This is because
is indeterminate.
This is an extremely common misconception. There is a vast difference between 
and the form of a limit, which may be labelled as "
" (note the quotes!), just as there is a difference between 
and the form "
" of some limits.
Here are the facts under standard mathematical conventions:
in contexts where the exponent is a natural number.
"
" is a label referring to an indeterminate form of some limits.
is undefined.
"
" is a label referring to another indeterminate form of some limits.
Limits with form "
" or "
" may have a value or may not. That is precisely why we call their form indeterminate, because we cannot determine the value so easily by their form alone.
is not a limit, and if the exponent is a natural number (like for rings or in combinatorics or in the binomial theorem or in power series or ...) then its value is always
.
If you do not believe this, see the conventional statement of the binomial theorem here and here (equation 4) and the definition of power series here and here.
Answer from user21820 on Stack ExchangeVideos
This is because
This is an extremely common misconception. There is a vast difference between and the form of a limit, which may be labelled as "" (note the quotes!), just as there is a difference between and the form "" of some limits.
Here are the facts under standard mathematical conventions:
"
"
Limits with form "
If you do not believe this, see the conventional statement of the binomial theorem here and here (equation 4) and the definition of power series here and here.
On pondering this good question further, I think that part of the problem is that we have no name for the functions 
. A clean way of getting around the difficulty might be the following:
Define functions 
for nonnegative integers 
inductively as follows:
for all 
, 
, and for 
, define $P_{n+1}(x)=xP_n(x)$. You see that this makes 
the constant function 
, and for 
, 
.
Then your function can be written 
.
I believe the issue here is the definition of negative numbers raised to rational powers that's used by Wolfram or your graphing software. A robust definition of exponentials that's used by most software because it generalises to complex numbers is: $$a^b := \exp({b\ln a}),$$ where $\exp x=e^x$, and in the case where $a$ is complex, $\ln$ is the principal logarithm. The issue now, is when you try to evaluate something like $(-1)^{2/3}$. If we use the definition we have $$(-1)^{2/3}=\exp\left(\frac23\ln(-1)\right),$$ which clearly makes no sense in $\mathbb R$ (the $\ln$ of a negative number is undefined). So to answer your question, the reason wolfram thinks $(-1)^{2/3}$ is undefined is because of the robust definition for exponentiation, which gives $(-1)^{2/3}$ as a complex number.
When dealing with fractional powers of negatives, Wolfram Alpha prefers to take the complex branch, so that neither (-1)^(1/3) nor (-1)^(2/3) are considered real.
Of course, ((-1)^2)^(1/3) will yield the expected result, but you can also force the real branch with (cubicroot(-1))^2.
https://www.wolframalpha.com/input/?i=(cubicroot(x%5E2-1))%5E2;(x%5E2-1)%5E(2%2F3)
Hint: The domain of a function is the set of input values for which the function is defined. Do you know of any values for which the square root is not defined?
The function $f(x)$ is not defined when $$x^2 - 4x - 45 \lt 0,\tag{1}$$ as the square root function is defined in the reals for non-negative reals only. The only valid "input" for the square root function is non-negative real numbers.
If a function $f(x)$ is not defined on an interval(s) = $I \subseteq \mathbb R$, or is not defined for any particular $c\in \mathbb R$, then the interval(s) and/or points at which $f(x)$ is not defined are excluded from the function's domain.
So what are you left with if you exclude all $x\in \mathbb R$ such that $x^2 - 4x - 45 < 0$? This amounts to excluding your solutions to the inequality given in $(1)$. What remains in $\mathbb R$ is then your domain.
Alternatively, your domain consists of all real $x$ such that $$x^2 - 4x - 45 \geq 0\tag{2}$$
Then your task would then be, essentially, to determine the solutions to the equality given by $(2)$. its solution will be your domain.