Solução_Calculo_Stewart_6e

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F. TX.10 SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 173 5. y = f(x) =x 4 +4x 3 = x 3 (x +4) A. D = R B. y-intercept: f(0) = 0; H. x-intercepts: f(x) =0 ⇔ x = −4, 0 C. No symmetry D. No asymptote E. f 0 (x) =4x 3 +12x 2 =4x 2 (x +3)> 0 ⇔ x>−3,sof is increasing on (−3, ∞) and decreasing on (−∞, −3). F. Local minimum value f(−3) = −27, no local maximum G. f 00 (x) =12x 2 +24x =12x(x +2)< 0 ⇔ −2 1,sof is CU on (1, ∞) and (x − 1) 3 11. y = f(x) =1/(x 2 − 9) A. D = {x | x 6=±3} =(−∞, −3) ∪ (−3, 3) ∪ (3, ∞) B. y-intercept = f(0) = − 1 9 ,no 1 x-intercept C. f(−x) =f(x) ⇒ f is even; the curve is symmetric about the y-axis. D. lim =0,soy =0 x→±∞ x 2 − 9 is a HA. 1 1 lim = −∞, lim x→3 − x 2 − 9 x→3 + x 2 − 9 = ∞, lim 1 x→−3 − x 2 − 9 = ∞, lim 1 = −∞, sox =3and x = −3 x→−3 + x 2 − 9 are VA. E. f 0 2x (x) =− (x 2 2 > 0 ⇔ x
F. 174 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION TX.10 decreasing on (0, 3) and (3, ∞). F. Local maximum value f(0) = − 1 9 . H. G. y 00 = −2(x2 − 9) 2 +(2x)2(x 2 − 9)(2x) (x 2 − 9) 4 = 6(x2 +3) (x 2 − 9) 3 > 0 ⇔ x 2 > 9 ⇔ x>3 or x 0 ⇔ −3
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Solução_Calculo_Stewart_6e

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