Solução_Calculo_Stewart_6e

F.
182 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
C. No symmetry D. lim f(x) =−∞ and lim
x→(1/2) −
lim
x→±∞
[f(x) − (−x +2)]= lim
E. f 0 (x) =−1 −
f(x) =∞,sox = 1 x→(1/2) + 2
is a VA.
1
=0,sotheliney = −x +2is a SA.
x→±∞ 2x − 1
2
(2x − 1) < 0 for x 6= 1 ,sof is decreasing on
−∞, 1 2 2 2
and 1
2 , ∞ . F. No extreme values G. f 0 (x) =−1 − 2(2x − 1) −2 ⇒
H.
f 00 (x) =−2(−2)(2x − 1) −3 8
(2) =
(2x − 1) 3 ,sof 00 (x) > 0 when x> 1 and 2
f 00 (x) < 0 when x< 1 . Thus, f is CU on 1
, ∞ and CD on
−∞, 1 2 2 2 .NoIP
63. y = f(x) =(x 2 +4)/x = x +4/x A. D = {x | x 6= 0} =(−∞, 0) ∪ (0, ∞) B. No intercept
C. f(−x) =−f(x) ⇒ symmetry about the origin D. lim (x +4/x) =∞ but f(x) − x =4/x → 0 as x → ±∞,
x→∞
so y = x is a slant asymptote.
lim (x +4/x) =∞ and
x→0 +
lim
x→0 − (x +4/x) =−∞,sox =0 is a VA. E. f 0 (x) =1− 4/x 2 > 0 ⇔
x 2 > 4 ⇔ x>2 or x 0 ⇔ x>0 so f is CU on
(0, ∞) and CD on (−∞, 0). NoIP
65. y = f(x) = 2x3 + x 2 +1
x 2 +1
=2x +1+ −2x
x 2 +1
A. D = R B. y-intercept: f(0) = 1; x-intercept: f(x) =0 ⇒
0=2x 3 + x 2 +1=(x + 1)(2x 2 − x +1) ⇒ x = −1 C. No symmetry D. No VA
lim
x→±∞
[f(x) − (2x + 1)] = lim
x→±∞
−2x
x 2 +1 =
lim
x→±∞
−2/x
=0, so the line y =2x +1is a slant asymptote.
1+1/x2 E. f 0 (x) =2+ (x2 +1)(−2) − (−2x)(2x)
= 2(x4 +2x 2 +1)− 2x 2 − 2+4x 2
= 2x4 +6x 2
(x 2 +1) 2 (x 2 +1) 2 (x 2 +1) = 2x2 (x 2 +3)
2 (x 2 +1) 2
so f 0 (x) > 0 if x 6= 0.Thus,f is increasing on (−∞, 0) and (0, ∞). Sincef is continuous at 0, f is increasing on R.
F. No extreme values
G. f 00 (x) = (x2 +1) 2 · (8x 3 +12x) − (2x 4 +6x 2 ) · 2(x 2 +1)(2x)
[(x 2 +1) 2 ] 2
= 4x(x2 +1)[(x 2 + 1)(2x 2 +3)− 2x 4 − 6x 2 ]
= 4x(−x2 +3)
(x 2 +1) 4 (x 2 +1) 3
so f 00 (x) > 0 for x