求迪厅痴汉链接
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发布时间：2022-04-28 21:27
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热心网友
时间：2022-06-23 06:43
迪厅痴汉链接
www.viy606.com设f(x)=2(logX)^2+2a log(1/x)+b,己知当x=1/2时,f(x)取得最小值为-8,求a-b
解：f(1/2) = 2 * (-1)^2 + 2a * 1 + b = 2 + 2a + b = -8, 2a + b = -10 -------------- (1)
f(x) = 2 (lnx / ln2)^2 + 2a (ln(1/x) / ln2) + b
= 2 (lnx)^2 / (ln2)^2 - 2a (lnx / ln2) + b
f'(x) = 2/(ln2)^2 * 2lnx * 1/x - 2a/ln2 * 1/x = 0
2/(ln2) * lnx - a = 0
x = 1/2
-2 - a = 0, a = -2, 带入(1)得
b = -6
a - b = -2 - (-6) = 4
