1. **Problem 3:** Given the function $h : x \to px - q$ with $h(-3) = -13$ and $h(2) = -3$, find: (a) The values of $p$ and $q$. (b) The values of $p$ and $q$ (repetition of (a), so same answer). (c) The value of $x$ such that $h(x) = x$ (the value of $x$ which maps onto itself). **Step 1:** Write the given conditions as equations: $$h(-3) = p(-3) - q = -3p - q = -13$$ $$h(2) = 2p - q = -3$$ **Step 2:** Solve the system of linear equations: From the two equations: $$-3p - q = -13$$ $$2p - q = -3$$ Subtract the second from the first to eliminate $q$: $$(-3p - q) - (2p - q) = -13 - (-3)$$ $$-3p - q - 2p + q = -13 + 3$$ $$-5p = -10$$ $$p = 2$$ **Step 3:** Substitute $p=2$ into one of the equations to find $q$: $$2p - q = -3 \implies 2(2) - q = -3 \implies 4 - q = -3 \implies q = 7$$ **Answer for (a) and (b):** $$p = 2, \quad q = 7$$ **Step 4:** Find $x$ such that $h(x) = x$: $$h(x) = px - q = x$$ Substitute $p=2$ and $q=7$: $$2x - 7 = x$$ $$2x - x = 7$$ $$x = 7$$ **Answer for (c):** $$x = 7$$ --- 2. **Problem 4:** Given functions $f$ and $g$ from an arrow diagram (not provided explicitly), find: (a) $g(7)$ (b) $g^2(x) = g(g(x))$ (c) $g(f(x))$ (d) $f(g(x))$ Since the explicit functions or mappings are not provided, we cannot compute exact values or expressions. Please provide the functions or mappings for $f$ and $g$ to proceed. --- 3. **Problem 5:** Given functions: $$g : x \to \frac{1}{x - 1}, x \neq 1$$ $$h : x \to 2x + 3$$ Find: (a) The functions $g(x)$, $h(g(x))$, $g^2(x) = g(g(x))$, and $h^2(x) = h(h(x))$. (b) Values of $x$ such that $g(h(x)) = h(g(x)) = \frac{31}{8}$. **Step 1:** Write the given functions: $$g(x) = \frac{1}{x - 1}$$ $$h(x) = 2x + 3$$ **Step 2:** Find $h(g(x))$: $$h(g(x)) = h\left(\frac{1}{x - 1}\right) = 2 \cdot \frac{1}{x - 1} + 3 = \frac{2}{x - 1} + 3$$ **Step 3:** Find $g^2(x) = g(g(x))$: $$g(g(x)) = g\left(\frac{1}{x - 1}\right) = \frac{1}{\frac{1}{x - 1} - 1} = \frac{1}{\frac{1 - (x - 1)}{x - 1}} = \frac{1}{\frac{2 - x}{x - 1}} = \frac{x - 1}{2 - x}$$ **Step 4:** Find $h^2(x) = h(h(x))$: $$h(h(x)) = h(2x + 3) = 2(2x + 3) + 3 = 4x + 6 + 3 = 4x + 9$$ **Answer for (a):** $$g(x) = \frac{1}{x - 1}, \quad h(g(x)) = \frac{2}{x - 1} + 3, \quad g^2(x) = \frac{x - 1}{2 - x}, \quad h^2(x) = 4x + 9$$ **Step 5:** Find $x$ such that: $$g(h(x)) = h(g(x)) = \frac{31}{8}$$ Calculate $g(h(x))$: $$g(h(x)) = g(2x + 3) = \frac{1}{(2x + 3) - 1} = \frac{1}{2x + 2} = \frac{1}{2(x + 1)}$$ Set equal to $\frac{31}{8}$: $$\frac{1}{2(x + 1)} = \frac{31}{8} \implies 2(x + 1) = \frac{8}{31} \implies x + 1 = \frac{4}{31} \implies x = \frac{4}{31} - 1 = -\frac{27}{31}$$ Calculate $h(g(x))$: $$h(g(x)) = \frac{2}{x - 1} + 3 = \frac{31}{8}$$ Solve for $x$: $$\frac{2}{x - 1} + 3 = \frac{31}{8}$$ $$\frac{2}{x - 1} = \frac{31}{8} - 3 = \frac{31}{8} - \frac{24}{8} = \frac{7}{8}$$ $$2 = \frac{7}{8}(x - 1)$$ $$2 = \frac{7}{8}x - \frac{7}{8}$$ $$\frac{7}{8}x = 2 + \frac{7}{8} = \frac{16}{8} + \frac{7}{8} = \frac{23}{8}$$ $$x = \frac{23}{8} \cdot \frac{8}{7} = \frac{23}{7}$$ **Answer for (b):** $$x = -\frac{27}{31} \quad \text{for} \quad g(h(x)) = \frac{31}{8}$$ $$x = \frac{23}{7} \quad \text{for} \quad h(g(x)) = \frac{31}{8}$$ --- 4. **Problem 6:** Given the inverse function: $$g^{-1}(x) = \frac{x - 4}{11}$$ Find the function $g$. **Step 1:** Recall that if $g^{-1}(x) = \frac{x - 4}{11}$, then $g$ is the inverse of this function. Let $y = g^{-1}(x) = \frac{x - 4}{11}$. To find $g$, solve for $x$ in terms of $y$: $$y = \frac{x - 4}{11} \implies 11y = x - 4 \implies x = 11y + 4$$ Since $y = g^{-1}(x)$, then $x = g(y)$. Replace $y$ by $x$ to write $g(x)$: $$g(x) = 11x + 4$$ **Answer:** $$g(x) = 11x + 4$$