Results 1 to 2 of 2

Math Help - Domain f o g

  1. #1
    Member
    Joined
    Sep 2007
    Posts
    81

    Domain f o g

    Hi all,

    First things first, what I've seen as the answer for these is described as follows.

    <br />
Dom (f o g) = {x : x \in Domg-and-g(x) \in Dom f}<br />

    I'm not sure what is meant by the second section though, g(x) in Dom f?

    For instance

    f(x) = \sqrt{4 - x^2}
    g(x) = \ln{x}

    So the dom for f(x) is [-2, 2]
    the dom for g(x) is (0, inf]

    How does it work fitting into this formula?

    Dom(fog) = x : x \in x > 0 and \ln{x} \in [-2, 2]

    So Dom(fog) = 0 < x < 2 ???

    I know this is incorrect, but this is how my logic of applying the formula works.

    Anyone who can point out my mistakes or explain it in english a bit better would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Domain of composite function

    Hello Peleus
    Quote Originally Posted by Peleus View Post
    Hi all,

    First things first, what I've seen as the answer for these is described as follows.

    <br />
Dom (f o g) = {x : x \in Domg-and-g(x) \in Dom f}<br />

    I'm not sure what is meant by the second section though, g(x) in Dom f?

    For instance

    f(x) = \sqrt{4 - x^2}
    g(x) = \ln{x}

    So the dom for f(x) is [-2, 2]
    the dom for g(x) is (0, inf]

    How does it work fitting into this formula?

    Dom(fog) = x : x \in x > 0 and \ln{x} \in [-2, 2]

    So Dom(fog) = 0 < x < 2 ???

    I know this is incorrect, but this is how my logic of applying the formula works.

    Anyone who can point out my mistakes or explain it in english a bit better would be appreciated.
    As I'm sure you know, when we 'feed' the value x through the composite function f \circ g, x is first 'fed into' g, and then the output of g (which is represented by g(x)) is then used as the input to the function f. To ensure a valid input into f, then, we must make sure that the output g(x) lies in the domain of f. That is what is meant by g(x) \in Dom(f).

    As far as the example you quote is concerned, g(x) = \ln(x), any value of x in the domain (0, +\infty) will produce an output from g, but not every output can then be used as an input to f: only those outputs that are themselves in [-2,2].

    So you need to ask yourself the question: What values of x can I use as inputs for \ln(x) that will produce outputs in the range [-2,2] ? The answer will be given by the solutions to the equations \ln(x_{min}) = -2 and \ln(x_{max}) = 2.

    Raising e to the power of both sides of each equation we get:

    x_{min} = e^{-2} and x_{max} = e^2

    There's the domain of f\circ g, then: [e^{-2},e^2]

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Converting from the S domain to the Z domain
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: February 24th 2011, 11:15 PM
  2. Domain
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 29th 2010, 01:20 AM
  3. Replies: 6
    Last Post: September 16th 2009, 07:25 AM
  4. Domain
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 14th 2009, 05:53 PM
  5. Domain of (26x-42)/(x^2 -2x - 63) ?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 4th 2009, 09:35 PM

Search Tags


/mathhelpforum @mathhelpforum