【高一数学】一道反函数化简题》》》

(x+1)/(x-1)=(x-1+2)/(x-1)=(x-1)/(x-1)+2/(x-1)=1+2/(x-1)
x>1,x-1>0
1/(x-1)>0
2/(x-1)>0
所以1+2/(x-1)>1
所以y>ln1=0
所以反函数定义域是x>0
e^y=1+2/(x-1)
2/(x-1)=e^y-1
x-1=2/(e^y-1)
x=1+2/(e^y-1)=(e^y-1+2)/(e^y-1)=(e^y+1)/(e^y-1)
所以反函数
y=(e^x+1)/(e^x-1),其中x>0

e^y=(x+1)/(x-1),
(x-1)e^y=x+1
(e^y-1)x=1+e^y
x=(1+e^y)/(e^y-1)
所以反函数为y=(1+e^x)/(e^x-1)