Question

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第59、61🙇

49-60 Find the absolute maximum and absolute minimum
values of f on the given interval.
49. f(x) = 12 + 4x - x², [0, 5]
50. f(x) = 5 + 54x2x³, [0, 4]
(51.) f(x) = 2x³ 3x² - 12x + 1, [-2,3]
52. f(x) = x³ - 6x² + 5, [-3,5]
53. f(x) = 3x4 - 4x³ - 12x² + 1, [-2,3]
54. f(t) = (t²-4)³, [2, 3]
1
55. f(x) = x + = ₁
X
56. f(x)
X
2
x² = x + 1’
-
57. f(t) = t - √√t, [-1,4]
√t
1 + t22
59. f(t) = 2 cost + sin 2t, [0, π/2] on ball
=
[0.2, 4]
58. f(t) =
=
[0, 3]
[0, 2]

HOWCAL
61. If a and b are positive numbers, find the maximum value
of f(x) = x^(1-x), 0≤x≤ 1,ood l

微積分

Answer 1

呵

極值找斜率為零的點，所以微分