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December 18, 2014

December 18, 2014

Posted by **Francesca** on Tuesday, February 15, 2011 at 2:53pm.

f(x) = {(1, 2), (2,1), (3,1), (4, 4)}

g(x) = {(1, 2), (2,4), (3, 1), (4, 3)}

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 5:27pmCan 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

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 6:31pmWell, 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 :)

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 6:33pmI read the the link you provided previous to posting this discussion, but I was still confused.

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 6:52pmSorry, 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.

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 7:00pmI'm using the search bar at the very top right-hand 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.

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 7:12pmYes, 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=Screenshot2011-02-15at61144PM.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!

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 7:27pmooOo I think I get it now. . .

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 7:31pmGreat!

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 7:56pmOk 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).

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 7:59pm"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. . .

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 8:18pmHere 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)}

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 8:54pmYes, 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 one-to-one 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.

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 8:57pmSome reading about surjection:

http://en.wikipedia.org/wiki/Surjective_function

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 9:29pmSo is this correct?

g ° f = {(1, 4), (2, 2), (3, 2), (4,3)}

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 9:44pmThere 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

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 9:46pmCorrect!

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

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 9:54pm"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.

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 9:56pmIs this correct?

g ° h ={(1,2), (3,1), (1,2), (4,3)}

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 9:59pmg ° h ={(1,2), (3,1), (1,2), (4,3)}

is correct.

Hope you get it now, and ready to try h²!

- Discrete Math-correction -
**MathMate**, Tuesday, February 15, 2011 at 10:04pmg ° 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!

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 10:14pmLol. . .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.

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 10:24pm(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).

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 10:29pmThey 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. . .

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 10:32pmGood!

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 10:41pmOh 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)

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 10:45pmHey, 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)}

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 10:48pmPerfect!

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.

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 11:01pmThe 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.

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 11:02pmHow 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?

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 11:03pmFor further reading, see:

http://en.wikipedia.org/wiki/Inverse_function

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 11:04pmOh 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!

- Discrete Math -
**Francesca**, Tuesday, February 15, 2011 at 11:06pmAlso thank you for the time you took out to help me :)

- Discrete Math -
**MathMate**, Tuesday, February 15, 2011 at 11:07pmThe 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 one-to-one (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.

- Discrete Math :) -
**MathMate**, Tuesday, February 15, 2011 at 11:08pmKeep up the good work, and post if you have other questions.

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