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Math Help - question on curve, points of inflection, stationary points and max and min points

  1. #1
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    question on curve, points of inflection, stationary points and max and min points

    For the function: y=2x^2/x^2-1

    (i) Find all stationary points.


    (ii) Find the intervals where y is increasing and the intervals where y is decreasing.


    (iii) Find all local minima and local maxima.


    (iv) Find all inflection points.


    (v) Sketch the graph, indicating all maxima and minima, and all in
    ection points.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: question on curve, points of inflection, stationary points and max and min points

    First, start differentiating the function.
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    Re: question on curve, points of inflection, stationary points and max and min points

    1st find \frac{dy}{dx}=-\frac{4x}{(x^2)-1}
    putting \frac{dy}{dx}=0 gives (0,0) as stationary point
    Now for x < 0, y is increasing and for x > 0, y is decreasing
    apply second derivative test to find (0,0) as maxima
    curve has no point of inflection
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: question on curve, points of inflection, stationary points and max and min points

    You forgot a square in the denominator:
    \frac{dy}{dx}\left(\frac{2x^2}{x^2-1}\right)=\frac{4x(x^2-1)-2x^2(2x)}{(x^2-1)^2}=\frac{-4x^3-4x-4x^3}{(x^2-1)^2}=\frac{-4x}{(x^2-1)^2}

    Which is offcourse don't make any difference in calculating the stationary points, but maybe for the inflection points it does.
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