 12y + 8z = −14  −6 + 14( − 12 ) − 4( − 74 ) = − 6  x + 14 y − 4z = − 6 7 4

1 2

47. There are an infinite number of linear systems that have ( a, a + 4, a) as their solution.

One possible system is:

 1a + 0( a + 4) − 1( a)  − 2a + 1( a + 4) + 3( a)  5a − 7 a + 4 + 2 a ( ) ( )  Let a = 1.

 1(1) + 0(5) − 1(1) = 0 − z = 0  x   6  −2 x + y + 3z = 6 − 2(1) + 1(5) + 3(1) =  5 1 − 7 5 + 2 1 = − 28  5 x − 7 y + 2 z = − 28 () ()   () 48. There are an infinite number of linear systems that have (3a, a, a + 2) as their solution.

One possible system is:

 1(3a) + 3a + 2( a + 2)  0(3a) − 1( a) + 1( a + 2) 2 3a + 2a − 5 a + 2 ( )  ( ) Let a = 2.

 1(3 ⋅ 2) + 3( 2) + 2( 2 + 2) = 20  x + 3 y + 2 z = 20   0 3 2 1 2 1 2 2 2 ⋅ − + + =  −y + z = 2 ( ) ( ) ( )   2 3 ⋅ 2 + 2 2 − 5 2 + 2 = − 4 2 x + 2 y − 5 z = − 4 ) ( ) ( )   ( 49. x + y + z = 8

50. x + 2 y + z = 4

Sample answer: (8, 0, 0), (0, 8, 0), (0, 0, 8), ( 2, 2, 4)

Sample answer: ( 4, 0, 0), (0, 2, 0), (0, 0, 4), ( 2, 1, 0)

4 2

4 x

4 8

Chapter 8

702

Linear Systems and Matrices

51. 3 x + 2 y + 2 z = 12

Sample answer: ( 4, 0, 0), (0, 6, 0), (0, 0, 6), ( 2, 3, 0)

61.

A B 1 1 = = + x + x x ( x + 1) x x + 1 2

1 = A ( x + 1) + Bx = ( A + B ) x + A

 A+ B = 0  A = 1  B = −1 

−1 1 1 1 1 = + = − x − x x x +1 x x +1

62.

1 = A ( 2 x − 3) + B ( 2 x + 3)

52. 5 x + y + 3 z = 15

3 1 Let x = − : 1 = −6 A  A = − 2 6 3 1 Let x = : 1 = 6 B  B = 2 6 1 1 1 1  =  −  4 x2 − 9 6  2x − 3 2 x + 3 

Sample answer: (3, 0, 0), (0, 15, 0), (0, 0, 5), (1, 7, 1) z

12 6

6 x

53.

63.

= ( A + 2B) x + ( A − B)

x−2 A B = + x + 4x + 3 x + 3 x + 1

55.

12 12 A B C = = + + x 3 − 10 x 2 x 2 ( x − 10 ) x x 2 x − 10

56.

x2 − 3x + 2 x2 − 3x + 2 A B C = = + + 4 x 3 + 11x 2 x 2 ( 4 x + 11) x x 2 4 x + 11

58.

59.

60.

( x − 5) 6x + 5

( x + 2)

x( x + 1)

x (3x − 1)

A B C D + + + 2 3 ( x + 2 ) ( x + 2 ) ( x + 2 ) ( x + 2 )4

x + 4 2

( x − 5 ) ( x − 5 ) 2 ( x − 5 )3

x −1 2

A B + 2x − 1 x + 1

5 − x = A ( x + 1) + B ( 2 x − 1)

4 x2 + 3

5− x 5− x = 2 x 2 + x − 1 ( 2 x − 1)( x + 1) =

7 7 A B = = + 2 x − 14 x x ( x − 14 ) x x − 14

54.

57.

1 A B = + 4 x2 − 9 2 x + 3 2 x − 3

A Bx + C Dx + E + 2 + 2 x x +1 ( x2 + 1)

A B C D + 2 + + x x 3x − 1 (3x − 1)2

 A + 2 B = −1  A = −1 − 2 B  A − B = 5 ( −1 − 2 B ) − B = 5  B = −2 and A = 3 5− x 3 −2 = + 2 x2 + x − 1 2 x − 1 x + 1 64.

x−2 A B = + x2 + 4 x + 3 x + 3 x + 1 A ( x + 1) + B ( x + 3) = x − 2

( A + B ) x + ( A + 3B ) = x − 2

 A+ B = 1   A + 3 B = −2 Solving for A and B, A = 25 , B = − 32

x−2 52 32 = − x2 + 4 x + 3 x + 3 x + 1

Section 8.3

65.

x 2 + 12 x + 12 x 2 + 12 x + 12 A B C = = + + 3 x − 4x x ( x − 2 )( x + 2 ) x x + 2 x − 2

69.

x 2 + 12 x + 12 = A ( x + 2 )( x − 2 ) + Bx ( x − 2 ) + Cx ( x + 2 )

 A+ B+ C = 1   − 2 B + 2C = 12  −4 A = 12  A = −3   B+C = 4   −B + C = 6

70.

6 x − 3 = A ( x − 1) + B ( x + 4 )

 A+B+ C = 1  3 B − 3C = 12   −9 A = −9  Solving, A = 1, B = 2 and C = −2

Let x = 1: 3 = 5B  B =

x + 12 x − 9 1 2 2 = + − x3 − 9 x x x −3 x +3

4 x 2 + 2 x − 1 = Ax ( x + 1) + B ( x + 1) + Cx 2 = ( A + C ) x2 + ( A + B) x + B

A +C = 4  = 2 A + B  B = −1  B = −1  A = 3  C = 1

68.

3 5

Let x = −4 : − 27 = −5 A  A =

4 x2 + 2 x − 1 A B C = + 2 + x 2 ( x + 1) x x x +1

x3 + 2 x2 − x + 1 6x − 3 = x −1+ + x2 + 3x − 4 x ( 4 )( x − 1) 6x − 3 A B = + + 4 − 1 + 4 −1 x x x x ( )( )

A x − 9 + Bx ( x + 3 ) + Cx ( x − 3 ) = x + 12 x − 9

67.

2 x3 − x 2 + x + 5 1 17 = 2x − 7 + + x2 + 3x + 2 x +1 x + 2

x + 12 x − 9 x + 12 x − 9 A B C = = + + x3 − 9x x ( x − 3)( x + 3) x x − 3 x + 3

)

A = 1  B = 17

(

 A + B = 18  2 A + B = 19

2C = 10  C = 5  B = −1 x 2 + 12 x + 12 −3 −1 5 = + + x3 − 4 x x x+2 x−2

2 x2 − x2 + x + 5 18 x + 19 = 2x − 7 + x2 + 3x + 2 ( x + 1)( x + 2 ) 18 x + 19

= ( A + B + C ) x + ( −2 B + 2C ) x + ( −4 A )

66.

703

( x + 1)( x + 2 ) x + 1 x + 2 18 x + 19 = A ( x + 2 ) + B ( x + 1) = ( A + B) x + (2 A + B)

Multivariable Linear Systems

27 5

x3 + 2 x2 − x + 1 27 3 = x −1+ + 2 x + 3x − 4 5 ( x + 4 ) 5 ( x − 1) 71.

( x − 1)

= x +3+

6 x2 − 8x + 3

( x − 1)

6 x2 − 8x + 3

( x − 1)

A B C + + x − 1 ( x − 1)2 ( x − 1)3

6 x2 − 8 x + 3 = A ( x − 1) + B ( x − 1) + C 2

= Ax2 + ( −2 A + B) x + ( A − B + C )

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

= 6  A  A B − + = −8 2   A− B+C = 3 

2x − 3

A = 6  B = −8 + 2 ( 6 ) = 4  C = 3 − 6 + 4 = 1

( x − 1)

= 2

A B + x − 1 ( x − 1)2

2 x − 3 = A ( x − 1) + B

( x − 1)

6 4 1 + + + x+3 x − 1 ( x − 1)2 ( x − 1)3

Let x = 1: − 1 = B Let x = 0 : − 3 = − A + B −3 = − A − 1

2=A 2 1 = − 2 2 ( x − 1) x − 1 ( x − 1) 2x − 3

704

72.

Chapter 8

4x4

( 2 x − 1)

Linear Systems and Matrices

x 3 24 x 2 − 16 x + 3 + + 3 2 4 4 ( 2 x − 1)

24x2 −16x + 3 1  A B C   =  + + 3 2 4  2x −1 ( 2x −1) ( 2x −1)3  4( 2x −1)   24x2 −16x + 3 = A( 2x −1) + B( 2x −1) + C 2

= A4x2 + ( −4A + 2B) x + ( A − B + C)

4 A = 24  A = 6 −4 A + 2 B = −16  2 B = 8  B = 4 A− B +C = 3 C =1 4 x4

( 2 x − 1) 73.

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

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

A Bx + C + 2 x −1 x − 2

x = A( x 2 − 2) + ( Bx + C )( x − 1) x = Ax 2 − 2 A + Bx 2 − Bx + Cx − C x = ( A + B) x 2 + (− B + C ) x + (− 2 A − C )

= 0  A + B  − B + C = 1  − 2 A −C = 0  = 0 A + B  − B + C = 1   2B − C = 0 

2 Eq. 1 + Eq. 3

= 0 A + B  −B + C = 1   C = 2 

2 Eq. 2 + Eq. 3

C = 2 − B + C = 1  − B + ( 2) = 1  B = 1 A + B = 0  A + (1) = 0  A = − 1

x (−1) + (1) x + 2 = x3 − x 2 − 2 x − 2 x −1 x2 − 2 x + 2 1 = − + x − 1 x2 − 2

Section 8.3

74.

Multivariable Linear Systems

705

2x2 + x + 8 2x2 + x + 8 = 2 4 2 x + 8 x + 16 ( x 2 + 4) Ax + B Cx + D + 2 x2 + 4 ( x 2 + 4)

2 x 2 + x + 8 = ( Ax + B )( x 2 + 4) + Cx + D = Ax 3 + Bx 2 + ( 4 A + C ) x + ( 4 B + D)  A  B   + 4 A   4B 

= 0 = 2 = 1

+ D = 8

A = 0 B = 2

4(0) + C = 1  C = 1 4( 2) + D = 8  D = 0

(1) x + 0 2x2 + x + 8 0x + 2 = 2 + 2 4 2 x + 8 x + 16 x + 4 ( x2 + 4) x 2 + x 2 + 4 ( x 2 + 4)2

= 75.

x − 12 A B = + x ( x − 4) x x − 4

76.

x − 12 = A ( x − 4 ) + Bx

x − 12 x ( x − 4)

2 ( 4 x − 3) 2

3 2 ,y = − x x−4

y= 3 x 2

8 10

−6

−8

Vertical asymptotes: x = 0 and x = 4

3 5 , y= x −3 x+3 y 8

6 4

2 y=− x−4 2

y= 3 x

3 5 + x −3 x +3

x −9 2 ( 4 x − 3) y= x2 − 9

−6

A B = + x2 − 9 x −3 x +3 2 ( 4 x − 3) = A ( x + 3) + B ( x − 3)

Let x = 3 : 18 = 6 A  A = 3 Let x = − 3: − 30 = − 6 B  B = 5

 A+ B = 1  A = 3, B = −2  = −12 −4 A x − 12 3 2 = − x ( x − 4) x x − 4 y=

2 ( 4 x − 3)

8 10

y=−

2 x−4

−8

Vertical asymptotes: x = 0 and x = 4

The combination of the vertical asymptotes of the terms of the decompositions are the same as the vertical asymptotes of the rational function.

−4

y= 4

−4 −6 −8

Vertical asymptotes: x = ±3

5 x+3

3 x−3

−4

5 x+3

−4 −6

3 x−3

−8

Vertical asymptotes: x = 3, x = −3

The combination of the vertical asymptotes of the terms of the decompositions are the same as the vertical asymptotes of the rational function.

706

Chapter 8

Linear Systems and Matrices

77. s = 12 at 2 + v0t + s0

(1, 128) , ( 2, 80 ) , ( 3, 0 ) 1  128 = 2 a + v0 + s0  a + 2v0 + 2 s0 = 256  80 = 2a + 2v0 + s0  2 a + 2 v0 + s0 = 80  9 0 = a + 3v0 + s0  9a + 6v0 + 2 s0 = 0  2 Solving the system, a = −32, v0 = 0, s0 = 144. 1 Thus, s = ( −32 ) t 2 + ( 0 ) t + 144 2 = −16t 2 + 144. 78. s = 12 at 2 + v0 t + s0

(1, 32 ) , ( 2, 32 ) , ( 3, 0 ) 32 = 12 a + v0 + s0  a + 2 v0 + 2 s0 = 64  32 = 2 a + 2 v0 + s0  2 a + 2 v0 + s0 = 32  0 = 9 a + 3v + s  9 a + 6 v + 2 s = 0 0 0 0 0 2  Solving the system, a = −32, v0 = 48, s0 = 0.

1 ( −32 ) t 2 + 48t 2 = −16t 2 + 48t.

Thus, s =

79. s = 12 at 2 + v0t + s0

(1, 352 ) , ( 2, 272 ) , ( 3, 160 ) 352 = 12 a + v0 + s0  a + 2 v0 + 2 s0 = 704  272 = 2 a + 2 v0 + s0  2a + 2 v0 + s0 = 272 160 = 9 a + 3v + s  9a + 6 v + 2 s = 320 0 0 0 0 2  Solving the system, a = −32, v0 = −32, s0 = 400. Thus, s = 12 ( −32 ) t 2 − 32t + 400 = −16t 2 − 32t + 400.

80. s = 12 at 2 + v0t + s0

(1, 132 ) , ( 2, 100 ) , ( 3, 36 ) 132 = 12 a + v0 + s0  a + 2 v0 + 2 s0 = 264  100 = 2 a + 2 v0 + s0  2 a + 2 v0 + s0 = 100  36 = 9 a + 3v + s  9a + 6 v + 2 s = 72 0 0 0 0 2 

81. y = ax 2 + bx + c passing through (0, 0), (3, 0), ( 4, 4)

c = 0    9a + 3b + c = 0 16a + 4b + c = 4 

16a + 4b + c = 4   9a + 3b + c = 0  c = 0  Substitute c = 0 into Equations 1 and 2. 16a + 4b = 4   9a + 3b = 0

 48a + 12b = 12  −36a − 12b = 0

(3) Eq. 1 (− 4) Eq. 2

48a + 12b = 12  = 12 12a

Eq. 1 + Eq. 2

12a = 12  a = 1 48(1) + 12b = 12  12b = − 36  b = − 3 Answer: a = −1, b = − 3, c = 0 The equation of the parabola is y = x 2 − 3 x. 82. y = ax 2 + bx + c passing through (0, 5), (1, 6), ( 2, 5)

c = 5   + + = 6 a b c  4a + 2b + c = 5 

4a + 2b + c = 5   a + b + c = 6  c = 5  Substitute c = 5 into Equations 1 and 2. 4a + 2b = 0   a + b = 1 4a + 2b = 0  = −2 2a

(− 2) Eq. 1 + Eq. 2

2a = − 2  a = −1

Solving the system a = −32, s0 = 16, s0 = 132.

4( −1) + 2b = 0  2b = 4  b = 2

Thus, s = 12 ( −32 ) t 2 + 16t + 132

Answer: a = −1, b = 2, c = 5

= −16t 2 + 16t + 132.

The equation of the parabola is y = − x 2 + 2 x + 5.

Section 8.3 83. y = ax 2 + bx + c passing through

85.

(−1, 1), (0, − 4), (1, −13)

Multivariable Linear Systems

707

x 2 + y 2 + Dx + Ey + F = 0 passing through

( 0, 0 ) , ( 5, 5) , (10, 0 )  ( 0, 0) : F =0  F= 0  5, 5 : 25 25 5 5 0 5 5 D E F D E =−50 + + + + =  + ( )  (10, 0) : 100 +10D + F = 0  10D =−100 

1 a − b + c =  c = −4  a + b + c = −13 

10 D = −100  D = −10

a + b + c = −13  1 a − b + c =  c = −4 

5 ( −10 ) + 5E = −50  E = 0 The equation of the circle is x 2 + y 2 − 10 x = 0. To graph, solve for y.

Substitute c = − 4 into Equations 1 and 2.

x 2 + y 2 − 10 x = 0

a + b = − 9  5 a − b =

y 2 = − x 2 + 10 x y = ± − x 2 + 10 x

 a + b = −9  = − 4 Eq.1 + Eq.2 2a

Let y1 = − x 2 + 10 x and y2 = − x 2 + 10 x . 6

2a = − 4  a = − 2

( − 2) + b = − 9  b = − 7

−5

Answer: a = − 2, b = − 7, c = − 4 The equation of the parabola is y = − 2 x 2 − 7 x − 4.

84. y = ax 2 + bx + c passing through

(− 2, 9), (−1, 0), (1, 6) 4a − 2b + c = 9   a − b+c = 0  a + b+c = 6 

86.

x 2 + y 2 + Dx + Ey + F = 0 passes through

( 0, 0 ) , ( 0, 6 ) , ( 3, 3) . ( 0, 0 ) : F =0  0, 6 : 36 6 E F + + = 0  E = −6 ) (  0  ( 3, 3 ) : 18 + 3 D + 3E + F = 0  D = The equation of the circle is x 2 + y 2 − 6 y = 0. To graph, complete the square first, then solve for y.

4a − 2b + c = 9   a − b+ c = 0 2a + 2c = 6 

Eq.2 + Eq.3

4a − 2b + c = 9  c = 9 2a − 2a + 2c = 6 

4a − 2b + c = 9  −c = 9  2a  − 3c = 3 

−6

x 2 + y2 − 6 y + 9 = 9 x 2 + ( y − 3) = 9 2

( y − 3) = 9 − x 2 2

Eq.1 + ( − 2)Eq.2

y − 3 = ± 9 − x2 y = 3 ± 9 − x2 Let y1 = 3 + 9 − x 2 and y2 = 3 − 9 − x 2 .

Eq.2 + ( −1)Eq.3

− 3c = 3  c = −1 2a − ( −1) = 9  a = 4 4( 4) − 2b + ( −1) = 9  − 2b = − 6  b = 3 Answer: a = 4, b = 3, c = −1

−9

9 −2

The equation of the parabola is y = 4 x 2 + 3 x − 1.

708

Chapter 8

Linear Systems and Matrices

87.

x 2 + y 2 + Dx + Ey + F = 0 passes through

89. Let x = amount borrowed at 4%,

( −3, − 1) , ( 2, 4 ) , ( −6, 8 ) . ( −3, − 1) : 10 − 3D − E + F = 0  10 = 3D + E − F ( 2, 4 ) : 20 + 2 D + 4 E + F = 0  20 = −2 D − 4 E − F ( −6, 8 ) : 100 − 6 D + 8 E + F = 0  100 = 6 D − 8E − F Answer: D = 6, E = −8, F = 0 10

−9

9 −2

The equation of the circle is x + y + 6 x − 8 y = 0. To graph, complete the squares first, then solve for y.

( x + 6 x + 9 ) + ( y − 8 y + 16) = 0 + 9 + 16 2

y − 4 = ± 25 − ( x + 3)

88.

and y2 = 4 − 25 − ( x + 3 ) . 2

91. Let C = amount in certificates of deposit. Let M = amount in muncipal bonds. Let B = amount in blue-chip stocks. Let G = amount in growth stocks.

3 13 2   x − 2  + ( y + 1) = 4  

  C+ M+ B + G = 500,000  0.03C + 0.05M + 0.08 B + 0.1G = 0.05 ( 500,000 )  1  B + G = ( 500,000 ) 4  The system has infinitely many solutions. Let G = s, then B = 125,000 − s

13  3 − x−  4  2

y = −1 ±

13  3 − x−  4  2

M = 187,500 − s C = 187,500 + s.

Let y1 = −1 +

Answer: (187,500 + s, 187,500 − s, 125,000 − s, s )

13  3 −  x −  and 4  2

y2 = −1 −

13  3 −x −  . 4  2

So, $20,000 at 4%, $2500 at 6%, and 7500 at 8%.

Solving the system, x = $300,000, y = $400,000, and z = $75,000.

)

y +1 = ±

x + ( 2500) + (7500) = 30,000  x = 20,000

z = amount invested at 10%.

9 9  2 2  x − 3x + 4  + y + 2 y + 1 = 0 + 4 + 1  

13  3 − x−  4  2

( − 4) Eq.1 + Eq.2

x+ y + z = 775,000   0.08 x + 0.09 y + 0.1z = 67,500  x − 4z = 0 

−2 E = −4  E = 2 3D = −9  D = −3 The equation of the circle is x 2 + y 2 − 3 x + 2 y = 0. To graph, complete the squares first, then solve for y.

y + z = 30,000 x +  + = 35,000 2 y 4 z   3y − z = 0 

90. Let x = amount invested at 8%, y = amount invested at 9%, and

( 0, 0 ) , ( 0, − 2 ) , ( 3, 0 )  ( 0, 0 ) : F =0 F = 0  − − + = 0  −2 E = −4 E F 0, 2 : 4 2 ( )   + F = 0  3 D = −9  ( 3, 0 ) : 9 + 3D

( y + 1) =

(100)Eq.2

2( 2500) + 4 z = 35,000  z = 7500

x 2 + y 2 + Dx + Ey + F = 0 passing through

(

y + z = 30,000  x +  4 x + 6 y + 8 z = 155,000  3y − z = 0 

14 y = 35,000  y = 2500

y = 4 ± 25 − ( x + 3)

Let y1 = 4 + 25 − ( x + 3 )

x + y + z = 30,000   1550 0.04 x + 0.06 y + 0.08 z =  − = 3 y z 0 

x + y + z = 30,000  2 y + 4 z = 35,000   14 y = 35,000 Eq.2 + ( 4) Eq.3 

( x + 3) + ( y − 4 ) = 25 2 2 ( y − 4 ) = 25 − ( x + 3) 2

y = amount borrowed at 6%, and z = amount borrowed at 8%.

−2

One possible solution: Let s = $100,000. Certification of deposit: $287,500 Municipal bonds: $87,500 Blue-chip stocks: $25,000 Growth stocks: $100,000

−4

Section 8.3 92. Let C = amount in certificates of deposit. Let M = amount in muncipal bonds. Let B = amount in blue-chip stocks. Let G = amount in growth stocks.

  C+ M + B+ G = 500,000  0.02C + 0.04 M + 0.1B + 0.14G = 0.06(500,000)  1  B+ G = (500,000) 4  The system has infinitely many solutions. Let G = s , then B = 125,000 − s M = 500,000 − 2 s C = 2 s − 125,000. Answer: ( 2 s − 125,000, 500,000 − 2 s, 125,000 − s, s )

One possible solution is: Let s = 100,000. Certificates of deposit: $75,000 Municipal bonds: $300,000 Blue-chip stocks: $25,000 Growth stocks: $100,000 93. Let x = number of 1-point free throws.

Let y = number of 2-point baskets. Let z = number of 3-point baskets.  x + 2 y + 3z  x − y x − z 

 3 x + 4 y + 5z = 72  y − 2z = 2  x −z= 0 

Solving the system, x = 4, y = 10, z = 4. 4 par-3 holes, 10 par-4 holes, and 4 par-5 holes 95.  I1 − I 2 + I 3 = 0 Equation 1  = 7 Equation 2  3 I1 + 2 I 2  I I 2 + 4 = 8 Equation 3 2 3   I1 − I 2 + I 3 = 0  5 I 2 − 3I 3 = 7 −3 Eq. 1 + Eq. 2   2 I 2 + 4 I3 = 8   I1 − I 2 + I 3 = 0  10 I 2 − 6 I 3 = 14 2 Eq. 2   10 I 2 + 20 I 3 = 40 5 Eq. 3   I1 − I 2 + I 3 = 0  10 I 2 − 6 I 3 = 14   26 I 3 = 26 

− Eq. 2 + Eq. 3

I 1 − 2 + 1 = 0  I1 = 1

−3 =

Answer: I1 = 1 ampere, I 2 = 2 amperes,

= −105 =

(−1) Eq.1 + Eq.2

−3

 x + 2 y + 3 z = 93  y + z = 35 − 13 Eq.2   2 y + 4 z = 96 Eq.1 + ( −1)Eq.2 

( )

x + 2 y  y   

94. Let x = number of par-3 holes. Let y = number of par-4 holes. Let z = number of par-5 holes.

10 I 2 − 6 (1) = 14  I 2 = 2

= −12

2 y + 3z x +  3 y − 3z −  x − z 

709

26 I 3 = 26  I 3 = 1

Multivariable Linear Systems

I 3 = 1 ampere

96.  t1 − 2t2 = 0   − = 128  2t2 − 2a = 128 t a 2  1  t2 + a = 32  −2t2 − 2a = −64  −4 a = 64

a = −16 t2 = 48

= 35

t1 = 96

= 26

2 z = 26  z = 13 y + ( 26) = 35  y = 9 x + 2(9) + 3(13) = 93  x = 10 The University of Connecticut scored 10 free throws, 22 two-point baskets, and 13 three-point baskets to score a total of 93 points.

Answer: t1 = 96 lb, t2 = 48 lb, a = −16 ft sec 2 97. Least squares regression parabola through ( −4, 5) , ( −2, 6 ) , ( 2, 6 ) , ( 4, 2 )  4c + 40a = 19  = −12 40b   40c + 544 a = 160 

Solving the system, a = − 245 , b = − 103 , and c = 416 . Thus, y = − 245 x 2 − 103 x + 416 .

710

Chapter 8

Linear Systems and Matrices

98. Least squares regression parabola through ( −2, 0 ) , ( −1, 0 ) , ( 0, 1) , (1, 2 ) ( 2, 5)  5c + 10 a = 8  10 b = 12  10 c + 34 a = 22 

Solving the system, a = 37 , b = 65 , c = 26 . 35

. Thus, y = 37 x 2 + 65 x + 26 35

102. (a) a (120 ) + b (120 ) + c = 68   2 a (140 ) + b (140 ) + c = 55  2 a (160 ) + b (160 ) + c = 30 Solving the system, a = −0.015, b = 3.25 and c = −106. 2

y = −0.015 x 2 + 3.25 x − 106

(b)

100

99. Least squares regression parabola through ( 0, 0 ) , ( 2, 2 ) , ( 3, 6 ) , ( 4, 12 )  4c + 9b + 29a = 20   9c + 29b + 99a = 70 29c + 99b + 353a = 254 

110

Solving the system, a = 1, b = −1, and c = 0. Thus, y = x 2 − x.

 4c + 6b + 14a = 25   6c + 14b + 36a = 21 14c + 36b + 98a = 33 

Solving the system, a = − 54 , b = 209 , and c = 199 . 20

. Thus, y = − 54 x 2 + 209 x + 199 20 101. (a)  a ( 30 ) + b ( 30 ) + c = 55   2  a ( 40 ) + b ( 40 ) + c = 105  2  a ( 50 ) + b ( 50 ) + c = 188 Solving the system, a = 0.165, b = −6.55, and c = 103. 2

y = 0.165 x 2 − 6.55 x + 103

(b)

500

100. Least squares regression parabola through ( 0, 10 ) , (1, 9 ) , ( 2, 6 ) , ( 3, 0 )

170

120 p 120 p , 0 ≤ p ≤ 100 = 10,000 − p2 (100 − p )(100 + p ) 120 p

A − B = 120   0 100 A + 100 B =

Hence, A = 60, B = −60 and 60 60 120 p − = . 100 − p 100 + p 10,000 − p 2 104. (a)

2000 ( 4 − 3 x )

, 0 ≤ x ≤1

(11 − 7 x )( 7 − 4 x ) (11 − 7 x ) ( 7 − 4 x ) 2000 ( 4 − 3 x ) = A ( 7 − 4 x ) + B (11 − 7 x )  −6000 = −4 A − 7 B    8000 = 7 A + 11B

2000 ( 4 − 3 x )

(11 − 7 x )( 7 − 4 x ) 60

(100 − p )(1000 + p ) 100 − p 100 + p A (100 + p ) + B (100 − p ) = 120 p

A = −2000 B = 2000

2000 −2000 + 11 − 7 x 7 − 4 x

2000 2000 − 7 − 4 x 11 − 7 x

2000 7 − 4x 2000 y2 = 11 − 7 x

(b) y1 =

700

2000 7 − 4x 2000 11 − 7x

105. False, The coefficient of y in the second equation is not 1. 106. True. A common point would be a solution.

Section 8.3 107. A = −1  B = 1. No, the problem was not worked correctly. You must first divide the improper fraction. 108.  x + y  y + z   + z  x ax + by + cz

= = = =

2 2 2 0

Multivariable Linear Systems

115. (a) f ( x ) = x 3 + x 2 − 12 x

(

711

)

= x x 2 + x − 12 = x ( x + 4 )( x − 3 )  x = 0, − 4, 3 y

(b) 25 20

Sample answers: (a) a = 1, b = 1, c = −2 (b) a = 1, b = 1, c = 2 (c) Not possible

−6

−2

−5

− 10

109. No, they are not equivalent. In the second system, the constant in the second equation should be −11 and the coefficient of z in the third equation should be 2.

− 15

116. (a) f ( x ) = −8 x 4 + 32 x 2

(

)

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

110. ( x, y ): (3,3), ( 4, 6), (5, 10)

 x = 0, 0, − 2, 2

 9a + 3b + c = 3  16a + 4b + c = 6 25a + 5b + c = 10 

(b) 35

111. Answers will vary. Sample answer:  2x + y − 5z = 3  − 4 x − 2 y + 10 z = 7

−5 −4 −3

112. When using Gaussian elimination to solve a system of linear equations, a system has no solution when there is a row representing a contradictory equation such as 0 = N , where N is a nonzero real number.  x + y = 3 Equation 1 For instance:   − x − y = 3 Equation 2

−1

3 4 5

117. (a) f ( x ) = 2 x 3 + 5 x 2 − 21x − 36

= ( 2 x + 3 )( x + 4 )( x − 3 )

 x = − 32 , − 4, 3 y

(b) 20 10

 x+y=3  0 = 6 Eq. 1 + Eq.2 

−6

−2

No solution

− 30

 y + λ = 0 113.    x = y = −λ λ = 0 x +   x + y − 10 = 0  2 x − 10 = 0 x=5

− 40 − 50 − 60

118. (a) f ( x ) = 6 x 3 − 29 x 2 − 6 x + 5

= ( x − 5 )( 2 x + 1)( 3 x − 1)

y=5

λ = −5  2x + λ = 0  x = y = −λ 2 114.   2y + λ = 0  x + y − 4 = 0  2x − 4 = 0 2x = 4 x=2 y=2 λ = −4

 x = 5, − 12 , 13

(b)

y 10

−4

−2

712

Chapter 8

Linear Systems and Matrices

119. 4 3 tan θ − 3 = 1

4 3 tan θ = 4 tan θ =

θ=

1 3

π 6

3 3

120. 6cos x − 2 = 1 6cos x = 3 1 cos x = 2

+ 2nπ , n is an integer 3 5π x= + 2nπ , n is an integer 3 x=

+ nπ , n is an integer

121. Answers will vary. (Make a Decision)

Section 8.4 Matrices and Systems of Equations 1.

matrix

reduced row-echelon form

Gauss-Jordan elimination

 −2 x + 3 y = 5 Yes, the coefficient matrix for the system   6x + 7y = 4  −2 3 is  .  6 7

 −2 x + 3 y = 5 No, the augmented matrix for the system   6x + 7y = 4  −2 3  5  is   is 2 × 3.  6 7  4

Yes, the augmented matrix is row-equivalent to its reduced row-echelon form because two matrices are row equivalent when one can be obtained from the other by a sequence of elementary row operations.

Since the matrix has one row and two columns, its dimension is 1 × 2.

Since the matrix has one row and four columns, its dimension is 1 × 4.

Since the matrix has three rows and one column, its dimension is 3 × 1.

10. Since the matrix has three rows and three columns, its dimension is 3 × 3. 11. Since the matrix has two rows and two columns, its dimension is 2 × 2. 12. Since the matrix has two rows and four columns, its dimension is 2 × 4. 13.  4 x − 3 y = −5  − x + 3 y = 12  4 −3  −5     −1 3  12  The dimension is 2 × 3.

14. 7 x + 4 y = 22  5 x − 9 y = 15 4  22  7   − 5 9  15   The dimension is 2 × 3.

15.  x + 10 y − 2 z = 2

  5x − 3 y + 4 z = 0 2 x + y =6 

 1 10 −2  2    5 −3 4  0 :  2 1 0  6 The dimension is 3 × 4.  1 −3 1  1 16.  0 4 0  0   0 0 7  −5

The dimension is 3 × 4. 17.  7 x − 5 y + z = 13  − 8 z = 10 19 x 1  13  7 −5   −  10  19 0 8  The dimension is 2 × 4.

18. 9 x + 2 y − 3z = 20   − 25 y + 11z = −5 2 −3  20  9   − 0 25 11  −5  The dimension is 2 × 4. 3 4  0  19.    1 −1  7  3 x + 4 y = 0   x− y= 7

Section 8.4  7 −5  2  20.   0  −2  8 7 x − 5 y = 2  8 x = −2 

R2 →  1 −1 − 3  R1 → 2 4 8 2 6 4  − 2 R1 + R2 − 2 R1 + R3

31. (a) 22. 6 x + 2 y − z − 5w = −25  + 7 z + 3w = 7 − x  4 x − y − 10 z + 6 w = 23   8 y + z − 11w = −21

(i)

(ii)

Multiply Row 1 by − 14 .

24. Add − 3 times Row 2 to Row 1. 25. 5 times Row 1 added to Row 3.

(iii) (iv)

26. Interchange Rows 1 and 2.  1 4 3    2 10 5 

27.

1  −2R1 + R2 → 0 6 8 3    4 −3 6 

28. 1 3

29.

3  2 −1

2  3 9

 1 −1 − 3  → 0 6 14 → 0 8 10

2  −1  5 

 −3 4  22     6 −4  −28 

R2 + R1 →  3 0  −6    6 −4  −28   3 0  −6    −2 R1 + R2 → 0 −4  −16   3 0  −6  − 14 R2 → 0 1  4  1 R →  1 0  −2  3 1   0 1  4 

Answer: x = −2, y = 4 (b)

−3 x + 4 y = 22 Equation 1   6 x − 4 y = −28 Equation 2 Add Equation 1 and Equation 2 to eliminate y: 3 x = −6 x = −2 Substitute x = −2 into Equation 1: −3 ( −2 ) + 4 y = 22 4 y = 16 y=4

R1 →  1 2 83     4 −3 6 

Answer: ( −2, 4 )

 1 1 4 −1    3 8 10 3  −2 1 12 6 

1  −3 R1 + R2 → 0 2 R1 + R3 → 0  1  1 R → 0 5 2  0

713

8 3 2 4    1 −1 − 3 2 2 6 4 9 

30.

 0 12 3  0   21.  −2 18 5  10  1 7 −8  43  12 y + 3z = 0   −2 x + 18 y = 10  x + 7 y − 8 z = 43 

23.

Matrices and Systems of Equations

1 5 3 1 1 3

4 −1  −2 6   20 4  4 −1  6 − 25 5  20 4 

714

Chapter 8

Linear Systems and Matrices

 7 13 1  −4    32. (a)  −3 −5 −1  −4   3 6 1  −2 

(i)

(ii)

(iii)

(iv) (v)

R2 + R1 →  4 8 0  −8     −3 −5 −1  −4   3 6 1  −2  1 R →  1 2 0  −2  4 1    −3 −5 −1  −4   3 6 1  −2 

 1 2 0  −2    R3 + R2 →  0 1 0  −6   3 6 1  −2 

1  0  −3R1 + R3 → 0 −2 R2 + R1 →  1  0 0

2 0  −2   1 0  −6  0 1  4  0 0  10   1 0  −6  0 1  4 

Answer: x = 10, y = −6, z = 4 (b)

 7 x + 13 y + z = −4 Equation 1  −3 x − 5 y − z = −4 Equation 2  3 x + 6 y + z = −2 Equation 3  Add Equation 2 and Equation 3: y = −6 Substitute y = −6 into Equations 1 and 2 and add the equations: 7 x + 13( −6) + z = −4 −3 x − 5( −6) − z = −4 4 x = 40 x = 10 Substitute x = 10 into Equation 2: −3(10) − 5( −6) − z = −4 z=4

33. (i)

(ii)

(iii)

(iv)

34. (i)

(ii)

(iii)

(iv)

(v)

Answer: (10, − 6, 4 )

(c) Answers will vary. 1 0 0 0  35.  0 1 1 5   0 0 0 0  This matrix is in reduced row-echelon form.  1 0 1 0 36.  0 1 0 2   0 0 1 0  The matrix is in row-echelon form, but not reduced row-echelon form. There is a one above the leading one of row three.

Section 8.4  3 0 3 7   37. 0 −2 0 4  0 0 1 5 The first nonzero entries in rows one and two are not one. The matrix is not in row-echelon form.

44.

1 3 0 0    38.  0 0 1 8   0 0 0 0 

R1 →  1 1 1    −1 0 −4   2 4 −2   1 1 1   R1 + R2 → 0 1 −3 −2 R1 + R3 → 0 2 −4      − R2 + R1 → 1 0 4   0 1 −3 −2 R2 + R3 → 0 0 2     1 0 4   0 1 −3 1 R → 0 0 1 2 3   −4 R3 + R1 →  1 0 0    3R3 + R2 → 0 1 0  0 0 1

 1 − 4 5   2 R1 + R2 → 0 − 2 4 1 − 4

5  1 − 2

1 − 3 2   5 0 7

42.

 1 − 3 2 (− 5) R1 + R2 → 0 15 − 3 2 1 − 3  1 R → 0 1 − 15   15 2

( ) 43.

 1 2 −1 3    3 7 −5 14   −2 −1 −3 8   1  −3R1 + R2 → 0 2 R1 + R3 → 0  1  0 −3R2 + R3 → 0 

−1

1 −2 3 −5 2

−1

1 −2 0 1

3  5 14   3  5 −1 

 3 3 3   −1 0 −4   2 4 −2    1 3

5  1 −4   6 − 6 − 2

(− 12 )R → 0

 1 −3 0 −7    −3 10 1 23  4 −10 2 −24     1 −3 0 −7    3 R1 + R2 → 0 1 1 2 −4 R1 + R3 → 0 2 2 4   

45.

1 1 0 0   40.  0 1 0 −1  0 0 0 2  The first nonzero entry in row three is two, not one. The matrix is not in row-echelon form.

41.

715

 1 −3 0 −7    1 1 2 0 −2 R2 + R3 → 0 0 0 0   

This matrix is in reduced row-echelon form.  1 0 2 1   39.  0 1 −3 10   0 0 1 0  This matrix is in row-echelon form, but not reduced row-echelon form.

Matrices and Systems of Equations

46.

 1 3 2    5 15 9   2 6 10 

 1 3 2   −5R1 + R2 → 0 0 −1   −2 R1 + R3 → 0 0 6  2 R2 + R1 →  1 3 0  0 0 −1   6 R2 + R3 → 0 0 0   1 3 0   −1R2 → 0 0 1 0 0 0   

716

Chapter 8

Linear Systems and Matrices  −4 6 1 0    1 −2 3 −4 

47.

→ R1 →  1 −2 3 −4   → R2 →  −4 1 0 6  1 −2 3 −4    4 R1 + R2 → 0 −7 12 −10   1 −2 3 −4   10  12 − 7 7  1 0 3 8 1 0 − 7 − 7   10  12 7  0 1 − 7

− 71 R2 → 2 R2 + R1 →

 4  R2 →  −1 5 10 −32  48.  5 1 2    4  −1 5 10 −32  R1 →  5 1 2  −1 5 10 −32    5 R1 + R2 →  0 26 52 −156  ( −1) R1 →  1 −5 −10 32    1 R → 0 1 2 −6  26 2

2 5 R2 + R1 →  1 0 0   0 1 2 6 − 

52. x − 2 z = − 7  9  y + z =  z = −3  y + ( − 3) = 9 y = 12 x − 2( − 3) = − 7 x = −13

Answer: ( −13, 12, − 3)  1 0  7 53.    0 1  −5  x=7 y = −5

Answer: ( 7, − 5)  1 0  −2  54.   0 1  4  x = −2 y=4

Answer: ( −2, 4 )

49.  x − 2 y = 4  y = −3 

x = 2 y + 4 = 2 ( −3 ) + 4 = −2

Answer: ( −2, − 3)

55.  1 0 0  −4    0 1 0  −8 0 0 1  2  x = −4

y = −8 z=2

50.  x + 8 y = 12  y=3 

Answer: ( −4, − 8, 2 )

x + 8 ( 3) = 12 x = −12

Answer: ( −12, 3 ) 51. x − y + 4 z = 0  y − z = 2   z = −2  y − ( − 2) = 2

56.  1 0 0  3   0 1 0  −1 0 0 1  0  x=3

y = −1 z=0 Answer: ( 3, − 1, 0 )

y = 0

x − 0 + 4( − 2) = 0 x = 8 Answer: (8, 0, − 2)

Section 8.4 57.  x + 2 y = 7  2 x + y = 8  1 2  7   2 1  8  1 2  7   −2 R1 + R2 → 0 −3  −6 

 1 2  7   0 1  2  y=2 x + 2 (2) = 7  x = 3 − 13 R2 →

Answer: ( −3, 2 ) 58. 2 x + 6 y = 16  2 x + 3 y = 7 2 6  16    2 3  7  2 6  16     0 −3  −9 

Matrices and Systems of Equations

60.  x + 2 y = 0   x+ y=6 3 x − 2 y = 8 

1 2   1  1 3 −2   1 2   − R1 + R2 → 0 −1  −3 R1 + R3 → 0 −8   1 2   − R2 → 0 −1  −8 R2 + R3 → 0 0   No solution, inconsistent

59. − x + y = −22  3 x + 4 y = 4 4 x − 8 y = 32 

 −1 1  −22    4  3 4   4 −8  32     −1 1  −22    3R1 + R2 → 7  −62   0  0 −4  −56  4 R1 + R3 →    −1 1  −22    → R2 →  0 −4  −56  → R3 →  0 7  −62     −1 1  −22    − 14 R2 → 14   0 1   0 0  −160  −7 R2 + R3 →   No solution, inconsistent

0  6 8  0  6 8  0  6 −40  

61.  x + 2 y − 3 z = − 28  4 y + 2z = 0  − x + −5 y − z =   1 2 − 3  − 28   2  0  0 4 −1 1 −1  − 5  

 2 0  −2    0 1  3 y = 3, x = −1

Answer: ( −1, 3)

717

1R 4 2

R1 + R3

 1 2 − 3  − 28   1  → 0 1 0 2 → 0 3 − 4  − 33

− 2 R2 + R1 →  1 0 − 4  − 28   1  0 0 1 2 − 3R2 + R3 → 0 0 − 11  − 33 2

2R − 11 3

 1 0 − 4  − 28   1  0 0 1 2 1  6 → 0 0

4 R3 + R1 →  1 0 0  − 4   − 12 R3 + R2 → 0 1 0  − 3 0 0 1  6  Answer: ( − 4, − 3, 6)

718

Chapter 8

Linear Systems and Matrices

62. − x + y − 2z = − 4  8  4 x + 2 y + 3z =  2x + 4 y − 7z = 1  −1 1 − 2  − 4   8  4 2 −3   2 4 −7  1  (−1) R1 →  1 −1 2  4   4 2 − 3  8 2 4 − 7  1   2  4  1 −1  (− 4) R1 + R2 → 0 6 −11  − 8 (− 2) R1 + R3 → 0 6 −11  − 7  1 −1 2  4   1 R → 0 1 − 11  − 43  6 2 6   1 R → 0 1 − 11  − 76  6 3 6 The corresponding system of equations is x − y + 2 z = 4  11 y − 6 z = − 34 .   y − 11 z = − 76 6 

z = − 43 and y − 11 z = − 76 , the system is Because y − 11 6 6 inconsistent and there is no solution. 63.  3 x + 2 y − z + w = 0   x − y + 4 z + 2w = 25  −2 x + y + 2 z − w = 2  x + y + z + w = 6

 3 2 −1 1   1 −1 4 2  −2 1 2 −1   1 1 1 1 1  0 0  0 1  0 0  0 1  0 0  0

0   25  2   6  

4 2  25 −1  5 −13 −5  −75 −1 10 3  52   2 −3 −1  −19  4 2 −1 1 −10 −3 0 37 10 0 17 5

 25   −52   185   85

−1



4 2 1 −10 −3 0 37 10 15 0 0 37

25   −52   185  0  

w = 0, z = 185 = 5, y = −52 + 10 ( 5 ) = −2 37 x = 25 + ( −2 ) − 4 ( 5 ) = 3

Answer: ( 3, − 2, 5, 0 )

64.  x − 4 y + 3 z − 2 w = 9  3 x − 2 y + z − 4 w = −13    −4 x + 3 y − 2 z + w = − 4  −2 x + y − 4 z + 3w = −10

 1 −4 3 −2  9   1 −4  13  3 −2  −4 3 −2 1  −4    3  −10  1 −4  −2  1 −4 3 −2  9   − − 0 10 8 40  2   0 −13 10 −7  32    8  0 −7 2 −1   1 −4 3 −2  1 − 45 0 5 1  2 22 0 0 − 5 − 5  18 2 0 0 − 5 5  1 −4 3 −2  1 − 45 0 5 1  0 0 1 11  0 40 0 0

9  −4   −20    −20   

9  −4  50    160    

w = 160 = 4, z = 50 − 4 (11) = 6 40 y = −4 + 54 ( 6 ) − 15 ( 4 ) = 0

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

Answer: ( −1, 0, 6, 4 )

− 3z = −2 65.  x  3 x + y − 2 z = 5  2x + 2y + z = 4  1  3 2  1  −3 R1 + R2 → 0 −2 R1 + R3 → 0  1  0 −2 R2 + R3 → 0  1  0 − 17 R3 → 0  3 R3 + R1 →  1  −7 R3 + R2 → 0 0

0 −3  −2   1 −2  5 2 1  4   0 −3  −2  1 7  11 2 7  8  0 −3  −2   1 7  11 0 −7  −14   0 −3  −2   1 7  11 0 1  2  0 0  4  1 0  −3 0 1  2 

Answer: ( 4, − 3, 2 )

Section 8.4 66. 2 x − y + 3 z = 24  2 y − z = 14  7 x − 5 y = 6 

Matrices and Systems of Equations

67.  − x − y − 3 z = −12  6 y − 4z =  2x − − 2 x + 4 y + 14 z = 19 

2 −1 3  24    0 2 −1  14   7 −5 0  6  14 −7 21  168    2 −1  14   0 14 −10 0  12 

 −1 −1 − 3  −12   6  2 −1 − 4  − 2 4 14  19  

(−1) R1

14 −7 21  168    14   0 2 −1   0 −3 −21  −156 

(− 2) R1 + R2

21  168  14 −7   0 1 7  52    0 0 −15  −90 

− 13 R2

R1 →  2  0 − 151 R3 →  0 R2 + R1 →  2  −7 R3 + R2 →  0  0 1 7

−1 3  24   1 7  52  0 1  6  0 10  76   1 0  10  0 1  6 

R1 →  1 0 5  38   0 1 0  10  0 0 1  6  −5R3 + R1 →  1 0 0  8   0 1 0  10 0 0 1  6 1 2

719

2 R1 + R3

( −1) R2 + R1 (− 6) R2 + R3

→ 1 1 3  12   − − 2 1 4  6  − 2 4 14  19   1 3  12 1   → 0 − 3 −10  −18 → 0 6 20  43

 1 1 3  12    6 → 0 1 10 3 0 6 20  43   1 →  1 0 − 3  6   0 1 10  6 3   → 0 0 0  7 

Because 0 ≠ 7, the system is inconsistent and there is no solution.

3z = 3 68. 2 x +  4 3 7 x y z=5 − +  8 x − 9 y + 15z = 9 

Answer: (8, 10, 6)

−2 R1 + R2 → −4 R1 + R3 →

−3 R2 + R3 → R1 → − R2 → 1 2 1 3

z=a

 2 0 3  3    4 −3 7  5  8 −9 15  9  2 0 3  3   0 −3 1  −1 0 −9 3  −3   2 0 3  3   0 −3 1  −1 0 0 0  0    3 3 1 0  2  2  1 1 0 1 − 3  3    0 0 0  0  

y = 13 a + 13 x = − 23 a + 23

Answer: ( − 23 a + 32 , 13 a + 13 , a )

720

Chapter 8

Linear Systems and Matrices 71.  x + 2 y + z = 8  3 x + 7 y + 6 z = 26

69.  x + y − 5z = 3  − 2z = 1 x 2 x − y − z = 0 

− R1 + R2 → −2 R1 + R3 →

1 1  1 0  2 −1 1 1  0 −1 0 −3

−5  3  −2  −1 −1  0  −5  3   3  −2  9  −6 

 1 1 −5  3   0 −1 3  −2  −3R2 + R3 → 0 0 0  0    R2 + R1 →  1 0 −2  1   − R2 → 0 1 −3  2  0 0 0  0    Let z = a , any real number. y − 3a = 2  y = 3a + 2 x − 2a = 1  x = 2a + 1 Answer: ( 2 a + 1, 3a + 2, a ) 70.  2 2 −1  2     1 −3 1  28  −1 1 0  14  1  0 0 1  0 0

−3 8 −2 −3 −2 0

28   −3  −54  1  42  1  28   1  42  1  114  1 

− R3 + R1 →

 1 −3 0  −86  − R3 + R2 → 0 −2 0  −72  0 0 1  114  3 R2 + R1 →  1 0 0  22    − 12 R2 → 0 1 0  36  0 0 1  114 

Answer: ( 22, 36, 114 )

1 2 1  8   3 7 6  26

(− 3) R1 + R2

 1 2 1  8   → 0 1 3  2

(− 2) R2 + R1

→  1 0 − 5  4   3  2 0 1

(− 2) R2 + R1

→  1 0 − 5  4   3  2 0 1

 x − 5z = 4 The resulting system is  .  y + 3z = 2

Let z = a, any real number. x = 5a + 4 y = − 3a + 2 Answer: (5a + 4, − 3a + 2, a )

x + y + 4 z = 5 72.  2 x + y − z = 9  1 1 4  5   2 1 −1  9

(− 2) R1 + R2 (−1) R2

4  5 1 1   → 0 −1 − 9  −1

 1 1 4  5   → 0 1 9  1

(−1) R2 + R1

→  1 0 − 5  4   0 1 9  1

 x − 5z = 4 The resulting system is  . y + 9z = 1

Let z = a, any real number.

x = 5a + 4 y = − 9a + 1 Answer: (5a + 4, − 9a + 1, a) 73.  x + 2 y + 4z =   2 x + 5 y + 20 z = 10 − x + 3 y + 8z = − 2  2 0  1 1 4  1 0 0        2 5 20 10 0 1 0 6  −     −1 3 8  − 2 0 0 1  − 2    

Answer: (0, − 6, − 2)

Section 8.4 80. (a)

74.  x + y + z = 0  2 x + 3 y + z = 0 3 x + 5 y + z = 0 

(b)

Answer: ( − 3, 6, − 4 )

Yes, the systems yield the same solution. 81.

6 1   6 0    3 0    −9   0

 

5 0  0  0 1  3 0 0  0  0 0  0 

−9 R1 + R3 →

z = 3, y = a, x = −5a

2  −6  1   0 4 0 1  1   0 5 2 6  −3    2 −1 −1  3  0

0 1 0 0

0 0 1 0

0 0 0 1

1 1 1   3 − R2 → 0 1 2  −3 R2 + R3 → 0 0 1   c=5 b + 23 ( 5 ) = 192  b = 2

 1   0  4   −2 

z = −3; y = 5 ( −3 ) + 16 = 1; x = 2 (1) − ( −3 ) − 6 = −1 Answer: ( −1, 1, − 3 ) z = −3, y = −3 ( −3 ) − 8 = 1, x = −1 + 2 ( −3 ) + 6 = −1 Answer: ( −1, 1, − 3 )

Yes, the systems yield the same solution. 78. (a)

z = 2, y = 2 − 4 = −2, x = 3 ( −2 ) − 4 ( 2 ) − 11 = −25. Answer: ( −25, − 2, 2 )

(b)

z = 2, y = −3 ( 2 ) + 4 = −2, x = −4 ( −2 ) − 11 = −3. Answer: ( −3, − 2, 2 )

No, the solutions are not the same. 79. (a)

z = 8, y = 7 ( 8 ) − 54 = 2, x = 4 ( 2 ) − 5 ( 8 ) + 27 = −5 Answer: ( −5, 2, 8 )

(b)

1 1  8  2 1  13 3 1  20  1 1  8  −2 −3  −19  −6 −8  −52 

1 2

a+2+5=8

z = 8, y = −5 ( 8 ) + 42 = 2, x = 6 ( 2 ) − 8 + 15 = 19 Answer: (19, 2, 8 )

No, the solutions are not the same.

8   5 

19 2

a =1

Answer: y = x + 2 x + 5

Answer: (1, 0, 4, − 2 )

(b)

1  4  9 1  0 0

−4 R1 + R2 →

Answer: (−5a, a, 3) 1 −1

f ( x ) = ax 2 + bx + c  f (1) = a + b + c = 8   f (2) = 4 a + 2b + c = 13  f (3) = 9a + 3b + c = 20 

75.  2 x + 10 y + 2 z = 6   x + 5y + 2z = 6   x + 5y + z = 3 −3 x − 15 y − 3z = −9 

z = −4, y = 2 ( −4 ) + 14 = 6, x = 6 − 3 ( −4 ) − 21 = − 3

x = −2 a

77. (a)

z = −4, y = −6 ( −4 ) − 18 = 6, Answer: ( −3, 6, − 4 )

Answer: ( −2 a, a, a )

2  3 76.  1   5

721

x = −3 ( 6 ) + ( −4 ) + 19 = −3

 1 1 1  0  1 0 2  0      2 3 1  0    0 1 −1  0   3 5 1  0   0 0 0  0  Let z = a , any real number. y=a

 2 10 2  5 2  1  1 5 1   −3 −15 −3

Matrices and Systems of Equations

82.

f ( x ) = ax 2 + bx + c  f (1) = a + b + c = 9   f (2) = 4 a + 2b + c = 8  f (3) = 9a + 3b + c = 5 

 1 1 1  9    4 2 1  8  9 3 1  5 1  −4 R1 + R2 → 0 −9 R1 + R3 → 0  1  − 12 R2 → 0 −3 R2 + R3 → 0  c=8

9 1   −2 −3  −28  −6 −8  −76   1 1  9  3 1 2  14  0 1  8  1

b + 32 ( 8 ) = 14  b = 2 a + 8 + 2 = 9  a = −1

Answer: y = − x 2 + 2 x + 8

722

Chapter 8

Linear Systems and Matrices

83.

f ( x ) = ax 2 + bx + c

85.

 f (1) = a + b + c = 2   f ( −2) = 4 a − 2b + c = 11  f (3) = 9a + 3b + c = 16 

1 1 1  2    4 −2 1  11 9 3 1  16     1 1 1  2   −4 R1 + R2 → 0 −6 −3  3   − 9 R1 + R3 → 0 −6 −8  −2   1 1 1  2   0 −6 −3  3   (−1) R2 + R3 → 0 0 −5  −5 −5c = −5  c = 1 −6b − 3c = 3  −6b = 3 + 3 = 6  b = −1 a + b + c = 2  a = 2 +1−1 = 2 Answer: y = 2 x 2 − x + 1 84.

f (−1) = a − b + c = 7 f (1) = a + b + c = − 3

Solving the system, a = −9, b = −5, c = 11. f ( x ) = −9 x 2 − 5 x + 11

86.

1 1 1  −1    4 −2 1  2  4 2 1  −6    1 1 1  −1   −4 R1 + R2 → 0 −6 −3  6    −4 R1 + R3 → 0 −2 −3  −2  1 1 1  −1   0 1 2 1  −2  − 3 R2 →  0 −2 −3  −2    1 1  1  −1   0 2 1  −2     R2 + R3 → 0 0 −2  −4  −2c = −4  c = 2 2b + c = −2  2b = −2 − 2 = −4  b = −2 a + b + c = −1  a = −1 + 2 − 2 = −1 Answer: y = − x 2 − 2 x + 2

f ( x ) = ax 2 + bx + c f (−2) = 4a − 2b + c = − 3 f (1) = a + b + c = − 3 f (2) = 4a + 2b + c = −11

Solving the system, a = −2, b = −2, c = 1. f ( x ) = −2 x 2 − 2 x + 1

87.

f ( x ) = ax 3 + bx 2 + cx + d f (−2) = −8a + 4b − 2c + d = − 7

f (−1) = − a + b − c + d =

f (1) =

a+ b+ c+d = −4

f (2) =

8a + 4b + 2c + d = − 7

Solving the system, a = 1, b = −2, c = −4, d = 1.

f ( x ) = ax 2 + bx + c  f (1) = a + b + c = −1   f ( −2) = 4 a − 2b + c = 2  f (2) = 4 a + 2b + c = −6 

f ( x ) = ax 2 + bx + c f (−2) = 4a − 2b + c = −15

f ( x) = x3 − 2 x 2 − 4 x + 1

88.

f ( x ) = ax 3 + bx 2 + cx + d f (−2) = −8a + 4b − 2c + d = − 17

f (−1) = − a + b − c + d = − 5 f (1) =

a+ b+ c+d =

f (2) =

8 a + 4 b + 2c + d =

Solving the system, a = 1, b = −1, c = 2, d = −1. f ( x) = x3 − x 2 + 2 x − 1

Section 8.4 89.  I1 − I 2 + I 3 = 0  =7  2 I1 + 2 I 2  2 I 4 I + 2 3 =8 

1  2 0  1  −2 R1 + R2 → 0 0  1  R3 → 0 R2 → 0 1  1 R → 0 2 2 0  1  0 −4 R2 + R3 → 0  1  0 − 101 R3 → 0

−1 1  0   2 0  7 2 4  8  −1 1  0   4 −2  7  2 4  8  −1 1  0   2 4  8 4 −2  7   −1 1  0   1 2  4 4 −2  7   −1

0   4  −9   0  4 9  10 

1 

1 2 0 −10 −1 1  1 2  0 1 

 1 5 10 20  95  1 0 0 0  15     1 1 1  26   0 1 0 0  8  1 0 1 −4 0  0  0 0 1 0  2       1 −2 0 0  −1 0 0 0 1  1 x = 15 y =8 z=2 w =1 The server has 15 $1 bills, 8 $5 bills, 2 $10 bills, and one $20 bill.

92. Let x = number of pounds of glossy. Let y = number of pounds of semi-glossy. Let z = number of pounds of matte.

y +  x + z = 100  4.25 y 3.75 z = 480 5.5 x + +   y + z = 50  Solving the system, x = 50, y = 35 and z = 15. 50 pounds of glossy, 35 pounds of semi-glossy, and 15 pounds of matte

13 I1 − 115 + 109 = 0  I1 = 10 amperes

 1 1 1  1,500,000 1 0 0  200,000     74,000  0 1 0  500,000 0.03 0.04 0.06   4 0 −1  0 0 0 1  800,000     Answer: ( 20,000, 500,000, 800,000 ) $200,000 at 3%, $500,000 at 4%, and $800,000 at 6%

723

91. Let x = number of $1 bills. Let y = number of $5 bills. Let z = number of $10 bills. Let w = number of $20 bills.  x + 5 y + 10 z + 20 w = 95   x + y + z + w = 26  = 0 y − 4z   x − 2 y = −1

I 3 = 109 amperes; I 2 + 2 ( 109 ) = 4  I 2 = 115 amperes;

90. Let x = amount at 3%. Let y = amount at 4%. Let z = amount at 6%. x+ y+ z = 1,500,000   0.03 x + 0.04 y + 0.06 z = 74,000  4x z=0 − 

Matrices and Systems of Equations

93.

8x2

( x − 1) ( x + 1) 2

A B C + + x + 1 x − 1 ( x − 1)2

8 x 2 = A ( x − 1) + B ( x − 1)( x + 1) + C ( x + 1) 2

(

)

(

)

8 x 2 = A x 2 − 2 x + 1 + B x 2 − 1 + C ( x + 1) 8 x = ( A + B ) x + ( −2 A + C ) x + ( A − B + C ) 2

 A + B  System of equations: −2 A  A − B   1 1 0  8  1 0 0    rref   −2 0 1  0  → 0 1 0   1 −1 1  0  0 0 1     A = 2, B = 6, C = 4 8x2

( x − 1) ( x + 1) 2

= 8 + C = 0 + C = 0 2  6 4 

2 6 4 + + x + 1 x − 1 ( x − 1)2

724

Chapter 8

94. (a)

Linear Systems and Matrices

f ( x ) = ax 2 + bx + c

96. (a)

f (1) =

c = 5.0  f (0) =   f (15) = 225a + 15b + c = 9.6   f (30) = 900 a + 30b + c = 12.4

y = −0.004 x 2 + 0.367 x + 5

+ b + c = 55.5

f ( 2) = 4a

+ 2b + c = 55.4

160

(c)

120

(c)

Maximum height ≈ 13 feet

2015: f (5) = 0.55(5) − 1.75(5) + 56.7 = $61,700

Strikes ground ( y = 0 ) when x ≈ 104 feet.

2020: f (10) = 0.55(10) − 1.75(10) + 56.7 = $94,200

(

)

−0.004 x − 91.75 x + 2104.5 + 5 + 8.418

Maximum height = 13.418 feet Range: y = 0  x=

−0.367 ± 0.367 2 + 4 ( 0.004 ) 5

95. (a) f (t ) =

−0.008

at 2 +

≈ 103.793 feet

c = 1831 a

+ b + c = 1922

f ( 2) = 4 a

+ 2b + c = 2061

24 0 0 1  1831 1 0 0      67  1 1 1  1922  0 1 0  4 2 1  2061 0 0 1  1831     y = 24t 2 + 67t + 1831 2800

0 1800

2020: f (10) = 24(10) + 67(10) + 1831 = $4901 2

2025: f (15) = 24(15) + 67(15) + 1831 = $8236 2

(d) Answers will vary.

2025: f (15) = 0.55(15) − 1.75(15) + 56.7 = $154,200

(d) Answers will vary. 97. (a)

bt + c

f ( 0) = f (1) =

(d) Complete the square:

(b)

+ 3b + c = 56.4

 1 1 1  55.5  1 0 0  0.55     4 2 1  55.4  0 1 0  −1.75 9 3 1  56.4 0 0 1  56.7     2 y = 0.55t − 1.75t + 56.7 (b)

at 2 + bt + c

f (3) = 9a

 0 0 1  5.0   1 0 0  −0.004       225 15 1  9.6   0 1 0  0.367 900 30 1  12.4  0 0 1  5     (b)

f (t ) =

 x1   x1      

+ x3 − x2 x2

− x4 + x5 + x6 − x6

x3 x4 x5

+ x7 + x7

= = = = = =

600 0 500 600 0 500

1 0 1 0 0 0 0  600   0 1 −1 0 −1 0 0 0  0 1 0 0 1 0 0  500   0 0 1 0 0 1 0  600 x1 − x6 = 0   0 0 0 0 0 − 1 1 1    x 2 − x7 = 0 0 0 0 0 1 0 1  500 x3 + x6 = 600   0 0 0 0 0 0 0 0  x 4 − x6 + x 7 = 0 1 0 0 0 0 −1 0  0 x5 + x7 = 500   0 0 1 0 0 0 0 −1  x6 = s 0 0 1 0 0 1 0  600 x7 = t    0 0 0 1 0 −1 1  0   0 0 0 0 1 0 1  500 0 0 0 0 0 0 0  0   0 0 0 0 0 0 0 0  x1 = s, x2 = t , x3 = 600 − s, x4 = s − t , x5 = 500 − t ,

x6 = s, x7 = t (b) When x6 = 0, x7 = 0. x1 = 0, x2 = 0, x3 = 600, x4 = 0, x5 = 500, x6 = 0, x7 = 0

x1 = 500, x2 = 100, x3 = 100, x4 = 400, x5 = 400, x6 = 500, x7 = 100

Section 8.4 98. (a)

600

x3 600

x5 700

− x2 x2

= 600 0 = = 700

− x4 + x5 + x6 − x6

x3 x4 x5

+ x7 + x7

= 600 0 = = 700

1 0 1 0 0 0 0  600   1 −1 0 −1 0 0 0  0 0 1 0 0 1 0 0  700   0 0 1 0 0 1 0  600   0 0 0 1 0 −1 1  0 0 0 0 0 1 0 1  700 1 0 0 0 0 −1 0  0   0 1 0 0 0 0 −1  0 0 0 1 0 0 1 0  600    0 0 0 1 0 −1 1  0   0 0 0 0 1 0 1  700 0 0 0 0 0 0 0  0   0 0 0 0 0 0 0  0 x1 − x6 = 0 x 2 − x7 = 0 x3 + x6 = 600 x 4 − x6 + x 7 = 0 x5 + x7 = 700 x6 = s x7 = t

x1 = s, x2 = t, x3 = 600 − s, x4 = s − t, x5 = 700 − t,

( Equation 1) + ( Equation 2 ) → new Equation 1 ( Equation 1) + 2 ( Equation 2 ) → new Equation 2 2 ( Equation 1) + ( Equation 2 ) → new Equation 3  x + y + 7 z = −1   x + 2 y + 11z = 0 2 x + y + 10 z = −3 

 1 0 102. (a)   is in reduced row-echelon form. 0 1  1 b (b)   , where b ≠ 0 is in row-echelon form. 0 1  1 0 (c)   , where c ≠ 0 is in neither form. c 1  1 b (d)   , where b ≠ 0 and c ≠ 0 is in neither form. c 1

103. No. Answers will vary.  1 2  −4  104.   0 a  b 

(a) For the system to have exactly one solution, a ≠ 0 and b can be any real number. (b) For the system to have infinitely many solutions a = b = 0. (c) For the system to have no solutions, a = 0, but b ≠ 0. 105. f ( x ) =

7 −x −1 y

x6 = s, x7 = t (c)

When x1 = 600, x7 = 200.

x1 = 600, x2 = 200, x3 = 0, x4 = 400, x5 = 500, x6 = 600, x7 = 200 (d)

725

101.  x + 3z = −2 Equation 1   y + 4 z = 1 Equation 2

700

+ x3

(b) x1 x1

Matrices and Systems of Equations

When x4 = 150, x5 = 350.

x1 = 500, x2 = 350, x3 =100, x4 =150, x5 = 350, x6 = 500, x7 = 350 99. True. If while in the process of using Gaussian elimination, a row is obtained with entries of zero except for the last row entry, then it can be concluded that the system is inconsistent.

− 12 − 8

−4

−4 −8

− 12

Vertical asymptote: x = −1 Horizontal asymptote: y = 0

100. False. Any matrix, including an augmented matrix for a dependent system of linear equations, can be written in reduced row-echelon form.

726

Chapter 8

106. f ( x ) =

Linear Systems and Matrices

4x 5x2 + 2

108. f ( x ) =

x 2 − 36 x +1

0.8 0.6 8

0.4

0.2

−4 −2

x − 12 − 8 − 4

−8 − 12

− 0.8

Horizontal asymptote: y = 0 107. f ( x ) =

Vertical asymptote: x = −1 Slant asymptote: y = x − 1

x2 − 2 x − 3 5 = x+2+ x−4 x−4 y 16 12 8 4

−8

−4

Vertical asymptote: x = 4 Slant asymptote: y = x + 2

Section 8.5 Operations with Matrices 1.

equal

scalars

zero, O

identity

(a) iii

(b) i

(d) v

(a) ii

(b) iv

(d) iii

No. In general AB ≠ BA; matrix multiplication is not commutative.

12.

x + 4 = 2 x + 9  x = −5 2 y = −8  y = −4 z + 2 = 11  z = 9  5 −2   3 A+B=  + 1  −2 3  5 −2   3 (b) A − B =  − 1  −2 3

13. (a)

(e) ii

Given A is a 2 × 3 matrix and B is a 3 × 4 matrix, than AB is a 2 × 4 matrix.

x = 5, y = −8

10.

x = 13, y = 12

11. 2 x + 7 = 5  x = − 1 3 y = 12  y = 4 2 z − 6 = 3 z − 14  z = 8

(c)

1 8 −1 =  6   1 7 1  2 −3 =  6   5 −5 

5 −2  3 ( 5 ) 3 ( −2 )  15 −6  3A = 3  = =  1 3 ( 3 ) 3 (1)   9 3 3

5 −2  3 1  (d) 3A − 2B = 3  − 2  3 1 −2 6 15 −6 −6 − 2  9 −8 = + =   9 3  4 −12  13 −9

Section 8.5

1 2 −3 −2  1− 3 2 − 2 −2 0 14. (a) A + B =  +  = =  2 1  4 2 2 + 4 1+ 2  6 3 1 2 −3 −2  1 + 3 2 + 2  4 4 (b) A − B =  − = =  2 1  4 2 2 − 4 1 − 2 −2 −1 1 2   3 (1) 3 ( 2 )  3 6  (c) 3 A = 3  = =  2 1  3 ( 2 ) 3 (1)  6 3  3 6   −3 −2  3 + 6 6 + 4  (d) 3 A − 2 B =   − 2 =  6 3    4 2  6 − 8 3 − 4   9 10  =   −2 −1  8 −1  1 6   8 + 1 −1 + 6        15. (a) A + B =  2 3 +  −1 −5 =  2 − 1 3 − 5  −4 5  1 10   −4 + 1 5 + 10  5  9   =  1 −2   −3 15  8 −1  1 6   8 − 1 −1 − 6        (b) A − B =  2 3 −  −1 −5 =  2 + 1 3 + 5  −4 5  1 10   −4 − 1 5 − 10   7 −7    =  3 8  −5 −5  8 −1  3 ( 8 ) 3 ( −1)   24 −3        (c) 3 A = 3  2 3 = 3 ( 2 ) 3 ( 3)  =  6 9    −4 5  3 −4    ( ) 3 ( 5 )   −12 15  12   22 −15  24 −3  2 (d) 3 A − 2 B =  6 9  −  −2 −10  =  8 19   −12 15  2 20   −14 −5

Operations with Matrices

727

1 −1 3 −2 0 −5 1− 2 −1− 0 3 − 5 16. (a) A+ B =   +  =  0 6 9 −3 4 −7 0 −3 6 + 4 9 − 7 −1 −1 −2 =  −3 10 2 1 −1 3 −2 0 −5 1+ 2 −1− 0 3 + 5 (b) A− B =   −  =  0 6 9 −3 4 −7 0 + 3 6 − 4 9 + 7 3 −1 8 =  3 2 16  1 −1 3  3 (1) 3 ( −1) 3 ( 3 )  (c) 3 A = 3   =  0 6 9  3 (10 ) 3 ( 6 ) 3 ( 9 )   3 −3 9  =  0 18 27  3 (d) 3 A − 2 B =  0 7 = 6

−3 9  −4 0 −10  −  18 27  −6 8 −14  −3 19  10 41

 4 5 −1 3 4  17. (a) A + B =    1 2 −2 −1 0   1 0 −1 1 0  +   −6 8 2 −3 −7   4 + 1 5 + 0 −1 − 1 3 + 1 4 + 0  =  1 − 6 2 + 8 −2 + 2 −1 − 3 0 − 7   5 5 −2 4 4  =   −5 10 0 −4 −7  4 5 −1 3 4  (b) A − B =    1 2 −2 −1 0   1 0 −1 1 0  −   −6 8 2 −3 −7  4 − 1 5 − 0 −1 + 1 3 − 1 4 − 0  =  1 + 6 2 − 8 −2 − 2 −1 + 3 0 + 7  3 5 0 2 4  =  7 −6 −4 2 7   4 5 −1 3 4  (c) 3 A = 3    1 2 −2 −1 0  3 ( 4 ) 3 ( 5 ) 3 ( −1) 3 ( 3 ) 3 ( 4 )  =   3 (1) 3 ( 2 ) 3 ( −2 ) 3 ( −1) 3 (10 )  12 15 −3 9 12  =   3 6 −6 −3 0  12 15 −3 9 12  (d) 3 A − 2 B =    3 6 −6 −3 0 

0  2 0 −2 2 −   −12 16 4 −6 −14  15 −1 7 12  10 =  − − 15 10 10 3 14   © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

728

Chapter 8

Linear Systems and Matrices

0   −3 5 1  −1 4     3 − 2 2 2 − 4 − 7     18. (a) A + B =  5 4 −1 +  10 −9 −1     8 −6   3 2 −4   0  −4 −1 0   0 1 −2     0 + 1 −4 9 1  −1 − 3 4 + 5     3 + 2 − 2 − 4 2 − 7 5 − 6 − 5     =  5 + 10 4 − 9 −1 − 1 =  15 −5 −2      0 + 3 8 + 2 −6 − 4  3 10 −10 −4 + 0 −1 + 1 0 − 2 −4 0 −2     0   −3 5 1  −1 4     2   2 −4 −7   3 −2 (b) A − B =  5 4 −1 −  10 −9 −1     8 −6   3 2 −4   0  −4 −1 0   0 1 −2     4−5 0 − 1  2 −1 −1  −1 + 3     − − + + 7  1 2 3 2 2 4 2 9  =  5 − 10 4 + 9 −1 + 1 =  −5 13 0      8 − 2 −6 + 4   −3 6 −2   0−3  −4 − 0 −1 − 1 0 + 2   −4 −2 2     

 −1 4 0   3 ( −1) 3 ( 4 ) 3 ( 0 )    3 ( 3 ) 3 ( −2 ) 3 ( 2 )   3 −2 2    (c) 3 A = 3  5 4 −1 =  3 ( 5 ) 3 ( 4 ) 3 ( −1)     8 −6  3(0) 3 ( 8 ) 3 ( −6 )   0   −4 −1 0     3 ( −4 ) 3 ( −1) 3 ( 0 )  0  −3 12   − 9 6 6  =  15 12 −3    0 24 −18   −12 −3 0    −3   9 (d) 3 A − 2 B =  15   0  −12 

0   −6 10 2    6   4 −8 −14  12 −3 −  20 −18 −2     24 −18   6 4 −8  −3 0   0 2 −4 

12 −6

 3 2 −2    20   5 2 =  −5 30 −1    −6 20 −10   −12 −5 4  

19. (a) A + B is not possible. (b) A − B is not possible.

 6 0 3  3 ( 6 ) 3 ( 0 ) 3 ( 3 )  (c) 3 A = 3   =  −1 −4 0  3 ( −1) 3 ( −4 ) 3 ( 0 )  0 9  18 =   −3 −12 0 

(d) 3 A − 2 B is not possible. 20. (a) A + B is not possible. (b) A − B is not possible.  3  3 ( 3 )   9        (c) 3 A = 3  2  = 3 ( 2 )  =  6   −1 3 −1   −3  ( ) 

(d) 3 A − 2 B is not possible.

−5 0  7 1 −10 −8 −5 0 −3 −7 21.  + + = +   3 −6 −2 −1  14 6  3 −6 12 5 −8 −7 =   15 −1   6 9  0 5  −13 −7  6   22.  −1 0 + −2 −1  +  4 −1 = −3   7 1  3 −6   −6 0 10        −7  = 1  4

23.

14 −13 −7    −1 +  4 −1 −5  −6 0 7  −2 −5

 −4 0

1  5 1 − 2  1  − 9 −1 3 −   = 3   − − − − 0 2 12 12 6 3 12 8 15      

1  3 

 − 3 − 13 =  8 3 − 4

1  − 5

24. 1 2

([3 − 2 4 0] − [10 − 6 −18 9]) = 12[− 7 4 22 − 9] = − 72

25.

26.

3 7

2 11 − 92 

5  2  −3 0   −17.143 2.143    + 6 ≈  − − 1 4    2 2   11.571 10.286 

− 22 20 −142.8 131.2 4  14 −11   + 7  =   5 − 22 19  13 6  73.4 57.2

Section 8.5

 3.211 6.829  1.630 − 3.090   1  4.914 −  5.256 8.335 27. − 5−1.004 4  0.055 − 3.889 − 9.768 4.251   

− 31.125 − 48   =  59.5 − 31  − 35 −15 

729

30. 2 X = 2 A + B

X = A + 12 B 3 − 2 −1  0 3 − 2 −1  0 2    1     0 + 2  2 0 =  1 0 +  1 0 =  1  3 − 4 − 4 −1  3 − 4 − 2 − 1         2

−16.463 − 33.373   =  3.706 − 26.654  2.167 18.382 

 12 11  − 3 13   10 15   1     28. − 3− 20 10 −   7 0 +  − 3 8  8  12 4         6 9 −14 15 

Operations with Matrices

1 − 2 2   =  2 0  1 − 9 2 

31. 2 A + 4 B = −2 X 3  −2 −1   0     X = − A − 2B = − 1 1 0 − 2  2 0  3 −4   −4 −1   2 1   0 −6   2 −5        0  =  −5 0 =  −1 0  +  −4  −3 4   8 2   5 6 

29. 3 X = A + 3B X = 13 A + B − 2 −1  0 3 − 23 − 13   0 3         = 13  1 0 −  2 0 =  13 0 +  2 0  3 − 4 − 4 −1  1 − 4  − 4 −1        3 8 − 2 3  73  0 =  3  −3 − 7  3 

32. − 3 X = −3 A = 9 B X = − A − 3B − 2 −1  0 3  2 1  0 − 9         0 − 3 2 0 =  −1 0 + − 6 0 = ( −1)  1  3 − 4 − 4 −1 − 3 4  12 3         2 − 8   0 = − 7  9 7  

1 3 −1  11   6 0   = − 11 5 33. AB = 4 − 5     1 7 −  2   54 − 6 6    

2 0 0  13 0 0 0 6 0      1 37. AB = 0 4 0 0 − 4 0 = 0 −1 0 0 0 − 1  0 0 − 2 0 0 16    3 

 −1 6   −2 51    2 3   AB =  −4 5   =  −8 33 0 9   0 27   0 3   

0 0  1 0 0  5 0 0   15      AB = 0 −8 0   0 − 18 0  = 0 1 0  0 0 7   0 0 12  0 0 27  

34.

35. A is 3 × 2 and B is 3 × 3  AB is not defined. 36. A is 2 × 4, B is 2 × 2  AB is not defined.

38.

 5  10 − 40 20     24 −12 39. AB = − 3[2 − 8 4] = − 6  4  8 − 32 16    

730

Chapter 8

Linear Systems and Matrices

5 −15 −5 −25 −45 40. AB =   −3 −1 −5 −9 =   6   −18 −6 −30 −54 41. (a)

(b) (c)

42. (a)

(b) (c)

43. (a)

(b)

(c)

44. (a)

(b)

(c)

45. (a)

 1 2   2 −1 0 15 AB =   =   4 2   −1 8 6 12   2 −1  1 2   −2 2  BA =   =   −1 8  4 2   31 14  1 2  1 2   9 6  A2 = AA =   =   4 2   4 2  12 12  3  −2 0   −6 12   6 AB =   =   −2 −4   2 4   −4 −16   −12 −6  BA =    4 −10 

3 −1 3 −1  9 − 1 −3 − 3 A2 = AA =   =   1 3  1 3 3 + 3 −1 + 9   8 −6  =  8 6 1 −1  1 3 AB =    1 1 −3 1 1(1) + ( −1)( −3) 1( 3) + ( −1)(1)   4 2 = =  1( 3) + 1(1)  −2 4  1(1) + 1( −3)  1 3 1 −1 BA =     −3 1 1 1  1(1) + ( 3 )1 1( −1) + 3 (1)   4 2  = =   −3 (1) + (1)(1) −3 ( −1) + 1(1)   −2 4  1 −1 1 −1 A2 = AA =    1 1 1 1 1(1) + ( −1)1 1( −1) + ( −1)(1)  0 −2 = =  1( −1) + 1(1)  2 0 1(1) + (1)(1)

(b)

 7   BA = 1 1 2   8  = 7 + 8 − 2  = 13  −1

(c)

A2 is not defined.

46. (a)

2    AB = 3 2 1 3  = 3 ( 2 ) + 2 ( 3) + 1( 0 )  = 12  0 

2( 3) 2( 2) 2(1)  6 4 2 2     (b) BA = 3 3 2 1 = 3( 3) 3( 2) 3(1)  = 9 6 3   0 0 0 0  0( 3) 0( 2) 0(1)   (c) The number of columns of A does not equal the number of rows of A; the multiplication is not possible.

3  6 3  30 6   6 A2 = AA =   =  − − − − 2 4 2 4     −4 10  3 −1  1 −3 3 − 3 −9 − 1 AB =   =  1 1 + 9 −3 + 3  1 3  3  0 −10  =  0 10  1 −3 3 −1 3 − 3 −1 − 9  BA =   =  1  1 3  9 + 1 −3 + 3 3  0 −10  =  0 10

 7  7 7 14      AB =  8 1 1 2  =  8 8 16   −1  −1 −1 −2 

47.

124 −70    AB = 228 452  192 −72 

48.

 151 25 48    AB = 516 279 387   47 −20 87 

49. A is 3 × 4, B is 3 × 2  AB is not defined. 50. A is 3 × 3, B is 4 × 2  AB is not defined.  3 1  1 0   1 0  1 2  1 0  5 8 51.   = =  0 −2 −2 2  2 4 4 −4 2 4 −4 −16         

3  0 5 −1  6  18 − 20  − 2 2 52.   −1 − 3   =   − − 1 2 0 0 1    − 4 − 5  4 1     4 0   −2 3   0 2 −2       53.     0 −1 +  −3 5   4 1 2    −1 2   0 −3       2 3  0 2 −2     −4 10  =   −3 4  =    4 1 2   −1 −1  3 14   

Section 8.5  3  3     −1 −1 54.   5 −6  + 7 −1 +  −8 9  =    4 2   5  5      7   7 

(

)

6  12   −4 −2  =  20 10     28 14 

1 2  4  55.   3 2  0  1 2  2   4  2  (a)     =      is not a solution. 3 2  1  8  1  1 2   −2   4   −2  (b)     =      is a solution. 3 2   3  0   3  1 2   −4   4   −4  (c)     =      is not a solution. 3 2   4   −4   4  1 2   2   −4   2  (d)     =      is not a solution. 3 2   −3  0   −3

5 −7 58.  1 3 5 (a)  3 5 (b)  3 (c) (d)

 6   −1  6   −1

(d)

2   3  0   3   =      is not a solution. 5  −9  −48  −9 

 6 2   −3  0   −3     =      is not a solution.  −1 5  9  48  9

 −2 −3 57.   4 2  −2 (a)   4  −2 (b)   4 (c)

2  2  0  2   =      is not a solution. 5  −6   −32   −6 

 −6    20  −3 3   −6  3     =      is not a solution. 2  0   12  0  −3  4   −14   4    =      is not a solution. 2   2   20  2 

731

 −15   17 −7  −4   15  −4    =      is not a solution. 1  −5   −17  −5  −7 5  11 5     =      is not a solution. 1 2  17 2 

5 −7   4   −15  4     =      is a solution. 1 5   17  5  3 5 −7   2   −67   2     =      is not a solution. 1 11  17  11 3

 x1   −1 1 4 A= , X =  , B =   2 1 x −   0   2 (b) By Gauss-Jordan elimination on  −1 1  4     −2 1  0 

59. (a)

− R1 → 2 R1 + R2 →

1 −1  −4    0 −1  −8 

− R2 + R1 → 1 0  4    , we have − R2 → 0 1  8  x 1 = 4 and x2 = 8.

 6 2  0 56.    −1 5  16   6 2   −1  0   −1 (a)     =      is a solution.  −1 5  3 16   3 (b)

Operations with Matrices

4 Answer: X =   8   x1  2 3  5  A= , X =  , B =   1 4  10   x2  (b) By Gauss-Jordan elimination on 1 4  10    2 3  5 

60. (a)

 1 4  10    −2 R1 + R2 → 0 −5  −15 −4 R2 + R1 →  1 0  −2  , − 15 R2 → 0 1  3 we have x 1 = −2 and x2 = 3.

 −2  Answer: X =    3

 −2 −3  −6   −6   −6     =      is not a solution.  4 2   6   −12   6   −2 −3  6   −6   6      =      is a solution.  4 2   −2   20   −2 

732

Chapter 8

Linear Systems and Matrices

 x1   −2 −3  −4  A= , X =  , B =   6 1 x    −36   2 (b) By Gauss-Jordan elimination on  −2 −3  −4    1  −36   6

61. (a)

 −2 −3  3R1 + R2 →  0 −8  −2 −3  1 R − → 1 ( 8) 2  0 3R1 + R2 → − 12 R1 →

 −4    −48

 1 −2 3  x1   9   A =  −1 3 −1 , X =  x2  , B =  −6       2 −5 5  x3   17  (b) By Gauss-Jordan elimination on  1 −2 3  9     −1 3 −1  −6   2 −5 5  17 

63. (a)

 −4    6  −2 0  14     0 1  6

R1 + R2 → −2R1 + R3 → 2R2 + R1 →

 1 0  −7    , we have 0 1  6 

R2 + R3 →

 −7 Answer: X =    6

 1   Answer: X =  −1  2 

62. (a)

1 −3  4R1 + R2 → 0 −3 ( −1) R2 + R1 →  1 0  0 −3

 12    35  −23   35

 1 0  −23   , we have − 13 R2 → 0 1  − 353  x 1 = −23, x2 = − 353 .  −23 Answer: X =  35   − 3 

−2 3  9   1 2  3 −1 −1  −1 0 7  15  1 2  3 0 1  2 

−7 R3 + R1 →  1 0 0  1   −2 R3 + R2 → 0 1 0  −1 , we have 0 0 1  2  x 1 = 1, x2 = −1, x3 = 2.

x1 = −7, x2 = 6.

 x1   −4 9   −13 A= , X =  , B =   1 3 x −    12   2 (b) By Gauss-Jordan elimination on  −4 9  −13 R2 →  1 −3  12       1 −3  12  R1 →  −4 9  −13

1  0 0 1  0 0

64. (a)

 1 1 −3   x1   9   A =  −1 2 0  , X =  x2  , B =  6       1 −1 1  x3   −5

(b) By Gauss-Jordan elimination on  1 1 −3  9   −1 2 0  6     1 −1 1  −5

1 0  0 1 0  2R2 − 3R3 → 0

1 −3  9  3 −3  15  2 −4  14  1 −3  9 3 −3  15  0 6  −12 

1 0  0 1 0  0

1 −3  9  1 −1  5  0 1  −2  0 −2  4  1 −1  5  0 1  −2 

R1 + R2 → R1 − R3 →

1 3 1 6

R2 → R3 →

− R2 + R1 →

 1 0 0  0 0 1 0  3 , we have   0 0 1  −2  x 1 = 0, x2 = 3, x3 = −2. 2 R3 + R1 → R3 + R2 →

 0 Answer: X =  3  −2 

Section 8.5  x1   1 −5 2   −20        A =  −3 1 −1 , X =  x2  , B =  8  x   0 −2 5  −16   3 (b) By Gauss-Jordan elimination on  1 −5 2  −20    1 −1  8  −3  0 −2 5  −16 

65. (a)

 x1  1 −1 4   17        A = 1 3 0  , X =  x2  , B =  −11 x  0 −6 5   40   3 (b) By Gauss-Jordan elimination on  1 −1 4  17     1 3 0  −11  0 −6 5  40 

( −1) R1 + R2 →

−5 2  −20   −2 5  −16  0 1  −2  −2 R3 + R1 →  1 −5 0  −16    − 5R3 + R2 → 0 −2 0  −6  0 0 1  −2   1 −5 0  −16  1 ( − 2 ) R2 → 0 1 0  3 0 0 1  −2 

−1 4  17   4 −4  −28  5  40  −6

R2 →

 1 0 0  4    0 1 0  −5 , we have  0 0 1  2  x 3 = 2, x2 = −5, x1 = 4. R2 + R1 →

 4   Answer: X =  −5  2 

5R2 + R1 →

 −1   Answer: X =  3  −2 

1  0  0 1  0  0

−1 4  17   1 −1  −7  −6 5  40   1 −1 4  17    1 −1  −7  0 6R2 + R3 →  0 0 −1  −2  4 R3 + R1 →  1 −1 0  9    − R3 + R2 →  0 1 0  −5 − R3 →  0 0 1  2  1 4

−5 2  −20   −2 5  −16  0 −30  60 

 1 0 0  −1   0 1 0  3 , we have 0 0 1  −2  x 1 = −1, x2 = 3, x3 = −2.

733

66. (a)

 1 −5 2  −20    3R1 + R2 → 0 −14 5  −52  0 −2 5  −16   1 −5 2  −20    R3 → 0 −2 5  −16  R2 → 0 −14 5  −52  1  0 0 −7R2 + R3 1  0 1 − 30 R3 → 0

Operations with Matrices

67. (a)

5  7 −2   A ( B + C ) =  −6 13 −8   16 11 −3

5  7 −2   (b) AB + AC =  −6 13 −8   16 11 −3 The answers are the same. 

68. (a)

0  5  −10 −8 13

( B + C ) A = 

4 −5

 9 5 0   BA + CA =  4 −5 5  −10 −8 13 The answers are the same. (b)

734

Chapter 8

Linear Systems and Matrices

 25 −34 28    69. (a) A ( BC ) =  −53 34 −7   −76 30 21   25 −34 28    (b) ( AB ) C =  −53 34 −7   −76 30 21  The answers are the same.  33 9 −12    12  70. (a) c ( AB ) =  −33 12  −18 39 18   33 9 −12    12  (b) ( cA ) B =  −33 12  −18 39 18  The answers are the same.

71. (a)

26 11 0  ( A + B ) =  11 20 −3  11 14 0  2

26 11 0    2 2 (b) A + AB + BA + B =  11 20 −3  11 14 0  The answers are the same.

75. (a) and (b) c( AB)  Not possible, A is a 2 × 3 matrix and B is a 2 × 3 matrix. 76. (a) and (b) A ( B + C )  Not possible because B and C

have different dimensions. 77. (a) and (b) CA − BC  Not possible because CA is 3 × 3, and BC is 2 × 2. 78. (a) and (b) dAB2  Not possible because AB is not

defined, nor is B2 . 79. (a) and (b)

 1 2 −2   6 12 −12  cdA = ( −2 )( −3)  =  0  −1 1 0   −6 6  1 80. (a) and (b) cA + dB = ( −2 )   −1  −2 −4 =  2 −2

 1 −16 7  =  1 0 8

81.

9 −6   0   −1 72. (a) ( A − B ) =  41 −12 25 0 −26 

The answers are the same. 73. (a) and (b)  1 2 −2   −1 4 −1 A + cB =   + ( −2 )    −1 1 0   −2 −1 0   1 2 −2   2 −8 2  = +   −1 1 0   4 2 0  3 −6 0  =  3 3 0   −1 4 −1  1 2 −2  74. (a) and (b) B + dA =   + ( −3 )   − 2 − 1 0    −1 1 0   −4 −2 5 =   1 −4 0 

2 0  A=   4 5

2 02 0 2 0 1 0 A2 − 5A + 2I2 =    − 5  + 2  4 54 5 4 5 0 1 −4 0 =   8 2

9 −6   0   −1 (b) A2 − AB − BA + B2 =  41 −12 25 0 −26 

2 −2   −1 4 −1  + ( −3 )   1 0  −2 −1 0  4   3 −12 3 +  0  6 3 0

5 3 82. B =   1 4 5 3 5 3 5 3  1 0  1 0 B2 − 7 B + 6I2 =    − 7  + 6   1 4 1 4 1 4 0 1 0 1 −1 6 =    2 − 3

83.

100 120 60 40 0.85 A = 0.85  140 160 200 80  85 102 51 34 =   119 136 170 68

100 90 70 30  90 81 63 27 84. 0.90  =   40 20 60 60  36 18 54 54  70 50 25  84 60 30  85. 1.20  =   35 100 70   42 120 84 

Section 8.5

615 86. 1.12 A = 1.12  995 688.80 = 1114.40

670 740 990   1030 1180 1105 750.40 828.80 1168.80   1153.60 1321.60 1237.60 

87. (a) B = 3.50 6.00

125 100 75  (b) BA = 3.50 6.00  100 175 125 = $1037.50 $1400.00 $1012.50 The entries represent the profit for both crops at the three outlets. 88. (a) A =  12

(b) AB =  12

1

1 4

325 387    1 236 281 = [575.5 685.75] 354 422  

The entries are the total calories burned by each person.

1.0 0.5 0.2 15 13 $23.20 $20.50      89. LW = 1.6 1.0 0.2 12 11 = $38.20 $33.80 2.5 2.0 0.4 11 10 $76.90 $68.50      The entries represent the labor costs at the two plants for the three boat sizes.

 840 1100    3 2 2 3 0  1200 1350    90. ST = 0 2 3 4 3  1450 1650     4 2 1 3 2  2650 3000  3050 3200    $15,770 $18,300    = $26,500 $29,250  $21,260 $24,150  The entries represent the wholesale and retail prices of the inventory at each outlet. 0.6 0.1 0.1 0.6 0.1 0.1    91. P 2 = 0.2 0.7 0.1 0.2 0.7 0.1 0.2 0.2 0.8  0.2 0.2 0.8  0.40 0.15 0.15   =  0.28 0.53 0.17  0.32 0.32 0.68  This product represents the proportion of changes in party affiliation after two elections.

Operations with Matrices

735

 0.4 0.15 0.15   0.6 0.1 0.1  92. P 3 = P 2 P =  0.28 0.53 0.17   0.2 0.7 0.1      0.32 0.32 0.68   0.2 0.2 0.8   0.300 0.175 0.175    =  0.308 0.433 0.217   0.392 0.392 0.608 

0.300 0.175 0.175  0.6 0.1 0.1    P 4 = P 3 P =  0.308 0.433 0.217  0.2 0.7 0.1 0.392 0.392 0.608  0.2 0.2 0.8  0.250 0.188 0.188    =  0.315 0.377 0.248   0.435 0.435 0.565 0.250 0.188 0.188  0.6 0.1 0.1    P 5 = P 4 P =  0.315 0.377 0.248  0.2 0.7 0.1  0.435 0.435 0.565  0.2 0.2 0.8   0.225 0.194 0.194    = 0.314 0.345 0.267   0.461 0.461 0.539   0.225 0.194 0.194  0.6 0.1 0.1    P = P P = 0.314 0.345 0.267   0.2 0.7 0.1  0.461 0.461 0.539   0.2 0.2 0.8   0.213 0.197 0.197    =  0.311 0.326 0.280  0.477 0.477 0.523  6

 0.213 0.197 0.197   0.6 0.1 0.1    P = P P =  0.311 0.326 0.280  0.2 0.7 0.1 0.477 0.477 0.523  0.2 0.2 0.8  0.206 0.198 0.198    =  0.308 0.316 0.288  0.486 0.486 0.514  7

0.206 0.198 0.198  0.6 0.1 0.1    P 8 = P 7 P =  0.308 0.316 0.288  0.2 0.7 0.1 0.486 0.486 0.514  0.2 0.2 0.8   0.203 0.199 0.199    =  0.305 0.309 0.292  0.492 0.492 0.508 

As P is raised to higher and higher powers, the resulting matrices appear to be approaching the matrix 0.2 0.2 0.2     0.3 0.3 0.3  .  0.5 0.5 0.5 

736

Chapter 8

Linear Systems and Matrices

93. True. To add two matrices, corresponding entries are added, therefore the matrices must have the same dimension.

1 0  2 2 2 106. ( A + B ) =   = A + AB + BA + B 2 1  

 −6 −2   4 0   −24 2  94. False.   =   2 −6   0 −1  8 6   4 0   −6 −2   −24 −8   =   0 −1  2 −6   −2 6  In general AB ≠ BA, matrix multiplication is not commutative.

0 1 2 3 2 107. AC =   = 0 1 2 3 2 1 0  2 3 2 BC =   = 1 0  2 3 2 AC = BC , but A ≠ B.

For 95–101, A is of dimension 2 × 3, B is of dimension 2 × 3, C is of dimension 3 × 2, D is of dimension 2 × 2. 95. A + 2C is not possible. A and C are not of the same dimension. 96. B − 3C is not possible. B and C are not of the same dimension. 97. CD is not possible. The resulting dimension is 3 × 2.

101. D ( A − 3 B ) is possible. The resulting dimension is 2 × 3. 102. C ( A + 2 B ) is possible. The resulting dimension is

3 × 3. 1 0  2 103. ( A + B ) =   2 1  0 0  A2 + 2 AB + B2 =   3 2 7 −16  2 104. ( A − B ) =   23 8 8 −16  A2 − 2 AB + B2 =   7 22 

3  3

3 3  1 −1 108. A =  , B =   4 4    −1 1  3 3   1 −1 0 0  AB =   =   4 4   −1 1 0 0  AB = O but A ≠ O and B ≠ O.  i 0   i 0   −1 0  2 109. (a) A2 =    =  and i = −1 0 i  0 i   0 −1 −1 0  i 0 −i 0 3 A3 = A2 A =     =   and i = −i  0 −1 0 i   0 −i 

99. CA − D is not possible. CA and D are not of the same dimension. 100. CB − D is not possible. CB and D are not of the same dimension.

3  3

−i 0  i 0 1 0 4 A4 = A3 A =    =  and i = 1 i i 0 − 0 0 1      

(b)

0 −i  0 −i  1 0  B2 =    = ,  i 0   i 0  0 1  The identity matrix

110. The product of two diagonal matrices of the same dimension is a diagonal matrix whose entries are the products of the corresponding diagonal entries of A and B. 111. Given A is a 3 × 2 matrix and B is a 2 × 2 matrix. (a) A ≠ B, A and B are of different dimensions. (b) A + B is not defined; A and B are of different dimensions.

 3 −2  105. ( A + B )( A − B ) =   3 4 2 −2  A2 − B 2 =   5 4

Section 8.6

The Inverse of a Square Matrix

1 2 113. 3 ln 4 − ln x + 3 3

4000 8000 5000 112. A =   6000 10,000 5000

(

) = ln 4 − ln ( x + 3) 3

  64  = ln  13   2 x + 3  

(a) a21 is the amount of Model B TVs sent to warehouse 1. (b) To find the shipment levels when the shipments are increased by 15% multiply matrix A by 1.15.

737

(

114. 2 ln 7t 4 −

)

2 35 3 ln t 5 = ln(7t 4 ) − ln (t 5 ) 5

= ln ( 49t 8 ) − ln (t 3 )  49t 8  = ln  3   t  = ln ( 49t 5 )

Section 8.6 The Inverse of a Square Matrix 1.

inverse

nonsingular, singular

No. A square matrix must be invertible or nonsingular to have an inverse.

Yes. If A and B are square matrices, and AB = I , then BA = I .

 (1)( − 2) + (3)(1) 3 − 2 − 3  1 5. AB =    =  1 −1 − 2  1 ( −1)( − 2) + ( − 2)(1)

(1)(− 3) + (3)(1)  (−1)(− 3) + (− 2)(1)

 1 0 =   0 1 ( − 2)(1) + ( − 3)( −1) 3 − 2 − 3  1 BA =    =  1 −1 − 2 (1)(1) + (1)(−1)   1

(− 2)(3) + (− 3)(− 2)  (1)(3) + (1)(− 2)

 1 0 =   0 1 ( − 4)( 2) + (3)(3) 3 2 3 − 4 6. AB =    =  (3)( 2) + ( − 2)(3)  3 − 2  3 4

(− 4)(3) + (3)(4)  (3)(3) + (− 2)(4)

 1 0 =   0 1 ( 2)( − 4) + (3)(3) 3 2 3 − 4 BA =    =  − 3 4 3 2 (3)( − 4) + ( 4)(3)   

(2)(3) + (3)(− 2)  (3)(3) + (4)(− 2)

 1 0 =   0 1 2 5 4 −1 − 5 + 6 10 − 10  1 0 7. AB =    3 − 5 =   =   6 − 5 3 2  2 − 3 + 3 0 1  2 2 5 4 −1  − 5 + 6 − 4 + 4  1 0 = 15 15 BA =  3  =    5  6 − 5  2 − 2  3 2  2 − 2 0 1

738

Chapter 8

Linear Systems and Matrices

− 1 − 5   8 − 4 + 5 − 5 + 5  5  1 0 4 2 2 = 8. AB =  2   =    − − 4 2 2  5 − 4  0 1  1  8 − 8 5 − 1 − 5   8 − 4 + 5 −10 + 10  1 0 4 = BA =   2   =   − − − − 4 2 2 2 5 4 1 2      0 1 

 2 −17 11  1 1 2     AB =  −1 11 −7   2 4 −3  0 3 −2   3 6 −5 4 + 51 − 55  2 − 34 + 33 2 − 68 + 66   =  −1 + 22 − 21 −1 + 44 − 42 −2 − 33 + 35  −9 + 10  6−6 12 − 12  1 0 0   = 0 1 0 0 0 1

2 0  1 0  11.  A  I  =   0 3  0 1 1 R →  1 0  12 0  2 1 −1   =  I  A  1 R →  0 1  0 13   3 2  1 0 3 0  A −1 =  2 1  = 61    0 3  0 2 

12.

 1 1 2   2 −17 11    BA =  2 4 −3  −1 11 −7   3 6 −5  0 3 −2   2 − 1 −17 + 11 + 6 11 − 7 − 4   1 0 0      =  4 − 4 −34 + 44 − 9 22 − 28 + 6  = 0 1 0   6 − 6 −51 + 66 − 15 33 − 42 + 10  0 0 1 4 6 − 4 1 5 − 2  1  10. AB =  −1 2 4 4  1 − 4 −11  0 −1 −1  −1 4 7    =

 8 − 4 + 0 − 2 + 8 − 6 −10 + 16 − 6  1 − 8 + 11 5 − 16 + 11  4 − 4 + 0 −1 + 8 − 7 − 5 + 16 − 7  

1 − 4 + 4 + 0 4

1 0 4

4 0 0  1 0 0    4 0 = 0 1 0 0 0 4 0 0 1    

2  1 0  1  −3 1 0  7 −2  −1  =  I  A  1  −3 1 

 1 −2  1 0   −2 R1 + R2 →  0 1  −2 1 2 R2 + R1 →  1 0  −3 2   0 1  −2 1

 8 + 1 − 5 −16 − 4 + 20 − 24 − 11 + 35  − 4 − 8 + 16 − 6 − 22 + 28  0 −1+1 0+ 4−4 0 + 11 − 7 

4 6 − 4 1 5 − 2    1 BA = 4  1 − 4 −11  −1 2 4  −1 4 7  0 −1 −1 

2  1 0  7  0 1

 1 −2  1 0  13.  A  I  =    2 −3  0 1

1 2 + 2 − 4 4

4 0 0  1 0 0     1 = 4 0 4 0 = 0 1 0 0 0 4 0 0 1    

1  A  I  =  3 1  −3 R1 + R2 → 0 −2 R2 + R1 →  1  0  7 −2  A−1 =   1  −3

 −3 2  A−1 =    −2 1

 −2 5   14. A =  6 −15  0 1   A has no inverse because it is not square. 15.

 2 7 1 A=   −3 −9 2  A has no inverse because it is not square.

Section 8.6

 −7 33  1 0  16.  A  I  =    4 −19  0 1 − 17 R1 →  1 − 337  − 17 0    0 1  4 −19   1 − 337  − 17 0    1 4 1 −4 R1 + R2 →  0 − 7  7

1  0 0  − R2 + R1 →  1  1 R → 0 2 2 −3R2 + R3 → 0  − R3 + R1 →  1  − R3 + R2 → 0 2 R3 → 0 

−3R1 + R2 → −3R1 + R3 →

 −19 −33 A −1 =    −4 −7 

0  0 1 

 1 2 2  1 0 0   18.  A  I  =  3 7 9  0 1 0   −1 −4 −7  0 0 1    1 2 2  1 0 0  −3R1 + R2 → 0 1 3  −3 1 0   R1 + R3 → 0 −2 −5  1 0 1   −2 R2 + R1 →  1 0 −4  7 −2 0    1 0 0 1 3  −3 2R2 + R3 → 0 0 1  −5 2 1   4 R3 + R1 →  1 0 0  −13 6 4  −3R3 + R2 → 0 1 0  12 −5 −3   0 0 1  −5 2 1   =  I  A −1 

 −13 6 4    A =  12 −5 −3  −5 2 1   −1

1 1  1 0 0  2 1  −3 1 0  3 2  −3 0 1  0 1 0

1 2 1 2 1 2

 2 3 −  2 3  2

− 12 −

1 2 3 2

0  0 1 

0 0  1 1 −1  1 0  −3 2 −1 0 1  3 −3 2  

=  I  A −1 

 1 1 −1   A =  −3 2 −1  3 −3 2    −1

0 0  1 0 0 1 

0 0  1 0 −1 1  Since the first 3 entries of row 3 are all zeros, the inverse does not exist.

739

 1 1 1  1 0 0     19.  A  I  = 3 5 4  0 1 0 3 6 5  0 0 1  

−33 R2 + R1 →  1 0  −19 −33   1 4 0 1  − 7 7   1 0  −19 −33   −7 R2 →  0 1  −4 −7 

1 0 0  1 0  17.  A  I  =  3 5 0  0 1 2 5 0  0 0  1 0 0  1 −3 R1 + R2 → 0 5 0  −3  −2R1 + R3 → 0 5 0  −2  1 0 0  1   0 5 0  −3 0 0 0  1 R2 − R3 → 

The Inverse of a Square Matrix

20. [ A  I ] =

 1 6 10  1 0 0   3 4 0  0 1 0 2 5 5  0 0 1  

− 2 R1 + R3

6 10  1 0 0 1   → 0 −14 − 30  − 3 1 0 → 0 − 7 −15  − 2 0 1

− 12 R1 + R3

1 6 10  1 0 0   − − −  0 14 30 3 1 0  → 0 0 0  − 72 − 12 1

− 3R1 + R2

A−1 does not exist.

21.

3  − 32 2 1  9  7 A =  2 − 2 −3  −1 1 1 

22.

−175 37 −13   A =  95 −20 7  14 −3 1  

23.

 −12 −5 −9   A =  −4 −2 −4   −8 −4 −6   

24.

A−1 does not exist.

25.

−1

A =

5 11

 0 −4 2     −22 11 11  22 −6 −8  

740

Chapter 8

Linear Systems and Matrices

26.

 3.75 0 −1.25   A =  3.4583 −1 −1.375  4.1667 0 −2.5  

27.

A−1 does not exist.

−1

28.

 −24 7 1 −2    −10 3 0 −1 A−1 =   −29 7 3 −2     12 −3 −1 1

4 − 3 29.   8 − 6

−1

− 6 3  4

(4)(− 6) − (− 3)(8)  − 8

1 − 6 − 3   0 8 4 Inverse does not exist. =

−1

   2 2  −1  2 2   − 12 1 30.  1 −2  =  =  = 1( 2 ) − ( −2 )( −3)  3 1 4  3 1  − 43  −3 2 

7 31.  21 5

− 1 32.  41  3

−1

 54 − 34  1 = 7 4 4 ( 2 )( 5 ) − ( − 34 )( 15 ) − 15 5 =

20  54  59  − 15

1  16 15   59  −4 70 

3 4 7 2

− 12   − 14 

  

−1

− 23   89 1 = 1 8 8 1 2  1 ( − 4 ) ( 9 ) − ( 3 )( − 3 )  − 3 9

  −  2 3

1 4

2  8 3 = 10  91 1 − 3 − 4  Inverse does not exist.

−1

  5 −3 1 5 −3  135 1 33.  2 3 =  =  = 2 ( 5 ) − ( 3)( −1)  1 2  13  1 2   131  −1 5  −1

   −5 −12  1 34.  7 12  =  ( 7 )( −5) − ( −8)(12 )  8 7   −8 −5  =

1  −5 −12    61  8 7

−1

 0 − 12   1 2 k=0 35.   = 1 1  −2 0  4 2 −1

 − 13  −1 1 36.  =   2  3  2 1

1 3 1 3

 2 k= 3 

 x   −3 2   5   5  37.   =    =    y   −2 1 10  0  Answer: ( 5, 0 )

− 133  2  13 

 x   −3 2  0  6  38.   =    =    y   −2 1  3  3  Answer: ( 6, 3)

 x   −3 2   4   −8  39.   =    =    y   −2 1 2   −6  Answer: ( −8, − 6 )

 x   −3 2   1   −7  40.   =    =    y   −2 1  −2   −4  Answer: ( −7, − 4 )

 x   1 1 −1 0   3        41.  y  =  −3 2 −1  5 =  8  z   3 −3 2  2   −11   Answer: ( 3, 8, − 11)

Section 8.6  x   1 1 −1  −1  1        42.  y  =  −3 2 −1  2  =  7   z   3 −3 2   0   −9 

49.

 −1 A −1 =   2

Answer: ( 2, 1, 0, 0 )

 5 50. A =  56 − 4 A−1 =

51.

 24 −12  1  4 −2  1  =   18  12  −5 3 18 ( 24 ) − 12 ( 30 )  −30

x 1  4 −2  13  12  −1  =A b=    =   3 23  13  12  −5  y

Answer: ( 12 , 13 ) 47.

 −4 0.8 A=   2 −0.4

A−1 =

 −4 −0.8  1   1.6 − 1.6  −2 −0.4 

A −1 does not exist. 48.

 −0.2 −0.6  A=   −1 1.4  −1 35 15 A −1 =   8 25 5 

1 2 1 3

  −2   −4   =    −12   −8 

−1  3 2 

3  52  5  3   5   4 − − − 1 ( )       6  2   4  1

3 1 1  =  52 0  4

5  6

1 

5  6

 4 −1 1   A =  2 2 3  5 −2 6 

−5  −5 x  18 4   1    y 3 19 10   10  = −   55   z   −14 3 10   1  −55  −1    1  = 55 165  =  3  110   2 

 18 12  A=  30 24  A −1 =

1 2 1 3

−5   18 4   A−1 = 551  3 19 −10   −14 3 10 

Answer: ( 2, − 2 ) 46.

  

A−1 does not exist.

Answer: ( −32, − 13, − 37, 15 )

3 4  A=  5 3 1  3 −4  A −1 =   3 9 − 20  −5 x 1  3 −4   −2  1  −22   2   =−    = −  =  3  4  11  −5 11  22   −2   y

741

Answer: ( −4, − 8 )

 x   −24 7 1 −2   1  −32         −10 y 3 0 −1  −2   −13 = 44.   =   z   −29 7 3 −2   0   −37         1  −3  15  w   12 −3 −1

45.

3 8 3 4

    −1 x −1  =A b= 2 y   

Answer: (1, 7, − 9 )  x1   −24 7 1 −2   0  2         x −10 3 0 −1  1  1 43.  2  =  =  x3   −29 7 3 −2   −1 0          x4   12 −3 −1 1  2  0 

− 1 A =  43  2

The Inverse of a Square Matrix

Answer: ( −1, 3, 2 )

52.

3  4 −2   A = 2 2 5  8 −5 −2  A −1 =

 −21 19 16  1   −44 32 14  82   26 −4 −12 

x  −21 19 16   −2   5   1      = y 14   16  =  8    82  −44 32  z   26 −4 −12   4   −2 

Answer: ( 5, 8, − 2 )

 x  −1  35 15   2.4   6   =   =   y  8  25 5   −8.8   −2 

Answer: ( 6, − 2 )

742

Chapter 8

Linear Systems and Matrices

53.

 5 −3 2  2     A =  2 2 −3 , B =  3  −1 7 −8   4  A−1 does not exist. The system actually has an infinite number of solutions of the form x = 0.3125t + 0.8125 y = 1.1875t + 0.6875 z=t where t is any real number.

54.

 2 3 5  4     A =  3 5 −9  , B =  7   5 9 17  13  166 1  −1 −96 A = 54   2 x  166   1  = y   54  −96  z   2

−6 −52   9 33 −3 1 −6 −52   4   −1     9 33  7  =  2  −3 1 13  0 

Answer: ( −1, 2, 0 )

55.

56.

5 2  1 4 A= 2 −2   1 0

1  11    2 −2  −7 , B=   3 5 1    0 −3  −1

0.141 0.394   0.338 −0.352   0.042 0.164 − 0.066 −0.117  A−1 ≈   −0.141 0.230 0.108 −0.164     0.113 −0.117 0.047 −0.202  0.141 0.394   11  x   0.338 −0.352      − −0.117   −7  0.042 0.164 0.066 y  =  z   −0.141 0.230 0.108 −0.164   3       w  0.113 −0.117 0.047 −0.202   −1  6.21   −0.77  =  −2.67     2.40  Answer: ( 6.21, − 0.77, − 2.67, 2.40 ) 1 1 1  10,000     57. Let A = 0.045 0.05 0.07 and B =  560 .   0  −1 0 2   

2  7 −3 0  41     −2 1 0 −1 −13 A= , B=  4 0 1 −2   12      − 1 1 0 − 1    −8 

Because AX = B, use A−1 to solve for X .

1  0 −1 0   − 1 − 5 0 3  A −1 =   −2 −4 1 −2    1  −1 −4 0 1  41  5  x   0 −1 0        y − − 1 5 0 3  =   −13 =  0   z   −2 −4 1 −2   12   −2         1  −8   3  w   −1 −4 0

$4000 in AAA-rated bonds, $2000 in A-rated bonds, and $4000 in B-rated bonds

Answer: ( 5, 0, − 2, 3)

 38 − 600

X = A

−1

1  −9 B = 11 

−18 

1 10,000 4000     200 5  560 = 2000 4000 400 −1  0  

1 1 1  20,000     58. Let A = 0.045 0.05 0.07 and B =  955.   0 2 0 −1   Because AX = B , use A−1 to solve for X . X = A

−1

 38 − 600 − 4 20,000 17,000     200 5  955  =  1000 −18  2000 400 −1  0    

1  −9 B = 11 

$17,000 in AAA-rated bonds, $1000 in A-rated bonds, and $2000 in B-rated bonds

Section 8.6 1 1 1  300,000     59. Let A = 0.045 0.05 0.07 and B =  18,450.   0 2 −1 0  

The Inverse of a Square Matrix

62. E1 = 10, E 2 = 10

Because AX = B , use A−1 to solve for X .

4 I3  2 I1 +  I2 + 4 I3   I1 + −I3 I2 

 38 − 600 − 4 300,000    −1 1 X = A B = 11  − 9 200 5  18,450  −18 400 −1  0  

2 0 4    A = 0 1 4   1 1 −1

 30,000   =  90,000 180,000  

−1

A =

Because AX = B , use A−1 to solve for X .  38 − 600 − 4 500,000    −1 1 X = A B = 11  − 9 200 5  28,000  −18 400 − 4  0   200,000   = 100,000 200,000  

4 I3  2 I1 +  4 I3 + I 2   I1 + −I3 I2 

y = bags for general planting z = bags for hardwood plants.  2 1 2   x  sand       1 2 2 y loam =       1 1 2   z  peat moss  A

Answer: I1 = 0.5 ampere, I 2 = 3 amperes, I 3 = 3.5 amperes

500  100      A−1 500  = 100   400  100  Answer: 100 bags for seedlings, 100 bags for general planting, and 100 bags for hardwood plants

= 0

 I1   5 −4 4  15 0.5   1     I 8  17  =  3 =  2  14  −4 6  I 3   1 2 −2   0  3.5

0 − 1  1   A −1 =  0 1 − 1  − 12 − 12 − 32 

= 15 = 17

 5 −4 4    A−1 = 141  −4 6 8  1 2 −2 

 5 −4 4    8  −4 6  1 2 −2 

For Exercises 63 and 64, let x = bags for seedlings

63.

2 0 4    A = 0 1 4   1 1 −1

= 0

Answer: I1 = 5 7 amps, I 2 = 10 7 amps, I 3 = 15 7 amps

$200,000 in AAA-rated bonds, $100,000 in A-rated bonds, and $200,000 in B-rated bonds 61. E1 = 15, E2 = 17

1 14

= 10 = 10

 I1   5 −4 4  10   5 7    1     6 8  10  = 10 7   I 2  = 14  −4  I 3   1 2 −2   0  15 7 

$30,000 in AAA-rated bonds, $90,000 in A-rated bonds, and $180,000 in B-rated bonds 1 1 1  500,000     60. Let A = 0.045 0.05 0.07 and B =  28,000.   0 2 −1 0  

743

64.

500   50      A−1 750  = 300   450   50  Answer: 50 bags for seedlings, 300 bags for general planting, and 50 bags for hardwood plants

65. (a) Let x = number of roses. Let y = number of lilies. Let z = number of irises.  x + y + z = 120  2.5 x + 4 y + 2 z = 300  x − 2 y − 2z = 0  (b)

1 1  x  120   1      2.5 4 2    y  = 300   1 −2 −2   z   0  A

744

Chapter 8

Linear Systems and Matrices

1  23  120  80  0 3  7    1 1  300  = 10  − (c) X = A B =  − 6 2 12    3 − 1 − 1   0  30     2 4  2 80 roses, 10 lilies, 30 irises −1

69. False. Only square matrices are non-singular. 70. Answers will vary. One possibility is if A is a 2 × 2 a b  matrix given by A =   , then A is invertible c d 

( A exists ) if and only if ad − bc ≠ 0. −1

66. y = at 2 + bt + c c = 3250   (a)  a + b + c = 3757 4a + 2b + c = 4416 

0 0 1 a 3250      1 1 1 b = 3757 (b)  4 2 1 c  4416      A X = B 1  12 −1 2   A−1 = − 32 2 − 12   1 0 0 

 76    X = A B =  431  3250   −1

y = 76t 2 + 431t + 3250

(c)

7500

71. Answers will vary. 72.

 a b   1   d −b  AA−1 =      c d   ad − bc   −c a  1  a b   d −b  =    ad − bc  c d   −c a  0  1 0 1 ad − bc  =  ad − bc 0 1 ad − bc  0 1  d −ba b  A−1A =    ad − bc −c a c d  0  1 0 1 ad − bc =  =  ad − bc 0 1 ad − bc  0 =

73. (a) Given a A =  11 0

 1 0  −1  a11  , A =  a12   0 

Given 5

0 3000

2013: t (13) = 76(13) + 431(13) + 3250 2

= 5277, or 5,277,000 travelers 2014: t (14) = 76(14) + 431(14) + 3250 2

= 6190, or 6,190,000 travelers 2015: t (15) = 76(15) + 431(15) + 3250 2

= 7305, or 7,305,000 travelers

(e) Answers will vary.

67. True. AA −1 = A −1 A = I 68. True. Let A be an n × n square matrix, and In be the identity matrix. If there exists a matrix A−1 such that AA−1 = I n = A−1 A, then A−1 is called the inverse of A.

 a11 0  A =  0 a22  0 0

 0   , a ≠ 0, a ≠ 0 11 22 1   a22 

 1   a11 0   0  , A−1 =  0  a33    0 

0 1 a22 0

 0    0 ,  1   a33 

a11 , a22 , a33 ≠ 0 (b) In general, the inverse of the diagonal matrix A is

 1 0 a  11  1 0 a 22   0 0      0 0 

0 0 1 a33  0

  0     0    (assuming aii ≠ 0 ).  0       1   ann 

Section 8.7

 x 0  −1 74. Given A =   . A exists (A is invertible) if 0 y  xy − 0 ≠ 0  xy ≠ 0.

(

)(

)

75. e 2 x + 2e x − 15 = e x + 5 e x − 3 = 0  e x = 3  x = ln 3 ≈ 1.099

The Determinant of a Square Matrix

(

)(

745

)

76. e 2 x − 10e x + 24 = e x − 6 e x − 4 = 0 e = 6  x = ln 6 ≈ 1.792 x

e x = 4  x = ln 4 ≈ 1.386

77. 7 ln 3 x = 12 ln 3 x = 127 3 x = e12 7 x = 13 e12 7 ≈ 1.851

78. ln ( x + 9 ) = 2 x + 9 = e2 x = e2 − 9 ≈ −1.611

Section 8.7 The Determinant of a Square Matrix 1. determinant

13.

2. minor 3. For a square matrix B, given the minor M 23 = 5, the cofactor is C23 = ( −1)

2+3

M23 = ( −1)( 5) = −5.

4. No. To find the determinant of a matrix using expansion by cofactors, you do not need to find all cofactors. The determinant of a square matrix A is the sum of the entries in any row or any column of A multiplied by their respective cofactors.

14.

1.9 − 0.3 5.6

3.2

7.2

0.8

3.6 − 0.4

= 7.76

= − 5.76

1.3 0.2 3.2

15.

0.2 6.2 0.2 = 11.998 − 0.4 4.4 0.3 5.1 0.2 7.3

16. − 6.3 0.2 0.2 = −159.632

5. 4 = 4

0.5 3.4 0.4

6. −12 = −12 7.

8 4

= 8 ( 3) − 4 ( 2 ) = 24 − 8 = 16

2 3 −5 2 6 3 −7 6 1 2

= ( −5)( 3) − ( 6 )( 2 ) = −27

3 4 17.   2 −5 (a)

M11 = −5

M12 = 2 M 21 = 4 M 22 = 3

= −7 ( 3) − 6 ( 12 ) = −21 − 3 = −24

(b) C11 = M11 = −5 C12 = − M12 = −2 C21 = − M 21 = −4

10.

11.

12.

4 −3 0

= ( 4 )( 0 ) − ( 0 )( −3) = 0

C22 = M 22 = 3

( 3)( 3) − (4)(3) = 3 − 12 = − 9

= (9)( 4) −

( 5 )( 5 ) = 36 − 5 = 31

746

Chapter 8

Linear Systems and Matrices

 11 6 18.   − 3 2

9 4  −2   20.  7 −6 0   6 7 −6 

(a) M 11 = 2 M12 = − 3

(a)

M 21 = 6 M 22 = 11

M12 =

(b) C11 = M 11 = 2 C12 = − M 12 = 3 C21 = − M 21 = − 6 C22 = M 22 = 11 6 3  −4   19.  7 −2 8   1 0 −5 (a)

M11 = M12 =

−2

0 −5 7

M 23 = M31 = M32 = M33 =

−4

1 −5 −4 6 1 0

−2

6 −6 −2 9 6 7

= −12

= −68

9 4 = 24 −6 0

M32 =

−2 4 = −28 7 0 −2

−6

= −51

(b) C11 = ( −1) M11 = 36 2

C12 = ( −1) M12 = 42 3

= 17

C13 = ( −1) M13 = 85 4

C21 = ( −1) M 21 = 82 3

C22 = ( −1) M 22 = −12 4

C23 = ( −1) M 23 = 68 5

−4 3 = −53 7 8 7

= −42

M31 =

M33 =

6 3 = 54 −2 8

6 −6

9 4 = −82 7 −6

= 10

= −6

−4

= 36

M 21 =

M 23 =

6 3 M 21 = = −30 0 −5

7 −6

7 −6 = 85 6 7

= −43

1 −5

−6

M13 =

M 22 =

7 −2 M13 = =2 1 0

M 22 =

M11 =

C31 = ( −1) M 31 = 24 4

C32 = ( −1) M 32 = 28 5

= −34

C33 = ( −1) M 33 = −51 6

(b) C11 = ( −1) M11 = 10 2

C12 = ( −1) M12 = 43 3

C13 = ( −1) M13 = 2 4

21. (a)

−3 4

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

C21 = ( −1) M21 = 30 3

C22 = ( −1) M22 = 17

= −3 ( 23 ) − 2 ( −8 ) − 22 = −75

C23 = ( −1) M23 = 6 5

C31 = ( −1) M31 = 54 4

C32 = ( −1) M32 = 53 5

(b)

−3 4

2 1 4 6 −3 1 −3 1 5 6 = −2 +5 +3 2 1 2 1 4 6 2 −3 1 = −2 ( −8 ) + 5 ( −5 ) + 3 ( −22 ) = −75

C33 = ( −1) M33 = −34 6

Section 8.7

22. (a)

747

The Determinant of a Square Matrix

23. (a)

6 0 −3 5 0 −3 5 6 −3 5 4 13 6 −8 = −4 0 7 4 + 13 −1 7 4 −1 0 7 4 6 0 2 8 0 2 8 6 0 2

−3 4 2 4 2 −3 2 −3 4 6 3 1 = −6 +3 −1 4 −8 4 −7 −7 −8 4 −7 −8 = −6 ( −18 ) + 3 (16 ) − 5 = 151 (b)

6 0 −3 6 0 −3 −6 −1 0 7 − 8 −1 0 7 8 6 0 8 6 0

−3 4 2 6 3 −3 4 −3 4 6 3 1 =2 − −8 4 −7 4 −7 6 3 4 −7 −8

= −4 ( −282 ) + 13 ( −298 ) − 6 ( −174 )

= 2 ( −54 ) − ( 5 ) − 8 ( −33) = 151

− 8 ( −234 )

(b)

= 170 6 0 −3 5 4 6 −8 6 −3 5 4 13 6 −8 = 0 −1 7 4 + 13 −1 7 4 −1 0 7 4 8 0 2 8 0 2 8 6 0 2

6 −3 5 6 + 0 4 6 −8 + 6 4 8 −0

−3 5 6 −8

−1

= 0 + 13 ( −298 ) + 0 + 6 ( 674 ) = 170

24. (a)

10 4 0 1

8 3 −7 8 3 −7 10 3 −7 10 8 −7 10 8 3 0 5 −6 5 −6 − 3 4 5 −6 + 2 4 0 −6 − 7 4 0 5 =0 0 3 2 7 0 −3 2 1 −3 2 1 0 2 1 0 −3 0 −3 2 = 0 ( −64 ) − 3 ( −3 ) + 2 ( −112 ) − 7 (136 )

(b)

10 4 0 1

= −1167 8 3 −7 0 5 −6 8 3 −7 8 3 −7 8 3 −7 0 5 −6 = 10 3 2 7 − 4 3 2 −7 + 0 0 5 −6 − 1 0 5 −6 3 2 7 0 −3 2 0 −3 2 0 −3 2 3 2 7 0 −3 2

= 10 ( 24) − 4 ( 245) + 0 ( −64) − 1( 427 ) = −1167

25. Expand along Column 3. 1 4 −2 3 2 1 4 3 2 0 = −2 +3 3 2 −1 4 3 −1 4 = −2 (14 ) + 3 ( −10 ) = −58

26. Expand along Row 1. −3

1 0

7 11 5 = − 3 1

2 5

11 5 2

−1

7 5 1 2

7 11 1

= − 3(12) − 1(9) = − 45

748

Chapter 8

Linear Systems and Matrices

27. Expand along Row 2. 6 3 −7 3 −7 6 −7 6 3 0 0 0 =0 −0 +0 =0 −6 3 4 3 4 −6 4 −6 3 28. Expand along Row 3. 1

1 2

3 −5 9 = 0 0

0 0

1 2 −5 9

−0

3 −9

3 −5

1 −1 8

35.

29. Expand along Column 1. −1 2 −5 3 −4 2 −5 2 −5 0 3 −4 = −1 −0 +0 0 3 0 3 3 −4 0 0 3

0 0

−1 0 1 5

− ( 0)

−1 0

+ ( 0)

4 5

36.

−1 −1 4

= ( −1)( 5 ) − 1( 0 )

37.

= −5

31. Expand along Column 3. 2 6 6 2 2 7 6 2 6 2 2 7 3 6 = 6 1 5 1 −3 1 5 1 1 5 0 1 3 7 7 3 7 7 3 7 0 7 38.

= 6 ( −20 ) − 3 (16 ) = −168

32. Expand along Column 4. 3 6 −5

−2 2

1 1

0 3

−1 −1

−2 2

3 6 −5

2 − ( −1) − 2 2

= 4 1 1

0 3 −1

1 1

= − 210

33. Expand along Column 2. 3 2 4 −1 5 −2 −2 0 1 3 2 1 1 0 0 4 0 = −2 6 6 0 2 −1 0 3 3 0 5 1 0

6 0 −4

2 0

0 2 2 8

6 2

0 4 0 2 −1 0 5

1 0

= −336

6 0 8

1 −1

−4 −7

6 0

0 9 0 14

= 7441

3 −2

4 3 1

−1

1 0

0 3 2

−1

7 −8 0 0 − 2

3 0 2

3 −5

1 2 3

−2 0

−3 3

0 −2

= 566

3 −2

−1

4 5 −1

2 4

1 3

2 −4

−3

− 5 2 −1

39. (a) A =

−2

(b) B =

2 6 3 2 6 3 = 5 ⋅1 3 4 1 3 4 1 0 0 2 0 0 2

0 −3

= −1( 9 ) − 0 ( 6 ) + 0 ( 7 ) = −9

−1 −1 0 = (1) 4 1 5

4 3 2

= 5 ( −20 ) = −100

= 0

30.

34. Expand along Column 1. 5 2 0 0 −2 1 0 1 4 3 2 0 0 0 2 6 3 =5 0 0 0 3 4 1 0 0 0 0 0 2

0 −1

= − 636

= 0 = −1

1  1 2 − 2 − 2 − 5 (c) AB =    =    4 − 2 0 −1  4 10

(d) AB =

− 2 −5 4

= 0; AB = A ⋅ B

= ( −2 )( −2 ) 6 2 −1 3 5 1 = 4 (103) = 412

Section 8.7

40. (a)

A =

= −8

3 −2 −1 1

B =

(c)

 4 0  −1 1   −4 4 AB =   =   3 −2  −2 2   1 −1 −4

AB =

41. (a) A =

(c)

−1 − 3 4 = − 23 0

(b) B =

− 2 −1

−9

44. (a)

6 = − 23; AB = A ⋅ B

−1 − 6

4 0

(c)

42. (a) A = 1 − 4 2 = 9 3

1 0

−1 4

(b) B = 0

1 3

−1 −28

−4 −14 −11

−4

−2

−1

3 −3 −1

7 −4 (d) AB = 8 − 9

= −46

4 −1

5 0 0

0 0

−3

2 5 0

= 89

 −1 5 2 0   1 5     0 0 1 1  10 −1  3 −3 −1 0   2 0     4 2 4 −1  −3 2

53 −10 10

2 0 1 2 −1 4 7 − 4 9      1 0 = 8 − 9 6 (c) AB =  1 −1 2 0 3 1 0 3 − 2 1 6 − 2 12     

= 5500; AB = A B

1 0 = −10

3 −2

= −220

0 0

10 −1 2 4

(b)

(d) AB = − 5 −10

0 −2

 6 4 0 1 0 −5 0 −2 −7 −16 −1 −28       2 −3 −2 −4−2 4 −1 −4 = −4 −14 −11 8  0 1 5 0 3 0 1 0  13 4 4 −4      3 2 2  1 −2 3 0 −2 −1 0 −1 1

(d)

4 1 − 9   = − 5 −10 6  4 −1 −1 

= −25

0 1

4 −1 −4

3 0 1 −2

 3 2 0  3 0 1    (c) AB =  −1 − 3 4  0 2 −1 − 2 0 1 − 2 −1 1 

749

1 5 0 −1

−7 −16

2 −1 = 1

−2

(b)

2 0

−2

0 −5

= 0; AB = A B

1 −1 3

0 −1

−2 2

2 −3 −2 −4

43. (a)

(b)

(d)

The Determinant of a Square Matrix

(d)

0 0  2 4 0 1  5 0

 53 −10 10 22    2 5 1 −1 =  −29 18 −6 −13    35 16 −1 12  22

−1

−29

18 −6 −13

5 −1

= −4094; AB = A B

9 6 = − 90; AB = A ⋅ B

6 − 2 12

45.

w x = wz − xy y z −

y z = − ( xy − wz ) = wz − xy w x

Thus,

w x y z =− . y z w x

750

Chapter 8

w cx

46.

y c

w x

w cx y

w x y

49.

= c ( wz − xy )

Thus,

47.

= cwz − cxy = c ( wz − xy )

cz y

Linear Systems and Matrices

(

48.

x2 y2

) (

= yz − xz − y z + x z + xy ( y − x )

)

(

)

= z 2 ( y − x ) − z y 2 − x 2 + xy ( y − x ) = z ( y − x ) − z ( y − x )( y + x ) + xy ( y − x ) 2

= wz − xy

= ( y − x )  z 2 − z ( y + x ) + xy  = ( y − x )  z 2 − zy − zx + xy 

w x

= ( y − x )  z 2 − zx − zy + xy 

= ( y − x )  z ( z − x ) − y ( z − x ) 

w x + cw . = z y z + cy

= ( y − x )( z − x )( z − y )

w x = cxw − cxw = 0 cw cx w x = 0. cw cx

Thus,

50.

x2 x + z2 y

) (

w x + cw = w ( z + cy ) − y ( x + cw ) = wz − xy y z + cy Thus,

y2 x − z2 z

= yz 2 − y 2 z − xz 2 − x 2 z + xy 2 − x 2 y

w x y

x2 y y2 = z z2

1 x 1 y 1 z

a+b a a a+b a a a a a −a +a a a+b a = (a + b) a a+b a a+b a+b a a a a+b 2 = ( a + b ) ( a + b ) − a 2  − a  a ( a + b ) − a 2  + a  a 2 − a ( a + b )   

= ( a + b ) − a 2 ( a + b ) − a 2 ( a + b ) + a3 + a3 − a 2 ( a + b ) 3

= ( a + b ) − 3a 2 ( a + b ) + 2a 3 3

= a 3 + 3a 2b + 3ab 2 + b3 − 3a 3 − 3a 2b + 2a 3 = 3ab 2 + b3 = b 2 ( 3a + b )

51.

52.

x 2 =2 1 x

53.

2 x −3 −2 2 x

x2 − 2 = 2

4 x2 − 6 = 3

x2 = 4

4 x2 = 9

x = ±2

x 2 = 94

x 4 −1 x

x = ± 32

= 20

x 2 + 4 = 20 x 2 = 16 x = ±4

54.

4 9x

9x2 − 8 = 8 9 x 2 = 16 16 9 4 x=± 3

x2 =

Section 8.7

55.

2 x −2

= −1

The Determinant of a Square Matrix

1 2 x −1 3 2 = 0 3 −2 1

61.

( x 2 − 2 x ) − 4 = −1 1

x2 − 2x − 3 = 0

( x − 3)( x + 1) = 0

3 2 −1 2 −1 3 −2 +x =0 −2 1 3 1 3 −2 7 − 2 ( −7 ) + x ( −7 ) = 0

x −3 = 0  x = 3

21 = 7 x x=3

x + 1 = 0  x = −1 56.

x +1 2 =4 −1 x

1 x −2 1 3 3 =0 0 2 −2

62.

x2 + x + 2 = 4 x2 + x − 2 = 0

( x + 2 )( x − 1) = 0

57.

−12 − ( −2 x + 4 ) = 0

x+3 2 =0 1 x+2

2 x = 16 x =8

x2 + 5x + 4 = 0

63.

4u −1 = 8uv − 1 −1 2v

64.

3x 2 1

−3 y 2 = 3 x 2 − −3 y 2 = 3 x 2 + 3 y 2 1

65.

e2 x 2e 2 x

e3 x = 3 e 5 x − 2e 5 x = e 5 x 3e3 x

66.

e− x −e − x

xe − x = (1 − x ) e −2 x − − xe −2 x (1 − x ) e− x

( x + 4 )( x + 1) = 0 x = −4, − 1

58.

x −1

3 x −2

= 0

( x2 − 3x + 2) − 6 = 0 x 2 − 3x − 4 = 0

( x − 4)( x + 1) = 0

59.

(

)

(

x − 4 = 0  x = 4

= e −2 x − xe −2 x + xe −2 x

x + 1 = 0  x = −1

= e −2 x

2x 1 =x −1 x − 1

67.

x ln x 1 = 1 − ln x 1 x

68.

x x ln x = x (1 + ln x ) − x ln x 1 1 + ln x

2 x2 − 2 x + 1 = x 2 x2 − 3x + 1 = 0

( x − 1)( 2 x − 1) = 0 x = 1,

60.

3 3 x −2 −1 = 0 ( Expand first column ) 2 −2 2 −2

x = −2, 1

( x + 3)( x + 2 ) − 2 = 0

1 2

x −1 x = −8 x +1 2 2

2 x − 2 − x − x = −8

751

)

= x + x ln x − x ln x = x 69. True. Expand along the row of zeros. 70. True. If a square matrix has two columns that are equal, then elementary column operations can be used to create a column with all zeros.

− x2 + x + 6 = 0 x2 − x − 6 = 0

( x − 3)( x + 2) = 0 x = 3, − 2

752

Chapter 8

Linear Systems and Matrices

71. Answers will vary. Possible answer  1 0 −3  3 1 5     A = 6 −2 7  , B =  −8 1 0  9  −7 6 −2  5 −1

 1 + 3 0 + 1 −3 + 5  4 1 2      A + B =  6 − 8 −2 + 1 7 + 0  =  −2 −1 7  9 + 7 5 + 6 −1 − 2   2 11 −3 Expand along Row 1 to find A + B . A+B =4

−1 7 −2 7 −2 −1 −1 +2 11 −3 2 −3 2 11

= 4 ( −74 ) − ( −8 ) + 2 ( −20 ) = −328

Expand along Row 1 to find A . A =1

−2 7 6 7 6 −2 −0 +3 5 −1 9 −1 9 5

= −33 − 0 − 3 ( 48 ) = −177

Expand along Column 3 to find B . B =5

−8 1 −7 6

−0

3 1 −7 6

−2

75. (a)

(b)

(c)

76. (a)

A =4

 −1.25 2 −2.25   (b) A =  0.25 0 0.25  −0.5 1 −1.5 1 (c) det A −1 = 4 1 (d) In general, det A −1 = . det A −1

( )

77. (a) Columns 2 and 3 are interchanged. (b) Rows 1 and 3 are interchanged.

−8 1

(b)

A + B = −177 − 227 = −404

( )

( ) det1 A .

78. (a)

= −227

 −4 −5 1.5   A−1 =  −1 −1 0.5  −1 −1 0  1 det A −1 = 2

(d) In general, det A −1 =

3 1

= 5 ( −41) − 2 (11)

A =2

( −5 ) times Row 1 is added to Row 2. ( −2 ) times Row 2 is added to Row 1.

79. (a) 3 is factored out of the first row of A. (b) 2 is factored out of the first column of A and 4 is factored out of the second column of A.

Therefore, A + B ≠ A + B . 72. Yes. Given A is a 3 × 3 matrix such that A = 5, then

2 A = 23 A = 8 ( 5 ) = 40, using cA = c n A , when A is an n × n matrix.

73. (a)

A =6

(b)

 A −1 =  

(c)

det A −1 =

1 3 1 3

−    1 3 1 6

( ) 61 ( )

A = −3

(b)

1 A −1 =  23  3

(c)

7 0 , A = 7 ( 4 ) − 0 = 28 0 4

(b)

−1 0 0 A = 0 5 0 , A = ( −1)( 5)( 2 ) = −10 0 0 2

(c)

2 0 0 −2 A= 0 0 0 0

0 0 1 0

0 0 0 3

Using cofactors and a11 ,

1 (d) In general, det A −1 = . det A

74. (a)

80. (a)

− 13   − 53  1 det A −1 = − 3

( )

( ) det1 A .

(d) In general, det A −1 =

A = 2 ⋅ C11 + 0 ⋅ C12 + 0 ⋅ C13 + 0 ⋅ C14 . −2 0 0 C11 = 0 1 0 0 0 3 A = 2C11 = 2 ( −2 ⋅ 1 ⋅ 3 ) = 2 ⋅ ( −6 ) = −12 In each case, the determinant of the matrix is the product of the diagonal entries. From this, you can conjecture that the determinant of a diagonal matrix is the product of the diagonal entries.

Section 8.7

81. (a)

(b)

 3 −2  3 −2 = 15  ; 5 0 5 0 1 3 −7 1  3 −7   0 5 9 ; 0 5 9 − − − − = −75   0 0 5 0 0 5

The Determinant of a Square Matrix

(c)

4 0   3 −3 3 6   2 −2

0 0 5 1

0 4 0  0  3 −3 ; 0 3 6  2  2 −2

0 0 5 1

753

0 0 = −120 0 2

The determinant of a triangular matrix is the product of the entries along the main diagonal.

82. Because Row 3 is all zeros, the determinant of the matrix is equal to zero. 83. Answers will vary.

84.

4 7

5 8

6 9 =0

10 11 12 13 14 15 = 0

10 11 12

16 17 18

33 34 35 36 37 38 = 0 39 40 41

−5 −4 −3 −2 −1 0 = 0 1 2 3

19 23 27 31

57 61 65 69

20 24 28 32

21 25 29 33

22 26 =0 30 34

58 62 66 70

59 63 67 71

60 64 =0 68 72

For an n × n matrix ( n > 2 ) with consecutive integer entries, the determinant appears to be 0.

x x +1 x + 2 x+4 x+5 x+3 x+5 − ( x + 1) x+3 x+4 x+5 = x x+7 x+8 x+6 x+8 x+6 x+7 x+8

+ ( x + 2)

x+3 x+4 x+6 x+7

= x ( x + 4 )( x + 8) − ( x + 7 )( x + 5)  − ( x + 1) ( x + 3)( x + 8) − ( x + 6 ) ( x + 5)  

(





)(

+ x + 2  x + 3 

(

) ( x + 7 ) − ( x + 6)(

) (



)

x + 4  

)

) (

(

)

= x  x2 + 12 x + 32 − x 2 + 12 x + 35  − ( x + 1)  x2 + 11x + 24 − x 2 + 11x + 30  

(

) (



)

+ ( x + 2 )  x 2 + 10 x + 21 − x 2 + 10 x + 24  

= −3x − ( x + 1)( −6 ) + ( x + 2 )( −3) = −3x + 6 x + 6 − 3x − 6 = 0 85. 4 y2 − 12 y + 9 = ( 2 y − 3)

86. 4 y2 − 28 y + 49 = ( 2 y − 7 )

87. 3 x − 10 y = 46   x + y = −2 y = −x − 2

3x − 10 ( − x − 2 ) = 46 13x = 26 x=2 y = −2 − 2 = −4

Answer: ( 2, − 4 )

Chapter 8

754

Linear Systems and Matrices

 5 x + 7 y = 23 88.  −4 x − 2 y = −4  2 x + y = 2  y = 2 − 2 x 5 x + 7 ( 2 − 2 x ) = 23 −9 x = 9 x = −1, y = 2 − 2 ( −1) = 4

Answer: ( −1, 4 )

Section 8.8 Applications of Matrices and Determinants 1.

Cramer’s Rule

cryptogram

The area of the triangle with vertices ( x1, y1 ) , ( x2 , y2 ) , ( x3 , y3 ) can be found by: Area = ±

1 2

x1 x2 x3

−3 2 1 1 2 1 = −3 ( 6 ) − 2 ( 2 ) + 1( −2 ) = −24 −1 −4 1 Area of rhombus = −24 = 24 square units

y1 1 y2 1 , y3 1

10. Vertices: ( −4, 4 ) , ( 6, 8 ) , ( 2, 1) , ( −8, − 3)

−4 4 1 6 8 1 = −4 ( 7 ) − 4 ( 4 ) + 1( −10 ) = −54 2 1 1

if the determinant is equal to −6, then multiply by − 12 to yield an area of 3 square units. 4.

If the value of the determinant is equal to 0, then the points are collinear.

Vertices: ( −2, 4 ) , ( 2, 3) , ( −1, 5) 1 2

−2 4 1  3 1 2 1 2 3  −4 + 2 3 1 = 12  −2  − − 5 1 1 1 1 5   −1 5 1

=  −2 ( −2 ) − 4 ( 3) + 13 = Area = square units 1 2

1 2

( 5) =

Area of rhombus = −54 = 54 square units −1 5 1 11. 4 = ± 12 −2 0 1 x 2 1 8 = ± ( −1)( −2 ) − 5 ( −2 − x ) + 1( −4 )  8 = ± 5 x + 8

5 2

5 x + 8 = 8 or 5 x + 8 = −8

5 2

−3 2 3

5 6 −5

1 1 = 12 ( −3 )11 − 5 ( −1) + 1( −28 )  = 12 ( −56 ) 1

Area = 28 square units

Vertices: ( 0, 12 ) , ( 25 , 0 ) , ( 4, 3 ) 0

1 1 5 0 1 = 12  − 12 ( − 23 ) + 152  = 12  334  = 338 2 2 4 3 1 1 2

Area = 338 square units

x = 0 or

Vertices: ( −3, 5 ) , ( 2, 6 ) , ( 3, − 5 ) 1 2

Vertices: ( 92 , 0 ) , ( 2, 6 ) , ( 0, − 32 ) 0 1 2 6 1 = 12  92 ( 6 + 32 ) + 1( −3)  0 − 32 1 9 2

1 2

= 12 ( 30.75 ) = 15.375 = 123 8

Vertices: ( −3, 2 ) , (1, 2 ) , ( −1, − 4 ) , ( −5, − 4 )

x = − 165

x = 0, − 165

12.

−4 2 1 4 = ± 12 −3 5 1 −1 y 1 8 = ±  −4 ( 5 − y ) − 2 ( −2 ) + 1( −3 y + 5 )  8 = ± −20 + 4 y + 4 − 3 y + 5 8 = ±  y − 11 y − 11 = 8

or y − 11 = −8

y = 19 or y = 3, 19

y =3

13. Points: ( 3, − 1) , ( 0, − 3 ) , (12, 5 )

3 −1 1 0 −3 1 = 3 ( −8) + 12 ( 2 ) = 0 12 5 1 The points are collinear.

Area = 123 square units 8

Section 8.8 14. Points: ( 3, − 5 ) , ( 6, 1) , ( 4, 2 ) 3 −5 1 6 1 3 −5 3 −5 6 1 1 = − + 4 2 4 2 6 1 4 2 1

15. Points: ( 2, − 12 ) , ( −4, 4 ) , ( 6, − 3 ) 2 − 12 1 −4 4 1 = 2 ( 7 ) + 12 ( −10 ) + 1( −12 ) −3 1

= −3 ≠ 0 The points are not collinear.

16. Points: ( 0, 12 ) , ( 2, − 1) , ( −4, 27 ) 0 12 1 2 −1 1 = − 12 ( 2 + 4 ) + 1( 7 − 4 )

−4

7 2

1 −2 1 x 2 1 =0 5 6 1

−3 5 1 −6 ( y − 5 ) − 2 ( −5 + 3 ) + ( −25 + 3 y ) = 0 −6 y + 30 + 4 − 25 + 3 y = 0 −3 y = −9 y=3

9 −9 −90 = = −3 −7 11 30 3 −9

−10 −3 12 9 −54 x= = = −1 4 −3 54 6 9 4 −10 6 12 108 y= = =2 4 −3 54 6 9

6 17 −13 −76 −235 y= = =5 6 −5 −47 −13 3

−6 2 1 −5 y 1 = 0

−1 11

21.  4 x − 3 y = −10  6 x + 9 y = 12

17 −5 3 −76 −329 x= = =7 6 −5 −47 −13 3

8 x − 24 = 0 x=3

19.  −7 x + 11y = −1   3x − 9y = 9

3 2 = 12 − 12 = 0 6 4

22.  6 x − 5 y = 17  −13 x + 3 y = −76

1( −4 ) + 2 ( x − 5 ) + 1( 6 x − 10 ) = 0

18.

20.  3 x + 2 y = −2  6 x + 4 y = 4

Answer: ( −1, 2 )

= −3 + 3 = 0 The points are collinear.

17.

755

Cramer’s Rule cannot be used because the determinant is zero.

= 8 − 26 + 33 = 15 ≠ 0 The points are not collinear.

Applications of Matrices and Determinants

Answer: ( 7, 5) 23. 4 x − y + z = − 5  2 x + 2 y + 3z = 10 6 x + y + 4 z = − 5 

4 −1 1 = 0 2 2 3 6

1 4

Cramer’s Rule cannot be used because the determinant is zero.

−7 −1 y=

3 9 −60 = = −2 −7 11 30 3 −9

Answer: ( −3, − 2 )

756

Chapter 8

Linear Systems and Matrices

24.  4 x − 2y + 3 z = −2 4  2 x + 2 y + 5 z = 16 D = 2  8x − 5 y − 2z = 4 8 

3 −2 −2 16 2 5 4 −5 −2 4 −2 3 2 16 5 8 4 −2 −82 4 −2 −2 2 2 16 8 −5 4

3 5 = −82 −2

2 x + 3 y − 5z = 1  3 x + 5 y + 9z = −16  5 x + 9 y + 17z = −30 

2 x + 3 y − 5 z = 1  1 y + 332 z = − 352  2  3 y + 592 z = − 652 2   2 x + 3 y − 5z = 1  y + 33z = −35   −20 z = 20 

−656 =8 −82

z = −1 y = −35 + 33 = −2 x = (1 − 5 + 6 ) 2 = 1

−82

Answer: (1, − 2, − 1)

164 = −2 −82

Answer: ( 5, 8, − 2 ) 25. (a)

26. (a)

−410 =5 −82

−82

−2 2 −5

(b)

 3 x + 3 y + 5z = 1  3 x + 5 y + 9 z = 2  5 x + 9 y + 17 z = 4 

2 3 −5 D = 3 5 9 = −20 5 9 17

1 3 −5 x=

3 x + 3 y + 5 z = 1  2 y + 4z = 1   4 y + 263 z = 37  3 x + 3 y + 5 z = 1  2 y + 4z = 1   2 z = 13 3  z = 12

2 y + 4 ( 12 ) = 1  y = − 12

3 x + 3 ( − 12 ) + 5 ( 12 ) = 1  x = 0

−16 5 9 −30 9 17 −20 2 1 −5 3 −16 9 5 −30 17 −20 2 3 1 3 5 −16 5 9 −30 −20

−20 =1 −20

40 = −2 −20

20 = −1 −20

Answer: (1, − 2, − 1)

Answer: ( 0, − 12 , 12 ) (b)

3 3 5 D= 3 5 9 =4 5 9 17

1 3

2 5 9 4 9 17 4 3 1 5 3 2 9 5 4 17 4 3 3 1 3 5 2 5 9 4 4

=−

1 2

Answer: ( 0, − 12 , 12 )

Section 8.8

Applications of Matrices and Determinants

49,853.8 6985 615

56,453.6 6985 615 5004.4 27. (a) a =

615

4436.8

450.8 55 5 ≈ 0.143 80,499 6985 615

28. (a) a =

6985

615

80,499 49,853.8 615

5004.4

6985

615 450.8 5 ≈ 1.417 80,499 6985 615

b =

6985

615

80,499 6985 49,853.8

5004.4

6985

615 55 450.8 ≈ 57 80,499 6985 615

c =

615

4436.8

615 55 401.2 ≈ − 56.220 80,499 6985 615

6985

615

6985

615

120

615

y = − 0.929t 2 + 22.789t − 56.220

y = 0.143t 2 + 1.417t + 57

4436.8

615 401.2 5 ≈ 22.789 80,499 6985 615

6985

(b)

401.2 55 5 ≈ − 0.929 80,499 6985 615

80,499 6985 56,453.6

c =

615

80,499 56,453.6 615

b =

615

6985

(b)

757

The model fits the data well. (c) Answers will vary.

≈ $110.43 billion

758

Chapter 8

Linear Systems and Matrices

29. (a) C O M E

H O M E

S O O N

[3 15] [13 5] [0 8] [15 13] [5 0] [19 15][15 14] 1 2 (b) [3 15] ⋅   = [48 81] 3 5 1 2  = [28 51] 3 5

[13 5] ⋅ 

1 2  = [24 40] 3 5

[0 8] ⋅ 

1 2  = [54 95] 3 5

[15 13] ⋅ 

1 2  = [5 10] 3 5

[5 0] ⋅ 

1 2  = [64 113] 3 5

[19 15] ⋅ 

1 2  = [57 100] 3 5 The encoded message:

[15 14] ⋅ 

48 81 28 51 24 40 54 95 5 10 64 113 57 100 30. (a) H E L P

T H E

W A Y

[8 5] [12 16][0 9][19 0] [15 14] [0 20] [8 5] [0 23] [1 25] − 2 3 (b) [8 5] ⋅   = [− 21 29]  −1 1 − 2 3 [12 16] ⋅   = [− 40 52]  −1 1 − 2 3 [0 9] ⋅   = [− 9 9]  −1 1 − 2 3 [19 0] ⋅   = [− 38 57]  −1 1 − 2 3  = [− 44 59]  −1 1

[15 14] ⋅ 

− 2 3  = [− 20 20]  −1 1

[0 20] ⋅ 

− 2 3  = [− 21 29]  −1 1

[8 5] ⋅ 

− 2 3  = [− 23 23]  −1 1

[0 23] ⋅ 

− 2 3  = [− 27 28]  −1 1 The encoded message:

[1 25] ⋅ 

− 21 29 − 40 52 − 9 9 − 38 57 − 44 59 − 20 20 − 21 29 − 23 23 − 27 28

Section 8.8 31. (a) C A L L

M E

T O M O R

Applications of Matrices and Determinants

759

O W

[3 1 12] [12 0 13] [5 0 20] [15 13 15][18 18 15] [23 0 0]  1 −1 0   (b) [3 1 12] ⋅  1 0 −1 = [− 68 21 35] − 6 2 3    1 −1 0 [12 0 13] ⋅  1 0 −1 = [− 66 14 39] − 6 2 3    1 −1 0 [5 0 20] ⋅  1 0 −1 = [−115 35 60] − 6 2 3    1 −1 0 [15 13 15] ⋅  1 0 −1 = [− 62 15 32] − 6 2 3    1 −1 0 ⋅ 18 18 15 [ ]  1 0 −1 = [− 54 12 27] − 6 2 3    1 −1

0  0 −1 = [23 − 23 0] 2 3

[23 0 0] ⋅  1 − 6 

The encoded message: − 68 21 35 − 66 14 39 −115 35 60 − 62 15 32 − 54 12 27 23 − 23 0 H

33.

32. (a) 8 1 16 16 25 0 D

2 9 18 20 8 4 1 25 0 (b)  8 1 16 A = −5 −41 −87  16 25 0 A = 91 207 257   2 9 18 A = 11 − 5 −41 20 8 4 A = 40 80 84  1 25 0 A = 76 177 227  The encoded message: −5 − 41 − 87 91 207 257 11 − 5 − 41 40 80 84 76 177 227

1 2  −1  −5 2  A= , A =   3 5   3 −1 11 21   8 1 H     64 112   16 16 P  25 50   25 0 Y    −5 2     29 53   3 −1 = 14 5 N   23 0 W  23 46        40 75   25 5 Y  55 92   1 18 A    

A P E E R

Message: HAPPY NEW YEAR

760

Chapter 8

34.

−1

Linear Systems and Matrices  11 2 −8   =  4 1 −3 −8 −1 6   2 −8  1 −3 = [8 1 22] −8 −1 6  

H A V

 11 2 −8 [19 −25 13] 4 1 −3 = [5 0 1] −8 −1 6  

E _ A

[72 −76 61] 4

2 −8  1 −3 = [0 7 18] −8 −1 6  

_ G R

 11 2 −8 − 95 118 71 [ ] 4 1 −3 = [5 1 20] −8 −1 6  

E A T

 11

[112 −140 83] 4

 11

 11 2 −8 [20 21 38] 4 1 −3 = [0 23 5] −8 −1 6  

 11 2 −8 [35 −23 36] 4 1 −3 = [5 11 5] −8 −1 6  

E K E

 11 2 −8 [42 −48 32] 4 1 −3 = [14 4 0] −8 −1 6  

N D _

W E

Message: HAVE A GREAT WEEKEND 35. Let A be the 2 × 2 matrix needed to decode the message. −19 −19  0 19 _ S  37  A = 21 5 U E 16     −1  16 −19 −19  0 19 399  = A−1 =     37 37 16 21 5      − 399

1 21  0  1  21 − 21 

2  5  3 1 C  25 14 3 N 11      −2 −7  5 12 E     −15 −15  1 1  0 15 _  32 14   = 18 4 R   −1 −2   8 13 − −    5 18 E  38 19 19 0 S     19 19 − −    0 19 _  37 21 5 U 16 Message: CANCEL ORDERS SUE

A C L O D R _ S E

19  1 1  =   5 −1 −2

Chapter 8 Review

761

36. Let A be the 2 × 2 matrix needed to decode the message. R  −18 −18   0 18   A =    1 16  15 14  O N −1

8  −18 −18   0 18   − 135 A=   = 1  1 16  15 14   270

− 151   0 18   1   15 14  15 

 −1 −2  =  1  1  8   −15  −13   5  5   5   −1  20   −18  1 

 13 21 5 M E    −10  5 20 E T  0 13 M −13    10 0 E  5 25  −1 −2  20 15 T O  = 19  1 1  14 9 N I    6 8 G H  7  20 40 0 T    R −18  0 18  15 14 O N 16   Message: MEET ME TONIGHT RON

37. True. Cramer’s Rule requires that the determinant of the coefficient matrix be nonzero.

 x + y =1 38. False. The system  has infinitely many 2 x + 2 y = 2  1 1   solutions, yet det   = 0.  2 2    

41.

4 y + x = 19 x + 4 y − 19 = 0

42.

39. Answers will vary. 40. Answers will vary. Sample answer: To find the equation of a line through two points, you could find the slope and then use the point-slope form of an equation, or use a matrix and evaluate the determinant. Using the point-slope form of an equation may be easier because the work is straightforward. However, it can be easy to make a mistake when working with fractions. Using the determinant of a matrix may be easier because the determinant of a 2 × 2 matrix is ad − bc. However, it may be difficult to remember how the determinant can be used to find the equation of a line.

5−3 −1 ( x + 1) = 4 ( x + 1 ) −1 − 7 4 y − 20 = − x − 1 y−5=

43.

y+6=

10 − ( −6 )

−2 − 0 y = −8 x − 6 8x + y + 6 = 0

( x − 0 ) = −8 x

−3 + 1 2 ( x − 3) = 7 ( x − 3) 3 − 10 7 y + 21 = 2 x − 6 y+3=

7 y − 2 x = −27 2 x − 7 y − 27 = 0

44.

12 − 2 5 ( x + 4) = − ( x + 4) −4 − 4 4 4 y − 48 = −5 x − 20 4 y + 5 x = 28 y − 12 =

5 x + 4 y − 28 = 0

Chapter 8 Review 1.

y=2−x  x + y = 2    x − y = 0  x − ( 2 − x ) = 0 2x − 2 = 0 x =1 y = 2 −1 = 1 Answer: (1, 1)

2 x − 3 y = 3   x− y=0

y=x

2 x − 3( x ) = 3 −x = 3 x = −3  y = −3 Answer: ( −3, − 3 )

762

Chapter 8

Linear Systems and Matrices

3.  4 x − y = 1  − 8 x + 2 y = 5 

y = 4 x − 52

(

7.  x 2 − y 2 = 9   x − y = 1  x = y + 1

( y + 1) − y2 = 9 2

)

4 x − 4 x − 52 = 1

2y + 1 = 9

4 x − 4 x + 52 = 1 5 2

≠ 1 Answer: ( 5, 4 )

No solution 4.

10 x + 6 y = −14   x + 9 y = −7 

x = −9 y − 7

10 ( −9 y − 7 ) + 6 y = −14 −90 y − 70 + 6 y = −14 −84 y = 56 y = − 23

Answer: ( −1, − 23 ) 5.

8.  x 2 + y 2 = 169  3x + 2 y = 39 

y = −0.5 x + 0.75

1.25 x − 4.5 ( −0.5 x + 0.75 ) = −2.5 1.25 x + 2.25 x − 3.375 = −2.5 3.5 x = 0.875 x = 0.25

x = 13 ( 39 − 2 y ) 2

 13 ( 39 − 2 y )  + y 2 = 169  

1 9

(1521 − 156 y + 4 y ) + y = 169 2

13 9 13 3

y 2 − 523 y =

y ( 13 y − 4 ) =

1 3

Answer: (13, 0 ) , ( 5, 12 )

9.  y = 2 x 2  4 2 2 4 2  y = x − 2 x  2 x = x − 2 x

0 = x4 − 4x2

(

0 = x2 x2 − 4

)

0 = x ( x + 2 )( x − 2 ) 2

x = 0, x = −2, x = 2

2 3 y− 5 5

y = 0, y = 8, y = 8 Answer: ( 0, 0 ) , ( −2, 8 ) , ( 2, 8 )

3 1 4 2 − y −  + y = − 5 5 5 5 2 3 1 4 − y+ + y=− 5 5 5 5 1 7 − y=− 5 5 y=7 x=  11  Answer:  , 7   5 

0  y = 0, 12

( ) y = 12 : x = ( 39 − 2 (12 ) ) = 5

Answer: ( 0.25, 0.625 ) 2 3   − x + 5 y = 5   −x + 1 y = − 4  5 5

y = 0 : x = 13 39 − 2 ( 0 ) = 13

y = −0.5 ( 0.25 ) + 0.75 = 0.625

169 − 523 y + 49 y 2 + y 2 = 169

x = −9 ( − 23 ) − 7 = −1

y = 0.75   0.5 x +  1.25 x 4.5 y = −2.5 − 

y=4 x=5

2 3 11 ( 7) − 5 = 5 5

10. x =  x =

y +3 y2 + 7

y + 3 = y2 + 7 0 = y2 − y + 4 y = y =

− ( −1) ± 1±

( −1)2 − 4(1)( 4) 2(1)

−15 2

No real solution

Chapter 8 Review 1  11.  5 x + 6 y = 7  y1 = ( 7 − 5 x )  6  − x − 4 y = 0  y = − x 2  4

763

y = 2(6 − x ) 15.   x −2  y = 2 6

6 −2 −9

9 −2

Point of intersection: ( 4, 4 )

−6

Point of intersection: ( 2, − 12 ) 12.  8 x − 3 y = −3  y = 13 ( 8 x + 3 ) = 83 x + 1  1 2 x + 5 y = 28  y = 5 ( 28 − 2 x ) 10

16.   y = ln ( x + 2 ) + 1   y = − x 3

−4

−3 −9

Point of intersection: ( −1, 1)

9 −2

Point of intersection: (1, 5) 13.  y 2 − 4 x = 0    x + y = 0 

y 2 = 4 x  y = ±2 x y = −x

17. Let x = number of plants.

C = 6.43 x + 5000 Cost equation  Revenue equation  R = 12.68 x R = C  6.43 x + 5000 = 12.68 x −6.25 x = −5000

−6

x = 800 plants 12

−6

Points of intersection: ( 0, 0 ) , ( 4, − 4 ) 14.  y = 2 x 2 − 4 x + 1  2  y = x − 4 x + 3

19. 2l + 2 w = 480  l = 1.50w  4

−2

Points of intersection:

I1 = I 2  0.015 x + 55,000 = 0.02 x + 52,000 −0.005 x = −3000 x = 600,000 Total annual sales of $600,000 or greater would make the second offer the better offer.

−2

18. Let x = amount of total sales.  I1 = 0.015 x + 55,000 Income option 1   I 2 = 0.02 x + 52,000 Income option 2

( 2, 5 − 4 2 ) , ( − 2, 5 + 4 2 )

or (1.41, − 0.66 ) , ( −1.41, 10.66 )

2 (1.50 w ) + 2 w = 480 5w = 480 w = 96 l = 144 The dimensions are 96 × 144 meters.

764

Chapter 8

Linear Systems and Matrices 22. 7 x + 2 y = 12   x − 2y = −4

20. 2l + 2 w = 68  w = 89 l 

8  2l + 2  l  = 68 9  34l = 68 9 l = 18 w = 16 The dimensions are 16 × 18 feet.

Add to solve for x.

8x = 8 x =1 7(1) + 2 y = 12  y = 52

( )

Answer: 1, 52

23. 2 x − y = 2  16 x − 8 y = 16  6 x + 8 y = 39  6 x + 8 y = 39

21.  x + 3 y = − 5  − x − 8 y = 0

22 x = 55

Add to solve for y.

55 x = 22 = 25

−5y = −5

y=3

y =1

Answer: ( ,3) 5 2

x + 3(1) = − 5  x = − 8 Answer: ( − 8, 1)

24. 7 x + 12 y = 63  −7x − 12 y = −63  2 x + 3 y = 15  8 x + 12 y = 60  x = −3

y=7 Answer: ( −3, 7 ) 3 y = 7  20 x + 30 y = 14  20 x + 30 y = 14 25.  54 x + 10 50 2 1 1  5 x + 2 y = 5  4 x + 5 y = 2  −20 x − 25 y = −10 5y = 4

y = 54 x = − 12

Answer: ( − 12 , 45 ) or ( −0.5, 0.8 ) 26. 0.2 x + 0.3 y = 0.14  0.4 x + 0.5 y = 0.20 −2 times Equation 1 and add to Equation 2 to eliminate x: −0.4 x − 0.6 y = −0.28

0.4 x + 0.5 y = 0.20 − 0.1 y = −0.08 y = 0.8 Substitute y = 0.8 into Equation 1: 0.2 x + 0.3 ( 0.8 ) = 0.14 0.2 x + 0.24 = 0.14 0.2 x = −0.1 x = −0.5 Answer: ( −0.5, 0.8 )

3x − 2 y = 0  3x − 2 y = 0 27.  3x + 2 y + 5 = 10  3 x + 2 y = 0 ( )  6x

=0 x=0 y=0

Answer: ( 0, 0 ) 28.  6 x − 4( y − 1) = 19  = 3 − 3x + 2 y

 6 x − 4 y = 15  6 x − 4 y = 15  − 3x + 2 y = 3  − 6 x + 4 y = 6 0 ≠ 21 No solution

Chapter 8 Review 29. 1.5 x + 2.5 y = 8.5  3x + 5 y = 17  6 x + 10 y = 24  −3x − 5 y = −12 

Inconsistent; no solution

7 (7 2) x + 1 = x + 1 34.  x − 7 y = −1  y = 2 7 2 7  x  −x + 2y = 4  y = + 2  2 3

30. 1.25 x − 2 y = 3.5  5x − 8 y = 14  5 x − 8 y = 14  −5 x + 8 y = −14 

0= 0 Infinite number of solutions Let y = a , then 5 x − 8a = 14  x = 145 + 85 a.

Answer: ( 145 + 85 a, a )

31. 3 x + 2 y = 0  y = − 23 x   x − y = 4  y = x − 4

765

−3

3 −1

Inconsistent; lines are parallel. 35.  8 x − 2 y = 10   − 6 x + 1.5 y = − 7.5 

y = 4x − 5 y = 4x − 5

2 −6 −4

−4

Consistent; All points on the line y = 4 x − 5

−6

Consistent Answer: (1.6, − 2.4 ) 32.  x + y =   −2 y =

 y =6− x

1  36.  − x + 3.2 y = 10.4  y = ( x + 10.4 )  3.2   −2 x − 9.6 y = 6.4  y = −1 ( 2 x + 6.4 )  9.6

− 12 + 2 x  y = 6 − x

7 −15

−3

−7

Consistent Answer: ( −7.52, 0.9 )

−1

Consistent; Infinite number of solutions of form ( x, 6 − x ) or ( 6 − y, y ) . All points on line y = 6 − x.

37.

15 = 0.00021x

33.  14 x − 15 y = 2  y = 54 x − 10  5 1 −5 x + 4 y = 8  y = 4 ( 8 + 5 x ) = 4 x + 2

500,000 159 , p= 7 7  500,000 159  Point of equilibrium:  ,  7 7   x=

−7

−6

Inconsistent; lines are parallel.

Demand = Supply 37 − 0.0002 x = 22 + 0.00001x

38.

Supply = Demand 45 + 0.0002 x = 120 − 0.0001x

0.0003 x = 75 x = 250,000 units p = $95.00 Point of equilibrium: ( 250,000, 95 )

766

Chapter 8

Linear Systems and Matrices

39. Let x = speed of the slower plane. Let y = speed of the faster plane. Then, distance of first plane + distance of second plane = 275 miles. (rate of first plane )(time) + (rate of second plane)(time) = 275 miles. 40 40  x ( 60 ) + y ( 60 ) = 275  y = x + 25  2 2 x + 3 ( x + 25 ) = 275 3

4 x + 50 = 825 4 x = 775 x = 193.75 mph y = x + 25 = 218.75 mph

40. Let x = amount invested in 6.75% bond. Let y = amount invested in 7.25% bond.

 x + y = 46,000  0.0675 x + 0.0725 y = 3245 Solving the system, x = 18,000, y = 28,000. At most $18,000 can be invested in the 6.75% bond. 41.  x − 4 y + 3z = 3  − y + z = −1   z = −5 

Substitute z = −5 into Equation 2: − y + ( −5 ) = −1 −y = 4 y = −4 Substitute y = −4 and z = −5 into Equation 1: x − 4 ( −4 ) + 3 ( −5 ) = 3 x + 16 − 15 = 3 x=2 Answer: ( 2, − 4, − 5 ) 42.  x − 7 y + 8 z = −14  y − 9 z = 26   z = −3 

Substitute z = −3 into Equation 2:

y − 9 z = 26  y = 9 ( −3 ) + 26 = −1

Substitute y = −1 and z = −3 into Equation 1: x − 7 y + 8 z = −14  x = −14 + 7 ( −1) − 8 ( −3 ) = 3 Answer: ( 3, − 1, − 3 )

43.  x − 2 y + 5 z = −12  + 6 z = − 28 2 x  x + y + 18 z = − 4 

x − 2 y + 5 z = −12  3 y + 13z = 8 −1 Eq.1 + Eq.3   4 y 4 z 4 − = − − 2 Eq.1 + Eq.2   x − 2 y + 5 z = −12  3 y + 13 z = 8   64 = 44 4 Eq.2 + ( − 3) Eq.3 z  11 64 z = 44  z = 16

( ) 5 11 x − 2( − 16 ) + 5(167 ) = −12  x = − 257 16 257 5 11 Answer: ( − 16 , − 16 , 16 )

15 5 11 3 y + 13 16 = 8  3 y = − 16  y = − 16

44.  x + y + z = 2  − x + 3 y + 2 z = 8  4x + y =4 

x + y + z = 2   4 y + 3z = 10 Eq.1 + Eq.2  −3y − 4 z = −4 −4 Eq.1 + Eq.3  x + y + z = 2   4 y + 3z = 10  − 74 z = 27 

3 4

Eq.2 + Eq.3

− 74 z = 72  z = −2

4 y + 3 ( −2 ) = 10  y = 4

x + 4 + ( −2 ) = 2  x = 0

Answer: ( 0, 4, − 2 ) 45.  x − 2 y + z = −6  = −7 2 x − 3 y − x + 3 y − 3z = 11 

 x − 2 y + z = −6  y − 2 z = 5 −2 Eq.1 + Eq. 2   y − 2 z = 5 Eq.1 + Eq. 3   x − 2 y + z = −6  y − 2z = 5   0 = 0 − Eq. 2 + Eq. 3 

Let z = a , then y = 2 a + 5. x − 2 ( 2 a + 5 ) + a = −6 x − 3a − 10 = −6 x = 3a + 4 Answer: ( 3a + 4, 2 a + 5, a ) , where a is any real number

Chapter 8 Review 46. 2 x + 6z = − 9  3 x − 2 y + 11z = −16 3 x − y + 7z = −11 

49. 5 x − 12 y + 7 z = 16  3 x − 7 y + 4 z = 9

− x + 2 y − 5 z = 7 − Eq. 2 + Eq. 1  3 x − 2 y + 11z = −16 3 x − y + 7 z = −11  − x + 2 y − 5 z = 7  4 y − 4 z = 5 3 Eq. 1 + Eq. 2   5 y − 8 z = 10 3 Eq. 1 + Eq. 3  − x + 2 y − 5 z = 7  4 y − 4z = 5   − 3y =0 

− 2 Eq. 2 + Eq. 3

−3 y = 0  y = 0 4 ( 0 ) − 4 z = 5  z = − 54

−x + 2 (0) − 5(−

)=7 x =− Answer: ( − , 0, − ) 5 4

3 4

3 times Eq.1 and ( −5 ) times Eq. 2:

 15 x − 36 y + 21z = 48  −15 x + 35 y − 20 z = −45 Adding, − y + z = 3  y = z − 3. 5 x − 12 ( z − 3 ) + 7 z = 16

5 x − 5z + 36 = 16 5 x = 5z − 20 x = z−4 Let z = a , then x = a − 4 and y = a − 3. Answer: ( a − 4, a − 3, a ) , where a is any real number 50. 2 x + 5 y − 19 z = 34  3 x + 8 y − 31z = 54 6 x + 15 y − 57 z = 102  6 x + 16 y − 62 z = 108

3 4

5 4

47.  x − 2 y + 3z = −5  2 x + 4 y + 5 z = 1  x + 2y + z = 0 

 x − 2 y + 3z = −5  8y − z = 11   4 y − 2z = 5   x − 2 y + 3z = −5  4y − 2 z = 5   3z = 1  1 z= 3

6 x + 15 y − 57 z = 102  y − 5z = 6 

Let z = a y = 5a + 6

x = 102 + 57a − 15 ( 5a + 6 )  6 = −3a + 2

−2 Eq.1 + Eq. 2 −Eq.1 + Eq. 3

Answer: ( 2 − 3a, 5a + 6, a ) , where a is any real number

Interchange Eq.2 and Eq.3

51. x + 5 y + z = 10

−2Eq. 2 + Eq. 3

z 10

17 4 y = 5 + 2 ( 13 ) = 173  y = 12

17 x = 2 ( 12 ) − 3 ( 13 ) − 5 = − 196

17 , 13 ) Answer: ( − 196 , 12

4 8

48.  x − 2 y + z = 5  2 x + 3 y + z = 5  x + y + 2z = 3 

Sample answer: (10, 0, 0), (0, 2, 0), (0, 0, 10), (1, 1, 4) 52. 2 x + 3 y + 4 z = 12 z

x − 2y + z = 5  7 y − z = −5   3 y + z = −2  Adding the second and third equations, 10 y = −7  y = − 107

z = 7 y + 5 = 7 ( − 107 ) + 5 = 101 x = 2y − z + 5 = 2(−

767

7 10

Answer: ( 72 , − 107 , 101 )

)−

1 10

6 4

2 x

+5=

7 2

2 6

Sample answer: (6, 0, 0), (0, 4, 0), (0, 0, 3), ( 2, 0, 2)

768

Chapter 8

Linear Systems and Matrices

53.

4−x A B = + x2 + 6 x + 8 x + 2 x + 4

58.

4 − x = A( x + 4) + B ( x + 2)

59.

y = ax 2 + bx + c

y = 2 x2 + x − 5

60.

3 2 3 2

( −1, 0 )  a − b + c = 0 (1, 4 )  a + b + c = 4 ( 2, 3)  4a + 2b + c = 3 y = − x2 + 2x + 3

x2 2 x − 15 =1− 2 x + 2 x − 15 x + 2 x − 15 −2 x + 15 A B = + ( x + 5)( x − 3) x + 5 x − 3 −2 x + 15 = A( x − 3) + B( x + 5)

Let x = −5: 25 = −8 A  A = −

y = ax 2 + bx + c

Solving the system, a = −1, b = 2, c = 3.

25 8

9 8

x2 25 9 =1− + x + 2 x − 15 8( x + 5) 8( x − 3) 2

x2 + 2x A Bx + C = + x3 − x2 + x − 1 x − 1 x2 + 1

(

)

Solving the system, a = 2, b = 1, c = −5.

9 1 3 3  =  −  x2 − 9 2 x − 3 x + 3

57.

)

( −1, − 4 )  a − b + c = −4 (1, − 2 )  a + b + c = −2 ( 2, 5)  4a + 2b + c = 5

9 A B = + x2 − 9 x −3 x +3 9 = A( x + 3) + B( x − 3)

Let x = 3: 9 = 8 B  B =

)

(

Let x = −1 : 1 = A Let x = −2 : 2 = − B  B = −2 −x 1 2 = − x2 + 3x + 2 x + 1 x + 2

56.

(

2 x2 − x + 7 2 −x − 1 = + 2 x 4 + 8 x 2 + 16 x 2 + 4 x2 + 4

− x = A ( x + 2 ) + B ( x + 1)

Let x = −3: 9 = −6 B  B = −

= 0  A  = 2 B   +C = −1 4 A  +D= 7 4B A = 0, B = 2, C = −1, D = −1

−x A B = + x2 + 3x + 2 x + 1 x + 2

Let x = 3: 9 = 6 A  A =

(

)

2 x − x + 7 = ( Ax + B ) x + 4 + Cx + D

 A + B = −1  A = 3, B = −4  4 A + 2 B = 4 4−x 3 4 = − x2 + 6 x + 8 x + 2 x + 4

55.

(

= ( A + B) x + ( 4 A + 2B)

54.

2 x2 − x + 7 2 x 2 − x + 7 Ax + B Cx + D = = 2 + 2 2 4 2 x + 8 x + 16 x +4 x2 + 4 x2 + 4

)

x 2 + 2 x = A x 2 + 1 + ( Bx + C )( x − 1) = ( A + B) x2 + (C − B) x + ( A − C )

61. Let x = number of par-3 holes. Let y = number of par-4 holes. Let z = number of par-5 holes.  x + y + z = 18 Equation 1  y − 2 z = 2 Equation 2  x − z = 0 Equation 3   x + y + z = 18  y − 2z = 2   − y − 2 z = −18 − Eq. 1 + Eq. 3   x + y + z = 18  y − 2z = 2   − 4 z = −16 Eq. 2 + Eq. 3  −4 z = −16  z = 4

y − 2 ( 4 ) = 2  y = 10 x + 10 + 4 = 18  x = 4 Answer: 4 par–3 holes, 10 par–4 holes, 4 par–5 holes

 A+ B =1  3 3 1 − B + C = 2  A = 2 , B = − 2 , C = 2  A−C = 0 

x2 + 2 x x −3  3 2 − (1 2 ) x + 3 2 1  3 = + =  − 2  3 2 2 x − x + x −1 x −1 x +1 2  x −1 x +1

Chapter 8 Review 62. Let x = amount invested at 7%. Let y = amount invested at 9%. Let z = amount invested at 11%. x+ y+ z = 40,000 Equation 1   x y z = 8000 Equation 2 2 − −  0.07 x + 0.09 y + 0.11z = 3500 Equation 3  y + z = 40,000 x +  −3y − 3z = −72,000 −2 Eq. 1 + Eq. 2   0.02 y + 0.04 z = 700 0.07 Eq. 1 + Eq. 3 −  y+ z = 40,000 x +  1  y+ z = −24,000 − Eq. 2  3   0.02 y + 0.04 z = 700 x + y + z = 40,000   y + z = −24,000   0.02z = 220 −0.02 Eq. 2 + Eq. 3  0.02 z = 220  z = 11,000 y + 11,000 = 24,000  y = 13,000 x + 13,000 + 11,000 = 40,000  x = 16,000 Answer: $16,000 at 7%, $13,000 at 9%, $11,000 at 11%

63. Dimension 3 × 1

68.  − x + y = 12  = −90  10 x

 −1 1  12   10 0  −90 69.  3 x − 5 y + z = 25  − 2 z = −14 −4 x  6x + y = 15 

1  25  3 −5   − − 4 0 2  −14    6 1 0  15 70. 8 x − 7 y + 4 z = 12  3 x − 5 y + 2 z = 20  5 x + 3 y − 3z = 26 

8 −7 4  12    3 −5 2  20  5 3 −3  26   5 1 7  −9    71.  4 2 0  10   9 4 2  3

64. Dimension 2 × 4

 5x + y + 7 z = −9  = 10 4 x + 2 y 9 x + 4 y +2 z = 3 

65. Dimension 1 × 1 66. Dimension 1 × 5 67.  6 x − 7 y = 11   −2 x + 5 y = −1

 6 −7  11    −2 5  −1

769

13 16 7 3  0  72.  0 21 0 5  12   4 10 −4 3  −1 13x + 16 y + 7 z + 3w = 0  21y + 5w = 12   4x + 10 y − 4 z + 3w = −1 

 1 − 4 0 −1 4 0  1 − 4 0 73.        0 1 97  7 9   3 − 5 9 0  1 3 − 7 7 6 −14 3 − 7 0 − 2 2 1 74.          0 1 − 7  2 6 14 0 2 7 0 2 7 − − −       2  

770

Chapter 8

Linear Systems and Matrices

0 1 1    1 2 3 2 2 2 

75.

R1 + R2 →  1 3 4    − R1 + R2 → 0 −1 −1 −2 R1 + R3 → 0 −4 −6  3 R2 + R1 →  1 0 1   − R2 → 0 1 1 −4 R2 + R3 → 0 0 −2 

 4 8 16     3 −1 2   −2 10 12 

76.

R1 →  1 2 4    3 R1 + ( −4 ) R2 → 0 28 40  2 R3 + R1 → 0 28 40  1 4

1 2  R2 → 0 1 R3 + ( −1) R2 → 0 0 1 28

4   0 

10 7

 3 −2  1 0 77.     4 −3 0 1  1 1 2  1 0 0 78.  −1 0 3  0 1 0  1 2 8 0 0 1

1.5 3.6 4.2  1 0 0      79. 0.2 1.4 1.8   0 1 0  2.0 4.4 6.4  0 0 1 

82. 2 x − 5 y = 2  3 x − 7 y = 1 2 −5  2    3 −7  1  1 −2  −1 R2 − R1 →   3 −7  1   1 −2  −1 −3 R1 + R2 →   0 −1  4  y = −4

x = 2 ( −4 ) − 1 = −9 Answer: ( −9, − 4 ) 83. 0.6 x − 0.9 y = − 0.3  0.3 0.4 x − 0.6 y = 0.6 − 0.9  − 0.3   0.3 0.4 − 0.6  5R 3 1

→

10 R2

→

1 − 3  − 1 2 2   3 4 − 6 

− 4 R1 + R2

1 − 3  − 1  2 2   → 0 0  5

Inconsistent; no solution 84. 0.2 x − 0.1y = 0.07  0.4 x − 0.5 y = −0.01

0.2 −0.1  0.07   0.4 −0.5  −0.01 5R1 → 1 −0.5  0.35   −2 R1 + R2 → 0 −0.3  −0.15 y = 0.5

 4.1 8.3 1.6  1 0 0      80.  3.2 −1.7 2.4   0 1 0   −2.3 1.0 1.2  0 0 1 

x = 0.5 ( 0.5 ) + 0.35 = 0.6 Answer: ( 0.6, 0.5)

81. 5 x + 4 y = 2   − x + y = −22

2  5 4    1 1 22 − −    4 R2 + R1 →  1 8  −86    R1 + R2 → 0 9  −108  9 y = −108 y = −12 x = −8 ( −12 ) − 86 = 10

Answer: (10, − 12 )

Chapter 8 Review 85.  2 x + 3 y + 3z = 3   6 x + 6 y + 12 z = 13  12 x + 9 y − z = 2 

 2 3 3  3    6 6 12  13 12 9 −1  2  3 3  3 2   3  4 −3R1 + R2 → 0 −3 −6 R1 + R3 → 0 −9 −19  −16  R2 + R1 → 2 0 6  7   0 3 3 4  −   −3R2 + R3 → 0 0 −28  −28  R1 → − R2 → − R3 → 1 2 1 3 1 28

y − 1 = − 43  y = − 13 x + 3 (1) = 72  x = 12 Answer: ( 12 , − 13 , 1)

1  2  3 1  − 2 R1 + R2 → 0 − 3R1 + R3 → 0

5 R2 + R3 →

 1 −1 4 −1   1 3 −2 1 0 1 −1 1  1 1 2 0

4  1    −4  0   −3 0    0  0 

1  1 0 0  −1 0 1 0  0  0 0 1  −2  0 0 0 

Answer: (1, − 1, 0, − 2 )  x + y + z = −3 89.  4 x + y − 2 x = −12 1  − 3 1 1   −  −12 4 1 2 

(− 4) R1 + R2 →

1 1  − 3 1   0 3 6 − −  0 

(− 13 ) R

 1 1 1  − 3   0 1 2  0

→

z = a

86.  x + 2 y + 6z = 1  2x + 5 y + 15z = 4 3 x + y + 3z = −6 

−2 R2 + R1 →

88.  x − y + 4 z − w = 4   x + 3 y − 2 z + w = −4  y − z + w = −3   2 x + z +w= 0

7 1 0 3   2  4  0 1 1 − − 3  0 0 1  1 

z =1

771

y + 2a = 0  y = − 2a x + y + z = − 3  x = − 3 + 2a − a

1  4 3  −6 

2 6  5 15  1

2 6  1  1 3  2 −5 −15  −9   1 0 0  −3   0 1 3  2  0 0 0  1

0 =1 Inconsistent, no solution 87.  x + 2 y − z = 1  y+z=0 

 1 2 −1  1  1 0 −3  1    0 1 1  0  0 1 1  0  y = −z x = 1 + 3z

Answer: (1 + 3a, − a, a ) , a is a real number

x = a −3

Answer: ( a − 3, − 2a, a )  2x + 2 y = 4 90.  − 3 x − y = −16 4  2 2    − 3 −1  −16

( 12 ) R

→

2  1 1    − 3 −1  −16

2 1 1    3R1 + R2 → 0 2  −10

1R 2 2

→

 1 1  2   0 1  − 5

y = −5 x + ( − 5) = 2  x = 7

Answer: (7, − 5)

772

Chapter 8

Linear Systems and Matrices 92. 4 x + 4 y + 4 z = 5   4 x − 2 y − 8z = 1  5 x + 3 y + 8z = 6 

91.  − x + y + 2z = 1   2 x + 3 y + z = −2  5 x + 4 y + 2z = 4 

− R1 → 2 R1 + R2 → 5 R1 + R3 → 1 5

 −1 1   2 3  5 4  1 −1  0 5 0 9

1  R2 → 0 0

2  1  1  −2  2  4  −2  −1  5  0 12  9  −1 −2  −1  1 1  0 9 12  9  0 −1  −1  1 1  0 0 3  9  0 −1  −1  1 1  0 0 1  3

1  0 −9 R2 + R3 → 0 1  0 1 0 R → 3 3 R3 + R1 →  1 0 0  2    − R3 + R2 → 0 1 0  −3 0 0 1  3 x = 2, y = −3, z = 3 R2 + R1 →

R3 − R1 → −4 R1 + R2 → −5R1 + R3 → R2 + R1 → 1 R → 2 2 −8 R2 + R3 → 8 R3 + R1 → 12R3 + R2 → 1 R → 84 3

 4 4 4  5    4 −2 −8  1  5 3 8  6  4  1  1 −1   0 2 24  −3 −  0 8 −12  1  1 0 −8  − 12   3 0 1 −12  − 2  0 0 84  13   31  1 0 0  42   5  0 1 0  14  0 0 1  13  84  

31 x = 42 , y = 145 , z = 13 84

31 , 145 , 13 Answer: ( 42 84 )

Answer: ( 2, − 3, 3 ) 93. x +  x −  x

y + 2z =

y + 4z =

+ 3z = − 5

2  4 2  4 1 1 1 1     R1 + ( −1) R2 → 0 2 − 2  3  3 0 2 − 2  R1 + ( −1) R3 → 0 1 −1  9 R2 + ( − 2) R3 →0 0 0  −15 Inconsistent, no solution 94.  x + y + 4 z = 0  2 x + y + 2z = 0 − x + y − 2 z = −1  1   1 1 4  0  1 0 0  5    3 2 1 2 0 0 1 0 −    5     −1 1 −2  −1 0 0 1  101   

Answer: ( 15 , − 35 , 101 )

Chapter 8 Review 95.  x + 2 y − z = 3   x − y − z = −3 2 x + y + 3z = 10   1 2 −1  3  1 −1 −1  −3    2 1 3  10  R1 − R2 → 2 R1 − R3 →

 1 2 −1  3 0 3 0  6    0 3 −5  −4 

 1 2 −1  3 R2 → 0 1 0  2    R2 − R3 → 0 0 5  10 

 1 2 −1  7   1 0 0  3     97.  0 −1 −1  4   0 1 0  0   4 0 −1  16  0 0 1  −4  Answer: ( 3, 0, − 4 )  3 0 6  0   1 0 0  −2      1 98.  −2 1 0  5  0 1 0   0 1 2  3 0 0 1  1 Answer: ( x, y , z ) = ( −2, 1, 1)

1 3

R1 − 2 R2 →  1 0 −1  −1 0 1 0  2    1 5 R3 → 0 0 1  2  R1 + R3 →

 1 0 0  1 0 1 0  2    0 0 1  2 

x = 1, y = 2, z = 2

Answer: (1, 2, 2)

3 R1 − R2 → R1 + R3 → − 81 R2 → R2 − 4 R3 →

1 12

1 2

R3 →

R3 + R2 →

1  2  1 −3  3 1  − − 0 1 2 2  0 0 1  1  1 −3 1  2    1 0  −1 0 0 0 1  1

1  0 0 − R3 + R1 →  1  0 0

x = − 2, y = − 1, z = 1

Answer:

Inconsistent, no solution  3 −1 5 −2  −44   1 0    1 6 4 −1  1 0 1  100.   5 −1 1 3  −15 0 0    0 4 −1 −8  58  0 0

0 0 1 0

0 0 0 1

2   6   −10    −3

( 2, 6, − 10, − 3)

101. x = 12 y = −7

1  2  1 −3   − − 3 1 1  −6    −1 1 −3  −2  1  2  1 −3   0 −8 4  12  0 −2 −2  0  1  2  1 −3   1 1 − 2  − 23  0 0 0 12  12 

3R2 + R1 →

99.

 1 0 − 3  0 2  20 4 12     7  0 4  12  0 1 1 6 6   5 6 − 8  − 4 0  1   0 0

Answer:

96.  x − 3 y + z = 2  3 x − y − z = −6 − x + y − 3z = −2 

773

0 1  −1  1 0  −1 0 1  1 0 0  −2   1 0  −1 0 1  1

( −2, − 1, 1)

102. x = 8 y=0

103. x + 3 = 5 x − 1  x = 1 −4 y = −44  y = 11 y + 5 = 16  y = 11 6x = 6  x = 1 Answer: x = 1, y = 11 104. 2 x = x − 10  x = −10

−4 = 2y  y = −2 − 5 = 12 x  x = −10 Answer: x = −10, y = − 2  7 3 10 −20  105. A =  , B =   − 1 5   14 −3  7 3 10 −20  17 −17 (a) A + B =  + =  2  −1 5 14 −3 13  7 3 10 −20   −3 23 (b) A − B =  − =   −1 5 14 −3  −15 8  7 3  14 6  (c) 2 A = 2  =   −1 5  −2 10   7 3 10 −20  37 −57  (d) A + 3B =   + 3 =   −1 5 14 −3  41 −4 

774

Chapter 8

Linear Systems and Matrices

6 0 7   0 5 1     106. A =  5 −1 2  , B =  −4 8 6   3 2 3  2 −1 1

6 0 7   0 5 1 6 5 8        (a) A + B =  5 −1 2  +  −4 8 6  =  1 7 8   3 2 3  2 −1 1  5 1 4  6 0 7   0 5 1 6 −5 6        (b) A − B =  5 −1 2  −  −4 8 6  = 9 −9 −4   3 2 3  2 −1 1  1 3 2  (c)

0 14  6 0 7  12     2 A = 2  5 −1 2  = 10 −2 4   3 2 3  6 4 6 

(d)

6 0 7   0 5 1  6 15 10        A + 3B =  5 −1 2  + 3  −4 8 6  =  −7 23 20   3 2 3  2 −1 1  9 −1 6 

 6 0  −11 16 19    , B = 107. A =    8 −4  − 7 − 2 1    −2 10  (a) Not possible (b) Not possible  −11 16 19  −22 32 38 (c) 2 A = 2 =  =   −7 −2 1  −14 −4 2  (d) Not possible 8  6 5 − 8     4, B =  3  108. A = −1 2  3 9 − 2 7     (a) A + B = Not possible (b) A − B = Not possible 8  6 5 12 10 16      4 = − 2 4 8  (c) 2 A = 2 −1 2  3 9 − 2  6 18 − 4     (d) A + 3 B = Not possible 3 −6  2 1 0 5 109.   − 3  0 5 4 0 2 5 − −     2 1 0  15 9 −18 = −   0 5 −4   0 −6 15  −13 −8 18 =   0 11 −19 

 1 2 7 1     110. −4  5 −4  + 8  1 2  6 0   1 4   −4 −8 56 8   52 0         −20 16  +  8 16  =  −12 32   −24 0   8 32   −16 32   − 2 0 7 − 8   8 −1 111. −1  + 5   +   3  − 2 4   3 −1 4 1 − 8 5 − 8 =   + 5  − 2 4 2   7 1 25 − 40 − 8 =   +   10  2 − 4 35 = 17 − 39   6 37

 − 4 −1 − 3 4 −1 1 13 − 7  112. 6   −  2 5 7 10 14 3 8 −1  − − −     − 3 − 2 −16 11 = 6  −12 − 2 −1 − 9 66  −18 −12 − 96 =   − 72 −12 − 6 − 54 21 − 45 8 − 2 5 4 − 2 − 3 4  + = 6 113. − 83     93 327 − 1 3 1 2 7 6      8 8  − 29 0.2 2.7  4.4 − 2.3  5     114. − 5  7.3 − 2.9 + 74  6.6 11.6 = − 499  20 8.6 −1.5 2.1 3.9  − 365   8

 − 159 8  291 8 

 − 201 40  174  5  − 147  40 

115. X = 3 A − 2 B  −4 0   −2    X = 3  1 −5 − 2  −2  −3 2   4 0  4  −12    X =  3 −15 +  4  −9 6   −8

2  1 4  −4   −8 −4     −2  =  7 −17  −8   −17 −2 

Chapter 8 Review 116. 6 X = 6 A + 3B

0 6 − 4 − 2 2 − 30       6 X = 6  1 − 5 + 3 − 2 1 =  0 − 27 −3  4 4  −6 2 24     − 30

X =

6 1 − 5    9 0 − 27 =  0 − 2   −6  −1 24 4  

1 6

117. 3 X + 2 A = B 3X = B − 2 A  −2 2   −4 0   6 2        3 X =  −2 1 − 2  1 −5 =  −4 11  4 4   −3 2   10 0   6 2   2 23     1  X = 3  −4 11 =  − 43 113   10 0   103 0   

2 A = 3 X + 5B

118.

0 − 4 − 2 2     2  1 − 5 = 3 X + 5 − 2 1 − 3  4 4 2     2 −10 − 2 2     − − 12 15 5   − 2 1 = 3 X − 26 −16  4 4      2 −10  1 12 −15 = X 3 − 26 −16    2 − 10  3  3   4 − 5 = X  26 16  − 3 − 3 

775

 1 2 6 −2 8 119. AB = 5 −4   4 0 0 6 0   1( 6) + 2( 4) 1( −2) + 2( 0) 1(8) + 2( 0)    = 5( 6) + ( −4)( 4) 5( −2) + ( −4)( 0) 5( 8) + ( −4)( 0)   6( 6) + ( 0)( 4) 6( −2) + ( 0)( 0) 6( 8) + ( 0)( 0)    14 −2 8   = 14 −10 40 36 −12 48

−1 2 1   120. AB = [6 − 5 7]  4 − 3 0 = [30 34 13]  8 1 1 

 1 5 6 7 5 2 121.    is undefined. 2 −4 0  0 1 0  2 4 − 3 2  3    AB 2 3 −1 is undefined. =  − 4 0 122. −1 3 0 6 2  1  4  14 −22 22     3 −5 6    123. 11 −7    =  19 −41 80  2 2 2 − −   42 −66 66  12 3     1 1 3 10    −2   13 24  124.    −5 2  =    4 −2 2   3 2  20 4    2 1   4 2 −2 4  2 1  2 6 125.     +   =    6 0  −3 1  0 4  6 0 −3 5 2 ( 2) + 1( −3) 2 ( 6) + 1( 5)  =  6 ( 2) + 0 6( 6) + 0   1 17 =  12 36

1 −1  0 31 0 1 −1 15 − 9  4 − 3 126.       =    =  4 2  1 25 − 3 4 2  11 − 6 82 − 48 127. (a) B = [$99 $112] 80 120 140 (b) BA = [99 112]   = [$12,400 $23,080 $22,820] 40 100 80

The entries represent the total prices of all tires at each factory.

776

Chapter 8

Linear Systems and Matrices

128. (a) B = [$79.99 $109.95 $189.99]

8200 7400   (b) BA = [$79.99 $109.95 $189.99] 6500 9800 = [$2,396,539 $2,581,388] 5400 4800   The entries represent the total prices of all headphones at each warehouse.

 −4 −1  −2 −1  1 0  129. AB =   =  ; BA = I 2  7 2   7 4  0 1  1 1 0   −2 −3 1  1 0 0       130. AB =  1 0 1  3 3 −1 = 0 1 0  ; BA = I 3 6 2 3  2 4 −1 0 0 1 131.

 −6 5  1 0  AI 2 =    −5 4  0 1 − 16 R1 → 1 − 56  − 16 0   − 15 R2 → 1 − 54  0 − 15  0  1 − 56  − 16  1 1 − R1 + R2 → 0 301  − 6 5 0 25 R2 + R1 →  1 0  4  1 1 1  0 30  6 − 5   1 0  4 0 −1  =I A 30 R2 → 0 1  5 −6  2 4 0 A−1 =    5 −6 

132.

AI 2 =

 2 0 3  1 0 0   AI 3 =  −1 1 1  0 1 0   2 −2 1  0 0 1

133.

R1 →  1 0 23  12 0 0    R1 + 2 R2 → 0 2 5  1 2 0  R1 − R3 → 0 2 2  1 0 −1 1 2

 1 0 32  12 0 0    R2 → 0 1 52  12 1 0  R2 − R3 → 0 0 3  0 2 1  1 0 32  12 0 0    5 1 0 1 2  2 1 0  1  0 1  0 23 13  3 R3 →  0 1 2

R1 − 32 R3 →  1 0 0  12 −1 − 12   −1 −5  R2 − 52 R3 → 0 1 0  12 − 23 6  = I3 A 2 1  0 0 1  0 3 3 1 1  2 −1 − 2    −1 A =  12 − 23 − 65  2 1 0 3 3 

− 3 − 9  1 0    4 12  0 1

− 13 R1 → 1 3  − 13 0   1R → 1 3  0 14  4 2  1 3  − 13 − R1 + R2 →  1 3 0 0 

0  1  4

Inverse does not exist.

Chapter 8 Review 1  1 0 0  0 −2   134. AI 3 =  −5 −2 −3  0 1 0   7 3 4  0 0 1 R2 →  −5 −2 −3  0 1 0    R1 →  0 −2 1  1 0 0  7 3 4  0 0 1 − R1 →  1  0 −2 3 1 R →  1 7 3 7 1 5

3 5

2 5

 0 −

1  1 4 7

 0

0  0 0 0 17  1 5

1  0 − 15 0    1  1 0 0 0 −2 1 1 − R1 + R3 → 0 351 − 351  0 5 7 3 2 1 1 5  0 − 5 0 5   − 12 R2 → 0 1 − 12  − 12 0 0 1 1 0 1 − 1  0 35 35 5 7  3 5

2 5

4 1  − 25 R2 + R1 →  1 0 5 5  1 1 0 1 − 2  − 2 − 351 R2 + R3 → 0 0 − 701  701

0  0 0 1 1 5 7

− 15

 1  4 139.   3  −  1

6  4 2 6 4 1 2  2 −1 − 2 3

8  1 11   A−1 =  −1 −7 −5  −1 −14 −10  −1

 15 2 6  135.   =1  10  3 −6  −1

  − 

 231  3 −10  136.   = − 2 2  23 4  1 −1 − 2   137. 0 3 − 2  1 2 − 4  

1 5 1 15

5 23 3 46

  

−1

 1 2 0  1 0 2     138.  −1 1 1 = 0 0 −1  0 −1 0   1 1 3

6 − 5.5 3.5 − 3   −1 1 −2 2  =  7 −15 14.5 − 9.5    −1 2.5 − 2.5 1.5

−1

3 7 − 2 − 2.5   − − 3 4 4.5 11 =   14.5 −16 − 40 12   1 3 −1  −1

−1

 −7 2  2 −2   1 −1 1 = 141.   = ( −7)( 2 ) − ( 2 )( −8) 8 −7 4 − 72   −8 2  −1

10 4   3 −4   23 1 = 142.   = (10 )( 3) − ( 4 )( 7) −7 10 − 27  7 3

−2   5

−1

−1 10  20 −10 1 143.   =   2 20 1 20 10 2 − − ( )( ) ( )( )   −2 −1 1  20 −10   −40 −2 −1 − 1 3  =  12 14   20 40  =

−1

5  −6 −5   3 1 144.   =   3 3 3 6 − − 6 3 5 3 − − − ( )( ) ( )( )     =

5 1  3   −3  −3 −6 

 −1 − 53  =  2  1

− 1 145.  32  10

20  − 6

−1

 − 6 − 20  3  (20) − 10 − 12 

( )(− 6) − ( ) 3 10

− 12

1  − 6 − 20 = −  3  3 − 10 − 12 

−1

Inverse does not exist.

−1

0 2 8  8   − − 4 2 0 2  140.   1 2 1 4   4 1 1 −1

4 1 1 0  − 15 0 5 5   1 1 0 1 − − 0 0  2 2   −70 R3 → 0 0 1  −1 −14 −10 

− 45 R3 + R1 →  1 0 0  1 11 8   1 + → 0 1 0 − 1 − 7 − 5 = I 3 A−1  R R 2 2 3  0 0 1  −1 −14 −10 

777

− 3 146.  44 − 5

5 2  − 83 

2 =  1 10

20  3  1 6

1 8 − 3 − − 54

−1

(− 34 )( ) ( )( ) 1 − 3  4  54

− 52   − 43 

− 2 =  31  5

− 85   3 − 16 

− 8 − 5  2  43  3 5  5 − 4  2 

778

Chapter 8

Linear Systems and Matrices

−1

153.  x + 2 y = −1  3 x + 4 y = −5

 −1 4  7 4 147.   =   2 −7 2 1  x   7 4   8  36   =   =    y  2 1  −5 11 

−1

1  x   −2 1  −1  −3  −2 1 2    =  3 − 1    =  3 − 1  =   3 4 y      2  1 2 2   −5  2

Answer : ( 36, 11) −1

 141  2 3 148.   = 2  7  4 −1  x   141  = 2  y   7

x = − 3, y = 1

Answer: ( −3, 1)

  −  3 14 1 7

154.  x + 3 y = 23  −6 x + 2 y = −18

  −10   − 12     −   1  −3 3 14 1 7

−1

 1 3 0.1 −0.15 x 0.1 −0.15 23 5   =    =    =   −6 2 0.3 0.05 y 0.3 0.05−18 6

Answer: ( − 12 , − 3)

x = 5, y = 6

Answer: ( 5, 6 )

−1

 −1 −1 1 3 2 −1     149.  1 −1 2  =  3 83 − 37  7  2 5 1 1 − 35  3  1  6   2   x   −1 −1       8 7 3 y = − −1 =  −1 3 3    7  z   2 − 35   7   −2  3 

155. −3 x − 3 y − 4 z = 2  y + z = −1   4 x + 3 y + 4 z = −1  −1

−3 −3 −4 1 0 1 x  1 0 1  2  1            0 1 1 = 4 4 3 y =  4 4 3 −1 =  1  4 3 4 4 −3 −3 z −4 −3 −3 −1 −2

Answer: ( 2, − 1, − 2 )

x = 1, y = 1, z = − 2

−1

 −1 4 −2   −11 −6 −2      2 9 5 − = 150.    −3 −2 −1  −1 5 −4   −1 −1 −1

Answer: (1, 1, − 2 ) 156. 2 x + 3 y − 4 z = 1   x − y + 2z = −4 3x + 7 y − 10 z = 0 

 x   −11 −6 −2   12   −2          y  =  −3 −2 −1  −25 =  4   z   −1 −1 −1  10   3

2 3 −4    2  1 −1  3 7 −10 

Answer: ( −2, 4, 3)

1 2  2 1 151.   1 −1  0 0

1 −1  1 1 0 2  1 1

−1

does not exist.

Inconsistent; no solution. 157. − 23 = − 23 158. 0 = 0

A−1 does not exist. So, it is not possible to solve the system using the inverse.

1 1  1 −1 152.  0 1   1 0

−1

 23 1 1  2  2 1 =  31  0 1  23  0 1  − 3

 x   23    2 y =  3  z   13    2  w   − 3

− −

1 3 1 3 1 3 1 3

−1 0 0 1

− 13 −

1 3 1 3 1 3

  0 −  0 −   1 

−1

2 3 1 3 2 3 1 3

  1  1  −   −3  1 = −   2   −2        2   1 2 3 1 3 2 3 1 3

159.

160.

2 −4 −9

= 8 ( −4 ) − 2 ( 5) = −42

= ( −9 )( −4 ) − (11)( 7 )

7 −4

= 36 − 77 = − 41

161.

162.

50 −30 10

14 −24 12 −15

= 50 ( 5) − ( −30 )(10 ) = 550 = 14 ( −15) − ( −24 )(12 ) = 78

Chapter 8 Review  8 3 4   166. A =  6 5 −9   −4 1 2 

2 −1 163. A =   7 4 (a) M11 = 4 M 21 = −1 M12 = 7 M 22 = 2

(a) M11 =

(b) C11 = 4 C21 = 1 C12 = −7 C22 = 2

M13 =

3 6  164. A =   5 −4  (a) M11 = −4 M12 = 5 M 21 = 6 M 22 = 3

(b) C11 = −4 C12 = −5 C21 = −6 C22 = 3

M 21 = M 23 = M31 =

1 8 5

3 4 8 4 = 2, M 22 = = 32, −4 2 1 2

M 23 =

8 3 = 20 −4 1

M31 =

3 4 8 4 = −47, M32 = = −96, 5 −9 6 −9 8 3 6 5

= 22

−2 4 1 4 1 −2 4 −6 0 2 = 6 −2 3 4 5 3 5 3 4

167.

= 6 (13) − 2 ( −26 ) = 130

= 22

2 −1

= 26

C31 = −47, C32 = 96, C33 = 22

2 −1 3 −1 = 20, M 22 = = 19, 8 6 1 6 3 2

C21 = −2, C22 = 32, C23 = −20

= −21

−4 1

(b) C11 = 19, C12 = 24, C13 = 26

−2 0 5 0 (a) M11 = = 30, M12 = = −12, 8 6 1 6 −2 5

5 −9 6 −9 = 19, M12 = = −24, −4 2 1 2

M 21 =

M33 =

 3 2 −1   165. A =  −2 5 0   1 8 6 

M13 =

779

= 5, M32 =

3 −1 −2

= −2,

7 −1

2 −3

168.

−5

4 = 4(3 − 0) − 7( − 2 + 20) − 1(0 − 15)

0 −1 = 12 − 126 + 15

3 2 M33 = = 19 −2 5

= − 99

(b) C11 = 30, C12 = 12, C13 = −21

(Expansion along Row 1)

C21 = −20, C22 = 19, C23 = −22

−2

C31 = 5, C32 = 2, C33 = 19

5 0

2 −1 0 = (3) ( − 2)( −1) − ( 2)(5) = − 24

169.

−1

1 0

(Expansion along Column 3)

0 1 −2 1 −2 0 1 2 = ( −1) = ( −1) (1)( 2) − (1)( −2)  = −4 1 2 −1 −1 3

170.

(Expansion along Column 1) 1 4 −2 171.

3 1

0 = ( − 2)

−2 2

3 1 −2 2

+ (1)

1 4 3 1

= ( − 2)(8) + (1)( −11) = − 27

(Expansion along Column 3)

780

Chapter 8

Linear Systems and Matrices

172.

0 3 1 5 −2 1 = −3 ( 5 − 1) + 1( 30 + 2 ) = −12 + 32 = 20 1 6 1 (Expansion along Row 1) 3 0 −4 0 0 8

1 2

0 8 2

= 3 1 8 2 + ( −4 ) 6 1 2 (Expansion along Row 1) 6 1 8 2 3 −4 1 0 3 1 0 3 −4 1

173.

(

)

= 3 8 8 − ( −8 ) − 1(1 − 6 ) + 2 ( −4 − 24 )  − 4 0 − 6 ( 8 − 6 ) + 0    = 3 128 + 5 − 56  − 4  −12  = 279

−5 6 174.

0 0

1 −1 2

−1 2

= −5 4 −5 1 − 6 −3 −5 1 (Expansion along Row 1) −3 4 −5 1 6 0 3 1 0 3 1 6 0 3 = 5 6 ( −1 + 10 ) + 3 ( −5 + 4 )  − 6 ( −1 + 10 ) + 3 ( 0 − 3 )  = −5 54 − 3 − 6 9 − 9  = −255

175. (1, 0 ) , ( 5, 0 ) , ( 5, 8)

1 0 1 1 2

5 0 1=

( 32 ) = 16

1 2

179. ( −2, − 1) , ( 4, 9) , ( −2, − 9) , ( 4, 1)

The figure is rhombus. −2 −1 1 4

5 8 1 Area = 16 square units

176. ( −4, 0 ) , ( 4, 0 ) , ( 0, 6 )

−4 0 1

Area = 12 4 0 1 = 12 ( 48 ) = 24 0 6 1 Area = 24 square units

(

) ( , 1)

177. ( 12 , 1) , 2, − 25 , 1 2

1 1

3 2

1 1

3 2

2 − 25 1 = 12 ( 27 ) = 74

1 2

Area = 74 square units

178.

( , 1) , ( 4, − ) , ( 4, 2) 3 2

Area =

1 2

3 2

1 1

2 1

9 1 = −48

−2 −9 1 Area = 48 square units

180. ( −4, 8) , ( 4, 0 ) , ( −4, 0 ) , ( 4, − 8)

The figure is a rhombus. −4 8 1 4 0 1 = −64 −4 0 1 Area = 64 square units

181. ( 2, 4 ) , ( 5, 6 ) , ( 4, 1)

2 4 1 1 2

5 6 1 = 12 ( −13 ) 4 1 1

Area = 132 square units

4 − 12 1 = 12 ( 254 ) = 258 square units

Chapter 8 Review

0 −5 1 −5 1 −5 1 1 1 = ( −2 ) + ( 4) 7 1 1 1 7 1

182. 2 4

−7

9 −5 3 −7

Collinear

9 37 174 y= = = −2 3 8 −87 9 −5

−1 7 1 −9 1 3 1 3 −9 − ( 7) + (1) 3 −9 1 = ( −1) −3 1 −3 15 15 1 −3 15 1

Answer: ( 3, − 2 )

Collinear

−11 3 −5 −3 −1 1

184. ( −3, 2 ) , ( 2, − 3) , ( −4, − 4 )

−3

2 1

2 −3 1 = 12 ( −35 )

1 2

189. x =

15 −4 −2

Area = 352 square units

5 2 1 1 3 = =1 185. x = 1 2 3

1 5 −1 1 6 = =2 1 2 3 −1 1

−14 = −1 14

56 =4 14

−3 15

70 =5 14

Answer: ( −1, 4, 5) 15 − 2

−10 −1

− 39

−1 2 −21 = = −3 2 −1 7

− 3 −1 − 7 90 = 5 −2 1 0

190. x =

3 −9

−8

3 −9

2 −10

−1 − 7

 Not possible

3 −1 28 = =4 2 −1 7 3

Answer: ( −3, 4 )

3 −4 187. x =

14 3 −11

4 −1 −1 −4

1 6

−2 −11 −5 4 −3 1 −1 15 6

Answer: (1, 2 )

3 −5

−2

−1 1

186. x =

4 −1 −1 −4

−4 −4 1

37 −5 −261 = =3 188. x = 3 8 −87

= −2 ( −12 ) + 4 ( −6 ) = 0

183.

−15 6 − 42 = 8 −4 0 −12

781

191.

1 −3 2 2 2 −3 = 20 1 −7 8 A1 = 0 A2 = −48 A3 = −52

 Not possible

48 x = 0, y = − 20 = −2.4, z = − 52 = −2.6 20

Answer: ( 0, − 2.4, − 2.6 )

782

Chapter 8

Linear Systems and Matrices

10 −21 −7 4

192. x =

5 −21

14 −21 −7 −4

1 −3

2 −2 =

1176 3 = 1568 4

(b) D = 2 2

2 −2

56 −21

5 −3

1 −4

4 66 5 2

−4

4 −2

2 −1 −4

1400 25 = = y= 14 −21 −7 1568 28 −4 2 −2 56 −21

14 −21 10 −4 z=

−1 x=

14 10 −7

1 −4 = 66

56 −21

14 −21 −7 −4

−4088 73 =− 1568 28

2 −2

56 −21

73   3 25 Answer:  , , −  28   4 28

193. (a)  x − 3 y + 2 z = 5  2 x + y − 4 z = −1 2 x + 4 y + 2 z = 3 

66 1 −3

1 −1

106 53 = 66 33

−34 −17 = 66 33

61 = 66 66 17 61   53 Answer:  , − ,  33 33 66   z=

194. (a)  x + 2 y − z = −3  2 x − y + z = −1 4 x − 2 y − z = 5 

 x + 2 y − z = −3  − 5 y + 3z = 5   − 10 y + 3z = 17 

 x − 3y + 2z = 5  7 y − 8z = −11   10 y − 2 z = −7 

 x + 2y − z = −3   − 5y + 3z = 5  − 3z = 7 

 x − 3y + 2z = 5  7 y − 8z = −11   66 z = 617 7  61 z= 66 17  61  7 y − 8   = −11  y = − 33  66 

z=−

53  17   61  x − 3 −  + 2   = 5  x = 33  33   66 

7 3

12  7 −5 y + 3  −  = 5  y = − 3 5   8  12   7  x + 2  −  −  −  = −3  x = − 15  5   3 12 7  8 Answer:  − , − , −  5 3  15

17 61   53 Answer:  , − ,  33 66   33

Chapter 8 Review 1

 12  197. A =  − 14 − 1  2

2 −1

−1

(b) D = 2 −1 1 = 15 4 −2 −1 −3 x=

5 −2 −1 15 1 −3 −1 2 −1

 32   9  3   −1  −8 

2 −1

−1 −1

−8 15

5 −1

15 2 −3

−1 −1

4 −2

−36 −12 = 15 5

195. L O O K _ O U T _ B E L O W (a) 12 15 15 11 0 15 21 20 0  2 5 12  15 23 0

(b) 12 15 15 A = −21 6 0 11 0 15 A =  −68 8 45 21 20 0  A = 102 −42 −60 

−

  −    1 4 1 8 1 4

− 14 1 8 5 16

  −    1 4 1 8 3 16

30 −7 30  3 1     5 10 80  0 25 1 1 1  37 34 16  4 − 4 4   21 0    3   1 − 18  =  5 1 40 −7 38  8 8 5 3  −3 8 36 − 1  0 13    16 16 16   16 −1 58  0 14  23 46 0 23 0    Message: CAN YOU HEAR ME NOW

14 C  15 − 8 U  18 E 5 −  15 − 0 W

A N Y O − H A R M E N O −

1  23 13  3  1 1 1 199. A =  − 3 3 3  1 1 −1 6  6 3  21 −11 14  20 8 1 T H A     − − 29 11 18   20 0 9 T − I 32 −6 −26  19 0 13 S − M 1      23 13  3 25 0 6  Y − F  31 −19 −12  − 1 1  1 = 3 10 6 26   13 31 9 14 1 I N A    6 3 − 16    13 −11 −2  12 0 1 L − A 37 28 −8 14 19 23 N S W      5 13 36   5 18 0  E R − Message: THAT IS MY FINAL ANSWER

−1

2 5 12  A =  −53 20 21 15 23 0  A = 99 −30 −69 

Cryptogram: − 21 6 0 − 68 8 45 102 − 42 − 53 20 21 99 − 30 − 60

−

3 8 1 4

−46 37   9 0 23 I − W    1  −48 15  12 − 14 4  9 12 12  I L L   3 1 1 −14 10   − 4 − 8 − 8  = 0 2 5 − B E    1  −6 2   − 12 − 14 0 2 1 − B A 4   3 11 0  C K − −22 −3   Message: I WILL BE BACK

 14  198. A−1 =  83 − 1  16

−35 −7 = 15 15 3 12 7  8 Answer:  − , − , −  15 5 3  z=

− 14

783

− 60

196. J U S T _ D O _ I T

(a)

20 0 4 15 0 9 20 0 0 10 21 19

(b)

10 21 19 A = −49 −78 −23

20 0 4 A = 52 28 4 15 0 9 A = 57 33 9 20 0 0 A = 40 20 0 Cryptogram: −49 −78 −23 52 28 4 57 33 9 40 20 0

−

784

Chapter 8

Linear Systems and Matrices 201. False. The solution may have irrational numbers.

200. (a) 4b + 46a = 75.9  46b + 534a = 873.5

75.9

202. True. Expansion by Row 3 gives a11 a12 a13 a21 a22 a23 a31 + c1 a32 + c2 a33 + c3

873.5 534 b = = 17.48 4 46

= ( a31 + c1 )

46 534 4 a =

75.9

46 873.5 = 0.13 4 46

= a31

46 534 y = 0.13t + 17.48 (b)

a13 a a a a − ( a32 + c2 ) 11 13 + ( a33 + c3 ) 11 12 a23 a21 a23 a21 a22

a13 a a a a a − a32 11 13 + a33 11 12 + c1 12 a23 a21 a23 a22 a21 a22

− c2

a11 a13 a a + c3 11 12 a21 a22 a21 a22

a11

a12

a13

a11

a12

a13

= a21

a22

a23 + a21

a22

a23

a31

a32

a33

a13 a23

Note: Expand each of these matrices by Row 3 to see the previous step.

a12 a22

203. The row operations on matrices are equivalent to the operations used in the method of elimination when solving a system of equations.

204. A square n × n matrix A has an inverse

A−1 if det ( A) ≠ 0.

2.52 = 0.13t t ≈ 19.38  2019

Chapter 8 Test 1.

x − y = 6  y = x − 6. Then

3x + 5( x − 6) = 2 

y = x − 1 = ( x − 1)  x = 1 or 3

1 = ( x − 1) = x 2 − 2 x + 1  x 2 − 2 x = 0. 2

8 x = 32  x = 4, y = 4 − 6 = −2.

Thus, x = 1 or x ( x − 2 ) = 0  x = 0, 1, 2.

Answer: ( 4, − 2 )

Answer: ( 0, − 1) , (1, 0 ) , ( 2, 1)

3. x − y = − 3  y = x + 3  4 x − ( x + 3)

= 7

4 x − ( x 2 + 6 x + 9) = 7 x 2 + 2 x + 16 = 0 x = x =

−2 ±

(2)2 − 4(1)(16) 2(1)

−2 ±

− 62

2 No real solution

Chapter 8 Test 4. 3x + 5 y = −11  y = 19 5 x −

Multiply equation 2 by 5.

28 x = 84 x = 3

Answer: (3, − 4) 5. 1.5 x + 3y = −9  5  2 x − 1.2 y =

Multiply Equation 1 by 2 and Equation 2 by 5.  3 x + 6 y = −18  25 10 x − 6 y = 7 13x = 7  x = 13

( )

7 3 13 + 6 y = −18  6 y = 255  y = − 85 13 26

(

7 Answer: 13 , − 85 26

)

x 3 + x 2 + x + 2 A B Cx + D = + 2 + 2 x4 + x2 x x x +1

(

) (

)

x 3 + x 2 + x + 2 = Ax x 2 + 1 + B x 2 + 1 + ( Cx + D ) x 2 = ( A + C ) x + ( B + D ) x 2 + Ax + B 3

=1 A + C  + D =1  B  A =1   B =2 A = 1, C = 0, B = 2, D = −1 x3 + x2 + x + 2 1 2 1 = + 2 − 2 x4 + x2 x x x +1

Answer: (1.5 − 2a, 1 + 2a, a ) , where a is any real number 3 1  10  2   11.  2 −3 −3  22  row reduces to  4 −2 3  −2 

z = −2 y = −2 − ( −2 ) = 0

x = 3 + 4 ( 0 ) + ( −2 ) = 1

Answer: (1, 0, − 2 ) 2

z = a , y = 2 a + 1, x = 1.5 − 2 a.

x − 4y − z = 3  y + z = −2   − 4z = 8 

6 = a (0) + b (0)

A B + x − 1 ( x − 1)2

 1 0 2  1.5 2 1 2  4      1 . 10. 2 2 0  5 row reduces to 0 1 −2  0 0 0  2 −1 6  2  0  Infinite number of solutions. Let

 x − 4y − z = 3   2 x − 5y + z = 0 3 x − 3 y + 2 z = −1  x − 4y − z = 3  3y + 3 z = − 6   9y + 5z = −10 

( x − 1) ( 5 x − 2 ) = A ( x − 1) + B = Ax + ( − A + B ) 2

= 5  A  − A + B = −2  B = 3 5x − 2 5 3 = + 2 2 x − 1 ( x − 1) ( x − 1)

 3 x + 5 y = −11  25 x − 5 y = 95

3(3) + 5 y = −11  y = − 4

5x − 2

785

 1 0 0  5   0 1 0  2  . 0 0 1  −6  Answer: ( 5, 2, − 6 )

+cc=6

2 = a ( −2 ) + b ( −2 ) + c 2

9 2

= a ( 3)

+ b ( 3) + c

4a − 2b + 6 = 2 or 2a − b = −2 Hence,  9 3 9a + 3b + 6 = 2 or 9a + 3b = − 2 . Solving this system for a and b, you obtain a = − 12 , b = 1. Thus, y = − 12 x 2 + x + 6.

786

Chapter 8

Linear Systems and Matrices 3 −2 −1 4 10 = =1 17. x = 2 −2 10 1 4

12. (a)

 1 0 4   A − B =  −7 −6 −1  0 4 0 

(b)

 15 12 12    3 A =  −12 −12 0   3 6 0 

(c)

4 12   7   3 A − 2 B =  −18 −16 −2   1 10 0 

(d)

 36 20 4    AB =  −28 −24 −4   10 8 2 

− 35 − 83 2 3

Upper right: x1 + x3 = x4 + 600

1  1 . 0 

 − 43 − 35  A =  − 43 − 83 2  1 3  3  7   −2      A −1  −5 =  3  −1  −1 −1

1  1 0 

Answer: ( −2, 3, − 1) 14.

−25 18 = ( −25)( −7 ) − 6 (18 ) = 67 6 −7 4

1  Answer:  1, −  2 

18. Upper left: 400 + x2 = x1

 −2 2 3  1 0 0    13.  1 −1 0  0 1 0  reduces to  0 1 4  0 0 1  1 0 0  − 43  4 0 1 0  − 3 1 0 0 1  3 

2 3 1 −1 −5 1 y= = =− 2 −2 10 2 1 4

Lower left: 300 = x2 + x3 + x5 Lower right: x5 + x4 = 100 = 400  x1 − x2  + x3 − x 4 = 600  x1  x + x x = 300 + 2 3 5   x4 + x5 = 100 

Solving the system:  1 −1 0 0 0    1 0 1 −1 0  0 1 1 0 1   0 0 0 1 1 

400   1   600  0 → 300  0   100  0

0 1 0 0

1 1 0 0

0 0 1 0

1 1 1 0

 700    300   100   0  

Letting x3 = a and x5 = b be real numbers, we have: x5 = b x4 = 100 − b x3 = a x2 = 300 − a − b x1 = 700 − b − a

0 3

15. det ( A ) = 1 −8 2 3 2 2

= 4 ( −16 − 4 ) − 0 + 3 ( 2 + 24 ) = −80 + 78 = −2 16.

x1 x2 x3

y1 1 −1 1 1 y2 1 = 4 11 1 y3 1 −1 −5 1 = −1(16 ) − 1( 4 + 1) + 1( −20 + 11) = −16 − 5 − 9 = −30

Area = −30 = 30 square units

C H A P T E R 9 Sequences, Series, and Probability Section 9.1

Sequences and Series..........................................................................788

Section 9.2

Arithmetic Sequences and Partial Sums ............................................804

Section 9.3

Geometric Sequences and Series .......................................................811

Section 9.4

The Binomial Theorem ......................................................................822

Section 9.5

Counting Principles ............................................................................832

Section 9.6

Probability ...........................................................................................836

Chapter 9 Review .......................................................................................................841 Chapter 9 Test ...........................................................................................................847

C H A P T E R 9 Sequences, Series, and Probability Section 9.1 Sequences and Series 1.

terms

recursively

index, upper limit, lower limit

series

(a) A finite sequence is a function whose domain is the set of the first n positive integers. (b) An infinite sequence is a function whose domain is the set of positive integers.

1 10. an =   2

1 1 a1 =   = 2 2 2

1 1 a2 =   = 4 2 3

1 1 a3 =   = 2 8 4

Since n! = 1 ⋅ 2 ⋅ 3( n − 1)( n ) , then 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 = 6!.

1 1 a4 =   = 2 16  

an = 2 n − 5

1 1 a5 =   = 32 2

a1 = 2 (1) − 5 = − 3

a2 = 2 ( 2 ) − 5 = − 1 a3 = 2 ( 3) − 5 = 1

a4 = 2 ( 4 ) − 5 = 3

 1 11. an =  −   2

a5 = 2 ( 5) − 5 = 5

1  1 a1 =  −  = − 2 2  

an = 4 n − 7

1  1 a2 =  −  = 4  2

a1 = 4 (1) − 7 = −3

a2 = 4 ( 2 ) − 7 = 1

1  1 a3 =  −  = − 8  2

a4 = 4 ( 4 ) − 7 = 9

1  1 a4 =  −  = 2 16  

a3 = 4 ( 3) − 7 = 5

a5 = 4 ( 5) − 7 = 13 9.

an = 3n

a1 = 31 = 3 a2 = 32 = 9 a3 = 33 = 27

1  1 a5 =  −  = − 32  2 12. an = ( −2 )

a1 = ( −2 ) = −2 1

a4 = 34 = 81

a2 = ( −2 ) = 4

a5 = 35 = 243

a3 = ( −2 ) = −8

a4 = ( −2 ) = 16 4

a5 = ( −2 ) = −32 5

788

Section 9.1 n +1 n 1+1 a1 = =2 1 2 +1 3 a2 = = 2 2 3 +1 4 a3 = = 3 3 4 +1 5 a4 = = 4 4 5 +1 6 = a5 = 5 5

13. an =

n n +1 1 1 a1 = = 1+1 2 2 2 a2 = = 2 +1 3 3 3 = a3 = 3 +1 4 4 4 a4 = = 4 +1 5 5 5 = a5 = 5 +1 6

17. an =

(a)

a1 = a2 = a3 =

2n n +1

2 (1)

1+1 2 (2) 2 +1 2 ( 3) 3 +1 2(4)

n a1 = 0

a3 = 0 a4 = 0.5 a5 = 0 (b)

18. an =

(a)

a1 =

1 + ( −1)

=0 1 1+1 2 = =1 a2 = 2 2 1 + ( −1) =0 a3 = 3 1+1 2 1 = = a4 = 4 4 2 1 + ( −1) =0 a5 = 5 1 + ( −1)

2n a1 = 0

a2 = 0.5 a3 = 0 a4 = 0.25

15. an =

16. an =

789

a2 = 1

14. an =

n n2 + 1 1 1 = a1 = 2 1 +1 2 2 2 a2 = 2 = 2 +1 5 3 3 a3 = 2 = 3 + 1 10 4 4 = a4 = 2 4 + 1 17 5 5 = a5 = 2 5 + 1 26

1 + ( −1)

Sequences and Series

a5 = 0 (b)

1−1 =0 2 1+1 1 a2 = = 2 (2) 2 a1 =

a3 =

1−1 =0 2 ( 3)

a4 =

1+1 1 = 2 (4) 4

a5 =

1−1 =0 2 ( 5)

1 2n a1 = 0.5

19. an = 1 −

(a)

4 3

a2 = 0.75

6 3 = 4 2

a4 = 0.9375

a3 = 0.875 a5 ≈ 0.9688

8 a4 = = 4 +1 5 2 ( 5 ) 10 5 a5 = = = 5 +1 6 3

790

Chapter 9

(b)

Sequences, Series, and Probability

1 1 = 21 2 1 1 3 a2 = 1 − 2 = 1 − = 2 4 4 1 7 a3 = 1 − 3 = 2 8 1 15 a4 = 1 − 4 = 2 16 1 31 a5 = 1 − 5 = 2 32

a1 = 1 −

22. an =

(a)

a3 ≈ 0.5774 a4 = 0.5 a5 ≈ 0.4472 (b)

a3 ≈ 0.1925 a4 = 0.125 a5 ≈ 0.0894 (b)

1 =1 1 1 a2 = 3 2 2 1 a3 = 3 2 3 1 1 a4 = 3 2 = 4 8 1 a5 = 3 2 5 a1 =

3 1 4 1

a5 =

a5 ≈ 0.2373

a2 ≈ 0.3536

2 1

a4 =

a4 ≈ 0.3164

32 9 a2 = 2 = 4 16 33 27 a3 = 3 = 4 64 34 81 a4 = 4 = 4 256 35 243 a5 = 5 = 4 1024 1 21. an = 3 2 n (a) a1 = 1

a3 =

a3 ≈ 0.4219

3 3 = 1 4 4

a1 = 1 a2 =

a2 ≈ 0.5625

a1 =

n a1 = 1 a2 ≈ 0.7071

3n 20. an = n 4 (a) a1 = 0.75

(b)

23. an =

(a)

( −1)

1 2

5 n

n2 a1 = −1

a2 = 0.25 a3 ≈ −0.1111 a4 = 0.0625 a5 = −0.04 (b)

−1 = −1 1 1 a2 = 4 1 a3 = − 9 1 a4 = 16 1 a5 = − 25 a1 =

n n  24. an = ( −1)    n +1 (a) a1 = −0.5

a2 ≈ 0.6667 a3 = −0.75 a4 = 0.8 a5 ≈ −0.8333

Section 9.1 1

(a)

a1 = 0 a2 = 0 a3 = 6 a4 = 24 a5 = 60

(b)

a1 = 1(1 − 1)(1 − 2 ) = 0 a2 = 2 ( 2 − 1)( 2 − 2 ) = 0 a3 = 3 ( 3 − 1)( 3 − 2 ) = 6

25. an = ( 2n − 1)( 2n + 1)

a4 = 4 ( 4 − 1)( 4 − 2 ) = 24

a1 = 3

(a)

791

26. an = n ( n − 1)( n − 2 )

1 1 =− 1+1 2 2 2 2 = a2 = ( −1) 1+ 2 3 3 3 3 =− a3 = ( −1) 3 +1 4 4 4 4 a4 = ( −1) = 4 +1 5 5 5 5 a5 = ( −1) =− 5 +1 6

a1 = ( −1)

(b)

Sequences and Series

a5 = 5 ( 5 − 1)( 5 − 2 ) = 60

a2 = 15 a3 = 35 a4 = 63 a5 = 99

a1 = (1)( 3 ) = 3

(b)

a2 = ( 3 )( 5 ) = 15

a3 = ( 5 )( 7 ) = 35

a4 = ( 7 )( 9 ) = 63

a5 = ( 9 )(11) = 99

27. an = 2 ( 3n − 1) + 5

28. an = 2n( n + 1) − 7

−3

105

137

173

213

n +1 n

29. an = 1 +

2.5

2.33 2.25

2.2

2.17 2.14 2.13 2.11

2.1

30. an =

4n2 (n + 2)

1.33

7.2

10.67

14.29

21.78

25.6

29.45

33.33

792

Chapter 9

Sequences, Series, and Probability

31. an = ( −1) + 1 n

32. an = ( −1)

n +1

33. a10 =

102 100 = 102 + 1 101

34. a5 =

52 25 = 2 ( 5 ) + 1 11

46.

47.

35. a25 = ( −1) 3 ( 25) − 2  = −73 25

38. a6 =

36 729 = 6 3 +1 730

39. 3, 8, 13, 18, 23,  an = 5n − 2

40. 3, 7, 11, 15, 19, 

an = 4 n − 1 41. 7, 13, 19, 25, 31,  an = 6 n + 1

42. 9, 11, 13, 15, 17 an = 2 n + 7

43. 0, 3, 8, 15, 24,  an = n 2 − 1

44. 4, 7, 12, 19, 28,  an = n 2 + 3

45.

2 3 4 5 6 , , , , , 3 4 5 6 7 n +1 an = n+2

1 −1 1 −1 , , , , 2 4 8 16 an =

36. a16 = ( −1) 16 (15)  = −240

26 64 = 37. a6 = 6 2 + 1 65

2 3 4 5 6 , , , , , 1 3 5 7 9 n +1 an = 2n − 1

48.

( − 1)

n +1

1 −2 4 −8 , , , , 3 9 27 81 an =

( −1)

n +1

2 n −1

( −2 ) 3

n −1

1 1 1 1 1 49. 1 + , 1 + , 1 + , 1 + , 1 + ,  1 2 3 4 5 1 an = 1 + n

1 1 1 1 1 ,1 + ,1 + ,1 + ,1 + , 3 6 11 18 27 1 an = 1 + 2 n + 2

50. 1 +

51. 1, 3, 1, 3, 1, 3, 

an = 2 + ( −1)

52. 1, − 1, 1, − 1, 1, − 1, 

an = ( −1)

n +1

53. a1 = 28, ak = ak −1 − 4 a1 = 28 a2 = a1 − 4 = 28 − 4 = 24 a3 = a2 − 4 = 24 − 4 = 20 a4 = a3 − 4 = 20 − 4 = 16 a5 = a4 − 4 = 16 − 4 = 12

Section 9.1 54. a1 = 15, ak = ak −1 + 3 a1 = 15

Sequences and Series

793

60. a1 = 25, ak +1 = ak − 5 a1 = 25

a2 = a1 + 3 = 15 + 3 = 18

a2 = a1 − 5 = 25 − 5 = 20

a3 = a2 + 3 = 18 + 3 = 21

a3 = a2 − 5 = 20 − 5 = 15

a4 = a3 + 3 = 21 + 3 = 24

a4 = a3 − 5 = 15 − 5 = 10

a5 = a4 + 3 = 24 + 3 = 27

a5 = a4 − 5 = 10 − 5 = 5

55. a1 = 3, ak +1 = 2 ( ak − 1)

a1 = 3 a2 = 2 ( a1 − 1) = 2 ( 3 − 1) = 4

a3 = 2 ( a2 − 1) = 2 ( 4 − 1) = 6

a4 = 2 ( a3 − 1) = 2 ( 6 − 1) = 10

a5 = 2 ( a4 − 1) = 2 (10 − 1) = 18 56. a1 = 32, ak +1 = 12 ak

a1 = 32 a2 = 12 a1 = 12 ( 32 ) = 16 a3 = 12 a2 = 12 (16 ) = 8

(8 ) = 4 a5 = a4 = ( 4 ) = 2 1 2

a4 = a3 =

1 2

57. a0 = 1, a1 = 3, ak = ak − 2 + ak −1 a0 = 1 a1 = 3 a2 = a0 + a1 = 1 + 3 = 4 a3 = a1 + a2 = 3 + 4 = 7 a4 = a2 + a3 = 4 + 7 = 11

58. ao = −1, a1 = 5, ak = ak − 2 + ak −1

In general, an = 30 − 5n. 61. a1 = 81, ak +1 =

a1 = 81 1 1 a1 = ( 81) = 27 3 3 1 1 a3 = a2 = ( 27 ) = 9 3 3 1 1 a4 = a3 = ( 9 ) = 3 3 3 1 1 a5 = a4 = ( 3 ) = 1 3 3 a2 =

1 In general, an = 81  3

a2 = a0 + a1 = −1 + 5 = 4 a3 = a1 + a2 = 5 + 4 = 9

a2 = ( −2 ) a1 = ( −2 )(14 ) = −28

a3 = ( −2 ) a2 = ( −2 )( −28 ) = 56

a4 = ( −2 ) a3 = ( −2 )( 56 ) = −112

a5 = ( −2 ) (a4 ) = ( −2 )( −112 ) = 224 In general, an = 14 ( −2 ) 63. an =

(a)

a5 = a4 + 2 = 12 + 2 = 14

In general, an = 2n + 4.

1 n! a0 = 1

a2 = 0.5 a3 ≈ 0.1667 a4 ≈ 0.0417

a1 = 6

a4 = a3 + 2 = 10 + 2 = 12

n −1

a1 = 1

59. a1 = 6, ak +1 = ak + 2

a3 = a2 + 2 = 8 + 2 = 10

243 1 = 81( 3 )   = n . 3 3

a1 = 14

a4 = a2 + a3 = 4 + 9 = 13

a2 = a1 + 2 = 6 + 2 = 8

n −1

62. a1 = 14, ak +1 = ( −2 ) ak

a0 = −1 a1 = 5

1 ak 3

(b)

1 =1 0! 1 a1 = = 1 1! 1 1 a2 = = 2! 2 1 1 a3 = = 3! 6 1 1 a4 = = 4! 24

a0 =

794

Chapter 9

64. an =

1 n + ( 1)!

(a)

a0 = 1

Sequences, Series, and Probability

a1 = 0.5

66. an =

(a) a0 = 0 a1 = 0.167

a2 ≈ 0.1667

a2 ≈ 0.333

a3 ≈ 0.0417

a3 = 0.225

a4 ≈ 0.0083 (b)

1 a0 = = 1 1! 1 1 a1 = = 2! 2 1 1 a2 = = 3! 6 1 1 a3 = = 4! 24 1 1 a4 = = 5! 120

a4 ≈ 0.089

(b) a0 =

03 0 = = 0 (0 + 2)! 2!

a1 =

13 1 1 = = 1 + 2 ! 3! 6 ( )

a2 =

23 23 8 1 = = = (2 + 2)! 4! 24 3

a3 =

33 33 27 9 = = = (3 + 2)! 5! 120 40

a4 =

43 43 64 4 = = = + 4 2 ! 6! 720 45 ( )

n2 65. an = ( n + 1)!

(a)

a0 = 0 a1 = 0.5 a2 ≈ 0.6667 a3 = 0.375

67. an =

( −1) ( 2n )!

(a)

a0 = 1

a1 = 0.5

a4 ≈ 0.1333 (b)

a2 ≈ 0.0417

a0 = 0 1 a1 = 2 22 2 a2 = = 3! 3 32 9 3 = = a3 = 4! 24 8 16 16 2 a4 = = = 5! 120 15

n3 ( n + 2) !

a3 ≈ 0.0014 a4 ≈ 0.0000248

( −1) = 1 a = 0

(b)

( −1) = 1 2

a1 =

( −1) = 1 a = 4

a3 =

( −1)

( −1) = 8

a4 =

1 720 1 40,320

Section 9.1

68. an =

( −1) ( 2n + 1)!

(a)

a0 = −1

2 n +1

a1 ≈ −0.1667

69.

2! 2! 1 = = 4! 4 ⋅ 3 ⋅ 2! 12

70.

5! 5! 1 = = 7! 7 ( 6 )( 5!) 42

a2 ≈ −0.0083 a3 ≈ −0.000198

71.

a4 ≈ −0.0000028

a0 =

(b)

−11 = −1 1!

( −1) = − 1 3

a1 =

72.

3! 6 −1 1 =− a2 = 5! 120 −1 1 =− a3 = 7! 5040 −1 1 =− a4 = 9! 362,880

73.

(n + 3)! = ( n + 3)(n + 2)(n + 1)n! = n + 3 n + 2 n + 1 ( )( )( )

74.

( n + 2 )! = ( n + 2 )( n + 1) n! = n + 2 n + 1 ( )( ) n! n!

75.

( 2n − 1)!

795

12! 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8! = 4!8! 4!8! 12 ⋅ 11 ⋅ 10 ⋅ 9 = = 495 4⋅3⋅2

10! 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 = 5! ⋅ 3! 3! 10 ⋅ 9 ⋅ 8 ⋅ 7 3⋅ 2 ⋅1 = 5040 =

( 2n − 1)!

( 2n + 1)! ( 2n + 1)( 2n )( 2n − 1)! 1 = 2 n ( 2 n + 1)

76.

Sequences and Series

( 2 n − 2) ! = ( 2 n − 2) ! 1 = ( 2 n )! (2n)(2n − 1)(2n − 2)! 2n(2n − 1) 8 n +1 an → 0 as n → ∞

77. an =

8 11 Matches graph (c).

79. an =

an → 0 as n → ∞ a1 = 4, a5 ≈ 0.013

Matches graph (d).

4n n! an → 0 as n → ∞

80. an =

4 4 256 2 = = 10 4! 24 3 Matches graph (a). a1 = 4, a4 =

81. an =

a1 = 4, a10 =

8n n +1 an → 8 as n → ∞

78. an =

2n + 2 2n!

2 n 3

8( 4)

32 = 5 5 Matches graph (b). a1 = 4, a4 =

Chapter 9

796

82. an =

Sequences, Series, and Probability

1 n +3 2

85. an = 3

2n n +1

83. an = 16( − 0.5)

n −1

86. an =

3n 2 n2 + 1

− 10

84. an = 8( − 0.75)

n −1

87.

 ( 2i +1) = ( 2 +1) + ( 4 +1) + ( 6 +1) + (8 +1) + (10 +1) = 35 i =1

88.

 ( 3i − 1) = ( 3 ⋅ 1 − 1) + ( 3 ⋅ 2 − 1) + ( 3 ⋅ 3 − 1) i =1

+ ( 3 ⋅ 4 − 1) + ( 3 ⋅ 5 − 1) + ( 3 ⋅ 6 − 1) = 57

−8

i =0

89. 4i 2 = 4i 2

= 4(02 + 12 + 22 + 32 + 42 + 52 + 62 ) = 364

k =0

i =0

90.  3i 2 = 3 i 2

(

)

= 3 02 + 12 + 22 + 32 + 42 + 52 = 165 91.

124

 j 2 − 3 = 9 − 3 + 16 − 3 + 25 − 3 = 6 + 13 + 22 = 429 j =3

92.

 j 2 + 1 = 3 + 1 + 4 + 1 + 5 + 1 = 4 + 5 + 6 = 60 j =3

93.

10 = 10 + 10 + 10 + 10 = 40 k =1

94.

 4 = 4 + 4 + 4 + 4 + 4 = 20 k =1

3 2 3 2 3 2 3 2 3 2 95.  (i − 1) + (i + 1)  = (1) + (3)  + ( 2) + ( 4)  + (3) + (5)  + ( 4) + (6)            i=2

= (1 + 9) + (8 + 16) + ( 27 + 25) + (64 + 36) = 186

Section 9.1 7

96.

Sequences and Series

797

 (k + 1) + (k − 3)  = (3) + (−1)  + (4) + (0)  + (5) + (1)  + (6) + (2)  + (7) + (3)  + (8) + (4)  2

k =2

= (3 + 1) + ( 4) + (5 + 1) + (6 + 4) + (7 + 9) + (8 + 16) = 64 4

  1 2    2 2    6 2  106. 1 −    + 1 −    +  + 1 −      6     6     6  

97.  2i = 20 + 21 + 2 2 + 23 + 2 4 = 31 i =0 4

2 6  k  =  1 −    ≈ 3.472 k =1    6  

98.  ( −2 ) = ( −2 ) + ( −2 ) + ( −2 ) + ( −2 ) + ( −2 ) = 11 0

j =0 6

99.  ( 24 − 3 j ) = 81

107. − 3 + 9 − 27 + 81 − 243 + 729 =  ( −1) 3i = 546

j =1

100.

i =1

 3 j + 1 ≈ 5.03

108. 1 −

j =1

1 1 1 1 1 1 1 1 1 + − + ... − = − + − + − 7 2 4 8 128 2 0 21 2 2 23 2 n

7  1 =   −  ≈ 0.664 2 n=0 

(−1)k

19 = = 0.63 101.  30 k = 0 ( k + 1)! 109.

20 ( −1) 1 1 1 1 1 − 2 + 2 − 2 + − 2 =  2 2 1 2 3 4 20 i i =1

9 1 1 1 1 1 103. + + + + =  ≈ 0.94299 3 (1) 3 ( 2 ) 3 ( 3 ) 3 ( 9 ) i =1 3!

110.

10 ( − 1) ≈ 0.246 1 1 1 1 − + − − = 1⋅ 3 2 ⋅ 4 3 ⋅ 5 10 ⋅12 k =1 k ( k + 2 )

15 5 5 5 5 5 + + + + = ≈ 11.904 104. 1+1 1+ 2 1+ 3 1 + 15 i =1 1 + i

111.

5 1 3 7 15 31 2 i − 1 129 + + + + =  i +1 = = 2.015625 4 8 16 32 64 j =1 2 64

 1    2   3   8  105. 2   + 3 + 2   + 3 + 2   + 3 +  + 2   + 3  8   8   8   8 

112.

6 1 2 6 24 120 720 k! + + + + + =  k = 18.25 2 4 8 16 32 64 k =1 2

( −1) = 3 = 0.375 k

102.  k =0

i +1

 0.82128 k +1

8  i  =  2   + 3 = 33 i =1   8  

i 4  1 1  1 2  1 3  1 4  1092 1 113.  7  = 7   +   +   +    = 625 5 5  5    5  i =1  5  5  1  242 1 1 1 1 1  114.  2  = 2   +   +   +   +    = 3 3 3 3 3 3 243          i =1     i

3 3  1 115.  4  −  = −1.5 = − 2 2 n =1   n

4 −51  1 116.  8  −  = ≈ −1.59375 4 32 n =1 

798

Chapter 9

117. (a)

Sequences, Series, and Probability

 10i = 101 + 102 + 103 + 104

120. (a)

i =1

1

2 10  = 2( 0.1) + 2( 0.01) + 2( 0.001) + 2( 0.0001) i=1

6 6 6 6 = + + + 10 100 1000 10,000

= 2 ( 0.1111) = 0.2222

= 0.6666

3333 = 5000 ∞

(b) 

i i =1 10

∞

= 6

∞  1  (b)  2   = 2 ( 0.1) + 2 ( 0.01) + 2 ( 0.001) +  i =1  10  = 0.2222 2 = 9

i i =1 10

= 6(0.1 + 0.01 + 0.001 + ) = 6(0.111) = 0.666

118. (a)

0.03   121. An = 5000  1 +  , n = 1, 2, 3,  4  

2 3

(a)

 10 = 10 + 10 + 10 + 10 k =1

= 0.4444 1111 = 2500 ∞

(b)  k =1

4 = 4 0.1 + 0.01 + 0.001 +  10k

(b)

= 4 0.111

119. (a)

4 9

1

0.035   (a) A1 = 10,0001 +  = 10,029.17 12  

 = 0.7777 7777 = 10,000

 1  7   = 7 0.1 + 0.01 + 0.001 +   k =1  10  = 0.777 7 = 9 ∞

A40 ≈ $6741.74 n



(b) A2 ≈ 10,058.42

A3 ≈ 10,087.76



A4 ≈ 10,117.18

A5 ≈ 10,146.69

A6 ≈ 10,176.28

A7 ≈ 10,205.96

 7  10  = 7 10 + 100 + 1000 + 10,000  k =1

(b)

 1 

0.03   A1 = 5000  1 +  = $5037.50 4   A2 ≈ $5075.28 A3 ≈ $5113.35 A4 ≈ $5151.70 A5 ≈ $5190.33 A6 ≈ $5229.26 A7 ≈ $5268.48 A8 ≈ $5307.99

0.035   122. An = 10,0001 +  , n = 1, 2, 3,  12  

= 0.444 =

1111 5000

A8 ≈ 10,235.73 A60 ≈ 11,909.43

Section 9.1

Sequences and Series

799

123. (a) p0 = 5500 pn = 0.75 pn −1 + 500

(b) In 2016, there will be p1 = 0.75 p0 + 500 = 4625 fish. In 2017, there will be p2 = 0.75 p1 + 500 ≈ 3969 fish. In 2018, there will be p3 = 0.75 p2 + 500 = 3477 fish. In 2019, there will be p4 = 0.75 p3 + 500 ≈ 3107 fish. (Answers will vary slightly.) ∞

 0.75 pn −1 + 500 = 2000 trout. As time passes, the population of trout decreases at a decreasing rate. Because the

(c)

p=0

population is growing smaller and still declines 25%, each time 25% is taken from a smaller number there is a smaller decline in the number of trout. 124. (a) and (c) 5000

(b) Linear sequence: Rn = 625.96n − 3841.5, r 2 ≈ 0.9779 Quadratic sequence: Rn = 34.7875n 2 − 104.58n − 107.675, r 2 ≈ 0.9843

(c) The quadratic sequence is the better fit, because its coefficient of determination is closer to 1. (d) 2017: Let n = 17  R17 = 34.7875(17) − 104.58(17) − 107.675 ≈ $8168.1 million 2

(e) 34.7875n 2 − 104.58n − 107.675 = 15,000

34.7875n 2 − 104.58n − 15,107.675 = 0 Using the Quadratic Formula: n ≈ 22.4, or 2022 13

 34.7875n 2 − 104.58n − 107.675 ≈ $16,386.1 million, which is equal to the sum of the revenues in the table.

(f)

n =8

125. True by the Properties of Sums. 126. True. The sums are equal because 4

2 = 2 + 2 + 2 + 2 = 2 j =1

j −2

j =3

800

Chapter 9

Sequences, Series, and Probability

127. a0 = 1, a1 = 1, ak + 2 = ak +1 + ak ; bn =

an +1 , n>0 an

a0 = 1

n +1

2n+1 5

2n 5

) − 2 (1 − 5 ) + 4 (1 + 5 ) − 4 (1 − 5 ) n +1

n +1

2n +2 5

a4 = 3 + 2 = 5

2n +2 5

a5 = 5 + 3 = 8

(1 + 5 ) (1 + 5 ) − (1 − 5 ) (1 − 5 ) = 2

a6 = 8 + 5 = 13 a7 = 13 + 8 = 21

2n +2 5

(1 + 5 ) =

a8 = 21 + 13 = 34 a9 = 34 + 21 = 55 a10 = 55 + 34 = 89

n+2

(

− 1− 5

2n + 2 5

)

n+2

= an + 2

a11 = 89 + 55 = 144

Yes, this is the recursive formula for the Fibonacci sequence.

b0 = 11 = 1 b1 = 21 = 2

xn n! x a1 = = x 1 x2 x2 = a2 = 2! 2 x3 x3 = a3 = 3! 6 x4 x4 = a4 = 4! 24 x5 x5 = a5 = 5! 120

133. an =

b2 = 23 b3 = 35 b4 = 85 b5 = 138 21 b6 = 13

b7 = 34 21 55 b8 = 34

b9 = 89 55 a n + 1 a n + a n −1 = an an a n −1 1 1 =1+ =1+ an an bn −1 a n −1

(1 + 5 ) − (1 − 5 ) 129. a = n

2n 5

(1 + 5 ) − (1 − 5 ) = 1 a = 1

21 5 a2 = 1, a3 = 2 a4 = 3, a5 = 5

x2 n2 x2 a1 = 1 x2 a2 = 4 x2 a3 = 9 x2 a4 = 16 x2 a5 = 25

134. an =

130. These are the first five terms of the Fibonacci sequence.

(1 + 5 ) − (1 − 5 ) 131. a = n +1

n +1

2 n +1 5

(1 + 5 ) − (1 − 5 ) = n+2

an + 2

(1 + 5 ) 2 (1 + 5 ) + 4 − (1 − 5 ) 2 (1 − 5 ) + 4 =

a3 = 2 + 1 = 3

n +1

( =

a2 = 1 + 1 = 2

=1+

n +1

2 1+ 5

a1 = 1

128. bn =

(1 + 5 ) − (1 − 5 ) + (1 + 5 ) − (1 − 5 ) 132. a + a =

n+2

2n + 2 5

Section 9.1

( −1) x 2 n +1 n

135. an =

a1 =

2n + 1 ( −1) x 2 +1

2 +1

− x3 3

( − 1) x n + 1 136. a =

− x3 x3 =− 3! 6 x5 x5 = a2 = 5! 120 x7 x7 a3 = − = − 7! 5040 x9 x9 a4 = = 9! 362,880 a5 =

a1 =

n +1

1 +1

1+1

−x 2

( −1) x 2 +1 = x 3 a = 2

2 +1

( −1) x 3 +1

a3 =

3 +1

( −1) x 4 +1

−x4 4

x5 5

a4 =

4 +1

( −1) x 5+1 = − x 6 5 +1

( −1) x 2 n 137. an = ( 2 n )! n

x2 a1 = − 2 x4 x4 a2 = = 4! 24 x6 − x6 a3 = =− 6! 720 x8 x8 a4 = = 8! 40,320 a5 =

( −1) x n n!

( −1) x 1!

= −x

( −1) x 2 = x 2 2

a2 =

2! 2 − x3 x3 =− a3 = 3! 6 4 4 x x = a4 = 4! 24 − x5 x5 =− a5 = 5! 120

a5 =

− x11 x11 =− 11! 39,916,800 n

139. an =

( −1) : x

( −1) x 2 n +1 ( 2n + 1)!

a1 =

801

138. an =

( −1) x 2(2) +1 = x 5 a2 = 2 (2) + 1 5 3 2 ( 3 ) +1 ( −1) x x7 =− a3 = 2 ( 3) + 1 7 4 2 ( 4 ) +1 ( −1) x x9 = a4 = 2 ( 4) + 1 9 5 2 ( 5 ) +1 ( −1) x − x11 a5 = = 2 ( 5) + 1 11 n

Sequences and Series

( −1) x n +1 ( n + 1)! n

140. an =

a1 = −

x2 2

x3 x3 = 3! 6 x4 −x4 a3 = =− 4! 24 x5 x5 a4 = = 5! 120 x6 − x6 a5 = =− 6! 720 a2 =

x10 − x10 =− 10! 3,628,800

802

Chapter 9

Sequences, Series, and Probability

( −1) ( x + 1) n +1

141. an =

a1 = x + 1 a2 = − a3 =

a3 = −

( x + 1)

120

a5 = −

( x − 1)

( x − 1) a =

( x + 1)

a4 = − a5 =

( x + 1)

( −1) ( x − 1) ( n + 1)! ( x + 1) a =−

142. an =

120

( x − 1)

720

1 1 − 2n 2n + 2 1 1 1 a1 = − = 2 4 4 1 1 1 a2 = − = 4 6 12 1 1 1 a3 = − = 6 8 24 1 1 1 a4 = − = 8 10 40 1 1 1 a5 = − = 10 12 60 1  1 1 1 1  1 nth partial sum =  −  +  −  +  +  −  2 4 4 6  2n 2 n + 2 

143. an =

1 1 − 2 2n + 2

1 1 − n n +1 1 1 a1 = 1 − = 2 2 1 1 1 a2 = − = 2 3 6 1 1 1 a3 = − = 3 4 12 1 1 1 a4 = − = 4 5 20 1 1 1 a5 = − = 5 6 30

144. an =

1  1 1   1 1 1 nth partial sum =  −  +  −  +  +  −  1 2 2 3 n n +1      1 =1− . n +1

Section 9.1

Sequences and Series

803

1 1 − n +1 n + 2 1 1 1 a1 = − = 2 3 6 1 1 1 a2 = − = 3 4 12 1 1 1 a3 = − = 4 5 20 1 1 1 a4 = − = 5 6 30 1 1 1 a5 = − = 6 7 42

145. an =

1  1 1 1 1  1 nth partial sum =  −  +  −  +  +  −  2 3 3 4  n +1 n + 2  1 1 = − 2 n+2 1 1 − n n+2 1 2 a1 = 1 − = 3 3 1 1 1 a2 = − = 2 4 4 1 1 2 a3 = − = 3 5 15 1 1 1 a4 = − = 4 6 12 1 1 2 a5 = − = 5 7 35

146. an =

 1 1 1 1 1 nth partial sum =  1 −  +  −  +  −  +   3 2 4 3 5 1  1 1   1 + − + −   n −1 n +1   n n + 2  1  1 1  3 1 1  = 1 +  −  + − = − n n n n 2 1 2 2 1 2 + + + +     147. Yes. If the sequence is finite and the terms are integer terms, then the sum can be found. 148. (a) a1 = 1 a2 = 2 a3 = 3 a4 = 4 a5 = 5 a6 = 6

(b)

an = n

(c)

n

6  4  2 149. (a) A − B =   −   =   3 −3 6 4 6 −10 (b) 2 B − 3 A = 2   − 3  =   3 3 −     −15 6  4  (c) AB =     = Not possible 3 − 3  4  6 (d) BA =     = Not possible − 3 3

n =1

804

Chapter 9

Sequences, Series, and Probability  −3 −7 4    151. (a) A − B =  4 4 1  1 4 3

 10 19  150. (a) A − B =    −12 −5  −30 −45 (b) 2 B − 3 A =   4  28  56 (c) AB =   48  48 (d) BA =   36

 8 17 −14    (b) 2 B − 3 A =  −12 −13 −9   −3 −15 −10 

−43  114  −72   122 

 −2 7 −16    45 (c) AB =  4 42  1 23 48  16 31 42    (d) BA = 10 47 31 13 22 25

−1 4    0 4 0  152. (a) A − B =  5 1  −   = Not possible  0 −1 3 1 − 2   −1 4  0 4 0    − 3 (b) 2 B − 3 A = 2    5 1  = Not possible 3 1 2 −      0 −1 −1 4  12 0 − 8   0 4 0    = (c) AB =  5 1     3 21 − 2 3 1 2 −   0 −1  − 3 −1 2      −1 4  0 4 0   20 4   (d) BA =   5 1 =   3 1 − 2   2 15   0 −1

Section 9.2 Arithmetic Sequences and Partial Sums 1. an = a1 + ( n − 1) d

8. 3.7, 3.1, 2.5, 1.9, 1.3,

Arithmetic sequence, d = − 0.6

2. nth partial sum 3. A sequence is arithmetic when the differences between consecutive terms are the same or constant, which is known as the common difference. 4. The arithmetic sequence, an = 4 n + 1, has a common

difference of 4, since it is in the form an = a1 + ( n − 1) d.

9. 12 , 2 2 , 32 , 4 2 , 52 , Not an arithmetic sequence 10. 3.7, 4.3, 4.9, 5.5, 6.1, Arithmetic sequence, d = 0.6 11. an = 8 + 13n

a1 = 8 + 13 (1) = 21

5. 10, 12, 14, 16, 18,

a2 = 8 + 13 ( 2 ) = 34

Arithmetic sequence, d = 2 6. 4, 9, 14, 19, 24, Arithmetic sequence, d = 5

a3 = 8 + 13 ( 3 ) = 47

5 3 , 2, , 1, 2 2

a5 = 8 + 13 ( 5 ) = 73

7. 3,

Arithmetic sequence, d = −

a4 = 8 + 13 ( 4 ) = 60

1 2

Arithmetic sequence, d = 13

Section 9.2 12. an = 150 − 7n

a1 = 150 − 7 (1) = 143

a2 = 150 − 7 ( 2 ) = 136 a3 = 150 − 7 ( 3 ) = 129

a4 = 150 − 7 ( 4 ) = 122 a5 = 150 − 7 ( 5 ) = 115

Arithmetic sequence, d = −7 13. an = 53 − 4( n + 6) = − 4n + 29

a1 = − 4(1) + 29 = 25

Arithmetic Sequences and Partial Sums

17. an = 3 + 2 ( −1)

a1 = 3 + 2 ( −1) = 1 1

a2 = 3 + 2 ( −1) = 5 2

a3 = 3 + 2 ( −1) = 1 3

a4 = 3 + 2 ( −1) = 5 4

a5 = 3 + 2 ( −1) = 1 5

Not an arithmetic sequence 18. an = ( −1)

a2 = − 4( 2) + 29 = 21

a1 = ( −1) = −1

a3 = − 4(3) + 29 = 17

a2 = ( −1) = 1

a4 = − 4( 4) + 29 = 13 a5 = − 4(5) + 29 = 9 Arithmetic sequence, d = − 4 14. an = 1 + ( n − 1) 4

a3 = ( −1) = −1 3

a4 = ( −1) = 1 4

a5 = ( −1) = −1 5

Not an arithmetic sequence

a1 = 1 + (1 − 1) 4 = 1

a2 = 1 + ( 2 − 1) 4 = 5

19. an =

a4 = 1 + ( 4 − 1) 4 = 13

a1 =

a3 = 1 + ( 3 − 1) 4 = 9

a5 = 1 + ( 5 − 1) 5 = 17

Arithmetic sequence, d = 4 15. an = 2 n + n

a2 = a3 =

a1 = 2 + 1 = 3 a2 = 22 + 2 = 6 a3 = 23 + 3 = 11 a4 = 2 4 + 4 = 20 5

a5 = 2 + 5 = 37 Not an arithmetic sequence 16. an = 2 n −1

a3 = 2

a4 =

(−1)2 n 4

(−1)2(1) = 1 4

(−1)

2( 2)

1 4

(−1)2(3) = 1 4

(−1)

2( 4)

1 4

(−1) ( ) = 1 25

a5 =

Arithmetic sequence, d = 0 20. an = ( −1)

2 n +1

a1 = ( −1) = −1 3

a1 = 21−1 = 1 2 −1

3 −1

a2 = 2

805

a4 = 2 4 −1 = 8

a2 = ( −1) = −1 5

a3 = ( −1) = −1 7

a4 = ( −1) = −1 9

a5 = 25 −1 = 16

a5 = ( −1) = −1

Not an arithmetic sequence

Arithmetic sequence, d = 0

21. a1 = 1, d = 6

an = a1 + ( n − 1)d = 1 + ( n − 1)(6) = 6n − 5

806

Chapter 9

Sequences, Series, and Probability

22. a1 = 15, d = 4

31. a1 = 5, d = 6

an = a1 + ( n − 1) d = 15 + ( n − 1) 4 = 11 + 4n 23. a1 = 43, d = − 7

an = a1 + ( n − 1)d = 43 + ( n − 1)( − 7) = − 7 n + 50

a2 = 5 + 6 = 11 a3 = 11 + 6 = 17 a4 = 17 + 6 = 23 a5 = 23 + 6 = 29

24. a1 = 100, d = −8

an = a1 + ( n − 1) d

32. a1 = 5, d = −

= 100 + ( n − 1)( −8 ) = 108 − 8n

9 3 3 25. 3, , , , 0,  4 2 4 a2 = a1 + ( n − 1)d 9 3 = 3 + ( 2 − 1)d  − = d 4 4

3 15  3 an = a1 + ( n − 1)d = 3 + ( n − 1) −  = − n + 4 4  4 26. 4,

a1 = 5

3 7 5 , − 1, − ,, d = − 2 2 2

 5  13 5 an = a1 + ( n − 1) d = 4 + ( n − 1)  −  = − n  2 2 2

27. a1 = − 5, a4 = 22

a4 = a1 + 3d  22 = − 5 + 3d  d = 9 an = a1 + ( n − 1)d = − 5 + ( n − 1)(9) = 9n − 14 28. a1 = −4, a5 = 16

3 4

a1 = 5 3 17 = 4 4 17 3 14 7 a3 = − = = 4 4 4 2 7 3 11 a4 = − = 2 4 4 11 3 8 a5 = − = =2 4 4 4 a2 = 5 −

33. a1 = − 2.6, d = 0.2 a1 = − 2.6 a2 = − 2.6 + 0.2 = − 2.4 a3 = − 2.4 + 0.2 = − 2.2 a4 = − 2.2 + 0.2 = − 2.0 a5 = − 2.0 + 0.2 = −1.8

34. a1 = −10, d = 9 a1 = −10 a2 = −10 + 9 = −1

an = a1 + ( n − 1) d

a3 = −1 + 9 = 8

16 = −4 + 4 d d =5

a4 = 8 + 9 = 17

an = −4 + ( n − 1) 5 = −9 + 5n 29. a3 = 94, a6 = 85

a6 = a3 + 3d  85 = 94 + 3d  d = −3 a1 = a3 − 2 d  a1 = 94 − 2 ( −3 ) = 100

an = a1 + ( n − 1) d

= 100 + ( n − 1)( −3) = 103 − 3n

30. a5 = 190, a10 = 115

a10 = a5 + 5d  115 = 190 + 5d  d = −15 a1 = a5 − 4 d  a1 = 190 − 4 ( −15 ) = 250

a5 = 17 + 9 = 26

35. a8 = 26, a12 = 42

26 = a8 = a1 + ( n − 1) d = a1 + 7d 42 = a12 = a1 + ( n − 1) d = a1 + 11d Answer: d = 4, a1 = −2 a1 = −2 a2 = −2 + 4 = 2 a3 = 2 + 4 = 6 a4 = 6 + 4 = 10 a5 = 10 + 4 = 14

an = a1 + ( n − 1) d = 250 + ( n − 1)( −15 ) = 265 − 15n

Section 9.2 36. a11 = a6 + 5d

−73 = −38 + 5d  d = −7

a6 = a1 + 5d  −38 = a1 + 5 ( −7 )  a1 = −3 a2 = −3 − 7 = −10 a3 = −10 − 7 = −17 a4 = −17 − 7 = −24 a5 = −24 − 7 = −31 37. a3 = 19, a15 = −1.7

a15 = a3 + 12 d −1.7 = 19 + 12 d  d = −1.725 a3 = a1 + 2 d  19 = a1 + 2 ( −1.725 )  a1 = 22.45 a2 = a1 − 1.725 = 20.725 a3 = 19 a4 = 19 − 1.725 = 17.275 a5 = 17.275 − 1.725 = 15.55 38. a14 = a5 + 9d

38.5 = 16 + 9d  d = 2.5

a5 = a1 + 4 d  16 = a1 + 4 ( 2.5)  a1 = 6 a2 = 6 + 2.5 = 8.5 a3 = 8.5 + 2.5 = 11 a4 = 11 + 2.5 = 13.5 a5 = 13.5 + 2.5 = 16

39. a1 = 15, ak +1 = ak + 4

a2 = a1 + 4 = 15 + 4 = 19

Arithmetic Sequences and Partial Sums 3 1 , a k +1 = − + a k 4 10 1 3 5 1 = a2 = − + = 10 5 10 2 1 1 4 2 = a3 = − + = 10 2 10 5 1 2 3 a4 = − + = 10 5 10 1 3 1 = a5 = − + 10 10 5 1 d=− 10 7 1 an = − n 10 10

41. a1 =

42. a1 = 1.5, ak +1 = ak − 2.5

a2 = 1.5 − 2.5 = −1.0 a3 = −1.0 − 2.5 = −3.5 a4 = −3.5 − 2.5 = −6.0 a5 = −6.0 − 2.5 = −8.5 d = −2.5 an = 4.0 − 2.5n 43. a1 = 5, a2 = 11  d = 6

a10 = a1 + 9d = 5 + 9 ( 6 ) = 59 44. a2 = a1 + d

13 = 3 + d  d = 10 a9 = a1 + 8d = 3 + 8 (10 ) = 83

a3 = 19 + 4 = 23 a4 = 23 + 4 = 27 a5 = 27 + 4 = 31 d = 4, an = 11 + 4 n 40. a1 = 6, ak +1 = ak + 5 a1 = 6 a2 = 6 + 5 = 11 a3 = 11 + 5 = 16

807

45. a1 = 4.2, a2 = 1.8  d = − 2.4

a7 = a1 + 6d = 4.2 + 6( − 2.4) = −10.2 46. d = a2 − a1 = −13.8 − ( −0.7 ) = −13.1

a8 = a1 + 7d = −0.7 + 7 ( −13.1) = −92.4 47. an = 15 − 23 n 16

a4 = 16 + 5 = 21 a5 = 21 + 5 = 26

d = 5, an = 5n + 1 0

48. an = −5 + 2 n 16

−6

Chapter 9

808

Sequences, Series, and Probability 51. an = 4 n − 5

49. an = 0.4 n − 2 2.4

− 2.0

−1

52. an = 17 + 3n

50. an = −1.3n + 7

an 20

26 29 32

−8

53. an = 20 − 34 n

17.75 17 16.25 15.5

14.75

13.25 12.5

an 19.25 18.5

54. an =

4 n −3 5

− 2.2

− 1.4

− 0.6

0.2

1.0

1.8

2.6

3.4

4.2

5.0

55. an = 1.5 + 0.05n

1.6

1.65 1.7 1.75 1.8 1.85

1.9

1.95

2.0

1.55

56. an = 8 − 12.5n

−4.5 −17 −29.5 −42 −54.5 −67 −79.5 −92 −104.5 −117

57. S10 =

10 ( 2 + 20 ) = 110 2

59. S5 =

5 −1 + ( −9 ) = −25 2

58. S7 =

7 (1 + 19 ) = 70 2

60. S6 =

6 ( − 2) + ( −17) = − 57 2

100

61. S100 =  n = 1 + 2 +  + 99 + 100 = n =1

)

100 (1 + 100 ) = 5050 2

62. S50 =  −n = ( −1) + ( −2 ) +  + ( −50 ) = n =1

(

50 ( −1) + ( −50 )   2 

= −1275

Section 9.2

63. S131 = 64. S61 =

131 ( −100 + 30 ) = −4585 2

74.

61 ( −10 + 50 ) = 1220 2

= 3775 − 1275 = 2500 500 75.  ( n + 8 ) = ( 9 + 508 ) = 129,250 2 n =1

65. 8, 20, 32, 44, , n = 10 a1 = 8, a2 = 20  d = 12

a10 = a1 + 9d = 8 + 9 (12 ) = 116 S10 =

Arithmetic Sequences and Partial Sums

10 (8 + 116 ) = 620 2

100

n = 51

n −1

809

 n −  n = 2 ( 51 + 100 ) − 2 (1 + 50 )

500

250

250 (999 + 750) = 218,625 2

76. (1000 − n) = n =1

77.

 ( 2n + 1) = 440 n =1

66. 2, 8, 14, 20, , n = 25 a1 = 2, a2 = 8  d = 6

a25 = 2 + ( 25 − 1)( 6 ) = 146 S25 =

25 ( 2 + 146 ) = 1850 2

78.

n =1 100

79.



n+5 101 = (2.5 + 52.5) = 2777.5 2 2

100

4−n

n=0

67. 7.2, 6.4, 5.6, 4.8,  ; n = 10 a1 = 7.2, a2 = 6.4  d = − 0.8

80.

10 (a1 + a10 ) = 5(7.2 + 0) = 36 2

68. 4.2, 3.7, 3.2, 2.7, , n = 12 a1 = 4.2, a2 = 3.7  d = −0.5

a12 = 4.2 + 11( −0.5 ) = −1.3 12 ( 4.2 − 1.3) = 17.4 2 69. a1 = 100, a25 = 220, n = 25 S12 =

25 ( a1 + a25 ) = 12.5 (100 + 220 ) = 4000 2 70. a1 = 15, a100 = 307, n = 100 S25 =

S100 =

100 (15 + 307 ) = 16,100 2

71. a1 = 1, a50 = 50, n = 50 50

 n = 2 (1 + 50 ) = 1275 n =1

72. an = 2 n

a1 = 2, a100 = 200, n = 100 100

 2n = n =1

100 ( 2 + 200 ) = 10,100 2

20 10 73.  n −  n = 11 + 30 ) − (1 + 10 ) ( 2 2 n =11 n =1 = 410 − 55 = 355

 4

= −1161.5



2 

n=0

a10 = a1 + 9d = 7.2 + 9( − 0.8) = 0 S10 =

(40 − 2n) = 2 38 + (− 60) = − 550

81.

  250 − 5 i  = 14,268 i =1

82.

200

 (10.5 + 0.025 j ) = 2602.5 j =1

83. a1 = 14, a18 = 31

S18 =

18 (14 + 31) = 405 bricks 2

84. a1 = 20 : first row

a2 = 21: second row a3 = 22 : third row  a1 = 20, d = 1, n = 20

an = 20 + ( n − 1)(1) = 19 + n a20 = 39 S20 =

20 ( 20 + 39 ) = 590 seats 2

85. a1 = 30,000

a2 = 30,000 + 5000 = 35,000 d = 5000 a5 = 30,000 + 4 ( 5000 ) = 50,000

S5 = 25 ( 30,000 + 50,000 ) = $200,000

810

Chapter 9

Sequences, Series, and Probability

(d) a9 = 772.925, a16 = 1028.875

86. a1 = 16 feet: first second

a2 = 48 feet: second second

10 (772.925 + 1028.875) = 9009 thousand 2 degrees, or 9,009,000 degrees S8 =

a3 = 80 feet: third second  a1 = 16, d = 32, n = 8

Answers will vary.

an = 16 + ( n − 1)( 32 ) = −16 + 32 n

89. True. If the nth term of an arithmetic sequence is an and

a8 = 240 S8 =

the common difference is d, the ( n + 1) th term can be

8 (16 + 240 ) = 1024 feet 2

found using the recursion formula, an +1 = an + d. 90. False. You need to know how many terms are in the sequence.

87. (a) an = 0.39n + 1.89

(b)

Year

2006

2007

2008

2009

Sales (in billions of dollars)

4.23

4.62

5.01

5.40

Year

2010

2011

2012

2013

Sales (in billions of dollars)

5.79

6.18

6.57

6.96

a2 = x + 2 x = 3 x a3 = 3 x + 2 x = 5 x a4 = 7 x a5 = 9 x a6 = 11x a7 = 13 x a8 = 15 x

The model fits the data well.

a9 = 17 x

a10 = 19 x

a1 = 4.23, a8 = 5.01 S8 =

91. a1 = x

8 ( 4.23 + 6.96) = $44.76 billion 2

92. a1 = − y

a2 = − y + 5 y = 4 y

(d) a9 = 7.35, a16 = 10.08

a3 = 9 y

8 S8 = (7.35 + 10.08) = $69.72 billion 2

a4 = 14 y a5 = 19 y

Answers will vary.

a6 = 24 y a7 = 29 y

88. (a) an = 25.595n + 440.19

a8 = 34 y

(b)

a9 = 39 y

Year

2005

2006

2007

2008

Master’s degrees conferred (in thousands)

568

594

619

645

Year

2009

2010

2011

2012

d = a2 − a1. So, an = a1 + ( n − 1) d

Master’s degrees conferred (in thousands)

671

696

722

747

(or an = a1 + ( n − 1)( a2 − a1 ) ).

a10 = 44 y 93. Given a1 and a2 , and because the sequence is

arithmetic, the common difference is

The model fits the data well. (c) an = 25.595n + 440.19

a1 = 568.165 a8 = 747.33 8 (568.165 + 747.33) = 5,261.98 thousand 2 degrees, or about 5,262,000 degrees S8 =

94. a20 = a1 + 19 ( 3) = a1 + 57 n ( a1 + a20 ) 2 20 = a1 + ( a1 + 57 ) = 650 2 10 ( 2 a1 + 57 ) = 650 S=

(

)

20 a1 = 80 a1 = 4

Section 9.3

n ( a1 + 5) + ( an + 5) 2 n = ( a1 + a2 + 10 ) 2 n = ( a1 + a2 ) + 5n 2 = S n + 5n

95. S =

(

)

a7 = a1 + (7 − 1)d 11 = − 7 + 6d

1 + 2 + 3 +  + 99 + 100 = x 100 + 99 +  + 2 + 1 = x Adding: 101 + 101 + ⋅ ⋅ ⋅ + 101 + 101 = 2 x 100 (101) 100 (101) = 2 x  x = = 5050 2 n ( n + 1) In general, 1 + 2 +  + n = . 2 98. (a) True. The differences between consecutive terms are the same. (b) False. The common difference is − 2.

18 = 6d 3 = d

A recursion formula is a1 = 7, an + 1 = an + 3. (b) No, an arithmetic sequence is not possible.

99.  2 −1 7  −10  row reduces to    3 2 −4  17   6 −5 1  −20 

 1 0 0  1    0 1 0  5 . 0 0 1  −1

Let a1 = 4, a8 = 28.5, and n = 8. a8 = a1 + (8 − 1)d

Answer: (1, 5, − 1)

28.5 = 4 + 7 d 24.5 = 7 d 3.5 = d

A recursion formula is a1 = 4, an + 1 = an + 3.5.

811

97. Gauss might have done the following:

96. (a) Yes. − 7, − 4, − 1, 2, 5, 8, 11

Let a1 = − 7, a7 = 11, and n = 7.

Geometric Sequences and Series

 −1 4 10  4    1  31 row reduces to 100.  5 −3  8 2 −3  −5  1 0 0  2   0 1 0  −6  . 0 0 1  3 Answer: ( 2, − 6, 3)

101. Answers will vary. (Make a Decision)

Section 9.3 Geometric Sequences and Series 1. geometric, common 2. an = a1r n −1 3. geometric series 4. No. A sequence is geometric when the ratios of a a consecutive terms are the same; 2 = 3 =  = r , r ≠ 0, a1 a2

where r is the common ratio. 5. In order for an infinite geometric series to have a sum, ∞ a r < 1 and S =  a1r i = 1 . 1−r i =0

6. (a) The sum of a finite geometric series is n

Sn =  air i −1 =

(

a1 1 − r n

)

. 1− r i =1 (b) The sum of an infinite geometric series is ∞ a S =  a1r i = 1 , r < 1. 1−r i =0 7. 5, 15, 45, 135, Geometric sequence r =3 8. 3, 12, 48, 192, Geometric sequence r = 123 = 4

812

Chapter 9

Sequences, Series, and Probability

9. 6, 18, 30, 42,

Not a geometric sequence (Note: It is an arithmetic sequence with d = 12.) 10. 4, 19, 34, 49,

Not a geometric sequence (Note: It is an arithmetic sequence with d = 15. )

1 1 1 11. 1, − , , − , 2 4 8 Geometric sequence 1 r=− 2 8 12. 9, − 6, 4, − ,  3 Geometric sequence 2 r=− 3

19. a1 = 1, r =

1 1 a2 = 1  = 2 2 11 1 a3 =   = 22 4 a4 =

r =

2 1 = = 0.1 20 10

14. 5, 1, 0.2, 0.04 Geometric sequence 1 r = = 0.2 5

1 1 1 , , , 2 3 4 Not a geometric sequence

15. 1,

16.

1 2 3 4 , , , , 5 7 9 11 Not a geometric sequence

11 1  = 42 8

1 1  1 a5 =   = 8  2  16 20. a1 = 2, r =

1 3

1 2 a2 = 2   = 3 3 21 2 a3 =   = 33 9 a4 =

21 2  = 9  3  27

a5 =

2 1 2  = 27  3  81

13. 20, 2, 0.2, 0.02,

Geometric sequence

1 2

21. a1 = 5, r = −

1 10

1  1  a2 = 5  −  = − 2  10   1  1  1 a3 =  −  −  =  2  10  20 a4 =

1  1  1 −  = − 20  10  200

1  1  1  a5 =  −  −  =  200  10  2000 22. a1 = 6, r = −

1 4

17. a1 = 6, r = 3

a2 = 6 ( 3) = 18

3  1 a2 = 6  −  = − 4 2   2

a3 = 18 ( 3) = 54

3  1 a3 = 6  −  =  4 8

a5 = 162 ( 3) = 486

3  1 a4 = 6  −  = − 32  4

a4 = 54 ( 3) = 162

18. a1 = 4, r = 2

a2 = 4 ( 2 ) = 8

3  1 a5 = 6  −  = 4 128  

a3 = 8 ( 2 ) = 16

a4 = 16 ( 2 ) = 32

a5 = 32 ( 2 ) = 64

Section 9.3

a2 = 9 ( e ) = 9e

a3 = ( 9e )( e ) = 9e

a2 = 3(5) = 15 a3 = 3(15) = 45

( ) ( e ) = 9e a = ( 9e ) ( e ) = 9e 2

a4 = 3( 45) = 135 a5 = 3(135) = 405

24. a1 = 7 r =

r = 3, an = 5(3)

( 5) = 7 5 a = (7 5 )( 5 ) = 35 a = (35)( 5 ) = 35 5 a = (35 5 )( 5 ) = 175 3

1 ak 2

3  81  243 a5 = −  −  = 2 4  8

1 a2 = ( 64 ) = 32 2 1 a3 = ( 32 ) = 16 2 1 a4 = (16 ) = 8 2 1 a5 = ( 8 ) = 4 2

1 1 , an = 64   2 2

26. a1 = 81, ak +1 =

n −1

3 29. a1 = 6, ak +1 = − ak 2 3 a2 = − ( 6 ) = −9 2 3 27 a3 = − ( −9 ) = 2 2 3  27  81 a4 = −   = − 2 2  4

a2 = 7

25. a1 = 64, ak +1 =

813

28. a1 = 5, ak + 1 = 3ak

23. a1 = 9, r = e

a4 = 9e

Geometric Sequences and Series

3  3 r = − , an = 6  −  2  2

n −1

1 = 128   2

1 ak 3

 3 = −4  −   2

2 30. a1 = 30, ak +1 = − ak 3 −2 2 a2 = − a1 = ( 30 ) = −20 3 3 2 40 a3 = − ( −20 ) = 3 3 2  40  80 a4 = −   = − 3 3  9 2  80  160 a5 = −  −  = 3  9  27

1 (81) = 27 3 1 a3 = ( 27 ) = 9 3 1 a4 = ( 9 ) = 3 3 1 a5 = ( 3 ) = 1 3 a2 =

1 1 r = , an = 243   3 3

n −1

2  2 r = − , an = 30  −  3  3

n −1

31. a1 = 11, r = 1.03, n = 12

an = a1r n −1 n

(a)

a12 = 11(1.03)

= 11(1.03) ≈ 15.227

(b)

a12 = 11(1.03)

= 11(1.03) ≈ 15.227

12 −1 12 −1

11 11

32. a1 = 24, r = 2.6, n = 8

27. a1 = 9, ak +1 = 2 ak

an = a1r n −1

a2 = 2 ( 9 ) = 18

a3 = 2 (18 ) = 36

a4 = 2 ( 36 ) = 72

(a)

a8 = 24 ( 2.6 )

(b)

a8 = 24 ( 2.6 ) ≈ 19,276.344

8−1

= 24 ( 2.6 ) ≈ 19,276.344 7

a5 = 2 ( 72 ) = 144 9 r = 2, an =   2 n = 9 2 n −1 2

(

)

814

Chapter 9

Sequences, Series, and Probability 39. 7, 21, 63, r =3

4 33. a1 = 8, r = − , n = 7 3

an = 7 ( 3 )

an = a1r n −1 7 −1

(a)

 4 a7 = 8  −   3

 4 = 8  −  ≈ 44.949  3

(b)

32,768  4 a7 = 8  −  = 3 729  

(a)

 3 a9 = 8  −   4

 3 = 8  −  ≈ 0.801  4

(b)

6561  3 a9 = 8  −  =  4  8192

(b) a14 = 7 38. a1 = 2, r =

an = a1r

2, n = 14 14 − 1

( 2)

14 − 1

≈ 633.568 = 7

( 2 ) ≈ 633.568

3, n = 11

( 3)

(

(b) a11 = 2 3

11 − 1

)

11 − 1

( 3 ) = 486 10

= 2

( 3) = 486

= 2

44. a1 = 5, a3 =

n −1

(a) a11 = 2

a4 = a1r 3 1 = 4r 3 2 1 = r3 8 1 =r 2 an = a1r n −1 1 1 a10 = 4   = 2 128  

n −1

( 2)

1 , n = 10 2

43. a1 = 4, a4 =

 1  12 −1  1  11 a12 =  − = −  (2)  2 = −16  128   128   1  11 a12 =  −  ( 2 ) = −16  128 

(a) a14 = 7

an = 4 ( 2 ) 22 −1 a22 = 4 ( 2 ) = 8,388,608 n −1

an = a1r n −1

an = a1r

= 50,388,480

42. 4, 8, 16, 8 r= =2 4

1 , r = 2, n = 12 128

37. a1 = 7, r =

= 8,957,952

n −1 10 −1

 1  1 (b) a6 =  − 86 −1 =  − 85 = 8192  4  4

(b)

7 −1

a10 = 5 ( 6 )

 1 (a) a6 =  − 86 −1 = 8192  4

(a)

n −1

41. 5, 30, 180, 30 r= =6 5

an = 5 ( 6 )

1 35. a1 = − , r = 8, n = 6 4 an = a1r n − 1

36. a1 = −

= 45,927

40. 3, 36, 432, 36 r= = 12 3

a7 = 3 (12 )

n −1

9 −1

an = 3 (12 )

3 34. a1 = 8, r = − , n = 9 4 an = a1r

a9 = 7 ( 3 )

n −1

45 , n =8 4

a3 = a1r 2 45 = 5r 2 4 9 = r2 4 3 =r 2 an = a1r n −1 7

 3  10,935 a8 = 5   = 128 2

Section 9.3

45. a2 = −18, a5 =

48. an = 20 ( 0.85)

2 , n=6 3

n −1

49. an = 2 (1.3 )

n −1

64 ,n = 6 27

a5 = a2r 3

50. an = 10 ( −1.2 )

64 = − 8r 3 27 8 − = r3 27 2 − = r 3 a6 = a5r

n −1

− 50

1 , 2

51. 8, − 4, 2, − 1,

128  64  2  a6 =   −  = − 27 3 81    47. an = 12 ( −0.75)

815

a5 = a2 r 3 2 = −18r 3 3 1 − = r3 27 1 − =r 3 a6 = a5r 2  2  1  a6 =   −  = − 9  3  3 

46. a2 = − 8, a5 =

Geometric Sequences and Series

S1 = 8 S2 = 8 + ( −4 ) = 4

n −1

S3 = 8 + ( −4 ) + 2 = 6

S4 = 8 + ( −4 ) + 2 + ( −1) = 5

S5 = 8 + ( −4 ) + 2 + ( −1) +

− 10

52. 8, 12, 18, 27,

1 11 = 2 2

81 , 2

S1 = 8 S2 = 8 + 12 = 20 S3 = 8 + 12 + 18 = 38 S4 = 8 + 12 + 18 + 27 = 65 S5 = 8 + 12 + 18 + 27 +

 1

n −1

31.5 31.75 31.875 31.9375 31.96875

∞

16  − 2 

53.

81 211 = 2 2

n =1

816

Chapter 9

54.

 4 ( 0.2 )

n −1

4.8

4.96 4.992 4.9984 4.99968

55.

2

∞

Sequences, Series, and Probability

n =1

S9 = 56.

(

1 1 − 29 1− 2

n −1

n =1

( S = 7

1 − ( −2 )

 1

1

63.

i −1

 a1 = 64, r = −

i −1

 a1 = 32, r =

1 − (1 4 )

n −1

21 6 6 12 5  = 12 5  n=0 n =1

1 2

64.

60.

7

10 6  = 10 6 

n=0

i =1

n −1

 a1 = 300, r = 1.06

 500 (1.04 ) =  500 (1.04 ) n

n −1

 a1 = 500, r = 1.04

n =1

 1 − (1.04 )7   ≈ 3949.15 S7 = 500   1 − 1.04    n −1

n=7

n −1

n =1

 a1 = 12, r =

6 5

66. 7 + 14 + 28 +  + 896

r = 2 and 896 = 7 ( 2 ) 8

 7(2)

n −1

n =8

n −1

n =1

67. 2 −

n −1

 a1 = 10, r =

i −1

 a1 = 8, r = −

1 1 1 + − + 2 8 2048

r=− 7 6

1 1  1 = 2 −  and 4 2048  4

 1 2 −   4 n =1  7

n −1

n=7

n −1

3 3 − − 5 625 n −1 3 r = −0.2 and − = 15 ( −0.2 )  n = 6 625

68. 15 − 3 +

 1

n =1

 5 ( 3)

 7  = − 60 1 −    ≈ 646.81  6    10

n =1

 8  − 4 

 300 (1.06 ) =  300 (1.06 )

r = 3 and 3645 = 5 ( 3)

1 − (7 6)16   S16 = 10   1 − (7 6) 

61.

1 3

65. 5 + 15 + 45 +  + 3645

1 4

21  6  = − 60 1 −    ≈ 2700.31  5   

 a1 = 5, r = −

n=0

1 − (6 5)21   S 21 = 12   1 − (6 5) 

i −1

 1 − (1.06 )6   ≈ 2092.60 S6 = 300   1 − 1.06   

 1 5 −   3 i =1 

n =0

(1 − (1 4 ) ) = 1365 S = 32

59.

≈5

) = 171

i =1

4.9999872 ≈ 5

 1 − ( −1 3 )10   ≈ 3.75 S10 = 5   1 − ( −1 3 )   

 1 − ( −1 2 )7  128   1 7  = S7 = 64  1 −  −   = 43 3   2    1 − ( −1 2 )   

 32  4 

i =1

58.

62.

 a1 = 1, r = −2

 64  − 2 

4.999936

) = 511

1 1 − ( −2 )

57.

 a1 = 1, r = 2

n −1

 ( −2 ) 9

1 4

15 ( −0.2 )

n −1

n =1

 1 − ( −1 4 )10  32   1 10  = S10 = 8  1 −  −   ≈ 6.4  1 − ( −1 4 )  5   4    

Section 9.3

4 5

69. a1 = 10, r =

∞

 1

1−

n −1

∞

n=0

∞

2 3

= 18

5 5 10 = =  1 3 3 1− −     2 2

1 4

7

 2  3 

n =0

79. 9 + 6 + 4 +

n =1

74.

∞

5

 8  3 

n −1

n =1

7  does not have a finite sum  > 1 . 3   5  does not have a finite sum  > 1  . 3  

75. a1 = 10, r = 0.11 ∞

n=0

−10 25 = − ≈ −8.333 1.2 3

∞ 8 2 +  =  9  3 n=0  3 

80. 8 + 6 +

9 9 = 1 = 27 2 1− 3 3

∞ 9 27 3 + +  = 8  2 8 n =0  4 

8 = 32 1 − 34 n

81. 3 +

∞ 15 75 375 5 + + +  =  3  does not have 2 4 8 n=0  2 

5 5 and > 1. 2 2

a finite sum because r =

a1 10 10 = = 1 − r 1 − 0.11 0.89 1000 = ≈ 11.236 89

10 ( 0.11) = n

−10

 −10 ( −0.2 ) = 1 − r = 1 − ( −0.2 )

a 5 5  1 = = 4  5 −  = 1 −1 r =  1 5 n=0  4  1 − −     4 4

73.

−3 30 =− ≈ −1.579 1.9 19

78. a1 = −10, r = −0.2 a1

72. a1 = 5, r = −

−3

n=0

 5  − 2  = 1 − r =

100

 −3 ( −0.9) = 1 − r = 1 − ( −0.9)

1 2

∞

≈ 9.091

∞

n=0

∞

77. a1 = −3, r = −0.9

 6  3  = 1 − r = 71. a1 = 5, r = −

n=0

2 70. a1 = 6, r = 3 2

 5 ( 0.45) = 1 − r = 1 − 0.45 = 0.55 = 11

a 10 4 10   = 1 = = 50  n=0  5  1− r 1− 4 5

∞

817

76. a1 = 5, r = 0.45

∞

Geometric Sequences and Series

82. 2 +

∞ 7 49 343 7 + + +  =  2  does not have a 3 18 108 n=0  6 

finite sum because r =

7 7 and > 1. 6 6

83. − 7 + 2 −

84. −6 + 5 −

∞ 4 8 a1 −7 49  2 7 + −  =  − 7 −  = = = ( − 7)  = − 7 49 1−r 1 − ( − 2 7) 9  7 9 n=0

∞ 25 125  5 + +  =  −6  −  6 36 n=0  6

= ∞

85. 0.36 =  0.36 ( 0.01)

−6 −6 36 = =− ≈ −3.2727 1 − ( − 5 6 ) 11 6 11

∞

86. 0.297 =  0.297 ( 0.001)

n=0

0.36 0.36 36 4 = = = 1 − 0.01 0.99 99 11

n =0

0.297 0.297 297 11 = = = 1 − 0.001 0.999 999 37

818

Chapter 9

Sequences, Series, and Probability

∞

87. 1.25 = 1.2 +  0.05 ( 0.1)

92.

n=0

6 0.05 + 5 1 − 0.1 6 0.05 = + 5 0.9 6 5 113 = + = 5 90 90

The series is geometric with common ratio 2 and first term a1 = 1. The sum of the first 15 terms is 15  1 − 215  S15 =  (1) 2 n −1 = (1)   = 32,767. n =1  1− 2 

∞

88. 1.38 = 1.3 +  0.08 ( 0.1)

93.

0.08 1 − 0.1 0.08 = 1.3 + 0.9 3 4 7 25 =1 + =1 = 10 45 18 18

∞

a2 30 1 = = . Because the a1 90 3

94.

∞

3n −1 n =1

The series is geometric with common ratio 3, and first term a1 = 1.

n −1

The sum of the first 15 terms is 15  1 − 315  S15 =  (1)3n −1 = (1)  = 7,174,453. n =1  1−3  nt

95.

5 7 9 11 + + + + 4 4 4 4 The series is arithmetic with common difference d = 12 .

r 0.03    A = P  1 +  = 1000  1 +  n n   

n = 1: A = 1000 (1 + 0.03) ≈ $1343.92

(b)

0.03   n = 2 : A = 1000  1 +  2  

(c)

0.03   n = 4 : A = 1000  1 +  4  

(d)

0.03   n = 12 : A = 1000  1 +  12  

(e)

0.03   n = 365 : A = 1000  1 +  365  

Because the first term is , the nth term is 5 33 1 . an = a1 + ( n − 1) d = + ( n − 1)   and a15 = 4 4 2

∞

 6n n =1

a15 = 90, the sum of the first 15 terms is 15 ( 6 + 90 ) = 720. 2

2 (10 )

4 (10 )

≈ $1346.86 ≈ $1348.35

12 (10 )

The sum of the first 15 terms is 15 5  1   15  5 33  285 S15 =   + ( n − 1)    =  +  = 4 n =1  4  2  2  4 4 

The series is arithmetic with common difference d = 6. Because the first term is a1 = 6 and the 15th term is

n (10 )

(a)

5 4

n =1

n −1

n =1

The sum of the first 15 terms is 15  1 − 0.815  n −1 S15 =  6 ( 0.8 ) = 6   ≈ 28.944. n =1  1 − 0.8 

The sum of the first 15 terms is  1 − ( 1 )15  15 n −1 3  = 135. S15 =  90 ( 13 ) = 90   1 − ( 13 )  n =1  

n =0

first term is a1 = 90, the nth term is

S15 =  6 n =

∞

 6 ( 0.8 ) =  6 ( 0.8 )

89. 90 + 30 + 10 + 103 + 

91.

n =0

= 1.3 +

The series is geometric, r =

∞

 6 ( 0.8 )

The series is geometric with common ratio r = 0.8.

n =0

90.

n −1

n =1

1 an = a1r n −1 = 90   3

∞

2

96.

≈ $1349.35

365 (10 )

≈ $1349.84

∞

n+8 8 n=0



The series is arithmetic with common difference d = ∞ ( n − 1) + 8 = ∞ 1 n + 7 n+8 =  )  ( 8 8 n=0 n =1 n =1 8 ∞



The sum of the first 15 terms is 15  11  225 S15 = = 28.125. 1 +  = 2 4 8

1 . 8

Section 9.3

97.

60 0.03   A = 100  1 +  12  n =1 

A = Per 12 + Pe2 r 12 +  + Pe Nr 12

  0.03  1 − (1 + 0.03 12 )   = 100  1 + ⋅  12  1 − (1 + 0.03 12 )    

=  Per 12 ⋅n n −1

(1 − (e ) ) = Pe r 12

  0.02  1 − (1 + 0.02 12)   = 501 + ⋅ 12  1 − (1 + 0.02 12)  72

99. Let N = 12 t be the total number of deposits.

N r    12   = P  + 1   −1 +  1 +    r    12   N    12  r  = P  1 +  − 1  1 +  r   12    12 t    12  r  = P  1 +  − 1  1 +  r   12   

( e − 1) ( e − 1)

r 12

r  r  r     A = P 1 +  + P 1 +  +  + P 1 +  12 12 12       N −1  r  r  r      = 1 +   P + P 1 +  +  + P 1 +    12    12   12  

N r  12    r    = P  1 +  −  1 −  1 +    12  r    12  

1 − er 12

r 12

(a) Compounded monthly: 12 ( 20 )   0.07  12  A = 50  1 + − 1  1 +   ≈ $26,198.27 12  0.07     (b) Compounded continuously:

≈ $3827.90

r 12

101. P = $50, r = 7%, t = 20 years

1 − 1.0017 72  ≈ 50(1.0017) ⋅    1 − 1.0017 

r   1− 1+ r   12   = P 1 +   12  1 −  1 + r   12   

12 t

r 12

= Pe

n −1

(1 − e ) (1 − (e ) ) r 12

r 12

 1 − 1.002560  = 100 (1.0025 ) ⋅    1 − 1.0025  ≈ $6480.83

r  N  r   = P 1 +   1 +  12 12 n 1 =    

819

100. Let N = 12 t be the total number of deposits. 60

72 0.02   98. A = 501 +  12   n =1

Geometric Sequences and Series

(

50e0.07 12 e e

0.07( 20 )

0.07 12

) ≈ $26,263.88

−1

102. P = 75, r = 0.04, t = 25 (a) Compounded monthly: 12 ( 25 )   0.04  12  A = 75  1 + − 1  1 +   ≈ $38,688.25 12 0.04      (b) Compounded continuously:

(

75e0.04 12 e e

0.04 ( 25)

0.04 12

) ≈ $38,725.81

−1

103. P = 100, r = 5% = 0.05, t = 40 (a) Compounded monthly: 12 ( 40 )   0.05  12  A = 100  1 + − 1  1 +   ≈ $153,237.86 12  0.05     (b) Compounded continuously:

(

100e0.05 12 e

0.05( 40 )

) ≈ $153,657.02

−1

0.05 12

e −1 104. P = $20, r = 6%, t = 50 years

(a) Compounded monthly: 12 ( 50 )   0.06  12  A = 20  1 + − 1  1 +   ≈ $76,122.54 12  0.06     (b) Compounded continuously:

(

20e0.06 12 e e

0.06 ( 50 )

0.06 12

−1

) ≈ $76,533.16

−1

820

Chapter 9

Sequences, Series, and Probability

105. First shaded area:

108. (a) an ≈ 1266.52(1.005)

16 2 4

(b) The population of China is growing at a rate of about 0.5% per year.

16 2 1 16 2 Second shaded area: + ⋅ 4 2 4

Third shaded area:

a18 = 1266.52(1.005)

16 2 1 16 2 1 16 2 + + , etc. 4 2 4 4 4

(d) an = 1.4 billion  an = 1400 million

 1 − (1 2 )6  16 1   = 64  4 n=0  2   1 − 1 2  n

1400 = 1266.52(1.005)

21 2  1  8  2  1  8  2  1  8  106. 27   + 27    + 27    + 27    9  9  9   9  9   9  9 

3 2465  1  8  =  272    = ≈ 273.89 square inches 9 n= 0  9  9 

107. (a) a0 = 70 degrees a1 = 0.8 ( 70 ) = 56 degrees 

an = ( 0.8 ) ( 70 ) n

(b) a6 = ( 0.8 ) ( 70 )  18.35°F 6

(c)

1400 n = (1.005) 1266.52  1400  ln   = n ln (1.005)  1266.52 

  1 6  = 128  1 −    = 126 square inches  2   

a12 = ( 0.8 )

≈ 1385.5 million

The value is close to the value predicted by the U.S. Census Bureau.

Total area of shaded region: 2

( 70 ) ≈ 4.81°F

 1400  ln    1266.52  = n ln 1.005 n ≈ 20.09 During the year 2020, China’s population will reach 1.40 billion people. 109. (a) Surface area of a sphere is 4π r 2 . The surface area of the sphere flake is   1 2    1 2  2 S = 4π (1) + 9  4π    + 92  4π    +   3   9      ∞

= 4π + 4π + 4π +  =  4π . n =1

(b) Volume of a sphere is 43 π r 2 . The volume of the

sphere flake is 0

a3 ≈ 35.8 a4 ≈ 28.7

Thus, the water freezes between 3 and 4 hours, about 3.5 hours.

 4  1 3   4  1 3  3 4 π (1) + 9  π    + 92  π    +   3 3  3 9  3     2

∞ 4 4 1 4 1 4 1 π + π   + π   + =  π   . 3 3 3 3 3 n=0 3 3

Section 9.3

110. (a) 8000 + 0.9 ( 8000 ) +  + ( 0.9 )

n −1

(8000 ) =  8000 ( 0.9 ) i =0

(b) After 10 years:

After 15 years: 14  1 − 0.9 14 i = 14, N15 =  8000 ( 0.9 ) = 8000   1 − 0.9 i =0  ≈ 63,529 units

  

After 20 years:  1 − 0.9 19 i = 19, N 20 =  8000 ( 0.9 ) = 8000   1 − 0.9 i =0  ≈ 70,274 units 19

∞

i =0

  

8000 = 80,000 1 − 0.9

If this trend continues indefinitely, the number of units will be 80,000. 111. The ball falls 6 feet, then rebounds to a height of

( 6 ) = 4.5 feet, then falls 4.5 feet, then rebounds 3 4.5) = 3.375 feet, and so on. The terms of this 4( 3 4

geometric series can be written as 2

∞ 3 3 3 9 ) +   ( 9 ) +  or 6 +  9   ( 4 n =1  4  4

So, the sum is 6 +

x 2

115. a1 = 3, r =

a4 =

3x 2  x  3x3  = 4 2 8

a5 =

3x3  x  3x 4  = 8  2  16

1 4 1 7x a2 = (7 x) = 4 4 7x 72 x2 49 x 2 a3 = = (7 x) = 4 4 4 73 x 3 343 x3 a4 = = 4 4 74 x 4 2401x 4 a5 = = 4 4

116. a1 =

117. a1 = 100, r = e x , n = 9

an = a1r n −1

( ) = 100e

a9 = 100 e x

6 + ( 4.5)( 2 ) + ( 3.375)( 2 ) + or 6+9+

821

 x  3x a2 = 3   = 2 2 3x  x  3x2 a3 =  = 2 2 4

9  1 − 0.9 10  9 i = 9, N10 =  8000 ( 0.9 ) = 8000   1 − 0.9  i =0   ≈ 52,106 units

Geometric Sequences and Series

n −1

9 = 42 feet. 1 − 34

(Note: Other possible solutions close yield 42 feet as an answer.) 112. False. If the sequence is geometric, then a7 = a5r 2 and if a7 = 16 then r 2 = 16 and r = ±4. a5

118. a1 = 4, r =

4x , n=6 3

an = a1r n −1 5

4096 5  4x  a6 = 4  x  = 243  3   1 − 0.5 x  119. (a) f ( x ) = 6    1 − 0.5  ∞

1

 6  2  = 1 − 1 2 = 12 n=0

113. True. If the sequence is geometric and the common ratio is r = 1, then all terms are equal to

a1 , an = a1r n −1  an = a1. If the sequence is arithmetic with common difference d = 0, then all terms are equal

to a1, an = a1 + ( n − 1) d  an = a1. Therefore both sequences have the same terms, an = a1. 114. False. You multiply the first term by the common ratio raised to the ( n − 1)th power.

The horizontal asymptote of f(x) is y = 12. This corresponds to the sum of the series. 28

−3

− 20

822

Chapter 9

(b)

Sequences, Series, and Probability n −1

 1 − 0.8 x  f ( x) = 2    1 − 0.8 

 4 122. (a) an = 20  is represented by graph (ii) because  3 the values of each term are increasing.

2 4 2  = = 10  1− 4 5 n=0  5  ∞

n −1

 3 an = 20  is represented by graph (i) because  4 the values of each term are decreasing.

The horizontal asymptote of f(x) is y = 10. This corresponds to the sum of the series. 14

(b) Because r =

−6

 3 an = 20   4

−6

120. Given a real number r between –1 and 1,

an = an −1r < an −1 , which shows that the terms decrease. 121. To use the first two terms of a geometric series to find the nth term, first divide the second term by the first term, to obtain the common ratio. The nth term is the first term multiplied by the common ratio raised to the

( n − 1) th power.

3 < 1, the terms of the sequence 4

n −1

can be summed.

 −1 3 4  123. det  −2 8 0  = 4 ( −10 − 0 ) − 1( −8 + 6 )  0 5 −1 = −40 + 2 = −38

 −1 0 4    124. det  −4 3 5 = −1( −9 − 10 ) + 4 ( −8 − 0 )  0 2 −3 = 19 − 32 = −13

a2 , an = a1r n −1 a1

125. Answers will vary. (Make a Decision)

Section 9.4 The Binomial Theorem 1.

n Cr or   r

10. 18 C2 =

2. expanding, binomial 3. Both the Binomial Theorem and Pascal’s Triangle can be used to find binomial coefficients.

(

)

4. The sum of the powers of the third term 3x 2 y in the

expression ( x + y ) is 3. 3

C5 =

6. 8 C6 =

7! 7 ⋅ 6 ⋅ 5! 42 = = = 21 2!5! 2 ⋅ 5! 2 8! 8⋅7 = = 28 2!6! 2 ⋅ 1

C15 =

20! 20 ⋅ 19 ⋅ 18 ⋅ 17 ⋅ 16 = = 15,504 15!5! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1

19 C12

9. 14 C1 =

19! 19 ⋅ 18 ⋅ 17 ⋅ 16 ⋅ 15 ⋅ 14 ⋅ 13 = = 50,388 7!12! 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1

14! 14 ⋅ 13! = = 14 13!1! 13!

18! 18 ⋅ 17 = = 153 16!2! 2 ⋅1

 12  12! =1 11.   = 12 C0 = 0!12! 0

 20  20! =1 12.   = 20 C20 = 20!0!  20   10  10! 10 ⋅ 9 ⋅ 8 ⋅ 7 = = 210 13.   = 10 C4 = 6!4! 4 ⋅ 3 ⋅ 2 ⋅ 1 4

 10  10! 10 ⋅ 9 ⋅ 8 ⋅ 7 = = 210 14.   = 10 C6 = 6 4!6! 4 ⋅ 3 ⋅ 2 ⋅1   100  100! 100 ⋅ 99 = = 4950 15.   = 100 C98 = 98 98!2! 2 ⋅1  

 100  100! 100 ⋅ 99 = = 4950 16.   = 100 C2 = 2 98!2! 2 ⋅1   17.

C36 = 749,398

18. 34 C 4 = 46,376

Section 9.4

C46 = 230,300

21.

250

C498 = 124,750

22.

1000

19.

20.

500

823

C2 = 31,125 C2 = 499,500

23. ( x + 1) = 5C0 x 5 + 5C1 x 4 (1) + 5C2 x 3 (1) + 5C3 x 2 (1) + 5C4 x(1) + 5C5 x(1) 5

The Binomial Theorem

= 1 ⋅ x 5 + 5 ⋅ x 4 + 10 ⋅ x 3 + 10 ⋅ x 2 + 5 ⋅ x + 1 = x 5 + 5 x 4 + 10 x 3 + 10 x 2 + 5 x + 1

24.

( x + 1) = 6 C0 x 6 + 6 C1 x 5 (1) + 6 C2 x 4 (1) + 6 C3 x3 (1) 4 5 6 + 6 C 4 x 2 (1) + 6 C5 x (1) + 6 C6 (1) 6

26.

25.

27.

( a + 3) = 3 C0 a + 3 C1a ( 3 ) + 3 C2a ( 3) + 3 C3 ( 3 ) 2 3 = a 3 + 3a 2 ( 3 ) + 3a ( 3 ) + ( 3 ) 3

= a4 + 8a3 + 24a2 + 32a + 16

= x 6 + 6 x 5 + 15 x 4 + 20 x 3 + 15 x 2 + 6 x + 1 3

( a + 2) = 4C0a4 + 4C1a3 ( 2) + 4C2a2 ( 2) + 4C3a( 2) + 4C4 ( 2)

( y − 4 ) = 3 C0 y 3 − 3 C1 y2 ( 4 ) + 3 C2 y ( 4 ) − 3 C3 ( 4 ) . 3

= y3 − 12 y 2 + 48 y − 64

= a 3 + 9a 2 + 27a + 27

28. ( y − 5)4 = 4C0 y 4 − 4C1 y 3 (5) + 4C2 y 2 (5)2 − 4C3 y (5)3 + 4C4 (5)4 = 1 ⋅ y 4 − 4 ⋅ y 3 ⋅ 5 + 6 ⋅ y 2 ⋅ 25 − 4 ⋅ y ⋅ 125 + 1 ⋅ 625 = y 4 − 20 y 3 + 150 y 2 − 500 y + 625

29.

( x + y ) = 5 C0 x 5 + 5 C1 x 4 y + 5 C2 x 3 y 2 + 5 C3x 2 y3 5

+ 5 C 4 xy 4 + 5 C5 y 5 = x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 xy 4 + y 5

30.

( x + y ) = 6 C0 x 6 + 6 C1x 5y + 6 C2 x 4 y2 + 6 C3 x 3 y3 + 6 C4 x 2 y4 + 6 C5xy5 + 6 C6 y6 6

= x 6 + 6 x 5 y + 15 x 4 y 2 + 20 x 3 y 3 + 15 x 2 y 4 + 6 xy 5 + y 6 31.

( 2x − y) = 5C0 ( 2x) − 5C1 ( 2x) y + 5C2 ( 2x) y2 − 5C3 ( 2x) y3 + 5C4 ( 2x) y4 − 5C5y5 = 32 x 5 − 5 (16 x 4 ) y + 10 ( 8 x 3 ) y 2 − 10 ( 4 x 2 ) y 3 + 5 ( 2 x ) y 4 − y 5 5

= 32 x 5 − 80 x 4 y + 80 x 3 y 2 − 40 x 2 y 3 + 10 xy 4 − y 5 32. (5 x − y ) = 4C0 (5 x) − 4C1 (5 x) y + 4C2 (5 x) y 2 − 4C3 (5 x) y 3 + 4C4 y 4 4

= 625 x 4 − 4(125 x3 ) y + 6( 25 x 2 ) y 2 − 4(5 x) y 3 + y 4 = 625 x 4 − 500 x3 y + 150 x 2 y 2 − 20 xy 3 + y 4 33. ( 4 y − 3) = 3C0 ( 4 y ) − 3C1 ( 4 y ) (3) + 3C2 ( 4 y )(3) = 3C3 (3) 3

= 64 y 3 − 3(16 y 2 )(3) + 3( 4 y )(9) − 27 = 64 y 3 − 144 y 2 + 108 y − 27 34. ( 2 y − 5) = 3C0 ( 2 y ) − 3C1 ( 2 y ) (5) + 3C2 ( 2 y )(5) − 3C3 (5) 3

= 8 y 3 − 3( 4 y 2 )(5) + 3( 2 y )( 25) − 125 = 8 y 3 − 60 y 2 + 150 y − 125

824

Chapter 9

Sequences, Series, and Probability

35. ( 2r + 3s ) = 6C0 ( 2r ) − 6C1 ( 2r ) (3s ) + 6C2 ( 2r ) (3s ) − 6C3 ( 2r ) (3s ) + 6C4 ( 2r ) (3s ) − 6C5 ( 2r ) (3s ) + 6C6 (3s ) 6

= ( 2r ) − 6( 2r ) (3s ) + 15( 2r ) (3s ) − 20( 2r ) (3s ) + 15( 2r ) (3s ) − 6( 2r ) (3s ) + (3s ) 6

= 64r 6 − 576r 5 s + 2160r 4 s 2 − 4320r 3 s 3 + 4860r 2 s 4 − 2916rs 5 + 729 s 6

36. ( 4 x − 3 y ) = 4C0 ( 4 x) − 4C1 ( 4 x) 3 y + 4C2 ( 4 x) (3 y ) − 4C3 ( 4 x)(3 y ) + 4C4 (3 y ) 4

= ( 4 x) − 4( 4 x) (3 y ) + 6( 4 x) (3 y ) − 4( 4 x)(3 y ) + 1(3 y ) 4

= 256 x 4 − 768 x3 y + 864 x 2 y 2 − 432 xy 3 + 81y 4 37.

( x + 2) = C ( x ) + C ( x ) (2) + C ( x ) 2 + C ( x ) 2 + C (2) 4

= x8 + 8 x 6 + 24 x 4 + 32 x 2 + 16 38. ( y 2 + 2) = 6C0 ( y 2 ) + 6C1 ( y 2 ) ( 2) + 6C2 ( y 2 ) ( 2) + 6C3 ( y 2 ) ( 2) + 6C4 ( y 2 ) ( 2) + 6C5 ( y 2 )( 2) + 6C6 ( 2) 6

= ( y 2 ) + 6( y 2 ) 2 + 15( y 2 ) 22 + 20( y 2 ) 23 + 15( y 2 ) 2 4 + 6( y 2 )25 + 26 6

= y12 + 12 y10 + 60 y 8 + 160 y 6 + 240 y 4 + 192 y 2 + 64

39. (5 − x 2 ) = 5C0 (5) − 5C1 (5) ( x 2 ) + 5C2 (5) ( x 2 ) − 5C3 (5) ( x 2 ) + 5C4 (5)( x 2 ) − 5C5 ( x 2 ) 5

= 3125 − 5(625)( x 2 ) + 10(125) x 4 − 10( 25) x 6 + 5(5) x8 − x10 = 3125 − 3125 x 2 + 1250 x 4 − 250 x 6 + 25 x8 − x10 = − x10 + 25 x8 − 250 x 6 + 1250 x 4 − 3125 x 2 + 3125

40.

( 3 − y ) = C ( 3) − C ( 3) y + C ( 3) ( y ) − C ( y ) 3

= 27 − 27 y 2 + 9 y 4 − y 6 = − y 6 + 9 y 4 − 27 y 2 + 27 41.

( x + y ) = C ( x ) + C ( x ) ( y ) + C ( x ) ( y ) + C ( x )( y ) + C ( y ) 2

= x 8 + 4 x 6 y 2 + 6 x 4 y 4 + 4 x 2 y 6 + y8 42.

( x +y ) = C ( x ) + C ( x ) ( y ) + C ( x ) ( y ) + C ( x ) ( y ) + C ( x ) ( y ) + C ( x )( y ) + C ( y ) 2

6 0

6 1

6 2

6 3

6 4

6 5

6 6

= x 12 + 6 x 10 y 2 + 15 x 8 y 4 + 20 x 6 y 6 + 15 x 4 y 8 + 6 x 2 y 10 + y 12 43. (3x3 − y ) = 6C0 (3 x3 ) − 6C1 (3 x3 ) y + 6C2 (3x3 ) y 2 − 6C3 (3x3 ) y 3 + 6C4 (3x3 ) y 4 − 6C5 (3 x3 ) y 5 + 6C6 y 6 6

= (3x3 ) − (6)(3 x3 ) y + (15)(3 x3 ) y 2 − ( 20)(3 x3 ) y 3 + (15)(3 x3 ) y 4 − (6)(3x3 ) y 5 + y 6 6

= 729 x18 − (6)( 243x15 ) y + (15)(81x12 ) y 2 − ( 20)( 27 x9 ) y 3 + (15)(9 x 6 ) y 4 − (6)(3x3 ) y 5 + y 6 = 729 x18 − 1458 x15 y + 1215 x12 y 2 − 540 x9 y 3 + 135 x 6 y 4 − 18 x3 y 5 + y 6 44.

(2 x − y) = C (2 x ) y − C (2 x ) y + C (2 x ) y + C (2 x ) y + C (2 x ) y − C y 3

= 32 x 5 − 80 x12 y + 80 x 9 y 2 − 40 x 6 y3 + 10 x 3 y 4 − y 5

Section 9.4 5

The Binomial Theorem

825

1  1 1 1 1 1 45.  + y  = 5C0   + 5C1   y + 5C2   y2 + 5C3   y3 + 5C4   y4 + 5C5y5 x x x x x           x 1 5 y 10 y 2 10 y 3 5 y 4 + + 3 + 2 + + y5 x5 x4 x x x

= 6

2 4 6 1  1 1 1 1 1 1 5 46.  + y  = 6C0   + 6C1   ( y ) + 6C2   ( y ) + 6C3   y 3 + 6C4   ( y ) + 6C5  ( y ) + 6C6 ( y ) x   x  x  x  x  x  x 6

( )

1 1 1 1 1 1 =   + 6  y + 15  y 2 + 20  y 3 + 15  y 4 + 6  y 5 + y 6  x  x  x  x  x  x =

6 y5 1 6 y 15 y 2 20 y 3 15 y 4 + + + y6 + 5 + + 6 4 3 2 x x x x x x

2 2 2 2 2 2 3 4 47.  − 2 y  = 4C0   − 4C1   ( 2 y ) + 4C2   ( 2 y ) + 4C3  ( 2 y ) + 4C4 ( 2 y ) x   x  x  x  x 4

2 8  4 2 =   − 4 3 ( 2 y ) + 6 2  4 y 2 + 4  8 y 3 + 16 y 4 x x x        x =

( )

( ) (

)

16 64 y 96 y 2 64 y 3 − 3 + − + 16 y 4 4 2 x x x x

2 3 4 5 2  2 2 2 2 2 48.  − 3y  = 5 C0   − 5 C1   ( 3y ) + 5 C2   ( 3y ) − 5 C3   ( 3y ) + 5 C4   ( 3y ) − 5 C5 ( 3y ) x x x x x x             32 240 720 2 1080 3 810 4 5 = 5 − 4 y+ 3 y − 2 y + y − 243 y x x x x x

49.

( 4 x − 1) − 2 ( 4 x − 1) = ( 64 x 3 − 48 x 2 + 12 x − 1) − 2 ( 256 x 4 − 256 x 3 + 96 x 2 − 16 x + 1) 3

= −512 x 4 + 576 x 3 − 240 x 2 + 44 x − 3

50.

( x + 3) − 4( x + 3) = ( x5 + 15x4 + 90x3 + 270x2 + 405x + 243) − 4( x4 + 12x3 + 54x2 + 108x + 81) 5

= x5 + 11x4 + 42x3 + 54x2 − 27x − 81

(

) (

51. 3 ( x + 1) + 4 ( x + 1) = 3 x 5 + 15 x 4 + 30 x 3 + 30 x 2 + 15 x + 3 + 4 x 3 + 12 x 2 + 12 x + 4 5

)

= 3 x + 15 x + 34 x + 42 x + 27 x + 7

( )

( )( )

( )

4 2 52. 2 ( x − 3) + 5( x − 3) = 2  x4 − 4 x3 ( 3) + 6 x2 32 − 4 ( x ) + 33 + 34  + 5  x2 − 2 ( x )( 3) + 32   

(

) (

= 2 x4 − 12 x3 + 54 x2 − 108x + 81 + 5 x2 − 6 x + 9 4

)

= 2 x − 24 x + 113x − 246 x + 207

(

) (

53. −3( x − 2) − 4 ( x + 1) = −3x3 + 18x2 − 36 x + 24 − 4 x6 + 24 x5 + 60 x 4 + 80 x3 + 60 x2 + 24 x + 4 3

)

= −4 x − 24 x − 60 x − 83x − 42 x − 60 x + 20

(

)

( ) = − (5 x5 + 50 x 4 + 200 x3 + 400 x 2 + 400 x + 160) + ( 2 x 2 − 4 x + 2) = −(5 x5 + 50 x 4 + 200 x3 + 402 x 2 + 396 x + 162)

54. − 5( x + 2) − 2( x − 1) = − 5 x5 + 50 x 4 + 200 x3 + 400 x 2 + 400 x + 160 − 2 x 2 − 2 x + 1 5

= − 5 x5 − 50 x 4 − 200 x3 − 402 x 2 − 396 x − 162 55.

( x + 8) , n = 4 3 10 − 3 (8 ) = 120 x 7 ( 512 ) = 61,440 x 7 10 C3 x 10

56.

( x + 6)6 , n = 7 6 C6 x

(6)

= (1)46,656 = 46,656

826

Chapter 9

57.

( x − 6y) , n = 3 2 5−2 ( −6 y ) = 10 x 3 ( 36 ) y2 = 360 x 3 y2 5 C2 x

58.

59.

Sequences, Series, and Probability 66. The term involving x 2 y8 in the expansion of ( 4x − y )

( x − 10 z ) , n = 4 3 7 −3 ( −10 z ) = −35,000 x 4 z3 7 C3 x 7

(10 − 8 )!8!

⋅ 16 x 2 y 8 = 720 x 2 y 8 .

67. The term involving x6 y3 in the expansion of

( 3x − 2 y ) is 6 3 6 3 6 3 9 C3 ( 3 x ) ( −2 y ) = 84 ( 3 ) ( −2 ) x y 9

( 4 x + 3y ) , n = 8 9−7 7 ( 3 y ) = 36 (16 ) x 2 ( 37 ) y7 9 C7 ( 4 x ) 9

= −489,888 x 6 y 3 . The coefficient is –489,888.

( 5a + 6 b ) , n = 5 5− 4 4 ( 6b ) = 32,400 a ⋅ b4 5 C 4 ( 5a ) 5

68. The term involving x 4 y 4 in the expansion of ( 2 x − 3 y )

(10 x − 3 y ) , n = 10 12 − 9 9 3 9 ( −3 y ) = 220 (10 ) ( −3) x 3 y9 12 C9 (10 x ) 12

( )

C 4 ( 2 x ) ( −3 y ) = 70 2 4 ( −3 ) x 4 y 4 4

61.

10!

The coefficient is 720.

= 1,259,712 x 2 y 7 60.

C8 ( 4 x ) ( − y ) = 2

The coefficient is 90,720.

(

69. The term involving x8 y6 in the expansion of x 2 + y

(7x − 2 y ) , n = 7 15 − 6 6 9 6 ( − 2 y ) = 5005 ( 7 ) ( − 2 ) x9 y 6 15 C6 ( 7 x ) 15

= 12,926,067,394, 240 x9 y 6 63.

) is 10

( ) y = 210 x y . The coefficient is 210.

C6 x 2

8 6

70. The term involving z14 y5 in the expansion of ( z 2 + y ) is 12

( x + 3) , ax 5 7 5 5 12 C7 x ( 3 ) = 1,732,104 x 12

12 C5

( z 2 ) y 5 = 792 z14 y 5. The coefficient is 792. 7

71. 4th entry of 7th row: 7 C4 = 35

a = 1,732,104 64.

= 90,720 x 4 y 4 .

= −4,330,260,000 x 3 y 9 62.

72. 3rd entry of 6th row: 6 C3 = 20

( x + 4 ) , ax 4 8 4 4 12 C8 x ( 4 ) = 32,440,320 x 12

73. 5th entry of 6th row: 6 C5 = 6

a = 32,440,320

74. 2nd entry of 5th row: 5 C2 = 10

65. The term involving x y in the expansion of ( x − 2 y )

8 2

10! ⋅ 4 x 8 y 2 = 180 x 8 y 2 . 2!8! The coefficient is 180. C2 x 8 ( −2 y ) = 2

75. 5th row of Pascal’s Triangle: 1 5 10 10 5 1

( 5 y + 2 ) = 1( 5 y ) + 5 ( 5 y ) 2 + 10 ( 5 y ) 22 + 10 ( 5 y ) 23 + 5 ( 5 y ) 24 + 25 5

= 3125 y 5 + 6250 y 4 + 5000 y 3 + 2000 y 2 + 400 y + 32

76. 6th row of Pascal’s Triangle: 1 6 15 20 15 6 1 6

( 2 v + 3 ) = 1 ( 2 v ) + 6 ( 2 v ) (3 ) + 15 ( 2 v ) (3 ) + 20 ( 2 v ) (3 ) 2 4 5 6 + 15 ( 2 v ) ( 3 ) + 6 ( 2 v )( 3 ) + ( 3 )

= 64 v 6 + 576 v 5 + 2160 v 4 + 4320 v 3 + 4860 v 2 + 2916 v + 729

Section 9.4

The Binomial Theorem

827

77. 5th row of Pascal’s Triangle: 1 5 10 10 5 1

(2 x + 3 y ) = 1(2 x) + 5(2 x) (3 y ) + 10(2 x) (3 y ) + 10(2 x) (3 y ) + 5(2 x)(3 y ) + (3 y ) 5

= 32 x 5 + 240 x 4 y + 720 x 3 y 2 + 1080 x 2 y 3 + 810 xy 4 + 243 y 5

78. 5th row of Pascal’s Triangle: 1 5 10 10 5 1

(3x + 4 y ) = 1(3x) + 5(3x) (4 y ) + 10(3x) (4 y ) + 10(3x) (4 y ) + 5(3x)( 4 y ) + 1( 4 y ) 5

= 243x 5 + 1620 x 4 y + 4320 x3 y 2 + 5760 x 2 y 3 + 3840 xy 4 + 1024 y 5

79. 4th row of Pascal’s Triangle: 1 4 6 4 1

( 3t − 2v ) = 1( 3t ) − 4 ( 3t ) ( 2v ) + 6 ( 3t ) ( 2v ) − 4 ( 3t )( 2v ) + 1( 2v ) 4

= 81t 4 − 216t 3v + 216t 2 v2 − 96tv3 + 16v 4 80. 4th row of Pascal’s Triangle: 1 4 6 4 1

( 5v − 2z ) = 1( 5v ) − 4 ( 5v ) ( 2z ) + 6 ( 5v ) ( 2z ) − 4 ( 5v )( 2z ) + 1( 2z ) 4

= 625v 4 − 1000v3 z + 600v2 z2 − 160vz3 + 16z 4 81.

(3 x + 5) = (3 x ) + 3(3 x ) (5) + 3(3 x )(5) + (5) 3

= 27 x 82.

x + 135 x + 225 x + 125

(2 t − 7) = (2 t ) − 3(2 t ) (7) + 3(2 t )(7) − (7) 3

= 8t t − 84t + 294 t − 343 83.

( x − y ) = ( x ) − 3 ( x ) ( y ) + 3 ( x )( y ) − ( y ) 23

= x 2 − 3 x 4 3 y1 3 + 3 x 2 3 y 2 3 − y 84.

85.

( u + v ) = u + 5u v + 10u v + 10u v + 5u v + v 35

f ( x + h) − f ( x ) h

12 5 1 5

95 25

65 35

35 45

( x + h ) − x3 3

h x 3 + 3 x 2h + 3 xh2 + h3 − x 3 = h

( =

h 3 x 2 + 3 xh + h 2

87.

f ( x +h) − f ( x) h

f ( x + h) − f ( x )

)

( x + h) − x4

h x 4 + 4 x 3h + 6 x 2h2 + 4 xh3 + h 4 − x 4 = h =

(

h 4 x 3 + 6 x 2 h + 4 xh2 + h3

4 2

3 3

2 4

(

h 6x5 +15x4h +20x3h2 +15x2h3 +6xh4 + h5

)

h = 6x5 +15x4h +20x3h2 +15x2h3 +6xh4 +h5, h ≠ 0

( x +6x h +15x h +20x h +15x h +6xh +h ) − x =

h = 3 x 2 + 3 xh + h2 , h ≠ 0 86.

( x +h) − x6 6

)

h = 4 x 3 + 6 x 2h + 4 xh 2 + h3 , h ≠ 0

88.

f ( x + h) − f ( x) h

( x + h ) − x8 8

h  x8 + 8 x 7 h + 28 x 6 h 2 + 56 x5 h3  8  − x 4 4 3 5 2 6 7 8 +70 x h + 56 x h + 28 x h + 8 xh + h  = h  8 x 7 + 28 x 6 h + 56 x 5 h 2  h  +70 x 4 h3 + 56 x 3h 4 + 28 x 2 h5 + 8 xh 6 + h7   =  h = 8 x 7 + 28 x 6 h + 56 x 5 h 2 + 70 x 4 h3 +56 x 3h 4 + 28 x 2 h5 + 8 xh6 + h7 , h≠0

828

Chapter 9

Sequences, Series, and Probability

89.

(1 + i ) =4 C014 +4 C1 (1) i +4 C2 (1) i2 +4 C31 ⋅ i3 +4 C4i4 4

= 1 + 4i − 6 − 4i + 1 = −4 90. (1 − i ) = 6C0 (1) − 6C1 (1) (i ) + 6C2 (1) (i ) − 6C3 (1) (i ) + 6C4 (1) (i ) − 6C5 (1)(i ) + 6C6 (i) 6

= 1 − 6i + 15i 2 − 20i 3 + 15i 4 − 6i 5 + i 6 = 1 − 6i − 15 + 20i + 15 − 6i − 1 = 8i

( )

( )( )

91. ( 4 + i ) = 4 C0 ( 4 ) + 4 C1 43 i + 4 C2 4 2 i 2 4

( )+ C i

+ 4 C3 ( 4 ) i

= 256 + 256i − 96 − 16i + 1 = 161 + 240i

92. ( 2 + i ) = 5C0 ( 2) + 5C1 ( 2) i + 5C2 ( 2) i 2 + 5C3 ( 2) i 3 + 5C4 ( 2)i 4 + 5C5i 5 5

= 32 + 80i − 80 − 40i + 10 + i = − 38 + 41i 93. ( 2 − 3i ) =6 C0 26 −6 C1 25 ( 3i ) +6 C2 24 ( 3i ) −6 C3 23 ( 3i ) +6 C4 22 ( 3i ) 6

−6 C5 2 ( 3i ) +6 C6 ( 3i ) 5

= 64 − 576i − 2160 + 4320i + 4860 − 2916i − 729 = 2035 + 828i 94. ( 3 − 2 i ) = 6 C 0 3 6 − 6 C 1 3 5 ( 2 i ) + 6 C 2 3 4 ( 2 i ) − 6 C 3 3 3 ( 2 i ) 6

+ 6 C 4 32 (2i ) − 6 C 5 3 (2i ) + 6 C 6 (2i ) 4

= 729 − 2916 i − 4860 + 4320 i + 2160 − 576 i − 64 = − 2035 + 828 i

95.

( 5 + −16 ) = ( 5 + 4i ) 3

98.

=3 C0 53 + 3 C1 52 ( 4i ) + 3 C2 5 ( 4i ) + 3 C3 ( 4i ) 2

= 125 + 300i − 240 − 64i = −115 + 236i 96.

( 5 + −9 ) = ( 5 + 3i ) 3

( 4 + 3i ) = C 4 + C 4 3i + C 4 ( 3i ) + C 4 ( 3i ) + C ( 3i ) 4

= −10 + 198i 0

= 184 − 440 3i

= 125 + 225i − 135 − 27i

=3 C0 53 + 3 C1 52 ( 3i ) + 3 C2 5 ( 3i ) + 3 C3 ( 3i )

= 625 − 500 3i − 450 + 60 3i + 9

97.

( 5 − 3i ) = C 5 − C 5 ( 3i ) + C 5 ( 3i ) − C 5 ( 3i ) + C ( 3i )

= 256 + 256 3i − 288 − 48 3i + 9

3  1 3  1 99.  − + i  = −1 + 3 i  2 2  8 

(

)

( ) ( )

 C −1 3 + C −1 2 3i  1 3 0 ( ) 3 1( )  =  2 3 8 + C −1 3i + C 3i   3 2( ) 3 3   1 = −1 + 3 3i + 9 − 3 3i   8

( )

= −23 + 208 3i

Section 9.4 3

3  1 3  1 i  = −1 + 3i 100.  − + 3  27  3 1 3 2 = 3i + 3C2 ( −1) 3 C0 ( −1) + 3C1 ( −1) 27  1 −1 + 3 3i + 9 − 3 3i  =  27  8 = 27

(

The Binomial Theorem

829

)

(

)

( 3i) + C ( 3i)  2

3 3

3 3 2 1  3  1 3 3 9 3 3  1  3  =− 101.  1 − 3 i  = C  1  − C  1   3 i  + 3 C2    − i− + i  − 3 C3  i  =    3 0   3 1     8 64 64 64 64 3 3 3       3  3  3  4 4 

3  3  1 3 3 9 3 3 3 2  1  3  i− + i = −1 i  − 3 C3  i = − 102.  1 − 3 i  = C  1  − C  1   3 i  + 3 C2      8 8 8 2  8  2  2   2 2  3 0  2  3 1  2   2      

103. (1.02 ) = (1 + 0.02 ) 8

= 1 + 8 ( 0.02 ) + 28 ( 0.02 ) + 56 ( 0.02 ) + 70 ( 0.02 ) + 56 ( 0.02 ) + 28 ( 0.02 ) + 8 ( 0.02 ) + ( 0.02 ) 2

= 1 + 0.16 + 0.0112 + 0.000448 +  ≈ 1.172 104. ( 2.005)

= ( 2 + 0.005)

= 210 + 10 ( 2 ) ( 0.005) + 45 ( 2 ) ( 0.005) + 120 ( 2 ) ( 0.005) + 210 ( 2 ) ( 0.005) + 252 ( 2 ) ( 0.005) + 210 ( 2 ) ( 0.005) 9

+ 120 ( 2 ) ( 0.005) + 45 ( 2 ) ( 0.005) + 10 ( 2 )( 0.005) + ( 0.005) 3

= 1024 + 25.6 + 0.288 + 0.00192 + 0.0000084 +  ≈ 1049.890

105. ( 2.99 )

= ( 3 − 0.01)

= 312 − 12 ( 3) ( 0.01) + 66 ( 3) 11

( 0.01) − 220 ( 3) ( 0.01)3 + 495 ( 3)8 ( 0.01)4 − 792 ( 3)7 ( 0.01)5 + 924 ( 3)6 ( 0.01)6 2

−792 ( 3) ( 0.01) + 495 ( 3) ( 0.01) − 220 ( 3) ( 0.01) + 66 ( 3) ( 0.01) − 12 ( 3)( 0.01) + ( 0.01) 5

≈ 510,568.785

106. (1.98) = ( 2 − 0.02) = 29 − 9 ( 2) ( 0.02) + 36 ( 2 ) ( 0.02 ) − 84 ( 2 ) ( 0.02 ) + 126 ( 2 ) ( 0.02 ) − 126 ( 2 ) ( 0.02 ) 9

+ 84 ( 2 ) ( 0.02 ) − 36 ( 2 ) ( 0.02 ) + 9 ( 2 )( 0.02) − ( 0.02 ) 3

= 512 − 46.08 + 1.8432 − 0.043008 + 0.00064512 ≈ 467.721

830

Chapter 9

Sequences, Series, and Probability

107. g is shifted two units to the right of f.

110. g is shifted three units to the right of f.

2 −7

f −4

−8

−3

f ( x ) = −x4 + 4 x2 − 1

f ( x) = x 4 − 5 x 2

g ( x ) = f ( x − 3)

g ( x ) = f ( x − 2) = ( x − 2) − 5( x − 2) 4

= − ( x − 3) + 4 ( x − 3) − 1 4

= − x 4 + 12 x 3 − 54 x 2 + 108 x − 81 + 4 x 2 − 24 x + 36 − 1

= x 4 − 8 x3 + 24 x 2 − 32 x + 16 − 5 x 2 + 20 x − 20

= − x 4 + 12 x 3 − 50 x 2 + 84 x − 46

= x 4 − 8 x3 + 19 x 2 − 12 x − 4 108. g is shifted four units to the left of f. 4

111. f ( x ) = (1 − x )

g ( x ) = 1 − 3x

−8

h ( x ) = 1 − 3x + 3x2

p ( x ) = 1 − 3x + 3x2 − x3 = f ( x )

Since p( x ) is the expansion of f ( x ), they have the same graph.

−4

f ( x ) = x3 − 4 x

g ( x ) = f ( x + 4)

= ( x + 4) − 4 ( x + 4)

−6

f=p

= x 3 + 12 x 2 + 48 x + 64 − 4 x − 16 −3

= x 3 + 12 x 2 + 44 x + 48 109. g is shifted five units to the left of f.

112. f ( x ) = (1 − 12 x )

g ( x) = 1 − 2x +

− 10

−6

f ( x) = − x3 + 3 x 2 − 4 g ( x) = f ( x + 5) = −( x + 5) + 3( x + 5) − 4 3

= − x 3 − 15 x 2 − 75 x − 125 + 3 x 2 + 30 x + 75 − 4 3

3 2 x 2 3 1 h ( x ) = 1 − 2 x + x2 − x3 2 2 3 2 1 3 1 4 p( x) = 1 − 2x + x − x + x = f ( x) 2 2 16 Since p( x ) is the expansion of f ( x ), they have the same graph.

= − x − 12 x − 45 x − 54

−4

f=p 8

h −3

Section 9.4

113.

1 1  1  1  C4     = 35    ≈ 0.273 2 2      16  8 

1 3  1   2187  114. 10 C3     = 120     ≈ 0.2503 4 4  64  16,384  4

1 2  1  16  115. 8 C4     = 70    ≈ 0.171 3 3  81  81  1 1  1  1  116. 8 C4     = 70    ≈ 0.2734 2 2  16  16 

7th row: 1 7 21 35 35 21 7 1

1 9 36 84 126 126 84 36 9 1 9th row: 10th row: 1 10 45 120 210 252 210 120 45 10 1 122. (a) The number of terms in the expansion is one greater than the power.

(b) There are n + 1 terms in the expansion of ( x + y ) . n

123.

n!  n − ( n − r )  !( n − r )!   n! n! = = = n Cr r !( n − r )! ( n − r )!r !

Cn − r =

= 0.005t 2 + 0.1t + 0.5 − 0.37t − 3.7 + 25.7 = 0.005t 2 − 0.27t + 22.5 (b)

124. 0 = (1 − 1)

= n C0 − n C1 + n C2 − n C3 +  ( ± n C n )

g 0

1 8 28 56 70 56 28 8 1

8th row:

g (t ) = f (t + 10) 2

118. (a) f (t ) = − 0.039t 2 + 1.30t + 17.7

125.

126.

= − 0.039t 2 − 0.39t − 0.975 + 1.3t + 6.5 + 17.7 = − 0.039t 2 + 0.91t + 23.225 g f

119. True. The coefficients from the Binomial Theorem can be used to find the numbers in Pascal’s Triangle. 120. False. Expanding binomials that represent differences is just as accurate as expanding binomials that represent sums, but for differences the coefficient signs are alternating.

( n − r )!r ! ( n − r + 1)!( r − 1)! n!( n − r + 1) n! r = + − − + − + n r ! r ! n r 1 n r 1 ( ) ( ) ( )!( r − 1)!r n!( n − r + 1) n! r = + ( n − r + 1)!r ! ( n − r + 1)!r ! n! ( n − r + 1 + r ) = ( n − r + 1)!r ! n!( n + 1) = ( n − r + 1)!r ! ( n + 1)! = C = ( n + 1 − r )!r ! n +1 r

= − 0.039(t + 5) + 1.30(t + 5) + 17.7

C r + n C r −1 =

g (t ) = f (t + 5)

(b)

831

121. The first and last numbers in each row are 1. Every other number in each row is found by adding the two numbers immediately above the number. In order to find rows 8– 10 of Pascal’s Triangle, row 7 is used to start:

117. (a) f (t ) = 0.005t 2 − 0.37t + 25.7

= 0.005(t + 10) − 0.37(t + 10) + 25.7

The Binomial Theorem

C0 + n C1 + n C2 + n C3 +  + n Cn = (1 + 1) = 2n n

127. (a) Second term of ( 2 x − 3y ) is 5

5 ( 2 x ) ( −3 y ) = −240 x 4 y. 4

1  (b) Fourth term of  x + 7 y  2 

3 1  C3  x  ( 7 y ) = 857.5 x 3 y 3 . 2  

832

Chapter 9

Sequences, Series, and Probability

 −1 −4  128. A =    1 2  2 4 1 1 2 1 4 − − − ( )( ) ( )( )  −1 −1 2   1 1  2 4   =  = 1 1  2  −1 −1  − −   2 2 

A −1 =

11 −12  A=   2 −2 

129.

A−1 =

 −2 12 

(11)( −2 ) − ( 2 )( −12 )  −2 11  −1 6  1  −2 12    11   = 2  −2 11 −1 2  

Section 9.5 Counting Principles 1. Fundamental Counting Principle 2. distinguishable permutations 3. A permutation is an ordering of n elements. 4. The number of permutations of n elements taken r at a n! time is given by the formula n Pr = . ( n − r )! 5. Odd integers: 1, 3, 5, 7, 9, 11, 13, 15 8 ways 6. Even integers 2, 4, 6, 8, 10, 12, 14, 7 ways 7. Prime integers: 1, 3, 5, 7, 11, 13 6 ways

13. Amplifiers: 4 choices

Stereo Receivers: 10 choices Speakers: 5 choices Total: 4 ⋅ 10 ⋅ 5 = 200 ways 14. Math courses: 2 Science courses: 3 Social sciences and humanities courses: 5 Total: 2 ⋅ 3 ⋅ 5 = 30 ways 15. 210 = 1024 ways 16. Earthworms: 9 Frogs: 4 Fetal pigs: 7 Total: 9 ⋅ 4 ⋅ 7 = 252 ways

8. Integers greater than 11: 12, 13, 14, 15 4 ways

17. (a) (b) (c)

9. Sum is 20: 5 + 15, 6 + 14, 7 + 13, 8 + 12, 9 + 11, 10 + 10, 11 + 9, 12 + 8, 13 + 7, 14 + 6, 15 + 5

9 ⋅ 10 ⋅ 10 = 900 numbers 9 ⋅ 9 ⋅ 8 = 648 numbers 9 ⋅ 10 ⋅ 2 = 180 numbers

18. (a) (b)

4 ⋅ 10 ⋅ 10 ⋅ 10 = 4000 numbers 9 ⋅ 10 ⋅ 10 ⋅ 5 = 4500 numbers

11 ways 10. Sum is 22: 7 + 15, 8 + 14, 9 + 13, 10 + 12, 11 + 11, 12 + 10, 13 + 9,

14 + 8, 15 + 7 9 ways 11. Distinct integers whose sum is 8:

1 + 7, 2 + 6, 3 + 5 3 ways 12. Distinct integers whose sum is 24:

9 + 15, 10 + 14, 11 + 13 3 ways

19. 3 ( 8 ⋅ 10 ⋅ 10 )(10 ⋅ 10 ⋅ 10 ⋅ 10 ) = 24, 000, 000 numbers 20. 26 ⋅ 26 ⋅ 26 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 = 175,760,000 license plates 21. (a) 10 5 = 100,000 zip codes

(b) There are 2 ⋅ 10 4 = 20,000 zip codes beginning with a one or a two. 22. (a) 10 9 = 1,000,000,000 nine-digit zip codes

(b) There are 2 ⋅ 108 = 200,000,000 nine-digit zip codes beginning with a one or a two. 23. (a) 10 4 = 10,000 ATM codes

(b) There are 9 ⋅ 103 = 9000 ATM codes that don’t begin with zero.

Section 9.5

24. (a)

263 + 263 = 35,152

35.

6 P5

36.

(b) There are 2 ⋅ 253 possibilities that don’t have Q. Hence, 2 ⋅ 263 − 2 ⋅ 253 = 3902 have at least one Q. 25. (a) (b) 26. (a)

(b)

6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 720 ways 6 ⋅ 1 ⋅ 4 ⋅ 1 ⋅ 2 ⋅ 1 = 48 ways 8! = 40,320 ways

( 5!)( 3!) = 120 ( 6 ) = 720 ways

P4 =

(6 − 5)! 7!

( 7 − 4 )!

37. 30 P6 =

27. 5! = 120 ways

6! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 720 1!

7! = 7 ⋅ 6 ⋅ 5 ⋅ 4 = 840 3!

30! 30! = = 427,518,000 − 30 6 ! ( ) 24!

P8 = 1,814,400

38.

29. 9! = 362,880 ways

39.

120

P4 = 197,149,680

30. 4! = 24 ways

40.

100

P3 = 970,200

41.

28. 8! = 40,320 ways

31.

P4 =

4! 4! = = 24 4 − 4 ! 0! ( )

32.

P5 =

5! 5! = = 120 ( 5 − 5 )! 0!

33.

P3 =

(8 − 3)!

43. 104 P3 =

44.

P3 =

104!

BACD BADC BCAD BCDA ADBC BDAC ADCB BDCA

42.

15! 6! = 15 ⋅ 14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7

P9 =

= 1,816,214,400 ways

104! = 104 ⋅ 103 ⋅ 102 = 1,092,624 orders of the top three places 101!

8! = 336 orders 5!

45. ABCD ABDC ACBD ACDB

12! 8! = 12 ⋅ 11 ⋅ 10 ⋅ 9

P4 =

= 11,880 ways

8! = 8 ⋅ 7 ⋅ 6 = 336 5!

(104 − 3)!

833

20! 20! = = 20 (19 ) = 380 20 − 2 ! 18! ( )

P2 =

34.

Counting Principles

CABD CADB CBAD CBDA CDAB CDBA

49.

DABC DACB DBAC DBCA DCAB DCBA

46. ABCD ACBD

DBCA DCBA

50. 51.

7! 7! = 2!1!1!1!1!1! 2! = 7⋅6⋅5⋅4⋅3 = 2520 11! 11! = = 11 ⋅ 10 ⋅ 9 ⋅ 7 ⋅ 5 = 34,650 1!4!4!2! 4!4!2! 5

5! 5⋅4 = = 10 2!3! 2

C2 =

6! 6⋅5⋅4 = = 20 3!3! 6 4! 53. 4 C1 = =4 1!3!

52.

C3 =

C1 =

7! =7 1!6!

47.

7! 7! 7⋅6 ⋅5⋅4 = = = 420 2!1!3!1! 2!3! 2 ⋅1

54.

48.

8! 8⋅7⋅6 = = 56 3!5! 3 ⋅ 2 ⋅ 1

55.

C0 =

25! =1 0!25!

56.

C0 =

20! =1 0!20!

Chapter 9

834

57. 33 C4 =

33! 33! = = 40,920 (33 − 4)! 29!4!

59.

42 C5 = 850,668

Select two of four cards for pair: 4 C2 13

62. ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF, BCD, BCE, BCF, BDE, BDF, BEF, CDE, CDF, CEF, DEF 6 C3 = 20 ways

72. There are 27 good sets and 3 defective sets. (a) 27 C4 = 17,550 ways

(b) (c)

67.

68.

C2 =

C2 ⋅ 3 C2 = 351 ⋅ 3 = 1053 ways

C4 + 27 C3 ⋅ 3 C1 + 27 C2 ⋅ 3 C2 = 17,550 + 8775 + 1053

= 23,378 ways 5

C2 − 5 = 10 − 5 = 5 diagonals

74.

C2 − 6 = 15 − 6 = 9 diagonals

100! 100 ( − 14)!14!

75.

C2 − 8 = 28 − 8 = 20 diagonals

100! ≈ 4.42 × 1016 committees 86!14!

76. 10 C2 − 10 =

20! 20! = = 190 ways (20 − 18)!18! 2!18!

77. 14 ⋅ n P3 = n + 2 P4

10! − 10 = 45 − 10 = 35 diagonals 8!2!

Note: n ≥ 3 for this to be defined.  n !  ( n + 2 )! 14  =  ( n − 3 )!  ( n − 2 )!

40! 40! = = 5,586,853,480 ways 65. 40 C12 = (40 − 12)!12! 28!12! 66.

73.

63. 100 C14 =

20 C18

C1 ⋅ 4 C3 ⋅ 12 C1 ⋅ 4 C2 = 13 ⋅ 4 ⋅ 12 ⋅ 6 = 3744 ways to get a

full house

C6 = 15,890,700

61. AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, EF 6 C2 = 15 ways

64.

Select three of four cards for three of a kind: 4 C3 Select type of card for pair: 12 C1

71. Select type of card for three of a kind: 13 C1

C7 = 120

58.

60.

Sequences, Series, and Probability

14n ( n − 1)( n − 2 ) = ( n + 2 )( n + 1) n ( n − 1) (We can divide here by n ( n − 1) since n ≠ 0, n ≠ 1. )

9! = 36 lines 7!2!

14 n − 28 = n 2 + 3n + 2

59! 39! ⋅ 54!5! 38!1! = 5,006,386 ⋅ 39 = 195,249,054 ways

0 = n 2 − 11n + 30

C5 ⋅ 39 C1 =

( 13 C7 )( 20 C3 ) = 1716 ⋅ 1140 = 1,956,240 ways

69. There are 7 administrators, 12 faculty, and 25 students. 7 C1 ⋅ 12 C3 ⋅ 25 C 2 = 7 ⋅ 220 ⋅ 300 = 462,000 committees 70. (a)

(b) (c) (d)

3! = 3 relationships 2!1! 8! 8⋅7 = = 28 relationships 8 C2 = 2!6! 2 12! 12 ⋅ 11 = = 66 relationships 12 C 2 = 2!10! 2 20! 20 ⋅ 19 = = 190 relationships 20 C 2 = 2!18! 2 3

C2 =

0 = ( n − 5 )( n − 6 ) n = 5 or n = 6 78.

P5 = 18 ⋅ n − 2 P4

Note: n ≥ 6 for this to be defined.  ( n − 2 )!  n! = 18   (We can divide by ( n − 5)!  ( n − 6 )! 

( n − 2 ) , ( n − 3) , ( n − 4 ) since n ≠ 2, n ≠ 3, and n ≠ 4. ) n ( n − 1)( n − 2 )( n − 3)( n − 4 ) = 18 ( n − 2 )( n − 3)( n − 4 )( n − 5 ) n 2 − n = 18n − 90 2

n − 19n + 90 = 0

( n − 9 )( n − 10 ) = 0 n = 9 or n = 10

Section 9.5 79.

P4 = 10 ⋅ n −1 P3

( n + 1)! = ( n + 2 )! ( n − 1)! ( n − 1)! 4 ( n + 1)! = ( n + 2 )!

n! = 10 ( n − 4 )! ( n − 4 )! n! = 10 ( n − 1)!

n = 10 80.

n=2

P6 = 12 ⋅ n −1 P5

5 ⋅ n −1 P1 = n P2

84.

( n − 1)! ( n − 6 )! n! = 12 ( n − 1)!

( n − 6 )!

( n − 1)! = n! ( n − 2 )! ( n − 2 )! 5 ( n − 1)! = n!

= 12

n = 12

n=5 85. False. Because order does not matter, it is an example of a combination.

P = 4 ⋅ n P2

81.

n +1 3

( n + 1)! = 4 n! ( n − 2 )! ( n − 2 )! ( n + 1)! = 4n!

86. True. the Fundamental Counting Principle determines the number of ways different events can occur by multiplying the number of ways each event occurs.

n=3 82.

n+2

835

4 ⋅ n +1 P2 = n + 2 P3

83.

( n − 1)!

Counting Principles

87.

P3 = 6 ⋅ n + 2 P1

P ≈ 3.836 × 10139

100 80

This number is too large for some calculators to evaluate.

( n + 2 )! = 6 ( n + 2 )! ( n − 1)! ( n + 1)! ( n + 1)! = 6 ( n − 1)! ( n + 1)( n ) = 6

88. (a) (b) (c) (d)

n=2

Permutation because order matters Combination because order does not matters Permutation because order matters Combination because order does not matters

89. The symbol n Pr means the number of ways to choose and

order r elements out of a set of n elements.

90.

—

No, Sample answer: 7 P0 =

7! 7! 7! 7! = = 1 and 7C 0 = = = 1 So, the value of n Pr is not always greater (7 − 0)! 7! (7 − 0)!0! 7!0!

than n C r .

Chapter 9

836

91.

92.

Pn −1 =

Cn =

Sequences, Series, and Probability

( n − ( n − 1) )!

n! n! = = n Pn 1! 0!

95.

( n − n )! n!

−5 −14 7 2 88 y= = = −8 3 −5 −11 7 −2

n! = 0! n! n! n! = = = n C0 n!0! ( n − 0 )!0! 93.

Answer: ( −2, − 8 )

n!  n − ( n − 1) !( n − 1)!   n! = (1)!( n − 1)!

C n −1 =

94.

3 −14 2 −2 22 x= = = −2 3 −5 −11 7 −2

Cr =

−1 − 4 96. x =

−4 5 − 21 = = −1 −3 − 4 21

n! = C ( n − 1)!1! n 1 y =

n! ( n − r )!r !

−3

−1

9 −4 21 = =1 −3 − 4 21 9

1  n!  =   r !  ( n − r )! 

Answer: ( −1, 1)

P = n r r!

Section 9.6 Probability 9. {(H, 1), (H, 2), (H, 3) (H, 4), (H, 5), (H, 6), (T, 1),

1. sample space 2. the number of outcomes in the event; the number of outcomes in the sample space 3. mutually exclusive

(T, 2) (T, 3), (T, 4), (T, 5), (T, 6)}

10. {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2) (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4),

4. independent 5. The probability of an event E is given by P( E ) =

n( E ) , n( S )

(5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

where 0 ≤ P ( E ) ≤ 1. 6. The probability of an impossible event is P ( E ) = 0. 7. The probability of a certain event is P ( E ) = 1. 8. (a)

Probability of the union of two events P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B ) Matches (ii).

(b) Probability of mutually exclusive events P ( A ∪ B ) = P ( A) + P ( B ) Matches (i). (c)

Probability of independent events P ( A and B) = P ( A) ⋅ P (B) Matches (iii).

11.

{ ABC , ACB, BAC , BCA, CAB, CBA}

( A, B), ( A, C ), ( A, D), ( A, E ), ( B, C ), 12.   ( B, D), ( B, E ), (C , D), (C , E ), ( D, E ) 13. E = {HTT , THT , TTH }

P( E ) =

n( E ) 3 = n( S ) 8

14. E = {HHH , HHT , HTH , HTT }

P( E ) =

n( E ) 4 1 = = n( S ) 8 2

Section 9.6 15. E = {HHH , HHT , HTH , HTT , THH , THT , TTH }

P( E ) =

16. E = {HHH , HHT , HTH , THH }

P( E ) =

n( E ) 4 1 = = n( S ) 8 2

17. E = {K , K , K , K , Q, Q, Q, Q, J , J , J , J }

P( E ) =

n( E ) n( S )

6 3 = 52 26

20. E = 4 suits ⋅ 9 cards (2–10) n( E ) 36 9 P( E ) = = = n(S ) 52 13 21. E = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

n( E ) 12 3 = = n(S ) 52 13

18. E = 26 black cards n( E ) 26 1 = = P( E ) = n(S ) 52 2

837

E = 6, red face cards

19.

n( E ) 7 = n( S ) 8

Probability

P( E ) =

n( E ) 5 = n(S ) 36

(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3),  22. E =   (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)   n( E ) 15 5 P( E ) = = = n(S ) 36 12 E = {(1, 1), (1, 2), (1, 3), ( 2, 1), ( 2, 2), (3, 1)}

23.

P( E ) =

n( E ) n( S )

6 1 = 36 6

(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4),  24. E =   (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)   n( E ) 19 P( E ) = = n(S ) 36 25. E = 66 ticket holders less than 19 years old n( E ) 66 3 P(E ) = = = n(S ) 2200 100

C ⋅ C + C ⋅ C + C ⋅ C 32. P ( E ) = 1 1 2 1 1 1 3 1 2 1 3 1 6 C2 =

2 + 3 + 6 11 = 15 15

26. E = 1342 ticket holders older than 39 years old n( E ) 1342 61 P(E ) = = = n(S ) 2200 100

33. P ( E ′) = 1 − P ( E ) = 1 − 0.75 = 0.25

27. E = 792 ticket holders 19 to 39 years old n( E ) 792 9 P( E ) = = = n(S ) 2200 25

35. P ( E ′) = 1 − P ( E ) = 1 − 23 = 13

28. E = 572 ticket holders younger than 19 years old or older than 59 years old

P( E ) =

n( E ) 572 13 = = n(S ) 2200 50 C2 3 1 = = 15 5 6 C2

29. P ( E ) = 3

34. P ( E ′) = 1 − P ( E ) = 1 − 0.36 = 0.64

36. P ( E ′) = 1 − P ( E ) = 1 − 87 = 18 37. P ( E ) = 1 − P ( E ′) = 1 − p = 1 − 0.12 = 0.88 38. P ( E ) = 1 − P ( E ′) = 1 − p = 1 − 0.84 = 0.16

= 207 39. P ( E ) = 1 − P( E ′) = 1 − 13 20 61 39 = 100 40. P ( E ) = 1 − P( E ′) = 1 − 100

C2 1 = 15 6 C2

30. P( E ) = 2

C2 6 2 = = C 15 5 6 2

31. P ( E ) = 4

838

Chapter 9

Sequences, Series, and Probability

41. E = The card is a club or a king. Event A = club

46.

P( E ) =

Event B = king

P( E ) = P( A) + P(B) − P( A ∩ B) 13 4 1 16 4 = + − = = 52 52 52 52 13 42. E = the card is a face card or black card

E = 22 to 30 years old

47.

n( E ) n( S )

2 1 = 32 16

E = 20 to 30 years old P( E ) =

n( E ) 20 5 = = n( S ) 32 8

Event B = black card

48. E = 18 to 21 years old n( E ) 29 P( E ) = = n( S ) 32

P ( E ) = P ( A) + P ( B ) − P ( A ∩ B ) 12 26 6 32 8 = + − = = 52 52 52 52 13

49. E = older than 21 years old n( E ) 3 P( E ) = = n(S ) 32

Event A = face card

43. E = the card is a 5 or 2

Event A = 5 Event B = 2

P( E ) = P( A) + P( B) − P( A ∩ B) 4 4 0 8 2 = + − = = 52 52 52 52 13 44. E = the card is a heart or a spade

Event A = heart

50. E = younger than 31 years old n( E ) 31 P( E ) = = n(S ) 32 51. E = all three numbers are even n( E ) 5⋅5⋅5 125 1 = = = P(E ) = n(S ) 10 ⋅ 10 ⋅ 10 1000 8 52. E = all three numbers are a factor of 8 n( E ) 4⋅4⋅4 64 8 = = = P(E ) = n(S ) 10 ⋅ 10 ⋅ 10 1000 125 53.

Event B = spade

P ( E ) = P ( A) + P ( B ) − P ( A ∩ B ) 13 13 0 26 1 = + − = = 52 52 52 52 2 In Exercises 45 − 50, use the following table. Age

Number of students

18 – 19

20 – 21

22 – 30

31 − 40

45. E = 18 to 19 years old n( E ) 11 P( E ) = = n( S ) 32

E = all three numbers less than or equal to 3 P( E ) =

54.

n( E ) n( S )

3⋅3⋅3 27 = 10 ⋅ 10 ⋅ 10 1000

E = all three numbers less than or equal to 9 P( E ) =

n( E ) n( S )

2⋅2⋅2 8 1 = = 10 ⋅ 10 ⋅ 10 1000 125

55. E = two numbers are less than 5 and the other number is 10 n( E ) 4 ⋅ 4 ⋅1 16 2 = = = P(E ) = n(S ) 10 ⋅ 10 ⋅ 10 1000 125 56. E = one number is 2, 4, or 6, and the other two numbers are odd n( E ) 3⋅5⋅5 3 = = P( E ) = n(S ) 10 ⋅ 10 ⋅ 10 40 57. p + p + 2 p = 1 p = 0.25 1 1 1 Taylor: 0.50 = , Moore: 0.25 = , Perez: 0.25 = 2 4 4

Section 9.6

Probability

839

58. (a) 12% of 11.46 million  0.12(11.45) ≈ 1.38 million unemployed workers

(b) 29% + 3% = 32% of 11.46 million  0.32(11.46) ≈ 3.67 million unemployed workers (c) P( 24 – 44 age group) = 0.39 (d) P( 25 – 64 age group) = 0.39 + 0.29 = 0.68 (e) P( 45 or older age group) = 0.29 + 0.03 = 0.32 59. (a) 11.6% of 206.9 million  0.116( 206.9) ≈ 24.00 million have advanced degrees

E = Bachelor's degree or higher

(b)

P( E ) = 0.201 + 0.116 = 0.317 E = high school diploma or gone on to post seconday education

(c)

P( E ) = 0.298 + 0.168 + 0.098 + 0.201 + 0.116 = 0.881 E = earned a degree

(d)

P( E ) = 0.098 + 0.201 + 0.116 = 0.415 60. (a)

(b) (c)

45 = 0.45 100 45 = 0.45 100 23 = 0.23 100

64. (a)

(b) (c)

61. (a) Number of ways to arrange the digits = 5! = 120 1 Probability of guessing each digit = 120 (b) Number of ways to arrange the digits if first digit is known = 4! = 24 1 Probability of guessing other 4 digits = 24 62. (a)

( 8 C2 )( 100 C5 ) = 0.0756 108

(b)

P( E ) = (0.32)(0.32) = 0.1024 (b)

E = At least one shopper paid in cash P( E ) = 0.32

P ( SS ) = (0.985)2 ≈ 0.9702

66. (a)

(b)

P (S ) = 1 − P( FF ) = 1 − (0.015)2 ≈ 0.9998

C4 126 14 = = (4 good units) 495 55 12 C4

(c)

P ( FF ) = (0.015)2 ≈ 0.0002

( 9 C2 )( 3 C2 ) = 108 = 12 12

(c)

E = Both shoppers paid only in cash

≈ 6.929 × 10 −4

108

63. (a)

65. (a)

P ( E ) = (1 − P ( E ′)) = (1 − 0.32) = (0.68) = 0.4624

( 8 C2 )( 25 C2 )( 25 C3 )

(d)

20 20 1 ⋅ = 40 40 4  20  20  1 P ( EO or OE ) = 2    =  40  40  2 29 29 841 P ( N1 < 30, N 2 < 30) = ⋅ = 40 40 1600 40 1 1 P ( N1 N1 ) = ⋅ = 40 40 40 P ( EE ) =

( 9 C3 )( 3 C1 ) 12 C4

495

(2 good units)

252 28 = (3 good units) 495 55

At least 2 good units:

12 28 14 54 + + = 55 55 55 55

67. (a)

1 1  5  = 15,625   6

(b)

4096 4  5  = 15,625 = 0.262144  

(c)

1 − 0.262144 = 0.737856 =

11,529 15,625

840

Chapter 9

Sequences, Series, and Probability 4

68. (a)

(d)

1 1 P ( BBBB ) =   = 2 16   4

(b) (c)

69. (a)

1 1 1 P( BBBB) + P(GGGG ) =   +   = 2 2 8 P(at least one boy) = 1 − P(no boys) = 1 − P(GGGG ) 1 15 =1− = 16 16 If the center of the coin falls within the circle of radius d 2 around a vertex, the coin will cover the vertex.

Area in which coin may fall so that it covers a vertex P(coin covers a vertex) = Total area   d 2  n π      2   1 =  = π nd 2 4

(e) n

0.88

0.75

0.59

0.49

0.29

0.11

0.03

0.12

0.25

0.41

0.51

0.71 0.89

0.97

(f)

70. 1 −

75. P ( A) = 0.76 and P ( B) = 0.58

(a)

A and B cannot be mutually exclusive, because P ( A) + P ( B) = 0.76 + 0.58 = 1.34 > 1. (b) A′ and B′ can be mutually exclusive, because A′ = 1 − 0.76 = 0.24 and B′ = 1 − 0.58 = 0.42  A′ + B′ = 0.24 + 0.42 = 0.66 < 1.

71. True. P ( E ) + P ( E ′) = 1 72. True. Two events are independent if the occurrence of one has no effect on the occurrence of the other. 73. (a)

(c)

As you consider successive people with distinct birthdays, the probabilities must decrease to take into account the birth dates already used. Since the birth dates of people are independent events, multiply the respective probabilities of distinct birthdays. 365 364 363 362 (b) ⋅ ⋅ ⋅ 365 365 365 365 365 (c) P1 = =1 365 365 364 364 365 − (2 − 1) ⋅ = P2 = P1 = P1 365 365 365 365 365 364 363 363 365 − (3 − 1) ⋅ ⋅ = P3 = P2 = P2 365 365 365 365 365 365 364 363 365 − ( n − 1) 365 − (n − 1) ⋅ ⋅ ⋅⋅⋅⋅ = Pn = Pn −1 365 365 365 365 365

0.76 ≤ P ( A ∪ B ) ≤ 1, where P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B ) .

(45)2 9 7  45  3 =1−   =1−   =1− = (60)2 16 16  60  4

23; see the table in part (e).

74. Answers will vary.

(b) Experimental results will vary. 2

Qn is the probability that the birthdays are not distinct, which is equivalent to at least 2 people having the same birthday.

76. No. They are not mutually exclusive because events A and B share outcomes.

77. 8 C4 =

(8 − 4)!4!

8! = 70 4!4!

C5 =

9! 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5! = = 126 4!5! 24 ⋅ 5!

C8 =

11! 11 ⋅ 10 ⋅ 9 ⋅ 8! = = 165 8!3! 8!6

78.

79.

80. 16 C13 =

16! 16 ⋅ 15 ⋅ 14 ⋅ 13! = = 560 3!13! 13!6

Chapter 9 Review

841

Chapter 9 Review 1.

2n 2n + 1

8. a1 = 49, ak +1 = ak + 6

21 2 a1 = 1 = 2 +1 3 22 4 a2 = 2 = 2 +1 5 23 8 a3 = 3 = 2 +1 9 24 16 a4 = 4 = 2 + 1 17 25 32 a5 = 5 = 2 + 1 33

a3 = 55 + 6 = 61

an =

a2 = a1 + 6 = 49 + 6 = 55 a4 = 61 + 6 = 67 a5 = 67 + 6 = 73 9.

18! 18! = 20! 20 ⋅ 19 ⋅ 18! 1 1 = = 20 ⋅ 19 380

10.

10! 10 ⋅ 9 ⋅ 8! = = 90 8! 8!

2n − 1

11.

a1 =

( −1) 5(1) = − 5 = − 5 2(1) − 1 1

(n + 1)! (n + 1)n(n − 1)! = = n(n + 1) (n − 1)! (n − 1)!

12.

a2 =

( −1) 5( 2) = 10 2( 2) − 1 3

2 n! 2 n! 2 = = (n + 1)! (n + 1)n! n + 1

13.

 5 = 6(5) = 30

( −1) 5n n

2. an =

a3 =

( −1) 5(3) = − 15 = − 3 2(3) − 1 3

a4 =

( −1) 5( 4) = 20 2( 4) − 1 7

14.

( −1) 5(5) = − 25 a5 = 2(5) − 1 9

15.

i =1

Denominators are successive odd numbers. 2 , n = 1, 2, 3, ... an = 2n − 1

an =

n+2 , n = 1, 2, 3, ... n +1

7. a1 = 9, ak +1 = 13 ak

a2 = 13 a1 = 13 (9) = 3 a3 = 13 a2 = 13 (3) = 1 1 3 1 a5 = 13 a4 = 13 ( 13 ) = 9

a4 = 13 a3 = 13 (1) =

 j =1 +2 +3 +4 j =1

=6+

16.

Common difference is −2. an = 52 − 2n, n = 1, 2, 3, ...

 4k = 8 + 12 + 16 + 20 = 56 k =2

Common difference is 5. an = 5n + 2, n = 1, 2, ...

3 2 3 205 + + = 2 3 8 24

 i +1 = 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 i =1

≈ 6.17 17.

20 1 1 1 1 1 + + + + = 2(1) 2(2) 2(3) 2(20) k =1 2 k ≈ 1.799

( )

18. 3 12 + 3 22 + 3 32 +  + 3 92 = 3k 2 = 855

19.

k =1

9 1 2 3 9 k + + + + = ≈ 7.071 2 3 4 10 k =1 k + 1 k

20. 1 −

∞ 1 1 1 3  1 + − +  =  −  = 3 9 27 3 4 k =0 

842

Chapter 9 4

21. (a) 

j =110

Sequences, Series, and Probability 8 8 8 8 + + + 10 100 1000 10,000

= 0.8 + 0.08 + 0.008 + 0.008

(b)

∞

= 0.8888 =

1111 1250

8 ∞ 1

j =1

22. (a)

10 j = 10 10 j = 10 ⋅ 1 − 1 10 = 10 ⋅ 9 = 9 4

j =1

3,030,303

100 j = 1001 + 1002 + 1003 + 1004 = 100,000,000 = 0.030303 j =1

(b)

∞

j =1

 1 

j =0

23. (a)

∞

100

 100 j = 100   100 j  = 100 ⋅ 1 − 1 100 = 99 = 100 ⋅ 99 = 33 100

10,101

4(0.01) = 4(0.01) + 4(0.01) + 4(0.01) + 4(0.01) = 250,000 = 0.040404 k

k =1

(b)

∞

k =1

24. (a)

∞

100

 4(0.01) = 100  (0.01) = 100 ⋅ 1 − 0.01 = 100 ⋅ 99 = 100 ⋅ 99 = 99 k

k =0

100

1111

2(0.1) = 2(0.1) + 2(0.1) + 2(0.1) + 2(0.1) = 5000 = 0.2222 k

k =1

(b)

∞

2 ∞

 2(0.1) = 10 (0.1) = 10 ⋅ 1 − 0.1 = 10 ⋅ 90 = 10 ⋅ 9 = 9

k =1

0.02   25. an = 3000  1 +  4  

100

26. (a) an = − 0.028n 2 + 2.31n + 57.6 n = 1: a1 = 58.982  $58,982 n = 2: a2 = 61.208  $61,208

(a)

0.02   a1 = 3000  1 +  = $3015 4  

n = 3: a3 = 63.378  $63,378 n = 4: a4 = 65.492  $65,492

0.02   a2 = 3000  1 +  = $3030.08 4  

n = 5: a5 = 67.550  $67,550 n = 6: a6 = 69.552  $69,552

0.02   a3 = 3000  1 +  = $3045.23 4  

n = 7: a7 = 71.498  $71,498

n = 8: a8 = 73.388  $73,388

n = 10: a10 = 77.000  $77,000

0.02   a4 = 3000  1 +  = $3060.45 4  

n = 9: a9 = 75.222  $75,222

0.02   a5 = 3000  1 +  = $3075.75 4  

n = 11: a11 = 78.222  $78,222 n = 12: a12 = 80.388  $80,388

0.02   a6 = 3000  1 +  = $3091.13 4  

(b)

0.02   a7 = 3000  1 +  = $3106.59 4   8

0.02   a8 = 3000  1 +  = $3122.12 4  

(b)

0.02   a40 = 3000  1 +  = $3662.38 4  

Chapter 9 Review (c) n = 13: a13 = 81.998  $81,998

34. a2 = 14, a6 = 22

n = 14: a14 = 83.552  $83,552

a6 = a2 + 4 d

n = 15: a15 = 85.050  $85,050

22 = 14 + 4 d 8 = 4d

n = 16: a16 = 86.492  $86,492 n = 18: a18 = 89.208  $89,208

2=d a1 = a2 − d

n = 19: a19 = 90.482  $90,482

a1 = 14 − 2 = 12

n = 17: a17 = 87.878  $87,878

a2 = 12 + 2 = 14

Answers will vary. Sample answer: In the near future the values seem reasonable. However, in the long term, since the model is quadratic, the average salaries will start to decrease around the year 2014. 27. Arithmetic d = 3 − 5 = −2

a3 = 14 + 2 = 16 a4 = 16 + 2 = 18 a5 = 18 + 2 = 20

35. a1 = 35, ak +1 = ak − 3

a1 = 35

28. Not arithmetic

a2 = a1 − 3 = 35 − 3 = 32 a3 = a2 − 3 = 32 − 3 = 29

29. Not arithmetic

a4 = a3 − 3 = 29 − 3 = 26

30. Arithmetic 8 9 1 d= − =− 9 9 9 31. a1 = 3, d = 4

a5 = a4 − 3 = 26 − 3 = 23 an = 35 + (n − 1)(−3) = 38 − 3n, d = −3 36. a1 = 15, ak +1 = ak + 25

a1 = 15

a1 = 3

a2 = 15 + 25 = 352

a2 = 3 + 4 = 7 a3 = 7 + 4 = 11

a3 = 352 + 25 = 402 = 20

a4 = 11 + 4 = 15

a4 = 20 + 25 = 452

a5 = 15 + 4 = 19

a5 = 452 + 25 = 502 = 25 an = 15 + 25 ( n − 1) = 252 + 25 n, d = 25

32. a1 = 8, d = −2 a1 = 8 a2 = 8 − 2 = 6 a3 = 6 − 2 = 4 a4 = 4 − 2 = 2 a5 = 2 − 2 = 0

37. an = 100 + (n − 1)(−3) = 103 − 3n 25

n =1

38. a3 = a1 + 2 d 28 = 10 + 2 d 18 = 2 d 9=d

a10 = a4 + 6 d

an = 10 + ( n − 1)9 = 1 + 9n

28 = 10 + 6 d 18 = 6 d 3=d a1 = a4 − 3d a1 = 1 a2 = 1 + 3 = 4 a3 = 4 + 3 = 7

 (25)(26)   = 1600 2 

 (103 − 3n) = 103 − 3 n = 25(103) − 3 

33. a4 = 10, a10 = 28

a1 = 10 − 3(3)

843

n =1

 (25)(26)   = 2950 2 

 (1 + 9n) = 1 + 9 n = 25(1) + 9  39.

j =1

(2.5 j − 5) = 2.5 j = 5 j =1

11(12)  = 2.5  − 11(5) = 110  2 

a4 = 7 + 3 = 10 a5 = 10 + 3 = 13

844

Chapter 9

Sequences, Series, and Probability

40.

  3 k + 4  = 3  k + 4

2

2 11

k =1



k =1

2 (11)(12) = ⋅ + 11(4) = 88 3 2

41. The sum of the first 50 positive multiples of 5 50 50 (5 + 250) S50 =  5n = 5 +  + 250 = 2 n =1 = 6375 42. The sum of the integers from 10 to 70

1 49. a1 = 4, r = − 4

a1 = 4

a2 = 4 ( − 14 ) = −1 a3 = −1( − 14 ) = 14

a4 = 14 ( − 14 ) = − 161

a5 = − 161 ( − 14 ) = 641

3 50. a1 = 2, r = 2

n =1

a1 = 2

S70 − S9 =  n − n = (1 + 2 +  + 70) − (1 + 2 +  + 9)

70 9 (1 + 70) − (1 + 9) 2 2 = 2485 − 45 = 2440 =

43. Starting salary: $53,300 with a salary increase of $1066 per year

an = 53,300 + ( n − 1)(1006) (a) a5 = 53,300 + (5 − 1)(1066) = $57,564

a2 = 2 ( 23 ) = 3 a3 = 3 ( 23 ) = 29

a4 = 92 ( 23 ) = 274

a5 = 274 ( 23 ) = 818 51. a1 = 9, a2 = 6 a2 = a1 r

6 = 9r = r  r = 23

6 9

n =1

a1 = 9

(b) S5 =  53,300 + ( n − 1)(1066) = 53,300 +  + 57,564 5 = (53,300 + 57,564) = $277,160 2

44. First trip around a field: 123 bales with 11 fewer bales each additional trip; there are 7 trips around the field. an = 123 + (n − 1)(−11) 7

S7 =  (123 + (n − 1)(−11)) = 123 +  + 57 n =1

6 (123 + 57) 2 = 630 bales =

45. Geometric

r = 2

a2 = 6 a3 = 6( 23 ) = 4

a4 = 4( 23 ) = 83

a5 = 83 ( 23 ) = 169

52. a1 = 2, a2 = 12 a2 = a1 r

12 = 2r 12 2

= r  r = 6

a1 = 2 a2 = 12 a3 = 12(6) = 72

46. Geometric

a4 = 72(6) = 432

r=−

a5 = 432(6) = 2592

1 3

47. Not geometric 48. Not geometric

1 53. a1 = 120, ak +1 = 3 ak

a1 = 120 a2 = 13 (120) = 40 a3 = 13 (40) = 403 a4 = 13 ( 403 ) = 409 a5 = 13 ( 409 ) = 40 27 an = 120 ( 13 )

n −1

, r = 13

Chapter 9 Review 54. a1 = 200, ak +1 = 0.1ak

65. Initial value: $130,000 Each year, value depreciates 0.30 of previous year’s value (a) at = 0.70 at −1 or a0 = 130,000

a1 = 200 a2 = 0.1(200) = 20 a3 = 0.1(20) = 2

a1 = 130,000(0.70)

a4 = 0.1(2) = 0.2

a2 = 130,000(0.70)2 

a5 = 0.1(0.2) = 0.02 an = 200(0.1)

n −1

at = 130,000(0.7)t

, r = 0.1 (b)

55. (a) and (b)

a2 −8 1 = =− a1 16 2

48 0.04   A = S48 =  80  1 +  12  n =1 

an = a1r n −1  1 a6 = 16  −   2

6 −1

a5 = 130,000(0.7)5 = $21,849.10

0.04  0.04    66. A = 80  1 +  +  + 80  1 +  12  12   

a1 = 16, a2 = −8, n = 6 r=

16 1  1 = 16  −  = − =− 32 2  2

a4 1 = a3 6

an = a1r n −1  a9 = a4r 5 5

1 1 a9 = (1)   = ≈ 0.000129 7776 6 7

57. 5i −1 = i =1

58.

3

i −1

3(− 4)

n −1

n =1

 1 60. 12  −  n =1  2 4

7 61.  4   i =1  8  ∞

62.

∞

1

 6  3 

i −1

i =1

63.

∞

3

 4 2 

n −1

 1 − ( − 4)   = 9831 = 3  1 − ( − 4)    7

 1  64.  1.3   k =1  10 

)

67.

C8 =

10! 10 ⋅ 9 ⋅ 8! = = 45 8!2! 8!2

68.

C5 =

12! 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7! = = 792 5!7! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ⋅ 7!

 14  14! 14 ⋅ 13 ⋅ 12! = = 91 70.   = 14 C12 = 12 12!2! 12! ⋅ 2   71. ( x + 5) 4 = 4 C0 x 4 + 4 C1 x 3 (5) + 4 C2 x 2 (5)2 + 4 C3 x(5)3 + 4 C4 (5)4

= x 4 + 20 x 3 + 150 x 2 + 500 x + 625

n −1

= 7.5

72. ( y − 3)3 = 3 C0 y 3 − 3 C1 y 2 (3) + 3 C2 y(3)2 − 3 C3 (3)3

= y 3 − 9 y 2 + 27 y − 27

4 7 =  4  = = 32 − 8 1 78 i =0   ∞

73. ( a − 4b )5 = 5 C0 a 5 − 5 C1a 4 (4b) + 5 C2 a 3 (4b) 2 − 5 C3 a 2 (4b)3

+ 5 C4 a (4b) 4 − 5 C5 (4b)5

6 1 = 6  = =9 1−1 3 i =0  3  ∞

k −1

k =1 ∞

= 3( − 4)

9 9! 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5! = = 126 69.   = 9 C4 = 4 4!5! 4 ⋅ 3 ⋅ 2 ⋅ 5!  

1 − 35 = 121 1− 3

i =1

59.

1 − 57 = 19,531 1−5

(

a1 1 − r n

1− r   0.04  48   1 − 1 +  12     = 80.26 ≈ $4170.62   0.04    1 − 1 +    12    

56. (a) and (b) a3 = 6, a4 = 1, n = 9

845

Not possible because r = 32 and 32 > 1.

= a 5 − 20a 4b + 160a 3b 2 − 640a 2b 3 + 1280ab 4 − 1024b 5 7

74. (3x + y) = 7 C0 (3x) + 7 C1(3x) y + 7 C2 (3x) ( y) + 7 C3 (3x) ( y) 3

+ 7 C4 (3x) ( y) + 7 C5 (3x) ( y) + 7 C6 (3x)( y) + 7 C7 ( y)7 k −1

1.3 13  1  = 1.3   = = − 10 1 1 10 ( ) 9 k =0   ∞

= 2187 x 7 + 5103 x 6 y + 5103 x 5 y 2 + 2835 x 4 y3 +945 x 3 y 4 + 189 x 2 y 5 + 21xy6 + y 7

Chapter 9

846

Sequences, Series, and Probability

75. The 2nd number in the 7th row is 7 C2 = 21.

8 ⋅ n P2 = n +1 P3

88.

76. The 8th entry in the 9th row is 9 C7 = 36.

77. The 5th number in the 8th row is 8 C4 = 70.

n! (n + 1)! = (n − 2)! ( n − 2)! 8n! = (n + 1)! n=7

78. The 6th entry in the 10th row is 10 C5 = 252. 79. Two numbers from 1 to 15 whose sum is 12 (without replacement) 10 ways: 1 + 11

2 + 10 3+9 4+8 5+7 7+5 8+4 9+3 10 + 2 11 + 1 80. Six speaker systems, five DVD players, and six flat screen televisions

C6 =

89.

90.

C48 =

92.

( 7 C2 )( 11C2 ) = 21 ⋅ 55 = 1155 ways

93.

10 1 1 ⋅ = 10 9 9

94. P( E ) =

95. (a)

(b) (4)(3)(6)(3) = 216 schedules (2)(3)(6)(3) = 108 schedules (2)(3)(2)(3) = 36 schedules

82. (a)

10 7 = 10,000,000 possible calls 2 ⋅ 106 = 2,000,000 calls

(b) 83.

12 P5 =

84.

85.

86. 87.

P4 =

(c)

n( E ) 1 1 = = n(S ) 5! 120

208 = 0.416 500 400 = 0.8 500 37 = 0.074 500

720 5  6  5  4  3  2  1  6! 96.        = 6 = = 6 6 6 6 6 6 6 46,656 324       

12! = 12 ⋅11 ⋅ 10 ⋅ 9 ⋅ 8 = 95,040 7!

97. True. ( n + 2)! (n + 2)(n + 1)n! = = ( n + 2)(n + 1) n! n!

6! = 360 2!

98. True by Properties of Sums.

8! 8! = = 7! = 5040 permutations 2!2!2!1!1! 8 9! 9! = = 90,720 permutations 2!2! 4 n +1

50! 50 ⋅ 49 = = 1225 2!48! 2

91. 12! = 479,001,600 ways

Total: 6 ⋅ 5 ⋅ 6 = 180 ways 81. (a) (b) (c)

8! 8⋅7 = = 28 2!6! 2

P2 = 4 ⋅ n P1

n! (n + 1)! = 4⋅ ( n − 1)! (n − 1)! ( n + 1)! = 4 ⋅ n!

3i = 31 + 32 +  + 36 = 1092 j =1 8

3 j − 2 = 33 − 2 + 34 − 2 +  + 38 − 2 = 1092 j =1

99. (a)

Each term is obtained by adding the same constant (common difference) to the preceding term. (b) Each term is obtained by multiplying the same constant (common ratio) by the preceding term.

n=3

Chapter 9 Test

847

Chapter 9 Test 1. an = 3( 23 )

n −1

a1 = 3( 23 ) = 3(1) = 3 0

a2 =

11! ⋅ 4! 11! = 4! ⋅ 7! 7! 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7! 7! = 11 ⋅ 10 ⋅ 9 ⋅ 8 =

() = ()=2 1

3 23

4 3 3 8 a4 = 3( 23 ) = 3( 278 ) = 9 4 16 16 2 a5 = 3( 3 ) = 3( 81 ) = 27 a3 = 3( 23 ) = 3( 94 ) = 2

= 7920 6.

2. a1 = 12, and ak +1 = ak + 4

a2 = 12 + 4 = 16

= 5000 − 100n + 100

b1 = − x

= 5100 − 100n

x2 2 x3 b3 = − 3

b2 =

10. Geometric: a1 = 4, ak +1 = 12 ak

an = a1 r n −1 a2 = a1 r

x4 4 x5 b5 = − 5

b4 =

( ) = 4r

1 4 2

1 2

2 n +1

x 2 n +1 11.

( −1)2(1) +1 x 2(1) +1 = ( −1)3 x3 = − x3 b = 1!

2( 2) +1

b4 =

( −1) ( )

x 2(3) +1

( −1) x 5

= −

x 2

(−1) x7 = − x7 7

( −1)2(4) +1 x 2(4) +1 = ( −1)9 x9 = − x9 4!

( −1) ( )

2 5 +1

b5 =

2! 2 3 +1

b3 =

2( 2) +1

x 2(5) +1

= r

an = 4( 12 )

( −1)

2n( n − 1)n! 2n! = = 2n − 1 ! n (n − 1)! ( )

= 5000 + ( n − 1)( −100)

b2 =

1 n +1

an = a1 + ( n − 1)d

( −1) x n 3. b =

9. Arithmetic: a1 = 5000, d = − 100

a5 = 24 + 4 = 28

( −1)

(n + 1)n!

an = n 2 + 1, n = 1, 2, 3, 

a4 = 20 + 4 = 24

4. bn =

8. 2, 5, 10, 17, 26, 

a3 = 16 + 4 = 20

(n + 1)!

( −1) x11 = − x11 11

2 3(1) + 1

12. 2 +

n −1

2 3( 2) + 1

++

12 2 2 =  +1 3(12) + 1 3 n n =1

1 1 1 1 + + + + 2 8 32 128 1 1 1   = 21 + + + +  4 16 64     1 0  1 1  1 2  1 3  = 2   +   +   +   +   4    4 4  4   n

∞ ∞ 1 1 = 2  =  2  n = 0 4  n =1  4 

n −1

120

Chapter 9

848 8

13.

1

 49 7 

n −1

n =1

Sequences, Series, and Probability  1 − (1 7)8   = 49  1−17   

15.

 ( −1)n 2n     n −1  n=0  5  ∞

n −1

 1 − 1 78  = 49  6 7    ≈ 57.17

5 = 5  7 25 = 7

16. (3a − 5b) = (3a ) − 4(3a ) (5b) + 6(3a ) (5b) − 4(3a )(5b) + (5b) 4

  1 = 5  − − 1 2 5 ( )  

7  7(8)  14.  (8n − 5) = 8  − 5(7) = 224 − 35 = 189 n =1  2 

∞  ( −1)2  =  5  5  n=0 

= 81a 4 − 540a 3b + 1350a 2b 2 − 1500ab 3 + 625b 4

17. 9 C3 = 84 18.

19.

20 C3

70 P3

= 1140

26. (a)

( 3060 )( 3060 ) = 12 ⋅ 12 = 14

(b)

11 ⋅ 11 60 60

(c)

1 60

121 ≈ 0.0336 = 3600

≈ 0.0167

70! 67! = 70 ⋅ 69 ⋅ 68 =

= 328,440 9! = 9 ⋅ 8 = 72 7!

20. 9 P2 =

4 ⋅ n P3 = n +1P4

21. 4

(n + 1)! n! = (n − 3)! ( n − 3)! 4n! = ( n + 1)! n = 3

22. 26 ⋅ 10 ⋅ 10 ⋅ 10 = 26,000 ways

23.

25 C4

25! 25 ⋅ 24 ⋅ 23 ⋅ 22 = = 12,650 ways 21!4! 24

24. There are three face cards in the suit of hearts, so 3 P( E ) = . 52 25. The probability of random selection of six teams from the twelve teams in the Big Ten conference is C 1 P( E ) = 6 6 = C 924 12 6

C H A P T E R 1 0 Topics in Analytic Geometry Section 10.1

Circles and Parabolas .........................................................................850

Section 10.2

Ellipses ................................................................................................863

Section 10.3

Hyperbolas ..........................................................................................873

Section 10.4

Parametric Equations..........................................................................884

Section 10.5

Polar Coordinates ...............................................................................892

Section 10.6

Graphs of Polar Equations .................................................................903

Section 10.7

Polar Equations of Conics ..................................................................912

Chapter 10 Review ....................................................................................................919 Chapter 10 Test .........................................................................................................940 Chapters 8–10 Cumulative Test ..............................................................................943

C H A P T E R 1 0 Topics in Analytic Geometry Section 10.1 Circles and Parabolas 1. conic section

(1 − 9) + − 7 − (− 5) 2

12. Diameter =

2. locus

3. circle, center

( x − h ) + ( y − k ) = r 2 , where the point ( h, k ) is the 2

6. A line that is tangent to a parabola at a point P creates or makes equal angles with the two lines, one line passing though P and the focus, and the axis of the parabola. 2

( x − 5) + ( y + 6) 2

( 11)

14. x 2 + y 2 = 121

Center: (0, 0) Radius: 11

15. ( x − 5) + y 2 = 9 2

Center: (5, 0)

= 4 + 49 = 53

( x − h) + ( y − k ) = r 2 2 ( x − 3) + ( y − 7) = 53 2

= 17

Radius: 6

( 3 − 1) + ( 7 − 0 )

Center: (0, 0)

x 2 + y 2 = 11 9. Radius =

13. x 2 + y 2 = 36

x 2 + y 2 = 16

8. ( x − 0) + ( y − 0) =

)

( x − h) 2 + ( y − k ) 2 = r 2

center of the circle and the radius is r.

7. ( x − 0 ) + ( y − 0 ) = ( 4 )

(

68 = 2 17

 1 + 9 − 7 + ( − 5)  Center =  ,  = (5, − 6) 2  2 

5. The standard equation of a circle is given by 2

1 2 17 = 2

Radius =

4. parabola, directrix, focus

64 + 4 =

Radius: 3

16. x 2 + ( y + 8) = 25 2

10. Radius = 6 − ( −2 )  +  −3 − 4 

Center: (0, − 8)

Radius: 5

= 64 + 49 = 113

17. ( x + 1) + ( y + 6) = 19 2

( x − h) + ( y − k ) = r2 2 2 ( x − 6 ) + ( y + 3) = 113 2

Center: ( −1, − 6) Radius:

36 + 4 =

40 = 2 10

(− 6 − 0) + 0 − (− 2)

11. Diameter =

1 Radius = 2 10 = 2

(

)

( x − h) 2 + ( y − k ) 2 = r 2 ( x + 3) + ( y + 1) = 10

850

18. ( x + 7) + ( y − 3) = 32 2

Center: ( − 7, 3)

 − 6 + 0 0 + ( − 2)  Center =  ,  = ( − 3, − 1) 2  2 

Radius:

19.

32 = 4 2

1 2 1 2 x + y =1 4 4 x2 + y2 = 4

Center: ( 0, 0 ) Radius: 2

Section 10.1

20.

1 2 1 2 x + y =1 9 9 2 x + y2 = 9

x + y 2 = 16

Center: ( 0, 0 ) Radius: 4

Radius: 3

22.

4 2 4 2 x + y =1 3 3 3 x2 + y2 = 4 Center: ( 0, 0 ) Radius:

5 3 2 1 −5

−3 −2 −1

3 2

x 2 + y 2 = 81

Center: ( 0, 0 ) Radius: 9 y

10 8 6 4 2

( x − 2 x + 1) + ( y + 6 y + 9 ) = −9 + 1 + 9 2

− 10

Radius: 1

( x − 10 x + 25) + ( y − 6 y + 9) = −25 + 25 + 9 2

( x − 5) + ( y − 3) = 9 Center: ( 5, 3 ) 2

Radius: 3

) (

)

25. 4 x 2 + 3 x + 94 + 4 y 2 − 6 y + 9 = −41 + 9 + 36

−6 −4 −2

2 4 6 8 10

x2 + 8x + y 2 + 2 y + 8 = 0

29.

( x 2 + 8 x + 16) + ( y 2 + 2 y + 1) = − 8 + 16 + 1 ( x + 4)2 + ( y + 1)2 = 9 Center: ( − 4, −1) Radius: 3

4 ( x + 32 ) + 4 ( y − 3 ) = 4 2

3 2

( x + ) + ( y − 3) = 1 Center: ( − , 3 ) 2

Radius: 1

) (

)

2 2 26. 9 x + 6 x + 9 + 9 y − 4 y + 4 = −17 + 81 + 36

9 ( x + 3 ) + 9 ( y − 2 ) = 100 2

4 3 2 1

3 2

−4 −6 −8 − 10

Center: (1, − 3 )

(

y 2 = 81 − x 2

28.

9 2 9 2 x + y =1 2 2 2 2 x + y2 = 9 Center: ( 0, 0 )

(

−5

( x − 1) + ( y + 3) = 1

24.

1 2 3

−2 −3

2 Radius: 3 23.

851

x 2 = 16 − y 2

27.

Center: ( 0, 0 )

21.

Circles and Parabolas

−9 −8

−6 −5

−3 −2

−3 −4 −5 −6

( x + 3) + ( y − 2 ) = 1009 Center: ( −3, 2 ) 2

Radius: 103

Chapter 10

852

Topics in Analytic Geometry

x 2 − 6 x + y2 + 6 y + 14 = 0

30.

33.

x 2 + 2 x + y2 − 35 = 0

( x − 6 x + 9) + ( y + 6 y + 9) = −14 + 9 + 9

( x + 2 x + 1) + y = 35 + 1

( x − 3) + ( y + 3) = 4

( x + 1) + y2 = 36

Center: ( 3, − 3 )

Center: ( −1, 0 )

Radius: 2

Radius: 6

1 −2 −1

1 2

4 5 6 7 8

10 8

−2 −3 −4 −5 −6 −7 −8 −9

4 2 −10 −8

2 4 6 8 10

−8 −10

( x − 14 x + 4 ) + ( y + 8 y + 16 ) = −40 + 49 + 16 2

( x − 7) + ( y + 4 ) = 25 2

Center: ( 7, − 4 )

34.

x 2 + y2 + 10 y + 9 = 0

(

)

x 2 + y2 + 10 y + 25 = −9 + 25

(

x 2 + y + 5 = 16 Radius: 4

6 4 2

4 6 8 10

14 16 18

−5 −4 −3 −2 −1

(

x 2 + 6 x + y2 − 12 y + 41 = 0

) (

)

x 2 + 6 x + 9 + y2 − 12 y + 36 = −41 + 9 + 36

( x + 3) + ( y − 6 ) = 4 2

Center: ( −3, 6 )

1 2 3 4 5

−2 −3 −4 −5 −6 −7 −8

−4 −6 −8 −10 −12 −14

32.

)

Center: ( 0, − 5 )

Radius: 5

−2

−4

x 2 − 14 x + y2 + 8 y + 40 = 0

31.

−4 −2

35. ( x + 5) + ( y − 3) = 25 2

y-intercepts: (0 + 5) + ( y − 3) = 25 2

25 + ( y − 3) = 25 2

( y − 3)

Radius: 2

= 0

y = 3

(0, 3)

9 8 7 6 5 4 3 2 1 −8 −7 −6 −5 −4 −3 −2 −1

x-intercepts: ( x + 5) + (0 − 3) = 25 2

( x + 5) + 9 = 25 2

1 2

( x + 5)

= 16

x + 5 = ±4 x = − 9 and x = −1

( − 9, 0) and ( −1, 0)

Section 10.1 2

853

Let x = 0.

39. y-intercepts:

36. ( x − 1) + ( y + 4) = 16

Circles and Parabolas

y − 6 y − 27 = 0

y-intercepts: (0 − 1) + ( y + 4) = 16 2

y 2 − 6 y + 9 = 27 + 9

1 + ( y + 4) = 16 2

( y + 4)

( y − 3) = 36 2

y − 3 = ±6 y = 9, − 3

= 15

y = −4 ±

(0, − 4 ±

)

( 0, 9 ) , ( 0, − 3) 2

x − 2 x − 27 = 0

x 2 − 2 x + 1 = 27 + 1

( x − 1) + 16 = 16 2

( x − 1)

Let y = 0.

x-intercepts:

x-intercepts: ( x − 1) + (0 + 4) = 16 2

( x − 1) = 28 2

= 0

x − 1 = ± 28

x −1 = 0

x =1± 2 7

x =1

(1 ± 2 7, 0 )

(1, 0) 37. y-intercepts: ( 0 − 6 ) + ( y + 3 ) = 16 2

( y + 3 ) = 16 − 36 2

= −20

No solution No y-intercepts

y + 2y + 9 = 0 No solution No y-intercepts x-intercepts: Let y = 0.

x2 + 8x + 9 = 0

x-intercepts: ( x − 6 ) + ( 0 + 3 ) = 16 2

Let x = 0.

40. y-intercepts:

x 2 + 8 x + 16 = −9 + 16

( x − 6) = 7

( x + 4) = 7

x −6 = ± 7

x+4=± 7

x =6± 7

x = −4 ± 7

(6 ± 7, 0 )

38. y-intercepts: ( 0 + 7 ) + ( y − 8 ) = 4 2

( y − 8 ) = 4 − 49 2

= −45

No solution No y-intercepts x-intercepts: ( x + 7 ) + ( 0 − 8 ) = 4 2

( x + 7 ) = 4 − 64 2

= −60

No solution No x-intercepts

( −4 ± 7, 0 )

41. (a) Radius: 52; center: ( 0, 0 ) x 2 + y 2 = 52 2  x 2 + y 2 = 2704

(b) The distance from ( −40, − 30 ) or 40 miles west and 30 miles south is

( 0 + 40 ) + ( 0 + 30 ) 2

= 402 + 302 = 2500 = 50 miles. Therefore, you would have felt the earthquake. (c) x 2 + y 2 = 52 2 y

60 40 20 −60 −40 −20 −20

−40 −60

You were 52 − 50 = 2 miles from the outer boundary.

854

Chapter 10

Topics in Analytic Geometry

42. (a) Area of a circle: A = π r 2

49. Vertex: ( 0, 0 )  h = 0, k = 0

π r 2 = 2000 2000

Graph opens upward. x 2 = 4 py

Point on graph: ( 3, 6 )

r2 =

2000

≈ 25.231 feet

9 = 24 p

2 (b) π R = 2500 2500 R2 =

3 8

2500

≈ 28.209 feet

The increase in radius is 28.209 − 25.231 = 2.978 feet. 43.

y 2 = −4 x

Vertex: ( 0, 0 ) Opens to the left since p is negative. Matches graph (e). 44.

x2 = 2 y

Vertex: ( 0, 0 ) p = 12 > 0

Opens upward. Matches graph (b). 45.

x 2 = −8 y

Vertex: ( 0, 0 ) Opens downward since p is negative. Matches graph (d). 46.

47.

y 2 = −12 x

Thus, x 2 = 4 ( 83 ) y  y = 23 x 2  x 2 = 32 y. 50. Point: ( −2, 6 ) x = ay 2 −2 = a ( 6 )

− 181 = a x = − 181 y 2 y 2 = −18 x

51. Vertex: (0, 0)  h = 0, k = 0

Focus: (0, 2)  p = 2

( x − h)

= 4 p( y − k )

x 2 = 4( 2) y x2 = 8 y 52. Vertex: (0, 0)  h = 0, k = 0

Focus: (0, 3)  p = 3

( x − h)

= 4 p( y − k )

Vertex: ( 0, 0 )

x = 4(3) y

p = −3 < 0 Opens to the left. Matches graph (f).

x 2 = 12 y

( y − 1) = 4 ( x − 3) 2

Vertex: ( 3, 1) Opens to the right since p is positive. Matches graph (a). 48.

32 = 4 p ( 6 )

( x + 3) = −2 ( y − 1) 2

Vertex: ( −3, 1)

53. Vertex: (0, 0)  h = 0, k = 0

1  1  Focus:  − , 0   p = − 2  2 

( y − k)

= 4 p( x − h)

 1 y 2 = 4 −  x  2 y2 = − 2x

p = − 12 < 0

Opens downward. Matches graph (c).

Section 10.1 54. Vertex: (0, 0)  h = 0, k = 0

3  3  Focus:  − , 0   p = − 2  2 

x 2 = 4 py

( −8) = 4 p ( −2 ) 2

64 = −8 p  p = − 8

 3 y = 4 −  x  2

x 2 = 4 ( − 8) y

x 2 = − 32 y

y2 = − 6x

61.

( x − h)

y = 12 x 2 x 2 = 2 y = 4 ( 12 ) y; p = 12

Directrix: y = 4  p = − 4

Vertex: ( 0, 0 )

= 4 p( y − k )

Focus: ( 0, 12 )

x = 4( − 4) y 2

Directrix: y = − 12

x = −16 y

56. Vertex: (0, 0)  h = 0, k = 0

Directrix: y = −1  p = 1

( x − h)

3 2

= 4 p( y − k )

−3

x = 4(1) y 2

−2

Directrix: x = − 2  p = 2

( y − k)

= 4 p( x − h)

y 2 = 4( 2) x

62.

Directrix: x = 5  p = − 5

( y − k)

= 4 p( x − h)

y = 4( − 5) x 2

y 2 = − 20 x

y = −2 x 2

x 2 = − 12 y = 4 ( − 18 ) y, p = − 81

Vertex: ( 0, 0 )

Focus: ( 0, − 18 )

Directrix: y = 81

y2 = 8x 58. Vertex: (0, 0)  h = 0, k = 0

−1

x2 = 4 y 57. Vertex : (0, 0)  h = 0, k = 0

855

60. Vertical axis Passes through ( −8, − 2 )

( y − k ) = 4 p( x − h)

55. Vertex: (0, 0)  h = 0, k = 0

Circles and Parabolas

y 1 −3

−2

−1

−2 −3 −4 −5

59. Vertex: ( 0, 0 )  h = 0, k = 0

Horizontal axis and passes through the point ( 3, 3)

( y − k ) = 4 p ( x − h) 2 ( y − 0) = 4 p ( x − 0) 2

y 2 = 4 px 32 = 4 p ( 3 ) 9 = 12 p  p = 34 y 2 = 4 ( 34 ) x y 2 = 3x

856

Chapter 10

63.

y 2 = −6 x

Topics in Analytic Geometry 66.

y = 4 ( − 23 ) x; p = − 23

y2 = − x y 2 = 4 ( − 14 ) x , p = − 14

Vertex: ( 0, 0 )

Directrix: x = 32

Directrix: x = 14

Focus: ( −32 , 0 )

Focus: ( − 14 , 0 ) y

−6 −5 −4 −3 −2 −1

−5

−4

−3

−2

−1 −2

−3

−4

64.

67. ( x + 1) − 8( y + 2) = 0 2

y2 = 3x

y 2 = 4 ( 34 ) x; p = 34

( x + 1)2 = 8( y + 2)

Vertex: ( 0, 0 )

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

Focus: ( 34 , 0 )

Vertex: ( −1, − 2)

Directrix: x = − 34

Focus: ( −1, 0)

Directrix: y = − 4 4

y 12

−2

8 6 4 2 −10 −8 −6

65.

x 2 = −6 y

x = 4(− 2

3 2

68.

Vertex: ( 0, 0 )

( x − 5) + ( y + 4 ) = 0 2 ( y + 4 ) = − ( x − 5) = 4 ( − 14 ) ( x − 5) 2

Vertex: ( 5, − 4 )

Focus: ( 0, − 23 )

Focus: ( 5 − 14 , − 4 ) = ( 194 , − 4 )

Directrix: y = 23

Directrix: x = 214

4 1

2 −4

−2 −4 −6 −8

−6

x2 + 6y = 0

−6

−2 −1 −1

−2 −3 −4 −5 −6 −7

Section 10.1

69.

( x + ) = 4 ( y − 2 )  h = − , k = 2, p = 1 Vertex: ( − , 2 ) 3 2

3 2

72.

Circles and Parabolas

857

y2 − 4 y − 4 x = 0

( y − 2 ) = 4 ( x + 1); p = 1 2

3 2

Vertex: ( −1, 2 )

Focus: ( − 32 , 2 + 1) = ( − 23 , 3 )

Focus: ( 0, 2 )

Directrix: y = 1

Directrix: x = −2

y 6

5 4

−5 −4 −3 −2 −1

−4

−2

70.

( x + ) = 4 ( y − 1)  p = 1 Vertex: ( − , 1) Focus: ( − , 1 + 1) = ( − , 2 ) 1 2

1 2

73.

x2 + 4 x + 6 y − 2 = 0 x 2 + 4 x + 4 = −6 y + 2 + 4 = −6 y + 6

1 2

( x + 2 ) = −6 ( y − 1) 2 ( x + 2 ) = 4 ( − 23 ) ( y − 1) Vertex: ( −2, 1) Focus: ( −2, 1 − ) = ( −2, − ) 2

Directrix: y = 0 y 4

3 2

1 2

Directrix: y = 25 y

−3

−2

−1

5 4

−1

71.

y + 6 y + 8 x + 25 = 0

( y + 3) = 4 ( −2 )( x + 2 ) ; p = −2 Vertex: ( −2, − 3 ) Focus: ( −4, − 3 ) 2

Directrix: x = 0

−6

−2 −4 −6

−3

x2 − 2 x + 8y + 9 = 0

( x − 1) = −8 ( y + 1) = 4 ( −2 )( y + 1) Vertex: (1, − 1) Focus: (1, − 3 ) 2

−4

−1

x 2 − 2 x + 1 = −8 y − 9 + 1

2 −8

−4 −3

−2

74. y

−10

−6

Directrix: y = 1 y

−8 2

−4 −3 −2 −1

2 3 4 5 6

−2 −3 −4 −5 −6 −7 −8

858

Chapter 10

75.

y2 + x + y = 0

Topics in Analytic Geometry 78. 4 x − y 2 − 2 y − 33 = 0

y + y + 14 = − x + 14

y 2 + 2 y + 1 = 4 x − 33 + 1

( y + ) = 4 ( − )( x − ) 2

1 2

1 4

h = 14 , k = − 12 , p = − 14

Vertex: ( 14 , − 12 )

Focus: ( 0, − 12 )

( y + 1) = 4 (1)( x − 8 ) Vertex: ( 8, − 1) Focus: ( 9, − 1) 2

Directrix: x = 7

Directrix: x = 12

y 6

4 2

−3

−2

−1

−2

−4 −6

−2

79. Vertex: ( 3, 1)  h = 3, k = 1 76.

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

Opens downward Passes through: ( 2, 0 ) , ( 4, 0 )

Vertex: ( −1, 0 )

y = − ( x − 2 )( x − 4 )

Focus: ( 0, 0 )

= − x2 + 6 x − 8 = − ( x − 3) + 1 2

Directrix: x = −2

( x − 3) = − ( y − 1) 2

y 6

80. Vertex: (8, 4)  h = 8, k = 4

4 2

−4

−2

−4 −6

(

y = 14 x 2 − 2 x + 5

77.

4 y − 4 = ( x − 1)

)

Opens to the left Passes through: (0, 0)

( y − k)

= 4 p ( x − h)

( y − 4)

= 4 p( x − 8)

(0 − 4)

= 4 p(0 − 8)

16 = − 32 p

( x − 1) = 4 (1)( y − 1) 2

p = −

h = 1, k = 1, p = 1

Vertex: (1, 1)

( y − 4)

Focus: (1, 2 )

1 2

 1 = 4 − ( x − 8)  2

( y − 4)2 = − 2( x − 8)

Directrix: y = 0 y 6

−2

Section 10.1 81. Vertex: ( −4, 0 )  h = −4, k = 0

Directrix: y = 2 Vertical axis p = 4−2 =2

16 = 16 p 1= p

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

88. Vertex: ( −2, 1)  h = −2, k = 1

Directrix: x = 1  p = −3

y2 = 4 ( x + 4 )

( y − k ) = 4 p ( x − h) 2 ( y − 1) = 4 ( −3) ( x − ( −2 ) ) 2 ( y − 1) = −12 ( x + 2 ) 2

82. Vertex: ( 3, − 3 )  h = 3, k = −3

Opens upward Passes through: ( 0, 0 )

( x − 3) = 4 p ( y + 3) 2 ( 0 − 3) = 4 p ( 0 + 3) 2

89. Focus: ( 2, 2 )

Directrix: x = −2 Horizontal axis Vertex: ( 0, 2 )

9 = 12 p 3 4

p =2−0 =2

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

( y − 2 ) = 4 ( 2 )( x − 0 ) 2 ( y − 2) = 8x 2

3 4

83. Vertex: ( −2, 0 )  h = −2, k = 0

90. Focus: ( 0, 0 )

Directrix: y = 8 Vertical axis Vertex: ( 0, 4 )

Opens to the right Focus: ( − 32 , 0 ) 1 2

859

87. Vertex: ( 0, 4 )

Opens to the right Passes through: ( 0, 4 )

( y − 0) = 4 p ( x + 4) y2 = 4 p ( x + 4 ) 2 ( 4) = 4 p (0 + 4)

Circles and Parabolas

p = 4 − 8 = −4

y = 4 ( 12 ) ( x + 2 ) 2

( x − 0 ) = 4 ( −4 )( y − 4 ) x 2 = −16 ( y − 4 ) 2

y2 = 2 ( x + 2 )

84. Vertex: ( 3, − 3 )  h = 3, k = −3

Focus: ( 3, − 94 )  p = 34

91.

y2 − 8 x = 0 y = 8x

( x − h) = 4 p ( y − k ) 2 ( x − 3) = 3 ( y + 3 ) 2

85. Vertex: ( 5, 2 )

and x − y + 2 = 0

y3 = x + 2

y1 = 8 x y2 = − 8 x

The point of tangency is ( 2, 4 ) .

Focus: ( 3, 2 )

Horizontal axis: p = 3 − 5 = −2

( y − 2 ) = 4 ( −2 )( x − 5) 2 ( y − 2 ) = −8 ( x − 5) 2

−6

−3

86. Vertex: ( −1, 2 )  h = −1, k = 2

Focus: ( −1, 0 )  p = −2

( x − h) = 4 p ( y − k ) 2 ( x + 1) = 4 ( −2 )( y − 2 ) 2 ( x + 1) = −8 ( y − 2 ) 2

860

Chapter 10

92.

x 2 + 12 y = 0 and

12 y = − x

Topics in Analytic Geometry x + y−3= 0

y2 = 3 − x

1 2 x 12 The point of tangency is ( 6, − 3 ) . y1 = −

1 95. x = − 2 y 2  y 2 = − x 2  1 = 4 −  x  8 

d1 = −4

1 + b 8 2

x 2 = 2 y, ( 4, 8 ) , p =

1  1 , focus:  0,  2  2

1 d1 = − b 2 2

1  17  d2 = ( 4 − 0 ) +  8 −  = 2 2   1 17  b = −8 d1 = d2  − b = 2 2 8 − ( −8 ) m= =4 4−0 Tangent line: y = 4 x − 8  4 x − y − 8 = 0 2

d2 =

1 17 2   − 2 +  + ( −1 − 0) = 8 8  

d1 = d 2 

1 17 +b =  b = 2 8 8

To find the slope, use the points ( − 2, −1) and ( 2, 0). m =

0 − ( −1)

2 − ( − 2)

96. x = − 2 y 2 

2 y = x2



1 4   y = x2 2 1 p= 2  1 Focus:  0,   2

d1 =

2 9 1 ( −3 − 0 ) +  2 − 2  = 5  

1 −b =5 2 9 2 − ( 9 2 ) − ( 9 2) 0+3

1 y2 = − x 2  1 = 4 −  x  8 p = −

1 8

1 +b 8

= −3

1 65 2   − 8 +  + ( 2 − 0) = 8 8  

d1 = d 2 

1 65 +b =  b = 8 8 8

To find the slope, use the points ( − 8, 2) and (8, 0). m =

b=− m=

1 1 x −  4y − x + 2 = 0 4 2

d2 = 2

d2 =

1 4

 1  Focus:  − , 0   8 

1 −b 2

d1 =

Tangent line: y =

Let y = 0  x = 2  x -intercept ( 2, 0 ) . 94.

1 8

 1  Focus:  − , 0   8 

−4

93.

p = −

0 − 2 1 = − 8 − ( − 8) 8

1 Tangent line: y = − x + 1  8 y + x − 8 = 0 8

9  6x + 2y + 9 = 0 2 3  3  Let y = 0  x = −  x -intercept  − , 0  . 2  2 

Tangent line: y = −3 x −

Section 10.1 97. Parabola: Vertex: (0, 4)

Circles and Parabolas

861

101. (a)

Passes through ( ±4, 0)

(− 640, 152)

(640, 152)

x 2 = 4 p ( y − 4)

16 = 4 p(0 − 4) 16 = −16 p −1 = p

x 2 = − 4( y − 4)

x 2 = 4 py

(b)

640 2 = 4 p (152 )

98. Vertex: (0, 16)  h = 0, k = 0

p = 12,800 19

Points on the parabola: ( ±2, 6) Vertical axis

( x − 0)

= 4 p( y − 16)

( 2 − 0)

= 4 p(6 − 16)

19 y = 51,200 x2

(c)

4 = 4 p( −10)

200

400

500

14.84

59.38

92.77

133.59

4 = −40 p 1 − 10 = p

( )( y − 16)

1 = 4 − 10

x 2 = − 52 ( y − 16) x 2 = − 52 (0 − 16)

1 x 2 = −640 y or y = − 640 x2

1 x 2  x = 8 feet (b) −0.1 = − 640

x 2 = 32 5 x = ±

256 = 4 p ( − 25 )  p = −160 x 2 = 4 ( −160 ) y

When y = 0, the lattice arch is at ground level.

32 5

102. (a) x 2 = 4 py passes through point (16, − 25 ) .

103. (a) x 2 = 4 py, p =

≈ ± 2.53

So, the lattice arch is about 2( 2.53) = 5.06 feet wide at ground level.

3 2

3 x2 = 4   y = 6y 2

( or y = 6 x ) 2

99. Vertex: ( 0, 48 )  h = 0, k = 48

(

Passes through 10 3, 0

)

6 5

Vertical axis

( x − 0 ) = 4 p ( y − 48 ) 2

(10 3 − 0 ) = 4 p ( 0 − 48)

8 in. 2

300 = −192 p

1 −4 −3 −2 −1 −1

x 2 = − 254 ( y − 48 ) 100. Vertex: ( 0, 0 ) y 2 = 4 px

Point: (1000, 800 )

−2

25 − 16 =p

25 x 2 = 4 ( − 16 ) ( y − 48 )

(0, 32 (

(b) When x = 4, 6 y = 16

Depth:

16 8 = . 6 3

8 ≈ 2.67 inches 3

800 2 = 4 p (1000 )  p = 160 y 2 = 4 (160 ) x y 2 = 640 x

862

Chapter 10

Topics in Analytic Geometry

104. (a) V = 17,500 2 mi hr ≈ 24,750 mi hr

(b) p = −4100, ( h, k ) = ( 0, 4100 )

( x − 0 ) = 4 ( −4100 )( y − 4100 ) x 2 = −16,400 ( y − 4100 ) 2

105. (a) x 2 = −

v2 ( y − s) 16

( 28 ) x =− 2

( y − 100 ) 16

x 2 = −49 ( y − 100 )

(b) The ball hits the ground when y = 0. x 2 = −49 ( 0 − 100 )

(

109. The slope of the line joining 2, − 2 2

(

is −2 2

) 2 = − 2 . The slope of the tangent line is

2 = 2 2. Thus, the tangent line is:

2 ( x − 2) 2 2y + 4 2 = 2 x − 2 2 y+2 2 =

2 x − 2 y = 6 2.

(

)

5. The slope of the tangent line is

2 −2 5 = −1

5. Thus, the tangent line is:

(

x = 4900

y−2 = 5 x+2 5

x = 70 The ball travels 70 feet.

y − 2 = 5 x + 10

x2 =

−v 2 ( y − s) 16

( 792 )

( 0 − 30,000 ) 16 x 2 = 1,176,120,000 x ≈ 34,294.6 The distance is about 34,249.6 feet. x2 = −

107. The slope of the line joining (3, − 4) and the center is − 43 . The slope of the tangent line at (3, − 4) is 34 .

Thus, the tangent line is: 3 y + 4 = ( x − 3) 4 4 y + 16 = 3 x − 9

3 x − 4 y = 25. 108. The slope of the line joining (−5, 12) and the center is − 125 . The slope of the tangent line at ( − 5, 12) is 125 .

Thus, the tangent line is: 5 y − 12 = ( x + 5 ) 12 12 y − 144 = 5 x + 25

5 x − 12 y + 169 = 0.

)

110. The slope of the line joining −2 5, 2 and the center is

540 mi 5280 ft 1 hr 1min ⋅ ⋅ ⋅ = 792 ft/s 106. 1 hr 1 mi 60 min 60 s s = 30,000 The crate hits the ground when y = 0.

) and the center

)

5 x − y + 12 = 0.

111. The slope of the line joining ( 4, 9) and the center

1−9 4 = . The slope of the tangent −2 − 4 3

( − 2, 1) is m =

3 line is m⊥ = − . Thus, the tangent line is: 4 3 ( x − 4) 4 3 y −9 = − x +3 4 3 y = − x + 12 4 y −9 = −

112. The slope of the line joining

(0, 3) is m =

( 7, 0) and the center

3−0 3 3 7 = − = − . The slope 7 0− 7 7

of the tangent line is m⊥ =

7 . Thus, the tangent line 3

is: y −0 =

7 x − 3

y =

7 7 x − 3 3

(

)

113. False. The center is ( 0, − 5 ) . 114. True. ( ± r , 0 ) and ( 0, ± r ) are on the graph. 115. False. A circle is a conic section. 116. True. The vertix is the closest point to the directrix or focus.

Section 10.2 117. False. The directrix y = − 14 is below the x-axis.

For the upper half of the parabola, y − 3 = 6 ( x + 1) y = 6 ( x + 1) + 3.

122. ( y + 1) = 2 ( x − 2 ) 2

For the lower half of the parabola,

5 4 3 2 1 −5 −4 −3 −2 −1

863

121. ( y − 3) = 6 ( x + 1)

118. True. If the vertex and focus of a parabola are on a horizontal line, then the directrix of the parabola is a vertical line. 119. The graph of x 2 + y 2 = 0 is a single point, ( 0, 0 ) .

Ellipses

y + 1 = − 2 ( x − 2) 1 2 3 4 5

y = −1 − 2 ( x − 2 ) .

−2 −3 −4 −5

The plane intersects the double-napped cone at the vertices of the cones. 120. (a) A circle is formed when a plane intersects the top or bottom half of a double-napped cone and is perpendicular to the axis of the cone. (b) An ellipse is formed when a plane intersects only the top or bottom half of a double-napped cone but is not perpendicular to the axis of the cone, is not parallel to the side of the cone, and does not intersect the vertex. (c) A parabola is formed when a plane intersects the top or bottom half of a double-napped cone, is parallel to the side of the cone, and does not intersect the vertex. (d) A hyperbola is formed when a plane intersects both halves of a double-napped cone, is parallel to the axis of the cone, and does not intersect the vertex.

123. f ( x ) = 3 x 3 − 4 x + 2

Relative maximum: ( −0.67, 3.78 ) Relative minimum: ( 0.67, 0.22 ) 124. f ( x ) = 2 x 2 + 3 x

Relative minimum: ( −0.75, −1.13 ) 125. f ( x ) = x 4 + 2 x + 2

Relative minimum: ( −0.79, 0.81) 126. f ( x ) = x 5 − 3 x − 1

Relative minimum: ( 0.88, − 3.11) Relative maximum: ( −0.88, 1.11)

Section 10.2 Ellipses 1.

ellipse, foci

major axis, center

minor axis

c a

8. The ellipse is elongated since its eccentricity is closer to 2 15 1, e = ≈ 0.968. 8 9.

In Exercises 5 − 8, use the equation of the ellipse

a = 3, b = 2 Vertical major axis Matches graph (b).

x 2 y2 + = 1. 22 82

The major axis is vertical since 8 > 2.

The length of the major axis is 2a or 2 ( 8 ) = 16 units.

The length of the minor axis is 2b or 2 ( 2 ) = 4 units.

x 2 y2 + =1 4 9 Center: ( 0, 0 )

10.

x 2 y2 + =1 9 4 Center: ( 0, 0 ) a = 3, b = 2 Horizontal major axis Matches graph (a).

864

Chapter 10

Topics in Analytic Geometry

11.

( x − 2) + y + 1 2 = 1 ( ) 16 2

Center: ( 2, − 1) a = 4, b = 1 Horizontal major axis Matches graph (c).

( x + 2) + ( y + 2) = 1 2

12.

9 4 Center: ( −2, − 2 )

17. Center: (0, 0)  h = 0, k = 0

Foci: (0, ± 5)  c = 5 Vertical major axis: 14 units  a = 7 a2 = b2 + c2 7 2 = b 2 + 52 24 = b 2 2 6 = b

( x − h) + ( y − k ) 2

a = 3, b = 2 Horizontal major axis Matches graph (d).

13. Center: ( 0, 0 )  h = 0, k = 0 a = 4, b = 2 Vertical major axis x 2 y2 + =1 4 16

Foci: ( ±2, 0 )  c = 2 Horizontal major axis: 10 units  a = 5

a 2 = b2 + c2 52 = b 2 + 2 2 21 = b 2

( x − h) + ( y − k ) = 1 2

x y2 + =1 25 21 19. Vertices: (0, ± 4)  a = 4

Center: (0, 0)  h = 0, k = 0

x 2 4 y2 + =1 4 9

Vertical major axis

15. Center: ( 0, 0 )  h = 0, k = 0 a = 3, c = 2  b = 9 − 4 = 5 Horizontal major axis x 2 y2 + =1 9 5

16. Vertices: ( 0, ± 8 )  a = 8

Foci: ( 0, ± 4 )  c = 4

b2 = a 2 − c2 = 64 − 16 = 48 Center: ( 0, 0 )  ( h, k )

( x − h) + ( y − k ) = 1 2

21 = b

18. Center: ( 0, 0 )  h = 0, k = 0

Vertices: ( ±2, 0 )  a = 2

x 2 y2 + =1 a 2 b2 x2 y2 + =1 2 2 2 (3)

x2 y2 + =1 24 49

14. Center: ( 0, 0 )  h = 0, k = 0 3 3  Endpoints of minor axis:  0, ±   b = 2 2 

a x 2 y2 + =1 48 64

( x − h) + ( y − k ) 2

x2 y2 + =1 2 b 16 Point: (3, 1) 32 12 + =1 2 16 b 9 15 = b2 16 144 = 15b 2 144 = b2 15 x2 y2 + =1 144 16 15 y2 15 x 2 + =1 144 16 y2 5x2 + =1 48 16

Section 10.2 20. Vertices: ( ± 8, 0)  a = 8

Center: (0, 0)  h = 0, k = 0 Horizontal major axis

( x − h) + ( y − k ) 2

Center: ( 3, 5 )  h = 3, k = 5 Minor axis of length 6  b = 3 Vertical major axis

( x − h) + ( y − k ) = 1

x2 y2 + 2 =1 64 b

( x − 3) + ( y − 5) = 1

Point: (5, − 3)

Major axis of length 36  a = 18 Vertical major axis

a 2 = b2 + c2

576 = b2 39

182 = b 2 + 42 324 − 16 = b 2

x2 y2 + =1 576 64 39

308 = b 2

( x − h)

x2 39 y 2 + =1 64 576

( x − 0) 308

Center: ( 2, − 1)  h = 2, k = −1 Endpoints of minor axis: ( 2, 0 ), ( 2, − 2 )  b = 1

( x − h) + ( y − k ) = 1 2

( x − 2 ) + ( y + 1) = 1 2

22. Center: ( 2, 3 )  h = 2, k = 3

( y − k)

( y − 4)

324

( y − 4) x2 + 308 324

21. Vertices: ( 0, − 1), ( 4, − 1)  a = 2

x2 13 y 2 + =1 64 192

Center: (0, 4)  h = 0, k = 4

576 = 39b 2

25. Foci: (0, 0), (0, 8)  c = 4

(− 3) 52 + 2 =1 64 b 9 39 = b2 64

26. Center: ( 2, 0 )  h = 2, k = 0 c = 2, a = 3  b 2 = a 2 − c 2 = 9 − 4 = 5 Horizontal major axis

( x − 2 ) + y2 = 1 2

27. Center: ( 3, 2 )  ( h, k )

a = 3, b = 1 Vertical major axis

a = 3c Foci: (1, 2 ) , ( 5, 2 )  c = 2, a = 6

( x − h) + ( y − k ) = 1

b2 = a 2 − c 2 = 36 − 4 = 32

( x − 2 ) + ( y − 3) = 1 2

865

24. Vertices: ( 3, 1), ( 3, 9 )  a = 4

Ellipses

23. Center: ( 4, 2 )  h = 4, k = 2

( x − h) + ( y − k ) = 1 2

( x − 3) + ( y − 2 ) = 1 2

a = 4, b = 1  c = 16 − 1 = 15 Horizontal major axis

( x − 4) + ( y − 2) = 1 2

866

Chapter 10

Topics in Analytic Geometry

28. Center: (0, 4)  h = 0, k = 4

Vertices: (0, −1), (0, 9)  a = 5

a = 5c  5 = 5c  c = 1

Foci: ( 4, − 1 ± 3 )  ( 4, − 4 ) , ( 4, 2 )

24 = b 2

Vertical major axis 2

( y − 4) +

( y − 4) x2 + 24 25

( x − 0) 24

29.

( y − k)

+ 2

6 4

2 −4 − 2 −2

−8

( x + 3) + ( y − 2 ) = 1 2

32.

c = 64 − 9 = 55

Vertices: ( ±8, 0 ) Foci: ± 55, 0

12 16 Center: ( −3, 2 )

a = 4, b = 2 3, c = 16 − 2 = 2

)

Foci: ( −3, 2 ± 2 )  ( −3, 0 ) , ( −3, 4 ) Vertices: ( −3, 2 ± 4 )  ( −3, − 2 ) , ( −3, 6 )

c 55 = a 8

10 8 6 4 2 −4 −2

−6

a = 8, b = 3,

− 10

−4

x 2 y2 + =1 64 9 Center: ( 0, 0 )

c 3 = a 5

(

16 25 Center: ( 4, − 1)

Vertices: ( 4, − 1 ± 5 )  ( 4, − 6 ) , ( 4, 4 )

52 = b 2 + 12

a = 5, b = 4, c = 3

a2 = b2 + c2

( x − h)

( x − 4 ) + ( y + 1) = 1 2

31.

c 2 1 = = a 4 2 y

2 4

3 2

−4 −6 −8 − 10

1 −7

−4 −3 −2 −1

−2

30.

x 2 y2 + =1 16 81 Center: ( 0, 0 )

a = 9, b = 4,

c = 81 − 16 = 65

6 4 2

Vertices: ( 0, ± 9 )

(

Foci: 0, ± 65 c 65 e= = a 9

)

−10 −8 − 6

−2

6 8 10

−4 −6 −10

Section 10.2

( x + 5) + y − 1 2 = 1 ( ) 9 2

33.

35. (a)

Center: ( −5, 1) 3 , b = 1, c = 2

9 5 −1 = 4 2

 5   5  Foci:  −5 + , 1  ,  −5 − , 1     2 2     Vertices: 3   7   3   13    −5 + , 1  =  − , 1  ,  −5 − , 1  =  − , 1  2 2 2        2 

x 2 + 9 y 2 = 36

(b)

a = 6, b = 2, c = 36 − 4 = 32 = 4 2

Center: ( 0, 0 ) Vertices: ( ±6, 0 )

(

Foci: ±4 2 , 0 e=

)

c 4 2 2 2 = = 6 3 a y

5 2 5 = 32 3 y

−10 −8

6 8 10

2 1

−7 −6 −5 −4 −3 −2 −1

36. (a) 16 x 2 + y 2 = 16

−2 −3

x2 +

−4

( x + 2) + 2

( y + 4) 1 4

(b)

y2 =1 16

a = 4, b = 1, c = 16 − 1 = 15

Center: ( 0, 0 )

Vertices: ( 0, ± 4 )

1 3 a = 1, b = , c = a 2 − b 2 = 2 2 Center: ( −2, − 4 )     3 3 Foci:  −2 + , − 4  ,  −2 − , − 4     2 2    

(

Foci: 0, ± 15 e=

c 3 = a 2

)

c 15 = a 4 y

Vertices: ( −3, − 4 ) , ( −1, − 4 ) e=

−4 −6 −8 −10

34.

867

x2 y2 + =1 36 4

Ellipses

1 −5 −4 −3 −2

2 3 4 5

−3

−2

−1

1 −1

−4 −5

−2 −3 −4 −5

868

Chapter 10

37. (a)

Topics in Analytic Geometry 39. (a)

49 x 2 + 4 y 2 − 196 = 0

(

) (

9 x 2 + 4 x + 4 + 4 y 2 − 6 y + 9 = −36 + 36 + 36

( x + 2 ) + ( y − 3) = 1

x y + =1 4 49 a = 7, b = 2, c = 49 − 4 = 45 = 3 5

Center: ( 0, 0 )

(b)

(

Foci: 0, ± 3 5

(

Foci: −2, 3 ± 5

)

c 5 = 3 a y

(c)

2 −8 −6 −4

−6 −5 −4 −3 −2 −1 −8

40. (a)

4 x 2 + 49 y 2 − 196 = 0

(

9 ( x − 3 ) + 4 ( y + 5 ) = 144 2

Center: ( 0, 0 )

(b)

(

Foci: 3, − 5 ± 2 5

)

a = 6, b = 4, c = 20 = 2 5

Center: ( 3, − 5 )

Vertices: ( ±7, 0 )

(

( x − 3) + ( y + 5) = 1

a = 7, b = 2, c = 49 − 4 = 45 = 3 5

Foci: ±3 5, 0

)

x2 y2 + =1 49 4

)

Vertices: ( 3, − 5 ± 6 )  ( 3, 1) , ( 3, − 11)

c 3 5 e= = a 7

c 2 5 5 = = a 6 3 y

(c)

6 4

−2

) (

4 x + 49 y = 196

−8

9 x 2 − 6 x + 9 + 4 y 2 + 10 y + 25 = −37 + 81 + 100

(c)

−2

(b)

Vertices: ( −2, 6 ) , ( −2, 0 )

c 3 5 e= = 7 a (c)

4 a = 3, b = 2, c = 5

Center: ( −2, 3 )

Vertices: ( 0, ± 7 )

38. (a)

)

49 x + 4 y = 196

(b)

9 x 2 + 4 y 2 + 36 x − 24 y + 36 = 0

−2 2

10 12

−4 −6 −8

− 10 − 12

Section 10.2 6 x 2 + 2 y 2 + 18 x − 10 y + 2 = 0

41. (a)

1  12  x 2 − 1 +  + 20 y 2 + 2 y + 1 = 37 + 3 + 20 4 

(

3 2

5 2

4 (b) a = 2 3, b = 2, c = 2 2

2 1  12  x −  + 20 ( y + 1) = 60 2 

(b)

a = 5, b = 3, c = 5 − 3 = 2 1  Center:  , − 1  2   1  Vertices:  ± 5, − 1 2  

 3 5 Center:  − ,   2 2  3 5  Foci:  − , ± 2 2   2 2   3 5  Vertices:  − , ± 2 3  2 2   c e= = a

1  Foci:  ± 2, − 1  2   e=

c = a

6 = 3 3

10 5

5 y

(c)

2 1

(c)

)

3 5   6  x +  + 2  y −  = 24 2 2  

(x + ) + (y − ) =1

869

12 x 2 + 20 y 2 − 12 x + 40 y − 37 = 0

43. (a)

9 25  27 25   6  x 2 + 3x +  + 2  y2 − 5y + +  = −2 + 4 4  2 2  

Ellipses

−3

−2

−1

−2

−3 2

−6

−4

42. (a)

(

36 x 2 + 9 y 2 + 48 x − 36 + 43 = 0

44. (a)

−2

4 4  36  x 2 + x +  + 9 y 2 − 4 y + 4 = −43 + 16 + 36 3 9 

(

25   x 2 − 6 x + 9 + 4  y2 + 5 y +  = 2 + 9 + 25 4  

)

2 2  36  x +  + 9 ( y − 2 ) = 9 3 

( x − 3) + 4  y + 2  = 36 5



( x + ) + ( y − 2) = 1



(b)

36 9 a = 6, b = 3, c = 36 − 9 = 27 = 3 3

(b)

−3

−1 −2 −3 −4 −6 −7 −8

a = 1, b =

1 1 3 , c = 1− = 2 4 2

 2 3 Foci:  − , 2 ±   3 2  

c 3 = a 2

c 3 = 2 a y

(c)

6 5 4 3 2 1

 2  Center:  − , 2   3   2   2   2  Vertices:  − , 2 ± 1    − , 1  ,  − , 3  3    3   3 

5  Foci:  3 ± 3 3, −  2  5  5  Vertices:  9, −  ,  −3, −  2 2    

(c)

1 4

5  Center:  3, −  2 

2 3

( x − 3 ) + ( y + 25 ) = 1 2

)

1 2 3 4 5 6 7

9 10

−2

−1

870

45.

Chapter 10

Topics in Analytic Geometry 50. Vertices: (0, ± 8), a = 8, h = 0, k = 0

x 2 y2 + =1 4 9 a = 3, b = 2,

Eccentricity: e =

c = 9−4 = 5 e=

c 5 = 3 a

b 2 = a 2 − c 2 = 64 − 16 = 48

x 2 y2 + =1 46. 25 49 a = 7, b = 5 2

Vertical major axis

a =b +c

x2 y2 + 2 =1 2 b a

x2 y2 + =1 48 64

49 = 25 + c 2 24 = c 2 2 6 =c

51. Foci: (1, 1) and (1, 13)  c = 6

c 2 6 e= = a 7

Eccentricity: e =

x 2 + 9 y 2 − 10 x + 36 y + 52 = 0

47.

( x − 10 x + 25) + 9 ( y + 4 y + 4 ) = −52 + 25 + 36 2

(

a = 9

a2 = b2 + c2 81 = b2 + 36 The center is the midpoint between the foci, (1, 7), and

4 x 2 + 3 y 2 − 8 x + 18 y + 19 = 0

( x − 1)

the major axis is vertical.

) (

)

4 ( x − 1) + 3 ( y + 3 ) = 12 2

( x − 1) + ( y + 3) = 1 2

3 4 a = 2, b = 3, c = 4 − 3 = 1

c 1 e= = a 2 49. Vertices: ( ± 5, 0), a = 5, h = 0, k = 0

Eccentricity: e =

3 c = 5 a

3 c = 5 5 3 = c

( x − 1)

( y − 7) a

( y − 7)

52. Foci: ( − 6, 5) and ( 2, 5)  c = 4

Eccentricity: e = e =

4 5

c 4 4  = a 5 a a = 5

a 2 = b2 + c2 25 = b 2 + 16 9 = b2

b = a − c = 25 − 9 = 16 Horizontal major axis 2

c 2 6  = a 3 a 2a = 18

c 2 2 = 3 a

4 x 2 − 2 x + 1 + 3 y 2 + 6 y + 9 = −19 + 4 + 27

2 3

45 = b2

a = 3, b = 1, c = 9 − 1 = 2 2

48.

e =

( x − 5) + 9 ( y + 2 ) = 9 2 2 ( x − 5) + ( y + 2 ) = 1

1 c = 2 a 1 c = 2 8 c = 4

x y + 2 =1 a2 b x2 y2 + =1 25 16

The center is the midpoint between the foci, ( − 2, 5), and the major axis is horizontal.

( x + 2)2 + ( y − 5)2 = 1 a2

( x + 2) 25

( y − 5) 9

Section 10.2 53. Center: ( 0, 0 ) , e = 0.97

x 2 y2 54. Let 2 + 2 = 1 be the equation of the ellipse. Then a b b = 2 and a = 3  c 2 = a 2 − b 2 = 9 − 4 = 5. Thus, the

π ab = 2π r 2

(

π a (10 ) = 2π (10 )

a = 20 Length of major axis: 2 a = 2 ( 20 ) = 40 units

58. For

x 2 y2 + = 1, we have c2 = a2 − b2 . a 2 b2

When x = c,

)

c2 y2 a 2 − b2  2 2 + = 1  y = b 1 −   a 2 b2 a2  

2 a = 6 feet.

b4 a2 2b 2  2y = . a

 y2 =

55. The length of the major axis and minor axis are 280 millimeters and 160 millimeters, respectively. Therefore, 2 a = 280  a = 140 and 2b = 160  b = 80. a 2 = b2 + c2 140 2 = 80 2 + c 2

59.

13,200 = c 2 13,200 = c

π a (10 ) = 200

tacks are placed at ± 5, 0 . The string has a length of

x2 y2 + =1 9 1 Horizontal major axis

20 33 = c The kidney stone and spark plug are each located at a focus, therefore they are 2c millimeters apart, or

)

a = 3, b = 1, c =

8 = 2 2

Points on the ellipse: ( ± 3, 0), (0, ±1)

2 20 33 = 40 33 ≈ 229.8 millimeters apart. Length of latus recta:

56. a + c = 947 + 6378 = 7325 a − c = 228 + 6378 = 6606 2 a = 13,931 a = 6965.5 c = 7325 − 6965.5 = 359.5 c e = ≈ 0.0516 a a−c b

2(1) 2b 2 = a 3

2 3

1  Additional points:  ± 2 2, ±  3  y 3 2

(− 2 2, 13 ) −2

−1

(− 2 2, − 13 ) − 2

−a

871

57. Area of ellipse = 2 ( area of circle )

2 a = 35.88  a = 17.94  a 2 ≈ 321.84 c c e =  0.97 =  c = 17.4018 a 17.94 c 2 = a 2 − b 2  b 2 = a 2 − c 2 ≈ 19.02 x2 y2 + =1 Ellipse: 321.84 19.02

(

Ellipses

(2 2, 13 ) 1

(2 2, − 13 )

−3

−b a+c

872

Chapter 10

60.

x2 y2 + =1 4 25

Topics in Analytic Geometry 62. 6 x 2 + 3 y 2 = 12

a = 5, b = 2, c =

x2 y2 + =1 2 4

Vertical major axis

a = 2, b =

Points on the ellipse: (0, ± 5), ( ± 2, 0)

Vertical major axis

2( 2) 2b 2 = a 5

Length of latus recta:

 4 Additional points:  ± , ±  5

(

)

(

4 , 5

−4

(

− 4, − 5

(

)

y 3

−2

)

( ) =2

−6

2, 0

2 2 2b 2 Length of latus recta: = a 2

 21  

Additional points: ±1, ±

(

Points on the ellipse: (0, ± 2), ±

8 5

− 4, 5

2, c =

)− 6 (

4 ,− 5

(− 1,

−3

)

−2

(− 1, −

(1,

−1

2) x

(1, −

−3

61. 16 x 2 + 4 y 2 = 64

63. True. If e ≈ 1, then the ellipse is elongated, not circular.

x2 y2 + =1 4 16

a = 4, b = 2, c =

64. True. The ellipse is inside the circle.

12 = 2 3

65. 16 x 2 + 25 y 2 − 32 x + 50 y + 16 = 0

Vertical major axis Points on the ellipse: (0, ± 4), ( ± 2, 0)

16( x 2 − 2 x + 1) + 25( y 2 + 2 y + 1) = −16 + 16 + 25

2( 2) 2b 2 Length of latus recta: = a 4

( x − 1)

(

Additional points: ±1, ± 2 3

16( x − 1) + 25( y + 1) = 25 2

= 2

25 16

)

(− 1, 2

(1, 2 3 )

66.

(1, − 2

−3

( y + 1)

9 x 2 + 25 y 2 − 36 x − 50 y + 61 = 0 9( x − 2) + 25( y − 1) = 0 2

−2

(− 1, − 2

9( x 2 − 4 x + 4) + 25( y 2 − 2 y + 1) = − 61 + 36 + 25

−1 −1

No. The equation represents an ellipse.

3) 3

−4 −3

Yes. The equation represents a point. 67.

x2 y2 + = 1, a = 328, b = 327 328 327 a 2 = b2 + c2 328 = 327 + c 2 1 = c2 1=c 1= c

c 1 = ≈ 0.055. a 328 Because the eccentricity is close to 0, the ellipse is nearly circular. The eccentricity is e =

Section 10.3 68. eB < e A < eC

71.

69. (a) The length of the string is 2 a.

(b) The path is an ellipse because the sum of the distances from the two thumbtacks is always the length of the string, that is, it is constant.

873

x 2 y2 + =1 a 2 b2 The sum of the distances from any point on the ellipse to the two foci is constant. Using the vertex ( a, 0 ) , you have

( a + c ) + ( a − c ) = 2 a. From the figure, 2 b2 + c 2 = 2a  a 2 = b2 + c 2 .

70. Center: ( 6, 2 )

Foci: ( 2, 2 ) , (10, 2 )  c = 4

( a + c ) + ( a − c ) = 2a = 36  a = 18

b2 = a 2 − b2  b = 182 − 16 = 308 Horizontal major axis

b2 + c2

c −a

( x − 6) + ( y − 2) = 1 2

Hyperbolas

−c

324

−b

308

72. Arithmetic: d = −11 73. Geometric: r = 12 74. Geometric: r = 2 75. Arithmetic: d = 1 76.

 3 = 1093 n

n=0

77.  4 ( 34 )

n −1

≈ 15.099

n =1

Section 10.3 Hyperbolas 1.

hyperbola

transverse axis, center

The center is ( h, k ).

The graph of a hyperbola has two distinct asymptotes.

Vertical transverse axis

The asymptotes intersect at the center of the hyperbola ( h, k ).

Matches graph (d).

Ellipse x2 y2 − =1 25 9

Center: (0, 0)

a = 3, b = 5, c =

x2 ( y + 3) − 4 9

Center: (0, − 3)

Center: (0, 0)

a = 5, b = 3, c =

y2 x2 − =1 9 25

a = 2, b = 3

Horizontal transverse axis

Horizontal transverse axis Matches graph (b).

Matches graph (a).

874

Chapter 10

10.

( y − 2) 2 − ( x − 2) 2 = 1

15. Vertices: ( 4, 1) , ( 4, 9 )  a = 4

Center: ( 2, 2); a = 2, b = 3

b 2 = c 2 − a 2  25 − 16 = 9

Topics in Analytic Geometry

Vertical transverse axis Matches graph (c). 11. Vertices: ( 0, ± 2 )  a = 2

Foci: ( 0, ± 4 )  c = 4 b 2 = c 2 − a 2 = 16 − 4 = 12

Foci: ( 4, 0 ) , ( 4, 10 )  c = 5 Center: ( 4, 5 ) = ( h, k )

( y − k ) − ( x − h) = 1 2

( y − 5) − ( x − 4 ) = 1 2

Center: ( 0, 0 )

16. Vertices: ( −2, 1) , ( 2, 1)  a = 2

y2 x 2 − =1 a 2 b2 y2 x 2 − =1 4 12

Foci: ( −3, 1) , ( 3, 1)  c = 3

Center: ( 0, 1) = ( h, k ) b2 = c2 − a 2  9 − 4 = 5

( x − h) − ( y − k ) = 1 2

12. Vertices: ( ±3, 0 )  a = 3

Foci: ( ±6, 0 )  c = 6

x 2 ( y − 1) − =1 4 5 2

b 2 = c 2 − a 2 = 36 − 9 = 27

Center: ( 0, 0 )

17. Vertices: ( 2, 3 ) , ( 2, − 3 )  a = 3

x 2 y2 − =1 a 2 b2 x 2 y2 − =1 9 27

Solution point: ( 0, 5 ) Center: ( 2, 0 ) = ( h, k )

( y − k ) − ( x − h) = 1 2

13. Vertices: ( 2, 0 ) , ( 6, 0 )  a = 2

Foci: ( 0, 0 ) , ( 8, 0 )  c = 4 b 2 = c 2 − a 2  16 − 4 = 12

Center: ( 4, 0 ) = ( h, k )

y2 ( x − 2 ) − =1 9 b2 2

b2 =

( x − h) − ( y − k ) = 1 2

( x − 4 ) − y2 = 1 2

14. Vertices: ( 2, 3 ) , ( 2, − 3 )  a = 3

9( x − 2)

y2 − 9

9 ( −2 )

25 − 9

36 9 = 16 4

y2 ( x − 2 ) − =1 9 9 4 2

Center: ( 2, 0 ) = ( h, k ) Foci: ( 2, 5 ) , ( 2, − 5 )  c = 5 b 2 = c 2 − a 2  25 − 9 = 16

( y − k ) − ( x − h) = 1 2

y2 ( x − 2 ) − =1 9 16 2

Section 10.3 18. Center: ( 0, 1) , a = 2

21.

( x − h) − ( y − k ) = 1 2

x 2 − y2 = 1 Center: ( 0, 0 ) ; Vertices: ( ±1, 0 )

(

( y − 1) = 1 x − 4 b2 2

Foci: ± 2 , 0

)

b Asymptotes: y = ± x = ± x a

Solution point: ( 5, 4 ) 25 9 − =1 4 b2 9 21 = b2 4 36 12 2 b = = 21 7

2 1

−2

2 −1

x 2 ( y − 1) − =1 4 12 7 2

−2

19. Vertices: ( 0, 4 ) , ( 0, 0 )

22. y 2 − x 2 = 1

Center: ( 0, 2 ) , a = 2

a = 1, b = 1, c =

( y − k ) − ( x − h) = 1

Center: (0, 0); Vertices: (0, ±1)

(

Foci: 0, ±

( y − 2) − x2 = 1 2

Passes through

(

)

9 5 −1 = 2 4 b b2 = 4  b = 2

−3

−2

( y − 2) − x2 = 1 2

( y − k ) − ( x − h) = 1 2

y 2 ( x − 1) − =1 4 b2 2

(

−3

20. Center: (1, 0 ) , a = 2

−

−1 −2

)

( −1 − 2 ) − 5 = 1

a Asymptotes: y = ± x = ± x b

5, − 1

875

a = 1, b = 1, c = 2

Hyperbolas

23.

y2 x2 − =1 1 16

a = 1, b = 4, c =

Center: (0, 0)

)

Vertices: (0, ±1); Foci: 0, ± 17

5 1 − =1 4 b2 1 1 = b=2 b2 4

a 1 Asymptotes: y = ± x = ± x b 4

Solution point: 0,

(

)

5 4 3 2

y 2 ( x − 1) − =1 4 4 2

−5 −4

4 5

−2 −3 −4 −5

876

Chapter 10

24.

x2 y2 − =1 9 1

Topics in Analytic Geometry

26.

a = 5, b = 6, c =

Center: (0, 0) 32 + 12 =

a = 3, b = 1, c =

(

Vertices: ( ± 5, 0)

)

(

61, 0

b 6 Asymptotes: y = ± x = ± x a 5

−1

10 8 6 4

−2 −3 −4 −5

25.

)

Foci: ±

b 1 Asymptotes: y = ± x = ± x a 3 5 4 3 2 1

Center: (0, 0)

Vertices: ( ± 3, 0); Foci: ± 10, 0

−5

x2 y2 − =1 25 36

− 10 − 8 − 6

4 6 8 10 −4 −6 −8 − 10

y2 x2 − =1 16 4 42 + 22 =

a = 4, b = 2, c =

20 = 2 5

27.

( x − 3)

−

( y − 1)

a = 3, b = 1, c =

Center: (0, 0)

(

Center: (3, 1)

Vertices: (0, ± 4) Foci: 0, ± 2 5

Vertices: (0, 1), (6, 1)

)

(

)

Foci: 3 ±

a Asymptotes: y = ± x = ± 2 x b

10, 1

Asymptotes: y = k ±

b 1 ( x − h) = 1 ± ( x − 3) a 3

8 6

−8 −6 −4 −2

−6

−2

−6

−4

−8

−6 −8

28.

( x − 2 ) 2 − ( y + 5) 2 = 1 4

Center: ( 2, − 5)

a = 2, b = 5, c =

Vertices: (0, − 5), ( 4, − 5)

(

Foci: 2 ±

)

−4 −6

29, − 5

Asymptotes: y = k ±

−4 −2 −2

−8

b 5 ( x − h) = − 5 ± ( x − 2) a 2

Section 10.3

29.

( y + 6)2 − ( x − 2)2 = 1

31. (a)

1 16

1 ,c = 4

4 x 2 − 9 y 2 = 36

1 = 16

a = 3, b = 2, c = 9 + 4 = 13

17 4

Vertices: ( ±3, 0 )

(

Foci: ± 13, 0

 Foci:  2, − 6 ± 

Asymptotes: y = ±

17   4 

a y = k ± ( x − h) b 1 = −6 ± ( x − 2) 14

−5 −4

−2

4 5

−2 −3 −4 −5

= − 6 ± 4( x − 2) y

32. (a)

1 2 3 4 5 6

25 x 2 − 4 y 2 = 100

x 2 y2 − =1 4 25 (b) Center: ( 0, 0 )

−2 −3 −4 −5 −6 −7 −8 −9

a = 2, b = 5, c = 4 + 25 = 29 Vertices: ( ±2, 0 )

(

Foci: ± 29, 0

( y + 4)2 − ( x + 3)2 = 1 1 9

1 4

(c)

1 1 ,b = ,c = 3 2

1 1 + = 9 4

13   6 

b 5 x=± x a 2

13 6

11   13   Vertices:  − 3, − ,  − 3, −  3  3 

 Foci:  − 3, − 4 ± 

)

Asymptotes: y = ±

Center: ( − 3, − 4) a =

b 2 x=± x a 3

Asymptotes:

30.

)

Vertices: ( 2, − 5), ( 2, − 7)

−4 −3 −2 −1

877

x 2 y2 − =1 9 4 (b) Center: ( 0, 0 )

Center: ( 2, − 6) a = 1, b =

Hyperbolas

6 4

−8 − 6 −4

−6 −8

Asymptotes: y = k ±

1 2 a ( x − h) = − 4 ± 13 ( x + 3) = − 4 ± ( x + 3) 3 b 2 y

1 −6 −5 −4 −3 −2 −1 −1

−3 −4 −5

878

Chapter 10

Topics in Analytic Geometry 35. (a)

33. (a) 6 y 2 − 3 x 2 = 24

9 x 2 − y 2 − 36 x − 6 y + 18 = 0

(

y2 x2 − =1 4 8

) (

( x − 2 ) − ( y + 3) = 1 2

(b) a = 2, b = 2 2, c = 2 3 Center: (0, 0)

(b)

(

Foci: 0, ± 2 3

1 a = 1, b = 3, c = 10

Vertices: (1, − 3 ) , ( 3, − 3 )

)

(

Foci: 2 ± 10, − 3

Asymptotes: y = ±

2 2

= ±

(c)

2 x 2

)

Asymptotes: y = k ±

b ( x − h) = −3 ± 3( x − 2 ) a

5 4 3

−6 −4 −2

−4 −6

1 −5 −4 −3 −2 −1

2 3 4 5

−8

−3 −4 −5

36. (a)

x 2 − 9 y 2 + 36 y − 72 = 0

(

)

x 2 − 9 y 2 − 4 y + 4 = 72 − 36

34. (a) 3 x 2 − 2 y 2 = 18

x 2 − 9 ( y − 2 ) = 36 2

x2 y2 − =1 6 9

(b) a =

Center: ( 2, − 3 )

Vertices: (0, ± 2)

(c)

)

9 x 2 − 4 x + 4 − y 2 + 6 y + 9 = −18 + 36 − 9

6, b = 3, c =

(b)

Center: (0, 0)

(

x2 ( y − 2) − =1 36 4 a = 6, b = 2, c = 36 + 4 = 2 10

Vertices: ±

6, 0

(

)

Center: ( 0, 2 )

)

Vertices: ( ±6, 2 )

(

Foci: ± 15, 0

Foci: ±2 10, 2

b 3 6 Asymptotes: y = ± x = ± x = ± x 2 a 6

Asymptotes: y = k ± (c)

12 8 4

1 2 3 4 5 −2 −3 −4 −5

b 1 x − h) = 2 ± x ( a 3

5 4 3 2

−5 −4 − 3

)

−8

−4

−8 − 12

Section 10.3 39. (a)

37. (a) 2 x 2 − 7 y 2 + 16 x + 18 = 0

(

Hyperbolas

9 y 2 − x 2 + 2 x + 54 y + 62 = 0

) (

)

2( x 2 + 8 x + 16) − 7 y 2 = −18 + 32

9 y 2 + 6 y + 9 − x 2 − 2 x + 1 = −62 − 1 + 81

2( x + 4) − 7 y 2 = 14

( y + 3) − ( x − 1) = 1 2

( x + 4) 2 − y 2 = 1 7

(b) a =

(b)

7, b =

(

7, 0

2 18 a = 2 , b = 3 2, c = 2 5

(

Vertices: 1, − 3 ± 2

Center: ( − 4, 0) Vertices: − 4 ±

Center: (1, − 3 )

2, c = 3

(

Foci: 1, − 3 ± 2 5

)

Asymptotes: y = k ±

Foci: ( − 7, 0), ( −1, 0)

)

a 1 ( x − h) = −3 ± 3 ( x − 1) b

(c)

Asymptotes:

879

b 14 y = k ± ( x − h) → y = ± ( x + 4) a 7

2 x

−6 −8 −10

−9 −8

−6

−4

40. (a)

−3 −4 −5

(

9 x 2 − y 2 + 54 x + 10 y + 55 = 0

) (

)

9 x 2 + 6 x + 9 − y 2 − 10 y + 25 = −55 + 81 − 25

( x + 3) − ( y − 5) = 1 2

3 y 2 − 5 x 2 + 6 y − 60 x − 192 = 0

38. (a)

3( y 2 + 2 y + 1) − 5( x 2 + 12 x + 36) = 192 + 3 − 180

(b)

( y + 1) − ( x + 6) 2

(b) a =

5, b =

1   8   10   Vertices:  −3 ± , 5    − , 5  ,  − , 5  3   3   3  

  10 Foci:  −3 ± , 5   3  

3, c = 2 2

Center: ( − 6, −1)

(

Vertices: − 6, −1 ±

(

Foci: − 6, −1 ± 2 2

Asymptotes: y = k ±

)

b ( x − h ) = 5 ± 3 ( x + 3) a

Asymptotes: a y = k ± ( x − h ) → y = −1 ± b (c)

1 10 a = , b = 1, c = 3 3

Center: ( −3, 5 )

3( y + 1) − 5( x + 6) = 15 2

10 8

15 ( x + 6) 3

6 4 2

−7 −6

−4 −3 −2

4 2 − 12 − 10 − 8

−4 −2 −2

−4

880

Chapter 10

Topics in Analytic Geometry

41. Vertices: ( ±1, 0 )  a = 1

45. Vertices: (1, 2 ) , ( 3, 2 )  a = 1

Center: ( 2, 2 ) = ( h, k )

Asymptotes:

b y = ±5 x  = 5 a b=5

Asymptotes: y = x , y = 4 − x

b =1 b =1 a

Center: ( 0, 0 )

( x − h) − ( y − k ) = 1 2

x 2 y2 − =1 a 2 b2 x 2 y2 − =1 1 25

( x − 2) − ( y − 2) = 1 2

Asymptotes: y = ±3 x 

46. Center: ( 3, − 3 ) , a = 3

42. Vertices: ( 0, ± 3 )  a = 3

Asymptotes: y = x − 6, y = − x

a = 3, b = 1 b

Center: ( 0, 0 )

a 3 = b=3 b b

( y − k ) − ( x − h) = 1 2

y2 x 2 − =1 a 2 b2 y2 − x2 = 1 9

( y + 3) − ( x − 3) = 1 2

Asymptotes: y = ±4 x 

47. Vertices: ( 0, 2 ) , ( 6, 2 )  a = 3

43. Foci: ( 0, ± 8 )  c = 8

a = 4  a = 4b b

c 2 = a 2 + b 2  64 = 16 b 2 + b 2

64 1024 = b2  a 2 = 17 17 y2 x 2 − =1 a 2 b2 y2 x2 − =1 1024 17 64 17

( x − 3) − ( y − 2 ) = 1 9

48. Vertices: ( 3, 0 ) , ( 3, 4 )  a = 2

Asymptotes: y =

3 b 3m Asymptotes: y = ± x  = 4 a 4m c 2 = a 2 + b 2  100 = ( 3m ) + ( 4 m ) 2

a = 4 ( 2 ) = 8, b = 3 ( 2 ) = 6

44. Foci: ( ±10, 0 )  c = 10

2=m

2 2 x, y = 4 − x 3 3

( x − h) − ( y − k ) = 1

17 y 2 17 x 2 − =1 1024 64

100 = 25m

Asymptotes: y =

b 2 = b=2 a 3 Center: ( 3, 2 ) = ( h, k )

Center: ( 0, 0 ) = ( h, k )

Center: ( 0, 0 )

2 2 x, y = 4 − x 3 3

a 2 = b=3 b 3 Center: ( 3, 2 ) = ( h, k )

( y − k ) − ( x − h) = 1 2

( y − 2 ) − ( x − 3) = 1 2

x 2 y2 − =1 a 2 b2 x 2 y2 − =1 64 36

Section 10.3 49. Foci: ( −1, 3), (9, 3)  c = 5

Asymptotes: y =

3 3 x, y = 6 − x 4 4

F2 : Your location (10,560, 0 )

P ( x, y ) : Location of lightning strike

19,800 = 9900  a 2 = 98,010,000 2 b2 = c 2 − a 2 = 13,503,600 c = 10,560, a =

3  25 = a 2 +  a  4  9 25 = a 2 + a 2 16 25 2 25 = a 16

x2 y2 − =1 98,010,000 13,503,600 y 10,000

Friend F1

25 = 16 + b 2 2

( x − h)

−

( x − 4)

−

( y − k)

−10,000

( y − 3)

2 a = 1 b a = 2b

52. The explosion occurred on the vertical line through ( 3300, 1100 ) and ( 3300, 0 ) . d2 − d1 = 4 (1100 ) = 4400

Hence, 2a = 4400

a = 2200 c = 3300 b2 = c 2 − a 2 . y

4000

c2 = a2 + b2

(3300, 1100)

3000

4 = ( 2b) + b 2 2

(−3300, 0)

4 = 5b 2

(3300, 0)

2000 1000

−4000

4 5

4 = a2 +

4 5

−4000

The explosion occurred on the hyperbola

16 = a2 5

( y − k ) − ( x − h)

( y − 4)

( x − 1) − 45

 x2   3300 2  y 2 = b 2  2 − 1 = 3300 2 − 2200 2  − 1  2 a   2200  y = −2750.

(

x 2 y2 − = 1. a 2 b2

Letting x = 3300,

Center: (1, 4) = ( h, k )

20,000

Asymptotes: y = 2 + 2 x , y = 6 − 2 x

16 5

You

(−10,560, 0) (10,560, 0)

50. Foci: (1, 2), (1, 6)  c = 2

b2 =

−20,000

Center: ( 4, 3) = ( h, k )

51. F1: Friend's location ( −10,560, 0 )

x 2 y2 − =1 a 2 b2

c2 = a2 + b2

9 = b

881

(1100 )(18 ) = 19,800

3 b = 4 a 3 b = a 4

16 = a

Hyperbolas

)

( 3300, − 2750 ) =1

882

Chapter 10

53. (a)

Topics in Analytic Geometry

x 2 y2 − =1 a 2 b2 a = 1; ( 2, 9 ) is on the curve, so 4 81 81 − 2 =1  2 = 3 1 b b 81  b2 =  b = 3 3. 3

55.

a = 5, b = 4, c = 25 + 16 = 41 The camera is 5 + 41 units from the mirror. 56. Center: ( 0, 0 )

Focus: ( 24, 0 )

x 2 y2 − = 1, − 9 ≤ y ≤ 9 1 27 (b) Because each unit is 12 foot, 4 inches is 23 of a unit. The base is 9 units from the origin, so 2 1 y =9− =8 . 3 3 25 When y = , 3

b 2 = c 2 − a 2 = 24 2 − a 2 = 576 − a 2

(

( 25 3 )  x ≈ 1.88998. x =1+

27 So the width is 2 x ≈ 3.779956 units, or 22.68 inches, or 1.88998 feet.

)

is approximately at (14.83, 0 ) . [Note: By the Quadratic Formula, the exact value of a is a = 12

Center: ( 0, 0 ) y

( 5 − 1). ]

57. 9 x 2 + 4 y 2 − 18 x + 16 y − 119 = 0 A = 9, C = 4 AC = 36 > 0, Ellipse

150

(x, 75)

58.

(−150, 0)

(150, 0) 75

150

x 2 + y 2 − 4 x − 6 y − 23 = 0 A = C = 1, Circle

d2 − d1 = (186,000 )( 0.001)

59. 16 x 2 − 9 y 2 + 32 x + 54 y − 209 = 0 A = 16, C = −9

AC = 16 ( −9 ) < 0, Hyperbola

= 186  2 a = 186  a = 93

b 2 = c 2 − a 2 = 150 2 − 932 = 13,851

x2 y2 − =1 2 93 13,851  752  x 2 = 932  1 +  ≈ 12,161.43  13,851  x ≈ 110.3 miles (b) 150 − 93 = 57 miles (c) Bay to Station 1: 30 miles Bay to Station 2: 270 miles ( 270 − 30 ) ≈ 0.00129 second 186,000 (d) In this case, d2 − d1 = 186,000 ( 0.00129) ≈ 239.94  a ≈ 120 and b 2 = c 2 − a 2 = 8100. The hyperbola

(

a ≈ ±38.83 or a ≈ ±14.83 Since a < c and c = 24, we choose a = 14.83. The vertex

54. Foci: ( ±150, 0 )  c = 150

(a)

)

a − 1728a + 331,776 = 0

−75

x2 y2 − =1 2 a 576 − a 2 2 2 24 24 − =1 2 a 576 − a 2 576 576 − =1 a2 576 − a 2

576 576 − a 2 − 576 a 2 = a 2 576 − a 2

−150

x 2 y2 − =1 25 16

x2 y2 − 2 = 1. 2 120 90

60.

x 2 + 4 x − 8 y + 20 = 0 A = 1, C = 0 AC = 0, Parabola

61.

y 2 + 12 x + 4 y + 28 = 0 C = 1, A = 0 AC = 0, Parabola

62. 4 x 2 + 25 y 2 + 16 x + 250 y + 541 = 0 A = 4, C = 25 AC = 100 > 0, Ellipse 63.

x 2 + y2 + 2 x − 6 y = 0 A = C = 1, Circle

For y = 60, x 2 = 20,800 and x ≈ 144.2. Position: (144.2, 60 )

Section 10.3 64. y 2 − x 2 + 2 x − 6 y − 8 = 0 A = −1, C = 1 AC < 0, Hyperbola 65.

At the point ( a, 0 ) , d2 − d1 = ( a + c ) − ( c − a ) = 2 a. 73. Center: ( 6, 2 )

Horizontal transverse axis Foci at ( 2, 2 ) and (10, 2 )  c = 4

AC = 0, Parabola

(c + a) − (c − a) = 6  a = 3

66. 9 x + 4 y − 90 x + 8 y + 228 = 0 A = 9, C = 4

b2 = c 2 − a 2 = 16 − 9 = 7

( x − 6) − ( y − 2) = 1 2

AC = 36 > 0, Ellipse

67. True. For a hyperbola, c 2 = a 2 + b 2 . The larger the ratio of b to a, the larger the eccentricity of the hyperbola, e = c a.

74. At the point ( a, 0 ) , the difference of the distances to the

foci ( ± c, 0 ) is ( c + a ) − ( c − a ) = 2 a. Let ( x , y ) be a point on the hyperbola.

68. False. b ≠ 0 because it is in the denominator.

2a +

( x − c ) + y2 2

4 a ( x − c ) + y 2 = 4cx − 4 a 2 2

a ( x − c ) + y 2 = cx − a 2 2

(

) a (c − a ) = (c − a ) x − a y

a 2 x 2 − 2cx + c 2 + y 2 = c 2 x 2 − 2 a 2 cx + a 4 2

71. Let ( x , y ) be such that the difference of the distances

from ( c, 0 ) and ( −c, 0 ) is 2a (again only deriving one of the forms).

( x + c ) + y2 − ( x − c ) + y2 2

( x − c ) + y2 = ( x + c ) + y2 2

4 a 2 + 4a ( x − c ) + y 2 + ( x − c ) + y 2 = ( x + c ) + y 2 2

70. True. The asymptotes are b y = ± x. a If they intersect at right angles, then −1 b a = =  a = b. a  b b − a   

( x − c ) + y2 = ( x + c ) + y2

4a 2 + 4a ( x − c ) + y2 + ( x − c ) + y2 = ( x + c ) + y2

is the graph of two intersecting lines.

2a +

−

( x − 1) − ( y − 1) = 0

2a =

( x + c ) + y2

2a =

69. False. For example, x2 − y2 − 2 x + 2 y = 0 2

883

72. d2 − d1 = constant by definition of hyperbola

x2 − 6x − 2 y + 7 = 0 A = 1, C = 0

Hyperbolas

x2 y2 − 2 2 a c − a2

Thus, c2 − a 2 = b2 . 75. If A = C ≠ 0, then by completing the square you obtain a circle. If A = 0 and C ≠ 0, then Cy 2 + Dx + Ey + F = 0 is a parabola (complete the square). Same for A ≠ 0 and C = 0. If AC > 0, then both A and C are positive (or both negative). By completing the square you obtain an ellipse. If AC < 0, then A and C have opposite signs. You obtain a hyperbola.

4a ( x − c ) + y2 = 4cx − 4a2 2

a ( x − c ) + y2 = cx − a2 2

(

) a (c − a ) = (c − a ) x − a y

a2 x 2 − 2cx + c2 + y2 = c2 x 2 − 2a2cx + a 4 2

2 2

Let b2 = c2 − a2 . Then a2b2 = b2 x 2 − a2 y2  1 =

2 2

x 2 y2 − . a 2 b2

884

76.

Chapter 10

Topics in Analytic Geometry

(

)

77. x 3 − 16 x = x x 2 − 16 = x ( x − 4 )( x + 4 )

x 2 y2 − =1 16 9 Center: ( 0, 0 )

78. x 2 + 14 x + 49 = ( x + 7 )

Horizontal transverse axis a = 4, b = 3

(

79. 2 x 3 − 24 x 2 + 72 x = 2 x x 2 − 12 x + 36

c = a + b = 16 + 9 = 25  c = 5 2

Vertices: ( ±4, 0 )

= 2x ( x − 6)

(

3 x 4

y2 x 2 − =1 9 16 Center: ( 0, 0 )

(

81. 16 x 3 + 54 = 2 8 x 3 + 27

(

)

82. 4 − x + 4 x 2 − x 3 = ( 4 − x ) + x 2 ( 4 − x )

(

)

= ( 4 − x ) x2 + 1

c = a + b = 9 + 16 = 25  c = 5 2

)

= x ( 3 x + 2 )( 2 x − 5 )

= 2 ( 2 x + 3) 4 x 2 − 6 x + 9

Vertical transverse axis a = 3, b = 4

)

80. 6 x 3 − 11x 2 − 10 x = x 6 x 2 − 11x − 10

Foci: ( ±5, 0 ) Asymptotes: y = ±

Vertices: ( 0, ± 3 ) Foci: ( 0, ± 5 )

3 x 4 Both hyperbolas have a center of ( 0, 0 ) and asymptotes

Asymptotes: y = ±

3 x. The first hyperbola has a horizontal 4 transverse axis, whereas the second hyperbola has a vertical transverse axis. The first hyperbola has vertices ( ±4, 0 ) and foci ( ±5, 0 ) . The second hyperbola has

of y = ±

vertices ( 0, ± 3 ) and foci ( 0, ± 5 ) . 6

−9

−6

Section 10.4 Parametric Equations 1.

plane curve, parametric equations, parameter

orientation

Given a set of parametric equations, the process of finding a corresponding rectangular equation is called eliminating the parameter.

The point on the plane curve x = t and y = t when t = 3 is ( 3, 3 ) .

x=t y=t+2 y = x + 2, line Matches (c).

x = t2 y =t −2t = y+2 x = ( y + 2)

Parabola opening to the right Matches (d). 7.

x= t y=t y = x 2 , parabola, x ≥ 0 Matches (b).

Section 10.4

1 1 t= t x

10. 1 +2 x

y=t+2 y=

Parametric Equations

x = 4 cos2 t , y = 4 sin t (a)

−

x = t, y = 2 − t (a)

−4

−2 2

−1

−2 −5 −4 −3 −2 −1

−1

(c)

2 2

−6

(d) −4

−3

y = 2 − t = 2 − x 2 , parabola The graph is an entire parabola rather than just the right half.

4 x 2 + y 2 = 16 cos2 t + 16 sin 2 t = 16 x y2 + = 1, parabola 4 16 The graph is an entire parabola rather than just the right portion. y

3 2 1 −5 −4 −3 − 2 −1

−2

1 2 3

−8

right half of the parabola.

−1

parabola to ( 0, 4 ) .

The curve starts at ( 0, 2 ) and moves along the

−1

The curve starts at ( 0, − 4 ) and moves along the

−2

−2 −3 −4 −5

−1

(d)

(c)

5 4 3 2 1

−2

(b)

−

Matches (a).

885

1 2 3

−2 −3 −5

11. The graph opens upward, contains (1, 0 ) , and is oriented

left to right. Matches (b). 12. The orientation of the graph is clockwise and the center is ( 2, 3 ). Matches (c).

886

Chapter 10

Topics in Analytic Geometry

13. x = t , y = −4t y = −4 x

17.

1 t, y = t 2 4

y = (4x)

y = 16 x 2

y 5

−2

−1

−1 −2

14.

−3

1 x = t, y = t 2 1 y = x or x − 2 y = 0 2

18.

−2

−1

x = t, y = t 3 y

6 6

−4

y = x3

−6

−2

x −6

−4

−2

−4 −6 −6

15.

x = 3t − 3, y = 2t + 1

19.

x+3 3  x + 3 y = 2 +1  3  t=

x = t + 2, y = t 2 t = x−2 y = ( x − 2)

2 x+3 3 y 4

5 4

1 −6

−4 −3 −2 −1 −1

−2 −1 −1

−2

20.

−3

16.

x= t y =1− t y = 1 − x2 , x ≥ 0

x = 3 − 2 t , y = 2 + 3t

3− x  y = 2 + 3   2  3 x + 2 y − 13 = 0 y

y 2

−2

−4 −6

−8

− 10

−4

−2

−4

Section 10.4 21.

24.

x = 2t , y = t − 2 t=

x  y = t −2 2 x = −2 2 =

Parametric Equations

887

x = cosθ , y = 4 sin θ 2

y x 2 +   = cos2 θ + sin 2 θ = 1 4 y2 x2 + = 1, ellipse 16

1 x−4 2

5 4

y 10 8 6

−5 −4 −3 −2

2 3 4 5

−2

−4 −5

25. 22.

x = t −1

x = e− t 

( )

y = e3 t  y = e t

y=t+2 t = y−2

= ( y − 2) − 1 = y−3

10 9 8 7 6 5 4 3 2 1

10 8 6 4 2

23.

−2

1 y=  x 1 y = 3 , x > 0, y > 0 x

x = t −1

−2

1 = et x

−2 −1

x = e2 t

26.

x = 2 cosθ , y = 3 sin θ

1 2 3 4 5 6 7 8

y = et  y2 = e2 t

 x  y 2 2  2  = cos θ ,  3  = sin θ

y 2 = x, y > 0; y = x , x > 0

x2 y2 + = cos2 θ + sin 2 θ = 1 4 9 x2 y2 + = 1, ellipse 4 9

4 3 2 1

2 1 −4 −3

−1 −1

−2 −4

888

Chapter 10

Topics in Analytic Geometry

27.

x = t 3  x1 3 = t

32.

y = 3ln t  y = ln t

( )

y = ln x1 3

y = ln x

−6

y 5 4 3 2 1

−4

−2 −1

2 3 4 5 6 7 8

33. x =

t 2

(

)

y = ln t 2 + 1

−2 −3 −4 −5

28.

x = sec θ y = tan θ

x = ln 2t  e x = 2t  t = 12 e x

( )= e

y = 2t 2 = 2 12 e x

1 2

− 12

2x −8

34.

x = 10 − 0.01et y = 0.4t 2

5 4

3 2 1

−3

29.

−2

−1

x = 4 + 3 cos θ , y = −2 + sin θ 1 −1

−5

30.

x = 4 + 3 cosθ , y = −2 + 2 sin θ 2

20 −5

35. Each curve represents a portion of the line y = 2 x + 1. (a) x = t y = 2t + 1 Domain: −∞ < x < ∞ Orientation: Left to right (b) x = cosθ y = 2 cosθ + 1 Domain: −1 ≤ x ≤ 1 Orientation: Depends on θ

(c)

x = e− t

y = 2e − t + 1 Domain: 0 < x < ∞ Orientation: Right to left (d) x = et

x = 4 sec θ , y = 2 tan θ

y = 2et + 1 Domain: 0 < x < ∞ Orientation: Left to right

−2

−6

31.

−40

− 12

−8

Section 10.4 36. Each curve represents a portion of the line 2 y + x − 8 = 0.

x=2 t

(a)

Line though (1, −1) and ( 4, 8) : x = 1 + t ( 4 − 1)  x = 1 + 3t y = −1 + t (8 − ( −1))  y = −1 + 9t

42. Circle: x = h + r cos θ y = k + r sin θ

Center: ( − 2, − 5); Radius: 7 x = − 2 + 7 cos θ y = − 5 + 7 sin θ

43. Ellipse: x = h + a cos θ y = k + b sin θ

Vertices: ( ± 5, 0); Foci: ( ± 4, 0) The center is the midpoint between the vertices, (0, 0). In addition, the distance from the center to a vertex is a = 5 and the distance from the center to a focus is c = 4. a2 = b2 + c2 25 = b 2 + 16 9 = b2 3 = b x = 0 + 5 cos θ  x = 5 cos θ y = 0 + 3 sin θ  y = 3 sin θ

( x − h) + ( y − k ) = 1 2

cos2 θ + sin 2 θ =

( x − h) + ( y − k ) = r 2

x = h + a cosθ y = k + b sin θ

x−h y−k = cosθ , = sinθ a b

( x − h) + ( y − k ) = 1 2

y = y1 + t ( y2 − y1 )

( x − x1 ) ( x2 − x1 )

x = x1 + t ( x2 − x1 )

x = h + r cosθ y = k + r sin θ

( x − h) − ( y − k ) = 1

41. Line through ( x1 , y1 ) and ( x2 , y2 ):

( x − h ) = cosθ , y − k = sinθ

39.

y = k + b tan θ

 x − x1  y = y1 +   ( y2 − y1 )  x2 − x1  y −y  y − y1 =  2 1  ( x − x1 )  x2 − x1 

38.

x = h + a sec θ

sec 2 θ − tan 2 θ =

x  x−2 y =3−t =3− =4− 2  2  Domain: −∞ < x < ∞ Orientation: Left to right 2 (d) x = −2t x y = 4 + t2 = 4 − 2 Domain: x ≤ 0 Orientation: Left to right for t ≤ 0 Right to left for t > 0 t=

889

x−h y−k = secθ , = tanθ a b

x y=4− t =4− ,y≤4 2 Domain: x ≥ 0 Orientation: Left to right (b) x = 2 3 t x y=4− 3 t =4− 2 Domain: −∞ < x < ∞ Orientation: Left to right (c) x = 2 ( t + 1)

37.

40.

Parametric Equations

44. Hyperbola: x = h + a sec θ y = k + b tan θ

Vertices: ( ± 2, 0); Foci: ( ± 3, 0) The center is the midpoint between the vertices, (0, 0). In addition, the distance from the center to a vertex is a = 2 and the distance from the center to focus is c = 3. c2 = a 2 + b2 9 = 4 + b2 5 = b2 5 = b x = 0 + 2 sec θ  x = 2 sec θ

y = 0+

5 tan θ  y =

5 tan θ

890

Chapter 10

Topics in Analytic Geometry

45. y = − 2 x + 1

51. y = e − x

(a) Let t = x  x = t

y = − 2t + 1

y = e −t

(b) Let t = 2 − x  x = 2 − t y = − 2( 2 − t ) + 1  y = 2t − 3

(b) Let t = 2 − x  x = 2 − t

y = e −(2 − t ) = et − 2

46. y = 4 x − 9

(a) Let t = x  x = t

52. y = ln ( x + 2)

y = 4t − 9

(a) Let t = x  x = t

(b) Let t = 2 − x  x = 2 − t

y = ln (t + 2)

y = 4( 2 − t ) − 9  y = − 4t − 1

47. y =

(b) Let t = 2 − x  x = 2 − t y = ln ( 2 − t + 2)

3 x

= ln ( 4 − t )

(a) Let t = x  x = t

53. x = 3 cos3 θ , y = 3 sin 3 θ

3 y = t

(b) Let t = 2 − x  x = 2 − t y =

−6

3 2−t

−4

48. y =

1 x2

54. x = 8θ − 4 sin θ , y = 8 − 4 cos θ

(a) Let t = x  x = t y =

1 t2

(b) Let t = 2 − x  x = 2 − t y =

(2 − t )

= 2

1 t 2 − 4t + 4

100

−6

55.

x = 2 cot θ , y = 2 sin 2 θ 4

49. y = 4 x + 5

(a) Let t = x  x = t

−6

y = 4t + 5 −4

(b) Let t = 2 − x  x = 2 − t y = 4( 2 − t ) + 5 2

= 4t − 16t + 21

3t 3t 2 , 3 1 + t 1 + t3

56.

57.

x = θ + sin θ , y = 1 − cosθ

50. y = x3 − x 2

(a) Let t = x  x = t y = t3 − t2

(b) Let t = 2 − x  x = 2 − t y = (2 − t ) − (2 − t ) 3

= 8 − 12t + 6t − t − 4 + 4t − t 3

= − t + 5t − 8t + 4

Section 10.4 58.

x = 2θ − 4 sin θ , y = 2 − 4 cosθ

891

(b) y = 5 + ( 225 sin 15°)t − 16t 2 = 0

t = 0

Parametric Equations

−225 sin 15° ±

( 225 sin 15°) − 4(−16)(5) 2( −16) 2

t ≈ −0.0839, 3.7236 Distance traveled before arrow hits ground:

−6

(225 cos 15°)(3.7236) ≈ 809.3 feet

59. Matches (b).

(c)

60. Matches (c).

61. Matches (d). 62. Matches (a). 63.

x = ( v0 cos θ ) t and y = h + ( v0 sin θ ) t − 16 t 2

(a)

Maximum height: 58.0 feet

v0 = 100 mph ≈ 146.67 ft/sec, h = 3

(d) Time arrow is in the air: ≈ 3.72 seconds

x = (146.67 cosθ ) t

y = 3 + (146.67 sin θ ) t − 16 t 2

x = (146.67 cos15° ) t

(b)

y = 3 + (146.67 sin15° ) t − 16t 2

900 0

(see part (b)) 65. True. x=t

first set

y = t 2 + 1 = x2 + 1 x = 3t

second set

y = 9t + 1 = ( 3t ) + 1 = x 2 + 1 2

450

The ball travels about 347 feet. So, it is not a home run. (c) x = (146.67 cos23° ) t y = 3 + (146.67 sin15° ) t − 16t 2

66. False. The graph of x = t 2 , y = t 2 represents the portion of the line y = x in the first quadrant. 67. False. For example, x = t 2 and y = t does not represent y as a function of x. 68. False. The equations represent a line. 69.

x = 6 cosθ , y = 6 sin θ (a) The orientation of the curve is counterclockwise starting at θ = 0 and the point ( 6, 0 ) .

(b) If the interval 0, 2π ) is used, the graph would be a 0

500

The ball travels about 490 feet. So, it is a home run. (d) Trying several values for θ , you’ll find that a minimum angle of about 19° is required for the hit to be a home run. 64. (a) x = (v0 cos θ )t and y = h + (v0 sin θ )t − 16t 2

h = 5, v0 = 225, θ = 15° x = ( 225 cos 15°)t y = 5 + ( 225 sin 15°)t − 16t 2

complete circle. (c) Answers will vary. Sample answer: The equations x = 6 cosθ and y = −6 sin θ would start at θ = 0 and the point ( 6, 0 ) , but trace the curve clockwise. (d) If the equations x = 6 sinθ and y = 6 cosθ were used, then the curve would start at θ = 0 and the point ( 6, 0 ) , and trace clockwise. 70. (a) The orientation is reversed.

(b) The graph is shifted one unit to the right. (c) The graph is shifted one unit up

892

Chapter 10

71.

f (−x) =

Topics in Analytic Geometry

4(−x)

(−x) + 1 2

4 x2 = f ( x) x2 + 1

Symmetric about the y-axis Even function 72.

73.

y = e x ≠ e − x ; e − x ≠ −e x No symmetry Neither even nor odd

74.

( x − 2) = y + 4  y = x2 − 4 x 2

No symmetry Neither even nor odd

f ( x) = x , x ≥ 0 No symmetry Neither even nor odd

Section 10.5 Polar Coordinates 7. (0, 0) matches C.

1. pole 2. directed distance, directed angle

x = 0 cos 0 = 0(1) = 0

3. The rectangular coordinates ( x , y ) and the polar

y = 0 sin 0 = 0(0) = 0

(0, 0)

coordinates ( r , θ ) are related by the following equations:

y and x r 2 = x 2 + y2 tan θ =

x = r cosθ y = r sin θ

5π   8.  − 2,  matches B. 4  

4. Yes, the polar coordinates (1, π ) and the rectangular

coordinates ( −1, 0 ) represent the same point. y

 5π 2 = − 2 −  = 4  2 

y = − 2 sin

 5π 2 = − 2 −  = 4  2 

( 2, 2 ) θ=π

(−1, 0) r=1

x = − 2 cos

( )

π 2

5π 3, 6

 π 5.  2,  matches A.  2 x = 2 cos y = 2 sin

π 2

= 2(0) = 0 = 2(1) = 2

(0, 2)  3π  6.  2,  matches D.  2

Three additional representations: 7π   5π   − 2π  =  3, −  3,  6 6     5π 11π     + π  =  −3,  −3,  6 6     5π π    − π  =  −3, −   −3, 6 6   

3π = 2(0) = 0 2 3π y = 2 sin = 2( −1) = − 2 2 (0, − 2) x = 2 cos

Section 10.5 10.

π 2

( ) 3π 2, 4

13.

 5π   11π  + π  =  4,  4,  6   6  

π  5π   − π  =  4, −   4, 6  6   14.

5π 7π     − 2π  =  − 4, −   − 4, 6 6    

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

Three additional representations:

 3π   −5π  − 2π  =  2,  2,  4 4     3π 7π     + π  =  −2,  −2,  4 4    

π 2

(−4, 56π )

Three additional representations:

11.

893

π 2

Polar Coordinates

(−5, − 116π )

π 2

1 2 3 4 5

(1, − π3 ) Three additional representations:

Three additional representations:

π    5π  1, − + 2π  = 1,  3    3 

11π π    + 2π  =  − 5,   − 5, − 6 6   

π 2π      −1, − + π  =  −1,  3 3    

11π    7π  − π  =  5,  5, −  6    6 

π 4π      −1, − − π  =  −1, −  3 3     12.

(3, − 76π )

π 2

11π 5π     + π  =  5, −   5, − 6 6     15.

(− 32 , − 74π )

π 2

Three additional representations:

7π    5π  + 2π  =  3,  3, −  6    6  7π 11π     − π  =  − 3,  − 3, −  6 6     7π π    + π  =  − 3, −   − 3, − 6 6   

Three additional representations:  3 7π   3 π + 2π  =  − ,  − , − 4  2   2 4  3 7π   3 5π  − π =  ,  ,−  4 2  2 4   3 7π   3 3π  + π =  , −   ,− 2 4 4    2

894

Chapter 10

Topics in Analytic Geometry

π 2

16.

19. Polar coordinates: (3, π ) x = 3 cos (π ) = − 3( −1) = − 3 y = 3 sin (π ) = 3(0) = 0 2

(

2π − 5 2, 3

Rectangular coordinates: ( − 3, 0)

)

π 2

Three additional representations:

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

2π 5π     + π  =  5 2,  5 2,  3 3     2π π    − π  =  5 2, −   5 2, 3 3    17.

π 2

x = 2 cos(0) = 2(1) = 2

( ) 0, −

20. Polar coordinates: ( 2, 0)

π 2

y = 2 sin (0) = 2(0) = 0

Rectangular coordinates: ( 2, 0) 1

π 2

Three additional representations:

3π   π   3π    0, ,  0, − ,  0,  2   2  2   18.

π 2

3π   21. Polar coordinates: 1, −  4  

(0, π6 ) 1

Three additional representations: 11π   5π   7π    0, ,  0, − ,  0, −  6 6 6      

2  3π  x = 1 cos −  = − 2  4  2  3π  y = 1 sin  −  = − 4 2  

 2 2 ,− Rectangular coordinates:  −  2   2 π 2

Section 10.5 5π   22. Polar coordinates:  −16,  2  

Polar Coordinates

895

3π   25. Polar coordinates:  − 5,  2  

 5π  x = −16 cos  = −16(0) = 0  2 

 3π  x = − 5 cos  = − 5(0) = 0  2 

 5π  y = −16 sin  = −16(1) = −16  2 

 3π  y = − 5 sin   = − 5( −1) = 5  2 

Rectangular coordinates: (0, −16)

Rectangular coordinates: (0, 5) π 2

π 2

7π   23. Polar coordinates:  0, −  6  

π  26. Polar coordinates:  − 3, −  6 

 3  7π  x = 0 ⋅ cos −  = 0  = 0  6   2 

 3 3 3  π x = − 3 cos −  = − 3  = − 2 2  6  

 7π   1 y = 0 ⋅ sin  −  = 0 −  = 0 6    2

3  π  1 y = − 3 sin  −  = − 3 −  = 6 2 2    

Rectangular coordinates: (0, 0)

 3 3 3 ,  Rectangular coordinates:  − 2 2  

π 2

 5π  24. Polar coordinates:  0,   4 

2π

 2  5π  x = 0 ⋅ cos  = 0 −  = 0  4   2 

27.

( r, θ ) =  2, 9   ( x, y ) = (1.53, 1.29 )

 2  5π  y = 0 ⋅ sin   = 0 −  = 0 4 2    

28.

( r, θ ) =  4,

Rectangular coordinates: (0, 0)

29.

( r , θ ) = ( −4.5, 1.3 )  ( x, y ) = ( −1.20, − 4.34 )

30.

( r , θ ) = (8.25, 3.5 )  ( x, y ) = ( −7.73, − 2.89 )

π 2





11π    ( x, y ) = ( −3.06, − 2.57 ) 9 

31. Polar coordinates: (1.5, − 2.82) 1

x = 1.5 cos( − 2.82) ≈ −1.42 y = 1.5 sin ( − 2.82) ≈ − 0.47

Rectangular coordinates: ( −1.42, − 0.47)

Chapter 10

896

Topics in Analytic Geometry

32. Polar coordinates: ( − 5.3, − 0.78)

36. Rectangular coordinates: ( −3, − 3 )

x = − 5.3 cos( − 0.78) ≈ − 3.77

r = 3 2 , tan θ = 1, θ = π 4

y = − 5.3 sin ( − 0.78) ≈ 3.73

Rectangular coordinates: ( − 3.77, 3.73)

5π   π  Polar coordinates:  3 2 ,  ,  −3 2 ,  4   4  y 2

33. Rectangular coordinates: ( −7, 0 )

r = 7, tan θ = 0, θ = 0

Polar coordinates: ( 7, π ) , ( −7, 0 )

−4

−3

−2

−1

−4

−9 −8 −7 −6 −5 −4 −3 −2 −1

r = 3 + 3 = 6, tan θ = 1, θ =

r = 5, tan θ = 0 is undefined, θ =

3 2 1

−3

y 2

−2

−1

−2

2 π  3π   Polar coordinates:  5,  ,  −5,  2   2 

5π   π  Polar coordinates:  6 , ,  − 6 ,  4 4   

34. Rectangular coordinates: ( 0, − 5 )

−1

( − 3, − 3 )

37. Rectangular coordinates:

−2 −3 −4 −5

−1

−3

5 4 3 2 1

−2

−3

−1

−3

−2 −3

38. Rectangular coordinates:

−4 −5 −6

35. Rectangular coordinates: (1, 1) r = 2 , tan θ = 1, θ = π 4

π  5π   Polar coordinates:  2 ,  ,  − 2 ,  4 4     y

−1 −2

−2 − 1 −1

−1

5π   11π   Polar coordinates:  2,  ,  −2,  6   6  

−2

r = 3 +1 = 2 −1 11π tan θ = ,θ= 6 3

−3

( 3, − 1)

−2 1

−3 −4 −5 −6

−3

Section 10.5 39. Rectangular coordinates: ( −3, 4 ) r=

44.

( −3) + ( 4) = 9 + 16 = 5 2

Polar Coordinates

( x, y ) = ( 3 2, 3 2 )  r = ( 3 2 ) + 3 ( 2 ) 2

897

= 36 ≈ 6

4 tan θ = , θ ≈ 2.214 −3 Polar coordinates: ( 5, 2.214 ) , ( −5, 5.356 )

3 2   3 2   

θ = arctan  =

y 5

( r, θ ) ≈ ( 6, 0.79 )

π 4

≈ 0.79

3 2

45.

1 −4

−3

−2

−1





40. Rectangular coordinates: ( 3, − 1) r=

(3) + ( −1) = 10 2

−1 ; θ ≈ 5.961 3

tan θ =

46.

( x, y ) =  − 

−1

−3

( x, y ) = ( 3, − 2)  r = 32 + ( −2) = 3 ≈ 3.61 47.

48.

= 29 ≈ 5.39

θ = arctan ( − ) ≈ 2.76

( r, θ ) ≈ ( 5.39, 2.76 )

)

49. 3 + 22 =



θ = arctan 

−

( r , θ ) ≈ (2.65, 2.28)

x 2 + y2 = 16 r 2 = 16 r=4

2 5

3, 2  r =

x 2 + y2 = 9 r =3

( x, y ) = ( −5, 2 )  r = ( −5) + ( 2 )

(

85 ≈ 2.30 16

r2 = 9

( r , θ ) ≈ ( 3.61, − 0.588 )

43. ( x, y ) = −

( r , θ ) ≈ ( 2.30, 3.85)

θ = arctan ( − 23 ) ≈ −0.588

42.

 7 −  θ = arctan  4  3  −   2 ≈ 3.85

−2

41.

    17 = ≈ 2.83 6 4   θ = arctan  3   5  2 ≈ 0.490

1 −1

7 3  7  3 , −   r = −  +−  4 2  4  2

−2

( r, θ ) ≈ ( 2.83, 0.490 )

( 10 , 5.961) , ( − 10, 2.820 )

Polar coordinates:

( x, y ) =  2 , 3   r =  2  +  3 

7 ≈ 2.65

2   + π ≈ 2.28 3 

y=x r sin θ = r cosθ sin θ = cosθ tan θ = 1

θ=

π 4

898

Chapter 10

50.

y = − y = − x tan θ = tan

Topics in Analytic Geometry

3x 3

(

2π = − 3

)

r cos θ + 8r cosθ + 16 = r 2 2

a sin θ

59.

60.

r = a secθ

r ( r − 8sin θ ) = 0

3 ( r cosθ ) − ( r sin θ ) + 2 = 0

r = 8sin θ

r ( 3cosθ − sin θ ) = −2 2 r=− 3cosθ − sin θ 2 = sin θ − 3cosθ 3x + 5y − 2 = 0

61.

r ( r − 2 a cosθ ) = 0 r = 2 a cosθ

62.

x 2 + y 2 − 2 ay = 0 r 2 − 2ar sin θ = 0

r ( r − 2 a sin θ ) = 0

r ( 3cosθ + 5sin θ ) = 2 2 3cosθ + 5sin θ

r = 2 a sin θ

y2 = x 3

63.

xy = 4

55.

x 2 + y 2 − 2 ax = 0 r 2 − 2 ar cosθ = 0

3 ( r cosθ ) + 5 ( r sin θ ) − 2 = 0

x 2 + y2 − 8 y = 0 r 2 − 8r sin θ = 0

3x − y + 2 = 0

54.

x 2 + y2 − 6 x = 0 r 2 = 6r cosθ r = 6 cosθ

x=a r cosθ = a

53.

4 −4 or r = 1 − cosθ 1 + cosθ

r 2 − 6r cosθ = 0

r = a cscθ

( r sinθ ) = ( r cosθ ) 2

( r cosθ )( r sin θ ) = 4 r cosθ sin θ = 4

sin 2 θ cos3 θ = tan 2 θ sec θ

r 2 ( 2 cosθ sin θ ) = 8

r 2 sin 2θ = 8 r 2 = 8csc 2θ

r cos θ = r 3 sin 3 θ 2

2r cosθ ⋅ r sin θ = 1 2r = sec θ csc θ

r 2 = 12 sec θ csc θ = csc ( 2θ )

( x + y ) = 9( x − y ) (r ) = 9 ( r cos θ − r sin θ ) r = 9 ( cos θ − sin θ ) 2

x 2 = y3

64.

2 xy = 1

56.

sin 2 θ = r cos3 θ

( r cosθ + 4 ) = r 2 r = ± ( r cosθ + 4 )

y = a

r =

57.

r 1 − cos2 θ − 8r cosθ − 16 = 0 2

r sin θ = a

52.

r sin θ − 8r cosθ − 16 = 0 2

2π 3

θ = 51.

y 2 − 8 x − 16 = 0

58.

65.

cos2 θ = cot 2 θ csc θ sin 3 θ r = 4sin θ

r 2 = 4r sin θ r 2 − 4r sin θ = 0 x 2 + y2 − 4 y = 0

r 2 = 9cos ( 2θ )

Section 10.5

66.

θ=

2π 3

 y  x  2 xy r 2 = 2    = 2  r  r  r r 4 = 2 xy

( x + y ) = 2 xy 2

y = − 3x

68.

r = 2sin 3θ

75.

(

r = 2 3sin θ − 4sin 3 θ

5π y = x 6

3 x 3

θ=

( x + y ) = 6 ( x + y ) y − 8y ( x + y ) = 6x y − 2y

77.

1 1 − cosθ r − r cosθ = 1 r=

x 2 + y2 − x = 1

, vertical line

x 2 + y2 = 1 + 2 x + x 2 y2 = 2 x + 1 78.

x 2 + y2 = ( 2 − y )

x=2

x2 + 4y − 4 = 0

79.

r 2 = cosθ r 3 = r cosθ

(x + y ) = x 32

x +y =x

(x + y ) = x 2

x 2 + y2 = 4 − 4 y + y2

r = 2sec θ

73.

2 1 + sin θ r + r sin θ = 2 r=

x 2 + y2 + y = 2

r cosθ = 2

r sinθ = −3 y = −3

r = − 3cos 2θ

r = −3cscθ

72.

( ) r = − 3 ( r cos θ − r sin θ ) ( x + y ) = 3( x − y ) or ( x + y ) = 9( x − y )

11π 6

70. θ = π , horizontal line y=0 71.

3 y=− x 3

2 x=0

r = − 3 cos 2 θ − sin 2 θ

− 3 y = 3 x

76.

11π y = tan θ = tan x 6

69. θ =

)

r 4 = 6r 3 sin θ − 8r 3 sin 3 θ

− 3 y = 3 x y=−

5π θ= 6 tan θ = tan

899

r 2 = sin 2θ = 2sin θ cosθ

74.

2π tan θ = tan 3 y =− 3 x

67.

Polar Coordinates

6 2 − 3sin θ r ( 2 − 3sin θ ) = 6 r=

(

2r = 6 + 3r sin θ

)

2 ± x 2 + y2 = 6 + 3y 23

(

)

4 x 2 + y2 = ( 6 + 3y )

4 x 2 + 4 y 2 = 36 + 36 y + 9 y 2 4 x 2 − 5 y 2 − 36 y − 36 = 0

900

80.

Chapter 10

Topics in Analytic Geometry

6 2 cosθ − 3sin θ 6 r= x  y 2   − 3  r r 6r r= 2x − 3y 6 1= 2x − 3y 2x − 3y = 6 r=

θ=

84.

7π 3 y = tan θ = tan = x 6 3 3y − 3 x = 0

Line through origin making angle of

π 6

with positive

x-axis y

81. The graph of r = 6 is a circle centered at the origin with a radius of 6 units. r =6

2 1 −3

−2

−1

r = 36

−2

x 2 + y 2 = 36

−3

y 8

r = 3secθ

85.

r cosθ = 3 x =3 x −3=0 Vertical line through ( 3, 0 )

4 2 −8

7π 6

−4 −2 −2

−4

−8 3

r =8

r 2 = 64

82.

x 2 + y 2 = 64

−2

Circle of radius 8 centered at origin

−1

−3

10 6 4 2

r = 2cscθ

86.

−6 −4 −2

−2

− 10

−1

2 4 6

−4 −6

r sin θ = 2 y=2 y−2 =0

Horizontal line through ( 0, 2 )

− 10

θ=

83.

y tan θ = tan = 1 = x 4 y=x The graph is the line y = x, which makes an angle of

θ=

π 4

with the positive x-axis.

−3

−2

−1

−2

y 3 2 1

−3

−2

−1

−2 −3

Section 10.5

Polar Coordinates

901

π 2

87. (a)

88. (a)

π 2

0 0

30, − π 2

(

(35, − π2 (

(

Since the passengers enter a car at the point π (r , θ ) =  30, − , r = 30 is the polar equation 2  for the model. (b) Since it takes 45 seconds for the Ferris wheel to complete one revolution clockwise, after 15 seconds a passenger car makes one-third of one revolution or an 2π angle of radians. 3 2π 7π = − , the passenger car 2 3 6 7π  5π    is at  30, −  =  30, . 6  6   

Because θ = −

−

(b) Since it takes 60 seconds for the Ferris wheel to complete one revolution clockwise, after 15 seconds a passenger car makes one quarter of one revolution or an angle of

π 2

radians.

Because θ = −

− = − π , the passenger car is 2 2 at (35, − π ) = (35, π ).

x = r cos θ = 35 cos π = 35( −1) = − 35

5π   (c) Polar coordinates:  30,  6   5π x = r cos θ = 30 cos = 15 3 ≈ 25.98 6 5π y = r sin θ = 30 sin = 15 6

Rectangular coordinates: (25.98, 15) The car is about 25.98 feet to the left of the center and 15 feet above the center. 91. (a)

Since the passengers enter a car at the point π (r , θ ) =  35, − , r = 35 is the polar equation for 2  the model.

y = r sin θ = 35 sin π = 35(0) = 0 Rectangular coordinates: ( − 35, 0) The car is directly left of the center. 89. True. The distances from the origin are the same. 90. False. For instance when r = 0, any value of θ gives the same point.

(r1, θ1 ) = ( x1 , y1 ) where x1 = r1 cos θ1 and y1 = r1 sin θ1. (r2 , θ 2 ) = ( x2 , y2 ) where x2 = r2 cos θ2 and y2 = r2 sin θ 2 . d =

( x1 − x2 ) + ( y1 − y2 ) 2

x12 − 2 x1 x2 + x2 2 + y12 − 2 y1 y2 + y2 2

( x12 + y12 ) + ( x22 + y22 ) − 2( x1x2 + y1 y2 )

r12 + r2 2 − 2( r1r2 cos θ1 cos θ 2 + r1r2 sin θ1 sin θ 2 )

r12 + r2 2 − 2r1r2 cos(θ1 − θ 2 )

(b) If θ1 = θ2 , then

d =

r12 + r22 − 2r1r2 =

(r1 − r2 )2 = r1 − r2 .

This represents the distance between two points on the line θ = θ1 = θ2 . (c) If θ1 − θ2 = 90°, then d =

r12 + r2 2 − 2r1r2 cos(90°) =

r12 + r2 2 .

902

Chapter 10

Topics in Analytic Geometry

 π   π   2π   5π   5π  92. (a) A 3, , B 4, , C  3, , D 2, , E  3,  4   3   6  4  3  

(b) Any point that lies three units from the pole lie on the graph of r = 3. So, points A, C, and E lie on r = 3. (c) The points B and D lie on the graph of θ =

π 4

93. For a coordinate ( r , θ ) , r is a directed distance and θ a

r = 2 ( h cosθ + k sin θ )

95.

directed angle, measured counterclockwise from the polar axis. For ( r , θ ± 2 nπ ) , you obtain the same point as

 x  y  r = 2 h  + k    r   r 2hx + 2 ky r= r r 2 = 2hx + 2 ky

( r, θ ) because the angle θ makes complete revolutions about the polar axis. For ( −r , θ ± ( 2 n + 1) π ) , you obtain the same point as ( r, θ ) because the angle θ makes one-

x 2 + y 2 = 2hx + 2 ky

half revolutions about the polar axis and because r is a directed distance, which must be positive, −r = r. So, the

x 2 − 2hx + y 2 − 2 ky = 0

( x − 2hx + h ) + ( y − 2ky + k ) = h + k 2

point ( r, θ ) has more than one representation.

x 3y r= + r r 2 r = x + 3y 2

x + y = x + 3y

Center: ( h, k ) Radius: h 2 + k 2 96. cos A =

b 2 + c 2 − a 2 192 + 252 − 132 = = 0.86 2bc 2 (19 )( 25 )

A ≈ 30.7°

x − x + y2 − 3 y = 0 2

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

cos B =

a 2 + c 2 − b2 132 + 252 − 192 = = 0.66615 2 ac 2 (13 )( 25 )

B ≈ 48.2° C ≈ 180° − 30.7° − 48.2° ≈ 101.1°

1 3 The graph is a circle, with center  ,  and radius 2 2 10 . 2

( x − h ) + ( y − k ) = h2 + k 2

r = cosθ + 3sinθ

94.

97.

A = 24°, a = 10, b = 6 b sin A ≈ 0.2440  B ≈ 14.1° a C = 180° − A − B ≈ 141.9° a sin C ≈ 15.17 c= sin A

sin B =

98. B = 180° − 56° − 38° = 86°

a c c sin A 12sin ( 56° ) = a= = ≈ 16.16 sin A sin C sin C sin ( 38° ) b c c sin B 12sin ( 86° ) = b= = ≈ 19.44 sin B sin C sin C sin ( 38° )

99. B = 71°, a = 21, b = 29

b2 = a 2 + c 2 − 2ac cos B = 885.458  b ≈ 29.76 sin B ≈ 0.9214  C ≈ 67.1° b A = 180° − B − C ≈ 41.9°

sin C = c

Section 10.6

Graphs of Polar Equations

903

Section 10.6 Graphs of Polar Equations 16. r = 5 + 4cosθ

1. convex limaçon 2. circle

θ=

4. cardioid

its because replacing (r, θ ) with

2 (r, π − θ ) or ( − r, − θ ) yields an equivalent equation.

6. The graph of r = 3 + 4cosθ is symmetric with respect to the polar axis because replacing (r, θ ) with (r, − θ ) or ( − r, π − θ ) yields an equivalent equation.

r = 3 + 4cosθ

Replace ( r , − θ ) : r = 3 + 4cos ( −θ ) Since cosθ = cos ( −θ ) : r = 3 + 4 cosθ 7. r = 3cos2θ is a rose curve. 8. r = 5 − 5sinθ is a cardioid.

10. r = 1 + 4cosθ is a limaçon with inner loop. 11. The graph has four leaves. Matches (c). 12. The graph has three leaves. Matches (d). 13. The graph is symmetric about the line θ = π 2, and

passes through ( r , θ ) = ( 3, 3π 2 ) . Matches (a). 14. The graph is symmetric about polar axis and passes through ( r , θ ) = ( 3, 0 ) . Matches (c). 15. r = 3

The polar curve is a circle with radius of 3, so it has symmetry to the line θ =

π 2

r = 5 − 4 cosθ Not an equivalent equation Polar axis: r = 5 + 4 cos ( −θ ) r = 5 + 4 cosθ Equivalent equation −r = 5 + 4cosθ Pole: Not an equivalent equation r = 5 + 4 cos (π + θ )

r = 5 + 4 cos ( cos π cosθ − sin π sin θ )

r = 5 − 4 cosθ Not an equivalent equation Answer: Symmetric with respect to the polar axis

17. r =

9. r 2 = 9cos2θ is a lemniscate.

, the polar axis, and the pole.

: −r = 5 + 4 cos ( −θ )

r = 5 + 4 ( cos π cosθ + sin π sin θ )

5. The graph of a polar equation is symmetric with respect to

−r = 5 + 4 cosθ Not an equivalent equation r = 5 + 4 cos (π − θ )

3. lemniscate

the line θ =

θ=

2 1 − cosθ

π 2

: −r =

2 1 − cos ( −θ )

2 1 − cosθ Not an equivalent equation 2 r= 1 − cos (π − θ ) −r =

2 1 − ( cos π cosθ + sin π sin θ )

2 1 + cos θ Not an equivalent equation 2 Polar axis: r = 1 − cos ( −θ ) r=

2 1 − cosθ Equivalent equation 2 Pole: −r = 1 − cosθ Not an equivalent equation 2 r= 1 − cos (π + θ ) r=

2 1 − (cos π cosθ − sin π sin θ ) 2 r= 1 + cosθ Not an equivalent equation Answer: Symmetric with respect to the polar axis r=

904

Chapter 10

18. r =

θ=

Topics in Analytic Geometry 20. r = 4 csc θ cosθ = 4cot θ π −r = 4cot ( −θ ) θ= : 2 r = 4cot θ Equivalent equation −r = 4cot (π − θ ) Polar axis:

2 1 + sin θ

π 2

: r=

2 1 + sin (π − θ )

2 1 + sin π cosθ − cos π sin θ 2 r= 1 + sin θ Equivalent equation 2 Polar axis: r = 1 + sin ( −θ ) r=

−r = 4cot ( −θ ) r = 4cot θ Equivalent equation r = 4cot (π + θ ) Pole: r = 4cot θ Equivalent equation

2 1 − sin θ Not an equivalent equation 2 −r = 1 + sin (π − θ ) r=

−r =

Answer: Symmetric with respect to θ =

21. r 2 = 16 sin 2θ π θ= : 2

( −r ) = 16 sin ( 2 ( −θ ) ) 2

r 2 = −16 sin 2θ Not an equivalent equation

(

r 2 = 16sin 2 (π − θ ) r 2 = 16sin ( 2π − 2θ )

)

r 2 = −16sin 2θ Not an equivalent equation

(

Polar axis: r 2 = 16sin 2 ( −θ )

1 + sin π cosθ + cos π sin θ 2 r= 1 − sin θ Not an equivalent equation Answer: Symmetric with respect to θ =

polar axis, and pole

1 + sin π cosθ − cos π sin θ 2 −r = 1 + sin θ Not an equivalent equation 2 Pole: −r = 1 + sin θ Not an equivalent equation 2 r= 1 + sin (π + θ ) r=

)

r = −16sin 2θ 2

Not an equivalent equation

( −r ) = 16sin ( 2 (π − θ ) ) 2

π 2

19. r = 6sin θ π θ = : −r = 6sin ( −θ ) 2 r = 6sin θ Equivalent equation Polar axis: r = 6sin ( −θ )

r 2 = −16sin 2θ

Not an equivalent equation Pole:

( −r ) = 16sin ( 2θ ) 2

r 2 = 16sin 2θ Equivalent equation Answer: Symmetric with respect to pole

r = −6sin (θ ) Not an equivalent equation − r = 6sin (π − θ )

− r = 6 ( sin π cos θ − cos π sin θ ) − r = 6sin θ Not an equivalent equation Pole: −r = 6sin θ Not an equivalent equation r = 6sin (π + θ )

r = −6sin θ Not an equivalent equation

Answer: Symmetric with respect to θ =

π 2

Section 10.6 25. r = 3sin θ

22. r 2 = 36 sin 2θ

θ=

( −r ) = 36sin ( 2 ( −θ ) ) 2

Graphs of Polar Equations

r 2 = −36sin 2θ Not an equivalent equation

(

r 2 = 36sin 2 (π − θ ) r = 36sin ( 2π − 2θ )

Symmetric with respect to θ =

)

905

π 2

3 Circle with radius of 2 π 2

r 2 = 36 ( sin 2π cos2θ + cos2π sin 2θ ) r 2 = −36sin 2θ Not an equivalent equation

(

Polar axis: r 2 = 16sin 2 ( −θ )

)

r = −16sin 2θ Not an equivalent equation

( − r ) = 16sin ( 2 (π − θ ) ) r 2 = 16sin ( 2π − 2θ ) r 2 = 16 ( sin 2π cos2θ − cos2π sin 2θ ) 2

26. r = 2cosθ Circle Radius: 1; center: (1, 0 ) π 2

r = −16sin 2θ Not an equivalent equation 2

Pole:

( −r ) = 16sin 2θ 2

r 2 = 16 sin 2θ Equivalent equation Answer: Symmetric with respect to the pole

23. r = 5 Circle

27. r = 3 (1 − cosθ )

π 2

Cardioid π 2

24. θ = −

5π 3

28. r = 4 (1 + sin θ )

Line

Cardioid

π 2

906

Chapter 10

Topics in Analytic Geometry

29. r = 4 + 5sin θ Limaçon with inner loop

32. r = sin 5θ

5-petal rose curve

π 2

Symmetry: θ =

π 2

r = sin 5θ r = sin 5(π − θ ) 4

r = sin (5π − 5θ ) r = (sin 5π cos5θ − cos5π sin 5θ ) r = sin 5θ Equivalent equation

30. r = 3 + 6cosθ

Zeros: r = 0

Limaçon with inner loop

sin 5θ = 0

π 2

sin 5θ = 0 5θ = 0, π , 2π , 3π , 4π , 5π

θ = 0, 2

31. r = 5cos3θ 3-petal rose curve

π 2π 3π 4π 5

,π

π 2

Symmetry: Polar axis r = 5cos3θ

(

r = 5cos 3 ( −θ ) r = 5cos ( −3θ )

)

r = 5cos3θ Equivalent equation r =0 Zeros: 5cos3θ = 0 cos3θ = 0 3π 5π , 2 2 π π 5π θ= , , 6 2 6

3θ =

π 2

Section 10.6 33. r = − 7sin 2θ

Symmetry: θ =

8-petal rose curve

π 2

, polar axis, and the pole

Symmetry: θ =

, polar axis, pole

r = 3cos 4θ

r = − 7 sin 2θ

907

34. r = 3cos 4θ

4-petal rose curve

θ =

Graphs of Polar Equations

: r = − 7sin ( 2(π − θ )) r = − 7sin[2π − 2θ ] r = − 7(sin 2π cos2θ − cos 2π sin 2θ )

r = 3cos 4( −θ ) r = 3cos 4θ Equivalent equation r = 0

Zeros:

r = − 7sin 2θ

3cos 4θ = 0

Equivalent equation

cos 4θ = 0

Polar axis: − r = − 7sin ( 2(π − θ ))

π 3π 5π 7π 9π 11π 13π 15π

4θ =

, , , , , , 2 2 2 2 2 2 2 π 3π 5π 7π 9π 11π 13π 15π , , θ = , , , , , 8 8 8 8 8 8 8 8

− r = − 7sin ( 2π − 2θ ) − r = − 7[sin 2π cos2θ − cos 2π sin 2θ ] − r = 7sin 2θ

π 2

r = − 7sin 2θ

Equivalent equation Pole: r = − 7sin ( 2(π − θ ))

r = − 7sin ( 2π − 2θ )

r = − 7(sin 2π cos2θ + cos 2π sin 2θ ) r = − 7sin 2θ

35. r = 1 + 2sin θ

Equivalent equation

Limaçon with inner loop

r = 0

Zeros:

− 7 sin 2θ = 0

Symmetry: θ =

sin 2θ = 0 2θ = 0, π , 2π , 3π , 4π

π 3π θ = 0, , π , , 2π 2

π 2

r = 1 + 2sin θ r = 1 + 2sin (π − θ ) r = 1 + 2 (sinπ cosθ − cos π sin θ ) r = 1 + 2 0 − ( −1) sin θ 

π 2

r = 1 + 2sin θ Equivalent equation 4 5 6

Zeros:

r = 0 1 + 2sin θ = 0 sin θ = −

θ =

1 2

7π 11π , 6 6

π 2

Chapter 10

908

36. r =

Topics in Analytic Geometry 40. r = 6 − 4sin θ

3 − 2cos θ

Cardioid

Symmetry: Polar axis r =

3 − 2cosθ

r =

3 − 2cos ( −θ )

r =

3 − 2cosθ

− 12

0 ≤ θ ≤ 2π

Equivalent equation

41. r =

r = 0

Zeros:

− 12

3 − 2cosθ = 0

3 π , 0 ≤θ ≤ sin θ − 2 cosθ 2 4

3 cosθ = 2 π 11π θ = , 6 6

−6

−4

π 2

42. r =

6 2sin θ − 3cosθ 4

0 6

−6

−4

0 ≤ θ ≤ 2π

37. r = 8cos2θ

43. r 2 = sin 2θ

−18

−3

−12

0 ≤ θ < 2π

−2

− 2π ≤ θ ≤ 2π

38. r = cos2θ 2

44. r 2 = 4 cos3θ r = ± 2 cos3θ

−3

−2 −3

0 ≤ θ ≤ 2π 39. r = 2 ( 5 − sin θ )

−2

0 ≤ θ ≤ 2π

45. r = 8sin θ cos2 θ − 18

− 14

0 ≤ θ < 2π

−3

3 −1

Section 10.6 46. r = 2cos ( 3θ − 2 )

Graphs of Polar Equations

909

52. r = 2 (1 − 2sin θ )

1 6

−6

−3

−2

−7

0 ≤θ ≤π

0 ≤ θ < 2π

47. r = 2 csc θ + 6

 3θ  53. r = 2cos   , 0 ≤ θ < 4π  2 

2 − 18

−3

− 10

0 ≤ θ < 2π

−2

48. r = 4 − sec θ

54. r = 3sin

5θ 2

−18

−6

−12

−4

0 ≤ θ ≤ 2π

0 ≤ θ < 4π

49. r = eθ

55. r = 9sin θ

100

− 40

560

−9

− 250

−2

0 ≤ θ ≤ 3π

0 ≤ θ < π

θ 2

50. r = e

56. r 2 = 25cos 2θ 15

r = ± 5 cos 2θ 4

− 15

30 −6

− 15

0 ≤ θ ≤ 3π

−4

51. r = 3 − 4cosθ

− 5

π 2

≤ θ <

π 2

−10

−5

0 ≤ θ < 2π

Chapter 10

910

Topics in Analytic Geometry

57. r = 2 − sec θ From the graph, as y → ±∞, x → −1. So, x = −1 is an asymptote.

60. r = 2 cos2θ sec θ 4

−6

−4

As y → ±∞, x → −2. So, x = −2 is a vertical asymptote. −4

61. (a) r = 2 + csc θ = 2 +

58.

r sin θ = 2sin θ + 1

π 2

1 sin θ

r ( r sin θ ) = 2r sin θ + r

2 4 6 8

(± x + y )( y) = 2y + (± x + y ) ( ± x + y ) ( y − 1) = 2 y ( ± x + y ) = y2−y1 2

The graph is a cardioid.

(b) Since r is at a maximum when r = 10 at θ = 0 radians, the microphone is most sensitive to sound when θ = 0.

4 y2

x 2 + y2 =

( y − 1)

(

y2 3 + 2 y − y2

x2 =

x=± =±

( y − 1)

(

)

y2 3 + 2 y − y2

( y − 1)

62. (a)

π 2

) 2

y 3 + 2 y − y2 y −1

The graph has an asymptote at y = 1. 5

(b) Since the area enclosed by the graph of a lemniscate 2 is a , the area of one loop is 1 a 2 . So, the area of one 2

−6

loop of r 2 = 16 cos 2θ is 12 (16) = 8 square units. 63. True. r = 6sin 5θ , n = 5 is odd. The rose curve has five

−3

petals. 59. r =

64. False. Because

r = a ± b cosθ

= 4 + 2cosθ , and 7

−2

From the graph, as x → ∞, y → 3. So, y = 3 is a horizontal asymptote.

a ≥ 2, the graph is a b convex Limaçon.

65. False. For example, let r = cos3θ . 66. (a) Because r = 5 sin θ is a circle, the graph matches (ii).

(b) Because r = 2 + 5 sin θ is a limaçon with inner loop, the graph matches (i). (c) Because r = 5 cos 2θ is a 4-petal rose curve, the graph matches (iii).

Section 10.6 67. The graph of r = f (θ ) is rotated about the pole through

(c)

an angle φ. Let ( r , θ ) be any point on the graph of

r = f (θ ) . Then ( r, θ + φ ) is rotated through the angle

φ , and since r = f ( (θ + φ )φ ) = f (θ ) , it follows that

(d)

( r, θ + φ ) is on the graph of r = f (θ − φ ) .

x2 = 9 x = ±3

(b) Rotation: φ = π

)

Rotated graph: r = f sin (θ − π ) = f ( − sin θ )

72.

3π 2

Original graph: r = f ( sin θ )

(d)

70. (a)

(b)

3π   r = 2 − sin  θ −  2   = 2 − cosθ   π  r = 2sin 2  θ −   6    π  π  = 4sin  θ −  cos  θ −  6  6 

3 =0 x−2

3 x−2 5( x − 2) = 3

2 ( sinθ − cosθ ) 2 π  r = 2 − sin  θ −  2 

= 2 + sin θ

3 x−2

π  r = 2 − sin  θ −  4 

= 2 + cosθ r = 2 − sin (θ − π )

y =5− 5−

5 x − 10 = 3 5 x = 13 13 x= 5

=2−

(c)

4 x2 + 4 No zeros y=6+

73.

  3π   Rotated graph: r = f  sin  θ −   = f ( cosθ ) 2   

(b)

x2 − 9 x +1

x2 − 9 =0 x +1 x2 − 9 = 0

= f ( − cosθ )

69. (a)

r = 2sin 2 (θ − π ) 

71.

  π  Rotated graph: r = f  sin  θ −   2   

  2π   r = 2sin 2  θ −  3    2π   2π   = 4sin  θ −  cos  θ −  3 3    

= 2sin 2θ = 4sin θ cosθ

Original graph: r = f ( sin θ )

(

911

= 2sin ( 2θ − 2π )

68. Use the result of Exercise 65. π (a) Rotation: φ = 2

Original graph: r = f ( sin θ )

Graphs of Polar Equations

74.

x 3 − 27 x2 + 4

x 3 − 27 =0 x2 + 4 x 3 − 27 = 0 x 3 = 27 x =3

  π  r = 2sin  2  θ −   2    = 2sin ( 2θ − π ) = −2sin 2θ = −4sin θ cosθ

912

Chapter 10

Topics in Analytic Geometry

Section 10.7 Polar Equations of Conics 1.

conic

(a) Matches (i). (b) Matches (iii). (c) Matches (ii).

2e 1 + e sin θ (a) Parabola (b) Ellipse (c) Hyperbola

8. r =

ep corresponds to a 1 + e cosθ conic with a vertical directrix.

An equation of the form r =

A conic with a horizontal directrix has a polar equation of ep below the pole. the form r = 1 − e sin θ

c −9

−4

9. r =

−4

10. r =

4 1 − cosθ e = 1  parabola Vertical directrix to left of pole Matches (b).

2e 1 + e cosθ (a) Parabola (b) Ellipse (c) Hyperbola r=

3 32 = 2 − cosθ 1 − (1 2 ) cosθ

1  ellipse 2 Vertical directrix to left of pole Matches (c). e=

−4

2e 1 − e cosθ (a) Parabola (b) Ellipse (c) Hyperbola r=

11. r =

1  ellipse 2 Vertical directrix to right of pole Matches (f). e=

−8

3 32 = 2 + cosθ 1 + (1 2 ) cosθ

4 1 − 3sin θ e = 3  hyperbola Horizontal directrix below the pole. Matches (e).

12. r = −4

2e 1 − e sin θ (a) Parabola (b) Ellipse (c) Hyperbola r=

b a 9

−9

c −8

3 1 + 2sin θ e = 2  hyperbola Horizontal directrix above the pole. Matches (d).

13. r =

4 1 + sin θ e = 1  parabola

14. r =

 π Vertex:  2,  2  Matches (a). 3 1 − cosθ e = 1  parabola

15. r =

Section 10.7

16. r = e =

17. r =

1 2 2 = 1 4 + sin θ 1 + 4 sin θ

1  ellipse 4

−1 −1 2 = 2 + 4sin θ 1 + 2sin θ e = 2  hyperbola 2

4 1 − 5cosθ

−3

−2

7 1 + sin θ e = 1  parabola

27. r =

18. r =

e= 20. r =

e= 21. r = e=

22. r =

14 1 = 14 + 17sin θ 1 + (17 14 ) sin θ 17  hyperbola 14

8 2 = 4 + 3sin θ 1 + ( 3 4 ) sin θ 3  ellipse 4

−9

9 3 = 3 − 2 cosθ 1 − ( 2 3 ) cosθ 2  ellipse 3

(1 2 )( 6 ) 6 = 2 + sin θ 1 + (1 2 ) sin θ

9 −3

12 2 − cosθ 6 = 1 − (1 2 ) cosθ

28. r =

1  ellipse 2

5 −5 = −1 + cos θ 1 − cosθ

− 15

− 10

e = 1  parabola

23. r =

3 −3 4 = − 4 − 8cos θ 1 + 2cos θ

−3 −4 + 2cos θ 34 = 1 − (1 2 ) cosθ

29. r =

e = 2  hyperbola

10 10 3 = 3 + 9sin θ 1 + 3sin θ e = 3  hyperbola

24. r =

5 1 − sin θ e = 1  parabola

25. r =

−2

−4

1  ellipse 2 2

−6

913

26. r =

e = 5  hyperbola

19. r =

Polar Equations of Conics

30. r =

−4 4 = −1 + cos θ 1 − cos θ

e = 1  parabola 9

− 12

−9

Chapter 10

914

31. r =

Topics in Analytic Geometry 37. e = 1, x = −1, p = 1 Vertical directrix to the left of the pole 1(1) 1 = r= 1 − 1cosθ 1 − cosθ

1 − cos (θ − π 4 ) 12

−6

18 −4

32. r =

7 1 + sin (θ − π 3)

− 16

−24

4 1 − 5 cos (θ + 3π 4 ) 2

−3

−2

34. r =

9 3 − 2cos (θ + π 2 ) 3

−11

35. r =

8 4 + 3sin (θ + π 6 ) 4

−9

10 3 + 9cos (θ + 2π 3) 2

−2

3 , y = −4, p = 4 4 Horizontal directrix below pole ( 3 4 ) 4 = 12 r= 1 − ( 3 4 ) sin θ 4 − 3sin θ

40. e =

41. e = 2, x = 1, p = 1 Vertical directrix to the right of the pole 2 (1) 2 = r= 1 + 2 cosθ 1 + 2 cosθ 3 , x = −1, p = 1 2 Vertical directrix to the left of the pole 3 2 (1) 3 = r= 1 − ( 3 2 ) cosθ 2 − 3cosθ

42. e =

π  43. Vertex:  1, −   e = 1, p = 2 2  Horizontal directrix below the pole 1( 2 ) 2 = r= 1 − 1sin θ 1 − sin θ 44. Parabola, e = 1, vertex: ( 8, 0 )

Vertical directrix to right of pole ep 16 r= = 1 + e cosθ 1 + cosθ

−8

36. r =

1 , y = 1, p = 1 2 Horizontal directrix above the pole (1 2 )(1) = 1 r= 1 + (1 2 ) sin θ 2 + sin θ

39. e =

33. r =

38. e = 1, y = −4, p = 4 Horizontal directrix below the pole 1( 4 ) 4 = r= 1 − (1) sin θ 1 − sin θ

45. Vertex: ( 5, π )  e = 1, p = 10

Vertical directrix to left of pole 1(10 ) 10 = r= 1 − 1cosθ 1 − cosθ

−2

Section 10.7

π  46. Vertex:  10,   e = 1, p = 20 2  Horizontal directrix above pole 1( 20 ) 20 r= = 1 + 1sin θ 1 + sin θ 2 3 Vertical directrix to the right of the pole ( 2 3) p = 2 p r= 1 + ( 2 3 ) cosθ 3 + 2 cosθ

2p 2p 2= =  p=5 3 + 2 cos0 5 10 r= 3 + 2 cosθ

p 3 + sin (π 2 )

p=8 8 3 + sin θ

5  π 49. Center:  5, , c = 5, a = 6, e = 6  2 Horizontal directrix below the pole r =

( 5 6) p

1 − (5 6) sin θ

(5 4) p = 5 p 1 − ( 5 4 ) sin θ 4 − 5sin θ

5p 4 − 5sin ( 3π 2 )

9 5

5 ( 9 5)

4 − 5sin θ

9 4 − 5sin θ

5 3 Vertical directrix to the right of the pole ( 5 3) p r= 1 + ( 5 3 ) cosθ

52. Center: ( 5, 0 ) , c = 5, a = 3, e =

1  3π  48. Center:  1,  , c = 1, a = 3, e = 2  3  Horizontal directrix above the pole (1 3) p = p r= 1 + (1 3 ) sin θ 3 + sin θ

5p 6 − 5sin θ

( 5 3) p  p = 16 1 + ( 5 3 ) cosθ 5

5  16  16 3 3  5  = r= 5 1 5 3cosθ + 1 + cosθ 3 16 r= 3 + 5cosθ

4 1 45 4  = 4 53. Center:  , π , c = , a = , e = 5 5 5 1 5   Vertical directrix to the left to the pole r =

4p 1 − 4cosθ

1 =

5p 6 − 5sin (3π 2)

−1 =

4p 1 − 4cos0

p =

11 5

p =

3 4

r =

11 6 − 5sin θ

r =

3 1 − 4cosθ

50. Center: ( 8, 0 ) , c = 8, a = 12, e =

Vertical directrix to left of pole ( 2 3) p = 2 p r= 1 − ( 2 3 ) cosθ 3 − 2 cosθ 2p = 2 p  p = 10 3−2 20 r= 3 − 2 cosθ

20 =

915

5  3π  51. Center:  5,  , c = 5, a = 4, e = 2  4  Horizontal directrix below the pole

47. Center: ( 4, π ) , c = 4, a = 6, e =

Polar Equations of Conics

c 2 = a 3

c 5 5 3 5 π  54. Center:  ,  , c = , a = , e = = a 3 2 2 2 2 Horizontal directrix above the pole ( 5 3) p = 5 p r= 1 + ( 5 3 ) sin θ 3 + 5sin θ 1=

5p 8  p= 3 + 5sin (π 2 ) 5

8 3 + 5sin θ

916

Chapter 10

Topics in Analytic Geometry

55. When θ = 0, r = c + a = ea + a = a (1 + e ) .

Therefore,

ep 1 − e cos0 a (1 + e )(1 − e ) = ep

)

1 − e2 a ep Thus, r = = . 1 − e cosθ 1 − e cosθ

57. Earth: a ≈ 9.295 × 10 7 , e ≈ 0.0167

(1 − ( 0.0167) ) 9.295 × 10 = 9.2930 × 10 r= 2

(1 − e ) a = (1 − e )(1 + e ) a = a (1 − e )

Aphelion: r = 9.295 × 10 7 (1 + 0.0167 ) = 9.4508 × 107 miles

1 − e cosθ

1− e

1 − 0.0167cosθ

= 9.1404 × 10 7 miles

1 − e cos π 1+ e Maximum distance occurs when θ = 0.

(1 − e ) a = (1 − e )(1 + e ) a = a (1 + e )

Perihelion: r = 9.295 × 10 7 (1 − 0.0167 )

1 − 0.0167cosθ

56. Minimum distance occurs when θ = π .

1 − e cos θ

Aphelion: r = a (1 + e )

a 1 − e2 = ep.

(

(1 − e ) a : Perihelion: r = a (1 − e ) 2

a (1 + e ) =

(

In Exercises 57 – 62, use the following result from Exercises 55 and 56:

58. Mercury: a ≈ 3.5983 × 10 7 , e ≈ 0.2056

(1 − ( 0.2056 ) ) 3.5983 × 10 = 3.4462 × 10 r= 2

1 − 0.2056 cosθ

Perihelion: r = 3.5983 × 10 7 (1 − 0.2056 ) = 2.8585 × 107 miles

Aphelion: r = 3.5983 × 10 7 (1 + 0.2056 ) = 4.3381 × 10 7 miles 59. Mars: a ≈ 1.4164 × 108 , e ≈ 0.0934 1 − (0.0934)2 (1.4163 × 108 ) 8  ≈ 1.4039 × 10 r =  1 − 0.0934 cos θ 1 − 0.0934 cos θ

Perihelion: r = 1.4163 × 108 (1 − 0.0934) ≈ 1.2840 × 108 miles Aphelion: r = 1.4163 × 108 (1 + 0.0934) ≈ 1.5486 × 108 miles 60. Saturn: a ≈ 1.4267 × 109 , e ≈ 0.0539 1 − (0.0542)2 (1.4267 × 109 ) 9  ≈ 1.4225 × 10 r =  1 − 0.0542 cos θ 1 − 0.0542 cos θ

Perihelion: r = 1.4267 × 109 (1 − 0.0542) ≈ 1.3494 × 109 km Aphelion: r = 1.4267 × 109 (1 + 0.0542) ≈ 1.5040 × 109 km 61. Venus: a ≈ 1.0821 × 108 , e ≈ 0.0068 1 − (0.0068)2 (1.0821 × 108 ) 8  ≈ 1.0820 × 10 r =  1 − 0.0068 cos θ 1 − 0.0068 cos θ

Perihelion: r = 1.0821 × 108 (1 − 0.0068) ≈ 1.0747 × 108 km Aphelion: r = 1.0821 × 108 (1 + 0.0068) ≈ 1.0895 × 108 km

Section 10.7 64. (a)

62. Jupiter: a ≈ 7.7841 × 108 ; e ≈ 0.0484

(1 − ( 0.0484) ) 7.7841× 10 2

1 − 0.0484cosθ 7.7659 × 108 = 1 − 0.0484 cosθ

126,800 + 4119 = 65,459.5 2 c = 65,459.5 − 4119 = 61,340.5 c 61,340.5 e= = ≈ 0.937 a 65,459.5 ep ep ep ep 2ep 2a = + = + = 1 − e cos0 1 − e cos (π ) 1 − e 1 + e 1 − e2

Aphelion: r = 7.7841 × 108 (1 + 0.0484 ) = 8.1609 × 108 km 63. a = 4.498 × 10 9 , e = 0.0086, Neptune

Thus, p =

a = 5.906 × 10 , e = 0.2488, Pluto

(1 − 0.0086 ) 4.498 × 10 2

1 − 0.0086 cosθ 4.4977 × 10 9 = 1 − 0.0086 cosθ

(1 − 0.2488 ) 5.906 × 10 2

1 − 0.2488 cosθ 5.5404 × 10 9 = 1 − 0.2488 cosθ (b) Neptune: Perihelion: 4.498 × 109 × (1 − 0.0086 ) ≈ 4.4593 × 109 km

Aphelion: 4.498 × 109 (1 + 0.0086 ) ≈ 4.5367 × 109 km Pluto: Perihelion: 5.906 × 109 × (1 − 0.2488 ) ≈ 4.4366 × 109 km Aphelion: 5.906 × 109 (1 + 0.2488 ) ≈ 7.3754 × 109 km (c)

7 × 10 9

− 5 × 10 9

Radius of Earth ≈ 4000 miles. Choose ep r= . 1 − e cosθ

= 7.4073 × 108 km

Pluto: r =

917

Vertices: (126,800, 0 ) and ( 4119, π )

Perihelion: r = 7.7841 × 108 (1 − 0.0484 )

(a) Neptune: r =

Polar Equations of Conics

(

a 1 − e2 e

) ≈ 8525.2. Thus,

ep 7988.1 . ≈ 1 − e cosθ 1 − 0.937cosθ (b) When θ = 60°, r ≈ 15,029 and the distance from the surface of Earth to the satellite is 15,029 − 4000 = 11,029 miles. r=

(c) When θ = 30°, r ≈ 42,370 and distance = 38,370 miles. 4 −4 3 = −3 − 3sin θ 1 + sin θ False. The directrix is below the pole.

65. r =

16 π  9 − 4 cos  θ +  4  False. The graph is not an ellipse. (It is two ellipses.)

66. r 2 =

e(−x) ex for e > 1, and r = 1 + e cosθ 1 − e cosθ represent the same hyperbola.

67. True. Both r = 8 × 10 9

− 7 × 10 9

(d) Yes. Pluto is closer to the sun for just a very short time. Pluto was considered the ninth planet because its mean distance from the sun is larger than that of Neptune. (e) Although the graphs intersect, the orbits do not. So, the planets won’t collide.

68. True. The graph r =

1 − sin (θ − π 4 )

rotating the graph of r =

can be obtained by

5 5π about the pole 1 + sin θ 4

radians clockwise.

918

Chapter 10

Topics in Analytic Geometry

x 2 y2 + =1 a 2 b2 r 2 cos2 θ r 2 sin 2 θ + =1 a2 b2

69.

(

72.

a = 3, b = 4, c = 5, e =

)

r 2 cos2 θ r 1 − cos θ + =1 a2 b2 2 2 2 2 2 2 2 2 r b cos θ + r a − r a cos θ = a 2 b 2 2

(

)

r 2 b 2 − a 2 cos2 θ + r 2 a 2 = a 2 b 2

r2 =

73.

For an ellipse, b 2 − a 2 = − c 2 . Hence, 2

c c −r   cos2 θ + r 2 = b 2 , e = a a −r 2e2 cos2 θ + r 2 = b 2

(

)

r2 =

74.

b2 . 1 − e cos2 θ

(

Vertices: ( 4, 0 ) , ( 4, π )  a = 4

)

c = a + b  25 = 16 + b 2  b 2 = 9 2

)

r =

c c r 2   cos2 θ − r 2 = b2 , e = a a r 2 e2 cos2 θ − r 2 = b2

)

r 2 e 2 cos2 θ − 1 = b2

71.

r2 =

76. Center: ( x, y ) = ( 0, 0 ) , c = 4, a = 5, e =

4 5

b 2 = a 2 − c 2 = 25 − 16 = 9  b = 3

x2 y2 + =1 169 144 a = 13, b = 12, c = 5, e =

5 1 −   cos2 θ 4 −9 r2 = 25 1 − cos2 θ 16 −144 2 r = 16 − 25cos2 θ 144 r2 = 25cos2 θ − 16

r 2 c 2 cos2 θ − r 2 a 2 = a 2 b 2

b2 e cos2 θ − 1 −b 2 = 1 − e2 cos2 θ

− ( 3)

a 2 + b2 = c2

r2 =

x 2 y2 −b 2 − 2 = 1  r2 = 2 a b 1 − e 2 cos2 θ

r 2 b 2 + a 2 cos2 θ − r 2 a 2 = a 2 b 2

(

x 2 y2 − =1 36 4

75. Hyperbola  One focus: ( 5, 0 )  c = 5

r 2 cos2 θ r 1 − cos θ − =1 a2 b2 r 2 b 2 cos2 θ − r 2 a 2 + r 2 a 2 cos2 θ = a 2 b 2 2

b2 16 400 = = 2 2 2 1 − e cos θ 1 − ( 9 25 ) cos θ 25 − 9cos2 θ

2 10 10 = 6 3 2 4 36 − − b = = r2 = 1 − e 2 cos2 θ 1 − (10 9 ) cos2 θ 10 cos2 θ − 9

70.

3 5

a = 6, b = 2, c = 40 = 2 10 , e =

x 2 y2 − =1 a2 b2 r 2 cos2 θ r 2 sin 2 θ − =1 a2 b2

x 2 y2 + =1 25 16

r2 =

5 3

144 −16 = 1 − ( 25 9 ) cos2 θ 25cos2 θ − 9

a = 5, b = 4, c = 3, e =

−r 2c 2 cos2 θ + r 2 a 2 = a 2 b2

r 2 1 − e2 cos2 θ = b 2

x 2 y2 − =1 9 16

5 13

144 24,336 = 1 − ( 25 169 ) cos2 θ 169 − 25cos2 θ

r2 =

b2 9 225 = = 1 − e cos2 θ 1 − (16 25 ) cos2 θ 25 − 16 cos2 θ 2

r = a sin θ + b cosθ

77.

r 2 = ar sin θ + br cosθ x 2 + y 2 = ay + bx Circle

Chapter 10 Review 78. (a) The value of e = 0.9 matches graph C because when e < 1, the graph is an ellipse.

(b) The value of e = 1.0 matches graph B because when e = 1, the graph is a parabola. (c) The value of e = 1.1 matches graph A because when e > 1, the graph is a hyperbola. 79. Answers will vary. Sample answers: The eccentricity of a conic is the constant ratio of the locus of a point that moves such that the distance from a fixed point (focus) to the distance from a fixed line (directrix). The eccentricity denoted by e, determines the type of conic: 0 < e < 1  ellipse, e = 1  parabola, and e > 1  hyperbola. a Given r = b + c sin θ ab , a ≠ 0, b ≠ 0, c ≠ 0 r= 1 + c b sin θ

If c b < 1  ellipse. If c b = 1  parabola. If c b > 1  hyperbola.

919

3 4 For Exercises 80 − 83: sin u = − , cos u = , 5 5 1 1 cos v = , sin v = − . 2 2

80. cos ( u + v ) = cos u cos v − sin u sin v

= =

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

2 10

5 2

81. sin ( u + v ) = sin u cos v + sin v cos u

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

−7

=−

5 2

7 2 10

82. sin ( u − v ) = sin u cos v − sin v cos u

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

1 5 2

2 10

83. cos ( u − v ) = cos u cos v + sin u sin v

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

5 2

7 2 10

Chapter 10 Review 1.

Hyperbola

Parabola

Radius =

6. Radius =

( −3 − 0 ) + ( −4 − 0 ) 2

( x + 1)2 + ( y − 2)2 = 32

Radius =

(8 − 0 ) + ( −15 − 0 ) 2

= 64 + 225 = 289 = 17 2

x + y = 289 1 5. Radius = 2 13 2

(

)

Center = ( −1, 2)

x + y = 25 4.

(

= 4 2

= 9 + 16 = 25 = 5 2

1 8 2 2

1 2 1 2 x + y = 18 2 2 x 2 + y 2 = 36 Center: ( 0, 0 ) Radius: 6

Center = ( 2, 4)

( x − 2)2 + ( y − 4)2 = 13

920

Chapter 10

Topics in Analytic Geometry

3 2 3 2 x + y =1 4 4 4 2 x + y2 = 3

13. ( x − 2) + ( y + 4) = 9 2

x-intercepts: ( x − 2) + (0 + 4) = 9 2

(

( x − 2)

Center: ( 0, 0 )

2 3 = 3 3

(

( y + 4)

(x − ) + (y + ) =1 Center: ( , − ) 2

1 2

(0, − 4 ±

3 4

) (

( x + 8)

)

( − 8, 0)

( x + 4 ) + ( y − 3) = 494 Center: ( −4, 3) 2

= 0

x = −8

4 ( x + 4 ) + 4 ( y − 3 ) = 49 2

11.

)

x-intercepts: ( x + 8) + (0 + 5) = 25

4 x 2 + 8 x + 16 + 4 y 2 − 6 y + 9 = −51 + 64 + 36

Radius:

4 x 2 + 4 y 2 + 32 x − 24 y + 51 = 0

(

14. ( x + 8) + ( y + 5) = 25

Radius: 1 10.

= 5

y = −4 ±

16 ( x − 12 ) + 16 ( y + 34 ) = 16 2

y + 4 = ± 5

)

16 x 2 − x + 14 + 16 y 2 + 23 y + 169 = 3 + 4 + 9

1 2

= −7

y-intercepts: (0 − 2) + ( y + 4) = 9

16 x 2 + 16 y 2 − 16 x + 24 y − 3 = 0

)

No real solutions, so no x-intercepts.

Radius:

y-intercepts: (0 + 8) + ( y + 5) = 25 2

( y + 5)

7 2

= − 41

No real solutions, so no y-intercepts.

( x + 4 x + 4 ) + ( y + 6 y + 9) = 3 + 4 + 9 2

( x + 2 ) + ( y + 3 ) = 16 Center: ( −2, − 3) 2

Radius: 4 y

15.

y = − 81 x 2 x 2 = 4 ( −2 ) y, p = −2 Vertex: ( 0, 0 ) Focus: ( 0, − 2 ) Directrix: y = 2

−7 −6

−4 −3 −2 −1

2 3

−2 −3 −4 −5 −6

− 12 − 9 − 6

9 12

−6 −9 −12 −15

−8

−18

12.

( x + 8 x + 16 ) + ( y − 10 y + 25) = 8 + 16 + 25 2

−21

( x + 4 ) + ( y − 5 ) = 49 Center: ( −4, 5) 2

Radius: 7 y

14 10 8 6 4 2 − 14

− 10

2 4 6

−4 −6

Chapter 10 Review 19. Vertex: (0, 0), focus: (0, 5)

16. 4 x − y 2 = 0

y 2 = 4 (1) x, p = 1

h = 0, k = 0, p = 5

Vertex: ( 0, 0 )

Vertical axis

Focus: (1, 0 )

x 2 = 4 py x 2 = 4(5)( y )

Directrix: x = −1

x 2 = 20 y

20. Vertex: ( − 3, 0), focus: (0, 0)

6 4

h = − 3, k = 0, p = 3

2 −2

−2

Horizontal axis

−4 −6

17.

1 2

921

= 4 p ( x − 4)

( y − 0)

= 4(3)( x + 3)

y 2 = 12( x + 3)

y 2 + 18 x = 0 1 2

( y − k)

21. Vertex: ( −6, 4 )

y 2 = −18 x y = −36 x = 4 ( −9 ) x, p = −9 2

Vertex: ( 0, 0 )

Passes through ( 0, 0 ) Vertical axis

( x + 6) = 4 p( y − 4) 2 (0 + 6) = 4 p (0 − 4) 2

Focus: ( −9, 0 ) Directrix: x = 9

36 = −16 p

− 94 = p

( x + 6 ) = 4 ( − 94 ) ( y − 4 ) 2 ( x + 6 ) = −9 ( y − 4 ) 2

4 − 20 − 16 − 12 − 8

−4

22. Vertex: ( 0, 5)

( y − 5) = 4 p ( x − 0 ) = 4 px ( 6, 0 ) on graph: 2 ( 0 − 5) = 4 p ( 6 )  p = 2425 2 ( y − 5) = 4 ( 2425 ) x 2 ( y − 5) = 256 x 2

18.

1 4

y − 8x2 = 0 8 x 2 = 14 y

1 x 2 = 321 y = 4 ( 128 ) y, p = 1281

Vertex: ( 0, 0 )

1 Focus: ( 0, 128 )

1 Directrix: y = − 128

y 5 16 1 4 3 16 1 8

−3

−1 − 1 8

)0, 1281 ) (0, 0)

1 8

3 16

922

Chapter 10

Topics in Analytic Geometry

23.

1  1 x 2 = −2 y = 4  −  y, p = − 2  2

26. (a) Parabola: Vertex: ( 0, 4 )

Passes through ( ±6, 0 )

1  Focus:  0, −  2  1 d1 = + b 2

x2 = 4 p ( y − 4)

36 = 4 p ( 0 − 4 ) 36 = −16 p −4= p

1 5  d2 = ( 2 − 0 ) +  −2 +  = 2 2   1 5 d1 = d2  + b =  b = 2 2 2 b+2 4 Slope of tangent line: = = −2 0 − 2 −2 2

x 2 = − 16 ( y − 4 ) Circle: Passes through ( ±6, 0 ) Radius: r = 14 Center: ( 0, k )

Equation: y + 2 = −2 ( x − 2 )

x 2 + ( y − k ) = 142 2

y = −2 x + 2

( ±6 ) + ( 0 − k ) = 142

x-intercept: (1, 0 ) 24.

(

−2 x = y

4 − 12 x

)= y

36 + k 2 = 196

k 2 = 160

k = − 160 = −4 10

p = − 12

(

x 2 + y + 4 10

 1  Focus:  − , 0  2 

(

x 2 + y + 4 10

25. x = 4 p ( y −12) 2

( 4, 10 ) on curve: 16 = 4 p (10 − 12 ) = −8 p  p = −2 x = 4 ( −2 )( y − 12 ) = −8 y + 96 2

) = 196  y = 196 − x − 4 10 2

) ( 196 − x − 4 10 )

(

d = − 161 x 2 + 4 −

x-intercept: ( 8, 0 )

Circle:

1 d1 = + b 2 2 1 17  d2 =  −8 +  + ( −4 − 0 ) = 2 2  1 17 +b= b=8 2 2 −4 − 0 1 m= = −8 − 8 4 1 y − 0 = ( x − 8) 4 1 y = x−2 4

) = 196

(b) Parabola: x 2 = − 16 ( y − 4 )  y = − 161 x 2 + 4

Let ( b, 0 ) be the x-intercepts of the tangent line.

= − 161 x 2 − 196 − x 2 + 4 + 4 10 x

2.65

2.57

2.35

1.97

1.45

27.

x2 y2 + =1 25 36 a = 6, b = 5, c =

36 − 25 =

Center: (0, 0) Vertices: (0, ± 6)

(

)

Eccentricity:

11 6

Foci: 0, ± 11

− x 2 + 96 8

y = 0 if x = 96  x = 4 6  width is 8 6 meters. 2

4 2 −8 −6

−2 −2

−4 −8

Chapter 10 Review

28.

y2 x2 + =1 49 2

31. (a)

a = 7, b =

2, c =

49 − 2 =

( x − 1) + ( y + 4 ) = 1 2

9 16 (b) Center: (1, − 4 )

)

47 7

Eccentricity:

a = 4, b = 3, c = 16 − 9 = 7 Vertices: (1, 0 ) , (1, − 8 )

(

Foci: 1, − 4 ± 7

e= −8 −6 −4 −2

)

16 ( x − 1) + 9 ( y + 4 ) = 144 2

Vertices: (0, ± 7)

(

) (

= −16 + 16 + 144

Center: (0, 0) Foci: 0, ±

(

16 x 2 − 2 x + 1 + 9 y 2 + 8 y + 16

923

)

c 7 = a 4 y

−2

−8

−3 −4

( x − 3)2 + ( y − 2)2 = 1 29. 36

−5 −6

a = 6, b = 2, c =

36 − 4 = 4 2

−8

Center: (3, 2)

32. (a)

Vertices: ( − 3, 2), (9, 2)

(

)

(b)

25 4 a = 5, b = 2, c = 21 Center: ( −2, 3)

Vertices: ( 3, 3) , ( −7, 3 )

2 −4 −2 −2

(

Foci: −2 ± 21, 3

−4

−6 −8

)

( x + 2 ) + ( y − 3) = 1

4 2 2 2 = 6 3

( x + 5)

(

4 ( x + 2 ) + 25 ( y − 3 ) = 100

30.

)

= −141 + 16 + 225

Foci: 3, ± 4 2, 2 Eccentricity:

(

4 x 2 + 4 x + 4 + 25 y 2 − 6 y + 9

( y − 1) 9

a = 3, b = 1, c =

6 4

9 −1 =

8 = 2 2

2 −8

Vertices: ( − 5, 4), ( − 5, − 2)

(

Eccentricity:

Center: ( − 5, 1) Foci: − 5, 1 ± 2 2

c 21 = a 5

(c)

)

−6

−4

−2

−4

)

−6

7 6 5 4 3 2 1

2 2 3

−9 −8 − 7

−5

−3 −2 −1

−2 −3

924

Chapter 10

Topics in Analytic Geometry y

(

33. (a)

) (

)

3 x 2 + 4 x + 4 + 8 y2 − 14 y + 49 = −403

20 16 12 8 4

(c)

+12 + 392 3 ( x + 2 ) + 8 ( y − 7) = 1 2

−12 −8 −4

( x + 2 ) + ( y − 7) = 1 2

3 2 , b= 3 4

35. Vertices: ( ±5, 0 )

1 1 5 30 − = c= 3 8 24 12  3  , 7 Vertices:  −2 ±   3   c2 = a 2 − b2 =

 30  , 7 Foci:  −2 ±   12   e=

c = a

30 12 3 3

6 4 2

−2

−1

15 x 2 + y 2 − 90 x + 5 y − 140 = 0

34. (a)

25  25  15( x 2 − 6 x + 9) +  y 2 + 5 y +  = 140 + 135 + 4 4  2

5 1125 2  15( x − 3) +  y +  = 2 4 

( x − 3) + 2

x 2 y2 + =1 25 9

Vertical major axis Center: ( 0, 0 ) , a = 6

−3

a = 5, c = 4  b = 3,

Passes through ( 2, 2 )

10 4

Foci: ( ±4, 0 )

36. Vertices: ( 0, ± 6 )

−4

−8 −12 −16 −20

(b) Center: ( −2, 7 ) a=

12 16 20 24

5  y +  2 

75 4

15 5 5 3 (b) a = ,b = ,c = 2 2 5  Center:  3, −  2 

1125 4

1125 75 5 42 − = 4 4 2

x 2 y2 + =1 b 2 36 22 22 + =1 b2 36 4 1 8 =1− = b2 9 9 36 9 b2 = = 8 2 x2 y2 + =1 9 2 36

37. Vertices: ( −3, 0 ) , ( 7, 0 )

Foci: ( 0, 0 ) , ( 4, 0 ) Horizontal major axis Center: ( 2, 0 )

a = 5, c = 2, b = 25 − 4 = 21

( x − h) + ( y − k ) = 1 2

( x − 2 ) + y2 = 1 2

 5 5 Vertices:  3, − ±  2 2  

 5 5 42  Foci:  3, − ±  2 2   e =

c 5 42 2 = = a 15 5 2

210 15

Chapter 10 Review 38. Vertices: ( 2, 0 ) , ( 2, 4 )

Center: ( 0, 0 )

Vertical major axis Center: ( 2, 2 )

Transverse axis vertical a = 1  a2 = 1

a = 2, c = 1,

c = 2  c2 = 4

b = 4 −1 = 3

c2 = a2 + b2

( x − h) + ( y − k ) = 1

4 = 1 + b2

( x − 2) + ( y − 2) = 1

( y − k ) − ( x − h) = 1

x2 y2 + = 1, a = 324 = 18, b = 196 = 14 324 196 Longest distance: 2a = 2 (18 ) = 36 feet Shortest distance: 2b = 2 (14 ) = 28 feet c 2 = a 2 − b 2 = 128

(

Foci: ±8 2, 0

)

Distance between foci: 16 2 ≈ 22.63 feet 41. a − c = 1.3495 × 109

a + c = 1.5045 × 10

3 = b2 2

( y − 0) − ( x − 0) = 1 2

39. a = 5, b = 4, c = a 2 − b2 = 25 − 16 = 3 The foci should be placed 3 feet on either side of the center and have the same height as the pillars. 40.

44. Vertices: ( 0, ± 1) ; foci: ( 0, ± 2 )

Foci: ( 2, 1) , ( 2, 3)

Adding, 2 a = 2.854 × 10 9  a = 1.427 × 10 9. Then c = 1.5045 × 10 9 − 1.427 × 10 9 = 0.0775 × 10 9 c e = ≈ 0.0543. a 72 = 36 2 c e = = 0.2056  c = ae = 7.4016 a b2 = a 2 − c 2 = 362 − 7.4016 2  1214.2 x2 y2 + =1 1296 1241.2

42. a =

925

y −

x2 =1 3

45. Foci: ( 3, ± 2 ) ; asymptotes: y = ±2 ( x − 3)

Vertical transverse axis Center: ( 3, 0 )  c = 2 a = 2  a = 2b b c2 = a2 + b2 4 = ( 2b ) + b2 = 5b 2  b 2 = 2

4 2 16 ,a = 5 5

5 y2 5 ( x − 3) − =1 16 4 2

46. Foci: ( 0, 0 ) , ( 8, 0 )  c = 4

Center: ( 4, 0 ) Asymptotes: y = ±2 ( x − 4 ) 

b = 2  b = 2a a

c 2 = a 2 + b2 16 = a 2 + ( 2 a ) = 5a 2  a = 2

( x − 4) − y 2

16 5

4 5

, b=

8 5

64 5

43. Vertices: ( ±4, 0 ); foci: ( ±6, 0 )

x 2 y2 − =1 a 2 b2 a=4 c 2 = a 2 + b2  36 = 16 + b2  b = 20 = 2 5 2

x y − =1 16 20

926

Chapter 10

47. (a)

Topics in Analytic Geometry

5 y 2 − 4 x 2 = 20 2

49. (a)

y x − =1 4 5

(b)

(

)

9 ( x − 1) − 16 ( y + 1) = 144 2

( x − 1) − ( y + 1) = 1 2

a = 2, b = 5,

c= 4+5 =3

(b) Center: (1, − 1) , a = 4, b = 3, c = 5

Center: ( 0, 0 )

Vertices: ( 5, − 1) , ( −3, − 1)

Vertices: ( 0, ± 2 )

Foci: ( 6, − 1) , ( −4, − 1)

Foci: ( 0, ± 3) Asymptotes: y = ±

Asymptotes: y = −1 ±

2 5 x 5 (c)

(c)

(

9 x 2 − 2 x + 1 − 16 y 2 + 2 y + 1 = 151 + 9 − 16

3 ( x − 1) 4

5 4 3

6 4 2

1 −5 −4 −3 −2 −1

1 2 3 4 5

−6 −4

−2

−4 −3 −4 −5

−6 −8

x 2 − y2 =

48. (a)

50. (a)

9 4

(

)

3 3 , b= , 2 2 9 9 c= + = 4 4

25 ( y + 3 ) − 4 ( x + 1) = 100 2

(b) Center: ( −1, − 3) , a = 2, b = 5, c = 29 Vertices: ( −1, − 1) , ( −1, − 5)

9 3 3 2 = = 2 2 2

(

Foci: −1, − 3 ± 29

Center: ( 0, 0 )

(c)

 3 2  , 0 Foci:  ±   2   Asymptotes: y = ± x

3 4 5

2 ( x + 1) 5

y 6 4 2

−8 −6

5 4 3 2 1

)

Asymptotes: y = − 3 ±

 3  Vertices:  ± , 0   2 

−5 −4 −3

( y + 3) − ( x + 1) = 1

(c)

)

x2 y2 − =1 ( 9 4) ( 9 4) (b)

(

25 y 2 + 6 y + 9 − 4 x 2 + 2 x + 1 = −121 + 225 − 4

−2 −4 −6 −8 − 10 − 12 − 14

4 6 8 10

−2 −3 −4 −5

Chapter 10 Review 51. (a)

( y − 2 y + 1) − 4 ( x + 12 x + 36 ) = −59 + 1 − 144 2

( y − 1) − 4 ( x + 6 ) = −202 2 2 ( x + 6 ) − ( y − 1) = 1 (101 2 ) 202 2

53.

x2 − y 2 − 4 = 0 x2 − y 2 = 4

b ( x − h) a 1 = 0 ± ( x − 0) 1 = ±x

Asymptotes: y = k ±

(b) Center: ( −6, 1) a2 =

101 2 101 505 , b = 202, c 2 = + 202 = 2 2 2

y 4

 101  , 1 Vertices:  −6 ±   2  

3 2 1

  505 , 1 Foci:  −6 ±   2   Asymptotes: y = 1 ± 2( x + 6) (c)

−4 −3

25 y 2 − 9 x 2 = 225

− 20 − 30

) (

)

9 x − 8 x + 16 − y − 8 y + 16 = −119 + 144 − 16

(

y2 x2 − =1 9 25 a Asymptotes: y = k ± ( x − h) b 3 = 0 ± ( x − 0) 5 3 = ± x 5

) − ( y − 4) = 9 ( y − 4) = 1 ( x − 4) − 2

9 x−4

y 8

4 2

(b)

54. 25 y 2 − 9 x 2 − 225 = 0

(

−4

52. (a)

−3

− 10

−1 −2

− 30 − 20

927

Center: ( 4, 4 )

−8 −6 −4

Vertices: ( 4 ± 1, 4 ) : ( 3, 4 ) , ( 5, 4 )

(

−4

a = 1, b = 3, c = 10

Foci: 4 ± 10, 4

−2 −6 −8

)

Asymptotes: y = 4 ± 3( x − 4) (c)

12 10 8 6 4 2 −2 −2

10 12

928

Chapter 10

Topics in Analytic Geometry

55.

5 y 2 − x 2 + 6 x − 34 = 0

57. d2 − d1 = 186,000 ( 0.0005 )

5 y − ( x − 6 x + 9) = 34 − 9 2

2 a = 93 a = 46.5 c = 100

5 y 2 − ( x − 3) = 25 2

y2 ( x − 3) − 5 25

b = c2 − a2

Asymptotes: y = k ±

x2 y 2 − =1 a 2 b2

a ( x − h) b

 x2   602  x = 60  y2 = b2  2 −1 = 1002 − 46.52  −1 2 46.5 a     ≈ 5211.57  y ≈ 72.2 72.2 miles north

(

5 = ± ( x − 3) 5 y 8

(−100, 0)

100 80 60 40 20

−60

−20

−4 −2

−4

9 x 2 − 4 y 2 − 36 x − 8 y − 4 = 0 9( x 2 − 4 x + 4) − 4( y 2 + 2 y + 1) = 4 + 36 − 4 9( x − 2) − 4( y + 1) = 36 2

( x − 2) − ( y + 1) 2

b Asymptotes: y = k ± ( x − h) a 3 = −1 ± ( x − 2) 2

(100, 0) 20

−40 −60 −80 −100

−6

56.

)

(60, 0)

58. Q : Your location ( 0, 0 )

F1 : Location of friend to the west ( −10,560, 0 ) F2 : Location of friend to the east (10,560, 0 ) P ( x, y ) : Location of lightning strike

The sound at F1 is heard 8 − 6 = 2 seconds after the sound at F2 .

(1100 )( 2 ) = 2200 = 2a x 2 y2 − =1 a2 b2

y 6

2200 = 1100  a 2 = 1,210,000 2 b2 = c 2 − a 2 = 110,303,600 The equation of the first hyperbola is x2 y2 − = 1. 1,210,000 110,303,600 c = 10,560, a =

−4 −2 −2

−8

y P(x, y)

F1 (−10,560, 0)

(0, 0)

Q (10,560, 0)

For the second hyperbola, place the center at ( 5280, 0 ) . Q : Your location ( 0, 0 )

F2 : Location of friend to the east (10,560, 0 ) P ( x, y ) : Location of lightning strike

Chapter 10 Review The sound at F2 is heard 6 seconds after the sound at Q.

64.

(1100 )( 6 ) = 6600 = 2a ( x − 5280 ) − y2 = 1 2

b 6600 c = 5280, a = = 3300  a 2 = 10,890,000 2 b 2 = c 2 − a 2 = 16,988,400 The equation of the second hyperbola is

( x − 5280 ) − 2

10,890,000

929

x = t, y = 8−t t

8 7

y2 = 1. 16,988,400

6 5 4 3

2 1

P(x, y)

−3 −2 −1 −1

(10,560, 0) Q

(5,280, 0)

65.

x = 2t y = 4t x . 2 x y = 4  2

Let t = 59. 6 x 2 − x − 2 y + 13 = 0

AC = 6(0) = 0  Parabola

y = 2x

60. 3x 2 + 5 y 2 − 10 x − 8 y − 90 = 0

AC = (3)(5) = 15 > 0  Ellipse

4 3

61. 8 x + 8 y − 7 x + 14 y + 1 = 0

A = C = 8 ≠ 0  Circle 62.

−4 −3 −2 −1

− x 2 + 4 y 2 + 4 x + 3 y + 12 = 0

66.

−2

−1

−8

−5

−2

−1

−5

x = 4t + 1 y = 2 − 3t 8 6 2

−4

−6

−8

−8 −6 −4 −2 −2

−4

x = 3t − 2, y = 7 − 4t

− 12 − 8

−3

AC = ( −1)( 4) = − 4 < 0  Hyperbola 63.

x −1 . 4  x −1  y = 2 − 3   4 

Let t =

y = 2 − 34 x + 34 y = − 34 x + 114

930

Chapter 10

Topics in Analytic Geometry

67.

x = t 2 + 2, y = 4t 2 − 3

71.

t = x−2 2

x=3 y=t

y = 4 ( x − 2 ) − 3 = 4 x − 11, x ≥ 2

y −3

10 8

−4

6 4

Not a function

−4 −2

10 12

72.

−4

68.

x=t y=2 7

x = 5t , y = t 2 t = 15 x  y = ( 15 x ) = 251 x 2 2

−6

−1

3 2

Function

−8 −6 −4 −2

73.

x = 2t y = 4t

−2 −3

y = 2 ( 2t ) = 2 x

−4

1 69. x = t , y = t 2 2 1 t = x1 3  y = x 2 3 2 3

−6

−4

Function

5 4

74.

3 2

x = 1 + 4t y = 2 − 3t

1 −4 −3 −2 −1

−2

3 3 x −1  x − 1  y = 2 − 3 =2− x+  4  4 4 4

11 3 − x 4 4 3 x + 4 y − 11 = 0 y=

−3

70.

4 , y = t2 − 1 t 4 16 t =  y = 2 −1 x x x=

−4

8 −2

Function

− 12 − 8

−4

Chapter 10 Review 75.

x = t2, y = t t=y

79.

( ) =y , y≥0

x = y2

1 t y=t

y=t=

y = 4 x, x ≥ 0

1 x 4

−6 −1

−4

−1

Function

Function 76.

931

80.

x = t + 4, y = t 2 t = x−4 y = ( x − 4)

x=t 1 1 y= = t x 4

7 −6

−2

−4

10 −1

Function

Function 77.

81.

x=3t y=t t = x3  y = t = x3 y = x3

x = 6 cosθ , y = 6sin θ x y cosθ = , sin θ = 6 6 x 2 y2 + =1 36 36 2 2 x + y = 36

−6

− 12

−4

−8

Function 78.

Not a function

x=t

82.

y = 3 t = 3 x = x1 3

x = 3 + 3cosθ , y = 2 + 5sin θ x −3 y−2 , sin θ = 3 5

cosθ =

( x − 3) + ( y − 2 ) = 1 2

−3

−2

Function

−4

Not a function

932

Chapter 10

83.

y = 6x + 2 (a) Let t = x. x=t

Topics in Analytic Geometry 92. Because 20 yards = 60 feet, (60, 5) is a point on the

graph.  3 x = v0  t  2 

y = 6t + 2 (b) Let t = 1 − x. x =1− t

 3 60 = v0  t  2  120 = v0 3t

y = 6 (1 − t ) + 2  y = −6t + 8 84.

y = 10 − x (a) Let t = x. x=t

1 y = 7 + v0  t − 16t 2  2

y = 10 − t (b) Let t = 1 − x. x =1− t

 120  1  2 5 = 7 +  t − 16t  3t  2  60 −2 = − 16t 2 3

y = 10 − (1 − t )  y = t + 9 85.

y = x2 + 2 (a) Let t = x.

16t 2 = 3 +

x=t

t 2 ≈ 2.29

y = t2 + 2

t ≈ ±1.51

(b) Let t = 1 − x. x =1− t

So, the ball is caught after 1.51 seconds.

v0 =

y = (1 − t ) + 2  y = t − 2t + 3 2

86.

60 3

120 3 (0.92)

≈ 45.9

y = 2 x3 + 5x (a) Let t = x. x=t

The speed of the football is about 45.9 feet per second when it is released. 93.

y = 2t 3 + 5t

(b) Let t = 1 − x. x =1− t y = 2 (1 − t ) + 5 (1 − t )  y 2

y = −2t + 6t − 11t + 7

88. x = x1 + t ( x2 − x1 ) = − 2 + t ( − 8 − ( − 2)) = − 2 − 6t y = y1 + t ( y2 − y1 ) = 0 + t (0 − 0) = 0

89. x = x1 + t ( x2 − x1 ) = 0 + t ( − 5 − 0) = − 5t

Using the zoom and trace features, the maximum height is approximately 15.19 feet.

87. x = x1 + t ( x2 − x1 ) = 3 + t (3 − 3) = 3 y = y1 + t ( y2 − y1 ) = − 6 + t (1 − ( − 6)) = − 6 + 7t

94.

1 5 = 7 + ( 45.8) t − 16t 2  2 16t 2 − 22.9t − 2 = 0 t ≈ 1.69

The receiver has 1.69 seconds to position himself.

y = y1 + t ( y2 − y1 ) = 4 + t (7 − 4) = 4 + 3t 90. x = x1 + t ( x2 − x1 ) = − 3 + t ( 2 − ( − 3)) = − 3 + 5t

y = y1 + t ( y2 − y1 ) = −1 + t (9 − ( −1)) = −1 + 10t 3 2 1 2 y = 7 + v0 (sin θ ) t − 16t = 7 + v0 t − 16t 2 2

91. x = v0 (cos θ ) t = v0 (cos 30°) t = v0 t

Chapter 10 Review π 2

95.

π 2

98.

(2, π4) 3

933

(− 3, 56π ) Three additional representations:

Three additional representations:

7π   π   − 2π  =  2, −  2,  4 4     π 5π     + π  =  −2,  −2,  4 4    

π  5π   − π  =  3, −   3, 6  6    5π   11π  + π  =  3,  3,  6 6     5π 7π     − 2π  =  − 3, −   − 3, 6 6    

π 3π     − π  =  −2, −  −2,  4 4     96.

(7, − 43π ) 1 2 3 4 5

(4, − π3 )

4π π    + π  =  − 7, −   7, − 3 3    4π 5π     − π  =  − 7,  − 7, −  3 3    

100.

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

π 2

(− 2, − 116π )

Three additional representations:

Three additional representations: π    5π   4, − + 2π  =  4,  3    3  π 2π      − 4, − + π  =  − 4,  3 3     π 4π      − 4, − − π  =  − 4, −  3 3     97.

π 2

99.

π 2

(− 3, − 56π ) 1

Three additional representations:

11π π    + 2π  =  −2,   −2, − 6 6    11π    7π  + 3π  =  2,  2, −  6 6    

5π 7π     + 2π  =  − 3,  − 3, −  6 6    

11π 5π     + π  =  2, −  2, −  6 6    

5π    π + π  =  3,   3, − 6    6 5π 11π     − π  =  3, −  3, −  6 6    

934

Chapter 10

Topics in Analytic Geometry

 π 101. ( r , θ ) =  3,   2

5π   104. ( r , θ ) = 1, −  6  

x = r cosθ = 3cos π2 = 3(0) = 0 y = r sin θ = 3sin 2 = 3(1) = 3 π

( x, y ) = (0, 3) π 2

3  5π  x = r cosθ = cos  −  = − 2  6  1  5π  y = r sin θ = sin  −  = − 2  6  

(3, π2 )



π 2

1 2 3

(

1, −

 7π  102. ( r , θ ) =  5,   4   2 7π 5 2 = 5  = 4 2  2 

x = r cosθ = 5cos y = r sin θ = 5sin

3 1 , −  2 2

( x, y ) =  −

 7π 2 5 2 = 5 −  = − 4 2  2 

5 2 5 2  ,−  2   2

( x, y ) = 

5π 6

)

11π   105. ( r , θ ) =  0, −  6    3  11π  x = r cosθ = 0 ⋅ cos  −  = 0  = 0 ⋅  6    2   11π  1 y = r sin θ = 0 ⋅ sin  −  = 0⋅  = 0  6   2

( x, y ) = (0, 0)

π 2

π 2 1

(

7π 4

(0, −116π )

)

2π   103. ( r , θ ) =  − 4,  3   x = r cosθ = − 4cos

2π  1 = − 4 −  = 2 3  2

 3 2π y = r sin θ = − 4sin = − 4  = − 2 3 3  2 

( x, y ) = (2, − 2

)

(− 4, 23π )

5π   106. ( r , θ ) =  − 2,  3   x = r cosθ = − 2 cos

5π 1 = − 2  = −1 3  2

y = r cosθ = − 2 sin

 5π 3 = − 2 −  = 3 2  

( x, y ) = (−1,

π 2

)

π 2

(−2, 53π ) 1

Chapter 10 Review 107. ( x, y ) = ( 0, − 9 )

r = 9, tanθ undefined, θ =

(

110. ( x, y ) = −3, − 3

tan θ = 1  θ =

2 π  3π   ( r, θ ) =  9, 2 ,  −9, 2     

7π



 

−3

3 2 1

(0, − 9)

−9

−4 −3 −2 − 1 −1

− 12

(−3, −

108. ( x, y ) = ( −3, 4 )

r = 5, tan θ =

−4 3

3 ) −2 −3

111. x 2 + y 2 = 81

or ( 5, 2.214 ) , ( −5, 5.356 )

r 2 = 81 r =9 112. x 2 + y 2 = 48

(radians)

r 2 = 48 r=4 3

(−3, 4)

113. x 2 + y 2 − 4 x = 0

3 2

r 2 − 4r cosθ = 0 r = 4 cosθ

1 −4 −3 − 2 −1 −1

114. x 2 + y 2 = 6 y

−2 −3

r 2 = 6r sin θ r = 6sin θ

−4

109. ( x, y ) = ( 5, − 5 )

3π r = 5 2, tan θ = −1  θ = 4 7π   3π  ( r, θ ) =  5 2, 4  ,  −5 2, 4     

xy = 5

115.

( r cosθ )( r sin θ ) = 5 r 2 = 5csc θ sec θ

xy = −2

116.

( r cosθ )( r sinθ ) = −2

r 2 = −2sec θ csc θ

2 1 1

−2 −3 −5

−4

( 5, 126.87°) , ( −5, 306.87°)

−4



−6

−2 −1 −1

( r, θ ) =  2 3, 6  ,  −2 3, 6 

3 −3

7π 6

−6

)

r 2 = ( −3 ) + 3 = 12  r = 2 3

−9

935

(5, − 5)

4x2 + y = 1

117.

4 ( r cosθ ) + ( r sin θ ) = 1 2

(

)

4r 2 cos2 θ + r 2 1 − cos2 θ = 1 r 3cos θ + 1 = 1 2

r2 =

1 3cos2 θ + 1

936

Chapter 10

Topics in Analytic Geometry 2 x 2 + 3 y2 = 1

118.

126. θ =

2 ( r cosθ ) + 3 ( r sin θ ) = 1 2

(

tan θ =

)

r 2 2 cos2 θ + 3sin 2 θ = 1

((

) ) r ( 2 + sin θ ) = 1

y = 3 x

y = 3x

r 2 1 − sin θ + 3sin θ = 1 2

4π 3

127. r = 5, circle

1 2 + sin 2 θ

r2 =

π 2

r =5

119. 2

x + y 2 = 52 2

x 2 + y 2 = 25

r = 12

120.

x 2 + y 2 = 144 128. r = 3, circle

r = 3cosθ

121.

π 2

r 2 = 3r cosθ x 2 + y2 = 3x r = 8sin θ

122.

r 2 = 8r sin θ

x 2 + y2 = 8 y r 2 = cos2θ

123.

r 2 = 1 − 2sin 2 θ r = r − 2r sin θ 4

( x + y ) = x + y − 2y 2

129. θ =

π 2

, y -axis π 2

(x + y ) − x + y = 0 2

r 2 = sin θ

124.

r 3 = r sin θ

( x + y ) = y or (x + y ) = y 2

5π θ= 125. 4 y tan θ = x 5π y tan = 4 x y 1= x y=x

130. θ = −

5π , line 6 π 2

Chapter 10 Review 131. r = − 5cosθ , circle

937

135. r = 3 − 5sin θ

Limaçon with loop

π 2

Symmetric: line θ =

π 2

Maximum r -value : r = 8 when θ = 1

3π 2

3  Zeros: r = 0 when θ ≈ 0.6435, 2.4981  sin θ =   5 π 2

132. r = 2sinθ , circle

1 2 3

π 2

136. r = 6 − 6cosθ

Cardioid Symmetry: polar axis

133. r = 5 + 2cosθ

Maximum: r = 12 when θ = π

Convex limaçon Symmetric with respect to polar axis

Zero: r = 0 when cosθ = 1  θ = 0 π 2

r is maximum at θ = 0: ( r , θ ) = (7, 0) Zeros of r: None π 2

137. r = −3cos2θ , 0 ≤ θ ≤ 2π Four-leaved rose curve

Symmetric with respect to θ =

134. r = 1 + 4sin θ Limaçon with inner loop

Symmetric with respect to θ =

r is maximum at θ =

π 2

 π :  5,  2  2

r = 0 when 4 sin θ = −1  sin θ = −

, polar axis, and pole

The value of r is a maximum ( 3) at θ = 0,

π 2

, π,

3π . 2

3π 5π 7π r = 0 for θ = , , , 4 4 4 4 π 2

1  θ ≈ 3.394, 6.031 4

π 2

938

Chapter 10

Topics in Analytic Geometry

138. r = cos5θ Five-leaved rose curve Symmetric with respect to polar axis nπ r is a maximum value of 1 at θ = , n = 0, 1, 2,  5 π 2 nπ r = 0 for θ = , n = 0, 1, 2,  + 10 10

1 1 + 2sin θ e = 2 > 1  hyperbola

141. r =

−3

π 2

−1

6 1 + sin θ e = 1  parabola

142. r = 0

−9

139. r = 5sin 2θ Lemniscate Symmetry with respect to pole

−4

5π Maximum r -value : 5 when θ = , 4 4 π 3π Zeros: r = 0 when θ = 0, , π , 2 2 π 2

4 5 − 3cosθ 45 = 1 − ( 3 5 ) cosθ

143. r =

3 < 1  ellipse 5 2

−2

6 −6 = −1 + 4 cosθ 1 − 4 cosθ e = 4 > 1  hyperbola

144. r = 140. r = cos 2θ Lemniscate 2

Symmetry: Pole, polar axis, and line θ = Maximum: r = 1 when θ = 0, π , 2π Zeros: r = 0 when θ = π 2

π 4

2 −6

3π 5π 7π , , 4 4 4

−4

3 34 = 4 − 4cosθ 1 − cosθ e = 1  parabola

145. r = 1

−4

Chapter 10 Review 5 6 + 2sin θ 56 = 1 + (1 3 ) sin θ

ep 1 + e cosθ Vertices: (1, 0 ) , ( − 7, π )  a = 3

146. r =

150. Hyperbola: r =

One focus: ( 0, 0 )  c = 4

1 e = < 1  ellipse 3

r= −3

939

c 4 4 3p 7 = , 1=  p= a 5 1 + ( 4 3) cosθ 4

( 4 3)( 7 4 ) = 73 7 = 1 + ( 4 3) cosθ 1 + ( 4 3) cosθ 3 + 4cosθ

151. e = 0.093 −2

147. Parabola

r =

ep ,e = 1 1 − e cosθ

Vertex: (3, π ) Vertical directrix to the left of the pole r =

p 1 − cosθ

3 =

p 1 − cos π

p = 6 r =

6 1 − cosθ

148. Parabola: r =

ep , e =1 1 + e sin θ

 π Vertex:  2,  2  Focus: ( 0, 0 )  p = 4 4 r= 1 + sin θ

ep 149. Ellipse: r = 1 − e cosθ  π   3π  Vertices:  5, , 1,   a = 3  2  2 

One focus: (0, 0)  c = 2 Horizontal directrix below the pole c 2 = a 3 2 3p 5  p = 5 = 2 1 − ( 2 3) sin ( π2 ) e =

r =

ep . 1 − e cosθ 0.093 p 0.093 p 2a = + 1 − 0.093cosθ 1 − 0.093cos π = 0.1876 p = 3.05  p ≈ 16.258, ep ≈ 1.512

Use r =

(2 3)(5 2) = 53 5 = 1 − ( 2 3) sin θ 1 − ( 2 3) sin θ 3 − 2sin θ

1.512 1 − 0.093cosθ 1.512 Perihelion: ≈ 1.383 astronomical units 1 + 0.093 1.512 Aphelion: ≈ 1.667 astronomical units 1 − 0.093 r=

ep (horizontal directrix below pole). 1 − e sin θ e = 1 (parabola) −π When θ = , r = 6,000,000. 2 p p r= = = 6,000,000  p = 12,000,000  −π  2 1 − sin    2 

152. Use r =

12,000,000 1 − sin θ

When θ = −

, the distance is approximately 3 6,430,781 miles.

153. False. The y 4 -term is not second degree. 154. False. There are infinitely many sets possible. For example, x = t , y = 3 − 2t

x = 3t , y = 3 − 6t. 155. (a) Vertical translation (b) Horizontal translation (c) Reflection in the y-axis (d) Vertical shrink 156. (a) Major axis horizontal (b) Circle (c) Longer major axis (d) Horizontal translation

940

Chapter 10

Topics in Analytic Geometry

157. The orientation of the graph would be reversed.

158. (a) The speed would double. (b) The elliptical orbit would be flatter. The length of the major axis is greater.

Chapter 10 Test y2 = 8 x = 2 ( 4 ) x

x 2 − 4 y2 − 4 x = 0

x 2 − 4 x + 4 − 4 y2 = 4

Parabola Vertex: ( 0, 0 )

( x − 2 ) − 4 y2 = 4 2 ( x − 2 ) − y2 = 1 2

Focus: ( 2, 0 )

4 6

−2

−4

−2

−4

−2

−4 −6

x 2 + 4 y 2 − 24 x + 32 = 0

Hyperbola Center: ( 2, 0 )

( x2 − 24 x + 144) + 4 y 2 = − 32 + 144 ( x − 12)2 + 4 y 2 = 112 ( x − 12)

112

a = 2, b = 1, c = 5 Vertices: ( 0, 0 ) , ( 4, 0 )

y =1 28

(

Foci: 2 ± 5, 0

Ellipse Center: (12, 0)

a =

112 = 4 7, b =

c =

84 = 2 21

(

Vertices: 12 ± 4 7 0

(

Foci: 12 ± 2 21 0

)

Vertex: ( 6, − 2 ) , p = 2

( y + 2 ) = 4 ( 2 )( x − 6 ) 2 ( y + 2) = 8( x − 6) 2

28 = 2 7,

)

y 8 4

y 10 8 6 4 2 −2 −4 −6 −8 −10

)

−4

6 9 12 15 18

24 27

−8

Center: ( −6, 3) a = 7, b = 4

( x + 6 ) + ( y − 3) = 1 2

a = 3,

3 a = b=2 2 b

y2 x2 − =1 9 4

Chapter 10 Test

x2 −

y2 =1 4 y2 = x2 − 1 4

x = 2 + 3cosθ y = 2sin θ 2

−9

y = ±2 x 2 − 1

 x−2  y   +  =1  3  2

( x − 2 ) + y2 = 1 2

−6

9 Ellipse

8. x = 1 − t  t = 1 − x

y = 5t  t =

10.

941

y 5

4 3

y = 5(1 − x)

y = − 5x + 5 −2

Line y

−2 −3 −4

6 5 4 3 2 1

−5 −4 −3 −2 −1

11. y = − 3 x + 5

(a) Let t = x. x = t 2 3 4 5

y = − 3t + 5

(b) Let t = 2 − x 9.

x = 2 −t

x = t +2 y=

y = −3( 2 − t ) + 5  y = 3t − 1

t  t = 4y 4

12. xy = 8

x = 16 y 2 + 2 2

x = 16 y + 2 Right-hand portion of hyperbola

(a) Let t = x. x = t y =

(b) Let t = 2 − x

6 4

x = 2−t

2 −4 −2 −2

8 t

8 10 12

y =

−4 −6 −8

8 2−t

13. x 2 + 2 y = 4

(a) Let t = x.

x = t t2 + 2y = 4  y =

4 − t2 2

Let t = 2 − x

(b)

x = 2−t

(2 − t ) + 2 y = 4 2

4 − (2 − t ) 2 4t − t 2 y = 2

y =

942

Chapter 10

Topics in Analytic Geometry

5π   14.  −2,  6  

20. r =

5π = 3 6 5π y = r sin θ  y = −2sin = −1 6

x = r cosθ  x = −2cos

Answer: 15.

3  Hyperbola 2 10

( 3, − 1)

− 12

7π

( x, y ) = ( 2, − 2 ) , r = 8 = 2 2 , θ = 4

7π π 3π ( r, θ ) =  2 2, 4  =  2 2, − 4  =  −2 2, 4        16.

4 2 = 3 2 + 3sin θ 1 + 2 sin θ

x 2 + y 2 − 3k = 0

−6

21. Because r = f (sin θ ) is symmetric to the line θ =

, 2 the rose curve r = a sin(7θ ), a ≠ 0 has seven petals and

is symmetric to the line θ =

r − 3r cosθ = 0 2

r 2 = 3r cosθ r = 3cosθ

17.

22. r = 8cos3θ 3-petal rose curve

r = 2 sin θ

Symmetry: Polar axis

r 2 = 2r sin θ x 2 + y2 = 2 y

r = 8cos3θ

x 2 + y2 − 2 y + 1 = 1

r = 8cos( − 3θ )

r = 8cos 3( −θ )

x 2 + ( y − 1) = 1 2

r = 8cos3θ

18. r = 2 + 3 sinθ Limaçon with inner loop

Equivalent equation r = 0

Zeros:

8cos3θ = 0

−6

cos3θ = 0 3θ =

θ =

−2

π 3π 5π 2

π π 5π , , 6 2 6

π 2

1 1 − cosθ e = 1  Parabola

19. r =

10 −4

−8

23. r =

( 14 ) ( 4 ) ep = 1 + e sin θ 1 + ( 14 ) sin θ

4 1 = 4 + sin θ 1 + ( 14 ) sin θ

24. r =

( 54 ) ( 2 ) = 10 ep = 1 + e sin θ 1 + ( 45 ) sin θ 4 + 5 sin θ

Chapters 8–10 Cumulative Test

943

Chapters 8–10 Cumulative Test 1.

 −1 −3  5   1 0  4   row reduces to  . 2  10   4  0 1  −3 

Answer: ( 4, − 3) 2.

−1

9. (a)

2 x − y2 = 0 x−y=4 x= y+4

−1  1 2  −175 37 −13     − = 3 7 10 7    95 −20  −5 −7 −15  14 −3 1 1

(b)

2 ( y + 4 ) − y2 = 0

= 1( −105 − 70) − 2 ( −45 − 50) − 1( −21 + 35)

y2 − 2 y − 8 = 0

( y − 4 )( y + 2 ) = 0

= −175 + 190 − 14 = 1

y = 4, x = 8 y = −2, x = 2 Answer: ( 2, − 2 ) , ( 8, 4 )

1  13  2 −3   1 −2  −6  row reduces to  −4  1 −3 3  12 

10.

0 0 1 6 2 1 = 44  Area = 12 ( 44 ) = 22 square units 8 10 1

a3 =

 −18 15 −14    5 A + 3 B =  28 11 34   −20 52 −1

 3 −31 2    AB =  22 18 6   52 −40 14 

31  5 36   BA =  −36 12 −36   16 0 18 

1 9

1 11 1 a5 = 13

(b)

 1 0 0  1 0 1 0  2.   0 0 1  0 Answer: (1, 2, 0)

1 7

a4 = −

 1 1 − 5  3  4 1 −11  6 row reduces to   −1 0 − 2  −1

 −7 −10 −16    3 A − B =  −6 18 9  −12 16 7 

a1 =

a2 = −

Answer: ( 35 , − 4, − 15 )

( −1) = 1 2 (1) + 3 5 1+ 1

11. (a)

3 1 0 0   5   0 1 0 4  −  . 0 0 1  − 1  5 

−1

det ( A) = 3 7 −10 −5 −7 −15

a1 = 3 ( 2 )

1 −1

a2 = 6 a3 = 12 a4 = 24 a5 = 48 6

12.

 ( 7k − 2 ) =

7 ( 6 )( 7 ) 2

k =1

13.

− 2 ( 6 ) = 135

 k + 4 = 1 + 4 + 4 + 4 + 9 + 4 + 16 + 4 k =1

≈

47 52

n  1 − ( 3 )11  3 4  ≈ 34.4795 14.  9   = 9   1 − ( 34 )  n=0 4   10

15.

 1

 100  − 2  =  100  − 2  n=0

n −1

n =1

= 100 ≈

1 − ( − 12 )

2 (100) ≈ 66.67 3

944

Chapter 10

16.

 3 − 5  = 1 − ( − ) =

∞

 3

Topics in Analytic Geometry

3 5

n =0 ∞

= 8 5

15 8

19. (a)

5 −0.1 =− 1.02 51

∞ 1 1  1 + −  = 4 −  2 4 2 n=0 

18. 4 − 2 + 1 −

= 21.

20.

C18 =

20! 20! = = 190 ( 20 − 18 )!18! 2!18!

20! 20  20   2  = 20 − 2 ! 2! = 18! 2! = 190 ( )

(b)

1 17.  5 ( −0.02) = 5 ( −0.02 ) 1 − − ( 0.02) n =1 n

( x + 3) = x 4 + 12 x 3 + 54 x 2 + 108 x + 81 4

4 8 = 1 1 − (− 2 ) 3

( 2 x + y ) = 32 x + 80 x y + 80 x y + 40 x y + 10 xy + y 2

4 2

3 4

2 6

22. ( x − 2 y) = x 6 − 12 x 5 y + 60 x 4 y2 − 160 x 3 y3 + 240 x 2 y 4 − 192 xy5 + 64 y6 6

23.

(3a − 4b) = 6561a − 69,698a b + 326,592a b − 870,912a b 8

6 2

5 3

+ 1,451,520 a 4 b 4 − 1,548,288a 3b5 + 1,032,192 a 2 b6 − 393,216 ab 7 + 65,536b8 24. Permutations of C, O, U, N, T 5

P5 =

( 5 − 5)!

5! = 120 0!

25. Permutations of P, R, I, N, C, I, P, L, E 9! 9! = = 362,880 9 P9 = (9 − 9)! 0!

However, because there are 2! permutations of two letter Is, and 2! permutations of two letter Ps, the number of distinguishable permutations is: 9! 362,880  = 90,720 permutations. 2!2! 2⋅2

27. Permutations of C, I, N, C, I, N, N, A, T, I 10! 10! = = 3,628,800 10 P10 = (10 − 10)! 0!

However, because there are 2! permutations of two letter Cs, 3! permutations of three letter Ns, and 3! permutations of three letter Is, the number of distinguishable 10! 3,628,800  = 50,400 permutations is: 3!3!2! 6⋅6⋅2 permutations. 28. Hyperbola with center ( 5, − 3) y 15

26. Permutations of A, T, L, A, N, T, A 7 P7

(7 − 7)!

7! = 5040 0!

However, because there are 3! permutations of three letter As, and 2! permutations of two letter Ts, the number of 7! 5040 distinguishable permutations is:  = 420 3!2! 6⋅2

− 10 − 5

−5

− 15

permutations.

Chapters 8–10 Cumulative Test 29. Ellipse with center ( 2, − 1)

34. Center: ( 0, − 4 ) ; a = 2

( y + 4) − x2 = 1 2

b2 16 16 16 ( 4, 0 ) : 4 − b2 = 1  b2 = 3  b2 = 3 4

−1

945

−1

( y + 4) − x2 = 1 2

−2

−3 −4

35. 30. Hyperbola with center ( 0, 0 )

16 3

x = 2t + 1, y = t 2 (a), (b) y

−8 −6

−2

−4 −6

−3 −2 −1 −1

−8

−2

31.

( x − 2 x + 1) + ( y − 4 y + 4 ) = −1 + 1 + 4 2

(c)

( x − 1) + ( y − 2 ) = 4 2

36.

Circle

x −1 x2 − 2x + 1  x −1   y = =  2 4  2 

x = cosθ , y = 2sin 2 θ (a), (b) y

−3 −2

32.

−1

−2

−1

(c)

y = 2sin 2 θ

( ) = 2 (1 − x ) , − 1 ≤ x ≤ 1 = 2 1 − cos2 θ

1 3

37.

x = 4 ln t, y = 12 t 2 (a), (b)

33. Center: (1, 4 )

a = 5, b = 2

( x − 1) + ( y − 4 ) = 1 2

−1

−2

( x − 2 ) = 4 p ( y − 3)

( 0, 0 ) : 4 = 4 p ( −3)  p = − 2 ( x − 2 ) = 4 ( − 13 ) ( y − 3 ) 2 ( x − 2 ) = − 43 ( y − 3 )

4 2

−4

(c)

−2

x 2

t = e x 4  y = 12 e , x ≥ 0

946

Chapter 10

Topics in Analytic Geometry

38.

y = 3x − 2 (a) Let t = x. x=t y = 3t − 2 x (b) Let t = . 2 x = 2t

42.





7π   π  11π    8, −  ,  −8, −  ,  −8,  6   6  6   π 2

(8, 56π )

y = 3 ( 2t ) − 2  y = 6t − 2

39.

5π

( r, θ ) =  8, 6 

x 2 − y = 16 (a) Let t = x.

x=t y = t 2 − 16 x . 2 x = 2t

(b) Let t =

y = ( 2t ) − 16  y = 4t 2 − 16 2

40.

π  43. ( r , θ ) =  5, −  2  3π  π   3π    5, ,  − 5, ,  − 5, −  2  2   2  

2 x (a) Let t = x. y=

π 2

x=t 2 y= t x (b) Let t = . 2 x = 2t 2 1 y= y= 2t t e2 x 41. y = 2 x e +1 (a) Let t = x. x=t

e e

2(2t )

(5, − π2 ) 44.

5π

( r, θ ) =  −2, 4  



3π   π   7π    −2, −  ,  2,  ,  2, −  4 4 4       π 2

e2 t y = 2t e +1 x (b) Let t = . 2 x = 2t 2(2t )

(− 2, 54π ) 1

y=

e e4t + 1

Chapters 8–10 Cumulative Test

45.

11π

( r, θ ) =  −3, − 6  

51. r =



947

5 6 + 2sin θ

Ellipse

π   7π   5π    −3,  ,  3,  ,  3, −  6  6   6  

π 2

−3

−2

52. 32,500 + 32,500(1.03) +  + 32,500(1.03)

(

− 3, −

46.

11π 6

= 32,500(1.03)

)

 1 − 1.0315  = 32,500   1 − 1.03  = $604,464.70

4x + 4y + 1 = 0 4r cosθ + 4r sin θ + 1 = 0 r  4cosθ + 4sin θ  = −1 −1 r= 4cosθ + 4sin θ r = 4 cosθ

47.

r 2 = 4r cosθ x 2 + y2 = 4 x

(6, 14) : 14 = − a (6) + 16 2

x 2 − 4 x + 4 + y2 = 4

36 a = 2

( x − 2 ) + y2 = 4 2

a = 181 y = − 181 x 2 + 16

2 4 − 5cos θ 4r − 5r cos θ = 2 r=

(

) − 5x = 2 16 ( x + y ) = ( 5 x + 2 ) = 25 x + 20 x + 4

4 x +y

53. Since the price is at least $600, there are two ways to select the first digit (6 or 7), two ways for the second digit, and three ways for the third digit. So, there are 2 ⋅ 2 ⋅ 1 = 4 different ways to arrange the digits. 1 Therefore, the probability is p = = 0.25. 4 54. Let y = −ax 2 + 16.

x 2 + y2 − 4 x = 0

48.

n −1

n =1

16 y 2 + 9 x 2 − 20 x = 4

y = 0:16 = 181 x 2 x 2 = 288 x = 12 2 Width: 24 2 meters

49. r = − 4, Circle 6

−9

−6

50. r = 3 − cosθ

Convex limaçon 4

−6

−4

CH EC KPOIN TS

Chapter P

.............................................................................................................950

Chapter 1

.............................................................................................................959

Chapter 2

.............................................................................................................971

Chapter 3

.............................................................................................................995

Chapter 4

...........................................................................................................1011

Chapter 5

...........................................................................................................1027

Chapter 6

...........................................................................................................1042

Chapter 7

...........................................................................................................1057

Chapter 8

...........................................................................................................1069

Chapter 9

...........................................................................................................1095

Chapter 10

...........................................................................................................1106

C H E C K P O I N T S Chapter P Checkpoints for Section P.1 1. (a) The inequality x > − 3 denotes all real numbers

5. Expression

greater than − 3.

−5

−4

−3

−2

−1

4x − 5

(b) The inequality 0 < x ≤ 4 means that x > 0 and x ≤ 4. This double inequality denotes all real numbers between 0 and 4, including 4 but not including 0. x 0

6. (a)

(b)

2. The inequality − 2 < x ≤ 4 can represent the statement “x is greater than − 2 and at most 4.” 3. The interval consists of real numbers greater than or equal to − 2 and less than 5. 4. (a) If x > − 3, then

(b) If x < − 3, then

x+3 x+3 = = 1. x+3 x+3

Value of Variable

Substitute

Value of Expression

x = 0

4(0) − 5

0 − 5 = −5

3 x 3x x ⋅ = = 5 6 30 10

x x 2x 2x 2 + = + ⋅ 10 5 10 5 2 x 4x = + 10 10 5x = 10 x = 2

x+3 − ( x + 3) = = −1. x+3 x+3

Checkpoints for Section P.2 1. (a)

(2 x− 2 y3 )(− x4 y) = (2)(−1)( x− 2 )( x4 )( y3 )( y) = − 2x2 y4

(b)

(4a2b3 ) = 1, a ≠ 0, b ≠ 0

(c)

(− 5 z )3 ( z 2 ) = (− 5)3 ( z )3 z 2

= −125 z 5 2. (a) 2a − 2 =

2 a2

Property 3

(b)

3a − 3b 4 3b 4 ⋅ b = −1 15ab 15a ⋅ a 3 =

x (c)    10 

950

−1

x −1 10−1

Property 7

10 x

Property 3

(d)

b5 5a 4

(− 2 x 2 ) (4 x3 ) 3

−1

Property 3 Property 1

= ( − 2) ( x 2 ) ⋅ 4−1 ⋅ ( x3 ) 3

−8x6 4 x3

= − 2 x3

−1

Property 5 Properties 3 and 6 Property 2

Solutions to Checkpoints 3. Expression: 134 + 5− 2

951

6. Because 184,000,000 = 1.84 × 108 ,

Graphing Calculator Keystrokes: 13 ∧ 4 + 5 ∧ − 2 Enter

51,000,000 = 5.1 × 107 , and 4.8 × 10−3 , you can evaluate the expression using

Display: 28,561.04

( 1.84 EE 8 − 5.1 EE 7 ) / 4.8 EE − 3 Enter . The display on the calculator is 2.771 E 10 or

4. 45,850 = 4.585 × 104

27708333333.3.

5. − 2.718 × 10− 3 = − 0.002718

So,

7. (a) − 144 = −12 because −

184,000,000 − 51,000,000 ≈ 2.771 × 1010. 0.0048

( 144 ) = ( 12 ) = − (12) = −12. 2

(b)

−144 is not a real number because no real number raised to the second power produces −144.

(c)

52 25 25 5 5 . = because   = 2 = 8 64 64 8 8

(d) − 3

 8   3 8  8 2  2 = − because −  3  = −  3  = −  .   27 3  3  27   27 

125 = 5 =

8. (a)

125 5

Property 3

Simplify.

= 5

(b)

1252 =

= (5)

= 25

(c)

= 10. (a) 3 8 +

−135 x = 3 ( − 27) ⋅ 5 ⋅ x 3 3

= 3 x3

Simplify.

= − 3x 3 5

= x

Property

Property 4 Simplify. 4⋅2 +

9⋅2

2 +3

Find square factors.

Find square roots.

Multiply.

= (6 + 3) 2 = 9 81x 5 −

= 2a 2 3

4a 4 ⋅ 6 a

( 2 a 2 ) ⋅ 6a

= 3 ( − 3 x) ⋅ 5

= 6

(b)

4 ⋅ 6 ⋅ a4 ⋅ a =

Property 2

18 = 3

24a 5 =

Simplify.

= 3⋅2

250 = 3 125 ⋅ 2 = 3 53 ⋅ 2 = 5 3 2

(d)

42 ⋅ 2 = 4 2

16 ⋅ 2 =

Property 1

x2 ⋅ 3 x = 3 x2 ⋅ x

(c)

Simplify.

x = 2⋅2 x

(d)

(b)

Simplify.

( 125 )

32 =

9. (a)

Combine like radicals.

24 x 2 =

Simplify. 3

27 x 3 ⋅ 3 x 2 −

8 ⋅ 3x 2

Find cube factors.

= 3x 3 3x 2 − 2 3 3x 2

Find cube roots.

= ( 3 x − 2)

Combine like radicals.

952

Solutions to Checkpoints

11. (a)

(b)

5 5 = ⋅ 3 2 3 2

2 is rationalizing factor.

5 2 3( 2)

Multiply.

5 2 6

Simplify.

1 = 25

3 1 5 ⋅ 3 25 5

5 5

8 6 −

Simplify. 8 6 −

5 is rationalizing factor.

Multiply.

(c)

2 2

6 + 6 +

⋅

( 6 + 2) ( 6) − ( 2) 8( 6 + 2 ) =

Multiply numerator and denominator by conjugate of denominator.

Multiply.

Simplify denominator.

( 6+

2− 2 2− 2 2+ = ⋅ 3 3 2+

2 2

Multiply numerator and denominator by conjugate of numerator.

2 −2

Multiply.

= 2

12.

2 2

= =

13. (a)

(b) (c)

4+ 2

(

2 −2

32+ 2

(

32+

Divide out common factor.

)

Simplify.

)

27 = 271 3

x3 y 5 z = ( x3 y 5 z )

( x 2 − 7)

−1 2

= x3 2 y 5 2 z1 2

( x − 7)

( )

(d) − 3b1 3c 2 3 = − 3 bc 2 14. (a)

)

(−125)− 2 3 = ( 3 −125 )

1 2

x −7

= − 3 3 bc 2 −2

= ( − 5)

−2

(− 5)2

1 25

(b)

(4 x2 y3 2 )(− 3x−1 3 )( y −3 5 ) = −12 x(2) − (1 3) y(3 2) − (3 5) = −12 x5 3 y9 10 , x ≠ 0, y ≠ 0

(c)

3 4

27 = 12 27 = 12 (3) = 33 12 = 31 4 = 4 3

(d) (3 x + 2)

(3 x + 2)−1 2 = (3 x + 2)(5 2) − (1 2) = (3 x + 2)2 , x ≠ −

2 3

Solutions to Checkpoints

953

Checkpoints for Section P.3 Standard Form

1. Polynomial

Degree

Leading Coefficient

−7

6 − 7 x3 + 2 x − 7 x3 + 2 x + 6

(

) (

)

2. 2 x3 − x + 3 − x 2 − 2 x − 3 3

= 2x − x + 3 − x + 2x + 3 = 2 x 3 − x 2 + ( − x + 2 x) + (3 + 3) = 2 x3 − x 2 + x + 6

(3x − 1)( x − 5) = 3x − 15 x − x + 5 2

= 3 x 2 − 16 x + 5 4.

x2 + 2x + 3 × x2 − 2x + 3

← x 2 ( x 2 + 2 x + 3)

x 4 + 2 x3 + 3x 2

← − 2 x( x 2 + 2 x + 3)

− 2 x3 − 4 x 2 − 6 x

3x + 6 x + 9 ← 3( x 2 + 2 x + 3) 2

x 4 + 0 x3 + 2 x 2 + 0 x + 9

(

)(

)

So, x 2 + 2 x + 3 x 2 − 2 x − 3 = x 4 + 2 x 2 + 9. 5. This product has the form (u + v)(u − v) = u 2 − v 2 .

( x − 2 + 3 y )( x − 2 − 3 y ) = ( x − 2) + 3 y ( x − 2) − 3 y  = ( x − 2) − ( 3 y ) 2

= x2 − 4x + 4 − 9 y2 = x2 − 9 y 2 − 4x + 4

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

5x 2 is a common factor.

= 5 x 2 ( x − 3)

(b) − 3 + 6 x − 12 x3 = −12 x3 + 6 x − 3

( ) = − 3( 4 x3 − 2 x + 1) (c) ( x + 1)( x 2 ) − ( x + 1)( 2) = ( x + 1)( x 2 − 2)

= − 3 4 x3 + ( − 3)( − 2 x) + ( − 3)(1)

7. 100 − 4 y 2 = 4( 25 − y 2 )

( x + 1) is a common factor. 9. 9 x 2 − 30 x + 25 = (3x) − 2(3x)(5) + 52 2

= (3x − 5)

2 2 = 4 (5) − ( y )    = 4(5 + y )(5 − y )

8. ( x − 1) − 9 y 4 = ( x − 1) − (3 y 2 ) 2

− 3 is a common factor.

10. 64 x3 − 1 = ( 4 x) − (1) 3

= ( 4 x − 1)(16 x 2 + 4 x + 1)

2 = ( x − 1) + 3 y 2  ( x − 1) − 3 y 

= ( x − 1 + 3 y 2 )( x − 1 − 3 y 2 )

954

Solutions to Checkpoints

11. (a) x3 + 216 = ( x) + (6) 3

13. (a) For the trinomial 2 x 2 − 5 x + 3, you have a = 2 and c = 3, which means that the factors of 3 must have like signs. The possible factorizations are

= ( x + 6)( x 2 − 6 x + 36) (b) 5 y 3 + 135 = 5( y 3 + 27)

( 2 x + 1)( x + 3), ( 2 x − 1)( x − 3), ( 2 x + 3)( x + 1), and ( 2 x − 3)( x − 1).

= 5( y ) + (3)    3

= 5( y + 3)( y 2 − 3 y + 9)

12. For the trinomial x 2 + x − 6, you have a = 1, b = 1, and c = − 6. Because b is positive

Testing the middle term, you will find the correct factorization to be 2 x 2 − 5 x + 3 = ( 2 x − 3)( x − 1).

and c is negative, one factor of − 6 is positive and one is negative. So, the possible factorizations of x 2 + x − 6 are

(b) For the trinomial 12 x 2 + 7 x + 1, you have a = 12, b = 7, and c = 1. Because a, b, and c are all positive, the factors of a and c are positive.

( x − 3)( x + 2), ( x + 3)( x − 2), ( x + 6)( x − 1), and ( x − 6)( x + 1).

So, the possible factorizations are

Testing the middle term, you will find the correct factorization to be x 2 + x − 6 = ( x + 3)( x − 2).

(

)

14. x 3 + x 2 − 5 x − 5 = ( x 3 + x 2 ) − (5 x + 5) = x ( x + 1) − 5( x + 1) 2

= ( x + 1)( x 2 − 5)

(12 x + 1)( x + 1), (6 x + 1)(2 x + 1), and (4 x + 1)(3x + 1). Testing the middle term, you will find the correct factorization to be 12 x 2 + 7 x + 1 = ( 4 x + 1)(3x + 1).

Group terms. Factor. Distributive Property

15. 2 x 2 + 5 x − 12 = 2 x 2 + 8 x − 3 x − 12 = ( 2 x + 8 x ) − (3 x + 12) 2

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

Rewrite middle term. Group terms. Factor. Distributive Property

Checkpoints for Section P.4 1. (a) The domain of the polynomial 4 x 2 + 3, x ≥ 0 is the set of all real numbers that are greater than or equal to 0. The domain is specifically restricted.

(b) The domain of the radical expression

x + 7 is the set of all real numbers greater than or equal to − 7,

because the square root of a negative number is not a real number. 1− x is the set of all real number except x = 0, which would x result in division by zero, which is undefined.

4 ( x + 3) 4 x + 12 = x − 3 x − 18 ( x − 6) ( x + 3) 2

4 , x ≠ −3 x−6

Factor completely. Divide out common factor.

Solutions to Checkpoints

(3x + 2) ( x − 1) 3x 2 − x − 2 3x 2 − x − 2 = = 2 2 5 − 4x − x − x − 4x + 5 − ( x + 5) ( x − 1)

Write in standard form.

3x + 2 ,x ≠ 1 x +5

Divide out common factor.

= −

5 x (3x + 1) ( x − 5) ( x + 3) 15 x 2 + 5 x x 2 − 2 x − 15 ⋅ 2 = ⋅ 2 x − 3 x − 18 x 3 x − 8 x − 3 x ( x − 6) ( x + 3) (3 x + 1) ( x − 3) 3

5( x − 5) 1 , x ≠ − 3, x ≠ − , x ≠ 0 3 ( x − 6)( x − 3)

x3 − 1 x2 + x + 1 x3 − 1 x 2 + 2 x + 1 ÷ 2 = 2 ⋅ 2 x − 1 x + 2x + 1 x − 1 x2 + x + 1 =

( x − 1) ( x 2 + x + 1) ( x + 1) ( x + 1) ⋅ x2 + x + 1 ( x + 1) ( x − 1)

= x + 1, x ≠ ±1

955

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

Factor completely. Divide out common factors.

Basic definition

x2 + 2x − 2 x + 1 (2 x − 1)( x + 2)

Distributive Property

x2 + 1 (2 x − 1)( x + 2)

Combine like terms.

7. The LCD of the rational expression

Invert and multiply.

4 x+5 4 − 2 + is x( x + 2)( x − 2). x x −4 x+2

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

Rewrite using the LCD.

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

Combine fractions.

4 x 2 − 16 − x 2 − 5 x + 4 x 2 − 8 x x( x + 2)( x − 2)

Distributive Property

7 x 2 − 13x − 16 x( x + 2)( x − 2)

Combine like terms.

 1 + 1( x + 2)   1  + 1    x+ 2  2 x +   =  8. x   x − 1(3)   − 1   3   3    x + 3   x + 2 =   x − 3    3  3 x +3 = ⋅ x + 2 x −3 3( x + 3) = ( x + 2)( x − 3)

Combine fractions.

Simplify.

Rewrite as multiplication. Multiply.

956

Solutions to Checkpoints

9. ( x − 1)

−1 3

− x( x − 1)

−4 3

= ( x − 1)

−4 3 

= ( x − 1)

−4 3

10.

x 2 ( x 2 − 2)

−1 2

( x − 1)

( −1 3) − (− 4 3)

( x − 1)1 − x  

( x − 1)

+ ( x 2 − 2)

x2 − 2

x 2 ( x 2 − 2)

−1 2

+ ( x 2 − 2)

x2 − 2 0

( x 2 − 2)

9+ h −3 = h

x2 + x2 − 2

= =

( x2 − 2) ⋅ 12 ( x 2 − 2)

x 2 ( x 2 − 2) + ( x 2 − 2)

11.

− x 

( x 2 − 2)

2x2 − 2

( x 2 − 2)

2( x + 1)( x − 1)

9+ h −3 ⋅ h

( 9 + h ) − (3) = h( 9 + h + 3) 2

( x 2 − 2)

9+ h +3 9+ h +3

12.

x + 2 − 2

h h

( 9 + h + 3)

( 9 + h + 3) 1

9+ h +3

x+ 2 + x+ 2 +

⋅

x+ 2 − 2

( x + 2) − ( x ) = 2( x + 2 + x )

(9 + h ) − 9

,h ≠ 0

x+ 2− x

( x+2+

(

)

2 x + 2 + 1 x+ 2 +

x x

Checkpoints for Section P.5 y

(−3, 2)

−4

−2

(3, 1) 2

(0, −2)

(−1, −2)

(4, −2)

700 600 500 400 300 200 100 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013

−4

2. To sketch a scatter plot of the data shown in the table, first draw a vertical axis to represent the number of employees E (in thousands) and a horizontal axis to represent the year. Then plot the resulting points. Note that the break in the t-axis indicates that the numbers 0 through 2004 have been omitted. Number of employees (in thousands)

Year

Solutions to Checkpoints 3. Let ( x1 , y1 ) = (3, 1) and ( x2 , y2 ) = ( − 3, 0).

Then apply the Distance Formula.

d =

( x2 − x1 ) + ( y2 − y1 )

(− 3 − 3) + (0 − 1)

(− 6) + (−1)

36 + 1

5. You can find the length of the pass by finding the distance between the points (10, 10) and ( 25, 32). y

2 35

(25, 32)

25 20

15 10

(10, 10)

5 5

≈ 6.08 So, the distance between the points is about 6.08 units. 4. The three points are plotted in the figure. y

(5, 5)

5 4 3

1 −1 −1 −2

(2, −1)

−3

(5 − 2) + (5 − (−1))

32 + 62

9 + 36

d2 =

(6 − 2) + (− 3 − (−1)) 2

42 + ( − 2)

16 + 4

d =

( x2 − x1 ) + ( y2 − y1 )

(25 − 10) + (32 − 10)

152 + 222

225 + 484

709

2 2

6. Let ( x1 , y1 ) = ( − 2, 8) and ( x2 , y2 ) = ( −4, −10).

 x + x2 y1 + y2  Midpoint =  1 ,  2   2  − 2 + 4 8 + ( −10)  =  ,  2  2 

(6 − 5)2 + (− 3 − 5)2

(1)2 + (− 8)2

1 + 64

= (1, −1) The midpoint of the line segment is (1, −1). 7. One solution to the problem is to assume the annual revenues followed a linear pattern. With this assumption, you can estimate the 2012 revenue to be the midpoint of the line segment connecting the points (2011, 37.9) and (2013, 59.8).  2011 + 2013 37.9 + 59.8  Midpoint =  ,  2 2   = ( 2012, 48.85)

Because ( d1 ) + ( d 2 ) = 45 + 20 = 65, you can 2

 2 2 =  ,−   2 2

d3 =

So, the pass is about 26.6 yards long.

(6, −3)

≈ 26.6 years

Using the Distance Formula, the lengths of the three sides are as follows. d1 =

957

conclude by the Pythagorean Theorem that the triangle must be a right triangle.

So, you would estimate the 2012 revenue for Google to have been about $48.85 billion.

958

Solutions to Checkpoints

8. The radius of the circle is the distance between

9. To shift the vertices two units to the left, subtract 2 from each of the x-coordinates. To shift the vertices four units down, subtract 4 from each of the y-coordinates.

(1, − 2) and (− 3, − 5). r =

( x − h) + ( y − k ) 2

1 − ( − 3) + − 2 − ( − 5)

42 + 32

16 + 9

Original point

Translated Point

(1, 4)

(1 − 2, 4 − 4) = (−1, 0)

(1, 0)

(1 − 2, 0 − 4) = ( −1, − 4)

(3, 2)

(3 − 2, 2 − 4) = (1, − 2)

(3, 6)

(3 − 2, 6 − 4) = (1, 2)

= 5

Using ( h, k ) = ( − 3, − 5) and r = 5, the equation of the

circle is

( x − h) 2 + ( y − k ) 2 = r 2 2

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

( x + 3) + ( y + 5) 2

(3, 6)

3 2

(1, 4)

(−1, 0) −4 −3 −2

= 25.

−1

(3, 2)

(1, −2)

(1, 0) 1

(1, 2)

(−1, −4)

−5

Checkpoints for Section P.6 1. (a) Begin by scanning the data to find the least and greatest numbers. For the data, the least number is 61 and the greatest is 95.

Next, draw a portion of a real number line that includes the interval [61, 95]. To create the line plots, start with the first number, 68, and record a • above 68 on the number line. Continue recording a • for each number in the list until you obtain the line plot.

100

Test Scores

(b) From the line plot, you can see that 82 occurs with the greatest frequency. (c) Because the range is the difference between the greatest and least data values, the range of scores is 95 − 61 = 34.

2. To begin constructing a frequency distribution, first decide on the number of intervals. There are several ways to group the data. However because the least number is 61 and the greatest is 95, it seems that eight intervals would be appropriate. By tallying the data into the eight intervals, you obtain the frequency distribution shown below.

Sample answer: Interval Tally

[60, 65) [65, 70) [70, 75) [75, 80) [80, 85) [85, 90) [90, 95) [95, 100)

|||| ||| ||||| |||| || ||||| | ||| || |

You can construct the histogram by drawing a vertical axis to represent the number of students who earned that score and a horizontal axis to represent the test scores. Then, for each interval, draw a vertical bar whose height is the total tally. Note that the break in the horizontal axis indicates that the numbers 0 through 59 have been omitted. Number of students

10 8 6 4 2 60 65 70 75 80 85 90 95 100

Test scores

Solutions to Checkpoints 3. To create a bar graph, begin by drawing a vertical axis to represent the number of stores and a horizontal axis to represent the year.

5. Begin by drawing a vertical axis to represent the number of stores. Then label the horizontal axis to represent the year and plot the points from the table in Example 3. Finally, connect the points with line segments.

12,000 10,000

6,000 4,000

2013

2008

Year

2012

2,000

2013

2012

2011

2010

2009

2,000

8,000

2011

4,000

10,000

2010

6,000

2009

Number of stores

12,000

8,000

2008

Number of stores

959

Year

From the graph, you can see that the number of stores has steadily increased from 2008 through 2013. 4. For the data, a horizontal bar graph seems appropriate. 2013 2011 2010

10,000

8,000

6,000

4,000

2,000

14,000

Dollar General Walmart

2009 2008

12,000

Year

2012

Number of stores

Chapter 1 Checkpoints for Section 1.1 1. (a)

y = 14 − 6 x ?

− 5 = 14 − 6(3)

Write original equation.

(b)

y = 14 − 6 x

Write original equation.

26 = 14 − 6( − 2)

Substitute 3 for x and − 5 for y.

Substitute − 2 for x and 26 for y.

− 5 = 14 − 18 −5 ≠ − 4

26 = 14 + 12 26 = 26

(3, − 5) is not a solution. 

(− 2, 26) is a solution. 

2. To graph y − 2 x = 1  y = 2 x + 1 construct a table of values that consists of several solution points. Then plot the points and connect them. y

y = 2x + 1

( x, y )

−2

y = 2( − 2 ) + 1 = − 3

( − 2, − 3)

−1

y = 2( −1) + 1 = −1

(−1, −1)

y = 2( 0 ) + 1 = 1

(0, 1)

y = 2(1) + 1 = 3

(1, 3)

y = 2( 2 ) + 1 = 5

(2, 5)

10 6 2 −3

−2

−4 −6

960

Solutions to Checkpoints

3. To graph y = 1 − x 2 , construct a table of values that consists of several solution points. Then plot the points and connect them with a smooth curve. y

( x, y )

y = 1 − x2

−2

y = 1 − ( − 2) = − 3

(− 2, − 3)

−1

y = 1 − ( −1) = 0

(−1, 0)

y = 1 − ( 0) = 1

(0, 1)

y = 1 − (1) = 0

(1, 0)

y = 1 − ( 2) = − 3

(2, − 3)

4. To graph y = ( x − 1) − 3, enter the equation in a 2

graphing utility. Then use a standard viewing window to obtain the graph.

4 3 2

−4 −3 −2

−1

−2 −3 −4

5. To graph x 2 + y 2 = 4, first solve the equation for y.

x2 + y2 = 4 y 2 = 4 − x2

4 − x2

y = ± −6

Enter y1 =

4 − x 2 and y2 = −

4 − x 2 in a

graphing utility. Because the graph is a circle, use a square viewing window to obtain the graph.

−4

−6

−4

6. (a) Algebraic Solution

(b) Algebraic Solution

To begin, find how far the runner can run in 3.1 hours.

To find how long it will take to run a 26.2-mile marathon, solve the equation below. d = rt

d = rt = (5.2)(3.1)

(26.2) = (5.2)t

≈ 16.1

26.2 5.2

So, the runner can run about 16.1 miles in 3.1 hours. Graphical Solution

Use a graphing utility to graph d = 5.2t  y = 5.2 x. 20

= t

5.0 ≈ t So, it will take the runner about 5 hours to run 26.2 miles. Graphical Solution

Adjust the viewing window from part (a) so it shows the graph at y = 26.2. 30

Use the value feature to estimate that when x = 3.1, the distance is y ≈ 16.12 miles.

Use the zoom and trace features to estimate that when y = 26.2, the time is x ≈ 5.04 hours.

Solutions to Checkpoints

961

7. (a) Numerical Solution

Enter y = 2075 + 0.08 x in a graphing utility. Then use the table feature of the graphing utility to create a table. Start the table at x = 1400 with a step of 10.

From the table, when x = 1480, the wages are y = $2193.40. (b) Numerical Solution Start the table from part (a) at x = 1750 with a step of 25.

From the table, wages of $2225, result in sales of $1875.

Checkpoints for Section 1.2 1. (a) m =

8 − ( − 6) y2 − y1 14 = = = 2 x2 − x1 2 − ( − 5) 7

(b) m =

y2 − y1 5−2 3 3 = = = − −2 2−4 2 x2 − x1

−1 − ( −1) y2 − y1 0 = = = 0 3−0 3 x2 − x1

y − y1 = m( x − x1 ) y − ( − 7) = 2( x − 3) y + 7 = 2x − 6 y = 2 x − 13

3. Let x = 12 represent 2012. Let (12, 156.508) and (13, 170.910) be two points on the line representing the sales. The slope of this line is

170.910 − 156.508 13 − 12 = 14.402.

m =

Next use the point-slope form to find the equation of the line. y − y1 = m( x − x1 ) y − 170.910 = 14.402( x − 13) y − 170.910 = 14.402 x − 187.226 y = 14.402 x − 16.316 Now, using this equation, you can predict the 2014 sales ( x = 14) to be y = 14.402(14) − 16.316 = $185.312 billion.

4. Begin by writing the equation in slope-intercept form.

x − 2y = 4 −2y = − x + 4 y = 12 x − 2 From the slope-intercept form of the equation, the slope is 12 and the y-intercept is (0, − 2). Because the slope is positive, you know that the graph of the equation is a line that rises one unit for every two units it moves to the right.

962

Solutions to Checkpoints

5. (a)

7. Begin by writing the equation of the line in slope-intercept form.

5x − 3 y = 8 −5

− 3 y = −5x + 8

y = 53 x − 83

−1

(b)

1 −2

Therefore, the given line has slope m = 53. Any line perpendicular to the given line will have a slope of − 53. So, the line passing through the point ( − 4, 1) has the

−4

(c)

following equation. y − y1 = m( x − x1 )

3 −5

y − 1 = − 53 ( x − ( − 4)) y − 1 = − 53 ( x + 4)

−7

The first graph does not show both intercepts. The third graph is best because it shows both intercepts and gives the most accurate view of the slope by using a square setting. 6. Begin by writing the equation of the line in slope-intercept form.

y − 1 = − 53 x − 12 5 y = − 53 x − 57 8. The lines y = 12 x and y = − 2 x are perpendicular. 10

y = 2x

5x − 3 y = 8 −3 y = − 5 x + 8

y =

−15

5x − 8 3 3

Therefore, the given line has slope m = 53. Any line

y = 1x 2

−10

y = −2x

parallel to the given line must also have a slope of 53. So, the line through ( − 4, 1) has the following equation. y − y1 = m( x − x1 ) y − 1 = 53 ( x − ( − 4)) y − 1 = 53 ( x + 4) y − 1 = 53 x + 20 3 y = 53 x + 23 3

Solutions to Checkpoints

963

Checkpoints for Section 1.3 1. The table does describe y as a function of x. Each input value is matched with exactly one output value.

4. Because x = − 2 is less than 0, use f ( x) = x 2 + 1

to obtain f ( − 2) = ( − 2) + 1 = 4 + 1 = 5. 2

2. (a) Solving for y yields

x2 + y 2 = 8 y 2 = 8 − x2 y = ±

Because x = 2 is greater than or equal to 0, use f ( x ) = x − 1 to obtain f ( 2) = 2 − 1 = 1.

Write original equation. Subtract x 2 from each side.

Because x = 3 is greater than or equal to 0, use f ( x ) = x − 1 to obtain f (3) = 3 − 1 = 2.

8 − x 2 . Solve for y.

The ± indicates that for a given value of x there correspond two values of y. So, y is not a function of x.

5. (a) The domain of f consists of all first coordinates in the set of ordered pairs. Domain = {− 2, −1, 0, 1, 2}

(b) Solving for y yields 2

y − 4 x = 36

Write original equation. 2

y = 36 + 4 x .

Add 4 x to each side.

Each value of x corresponds to exactly one value of y. So, y is a function of x. 3. (a) Replacing x with 2 in f ( x) = 10 − 3x 2 yields the

following.

f ( 2) = 10 − 3( 2)

(b) Excluding x-values that yield zero in the denominator, the domain of g is the set of all real numbers x except x = 3.

6. (a) Because the function represents the circumference of a circle, the values of the radius r must be positive. So, the domain is the set of real numbers r such that r > 0.

(b) This function is defined only for x-values for which x − 16 ≥ 0. You can conclude that x ≥ 16. So, the domain of h is all real numbers that are greater than or equal to 16.

= 10 − 12 = −2 (b) Replacing x with −4 yields the following.

f ( − 4) = 10 − 3( − 4)

= 10 − 48

−6

= − 38 (c) Replacing x with x − 1 yields the following. f ( x − 1) = 10 − 3( x − 1)

= 10 − 3( x 2 − 2 x + 1) = 10 − 3 x 2 + 6 x − 3 = − 3x 2 + 6 x + 7

−4

Using the trace feature of a graphing utility, you can determine that the x-values extend from − 2 to 2 and the y-values extend from 0 to 2. So, the domain of the function f is all real numbers such that − 2 ≤ x ≤ 2 and the range of the function f is all real numbers such that 0 ≤ y ≤ 2.

8. For 2010; t = 10, so use N = − 4.64t + 76.2. N (10) = − 4.64(10) + 76.2 = − 46.4 + 76.2 = 29.8 thousand employees

For 2012; t = 12, so use N = 0.94t + 20.0. N (12) = 0.9(12) + 20.0 = 10.8 + 20.0 = 30.8 thousand employees

Solutions to Checkpoints

964

9. Algebraic Solution

When x = 60, you can find the height of the baseball as follows. f ( x) = − 0.004 x 2 + 0.3x + 6

Write original function.

f (60) = − 0.004(60) + 0.3(60) + 6

Substitute 60 for x.

= 9.6

Simplify.

When x = 60, the height of the ball thrown by the second baseman is 9.6 feet. So, the first baseman cannot catch the baseball without jumping. Graphical Solution 15

When x = 60, the height of the ball thrown by the second baseman is 9.6 feet. So, the first baseman cannot catch the ball without jumping. ( x + h)2 + 2( x + h) − 3 − ( x 2 + 2 x − 3) f ( x + h) − f ( x)   10. = h h x 2 + 2 xh + h 2 + 2 x + 2h − 3 − x 2 − 2 x + 3 = h 2 xh + h 2 + 2h = h 2( 2 x + h + 2) = h = 2 x + h + 2, h ≠ 0

Checkpoints for Section 1.4 1. (a) The open dot at ( − 3, − 6) indicates that x = − 3 is not

in the domain of f. So, the domain of f is all real numbers, except x = − 3, or ( −∞, − 3) ∪ ( − 3, ∞ ).

3. This is not a graph of y as a function of x because you can find a vertical line that intersects the graph twice. 4.

(b) Because (0, 3) is a point on the graph of f, it follows

(− 2, 3)

that f (0) = 3. Similarly, because the point (3, − 6) is a point on the graph of f, it follows that f (3) = − 6. (c) Because the graph of f does not extend above f (0) = 3, the range of f is the interval ( −∞, 3]. 2. Because the expression under a radical cannot be negative, the domain of f ( x) = x − 1 is the set of all

real numbers such that x − 1 ≥ 0. Solve this inequality for x, as follows.

3 2 1

−4

−2

−1

(0, − 1)

−2

This function is increasing on the interval ( − ∞, − 2), decreasing on the interval ( − 2, 0), and increasing on the interval (0, ∞ ).

x −1 ≥ 0 x ≥1 So, the domain is the set of all real numbers greater than or equal to 1. Because the value of the radical expression is never negative, the range of f ( x) = x − 1 is the set of all real nonnegative numbers, y ≥ 0 or f ( x ) ≥ 0.

Solutions to Checkpoints 5.

965

8. This piecewise-defined function consists of two linear functions. At x = − 4 and to the left of x = − 4, the

−4

graph is the line y = − 12 x − 6, and to the right of

x = − 4, the graph is the line y = x + 5. The point

( − 4, − 2) is a solid dot and ( − 4, 1) is an open dot. This is because f ( − 4) = − 2.

−5

By using the zoom and the trace features or the maximum feature of a graphing utility, you can estimate that the function has a relative maximum at the point ( − 0.875, 6.0625). 6.

y 6

2 −8

−6

−4

−2

−4 −6

−8

By using the minimum and maximum features of a graphing utility, you can estimate that the function has a relative minimum at (1, − 7) and a relative maximum

2 −6

at ( − 2, 20). 7. Using the minimum feature of a graphing utility, you can determine that the minimum temperature during the 24-hour period was approximately 34°F, which

occurred at about 1:48 A.M. ( x ≈ 19.8).

−6

The graph is symmetric with respect to the y-axis. So, the function is even.

10. (a) The function f ( x ) = 5 − 3 x is neither odd nor even because f ( − x ) ≠ − f ( x) and f ( − x ) ≠ f ( x ) as follows.

f ( − x) = 5 − 3( − x) = 5 + 3 x ≠ − f ( x) ≠ f ( x) So, the graph of f is neither symmetric with respect to the origin nor with respect to the y-axis. (b) The function g ( x) = x 4 − x 2 − 1 is even because g ( − x ) = g ( x ). g ( − x) = (− x) − (− x) − 1 4

= x4 − x2 − 1 = g ( x)

So, the graph of g is symmetric with respect to the y-axis. (c) The function h( x) = 2 x3 + 3x is odd because h( − x ) = − h( x ), h( − x) = 2( − x) + 3( − x) 3

= − 2 x3 − 3x = −( 2 x 3 + 3 x ) = − h( x )

So, the graph of h is symmetric with respect to the origin.

966

Solutions to Checkpoints

Checkpoints for Section 1.5 1. (a) The graph of h( x) = x3 + 5 is an upward shift of

five units of the graph of f ( x) = x3.

5. (a) Relative to the graph of f ( x) = x 2 , the graph of

g ( x) = 4 x 2 = 4 f ( x) is a vertical stretch (each y-value is multiplied by 4) of the graph of f.

y 10

y g(x) = 4x 2

2 −3

−1

2 1

f(x) = x 2

(b) The graph of g ( x) = ( x − 3) + 2 is a right shift of 3

three units and an upward shift of two units of the graph of f ( x) = x3. y

−3

−2

−1

(b) Relative to the graph of f ( x) = x 2 , the graph of

h( x) = 14 x 2 = 14 f ( x) is a vertical shrink (each y-value is multiplied by 14 ) of the graph of f.

f(x) = x 2

h(x) = 1 x 2 4

6 2

4 3

−2

2. The graph of k is a vertical shift of one unit downward of the graph of f ( x) = x 2 . So, the equation for k is

k ( x) = x 2 − 1.

1 −4 −3 −2 −1 −1

6. (a) Relative to the graph of f ( x) = x 2 + 3, the graph

3. The graph of k is a horizontal shift of two units to the left followed by a reflection in the x-axis of the graph of f ( x) = x 2 . So, the equation for k is

of g ( x) = f ( 2 x) = ( 2 x) + 3 = 4 x 2 + 3 is a 2

horizontal shrink (c > 1) of the graph of f. y g(x) = 4x 2 + 3

k ( x) = − ( x + 2) . 2

6 5

4. Algebraic Solution

Relative to the graph of f ( x ) = x , the graph of g ( x ) = − x is a reflection of the graph of f in the x-axis.

f(x) = x 2 + 3

2 1

f ( x) = x −3

= −x

−2

−1

(b) Relative to the graph of f ( x) = x 2 + 3, the graph

= g ( x)

( ) ( 12 x) + 3 = 14 x2 + 3 is a 2

Graphical Solution

of h( x) = f 12 x =

From the graph, you can see that the graph of g is a reflection of the graph of f in the x-axis.

horizontal stretch (0 < c < 1) of the graph of f.

f(x) = 兩 x 兩

f (x) = x 2 + 3

6 5

−6

−4

−2

g(x) = −兩 x 兩

−4 −6

h(x) = 1 x 2 + 3 4

1 −4 − 3 − 2 − 1

Solutions to Checkpoints

967

Checkpoints for Section 1.6 1. The sum of f and g is

6. Algebraic Solution

( f + g )( x) = f ( x) + g ( x) = ( x 2 ) + (1 − x)

(a) The composition of f with g is as follows.

( f  g )( x) = f ( g ( x)) = f ( 4 x 2 + 1)

= x 2 − x + 1. When x = 2, the value of this sum is

= 2( 4 x 2 + 1) + 5

( f + g )(2) = ( 2)2 − ( 2) + 1

= 8x2 + 2 + 5 = 8x2 + 7

= 3.

(b) The composition of g with f is as follows.

2. The difference of f and g is

( g  f )( x) = g ( f ( x)) = g ( 2 x + 5)

( f − g )( x) = f ( x) − g ( x) = ( x 2 ) − (1 − x)

= 4( 2 x + 5) + 1 2

= x 2 + x − 1. When x = 3, the value of the difference is

= 4( 4 x 2 + 20 x + 25) + 1

( f − g )(3) = (3)2 + (3) − 1

= 16 x 2 + 80 x + 100 + 1 = 16 x 2 + 80 x + 101

= 11.

3. The product of f and g is

( f  g )(− 12 ) = 8( − 12 ) + 7 2

( fg ) = f ( x) g ( x) = ( x 2 )(1 − x)

()

= 8 14 + 7 = 2 + 7

= x 2 − x3

= 9

= − x3 + x 2 .

Numerical Solution

When x = 3, the value of the product is

(a), (b) and (c)

( fg )(3) = −(3)3 − (3)2 = − 27 + 9 = −18.

Enter y1 = f ( x ), y2 = g ( x), and y3 = ( f  g )( x ), and y4 = ( g  f )( x).

4. The quotient of f and g is

f ( x) f =  ( x) = g ( x) g

x −3 16 − x 2

( )

Then use the table feature to find ( f  g ) − 12 .

The quotient of g and f is g ( x) g =  ( x ) = f ( x) f

16 − x 2 . x −3

The domain of f is [3, ∞ ) and the domain of g is [− 4, 4]. The intersection of these two domains is [3, 4]. So, the domain of f g is [3, 4) and the domain of g f is (3, 4]. 5. The composition of f ( x) = x 2 with g ( x ) = x − 1

is ( f  g )( x) = f ( g ( x)) = f ( x − 1) = ( x − 1) . 2

Then using ( f  g )( x) = ( x − 1) , you can find 2

( f  g )(0) = (0 − 1)2 = 1.

968

Solutions to Checkpoints

7. Algebraic Solution

8. (a)

The composition of f with g is as follows.

( f  g )( x) = f ( g ( x)) = f ( x6 )

( f  g )( x) = f ( g ( x))

= ( x6 )

= f ( x 2 + 4)

= x2

x2 + 4

The domain of f is [0, ∞) and the domain of g is the set

(b) ( g  f )( x) = g ( f ( x))

of all real numbers.

= g ( x1 3 )

The range of g is [4, ∞ ), which is in the range of f,

= ( x6 )

[0, ∞). Therefore, the domain of f  g is all real numbers.

9. Let the inner function be g ( x) = 8 − x and the outer

Graphical Solution 7

= x2

function be f ( x) =

x 2+ 4

6 −1

The x-coordinates of the points on the graph appear to be all real numbers. So, the domain of f  g is ( − ∞ , ∞ ). 10. (a)

x . 5

8− x 5 = f (8 − x)

h( x ) = −6

= f ( g ( x ))

( N  T )(t ) = N (T (t )) = 8( 2t + 2) − 14( 2t + 2) + 200 2

= 8( 4t 2 + 8t + 4) − 28t − 28 + 200 = 32t 2 + 64t + 32 − 28t − 28 + 200 = 32t 2 + 36t + 204 The composite function ( N  T )(t ) represents the number of bacteria in the food as a function of the amount of time the food has been out of refrigeration. (b) To find the time when the bacteria count reaches 1000, set N (T (t )) = 1000 and solve for t. N (T (t )) = 1000 32t 2 + 36t + 204 = 1000 32t 2 + 36t − 796 = 0

Using the Quadratic Formula, you can approximate the time to be t ≈ 4.46 hours.

Checkpoints for Section 1.7 1. The function f multiplies each input by 15. To “undo” this function, you need to multiply each input by 5.

So, the inverse function of f ( x) = 15 x is f −1( x) = 5x.

(

)

To verify this, show that f f −1 ( x) = x and f −1 ( f ( x)) = x.

f ( f −1 ( x)) = f (5 x) = 15 (5 x) = x

( )

f −1( f ( x)) = f −1 15 x = 5 15 x = x So, the inverse function of f ( x) = 15 x is f −1( x) = 5x.

Solutions to Checkpoints

969

2. The function f adds 7 to each input. To “undo” this function, you need to subtract 7 from each input. So, the inverse function of f ( x ) = x + 7 is f −1 ( x) = x − 7.

(

)

To verify this, show that f f −1 ( x) = x and f −1 ( f ( x)) = x. f ( f −1 ( x )) = f ( x − 7) = ( x − 7) + 7 = x f −1 ( f ( x )) = f −1 ( x + 7) = ( x + 7) − 7 = x

So, the inverse function of f ( x ) = x + 7 is f −1 ( x) = x − 7. 3. By forming the composition of f ( x) = x5 with g ( x) = 5 x , you have f ( g ( x)) = f

( x ) = ( x ) = x. 5

( )

5 By forming the composition of g ( x) = 5 x with f ( x) = x5 , you have g ( f ( x)) = g x5 = x5 = x.

Because f ( g ( x)) = g ( f ( x)) = x, f ( x) = x5 and g ( x) = 5 x are inverse functions of each other. 4. By forming the composition of f ( x ) =

f ( g ( x)) = f (7 x + 4) =

x − 4 with g ( x ) = 7 x + 4, you have 7

(7 x + 4) − 4 = 7 x = x. 7

Then by forming the composition of g ( x ) = 7 x + 4 and f ( x ) =

x − 4 , you have 7

 x − 4  x − 4 g ( f ( x )) = g   = 7  + 4 = x − 4 + 4 = x.  7   7 

Because f ( g ( x )) = g ( f ( x )) = x, f ( x ) = By forming the composition of f ( x) =

x − 4 and g ( x ) = 7 x + 4 are inverse functions of each other. 7

x − 4 7 with h( x ) = , you have 7 x − 4

 7   − 4 − 4 x + 23  7   x − 4 f ( g ( x)) = f  = = ≠ x.  7 7( x − 4)  x − 4 Because this composition is not equal to the identity function x, it follows that h( x ) = inverse function of f ( x) = 5.

x − 4 . 7

6. You can verify that f ( x) = x5 and g ( x) = 5 x are

f(x) = x5

−6

7 is not the x − 4

g(x) = 5 x −4

From the graph, it appears that f ( x) = x5 and

inverse functions of each other numerically by using a graphing utility. Enter y1 = x5 , y2 = 5 x , y3 = y1 ( y2 ), and y4 = y2 ( y1 ). Then use the table feature to create a table.

g ( x) = 5 x are inverse functions of each other.

Note that the entries for x, y3 , and y4 are the same. So, it appears that f ( g ( x)) = g ( f ( x)) = x and you can conclude that f ( x) = x5 and g ( x) = 5 x are inverse functions of each other. © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

970

Solutions to Checkpoints 8. Because f ( −1) = −1 = 1 and f (1) = 1 = 1, you

7. Algebraic Solution

Let a and b be real numbers with f ( a) = f (b). 3

a = 3 b

( a) = ( b) 3

have two inputs matched with the same output. So, f is not one-to-one and does not have an inverse function. You can confirm this graphically by noticing that the horizontal line y = 1 intersects the graph of f ( x ) = x

twice.

a = b

So, f ( a ) = f (b) implies that a = b. You can

y = 兩 x兩

conclude that f is one-to-one and does have an inverse function.

−6

Graphical Solution

−1

From the graph of f ( x) =

x , you can see that a 9. The graph of f ( x ) = 2 x − 3 passes the Horizontal

horizontal line will intersect the graph at most once. Therefore, you can conclude that f is one-to-one and does have an inverse function. 3

Line Test. So, you know that f is one-to-one and has an inverse function. f ( x) = 2 x − 3

y= 3 x

y = 2x − 3 −6

x = 2y − 3 x +3 2 x +3 f −1 ( x) = 2 y =

−3

1 −3

f (x) = 2 x − 3 −4

10. The graph of f ( x) = 3 x + 10 passes the Horizontal Line Test. So, you know that f is one-to-one

and has an inverse function. f ( x ) = 3 x + 10

f(x) = 3 10 + x

y = 3 x + 10 x = 3 y + 10

−5

x 3 = y + 10

− 10

y = x 3 − 10 f −1 ( x ) = x 3 − 10

f −1(x) = x 3 − 10

The graphs of f and f −1 are reflections of each other in the line y = x. So, the inverse function of

f ( x) = 3 x + 10 is f −1( x) = x3 − 10.

(

)

To verify, check that f f −1 ( x) = x and f −1 ( f ( x)) = x.

f ( f −1 ( x)) = f ( x3 − 10)

(

f −1 ( f ( x)) = f −1 3 10 + x

= 3 10 + ( x3 − 10)

= 3 x3

= 10 + x − 10

= x

)

( 10 + x ) − 10 3

Solutions to Checkpoints

971

11. The graph of f ( x) = ( x − 1) , x ≥ 1, passes the Horizontal Line Test. So, you know that 2

f is one-to-one and has an inverse function. f ( x) = ( x − 1)

y = ( x − 1)

x = ( y − 1)

y 5 4

−1

x +1

( x) =

x +1

x+1

y=x

3 2

x = y −1 y =

y = (x − 1)2, x ≥ 1 −1

−1

The graphs of f and f −1 are reflections of each other in the line y = x. So, the inverse function of f ( x) = ( x − 1) is f −1 ( x ) = 2

(

x + 1.

)

To verify, check that f f −1( x) = x and f −1 ( f ( x )) = x.

( x + 1) = ( x + 1) − 1   = ( x)

f ( f −1 ( x)) = f

f −1 ( f ( x )) = f −1 ( x − 1) 2

( x − 1)

= x −1+1

= x

Chapter 2 Checkpoints for Section 2.1 1.

4x 1 5 − = x + 9 3 3 5  4x  1 (9)  − (9)  = (9) x + 9   9   3  3 4 x − 3 = 9 x + 15 − 5 x = 18 18 x = − 5

Write original equation. Multiply each term by the LCD. Simplify. Combine like terms. Divide each side by − 5.

3x 12 = 5+ x −4 x −4 3x   12  ( x − 4)  = ( x − 4)5 + ( x − 4)   x − 4  x − 4 3x = 5 x − 20 + 12, x ≠ 4 − 2 x = −8 x = 4

Write original equation. Multiply each term by LCD. Simplify. Divide each side by − 2. Extraneous solution

In the original equation, x = 4 yields a denominator of zero. So, x = 4 is an extraneous solution, and the original equation has no solution.

972

Solutions to Checkpoints

3. Draw a diagram, 3w w

Verbal Model:

2 ⋅ Length + 2 ⋅ Width = Perimeter

Labels:

Perimeter = 112 Width = w

(feet) (feet) (feet)

Length = l = 3w

Equation:

2(3w) + 2( w) = 112 6w + 2w = 112

Write equation. Simplify.

8w = 12

Combine like terms.

w = 14

Divide each side by 8.

Because the length is three times the width, l = 3w = 3(14) = 42.

So, the dimensions of the family room are 14 feet by 42 feet. 4. To solve this problem, use the result from geometry that the ratios of the corresponding sides of similar triangles are equal.

x ft

4 ft

1.8 ft

55 ft Not drawn to scale

Verbal Model: Labels:

Equation:

Height of building Height of post = Length of building’s shadow Length of posts, shadow Height of building = x Length of building’s shadow = 55 Height of post = 4 Length of post ’s shadow = 1.8

(feet) (feet) (feet) (feet)

x 4 = 55 1.8 4 x = ⋅ 55 1.8 x ≈ 122.2

So, the building is about 122.2 feet high.

Solutions to Checkpoints

(dollars) Inventory of 46-inch sets = 50,000 − x (dollars) Profit from 32-inch sets = 0.22 x (dollars) Profit from 46-inch sets = 0.40(50,000 − x) (dollars) Total profit = 0.35(50,000) = 17,500 (dollars)

5. Inventory of 32-inch sets = x

The equation is as follows.

0.22 x + 0.40(50,000 x) = 17,500 0.22 x + 20,000 − 0.40 x = 17,500 − 0.18 x = − 2500 x ≈ 13,888.89

973

7. First solve for t in the formula for distance.

Then use d = 5 and r = 6 to find the time.

d = rt d = t r 5 = t 6

Formula for distance. Divide each side by r. Substitute d = 5 and r = 6.

5   60 min  So, it will take  hour  ⋅   = 50 minutes to 6    1 hour  jog 5 miles.

So, $13,888.89 was invested in 32-inch sets and 50,000 − 13,888.89 = $36,111.11, was invested in 46-inch sets. 6. The formula for the volume of a cylindrical container is V = π r 2h. To find the height of the container, solve for h. h =

π r2

Then, using V = 84 and r = 3, find the height. h =

π (3)

84 9π h ≈ 2.97 h =

So, the height of the container is about 2.97 inches. You can use unit analysis to check that your answer is reasonable.

84 in.3 84 in. ⋅ in. ⋅ in. 84 = = in. ≈ 2.97 in. 9π in.2 9π 9π in. ⋅ in.

Checkpoints for Section 2.2 1. To find the x-intercept, let y = 0, and solve for x. This produces

0 = 2x − 3 3 = 2x 3 2

= x

which implies that the graph has one x-intercept at

( 32 , 0).

To find the y-intercept, let x = 0, and solve for y. This produces y = 2(0) − 3 y = −3 which implies that the graph has one y-intercept at (0, − 3).

2. Algebraic Solution

To verify that 4 is a zero of f, check that f ( 4) = 0. f ( x) = 24 − 6 x

Write original function.

f ( 4) = 24 − 24

Simplify.

f ( 4) = 24 − 6( 4) Substitute 4 for x. f ( 4) = 0

4 is a zero. 

Graphical Solution

Use a graphing utility to graph y = 24 − 6 x. From the figure, it appears that the graph of f has an x-intercept when x = 4. So, 4 is a zero of f. Use the zero or root feature of the graphing utility to verify this zero. 2

−1

(4, 0)

−2

974

Solutions to Checkpoints

3. Graph the function y = − x 3 + x + 3. You can see

from the graph that there is one x-intercept. It lies between 1 and 2 and is approximately 1.5. 5

5. Algebraic Solution

To begin, solve each equation for y to obtain y = − 14 x + 12 and y = x. Next, set the two expressions for y equal to each other and solve the resulting equation for x, as follows.

y = −x3 + x + 3 −6

− 14 x + 12 = x

− 54 x = − 12 x = 52

−1

By using the zero or root feature of a graphing utility, you can improve the approximation. Choose a left bound of x = 1 and a right bound of x = 2. To four-decimalplace accuracy, the solution is x ≈ 1.617. Check this approximation on your calculator. You will find that the value of y is y = − (1.6717) + (1.6717) + 3 ≈ 0. 3

4. In general form, this equation is 3x − 3x − 3 = 0. So, you can begin by graphing y = 3 x 4 − 3 x3 − 3 = 0. 4

Subtract x and 12 from each side. Multiply each side by − 54 .

When x = 52 , the y-value of each of the original equations is 52 . So, the point of intersection is

( 52 , 52 ).

Graphical Solution

To begin, solve each equation for y to obtain y1 = − 14 x + 12 and y2 = x. Then use a graphing utility to graph both equations in the same viewing window, as shown. Use the intersect feature to approximate the point of intersection. y1 = − 14 x + 12

−6

Equate equations for y.

y2 = x

−6

−4

This graph has two x-intercepts, and by using the zoom and trace features you can approximate the corresponding solutions to be x ≈ − 0.82 and x ≈ 1.38.

−5

The point of intersection is

( 52 , 52 ).

6. Begin by using the graphing utility to graph both functions, as shown. From this display, you can see that the two graphs have only one point of intersection. Then, using the zoom and trace features, approximate the points of intersection to be (−1.871, 10.5), (0, 0) and (1.871, 10.5). 18

y2 = 2x4 − 4x2 y1 = 3x2

−3

3 −6

To test the reasonableness of these approximations, you can evaluate both functions at x = −1.871, x = 0, and x = 1.871. Quadratic Function

y = 3( −1.871) ≈ 10.50 2

y = 3(0) = 0 2

y = 3(1.871) ≈ 10.50 2

Quartic Function

y = 2( −1.871) − 4( −1.871) ≈ 10.51 4

y = 2(0) − 4(0) = 0 4

y = 2(1.871) − 4(1.871) ≈ 10.51 4

Because both functions yield approximately the same y-values, you can conclude that the approximate coordinates of the points of intersection are ( −1.871, 10.5), (0, 0), and (1.871, 10.5).

Solutions to Checkpoints

975

Checkpoints for Section 2.3 1. (a)

(7 + 3i) + (5 − 4i) = 7 + 3i + 5 − 4i = (7 + 5) + (3 − 4)i

Remove parentheses. Group like terms.

= 12 − i (b)

(−1 +

) (

Write in standard form.

)

−8 + 8 −

− 50 = −1 +

−8 + 8 −

− 50

Remove parentheses.

= −1 + 2 2i + 8 − 5 2i

(

Write in i -form.

)

= ( −1 + 8) + 2 2 − 5 2 i

Group like terms.

= 7 − 3 2i

Write in standard form.

Remove parentheses.

= ( − 3 + 3) + ( 2 − 4 + 3)i

Group like terms.

= i

Write in standard form.

(d) (5 − 3i ) + (3 + 5i ) − (8 + 2i ) = 5 − 3i + 3 + 5i − 8 − 2i

2. (a)

− 25 = (3i )(5i )

−9 ⋅

= 15i

Remove parentheses.

= (5 + 3 − 8) + ( − 3 + 5 − 2)i

Group like terms.

= 0 + 0i

Simplify.

= 0

Write in standard form.

Write each factor in i -form. Multiply.

= 15( −1)

i 2 = −1

= −15

Simplify.

(b) ( 2 − 4i )(3 + 3i ) = 2(3 + 3i ) − 4i(3 + 3i ) = 6 + 6i − 12i − 12i

Distributive Property Distributive Property

= 6 + 6i − 12i − 12( −1)

i 2 = −1

= (6 + 12) + (6 − 12)i

Group like terms.

= 18 − 6i

Write in standard form.

Square of a binomial

= 4( 4 + 2i ) + 2i( 4 + 2i ) = 16 + 8i + 8i + 4i

= 16 + 8i + 8i + 4( −1)

Distributive Property Distributive Property i 2 = −1

= (16 − 4) + (8i + 8i )

Group like terms.

= 12 + 16i

Write in standard form.

3. (a) The complex conjugate of 3 + 6i is 3 − 6i.

(3 + 6i)(3 − 6i) = (3) − (6i) 2

= 9 − 36i 2 = 9 − 36( −1) = 45 (b) The complex conjugate of 2 − 5i is 2 + 5i.

(2 − 5i)(2 + 5i) = (2) − (5i) 2

= 4 − 25i 2 = 4 − 25( −1) = 29

976

Solutions to Checkpoints

4. 2 + i = 2 + i ⋅ 2 + i 2 −i 2 −i 2 + i

Multiply numerator and denominator by complex conjugate of the denominator.

4 + 2i + 2i + i 2 4 − i2 4 − 1 + 4i = 4 − ( −1) =

Expand. i 2 = −1

3 + 4i 5 3 4 = + i 5 5 =

Simplify. Write in standard form.

Checkpoints for Section 2.4 1.

2 x 2 − 3x + 1 = 6

Write original equation.

2 x − 3x − 5 = 0

Write in general form.

(2 x − 5)( x + 1) = 0

Factor.

2 x − 5 = 0  x = 52

Set 1st factor equal to 0.

x + 1 = 0  x = −1

The solutions are x = −1 and x = Check:

Set 2nd factor equal to 0. 5. 2

x = −1 2 x 2 − 3x + 1 = 6 ?

2( −1) − 3( −1) + 1 = 6 2

2(1) + 3 + 1 = 6 6 = 6 5 2

x =

2 x 2 − 3x + 1 = 6

( ) − 3( 52 ) + 1 = 6 2( 25 − 15 +1= 6 4) 2

2 52

6 = 6

Solutions to Checkpoints 2. (a) 3x 2 = 36

Write original equation.

x = 12

Divide each side by 3.

x = ± 12

Extract square roots.

x = ±2 3 The solutions are x = ± 2 Check:

x = −2

(

x −1 = ± x =1±

10 10

x = 1−

(

1 − 

) = 36 ?

)

x =1+

(

1 + 

) = 36 ?

−6

Extract square roots.

3x − 2 = ±

Write in i -form.

3x = 2 ±

Add 2 to each side.

2 6 i ± 3 3

The solutions are x = x2 − 4 x − 1 = 0 x − 4x = 1

2 6 ± i. 3 3 Add 1 to each side.

x − 4 x + ( 2) = 1 + ( 2) 2

= 5

Add 22 to each side.

Simplify.

x − 2 = ±

x = 2±

The solutions are x = 2 ± Check: x = 2 −

Divide each side by 3.

Write original equation.

3x − 2 = ±

( x − 2)

10 = 10 

36 = 36 

)

= 10

10 − 1 = 10  2

( 10 ) = 10

3(12) = 36

x =

( x − 1)

3 x 2 = 36 2

10 = 10 

36 = 36 

(

= 10

(− 10 ) = 10

3(12) = 36

10 − 1 = 10 

x = 2

10.

( x − 1)

3 x 2 = 36 3 −2

Check:

(b) ( x − 1) = 10

The solutions are x = 1 ±

977

Extract square roots. 5

Add 2 to each side.

5 x2 − 4 x − 1 = 0

(2 −

) − 4(2 − 5 ) − 1 = 0 (4 − 4 5 + 5) − 8 + 4 5 − 1 = 0 5

4 + 5 −8 −1 = 0 0 = 0

x = 2+

5 also checks. 

978 4.

Solutions to Checkpoints 2 x2 − 4x + 1 = 0

Write original equation.

2x 2 − 4 x = −1

Subtract 1 from each side.

1 Divide each side by 2. 2 1 2 2 x 2 − 2 x + (1) = − + (1) Add 12 to each side. 2 1 2 Simplify. ( x − 1) = 2 x2 − 2 x = −

x −1 = ±

1 2

Extract square roots.

x −1 = ±

2 2

Rationalize the denominator.

x = 1±

2 2

Add 1 to each side.

The solutions are x = 1 ± Check: x = 1 −

2 . 2

2 2 2 x2 − 4 x + 1 = 0 2

  ? 2 2 21 −  − 41 −  + 1 = 0 2  2    ? 1  21 − 2 +  − 4 + 2 2 + 1 = 0 2  2− 2

2 +1− 4+ 2

2 +1= 0 ?

2+1− 4 +1= 0 0 = 0 x =1+ 5.

2 also checks.  2

3 x 2 − 10 x − 2 = 0

Original equation

3 x − 10 x = 2

Add 2 to each side.

10 2 x2 − x = 3 3

Divide each side by 3.

x2 −

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

5 Add   to each side.  3

5 31  x −  = 3 9  x −

Simplify.

5 31 = ± 3 3 x =

The solutions are x =

Extract square roots.

5 31 ± 3 3

5 ± 3

Add

5 to each side. 3

31 . 3

Solutions to Checkpoints 6. x 2 + 2 x + 17 = 0

979

Write original equation.

x 2 + 2 x = −17

Subtract 17 from each side.

x 2 + 2 x + 22 = −17 + 22

Add 22 to each side.

( x + 2)2 = −13

Simplify.

x + 2 = ± 13i

Extract square roots.

x = − 2 ± 13i

Subtract 2 from each side.

The solutions are x = − 2 ± 13i. 7. 3x 2 + 2 x − 10 = 0

x = x = x = x = x = x =

Write original equation.

−b ±

b 2 − 4ac 2a

Quadratic Formula

−2 ±

(2) − 4(3)(−10) 2(3)

Substitute a = 3, b = 2, and c = −10.

−2 ±

4 + 120 6

Simplify.

−2 ±

124

− 2 ± 2 31 6

(

2 −1 ±

Simplify.

)

Factor out common factor.

x =

−1 ± 31 3

The solutions are

−1 ± 31 . 3

Check: x =

Simplify.

−1 + 31 3 3 x 2 + 2 x − 10 = 0 2

?  −1 + 31   −1 + 31  3  + 2  − 10 = 0 3 3     ? 1 2 31 31 2 31 + + + − 10 = 0 3 3 3 3 ?

10 − 10 = 0 0 = 0

The solution x =

−1 − 31 also checks.  3

980

Solutions to Checkpoints

8. Algebraic Solution

This equation has a common factor of 3. You can simplify the equation by dividing each side of the equation by 3. 3x 2 − 6 x + 3 = 0

Write original equation.

x2 − 2x + 1 = 0

Divide each side by 3.

x = x = x =

b 2 − 4ac 2a

−b ±

− ( − 2) ± 2±

(− 2)2 − 4(1)(1) 2(1)

4−4 2

2± 0 2 x =1 x =

Quadratic Formula Substitute. Simplify. Simplify. Repeated solution

The equation has only one solution, x = 1. Check this solution in the original equation. Graphical Solution

Use a graphing utility to graph y = 3 x 2 − 6 x + 3. Use the zero feature of the graphing utility to approximate any values of x for which the function is equal to zero. From the figure, it appears that the function is equal to zero when x = 1. This is the only solution of the equation 3x 2 − 6 x + 3 = 0. 3

y = 3x2 − 6x + 3 −2

−2

9. Algebraic Solution

The equation x 2 − 2 x + 2 = 0 is in general form, so use the Quadratic Formula with a = 1, b = − 2, and c = 2. x2 − 2x + 2 = 0 x = x = x =

−b ±

b 2 − 4ac 2a

− ( − 2) ± 2± 2±

4−8 2

−4 2 2 ± 2i x = 2 x =1±i x =

(− 2)2 − 4(1)(2) 2(1)

Quadratic Formula Substitute. Simplify. Simplify. Simplify radical. Solutions.

The equation has no real solution, but it has two imaginary solutions, x = 1 ± i. Graphical Solution

Use a graphing utility to graph y = x 2 − 2 x + 2. Note in the figure that the graph of the function does not appear to have an x-intercept. From this, you can conclude that the equation x 2 − 2 x + 2 = 0 has no real solution.

y = x2 − 2x + 2 −3

5 −1

Solutions to Checkpoints

981

10. To set up a mathematical model for the height of the rock, use the position equation s = −16t 2 + v0t + s0 .

Because the object is dropped, the initial velocity is v0 = 0 feet per second. Because the initial height is s0 = 196 feet, you have s = −16t 2 + 196. To find the time the rock hits the ground, let the height s be zero and solve the equation for t. s = −16t 2 + 196 Write positive equation. 0 = −16t 2 + 196 Substitute 0 for height. 16t 2 = 196 t2 =

Add 162 to each side.

196 16

Divide each side by 16.

196 16 14 t = 4 t = 3.5 t =

Extract positive square root. Simplify. Simplify.

The rock will take 3.5 seconds to hit the ground. 11. To find the year in which the number of Internet users reached 188 million, you can solve the equation as follows. − 0.25t 2 + 16.9t − 50 = 188 − 0.25t 2 + 16.9t − 238 = 0 t =

− (16.9) ±

(16.9) − 4(− 0.25)(− 238) 2( − 0.25) 2

t =

−16.9 ± 47.61 − 0.5

t =

−16.9 ± 6.9  t = 20 or 47.6 − 0.5

The domain of the model is 5 ≤ t ≤ 24, so choose t = 20. Because t = 5 corresponds to 1995, it follows that t = 20 must correspond to 2010. So, the number of U.S. adults who use the Internet reached 188 million during 2010. 12. Using the Pythagorean Theorem, you have the following.

a 2 + b2 = c2 x + (3 x) = 102 2

Pythagorean Theorem 2

Subsitute for a, b, and c.

x + 9 x = 10,404 Simplify. 10 x 2 = 10,404 Combine like terms. x 2 = 1040.4 Divide each side by 10. x ≈ 32.26

Extract positive square root.

The total distance covered by walking on the L-shaped sidewalk is x + 3x = 4 x = 4 1040.4 ≈ 129.02 feet. Walking on the diagonal sidewalk saves a person about 129.02 − 102 = 27.02 feet.

982

Solutions to Checkpoints

Checkpoints for Section 2.5 1.

9 x 4 − 12 x 2 = 0

Write original equation.

3x (3 x − 4) = 0

Factor out common factor.

9 x 4 − 12 x 2 = 0

x = 0

Set 1st factor equal to 0.

3x − 4 = 0  3x = 4

Set 2nd factor equal to 0.

9(0) − 12(0) = 0

3x 2 = 0  2

4 x = 3

Check: x = 0 4

x = 4 3

3 3 9 x 4 − 12 x 2 = 0 4

2 3 2 3 ? 9  − 12  = 0 3    3   16   4 ? 9  − 12  = 0 9    3

±2 3 x = 3 The solutions are x = 0 and x = ±

0 = 0

x = ±

2 3 . 3

16 − 16 = 0 0 = 0

The solution x = − 2. (a)

x3 − 5 x 2 − 2 x + 10 = 0

Write original equation.

x ( x − 5) − 2( x − 5) = 0

Factor by grouping.

( x − 5)( x − 2) = 0

Distributive Property

x −5 = 0  x = 5

Set 1st factor equal to 0.

x −2 = 0  x = 2 2

= ± The solutions are x = − Check:

2 also checks. 

Set 2nd factor equal to 0. 2

2, and 5.

x = 5

x3 − 5 x 2 − 2 x + 10 = 0 ?

(5) − 5(5) − 2(5) + 10 = 0 3

125 − 125 − 10 + 10 = 0 0 = 0

x =

2 x3 − 5 x 2 − 2 x + 10 = 0

( 2 ) − 5( 2 ) − 2( 2 ) + 10 = 0 3

2 2 − 10 − 2 2 + 10 = 0 0 = 0 The solution x = −

2 also checks. 

Solutions to Checkpoints (b) 6 x3 − 27 x 2 − 54 x = 0

983

Write original equation.

3x( 2 x − 9 x − 18) = 0 2

Factor out common factor.

3 x( 2 x + 3)( x − 6) = 0

Factor quadratic factor.

3x = 0  x = 0

Set 1st factor equal to 0.

2 x + 3 = 0  x = − 32

Set 2nd factor equal to 0.

x −6 = 0  x = 6

Set 3rd factor equal to 0.

3 The solutions are x = − , 0, and 6. 2

Check:

x = 0

6 x3 − 27 x 2 − 54 x = 0 ?

6(0) − 27(0) − 54(0) = 0 3

0 = 0

x = − 32 6 x3 − 27 x 2 − 54 x = 0

( ) − 27(− 32 ) − 54(− 32 ) = 0 6( − 27 − 27( 94 ) + 27(3) = 0 8 )

6 − 32

− 81 − 243 + 81 = 0 4 4 0 = 0

x = 6 6 x3 − 27 x 2 − 54 x = 0 ?

6(6) − 27(6) − 54(6) = 0 3

6( 216) − 27(36) − 324 = 0 0 = 0

984

Solutions to Checkpoints

3. (a) This equation is of quadratic type with u = x 2 .

x 4 − 7 x 2 + 12 = 0

Write original equation.

( x2 ) − 7( x2 ) + 12 = 0 2

Quadratic form

u = x2

u − 7u + 12 = 0

(u − 3)(u − 4) = 0

Factor.

u −3 = 0  u = 3

Set 1st factor equal to 0.

u − 4 = 0  u = 4

Set 2nd factor equal to 0.

Replace u with x and solve for x in each equation. u = 3

u = 4

x2 = 3

x2 = 4

x = ±

x = ±2

The solutions are x = − x = −

Check:

3, − 2, and 2.

x = −2

x 4 − 7 x 2 + 12 = 0

(− 3) − 7(− 3) + 12 = 0 ?

(− 2) − 7( − 2) + 12 = 0

16 − 28 + 12 = 0

9 − 7(3) + 12 = 0

0 = 0

− 21 + 21 = 0 0 = 0

The solution x = 2 also checks. 

3 also checks. 

The solution x =

(b) This equation is of quadratic type with u = x 2 . 9 x 4 − 37 x 2 + 4 = 0

Write original equation.

9( x 2 ) − 37( x 2 ) + 4 = 0 2

Quadratic form

9u − 37u + 4 = 0

u = x2

(9u − 1)(u − 4) = 0

Factor.

9u − 1 = 0  u = 19

Set 1st factor equal to 0.

u − 4 = 0  u = 4

Set 2nd factor equal to 0.

Replace u with x 2 and solve for x in each equation.

u = 19

u = 4

x = 19 x = ± 13

x2 = 4

x = ±2

The solutions are − 13, 13 , − 2, and 2. Check:

x = − 13

x = −2

9 x 4 − 37 x 2 + 4 = 0

( ) − 37(− 13 ) + 4 = 0 1 − 37 1 + 4 = 0 9( 81 ) (9) ?

9( − 2) − 37( − 2) + 4 = 0

9(16) − 37( 4) + 4 = 0

9 − 13

1 − 37 + 4 = 0 9 9

−4 + 4 = 0 

? ?

144 − 148 + 4 = 0  The solution x = 2 also checks. 

Solutions to Checkpoints 4. −

40 − 9 x + 2 = x

Write original equation.

−

Isolated radical.

40 − 9 x = x − 2

(− 40 − 9 x ) = ( x − 2) 2

985

Square each side.

40 − 9 x = x 2 − 4 x + 4 2

0 = x + 5 x − 36 0 = ( x − 4)( x + 9)

Simplify. Write in general form. Factor.

x −4 = 0  x = 4

Set 1st factor equal to 0.

x + 9 = 0  x = −9

Set 2nd factor equal to 0.

Check: x = 4

− −

40 − 9 x + 2 = x ?

40 − 9( 4) + 2 = 4 ?

−

4 + 2 = 4 ?

−2 + 2 = 4 0 ≠ 4

x = 4 is an extraneous solution. x = −9 −

40 − 9( − 9) + 2 = − 9 ?

− 121 + 2 = − 9 ?

−11 + 2 = − 9 ?

−9 = −9 

The only solution is x = − 9. 5. 3

( x − 5)2 3 = 16

Write original equation.

( x − 5) = 16 2 ( x − 5) = 4096

Rewrite in radical form.

Cube each side.

x − 5 = ± 64

Extract square roots.

x = 5 ± 64

Add 5 to each side.

x = − 59, x = 69

x = 69

Check: x = − 59

( x − 5)2 3 = 16 ?

(− 59 − 5)2 3 = 16 (− 64)

= 16 ?

(− 4)2 = 16 16 = 16 

( x − 5)2 3 = 16 ?

(69 − 5)2 3 = 16 (64)

= 16 ?

(4)2 = 16 16 = 16 

The solutions are x = − 59 and 69.

986

Solutions to Checkpoints

6. The LCD of the three terms is x( x + 3). 4 2 + = −3 x x +3 4 2 = x( x + 3)( − 3) x( x + 3) + x( x + 3) x x +3 4( x + 3) + 2 x = − 3 x( x + 3), x ≠ − 3, 0

Write original equation. Multiply each term by LCD. Simplify.

6 x + 12 = − 3 x 2 − 9 x

Simplify.

3 x 2 + 15 x + 12 = 0

Write in general form.

3( x + 5 x + 4) = 0

Factor.

x + 5x + 4 = 0

Divide each side by 3.

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

Factor.

x + 1 = 0  x = −1

Set 1st factor equal to 0.

x + 4 = 0  x = −4

Set 2nd factor equal to 0.

Both x = − 4 and x = −1 are possible solutions. Check: x = − 4

x = −1

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

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

−1 + ( − 2) = − 3

− 4 + (1) = − 3

−3 = −3 

So, the solutions are x = − 4 and x = −1. 7. x 2 + 4 x = 5x + 12 First Equation x 2 + 4 x = 5 x + 12 2

x − x − 12 = 0

( x + 3)( x − 4) = 0

Use positive expression. Write in general form. Factor.

x + 3 = 0  x = −3

Set 1st factor equal to 0.

x − 4 = 0  x = 4

Set 2nd factor equal to 0.

Second Equation

− ( x 2 + 4 x) = 5 x + 12 2

− x − 4 x = 5 x + 12 2

0 = x + 9 x + 12

Use negative expression. Distributive Property Write in general form.

Use the Quadratic Formula to solve the equation 0 = x 2 + 9 x + 12. x = x = x =

−b ±

b 2 − 4ac 2a

−9 ±

92 − 4(1)(12) 2(1)

− 9 ± 33 2

The possible solutions are x = − 3, x = 4, and x =

− 9 ± 33 . 2

Solutions to Checkpoints

987

Check: x = − 3

x 2 + 4 x = 5 x + 12 ?

(− 3)2 + 4(− 3) = 5(− 3) + 12 ?

− 3 = −15 + 12 3 ≠ −3

x = 4 ?

(4) + 4(4) = 5(4) + 12 2

32 = 32 32 = 32 

x =

− 9 ± 33 2

(− 9 +

) + 4  − 9 + 2

33  ?  − 9 + 33   = 5  + 12 2 2   

 

21 5 33 ? − 45 5 33 − + + 12 = 2 2 2 2 5

x =

− 21 5 33 33 21 − = +  2 2 2 2

− 9 ± 33 2

(− 9 −

33 2

) + 4 − 9 − 2

 

33  ?  = 2 

21 5 33 ? + = 2 2

 − 9 − 33  5  + 12 2   − 21 5 33 − 2 2

21 5 33 21 5 33 + ≠ − −  2 2 2 2 x = − 3 and x =

− 9 − 33 − 9 + 33 are extraneous solutions. So, the solutions are x = 4 and x = . 2 2

988

Solutions to Checkpoints

8. Verbal Model:

Labels:

Cost per student ⋅ Number of students

Cost of trip = 560 Original number of students = x

(dollars) (people)

New number of students = x + 8

(people)

560 Original cost per student = x 560 New cost per student = − 3.50 x

Equation:

= Cost of trip

(dollars per person) (dollars per person)

 560  − 3.50 ( x + 8) = 560  x  

Check:

 560 − 3.50 x   ( x + 8) = 560 x  

?  560  − 3.50 ( x + 8) = 560  x   ?  560  − 3.50 (32 + 8) = 560   32 

(560 − 3.50 x)( x + 8) = 560 x, x ≠ 0

(14)( 40) = 560

560 x + 4480 − 3.5 x 2 − 28 x = 560 x

560 = 560 

− 3.5 x − 28 x + 4480 = 0 x 2 + 8 x − 1280 = 0

( x − 32)( x + 40) = 0 x − 32 = 0  x = 32 x + 40 = 0  x = − 40 Choosing the positive value of x, the number of students in the original group is 32. 9. Formula:

Labels:

r  A = P 1 +  n 

Balance = A = 3544.06

(dollars) (dollars)

Principal = P = 2500

Time = t = 5

Equation:

Compounding periods per year = n = 12

(years) (compoundings per year)

Annual interest rate = r

(percent in decimal form)

r   3544.06 = 25001 +  12   3544.06 r   = 1 +  2500 12  

r   1.4176 ≈ 1 +  12   r 1 60 (1.4176) = 1 + 12 r 1.0058 ≈ 1 + 12 r 0.0058 ≈ 12 0.0696 ≈ r

(12)(5)

Check:

r  A = P1 +  n 

? 0.07   3544.06 = 25001 +  12  

(12)(5)

3544.06 = 3544.06 

0.07 ≈ r

The annual interest rate is about 0.07 or 7%.

Solutions to Checkpoints

989

Checkpoints for Section 2.6 1. 7 x − 3 ≤ 2 x + 7

Write original inequality.

5 x ≤ 10

Subtract 2 x and add 3 to each side.

x ≤ 2

Divide each side by 5.

The solution set is all real numbers less than or equal to 2. x 0

2. (a) Algebraic Solution 2 − 53 x > x − 6

Write original inequality.

6 − 5 x > 3x − 18

Multiply each side by 3.

− 8 x > − 24

Subtract 3x and subtract 6 from each side.

x < 3

Divide each side by − 8 reverse the inequality symbol.

The solution set is all real numbers that are less than 3. x 0

(b) Graphical Solution Use a graphing utility to graph y1 = 2 − 53 x and y2 = x − 6 in the same viewing window. Use the intersect feature to determine that the graphs intersect at (3, − 3). The graph of y1 lies above the graph of y 2 and to the left of their point of intersection, which implies that y1 > y 2 for all x < 3. 3

y1 = 2 − 53 x

−4

y2 = x − 6 −7

1 < 2 x + 7 < 11

Write original inequality.

1 − 7 < 2 x + 7 − 7 < 11 − 7 −6 < 2x < 4

Subtract 7 from each part. Simplify.

6 2x 4 < < 2 2 2 −3 < x < 2

−

Divide each part by 2. Simplify.

The solution set is all real numbers greater than − 3 and less than 2, which is denoted by ( − 3, 2). x

−4 −3 −2 −1

990

Solutions to Checkpoints

4. x − 20 ≤ 4

Write original inequality.

− 4 ≤ x − 20 ≤ 4

Write equivalent inequalities.

− 4 + 20 ≤ x − 20 + 20 ≤ 4 = 20

Add 20 to each part.

16 ≤ x ≤ 24

Simplify.

The solution set is all real numbers that are greater than or equal to 16 and less than or equal to 24, which is denoted by [16, 24]. x 12 14 16 18 20 22 24 26 28

5. To determine the intervals on which x 2 − 2 x − 3 is entirely negative and those on which it is entirely positive, factor the quadratic as x 2 − 2 x − 3 = ( x + 1)( x − 3). The key numbers occur at x = −1 and x = 3. So, the test intervals for the

quadratic are ( −∞, −1), ( −1, 3), and (3, ∞ ). In each test interval, choose a representative x-value and evaluate the polynomial, shown in the table. Interval

x-Value

Value of Polynomial

Sign of Polynomial

( − ∞, −1)

x = −2

(− 2) − 2(− 2) − 3 = 5

Positive

(−1, 3)

x = 0

(0) − 2(0) − 3 = − 3

Negative

(3, ∞)

x = 5

(5) − 2(5) − 3 = 12

Positive

The polynomial has negative values for every x in the interval ( −1, 3) and positive values for every x in the intervals

(−∞, −1) and (3, ∞ ). The graph of y = x 2 − 2 x − 3 confirms this result. 8

−4

y = x2 − 2x − 3 −8

6. (a) Algebraic Solution

2 x 2 + 3x < 5

Write original inequality.

2 x + 3x − 5 < 0

Write in general form.

(2 x + 5)( x − 1) < 0

Factor.

Key numbers: x = − 52 and x = 1

(

)(

)

Test intervals: −∞, − 52 , − 52 , 1 , (1, ∞) Test: Is ( 2 x + 5)( x − 1) < 0?

(

)

After testing the intervals, you can see that the polynomial 2 x 2 + 3x − 5 is negative on the open interval − 52 , 1 .

(

)

So, the solution set of the inequality is − 52 , 1 . (b) Graphical Solution First write the polynomial inequality 2 x 2 + 3x < 5 as 2 x 2 + 3x − 5 > 0. Then use a graphing utility to graph y = 2 x 2 + 3x − 5. You can see that the graph is below the x-axis when x is greater than − 52 and when x is less than 1.

(

)

So, the solution set is − 52 , 1 . 1 −7

( − 52 , 0 )

(1, 0)

−7

Solutions to Checkpoints 3x3 − x 2 − 12 x > − 4

7. 2

3 x − x 2 − 12 x + 4 > 0 x 2 (3 x − 1) − 4(3 x − 1) > 0

(3x − 1)( x − 4) > 0 (3x − 1)( x + 2)( x − 2) > 0 2

991

Write original inequality. Write in general form. Factor. Factor. Factor.

(

) ( 13, 2), and (2, ∞).

The key numbers are x = − 2, x = 13 , and x = 2, and the test intervals are ( −∞, − 2), − 2, 13 , Test interval

( − ∞ , − 2)

x-value

Polynomial value

Conclusion

x = −3

3( − 3) − ( − 3) − 12( − 3) + 4 = − 50

Negative

(− 2, 13 ) ( 13, 2)

x = 0

3(0) − (0) − 12(0) + 4 = 4

Positive

x =1

3(1) − (1) − 12(1) + 4 = − 6

Negative

(2, ∞ )

x = 3

3(3) − (3) − 12(3) + 4 = 40

Positive

(

)

From this, you can conclude that the inequality is satisfied on the open intervals − 2, 13 and ( 2, ∞).

(

)

So, the solution set is − 2, 13 ∪ ( 2, ∞). 1 3 −4 −3 −2 −1

x 1

8. (a) The solution set of x 2 + 6 x + 9 < 0 is empty. In other words, the quadratic x 2 + 6 x + 9 is not less than 0 for any value of x.

(b) The solution set of x 2 + 4 x + 4 ≤ 0 consists of the single real number − 2, because the quadratic x 2 + 4 x + 4 has only one key number, x = − 2, and it is the only value that satisfies the inequality. (c) The solution set of x 2 − 6 x + 9 > 0 consists of all real numbers except x = 3. In interval notation, the solution set can be written as ( − ∞, 3) ∪ (3, ∞ ). (d) The solution set of x 2 − 2 x + 1 ≥ 0 consists of the entire set of real numbers ( − ∞, ∞ ). In other words, the value of the quadratic x 2 − 2 x + 1 is non-negative for every real value of x.

992

Solutions to Checkpoints

4x − 1 > 3 x −6 4x − 1 −3 > 0 x −6 4 x − 1 − 3( x − 6) > 0 x −6 4 x − 1 − 3x + 18 > 0 x −6 x + 17 > 0 x −6

Write original inequality. Write in general form. Combine fractions with LCD. Simplify. Simplify.

Key numbers: x = −17, x = 6 Test intervals: ( − ∞, −17), ( −17, 6), and (6, ∞ ) Test: Is

x + 17 > 0? x −6

Test interval

x-value

Polynomial value

Conclusion

(− ∞, −17)

− 20

( 20) + 17 = 3 ( 20) − 6 14

Positive

(−17, 6)

(0) + 17 = − 17 6 (0) − 6

Negative

(6, ∞ )

(8) + 17 = 25 2 (8) − 6

Positive

After testing these intervals, you can see that the inequality is satisfied on the open intervals ( − ∞, −17) and (6, ∞ ). So, you can conclude that the solution set consists of all real numbers in the intervals ( − ∞, −17) ∪ (6, ∞ ). − 17 − 24

− 12

6 0

10. Algebraic Solution

Graphical Solution

Recall that the domain of an expression is the set of all x-values for which the expression is defined. Because 2

x − 7 x + 10 is defined only of x − 7 x + 10 is non-negative, the domain is given by x 2 − 7 x + 10 ≥ 0. 2

x 2 − 7 x + 10 ≥ 0

Write in general form.

( x − 2)( x − 5) ≥ 0

Factor.

Begin by sketching the graph of the equation y = x 2 − 7 x + 10. From the graph, you can determine that the x-values extend up to 2 (including 2) and from 5 and beyond (including 5). So, the domain of

x 2 − 7 x + 10 is ( − ∞, 2] ∪ [5, ∞ ).

the expression y 8

So, the inequality has two key numbers:

x = 2 and x = 5.

Key numbers: x = 2, x = 5 Test intervals: ( − ∞, 2), ( 2, 5), (5, ∞ )

−4 −2 −2

−4

Test: Is ( x − 2)( x − 5) ≥ 0? A test shows that the inequality is satisfied in the unbounded half-closed intervals ( − ∞, 2] or [5, ∞ ). So, the domain of the expression ( − ∞, 2] ∪ [5, ∞).

x 2 − 7 x + 10 is

Solutions to Checkpoints

993

11. The position of an object moving vertically can be modeled by the position equation s = −16t 2 + v0t + s0 , where s is the

height in feet and t is the time in seconds. In this case, s0 = 0 and v0 = 208. So, you need to solve the inequality −16t 2 + 208t > 640. Using a graphing utility, graph y1 = −16t 2 + 208t and y2 = 640, as shown. From the graph, you can

determine that −16t 2 + 208t > 640 for t between 5 and 8. You can verify this result algebraically. −16t 2 + 208t > 640 t 2 − 13t < − 40 t 2 − 13t + 40 < 0

(t − 8)(t − 5) < 0 The key numbers are t = 5 and t = 8. A test will verify that the height of the projectile will exceed 640 feet when 5 < t < 8; that is, during the time interval (5, 8) seconds. 700

y1 = −16t2 + 208t

y2 = 640 600

Checkpoints for Section 2.7 1. Begin by representing the data with a set of ordered pairs. Let t represent the year with t = 6 corresponding to 2006.

(6, 179.4), (7, 185.4), (8, 191.0), (9, 196.7), (10, 202.6), (11, 208.7), (12, 214.9), (13, 221.4) Then plot each point in a coordinate plane as shown. Median sales price (in thousands of dollars)

230 220

(b) The scatter plot relating the number of hours watching television and test scores has a negative correlation. This means that the more time a student spent watching television, the lower his or her test score tended to be. 3. Sample answer: Let t represent the year, with t = 6 corresponding to 2006. After plotting the data points, draw the line that you think best fits the data. Two points that lie on the line are

(6, 179.4) and (13, 221.4).

210 200

You can find the equation of the line through these two points. First find the slope.

190 180 170 6 7 8 9 10 11 12 13 14

m =

Year (0 ↔ 2006)

2. (a) Construct a scatter plot of the data with the horizontal axis representing the number of hours spent watching television and the vertical axis representing the score obtained on the test.

221.4 − 179.4 42 = = 6 13 − 6 7

Then using the point-slope form, you obtain S − 179.4 = 6(t − 6) S − 179.4 = 6t − 36 S = 6t + 143.4. S

100 90 80 70 60 50 40 30 20 10 2 4 6 8 10 12 14 16 18 20

TV hours

Median sales price (in thousands of dollars)

Test scores

230 220 210 200 190 180 170 6 7 8 9 10 11 12 13 14

Year (6 ↔ 2006)

994

Solutions to Checkpoints

4. Graphical Solution

(a) Use the linear regression feature of a graphing utility to approximate the model to be N = 0.7069t + 4.761.

(b) Graph the actual data and the model. From the graph, it appears that the model is not a good fit for the actual data. 15

Numerical Solution

(a) Using the linear regression feature of a graphing utility, you can find that a linear model for the data is

N = 0.7069t + 4.761. (b) You can see how well the model fits the data by comparing the actual values of N with the values of N given by the model, which are labeled N* in the table below. From the table, it appears that the model is not a good fit for the actual data. Year

2007

7.078

9.709

2008

8.924

10.416

2009

14.625

11.123

2010

14.825

11.830

2011

13.747

12.537

2012

12.506

13.244

2013

11.46

13.951

5. After entering the data into a graphing utility and using the linear regression feature of a graphing utility, you can find the least squares regression line for the data to be e = 399.9 g − 653.

Solutions to Checkpoints

995

Chapter 3 Checkpoints for Section 3.1 1. (a) With respect to the graph of f ( x) = x 2 , the graph of g ( x) = x 2 − 4 is obtained by a vertical shift four units downward. y

4 3 2 1 −5 −4 −3

−1

3 4 5

−2

−5 −6

(b) With respect to the graph of f ( x) = x 2 , the graph of h( x) = ( x − 3) + 1 is obtained by a horizontal shift three units to 2

the right and a vertical shift one unit upward. y

6 5 4 3 2 1 − 2 −1 −1

−2

2. Write the quadratic function in standard form by completing the square. The first step is to factor out any coefficient of x 2 that is not 1. f ( x) = 3x 2 − 6 x + 4 = 3( x 2 − 2 x ) + 4

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

From the standard form, you can see that the graph of f is a parabola that opens upward with vertex (1, 1).

3. f ( x) = x 2 − 4 x + 3 = ( x 2 − 4 x + 4 − 4) + 3 = ( x 2 − 4 x + 4) − 4 + 3 = ( x − 2) − 1 2

The graph of f is a parabola that opens upward with vertex ( 2, −1). The x-intercepts are determined as follows. x2 − 4 x + 3 = 0

( x − 3)( x − 1) = 0 x −3 = 0 x = 3 x −1 = 0 x =1 So, the x-intercepts are (3, 0) and (1, 0).

996

Solutions to Checkpoints

4. Because the vertex of the parabola is ( h, k ) = ( − 4, 11),

5. Algebraic Solution

the equation has the form

For this quadratic function,

f ( x ) = a( x − ( − 4)) + 11

f ( x) = ax 2 + bx + c = − 0.007 x 2 + x + 4

which implies that a = − 0.007 and b = 1. Because

= a( x + 4) + 11. 2

Because the parabola passes through the point ( − 6, 15), it follows that f ( − 6) = 15. So you obtain

f ( x) = a( x + 4) + 11 2

15 = a( − 6 + 4) + 11 2

15 = 4a + 11

b . So, 2a the baseball reaches its maximum height when it is b x = − 2a 1 = − 2( − 0.007)

a < 0, the function has a maximum at x = −

4 = 4a 1 = a.

At this distance, the maximum height is

The equation in standard form is

f (71.4) = − 0.007(71.4) + (71.4) + 4 ≈ 39.7 feet. 2

f ( x) = 1( x + 4) + 11 = ( x + 4) + 11. 2

1 ≈ 71.4 feet from home plate. 0.014

Graphical Solution

Use the maximum feature of a graphing utility to find the maximum height reached by the baseball. 50

200

The maximum height is y ≈ 39.7 feet at x ≈ 71.4 feet.

Checkpoints for Section 3.2 1. (a) With respect to the graph of f ( x) = x3 , the graph of g ( x) = − x3 + 2 is obtained by a reflection in the x-axis and a

vertical shift two units upward. y

5 4 3 1 −4 −3 −2 −1 −1

−2 −3

(b) With respect to the graph of f ( x) = x3 , the graph of h( x) = x3 − 5 is obtained by a vertical shift five units downward. y

1 −4 −3 −2 −1 −1

−2 −3 −4

−7

Solutions to Checkpoints

997

(c) With respect to the graph of f ( x) = x3 , the graph of k ( x) = ( x + 2) is obtained by a horizontal shift two units to the 3

left. y

1 −6 −5 −4

−2 −1 −1

−2 −3 −4

2. Because the degree is odd and the leading coefficient is positive, the graph of f ( x) = 2 x3 − 3x 2 + 5 falls to the

5. f ( x) = x 3 + 2 x 2 − 3x 0 = x3 + 2 x 2 − 3x

left and rises to the right.

0 = x( x 2 + 2 x − 3)

3. (a) Because the degree is even and the leading coefficient is negative, the graph of f ( x) = − x 4 + 2 x 2 + 6 falls to the left

0 = x( x + 3)( x − 1)

x = 0 x+3 = 0

and right.

x = −3

(b) Because the degree is odd and the leading coefficient is negative, the graph of f ( x) = − x5 + 3x 4 − x

x −1 = 0 x =1

rises to the left and falls to the right.

So, the real zeros are x = − 3, x = 0, and x = 1 and

4. Algebraic Solution

the corresponding x-intercepts are (0, 0), ( − 3, 0), and

f ( x) = x3 + x 2 − 6 x

(1, 0). Using the maximum and minimum features of a

0 = x3 + x 2 − 6 x

graphing utility, you can approximate the two relative extrema to be (0.54, − 0.88) and ( −1.87, 6.06).

0 = x( x 2 + x − 6) 0 = x( x + 3)( x − 2)

(−1.87, 6.06) 7

x = 0

−7

x +3 = 0

(−3, 0) (0, 0)

(1, 0)

−2 (0.54, −0.88)

x = −3 x − 2 = 0 6.

x = 2 So, the real zeros are x = − 3, x = 0, and x = 2.

f ( x) = x5 − 2 x 4 + x3 − 2 x 2 0 = x5 − 2 x 4 + x3 − 2 x 2

Graphical Solution

0 = x 2 ( x3 − 2 x 2 + x − 2)

The graph of f has three x-intercepts

0 = x 2  x 2 ( x − 2) + ( x − 2)

(− 3, 0), (0, 0), and ( 2, 0) as shown. 10

y = x 3 + x 2 − 6x

(− 3, 0) −6

0 = x 2 ( x − 2)( x 2 + 1) 0 = x 2 ( x − 2)( x 2 + 1) x2 = 0

(0, 0)

(2, 0)

−6

So, the real zeros of f are x = − 3, x = 0, and x = 2. Use the zero or root feature of a graphing utility to verify these zeros.

x = 0 x−2 = 0 x = 2 The factor x 2 + 1 has no real zeros, only imaginary zeros. So, the real zeros are x = 0 and x = 2.

998

Solutions to Checkpoints

7. Sample answer: The zeros x = − 2, x = − 1, x = 1,

and x = 2 correspond to the factors ( x + 2), ( x + 1),

( x − 1), and ( x − 2). So, a possible function is f ( x) = ( x + 2)( x + 1)( x − 1)( x − 2) = x 4 − 5 x 2 + 4. 8. 1. Apply the Leading Coefficient Test. Because the leading coefficient is negative and the degree is even, you know that the graph eventually falls to the left and right.

2. Find the Real Zeros of the Polynomial. By factoring 1 3 9 f ( x) = − x 4 + x3 − x 2 4 2 4 1 2 2 = − x ( x − 6 x + 9) 4 1 2 2 = − x ( x − 3) 4

9. Graphical Solution

From the graph, you can see that the graph of f ( x) = x3 − 4 x 2 + 1 crosses the x-axis three times, first between −1 and 0, again between 0 and 1, and finally between 3 and 4. So, you can conclude that the function has zeros in the intervals ( −1, 0), (0, 1), and (3, 4). 2 −3

− 10

Numerical Solution

From the table, you can see that f ( −1) and f (0) differ in

you can see that the real zeros of f are x = 0 (of even multiplicity 2) and x = 3 (of even multiplicity 2). So, the x-intercepts occur at (0, 0) and (3, 0). Add these points to your graph. 3. Plot a Few Additional Points. To sketch the graph by hand, find a few additional points, as shown in the table. Be sure to choose points between the zeros and to the left and right of the zeros. Then plot the points. x

−1

f(x)

−4

−1

−4

sign. So, you can conclude from the Intermediate Value Theorem that the function has a zero between −1 and 0. Similarly, f (0) and f (1) differ in sign, so the function has a zero between 0 and 1. Likewise, f (3) and f ( 4) differ in sign, so the function has a zero between 3 and 4. So, you can conclude that the function has zeros in the intervals ( −1, 0), (0, 1), and (3, 4).

4. Draw the Graph. Draw a continuous curve through the points. Because the zero x = 0 is of even multiplicity, you know that the graph only touches the x-axis, and because the zero x = 3 is also of even multiplicity, you know that the graph only touches the x-axis at x = 3. When you are unsure of the shape of a portion of the graph, plot some additional points. y

2 1 −2 −1

−6

Solutions to Checkpoints

999

Checkpoints for Section 3.3 1. To divide 3 x 2 + 19 x + 28 by x + 4 using long division, you can set up the operation as shown.

3x + 7 x + 4 3 x 2 + 19 x + 28 3 x 2 + 12 x

Multiply 3 x by x + 4.

7 x + 28 7 x + 28 0

2. To divide x 3 − 2 x 2 − 9 by x − 3 using long division, you can set up the operation as shown. Because there is no x-term in the dividend, rewrite the dividend as x 3 − 2 x 2 + 0 x − 9 before you apply the Division Algorithm.

Subtract. Multiply 7 by x + 4. Subtract.

x2 + x + 3 3

x − 3 x − 2x2 + 0 x − 9 x3 − 3x 2 x + 0x − 9

Multiply x 2 by x − 3. Subtract.

x 2 − 3x 3x − 9 3x − 9 0

Multiply x by x − 3. Subtract. Multiply 3 by x − 3. Subtract.

From this division, you can conclude that

3 x 2 + 19 x + 28 = ( x + 4)(3x + 7).

So, x − 3 divides evenly into x3 − 2 x 2 − 9, and you

x3 − 2 x 2 − 9 = x 2 + x + 3, x ≠ 3. x −3 You can check this result by multiplying ( x − 3)( x 2 + x + 3) = x3 − 3x 2 + x 2 − 3x + 3x − 9 can write

= x 3 − 2 x 2 − 9.

3. To divide − x 3 + 9 x + 6 x 4 − x 2 − 3 by 1 + 3x using long division, begin by rewriting the dividend and divisor in

descending powers of x. 2 x3 − x 2 4

3x + 1 6 x −

x − x + 9x − 3

6 x 4 + 2 x3 − 3x − x + 9 x − 3

Multiply 2 x 3 by 3x + 1. Subtract.

− 3x3 − x 2

Multiply − x 2 by 3 x + 1.

9x − 3 9x + 3 −6

So, you have

Subtract. Multiply 3 by 3 x + 1. Subtract.

6 x 4 − x3 − x 2 + 9 x − 3 6 = 2 x3 − x 2 + 3 − . 3x + 1 3x + 1

1000 Solutions to Checkpoints 4. To divide 5 x 3 + 8 x 2 − x + 6 by x + 2 using synthetic division, you can set up the array as shown. −2

−1

Then, use the synthetic division pattern by adding terms in columns and multiplying the results by − 2. Dividend: 5x3 + 8x2 − x + 6 8

−1

− 10

−6

−2

(−

(5 −2

)

−2

Divisor: x+2

Remainder: 0

Quotient: 5x 2 − 2x + 3

So, you have 5. (a) f ( −1)

5 x3 + 8 x 2 − x + 6 = 5 x 2 − 2 x + 3, x ≠ − 2. x+2 −1

−3

−8

−4 6

−6

−9

Because the remainder is r = 1, f ( −1) = 1. Check: f ( −1) = 4( −1) + 10( −1) − 3( −1) − 8 3

= 4( −1) + 10(1) + 3 − 8

()

4 4

−3

−8

104

404

101

396

Because the remainder is r = 396, f ( 4) = 396. Check: f ( 4) = 4( 4) + 10( 4) − 3( 4) − 8 3

= 4(64) + 10(16) − 12 − 8 = 396

−3

−8

3 2

− 13 2

()

Because the remainder is r = − 13 , f 12 = − 13 . 2 2

()

( ) + 10( 12 ) − 3( 12 ) − 8 = 4( 18 ) + 10( 14 ) − 32 − 8

Check: f 12 = 4 12

(b) f ( 4)

1 2

= 12 + 52 − 32 − 8 = − 13 2 (d) f ( − 3)

−3

−8

− 12

−9

−2

− 17

4 4

Because the remainder is r = −17, f ( − 3) = −17. Check: f ( − 3) = 4( − 3) + 10( − 3) − 3( − 3) − 8 3

= 4( − 27) + 10(9) + 9 − 8 = −17

Solutions to Checkpoints 1001 6. Algebraic Solution

Using synthetic division with the factor ( x + 3), you obtain the following. −3

− 19

− 30

−3

− 10

1 1

→

0 remainder, so f ( − 3) = 0 and ( x + 3) is a factor.

Because the resulting quadratic expression factors as x 2 − 3x − 10 = ( x − 5)( x + 2), the complete factorization of f ( x) is f ( x) = x3 − 19 x − 30 = ( x + 3)( x − 5)( x + 2). Graphical Solution

From the graph of f ( x) = x3 − 19 x − 30, you can see there are three x-intercepts. These occur at x = − 3, x = − 2, and x = 5. This implies that ( x + 3), ( x + 2), and ( x − 5) are factors of f ( x ). y = x 3 − 19x − 30 10 −6

(5, 0)

(−2, 0) (−3, 0) −70

7. Because the leading coefficient is 1, the possible rational zeros of f ( x) = x3 − 5 x 2 + 2 x + 8 are the factors of the

constant term. Possible rational zeros: ± 1, ± 2, ± 4, ± 8 By testing these zeros, f ( −1) = ( −1) − 5( −1) + 2( −1) + 8 = 0 3

f (1) = (1) − 5(1) + 2(1) + 8 = 6 3

f ( − 2) = ( − 2) − 5( − 2) + 2( − 2) + 8 = − 24 3

f ( 2) = ( 2) − 5( 2) + 2( 2) + 8 = 0 3

f ( − 4) = ( − 4) − 5( − 4) + 2( − 4) + 8 = −144 3

f ( 4) = ( 4) − 5( 4) + 2( 4) + 8 = 0 3

f ( − 8) = ( − 8) − 5( − 8) + 2( − 8) + 8 = − 840 3

f (8) = (8) − 5(8) + 2(8) + 8 = 216 3

you can conclude that the polynomial function f ( x) = x3 − 5 x 2 + 2 x + 8 has three rational zeros x = −1, x = 2, and x = 4.

1002 Solutions to Checkpoints 8. The leading coefficient of f ( x) = 2 x 4 − 9 x3 − 18 x 2 + 71x − 30 is 2 and the constant term is − 30.

Possible rational zeros:

Factors of − 30 ± 1, ± 2, ± 3, ± 5, ± 6, ± 10, ± 15, ± 30 = Factors of 2 ±1, ± 2

1 3 5 15 = ±1, ± 2, ± 3, ± 5, ± 6, ±10, ±15, ± 30, ± , ± , ± , ± 2 2 2 2 Choose a value of x and use synthetic division. x = −3 −3

−9

− 18

− 30

−6

− 81

− 15

− 10

2 2

→

0 remainder, so x + 3 is a factor.

Choose another value of x and use synthetic division. x = 2

2 2

− 15

− 10

− 22

− 11

→

0 remainder, so x + 3 is a factor.

So, f ( x) = 2 x 4 − 9 x3 − 18 x 2 + 71x − 30 factors as

f ( x) = ( x + 3)( x − 2)( 2 x 2 − 11x + 5) = ( x + 3)( x − 2)( x − 5)( 2 x − 1) and

you can conclude that the rational zeros of f are x = − 3, x =

1 , x = 2, and x = 5. 2

9. The original polynomial has three variations in sign.

− to +

−

to +

↓

f ( x) = − 2 x3 + 5 x 2 − x + 8 ↑

↑

+ to − The polynomial f ( − x) = − 2( − x) + 5( − x) − ( − x) + 8 = 2 x3 + 5 x + x + 8 has no variations in sign. 3

So, from Descartes’s Rule of Signs, the polynomial f ( x) = − 2 x3 + 5 x 2 − x + 8 has either three positive real zeros or one positive real zero and no negative real zeros. From the graph, you can see that the function has only one real zero. 14

−1

y = −2x 3 + 5x 2 − x + 8

3 −2

Solutions to Checkpoints 1003 10. The possible real zeros of f ( x) = 8 x3 − 4 x 2 + 6 x − 3 are as follows.

±1, ± 3 Factors of − 3 1 1 3 1 3 3 = = ± , ± , ± , ± , ± , ±1, ± , ± 3 ±1, ± 2, ± 4, ± 8 Factors of 8 8 4 8 2 4 2 The original polynomial f ( x) has three variations in sign. The polynomial f ( − x) = 8( − x) − 4( − x) + 6( − x) − 3 = − 8 x3 − 4 x 2 − 6 x − 3 3

has no variations in sign. So, you can apply Descartes’s Rule of Signs to conclude that there are either three positive real zeros or one positive real zero, and no negative real zeros. Using x = 1, synthetic division produces the following.

8 8

−4

−3

→

1 is not a zero.

So, x = 1 is not a zero, but because the last row has all positive entries, you know that x = 1 is an upper bound for the real zeros. So, you can restrict the search to real zeros between 0 and 1. Using x = 12 , synthetic division produces the following. 1 2

−4

−3

→

1 2

is a real zero.

f ( x) = 8 x3 − 4 x 2 + 6 x − 3 1  =  x − (8 x 2 + 6) 2 

Because 8 x 2 + 6 has no real zeros, it follows that x = 12 is the only real zero. 11. 3 x 4 − 14 x 2 − 4 x = 0 x (3 x 3 − 14 x − 4) = 0 So, x = 0 is a real zero.

1 2 4 Possible rational zeros : ± 1, ± 2, ± 4, ± , ± , ± 3 3 3

−2 3 3

− 14

−4

−6

−2

x = −2 is a real zero.

Use the Quadratic Formula. 3x 2 − 6 x − 2 = 0 x= x=

− ( − 6) ±

( − 6 ) − 4 ( 3)( − 2 ) 2 ( 3) 2

3 ± 15 3

So, the real zeros are x = 0, − 2, and

3 ± 15 . 3

1004 Solutions to Checkpoints

Checkpoints for Section 3.4 1.

f ( x ) = x 4 − 1 = ( x 2 + 1)( x 2 − 1) = ( x 2 + 1)( x + 1)( x − 1)

4. Because 7i is a zero and the polynomial is stated to have real coefficients, you know that the conjugate − 7i

must also be a zero.

x2 + 1  x = ± i

So, the four zeros are 2, − 2, 7i , and –7i.

x + 1  x = −1

Then, using the Linear Factorization Theorem, f ( x)

x −1 x =1

The fourth-degree polynomial function f ( x) = x 4 − 1 has exactly four zeros, x = ± i and x = ±1. 2. Factor the polynomial completely as follows.

f ( x) = x 4 − 1 = ( x 2 − 1)( x 2 + 1)

can be written as f ( x ) = a( x − 2)( x + 2)( x − 7i )( x + 7i ).

For simplicity, let a = 1. Then multiply the factors with real coefficients to obtain

( x + 2)( x − 2) = x 2 − 4 and multiply the complex conjugates to obtain

= ( x + 1)( x − 1)( x + i )( x − i )

( x − 7i)( x + 7i) = x 2 + 49.

Set f ( x ) = 0: ( x + 1)( x − 1)( x + i )( x − i ) = 0

So, you obtain the following fourth-degree polynomial function.

Set 1st factor equal to 0: x + 1 = 0

f ( x) = ( x 2 − 4)( x 2 + 49) = x 4 + 49 x 2 − 4 x 2 − 196

x = −1 Set 2nd factor equal to 0: x − 1 = 0

x =1 Set 3rd factor equal to 0: x + i = 0

x = −i Set 4th factor equal to 0: x − i = 0

x = i So, the zeros are x = ± 1 and x = ± i. 3. Factor the polynomial completely as follows.

f ( x) = x 4 − 256

= x 4 + 45 x 2 − 196

5. Because 2i is a zero and the polynomial is stated to have real coefficients, you know that the conjugate − 2i must

also be a zero.

f ( x) = a( x − 1)( x + 2)( x − 2i)( x + 2i) = a( x 2 + x − 2)( x 2 + 4)

= a( x 4 + 4 x 2 + x3 + 4 x − 2 x 2 − 8) = a( x 4 + x3 + 2 x 2 + 4 x − 8)

To find the value of a, use the fact that f ( −1) = 10

= ( x 2 − 16)( x 2 + 16)

4 3 2 to obtain a ( −1) + ( −1) + 2( −1) + 4( −1) − 8 = 10.  

= ( x + 4)( x − 4)( x + 4i )( x − 4i )

Because a = −1,

Set f ( x ) = 0: ( x + 4)( x − 4)( x + 4i )( x − 4i ) = 0 Set 1st factor equal to 0: x + 4 = 0

f ( x ) = −1( x 4 + x 3 + 2 x 2 + 4 x − 8) = − x 4 − x 3 − 2 x 2 − 4 x + 8.

x = −4 Set 2nd factor equal to 0: x − 4 = 0

x = 4 Set 3rd factor equal to 0: x + 4i = 0

x = − 4i Set 4th factor equal to 0: x − 4i = 0

x = 4i So, the zeros are x = ± 4 and x = ± 4i.

Solutions to Checkpoints 1005 6. (a) Begin by factoring the polynomial as the product of two quadratic polynomials.

x4 − 2 x2 − 3 = ( x2 − 3)( x2 + 1) Both of these factors are irreducible over the rationals.

(

3 x +

)(

3 ( x 2 + 1), where the quadratic factor is

(

3 x−

)(

3 ( x + i )( x − i).

(b) By factoring over the reals, you have x 4 − 2 x 2 − 3 = x −

)

irreducible over the reals. (c) In completely factored form, you have x 4 − 2 x 2 − 3 = x +

)

7. Because complex zeros occur in conjugate pairs you know that if 4i is a zero of f ( x ) = 3 x 3 − 2 x 2 + 48 x − 32, so is − 4i.

This means that both ( x − 4i) and ( x + 4i) are factors of f.

( x − 4i)( x + 4i) = x2 − 16i 2 = x2 + 16 Using long division, you can divide x 2 + 16 into f ( x) to obtain the following. 3x − 2 2

x + 16 3 x − 2 x + 48 x − 32 3x3 + 48 x − 2 x2 − 2 x2

− 32 − 32 0

(

)

So, you have f ( x) = x2 + 16 (3x − 2) and you can conclude that the real zeros of f are x = − 4 i , x = 4 i , and x = 23.

Checkpoints for Section 3.5 1. Because the denominator is zero when x = 1, the domain of f ( x ) =

3x is all real numbers except x = 1. x −1

To determine the behavior of f at this excluded value, evaluate f ( x) to the left and to the right of x = 1. x

0.5

0.9

0.99

0.999

→ 1

f ( x)

−3

− 27

− 297

− 2997

→ −∞

1 ←

1.001

1.01

1.1

1.5

f ( x)

∞ ←

3003

303

f (x) = x 3x −1

−6

10 −4

As x approaches 1 from the left, f ( x) decreases without bound. As x approaches 1 from the right, f ( x) increases without bound.

1006 Solutions to Checkpoints

2. For f ( x ) =

5x2 , the degree of the numerator is equal x2 − 1

to the degree of the denominator. The leading coefficient of the numerator is 5 and the leading coefficient of the denominator is 1, so the graph has the line y = 5 as a horizontal asymptote. To find any vertical asymptotes, set the denominator equal to zero and solve the resulting equation for x.

x2 − 1 = 0

( x + 1)( x − 1) = 0 x + 1 = 0  x = −1 x −1 = 0  x =1 This equation has two real solutions, x = − 1 and x = 1, so the graph has the lines x = − 1 and x = 1 as vertical asymptotes. 9

5x 2 x2 − 1

x = −1

x=1

−3

3. For f ( x ) =

x 2 − 25 , the degree of the numerator x2 + 5x

is equal to the degree of the denominator. The leading coefficients of the numerator and denominator are both 1, so the graph has the line y = 1 as a horizontal asymptote. To find any vertical asymptotes, first factor the numerator and denominator as follows. f ( x) =

x 2 − 25 = x2 + 5x

3x 2 + x − 5 x2 + 1

(a) Because the denominator is zero when x 2 + 1 = 0, solve this equation to determine that the domain of f is all real numbers. (b) Because the denominator of f is not factorable over the reals, it does not have a real zero, and therefore, the graph of f has no vertical asymptotes. (c) Because the degrees of the numerator and denominator are the same, and the leading coefficient of the numerator is 3 and the leading coefficient of the denominator is 1, the horizontal asymptote of f is y = 3. 5. (a) The cost to remove 20% of the pollutants is 255(20) C = = $63.75 million. 100 − (20)

The cost to remove 45% of the pollutants is 255(45) C = ≈ $208.64 million. 100 − 45

y=5 −9

4. f ( x ) =

The cost to remove 80% of the pollutants is 255(80) C = = $1020 million. 100 − 80 (b) The cost to remove 100% of the pollutants is 255(100) C = which is undefined. 100 − (100) So, it would not be possible to remove 100% of the pollutants.

( x + 5) ( x − 5) x − 5 = , x ≠ −5 x x ( x + 5)

By setting the denominator x (of the simplified function) equal to zero, you can determine that the graph has the line x = 0 as a vertical asymptote. To find any holes in the graph, note that the function is undefined at x = 0 and x = − 5. Because x = − 5 is not a vertical asymptote of the function, there is a hole in the graph at x = − 5. To find the y-coordinate of the hole, substitute x = − 5 into the simplified form of the function. y =

x −5 −5 − 5 −10 = = = 2 x −5 −5

So, the graph of the rational function has a hole at ( − 5, 2). 2

y = x 2 − 25

x − 5x 9

y=1 −9

x=0 −3

Solutions to Checkpoints 1007

Checkpoints for Section 3.6 1. (a) With respect to the graph of f ( x ) =

1 , the graph x

3. f ( x ) =

2x + 3 x +1

y-intercept: (0, 3), because C (0) = 3

1 is obtained by a horizontal shift x − 4 four units to the right.

of g ( x ) =

(

)

( )

x-intercept: − 32 , 0 , because C − 32 = 0 Vertical asymptote: x = − 1, zero of denominator

Horizontal asymptote: y = 2, because degree of

4 3

N ( x) = degree of D( x)

2 1 −1 −1

Additional points:

−2 −3 −4

(b) With respect to the graph of f ( x ) =

−3

−2

−1

C( x)

3 2

Undefined

5 2

9 4

1 , the graph x

1 − 1 is obtained by a horizontal x + 3 shift three units to the left and a vertical shift one unit downward.

of h ( x ) =

3 1 − 5 − 4 −3 −2

4 3 2

4. f ( x) =

1 −6 −5 −4

−2

y-intercept: (0, 0), because f (0) = 0

−2 −3

x-intercept: (0, 0), because f (0) = 0

−4

2. f ( x ) =

Vertical asymptotes: x = − 2, x = 1, zeros of

1 x + 3

denominator Horizontal asymptote: y = 0, because degree of

( )

y-intercept: 0, 13 , because f (0) = 13

N ( x) < degree of D( x)

x-intercept: none, because 1 ≠ 0 Vertical asymptote: x = − 3, zero of denominator

Additional points:

Horizontal asymptote: y = 0, because degree of

N ( x) < degree of D( x)

−3

−2

−1

f ( x)

−

9 4

Undefined

3 2

Undefined

3 2

9 10

Additional points: x

f ( x)

−5

3x 3x = x2 + x − 2 ( x + 2)( x − 1)

−4

−3

−2

−1

2 4

− 12

−1

Undefined

1 2

1 4

1 5

3 2

−4 −3

4 3

−2

−3

1 − 2 −1 −1

−1 −1

−4

−2 −3 −4

1008 Solutions to Checkpoints

5. f ( x) =

x2 − 4 x − x−6 2

( x + 2) ( x − 2) ( x − 3) ( x + 2)

x−2 , x ≠ −2 x−3

2  2 y-intercept:  0, , because f (0) = 3  3

x-intercept: ( 2, 0), because f ( 2) = 0 Vertical asymptote: x = 3, zero of (simplified) denominator

4  Hole:  − 2, , f is not defined at x = − 2 5 

Horizontal asymptote: y = 1, because degree of N ( x) = degree of D( x)

Additional points:

−7

−5

−2

−1

f ( x)

9 10

7 8

Undefined

3 4

1 2

Undefined

3 2

6. f ( x) =

− 2 −1 −1

−2 −3 −4

3x 2 + 1 x

First divide 3 x 2 + 1 by x, either by long division:

3x x 3x 2 + 0 x + 1 3x 2 1 2

3x + 1 1 = 3x + x x

or by separating the numerator and simplifying:

3x 2 + 1 3x 2 1 1 = + = 3x + . x x x x So, the slant asymptote is y = 3 x, since

3x 2 + 1 1 = 3x + . x x

y-intercept: none, since f (0) is undefined. x-intercept: none, since 3 x 2 + 1 ≠ 0 for real numbers. Vertical asymptote: x = 0, zero of denominator Slant asymptote: y = 3x

Additional points:

−2

−1

− 0.5

0.5

f ( x)

−

13 2

−4

−

7 2

Undefined

7 2

13 2

6 4 2 −8 −6 −4 −2

−8

Solutions to Checkpoints 1009 7. Graphical Solution

Numerical Solution 1 in.

2 in.

1 in.

2 in.

x 1 in.

1 in.

Let A be the area to be minimized.

A = ( x + 4)( y + 2)

A = ( x + 4)( y + z)

The printed area inside the margins is modeled by 40 40 = xy or y = . x

To find the minimum area, rewrite the equation for A in 40 terms of just one variable by substituting for y. x

 40  A = ( x + 4) + 2  x 

 40  + 2 A = ( x + 4)  x 

 40 + 2 x  = ( x + 4)  x  

( x + 4)(40 + 2 x) , x > 0 x

The graph of this rational function is shown below. Because x represents the width of the printed area, you need to consider only the portion of the graph for which x is positive. Using a graphing utility, you can approximate the minimum value of A to occur when x ≈ 8.9 inches. The corresponding value of y is 40 ≈ 4.5 inches. 8.9

Use the table feature of a graphing utility to create a table of values for the function

So, the dimensions should be 8.9 + 4 = 12.9 inches by 4.5 + 2 = 6.5 inches.

To approximate the minimum value of y1 to one decimal place, change the table so that it starts at x = 8 and increases by 0.1. The minimum value of y1 occurs when x ≈ 8.9 as shown.

200

(x + 4)(40 + 2x) x

y1 =

( x + 4)(40 + 2 x) , x > 0 x

beginning at x = 6. From the table, you can see that the minimum value of y1 occurs when x is somewhere between 8 and 9, as shown.

40 ≈ 4.5 inches. 8.9 So, the dimensions should be 8.9 + 4 = 12.9 inches by 4.5 + 2 = 6.5 inches.

The corresponding value of y is

0 − 10

1010 Solutions to Checkpoints

Checkpoints for Section 3.7 1. Enter the data into a graphing utility and then display the scatter plot. From the scatter plot, it appears the data follow a parabolic pattern. So, the data can be better modeled by a quadratic function.

3. Begin by entering the data into a graphing utility and displaying the scatter plot. From the scatter plot, you can see that the data show a parabolic trend. 225

0 210

5100

2. (a) Begin by entering the data into a graphing utility and displaying the scatter plot. From the scatter plot, you can see that the data appear to follow a parabolic pattern.

Using the regression feature of a graphing utility, you can find the quadratic model. The quadratic model that best fits the data is given by y = − 0.61x 2 + 2.3x + 219.

(b) Using the regression feature of a graphing utility, you can find the quadratic model. So, the quadratic equation that best fits the data is given by y = 0.0008 x 2 + 0.06 x.

To estimate the first year when the annual sales will be less than $190 million, set the regression equation equal to 190 and solve for x using the Quadratic Formula. 190 = − 0.607 x 2 + 2.28 x + 219 0 = − 0.607 x 2 + 2.28 x + 29 x = x =

−b ±

b 2 − 4ac 2a

−( 2.28) ±

(2.28)2 − 4(− 0.607)(29) 2( − 0.607)

x ≈ 9.04

(c) Graph the data and the model in the same viewing window. Using the value or trace features of the graphing utility, you can conclude that it takes approximately 5.7 seconds for an automobile, from a standing start, to reach a speed of 55 miles per hour.

So, the first year that annual sales will be less than $190 million is year 10.

Solutions to Checkpoints 1011 4. Let x represent the year, with x = 5 corresponding to 2005. Begin by entering the data into a graphing utility. Using the regression feature, a linear model for the data is y = 10.26 x − 28.8 and a quadratic model for the data is

y = − 0.762x2 + 22.45x − 74.5. Plot the data and the linear model in the same viewing window. 85

Then plot the data and the quadratic model in the same viewing window. 85

To determine which model fits the data better, compare the y-values given by each model with the actual y-values. The model whose y-values are closet to the actual values is the better fit. In this case, the better-fitting model is the quadratic model.

Chapter 4 Checkpoints for Section 4.1 1. Function Value

( 2) = 8

−

3. The table lists some values for each function and the graph shows a sketch for each function. Note that both graphs are decreasing and the graph of g ( x) = 9− x is

Graphing Calculator Keystrokes 8 ∧ ( ( −)

decreasing more rapidly than the graph of f ( x) = 3− x.

2 ) Enter

Display 0.052824803759 2. The table lists some values for each function and the graph shows a sketch of the two functions. Note that both graphs are increasing and the graph of g ( x) = 9 x is

–2

–1

9− x

1 8

1 64

1 512

3− x g ( x)

1 3

1 9

1 27

g(x) = 9 − x

increasing more rapidly than the graph of f ( x) = 3x. x

–3

–2

–1

1 27

1 9

1 3

1 729

1 81

1 9

f (x) = 3− x −4

−3

−2

−1

g(x) = 9 x

f(x) = 3x 1

1012 Solutions to Checkpoints 4. (a) Because g ( x) = 4 x − 2 = f ( x − 2), the graph of g

5. Graphical Solution 24

can be obtained by shifting the graph of f two units to the right. y

y2 y1

f(x) = 4 x

2 1 −1

2 As x increases, the graph of y1 = 1 +  gets closer x 

−2

100 0

and closer to the graph of the line y2 = e 2 .

g(x) = 4 x − 2 1

−1

Numerical Solution Use the table feature (in ask mode) to evaluate y1 for increasing values of x.

(b) Because h( x) = 4 x + 3 = f ( x) + 3 the graph of h can be obtained by shifting the graph of f upward three units. y 6

From the table, it seems reasonable to conclude that

h(x) = 4 x + 3 5

2  2 1 +  → e as x → ∞. x 

3 2

f(x) = 4 x

1 −3

−2

−1

(c) Because k ( x) = 4− x − 3 = f ( − x) − 3, the graph of k can be obtained by reflecting the graph of f in the y-axis and shifting the graph of f downward three units. y 3 2 1 −3

−2

−1

f(x) = 4 x 1

k(x) = 4−x − 3 −2 −3

6. Function Value

Graphing Calculator Keystrokes

Display

(a) f (0.3) = e0.3

e x 0.3 Enter

1.3498588

(b) f ( −1.2) = e−1.2

e x ( − ) 1.2 Enter

0.3011942

e x 6.2 Enter

492.74904

7. To sketch the graph of f ( x) = 5e0.17 x , use a graphing utility to construct a table of values. After constructing the table, plot the points and draw a smooth curve. x

–3

–2

–1

f ( x)

3.002

3.559

4.218

5.000

5.927

7.025

8.326

y 25 20 15 10

−8 −6 −4 −2

2 4 6 8 10

Solutions to Checkpoints 1013 8. Algebraic Solution From Example 8, let P = 9000, r = 2.5% = 0.025, n = 1, and t = 10. Using the formula for compound

A = Pert

interest with n compoundings per year, you have r  A = P1 +  n 

0.025   = 90001 +  1  

(1)(10)

Formula for continuous compounding

= 6000e0.04(7)

Substitute P, r , and t.

≈ $7938.78

Use a calculator.

10. Algebraic Solution 1 y = 10  2

= 9000(1.025)

≈ $11,520.76.

t 29

160 29

1 = 10  2

So, the balance in the account after 10 years is about $11,520.76.

1 = 10  2 ≈ 0.218

Graphical Solution Substitute the values for P, r, and n into the formula for compound interest with n compoundings per year and simplify to obtain A = 9000(1.025) . Use a graphing t

utility to graph A = 9000(1.025) . Then use the value t

feature to approximate the value of A when t = 10, as shown. 15,000

5.517

So, about 0.218 gram is present after 160 years. Graphical Solution t 29

1 Use a graphing utility to graph y = 10  . Then use 2 the value feature to approximate the value of y when t = 160, as shown. 12

The balance in the account after 10 years is $11,520.80. 9. (a) For quarterly compounding, you have n = 4. So, in 7 years at 4%, the balance is as follows.

r  A = P1 +  n 

Formula for compound interest

0.04   = 60001 +  4   ≈ $7927.75

4(7)

Use a calculator.

200

When t = 160, y ≈ 0.218. So, about 0.218 gram is present after 160 years. 11. (a) To find the initial population, evaluate Q(t ) when t = 0.

Q(0) = 10e0.02(0) = 10e0 = 10 flies

Substitute P, r , n, and t.

(b) For monthly compounding, you have n = 12. So, in 7 years at 4%, the balance is as follows. r  A = P1 +  n 

(b) After 72 hours, the population size is Q(72) = 10e0.02(72) = 20e1.44 ≈ 42 flies. (c)

Formula for compound interest 12(7)

0.04   = 60001 +  12   ≈ $7935.08

Substitute P, r , n, and t.

Use a calculator.

Checkpoints for Section 4.2 1. (a) f (1) = log6 1 = 0 because 60 = 1.

( )

1 1 = − 3 because − 3 1 . (b) f 125 = log 5 125 5 = 125

1014 Solutions to Checkpoints 2. Function Value

Graphing Calculator Keystrokes

Display

(a) f ( 275) = log10 275

LOG 275 ENTER

2.4393327

(b) f (0.275) = log10 0.275

LOG 0.275 ENTER

– 0.5606673

( ) (d) f ( 12 ) = log 12

(−) ( 1 ÷ 2 ) ) ENTER

LOG (

ERROR

LOG ( 1 ÷ 2 ) ENTER

– 0.3010300

3. (a) Using Property 4, log10 x = log10 2, so x = 2.

(b) Using Property 3, 20

log

20 3

= 3.

4. (a) For f ( x) = 8x , construct a table of values. Then plot the points and draw a smooth curve. x

–2

–1

f ( x) = 8x

1 64

1 8

(b) Because g( x) = log8 x is the inverse function of f ( x) = 8x ,

the graph of g is obtained by plotting the points ( f ( x), x) and

5 4

connecting them with a smooth curve. The graph of g is a reflection of the graph of f in the line y = x. x

g ( x ) = log 8 x

1 64

1 8

−2

−1

f (x) = 8x

3 2

g(x) = log 8x

1 −2 −1 −1

−2

5. Construct a table of values by using the Inverse Property of Logarithms. Plot the points and connect them with a smooth curve. Note that x = 0 (the y-axis) is a vertical asymptote of the graph.

1 9

1 3

f ( x) = log3 x

−2

−1

5 4 3 2 1 −1 −2 −3 −4 −5

1 2 3 4 5 6 7 8 9

6. (a) Because g( x) = −1 + log10 x = −1 + f ( x), the

(b) Because g( x) = log10 ( x + 3) = f ( x + 3), the

graph of g can be obtained by shifting the graph of f one unit downward.

graph of g can be obtained by shifting the graph of f three units to the left. y

2 −2 −2 −4 −6 −8

8 10 12 14

−6 −4

−2

−4 −6 −8

Solutions to Checkpoints 1015 7. Function Value

Graphing Calculator Keystrokes

Display

f (0.01) = ln 0.01

LN 0.01 ENTER

− 4.6051702

f ( 4) = ln 4

LN 4 ENTER

1.3862944

( 3 + 2) = ln( 3 + 2) f ( 3 − 2) = ln ( 3 − 2) f

8. (a) ln e1 3 = 13

3 ln e 4

3 ) + 2 ) ENTER

1.3169579

LN ( (

3 ) − 2 ) ENTER

ERROR

Inverse Property

(b) 5 ln 1 = 5(0) = 0 (c)

LN ( (

Property 1

= 34 (1) = 34 Property 2

(d) eln 7 = 7

Inverse Property

9. Algebraic Solution

Because ln( x + 3) is defined only when x + 3 > 0, it follows that the domain of f is ( −3, ∞). Graphical Solution 3

−5

−3

The x-coordinates of points on the graph appear to extend from the right of − 3 to ∞ . So, you can estimate the domain to be ( − 3, ∞). 10. Algebraic Solution

(a) After 1 month, the average score was the following. f (1) = 75 − 6 ln (1 + 1) = 75 − 6 ln 2 ≈ 75 − 6(0.6931) ≈ 70.84

(b) After 9 months, the average score was the following. f (9) = 75 − 6 ln (9 + 1) = 75 − 6 ln 10 ≈ 75 − 6( 2.3026) ≈ 61.18

(c) After 12 months, the average score was the following. f (12) = 75 − 6 ln (12 + 1) = 75 − 6 ln 13 ≈ 75 − 6( 2.5649) ≈ 59.61

1016 Solutions to Checkpoints Graphical Solution

(a)

100

When t = 1, f (1) ≈ 70.84. So, the average score after 1 month was about 70.84. (b)

100

When t = 9, f (9) ≈ 61.18. So, the average score after 9 months was about 61.18. (c)

100

When t = 12, f (12) ≈ 59.61. So, the average score after 12 months was about 59.61.

Checkpoints for Section 4.3 1. log 2 12 =

log 12 log 2

log x log a

2. log 2 12 =

Use a calculator.

≈

log a x =

1.07918 0.30103 ≈ 3.5850 ≈

Simplify.

3. (a) log10 75 = log10 (3 ⋅ 25)

2.48491 0.69315 ≈ 3.5850

log a x =

ln x ln a

Use a calculator. Simplify.

Rewrite 75 as 3 ⋅ 25.

= log10 3 + log10 25

Product Property

Rewrite 25 as 52 .

= log10 3 + log10 5

ln 12 ln 2

= log10 3 + 2 log10 5 Power Property 9 = log 9 − log 125 (b) log10 125 10 10

= log10 32 − log10 53

Rewrite 9 as 32 and 125 as 52 .

= 2 log10 3 − 3 log10 5

Power Property

4. ln e5 = 5 ln e = 5(1) = 5

5. log 3

Quotient Property

Power Property Inverse Property

4x2 4 x2 = log 3 1 2 y y

Rewrite radical using rational exponent.

= log 3 4 x 2 − log 3 y1 2 2

Quotient Property 12

= log 3 4 + log 3 x − log 3 y

Product Property

1 = log 3 4 + 2 log 3 x − log 3 y 2

Power Property

Solutions to Checkpoints 1017 6. 2 log10 ( x + 3) − 2 log10 ( x − 2) = 2 log10 ( x + 3) − log10 ( x − 2)    2

  x + 3   = 2 log10  2    ( x − 2)    x+3   = log10   ( x − 2)2   

Power Property Quotient Property

Power Property

( x + 3) 4 ( x − 2) 2

= log10

Simplify.

7. To solve this problem, take the natural logarithm of each of the x- and y-values of the ordered pairs.

(ln x, ln y): (− 0.994, − 0.673), (0.000, 0.000), (1.001, 0.668), (2.000, 1.332), (3.000, 2.000) ln y 4 3 2 1 −2

−1

ln x

−2

By plotting the ordered pairs, you can see that all five points appear to lie in a line. Choose any two points to determine the slope of the line. Using the points (0, 0) and (1.001, 0.668), the slope of the line is m =

0.668 − 0 2 = 0.668 ≈ . 1−0 3

By the point-slope form, the equation of the line is y = 23 x, where y = ln y and x = ln x. So, the logarithmic equation is ln y = 23 ln x.

Checkpoints for Section 4.4 1. Original Equation

Rewritten Equation

Solution

Property

(a) 2x = 512

2 x = x9

x = 9

One-to-One

(b) log 6 x = 3

log 6 x

= 63

x = 216

Inverse

ln 5 = ln ex

ln 5 = x

Inverse

32 x = 3−1

2 x = −1

One-to-One

5 = e

(d) 9 = 13 x

x = − 12

2. Algebraic Solution

(a)

e x = 10

Write original equation.

Take ln of each side.

x = ln 10

Inverse Property

x ≈ 2.30

Use a calculator.

ln e = ln 10

The solution is x = ln 10 ≈ 2.30.

1018 Solutions to Checkpoints

(b)

2(5x ) = 32

Write original equation.

Divide each side by 2.

Take log (base 5) of each side.

x = log5 16

Inverse Property

5 = 16 log5 5 = log5 16

x =

ln 16 ≈ 1.72 Change-of-base formula ln 5

The solution is x = log5 16 ≈ 1.72. Graphical Solution

(a) Graph both y1 = e x and y2 = 10 in the same viewing window. Then use the intersect feature to find the point of intersection. 20

The intersection point is about ( 2.30, 10). So, the solution is x ≈ 2.30. (b) Graph both y1 = 2(5 x ) and y = 32 in the same viewing window. Then use the intersect feature to find the point of intersection. The intersection point is about (1.72, 32). So, the solution is x ≈ 1.72 40

3. Algebraic Solution

2e3 x + 5 = 29

Write the original equation.

Subtract 5 from each side.

= 24

e3 x = 12 ln e

= ln 12

3 x = ln 12

Divide each side by 2. Take ln of each side. Inverse Property

x = 13 ln 12 Multiply each side by 13. x ≈ 0.83

Use a calculator.

Graphical Solution

First, rewrite the equation as 2e3 x − 24 = 0. Use a graphing utility to graph y = 2e3 x − 24. Then use the zero or root feature to approximate the value of x for which y = 0, that is, find the x-intercept. 10 − 0.5

1.5

− 30

The zero occurs at x ≈ 0.83. So, the solution of the original equation is x ≈ 0.83.

Solutions to Checkpoints 1019 4. 6( 2t + 5 ) + 4 = 11 6( 2

t +5

Write original equation.

)=7

2t + 5 =

Subtract 4 from each side.

7 6

Divide each side by 6.

7 log 2 2t + 5 = log 2   6 7 t + 5 = log 2   6 7 t = log 2   − 5 6 1 ln   6 t =   −5 ln 2 t ≈ − 4.78

Take log ( base 2) of each side. Inverse Property Subtract 5 from each side.

Change-of-base formula. Use a calculator.

7 The solution is t = log 2   − 5 ≈ − 4.78. 6 Check: t ≈ − 4.78 6( 2t + 5 ) + 4 = 11 ?

6 2(− 4.78 + 5)  + 4 = 11   ?

6(1.16) + 4 = 11 10.96 ≈ 11 

5. Algebraic Solution e 2 x − 7e x + 12 = 0

Write original equation.

(e x ) − 7e x + 12 = 0 (e x − 3)(e x − 4) = 0

Write in quadratic form. Factor.

e −3 = 0  e = 3

Set 1st factor equal to 0.

x = ln 3 ex − 4 = 0  ex = 4 x = ln 4

Solution Set 2nd factor equal to 0. Solution

The solutions are x = ln 3 ≈ 1.10 and x = ln 4 ≈ 1.39. Check: x = ln 3

Check: x = ln 4

e 2 x − 7e x + 12 = 0 ?

e 2(ln 4) − 7e(ln 4) + 12 = 0 2

e 2(ln 3) − 7e(ln 3) + 12 = 0

eln 4 − 7eln 4 + 12 = 0

( ) − 7eln 3 + 12 =? 0

42 − 7( 4) + 12 = 0

ln 32

32 − 7(3) + 12 = 0

0 = 0 

1020 Solutions to Checkpoints Graphical Solution

Use a graphing utility to graph y = e2 x − 7e x + 12 and then find the zeros. y = e2x − 7e x + 12

Zeros occur at x ≈ 1.099 and x ≈ 1.386. 3

−1

So, you can conclude that the solutions are x ≈ 1.10 and x ≈ 1.39. 6. (a) ln x =

2 3

Write original equation.

e ln x = e 2 3

(b) log 2 ( 2 x − 3) = log 2 ( x + 4)

Exponentiate each side.

x = e2 3

Inverse Property

2x − 3 = x + 4 x = 7

Write original equation. One-to-One Property Solution

7. Algebraic Solution 7 + 3 ln x = 5

Write original equation.

3 ln x = − 2

Subtract 7 from each side.

ln x = − 23

Divide each side by 3.

eln x = e − 2 3

Exponentiate each side.

−2 3

Inverse Property

x ≈ 0.51

Use a calculator.

x = e

Graphical Solution

Use a graphing utility to graph y1 = 7 + 3 ln x and y2 = 5. Then find the intersection point. y2 = 5

−1

y1 = 7 + 3 lnx

The point of intersection is about (0.513, 5). So, the solution is x ≈ 0.51. 8. 3 log 4 6 x = 9

Write original equation.

log 4 6 x = 3 4

log 4 6 x

= 4

Divide each side by 3. 3

6 x = 64 32 3

x =

Exponentiate each side ( base 4). Inverse Property Divide each side by 6 and simplify.

Check: x = 32 3

3 log 4 6 x = 9

( )

3 log 4 6 32 =9 3 ?

3 log 4 64 = 9 ?

3 log 4 43 = 9 ?

3⋅ 3=9 9 = 9 

Solutions to Checkpoints 1021 9. Algebraic Solution

log10 x + log10 ( x − 9) = 1 log10  x( x − 9) = 1 10

log10  x( x − 9)

= 101

x( x − 9) = 10 x 2 − 9 x − 10 = 0

( x − 10)( x + 1) = 0 x − 10 = 0  x = 10 x + 1 = 0  x = −1

Write original equation. Product Property of Logarithms Exponentiate each side ( base 10). Inverse Property Write in general form. Factor. Set 1st factor equal to 0. Set 2nd factor equal to 0.

Check: x = 10 log10 x + log10 ( x − 9) = 1 ?

log10 (10) + log10 (10 − 9) = 1 ?

log10 10 + log10 1 = 1 ?

1+ 0 =1 1 =1 

Check: x = −1 log10 x + log10 ( x − 9) = 1 ?

log10 ( −1) + log10 ( −1 − 9) = 1 ?

log10 ( −1) + log10 ( −10) = 1

x = −1 and x = −10 are not in the domain of log x. So, it does not check. The solutions appear to be x = 10 and x = −1. But when you check these in the original equation, you can see that x = 10 is the only solution. Graphical Solution

First, rewrite the original solution as

5 y = log10 x + log10 (x − 9) − 1

log10 x + log10 ( x − 9) − 1 = 0. Then use a graphing utility to graph the equation y = log10 x + log10 ( x − 9) − 1 and find the zeros.

−5

1022 Solutions to Checkpoints

10. log a

1 = log a 1 − log a x x = 0 − log a x

Quotient Property

12. Using the formula for continuous compounding, the balance is

A = Pert

= − log a x

= 500e0.0525t . To find the time required for the balance to double, let A = 1000 and solve the resulting equation for t.

11. First, rewrite the original equation as 5e0.3 x − 17 = 0.

Then use a graphing utility to graph y = 5e0.3 x − 17. Use the zero or root feature to find the value(s) of x such that y = 0.

500e0.0525t = 1000

5 −1

= 2

Divide each side by 500.

ln e

0.0525t

= ln 2

Take natural log of each side.

0.0525t = ln 2

ln 2 0.0525 t ≈ 13.20 t =

−15

So, the solution is x ≈ 4.079.

Let A = 1000.

0.0525t

Inverse Property Divide each side by 0.0525. Use a calculator.

The balance in the account will double after approximately 13.20 years.

Check:

Because the interest rate is lower than the interest rate in Example 12, it will take more time for the account balance to double.

5e 0.3 x = 17 ?

5e 0.3(4.079) = 17 ?

5e1.2237 = 17 16.9987 ≈ 17 

So, the solution x ≈ 4.079 seems reasonable. 13. 33.374 + 9.162 ln t = y

33.374 + 9.162 ln t = 52.5 9.162 ln t = 19.126 19.126 9.162 eln t = e19.126 9.162 ln t =

t = e

19.126 9.162

t ≈ 8.07

Write original equation. Substitute 52.5 for y. Subtract 33.374 from each side. Divide each side by 9.162. Exponentiate each side. Inverse Property Use a calculator.

Because t = 4 represents 2004, it follows that the average salary for public school teachers reached $52.5 thousand in 2008.

Solutions to Checkpoints 1023

Checkpoints for Section 4.5 1. Algebraic Solution

To find when the world population will reach 9 billion, let P = 9000 in the model and solve for t. 6128e0.0113t = P 6128e

0.0113t

Write original equation.

= 9000

Substitute 9000 for P.

e0.0113t ≈ 1.46867 ln e

0.0113t

Divide each side by 6128.

≈ ln 1.46867 Take natural log of each side.

0.0113t ≈ 0.38436 t ≈ 34.0

Inverse Property Divide each side by 0.0113.

According to the model, the world population will reach 9 billion in 2034. Graphical Solution 10,000

The intersection point of the model y = 6128e0.0113t and the line y = 9000 is about (34.0, 9000). So, according to the model, the world population will reach 9 billion in 2034. 2. Let y be the number of bacteria at time t. From the given information you know that y = 100 when t = 1 and y = 200 when t = 2. Substituting this information into the model y = aebt produces 100 = ae(1)b and 200 = ae(2)b .

To solve for b, solve for a in the first equation.

100 = aeb

Write first equation.

100 = a Solve for a. eb Then substitute the result into the second equation. 200 = ae 2b

Write second equation.

 100  200 =  b e 2b  e  200 = eb 100 2 = eb

Simplify.

ln 2 = ln eb

Take natural log of each side.

ln 2 = b

Inverse Property

Substitute

100 for a. eb

Simplify and divide each side by 100.

Use b = ln 2 and the equation you found for a. 100 eln 2 100 = 2 = 50

a =

Substitute ln 2 for b. Inverse Property Simplify.

So, with a = 50 and b = ln 2, the exponential growth model is y = 50e(ln 2)t . ln 2 3 After 3 hours, the number of bacteria will be y = 50e ( ) = 400 bacteria.

1024 Solutions to Checkpoints 3. Algebraic Solution

In the carbon dating model, substitute the given value of R to obtain the following. 1 − t 8223 e = R 1012

Write original model.

e − t 8223 1 = 14 1012 10 1 e − t 8223 = 2 10 1 e − t 8223 = 100 1 ln e − t 8223 = ln 100 t − ≈ − 4.6052 8223 t ≈ 37,869

Substitute

1 for R. 1014

Multiply each side by 102. Simplify. Take natural log of each side. Inverse Property Multiply each side by − 8223.

So, to the nearest thousand years, the age of the fossil is about 38,000 years. Graphical Solution

Use a graphing utility to graph the formula for the ratio of carbon 14 to carbon 12 at any time t as y1 =

1 − x 8223 e . 1012

In the same viewing window, graph y2 =

1 . 1014

Use the intersect feature to estimate that x ≈ 18,934 when y = 10−13

y1 = 112 e−x/8223 10

y2 = 114 10 10,000

1 . 1013

Use the intersect feature to estimate that x ≈ 37,868 when y = 1/10 14. 45,000

− 10−13

So, to the nearest thousand years, the age of the fossil is about 38,000 years. 4. Use a graphing utility to graph the function. On this bell-shaped curve, the maximum value of the curve represents the average score. Using the maximum feature of the graphing utility, you can see that the average SAT critical reading score for college-bound seniors in 2013 was 496. 0.004

200

y = 0.0035e−(x − 496)2 / 26,450

800

− 0.001

Solutions to Checkpoints 1025 5. Algebraic Solution

To find the number of days that 250 students are infected, let y = 250 and solve for t. 5000 = y 1 + 4999e − 0.8t 5000 = 250 1 + 4999e − 0.8t 5000 = 1 + 4999e − 0.8t 250

Write original model. Substitute 250 for y. Divide each side by 250 and multiply each side by 1 + 4999e − 0.8t .

20 = 1 + 4999e − 0.8t 19 = 4999e

Simplify.

− 0.8t

Subtract 1 from each side.

19 = e− 0.8t 4999  19  − 0.8t ln   = ln e  4999 

Divide each side by 4999. Take natural log of each side.

 19  ln   = − 0.8t  4999  − 5.5726 ≈ − 0.8t

Inverse Property Use a calculator.

t ≈ 6.97

Divide each side by − 0.8.

So, after about 7 days, 250 students will be infected. Graphical Solution

To find the number of days that 250 students are infected, use a graphing utility to graph. y1 =

5000 and y2 = 250 1 + 4999e − 0.8 x

in the same viewing window. Use the intersect feature of the graphing utility to find the point of intersection of the graphs. The point of intersection 2000 occurs near x ≈ 6.96. So, after about 7 days, at least 250 students 0 will be infected.

y1 =

5000 1 + 4999e− 0.8x y2 = 250 16

−1500

6. (a) Because I 0 = 1 and R = 6.0, you have the following.

R = log

I I0

6.0 = log

I 1

Substitute 1 for I 0 and 6.0 for R.

106.0 = 10log I

Exponentiate each side ( base 10).

6.0

= I

Inverse Property

1,000,000 = I

Simplify.

(b) Because I 0 = 1 and R = 7.9, you have the following. I 1

Substitute 1 for I 0 and 7.9 for R.

107.9 = 10log I

Exponentiate each side ( base 10).

7.9 = log

7.9

= I

79,432,823 ≈ I

Inverse Property Simplify.

1026 Solutions to Checkpoints

Checkpoints for Section 4.6 1.

3. Begin by entering the data into a graphing utility. Then use the regression feature of the graphing utility to find logarithmic and linear models for the data.

From the figure, it appears that the data can best be modeled by an exponential function. Logarithmic Model

2. Begin by entering the data into a graphing utility. Then use the regression feature of the graphing utility to find exponential and power models for the data.

Exponential Model

Power Model

r 2 ≈ 0.9023. So, the logarithmic model is the better fit since its r 2 -value is closer to 1.

So, an exponential model for the data is x y = 16.639(0.500) , and a power model for the data is

y = 16.639(0.500) x

y = 9.578 x −1.957

y = 0.346x + 2.323

y = 2.460 + 1.090ln x

y = 9.578 x . Plot the data and each model in the same viewing window. To determine which model fits the data better, compare the coefficients of determination for each model. For exponential model, r 2 ≈ 0.9986

−1.957

and for the power model, r 2 ≈ 0.8739. So, the exponential model is the better fit since its r 2 -value is closer to 1.

Linear Model

So, a logarithmic model for the data is y = 2.460 + 1.090 ln x, and a linear model for the data is y = 0.346 x + 2.323. Plot the data and each model in the same viewing window. To determine which model fits the data better, compare the coefficients of determination for each model. For the logarithmic model, r 2 ≈ 0.9826 and for the linear model,

Logarithmic Model

Linear Model

4. To find the time required for the initial 10.08 grams to decay to 5 grams, set y = 5 and solve for t. This value is the half-life of the radioactive substance.

y = 10.08e− 0.131x 0

Exponential Model

5 = 10.08e− 0.131x

5 = e − 0.131x 10.08  5  − 0.131x ln   = ln e  10.08 

Power Model

 5  ln   = − 0.131x  10.08  1  5  ln   = x − 0.131  10.08  x ≈ 5.4 days 5. Use a graphing utility to graph the logistic model y =

100 . 1 + 7e − 0.069 x

120

The horizontal asymptotes are y = 0 to the left as x → −∞ and y = 100 to the right as x → ∞. As the number of egg masses increases, the percent of defoliation approaches 100%. Because the domain is x ≥ 0, there is no context for the asymptote y = 0.

Solutions to Checkpoints 1027

Chapter 5 Checkpoints for 5.1 1. (a) Sample answers: 55° + 360° = 415° 55° − 360° = − 305°

(b) Sample answers: − 28° + 360° = 332° − 28° − 360° = − 388°

2. (a) The complement of 23° is 90° − 23° = 67°.

The supplement of 23° is 180° − 23° = 157°. (b) Because − 28° < 0, the angle does not have a complement nor supplement.  π rad  π 3. (a) 60° = 60 deg  radians  =  180 deg  3  

(

)

 π rad  16π (b) 320° = 320 deg  radians  =  180 deg  9  

(

)

4. (a)

π π  180 deg  =  rad   = 30° 6 6  π rad 

(b)

5π  180 deg  5π =  rad   = 300° 3  3  π rad 

5. (a) The complement of

π 2

−

3π 8π 3π 5π . = − = 16 16 16 16

(b) The supplement of

π −

3π is 16

5π is 12

5π 12π 5π 7π . = − = 12 12 12 12

4π is 3

4π 4π 6π 2π . + 2π = − + = 3 3 3 3

6. To use the formula s = r θ , first convert 160° to radian

measure.  π rad  8π 160° = 160 deg  radians  =  180 deg  9  

(

)

Then, using a radius of r = 27 inches, you can find the arc length to be s = rθ  8π  = ( 27)   9  = 24π ≈ 75.40 inches.

7. In one revolution, the arc length traveled is

s = 2π r = 2π (8) = 16π centimeters. The time required for the second hand to travel this distance is

t = 1 minute = 60 seconds. So, the linear speed of the tip of the second hand is

s t 16π centimeters = 60 seconds

Linear speed =

≈ 0.838 centimeter per second. 8. (a) Because each revolution generates 2π radians, it follows that the saw blade turns (2400)(2π ) = 4800π radians per minute.

In other words, the angular speed is

Angular speed =

t 4800π radians = 1 minute = 4800π radians per minute.

(b) The radius is r = 4. The linear speed is s t rθ = t 4)( 4800π ) inches ( = 1 minute

Linear speed =

= 60,319 inches per minute.

1028 Solutions to Checkpoints

Checkpoints for Section 5.2 1. 4

By the Pythagorean Theorem, ( hyp ) = (opp ) + (adj) , it follows that 2

adj =

42 − 22 =

12 = 2 3.

So, the six trigonometric functions of θ are as follows. sin θ =

opp 2 1 = = hyp 4 2

cos θ =

adj 2 3 = = hyp 4

3 2

tan θ =

opp 2 = = adj 2 3

1 = 3

3 3

csc θ =

hyp 4 = = 2 opp 2

sec θ =

hyp 4 = = adj 2 3

2 2 3 = 3 3

cot θ =

adj 2 3 = = opp 2

3. 45° 2

30° 2

60° 1

45° 1

For θ = 60°, you have adj = 1, opp =

adj 1 = =1 cot 45° = opp 1

3 and

hyp = 2.

sec 45° =

hyp = adj

2 = 1

csc 45° =

hyp = adj

2 = 1

So, tan 60° =

opp = adj

3 = 1

For θ = 30°, you have adj =

3. 3, opp = 1 and

hyp = 2. So, tan 30° =

opp = adj

1 = 3

3 . 3

Function

Mode

Graphing Calculator Keystrokes

Display

(a) sin 32.8°

Degree

SIN ( 32.8 ) ENTER

0.5417082

(b) csc 1.2

Radian

( SIN ( 1.2 ) ) x −1 ENTER

1.0729164

5. (a) To find the value of sin θ , use the Pythagorean

identity sin θ + cos θ = 1. So, you have 2

tan θ =

(0.28) + cos2 θ = 1 2

cos 2 θ = 1 − (0.28)

cos 2 θ = 1 − 0.0784 cos 2 θ = 0.9216 cosθ =

(b) Now, knowing the sine and cosine of θ , you can find the tangent of θ to be

0.9216

cosθ = 0.96.

sin θ cosθ

0.28 0.96 7 tan θ = 24 tan θ ≈ 0.2917. tan θ =

Solutions to Checkpoints 1029

6. (a) cot θ = =

1 tan θ

Reciprocal Identity

1 2

(b) sec2 θ = 1 + tan 2 θ

sec 2θ = 1 + ( 2)

Pythagorean identity

sec 2θ = 5 sec θ =

5 Use the definitions of cot θ and sec θ and the triangle to check these results.

θ 1

 sin θ  1    7. (a) tan θ csc θ =   cos θ  sin θ     1 = cos θ

Use a quotient identity and a reciprocal identity. Simplify.

= secθ

Reciprocal identity

(b) (csc θ + 1)(csc θ − 1) = csc2 θ − csc θ + csc θ − 1

Simplify.

= cot θ

Pythagorean identity

8. From the figure you can see that

tan 64.6° =

opp y = x adj

where x = 19 and y is the height of the flagpole. So, the height of the flagpole is y = 19 tan 64.6° ≈ 19( 2.106)

9. From the figure, you can see that the sine of the angle θ opp 200 3 = = hyp 400

10. From the figure, you can see that opp 3.5 sin 11.5° = = . hyp c

Solve the equation to find c. 3.5 sin 11.5° = c c sin 11.5° = 3.5

c =

≈ 40 feet.

is sin θ =

FOIL Method

= csc θ − 1 2

3 . 2

Now you can recognize that θ = 60°.

3.5 sin 11.5°

So, the length c of the loading ramp is 3.5 3.5 c = ≈ ≈ 17.6 feet. sin 11.5 0.1994 Also from the figure, you can see that opp 3.5 tan 11.5° = . = adj a So, the length a of the ramp is 3.5 3.5 a = ≈ ≈ 17.2 feet. tan 11.5° 0.2034

1030 Solutions to Checkpoints

Checkpoints for Section 5.3 1. Referring to the figure shown, you can see that x = − 2, y = 3, and

r =

x2 + y 2 =

(− 2) + (3) 2

13.

So, you have the following. sin θ =

y = r

3 3 13 = 13 13

(−2, 3)

x 2 2 13 cos θ = = − = − r 13 13 y 3 tan θ = = − x 2

2. Note that θ lies in Quadrant II because that is the only quadrant in which the sine is positive and the tangent is negative. Moreover, using

θ x

4. (a) Because 213° lies in Quadrant III, the angle it makes with the x-axis is θ ′ = 213° − 180° = 33°. y

4 y sin θ = = 5 r

and the fact that y is positive in Quadrant II, you can let y = 4 and r = 5. 2

x + y

5 =

x 2 + 42

θ ′ = 33°

r =

θ = 213°

25 = x 2 + 16 ±3 = x

14π lies in Quadrant IV, the angle it makes 9 with the x-axis is

Since x is negative in Quadrant II, x = − 3.

θ ′ = 2π −

(b) Because

9 = x2

So, cos θ =

x y 3 4 = − and tan θ = = − . r 5 x 3

3. To begin, choose a point on the terminal side of the angle 3π . 2 π 2

(0, − 1) 3π 2

θ=

14 π 9

4π θ ′= 9

4π lies in Quadrant II, the angle it makes 5 with the x-axis is

θ′ = π −

For the point (0, −1), r = 1 and you have the following. 3π y −1 = = = −1 2 1 r 3π 0 x cot = = = 0 2 −1 y

14π 18π 14π 4π . = − = 9 9 9 9

4π π = . 5 5 y

sin

θ ′=

π 5

θ=

4π 5 x

Solutions to Checkpoints 1031

5. (a) Because θ =

7π 7π π lies in Quadrant IV, the reference angle is θ ′ = 2π − as shown. = 4 4 4

π θ=7 4

θ ′=

π 4

Because the sine is negative in Quadrant IV, you have sin

7π π = ( −)sin 4 7 = −

2 . 2

(b) Because −120° + 360° = 240°, it follows that − 120° is coterminal with the third-quadrant angle 240°. So, the reference angle is θ ′ = 240° − 180° = 60° as shown. y

θ = − 120° θ ′ = 60°

1 Because the cosine is negative in Quadrant III, you have cos( −120°) = ( −)cos 60° = − . 2

11π 11π π lies in Quadrant IV, the reference angle is θ ′ = 2π − as shown. = 6 6 6

π θ = 11 6

θ ′=

π 6

Because the tangent is negative in Quadrant IV, you have tan

11π π 3 = ( −) tan = − . 6 6 3

1032 Solutions to Checkpoints 6. (a) Using the Pythagorean Identity sin 2 θ + cos 2 θ = 1, you obtain the following. sin 2 θ + cos 2 θ = 1

Write identity.

 4 2  −  + cos θ = 1  5 16 + cos 2 θ = 1 25 cos 2 θ = 1 − cos 2 θ =

9 25

Substitute −

4 for sin θ . 5

Simplify. 16 25

Subtract

16 from each side. 25

Simplify.

Because cos θ < 0 in Quadrant III, you can use the negative root to obtain 9 3 = − . 25 5

cos θ = −

(b) Using the trigonometric identity tan θ = tan θ =

sin θ cos θ

−4 5 −3 5

4 3

sin θ , you obtain the following. cos θ

Function

Mode

Graphing Calculator Keystrokes

Display

(a) tan 119°

Degrees

TAN ( 119 ) ENTER

−1.8040478

(b) csc 5

Radian

( SIN ( 5 ) ) x −1 ENTER

−1.0428352

Radian

COS ( π ÷ 5 )

0.8090170

π 5

8. Algebraic Solution

Graphical Solution

Use a graphing utility set in parametric and radian modes to graph X IT = sin T and YIT = cos T. (a)

f (t ) = cos t f ( 2) = cos 2 ≈ 0.4161

1.5

− 1.5

Use the trace feature to estimate that sin 2 ≈ 0.4161.

−1

(b)

f (t ) = cos t

π π  f   = cos 3 3 1 = 2

1.5

−1.5

−1

Use the trace feature to estimate that cos

π 3

≈ cos 1.047 = 0.5.

Solutions to Checkpoints 1033

Checkpoints for Section 5.4 1. Note that y = 2 cos x = 2(cos x ) indicates that the y-values for the key points will have twice the magnitude of those on the graph of y = cos x. Divide the period 2π into four equal parts to get the key points. Maximum

Intercept

Minimum

Intercept

Maximum

(0, 2)

π   , 0 2 

(π , − 2)

 3π   , 0  2 

(2π , 2)

By connecting these key points with a smooth curve and extending the curve in both directions over the  π 9π  interval − ,  , you obtain the graph shown.  2 2

y 3 2

2π

3π

4π

−2 −3

2. (a) Because the amplitude of y =

1 1 1 1 sin x is , the maximum value is and the minimum value is − . 3 3 3 3

Divide one cycle, 0 ≤ x ≤ 2 π , into for equal parts to get the key points. Intercept

Maximum

Intercept

Minimum

Intercept

(0, 0)

π 1  ,   2 3

(π , 0)

 3π 1   ,−  3  2

(2π , 0)

(b) A similar analysis shows that the amplitude of y = 3 sin x is 3, and the key points are as follows. Intercept

Maximum

Intercept

Minimum

Intercept

(0, 0)

π   , 3 2 

(π , 0)

 3π   , − 3 2  

(2π , 0)

y 4

y = 3sin x y = 1 sin x 3

π 2

−4

5π 2

1034 Solutions to Checkpoints

3. The amplitude of y = cos

x 1 is 1. Moreover, because b = , the period is 3 3

2π 2π 1 = = 6π . Substitute for b. 1 b 3 3 Now, divide the period-interval [0, 6π ] into four equal parts using the values

3π 9π to obtain the key points. , 3π , and 2 2

Maximum

Intercept

Minimum

Intercept

Maximum

(0, 1)

 3π   , 0  2 

(3π , −1)

 9π   , 0  2 

(6π , 1)

y 2

−π

3π

5π

−1 −2

π  4. y = 2 cos x −  2  Algebraic Solution

The amplitude is 2 and the period is 2π. By solving the equations x −

and

x −

π 2

= 0  x =

π 2

= 2π  x =

5π 2

π 5π  you see that the interval  ,  corresponds to one cycle of the graph. Dividing this interval into four equal parts produces 2 2  the key points.

Maximum

Intercept

Minimum

Intercept

Maximum

π   , 2 2 

(π , 0)

 3π   , − 2 2  

(2π , 0)

 5π   , 2  2 

Graphical Solution

π  Use a graphing utility set in radian mode to graph y = 2 cos  x −  as shown. 2  Use the minimum, maximum, and zero or root features of the graphing utility to approximate the key points

(1.57, 2), (3.14, 0), ( 4.71, − 2), (6.28, 0) and (7.85, 2). 3

−π 2

3π

−3

( 2(

y = 2cos x − π

Solutions to Checkpoints 1035

5. y = −

1 sin (π x + π ) 2

Algebraic Solution 1 2π 2π and the period is = = 2. 2 b π

The amplitude is

By solving the equations

πx + π = 0 π x = −π x = −1

π x + π = 2π

and

πx = π x =1 you see that the interval [−1, 1] corresponds to one cycle of the graph. Dividing this into four equal parts produces the key points. Intercept

Minimum

( −1, 0)

 1 1 − , −   2 2

Intercept

Maximum

Intercept

(0, 0)

1 1  ,  2 2

(1, 0)

Graphical Solution 1 Use a graphing utility set in radian mode to graph y = − sin (π x + π ). 2

Use the minimum, maximum, and zero or root features of the graphing utility to approximate the key points

(−1, 0),  − , − , (0, 0),  , 1  2

−2

1 1 , and (1, 0). 2 2

1 2

y = − 1 sin(π x + π) 2

−2

6. The amplitude of y = 2 cos x − 5 is 2 and the period is 2π . The key points over the interval [0, 2π ] are

π 3π  − 5 , and ( 2π , − 3). Compared with the graph of f ( x) = 2 cos x, the graph 2   2  of y = 2 cos x − 5 is shifted downward five units.

(0, − 3),  , − 5, (π , − 7),  2 2π

−2π

−8

y = 2cos x − 5

7. Sample answer: The amplitude of this cosine curve is 2. The period is 2π , and there is a right phase shift of π . So, you can

write y = 2cos ( x − π ).

1036 Solutions to Checkpoints 8. Use a sine model of the form y = a sin (bt − c ) + d .

The difference between the maximum value and minimum value is twice the amplitude of the function. So, the amplitude is

1 ( maximum depth ) − ( minimum depth ) 2 1 = (11.3 − 0.1) = 5.6. 2 The sine function completes one half cycle between the times at which the maximum and minimum depths occur. So, the period p is p = 2 ( time of min. depth ) − ( time of max. depth ) a =

= 2(10 − 4) = 12

which implies that b = bt − c =

π 2

2π ≈ 0.524. Because high tide occurs 4 hours after midnight, consider the maximum to be p

≈ 1.571.

So, (0.524)( 4) − c ≈ 1.571

c ≈ 0.525. Because the average depth is

1 (11.3 + 0.1) = 5.7, it follows that d = 5.7. So, you can model the depth with the function 2

y = a sin (bt − c) + d = 5.6 sin (0.524t − 0.525) + 5.7. 12

y = 5.6 sin(0.524t − 0.525) + 5.7

Checkpoints for Section 5.5 2. y = tan 2 x

x 4 By solving the equations

1. y = tan

By solving the equations 2x = −

π π x x = − and = 4 2 4 2 x = − 2π x = 2π

x = −

you can see that two consecutive vertical asymptotes occur at x = − 2π and x = 2π . Between these two asymptotes, plot a few points including the x-intercept. − 2π

f ( x)

Undef.

−π

−1

2π Undef.

and 2 x =

x = 4 4 you can see that two consecutive vertical asymptotes π π occur at x = − and x = . Between these two 4 4 asymptotes, plot a few points including the x-intercept. −

tan 2x

Undef.

−

−1

−4

−3π

5π

π 2

Undef.

Solutions to Checkpoints 1037

3. y = cot

4 By solving the equations x x = 0 and = π 4 4 x = 0 x = 4π you can see that two consecutive vertical asymptotes occur at x = 0 and x = 4π . Between these two asymptotes, plot a few points, including the x-intercept. x

cot

x 4

2π

3π

4π

Undef.

−1

Undef.

5. y = sec

x as indicated 2 x by the gray curve. Then, form the graph of y = sec as 2 the black curve. Note that the x-intercepts of x y = cos , (π , 0), (3π , 0), (5π , 0),  correspond to the 2 vertical asymptotes x = π , x = 3π , x = 5π ,  of the

Begin by sketching the graph of y = cos

x graph of y = sec . Moreover, notice that the period of 2 2π x x = 4π . y = cos and y = sec is 1 2 2 2

(2(

y = sec π

4 3

− 2π

x 2

5π

2π

−1

−4

= 2π 2 π 3π x = − x = 2 2 you can see that one cycle of the sine function π 3π corresponds to the interval from x = − to x = . 2 2 The graph of this sine function is represented by the gray curve. Because the sine function is zero at the midpoint and endpoints of this interval, the corresponding cosecant function π  y = 2 csc  x +  y = 2csc(x + π ( y 2 2    3   1 y = 2sin (x + π ( 2  = 2 1 π    + sin x    x 2   π   −1 2

= 0 and x +

has vertical asymptotes at π π 3π x = − ,x = ,x = , 2 2 2 and so on. The graph of the cosecant curve is represented by the black curve.

((

−3

π  Begin by sketching the graph of y = 2 sin  x + . For 2  this function, the amplitude is 2 and the period is 2π. By solving the equations x+

y = cos x 2

−2

π  4. y = 2 csc x +  2 

4π

6. f ( x ) = e x sin 4 x

Consider f ( x) as the product of these two functions y = e x and y = sin 4 x

each of which has a set of real numbers as its domain. For any real number x, you know that e x sin 4 x ≤ e x which means that − e x ≤ e x sin 4 x ≤ e x . Furthermore, because f ( x ) = e x sin 4 x = ± e x at x = sin 4 x = ± 1 at 4 x =

π 2

π 8

nπ since 4

+ nπ

and f ( x) = e x sin 4 x = 0 at x =

nπ since sin 4 x = 0 at 4

4x = nπ the graph of f touches the curve y = − e x or y = e x at x =

π 8

nπ nπ and has x-intercepts at x = . 4 4 8

y = ex

−3

7 π 8

− 7π 8

y = −e x −8

f(x) = e xsin 4x

1038 Solutions to Checkpoints

Checkpoints for Section 5.6 1. (a) Because sin

π 2

π  π π lies in − , , it follows that arcsin 1 = . 2 2  2 2

= 1, and

(b) It is not possible to evaluate y = sin −1 x when x = − 2 because there is no angle whose sine is − 2. Remember that the domain of the inverse sine function is [−1, 1]. 2. Using a graphing utility, you can graph the three functions with the following keystrokes.

Function

Keystroke

Display

f ( x) = sin x

y = SIN ( x )

y1 = sin ( x)

g ( x) = arcsin x

y = 2ND SIN ( x )

y2 = sin −1 ( x )

y = x

y3 = x

Remember to check the mode to make sure the angle measure is set to radian mode. Although the graphing utility will  π π graph the sine function for all real values of x, restrict the viewing window to values of x to be the interval − , .  2 2 1

−π

 π π Notice that the graphs of y1 = sin x,  − ,  and y2 = sin −1 x are  2 2 reflections of each other in the line y3 = x. So, g is the inverse of f.

−1

 2 3π 2 3π  3π  and 3. (a) Because cos   = − lies in [0, π ], it follows that cos −1  − .  = 4 2 4 2 4     π  (b) Because tan   = 3

3 and

π 3

 π π lies in − ,  , it follows that tan −1  2 2

( 3) = π3 .

Function

Mode

Graphing Calculator Keystrokes

(a) arctan 4.84

Radian

TAN −1 ( 4.84 ) ENTER

From the display, it follows that arctan 4.84 ≈ 1.3670516. (b) arcsin ( −1.1)

Radian

SIN −1 ( ( −) 1.1 ) ENTER

In radian mode the calculator should display an error message because the domain of the inverse sine function is [−1, 1]. (c) arccos ( − 0.349)

Radian

COS−1 ( ( −) 0.349 ) ENTER

From the display, it follows that arccos ( − 0.349) ≈ 1.9273001. 5. Because g ( x ) = arccos ( x + 1) = f ( x + 1), the graph of g can be obtained by shifting the graph of f one unit to the left. f(x) = arccos x π

g(x) = arccos (x + 1) −2

Solutions to Checkpoints 1039 6. (a) Because −14 lies in the domain of the arctangent

function, the inverse property applies, and you have tan tan −1 ( −14) = −14. (b) In this case,

7π does not lie in the range of the 4

arcsine function, −

π 2

≤ y ≤

π 2

7. Algebraic Solution 3  3 If you let u = arctan  − , then tan u = − . Because 4 4   the range of the inverse tangent function is the first and fourth quadrants and tan u is negative, u is a fourth-quadrant angle. You can sketch and label angle u.

7π 7π π is coterminal with − 2π = − 4 4 4 which does lie in the range of the arcsine function, and you have

3 Angle whose tangent is − . 4

However,

4 u

7π    π  −1  sin −1  sin  = sin sin  −  4     4 

−3

= − . 4 (c) Because 0.54 lies in the domain of the arccosine function, the inverse property applies and you have cos(arccos 0.54) = 0.54.

42 + ( − 3)

= 5

 4  3  So, cos arctan  −  = cos u = . 5  4  

Graphical Solution

Use a graphing utility set in radian mode to graph y = cos(arctan x ). y = cos (arctan x) 2

−3

−2

When x = −

3 4 = − 0.75, y = 0.8 = . 4 5

opp x = , you can sketch a right adj 1 triangle with acute angle u as shown. From this triangle, you can convert to algebraic form.

8. If you let u = arctan x, then tan u = x, where x is any real number. Because tan u = sec(arctan x) = sec u 1 2 + x

x2 + 1 1

x2 + 1

u = arctan x 1

Checkpoints for Section 5.7 1. Because c = 90°, it follows that A + B = 90° and B° = 90° − 20° = 70°.

To solve for a, use the fact that opp a tan A = =  a = b tan A. adj b So, a = 15 tan 20° ≈ 5.46. Similarly, to solve for c, use the fact that cos A = So, c =

15 ≈ 15.96. cos 20°

adj b b  c = = . hyp c cos A

1040 Solutions to Checkpoints

1600 ft 100 ft

16 feet

Not drawn to scale

Using the tangent function, you can see that 80°

tan A =

a From the equation sin A = , it follows that c

So, the angle of depression is

a = c sin A

A = arctan (0.0625) radian

= 16 sin 80°

≈ 0.06242 radian

≈ 15.8. So, the height from the top of the ladder to the ground is about 15.8 feet. 3.

opp 100 = = 0.0625 adj 1600

≈ 3.58°.

5. D

(1260 ((10nm( = 2nm

b C

16°

( 41 ((8nm( = 2nm

For triangle BCD, you have B = 90° − 16° = 74°.

43°

The two sides of this triangle can be determined to be

35° 65 ft

b = 2 sin 74° and d = 2 cos 74°.

Note that this problem involves two right triangles. For the smaller right triangle, use the fact that a to conclude that the height of the church tan 35° = 65 is a = 65 tan 35°. For the larger right triangle use the equation a + s to conclude that a + s = 65 tan 43°. tan 43° = 65 So, the height of the steeple is

For triangle ACD, you can find angle A as follows. tan A =

= 65 tan 43° − (65 tan 35°)

The angle with the north south line is 90° − 37° = 53°. So, the bearing of the ship is N 53° W. Finally, from triangle ACD you have

c =

≈ 15.1 feet.

b 2 sin 74° = ≈ 0.7535541 d + 2 2 cos 74° + 2

A = arctan A ≈ arctan 0.7535541 radian ≈ 37°

sin A =

s = 65 tan 43° − a

b , which yields c

b 2 sin 74° = ≈ 3.2 nautical miles. sin A sin 37°

6. Because the spring is at equilibrium ( d = 0) when t = 0, use the equation d = a sin wt .

Because the maximum displacement from zero is 6 and the period is 3, you have the following. Amplitude = a = 6 Period =

2π 2π = 3  w = w 3

So, an equation of motion is d = 6 sin

2π t. 3

The frequency is

2π w 1 Frequency = = 3 = cycle per second. 2π 2π 3 .

Solutions to Checkpoints

1041

7. Algebraic Solution

The given equation has the form d = 4 cos 6π t , with a ≈ 4 and w = 6π . (a) The maximum displacement is given by the amplitude. So, the maximum displacement is 4. (b) Frequency =

6π w = = 3 cycles per unit of time 2π 2π

(c) d = 4 cos 6π ( 4) = 4 cos 24π = 4(1) = 4 (d) To find the least positive value of t, for which d = 0, solve the equation 4 cos 6π t = 0. First divide each side by 4 to obtain cos 6π t = 0. This equation is satisfied when 6π t =

π 3π 5π 2

Divide each of these values by 6π to obtain t = So, the least positive value of t is t =

,.

1 1 5 , , ,. 12 4 12

1 . 12

Graphical Solution

(a) Use a graphing utility set in radian mode. d = 4cos 6π t

−0.25

−6

The maximum displacement is from the point of equilibrium ( d = 0) is 4. 6

d = 4cos 6π t

− 0.25

0.5

−6

t ≈ 0.333

(b) The period is the time for the graph to complete one cycle, which is t ≈ 0.333. So, the frequency is about 1 ≈ 3 per unit of time. 0.333 (c)

d = 4cos 6π t

3.5

4.5

−6

The value of d when t = 4 is d = 4. (d)

t ≈ 0.083

0.5

−0.25

−6

The least positive value of t for which d = 0 is t ≈ 0.083.

1042 Solutions to Checkpoints

Chapter 6 Checkpoints for Section 6.1 1. Using a reciprocal identity, you have cot x =

1 1 = = 3. 1 tan x 3

Using a Pythagorean identity, you have 2

1 10 1 sec 2 x = 1 + tan 2 x = 1 +   = 1 + . = 3 9 9  

Because tan x > 0 and cos x < 0, you know that the angle x lies in Quadrant III. Moreover, because sec x is negative when x is in Quadrant III, choose the negative root and obtain sec x = −

10 10 = − . 9 3

Using a reciprocal identity, you have cos x =

1 = − sec x

1 3 3 10 . = − = − 10 10 3 10

Using a quotient identity, you have tan x =

 3 10  1  sin x 10  sin x = cos x tan x =  −  3  = − 10 . cos x 10  

Using a reciprocal identity, you have csc x =

1 = − sin x

1 10 = − = − 10. 10 10 10

sin x = −

10 10

csc x = − 10

cos x = −

3 10 10

sec x = −

tan x =

1 3

10 3

cot x = 3

2. First factor out a common monomial factor then use a fundamental identity. cos 2 x csc x − csc x = csc x(cos 2 x − 1)

Factor out a common monomial factor.

= − csc x(1 − cos x)

Factor out −1.

= − csc x sin 2 x

Pythagorean identity

 1  2 = − sin x  sin x  = − sin x

Reciprocal identity Multiply.

3. (a) This expression has the form u 2 − v 2 , which is the difference of two squares. It factors as 1 − cos 2 θ = (1 − cos θ )(1 + cos θ ).

(b) This expression has the polynomial form ax 2 + bx + c, and it factors as 2csc 2 θ − 7csc θ + 6 = ( 2csc θ − 3)(csc θ − 2).

Solutions to Checkpoints

1043

4. Use the identity sec 2 x = 1 + tan 2 x to rewrite the expression.

sec 2 x + 3tan x + 1 = (1 + tan 2 x) + 3tan x + 1

Pythagorean identity

= tan x + 3tan x + 2

Combine like terms.

= ( tan x + 2)( tan x + 1)

Factor.

 cos x  1 − cos x  sin x  sin x 

Quotient and reciprocal identities

1 cos 2 x − sin x sin x

Multiply.

1 − cos 2 x sin x

Add fractions.

sin 2 x sin x

Pythagorean identity

5. csc x − cos x cot x =

= sin x 6.

Simplify.

cos 2 θ 1 − sin 2 θ = 1 − sin θ 1 − sin θ =

Pythagorean identity

(1 + sin θ )(1 − sin θ ) 1 − sin θ

= 1 + sin θ

Factor the numerator as the difference of squares. Simplify.

7. Begin by letting x = 3sin x, then you obtain the following.

9 − x2 =

9 − (3sin θ )

9 − 9sin 2 θ

Rule of exponents

9(1 − sin 2 θ )

Factor.

9cos 2 θ

Pythagorean identity

= 3cos θ

Substitute 3sin θ for x.

cos θ > 0 for 0 < θ =

π 2

Checkpoints for Section 6.2 1. Start with the left side because it is more complicated.

sin 2 θ + cos 2 θ 1 = cos 2 θ sec 2 θ cos 2 θ sec 2 θ =

1  1  cos 2 θ   2  cos θ 

Pythagorean identity Reciprocal identity

Simplify.

1044 Solutions to Checkpoints 2. Algebraic Solution

Start with the right side because it is more complicated. 1 1 1 + cos β + 1 − cos β + = 1 − cos β 1 + cos β (1 − cos β )(1 + cos β )

Add fractions.

2 1 − cos 2 β

Simplify.

2 sin 2 β

Pythagorean identity

= 2csc 2 β

Reciprocal identity

Numerical Solution

Use a graphing utility to create a table that shows the values of y1 = 2csc 2 x and y2 =

1 1 for different values of x. + 1 − cos x 1 + cos x

The values for y1 and y2 appear to be identical, so the equation appears to be an identity. 3. Algebraic Solution

By applying identities before multiplying, you obtain the following.

(sec2 x − 1)(sin 2 x − 1) = (tan 2 x)(− cos2 x)

Pythagorean identities

 sin x  2 =   ( − cos x)  cos x 

Quotient identity

 sin 2 x  2 =  ( − cos x) 2  cos x 

Property of exponents

= − sin 2 x

Multiply.

Graphical Solution

Using a graphing utility, let y1 = (sec 2 x − 1)(sin 2 x − 1) and y2 = − sin 2 x. y1 =

( cos1 x − 1( (sin x − 1) 2

1 −2π

2π

−3

y2 = −sin 2 x

Because the graphs appear to coincide, the given equation (sec 2 x − 1)(sin 2 x − 1) = − sin 2 x appears to be an identity.

Solutions to Checkpoints

1045

4. Convert the left into sines and cosines.

csc x − sin x =

1 − sin x sin x

1 − sin 2 x sin x

Add fractions.

cos 2 x sin x

Pythagorean identity

 cos x  cos x  =   Product of fractions   1  sin x  = cos x cot x

Quotient identity

5. Algebraic Solution Begin with the right side and create a monomial denominator by multiplying the numerator and denominator by 1 + cos x.

sin x sin x  1 + cos x  =   1 − cos x 1 − cos x  1 + cos x 

Multiply numerator and denomintor by 1 + cos x.

sin x + sin x cos x 1 − cos 2 x

Multiply.

sin x + sin x cos x sin 2 x

Pythagorean identity

sin x sin x cos x + sin 2 x sin 2 x

Write as separate functions.

1 cos x + sin x sin x

Simplify.

= csc x + cot x

Identities

Graphical Solution

Using a graphing utility, let y1 = csc x + cot x and y2 =

sin x . 1 − cos x

y1 = csc x + cot x 4

−2

−4

y2 =

sin x 1 − cos x

Because the graphs appear to coincide, the given equation appears to be an identity.

1046 Solutions to Checkpoints 6. Algebraic Solution

Working with the left side, you have the following. tan 2 θ sec 2 θ − 1 = 1 + sec θ sec θ + 1 =

Pythagorean identity

(sec θ + 1)(sec θ − 1) sec θ + 1

= sec θ − 1

Factor. Simplify.

Now, working with the right side, you have the following. 1 − cos θ 1 cos θ = − cos θ cos θ cos θ = sec θ − 1

Write as separate fractions. Identity and simplify.

This verifies the identity because both sides are equal to sec θ − 1. Numerical Solution

Use a graphing utility to create a table that shows the values of

y1 =

tan 2 x 1 − cos x and y2 = for different values of x. 1 + sec x cos x

The values of y1 and y2 appear to be identical, so the equation appears to be an identity. 7. tan x sec 2 x − tan x = tan x(sec 2 x − 1) = tan x tan 2 x 3

= tan x

Factor. Pythagorean identity Multiply.

Checkpoints for Section 6.3 1. sin x + 1 = 0

sin x = −1

Write original equation. Subtract 1 from each side.

To solve for x, note in the figure that the equation sin x = −1 has the solution x =

3π in the interval [0, 2π ). 2

Since sin x has a period of 2π , there are infinitely many other solutions, which can be written as x =

3π + 2 nπ , where n is 2

an integer. y

y = sin x 1

−

π 2

3π 2

7π 2 y = −1

Solutions to Checkpoints

1047

2. Algebraic Solution

Begin by isolating sin x on one side of the equation. sin x − sin x + sin x −

2 = − sin x

Write original equation.

2 = 0

Add sin x to each side.

sin x + sin x =

Add

2sin x =

Combine like terms.

sin x =

2 2

Divide each side by 2.

The solutions in [0, 2π ) are x =

π 4

2 each side.

and x =

3π . 4

Numerical Solution

Use a graphing utility set in radian mode to create a table that shows the values of y1 = sin x −

2 and y2 = − sin x for

different values of x. Your table should go from x = 0 to x = 2π using increments of π 8.

The values of y1 and y2 appear to be identical when x ≈ 0.7854 ≈ π 4 and x ≈ 2.3562 ≈ 3π 4. These values are the approximate solutions of sin x −

2 = − sin x.

3. Begin by isolating sin x on one side of the equation.

4sin 2 x − 3 = 0

Write original equation.

4sin x = 3 3 sin 2 x = 4

Add 3 to each side. Divide each side by 4.

sin x = ±

3 4

Extract square roots.

sin x = ±

3 2

Simplify.

Because sin x has a period of 2π , first find all solutions in the interval [0, 2π ). These solutions are x = x =

π 3

,x =

2π , 3

4π 5π , and x = . 3 3

Finally, add multiples of 2π to each of these solutions to obtain the general form. π 2π 4π 5π x = + 2 nπ , x = + 2 nπ , x = + 2 nπ , and x = + 2 nπ where n is an integer. 3 3 3 3

1048 Solutions to Checkpoints 4. Begin by collecting all terms on one side of the equation and factoring.

sin 2 x = 2sin x

Write original equation.

sin x − 2sin x = 0

Subtract 2sin x from each side.

sin x(sin x − 2) = 0

Factor.

By setting each of these factors equal to zero, you obtain sin x = 0 and sin x − 2 = 0

sin x = 2. In the interval [0, 2π ), the equation sin x = 0 has solutions x = 0 and x = π . Because sin x has a period of 2π , you would obtain the general forms x = 0 + 2 nπ and x = π + 2 nπ where n is an integer by adding multiples of 2π . No solution exists for sin x = 2 because 2 is outside the range of the sine function, [−1, 1]. Confirm this graphically by graphing y = sin 2 x − 2sin x. y

y = sin 2 x − 2 sin x

Notice that the x-intercepts occur at − 2π , − π , 0, π , 2π and so on.

These x-intercepts correspond to the solutions of sin 2 x − 2sin x = 0.

−π 2

3π 2

−2

5. Algebraic Solution

Treat the equation as a quadratic in sin x and factor. 2sin 2 x − 3sin x + 1 = 0

Write original equation.

(2sin x − 1)(sin x − 1) = 0

Factor.

Graphical Solution The x-intercepts are x ≈ 0.524, x = 2.618, and x = 1.571. 6

Setting each factor equal to zero, you obtain the following solutions in the interval [0, 2π ). 2sin x − 1 = 0 sin x = x =

sin x − 1 = 0

and

1 2 6

x =

3 −0.5

From the graph, you can conclude that the approximate solutions of 2sin 2 x − 3sin x + 1 = 0 in the interval

sin x = 1

π 5π

−

[0, 2π ) are x ≈ 0.524 =

π 2

x ≈ 1.571 =

π 2

π 6

, x ≈ 2.618 =

5π , and 6

6. This equation contains both tangent and secant functions. You can rewrite the equation so that it has only tangent functions by using the identity sec 2 x = tan 2 x + 1.

3sec 2 x − 2 tan 2 x − 4 = 0 3( tan x + 1) − 2 tan x − 4 = 0

Write original equation.

Pythagorean identity

Distributive property

3tan x + 3 − 2 tan x − 4 = 0 2

tan x − 1 = 0 2

Simplify.

tan x = 1

Add 1 to each side.

tan x = ±1

Extract square roots.

Because tan x has a period of π , you can find the solutions in the interval [0, π ) to be x = The general solution is x =

π 4

+ nπ and x =

π 4

and x =

3π . 4

3π + nπ where n is an integer. 4

Solutions to Checkpoints

1049

7. It is not clear how to rewrite this equation in terms of a single trigonometric function. Notice what happens when you square each side of the equation.

sin x + 1 = cos x 2

Write original equation.

sin x + 2sin x + 1 = cos x 2

Square each side. 2

sin x + 2sin x + 1 = 1 − sin x 2

Pythagorean identity

sin x + sin x + 2sin x + 1 − 1 = 0

Rewrite equation.

2sin x + 2sin x = 0

Combine like terms.

2sin x(sin x + 1) = 0

Factor.

Setting each factor equal to zero produces the following.

2sin x = 0

sin x + 1 = 0

and

sin x = 0

sin x = −1

x = 0, π

x =

3π 2

Because you squared the original equation, check for extraneous solutions. ?

Check x = 0:

Substitute 0 for x.

0 +1=1

Solution checks. 

Check x = π :

Check x =

sin 0 + 1 = cos 0

sin π + 1 = cos π

3π : 2

Substitute π for x.

0 + 1 ≠ −1

Solution does not check.

? 3π 3π + 1 = cos 2 2

Substitute

−1 + 1 = 0

Solution checks. 

sin

3π for x. 2

Of the three possible solutions, x = π is extraneous. So, in the interval [0, 2π ), the two solutions are x = 0 and x = 8. 2sin 2t −

3 = 0

Write original equation.

2sin 2t =

Add

sin 2t =

3 2

Divide each side by 2.

9. 2 tan

3 to each side.

In the interval [0, 2π ), you know that 2t =

π 3

and 2t =

2π are the only solutions. 3

π 3

+ 2nπ and 2t =

2π + 2 nπ . 3

Dividing these results by 2, you obtain the general solution t =

π 6

+ nπ and t =

π 3

Write original equation. Add 2 to each side. Divide each side by 2.

In the interval [0, π ), you know that

π x is the only = 2 4

solution. So, in general, you have

So, in general you have 2t =

x − 2 = 0 2 x 2 tan = 2 2 x tan = 1 2

3π . 2

+ nπ .

π x = + nπ . 2 4 Multiplying this result by 2, you obtain the general solution x =

π 2

+ 2nπ , where n is an integer.

1050 Solutions to Checkpoints 4 tan 2 x + 5 tan x − 6 = 0

10.

Write original equation.

(4 tan x − 3)( tan x + 2) = 0

Factor.

4 tan x − 3 = 0 and tan x + 2 = 0 3 4

tan x =

Set each factor equal to zero.

tan x = − 2

 3 x = arctan   4

x = arctan ( − 2)

Use inverse tangent function to solve for x.

 π π  π π These two solutions are in the interval  − , . Recall that the range of the inverse tangent function is  − , .  2 2  2 2

Finally, because tan x has a period of p, you add multiples of p to obtain 3 x = arctan   + nπ and x = arctan ( − 2) + nπ , where n is an integer. 4  3 You can use a calculator to approximate the values of x = arctan  ≈ 0.6435 and x = arctan ( − 2) ≈ −1.1071.  4

11. Use a graphing utility to graph y = x + 4sin 2 x in the interval [−π , π ]. Using the zero or root feature, you can see that the

solutions are x ≈ − 2.7607, x = 0, x ≈ −1.8049, x ≈ 1.8049, and x ≈ 2.7607. 10

−2

−10

−2

−10

−2

−10

12. (a) Let h = 3.2, s = 0.75, and S = 15.8. S = 6hs +

3 2 s  2 

3 − cosθ   sin θ 

15.8 = 6(3.2)(0.75) +

3 2 (0.75)  2 

 15.8 = 14.4 + (0.84375)   0 = (0.84375) 

3 − cosθ   sin θ 

3 − cosθ   − 1.4 sin θ 

 Use a graphing utility set in degree mode to graph the function y = (0.84375)  0.06 0

−0.24

3 − cosθ   − 1.4. sin θ 

0.06 100

100

−0.24

Using the zero or root feature, you can determine that the surface area is 15.8 square inches when θ ≈ 32.3° and θ ≈ 85.5°.

Solutions to Checkpoints

1051

 3 − cosθ  (b) From part (a), let h = 3.2 and s = 0.75 to obtain S = 14.4 + (0.84375) . Graph this function using a sin θ   graphing utility set in degree mode. Use the minimum feature to approximate the minimum point on the graph, which  1  occurs at θ ≈ 54.7 ° . By using calculus, it can be shown that the exact minimum value is θ = arccos   ≈ 54.7356°.  3 30

100

Checkpoints for Section 6.4 1. (a) To find the exact value of cos

π 12

−

π 4

π 12

, use the fact that

2. Because sin u =

5 as shown. 13

cos u =

The formula for cos (u − v) yields the following. cos

π 12

π π = cos  −  4 3 = cos

π 3

cos

π 4

+ sin

π 3

sin

2 + 4

6 4

2 + 4

(b) Using the fact that 75° = 30° + 45°, together with the formula for sin (u + v), you obtain the following.

 1  2   3  2  =    +    2  2   2   2  =

12 and u is in Quadrant I, 13

Because cos v = −

3 4 and v is in Quadrant II, sin v = 5 5

as shown. y

5 v −3

sin 75° = sin (30° + 45°) = sin 30° cos 45° + cos 30° sin 45°  1  2   3  2  =   +    2  2   2   2  =

2 6 + 4 4

2 + 4

You can find cos (u + v) as follows.

cos (u + v) = cos u cos v − sin u sin v  5  3   12  4  =   −  −     13  5   13  5  63 = − 65

1052 Solutions to Checkpoints 3. This expression fits the formula for sin (u + v). The figures show the angles u = arctan 1 and v = arccos x.

1 − x2

v 1

sin (u + v) = sin u cos v + cos u sin v = sin (arctan 1) cos (arccos x) + cos (arctan 1) sin (arccos x)

( 1− x )

 1   1  =  ( x ) +    2  2 = =

1 − x2 2

x + 2 x +

1 − x2 2

4. Using the formula for sin (u − v), you have

π π π  sin  x −  = sin x cos − cos x sin 2 2 2  = (sin x )(0) − (cos x)(1) = − cos x.

5. (a) Using the formula for sin (u − v), you have

6. Algebraic Solution

Using sum and difference formulas, rewrite the equation.

π 3π    sin  x +  + sin  x −  =1 2 2    π π 3π 3π sin x cos + cos x sin + sin x cos − cos x sin =1 2 2 2 2 (sin x)(0) + (cos x)(1) + (sin x)(0) − (cos x)(1) = 1 cos x + cos x = 1

π π  π  sin  3 − θ  = sin 3 cos θ − cos 3 sinθ 2 2  2 

2cos x = 1 cos x =

= ( −1)(cos θ ) − (0)(sin θ ) = − cos θ .

(b) Using the formula for tan (u − v), you have

tan θ − tan π  4 tan θ −  = π 4  1 + tan θ tan 4 tan θ − 1 = 1 + ( tan θ )(1)

1 2

So, the only solutions in the interval [0, 2π ) are x =

π 3

and x =

5π . 3

Graphical Solution

( )

(

)

y = sin x + π + sin x − 3π −1 2 2 2

2π

tan θ − 1 = . 1 + tan θ −4

The x-intercepts are x ≈ 1.047198 and x ≈ 5.235988. From the above figure, you can conclude that the approximate solutions in the interval [0, 2π ) are x ≈ 1.047198 =

π 3

and x ≈ 5.235988 =

5π . 3

Solutions to Checkpoints

1053

7. Using the formula for cos ( x + h), you have the following. cos ( x + h) − cos x h

cos x cos h − sin x sin h − cos x h cos x cos h − cos x − sin x sin h = h cos x (cos h − 1) − sin x sin h = h  cos h − 1   sin h  = cos x   − sin x   h    h  =

Checkpoints for Section 6.5 1. Begin by rewriting the equation so that it involves functions of x (rather than 2x). Then factor and solve. cos 2 x + cos x = 0

Write original equation.

Double-angle formula

2cos x − 1 + cos x = 0 2

2cos x + cos x − 1 = 0

Rearrange terms.

(2cos x − 1)(cos x + 1) = 0

Factor.

2cos x − 1 = 0

cos x + 1 = 0

1 2

cos x = −1

cos x = x =

π 5π 3

So, the general solution is x =

x = π

+ 2nπ , x =

2. Begin by drawing the angle θ, 0 < θ <

π 2

Set factors equal to zero. Solve by cos x. Solutions in [0, 2π ).

5π + 2 nπ , and x = π + 2 nπ , where n is an integer. 3

given sin θ =

3 . 5

5 4

(4, 3)

3 2

1 −1

−1

θ 1

From the sketch, you know that sin θ = Because x = 4, you know sin θ =

3 y = . r 5

3 4 3 , cos θ = , and tan θ = . 5 5 4

Using the double angle formulas, you have the following.

24  3  4  sin 2θ = 2sin θ cos θ = 2   = 25  5  5  2

7  4  3 cos2θ = cos 2 θ − sin 2 θ =   −   = 25 5 5 tan 2θ =

() ( 4)

3 2 34 2tan θ 24 = = 72 = 2 2 3 1 − tan θ 7 1− 16

1054 Solutions to Checkpoints 3. cos3 x = cos ( 2 x + x)

Rewrite 3x as sum of 2x and x.

= cos 2 x cos x − sin 2 x sin x

Sum formula

= ( 2cos 2 x − 1)(cos x) − ( 2sin x cos x)(sin x)

Double-angle formulas

= 2cos3 x − cos x − 2sin 2 x cos x

Distributive property and simplify.

= 2cos x − cos x − 2(1 − cos x)(cos x) 3

= 2cos x − cos x − 2cos x + 2cos x

Pythagorean identity Distributive property

= 4cos x − 3cos x

Simplify.

4. You can make repeated use of power-reducing formulas. tan 4 x = ( tan 2 x )

Property of exponents

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

Power-reducing formula

1 − 2 cos 2 x + cos 2 2 x 1 + 2 cos 2 x + cos 2 2 x

Expand.

 1 + cos 4x  1 − 2 cos 2 x +   2   =  1 + cos 4x  1 + 2 cos 2 x +   2   2 − 4 cos 2 x + 1 + cos 4 x 2 = 2 + 4 cos 2 x + 1 + cos 4 x 2 3 − 4 cos 2 x + cos 4 x = 3 + 4 cos 2 x + cos 4 x

Power-reducing formula

Simplify.

Collect like terms, invert, and multiply.

You can use a graphing utility to check this result. Notice that the graphs coincide. 4

y = tan 4 x

−2

y = 3 − 4cos 2x + cos 4x 3 + 4cos 2x + cos 4x

u 5. Begin by noting 105° is one-half of 210°. Then using the half-angle formula for cos   and the fact that 105° lies in 2 Quadrant II, you have the following.

cos 105° = −

1 + cos 210° = − 2

 3 1 +  −  2   = − 2

2− 3 2 = − 2

2− 3 = − 4

2− 2

The negative square root is chosen because cos θ is negative in Quadrant II.

Solutions to Checkpoints

1055

6. Algebraic Solution

cos 2 x = sin 2

x 2

Write original equation.

 1 − cos x  cos x =  ±  2   − x 1 cos cos 2 x = 2 2 cos 2 x = 1 − cos x

Half-angle formula Simplify. Multiply each side by 2.

2 cos x + cos x − 1 = 0

Simplify.

(2 cos x − 1)(cos x + 1) = 0

Factor.

2 cos x − 1 = 0 cos x = x =

cos x + 1 = 0

1 2

cos x = −1

π 5π 3

Set each factor equal to zero.

Solve each equation for cos x. Solutions in [0, 2π )

x = π

The solutions in the interval [0, 2π ) are x =

π 3

, x = π , and x =

5π . 3

Graphical Solution

Use a graphing utility to graph y = cos 2 x − sin 2

x in the interval [0, 2π ). Determine the approximate value of the 2

x-intercepts. 2

(2 )

y = cos 2 x − sin 2 x

The x-intercepts are x ≈ 1.04720, x ≈ 3.14159, and x ≈ 5.23599. 2π

−1

From the graph, you can conclude that the approximate solutions of cos 2 x = sin x ≈ 1.04720 =

π 3

, x ≈ 3.14159 = π , and x ≈ 5.23599 =

2x in the interval [0, 2π ) are 2

5π . 3

7. Using the appropriate product-to-sum formula sin u cos v =

1 sin (u + v ) + sin (u − v ) , you obtain the following. 2

1 sin (5 x + 3x ) + sin (5 x − 3 x ) 2 1 = (sin 8 x + sin 2 x) 2 1 1 = sin 8 x + sin 2 x 2 2

sin 5 x cos 3 x =

1056 Solutions to Checkpoints 8. Using the appropriate sum-to-product formula, u + v u − v sin u + sin v = 2sin   cos  , you obtain the following.  2   2 

 195° + 105°   195° − 105°  sin195° + sin105° = 2sin   cos   2 2     = 2sin150° cos 45°  1  2  = 2    2  2  2 2

sin 4 x − sin 2 x = 0

Write orignal equation.

 4x + 2x   4x − 2x  2cos   sin   = 0 2 2    

Sum-to-product formula

2cos3x sin x = 0

Simplify.

cos3 x sin x = 0

Divide each side by 2.

cos3 x = 0

sin x = 0 x = 0, π

cos3 x = 0

3π + 2nπ 2 π 2nπ π 2nπ and x = x = + + 6 3 2 3 π π 5π 7π 3π 11π , , , and x = , , 6 2 6 6 2 6

3x =

Set each factor equal to zero.

+ 2nπ and 3x =

So, all the solutions in the interval [0, 2π ) are x = 0, are x =

π 6

π π 5π

7π 3π 11π , , ,π, , , and . Notice that the general solutions 6 2 6 6 2 6

2nπ π 2nπ ,x = , and x = nπ , where n is an integer. + 3 2 3

To verify these solutions you can graph y = sin 4 x − sin 2 x and approximate the x-intercepts. y = sin 4x − sin 2x 3

−2π

2π

−2

The x-intercepts occur at 0,

π π 5π

7π , , , π, , 6 2 6 6

Solutions to Checkpoints

1057

Chapter 7 Checkpoints for Section 7.1 1. The third angle of the triangle is

3. Sketch and label the triangle as shown. B

C = 180° − A − B = 180° − 30° − 45° = 105°. c

By the Law of Sines, you have

a b c = = . sin A sin B sin C

Using a = 32 produces

By the Law of Sines, you have

b =

a 32 (sin B) = (sin 45°) ≈ 45.3 units sin A sin 30°

and a 32 c = (sin C ) = (sin 105°) ≈ 61.8 units. sin A sin 30°

31° 5 C

sin B sin A = b a  sin A  sin B = b   a   sin 31°  sin B = 5   12  B ≈ 12.39°

Reciprocal form Multiply each side by b. Substitute for A, a, and b.

Now, you can determine that 8°

C = 180° − A − B ≈ 180° − 31° − 12.39° ≈ 136.61°.

Then, the remaining side is 43°

c = 22 ft

c a = sin C sin A c =

In the figure, A = 43° and B = 90° − 8° = 82°. So, the third angle is C = 180° − A − B = 180° − 43° − 82° = 55°. Using the Law of Sines, you have

a 12 (sin C ) ≈ (sin 136.61°) sin A sin 31°

≈ 16.01 units.

4. Begin by making a sketch as shown.

a c . = sin A sin C

a=4

Because c = 22 feet, the length of pole is

b = 14

a 22 = sin 43° sin 55°

60°

22sin 43° a = sin 55°

It appears that no triangle is formed.

a ≈ 18.32 feet.

You can verify this using the Law of Sines.

sin B sin A = b a  sin A  sin B = b   a   sin 60°  sin B = 14  ≈ 3.0311 > 1  4  This contradicts the fact that sin B ≤ 1. So, no triangle can be formed having sides a = 4 and b = 14 and angle A = 60°.

1058 Solutions to Checkpoints 5. By the Law of Sines, you have

sin B sin A = b a  sin A   sin 58°  sin B = b  = 5  ≈ 0.9423.  a   4.5  There are two angles B1 ≈ 70.4° and B2 ≈ 180° − 70.4° = 109.6° between 0° and 180° whose sine is approximately 0.9423. For B1 ≈ 70.4°, you obtain the following. C = 180° − A − B1 = 180° − 58° − 70.4° = 51.6° 4.5 a (sin C ) = (sin 51.6°) ≈ 4.16 feet sin A sin 58°

c =

For B2 = 109.6°, you obtain the following. C = 180° − A − B2 = 180° − 58° − 109.6° = 12.4° a 4.5 (sin C ) = (sin 12.4°) ≈ 1.14 feet sin A sin 58°

c =

The resulting triangles are shown. C

C a = 4.5 ft

b = 5 ft

58° 70.4°

b = 5 ft 58° A

a = 4.5 ft 109.6° B2

6. Consider a = 24 yards, b = 18 yards, and angle C = 80° as shown. Then, the area of the triangle is

A =

1 1 ab sin C = ( 24)(18) sin 80° ≈ 212.7 square yards. 2 2

b = 18 yd 80° C

a = 24 yd

7. Because lines AC and BD are parallel, it follows that ∠ ACB ≅ ∠ CBD.

So, triangle ABC has the following measures as shown. The measure of angle B is 180° − A − C = 180° − 28° − 58° = 94°. D

C 58°

B 800 m 28°

Using the Law of Sines,

a b c = = . sin 28° sin 94° sin 58°

Because b = 800, C =

800 800 (sin 58°) ≈ 680.1 meters and a = (sin 28°) ≈ 376.5 meters sin 94° sin 94°

The total distance that you swim is approximately Distance = 680.1 + 376.5 + 800 = 1856.6 meters.

Solutions to Checkpoints

1059

Checkpoints for Section 7.2 1.

B c = 12 a=6 C

b=8

First, find the angle opposite the longest side – side c in this case. Using the alternative form of the Law of Cosines, you find that cos C =

a2 + b2 − c2 6 2 + 82 − 12 2 = ≈ − 0.4583. 2ab 2(6)(8)

Because cos C is negative, C is an obtuse angle given by C ≈ cos −1 ( − 0.4583) ≈ 117.28°.

At this point, it is simpler to use the Law of Sines to determine angle B.  sin C  sin B = b   c 

 sin 117.28°  sin B = 8   ≈ 0.5925 12  

Because C is obtuse and a triangle can have at most one obtuse angle, you know that B must be acute. So, B ≈ sin −1 (0.5925) ≈ 36.34° So, A = 180° − B − C ≈ 180° − 36.34° − 117.28° ≈ 26.38°. 2.

C b = 16 80° A c = 12

Use the Law of Cosines to find the unknown side a in the figure.

a 2 = b 2 + c 2 − 2bc cos A a 2 = 16 2 + 122 − 2(16)(12) cos 80° a 2 ≈ 333.3191 a ≈ 18.2570 Use the Law of Sines to find angle B.

sin B sin A = b a  sin A  sin B = b   a   sin 80°  sin B = 16   18.2570  sin B ≈ 0.8631 There are two angles between 0° and 180° whose sine is 0.8631. The two angles are B1 ≈ 59.67° and B2 ≈ 180° − 59.67° ≈ 120.33°. Because side a is the longest side of the triangle, angle A must be the largest angle, therefore B must be less than 80°. So, B ≈ 59.67°. Therefore, C = 180° − A − B ≈ 180° − 80° − 59.67° ≈ 40.33°.

1060 Solutions to Checkpoints 3. C

In triangle HCT, H = 45° (line HC bisects the right angle at H), t = 240, and c = 60. Using the Law of Cosines for this SAS case, you have h 2 = c 2 + t 2 − 2 ct cos H h 2 = 602 + 2402 − 2 (60)( 240) cos 45° h 2 ≈ 40835.3 h ≈ 202.1 So, the center fielder is approximately 202.1 feet from the third base.

t = 240 ft

60 ft

60 ft c

p = 60 ft

45° H

4. You have a = 30, b = 56, and c = 40.

So, using the alternative form of the Law of Cosines, you have cos B =

a +c −b 2 ac

30 + 40 − 56 = − 0.265. 2 (30)( 40)

So, B = cos −1 ( − 0.265) ≈ 105.37°, and thus the bearing from due north from point B to point C is 105.37° − 90° = 15.37°, or N 15.37° E.

5. Because s =

a +b + c 5+9+8 22 = = = 11, 2 2 2

Heron’s Area Formula yields Area =

s( s − a )( s − b)( s − c)

11(11 − 5)(11 − 9)(11 − 8)

(11)(6)( 2)(3)

396

≈ 19.90 square feet.

Checkpoints for Section 7.3   1. From the Distance Formula, it follows that PQ and RS have the same magnitude.

 PQ =

(3 − 0) + (1 − 0)

 RS =

(5 − 2) + (3 − 2)

Moreover, both line segments have the same direction because they are both directed toward the upper right on lines having a slope of 1−0 3− 2 1 = = . 3−0 5− 2 3   Because, PQ and RS have the same magnitude and direction, u and v are equivalent.

Solutions to Checkpoints

1061

2. Algebraic Solution

Let P( − 2, 3) = ( p1 , p2 ) and Q( − 7, 9) = ( q1 , q2 ). Then, the components of v = (v1 , v2 ) are

v1 = q1 − p1 = − 7 − ( − 2) = − 5 v2 = q2 − p2 = 9 − 3 = 6.

( − 5) + ( 6 ) 2

So, v = − 5, 6 and the magnitude of v is v =

61.

Graphical Solution

Use centimeter graph paper to plot the points P( − 2, 3) and Q( − 7, 9). Carefully sketch the vector v. Use the sketch to find the components of v = (v1 , v2 ). Then use a centimeter ruler to find the magnitude of v. The figure shows that the components of v are v1 = − 5 and v2 = 6, so v = − 5, 6 . The figure also shows that the magnitude of v is v =

61.

y 7

Q(−7, 9)

||v|| = √— 61

v2 = 6

P(−2, 3) v1 = −5 x

3. (a) The sum of u and v is

2u − 3v = 2 1, 4 − 3 3, 2

u + v = 1, 4 + 3, 2 = 1 + 3, 4 + 2

= 2, 8 − 9, 6

= 4, 6 .

= 2 − 9, 8 − 6

= − 7, 2 . y

8 6

(1, 4)

(4, 6)

u+v

(−7, 2)

−10 −8 −6 −4 −2

(0, 0)

(2, 8)

− 3v

2u − 3v

2u (0, 0) 2

−4 −6 −8

(b) The difference of u and v is

u + v = 1, 4 − 3, 2

= 1 − 3, 4 − 2

= − 2, 2 .

(1, 4)

−v (− 2, 2) u− v −4

−2

(0, 0) 2

1062 Solutions to Checkpoints 4. The unit vector in the direction of v is v = v

7. (a) The direction angle is determined from

6, −1

(6) + (−1) 2

b 6 = = −1. a −6

tan θ =

Because v = − 6i + 6 j lies in Quadrant II, θ lies in

1 6, −1 37

6 1 ,− 37 37

6 37 37 ,− . 37 37

Quadrant II and its reference angle is π θ ′ = arctan ( −1) = − = 45°. 4 So, it follows that the direction angle is θ = 180° − 45° = 135°. y

This vector has a magnitude of 1 because 2

1   6     + −  37 37    

(− 6, 6)

36 1 + = 37 37

−8

(−2, 6)

The result is shown graphically.

u = − 6i − 3j

(−8, 3)

−4

tan θ =

6 4

−2

b −4 4 = = . a −7 7

Because v = − 7 i − 4 j lies in Quadrant III, θ lies

− 12 − 10 − 8 − 6 − 4 − 2 −2

−6

(b) The direction angle is determined from

= − 6i − 3j

θ = 135°

−2

= − 6, − 3

θ ′ = 45°

5. Begin by writing the component form of vector u. u = − 8 − ( − 2), 3 − 6

37 = 1. 37

6. Perform the operations in unit vector form. 5u − 2 v = 5(i − 2 j) − 2( − 3i + 2 j) = 5i − 10 j + 6i − 4 j

in Quadrant III and its reference angle is  4 θ ′ = arctan   ≈ 0.51915 radian ≈ 29.74°. 7

So, it follows that the direction angle is θ = 180° + 29.74° = 209.74°. y

= 11i − 14 j

θ = 209.74° −7 −6 −5 −4 −3

θ ′ = 29.74° v

(−7, −4)

−1

−2 −3 −4 −5 −6 −7

8. The velocity vector v has a magnitude of 100 and a direction angle of θ = 225°.

v = v (cos θ )i + v (sin θ ) j = 100(cos 225°)i + 100(sin 225°) j   2 2 = 100 − i + 100 −  j  2   2  = − 50 2i − 50 2 j ≈ − 70.71i − 70.71j ≈ − 70.71, − 70.71 You can check that v has a magnitude of 100, as follows. v =

(− 50 2 ) + (− 50 2 ) 2

5000 + 5000

10,000 = 100

Solutions to Checkpoints

1063

B 12°

D 12° A

Solution Based in the figure, you can make the following observations.  BA = force of gravity = combined weight of boat and trailer  BC = force against ramp  AC = force required to move boat up ramp = 500 pounds By construction, triangles BWD and ABC are similar. So, angle ABC is 12°. In triangle ABC, you have  AC sin 12° =  BA 500 sin 12° =  BA

 BA =

500 sin 12°

 BA ≈ 2405.



So, the combined weight is approximately 2405 pounds. (In the figure, note that AC is parallel to the ramp.) y

150° x

Wi nd

10.

Using the figure, the velocity of the airplane (alone) is v1 = 450 cos 150°, sin 150° = − 225 3, 225 and the velocity of the wind is v 2 = 40 cos 60°, sin 60° = 20, 20 3 . So, the velocity of the airplane (in the wind) is v = v1 + v 2 = − 225 3 + 20, 225 + 20 3 ≈ − 369.7, 259.6 and the resultant speed of the airplane is v ≈

( − 369.7) + (259.6) 2

≈ 451.8 miles per hour.

Finally, given that θ is the direction angle of the flight path, you have tan θ ≈

259.6 ≈ 0.7022 − 369.7

which implies that θ ≈ 180° − 35.1° = 144.9°. So, the true direction of the airplane is approximately 270° + (180° − 144.9°) = 305.1°.

1064 Solutions to Checkpoints

Checkpoints for Section 7.4 1. u ⋅ v = 3, 4 ⋅ 2, − 3

3. cos θ =

= 3( 2) + 4( − 3) = 6 − 12

= −6

2. (a) Begin by finding the dot product of u and v. u ⋅ v = 3, 4 ⋅ − 2, 6 = 3( − 2) + 4(6) = − 6 + 24 = 18

(u ⋅ v) v = 18 − 2, 6 = − 36, 108 u + v = 3, 4 + − 2, 6 = 1, 10

u ⋅ (u + v ) = 3, 4 ⋅ 1, 10

= 43

(c) Begin by finding the dot product of v and v. v ⋅ v = − 2, 6 ⋅ − 2, 6 = − 2( − 2) + 6(6) = 40 = v ⋅ v = 40, it follows that

2 + 12

12 + 32

5 5 10 5 = 50 5 = 5 2 1 = 2 2 2

 1 1 = 6 −  + 10   3 5 = −2 + 2 = 0

= 4 + 36

2(1) + 1(3) 2

1 1 u ⋅ v = 6, 10 ⋅ − , 3 5

= 3 + 40

v⋅v

1, 3

4. Find the dot product of the two vectors.

= 3(1) + 4(10)

v =

2, 1

This implies that the angle between the two vectors is  2 π = 45°. θ = cos −1   = 2 4  

= 3 + ( − 2), 4 + 6

2, 1 ⋅ 1, 3

(b) Begin by finding u + v.

Because v

u⋅v = u v

Because the dot product is 0, the two vectors are orthogonal.

6 4 2 −2

−2

v −

1 5

1 3

= 2 10.

5. The projection of u onto v is u ⋅ v  3, 4 ⋅ 8, 2   3(8) + 4( 2)  64 16  32  8 v =  w1 = projv u =  8, 2 =   8, 2 =  68  8, 2 =  17  8, 2 = 17 , 17 .  8, 2 ⋅ 8, 2   v2  8 8 2 2 + ( ) ( )      

The other component, w2 is w 2 = u − w1 = 3, 4 −

So, u = w1 + w 2 =

64 16 , 17 17

= −

13 52 , . 17 17

64 16 13 52 , + − , 17 17 17 17

= 3, 4 .

Solutions to Checkpoints 6. Because the force due to gravity is vertical and downward, you can represent the gravitational force by the vector F = − 150 j.

Force due to gravity

30°

To find the force required to keep the cart from rolling down the ramp, project F onto a unit vector v in the direction of the ramp, as follows.

v = (cos 15°)i + (sin 15°) j = 0.966 i + 0.259 j

1065

Unit vector along ramp

So, the projection of F onto v is

w1 = projv F F ⋅ v v =   v2    ≈ ( − 38.85) v

Q not drawn to scale

Using a projection, you can calculate the work as follows.  W = proj ⎯ ⎯ →F PQ PQ  = (cos 30°) F PQ =

= (F ⋅ v) v ≈ ( 0, −150 ⋅ 0.966, 0.259 ) v

40 ft

3 (35)(40) 2

= 700 3 ≈ 1212.436

So, the work done is 1212.436 foot-pounds.

≈ − 37.53i − 10.06 j. The magnitude of this force is approximately 38.85. So, a force of approximately 38.85 pounds is required to keep the cart from rolling down the ramp.

Checkpoints for Section 7.5 1. The number z = 3 − 4i is plotted in the complex plane.

−1 −1

0 2 + (3)

r = 0 + 3i =

Real axis

−2

9 = 3

b . a Because a = 0 and because z = 0 + 3i = 3i lies on

and the argument θ is determined from tan θ =

−3 −4

2. z = 3i = 0 + 3i

The absolute value of z = 3i is

Imaginary axis

3 − 4i

−5 −6

It has an absolute value of z =

3 + ( − 4) 2

9 + 16

= 5.

the positive imaginary axis, θ =

π 2

So, the trigonometric form is π π  z = r (cos θ + i sin θ ) = 3  cos + i sin . 2 2  Imaginary axis

z = 3i

π 2

−3 −2 −1 −1

Real axis

−2 −3

1066 Solutions to Checkpoints 3. z = 6 − 6i

5π 5π   7π 7π   + i sin + i sin 5. z1 z2 = 2 cos  ⋅ 5 cos  6 6   6 6  

The absolute value of z = 6 − 6i is 6 2 + ( − 6)

r = 6 − 6i = =

36 + 36 =

 7π  7π   5π  5π = ( 2)(5) cos  + +  + i sin   6  6   6  6 

72 = 6 2

= 10(cos 2π + i sin 2π )

and the argument θ is determined from tan θ =

= 10

Because z = 6 − 6i lies in Quadrant IV, Imaginary axis

θ 1 −1 −1

= 10 1 + i (0)

b −6 = = −1. a 6

π π  π π  6. z1 z2 = 3 cos + i sin  ⋅ 4 cos + i sin  3 3  6 6  π π   π π = (3)( 4) cos  +  + i sin  +  6 6  3 3 

Real axis

π π  = 12 cos + i sin  2 2 

−2 −3 −4

= 12 0 + i (1)

−5 −6

6 − 6i

θ = 2π − arctan ( −1) = 2π −

π 4

= 12 i

7π . 4

You can check this by first converting the complex numbers to their standard forms and then multiplying algebraically.

So, the trigonometric form is 7π 7π   z = r (cos θ + i sin θ ) = 6 2  cos + i sin . 4 4  

1 π π 3  3 3 3  z1 = 3 cos + i sin  = 3 + i  = + i 3 3 2 2 2     2

  2π   2π  4. To write z = 8cos   + i sin   in standard  3   3   form, first find the trigonometric ratios. Because

−1  2π   2π  cos   = and sin   = 2  3   3 

 3 1  π π  + i  = 2 3 + 2 i z2 = 4 cos + i sin  = 4 6 6 2     2

3 3 3  i 2 3 + 2i So, z1 z2 =  + 2  2

(

3 , you can write 2

)

= 3 3 + 3i + 9i + 3 3 i 2

  2π   2π  z = 8cos   + i sin    3   3  

= 3 3 + 12i + 3 3 ( −1)

 1 3  = 8 − + i  2 2  

= 3 3 + 12i − 3 3 = 12 i.

= − 4 + 4 3 i.

z1 cos 40° + i sin 40° = z2 cos 10° + i sin 10° = cos ( 40° − 10°) + i sin ( 40° − 10°) = cos 30° + i sin 30° =

3 1 − i 2 2

Solutions to Checkpoints

( −1) + ( −1) 2

8. The absolute value of z = − 1 − i is r = −1 − i = and the argument θ given by tan θ =

1+1 =

5π . 4

1067

b −1 = = 1. −1 a

Because z = − 1 − i lies in Quadrant III, θ = π + arctan 1 = π + So, the trigonometric form is z = −1 − i =

π 4

5π 5π   2  cos + i sin . 4 4  

 5π 5π  4  + i sin Then, by DeMoivre’s Theorem, you have ( −1 − i ) =  2  cos  4 4    =

( 2 )  cos  ( 4 )  + i sin  ( 4 )   4

 4 5π 



 4 5π  



= 4(cos 5π + i sin 5π ) = 4 −1 + i (0) = − 4. 9. First, write 1 in trigonometric form z = 1(cos 0 + i sin 0). Then, by the nth root formula, with n = 4 and r = 1,

the roots are of the form 0 + 2π k 0 + 2π k  πk πk  πk πk   zk = 4 1 cos + i sin + i sin + i sin .  = (1) cos  = cos 4 4 2 2 2 2    

So, for k = 0, 1, 2 and 3, the fourth roots are as follows. z0 = cos 0 + i sin 0 = 1 + i(0) = 1 z1 = cos

π 2

+ i sin

π 2

= 0 + i(1) = i

z2 = cos π + i sin π = −1 + i(0) = −1 z3 = cos

3π 3π + i sin = 0 + i( −1) = −i 2 2 Imaginary axis

z1 = 0 + i = i z2 = −1 + 0i = −1

Real axis

z 0 = 1 + 0i = 1 z3 = 0 − i = −i

1068 Solutions to Checkpoints 10. The absolute value of z = − 6 + 6 i is r = −6 + 6i =

( − 6) + 6 2 = 2

36 + 36 =

and the argument θ is given by tan θ =

72 = 6

b 6 = = −1. a −6

Because z = − 6 + 6 i lies in Quadrant II, the trigonometric form of

z is z = − 6 + 6 i = 6 2 (cos 135° + i sin 135°). By the formula for nth roots, the cube roots have the form   135° + 360°k   135° + 360°k  zk = 3 6 2 cos   + i sin  . 3 3     

Finally, for k = 0, 1 and 2, you obtain the roots   135° + 360°(0)   135° + 360°(0)  z0 = 3 6 2 cos   + i sin  . 3 3      = 3 6 2 (cos 45° + i sin 45°)  2 = 3 6 2  +  2

2  i  ≈ 1.4422 + 1.4422 i 2 

  135° + 360°(1)   135° + 360°(1)  z1 = 3 6 2 cos   + i sin  . 3 3      = 3 6 2 (cos 165° + i sin 165°) ≈ −1.9701 + 0.5279 i

  135° + 360°( 2)   135° + 360°( 2)  z2 = 3 6 2 cos   + i sin  . 3 3      = 3 6 2 (cos 285° + i sin 285°) ≈ 0.5279 − 1.9701i Imaginary axis 3

z1 ≈ − 1.9701 + 0.5279i −3

z 0 ≈ 1.4422 + 1.4422i

1 −1

−1

−3

Real axis

z2 ≈ 0.5279 − 1.9701i

Solutions to Checkpoints

1069

Chapter 8 Checkpoints for Section 8.1 1. Algebraic Solution

 x − y = 0  5 x − 3 y = 6

Equation 1 Equation 2

Begin by solving for y in Equation 1.

x− y = 0 y = x Next substitute this expression for y into Equation 2 and solve the resulting single-variable equation for x. 5x − 3 y = 6

Write Equation 2.

5 x − 3( x) = 6

Substitute x for y.

2x = 6

Collect like terms.

x = 3

Divide each side by 2.

Finally, solve for y by back-substituting x = 3 into equation y = x, to obtain the corresponding value for y.

y = x

Write revised Equation 1.

y = 3

Substitute 3 for x. Check (3, 3) in Equation 2:

Check (3, 3) in Equation 1: x − y =0

5x − 3 y = 6

Write Equation 1.

3 − 3=0

5(3) − 3(3) = 6

Substitute for x and y.

0 = 0

Solution checks in Equation 1.

Write Equation 2. Substitute for x and y.

15 − 9 = 6 6 = 6

Solution checks in Equation 2.

Because (3, 3) satisfies both equations in the system, it is a solution of the system of equations. Graphical Solution

Begin by solving both equations for y. Then use a graphing utility to graph the equations y1 = x and y2 =

5 x − 2 in the 3

same viewing window. Use the intersect feature to approximate the point of intersection of the graphs. 5

−2

7 −1

The point of intersection is (3, 3). Check that (3, 3) is the exact solution as follows. Check (3, 3) in Equation 1:

x− y = 0 ?

3−3= 0 0 = 0

Check (3, 3) in Equation 2:

5x − 3 y = 6 ?

5(3) − 3(3) = 6 ?

15 − 9 = 6 6 = 6

1070 Solutions to Checkpoints Amount in 6.5% fund + Amount in 8.5% fund = Total investment

2. Verbal Model:

Interest for 6.5% fund + Interest for 8.5% fund = Total interest Amount in 6.5% fund = x

Labels:

Interest for 6.5% fund = 0.065 x Amount in 8.5% fund = y Interest for 8.5% fund = 0.085 y Total investment = 25,000 Total interest = 2600 x + y = 25,000   0.065 x + 0.085 y = 2000

System:

(dollars) (dollars) (dollars) (dollars) (dollars) (dollars) Equation 1 Equation 2

To begin, it is convenient to multiply each side of Equation 2 by 1000. This eliminates the need to work with decimals. 1000(0.065 x + 0.085 y ) = 1000( 2000) 65 x + 85 y = 2,000,000

Multiply each side of Equation 2 by 1000. Revised Equation 2

To solve this system, you can solve for x in Equation 1.

x = 25,000 − y

Revised Equation 1

Then, substitute this expression for x into revised Equation 2 and solve the resulting equation for y. 65 x + 85 y = 2,000,000 Write revised Equation 2. 65( 25,000 − y ) + 85 y = 2,000,000

Substitute 1 25000 − y for x.

1,625,000 − 65 y + 85 y = 2,000,000 20 y = 375,000 y = 18,750

Distributive Property Combine like terms. Divide each side by 20.

Next, back-substitute y = 18,750 to solve for x. x = 25,000 − y

Write revised Equation 1.

x = 25,000 − (18,750)

Substitute 18750 for y.

x = 6250

Subtract.

The solution is (6250, 18,750). So, $6250 is invested at 6.5% and $18,750 is invested at 8.5%.

Solutions to Checkpoints

1071

3. Algebraic Solution − 2x + y = 5   2 x − y + 3 x = 1

Equation 1 Equation 2

Begin by solving for y in Equation 1 to obtain y = 2 x + 5. Next, substitute this expression for y into Equation 2 and solve for x. x 2 − y + 3x = 1

Write Equation 2.

x − ( 2 x + 5) + 3x = 1 2

Substitute 2 x + 5 for y into Equation 2.

x − 2 x − 5 + 3x = 1

Simplify.

x + x −6 = 0

Write in standard form.

( x + 3)( x − 2) = 0

Factor.

x + 3 = 0  x = −3

Solve for x.

x − 2 = 0  x = 2 Back-substitute these values of x and solve for the corresponding values of y to produce the following solutions.

y = 2x + 5 y = 2( − 3) + 5 = −1 y = 2( 2) + 5 = 9 So, the solutions of the system are ( − 3, −1) and ( 2, 9).

Graphical Solution Solve each equation for y and use a graphing utility to graph the equations in the same viewing window. 12

− 12

12 −4

−4

One of the points of intersection is ( − 3, −1) and the other is ( 2, 9). So, the solutions of the system are ( − 3, −1) and ( 2, 9).

4.  2x − y = − 3  2 2 x + 4 x − y 2 = 0

Equation 1 Equation 2

Begin by solving for y in Equation 1 to obtain y = 2 x + 3. Next, substitute this expression for y into Equation 2 and solve for x. 2x2 + 4 x − y 2 = 0

Write Equation 2.

2 x + 4 x − ( 2 x + 3) = 0 2

Substitute 2 x + 3 for y into Equation 2.

2 x + 4 x − ( 4 x + 12 x + 9) = 0 2

Simplify.

− 2 x2 − 8x − 9 = 0

Combine like terms.

2x + 8x + 9 = 0 x = x =

−(8) ±

Write in standard form.

(8) − 4(2)(9) 2( 2)

−8 ± −8 4

Use the Quadratic Formula. Simplify.

Because the discriminant is negative, the equation 2 x 2 + 8 x + 9 = 0 has no real solution. So, the original system of equations has no real solution.

1072 Solutions to Checkpoints y

y = 2x + 1

6 5

y = 3 − log10 x

4 3 2 1 1

There is only one point of intersection of the graphs of the two equations, and (1, 3) is the solution point. Check (1, 3) in Equation 1:

Check (1, 3) in Equation 2:

y = 3 − log10 x

Write Equation 1.

− 2x + y = 1

Substitute for x and y.

− 2(1) + 3 = 1

3 = 3 − log10 1 ?

Write Equation 2. Substitute for x and y.

3=3 − 0

−2 + 3 = 1

3 = 3

Solution checks in Equation 1.

1 = 1

Solution checks in Equation 2.

6. The total cost of producing x bottles is Total cost = Cost per bottle ⋅ Number of bottles + Initial cost C = 0.50 x + 10,000.

The revenue obtained by selling x bottles is Total revenue = Price per bottle ⋅ Number of bottles R = 1.20 x.

Because the break-even point occurs when R = C , you have C = 1.20 x and the system of equations to solve is as follows.

C = 0.50 x + 10,000  C = 1.20 x Now you can solve by substitution. C = 0.50 x + 10,000 1.20 x = 0.50 x + 10,000 0.70 x = 10,000 10,000 0.7 x ≈ 14,286 bottles x =

Solutions to Checkpoints

1073

Checkpoints for Section 8.2 1. Because the coefficients of y differ only in sign, eliminate the y-terms by adding the two equations. 2x + y = 4

Write Equation 1.

2 x − y = −1

Write Equation 2.

= 3

3 4

Add equations. Solve for x.

Solve for y by back-substituting x = 34 into Equation 1.

()

2 34 + y = 4 3 + y = 4 2

y = 52 The solution is

( 34 , 52 ).

Check this in the original system. ?

( ) ( 52 ) = 4

2 34 +

3 + 52 2

=4 ?

( ) ( 52 ) = −1

2 34 −

3 − 52 2

= −1

Write Equation 1. Solution checks in Equation 1.  Write Equation 2. Solution checks in Equation 2. 

2. Algebraic Solution You can obtain coefficients that differ only in sign by multiplying Equation 1 by 2 and multiplying Equation 2 by − 3. 3x + 2 y = 7 

6 x + 4 y = 14

2 x + 5 y = 1  − 6 x − 15 y = − 3 −11 y = 11 y = −1

Multiply Equation 1 by 2. Multiply Equation 2 by − 3. Add Equations. Solve for y.

Solve for x by back-substituting y = 1 into Equation 1.

3x + 2 y = 7

Write Equation 1.

3x + 2( −1) = 7

Substitute −1 for y

3x − 2 = 7 3x = 9 x = 3 The solution is (3, −1). Graphical Solution Solve each equation for y and use a graphing utility to graph the equations in the same viewing window. From the graph, the solution is (3, −1). Check this in the original system. ?

3(3) + 2( −1) = 7 9 − 2 =7 ?

2(3) + 5( −1) = 1 6−5 =1

Write Equation 1. Solution checks in Equation 1. 

y1 = − 32 x + 72

−2

−4

y2 = − 25 x + 15

Write Equation 2. Solution checks in Equation 2. 

1074 Solutions to Checkpoints 3. First, write each equation in slope-intercept form.

2 x + 3 y = 6  y = 2 x + 2  3  4 x − 6 y = − 9  y = 23 x + 32 y 5 4

− 2x + 3y = 6

4x − 6y = − 9 −2

−1

The graph of the system is a pair of parallel lines. The lines have no points of intersection, so the system has no solution. The system is inconsistent.

4. Algebraic Solution To obtain coefficients that differ only in sign, multiply Equation 1 by 2. 6x − 5 y = 3 

12 x − 10 y = 6

Multiply Equation by 2.

−12 x + 10 y = 5  −12 x + 10 y = 5

Write Equation 2.

0 = 11

Add equations.

Because there are no values of x and y for which 0 = 11, you can conclude that the system is inconsistent and has no solution.

Graphical Solution Solve each equation for y and use a graphing utility to graph the equations in the same viewing window. From the figure, it appears that the system has no solution. When you use the intersect feature, the graphing utility cannot find a point of intersection and you will get an error message. The lines have the same slope but different y-intercepts, so they are parallel. 4

−12x + 10y = 5 6x −5y = 3

−6

−4

5. To obtain coefficients that differ only in sign, multiply Equation 1 by 8. 1x − 1y 2 8

= − 83 

− 4x + y =

4x − y = −3

3  − 4x + y = 3 0 = 0

Multiply Equation by 8. Write Equation 2. Add equations.

Because the two equations are equivalent (have the same solution set), the system has infinitely many solutions. The solution set consists of all points ( x, y ) lying on the line − 4 x + y = 3 as shown. Letting x = a, where a is any real number, the solutions of the system are ( a, 4a + 3) 6

−4x + y = 3 −6

6 −2

Solutions to Checkpoints

1075

6. The two unknown quantities are the speeds of the wind and of the plane. If r1 is the speed of the plane and r2 is the speed of the wind, then

r1 − r2 = speed of the plane against the wind r1 + r2 = speed of the plane with the wind. Using the formula distance = ( rate)( time)

for these two speeds, you obtain the following equations.

24   2000 = ( r1 − r2 ) 4 +  60   6  2000 = ( r1 − r2 ) 4 +  60   These two equations simplify as follows.  5000 = 11r1 − 11r2  20,000 = 41r1 + 41r2

Equation 1 Equation 2

To solve this system by elimination, multiply Equation 1 by 41 and Equation 2 by 11. 250,000 = 451r1 − 451r2

Multiply Equation 1 by 41.

220,000 = 451r1 + 451r2

Multiply Equation 2 by 11.

425,000 = 902r1

Add equations.

So, r1 =

425,000 ≈ 471.18 miles per hour 902

1 11r − 5000 and r2 = 11 ( 1 ) 1 11 ⋅ 425,000 − 5000  ≈ 16.63 miles per hour. r2 = 11   902  

Check this solution in the original system of equations. 24   2000 ≈ ( 471.18 − 16.63) 4 +  60   6  2000 ≈ ( 471.18 + 16.63) 4 +  60  

1076 Solutions to Checkpoints

Checkpoints for Section 8.3 1. From Equation 3, you know the value of z. To solve for y, back-substitute z = 3 into Equation 2 to obtain the following. Write Equation 2. y + 3z = 6 y + 3(3) = 6

Substitute 3 for z.

y = −3

Solve for y.

Then back-substitute y = − 3 and z = 3 into Equation 1 to obtain the following. 2 x − y + 5 z = 22

Write Equation 1.

2 x − ( − 3) + 5(3) = 22

Substitute − 3 for y and 3 for z.

2x = 4

Combine like terms.

x = 2

Solve for x.

The solution is x = 2, y = − 3, and z = 3, which can be written as the ordered triple ( 2, − 3, 3). Check this in the original system of equations. Check Equation 1: 2 x − y + 5 z = 22 ?

2( 2) − ( − 3) + 5(3) = 22 4 + 3 + 15 = 22 

Equation 2:

y + 3z = 6 ?

( − 3) + 3(3) = 6 −3 + 9 = 6  Equation 3:

z = 3

(3) = 3 

Solutions to Checkpoints

1077

2. Because the leading coefficient of the second equation is 2, multiply the first equation by − 2. − 2 x − 2 y − 2 z = −12

Multiply Equation 1 by − 2.

2x − y + z =

Write Equation 2.

−3y − z = −9  x + y + z = 6   −3y − z = −9 3 x + y − z = 2 

Add revised Equation 1 to Equation 2. Adding − 2 times the first equation to the second equation produces a new second equation.

− 3 x − 3 y − 3 z = −18

Multiply Equation 1 by − 3.

3x + y −

Write Equation 3.

z =

− 2 y − 4 z = −16 x + y + z = 6   −3y − z = −9  − 2 y − 4 z = −16 

Add revised Equation 1 to Equation 3. Adding − 3 times the first equation to the third equation produces a new third equation.

Now that you have eliminated all but the x in the upper position of the first column, work on the second column. x + y + z = 6   −3y − z = −9  − y − 2z = −8 

Multiplying the third equation by 2, produces a new third equation.

−3y −

Write Equation 2.

z = −9

3 y + 6 z = 24 5 z = 15

Multiply Equation 3 by − 3. Add equations.

x + y + z = 6   −3y − z = −9  5 z = 15 

Adding the second equation to − 3 times the third equation produces a new third equation.

x + y + z = 6  y + 13 z = 3   5 z = 15 

Multiplying the second equation by − 13 produces a new second equation.

x + y + z = 6  y + 13 z = 3   z = 3 

Multiplying the thrid equation by 15 produces a new third equation.

To solve for y, back-substitute z = 3 into Equation 2 to obtain the following. y + 13 (3) = 3 y = 2

Then back-substitute y = 2 and z = 3 into Equation 1 to obtain the following.

x + ( 2) + (3) = 6 x =1 The solution is x = 1, y = 2, and z = 3, which can be written as (1, 2, 3).

1078 Solutions to Checkpoints  x + y − 2z = 3  3. 3 x − 2 y + 4 z = 1 2 x − 3 y + 6 z = 8   x + y − 2z = 3  − 5 y + 10 z = − 8  2 x − 3 y + 6 z = 8 

Adding − 3 times the first equation to the second equation produces a new second equation.

x + y − 2 z = 3   − 5 y + 10 z = − 8  − 5 y + 10 z = 2 

Adding − 2 times the first equation to the third equation produces a new third equation.

x + y − 2 z = 3   − 5 y + 10 z = − 8  0 = 10 

Adding −1 times the second equation to the third equation produces a new third equation.

Because 0 = 10 is a false statement, this is an inconsistent system and has no solution. Moreover, because this system is equivalent to the original system, the original system has no solution. 4.  x + 2 y − 7 z = − 4  5 2 x + 3 y + z =  3 x + 7 y − 36 z = − 25  x + 2 y − 7 z = − 4  − y + 15 z = 13  3 x + 7 y − 36 z = − 25 

Adding − 2 times the first equation to the second equation produces a new second equation.

x + 2 y − 7 z = − 4  − y + 15 z = 13   y − 15 z = −13 

Adding − 3 times the first equation to the third equation produces a new third equation.

x + 2 y − 7 z = − 4  − y + 15 z = 13   0 = 0 

Adding the second equation to the third equation to produces a new third equation.

This result means that Equation 3 depends on Equations 1 and 2 in the sense that it gives no additional information about the variables. Because 0 = 0 is a true statement, this system has infinitely many solutions. However, it is incorrect to say that the solution is “infinite.” You must also specify the correct form of the solution. So, the original system is equivalent to the system. x + 2 y − 7 z = − 4  − y + 15 z = 13.  In the second equation, solve for y in terms of z to obtain the following. − y + 15 z = 13 − y = −15 z + 13 y = 15 z − 13 Back-substituting in the first equation produces the following. x + 2 y − 7z = − 4 x + 2(15 z − 13) − 7 z = − 4 x + 30 z − 26 − 7 z = − 4 x = − 23 z + 22

Finally, letting z = a where a is a real number, the solutions of the given system are all of the form x = − 23a + 22,

y = 15a − 13, and z = a. So, every ordered triple of the form ( − 23a + 22, 15a − 13, a) is a solution of the system.

Solutions to Checkpoints

1079

 x − y + 4z = 3 5.  −z = 0 4 x x − y + z = 3   4 y − 17 z = −12  x − y + z = 3  y − 17 z = −3  4

Adding − 4 times the first equation to the second equation produces a new second equation. Multiplying the second equation by 14 produces a new second equation.

Solve for y in terms of z to obtain the following. y − 17 z = −3 4 y = 17 z −3 4 Solve for x by back-substituting y = 17 z − 3 into Equation 1. 4 x − y + 4z = 3 x −

(

) + 4z = 3

17 z −3 4

x − 17 z + 3 + 4z = 3 4 x = 14 z

Finally, by letting z = a, where a is a real number, you have the solution x = 14 a, y = 17 a − 3, and z = a. 4

So, every ordered triple of the form

( 14 a, 174 a − 3, a) is a solution of the system. Because the original system had three

variables and only two equations, the system cannot have a unique solution.

1080 Solutions to Checkpoints 6. The expression is proper, so you should begin by factoring the denominator. Because 2 x 2 − x − 1 = ( 2 x + 1)( x − 1)

you should include one partial fraction with a constant numerator for each linear factor of the denominator. Write the form of the decomposition as follows. x +5 A B = + 2x2 − x − 1 2x + 1 x − 1 Multiplying each side of this equation by the least common denominator, ( 2 x + 1)( x − 1), leads to the basic equation x + 5 = A( x − 1) + B( 2 x + 1).

Because this equation is true for all x, substitute any convenient values of x that will help determine the constants A and B. Values of x that are especially convenient are those that make the factors x − 1 and 2 x + 1 equal to zero. For instance, to solve for B, let x = 1. Then 1 + 5 = A(1 − 1) + B 2(1) + 1

Substitute 1 for x.

6 = A(0) + B(3) 6 = 3B 2 = B. 1 To solve for A, let x = − and then 2 −

1  1  1  + 5 = A − 1 + B 2  + 1 2 2 2      9  3 = A −  + B(0) 2  2 9 3 = − A 2 2 − 3 = A.

So, the partial fraction decomposition is

x +5 −3 2 = + . 2x2 − x − 1 2x + 1 x − 1 Check this result by combining the two partial fractions on the right side of the equation, or by using your graphing utility. 10

−2

x+5 2x 2 − x − 1

− 10

Solutions to Checkpoints

1081

7. The expression is proper, so factor the denominator as x 3 + x 2 = x 2 ( x + 1). You should include one partial fraction with a

constant numerator for each power of x and x + 1, and write the form of the decomposition as follows. x + 4 A B C = + 2 + x3 + x 2 x x x +1

Multiplying each side by the LCD x 2 ( x + 1) leads to the basic equation x + 4 − Ax( x + 1) + B ( x + 1) = Cx 2 .

Letting x = − 1 eliminates the A- and B-terms and yields the following. −1 + 4 = A( −1)( −1 + 1) + B( −1 + 1) + C ( −1)

3 = 0 + 0 + C 3 = C

Letting x = 0, eliminates the A- and C-terms. 0 + 4 = A(0)(0 + 1) + B (0 + 1) + C (0)

4 = 0 + B + 0 4 = B

At this point, you have exhausted the most convenient values of x, so to find the value of A, use any other value of x along with the known values of B and C. So, using x = 1, B = 4, C = 3 1 + 4 = A(1)(1 + 1) + 4(1 + 1) + 3(1)

5 = 2A + 8 + 3 −6 = 2A − 3 = A. So, the partial fraction decomposition is x + 4 −3 4 3 = + 2 + . x3 + x 2 x x x +1

1082 Solutions to Checkpoints 8. By substituting the three values of t and s into the position equation, you can obtain three linear equations in a, v0 and s0 . When t = 1: 12 a (1) + v0 (1) + s0 = 104  a + 2v0 + 2 s0 = 208 2

When t = 2: 12 a ( 2) + v0 ( 2) + s0 = 76  2a + 2v0 + s0 = 76 2

When t = 3: 12 a (3) + v0 (3) + s0 = 16  9 a + 6v0 + 2 s0 = 32 2

This produces the following system of linear equation.  a + 2v0 + 2 s0 = 208  2a + 2v0 + s0 = 76 9a + 6v + 2 s = 32 0 0 

Now solve the system using Gaussian Elimination.  a + 2v0 + 2 s0 = 208  − 2v0 − 3s0 = − 340  9a + 6v + 2 s = 32 0 0 

Adding − 2 times the first equation to the second equation produces a new second equation.

208 a + 2v0 + 2 s0 =   − 2v0 − 3s0 = − 340  −12v − 16 s = −1840 0 0 

Adding − 9 times the first equation to the third equation produces a new third equation.

a + 2v0 + 2 s0 = 208   − 2v0 − 3s0 = − 340  2 s0 = 200 

Adding − 6 times the second equation to the third equation produces a new third equation.

a + 2v + 2 s = 208 0 0   3 v s + = 170  0 2 0  s0 = 100 

Multiplying the second equation by − 12 produces a new second equation and multiplying the third equation by 12 produces a new third equation.

So, s0 = 100. Find v0 by back-substituting s0 = 100 into Equation 2. v0 + 32 (100) = 170 v0 = 20

Find a by back-substituting s0 = 100 and v0 = 20 into Equation 1. a + 2( 20) + 2(100) = 208 a = − 32 So, the solution of this system is a = − 32, v0 = 20, and s0 = 100, which can be written as ( − 32, 20, 100). This results in a position equation of s = 12 ( − 32)t 2 + 20t + 100 = −16t 2 + 20t + 100

and implies that the object was thrown upward at a velocity of 20 feet per second from a height of 100 feet.

Solutions to Checkpoints

1083

9. Because the graph of y = ax 2 + bx + c passes through the points (0, 0), (3, − 3), and (6, 0), you can write the following.

When x = 0, y = 0: a (0) + b(0) + c = 0 2

When x = 3, y = − 3: a (3) + b(3) + c = − 3 2

When x = 6, y = 0: a (6) + b(6) + c = 0 2

This produces the following system of linear equations. c = 0    9a + 3b + c = − 3 36a + 6b + c = 0 

Equation 1 Equation 2 Equation 3

You can reorder these equations as shown. 36a + 6b + c = 0   9a + 3b + c = − 3  c = 0  36a + 6b + c = 0  − 6b − 3c = 12   c = 0 

Adding − 4 times the second equation to the first equation produces a new second equation.

1c = 0 a + 16 b + 36   b + 12 c = − 2  c = 0 

1 produces a new first equation and Multiplying the first equation by 36

multiplying the second equation by − 16 produces a new second equation.

So, c = 0, b + 12 (0) = − 2 b = − 2, 1 0 = 0 and a + 16 ( − 2) + 36 ()

a =

1. 3

The solution of this system is a = 13 , b = − 2, and c = 0. So, the equation of the parabola is y = 13 x − 2 x.

−4

y = 1 x 2 − 2x 3

(6, 0) (0, 0)

(3, −3)

−4

Checkpoints for Section 8.4 1. The matrix has two rows and three columns. The dimension of the matrix is 2 × 3.  x + y + z = 2  2.  2 x − y + 3 z = −1 − x + 2 y − z = 4  All of the variables are aligned in the system. Next, use the coefficients and constant terms as the matrix entries.  1 1 1  2   R2  2 −1 3  −1 −1 2 −1  4 R3   The augmented matrix has three rows and four columns, so it is a 3 × 4 matrix. R1

3. Add − 3 times the first row of the original matrix to the second matrix.

OriginalMatrix

New Row -Equivalent Matrix

1 2 3   3 12 5

3 1 2   − 3R1 + R2 → 0 6 − 4

1084 Solutions to Checkpoints 4. Linear System

Associated Augmented Matrix

y − z  2x +  x − y + z 4 2 2  − 6 x + 5 y + 4 z 

−3

1 −1 : − 3  2   4 2 2 : − 2 −  − 6 5 4 : 10 

= −2 =

Multiply the first equation by 12 .

 x + 1y − 1z = −3 2 2 2   4x − 2 y + 2z = − 2 − 6 x + 5 y + 4 z = 10 

1 R → 2 1

1  1 − 12 : − 32  2    4 −2 2 : − 2   5 4 : 10 − 6 

Add − 4 times the first equation to the second equation.

 x + 1y − 1z = −3 2 2 2  − 4 y + 4z = 4  − 6 x + 5 y + 4 z = 10 

1  1 − 12 : − 32  2   4 : 4 − 4 R1 + R2 →  0 − 4   5 4 : 10 − 6 

Multiply the second equation by − 14 .

 x +   − 6 x + 

−

1 y 2

1 z 2

y −

= −3 2

z =

−1

5 y + 4z =

 1 1 − 1 : − 3 2 2 2   − 14 R2 →  0 1 −1 : −1   4 : 10 − 6 5 

Add 6 times the first equation to the third equation.

x +    

1 y 2

−

1 z 2

= −3 2

y −

z =

−1

8y +

z =

1 1 − 1 : − 3  2 2  2 0 1 −1 : −1   6 R1 + R3 → 0 8 1 : 1  

Add − 8 times the second equation to the third equation.

x +    

1 y 2

−

y −

1 z 2

= −3 2

z =

−1

9z =

1 1 − 1 : − 3  2 2  2 0 1 −1 : −1   9 : 9 − 8R2 + R3 → 0 0  

Multiply the third equation by 19 .

x +    

1 y 2

−

y −

1 z 2

= −3 2

z =

−1

z =

1 1 − 1 : − 3  2 2  2 0 1 −1 : −1  1 R → 0 0 1 : 1  9 3 

At this point, you can use back-substitution to find x and y. y − z = −1 y − (1) = −1 y = 0 x + 12 y − 12 z

= − 23

x + 12 (0) − 12 (1) = − 23 x = −1 The solution is x = −1, y = 0, and z = 1.

Solutions to Checkpoints

1085

5. The matrix is in row-echelon form. The row consisting entirely of zeros occurs at the bottom of the matrix, and for each row that does not consist entirely of zeros, the first nonzero entry is 1. Furthermore, the matrix is in reduced row-echelon form, since every column that has a leading 1 has zeros in every position above and below its leading 1. 5 3  −19 − 3   4  8  3 4  4 − 8 − 6  26  

R3 + R1 →  1 − 3 − 3  7   4 4  8 3 4 − 8 − 6  26  

7 1 − 3 − 3    − 3R1 + R2 → 0 13 13  −13 − 4 R1 + R3 → 0 4 6  − 2 5 1 − 3 − 3    R2 + ( −3) R3 → 0 1 − 5  − 7 0 4 6  − 2  7 1 − 3 − 3    − − 0 1 5 7    − 4 R2 + R3 → 0 0 26  26

7 1 − 3 − 3    1 − 5  − 7 0 1 R → 0 0 1  1 26 3

Write augmented matrix.

Add R3 to R1 so first column has leading 1 in upper left corner.

Perform operations on R2 and R3 so first column has zeros below its leading 1.

Perform operations on R2 so second column has a leading 1.

Perform operations on R3 so second column has a zero below its leading 1.

Perform operations on R3 so third column has a leading 1.

The matrix is now in row-echelon form, and the corresponding system is as follows. x − 3 y − 3z = 7  y − 5z = − 7   z = 1 

Using back-substitution, you can determine that the solution is x = 4, y = −2 and z = 1 . 1 1 1   7. 1 2 2  1 −1 −1   1 1  1  − R1 + R2 → 0 1 1  − R1 + R3 → 0 −2 −2   1 1 1  1   0 1 1  1 2 R2 + R3 → 0 0 0  2

1  2 1 1  1 0

Write augmented matrix.

Perform row operations.

Note that the third row of this matrix consists entirely of zeros except for the last entry. This means that the original system of linear equations is inconsistent. You can see why this is true by converting back to a system of linear equations. x + y + z = 1  y + z =1   0 = 2  Because the third equation is not possible, the system has no solution.

1086 Solutions to Checkpoints 2  1 − 3 7   3 − 5  − 8 − 5  2 −2 − 3  15  

R2 + R1 →

− R1 →

5 −1  16  −1   3 − 5  − 8 − 5  2 − 2 − 3  15   1  −16  1 −5   − 5 3 − 5  − 8   2 − 2 − 3  15  

1  −16 1 −5   5 R1 + R2 → 0 − 22 0  − 88 2 − 2 − 3  15  1  −16 1 − 5   − 22 0  − 88 0 − 2 R1 + R3 → 0 8 − 5  47 1  −16 1 − 5   0 1 0 4   0 8 − 5  47  1  −16 1 − 5   1 0  4 0 − 8 R2 + R3 → 0 0 − 5  15  1 − 5 1  −16   1 0  4 0 − 15 R3 → 0 0 1  −3 1 − 22 R2 →

At this point, the matrix is in row-echelon form. Now, apply elementary row operations until you obtain zeros above each of the leading is, as follows.

2 − 6 6  46   2 − 3 0  31

1 − 3 3  23   2 − 3 0  31

− 2 R1 + R2 →

3  23 1 − 3    −15 − 0 3 6 

1R 3 2

3  23 1 − 3   0 1 2  − 5 − 

→

R1 + 3R2 →

8 1 0 − 3    0 1 − 2  − 5

The corresponding system of equations is  x − 3z = 8   y − 2 z = − 5. Solving for x and y in terms of z, you have x = 3z + 8 and y = 2 z − 5. To write a solution of the system that does not use any of the three variables of the system, let a represent any real number and let z = a. Substituting a for z in the equations for x and y, you have

x = 3z + 8 = 3a + 8 and y = 2 z − 5 = 2a − 5. So, the solution set can be written as an ordered triple of the form

(3a + 8, 2a − 5, a), where a is any real number. Remember that a solution set of this form represents an infinite number of solutions. Try substituting values for a to obtain a few solutions. Then check each solution in the original system of equations.

5 R2 + R1 →  1 0 1  4   4 0 1 0  0 0 1  − 3   − R3 + R1 →  1 0 0  7   4 0 1 0  0 0 1  − 3  

The matrix is now in reduced row-echelon form. Converting back to a system of linear equations, you have x = 7   y = 4. z = − 3 

So, the solution is x = 7, y = 4, and z = − 3, which can be written as the ordered triple (7, 4, − 3).

Solutions to Checkpoints

1087

Checkpoints for Section 8.5 a11 a12   6 3 1.   =   a21 a22  − 2 4 Because two matrices are equal when their corresponding entries are equal, you can conclude that a11 = 6, a12 = 3, a21 = −2, and a22 = 4. 4 + 2 −1 + ( −1) 4 −1 2 −1 6 − 2 2.   =   +   =   2 3 0 6 3 −     2 2 + 0 − 3 + 6  3. 3 A = 3  4 −1  12 − 3      0 4 =  0 12 − 3 8 − 9 24     4. 3 A − 2 B = 3  4 −1 − 2  0 4  12 − 3  0 8  12 −11           6  0 4 −1 3 =  0 12 − − 2 6 =  2 − 3 8  1 7 − 9 24  2 14 −11 10             1 3 − 4 0   1 3 − 4 0 5. 2   − 2 2 + − 3 1  = 2 − 2 2 + 2  − 3 1          2 6 − 8 0 =   +   − 4 4 − 6 2  − 6 6 =   −10 6

6. Begin by solving the matrix equation for X to obtain

2X − A = B 2X = B + A X = 12 ( B + A). Now, using the matrices A and B you have the following   4 −1 6 1  10 0  5 0 1 X = 12    +    = 2   =   − 2 8 −1 4  − 2 5 0 3 

( −1)(1) + ( 4)(0) −1 4     1 − 2 7. AB =  2 0   =  ( 2)(1) + (0)(0) 0 7   (1)(1) + ( 2)(0)  1 2    

(−1)(− 2) + (4)(7) −1 30  (2)(− 2) + (0)(7) =  2 − 4 (1)(− 2) + (2)(7)  1 12

4 − 3 − 2 0 (0)( − 2) + ( 4)(0) + ( − 3)(1) 0     8. AB = 2 1 7  0 − 4 = ( 2)( − 2) + (1)(0) + (7)(1) (3)( − 2) + ( − 2)(0) + (1)(1) 3 − 2 1  1 2  

(0)(0) + (4)( − 4) + (− 3)(2) − 3 − 22  (2)(0) + (1)( − 4) + (7)(2)  =  3 10 (3)(0) + (− 2)( − 4) + (1)( 2) − 5 10

9. Note that the dimension of A is 3 × 3 and the dimension of B is 3 × 3. So, the product will have dimension 3 × 3. Use the matrix feature of a graphing utility to find the product of A and B.

1088 Solutions to Checkpoints

− 2 x1 − 3x2 = − 4 10.   6 x1 + x2 = − 36 (a) In matrix form Ax = B , the system can be written as follows. − 2 − 3  x1   − 4    =   1  x2   6 − 36 (b) The augmented matrix is formed by adjoining matrix B to matrix A.

[ A  B] =

− 2 − 3  − 4   1  − 36  6

Use Gauss-Jordan elimination to rewrite the matrix. − 12 R1 →  1  6

− 6 R1 + R2

− 18 R2

2  1  − 36

3 2



3  2 = 1 2   → 0 − 8  − 48

= 1  → 0 

 − 2  1  6

3 2

− 32 R2 + R1 → = 1 0  7   0 1  6 = [I  X ]  x1  − 7 So, the solution of the matrix equation is X =   =   .  x2   6 11. The equipment lists E and the costs per item C can be written in matrix form as 12 15   E = 45 38 and C = [100 7 65]. 15 17   

The total cost of equipment for each team is given by the following product. 12 15   CE = [100 7 65] 45 38 = (100)(12) + (7)( 45) + (65)(15) 15 17   

(100)(15) + (7)(38) + (65)(17) = [2490

2871]

So, the total cost of equipment for the women’s team is $2490 and the total cost of equipment for the men’s team is $2871.

Checkpoints for Section 8.6 1. To show that B is the inverse of A, show that AB = I = BA, as follows.

 2 −1  −1 −1 − 2 + 3 − 2 + 2 1 0 AB =    =   =   − 3 1 − 3 − 2 3 − 3 3 − 2      0 1  −1 −1  2 −1 − 2 + 3 1 − 1  1 0 BA =    =   =   − 3 − 2 − 3 1 − 6 + 6 3 − 2 0 1 Because AB = I = BA, B is the inverse of A.

Solutions to Checkpoints 2. To find the inverse of A, solve the matrix equation AX = I for X. A X = I  1 − 2  X 11 X 12  1 0    =   3  X 21 X 22  −1 0 1  X 11 − 2 X 21  − X 11 + 3 X 21

3. Begin by adjoining the identity matrix to A to form the matrix  1 − 2 −1  1 0 0 [ A  I ] = 0 −1 2  0 1 0 .  1 − 2 0  0 0 1  

X 12 − 2 X 22  1 0  =   − X 12 + 3 X 22  0 1

Equating corresponding entries, you obtain two system of linear equations.  X 11 − 2 X 21 = 1  − X 11 + 3 X 21 = 0

1089

 X 12 − 2 X 22 = 0  − X 12 + 3 X 22 = 1

Use elementary row operations to obtain the form I  A−1   1 − 2 −1  1 0 0   1 − 2  0 −1 0 − R2 → 0 0 1  −1 0 1 − R1 + R3 → 0 2 R2 + R1 →  1 0 − 5  1 − 2 0   2 R3 − R2 → 0 1 0  − 2 −1 2 0 0 1  −1 0 1 

Solving the first system yields

X11 = 3 and X 21 = 1. Solving the second system yields.

5R3 + R1 →  1 0 0  − 4 − 2 5   −1 0 1 0  − 2 −1 2 = I  A  0 0 1  −1 0 1  

X12 = 2 and X 22 = 1. 3 2 So, the inverse of A is X = A−1 =  . 1 1

So, the matrix A is invertable and its inverse is A−1 = − 4 − 2 5   − 2 −1 2 .  −1 0 1 

You can check this by finding AA −1 and A −1 A. Check:

 1 − 2 3 2 1 0 AA−1 =    =   = I 1 3 1 1 −    0 1

Confirm this result by multiplying AA−1 to obtain I.

3 2  1 − 2 1 0 A−1 A =    =   = I 3 1 1 −1 0 1

Check:  1 − 2 −1 − 4 − 2 5 1 0 0      AA−1 = 0 −1 2 − 2 −1 2 = 0 1 0 = I  1 − 2 0  −1 0 0 1 0 1    

4. For the matrix A, apply the formula for the inverse of a 2 × 2 matrix to obtain ad − bc = (5)( 4) − ( −1)(3) = 23.

Because this quantity is not zero, the matrix is invertible. The inverse is formed by interchanging the entries on the 1 , as follows. main diagonal, changing the signs of the other two entries, and multiplying by the scalar 23 A−1 = =

 d − b 1   ad − bc − c a

Formula for the inverse of a 2 × 2 matrix

1  4 1   23 − 3 5

Substitute for a, b, c, d , and the determinant

4 23 =  −3  23

1 23



5 23 

Multiply by the scalar

1 . 23

1090 Solutions to Checkpoints 5. Begin by writing the system as AX = B. 1 1 1  x   6      3 5 4  y = 27 3 6 5  z  35     

1 0 0 1 1 1  1 0 0 1 1 1     1 0 − 3R1 + R2 → 0 2 1  − 3 1 0 3 6 5  0 0 1 − 3R1 + R3 → 0 3 2  − 3 0 1    

[ A  I ] = 3 5 4  0

1 R 2 2

1 1 1  1 0 0 1 0 0 1 1 1      3 1 1 → 0 1 12  − 32 12 0 − 0 1 0   2 2 2 3 0 3 2  − 3 0 1 − 3R2 + R3 → 0 0 1  − 23 1   2 2 

5 − 12 0 1 0 0 − R2 + R3 →  1 0 12  1 1 1  2     3 1 1 0 − 12 R3 + R2 →0 1 0  − 3 2 −1 2 0 1 2  − 2 0 0 1  3 − 3 2 2 R3 →0 0 1  3 − 3 1  

− 12 R3 + R1 → 1 0 0  1 1 −1   −1 0 1 0  − 3 2 −1 = I  A  0 0 1  3 − 3 2  1 −1  1   A−1 = − 3 2 −1  3 − 3 2  

Finally, multiply B by A−1 on the left to obtain the solution. X = A−1B 1 −1  6   1    = − 3 2 −1 27  3 − 3 2 35    − 2   = 1 7  

So, the solution is x = − 2, y = 1, and z = 7. Use a graphing utility to verify A−1 for the system of equations.

Checkpoints for Section 8.7 1. (a) det ( A) =

3 −1

(b) det ( B ) =

5 0 −4 2

3 6 2 4

= 1( −1) − 3( 2)

= 5( 2) − ( − 4)(0)

= 3( 4) − ( 2)(6)

= −1 − 6

= 10 + 0

= 12 − 12

= −7

= 10

= 0

2. Use the matrix editor to enter the matrix as [ A]. Then choose the determinant feature.

So, det ( A) = 0.64.

Solutions to Checkpoints

1091

3. To find the minor M11, delete the first row and first column of A and evaluate the determinant of the resulting matrix.  1 2 3   0 −1 5 , 2 1 4 

M 11 =

−1 5 1 4

= −1( 4) − 1(5) = − 9

Continuing this pattern, you obtain the minors. M 12 = M 13 = M 21 = M 22 = M 23 = M 31 = M 32 = M 33 =

0 5

= 0( 4) − 2(5) = −10

2 4 0 −1 2

2 3 1 4 1 3 2 4 1 2 2 1 2 3 −1 5 1 3 0 5 1

0 −1

= 0(1) − 2( −1) = 2 = 2( 4) − 1(3) = 5 = 1( 4) − 2(3) = −2 = 1(1) − 2( 2) = −3 = 2(5) − ( −1)(3) = 13 = 1(5) − 0(3) = 5 = 1( −1) − 0( 2) = −1

+ − +   Now, to find the cofactors, combine these minors with the checker board pattern of signs for a 3 × 3 matrix, − + − . + − +  

C11 = −9

C12 = 10

C13 = 2

C21 = −5

C22 = − 2

C23 = 3

C31 = 13

C32 = − 5

C33 = −1

4. The cofactors of the entries in the first row are as follows. C11 = + M 11 = C12 = − M 12 = C13 = + M 13 =

5 0 4 1 3 0 −1 1 3 5 −1 4

= 5(1) − 4(0) = 5 = − (3(1) − ( −1)(0)) = − 3 = 3( 4) − ( −1)(5) = 17

So, by the definition of a determinant, you have the following. A = a11C11 + a12C12 + a13C13 = 3(5) + 4( − 3) + ( − 2)(17) = − 31

1092 Solutions to Checkpoints

Checkpoints for Section 8.8 y

1. 6

(2, 5)

5 4 3 2

(4, 1)

(0, 0)

−1

Let ( x1 , y1 ) = (0, 0), ( x2 , y2 ) = ( 4, 1), and ( x3 , y3 ) = ( 2, 5). Then, to find the area of the triangle, evaluate the determinant. x1

y1 1

y2 1 = 4 1 1

0 0 1

y3 1

2 5 1

= (0)( −1)

1 1

5 1

+ (0)( −1)

4 1

+ (1)( −1)

2 1

4 1 2 5

= 0 + 0 + (1)(18) = 18

Using this value, you can conclude that the area of the triangle is 0 0 1 1 1 Area = 4 1 1 = (18) = 9 square units. 2 2 2 5 1 y

2. 6

(− 2, 4) 4 2

−4

−2

(3, − 1)

−2

(6, − 4)

−4

To determine if the points are collinear, let ( x1 , y1 ) = ( − 2, 4), ( x2 , y2 ) = (3, −1), and ( x3 , y3 ) = (6, − 4). Then, evaluate the determinant as follows. −2

4 1

y2 1 =

−1 1

y3 1

6 −4 1

y1 1

x2 x3

= ( −2)( −1)

−1 1 −4 1

+ ( 4)( −1)

3 3

6 1

+ (1)( −1)

4 3

−1

6 −4

= ( − 2)(3) + ( − 4)( − 3) + (1)( − 6) = 0

Because the value of this determinant is equal to zero, you can conclude that the three points are collinear.

Solutions to Checkpoints

1093

3. To begin, find the determinant of the coefficient matrix.

3 4

D =

= 9 − 20 = −11

5 3

Because this determinant is not zero, you can apply Cramer’s Rule.

1 4 x =

9 3 Dx = −11 D

− 33 3 − 36 = = 3 −11 −11

27 − 5 22 = = −2 −11 −11

3 1 Dy

y =

5 9 −11

So, the solution is x = 3 and y = −2 . 4. To find the determinant of the coefficient matrix, expand along the first row, as follows. 4 −1 1   2 3 2 5 − 2 6  

D = 4( −1)

2 3

−2 6

+ ( −1)( −1)

2 3 5 6

+ (1)( −1)

= 4(18) + (1)( − 3) + (1)( −14) = 55

5 −2

Because this determinant is not zero, you can apply Cramer’s Rule. Next, find Dx , Dy , and Dz . 12

−1 1

2 3 = (12)( −1)

Dx =

22 − 2 6 4 12

1 3 = ( 4)( −1)

Dy = 2

22 6

−1 12

Dz = 2

−2 6

1 3

5 22 6 4

2 3

1 = ( 4)( −1)

5 − 2 22

+ ( −1)( −1)

+ (12)( −1)

− 2 22

1 3

22 6

2 3 5 6

+ ( −1)( −1)

+ (1)( −1)

5 22

22 −2

5 22

+ (12)( −1)

= (12)(18) + (1)( − 60) + (1)( − 46) = 110

= ( 4)( − 60) + ( −12)( − 3) + (1)(39) = −165

5 −2

= ( 4)( 46) + (1)(39) + (12)( −14) = 55

Finally, you can determine the values of x, y , and z as follows. Dx 110 = = 2 D 55 Dy −165 = = −3 y = D 55 D 55 z = z = =1 D 55

x =

So, the solution is x = 2, y = −3, and z = 1. 5. Partioning the message (including blank spaces, but ignoring any punctuation) into groups of three produces the following uncoded 1 × 3 row matrices.

[15 23 12] [19 0

1] [18 5 0] [14 15

3] [20 21 18] [14

1 12]

1094 Solutions to Checkpoints 6. The coded row matrices are obtained by multiplying each of the uncoded row matrices found in Checkpoint Example 5 by the matrix A, as follows. Uncoded Matrix

Encoding Matrix A

[15 23 12]

 1 −1 0    1 0 −1 6 −2 −3  

Coded Matrix

[110 −39 −59]

[19 0 1]

 1 −1 0    1 0 −1 6 −2 −3  

[25 −21 −3]

[18 5 0]

 1 −1 0    1 0 −1 6 −2 −3  

[23 −18 −5]

[14 15 3]

 1 −1 0    1 0 −1 6 −2 −3  

[47

[20 21 18]

 1 −1 0    1 0 −1 6 −2 −3  

[149 −56 −75]

[14 1 12]

 1 −1 0    1 0 −1 6 −2 −3  

[87

−20 −24]

−38 −37]

So, the cryptogram is 110 − 39 − 59 25 − 21 − 3 23 − 18 − 5 47 − 20 − 24 149 − 56 − 75 87 − 38 − 37 . 7. First find the decoding matrix A−1 using matrix A from Checkpoint Example 6.

[ A  I ] = 1

−1 0  1 0 0   1 0 −1  0 1 0  6 − 2 − 3  0 0 1  

R2 →  1 0 −1  0 1 0   R1 →  1 −1 0  1 0 0 − 6 R1 + R3 → 0 4 − 3  − 6 0 1 0 1 0  1 0 −1    1  1 −1 0 − R1 + R2 → 0 −1 0 4 − 3  − 6 0 1   0 1 0  1 0 −1     0 1 1 1 1 0 − −  4 R2 + R3 → 0 0 1  − 2 − 4 1 0 1 0  1 0 −1    1 0 − R2 → 0 1 −1  −1 0 0 1  − 2 − 4 1  R1 + R3 →  1 0 0  − 2 − 3 1   R2 + R3 → 0 1 0  − 3 − 3 1 = I  A−1  0 0 1  − 2 − 4 1   Partition the message into groups of three to form the coded row matrices. Finally, multiply each coded row matrix by A−1 (on the right).

Solutions to Checkpoints Coded Matrix

Decoding Matrix A−1

[110 −39 −59]

−2 −3 1   −3 −3 1 −2 −4 1  

[15 23 12]

[25 −21 −3]

−2 −3 1   −3 −3 1 −2 −4 1  

[19 0 1]

[23 −18 −5]

−2 −3 1   −3 −3 1 −2 −4 1  

[18 5 0]

[47

−20 −24]

−2 −3 1   −3 −3 1 −2 −4 1  

[14 15 3]

[149 −56 −75]

−2 −3 1   −3 −3 1 −2 −4 1  

[20 21 18]

[87

−2 −3 1   −3 −3 1 −2 −4 1  

[14 1 12]

−38 −37]

1095

Decoded Matrix

So, the message is as follows.

[15 23 12] [19 0 1]

[18 5 0] [14 15 3] [20 21 18] [14

1 12]

Chapter 9 Checkpoints for Section 9.1 1. The first four terms of the sequence given by an = 2n + 1 are as follows. a1 = 2(1) + 1 = 3

1st term

a2 = 2( 2) + 1 = 5

2nd term

a3 = 2(3) + 1 = 7

3rd term

a4 = 2( 4) + 1 = 9

4th term

2. Algebraic Solution

The first four terms of the sequence given by an = 2 + ( −1)

a1 = a2 = a3 = a4 =

1 2 + ( −1)

2 +1 3 = 2 2

2 −1 1 = 3 3

2 +1 3 = 4 4

3 2 + ( −1) 4

2 −1 =1 1

2 2 + ( −1)

2 + ( −1) n

are as follows.

1096 Solutions to Checkpoints Numerical Solution

Set your graphing utility to sequence mode and enter the sequence. Use the table feature in ask mode to create a table showing the terms of the sequence.

So, you can estimate the first four terms of the sequence as follows. u1 = 1, u 2 = 1.5 =

3 1 3 , u3 = 0.33333 ≈ , and u 4 = 0.75 = 2 3 4

n : 1 2 3

3. (a)

5 

Terms : 1 5 9 13 17  an Apparent Pattern: Each term is 3 less than 4 times n. So, the apparent nth term is an = 4n − 3.

n : 1 2 3 4

(b)

5 

Terms : 3 5 7 9 11  an Apparent Pattern: Each term is 1 more than 2 times n, or each term is an odd number. So, the apparent nth term is an = 2n + 1. 4. The first five terms of the sequence are as follows.

a1 = 6

1st term is given.

a2 = a1 + 1 = a1 + 1 = 6 + 1 = 7

Use recursion formula.

a3 = a2 + 1 = a2 + 1 = 7 + 1 = 8

Use recursion formula.

a4 = a3 + 1 = a3 + 1 = 8 + 1 = 9

Use recursion formula.

a5 = a4 +1 = a4 + 1 = 9 + 1 = 10

Use recursion formula.

5. The first five terms of the sequence are as follows,

a0 = 1

0th term is given.

a1 = 3

1st term is given.

a2 = a2 − 2 + a2 −1 = a0 + a1 = 1 + 3 = 4

Use recursion formula.

a3 = a3 − 2 + a3 −1 = a1 + a2 = 3 + 4 = 7

Use recursion formula.

a4 = a4 − 2 + a4 −1 = a2 + a3 = 4 + 7 = 11

Use recursion formula.

6. Algebraic Solution

Graphical Solution

a0 =

3 +1 1+1 = = 2 0! 1

Using a graphing utility set to dot and sequence modes, enter the sequence. Next, graph the sequence.

a1 =

31 + 1 3+1 = = 4 1! 1

You can estimate the first five terms of the sequence as follows.

a2 =

3 +1 9 + 1 10 = = = 5 2! 2 2

a3 =

33 + 1 27 + 1 28 14 = = = 3! 6 6 3

a4 =

34 + 1 81 + 1 82 41 = = = 4! 24 24 12

u0 = 2

Use the trace feature to approximate the first five terms.

u1 = 4 u2 = 5 0

14 3 41 u4 ≈ 3.417 = 12 u3 ≈ 4.667 =

Solutions to Checkpoints

4 !( n + 1) ! 3!n !

1097

( 1 ⋅ 2 ⋅ 3 ⋅ 4)1 ⋅ 2 ⋅ 3  n ⋅ (n + 1) ( 1 ⋅ 2 ⋅ 3 )( 1 ⋅ 2 ⋅ 3  n )

= 4( n + 1) 4

8. ( 4i + 1) = 4(1) + 1 + 4( 2) + 1 + 4(3) + 1 + 4( 4) + 1 i =1

= 5

= 44 9. (a) The fourth partial sum is as follows. 4

(b) The sum of the series is as follows. ∞

 102 = 101 + 102 + 103 + 104

 10i = 101 + 102 + 103 + 104 + 105 + 

i =1

= 0.5 + 0.05 + 0.005 + 0.0005

= 0.5 + 0.05 + 0.005 + 0.0005 + 0.00005 + 

= 0.5555

= 0.55555  =

5 9

10. (a) The first three terms of the sequence are as follows. 0

0.03   A0 = 10001 +  = $1000 12  

Original deposit

0.03   A1 = 10001 +  = $1002.50 12  

First-month balance

0.03   A2 = 10001 +  ≈ $1005.01 12  

Second-month balance

(b) The 48th term of the sequence is as follows.

0.03   A48 = 10001 +  12  

≈ $1127.33

Four-year balance

Checkpoints for Section 9.2 1. The sequence whose nth term is 3n − 1 is arithmetic. The first four terms are as follows.

3(1) − 1 = 2 3( 2) − 1 = 5

2. You know that the formula for the nth term is of the form an = a1 + ( n − 1)d . Because the common difference is d = 5 and the first term is a1 = −1, the formula must

have the form

3(3) − 1 = 8

an = a1 + ( n − 1)d = −1 + 5( n − 1).

3( 4) − 1 = 11

So, the formula for the nth term is an = 5n − 6.

For this sequence, the common difference between consecutive terms is 3.

The sequence therefore has the following form.

2, 5 , 8, 11, 

The figure below shows a graph of the first 15 terms of the sequence. Notice that the points lie on a line.

5− 2=3

−1, 4, 9, 14, , 5n − 6, 

an = 5n − 6

−5

1098 Solutions to Checkpoints 3. You know that a8 = 25 and a12 = 41. So, you must add the common difference d four times to the eighth term to obtain the 12th term. Therefore, the eighth term and the 12th terms of the sequence are related by

4. For this arithmetic sequence, the common difference is d = 15 − 7 = 8.

There are two ways to find the tenth term. One way is to write the first ten terms (by repeatedly adding 8).

a12 = a8 + 4 d .

7, 15, 23, 31, 39, 47, 55, 63, 71, 79

Using a8 = 25 and a12 = 41, solve for d.

So, the tenth term is 79.

a12 = a8 + 4d

Another way to find the tenth term is to first find a formula for the nth term. Because the common difference is d = 8 and the first term is a1 = 7, the formula must have the form an = a1 + ( n − 1)d = 7 + ( n − 1)(8).

41 = 25 + 4d 16 = 4d 4 = d

Use the formula for the nth term of an arithmetic sequence to find a1.

an = a1 + ( n − 1)d a8 = a1 + (8 − 1)( 4)

Therefore, a formula for the nth term is an = 8n − 1 which implies that the tenth term is a10 = 8(10) − 1 = 79.

25 = a1 + (7)( 4) − 3 = a1 So, the formula for the nth term of the sequence is an = − 3 + ( n − 1)( 4) = − 3 + 4n − 4 = 4n − 7.

The sequence is as follows. a1

a10

a11 

−3

13 17 21 25 29

37 

5. To begin notice that the sequence is arithmetic (with a common difference of d = 37 − 40 = − 3 ).

Moreover, the sequence has 7 terms. So, the sum of the sequence is as follows. n (a1 + an ) 2 7 = ( 40 + 22) 2 = 217

Sn =

Sum of a finite arithmetic sequence Substitute 7 for n, 40 for a1 , and 22 for an . Simplify.

6. For this arithmetic sequence, a1 = 6 and d = 12 − 6 = 6.

So, an = a1 + ( n − 1)d = 6 + 6( n − 1) and the nth term is an = 6 n. Therefore, a120 = 6(120) = 720, and the sum of the first 120 terms is

n (a1 + a120 ) 2 120 = (6 + 720) 2 = 60(726)

S120 =

= 43,560.

Solutions to Checkpoints 7. Algebraic Solution

1099

Numerical Solution

The annual sales form an arithmetic sequence in which a1 = 160,000 and d = 20,000. So, an = 160,000 + 20,000( n − 1) and the nth term of the sequence is an = 20,000 n + 140,000. Therefore, the 10th term of the sequence is

a10 = 20,000(10) + 140,000 = 340,000. The sum of the first 10 terms of the sequence is

n (a1 + a10 ) 2 10 = (160,000 + 340,000) 2 = 5(500,000)

S10 =

The annual sales form an arithmetic sequence in which a1 = 160,000 and d = 20,000. So, an = 160,000 + 20,000( n − 1) and nth term of the

sequence is an = 20,000 n + 140,000. Use the list editor of a graphing utility to create a table that shows the sales for each of the first 10 years and the total sales for the first 10 years, as shown in the figure. Enter 1 through 10 for L1.

Enter 20000L1 + 140000 Enter cumSum(L2) for L2. for L3.

Total sales for the first ten years.

So, the total sales for the first 10 years are $2,500,000.

= 2,500,000. So, the total sales for the first 10 years will be $2,500,000.

Checkpoints for Section 9.3 1. The sequence whose nth term is 6( − 2) is geometric. For this sequence, the common ratio of consecutive terms is − 2. n

The first four terms, beginning with n = 1 are as follows. a1 = 6( − 2) = 6( − 2) = −12 1

a2 = 6( − 2) = 6( 4) = 24 2

a3 = 6( − 2) = 6( − 8) = − 48 3

a4 = 6( − 2) = 6(16) = 96 4

So, the sequence of terms are

−12, 24, − 48, 96, , 6( − 2) , .    n

24 =− 2 −12

2. Starting with a1 = 2, repeatedly multiply by 4 to obtain the following. a1 = 2

1st term

a2 = 2( 41 ) = 8

2nd term

a3 = 2( 42 ) = 32

3rd term

a4 = 2( 43 ) = 128

4th term

a5 = 2( 4) = 512

5th term

1100 Solutions to Checkpoints 4. The common ratio of this geometric sequence is r = 20 = 5. Because the first term is a1 = 4, the 4

3. Algebraic Solution

Use the formula for the nth term of a geometric sequence.

formula must have the form

an = a1r n −1

an = a1r n −1 12 − 1

a12 = 14(1.2)

= 4(5)

= 14(1.2)

The 12th term ( n = 12) of the sequence is

≈ 104.02

12 −1

a12 = 4(5)

Numerical Solution

For this sequence, r = 1.2 and a1 = 14. So an = 14(1.2)

n −1

= 4(5)

= 195,312,500.

5. The fifth term is related to the second term by the equation a5 = a2 r 3 .

. Use a graphing utility to create a table

that shows the terms of the sequence.

Because a5 = 81 and a2 = 6, you can solve for r as 4 follows.

The number in the 12th row is the 12th term of the sequence. So, a12 ≈ 104.02.

a5 = a2 r 3

Multiply the second term by r 5 − 3 .

81 = 6r 3 4

Substitute 81 for a5 and 6 for a2 . 4

27 8

= r3

Divide each side by 6.

3 2

= r

Take the cube root of each side.

You can obtain the eighth term by multiplying the fifth term by r 3. Multiply the fifth term by r 8 − 5 .

a8 = a5r 3

() 81 = 4 ( 27 8 )

Substitute 81 for a5 and 32 for r . 4

= 2187 32

Multiply fractions.

3 = 81 4 2

6. You have  2(0.25) i =1

i −1

Evalutate power.

= 2(0.25) + 2(0.25) + 2(0.25) +  + 2(0.25) . 0

Now, a1 = 2, r = 0.25, and n = 10, so applying the formula for the sum of a finite geometric sequence, you obtain the following. 1 − r n  S n = a1   1− r  10

2(0.25) i =1

i −1

Sum of a finite geometric series

1 − (0.25) = 2  1 − 0.25 ≈ 2.667

  

Substitute 2 for a1, 0.25 for r , and 10 for n. Use a calculator.

Solutions to Checkpoints

7. (a)

∞

1101

 5(0.5) = 5 + 5(0.5) + 5(0.5) +  + 5(0.5) +  n

n=0

5 1 − 0.5 5 = 0.5 = 10 =

Use

a1 and Subsitute 5 for a1 and 0.5 for r. 1− r

(b) To find the common ratio, divide any term by the preceding term. So, r = 15 = 0.2. The sum of the infinite geometric series is as follows. 5 + 1 + 0.2 + 0.04 +  = 5(0.2) + 5(0.2) + 5(0.2) + 5(0.2) +  0

a1 1− r 5 = 1 − 6.2 = 6.25 =

8. To find the balance in the account after 48 months, consider each of the 48 deposits separately. The first deposit will gain interest for 48 months, and its balance will be

0.02   A48 = 701 +  12  

= 70(1.0017) . 48

The second deposit will gain interest for 47 months, and its balance will be

0.02   A47 = 701 +  12  

= 70(1.0017) . 47

The last deposit will gain interest for only 1 month, and its balance will be 1

0.02   A1 = 701 +  = 70(1.0017). 12   The total balance in the annuity will be the sum of the balances of the 48 deposits. Using the formula for the sum of a finite geometric sequence, with A1 = 70(1.0017), r = 1.0017, and n = 48 you have

1 − r n  Sn = A1   1− r  1 − (1.0017) 48   ≈ $3500.85. S48 = 70(1.0017)   1 − 1.0017 

Checkpoints for Section 9.4 (11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7) ⋅ 6 ! 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 11 ! 1. (a) 11 = = = 462 = 5 5 ⋅ 4 ⋅ 3⋅ 2 ⋅1 6!5!   6! 5! (b) 9 C2 =

( 9 ⋅ 8) ⋅ 7 ! 9 ⋅ 8 9! = = 36 = 2 ⋅1 7!2! 7! 2!

5 ! (c)  5  = = 1  0 5! 0!

(d) 15 C15 =

15 ! 0 ! 15 !

= 1

1102 Solutions to Checkpoints

2. (a)

7 C5

7! 7⋅6⋅ 5 ⋅ 4 ⋅ 3 = = 21 2!5! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1

7! 7⋅6 7 (b)   = = = 21 2 5!2! 2 ⋅1   14 ! 14 = = 14 1 !13 ! 1

14 ! 14 14 (d)   = = = 14 1  1  13 !1 !

3. The binomial coefficients are n Cr

( n − r )!r !

 4C0 =

4! 4! 4! 4! 4! = 1, 4C1 = = 4, 4C2 = = 6, 4C3 = = 4, and 4C4 = = 1. 4!0! 3!1! 2!2! 1!3! 0!4!

So, the expansion is as follows.

( x + 1)2 = (1) x 4 + (4) x3 (1) + (6) x 2 (1)2 + ( 4) x(1)3 + (1)(1)4 = x 4 + 4 x3 + 6 x 2 + 4 x + 1

4. The binomial coefficients are n Cr

n! 4! 4! 4! 4! 4!  4C0 = = 1, 4C1 = = 4, 4C2 = = 6, 4C3 = = 4, and 4C4 = = 1. n − r ! r ! 4!0! 3!1! 2!2! 1!3! 0!4! ( )

So, the expansion is as follows.

( x − 1) = x + (−1) 4

= (1) x 4 + ( 4) x3 ( −1) + (6) x 2 ( −1) + ( 4) x( −1) + (1)( −1) 2

= x 4 − 4 x3 + 6 x 2 − 4 x + 1 5. (a) The binomial coefficients are n Cr

( n − r )!r !

 4C0 =

4! 4! 4! 4! 4! = 1, 4C1 = = 4, 4C2 = = 6, 4C3 = = 4, and 4C4 = = 1. 4!0! 3!1! 2!2! 1!3! 0!4!

So, the expansion is as follows.

(3 y − 1)

= (3 y ) + ( −1)

= (1)(3 y ) + ( 4)(3 y ) ( −1) + (6)(3 y ) ( −1) + ( 4)(3 y )( −1) + (1)( −1) 4

= 81 y 4 − 108 y 3 + 54 y 2 − 12 y + 1

(b) The binomial coefficients are n Cr

5 C5

(n − r )!r !

 5C0 =

5! 5! 5! 5! 5! = 1, 5C1 = = 5, 5C2 = = 10, 5C3 = = 10, 5C4 = = 5, and 5!0! 4!1! 3!2! 2!3! 1!4!

5! = 1. 0!5!

So, The expansion is as follows.

(2 x − y ) = (1)( 2 x) − (5)( 2 x) y + (10)(2 x) y 2 − (10)( 2 x) y 3 + (5)( 2 x) y 4 − (1) y 5 5

= 32 x 5 − 80 x 4 y + 80 x 3 y 2 − 40 x 2 y 3 + 10 xy 4 − y 5

Solutions to Checkpoints

1103

6. The binomial coefficients are n Cr

( n − r )!r !

3! 3! 3! 3! = 1, 3C1 = = 3, 3C2 = = 3, and 3C3 = = 1. So, the expansion is as 3!0! 2!1! 0!2! 0!3!

 3C0 =

follows.

(5 + y 2 ) = ( y 2 + 5) 3 = ( y 2 ) + (5) 3 2 2 3 = (1)( y 2 ) + (3)( y 2 ) (5) + (3)( y 2 )(5) + (1)(5) 3

= y 6 + 15 y 4 + 75 y 2 + 125

7. (a) Because the formula is for the ( r + 1) th term, r is one less than the number of the term you need. So, to find the fifth term

in this binomial expansion, use r = 4, n = 8, x = a, and y = 2b. n Cr x

n−r

y r = 8C4a8 − 4 ( 2b) = (70)( a 4 )( 2b) 4

= 70( 24 )a 4b 4 = 1120a 4b 4

(b) Note that ( 2a − 5b)

= ( 2a) + ( − 5b) . So, n = 11, r = 7, x = 2a, and y = − 5b. Substitute these values to obtain

the following. n Cr x

n−r

y r = 11C7 ( 2a ) ( − 5b) 4

= (330)(32a 4 )( − 78,125b 7 ) = − 412,500,000a 4b 7

So, the coefficient is − 412,500,000. 8.

1 1 1 1 1 1 1 1 1 C

9−0

8 9 C

9−1

28 36 C

9−2

10 20

1 5

9−4

1 6

9−5

1 56 70 56 28 8 1 84 126 126 84 36 9 1

9−3

1 4

10 15

3 6

5 6

3 4

1 2

9−6

9−7

9−8

8th row

9−9

Checkpoints for Section 9.5 1. To solve this problem, count the different ways to obtain a sum of 14 using two numbers from 1 to 8. First number :

6 7 8

Second number :

8 7 6

So, a sum of 14 can occur in three different ways.

1104 Solutions to Checkpoints 2. To solve this problem, count the different ways to obtain a sum of 12 using three different numbers from 1 to 8. First number: Second number: Third number:

1 1 3 3 8 8 1 1 4 4 4 7 1 1 5 5 6 6 2 2 3 3 7 7 2 2 4 4 6 6 3 3 4 4 5 5 3 8 8 1 1 3 4 7 1 7 1 4 5 6 6 1 1 5 3 7 2 7 2 3 4 6 2 6 2 4 4 5 5 3 3 4 8 3 1 8 3 1 7 4 7 1 4 1 6 5 1 6 5 1 7 3 7 2 3 2 6 4 6 2 4 2 5 4 3 5 4 3

So, a sum of 12 can occur in 36 ways. 3. There are three events in this situation. The first event is the choice of the first number, the second event is the choice of the second number, and the third event is the choice of the third number. Because there is a choice of 30 numbers for each event, it follows that the number of different lock combinations is

4. Because the product’s catalog number is made up of one letter from the English alphabet followed by a five-digit number, there are 26 choices for the first digit and 10 choices for each of the other 5 digits.

So, the number of possible catalog numbers is 26 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 = 2,600,000. 5. First position:

Any of the four letters

Second position:

Any of the remaining three letters

Third position:

Either of the remaining two letters

Fourth position:

The one remaining letter

So, the numbers of choices for the four positions are as follows. Permutations of four letters  4

President (first position): Five choices Vice-president (second position): Four choices Using the Fundamental Counting Principle, multiply these two numbers to obtain the following. Different orders of offices 

30 ⋅ 30 ⋅ 30 = 27,000.

6. Here are the different possibilities.

The total number of permutations of the four letters is 4! = 4 ⋅ 3 ⋅ 2 ⋅ 1

= 24.

President Vice-president 5

So, there are 5 ⋅ 4 = 20 different ways there can be a president and vice-president. 7. The word M I T O S I S has seven letters, of which there are two I’s, two S’s, and one M, T, and O. So, the number of distinguishable ways the letters can be written is 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2! n! 7! = 1260. = = n1 ! n2 ! 2 ! 2! 2 ! 2! 8. The following subsets represent the different combinations of two letters that can be chosen from the seven letters.

{A, B} {A, C} {A, D} {A, E} {A, F} {A, G}

{B, C} {C, D} {D, E} {E, F} {F, G} {B, D} {C, E} {D, F} {E, G} {B, E} {C, F} {D, G} {B, F} {C, G} {B, G}

From this list, you can conclude that there are 21 different ways that two letters can be chosen from seven letters.

Checkpoints for Section 9.6 1. Either coin can land heads up or tails up, and the six-sided die can land with a 1 through 6 up.

So, the sample space is as follows. S = {HH 1, HH 2, HH 3, HH 4, HH 5, HH 6 HT 1, HT 2, HT 3, HT 4, HT 5, HT 6, TH 1, TH 2, TH 3, TH 4, TH 5, TH 6, TT 1, TT 2, TT 3, TT 4, TT 5, TT 6}

Solutions to Checkpoints 2. (a) Let E = {TTT} and S = {HHH , HHT , HTH , HTT , TTT , THT , TTH , THH }.

The probability of getting three tails is

P( E ) =

n( E )

n( S )

1 . 8

(b) Because there are 52 cards in a standard deck of playing cards and there are 13 diamonds, the probability of drawing a diamond is P( E ) =

n( E ) n( S )

5. Because the deck has 4 aces, the probability of drawing an ace (event A) is P ( A) =

Similarly, because the deck has 13 spades, the probability of drawing a spade (event B) is P( B) =

Because one of the cards is an ace and a spade, the ace of spades, it follows that

2♠ 3♠ 7♠ 4♠ J♠ 5♠ 8♠ Q♠ 6♠ 9♠ K♠ 10♠ A♠

13 1 P( E ) = = = . n( S ) 52 4 For a set consisting of the aces, the sample space is 4 cards. So, the probability of drawing the ace at hearts is

n( S )

1 . 4

So, the probability of drawing a club from a standard deck of cards is the same as drawing the ace of hearts from the set of aces.

n( E ) n( S )

P(A ∩ B) Aces

A♣

A♥

Finally, applying the formula for the probability of the union of two events, the probability of drawing an ace or spade is as follows.

P( A ∪ B ) = P( A) + P( B) − P( A ∩ B )

4. The total number of colleges and universities is 4706. Because there are 640 colleges and universities in the Pacific region, the probability that the institution is in that region is

P( E ) =

1 . 52

A♦

n( E )

13 . 52

Spades

3. For a standard deck of 52 playing cards, there are 13 clubs. So, the probability of drawing a club is

P( E ) =

4 . 52

P( A ∩ B) =

13 = 52 1 = . 4

1105

4 13 1 + − 52 52 52 16 = ≈ 0.308 52 =

640 ≈ 0.136. 4706

6. To begin, add the number of employees to find that the total is 529. Next, let event A represent choosing an employee with 30-34 years of service, event B with 35-39 years of service, event C with 40-44 years of service, and event D with 45 or more years of service.

Because events A, B, C, and D have no outcomes in common, these four events are mutually exclusive and 35 21 8 2 66 + + + = ≈ 0.125. 529 529 529 529 529 So, the probability of choosing and employee who has 30 or more years of service is about 0.125. P ( A ∪ B ∪ C ∪ D ) = P ( A) + P ( B ) + P (C ) + P ( D ) =

7. The probability of selecting a number from 1 to 11 from a set of numbers from 1 to 30 is P ( A) =

11 . 30

So, the probability that both numbers are less than 12 is P ( A) ⋅ P ( A) =

11 11 121 ⋅ = ≈ 0.134. 30 30 900

8. Let A represent choosing an adult who says that his or her television would be very hard to give up. The probability of choosing an adult who says that his or her television would be very hard to give up is 0.35, the probability of choosing a second adult who says that his or her television would be very hard to give up is 0.35, and so on. Because these events are independent, the probability that all eight people say that his or her television would be very hard to give up is 8

P( A) = (0.35) = 0.000225. 8

1106 Solutions to Checkpoints

Chapter 10 Checkpoints for 10.1 1. The radius of the circle is the distance between (3, 1) and (1, −1).

r =

(3 − 1) + (1 − (−1)) 2

22 + 2 2 =

8 = 2 2

3. To find any x-intercept(s), let y = 0 and solve for x.

( x − 2) + (0 + 3) = 9 2 ( x − 2) = 0 2

x = 2

4+4

(2, 0) To find any y-intercept(s), let x = 0 and solve for y.

The equation of the circle with center ( h, k ) = (1, −1) and radius r = 2 2 = 2

= r2

( x − 1) + ( y − (−1))

( x − 1) + ( y + 1)

(0 − 2) + ( y + 3) = 9 2 4 + ( y + 3) = 9 2 ( y + 3) = 5 2

8 is as follows.

( x − h) + ( y − k )

( 8)

So, the x-intercept is ( 2, 0) and the y-intercepts are

(0, − 3 +

x + 4x + y + 4 y − 1 = 0

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

= 9

Then use the fact that the center is ( − 2, − 2) and the radius is r = 9 = 3 to plot a few additional points. The points ( − 5, − 2), ( − 2, 1), (1, − 2), and ( − 2, − 5) are convenient. Draw a circle that passes through the points. y

2 1 x

(0, − 3 ± 5 )

= 8

y = −3 ±

2. First, write the equation in standard form.

−6 −5 −4 −3 −2 −1 −1

−2

)

(

5 and 0, − 3 −

)

4. The axis of the parabola is vertical, passing through 3 (0, 0) and  0, . The standard form is x 2 = 4 py,  8 3 where p = . 8  3 So, the equation is x 2 = 4  y 8 3 2 x = y. 2 You can use a graphing utility to confirm this equation. 2 To do this, graph y1 = x 2 . 3 2

−3

−4 −5

x2 = 3 y

Focus 0, 3

) 8)

−6 −1

(0, 0)Vertex

−1

5. Convert to standard form by completing the square.

1 2 3 13 y + y + 4 2 4 4 x = y 2 + 6 y + 13 x =

Write orginal equation. Multiply each side by 4.

4 x − 13 = y + 6 y

Subtract 13 from each side.

4 x − 13 + 9 = y 2 + 6 y + 9

Add 9 to each side.

4x − 4 = y2 + 6 y + 9

Combine like terms.

4( x − 1) = ( y + 3)

Standard form

Comparing this equation with ( y − k ) = 4 p( x − h), you can conclude that h = 1, k = − 3, and p = 1. Because p is 2

positive, the parabola opens to the right. So, the focus is( h + p, k ) = (1 + 1, − 3) = ( 2, − 3). .

Solutions to Checkpoints

1107

6. Because the axis of the parabola is horizontal, and passes through ( 2, − 3) and ( 4, − 3), consider the equation

( y − k ) = 4 p( x − h), 2

where h = 2, k = − 3, and p = 4 − 2 = 2. So, the standard form is 2

 y − ( − 3) = 4( 2)( x − 2)

( y + 3) 2

= 8( x − 2).

(y + 3)2 = 8(x − 2)

−2

Focus (4, − 3)

Vertex (2, − 3) −8

7. For this parabola, p = y = 3x 2

1  1 and the focus is  0,  as shown in the figure below. 12  12 

( ( 0, 1 12

(1, 3)

α −1

α (0, b)

You can find the y-intercept (0, b) of the tangent line by equating the lengths of the two sides of the isosceles triangle shown in the figure: d1 =

1 − b 12

and d2 = =

(1 − 0) +  3 − 2



1  12 

 35  12 +    12 

37 . 12

1 1 . The order of subtraction for the distance is important because the distance must − b rather than b − 12 12 be positive.) Setting d1 = d 2 produces

(Note that d1 =

1 37 −b = 12 12 b = − 3. So, the slope of the tangent line through (0, − 3) and (1, 3) is

m =

3 − ( − 3) 1−0

= 6

and the equation of the tangent line in slope-intercept form is y = 6 x − 3.

1108 Solutions to Checkpoints

Checkpoints for Section 10.2 1. Because the foci occur at ( 2, 0) and ( 2, 6), the center of the ellipse is ( 2, 3) and the distance from the center to one of the foci

is c = 3. Because 2 a = 8, you know that a = 4. Now, from c 2 = a 2 − b 2 , you have a2 − c2 =

b =

42 − 32 =

Because the major axis is vertical, the standard equation is as follows.

( x − h) + ( y − k )

( x − 2) + ( y − 3)

( 7)

(2, 6)

(2, 3) (2, 0)

−2

( x − 2) + ( y − 3) 2

a=4

−1

b= 7

2. Algebraic Solution x 2 + 4 y 2 = 16 x2 y2 + =1 16 4

The center of the ellipse is (0, 0). Because the denominator of the x 2 -term is greater than that of the y 2 -term, you can conclude that the major axis is horizontal. Additionally, because a = 4, the vertices are ( − 4, 0) and ( 4, 0), and because b = 2, the endpoints of the minor axis are (0, − 2) and (0, 2). y

5 4 3

(−4, 0) −5

x2 y2 + =1 42 22

(0, 2) (4, 0)

−2 −1

2 3

−3 −4 −5

(0, −2)

Graphical Solution

Solve the equation of the ellipse for y. x 2 + 4 y 2 = 16 4 y 2 = 16 − x 2 y2 =

16 − x 2 4

y = ±

16 − x 2 4

Then use a graphing utility to graph

y1 = 12 16 − x2 and y2 = − 12 16 − x2 in the same viewing window. Be sure to use a square setting. 4

−6

−4

The center of the ellipse is( 0, 0) and the major axis is horizontal. The vertices are( − 4, 0) and( 4, 0). .

Solutions to Checkpoints

1109

3. Begin by writing the original equation in standard form. 9 x 2 + 4 y 2 + 36 x − 8 y + 4 = 0 2

(

Write original equation.

9 x + 36 x + 4 y − 8 y = − 4

Group terms.

) + 4( y − 2 y + ) = − 4

Factor out leading coefficients.

9 x + 4x +

9( x 2 + 4 x + 4) + 4( y 2 − 2 y + 1) = − 4 + 36 + 4

Complete the square.

9( x + 2) + 4( y − 1) = 36 2

( x + 2)

( y − 1) 9

( y − 1)

Write in completed square form.

Divide each side by 36.

Write in standard form.

The center is ( h, k ) = ( − 2, 1). Because the denominator of the y-term is a 2 = 32 , the endpoints of the major axis lie three units above and below the center. So, the vertices are ( − 2, 4) and ( − 2, − 2). Similarly, because the denominator of the x-term is b 2 = 22 , the endpoints of the minor axis lie two units to the right and left of the center at ( − 4, 1) and (0, 1). (y − 1)2 (x + 2)2 + 22 = 1 32 4

(− 2, 4)

−8

(− 4, 1)

(0, 1)

(− 2, 2) −4

4. By completing the square, you can write the original equation in standard form. 5 x 2 + 9 y 2 + 10 x − 54 y + 41 = 0 2

(

Write original equation.

5 x + 10 x + 9 y − 54 y = − 41 2

5 x + 2x +

Group terms.

) + 9( y − 6 y + ) = − 41

Factor out leading coefficients.

5( x + 1) + 9( y − 3) = 45

Write in completed square form.

5( x 2 + 2 x + 1) + 9( y 2 − 6 y + 9) = − 41 + 5 + 81 2

( x + 1)

( y − 3)

( 5)

Divide each side by 45.

Write in standard form.

The major axis is horizontal, where h = −1, k = 3, a = 3, b =

5 and c = =

So, you have the following. Center: ( −1, 3) Vertices: ( − 4, 3), ( 2, 3) Foci: ( − 3, 3), (1, 3)

(x + 1)2 (y − 3)2 + ( 5 )2 = 1 32 6

a 2 − b2 32 −

( 5) = 2

focus

4 = 2. −6

center vertex −2

vertex 6

1110 Solutions to Checkpoints 5.

aphelion Encke’s coment Sun

2.345 astronomical units ≈ 218.885 million miles

4.429 astronomical units ≈ 411.897 million miles

perihelion

Because 2 a = 411.897 and 2b = 218.885, you have a ≈ 205.95 and b ≈ 109.44 a2 − b2

which implies that c =

205.952 − 109.44 2

≈ 174.47.

So, the greatest distance, the aphelion, from the sun’s center to the comet’s center is

a + c ≈ 205.95 + 174.47 = 380.42 million miles and the least distance, the perihelion, is a − c = 205.95 − 174.47 = 31.48 million miles.

Checkpoints for Section 10.3 1. By the Midpoint Formula, the center of the hyperbola occurs at the point

(h, k ) =  

2 + 2 4 + 2 ,−  = ( 2, −1). 2 2 

Furthermore, a = 2 − ( −1) = 3 and c = 3 − ( −1) = 4, and it follows that b =

c2 − a2 =

So, the hyperbola has a vertical transverse axis and the standard form of the equation is

( y − k)

−

( x − h)

( y + 1)2 − ( x − 2)2 = 1.

( 7)

This equation simplifies to

( y + 1)2 − ( x − 2)2 = 1 9

(y + 1)2 (x − 2)2 =1 − 9 7

(2, 3) (2, 2) 4 (2, − 1) (2, − 4) (2, − 5)

−8

−4

42 − 32 =

Solutions to Checkpoints

1111

2. Algebraic Solution

Divide each side of the equation by 36, and write the equation in standard form.

4 y 2 − 9 x 2 = 36 y2 x2 − =1 9 4 y2 x2 − 2 =1 2 3 2 From this, you can conclude that a = 3, b = 2, and the transverse axis is vertical. Because the center is (0, 0), the vertices occur at (0, 3) and (0, − 3). The endpoints of the conjugate axis occur at ( 2, 0) and ( − 2, 0). Using these four points, sketch a rectangle that is 2a = 6 units tall and 2b = 4 units wide. Finally, by drawing the asymptotes through the corners of this rectangle, you can complete the sketch. Note that the 3 3 asymptotes are y = x and y = − x. 2 2 y

(0, 3)

(− 2, 0)

(2, 0)

−4

y2 x2 − =1 9 4

−2 −4

(0, − 3)

Graphical Solution

Solve the equation of the hyperbola for y as follows.

4 y 2 − 9 x 2 = 36 4 y 2 = 9 x 2 + 36 y2 =

9 x 2 + 36 4 9 x 2 + 36 4

y = ± y = ±

3 2

x2 + 4

Then use a graphing utility to graph y1 = From the graph, you can see that the traverse axis is vertical and the vertices are − 8 (0, 3) and (0, − 3).

y1 = 32 x2 + 4 8

y2 = − 32 −8

3 2

x2 + 4

x 2 + 4 and y2 = −

3 2

x 2 + 4 in the same viewing window.

1112 Solutions to Checkpoints 3.

9 x 2 − 4 y 2 + 8 y − 40 = 0 2

Write original equation.

9 x − 4 y + 8 y = 40

Group terms.

(

Factor out leading coefficients.

9x − 4 y − 2 y +

) = 40

9 x 2 − 4( y 2 − 2 y + 1) = 40 − 4 9 x 2 − 4( y − 1) = 36 2

( y − 1) x2 − 4 9

( y − 1) x2 − 22 32

Complete the square. Write in completed square form. y

(0, 4)

Divide each side by 36.

Write in standard form.

(− 13, 1) (− 2, 1)

From this equation you can conclude that the hyperbola has a horizontal transverse axis centered at (0, 1). The vertices are at ( − 2, 1) and ( 2, 1), and has a conjugate axis with

−4 −3

endpoints (0, 4) and (0, − 2). To sketch the hyperbola, draw a rectangle through these Using a = 2 and b = 3, you can conclude that the equations of the asymptotes are y =

( 13, 1)

(0, 1) (2, 1) 3

−1

(0, − 2)

y = 32 x + 1 y = − 32 x + 1

3 3 x + 1 and y = − x + 1. 2 2

Finally, you can determine the foci by using the equation c 2 = a 2 + b 2 . So, you have c =

( 13, 1) and (− 13, 1).

−3

four points. The asymptotes are the lines passing through the corners of the rectangle.

are

x 2 (y − 1)2 =1 − 4 9

22 + 32 =

13, and the foci

3 + 9 2 + 2 4. Using the Midpoint Formula you can determine the center is ( h, k ) =  ,  = (6, 2). 2   2

Furthermore, the hyperbola has a horizontal transverse axis with a = 3. From the original equations, you can determine the b 2 b 2 slopes of the asymptotes to be m1 = = and m2 = − = − , and because a = 3, you can conclude that b = 2. a 3 a 3 So, the standard form of the equation of the hyperbola is

( x − 6)2 − ( y − 2)2 = ( x − 6)2 − ( y − 2)2 = 1. 32

5. Assuming sound travels at 1100 feet per second, the explosion took place 4400 feet farther from B than from A. The locus of all points that are 4400 feet closer to A than to B is one branch of a hyperbola with foci at A and B. Because the hyperbola is centered at the origin and has a horizontal transverse axis, the standard form of its equation is

x2 y2 − = 1. a2 b2 From Example 5, it was determined that the foci are 2640 units from the center, c = 2640. Let d A and d B be the distances of any point on the hyperbola from the foci at A and B, respectively. So, you have

4400 A B

d B − d A = 2a x

4400 = 2 a

The points are 4400 feet closer to A than to B.

2200 = a 2

4400

2000

Divide each side by 2. 2

So, b = c − a = 26402 − 22002 = 2,129,600 and you can conclude that the explosion occurred somewhere on the right branch of the hyperbola x2 y2 − = 1. 4,840,000 2,129,600

Solutions to Checkpoints

1113

6. (a) For the equation 3 x 2 + 3 y 2 − 6 x + 6 y + 5 = 0, you have A = C = 3. So, the graph is a circle.

(b) For the equation 2 x 2 − 4 y 2 + 4 x + 8 y − 3 = 0, you have AC = 2( − 4) < 0. So, the graph is a hyperbola. (c) For the equation 3 x 2 + y 2 + 6 x − 2 y + 3 = 0, you have AC = 3(1) > 0. So, the graph is an ellipse. (d) For the equation 2 x 2 + 4 x + y − 2 = 0, you have AC = 2(0) = 0. So, the graph is a parabola.

Checkpoints for Section 10.4 1. Using values of t in the specified interval, − 2 ≤ t ≤ 2, 2

the parametric equations x = 2t and y = 4t + 2 yield the points ( x, y ) shown. t

−2

−4

−1

−2

2. x = (t − 2) and y = 3t 2 2

First, set the graphing utility to parametric mode. Enter the parametric equations for x and y. From the graph, you can see that y is a function of x since it passes the Vertical Line Test.

By plotting these points in the order of increasing values of t, you obtain the curve shown. The arrows on the curve indicate its orientation as t increases from − 2 to 2.

18 16 14 12 10 8 6 4

−5 −4 −3 −2 −1

−1

8 −1

1 2 3 4 5

1114 Solutions to Checkpoints

3. x =

t +1 1 and y = t −1 t −1

Solving for t in the equation for x produces the following. 1 x = t −1 1 2 x = t −1 1 t −1 = 2 x 1 t = 2 +1 x 1 + x2 t = x2

1 + x2 +1 x2 y = 2 1+ x −1 x2 1 + x2 + x2 x2 = 1 + x2 − x2 x2 2x2 + 1 x2 = 1 x2 2x2 + 1 x2 ⋅ = x2 1 = 2x2 + 1

y = 2x 2 + 1 −4

−1

t+1

y= t−1 x= t=2 t=3 t=4

1 t−1

4 −1 5

y = 2x 2 + 1, x > 0

t = −1 x=t y = t2 + 2

t=1 t=0

−3

3 −1

(b) Letting t = 2 − x , you obtain the parametric equations

t = 2 − x  x = 2 − t and The curve represented by the parametric equations is shown. 5

t=1 t=2 3 −1

4 −1

−3

for all values of x. The parametric equation for x is defined only when t > 1. This implies that you should restrict the domain of x to positive values.

−4

The curve represented by the parametric equations is shown.

t=3 x=2− t 2 y = t − 4t + 6

From this rectangular equation, you can recognize that the curve is a parabola that opens upward and has its vertex at (0, 1). Also, this rectangular equation is defined

−4

x = t and y = x 2 + 2 = t 2 + 2.

y = x 2 + 2 = t 2 − 4t + 6.

Now, substituting in the equation for y, you obtain the following rectangular equation.

4. (a) Letting t = x, you obtain the parametric equations

Solutions to Checkpoints

1115

Checkpoints for Section 10.5  π 1. (a) The point ( r , θ ) =  3,  lies three units  4 from the pole on the terminal side of the angle π θ = . 4

π  (b) The point ( r , θ ) =  2, −  lies two 3  units from the pole on the terminal side π of the angle θ = − 3

π 2

3π 2

5π  5π  (c) The point ( r , θ ) =  2, , which coincides with  lies two units from the pole on the terminal side of the angle θ = 3 3   π  the point  2, − . π 2 3 

3π 2

2. (a) The point is shown. Three other representations are as follows.

 4π   10π  + 2π  =  3,  3,  3   3  

Add 2π to θ .

4π π    − π  =  − 3,   − 3, 3 3   

Replace r with − r and subtract π from θ .

4π 7π     + π  =  − 3,  − 3,  3 3    

Replace r with − r and add π to θ .

π 2

3π 2

π  7π   4π   10π    3,  =  3,  =  − 3,  =  − 3,  = 3   3  3   3  

1116 Solutions to Checkpoints

(b) The point is shown. Three other representations are as follows: 5π    7π  + 2π  =  2,  2, −  6    6 

Add 2π to θ .

5π 11π     − π  =  − 2, −  − 2, −  6 6    

Replace r with − r and subtract π from θ .

5π π    + π  =  − 2,   − 2, − 6 6   

Replace r with − r and add π to θ .

π 2

3π 2

5π  11π  π   7π     2, −  =  2,  =  − 2, −  =  − 2,  =  6  6  6  6    

(c) The point is shown. Three other representations are as follows. 3π    11π  + 2π  =  −1,  −1,  Add 2π to θ . 4 4    

π 2

π  3π   − π  = 1, −  1, 4  4  

Replace r with − r and subtract π from θ .

 3π   7π  + π  = 1, 1,  4    4 

Replace r with − r and add π to θ .

3π   11π   π   7π    −1,  =  −1,  = 1, −  = 1,  = 4   4   4  4    π 3. For the point ( r , θ ) =  4,  , you have the following.  3

x = r cos θ = 4 cos

π 3

1 = ( 4)  = 2 2

4 2

) )

π 3 (x, y) = (2, 2 4

y-axis, you have 2 π is undefined  θ = . 2 0

)

x2 + y 2 =

 π So, one set of polar coordinates is ( r , θ ) =  2,  as  2 shown.

3) x

4 2

) π2 )

(r, θ) = 2,

(x, y) = (0, 2) 2

02 + 22 = 2.

(r, θ) = 4,

4. For the point ( x, y ) = (0, 2), which lies on the positive

r =

The rectangular coordinates are ( x, y ) = 2, 2 3 . y

Choosing a positive value for r,

 3 = ( 4)  = 2 3 3  2 

(

3π 2

tan θ =

y = r sin θ = 4 sin

Solutions to Checkpoints

1117

5. The graph of the polar equation r = 6 sin θ is not evident by simple inspection, so convert to rectangular form.

Because r = 6 sin θ  r 2 = 6r sin θ , use the relationships r 2 = x 2 + y 2 and y = r sin θ .

r 2 = 6r sin θ x2 + y2 = 6 y Now you see that the graph is a circle. The graph of x 2 + y 2 = 6 y in standard form is x2 + y2 − 6 y = 0

x2 + ( y − 3) = 9, which is a circle centered at (0, 3) with a radius of 3. 2

Checkpoints for Section 10.6 1. The cosine function is periodic, so you can get a full range of r-values by considering values of θ in the interval 0 ≤ θ ≤ 2π , as shown.

2π 3

3π 4

5π 6

3 3

3 2

−3

−3 2

−3 3

−6

7π 6

5π 4

4π 3

3π 2

5π 3

7π 4

11π 6

2π

−3 3

−3 2

−3

3 2

3 3

By plotting these points, it appears that the graph is a circle of radius 3 whose center is at the point ( x, y ) = (3, 0). π 2 3 2 1

3π 2

1118 Solutions to Checkpoints 2. Replacing ( r , θ ) by ( r , π − θ ) produces the following. r = 3 + 2 sin (π − θ ) = 3 + 2 (sin π cos θ − cos π sin θ ) = 3 + 2 (0) cos θ − ( −1) sin θ  = 3 + 2 (0 + sin θ ) = 3 + 2 sin θ

So, you can conclude that the curve is symmetric with respect to the line θ = π 2. Plotting the points in the table and using symmetry with respect to the line θ = π 2, you obtain the graph shown, called a dimpled limaçon.

2π 3

7π 6

3+ 5π 4

3−

2 4π 3

3−

3 3π 2

5π 3

3−

3π 4

11π 6

2π

3 7π 4

5π 6

3−

π 2 6 4 3 2 1

3 4 5 3π 2

3. From the equation r = 4 sin 2θ , you can obtain the following.

Symmetry: With respect to the line θ =

π 2

π 3π 5π 7π

Maximum value of r : r = 2, when 2θ =

, , , 2 2 2 2 π 3π 5π 7π or θ = , , , 4 4 4 4

Zeros of r : r = 0 when 2θ = 0, π , 2π , 3π

or θ = 0,

π 2

, π,

3π 2

By plotting these points using symmetry with respect to θ =

π 2

3π 4

4 4

−4

π 2

, zeros, and maximum values, you obtain the graph shown. π 2

5π 4

3π 2

7π 4

2π

−4

4 3

3π 2

Solutions to Checkpoints

r = 3 sin 3θ

4. Equation:

5. Equation: r 2 = 4 cos 2θ

Type of curve: Lemniscate Symmetry : With respect to the polar axis,

Type of curve: Rose curve with n = 3 petals Symmetry: With respect to the line θ =

π 2

the line θ =

π 2π 3

, and the pole 2 Maximum value of r : r = 2 when θ = 0, π

Zeros of r: r = 0 when 3θ = 0, π , 2π , 3π

or θ = 0,

,π

Zeros of r : r = 0 when θ =

Using a graphing utility in polar mode, enter the equation ( with 0 ≤ θ ≤ π ). You obtain the graph shown.

−6

π 3π

, 4 4 Use a graphing utility in polar mode, enter the equations r1 = 4 cos 2θ and 3π   4 cos 2θ  with − 2π ≤ θ ≤ . 2   You obtain the graph shown. r2 = −

−4

−6

−4

Checkpoints for Section 10.7 1. Algebraic Solution

To identify the type of conic, rewrite the equation in the form r = r = r =

8 2 − 3 sin θ

ep . 1 ± e sinθ

Write original equation.

Divide numerator and denominator by 2.

1 − 32 sin θ

3 > 1, you can conclude the graph is a hyperbola. 2

Because e =

Graphical Solution

Use a graphing utility in polar mode and be sure to use a square setting as shown. 5 − 12

) 85 , 32π )

8 2 − 3sin θ

)− 8, π2) − 15

The graph of the conic appears to be a hypberbola.

1119

1120 Solutions to Checkpoints 2. Dividing the numerator and denominator by 2, you have r =

32 . 1 − 2 sin θ

Because e = 2 > 1, the graph is a hyperbola. The transverse axis of the hyperbola lies on the line θ =

 3 π  1 3π  , and the vertices occur at  − ,  and  , . 2  2 2 2 2 

Because the length of the transverse axis is 1, you can see that a =

1 . To find b, write 2

3 2 1 b 2 = a 2 (e 2 − 1) =   ( 2) − 1 = .   2 4   So, b =

3 3 . You can use a and b to determine that the asymptotes of the hyperbola are y = −1 ± x. 2 3 r=

3 2 − 4sin θ

) π) 1 3 , 2 2

−3

)

3 π − , 2 2

)

−4

3. Because the directrix is vertical and to the left of the pole, use an equation of the form r =

ep . 1 + e cos θ

Moreover, because the eccentricity of a parabola is e = 1 and the distance between the pole and the directrix is p = 2, you have the equation r =

ep 2 . = 1 + e cos θ 1 − cos θ

4. Using a vertical major axis as shown, choose an equation of the form r =

Because the vertices of the ellipse occur when θ =

π 2

3π , you can determine the length of the major axis to be the 2 π

and θ =

sum of the r-values of the vertices. 2a =

ep . 1 + e sin θ

Sun 2

0.848 p 0.848 p + ≈ 6.038 p ≈ 4.429 1 + 0.848 1 − 0.848

Encke’s comet

So, p ≈ 0.734 and ep ≈ (0.848)(0.734) ≈ 0.622. Using this value of ep in the equation, you have r =

ep 0.622 = , 1 + e sin θ 1 + 0.848 sin θ

where r is measured in astronomical units. To find the closest point to the sun (the focus), substitute θ = r =

π 2

into this equation.

0.622 0.622 = 1 + 0.848 sin π 2 1 + 0.848(1) ≈ 0.337 astronomical unit ≈ 31,400,000 miles

3π 2