Write the points of non-differentiability
Question:

Write the points of non-differentiability of $f(x)=|\log | x||$.

Solution:

We have,
f (x) = |log |x||

$|x|=\left\{\begin{array}{cl}-x & -\infty<x<-1 \\ -x & -1<x<0 \\ x & 0<x<1 \\ x & 1<x<\infty\end{array}\right.$

$\log |x|=\left\{\begin{array}{cc}\log (-x) & -\infty<x<-1 \\ \log (-x) & -1<x<0 \\ \log (x) & 0<x<1 \\ \log (x) & 1<x<\infty\end{array}\right.$

$|\log | x||=\left\{\begin{array}{lc}\log (-x) & -\infty<x<-1 \\ -\log (-x) & -1<x<0 \\ -\log (x) & 0<x<1 \\ \log (x) & 1<x<\infty\end{array}\right.$

$(\mathrm{LHD}$ 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