1. **Problem Statement:** We are given cost functions $f(x)$ representing the cost of producing $x$ cell phones during summer and winter periods. We need to interpret the meaning of the derivative $f'(x)$ and calculate $f'(2)$ for both given cost functions. 2. **Meaning of the derivative $f'(x)$:** The derivative $f'(x)$ represents the rate of change of the cost with respect to the number of cell phones produced. In other words, $f'(x)$ tells us how much the cost changes when we produce one additional phone at production level $x$. It is the marginal cost at $x$. 3. **Summer cost function:** $$f(x) = (2x^3 + 2x)(2x + 8)^3$$ We want to find $f'(2)$. 4. **Find $f'(x)$ for summer:** Use the product rule: If $f(x) = u(x)v(x)$, then $$f'(x) = u'(x)v(x) + u(x)v'(x)$$ Let $$u(x) = 2x^3 + 2x$$ $$v(x) = (2x + 8)^3$$ Calculate derivatives: $$u'(x) = 6x^2 + 2$$ $$v'(x) = 3(2x + 8)^2 \cdot 2 = 6(2x + 8)^2$$ So, $$f'(x) = (6x^2 + 2)(2x + 8)^3 + (2x^3 + 2x) \cdot 6(2x + 8)^2$$ 5. **Evaluate $f'(2)$:** Calculate each part: $$u'(2) = 6(2)^2 + 2 = 6 \cdot 4 + 2 = 24 + 2 = 26$$ $$v(2) = (2 \cdot 2 + 8)^3 = (4 + 8)^3 = 12^3 = 1728$$ $$u(2) = 2(2)^3 + 2(2) = 2 \cdot 8 + 4 = 16 + 4 = 20$$ $$v'(2) = 6(2 \cdot 2 + 8)^2 = 6(4 + 8)^2 = 6 \cdot 12^2 = 6 \cdot 144 = 864$$ Therefore, $$f'(2) = 26 \cdot 1728 + 20 \cdot 864 = 44928 + 17280 = 62208$$ 6. **Winter cost function:** $$f(x) = x^2 + \frac{1}{x} + 1$$ Find $f'(2)$. 7. **Find $f'(x)$ for winter:** $$f'(x) = 2x - \frac{1}{x^2} + 0 = 2x - \frac{1}{x^2}$$ 8. **Evaluate $f'(2)$:** $$f'(2) = 2 \cdot 2 - \frac{1}{2^2} = 4 - \frac{1}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} = 3.75$$ **Final answers:** - The derivative $f'(x)$ represents the marginal cost, or how cost changes with producing one more phone. - $f'(2)$ during summer is $62208$. - $f'(2)$ during winter is $\frac{15}{4} = 3.75$.