Find the inflection points of the curve y = x e^(1-x)
Find the inflection points of the curve $y = x\,e^{1-x}$.
Explanation of Method
One sign change, one inflection.
Step 1: The derivatives
\[ y' = e^{1-x}-x\,e^{1-x} = e^{1-x}\bigl(1-x\bigr), \]
\[ y'' = -e^{1-x}-\bigl(1-x\bigr)e^{1-x} = e^{1-x}\left[-1-\bigl(1-x\bigr)\right] = e^{1-x}\bigl(x-2\bigr). \]
Step 2: The candidate and its sign
As $e^{1-x}\gt 0$ everywhere, $y'' = 0$ only at $x = 2$, and
\[ x\lt 2:\ y''\lt 0 \ \text{(concave down)}; \qquad x\gt 2:\ y''\gt 0 \ \text{(concave up)}. \]
Step 3: The point
\[ y(2) = 2e^{-1} = \frac{2}{e}. \]
Step 4: Verification
\[y''(1) = -e^{0} = -1\lt 0; \qquad y''(3) = e^{-2} = 0.1353\gt 0 \;\;\checkmark\]
\[\text{The maximum is at } x = 1,\ y(1) = 1,\ \text{to the left of the inflection} \;\;\checkmark\]
\[y\to0^{+} \text{ as } x\to+\infty \text{ and } y\to-\infty \text{ as } x\to-\infty \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\left(2,\;\frac{2}{e}\right) \approx \bigl(2,\ 0.7357589\bigr)\,}\]
Remark
Remark. The pattern $x^{n}e^{-x}$ always places its maximum at $x = n$ and its inflections where $x^{2}-2nx+n(n-1) = 0$; for $n = 1$ that reduces to the single point $x = 2$ found here. The factor $e^{1}$ merely scales the curve and moves nothing.