#### Answer

$\left(- \infty, 2 \right]$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
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]
.$