College Algebra, Chapter 3, 3.6, Section 3.6, Problem 38

Determine the functions $f \circ g, \quad g \circ f, \quad f \circ f$ and $g \circ g$ and their domains if $f(x) = x^2$ and $g(x) = \sqrt{x-3}$
For $f \circ g$

$\begin{equation} \begin{aligned} f \circ g &= f(g(x)) && \text{Definition of } f \circ g\\ \\ f \circ g &= \left( \sqrt{x-3} \right)^2 && \text{Definition of } g\\ \\ f \circ g &= x -3 && \text{Definition of } f \end{aligned} \end{equation}$

The domain of the function is $(-\infty, \infty)$

For $g \circ f$

$\begin{equation} \begin{aligned} g \circ f &= g(f(x)) && \text{Definition of } g \circ f\\ \\ g \circ f &= \sqrt{(x^2)-3} && \text{Definition of } f\\ \\ g \circ f &= \sqrt{x^2 - 3} && \text{Definition of } g \end{aligned} \end{equation}$

Since the function involves square root we want

$\begin{equation} \begin{aligned} x^2 -3 &\geq 0\\ \\ x^2 &\geq 3\\ \\ x &\geq \pm \sqrt{3} \end{aligned} \end{equation}$

Thus, the domain is $[-\sqrt{3},\infty)$

For $f \circ f$

$\begin{equation} \begin{aligned} f \circ f &= f(f(x)) && \text{Definition of } f \circ f\\ \\ f \circ f &= (x^2)^2&& \text{Definition of } f\\ \\ f \circ f &= x^4 && \text{Definition of } f \end{aligned} \end{equation}$

The domain of the function is $(-\infty,\infty)$

For $g \circ g$

$\begin{equation} \begin{aligned} g \circ g &= g(g(x)) && \text{Definition of } g \circ g\\ \\ g \circ g &= \sqrt{\sqrt{x-3}-3} && \text{Definition of } g\\ \end{aligned} \end{equation}$

Since the equation involves square root, we want

$\begin{equation} \begin{aligned} \sqrt{x-3} - 3 &\geq 0 && \text{Add }3 \\ \\ \sqrt{x-3} &\geq 3 && \text{Square both sides}\\ \\ x- 3 &\geq 9 && \text{Add } 3\\ \\ x &\geq 12 \end{aligned} \end{equation}$

Thus, the domain of $g \circ g$ $[12,\infty)$