f(x)在[0,1]上连续,在(0,1)内可导,且f(0)=1,f(1)=0,证明在(0,1)内至少存在一点ξ使得f'(ξ)=f(ξ)/ξ
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![f(x)在[0,1]上连续,在(0,1)内可导,且f(0)=1,f(1)=0,证明在(0,1)内至少存在一点ξ使得f'(ξ)=f(ξ)/ξ](/uploads/image/z/11504302-70-2.jpg?t=f%EF%BC%88x%EF%BC%89%E5%9C%A8%EF%BC%BB0%2C1%EF%BC%BD%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8%EF%BC%880%2C1%EF%BC%89%E5%86%85%E5%8F%AF%E5%AF%BC%2C%E4%B8%94f%EF%BC%880%EF%BC%89%3D1%2Cf%EF%BC%881%EF%BC%89%3D0%2C%E8%AF%81%E6%98%8E%E5%9C%A8%EF%BC%880%2C1%EF%BC%89%E5%86%85%E8%87%B3%E5%B0%91%E5%AD%98%E5%9C%A8%E4%B8%80%E7%82%B9%CE%BE%E4%BD%BF%E5%BE%97f%27%EF%BC%88%CE%BE%EF%BC%89%3Df%EF%BC%88%CE%BE%EF%BC%89%2F%CE%BE)
f(x)在[0,1]上连续,在(0,1)内可导,且f(0)=1,f(1)=0,证明在(0,1)内至少存在一点ξ使得f'(ξ)=f(ξ)/ξ
f(x)在[0,1]上连续,在(0,1)内可导,且f(0)=1,f(1)=0,证明在(0,1)内至少存在一点ξ使得f'(ξ)=f(ξ)/ξ
f(x)在[0,1]上连续,在(0,1)内可导,且f(0)=1,f(1)=0,证明在(0,1)内至少存在一点ξ使得f'(ξ)=f(ξ)/ξ
令g(x)=xf(x),g'(x)=f(x)-xf(x).根据罗尔定理,存在ζ使上式成立.
罗尔定理