Solução_Calculo_Stewart_6e

F.
218 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
23. y = f(x) =
1
x(x − 3) 2 A. D = {x | x 6= 0, 3} =(−∞, 0) ∪ (0, 3) ∪ (3, ∞) B. No intercepts. C. No symmetry.
1
D. lim
=0,soy =0 is a HA.
x→±∞ x(x − 3) lim
2
1
1
1
= ∞, lim
= −∞, lim
x→0 + x(x − 3)
2
x→0 − x(x − 3)
2 x→3 x(x − 3) = ∞, 2
so x =0and x =3are VA. E. f 0 (x) =− (x − 3)2 +2x(x − 3) 3(1 − x)
= ⇒ f 0 (x) > 0 ⇔ 1 0 ⇔ x>0 ⇒ f is CU on (0, 3) and (3, ∞) and
CD on (−∞, 0). NoIP
25. y = f(x) = x2
64
= x − 8+
x +8 x +8
A. D = {x | x 6=−8} B. Intercepts are 0 C. No symmetry
x 2
64
D. lim = ∞, butf(x) − (x − 8) = → 0 as x →∞,soy = x − 8 is a slant asymptote.
x→∞ x +8 x +8
x 2
lim
x→−8 + x +8 = ∞ and
lim x 2
x→−8 − x +8 = −∞,sox = −8 is a VA. E. f 0 (x) =1−
64 x(x +16)
=
(x +8)
2
(x +8) > 0 ⇔
2
x>0 or x 0 ⇔ x>−8,sof is CU on (−8, ∞) and
CD on (−∞, −8). NoIP
27. y = f(x) =x √ 2+x A. D =[−2, ∞) B. y-intercept: f(0) = 0; x-intercepts: −2 and 0 C. No symmetry
D. No asymptote E. f 0 x
(x) =
2 √ 2+x + √ 1
2+x =
2 √ 3x +4
[x +2(2+x)] =
2+x 2 √ 2+x =0when x = − 4 ,sof is
3
decreasing on
−2, − 4 3 and increasing on −
4
, ∞ . F. Local minimum value f
− 4 3 3 = −
4 2
= − 4√ 6
≈−1.09,
3 3 9
no local maximum
2 √ 1
2+x · 3 − (3x +4) √
G. f 00 2+x
(x) =
4(2 + x)
=
3x +8
4(2 + x) 3/2
=
6(2 + x) − (3x +4)
4(2 + x) 3/2
H.
f 00 (x) > 0 for x>−2,sof is CU on (−2, ∞). NoIP