Solução_Calculo_Stewart_6e

F.
210 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION TX.10
⎧
10x
11. f(x) = 10
⎪⎨
−8
+ C 1 = − 5
x 9 =10x−9 has domain (−∞, 0) ∪ (0, ∞),soF (x) =
−8
4x + C 8 1 if x0
See Example 1(b) for a similar problem.
13. f(u) = u4 +3 √ u
u 2
= u4
u 2 + 3u1/2
u 2 = u 2 +3u −3/2 ⇒
F (u) = u3
3
+3
u−3/2+1
−3/2+1 + C = 1 3 u3 +3 u−1/2
−1/2 + C = 1 3 u3 − 6 √ u
+ C
15. g(θ) =cosθ − 5sinθ ⇒ G(θ) =sinθ − 5(− cos θ)+C =sinθ +5cosθ + C
17. f(x) =5e x − 3coshx ⇒ F (x) =5e x − 3sinhx + C
19. f(x) = x5 − x 3 +2x
= x − 1 x 4 x + 2 x = x − 1 3 x +2x−3 ⇒
F (x) = x2
x
−3+1
2 − ln |x| +2 + C = 1 2
−3+1
x2 − ln |x| − 1 x + C 2
21. f(x) =5x 4 − 2x 5 ⇒ F (x) =5· x5
5 − 2 · x6
6 + C = x5 − 1 3 x6 + C.
F (0) = 4 ⇒ 0 5 − 1 3 · 06 + C =4 ⇒ C =4,soF (x) =x 5 − 1 3 x6 +4.
The graph confirms our answer since f(x) =0when F has a local maximum, f is
positive when F is increasing, and f is negative when F is decreasing.
23. f 00 (x) =6x +12x 2 ⇒ f 0 (x) =6· x2
2
f(x) =3· x3
3
25. f 00 (x) = 2 3 x2/3 ⇒ f 0 (x) = 2 3
+12·
x3
3 + C =3x2 +4x 3 + C ⇒
+4·
x4
4 + Cx + D = x3 + x 4 + Cx + D [C and D are just arbitrary constants]
x
5/3
5/3
+ C = 2 5 x5/3 + C ⇒ f(x) = 2 5
x
8/3
8/3
+ Cx + D = 3
20 x8/3 + Cx + D
27. f 000 (t) =e t ⇒ f 00 (t) =e t + C ⇒ f 0 (t) =e t + Ct + D ⇒ f(t) =e t + 1 2 Ct2 + Dt + E
29. f 0 (x) =1− 6x ⇒ f(x) =x − 3x 2 + C. f(0) = C and f(0) = 8 ⇒ C =8,sof(x) =x − 3x 2 +8.
31. f 0 (x) = √ x(6 + 5x) =6x 1/2 +5x 3/2 ⇒ f(x) =4x 3/2 +2x 5/2 + C.
f(1) = 6 + C and f(1) = 10 ⇒ C =4,sof(x) =4x 3/2 +2x 5/2 +4.
33. f 0 (t) =2cost +sec 2 t ⇒ f(t) =2sint +tant + C because −π/2