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

Question

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 $\mathbb{R}$

Answer 1

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 $\mathbb{R}$.

Answer 2

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 $\mathbb R$.