1. **Problem statement:** We need to find the cost function $K(x)$ given the derivative(s) and initial conditions for each part. --- ### a) Given: $$K'(x) = 0.06x^2 - x + 21, \quad K(0) = -600$$ **Step 1:** Integrate $K'(x)$ to find $K(x)$: $$K(x) = \int (0.06x^2 - x + 21) \, dx = 0.06 \cdot \frac{x^3}{3} - \frac{x^2}{2} + 21x + C$$ Simplify coefficients: $$K(x) = 0.02x^3 - 0.5x^2 + 21x + C$$ **Step 2:** Use initial condition $K(0) = -600$: $$K(0) = 0.02 \cdot 0 - 0.5 \cdot 0 + 21 \cdot 0 + C = C = -600$$ **Final function:** $$\boxed{K(x) = 0.02x^3 - 0.5x^2 + 21x - 600}$$ --- ### b) Given: $$K'(x) = 0.2x + 4, \quad K(10) = 15$$ **Step 1:** Integrate $K'(x)$: $$K(x) = \int (0.2x + 4) \, dx = 0.2 \cdot \frac{x^2}{2} + 4x + C = 0.1x^2 + 4x + C$$ **Step 2:** Use $K(10) = 15$: $$15 = 0.1 \cdot 10^2 + 4 \cdot 10 + C = 0.1 \cdot 100 + 40 + C = 10 + 40 + C = 50 + C$$ Solve for $C$: $$C = 15 - 50 = -35$$ **Final function:** $$\boxed{K(x) = 0.1x^2 + 4x - 35}$$ --- ### c) Given: $$K''(x) = 0.18x - 2, \quad K'(10) = 9, \quad K(10) = 280$$ **Step 1:** Integrate $K''(x)$ to find $K'(x)$: $$K'(x) = \int (0.18x - 2) \, dx = 0.18 \cdot \frac{x^2}{2} - 2x + C_1 = 0.09x^2 - 2x + C_1$$ **Step 2:** Use $K'(10) = 9$: $$9 = 0.09 \cdot 10^2 - 2 \cdot 10 + C_1 = 0.09 \cdot 100 - 20 + C_1 = 9 - 20 + C_1 = -11 + C_1$$ Solve for $C_1$: $$C_1 = 9 + 11 = 20$$ So, $$K'(x) = 0.09x^2 - 2x + 20$$ **Step 3:** Integrate $K'(x)$ to find $K(x)$: $$K(x) = \int (0.09x^2 - 2x + 20) \, dx = 0.09 \cdot \frac{x^3}{3} - x^2 + 20x + C_2 = 0.03x^3 - x^2 + 20x + C_2$$ **Step 4:** Use $K(10) = 280$: $$280 = 0.03 \cdot 10^3 - 10^2 + 20 \cdot 10 + C_2 = 0.03 \cdot 1000 - 100 + 200 + C_2 = 30 - 100 + 200 + C_2 = 130 + C_2$$ Solve for $C_2$: $$C_2 = 280 - 130 = 150$$ **Final function:** $$\boxed{K(x) = 0.03x^3 - x^2 + 20x + 150}$$