设f(x)在[0,+∞)上连续,在(0,+∞)内可导且满足f(0)=0,f(x)≥0,f(x)≥f’(x)(x>0),求证:f(x)≡0.

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问题 设f(x)在[0,+∞)上连续,在(0,+∞)内可导且满足f(0)=0,f(x)≥0,f(x)≥f’(x)(x>0),求证:f(x)≡0.

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答案由f’(x)-f(x)≤0, 得 e-x[f’(x)-f(x)]=[e-xf(x)]’≤0. 又 f(x)e-xx=0=0, 则 f(x)e-x≤f(x)e-xx=0=0.进而 f(x)≤0(x∈[0,+∞)),因此 f(x)≡0([*]x∈[0,+∞)).

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