# What is the domain and range of y= 1/(x^2-25)?

Mar 18, 2018

The domain of $y$ is $x \in \mathbb{R} - \left\{- 5 , 5\right\}$.
The range is $y \in \left[- \frac{1}{25} , 0\right) \cup \left(0 , + \infty\right)$

#### Explanation:

As you cannot divide by $0$, the denominator is $\ne 0$

Therefore,

${x}^{2} - 25 \ne 0$, $\implies$ $x \ne - 5$ and $x \ne 5$

The domain of $y$ is $x \in \mathbb{R} - \left\{- 5 , 5\right\}$

To calculate the range, proceed as follows

$y = \frac{1}{{x}^{2} - 25}$

$y \left({x}^{2} - 25\right) = 1$

$y {x}^{2} - 1 - 25 y = 0$

${x}^{2} = \frac{1 + 25 y}{y}$

$x = \sqrt{\frac{1 + 25 y}{y}}$

Therefore,

$y \ne 0$

and

$1 + 25 y \ge 0$

$y \ge - \frac{1}{25}$

The range is $y \in \left[- \frac{1}{25} , 0\right) \cup \left(0 , + \infty\right)$

graph{1/(x^2-25) [-6.24, 6.244, -3.12, 3.12]}