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Math Help - solution of equation/ rolle's theorem

  1. #1
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    solution of equation/ rolle's theorem

    For c > 0, prove that the following equation does not have two solutions:

    x^3 - 3x + c = 0 ;  0 < x < 1
    I think we have to show this using Rolle's theorem, but I am not being able to use the theorem properly on this question. Can anyone give any suggestions?
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  2. #2
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    The LHS is decreasing (differentiate). Can a decreasing function have two zeros?
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  3. #3
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    Quote Originally Posted by maddas View Post
    The LHS is decreasing (differentiate). Can a decreasing function have two zeros?
    hi maddas,

    How is the function decreasing? Its derivative is positive, doesn't it mean that it is strictly increasing?
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  4. #4
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    Well, if its increasing, it can't have two zeros either. Is the derivative really positive?

    edit: no, I'm pretty sure its negative.
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  5. #5
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    Quote Originally Posted by maddas View Post
    Well, if its increasing, it can't have two zeros either. Is the derivative really positive?

    edit: no, I'm pretty sure its negative.
    Thanks maddas, it is negative...

    so differentiating gives 3x^2 - 3 = 0. But then, f'(x) = 0 for x= \pm 1.
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  6. #6
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    Fortunately you're only considering 0<x<1.
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  7. #7
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    Quote Originally Posted by maddas View Post
    Fortunately you're only considering 0<x<1.
    So , I would do the same for x^5 + 5x + 1 = 0 ; -1<x<0 to show that the equation has only one solution..?
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  8. #8
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    Pretty sure.
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