Answer
[-3, 8/3]
Work Step by Step
The question asks for the domain of the function given in the problem
Given $f(x) = \sqrt {24 - x - 3x^2}$
The domain is such that $24 - x - 3x^2$ is greater than or equal to 0.
$24 - x - 3x^2 \geq 0$
$3x^2 + x - 24 \leq 0$
$(3x - 8) (x + 3) \leq 0$
$x = 8/3, -3$
(-∞, -3] $(-)(-) = (+)$
[-3, 8/3] $(+)(-) = (-)$
[8/3, ∞) $(+)(+) = (+)$
Thus the Domain is [-3, 8/3]