# How do you find the domain of f(x)=(3x-1)/(x^2+9)?

Apr 28, 2018

The domain of $f \left(x\right)$ is $\mathbb{R}$

#### Explanation:

The function is

$f \left(x\right) = \frac{3 x - 1}{{x}^{2} + 9}$

$\forall x \in \mathbb{R} , \text{ the denominator is } {x}^{2} + 9 > 0$ and is $\ne 0$

Therefore,

The domain of $f \left(x\right)$ is $\mathbb{R}$ or in interval notation $x \in \left(- \infty , + \infty\right)$

graph{(3x-1)/(x^2+9) [-7.02, 7.03, -3.51, 3.51]}

Apr 28, 2018

$x \in \left(- \infty , \infty\right)$

#### Explanation:

$\text{the denominator of "f(x)" cannot be zero as this would}$
$\text{make "f(x)" undefined}$

${x}^{2} + 9 \text{ is always positive for all } x \in \mathbb{R}$

$\Rightarrow \text{domain } x \in \left(- \infty , \infty\right)$
graph{(3x-1)/(x^2+9) [-10, 10, -5, 5]}