Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 4 REVIEW ¤ 219
29. y = f(x) =sin 2 x − 2cosx A. D = R B. y-intercept: f(0) = −2 C. f(−x) =f(x),sof is symmetric with respect
to the y-axis. f has period 2π. D. No asymptote E. y 0 =2sinx cos x +2sinx =2sinx (cos x +1). y 0 =0 ⇔
sin x =0or cos x = −1 ⇔ x = nπ or x =(2n +1)π. y 0 > 0 when sin x>0,sincecos x +1≥ 0 for all x.
Therefore, y 0 > 0 [and so f is increasing] on (2nπ, (2n +1)π); y 0 < 0 [and so f is decreasing] on ((2n − 1)π, 2nπ).
F. Local maximum values are f((2n +1)π) =2; local minimum values are f(2nπ) =−2.
G. y 0 =sin2x +2sinx ⇒ y 00 =2cos2x +2cosx =2(2cos 2 x − 1) + 2 cos x =4cos 2 x +2cosx − 2
=2(2cos 2 x +cosx − 1) = 2(2 cos x − 1)(cos x +1)
y 00 =0 ⇔ cos x = 1 or −1 ⇔ x =2nπ ± π or x =(2n +1)π. H.
2 3
y 00 > 0 [and so f is CU] on 2nπ − π , 2nπ + π
3 3 ; y 00 ≤ 0 [and so f is CD]
on 2nπ + π , 2nπ + 5π
3 3 .Thereareinflection points at 2nπ ±
π
, − 1
3 4 .
31. y = f(x) =sin −1 (1/x) A. D = {x | −1 ≤ 1/x ≤ 1} =(−∞, −1] ∪ [1, ∞) . B. No intercept
C. f(−x) =−f(x), symmetric about the origin D. lim
x→±∞ sin−1 (1/x) =sin −1 (0) = 0, soy =0is a HA.
E. f 0 (x) =
1
− 1
−1
= √ < 0,sof is decreasing on (−∞, −1) and (1, ∞) .
1 − (1/x)
2 x 2 x4 − x2 F. No local extreme value, but f(1) = π is the absolute maximum value
2
H.
and f(−1) = − π 2
is the absolute minimum value.
G. f 00 (x) = 4x3 − 2x
2(x 4 − x 2 ) 3/2 = x 2x 2 − 1
(x 4 − x 2 ) 3/2 > 0 for x>1 and f 00 (x) < 0
for x 0 ⇔ −2x +1> 0 ⇔ x< 1 2 and f 0 (x) < 0 ⇔ x> 1 2 ,
so f is increasing on −∞, 1 2
no local minimum value
and decreasing on
1 , ∞ . F. Local maximum value f
1
2 2 =
1
2 e−1 =1/(2e);
G. f 00 (x)=e −2x (−2) + (−2x +1)(−2e −2x )
H.
=2e −2x [−1 − (−2x +1)]=4(x − 1) e −2x .
f 00 (x) > 0 ⇔ x>1 and f 00 (x) < 0 ⇔ x