由f(x)>=f'(x)得,e^(-x)f(x)-f'(x)e^(-x)=[e^(-x)f(x)]'>=0,所以e^(-x)f(x)
证明函数恒等式设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
=0,所以e^(-x)f(x)"}}}'>
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