The mean value theorem is stated as follows
Let there be a function $f$ which is continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists $c$ belonging to $(a,b)$ such that $f′(c)=\frac{f(b)−f(a)}{b−a}$.
Now, here it is assumed that the function is continuous on a closed interval. What I don't understand is the use of making the function continuous on a closed interval. I understand that then $f(a)$ and $f(b)$ will not be defined if we use rather an open interval (a,b), but then we can take into account the limits as x approaches a and b of the function (sometimes one-sided limits) instead of the values of the function at $a$ and $b$.
I know this would be tedious and would make the theorem complicated, but what I want to ask is that whether there is any other reason for proposing a closed continuous function rather than an open or half-open one.
Best Answer
If you just consider continuity of $f(x)$ on $(a, b)$, then the limits at endpoints could be infinity or do not exist at all.
For example $f(x)= \tan (x)$ on $(-\pi /2, \pi /2)$.
In that case the statement of the theorem does not make sense as it stands.