Solução_Calculo_Stewart_6e

F.
82 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
TX.10
35. (a) The slope of the tangent line at (2, 1) is
f(x) − f(2) 9 − 2x 2 − 1 8 − 2x 2
lim
=lim
= lim
x→2 x − 2 x→2 x − 2 x→2 x − 2
= lim −2(x 2 − 4) −2(x − 2)(x +2)
= lim
x→2 x − 2 x→2 x − 2
=lim
x→2
[−2(x +2)]=−2 · 4=−8
(b) An equation of this tangent line is y − 1=−8(x − 2) or y = −8x +17.
37. (a) s = s(t) =1+2t + t 2 /4. The average velocity over the time interval [1, 1+h] is
v ave =
s(1 + h) − s(1)
(1 + h) − 1
= 1+2(1+h)+(1+h)2 4 − 13/4
h
=
10h + h2
4h
= 10 + h
4
So for the following intervals the average velocities are:
(i) [1, 3]: h =2, v ave =(10+2)/4 =3m/s
(iii) [1, 1.5]: h =0.5, v ave =(10+0.5)/4 =2.625 m/s
(ii) [1, 2]: h =1, v ave =(10+1)/4 =2.75 m/s
(iv) [1, 1.1]: h =0.1, v ave =(10+0.1)/4 =2.525 m/s
s(1 + h) − s(1) 10 + h
(b) When t =1, the instantaneous velocity is lim
= lim = 10
h→0 h
h→0 4 4
=2.5 m/s.
39. (a) f 0 f(x) − f(2) x 3 − 2x − 4
(2) = lim
=lim
x→2 x − 2 x→2 x − 2
(x − 2) x 2 +2x +2
= lim
x→2 x − 2
= lim
x→2
x 2 +2x +2 =10
(c)
(b) y − 4=10(x − 2) or y =10x − 16
41. (a) f 0 (r) is the rate at which the total cost changes with respect to the interest rate. Its units are dollars/(percent per year).
(b) The total cost of paying off the loan is increasing by $1200/(percent per year) as the interest rate reaches 10%. Soifthe
interest rate goes up from 10% to 11%, the cost goes up approximately $1200.
(c) As r increases, C increases. So f 0 (r) will always be positive.
43.