设函数y=f(x)对任意x∈R都有f(x+1)=af(x)(a>0).
设函数y=f(x)对任意x∈R都有f(x+1)=af(x)(a>0).
已知函数y=f(x),满足:对任意a,b∈R,都有af(a)+bf(b)>af(b)+bf(a).(1)试证明:f(x)
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数列&函数·问题设函数f(x)满足f(0)=1,且对任意x,y∈R,都有f(xy+1)=f(x)f(y)-f(y)-x+
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设函数f(x)满足f(0)=1,且对任意x,y属于R,都有f(xy+1)=f(x)乘f(y)减f(y)减x加2.求f(x
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