Timeline for The role of the mean value theorem (MVT) in first-year calculus

Current License: CC BY-SA 3.0

9 events

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Jun 15, 2020 at 7:27	history	edited	CommunityBot		Commonmark migration
Dec 15, 2017 at 20:07	comment	added	LSpice		(For anyone who shares my naïveté, Bers explains why it is true that "everywhere-positive derivative implies strictly increasing" implies "everywhere-non-negative derivative implies non-decreasing": namely, if $f' \ge 0$ everywhere, then, for all $\epsilon > 0$, we have that $f' + \epsilon > 0$ everywhere, so $f + \epsilon\operatorname{id}$ is strictly increasing.)
Dec 12, 2017 at 17:18	comment	added	LSpice		@AnnaTaurogenireva, is it obvious that "a function with everywhere-positive derivative is strictly increasing" implies that "a function with everywhere-$0$ derivative is constant"? One can weaken the former statement to "a function with everywhere-non-negative derivative is non-decreasing", in which case constancy follows, but sometimes one wants to be able to conclude strict-increasing-ness. It seems incredibly awkward and unintuitive to have two such very similar, but not identical, statements, but I think elementary calculus really does use both.
Dec 10, 2017 at 22:38	history	edited	Martin Sleziak	CC BY-SA 3.0	added links to papers
Dec 10, 2017 at 22:31	history	edited	Martin Sleziak	CC BY-SA 3.0	added links to papers
Dec 10, 2017 at 22:23	history	edited	Martin Sleziak	CC BY-SA 3.0	added links to papers
Jul 15, 2010 at 13:27	history	edited	Pete L. Clark	CC BY-SA 2.5	added 1451 characters in body
Jul 13, 2010 at 12:02	comment	added	O.R.		I read the ones that I had access and I think one can conclude that a course will do fine with just $f'>0$ implies $f$ increasing and that one can prove as many consequence of this as is appropriate for the audience, among them MVT. I guess that answers the first question.
Jul 13, 2010 at 11:03	history	answered	Pete L. Clark	CC BY-SA 2.5