# How do you find the domain and range of f(x)=sqrt(9-x^2)?

Refer to explanation

#### Explanation:

For the domain we have that

$9 - {x}^{2} \ge 0 \implies \left(3 - x\right) \cdot \left(3 + x\right) \ge 0 \implies - 3 \le x \le 3$ hence

$D \left(f\right) = \left[- 3 , 3\right]$

and the range is

$R \left(f\right) = \left[0 , 3\right]$

The graph of the function is

graph{(9-x^2)^(1/2) [-10, 10, -5, 5]}

Actually the function represents the upper half of the circle with
equation

${x}^{2} + {y}^{2} = {3}^{2}$