What is the domain of $(\arctan(x-1))^{1/(x-3)}$ I would like to know how to solve the domain for this function in real numbers:
$(\arctan(x-1))^{1/(x-3)}$
I usually know how to get to the domain, but this one caught me by surprise, where I don't even know where to begin or how to deal with it.
 A: Exponentiating with an arbitrary real exponent is only defined when the base is positive.
Thus you need $\arctan(x-1)>0$, besides the obvious $x-3\ne0$.
The first inequality is equivalent to $x>1$, so the domain is $(1,3)\cup(3,\infty)$.
Exponentiation $0^t$ might be considered defined, but only with positive exponent (pretty useless anyway). Not this case.
Some people also consider exponentiation defined for negative base, provided the exponent is rational, with a fairly complicated rule based on the expression of the rational with coprime numerator and denominator. I hope it's not your case.
A: This is a comment about your problem in which I have some doubts. Putting $$y=(\arctan(x-1))^{1/(x-3)}\tag 1$$ you can consider the equivalent expression (where $y\ne 0$) $$y^{x-1}=y^2\arctan(x-1)\tag 2$$
 From $(1)$ it easily follows that  $x=1$ and $x=3$ are not in the domain (you have $\arctan 2\approx 63,434949^{\circ}$ and $y$ becomes infinite in both cases). Besides, no problem for all element of $D_1=]1,3[\space\cup\space]3, +\infty[$ and plotting with Desmos the corresponding curve $(1)$ (or $(2)$ if you want because both have the same graphic), this $D_1$ is apparently given as the required domain.

I have put expression $(2)$ in case it could give some help to really finish because $D_1$ above is not the whole domain. We have for example:
$x=0$ gives $$(\arctan (-1))^{-\frac13}=\dfrac{1}{\sqrt[3]{\arctan(- 1)}}=\dfrac{-1}{\sqrt[3]{\arctan( 1)}}=\dfrac{-1}{\sqrt[3]{\frac{\pi}{4}}}\approx -1.083852$$ so $x=0$ belongs to the domain but $0\notin D_1$ There is a lot of other values of $x$ for which the function is well defined, for example:
$$f(0.1)\approx 1.113149\\f(0.5)\approx 1.359955\\f(-0.9)\approx 0.978994\\f(-6.3)\approx-0.961934\\f(-6.2)\text { is undefined }$$ For $x=-1$ we have $f(-1)=\dfrac{1}{\sqrt[4]{-\arctan 2}}\notin\mathbb R$ and similarly the function is undefined for $x=-(2n+1)$ but for example for $x=-2$ one has $f(-2)\approx -0.956498$.
We can see that the domain $D_1$ apparently given by Desmos is not complete. But then what is this domain? Here I stop this comment.
