The function

Solution:

(a) continuous everywhere but not differentiable at x = 0

Given: $f(x)=e^{-|x|}= \begin{cases}e^{x}, & x \geq 0 \\ e^{-x}, & x<0\end{cases}$

Continuity :

$\lim _{x \rightarrow 0^{-}} f(x)$

$=\lim _{h \rightarrow 0} f(0-h)$

$=\lim _{h \rightarrow 0} e^{-(0-h)}$

$=\lim _{h \rightarrow 0} e^{h}$

$=1$

RHL at x = 0

$\lim _{x \rightarrow 0^{+}} f(x)$

$=\lim _{h \rightarrow 0} f(0+h)$

$=\lim _{h \rightarrow 0} e^{(0+h)}$

$=1$

and $f(0)=f(0)=e^{0}=1$

Thus, $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f$

Hence, function is continuous at x = 0

Differentiability at x = 0

(LHD at x = 0)

$\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$

$=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0}$

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

$=\lim _{h \rightarrow 0} \frac{e^{h}}{h}$

$=\infty$

Therefore, left hand derivative does not exist.
Hence, the function is not differentiable at x = 0.