Is f(x) differentiable at x=-1 ?

Question

f(x)=|x+1|e^x

(that's abs(x+1)*e^x)

it has to be done with the definition of the derivative.

i see that approaching x=-1 from one side the limit is e^(-1) and from the other side it's -e^(-1) does that mean its not differentiable at this point ? or is it enough that the limit is constant at the questionable point to be differentiable ?

thanks

Answer 1

you have to calculate:

lim x--> - 1  [f(x)- f(-1) ]/ (x+1).  that is the definition of the derivative.

f( -1) = |-1+1|e ^ (-1) = 0

first you calculate lim x-->-1, x <-1 of [f(x)- f(-1) ]/ (x+1).  

then you calculate lim x-->-1, x >-1 of [f(x)- f(-1) ]/ (x+1).  

those two have to be equal.

because of the module, the formula for f(x) changes depending on x<-1 or x>-1.

if x<-1 => x+1<0 => |x+1|= -(x+1).

if x>-1 then |x+1|= x+1.

so:

lim x-->-1, x <-1 of [f(x)- f(1) ]/ (x-1) = lim x-->-1, x <-1 of [-(x+1) e^x] / (x+1) =  - lim x-->-1, x <-1 e^ x = -e^ -1 = -1/e.

then you calculate lim x-->-1, x >-1 of [f(x)- f(1) ]/ (x+1) =

lim x-->-1, x >1- of [(x+1) e^x] / (x+1) = lim x-->-1, x >1- of e^x = e ^ -1 = 1/ e

since the two limits are different,  f(x) is NOT differentiable at x=- 1

Answer 2

The answer is 'no.'

Search 'Yahoo answers' for the following question:

Prove that the absolute value function is not differentiable at the origin.

You have an answer.

Here, instead of |x| at x=0, you have |x+1| at x=-1.