1. **Problem statement:** Given the function $F(x)=3x^5-20x^3$, find its domain, parity, limits at domain boundaries, asymptotes, first and second derivatives with sign tables, intercepts, intersections with asymptotes, other points, max/min points, intervals of increase/decrease, concavity, and inflection points. 2. **Domain:** Polynomial functions are defined for all real numbers, so the domain is $(-\infty, \infty)$. 3. **Parity:** Check if $F(-x)=F(x)$ (even) or $F(-x)=-F(x)$ (odd). Calculate: $$F(-x)=3(-x)^5 - 20(-x)^3 = -3x^5 + 20x^3 = - (3x^5 - 20x^3) = -F(x)$$ Since $F(-x) = -F(x)$, the function is **odd**. 4. **Limits at domain boundaries:** $$\lim_{x \to \infty} F(x) = \lim_{x \to \infty} (3x^5 - 20x^3) = \infty$$ $$\lim_{x \to -\infty} F(x) = \lim_{x \to -\infty} (3x^5 - 20x^3) = -\infty$$ 5. **Asymptotes:** Polynomial functions have no vertical or horizontal asymptotes. 6. **First derivative:** $$F'(x) = \frac{d}{dx}(3x^5 - 20x^3) = 15x^4 - 60x^2$$ Factor: $$F'(x) = 15x^2(x^2 - 4) = 15x^2(x-2)(x+2)$$ 7. **Second derivative:** $$F''(x) = \frac{d}{dx}(15x^4 - 60x^2) = 60x^3 - 120x$$ Factor: $$F''(x) = 60x(x^2 - 2) = 60x(x - \sqrt{2})(x + \sqrt{2})$$ 8. **Sign table for $F'(x)$:** - Critical points at $x = -2, 0, 2$ - Test intervals: - $(-\infty, -2)$: $F'(x) > 0$ - $(-2, 0)$: $F'(x) < 0$ - $(0, 2)$: $F'(x) < 0$ - $(2, \infty)$: $F'(x) > 0$ 9. **Sign table for $F''(x)$:** - Possible inflection points at $x = -\sqrt{2}, 0, \sqrt{2}$ - Test intervals: - $(-\infty, -\sqrt{2})$: $F''(x) < 0$ - $(-\sqrt{2}, 0)$: $F''(x) > 0$ - $(0, \sqrt{2})$: $F''(x) < 0$ - $(\sqrt{2}, \infty)$: $F''(x) > 0$ 10. **Intercepts:** - $y$-intercept at $x=0$: $F(0) = 0$ - $x$-intercepts where $F(x) = 0$: $$3x^5 - 20x^3 = x^3(3x^2 - 20) = 0$$ So $x=0$ or $3x^2 - 20=0 \Rightarrow x= \pm \sqrt{\frac{20}{3}} = \pm \frac{2\sqrt{15}}{3}$ 11. **Intersection with asymptotes:** None, since no asymptotes. 12. **Other points:** For example, $F(1) = 3(1)^5 - 20(1)^3 = 3 - 20 = -17$. 13. **Max and min points:** - From sign of $F'(x)$: - Local max at $x = -2$ (derivative changes from positive to negative) - Local min at $x = 2$ (derivative changes from negative to positive) - Calculate values: $$F(-2) = 3(-2)^5 - 20(-2)^3 = 3(-32) - 20(-8) = -96 + 160 = 64$$ $$F(2) = 3(2)^5 - 20(2)^3 = 3(32) - 20(8) = 96 - 160 = -64$$ 14. **Increasing and decreasing intervals:** - Increasing where $F'(x) > 0$: $(-\infty, -2) \cup (2, \infty)$ - Decreasing where $F'(x) < 0$: $(-2, 0) \cup (0, 2)$ 15. **Concavity and inflection points:** - Concave up where $F''(x) > 0$: $(-\sqrt{2}, 0) \cup (\sqrt{2}, \infty)$ - Concave down where $F''(x) < 0$: $(-\infty, -\sqrt{2}) \cup (0, \sqrt{2})$ - Inflection points at $x = -\sqrt{2}, 0, \sqrt{2}$ - Calculate $F(x)$ at inflection points: $$F(-\sqrt{2}) = 3(-\sqrt{2})^5 - 20(-\sqrt{2})^3 = 3(-4\sqrt{2}) - 20(-2\sqrt{2}) = -12\sqrt{2} + 40\sqrt{2} = 28\sqrt{2}$$ $$F(0) = 0$$ $$F(\sqrt{2}) = 3(\sqrt{2})^5 - 20(\sqrt{2})^3 = 3(4\sqrt{2}) - 20(2\sqrt{2}) = 12\sqrt{2} - 40\sqrt{2} = -28\sqrt{2}$$