I have the following statement:
Determine the domain and range of $\large{f(x) = \frac{x}{x^2 - 1}}$
The domain are allowed input values, in this case the function is undeterminated in reals when $x \in \{-1, 1\}$ hence the domain is $\mathbb{R} - \{-1, 1\}.$
But, get the range is harder to me.
My attempt was:
Let $f(x) = y$, that is $y = \frac{x}{x^2 -1} \iff yx^2 -x-y=0$
In the case that $y = 0$ i have: $-x = 0 \iff x = 0$ and since $x \in Dom_f \to y \in Rec_f$.
In other case, $ y\neq 0$ i have:
$\large{x = \frac{1 \pm \sqrt{1+4y^2} }{2y}}$ and from here i need to get $\frac{1 \pm \sqrt{1+4y^2} }{2y} \in \mathbb{R} - \{-1, 1\}$.
so there shouldn't be a $y$ related to $x = \pm 1$.
here i don't know how to continue. Any help is really appreciated.