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Math Help - composite function for given domain

  1. #1
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    composite function for given domain

    Hi:
    Question: Find the composite function gf for the domain 3<=x<=7.
    f:x>12/x-1, 3<=x<=7
    g:x>x+2, xER

    If it were a value, say (2), I understand that I need to input value into f before g, like in a computer. But I can't understand part of question 'for the domain', and can't get the required answer.

    Answer: gf:x>2x+10/x-1, 3<=x<=7

    thank you.
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  2. #2
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    Quote Originally Posted by lemonz View Post
    Question: Find the composite function gf for the domain 3<=x<=7.
    f:x>12/x-1, 3<=x<=7
    g:x>x+2, xER

    If it were a value, say (2), I understand that I need to input value into f before g, like in a computer. But I can't understand part of question 'for the domain', and can't get the required answer.
    In what follows, I will be guessing that you mean the first function to be f(x)\, =\, \frac{12}{x\, -\, 1}.

    The function f(x), without the restriction, is defined everywhere but at x = 1. The restriction allows you to ignore the division-by-zero problem at x = 1.

    Then you're plugging f(x) into g(x). But g(x) is defined everywhere, while the restricted f(x) is defined only for 3 < x < 7. So you can't say that g(f(x)) is defined everywhere, because you're stuck with that restriction on f(x). If you can't plug, say, x = -5 into f(x), then you can't plug x = -5 into g(f(x)); it's still not an allowed input value.

    So you need to look at the restriction explicitly placed on f(x), and then also check to see if, by composing functions, any additional restrictions are added.

    . . . . . g(f(x))\, =\, \frac{12}{(x\, +\, 2)\, -\, 1}\, =\, \frac{12}{x\, +\, 1}

    The only additional restriction, to avoid dividing by zero, would be x = -1, but that's not in the domain of f(x) anyway. So the domain of g(f(x)) remains just the domain of f(x).

    Hope that helps!
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  3. #3
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    So,
    the asymptote f(x):12/x-1, 3<=x<=7
    can be drawn on the graph in all places within the domain 3<=x<=7, except x-1 - because 1/x where x cannot = 0
    g(x): has no restrictions as xER = all numbers -minus to infinity and all +numbers to infinity
    So, g(f(x)) has the same restrictions as f(x) and is limited to the same domain.

    Hmmm! I hope I have understood that. But not sure about last bit, as the answer I am required to find is:
    gf:x>2x+10/x-1, 2<=x<=7
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