f(x)=(x-3)^3 [0,4]
f(x)=x^3-12X [1,4]
1.$\displaystyle f'(x)=[(x-3)^3]'=[x^3-9x^2+27x-27]'=3x^2-18x+27$
$\displaystyle f'(x)=0 \Longrightarrow x=3$
Notice, that the derivative has a double root from what results, that the $\displaystyle f$ function is a hyperbola with focus in $\displaystyle x=3 \in [0,4]$,
the extrema are then $\displaystyle f(0)$ and $\displaystyle f(4)$:
$\displaystyle f(0)=-27$
$\displaystyle f(4)=1$
2.$\displaystyle f'(x)=[x^3-12x]'=3x^2-12=3(x+2)(x-2)$, $\displaystyle x=2$ belongs to the interval,
all we have to do now is to check the $\displaystyle f$ function values for the variable of $\displaystyle 1$, $\displaystyle 2$ and $\displaystyle 4$ and pick the highest and the lowest one.