设f(x)是定义在R上的函数,且对于任意x.y∈R恒有f(x+y)=f(x)f(y),且x>0时,f(x)>1.证明
设f(x)是定义在R上的函数,且对于任意x.y∈R恒有f(x+y)=f(x)f(y),且x>0时,f(x)>1.证明
(1)当f(0)=1时,且x
(1)当f(0)=1时,且x
数学人气:125 ℃时间:2019-10-11 17:35:13
优质解答
f(0)=f(x)+f(-x),因为x>0时,f(x)>1,f(0)=1,根据反函数图像易知,0
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