I'm having difficulty understanding what you are trying to say.
To be clear, we say a function
f(x) has a stationary point when
f′(x)=0, and a point of inflection when
f′′(x)=0 (
and sign is swapped)
A stationary P.O.I is therefore one which satisfies
f′(x)=0 and f′′(x)=0.
For
f which are polynomials,
f′′ has lower degree than
f′ so it will have fewer solutions when put to 0. If
f has degree
n then
f′′=0 has at most
n−2 solutions. However, not all of them will be stationary points ...
We also need to take into account that if we want
x=x0 to be a sationary point of inflection, then
(x−x0) must be a factor of
f′ and
f′′.
So if
f(x)=Pn(x) (poly of n degree) then
f′(x)=Pn−1(x)=(x−x0)Q(x)and we also need
f′′(x)=Pn−2(x)=Q(x)+(x−x0)Q′(x)=(x−x0)R(x)This implies that
Q(x) must have factor
(x−x0).
Meaning that
(x−x0)2 is a factor of
f′.
All this says that a single stationary point of inflection
x0 must have its corresponding factor
(x−x0) occur at least once in factorised
f′′ and at least twice in factorised
f′.
Now take a polynomial of degree 9 ... its first derivative is degree 8 so it can have at most 4 different roots (repeated twice), meaning the second derivative of order 7 will contain all these four roots at least once.
One such example is the function
f(x)=70x9−315x8+180x7+840x6−882x5−630x4+840x3https://www.desmos.com/calculator/me2brwvwc2edit: others beat me to it after i left my reply for 30 mins before posting!