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Show that the equation x^3-15x+c=0 has at most one root in the interval [-2,2].

  • calc -

    Hint:
    For f(x) to have at most one root on [-2,2], it must be strictly increasing or decreasing on that interval.

    So let's examine f'(x) on the interval [-2,2].

    f(x)=x^3-15x+c
    f'(x)=3x²-15
    Absolute maximum on [-2,2] is when x=±2, f'(x)=4-15=-11
    Absolute minimum on [-2,2] is when x=0, f'(x)=-15.

    Therefore f'(x) is negative on [-2,2], therefore strictly decreasing. Under these circumstances, f(x) can have at most one zero (root).

  • calc -

    Thank you for your answer!!! My solution manual use Roll's theorem to solve it and I cannot understand it at all.
    Your f'(2) and f'(-2) should equal -3 tho.

  • calc -

    yo, you plugged the numbers wrong bruh...
    + or - 2 when you plug it into f prime of x, you get -3 bruh not -11...

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