设函数y=f(x)定义在R上,对与任意实数m;n,恒有f(m+n)=f(m)f(n).当x>0时,0<f(x)<1.①求证:f(0)=1且当x<0时,f(x)>1;②求证;f(x)在R上递减.

问题描述:

设函数y=f(x)定义在R上,对与任意实数m;n,恒有f(m+n)=f(m)f(n).当x>0时,0<f(x)<1.
①求证:f(0)=1且当x<0时,f(x)>1;
②求证;f(x)在R上递减.

f(0+0)=f(0)*f(0) 所以:f(0)=1或f(0)=0 f(1+0)=f(1)*f(0),如果f(0)=0,等式不成立.所以f(0)=1.--------------------------------------------------f(0)=f(-x)f(x)=1取a1所以:当x<0时,f(x...