Discuss the continuity and differentiability of f (x) = |log |x||.

Question:

Discuss the continuity and 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
At 0 function is not defined.
So, at 0 function is not continuous.

$(\mathrm{LHL}$ at $x=-1)=\lim _{x \rightarrow-1^{-}} f(x)$

$=\lim _{x \rightarrow-1^{-}} \log (-x)$

$=\log (1)=0$

$(\mathrm{RHL}$ at $x=-1)=\lim _{x \rightarrow-1^{+}} f(x)$

$=\lim _{x \rightarrow-1^{+}}-\log (-x)$

$=-\log 1=0$

$f(-1)=0$

Therefore, $f(x)=|\log | x||$ is continuous at $x=-1$'

$(\mathrm{LHL}$ at $x=1)=\lim _{x \rightarrow 1^{-}} f(x)$

$=\lim _{x \rightarrow 1^{-}}-\log (x)$

$=-\log (1)=0$

$(\mathrm{RHL}$ at $x=1)=\lim _{x \rightarrow 1^{+}} f(x)$

$=\lim _{x \rightarrow 1^{+}} \log (x)$

$=\log 1=0$

$f(1)=0$

Therefore, at $x=1, f(x)=|\log | x||$ is continuous.

Hence, function $f(x)=\| \log |x| \mid$ is not continuous at $x=0$

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