Yhyjyk

Do the end points of a domain come under critical points ?
I know we say critical point is a point where the derivative is zero or the derivative doesn't exist.

For ex:-
$$ f:[0,pi] rightarrow [-1,1] , f(x) = sin(x)$$
Does this have 1 critical point or 3 critical points(0 and $pi$ included) ?

NOTE:- This question is limited to only Single Variable Functions.Although I really would love an insight to this for Multivariable as well.

Yes, the function has 3 critical numbers. One is where the derivative of the function $f(x)=sinx, xin[0,pi]$ is zero and the other two happen to be the endpoints $x=0$ and $x=pi$ because the function $f(x)=sinx, xin[0,pi]$ is non-differentiable at those points.

Do you remember what it means for a function to be differentiable at a point? The function has to have a derivative at that point. What is the derivative of the function $f(x)=sinx, xin[0,pi]$ at $x=0$? Well, it should be:

$$
lim_xto0fracsinx-sin0x-0=lim_xto0fracsinxx
$$

Which is nothing more than two one-sided limits (if those two limits exist and are equal to each other, the limit itself exists):

$$
lim_xto0^-fracsinxx, lim_xto0^+fracsinxx
$$
But the first of those two limits for all intents and purposes is nonexistent because all x values that lie to the left of $0$ are not in the domain of the function $f(x)=sinx, xin[0,pi]$. For a limit to exist, you need two one-sided limits. But you've got only one! Thus, the derivative at $x=0$ does not exist which makes it a critical number. The exact same idea applies to the other endpoint.

$$f'(x) = cos(x) = 0 iff x = fracpi2$$
The function $f$ has only one critical point, since the derivative ($f'(x) = cos(x)$) exists at the endpoints. ($0$ and $pi$.)