Points of inflection on a function of 9 degree

I am recapping my whole A-Level from the beginning via an online course.

I have learned something that I previously overlooked regarding stationary points, or perhaps I forgot it.

The most obvious is that turning points are clearly stationary points but I recently learned that some inflection points can also be considered stationary.

I’m told that inflection stationary points count as two points due to there being two hidden turning points or something.

So, the max turning points on a graph with degree 9, would be 8.

Max inflection stationary points would be 4?

As I totally overlooked these inflections, I’m not 100% certain.

The question is asking for the number of stationary inflexion points (inflexion, NOT inflection. Are you taking an American course?)

Try to think about the maths involved - in order to be a point of inflexion, the second derivative of a point must be zero. Essentially, you reduce the order of your original function to give you a new one whose roots give you the points of inflexion.
Therefore, the answer would be 7.

A stationary inflexion point requires the first derivative to also be zero, surely?

My take (@mqb2766 - slightly different from yours)

If f is your original poly, and f' its derivative (so degree 8), then for $a_i$ to be a stationary inflexion point we must have $f'(a_i) = f''(a_i) = 0$ and so $(x-a_i)^2$ divides f'.

So you can have at most 4 (distinct) stationary inflexion points.

e.g. $\int x^2(x-1)^2(x-2)^2(x-3)^2 dx$ (cut paste version: \int x^2(x-1)^2(x-2)^2(x-3)^2 dx)

which you can get Wolfram to plot directly without needing to do the integration (fortunately!).

Edit: Debatable if this is "beyond A-level" but it's certainly more a STEP-type question than an A-level one (at least if you want the answer to include justification).

I am recapping my whole A-Level from the beginning via an online course.

I have learned something that I previously overlooked regarding stationary points, or perhaps I forgot it.

The most obvious is that turning points are clearly stationary points but I recently learned that some inflection points can also be considered stationary.

I’m told that inflection stationary points count as two points due to there being two hidden turning points or something.

So, the max turning points on a graph with degree 9, would be 8.

Max inflection stationary points would be 4?

As I totally overlooked these inflections, I’m not 100% certain.

The question is asking for the number of stationary inflexion points (inflexion, NOT inflection. Are you taking an American course?)

Try to think about the maths involved - in order to be a point of inflexion, the second derivative of a point must be zero. Essentially, you reduce the order of your original function to give you a new one whose roots give you the points of inflexion.
Therefore, the answer would be 7.

Was in the middle of making tea and was coming back to say somehting like this. So zero first and second derivative at each point (assume a sign change) (so two constraints for each statoinary point of inflection) and the vallue of the constant in the polynomial is irrelevant so (n-1)/2 which is 4 as you say.

A stationary inflexion point requires the first derivative to also be zero, surely?

My take (@mqb2766 - slightly different from yours)

If f is your original poly, and f' its derivative (so degree 8), then for $a_i$ to be a stationary inflexion point we must have $f'(a_i) = f''(a_i) = 0$ and so $(x-a_i)^2$ divides f'.

So you can have at most 4 (distinct) stationary inflexion points.

e.g. $\int x^2(x-1)^2(x-2)^2(x-3)^2 dx$ (cut paste version: \int x^2(x-1)^2(x-2)^2(x-3)^2 dx)

which you can get Wolfram to plot directly without needing to do the integration (fortunately!).

Edit: Debatable if this is "beyond A-level" but it's certainly more a STEP-type question than an A-level one (at least if you want the answer to include justification).

Okay, great.

So, I was right to say four, due to stationary inflection points having a gradient=0 hence, the concept of it being 0.

The course is okay but it’s not great. Their answer suggested 8 but their answer referred to stationary points as a whole ignoring the fact that the question specifically asked for inflection stationary points.

Thank you for confirmation. I have reported their error

Okay, great.

So, I was right to say four, due to stationary inflection points having a gradient=0 hence, the concept of it being 0.

The course is okay but it’s not great. Their answer suggested 8 but their answer referred to stationary points as a whole ignoring the fact that the question specifically asked for inflection stationary points.

Thank you for confirmation. I have reported their error

Yeah... Not only were they wrong, they couldn't even spell inflexion right... Doesn't inspire confidence...

I think the quesiton is wrong. Asking about the number of stationary points would be more a level.

Do you mean, it was likely meant to be the maximum number of stationary points? Do you think inflexion points are not likely to be A-Level material then? It could explain why I don’t recall it from the book I studied

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 $x=x_0$ to be a sationary point of inflection, then $(x-x_0)$ must be a factor of $f'$ and $f''$.

So if $f(x) = P_n(x)$ (poly of n degree) then

$f'(x) = P_{n-1}(x) = (x-x_0) Q(x)$

and we also need

$f''(x) = P_{n-2}(x) = Q(x) + (x-x_0)Q'(x) = (x-x_0)R(x)$

This implies that $Q(x)$ must have factor $(x-x_0)$.

Meaning that $(x-x_0)^2$ is a factor of $f'$.

All this says that a single stationary point of inflection $x_0$ must have its corresponding factor $(x-x_0)$ 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

$f(x) = 70 x^9 - 315 x^8 + 180 x^7 + 840 x^6 - 882 x^5 - 630 x^4 + 840 x^3$

https://www.desmos.com/calculator/me2brwvwc2


edit: others beat me to it after i left my reply for 30 mins before posting!

I think the quesiton was just asking about the number of stationary points and the extra "inflection" was a typo. Fairly obviously thats just the number of roots of the n-1 (derivative) polynomial so 8.

Inflexion is certainly an a level topic, but reasoning about it (stationary inflexion) in factorised polynomial form is probably beyond what would be expected.

Yeah... Not only were they wrong, they couldn't even spell inflexion right... Doesn't inspire confidence...

Do you have the instruction manual for using desmos lol. I can just about figure out how to add range values for x lol.

Seen this?

https://help.desmos.com/hc/en-us/articles/4406040715149-Getting-Started-Desmos-Graphing-Calculator

I just checked out of interest and AQA, Edexcel, OCR (A and MEI) and WJEC all spell it "inflection" in their A level specifications, so I'm assuming the course has just copied from there. Interesting to see the common usage seems to be shifting away from the British spelling, at least within school courses.

(Of course, the question being wrong is a whole other issue!)