$\therefore {\mathit{f}}^{\prime }\left(\mathit{x}\right)=4-\frac{1}{2}\left(2\mathit{x}\right)=4-\mathit{x}$

$\Rightarrow 4-\mathit{x}=0$
$\Rightarrow \mathit{x}=4$
Then, we evaluate the $\mathit{f}$ at critical point $\mathit{x}=4$ and at the end points of the interval $\left[-2,\frac{9}{2}\right]$.
$\mathit{f}\left(4\right)=16-\frac{1}{2}\left(16\right)=16-8=8$
$\mathit{f}\left(-2\right)=-8-\frac{1}{2}\left(4\right)$
$=-8-2=-10$
$\mathit{f}\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}{\left(\frac{9}{2}\right)}^2$
$=18-\frac{81}{8}=7.875$
Thus, the absolute maximum value of $\mathit{f}$ on $\left[-2,\frac{9}{2}\right]$ is 8 occurring at $\mathit{x}=4$.