Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 126

Illustrate the $f$ and $f'$ of the function $f(x) = x^2 (x - 2) (x + 2)$.
Then, estimate points at which the tangent line to $f$ is horizontal. If no such points exists, state that fact.


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


Then, by using product rule,

$
\begin{equation}
\begin{aligned}
f'(x) &= x^2 \cdot \frac{d}{dx} (x^2 - 4) + (x^2 - 4) \cdot \frac{d}{dx} (x^2)\\
\\
&= x^2(2x) + (x^2 - 4)(2x)\\
\\
&= 2x^3 + 2x^3 - 8x\\
\\
&= 4x^3 - 8x
\end{aligned}
\end{equation}
$




Based from the graph, the points at which the tangent line to $f$ is horizontal (slope = 0) are
at $x \approx -1.40, x= 0$ and $x \approx 1.40$