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•  » Polynomials and Stationary Points - much harder than it seems!

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SteveB
2013-06-11 06:21:27

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.

anonimnystefy
2013-06-11 04:28:04

Discrete and discreet are not the same.

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

bob bundy
2013-06-11 03:32:24

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
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

anonimnystefy
2013-06-10 23:09:46

That is the same one.

bob bundy
2013-06-10 22:02:47

you haven't yet replied to the Discrete Calculus topic

What is this topic?

url?

Bob

anonimnystefy
2013-06-10 21:37:52

The other what?

bob bundy
2013-06-10 17:33:03

#### Stefy wrote:

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

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

Bob

anonimnystefy
2013-06-10 06:48:52

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.

bob bundy
2013-06-10 04:50:28

hi Stefy

You did that on paper of course.

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

Thanks.

Bob

anonimnystefy
2013-06-10 03:53:40

Hi Bob

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

bob bundy
2013-06-09 04:33:51

Thank you.

Bob

anonimnystefy
2013-06-09 04:32:23

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

bob bundy
2013-06-09 04:14:43

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

anonimnystefy
2013-06-09 03:56:54

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

bob bundy
2013-06-09 03:39:09

hi Stefy,

Well I thought I had.  See post 4

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

Bob