是問(wèn)奇偶性嗎?
f(x)=lg[(x+1)/(x-1)]
由(x+1)/(x-1)>0得定義域
(-∞,-1)U(1,+∞)
f(-x)+f(x)=lg[(-x+1)/(-x-1)]+lg[(x+1)/(x-1)]
=lg[(x-1)/(x+1)*(x+1)/(x-1)]
=lg1=0
∴f(-x)=-f(x)
∴f(x)是奇函數(shù)
g(x)=e^x+1/e^x定義域?yàn)镽
g(-x)=e^(-x)/1/e^(-x)=1/e^x+e^x=g(x)
∴g(x)是偶函數(shù)