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

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(A) Given,$f(x) = \begin{cases} e^{-x}, & x \geq 0 \ e^{x}, & x < 0 \end{cases}$ $LHL = \lim_{x 	o 0^{-}} f(x) = \lim_{x 	o 0} e^{x} = 1$ $RHL = \l

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continuous everywhere but not differentiable at $x = 0$

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(A) Given,$f(x) = \begin{cases} e^{-x}, & x \geq 0 \ e^{x}, & x $LHL = \lim_{x 	o 0^{-}} f(x) = \lim_{x 	o 0} e^{x} = 1$ $RHL = \lim_{x 	o 0^{+}} f(x) = \lim_{x 	o 0} e^{-x} = 1$ Also,$f(0) = e^{0} = 1$ Since $LHL = RHL = f(0)$,the function is continuous for every value of $x$. Now,we check for differentiability at $x = 0$: $LHD = \left(\frac{d}{dx} e^{x}ight)_{x = 0} = \left[e^{x}ight]_{x = 0} = 1$ $RHD = \left(\frac{d}{dx} e^{-x}ight)_{x = 0} = \left[-e^{-x}ight]_{x = 0} = -1$ Since $LHD 
eq RHD$,$f(x)$ is not differentiable at $x = 0$. Therefore,$f(x) = e^{-|x|}$ is continuous everywhere but not differentiable at $x = 0$.

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

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