设函数y=f(x)定义在R上,当x>0时f(x)〉1,且对于任意实数a,b∈R,有f(a+b)=f(a)f(b),判断f（x）在R上的单调

要学会透视这类抽象函数,估计是y=a^x模型,且a>1x2>x1,x2-x1>1f(1+0)=f(1)f(0),f(1)>1,:.f(0)=1当x>0,f(x-x)=f(x)f(-x)=f(0)=1,:.f(-x)=1/f(x)>0.:.x∈R,f(x)>0.f(x2)=f(x2-x1+x1)=f(x2-x1)f(x1)又 f(x2-x1)>1,:.f(x2...