= Solution
{c}
For $r>\delta$, the sign of $f(N)$ is positive on $0<N<N^*$ and negative on $N>N^*$. Its graph passes through $0,N^*$, has positive initial slope $r-\delta$, and eventually behaves like $-\delta N$. The <phase line> therefore points toward $N^*$ from both sides.
The vector field is continuously differentiable, so solutions are unique. No positive solution crosses either equilibrium; one starting below $N^*$ increases and stays below it, and one starting above it decreases and stays above it. These bounded solutions exist for all forward time and have a limit. By continuity of $f$, any such limit must be a zero of $f$: otherwise the derivative would stay bounded away from zero near the limit. The increasing solution is bounded below by its strictly positive initial value, excluding zero; the decreasing one stays above $N^*$. Consequently
$$
\boxed{N(0)>0\quad\Longrightarrow\quad N(t)\longrightarrow N^*.}
$$
The solution starting exactly at $N^*$ is constant. The zero initial condition remains at zero, which is why positivity is stipulated.
\Image[/past-exam-of-the-short-options-course-of-the-university-of-oxford/2021-population-bifurcation.png]
{title=Population growth curve and nonnegative steady-state bifurcation diagram}
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