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

y = sin x for x <= 0

y = 1/3 x^3 +x for x >0.

I'm typing this late at night and may be missing something from fatigue, but

I think these bits patch so that y (0) = 0, y ' (0) = 1, y " (0) =0, and

there is neither an inflection nor an extreme value at the origin. It's

concave up to both sides. There are also no "adjacent points" (no interval

/ neighborhood) where the second derivative is zero as you thought would be

required.

The function is only twice differentiable at x = 0 because the patch

produces a discontinuity in the third derivative (-1 from one side and 2

from the other).

Lenore Horner

I don't think your example works. For x<0 the second derivative is negative

and for x>0 the second derivative is positive, thus this is a place where

the second derivative changes sign.

For continuous functions, in order for a point where the second derivative

is zero to be neither an extremum nor an inflection point, the second

derivative must be zero for at least one adjacent point on at least one

side.

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