Question

非零函数F(X),任意a b 函数F[a+b]=F[a]*F[b],当X小于0时,F（x）大于1 求证f（x)为减函数.

Answer 1

a=b=0时

f(0)=f(0)*f(0)

f(0)[1-f(0)]=0

f(x)非0,所以 1-f(0)=0,即 f(0)=1

a=-b≠0时

f(0)=f(a)* f(-a)=1

f(x)*f(-x)=1

f(x)=1/f(-x)

故f(x)与 f(-x)同号

因为 f(x)非0,x1

当 x>0时,-x1,则 f(x) >0

故 f(x) 恒大于0

令 x1>x2

f(x2-x1)=f(x2)*f(-x1)

=f(x2)*1/f(x1)

=f(x2)/f(x1)

x2-x11

即 f(x2)/f(x1) > 1

f(x2)>f(x1) 结论得证