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Math Help - Rolle's Theorem

  1. #1
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    Rolle's Theorem

    Hello I have some question about the Rolle's theorem

    Here is the question:
    Use Rolle's theorem to show that f(x)=x^3+3x-5 has exactly one real root.

    what is my limit!?I need some limit right!?

    Thanks
    NaNa
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  2. #2
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    Quote Originally Posted by NaNa View Post
    Hello I have some question about the Rolle's theorem

    Here is the question:
    Use Rolle's theorem to show that f(x)=x^3+3x-5 has exactly one real root.

    what is my limit!?I need some limit right!?

    Thanks
    NaNa
    No note quite.

    Suppose by way of contradiction that

    f(a)=f(b)=0,a \ne b note that f'(x)=3x^2+3=3(x^2+1)

    Now if you apply Rolle's theorem

    f(a)-f(b)=f'(c)(b-a) \iff 0=f'(c)(b-a)

    Now we have a problem since b \ne a \implies b-a \ne 0

    That mean that f'(c)=0 \implies 0=3(c^2+1) this is a contradiction because there is no real number c that fits the bill.

    So this is Half of the proof. We have show that there cannot be more than two roots, it is still up to you to show that there is at least one.

    Good luck.
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  3. #3
    MHF Contributor Calculus26's Avatar
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    Not to put too fine of a point on it but




    is a statement of the mean value theorem which granted is prooved using Rolle's Thm

    But we don't need the MVT for this problem

    Rolles theorem states if f(a) = f(b) = 0 under the proper continuity and differentiability conditions then there is at least one point c where
    f ' (c) = 0

    As stated assume f(a) = f(b) = 0

    Therefore and the results follows as stated
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by Calculus26 View Post
    Not to put too fine of a point on it but




    is a statement of the mean value theorem which granted is prooved using Rolle's Thm

    But we don't need the MVT for this problem

    Rolles theorem states if f(a) = f(b) = 0 under the proper continuity and differentiability conditions then there is at least one point c where
    f ' (c) = 0

    As stated assume f(a) = f(b) = 0

    Therefore and the results follows as stated
    Thanks I got a little a head of my self . Thanks again
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  5. #5
    MHF Contributor Calculus26's Avatar
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    When I have time I read your posts because I've always been impressed
    with you replies --- I probably wouldn't even have read it if you weren't so damn good.
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