Explanation

(a)
We have, $\ f(x) = x^{3} - 3x^{2} + 2x + 10$
$f^{'}(x) = 3x^{2} - 6x + 2$
Put $f^{'}(x) = 0,\ \ \ \ \ \ \ \ \ \ \ \ \ 3x^{2} - 6x + 2 = 0$
$\Rightarrow \ \ x = \frac{6 \pm \sqrt{36 - 24}}{2 \times 3}\$
$= \frac{3 + \sqrt{3}}{3}$ $\begin{bmatrix}
\because x = \frac{3 - \sqrt{3}}{3}\ does\ not\ lie \\
in\ the\ given\ interval\ \ \ \ \ \ \ \ \ \
\end{bmatrix}$