This doesn't make any sense. A local minimum is just the lowest value of a function on a given interval. A global minimum is a local minimum where the given interval is the domain of the function. The first and second derivative tests tell you if something is a local max or min on some interval, but they don't tell you which interval. It just depends on where you're looking. For example:

This shows f(x) = sin(x) + x/2 on the interval x = [0, 2π]. The local minimum on this interval is clearly at (0,0)--just look at the graph--but f''(0) = 0.

Post has been edited 1 time(s), last time on Oct 30 2011, 5:43 pm by FatalException.

I really don't have the time or willingness to teach every uneducated person on the internet basic calculus.

http://www.staredit.net/?p=shoutbox&view=1499

That's the most effort I'm willing to put into responding to people who are still misunderstanding such a fundamental concept.

Local minimums don't exist on intervals; those are global minimums. Local minimums exist within neighborhoods, which do not have boundaries. This means that the local minimum on this graph on this interval is right around where x is about 4.2

I'm sorry that basic calculus was a few years ago for me so I remember a working definition for what I have to use it for. No need be an asshole about it.

I don't ever recall using any of this outside of precalculus. Then again, why simply look at graphs and see behavior of curves when you can use wordy explanations to convey a much more convoluted meaning and confuse the shit out of people?

You can't explain things easily via words or graphs to people who insist on quoting anything you say and going "Nuh-uh!"

The problem isn't that they misunderstood what was said, it was that they didn't think it was true due to lack of knowledge on the subject.

Guys guys.

What the fuck, I expected more from you. Like this:



Now THAT is "Actual Troll Mathmatics."

I don't think this has been covered specifically, but I'll go over it if you're still curious why it isn't working.

The test for concavity failed; that is, it didn't show anything. For future reference (summations), tests can and will fail, so now is the time to let go of the idea that some formula or procedure will always work. In this case, the reason can be explained easily.

The test for concavity is so named (inappropriately, imo) because of the way a particular curve appears to be. If we look at x³, we can see that (-∞, 0) is concave down; while (0, ∞+) is concave up.



But what about when x = 0? Is it concave up, down, both, or neither? The answer lies in the test itself.

The test for concavity doesn't really test concavity, it tests whether the point has a negative or positive acceleration. From this perspective, the test makes more sense, and its failings even more. It didn't so much as tell you that the point is a minimum or not, but rather that the instantaneous acceleration at that point is 0.

Just fyi, you can't define concavity on a point. There is no answer to a question you cannot ask.
And, the test for concavity tells a variety of things even without such physical explanations. In modern mathematics, no such thing is required. It's no acceleration. It is meaningful in itself.

Here's a very textbook example of what you can find out through convex functions.

f is a continuous function in the given interval [0,1].
f''(x) < 0, f'(x) > 0, f(0)=0, f(1)>0
Compare A, B, and C.
A= f'(1), B= f(1), C= 2(integral 0->1)f(x)dx

Post has been edited 1 time(s), last time on Oct 31 2011, 2:24 pm by BeDazed.

I was expecting this as well!

Yes it always is. Just because DTBK happens to arbitrarily decide that you need it to be in an open interval and not a closed one doesn't mean it isn't.

http://mathworld.wolfram.com/LocalMinimum.html

I win. Since we all pray to Wolfram, God of Math, and He says nothing about the neighborhood being open or closed, therefore my definition is permissible, and you're all imposing arbitrary unnecessary restrictions.

It's like you guys arguing that squares aren't rectangles. THEY ARE.

Edit: Troll Math Accomplished.

http://www.britannica.com/EBchecked/topic/199172/extremum

You lose, kthx

I've already established that Wolfram is the God of Math. Therefore, you still lose, for my God is greater than your false gods.

Neighborhoods always are on open intervals ;_;

Wolfram, master of incandescent lightbulbs, we hail thee.

They're not always intervals and they're not always open.

Thats the right idea.

Seriously, you guys gotta start thinking outside the epsilon.

I'm still struggling with ±δ

Wolfram's definition is equivalent; a function cannot have a local extremum on the boundary point of its definition on a closed interval.

A function f : A c R → B c R has a local minimum at x₀ c B iff there exists an open interval I c B s.t. x₀ c I and f(x₀) <= f(x) for all x in I.

This definition is equivalent to every one which has been offered by a reliable source.

It says nothing about your open interval "I" being contained in B, or about the neighborhood being necessarily open, Mr. Add Arbitrary Restrictions

But seriously, stop being a stick in the mud and HAIL WOLFRAM, GOD OF MATHEMATICS. This is a troll math thread, not an actual math thread.

It's an "actual troll math" thread actually.