Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 2.8 THEDERIVATIVEASAFUNCTION ¤ 73
17. (a) By zooming in, we estimate that f 0 (0) = 0, f 0 1
2
=1, f 0 (1) = 2,
and f 0 (2) = 4.
(b) By symmetry, f 0 (−x) =−f 0 (x). Sof 0 − 1 2
= −1, f 0 (−1) = −2,
and f 0 (−2) = −4.
(c) It appears that f 0 (x) is twice the value of x, so we guess that f 0 (x) =2x.
(d) f 0 f(x + h) − f(x) (x + h) 2 − x 2
(x) =lim
=lim
h→0 h
h→0 h
x 2 +2hx + h 2 − x 2 2hx + h 2 h(2x + h)
=lim
=lim =lim
=lim(2x + h) =2x
h→0 h
h→0 h h→0 h
h→0
1
19. f 0 f(x + h) − f(x)
(x)= lim
= lim
(x + h) − 1
2 3 − 1 x −
1
2 3
h→0 h
h→0 h
=lim
1
h 2
h→0
h =lim
h→0
1
2 = 1 2
Domain of f = domain of f 0 = R.
21. f 0 f(t + h) − f(t)
5(t + h) − 9(t + h)
2
− (5t − 9t 2 )
(t) = lim
=lim
h→0 h
h→0 h
1
= lim
x + 1 h − 1 − 1 x + 1 2 2 3 2 3
h→0 h
5t +5h − 9(t 2 +2th + h 2 ) − 5t +9t 2 5t +5h − 9t 2 − 18th − 9h 2 − 5t +9t 2
= lim
= lim
h→0 h
h→0 h
5h − 18th − 9h 2 h(5 − 18t − 9h)
= lim
= lim
=lim(5 − 18t − 9h) =5− 18t
h→0 h
h→0 h
h→0
Domain of f = domain of f 0 = R.
23. f 0 f(x + h) − f(x) (x + h) 3 − 3(x + h)+5 − (x 3 − 3x +5)
(x) = lim
=lim
h→0 h
h→0 h
x 3 +3x 2 h +3xh 2 + h 3 − 3x − 3h +5 − x 3 − 3x +5
3x 2 h +3xh 2 + h 3 − 3h
= lim
= lim
h→0 h
h→0 h
h 3x 2 +3xh + h 2 − 3
= lim
=lim 3x 2 +3xh + h 2 − 3 =3x 2 − 3
h→0 h
h→0
Domain of f = domain of f 0 = R.
√ √
25. g 0 g(x + h) − g(x) 1+2(x + h) − 1+2x 1+2(x + h)+ 1+2x
(x) =lim
= lim
√
h→0 h
h→0 h
1+2(x + h)+ 1+2x
(1 + 2x +2h) − (1 + 2x)
2
2
=lim
h→0
1+2(x √ =lim√ √ =
h
+ h)+ 1+2x h→0 1+2x +2h + 1+2x 2 √ 1+2x = 1
√ 1+2x
Domain of g = − 1 2 , ∞ , domain of g 0 = − 1 2 , ∞ .