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Math Help - how to prove that gradient of curve is never less than something

  1. #1
    Member helloying's Avatar
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    how to prove that gradient of curve is never less than something

    the eqn of curve is x - x - 5x-3
    dy/dx=3x -2x-5
    dy/dx=6x-2

    how to show that gradient of the curve is never less than -16/3
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  2. #2
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    Quote Originally Posted by helloying View Post
    the eqn of curve is x - x - 5x-3
    dy/dx=3x -2x-5
    dy/dx=6x-2

    how to show that gradient of the curve is never less than -16/3
    Assume that the gradient of the curve COULD be less than -\frac{16}{3}.

    Then \frac{dy}{dx} = 3x^2 - 2x - 5 < -\frac{16}{3}

    3x^2 - 2x + \frac{1}{3} < 0

    3\left(x^2 - \frac{2}{3}x + \frac{1}{9}\right) < 0

    3\left[x^2 - \frac{2}{3}x + \left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2 + \frac{1}{9}\right] < 0

    3\left[\left(x - \frac{1}{3}\right)^2 - \frac{1}{9} + \frac{1}{9}\right] < 0

    3\left(x - \frac{1}{3}\right)^2 < 0


    This is clearly a contradiction, as \left(x - \frac{1}{3}\right)^2 > 0.

    Therefore the gradient can never be less than -\frac{16}{3}.
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