1. **State the problem:** We need to plot the function $f(x) = x e^{-x}$ and identify key points: the origin $(0,0)$, the local maximum at $(1, \frac{1}{e})$, and the inflection point at $(2, \frac{2}{e^2})$. 2. **Formula and important rules:** The function is $f(x) = x e^{-x}$. The exponential decay $e^{-x}$ causes the function to approach zero as $x \to +\infty$. As $x \to -\infty$, $x$ becomes large negative and $e^{-x}$ grows exponentially, so $f(x) \to -\infty$. 3. **Key points:** - At $x=0$, $f(0) = 0 \cdot e^0 = 0$. - Local maximum at $x=1$: $f(1) = 1 \cdot e^{-1} = \frac{1}{e} \approx 0.37$. - Inflection point at $x=2$: $f(2) = 2 \cdot e^{-2} = \frac{2}{e^2} \approx 0.27$. 4. **Behavior:** - As $x \to -\infty$, $f(x) \to -\infty$. - As $x \to +\infty$, $f(x) \to 0$ from above. 5. **Graph features:** The curve starts from negative infinity on the left, passes through the origin, rises to a local maximum at $(1, \frac{1}{e})$, then decreases, passing through the inflection point at $(2, \frac{2}{e^2})$, and approaches zero asymptotically. Final answer: The function $f(x) = x e^{-x}$ has the described behavior and key points as above, plotted with a professional blue curve.