1. **State the problem:** We are given two functions: $$f(x) = \frac{6}{x+4}$$ and $$g(x) = \frac{3}{x}$$ We need to find: a. The composition $(f \circ g)(x) = f(g(x))$. b. The domain of $f \circ g$. 2. **Find the composition $(f \circ g)(x)$:** By definition, $(f \circ g)(x) = f(g(x))$. Substitute $g(x)$ into $f$: $$ (f \circ g)(x) = f\left(\frac{3}{x}\right) = \frac{6}{\frac{3}{x} + 4} $$ 3. **Simplify the expression:** Combine the denominator: $$ \frac{3}{x} + 4 = \frac{3}{x} + \frac{4x}{x} = \frac{3 + 4x}{x} $$ So, $$ (f \circ g)(x) = \frac{6}{\frac{3 + 4x}{x}} = 6 \times \frac{x}{3 + 4x} = \frac{6x}{3 + 4x} $$ 4. **Find the domain of $f \circ g$:** The domain consists of all $x$ values for which the expression is defined. - First, $g(x) = \frac{3}{x}$ is defined for $x \neq 0$. - Next, $f(g(x)) = \frac{6}{g(x) + 4} = \frac{6}{\frac{3}{x} + 4}$ requires the denominator $\frac{3}{x} + 4 \neq 0$. Set denominator not equal to zero: $$ \frac{3}{x} + 4 \neq 0 \implies \frac{3 + 4x}{x} \neq 0 $$ This fraction is zero when numerator is zero and denominator is not zero. So, $$ 3 + 4x \neq 0 \implies 4x \neq -3 \implies x \neq -\frac{3}{4} $$ Also, denominator $x \neq 0$ because it is in the denominator of $g(x)$. 5. **Domain in interval notation:** Exclude $x=0$ and $x = -\frac{3}{4}$ from all real numbers: $$ (-\infty, -\frac{3}{4}) \cup (-\frac{3}{4}, 0) \cup (0, \infty) $$ **Final answers:** $\text{a. } (f \circ g)(x) = \frac{6x}{3 + 4x}$ $\text{b. Domain} = (-\infty, -\frac{3}{4}) \cup (-\frac{3}{4}, 0) \cup (0, \infty)$