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Show a function has at most one Root (IVT & Rolles [quick question])

  1. Nov 25, 2008 #1
    show x^3-15x+c=0 has at most one root on the interval [-2, 2].

    My work/attempted solution.

    Using the IVT i determined that there is a possible root (N=c) within the interval of [-2,2]. because c is my constant i decided to Use [d] as my letter for explanation.

    c=-22 when for f(-2)
    c=22 for f(2)

    f is continuous on [-2,2] because it is a polynomial. Let -c be a number b/w f(-2) and f(2) where f(-2) does not equal f(2). Then there exist a root d on (-2,2) such that F(d)=-c

    I am stuck there. I feel like i have done a bunch of stuff wrong so a little nudge in the right direction would be great.

  2. jcsd
  3. Nov 25, 2008 #2


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    Homework Helper
    Gold Member

    Hard to follow what you are doing especially when not using LaTex (in fact I'm not sure you are doing anything :biggrin:).

    However, just ask yourself where are the turning points (the maximum and minimum) of your polynomial and you will soon see.

    When they told you 'Rolle's Theorem' they were pretty well telling you that.
  4. Nov 25, 2008 #3


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    "when for f(-2)"? f(-2)= c+ 22, not "22= -c".

    f(2)= 8- 30+ c= c- 22. That does not say c= 22 unless you are assuming f(2)= 0- which you have no right to assume.

    Don't use "c" to mean two different things. Did you mean d? But f(d)= -c is not f(d)= 0 which is what you want.

    You cannot prove that x3- 15x2+ c= 0 has a root in [-2, 2] because it is not true: x3- 15x+ 50= 0 has NO roots in that interval. And you are only asked to prove that there are not MORE than one root.

    Rolle's theorem, which you mention in your title, says that if there are two points at which f(x)= 0, then there must be an x between them where f'(x)= 0. Look at the derivative of f, not f.
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