for the function on the interval of

I don't need to find the inflection values. Just the number of inflections points for the graph over the interval.

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- May 4th 2009, 09:06 PMswatpup32How do you determine the number of inflection points
for the function on the interval of

I don't need to find the inflection values. Just the number of inflections points for the graph over the interval. - May 4th 2009, 10:09 PMredsoxfan325
Solve on

Let (this extends the domain to )

.

Clearly, a solution is . There are eight others, but I only know that because I graphed the equation and counted them. I honestly don't know how to solve for the other roots analytically.

But there are 9 zeroes and thus 9 inflection points. - May 5th 2009, 06:27 AMKaitosan
Not neccessarily true. An inflection point is a point in which (1.) the slope of f' is zero or undefined vertically and (2.) there is a change of the sign of the slope of f' from - to + or from + to - . A zero can exist without the slope sign changing, therefore it'd not be an inflection point.

- May 5th 2009, 07:42 AMredsoxfan325
Ok, but this is a continuous function defined on a compact set, so it can't be unbounded, and whenever it changes from positive to negative there will be a corresponding zero by the IVT.

- May 5th 2009, 02:09 PMTheEmptySet
I have an indeterminate form ie this is not well defined, and I checked if the limits go to zero and they do not.

The zero factor principle does not apply becuase

Thanks for the graph skeeter.

So the 2nd derivative is

This can be factored as follows

The 2nd factor only has x=0 as a solution so the solutions are

Note that in the factorization that for all

Since changes sign at each of it zero all of the above zero's are critical points - May 5th 2009, 02:31 PMskeeter
graph of the 2nd derivative ...