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Math Help - Showing gradient of a curve (Q10)

  1. #1
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    Showing gradient of a curve (Q10)

    A curve is such that \frac{d^2y}{dx^2}=5x-4. The gradient of the curve at the point (-2,4) is 3.

    Show that the gradient is never less than -\frac{83}{5}
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  2. #2
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    Quote Originally Posted by Punch View Post
    A curve is such that \frac{d^2y}{dx^2}=5x-4. The gradient of the curve at the point (-2,4) is 3.

    Show that the gradient is never less than -\frac{83}{5}
    The gradient of the curve is  \frac{dy}{dx}=\int (5x-4)dx=\frac{5x^2}{2}-4x+c

    At (-2,4), dy/dx=3 so c= -15

    dy/dx=\frac{5x^2}{2}-4x-15

    Complete the square and you will notice that the minimum value of the gradient is -83/5.
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  3. #3
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    \frac{dy}{dx}=\frac{5x^2}{2}-4x-15
    =\frac{5}{2}x^2-4x-15+(\frac{4}{2})^2-(\frac{4}{2})^2
    =(\frac{5}{2}x-2)^2-15-(\frac{4}{2})^2
    =(\frac{5}{2}x-2)^2-19

    minimum value=-19 which doesnt tally with the question of -\frac{83}{5}
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