Explanation

(b)
$\because\ \ f^{'}(x) = 2\left( \frac{1}{3} \right)\sin\left( \frac{x}{6} \right)\cos\left( \frac{x}{6} \right) + \left( \frac{1}{3} \right)\cos\frac{x}{3} - \left( \frac{1}{3} \right)$
$= \left( \frac{1}{3} \right)\left\lbrack 2\sin\left( \frac{x}{6} \right)\cos\left( \frac{x}{6} \right) - 2\sin^{2}\left( \frac{x}{6} \right) \right\rbrack$
$= \left( \frac{2}{3} \right)\sin\left( \frac{x}{6} \right)\cos{\left( \frac{x}{6} \right) - 2\sin\left( \frac{x}{6} \right)}$
Put $f^{'}(x) = 0\ \ \ \Leftrightarrow \ \ \sin\left( \frac{x}{6} \right) = 0$
$\Rightarrow \ \tan\left( \frac{x}{6} \right) = 1$
$\Rightarrow \ \frac{x}{6} = k\pi,\ k \in I\ \ or\ \frac{x}{6} = n\pi + \frac{\pi}{4},\ \ n \in I$
$x^{2} - 10 < - 19.5\ x$
$\Leftrightarrow \ \ \ (x + 9.75)^{2} < 105.0625$
$\Leftrightarrow \ \ (x - 0.5)(x + 20) < 0$
$\Leftrightarrow \ - 20 < x < 0.5$
So, the critical points satisfying the last inequality will be $0,\ 6\pi,\ - \frac{9\pi}{2}$