Sure, let's fill in the table by finding the corresponding [tex]\( x \)[/tex] and [tex]\( g(x) \)[/tex] for the given function [tex]\( g(x) = 3 - 8x \)[/tex].
1. Finding [tex]\( x \)[/tex] when [tex]\( g(x) = 0 \)[/tex]:
To find [tex]\( x \)[/tex] when [tex]\( g(x) = 0 \)[/tex], we need to solve the equation:
[tex]\[
3 - 8x = 0
\][/tex]
Rearranging the equation to solve for [tex]\( x \)[/tex]:
[tex]\[
8x = 3 \implies x = \frac{3}{8} \implies x = 0.375
\][/tex]
2. Finding [tex]\( g(x) \)[/tex] when [tex]\( x = 0 \)[/tex]:
To find [tex]\( g(x) \)[/tex] when [tex]\( x = 0 \)[/tex]:
[tex]\[
g(0) = 3 - 8 \cdot 0 \implies g(0) = 3
\][/tex]
3. Finding [tex]\( x \)[/tex] when [tex]\( g(x) = -5 \)[/tex]:
To find [tex]\( x \)[/tex] when [tex]\( g(x) = -5 \)[/tex], we need to solve the equation:
[tex]\[
3 - 8x = -5
\][/tex]
Rearranging the equation to solve for [tex]\( x \)[/tex]:
[tex]\[
3 + 5 = 8x \implies 8 = 8x \implies x = 1.0
\][/tex]
4. Finding [tex]\( g(x) \)[/tex] when [tex]\( x = 3 \)[/tex]:
To find [tex]\( g(x) \)[/tex] when [tex]\( x = 3 \)[/tex]:
[tex]\[
g(3) = 3 - 8 \cdot 3 \implies g(3) = 3 - 24 \implies g(3) = -21
\][/tex]
Now that we have the necessary values, we can fill in the table:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $g(x)$ \\
\hline
0.375 & 0 \\
\hline
0 & 3 \\
\hline
1.0 & -5 \\
\hline
3 & -21 \\
\hline
\end{tabular}
\][/tex]
So, the completed table is:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $g(x)$ \\
\hline
0.375 & 0 \\
\hline
0 & 3 \\
\hline
1.0 & -5 \\
\hline
3 & -21 \\
\hline
\end{tabular}
\][/tex]
