# Solution Manual for College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14th Edition

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2 FUNCTIONS EXERCISE 2-1 2. 4. 6. 8. 10. The table specifies a function, since for each domain value there corresponds one and only one range value. 12. The table does not specify a function, since more than one range value corresponds to a given domain value. (Range values 1, 2 correspond to domain value 9.) 14. This is a function. 16. The graph specifies a function; each vertical line in the plane intersects the graph in at most one point. 18. The graph does not specify a function. There are vertical lines which intersect the graph in more than one point. For example, the y-axis intersects the graph in two points. 20. The graph does not specify a function. 22. y ๏ฝ 4x ๏ซ 26. x ๏ซ xy ๏ซ 1 ๏ฝ 0 is neither linear nor constant. 1 x is neither linear nor constant. 24. 28. 2 x ๏ญ 4 y ๏ญ 6 ๏ฝ 0 is linear. y ๏ญ x 3 ๏ซ 2x 1 ๏ซ ๏ฝ 1 simplifies to y ๏ฝ 2 4 2 constant. Copyright ยฉ 2019 Pearson Education, Inc. 2-1 2-2 CHAPTER 2: FUNCTIONS 30. 32. 34. 36. 38. 3x 2 . Since the denominator is bigger than 1, we note that the values of f are between 0 and 3. x2 ๏ซ 2 Furthermore, the function f has the property that f(โx) = f(x). So, adding points x = 3, x = 4, x = 5, we have: f(x) = x F(x) โ5 โ4 โ3 โ2 โ1 0 1 2 3 4 5 2.78 2.67 2.45 2 1 0 1 2 2.45 2.67 2.78 The sketch is: 40. y = f(4) = 0 42. y = f(โ2) = 3 44. f ( x ) ๏ฝ 4 at x ๏ฝ 5. 46. f ( x ) ๏ฝ 0 at x ๏ฝ ๏ญ5, 0, 4. 48. Domain: all real numbers. 50. Domain: all real numbers except x = 2. 52. Domain: x ๏ณ ๏ญ5 or [ ๏ญ5, ๏ฅ ). 54. Given 6 x ๏ญ 7 y ๏ฝ 21 . Solving for y we have: ๏ญ7 y ๏ฝ 21 ๏ญ 6 x and y ๏ฝ 6 7 x ๏ญ 3. This equation specifies a function. The domain is R, the set of real numbers. Copyright ยฉ 2019 Pearson Education, Inc. EXERCISE 2-1 56. Given x ( x ๏ซ y ) ๏ฝ 4 . Solving for y we have: xy ๏ซ x ๏ฝ 4 and y ๏ฝ 58. Given x ๏ซ y ๏ฝ 9. Solving for y we have: y ๏ฝ 9 ๏ญ x 2 4๏ญx x This equation specifies a function. The domain is all real numbers except 0 2 2 2 2 2 . and y ๏ฝ ๏ฑ 9 ๏ญ x 2 . This equation does not define y as a function of x. For example, when x = 0, y = ๏ฑ 3. 60. Given x ๏ญ y ๏ฝ 0. . Solving for y we have: y ๏ฝ 3 3 1/6 x and y ๏ฝ x . This equation specifies a function. The domain is all nonnegative real numbers, i.e., x ๏ณ 0 . 62. f ( ๏ญ3 x ) ๏ฝ ( ๏ญ3 x ) ๏ญ 4 ๏ฝ 9 x ๏ญ 4 64. f ( x ๏ญ 1) ๏ฝ ( x ๏ญ 1) ๏ญ 4 ๏ฝ x ๏ญ 2 x ๏ซ 1 ๏ญ 4 ๏ฝ x ๏ญ 2 x ๏ญ 3 66. f (x ) ๏ฝ (x ) ๏ญ 4 ๏ฝ x ๏ญ 4 68. f ( 4 x ) ๏ฝ x1/ 4 70. f ( ๏ญ3) ๏ซ f ( h ) ๏ฝ ( ๏ญ3)2 ๏ญ 4 ๏ซ h 2 ๏ญ 4 ๏ฝ 5 ๏ซ h 2 ๏ญ 4 ๏ฝ h 2 ๏ซ 1 72. f ( ๏ญ3 ๏ซ h ) ๏ฝ ( ๏ญ3 ๏ซ h ) 2 ๏ญ 4 ๏ฝ 9 ๏ญ 6 h ๏ซ h 2 ๏ญ 4 ๏ฝ 5 ๏ญ 6 h ๏ซ h 2 74. f ( ๏ญ3 ๏ซ h ) ๏ญ f ( ๏ญ3) ๏ฝ ๏ฉ( ๏ญ3 ๏ซ h )2 ๏ญ 4 ๏น ๏ญ ๏ฉ( ๏ญ3)2 ๏ญ 4 ๏น ๏ฝ (9 ๏ญ 6h ๏ซ h 2 ๏ญ 4) ๏ญ (9 ๏ญ 4) ๏ฝ ๏ญ6h ๏ซ h 2 ๏ซ ๏ป ๏ซ ๏ป 76. (A) f ( x ๏ซ h) ๏ฝ ๏ญ3( x ๏ซ h) ๏ซ 9 ๏ฝ ๏ญ3 x ๏ญ 3h ๏ซ 9 (B) f ( x ๏ซ h) ๏ญ f ( x) ๏ฝ ๏จ ๏ญ3x ๏ญ 3h ๏ซ 9๏ฉ ๏ญ ๏จ ๏ญ3 x ๏ซ 9๏ฉ ๏ฝ ๏ญ3h (C) f ( x ๏ซ h) ๏ญ f ( x) ๏ญ3h ๏ฝ ๏ฝ ๏ญ3 h h (A) f ( x ๏ซ h) ๏ฝ 3( x ๏ซ h) 2 ๏ซ 5( x ๏ซ h) ๏ญ 8 78. 2 2 2 3 3 2 2 2 6 ๏จ ๏ฉ ๏ญ4 ๏ฝ x 2 1/ 2 ๏ญ4๏ฝ x ๏ญ4 ๏ฝ 3( x 2 ๏ซ 2 xh ๏ซ h 2 ) ๏ซ 5 x ๏ซ 5h ๏ญ 8 ๏ฝ 3 x 2 ๏ซ 6 xh ๏ซ 3h 2 ๏ซ 5 x ๏ซ 5h ๏ญ 8 (B) ๏จ ๏ฉ ๏จ f ( x ๏ซ h) ๏ญ f ( x) ๏ฝ 3 x 2 ๏ซ 6 xh ๏ซ 3h 2 ๏ซ 5 x ๏ซ 5h ๏ญ 8 ๏ญ 3 x 2 ๏ซ 5 x ๏ญ 8 ๏ฉ ๏ฝ 6 xh ๏ซ 3h 2 ๏ซ 5h 80. (C) f ( x ๏ซ h) ๏ญ f ( x) 6 xh ๏ซ 3h 2 ๏ซ 5h ๏ฝ ๏ฝ 6 x ๏ซ 3h ๏ซ 5 h h (A) f(x + h) = x2 + 2xh + h2 + 40x + 40h (B) f(x + h) โ f(x) = 2xh + h2 + 40h (C) f ( x ๏ซ h) ๏ญ f ( x ) = 2x + h + 40 h Copyright ยฉ 2019 Pearson Education, Inc. 2-3 2-4 82. CHAPTER 2: FUNCTIONS Given A = l w = 81. 81 162 81 ๏ฝ 2l ๏ซ . . Now P ๏ฝ 2l ๏ซ 2 w ๏ฝ 2l ๏ซ 2 Thus, w = l l l The domain is l > 0. 84. Given P = 2 ๏ฌ + 2w = 160 or ๏ฌ + w = 80 and ๏ฌ = 80 โ w. Now A = ๏ฌ w = (80 โ w)w and A = 80w โ w2 . The domain is 0 โค w โค 80. [Note: w โค 80 since w > 80 implies ๏ฌ ๏ผ 0.] 86. (A) 88. (A) (B) p(11) = 1,340 dollars per computer p(18) = 920 dollars per computer R(x) = xp(x) = x(2,000 โ 60x) thousands of dollars Domain: 1 โค x โค 25 (C) (B) Table 11 Revenue x(thousands) R(x)(thousands) 1 5 10 15 20 25 90. (A) \$1,940 8,500 14,000 16,500 16,000 12,500 P(x) = R(x) โ C(x) = x(2,000 โ 60x) โ (4,000 + 500x) thousand dollars = 1,500x โ 60×2 โ 4,000 Domain: 1 โค x โค 25 (C) (B) Table 13 Profit x (thousands) 1 5 10 15 20 25 P(x) (thousands) โ\$2,560 2,000 5,000 5,000 2,000 โ4,000 Copyright ยฉ 2019 Pearson Education, Inc. EXERCISE 2-2 92. (A) Given 5v โ 2s = 1.4. Solving for v, we have: v = 0.4s + 0.28. If s = 0.51, then v = 0.4(0.51) + 0.28 = 0.484 or 48.4%. (B) Solving the equation for s, we have: s = 2.5v โ 0.7. If v = 0.51, then s = 2.5(0.51) โ 0.7 = 0.575 or 57.5%. EXERCISE 2-2 Domain: [0, ๏ฅ) ; range: [1, ๏ฅ). 2. f ( x) ๏ฝ 1 ๏ซ x 4. f ( x) ๏ฝ x 2 ๏ซ 10 Domain: all real numbers; range: [10, ๏ฅ ). 6. f ( x ) ๏ฝ 5 x ๏ซ 3 Domain: all real numbers; range: all real numbers. 8. f ( x) ๏ฝ 15 ๏ญ 20 x 10. f ( x) ๏ฝ ๏ญ8 ๏ซ 3 x Domain: all real numbers; range: all real numbers. 12. 14. 16. 18. 20. 22. Domain: all real numbers; range: ( ๏ญ๏ฅ,15]. Copyright ยฉ 2019 Pearson Education, Inc. 2-5 2-6 CHAPTER 2: FUNCTIONS 24. 26. 28. The graph of h(x) = โ|x โ 5| is the graph of y = |x| reflected in the x axis and shifted 5 units to the right. 30. The graph of m(x) = (x + 3)2 + 4 is the graph of y = x2 shifted 3 units to the left and 4 units up. 32. The graph of g(x) = โ6 + 3 x is the graph of y = 3 x 34. The graph of m(x) = โ0.4×2 is the graph of y = x2 reflected in the x axis and vertically shifted 6 units down. contracted by a factor of 0.4. 36. The graph of the basic function y = |x| is shifted 3 units to the right and 2 units up. y = |x โ 3| + 2 38. The graph of the basic function y = |x| is reflected in the x axis, shifted 2 units to the left and 3 units up. Equation: y = 3 โ |x + 2| 40. The graph of the basic function 3 x is reflected in the x axis and shifted up 2 units. Equation: y = 2 โ 3 x 42. The graph of the basic function y = x3 is reflected in the x axis, shifted to the right 3 units and up 1 unit. Equation: y = 1 โ (x โ 3)3 Copyright ยฉ 2019 Pearson Education, Inc. EXERCISE 2-2 44. g(x) = 3 x ๏ซ 3 + 2 46. g(x) = โ|x โ 1| 48. g(x) = 4 โ (x + 2)2 50. ๏ฌ x ๏ซ 1 if x ๏ผ ๏ญ1 g ( x) ๏ฝ ๏ญ ๏ฎ2 ๏ซ 2 x if x ๏ณ ๏ญ1 52. ๏ฌ 10 ๏ซ 2 x if h( x ) ๏ฝ ๏ญ ๏ฎ40 ๏ซ 0.5 x if 54. ๏ฌ 4 x ๏ซ 20 if ๏ฏ h( x ) ๏ฝ ๏ญ 2 x ๏ซ 60 if ๏ฏ๏ญ x ๏ซ 360 if ๏ฎ 0 ๏ฃ x ๏ฃ 20 x ๏พ 20 2-7 0 ๏ฃ x ๏ฃ 20 20 ๏ผ x ๏ฃ 100 x ๏พ 100 56. The graph of the basic function y = x is reflected in the x axis and vertically expanded by a factor of 2. Equation: y = โ2x 58. The graph of the basic function y = |x| is vertically expanded by a factor of 4. Equation: y = 4|x| 60. The graph of the basic function y = x3 is vertically contracted by a factor of 0.25. Equation: y = 0.25×3. 62. Vertical shift, reflection in y axis. Reversing the order does not change the result. Consider a point (a, b) in the plane. A vertical shift of k units followed by a reflection in y axis moves (a, b) to (a, b + k) and then to (โa, b + k). In the reverse order, a reflection in y axis followed by a vertical shift of k units moves (a, b) to (โa, b) and then to (โa, b + k). The results are the same. Copyright ยฉ 2019 Pearson Education, Inc. 2-8 CHAPTER 2: FUNCTIONS 64. Vertical shift, vertical expansion. Reversing the order can change the result. For example, let (a, b) be a point in the plane. A vertical shift of k units followed by a vertical expansion of h (h > 1) moves (a, b) to (a, b + k) and then to (a, bh + kh). In the reverse order, a vertical expansion of h followed by a vertical shift of k units moves (a, b) to (a, bh) and then to (a, bh + k); (a, bh + kh) โ  (a, bh + k). 66. Horizontal shift, vertical contraction. Reversing the order does not change the result. Consider a point (a, b) in the plane. A horizontal shift of k units followed by a vertical contraction of h (0 < h 700, the charge is 54 + .053(x โ 700) = 16.9 + 0.053x. (B) Thus, ๏ฌ 8.5 ๏ซ .065 x if 0 ๏ฃ x ๏ฃ 700 W ( x) ๏ฝ ๏ญ ๏ฎ16.9 ๏ซ 0.053 x if x ๏พ 700 74. (A) Let x = taxable income. If 0 โค x โค 12,500, the tax due is \$.02x. At x = 12,500, the tax due is \$250. For 12,500 50,000, the tax due is 1,250 + .06(x โ 50,000) = .06x โ 1,250. Thus, 0.02 x if 0 ๏ฃ x ๏ฃ 12,500 ๏ฌ ๏ฏ T ( x) ๏ฝ ๏ญ 0.04 x ๏ญ 250 if 12,500 ๏ผ x ๏ฃ 50, 000 ๏ฏ0.06 x ๏ญ 1, 250 if x ๏พ 50, 000 ๏ฎ Copyright ยฉ 2019 Pearson Education, Inc. EXERCISE 2-3 (B) 76. (A) The graph of the basic function y = x3 is vertically expanded by a factor of 463. 78. (C) T(32,000) = \$1,030 T(64,000) = \$2,590 (A) The graph of the basic function y = 3 x is reflected in the x axis and shifted up 10 units. (B) (B) EXERCISE 2-3 2. x 2 ๏ซ 16 x (standard form) 2 x ๏ซ 16 x ๏ซ 64 ๏ญ 64 (completing the square) 2 ( x ๏ซ 8) ๏ญ 64 4. x 2 ๏ญ 12 x ๏ญ 8 (standard form) 2 ( x ๏ญ 12 x) ๏ญ 8 ( x 2 ๏ญ 12 x ๏ซ 36) ๏ซ 8 ๏ญ 36 (vertex form) (completing the square) ( x ๏ญ 6) 2 ๏ญ 44 (vertex form) 6. 3 x 2 ๏ซ 18 x ๏ซ 21 (standard form) 3( x 2 ๏ซ 6 x ) ๏ซ 21 3( x 2 ๏ซ 6 x ๏ซ 9 ๏ญ 9) ๏ซ 21 (completing the square) 3( x ๏ซ 3) 2 ๏ซ 21 ๏ญ 27 3( x ๏ซ 3) 2 ๏ญ 6 (vertex form) 8. ๏ญ5 x 2 ๏ซ 15 x ๏ญ 11 2-9 (standard form) ๏ญ5( x 2 ๏ญ 3x ) ๏ญ 11 ๏ญ5( x 2 ๏ญ 3x ๏ซ 94 ๏ญ 94 ) ๏ญ 11 (completing the square) ๏ญ5( x ๏ญ 23 ) 2 ๏ญ 11 ๏ซ 454 ๏ญ5( x ๏ญ 23 ) 2 ๏ซ 14 (vertex form) Copyright ยฉ 2019 Pearson Education, Inc. 2-10 CHAPTER 2: FUNCTIONS 10. The graph of g(x) is the graph of y = x2 shifted right 1 unit and down 6 units; g ( x ) ๏ฝ ( x ๏ญ 1) 2 ๏ญ 6. 12. The graph of n(x) is the graph of y = x2 reflected in the x axis, then shifted right 4 units and up 7 units; n ( x ) ๏ฝ ๏ญ( x ๏ญ 4) 2 ๏ซ 7. 14. (A) g (B) m (C) n (D) f 16. (A) x intercepts: โ5, โ1; y intercept: โ5 (B) Vertex: (โ3, 4) (C) Maximum: 4 (D) Range: y โค 4 or (โโ, 4] 18. (A) x intercepts: 1, 5; y intercept: 5 (B) Vertex: (3, โ4) (C) Minimum: โ4 (D) Range: y โฅ โ4 or [โ4, โ) 20. g(x) = โ(x + 2)2 + 3 (A) x intercepts: โ(x + 2)2 + 3 = 0 (x + 2)2 = 3 x+2=ยฑ 3 x = โ2 โ 3 , โ2 + 3 y intercept: โ1 (B) 22. Vertex: (โ2, 3) (C) Maximum: 3 (D) Range: y โค 3 or (โโ, 3] n(x) = (x โ 4)2 โ 3 (A) x intercepts: (x โ 4)2 โ 3 = 0 (x โ 4)2 = 3 xโ4=ยฑ 3 x=4โ 3,4+ 3 y intercept: 13 (B) Vertex: (4, โ3) (C) Minimum: โ3 (D) Range: y โฅ โ3 or [โ3, โ) 24. y = โ(x โ 4)2 + 2 26. y = [x โ (โ3)]2 + 1 or y = (x + 3)2 + 1 28. g(x) = x2 โ 6x + 5 = x2 โ 6x + 9 โ 4 = (x โ 3)2 โ 4 (A) x intercepts: (x โ 3)2 โ 4 = 0 (x โ 3)2 = 4 x โ 3 = ยฑ2 x = 1, 5 y intercept: 5 (B) Vertex: (3, โ4) (C) Minimum: โ4 (D) Range: y โฅ โ4 or [โ4, โ) Copyright ยฉ 2019 Pearson Education, Inc. EXERCISE 2-3 30. 3๏น 1๏น ๏ฉ ๏ฉ s(x) = โ4×2 โ 8x โ 3 = โ4 ๏ช x 2 ๏ซ 2 x ๏ซ ๏บ = โ4 ๏ช x 2 ๏ซ 2 x ๏ซ 1 ๏ญ ๏บ 4๏ป 4๏ป ๏ซ ๏ซ 1 ๏ฉ ๏น = โ4 ๏ช ( x ๏ซ 1) 2 ๏ญ ๏บ = โ4(x + 1)2 + 1 4๏ป ๏ซ (A) โ4(x + 1)2 + 1 = 0 4(x + 1)2 = 1 x intercepts: 1 4 1 x+1=ยฑ 2 3 1 x=โ ,โ 2 2 (x + 1)2 = y intercept: โ3 (B) 32. Vertex: (โ1, 1) (C) Maximum: 1 (D) Range: y โค 1 or (โโ, 1] v(x) = 0.5×2 + 4x + 10 = 0.5[x2 + 8x + 20] = 0.5[x2 + 8x + 16 + 4] = 0.5[(x + 4)2 + 4] = 0.5(x + 4)2 + 2 (A) x intercepts: none y intercept: 10 (B) Vertex: (โ4, 2) (C) Minimum: 2 (D) Range: y โฅ 2 or [2, โ) 34. g(x) = โ0.6×2 + 3x + 4 (A) g(x) = โ2: โ0.6×2 + 3x + 4 = โ2 0.6×2 โ 3x โ 6 = 0 (B) g(x) = 5: โ0.6×2 + 3x + 4 = 5 โ0.6×2 + 3x โ 1 = 0 0.6×2 โ 3x + 1 = 0 x = โ1.53, 6.53 x = 0.36, 4.64 (C) g(x) = 8: โ0.6×2 + 3x + 4 = 8 โ0.6×2 + 3x โ 4 = 0 0.6×2 โ 3x + 4 = 0 No solution Copyright ยฉ 2019 Pearson Education, Inc. 2-11 2-12 CHAPTER 2: FUNCTIONS 36. Using a graphing utility with y = 100x โ 7×2 โ 10 and the calculus option with maximum command, we obtain 347.1429 as the maximum value. 38. m(x) = 0.20×2 โ 1.6x โ 1 = 0.20(x2 โ 8x โ 5) = 0.20[(x โ 4)2 โ 21] = 0.20(x โ 4)2 โ 4.2 (A) x intercepts: 0.20(x โ 4)2 โ 4.2 = 0 (x โ 4)2 = 21 x โ 4 = ยฑ 21 x = 4 โ 21 = โ0.6, 4 + 21 = 8.6; y intercept: โ1 (B) Vertex: (4, โ4.2) 40. (C) Minimum: โ4.2 (D) Range: y โฅ โ4.2 or [ ๏ญ4.2, ๏ฅ ) n(x) = โ0.15×2 โ 0.90x + 3.3 = โ0.15(x2 + 6x โ 22) = โ0.15[(x + 3)2 โ 31] = โ0.15(x + 3)2 + 4.65 (A) x intercepts: โ0.15(x + 3)2 + 4.65 = 0 (x + 3)2 = 31 x + 3 = ยฑ 31 x = โ3 โ 31 = โ8.6,โ3 + 31 = 2.6; y intercept: 3.30 (B) Vertex: (โ3, 4.65) 42. (C) Maximum: 4.65 (D) Range: x โค 4.65 or ( ๏ญ๏ฅ, 4.65] ( x ๏ซ 6)( x ๏ญ 3) ๏ผ 0 Therefore, either ( x ๏ซ 6) ๏ผ 0 and (x ๏ญ 3) ๏พ 0 or ( x ๏ซ 6) ๏พ 0 and ( x ๏ญ 3) ๏ผ 0. The first case is impossible. The second case implies ๏ญ6 ๏ผ x ๏ผ 3. Solution set: (๏ญ6,3). 44. x 2 ๏ซ 7 x ๏ซ 12 ๏ฝ ( x ๏ซ 3)( x ๏ซ 4) ๏ณ 0 Therefore, either ( x ๏ซ 3) ๏ณ 0 and (x ๏ซ 4) ๏ณ 0 or ( x ๏ซ 3) ๏ฃ 0 and ( x ๏ซ 4) ๏ฃ 0. The first case implies x ๏ณ ๏ญ3 and the second case implies x ๏ฃ ๏ญ4. Solution set: (๏ญ๏ฅ, ๏ญ 4] ๏ [๏ญ3, ๏ฅ). 46. 48. 50. โ0.88 โค x โค 3.52 x = โ1.27, 2.77 52. f is a quadratic function and max f(x) = f(โ3) = โ5 Axis: x = โ3 Vertex: (โ3, โ5) Range: y โค โ5 or (โโ, โ5] x intercepts: None Copyright ยฉ 2019 Pearson Education, Inc. x 2.72 EXERCISE 2-3 54. (A) (B) f(x) = g(x): โ0.7x(x โ 7) = 0.5x + 3.5 โ0.7×2 + 4.4x โ 3.5 = 0 x= ๏ญ4.4 ๏ฑ (4.4) 2 ๏ญ 4(0.7)(3.5) ๏ญ1.4 = 0.93, 5.35 (C) f(x) > g(x) for 0.93 < x < 5.35 (D) f(x) < g(x) for 0 โค x < 0.93 or 5.35 g(x) for 1.08 < x < 6.35 (D) f(x) < g(x) for 0 โค x < 1.08 or 6.35 < x โค 9 58. The graph of a quadratic with no real zeros will not intersect the xโaxis. 60. Such an equation will have b 2 ๏ญ 4ac ๏ฝ 0. 62. Such an equation will have k ๏ผ 0. a Copyright ยฉ 2019 Pearson Education, Inc. 2-13 2-14 64. CHAPTER 2: FUNCTIONS ax 2 ๏ซ bx ๏ซ c ๏ฝ a ( x ๏ญ h) 2 ๏ซ k ๏ฝ a ( x 2 ๏ญ 2hx ๏ซ h 2 ) ๏ซ k ๏ฝ ax 2 ๏ญ 2ahx ๏ซ ah 2 ๏ซ k Equating constant terms gives k ๏ฝ c ๏ญ ah 2 . Since h is the vertex, we have h ๏ฝ ๏ญ b . Substituting then gives 2a 2 ๏ฆ b2 ๏ถ b ๏ฝc๏ญ 2 ๏ท 4a ๏จ 4a ๏ธ 2 k ๏ฝ c ๏ญ ah ๏ฝ c ๏ญ a ๏ง ๏ฝ 66. 4ac ๏ญ b 2 4a f(x) = โ0.0117×2 + 0.32x + 17.9 (B) (A) x Mkt Share f ( x) 5 10 15 18.8 20.0 20.7 19.2 19.9 20.1 20 25 30 20.2 17.4 16.4 19.6 18.6 17 35 15.3 14.8 (C) For 2025, x = 45 and f(45) = โ0.0117(45)2 + 0.32(45) + 17.9 = 8.6% For 2028, x = 48 and f(48) = โ0.0117(48)2 + 0.32(48) + 17.9 = 6.3% (D) Market share rose from 18.8% in 1985 to a maximum of 20.7% in 1995 and then fell to 15.3% in 2010. 68. Verify 70. (A) (B) R(x) = 2,000x โ 60×2 100 ๏ถ ๏ฆ x๏ท = ๏ญ60 ๏ง x 2 ๏ญ 3 ๏ธ ๏จ 100 2500 2500 ๏น ๏ฉ = ๏ญ60 ๏ช x 2 ๏ญ x๏ซ ๏ญ 3 9 9 ๏บ๏ป ๏ซ ๏ฉ๏ฆ 50 ๏ถ 2500 ๏น = ๏ญ60 ๏ช๏ง x ๏ญ ๏ท ๏ญ ๏บ 3 ๏ธ 9 ๏บ๏ป ๏ช๏ซ๏จ 2 2 50 ๏ถ 50,000 ๏ฆ = ๏ญ60 ๏ง x ๏ญ ๏ท + 3 ๏ธ 3 ๏จ 16.667 thousand computers (16,667 computers); 16,666.667 thousand dollars (\$16,666,667) (C) 2000 ๏ญ 60(50 / 3) ๏ฝ \$1,000 Copyright ยฉ 2019 Pearson Education, Inc. EXERCISE 2-3 72. 2-15 (A) ๏ฆ 50 ๏ถ ๏ฆ 50 ๏ถ p ๏ง ๏ท = 2,000 โ 60 ๏ง ๏ท = \$1,000 ๏จ 3 ๏ธ ๏จ 3 ๏ธ (B) R(x) = C(x) x(2,000 โ 60x) = 4,000 + 500x 2,000x โ 60×2 = 4,000 + 500x 60×2 โ 1,500x + 4,000 = 0 6×2 โ 150x + 400 = 0 x = 3.035, 21.965 Break-even at 3.035 thousand (3,035) and 21.965 thousand (21,965) (C) 74. Loss: 1 โค x < 3.035 or 21.965 < x โค 25; Profit: 3.035 < x 0 for x < 5 Domain: x < 5 or (โโ, 5) (2-1) f(x) = 4×2 + 4x โ 3 = 4(x2 + x) ๏ญ 3 1๏ถ ๏ฆ = 4 ๏ง x2 ๏ซ x ๏ซ ๏ท ๏ญ 3 ๏ญ 1 4๏ธ ๏จ 2 1๏ถ ๏ฆ = 4 ๏ง x ๏ซ ๏ท ๏ญ 4 (vertex form) 2๏ธ ๏จ Intercepts: y intercept: f(0) = 4(0)2 + 4(0) ๏ญ 3 = ๏ญ3 x intercepts: f(x) = 0 2 1๏ถ ๏ฆ 4๏งx๏ซ ๏ท ๏ญ 4 = 0 2๏ธ ๏จ 2 1๏ถ ๏ฆ ๏งx๏ซ 2๏ท =1 ๏จ ๏ธ 1 = ยฑ1 x+ 2 1 3 1 x=๏ญ ยฑ1=๏ญ , 2 2 2 ๏ฆ 1 ๏ถ Vertex: ๏ง ๏ญ , ๏ญ4 ๏ท ; minimum: ๏ญ4; range: y โฅ ๏ญ4 or [๏ญ4, โ) 2 ๏จ ๏ธ Copyright ยฉ 2019 Pearson Education, Inc. (2-3) 2-37 2-38 CHAPTER 2: FUNCTIONS 44. f(x) = ex โ 1, g(x) = ln(x + 2) 2 f Points of intersection: (โ1.54, โ0.79), (0.69, 0.99) g -3 (2-5, 2-6) 3 -2 45. f(x) = 50 x2 ๏ซ 1 : x ๏ญ3 ๏ญ2 ๏ญ1 f ( x) 0 1 2 3 5 10 25 50 25 10 5 (2-1) 46. f(x) = ๏ญ66 2 ๏ซ x2 x ๏ญ3 f ( x ) ๏ญ6 : ๏ญ2 ๏ญ11 ๏ญ1 ๏ญ22 0 ๏ญ66 1 ๏ญ22 2 ๏ญ11 3 ๏ญ6 (2-1) For Problems 47โ50, f(x) = 5x + 1. 47. f(f(0)) = f(5(0) + 1) = f(1) = 5(1) + 1 = 6 (2-1) 48. f(f(โ1)) = f(5(โ1) + 1) = f(โ4) = 5(โ4) + 1 = โ19 (2-1) 49. f(2x โ 1) = 5(2x โ 1) + 1 = 10x โ 4 (2-1) 50. f(4 โ x) = 5(4 โ x) + 1 = 20 โ 5x + 1 = 21 โ 5x (2-1) 51. f(x) = 3 โ 2x (A) f(2) = 3 โ 2(2) = 3 โ 4 = โ1 (B) f(2 + h) = 3 โ 2(2 + h) = 3 โ 4 โ 2h = โ1 โ 2h (C) f(2 + h) โ f(2) = โ1 โ 2h โ (โ1) = โ2h f (2 ๏ซ h) ๏ญ f (2) 2h =โ = โ2 (D) h h 52. f(x) = x2 โ 3x + 1 (A) f(a) = a2 โ 3a + 1 Copyright ยฉ 2019 Pearson Education, Inc. (2-1) CHAPTER 2 REVIEW 2-39 (B) f(a + h) = (a + h)2 โ 3(a + h) + 1 = a2 + 2ah + h2 โ 3a โ 3h + 1 (C) f(a + h) โ f(a) = a2 + 2ah + h2 โ 3a โ 3h + 1 โ (a2 โ 3a + 1) = 2ah + h2 โ 3h (D) f ( a ๏ซ h) ๏ญ f ( a ) 2ah ๏ซ h 2 ๏ญ 3h h(2a ๏ซ h ๏ญ 3) = = = 2a + h โ 3 h h h (2-1) 53. The graph of m is the graph of y = |x| reflected in the x axis and shifted 4 units to the right. 54. The graph of g is the graph of y = x3 vertically contracted by a factor of 0.3 and shifted up 3 units. (2-2) 55. 56. (2-2) The graph of y = x2 is vertically expanded by a factor of 2, reflected in the x axis and shifted to the left 3 units. Equation: y = โ2(x + 3)2 (2-2) Equation: f(x) = 2 x ๏ซ 3 โ 1 (2-2) 57. f ( x) ๏ฝ n( x ) 5x ๏ซ 4 . Since degree n(x) = 1 1 = degree d(x), there is no horizontal asymptote. d ( x ) 100 x ๏ซ 1 (2-4) 60. 61. n( x ) x 2 ๏ซ 100 x 2 ๏ซ 100 ๏ฝ 2 ๏ฝ . Since n( x ) ๏ฝ x 2 ๏ซ 100 has no real zeros and d ( x ) x ๏ญ 100 ( x ๏ญ 10)( x ๏ซ 10) d (10) ๏ฝ d ( ๏ญ10) ๏ฝ 0, x = 10 and x = ๏ญ10 are the vertical asymptotes of the graph of f. f ( x) ๏ฝ f ( x) ๏ฝ (2-4) n( x ) x 2 ๏ซ 3x x ( x ๏ซ 3) x ๏ซ 3 ๏ฝ ๏ฝ ๏ฝ , x ๏น 0. x = ๏ญ2 is a vertical asymptote of the graph of f. d ( x ) x 2 ๏ซ 2 x x ( x ๏ซ 2) x ๏ซ 2 (2-4) 62. True; p(x) = p ( x) is a rational function for every polynomial p. 1 Copyright ยฉ 2019 Pearson Education, Inc. (2-4) 2-40 CHAPTER 2: FUNCTIONS 1 = xโ1 is not a polynomial function. x 63. False; f(x) = 64. False; f(x) = 65. True: let f(x) = bx, (b > 0, b โ  1), then the positive x-axis is a horizontal asymptote if 0 < b 1. (2-5) 66. True: let f(x) = logbx (b > 0, b โ  1). If 0 < b 1, then the negative y-axis is a vertical asymptote. 67. True; f(x) = (2-6) x has vertical asymptote x = 1 and horizontal asymptote y = 1. x ๏ญ1 68. 69. (2-2) (2-2) 70. 71. (2-4) y = โ(x โ 4)2 + 3 (2-2, 2-3) f(x) = โ0.4×2 + 3.2x + 1.2 = โ0.4(x2 โ 8x + 16) + 7.6 = โ0.4(x โ 4)2 + 7.6 (A) y intercept: 1.2 x intercepts: โ0.4(x โ 4)2 + 7.6 = 0 (x โ 4)2 = 19 x = 4 + 19 โ 8.4, 4 โ 19 โ โ0.4 (B) Vertex: (4.0, 7.6) (C) Maximum: 7.6 (D) Range: y โค 7.6 or (โโ, 7.6] (2-3) (A) y intercept: 1.2 x intercepts: โ0.4, 8.4 72. (B) Vertex: (4.0, 7.6) (C) Maximum: 7.6 (D) Range: y โค 7.6 or (โโ, 7.6] 73. (2-3) log 10ฯ = ฯ log 10 = ฯ 10log 2 = y is equivalent to log y = log 2 which implies y = 2 Similarly, ln e ฯ = ฯ ln e = ฯ (Section 2-5, 4.b & g) and eln 2 = y implies ln y = ln 2 and y= 2 . (2-6) Copyright ยฉ 2019 Pearson Education, Inc. CHAPTER 2 REVIEW 74. 2-41 log x ๏ญ log 3 = log 4 ๏ญ log (x + 4) 4 x log ๏ฝ log 3 x๏ซ4 4 x ๏ฝ 3 x๏ซ4 x ( x ๏ซ 4) ๏ฝ 12 x 2 ๏ซ 4 x ๏ญ 12 ๏ฝ 0 ( x ๏ซ 6)( x ๏ญ 2) ๏ฝ 0 x ๏ฝ ๏ญ6, 2 Since log(๏ญ6) is not defined, ๏ญ6 is not a solution. Therefore, the solution is x = 2. 75. ln(2x โ 2) โ ln(x โ 1) = ln x ๏ฆ 2x ๏ญ 2 ๏ถ = ln x ln ๏ง ๏จ x ๏ญ 1 ๏ท๏ธ 76. ๏ฉ 2( x ๏ญ 1) ๏น ln ๏ช ๏บ = ln x ๏ซ x ๏ญ1 ๏ป ln 2 = ln x x=2 77. ln(x + 3) โ ln x = 2 ln 2 ๏ฆ x ๏ซ3๏ถ ln ๏ง = ln(22) ๏จ x ๏ท๏ธ x๏ซ3 =4 x x + 3 = 4x 3x = 3 x=1 (2-6) log 3×2 = 2 + log 9x ln y โ ln c = โ5t ๏ฆx๏ถ log ๏ง ๏ท = 2 ๏จ3๏ธ y = eโ5t c y = โ5t c ln y = ceโ5t (2-6) Let x be any positive real number and suppose log1x = y. Then 1y = x. But, 1y = 1, so x = 1, i.e., x = 1 for all positive real numbers x. This is clearly impossible. 80. (2-6) ln y = โ5t + ln c 78. log 3×2 โ log 9x = 2 ๏ฆ 3×2 ๏ถ log ๏ง ๏ท =2 ๏จ 9x ๏ธ x = 102 = 100 3 x = 300 (2-6) 79. (2-6) (2-6) The graph of y = 3 x is vertically expanded by a factor of 2, reflected in the x axis, shifted 1 unit to the left and 1 unit down. Equation: y = โ2 3 x ๏ซ 1 โ 1 (2-2) Copyright ยฉ 2019 Pearson Education, Inc. 2-42 81. CHAPTER 2: FUNCTIONS G(x) = 0.3×2 + 1.2x โ 6.9 = 0.3(x2 + 4x + 4) โ 8.1 = 0.3(x + 2)2 โ 8.1 (A) y intercept: โ6.9 x intercepts: 0.3(x + 2)2 โ 8.1 = 0 (x + 2)2 = 27 x = โ2 + (B) Vertex: (โ2, โ8.1) 27 โ 3.2, โ2 โ 27 โ โ7.2 (C) Minimum: โ8.1 (D) Range: y โฅ โ8.1 or [โ8.1, โ) (2-3) (A) y intercept: โ6.9 x intercept: โ7.2, 3.2 82. (B) Vertex: (โ2, โ8.1) (C) Minimum: โ8.1 (D) Range: y โฅ โ8.1 or [โ8.1, โ) 83. (A) (2-3) S(x) = 3 if 0 โค x โค 20; S(x) = 3 + 0.057(x โ 20) = 0.057x + 1.86 if 20 < x โค 200; S(200) = 13.26 S(x) = 13.26 + 0.0346(x โ 200) = 0.0346x + 6.34 if 200 1000 if ๏ฌ3 ๏ฏ0.057 x ๏ซ 1.86 if ๏ฏ Therefore, S(x) = ๏ญ ๏ฏ0.0346 x ๏ซ 6.34 if ๏ฏ๏ฎ0.0217 x ๏ซ 19.24 if 0 ๏ฃ x ๏ฃ 20 20 ๏ผ x ๏ฃ 200 200 ๏ผ x ๏ฃ 1000 x ๏พ 1000 (B) (2-2) mt 84. r๏ถ ๏ฆ A ๏ฝ P ๏ง 1 ๏ซ ๏ท ; P = 5,000, r = 0.0125, m = 4, t = 5. ๏จ m๏ธ 4(5) 20 ๏ฆ 0.0125 ๏ถ ๏ฆ 0.0125 ๏ถ A ๏ฝ 5000 ๏ง 1 ๏ซ ๏ท ๏ฝ 5000 ๏ง 1 ๏ซ ๏ท ๏ป 5321.95 4 ๏ธ 4 ๏ธ ๏จ ๏จ After 5 years, the CD will be worth \$5,321.95 Copyright ยฉ 2019 Pearson Education, Inc. (2-5) CHAPTER 2 REVIEW 2-43 mt 85. r๏ถ ๏ฆ A ๏ฝ P ๏ง 1 ๏ซ ๏ท ; P = 5,000, r = 0.0105, m = 365, t = 5 m ๏จ ๏ธ 365(5) 1825 ๏ฆ 0.0105 ๏ถ ๏ฆ 0.0105 ๏ถ A ๏ฝ 5000 ๏ง 1 ๏ซ ๏ฝ 5000 ๏ง 1 ๏ซ ๏ท 365 365 ๏ท๏ธ ๏จ ๏ธ ๏จ After 5 years, the CD will be worth \$5,269.51. ๏ป 5269.51 (2-5) mt 86. r๏ถ ๏ฆ A = P ๏ง 1 ๏ซ ๏ท , r = 0.0659, m = 12 ๏จ m๏ธ 12 t ๏ฆ 0.0659 ๏ถ = 3P or (1.005492)12t = 3 Solve P ๏ง 1 ๏ซ 12 ๏ท๏ธ ๏จ for t: 12t ln(1.005492) = ln 3 ln 3 โ 16.7 year. t= 12 ln(1.005492) 87. (2-5) A ๏ฝ Pe rt , r ๏ฝ 0.0739 . Solve 2 P ๏ฝ Pe0.0739t for t. 2 P ๏ฝ Pe0.0739t e0.0739t ๏ฝ 2 0.0739t ๏ฝ ln 2 t๏ฝ 88. ln 2 ๏ป 9.38 years. 0.0739 (2-5) p(x) = 50 โ 1.25x Price-demand function C(x) = 160 + 10x Cost function R(x) = xp(x) = x(50 โ 1.25x) Revenue function (A) (B) R = C x(50 โ 1.25x) = 160 + 10x โ1.25×2 + 50x = 160 + 10x โ1.25×2 + 40x = 160 โ1.25(x2 โ 32x + 256) = 160 โ 320 โ1.25(x โ 16)2 = โ160 (x โ 16)2 = 128 x = 16 + 128 โ 27.314, 16 โ 128 โ 4.686 R = C at x = 4.686 thousand units (4,686 units) and x = 27.314 thousand units (27,314 units) R < C for 1 โค x < 4.686 or 27.314 C for 4.686 < x < 27.314 Copyright ยฉ 2019 Pearson Education, Inc. 2-44 CHAPTER 2: FUNCTIONS (C) Max Rev: 50x โ 1.25×2 = R โ1.25(x2 โ 40x + 400) + 500 = R โ1.25(x โ 20)2 + 500 = R Vertex at (20, 500) Max. Rev. = 500 thousand (\$500,000) occurs when output is 20 thousand (20,000 units) Wholesale price at this output: p(x) = 50 โ 1.25x p(20) = 50 โ 1.25(20) = \$25 (2-3) 89. (A) P(x) = R(x) โ C(x) = x(50 โ 1.25x) โ (160 + 10x) = โ1.25×2 + 40x โ 160 (B) P = 0 for x = 4.686 thousand units (4,686 units) and x = 27.314 thousand units (27,314 units) P < 0 for 1 โค x < 4.686 or 27.314 0 for 4.686 < x < 27.314 Maximum profit is 160 thousand dollars (\$160,000), and this occurs at x = 16 thousand units(16,000 units). The wholesale price at this output is p(16) = 50 โ 1.25(16) = \$30, which is \$5 greater than the \$25 found in 88(C). (2-3) (C) 90. (A) The area enclosed by the storage areas is given by A = (2y)x Now, 3x + 4y = 840 3 so y = 210 โ x 4 3 ๏ถ ๏ฆ Thus A(x) = 2 ๏ง 210 ๏ญ x ๏ท x 4 ๏ธ ๏จ 3 = 420x โ x2 2 (B) Clearly x and y must be nonnegative; the fact that y โฅ 0 implies 3 210 โ x โฅ 0 4 3 and 210 โฅ x 4 840 โฅ 3x 280 โฅ x Thus, domain A: 0 โค x โค 280 (C) Copyright ยฉ 2019 Pearson Education, Inc. CHAPTER 2 REVIEW (D) Graph A(x) = 420x โ 3 2 x and y = 25,000 together. 2 There are two values of x that will produce storage areas with a combined area of 25,000 square feet, one near x = 90 and the other near x = 190. 2-45 30,000 0 3 0 (E) (F) x = 86, x = 194 3 2 3 x = โ (x2 โ 280x) 2 2 Completing the square, we have 3 A(x) = โ (x2 โ 280x + 19,600 โ 19,600) 2 3 = โ [(x โ 140)2 โ 19,600] 2 3 = โ (x โ 140)2 + 29,400 2 A(x) = 420x โ The dimensions that will produce the maximum combined area are: x = 140 ft, y = 105 ft. The maximum area is 29,400 sq. ft. 91. (A) Quadratic regression model, Table 1: (2-3) To estimate the demand at price level of \$180, we solve the equation ax2 + bx + c = 180 for x. The result is x โ 2,833 sets. (B) Linear regression model, Table 2: (C) The condition is not stable; the price is likely to decrease since the supply at the price level of \$180 exceeds the demand at this level. (D) Equilibrium price: \$131.59 Equilibrium quantity: 3,587 cookware set. To estimate the supply at a price level of \$180, we solve the equation ax + b = 180 for x. The result is x โ 4,836 sets. Copyright ยฉ 2019 Pearson Education, Inc. (2-3) 2-46 CHAPTER 2: FUNCTIONS (A) 92. Cubic Regression y ๏ฝ 0.30395 x3 ๏ญ 12.993 x 2 ๏ซ 38.292 x ๏ซ 5, 604.8 y ๏ฝ 0.30395(38)3 ๏ญ 12.993(38) 2 ๏ซ 38.292(38) ๏ซ 5,604.8 ๏ป 4,976 (B) The predicted crime index in 2025 is 4,976. 93. (A) N(0) = 1 ๏ฆ1๏ถ N๏ง ๏ท = 2 ๏จ2๏ธ (B) log 22t = log 109 = 9 2t log 2 = 9 9 โ 14.95 t= 2 log 2 N(1) = 4 = 22 ๏ฆ 3๏ถ N ๏ง ๏ท = 8 = 23 ๏จ2๏ธ N(2) = 16 = 24 ๏ Thus, we conclude that N(t) = 22t or N = 4t 94. We need to solve: 22t = 109 Thus, the mouse will die in 15 days. (2-6) 1 Given I = I0eโkd. When d = 73.6, I = I0. Thus, we have: 2 1 โk(73.6) I =I e 2 0 0 1 eโk(73.6) = 2 1 2 ln(0.5) โ 0.00942 k= ๏ญ73.6 โk(73.6) = ln Thus, k โ 0.00942. To find the depth at which 1% of the surface light remains, we set I = 0.01I0 and solve 0.01I0 = I0eโ0.00942d for d: 0.01 = eโ0.00942d โ0.00942d = ln 0.01 ln 0.01 โ 488.87 d= ๏ญ0.00942 Thus, 1% of the surface light remains at approximately 489 feet. Copyright ยฉ 2019 Pearson Education, Inc. (2-6) CHAPTER 2 REVIEW 95. (A) Logarithmic regression model: Year 2023 corresponds to x = 83; y(83) โ 6,134,000 cows. (B) 96. ln (0) is not defined. (2-6) Using the continuous compounding model, we have: 2P0 = P0e0.03t 2 = e0.03t 0.03t = ln 2 ln 2 t= โ 23.1 0.03 Thus, the model predicts that the population will double in approximately 23.1 years. 97. (2-5) (A) The exponential regression model is y = 47.194(1.0768)x. To estimate for the year 2025, let x = 45 ๏ y = 47.19368975(1.076818175)45 ๏ป 1,319.140047. The estimated annual expenditure for Medicare by the U.S. government, rounded to the nearest billion, is approximately \$1,319 billion. (This is \$1.319 trillion.) (B) To find the year, solve 47.194(1.0768)x = 2,000. Note: Use 2,000 because expenditures are in billions of dollars, and 2 trillion is 2,000 billion. 47.194(1.0768)x = 2,000 2,000 1.0768x = 47.194 ๏ฆ 2,000 ๏ถ ln(1.0768x) = ln ๏ง ๏ท ๏จ 47.194 ๏ธ ๏ฆ 2,000 ๏ถ xln1.0768 = ln ๏ง ๏ท ๏จ 47.194 ๏ธ ๏ฆ 2,000 ๏ถ ln ๏ง ๏ท 47.194 ๏ธ ๏ป 50.6 years x= ๏จ ln1.0768 1,980 + 50.63 = 2,030.63 Annual expenditures exceed two trillion dollars in the year 2031. (2-5) Copyright ยฉ 2019 Pearson Education, Inc. 2-47

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### Solution Manual for College Mathematics for Business, Economics, Life Sciences, and Social Sciences, 14th Edition

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