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

Options

A. continuous and differentiable everywhere not continuous at x=0
B. ontinuous and differentiable everywhere
C. not continuous at x=0
D. None of the above

Answer

A

Explanation

Given `f(x)={{:(,e^(-x),x ge 0),(,e^(x),xlt 0):}`
`LHL=underset(x to 0^(+))lim f(x)=underset(x to 0)lim e^(x)=1`
`RHL=underset(x to 0^(+))lim f(x)=underset(x to 0)lim e^(-x)=1`
`"Also"f(0)=e^(0)=1`
`therefore LHL=RHL=f(0)`
`therefore` It is continuous for every value of x.
Now, LHD at x=0
`((d)/(dx)e^(x))_(x=0)=[e^(x)]_(x)=e^(0)=1`
RHD at x=0
`((d)/(dx)d^(-x))_(x=0)=[-e^(-x)]_(x=0)=-1`
So, f(x) is not differentiable at x=0
Hence `f(x)=e^(-|x|)` is continuous everywhere but not differentiable at x=0