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Topic: [ap-calculus] Double Derivatives
Replies: 1   Last Post: Oct 25, 2012 5:49 AM

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Lenore Horner

Posts: 74
Registered: 7/11/10
Re: [ap-calculus] Double Derivatives
Posted: Oct 25, 2012 5:49 AM
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If the first derivative isn't zero, then my comments about the second derivative aren't relevant.
(Which wasn't true for the original function either, but somehow I thought that was what was happening.)

On Oct 25, 2012, at 3:49 AM, Jon Stark wrote:

> 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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