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Homework Help: What is the condition for fucntion fg^(-1) to exist ?

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data

    what is the condition for fucntion fg^(-1) to exist ?

    2. Relevant equations



    3. The attempt at a solution

    i understand that g^(-1) only happens when its one-one , how about fg^(-1) ?
     
  2. jcsd
  3. Apr 7, 2010 #2
    Re: function

    Are you thinking of [tex](f \circ g)^{-1} = g^{-1} \circ f^{-1}[/tex] ?

    Then I think this is what you are looking for
    http://planetmath.org/encyclopedia/InverseOfCompositionOfFunctions.html [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Apr 7, 2010 #3

    HallsofIvy

    User Avatar
    Science Advisor

    Re: function

    [itex]fg^{-1}[/itex] is only defined on the intersection of the domains of f and g.
     
  5. Apr 7, 2010 #4
    Re: function

    We know that HallsoftIvy, but you think since this is a pre-Calculus question that the original poster is simply looking for and try to understand the definition of

    [tex]f \circ g^{-1}[/tex]?
     
  6. Apr 7, 2010 #5
    Re: function

    thanks for helping , i mean [tex]fg^{-1}[/tex] , because i had this previous example , where i am supposed to prove that function gf exist , so the range of f is a subset or equal the domain of g . I wonder if i could applying the same reasoning for this function .
     
  7. Apr 7, 2010 #6

    Mark44

    Staff: Mentor

    Re: function

    You aren't being clear in what you're asking. Which of the following are you asking about?
    • fg-1, the product of f and g-1
    • (fg)-1, the inverse of the product of f and g
    • f [itex]\circ[/itex] g-1, the composition of f and g-1
    • (f [itex]\circ[/itex] g)-1, the inverse of the composition of f and g
     
  8. Apr 7, 2010 #7
    Re: function

    sorry, i meant number 3 .
     
  9. Apr 7, 2010 #8

    Mark44

    Staff: Mentor

    Re: function

    When who is one-to-one? See if you can restate your question in a more precise way.
     
  10. Apr 7, 2010 #9
    Re: function

    ok i will look for an example .

    The function f and g are defined by f(x)= ln x , x are all positive real numbers and g(x)=x^2-1 , x is all positive real numbers . Determine whether the composite function g^(-1) o f exists . Find the restricted domain of f such that the function exists .
     
  11. Apr 7, 2010 #10

    Mark44

    Staff: Mentor

    Re: function

    For this example, the restricted domain for g makes it a one-to-one function, so the inverse, g-1, exists.

    For g-1 o f, you need to look at the domain of f (x such that x > 0), and the range of f. The outputs from f are going to be the inputs to g-1. Are there any numbers in the range of the ln function that are not allowed in the domain for g-1?
     
  12. Apr 8, 2010 #11
    Re: function

    the range of f is y such that [tex]y\in R [/tex]

    the domain of g^(-1) is the range of g which is (-1 , infinity)

    since the [tex]R_f[/tex] is not equal or a subset of [tex]D_{g^{-1}}[/tex] , [tex]g^{-1} \circ f[/tex] doesn't exist .

    for the function g^(-1) o f to exist , range of f has to be (-1 , infinity)

    Am i correct ?
     
  13. Apr 8, 2010 #12

    Mark44

    Staff: Mentor

    Re: function

    Yes. Now, can you figure out how to restrict the domain of f so that the range of f is {y | y > -1}?

    IOW, what restrictions can you place on the domain of ln(x) so that the range is {y | y > -1}?
     
  14. Apr 9, 2010 #13
    Re: function

    domain of f is (1/e , infinity) ?
     
  15. Apr 9, 2010 #14

    Mark44

    Staff: Mentor

    Re: function

    Yes, good!
     
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