Is Mathway wrong about the domain of $\sqrt{x-13}\sqrt{x-13}$? Say $f(x) = \sqrt{x-13}$. You're trying to find the domain of $(ff)(x)$ where $$(ff)(x) = \sqrt{x-13}\sqrt{x-13}.$$
However, $\sqrt{x-13}\sqrt{x-13} = x-13 $, right? 
The domain of $x-13$ in interval notation: $ (-\infty, +\infty) $
The domain of $\sqrt{x-13}$ in interval notation: $[13, +\infty) $
The domain of $\sqrt{x-13}\sqrt{x-13}$ in interval notation: 

The expression should first simplify to $x-13 $,  and we already found
  out that   "the domain of $x-13$ in interval notation: $(-\infty,+\infty)$"

So, then, why does Mathway say that the domain of $\sqrt{x-13}\sqrt{x-13}$ is $[13, +\infty)$? 
 A: My guess is that the function $\sqrt{x}:\mathbb{R}_+\to\mathbb{R}$ is defined only for $x\geq 0$. Thus, $x<13$,  the function $\sqrt{x-13}$  is undefined. Therefore, $\sqrt{x-13}\sqrt{x-13}$ is undefined. Hence, $\sqrt{x-13}\sqrt{x-13}=x-13$ only if $x\geq 13$.
It is just a matter of definition. In general, $\sqrt{\cdot}:\mathbb{R}_+\to\mathbb{R}_+$ and $(\cdot)^{1/2}:\mathbb{C}\to\mathbb{C}$. So,$ (x−13)^{1/2}(x−13)^{1/2}=x−13$ always. 
A: I know nothing about Mathway but I don't think that it is necessary.  It is being a little pedantic but that is necessary in maths.  If you are not careful in this way then you can "prove" all sorts of nonsense such as $1 = 0$.  Incorrect handling of square roots is common in fake proofs.  
I am assuming that you are dealing with the real numbers $\mathbb{R}$.  The complex numbers $\mathbb{C}$ add different complications.  Square roots need careful handling there as well.  
$$\sqrt{x - 13}\sqrt{x - 13} = x - 13$$
This is true whenever the left hand side is defined so when $x \ge 13$.  It is not true when $x < 13$ as the left hand side is not defined.  
When expressions like this arise in real problems, you need to note these assumptions because they might not be apparent in the final result.  You need to treat these cases separately.  Maybe the final result is true for them as well but it requires a different proof.  Maybe it isn't even though the final result does not appear to have the restriction.  
A: If I understand the problem correctly, $(ff)(x)$ means $f(x)\cdot f(x)$ in which case I think Mathway is correct. In effect you are multiplying the 'output' of a function by the 'output' of another function. You are not multiplying two expressions to make a third expression.So, if one of the two functions has an undefined output for a given value of $x$, then the result of the multiplication must be undefined. Obviously in this case, both will have the same output for whatever value of $x$, and that will only be a defined output for $x\ge 13$.
