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

Option: 1

continuous everywhere but not differentiable at x=0


Option: 2

continuous and differentiable everywhere


Option: 3

not continuous at x=0


Option: 4

none of these


Answers (1)

best_answer

We have,   \mathrm{ f(x)=\left\{\begin{array}{c}e^{-x}, x \geq 0 \\ e^x, x<0\end{array}\right.}
Clearly, f(x) is continuous and differentiable for all non-zero x.
Now,   \mathrm{ \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} e^x=1~ and, ~\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} e^{-x}=1.}
Also, \mathrm{ f(0)=e^0=1.}
So, f(x) is continuous for all x.

\mathrm{ ( ~LHD ~at x=0)=\left(\frac{d}{d x}\left(e^x\right)\right)_{x=0}=\left[e^x\right]_{x=0}=e^0=1 .}

\mathrm{ (( RHD ~at ~x=0)=\left(\frac{d}{d x}\left(e^{-x}\right)\right)_{x=0}=\left[-e^{-x}\right]_{x=0}=-1 .}

So, f(x) is not differentiable at x=0.
Hence,  \mathrm{f(x)=e^{-|x|} .}  is everywhere continuous but not differentiable at x=0. This fact is also evident from the graph of the function.

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Divya Prakash Singh

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