# What is the domain and range of g(x) =(5x)/(x^2-36)?

Feb 10, 2018

$x \in \mathbb{R} , x \ne \pm 6$
$y \in \mathbb{R} , y \ne 0$

#### Explanation:

The denominator of g(x) cannot be zero as this would make g(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be.

$\text{solve } {x}^{2} - 36 = 0 \Rightarrow \left(x - 6\right) \left(x + 6\right) = 0$

$\Rightarrow x = \pm 6 \leftarrow \textcolor{red}{\text{ are excluded values}}$

$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne \pm 6$

$\text{or in interval notation as}$

$\left(- \infty , - 6\right) \cup \left(- 6 , 6\right) \cup \left(6 , + \infty\right)$

$\text{for range divide terms on numerator/denominator by the}$
$\text{highest power of x that is } {x}^{2}$

$g \left(x\right) = \frac{\frac{5 x}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{36}{x} ^ 2} = \frac{\frac{5}{x}}{1 - \frac{36}{x} ^ 2}$

$\text{as } x \to \pm \infty , g \left(x\right) \to \frac{0}{1 - 0}$

$\Rightarrow y = 0 \leftarrow \textcolor{red}{\text{is an excluded value}}$

$\Rightarrow \text{range is } y \in \mathbb{R} , y \ne 0$

$\left(- \infty , 0\right) \cup \left(0 , + \infty\right) \leftarrow \textcolor{b l u e}{\text{in interval notation}}$
graph{(5x)/(x^2-36) [-10, 10, -5, 5]}