# 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||

$|x|=\left\{\begin{array}{cl}-x & -\infty$\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