1. **State the problem:** We have a cost function $C(x) = x^3 - 3x^2 + 4x$ where $x$ is in hundreds of micro-components and $C(x)$ is in hundreds of dollars. (a) Find the derivative $C'(x)$. 2. **Find $C'(x)$:** Using the power rule, $C'(x) = 3x^2 - 6x + 4$. 3. **Marginal cost at 200 micro-components:** Since $x$ is in hundreds, $200$ micro-components correspond to $x=2$. Calculate $C'(2) = 3(2)^2 - 6(2) + 4 = 12 - 12 + 4 = 4$. Interpretation: The marginal cost of producing one additional hundred micro-components at $x=2$ is $4$ hundred dollars, or $400$ dollars. 4. **Revenue function:** $R(x) = 0.6x^3 + x^2 + 10x - 2$. 5. **Profit function:** $P(x) = R(x) - C(x) = (0.6x^3 + x^2 + 10x - 2) - (x^3 - 3x^2 + 4x) = -0.4x^3 + 4x^2 + 6x - 2$. 6. **Derivative of profit function $P'(x)$:** $P'(x) = \frac{d}{dx}(-0.4x^3 + 4x^2 + 6x - 2) = -1.2x^2 + 8x + 6$. 7. **Intervals where $P(x)$ is increasing or decreasing:** Find critical points by solving $P'(x) = 0$: $$-1.2x^2 + 8x + 6 = 0$$ Multiply both sides by $-5/6$ to clear decimals: $$2x^2 - \frac{40}{3}x - 5 = 0$$ Or solve directly: Using quadratic formula: $$x = \frac{-8 \pm \sqrt{8^2 - 4(-1.2)(6)}}{2(-1.2)} = \frac{-8 \pm \sqrt{64 + 28.8}}{-2.4} = \frac{-8 \pm \sqrt{92.8}}{-2.4}$$ Calculate $\sqrt{92.8} \approx 9.63$: $$x_1 = \frac{-8 + 9.63}{-2.4} = \frac{1.63}{-2.4} \approx -0.68$$ $$x_2 = \frac{-8 - 9.63}{-2.4} = \frac{-17.63}{-2.4} \approx 7.35$$ Test intervals: - For $x < -0.68$, pick $x=-1$: $P'(-1) = -1.2(1) + 8(-1) + 6 = -1.2 -8 +6 = -3.2 < 0$ decreasing. - For $-0.68 < x < 7.35$, pick $x=0$: $P'(0) = 6 > 0$ increasing. - For $x > 7.35$, pick $x=8$: $P'(8) = -1.2(64) + 64 + 6 = -76.8 + 70 = -6.8 < 0$ decreasing. So $P(x)$ is decreasing on $(-\infty, -0.68)$, increasing on $(-0.68, 7.35)$, and decreasing on $(7.35, \infty)$. 8. **Optimal production level:** Optimal production occurs where $P'(x) = 0$ and $P(x) > 0$. Check $P(-0.68)$: $$P(-0.68) = -0.4(-0.68)^3 + 4(-0.68)^2 + 6(-0.68) - 2 \approx -0.4(-0.314) + 4(0.462) - 4.08 - 2 = 0.126 + 1.848 - 4.08 - 2 = -4.106 < 0$$ Not positive. Check $P(7.35)$: $$P(7.35) = -0.4(7.35)^3 + 4(7.35)^2 + 6(7.35) - 2$$ Calculate: $7.35^3 \approx 397.5$, $7.35^2 \approx 54.02$ $$P(7.35) = -0.4(397.5) + 4(54.02) + 44.1 - 2 = -159 + 216.08 + 44.1 - 2 = 99.18 > 0$$ Thus, optimal production level is approximately $x = 7.35$ hundred micro-components (735 units) with expected profit about $99.18$ hundred dollars or $9918$ dollars. --- 9. **For function $f(x) = -\frac{1}{x} + \frac{a}{2x^2}$, find $f'(x)$:** Use power rule rewriting terms: $$f(x) = -x^{-1} + \frac{a}{2} x^{-2}$$ Derivative: $$f'(x) = (-1)(-x^{-2}) + \frac{a}{2}(-2)x^{-3} = x^{-2} - a x^{-3} = \frac{1}{x^2} - \frac{a}{x^3}$$ Final answer: $$f'(x) = \frac{1}{x^2} - \frac{a}{x^3}$$