※ 引述《smart55 (大東)》之銘言：
: 請教一下4cos(x-pi/9)+2sin(2x)的最小值要怎麼用微分導數為零求出？謝謝。
f(x) = 4cos(x - pi/9) + 2sin(2x)
f'(x) = -4sin(x - pi/9) + 4cos(2x)
= 4[sin(pi/2 - 2x) - sin(x - pi/9)]
= 8cos(7pi/36 - x/2)sin(11pi/36 - 3x/2)
= -8sin(3x/2 - 11pi/36)cos(x/2 - 7pi/36)
極值點x = 11pi/54, 47pi/54, 75pi/54, 83pi/54
極大值 極小值 極大值 極小值
f(47pi/54) = 4cos(41pi/54) + 2sin(47pi/27)
= -4cos(13pi/54) - 2sin(7pi/54)
f(83pi/54) = 4cos(77pi/54) + 2sin(83pi/27)
= -4cos(23pi/54) - 2sin(2pi/27)
f(47pi/54) - f(83pi/54) = -8sin(pi/3)sin(5pi/54) - 4cos(11pi/108)sin(3pi/108)
= -2sqrt(3)sin(5pi/54) - 4cos(11pi/108)sin(pi/36) < 0
所以f(x)最小值 = -4cos(13pi/54) - 2sin(7pi/54)
發生在x = 47pi/54 + 2kpi, k = 整數
當然這個函數是無窮延展，我這邊只是要算出所有的極大值和極小值
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※ 編輯: Honor1984 (61.56.10.112), 11/28/2017 12:41:03