# Mean Tables

The AP calculus exams always seem to have a multiple-choice table question in which the stem describes function in words and students are asked which of 5, now 4, tables could be a table of values for the function.  Could be because you can never be sure without other information what happens between values in the table. So, the way to solve the problem is to eliminate choices that are at odds with the description.

The question style nicely makes students relate a verbal description with the numerical information in the tables. This uses two parts of the Rule of Four.

In 2003, question AB 90 told students that a function f had a positive first derivative and a negative second derivative on the closed interval [2, 5]. There were five tables to choose from.

There is a fairly quick way to solve the problem, but I want to go a little slower and discuss the theorems that apply.

First, since the function has first and second derivatives on the interval, the function and its first derivative are continuous on the interval. This is important since, if they were not continuous, there would be no way to solve the problem.

Next, since the first derivative is positive, the function must be increasing. This allowed students to quickly eliminate three choices where the function was obviously decreasing. The remaining tables showed increasing values and thus could not be eliminated based on the first derivative.

The two remaining tables were Notice that in table A the values are increasing at an increasing rate, and in table B the values are increasing at a decreasing rate. Thus, table B is the correct choice. By the end of the year that kind of reasoning is enough for students to determine the correct answer.

Students could also draw a quick graph and see that table A was concave up and B was concave down. This will give them the correct answer, but technically it is a wrong approach since, once again, there is no way to know what happens between the values; we should not just connect the points and draw a conclusion.

The correct reasoning is based on the Mean Value Theorem (MVT).

In table A, by the MVT there must be a number c1 between 2 and 3 where ${f}'\left( {{c}_{1}} \right)=2$ the slope between the points (2,7) and (3, 9). Also, there must be a number c2 between 3 and 4 where ${f}'\left( {{c}_{2}} \right)=3$ the slope between (3, 9) and (4, 12), and likewise a number c3 between 4 and 5 where ${f}'\left( {{c}_{3}} \right)=4$.

Then applying the MVT to these values of ${f}'\left( x \right)$, there must be a number, say d1 between c1 and c2 where ${{f}'}'\left( {{d}_{1}} \right)=\frac{3-2}{{{c}_{2}}-{{c}_{1}}}>0$. Since the second derivative should be negative everywhere, table A is eliminated making the remaining table B the correct choice.

If we do a similar analysis of Table B, we find that the MVT values for the second derivative are all negative. However, we cannot be sure this is true for all values of f in table B, since we can never be sure what happens between the values in a table. But table B is the only one that could be the one described since the others clearly are not.

In this post we saw how the MVT can be used in a numerical setting. I discussed the MVT in an analytic setting on September 28, 2012 and graphically on October 1, 2012.