Why Rolle's theorem considers differentiability on open $(a,b)$ rather than closed $[a,b]$?

Question

This question is a duplicate,but still I'm posting it because the answers given in the previously asked questions(by other users) are not satisfactory and since my reputation in the website isn't high enough, yet, I couldn't post any comments in the previous questions to get doubts clarified.

So, why is it defined in open $(a,b)$ rather than closed $[a,b]$.

Could you explain with an example?

Note: preiously asked questions(whose answers are not satisfactory) are:

Continuous and differentiable function in Rolle's theorem

Reason for the diferentiability of function on open interval in Rolle's Theorem

Answer 1

Let $f:[a,b] \to \mathbb R$ a continuous function, which is differentiable on $(a,b)$. If $f(a)=f(b)$, then Rolle's theorem says: there is $s \in (a,b)$ such that $f'(s)=0$.

Look at a proof of this theorem, then you will see:

1. you need that $f$ is continuous on $[a,b]$

and

2. you only(!) need that $f$ is differentiable on $(a,b)$.

Answer 2

A function is called differentiable if it is differentiable at each point in its domain.

You basically want to talk about function being differentiable at every point in the domain...

You can not talk about a function being differentiable at a point which is not an interior point...

If your function is $f:[a,b]\rightarrow \mathbb{R}$ then you have to say about differentiability of $f$ at $a$ but as that notion is not there, you just ignore $a$ (as well as $b$) and considered $f:(a,b)\rightarrow \mathbb{R}$.

In the new domain $(a,b)$, every point is interior point, so you can say function is differentiable on $(a,b)$.