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Curver sketching

  1. Apr 16, 2009 #1
    in trying to sketch the curve of y=(2-x^2)/(1+x^4)



    y'=(-2x-8x^3+2x^5)/(1+x^4)^2

    y''=[ (1+x^4)^2(-2-24x^2+10x^4)-(-2x-8x^3+2x^5)8x^3(1+x^4) ] / (1+x^4)^4

    =[ -2(1+x^4)(1+12x^2-5x^4)+16x^4(1+4x^2-x^4) ] /(1+x^4)^3

    dead end
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 17, 2009 #2

    lanedance

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    to get started try plotting 2-x^2, (1+x^4) and 1/(1+x^4), this should give you a feel for the general shape of the function

    noticing that the denominator is always > 0 and both the numerator & denominator are symmetric should also help
     
  4. Apr 17, 2009 #3
    i need to calculate the exact points of inflections
     
  5. Apr 17, 2009 #4

    lanedance

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    hmmm... how about considering a substitution u = x^2 then

    this should hopefully lead to a simpler calculation of the zeroes of the derivatives w.r.t. x

    then if ' denotes dervative w.r.t. x, using the chain rule

    [tex] y' = \frac{dy}{dx} = \frac{d}{dx} y(u(x)) = \frac{d y(u)}{du} \frac{du(x)}{dx} = \frac{dy}{du} u'[/tex]

    and so on for the next one, where you'll need the product rule as well
     
    Last edited: Apr 17, 2009
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