# What is the domain of g(x) = (x+5)/(3x^2+23x-36) in set notation?

Mar 11, 2017

$\left\{x \in \mathbb{R} | x < - 9 \mathmr{and} - 9 < x < \frac{4}{3} \mathmr{and} x > \frac{4}{3}\right\}$

#### Explanation:

The domain of a function represents the possible input values, i.e. values of $x$, for which the function is defined.

Notice that your function is actually a fraction that has two rational expressions as its numerator and denominator, respectively.

As you know, a fraction that has a denominator equal to $0$ is undefined. This implies that any value of $x$ that will make

$3 {x}^{2} + 23 x - 36 = 0$

will not be part of the domain of the function. This quadratic equation can be solved by using the quadratic formula, which for a generic quadratic equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{a {x}^{2} + b x + c = 0}}}$

looks like this

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{x}_{1 , 2} = \frac{- b + - \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a}}}} \to$ the quadratic formula

$\left\{\begin{matrix}a = 3 \\ b = 23 \\ c = - 36\end{matrix}\right.$

Plug in your values to find

${x}_{1 , 2} = \frac{- 23 \pm \sqrt{{23}^{2} + 4 \cdot 3 \cdot \left(- 36\right)}}{2 \cdot 3}$

${x}_{1 , 2} = \frac{- 23 \pm \sqrt{961}}{6}$

${x}_{1 , 2} = \frac{- 23 \pm 31}{6} \implies \left\{\begin{matrix}{x}_{1} = \frac{- 23 - 31}{6} = - 9 \\ {x}_{2} = \frac{- 23 + 31}{6} = \frac{4}{3}\end{matrix}\right.$

So, you know that when

$x = - 9 \text{ }$ or $\text{ } x = \frac{4}{3}$

the denominator is equal to $0$ and the function is undefined. For any other value of $x$, $f \left(x\right)$ will be defined.

This means that the domain of the function in set notation will be

$\left\{x \in \mathbb{R} | x < - 9 \mathmr{and} - 9 < x < \frac{4}{3} \mathmr{and} x > \frac{4}{3}\right\}$

graph{(x+5)/(3x^2 + 23x - 36) [-14.24, 14.23, -7.12, 7.12]}

As you can see from the graph, the function is not defined for $x = - 9$ and $x = \frac{4}{3}$, i.e. the function ahs two vertical asymptotes in those two points.

Alternatively, you can write the domain as

$x \in \mathbb{R} \text{\} \left\{- 9 , \frac{4}{3}\right\}$

In interval notation, the domain would look like this

$x \in \left(- \infty , - 9\right) \cup \left(- 9 , \frac{4}{3}\right) \cup \left(\frac{4}{3} , + \infty\right)$