Talk:Fermat's theorem (stationary points)

Notation

IMO in the Fermat's theorem section the notation should be changed back from ${\frac {df}{dx}}(x_{0})=0$ 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)[reply]

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)[reply]

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)[reply]

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

"Local" always is in reference to the topology. That is why the correct statement of the theorem is that f is defined on U which is an open subset of the reals -- taking the topology of the reals (and not something more interesting). — Preceding unsigned comment added by 129.110.242.24 (talk) 23:58, 6 May 2016 (UTC)[reply]

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)[reply]

In fact, it uses Darboux's theorem, which is again proved by Fermat's theorem, so we have a circular proof. --Roentgenium111 (talk) 15:24, 20 March 2009 (UTC)[reply]

Merge proposal

This article was tagged to merge First derivative test into this article. However, I have an alternative proposal: we merge First derivative test, Second derivative test, Second partial derivative test, and Higher-order derivative test into one article, and leave this theorem article separate. They're all very short, and essentially extensions of the same idea. Name could be something like Derivative tests. Thoughts anyone? Forbes72 (talk) 05:06, 9 March 2015 (UTC)[reply]

@Forbes72 Support. Lbertolotti (talk) 17:24, 4 September 2015 (UTC)[reply]

@User talk:Lbertolotti That's only two people for consensus, but since its been proposed for 5 months, I've just gone ahead and done some BOLD merging. I didn't merge second partial derivative test because that seemed the most different of the four articles, but the other three fit together quite nicely I think. Forbes72 (talk) 22:12, 5 September 2015 (UTC)[reply]

"Infinitesimal"

The word "infinitesimal" is used in this article. What does this formally mean? Are we talking about non-standard analysis? SomePseudonym (talk) 16:10, 16 March 2016 (UTC)[reply]

"Intuition"

None of this section should be in the article and I suggest deleting it. Darboux's theorem is a consequence of the Extreme Value Theorem and Fermat's theorem. The talk about "Increasing" and "Decreasing" are really consequences of the Mean Value Theorem which is a consequence Rolle's theorem which is a consequence of Fermat's theorem.

(EVT, FT) implies DT

(EVT, FT) implies RT implies MVT

It makes no sense to make arguments for FT from DT or from MVT, that is arguing from the conclusion. — Preceding unsigned comment added by 129.110.242.24 (talk) 22:30, 6 May 2016 (UTC)[reply]