# How do you find the domain of h(x)= sqrt( 4-x^2)/( x-3)?

Jun 11, 2015

The domain is restricted by the numerator to $\left[- 2 , 2\right]$ and by the denominator to $\left(- \infty , 3\right) \cup \left(3 , \infty\right)$, which is no additional restriction. So the domain of $h \left(x\right)$ is $\left[- 2 , 2\right]$

#### Explanation:

In order for the numerator of $h \left(x\right)$ to be defined, $4 - {x}^{2} \ge 0$.

Adding ${x}^{2}$ to both sides we get ${x}^{2} \le 4 = {2}^{2}$

hence $- 2 \le x \le 2$

In order for the denominator to be defined and non-zero, we just need $x \ne 3$

This condition only affects values of $x$ outside the $\left[- 2 , 2\right]$ range we already know the domain is restricted to.

So $h \left(x\right)$ is well defined for all $x \in \left[- 2 , 2\right]$.