1. **Stating the problem:** We are given a function $$F(x) = -x + (x - 2)\ln(2 - x) + e^{x-1}$$ and asked to evaluate the average value of its derivative $$f'(x) = \frac{n(x)}{x-2}$$ over the interval $$[0,1]$$ using the formula for the average value of a function: $$M = \frac{1}{b - a} \int_a^b f(x) \, dx$$ where $$a=0$$ and $$b=1$$. 2. **Understanding the formula:** The average value $$M$$ of a function $$f(x)$$ on $$[a,b]$$ is given by: $$M = \frac{1}{b - a} \int_a^b f(x) \, dx$$ This means we integrate the function over the interval and then divide by the length of the interval. 3. **Relating $$f'(x)$$ and $$F(x)$$:** Since $$F(x)$$ is an antiderivative of $$f(x)$$, by the Fundamental Theorem of Calculus: $$\int_a^b f'(x) \, dx = F(b) - F(a)$$ 4. **Calculate $$F(1)$$ and $$F(0)$$:** $$F(1) = -1 + (1 - 2)\ln(2 - 1) + e^{1-1} = -1 + (-1)\ln(1) + e^0 = -1 + 0 + 1 = 0$$ $$F(0) = -0 + (0 - 2)\ln(2 - 0) + e^{0-1} = 0 - 2\ln(2) + e^{-1} = -2\ln(2) + \frac{1}{e}$$ 5. **Calculate the integral:** $$\int_0^1 f'(x) \, dx = F(1) - F(0) = 0 - \left(-2\ln(2) + \frac{1}{e}\right) = 2\ln(2) - \frac{1}{e}$$ 6. **Calculate the average value $$M$$:** $$M = \frac{1}{1 - 0} \int_0^1 f'(x) \, dx = 2\ln(2) - \frac{1}{e}$$ **Final answer:** $$\boxed{M = 2\ln(2) - \frac{1}{e}}$$