Scribe Post 9/14 (FTS)

REMEMBER: Changes will be made. Yay.

     On Friday, IWs #1-4 were due (though we only passed in three IWs as the first one was just to rememorize the unit circle). However, instead of passing them in immediately and starting the lesson, O'Brien asked if anyone had any questions on any of the IWs.

-----Review---------     
     We first addressed how to solve #31 on page 66 algebraically, which read:

O'Brien explained that this was a really good question to ask as we will probably have one similar to it during our quiz (which will cover IW #1-6) on Tuesday. He first confirmed that this is an indeterminate function for when we substitute in 0 for x, we have division by 0. He then reminded us of how we proved that  (see our proof from Anna's blog post here). When we factor out the
expression, we are left with: 

.

By the properties of limits (to review, click here and see property 3), we know that we can simply multiply the limits and of both of these factors to get the limit of our original function. We know that  and that when we substitute 0 in for x. , therefore . 

     We also looked at #33 on page 66, which read:. Here, we can separate the function into two functions --  and. We know that the first is 1 and the limit of the second is 0. 1 x 0=0=limit of the original problem. 

     Then we looked at #72, which read : . This problem was true -- to prepare for the quiz, just make sure you know that on questions like this, you are looking to break apart the fraction in order to get  so that you won't have an indeterminate function to work with. 

     

     The last problem that we looked at was #43, which was a picture of a graph and various true or false questions regarding right and left hand limits. To prepare for a question like this, make sure you understand that with limits, it does not matter what the function is at the value that x is approaching. What matters is what is around that area. Also, make sure that you know your left from your right. 

-------------------------------

     After this review, O'Brien started the lesson by talking about "Happy Limits". These are limits that are easily found because you can simply substitute the value that x is approaching into the equation, solve, and this number will be the limit of the function. Why is this, you ask?

BECAUSE THE FUNCTION IS CONTINUOUS AT THAT POINT.

This means that there are no bottomless holes, no sketchy gaps, no creepy asymptotes, no division-by-0. NOTHING. They are, in a word, beautiful.

Look:                                

I mean...who doesn't freak out when they see that? BUT because the function is continuous, we can just substitute in 3 and our final answer is:

Good stuff. 

     At this point, O'Brien explained that one of our objectives today was to look at continuity. From last year, we know that a function is continuous if we can draw it without picking up our writing instrument. He also provided us with the formal definition:

A function is continuous at a point c if 

as long as:

a) c must be in the domain of f(x) 

b)must exist

Along with this definition, O'Brien added:
To be considered continuous, a function must be continuous at every point on its domain. 

"Well what about asymptotes? And holes? And gaps?"

What about them? They aren't in the domain. It doesn't matter that there aren't points there.

::::UPDATE::::
Here is an example of another continuous function that we went over during the 40 minute class period.

Here f(x) is continuous. Yes, I realize that you cannot draw f(x) without picking up your pencil, which is how we defined continuous functions last year. The point is that there is an asymptote at x=3, meaning that 3 is not in the domain, meaning that it's okay that there are no points there.  On the other hand, if we were to define a point on the asymptote, such as (3,4), the function suddenly becomes discontinuous because it does not follow the rule stated in blue above.

As x approaches c (which, in this case, is 3), the limit is infinity. Yet f(3)=4. Because 4 is not equivalent to infinity (and because 4 never could be equivalent with infinity as infinity is a concept while 4 is a number), this function with a redefined point is not continuous.
----------------------------------------------------------------------------------------------------------------------------------

In this diagram, you can see that . As this fits the rule that we defined above, the function must be continuous. But what about at point b? Here there is a hole . . . is that an issue?
O'Brien explained that because the hole is at the endpoint of the function, it is sufficient to just have the one-sided limit (in this case, the left-sided limit) equal to f(c). However, this graph is not continuous at x=b. There is no point there. There is no graph. Along with this, point b is not actually in the domain of the function, so it doesn't matter that there is no point there. Of course, it is also not continuous at a plethora of other points such as any point to the right of b and any point to the left of a.


     Here's another example:

     This function is not continuous, as there is clearly a hole at point b. The limit of the function as x approaches b appears to be a few units below M. However, f(b) = M. Because this does not fit the rule, this function is not continuous. If you still don't really understand the rules of continuous functions, click here for a video by a very beautiful man and watch until 3:20.

     O'Brien said that all of the functions that we learned last year were continuous (quadratic, cubic, exponential, rational, logarithmic, etc) except for one function, which was the floor function (below):

BEEP BEEP BEEP BEEP BEEP BEEP *FIRE DRILL* BEEP BEEP BEEP BEEP BEEP BEEP



     When we came back from an uneventful fire drill (there wasn't even a firetruck), O'Brien drew this graph on the board:

We identified that the discontinuous points were at x=4, 3, 1, 2, and -1. Then O'Brien classified each discontinuity for us:

1) Removable Discontinuities: Removable discontinuities are where you can define the function at that point without breaking the vertical line test. X=-1, 2, and 4 are all removable because if we redefine those points, we can fill the holes. We are able to do this with all of these points because they have a limit i.e. the limit as x approaches 2 is 2.

2) Jump Discontinuities: Jump discontinuities are also known as non-removable discontinuities. We cannot redefine the function at these points because then it would not pass the vertical line test. Along with this, the right-hand and left-hand limits do not approach the same values at this point. X=1 is a jump discontinuity. i.e. At X=1, the left-hand limit is 2.5 and the right-hand limit is 1. If we were to redefine x=1 to equal 1, we would fail the vertical line test.

3) Infinite Discontinuities: An infinite discontinuity is also known as an asymptote, where the function is always discontinuous at one point. X=3 is an infinite discontinuity.

***If any of this is unclear, go to this page 80 in our textbook, which gives a pretty good overview of the different types of discontinuities. Also, if you watch this video (same as the previous one) from 3:20, you will get a pretty good idea as to how to recognize the different types of discontinuities.

     Before we moved on, O'Brien showed us an "oscillating discontinuity" by graphing

 in Geogebra. It looked like so:

As we zoomed in, we saw that there was an infinite number of oscillations as we got closer to 0.

Cool! I guess...if you're into that kind of stuff...

     It was at this time that we transitioned toward derivatives. Just for review, a derivative is the slope of a tangent line to a function. However, we'll revise this blog post in upcoming weeks when we learn more about derivatives. Instead, for this class, we directed our attention to our other objective which pertained to local linearity and limits. 

     We began by looking at this limit problem: .  We saw that this was an indeterminate function as when we substitute in 1 for x, we get division by 0. O'Brien asked us to type this equation into Geogebra. When asked how to do a limit in Geogebra, he instructed us to just type in the function because the limit command in Geogebra is weird. 

We typed in and got the following picture:

We created the slider k and then defined point A as (k,f(k)). We used a spreadsheet in Geogebra in order to see what happened as we got really close to 1. At x=1, Geogebra tells us that the limit of the graph is  . However, around 1, the limit is approximately -1.359. 
     O'Brien then showed us that we could find this without technology by looking at the numerator and the denominator separately. This is what the graphs (red)  and    (black) look like near x=1:

Notice both items look nearly linear (this explains local linearity -- as we zoom in close enough on a given point on any function, even if it is curvy, we will eventually see a straight line). Also notice that both functions go through (1,0). If we can find the slope of both of these lines, we can put the equations into point-slope form. We know that the slope of 2x-2 = 2. Therefore, the new equation of this is y=2(x-1). However, the red curve is harder to find the slope of. If we imagine that it's a bit more linear, we can see that we go over about 1 and go up 2 (almost 3 units). After asking Duncan, Cole, and Autumn for help, I finally realized how a slope of e (which is about 2.72) was a reasonable slope to start out with. When we plug this into the point-slope form, we get . To check that this is the right linear equation for  (red), we graphed both this and  (black) on the same graph.

These are obviously not the same line. Obviously. The line that we created has the wrong slope -- let's make it negative:

Much better! Now we know that our numerator is and that our denominator is . When we combine them, we'll get . We can cancel out the (x-1) from both the numerator and the denominator and we will be left with  which is the limit of our original function. In summation, we basically separated the numerator and the denominator into two fractions, zoomed in on the graph at x=1, found the slope of these straight lines, and used the slopes to get the limit of an indeterminate function.


Let's look at another example:
We are given:. We know that .  If we look at the graph of and compare it to the graph of , we can see that we have compressed  in by 2, thus making the slope steeper.

Note: sin(x)=blue

sin(2x)=black

If the slope of the numerator is 2 and the slope of the denominator is 1 (as y=x has a slope of 1), then the slope of the original function is 2, thus making the limit of the function (as x approaches 0) 2. 

     We can also look at the function:, we will see something very surprising. 

Althoughand  are not the same graph, they do have the same slope when we look at the graphs around 0. This is because we are only looking at the LOCAL LINEARITY of graphs, not the graph as a whole. In both cases, the limit is 2. 

     This method of finding the slope of an indeterminate function and using this as its limit is known as
L'Hôpital's Rule (also known as L'Hospítal's Rule). The official definition of this is as follows (from page 448 in our textbook):

Suppose that f(a)=g(a)=0, and that f '(a) and g '(a) exist, and that g '(a) does not equal 0, then 

.

For further explanations, go to this video.


     We ended the class by talking about why radians rock. (I'll admit, I was always skeptical for the fact that I learned the unit of degrees first and always found that to be rather simple. However, after this explanation, I was sold. Radians rock.)
We graphed (in radian mode) and went to the table. We wanted to see what happened to this graph as x approached 0. When x=0.000001 and when x= -0.000001, y=1. However, in degree mode, when x=0.000001 and when x=-0.000001, y=0.0174532... Luckily, Cal miraculously figured out that 0.0174532 was the same as . What would you prefer... or 1?

That's what I thought. 

During our next 80 minute class, we will also be talking about the Intermediate Value Theorem. For a brief intro, check out this beautiful man.

The homework for the forty minute class is to do IW #6 (see full assignment below) and to study for a QUIZ ON TUESDAY that will cover IW #1-6.
IW #6: p. 84/7, 11, 13, 14, 15, 23, 41, 43, 47, 54, 57, 59

Please note that Dunks, Cole, and Auts are all beautiful people. I probably would have died without them.

Update:::::::

The next scribe ended up being Coleson, as Crockett was "sick"...