Concavity intervals and inflection points of f(x) = x^2 e^(-x)

Using the product rule, $f'(x) = 2xe^{-x} - x^2e^{-x} = e^{-x}(2x - x^2)$. Differentiating again with the product rule: $f''(x) = e^{-x}(2 - 2x) - e^{-x}(2x - x^2) = e^{-x}(x^2 - 4x + 2)$. Since $e^{-x} > 0$ for all $x$, the sign of $f''(x)$ is determined entirely by the quadratic factor $x^2 - 4x + 2$.

Setting $x^2 - 4x + 2 = 0$ and using the quadratic formula gives $x = \frac{4 \pm \sqrt{16-8}}{2} = 2 \pm \sqrt{2}$. Testing values shows the quadratic (and hence $f''$) is positive for $x < 2-\sqrt{2}$, negative for $2-\sqrt{2} < x < 2+\sqrt{2}$, and positive again for $x > 2+\sqrt{2}$, so the sign of $f''$ changes at both points. Thus $f$ is concave up on $(-\infty, 2-\sqrt{2})$ and $(2+\sqrt{2}, \infty)$, concave down on $(2-\sqrt{2}, 2+\sqrt{2})$, with inflection points at $x = 2-\sqrt{2}$ and $x = 2+\sqrt{2}$.

The two inflection points are at two minus square root two and two plus square root two, and x equals two lies between them. On that middle interval, the second derivative is negative, so the graph is concave down. This matches the earlier finding that f is concave down on the interval between the two inflection points.