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:

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