For any generic polynomial function, it seems that a point of inflection is always half-way between the nearest minimum and maximum (if they exist).
Is there a way to prove that midpoints of adjacent minima and maxima are POIs? Or might I be missing a simple counterexample?
You are probably only trying cubics. If you have the function $f(x)=ax^3+bx^2+cx+d$ the local extrema come at the roots of $3ax^2+2bx+c=0$, which are $\frac {-b\pm \sqrt{b^2-3ac}}{3a}$. The midpoint of these is $\frac {-b}{3a}$. The inflection point is at the root of $6ax+2b=0$, which is $\frac {-b}{3a}$
If you try higher degree polynomials it fails. Take $f(x)=x^4-2x^2+1$ The extrema are at roots of $4x^3-4x=0$, so at $0,\pm 1$. The inflection points are at the roots of $12x^2-4=0$, which are $\pm \frac 1{\sqrt 3}$, not $\pm \frac12$
In general, it is not.
$$x^4-2x^2$$ has a maximum at $x=0$ and a minimum $x=1$ and an inflection point at $x=\dfrac1{\sqrt3}.$
Having a de-centered inflection is tantamount to having a de-centered extremum between two roots in the first derivative, which is also quite possible.
Consider the function
$$y'(x)=(x-a)(x-b)(x-d).$$
If has an extremum where
$$(x-a)(x-b)+(x-a)(x-d)+(x-b)(x-d)=0.$$
If we want this to occur at $x=c$, we can set
$$d=c+\frac{(c-a)(c-b)}{2c-a-b}.$$
So the antiderivative of
$$(x-a)(x-b)\left(x-c-\frac{(c-a)(c-b)}{2c-a-b}\right)$$
has two extrema and an inflection point wherever you want (except in the middle !)
This is true for cubic polynomials $p$: If $p$ has local extrema, then the quadratic function $p'$ has roots and these roots are symmetric about $x = r$, where $r$ is the unique root of $p''$ (so that $(r, p(r))$ is an inflection point). Of course, the claim is vacuous for linear and quadratic polynomials.
On the other hand, the claim is in general false for polynomials of degree $> 3$: If a quartic polynomial has more than one extremum, it has three, and so $p'$ has three real roots. By translating, dilating, and reflecting (all of which preserve midpoints) we may as well assume that $$p'(x) = x (x - r) (x - 1),$$ so that the extrema of $p$ occur at $0, r, 1$, where $0 < r < 1$. At the midpoints $\frac{1}{2} r$, $\frac{1}{2} (r + 1)$ of consecutive roots, we compute that $$p''\left(\tfrac{1}{2}r\right) = -\tfrac{1}{4} r^2 \qquad \textrm{and} \qquad p''\left(\tfrac{1}{2}(r + 1)\right) = -\tfrac{1}{4} (1 - r)^2,$$ which by hypothesis are never zero, so these are not inflection points. Thus, we can conclude for quartic polynomials that the claim never holds (at least in the nonvacuous case, that is, when the polynomial has multiple extrema).
The simple example $p(x) = x^5 - x$ shows that the claim can hold for specific quintic polynomials, but it fails in general in the degree $5$ case.
I think your intuition comes from a cubic polynomial, which when you look at its derivative, the zeros of the derivative correspond to the two local extrema of the cubic, and the inflexion point corresponds to the extremum of the derivative, a quadratic, which is indeed the midpoint between the two roots because a quadratic is symmetric in its extremum. However, polynomials of higher degree can skew this symmetry and prevent the inflexion point to lie in the middle. Examples have been given above.
Hope that helps,
It's easy to prove that this cannot be the case, by substituting $x=u^2$. Clearly your polynomial is still a polynomial, although now with only even powers of $u$. This substitution is non-linear, so a point $x_{mid} = (x_a+x_b)/2$ will not correspond to $u_{mid} = (u_a+u_b)/2$.
So if $x_{mid}$ is an inflection point, then $u_{mid}$ isn't, and vice versa. And that means there's at least one polynomial for which the assertion doesn't hold.
