The Mean Value Theorem

Posted on October 23, 2018 by Lin McMullin

Another application of the derivative is the Mean Value Theorem (MVT). This theorem is very important. One of its most important uses is in proving the Fundamental Theorem of Calculus (FTC), which comes a little later in the year.

See last Fridays post Foreshadowing the MVT for an a series of problems that will get your students ready for the MVT.

Here are some previous post on the MVT:

Fermat’s Penultimate Theorem A lemma for Rolle’s Theorem: Any function extreme value(s) on an open interval must occur where the derivative is zero or undefined.

Rolle’s Theorem A lemma for the MVT: On an interval if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b) and f(a) = f(b), there must exist a number c in the open interval (a, b) where f ‘(c) = 0.

Mean Value Theorem I Proof

Mean Value Theorem II Graphical Considerations

Darboux’s Theorem The Intermediate Value Theorem for derivatives.

Mean Tables

Revised from a post of October 31, 2017

Foreshadowing the MVT

Posted on October 19, 2018 by Lin McMullin

The Mean Value Theorem (MVT) is proved by writing the equation of a function giving the (directed) length of a segment from the given function to the line between the endpoints as you can see here. Since the function and the line intersect at the endpoints of the interval this function satisfies the hypotheses of Rolle’s theorem and so the MVT follows directly. This means that the derivative of the distance function is zero at the points guaranteed by the MVT. Therefore, these values must also be the location of the local extreme values (maximums and minimums) of the distance function on the open interval. *

Here is an exploration with three similar examples that use this idea to foreshadow the MVT. You, of course, can use your own favorite function. Any differentiable function may be used, in which case a CAS calculator may be helpful. Answers are at the end.

First example:

Consider the function $f(x)=x+2\sin (\pi x)$ defined on the closed interval [–1,3]

Write the equation of the line through the endpoints of the function.
Write an expression for h(x) the vertical distance between f(x) and the line found in part 1.
Find the x-coordinates of the local extreme values of h(x) on the open interval (–1,3).
Find the slope of f(x) at the values found in part 3.
Compare your answer to part 4 with the slope of the line. Is this a coincidence?

Second example: slightly more difficult than the first.

Consider the function $f\left( x \right)=1+x+2\cos \left( x \right)$ defined on the closed interval $\left[ {\tfrac{\pi }{2},\tfrac{{9\pi }}{2}} \right]$

Write the equation of the line through the endpoints of the function.
Write an expression for h(x) the vertical distance between f(x) and the line found in part 1.
Find the x-coordinates of the local extreme values of h(x) on the open interval $\left( {\tfrac{\pi }{2},\tfrac{{9\pi }}{2}} \right)$
Find the slope of f(x) at the values found in part 3.
Compare your answer to part 4 with the slope of the line. Is this a coincidence?

Third example: In case you think I cooked the numbers

Consider the function $\displaystyle f(x)={{x}^{3}}$ defined on the closed interval $\displaystyle [-4.5]$

Write the equation of the line thru the endpoints of the function.
Write an expression for h(x) the vertical distance between f(x) and the line found in part 1.
Find the x-coordinates of the local extreme values of h(x) on the open interval $\displaystyle (-4,5)$
Find the slope of f(x) at the values found in part 3.
Compare your answer to part 4 with the slope of the line. Is this a coincidence?

Answers

First example:

y = x
$\displaystyle h(x)=f(x)-y(x)=\left( {x+2\sin (\pi x)} \right)-\left( x \right)=2\sin (\pi x)$
${h}'\left( x \right)=2\pi \cos \left( {\pi x} \right)=0$ when x = –1/2, ½, 3/2 and 5/2
$\displaystyle {f}'\left( x \right)=1+2\pi \cos \left( {\pi x} \right)$ , the slope = 1 at all four points
They are the same. Not a coincidence.

Second example:

The endpoints are $\left( {\tfrac{\pi }{2},1+\tfrac{\pi }{2}} \right)$ and $\left( {\tfrac{{9\pi }}{2},1+\tfrac{{9\pi }}{2}} \right)$ ; the line is $y=x+1$
$h\left( x \right)=f\left( x \right)-y\left( x \right)=\left( {1+x+2\cos (x)} \right)-\left( {x+1} \right)=2\cos \left( x \right)$
${h}'\left( x \right)-2\sin (x)=0$ when $x=\pi ,2\pi ,3\pi ,\text{ and }4\pi$
${f}'\left( x \right)=1-2\sin \left( x \right)$ , at the points above the slope is 1.
They are the same. Not a coincidence.

Third example:

The endpoints are (-4, -64) and (5, 125), the line is $\displaystyle y=125+21(x-5)=21x+20$
$\displaystyle h(x)={{x}^{3}}-21x-20$
$\displaystyle {h}'(x)=3{{x}^{2}}-21=0$ when $\displaystyle x=\sqrt{7},-\sqrt{7}$
$\displaystyle {f}'\left( {\pm \sqrt{7}} \right)=3{{\left( {\pm \sqrt{7}} \right)}^{2}}=21$
They are the same. Not a coincidence.

* It is possible that the derivative is zero and the point is not an extreme value. This is similar to the situation with a point of inflection when the first derivative is zero but does not change sign.

Teaching Concavity

Posted on October 17, 2018 by Lin McMullin

As you’ve probably noticed, different authors use different definitions based on how they plan to present topics later in their texts. So, the same concept seems to have different definitions. A definition in one book is a theorem in another. Students should be aware of this; looking at different definitions and related theorems about the same idea helps their mathematical education.

Here is a thought on exploring how concepts may be defined and learning about concavity are the same time.

Start by drawing the 4 shapes of (non-linear) graphs (See Concepts Related to Graphs and The Shapes of a Graph.)

Increasing, concave up
Increasing concave down
Decreasing concave up
Decreasing concave down

Look at the sine or cosine graphs which show all four and the tangent graph that shows only two. Look at other functions as well and identify which parts (intervals) exhibit each shape.

Next, challenge the students, in groups, individually, in class, or for homework to find analytic ways to say what concavity is or to identify it from equations. If they need hints suggest they look at derivatives (first and second) and tangent lines. Don’t limit them – explain that there are other ways. They should try to find several ways.

Hopefully, they will come up with some of these. (I have purposely listed these in a rough preliminary form. That’s how math is done – get an idea and then develop and formalize it)

A function is concave up (down) when:

The slope (derivative) is increasing (decreasing)
The second derivative is positive (negative).
The tangent line lies below (above) the graph
All (any, every) segments joining points in the interval lie above (below) the graph.
Others ???

Finally, clean these up. Help the students:

State them as “if, then” or “if, and only if” statements giving the hypotheses for each.
Consider how one can be used to imply the others – in either direction.
Determine which implies the inclusion of endpoints (1 and maybe 3).
Discuss which the class thinks would make the best definition, and which should become theorems.
Taking whichever definition your book uses, show how the others can be proved.

The point here is the thinking, forming ideas, doing the mathematics; the understanding of concavity will follow.

L’Hospital’s Rule

Posted on October 16, 2018 by Lin McMullin

Another application of the derivative

L’Hospital’s Rule

Locally Linear L’Hospital’s Demonstration of the proof

L’Hospital Rules the Graph

Good Question An AP Exam question that can be used to delve deeper into L’Hospital’s Rule (2008 AB 6)

Guillaume de l’Hospital
1661 – 1704

Revised from a post of November 7, 2017

There will be two extra posts this week! Check tomorrow for some suggestions on “Teaching Concavity” and on Friday for “Foreshadowing the MVT.”

I made a major update to last Friday’s post On Scaling. It includes a suggestion from a reader of this blog with a Desmos graph that will calculate the Kennedy scale scores for you.

On Scaling

Posted on October 12, 2018 by Lin McMullin

Why “scaling” is necessary

No teacher can make two tests on the same topics equal in difficulty. No two teachers, even if they collaborate, can make two tests on the same topic equal in difficulty. No two teachers in different schools, districts, or states can make two tests on the same subject equal in difficulty. Even professional testing companies, such as the Educational Testing Service (ETS) that writes the AP exams, cannot write two tests on the same courses of equal difficulty.

Scaling is needed to account for the difference in difficulty. Scaling attempts to make the scores on different forms of a test indicate that a student writing the test has the same amount of knowledge as another student with a similar score.

The ETS does this by pre-testing its items on college students and including several questions from previous years to help judge the difficulty from year to year. They do a great deal of statistics on each item each year. But they do not pretend that this year’s test is the same difficulty as last year’s test. After their computations and consultations with colleges are done, they scale the test. Their goal is to make the score indicate the same amount of knowledge from test to test and year to year.

A teacher cannot do that in his or her class. They don’t have the resources or the time. Yet, there are ways to even out the difficulty of your classroom tests and quizzes. .

Some poor ways to scale

In what follows, P will represent the percentage of the total points available on a test that a student earns, and S will equal the score the student is given for that percentage.

Percentage scaling (S = P): For many years I, and I expect most teachers, simply let S = P. But sometimes the scores were kind of low: the test was too hard, or the students didn’t do well (or maybe the teacher didn’t do well). What to do? Among the usual solutions are (1) give a make-up test, (2) let the students make corrections to earn back some of the points, (3) scale the test by raising all the grades arbitrarily, or (4) make sure the next test is “easy.” I’ve tried all of them.

Doesn’t make too much sense, does it?

Categories: For quite a few years, I listed the percentages from highest to lowest and looked for natural breaks to separate the scores into 90, 80, 70, etc. Intermediate scores were spread between the cut points. If you don’t need a number to put on the report cards, the categories become A, B, C, etc. with perhaps a “+” or a “–“ attached.

Comic Interlude – the “Square Root Scale”

The “square root scale” is $S=10\sqrt{P}$ . So, a 36 is scaled to a 60, an 81 to a 90, and a 70 to an 84. What this accomplishes is to raise everyone score for no reason other than to raise the score. See the graph below.

The Square Root Curve, $S=10\sqrt{P}$ ., in red and the Percentage Curve, S = P, in blue

Compared to the percentage grade, the low scores get raised more than the higher scores. Everyone wins big time, but what does it tell you? I can see no justification for this, except maybe the “complicated” algebra involved fools the students, administrators, and parents into thinking that something really scientific is going on. It’s not.

(Since this is a calculus blog, there is a calculus exercise in the appendix below that analyzes this scheme.)

A Better Choice for Scaling – the Kennedy Scale

While no method is perfect, this method suggested in Assessing True Academic Success by Dan Kennedy [1] is a reasonable and easy one. The entire article is worth reading every year and discusses a lot about assessment, besides just scaling.

He writes of his method, “Mathematically, the effect of scaling is to adjust the mean, a primary goal, and reduce the standard deviation, a secondary effect that helps me keep the entire class engaged.” “[Teachers] can challenge [their] students to do just about anything, then see how far they can go. …[Students] are freed from the burden of getting a certain percent right, so they can concentrate on doing as much as they can as well as they can.”

I used this method for BC Calculus and 8^th grade Algebra 1 in the year I came out of retirement and was happy with the results.

Here’s how the method works. First, determine the class mean you desire. Kennedy suggests a class average of 82 for regular classes, 85 for electives, and 90 for advanced. These are based on his school wide empirical (historical) data. You may use your own data or just what you think is reasonable.

Using two data points (class mean, desired mean) and (highest score, 99). (The 99 could be adjusted as you see fit.} Write the equation of the line through these points (P, S) expressing S as a function of P. Use this function to scale the test.

This TI-8x program, from the same article, will easily compute the scores for you. (There is a typo in the fourth line; it should read 0->Ymin:126->Ymax.)

Update Excel Spread Sheet for Kennedy Scale.

At the suggestion of a reader, here is an Excel spreadsheet for you may download for the Kennedy Curve. Enter the four values at the top left and the scores w ill be calculated.

Updated December 8, 2020

Update Desmos Program for Kennedy Score

Dan Anderson sent a comment (see below) with a link to a Desmos graph he made that will calculate the Kennedy scale for your tests. You can access the graph here. Once you’ve opened it, save it to your Desmos files.

It works like this: enter the 4 numbers in the left column AverageRawScore, DesiredAverage, MaxRawScore, and DesiredMax as they apply to your test. The scaled scores will appear in the table in the lower left.

To scale your exam, delete everything in the x₁ column and enter your scores (in any order, with duplicates). The scaled scores appear in the second column of the table and the pairs are graphed.

The two highlighted points are (AverageRawScore, DesiredAverage) and (MaxRawScore, DesiredMax). These may be dragged to see the effect of changing them.

A final caution: If the AverageRawScore is greater then or equal to the DesiredAverage (or even close), then some scores may be scaled down. You probably want to avoid this (although, it is consistent with the idea).

Updated October 13, 2018

Update October 19, 2020

Remember, by scaling, you are not giving away free points; you are trying to account for the difference in difficulty from one test to the next.

Scaling Different Versions of the Same Test How to adapt the Kennedy method when using different versions of the same test in your class.

Update August 24, 2021

Appendix: An analysis of the Square Root Curve – A Calculus Exercise

For the function $S=10\sqrt{P}$ .

Determine the percentage score(s), P, which receives the least points using this method. Justify your answer.
Determine the percentage score(s), P, which receives the most points using this method. Justify your answer.
At the value found in 2, what is the slope of the line tangent to the graph of $S=10\sqrt{P}$ ?
Compare your answer for 3 to the slope of S = P. Why must this be so? Is it related to the MVT?

Solution

Since the Square Root curve lies above the percentage curve all the values receive some increase except the end points (P = 0 and P = 100) which receive no increase.
Let I = the increase in the score, then

$I=10\sqrt{P}-P$

$\displaystyle \frac{{dI}}{{dP}}=\frac{{10}}{{2\sqrt{P}}}-1$

$\displaystyle \frac{{10}}{{2\sqrt{P}}}-1=0,\text{ when }P=25$

This is the maximum since it is the only place where P’ changes from positive to negative. At P = 25 the score is raised by 25 points to a 50.

3. $\displaystyle \frac{{dS}}{{dP}}=\frac{{10}}{{2\sqrt{P}}}$ . At P = 25, dS/dP = 1. The slope of the tangent line is 1.

4. At P = 25 the slope of the tangent line to the square root scale is 1: the tangent is parallel to the percentage graph. The square root scale to the left of P =25 is raising faster then S = P therefore its slope is greater. After P = 25 the slope of the square root scale decreases and drops faster than the slope of S = P. P = 25 is the place where the slope changes from steeper to less steep and thus where the slopes are equal. This is the farthest point vertically above the percentage graph. This is also the point guaranteed by the MVT on the interval [0, 100].

[1] Assessing True Academic Success by Dan Kennedy, The Mathematics Teacher, September 1999, page 462 – 466).

Graphing – an Application of the Derivative.

Posted on October 9, 2018 by Lin McMullin

Graphing and the analysis of graphs given (1) the equation, (2) a graph, or (3) a table of values of a function and its derivative(s) makes up the largest group of questions on the AP exams. Most of the other applications of the derivative depend on understanding the relationship between a function and its derivatives.