设f(x)在[a,b]上连续,在(a,b)内可导,f(a)f(b)>0,f(a)f[(a+b)/2]

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设f(x)在[a,b]上连续,在(a,b)内可导,f(a)f(b)>0,f(a)f[(a+b)/2]
设f(x)在[a,b]上连续,在(a,b)内可导,f(a)f(b)>0,f(a)f[(a+b)/2]

设f(x)在[a,b]上连续,在(a,b)内可导,f(a)f(b)>0,f(a)f[(a+b)/2]
不妨设f(a)>0,f(b)>0,则f((a+b)/2)<0.令F(x)=e^(-x)f(x),记c=(a+b)/2
则F(a)>0,F(b)>0,F(c)<0,分别在【a,c】【c,b】上用
零点定理知道,存在c1再在【c1,c2】上用Rolle定理,存在α位于(a,b),使得
F'(α)=0,即e^(-α)(f'(α)-f(α))=0,于是
f'(α)=f(α).

楼主应该是（a,b）不是(0,1)吧

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