Answer
Absolute maximum: $f(2)=16$
Absolute minimum: $f(5) = 7$
Work Step by Step
Using closed interval method, as $f$ in continuous over the given range, let's first calculate the value of $f$ at critical numbers followed by values of $f$ at endpoints of the given range.
For critical numbers, $f'(x)=0$ or $f'(x)$ should not exist. Let's differentiate $f(x)=12+4x-x^2$ with respect to $x$ keeping $[0,5]$ as interval.
$$\frac{df}{dx}=4-2x$$ $\frac{df}{dx}$ exists on the given interval. Thus, $\frac{df}{dx}=0$, $$4-2x=0\implies x=2.$$
Thus, $f(2)=12+4(2)-2^2=16$. At endpoints,
$f(0)=12$ & $f(5)=7$. Thus, absolute maximum $=f(2)=16$ and absolute minimum $=f(5)=7$.
