The average rate of change of a function $f(x)$ on the interval $[a, b]$ is given by the formula:

$\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}$

In this case, $f(x) = \sqrt{x}$, and the interval is $[9, 9.06]$.

1. Compute $f(9) = \sqrt{9} = 3$.
2. Compute $f(9.06) = \sqrt{9.06}$.

Now, let's calculate:

$f(9.06) = \sqrt{9.06} \approx 3.0100$

Now apply the formula for the average rate of change:

$\text{Average rate of change} = \frac{f(9.06) - f(9)}{9.06 - 9}$ $\text{Average rate of change} = \frac{3.0100 - 3}{9.06 - 9} = \frac{0.0100}{0.06} \approx 0.1667$

Thus, the average rate of change is approximately:

$\boxed{0.1667}$

Would you like more details, or do you have any questions?

Here are 5 related questions:

1. What is the general formula for the average rate of change of any function?
2. How does the average rate of change relate to the slope of a secant line?
3. What happens to the average rate of change as the interval shrinks?
4. Can we find the instantaneous rate of change at a point using limits?
5. How is the average rate of change different from the derivative?

Tip: The average rate of change over an interval gives you the slope of the secant line through the points on the graph at the interval's endpoints.