1. **Problem Statement:** Find the extrema (maximum and minimum points) of the function $$y = x^3 - 3x^2 + 4$$. 2. **Formula and Rules:** To find extrema, we use the first derivative test. Extrema occur where the first derivative $$y'$$ is zero or undefined. 3. **Find the first derivative:** $$y' = \frac{d}{dx}(x^3 - 3x^2 + 4) = 3x^2 - 6x$$ 4. **Set the first derivative equal to zero to find critical points:** $$3x^2 - 6x = 0$$ Factor out $$3x$$: $$3x(x - 2) = 0$$ So, $$x = 0$$ or $$x = 2$$. 5. **Find the second derivative to determine the nature of critical points:** $$y'' = \frac{d}{dx}(3x^2 - 6x) = 6x - 6$$ 6. **Evaluate the second derivative at each critical point:** - At $$x=0$$: $$y''(0) = 6(0) - 6 = -6 < 0$$, so $$x=0$$ is a local maximum. - At $$x=2$$: $$y''(2) = 6(2) - 6 = 12 - 6 = 6 > 0$$, so $$x=2$$ is a local minimum. 7. **Find the corresponding $$y$$ values:** - $$y(0) = 0^3 - 3(0)^2 + 4 = 4$$ - $$y(2) = 2^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0$$ **Final answer:** The function has a local maximum at $$(0, 4)$$ and a local minimum at $$(2, 0)$$.