To determine which exponential function is represented by the given table, we need to evaluate the functions provided for each [tex]\( x \)[/tex] value in the table and compare the results to the [tex]\( f(x) \)[/tex] values listed.
The table is as follows:
[tex]\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-2 & 12.5 \\
-1 & 2.5 \\
0 & 0.5 \\
1 & 0.1 \\
2 & 0.02 \\
\hline
\end{array}
\][/tex]
The exponential functions to evaluate are:
1. [tex]\( f(x) = 0.2(0.5^x) \)[/tex]
2. [tex]\( f(x) = 0.5(5^x) \)[/tex]
3. [tex]\( f(x) = 0.5(0.2^x) \)[/tex]
4. [tex]\( f(x) = 0.2(0.2^x) \)[/tex]
We proceed by evaluating each function at the given [tex]\( x \)[/tex] values.
### Function 1: [tex]\( f(x) = 0.2(0.5^x) \)[/tex]
[tex]\[
\begin{array}{c|c}
x & f(x) = 0.2(0.5^x) \\
\hline
-2 & 0.2(0.5^{-2}) = 0.2 \left(\frac{1}{0.25}\right) = 0.2 \cdot 4 = 0.8 \\
-1 & 0.2(0.5^{-1}) = 0.2 \left(\frac{1}{0.5}\right) = 0.2 \cdot 2 = 0.4 \\
0 & 0.2(0.5^{0}) = 0.2(1) = 0.2 \\
1 & 0.2(0.5^{1}) = 0.2(0.5) = 0.1 \\
2 & 0.2(0.5^{2}) = 0.2(0.25) = 0.05 \\
\end{array}
\][/tex]
The calculated values do not match the table values, so function 1 is not the correct one.
### Function 2: [tex]\( f(x) = 0.5(5^x) \)[/tex]
[tex]\[
\begin{array}{c|c}
x & f(x) = 0.5(5^x) \\
\hline
-2 & 0.5(5^{-2}) = 0.5 \left(\frac{1}{25}\right) = 0.5 \cdot 0.04 = 0.02 \\
-1 & 0.5(5^{-1}) = 0.5 \left(\frac{1}{5}\right) = 0.5 \cdot 0.2 = 0.1 \\
0 & 0.5(5^{0}) = 0.5(1) = 0.5 \\
1 & 0.5(5^{1}) = 0.5(5) = 2.5 \\
2 & 0.5(5^{2}) = 0.5(25) = 12.5 \\
\end{array}
\][/tex]
The calculated values match but are in the opposite order, indicating a potential inverse relationship but not a direct one suitable for this table, so function 2 is not the correct one.
### Function 3: [tex]\( f(x) = 0.5(0.2^x) \)[/tex]
[tex]\[
\begin{array}{c|c}
x & f(x) = 0.5(0.2^x) \\
\hline
-2 & 0.5(0.2^{-2}) = 0.5 \left(\frac{1}{0.04}\right) = 0.5 \cdot 25 = 12.5 \\
-1 & 0.5(0.2^{-1}) = 0.5 \left(\frac{1}{0.2}\right) = 0.5 \cdot 5 = 2.5 \\
0 & 0.5(0.2^{0}) = 0.5(1) = 0.5 \\
1 & 0.5(0.2^{1}) = 0.5(0.2) = 0.1 \\
2 & 0.5(0.2^{2}) = 0.5(0.04) = 0.02 \\
\end{array}
\][/tex]
The values calculated from function 3 match perfectly with the values provided in the table.
### Function 4: [tex]\( f(x) = 0.2(0.2^x) \)[/tex]
[tex]\[
\begin{array}{c|c}
x & f(x) = 0.2(0.2^x) \\
\hline
-2 & 0.2(0.2^{-2}) = 0.2 \left(\frac{1}{0.04}\right) = 0.2 \cdot 25 = 5 \\
-1 & 0.2(0.2^{-1}) = 0.2 \left(\frac{1}{0.2}\right) = 0.2 \cdot 5 = 1 \\
0 & 0.2(0.2^{0}) = 0.2(1) = 0.2 \\
1 & 0.2(0.2^{1}) = 0.2(0.2) = 0.04 \\
2 & 0.2(0.2^{2}) = 0.2(0.04) = 0.008 \\
\end{array}
\][/tex]
The values calculated from function 4 do not match the values in the table, so function 4 is not the correct one.
Therefore, the exponential function represented by the table is:
[tex]\[ f(x) = 0.5(0.2^x) \][/tex]
