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Values of concavity. Please help!

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data

    For what intervals is g(x) = 1/(x^2+1) concave down?


    2. Relevant equations

    Quotient Rule

    3. The attempt at a solution

    OK, so I have found the first and second derivative, but I am confused as to how I should use them to find where the function is concave down. Do I set the second derivative equal to zero? Use the quadratic formula? I know the answer from the back of the book, but my attemps to get the same answer have failed. Please point me in the right direction! Thank you :)
     
  2. jcsd
  3. Nov 9, 2008 #2

    gabbagabbahey

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    Setting the second derivative equal to zero will give you the inflection points, but you also need to determine whether the second derivative is (+) or (-) on each side of these inflection points....what are you getting for your solution?
     
  4. Nov 9, 2008 #3
    Ok, so for the 1st derivative I have:

    -2x/(x^2+1)^2

    Second:

    -6x^2-2/ (x^2+1)^2

    Do I need to factor the denom? Also, you stated that I need to determine the values on either side of the inflection points, I should do that with my calculator, right? So, if I have this correct, determining the value (+ or -) of the function on both sides of the inflection points will tell if the original funtion is concave up or down?

    Thank you!
     
  5. Nov 9, 2008 #4

    gabbagabbahey

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    Close, I get the numerator as being +6x^2-2... and you don't need to factor the denominator; the expression will only be zero when the numerator is zero....what does that give you for you inflection points?
     
  6. Nov 9, 2008 #5
    WOW! You have been so much help!

    I got


    X= sqrt of (1/3)

    So the function is concave down for all X values between -< X < + of sqrt (1/3)

    I have been going crazy trying to figure this is out! Thank you for your time :)
     
  7. Nov 9, 2008 #6

    gabbagabbahey

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    No problem :smile:
     
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