# What is the domain and range of 3sqrt( x^2 - 9)?

Domain: $x \le - 3$ or $x \ge 3$ also Domain: $\left(- \infty , - 3\right] \cup \left[3 , \infty\right)$
Range: $\left[0 , + \infty\right)$

#### Explanation:

x can take on values -3 or less up to $- \infty$
also x can take on values 3 or higher up to $+ \infty$

that is why Domain :$x \le - 3$ or $x \ge 3$

The lowest possible value is 0 up to $+ \infty$ and that is the range .
That is if we let

$y = 3 \cdot \sqrt{{x}^{2} - 9}$

when $x = \pm 3$ the value of $y = 0$ and when $x$ approaches a very high value, the value of $y$ approaches a very high value also.

So the Range: $\left[0 , + \infty\right)$