# How do you find the domain and range of sqrt(16-x^2)?

Aug 6, 2017

Domain : $- 4 \le x \le 4$, in interval notation : $\left[- 4 , 4\right]$
Range: $0 \le f \left(x\right) \le 4$, in interval notation : $\left[0 , 4\right]$

#### Explanation:

$f \left(x\right) = \sqrt{16 - {x}^{2}}$ , for domain under root should not be

negative quantity. $16 - {x}^{2} \ge 0 \mathmr{and} 16 \ge {x}^{2} \mathmr{and} {x}^{2} \le 16$

$\therefore x \le 4 \mathmr{and} x \ge - 4$ . Domain : $- 4 \le x \le 4 \mathmr{and} \left[- 4 , 4\right]$

Range : $f \left(x\right)$ is maximum at $x = 0 , f \left(x\right) = 4$ and

$f \left(x\right)$ is minimum at $x = 4 , f \left(x\right) = 0$

Range : $0 \le f \left(x\right) \le 4 \mathmr{and} \left[0 , 4\right]$

Domain : $- 4 \le x \le 4$, in interval notation : $\left[- 4 , 4\right]$

Range: $0 \le f \left(x\right) \le 4$, in interval notation : $\left[0 , 4\right]$

graph{(16-x^2)^0.5 [-10, 10, -5, 5]}