1. Problem: Find the product function $(fg)(x)$ where $f(x) = 4x - 2$ and $g(x) = 2x + 2$. Step 1: Recall that $(fg)(x) = f(x) \cdot g(x)$. Step 2: Substitute the given functions: $$ (fg)(x) = (4x - 2)(2x + 2) $$ Step 3: Expand the product: $$ = 4x \cdot 2x + 4x \cdot 2 - 2 \cdot 2x - 2 \cdot 2 $$ $$ = 8x^2 + 8x - 4x - 4 $$ Step 4: Simplify like terms: $$ 8x^2 + 4x - 4 $$ Answer: $(fg)(x) = 8x^2 + 4x - 4$ 2. Problem: Find $(2f + 3g)(1)$ where $f(x) = 5x - 2$ and $g(x) = 4 - 3x$. Step 1: Compute $2f(x) + 3g(x)$: $$ 2f(x) + 3g(x) = 2(5x - 2) + 3(4 - 3x) $$ $$ = 10x - 4 + 12 - 9x $$ $$ = (10x - 9x) + (-4 + 12) $$ $$ = x + 8 $$ Step 2: Evaluate at $x=1$: $$ 1 + 8 = 9 $$ Answer: $(2f + 3g)(1) = 9$ 3. Problem: Given $f(x) = \sqrt{25 - x^2}$ and $g(x) = \sqrt{x} + 4$, find: A. Domain of $(fg)(x)$ B. Domain of $\left(\frac{f}{g}\right)(x)$ C. Domain of $\left(\frac{g}{f}\right)(x)$ Step 1: Find domain of $f(x)$: $$ 25 - x^2 \geq 0 \implies x^2 \leq 25 \implies -5 \leq x \leq 5 $$ Step 2: Find domain of $g(x)$: $$ x \geq 0 $$ (since $\sqrt{x}$ requires $x \geq 0$) Step 3: Domain of $(fg)(x) = f(x) \cdot g(x)$ is intersection of domains of $f$ and $g$: $$ [0,5] $$ Step 4: Domain of $\left(\frac{f}{g}\right)(x)$ requires $g(x) \neq 0$ and $x$ in domain of both $f$ and $g$. Since $g(x) = \sqrt{x} + 4 > 0$ for all $x \geq 0$, no zeros. So domain is $[0,5]$. Step 5: Domain of $\left(\frac{g}{f}\right)(x)$ requires $f(x) \neq 0$ and $x$ in domain of both. $f(x) = 0$ when $25 - x^2 = 0 \implies x = \pm 5$. Within $[0,5]$, $f(5) = 0$, so exclude $x=5$. Domain is $[0,5)$. Answer: A. Domain of $(fg)(x)$ is $[0,5]$ B. Domain of $\left(\frac{f}{g}\right)(x)$ is $[0,5]$ C. Domain of $\left(\frac{g}{f}\right)(x)$ is $[0,5)$ 4. Problem: Given relation $R = \{(x,y): y \leq x - 2 \text{ and } y \geq 2x - 4\}$ A. Sketch the graph of $R$ (description only) B. Find domain of $R$ C. Find range of $R$ Step 1: The relation is the set of points between two lines: $$ y \leq x - 2 $$ $$ y \geq 2x - 4 $$ Step 2: Find intersection of the lines: $$ x - 2 = 2x - 4 $$ $$ -2 + 4 = 2x - x $$ $$ 2 = x $$ $$ y = 2 - 2 = 0 $$ Intersection point is $(2,0)$. Step 3: Domain is all $x$ values where the region exists. Since $y$ is between two lines, and the lines intersect at $x=2$, check behavior: For $x < 2$, $2x - 4$ is less than $x - 2$, so region exists. For $x > 2$, $2x - 4$ is greater than $x - 2$, so no points satisfy both inequalities. Therefore, domain is $(-\infty, 2]$. Step 4: Range is $y$ values between the two lines over the domain. At $x=2$, $y=0$. As $x \to -\infty$, $x - 2 \to -\infty$ and $2x - 4 \to -\infty$, so range is $(-\infty, 0]$. Answer: A. Region between lines $y \leq x - 2$ and $y \geq 2x - 4$ for $x \leq 2$. B. Domain: $(-\infty, 2]$ C. Range: $(-\infty, 0]$ 5. Problem: For $g(x) = -2x^2 + 6x$, find vertex, line of symmetry, max/min value, and range. Step 1: Identify $a = -2$, $b = 6$, $c = 0$. Step 2: Vertex $x$-coordinate: $$ x = -\frac{b}{2a} = -\frac{6}{2 \times -2} = -\frac{6}{-4} = 1.5 $$ Step 3: Vertex $y$-coordinate: $$ g(1.5) = -2(1.5)^2 + 6(1.5) = -2(2.25) + 9 = -4.5 + 9 = 4.5 $$ Vertex is $(1.5, 4.5)$. Step 4: Since $a = -2 < 0$, parabola opens downward, so vertex is maximum. Step 5: Line of symmetry is $x = 1.5$. Step 6: Maximum value is $4.5$. Step 7: Range is $(-\infty, 4.5]$. Answer: Vertex: $(1.5, 4.5)$ Line of symmetry: $x = 1.5$ Maximum value: $4.5$ Range: $(-\infty, 4.5]$ 6. Problem: Given vertex of $g(x) = x^2 + bx + c$ is $(6, -1)$, find $b$. Step 1: Vertex formula: $$ x = -\frac{b}{2a} $$ Since $a=1$, $$ 6 = -\frac{b}{2} \implies b = -12 $$ Answer: $b = -12$ 7. Problem: Rewrite $g(x) = x^2 - 12x + 5$ in vertex form $a(x - h)^2 + k$ and sketch. Step 1: Complete the square: $$ g(x) = x^2 - 12x + 5 $$ $$ = (x^2 - 12x + 36) - 36 + 5 $$ $$ = (x - 6)^2 - 31 $$ Answer: $g(x) = (x - 6)^2 - 31$ Vertex at $(6, -31)$, parabola opens upward. Slug: "function operations" Subject: "algebra" Desmos: {"latex":"y=(4x-2)(2x+2)","features":{"intercepts":true,"extrema":true}} q_count: 7