To answer this question, let's analyze the given table of values for the functions [tex]\( f(x) = 2^x \)[/tex] and [tex]\( g(x) = \left(\frac{1}{2}\right)^x \)[/tex]:
[tex]\[
\begin{array}{|c|c|c|}
\hline
x & f(x)=2^x & g(x)=\left(\frac{1}{2}\right)^x \\
\hline
2 & 4 & \frac{1}{4} \\
\hline
1 & 2 & \frac{1}{2} \\
\hline
0 & 1 & 1 \\
\hline
-1 & \frac{1}{2} & 2 \\
\hline
-2 & \frac{1}{4} & 4 \\
\hline
\end{array}
\][/tex]
First, observe that [tex]\( f(x) = 2^x \)[/tex] is an increasing function:
- When [tex]\( x = 2 \)[/tex], [tex]\( f(x) = 4 \)[/tex]
- When [tex]\( x = 1 \)[/tex], [tex]\( f(x) = 2 \)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( f(x) = 1 \)[/tex]
- When [tex]\( x = -1 \)[/tex], [tex]\( f(x) = \frac{1}{2} \)[/tex]
- When [tex]\( x = -2 \)[/tex], [tex]\( f(x) = \frac{1}{4} \)[/tex]
Next, observe that [tex]\( g(x) = \left(\frac{1}{2}\right)^x = 2^{-x} \)[/tex], which is a decreasing function:
- When [tex]\( x = 2 \)[/tex], [tex]\( g(x) = \frac{1}{4} \)[/tex]
- When [tex]\( x = 1 \)[/tex], [tex]\( g(x) = \frac{1}{2} \)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( g(x) = 1 \)[/tex]
- When [tex]\( x = -1 \)[/tex], [tex]\( g(x) = 2 \)[/tex]
- When [tex]\( x = -2 \)[/tex], [tex]\( g(x) = 4 \)[/tex]
If you look closely at the table, you can see that [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] have the same values but in reverse order with respect to the [tex]\( y \)[/tex]-axis:
- [tex]\( f(2) = 4 \)[/tex] and [tex]\( g(-2) = 4 \)[/tex]
- [tex]\( f(1) = 2 \)[/tex] and [tex]\( g(-1) = 2 \)[/tex]
- [tex]\( f(0) = 1 \)[/tex] and [tex]\( g(0) = 1 \)[/tex]
- [tex]\( f(-1) = \frac{1}{2} \)[/tex] and [tex]\( g(1) = \frac{1}{2} \)[/tex]
- [tex]\( f(-2) = \frac{1}{4} \)[/tex] and [tex]\( g(2) = \frac{1}{4} \)[/tex]
This relationship shows that [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are reflections of each other over the [tex]\( y \)[/tex]-axis. Hence, the correct conclusion is:
The functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are reflections over the [tex]\( y \)[/tex] axis.
