Answer
$\left[-3, \frac{8}{3}\right] $
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a polynomial function.
$f(x)=\sqrt {24-x-3x^2},$
$\sqrt {24-x-3x^2} \geq 0,$
$24-x-3x^2\geq 0,$
$-3x^2-9x+8x+24,$
$-3x(x+3)+8(x+3),$
$(-3x+8)(x+3),$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(-3x+8)(x+3)=0$
$x=-3$ or $x=\frac{8}{3}$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $g(x)=24-x-3x^2$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & factors,signs & g(a),signs \\
& &(-3a+8)(a+3)& \\
(-\infty, -3) & -5 & (+)(-)=(-) & Undefined\\
(-3, \frac{8}{3}) & 0 & (+)(+)=(+) & (+)\\
(\frac{8}{3},\infty) & 5 & (-)(+)=(-) & Undefined
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
The domain is: $\left[-3, \frac{8}{3}\right] $
