1. Problem: Given sets A = {2,3,4}, B = {1,3,5,7}, and C = {3,4,5}, find: a) $B \cup C$ Formula: Union of two sets includes all elements from both sets without repetition. Calculation: $$B \cup C = \{1,3,5,7\} \cup \{3,4,5\} = \{1,3,4,5,7\}$$ b) $A \cap (B \cup C)$ Formula: Intersection includes elements common to both sets. Calculation: $$A \cap (B \cup C) = \{2,3,4\} \cap \{1,3,4,5,7\} = \{3,4\}$$ c) $A \cup B$ Calculation: $$A \cup B = \{2,3,4\} \cup \{1,3,5,7\} = \{1,2,3,4,5,7\}$$ d) $A \cup C$ Calculation: $$A \cup C = \{2,3,4\} \cup \{3,4,5\} = \{2,3,4,5\}$$ e) $(A \cap B) \cup (A \cap C)$ Calculate each intersection: $$A \cap B = \{3\}$$ $$A \cap C = \{3,4\}$$ Then union: $$(A \cap B) \cup (A \cap C) = \{3\} \cup \{3,4\} = \{3,4\}$$ 2. Problem: Universal set $U$ contains natural numbers between 12 and 21. Given: $$A = \{\text{multiples of 2 between 12 and 19 inclusive}\}$$ $$B = \{\text{positive integers between 15 and 19 inclusive}\}$$ a) List elements: $$U = \{12,13,14,15,16,17,18,19,20,21\}$$ $$A = \{12,14,16,18\}$$ $$B = \{15,16,17,18,19\}$$ b) Venn diagram: (Not shown here, but sets overlap at 16 and 18) c) Find $A' \cap B'$ Complement sets relative to $U$: $$A' = U \setminus A = \{13,15,17,19,20,21\}$$ $$B' = U \setminus B = \{12,13,14,20,21\}$$ Intersection: $$A' \cap B' = \{13,20,21\}$$ 3. Solve inequalities: a) $2(2x + 3) - 10 < 6(x - 2)$ Step 1: Expand $$4x + 6 - 10 < 6x - 12$$ Step 2: Simplify $$4x - 4 < 6x - 12$$ Step 3: Subtract $4x$ both sides $$\cancel{4x} - 4 < \cancel{4x} + 6x - 12 \Rightarrow -4 < 2x - 12$$ Step 4: Add 12 both sides $$-4 + 12 < 2x - 12 + 12 \Rightarrow 8 < 2x$$ Step 5: Divide both sides by 2 $$\frac{8}{\cancel{2}} < \frac{2x}{\cancel{2}} \Rightarrow 4 < x$$ Solution: $x > 4$ b) $\frac{x-3}{4} + 6 \geq 2 + \frac{4x}{3}$ Step 1: Multiply entire inequality by 12 (LCM of 4 and 3) to clear denominators $$12 \times \left(\frac{x-3}{4} + 6\right) \geq 12 \times \left(2 + \frac{4x}{3}\right)$$ Step 2: Distribute $$3(x-3) + 72 \geq 24 + 16x$$ Step 3: Expand $$3x - 9 + 72 \geq 24 + 16x$$ Step 4: Simplify $$3x + 63 \geq 24 + 16x$$ Step 5: Subtract $3x$ both sides $$63 \geq 24 + 13x$$ Step 6: Subtract 24 both sides $$39 \geq 13x$$ Step 7: Divide both sides by 13 $$\frac{39}{\cancel{13}} \geq \frac{13x}{\cancel{13}} \Rightarrow 3 \geq x$$ Solution: $x \leq 3$ c) $\frac{2x}{5} - \frac{x-3}{2} \leq \frac{2x}{3} - \frac{3}{10}(x + 2)$ Step 1: Multiply entire inequality by 30 (LCM of 5,2,3,10) to clear denominators $$30 \times \left(\frac{2x}{5} - \frac{x-3}{2}\right) \leq 30 \times \left(\frac{2x}{3} - \frac{3}{10}(x + 2)\right)$$ Step 2: Distribute $$6 \times 2x - 15(x - 3) \leq 10 \times 2x - 9(x + 2)$$ Step 3: Simplify $$12x - 15x + 45 \leq 20x - 9x - 18$$ Step 4: Combine like terms $$-3x + 45 \leq 11x - 18$$ Step 5: Add $3x$ both sides $$45 \leq 14x - 18$$ Step 6: Add 18 both sides $$63 \leq 14x$$ Step 7: Divide both sides by 14 $$\frac{63}{\cancel{14}} \leq \frac{14x}{\cancel{14}} \Rightarrow \frac{9}{2} \leq x$$ Solution: $x \geq \frac{9}{2}$ 4. Function $f(x) = x^2 + 3x$ a) Find $f(-2)$ $$f(-2) = (-2)^2 + 3(-2) = 4 - 6 = -2$$ b) Find $f(1 + h)$ $$f(1 + h) = (1 + h)^2 + 3(1 + h) = 1 + 2h + h^2 + 3 + 3h = h^2 + 5h + 4$$ 5. Given $f(x) = \frac{x+1}{x-1}$, find $$\frac{f(x+h) - f(x)}{h}$$ Step 1: Compute $f(x+h)$ $$f(x+h) = \frac{(x+h)+1}{(x+h)-1} = \frac{x + h + 1}{x + h - 1}$$ Step 2: Compute difference quotient $$\frac{\frac{x + h + 1}{x + h - 1} - \frac{x + 1}{x - 1}}{h} = \frac{\frac{(x + h + 1)(x - 1) - (x + 1)(x + h - 1)}{(x + h - 1)(x - 1)}}{h}$$ Step 3: Simplify numerator Expand: $$(x + h + 1)(x - 1) = x^2 - x + hx - h + x - 1 = x^2 + hx - h - 1$$ $$(x + 1)(x + h - 1) = x^2 + xh - x + x + h - 1 = x^2 + xh + h - 1$$ Difference: $$x^2 + hx - h - 1 - (x^2 + xh + h - 1) = x^2 + hx - h - 1 - x^2 - xh - h + 1 = (hx - xh) + (-h - h) + (-1 + 1) = 0 - 2h + 0 = -2h$$ Step 4: Substitute back $$\frac{-2h}{(x + h - 1)(x - 1)} \times \frac{1}{h} = \frac{-2h}{h (x + h - 1)(x - 1)} = \frac{-2}{(x + h - 1)(x - 1)}$$ Final answer: $$\frac{f(x+h) - f(x)}{h} = \frac{-2}{(x + h - 1)(x - 1)}$$ 6. Given $f(x) = \sqrt{x}$ and $g(x) = \sqrt{10 - x}$ a) $(f + g)(x) = \sqrt{x} + \sqrt{10 - x}$ Domain: $x \geq 0$ and $10 - x \geq 0 \Rightarrow x \leq 10$, so domain is $[0,10]$ b) $(f - g)(x) = \sqrt{x} - \sqrt{10 - x}$ Domain same as above: $[0,10]$ c) $(fg)(x) = \sqrt{x} \times \sqrt{10 - x} = \sqrt{x(10 - x)}$ Domain: $x \geq 0$, $10 - x \geq 0$, and $x(10 - x) \geq 0$ which is true for $x \in [0,10]$ d) $\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x}}{\sqrt{10 - x}} = \sqrt{\frac{x}{10 - x}}$ Domain: $x \geq 0$, $10 - x > 0$ (denominator not zero), so $x \in [0,10)$ 7. Given $f(x) = \frac{x}{x-1}$ and $g(x) = \frac{x-4}{x+2}$ a) Find $f(g(x))$ $$f(g(x)) = \frac{g(x)}{g(x) - 1} = \frac{\frac{x-4}{x+2}}{\frac{x-4}{x+2} - 1} = \frac{\frac{x-4}{x+2}}{\frac{x-4 - (x+2)}{x+2}} = \frac{\frac{x-4}{x+2}}{\frac{x-4 - x - 2}{x+2}} = \frac{\frac{x-4}{x+2}}{\frac{-6}{x+2}} = \frac{x-4}{x+2} \times \frac{x+2}{-6} = \frac{x-4}{-6} = -\frac{x-4}{6}$$ Domain: $x \neq -2$ (denominator of $g(x)$), and $g(x) \neq 1$ (denominator of $f(g(x))$) Check $g(x) = 1$: $$\frac{x-4}{x+2} = 1 \Rightarrow x - 4 = x + 2 \Rightarrow -4 = 2$$ No solution, so no restriction from this. Final domain: $x \neq -2$ b) Find $g(f(x))$ $$g(f(x)) = \frac{f(x) - 4}{f(x) + 2} = \frac{\frac{x}{x-1} - 4}{\frac{x}{x-1} + 2} = \frac{\frac{x - 4(x-1)}{x-1}}{\frac{x + 2(x-1)}{x-1}} = \frac{\frac{x - 4x + 4}{x-1}}{\frac{x + 2x - 2}{x-1}} = \frac{\frac{-3x + 4}{x-1}}{\frac{3x - 2}{x-1}} = \frac{-3x + 4}{3x - 2}$$ Domain: $x \neq 1$ (denominator of $f(x)$), and $3x - 2 \neq 0 \Rightarrow x \neq \frac{2}{3}$ 8. Given $f(g(x)) = \frac{x^2}{x^2 + 1}$ and $g(x) = \sqrt{1 - x^2}$, find $f(x)$ Let $y = g(x) = \sqrt{1 - x^2}$, then $y^2 = 1 - x^2 \Rightarrow x^2 = 1 - y^2$ Given: $$f(g(x)) = f(y) = \frac{x^2}{x^2 + 1} = \frac{1 - y^2}{(1 - y^2) + 1} = \frac{1 - y^2}{2 - y^2}$$ Therefore: $$f(x) = \frac{1 - x^2}{2 - x^2}$$ 9. Given $f(x) = A(2x - 3) - B(x - 2)$ and $g(x) = x - 1$, find $A$ and $B$ such that $f(x) = g(x)$ Step 1: Expand $f(x)$ $$f(x) = 2Ax - 3A - Bx + 2B = (2A - B)x + (-3A + 2B)$$ Step 2: Equate coefficients with $g(x) = x - 1$ Coefficient of $x$: $$2A - B = 1$$ Constant term: $$-3A + 2B = -1$$ Step 3: Solve system From first: $$B = 2A - 1$$ Substitute into second: $$-3A + 2(2A - 1) = -1 \Rightarrow -3A + 4A - 2 = -1 \Rightarrow A - 2 = -1 \Rightarrow A = 1$$ Then: $$B = 2(1) - 1 = 1$$ 10. Given $$f(x) = A(x^2 + 1) + B(x^2 - 2x + 4) + C(x^2 + x - 1)$$ and $$g(x) = 7x^2 + 5x - 1$$ Find $A, B, C$ such that $f(x) = g(x)$ Step 1: Expand $f(x)$ $$f(x) = A x^2 + A + B x^2 - 2 B x + 4 B + C x^2 + C x - C$$ Group terms: $$f(x) = (A + B + C) x^2 + (-2 B + C) x + (A + 4 B - C)$$ Step 2: Equate coefficients with $g(x)$ $$A + B + C = 7$$ $$-2 B + C = 5$$ $$A + 4 B - C = -1$$ Step 3: Solve system From second: $$C = 5 + 2 B$$ Substitute into first: $$A + B + 5 + 2 B = 7 \Rightarrow A + 3 B = 2$$ Substitute into third: $$A + 4 B - (5 + 2 B) = -1 \Rightarrow A + 4 B - 5 - 2 B = -1 \Rightarrow A + 2 B - 5 = -1 \Rightarrow A + 2 B = 4$$ Step 4: Subtract equations $$(A + 3 B) - (A + 2 B) = 2 - 4 \Rightarrow B = -2$$ Step 5: Find $A$ $$A + 3(-2) = 2 \Rightarrow A - 6 = 2 \Rightarrow A = 8$$ Step 6: Find $C$ $$C = 5 + 2(-2) = 5 - 4 = 1$$ 11. Line passes through $P(-2,1)$ and $Q(4,k)$ with slope $-\frac{2}{3}$ Formula for slope: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Substitute: $$-\frac{2}{3} = \frac{k - 1}{4 - (-2)} = \frac{k - 1}{6}$$ Multiply both sides by 6: $$6 \times -\frac{2}{3} = k - 1 \Rightarrow -4 = k - 1$$ Solve for $k$: $$k = -3$$ 12. Points $(1,m)$ and $(-2,-1)$ are 5 units apart Distance formula: $$\sqrt{(1 + 2)^2 + (m + 1)^2} = 5$$ Simplify: $$\sqrt{3^2 + (m + 1)^2} = 5$$ Square both sides: $$9 + (m + 1)^2 = 25$$ Subtract 9: $$(m + 1)^2 = 16$$ Take square root: $$m + 1 = \pm 4$$ Solve: $$m = 3 \text{ or } m = -5$$ 13. Points $B(3,1)$ and $C(x,y)$ have midpoint $(-1,2)$ Midpoint formula: $$\left(\frac{3 + x}{2}, \frac{1 + y}{2}\right) = (-1, 2)$$ Equate components: $$\frac{3 + x}{2} = -1 \Rightarrow 3 + x = -2 \Rightarrow x = -5$$ $$\frac{1 + y}{2} = 2 \Rightarrow 1 + y = 4 \Rightarrow y = 3$$ 14. Write $f(x) = 3x^2 + 6x + 5$ in vertex form Formula: $$f(x) = a(x - h)^2 + k$$ Step 1: Factor out 3 from first two terms $$f(x) = 3(x^2 + 2x) + 5$$ Step 2: Complete the square inside parentheses Take half of 2, square it: $(\frac{2}{2})^2 = 1$ Add and subtract 1 inside parentheses: $$f(x) = 3(x^2 + 2x + 1 - 1) + 5 = 3((x + 1)^2 - 1) + 5$$ Step 3: Simplify $$f(x) = 3(x + 1)^2 - 3 + 5 = 3(x + 1)^2 + 2$$ 15. Graph quadratic equations (description only): a) $y = x^2 + 2x - 8$ b) $f(x) = 4x - x^2$ 16. Graph rational functions (description only): a) $f(x) = \frac{1}{x^2 - 4}$ b) $f(x) = \frac{2x}{x - 2}$