# How do you find the domain and range of (x^2-64)/(x-8)?

Apr 1, 2016

Domain: $\mathbb{R} \text{\} \left\{8\right\} = \left(- \infty , 8\right) \cup \left(8 , \infty\right)$

Range: $\mathbb{R} \text{\} \left\{16\right\} = \left(- \infty , 16\right) \cup \left(16 , \infty\right)$

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} - 64}{x - 8} = \frac{\left(x - 8\right) \left(x + 8\right)}{x - 8} = x + 8$

with exclusion $x \ne 8$

When $x = 8$, both the numerator and denominator are $0$ resulting in an undefined value. So $x = 8$ is not in the domain.

$f \left(x\right)$ is well defined for all other values of $x$

So the domain is $\mathbb{R} \text{\} \left\{8\right\} = \left(- \infty , 8\right) \cup \left(8 , \infty\right)$

The range is the set of values that $f \left(x\right)$ takes for $x \in \mathbb{R} \text{\} \left\{8\right\}$, namely $\left(- \infty , 16\right) \cup \left(16 , \infty\right)$.

Note that $f \left(x\right)$ can take any value except $16$