# 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 $e^y>0$ (assuming here that we are just talking about real numbers). So $$y = \ln(e^y) = \dots$$

• Wow, what a dopey mistake! Thanks.
– ivan
Mar 24 '13 at 15:00
• @ivan: No problem. Glad to help. Mar 24 '13 at 15:03

$$\text{If }x=\frac{e^y-e^{-y}}2,2xe^y=e^{2y}-1$$

$$\implies e^{2y}-2xe^y-1=0$$ which is a Quadratic Equation in $e^y$

$$\text{ So,}e^y=\frac{2x\pm\sqrt{(2x)^2-4\cdot1\cdot(-1)}}{2\cdot1}=x\pm\sqrt{x^2+1}$$

For real $y,e^y>0\implies e^y=x+\sqrt{x^2+1}$ as $x<\sqrt{x^2+1}$

enter image description here](https://i.stack.imgur.com/IxpIA.jpg) Some parts of ans has been done directly without detailed steps