# Point of inflection

1. Oct 29, 2009

### FedEx

Hi

Does sine have a point of inflexion?

Well there are many interpretations for the point if infelxion....

I saw on wiki that sin2x has point of inflexion. Well the second derivatie does change its sign at pi by 2, but the first derivative is not eqaul to zero. Can we say that sine has a point if infelxion at pi by two. WIKI says yes...

2. Oct 29, 2009

### fourier jr

it has nothing to do with the first derivative. an inflection point is where the second derivative = 0 so sin(x) has inflection pts at integer multiples of pi

3. Oct 29, 2009

### l'Hôpital

Untrue,consider f(x) = x4. f ''(x) = 12x2. f '' (0) = 0, but that is not an inflection point.

Inflection point is when the concavity changes.

4. Oct 29, 2009

### leright

When the function is concave up it has a positive second derivative and when it is concave down it has a negative second derivative. Thus, at the point where the function switches from concave up to concave down (the inflection point) the second derivative would be zero. However, as you point out, some functions have points that have second derivatives that are zero but they are not inflection points, such as straight lines. It seems the function x^4 is one of these odd exceptions. I have not realized this before. Is the function x^4 briefly a flat line at x=0?

Straight lines obviously have second derivatives that are zero but have no points of inflection. f(x) = x^4 is strange though. Does it have a region around x=0 where it is perfectly straight? The same 'problem' occurs for x^6, x^8, etc...not for x^2 though. The problem doesn't occur for x^3, x^5, x^7, etc since x=0 is indeed an inflection point for these functions.

Last edited: Oct 29, 2009
5. Oct 29, 2009

### leright

deleted

6. Oct 29, 2009

### l'Hôpital

f(x) = ax2n is always a parabola for any positive integer n, so concavity is always the same sign of a.

Last edited: Oct 29, 2009
7. Oct 29, 2009

### leright

I see that. This is very strange to me.

8. Oct 29, 2009

### l'Hôpital

Why?

You believe y = x2 is a parabola.

Let x = un, for some positive integer n.

then y = u2n. Still a parabola. : )

You can think of it in quadratic form.

As long as you have y = ax2n+ bxn + c, you have a parabola.

9. Oct 29, 2009

### leright

I don't think it is strange that it is a parabola. I think it is strange that its second derivative is zero at x=0.

10. Oct 30, 2009

### FedEx

A parabola only if we plot y versus x^n(ofcourse)

11. Oct 30, 2009

### FedEx

Well let me write down all the conditions which i know for a point to be a point if inflection

1)Concavity changes. That is if concave upwards than changes to concave downwards

2)It should be a critical point. dy/dx = zero.

3) The lowest order non zero derivative should be of odd order( ie third,fifth...)

I dont which of them is the necessary and the sufficient condition

12. Oct 30, 2009

### leright

a point doesn't need to be a critical point to be an inflection point.

13. Nov 16, 2009