= Solution
Put $X=z/\epsilon$. The travelling equations become $cv_X=f(v)-w+v_{XX}$ and $cw_X=\epsilon(\gamma v-w)$. Hence, for nonzero $O(1)$ speed and $\epsilon\gamma\ll1$, $w$ is constant to leading order across the front.
Writing $p=v_X$ gives the planar system $v_X=p$, $p_X=cp-f(v)+w$. At the outer cubic roots $A=(V_-,0)$ and $B=(V_+,0)$, $f'(V_\pm)<0$. The <Jacobian determinant> is $f'(V_\pm)<0$, so both are <saddle points>. The increasing front is a <heteroclinic orbit> from the <unstable manifold> of $A$ to the <stable manifold> of $B$. Varying $c$ supplies the shooting parameter required to match these manifolds. For this cubic one can also construct it explicitly, as in the next solution.
Multiply the second-order equation by $v_X$ and integrate across the front. Because $v_X\to0$ at both ends, the term $\int v_{XX}v_X\,dX=[v_X^2/2]_{-\infty}^\infty$ vanishes. The remaining identity is
$$
\boxed{c(w)=\frac{\displaystyle\int_{V_-(w)}^{V_+(w)}[f(v)-w]\,dv}{\displaystyle\int_{-\infty}^\infty v_X^2\,dX}.}
$$
The denominator is positive for a nonconstant front, so its direction is determined by the signed <area> between the cubic and the horizontal recovery line: <Maxwell balance for a scalar reaction-diffusion front> gives zero speed when that <area> vanishes.
The <derivative> in the denominator must use the inner coordinate $X$. If one instead uses the source's original $z$ and denotes $J_z=\int v_z^2\,dz$, then $J_z=\epsilon^{-1}\int v_X^2\,dX$ and the identity is $c=\int(f-w)\,dv/(\epsilon J_z)$. Thus the printed $v'(z)$ is valid only if $z$ has silently been reused for the rescaled inner coordinate; with the original coordinate it misses a factor $\epsilon$.
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