1. **Evaluate the limits involving absolute value:** Given the function $$f(x) = \frac{1}{x} - \frac{1}{|x|}$$ - For $$x \to 0^-$$ (approaching zero from the left), $$|x| = -x$$ because $$x$$ is negative. So, $$f(x) = \frac{1}{x} - \frac{1}{-x} = \frac{1}{x} + \frac{1}{x} = \frac{2}{x}$$ As $$x \to 0^-$$, $$\frac{2}{x} \to -\infty$$. - For $$x \to 0^+$$ (approaching zero from the right), $$|x| = x$$. So, $$f(x) = \frac{1}{x} - \frac{1}{x} = 0$$ Therefore, $$\lim_{x \to 0^-} f(x) = -\infty$$ and $$\lim_{x \to 0^+} f(x) = 0$$. 2. **Find the tangent and normal lines to the parabola $$y = x^2 - 8x + 9$$ at $$(3, -6)$$:** - Compute derivative: $$y' = 2x - 8$$ - Slope at $$x=3$$: $$m = 2(3) - 8 = 6 - 8 = -2$$ - Equation of tangent line: $$y - (-6) = -2(x - 3) \Rightarrow y + 6 = -2x + 6 \Rightarrow y = -2x$$ - Slope of normal line is negative reciprocal: $$m_{normal} = \frac{1}{2}$$ - Equation of normal line: $$y + 6 = \frac{1}{2}(x - 3) \Rightarrow y = \frac{1}{2}x - \frac{3}{2} - 6 = \frac{1}{2}x - \frac{15}{2}$$ 3. **Interpretation of derivative $$f'(x)$$ for cost function $$C = f(x)$$:** (a) $$f'(x)$$ represents the rate of change of cost with respect to yards produced, i.e., the marginal cost. Its units are dollars per yard. (b) $$f'(1000) = 9$$ means producing the 1001st yard will increase cost by approximately 9 dollars. (c) Generally, marginal cost may increase with production due to factors like resource limits, so likely $$f'(50) < f'(500) < f'(5000)$$, but exact values depend on $$f$$. 4. **Rock thrown upward on Mars with height $$H = 10t - 1.86t^2$$:** (a) Velocity is derivative of height: $$v(t) = H'(t) = 10 - 3.72t$$ At $$t=1$$: $$v(1) = 10 - 3.72 = 6.28 \text{ m/s}$$ (b) Velocity at $$t=a$$: $$v(a) = 10 - 3.72a$$ (c) Rock hits surface when $$H=0$$: $$10t - 1.86t^2 = 0 \Rightarrow t(10 - 1.86t) = 0$$ Nonzero solution: $$t = \frac{10}{1.86} \approx 5.38 \text{ seconds}$$ (d) Velocity at impact: $$v(5.38) = 10 - 3.72 \times 5.38 = 10 - 20 = -10 \text{ m/s}$$ (negative means downward). 5. **Tangent line to $$y = g(x)$$ at $$x=5$$ with $$g(5) = -3$$ and $$g'(5) = 4$$:** Equation: $$y - (-3) = 4(x - 5) \Rightarrow y + 3 = 4x - 20 \Rightarrow y = 4x - 23$$ 6. **Tangent line to $$y = f(x)$$ at $$(4,3)$$ passes through $$(0,2)$$:** Slope: $$m = \frac{3 - 2}{4 - 0} = \frac{1}{4}$$ So, $$f(4) = 3$$ and $$f'(4) = \frac{1}{4}$$ 7. **Find points where tangent line to $$y = x^4 - 6x^2 + 4$$ is horizontal:** Derivative: $$y' = 4x^3 - 12x$$ Set $$y' = 0$$: $$4x^3 - 12x = 0 \Rightarrow 4x(x^2 - 3) = 0$$ Solutions: $$x = 0, x = \pm \sqrt{3}$$ Corresponding points: $$y(0) = 4$$ $$y(\pm \sqrt{3}) = (\sqrt{3})^4 - 6(\sqrt{3})^2 + 4 = 9 - 18 + 4 = -5$$ Points: $$(0,4), (\sqrt{3}, -5), (-\sqrt{3}, -5)$$ 8. **Rolle's Theorem verification and finding $$c$$:** Since no specific function or interval is given, this cannot be answered here.