The value of the derivative of |x−1|+|x−3| at x=2 is:

A
2
B
-2
C
0
D
not defined

Solution:

Let f(x) = I x- 1l + I x-3 I
At x= 2,
$\begin{array}{l} | x - 1 | = x - 1 and | x - 3 | = - x + 3 \\ \Rightarrow f (x) = x - 1 - x + 3 = 2 \end{array}$
which is constant function. $\Rightarrow$ f ' (2) = 0

The value of the derivative of $| x - 1 | + | x - 3 |$ at x=2 is:

Solution:

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