Solução_Calculo_Stewart_6e

F.
150 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
47. f(x) =3x 2 − 12x +5, [0, 3]. f 0 (x) =6x − 12 = 0 ⇔ x =2. Applying the Closed Interval Method, we find that
f(0) = 5, f(2) = −7,andf(3) = −4. Sof(0) = 5 is the absolute maximum value and f(2) = −7 is the absolute minimum
value.
49. f(x) =2x 3 − 3x 2 − 12x +1, [−2, 3]. f 0 (x) =6x 2 − 6x − 12 = 6(x 2 − x − 2) = 6(x − 2)(x +1)=0 ⇔
x =2, −1.
f(−2) = −3, f(−1) = 8, f(2) = −19,andf(3) = −8. Sof(−1) = 8 is the absolute maximum value and
f(2) = −19 is the absolute minimum value.
51. f(x) =x 4 − 2x 2 +3, [−2, 3]. f 0 (x) =4x 3 − 4x =4x(x 2 − 1) = 4x(x +1)(x − 1) = 0 ⇔ x = −1, 0, 1.
f(−2) = 11, f(−1) = 2, f(0) = 3, f(1) = 2, f(3) = 66. Sof(3) = 66 is the absolute maximum value and f(±1) = 2 is
the absolute minimum value.
53. f(x) = x , [0, 2].
x 2 +1 f
0 (x) = (x2 +1)− x(2x)
= 1 − x2
=0 ⇔ x = ±1,but−1 is not in [0, 2]. f(0) = 0,
(x 2 +1) 2 (x 2 +1)
2
f(1) = 1 , f(2) = 2 .Sof(1) = 1 2 5 2
is the absolute maximum value and f(0) = 0 is the absolute minimum value.
55. f(t) =t √ 4 − t 2 , [−1, 2].
f 0 (t) =t · 1 (4 − 2 t2 ) −1/2 (−2t)+(4− t 2 ) 1/2 · 1= √ −t2 + √ 4 − t 2 = −t2 +(4− t 2 )
√ = √ 4 − 2t2 .
4 − t
2 4 − t
2 4 − t
2
f 0 (t) =0 ⇒ 4 − 2t 2 =0 ⇒ t 2 =2 ⇒ t = ± √ 2,butt = − √ 2 is not in the given interval, [−1, 2].
f 0 (t) does not exist if 4 − t 2 =0 ⇒ t = ±2, but−2 is not in the given interval. f(−1) = − √ 3, f √ 2 =2,and
f(2) = 0. Sof √ 2 =2is the absolute maximum value and f(−1) = − √ 3 is the absolute minimum value.
57. f(t) =2cost +sin2t, [0, π/2].
f 0 (t) =−2sint +cos2t · 2=−2sint +2(1− 2sin 2 t)=−2(2 sin 2 t +sint − 1) = −2(2 sin t − 1)(sin t +1).
f 0 (t) =0 ⇒ sin t = 1 or sin t = −1 ⇒ t = π . f(0) = 2, f( π )=√ √ √
2 6 6
3+ 1 2 3=
3
2 3 ≈ 2.60,andf(
π
)=0. 2
So f( π )= √ 3
6 2 3 is the absolute maximum value and f(
π
2
)=0is the absolute minimum value.
59. f(x) =xe −x2 /8 , [−1, 4]. f 0 (x) =x · e −x2 /8 · (− x 4 )+e−x2 /8 · 1=e −x2 /8 (− x2
4 +1).Sincee−x2 /8 is never 0,
f 0 (x) =0 ⇒ −x 2 /4+1=0 ⇒ 1=x 2 /4 ⇒ x 2 =4 ⇒ x = ±2,but−2 is not in the given interval, [−1, 4].
f(−1) = −e −1/8 ≈−0.88, f(2) = 2e −1/2 ≈ 1.21,andf(4) = 4e −2 ≈ 0.54. Sof(2) = 2e −1/2 is the absolute maximum
value and f(−1) = −e −1/8 is the absolute minimum value.
61. f(x) =ln(x 2 + x +1), [−1, 1]. f 0 1
(x) =
x 2 + x +1 · (2x +1)=0 ⇔ x = − 1 2 .Sincex2 + x +1> 0 for all x,the
domain of f and f 0 is R. f(−1)=ln1=0, f
− 1 2 =ln
3
≈ −0.29,andf(1) = ln 3 ≈ 1.10. Sof(1) = ln 3 ≈ 1.10 is
4
the absolute maximum value and f
− 1 2 =ln
3
4
≈ −0.29 is the absolute minimum value.