1. **State the problem:** Find the points of inflection of the function $$f(x) = - \frac{x^3 - 2x^2 + x - 1}{x - 2}, \quad x \neq 2.$$ 2. **Recall the definition:** Points of inflection occur where the second derivative changes sign, i.e., where $$f''(x) = 0$$ or is undefined, and the concavity changes. 3. **Simplify the function:** First, perform polynomial division on the numerator by the denominator to simplify $$f(x)$$. Divide $$x^3 - 2x^2 + x - 1$$ by $$x - 2$$: $$x^3 - 2x^2 + x - 1 = (x - 2)(x^2 + 1) + 1$$ So, $$f(x) = - \frac{(x - 2)(x^2 + 1) + 1}{x - 2} = - \left(x^2 + 1 + \frac{1}{x - 2}\right) = -x^2 - 1 - \frac{1}{x - 2}.$$ 4. **Find the first derivative:** $$f'(x) = -2x - 0 - \left(-\frac{1}{(x - 2)^2}\right) = -2x + \frac{1}{(x - 2)^2}.$$ 5. **Find the second derivative:** $$f''(x) = -2 - \frac{d}{dx} \left(-\frac{1}{(x - 2)^2}\right) = -2 - \left(-2 \cdot \frac{1}{(x - 2)^3}\right) = -2 - \left(-\frac{2}{(x - 2)^3}\right) = -2 + \frac{2}{(x - 2)^3}.$$ 6. **Set the second derivative equal to zero to find candidates for inflection points:** $$-2 + \frac{2}{(x - 2)^3} = 0$$ $$\frac{2}{(x - 2)^3} = 2$$ $$\frac{\cancel{2}}{(x - 2)^3} = \cancel{2}$$ $$\frac{1}{(x - 2)^3} = 1$$ 7. **Solve for $$x$$:** $$(x - 2)^3 = 1$$ $$x - 2 = 1$$ $$x = 3.$$ 8. **Check the domain:** $$x = 3$$ is in the domain since $$x \neq 2$$. 9. **Verify concavity change:** For $$x < 3$$, pick $$x = 2.5$$: $$(2.5 - 2)^3 = 0.5^3 = 0.125 > 0$$ $$f''(2.5) = -2 + \frac{2}{0.125} = -2 + 16 = 14 > 0$$ (concave up) For $$x > 3$$, pick $$x = 4$$: $$(4 - 2)^3 = 2^3 = 8 > 0$$ $$f''(4) = -2 + \frac{2}{8} = -2 + 0.25 = -1.75 < 0$$ (concave down) Concavity changes from up to down at $$x=3$$, confirming an inflection point. **Final answer:** The function has a point of inflection at $$x = 3$$.