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Let's work through the problems shown in the image step by step.

### (a) Average Rate of Change over \( 6 \leq t \leq 12 \)

The average rate of change of a function \( f(t) \) over an interval \( [a, b] \) is given by:

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

Here, the function to be analyzed is \( g(t) \) (as we are working with \( 6 \leq t \leq 12 \), where \( f(t) = g(t) \)) from the table. We have:

- \( g(12) = 474 \) (from the table)
- \( g(6) = 306 \) (from the table)

Now, compute the average rate of change:

\[
\frac{g(12) - g(6)}{12 - 6} = \frac{474 - 306}{12 - 6} = \frac{168}{6} = 28 \, \text{cubic meters per hour}
\]

### (b) Approximation of \( g'(9) \)

To approximate \( g'(9) \), we can use a symmetric difference quotient based on the table values around \( t = 9 \). Using the values at \( t = 8 \) and \( t = 10 \), we apply:

\[
g'(9) \approx \frac{g(10) - g(8)}{10 - 8}
\]

From the table:
- \( g(10) = 428 \)
- \( g(8) = 376 \)

Now, compute the approximation:

\[
g'(9) \approx \frac{428 - 376}{10 - 8} = \frac{52}{2} = 26 \, \text{cubic meters per hour}
\]

### (c) Continuity of \( f \) for \( 0 \leq t \leq 12 \)

To determine whether \( f(t) \) is continuous for \( 0 \leq t \leq 12 \), we need to check continuity at the transition point \( t = 6 \), where the definition of \( f(t) \) changes from the formula \( 48t + t^2 - \frac{t^3}{12} \) for \( 0 \leq t < 6 \) to the function \( g(t) \) for \( 6 \leq t \leq 12 \).

For \( f(t) \) to be continuous at \( t = 6 \), we need:

\[
\lim_{t \to 6^-} f(t) = \lim_{t \to 6^+} f(t)
\]

- From the function \( f(t) = 48t + t^2 - \frac{t^3}{12} \) for \( t \leq 6 \), we compute:

\[
f(6) = 48(6) + 6^2 - \frac{6^3}{12} = 288 + 36 - 18 = 306
\]

- From the table, \( g(6) = 306 \), so:

\[
\lim_{t \to 6^+} g(t) = 306
\]

Thus, \( f(t) \) is continuous because the left-hand and right-hand limits at \( t = 6 \) are equal to \( f(6) \).

### (d) Instantaneous Rate of Change of \( f \) at \( t = 2 \)

To find \( f'(2) \), we differentiate the expression for \( f(t) \) for \( 0 \leq t < 6 \):

\[
f(t) = 48t + t^2 - \frac{t^3}{12}
\]

Differentiate term-by-term:

\[
f'(t) = 48 + 2t - \frac{3t^2}{12} = 48 + 2t - \frac{t^2}{4}
\]

Now, substitute \( t = 2 \):

\[
f'(2) = 48 + 2(2) - \frac{2^2}{4} = 48 + 4 - \frac{4}{4} = 52 - 1 = 51 \, \text{cubic meters per hour}
\]

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Do you want more details on any part or have any questions?

### Follow-up Questions:
1. How do you find instantaneous rates of change using limits?
2. What does the average rate of change represent in practical terms?
3. Why is continuity important in real-world models?
4. How do you handle non-differentiable points in piecewise functions?
5. What are common techniques for approximating derivatives from tabular data?

### Tip: 
When dealing with piecewise functions, always check for continuity at the points where the function definition changes, especially in applied problems!

At an excavation site, the amount of dirt removed in cubic meters is modeled by a piecewise function. The problem involves finding the average rate of change, approximating the instantaneous rate of change using table data, checking the continuity of the function, and calculating the derivative for a given time.

This problem covers the analysis of a piecewise function representing dirt removal at an excavation site. Topics include finding average and instantaneous rates of change, checking continuity at transition points, and differentiating the function. Suitable for high school calculus students.

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