EXERCISE 3-7 Tings to remember: 1. MARGINAL COST, REVENUE, AND PROFIT If is te number of units of a product produced in some time interval, ten: Total Cost C() Marginal Cost C'() Total Revenue R() Marginal Revenue R'() Total Profit P() R() - C() Marginal Profit P'() R'() - C'() (Marginal Revenue) - (Marginal Cost) Marginal cost (or revenue or profit) is te instantaneous rate of cange of cost (or revenue or profit) relative to production at a given production level.. MARGINAL COST AND EXACT COST If C() is te cost of producing items, ten te marginal cost function approimates te eact cost of producing te ( + 1)st item: Marginal Cost Eact Cost C'() C( + 1) - C() Similar interpretations can be made for total revenue and total profit functions. 3. BREAK-EVEN POINTS Te BREAK-EVEN POINTS are te points were total revenue equals total cost. 4. MARGINAL AVERAGE COST, REVENUE, AND PROFIT If is te number of units of a product produced in some time interval, ten: Average Cost C () C() Cost per unit Marginal Average Cost C '() Average Revenue R () R() Revenue per unit Marginal Average Revenue R '() Average Profit P () P() Profit per unit Marginal Average Profit P '() EXERCISE 3-7 135
1. C() 000 + 50-0.5 (A) Te eact cost of producing te 1st food processor is: C(1) - C(0) 000 + 50(1) - (1) # - 000 + 50(0) " (0) & % ( $ % '( 89.50-800 9.50 or $9.50 (B) C'() 50 - C'(0) 50-0 30 or $30 3. C() 60,000 + 300 60,000 + 300 (A) C () 60,000 + 300 60,000-1 + 300 60,000 + 300(500) C (500) 10,000 40 or $40 500 500 (B) C '() -60,000 - "60,000 C '(500) "60,000 (500) -0.4 or -$0.4 Interpretation: At a production level of 500 frames, average cost is decreasing at te rate of 4 per frame. (C) Te average cost per frame if 501 frames are produced is approimately $40 - $0.4 $419.76. 5. P() 30-0.3-50, 0 100 (A) Te eact profit from te sale of te 6t skateboard is: P(6) - P(5) 30(6) - 0.3(6) - 50 - [30(5) - 0.3(5) - 50] 37.0-31.50 $14.70 (B) Marginal profit: P'() 30-0.6 P'(5) $15 7. P() 5-00 - 450 P'() 5-100 (A) P'(450) 5-450 0.5 or $0.50 100 Interpretation: At a production level of 450 cassettes, profit is increasing at te rate of 50 per cassette. (B) P'(750) 5-750 -.5 or -$.50 100 Interpretation: At a production level of 750 cassettes, profit is decreasing at te rate of $.50 per cassette. 136 CHAPTER 3 LIMITS AND THE DERIVATIVE
9. P() 30-0.03-750 Average profit: P () P() 30-0.03-750 (A) At 50, P (50) 30 - (0.03)50-750 50 30-0.03-750 -1 13.50 or $13.50. (B) P '() -0.03 + 750 - -0.03 + 750 P '(50) -0.03 + 750-0.03 + 0.3 0.7 or $0.7; at a (50) production level of 50 mowers, te average profit per mower is INCREASING at te rate of $0.7 per mower. (C) Te average profit per mower if 51 mowers are produced is approimately $13.50 + $0.7 $13.77. 11. 4,000-40p (A) Solving te given equation for p, we get 40p 4,000 - and p 100-1 or p 100-0.05 40 Since p 0, te domain is: 0 4,00 (B) R() p 100-0.05, 0 4,000 (C) R'() 100-0.05; R'(1,600) 100-80 0 At a production level of 1,600 radios, revenue is INCREASING at te rate of $0 per radio. (D) R'(,500) 100-15 -5 At a production level of,500 radios, revenue is DECREASING at te rate of $5 per radio. 13. Price-demand equation: 6,000-30p Cost function: C() 7,000 + 60 (A) Solving te price-demand equation for p, we get p 00-1 ; domain: 0 6,000 30 (B) Marginal cost: C'() 60 (C) Revenue function: R() 00-1 30 ; domain: 0 6,000 (D) Marginal revenue: R'() 00-1 15 (E) R'(1,500) 100; at a production level of 1,500 saws, revenue is INCREASING at te rate of $100 per saw. R'(4,500) -100; at a production level of 4,500 saws, revenue is DECREASING at te rate of $100 per saw. EXERCISE 3-7 137
Hundreds of Tousands Tousands (G) Profit function: P() R() - C() 00-1 30 - [7,000 + 60] 140-1 30-7,000 (H) Marginal profit: P'() 140-1 15 (I) P'(1,500) 140-100 40; at a production level of 1,500 saws, profit is INCREASING at te rate of $40 per saw. P'(3000) 140-00 -60; at a production level of 3,000 saws, profit is DECREASING at te rate of $60 per saw. 15. (A) Assume p m + b. We are given 16 m 00 + b and 14 m 300 + b Subtracting te second equation from te first, we get -100m so m - 1 50-0.0 Substituting tis value into eiter equation yields b 0. Terefore, P 0-0.0; domain: 0 1,000 (B) Revenue function: R() p 0-0.0, domain: 0 1,000. (C) C() m + b. From te finance department's estimates, m 4 and b 1,400. Tus, C() 4 + 1,400. Tousands of dollars Hundreds of units (E) Profit function: P() R() - C() 0-0.0 - [4 + 1,400] 16-0.0-1,400 138 CHAPTER 3 LIMITS AND THE DERIVATIVE
(F) Marginal profit: P'() 16-0.04 P'(50) 16-10 6; at a production level of 50 toasters, profit is INCREASING at te rate of $6 per toaster. P'(475) 16-19 -3; at a production level of 475 toasters, profit is DECREASING at te rate of $3 per toaster. 17. Total cost: C() 4 + 1,900 Total revenue: R() 00-0., 0 1,000 (A) R'() 00-0.4 Te grap of R as a orizontal tangent line at te value(s) of were R'() 0, i.e., 00-0.4 0 or 500 (B) P() R() - C() 00-0. - (4 + 1,900) 176-0. - 1,900 (C) P'() 176-0.4. Setting P'() 0, we ave 176-0.4 0 or 440 (D) Te graps of C, R and P are sown below. Break-even points: R() C() 00-0. 4 + 1,900 0. - 176 + 1,900 0 176 ± (176) " (4)(0.)(1,900) (quadratic formula) (0.) 176 ± 30,976 " 17,50 0.4 176 ± 13,456 176 ± 116 730, 150 0.4 0.4 Tus, te break-even points are: (730, 39,40) and (150, 5,500). -intercepts for P: -0. + 17.6-1,900 0 or 0. - 176 + 1,900 0 wic is te same as te equation above. Tus, 150 and 730. EXERCISE 3-7 139
19. Demand equation: p 0-0 - 1/ Cost equation: C() 500 + 100 C() (A) Revenue R() p (0-1/ ) or R() 0-3/ 0 R() 400 (B) Te graps for R and C for 0 400 are sown at te rigt. 0 Break-even points (44, 588) and (58, 1,016). 1. (A) (B) Fied costs $71,680; variable costs $11 (C) Let y p() be te quadratic regression equation found in part (A) and let y C() be te linear regression equation found in part (B). Ten revenue R() p(), and te break-even points are te points were R() C(). break-even points: (713, 807,703), (5,43, 1,376,7) 0,000,000 (D) Te company will make a profit wen 713 5,43. From part (A), p(713) 1,133 and p(5,43) 54. Tus, te company will make a profit for te price range $54 p $1,133. 0 8,000 CHAPTER 3 REVIEW 1. (A) f(3) - f(1) (3) + 5 - [(1) + 5] 16 f(3) " f(1) (B) Average rate of cange: 16 3 " 1 8 f(3) " f(1) (C) Slope of secant line: 16 3 " 1 8 (D) Instantaneous rate of cange at 1: Step 1. f(1 + ) " f(1) (1 + ) + 5 " [(1) + 5] (1 + + ) + 5 " 7 f(1 + ) " f(1) Step. (4 + ) 4 0 (E) Slope of te tangent line at 1: 4 4 + 4 + (F) f'(1) 4 (3-1, 3-4) 140 CHAPTER 3 LIMITS AND THE DERIVATIVE
. f() -3 + f( + )! f() Step 1. Simplify f( + )! f()!3( + ) +! (!3 + )!3! 3 + + 3! -3 f( + )! f() Step. Evaluate.!0 f( + )! f() (-3) -3 (3-3)!0!0 3. (A) (5f() 3g()) 5 f() + 3 g() 5 + 3 4!1!1!1 g()] 4 8 (B)!1 g() (C)!1 (D)!1 [f()g()] [ f()][!1!1 g() f() "1 f() 4 "1 [5 + - 3g()] 5 + - 3 g()!1!1!1 5 + - 3(4) -5 (3-1) 4. (A) f() 1 (B) f() 1 (C) f() 1!1!1 +!1 (D) f(1) 1 (3-1) 5. (A)! (B)! + (C) f() does not eist! (D) f() 3 (3-1) 6. (A) f() 4 (B) f() 4 (C) f() 4!3!3 +!3 (D) f(3) does not eist (3-1) 7. (A) From te grap, f() does not eist since!1!1 f()!1 + f() 3. (B) f(1) 3 (C) f is NOT continuous at 1, since f() does not eist. (3-)!1 8. (A)! f() (B) f() is not defined (C) f is NOT continuous at since f() is not defined. (3-) CHAPTER 3 REVIEW 141
9. (A) f() 1!3 (B) f(3) 1 (C) f is continuous at 3 since f() f(3).!3 (3-) f() 5 (3-3) 10. "# f() 5 (3-3) 11. "#$ 1. f() (3-3) 13. f() - (3-3) " + # " 14. f() 0 (3-1) 15. f() 0 (3-1) # "0 "0 + 16. f() 0 (3-1) "0 17. is a vertical asymptote (3-3) 18. y 5 is a orizontal asymptote (3-3) 19. f is discontinuous at (3-) 0. f() 5 Step 1. Find f( + ): f( + ) 5( + ) 5( + + ) 5 + 10 + 5 Step. Find f( + ) f(): f( + ) f() 5 + 10 + 5 5 10 + 5 Step 3. Find f( + ) " f() : f( + ) " f() 10 + 5 10 + 5 f( + ) " f() Step 4. Find "0 f( + ) " f() (10 + 5) 10 "0 "0 Tus, f () 10 (3-4) 1. (A) () (3f()) 3f (); (5) 3f (5) 3(-1) -3 (B) () (-g()) -g (); (5) -g (5) -(-3) 6 (C) () f (); (5) (-1) - (D) () -g (); (5) -(-3) 3 (E) () f () + 3g (); (5) (-1) + 3(-3) -11 (3-5). f() 1 3 3-5 + 1; f'() - 10 (3-5) 3. f() 1/ - 3 f'() 1-1/ - 3 1 1-3 (3-5) 14 CHAPTER 3 LIMITS AND THE DERIVATIVE
4. f() 5 f'() 0 (3-5) 5. f() 3 + 53 4 6. f() 0.5 4 + 0.54 0.5-4 + 0.5 4 3-1 + 5 4 3 ; f'() - 3 - + 15 4-3 + 15 4 (3-5) f'() 0.5(-4) -5 + 0.5(4 3 ) - -5 + 3-5 + 3 (3-5) 7. f() (3 3 )( + 1) 3 4 + 3 3 f () 1 3 + 9 (3-5) For Problems 8 31, f() +. 8. 1 3 1, y f( ) f( 1 ) 1 10, "y " 10 5. (3-6) 9. f( 1 + ") # f( 1 ) " f(1 + ) " f(1) f(3) " f(1) 1 " 5 (3-6) 30. dy f ()d ( + 1)d. For 1 1, 3, d 3 1, dy ( 1 + 1) 3 6 (3-6) 31. y f( + ) f(); at 1, 0., y f(1.) f(1) 0.64 dy f ()d were f () + 1; at 1 dy 3(0.) 0.6 (3-6) 3. From te grap: (A) " f() 4 (B) " + f() 6 (C) f() does not eist since f()! " " + f() (D) f() 6 (E) No, since f() does not eist. (3-)! 33. From te grap: (A) f() 3 (B) f() 3 (C) f() 3 (D) f(5) 3 "5 "5 +!5 (E) Yes, since f() f(5) 3. (3-)!5 34. (A) f() < 0 on (8, ) (B) f() 0 on [0, 8] (3-) CHAPTER 3 REVIEW 143
35. - < 1 or - - 1 < 0 Let f() - - 1 ( + 3)( - 4). Ten f is continuous for all and f(-3) f(4) 0. Tus, -3 and 4 are partition numbers. f() + + + + + - - - - - - + + + + -4-3 0 4 5 Test Numbers f() -4 8 (+) 0-1 (-) 5 8 (+) Tus, - < 1 for: -3 < < 4 or (-3, 4). (3-) 36.! 5 + 3 > 0 or! 5 ( + 3) > 0 Let f( )! 5. Ten f is discontinuous at 0 and -3, ( + 3) and f(5) 0. Tus, -3, 0, and 5 are partition numbers. Test Numbers f() - - - + + - - - - + + + Tus,! 5 + 3-4 -3-1 0 1 5 6 > 0 for -3 < < 0 or > 5 f()!4! 9 (!) 4!1 3(+) 1!1(!) 1 6 (+) 54 or (-3, 0) (5, ). (3-) 37. 3 + - 4 - > 0 Let f() 3 + - 4 -. Te f is continuous for all and f() 0 at -.349, -0.4707 and 1.8136. f() - - - 0 + + + 0 - - - - 0 + + + -.34-0.47 0 1.81 Tus, 3 + - 4 - > 0 for -.349 < < -0.4707 or 1.8136 < <, or (-.349, -0.4707) (1.8136, ). (3-) 38. f() 0.5-5 f(4)! f() (A) 0.5(4)! 5! [0.5()! 5] 4! 8! 3 144 CHAPTER 3 LIMITS AND THE DERIVATIVE
(B) f( + )! f() f( + )! f() (C)!0 39. y 1 3-3 5 - + 1; 0.5( + )! 5! [0.5()! 5] 0.5(4 + 4 + )! 5 + 3 + 0.5 ( + 0.5) + 0.5 ( + 0.5) (3-4)!0 dy d 1 3 (-3)-4 5(-) -3 - -4 + 10-3 (3-5) 40. y 3 + 5 3 3 1/ + 5 3-1/ ; y' 3 # 1 "1 & % $ ' ( + 5 # 3 " 1 "3 & % ( 3 $ ' 4 1 " 5 6 3 3 4 " 5 (3-5) 6 3 41. g() 1.8 3 # g'() 1.8% 1 3 " 3 $ + 0.9 3 1.8 1/3 + 0.9-1/3 & ' ( + 0.9 # " 1 3 "4 3 & % ( $ ' 0.6 -/3-0.3-4/3 0.6 3 " 0.3 4 3 (3-5) 4. y 3! 3 5 3 5-3 5-3 ; y' - 3 5 (-3-4 ) 9 5 4 (3-5) 43. f() + 4 f'() (A) Te slope of te grap at 1 is m f'(1). (B) f(1) 1 + 4 5 Te tangent line at (1, 5), were te slope m, is: (y - 5) ( - 1) [Note: (y - y 1 ) m( - 1 ).] y 5 + - y + 3 (3-4, 3-5) 44. f() 10 - f'() 10 - Te tangent line is orizontal at te values of suc tat f'() 0: 10-0 5 (3-4) CHAPTER 3 REVIEW 145
45. f() ( + 3)( - 45) f'() ( + 3)() + ( - 45)(1) 3 + 6-45 Set f'() 0: 3 + 6-45 0 + - 15 0 ( - 3)( + 5) 0 3, -5 (3-5) 46. f() 4-3 - 5 + 7 f'() 4 3-6 - 10 + 7 Set f'() 4 3-6 - 10 + 7 0 and solve for using a rootapproimation routine on a graping utility: f'() 0 at -1.34, 0.58,.6 (3-5) 47. f() 5 10 3 5 + 10 f () 5 4 30 5 5( 4 6 1) Let f () 5( 4 6 1) 0 and solve for using a rootapproimation routine on a graping utility; tat is, solve 4 6 1 0 for. f () 0 at ±.484 (3-5) 48. y f() 8-4 + 1 (A) Instantaneous velocity function; v() f'() 16-4. (B) v(3) 16(3) - 4 44 ft/sec. (3-5) 49. y f() -5 + 16 + 3 (A) Instantaneous velocity function: v() f'() -10 + 16. (B) v() 0 wen -10 + 16 0 10 16 1.6 sec (3-5) 50. (A) f() 3, g() ( - 4) 3, () ( + 3) 3 Te grap of g is te grap of f sifted 4 units to te rigt; te grap of is te grap of f sifted 3 units to te left. 5 5 5 f 5 g ' f' g' (B) f'() 3, g'() 3( - 4), '() 3( + 3) 5 Te grap of g' is te grap of f' sifted 4 units to te rigt; te grap of ' is te grap of f' sifted 3 units to te left. 5 5 5 (1-, 3-5) 146 CHAPTER 3 LIMITS AND THE DERIVATIVE
51. (A) g() kf(), k > 1; g () kf () Te grap of g is a vertical stretc of te grap of f; te grap of g is a vertical stretc of te grap of f. (B) g() f() + k, g () f () Te grap of g is a vertical translation of te grap of f; te grap of g is te same as te grap of f. (3-5) 5. f() 1, f () 6. y f( + ) f(); at 3, 0. y f(3.) f(3) 1 3. - 1 3 0.6816. dy f ()d 6 d; at 3, d 0. dy 6 (0.) 0.698 (3-6) 3 53. f() 1/3, f () 1 3 -/3 1 3 3 3 8.1 (8.1) 1/3 f(8.1) f(8) + df + df df f ()d; at, d 0.1 1 df (0.1) 0.00833 (5 decimal places) 3 3(8) Terefore: 3 8.1.00833 Calculator value:.00830 (5 decimal places) (3-6) 54. f() - 4 is a polynomial function; f is continuous on (-, ). (3-) 55. f() + 1! is a rational function and te denominator - is 0 at. Tus f is continuous for all suc tat, i.e., on (-, ) (, ). (3-) + 4 56. f() is a rational function and te denominator + 3! 4 + 3-4 ( + 4)( - 1) is 0 at -4 and 1. Tus, f is continuous for all ecept -4 and 1, i.e., on (-, -4) (-4, 1) (1, ). (3-) 3 57. f() 4 " ; g() 4 - is continuous for all since it is a polynomial function. Terefore, f() 3 g() is continuous for all, i.e., on (-, ). (3-) 58. f() 4 " ; g() 4 - is continuous for all and g() is nonnegative for -. Terefore, f() g() is continuous for -, i.e., on [-, ]. (3-) CHAPTER 3 REVIEW 147
59. f()! 3 (! 3)! 3, 0 (A) f()!1!1 (B) f()!3!3 ( - 3) 0 (C) f()!0!0! 3!3! 3! 3-3!1 ( " 3)!1! -1 does not eist since and!3 (3-1) 60. f() + 1 (3! ) ( + 1) + 1 (A)!1(3! )!1 (3 " ) 1!1 ( + 1) + 1!"1(3! )! "1!"1 (B) (3 " ) 0 4 0 + 1 (C) (3! ) does not eist since ( + 1) 4!3!3 and!3 (3 - ) 0 (3-1) 61. f()! 4! 4 (A) f() -1!4 (C)!4 # "1 if < 4 $ % 1 if > 4 (B) f() 1!4 + f() does not eist. (3-1) 6. f()! 3 9!! 3 (3 + )(3! )!(3! ) (3 + )(3! )!1 3 +, 3!1 (A) f()!3!3 3 + -1 6!1 (B) f() does not eist!"3!"3 3 +!1 (C) f()!0!0 3 + - 1 3 (3-1) 148 CHAPTER 3 LIMITS AND THE DERIVATIVE
63. f()!!! 7 + 10 (A) f()!"1!"1 (B) f()!! (C) f()!5 (! )( + 1) (! )(! 5) + 1! 5, + 1! 5 0 64. f() 3 " 6 3( " ) (A) 3 " 6 3 3 "# (B) + 1! 5 3!3-1 + 1 does not eist (3-1)!5! 5 "# "#$ 3 " 6 "#$ (C) " 65. f() " + (A) "# (B) 3 " 6 3( " ) " ; 3 3 3( " ) -! 3 " 6 3 3( " ) 3 3 " 1 + 1 3 3 3 " 1 + 1 "# 3 "#$ 3 " 1 + 1 "#$ 3 3 3 does not eist. (3-3) "# 3 "#$ 3-3 (C) " 3( " ) 3 3 " + ; 3( " ) " 3( " ) (3-3) 66. f() 3( " ) 3 (A) "# 3( " ) 3 "# 3 3 "# 3 0 (B) "#$ 3( " ) 3 "#$ 3 3 "#$ 3 0 (C) -, " 3 3( " ) " + 3( " ) 3 ; does not eist. (3-3) " 3 3( " ) CHAPTER 3 REVIEW 149
67. f() + 4 f( + ) " f() "0 68. Let f() 1 + f( + ) " f() "0 [( + ) + 4] - [ + 4] 0 0 4 + 4 + + 4-8 0 4 + (4 + ) 4 (3-1) 0 0 1 ( + ) + - 1 + + - ( + + ) ( + + )( + ) 0 - ( + + )( + ) 0-1 ( + + )( - ) -1 ( + ) (3-1) 0 69. (A) f() -6, f() 6; "# "# + (B)!0 f() 4!" (C) f(), f() -; "# "# + 10! f() does not eist f() does not eist -10 10 70. f() - Step 1. Simplify f( + ) " f() f( + ) " f(). -10 [( + ) - ( + )] - ( - ) + + - - - + + - + - 1 (3-1) 150 CHAPTER 3 LIMITS AND THE DERIVATIVE
f( + )! f() Step. Evaluate.!0 f( + )! f() ( + - 1) - 1!0!0 Tus, f'() - 1. (3-4) 71. f() - 3 Step 1. Simplify f( + )! f() f( + )! f(). [ + " 3] " ( " 3) + " 1 + + f( + )! f() Step. Evaluate.!0 f( + )! f() 1!0!0 + + 1 + + + + + " [ + + ] (3-4) 7. f is not differentiable at 0, since f is not continuous at 0.(3-4) 73. f is not differentiable at 1; te curve as a vertical tangent line at tis point. (3-4) 74. f is not differentiable at ; te curve as a "corner" at tis point. (3-4) 75. f is differentiable at 3. In fact, f'(3) 0. (3-4) 76. f() 5 ; f is discontinuous at 7 " 7 5 5 -, " 7 " 7 "7 f() "# "# "7 + 5 " 7 "# ; 7 is a vertical asymptote 5 5; y 5 is a orizontal asymptote. 77. f() " + 5 ; f is discontinuous at 4. ( " 4) " + 5 " + 5 "4 -, ( " 4) "4 + - ; 4 is a vertical asymptote. ( " 4) " + 5 "# ( " 4) "# " "# (3-3) " ; y 0 is a orizontal asymptote. (3-3) CHAPTER 3 REVIEW 151
78. f() + 9 ; f is discontinuous at 3. " 3 + 9 + 9 "3 -, " 3 "3 + " 3 "# + 9 " 3 "# ; 3 is a vertical asymptote. ; no orizontal asymptotes. (3-3) "# " 9 79. f() + " " 9 ; f is discontinuous at -, 1. ( + )( " 1) At -: " 9 " 9 "# # -, ( + )( " 1) "# + ( + )( " 1) ; - is a vertical asymptote. At 1 " 9 " 9 "1, ( + )( " 1) "1 + ( + )( " 1) - ; 1 is a vertical asymptote. " 9 "# + " "# 1 1; "# y 1 is a orizontal asymptote. (3-3) 80. f() 3 " 1 3 " " + 1 ( " 1)( + + 1) ( " 1)( " 1) ( " 1)( + + 1) ( " 1) ( + 1) + + 1, 1. f is discontinuous at 1, -1. ( " 1)( + 1) At 1: + + 1 f() "1 "1 ( " 1)( + 1) asymptote. At -1: + + 1 "#1 ( " 1)( + 1) asymptote. -, f() ; 1 is a vertical "1 +, "#1 + + + 1 ( " 1)( + 1) - ; -1 is a vertical 3 " 1 "# 3 " " + 1 3 "# 3 1 1; y 1 is a orizontal "# asymptote. (3-3) 81. f() 1/5 ; f'() 1 5-4/5 1 5 4/5 Te domain of f' is all real numbers ecept 0. At 0, te grap of f is smoot, but te tangent line to te grap at (0, 0) is vertical. (3-4) 15 CHAPTER 3 LIMITS AND THE DERIVATIVE
$ 8. f() % " m if # 1 &" + m if > 1 f() 5 f() 5 5 5 5 5 5 f() 1, f() -1 "1 # "1 + (C) f() 1 - m, f() -1 + m "1 # "1 + We want 1 - m -1 + m wic implies m 1. f() 5 5 f() -1, "1 # "1 + f() 1 5 5 5 (D) Te graps in (A) and (B) ave jumps at 1; te grap in (C) does not. (3-) 83. f() 1 - - 1, 0 f(1 + )! f(1) 1! 1 +! 1! 1 (A) "0 # "0 # "0 #! "0 # 1 ( - if < 0) f(1 + )! f(1) 1! 1 +! 1! 1 (B) "0 + "0 + "0 +!! "0 + -1 ( if > 0) f(1 + )! f(1) (C) does not eist, since te left it and te!0 rigt it are not equal. (D) f'(1) does not eist. (3-4) CHAPTER 3 REVIEW 153
84. (A) S() 7.47 + 0.4000 for 0 90; S(90) 43.47; S() 43.47 + 0.076 ( - 90) 4.786 + 0.076, > 90 Terefore, " S() 7.47 + 0.4000 if 0 % % 90 # $ 4.786 + 0.076 if > 90 (B) $80 $60 $40 S() (C) S() S() 43.47 S(90); "90 # "90 + S() is continuous at 90. (3-) $0 90 180 85. C() 10,000 + 00-0.1 (A) C(101) - C(100) 10,000 + 00(101) - 0.1(101) - [10,000 + 00(100) - 0.1(100) ] 9,179.90-9,000 $179.90 (B) C'() 00-0. C'(100) 00-0.(100) 00-0 $180 (3-7) 86. C() 5,000 + 40 + 0.05 (A) Cost of producing 100 bicycles: C(100) 5,000 + 40(100) + 0.05(100) 9000 + 500 9500 Marginal cost: C'() 40 + 0.1 C'(100) 40 + 0.1(100) 40 + 10 50 Interpretation: At a production level of 100 bicycles, te total cost is $9,500 and is increasing at te rate of $50 per additional bicycle. (B) Average cost: C () C() C (100) 5000 100 5000 + 40 + 0.05 Marginal average cost: C '() - 5000 + 0.05 + 40 + 0.05(100) 50 + 40 + 5 95 and C '(100) - 5000 (100) + 0.05-0.5 + 0.05-0.45 Interpretation: At a production level of 100 bicycles, te average cost is $95 and te marginal average cost is decreasing at a rate of $0.45 per additional bicycle. (3-7) 154 CHAPTER 3 LIMITS AND THE DERIVATIVE
87. Te approimate cost of producing te 01st printer is greater tan tat of producing te 601st printer (te slope of te tangent line at 00 is greater tan te slope of te tangent line at 600). Since te marginal costs are decreasing, te manufacturing process is becoming more efficient. (3-7) 88. p 5-0.1, C() + 9,000 (A) Marginal cost: C'() Average cost: C () C() Marginal cost: C '() - 9,000 + 9,000 (B) Revenue: R() p 5-0.01 Marginal revenue: R'() 5-0.0 Average revenue: R () R() 5-0.01 Marginal average revenue: R '() -0.01 (C) Profit: P() R() - C() 5-0.01 - ( + 9,000) 3-0.01-9,000 Marginal profit: P'() 3-0.0 Average profit: P () P() 3-0.01-9,000 Marginal average profit: P '() -0.01 + 9,000 (D) Break-even points: R() C() 5-0.01 + 9,000 0.01-3 + 9,000 0 -,300 + 900,000 0 ( - 500)( - 1,800) 0 Tus, te break-even points are at 500, 1,800; break-even points: (500, 10,000), (1,800, 1,600). (E) P'(1,000) 3-0.0(1000) 3; profit is increasing at te rate of $3 per umbrella. P'(1,150) 3-0.0(1,150) 0; profit is flat. P'(1,400) 3-0.0(1,400) -5; profit is decreasing at te rate of $5 per umbrella. CHAPTER 3 REVIEW 155
(F) R C $16,000 $1,000 R Profit C Loss $8,000 Loss $4,000 500 1,800,500 (3-7) 40t " 80 89. N(t) 40-80 t t, t (A) Average rate of cange from t to t 5: 40(5) " 80 40() " 80 " 5 N(5) " N() 5 " 3 10 15 8 components per day. (B) N(t) 40-80 t 40 80t-1 ; N (t) 80t - 80 t. N () 80 0 components per day. (3-5) 4 90. N(t) t + 1 3 t3/, N (t) + 1 t1/ 4 + t N(9) 18 + 1 3 (9)3/ 7, N (9) 4 + 9 7 3.5 After 9 monts, 7,000 pools ave been sold and te total sales are increasing at te rate of 3,500 pools per mont. (3-5) 91. (A) (B) N(50) 38.6, N'(50).6; in 100, natural gas consumption will be 38.6 trillion cubic feet and will be INCREASING at te rate of.6 trillion cubic feet per year. (3-4) 9. (A) (B) Fied costs: $484.1; variable cost per kringle: $.11. 156 CHAPTER 3 LIMITS AND THE DERIVATIVE
(C) Let p() be te linear regression equation found in part (A) and let C() be te linear regression equation found in part (B). Ten revenue R() p() and te break-even points are te points were R() C(). Using an intersection routine on a graping utility, te break-even points are: (51, 591.15) and (48, 1,007.6). 0 1500 (D) Te bakery will make a profit wen 51 < < 48. From te regression equation in part (A), p(51) 11.64 and p(48) 4.07. Tus, te bakery will make a profit for te price range $4.07 < p < $11.64. (3-7) 93. C() 500 500 -, 1. Te instantaneous rate of cange of concentration at meters is: C () 500(-) -3 "1000 3 Te rate of cange of concentration at 10 meters is: C (10) "1000 10 3-1 parts per million per meter Te rate of cange of concentration at 100 meters is: C (100) "1000 (100) 3 "1000 100,000-1 -0.001 parts per million per 1000 meter. (3-5) 94. F(t) 0.16t 1.6t + 10, F (t) 0.3t 1.6 F(4) 98.16, F (4) -0.3. After 4 ours te patient s temperature is 98.16 F and is decreasing at te rate of 0.3 F per our. (3-5) 95. N(t) 0 t 0t 1/ " Te rate of learning is N'(t) 0$ 1% ' t -1/ 10t -1/ 10 # & t. (A) Te rate of learning after one our is N'(1) 10 1 10 items per our. 0 400 (B) Te rate of learning after four ours is N'(4) 10 4 10 5 items per our. (3-5) CHAPTER 3 REVIEW 157
96. (A) 14 1 C! C ma 1 (B) C(T) 1T 150 + T 10 C 8 CTE 6 C ma 6 9 4 6 3 0 0 00 400 600 800 1000 100 M 150 Kelvins T 500 1000 T (C) C(600) 1(600) 150 + 600 9.6 To find T wen C 10, solve 1T 150 + T 10 for T. 1T 150 + T 10 1T 1500 + 10T T 1500 T 750 T 750 wen C 10. (3-3) 158 CHAPTER 3 LIMITS AND THE DERIVATIVE