1. **State the problem:** We need to find the local maxima and minima of the function $$f(x)=x^3-6x^2+9x+1$$ using the second derivative test. 2. **Find the first derivative:** The first derivative $$f'(x)$$ gives the slope of the function and helps find critical points. $$f'(x) = \frac{d}{dx}(x^3-6x^2+9x+1) = 3x^2 - 12x + 9$$ 3. **Find critical points:** Set $$f'(x) = 0$$ to find critical points. $$3x^2 - 12x + 9 = 0$$ Divide both sides by 3: $$x^2 - 4x + 3 = 0$$ Factor: $$(x-3)(x-1) = 0$$ So, critical points are $$x=1$$ and $$x=3$$. 4. **Find the second derivative:** $$f''(x) = \frac{d}{dx}(3x^2 - 12x + 9) = 6x - 12$$ 5. **Apply the second derivative test:** - At $$x=1$$: $$f''(1) = 6(1) - 12 = 6 - 12 = -6 < 0$$, so $$f$$ has a local maximum at $$x=1$$. - At $$x=3$$: $$f''(3) = 6(3) - 12 = 18 - 12 = 6 > 0$$, so $$f$$ has a local minimum at $$x=3$$. 6. **Find the function values at critical points:** - $$f(1) = 1^3 - 6(1)^2 + 9(1) + 1 = 1 - 6 + 9 + 1 = 5$$ - $$f(3) = 3^3 - 6(3)^2 + 9(3) + 1 = 27 - 54 + 27 + 1 = 1$$ **Final answer:** - Local maximum at $$x=1$$ with value $$f(1)=5$$. - Local minimum at $$x=3$$ with value $$f(3)=1$$.