Question

Write the points of non-differentiability of 

\[f \left( x \right) = \left| \log \left| x \right| \right| .\]

Answer 1

We have,
f (x) = |log |x|| 

`|x| = {(-x,-infty<x<-1),(-x, -1<x<0),(x, 0<x <1),(x, 1< x <infty):}`

`log |x| ={ (log(-x), - infty<x<-1),(log-x ,-1<x<0),(log(x), 0 <x<1),(log(x) , 1<x<infty):}`

`|log|x|| = {(log (-x), -infty<x<-1),(-log(-x),-1<x<0),(-log(x), 0<x<1),(log (x), 1 <x<infty):}`

\[\left( \text { LHD at x = } - 1 \right) = \lim_{x \to - 1^-} \frac{f\left( x \right) - f\left( - 1 \right)}{x + 1}\]
\[ = \lim_{x \to - 1^-} \frac{\log \left( - x \right) - 0}{x + 1}\]
\[ = \lim_{h \to 0} \frac{\log \left( 1 + h \right)}{- 1 - h + 1}\]
\[ = - \lim_{h \to 0} \frac{\log \left( 1 + h \right)}{h} = - 1\]

\[\left( \text { RHD at x } = - 1 \right) = \lim_{x \to - 1^+} \frac{f\left( x \right) - f\left( - 1 \right)}{x + 1}\]
\[ = \lim_{x \to - 1^+} \frac{- \log \left( - x \right) - 0}{x + 1}\]
\[ = \lim_{h \to 0} \frac{- \log \left( 1 - h \right)}{- 1 + h + 1}\]
\[ = \lim_{h \to 0} \frac{- \log \left( 1 - h \right)}{h} = 1\]

Here, LHD ≠ RHD
So, function is not differentiable at x = − 1

At 0 function is not defined.

\[\left( \text { LHD at x } = 1 \right) = \lim_{x \to 1^-} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1}\]
\[ = \lim_{x \to 1^-} \frac{- \log \left( x \right) - 0}{x - 1}\]
\[ = \lim_{h \to 0} \frac{- \log \left( 1 - h \right)}{1 - h - 1}\]
\[ = - \lim_{h \to 0} \frac{\log \left( 1 - h \right)}{h} = - 1\]

\[\left( \text { RHD at x } = 1 \right) = \lim_{x \to 1^+} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1}\]
\[ = \lim_{x \to 1^+} \frac{\log \left( x \right) - 0}{x - 1}\]
\[ = \lim_{h \to 0} \frac{\log \left( 1 + h \right)}{1 + h - 1}\]
\[ = \lim_{h \to 0} \frac{\log \left( 1 + h \right)}{h} = 1\]

Here, LHD ≠ RHD
So, function is not differentiable at x = 1
Hence, function is not differentiable at x = 0 and ± 1