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

Options

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

Answer

A

Explanation

Given , `f(x) = {{:(e^(-x) , x ge 0 ),(e^(x), x lt 0 ):}`
LHL`= underset(xto0^(-))limf(x) = underset(xto0)lim e^(x) = 1`
RHL`= underset(xto0^(+))limf(x) = underset(xto0)lim e^(-x) = 1`
Also , ` f(0) = e^(0) = 1`
`because ` 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=0) = e^(0) = 1`
RHD at x= 0
`((d)/(dx)e^(-x))_(x=0) = [-e^(x)]_(x=0) = e^(0) = -1`
So, f(x) is not differentiable at x = 0
Hence , ` f(x) = ^(-|x|)` is continuous everywhere but not
differentiable at x = 0