Solução_Calculo_Stewart_6e

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F. TX.10 SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 181 55. y = − W 24EI x4 + WL 12EI x3 − WL2 24EI x2 = − W 24EI x2 x 2 − 2Lx + L 2 = −W 24EI x2 (x − L) 2 = cx 2 (x − L) 2 where c = − W isanegativeconstantand0 ≤ x ≤ L. Wesketch 24EI f(x) =cx 2 (x − L) 2 for c = −1. f(0) = f(L) =0. f 0 (x) =cx 2 [2(x − L)] + (x − L) 2 (2cx) =2cx(x − L)[x +(x − L)] = 2cx(x − L)(2x − L). Sofor0
F. 182 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION TX.10 C. No symmetry D. lim f(x) =−∞ and lim x→(1/2) − lim x→±∞ [f(x) − (−x +2)]= lim E. f 0 (x) =−1 − f(x) =∞,sox = 1 x→(1/2) + 2 is a VA. 1 =0,sotheliney = −x +2is a SA. x→±∞ 2x − 1 2 (2x − 1) < 0 for x 6= 1 ,sof is decreasing on −∞, 1 2 2 2 and 1 2 , ∞ . F. No extreme values G. f 0 (x) =−1 − 2(2x − 1) −2 ⇒ H. f 00 (x) =−2(−2)(2x − 1) −3 8 (2) = (2x − 1) 3 ,sof 00 (x) > 0 when x> 1 and 2 f 00 (x) < 0 when x< 1 . Thus, f is CU on 1 , ∞ and CD on −∞, 1 2 2 2 .NoIP 63. y = f(x) =(x 2 +4)/x = x +4/x A. D = {x | x 6= 0} =(−∞, 0) ∪ (0, ∞) B. No intercept C. f(−x) =−f(x) ⇒ symmetry about the origin D. lim (x +4/x) =∞ but f(x) − x =4/x → 0 as x → ±∞, x→∞ so y = x is a slant asymptote. lim (x +4/x) =∞ and x→0 + lim x→0 − (x +4/x) =−∞,sox =0 is a VA. E. f 0 (x) =1− 4/x 2 > 0 ⇔ x 2 > 4 ⇔ x>2 or x 0 ⇔ x>0 so f is CU on (0, ∞) and CD on (−∞, 0). NoIP 65. y = f(x) = 2x3 + x 2 +1 x 2 +1 =2x +1+ −2x x 2 +1 A. D = R B. y-intercept: f(0) = 1; x-intercept: f(x) =0 ⇒ 0=2x 3 + x 2 +1=(x + 1)(2x 2 − x +1) ⇒ x = −1 C. No symmetry D. No VA lim x→±∞ [f(x) − (2x + 1)] = lim x→±∞ −2x x 2 +1 = lim x→±∞ −2/x =0, so the line y =2x +1is a slant asymptote. 1+1/x2 E. f 0 (x) =2+ (x2 +1)(−2) − (−2x)(2x) = 2(x4 +2x 2 +1)− 2x 2 − 2+4x 2 = 2x4 +6x 2 (x 2 +1) 2 (x 2 +1) 2 (x 2 +1) = 2x2 (x 2 +3) 2 (x 2 +1) 2 so f 0 (x) > 0 if x 6= 0.Thus,f is increasing on (−∞, 0) and (0, ∞). Sincef is continuous at 0, f is increasing on R. F. No extreme values G. f 00 (x) = (x2 +1) 2 · (8x 3 +12x) − (2x 4 +6x 2 ) · 2(x 2 +1)(2x) [(x 2 +1) 2 ] 2 = 4x(x2 +1)[(x 2 + 1)(2x 2 +3)− 2x 4 − 6x 2 ] = 4x(−x2 +3) (x 2 +1) 4 (x 2 +1) 3 so f 00 (x) > 0 for x
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Solução_Calculo_Stewart_6e

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