Explanation

(c)
Let $f(x) = |x - 1| + |x - 3|$
$= \left\{ \begin{array}{r}
x - 1\ \ \ \ \ + x - 3\ \ \ \ ,\ x \geq 3 \\
x - 1 + 3 - x,\ \ \ \ 1 \leq x < 3 \\
1 - x\ \ \ \ + 3 - x,\ \ \ x \leq 1
\end{array} \right.\$
$= \left\{ \begin{array}{r}
2x - 4,\ x \geq 3 \\
2,\ 1 \leq x < 3 \\
4 - 2x,\ x \leq 1
\end{array} \right.\$
At $x = 2$, function is
$f(x) = 2$
$\Rightarrow \ \ f^{'}(x) = 0$