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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)?

Question

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Solution

To find a function rule for f(t) f(t) that increases by a factor of 9 for every unit interval and starts with f(0)=1 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)rt f(t) = f(0) \cdot r^t where r r is the growth factor and f(0) f(0) is the initial value. In this case, f(0)=1 f(0) = 1 and the growth factor r r is 9.

3. Analysis and Detail

Given:

  • f(0)=1 f(0) = 1
  • The function increases by a factor of 9, meaning: f(1)=f(0)9=19=9 f(1) = f(0) \cdot 9 = 1 \cdot 9 = 9 f(2)=f(1)9=99=92 f(2) = f(1) \cdot 9 = 9 \cdot 9 = 9^2 f(3)=f(2)9=929=93 f(3) = f(2) \cdot 9 = 9^2 \cdot 9 = 9^3

From this pattern, we can see that: f(t)=19t=9t f(t) = 1 \cdot 9^t = 9^t

4. Verify and Summarize

Verify with f(0) f(0) : f(0)=90=1 f(0) = 9^0 = 1 This condition is satisfied, confirming the function f(t)=9t f(t) = 9^t increases by a factor of 9 for every unit increase in t t .

Final Answer

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

This problem has been solved

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