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Solution:

(b) f (x) is continuous for all x in its domain but not differentiable at x = ± 1

We have, $f(x)=\left|\log _{e}\right| x||$

We know that $\log$ function is defined for positive value.

Here, $|x|$ is positive for all non zero $x$.

Therefore, domain of function is $R-\{0\}$

And we know that logarithmic function is continuous in its domain.

Therefore, $\left|\log _{\mathrm{e}}\right| x||$ is continuous in its domain.

 

We will check the differentiability at its critical points.

$\left|\log _{\mathrm{e}}\right| x||=\left\{\begin{array}{lc}\log _{e}(-x) & -\infty

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

$=\lim _{x \rightarrow-1^{-}} \frac{\log _{\mathrm{e}}(-x)-0}{x+1}$

$=\lim _{h \rightarrow 0} \frac{\log _{\mathrm{e}}[-(-1-h)]}{-1-h+1}$

$=\lim _{h \rightarrow 0} \frac{\log _{\mathrm{e}}(1+h)}{-h}$

 

$=-1$

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

$=\lim _{x \rightarrow-1^{+}} \frac{-\log _{e}(-x)-0}{x+1}$

$=\lim _{h \rightarrow 0} \frac{-\log _{e}[-(-1+h)]}{-1+h+1}$

$=\lim _{h \rightarrow 0} \frac{-\log _{e}(1-h)}{h}$

$=-\lim _{h \rightarrow 0} \frac{\log _{e}(1-h)}{h}$

 

$=-1 \times-1=1$

Here, $\mathrm{LHD} \neq \mathrm{RHD}$

Therefore, the given function is not differentiable at $x=-1$.

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

$=\lim _{x \rightarrow 1^{-}} \frac{-\log _{\mathrm{e}}(x)-0}{x-1}$

$=\lim _{h \rightarrow 0} \frac{-\log _{\mathrm{e}}[(1-h)]}{1-h-1}$

$=\lim _{h \rightarrow 0} \frac{\log _{\mathrm{e}}(1-h)}{h}$

 

$=-1$

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

$=\lim _{x \rightarrow 1^{+}} \frac{\log _{e}(x)-0}{x-1}$

$=\lim _{h \rightarrow 0} \frac{\log _{e}[(1+h)]}{1+h-1}$

$=\lim _{h \rightarrow 0} \frac{\log _{e}(1+h)}{h}$

 

$=1$

Here, LHD $\neq$ RHD

Therefore, the given function is not differentiable at $x=1$.

 

Therefore, given function is continuous for all $x$ in its domain but not differentiable at $x=\pm 1$