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Math Help - Set Problems

  1. #1
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    Set Problems

    Using ordered-pair notation, let g be the function defined by
    g = {(5n, 3n) | n element of Z}
    a) State the domain and a reasonable codomain of g.
    b) Evaluate g(10), g(g(50)), and g^-1(-6).

    Isn't the domain all real numbers?

    How do I go about evaluating and finding the domain and codomain
    Last edited by xmathlover; April 21st 2013 at 09:45 AM.
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  2. #2
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    Re: Set Problems

    Quote Originally Posted by xmathlover View Post
    Using ordered-pair notation, let g be the function dened by
    g = f(5n, 3n) | n element of Z
    a) State the domain and a reasonable codomain of g.
    b) Evaluate g(10), g(g(50)), and g^-1(-6).
    Isn't the domain all real numbers?
    How do I go about evaluating and finding the domain and codomain

    What you have posted leaves a great many questions.

    What in the world is the definition of f(5n,3n)~?

    Do you mean g(n)=f(5n,3n)~?.

    Where do ordered pairs come in to it at all?

    Please repost a meaningful statement.
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  3. #3
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    Re: Set Problems

    I fixed it.

    {5n, 3n | n elements of Z}
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    Re: Set Problems

    Quote Originally Posted by xmathlover View Post
    I fixed it.

    {5n, 3n | n elements of Z}
    If g=\{(5n,3n):n\in\mathbb{Z}\} then the domain of g is the set of multiples of five.

    The image set (final set or codomain) is the set of multiples of three.

    EX: g(10)=g(5\cdot 2)=3\cdot 2)=6
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  5. #5
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    Re: Set Problems

    How did you come to that conclusion? do you look at the ordered pairs as x input, y output?

    How would I evaluate g(g(50)) and g^-1(-6)?

    g(50) = g(5*10) = 3 * 10 = 30, now how do I run it through the outer g?
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  6. #6
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    Re: Set Problems

    Quote Originally Posted by xmathlover View Post
    How did you come to that conclusion? do you look at the ordered pairs as x input, y output?

    How would I evaluate g(g(50)) and g^-1(-6)?

    g(50) = g(5*10) = 3 * 10 = 30, now how do I run it through the outer g?


    g(30)=g(5\cdot 6)=3\cdot 6=18 and -6=3\cdot ? so ?
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  7. #7
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    Re: Set Problems

    what do you do with the g^-1?
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  8. #8
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    Re: Set Problems

    Quote Originally Posted by xmathlover View Post
    what do you do with the g^-1?

    From your questions, it seems that you need a good review of function notation.

    g(g^{-1}(t))=t. That is if g^{-1}(t)=s then g(s)=t

    Thus g^{-1}(-21)=-35, now can you explain why?
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  9. #9
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    Re: Set Problems

    you lost me! lol...
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  10. #10
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    Re: Set Problems

    Quote Originally Posted by xmathlover View Post
    you lost me! lol...

    I really don't think that you are ready to do these questions.
    Why not give your notes and/or textbook a good review?

    Asking for g^{-1}(-21) is the same as asking g(?)=-21, fill in the blank?
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  11. #11
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    Re: Set Problems

    for this set {5n, 3n}?

    g(7) = -21

    wouldn't g^-1(-6) = 2?
    Last edited by xmathlover; April 21st 2013 at 11:11 AM.
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  12. #12
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    Re: Set Problems

    Quote Originally Posted by xmathlover View Post
    for this set {5n, 3n}?

    g(7) = -21
    No. g(7) does not exist, because seven is not in the domain of g. WHY?
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  13. #13
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    Re: Set Problems

    because it isn't a multiple or 5 or 3?

    But by that assumption nothing will work.
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  14. #14
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    Re: Set Problems

    Quote Originally Posted by xmathlover View Post
    because it isn't a multiple or 5 or 3?

    But by that assumption nothing will work.

    See, as I told you: You do not understand the notation.

    g^{-1}(-21)=-35. It does work: g(-35)=-21.
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  15. #15
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    Re: Set Problems

    oh so then, g^1(-6) = -10

    Is the domain of g express liked this {x | x +- 5} or should it be expressed different? My notes suck...they don't explain anything about domain and codomain in orderpair notation
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