1. **State the problem:** Simplify the composite functions given by $m(x) = x^3 - x$, $n(x) = \frac{5}{x}$, $p(x) = 3x - 6$, and $q(x) = \frac{x^2}{x - 5}$. Find $(p \circ m)(x)$, $(q \circ p)(x)$, $(p \circ p)(x)$, $(n \circ m)(x)$, $(n \circ q)(x)$, and $(p \circ q)(x)$. Also state the domain of each. 2. **Recall the composite function formula:** $(f \circ g)(x) = f(g(x))$. We substitute the inner function into the outer function. 3. **Calculate each composite:** - $(p \circ m)(x) = p(m(x)) = p(x^3 - x) = 3(x^3 - x) - 6 = 3x^3 - 3x - 6$ - $(q \circ p)(x) = q(p(x)) = q(3x - 6) = \frac{(3x - 6)^2}{(3x - 6) - 5} = \frac{(3x - 6)^2}{3x - 11}$ - $(p \circ p)(x) = p(p(x)) = p(3x - 6) = 3(3x - 6) - 6 = 9x - 18 - 6 = 9x - 24$ - $(n \circ m)(x) = n(m(x)) = n(x^3 - x) = \frac{5}{x^3 - x} = \frac{5}{x(x^2 - 1)} = \frac{5}{x(x - 1)(x + 1)}$ - $(n \circ q)(x) = n(q(x)) = n\left(\frac{x^2}{x - 5}\right) = \frac{5}{\frac{x^2}{x - 5}} = 5 \cdot \frac{x - 5}{x^2} = \frac{5(x - 5)}{x^2}$ - $(p \circ q)(x) = p(q(x)) = p\left(\frac{x^2}{x - 5}\right) = 3 \cdot \frac{x^2}{x - 5} - 6 = \frac{3x^2}{x - 5} - 6 = \frac{3x^2 - 6(x - 5)}{x - 5} = \frac{3x^2 - 6x + 30}{x - 5}$ 4. **State the domains:** - For $(p \circ m)(x)$, domain is all real numbers since $m(x)$ and $p(x)$ are polynomials. - For $(q \circ p)(x)$, denominator $3x - 11 \neq 0 \Rightarrow x \neq \frac{11}{3}$. - For $(p \circ p)(x)$, domain is all real numbers (polynomial). - For $(n \circ m)(x)$, denominator $x(x - 1)(x + 1) \neq 0 \Rightarrow x \neq 0, \pm 1$. - For $(n \circ q)(x)$, denominator $x^2 \neq 0 \Rightarrow x \neq 0$, and also $x - 5 \neq 0 \Rightarrow x \neq 5$ from inside $q(x)$. - For $(p \circ q)(x)$, denominator $x - 5 \neq 0 \Rightarrow x \neq 5$. **Final answers:** $$ (p \circ m)(x) = 3x^3 - 3x - 6, \quad \text{domain: } (-\infty, \infty) $$ $$ (q \circ p)(x) = \frac{(3x - 6)^2}{3x - 11}, \quad \text{domain: } (-\infty, \frac{11}{3}) \cup (\frac{11}{3}, \infty) $$ $$ (p \circ p)(x) = 9x - 24, \quad \text{domain: } (-\infty, \infty) $$ $$ (n \circ m)(x) = \frac{5}{x(x - 1)(x + 1)}, \quad \text{domain: } (-\infty, -1) \cup (-1, 0) \cup (0, 1) \cup (1, \infty) $$ $$ (n \circ q)(x) = \frac{5(x - 5)}{x^2}, \quad \text{domain: } (-\infty, 0) \cup (0, 5) \cup (5, \infty) $$ $$ (p \circ q)(x) = \frac{3x^2 - 6x + 30}{x - 5}, \quad \text{domain: } (-\infty, 5) \cup (5, \infty) $$