The value of frac{d}{dx}[|x - 1| + |x - 5|] a | vedclass

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(B) Let $f(x) = |x - 1| + |x - 5|$.We define the function based on the intervals of $x$:$f(x) = \begin{cases} -(x - 1) - (x - 5), & x < 1 \ (x - 1) -

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$0$

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(B) Let $f(x) = |x - 1| + |x - 5|$.We define the function based on the intervals of $x$:$f(x) = \begin{cases} -(x - 1) - (x - 5), & x  5 \end{cases}$Simplifying the expressions:$f(x) = \begin{cases} 6 - 2x, & x  5 \end{cases}$Since $x = 3$ lies in the interval $(1, 5)$,we have $f(x) = 4$ for all $x$ in this interval.Therefore,the derivative $f'(x) = \frac{d}{dx}(4) = 0$ for $x \in (1, 5)$.Thus,at $x = 3$,the value is $0$.

The value of $\frac{d}{dx}[|x - 1| + |x - 5|]$ at $x = 3$ is

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