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    I don't get how we step from the stationary point expression equalling zero, to the inequality. Maybe it is the discriminant, but it is two stationary points!! :confused:
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    (Original post by Gmart)
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    I don't get how we step from the stationary point expression equalling zero, to the inequality. Maybe it is the discriminant, but it is two stationary points!! :confused:
    Not quite sure what you are saying but this is indeed a use of the discriminant.

    We require 2 real roots of the equation and for this to occur b^2-4ac>0 .
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    (Original post by Gmart)
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    I don't get how we step from the stationary point expression equalling zero, to the inequality. Maybe it is the discriminant, but it is two stationary points!! :confused:
    At the stationary points dy/dx = 0. From this we get the quadratic which equals zero. For there to be two stationary points, this equation must have two solutions, this is where the discriminant comes in.
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    But the discriminant just has one inequality sign, the answer has two!!!

    And the discriminant is about having two roots, not two stationary points, and a quadratic only has one stationary point anyway.
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    (Original post by Gmart)
    But the discriminant just has one inequality sign, the answer has two!!!

    And the discriminant is about having two roots, not two stationary points, and a quadratic only has one stationary point anyway.
    You don't understand.

    Once you differentiated the numerator of the expression must be equal to zero.

    ie. x^2 +6x +3\lambda +6\lambda^2 = 0

    This determines the x-coordinates of the points at which you have stationary points. You want 2 stationary points, therefore you want this equation to give you two values of x. The equation must have two distinct roots. We use the discriminant.

    36 - 4(3\lambda + 6\lambda^2) = 36 - 12\lambda - 24\lambda^2 >0

    2\lambda^2 + \lambda - 3 < 0

    (2\lambda + 3)(\lambda -1) < 0

    Sketch the curve on the LHS, and determine the points at which it intersects the x-axis. Now look for the region where the curve is below the x-axis. Does that help?
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    (Original post by Ateo)
    You don't understand.

    Once you differentiated the numerator of the expression must be equal to zero.

    ie. x^2 +6x +3\lambda +6\lambda^2 = 0

    This determines the x-coordinates of the points at which you have stationary points. You want 2 stationary points, therefore you want this equation to give you two values of x. The equation must have two distinct roots. We use the discriminant.

    36 - 4(3\lambda + 6\lambda^2) = 36 - 12\lambda - 24\lambda^2 >0

    2\lambda^2 + \lambda - 3 < 0

    (2\lambda + 3)(\lambda -1) < 0

    Sketch the curve on the LHS, and determine the points at which it intersects the x-axis. Now look for the region where the curve is below the x-axis. Does that help?
    That sure does!!

    Thank you so much, I have been working myself up into a tizz about that for an embarrassing amount of time :O
 
 
 
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