critical point

[synx]

Are there any critical points on e^(2/x) ? its derivative is [-2e^(2/x)]/x^2
if you solve (set f'(x) = to 0) you get 0, but x=0 is not in the domain, so im not sure if it can be a critical point. Would x=0 be a critical point or did i mess up and is there a different way to find critical points?
the domain of this function is just x= cant be 0, right?

 

[galactus]

Continuity requires that:

#1. f(c) is defined.

#2. \(\displaystyle \L\\\lim_{x\to\c}f(x)\) exists

#3. \(\displaystyle \L\\\lim_{x\to\c}f(x)=f(c)\)

(x approaches c. It won't display for some reason).

A function can not be differentiable at a point of discontinuity.

Is your function defined at 0.

Does \(\displaystyle \L\\\lim_{x\to\0}e^{\frac{2}{x}}\) exist?.