1. **Problem statement:** Find and classify all stationary points of the function $$f(x) = x^3 - 6x^2 + 9x + 2$$. 2. **Find stationary points:** Stationary points occur where the first derivative $$f'(x)$$ equals zero. Given $$f'(x) = 3x^2 - 12x + 9$$, set it to zero: $$3x^2 - 12x + 9 = 0$$ 3. **Solve the quadratic equation:** Divide both sides by 3: $$x^2 - 4x + 3 = 0$$ Factor: $$(x - 3)(x - 1) = 0$$ So, $$x = 1$$ or $$x = 3$$. 4. **Classify stationary points using the second derivative:** Calculate $$f''(x)$$: $$f''(x) = \frac{d}{dx}(3x^2 - 12x + 9) = 6x - 12$$ Evaluate at $$x=1$$: $$f''(1) = 6(1) - 12 = -6 < 0$$, so $$x=1$$ is a local maximum. Evaluate at $$x=3$$: $$f''(3) = 6(3) - 12 = 6 > 0$$, so $$x=3$$ is a local minimum. 5. **Find the function values at stationary points:** $$f(1) = 1 - 6 + 9 + 2 = 6$$ $$f(3) = 27 - 54 + 27 + 2 = 2$$ --- 6. **Second-order partial derivatives:** The function $$f(x)$$ is a single-variable function, so second-order partial derivatives do not apply here. **Final answer:** - Stationary points at $$x=1$$ (local maximum, $$f(1)=6$$) and $$x=3$$ (local minimum, $$f(3)=2$$). - No second-order partial derivatives since $$f$$ is a function of one variable.