# Probability that a cubic has two turning points

Most cubic polynomials with real coefficients have two turning points, a local maximum and a local minimum. But how do you quantify “most”?

Here’s how one author did it [1]. Start with the cubic polynomial

x³ + ax² + bx + c

Since multiplying a polynomial by a nonzero constant doesn’t change how many turning points it has, we might as well assume the leading coefficient is 1.

In his paper, Robert Fakler assumes a, b, and c are chosen randomly from an interval [−k, k]. He shows that for k ≤ 3, the probability that the polynomial has two turning points is

p = (9 + k)/18.

For k ≥ 3, the probability is

p = 1 − √(3/k) / 3

and so as k → ∞,  p → 1.

[1] Robert Fakler. Do Most Cubic Graphs Have Two Turning Points? The College Mathematics Journal, Vol. 30, No. 5 (Nov., 1999), pp. 367-369

## 7 thoughts on “Probability that a cubic has two turning points”

1. By randomly chosen in [-k, k], I assume you mean uniformly?

2. They should be equal at 3, but aren’t. Typo?

3. Nathan Hannon

The limiting result is not surprising when you consider what the limit is actually describing: a cubic that is very nearly a quadratic. If you assume that the x^3 term can be ignored except for very large x, you have something that looks almost like a parabola with one local extremum, and then a second local extremum as the tiny x^3 term finally takes over for either large negative or large positive x.

This also raises the question of whether some other way to choose a random cubic might be more natural. I would argue that having the lower order terms be much larger than the x^3 term is not what most people would have in mind.

4. Marco T

One can also use the transformation x = (x’-a/3) which is just a horizontal translation (and thus doesn’t change the number of turning points) and gets rid of the x^2 term. Similarly, we can get rid of c since doing so is just a vertical translation. We end up with x^3+bx. Studying the number of turning points of that cubic is simple. It has two if b0. This means that for b in [-k,k] the chance that it has two turning points is 1/2 and that it has none is 1/2.

5. There are many ways to approach this, but in my opinion, the change of variables eliminating the quadratic term departs from common experience. Most cubics that you run into in the wild have two turning points, though there’s a good chance that a change of variables may turn them into a cubic without a turning point.

6. @BobC: I believe they both reduce to p = 2/3 when k = 3.

7. @Kenneth: Yes, that’s the default assumption.