设f(x)在[0,1]上二阶可导,且f(0)=f(1),试证:至少存在一个§属于(0,1),使f''(§)=2f'(§)/(1-§)
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设f(x)在[0,1]上二阶可导,且f(0)=f(1),试证:至少存在一个§属于(0,1),使f''(§)=2f'(§)/(1-§)
设f(x)在[0,1]上二阶可导,且f(0)=f(1),试证:至少存在一个§属于(0,1),使f''(§)=2f'(§)/(1-§)
设f(x)在[0,1]上二阶可导,且f(0)=f(1),试证:至少存在一个§属于(0,1),使f''(§)=2f'(§)/(1-§)
构造函数F(x)=(x^2-x)f'(x)+f(x)
F(0)-F(1)=F'(ξ)=f''(ξ)(ξ^2-ξ)+2ξf'(ξ)=0
即f''(ξ)(ξ-1)+2f'(ξ)=0
所以f''(ξ)=2f'(ξ)/(1-ξ)
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