
Polynomials and Stationary Points  much harder than it seems!
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. TI89 or TInspire) is allowed.
I am currently advanced high school level.
Thank you!!!!
 bob bundy
 Moderator
Re: Polynomials and Stationary Points  much harder than it seems!
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
Re: Polynomials and Stationary Points  much harder than it seems!
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 + (3k54)x^2 + (6412k)x + (8K  48)
second derivative = 36x^2 + 6(k18)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 (20130608 14:11:11)
 bob bundy
 Moderator
Re: Polynomials and Stationary Points  much harder than it seems!
hi robertpatch
I've been playing with values of k using
http://www.mathsisfun.com/data/function … amp;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
 bob bundy
 Moderator
Re: Polynomials and Stationary Points  much harder than it seems!
improved accuracy and graphs.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Polynomials and Stationary Points  much harder than it seems!
thanks bob that means the teacher's answers are wrong, haha. i better point it out. thanks for all of your help
Re: Polynomials and Stationary Points  much harder than it seems!
Hi Bob
Have you actually found such a quartic with 2 stationary points, because I have not?
The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 bob bundy
 Moderator
Re: Polynomials and Stationary Points  much harder than it seems!
hi Stefy,
Well I thought I had. See post 4
By all means tell me where I'm going wrong.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Polynomials and Stationary Points  much harder than it seems!
Well, I have checked all k between 10 000 and 10 000 and in no case was there 2 stationary points.
The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 bob bundy
 Moderator
Re: Polynomials and Stationary Points  much harder than it seems!
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
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Polynomials and Stationary Points  much harder than it seems!
Oh, you are right. I will try getting the exact answer if possible.
The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 bob bundy
 Moderator
Re: Polynomials and Stationary Points  much harder than it seems!
Thank you.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Polynomials and Stationary Points  much harder than it seems!
Hi Bob
I don't think you'll like seeing this one:
The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 bob bundy
 Moderator
Re: Polynomials and Stationary Points  much harder than it seems!
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Polynomials and Stationary Points  much harder than it seems!
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.
The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 bob bundy
 Moderator
Re: Polynomials and Stationary Points  much harder than it seems!
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
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Polynomials and Stationary Points  much harder than it seems!
The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 bob bundy
 Moderator
Re: Polynomials and Stationary Points  much harder than it seems!
you haven't yet replied to the Discrete Calculus topic
What is this topic?
url?
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Polynomials and Stationary Points  much harder than it seems!
The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 bob bundy
 Moderator
Re: Polynomials and Stationary Points  much harder than it seems!
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 [dihskreet] 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 [kalkyuhluhs] noun, plural cal·cu·li [kalkyuhlahy] 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
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Polynomials and Stationary Points  much harder than it seems!
Discrete and discreet are not the same.
I was referring to the topic where we were having the antonym discussion.
Last edited by anonimnystefy (20130611 04:37:26)
The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Polynomials and Stationary Points  much harder than it seems!
The cubic equation for dy/dx=0 was:
0 = 12x^3 + 3(k18)x^2 + (6412kx) + 8k48
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 (20130611 06:48:39)
