The value of the derivative of |x - 1| + |x - | vedclass

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(B) Let $f(x) = |x - 1| + |x - 3|$.For $x$ in the interval $(1, 3)$,we have $x - 1 > 0$ and $x - 3 < 0$.Thus,$|x - 1| = x - 1$ and $|x - 3| = -(x - 3)

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$0$

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(B) Let $f(x) = |x - 1| + |x - 3|$.For $x$ in the interval $(1, 3)$,we have $x - 1 > 0$ and $x - 3 Thus,$|x - 1| = x - 1$ and $|x - 3| = -(x - 3) = -x + 3$.Substituting these into the function,we get $f(x) = (x - 1) + (-x + 3) = 2$.Since $f(x) = 2$ for all $x \in (1, 3)$,the function is constant in this neighborhood of $x = 2$.The derivative of a constant function is $0$.Therefore,$f'(2) = 0$.

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

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