To determine which of the given exponential functions matches the table of values, we will evaluate each function at the specified [tex]\(x\)[/tex] values and compare the results to the given [tex]\(f(x)\)[/tex] values.
The functions to consider are:
1. [tex]\(f(x)=2\left(2^x\right)\)[/tex]
2. [tex]\(f(x)=0.8\left(0.8^x\right)\)[/tex]
3. [tex]\(f(x)=2\left(0.8^x\right)\)[/tex]
4. [tex]\(f(x)=0.8\left(2^x\right)\)[/tex]
Let's evaluate each function at the given [tex]\(x\)[/tex] values: [tex]\(-2, -1, 0, 1, 2\)[/tex].
### Evaluating [tex]\(f(x)=2\left(2^x\right)\)[/tex]
[tex]\[
\begin{aligned}
f(-2) &= 2(2^{-2}) = 2 \left(\frac{1}{4}\right) = 0.5, \\
f(-1) &= 2(2^{-1}) = 2 \left(\frac{1}{2}\right) = 1, \\
f(0) &= 2(2^0) = 2(1) = 2, \\
f(1) &= 2(2^1) = 2(2) = 4, \\
f(2) &= 2(2^2) = 2(4) = 8.
\end{aligned}
\][/tex]
### Evaluating [tex]\(f(x)=0.8\left(0.8^x\right)\)[/tex]
[tex]\[
\begin{aligned}
f(-2) &= 0.8 (0.8^{-2}) = 0.8 \left(\frac{1}{(0.8)^2}\right) = 0.8 \left(\frac{1}{0.64}\right) \approx 1.25, \\
f(-1) &= 0.8 (0.8^{-1}) = 0.8 \left(\frac{1}{0.8}\right) = 1, \\
f(0) &= 0.8 (0.8^0) = 0.8, \\
f(1) &= 0.8 (0.8^1) = 0.8 \cdot 0.8 = 0.64, \\
f(2) &= 0.8 (0.8^2) = 0.8 \cdot 0.64 = 0.512.
\end{aligned}
\][/tex]
### Evaluating [tex]\(f(x)=2\left(0.8^x\right)\)[/tex]
[tex]\[
\begin{aligned}
f(-2) &= 2 (0.8^{-2}) = 2 \left(\frac{1}{(0.8)^2}\right) = 2 \left(\frac{1}{0.64}\right) \approx 3.125, \\
f(-1) &= 2 (0.8^{-1}) = 2 \left(\frac{1}{0.8}\right) = 2.5, \\
f(0) &= 2 (0.8^0) = 2, \\
f(1) &= 2 (0.8^1) = 2 \cdot 0.8 = 1.6, \\
f(2) &= 2 (0.8^2) = 2 \cdot 0.64 = 1.28.
\end{aligned}
\][/tex]
### Evaluating [tex]\(f(x)=0.8\left(2^x\right)\)[/tex]
[tex]\[
\begin{aligned}
f(-2) &= 0.8 (2^{-2}) = 0.8 \left(\frac{1}{4}\right) = 0.2, \\
f(-1) &= 0.8 (2^{-1}) = 0.8 \left(\frac{1}{2}\right) = 0.4, \\
f(0) &= 0.8 (2^0) = 0.8, \\
f(1) &= 0.8 (2^1) = 0.8 \cdot 2 = 1.6, \\
f(2) &= 0.8 (2^2) = 0.8 \cdot 4 = 3.2.
\end{aligned}
\][/tex]
### Conclusion
Comparing the results of each function evaluation to the given values [tex]\(f(x)\)[/tex]:
[tex]\[
\begin{array}{cc}
x & f(x) \\
-2 & 0.2 \\
-1 & 0.4 \\
0 & 0.8 \\
1 & 1.6 \\
2 & 3.2 \\
\end{array}
\][/tex]
We observe that the evaluations of [tex]\(f(x) = 0.8(2^x)\)[/tex]:
[tex]\[
\begin{aligned}
f(-2) &= 0.2, \\
f(-1) &= 0.4, \\
f(0) &= 0.8, \\
f(1) &= 1.6, \\
f(2) &= 3.2,
\end{aligned}
\][/tex]
Exactly match the table values. Hence, the exponential function represented by the table is:
[tex]\[
\boxed{f(x) = 0.8\left(2^x\right)}
\][/tex]
