Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 4.1 MAXIMUM AND MINIMUM VALUES ¤ 149
29. f(x) =5x 2 +4x ⇒ f 0 (x) =10x +4. f 0 (x) =0 ⇒ x = − 2 ,so− 2 5 5
is the only critical number.
31. f(x) =x 3 +3x 2 − 24x ⇒ f 0 (x) =3x 2 +6x − 24 = 3(x 2 +2x − 8).
f 0 (x) =0 ⇒ 3(x +4)(x − 2) = 0 ⇒ x = −4, 2. These are the only critical numbers.
33. s(t) =3t 4 +4t 3 − 6t 2 ⇒ s 0 (t) =12t 3 +12t 2 − 12t. s 0 (t) =0 ⇒ 12t(t 2 + t − 1) ⇒
t =0 or t 2 + t − 1=0. Using the quadratic formula to solve the latter equation gives us
t = −1 ± 1 2 − 4(1)(−1)
2(1)
= −1 ± √ 5
2
≈ 0.618, −1.618. The three critical numbers are 0, −1 ± √ 5
.
2
35. g(y) =
y − 1
y 2 − y +1
⇒
g 0 (y) = (y2 − y +1)(1)− (y − 1)(2y − 1)
= y2 − y +1− (2y 2 − 3y +1)
= −y2 +2y y(2 − y)
=
(y 2 − y +1) 2 (y 2 − y +1) 2 (y 2 − y +1)
2
(y 2 − y +1) . 2
g 0 (y) =0 ⇒ y =0, 2. The expression y 2 − y +1is never equal to 0,sog 0 (y) exists for all real numbers.
The critical numbers are 0 and 2.
37. h(t) =t 3/4 − 2t 1/4 ⇒ h 0 (t) = 3 4 t−1/4 − 2 4 t−3/4 = 1 4 t−3/4 (3t 1/2 − 2) = 3 √ t − 2
4 4√ t 3 .
h 0 (t) =0 ⇒ 3 √ t =2 ⇒ √ t = 2 3
⇒ t = 4 9 . h0 (t) does not exist at t =0, so the critical numbers are 0 and 4 9 .
39. F (x) =x 4/5 (x − 4) 2 ⇒
F 0 (x) =x 4/5 · 2(x − 4) + (x − 4) 2 · 4
5 x−1/5 = 1 5 x−1/5 (x − 4)[5 · x · 2+(x − 4) · 4]
=
(x − 4)(14x − 16) 2(x − 4)(7x − 8)
=
5x 1/5 5x 1/5
F 0 (x) =0 ⇒ x =4, 8 7 . F 0 (0) does not exist. Thus, the three critical numbers are 0, 8 7 ,and4.
41. f(θ) =2cosθ +sin 2 θ ⇒ f 0 (θ) =−2sinθ +2sinθ cos θ. f 0 (θ) =0 ⇒ 2sinθ (cos θ − 1) = 0 ⇒ sin θ =0
or cos θ =1 ⇒ θ = nπ [n an integer] or θ =2nπ. The solutions θ = nπ include the solutions θ =2nπ, so the critical
numbers are θ = nπ.
43. f(x) =x 2 e −3x ⇒ f 0 (x) =x 2 (−3e −3x )+e −3x (2x) =xe −3x (−3x +2). f 0 (x) =0 ⇒ x =0, 2 3
[e −3x is never equal to 0]. f 0 (x) always exists, so the critical numbers are 0 and 2 3 .
45. The graph of f 0 (x) =5e −0.1|x| sin x − 1 has 10 zeros and exists
everywhere, so f has 10 critical numbers.