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Show y = (1+x)/(1+x^2) has three inflection points

  1. May 3, 2012 #1
    1. The problem statement, all variables and given/known data

    We are given the curve y = (1+x)/(1+x^2)


    2. Relevant equations

    y' and y''

    3. The attempt at a solution

    I know the inflection points of y are the local minimum and maximum of y'; this can also be restated as the critical points of y''. My attempt is to find the zero's of y'' and show there are only three critical points that satisfy as being the local extremes of y'. However, I end up getting a huge fifth degree polynomial for y''. This thing is a monster to solve; it does not seem to simplify either.

    y'' = (2x^5+6x^4-4x^3+5x^2-8x-3)/(x^4+2x^2+1)^2

    Perhaps I can show y'' has three zero's by IVT but there is a second part that requires the actual x values that make those zero's.

    Anyone have any other ideas?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 3, 2012 #2
    I can't be sure, but it looks like you may have made an algebraic mistake when simplifying the second derivative.

    Edit: I am sure. Your second derivative is incorrect.
     
  4. May 5, 2012 #3
    You must excuse me, it seems I have been having quite a few brain malfunctioning moments these past couple of days.

    The proper second derivative to y is y'' = (2(1+x^2)(x^3+3x^2-3x-1))/(1+x^2)^4

    I actually figured it out. It was just a matter of applying algebra. Since (1+x^2) > 0 it is not a critical value. But x^3+3x^2-3x-1 factors out as (x-1)(x^2+4x+1). So x=1 is a critical value for y'. And x=-2+/-sqrt(3) by the quadratic formula. So at those values for x we get the local extremes of y'. Thus these points represent the inflection points of y!

    Thank you for double checking my work. I appreciate it!
     
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