Deriving the inverse of $\cosh$ I am trying to derive the inverse of $f(x)=\cosh x$ when I restrict the domain so that I have $[0,\infty)$ as the domain and $[1,\infty)$ as the range of $f(x)$, respectively.
In my algebra, I got to the point where
$$e^x=y\pm\sqrt{y^2-1}$$
I noted that


*

*$y\pm\sqrt{y^2-1}>0$ so the fact that $e^x$ should always be positive did not help.

*Since $x\geq0$, I must have $e^x\geq1$ and either of the inequalities $y\pm\sqrt{y^2-1}\geq1$ yields $y\leq1$ but this contradicts the range of $f(x)$.


Therefore I have no idea what I should do next and what reasoning I am supposed to give so that $e^x=y+\sqrt{y^2-1}$.
 A: You want something of the form $x = f(y)$. Now look at $f(y) = \ln\left(y \pm \sqrt{y^2 - 1}\right)$. Do you want that function to be positive or negative?
A: $$y+\sqrt{y^2-1}\ge1\iff\sqrt{y^2-1}\ge1-y\ \ \ \  (1)$$
As $\sqrt{y^2-1}\ge0,$
$(1)$ is always true if $1-y\le0\iff y\ge1$
Else i.e, for $y<1,$
Squaring we get $y^2-1\ge(1-y)^2\iff y\ge1$ which is contradictory 
A: Ok, I think that I found a way to make sense of this. If we have
$$e^x=y\pm\sqrt{y^2-1}$$
Then taking natural logarithm of both sides of the equality gives
$$x=\log\left(y\pm\sqrt{y^2-1}\right)$$
where $x\geq0$ and $y\geq1$ but we note that
$$
\begin{align}
y-\sqrt{y^2-1}&=\left(y-\sqrt{y^2-1}\right)\left(
\frac{y+\sqrt{y^2-1}}{y+\sqrt{y^2-1}}\right)\\
&=\frac{y^2-y^2+1}{y+\sqrt{y^2-1}}\\
&=\frac{1}{y+\sqrt{y^2-1}}
\end{align}
$$
Therefore we must have
$$x=\pm\log\left(y+\sqrt{y^2-1}\right)$$
When $y\geq1$, we have $y+\sqrt{y^2-1}\geq1$; that is $\log\left(y+\sqrt{y^2-1}\right)\geq0$ but $x\geq0$ so we conclude that
$$x=\log\left(y+\sqrt{y^2-1}\right)$$
