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

Domain: $\left(- \infty , - \sqrt{2}\right) \cup \left(- \sqrt{2} , \sqrt{2}\right) \cup \left(\sqrt{2} , + \infty\right)$
Range: $\left(- \infty , - \frac{1}{2}\right] \cup \left(0 , + \infty\right)$

Explanation:

The domain can be determined by equating ${x}^{2} - 2 = 0$

solving for x:
$x = \pm \sqrt{2}$ and y is not defined when $x = \pm \sqrt{2}$

therefore the domain excludes $x = \pm \sqrt{2}$

For the Range:

$y$ can have a value from -1/2 and lower and values higher than zero.

So that Range: $\left(- \infty , - \frac{1}{2}\right] \cup \left(0 , + \infty\right)$

graph{y=1/(x^2-2)[-20,20,-10,10]}

have a nice day!