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Math Help - Turning point

  1. #1
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    dfsdfdf
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    Turning point

    Dear Sir,
    I would be most grateful if some friends can help to solve the attached question.
    Thanks
    Attached Thumbnails Attached Thumbnails Turning point-turning-point.jpg  
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  2. #2
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    Hello, kingman!

    I'm coming up with a strange result . . .



    A curve has the equation: . y \:=\:\dfrac{2x^2+px+2}{x+1} where p is a constant.

    By considering \dfrac{dy}{dx} .
    . . . I don't see how
    . . find the range of p such that the equation: . \dfrac{2x^2+px+2}{x+1} \:=\:k
    . . has exactly 2 real roots for all real values of k.

    The equation is: . 2x^2 + px + 2 \:=\:k(x+1)

    . . which simplifies to: . 2x^2 + (p-k)x + (2-k) \:=\:0


    A quadratic has two real roots if its discriminant is positive.

    The discriminant is: . b^2 - 4ac \;=\;(p-k)^2 - (4)(2)(2-k)

    . . And we want: . p^2 - 2kp + k^2 + 8x - 16 \;>\;0 .[1]


    This is an up-opening parabola.
    . . It will be positive if its vertex is above the p-axis.

    Its vertex is found at: . p \:=\:\dfrac{\text{-}b}{2a} \:=\:\dfrac{\text{-}(\text{-}2k)}{2(1)} \:=\:k

    . . Hence, the vertex is: . (k,\;8k-16)

    The parabola is positive if: . 8k-16 \:>\:0 \quad\Rightarrow\quad k \:>\:2


    We have a constaint on k.

    Yet we must find values of p which satisfy [1] for all values of k.


    Is there a flaw in my reasoning?

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  3. #3
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    If eq 1 is written in powers of k instead of p we can get the answer but I still wonder why.But question required one to start from dy/dy with the hint that if dy/dy =0 has no turning point than the answer follows and again I wonder why this is so.
    Thanks
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