1. **State the problem:** Find the absolute maximum value of the function $f(x) = xe^{-x}$ on the interval $[0,2]$. 2. **Formula and rules:** To find absolute extrema on a closed interval, evaluate the function at critical points inside the interval and at the endpoints. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(xe^{-x}) = e^{-x} - xe^{-x} = e^{-x}(1 - x)$$ 4. **Find critical points by setting $f'(x) = 0$:** $$e^{-x}(1 - x) = 0$$ Since $e^{-x} \neq 0$ for all $x$, we have: $$1 - x = 0 \implies x = 1$$ 5. **Evaluate $f(x)$ at critical points and endpoints:** - At $x=0$: $$f(0) = 0 \cdot e^{0} = 0$$ - At $x=1$: $$f(1) = 1 \cdot e^{-1} = e^{-1}$$ - At $x=2$: $$f(2) = 2 \cdot e^{-2}$$ 6. **Compare values:** - $f(0) = 0$ - $f(1) = e^{-1} \approx 0.3679$ - $f(2) = 2e^{-2} \approx 2 \times 0.1353 = 0.2706$ 7. **Conclusion:** The absolute maximum value on $[0,2]$ is $f(1) = e^{-1}$. **Answer: A) $e^{-1}$**