The image presents problem 71 related to the average rate of change and average value of a function. Here's a summary of the given problem:

• Given: A function $f$ is differentiable on the interval $[a, b]$.

• The average rate of change of $f$ over $[a, b]$ is: $\frac{f(b) - f(a)}{b - a}$

• The instantaneous rate of change at any point $x$ is $f'(x)$.

• The chapter defines the average value of a function $f'$ over the interval $[a, b]$, and for consistency, it is required that: $\frac{f(b) - f(a)}{b - a} = \text{average value of } f' \text{ on } [a, b].$

This leads to the idea that the average value of $f'(x)$ over the interval $[a, b]$ is the same as the average rate of change of $f$ over $[a, b]$.

This can be mathematically expressed using the mean value theorem for integrals: $\frac{1}{b-a} \int_a^b f'(x) \, dx = \frac{f(b) - f(a)}{b - a}.$

This shows that the average value of the derivative $f'(x)$ over $[a, b]$ is equivalent to the difference quotient, which is the average rate of change of the function $f$ itself.

Would you like further elaboration on how this connects to the mean value theorem, or a detailed example? Here are five related questions to explore this further:

1. What is the geometric interpretation of the average rate of change of a function?
2. How does the Mean Value Theorem relate to the concept of the average rate of change?
3. Can the equality between the average value of $f'(x)$ and the average rate of change fail for any differentiable function?
4. How can you apply this concept to real-world problems involving average and instantaneous rates of change?
5. How do you calculate the average value of a general continuous function over an interval?

Tip: Remember that the average value of a function over an interval is found by integrating the function and dividing by the length of the interval.