1. **State the problem:** We need to sketch the curve of the function $$f(x) = x^3 - 3x^2 + 2$$ and analyze its key features. 2. **Formula and rules:** To sketch a cubic function, we find: - The first derivative $$f'(x)$$ to locate critical points (where slope is zero). - The second derivative $$f''(x)$$ to determine concavity and inflection points. - The intercepts with axes. 3. **Find the first derivative:** $$f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 2) = 3x^2 - 6x$$ 4. **Find critical points by setting $$f'(x) = 0$$:** $$3x^2 - 6x = 0$$ $$3x\cancel{(x - 2)} = 0$$ $$\cancel{3}x(x - 2) = 0$$ So, $$x = 0$$ or $$x = 2$$. 5. **Find the second derivative:** $$f''(x) = \frac{d}{dx}(3x^2 - 6x) = 6x - 6$$ 6. **Determine concavity at critical points:** - At $$x=0$$: $$f''(0) = 6(0) - 6 = -6 < 0$$, so local maximum. - At $$x=2$$: $$f''(2) = 6(2) - 6 = 6 > 0$$, so local minimum. 7. **Find function values at critical points:** - $$f(0) = 0^3 - 3(0)^2 + 2 = 2$$ - $$f(2) = 2^3 - 3(2)^2 + 2 = 8 - 12 + 2 = -2$$ 8. **Find y-intercept:** At $$x=0$$, $$f(0) = 2$$ (already found). 9. **Find x-intercepts by solving $$f(x) = 0$$:** $$x^3 - 3x^2 + 2 = 0$$ Try possible rational roots: $$x=1$$: $$1 - 3 + 2 = 0$$, so $$x=1$$ is a root. Divide polynomial by $$x-1$$: $$\frac{x^3 - 3x^2 + 2}{x-1} = x^2 - 2x - 2$$ Solve quadratic: $$x^2 - 2x - 2 = 0$$ $$x = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$$ 10. **Summary of intercepts:** - x-intercepts: $$x=1$$, $$x=1+\sqrt{3}$$, $$x=1-\sqrt{3}$$ - y-intercept: $$2$$ 11. **Find inflection point by setting $$f''(x) = 0$$:** $$6x - 6 = 0$$ $$6(x - 1) = 0$$ $$x = 1$$ Calculate $$f(1)$$: $$1 - 3 + 2 = 0$$ So inflection point at $$(1,0)$$. 12. **Conclusion:** - Local max at $$(0,2)$$ - Local min at $$(2,-2)$$ - Inflection point at $$(1,0)$$ - x-intercepts at $$1$$, $$1+\sqrt{3}$$, $$1-\sqrt{3}$$ - y-intercept at $$2$$ These points help sketch the curve accurately.