= Solution
For the shifted beam <Maxwell-Boltzmann distribution>,
$$
f_b'(v)=\frac{n_b(V-v)}{\sqrt{2\pi}\sigma_b^3}e^{-(v-V)^2/(2\sigma_b^2)}.
$$
On the rising flank put $u=(V-v)/\sigma_b>0$. The positive slope is proportional to $u e^{-u^2/2}$, whose derivative is $(1-u^2)e^{-u^2/2}$. Its unique positive maximum is at $u=1$, giving
$$
\boxed{v_\phi\simeq V-\sigma_b=V-\sqrt{\frac{k_{\rm B}T_b}{m_e}}.}
$$
This is the region where beam drive can most readily overcome the negative plasma slope when that background changes little over the narrow beam width. It is an estimate for the first resonant unstable wave, not a proof that every beam-plasma dispersion relation has its exact marginal phase velocity at this value. Both the total slope and the real dispersion relation determine the exact marginal mode.
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