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Let's work through this problem based on the graph and questions you've provided. The task is to find the **average rate of change** of the bacterial population with respect to time over different time intervals. The rate of change is essentially the slope of the secant line between two points on the graph, calculated using the formula:

\[
\text{Average Rate of Change} = \frac{\Delta P}{\Delta t} = \frac{P(t_2) - P(t_1)}{t_2 - t_1}
\]

Where:
- \(P(t)\) is the population at time \(t\).
- \(t_1\) and \(t_2\) are the time intervals.

### Part a: Interval [1, 2]
At \(t = 1\), the population is 3 million, and at \(t = 2\), the population is 5 million. So, the rate of change is:

\[
\text{Rate of Change} = \frac{5 - 3}{2 - 1} = \frac{2}{1} = 2 \text{ million per minute}
\]

### Part b: Interval [2, 3]
At \(t = 2\), the population is 5 million, and at \(t = 3\), the population is also 5 million. Thus, the population did not change during this time:

\[
\text{Rate of Change} = \frac{5 - 5}{3 - 2} = \frac{0}{1} = 0 \text{ million per minute}
\]

### Part c: Interval [3, 4]
At \(t = 3\), the population is 5 million, and at \(t = 4\), the population decreases to 4.2 million. So the rate of change is:

\[
\text{Rate of Change} = \frac{4.2 - 5}{4 - 3} = \frac{-0.8}{1} = -0.8 \text{ million per minute}
\]

### Part d: Interval [4, 5]
At \(t = 4\), the population is 4.2 million, and at \(t = 5\), the population is 1 million. So the rate of change is:

\[
\text{Rate of Change} = \frac{1 - 4.2}{5 - 4} = \frac{-3.2}{1} = -3.2 \text{ million per minute}
\]

---

### Part e: When did the population begin to decrease?
Looking at the graph, the population begins to decrease after \(t = 3\) minutes. Before this, the population increases, peaking at 5 million at \(t = 3\).

### Part f: When did the rate of decrease of the population slow down?
The rate of decrease slows down between \(t = 4\) and \(t = 5\). Prior to that, between \(t = 3\) and \(t = 4\), the population decreases by 0.8 million, but between \(t = 4\) and \(t = 5\), it decreases more sharply by 3.2 million.

Would you like more details on any of these parts? Let me know!

Here are some related questions for you:
1. How is the average rate of change related to the slope of a function?
2. What does a positive rate of change signify about the bacterial population?
3. How can we use the first derivative to interpret the instantaneous rate of change?
4. How could you approximate the instantaneous rate of change at \(t = 3\)?
5. What does it mean when the rate of change is zero?

**Tip**: When interpreting graphs, always pay close attention to points where the slope changes direction—these often indicate important behaviors such as population peaks or declines!

Find and interpret the average rate of change of the bacterial population with respect to time for the given time intervals: 1 to 2, 2 to 3, 3 to 4, 4 to 5. Also, determine when the population began to decrease and when the rate of decrease slowed down.

Learn how to find and interpret the average rate of change in a bacterial population over time intervals using the given data. Discover when the population began to decrease and when the rate of decrease slowed down. Suitable for high school students (Grades 9-12).

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