1. **State the problem:** Find the minimum possible value of the function $$f(x) = \frac{5 - 4x + x^2}{2 - x}$$ for all real numbers $$x < 2$$. 2. **Rewrite the function:** The numerator is $$x^2 - 4x + 5$$ and the denominator is $$2 - x$$. 3. **Domain:** Since the denominator cannot be zero, $$x \neq 2$$, and the problem restricts $$x < 2$$. 4. **Simplify the expression:** Perform polynomial division or rewrite the numerator in terms of the denominator. Note that $$x^2 - 4x + 5 = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1$$. So, $$f(x) = \frac{(x - 2)^2 + 1}{2 - x}$$. 5. **Rewrite denominator:** $$2 - x = -(x - 2)$$, so $$f(x) = \frac{(x - 2)^2 + 1}{-(x - 2)} = -\frac{(x - 2)^2 + 1}{x - 2}$$. 6. **Set $$t = x - 2$$, then $$t < 0$$ (since $$x < 2$$). The function becomes: $$f(t) = -\frac{t^2 + 1}{t} = -\left(t + \frac{1}{t}\right)$$. 7. **Analyze $$g(t) = t + \frac{1}{t}$$ for $$t < 0$$:** - For $$t < 0$$, $$t + \frac{1}{t}$$ has a maximum at $$t = -1$$ because the derivative $$1 - \frac{1}{t^2} = 0$$ at $$t = \pm 1$$. - At $$t = -1$$, $$g(-1) = -1 + \frac{1}{-1} = -1 - 1 = -2$$. - Since $$g(t)$$ has a minimum at $$t = -1$$ for $$t < 0$$, the maximum value of $$g(t)$$ on $$t < 0$$ is $$-2$$. 8. **Recall $$f(t) = -g(t)$$, so the minimum value of $$f(t)$$ is the negative of the maximum of $$g(t)$$. Therefore, minimum of $$f(t) = -(-2) = 2$$. 9. **Answer:** The minimum possible value of $$f(x)$$ for $$x < 2$$ is $$2$$. **Final answer: (C) 2**