What are the absolute extrema of f ( x ) = x 3 - 3...

There are two potential candidates for an interval's absolute extrema. They are the endpoints of the interval (here, ${\displaystyle 0}$ and ${\displaystyle 3}$) and the critical values of the function located within the interval.
The critical values can be found by finding the function's derivative and finding for which values of ${\displaystyle x}$ it equals ${\displaystyle 0}$.
We can use the power rule to find that the derivative of ${\displaystyle f\left(x\right)=x^3-3x+1}$ is ${\displaystyle f\prime \left(x\right)=3x^2-3}$.
The critical values are when ${\displaystyle 3x^2-3=0}$, which simplifies to be ${\displaystyle x=\pm 1}$. However, ${\displaystyle x=-1}$ is not in the interval so the only valid critical value here is the one at ${\displaystyle x=1}$. We now know that the absolute extrema could occur at ${\displaystyle x=0,x=1,}$ and ${\displaystyle x=3}$.
To determine which is which, plug them all into the original function.
${\displaystyle f\left(0\right)=1}$
${\displaystyle f\left(1\right)=-1}$
${\displaystyle f\left(3\right)=19}$
From here we can see that there is an absolute minimum of ${\displaystyle -1}$ at ${\displaystyle x=1}$ and an absolute maximum of ${\displaystyle 19}$ at ${\displaystyle x=3}$.
Check the function's graph:
graph{x^3-3x+1 [-0.1, 3.1, -5, 20]}