#### Answer

$(-\infty,11]$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
The domain of the given function, $
f(x)=\sqrt{11-x}
,$ 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}
11-x\ge0
\\\\
-x\ge-11
.\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}
-x\ge-11
\\\\
\dfrac{-x}{-1}\ge\dfrac{-11}{-1}
\\\\
x\le11
.\end{array}
Hence the domain is $
(-\infty,11]
.$