f(x) = ||x| - 1| is not differentiable at | vedclass

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(B) The function is defined as $f(x) = ||x| - 1|$.We can break this down based on the sign of the inner expression:$f(x) = \begin{cases} |x| - 1, & |x

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$\pm 1, 0$

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(B) The function is defined as $f(x) = ||x| - 1|$.We can break this down based on the sign of the inner expression:$f(x) = \begin{cases} |x| - 1, & |x| - 1 \ge 0 \ -(|x| - 1), & |x| - 1 $f(x) = \begin{cases} |x| - 1, & |x| \ge 1 \ 1 - |x|, & |x| This can be further expanded for different intervals of $x$:$f(x) = \begin{cases} -x - 1, & x \le -1 \ x + 1, & -1 $A$ function is not differentiable at points where there are sharp corners (cusps) in its graph.By observing the graph,the function has sharp corners at $x = -1$,$x = 0$,and $x = 1$.Therefore,$f(x)$ is not differentiable at $x \in \{-1, 0, 1\}$.

$f(x) = ||x| - 1|$ is not differentiable at

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