Inverse Function of $ \frac{1}{2} ( e^x - e^{-x} ) $

Question

Should find the inverse of:

$$ f(x) = \frac{1}{2} ( e^x - e^{-x} ) $$

I tried a lot. But I don't know how to proceed on $$ 2x = \frac{(e^y)^2 - 1}{e^y} $$

Writing $ e^{-y} $ as $ \frac{1}{e^y} $ is right? I know somewhere I need to use the $ ln $ but I dont know when

Answer 1

The quantity $z=\mathrm e^y$ solves the quadratic $z^2-2xz-1=0$. Surely you can write down the roots of the quadratic and from there, deduce $y$ (just remember that $z\gt0$).

Answer 2

Let $e^x=z$. Thus, you have:

$$y=\frac{1}{2} (z-\frac{1}{z})$$

Solve the resulting quadratic for $z$ and then set $x = ln(z)$.