1. **State the problem:** Find the second derivative $f''(x)$ of the function $$f'(x) = (-3x^2 + 9x - 3)(x^2 - x + 1) - 3$$ using the product rule. 2. **Recall the product rule:** For two functions $u(x)$ and $v(x)$, $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$ 3. **Identify functions:** Let $$u(x) = -3x^2 + 9x - 3$$ $$v(x) = x^2 - x + 1$$ 4. **Find first derivatives:** $$u'(x) = -6x + 9$$ $$v'(x) = 2x - 1$$ 5. **Find $f'(x)$ without the constant -3:** $$f'(x) = u(x)v(x) - 3$$ 6. **Find $f''(x)$:** Differentiate $f'(x)$: $$f''(x) = \frac{d}{dx}[u(x)v(x)] - 0 = u'(x)v(x) + u(x)v'(x)$$ 7. **Substitute:** $$f''(x) = (-6x + 9)(x^2 - x + 1) + (-3x^2 + 9x - 3)(2x - 1)$$ 8. **Expand terms:** $$(-6x + 9)(x^2 - x + 1) = -6x^3 + 6x^2 - 6x + 9x^2 - 9x + 9 = -6x^3 + 15x^2 - 15x + 9$$ $$(-3x^2 + 9x - 3)(2x - 1) = -6x^3 + 3x^2 + 18x^2 - 9x - 6x + 3 = -6x^3 + 21x^2 - 15x + 3$$ 9. **Add the two expansions:** $$f''(x) = (-6x^3 + 15x^2 - 15x + 9) + (-6x^3 + 21x^2 - 15x + 3)$$ $$= -12x^3 + 36x^2 - 30x + 12$$ **Final answer:** $$f''(x) = -12x^3 + 36x^2 - 30x + 12$$