设f(x)在[0,1]上二阶可导,且|f"(x)|≤1(x∈[0,1]),又f(0)=f(1),证明: |f’(x)|≤(x∈[0,1]).

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问题 设f(x)在[0,1]上二阶可导,且|f"(x)|≤1(x∈[0,1]),又f(0)=f(1),证明:
|f’(x)|≤(x∈[0,1]).

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答案由泰勒公式得, f(0)=f(x)-f’(x)x+[*]f"(ξ1)x21∈(0,x), f(1)=f(x)+f’(x)(1-x)+[*]f"(ξ2)(1-x)22∈(x,1) 两式相减,得f’(x)=[*]f"(ξ1)x2-[*]f"(ξ2)(1-x)2 两边取绝对值,再由|f"(x)|≤1,得 [*]

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