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)|≤1/2(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)|≤1/2(x∈[0,1]).

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答案由泰勒公式得f(0)=f(x)-f’(x)x+1/2f"(ξ1)x2,ξ1∈(0,x),f(1)=f(x)+f’(x)(1-x)+1/2f"(ξ2)(1-x)2,ξ2∈(x,1),两式相减,得f’(x)=1/2f"(ξ1)x2-1/2f"(ξ2)(1-x)2.两边取绝对值,再由|f"(x)|≤1,得|f’(x)|≤1/2[x2+(1-x)2]=(x-1/2)2+1/4≤1/2.

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