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Topic: [ap-calculus] Double Derivatives
Replies: 1   Last Post: Oct 25, 2012 3:49 AM

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Jon Stark

Posts: 914
Registered: 6/10/06
RE: [ap-calculus] Double Derivatives
Posted: Oct 25, 2012 3:49 AM
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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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