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Show that the equation has at most one root

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that x^5 +10x +c=0 has at most one real root on the interval [-1, 1]


    2. Relevant equations

    3. The attempt at a solution

    I first try to find an x value inorder to find c so i take the end points of the interval and solve for for c

    -1 +6=-c
    -5=c

    1-6=-c
    -5=c


    I get two c's which isn't right so I'm guessing depending on which c i choose there is only 1 root?
     
  2. jcsd
  3. Oct 15, 2012 #2

    LCKurtz

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    Hint: Look at the derivative.
     
  4. Oct 15, 2012 #3

    HallsofIvy

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    Blast you, LCKurtz!

    I was going to say that!
     
  5. Oct 15, 2012 #4
    So i took the derivative and got:
    f'(x)=5X^4 -6

    So i plug in the end points and the outputs seem to be the same which means that f'(x)<0 on [-1,1] meaning that it can only cross the x axis once along that interval, correct?
     
  6. Oct 15, 2012 #5

    LCKurtz

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    Heh heh. Go ahead and say it. :smile: It's OK to repeat, like Erdos did when he gave a "silver spoon" argument: "Chebyshev said it, and I say it again, There is always a prime between n and 2n".
     
  7. Oct 15, 2012 #6

    LCKurtz

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    The derivative of 10x is -6??????
     
  8. Oct 15, 2012 #7
    woops it seems i wrote the wrong equation in the original post. The equation in question is

    x^5 -6 +c =0

    sorry about that
     
  9. Oct 16, 2012 #8

    LCKurtz

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    So the derivative is...., and therefore.....?
     
  10. Oct 16, 2012 #9
    f'(x)=5X^4 -6

    f'(x)<0 on [-1,1]

    therefore there is only 1 root?
     
  11. Oct 16, 2012 #10

    HallsofIvy

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    Between any two roots of a polynomial, there is a root of its derivative. Since f' has no root in [-1, 1], f can has no more than one root in that interval.

    Now, what can you say about c so that there exists exactly one root in that interval?
     
  12. Oct 16, 2012 #11

    LCKurtz

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    The derivative of -6 is not -6.
     
  13. Oct 16, 2012 #12
    Wow whats wrong with me I meant to say

    x^5 -6x +c =0

    Please forgive my stupidity. :P
     
  14. Oct 16, 2012 #13

    Mark44

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    Aside from the fact that this isn't the right approach, you are making two tacit assumptions.
    1. When you substitute x = -1, you are assuming that the root is at x = -1, and looking for the value of c that makes that happen.
    2. When you substitute x = 1, you are now assuming that the root is at x = +1, and looking for the almost certainly different value of c that makes that happen.

    It would be extremely unusual for the same value of c to work in both cases.
     
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