Thank you for the clarification. The function is \( s(t) = \frac{1}{t^{-3}} \), which simplifies to:

\[
s(t) = t^3
\]

Now, to find the average rate of change of the function \( s(t) = t^3 \) on the interval \([0, 1]\), we use the formula:

\[
\text{Average Rate of Change} = \frac{s(b) - s(a)}{b - a}
\]

Substituting \( a = 0 \) and \( b = 1 \):

\[
s(1) = 1^3 = 1
\]

\[
s(0) = 0^3 = 0
\]

Now, compute the average rate of change:

\[
\text{Average Rate of Change} = \frac{1 - 0}{1 - 0} = \frac{1}{1} = 1
\]

So, the average rate of change of the function \( s(t) = t^3 \) on the interval \([0, 1]\) is **1.000** (to three decimal places).

Would you like further details or have any other questions?

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Here are 5 related questions:
1. How does the average rate of change relate to the derivative of a function?
2. What is the significance of the average rate of change being constant over an interval?
3. How would the average rate of change differ if we changed the interval to \([1, 2]\)?
4. Can the average rate of change of a cubic function ever be zero?
5. How does the shape of the graph of a cubic function affect its average rate of change over different intervals?

**Tip:** When computing the average rate of change, always carefully check the interval boundaries and ensure that the function is well-defined at those points.

Learn how to calculate the average rate of change of a cubic function using a step-by-step example. Understand the relationship between the function's values at different points and its rate of change over an interval. Suitable for Grades 11-12 students.

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