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Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition is here to help you with textbook problems, offering clear solutions and helpful explanations.

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     S OLUTIONS M ANUAL C ALCULUS FOR B USINESS , E CONOMICS , L IFE S CIENCES , AND S OCIAL S CIENCES F OURTEENTH E DITION AND C ALCULUS FOR B USINESS , E CONOMICS , L IFE S CIENCES , AND S OCIAL S CIENCES B RIEF V ERSION F OURTEENTH E DITION Raymond A. Barnett Merritt College Michael R. Ziegler Marquette University Karl E. Byleen Marquette University Christopher J. Stocker Marquette University

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    1-1 1 FUNCTIONS AND GRAPHS EXERCISE 1-1 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. 1 4 y x x   is neither linear nor constant. 24. 2 4 6 0 x y    is linear. 26. 1 0 x xy    is neither linear nor constant. 28. 3 2 1 simplifies to 2 4 1 constant. 2 y x x y     

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    1-2 CHAPTER 1: FUNCTIONS AND GRAPHS 30. 32. 34. 36. 38. f ( x ) = 2 2 3 2 x x  . Since the denominator is bigger than 1, we note that the values of f are between 0 and 3. Furthermore, the function f has the property that f (– x ) = f ( x ). So, adding points x = 3, x = 4, x = 5, we have: x –5 –4 –3 –2 –1 0 1 2 3 4 5 F ( x ) 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 (–2) = 3 44. ( ) 4 at 5. f x x   46. ( ) 0 at 5, 0, 4. f x x    48. Domain : all real numbers. 50. Domain: all real numbers except x = 2 . 52. Domain: 5 or [ 5, ). x     54. Given 6 7 21 x y   . Solving for y we have: 6 7 21 6 and 3. 7 y x y x      This equation specifies a function. The domain is R , the set of real numbers.

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    EXERCISE 1-1 1-3 56. Given ( ) 4 x x y   . Solving for y we have: 2 2 4 4 and . x xy x y x     This equation specifies a function. The domain is all real numbers except 0 58. Given 2 2 . 9 x y   Solving for y we have: 2 2 2 9 and 9 . y x y x      This equation does not define y as a function of x . For example, when x = 0, y =  3. 60. Given 3 0 . x y   . Solving for y we have: 3 1/6 and . y x y x   This equation specifies a function. The domain is all nonnegative real numbers, i.e., 0 x  . 62. 2 2 ( 3 ) ( 3 ) 4 9 4 f x x x       64. 2 2 2 ( 1) ( 1) 4 2 1 4 2 3 f x x x x x x            66. 2 6 3 3 ( ) ( ) 4 4 f x x x     68.   2 1/ 4 1/ 2 4 ( ) 4 4 4 f x x x x       70. 2 2 2 2 ( 3) ( ) ( 3) 4 4 5 4 1 f f h h h h             72. 2 2 2 ( 3 ) ( 3 ) 4 9 6 4 5 6 f h h h h h h              74. 2 2 2 2 ( 3 ) ( 3) ( 3 ) 4 ( 3) 4 (9 6 4) (9 4) 6 f h f h h h h h                             76. (A) ( ) 3( ) 9 3 3 9 f x h x h x h          (B)     ( ) ( ) 3 3 9 3 9 3 f x h f x x h x h            (C) ( ) ( ) 3 3 f x h f x h h h       78. (A) 2 2 2 2 2 ( ) 3( ) 5( ) 8 3( 2 ) 5 5 8 3 6 3 5 5 8 f x h x h x h x xh h x h x xh h x h                   (B)     2 2 2 2 ( ) ( ) 3 6 3 5 5 8 3 5 8 6 3 5 f x h f x x xh h x h x x xh h h               (C) 2 ( ) ( ) 6 3 5 6 3 5 f x h f x xh h h x h h h         80. (A) f ( x + h ) = x 2 + 2 xh + h 2 + 40 x + 40 h (B) f ( x + h ) – f ( x ) = 2 xh + h 2 + 40 h (C) ( ) ( ) f x h f x h   = 2 x + h + 40

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    1-4 CHAPTER 1: FUNCTIONS AND GRAPHS 82. Given A = l w = 81. Thus, w = 81 l . Now 81 162 2 2 2 2 2 . P l w l l l l       The domain is l > 0. 84. Given P = 2  + 2 w = 160 or  + w = 80 and  = 80 – w . Now A =  w = (80 – w ) w and A = 80 w – w 2 . The domain is 0 ≤ w ≤ 80. [Note: w ≤ 80 since w > 80 implies  0.]  86. (A) (B) p (11) = 1,340 dollars per computer p (18) = 920 dollars per computer 88. (A) R ( x ) = xp ( x ) = x (2,000 – 60 x ) thousands of dollars Domain: 1 ≤ x ≤ 25 (B) Table 11 Revenue x (thousands) R ( x )(thousands) 1 $1,940 5 8,500 10 14,000 15 16,500 20 16,000 25 12,500 (C) 90. (A) P ( x ) = R ( x ) – C ( x ) = x (2,000 – 60 x ) – (4,000 + 500 x ) thousand dollars = 1,500 x – 60 x 2 – 4,000 Domain: 1 ≤ x ≤ 25 (B) Table 13 Profit x (thousands) P ( x ) (thousands) 1 –$2,560 5 2,000 10 5,000 15 5,000 20 2,000 25 –4,000 (C)

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    EXERCISE 1-2 1-5 92. (A) Given 5 v – 2 s = 1.4. Solving for v , we have: v = 0.4 s + 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 = 2.5 v – 0.7. If v = 0.51, then s = 2.5(0.51) – 0.7 = 0.575 or 57.5%. EXERCISE 1-2 2. ( ) 1 f x x   Domain: [0, )  ; range: [1, ).  4. 2 ( ) 10 f x x   Domain: all real numbers; range: [10, ).  6. ( ) 5 3 f x x   Domain: all real numbers; range: all real numbers. 8. ( ) 15 20 f x x   Domain: all real numbers; range: ( ,15].  10. 3 ( ) 8 f x x    Domain: all real numbers; range: all real numbers. 12. 14. 16. 18. 24. 20. 26. 22.

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    1-6 CHAPTER 1: FUNCTIONS AND GRAPHS 28. The graph of h ( x ) = –| x – 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 = x 2 shifted 3 units to the left and 4 units up. 32. The graph of g ( x ) = –6 + 3 x is the graph of y = 3 x shifted 6 units down. 34. The graph of m ( x ) = –0.4 x 2 is the graph of y = x 2 reflected in the x axis and vertically 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 – 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 – | 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 – 3 x 42. The graph of the basic function y = x 3 is reflected in the x axis, shifted to the right 3 units and up 1 unit. Equation: y = 1 – ( x – 3) 3

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    EXERCISE 1-2 1-7 44. g ( x ) = 3 3 x  + 2 46. g ( x ) = –| x – 1| 48. g ( x ) = 4 – ( x + 2) 2 50. 1 if 1 ( ) 2 2 if 1 x x g x x x           52. 10 2 if 0 20 ( ) 40 0.5 if 20 x x h x x x          54. 4 20 if 0 20 ( ) 2 60 if 20 100 360 if 100 x x h x x x x x                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 = –2 x 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 = x 3 is vertically contracted by a factor of 0.25. Equation: y = 0.25 x 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 (– a , b + k ). In the reverse order, a reflection in y axis followed by a vertical shift of k units moves ( a, b ) to (– a, b ) and then to (– a, b + k ). The results are the same. 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 ) ≠ ( a, bh + k ).

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    1-8 CHAPTER 1: FUNCTIONS AND GRAPHS 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 < 1) moves ( a, b ) to ( a + k, b ) and then to ( a + k, bh ). In the reverse order, a vertical contraction of h followed by a horizontal shift of k units moves ( a, b ) to ( a, bh ) and then to ( a + k, bh ). The results are the same. 68. (A) The graph of the basic function y = x is vertically expanded by a factor of 4. 70. (A) The graph of the basic function y = x 2 is reflected in the x axis, vertically contracted by a factor of 0.013, and shifted 10 units to the right and 190 units up. (B) (B) 72. (A) Let x = number of kwh used in a winter month. For 0 ≤ x ≤ 700, the charge is 8.5 + .065 x . At x = 700, the charge is $54. For x > 700, the charge is 54 + .053( x – 700) = 16.9 + 0.053 x . Thus, 8.5 .065 if 0 700 ( ) 16.9 0.053 if 700 x x W x x x          (B) 74. (A) Let x = taxable income. If 0 ≤ x ≤ 12,500, the tax due is $.02 x . At x = 12,500, the tax due is $250. For 12,500 < x ≤ 50,000, the tax due is 250 + .04( x – 12,500) = .04 x – 250. For x > 50,000, the tax due is 1,250 + .06( x – 50,000) = .06 x – 1,250. Thus, (B) 0.02 if 0 12,500 ( ) 0.04 250 if 12,500 50, 000 0.06 1, 250 if 50, 000 x x T x x x x x              (C) T (32,000) = $1,030 T (64,000) = $2,590

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    EXERCISE 1-3 1-9 76. (A) The graph of the basic function y = x 3 is vertically expanded by a factor of 463. 78. (A) The graph of the basic function y = 3 x is reflected in the x axis and shifted up 10 units. (B) (B) EXERCISE 1-3 2. y = 2 x + 1 0 1 2 2 4 3 x y 4. 8 x – 3 y = 24 0 8 3 0 6 8 x y  6. Slope: 3 m  8. Slope: 10 3 m   y- intercept: b = 2 y- intercept: b = 4 10. Slope: 4 ; m   x -intercept: 3 12. Slope: 3; m   x -intercept: 2 14. Slope: 9 / 2 m   ; x -intercept: 4/9 16. 5 y x   18. 6 9 6 9 , ; 7 2 7 2 m b y x      20. x- intercept: 1; y- intercept: 3; 3 3 y x    22. x- intercept: 2, y- intercept: – 1 ; 1 1 2 y x   24. (A) g (B) m (C) n (D) f 26. (A) x -intercepts: – 5, – 1; y- intercept: – 5 (B) Vertex: ( – 3, 4) (C) Maximum: 4 (D) Range: y ≤ 4 or ( – ∞, 4] 28. (A) x- intercepts: 1, 5; y- intercept: 5 (B) Vertex: (3, – 4) (C) Minimum: – 4 (D) Range: y ≥ – 4 or [ – 4, ∞)

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    1-10 CHAPTER 1: FUNCTIONS AND GRAPHS 30. g ( x ) = – ( x + 2) 2 + 3 (A) x- intercepts: – ( x + 2) 2 + 3 = 0 ( x + 2) 2 = 3 x + 2 = ± 3 x = – 2 – 3 , – 2 + 3 y- intercept: – 1 (B) Vertex: ( – 2, 3) (C) Maximum: 3 (D) Range: y ≤ 3 or ( – ∞, 3] 32. n ( x ) = ( x – 4) 2 – 3 (A) x- intercepts: ( x – 4) 2 – 3 = 0 ( x – 4) 2 = 3 x – 4 = ± 3 x = 4 – 3 , 4 + 3 y- intercept: 13 (B) Vertex: (4, – 3) (C) Minimum: – 3 (D) Range: y ≥ – 3 or [ – 3, ∞) 34. (A) Slope: 5 2 3 3 1 2 m     . (B) Point-slope form: 3 2 ( 1) 2 y x    . (C) Slope-intercept form: 3 1 2 2 y x   . (D) Standard form: 3 2 1 x y    . 36. (A) Slope: 7 3 4 3 2 5 m       . (B) Point-slope form: 4 3 ( 2) 5 y x     . (C) Slope-intercept form: 4 23 5 5 y x    . (D) Standard form: 4 5 23 x y   . 38. (A) Slope: 4 4 0 0 1 m     . (B) Point-slope form: 4 0 y   . (C) Slope-intercept form: 4 y  . (D) Standard form: 4 y  . 40. (A) Slope: 3 0 m   not defined (B) Point-slope form: none. (C) Slope-intercept form: none. (D) Standard form: 2 x  . 42. g ( x ) = x 2 – 6 x + 5 = x 2 – 6 x + 9 – 4 = ( x – 3) 2 – 4 (A) x- intercepts: ( x – 3) 2 – 4 = 0 ( x – 3) 2 = 4 x – 3 = ±2 x = 1, 5 y- intercept: 5 (B) Vertex: (3, – 4) (C) Minimum: – 4 (D) Range: y ≥ – 4 or [ – 4, ∞) 44. s ( x ) = – 4 x 2 – 8 x – 3 = – 4 2 3 2 4 x x         = – 4 2 1 2 1 4 x x          = – 4 2 1 ( 1) 4 x         = – 4( x + 1) 2 + 1

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    EXERCISE 1-3 1-11 (A) x- intercepts: – 4( x + 1) 2 + 1 = 0 4( x + 1) 2 = 1 ( x + 1) 2 = 1 4 x + 1 = ± 1 2 x = – 3 2 , – 1 2 y- intercept: – 3 (B) Vertex: ( – 1, 1) (C) Maximum: 1 (D) Range: y ≤ 1 or ( – ∞, 1] 46. v ( x ) = 0.5 x 2 + 4 x + 10 = 0.5[ x 2 + 8 x + 20] = 0.5[ x 2 + 8 x + 16 + 4] = 0.5[( x + 4) 2 + 4] = 0.5( x + 4) 2 + 2 (A) x- intercepts: none y- intercept: 10 (B) Vertex: ( – 4, 2) (C) Minimum: 2 (D) Range: y ≥ 2 or [2, ∞) 48.   54 9 54 0, 6; solution set: 6, 6 x x         50.   4 44 0, 4 44, 11; solution set: 11, x x x         52. ( 6)( 3) 0 x x    Therefore, either ( 6) 0 and ( 3) 0 or ( 6) 0 and ( 3) 0. x x x x         The first case is impossible. The second case implies 6 3. x    Solution set: ( 6,3).  54. 2 7 12 ( 3)( 4) 0 x x x x       Therefore, either ( 3) 0 and ( 4) 0 or ( 3) 0 and ( 4) 0. x x x x         The first case implies 3 x   and the second case implies 4. x   Solution set: ( , 4] [ 3, ).      56. 58. (A) (B) Set f ( x ) = 0, – 0.8 x + 5.2 = 0, x = 6.5. Set x = 0, y = 5.2. (C) (D) x -intercept: x = 6.5, y -intercept: y = 5.2 (E) – 0.8 x + 5.2 < 0 x > 6.5

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    1-12 CHAPTER 1: FUNCTIONS AND GRAPHS 60. g ( x ) = – 0.6 x 2 + 3 x + 4 (A) g ( x ) = – 2: – 0.6 x 2 + 3 x + 4 = – 2 0.6 x 2 – 3 x – 6 = 0 x = – 1.53, 6.53 (B) g ( x ) = 5: – 0.6 x 2 + 3 x + 4 = 5 – 0.6 x 2 + 3 x – 1 = 0 0.6 x 2 – 3 x + 1 = 0 x = 0.36, 4.64 (C) g ( x ) = 8: – 0.6 x 2 + 3 x + 4 = 8 – 0.6 x 2 + 3 x – 4 = 0 0.6 x 2 – 3 x + 4 = 0 No solution 62. Using a graphing utility with y = 100 x – 7 x 2 – 10 and the calculus option with maximum command, we obtain 347.1429 as the maximum value. 64. The slope of the line through 1 1 2 2 ( , ) and ( , ) x y x y is 2 1 2 1 y y x x   ; the slope of the line through 1 1 3 3 ( , y ) and ( , ) x x y is 3 1 3 1 y y x x   . By Problem 57, 2 1 3 1 2 1 3 1 . y y A y y x x B x x        66. We have two representations of ( d, P ): (0, 14.7) and (34, 29.4). (A) A line relating P to d passes through the above two points. Its equation is: P – 14.7 = (29.4 14.7) (34 0)   ( d – 0) or 0.432 14.7 P d   (B) The rate of change of pressure with respect to depth is approximately 0.432 lbs/in 2 per foot. (C) For d = 50, P = 0.432 (50) + 14.7 ≈ 36.3 lbs/in 2

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    EXERCISE 1-3 1-13 (D) For P = 4 atmospheres, we have P = 4(14.7) = 58.8 lbs/in 2 and hence 58.8 = 0.432 d + 14.7 or d = 58.8 14.7 0.432 102   ft. 68. We have two representations of ( t, a ): (0, 2,880) and (180, 0). (A) The linear model relating altitude a to the time in air t has the following equation: a – 2,880 = (0 2,880) (180 0)   ( t – 0) or a = – 16 t + 2,880 (B) The rate of descent for an ATPS system parachute is 16 ft/sec. (C) It is 16 ft/sec. 70. Let y be daily cost of producing x tennis rackets. Then we have two points for ( x, y ): (50, 3,855) and (60, 4,245). (A) Since x and y are linearly related, then the two points (50, 3,855) and (60, 4,245) will lie on the line expressing the linear relationship between x and y . Therefore y – 3,855 = (4, 245 3,855) (60 50)   ( x – 50) or y = 39 x + 1,905 (B) (C) The y intercept, $1,905, is the fixed cost and the slope, $39, is the cost per racket. 72. We observe that for ( t, V ) two points are given: (0, 224,000) and (16, 115,200) (A) A linear model will be a line passing through the two points (0, 224,000) and (16, 115,200). The equation of this line is: V – 115,200 = (224, 000 115, 200) (0 16)   ( t – 16) or V = – 6,800 t + 224,000 (B) For t = 10 V = – 6,800(10) + 224,000 = $156,000

  16. 1-14 CHAPTER 1: FUNCTIONS AND GRAPHS (C) For V = $100,000 100,000 = – 6,800 t + 224,000 or t = (224, 000 100, 000) 6,800 18.24   So, during the 19 th year, the depreciated value falls below $100,000. (D) 74. Let T be the true airspeed at the altitude A (thousands of feet). Since T increases 1.6% , (1000) 200(1.016) 203.2. T   (A) A linear relationship between A and T has slope (203.2 200) 3.2 1 m    . Therefore, 3.2 200. T A   (B) For A = 6.5 (6,500 feet), T = 3.2(6.5) + 200 = 20.8+200 = 220.8 mph 76. (A) For ( x, p ) we have two representations: (9,800, 1.94) and (9,400, 1.82). The slope is m = (1.94 1.82) (9,800 9, 400)   = 0.0003 Using one of the points, say (9,800, 1.94), we find b : 1.94 = (0.0003)(9,800) + b or b = – 1 So, the desired equation is: p = 0.0003 x – 1. (B) Here the two representations of ( x, p ) are: (9,300, 1.94) and (9,500, 1.82). The slope is (1.94 1.82) 0.0006 (9,300 9,500) m      Using one of the points, say (9,300, 1.94) we find b : 1.94 = – 0.0006(9,300) + b or b = 7.52 So, the desired equation is: p = – 0.0006 x + 7.52. (C) To find the equilibrium point, we need to solve 0.0003 x – 1 = – 0.0006 x + 7.52 for x . Observe that 0.0009 x = 8.52 or x = 8.52 0.0009 = 9,467 Substituting x = 9,467 in either of equations in (A) or (B) we obtain .0003(9, 467) 1 1.84 p    So, the desired point is (9,467, 1.84).

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