# What is the domain and range of y = 5 - (sqrt(9-x^2))?

Feb 26, 2018

Donain: $\left[- 3 , + 3\right]$ Range: $\left[2 , 5\right]$

#### Explanation:

$f \left(x\right) = 5 - \left(\sqrt{9 - {x}^{2}}\right)$

$f \left(x\right)$ is defined for $9 - {x}^{2} \ge 0 \to {x}^{2} \le 9$

$\therefore f \left(x\right)$ is defned for $\left\mid x \right\mid \le 3$

Hence the domain of $f \left(x\right)$ is $\left[- 3 , + 3\right]$

Consider, $0 \le \sqrt{9 - {x}^{2}} \le 3$ for $x \in \left[- 3 , + 3\right]$

$\therefore {f}_{\max} = f \left(\left\mid 3 \right\mid\right) = 5 - 0 = 5$

and, ${f}_{\min} = f \left(0\right) = 5 - 3 = 2$

Hence, the range of $f \left(x\right)$ is $\left[2 , 5\right]$

We can see these results from the graph of $f \left(x\right)$ below.

graph{5-(sqrt(9-x^2)) [-8.006, 7.804, -0.87, 7.03]}