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To solve the problem, we need to find the average rate of change of the function \( f(x) = 5x^2 - 6 \) over the interval \([4, b]\).

The average rate of change of a function \( f(x) \) over the interval \([x_1, x_2]\) is given by:

\[
\text{Average Rate of Change} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}
\]

In this case:
- \( f(x) = 5x^2 - 6 \)
- \( x_1 = 4 \)
- \( x_2 = b \)

Let's calculate \( f(4) \) and \( f(b) \):

\[
f(4) = 5(4)^2 - 6 = 5(16) - 6 = 80 - 6 = 74
\]

\[
f(b) = 5b^2 - 6
\]

Now, substituting these into the formula for the average rate of change:

\[
\text{Average Rate of Change} = \frac{f(b) - f(4)}{b - 4} = \frac{(5b^2 - 6) - 74}{b - 4} = \frac{5b^2 - 80}{b - 4}
\]

So, the expression for the average rate of change of \( f(x) \) on the interval \([4, b]\) is:

\[
\frac{5b^2 - 80}{b - 4}
\]

Would you like more details on any part of this solution?

Here are five related questions to deepen your understanding:
1. How would the average rate of change differ if the interval were \([a, b]\) instead of \([4, b]\)?
2. What is the physical interpretation of the average rate of change for this quadratic function?
3. How does the average rate of change relate to the derivative of the function at a point?
4. What happens to the average rate of change as \( b \) approaches 4?
5. How can you use the average rate of change to approximate the slope of the tangent line at a point?

**Tip:** The average rate of change is equivalent to the slope of the secant line between two points on the graph of the function.

Find the average rate of change of f(x) = 5x^2 - 6 on the interval [4, b].

Learn how to calculate the average rate of change for the quadratic function f(x) = 5x^2 - 6 over the interval [4, b]. This detailed solution covers algebraic concepts and provides a step-by-step approach for high school students.

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