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

May 6, 2018

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

#### Explanation:

The denominator is $\ne 0$

${x}^{2} - 1 \ne 0$

$\left(x + 1\right) \left(x - 1\right) \ne 0$

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

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

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

Therefore,

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

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

This is a quadratic equation in $x$

The real solutions are when the discriminant is

$\Delta \ge 0$

$0 - 4 \cdot y \left(- \left(y + 1\right)\right) \ge 0$

$4 y \left(y + 1\right) \ge 0$

The solutions to this equation is obtained with a sign chart.

$y \in \left(- \infty , - 1\right] \cup \left(0 , + \infty\right)$

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

graph{1/(x^2-1) [-7.02, 7.024, -3.51, 3.51]}