1. 2 - 1 © 2012 Pearson Education, Inc.. All rights reserved. Chapter 2 Nonlinear Functions

2. 2 - 2 © 2012 Pearson Education, Inc.. All rights reserved. Figure 1

3. 2 - 3 © 2012 Pearson Education, Inc.. All rights reserved. Section 2.1 Properties of Functions

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8. 2 - 8 © 2012 Pearson Education, Inc.. All rights reserved. Figure 5a - 5b

9. 2 - 9 © 2012 Pearson Education, Inc.. All rights reserved. Figure 5c - 5d

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11. 2 - 11 © 2012 Pearson Education, Inc.. All rights reserved. Figure 6

12. 2 - 12 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 Find the domain and range for the function Solution: The domain includes only those values of x satisfying since the denominator cannot be zero. Using the methods for solving a quadratic inequality produces the domain Because the numerator can never be zero, the denominator can take on any positive real number except for 0, allowing y to take on any positive value except for 0, so the range is

13. 2 - 13 © 2012 Pearson Education, Inc.. All rights reserved. Figure 7

14. 2 - 14 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 Given the function find each of the following. (a) (b) All values of x such that (a) Solution: Replace x with the expression x + h and simplify. (b) Solution: Set f (x) equal to − 5 and then add 5 to both sides to make one side equal to 0. Continued

15. 2 - 15 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 continued This equation does factor as Set each factor equal to 0 and solve for x.

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19. 2 - 19 © 2012 Pearson Education, Inc.. All rights reserved. Figure 10

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23. 2 - 23 © 2012 Pearson Education, Inc.. All rights reserved. Section 2.2 Quadratic Functions; Translation and Reflection

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25. 2 - 25 © 2012 Pearson Education, Inc.. All rights reserved. Figure 14

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27. 2 - 27 © 2012 Pearson Education, Inc.. All rights reserved. Figure 16

28. 2 - 28 © 2012 Pearson Education, Inc.. All rights reserved. Figure 17

29. 2 - 29 © 2012 Pearson Education, Inc.. All rights reserved. Figure 18

30. 2 - 30 © 2012 Pearson Education, Inc.. All rights reserved.

31. 2 - 31 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 For the function (a) complete the square, (b) find the y-intercept, (c) find the x intercepts, (d) find the vertex, and (e) sketch the graph. Solution (a): To begin, factor 2 from the x-terms so the coefficient of x 2 is 1: Next, we make the expression inside the parentheses a perfect square by adding the square of one-half of the coefficient of x, Continued

32. 2 - 32 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 continued Solution (b):The y-intercept (where x = 0) is − 1. Solution (c): To find the x-intercepts, solve Use the quadratic formula to verify that the x-intercepts are at Solution (d): The function is now in the form Continued

33. 2 - 33 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 continued

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37. 2 - 37 © 2012 Pearson Education, Inc.. All rights reserved. Figure 23-25

38. 2 - 38 © 2012 Pearson Education, Inc.. All rights reserved. Figure 26 - 27

39. 2 - 39 © 2012 Pearson Education, Inc.. All rights reserved. Figure 28 - 30

40. 2 - 40 © 2012 Pearson Education, Inc.. All rights reserved. Figure 31

41. 2 - 41 © 2012 Pearson Education, Inc.. All rights reserved. Figure 32

42. 2 - 42 © 2012 Pearson Education, Inc.. All rights reserved. Section 2.3 Polynomial and Rational Functions

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44. 2 - 44 © 2012 Pearson Education, Inc.. All rights reserved. Figure 33

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46. 2 - 46 © 2012 Pearson Education, Inc.. All rights reserved. Figure 35

47. 2 - 47 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 Graph Solution: Using the principles of translation and reflection, we recognize that this is similar to the graph of but reflected vertically (because of the negative in front of x 6 ) and 64 units up.

48. 2 - 48 © 2012 Pearson Education, Inc.. All rights reserved. Figure 36 - 37

49. 2 - 49 © 2012 Pearson Education, Inc.. All rights reserved. Figure 38

50. 2 - 50 © 2012 Pearson Education, Inc.. All rights reserved. Figure 39

51. 2 - 51 © 2012 Pearson Education, Inc.. All rights reserved. Figure 40

52. 2 - 52 © 2012 Pearson Education, Inc.. All rights reserved. Figure 41

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56. 2 - 56 © 2012 Pearson Education, Inc.. All rights reserved. Figure 42

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58. 2 - 58 © 2012 Pearson Education, Inc.. All rights reserved. Figure 43

59. 2 - 59 © 2012 Pearson Education, Inc.. All rights reserved. Figure 44

60. 2 - 60 © 2012 Pearson Education, Inc.. All rights reserved. Figure 45

61. 2 - 61 © 2012 Pearson Education, Inc.. All rights reserved. Section 2.4 Exponential Functions

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63. 2 - 63 © 2012 Pearson Education, Inc.. All rights reserved. Figure 46

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65. 2 - 65 © 2012 Pearson Education, Inc.. All rights reserved. Figure 48

66. 2 - 66 © 2012 Pearson Education, Inc.. All rights reserved.

67. 2 - 67 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 Solve Solution: Since the bases must be the same, write 25 as 5 2 and 125 as 5 3.

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69. 2 - 69 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 Find the interest earned on $4400 at 3.25% interest compounded quarterly for 5 years. Solution: Use the formula for compound interest with P = 4400, r = 0.0325, m = 4, and t = 5. The investment plus the interest is $5172.97. The interest amounts to $5172.97 − $4400 = $772.97.

70. 2 - 70 © 2012 Pearson Education, Inc.. All rights reserved. Figure 49

71. 2 - 71 © 2012 Pearson Education, Inc.. All rights reserved. Figure 50

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73. 2 - 73 © 2012 Pearson Education, Inc.. All rights reserved.

74. 2 - 74 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 3 Find the amount after 4 years if $800 is invested in an account earning 3.15% compounded continuously. Solution: In the formula for continuous compounding, let P = 800, t = 4 and r = 0.0315 to get or $907.43.

75. 2 - 75 © 2012 Pearson Education, Inc.. All rights reserved. Figure 51- 52

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77. 2 - 77 © 2012 Pearson Education, Inc.. All rights reserved. Section 2.5 Logarithmic Functions

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79. 2 - 79 © 2012 Pearson Education, Inc.. All rights reserved. Example 1

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81. 2 - 81 © 2012 Pearson Education, Inc.. All rights reserved. Figure 54

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83. 2 - 83 © 2012 Pearson Education, Inc.. All rights reserved. Example 3 If all the following variable expressions represent positive numbers, then for a > 0, a ≠ 1, the statements in (a)–(c) are true.

84. 2 - 84 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 3 Write the expression as a sum, difference, or product of simpler logarithms. Solution: Using the properties of logarithms,

85. 2 - 85 © 2012 Pearson Education, Inc.. All rights reserved. Figure 55

86. 2 - 86 © 2012 Pearson Education, Inc.. All rights reserved.

87. 2 - 87 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 4 Evaluate Solution: Using the change-of-base theorem for logarithms with x = 50 and a = 3 gives

88. 2 - 88 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 5 Solve for x: Solution: This leads to two solutions: x = − 4 and x = 2. But notice that x = − 4 is not a valid value for x in the original equation, since the logarithm of a negative number is undefined. The only solution is, therefore, x = 2.

89. 2 - 89 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 6 Solve for x: Solution: Taking natural logarithms on both sides gives

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91. 2 - 91 © 2012 Pearson Education, Inc.. All rights reserved. Figure 56

92. 2 - 92 © 2012 Pearson Education, Inc.. All rights reserved. Figure 57

93. 2 - 93 © 2012 Pearson Education, Inc.. All rights reserved. Section 2.6 Applications: Growth and Decay; Mathematics of Finance

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95. 2 - 95 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 Yeast in a sugar solution is growing at a rate such that 5 g grows exponentially to 18 g after 16 hours. Find the growth function, assuming exponential growth. Solution: The values of y 0 and k in the exponential growth function y = y 0 e kt must be found. Since y 0 is the amount present at time t = 0, y 0 = 5. To find k, substitute y = 18, t = 16, and y 0 = 5 into the equation y = y 0 e kt. Now take natural logarithms on both sides and use the power rule for logarithms and the fact that Continued

96. 2 - 96 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 continued The exponential growth function is where y is the number of grams of yeast present after t hours.

97. 2 - 97 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 Estimate the age of a sample with 1/10 the amount of carbon- 14 as a live sample. Solution: Continued

98. 2 - 98 © 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 Continued

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100. 2 - 100 © 2012 Pearson Education, Inc.. All rights reserved. Figure 58

101. 2 - 101 © 2012 Pearson Education, Inc.. All rights reserved. Graphs of Basic Functions

102. 2 - 102 © 2012 Pearson Education, Inc.. All rights reserved. Chapter 2 Extended Application

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105. 2 - 105 © 2012 Pearson Education, Inc.. All rights reserved. Figure 61