# What is the domain and range of F(x) = 1/ sqrt(4 - x^2)?

##### 1 Answer
Jul 8, 2018

The domain is $x \in \left(- 2 , 2\right)$. The range is $\left[\frac{1}{2} , + \infty\right)$.

#### Explanation:

The function is

$f \left(x\right) = \frac{1}{\sqrt{4 - {x}^{2}}}$

What'under the sqrt sign must be $\ge 0$ and we cannot divide by $0$

Therefore,

$4 - {x}^{2} > 0$

$\implies$, $\left(2 - x\right) \left(2 + x\right) > 0$

$\implies$, $\left\{\begin{matrix}2 - x > 0 \\ 2 + x > 0\end{matrix}\right.$

$\implies$, $\left\{\begin{matrix}x < 2 \\ x > - 2\end{matrix}\right.$

Therefore,

The domain is $x \in \left(- 2 , 2\right)$

Also,

${\lim}_{x \to {2}^{-}} f \left(x\right) = {\lim}_{x \to {2}^{-}} \frac{1}{\sqrt{4 - {x}^{2}}} = \frac{1}{O} ^ + = + \infty$

${\lim}_{x \to - {2}^{+}} f \left(x\right) = {\lim}_{x \to - {2}^{+}} \frac{1}{\sqrt{4 - {x}^{2}}} = \frac{1}{O} ^ + = + \infty$

When $x = 0$

$f \left(0\right) = \frac{1}{\sqrt{4 - 0}} = \frac{1}{2}$

The range is $\left[\frac{1}{2} , + \infty\right)$

graph{1/sqrt(4-x^2) [-9.625, 10.375, -1.96, 8.04]}