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Math Help - 14th Degree Polynomial Help!

  1. #1
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    14th Degree Polynomial Help!

    I need help find all the turning points of this function:

    f(x) = x^(14)-12x^(13)-21x^(12)+744x^(11)-1326x^(10)-16800x^(9)+56862x^(8)+153768x^(7)-817347x^(6)-253652x^(5)+4829511x^(4)-3026448x^(3)-9283680x^(2)+5702400x+7776000

    the 14th degree polynomial is the expanded one. Here is the original one:

    f(x) = (x+1)^(2)(x-5)^(3)(x+4)^(4)(x-3)^(5)

    What i want to know is that, how can i find the turning points. I want to know how you can find all the turning points of this function. If you could show, can you please list all the working out, please!

    Thanks!
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  2. #2
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    Lightbulb Re: 14th Degree Polynomial Help!

    Attached Thumbnails Attached Thumbnails 14th Degree Polynomial Help!-14-degree-polynomial.png  
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  3. #3
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    Re: 14th Degree Polynomial Help!

    What does the graph show?
    I don't really need the graph.
    What I want to know is that, how can i find all the turning points of the function

    And by the way, what program did you use to graph it?
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  4. #4
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    Re: 14th Degree Polynomial Help!

    Quote Originally Posted by Jeka View Post
    What does the graph show?
    I don't really need the graph.
    What I want to know is that, how can i find all the turning points of the function

    And by the way, what program did you use to graph it?
    Actually the graph turns out to be very useful. Turning points occur when the derivative is zero, and changes signs (from + to -, or vice versa). The dashed line represents the derivative of f(x) (denoted f'(x) or \frac{d}{dx} f(x)) and intersects the x-axis at -4, -2.4916, -1, etc.
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  5. #5
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    Re: 14th Degree Polynomial Help!

    Hello, Jeka!

    Are you familiar with the Extended Product Rule?

    For three functions: . y \;=\;f(x)\!\cdot\!g(x)\!\cdot\!h(x)

    . . we have: . y' \;=\;{\color{blue}f'(x)}\!\cdot\!g(x)\!\cdot\!h(x) + f(x)\!\cdot\!{\color{blue}g'(x)}\!\cdot\!h(x) + f(x)\!\cdot\!g(x)\!\cdot\!{\color{blue}h'(x)}

    Do you see the pattern?



    I need help find all the turning points of this function:

    . . f(x) \:=\: (x+1)^2(x-5)^3(x+4)^4(x-3)^5

    The derivative is: . f'(x) \;=\;\begin{Bmatrix}{\color{blue}2(x+1)}\cdot(x-5)^3(x+4)^4(x-3)^5 \\ + (x+1)^2\cdot {\color{blue}3(x-5)^2}\cdot(x+4)^4(x-3)^5 \\ +(x+1)^2(x-5)^3\cdot {\color{blue}4(x+4)^3}\cdot (x-3)^5 \\ +(x+1)^2(x-5)^3(x+4)^4\cdot {\color{blue}5(x-3)^4} \end{Bmatrix} \;=\;0

    Factor: . f'(x) \;=\;(x+1)(x-5)^2(x+4)^3(x-3)^4\cdot\begin{Bmatrix}2(x-5)(x+4)(x-3) \\ + 3(x+1)(x+4)(x-3) \\ + 4(x+1)(x-5)(x-3) \\ + 5(x+1)(x-5)(x+4) \end{Bmatrix} \;=\;0

    Some turning points are at: . x \:=\:\text{-}1,5,\text{-}4,3

    I found no zeros for the cubic polynomial in the braces.
    (But I see that MaxJasper found them.)
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  6. #6
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    Re: 14th Degree Polynomial Help!

    I don't really get it...
    What i know right now is that, there are turning points at
    (-1,0) & (5,0) & (-4,0) & (3,0)

    There should be 3 or 4 more left
    But i can't seem to find it.
    And i'm talking about this function:
    f(x) = (x+1)^(2)(x-5)^(3)(x+4)^(4)(x-3)^(5)
    the above function, there are 4 four graphs (two types), combined together.
    I want to know how i can find the other 3 or 4 turning points.
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  7. #7
    MHF Contributor MarkFL's Avatar
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    Re: 14th Degree Polynomial Help!

    You should ultimately be able to find:

    f'(x)=2(x+1)(x-5)^2(x+4)^3(x-3)^4(7x^3-15x^2-72x+22)

    Obviously, the roots from the linear factors are easy enough to find, but the roots of the cubic factor are not, and I would simply use Newton's method to approximate these roots. Let:

    g(x)\equiv7x^3-15x^2-72x+22

    Thus: g'(x)=21x^2-30x-72

    So, we use the recursion:

    x_{n+1}=x_n-\frac{7x_n^3-15x_n^2-72x_n+22}{21x_n^2-30x_n-72}

    Looking at a graph of the cubic, we can then find reasonable guesses for the roots.

    Root 1: x_0\approx-2

    x_1\approx-2.69\bar{4}

    x_2\approx-2.50948582734

    x_3\approx-2.49174430831

    x_4\approx-2.49158454787

    x_5\approx-2.49158453496

    x_6\approx-2.49158453496

    Root 2: x_0\approx0

    x_1\approx0.30\bar5

    x_2\approx0.290395504076

    x_3\approx0.290370205766

    x_4\approx0.290370205694

    x_5\approx0.290370205694

    Root 3: x_0\approx4

    x_1\approx4.402\bar{7}

    x_2\approx4.34537960674

    x_3\approx4.34407214407

    x_4\approx4.34407147212

    x_5\approx4.34407147212

    So, we find the 3 roots of the cubic factor may be approximated by:

    x\approx-2.49158453496,0.290370205694,4.34407147212
    Last edited by MarkFL; October 3rd 2012 at 12:11 AM.
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  8. #8
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    Re: 14th Degree Polynomial Help!

    Sorry Guys, i seriously don't know where you guys are heading.
    I don't know what rule you're using, i don't know how you guys got that.
    It would be really really really helpful if you guys have some good explaination.

    Soroban, i'm really happy that you're helping me.
    But i don't know how you factored the first derivation of the function

    MarkFL2, you too, i appreciate your help.
    But i don't really know what you're doing. I don't know what rule. I know norhing.
    You're just like, write so much stuff. That looks so pro. But i don't know what its say

    Sorry, guys. Could you please have some explanation? Please!

    Thanks
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  9. #9
    MHF Contributor MarkFL's Avatar
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    Re: 14th Degree Polynomial Help!

    Maybe things have changed since I was a student back in the days of stone-washed jeans and mullets, but we were taught Newton's method in Chapter 2 of Calc I.
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  10. #10
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    Re: 14th Degree Polynomial Help!

    MarkFL2, well you can teach me the way you were taught. I just want it solved
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  11. #11
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    Re: 14th Degree Polynomial Help!

    If you read the link I provided, it will tell you how Newton's method works to converge to the root of a function.

    If you just want it solved, then plug the values I gave for the roots into the original function to find the approximate values of the turning points resulting from the cubic factor of the derivative. If you wish for exact values, then read this article on how to obtain them:

    Cubic function - Wikipedia, the free encyclopedia
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  12. #12
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    Re: 14th Degree Polynomial Help!

    But i'm not actually trying to find the x-intercepts on the x-axis. I'm trying to find the turning points. I already know all the x and y intercepts. What I am trying to find is the turning points, that aren't touching the x or y axis.
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  13. #13
    MHF Contributor MarkFL's Avatar
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    Re: 14th Degree Polynomial Help!

    The roots of the derivative give you the critical values. Roots of odd multiplicity will give you turning points, when plugged into the original function.
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  14. #14
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    Re: 14th Degree Polynomial Help!

    Sorry, I don't get it. Sorry.

    Can you PLEASE do an example, on my function.
    PLEASEEEEEEEEEEEEEEEEEEEEE!
    I beg you!
    Please!!!!!!!!
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  15. #15
    MHF Contributor MarkFL's Avatar
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    Re: 14th Degree Polynomial Help!

    I suggest working with simpler functions to get a feel for this first.
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