= Solution
Differentiation and factorization give $f'(x)=e^{-x^2}(1+x^2-2x^4)=e^{-x^2}(1-x^2)(1+2x^2)$. The exponential and $1+2x^2$ are positive, so the only <stationary points> are at $x=\pm1$. The derivative is positive for $|x|<1$ and negative for $|x|>1$, giving
$$
\boxed{(-1,-2/e)\text{ a global minimum},\qquad(1,2/e)\text{ a global maximum}.}
$$
The function is <odd>, crosses the origin with slope one, and tends to zero from below as $x\to-\infty$ and from above as $x\to\infty$. These facts determine the sketch.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-1-odd-gaussian-profile.png]
{title=Stationary points and Gaussian decay of the odd polynomial-exponential profile}
{description=Original plot of $(x+x^3)e^{-x^2}$, with its global extrema marked. Shading indicates the equal and opposite signed areas on the two half-lines.}
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