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Math Help - Function question

  1. #1
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    Function question

    Hello! My question is as follows:

    f(x) = (x-3)^3 , x > 3

    Find f^-1(f(x)) and f(f^-1(x)), stating clearly the rules and domains.

    I know that the rules of both are the same: f^-1(f(x)) = x and f(f^-1(x)) = x. However, I don't understand why are the domains different? Please explain to me! Thank you!
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    Quote Originally Posted by Tangera View Post
    Hello! My question is as follows:

    f(x) = (x-3)^3 , x > 3

    Find f^-1(f(x)) and f(f^-1(x)), stating clearly the rules and domains.

    I know that the rules of both are the same: f^-1(f(x)) = x and f(f^-1(x)) = x. However, I don't understand why are the domains different? Please explain to me! Thank you!
    You need reasoning similar to that used here.
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  3. #3
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    I understand why f^-1(f(x)) has a domain of x>3, but I don't know why the domain of f(f^-1(x)) is x > 0. Is it because since f(x) has a domain of x >3, so the range of f inverse is also [3, infinity). f inverse = (cube root x) + 3. For the range of f inverse to be [3, infinity), x > 0? Is my reasoning correct?
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    Quote Originally Posted by Tangera View Post
    I understand why f^-1(f(x)) has a domain of x>3, but I don't know why the domain of f(f^-1(x)) is x > 0. Is it because since f(x) has a domain of x >3, so the range of f inverse is also [3, infinity). f inverse = (cube root x) + 3. For the range of f inverse to be [3, infinity), x > 0? Is my reasoning correct?
    There is a rule that says:

    \text{dom} \, f^{-1} = \text{ran} \, f

    \text{ran} \, f^{-1} = \text{dom} \, f

    These rules can be thought of as a consequence of swapping y and x around in y = f(x) to get the rule for the inverse function ..... y's and x's swap --> domain and range swap .....

    In your example,

    \text{dom}\, f = [3, \infty) \, and \text{ran}\, f = [0, \infty) \, .

    Therefore:

    \text{dom} \, f^{-1} = \text{ran} \, f = [0, \infty).

    \text{ran} \, f^{-1} = \text{dom} \, f = [3, \infty).


    The composite function g(f(x)) exists iff \text{ran} f = \text{dom} \, g \, . When that condition is met, the domain of g(f(x)) is \text{dom} f. In you present question, this condition is met for both f(f^{-1}(x)) \, and f^{-1}(f(x)).

    So:

    The domain of f(f^{-1}(x)) = x\, is \text{dom} \, f^{-1} = [0, \infty).

    The domain of f^{-1}(f(x)) = x\, is \text{dom} \, f = [3, \infty).
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