A quick post here reflecting on a great solution presented by one of my AP Calculus BC kiddos. They had their first assessment on Friday. At this point, we are doing a quick and deep review of last year’s work. In our school, BC students have already completed AP Calculus AB and we spend this year digging deep and moving into the BC only topics. So, one of the questions I posed was in two parts:

- For what value(s) of
*x* is *x^10< x^6*
- For what value(s) of
*x* is *x^7 < x^3*

This, by the way, was a non-calculator assessment. I will be writing soon about my wavering on this issue. One of my students presented the following work:

^{x^10} < x^6 becomes x^4 < 1 and this is true whenever |x| < 1 (other than x = 0) so the intervals are (-1, 0) and (0, 1)
*x^*^{7} < x^^{3 }becomes x^^{7} – x^^{3} < 0 which becomes x^^{3} (x^^{4} – 1) < 0 since x^^{3} < 0 for all x < 0 we need x^^{4} – 1 to be positive. This is true when 1 < |x|. So, the overlap here is x < – 1. If x^^{4} – 1 is negative while x^^{3} is positive, then 0 < x < 1.

What knocked me out here was that he divided in one case (when it was safe with even powers) while he subtracted in the other case with odd powers. Now, I have not had the opportunity to ask him about this yet, but I have to imagine that this was not just luck. I think he had some instinct, and I want to gauge how conscious this instinct is, that there is a problem with dividing by x^^{3} which can, of course be negative. I get to labor on labor day, we have classes here on this holiday, so I will quiz him a bit about this. I’ll report back.

Thanks to Sam Shah for catching a mistake in my earlier version of this post!

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