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)一f(x)≤0,得e-x[f’(x)一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,+∞)).

解析 因f(x)≥0,若能证f(x)≤0,则f(x)≡0.因f(0)=0,若能证f(x)单调不增或对某正函数R(x),R(x)f(x)是单调不增的,这只需证f’(x)≤0或[R(x)f(x)]’≤0.由所给条件及积分因子法的启发,应采取后一种方法.
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