As we walked into class today, the daily problem on the board was a bit different from the ones we have had in the past. It was different because in the problem, two new operations were created: up arrow(ua) and down arrow(da). The problem was if a ua b means a^b and if a da b means ^b(radical symbol) a, what is the value of [(2 ua 6) da 3] ua 2?

To simplify this equation, one would replace the arrows with the given operation. By replacing the arrows with the operation they represent, the expression reads (³(radical symbol)2⁶)². Now we simplify everything in the parentheses first. 2⁶=64. ³(radical symbol)64 = 4. This just leaves us with 4² which equals 16. So the final answer to this problem was 16. The class worked well and efficiently and even put the problem up on the board to explain to the students who did not understand the problem.

The addition of two new operations made this problem confusing to most of the class. However, most people understood the problem after it was explained and drawn on the board.

After the practice problem was explained and talked about in a class discussion, Mr. Bieniek passed back our assessments that were taken at the end of yesterday’s class. The assessment was about whether or not a given set of inputs and outputs was a function.  The function’s inputs, a, b, and c all led to the same output, 1. We eventually came to the conclusion that the set of inputs and outputs was, in fact, a function. We referred again to the soda machine example. Mr. Bieniek explained that a soda machine can have multiple buttons(inputs) that will get you the same soda(output) and it will still “function” properly. However, if  a button has two outputs or two possible soda outcomes, the soda machine will not function. A question that came up was that if one input did not have an output, could the system be a function. The answer to this was no. In the soda machine example, it was said that if a button does not work and doesn’t give you an output or soda, the soda machine is “not functioning” properly. It was also noted that this function would create a horizontal line on a graph.

We also discussed what domain and ranges are. A domain is all the possible inputs or x-values. A range is all the possible outputs or y-values.

Next we were introduced to piecewise functions. When Mr. Bieniek put the first piecewise function on the board, much of the class was intimidated by it. The first piecewise function we received was:

f(x)= (x+4)²-2       If x is less than or equal to -2

-x                    If x is greater than -2 and less than or equal to 2

-(x-4)²+2      If x is greater than 2

Mr. Bieniek said that the best method to solve this problem was by making a table and then graphing it. He stressed that this was not the best way necessarily to graph a piecewise function all the time. The table that we made in class can be viewed by clicking on the link below

http://public.iwork.com/document/?a=p1017344002&d=Untitled.numbers

So now all what was left to do was to graph the points here and connect the points depending on the inequalities.

Next, we received another piecewise function. However, this time the class was told by Mr. Bieniek to try and graph this function without making a table. The piecewise function was as shown below.

f(x)=x² If x is less than 2

f(x)=6               If x=2

f(x)=10-x         If x is greater than 2 and less than or equal t0 6

The class graphed the equation fairly easily. After, we participated in a class wide discussion about the function. It was found that the f(x)=6 part of the function was confusing but after it was discussed, everybody understood it meant to graph the point (2,6). Also, there was controversy on whether to connect the parabola created by f(x)=x² to the line created by f(x)=10-x. The line would go through the point (2,6). After a discussion, it was determined there should not be line because it would be a vertical line, meaning that two or more inputs would have the same output. This is not possible in a function so it would be illogical to connect the parabola and the line

After this, we received two piecewise function problems for homework and class was dimissed.

For help with piecewise function, see the following sites

http://www.analyzemath.com/Graphing/piecewise_functions.html- A site that gives tutorials for piecewise functions

http://www.youtube.com/watch?v=Kac2I5ao49E- A video about piecewise functions

http://www.prenhall.com/divisions/esm/app/graphing/ti83/Graphing/Special_graphs/piece_wise/piece_wise.html- A site that shows how to graph piecewise functions on your TI calculators