If the solve the problem

(i)

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

$\Rightarrow f^{\prime}(x)=4-x$

For a local maximum or a local minimum, we must have

$f^{\prime}(x)=0$

$\Rightarrow 4-x=0$

$\Rightarrow x=4$

Thus, the critical points of $f$ are $-2,4$ and $4.5$.

Now,

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

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

$f(4.5)=4(4.5)-\frac{(4.5)^{2}}{2}=18-10.125=7.875$

Hence, the absolute maximum value when $x=4$ is 8 and the absolute minimum value when $x=-2$ is $-10$.