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The function $y=e^{-|x|}$ is
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Continuous but not differentiable at $x=0$
We have,
$f(x)= \begin{cases}e^{-x}, & x \geq 0 \\ e^x, & x \lt 0\end{cases}$
Clearly, $f(x)$ is continuous and differentiable for all non zero $x$.
Now $\lim _{x \rightarrow 0-} f(x)=\lim _{x \rightarrow 0} e^x=1 \quad \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) e^{-x}=1$
Also, $f(0)=e^0=1$. So, $f(x)$ is continuous for all $x$.
$($ LHD at $x=0)=\left(\frac{d}{d x}\left(e^x\right)\right)_{x=0}=1$
(RHD at $x=0)=\left(\frac{d}{d x}\left(e^{-x}\right)\right)_{x=0}=-1$
So, $f(x)$ is not differentiable at $x=0$.
Hence $f(x)=e^{-|x|}$ is everywhere continuous but not differentiable at $x=0$.
$f(x)= \begin{cases}e^{-x}, & x \geq 0 \\ e^x, & x \lt 0\end{cases}$
Clearly, $f(x)$ is continuous and differentiable for all non zero $x$.
Now $\lim _{x \rightarrow 0-} f(x)=\lim _{x \rightarrow 0} e^x=1 \quad \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) e^{-x}=1$
Also, $f(0)=e^0=1$. So, $f(x)$ is continuous for all $x$.
$($ LHD at $x=0)=\left(\frac{d}{d x}\left(e^x\right)\right)_{x=0}=1$
(RHD at $x=0)=\left(\frac{d}{d x}\left(e^{-x}\right)\right)_{x=0}=-1$
So, $f(x)$ is not differentiable at $x=0$.
Hence $f(x)=e^{-|x|}$ is everywhere continuous but not differentiable at $x=0$.
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