The function $f(x)=e^{-|x|}$ is

Given, $f(x)=\left\{\begin{array}{r}e^{-x}, x \geq 0 \\ e^{x}, x < 0\end{array}\right.$ $$ \begin{array}{l} \text { LHL }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} e^{x}=1 \\ \text { RHL }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} e^{-x}=1 \\ \text { Also, } \quad f(0)=e^{0}=1 \end{array} $$ $\because$ $$ \mathrm{LHL}=\mathrm{RHL}=f(0) $$ $\therefore$ It is continuous for every value of $x$. Now, LHD at $x=0$ $$ \left(\frac{d}{d x} e^{x}\right)_{x=0}=\left[e^{x}\right]_{x=0}=e^{0}=1 $$ RHD at $x=0$ $$ \left(\frac{d}{d x} e^{-x}\right)_{x=0}=\left[-e^{-x}\right]_{x=0}=-1 $$ So, $f(x)$ is not differentiable at $x=0$. Hence, $f(x)=e^{-\dashv x \mid}$ is continuous everywhere but not differentiable at $x=0$. Alternative
Hence, it is clear from the figure that $f(x)$ is continuous everywhere and not differentiable at $x=0 .$ $$