= Solution
With $s=0$, the <equilibrium of an autonomous differential equation>[equilibria] satisfy
$$
g\left(\frac{kg}{1+g^2}-1\right)=0.
$$
Besides $g=0$, the two positive equilibria are
$$
g_\pm=\frac{k\pm\sqrt{k^2-4}}2,
\qquad
g_-<1<g_+,
$$
where $g_-g_+=1$. Since
$$
\partial_gf(g,0)=\frac{2kg}{(1+g^2)^2}-1,
$$
the nonzero equilibrium relation $k=(1+g^2)/g$ gives
$$
\partial_gf(g,0)=\frac{1-g^2}{1+g^2}.
$$
The <fixed point stability for an autonomous differential equation> therefore shows that $g=0$ and $g_+$ are stable, while $g_-$ is unstable. The graph of $f(g,0)$ starts at zero with negative slope, crosses upward at $g_-$, crosses downward at $g_+$, and tends to $-\infty$ as $g\to\infty$.
For a constant input $s$, write the equilibrium equation as
$$
s=h(g),
\qquad
h(g)=g-\frac{kg^2}{1+g^2}.
$$
The low stable equilibrium and the intervening unstable equilibrium coalesce in a <saddle-node bifurcation>. More precisely, let $g_c$ be the first positive solution of
$$
h'(g_c)=1-\frac{2kg_c}{(1+g_c^2)^2}=0
$$
and define
$$
\boxed{s_c(k)=h(g_c)
=g_c-\frac{kg_c^2}{1+g_c^2}.}
$$
Equivalently, $f(g_c,s_c)=0$ and $\partial_gf(g_c,s_c)=0$, with $g_c$ on the low-concentration branch. If $s_1>s_c(k)$, that branch no longer exists. Holding the input long enough carries the trajectory into the <basin of attraction> of the high state. When the input returns to zero, the concentration converges to
$$
\boxed{g^*=g_+=\frac{k+\sqrt{k^2-4}}2.}
$$
This is the <saturating autocatalytic switch>.
For $k\gg1$, the threshold occurs at $g=O(k^{-1})$, so
$$
f(g,s)=s+kg^2-g+O(kg^4).
$$
The approximate double-root conditions are
$$
s+kg^2-g=0,
\qquad
2kg-1=0.
$$
Thus $g_c\simeq1/(2k)$ and
$$
\boxed{s_c(k)\simeq\frac1{4k}.}
$$
Hence the constant in $s_c(k)\simeq Ck^{-1}$ is $\boxed{C=1/4}$, as recorded by the <strong-autocatalysis switching threshold>.
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