Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 179
41. y =1/(1 + e −x ) A. D = R B. No x-intercept; y-intercept = f(0) = 1 2
. C. No symmetry
D. lim 1/(1 +
x→∞ e−x )= 1 =1and lim 1/(1 + 1+0 x→−∞ e−x )=0 since
lim
x→−∞ e−x = ∞], so f has horizontal asymptotes
y =0and y =1. E. f 0 (x) =−(1 + e −x ) −2 (−e −x )=e −x /(1 + e −x ) 2 . This is positive for all x,sof is increasing on R.
F. No extreme values G. f 00 (x) = (1 + e−x ) 2 (−e −x ) − e −x (2)(1 + e −x )(−e −x )
= e−x (e −x − 1)
(1 + e −x ) 4 (1 + e −x ) 3
The second factor in the numerator is negative for x>0 and positive for x 0 ⇒ 1 > 1/x ⇒ x>1 and
H.
f 0 (x) < 0 ⇒ 0 0]
x→−∞
f(x) =1,soy =0and y =1are HA; no VA
E. f 0 (x) =−2(1 + e x ) −3 e x = −2ex < 0,sof is decreasing on R F. No local extrema
(1 + e x )
3
G. f 00 (x)=(1+e x ) −3 (−2e x )+(−2e x )(−3)(1 + e x ) −4 e x
H.
= −2e x (1 + e x ) −4 [(1 + e x ) − 3e x ]= −2ex (1 − 2e x )
(1 + e x ) 4 .
f 00 (x) > 0 ⇔ 1 − 2e x < 0 ⇔ e x > 1 2
⇔ x>ln 1 2 and
f 00 (x) < 0 ⇔ x0} = ∞
n=−∞
(2nπ, (2n +1)π) =···∪ (−4π, −3π) ∪ (−2π, −π) ∪ (0,π) ∪ (2π, 3π) ∪ ···
B. No y-intercept; x-intercepts: f(x) =0 ⇔ ln(sin x) =0 ⇔ sin x = e 0 =1 ⇔ x =2nπ + π for each
2
integer n. C. f is periodic with period 2π. D. lim f(x) =−∞ and lim
x→(2nπ) +
x = nπ are VAs for all integers n.
x→[(2n+1)π]
−
f(x) =−∞, so the lines
E. f 0 (x) = cos x
sin x =cotx,sof 0 (x) > 0 when 2nπ < x < 2nπ + π for each
2
integer n,andf 0 (x) < 0 when 2nπ + π 2