Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 183
67. y = f(x) =x − tan −1 x, f 0 (x) =1− 1
1+x 2 = 1+x2 − 1
1+x 2 = x2
1+x 2 ,
f 00 (x) = (1 + x2 )(2x) − x 2 (2x)
= 2x(1 + x2 − x 2 ) 2x
=
(1 + x 2 ) 2 (1 + x 2 ) 2 (1 + x 2 ) . 2
lim f(x) − x −
π
x→∞
2 = lim π −
x→∞
2 tan−1 x = π − π =0,soy = x − π is a SA.
2 2 2
Also,
lim
x→−∞
f(x) −
x +
π
2
= lim
x→−∞
−
π
2 − tan−1 x = − π 2 − − π 2
=0,
so y = x + π 2 is also a SA. f 0 (x) ≥ 0 for all x, with equality ⇔ x =0,sof is
increasing on R. f 00 (x) has the same sign as x,sof is CD on (−∞, 0) and CU on
(0, ∞). f(−x) =−f(x),sof is an odd function; its graph is symmetric about the
origin. f has no local extreme values. Its only IP is at (0, 0).
69.
x 2
a 2 − y2
lim
x→∞
b
a
b =1 ⇒ y = ± b √
x2 − a 2 a
2 .Now
√
x2 − a 2 − b
a x = b a · lim
x→∞
which shows that y = b x is a slant asymptote. Similarly,
a
lim − b √
x2 − a
x→∞ a 2 −
− b
a x = − b a · lim
x→∞
√
x2 − a 2 − x √ x 2 − a 2 + x
√
x2 − a 2 + x = b a · lim
x→∞
−a 2
√
x2 − a 2 + x =0,
−a 2
√
x2 − a 2 + x =0,soy = − b x is a slant asymptote.
a
71. lim
x→±∞
f(x) − x
3 =
lim
x→±∞
x 4 +1
x
− x4
x = lim 1
x→±∞ x =0,sothegraphoff is asymptotic to that of y = x3 .
= −∞ and
A. D = {x | x 6= 0} B. No intercept C. f is symmetric about the origin. D. lim
x 3 + 1
x→0 − x
lim
x 3 + 1
= ∞,sox =0is a vertical asymptote, and as shown above, the graph of f is asymptotic to that of y = x 3 .
x→0 + x
E. f 0 (x) =3x 2 − 1/x 2 > 0 ⇔ x 4 > 1 3
⇔ |x| > 1 4 √ 3 ,sof is increasing on
−∞, − 1 4 √ 3
decreasing on − 4√ 1
, 0 and 0,
3
1
4√
3
. F. Local maximum value
f − 1
1
√ = −4 · 3 −5/4 , local minimum value f
4
3
4√ =4· 3 −5/4
3
G. f 00 (x) =6x +2/x 3 > 0 ⇔ x>0,sof is CU on (0, ∞) and CD
on (−∞, 0). NoIP
H.
1
and √ 4
3 , ∞ and