need help!!!!!!!!!! Find the inverse function of y=[e^(2x)-1] / [e^(2x)+1]?

Answer 1

x=(e^2y-1)/(e^2y+1)
x(e^2y+1)=e^2y-1
xe^2y+x=
xe^2y-e^2y=-x-1
e^2y(x-1)=
e^2y=-(x+1)/(x-1)
lne^2y=ln(x+1)(x-1)
2y=ln(x+1) + ln (x-1)
y=1/2[ln(x+1)+ln(x-1)

Answer 2

Your function is y = tanh(x). Solve for x to get

x = arctanh(y)

then interchange x and y to get the inverse function

y = arctanh(x)

Answer 3

let z = e^2x

y = (z - 1)/(z + 1)
y = 1 - 2/(z + 1)
2/(z + 1) = 1 - y
2/(1 - y) = z + 1
z = 2/(1 - y) - 1
e^2x = 2/(1 - y) - 1
2x = ln(2/(1 - y) - 1)
x = ln(2/(1 - y) - 1) / 2