Question

已知函数f(x)=lg(x+1/x-1)求值域还有奇偶性！急求~~~要详细解释！！

Answer 1

(x+1)/(x-1)>0
x<-1或x>1
定义域关于原点对称
设u=(x+1)/(x-1)
=(x-1+2)/(x-1)
=1+2/(x-1)
∴u≠1
∴lgu≠0
值域是(-∞,0)∪(0,+∞)
f(-x)
=lg[(-x+1)/(-x-1)]
=lg[(x-1)/(x+1)]
=lg[(x+1)/(x-1)]^(-1)
=-lg[(x+1)/(x-1)]
=-f(x)
∴f(x)是奇函数
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