Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 48

Determine the absolute maximum and absolute minimum values of $f(x) = x^3 - 6x^2 + 9x + 2$ on the interval $[-1,4]$.

Taking the derivative of $f(x)$

$
\begin{equation}
\begin{aligned}
f'(x) &= 3x^2 - 12x + 9\\
\\
f'(x) &= 3\left(x^2-4x+3\right)
\end{aligned}
\end{equation}
$

Solving for critical numbers, when $f'(x) = 0$

$
\begin{equation}
\begin{aligned}
0 &= 3(x^2 - 4x + 3)\\
\\
0 &= x^2 - 4x +3
\end{aligned}
\end{equation}
$


Using Quadratic formula, we have $x = 3 $ and $x = 1$

We have either absolute maximum and minimum values at $x = 3$ and $x = 1$
So,

$
\begin{equation}
\begin{aligned}
\text{when } x &= 3\\
\\
f(3) &= (3)^3 - 6(3)^2 + 9(3) +2 \\
\\
f(3) &= 2\\
\\
\\
\\
\text{when } x &= 1\\
\\
f(1) &= (1)^3 - 6(1)^2 + 9(1) +2 \\
\\
f(1) &= 6\\
\end{aligned}
\end{equation}
$

Evaluating $f(x)$ at end points $x = -1$ and $x = 4$

$
\begin{equation}
\begin{aligned}
\text{when } x &= -1\\
\\
f(-1) &= (-1)^3 - 6(-1)^2 + 9(-1) +2 \\
\\
f(-1) &= -14\\
\\
\\
\\
\text{when } x &= 4\\
\\
f(4) &= (4)^3 - 6(4)^2 + 9(4) +2 \\
\\
f(4) &= 6\\
\end{aligned}
\end{equation}
$

Therefore, we have absolute maximum value at $f(1) = f(4) = 6$ and absolute minimum value at $f(-1) = -14$ on the interval $[-1,4]$