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AP Calculus

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Show that the equation x^3 - 15x + c = o has exactly one real root.

All I know is that it has something to do with the Mean Value Theorem/Rolle's Theorem.

  • AP Calculus - ,

    You must have a typo.
    There are all kinds of values of c for which the above equation has 3 roots.
    e.g. if c=0
    x^3 - 15x = 0
    x(x^2 - 15) = 0
    x = 0 or x = ± √15

    are you sure it wasn't x^3 + 15x + c = 0 ?
    I will assume it was

    It's been 50 years but if I recall, Rolle's theorem says that if there are two x-intercepts , then the derivative must be zero between those two points

    so let f(x) = x^3 + 15x + c
    f '(x) = 3x^2 +15 = 0

    notice this is always positive, since x^2 = -5 has no real solution,
    so there are no turning points, and the curve cuts only once.
    You do know that every cubic must cut he x-axis at least once?

  • AP Calculus - ,

    The missing info it the interval [-2,2]. In this interval there could be just one root.

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