I have moved to a new server

Hi everyone. I want you to know that I’m still alive (lol) despite being inactive for a few months. I have also moved to a new server and sadly I have lost all the comments and I don’t know if I could still recover them.

Also, the new server which is supposed to be an upgrade has also been erratic. From time to time, the site goes offline and receiving an Error 503 or 504. Anyway, hopefully, It will improve over time.

Anyway, I’m wishing everyone a Happy New Year (I know it’s a bit late) and hopefully I can be more active this year,

Be safe.

I’m Still an Active GeoGebra User

GeoGebra has improved a lot since I was first acquainted with it in 2006. For the subscribers of this blog, you have probably come across my series of tutorials on how to use the app. In recent years, GeoGebra has evolved into several apps that can be used for specific tasks. It has also become compatible with major operating systems including Windows, Mac, Linux, iOS, and Android. The different GeoGebra apps are the following:

  • Scientific Calculator
  • Graphing Calculator
  • Geometry
  • 3D
  • CAS
  • Classic

The Classic has two versions and is the most comprehensive among the apps. The 3D app has also support from Augmented Reality. In addition, one of the recent updates of GeoGebra is its support for examinations.  

My Current Work on GeoGebra

Personally, I have been involved in developing applets and lessons for the GeoGebra Institute in my country (see example below). You can view my applets at the official GeoGebra website and some short videos demonstrating the capabilities of GeoGebra at our GeoGebra Institute Facebook page. The official website of our GeoGebra Institute is currently being updated here and hopefully, we can upload lessons that are readily available for teachers soon.

At the Institute where I work, we have also integrated GeoGebra in several of our teacher trainings. Last year, we have obtained funding to train potential trainers. The project includes developing a module for future trainers and a 2-phase training.  The training was supposed to be start in April but was postponed due to the current pandemic.

Some Resource on GeoGebra

There are so many tutorials if you want to learn GeoGebra. You can see my tutorials here or read the GeoGebra Manual. As for the applets, there are millions of applets uploaded in the GeoGebra website. Some of the excellent GeoGebraists are Daniel Mentrard, Tim Bzherzhinski. The GeoGebra Institute of Bogota is also very active.

Solving Rational Inequalities and the Sign Analysis Test

In this post, we are going to learn the steps in solving rational inequalities and see why we are doing them. For discussion’s purposes, let’s take the inequality

\dfrac{x - 1}{x^2 - 25} \geq 0

as an example.  In solving rational inequalities, first we get its critical values, then use the values to determine the intervals, and finally, test the values in each interval to see if they satisfy the inequality or not. Those that satisfy the inequality are included in its solution set.

Getting the Critical Values

To get the critical values, we equate the numerator and the denominator of the inequality to 0 (if applicable). If we think of the inequality as a function, that is, if we let

f(x) = \dfrac{x - 1}{x^2 - 25},

then equating the numerator to 0 means getting the zeroes of that function. Graphically, the zeroes are the intersections of the graph of the function and the x-axis. In the example above, when we equate x - 1 = 0, we get x = 1. This is where the graph will intersect the x-axis.

In equating the denominator to 0, we determine the values of x where there are holes or vertical asymptotes. If we factor both the numerator and denominator completely, the zeroes of the factors that we can cancel are the holes, otherwise vertical asymptotes. In the given above, equating the denominator to 0 only gives us asymptotes.

x^2 - 25 = 0
(x + 5)(x - 5) = 0
x = 5, = -5.

From the equation above, the vertical asymptotes of the function are at x = 5 and x =-5. Now that we found all the three critical points, we plot them on the number line. These points divide the number line into four intervals.

The criticial points divide the number line into four intervals.

Sign Analysis Test

Now that we identified the critical values, we do the sign analysis test. In this test, we choose a number from each interval and substitute it to the inequality to see if the resulting value satisfies the inequality. Since \frac{x - 1}{x^2 - 25} \geq 0, we are looking for values of x that will make the expression \frac{x-1}{x^2-25} either positive or 0.

In testing the intervals, if we choose x = -6, x = 0, x = 2, and x = 6. Substituting these values to the inequality give us

-\dfrac{7}{11}, \dfrac{1}{25}, -\dfrac{1}{21}, \dfrac{5}{11},

respectively.  This means that the intervals containing \frac{1}{25} and \frac{5}{11} are the solutions to the inequality.

We now test the critical values and see if they are included in the solution set. In the inequality above, since x = 1 will make the inequality 0, it is therefore included in the solution. However, x = 5 and x = -5 are not included as solutions because they will make the denominator 0 and will make the rational expression undefined. Therefore, the solutions are the intervals that contain positive values, includes x = 1, and exclude x = 5 and x = -5. In interval notation, the solution set is (5,1] \cup (5, \infty).

Interpreting the Solution

We are looking for positive values of x that will make the inequality

\dfrac{x - 1}{x^2 - 25} \geq 0,

true and obtained the solution (5,1] \cup (5, \infty). In the function

f(x) = \dfrac{x - 1}{x^2 - 25},

the values which are greater than 0 are the parts of the graph above the x-axis. As we can see, these are the intervals (5,1] and (5, \infty).

We can also verify the critical values x = 5 and x = -5 from graph. They are vertical asymptotes. Lastly, the value x = 1 is the intersection of the graph and the x-axis.

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