由于:f(0+0)=f(0)*f(0)得:f(0)=[f(0)]^2得:f(0)=0,或f(0)=1若f(0)=0,则对任何x,有:f(x)=f(x+0)=f(x)*f(0)=0因而对任何x:f'(x)=0命题成立.若f(0)=1,则:[f(x+h)-f(x)]/h=[f(x)*f(h)-f(x)]/h=f(x)*[f(h)-1]/h=f(x)*[f(h...