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 gives the critical points where the slope is zero or undefined. $$f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + 1) = 3x^2 - 12x + 9$$ 3. **Find critical points:** Set the first derivative equal to zero and solve for $x$. $$3x^2 - 12x + 9 = 0$$ Divide both sides by 3: $$x^2 - 4x + 3 = 0$$ Factor: $$(x - 3)(x - 1) = 0$$ So, critical points are at $$x = 1$$ 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 $$f(3) = 1$$.