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Consider the differential equation x'(t)=f(x(t)). The line intersects the x-axis at coordinates (-3.5,0), touches the x-axis at(-2.5,0), intersects the x-axis at (-1.5,0), and intersects the x-axis at (-0.5,0). List the constant or equilibrium solutions to this differential equation in increasing order  and indicate whether or not this equilibria are stable, semi-stable (stable from one side, unstable from the other), or unstable.

The given graph represents the function \( f(x) \) and the differential equation is \( x'(t) = f(x(t)) \). Equilibrium solutions occur where \( x'(t) = 0 \), which corresponds to \( f(x) = 0 \). These are the points where the graph of \( f(x) \) intersects the x-axis.

Based on the graph and the provided coordinates, the function intersects or touches the x-axis at the following points:

- \( x = -3.5 \)
- \( x = -2.5 \) (touches the x-axis, meaning it's semi-stable)
- \( x = -1.5 \)
- \( x = -0.5 \)

### Stability of each equilibrium point:
1. **At \( x = -3.5 \)**:
   - The graph of \( f(x) \) crosses the x-axis from positive to negative, indicating that the equilibrium is **stable** (since the flow on both sides moves towards the equilibrium).

2. **At \( x = -2.5 \)**:
   - The graph touches the x-axis and doesn't cross it. On one side, the function stays positive and on the other side, it stays negative. This is a **semi-stable** equilibrium (stable from one side and unstable from the other).

3. **At \( x = -1.5 \)**:
   - The graph crosses the x-axis from negative to positive, indicating that the equilibrium is **unstable** (since the flow on both sides moves away from the equilibrium).

4. **At \( x = -0.5 \)**:
   - The graph crosses the x-axis from positive to negative, indicating that the equilibrium is **stable** (since the flow on both sides moves towards the equilibrium).

### Final Answer:
- \( x = -3.5 \), **stable**
- \( x = -2.5 \), **semi-stable**
- \( x = -1.5 \), **unstable**
- \( x = -0.5 \), **stable**

Would you like further details on this? Here are five related questions:

1. How do you determine the stability of an equilibrium in a differential equation?
2. Can an equilibrium point be both stable and unstable simultaneously?
3. What role does the slope of the function at equilibrium points play in determining stability?
4. How would the graph change if the equilibrium points were shifted horizontally?
5. What is the significance of the differential equation \( x'(t) = f(x(t)) \) in physical systems?

**Tip**: When analyzing stability, remember that if a function crosses the x-axis, the stability is determined by whether it crosses from positive to negative (stable) or negative to positive (unstable).

Consider the differential equation x'(t)=f(x(t)). The line intersects the x-axis at coordinates (-3.5,0), touches the x-axis at(-2.5,0), intersects the x-axis at (-1.5,0), and intersects the x-axis at (-0.5,0). List the constant or equilibrium solutions to this differential equation in increasing order and indicate whether or not this equilibria are stable, semi-stable (stable from one side, unstable from the other), or unstable.

Analyze the stability of equilibrium solutions for the differential equation x'(t) = f(x(t)). Identify where the function intersects or touches the x-axis and determine if each equilibrium point is stable, semi-stable, or unstable. A detailed examination for college-level mathematics.

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