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Math Help - Either I'm wrong or the solution manual is: f(g(x))

  1. #1
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    Either I'm wrong or the solution manual is: f(g(x))

    I'm looking for either a confirmation or explanation as to why I'm or the manual is wrong.

    This is what the book gives as a solution:
    For f(g(x)) = (1-sqrt(x))^2 Domain0,infinity), Range: (1, infinity)
    For g(f(x)) = 1-sqrt(x^2) Domain: (0, infinity), Range (-infinity,1)

    Now I understand you can simplify to: 1-2(sqrt(x)) + x and 1-abs(x)

    What I don't get is why it gives the domain and ranges as those values. It seems perfectly reasonable to me for the domain to be [0, infinity) and range to be [0, infinity) for the first equation. I can put 0 in and the universe won't fold in on itself. For the second one I can get (-infinity,infinity) and [1,infinity].

    Is the solution manual incorrect or am I missing something?
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  2. #2
    MHF Contributor
    Joined
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    For f(g(x)), the range should be [0, infinity).

    For g(f(x)), the domain should be (-infinity, infinity).

    Otherwise, the book is correct.
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