## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\left(- \infty, 2 \right]$
$\bf{\text{Solution Outline:}}$ The domain of the given function, $f(x)=\sqrt{6-3x} ,$ is all the values of $x$ for which the radicand is greater than or equal to $0.$ Express the answer in the interval notation. $\bf{\text{Solution Details:}}$ Since the radicand should be greater than or equal to zero, then \begin{array}{l}\require{cancel} 6-3x\ge0 \\\\ -3x\ge-6 .\end{array} Dividing both sides by a negative number (and consequently reversing the inequality symbol), the inequality above is equivalent to \begin{array}{l}\require{cancel} -3x\ge-6 \\\\ x\le\dfrac{-6}{-3} \\\\ x\le2 .\end{array} Hence, the domain is $\left( -\infty, 2 \right] .$