# What is the domain and range of f(x)=sqrt(16-x^3)?

May 13, 2017

Domain: $x \le \sqrt[3]{16} \mathmr{and} \left(- \infty , \sqrt[3]{16}\right]$
Range: $f \left(x\right) \ge 0 \mathmr{and} \left[0 , \infty\right)$

#### Explanation:

$f \left(x\right) = \sqrt{16 - {x}^{3}}$ Domain: under root should not be negative , so
$16 - {x}^{3} \ge 0 \mathmr{and} 16 \ge {x}^{3} \mathmr{and} {x}^{3} \le 16 \mathmr{and} x \le \sqrt[3]{16}$

Domain: $x \le \sqrt[3]{16} \mathmr{and} \left(- \infty , \sqrt[3]{16}\right]$

Range: $f \left(x\right)$ is any real value $\ge 0$

Range: $f \left(x\right) \ge 0 \mathmr{and} \left[0 , \infty\right)$
graph{(16-x^3)^0.5 [-10, 10, -5, 5]}