# What is the domain and range of f(x) = sqrt((x^2) - 3)?

Domain: $x < - \sqrt{3} , x > \sqrt{3}$

Range: $f \left(x\right) \ge 0$

#### Explanation:

I'm going to assume for this question that we are staying within the realm of Real Numbers (and so things like $\pi$ and $\sqrt{2}$ are allowed but $\sqrt{- 1}$ is not).

The Domain of an equation is the list of all allowable $x$ values.

Let's look at our equation:

$f \left(x\right) = \sqrt{{x}^{2} - 3}$

Ok - we know that square roots can't have negative numbers in them, so what will make our square root term negative?

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

${x}^{2} < 3$

$x < \left\mid \sqrt{3} \right\mid \implies - \sqrt{3} < x < \sqrt{3}$

Ok - so we know that we can't have $- \sqrt{3} < x < \sqrt{3}$. All other $x$ terms are ok. We can list the domain in a few different ways. I'll use:

$x < - \sqrt{3} , x > \sqrt{3}$

The Range is the list of resulting values coming from the domain.

We already know that the smallest number the range will be is 0. As $x$ gets larger and larger (both in a positive and negative sense), the range will increase. And so we can write:

$f \left(x\right) \ge 0$

We can see this in the graph:

graph{sqrt(x^2-3) [-10,10,-2,7]}