f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证:存在ξ∈(0,1),使f``(ξ)=2f`(ξ)/1-ξ.f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证:存在ξ∈(0,1),使f``(ξ)=2f(ξ)/1-ξ.
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![f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证:存在ξ∈(0,1),使f``(ξ)=2f`(ξ)/1-ξ.f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证:存在ξ∈(0,1),使f``(ξ)=2f(ξ)/1-ξ.](/uploads/image/z/11616887-47-7.jpg?t=f%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%3D0%2C%E8%AF%95%E8%AF%81%EF%BC%9A%E5%AD%98%E5%9C%A8%CE%BE%E2%88%88%EF%BC%880%2C1%EF%BC%89%2C%E4%BD%BFf%60%60%28%CE%BE%29%3D2f%60%28%CE%BE%29%2F1-%CE%BE.f%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%3D0%2C%E8%AF%95%E8%AF%81%EF%BC%9A%E5%AD%98%E5%9C%A8%CE%BE%E2%88%88%EF%BC%880%2C1%EF%BC%89%2C%E4%BD%BFf%60%60%28%CE%BE%29%3D2f%28%CE%BE%29%2F1-%CE%BE.)
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f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证:存在ξ∈(0,1),使f``(ξ)=2f`(ξ)/1-ξ.f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证:存在ξ∈(0,1),使f``(ξ)=2f(ξ)/1-ξ.
f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证:存在ξ∈(0,1),使f``(ξ)=2f`(ξ)/1-ξ.
f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证:存在ξ∈(0,1),使f``(ξ)=2f(ξ)/1-ξ.
f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证:存在ξ∈(0,1),使f``(ξ)=2f`(ξ)/1-ξ.f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证:存在ξ∈(0,1),使f``(ξ)=2f(ξ)/1-ξ.
由条件,存在η∈(0,1),满足f'(η)=0.
令G(x) = (1-x)²f'(x),则G(η) = G(1) = 0
所以,存在ξ∈(η,1),使G'(ξ)=0,即(1-ξ)²f''(ξ)-2(1-ξ)f'(ξ)=0
由于ξ<1,所以(1-ξ)f''(ξ)-2f'(ξ)=0,即f''(ξ)=2f'(ξ)/(1-ξ).
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