Monthly Archives: September 2013

Mean Numbers

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Here is a problem for you and your students. The numbers are mean until you get to the end when they all become very nice and well-behaved.  

You could give this to your students individually or as a group exploration. Give each person or group a different function and/or different intervals. Choose a function that has several (3 – 5) turning points in the interval. The function should be differentiable on the open interval and continuous on the closed interval.

It is intended that the work be done on a graphing calculator; you will need to carry 6 or 7 decimal places in their work.

Here is a typical problem. A link to the solution is given at the end.

Consider the function f\left( x \right)=\sin \left( x \right) on the closed interval [1, 12].

  1. Write an equation of a line, y\left( x \right),  between the endpoints of the function. Give the decimal value of its slope and give a graph of the function and the line.
  2. Write the equation of a function h\left( x \right) that gives the vertical distance between f\left( x \right) and y\left( x \right). Since f may be both above and below y this function may have positive and negative values.
  3. Graph h and find its critical values. What are these places with respect to the graph of h?
  4. Calculate the derivative of f at the critical values of h.
  5. Interpret your result graphically.

Click here for the solution.

Foreshadowing the Chain Rule

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I assigned another very easy but good problem this week. It was simple enough, but it gave a hint of things to come.

Use the Product Rule to find the derivative of {{\left( f\left( x \right) \right)}^{2}}.

Since we have not yet discussed the Chain Rule, the Product Rule was the only way to go.

\frac{d}{dx}{{\left( f \right)}^{2}}=\frac{d}{dx}\left( f\cdot f \right)=f\cdot {f}'+{f}'\cdot f=2f\cdot f'

 And likewise for higher powers:

\frac{d}{dx}{{f}^{3}}=\frac{d}{dx}\left( f\cdot f\cdot f \right)=f\cdot f\cdot {f}'+f\cdot {f}'\cdot f+{f}'\cdot f\cdot f=3{{f}^{2}}{f}'

If you just look at the answer, it is not clear where the {f}' comes from. But the result foreshadows the Chain Rule.

Then we used the new formula to differentiate a few expressions such as {{\left( 4x+7 \right)}^{2}} and {{\sin }^{2}}\left( x \right) and a few others.

Regarding the Chain Rule: I have always been a proponent of the Rule of Four, but I have never seen a good graphical explanation of the Chain Rule. (If someone has one, PLEASE send it to me – I’ll share it.)

Here is a rough verbal explanation that might help a little.

Consider the graph of y=\sin \left( x \right). On the interval [0,2\pi ] it goes through all its value in order once – from 0 to 1 to 0 to -1 and back to zero. Now consider the graph of y=\sin \left( 3x \right). On the interval \left[ 0,\tfrac{2\pi }{3} \right] it goes through all the same values in one-third of the time. Therefore, it must go through them three times as fast. So the rate of change of y=\sin \left( 3x \right) between 0 and \tfrac{2\pi }{3} must be three times the rate of change of y=\sin \left( x \right). So the rate of change of  must be 3\cos \left( 3x \right). Of course this rate of change is the slope and the derivative.

At Just the Right Time

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This is about a little problem that appeared at just the right time. My class had just learned about derivatives (limit definition) and the fact that the derivative is the slope of the tangent line. But none of that was really firm yet. I had assigned this problem for homework:1

Find f (3) and f ‘ (3), assuming that the tangent line to y = f (x) at a = 3 has equation y = 5x + 2

To solve the problem, you need to realize that the tangent line and the function intersect at the point where x = 3. So, f (3) was the same as the point on the line where x = 3. Therefore, f (3) = 5(3) + 2 = 17.

Then you have to realize that the derivative is the slope of the tangent line, and we know the tangent line’s equation and we can read the slope. So f ‘ (3) = 5

In my previous retired years, I wrote a number of questions for several editions of a popular AP Calculus exam review book.2 I found it easy to write difficult questions. But what I was after was good easy questions; they are more difficult to write. One type of good easy question is one that links two concepts in a way that is not immediately obvious such as the question above. I am always amazed at the good easy questions on the AP calculus exams. Of course, they do not look easy, but that’s what makes them good.

Now a month from now this question will not be a difficult at all – in fact it did not stump all of my students this week. Nevertheless, appearing at just the right time, I think it did help those it did stump, and that’s why I like it.

______________________

1From Calculus for AP(Early Transcendentals) by Jon Rogawski and Ray Cannon. © 2012, W. H. Freeman and Company, New York  Website p. 126 #20

2 These review books are published by D&S Marketing Systems, Inc. Website

Right Answer – Wrong Question

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About this time every year the AP Calculus Community discussion turns to the sentence, “A function is continuous on its domain.” Functions such as f\left( x \right)=\frac{1}{x} cause confusion – is it continuous or not?  The confusion comes, I think, from the way we introduce continuity to new calculus students.

We say – and I did say this myself just last week – that the graph of a continuous function can be drawn without taking your pencil off the paper. That idea helps students get a start on understanding what continuity means, but it is not quite correct.

The definition of continuity requires that for a function to be continuous at a point, the limit at that point equals the value there (and that both the limit and value be finite). The only way a function can have a value at a point is if the point is in the domain. So, the definition of continuity can be applied only at points in the domain. If the domain of the function is not all Real numbers, then the function cannot be continuous “everywhere;” rather it can only be continuous on its domain. (And, of course, there are many examples of functions that are not continuous at all points in their domains.)

So what do you say about a function like f\left( x \right)=\frac{1}{x}?

Its domain is all Real numbers x\ne 0. The function is continuous at all the points in its domain and so it is continuous on its domain.

But that statement does not tell the whole story. We asked the wrong question. We should ask where the function is not continuous. If we ask where this function is not continuous, the answer is that the function is not continuous at x = 0. Asking where a function is not continuous requires that we consider the entire number line, all Real numbers. The answer often provides better information.

So then, obviously a function is not continuous at any and all the points not in its domain (plus perhaps some other points in its domain). Accepting that a function is continuous on its domain, even if correct, does give us as much information as asking where a function is not continuous.

Ask the right question!