To find the average rate of change of the function $f(x) = 9x^2 - 3$ over the interval $[1, t]$, we use the formula for the average rate of change:

$\text{Average rate of change} = \frac{f(t) - f(1)}{t - 1}$

1. Evaluate $f(t)$: $f(t) = 9t^2 - 3$

2. Evaluate $f(1)$: $f(1) = 9(1)^2 - 3 = 9 - 3 = 6$

3. Substitute $f(t)$ and $f(1)$ into the formula: $\text{Average rate of change} = \frac{(9t^2 - 3) - 6}{t - 1}$

4. Simplify the expression: $\text{Average rate of change} = \frac{9t^2 - 9}{t - 1}$ $= \frac{9(t^2 - 1)}{t - 1}$ $= \frac{9(t-1)(t+1)}{t-1}$

5. Cancel the common factor $t-1$ (assuming $t \neq 1$): $\text{Average rate of change} = 9(t + 1)$

Thus, the average rate of change of $f(x)$ over the interval $[1, t]$ is:

$9(t + 1)$

Would you like more details or have any questions about this solution?

Here are five related questions to deepen your understanding:

1. How would the solution change if the function were $f(x) = 9x^2 + 3$?
2. What is the significance of the average rate of change in the context of this function?
3. Can you find the instantaneous rate of change at $x = 1$?
4. How would the average rate of change be interpreted if $t \to 1$?
5. What is the average rate of change for a linear function $f(x) = mx + b$?

Tip: When calculating the average rate of change, it's essentially finding the slope of the secant line connecting two points on the graph of the function.