= Solution
<Derivative>[Differentiate] the positive-frequency branch on either side of the origin:
$$
\boxed{v_g(k)=\frac{d\omega}{dk}=v_s\,\operatorname{sgn}(k)\cos(ka/2),\qquad
0<|k|<\pi/a.}
$$
It tends to $+4.0$ km/s as $k\to0^+$ and to $-4.0$ km/s as $k\to0^-$, and vanishes at either <Brillouin zone>[zone boundary]. A positive-$k$ <Wave packet>[wavepacket] therefore transports <energy> to the right, while a negative-$k$ one transports it to the left. At the zone edge, neighboring <Chemical atom>[atoms] oscillate out of phase and the dispersion is flat, giving zero <group velocity> even though the frequency is nonzero.
The graph has separate one-sided limits at zero because $\omega(k)$ has a cusp there. Assigning one differentiable slope at exactly $k=0$ would erase the distinction between the two propagation directions; a precisely uniform translation has no preferred propagation direction. Plotting only $0\leq k\leq\pi/a$ is also legitimate if the missing negative branch is stated. <Group velocity> and <phase velocity> $\omega/k$ are different except in the long-wavelength linear limit.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2016-monatomic-chain-group-velocity.png]
{title=Signed group velocity across the first Brillouin zone}
{description=Original derivative of the positive-frequency acoustic branch. Open markers show the distinct limits at zero wavevector; the derivative is not defined at the cusp itself.}
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