# 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

## 1 Answer

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.

• "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
• @nextpuzzle Fair point. By "this" I meant this particular function is lost as an example. – Randall Jul 27 '18 at 1:53
• do you mean "if you try to get cute and give it domain (0,1], then it has a MAX but still no min? – MinYoung Kim May 12 at 2:16
• @MinYoungKim No. That function definitely has no max on $(0,1]$. – Randall May 12 at 2:19