Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 94

Illustrate the $f$ and $f'$ of the function $f(x) = 1.68x \sqrt{9.2 - x^2}$ over the
given interval $[-3,3]$. Then estimate points at which the tangent line is horizontal.

If $f(x) = 1.68x(9.2 - x^2)^{\frac{1}{2}}$, then by using Chain Rule and Product Rule


$
\begin{equation}
\begin{aligned}
f'(x) &= 1.68x \cdot \frac{d}{dx} \left[(9.2 - x^2)^{\frac{1}{2}} \right] + (9.2 - x^2)^{\frac{1}{2}} \cdot \frac{d}{dx} (1.68x)\\
\\
f'(x) &= 1.68x \cdot \frac{1}{2} (9.2 - x^2)^{\frac{1}{2} - 1} \cdot \frac{d}{dx} (9.2 - x^2) + (9.2 - x^2)^{\frac{1}{2}} (1.68)\\
\\
f'(x) &= \frac{1.68x}{2} (9.2 - x^2)^{-\frac{1}{2}} ( -2x) + (9.2 - x^2)^{\frac{1}{2}} ( 1 .68)\\
\\
f'(x) &= \frac{-1.68 x^2}{(9.2 - x^2)^{\frac{1}{2}}} + 1.68 (9.2 - x^2)^{\frac{1}{2}}\\
\\
f'(x) &= 1.68 \left[ \frac{-x^2 + 9.2 - x^2}{(9.2 - x^2)^{\frac{1}{2}}} \right]\\
\\
f'(x) &= \frac{15.456 - 3.36x^2}{(9.2- x^2)^{\frac{1}{2}}}
\end{aligned}
\end{equation}
$


Thus, the graph of $f$ and its derivative is



Based from the graph, the points at which the tangent line is horizontal (slope = 0) are
$x \approx -2.15$ and $x \approx 2.15$