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Thread: Domain of Composite Function

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

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

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

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

    My question is, what is the domain of the composite function $\displaystyle (f-of-g)(x)=f(g(x))$? I believe that it would be $\displaystyle x\ge\sqrt5$ since you can't plug a negative into $\displaystyle f$ and you can't plug a number between $\displaystyle 0$ and $\displaystyle \sqrt5$ 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 $\displaystyle f(x)\sqrt x \;\& \,g(x) = x^2 - 5$ then the domain of $\displaystyle f \circ g(x)$ is $\displaystyle \{x:x^2\ge 5\}=\left( { - \infty , - \sqrt 5 } \right] \cup \left[ {\sqrt 5 ,\infty } \right)$

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