Tag Archives: absolute value

CAS Derivatives

Here’s a fun problem from my calculus class today, enhanced by CAS. As a set-up, our last unit focused on interpreting the meaning of derivatives with multiple interpretations of the definition of the derivative as the only algebraic work they’ve done.  In that unit, the students discovered that vertical translations on functions didn’t change their derivatives, and horizontal translations on functions changed the corresponding derivatives by the same horizontal translation.  From their work with derivatives of power functions using a definition of the derivative, they hypothesized and proved \frac{d}{dx}(x^n)=n*x^{n-1} for natural (and a few other) values of n.  Knowing nothing else, I posed this.

Use your CAS, determine the derivatives of y=ln(x), y=ln(2x), y=ln(3x), and y=ln(4x).  Use your results to hypothesize the derivative of y=ln(n*x).  Justify your claim.

The following image from the first part shows that the pattern is easy to spot.

Unfortunately, I posed the problem with only 10 minutes remaining in class, but the students clearly knew \frac{d}{dx}(ln(n*x))=\frac{1}{x}, but the looming question wasWHY.  With a couple minutes to spare, one guessed rules of logarithms might apply, but not having used them since their first semester exam, he didn’t recall the property.  Some colleagues may argue that I should have insisted on my students having those rules memorized in advance, but I firmly believe that this particular problem actually gave a reason for my students to relearn their logarithm properties.

I let the awkward moment hang there until another called out with glee,
ln(n*x)=ln(n)+ln(x)“, to which a third exclaimed, “and ln(n) is a constant, making the derivative of ln(n*x) the same as the derivative of
ln(x),” clearly using her understanding of the effect of translations on functions and their derivatives that she learned in the last unit.

Two other nice ideas emerged:

  1. They thought it convenient that \frac{d}{dx}(ln(n*x))=\frac{1}{x}, but now really want to know why.
  2. A few observant ones noticed that ln(x) and \frac{1}{x} have different domains.  To these, I pointed out the warning at the bottom of the CAS screen above that most had completely overlooked when getting their initial answers.

These questions still linger for the class, but I argue that the use of CAS in my calculus class this afternoon

  • granted everyone equal access to the same mathematics,
  • emphasized the importance of logarithm properties in a more meaningful way than any pleas from me could have done,
  • left a need in many to discover why \frac{d}{dx}(ln(n*x))=\frac{1}{x}, and
  • raised domain issues that ultimately will lead to a deeper understanding for the existence of the absolute value in \int{\frac{1}{x}dx}=ln(|x|).

It was a good end to our first day back after Spring Break and left my students wanting to know more.  What a great place to leave a group of students, thanks to CAS.

Transforming inverse trig graphs

It all started when I tried to get an interesting variation on graphs of inverse trigonometric functions.  Tiring of constant scale changes and translations of inverse trig graphs, I tried i(x)=x*tan^{-1}x , thinking that this product of odd functions leading to an even function would be a nice, but minor, extension for my students.  

I reasoned that because the magnitude of arctangent approached \frac{\pi}{2} as x\rightarrow\infty, the graph of i(x)=x*tan^{-1}x must approach y=\frac{\pi}{2}|x| .  As shown and to my surprise, y=i(x) seemed to parallel the anticipated absolute value function instead of approaching it.  Hmmmm…..

If this is actually true, then the gap between i(x)=x*tan^{-1}x and y=\frac{\pi}{2}|x| must be constant.  I suspected that this was probably beyond the abilities of my precalculus students, but with my CAS in hand, I (and they) could compute that limit anyway. 

Now that was just too pretty to leave alone.  Because the values of x are positive for the limit, this becomes \displaystyle y=\frac{\pi}{2}|x|-x*tan^{-1}x=\frac{\pi}{2}x-x*tan^{-1}x=x*(\frac{\pi}{2}-tan^{-1}x) .

So, four things my students should see here (with guidance, if necessary) are

  1. i(x)=x*tan^{-1}x actually approaches y=\frac{\pi}{2}|x|-1,
  2. the limit can be expressed as a product,
  3. each of the terms in the product describes what is happening to the individual terms of the factors of i(x) as x approaches infinity, and
  4. (disturbingly) this limit seems to approach \infty*0.  A less-obvious recognition  is that as x\rightarrow\infty, \frac{\pi}{2}-tan^{-1}x  must behave exactly like \frac{1}{x} because its product with x becomes 1,

But what do I do with this for my precalculus students? 

NOTE:  As a calculus teacher, I immediately recognized the \infty*0 product as a precursor to L’Hopital’s rule.

\displaystyle\lim_{x\to\infty} [x*(\frac{\pi}{2}-tan^{-1}x)]\rightarrow\infty*0
=\displaystyle\lim_{x\to\infty}\frac{\frac{\pi}{2}-tan^{-1}x}{\frac{1}{x}}\rightarrow\frac{0}{0}
and this form permits L’Hopital’s rule
=\displaystyle\lim_{x\to\infty}\frac{\frac{-1}{\displaystyle 1+x^2}}{\frac{-1}{\displaystyle x^2}}
\displaystyle=\lim_{x\to\infty}\displaystyle\frac{x^2}{1+x^2}=1

OK, that proves what the graph suggests and the CAS computes.  Rather than leaving students frustrated with a point in a problem that they couldn’t get past (determining the gap between the suspected and actual limits), the CAS kept the problem within reach.  Satisfying enough for some, I suspect, but I’d love suggestions on how to make this particular limit more attainable for students without invoking calculus.  Ideas, anyone?

Quadratic Approximations to Sums of Absolute Values

I recently was notified that a former student’s idea is expected to be published in the Mathematics Teacher in early Fall, 2011.  Ian discovered a way to use a single quadratic function with a variable y-intercept to capture a limited number of values of an increasingly long sum of absolute value functions.  I’ll link a copy of the brief article once it is published.

The coolest part of the development of this idea is that Ian’s access to a CAS and the encouragement of his class to explore their ideas made this problem possible.  When Ian needed to learn additional mathematics (induction) to prove his theorem, he was receptive and confident.