Discuss the continuity and differentiability of `f(x)=e^(|x|)` .

Explanation

Given,
`f(x)=e^(|x|)`
`f(x)={{:(e^-x,xlt0),(e^x,xge0):}`
for the continuity at `x=0`
`f(0)=e^0=1`
`LHL=lim_(x->0^-)f(x)`
`=lim_(h->0)e^-(0-h)`
`=lim_(h->0)e^h`
`=1
Now,
`RHL=lim_(x->0^+)f(x)`
`=lim_(h->0)e^(0+h)`
`=lim_(h->0)e^h`
`=1
since,`LHL=RHL=f(0)`
So, f(x) is continuous at` x=0`

For differentiablility at `x=0`
`LHD=lim_(h->0)(f(0-h)-f(0))/(0-h)`
`=lim_(h->0)(e^h-1)/-h`
`=-lim_(h->0)(e^h-1)/h`
`=-1`
Now,
`RHD=lim_(h->0)(f(0+h)-f(0))/(0+h)`
`=lim_(h->0)(e^h-1)/h`
`=lim_(h->0)(e^h-1)/h`
`=1`
since,
`LHD ne RHD`
f(x) is not differentiable at `x=0`.