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设F(x)在【0,1】上连续,在(0,1)内可导,且F(0)=F(1)=0,F(0.5)=1,试证至少有一点W,使F'(W)=1
人气:156 ℃ 时间:2020-05-16 22:43:07
解答
设 G(x) = F(x) - x
G(0) = 0
G(1) = -1
G(0.5) = 0.5
所以 G(x) 的最大值必在开区间(0,1)中一点 w 达到.于是 G'(w) = 0,
即:F'(w) - 1 = 0 =====> F'(w) = 1
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