# How to determine the domain and range of the following function?

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.

We can find the range of a function by finding the inverse map of the function; the range of the function is the domain of its inverse map.

So, let us find the inverse map of the function $$f(x)=\frac{x}{x^2-1}$$ by finding its inverse map as follows.$$y=\frac{x}{x^2-1} \quad \Rightarrow \quad yx^2-x-y=0$$$$\Rightarrow \quad \begin{cases}x=0, & \text{if } y=0 \\ x=\frac{1 \pm \sqrt{1+4y^2}}{2y}, & \text{if } y\neq 0 \end{cases}$$(The last conclusion was obtained by using the quadratic formula: the solutions of the quadratic equation $$ax^2+bx+c=0$$, $$a\neq 0$$, are $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$).

The domain of the first case is $$\{ 0 \}$$.

To find the domain of the second case, we must exclude real numbers vanishing the denominator or making the radicand negative. Since $$1+4y^2 \gt 0$$ for any $$y \in \mathbb{R}$$, the domain of the second case is $$\mathbb{R} -\{ 0 \}$$.

Please note that the domain of a piecewise-defined function equals the union of the domain of the pieces. So the domain of the inverse map is$$\{ 0 \} \cup ( \mathbb{R}- \{ 0 \} ) = \mathbb{R}.$$Thus, we conclude that the range of the function $$f(x)=\frac{x}{x^2-1}$$ is$$R_f= \mathbb{R}.$$

You've discovered that, to get a given output $$y \neq 0$$, then the input you should choose is

$$x = \frac{1 \pm \sqrt{1+4y^2}}{2y}. \tag{*}$$

Thus the only (nonzero) $$y$$'s that we won't be able to get as an output are the $$y$$'s that would require us to choose $$x=-1$$ or $$x=1$$. So let's find these $$y$$'s. Assume that

$$-1 = \frac{1 \pm \sqrt{1+4y^2}}{2y}.$$

Multiply both sides by $$2y$$, subtract $$1$$, and square both sides to get $$(-2y-1)^2 = 1+4y^2$$. Solving this gives $$y=0$$, which isn't allowed since we assumed that $$y \neq 0$$ in order to obtain formula ($$*$$). If you had chosen $$x=1$$ instead of $$x=-1$$, then you'd also get $$y=0$$. What this tells us is that you can obtain all $$y$$'s using your formula other than $$y=0$$, but that's fine because we know that we can get $$y=0$$ anyway by choosing $$x=0$$. So actually, we can get all $$y$$'s; $$y = 0$$ we get from $$x=0$$, and $$y \neq 0$$ we get from ($$*$$). Therefore, the range is all of $$\mathbb{R}$$.