the function is *f*(*x*)=3000+15*x*+0.2x^2

Thank you for the correction. Let's work with the function \( f(x) = 3000 + 15x + 0.2x^2 \) and redo the calculations based on this function.

### 1. Average Rate of Change
To find the average rate of change of \( f(x) \) between \( x = 10 \) and \( x = 15 \):

\[
\text{Average Rate of Change} = \frac{f(15) - f(10)}{15 - 10}
\]

First, calculate \( f(10) \) and \( f(15) \):

\[
f(10) = 3000 + 15(10) + 0.2(10)^2 = 3000 + 150 + 20 = 3170
\]
\[
f(15) = 3000 + 15(15) + 0.2(15)^2 = 3000 + 225 + 45 = 3270
\]

Then, compute the average rate of change:

\[
\text{Average Rate of Change} = \frac{3270 - 3170}{5} = \frac{100}{5} = 20
\]

### 2. Instantaneous Rate of Change
The instantaneous rate of change is found by taking the derivative of \( f(x) \):

\[
f'(x) = \frac{d}{dx}[3000 + 15x + 0.2x^2] = 15 + 0.4x
\]

Now, evaluate the derivative at \( x = 10 \) and \( x = 15 \):

At \( x = 10 \):

\[
f'(10) = 15 + 0.4(10) = 15 + 4 = 19
\]

At \( x = 15 \):

\[
f'(15) = 15 + 0.4(15) = 15 + 6 = 21
\]

### 3. Rate of Change Equal to 0 or 1000
To find when the rate of change equals 0 or 1000:

1. **When \( f'(x) = 0 \):**
   \[
   15 + 0.4x = 0
   \]
   \[
   0.4x = -15 \Rightarrow x = \frac{-15}{0.4} = -37.5
   \]

2. **When \( f'(x) = 1000 \):**
   \[
   15 + 0.4x = 1000
   \]
   \[
   0.4x = 985 \Rightarrow x = \frac{985}{0.4} = 2462.5
   \]

### Summary:
- **Average rate of change** between \( x = 10 \) and \( x = 15 \) is **20**.
- **Instantaneous rate of change** at \( x = 10 \) is **19** and at \( x = 15 \) is **21**.
- Rate of change equals 0 at \( x = -37.5 \) and equals 1000 at \( x = 2462.5 \).

Would you like any more details or have any questions?

### Related Questions:
1. How does changing the coefficients of a quadratic function affect its rate of change?
2. What are practical examples of interpreting the instantaneous rate of change in real life?
3. How do you graphically interpret the average rate of change on the curve?
4. What does the point where the rate of change equals zero represent on the graph?
5. How would these calculations change if \( f(x) \) had a cubic term?

### Tip:
Always verify the units and context when interpreting the rate of change, especially in applied problems, as it gives meaning to the numbers calculated.

Explore how to calculate the average and instantaneous rates of change for a quadratic function f(x) = 3000 + 15x + 0.2x^2. Understand the steps involved in finding these rates at specific points and their significance. Suitable for advanced high school students.

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