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.
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.
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.
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