1. **State the problem:** Find the values of $x$ where the function $f(x) = x^4 - 18x^2$ has a relative minimum. 2. **Find the first derivative:** $$f'(x) = \frac{d}{dx}(x^4 - 18x^2) = 4x^3 - 36x$$ 3. **Set the first derivative equal to zero to find critical points:** $$4x^3 - 36x = 0$$ 4. **Factor the equation:** $$4x(x^2 - 9) = 0$$ 5. **Solve for $x$:** $$4x = 0 \Rightarrow x = 0$$ $$x^2 - 9 = 0 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$$ 6. **Find the second derivative to determine concavity:** $$f''(x) = \frac{d}{dx}(4x^3 - 36x) = 12x^2 - 36$$ 7. **Evaluate $f''(x)$ at each critical point:** - At $x=0$: $$f''(0) = 12(0)^2 - 36 = -36 < 0$$ (concave down, so relative maximum) - At $x=3$: $$f''(3) = 12(3)^2 - 36 = 12(9) - 36 = 108 - 36 = 72 > 0$$ (concave up, so relative minimum) - At $x=-3$: $$f''(-3) = 12(-3)^2 - 36 = 72 > 0$$ (concave up, so relative minimum) **Final answer:** The function has relative minima at $x = 3$ and $x = -3$. --- 1. **State the problem:** Evaluate the limit $$\lim_{x \to \infty} \frac{5x^3 - 3}{2x^3 + 1}$$ 2. **Divide numerator and denominator by $x^3$ (highest power in denominator):** $$\lim_{x \to \infty} \frac{\frac{5x^3}{x^3} - \frac{3}{x^3}}{\frac{2x^3}{x^3} + \frac{1}{x^3}} = \lim_{x \to \infty} \frac{5 - \frac{3}{x^3}}{2 + \frac{1}{x^3}}$$ 3. **As $x \to \infty$, terms with $\frac{1}{x^3} \to 0$: ** $$\lim_{x \to \infty} \frac{5 - 0}{2 + 0} = \frac{5}{2}$$ **Final answer:** The limit is $\frac{5}{2}$. (Note: The user answer was 1/2, but the correct limit is $\frac{5}{2}$.)