# How do you find the domain and range of f(x) = sqrt(x+6) /( 6+x)?

Apr 23, 2017

Domain : $x \ne - 6$ from the denominator.
$x \ge - 6$ as an argument for a square root.

#### Explanation:

Together we get domain $x > - 6$

Range:
Both the numerator and the denominator are now always positive.

So the range is $0 < f \left(x\right) < + \infty$
graph{1/(sqrt(x+6)) [-10, 10, -5, 5]}
Note:
You may have noticed that $6 + x = x + 6 = {\left(\sqrt{x + 6}\right)}^{2}$, so under the given conditions we may rewrite:
$f \left(x\right) = \frac{\cancel{\sqrt{x + 6}}}{\sqrt{x + 6}} ^ \cancel{2} = \frac{1}{\sqrt{x + 6}}$