Solução_Calculo_Stewart_6e

F.
74 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
TX.10
4(t + h)
27. G 0 G(t + h) − G(t) (t + h)+1 − 4t 4(t + h)(t +1)− 4t(t + h +1)
t +1
(t + h +1)(t +1)
(t) =lim
=lim
=lim
h→0 h
h→0 h
h→0 h
4t 2 +4ht +4t +4h − 4t 2 +4ht +4t
4h
=lim
= lim
h→0 h(t + h +1)(t +1)
h→0 h(t + h +1)(t +1)
=lim
h→0
4
(t + h +1)(t +1) = 4
(t +1) 2
Domain of G = domain of G 0 =(−∞, −1) ∪ (−1, ∞).
29. f 0 f(x + h) − f(x) (x + h) 4 − x 4 x 4 +4x 3 h +6x 2 h 2 +4xh 3 + h 4 − x 4
(x) =lim
=lim
=lim
h→0 h
h→0 h
h→0 h
4x 3 h +6x 2 h 2 +4xh 3 + h 4
=lim
=lim 4x 3 +6x 2 h +4xh 2 + h 3 =4x 3
h→0 h
h→0
Domain of f = domain of f 0 = R.
31. (a) f 0 f(x + h) − f(x) [(x + h) 4 +2(x + h)] − (x 4 +2x)
(x)= lim
=lim
h→0 h
h→0 h
=lim
h→0
x 4 +4x 3 h +6x 2 h 2 +4xh 3 + h 4 +2x +2h − x 4 − 2x
h
4x 3 h +6x 2 h 2 +4xh 3 + h 4 +2h h(4x 3 +6x 2 h +4xh 2 + h 3 +2)
=lim
=lim
h→0 h
h→0 h
=lim
h→0
(4x 3 +6x 2 h +4xh 2 + h 3 +2)=4x 3 +2
(b) Notice that f 0 (x) =0when f has a horizontal tangent, f 0 (x) is
positive when the tangents have positive slope, and f 0 (x) is
negative when the tangents have negative slope.
33. (a) U 0 (t) is the rate at which the unemployment rate is changing with respect to time. Its units are percent per year.
(b) To find U 0 U(t + h) − U(t) U(t + h) − U(t)
(t),weuse lim
≈ for small values of h.
h→0 h
h
For 1993: U 0 (1993) ≈
U(1994) − U(1993)
1994 − 1993
=
6.1 − 6.9
1
= −0.80
For 1994: We estimate U 0 (1994) by using h = −1 and h =1, and then average the two results to obtain a final estimate.
h = −1 ⇒ U 0 (1994) ≈
h =1 ⇒ U 0 (1994) ≈
U(1993) − U(1994)
1993 − 1994
U(1995) − U(1994)
1995 − 1994
=
=
6.9 − 6.1
−1
5.6 − 6.1
1
So we estimate that U 0 (1994) ≈ 1 [(−0.80) + (−0.50)] = −0.65.
2
= −0.80;
= −0.50.
t 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002
U 0 (t) −0.80 −0.65 −0.35 −0.35 −0.45 −0.35 −0.25 0.25 0.90 1.10