Damion Ellis

2023-02-18

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

Cheyenne Lynn

Beginner2023-02-19Added 10 answers

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 $x=\pm 1$. 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]}

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 $x=\pm 1$. 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]}