Precal Blog

On Friday, September 28th our sixth hour precalculus class went over the claimed problems from the previous nights’ homework. Section 1.4 Problems 3 and 8

Jessica took a stab at explaining to the class the process of number 3. She did an excellent job going over everything we really have learned so far about functions and their domains. This problem seemed to entail a lot of the material we have been learning, and Jessica really helped to point out a few important things we need to remember for our upcoming test. (THIS TUESDAY) I especially thought she did a good job of reminding us the two circumstances we must remember when working with the Composition of functions…

That not only must the values of the first function fit in the domain of the second, but the x values of the first equation must also fit into its’ domain.

A comosition function can not…”funtion” without values that fit into the given domains.

Next, Justin helped the class with number 8.

His presentation I also found to be beneficial in that in really went over everything in class that we have been learning. He did a good job of telling how the different functions really related to each other..and prooved it. He showed us the graphical, numerical and alegraic proof of his findings and the process we must go through to relate functions.

Our class was then cut short, due to homecoming activities..

Thank you Jessica and Justin for the effort and help you put into your presentations

Mondays Scribe will be………Stephanie M

[Note: When I typed this post originally I forgot that Internet Explorer stinks. Everyone should be using Firefox - it is superior to IE in every way. If you insist on using IE then there are two files you need to install. One is Adobe's SVG Viewer (which will no longer be supported after 1/1/08, and the other is Design Science's MathPlayer. Even with those files installed, I still can't get IE to display correctly. Download Firefox!

[Yet another note: Even though I had the most current version of Firefox (2.0.0.7) I still needed to install some fonts to get all of the notation to show up correctly. It is quick and easy to do and you can get them here.]

`f(x)=(x-2)(x-1)x(x+1)(x+2)/2`

agraph
width=300; height=200; xmin=-5; xmax=5; xscl=1;
plot((x-2)*(x-1)*x*(x+1)*(x+2)/2,-1.5,2);
endagraph

Typing in this function and creating its graph are extremely easy thanks to ASCIIMath. Double click on the graph and study the code you see. Do you see the simple, intuitive commands that were used to plot the function? Do you see how to restrict the domain? Play around with it and just refresh the page to start over.

To include mathematical notation in your post, just type it in like you “think” you should – but surround it with left-quotes. On my keyboard the left quote is with the tilde (~). So if I wanted to type in the quadratic formula normally it would look terrible: (-b +- sqrt(b^2-4ac))/(2a).
But use the left quotes around it and get: `(-b+-sqrt(b^2-4ac))/(2a)`.

Beautiful.

Today we talked a lot about the activity from the lab yesterday. Mr. B. focused mainly on the second page dynagraph. The first page dynagraph showed a more real world example of how the dynagraphs work. The function of CO2 in the atmosphere was the input whereas the rising temperature was the output. This output temperature was then “plugged in” to the second function, showing the increase in melting of Earth’s icecaps. The dynagraph from the second page worked very much in this same way. Function f(x)’s output would be plugged in to g(x) , giving another output.

We then focused on the domain and range of this dynagraph. The domain of f(x) was [-2,10]. When we merged the functions together, the dynagraphs did not keep the same domains and ranges. The values from f(x) must fit into the domain of g(x) which we found to be [2,2.5)(2.5,4.5]. The domain is basically [2,4.5], but 2.5 cannot be included. Since the equation for g(x) was 1/x, we knew that when x was 0, we’d have a problem. Sure enough, when x was 2.5, we had a problem. In f(x), the equation “2x-5″ would become 0 when 2.5 was plugged in. Plugging 0 into 1/x of course cannot be done because it is division by 0.

To figure the domain out algebraically, we used the technique we learned two days ago. We used the domain for g(x) and plugged f(x) in for the domain. Since that’s kind of confusing, it’s easier to just look at this. -4 < or equal to 2x-5 < or equal to 4. Using order of operations, we add 5 to all sides to get rid of it. This gives us 1 < or equal to 2x < or equal to 9. After dividing all sides by 2, we get a domain of .5 < or equal to x < or equal to 4.5. BUT WAIT A MINUTE! I THOUGHT THE DOMAIN WAS [2,2.5)(2.5,4.5]! Well that is a compelling point. Any number smaller than 2 will not work because it doesn’t fit the domain of g(f(x)). [Actually, numbers less than 2 don't fit in the domain of f. Remeber f(x) has to be in the domain of g, and x has to be in the domain of f ] So therefore, our original domain is correct.

So that was kinda difficult. That’s OK, Mr. B. said today that we will only be responsible for linear functions on the test. We won’t encounter any crazy situations where some number will make the graph go out of whack. So remember, TEST TUESDAY. And there problems 3 and 8 from section 1-4 are up for claims tomorrow.

Thine scribeth for the day of next shalt be Britney.

Today in class we went over the Composition of Functions worksheet that we worked on in the computer lab yesterday. Mr. Bieniek started off by describing the first page that had the f(x), g(x), and h(x) function dynapgraph. He described that the output of f(x) in the dynagraph is the input of g(x) and that the output of g(x) is the input of h(x). This is because the value f(x) produces has to produce for the next inside function. The next inside function then produces a value for the next function which is an outside function which will lead to the final result.

Alright, just to warn you, my post won’t be as amusing as Kaitlyn’s or Shelby’s but I’ll try my hardest.

Today in class, was spent in the computer lab. We had an assignment to do in Geometer’s Sketchpad exploring different composite functions. In this assignment, we were given numerous different composite functions and had to answer questions about them. Just as a reminder, composite functions are when the output of one function is used as the input of another function.

Our first exploration was using composite functions to observe how CO2, temperatures and melting were related to ice caps. First, what we had to do was combine the two fuctions ( CO2 and Temperature; Temperature and Melting Rate ) in a dynagraph, which is basically a visual representation of functions. After we did that, we were able to see clearly how the two functions acted together. We also had to figure out if the dynagraphs showed an f(g(x)) or g(f(x)) function.

The second exploration was about domains and ranges of composite functions. We were given yet another dynagraph and were told to combine the two functions. But, this time, after they were connected we were told to try and find the domains and ranges. To find the domain of a composite function, we had to find which numbers worked for both of the functions (what numbers where in both of the functions domains). The range was the ‘B’ values on the dynagraph. We were also told to manipulate the functions to get different results for the domain and range.

Mr. B then reminded us that section 1-4 problems were due for tomorrow. Tomorrow’s scribe will be Adam.

As the bell rang for class to begin, Mr. B gave us brief instuructions and told us to find our 2 o’clock partners. We then enjoyed a change of scenery by moving to the lab to complete another sketchpad activity.

Once we logged on and pulled up the sketch pad, there were instructions and explorations awaiting us. Today’s theme was composite functions, in an attempt to use viuals to better understand them. In case a few of you forgot, a composite function is when the output of one function is used as an input for another function; the result, a composite function.

We used a dynagraph as a visual to see how functions work for certain values. The dynagraph was used as a visual aid to decide whether graphs were (f(x)) or f(g(x)). More of these graphs were used to figure out similar problems. We did this to see how the graphs relate when put together.

The next part of the activity was about the domain and the range when dealing with composite functions. From this we learned that the function inside the parenthases will affect the outcome of the domain and range of the outside parentheses. The numbers that fit within both domains of the two or more functions will be the given domain for the composite function. The same principles apply for the range. Anything outside of this domain or range will leave the function undefined.

Though may of us did not complete the assignment, it still provided a better insight on composite functions and how they work. The scribe for tomorrow will be Emily.

I wanted to hand these out “live” but it slipped my mind. Here they are… MQ1 Learning Targets

Today we looked into comosite functions in depth. Many people I think found this a confusing topic but after a while everyone seemed to be on the same page. First we began with a white handout that gave us two functions both g(x)=10-2x with a domain of 1<x<4 and f(x)=x+2 with a domain of 3<x<7. We had to find what values of x would be able to meet both domain restrictions. The issues we had were 1. g(x) has to be in the domain of f(x) and 2. x has to be in the domain of g(x). After looking at this in various circumstances we were able to draw a line on the graph on the white handout that meet all of the restictions called f o g.

~The next scribe will be Juliana

Today, we started off by looking at a worksheet with two functions on it. These functions were also graphed. Our objective was to find f(g(3)). To find this, you must plug 3 into g(x) first, and then plug the answer, 4, into f(x) which gives you the answer of 6. Then we looked at other points that could fit the domain of both the functions. The first test a number must pass to become defined is whether or not it fits the domain of g. 3 happens to fit into the domain of greater than or equal to 1, and less than or equal to 4. Once three has been transformed into 4, it must then fit the domain of f(x). 4 also fits into the domain of x is greater than or equal to 3 and less than or equal to 7. We then created a table to find other points that would fit into these restrictions. Though we found that 1.5 through 3.5 also worked, we decided that we could be missing points, and the best way to find all the points was to figure out the domain and range algebraically. We came up with 3 ≤ g(x) ≤7, which equals 3≤ 10-2x≤7. We solved this by subtracting 10 from both 3 and 7, and then by dividing both sides by 2 giving us the equation 7/2 ≤ x ≤ 3/2. This also proves our t-chart from before correct, as 7/2 is 3.5 and 3/2 is 1.5. After a thorough understanding of this subject, we went through more examples using the same functions. We also learned that a function can be taken to itself as in g(g(x)) or f(f(x)). Our homework was to read 1.4 if we hadn’t already, and do the section problems for thursday.

The next scribe is Hannah.

Today we started off discussing our homework from this weekend, which was to find different ways to make a curved graph into a straight line. We only looked at two: (x squared,y) and (x, square root of y). Both of these worked. We learned how to tell if they were really straight lines, not just by looking at them but by picking a couple points on the line and tell if the slope is constant. We did this by finding the change in x over the change in y. People believe that those two worked because they both have a two to one ratio.

We also started talking about compositions of functions. They are kind of hard to explain , but here is an example(the one from class):

Lets call the original data f(x) and the square root function g(x)

When you apply the square root functionto the y values of f(x), then you write it like this: g(f(x))

These can be written and said different ways:

g(f(x)) – “g of f of x” , g o f – ” g of f”, or g o f(x) – “g of f of x”

We also tried to change a direct variation function(f) into a negative parabala with a y intercept of two(g). Translations and dialations wouldn’t work. So, we used different rules:

-Rule for vertical translation( to move the line up 2): v(x)

-Rule for squaring(to get a u shape): s(x)

-Rule for negating(making the slope negative): n(x)

So f(x)= x, s(f(x))= x squared, n(s(f(x)))= -x squared, and finally g(x)=v(n(s(f(x))))= -x squared + 2

This is a transformation, and that is as far as we got today. The next scribe is Tyler.