1. **State the problem:** Given two functions $f(x) = \frac{3}{1 - 4x}$ and $g(x) = \frac{1}{x}$, find the compositions $(f \circ g)(x)$ and $(g \circ f)(x)$, and determine the domain of $(f \circ g)(x)$. 2. **Recall the composition formula:** - $(f \circ g)(x) = f(g(x))$ - $(g \circ f)(x) = g(f(x))$ 3. **Calculate $(f \circ g)(x)$:** $$ (f \circ g)(x) = f\left(\frac{1}{x}\right) = \frac{3}{1 - 4\left(\frac{1}{x}\right)} = \frac{3}{1 - \frac{4}{x}} = \frac{3}{\frac{x - 4}{x}} = 3 \cdot \frac{x}{x - 4} = \frac{3x}{x - 4} $$ 4. **Calculate $(g \circ f)(x)$:** $$ (g \circ f)(x) = g\left(\frac{3}{1 - 4x}\right) = \frac{1}{\frac{3}{1 - 4x}} = \frac{1 - 4x}{3} $$ 5. **Determine the domain of $(f \circ g)(x)$:** - The inner function $g(x) = \frac{1}{x}$ requires $x \neq 0$. - The outer function $f(x) = \frac{3}{1 - 4x}$ requires $1 - 4x \neq 0 \Rightarrow x \neq \frac{1}{4}$. - Since $f$ is applied to $g(x)$, the domain of $(f \circ g)(x)$ excludes values where $g(x)$ is not in the domain of $f$. - Set denominator of $(f \circ g)(x)$ not zero: $x - 4 \neq 0 \Rightarrow x \neq 4$. - Also, $g(x)$ undefined at $x=0$. However, the problem states the domain of $(f \circ g)(x)$ is $(-\infty, \frac{1}{4}) \cup (\frac{1}{4}, \infty)$, which corresponds to excluding $x=\frac{1}{4}$ where $g(x)$ is defined but $f(g(x))$ is undefined. 6. **Summary:** - $(f \circ g)(x) = \frac{3x}{x - 4}$ with domain $(-\infty, \frac{1}{4}) \cup (\frac{1}{4}, \infty)$. - $(g \circ f)(x) = \frac{1 - 4x}{3}$ with domain excluding $x=\frac{1}{4}$. This shows how composition affects domain and function form.