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.
Best Answer
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$$R_f= \mathbb{R}.$$