I don't understand how to solve this question below:

Suppose g:A-->B and f:B-->C where A=B=C={1,2,3,4}, g={(1,4),(2,1),(3,1),(4,2)} and f={(1,3),(2,2),(3,4),(4,2)}.

Find g of f

Any help with this would be very much appreciated.

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- Jul 30th 2009, 10:51 PMkro{Solved} Need help finding g of f
I don't understand how to solve this question below:

Suppose g:A-->B and f:B-->C where A=B=C={1,2,3,4}, g={(1,4),(2,1),(3,1),(4,2)} and f={(1,3),(2,2),(3,4),(4,2)}.

Find g of f

Any help with this would be very much appreciated. - Jul 30th 2009, 11:49 PMSwlabr
Basically, here you have to follow a trail. Take 1 and see where it is mapped to, then look at 2, then 3 then 4. In g(f(x)) we apply f to x and then we apply g to the result. So, if x is 1 then f(x) is 3 and g(f(x)) is 1. Similarly, 2 is sent to 1. Thus, $\displaystyle g \circ f = \{(1,1)(2,1)(3,a)(4,b)\}$ ($\displaystyle (g \circ f)(x) = g(f(x))$). Can you see how to find $\displaystyle a$ and $\displaystyle b$?

- Jul 31st 2009, 12:17 AMkroI think I got it?
Swlabr,

Here's what I have so please let me know if this is correct:

f(1)=3 so g(f(x))=1

f(2)=2 so g(f(x))=1

f(3)=4 so g(f(x))=2

f(4)=2 so g(f(x))=1

so......f of g = {(1,1),(2,1),(3,2),(4,1)}

Is this correct? - Jul 31st 2009, 12:30 AMSwlabr
- Jul 31st 2009, 12:37 AMkroThanks alot!
Ohhh yeah....I meant g of f.

Thanks alot for you help......FYI, I'm about to post another question that's kinda similar to this one except I have to find: f of f^-1