Talk:Rolle's theorem

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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)