A function f(t) increases by a factor of 9 over every unit interval in t and f(0)=1.Which could be a function rule for f(t)?

To find a function rule for $f(t)$ that increases by a factor of 9 for every unit interval and starts with $f(0) = 1$, we can follow these steps:

1. Break Down the Problem

We need to determine a function that represents exponential growth, given the conditions provided (growing by a factor of 9 over each unit interval).

2. Relevant Concepts

The general form of an exponential function is: $f(t) = f(0) \cdot r^t$ where $r$ is the growth factor and $f(0)$ is the initial value. In this case, $f(0) = 1$ and the growth factor $r$ is 9.

3. Analysis and Detail

Given:

• $f(0) = 1$
• The function increases by a factor of 9, meaning: $f(1) = f(0) \cdot 9 = 1 \cdot 9 = 9$ $f(2) = f(1) \cdot 9 = 9 \cdot 9 = 9^2$ $f(3) = f(2) \cdot 9 = 9^2 \cdot 9 = 9^3$

From this pattern, we can see that: $f(t) = 1 \cdot 9^t = 9^t$

4. Verify and Summarize

Verify with $f(0)$: $f(0) = 9^0 = 1$ This condition is satisfied, confirming the function $f(t) = 9^t$ increases by a factor of 9 for every unit increase in $t$.

Final Answer

The function rule for $f(t)$ is: $f(t) = 9^t$