Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 50

Determine the absolute maximum and absolute minimum values of $f(x) = \left(x^2 -1\right)^3$ on the interval $[-1,2]$.

Taking the derivative of $f(x)$

$
\begin{equation}
\begin{aligned}
f'(x) &= 3\left(x^2 -1\right)^2(2x)\\
\\
f'(x) &= 6x\left(x^2 -1\right)^2
\end{aligned}
\end{equation}
$

Solving for critical numbers, when $f'(x) = 0$
$0 = 6x \left(x^2 -1\right)^2$

Hence,

$
\begin{equation}
\begin{aligned}
x &= 0 \text{ and } \left(x^2 -1\right) = 0\\
\\
x &= 0 \text{ and } x = \pm 1
\end{aligned}
\end{equation}
$



We have either absolute maximum and minimum values at $x = 0$ and $x = \pm 1$
So,

$
\begin{equation}
\begin{aligned}
\text{when } x &= 0\\
\\
f(0) &= \left(0^2 -1\right)^3\\
\\
f(0) &= -1\\
\\
\\
\\
\text{when } x &= 1\\
\\
f(1) &= \left(1^2 -1\right)^3 \\
\\
f(1) &= 0\\
\\
\\
\\
\text{when } x &= -1\\
\\
f(-1) &= \left((-1)^2 -1\right)^3 \\
\\
f(-1) &= 0\\
\end{aligned}
\end{equation}
$

Evaluating $f(x)$ at end points $x = -1$ and $x = 2$

$
\begin{equation}
\begin{aligned}
\text{when } x &= -1\\
\\
f(-1) &= \left((-1)^2 -1\right)^3 \\
\\
f(-1) &= 0\\
\\
\\
\\
\text{when } x &= 2\\
\\
f(4) &= \left((2)^2 -1\right)^3 \\
\\
f(4) &= 27\\
\end{aligned}
\end{equation}
$

Therefore, we have absolute maximum value at $f(2) = 27$ and the absolute minimum value at $f(0) = -1 $ on the interval $[-1,2]$