Talk:Rolle's theorem

The statement that "Rolle's Theorem is used in proving the mean value theorem, which eliminates the requirement that f(a)=f(b)." carries the incorrect suggestion the requirement f(a)=f(b) is eliminated but nothing else differs between the two theorems.

mikeliuk 13:30, 22 October 2006 (UTC)

That Bhaskara stated this before Rolle needs citation.

"For example, if f(x) = |x|, the absolute value of x, then we have that f(-1) = f(1), but there is no x between -1 and 1 for which f '(x) = 0."

This is invoked as an example of why the stated assumptions are necessary for the theorem to work. But shouldn't we state explicitly which of the assumption(s) this proposed function violates? Its my understanding that the absolute value function is continuous but not differentiable. I speak subject to correction in these matters, though. --Christofurio 12:23, Apr 11, 2005 (UTC)

You are correct about f(x) = |x| being continuous on [-1,1], but not differentiable on (-1,1).

My question is the "generalization" portion. When would a function be continuous on [a,b] but also have some c on (a,b) for which f'(c) = ±$$\infty$$? And without that addition, how are the assumptions "slightly broader" than the standard Rolle's theorem assumptions?--YLlama 06:02, Jul 11, 2005 (UTC)

For instance, let f: [-1, 1] --> R be defined by f(x) = x^(1/3) - x. This is the sort of function that satisfies the slightly relaxed hypotheses. Notice that f is not differentiable at f = 0. However, lim(h --> 0) (f(0 + h) - f(0))/h = + infinity. However, a function like absolute value does not satisfy these conditions -- we cannot say its derivative is + infinity or - infinity or anything else at the origin, because it does something different from the right and the left. I'm changing the page a little bit, b/c it shouldn't say that "all the assumptions are necessary" and later note that an assumption can be relaxed. Kier07 22:54, 13 December 2006 (UTC)


 * The generalization is wrong as stated, since the absolute value function f(x) = |x| satisfies the conditions -- the limit of the difference quotient, as defined there, is 1 for x in [0,infty) and -1 for x in (-infty,0) and thus is finite everywhere. Tesseran 15:10, 3 March 2007 (UTC)


 * If you're referring to the "Relaxed assumptions" section, then no, you're off a bit: the limit of the difference quotient for the absolute value function does not exist. Like you mention yourself, the limit on the left is -1 whereas the limit on the right is +1 (at the origin).  That is, the absolute value function has no well-defined slope at the origin.  On the other hand, the cube root function does have a well-defined slope (it has a vertical tangent).  The difference quotient has a limit of +infinity (from both the right and the left).
 * If you're referring to the "Generalizations" section, I don't see what you're saying. If this is the case, could you elaborate?  Lunch 01:50, 5 March 2007 (UTC)
 * My apologies--for some reason, my brain was seeing a + in the limit, which would mean that we were just taking the limit from the right. There is no problem with what's in the article. Sorry. Tesseran 08:21, 6 March 2007 (UTC)

formatting changes
Rbb, if you're interested in changing the formatting to improve readability in some browsers, well, OK, I'll go along. But please do not change prose to symbols where the prose is clear. Unnecessary symbols make the text more difficult to read -- especially for an elementary topic like this. Thank you, Lunch 01:35, 9 April 2007 (UTC)


 * Again, please stop changing prose into symbols. Also, please do not make any "corrections" to the article.  In the "Relaxed assumptions" section, you've reintroduced an error that has been the source of confusion for other readers before.  Clarify if you will, but don't keep the error.


 * Edit summaries would be nice, too. And it would be better to make all these formatting changes in one edit rather than making many small edits.  Lunch 01:56, 9 April 2007 (UTC)

"Relaxed assumptions" and vertical tangents
A number of contributors have "corrected" this section. Before doing so, please stop and think about the theorem and whether it applies to functions with a vertical tangent. Consider that the theorem can be applied (locally) to the inverse of the function. That is, the theorem is more a geometric statement independent of coordinate system than one that depends on a particular coordinate system.

I will be the first to admit that the section could be worded better, and the notation $$[-\infty,+\infty]$$ is non-standard. Feel free to improve these. However, please discuss technical changes here before making them to the article. Thank you, Lunch 19:42, 15 April 2007 (UTC)


 * A more standard notation is perhaps R . Thomas T Howard 03:11, 7 May 2007 (UTC)

Style in the History section
The following is a description of my reasons for making the disputed edit on 19 November.

My intention was to introduce a more formal tone (as recommended in the style guide), and to remove the artificial sense of excitement and suspense from the existing text (which reads like and excerpt from a piece of advertising copy promoting a gripping weekend course which offers a unique insight into the fascinating history of mathematical analysis - a tale of deception, betrayal and human ambition, fraught with drama suspense from beginning to end).

My revision from:

"A proof of the theorem had to wait until centuries later when ..."

to:

"The first known formal proof of the theorem was offered by Michel Rolle in 1691 ..."

avoids patronising the reader by refraining from making unverifiable assertions. How can it possibly be proven that Rolle's proof was the very first one offered? The word known acknowledges this limitation. reetep (talk) 09:42, 20 November 2007 (UTC)

Third example
The third example exhibits the following function:



f(x)=\begin{cases} \sqrt{1-x^2}&\text{for }x\in[0,1],\\ -\sqrt{1-x^2}&\text{for }x\in[-1,0]. \end{cases} $$


 * This function is not well defined since it takes more than one value at zero.
 * The assertion: "there is (even more than) one point in (−1,1) where the derivative is zero" is false.
 * The description "two semicircles put together is a smooth way at the origin" is inaccurate. It is more like a quarter circle centred at the origin with radius r in the quadrant {x>0, y>0}, relected in the diagonal line y=-x, as illustrated below.

The function is a very interesting one, and very worthy of exhibition in this context. I'm just not sure that the function is suitable for making the point that the author intended. I will contact the author and see if they wish to amend the article according to his/her original intention, before diving in myself. reetep (talk) 11:43, 20 November 2007 (UTC)

[[Image:Part_circle.svg|thumb|300px|True graph of the function on the intervals [-1,0) and [0,1]]]


 * Sorry, I didn't shift the semicircles. Thank you for pointing this out. It should work now. Schmock (talk) 20:36, 20 November 2007 (UTC)

If the function isn't differentiable at x=0, then the hypothesis of Rolle's theorem isn't satisfied. I guess that means that this example doesn't really illustrate Rolle's theorem; it's more of a coincidence than anything. I thought it was kind of confusing the first time I read through this article, so I just wanted to point it out to you guys. Thelittlestspoon (talk) 07:45, 9 May 2008 (UTC)


 * That is completely true - the given function is not differentiable on the interval, so the Rolle's theorem can not be applied. But you should have removed the reference of the third example as well. In the very next section, there is a mention of some third example (and there are only two), so I got confused, and had to search through history to find out what happened with that example :) -- Obradovi&#263; Goran ( t al k  02:01, 15 July 2008 (UTC)

"Smooth" function
"Rolle's theorem essentially states that a smooth function [...]"

Seeing as smooth function has a precise meaning in mathematics other than just continuous function, maybe it would be better to change smooth to something else ?

Does anyone have any ideas ? -- Xedi (talk) 04:07, 22 February 2008 (UTC)


 * Well, I think that differentiable is better term. I considered linking that to smooth function, but from the beginning of that article it is not clear what it means, and it means (in the context of Rolle's theorem) that function has first derivative. On the other hand, smooth is more intuitive, since a lot of people might not understand the concept of derivative. -- Obradovi&#263; Goran ( t al k  01:56, 15 July 2008 (UTC)

On second thought, I think that it is fine this way. The first sentence gives some insight to a reader not very familiar with calculus, and second sentence gives the formal mathematical definition. As it should be :) -- Obradovi&#263; Goran ( t al k  02:05, 15 July 2008 (UTC)

I believe that the appropriate requirement is that the function be continuously differentiable. — Preceding unsigned comment added by Brydustin (talk • contribs) 20:20, 6 May 2012 (UTC)

Second example deleted
How on earth can anyone say that the standard version of Rolle's theorem can fail, when, in fact, it doeas not; this is an implication, and if the antecedent does not hold, then how can anyone whine that neither the consequence does?! --84.47.117.130 (talk) 12:07, 3 January 2010 (UTC)
 * I've restored a reworded version of the example. Paul August &#9742; 19:34, 14 January 2010 (UTC)

Quaint notation
I cannot find a reason for the arrow notation for the one-sided limits in the proof, especially when the statement used modern (0+ and 0-) notation. I shall change it.

Tebello TheWHAT!!?? 13:56, 14 January 2011 (UTC)

fixed mistake in history
The history section incorrectly stated that Rolle proved the theorem, and did it using calculus. I've corrected it to this: "Although the theorem is named after Michel Rolle, Rolle's 1691 proof covered only the case of polynomial functions, and did not use the methods of differential calculus." More info: http://mathoverflow.net/questions/184358/what-did-rolle-prove-when-he-proved-rolles-theorem — Preceding unsigned comment added by 76.169.116.244 (talk) 17:55, 21 June 2016 (UTC)

Question regarding this excerpt
I wouldn't risk making an edit as all the participants seem more sophisticated in understanding than me. I just have a question regarding this excerpt:

"As the induction hypothesis, assume the generalization is true for n − 1. We want to prove it for n > 1."

Souldn't that be changed to say

"As the induction hypothesis, assume the generalization is true for n − 1. We want to prove it for n = 1." — Preceding unsigned comment added by 2601:19B:C403:1D80:85D9:5E27:8EDC:8355 (talk) 20:44, 24 December 2019 (UTC)


 * It was very awkwardly phrased, I have cleaned it up. (Your proposed phrasing was completely wrong, but hopefully now it is clear.)  --JBL (talk) 15:45, 26 December 2019 (UTC)

Multivariate version of Rolle's theorem
Shouldn't this result by Rai & Van Ryzin be mentioned in the article? --bender235 (talk) 01:18, 29 May 2020 (UTC)

Add informative infographic
I vote we add this infographic to the page https://xkcd.com/2042/