Quick math question.

[EpsiIon]

Can a cusp be an inflection point?
Or does the point have to be differentiable, as well?

I guess it depends on how you define an inflection point, but my calculus book says nothing specific about its definition.

Thanks for the help,
Epsilon

 

[EpsiIon]

Common, people. Just yes or no.

Epsilon

 

[EpsiIon]

Bump, buhdha bump bump -- bump BUMP.

 

[Scrapster]

You know I heard my math teacher use the term cusp just the other day. First time I heard it in a math setting and it doesn't fit into anything you just said. What kind of problem are you working on? Topic?

 

[EpsiIon]

Calculus. Second derivate. I need to know if a cusp can be an inflection point (if the second derivate changes sign on either side of it, of course). Or does an inflection point have to be differentiable, too? I think it does, but I'm not sure...

Epsilon

 

[Handle]

Wouldn't (0,0) of y=x^3 be a cusp and inflection point?

 

[Mister T]

what do you mean by cusp?

 

[Handle]

I'm not too familiar with the terminology, but I get the impression here that cusp is where the second derivative is zero (point at which concavity changes).

 

[js1973]

I wish there was a "clueless" emoticon. I'd probably have to bind it to one of the keys on my keyboard.

 

[EpsiIon]

No, not x^3. Maybe x^(1/3), though... Except I don't know if that's technically a cusp... Hmmm, that's a really interesting question...

lol Both ways of thinking about it make sense to me, so if you know the answer, share directly, will ya?

Epsilon

 

[xtreme2k]

I think the point has to be differentiable by both side with the same 2nd deriative. In another word the line must be smooth.
An inflection point is just a change of concavity. It has a 2nd deriative of 0.

 

[EpsiIon]

Ok... Here's what a graph with a cusp might look like (ignore the periods -- they're just spacers):

y
|
|.\ . ./
|..\../
|.. \/
|-------------------x

The cusp would be the point at the bottom of the graph... Notice that the first derivative of that point doesn't exist because you can draw almost any tangent line you'd like to. So the second derivative doesn't exist, either.

What I want to know is: If the sign of the second derivative on the RIGHT side of the cusp is the opposite of the sign of the second derivative on the LEFT side of the cusp, could it still be considered an inflection point (point where the sign of the second derivative changes) EVEN THOUGH the second derivative doesn't exist at that point (the cusp)...

Epsilon

 

[xtreme2k]

The line is not smooth. IE: Left side derivative is not the same as Right side derivative.
It is not considered a valid inflection point. I remember in first year uni that, to be differentiable at a point, the line must be SMOOTHat that point. That line is not smooth. Therefore it is not an inflection point.

 

[RossGr]

Ok Epsilon let me try to remeber.

First of all, the 1st deritive at the max and min points is not undefined it is zero, this is fundamental to solving max/min problems. Take the first deritive and set it equal to zero. Then take the 2nd derivitive and check the sign. IF it is positive the function is concave up at that point if it is negitive the function is concave down. The points of inflection are sign changes in the 2nd derivitive.

Hope this helps.
Ross

 

[EpsiIon]

<< An inflection point is just a change of concavity. It has a 2nd deriative of 0. >>



Not true. :S Here's a direct quote from my book:

&quot;Note: A zero value of the second derivative does not necessarily indicate an inflection point. A zero value of the second derivativve [f&quot;(c) = 0] tells us to go back and use the first derivative test.&quot;

But the first derivative test doesn't apply to cusps, because they have no first derivative. But that doesn't eliminate the possibility of a cusp being a point of inflection... Oh well...

Epsilon

 

[xtreme2k]

It is true that 2nd derivative = 0 does NOT guarentee inflection point. Sorry what I said was misleading

BUt then for the inflection point part, i will again say, the line MUST be smooth to be differentiable. Your line is NOT SMOOTH at that point, therefore it is NOT differentiable. Therefore will Have no first derivative. And 2nd derivative.

 

[RossGr]

The graph you have drawn there is for ABS(X) In this case because the derivitive from the right is different from the one from the left then it is indeed undefined at the point. But the 2nd derivitive is 0 on both sides, (1st = +/- 1 depending on the side, 2nd = 0) so the cusp is not a point of inflection.