```Date: Oct 25, 2012 3:49 AM
Author: Jon Stark
Subject: RE: [ap-calculus] Double Derivatives

NOTE: This ap-calculus EDG will be closing in the next few weeks. Please sign up for the new AP CalculusTeacher Community Forum at https://apcommunity.collegeboard.org/getting-startedand post messages there. ------------------------------------------------------------------------------------------------Perhaps this one works:y = sin x for x <= 0y = 1/3 x^3 +x for x >0.I'm typing this late at night and may be missing something from fatigue, butI think these bits patch so that y (0) = 0, y ' (0) = 1, y " (0) =0, andthere is neither an inflection nor an extreme value at the origin.  It'sconcave up to both sides.  There are also no "adjacent points" (no interval/ neighborhood) where the second derivative is zero as you thought would berequired.The function is only twice differentiable at x = 0 because the patchproduces a discontinuity in the third derivative (-1 from one side and 2from the other). Lenore HornerI don't think your example works.  For x<0 the second derivative is negativeand for x>0 the second derivative is positive, thus this is a place wherethe second derivative changes sign. For continuous functions, in order for a point  where the second derivativeis zero to be neither an extremum nor an inflection point, the secondderivative must be zero for at least one adjacent point on at least oneside.  ---To search the list archives for previous posts go tohttp://lyris.collegeboard.com/read/?forum=ap-calculus
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