函数f(x)=lg【2/(x+1) —1】的图像关于（）对称?

f(x)=lg{（2/x+1）-1} =lg[(2-x-1)/(x+1)] =lg[(1-x)/(x+1)] f(-x)=lg[(1-(-x))/(-x+1)] =lg[(x+1)/(1-x)] =lg[(1-x)/(1+x)]^-1 =-lg[(1-x)/(x+1)] -f(-x)=lg[(1-x)/(x+1)] 所以f(x)=-f(-x) 所以函数y=lg{（2/x+1）-1}是奇函数,关于原点对称