Damion Ellis

2023-02-18

What are the absolute extrema of $f\left(x\right)={x}^{3}-3x+1\in \left[0,3\right]$?

Cheyenne Lynn

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

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