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lfdahl

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\[\frac{\mathrm{d} x}{\mathrm{d} t} = x\left ( 1-\frac{x}{x_0} \right )-R_f\]

where $x_0$ is the equilibrium amount with no fishing and $R_f > 0$ is the constant rate of removal due to fishing. Assume $x(0) = \frac{x_0}{2}$.

(a). If $R_f < \frac{x_0}{4}$, solve for $x(t)$ and show that it tends to an equilibrium amount between $\frac{x_0}{2}$ and $x_0$.

(b). What happens if $R_f \geq \frac{x_0}{4}$?