Solução_Calculo_Stewart_6e

F.
166 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
77. Let the cubic function be f(x) =ax 3 + bx 2 + cx + d ⇒ f 0 (x) =3ax 2 +2bx + c ⇒ f 00 (x) =6ax +2b.
So f is CU when 6ax +2b>0 ⇔ x>−b/(3a),CDwhenx0,so(0, 0) is an inflection point. But g 00 (0) does not
exist.
83. (a) f(x) =x 4 sin 1 x ⇒ f 0 (x) =x 4 cos 1 − 1
+sin 1 x x 2 x (4x3 )=4x 3 sin 1 x − x2 cos 1 x .
g(x) =x 4 2+sin 1
=2x 4 + f(x) ⇒ g 0 (x) =8x 3 + f 0 (x).
x
h(x) =x 4 −2+sin 1
= −2x 4 + f(x) ⇒ h 0 (x) =−8x 3 + f 0 (x).
x
It is given that f(0) = 0, sof 0 f(x) − f(0)
x 4 sin 1
(0) = lim
=lim x − 0
=limx 3 sin 1
x→0 x − 0 x→0 x
x→0 x .Since
− x 3 ≤ x 3 sin 1 x ≤ x 3 and lim
x→0
x 3 =0,weseethatf 0 (0) = 0 by the Squeeze Theorem. Also,
g 0 (0) = 8(0) 3 + f 0 (0) = 0 and h 0 (0) = −8(0) 3 + f 0 (0) = 0,so0 is a critical number of f, g,andh.
For x 2n = 1
2nπ
For x 2n+1 =
1
1
[n a nonzero integer], sin =sin2nπ =0and cos =cos2nπ =1,sof 0 (x 2n )=−x 2 2n < 0.
x 2n x 2n
1
(2n +1)π , sin 1
1
=sin(2n +1)π =0and cos =cos(2n +1)π = −1, so
x 2n+1 x 2n+1
f 0 (x 2n+1 )=x 2 2n+1 > 0. Thus,f 0 changes sign infinitely often on both sides of 0.
Next, g 0 (x 2n) =8x 3 2n + f 0 (x 2n) =8x 3 2n − x 2 2n = x 2 2n(8x 2n − 1) < 0 for x 2n < 1 8 ,but
g 0 (x 2n+1 )=8x 3 2n+1 + x 2 2n+1 = x 2 2n+1(8x 2n+1 +1)> 0 for x 2n+1 > − 1 8 ,sog0 changes sign infinitely often on both
sides of 0.
Last, h 0 (x 2n) =−8x 3 2n + f 0 (x 2n) =−8x 3 2n − x 2 2n = −x 2 2n(8x 2n +1)< 0 for x 2n > − 1 8 and
h 0 (x 2n+1 )=−8x 3 2n+1 + x 2 2n+1 = x 2 2n+1(−8x 2n+1 +1)> 0 for x 2n+1 < 1 8 ,soh0 changes sign infinitely often on both
sides of 0.