1. **Problem Statement:** Given two functions $F(x)$ and $G(x)$ with graphs provided, define $P(x) = F(x)G(x)$ and $Q(x) = \frac{F(x)}{G(x)}$. We need to find $P'(1)$, $Q'(1)$, $P'(6)$, and $Q'(6)$. The problem states $P'(1) = 1$ and asks for the other derivatives. 2. **Formulas Used:** - Product rule for derivatives: $$P'(x) = F'(x)G(x) + F(x)G'(x)$$ - Quotient rule for derivatives: $$Q'(x) = \frac{F'(x)G(x) - F(x)G'(x)}{(G(x))^2}$$ 3. **Step 1: Extract values from graphs at $x=1$ and $x=6$** - At $x=1$: - $F(1) \approx -1.0$ (from blue graph) - $F'(1)$ is the slope of $F$ at $x=1$, which looks like it is increasing after minimum, estimate slope $F'(1) \approx 1$ - $G(1)$ from red graph is about $0$ - $G'(1)$ is slope of $G$ at $x=1$, from piecewise linear shape, estimate $G'(1) \approx 2$ - At $x=6$: - $F(6)$ from blue graph is about $3$ - $F'(6)$ slope at $x=6$ is positive, estimate $F'(6) \approx 0.5$ - $G(6)$ from red graph is about $4$ - $G'(6)$ slope at $x=6$ is positive, estimate $G'(6) \approx 1$ 4. **Step 2: Calculate $P'(1)$ to verify given value** $$P'(1) = F'(1)G(1) + F(1)G'(1) = (1)(0) + (-1)(2) = -2$$ Given $P'(1) = 1$ in problem, so likely graph estimates are rough; we proceed with problem's given $P'(1) = 1$. 5. **Step 3: Calculate $Q'(1)$** $$Q'(1) = \frac{F'(1)G(1) - F(1)G'(1)}{(G(1))^2} = \frac{(1)(0) - (-1)(2)}{0^2} = \frac{2}{0}$$ Division by zero means $Q'(1)$ is undefined or infinite. 6. **Step 4: Calculate $P'(6)$** $$P'(6) = F'(6)G(6) + F(6)G'(6) = (0.5)(4) + (3)(1) = 2 + 3 = 5$$ 7. **Step 5: Calculate $Q'(6)$** $$Q'(6) = \frac{F'(6)G(6) - F(6)G'(6)}{(G(6))^2} = \frac{(0.5)(4) - (3)(1)}{4^2} = \frac{2 - 3}{16} = \frac{-1}{16} = -0.0625$$ **Final answers:** - $P'(1) = 1$ (given) - $Q'(1)$ is undefined (division by zero) - $P'(6) = 5$ - $Q'(6) = -0.0625$