Question

求这道数学涵数题怎么解？

Answer 1

令a=b=0
则f(0)=[f(0)]^2
f(0)[f(0)-1]=0
因为f(x)为非零函数
则f(0)=1
令b=-a>0
f(0)=f(a)f(-a)=1
a<0，则f(a)>1
f(-a)=1/f(a)
则0<f(-a)<1
即当x>0时，0<f(x)<1
所以f(x)>0
令a=x1-x2
b=x2
且x1<x2
f(x1-x2+x2)=f(x1-x2)*f(x2)
f(x1)=f(x1-x2)*f(x2)
[x1-x2<0
f(x1-x2)>1
且f(x1)>0
f(x2)>0]
f(x1)/f(x2)>1
f(x1)>f(x2)
即得当x1<x2
f(x1)>f(x2)
所以f(x)为减函数