To find the average rate of change of the function $g(t)$ over the interval $[0, 3]$, we can apply the formula for the average rate of change between two points on a function:

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

Here, $a = 0$ and $b = 3$.

1. From the graph:
   • At $t = 0$, $g(0) = 1$.
   • At $t = 3$, $g(3) = 4$.

Now we apply the values into the formula:

$\text{Average rate of change} = \frac{g(3) - g(0)}{3 - 0} = \frac{4 - 1}{3 - 0} = \frac{3}{3} = 1$

So, the average rate of change of the function $g(t)$ on the interval $[0, 3]$ is 1.

Would you like more details or have any questions on this?

Here are 5 related questions you might find helpful:

1. What is the significance of the average rate of change for a function?
2. How does the average rate of change differ from the instantaneous rate of change?
3. Can the average rate of change be negative? What would that indicate?
4. How is the slope of the secant line related to the average rate of change?
5. How can you approximate the average rate of change if given a table of values?

Tip: The average rate of change over an interval is the same as the slope of the secant line passing through the two points of the function at the ends of the interval.