Composition Functions: Finding f(g(x)) for F(x)=1/(1+x) and G(x)=1/(2+x)

AI Thread Summary
To find the composition f(g(x)) for F(x) = 1/(1+x) and G(x) = 1/(2+x), the expression simplifies to (2+x)/(3+x). The discussion highlights that x = -3 and x = -2 are points where the function is undefined due to the domains of F and G. There is confusion regarding why x = -1 is also considered forbidden, but it is clarified that only G(x) = -1 is problematic. The conversation emphasizes the importance of considering the domains of both functions in composition. Understanding the domain and range of composition functions is crucial for solving these types of problems effectively.
Coco12
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Homework Statement



If F(x)= 1/(1+x)
G(x)= 1/(2+x)
determine f(g(x))

Homework Equations



No equations for composition functions

The Attempt at a Solution


F(g(x)) = 1/1+ 1/(2+x)

Then do you try to eliminate the fraction in the denominator by multiplying it by (2+x)?

They said that the x values that would not work are -3,-2,-1. I understand where they got -3 and -2 but where did they get -1?
 
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Hi Coco12! :smile:
Coco12 said:
They said that the x values that would not work are -3,-2,-1. I understand where they got -3 and -2 but where did they get -1?

Let's see … you got 1/(1+ 1/(2+x))

(pleeease use more brackets … you confused me at first :redface:)

= 1/((3+x)/(2+x))

= (2+x)/(3+x) which is not defined if x = -3
(and G(x) was not defined anyway at x = -2 )

Yes i agree :smile:, i don't see why x = -1 is forbidden.
 
tiny-tim said:
Hi Coco12! :smile:Let's see … you got 1/(1+ 1/(2+x))

(pleeease use more brackets … you confused me at first :redface:)

= 1/((3+x)/(2+x))

= (2+x)/(3+x) which is not defined if x = -3
(and G(x) was not defined anyway at x = -2 )

Yes i agree :smile:, i don't see why x = -1 is forbidden.
Wait how did you get 1/(3+x)(2+x)??
 
Coco12 said:
Wait how did you get 1/(3+x)(2+x)??

(what is it with you and brackets? :redface:)

i didn't, i got 1/((3+x)/(2+x))

anyway, what did you get?​
 
Never mind I got it. Thank you
 
If F(x)= 1/(1+x)

This would be why -1 is forbidden.
 
where did you get (1+x)??
 
Student100 said:
If F(x)= 1/(1+x)

This would be why -1 is forbidden.

But we're not interested in F(x), only in F(G(x)),

and only G(x) = -1 is forbidden (ie x = -3)
 
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Coco12 said:
where did you get (1+x)??

1. Homework Statement

If F(x)= 1/(1+x)
G(x)= 1/(2+x)
determine f(g(x))

Is this the problem? That's your f function. There is a thoerm that states the domains of both functions need to be considered in a compoistion. Unless I'm making this up in my head.
 
  • #10
I thought you only need to consider the domain of the inner function and the overall function.

Look at the above video
 
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  • #11
Yeah I just looked it up; I guess I was thinking about something else. Sorry for the confusion, I'll slump back to my little corner now. :)
 
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  • #12
no problem. Do you know anything about the in and out method of solving for the domain and ranges?
 
  • #13
Never heard of it, I'm probably a bit rustier than I should be with all this.
 
  • #14
Coco12 said:
no problem. Do you know anything about the in and out method of solving for the domain and ranges?
I haven't heard of it being called that, but for the composition f(g(x)), x has to be in the domain of g, and the range of g is the domain of f.
 
  • #15
I have another question related to that in the forums. Can you please see if you can help me? How do u find the domain and ranges of composition function
 
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