To determine which exponential function fits the given data table, we need to compare the provided data points with the candidate functions and see which function matches the data points exactly.
The given data points are:
- For [tex]\( x = -2 \)[/tex], [tex]\( f(x) = 12.5 \)[/tex]
- For [tex]\( x = -1 \)[/tex], [tex]\( f(x) = 2.5 \)[/tex]
- For [tex]\( x = 0 \)[/tex], [tex]\( f(x) = 0.5 \)[/tex]
- For [tex]\( x = 1 \)[/tex], [tex]\( f(x) = 0.1 \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( f(x) = 0.02 \)[/tex]
We have four candidate functions:
1. [tex]\( f(x) = 0.2 \left( 0.5^x \right) \)[/tex]
2. [tex]\( f(x) = 0.5 \left( 5^x \right) \)[/tex]
3. [tex]\( f(x) = 0.5 \left( 0.2^x \right) \)[/tex]
4. [tex]\( f(x) = 0.2 \left( 0.2^x \right) \)[/tex]
Next, we examine each function:
1. Checking [tex]\( f(x) = 0.2 \left( 0.5^x \right) \)[/tex]:
- [tex]\( f(-2) = 0.2 \left( 0.5^{-2} \right) = 0.2 \times 4 = 0.8 \)[/tex]
- [tex]\( f(-1) = 0.2 \left( 0.5^{-1} \right) = 0.2 \times 2 = 0.4 \)[/tex]
- [tex]\( f(0) = 0.2 \left( 0.5^0 \right) = 0.2 \times 1 = 0.2 \)[/tex]
- [tex]\( f(1) = 0.2 \left( 0.5^1 \right) = 0.2 \times 0.5 = 0.1 \)[/tex]
- [tex]\( f(2) = 0.2 \left( 0.5^2 \right) = 0.2 \times 0.25 = 0.05 \)[/tex]
Clearly, this function does not match most of the given data points.
2. Checking [tex]\( f(x) = 0.5 \left( 5^x \right) \)[/tex]:
- [tex]\( f(-2) = 0.5 \left( 5^{-2} \right) = 0.5 \times 0.04 = 0.02 \)[/tex]
- [tex]\( f(-1) = 0.5 \left( 5^{-1} \right) = 0.5 \times 0.2 = 0.1 \)[/tex]
- [tex]\( f(0) = 0.5 \left( 5^0 \right) = 0.5 \times 1 = 0.5 \)[/tex]
- [tex]\( f(1) = 0.5 \left( 5^1 \right) = 0.5 \times 5 = 2.5 \)[/tex]
- [tex]\( f(2) = 0.5 \left( 5^2 \right) = 0.5 \times 25 = 12.5 \)[/tex]
This function does not match the given data points in the correct order, so it is incorrect.
3. Checking [tex]\( f(x) = 0.5 \left( 0.2^x \right) \)[/tex]:
- [tex]\( f(-2) = 0.5 \left( 0.2^{-2} \right) = 0.5 \times 25 = 12.5 \)[/tex]
- [tex]\( f(-1) = 0.5 \left( 0.2^{-1} \right) = 0.5 \times 5 = 2.5 \)[/tex]
- [tex]\( f(0) = 0.5 \left( 0.2^0 \right) = 0.5 \times 1 = 0.5 \)[/tex]
- [tex]\( f(1) = 0.5 \left( 0.2^1 \right) = 0.5 \times 0.2 = 0.1 \)[/tex]
- [tex]\( f(2) = 0.5 \left( 0.2^2 \right) = 0.5 \times 0.04 = 0.02 \)[/tex]
This function matches all the given data points exactly:
[tex]\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-2 & 12.5 \\
\hline
-1 & 2.5 \\
\hline
0 & 0.5 \\
\hline
1 & 0.1 \\
\hline
2 & 0.02 \\
\hline
\end{array}
\][/tex]
Thus, the correct exponential function represented by the table is:
[tex]\[ f(x) = 0.5 (0.2^x) \][/tex]
