Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 130

Illustrate the $f$ and $f'$ of the function $\displaystyle \frac{0.01x^2}{x^4 + 0.0256}$.
Then, estimate points at which the tangent line to $f$ is horizontal. If no such points exists, state that fact.

If $\displaystyle f(x) = \frac{0.01x^2}{x^4 + 0.0256}$, then by applying Quotient Rule

$
\begin{equation}
\begin{aligned}
f'(x) &= \frac{(x^4 + 0.0256) \cdot \frac{d}{dx} (0.01x^2) - (0.01x^2) \cdot \frac{d}{dx} (x^4 + 0.0256) }{(x^4 + 0.0256)^2}\\
\\
f'(x) &= \frac{(x^4 + 0.0256)(0.02x) - (0.01x^2)(4x^3)}{(x^4 + 0.0256)^2}\\
\\
f'(x) &= \frac{0.02x^5 + 0.000512x - 0.04x^5}{(x^4 + 0.0256)^2}\\
\\
f'(x) &= \frac{-0.02x^5 + 0.000512x}{(x^4 + 0.0256)^2}
\end{aligned}
\end{equation}
$




Based from the graph, the points at which the tangent line to $f$ is horizontal (slope = 0) are
at $x = -0.4, x = 0$ and $x = 0.4$