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Homework Help: Composite functions and domains

  1. Feb 26, 2010 #1
    1. The problem statement, all variables and given/known data

    The forward-back function is
    [tex]f (t) = 2t[/tex] for [tex]0\leq{t}\leq{3}[/tex] ,
    [tex]f(t)= 12-2t[/tex] for [tex]3\leq{t}\leq{6}[/tex]. Graph [tex]f(f(t))[/tex] and find
    its four-part formula. First try t = 1.5 and 3.

    3. The attempt at a solution
    There are four possible composite functions from the two given functions:





    Since the domain of [tex]f (t) = 2t[/tex] is from 0 to 3 and the domain of [tex]f(t)= 12-2t[/tex] is from 3 to 6 it seems to me that only the first and fourth functions are valid. Functions two and three have a part of the composite function applied outside its defined domain. Instead of a four part formula I get a two part one:

    [tex]2(2(t))[/tex] for [tex]0\leq{t}\leq{3}[/tex]

    [tex]12-2(12-2t)[/tex] for [tex]3\leq{t}\leq{6}[/tex]

    The answer in the back of the book gives:

    [tex]2(2t)[/tex] for [tex]0\leq{t}\leq{1.5}[/tex]

    [tex]12-4t[/tex] for [tex]1.5\leq{t}\leq{3}[/tex]

    [tex]12-2(12-2t)[/tex] for [tex]3\leq{t}\leq{4.5}[/tex]

    [tex]2(12-2t)[/tex] for [tex]4.5\leq{t}\leq{6}[/tex]

    I also don't see the relevance of "First try t = 1.5 and 3".
    Any help will be appreciated.
  2. jcsd
  3. Feb 26, 2010 #2


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    Homework Helper
    Gold Member

    The relevance is just so that you see whether inside its domain it's a continuous function. though maybe with sharp corners (if the slope of the function is not an everywhere continuous function). You have a formula f(t) = one thing for a stretch up to t=3 but another thing for [tex]t\leq{3}[/tex]. Can you see intuitively that if the two formulae give the same result for t=3 the function is continuous, and if not there is a discontinuity?

    And what is the relevance of that? Well in dealing with a function you'll almost always want to know what it looks like - you would be naturally led a the start to look at f(x) at the break point t=3. As for 1.5 - well you'll see when you have calculated f(f(t)).
    Last edited: Feb 27, 2010
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