# How do you find the domain of f(x)=(x^3+x^2-22x-40)/(x^4-x^3-7x^2+x+6)?

Jul 2, 2018

Domain: $\left(- \infty , - 2\right) \cup \left(- 2 , - 1\right) \cup \left(- 1 , 1\right) \cup \left(1 , 3\right) \cup \left(3 , \infty\right)$

#### Explanation:

Given: $\frac{{x}^{3} + {x}^{2} - 22 x - 40}{{x}^{4} - {x}^{3} - 7 {x}^{2} + x + 6}$

This type of function is called a rational function.

$\frac{N \left(x\right)}{D \left(x\right)} = \frac{{a}_{n} {x}^{n} + \ldots}{{b}_{b} {x}^{m} + \ldots}$, where ${a}_{n} \text{ & } {b}_{m}$ are the leading coefficients and $n$ and $m$ are the degrees of the two functions.

$\textcolor{b l u e}{\text{Factor}}$ both the numerator and the denominator need to be factored.

The easiest way is to graph each function individually and find the zeros ($x$-intercepts).

This is the graph of $N \left(x\right) = {x}^{3} + {x}^{2} - 22 x - 40$
graph{x^3 + x^2 - 22x - 40 [-5, 6, -80, 10]}

It has $x$-intercepts at $x = - 4 , - 2 \mathmr{and} 5$

This means $\textcolor{red}{N \left(x\right)} = {x}^{3} + {x}^{2} - 22 x - 40 = \textcolor{red}{\left(x + 4\right) \left(x + 2\right) \left(x - 5\right)}$

This is the graph of $D \left(x\right) = {x}^{4} - {x}^{3} - 7 {x}^{2} + x + 6$:

graph{x^4 - x^3 - 7x^2 + x + 6 [-5, 6, -15, 10]}

It has $x$-intercepts at $x = - 2 , - 1 , 1 \mathmr{and} 3$

This means $\textcolor{red}{D \left(x\right)} = {x}^{4} - {x}^{3} - 7 {x}^{2} + x + 6 = \textcolor{red}{\left(x + 2\right) \left(x + 1\right) \left(x - 1\right) \left(x - 3\right)}$

$\textcolor{b l u e}{\text{Factored function:}}$
$f \left(x\right) = \frac{\left(x + 4\right) \cancel{\left(x + 2\right)} \left(x - 5\right)}{\cancel{\left(x + 2\right)} \left(x + 1\right) \left(x - 1\right) \left(x - 3\right)}$

$\textcolor{m a \ge n t a}{\text{Finding the Domain depends on holes, and vertical asymptotes.}}$

If you can cancel a factor that is both in the numerator and the denominator, that is where you have a hole; a removable discontinuity.

color (magenta)("Hole at " x = -2

Vertical asymptotes are found by setting $D \left(x\right) = 0$ after the holes are eliminated.

color (magenta)("Vertical asymptotes are at " x = -1, 1, x = 3

Domain: $\left(- \infty , - 2\right) \cup \left(- 2 , - 1\right) \cup \left(- 1 , 1\right) \cup \left(1 , 3\right) \cup \left(3 , \infty\right)$