Hi,

My problem is to find the inverse of $\displaystyle y = f(x) = \frac{1}{2}(e^x - \frac{1}{e^x}) $.

I can get up to using $\displaystyle v$ for $\displaystyle e^x$, and that $\displaystyle v > 0$

$\displaystyle 2yv = v^2 -1$

$\displaystyle 0 = v^2 - 2yv - 1$

The using the quadratic formula I get $\displaystyle v = \frac{2y \pm \sqrt{(-2y)^2 + 4}}{2}.$

Here I am stuck

Sorry for bad format.