Prove that the function $f$ given by $f(x) = |x - 1|, x \in R$ is not differentiable at $x = 1$.
(N/A) The given function is $f(x) = |x - 1|, x \in R$.
$A$ function $f$ is differentiable at $x = c$ if the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$ exist and are equal.
$LHD$ at $x = 1$:
$\mathop {\lim }\limits_{h \to {0^-}} \frac{f(1 + h) - f(1)}{h} = \mathop {\lim }\limits_{h \to {0^-}} \frac{|1 + h - 1| - |1 - 1|}{h} = \mathop {\lim }\limits_{h \to {0^-}} \frac{|h|}{h}$.
Since $h < 0$,$|h| = -h$,so the limit is $\mathop {\lim }\limits_{h \to {0^-}} \frac{-h}{h} = -1$.
$RHD$ at $x = 1$:
$\mathop {\lim }\limits_{h \to {0^+}} \frac{f(1 + h) - f(1)}{h} = \mathop {\lim }\limits_{h \to {0^+}} \frac{|1 + h - 1| - |1 - 1|}{h} = \mathop {\lim }\limits_{h \to {0^+}} \frac{|h|}{h}$.
Since $h > 0$,$|h| = h$,so the limit is $\mathop {\lim }\limits_{h \to {0^+}} \frac{h}{h} = 1$.
Since $LHD = -1$ and $RHD = 1$,$LHD \neq RHD$.
Therefore,the function $f(x) = |x - 1|$ is not differentiable at $x = 1$.
$A$ function $f$ is differentiable at $x = c$ if the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$ exist and are equal.
$LHD$ at $x = 1$:
$\mathop {\lim }\limits_{h \to {0^-}} \frac{f(1 + h) - f(1)}{h} = \mathop {\lim }\limits_{h \to {0^-}} \frac{|1 + h - 1| - |1 - 1|}{h} = \mathop {\lim }\limits_{h \to {0^-}} \frac{|h|}{h}$.
Since $h < 0$,$|h| = -h$,so the limit is $\mathop {\lim }\limits_{h \to {0^-}} \frac{-h}{h} = -1$.
$RHD$ at $x = 1$:
$\mathop {\lim }\limits_{h \to {0^+}} \frac{f(1 + h) - f(1)}{h} = \mathop {\lim }\limits_{h \to {0^+}} \frac{|1 + h - 1| - |1 - 1|}{h} = \mathop {\lim }\limits_{h \to {0^+}} \frac{|h|}{h}$.
Since $h > 0$,$|h| = h$,so the limit is $\mathop {\lim }\limits_{h \to {0^+}} \frac{h}{h} = 1$.
Since $LHD = -1$ and $RHD = 1$,$LHD \neq RHD$.
Therefore,the function $f(x) = |x - 1|$ is not differentiable at $x = 1$.
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