Category: 8-3

Compostition of Functions and how to solve algebraically

Student 08_09
Sep 25 |22:28

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.

 

Gray (grey?) area

Reversearp
Mar 9 |20:52

The oak tree data from today is not clearly modeled by any particular function. We have to do the best we can to choose the most appropriate model. Is the coefficient of determination acceptable? Is the end behavior appropriate? If we extrapolate, do we get a reasonable value? You may not find a function that is perfect in all areas but taking the context of the data into consideration should help you weigh the pros and cons of each model you consider.

By the way, did anyone try squaring the y-values…?

 

Power function = log, log

Reversearp
Mar 8 |20:30

Today we looked at an exponential function, and like yesterday it was linearized by taking the log of the y-values. Working through the algebra is really as simple as using the definition of logarithm to rewrite the log y = ax + b as a power of 10 exponential equation and simplifying.

The second problem was a power function that was linearized by taking the log of both x and y. This gave us a messy equation: log(y) = a⋅log(x) + b. Our (my) first thought was to raise both sides to the power of 10. This left us with logs on both sides of the equation – sort of like leaving yourself with x’s on both sides of an equation. With that said, our next attempt should probably be to get both logs together on one side and go from there. Can anyone do it for tomorrow?

 

Massaging the data

Reversearp
Mar 7 |11:17

We came to some important conclusions today. First, we discovered that for anything other than linear regression your calculator computes an R2 value instead of the normal r2. The difference is that R2 is calculated using the line of least squares that fits the transformed, “linearized” data – not the actual curve that best fits the original data. If you want the r2 value for the original data using the curve of best fit you must calculate by hand.

So, how do you find the best fit curve for non-linear data? You must transform one or both of the variables by squaring, or taking the the log, or the square root, or….? The goal is to plot the transformed points and see a line. Then running linear regression we can find the slope and y-intercept of the best fit line for the transformed data. Then re-write your y = mx + b form substiuting log(y) or x2 (or whatever transformation you made) in for x and y. Solve for y and we have the curve.

It is not a random process deciding on the transformation to use. Today we looked at exponential data and wound up taking the log of the y-values. That is not a coincidence that we “un-did” exponents with logs. Work with the problems for tomorrow and you will see some other techniques and where to use them.