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FedEx
Oct29-09, 09:59 PM
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...

fourier jr
Oct29-09, 10:06 PM
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

l'Hôpital
Oct29-09, 10:10 PM
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

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.

Matt1
Oct29-09, 10:49 PM
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.

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.

Matt1
Oct29-09, 11:09 PM
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.


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l'Hôpital
Oct29-09, 11:36 PM
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.

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

Matt1
Oct29-09, 11:47 PM
f(x) = ax2n is always a parabola for any positive integer n, so concavity is always the same sign of a.

I see that. This is very strange to me.

l'Hôpital
Oct29-09, 11:53 PM
I see that. This is very strange to me.

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.

Matt1
Oct30-09, 12:08 AM
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.

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.

FedEx
Oct30-09, 09:43 AM
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.

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

FedEx
Oct30-09, 09:49 AM
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

Matt1
Oct30-09, 01:51 PM
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

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

adrianne42
Nov16-09, 04:11 PM
Nicely salved ur equation....
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