Let's go over the image step by step to verify the results:

1. Part (a) - Average Rate of Change on the Interval [10, 60]

   • The average rate of change of a function $f$ on an interval $[a, b]$ is calculated as: $\frac{f(b) - f(a)}{b - a}$ From the graph, at $x = 10$, the function $f(10) \approx 480$, and at $x = 60$, $f(60) \approx 780$. Therefore, the average rate of change is: $\frac{f(60) - f(10)}{60 - 10} = \frac{780 - 480}{60 - 10} = \frac{300}{50} = 6$ The answer provided is correct: 6.

2. Part (b) - Interval with Average Rate of Change of 0

   • The average rate of change is zero on an interval where the function is constant, meaning the graph is flat.
   • Looking at the graph, the function appears to be flat between $x = 10$ and $x = 50$, suggesting the correct interval is $[10, 50]$. This choice is correct.

3. Part (c) - Compute the Difference and Explain Geometrically

   • The expression to compute is: $\frac{f(40) - f(0)}{40 - 0}$ Based on the graph, we estimate $f(40) \approx 780$ and $f(0) \approx 600$. Thus: $\frac{f(40) - f(0)}{40 - 0} = \frac{780 - 600}{40} = \frac{180}{40} = 4.5$ The provided answer appears incorrect, but geometrically, this value represents the slope of the line segment from $(0, f(0))$ to $(40, f(40))$. The correct interpretation is selected in the multiple-choice answer.

Would you like further details or explanations on any of these steps?

Here are some related questions you might explore:

1. What is the significance of the average rate of change for different types of functions?
2. How can you interpret intervals where the average rate of change is negative?
3. What are some common applications of calculating the average rate of change in real-world problems?
4. How would you calculate the instantaneous rate of change at a point using a derivative?
5. How can a secant line help in approximating the slope of a tangent line?

Tip: The average rate of change is closely related to the concept of a secant line. Understanding how secant lines approximate tangent lines will help in understanding derivatives.