Range of $ f(x)=\ln(x+\sqrt{x^2+1}) $ 
Find the range of $ f(x)=\ln(x+\sqrt{x^2+1}) $

We already know that the domain of $f$ is $\mathbb{R}$ so to find the range of $f$ we solve $y=f(x)$ in terms of $x$ and we try to limit the values of $y$.
$y=f(x) \Rightarrow y=\ln(x+\sqrt{x^2+1}) \Rightarrow e^y=x+\sqrt{x^2+1} \Rightarrow e^y-x=\sqrt{x^2+1}$
But we know that $\sqrt{x^2+1} > 0 \Rightarrow e^y-x>0 \Rightarrow e^y>x\quad (1)$
Then by squaring both sides we get $e^{2y}-2xe^y+x^2=x^2+1 \Rightarrow x=\dfrac{e^y-e^{-y}}{2}$
We know that $x \in \mathbb{R} \Rightarrow \dfrac{e^y-e^{-y}}{2}\in \mathbb{R}$ which does not provide any limitation for $y$.
Does $(1)$ provide any limitation for $y$? if not why?
I understand that $g(x)=\dfrac{e^x-e^{-x}}{2}$ has $\mathbb{R}$ as its domain and $g$ is essentially the inverse function of $f$ but I dont know how to formally show that the range of $f$ is $\mathbb{R}$
 A: You can easily evaluate
$$
\lim_{x\to\infty}f(x)=\infty
$$
but also, with $x=-t$,
$$
\lim_{x\to-\infty}f(x)=
\lim_{t\to\infty}\ln(-t+\sqrt{t^2+1})=
\lim_{t\to\infty}\ln\frac{1}{t+\sqrt{t^2+1}}=-\infty
$$
The intermediate value theorem ends the task.

Without calculus, one can find the inverse function: if $y=\ln(x+\sqrt{x^2+1})$, then
\begin{align}
e^y&=x+\sqrt{x^2+1},
\\
e^{-y}&=\frac{1}{\sqrt{x^2+1}+x}=\frac{\sqrt{x^2+1}-x}{x^2+1-x^2}=
\sqrt{x^2+1}-x
\end{align}
and so
$$
2x=e^{y}-e^{-y}
$$
which is usually written $x=\sinh y$.
Now this function is defined with no restriction on $y$:
$$
g(y)=\frac{e^{y}-e^{-y}}{2}=\sinh y
$$
This precisely means that the range of $f$ is $\mathbb{R}$.
A: The domain of a function is the set of input or argument values for which the function is real and defined.
Solving
$$x+\sqrt{x^2+1}>0$$
it is true $\forall x\in \mathbb{R}$. The function has no undefined points nor domain constraints. Therefore, the domain of $f$ is $\mathbb{R}$.
For the range you should find the codomain=range. In fact if a function $f(x)$ is mapping $x\mapsto y$, then the inverse function of $f(x)$ maps $y$ back to $x$.
If 
$$x=\ln \left(y+\sqrt{y^2+1}\right), \quad x\leftrightarrow y$$
solving to $x$ we have
$$\ln \left(e^x\right)=\ln \left(y+\sqrt{y^2+1}\right)$$
hence for the propriety $\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\quad \Rightarrow \quad f\left(x\right)=g\left(x\right)$, we have
$$e^x=y+\sqrt{y^2+1}$$
After squaring 
$$e^{2x}=y^2+2y\sqrt{y^2+1}+y^2+1$$
$$e^{2x}-1=\color{red}{2y^2+2y\sqrt{y^2+1}}=2y(y+\sqrt{y^2+1}) \tag{*}$$
Hence being $x=\ln \left(y+\sqrt{y^2+1}\right)\rightarrow y+\sqrt{y^2+1}=e^x$ we have
$$y=\frac{1}{2}e^{-x}\left(e^{2x}-1\right)$$
and the domain of this function (i.e. range) is $\mathbb R$.
