How do you find the domain and range of root3( (x^2-x-12))?

Aug 27, 2017

Domain: $\left(- \infty , \infty\right)$

Range: $\left[- \frac{\sqrt[3]{98}}{2} , \infty\right)$

Explanation:

Given:

$f \left(x\right) = \sqrt[3]{{x}^{2} - x - 12}$

We can convert the quadratic radicand into vertex form by completing the square:

${x}^{2} - x - 12 = {x}^{2} - 2 \left(\frac{1}{2}\right) x + {\left(\frac{1}{2}\right)}^{2} - {\left(\frac{1}{2}\right)}^{2} - 12$

$\textcolor{w h i t e}{{x}^{2} - x - 12} = {\left(x - \frac{1}{2}\right)}^{2} - \left(\frac{1}{4} + \frac{48}{4}\right)$

$\textcolor{w h i t e}{{x}^{2} - x - 12} = {\left(x - \frac{1}{2}\right)}^{2} - \frac{49}{4}$

So the minimum value $- \frac{49}{4}$ of the radicand occurs when $x = \frac{1}{2}$

Then:

$f \left(\frac{1}{2}\right) = \sqrt[3]{- \frac{49}{4}} = - \sqrt[3]{\frac{98}{8}} = - \frac{\sqrt[3]{98}}{2}$

Hence the range of $f \left(x\right)$ is $\left[- \frac{\sqrt[3]{98}}{2} , \infty\right)$.

Its domain is the whole of $\mathbb{R} = \left(- \infty , \infty\right)$, since it is well defined for any value of $x$.