Find the absolute maximum value and the absolute minimum value of the function given by $f(x) = 4x - \frac{1}{2}x^2$ for $x \in \left[-2, \frac{9}{2}\right]$.

(N/A) The given function is $f(x) = 4x - \frac{1}{2}x^2$.
First,we find the derivative of the function:
$f'(x) = \frac{d}{dx}(4x - \frac{1}{2}x^2) = 4 - x$.
To find the critical points,we set $f'(x) = 0$:
$4 - x = 0 \implies x = 4$.
Since $x = 4$ lies within the interval $\left[-2, \frac{9}{2}\right]$,we evaluate the function at the critical point and the endpoints of the interval:
$f(4) = 4(4) - \frac{1}{2}(4)^2 = 16 - 8 = 8$.
$f(-2) = 4(-2) - \frac{1}{2}(-2)^2 = -8 - 2 = -10$.
$f\left(\frac{9}{2}\right) = 4\left(\frac{9}{2}\right) - \frac{1}{2}\left(\frac{9}{2}\right)^2 = 18 - \frac{81}{8} = 18 - 10.125 = 7.875$.
Comparing these values,the absolute maximum value is $8$ at $x = 4$ and the absolute minimum value is $-10$ at $x = -2$.