Solução_Calculo_Stewart_6e

F.
13. f(x) =
lim
x→1
TX.10SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS ¤ 187
(x +4)(x − 3)2
x 4 (x − 1)
f(x) =−∞ and lim f(x) =∞.
− +
f(x) =
x +4
·
x
x 4
x→1
(x − 3)2
x 2
· (x − 1)
x3 as x → ±∞,sof is asymptotic to the x-axis.
has VA at x =0and at x =1since lim
x→0
f(x) =−∞,
dividing numerator
and denominator by x 3
=
(1 + 4/x)(1 − 3/x)2
x(x − 1)
Since f is undefined at x =0,ithasnoy-intercept. f(x) =0 ⇒ (x +4)(x − 3) 2 =0 ⇒ x = −4 or x =3,sof has
x-intercepts −4 and 3. Note, however, that the graph of f is only tangent to the x-axis and does not cross it at x =3,sincef is
positive as x → 3 − and as x → 3 + .
→ 0
From these graphs, it appears that f has three maximum values and one minimum value. The maximum values are
approximately f(−5.6) = 0.0182, f(0.82) = −281.5 and f(5.2) = 0.0145 and we know (since the graph is tangent to the
x-axis at x =3) that the minimum value is f(3) = 0.
15. f(x) =
x 2 (x +1) 3
(x − 2) 2 (x − 4) 4 ⇒ f 0 (x) =− x(x +1)2 (x 3 +18x 2 − 44x − 16)
(x − 2) 3 (x − 4) 5 [from CAS].
From the graphs of f 0 , it seems that the critical points which indicate extrema occur at x ≈−20, −0.3,and2.5, as estimated
in Example 3. (There is another critical point at x = −1,butthesignoff 0 does not change there.) We differentiate again,
obtaining f 00 (x) =2 (x +1)(x6 +36x 5 +6x 4 − 628x 3 + 684x 2 +672x +64)
(x − 2) 4 (x − 4) 6 .
From the graphs of f 00 , it appears that f is CU on (−35.3, −5.0), (−1, −0.5), (−0.1, 2), (2, 4) and (4, ∞) and CD