Singularities of $\operatorname{arctanh}(x)$

Question

The function $\operatorname{arctanh}(x)$ is undefined for all $\vert x\vert\ge 1$.

Does this mean, because the graph shoots up to infinity at $1$ and $-1$, that these are the singularities of the function,

or would this not be true since it actually breaks down for every point after as well?

For example, $\tan(x)$ has singularities at $\pi\over 2$, but is also defined before and after this point, unlike $\operatorname{arctanh}$.

Answer 1

$$\tanh x=\frac{e^{2 x}-1}{e^{2 x}+1}$$ The range of $\tanh x$ is $(-1,1)$ therefore the domain of $\text{arctanh }x$ is $(-1,1)$.

Solving $\frac{e^{2 x}-1}{e^{2 x}+1}=y$ we get $x=\frac{1}{2} \log \left(\frac{1+y}{1-y}\right);\;|y|<1$.

The graph of arctanh x is the symmetric of the graph of $\tanh x$ wrt the line $y=x$. Domain and range of the function and its inverse swap.

See the picture below

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Answer 2

The hyperbolic tangent is a bijective (strictly increasing) function $\tanh:\Bbb{R}\to (-1,1)$. Note that here the domain is all of $\Bbb{R}$, and the image is $(-1,1)$. This is in contrast to the tangent function whose domain is not all of $\Bbb{R}$, and the image is all of $\Bbb{R}$, and the tangent function is only injective once you restrict the domain appropriately (for example to $(-\frac{\pi}{2},\frac{\pi}{2})$).

The reason for this is that $\tanh$ is defined as $\frac{\sinh}{\cosh}$, and $\cosh(x)\geq 1$ for all $x\in\Bbb{R}$. Hence, the quotient never goes off to $\pm\infty$, unlike for $\tan$. So, yes, the inverse $\text{argtanh}: (-1,1)\to\Bbb{R}$ is also a strictly increasing function with $\lim\limits_{x\to -1^-}\text{argtanh}(x)=-\infty$ and $\lim\limits_{x\to 1^+}\text{argtanh}(x)=\infty$. I suggest using wolfram alpha/desmos to sketch the graphs of $\tanh$ and $\text{argtanh}$. So, sure if you wish, you can call $-1$ and $1$ singularities of $\text{argtanh}$, because it "blows up" there, and hence there is no way of continuously extending the definition outside of $(-1,1)$.