Solução_Calculo_Stewart_6e

F.
TX.10SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS ¤ 189
21. y = f(x) = 1 − e1/x
1+e . 2e 1/x
1/x FromaCAS,y0 =
x 2 (1 + e 1/x ) and 2 y00 = −2e1/x (1 − e 1/x +2x +2xe 1/x )
.
x 4 (1 + e 1/x ) 3
f is an odd function defined on (−∞, 0) ∪ (0, ∞). Its graph has no x-ory-intercepts. Since
lim
x→±∞
f(x) =0,thex-axis
is a HA. f 0 (x) > 0 for x 6= 0,sof is increasing on (−∞, 0) and (0, ∞). It has no local extreme values.
f 00 (x) =0for x ≈ ±0.417,sof is CU on (−∞, −0.417),CDon(−0.417, 0),CUon(0, 0.417), and CD on (0.417, ∞).
f has IPs at (−0.417, 0.834) and (0.417, −0.834).
23. (a) f(x) =x 1/x
(b) Recall that a b = e b ln a .
lim x1/x = lim e(1/x)lnx .Asx → 0 + , ln x
x→0 + x→0 + x
→−∞,sox1/x = e (1/x)lnx → 0. This
indicates that there is a hole at (0, 0). Asx →∞, we have the indeterminate form ∞ 0 . lim
x→∞ x1/x = lim
x→∞ e(1/x)lnx ,
ln x H 1/x
but lim = lim =0,so lim
x→∞ x x→∞ 1 x→∞ x1/x = e 0 =1. This indicates that y =1is a HA.
(c) Estimated maximum: (2.72, 1.45). No estimated minimum. We use logarithmic differentiation to find any critical
numbers. y = x 1/x ⇒ ln y = 1 y0
lnx ⇒
x y = 1 x · 1
x +(lnx) − 1
1 − ln x
⇒ y 0 = x 1/x =0 ⇒
x 2 x 2
ln x =1 ⇒ x = e. For0 e, y 0 < 0,sof(e) =e 1/e is a local maximum value. This
point is approximately (2.7183, 1.4447), which agrees with our estimate.
(d) From the graph, we see that f 00 (x) =0at x ≈ 0.58 and x ≈ 4.37. Sincef 00
changes sign at these values, they are x-coordinates of inflection points.