已知函数f(x)的定义域是(0,正无穷),且满足f(xy)=f(x)+f(y),f(1/2)=1,如果对于0<x<y,都有f(x)>f(y) .解不等式f(-x)+f(3-x)≥-2
人气:427 ℃ 时间:2019-08-19 17:53:39
解答
f(xy)=f(x)+f(y)中,令x=y=1得
f(1)=0,
f(xy)=f(x)+f(y)中,再令y=1/x
得f(1)=f(x)+f(1/x)=0
f(1/2)=1,所以f(2)=-1,f(4)=f(2)+f(2)=-2
f(-x)+f(3-x)≥-2中,先要有x
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