1. **State the problem:** We analyze the function f represented by curve (C) and line (d) in the given graphs. 2. **Domain of definition of f:** From the graph, the curve (C) is defined for all real x from about -5 to 7, so the domain is $[-5,7]$. 3. **Find $f(0)$, $f(1)$, $f(2)$, $f(-1)$, $f(-2)$ graphically:** - $f(0)$ is the y-value at $x=0$, approximately $f(0) = 2$ - $f(1)$ is about $3$ - $f(2)$ is about $0$ - $f(-1)$ is about $3$ - $f(-2)$ is about $4$ 4. **Find antecedents of 0 and 4 by f:** - $f(x) = 0$ at $x \\approx -3$ and $x \\approx 2$ - $f(x) = 4$ at $x = -2$ 5. **Solve:** a) $f(x) = 0$ solutions: $x = -3, 2$ b) $f(x) = 4$ solution: $x = -2$ 6. **Solve inequalities:** - $f(x) \\geq 0$ for $x \\in [-3,2]$ approximately - $f(x) > 0$ for $x \\in (-3,2)$ - $f(x) \\leq 0$ outside $[-3,2]$ - $f(x) < 0$ outside $(-3,2)$ - $f(x) > 4$ only at $x = -2$ - $f(x) \\leq 4$ for all $x$ except possibly at $x=-2$ 7. **Equation of (d):** Given as $y = x + 2$. 8. **Solve:** a) $f(x) = x + 2$ find intersection points A and B at $x = -3$ and $x = 3$. b) $f(x) > x + 2$ between $x = -3$ and $x = 3$. c) $f(x) - x - 2 < 0$ outside $[-3,3]$. 9. **Define $g(x) = \frac{1}{f(x) - 4}$ and $h(x) = \sqrt{f(x) - x - 2}$:** - Domain of $g$: $f(x) \neq 4$, so exclude $x = -2$. - Domain of $h$: $f(x) - x - 2 \geq 0$. 10. **Suppose $f(x) = ax^3 + bx + c$:** Using points: - $f(-3) = -1$ gives $-27a - 3b + c = -1$ - $f(-2) = 4$ gives $-8a - 2b + c = 4$ - $f(3) = 5$ gives $27a + 3b + c = 5$ Solving system: - Subtract second from first: $-19a - b = -5$ - Add first and third: $0a + 0b + 2c = 4$ so $c=2$ - Substitute $c=2$ into second: $-8a - 2b + 2 = 4$ so $-8a - 2b = 2$ From $-19a - b = -5$ multiply by 2: $-38a - 2b = -10$ Subtract from $-8a - 2b = 2$: $(-8a - 2b) - (-38a - 2b) = 2 - (-10)$ gives $30a = 12$ so $a = \frac{2}{5}$ Then $-19(\frac{2}{5}) - b = -5$ gives $-\frac{38}{5} - b = -5$ so $b = -\frac{38}{5} + 5 = -\frac{38}{5} + \frac{25}{5} = -\frac{13}{5}$ Final coefficients: $a = \frac{2}{5}$, $b = -\frac{13}{5}$, $c = 2$. --- **Second graph (red curve):** 1. Domain of $f$: all real except $x=0$ (vertical asymptote). 2. $f(1) = 0$, $f(-1) = 2$ from graph. 3. Antecedent of 1: no $x$ such that $f(x) = 1$ (horizontal asymptote). 4. Solve: a) $f(x) > 0$ for $x > 0$ and $x < 0$ where curve is above x-axis. b) $f(x) > 2$ for $x < 0$ near $-1$. 5. Suppose $f(x) = a + \frac{b}{x}$. - From $f(1) = 0$: $a + b = 0$ - From $f(-1) = 2$: $a - b = 2$ Adding: $2a = 2 \Rightarrow a = 1$ Then $b = -a = -1$ 6. $g(x) = \frac{1}{f(x)}$ domain excludes $x=0$ and points where $f(x) = 0$ (i.e., $x=1$). **Desmos latex:** $y=\frac{2}{5}x^3 - \frac{13}{5}x + 2$