Solução_Calculo_Stewart_6e

F.
186 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
9. f(x) =1+ 1 x + 8 x 2 + 1 x 3 ⇒ f 0 (x) =− 1 x 2 − 16
x 3 − 3 x 4 = − 1 x 4 (x2 +16x +3) ⇒
f 00 (x) = 2 x 3 + 48
x 4 + 12
x 5 = 2 x 5 (x2 +24x +6).
From the graphs, it appears that f increases on (−15.8, −0.2) and decreases on (−∞, −15.8), (−0.2, 0),and(0, ∞);thatf
has a local minimum value of f(−15.8) ≈ 0.97 and a local maximum value of f(−0.2) ≈ 72;thatf is CD on (−∞, −24)
and (−0.25, 0) and is CU on (−24, −0.25) and (0, ∞);andthatf has IPs at (−24, 0.97) and (−0.25, 60).
To find the exact values, note that f 0 =0 ⇒ x = −16 ± √ 256 − 12
2
= −8 ± √ 61 [≈ −0.19 and −15.81].
f 0 is positive (f is increasing) on −8 − √ 61, −8+ √ 61 and f 0 is negative (f is decreasing) on −∞, −8 − √ 61 ,
−8+
√
61, 0
,and(0, ∞). f 00 =0 ⇒ x = −24 ± √ 576 − 24
2
= −12 ± √ 138 [≈ −0.25 and −23.75]. f 00 is
positive (f is CU) on −12 − √ 138, −12 + √ 138 and (0, ∞) and f 00 is negative (f is CD) on −∞, −12 − √ 138
and −12 + √ 138, 0 .
11. (a) f(x) =x 2 ln x. The domain of f is (0, ∞).
(b)
ln x
lim x2 ln x = lim
H 1/x
= lim
x→0 + x→0 + 1/x 2 x→0 + −2/x = lim
3 x→0 +
There is a hole at (0, 0).
− x2
=0.
2
(c) It appears that there is an IP at about (0.2, −0.06) and a local minimum at (0.6, −0.18). f(x) =x 2 ln x ⇒
f 0 (x) =x 2 (1/x)+(lnx)(2x) =x(2 ln x +1)> 0 ⇔ ln x>− 1 2
⇔ x>e −1/2 ,sof is increasing on
1/ √
e, ∞ , decreasing on 0, 1/ √
e .BytheFDT,f 1/ √
e = −1/(2e) is a local minimum value. This point is
approximately (0.6065, −0.1839), which agrees with our estimate.
f 00 (x) =x(2/x)+(2lnx +1)=2lnx +3> 0 ⇔ ln x>− 3 2
⇔ x>e −3/2 ,sof is CU on (e −3/2 , ∞)
and CD on (0,e −3/2 ). IPis(e −3/2 , −3/(2e 3 )) ≈ (0.2231, −0.0747).