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Composition functions

  1. Dec 13, 2013 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    No equations for composition functions

    3. 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?
     
  2. jcsd
  3. Dec 13, 2013 #2

    tiny-tim

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    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.
     
  4. Dec 13, 2013 #3

    Wait how did you get 1/(3+x)(2+x)??
     
  5. Dec 13, 2013 #4

    tiny-tim

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

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

    anyway, what did you get?​
     
  6. Dec 13, 2013 #5
    Never mind I got it. Thank you
     
  7. Dec 13, 2013 #6

    Student100

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    If F(x)= 1/(1+x)

    This would be why -1 is forbidden.
     
  8. Dec 13, 2013 #7
    where did you get (1+x)??
     
  9. Dec 13, 2013 #8

    tiny-tim

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

    and only G(x) = -1 is forbidden (ie x = -3)
     
    Last edited: Dec 13, 2013
  10. Dec 13, 2013 #9

    Student100

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    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.
     
  11. Dec 13, 2013 #10
    I thought you only need to consider the domain of the inner function and the overall function.

    Look at the above video
     
    Last edited by a moderator: Sep 25, 2014
  12. Dec 13, 2013 #11

    Student100

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    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. :)
     
    Last edited: Dec 13, 2013
  13. Dec 13, 2013 #12
    no problem. Do you know anything about the in and out method of solving for the domain and ranges?
     
  14. Dec 13, 2013 #13

    Student100

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    Never heard of it, I'm probably a bit rustier than I should be with all this.
     
  15. Dec 13, 2013 #14

    Mark44

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    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.
     
  16. Dec 13, 2013 #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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