Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 128

Illustrate the $f$ and $f'$ of the function $\displaystyle f(x) = \frac{x^3 -1}{x^2 + 1}$.
Then, estimate points at which the tangent line to $f$ is horizontal. If no such points exists, state that fact.

If $\displaystyle f(x) = \frac{x^3 - 1}{x^2 + 1}$, then by applying Quotient Rule

$
\begin{equation}
\begin{aligned}
f'(x) &= \frac{(x^2 + 1) \cdot \frac{d}{dx} (x^3 - 1) - (x^3 - 1) \cdot \frac{d}{dx} (x^2 + 1) }{(x^2 + 1)^2}\\
\\
&= \frac{(x^2 + 1)(3x^2) - (x^3 - 1)(2x)}{(x^2 + 1)^2}\\
\\
&= \frac{3x^4 + 3x^2 - 2x^4 + 2x}{(x^2 + 1)^2}\\
\\
&= \frac{x^4 + 3x^2 + 2x}{(x^2 + 1)^2}
\end{aligned}
\end{equation}
$




Based from the graph, the points at which the tangent line to $f$ is horizontal (slope = 0) are
at $x \approx -0.6$ and $x = 0$