# How do you find the domain and range for f(x)= (x-3)/( x^2- 9)?

##### 1 Answer
Feb 6, 2016

$\frac{x - 3}{{x}^{2} - 9}$

$= \frac{x - 3}{\left(x - 3\right) \left(x + 3\right)} , x \ne 3 , x \ne - 3$

$= \frac{1}{x + 3} , x \ne 3 , x \ne - 3$

The domain will be $\mathbb{R}$ \ $\left\{- 3 , 3\right\}$

Remember that $x \ne 3$, there $\frac{1}{3 + 3} = \frac{1}{6}$ is not valid

therefore the range will be $\mathbb{R}$\ $\left\{\frac{1}{6}\right\}$