Domain of Composite Function

**pflo** · November 23rd 2010, 01:03 PM

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"?

**Plato** · November 23rd 2010, 01:21 PM

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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