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Math Help - Using Rolle's theorem to prove at most one solution of quadratic

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    Using Rolle's theorem to prove at most one solution of quadratic

    How do we use Rolle's theorem to prove that there is at most one solution of

    x^5 -5x +b=0

    in [-1,1] for any b \in R

    and for what values of b does this have a solution in [-1,1]?
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    MHF Contributor FernandoRevilla's Avatar
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    If -1\leq \alpha < \beta\leq 1 were roots of f(x)=x^5-5x+b then, would exist \xi\in (\alpha,\beta) such that f'(\xi)=5\xi^4-5=0 , but \xi=\pm 1 (contradiction). Try the second question.


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    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by maximus101 View Post
    How do we use Rolle's theorem to prove that there is at most one solution of

    x^5 -5x +b=0

    in [-1,1] for any b \in R
    Let f(x)=x^5-5x+b

    f'(x)=5x^4-5=5(x^4-1)\leq 0 in the interval [-1, 1].

    So f(x) is decreasing in the interval [-1, 1].

    Therefore, there is at most one solution of x^5-5x+b=0 in [-1, 1] for any b\in R.

    PS: This method does not use Rolle's theorem, though.
    Last edited by alexmahone; February 11th 2011 at 06:18 AM. Reason: Added postscript
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by maximus101 View Post
    and for what values of b does this have a solution in [-1,1]?
    f(-1)=4+b

    f(1)=-4+b

    For the conditions of the intermediate value theorem to be satisfied, f(-1) and f(1) should have opposite signs.

    4+b\geq0\geq-4+b (Note that f is a decreasing function.)

    -4\leq b\leq 4
    Last edited by alexmahone; February 11th 2011 at 06:42 AM. Reason: Stupid mistake
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