 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# The Transformation y = f(bx) - Problem 3

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Let’s graph another transformation. This time let’s graph y equals -2 times the greatest integer, less than or equals to 1/3x plus 2. This is a transformation of a new parent function, y is equals greatest integer less than or equal to x, whose graph looks like this.

It’s composed entirely of this 1 unit long segments, half closed, half open. And so this is a really interesting function to apply transformations to. Let me start with the table of values.

I deal with this function in sort of a special way because there are many values that will give this individual output. For example, to get the output of 0, I could use 0 I could use a ½, I could use 0.9, I could use any number between 0 and 1 except 1.

So I can take the entire interval from 0 to 1 not including 1 and I will get 0. And likewise, I could take the numbers from 1 to 2 and I’ll always get 1. From 1 to 2, but not including 2. I get a greatest of integer 1. And from 2 to 3 I get 2 and so on. Let me start with these values and I’ll transform them and then plot the results.

The transformation is minus 2 greatest integer of 1/3x plus 2. If I make this substitution u for 1/3x, then x equals 3 times u. That helps me get these values here. Then notice I’m using sort of a strange notation here and I’m actually using as input entire intervals. I want to see what happen so to this entire interval as it goes through this transformation x equals 3u.

So I’m going to take each of these values and multiply, them by 3. So 0 times 3 is 0, 1 times 3 is 3 then I get the interval 0, 3. I do the same thing to this one 1 times 3 is 3, 2 times 3 is 6 I get 3, 6. Same thing here and I get 6 9.

Let’s see what happens to the y values. I have -2 times the greatest integer of u plus 2. On my greatest integer of u values are right here so I’ll just have to take these values multiply them by -2 and add 2. So 0 times -2, 0 plus 2 is 2.

1 times -2 is -2 plus 2 is 0, 2 times -2 is -4 plus 2, -2. So let me see if I can get an idea of what this graph looks like just form these three points. From 0 to 3, I get an output of 2. So here is the interval from 0 to 3 when my output is 2. So it’s going to looks something like this.

From 3 to 6, I get 0. So here’s 3 and 6. From 3 to 6 I get 0. From 6 to 9, I get -2 and you can begin to see the pattern here. This is -2, 6, 9 and I’m going to get these segments which are 3 units long and they seem to be descending. So I’ll get -4 for 9 to 12 and so on. I’ll get 4 for -3 to 0 not including 0, and so and backwards.

And so that’s how I transform the greatest integer function. I use this special trick of using as input entire intervals because that tells me what’s going to happen to the left and right endpoints. And this is especially important when the transformation includes some kind of reflection and these end points switch. But feel free to use this trick, whenever you are transforming the greatest integer function.