设f(x)在[0,1]可导,f(0)=0,|f’(x)|≤|f(x)|,则f(1)=__________.

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问题 设f(x)在[0,1]可导,f(0)=0,|f’(x)|≤|f(x)|,则f(1)=__________.

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答案0

解析 当0<x<1时,
|f(x)|=|f(x)-f(0)|=|f’(ξ1)|x≤|f(ξ1)|·x
=|f(ξ1)-f(0)|·x=|f’(ξ2)|·ξ1·x
≤|f(ξ2)|·x2≤…≤|f(ξn)|·xn
≤M·xn0
其中0<ξn<...<ξ2<ξ1,M为|f(x)|在[0,1]上的最大值,故0<x<1时,f(x)=0,又由f(x)在x=1处左连续,有f(1)==0.
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