= Solution
Write $\Delta V=V_2-V_1>0$ and $\gamma_i=(1-v_i^2)^{-1/2}$, the initial <Lorentz factor>. The conserved <energy> from part (a) is $E=m\gamma_i+V_1$. To emerge on the higher plateau with positive speed, the particle needs $E>m+V_2$. Hence
$$
\gamma_i>1+\frac{\Delta V}{m},\qquad\boxed{v_i>v_{\mathrm{th}}=\sqrt{1-\left(1+\frac{\Delta V}{m}\right)^{-2}}.}
$$
Above this threshold, <conservation of energy> gives $\gamma_f=\gamma_i-\Delta V/m>1$. The transmitted <velocity> is positive, so
$$
\boxed{v_f=+\sqrt{1-\left[\frac1{\sqrt{1-v_i^2}}-\frac{\Delta V}{m}\right]^{-2}},\qquad v_i>v_{\mathrm{th}}.}
$$
Below threshold there is a turning point within the steep transition, where $\dot x=0$ and $V=E-m<V_2$. The particle returns to the lower plateau. Its final <potential energy> equals its initial value, so its final <Lorentz factor> equals $\gamma_i$ but its direction is reversed:
$$
\boxed{v_f=-v_i,\qquad 0<v_i<v_{\mathrm{th}}.}
$$
At exact threshold the limiting speed on the upper plateau is zero. This is the marginal case, not passage with positive outgoing speed. The ideal discontinuous potential does not specify the detailed trajectory at equality; a smooth transition can make the particle approach the top asymptotically. The final-speed formulas above cover the strictly transmitted and reflected regimes requested.
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