Functions interval question I have a question regarding the interval of a function.
The following function
$$f(x)=\frac{2x^2}{x^2-1}$$
$$f'(x)=\frac{-4x}{(x^2-1)^2}$$
I know that the point -1 and 1 in the number line would be gaps in the number but would this because of domain restriction in the original function?
Because I also know that if I the derivative has point that DNE then they can be included in the number line to determine whether the function increase or decrease.
 A: Following up on the discussion in the comments, it was clarified that the OP is having trouble understanding why the domain of the derivative of a function might not be the same as the domain of the initial function.
Answer: Read this definition of derivative of a function. As you can see, given a function $f$ with domain $D$, the derivative of $f$ at a point $a\in D$, which is denoted by $f'(a)$ is defined by:
$$f'(a):=\lim \limits_{x\to a}\left(\frac{f(x)-f(a)}{x-a}\right)$$
if the limit exists and is finite, otherwise $f'$ is not defined.
Since $a\in D$, then the domain of $f'$ will be a subset of $D$, but it can happen that the domain of $f'$ isn't $D$. For that to happen it suffices that the limit in the definition of $f'$ doesn't exist or isn't finite. 
Here's an example: let $g(x)=\sqrt x$, for all $x\in \Bbb R_0^+$.
Let's try to find $g'$. Let $a\in \Bbb R_0^+$.
$$\lim \limits_{x\to a}\left(\frac{g(x)-g(a)}{x-a}\right)=\lim \limits_{x\to a}\left(\frac{\sqrt x -\sqrt a}{x-a}\right)=\lim \limits_{x\to a}\left(\frac{\sqrt x -\sqrt a}{(\sqrt x-\sqrt a)(\sqrt x + \sqrt a)}\right)=\lim \limits_{x\to a}\left(\frac{1}{\sqrt x + \sqrt a}\right)$$
If $a\neq 0$, then you get $\displaystyle g'(a)=\frac{1}{2\sqrt a}$, as expected.
But if $a=0$, then the limit won't be finite, therefore $0$ is not in the domain of $g'$. It should be noted that the limit for $a=0$ is only taken for $x$ to the right of $0$.
A: The function is not defined (not in the domain of) at $x = 1, x=-1$
You must check the limit of the derivative as $x \to -1$ and $x \to 1$ approached from each direction, to understand or determine whether an asymptote, cusp, or vertical tangent e.g. exists there. For example, see the graph of your function is shown below:
By testing the limit of the derivative as $x \to -1^+, x \to -1^-, x\to 1^-, x\to 1^+$, you will see the $f'(x) \to -\infty, f'(x) \to +\infty, f'(x) \to -\infty, f'(x)\to +\infty,$ respectively. So it turns out that we have two vertical asymptotes, one at $x = 1$, and the other at $x = -1$.
Hence, the neither the derivative nor the function is defined at $x = -1, x = 1$. The only critical point of the function is the local maximum at $(0, 0)$. 

