College Algebra, Chapter 3, Review Exercises, Section Review Exercises, Problem 78

Suppose that $f(x) = \sqrt{x}$ and $\displaystyle g(x) = \frac{2}{x-4}$ find $f \circ g$, $g \circ f$, $f \circ f$ and $g \circ g$ and their domains

$\begin{equation} \begin{aligned} f \circ g &= f(g(x)) = f\left( \frac{2}{x - 4} \right) = \sqrt{\frac{2}{x-4}}\\ \\ g \circ f &= g(f(x)) = g(\sqrt{x}) = \frac{2}{\sqrt{x}-4}\\ \\ f \circ f &= f(f(x)) = f(\sqrt{x}) = \sqrt{\sqrt{x}} = \sqrt[4]{x}\\ \\ g \circ g &= g(g(x)) = g\left( \frac{2}{x-4} \right) = \frac{2}{\frac{2}{x-4}-4}\\ \\ &= \frac{2(x-4)}{2-4(x-4)} = \frac{2(x-4)}{2-4x+16} = \frac{2(x-4)}{-4x+18} = \frac{2(x-4)}{-2(2x-9)} = \frac{-(x-4)}{2x-9} \end{aligned} \end{equation}$


To find the domain of $f \circ g$, we want...

$\begin{equation} \begin{aligned} x - 4 &> 0 && \text{Add } 4\\ \\ x &> 4 \end{aligned} \end{equation}$

Thus, the domain of $f \circ g$ is $(4, \infty)$

To find the domain of $g \circ f$, we want $\sqrt{x } - 4 \neq 0$. The denominator is zero when...

$\begin{equation} \begin{aligned} \sqrt{x} - 4 &= 0 && \text{Add } 4\\ \\ \sqrt{x} &= 4 && \text{Square both sides}\\ \\ x &= 16 \end{aligned} \end{equation}$


Also, the function involves square root that is defined for only positive values. Thus, the domain of $g \circ f$ is $[0,16)\bigcup(16,\infty)$

To find the domain of $f \circ f$, recall that the function with even roots is defined only for positive values of $x$. Thus, the domain of $f \circ f$ is $[0, \infty)$

To find the domain of $g \circ g$, we want $2x - 9 \neq 0$ the denominator is zero when...

$\begin{equation} \begin{aligned} 2x - 9 &= 0 && \text{Add } 9 \\ \\ 2x &= 9 && \text{Divide } 2\\ \\ x &= \frac{9}{2} \end{aligned} \end{equation}$

Thus, the domain of $g \circ g$ is $\displaystyle \left( -\infty, \frac{9}{2} \right)\bigcup \left( \frac{9}{2}, \infty \right)$