From the previous problem show that $g(k)$ is given by
$$g(k)=\frac{2}{\Delta \bar{v}} \sqrt{\left(S \alpha_C / \pi k\right)-\alpha_C^2}-1$$
and the inverse is given by
$$k(g)=g^{-1}(k)=\frac{S \alpha_C / \pi}{\left[(\Delta \tilde{v} / 2)(1+g)^2\right]+\alpha_c^2}$$