# Thread: Domain of Composite Function

1. ## 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"?

2. 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.