Given the function $N(t) = 2(3^t) + C$, where $N$ represents the number of people infected and $t$ is the time in days, let's determine the average rate of change between day 1 and day 2.

Average Rate of Change

The average rate of change of a function $N(t)$ over an interval $[t_1, t_2]$ is calculated using the formula: $\text{Average Rate of Change} = \frac{N(t_2) - N(t_1)}{t_2 - t_1}$ For the interval from day 1 to day 2, we have:

• $t_1 = 1$
• $t_2 = 2$

The function is $N(t) = 2(3^t) + C$.

First, calculate $N(1)$ and $N(2)$:

$N(1) = 2(3^1) + C = 2 \cdot 3 + C = 6 + C$

$N(2) = 2(3^2) + C = 2 \cdot 9 + C = 18 + C$

Now, substitute these values into the formula for the average rate of change:

$\text{Average Rate of Change} = \frac{(18 + C) - (6 + C)}{2 - 1} = \frac{18 + C - 6 - C}{1} = \frac{12}{1} = 12$

Answer:

The exact form of the average rate of change between day 1 and day 2 is 12. This is the constant difference in the number of infected people per day between the two days.

Would you like any further details or explanations? Here are some related questions:

1. How would the average rate of change be affected if $C$ were negative or positive?
2. What does this rate of change suggest about the speed of the virus spread between days 1 and 2?
3. How would the average rate of change differ if the interval was between day 0 and day 2?
4. If the function had an additional term (e.g., $+5t$), how would that affect the average rate of change?
5. Could the average rate of change be zero for any interval? Under what circumstances?

Tip: Always remember that the constant $C$ in the function does not affect the rate of change; it only shifts the function vertically.