Talk:Inflection point

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I have expanded this, and corrected a few errors including that the second derivative at a point of inflection can be non-zero. Also added the term 'saddle-point' for 'non-stationary point of inflection'. And made clear that in principle there is no limit to how high an order of derivative you might need to go to if you are using that method. 158-152-12-77 01:38, 24 August 2005 (BST)

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[edit] Inconsistent definition of Saddle Point

The definition in this article differs substantially from the saddle point article. One or both need to be fixed to show the correct definition (or both definitions, if both are correct). I have never heard of the def in this article, but am familiar with the surface def in the separate saddle point article. However, that doesn't mean this def isn't also correct, as many terms have multiple meanings. StuRat 19:16, 24 August 2005 (UTC)

Your graphs are excellent - they add a lot to the article. The definition of 'saddle point' in the saddle point article is incomplete, because it does not cover the case of a function of a single variable. This is covered properly in the Mathworld definition, namely "A point of a function or surface which is a stationary point but not a extremum" (i.e. not a local extremum).
158-152-12-77 21:41, 25 August 2005 (BST)
I have amended saddle point article.
158-152-12-77 00:11, 26 August 2005 (BST)
Thanks. StuRat 23:18, 25 August 2005 (UTC)

[edit] Font problem

In four places towards then end of the article, when the first derivative is mentioned, it's impossible for me to see the apostrophe after the "f", so I thought the function itself was being referred to, and I got really confused. I really wish I could fix this myself, but I have no idea how.

The only way I can see to make it more visible is to add spaces:

HOW IT NOW APPEARS: f'(x)

WITH SPACES ADDED  : f ' (x)

Should we do that ? StuRat 21:34, 2 December 2006 (UTC)

[edit] Definition vs. Property

From the definitions:

a point on a curve at which the tangent crosses the curve itself.

This is a property of inflection points, but in no way defines them: tangents can cross the curve in many points, most of which, in some cases, are not inflection points. See Tangent.

Diego.pereira 20:49, 10 March 2007 (UTC)

I think you're misreading it: if it's an inflection point x=a, then we want the tangent line at the point a to cross the curve at the point a. --Cheeser1 (talk) 05:00, 19 November 2007 (UTC)

[edit] Graph of y=x^3 is wrong

The graph here resembles y=x^3 near the origin, but it doesn't pass the vertical line test like the x^3 function does. This doesn't affect what is being demonstrated, however, but nonetheless for the purposes of accuracy I believe the either figure should be replaced with a true x^3 graph, or that its caption contain the correct equation (unknown to me) for the graph shown. --unsigned

The first graph in the article is y=x^3 and is correct, it should be tangent to the x axis at 0. You may be talking about the cubic root. Oleg Alexandrov (talk) 03:42, 17 August 2007 (UTC)
I believe he's talking about how the second graph is not a function, Oleg. At x=4, notice how there are two y-values. I was about to say the same myself but I saw it was already here. --unsigned

Just plot x^3 + x instead of the rotated x^3 graph, which is not a function. --unsigned —Preceding unsigned comment added by 198.240.130.75 (talk) 14:39, 6 March 2008 (UTC)

[edit] Important Change Needed

A location where the "tangent line at the point a crosses the curve at point a" is not necessarily an inflection point.

The definition of inflection point is that the curvature changes from positive to negative or negative to positive at the point. That is, there exist c,d such that f''(x) has opposite signs on the intervals (c,a) and (a,d), Consider the function f(x) = \left\{ \begin{array}{c l} x^{5}\sin(\frac{1}{x}) & x\ne0\\ 0 & x=0 \end{array} \right.

This function is twice differentiable on (-\infty, \infty). The tangent line at x = 0 is y = 0 and crosses the curve there. However, the sign of f''(x) changes infinitely often in all intervals of the form (c,0) or (0,d).

To correct the page, the statement should be taken out of the definition section and restated later as "If f has an inflection point at a, then the tangent line at a crosses the graph at a." —Preceding unsigned comment added by 64.59.248.162 (talk) 20:31, 10 March 2008 (UTC)

How do you come up with f' (0) = 0? Oli Filth(talk) 20:40, 10 March 2008 (UTC)
f'(0)=\lim\limits_{h\to 0} \frac{f(h)-f(0)}{h}=\lim\limits_{h\to 0}\frac{h^5\sin(1/h)}{h}=\lim\limits_{h\to 0}h^4\sin(1/h)=0 since |\sin(1/h)|\le1.
Ah yes.
However, presumably your example should be f(x) = x4sin(1 / x), as in your current example, we have f(x) = f( − x), hence the tangent can only touch the curve at x = 0, not cross it.
However, with f(x) = x4sin(1 / x), we have that f^{\prime\prime}(x) = -f^{\prime\prime}(-x). Therefore, this example also satisfies a further test for an inflection point (namely that f^{\prime\prime}(x + \epsilon) and f^{\prime\prime}(x-\epsilon) should have opposite signs in the region of an inflection point at x).
Oli Filth(talk) 21:50, 10 March 2008 (UTC)
In any neighborhood of zero there are points a<0<b such that f''(a)f''(b)<0. That's good enough for me. --Cheeser1 (talk) 21:36, 10 March 2008 (UTC)
Actually, I used the fifth power so that the function is twice differentiable. The comment that in any neighborhood of zero, there are points a < 0 < b such that f''(x)f''(b)<0 is very interesting, but does not make an inflection point according to the accepted definition. —Preceding unsigned comment added by 64.59.248.162 (talk) 23:08, 10 March 2008 (UTC)
Which is...? --Cheeser1 (talk) 20:28, 11 March 2008 (UTC)