1. **State the points at which an extremum can occur, as well as the necessary condition for an extremum.** An extremum occurs where the first derivative $f'(x)$ is zero or undefined. Given: $$f(x) = 0.15x^3 - 0.2x^2 - 1x$$ Calculate the first derivative: $$f'(x) = \frac{d}{dx}(0.15x^3) - \frac{d}{dx}(0.2x^2) - \frac{d}{dx}(1x) = 0.45x^2 - 0.4x - 1$$ Set $f'(x) = 0$ to find critical points: $$0.45x^2 - 0.4x - 1 = 0$$ 2. **Examine the curvature of the function, i.e., the second derivative $f''$, and the type of extremum, i.e., the original function.** Calculate the second derivative: $$f''(x) = \frac{d}{dx}(f'(x)) = \frac{d}{dx}(0.45x^2 - 0.4x - 1) = 0.9x - 0.4$$ The type of extremum at a critical point $x_c$ depends on the sign of $f''(x_c)$: - If $f''(x_c) > 0$, $f$ has a local minimum at $x_c$. - If $f''(x_c) < 0$, $f$ has a local maximum at $x_c$. 3. **Determine a sufficient condition to formulate this relationship.** The sufficient condition for an extremum at $x_c$ is: $$f'(x_c) = 0 \quad \text{and} \quad f''(x_c) \neq 0$$ - If $f''(x_c) > 0$, then $x_c$ is a local minimum. - If $f''(x_c) < 0$, then $x_c$ is a local maximum. --- **Now, solve the quadratic for critical points:** $$0.45x^2 - 0.4x - 1 = 0$$ Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=0.45$, $b=-0.4$, $c=-1$. Calculate discriminant: $$\Delta = (-0.4)^2 - 4 \times 0.45 \times (-1) = 0.16 + 1.8 = 1.96$$ Calculate roots: $$x = \frac{0.4 \pm \sqrt{1.96}}{2 \times 0.45} = \frac{0.4 \pm 1.4}{0.9}$$ So, $$x_1 = \frac{0.4 + 1.4}{0.9} = \frac{1.8}{0.9} = 2$$ $$x_2 = \frac{0.4 - 1.4}{0.9} = \frac{-1}{0.9} \approx -1.111$$ Evaluate $f''$ at these points: $$f''(2) = 0.9 \times 2 - 0.4 = 1.8 - 0.4 = 1.4 > 0 \Rightarrow \text{local minimum at } x=2$$ $$f''(-1.111) = 0.9 \times (-1.111) - 0.4 = -1 - 0.4 = -1.4 < 0 \Rightarrow \text{local maximum at } x \approx -1.111$$ --- **Summary:** - Extremum points occur where $f'(x) = 0$, here at $x=2$ and $x \approx -1.111$. - The second derivative test determines the type: $f''(2) > 0$ means local minimum, $f''(-1.111) < 0$ means local maximum. - The sufficient condition for extrema is $f'(x_c) = 0$ and $f''(x_c) \neq 0$.