The absolute maximum value of the function $f(x) = 4x - \frac{1}{2}x^2$ in the interval $[-2, \frac{9}{2}]$ is:

$8$

The correct answer is Option (1) → $8$ ##

Given, $f(x) = 4x - \frac{1}{2}x^2$

$∴f'(x) = 4 - \frac{1}{2}(2x) = 4 - x$

Put $f'(x) = 0$

$\Rightarrow 4 - x = 0$

$\Rightarrow x = 4$

Then, we evaluate the $f$ at critical point $x = 4$ and at the end points of the interval $[-2, \frac{9}{2}]$.

$f(4) = 16 - \frac{1}{2}(16) = 16 - 8 = 8$

$f(-2) = -8 - \frac{1}{2}(4) = -8 - 2 = -10$

$f\left(\frac{9}{2}\right) = 4\left(\frac{9}{2}\right) - \frac{1}{2}\left(\frac{9}{2}\right)^2$

$= 18 - \frac{81}{8} = 7.875$

Thus, the absolute maximum value of $f$ on $[-2, \frac{9}{2}]$ is 8 occurring at $x=4$.