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 $[-1,2]$