To solve the problem, let's analyze the graph and the provided options.

The graph depicts a curve with some notable points, which are:

• A local minimum around $(0, 1)$
• A local maximum around $(-1.3, 2.2)$
• Another point labeled $(1, 4)$

Now let's evaluate the intervals and their potential local minima:

1. Over the interval $[-3, -2]$: This interval is on the far left side of the graph, where the function seems to decrease to negative values. This does not match any minimum of the function.

2. Over the interval $[-2, -1]$: The graph reaches a local maximum at about $(-1.3, 2.2)$, so this is a local maximum, not a local minimum.

3. Over the interval $[-1, 0.5]$: The graph has a local minimum at $(0, 1)$, which lies within this interval. Therefore, the correct statement is that the local minimum is 1 over this interval.

4. Over the interval $[0.5, 2]$: The graph is increasing in this region, reaching a value of 4 at $(1, 4)$, so this is not a local minimum.

Conclusion:

The correct answer is: "Over the interval $[-1, 0.5]$, the local minimum is 1."

Would you like further clarification on any part of this problem?

Related Questions:

1. What are the conditions for a point to be considered a local minimum in a graph?
2. How can you use the first derivative to identify local minima in a function?
3. What is the difference between a local minimum and a global minimum?
4. How do changes in intervals affect the identification of local extrema?
5. What role does the second derivative play in confirming whether a point is a minimum?

Tip:

A local minimum is a point where the function changes from decreasing to increasing, and the value of the function at that point is lower than the values in its immediate vicinity.