1. **Problem Statement:** We are given a differential equation $$\frac{dP}{dt} = f(P)$$ with initial value $$P(t_0) = P_0$$. We want to find for which positive initial values $$P_0$$ the solutions are strictly decreasing for all time. 2. **Key Concept:** A solution $$P(t)$$ is strictly decreasing if its derivative is negative for all time, i.e., $$\frac{dP}{dt} = f(P) < 0$$ for all $$t$$. 3. **Interpretation of the slope field:** The slope field shows the sign of $$f(P)$$ at different values of $$P$$. Since $$f$$ depends only on $$P$$, the sign of $$f(P)$$ determines whether the solution increases or decreases at that value. 4. **From the description:** The slope field lines decrease as $$P$$ increases, indicating that $$f(P)$$ transitions from positive to negative as $$P$$ increases. 5. **Conclusion:** For the solution to be strictly decreasing for all time starting at $$P_0 > 0$$, we need $$f(P_0) < 0$$ and also for all subsequent values of $$P(t)$$ along the solution. 6. **Since $$f(P)$$ depends only on $$P$$, the solution will move according to the sign of $$f(P)$$. If $$f(P) < 0$$ for all $$P > a$$ for some $$a > 0$$, then starting at any $$P_0 > a$$, the solution will decrease. 7. **From the slope field description, the slopes are negative for larger $$P$$ values (above some positive threshold) and positive below it. Thus, the solutions are strictly decreasing for all time if $$P_0$$ is in the interval where $$f(P) < 0$$, which is for $$P_0 > c$$ for some positive constant $$c$$. 8. **Therefore, the interval of positive initial values for which solutions are strictly decreasing is $$\boxed{(c, \infty)}$$ where $$c$$ is the positive value where $$f(P)$$ changes sign from positive to negative. Since the exact function $$f(P)$$ is unknown but the slope field indicates the sign change near a positive value, the answer is: $$P_0 \in (c, \infty)$$ for some positive $$c$$. This means all positive initial values greater than $$c$$ yield strictly decreasing solutions.