Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 46

Determine the absolute maximum and absolute minimum values of $f(x) = x^3 - 3x + 1$ on the interval $[0,3]$.

Taking the derivative of $f(x)$
$f'(x) = 3x^2 - 3$
Solving for critical numbers, when $f'(x) = 0$

$
\begin{equation}
\begin{aligned}
0 &= 3x^2 - 3\\
\\
3x^2 &= 3 \\
\\
x &= \pm 1
\end{aligned}
\end{equation}
$


We have either absolute maximum or minimum at $x = \pm 1$,
So,

$
\begin{equation}
\begin{aligned}
\text{when } x &= 1\\
\\
f(1) &= (1)^3 - 3(1) +1\\
\\
f(1) &= -1\\
\\
\\
\\
\text{when } x &= -1\\
\\
f(-1) &= (-1)^3 - 3(-1) + 1\\
\\
f(-1) &= 3
\end{aligned}
\end{equation}
$

Evaluating $f(x)$ at end points $x = 0$ and $x = 3$.

$
\begin{equation}
\begin{aligned}
\text{when } x &= 0\\
\\
f(0) &= (0)^3 - 3(0) +1\\
\\
f(0) &= 1\\
\\
\\
\\
\text{when } x &= 3\\
\\
f(3) &= (3)^3 - 3(3) + 1\\
\\
f(3) &= 19
\end{aligned}
\end{equation}
$


Therefore, we have absolute maximum value at $f(3) = 19$ and absolute minimum value $f(1) = -1$ on the interval $[0,3]$