Solução_Calculo_Stewart_6e

F.
70 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
TX.10
33. By Equation 3, lim
x→5
2 x − 32
x − 5 = f 0 (5),wheref(x) =2 x and a =5.
cos(π + h)+1
35. By Definition 2, lim
= f 0 (π),wheref(x) =cosx and a = π.
h→0 h
cos(π + h)+1
Or: ByDefinition 2, lim
= f 0 (0),wheref(x) =cos(π + x) and a =0.
h→0 h
37. v(5) = f 0 f(5 + h) − f(5) [100 + 50(5 + h) − 4.9(5 + h) 2 ] − [100 + 50(5) − 4.9(5) 2 ]
(5) = lim
=lim
h→0 h
h→0 h
(100 + 250 + 50h − 4.9h 2 − 49h − 122.5) − (100 + 250 − 122.5) −4.9h 2 + h
=lim
= lim
h→0 h
h→0 h
h(−4.9h +1)
=lim
=lim(−4.9h +1)=1m/s
h→0 h
h→0
The speed when t =5is |1| =1m/s.
39. Thesketchshowsthegraphforaroomtemperatureof72 ◦ and a refrigerator
temperature of 38 ◦ . The initial rate of change is greater in magnitude than the
rate of change after an hour.
41. (a) (i) [2000, 2002]:
(ii) [2000, 2001]:
(iii) [1999, 2000]:
P (2002) − P (2000)
2002 − 2000
P (2001) − P (2000)
2001 − 2000
P (2000) − P (1999)
2000 − 1999
(b) Using the values from (ii) and (iii), we have
=
=
=
77 − 55
2
68 − 55
1
55 − 39
1
13 + 16
2
= 22
2 =11percent/year
=13percent/year
=16percent/year
=14.5 percent/year.
(c) Estimating A as (1999, 40) and B as (2001, 70), the slope at 2000 is
70 − 40
2001 − 1999 = 30
2 =15percent/year.
43. (a) (i) ∆C
∆x
(ii) ∆C
∆x
(b)
=
C(105) − C(100)
105 − 100
=
6601.25 − 6500
5
6520.05 − 6500
1
=$20.25/unit.
C(101) − C(100)
= = =$20.05/unit.
101 − 100
5000 + 10(100 + h)+0.05(100 + h)
2
− 6500
=
=
h
=20+0.05h, h 6= 0
C(100 + h) − C(100)
h
20h +0.05h2
h
C(100 + h) − C(100)
So the instantaneous rate of change is lim
= lim (20 + 0.05h) =$20/unit.
h→0 h
h→0
45. (a) f 0 (x) is the rate of change of the production cost with respect to the number of ounces of gold produced. Its units are
dollars per ounce.