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-§)](/uploads/image/z/11616886-46-6.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%5B0%2C1%5D%E4%B8%8A%E4%BA%8C%E9%98%B6%E5%8F%AF%E5%AF%BC%2C%E4%B8%94f%280%29%3Df%281%29%2C%E8%AF%95%E8%AF%81%EF%BC%9A%E8%87%B3%E5%B0%91%E5%AD%98%E5%9C%A8%E4%B8%80%E4%B8%AA%C2%A7%E5%B1%9E%E4%BA%8E%280%2C1%29%2C%E4%BD%BFf%27%27%28%C2%A7%29%3D2f%27%28%C2%A7%29%2F%281-%C2%A7%29)
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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