The function y = e^{-|x|} is | vedclass

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(C) We have,$f(x) = \begin{cases} e^x, & x < 0 \ e^{-x}, & x \ge 0 \end{cases}$Clearly,$f(x)$ is continuous and differentiable for all $x 
eq 0$.Now

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Continuous but not differentiable at $x = 0$

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(C) We have,$f(x) = \begin{cases} e^x, & x Clearly,$f(x)$ is continuous and differentiable for all $x 
eq 0$.Now,$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0} e^x = 1$ and $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} e^{-x} = 1$.Also,$f(0) = e^0 = 1$. Since $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$,the function is continuous at $x = 0$.Now,check for differentiability at $x = 0$:Left-hand derivative $(LHD)$ at $x = 0$ is $\frac{d}{dx}(e^x) |_{x=0} = e^0 = 1$.Right-hand derivative $(RHD)$ at $x = 0$ is $\frac{d}{dx}(e^{-x}) |_{x=0} = -e^0 = -1$.Since $LHD 
eq RHD$ at $x = 0$,the function is not differentiable at $x = 0$.Thus,$f(x) = e^{-|x|}$ is continuous but not differentiable at $x = 0$.

The function $y = e^{-|x|}$ is

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