85. f() = 4 ( - 6) 2 f'() = 4 (2)( - 6)(1) + ( - 6) 2 (4 3 ) = 2 3 ( - 6)[ + 2( - 6)] = 2 3 ( - 6)(3-12) = 6 3 ( - 4)( - 6) Thus, the critical values are = 0, = 4, and = 6. Now we construct the sign chart for f' ( = 0, = 4, = 6 are partition numbers). f'() - - - - 0 + + + + + + + + + 0 - - - - 0 + + + f'()!1!210(!) -1 0 1 4 5 6 7 1 90(+) f() Decreasing Increasing Decreasing Increasing 5!750(!) 7 + Therefore, f is increasing on (0, 4) and on (6, ), f is decreasing on (-, 0) and on (4, 6); f has a local maimum at = 4 and local minima at = 0 and = 6. 87. f() = 3( - 2) 2/3 + 4 f'() = 3 2 3 ( - 2) -1/3 = 2 ( - 2) 1/3 Critical values: f' is not defined at = 2. [Note: f(2) is defined, f(2) = 4.] f'() 0 for all. Thus, the critical value for f is = 2; = 2 is also a partition number for f'. f'() f() - - - - ND + + + + + + 0 1 2 3 Decreasing Increasing f'() 1!2(!) 3 2(+) Therefore, f is increasing on (2, ) and decreasing on (-, 2); f has a local minimum at = 2. 89. f() = 22 2 ; domain : (-, ) + 1 f'() = (2 + 1)4! 2 2 (2) 4 ( 2 + 1) 2 = ( 2 + 1) 2 Critical values: 4 f'() = ( 2 + 1) 2 = 0 4 = 0 = 0 Thus, the critical value is = 0. 228 CHAPTER 5 GRAPHING AND OPTIMIZATION
The sign chart for f' is: - - - - 0 + + + + f'() f() -1 0 1 Decreasing Increasing minimum f'()!1!1(!) 1 1(+) Therefore, f is increasing on (0, ); f is decreasing on (-, 0); f(0) = 0 is a local minimum. 91. Let f() = 3 + k (A) k > 0 f'() = 3 2 + k > 0 for all. There are no critical values and no local etrema; f is increasing on (-, ). (B) k < 0 f'() = 3 2 + k;3 2 + k = 0 2 = - k 3 = ± " k 3 Critical values: = - " k 3, = " k 3 ; f'() + + + + + 0 - - - - - - - - 0 + + + + + - - k 3 0 - k 3 $ f is increasing on "#, " " k ' $ & ) and on & " k % 3 ( % 3, # ' ); f is decreasing ( # on " " k 3, " k & % (; f has a local maimum at = - " k and a local $ 3 ' 3 minimum at = " k 3. (C) k = 0 The only critical value is = 0. There are no etrema, the function is increasing for all. 93. (A) The marginal profit function, P', is positive on (0, 600), zero at = 600, and negative on (600, 1,000). (B) P'() 0 500 1,000 EXERCISE 5-1 229
95. (A) The price function, B(t), decreases for the first 15 months to a local minimum, increases for the net 40 months to a local maimum, and then decreases for the remaining 15 months. (B) B(t) 10 30 50 70 t 97. C() = 2 20 (A) C () = C() + 20 + 320 = 20 (B) Critical values: C '() = 1 20-320 2 = 0 C'() C() + 20 + 320 2-320(20) = 0 2-6400 = 0 ( - 80)( + 80) = 0 Thus, the critical value of C on the interval (0, 150) is = 80. Net, construct the sign chart for C ' ( = 80 is a partition number for C '). - - - - - - - 0 + + + + C "() 40 # 3 20 (#) 0 20 40 60 80 100 Decreasing Increasing 100 $ 1 60 (+) Therefore, C is increasing for 80 < < 150 and decreasing for 0 < < 80; C has a local minimum at = 80. 99. P() = R() - C() P'() = R'() - C'() Thus, if R'() > C'() on the interval (a, b), then P'() = R'() - C'() > 0 on this interval and P is increasing. 101. C(t) = 0.28t t 2 + 4, 0 < t < 24 C'(t) = (t2 + 4)(0.28)! 0.28t(2t) (t 2 + 4) 2 = 0.28(4! t2 ) (t 2 + 4) 2 Critical values: C' is continuous for all t on the interval (0, 24). C'(t) = 0.28(4 " t2 ) (t 2 + 4) 2 = 0 4 - t 2 = 0 (2 - t)(2 + t) = 0 Thus, the critical value of C on the interval (0, 24) is t = 2. 230 CHAPTER 5 GRAPHING AND OPTIMIZATION
The sign chart for C' (t = 2 is a partition number) is: + + + 0 - - - C'(t) t t C'(t) C(t) 0 1 2 3 1 (+) 3 (!) Increasing Decreasing maimum Therefore, C is increasing on (0, 2) and decreasing on (2, 24); C(2) = 0.07 is a local maimum. 103. P(t) = 8.4t t 2 + 49 + 0.1, 0 < t < 24 P'(t) = (t2 + 49)(8.4) " 8.4t(2t) (t 2 + 49) 2 = 8.4(49 " t2 ) (t 2 + 49) 2 Critical values: P is continuous for all t on the interval (0, 24): P'(t) = 8.4(49 " t2 ) (t 2 + 49) 2 = 0 49 - t 2 = 0 (7 - t)(7 + t) = 0 Thus, the critical value of P on (0, 24) is t = 7. The sign chart for P' (t = 7 is a partition number for P') is: P'(t) + + + + + + + + + + 0 - - - - - P'(t) t 6 + 8! P(t) 0 5 6 7 8 Increasing Decreasing Therefore, P is increasing for 0 < t < 7 and decreasing for 7 < t < 24; P has a local maimum at t = 7. EXERCISE 5-2 Things to remember: 1. CONCAVITY The graph of a function f is CONCAVE UPWARD on the interval (a, b) if f'() is increasing on (a, b) and is CONCAVE DOWNWARD on the interval (a, b) if f'() is decreasing on (a, b). 2. SECOND DERIVATIVE For y = f(), the SECOND DERIVATIVE, provided it eists, is: f"() = d d f'() Other notations for f"() are: d 2 y d 2 and y". EXERCISE 5-2 231
3. SUMMARY For the interval (a, b): f () f () of y = f() Eample + Increasing upward - Decreasing downward 4. INFLECTION POINT An INFLECTION POINT is a point on the graph of a function where the concavity changes (from upward to downward, or from downward to upward). If f is continuous on (a, b) and has an inflection point at = c, then either f (c) = 0 or f (c) does not eist. 5. GRAPHING STRATEGY Step 1. Analyze f(). Find the domain and the intercepts. The intercepts are the solutions to f() = 0 and the y intercept is f(0). Step 2. Analyze f'(). Find the partition numbers '() and the critical values of f(). Construct a sign chart for f'(), determine the intervals where f is increasing and decreasing, and find the local maima and minima. Step 3. Analyze f"(). Find the partition numbers "(). Construct a sign chart for f"(), determine the intervals where the graph is concave upward and concave downward, and find the inflection points. Step 4. Sketch the graph. Locate intercepts, local maima and minima, and inflection points. Sketch in what you know from steps 1 3. Plot additional points as needed and complete the sketch. 1. (A) The graph is concave upward on (a, c), (c, d), and (e, g). (B) The graph is concave downward on (d, e) and (g, h). (C) f"() < 0 on (d, e) and (g, h). (D) f"() > 0 on (a, c), (c, d), and (e, g). (E) f'() is increasing on (a, c), (c, d), and (e, g). (F) f'() is decreasing on (d, e) and (g, h). (G) Inflection points occur at = d, = e, and = g. (H) The local etrema occur at = d, = e, and = g. 3. f'() > 0, f"() > 0; (C) 5. f'() < 0, f"() > 0; (D) 232 CHAPTER 5 GRAPHING AND OPTIMIZATION
7. f() = 2 3-4 2 + 5-6 f'() = 6 2-8 + 5 f"() = 12-8 11. y = 2-18 1/2 dy = 2-9-1/2 d d 2 y d 2 = 2 + 9 2-3/2 15. f()= e -2 f () = e -2 (-2) = -2e -2 f () = -2e -2 (-2) 2e -2 = 2(2 2 1)e -2 17. y = ln 2 2 " 1 % $ ' ( ln (2) # & y = y = 4 3 # " 2 & % ( " (1 " 2 ln )3 2 $ ' 6 19. f() = 4 + 6 2 13. y = ( 2 + 9) 4 = 9. h() = 2-1 - 3-2 h'() = -2-2 + 6-3 h"() = 4-3 - 18-4 y' = 4( 2 + 9) 3 (2) = 8( 2 + 9) 3 y" = 24( 2 + 9) 2 (2) + 8( 2 + 9) 3 = 48 2 ( 2 + 9) 2 + 8( 2 + 9) 3 = 8( 2 + 9) 2 (7 2 + 9) " 2 ln 4 = 1 " 2 ln 3 = "52 + 6 2 ln 6 = 6 ln " 5 4 f'() = 4 3 + 12 f"() = 12 2 + 12 12 > 0 The graph is concave upward for all ; there are no inflection points. 21. f() = 3-4 2 + 5-2 f'() = 3 2-8 + 5 f"() = 6-8 f"() = 0: 6-8 = 0 = 4 " 3 Sign chart for f" partition number is 4 % $ ': # 3& f"() - - - - - 0 + + + + + 0 1 2 4 3 f"() 0!8(!) 2 4(+) $ Therefore, the graph is concave downward on "#, 4 ' & ) and concave % 3( # upward on % 4 $ 3, " & (; there is an inflection point at = 4 ' 3. EXERCISE 5-2 233
23. f() = - 4 + 12 3-12 + 24 f'() = -4 3 + 36 2-12 f"() = -12 2 + 72 f"() = 0: -12 2 + 72 = 0-12( - 6) = 0 = 0, 6 Sign chart for f" (partition numbers 0, 6): f"() - - - 0 + + + + + 0 - - - -1 0 1 6 7 f"()!1!84(!) 1 60(+) 7!84(!) Therefore, the graph is concave downward on (-, 0) and (6, ); concave upward on (0, 6); there are inflection points at = 0 and = 6. 25. f() = ln( 2 2 + 10) (Note: ( 2 2 + 10 = ( 1) 2 + 9 > 0 for all ) 1 f () = 2 " 2 + 10 (2 2) = 2 " 2 2 " 2 + 10 f () = (2 " 2 + 10)(2) " (2 " 2)(2 " 2) ( 2 " 2 + 10) 2 = 22 " 4 + 20 " [4 2 " 8 + 4] ( 2 " 2 + 10) 2 "2( " 4)( + 2) f () = ( 2 " 2 + 10) 2 f () = 0: -2( 4)( + 2) = 0 = 4, = -2 Sign chart for f" (partition numbers 4, -2): f () - - + + + + + + + + - - - -2 0 4 = "22 + 4 + 16 ( 2 " 2 + 10) 2 = "2(2 " 2 " 8) ( 2 " 2 + 10) 2 ; f"() "3 " 14 (") (25) 2 0 16 100 (+) 5 " 54 (") (45) 2 The graph is concave downward on (-, -2) and (4, ); the graph is concave upward on (-2, 4); there are inflection points at = -2 and = 4. 234 CHAPTER 5 GRAPHING AND OPTIMIZATION
27. f() = 8e e 2 f () = 8e 2e 2 f () = 8e 4e 2 f () = 0: 8e 4e 2 = 0 4e (2 e ) = 0 e = 2 = ln 2 Sign chart for f (partition number ln 2 0.69): f () + + + + - - - - 0 0.69 1 f"() 0 4(+) 1 8e " 4e 2 (") The graph is concave upward on (-, ln 2) and concave downward on (ln 2, ); there is an inflection point at = ln 2. 29. f"() - - - 0 + + + + 0 - - - -1 2 Inflection point Inflection point f'() + + + + 0 - - - - - 0 - - - - -2 2 f() Increasing Decreasing Decreasing maimum Neither maimum nor minimum Using this information together with the points (-4, 0), (-2, 3), (-1, 1.5), (0, 0), (2, -1), (4, -3) on the graph, we have f() 3-4 0 4-3 EXERCISE 5-2 235
31. f"() + + + + ND - - - - 0 + + + + 0 2 Point of Inflection Point of Inflection f'() + + + + + ND ++ + 0 -- - - - - 0 ++ + + f() Increasing Incr. 0 1 4 maimum Decreasing Increasing minimum f() 2 Using this information together with the points (-3, -4), (0, 0), (1, 2), (2, 1), (4, -1), (5, 0) on the graph, we have -3 0 5-4 33. f"() - - - - - 0 + + + + + 0 1 2 f'() + + + 0 - - - - - 0 + + + f() Increasing 0 1 2 maimum Decreasing 0 1 2 f() 2 0-2 minimum Increasing f() 5 5 5 5 236 CHAPTER 5 GRAPHING AND OPTIMIZATION