Precal Blog

Today’s class we mainly just listened to the presentations of the people who claimed problems 3 and 5 from fridays homework. Zach presented number 3 and Matt presented number 5. Then after that we had a contest in the last 15 minutes of class using some problems that Mr. Bieniek made or picked from the ACT

Tuesdays scribe will be Monica

Today in class (9/26) we continued to elaborate on composite functions.

 We discussed the composite function we were using the previous day in class, f(g(x)), in which g(x)=10-2x with the domain of `1<= x<=4`, and f(x)=x+2 with the domain of `3<=x<=7`. As Allison and Kevin explained, we determined that to get an output for the function f(g(x));

 -The value of g(x) must in the domain of f(x), and

-The value of x must be in the domain of g(x).

 We then better explained these statements by recognizing g(x) as the inside function and f(x) as the outside function, and rephrasing the statements as;

-The outputs of the inside function must be in the domain of the outside function, and

- X must be in the domain of the inside function.

 After this discussion, the class then worked on the problems for 1-4, with #3 and #5 being claimable.

 Lauren will post the next blog entry.

Today’s class was dedicated to working on problems in Section 1-4.  Claimable problems were numbers 3 and 5. 

Monday’s Scribe will be Dominic.

-Note from the Scribe: My previous blogs were late due to my inability to log on to the blog site.  I apologize for the inconvenience.

Today began with a continued discussion about compositions, and how they are not commutative.  In my last blog entry, I used an example of transforming a linear function into a quadratic function.

f(x)=x
s(f(x))=`x^2`
n(s(f(x)))=`-x^2`
v(n(s(f(x))))=`-x^2+2`

In this example, the vertical function has to be the last composition change.  If v(f(x)) were to be the first change made, the squaring function would create an entirely different graph, because (x+2) would be squared, as opposed to just x.  Also, if the vertical function change preceded the negative function change, the graph would actually be lowered by two units, as opposed to being raised two units.

Next, we went over another example of compositions using two piecewise functions.  They were f(x)=x+2: 3<=x<=7  and  g(x)=10-2x: 1<=x<=4.  We were told to graph the compound function f(g(x)), which can also be written as f o g(x), where the the o stands for “of.”  To get results for f(g(x)), the first thing is to plug in x values in g(x).  Let’s use 3 as an example for an x value.  This would yield a result of 4 when plugged into g(x).  Four is then used as the x value for finding the value of f(x), which would yield 6 as a final result.

To find the domain of f(g(x)), you must apply two rules:

1: All g(x) values must be in the domain of f(x).

2: All x values must be in the domain of g(x).

Because of these rules, you can find the exact domain by setting up the equation `3<=10-2x<=7`.  Note that 10-2x is g(x) and the 3 and 7 come from the domain of f(x).  If you were to work through this equation, you would get 1.5<=x<=3.5.  The next step is to check this answer and compare it to the domain of g(x) because of rule#2.  If g(x) stated 1<=x<=3, the domain of f(g(x)) would be 1.5<=x<=3.

Class began today with presentations on problems from section 1-3.  Claims for the section were number’s 6,11,and 20, and were presented by Curtis, Rick, and Francesca, respectively.  There was further questioning on the homework assignment, but all questions were answered very clearly.  After the questioning, the idea of an educational function game was suggested, and Mr. B relished the idea.

Next, we learned about compositions, another way to control functions. Mr. B used an example of combining four functions to make a linear function quadratic.

To change a linear function, specifically `y=x`, to a quadratic function, specifically `y=-x^2+2`, it must undergo three function changes: a squaring function, a negative function, and vertical function.

f(x)=x
s(f(x))=`x^2`
n(s(f(x)))=`-x^2`
v(n(s(f(x))))=`-x^2+2`

The final function will produce the desired result.

Take a look at this article from the NY Times. Very interesting with implications for mathematics students and teachers alike.

Then take the test to see where you stand.

Today in honors precalculus our class discussed section 1-4 of our textbook, which delt with the compostition of functions.  This section was aimed at helping us graph and evaluate the compostition of one function with another, given two functions.

We were originally given a sheet that had the functions g(x)=10-2x with domain `1<= x<=4` and f(x)=x+2 with the domain `3<=x<=7`.  These two funtions were already graphed on a graph and, so we were supposed to find f(g(3)).  To do this we plugged 3 into the function g(x)=10-2x to get g(x)=10-2(3), which equals 4.  We used the number 4 and plugged it into the function f(x)=x+2 to get f(x)=(4)+2, which is 6.  This means that the f(g(3)) is (4,6).  However when you plot f of g you use the origianl x value, which was 3, so your coordinates are (3,6).

After plugging 3 into this function, we had to find the domain and range for f of g.  To do this you have to remember two rules about composite functions.

1.  g(x) must be in the domain of f(x).

2.  x must be in the domain of g.

Following these rules we set up the equation `3<=10-2x<=7` because we are plugging the function g(x) in for the x in the domain of f(x).  Then we seperated this into two equations `3<=10-2x`, which equals `x<=3.5` because you have to flip the inequality sign when you divide by a negative number.  Our other equation was `10-2x<=7`, which equals `x>=1.5`.  This means that the domain for f of g is `1.5<=x<=3.5` or [1.5,3.5].  Then to find the range we plugged 1.5 and 3.5 into the function f(g(x)).  First we plugged 1.5 into the g(x) function to get 7 and then we plugged that into the f(x) equation to get 9.  For the other part of the domain  we plugged 3.5 into the function g(x) to get 3 and then we used that in the function f(x) to get 5.  This means that the range is [5,9].

Once we were finished with this composition function we treid the inverse function, which is g(f(x)).  We started by putting the f(x) function into the g(x) function to get 10-2(x+2).  However, this function did not work because our x values were out of the domain.  For instance, the domain of f(x) is `3<=x<=7`, so we plugged in 3 into our new function, but this gave us 0, which is not in the domain of g(x).  We tried the other numbers of the function f(x), but they all were out of the domain of g(x).

Through this class we learned that funtions are not the same when you change their order, so it is imprtant that you get the right function to graph.

Today in class we began by going over the claims for section 1-3:
First, Zak presented number 6, and explained how he  got the equation `g(x)=(sqrt(9-(x+3)^2))/2` by inserting the equation `f(x)=sqrt(9-x^2)` into the equation `g(x)=(1/2)f(x+3)`.  The original graph was horizontally translated 3 to the left (it is “opposite world” because x+3 is in parentheses).  The image was also vertically dilated by `1/2` because the whole equation was over 2.
Next, Brandon explained how the image in number 11 was vertically dilated by 3 because -1*3=-3, which is a point on the graph, and the vertex changes from 2 to 6, which is also by 3.  The graph was also moved 6 to the right, a horizontal translation.  Brandon then came up with the graph `g(x)=3f(x-6)`.  The 3 outside is altering the y variable, dilating it by 3, and the 6 inside translates the x values to the right 6 places (in “opposite world” again).
Finally, Sawyer explained problem 20.  He created a new graph with a horizontal dilation by 2 and a vertical translation by 4.  There was a class dispute over whether the graph also widened vertically, but planting points on the graph proved that the graph was only dilated horizontally.

Mr. B. explained a little more about “opposite world,” relating it to factoring we have done with quadratic equations.  For instance, if we came up with (x-2), x would be equal to (2,0), not -2.  We need to find the value that completes the task rather than using the number inside the parentheses.

Next, Mr. B. gave us an introduction to section 1-4 with a story of Damian and Karla.  Damian’s mood depended on his hunger and Karla’s mood depended on Damian’s mood.  We made a rough graph relating time of day as the independent variable versus Karla’s mood as the dependent variable.  The graph ended up with a lot of peaks and valleys, peaking at Damian’s meal times and decreasing until Damian’s next meal time.  This is called a periodic graph because the graph repeats in regular intervals.

We next decided to create variables from the information given: t=time of day, h=Damian’s hunger level, g=the amount of grouchiness from Damian, and k=the amount of positiveness from Karla.  Since the time of day affected Damian’s hunger, d(t).  Damian’s hunger then affected his grouchiness, so his hunger is imputed with grouchiness as an output to get g(h(t)).  Finally, Karla’s positiveness depends on Damian’s grouchiness, so k(g(h(t))).  This is called a composition of functions because one rule becomes the imput of a new rule.  This can also be called “k of g of h of t,” written as “kogoh(t)” with small circles in between the variables.

Finally, we tried to transform a graph of a parabola opening down through the point (0,2) into a straight line through the origin with a slope of 1.  We decided to transform, we needed: v=a vertical shift +2, s=square, and n=negative 1 function.  We attempted to make a composition of functions in this order, coming up with n(s(v(f(x)))), which gave us the equation `-(x+2)^2`.  This was wrong, because we didn’t want to square the 2, just the x.  Therefore we determined that the order matters in a composition of functions.  We redid the order and found that:
f(x)=x
s(f(x))=`x^2`
n(s(f(x)))=`-x^2`
v(n(s(f(x))))=`-x^2+2`
This equation gave us the right graph!

Finally, our homework for tonight was to read section 1.4! Tomorrow’s class scribe will be Allison!

Today September 23rd, we discussed section 1.3.  Several people had questions that needed clarifying. We went over how to make windows “friendly”. Each window on our calculator has 94 pixels. So in order to make each pixel show a friendly number when we trace a graph, we can use the equation `94*Y=X` when Y is equal to the number we want each pixel to be.  We then take `X` and divide it by 2, so we get what the maximum and minimum numbers for the window should be. For example `94*.5=47`, `47/2=23.5`  `X` maximum should be 23.5 and minimum should be -23.5. This will make the X axis pixels equal to .5.  We also discussed how to use functions on our calculator.  Type the function into the `Y=` window for Y1, and then press the VARS button.  Then go over to Y-VARS, then it is #1 “Function…”. Then select Y1.  Then you can go Y1(4) and it will display what the output of a function is when the input is 4. After that, we had class time to start working on the problems in 1.3. The next scribe will be Sam.

Yeah– Sorry about that. I think the scribe list is fully updated now. If you’re not crossed off and you posted a class period as the scribe (NOT something in addition to that- while if you do that it’s appreciated and will help you in the long run, it sadly doesn’t count), please let me know and I’ll look into it. Thanks ^_^

~~~~~~Andy~~~~~~