AP Calculus 09/07/10

Okay, so this took me about 15 minutes to figure out how to post a blog, and after all that, the 'New Post' thing is appearing at the top corner of the screen now. But it wasn't before. So if anyone else gets confused, apparently all you have to do is go to your dashboard to click 'Manage Blogs' and then just go to 'New Post' for the 2010-2011 blog, and you should be all set.

To start off class today, we took a three minute quizlet quiz on the trig functions/values. Some of us didn't do so hot, so if you can get to Mr. O'Brien's room before the end of the day 09/08, then you can retake the 10 question quiz. If you got a 90%, you can make it up to a 100%; lower than 90%, up to 90%. It would probably be best to study so here the link. (Hint: click 'link')

Then, as usual, Mr. O'Brien put a practice problem on the board for us to try to work through with the person sitting at our table.

Before we went over the problem, however, we went back to last class's link about What Is A Limit?. This included the Intuitive Definition of a Limit, which Nick wrote about in his last blog post.

"An Intuitive Definition of Limit is:
If f is a function and a and L are numbers such that
1. if x is close to a but not equal to a, then f(x) is close to L
2. As x gets closer and closer to a but not equal to a then f(x) gets closer and closer to L
3. We can make f(x) as close to L by making x close to a but not equal to a."

But back to the practice problem.
It was as follows:





As we tried to solve for the limit, we noticed that if x was greater than 1 then the limit was approaching 3, but if it was less than 1, the limit was approaching 1. This led us to Non-Existent Limits and One Sided Limits.

Non-Existent Limits are just like what we found for the practice problem. If the function does not have one consistent limit, then it is non-existent.
You would write this as:



The other part of this is what is called One Sided Limits. These are 1 and 3 values that we found. As x approached 1 from the negative (or lesser) side, then the one sided limit is 1, written as:




As x approached 1 from the positive (or greater) side, then the one sided limit is 3, written as:



(Notice the positive and negative signs next to the number x is approaching. It is what tells us that it is a one sided limit.)

This also lead us to the very basic rule that

if then exists.

Next, we ventured into the idea of how to find limits both graphically and algebraically. However, we mostly focused this class on the graphing side of that, so be prepared for a lot of screenshots of Geogebra coming your way. Sorry. I felt it was necessary.

Yes. So, Mr. O'Brien wanted everyone to follow along in Geogebra using their own computers, so they would know how to do it themselves in the future.

We started off with the basics- putting the practice problem/function into the graph:


Okay, so the screenshots turned out much blurrier than I would have liked. But you'll get the gist of it. The hard-to-read writing is the function, which was .

Next, we had to put in sliders (our slider was labeled with 'k'). However, that didn't change or impact the function at all when we put it in. It had no meaning. So we had to plot the point (k, f(x)). That made the slider applicable to the function. So, as we moved the slider the point moved and vice versa. We can track the point as it moves.

(You can't read it, but it says that k = -2.4)



(You also can't see this that well, but it says that k=2.4. The point is that you can see A -the point on the function- move. Just estimate where you think 2.4 or -2.4 should be on the x-axis and follow that to the point.)

Obviously, since our limit was supposed to be as x approached 1. We had to move the slider to see what would happen when k was equal to 1.


As you can hopefully see, the point disappeared. It vanished. Like magic. Like a magician took it. With magic. Magically.
BUT IT IS NOT MAGIC! Surprisingly, this is what is called the Non-Removable Discontinuity.



That blue box is where the Non-Removable Discontinuity is. It cannot be fixed easily, hence it is non-removable. There are in fact little gaps where x = 1. We will touch on this a bit later, but all in all, x cannot equal 1 because in our function -
- if x equaled 1, then the fraction at the end of the function would be undefined. So it cannot exist. Even though it may look like there are two values for 1 on that graph, there are little holes in the function.

We then went on the look at how to find limits graphically. Let's say that then using a graph above, you can trace where x is 2 and look around the point on the function where x equals 2 to see what happens to the function as x approaches 2. In this case, the limit is, in fact, 2.

The rule of thumb is that when looking graphically at a function to find the limit, it is not what is happening at the x value, but what is happening around it that will help you determine limit.


Now, we move on to the more advanced Geogebra workings- spreadsheets. DUN DUN DUN!
But really. It's not that bad, as I explained to Andy several times over (I love you Andy. I mean this lovingly. Love with a bit of teasing ), all you need to do is put your big kid pants on and go to 'View', then click 'Spreadsheet'. And for this problem, we went to 'Option' and changed it so that our decimals were rounded to 15 places. Or something around there. We just want to see far into the decimals and not have them rounded to the 10th power.

Once, we got to the Spreadsheet attachment, all you need to do was input several values approaching 1 into the A column.


You can mostly see the A column values. So the next step was to input =f(A1) into the B1 slot. I'm not sure if you can see that all that well. But once you do that and press enter. You'll need to highlight B1 and that will give you a blue box with a square in the corner. Grab that box and pull it down to end of the A values. That should give you something that looks a bit like this:

If you squint really hard and rub your tummy, you can almost make out those values. Good luck.

Mr. O'Brien then posed that question as to why we have that gap. I sort of touched on that briefly with the whole undefined fraction bit. However, we figured out that if you simplify the function with a positive x value, you will get . But if you simplify it with a negative x value you will get . Every time. With the exception of positive 1.

We plotted both of those simplified functions with the original function to look a little something like this:
(The upper red line is , and the bottom red line is . The blue overshadows the red.)

We played around a little bit with what would happen/how the graph would change if we changed the 1's in the fraction of the function to 2's (the first graph) or to -1's (the second graph).



Someone also brought up the possibility that it may not pass the vertical line test, and therefore it would not be a function. But seeing as there are gaps, like we discussed earlier, little holes, it would pass no problem.

We looped back around to if we changed to fraction 1's to negative 1's. That would make the function look like .

Using this we can simplify that so that when x is greater than -1, . And when x is less than -1, .

We can apply those simplified equations next to like we did to the original function to see how they compare.

It should look a bit like that. (Which is easier to read. Which I like.)
We can see that those two simplifications still line up with the function.

Next, we went to see what would happen if we multiplied the fraction of the function by 2, giving us . Graphically, it looked like this:


You can see that it stretches the graph a little bit, but the gap (or the Non-Removable Discontinuity) at -1 is still the same.


Going back to our altered function of , which we just graphed, we can established that . However if you were trying to find , the answer would be 7.
This proves that sometimes you can plug in what x is approaching to find the limit and sometimes you can't, so the best decision is to check your answer or find it by looking at what is around/approaching the value, instead of the value itself.

Then, we looped back to our original function of . This is when we derived a new function.
We decided to let if and to also let g(x) = 5 if .
The way to write this into Geogebra looks like this: if [x≠1, f, 5].
To break that down, that means that if x doesn't equal 1 then have g(x) equal f(x), and if not (if it does equal 1) then have it equal 5.

When graphed, it should look like this:
The next step was to insert g(x) into the spreadsheet to compare the differences. That should look like this:



If you can see that, when x equals 1, f(x) is undefined, but at g(x) it equals 5, like we specified. Other than that f(x) and g(x) are the same.

We then went on to plot (1, g(1)). This makes this point B at (1, 5). We had to change our slider to make it equal (k, g(k)) now instead of (k, f(k)) so that it would be applicable to our current problem.
This now means that when the the slider is at 1, instead of disappearing like it did before, it will overlap with point B at (1, 5).


(It's hard to see but the slider point is A and it is green. It's located at (1, 5) with point B, which is black.)


Mr. O'Brien then had us take away the absolute value portion of our function which made it which would simplify to if x≠1. When it does equal 1, the function become undefined.

All in all, the graph would look like this:

Just remember that it does not matter what is happening at 1 (or, in general, what x is approaching), it matters what is happening around it.

Now after all that, we took a break from Geogebra with a few minutes left in class and went over the previous homework (HW #2) and did it algebraically on the board.

And then the bell rang.
Homework for next class:
Assignment #3:
* p. 72/5, 7, 9, 13, 15, 17, 19, 39, 43, 63, 67 without using the calculator
* Be ready for Thursday's quiz.

Next scribe: Caitlin

Sorry I posted this so late, I got home late, and this took me forever. I really hope you were able to follow this, whoever you are.