Help finding inverse of $f(x)=\frac{e^x-e^{-x}}{2}$ I'm trying to find the inverse of $f(x)=\frac{e^x-e^{-x}}{2}$. My textbook says $f^{-1}(x)=\ln(x+\sqrt{x^2+1})$, but I haven't been able to get that answer. Switching $x$ and $y$, I tried solving for $y$ a few different ways, with limited success. Initially, I tried:

$$x=\frac{e^y-e^{-y}}{2}\\
2x=e^y-e^{-y}\\
2x=e^{-y}(e^{2y}-1)\\
\ln(2x)=\ln(e^{-y}(e^{2y}-1))\\
\ln(2x)=\ln(e^{-y})+\ln(e^{2y}-1)\\
\ln(2x)=-y+\ln(e^{2y}-1)\\
\ln(2x)=-y+\ln((e^y-1)(e^y+1))\\
\ln(2x)=-y+\ln(e^y-1)+\ln(e^y+1)$$

I can't see anything else I could do with that. I also tried squaring both sides back on that first line, but it lead me down another dead-end. The only thing that seemed to yield an answer was:

$$x=\frac{e^y-e^{-y}}{2}\\
2x=e^y-e^{-y}\\
2x=e^y-\frac{1}{e^y}\\
2x=\frac{e^{2y}-1}{e^y}\\
2xe^y=e^{2y}-1\\
0=e^{2y}-2xe^y-1\\
(e^y)^2-2x(e^y)-1=0$$
Then, the quadratic equation yields,
$$e^y=\frac{2x\pm \sqrt{4+4}}{2}=\frac{2x\pm 2\sqrt{2}}{2}\\
e^y=x\pm \sqrt{2}\\
y=\ln(x\pm \sqrt{2})$$

I'm not sure what to do about the $\pm$. I imagine one would be eliminated to make the domain of $f^{-1}$ match the range of $f$, but I'm not sure how to determine which (actually, the textbook says the domains and ranges of both $f$ and $f^{-1}$ are $(-\infty ,\infty)$. More importantly, the book's answer, $f^{-1}(x)=\ln(x+\sqrt{x^2+1})$ doesn't match mine! Help! :)
You have
$$
(e^y)^2 - 2x(e^y) - 1 = 0
$$
From this you get
$$
e^y = \frac{2x\pm \sqrt{{\color{red}{4x}^2} + 4}}{2}= x\pm \sqrt{x^2 + 1}
$$
Here you only get the $+$ in the $\pm$ because otherwise the numeerator is negative and you have $e^y&gt;0$ (assuming here that we are just talking about real numbers). So
$$
y = \ln(e^y) = \dots
$$
enter image description here](https://i.stack.imgur.com/IxpIA.jpg)
Some parts of ans has been done directly without detailed steps