A cyclist travels between point A and point B with a speed represented by the following equation; f'(x)=-3x²+18x What is the maximum speed traveled by the cyclist and what is the distance between the 2 points

To find the maximum speed of the cyclist represented by the function \( f'(x) = -3x^2 + 18x \), we can follow these steps:

1. **Find the critical points**: Set the derivative equal to zero to find the critical points.
   \[
   -3x^2 + 18x = 0
   \]
   Factor out \( -3x \):
   \[
   -3x(x - 6) = 0
   \]
   This gives us \( x = 0 \) and \( x = 6 \).

2. **Determine the maximum speed**: We evaluate \( f'(x) \) at the critical points and at the endpoints of the domain (if defined). Since no specific interval is given, we consider \( x = 0 \) and \( x = 6 \):
   \[
   f'(0) = -3(0)^2 + 18(0) = 0
   \]
   \[
   f'(6) = -3(6)^2 + 18(6) = -3(36) + 108 = -108 + 108 = 0
   \]

3. **Check the second derivative to confirm maxima**:
   \[
   f''(x) = -6x + 18
   \]
   Evaluate \( f''(6) \):
   \[
   f''(6) = -6(6) + 18 = -36 + 18 = -18 \quad (\text{concave down, maximum at } x=6)
   \]

4. **Calculate maximum speed**:
   Since \( f'(0) = 0 \) and \( f'(6) = 0 \) are both not maximum speeds, we need to confirm the nature of these values:
   By analyzing values around the critical points, we find that the function increases until \( x=6 \) and decreases afterward.

   For practical purposes, if the speed is bounded, \( x \) values should be considered in a defined range, typically from 0 to 6 for practical cycling distances.

5. **Determine the distance traveled**:
   Assuming the speed function represents the cyclist's speed over time, the distance can be calculated using:
   \[
   \text{Distance} = \int_{0}^{6} f'(x) \, dx
   \]

   Integrate \( f'(x) \):
   \[
   \int (-3x^2 + 18x) \, dx = -x^3 + 9x^2 + C
   \]
   Evaluating from \( 0 \) to \( 6 \):
   \[
   \left[-x^3 + 9x^2\right]_{0}^{6} = [-(6)^3 + 9(6)^2] - [0] = [-216 + 324] = 108
   \]

Thus, the maximum speed traveled by the cyclist is 0 (the maximum speed occurs at endpoints in this context without a defined speed function), and the distance between points A and B is **108 units**.

### Summary:
- **Maximum Speed**: The function's maximum value needs endpoints to assess speed; both evaluated speeds at critical points were zero.
- **Distance**: **108 units**

Would you like more details on any step? Here are some related questions to consider:

1. What is the significance of the second derivative in this context?
2. How would you evaluate speed at different intervals?
3. Could the function \( f'(x) \) be modified to provide a maximum speed at a different point?
4. How can you graph this function to visualize the speed?
5. What implications does this have for the cyclist's performance over different terrains? 

**Tip**: Always check both the critical points and endpoints when dealing with optimization problems in calculus!

A cyclist travels between point A and point B with a speed represented by the following equation; f'(x)=-3x²+18x What is the maximum speed traveled by the cyclist and what is the distance between the 2 points?

Explore how to determine the maximum speed of a cyclist using the derivative f'(x) = -3x² + 18x. This problem involves finding critical points, evaluating speed, and calculating distance traveled, yielding a maximum speed of 0 and a distance of 108 units. Ideal for advanced high school calculus students.

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