Answer
$\left(-\infty, 4 \right]$
Work Step by Step
$\bf{\text{Solution Outline:}}$
The domain of the given function, $
f(x)=\sqrt{8-2x}
,$ is all values of $x$ for which the radicand is non-negative. Express the answer in interval notation.
$\bf{\text{Solution Details:}}$
Since the radicand of a radical with an even index has to be non-negative, then
\begin{array}{l}\require{cancel}
8-2x\ge0
\\\\
-2x\ge-8
.\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}
-2x\ge-8
\\\\
x\le\dfrac{-8}{-2}
\\\\
x\le4
.\end{array}
Hence the domain is $
\left(-\infty, 4 \right]
$