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Math Help - Domain of Composite Function

  1. #1
    Member pflo's Avatar
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    Domain of Composite Function

    Suppose f(x)=\sqrt(x) and g(x)=x^2-5.
    Note that the domain of f is x\ge 0, the range of f is y\ge 0, the domain of g is all real numbers, and the range of g is also all real numbers.

    The composite function (g-of-f)(x)=g(f(x))=(\sqrt{x})^2-5=x-5. For this function the domain would be all real numbers, but if f does not exist then you can plug it in, and x\ge 0 for f to exist. So for this composite function the domain is x\ge 0.

    The composite function (f-of-g)(x)=f(g(x))=\sqrt{(x^2-5)}=\sqrt{x^2-5}. For this function the domain would be -\sqrt5\ge x\ge \sqrt5. Since all of these numbers are in the range of g, they can be plugged in to f.

    My question is, what is the domain of the composite function (f-of-g)(x)=f(g(x))? I believe that it would be x\ge\sqrt5 since you can't plug a negative into f and you can't plug a number between 0 and \sqrt5 into the resultant composite function. Is this correct? Is there a better way to think about this that does not involve "unions" and "intersections"?
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    Given that f(x)\sqrt x \;\& \,g(x) = x^2  - 5 then the domain of f \circ g(x) is \{x:x^2\ge 5\}=\left( { - \infty , - \sqrt 5 } \right] \cup \left[ {\sqrt 5 ,\infty } \right)

    The domain of g\circ f(x) is the set of non-negative numbers.
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