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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Yitzi_K on April 08, 2010, 08:08:18 pm

Title: Proving points of inflection
Post by: Yitzi_K on April 08, 2010, 08:08:18 pm: 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 $x^3-3x^2+x+7$

the first derivative is $3x^2-6x+1$

and the second is $6x-6$

so the second derivative equals $0$ when $x=1$

therefore the point of inflection is at $(1,6)$ .

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

as when $x>1 , \frac{d^2y}{dx^2}>0$ and when $x<1 , \frac{d^2y}{dx^2}<0$ ?
Title: Re: Proving points of inflection
Post by: brightsky on April 08, 2010, 08:14:35 pm: Better safe than sorry. :p
Title: Re: Proving points of inflection
Post by: brightsky on April 08, 2010, 08:20:00 pm: btw, I don't think that that is the point of inflexion. :D Point of inflexions ain't relevant to all cubics.
Title: Re: Proving points of inflection
Post by: Yitzi_K on April 08, 2010, 08:21:33 pm: No? So what is?
Title: Re: Proving points of inflection
Post by: brightsky on April 08, 2010, 08:22:02 pm: 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.
Title: Re: Proving points of inflection
Post by: brightsky on April 08, 2010, 08:25:58 pm: I think what you have found using the second derivative is the point where the graph turns (bad expression :buck2:).

(http://www4c.wolframalpha.com/Calculate/MSP/MSP6719a5644g8ie869eh0000119e6h69378iha31?MSPStoreType=image/gif&s=61&w=299&h=144)
Title: Re: Proving points of inflection
Post by: brightsky on April 08, 2010, 08:29:24 pm: Someone please confirm...:p
Title: Re: Proving points of inflection
Post by: superflya on April 08, 2010, 08:31:14 pm: yea u have a max and min :S

fail.
Title: Re: Proving points of inflection
Post by: Yitzi_K on April 08, 2010, 08:33:52 pm: 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.
Title: Re: Proving points of inflection
Post by: Yitzi_K on April 08, 2010, 08:37:58 pm: 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
Title: Re: Proving points of inflection
Post by: brightsky on April 08, 2010, 08:38:26 pm: Oh ok, gotcha. :p In that case its right.
Title: Re: Proving points of inflection
Post by: the.watchman on April 08, 2010, 08:39:09 pm: Quote from: brightsky on April 08, 2010, 08:36:21 pm
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.
Title: Re: Proving points of inflection
Post by: brightsky on April 08, 2010, 08:42:06 pm: Quote from: the.watchman on April 08, 2010, 08:39:09 pm
Quote from: brightsky on April 08, 2010, 08:36:21 pm
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. ;)
Title: Re: Proving points of inflection
Post by: shinny on April 08, 2010, 08:45:22 pm: Try your method (without a gradient check on both sides) on $y=x^{4}$ . 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)
Title: Re: Proving points of inflection
Post by: the.watchman on April 08, 2010, 08:50:17 pm: Just a good example for you brightsky, take the graph $y=x^3 + x$
It has a point of inflection at the origin, but it is not stationary, as the gradient at the point is not zero.
Title: Re: Proving points of inflection
Post by: brightsky on April 08, 2010, 08:52:18 pm: Yeah, does the second derivative method work with graphs like $y = x^2, y = x^4, \text{etc}$ ? The (0,0) won't be a point of inflection right?
Title: Re: Proving points of inflection
Post by: brightsky on April 08, 2010, 08:54:29 pm: Quote from: the.watchman on April 08, 2010, 08:50:17 pm
Just a good example for you brightsky, take the graph $y=x^3 + x$
It has a point of inflection at the origin, but it is not stationary, as the gradient at the point is not zero.

Yeah, we've been learning all about stationary point of inflections at Dr He, so I've become a bit disillusioned by the general definition of point of inflection. :p
Title: Re: Proving points of inflection
Post by: the.watchman on April 08, 2010, 08:54:39 pm: Quote from: brightsky on April 08, 2010, 08:52:18 pm
Yeah, does the second derivative method work with graphs like $y = x^2, y = x^4, \text{etc}$ ? The (0,0) won't be a point of inflection right?

For $y=x^2$ etc., the origin is a point of inflection, but it is a special type, a stationary point of inflection :)
Title: Re: Proving points of inflection
Post by: Yitzi_K on April 08, 2010, 08:56:24 pm: Quote from: shinny on April 08, 2010, 08:45:22 pm
Try your method (without a gradient check on both sides) on $y=x^{4}$ . 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)

2nd derivative gives $12x^2$ therefore PoI is 0... but this quartic has no PoI, it's a stationary point.

I get what you're saying. Thanks. I'd karma you if I could :)
Title: Re: Proving points of inflection
Post by: brightsky on April 08, 2010, 08:56:54 pm: Quote from: the.watchman on April 08, 2010, 08:54:39 pm
Quote from: brightsky on April 08, 2010, 08:52:18 pm
Yeah, does the second derivative method work with graphs like $y = x^2, y = x^4, \text{etc}$ ? The (0,0) won't be a point of inflection right?

For $y=x^2$ etc., the origin is a point of inflection, but it is a special type, a stationary point of inflection :)
Really? The concavity doesn't really change though?
Title: Re: Proving points of inflection
Post by: the.watchman on April 08, 2010, 08:59:48 pm: Quote from: brightsky on April 08, 2010, 08:56:54 pm
Quote from: the.watchman on April 08, 2010, 08:54:39 pm
Quote from: brightsky on April 08, 2010, 08:52:18 pm
Yeah, does the second derivative method work with graphs like $y = x^2, y = x^4, \text{etc}$ ? The (0,0) won't be a point of inflection right?

For $y=x^2$ etc., the origin is a point of inflection, but it is a special type, a stationary point of inflection :)
Really? The concavity doesn't really change though?

True, now I'm confused, is a stationary point of inflection a point of inflection? :idiot2:
Title: Re: Proving points of inflection
Post by: brightsky on April 08, 2010, 09:00:37 pm: Nah, don't think it's a point of inflection at all. As Yitzi_K said, it's just a stationary point.
Title: Re: Proving points of inflection
Post by: shinny on April 08, 2010, 09:01:44 pm: Quote from: the.watchman on April 08, 2010, 08:59:48 pm
Quote from: brightsky on April 08, 2010, 08:56:54 pm
Quote from: the.watchman on April 08, 2010, 08:54:39 pm
Quote from: brightsky on April 08, 2010, 08:52:18 pm
Yeah, does the second derivative method work with graphs like $y = x^2, y = x^4, \text{etc}$ ? The (0,0) won't be a point of inflection right?

For $y=x^2$ etc., the origin is a point of inflection, but it is a special type, a stationary point of inflection :)
Really? The concavity doesn't really change though?

True, now I'm confused, is a stationary point of inflection a point of inflection? :idiot2:

It is, when it's actually a point of inflection. It's not in the case of $y=x^{4}$ though; it's clearly just a turning point. Just a nice, simple example to show why you must do a gradient check I guess.
Title: Re: Proving points of inflection
Post by: the.watchman on April 08, 2010, 09:03:03 pm: Quote from: brightsky on April 08, 2010, 09:00:37 pm
Nah, don't think it's a point of inflection at all. As Yitzi_K said, it's just a stationary point.

But it is called a stationary ... oh crap you're right, it's just a stationary point / turning point
I'm an idiot =='

I got stationary points and SPOIs muddled :uglystupid2: