问答题：设f(x)在[0，1]上连续，在(0，1)内可导，且满足证明至少存在一点ξ∈

题目：

设f(x)在[0，1]上连续，在(0，1)内可导，且满足

证明至少存在一点ξ∈(0，1)，使得f’(ξ)=(1-ξ^-1)f(ξ)．

答案：

参考答案：

[分析]：注意积分

的被积函数xe^1-xf(x)的导数
[ze^1-xf(x)]’=xe^1-xf’(x)+e^1-xf(x)-xe^1-xf(x)．
=xe^1-x[f’(x)-(1-x^-1)f(x)]
注意

就是本题要证的结论，所以，令F(x)=xe^1-xf(x)，只要证明F(x)满足罗尔定理条件，本题就可得以证明．
[证] 令F(x)=e^1-xf(x)，则F(1)=f(1)，由积分中值定理知，存在

，使

由题设

F(1)=F(c)
从而F(x)在[c，1]上满足罗尔定理条件，则存在ξ∈(c，1)，使F’(ξ)=0，
即ξe^1-ξ[f’(ξ)-(1-ξ^-1)f(ξ)]=0
而ξe^1-ξ≠0，故f’(ξ)-(1-ξ^-1)f(ξ)=0
即f’(ξ)=(1-ξ^-1)f(ξ)

完形填空
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