Explanation

(b)
Let $f(x) = |x - 1| + |x - 5|$
$f(x) = \left\{ \begin{matrix}
- (x - 1) - (x - 5), & x < 1 \\
(x - 1) - (x - 5), & 1 \leq x \leq 5 \\
x - 1 + x - 5, & x > 5
\end{matrix} \right.\$
$f(x) = \left\{ \begin{array}{r}
\begin{matrix}
6 - 2x, & x < 1 \\
4, & x \leq 5
\end{matrix} \\
2x - 6,\ \ \ x > 5
\end{array} \right.\$
$\because x = 3 \in (1,\ 5)$
$\therefore For\ x = 3$,
$f(x) = 4 \Rightarrow f^{'}(x) = 0$