f'(c) f"(c) GRAPH OF f IS: f(c) EXAMPLE

Transcription

1 4. SECOND DERIVATIVE TEST Let c be a critical value for f(). f'(c) f"(c) GRAPH OF f IS: f(c) EXAMPLE 0 + Concave upward minimum 0 - Concave downward maimum 0 0? Test does not apply 5. SECOND DERIVATIVE TEST FOR ABSOLUTE EXTREMUM Let f be continuous on an interval I with only one critical value c on I: If f'(c) = 0 and f"(c) > 0, then f(c) is the absolute minimum of f on I. ( ) I If f'(c) = 0 and f"(c) < 0, then f(c) is the absolute maimum of f on I. ( I ) 1. Interval [0, 10]; absolute minimum: f(0) = 0; absolute maimum: f(10) = Interval [0, 8]; absolute minimum: f(0) = 0; absolute maimum: f(3) = 9 5. Interval [1, 10]; absolute minimum: f(1) = f(7) = 5; absolute maimum: f(10) = Interval [1, 9]; absolute minimum: f(1) = f(7) = 5; absolute maimum: f(3) = f(9) = 9 9. Interval [2, 5]; absolute minimum: f(5) = 7; absolute maimum: f(3) = f() = , I = (-, ) f'() = 2-2 = 2( - 1) f'() = 0: 2( - 1) = 0 = 1 = 1 is the ONLY critical value on I, and f(1) = 1 2-2(1) + 3 = 2 f"() = 2 and f"(1) = 2 > 0. Therefore, f(1) = 2 is the absolute minimum. The function does not have an absolute maimum since lim ±" f() =. 13. f() = , I = (-, ) f'() = -2-6 = -2( + 3) f'() = 0: -2( + 3) = 0 = -3 = -3 is the ONLY critical value on I, and f(-3) = -(-3) 2-6(-3) + 9 = 18 f"() = -2 and f"(-3) = -2 < 0. Therefore, f(-3) = 18 is the absolute maimum. The function does not have an absolute minimum since lim ±" f() = CHAPTER 5 GRAPHING AND OPTIMIZATION

2 15. f() = 3 +, I = (-, ) f'() = on I; f is increasing on I and lim "# f() = -, lim f() =. Therefore, f does not have any absolute etrema. " 17. f() = ; domain: all real numbers f'() = = 8 2 (3 - ) f"() = = 24(2 - ) Critical values: = 0, = 3 f"(0) = 0 (second derivative test fails) f"(3) = -72 f has a local maimum at = 3. Sign chart for f'() = 8 2 (3 - ) f'() (0 and 3 are partition numbers) From the sign chart, f does not have a local etremum at = 0; f has a local maimum at = 3 which must be an absolute maimum f() 0 3 Increasing Increasing Decreasing maimum since f is increasing on (-, 3) and decreasing on (3, ); f(3) = 54 is the absolute maimum of f. f does not have an absolute minimum since lim f() = lim f() = -. " "# 19. f() = + 16 ; domain: all real numbers ecept = 0. f'() = = ( 4)( + 4) = 2 f"() = 32 3 Critical values: = -4, = 4 f"(-4) = < 0; f has a local maimum at = -4 f"(4) = 1 > 0; f has a local minimum at = 4 2 lim f() = lim + 16 # " " " $ = ; lim f() = lim + 16 # "# "# " $ f has no absolute etrema. 21. f() = 2 2 ; domain: all real numbers + 1 f'() = (2 + 1)2 2 (2) ( 2 + 1) 2 = 2 ( 2 + 1) 2 = - ; f"() = (2 + 1) 2 (2) 2(2)( 2 + 1)(2) ( 2 + 1) 4 = 2 62 ( 2 + 1) 3 Critical value: = 0 Since f has only one critical value and f"(0) = 2 > 0, f(0) = 0 is 2 the absolute minimum of f. Since lim f() = lim " " 2 = 1, f has no + 1 absolute maimum; y = 1 is a horizontal asymptote for the graph of f. EXERCISE

3 23. f() = 2 2 ; domain: all real numbers + 1 f'() = (2 + 1)2 2(2) ( 2 + 1) 2 = 2 22 ( 2 + 1) 2 = 2(1 2 ) ( 2 + 1) 2 f"() = (2 + 1) 2 (4) 2(1 2 )(2)( 2 + 1)(2) ( 2 + 1) 4 = 4(2 3) ( 2 + 1) 3 Critical values: = -1, = 1 f"(-1) = 8 > 0; f has a local minimum at = -1 f"(1) = -8 < 0; f has a local maimum at = 1 Sign chart for f'() (partition numbers are -1 and 1) 2 lim f() = lim ±" ±" = 0 (the -ais is a horizontal asymptote) f'() f() Decreasing Increasing Decreasing minimum maimum We can now conclude that f(1) = 1 is the absolute maimum of f and f(-1) = -1 is the absolute minimum of f. 25. f() = ; domain: all real numbers + 1 f'() = (2 + 1)(2) ( 2 1)(2) 4 ( 2 + 1) 2 = ( 2 + 1) 2 f"() = (2 + 1) 2 (4) 4(2)( 2 + 1)2 ( 2 + 1) 4 = 4(3 2 ) ( 2 + 1) 3 Critical value: = 0 f"(0) = 12 > 0; f has a local minimum at = 0 4 Sign chart for f'() = ( 2 + 1) 2 (0 is the partition number) f'() f() lim f() = lim ±" ±" = 1 (y = 1 is a horizontal asymptote) 0 Decreasing Increasing minimum We can now conclude that f(0) = -1 is the absolute minimum and f does not have an absolute maimum. 27. f() = on I = [0, ) f'() = 4-8 = 4( - 2) f"() = 4 Critical value: = 2 f"(2) = 4 > 0; f has a local minimum at = 2 Since = 2 is the only critical value of f on I, f(2) = -2 is the absolute minimum of f on I. 312 CHAPTER 5 GRAPHING AND OPTIMIZATION

4 29. f() = on I = [0, ) f'() = = 3(2 - ) f"() = 6-6 Critical value (in (0, )): = 2 f"(2) = -6 < 0; f has a local maimum at = 2 Since f(0) = 0 and = 2 is the only critical value of f in (0, ), f(2) = 4 is the absolute maimum value of f on I. 31. f() = ( + 4)( - 2) 2 on I = [0, ) f'() = ( + 4)(2)( - 2) + ( - 2) 2 = ( - 2)[ ] = ( - 2)(3 + 6) = f"() = 6 Critical value in I: = 2 f"(2) = 12 > 0; f has a local minimum at = 2 Since f(0) = 16 and = 2 is the only critical value of f in (0, ), f(2) = 0 is the absolute minimum of f on I. 33. f() = on I = (0, ) Since lim f() = lim " " ( ) =, f does not have an absolute maimum on I. 35. f() = , > 0; I = (0, ) f'() = f'() = 0: = = 12 2 = 4 = 2 (-2 is not in I) = 2 is the only critical value of f on I, and f(2) = 20-3(2) = 8. f"() = ; f"(2) = - = -6 < 0. Therefore, f(2) = 8 is the 4 absolute maimum of f. The function does not have an absolute minimum since lim f() = -. (Also, lim f() = -.) + " 0 EXERCISE

5 37. f() = , > 0; I = (0, ) f'() = f'() = 0: = = = 64 = 4 = 4 is the only critical value of f on I and f(4) = (4) = 22 f"() = ; f"(4) = 4 4 = 3 > 0. Therefore, f(4) = 22 is the 2 absolute minimum of f. The function does not have an absolute maimum since lim f() =. (Also, lim f() =.) + " f() = on I = (0, ) f'() = = = (2 10)( 2 + 9) 4 f"() = Critical value (in (0, )): = 10 2 f"( 10) = (10) (10) 5 2 > 0; f has a local minimum at = Since 10 is the only critical value of f on I, f( absolute minimum of f on I. 10) = 41. f() = e 2, > 0 2 d f'() = d e e d d 2 4 = 2 e 2e 4 = e ( 2) 4 = e ( 2) 3 Critical values: f'() = e ( 2) 3 = 0 e ( - 2) = 0 = 2 [Note: e 0 for all.] Thus, = 2 is the only critical value of f on (0, ). Sign chart for f' [Note: This approach is a little easier than calculating f"()]: Test Numbers f'() f() Decr. Incr. f'() 1 e () 3 e 3 27 (+) is the 314 CHAPTER 5 GRAPHING AND OPTIMIZATION

6 By the first derivative test, f has a minimum value at = 2; f(2) = e2 2 2 = e f() = 3 e f'() = " d $ # d is the absolute minimum value of f. % ' e " d % ( $ & # d e ' 3 & (e ) 2 = 32 e 3 e e 2 = 2 (3 )e e 2 = 2 (3 ) e Critical values: f'() = 2 (3 ) e = 0 2 (3 - ) = 0 = 0 and = 3 Sign chart for f' [Note: This approach is a little easier than calculating f"()]: 0 0 Test Numbers f'() f'() e 1 (+) f() e (+) Increasing Increasing Decreasing 4 16 e () 4 By the first derivative test, f has a maimum value at = 3; f(3) = is the absolute maimum value of f. e 45. f() = 5-2 ln, > 0 f'() = 5-2 d d (ln ) - ln d d (2) " 1 % = 5-2$ ' - 2 ln = 3-2 ln, > 0 # & Critical values: f'() = 3-2 ln = 0 ln = 3 = 1.5; = e1.5 2 Thus, = e 1.5 is the only critical value of f on (0, ). Now, f"() = d d (3-2 ln ) = - 2 and f"(e 1.5 ) = - 2 e 1.5 < 0. Therefore, f has a maimum value at = e 1.5, and f(e 1.5 ) = 5e 1.5-2e 1.5 ln(e 1.5 ) = 5e 1.5-2(1.5)e 1.5 = 2e is the absolute maimum of f. EXERCISE

7 47. f() = 2 (3 - ln ), > 0 f'() = 2 d d (3 - ln ) + (3 - ln ) d d 2 = 2 1 Critical values: f'() = 5-2 ln = 0 (5-2 ln ) = ln = 0 Now ( ) + (3 - ln )2 = ln = 5-2 ln f"() = 5-2( 1 ) - 2 ln = 3-2 ln ln = 5 2 = 2.5 = e 2.5 [Note: 0 on (0, )] and f"(e 2.5 ) = 3-2 ln(e 2.5 ) = 3-2(2.5) = 3-5 = -2 < 0 Therefore, f has a maimum value at = e 2.5 and f(e 2.5 ) = (e 2.5 ) 2 (3 - ln e 2.5 ) = e 5 (3-2.5) = e is the absolute maimum value of f. 49. f() = ln(e - ), > 0 f'() = 1 e d d (e- ) = 1 e [e- - e - ] = 1 Critical values: f'() = 1 Sign chart for f'(): f'() f() = 0; = Increasing Decreasing maimum Test Numbers f() 1 2 1(+) () By the first derivative test, f has a maimum value at = 1; f(1) = ln(e -1 ) = -1 is the absolute maimum value of f. 51. f() = f'() = = 3( ) = 3( - 3)( - 1) Critical values: = 1, 3 (A) On the interval [-1, 5]: f(-1) = = -22 f(1) = = -2 f(3) = = -6 f(5) = = 14 Thus, the absolute maimum of f is f(5) = 14, and the absolute minimum of f is f(-1) = -22. (B) On the interval [-1, 3]: f(-1) = -22 f(1) = -2 f(3) = -6 Absolute maimum of f: f(1) = -2 Absolute minimum of f: f(-1) = CHAPTER 5 GRAPHING AND OPTIMIZATION

8 (C) On the interval [2, 5]: f(2) = = -4 f(3) = -6 f(5) = 14 Absolute maimum of f: f(5) = 14 Absolute minimum of f: f(3) = f() = ( - 1)( - 5) f'() = ( - 1)3( - 5) 2 + ( - 5) 3 = ( - 5) 2 ( ) = ( - 5) 2 (4-8) Critical values: = 2, 5 (A) Interval [0, 3]: f(0) = (-1)(-5) = 126 f(2) = (2-1)(2-5) = -26 f(3) = (3-1)(3-5) = -15 Absolute maimum of f: f(0) = 126 Absolute minimum of f: f(2) = -26 (B) Interval [1, 7]: f(1) = 1 f(2) = -26 f(5) = 1 f(7) = (7-1)(7-5) = = 49 Absolute maimum of f: f(7) = 49 Absolute minimum of f: f(2) = -26 (C) Interval [3, 6]: f(3) = (3-1)(3-5) = -15 f(5) = 1 f(6) = (6-1)(6-5) = 6 Absolute maimum of f: f(6) = 6 Absolute minimum of f: f(3) = f() = f'() = = 4 2 ( - 3) Critical values: = 0, = 3 (A) On the interval [-1, 2]: f(-1) = 10 f(0) = 5 f(2) = -11 Thus, the absolute maimum of f is f(-1) = 10; the absolute minimum of f is f(2) = -11. (B) On the interval [0, 4]: f(0) = 5 f(3) = -22 f(4) = 5 Absolute maimum of f: f(0) = f(4) = 5 Absolute minimum of f: f(3) = -22 (C) On the interval [-1, 1]: f(-1) = 10 f(0) = 5 f(1) = 2 Absolute maimum of f: f(-1) = 10 Absolute minimum of f: f(1) = 2 EXERCISE

9 57. f has a local minimum at = Unable to determine from the given information (f'(-3) = f"(-3) = 0). 61. Neither a local maimum nor a local minimum at = 6; = 6 is not a critical value of f. 63. f has a local maimum at = 2. EXERCISE 5-6 Things to remember: STRATEGY FOR SOLVING OPTIMIZATION PROBLEMS Step 1. Introduce variables, look for relationships among these variables, and construct a mathematical model of the form: Maimize (or minimize) f() on the interval I Step 2. Find the critical values of f(). Step 3. Use the procedures developed in Section 5-5 to find the absolute maimum (or minimum) value of f() on the interval I and the value(s) of where this occurs. Step 4. Use the solution to the mathematical model to answer all the questions asked in the problem. 1. Let one length = and the other = Since neither length can be negative, we have 0 and 10-0, or 10. We want the maimum value of the product (10 - ), where Let f() = (10 - ) = 10-2 ; domain I = [0, 10] f'() = 10-2; = 5 is the only critical value f"() = -2 f"(5) = -2 < 0 Thus, f(5) = 25 is the absolute maimum; divide the line in half. 3. Let one number =. Then the other number = f() = ( + 30) = ; domain I = (-, ) f'() = ; = -15 is the only critical value f"() = 2 f"(-15) = 2 > 0 Thus, the absolute minimum of f occurs at = -15. The numbers, then, are -15 and = CHAPTER 5 GRAPHING AND OPTIMIZATION

10 5. Let = the length of the rectangle and y = the width of the rectangle. Then, 2 + 2y = y = 50 y = 50 - We want to find the maimum of the area: y y A() = y = (50 - ) = Since 0 and y 0, we must have [Note: A(0) = A(50) = 0.] A'() = da d = 50-2; = 25 is the only critical value. Now, A" = -2 and A"(25) = -2 < 0. Thus, A(25) is the absolute maimum. The maimum area is A(25) = 25(50-25) = 625 cm 2, which means that the rectangle is actually a square with sides measuring 25 cm each. 7. Let the rectangle of fied area A have dimensions and y. Then A = y and y = A. The cost of the fence is C = 2B + 2By = 2B + 2AB y, > 0 Thus, we want to find the absolute minimum of C() = 2B + 2AB, > 0 Since lim C() = lim C() =, and C() > 0 for all > 0, we can 0 " conclude that C has an absolute minimum on (0, ). This agrees with our intuition that there should be a cheapest way to build the fence. 9. Let and y be the dimensions of the rectangle and let C be the fied amount which can be spent. Then C = 2B + 2By and y = C 2B 2B The area enclosed by the fence is: # C " 2B & A = y = $ % 2B ' ( Thus, we want to find the absolute maimum value of A() = C 2B - 2, 0 C 2B " Since A() is a continuous function on the closed interval 0, C % # $ 2B & ', it has an absolute maimum value. This agrees with our intuition that there should be a largest rectangular area that can be enclosed with a fied amount of fencing. EXERCISE

11 11. Price-demand: p() = ; cost: C() = 20, (A) Revenue: R() = p() = , 0 < R'() = R'() = = 0 implies = 500 R"() = -1; R"(500) = -1 < 0 R has an absolute maimum at = 500. p(500) = (500) = 250; R(500) = (500) 2-0.5(500) 2 = 125,000 The company should produce 500 phones each week at a price of $250 per phone to maimize their revenue. The maimum revenue is $125,000 (B) Profit: P() = R() - C() = (20, ) = ,000 P'() = P'() = = 0 implies = 365 P"() = -1; P"(365) = -1 < 0 P has an absolute maimum at = 365 p(365) = (365) = ; P(365) = (365) 2-0.5(365) 2-20,000 = 46, To maimize profit, the company should produce 365 phones each week at a price of $ per phone. The maimum profit is $46, # 13. (A) Revenue R() = p() = 200 " & % ( = $ 30' 30, 0 6,000 R'() = = Now R'() = = 0 implies = R"() = < Thus, R"(3000) = - 1 < 0 and we conclude that R has an absolute 15 maimum at = The maimum revenue is R(3000) = 200(3000) - (3000)2 = $300, (B) Profit P() = R() - C() = (72, ) 30 = ,000 P'() = Now P"() = = 0 implies = 2, and P"(2,100) = - < 0. Thus, the maimum profit 15 occurs when 2,100 television sets are produced. The maimum profit is P(2,100) = 140(2,100) - (2,100)2-72,000 = $75, the price that the company should charge is p(2,100) = 200-2, = $130 for each set. 320 CHAPTER 5 GRAPHING AND OPTIMIZATION

12 (C) If the government taes the company $5 for each set, then the profit P() is given by P() = (72, ) - 5 = ,000. P'() = Now = 0 implies = 2, P"() = and P"(2,025) = - < 0. Thus, the maimum profit in this case occurs when 2,025 television sets are produced. The maimum profit is P(2,025) = 135(2,025) - (2,025)2-72,000 = $64, and the company should charge p(2,025) = 200-2,025 = $132.50/set (A) (B) (C) The revenue at the demand level is: R() = p() where p() is the quadratic regression equation in (A). The cost at the demand level is C() given by the linear regression equation in (B). The profit P() = R() - C(). The maimum profit is $118,996 at the demand level = The price per saw at the demand level = 1422 is $ (A) Let = number of 10 reductions in price. Then = number of sandwiches sold at 10 reductions = price per sandwich 0 80 Revenue: R() = ( )(8-0.1) = , 0 80 R'() = R'() = = 0 implies = 32 R(0) = 5120, R(32) = 9216, R(80) = 0 Thus, the deli should charge = $4.80 per sandwich to realize a maimum revenue of $9216. EXERCISE

13 (B) Let = number of 20 reductions in price. Then = number of sandwiches sold = price per sandwich 0 40 Revenue: R() = ( )(8-0.2) = , 0 40 R'() = -8-6 R'() = -8-6 = 0 has no solutions in (0, 40) Now, R(0) = = $5120 R(40) = 0 Thus, the deli should charge $8 per sandwich to maimize their revenue under these conditions. 19. Let = number of dollar increases in the rate per day. Then = total number of cars rented and 30 + = rate per day. Total income = (total number of cars rented)(rate) y() = (200-5)(30 + ), 0 40 y'() = (200-5)(1) + (30 + )(-5) = = = 10(5 - ) Thus, = 5 is the only critical value and y(5) = (200-25)(30 + 5) = y"() = -10 y"(5) = -10 < 0 Therefore, the absolute maimum income is y(5) = $6125 when the rate is $35 per day. 21. Let = number of additional trees planted per acre. Then 30 + = total number of trees per acre and 50 - = yield per tree. Yield per acre = (total number of trees per acre)(yield per tree) y() = (30 + )(50 - ), 0 20 y'() = (30 + )(-1) + (50 - ) = 20-2 = 2(10 - ) The only critical value is = 10, y(10) = 40(40) = 1600 pounds per acre. y"() = -2 y"(10) = -2 < 0 Therefore, the absolute maimum yield is y(10) = 1600 pounds per acre when the number of trees per acre is Volume = V() = (12-2)(8-2), 0 4 = V'() = = 4( ) We solve = 0 by using the quadratic formula: 20 ± 400 " 4 # 24 # 3 10 ± 2 7 = = CHAPTER 5 GRAPHING AND OPTIMIZATION

14 Thus, = 10 " [0, 4]. V"() = V"(1.57) = (1.57) < is the only critical value on the interval Therefore, a square with a side of length = 1.57 inches should be cut from each corner to obtain the maimum volume. 25. Area = 800 square feet = y (1) Cost = (2y + ) From (1), we have y = 800. y Hence, cost C() = " + # $, or C() = , > 0, C'() = = 24(2 400) 24( 20)( + 20) 2 = 2. Therefore, = 20 is the only critical value. 19, 200 C"() = 3 19, 200 C"(20) = > 0. Therefore, = 20 for the 8000 minimum cost. The dimensions of the fence are shown in the diagram at the right (Epensive side) 27. Let = number of cans of paint produced in each production run. Then, number of production runs: 16,000, 1 16,000 Cost: C() = cost of storage + cost of set up = 16,000 (4) + (500) 2 [Note: 2 is the average number of cans of paint in storage per day.] Thus, C() = 2 + 8,000,000, 1 16,000 C'() = 2-8,000,000 2 = 22 8,000,000 2 = 2(2 4,000,000) 2 Critical value: = 2000 C"() = 16,000,000 3 ; C"(2000) > 0 Thus, the minimum cost occurs when = 2000 and the number of production runs is 16,000 2,000 = 8. EXERCISE

15 29. Let = number of books produced each printing. Then, the number of printings = 50,000. Cost = C() = cost of storage + cost of printing = ,000 (1000), > 0 [Note: 2 is the average number in storage each day.] C'() = ,000,000 2 = 2 100,000,000 ( + 10,000)( 10,000) 2 2 = 2 2 Critical value: = 10,000 C"() = 100,000,000 3 C"(10,000) = 100,000,000 (10,000) 3 > 0 Thus, the minimum cost occurs when = 10,000 and the number of printings is 50,000 10,000 = (A) Let the cost to lay the pipe on the land be 1 unit; then the cost to lay the pipe in the lake is 1.4 units. C() = total cost = (1.4) (1)(10 - ), 0 10 = (1.4)( ) 1/ C'() = (1.4) 1 2 (2 + 25) -1/2 (2) - 1 = (1.4)( ) -1/2-1 = 1.4 " C'() = 0 when = 0 or = = 25 2 = = = ±5.1 Thus, the critical value is = 5.1. C"() = (1.4)( ) -1/2 # + (1.4) " 1 & % (( ) -3/2 2 $ 2' 1.4 = ( ) (1.4)2 ( ) 3 2 = 35 ( ) C"(5.1) = [(5.1) ] 3 2 > 0 Thus, the cost will be a minimum when = 5.1. Note that: C(0) = (1.4) = 17 C(5.1) = (1.4) (10-5.1) = 14.9 C(10) = (1.4) 125 = Thus, the absolute minimum occurs when = 5.1 miles. 324 CHAPTER 5 GRAPHING AND OPTIMIZATION

16 (B) C() = (1.1) (1)(10 - ), 0 10 (1.1) " C'() = C'() = 0 when = 0 or (1.21) 2 = = 25 2 = = = ±10.91 Critical value: = > 10, i.e., there are no critical values on the interval [0, 10]. Now, C(0) = (1.1) = 15.5, C(10) = (1.1) Therefore, the absolute minimum occurs when = 10 miles. 33. C(t) = 30t 2-240t + 500, 0 t 8 C'(t) = 60t - 240; t = 4 is the only critical value. C"(t) = 60 C"(4) = 60 > 0 Now, C(0) = 500 C(4) = 30(4) 2-240(4) = 20, C(8) = 30(8) 2-240(8) = 500. Thus, 4 days after a treatment, the concentration will be minimum; the minimum concentration is 20 bacteria per cm Let = the number of mice ordered in each order. Then the number of orders = 500. C() = Cost = [Note: Cost = cost of feeding + cost of order, (10) 2 is the average number of mice at any one time.] C() = , 0 < 500 C'() = = = 2(2 2500) 2( + 50)( 50) 2 = 2 Critical value: = 50 (-50 is not a critical value, since the domain of C is > 0. C"() = 10,000 3 and C"(50) = 10, > 0 Therefore, the minimum cost occurs when 50 mice are ordered each time. The total number of orders is = 10. EXERCISE

17 37. H(t) = 4t 1/2-2t, 0 t 2 H'(t) = 2t -1/2-2 Thus, t = 1 is the only critical value. Now, H(0) = 4 0 1/2-2(0) = 0, H(1) = 4 1 1/2-2(1) = 2, H(2) = 4 2 1/ Therefore, H(1) is the absolute maimum, and after one month the maimum height will be 2 feet. 39. N(t) = t 2 - t 3, 0 t 8 The rate of increase = R(t) = N'(t) = 24t - 3t 2, and R'(t) = N"(t) = 24-6t. Thus, t = 4 is the only critical value of R(t). Now, R(0) = 0, R(4) = = 48, R(8) = = 0. Therefore, the absolute maimum value of R occurs when t = 4; the maimum rate of increase will occur four years from now. CHAPTER 5 REVIEW 1. The function f is increasing on (a, c 1 ), (c 3, c 6 ). (5-1, 5-2) 2. f'() < 0 on (c 1, c 3 ), (c 6, b). (5-1, 5-2) 3. The graph of f is concave downward on (a, c 2 ), (c 4, c 5 ), (c 7, b). (5-1, 5-2) 4. A local minimum occurs at = c 3. (5-1) 5. The absolute maimum occurs at = c 1, c 6. (5-1, 5-5) 6. f'() appears to be zero at = c 1, c 3, c 5. (5-1) 7. f'() does not eist at = c 4, c 6. (5-1) 8. = c 2, c 4, c 5, c 7 are inflection points. (5-2) 9. f"() ND Graph of f Concave Downward Concave Upward Concave Downward Concave Downward Point of Inflection Point of Inflection 326 CHAPTER 5 GRAPHING AND OPTIMIZATION

18 f'() ND + + f() Incr. Decr. Decr. Incr. maimum minimum Using this information together with the points (-3, 0), (-2, 3), (-1, 2), (0, 0), (2, -3), (3, 0) on the graph, we have f() Domain: all real numbers Intercepts: y-intercept: f(0) = 0 -intercepts: = 0 Asymptotes: Horizontal asymptote: y = 2 no vertical asymptotes Critical values: = (5-2) f'() f() Decreasing 0 minimum Increasing f"() f() -2 2 Concave Concave Concave 1 Downward Upward Downward Inflection Inflection point point 5 5 f() 2 (5-1, 5-2) 11. f() = y = f'() = y' = f"() = (5-2) y" = 8 3 (5-2) CHAPTER 5 REVIEW 327

19 13. f() = Step 1. Analyze f(): (A) Domain: All real numbers, (-, ) (B) Intercepts: y-intercept: f(0) = (0) (0) = 0 -intercepts: = 0 ( ) = 0 ( - 9) 2 = 0 = 0, 9 (C) Asymptotes: No horizontal or vertical asymptotes. Step 2. Analyze f'(): f'() = = 3( ) = 3( - 3)( - 9) Critical values: = 3, = 9 Partition numbers: = 3, = 9 Sign chart for f': f'() f() Increasing Decreasing Increasing Maimum Minimum Test Numbers f'() 0 81 (+) 5-24 (-) (+) Thus, f is increasing on (-, 3) and on (9, ); f is decreasing on (3, 9). There is a local maimum at = 3 and a local minimum at = 9. Step 3. Analyze f"(): f"() = 6-36 = 6( - 6) Thus, = 6 is a partition number for f". Sign chart for f": f"() Test Numbers Graph of f 0 Concave Downward 6 7 Concave Upward f"() 0-36 (-) 7 6 (+) Inflection Point Thus, the graph of f is concave downward on (-, 6) and concave upward on (6, ). The point = 6 is an inflection point. 328 CHAPTER 5 GRAPHING AND OPTIMIZATION