Talk:Fermat's theorem (stationary points)
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[edit] Notation
IMO in the Fermat's theorem section the notation should be changed back from 
to fˈ(x0) = 0, as this way it would probably appeal to a wider audience. For instance, this is a high school level calculus theorem, and high school students are probably much more confortable with the derivative notation. It also looks much better inline. AdamSmithee 07:38, 3 July 2006 (UTC)
[edit] Continuous?
Is it really necessary to assume that, f is continuous? I think, the proof works just with: "f is differentiable at point x0". What do you think? I've found the same assumption on PlanetMath – Fermat's Theorem (stationary points) (Sorry about my poor English!) Mozó 80.99.182.157 19:48, 2 October 2006 (UTC)
- You are right. You don't actually need that condition - I lifted it wholesale from PlanetMath and didn't notice. The function would be continuous at the point, but that just follows from it being differentiable AdamSmithee 08:56, 4 October 2006 (UTC)
[edit] Boundary?
According to what I think is the most commonly accepted definition, a function whose domain has a boundary (for example, if the domain is a closed interval) cannot have a LOCAL extremum at a boundary point. In the proof given here, there is talk of a region (neighborhood of a given point) throughout which the function is everywhere larger (resp smaller) than at the given point. This cannot apply at a boundary point. L P Meissner L P Meissner
[edit] Easy Proof?
The formal proof that is given is much more elegant then the informal one because it does not rely on the intermediate value theorem for derivatives. That's a quite nasty one to prove, and so should not be in a supposedly 'easy' section. Watson Ladd (talk) 00:31, 5 December 2007 (UTC)
