 ###### 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 Greatest Integer Function - Problem 2

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 the transformation of the greatest integer function. y equals -2 times the greatest integer less than or equal to x plus 1 plus 3. The parent function here is y equals the greatest integer less than or equal to x. I’ve got a graph of the function right here. Usually what I do when I’m graphing transformations is, I start with a table of values for my parent function. And I use the letter u, especially when there are horizontal transformations instead of x.

This one’s interesting though because, there is a whole interval of numbers that have an output of zero, and a whole interval of numbers that have an output of 1 and so on. And so, the way I want to deal with that is that I actually want to, on my table, I’m actually going to put the entire interval down that gives me zero. That’s the numbers from zero to 1 not including 1. So this interval, every number in this interval will give me an output of zero.

Every number from 1 to 2 not including 2, will give me an output of 1. Every number from 2 to 3, not including 3 will give me an output of 2 and so on. I can go in the negative direction. Every number from 1 to zero not including zero will give me -1. Now my transformation is -2 times the greatest integer less than or equal to x plus 1 plus 3. I need to take care of my horizontal transformations first.

The way I usually do that is I substitute u for x plus 1. If u equals x plus 1, then x equals u minus 1. To get my x values here I take my u values and subtract 1. Now since I have whole intervals, what I’m going to do is subtract 1 from each endpoint. The result is this becomes [-1, 0). This becomes [0, 1). This becomes [1, 2). This becomes [-2, -1).

What happens to the y values? Remember that x plus 1 is u so this is minus 2 times the greatest integer of u plus 3. So minus 2 times this column plus 3. Minus 2 times zero plus 3 is 3. Minus 2 times 1, -2 plus 3, is 1. Minus 2 times 2 is -4, plus 3 is -1 and minus 2 times -1 is 2 plus 3 is 5. Let me start by graphing all these, and they’re not just points, they are whole segments, from -1 to zero, I have 3. So -1’s here, zero’s here and I get the value 3. So my graph will look like that. That little segment there.

From zero to 1 I have 1. So over here 1. From 1 to 2 I get -1, so that’s down here. From -1 to -2 I get 5,that’s over here so I want to get 5. And this is going to continue up this way and down this way. I’ll draw one more though. You can indicate that it continues with an ellipsis in each direction. That’s our final graph; this is our transformed greatest integer function.