Closed or Open Intervals in Extreme Value Theorem, Rolle's Theorem, and Mean Value Theorem I am a little confused by the openness of intervals in Extreme Value Theorem (EVT), Rolle's Theorem (RT), and Mean Value Theorem (MVT):
EVT: Given a function $f:[a,b]\rightarrow R$ is continuous, $f$ attains its maximum and minimum at some $c\in [a,b]$. (closed intervals)
RT: Given $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, if $f(a)=f(b)$, then there exists $c\in (a,b)$ such that $f'(c)=0$. (I believe if $f$ is differentiable on $[a,b]$, then, there exists $c\in [a,b]$ such that $f'(c)=0$. Why do we need an open interval in this statement?)
MVT:  Given $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in (a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$. (Obviously, MVT goes with RT. Again, I believe we can just replace "differentiable on $(a,b)$ by "differentiable on $[a,b]$", and then get "there exists $c\in [a,b]$ such that $\dots$." Do we use an open interval for some specific reason?)
Thanks in advance for any explanation.
 A: For the EVT, the statement is clearly potentially false if you delete endpoints.  E.g., the function $f(x) = \frac{1}{x}$ on $(0,1)$ has neither a max nor a min.  If you try to get cute and give it domain $(0,1]$ then it has a min but still no max.  If you want to be guaranteed both, you need a closed and bounded interval.  
For RT, consider the upper half of a circle.  Let's take $f(x) =\sqrt{1-x^2}$ for definiteness.  For the "right" RT, note that this is continuous on $[-1,1]$ and differentiable on $(-1,1)$.  Connecting the dots at the ends, RT says there should be a horizontal tangent somewhere.  Well, sure, it's at the north pole $(0,1)$.  However, you cannot demand a version of RT that requires differentiability on $[-1,1]$ and retain this example:  the function is not differentiable on the closed interval $[-1,1]$ as the derivative blows up at the endpoints (vertical tangents).  This is what user nextpuzzle was getting at when they said the original version is stronger. 
As you've noted, RT is roughly equivalent to the MVT, so the same reason suffices.  
