# How do you find the domain and range of 2/ root4(9-x^2)?

Dec 16, 2017

The domain is $x \in \left(- 3 , 3\right)$. The range is $y \in \left[1.155 , + \infty\right)$

#### Explanation:

Let $y = \frac{2}{\sqrt[4]{9 - {x}^{2}}}$

The denominator must be $\ne 0$ and $> 0$

Therefore,

$9 - {x}^{2} > 0$

${x}^{2} < 9$

$x \in \left(- 3 , 3\right)$

The domain is $x \in \left(- 3 , 3\right)$

To find the range, proceed as follows

${y}^{4} = \frac{16}{9 - {x}^{2}}$

$9 - {x}^{2} = \frac{16}{y} ^ 4$

${x}^{2} = 9 - \frac{16}{y} ^ 4$

$x = \sqrt{\frac{9 {y}^{4} - 16}{{y}^{4}}}$

Therefore,

$y \ne 0$

and

$9 {y}^{4} - 16 \ge 0$

${y}^{4} \ge \frac{16}{9}$

$y \ge \frac{2}{\sqrt{3}}$

The range is $y \in \left[\frac{2}{\sqrt{3}} , + \infty\right)$

graph{root(4)(16/(9-x^2)) [-7.02, 7.03, -2.197, 4.827]}