# What is the domain and range of g(x)= (x^2 - 16)^(1/2)?

Jun 9, 2018

$x \setminus \in \left(- \setminus \infty , - 4\right] \setminus \cup \left[4 , \setminus \infty\right)$
$y \setminus \in \left[0 , \setminus \infty\right]$

#### Explanation:

$g \left(x\right) = \setminus \sqrt{{x}^{2} - 16}$

Anything underneath a square root must be greater than zero for the square root to be defined for real numbers.

$\setminus \therefore {x}^{2} - 16 > 0 \setminus R i g h t a r r o w {x}^{2} > 16 \setminus R i g h t a r r o w | x | > 4$

$\setminus \therefore x \setminus \in \left(- \setminus \infty , - 4\right] \setminus \cup \left[4 , \setminus \infty\right)$

The range will be $\left[0 , \setminus \infty\right)$ since the expression inside the sqrt ranges from 0 to $\setminus \infty$

$\setminus \therefore y \setminus \in \left[0 , \setminus \infty\right]$