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

Aug 3, 2015

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

#### Explanation:

The only restriction to the domain of the function will occur when the denominator is equal to zero. More specifically,

${x}^{2} - 2 = 0$

$\sqrt{{x}^{2}} = \sqrt{2} \implies x = \pm \sqrt{2}$

These two values of $x$ will make the function's denominator equal to zero, which means that they will be excluded from the function's domain.

No other restrictions apply, so you can say that the domain of the function is $\mathbb{R} - \left\{\pm \sqrt{2}\right\}$, or (-oo, -sqrt(2)) uu (-sqrt(2), sqrt(2)) uu (sqrt(2), +oo)#.

This restriction on the possible values $x$ can take will impact the function's range as well.

Because you don't have a value of $x$ that can make $y = 0$, the range of the function will not include this value, i.e. zero.

Simply put, because you have

$\frac{1}{{x}^{2} - 2} \ne 0 , \left(\forall\right) x \ne \pm \sqrt{2}$

the range of the function will be $\mathbb{R} - \left\{0\right\}$, or $\left(- \infty , 0\right) \cup \left(0 , + \infty\right)$.

In other words, the function's graph will have two vertical asymptotes at $x = - \sqrt{2}$ and $x = \sqrt{2}$, respectively.

graph{1/(x^2-2) [-10, 10, -5, 5]}