y=f(x)=[e^x-e^(-x)]/[e^x+e^(-x)]
=[e^(2x)-1]/[e^(2x)+1]
于是e^(2x)-1=ye^(2x)+y
即 e^(2x)=(y+1)/(1-y)
兩邊取對數(shù)有
2x=ln[(y+1)/(1-y)]
即 x=0.5ln[(y+1)/(1-y)]=0.5[ln(y+1)-ln(1-y)]
把自變量換成x即為
f(x)的反函數(shù)為 y=0.5[ln(x+1)-ln(1-x)]