设f(x)在[0,1]上二阶连续可导且f(0)=f(1),又|f"(x)|≤M,证明:|f’(x)|≤M/2.

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问题 设f(x)在[0,1]上二阶连续可导且f(0)=f(1),又|f"(x)|≤M,证明:|f’(x)|≤M/2.

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

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