Solução_Calculo_Stewart_6e

F.
76 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
47. f 0 f(x + h) − f(x) 2(x + h) 2 − (x + h) 3 − (2x 2 − x 3 )
(x) = lim
= lim
h→0 h
h→0 h
h(4x +2h − 3x 2 − 3xh − h 2 )
=lim
=lim(4x +2h − 3x 2 − 3xh − h 2 )=4x − 3x 2
h→0 h
h→0
f 00 f 0 (x + h) − f 0 (x)
4(x + h) − 3(x + h)
2
− (4x − 3x 2 ) h(4 − 6x − 3h)
(x) = lim
= lim
= lim
h→0 h
h→0 h
h→0 h
=lim(4 − 6x − 3h) =4− 6x
h→0
f 000 f 00 (x + h) − f 00 (x) [4 − 6(x + h)] − (4 − 6x) −6h
(x) = lim
=lim
=lim
h→0 h
h→0 h
h→0 h
=lim (−6) = −6
h→0
f (4) f 000 (x + h) − f 000 (x) −6 − (−6) 0
(x) = lim
=lim
=lim
h→0 h
h→0 h
h→0 h =lim(0)
= 0
h→0
The graphs are consistent with the geometric interpretations of the
derivatives because f 0 has zeros where f has a local minimum and a local
maximum, f 00 has a zero where f 0 has a local maximum, and f 000 is a
constant function equal to the slope of f 00 .
49. (a)Notethatwehavefactoredx − a as the difference of two cubes in the third step.
f 0 f(x) − f(a) x 1/3 − a 1/3
x 1/3 − a 1/3
(a) =lim
=lim
= lim
x→a x − a x→a x − a x→a (x 1/3 − a 1/3 )(x 2/3 + x 1/3 a 1/3 + a 2/3 )
1
=lim
x→a x 2/3 + x 1/3 a 1/3 + a = 1
2/3 3a or 1 2/3 3 a−2/3
(b) f 0 f(0 + h) − f(0)
√ 3
h − 0
(0) = lim
= lim = lim
h→0 h
h→0 h
exist, and therefore f 0 (0) does not exist.
(c) lim
x→0
|f 0 (x)| =lim
51. f(x) =|x − 6| =
x→0
1
TX.10
1
h→0 h
x − 6 = lim
x→6
|x − 6| .
2/3
. This function increases without bound, so the limit does not
= ∞ and f is continuous at x =0(root function), so f has a vertical tangent at x =0.
3x2/3
x − 6 if x − 6 ≥ 6 x − 6 if x ≥ 6
−(x − 6) if x − 6 < 0 = 6 − x if x6
However, a formula for f 0 is f 0 (x) =
−1 if x