Absolutely Extreme

The mantra of this year's AP calculus exams was "communication."

Many of the free-response questions required students to communicate to the reader not just how they found the answer, but also how they knew their answer was correct. "Bald" answers (a correct answer with nothing else) were not accepted. The directions "justify your answer", "show the work", "explain the meaning of" and like appear eleven times on the AB exam alone.

Students were asked to justify a maximum value of a function tree times on the AB exam. I'd like to discuss how to do that here. Students who found the correct maximum value or its location often, far too often, wrote poor justifications. And sometimes they missed earning the point by a single word.

The "maximum (minimum) value of a function" means the absolute maximum (minimum) value; not the local extreme value.

To justify an absolute extreme value it is necessary to consider the entire domain of the function, not just the local region near the critical point.

There are three main ways to justify an extreme value:

1. The Candidates' Test: Compute the value of the function at the critical point(s) and the endpoints of the domain and choose the maximum or minimum.
2. Discuss the sign of the first derivative on the entire domain. Say that the derivative is positive for all values on one side of the critical point and negative for all value on the other.
3. Discuss the sign of the second derivative on the entire domain.

The examples from the 2009 AB Exam.

2009 AB 2 / BC 2 (b) The Candidates' test works easily here. Setting the derivative equal to zero and finding the interior critical point earned the first 2 points in this question. The function had 2 critical points one of which was also an endpoint. So evaluating the function at 3 points and comparing the values easily justified the maximum and earned the last point.

Derivatives could also have been used.

2009 AB 3 (d) The domain in this questions was k ≥ 0 so, with no endpoint, the Candidates' Test cannot be used. So after earning 3 points (derivative = 0, solution k = 400, and value at the solution = $16,000) the justification point can be earned by writing that

• this is the only critical point and that the derivative changes sign from positive to negative here.
  
• or the student could claim that for 0 < k < 400 the derivative is negative and for k > 400 the derivative is positive.

This is where the mistakes were made. Omitting the word "only" or otherwise not saying there was only one critical point and the final point was not earned. Saying that the derivative was positive to the left of 400 and negative to the right was not enough. Saying the derivative changed sign here (without saying "only here") didn't do it either -- the sign could have changed elsewhere.

Finding the value of the derivative at particular points to the left and right is even worse. Now, of course, checking single points is how you determine what the sign of the derivative is, so the temptation is to report this as the justification. That doesn't do it. The justification must consider all the values on each side.

The second derivative was easy enough to compute in both these examples (although few used this approach). A justification here would need to cover the entire domain as well: "k = 400 is the only critical point and P''(k) = -4/8000 so k = 400 is the location of the maximum."

2009 AB 6 (c) gave the graph of the derivative of a function on a closed interval [-4, 4] and showed two critical points at x = -2 and x = 3ln(5/3).

The Candidates' Test could be used here, but the relative sizes of the 4 values of the function

were not clear, and this was a no calculator question. (See how this could be attempted without calculating the values at the end of the previous post in this blog.)

So the first derivative was the way to go here. Again students missed the justification point for not considering the entire domain. Having two critical points required some care (x = -2 was a critical point, but not an extreme value). Students made the same mistakes as in the previous example: not saying this was the only place the derivative changed from positive to negative, discussing the sign of the derivative at particular points to the left and right, not saying that f '(x) ≥0 for all x in the interval (-4, 3ln(5/3)) and so on.

So the lessons here are teach your students the meaning and use of "only." Practice writing justifications of absolute extremes that consider the entire domain and not just the local region near the critical point.