f(x)在【0,1】上连续,(0,1)内可导,f(1)=0,证至少存在一点ξ属于(0,1),使f'(ξ)=-2f(ξ)/ξ
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![f(x)在【0,1】上连续,(0,1)内可导,f(1)=0,证至少存在一点ξ属于(0,1),使f'(ξ)=-2f(ξ)/ξ](/uploads/image/z/12624753-57-3.jpg?t=f%EF%BC%88x%EF%BC%89%E5%9C%A8%E3%80%900%2C1%E3%80%91%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%EF%BC%880%2C1%EF%BC%89%E5%86%85%E5%8F%AF%E5%AF%BC%2Cf%EF%BC%881%EF%BC%89%3D0%2C%E8%AF%81%E8%87%B3%E5%B0%91%E5%AD%98%E5%9C%A8%E4%B8%80%E7%82%B9%CE%BE%E5%B1%9E%E4%BA%8E%EF%BC%880%2C1%EF%BC%89%2C%E4%BD%BFf%27%28%CE%BE%29%3D-2f%28%CE%BE%29%2F%CE%BE)
f(x)在【0,1】上连续,(0,1)内可导,f(1)=0,证至少存在一点ξ属于(0,1),使f'(ξ)=-2f(ξ)/ξ
f(x)在【0,1】上连续,(0,1)内可导,f(1)=0,证至少存在一点ξ属于(0,1),使f'(ξ)=-2f(ξ)/ξ
f(x)在【0,1】上连续,(0,1)内可导,f(1)=0,证至少存在一点ξ属于(0,1),使f'(ξ)=-2f(ξ)/ξ
令F(x)=x²f(x)
则 F(1)=F(0) =0
所以存在ξ属于(0,1),使得
F'(ξ) = 2ξf(ξ)+ξ²f'(ξ) = 0
整理有f'(ξ)=-2f(ξ)/ξ
证毕
f(∫)=19-1
∫ =18