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I'm having trouble finding the Points of Inflection for this function

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data
    It asks for any Points of Inflection
    The second derivative is given: y'' = (2x^5 + 4x^3 - 6x) / (x^2-1)^4

    2. Relevant equations
    The original function was x / (X^2 -1)


    3. The attempt at a solution
    I have set it equal to zero, and I get 2x^5 + 4x^3 - 6x = 0

    I then simplified, taking out the 2, as well as factoring out an x

    leaving me with

    x(x^2 + 3)(x^2 -1)....

    I get a weird answer though...x^2=-3 does not exist....and x = =+-1, are vertical asymptotes in the original function

    help?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Dec 4, 2011
  2. jcsd
  3. Dec 4, 2011 #2
    help....anyone ;_;
     
  4. Dec 4, 2011 #3

    SammyS

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    Hi agv567. Welcome to PF.

    Sometimes it takes a while for those of us who volunteer on the site to get around to answering questions. That's why there is a rule which forbids "bumping" your post until it has been here 24 hours.

    I'll look at your question & get back to you soon.
     
  5. Dec 4, 2011 #4

    eumyang

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    This can be simplified further. Also, are you sure the numerator isn't
    2x5 + 4x3 - 6x?

    There's one more factor to check, the one in bold above.
     
  6. Dec 4, 2011 #5
    EDIT: I'm an idiot

    And POI can't occure at asymptotes, right? so -1 or 1 are not POI?

    and x = 0...so the only POI occurs at (0,0), right? I checked with my calculator and it seems so.


    But what about the x^2 = -3 equation? Since it does not exist...I do nothing with it?
     
    Last edited: Dec 4, 2011
  7. Dec 4, 2011 #6

    SammyS

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    You have a typo in that second derivative. It's y'' = (2x5 + 4x3 - 6x) / (x^2-1)^4

    That should fix-up your problem!

    Also, please use proper parentheses. Otherwise what you would have is equivalent to: [itex]\displaystyle 2x^5 + 4x^3 -\frac{6x}{(x^2-1)^4}[/itex]
     
  8. Dec 4, 2011 #7
    Thank you so much. I see my mistake now and how it confused everyone haha. I still don't understand what to do with the x^2 = -3, though.
     
  9. Dec 4, 2011 #8

    SammyS

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    x2=-3 doesn't have any real number solutions !
     
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