What is a Function?

A Function is a special type of relation where each x-value in the domain corisponds to

**1 y-value in the range.**

*only*EG.

2 ==>8

0 ==>4

==>6

6 ==>3

Ordered Pairs: (2,8),(0,4),(0,6),(6,3)

As you can see in the ordered pairs the x-value of 0 has two different y-values. THIS IS AN EXAMPLE OF A RELATION,

**A FUCTION.**

*NOT*That was an example of using the ordered pairs. What if we were given a graph? Why use the

*Vertical Line Test!*What is the vertical line test? Well its the test used on graphs to see if it is a function or a relation. If the line passes through two points or more on the graph then it is a relation. If it passes through only one point then it is a Function.

Here are a few Examples from class.

You know that this is a function right? The vertical line test shows it is. Anywhere on the graph if you drew a vertical line there it would only go through one point.

On to the next graph.

This is also a function. No line going through two points

This is a relation. Not a function. As you can see there is a vertical line in the graph. If you put a vertical line through that it over laps. Which means that its going through more then one point. If you drew it anywhere else it would not hit more then two points, but because of that vertical line the graph made it is considered a relation.

This one is in the same boat as the last one. It is a relation becaues there are three points in a line. If you drew a vertical line there it would hit all three. Therefore it is a relation.

Okay I know that this is gettign repetitive right about now, but now moving on to the other topic that was covered on wednesday.

*Function Notation*Functions are usually represented with the letters "

**F**" or "

**G**"

Eg. f (x) = x + 2

This reads "The value of the function at this value of x is x +2"

Lets try doing a question from class , the one above. Well i think there is suppose to be a stament about what you switch f (x) with...but it seems i dont have it, I am very sorry for this. I do have the work though

f (x) = x + 2

f (3) = 3 + 2 ====> What ever you switch the x with on the right side you switch all the x's on the left side with the same thing and just evaluate left side. I am really sorry for how i said that and i know it should be something else mathematical, but i really dont know any other way to day it. ANYWAYS...continuing on.

f (3) = 5

Okay that was the easy one. Now we go into Compound Functions.

Let f (x) = x

^{2}/ x + 1 and g (x) = 4 / x

^{2}

1.f (g

^{(2)})

This may seem confusing at first but it really isnt that hard. You work your way inside out. You start with the g (2) function and find that. the you use that in the f function. Heres the work.

Oh we are replacing the x value with 2

g (x) = 4 / x

^{2}

g (2) = 4 / 2

^{2}

g (2) = 4 / 4

g (2) = 1

After you get that done just your answer you got from the g function and use it in the in function.

so instead of this.

f (g

^{(2)})

It will be this.

f (1)

To finish the problem your going finish with the f function

f (x) = x

^{2}/ x + 1

f (1) = 1

^{2}/ 1 + 1

f (1) = 1 / 2

f (g

^{(2)}) = 1 / 2

And thats Wednesday's class. I picked don already to scribe and it should be up today too so i dont need to pick one again.