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The function f(x)=e^(-|x|) is continuous...

The function `f(x)=e^(-|x|)` is continuous everywhere but not differentiable at `x=0` continuous and differentiable everywhere not continuous at `x=0` none of these

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
D
We have `f(x)=e^(-[x])={{:(,e^(x),x lt 0),(,e^(-x),x ge 0):}`
Clearly, it is everywhere continuous. So, statement-2 is true,
Now, `("LHD at x=0")=underset(x to 0^(-))lim (f(x)-f(0))/(x-0)=underset(x to 0^(-))lim (e^(x)-1)/(x)=1`
`("RHD at x=0")=underset(x to 0^(+))lim (f(x)-f(0))/(x-0)=underset(x to 0^(+))lim f(e^(x)-1)/(x)=-1`
`therefore` f(X) is not differentiable at x=0.
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