To determine the function that best represents the relationship between [tex]\( n \)[/tex] (the number of time intervals) and [tex]\( f(n) \)[/tex] (the amount of chlorine remaining), we need to examine the given data. The data provides the following values:
[tex]\[
\begin{tabular}{|l|l|}
\hline
n & f(n) \\
\hline
1 & 16 \\
\hline
2 & 8 \\
\hline
3 & 4 \\
\hline
4 & 2 \\
\hline
\end{tabular}
\][/tex]
From this data, we can infer the pattern and check if a potential function fits the given values.
The values of [tex]\( f(n) \)[/tex] decrease by a factor of 2 as [tex]\( n \)[/tex] increases by 1. This suggests that the amount of chlorine is halved each time period.
We propose a function of the form:
[tex]\[ f(n) = 16 \cdot (0.5)^{n-1} \][/tex]
Let's verify this function step-by-step with the given values of [tex]\( n \)[/tex]:
1. For [tex]\( n = 1 \)[/tex]:
[tex]\[
f(1) = 16 \cdot (0.5)^{1-1} = 16 \cdot (0.5)^0 = 16 \cdot 1 = 16
\][/tex]
This matches the value [tex]\( f(1) = 16 \)[/tex].
2. For [tex]\( n = 2 \)[/tex]:
[tex]\[
f(2) = 16 \cdot (0.5)^{2-1} = 16 \cdot (0.5)^1 = 16 \cdot 0.5 = 8
\][/tex]
This matches the value [tex]\( f(2) = 8 \)[/tex].
3. For [tex]\( n = 3 \)[/tex]:
[tex]\[
f(3) = 16 \cdot (0.5)^{3-1} = 16 \cdot (0.5)^2 = 16 \cdot 0.25 = 4
\][/tex]
This matches the value [tex]\( f(3) = 4 \)[/tex].
4. For [tex]\( n = 4 \)[/tex]:
[tex]\[
f(4) = 16 \cdot (0.5)^{4-1} = 16 \cdot (0.5)^3 = 16 \cdot 0.125 = 2
\][/tex]
This matches the value [tex]\( f(4) = 2 \)[/tex].
Since the proposed function [tex]\( f(n) = 16 \cdot (0.5)^{n-1} \)[/tex] correctly produces the given values for [tex]\( f(n) \)[/tex] at [tex]\( n = 1, 2, 3, \)[/tex] and [tex]\( 4 \)[/tex], we can confirm that this is the function that best describes the relationship.
Thus, the function that best shows the relationship between [tex]\( n \)[/tex] and [tex]\( f(n) \)[/tex] is:
[tex]\[ f(n) = 16 \cdot (0.5)^{n-1} \][/tex]
