1. **Problem 1:** Find $f'(2)$ for $f(x) = \sqrt{6 - x}$. - The function can be rewritten as $f(x) = (6 - x)^{\frac{1}{2}}$. - Using the chain rule, the derivative is $$f'(x) = \frac{1}{2}(6 - x)^{-\frac{1}{2}} \cdot (-1) = -\frac{1}{2\sqrt{6 - x}}.$$ - Evaluate at $x=2$: $$f'(2) = -\frac{1}{2\sqrt{6 - 2}} = -\frac{1}{2\sqrt{4}} = -\frac{1}{2 \cdot 2} = -\frac{1}{4}.$$ 2. **Problem 2:** Solve the equation $$\frac{1}{x-2} = \frac{3}{x+2} - \frac{6x}{x^2 - 4}.$$ - Note that $x^2 - 4 = (x-2)(x+2)$. - Rewrite the right side with common denominator $(x-2)(x+2)$: $$\frac{3}{x+2} = \frac{3(x-2)}{(x+2)(x-2)}, \quad \frac{6x}{x^2 - 4} = \frac{6x}{(x-2)(x+2)}.$$ - So the equation becomes $$\frac{1}{x-2} = \frac{3(x-2) - 6x}{(x-2)(x+2)} = \frac{3x - 6 - 6x}{(x-2)(x+2)} = \frac{-3x - 6}{(x-2)(x+2)}.$$ - Multiply both sides by $(x-2)(x+2)$ (excluding $x=\pm 2$): $$(x+2) = -3x - 6.$$ - Solve for $x$: $$x + 2 = -3x - 6 \implies x + 3x = -6 - 2 \implies 4x = -8 \implies x = -2.$$ - Check domain restrictions: $x \neq \pm 2$, so $x = -2$ is excluded. - No solution remains, so the solution set is empty $\emptyset$. 3. **Problem 3:** Identify the local minimum point of $$f(x) = -3x^4 + 4x^3 + 12x^2 + 1.$$ - Find the first derivative: $$f'(x) = -12x^3 + 12x^2 + 24x.$$ - Set $f'(x) = 0$: $$-12x^3 + 12x^2 + 24x = 0 \implies -12x^3 + 12x^2 + 24x = 0.$$ - Factor out $-12x$: $$-12x(x^2 - x - 2) = 0.$$ - Solve factors: $$-12x = 0 \implies x=0,$$ $$x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0 \implies x=2 \text{ or } x=-1.$$ - Find second derivative: $$f''(x) = -36x^2 + 24x + 24.$$ - Evaluate $f''(x)$ at critical points: - At $x=0$: $$f''(0) = 24 > 0,$$ so local minimum at $x=0$. - At $x=-1$: $$f''(-1) = -36(1) + 24(-1) + 24 = -36 - 24 + 24 = -36 < 0,$$ local maximum. - At $x=2$: $$f''(2) = -36(4) + 24(2) + 24 = -144 + 48 + 24 = -72 < 0,$$ local maximum. - Find $f(0)$: $$f(0) = 1.$$ - So the local minimum point is $(0, 1)$. **Final answers:** 1. $f'(2) = -\frac{1}{4}$. 2. Solution set is $\emptyset$. 3. Local minimum point is $(0, 1)$.