# What is the domain of f(x)=(x+3)/sqrt(x^2-9) ?

Sep 6, 2015

Domain: $\left(- \infty , - 3\right) \cup \left(3 , + \infty\right)$

#### Explanation:

The domain of the function will include any value of $x$ that does not make the denominator equal to zero and that does not make the expression under the radical negative.

For real numbers, you can only take the square root of positive numbers, which means that

${x}^{2} - 9 \ge 0$

SInce you also need this expression to be different from zero, you get

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

${x}^{2} - {3}^{2} > 0$

$\left(x - 3\right) \left(x + 3\right) > 0$

This inequality is true when you have both terms negative or both terms positive. For values of $x < - 3$ you have

$\left\{\begin{matrix}x - 3 < 0 \\ x + 3 < 0\end{matrix}\right. \implies \left(x - 3\right) \left(x + 3\right) > 0$

For values of $x > 3$ you get

$\left\{\begin{matrix}x - 3 > 0 \\ x + 3 > 0\end{matrix}\right. \implies \left(x - 3\right) \left(x + 3\right) > 0$

This means that any value of $x$ that is smaller than $\left(- 3\right)$ or greater than $3$ will be a valid solution to this inequality. On the other hand, any value of $x \in \left[- 3 , 3\right]$ will not satisfy this inequality.

This means that the domain of the function will be $\left(- \infty , - 3\right) \cup \left(3 , + \infty\right)$.