# How do you find the domain and range of g(z) = 1 / sqrt(4-z^2)?

Aug 9, 2017

Here the function is g(z) = 1/(sqrt(4-z^2). But sqrt(4-z^2 can't be $0$ because denominator cannot be $0$. That is because dividing by $0$ is not defined.

From this, we can say that ${z}^{2}$ can't be $4$. That means $z \ne \sqrt{4}$ $\implies$ $z \ne \pm 2$.

And the another thing to be noticed is that we can't have a negative number inside a square root. So $z$ must be in the range of $2$ & $- 2$.
Now we can say that Domain is all the real numbers between $- 2$ & $2$.

So Domain $= - 2$$<$$z$$<$$2 , z \in \mathbb{R} \mathmr{and} \left(- 2 , 2\right)$

From this Domain set, we can easily find out Range of the Function.

The Range of the Function is in between $0$ & $\frac{1}{\sqrt{3}}$ including $\frac{1}{\sqrt{3}}$.

So Range $= \left(0 , \frac{1}{\sqrt{3}}\right]$