1. **Problem Statement:** We consider the differential equation $\dot{x} = f(x)$ where $f(x)$ is given by the graph described. 2. **Phase Line and Equilibria:** Equilibrium points occur where $f(x) = 0$. From the graph, these points are approximately at $x \approx 0.6$ and $x \approx 3$. 3. **Classification of Equilibria:** - At $x \approx 0.6$, $f(x)$ changes from negative to positive, so the flow moves from left to right, indicating a source. - At $x \approx 3$, $f(x)$ changes from positive to negative, so the flow moves towards this point, indicating a sink. 4. **Slope Field Sketch:** The slope field shows the direction of $\dot{x}$ at each $x$. Near $x=0.6$, slopes are zero and increase moving right; near $x=3$, slopes are zero and decrease moving right. 5. **Solutions with Initial Conditions:** - For $x(0) = 2$, since $2$ is between the source at $0.6$ and sink at $3$, the solution moves towards the sink at $3$ as $t \to \infty$. So, $\lim_{t \to \infty} x(t) = 3$. - For $x(0) = 1$, which is less than the source at $0.6$, the solution moves away from the source towards $0.6$ but since $1 > 0.6$, it moves away from the source towards the sink at $3$. However, the graph suggests $f(x)$ is positive near $1$, so the solution moves right towards $3$. Thus, $\lim_{t \to \infty} x(t) = 3$. **Final answers:** - Equilibria: $x \approx 0.6$ (source), $x \approx 3$ (sink). - $\lim_{t \to \infty} x(t) = 3$ for both initial conditions $x(0) = 2$ and $x(0) = 1$.