Solução_Calculo_Stewart_6e

F.
TX.10SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS ¤ 191
becomes steeper as c increases. For c =0, the graph is the simple curve y = x 4 . For c0, we see that there is a HA at y =1, and that the graph
spreads out as c increases. At first glance there appears to be a minimum at (0, 0),butf(0) is undefined, so there is no
minimum or maximum. For c0,theinflection points spread out as c increases, and if c0, there are IP at
±
2c/3,e −3/2 . Note that the y-coordinate of the IP is constant.
31. Note that c =0is a transitional value at which the graph consists of the x-axis. Also, we can see that if we substitute −c for c,
cx
the function f(x) = will be reflected in the x-axis, so we investigate only positive values of c (except c = −1,asa
1+c 2 x2 demonstration of this reflective property). Also, f is an odd function. lim f(x) =0,soy =0is a horizontal asymptote for
x→±∞
all c. We calculate f 0 (x) = (1 + c2 x 2 )c − cx(2c 2 x)
= − c(c2 x 2 − 1)
(1 + c 2 x 2 ) 2 (1 + c 2 x 2 ) . f 0 (x) =0 ⇔ c 2 x 2 − 1=0 ⇔ x = ±1/c.
2
So there is an absolute maximum value of f(1/c) = 1 and an absolute minimum value of f(−1/c) =− 1 .Theseextrema
2 2
have the same value regardless of c, but the maximum points move closer to the y-axis as c increases.
f 00 (x) = (−2c3 x)(1 + c 2 x 2 ) 2 − (−c 3 x 2 + c)[2(1 + c 2 x 2 )(2c 2 x)]
(1 + c 2 x 2 ) 4
= (−2c3 x)(1 + c 2 x 2 )+(c 3 x 2 − c)(4c 2 x)
= 2c3 x(c 2 x 2 − 3)
(1 + c 2 x 2 ) 3 (1 + c 2 x 2 ) 3
f 00 (x) =0 ⇔ x =0or ± √ 3/c,sothereareinflection points at (0, 0) and
at ± √ 3/c, ± √ 3/4 .Again,they-coordinate of the inflection points does not
depend on c,butasc increases, both inflection points approach the y-axis.