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An Average Cubic

Consider the general cubic function f(x) = ax3 + bx2 + cx + d.

Two things you perhaps know:

  1. The general shape of this cubic, especially ideas of graphs where this cubic could have 1, 2, or 3 distinct real roots
  2. This cubic could have both local maximums and minimums.
Pepso Dent claims that she has created a new theorem, and wants it named after herself. Her claim: To find the location of local maximums or minimums of a cubic function, the x-coordinate is the average of two adjacent roots (assuming such exist).

Is Pepso correct with her claim?


Hint: Pick a cubic and graph it on a graphic calculator....test Pepso's conjecture. Repeat....does it seem true?

Now, what to do...maybe assume....


Solution Commentary: Start with a cubic function that has two roots...translate it vertically d-units to form another "graphically equivalent" cubic function where one of its roots is now at x = 0.

Assume that the second known root is at x = 2n (where n ≠ 0).

The cubic function is now f(x) = ax3 + bx2 + cx and also by taking the derivative, f'(x) = 3ax2 + 2bx + c.

Then, because of the nature of roots and min/max values, we have f(2n) = f'(n) = 0. But, this means that f(n) = 8an3 + 4bn2 + 2cn = 0 and f'(n) 3an2 + 2bn + c = 0.

Dividing the first equation by 2 and multiplying the second by n, we obtain 4an3 + 2bn2 + cn = 0 and 3an3 + 2bnn + cn = 0.

Solving this system, we discover that 0 = an3 or a = 0, as by assumption n ≠ 0.

But, this means that the original equation was a quadratic, not a cubic. That is, Pepso's claim has no merit....i.e. no cubic function has a min/max whose x-coordinate is the average of its two adjacent roots!

Reader T.R. (Bellingham) responds: "No, Pepso, but your sister Polly knows that for a cubic with three zeros, the x-coordinate of the point of inflection is their average."