Answer
$[-2, 3] $
Work Step by Step
Given $f(x) = \sqrt {6 + x - x^2}$
In a even root function, the domain is defined such that the function is always at (if defined) or above 0, so $f(x) \geq 0$
$\sqrt {6 + x - x^2} \geq 0$
$6 + x - x^2 \geq 0$
$x^2 - x - 6 \leq 0$
$(x-3)(x+2) \leq 0$
Find the zeros of the expressions in the numerator AND the denominator
$x = 3, -2$
Test numbers in between those zero values to determine if the function is negative or positive
(-∞, -2] $(-)(-) = (+)$
[-2, 3] $(-)(+) = (-)$
[3,∞) $(+)(+) = (+)$
Thus the solution is $[-2, 3] $
