Solução_Calculo_Stewart_6e

F.
176 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
23. y = f(x) =x/ √ x 2 +1 A. D = R B. y-intercept: f(0) = 0; x-intercepts: f(x) =0 ⇒ x =0
C. f(−x) =−f(x),sof is odd; the graph is symmetric about the origin.
D. lim f(x) = lim
x→∞ x→∞
and
lim f(x) =
x→−∞
=
lim
x→−∞
x
√
x2 +1 = lim
x→∞
x
√
x2 +1 =
lim
x→−∞
x/x
√
x2 +1/x = lim
x→∞
x/x
√
x2 +1/x =
1
− √ = −1 so y = ±1 are HA.
1+0
lim
x→−∞
x/x
√
x2 +1/ √ x 2 = lim
x→∞
x/x
√
x2 +1/− √ x 2 = lim
x→−∞
1
1+1/x
2 = 1
√ 1+0
=1
1
− 1+1/x 2
No VA.
√ 2x
x2 +1− x ·
E. f 0 2 √ x
(x) =
2 +1
= x2 +1− x 2
[(x 2 +1) 1/2 ] 2 (x 2 +1) = 1
> 0 for all x,sof is increasing on R.
3/2 (x 2 3/2
+1)
F. No extreme values
G. f 00 (x) =− 3 2 (x2 +1) −5/2 · 2x =
−3x
(x 2 +1) 5/2 ,sof 00 (x) > 0 for x0. Thus,f is CU on (−∞, 0) and CD on (0, ∞).
IP at (0, 0)
25. y = f(x) = √ 1 − x 2 /x A. D = {x ||x| ≤ 1, x 6= 0} =[−1, 0) ∪ (0, 1] B. x-intercepts ±1, noy-intercept
√
1 − x
2
C. f(−x) =−f(x), so the curve is symmetric about (0, 0) . D. lim
x→0 + x
−x 2 / √ 1 − x 2 − √ 1 − x 2
so x =0is a VA. E. f 0 (x) =
on (−1, 0) and (0, 1).
G. f 00 (x) =
f is CU on −1, −
2
IP at ± , ± √ 1
3 2
F. No extreme values
2 − 3x 2
x 3 (1 − x 2 ) > 0 3/2 ⇔
−1 1 and f 0 (x) < 0 when 0 < |x| < 1,sof is increasing on (−∞, −1) and (1, ∞),and
decreasing on (−1, 0) and (0, 1) [hence decreasing on (−1, 1) since f is
H.
H.
continuous on (−1, 1)].
F. Local maximum value f(−1) = 2, local minimum
value f(1) = −2 G. f 00 (x) = 2 3 x−5/3 < 0 when x 0
when x>0,sof is CD on (−∞, 0) and CU on (0, ∞). IPat(0, 0)