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**robertpatch****Member**- Registered: 2013-06-07
- Posts: 3

Hi,

I have a question from my recent test:

A quartic function has the equation y = (x - 2) (x - 4) (3x^2 + kx + 8), where k is any real number. How many stationary points can the quartic have?

I have tried differentiating and equating to zero. However, the cubic equation that this yields does not readily allow the number of solutions (and hence stationary points) to be deduced. The answer is apparently 1 or 3, but I do not know how to show that 2 can't be the answer too. I recognise that we don't need to necessarily find the stationary points....we simply need to find how many there are. However, I don't know how else to proceed. I would appreciate any help on how to approach this. A graphics calculator (eg. TI-89 or TI-nspire) is allowed.

I am currently advanced high school level.

Thank you!!!!

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,536

hi robertpatch

Welcome to the forum.

A cubic can only have two roots if one is also a turning point. (Would normally cross the x axis three times or once if a turning point/point of inflexion causes the graph to avoid before crossing for time two and three => to have two, one must be 'on the axis' ie a turning point there)

LATER EDIT:

Sorry, that's a bit garbled so I've made a diagram (see below)

A cubic may have one, two or three zeros.

If you differentiate the cubic again, then you can begin to test which of those cubics are possible.

The characteristic of the quadratic, b^2 - 4ac, will tell you if there are turning points. Only if such a turning point coincides with one of the zeros of the cubic, will you get two.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**robertpatch****Member**- Registered: 2013-06-07
- Posts: 3

thank you for your reply, bob. i am still confused because of the effect of the k, which prevents us from solving the cubic to get three solutions. so far i have:

dy/dx = 12x^3 + (3k-54)x^2 + (64-12k)x + (8K - 48)

second derivative = 36x^2 + 6(k-18)x +64 - 12k

For this second derivative, b^2 - 4ac = 36k^2+432k+2448, which is always positive since discriminant <0 for this quadratic involving k and the quadratic involving k is an upright parabola.

So, this means that we have two values at which the second derivative equals zero. However, i do not know how to check if these coincide with stationary points.

in short, with the original quartic, there must be a stationary point between x=2 and x=4. How do we work out whether there could be a stationary point of inflection for x<2?? my issue is still proving whether or not we can have 2 stationary points for the quartic graph

thanks again for your help everyone!

*Last edited by robertpatch (2013-06-07 16:11:11)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,536

hi robertpatch

I've been playing with values of k using

http://www.mathsisfun.com/data/function … =-8&ymax=8

All graphs go through (0,64) (2,0) and (4,0)

So there is always a minimum between 2 and 4.

At approximately k = -9.73 there appears to be a point of inflexion (with zero gradient) at about x = 1.75.

At approximately k = 5.95 there appears to be a point of inflexion (with zero gradient) somewhere between x = -0.2 and x = 0.

With k = 1 there is only the minimum between 2 and 4.

With k = 10 there are three turning points.

So the answer to the question "How many stationary points can the quartic have?" would appear to be 1,2 or 3.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,536

improved accuracy and graphs.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**robertpatch****Member**- Registered: 2013-06-07
- Posts: 3

thanks bob that means the teacher's answers are wrong, haha. i better point it out. thanks for all of your help

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Hi Bob

Have you actually found such a quartic with 2 stationary points, because I have not?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,536

hi Stefy,

Well I thought I had. See post 4

By all means tell me where I'm going wrong.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Well, I have checked all k between -10 000 and 10 000 and in no case was there 2 stationary points.

Here lies the reader who will never open this book. He is forever dead.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,536

But k is any real number so you must have checked an infinite number of cases. I'm impressed.

Did you try k= 5.95361523..... ?

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Oh, you are right. I will try getting the exact answer if possible.

Here lies the reader who will never open this book. He is forever dead.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,536

Thank you.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Hi Bob

I don't think you'll like seeing this one:

Here lies the reader who will never open this book. He is forever dead.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,536

hi Stefy

You did that on paper of course.

But it's nice to know there's an exact solution.

Thanks.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

What's even more interesting is that it seems there are more than one values of k which do that. If I remember correctly, something similar should happen at +/- 2sqrt(33).

By the way, you haven't yet replied to the Discrete Calculus topic and our "interesting" antonym discussion.

Here lies the reader who will never open this book. He is forever dead.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,536

Stefy wrote:

By the way, you haven't yet replied to the Discrete Calculus topic and our "interesting" antonym discussion.

Antonym reply done.

Cannot find the other ??? It is being very discreet.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

The other what?

Here lies the reader who will never open this book. He is forever dead.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,536

you haven't yet replied to the Discrete Calculus topic

What is this topic?

url?

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

That is the same one.

Here lies the reader who will never open this book. He is forever dead.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,536

The same one as what?

You want me to respond to a topic called 'Discrete Calculus'. This didn't ring any bells in my brain so I did a search. No topic found.

So I looked through the recent topics list. Still nothing.

So I asked for the url and you have said it is the same one. Same as what. Is this some sort of test? Can I see the invisible words? Am I expected to read your mind? Is there a secret code?

dictionary.com wrote:

dis·creet

[dih-skreet] adjective

1.

judicious in one's conduct or speech, especially with regard to respecting privacy or maintaining silence about something of a delicate nature; prudent; circumspect............................

ibid wrote:

cal·cu·lus

[kal-kyuh-luhs] noun, plural cal·cu·li [kal-kyuh-lahy]

1.

.....................2.

Pathology . a stone, or concretion, formed in the gallbladder, kidneys, or other parts of the body.

I get it. You are worried about my kidney stone but are trying to be discreet about it. Too late! I've told everyone now.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Discrete and discreet are not the same.

I was referring to the topic where we were having the antonym discussion.

*Last edited by anonimnystefy (2013-06-10 06:37:26)*

Here lies the reader who will never open this book. He is forever dead.

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**SteveB****Member**- Registered: 2013-03-07
- Posts: 557

The cubic equation for dy/dx=0 was:

0 = 12x^3 + 3(k-18)x^2 + (64-12kx) + 8k-48

The DELTA discriminant for whether this cubic equation has:

3 distinct real roots when DELTA > 0

multiple root with all roots real when DELTA = 0

or one real root and two complex conjugate roots when DELTA < 0

(Source: Wikipedia entry for cubic polynomial equations)

Using an algebraic package you can solve this for k where DELTA = 0

So DELTA = 432k^4 + 12096k^3 + 83520k^2 - 345600k - 3998208

I then used a grapical calculator with a polynomial of order 4 solver to get numerical solutions:

Two of them were complex conjugates of (-12.1161977552, +/- 3.62069124665) these are irrelevant for this analysis.

The real solutions are relevant, one of them was Bob's k = 5.9536152399

The other was also mentioned by Bob earlier to less accuracy and I am getting: k = -9.72121972948

These solutions for k indicate that there is a multiple root with all roots real and a graph shows that they have two stationary points

in terms of the original quartic expression involving x and k. One is an inflection and the other a minimum.

You could work out whether there are 3 turning points or 1 turning point for the ranges either side and in between these

if you wanted to give a full analysis of how many turning points you get for all values of real k.

(Perhaps draw a graph on a graphics calulator for the function of DELTA in terms of k and see where it is above zero

and where it is below zero. Then use the wiki quote that I gave above.)

If I am understanding this correctly there is one stationary point inbetween the two (k>-9.7212197... and k<5.9536152...)

and there are three stationary points for k < -9.721297... and for k > 5.9536152...

The exact formula for calculating the roots of k is extremely complicated and I would not like to attempt that one

without a something like Wolfram or another computer algebra package.

*Last edited by SteveB (2013-06-10 08:48:39)*

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