Discrete Math
posted by Francesca .
Find g ° f
f(x) = {(1, 2), (2,1), (3,1), (4, 4)}
g(x) = {(1, 2), (2,4), (3, 1), (4, 3)}

Can you check if there is a typo?
As it is, f(x) is not onto, i.e. f(2)=f(3)=1.
To find g°f, complete the following table:
x u=f(x) g(u) = g ° f(x)
1 2 4 = g°f(x)
2 1 2
3 1 2
4 4 3
Therefore
g°f : {(1,4),(2,2) ....}
As supplementary information, read up
http://en.wikipedia.org/wiki/Function_composition 
Well, I am suppose to find the composition of functions from a figure. If you don't mind I uploaded a photo of it on photobucket. Can you take a look, and offer any suggestions for the first problem, so that I can get an idea? Since this forum will not let me post a link I can give directions on how to find the photo. First, go to search bar and type: flutegirl516. Then, this message will appear: Are you looking for the Photobucket user flutegirl516? Click on this. Then you will see the photo album. There is only one picture.
Also, click "View as slide show" to make it larger. Tell me if you can view it okay. Thanks for any helpful replies :) 
I read the the link you provided previous to posting this discussion, but I was still confused.

Sorry, I am not able to find the link according to your directions, probably we have different search bars.
You can try posting the link that following the prefix
http://media.photobucket.com/
and I can append it myself. 
I'm using the search bar at the very top righthand corner, then I enter: flutegirl516. It will say no matches found, but it will say: Are you looking for the Photobucket user flutegirl516?
You can also try using the search bar drop down menu and clicking on "Users," and then enter flutegirl516. The album should pop up. Try it, hopefully you will be able to view it. 
Yes, indeed, I found it when I used Photobucket's search bar instead of the browser's search bar.
For reference, here's the link:
http://s1193.photobucket.com/albums/aa348/flutegirl516/?action=view¤t=Screenshot20110215at61144PM.png
You have correctly represented the two functions, so there is no typo.
In case it is not clear how composition works, you only have to imagine that the range of f(x), i.e. the right part of f(x), is merged with the domain of g(x), or the left part of g(x).
You can then follow the arrows from the domain of f(x) to the codomain of g(x). For example, f(1)=2, g(2)=4, which is exactly what is in the table I posted earlier, namely
g°f=g(f(x))=g(f(1))=g(2)=4
It is easier than you can imagine! 
ooOo I think I get it now. . .

Great!

Ok so
g ° f = {(1, 4), (2, 2), (3, 2), (4,3)}
There are two 1s in the range of f(x) though (1,2) and (3,1). . .Does that mean anything?
Is this correct:
Does g ° h = {(1, 2), (2?, 2?), (3, 2), (4, 1)}? For this one there was no 2 in the range of h.
Also, for h^2 = h ° h, what am I suppose to square? Domain (x) or Range (y). 
"There are two 1s in the range of f(x) though (1,2) **I meant to say (2,1)** and (3,1). . .Does that mean anything?" Anyway, scratch this statement of the previous post. . .

Here are the functions from the figures:
• f(x) = {(1, 2), (2,1), (3,1), (4, 4)}
• g(x) = {(1, 2), (2,4), (3, 1), (4, 3)}
• h(x) = {(1, 1), (2, 3), (3, 1), (4, 3)} 
Yes, the (2,1) and (3,1) mean something.
You will notice that f(x) does not map to the number 3. This makes f(x) not surjective (or "onto"). If it were, all the elements in the codomain {1,2,3,4} would be mapped by f(x).
As another example, g(x) is bijective, because it is onetoone and onto (or injective and surjective).
For g°h, you just have to create a similar table as in f°g:
x h(x) g°h=g(h(x))
1 h(1)=1 g(h(1))=g(1)=2
2 h(2)=3 g(h(2))=g(3)=1
3 h(3)=1 g(h(3))=g(1)=2
4 h(4)=4 g(h(4))=g(4)=3
Note: It is OK that there is no element 2 in the range of h, just like there is no element 3 in the range of f.
You can give it a try for h°h in a similar way. 
Some reading about surjection:
http://en.wikipedia.org/wiki/Surjective_function 
So is this correct?
g ° f = {(1, 4), (2, 2), (3, 2), (4,3)} 
There is something not connecting in my thought process. I'm still confused with g ° h does it = {(1,2), (3,1), (1,2), (4,3)}? I feel way off. . .I am doing something terribly wrong. IDK

Correct!
WOuld you like to try g°h again?
This one was not correct:
"Does g ° h = {(1, 2), (2?, 2?), (3, 2), (4, 1)}? For this one there was no 2 in the range of h. "
Hint: (g°h)(2)=g(h(2))=g(3)=1 
"Correct!" applied to g ° f = {(1, 4), (2, 2), (3, 2), (4,3)}
not g°h, sorry.
FOR G°h:
To do g°h, you can draw it graphically (at least for now), with the domain of h on the left, the range of h in the middle, and the range of g on the right.
Between the left and the middle, draw arrows according to h(x), and between the middle and the right, draw arrows according to g(x).
Start from any value x0 on the left, travel to the right through the middle according to the arrows, where you arrive is (g°h)(x0), or g(h(x0)).
Your previous attempt was almost correct, just (2,2) was not. 
Is this correct?
g ° h ={(1,2), (3,1), (1,2), (4,3)} 
g ° h ={(1,2), (3,1), (1,2), (4,3)}
is correct.
Hope you get it now, and ready to try h²! 
g ° h ={(1,2), (3,1), (1,2), (4,3)}
is not correct. I went too fast.
(g°h)(1)=g(h(1))=g(1)=2
(g°h)(2)=g(h(2))=g(3)=1
(g°h)(3)=g(h(3))=g(1)=2
(g°h)(4)=g(h(4))=g(3)=1
So g°h : {(1,2),(2,1),(3,2),(4,1)}
Hope you're ready to try h²! 
Lol. . .But really Thank You! You are really helping me to understand. I know you are probably annoyed by my silly questions, but I am really starting to get a better understanding.
Ok so would h ° h = {(1,1), (2,1), (3,1), (4,1)}? But the h² is throwing me off. 
(h°h)(1)=h(h(1))=h(1)=1
(h°h)(2)=h(h(2))=h(3)=1
(h°h)(3)=h(h(3))=h(1)=1
(h°h)(4)=h(h(4))=h(3)=1
Great,you got the correct answers!
(Even though they appear a little strange). 
They appear a little strange. . .how so? The h² does not effect the answer at all?
OK so last one f ° g ° h. . .I will attempt now, and check back and see if I am on the right track. . . 
Good!

Oh it looks strange b/c they all end in zero. Just noticed. . .
Is this correct?
f ° g ° h = {(1,1), (2,2), (3,1), (4,2) 
Hey, I know I said the previous would be the last one but can you check this one too. . .
Let f: A→B be a function from A to B. f = {(w, 1), (x, 2), (y, 3), (z, 2)}. Find f^1.
Answer: f^1 = {(1, w), (2, x), (3, y), (2, z)} 
Perfect!
I forgot to mention that the expression f°g°h is associative, which means that you can evaluate it as f°(g°h) or (f°g)°h. This is why the parentheses have been left out.
If you want extra practice, try both and verify the associativity. 
The function A>B is surjective, but not injective. The domain has more elements than codomain.
The inverse relation B>A is not a function, since f^{1}(x) is undefined when x=2, namely f^{1}(2)=x or z?
Under these circumstances, the inverse does not exist. 
How about this?
Let f: A→B be a function from A to B. f = {(w, 1), (x, 2), (y, 3), (z, 2)}. Find f^1.
Answer: f^1 = {(1, w), (2, x), (3, y), (2, z)}
So would this one not have an inverse? 
For further reading, see:
http://en.wikipedia.org/wiki/Inverse_function 
Oh okay we posted at the same time. . .You were such a big help! I really understand this stuff a lot better. Thank you a thousands times!

Also thank you for the time you took out to help me :)

The basic definition of a function is f(x) must have a unique value for a given x.
A> satisfies this requirement, but since it is not onetoone (injective), it maps both x and z to 2, which does not stop f(x) from being a function.
However, if you draw the diagram, you will see that the inverse function, if it exists, will have f^{1} pointing to both x and z, which makes f^{1} not a function. Therefore under these circumstances, f^{1} is not defined. 
Keep up the good work, and post if you have other questions.
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