Two exponential functions are shown in the table.

\begin{tabular}{|c|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)=2^x$[/tex] & [tex]$g(x)=\left(\frac{1}{2}\right)^x$[/tex] \\
\hline
2 & 4 & [tex]$\frac{1}{4}$[/tex] \\
\hline
1 & 2 & [tex]$\frac{1}{2}$[/tex] \\
\hline
0 & 1 & 1 \\
\hline
-1 & [tex]$\frac{1}{2}$[/tex] & 2 \\
\hline
-2 & [tex]$\frac{1}{4}$[/tex] & 4 \\
\hline
\end{tabular}

Which conclusion about [tex]$f(x)$[/tex] and [tex]$g(x)$[/tex] can be drawn from the table?

A. The functions [tex]$f(x)$[/tex] and [tex]$g(x)$[/tex] are reflections over the [tex]$x$[/tex] axis.
B. The functions [tex]$f(x)$[/tex] and [tex]$g(x)$[/tex] are reflections over the [tex]$y$[/tex] axis.
C. The function [tex]$f(x)$[/tex] is a decreasing function, and [tex]$g(x)$[/tex] is an increasing function.
D. The function [tex]$f(x)$[/tex] has a greater initial value than [tex]$g(x)$[/tex].



Answer :

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.