Explanation

(b)
$f(x) = x(x - 1)^{2}$
$f^{'}(x) = 2x(x - 1) + (x - 1)^{2}$
$= (x - 1)(2x + x - 1) = (x - 1)(3x - 1)$
$\therefore\ \ \ \ \ \ f^{'}(c) = \frac{f(2) - f(0)}{2 - 0}$
$\Rightarrow \ \ \ \ (c - 1)(3c - 1) = \frac{2 - 0}{2} = 1$
$\Rightarrow \ \ \ 3c^{2} - 4c = 0$
$\Rightarrow \ \ \ c(3c - 4) = 0$
$\Rightarrow \ \ \ c = 0$ or $c = \frac{4}{3}$
$\therefore$ The value of $c$ in $(0,\ 2)$ is $\frac{4}{3}$