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If $f: R \rightarrow R$ is defined as $f(x)=|x+1|+|x-1|$, then $f(x)$ is
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not differentiable at - 1and 1 only
Given, $f(x)=|x+1|+|x-1|$
$f(x)$ has a corner point only at $x=1,-1$ by the graph.
So, $f(x)$ is not differentiable at $x=-1,1$ only.
$f(x)$ has a corner point only at $x=1,-1$ by the graph.
So, $f(x)$ is not differentiable at $x=-1,1$ only.
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