Consider the following statements for f(x)=e^ -|x| : 1. The...

(1) $f(x)=e^{-|x|}, f(0)=e^{-|^{0}|}=1$. $\begin{gathered} f(x)=e^{-x}, \text{\text{ for } } x \geq 0 \\ =e^x, \quad \text{ for } x < 0 \end{gathered}$$\begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} e^x=1 \\ & \mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} e^{-x}=1\end{aligned}$$$As,$f(0)=$\mathrm{LHL}=\mathrm{RHL}$ Thus $f(x)$ is continuous at $x=0$. (2) \mathrm{LHD}=\lim _{x \rightarrow0^{-}} f^{\prime}(x)=\lim _{x \rightarrow0^{-}} e^x=e^0=1\mathrm{RHD}=\lim _{x \rightarrow0^{+}} f^{\prime}(x)=\lim _{x \rightarrow0^{+}}-e^{-x}=-e^0=-1 As LHD ≠ RHD Hence, $f(x)$ is not differentiable at $x=0$.$