#### Answer

$[-3,2)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
The domain of the given function, $
f(x)=\dfrac{\sqrt{x+3}}{\sqrt[4]{2-x}}
,$ are the combined restrictions of the numerator and the denominator.
$\bf{\text{Solution Details:}}$
In the numerator, since the radicand of a radical with an even index (index equals $2$), should be nonnegative, then
\begin{array}{l}\require{cancel}
x+3\ge0
\\\\
x\ge-3
.\end{array}
In the denominator, since the radicand of a radical with an even index (index equals $4$), should be nonnegative, and that the denominator cannot be zero, then
\begin{array}{l}\require{cancel}
2-x\gt0
\\\\
-x\gt-2
\\\\
x\lt\dfrac{-2}{-1}
\\\\
x\lt2
.\end{array}
Combining the two restrictions above, then the domain is the interval $
[-3,2)
.$