# How do you find the domain and range of g(x) = x/(x^2 - 16)?

Mar 17, 2018

The domain of $g \left(x\right)$ is $x \in \mathbb{R} - \left\{- 4 , 4\right\}$.
The range is $g \left(x\right) \in \mathbb{R}$

#### Explanation:

As you cannot divide by $0$, the denominator is $\ne 0$

Therefore,

${x}^{2} - 16 \ne 0$, $\implies$ $x \ne - 4$ and $x \ne 4$

The domain of $g \left(x\right)$ is $x \in \mathbb{R} - \left\{- 4 , 4\right\}$

To calculate the range, proceed as follows

Let $y = \frac{x}{{x}^{2} - 16}$

$y \left({x}^{2} - 16\right) = x$

$y {x}^{2} - x - 16 y = 0$

This is a quadratic equation in $x$, and in order to have solutions,
the discriminant $\ge 0$

$a = y$

$b = - 1$

$c = - 16 y$

$\Delta = {b}^{2} - 4 a c = {\left(- 1\right)}^{2} - 4 \left(y\right) \left(- 16 y\right) = 1 + 64 {y}^{2}$

$\forall y \in \mathbb{R}$, $\implies$, $\Delta \ge 0$

Therefore,

The range is $g \left(x\right) \in \mathbb{R}$

graph{x/(x^2-16) [-10, 10, -5, 5]}

Mar 17, 2018

Domain: $\left(- \infty , - 4\right) \cup \left(- 4 , 4\right) \cup \left(4 , \infty\right)$
Range: $\left(- \infty , \infty\right)$

#### Explanation:

Given: $\frac{x}{{x}^{2} - 16}$

First factor the denominator since $\left({x}^{2} - 16\right)$ is the difference of squares:

$\frac{x}{{x}^{2} - 16} = \frac{x}{\left(x - 4\right) \left(x + 4\right)}$

Find the Domain - valid input - usually $x$
For most functions, the domain is $\left(- \infty , \infty\right)$, the set of all reals. There are a number of factors that can cause this domain to be limited. Here are a few possibilities:

• a radical such as a square root - limits the domain
• a denominator - can produce holes and/or vertical asymptotes
• inverse trigonometry functions
• natural log function $\left(y = \ln x\right)$

In your example, the vertical asymptotes are the cause. When the denominator function $D \left(x\right) = 0$, the vertical asymptotes are found to be at $x = \pm 4$

Domain: $\left(- \infty , - 4\right) \cup \left(- 4 , 4\right) \cup \left(4 , \infty\right)$

Find the Range - valid output - usually $y$
For most functions, the range is also $\left(- \infty , \infty\right)$, the set of all reals. There are a number of factors that can cause this range to be limited. Here are a few possibilities:

• a radical such as a square root - limits the range
• a quadratic or even powered function can limit the range. The vertex will be a minimum or a maximum
• absolute value functions can have a vertex
• a rational function (has a numerator and denominator) can have a horizontal asymptote
• a natural exponential function ($y = {e}^{x}$)

In your example, w have a rational function. The degree of the numerator function = 1 $\left(n = 1\right)$and the degree of the denominator function = 2, $\left(m = 2\right)$. When $n < m$ there is a horizontal asymptote at $y = 0$.

Range: $\left(- \infty , 0\right) \cup \left(0 , \infty\right)$

But you can see from the graph below that the point $\left(0 , 0\right)$ exists. This means the domain is actually $\left(- \infty , \infty\right)$

How would you know this without graphing the function? Create a table of values.
$\underline{x | - 8 \text{ "|-4" "|-2" "|0" "|2" "|4" "|8 " }}$
$y | - \frac{1}{6} \text{ "|un" "|1/6" "|0" "|-1/6""|un" } | \frac{1}{6}$

$u n$ = undefined

graph{x/(x^2-16) [-10, 10, -5, 5]}