## Tricky Calculus Problem

Here’s a cute problem I heard from Moor Xu:

Let $f(x) = x^6 - 9x^2 - 6x$. $f$ has exactly three critical points. Find the parabola that passes through these critical points.

I’ve been doing a daily problem of the day for my physics camp students. Questions and answers are posted here the day after we give them to the students. I haven’t been copying them over to this blog because many are repeats, and none are original. They might still be entertaining if you’re in to that sort of thing.

### 3 Responses to “Tricky Calculus Problem”

1. htam Says:

This is pretty cool. Any sane person would probably take the derivative and get $f'(x)=6x^5-18x-6$.

Then I was thinking, okay, I definitely can’t solve that. But oh, I can consider $f$ in the polynomial ring with one variable modulo the ideal generated by $f'(x)$, and pick a quadratic in the equivalence class.

Namely, if $c$ is a critical point, it would satisfy $f'(c)=6c^5-18c-6=0$, and so any multiple of that would be satisfied. In particular, $c^6-3c^2-c=0$. Using that to simplify $f(c)$ gives $-6c^2-5c$, and thus $-6x^2-5x$ should work.

I don’t know if I did the calculations right, but I think the idea should be.

2. John Burk Says:

What is physics camp?

3. Mark Eichenlaub Says:

http://epgy.stanford.edu/summer/