The function [tex]$f(x)$[/tex] is defined as [tex]$f(x)=\frac{1}{3}(6)^x$[/tex]. Which table of values could be used to graph [tex]$g(x)$[/tex], a reflection of [tex]$f(x)$[/tex] across the [tex]$x$[/tex]-axis?

\begin{tabular}{|c|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)$[/tex] & [tex]$g(x)$[/tex] \\
\hline
-2 & [tex]$\frac{1}{188}$[/tex] & 12 \\
\hline
-1 & [tex]$\frac{1}{18}$[/tex] & 2 \\
\hline
0 & [tex]$\frac{1}{3}$[/tex] & [tex]$\frac{1}{3}$[/tex] \\
\hline
1 & 2 & [tex]$\frac{1}{18}$[/tex] \\
\hline
2 & 12 & [tex]$\frac{1}{108}$[/tex] \\
\hline
\end{tabular}

\begin{tabular}{|c|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)$[/tex] & [tex]$g(x)$[/tex] \\
\hline
-2 & [tex]$\frac{1}{108}$[/tex] & 12 \\
\hline
-1 & [tex]$\frac{1}{18}$[/tex] & 2 \\
\hline
0 & 0 & 0 \\
\hline
1 & 2 & [tex]$\frac{1}{18}$[/tex] \\
\hline
2 & 12 & 1 \\
\hline
\end{tabular}



Answer :

To determine which of the given tables represents the values of [tex]\( g(x) \)[/tex], the reflection of [tex]\( f(x) \)[/tex] across the [tex]\( x \)[/tex]-axis, we need to understand that reflecting [tex]\( f(x) \)[/tex] across the [tex]\( x \)[/tex]-axis results in [tex]\( g(x) = -f(x) \)[/tex].

Given the function [tex]\( f(x) = \frac{1}{3} \left( 6 \right)^x \)[/tex], we will perform the reflection to find [tex]\( g(x) \)[/tex].

Step-by-step calculation:

1. Define [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \frac{1}{3} \left( 6 \right)^x \][/tex]

2. Calculate values of [tex]\( f(x) \)[/tex] for given [tex]\( x \)[/tex] values:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = \frac{1}{3} \left( 6 \right)^{-2} = \frac{1}{3} \cdot \frac{1}{36} = \frac{1}{108} \][/tex]
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = \frac{1}{3} \left( 6 \right)^{-1} = \frac{1}{3} \cdot \frac{1}{6} = \frac{1}{18} \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{1}{3} \left( 6 \right)^0 = \frac{1}{3} \][/tex]

- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{1}{3} \left( 6 \right)^1 = \frac{6}{3} = 2 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = \frac{1}{3} \left( 6 \right)^2 = \frac{36}{3} = 12 \][/tex]

The corresponding values of [tex]\( f(x) \)[/tex] are:

[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -2 & \frac{1}{108} \\ -1 & \frac{1}{18} \\ 0 & \frac{1}{3} \\ 1 & 2 \\ 2 & 12 \\ \hline \end{array} \][/tex]

3. Calculate [tex]\( g(x) = -f(x) \)[/tex]:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ g(-2) = -f(-2) = -\frac{1}{108} \][/tex]
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ g(-1) = -f(-1) = -\frac{1}{18} \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = -f(0) = -\frac{1}{3} \][/tex]

- For [tex]\( x = 1 \)[/tex]:
[tex]\[ g(1) = -f(1) = -2 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ g(2) = -f(2) = -12 \][/tex]

4. Check the two given tables to see which matches [tex]\( g(x) \)[/tex]:

Comparing the values of [tex]\( g(x) \)[/tex] in the two tables:
- Table 1 does not match as the values in [tex]\( g(x) \)[/tex] column are inverted and not the ones calculated.
- Table 2 also does not match because it has incorrect values including a zero.

Given these observations, the corresponding [tex]\( g(x) \)[/tex] values do not match with either table presented.