# What is the domain and range of 1 / (x^2 + 5x + 6)?

May 24, 2018

The domain is $x \in \left(- \infty , - 3\right) \cup \left(- 3 , - 2\right) \cup \left(- 2 , + \infty\right)$. The range is $y \in \left(- \infty , - 4\right] \cup \left[0 , + \infty\right)$

#### Explanation:

The denominator is

${x}^{2} + 5 x + 6 = \left(x + 2\right) \left(x + 3\right)$

As the denominator must be $\ne 0$

Therefore,

$x \ne - 2$ and $x \ne - 3$

The domain is $x \in \left(- \infty , - 3\right) \cup \left(- 3 , - 2\right) \cup \left(- 2 , + \infty\right)$

To find the range, proceed as follows :

Let $y = \frac{1}{{x}^{2} + 5 x + 6}$

$y \left({x}^{2} + 5 x + 6\right) = 1$

$y {x}^{2} + 5 y x + 6 y - 1 = 0$

This is a quadratic equation in $x$ and the solutions are real only if the discriminant is $\ge 0$

$\Delta = {b}^{2} - 4 a c = {\left(5 y\right)}^{2} - 4 \left(y\right) \left(6 y - 1\right) \ge 0$

$25 {y}^{2} - 24 {y}^{2} + 4 y \ge 0$

${y}^{2} + 4 y \ge 0$

$y \left(y + 4\right) \ge 0$

The solutions of this inequality is obtained with a sign chart.

The range is $y \in \left(- \infty , - 4\right] \cup \left[0 , + \infty\right)$

graph{1/(x^2+5x+6) [-16.26, 12.21, -9.17, 5.07]}