# How do you find the domain of f(x)=(sqrt(x-9))/(x^2-16)?

Apr 2, 2017

{x|x≥9} or [9, oo[

#### Explanation:

The domain is the set of numbers that $x$ can be equal to.

Let's look at the numerator first. You can only take the square root of a number greater than or equal to $0$. That means that x-9≥0 (else you will be taking the square root of a negative number). Solving this gives x≥9.

Let's then look at the denominator. You cannot divide by zero. This means that ${x}^{2} - 16 \ne 0$. Solving this gives $x \ne 4 \vee x \ne - 4$.

Combining these gives x≥9 ^^ xne4 ^^ xne-4. However, since x≥9 implies that $x \ne - 4$ and $x \ne 4$, we can remove both.

Our final answer is x≥9. In set notation, we write {x|x≥9}. In interval notation, we write [9, oo[.