# What is the domain and range of y = (2x^2)/( x^2 - 1)?

Aug 7, 2017

The domain is $x \in \left(- \infty , - 1\right) \cup \left(- 1 , 1\right) \cup \left(1 , + \infty\right)$
The range is $y \in \left(- \infty , 0\right] \cup \left(2 , + \infty\right)$

#### Explanation:

The function is

$y = \frac{2 {x}^{2}}{{x}^{2} - 1}$

We factorise the denominator

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

Therefore,

$x \ne 1$ and $x \ne - 1$

The domain of y is $x \in \left(- \infty , - 1\right) \cup \left(- 1 , 1\right) \cup \left(1 , + \infty\right)$

Let's rearrage the function

$y \left({x}^{2} - 1\right) = 2 {x}^{2}$

$y {x}^{2} - y = 2 {x}^{2}$

$y {x}^{2} - 2 {x}^{2} = y$

${x}^{2} = \frac{y}{y - 2}$

$x = \sqrt{\frac{y}{y - 2}}$

For $x$ to a solution, $\frac{y}{y - 2} \ge 0$

Let $f \left(y\right) = \frac{y}{y - 2}$

We need a sign chart

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a a}$$2$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$y - 2$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(y\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

Therefore,

$f \left(y\right) \ge 0$ when $y \in \left(- \infty , 0\right] \cup \left(2 , + \infty\right)$

graph{2(x^2)/(x^2-1) [-16.02, 16.02, -8.01, 8.01]}