Problem with defining the range of $f(x)=\ln(x+\sqrt{x^2+1})$

Question

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

Personal work: We have to solve the inequality: $x+\sqrt{x^2+1}>0$

I've thought of raising both sides powers' to $^2$ so the root can go away. So, we have: $$x^2+x^2+1>0 \iff 2x^2+1>0$$ And it's always positive. Problem? I need to find the $D_f$ of $f(x).$

Answer 1

In your title you are asking for the range of $f(x)$, however it seems you are attempting to solve for the domain of $f(x)$, so I will provide such a solution. If it is not the case, please clarify.

The logarithm function is defined for $x>0$, so we want to solve the inequality $x+\sqrt{x^2+1}> 0$. This is equivalent to solving \begin{align}\sqrt{x^2+1}> -x \tag{1} \end{align} and we have to consider two cases: when $x$ is positive and when $x$ is negative. So

• Case $x > 0: $ notice the right hand side of $(1)$ becomes negative, and the left hand side is positive, since $x^2+1$ is always positive. Hence it is true for all $x>0$.

• Case $x\leq0: $ Then we have that both the RHS and LHS are non-negative, so we have that \begin{align}\sqrt{x^2+1}> -x \iff\left(\sqrt{x^2+1}\right)^2> (-x)^2 \iff x^2+1>x^2 \iff1>0,\end{align} which is true for all $x$.

Hence the domain of $f(x)=\ln(x+\sqrt{x^2+1})$ is all the real numbers, $\mathbb{R}$.

Answer 2

HINT: $\sqrt{x^2+1}>\sqrt{x^2}=|x|$, so the domain (against the title, as suggest notation and your attempt) is...

(Your method assumes nonnegative $x$).

Answer 3

WLOG $x=\cot2y$ where $0<2y\le\pi$

$$x+\sqrt{x^2+1}=\cot2y+\csc2y=\cot y$$ for $1+\cos2y\ne0\iff2y\ne\pi$

$0<\cot y<\infty$

Answer 4

No issues if $x\geq 0$. If $x<0$ then $\sqrt{x^2+1}+x$ is positive anyway, since $\sqrt{x^2+1}>|x|$, hence $f(x)=\log(x+\sqrt{x^2+1})$ is defined for any $x\in\mathbb{R}$. We may also notice that $$ f(\sinh u) = \log(\sinh u+\cosh u) = \log(e^u) = u $$ hence $f(x)=\text{arcsinh}(x)$ is the inverse function of $\sinh$, which is bijective over $\mathbb{R}$.