1. **Problem statement:** Given the function $$f(x) = 0.15x^3 - 0.2x^2 - 1x,$$ we need to analyze its extrema and curvature. 2. **Formula and rules:** Extrema occur where the first derivative $$f'(x)$$ equals zero (necessary condition). The nature of extrema (maxima or minima) is determined by the second derivative $$f''(x)$$: if $$f''(x) > 0$$, it's a minimum; if $$f''(x) < 0$$, it's a maximum. 3. **Find first derivative:** $$f'(x) = \frac{d}{dx}(0.15x^3 - 0.2x^2 - 1x) = 0.45x^2 - 0.4x - 1$$ 4. **Find critical points by solving $$f'(x) = 0$$:** $$0.45x^2 - 0.4x - 1 = 0$$ Divide both sides by 0.05 to simplify: $$\cancel{0.45/0.05}9x^2 - \cancel{0.4/0.05}8x - \cancel{1/0.05}20 = 0$$ Use quadratic formula: $$x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 9 \cdot (-20)}}{2 \cdot 9} = \frac{8 \pm \sqrt{64 + 720}}{18} = \frac{8 \pm \sqrt{784}}{18} = \frac{8 \pm 28}{18}$$ So, $$x_1 = \frac{8 + 28}{18} = 2$$ $$x_2 = \frac{8 - 28}{18} = -\frac{10}{18} = -\frac{5}{9}$$ 5. **Find second derivative:** $$f''(x) = \frac{d}{dx}f'(x) = \frac{d}{dx}(0.45x^2 - 0.4x - 1) = 0.9x - 0.4$$ 6. **Evaluate $$f''(x)$$ at critical points:** - At $$x=2$$: $$f''(2) = 0.9 \cdot 2 - 0.4 = 1.8 - 0.4 = 1.4 > 0$$, so minimum. - At $$x = -\frac{5}{9}$$: $$f''\left(-\frac{5}{9}\right) = 0.9 \cdot \left(-\frac{5}{9}\right) - 0.4 = -0.5 - 0.4 = -0.9 < 0$$, so maximum. 7. **Summary:** - There are two extrema. - At $$x=2$$, a minimum. - At $$x=-\frac{5}{9}$$, a maximum. 8. **Curvature and conjecture:** - The sign of $$f''(x)$$ at critical points determines the type of extremum. - Positive $$f''(x)$$ means minimum, negative means maximum. 9. **Sufficient condition:** - If $$f'(x_0) = 0$$ and $$f''(x_0) \neq 0$$, then $$x_0$$ is an extremum. - If $$f''(x_0) > 0$$, minimum; if $$f''(x_0) < 0$$, maximum. 10. **Function with parameter $$a$$:** $$h(x) = \frac{a}{8}x^3 - 0.2x^2 - a x$$ - Similar analysis applies by computing $$h'(x)$$ and $$h''(x)$$ and studying their zeros and signs.