(1)由题意f(0+3)=f(0)+f(3)得f(0)=0
设s=-t,得f(s+t)=f(0)=0=f(s)+f(t)
得-f(t)=f(s)=f(-t)
所以f(x)为奇函数
设s,t>0
则有s+t>s
f(s+t)=f(s)+f(t)得f(s+t)-f(s)=f(t)
又因为x>0时f(x)
(1)由题意f(0+3)=f(0)+f(3)得f(0)=0
设s=-t,得f(s+t)=f(0)=0=f(s)+f(t)
得-f(t)=f(s)=f(-t)
所以f(x)为奇函数
设s,t>0
则有s+t>s
f(s+t)=f(s)+f(t)得f(s+t)-f(s)=f(t)
又因为x>0时f(x)