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Math Help - gradient of a curve

  1. #1
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    gradient of a curve

    Find the gradient of the curve y=2x^3 - 5x^2 + 46x + 87 at the point where it crosses the x axis.

    I'm having real problems with this one as all the examples i've done before have given me a full set of co-ordinates.

    I know that it will cross the x axis when y equals 0 so i have at least one co-ordinate. I differentiated it to 6x^2 - 10x + 46 I then thought about making 6x^2 - 10x = -46 and trying to get a value of x from that but i'm getting nowhere.

    Any help on this is apprecciated
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  2. #2
    Senior Member DeMath's Avatar
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    \left. {2{x^3} - 5{x^2} + 46x + 87 = 0{\text{ }}} \right| \times 4 \Leftrightarrow

    \Leftrightarrow 8{x^3} - 20{x^2} + 184x + 348 = 0 \Leftrightarrow

    \Leftrightarrow 8{x^3} + 27 - 20{x^2} + 45 + 184x + 276 = 0 \Leftrightarrow

    \Leftrightarrow 8{x^3} + 27 - 5\left( {4{x^2} - 9} \right) + 92\left( {2x + 3} \right) = 0

    \Leftrightarrow \left( {2x + 3} \right)\left( {4{x^2} - 6x + 9} \right) - 5\left( {2x - 3} \right)\left( {2x + 3} \right) + 92\left( {2x + 3} \right) = 0.
    Last edited by DeMath; January 9th 2009 at 05:56 PM.
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  3. #3
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    no idea what you've done or how it helps me,

    sorry
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by ally79 View Post
    no idea what you've done or how it helps me,

    sorry
    \left. {2{x^3} - 5{x^2} + 46x + 87 = 0{\text{ }}} \right| \times 4 \Leftrightarrow

    \Leftrightarrow 8{x^3} - 20{x^2} + 184x + 348 = 0 \Leftrightarrow

    \Leftrightarrow 8{x^3} + 27 - 20{x^2} + 45 + 184x + 276 = 0 \Leftrightarrow

    \Leftrightarrow 8{x^3} + 27 - 5\left( {4{x^2} - 9} \right) + 92\left( {2x + 3} \right) = 0\Leftrightarrow

    \Leftrightarrow\left( {2x + 3} \right)\left( {4{x^2} - 6x + 9} \right) - 5\left( {2x - 3} \right)\left( {2x + 3} \right) + 92\left( {2x + 3} \right) = 0 \Leftrightarrow

    \Leftrightarrow \left( {2x + 3} \right)\left( {4{x^2} - 6x + 9 - 5\left( {2x - 3} \right) + 92} \right) = 0 \Leftrightarrow

    \Leftrightarrow \left( {2x + 3} \right)\left( {4{x^2} - 16x + 116} \right) = 0 \Leftrightarrow

    \Leftrightarrow \left[ \begin{gathered}2x + 3 = 0, \hfill \\{x^2} - 4x + 29 = 0; \hfill \\ \end{gathered}  \right. \Leftrightarrow \left[ \begin{gathered}{x_1} =  - \frac{3}{2}, \hfill \\{x_{2,3}} = 2 \pm i\sqrt 5 . \hfill \\ \end{gathered}  \right.

    {x_1} =  - \frac{3}{2} So \left( { - \frac{3}{2};y\left( { - \frac{3}{2}} \right)=0} \right) it's the point where {y\left( x \right) = 2{x^3} - 5{x^2} + 46x + 87} crosses the x axis.

    Or I do not understand you?
    Last edited by DeMath; January 9th 2009 at 03:56 PM. Reason: Sorry, x1 = - 3/2
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  5. #5
    Senior Member DeMath's Avatar
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    ally79, I wrote this post, if you have not yet realized that you need to do.

    First, remember the definition of a gradient \nabla F of the function F\left( {{x_1},{x_2}, \ldots ,{x_n}} \right):

    The gradient \nabla F of the function F\left( {{x_1},{x_2}, \ldots ,{x_n}} \right) is a vector, whose projection on the axises of coordinates are partial derivatives of functions F\left( {{x_1},{x_2}, \ldots ,{x_n}} \right), i.e.

    \nabla F = \left( {\frac{{\partial F}}{{\partial {x_1}}},\frac{{\partial F}}{{\partial {x_2}}}, \ldots ,\frac{{\partial F}}{{\partial {x_n}}}} \right).

    You have to find \left|{\nabla y(x)}\right| of y=2{x^3} - 5{x^2} + 46x + 87 at the point where it crosses the x axis, i.e. you need to find the absolute value of the derivative function y=2{x^3} - 5{x^2} + 46x + 87 at the point where it crosses the x axis.

    Finally, you need to find this value: \left| {\nabla y(x)} \right| = \left| {y'\left( { - \frac{3}{2}} \right)} \right|.
    Last edited by DeMath; January 11th 2009 at 04:25 AM.
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  6. #6
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    Find the gradient of the curve y=2x^3 - 5x^2 + 46x + 87 at the point where it crosses the x axis.
    using the rational root theorem and the intermediate value theorem, you can determine that there exists one real root at x = -\frac{3}{2}.

    (why teachers still make students do this is beyond me)

    now ... evaluate \frac{dy}{dx} at x = -\frac{3}{2}
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