Write the points of non-differentiability of $f(x)=|\log | x||$.
We have,
f (x) = |log |x||
$|x|=\left\{\begin{array}{cl}-x & -\infty
$\log |x|=\left\{\begin{array}{cc}\log (-x) & -\infty
$|\log | x||=\left\{\begin{array}{lc}\log (-x) & -\infty
$(\mathrm{LHD}$ at $x=-1)=\lim _{x \rightarrow-1^{-}} \frac{f(x)-f(-1)}{x+1}$
$=\lim _{x \rightarrow-1^{-}} \frac{\log (-x)-0}{x+1}$
$=\lim _{h \rightarrow 0} \frac{\log (1+h)}{-1-h+1}$
$=-\lim _{h \rightarrow 0} \frac{\log (1+h)}{h}=-1$
$(\mathrm{RHD}$ at $x=-1)=\lim _{x \rightarrow-1^{+}} \frac{f(x)-f(-1)}{x+1}$
$=\lim _{x \rightarrow-1^{+}} \frac{-\log (-x)-0}{x+1}$
$=\lim _{h \rightarrow 0} \frac{-\log (1-h)}{-1+h+1}$
$=\lim _{h \rightarrow 0} \frac{-\log (1-h)}{h}=1$
Here, LHD ≠ RHD
So, function is not differentiable at x = − 1
At 0 function is not defined.
$(\mathrm{LHD}$ at $x=1)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}$
$=\lim _{x \rightarrow 1^{-}} \frac{-\log (x)-0}{x-1}$
$=\lim _{h \rightarrow 0} \frac{-\log (1-h)}{1-h-1}$
$=-\lim _{h \rightarrow 0} \frac{\log (1-h)}{h}=-1$
$(\mathrm{RHD}$ at $x=1)=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}$
$=\lim _{x \rightarrow 1^{+}} \frac{\log (x)-0}{x-1}$
$=\lim _{h \rightarrow 0} \frac{\log (1+h)}{1+h-1}$
$=\lim _{h \rightarrow 0} \frac{\log (1+h)}{h}=1$
Here, LHD ≠ RHD
So, function is not differentiable at x = 1
Hence, function is not differentiable at x = 0 and ± 1
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