# 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.

• RT: Your second version, asks for more (differentiability at more points), and gives less (the point $c$ is less clear where it is since now it could also be at $a$ or at $b$). The original version is stronger. The second version is true, though. – user578878 Jul 27 '18 at 1:32
• For RT, to see why open intervals are "right" for differentiability, think about the upper half of a circle. – Randall Jul 27 '18 at 1:36
• The extreme value theorem requires a closed interval. The max / min may be at an endpoint. Over an open interval there may not be a max or a min. Rolles theorem / MVT still hold over closed intervals, but they telll you that there will be special points in the interior of the interval, i.e. not at the end points. – Doug M Jul 27 '18 at 1:50
• AH, FINALLY SOMEONE ASKS THIS! – Nick Jul 27 '18 at 1:57
• I put questions like this on topology exams all the time. It's good for students to think about why these theorems are written this way. – Randall Jul 27 '18 at 1:57

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.
• "you cannot ask this to work with differentiability on $[-1,1]$" can be understood as saying that the theorem wouldn't be true. You want to say that asking for differentiability on $[-1,1]$ would give a theorem that, while true, couldn't be applied to that particular $f$. – user578878 Jul 27 '18 at 1:52
• @MinYoungKim No. That function definitely has no max on $(0,1]$. – Randall May 12 at 2:19