# What is the domain and range of  f(x) = sqrt(x^2-36)?

Dec 19, 2015

Domain: $\mathbb{R} , \left(- \infty , - 6\right] \cup \left[+ 6 , + \infty\right]$
Range: $\mathbb{R} , \left[0 , + \infty\right)$

#### Explanation:

$f \left(x\right)$ is defined in the Real numbers provided
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 36 \ge 0$
This will be true provided
$\textcolor{w h i t e}{\text{XXX")x >= 6color(white)("XX}}$or$\textcolor{w h i t e}{\text{XX}} x \le - 6$
Therefore the domain (in $\mathbb{R}$) is $x \in \left(+ \infty , - 6\right] \cup \left[+ 6 , + \infty\right)$

Since $\sqrt{\text{any real number}}$ is defined as the principle square root (i.e. >= 0)
the range is $f \left(x\right) \in \left[0 , + \infty\right)$