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Topic: Defining "f o g" with Functions
Replies: 4   Last Post: Dec 31, 2012 8:19 AM

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Angela Richardson

Posts: 42
From: UK
Registered: 6/22/11
Re: Defining "f o g" with Functions
Posted: Oct 15, 2012 3:17 AM
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f={(a,b)} implies f(a)=b
f o g (x)=f(g(x))
f o g =(1,6), (2,4),(3,7),(4,4),(5,3) since g maps 1 to 2, which f maps to 6, and g maps 2 to 3, which f maps to 4 etc.
g is not one-to-one since both 2 and 4 are mapped to 3. f is one-to-one since no two ordered pairs have the same number as either their first or second element. A function has an inverse if and only if it is one-to-one. f^-1={(6,2),(7,1),(4,3),(3,5),(5,4)} which is the ordered pairs with the first and second elements swapped.


________________________________
From: Edward <discussions@mathforum.org>
To: discretemath@mathforum.org
Sent: Sunday, 14 October 2012, 18:32
Subject: Defining "f o g" with Functions

So I've been learning about functions. Ideas such as one-to-one, onto, bijections....and so on. Things have been going smoothly and I've been going on my own through the text trying to do all of the exercises to prove to myself I understand the concepts on my own but I came into a problem where I don't know where to begin. I'm not really familiar with "g o f" and "f o g". Can somebody show me how to figure this out:

Let S = {1,2,3,4,5} and let T = {3,4,5,6,7}. Define function f : S --> T and g : S --> S as follows:
f = {(2,6), (1,7), (3,4), (5,3), (4,5) and
g = {(2,3), (1,2), (4,3), (5,5), (3,1)}.

(a.) Find "f o g" or explain why "f o g" is not defined, Repeat for "g o f", "f o f" and "g o g".
(b.) Which (if any) of f, g are one-to-one? Which (if any) are onto? Explain
(c.) Find f^-1 if it exists. If not explain why.
(d.) Find g^-1 if it exists. If not explain why.

*** ^ represents exponent ***

Anybody who understands these concepts better than I do?



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