The function in the image appears to involve $f(x-y)$, and the expression is $(x+2y)(e^{-x+1}) + (x-y)e^{-x+2}$. Let's break it down and solve or simplify it further, depending on the requirement.

Step-by-step solution or simplification:

1. Rewrite the function: $f(x-y) = (x+2y)e^{-x+1} + (x-y)e^{-x+2}$

2. Simplify exponents:

   • $e^{-x+1} = e^{-x} \cdot e^1 = e^{-x}e$
   • $e^{-x+2} = e^{-x} \cdot e^2 = e^{-x}e^2$

   Substituting these values back: $f(x-y) = (x+2y)(e^{-x}e) + (x-y)(e^{-x}e^2)$

3. Factorize $e^{-x}$: $f(x-y) = e^{-x} \left[ (x+2y)e + (x-y)e^2 \right]$

4. Expand terms inside the brackets: $(x+2y)e + (x-y)e^2 = xe + 2ye + xe^2 - ye^2$

   Combine like terms: $= x(e + e^2) + y(2e - e^2)$

5. Final simplified form: $f(x-y) = e^{-x} \left[ x(e + e^2) + y(2e - e^2) \right]$

If this is the desired simplification or solution, it is now ready.

Would you like me to evaluate it for specific values of $x$ and $y$, or expand further? Let me know!

Here are five related questions to extend your understanding:

1. What is the significance of factoring $e^{-x}$ in this expression?
2. How does the behavior of $f(x-y)$ change as $x \to \infty$?
3. Can you find the partial derivatives of $f(x-y)$ with respect to $x$ and $y$?
4. How can this function be applied in real-world problems involving exponential decay?
5. What happens if $y = 0$? How does $f(x-y)$ simplify?

Tip: Always check if you can factor out common exponential terms to simplify expressions efficiently!