Answer
Absolute maximum: $f(3)=113$
Absolute minimum: $f(0) = 5$
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 $f(x)=5+54x-2x^3$ with respect to $x$ keeping $[0,4]$ as interval.
$$\frac{df}{dx}=54-6x^2$$ $\frac{df}{dx}$ exists on the given interval. Thus, $\frac{df}{dx}=0$, $$54-6x^2=0\implies x^2=9.$$ $$x=-3,3$$
Thus, $f(3)=5+54(3)-2(3)^3=113$. Not considering $x=-3$ because not in given interval. At endpoints,
$f(0)=5$ & $f(4)=93$. Thus, absolute maximum $=f(3)=113$ and absolute minimum $=f(0)=5$.
