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!