设f(x)在[0,1]上阶连续可导且f(0)=f(1),又|f’’(x)|≤M,证明:

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问题 设f(x)在[0,1]上阶连续可导且f(0)=f(1),又|f’’(x)|≤M,证明:

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答案由泰勒公式得 f(0)=f(x)+f’(x)(0-x)+[*](0-x)2,ξ∈(0,x), f(1)=f(x)+f’(x)(1-x)+[*](1-x)2,η∈(x,1), 两式相减得 f’(x)=[*][f"(ξ)x2-f’’(η)(1-x)2], 取绝对值得 |f’(x)|≤[*][x2+(1-x)2], 因为x2≤x,(1-x)2≤1-x,所以x2+(1-x)2≤1,故|f’(x)|≤[*]

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