# How do you find the domain and range of f(x)=x/(x^2 - 2x -3) ?

Mar 18, 2016

Domain : $\mathbb{R} - \left\{- 1 , 3\right\}$
Range: $\mathbb{R}$

#### Explanation:

To Find the domain

Equate the denominator(${x}^{2} - 2 x - 3$) to zero, then solve the equation for $x$

$\rightarrow {x}^{2} - 2 x - 3 = 0$
$\rightarrow x = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \cdot \left(1\right) \cdot \left(- 3\right)}}{2 \cdot 1}$
$\rightarrow x = 1 \pm 2$

$\implies x = - 1$ and $x = 3$

This means that, when $x = - 1$ or $3$, we have the ${x}^{2} - 2 x - 3 = 0$
Implying that $f \left(x\right) = \textcolor{red}{\frac{x}{0}}$ which is undefined.

Hence, the domain is all real numbers except $- 1$
and $3$

Also written as ${D}_{f} = \left(- \infty , - 1\right) \cup \left(- 1 , 3\right) \cup \left(3 , + \infty\right)$

To Find the Range

Step 1
say $f \left(x\right) = y$ and rearrange the function as a quadratic equation

$y = \frac{x}{{x}^{2} - 2 x - 3}$
$\rightarrow y \left({x}^{2} - 2 x - 3\right) = x$
$\rightarrow y {x}^{2} - 2 y x - 3 y - x = 0$
$\rightarrow y {x}^{2} + \left(- 2 y - 1\right) x - 3 y = 0$

Step 2
We know from the quadratic formula,
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
that the solutions of $x$ are real when ${b}^{2} - 4 a c \ge 0$

So likewise, we say, ${\left(- 2 y - 1\right)}^{2} - 4 \cdot \left(y\right) \cdot \left(- 3 y\right) \ge 0$
$\textcolor{w h i t e}{\text{ }} \textcolor{red}{b}$$\textcolor{w h i t e}{\text{ }} \textcolor{red}{a}$$\textcolor{w h i t e}{\text{ }} \textcolor{red}{c}$

Step 3
We solve the inequality for the values set of values of $y$

$\implies 4 {y}^{2} + 4 y + 1 + 12 {y}^{2} \ge 0$

$\rightarrow 16 {y}^{2} + 4 y + 1 \ge 0$

$\rightarrow 16 \left[{y}^{2} + \frac{1}{4} y + \frac{1}{16}\right] \ge 0$

$\rightarrow 16 \left[{\left(y + \frac{1}{8}\right)}^{2} - \frac{1}{64} + \frac{1}{16}\right] \ge 0$

$\rightarrow 16 \left[{\left(y + \frac{1}{8}\right)}^{2} + \frac{3}{64}\right] \ge 0$

Notice that for all values of $y$ the left hand side of the inequality be greater than (but not equal) to zero.

We then conclude that, $y$ can take all real values.

$y \in \mathbb{R} \iff f \left(x\right) \in \mathbb{R}$

So the Range is $\mathbb{R}$