(1)以m=n=0代入,得:
f(0)=f(0)+f(0)
f(0)=0
(2)以m=x、n=-x代入,得:
f(0)=f(x)+f(-x)
f(x)+f(-x)=0
即:f(-x)=-f(x)
所以函数f(x)是奇函数
(3)设:x1>x2,则:
f(x1)-f(x2)
=f[(x1-x2)+x2]-f(x2)
=[f(x1-x2)+f(x2)-f(x2)
=f(x1-x2)
因为x1-x2>0,则:
f(x1-x2)
(1)以m=n=0代入,得:
f(0)=f(0)+f(0)
f(0)=0
(2)以m=x、n=-x代入,得:
f(0)=f(x)+f(-x)
f(x)+f(-x)=0
即:f(-x)=-f(x)
所以函数f(x)是奇函数
(3)设:x1>x2,则:
f(x1)-f(x2)
=f[(x1-x2)+x2]-f(x2)
=[f(x1-x2)+f(x2)-f(x2)
=f(x1-x2)
因为x1-x2>0,则:
f(x1-x2)