07 | October | 2008 | Continuous Everywhere but Differentiable Nowhere

Teaching seniors is a mixed bag.

With college admissions, you often get students suffering from a Dr. Jekyll-and-Mr. Hyde complex. The first semester has seniors constantly stressed out, with SATs and SAT IIs, writing college admissions essays, with soliciting recommendations, and deciding where to apply. Many seniors also — at least in my school — tend to fill their plates to the edge. Some have crud oozing off the edges of their plates.

They’re overextended and stressed. And I feel for them.

And then, once college admission decisions have been made, you get seniors falling victim of the most ugly and heinous of all diseases: senioritis. Grades, the traditional motivator, have little curative effects. Their corporeal bodies might be in chairs, but their spirits have flown yonder. The stressed out blobs of nervous energy have become sedate lumps.

Right now I’m dealing with the blobs of nervous energy.

And quite a few of them are even more nervous.

A good number of seniors who sign up for regular (non-AP) calculus specifically for colleges. And this then becomes the rub. Because it’s about at this time of the year when students start to realize that they actually are getting graded in calculus. That it’s more than just the name of a course for college admissions. It’s — you won’t believe this — actually a course.

All I can say to them is that I’m here for them. And that it’s never Me Vs. Them, but always Us Vs. Calculus. But I suspect that this distinction will get blurred in their minds as the admission season gets underway.

Coming up with math explanations for students that don’t always “get it” can be tough. You have to be thinking on your toes, and you have to make your explanations understandable. It’s not always easy to succeed.

Two Recent Examples:

How would you explain to a student in a non-accelerated Algebra II (or even non-AP Calculus class) why it is that when you solve $\log_(x+2)+\log(x-2)=\log(5)$ in the standard algebraic way, you get $x=3$ and $x=-3$ ? Why is it that you generate the extraneous solution, $x=-3$ that you then have to eliminate, because it can’t “plug into” the original equation? Where does that extra solution come from?

How would you explain to a student — without using the formal definition of a limit — what a limit is? So that they understand it intuitively. Now ask yourself: would your explanation work for the constant equation $f(x)=6$ ? What would you say to a student who says “Why is the limit of f(x) as x approaches 0 equal to 6? The function has already reached 6. It isn’t approaching 6.”

And for those of you who want to prove your mathematical mettle, here’s a question that recently circulated through our math department. What’s the answer, and how would you explain it to a stuck student?:

Suppose that a function $f$ is differentiable at $x=1$ and $\lim_{h\rightarrow 0} \frac{f(1+h)}{h}=5$ . Find $f(1)$ and $f'(1)$ .