设f(x)是定义在R上的函数,且对于任意x,y∈R,恒有f(x+y)=f(x)f(y),且当x>0时,f(x)>1.证明:(1)当f(0)=1
问题描述：
设f(x)是定义在R上的函数,且对于任意x,y∈R,恒有f(x+y)=f(x)f(y),且当x>0时,f(x)>1.证明:(1)当f(0)=1
（1）当f(0)=1，且x
答
f(x+y) = f(x)f(y)for x f(x) ( -x > 0,=> f(-x)>1 )put y=xf(2x) = {f(x)}^2 > 0ie f(2x) >0 for all xf(x) > 0 for all xx => y = x+a (a>0)=> f(y) = f(x+a)= f(x)f(a)> f(x) (f(a) > 1 )f is increasing