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.