1. **State the problem:** We are given the function $$f(x) = -2x^4 + 36x^2$$ and need to find all intervals where $$f$$ is both increasing and concave down. 2. **Recall definitions:** - A function is **increasing** where its first derivative $$f'(x) > 0$$. - A function is **concave down** where its second derivative $$f''(x) < 0$$. 3. **Find the first derivative:** $$f'(x) = \frac{d}{dx}(-2x^4 + 36x^2) = -8x^3 + 72x$$ 4. **Find the second derivative:** $$f''(x) = \frac{d}{dx}(-8x^3 + 72x) = -24x^2 + 72$$ 5. **Determine where $$f'(x) > 0$$:** Factor $$f'(x)$$: $$f'(x) = -8x^3 + 72x = -8x(x^2 - 9) = -8x(x-3)(x+3)$$ Analyze sign intervals: - For $$x < -3$$: $$x$$ negative, $$x-3$$ negative, $$x+3$$ negative, so $$f'(x) = -8 \times (negative) \times (negative) \times (negative) = -8 \times negative = positive$$ - For $$-3 < x < 0$$: $$x$$ negative, $$x-3$$ negative, $$x+3$$ positive, so $$f'(x) = -8 \times (negative) \times (negative) \times (positive) = -8 \times positive = negative$$ - For $$0 < x < 3$$: $$x$$ positive, $$x-3$$ negative, $$x+3$$ positive, so $$f'(x) = -8 \times (positive) \times (negative) \times (positive) = -8 \times negative = positive$$ - For $$x > 3$$: all positive, so $$f'(x) = -8 \times positive = negative$$ Thus, $$f'(x) > 0$$ on intervals $$(-\infty, -3) \cup (0, 3)$$. 6. **Determine where $$f''(x) < 0$$:** $$f''(x) = -24x^2 + 72 < 0 \implies -24x^2 < -72 \implies x^2 > 3$$ So $$f''(x) < 0$$ for $$x < -\sqrt{3}$$ or $$x > \sqrt{3}$$. 7. **Find intervals where both conditions hold:** - Increasing and concave down means $$f'(x) > 0$$ and $$f''(x) < 0$$. - From step 5, increasing intervals: $$(-\infty, -3) \cup (0, 3)$$. - From step 6, concave down intervals: $$(-\infty, -\sqrt{3}) \cup (\sqrt{3}, \infty)$$. Intersect these: - On the left: $$(-\infty, -3) \cap (-\infty, -\sqrt{3}) = (-\infty, -3)$$ since $$-3 < -\sqrt{3}$$. - On the right: $$(0, 3) \cap (\sqrt{3}, \infty) = (\sqrt{3}, 3)$$. 8. **Final answer:** $$f$$ is increasing and concave down on intervals $$(-\infty, -3)$$ and $$(\sqrt{3}, 3)$$.