(Solution Manual) College Mathematics for Business Economics Life Sciences and Social Sciences

Barnett, Ziegler, Byleen, 13

Math & Statistics

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13th Edition

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Solution Manual for College Mathematics for Business Economics Life Sciences and Social Sciences, The downloadable file consists of 618 pages, and the total size of the file is 9.63 MB. Purchase now only for highly discounted price.

Author(s): Barnett, Ziegler, Byleen

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EXERCISE 1-1 1-1 Copyright © 2015 Pearson Education, Inc. 1 LINEAR EQUATIONS AND GRAPHS EXERCISE 1-1 2. 3– 4 6 1 9 3 6 1 5 3 6 1 5 3 1 5 5yy yy yy y y      4. 5x + 2 > 1 5x > −1 x > −1 5 6. −4x ≤ 8 4 4x ≥ 8 4 (Dividing by a negative number) x ≥ −2 8. −2x + 8 < 4 −2x + 8 – 8 < 4 − 8 −2x < −4 2 2x  > 42 (Dividing by a negative number) x > 2 or (2, ∞) 10. −4 < 2 y − 3 < 9 −1 < 2 y < 12 −1 2 < y < 6 or (1 / 2 , 6 ) . 12. 2433m Multiply both sides of the equation by 3 to obtain: m − 12 = 2 m = 14 14. 5 46x Multiply both sides by ( −4) which will result in changing the direction of the inequality as well. 20 6x and simplified we have 10.3x 16. 39 1 3 8 29 1 3 8 64 42 63y yy y y y y     18. −3(4 − x) = 5 − (x + 1) −12 + 3 x = 5 − x − 1 −12 + 3 x = 4 − x 12 − 12 + 3 x = 12 + 4 − x 3 x = 16 − x 4 x = 16 x = 4 1-2 CHAPTER 1: LINEAR EQUATIONS AND GRAPHS Copyright © 2015 Pearson Education, Inc. 20. x − 2 ≥ 2(x − 5) x − 2 ≥ 2x − 10 x − 2 + 2 ≥ 2x − 10 + 2 x ≥ 2x − 8 8 x  x ≤ 8 22. 4y − 3y = 1 2 Multiply both sides by 12: 3y − 4y = 6 −y = 6 y = −6 24. 2u − 2 3 < 3u + 2 2u − 3u < 2 + 2 3 6u < 8 3 u < 16 26. −4 ≤ 5x + 6 < 21 −6 − 4 ≤ 5x < 21 − 6 −10 ≤ 5x < 15 −2 ≤ x < 3 or [−2, 3) 28. −1 ≤ 2 3t + 5 ≤ 11 −5 − 1 ≤ 2 3t ≤ 11 − 5 −6 ≤ 2 3t ≤ 6 −18 ≤ 2t ≤ 18 −9 ≤ t ≤ 9 or [9 , 9 ] . 30. y = −2 3x + 8 y – 8 = −2 3x + 8 − 8 −2 3x = y − 8 −2x = 3y − 24 x = 32 4 2y  = −3 2y + 12 32. y = mx + b y − b = mx + b − b mx = y − b m = yb x 34. C = 5( 32)9F 9 5C = F − 32 32 + 9 5C = F F = 9 5C + 32 36. 10 8 3 6 18 3 14 18 3 14 146 3 14 36o r [ 1 4 / 3 , 6 ]u u u uu         38. (A) Two must be negative and one positive or all three must be positive. (B) Two must be positive and one negative or all three must be negative. (C) Two must be negative and one positive or all three must be positive. (D) a ≠ 0 and b and c must have opposite signs. EXERCISE 1-1 1-3 Copyright © 2015 Pearson Education, Inc. 40. If a and b are negative and b a > 1, then multiplying both sides by the negative number a we obtain b < a and hence a − b > 0. 42. False. Consider the two closed intervals [1, 2] and [2, 3]. Their intersection is {2} which is not an interval. 44. False. Consider the two closed intervals [ −1, 0] and [1, 2]. Their union is [ −1, 0]  [1, 2] which is not an interval. 46. True. Let A = [a, b], B = [c, d], where acb , so that AB  . Then A  B = [c, b] if bd and [, ]BdAc if db. In either case, the inters ection is a closed interval. 48. Let x = number of quarters in the meter. Then 100 − x = number of dimes in the meter. Now, 0.25 x + 0.10(100 − x) = 14.50 or 0.25 x + 10 − 0.10 x = 14.50 0.15 x = 4.50 x = 4.50 0.15 = 30 Thus, there will be 30 quarters and 70 dimes. 50. Let x be the amount invested in “Fund A” and (500,000 – x) the amount invested in “Fund B”. Then 0.052 x + 0.077(500,000 – x) = 30,000. Solving for x: (0.077)(500,000) – 30,000 = (0.077 – 0.052) x 8,500 = 0.025 x x = 8,500 0.025 = $340,000 So, $340,000 should be invested in Fund A and $160,000 in Fund B. 52. Let x be the price of the house in 1960. Then 29.6 200,000 229.6x (refer to Table 2, Example 10) 29.6200,000229.6x ≈ $25,784 To the nearest dollar, the house would be valued $25,784 in 1960. 54. (A) It is 60 – 0.15(60) = $51 (B) Let x be the retail price. Then 136 = x – 0.15 x = 0.85 x So, x = 136 0.85 = $160. 1-4 CHAPTER 1: LINEAR EQUATIONS AND GRAPHS Copyright © 2015 Pearson Education, Inc. 56. Let x be the number of times you must clean the living room carpet to make buying cheaper than renting. Then (20 + 2(16)) x = 300 + 3(9) x Solving for x 52 x = 300 + 27 x 25 x = 300 x = 300 25 = 12 58. Let x be the amount of the second employee’s sales during the month. Then (A) 3,000 + 0.05 x = 4,000 or x = 4,000 3,000 0.05 = $20,000 (B) In view of Problem 57 we have: 2,000 + 0.08( x – 7,000) = 3,000 + 0.05 x Solving for x: 2,000 – (0.08)7,000 – 3,000 = 0.05 x – 0.08 x −1,560 = −0.03x x = 1, 560 0.03 = $52,000 (C) An employee who chooses (A) will earn more than he or she would with the other option until $52,000 in sales is achieved, after which the other option would earn more. 60. Let x = number of books produced. Then Costs: C = 2.10 x + 92,000 Revenue: R = 15 x To find the break-even point, set R = C: 15 x = 2.10 x + 92,000 12.9 x = 92,000 x = 92,000 12.9 ≈ 7,132 Thus, 7,132 books will have to be so ld for the publisher to break even. 62. Let x = number of books produced. Costs: C(x) = 92,000 + 2.70 x Revenue: R(x) = 15 x (A) The obvious strategy is to raise the price of the book. (B) To find the break-even point, set R(x) = C(x): 15 x = 92,000 + 2.70 x 12.30 x = 92,000 x = 7,480 The company must sell more than 7,480 books to make a profit. (C) From Problem 60, the production level at the break-even point is: 7,132 books. At this production level, the costs are C(7,132) = 92,000 + 2.70(7,132) = $111,256.40 EXERCISE 1-2 1-5 Copyright © 2015 Pearson Education, Inc. If p is the new price of the book, then we need 7,132 p = 111,256.40 and p ≈ $15.60 The company should sell the book for at least $15.60. 64. −49 ≤ F ≤ 14 −49 ≤ 9 5C + 32 ≤ 14 −32 – 49 ≤ 9 5C ≤ 14 – 32 −81 ≤ 9 5C ≤ −18 (−81) · 5 ≤ 9C ≤ (−18) · 5 (8 1 )5 9 ≤ C ≤ (1 8 )5 9 −45 ≤ C ≤ −10 66. Note that IQ = MA CA  100 (see problem 65). Thus 80 < IQ < 140 80 < MA 12  100 < 140 or (80)(12) 100 < MA < (140)(12) 100 or 9.6 < MA < 16.8 EXERCISE 1-2 2. (A) 4. (B); slope is not defined for a vertical line 6. y = 2x+ 1 01 2243xy 8. 8x − 3y = 24 08 30 68x y  10. Slope: 3m 12. Slope: 10 3m y intercept: b = 2 y intercept: b = 4 14. Slope: 1 5m, y intercept: 1 2b 16. m = 1, b = 5 so 5 yx . 18. 6 7m, y intercept: 9 2b so 69 72yx 20. x intercept: 1; y intercept: 3; 33 yx  22. x intercept: 2, y intercept: −1; 112yx 1-6 CHAPTER 1: LINEAR EQUATIONS AND GRAPHS Copyright © 2015 Pearson Education, Inc. 24. y = −3 2x + 1 m = −3 2, b = 1 01 2224x y   26. 5x − 6y = 15 02 . 5 3035x y   28. 30. 5x − y = −2 −y = −5x − 2 Multiply both sides by ( −1); y = 5x + 2 m = 5 32. 2x − 3y = 18 −3y = −2x + 18 Divide both sides by (−3); y = 2 3x − 6 m = 2 3 34. 84 84 11 82xy yx yx    Slope = 1 8 36. (A) x = 4 (B) y = 3 (C) y = −2 3x + 8 38. 40. (A) (B) Set y = 0, −0.8x + 5.2 = 0, x = 6.5. Set x = 0, y = 5.2. EXERCISE 1-2 1-7 Copyright © 2015 Pearson Education, Inc. (C) (D) x intercept: 6.5; y intercept: 5.2 (E) 6.5x 42. The equation of the vertical line is x = −5 and the equation of the horizontal line is y = 6. 44. The equation of the vertical line is x = 2.6 and the equation of the horizontal line is y = 3.8. 46. Slope: 4m; point: (0,6). Using the point-slope form: 64 ( 0 ) 46yx yx   48. Slope: 10 m ; point: (2, 5). Using the point-slope form: (5 ) 1 0 ( 2 ) 51 02 0 10 15yx yx yx         50. Slope: 2/7m ; point: (7,1). Using the point-slope form: 21( 7 ) 7 212 7 21 7yx yx yx     52. Slope: 0.9m ; point: (2.3, 6.7). Using the point-slope form: 6.7 0.9( 2.3) 6.7 0.9 2.07 0.9 4.63yxyx yx    1-8 CHAPTER 1: LINEAR EQUATIONS AND GRAPHS Copyright © 2015 Pearson Education, Inc. 54. (A) m = 52 3 31 2 (B) Using y − y1 = m(x − x1), where m = 3 2 and ( x1, y1) = (1,2), we obtain: 3 2( 1 ) 2yx  or 32 1xy  (C) Slope-intercept form: 31 22yx 56. (A) m = 73 32  = −4 5 (B) Using y − y1 = m(x − x1), where m = −4 5 and ( x1, y1) = (−3,7), we obtain: y − 7 = −4 5(x + 3) or 4 x + 5y = 23. (C) Slope-intercept form: 42 3 55yx  58. (A) m = 44 01  = 0 1 = 0 (B) The line through (1, 4) and (0, 4) is horizontal; y = 4. (C) Slope-intercept form is the same: 4y. 60. (A) m = 30 22  = 3 0 which is not defined. (B) The line through (2, 0) and (2, −3) is vertical; x = 2. (C) No slope-intercept form 62. The graphs are parallel lines with slope −0.5. 64. Let C be the total weekly cost of producing x picnic tables. Then C = 1,200 + 45 x For C = $4,800, we have 1,200 + 45 x = 4,800 Solving for x we obtain x = 4,800 1,200 45 = 80 66. 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 EXERCISE 1-2 1-9 Copyright © 2015 Pearson Education, Inc. 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. 68. Let R and C be retail price and cost respectively. Then two points for ( C, R) are (20, 33) and (60, 93). (A) If C and R are linearly related, then the line expressing their relationship passes through the points (20, 33) and (60, 93). Therefore, R – 33 = (93 33) (60 20) (C – 20) or R = 1.5 C + 3 (B) For R = $240 we have 240 = 1.5 C + 3 or C = 240 3 1.5 = $158 70. 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, 224,0000) 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 (D) (B) For t = 10 V = −6,800(10) + 224,000 = $156,000 (C) For V = $100,000 100,000 = −6,800 t + 224,000 or t = (224,000 100,000) 6,80018.24 So, during the 19th year, the depreciated value falls below $100,000. 1-10 CHAPTER 1: LINEAR EQUATIONS AND GRAPHS Copyright © 2015 Pearson Education, Inc. 72. We have two representations for ( x, T) namely: (29.9, 212) and (28.4, 191). (A) The line of the form T = mx + b has slope: m = (212 191) (29.9 28.4)  = 14 Using, say (29.9, 212), will give the value for b: b = –206.6 T(B) For x = 31, we have T = 14(31) − 206.6 = 227.4 ˚F (C) For T = 199 ˚F, we have 199 = 14 x – 206.6 or x = 199 206.62148.97in Hg . (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)3.21m . Therefore, 3.2 200.TA (B) For A = 6.5 (6,500 feet), T = 3.2(6.5) + 200 = 20.8+200 = 220.8 mph 76. (A) I = mt + b At t = 0, I = 30,000. Therefore, I = mt + 30,000. At t = 20, I = 53,000 = m(20)+30,000 20 53,000 30,000 20 23,000 1150m mm Therefore, 1150 30,000It . (B) At t = 40, 1150(40) 30,000 76,000I  The median income in 2030 will be $76,000. 78. (A) f = mt + b At t = 0, f = 25.7. Therefore, f = mt + 25.7. EXERCISE 1-2 1-11 Copyright © 2015 Pearson Education, Inc. At t =10, 21.5 10 25.7 10 4.2 0.42m m m Therefore, 0.42 25.7 ft  . (B) Solve 0.42 25.7 12t for t 0.42 13.7 32.6t t   : The percentage of male smokers will fall below 12% in 2032. 80. (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 poi nts, 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 the equations in (A) or (B) we obtain .0003(9,467) 1 1.84p  So, the desired point is (9,467, 1.84). (D) 1-12 CHAPTER 1: LINEAR EQUATIONS AND GRAPHS Copyright © 2015 Pearson Education, Inc. 82. We have two re presentations of ( w, d): (3, 18) and (5, 10). (A) The line through these two points has a slope (18 10) (3 5)  = −4. So, the equation of the line is d − 10 = −4(w – 5) or d = −4w + 30 (B) For w = 0, d = 30 in. (C) For d = 0, −4w + 30 = 0 or w = 30 4 = 7.5 lbs. EXERCISE 1-3 2. (A) w = 52 + 1.9 h (B) The rate of change of weight with respect to height is 1.9 inches per kilogram. (C) 5'8 ꞌ' is 8 inches over 5 feet and the model predicts the weight to be w = 52 + 1.9(8) = 67.2 kg. (D) For w = 70, we have 70 = 52 + 1.9 h or h = 70 52 1.9 ≈ 9.5 So, the height of this man is predicted to be 5 ꞌ9.5'ꞌ. 4. We have two re presentations 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 14.7Pd (B) The rate of change of pressure w ith respect to depth is approximately 0.432 lbs/in2 per foot. (C) For d = 50, P = 0.432 (50) + 14.7 ≈ 36.3 lbs/in2 (D) For P = 4 atmospheres, we have P = 4(14.7) = 58.8 lbs/in2 and hence 58.8 = 0.432 d + 14.7 or d = 58.8 14.7 0.432102 ft. 6. We have two re presentations of ( t, a): (0, 2,880) and (180, 0). EXERCISE 1-3 1-13 Copyright © 2015 Pearson Education, Inc. (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 = −16t + 2,880 (B) The rate of descent for an ATPS system parachute is 16 ft/sec. (C) It is 16 ft/sec. 8. We have two re presentations of ( t, s): (0, 1,403) and (20, 1,481). So, the line passing through these poi nts has the following equation: s – 1,403 = (1,481 1,403) (20 0) (t – 0) or s = 3.9 t + 1,403 The slope of this line (model) is the rate of change of the speed of sound with respect to temperature; 3.9 m/s per ˚C. 10. (A) (B) The percent rate of change of fossil fuel consumption is −0.09% per year. (C) For x = 40 (2025 is 40 years from 1985), we have 0.09 40 85.8 82.2. y   i.e. 82.2% of total consumption. (D) Solve 0.09 85.8 80x  for x: 0.09 85.8 80 0.09 80 85.8 5.8 64.4x x x       Fossil fuel consumption will be less than 80% of total energy consumption in 2050. 1-14 CHAPTER 1: LINEAR EQUATIONS AND GRAPHS Copyright © 2015 Pearson Education, Inc. 12. (B) Solve 0.44 27.28 15t  for t: 0.44 15 27.28 12.28 27.91t t     The first year in which the percentage of male smokers will be less than 15% is 2025. 14. (B) For 2025, x = 45 and (45) 0.048(45) 1.14 3.3y  , so there will be about 3,300,000 gradua te students enrolled in 2016. (C) Graduate student enrollment is increasing at a rate of 0.048 million (or 48,000) students per year. 16. (B) At x = 5.3, y = 0.63(5.3) + 0.31= 3.649. There were approximately 3, 649,000 licensed drivers in Minnesota in 2010. (C) Solve 0.63 x + .031 = 4.1 for x: 0.63 3.79, 6.016xx The population of Wisconsin 2010 was approximately 6,016,000. 18. (B) For 2022, t = 24 and from the model 1.21(24) 12. 41. 06 1 I  . Wal-Marts operating income in 2024 will be approximately $41.1 billion. EXERCISE 1-3 1-15 Copyright © 2015 Pearson Education, Inc. 20. (B) For P = 30, we have: 30 = −0.54T + 34 or T = 34 30 0.54 ≈ 7˚F (C) For T = 15, we have: P = −0.54(15) + 34 = 25.9, i.e., the estimated percentage of propylene glycol is 25.9%. 22. (A) The rate of change of height with respect to Dbh is 1.66 ft/in. (B) One inch increase in Dbh produces a height increase of approximately 1.66 ft. (C) For x = 12, we have: y = 1.66(12) – 5.14 ≈ 15 ft. (D) For y = 25, we have: 25 = 1.66 x – 5.14 or x = 25 5.14 1.66 ≈ 18 in. 24. 5.85 36.32, slope: 5.85yx m  (A) Annual revenue is increasing at a rate of $5.85 billion per year. (B) For 2024, x = 24 and y = 5.85(24) + 36.32 ≈ 176.72. So, the predicted annual reve nue is approximately $177 billion. 26. Residential service: 32.1 687 yx  , cellular service: 64.8 201yx (A) Annual expenditure per consumer unit on residential telephone service is decreasing at a rate of $32.10 per year. Annual expenditure per consumer unit on cellular service is increasing at a rate of $64.80 per year. (B) For 2020, x = 20. For residential service we have: 32.1(20) 719 $77.00 y   For cellular service we have, 64.8(20) 136 $1432y  (C) For 2025, the models predict annua l expenditure per consumer unit to be −$83.50 on residential service and $1,756 on cellular service. The residentia l model clearly gives an unreasonable prediction for 2025. The cellular model pre diction seems overly high. 1-16 CHAPTER 1: LINEAR EQUATIONS AND GRAPHS Copyright © 2015 Pearson Education, Inc. 28. Men: 0.294 119.503 yx  Women: 0.156 128.473 yx  The graphs of these lines indicate that the women will not catch up with the men. To see this algebraically, if we set the equations equal to each other and solve, then we obtain x = −65, so the lines intersect at a point outside of the domain of our functions. Also, the men’s slope is st eeper so their times, already lower, are decreasing more rapidly. 30. Supply: y = 1.53 x + 2.85; Demand: y = −2.21x + 10.66 To find equilibrium price we solve the following equation for x and then use that to find y: 1.53 x + 2.85 = −2.21x + 10.66 or x = (10.66 2.85) (1.53 2.21)  ≈ 2.09, and y = 1.53(2.09) + 2.85 ≈ $6.05. EXERCISE 2-1 2-1 Copyright © 2015 Pearson Education, Inc. 2 FUNCTIONS AND GRAPHS EXERCISE 2-1 2. 4. 6. 8. 10. The table specifies a function, since for each dom ain 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 intersect s the graph in at most one point. 18. The graph does not specify a function. There are verti cal 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. 10 3y x is linear. 24. 28 xy is neither linear nor constant. 26. 22 2 2 3 3 3333 which is consta .3t4nx xx xy     28. 926 0xy  is linear. 30. 32. 2-2 CHAPTER 2 FUNCTIONS AND GRAPHS Copyright © 2015 Pearson Education, Inc. 34. 36. 38. f(x) = 2 23 2x 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. f(x) = 3, x < 0 at x = –4, –2 46. f(x) = 4 at x = 5 48. All real numbers 50. All real numbers except x = 2 52. 5 x 54. Given 67 2 1xy . Solving for y we have: 672 1 6 a n d 3 . 7yx y x   This equation specifies a function. The domain is R, the set of real numbers. 56. Given () 4xx y . Solving for y we have: 2 2 44a n d .xxy x y x This equation specifies a function. The domain is all real numbers except 0. 58. Given 22.9 xy Solving for y we have: 22 29a n d 9 . yxy x    This equation does not define y as a function of x. For example, when x = 0, y =  3. 60. Given 30. xy . Solving for y we have: 31 / 6and . yxy x   This equation specifies a function. The domain is all nonnegative real numbers, i.e., 0x. 62. 2(5 ) (5 ) 4 2 5 4 2 1f    64. 22 2(2 ) (2 ) 4 44 4 4fxx x x x x     66. 22( 1 0) ( 1 0) 4 1 0 0 4fx x x  68. 2 () 4 4fx x x   70. 22 2 2(3 ) () (3 ) 4 4 5 4 1ff h h h h       72. 222(3 ) (3 ) 4 9 6 4 5 6f h h hh hh         74. 22 2 2(3 ) (3 ) (3 ) 4 (3 ) 4 ( 9 6 4 ) ( 9 4 ) 6f hf h h h h h              EXERCISE 2-1 2-3 Copyright © 2015 Pearson Education, Inc. 76. (A) ( ) 3 ( ) 9 339fx h x h x h    (B)   () ( )3 3 9 3 93fxh f x x h x h     (C) () ( ) 33fx h fx h hh   78. (A) 2 22 22() 3 () 5 () 8 3( 2 ) 5 5 8 3635 5 8fx h x h x h xx h h x h xx h hx h            (B)  22 2 2() ( ) 3635 5 8 35 8 635fx h fx x x h h x h x x xh h h       (C) 2() ( ) 635635fx h fx x h h hxhhh     80. (A) f(x + h) = x2 + 2xh + h2 + 40 x + 40 h (B) f(x + h) – f(x) = 2 xh + h2 + 40 h (C) () ( )fxh f x h = 2x + h + 40 82. Given A = w = 81. Thus, w = 81 . Now P = 2 + 2w = 2 + 281  = 2 + 162 . The domain is  > 0. 84. Given P = 2 + 2w = 160 or  + w = 80 and  = 80 – w. Now A = w = (80 – w)w and A = 80 w – w2. 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) 2-4 CHAPTER 2 FUNCTIONS AND GRAPHS Copyright © 2015 Pearson Education, Inc. 90. (A) P(x) = R(x) – C(x) = x(2,000 – 60 x) – (4,000 + 500 x) thousand dollars = 1,500 x – 60 x2 – 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) 92. (A) 1.2 inches (B) Evaluate the volume function for x = 1.21, 1.22, …, and choose the value of x whose volume is closest to 65. (C) x = 1.23 to two decimal places 94. (A) V(x) = x2(108 4 ) x (B) 0 < x < 27 (C) Table 16 Volume x V(x) 5 2,200 10 6,800 15 10,800 20 11,200 25 5,000 (D) 96. (A) Given 5 v – 2s = 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 2-2 2. () 4 1 2fx x  Domain: all real numbers; range: all real numbers. 4. () 3fxx Domain: [0, ) ; range: [3, ). 6. () 5 2fx x  Domain: all real numbers; range: (, 2 ] . EXERCISE 2-2 2-5 Copyright © 2015 Pearson Education, Inc. 8. 3( ) 20 10 x fx Domain: all real numbers; range: all real numbers. 10. 12. 14. 16. 18. 20. 22. 26. 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. 24. 28. 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. 2-6 CHAPTER 2 FUNCTIONS AND GRAPHS Copyright © 2015 Pearson Education, Inc. 30. The graph of g(x) = –6 + 3x is the graph of y = 3x shifted 6 units down. 32. The graph of m(x) = –0.4 x2 is the graph of y = x2 reflected in the x axis and vertically contracted by a factor of 0.4. 34. The graph of the basic function y = |x| is shifted 3 units to the right and 2 units up. y = |x – 3| + 2 36. 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| 38. The graph of the basic function 3x is reflected in the x axis and shifted up 2 units. Equation: y = 2 – 3x 40. 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 – ( x – 3)3 42. g(x) = 33x+ 2 44. g(x) = –| x – 1| 46. g(x) = 4 – ( x + 2)2 48. g(x) = 1i f 1 22 i f 1xx xx    50. h(x) = 10 2 if 0 20 40 0.5 if 20xx xx   EXERCISE 2-2 2-7 Copyright © 2015 Pearson Education, Inc. 52. h(x) = 42 0 i f0 2 0 26 0 i f 2 0 1 0 0 360 if 100xx xx xx      54. 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 56. The graph of the basic function y = |x| is vertically expanded by a factor of 4. Equation: y = 4|x| 58. The graph of the basic function y = x3 is vertically contracted by a factor of 0.25. Equation: y = 0.25 x3. 60. 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. 62. 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). 64. Horizontal shift, ve rtical 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. 66. (A) The graph of the basic function y = x is vertically expanded by a factor of 4. 68. (A) The graph of the basic function y = x2 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) 2-8 CHAPTER 2 FUNCTIONS AND GRAPHS Copyright © 2015 Pearson Education, Inc. 70. (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, W(x) = 8.5 .065 if 0 700 16.9 0.053 if 700xx xx   (B) 72. (A) Let x = taxable income. If 0 ≤ x ≤ 15,000, the tax due is $.035 x. At x = 15,000, the tax due is $525. For 15,000 < x ≤ 30,000, the tax due is 525 + .0625( x – 15,000) = .0625 x – 412.5. For x > 30,000, the tax due is 1,462.5 + .0645( x – 30,000) = .0645 x – 472.5. Thus, (B) T(x) = 0.035 if 0 15,000 0.0625 412.5 if 15,000 30,000 0.0645 472.5 if 30,000xx xxxx     (C) T(20,000) = $837.50 T(35,000) = $1,785 74. (A) The graph of the basic function y = x3 is vertically expanded by a factor of 463. 76. (A) The graph of the basic function y = 3x is reflected in the x axis and shifted up 10 units. (B) (B) EXERCISE 2-3 2. 216x x (standard form) 4. 212 8xx (standard form) 264 64 16xx (completing the square) 2(1 2 ) 8xx 2(8 ) 6 4x (vertex form) 2(1 2 3 6 ) 8 3 6xx  (completing the square) 2(6 ) 4 4x (vertex form) EXERCISE 2-3 2-9 Copyright © 2015 Pearson Education, Inc. 6. 231 8 2 1xx (standard form) 2 2 2 23( 6 ) 21 3( 6 9 9) 21 (completing the square)3( 3) 21 273( 3) 6 (vertex form)xxxxxx    8. 251 5 1 1xx (standard form) 2 2 299 44 34 5 24 23 1 245( ) 11 5( ) 11 (completing the square) 5( ) (vertex form)3 3 11 5( )x xx xx x        10. The graph of g(x) is the graph of y = x2 shifted right 1 unit and down 6 units; 2() ( 1 ) 6 .gx x 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; 2() ( 4 ) 7 .nx x   14. (A) g (B) m (C) n (D) f 16. (A) x intercepts: –5, –1; y intercept: –5 (B) Vertex: (–3, 4) (C) Maximum: 4 (D) Range: y ≤ 4 or (–∞, 4] 18. (A) x intercepts: 1, 5; y intercept: 5 (B) Vertex: (3, –4) (C) Minimum: –4 (D) Range: y ≥ –4 or [–4, ∞) 20. 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] 22. 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, ∞)