Talk:Rolle's theorem
From Wikipedia, the free encyclopedia
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) = ±
? 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)
