1. **Problem statement:** Perform a full curve discussion (Kurvendiskussion) for the function $$f(x) = (x - 1) \cdot (x + 2)^2$$. 2. **Derivative calculation:** Use the product rule: $$f'(x) = (x - 1)' \cdot (x + 2)^2 + (x - 1) \cdot 2(x + 2)$$. 3. Calculate derivatives: $$ (x - 1)' = 1 $$ $$ f'(x) = 1 \cdot (x + 2)^2 + (x - 1) \cdot 2(x + 2) $$ 4. Expand and simplify: $$ f'(x) = (x + 2)^2 + 2(x - 1)(x + 2) $$ $$ = (x + 2)^2 + 2(x^2 + 2x - x - 2) $$ $$ = (x + 2)^2 + 2(x^2 + x - 2) $$ 5. Expand $(x + 2)^2$: $$ (x + 2)^2 = x^2 + 4x + 4 $$ 6. Substitute back: $$ f'(x) = x^2 + 4x + 4 + 2x^2 + 2x - 4 $$ 7. Combine like terms: $$ f'(x) = (x^2 + 2x^2) + (4x + 2x) + (4 - 4) = 3x^2 + 6x + 0 = 3x^2 + 6x $$ 8. Factor derivative: $$ f'(x) = 3x(x + 2) $$ 9. **Symmetry:** The function is neither even nor odd because of the mixed powers and terms. 10. **Zeros of the function:** Solve $$f(x) = 0$$: $$ (x - 1)(x + 2)^2 = 0 $$ Zeros are at $$x = 1$$ and $$x = -2$$ (with multiplicity 2). 11. **Behavior at infinity:** As $$x \to \infty$$, leading term dominates: $$f(x) \approx x \cdot x^2 = x^3 \to \infty$$. As $$x \to -\infty$$, $$f(x) \approx x^3 \to -\infty$$. 12. **Extrema:** Set $$f'(x) = 0$$: $$3x(x + 2) = 0$$ Solutions: $$x = 0$$ and $$x = -2$$. 13. Determine nature of extrema using second derivative: Calculate $$f''(x)$$: $$f''(x) = (3x^2 + 6x)' = 6x + 6$$ Evaluate at critical points: - At $$x=0$$: $$f''(0) = 6(0) + 6 = 6 > 0$$, so minimum. - At $$x=-2$$: $$f''(-2) = 6(-2) + 6 = -12 + 6 = -6 < 0$$, so maximum. 14. **Wendepunkte (inflection points):** Solve $$f''(x) = 0$$: $$6x + 6 = 0 \Rightarrow x = -1$$. 15. Calculate $$f(-1)$$ for inflection point: $$f(-1) = (-1 - 1) \cdot (-1 + 2)^2 = (-2) \cdot (1)^2 = -2$$. 16. **Summary:** - Zeros: $$x = -2$$ (double root), $$x = 1$$ - Extrema: max at $$x = -2$$, min at $$x = 0$$ - Inflection point at $$(-1, -2)$$ - Behavior at infinity: $$f(x) \to \infty$$ as $$x \to \infty$$, $$f(x) \to -\infty$$ as $$x \to -\infty$$.