设函数f(x)f(x)f(x)在闭区间[0,1][0,1][0,1]上连续,在(0,1)(0,1)(0,1)内可导,且f(0)=0f(0)=0f(0)=0,f(1)=2f(1)=2f(1)=2,证明在(0,1)(0,1)(0,1)内至少存在一点ξ\xiξ,使得f′(ξ)=2ξ+1f'(\xi)=2\xi + 1f′(ξ)=2ξ+1成立.
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