1. **Problem Statement:** Given two functions $F(x)$ and $G(x)$ with graphs described, define $P(x) = F(x)G(x)$ and $Q(x) = \frac{F(x)}{G(x)}$. We need to find the derivatives $P'(1)$, $Q'(1)$, $P'(6)$, and $Q'(6)$. 2. **Formulas:** - Product rule for $P(x)$: $$P'(x) = F'(x)G(x) + F(x)G'(x)$$ - Quotient rule for $Q(x)$: $$Q'(x) = \frac{F'(x)G(x) - F(x)G'(x)}{(G(x))^2}$$ 3. **Evaluate at $x=1$:** From the graphs (approximate values): - $F(1) \approx 1$ - $G(1) \approx 2$ - $F'(1)$ is slope of $F$ at $x=1$, approximate from graph: $F$ is increasing, slope $\approx 1$ - $G'(1)$ is slope of $G$ at $x=1$, $G$ is decreasing linearly between (0,-2) and (2,4), slope $= \frac{4 - (-2)}{2 - 0} = 3$ Calculate: $$P'(1) = F'(1)G(1) + F(1)G'(1) = (1)(2) + (1)(3) = 2 + 3 = 5$$ $$Q'(1) = \frac{F'(1)G(1) - F(1)G'(1)}{(G(1))^2} = \frac{(1)(2) - (1)(3)}{2^2} = \frac{2 - 3}{4} = \frac{-1}{4} = -0.25$$ 4. **Evaluate at $x=6$:** From the graphs (approximate values): - $F(6) \approx 5$ - $G(6) \approx 6$ - $F'(6)$ slope of $F$ at $x=6$, $F$ is increasing steeply, approximate slope $\approx 1$ - $G'(6)$ slope of $G$ at $x=6$, $G$ is increasing steeply from (5,0) to (7,7), slope $= \frac{7 - 0}{7 - 5} = \frac{7}{2} = 3.5$ Calculate: $$P'(6) = F'(6)G(6) + F(6)G'(6) = (1)(6) + (5)(3.5) = 6 + 17.5 = 23.5$$ $$Q'(6) = \frac{F'(6)G(6) - F(6)G'(6)}{(G(6))^2} = \frac{(1)(6) - (5)(3.5)}{6^2} = \frac{6 - 17.5}{36} = \frac{-11.5}{36} \approx -0.3194$$ **Final answers:** - (A) $P'(1) = 5$ - (B) $Q'(1) = -0.25$ - (C) $P'(6) = 23.5$ - (D) $Q'(6) \approx -0.3194$