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 \(x=0 .\)
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