Proving points of inflection

[Yitzi_K]

When finding a point of inflection (by finding when the second derivative is equal to zero) is it ok to leave it as that or do we have to prove that it's a point of inflection?

eg for

the first derivative is

and the second is

so the second derivative equals when

therefore the point of inflection is at .

Is it good enough to leave the answer like that or do I have to then say:

as when and when ?

[brightsky]

Better safe than sorry. :p

[brightsky]

btw, I don't think that that is the point of inflexion. Point of inflexions ain't relevant to all cubics.

[Yitzi_K]

No? So what is?

[brightsky]

Point of inflexions ain't relevant to all cubics. No point of inflexion for this one. If you solve the original "first" derivative, then you'll find that there are two "stationary points", but none of them are points of inflexion.

[brightsky]

I think what you have found using the second derivative is the point where the graph turns (bad expression  :buck2:).

[brightsky]

Someone please confirm...:p

[superflya]

yea u have a max and min :S

fail.

[Yitzi_K]

On the graph you've got there, it looks like I'm right? x=1 is the point where the rate of change of the gradient changes from positive to negative, which is a point of inflection (inflexion).

P of I =/= stationary point.

[Yitzi_K]

I don't think so...

take a look at this http://en.wikipedia.org/wiki/File:Animated_illustration_of_inflection_point.gif

that's pretty much exactly what I have

[brightsky]

Oh ok, gotcha. :p In that case its right.

[the.watchman]

Hmmm..seems like our definition of the inflection point is different. I thought all inflection points were stationary points, e.g. stationary point of inflection? :S

No, an inflection point doesn't have to be stationary, its definition is when the second derivative changes from positive to negative or vice versa.

[brightsky]

Hmmm..seems like our definition of the inflection point is different. I thought all inflection points were stationary points, e.g. stationary point of inflection? :S

No, an inflection point doesn't have to be stationary, its definition is when the second derivative changes from positive to negative or vice versa.

k, I get it. Thanks.

[shinny]

Try your method (without a gradient check on both sides) on . I (think) this gives the answer to the thread =T (note: have not done any Maths whatsoever in 2 years, might be wrong with the definitions here)

[the.watchman]

Just a good example for you brightsky, take the graph
It has a point of inflection at the origin, but it is not stationary, as the gradient at the point is not zero.