How do you find the domain and range of  sqrt(x+4)/(x+2)?

Jul 1, 2018

The domain is $x \in \left[- 4 , 2\right) \cup \left(2 , + \infty\right)$. The range is $y \in \mathbb{R}$

Explanation:

For the domain, there are $2$ conditions :

The denominator must be $\ne 0$

$\implies$, $x + 2 \ne 0$

$\implies$, $x \ne - 2$

And

What's under the sqrt sign must be $\ge 0$

$\implies$, $x + 4 \ge 0$

$\implies$, $x \ge - 4$

Therefore,

The domain is $x \in \left[- 4 , - 2\right) \cup \left(- 2 , + \infty\right)$

For the range, let $y = \frac{\sqrt{x + 4}}{x + 2}$

When $x = - 4$, $\implies$, $y = 0$

And

${\lim}_{x \to - {2}^{-}} y = - \infty$

${\lim}_{x \to - {2}^{+}} y = + \infty$

${\lim}_{x \to + \infty} y = {0}^{+}$

The range is $y \in \mathbb{R}$

graph{sqrt(x+4)/(x+2) [-11.44, 24.6, -8.1, 9.92]}