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If $f(x) = |x - 1| + |x - 3|$, then $f'(2)$ equals:

A$0$
B$2$
C$1$
D$-1$
Answer & Solution
Correct answer: A. $0$
For $1 < x < 3$: $|x - 1| = x - 1$, $|x - 3| = 3 - x$. So $f(x) = (x-1) + (3-x) = 2$ (constant!). **Hence $f'(2) = 0$.**
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