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With constant-effort harvesting, the constant-effort model isThe extinction multiplier becomes $g^{\prime }(0)-E$, so extinction is linearly stable when $g^{\prime }(0)-1<E<g^{\prime }(0)+1$. At $E_c=g^{\prime }(0)-1$ its multiplier crosses $+1$. Values large enough to make the next population negative lie outside the physical validity of this untruncated harvest model.

Any positive fixed point $N_E$ solves $g(N_E)/N_E=1+E$. Since that ratio is strictly decreasing, there is at most one such point, and it decreases with effort. Differentiating the fixed-point equation yieldsIf the unharvested positive branch exists, it continues for $0\le E<E_c$ and approaches zero as $E\uparrow E_c$. This follows because its required ratio lies between $1$ and $g^{\prime }(0)$; its intersection lies between zero and the unharvested fixed point. There is no positive fixed point for $E\ge E_c$.

The positive-state multiplier is $m_E=g^{\prime }(N_E)-E<1$ by concavity. It is stable precisely whenCrossing $-1$ can change stability through a flip, but the direction of change is not fixed by the three assumptions: $d{m}_E/dE=g^{\prime \prime }(N_E)N_E^{\prime }-1$ has no prescribed sign. In the illustrated saturating example, increasing effort eventually removes the positive branch at extinction. The finite upper stability bound for the algebraic extinction map should not be interpreted as a physically meaningful recovery of a population at unrealistically large harvest effort.