Messed up on the inverse of $\cosh(x)$ but I don't know where I went wrong

Question

If someone can tell me where I went wrong, that would be extremely appreciated :)

The function $f:[0,\infty)\mapsto[1,\infty)$ defined by $f(x)=\cosh x$ is bijective. Find a formula for $f^{-1}.$

If $y=\cosh x$, then \begin{align} & y = \frac{1}{2}\left(e^x+e^{-x}\right) \\[5pt] \implies & 2y = e^x+e^{-x} \\[5pt] \implies & 2ye^x = e^x(e^x+e^{-x}) \\[5pt] \implies & 2ye^x = e^{2x} + e^0 \\[5pt] \implies & 2ye^x = e^{2x}+1 \\[5pt] \implies & -e^{2x}+2ye^x - 1 = 0 \\[5pt] \implies & e^x = \frac{-2y \pm \sqrt{4y^2-4(-1 \times -1)}}{-2} \\[5pt] \implies & e^x = y + \sqrt{-2y+2} \text{ as $e^x>0$} \\[5pt] \implies & x = \ln(y+\sqrt{-2y+2}) \\[5pt] \implies & \cosh^{-1}x = \ln(y+\sqrt{-2y+2}) \end{align}

Answer 1

We have \begin{align} e^x &= \frac{-2y\pm\sqrt{(2y)^2-4}}{-2} \\[5pt] &= \color{red}{y\pm\frac{\sqrt{4y^2-4}}{-2}} \\[5pt] &= y\pm\frac{\sqrt{4}\sqrt{y^2-1}}{-2} \\[5pt] &= y\pm \frac{2\sqrt{y^2-1}}{-2} \\[5pt] &= y \mp\sqrt{y^2-1} \, . \end{align} Since $x$ is nonnegative, we know that $e^x \geq 1$. Hence, $y\mp\sqrt{y^2-1} \geq 1$. Notice that it has to be greater than or equal to $1$, not $0$ as you wrote in your question. Since $0<y-\sqrt{y^2-1}\leq 1$ for $y\geq1$, the negative branch doesn't work. Hence, $e^x=y\color{red}{+}\sqrt{y^2-1}$, and the result follows.

Answer 2

If $y=\cosh x$, then \begin{align} & y = \frac{1}{2}\left(e^x+e^{-x}\right) \\[5pt] \implies & 2y = e^x+e^{-x} \\[5pt] \implies & 2ye^x = e^x(e^x+e^{-x}) \\[5pt] \implies & 2ye^x = e^{2x} + e^0 \\[5pt] \implies & 2ye^x = e^{2x}+1 \\[5pt] \implies & -e^{2x}+2ye^x - 1 = 0 \\[5pt] \implies & e^x = \frac{-2y \pm \sqrt{4y^2-4(-1 \times -1)}}{-2} \\[5pt] \color{red} \implies & \color{red}{e^x = y + \sqrt{-2y+2} \text{ as $e^x>0$}}\\[5pt] \end{align}

Indeed you should get:

\begin{align} \implies & e^x = y ± (-1) \sqrt{y^2 -1} \\[5pt] \end{align}

Then, following the justification given in the comments by Soheil, you can conclude that since $x \in [0,\infty)$, you end $$\cosh^{-1}(x)= \log(x + \sqrt{x^2-1})$$