# What is the domain and range of p(x)= root3(x-6)/sqrt(x^2 - x - 30)?

May 18, 2018

The domain of $p$ can be defined as $\left\{x \in \mathbb{R} : x > 6\right\}$
and the range as $\left\{y \in \mathbb{R} : y > 0\right\}$.

#### Explanation:

First, we can simplify $p$ as given thusly:

$\frac{\sqrt[3]{x - 6}}{\sqrt[]{{x}^{2} - x - 30}} = \frac{\sqrt[3]{x - 6}}{\sqrt[]{\left(x - 6\right) \left(x + 5\right)}}$.

Then, further simplifying, we discern that
$\frac{\sqrt[3]{x - 6}}{\sqrt[]{\left(x - 6\right) \left(x + 5\right)}} = \frac{{\left(x - 6\right)}^{\frac{1}{3}}}{{\left(x - 6\right)}^{\frac{1}{2}} {\left(x + 5\right)}^{\frac{1}{2}}}$,

which, by means of dividing exponents, we deduce

$p \left(x\right) = \frac{1}{\sqrt[6]{x - 6} \sqrt[]{x + 5}}$ .

By seeing $p$ like this, we know that no $x$ can make $p \left(x\right) = 0$, and indeed $p \left(x\right)$ cannot be negative because the numerator is a positive constant and no even root (i.e. $2$ or $6$) can yield a negative number. Therefore the range of $p$ is $\left\{y \in \mathbb{R} : y > 0\right\}$.

Finding the domain is no more difficult. We know that the denominator cannot equal $0$, and by observing which values for $x$ would lead to thus, we find that $x$ must be greater than $6$. Thereby the domain of $p$ is $\left\{x \in \mathbb{R} : x > 6\right\}$.