College Algebra, Chapter 3, 3.6, Section 3.6, Problem 42

Determine the functions $f \circ g, \quad g \circ f, \quad f \circ f$ and $g \circ g$ and their domains if $\displaystyle f(x) = \frac{1}{\sqrt{x}}$ and $g(x) = x^2 - 4x$
For $f \circ g$,

$\begin{equation} \begin{aligned} f \circ g &= f(g(x)) && \text{Definition of } f\circ g\\ \\ f \circ g &= \frac{1}{\sqrt{x^2 - 4x}} && \text{Definition of } f \end{aligned} \end{equation}$

Since the function involves square root in the denominator, we want

$\begin{equation} \begin{aligned} x^2 - 4x &> 0\\ \\ x(x-4) &> 0 \end{aligned} \end{equation}$

The factors on the left hand side are $x$ and $x-4$. These factors are zero when $x$ is and $4$, respectively. These numbers divide the number line into interval
$(-\infty,0), (0,4), (4, \infty)$
By testing some points on the interval...



Thus, the domain where $x(x-4) > 0$ is $(-\infty,\infty) \bigcup (4, \infty)$

For $g \circ f$,

$\begin{equation} \begin{aligned} g \circ f &= g (g(x)) && \text{Definition of } g \circ f\\ \\ g \circ f &= \left( \frac{1}{\sqrt{x}} \right)^2 - 4 \left( \frac{1}{\sqrt{x}} \right) && \text{Definition of } f\\ \\ g \circ f &= \frac{1}{x} - \frac{4}{\sqrt{x}} && \text{Definition of } g \end{aligned} \end{equation}$

Since the function is a rational function that invovles square root, so the domain of $g \circ f$ is $(0, \infty)$

For $f \circ f$,

$\begin{equation} \begin{aligned} f \circ f &= f(f(x)) && \text{Definition of } f\circ f\\ \\ f \circ f &= \frac{1}{\sqrt{\frac{1}{\sqrt{x}}}} && \text{Definition of } f\\ \\ f \circ f &= \frac{1}{\frac{\sqrt{1}}{\sqrt{x}}} && \text{Simplify}\\ \\ f \circ f &= \sqrt[4]{x} \end{aligned} \end{equation}$

We know that if the index is any even number, the radicand can't have a negative value. So the domain if $f \circ f$ is $[0, \infty)$

For $g \circ g$,

$\begin{equation} \begin{aligned} g \circ g &= g(g(x)) && \text{Definition of } g \circ g\\ \\ g \circ g &= \left( x^2 - 4x \right)^2 -4 \left( x^2 - 4x \right) && \text{Definition of } g \\ \\ g \circ g &= x^4 - 8x^3 + 16x^2 - 4x^2 + 16x && \text{Simplify}\\ \\ g \circ g &= x^4 - 8x^3 + 12x^2 + 16x && \text{Definition of } g \end{aligned} \end{equation}$

The domain of $g \circ g$ is $(-\infty,\infty)$