[tex]\[
f(x) = -\frac{2}{3} x + 3 \quad \text{and} \quad g(x) = 2x^2 - 5
\][/tex]

\begin{tabular}{|c|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)$[/tex] & [tex]$g(x)$[/tex] \\
\hline
3 & [tex]$f(3)$[/tex] & 1 \\
\hline
-5 & [tex]$f(-5)$[/tex] & 45 \\
\hline
-3 & [tex]$g(-3)$[/tex] & 13 \\
\hline
-7 & [tex]$g(-7)$[/tex] & 93 \\
\hline
8 & [tex]$g(8)$[/tex] & 123 \\
\hline
9 & [tex]$g(9)$[/tex] & 157 \\
\hline
\end{tabular}



Answer :

To solve the question, we need to evaluate the functions [tex]\( f(x) = -\frac{2}{3} x + 3 \)[/tex] and [tex]\( g(x) = 2 x^2 - 5 \)[/tex] for the given values of [tex]\( x \)[/tex]. Here are the detailed steps:

1. For [tex]\( f(x) \)[/tex]:

- Calculate [tex]\( f(3) \)[/tex]:
[tex]\[ f(3) = -\frac{2}{3} \cdot 3 + 3 \][/tex]
The value of [tex]\( f(3) \)[/tex] is 1.

- Calculate [tex]\( f(-5) \)[/tex]:
[tex]\[ f(-5) = -\frac{2}{3} \cdot (-5) + 3 \][/tex]
The value of [tex]\( f(-5) \)[/tex] is approximately 6.333333333333333 which is close to 6.33 (infinite repeating decimal).

2. For [tex]\( g(x) \)[/tex]:

- Calculate [tex]\( g(-3) \)[/tex]:
[tex]\[ g(-3) = 2 \cdot (-3)^2 - 5 \][/tex]
The value of [tex]\( g(-3) \)[/tex] is 13.

- Calculate [tex]\( g(-7) \)[/tex]:
[tex]\[ g(-7) = 2 \cdot (-7)^2 - 5 \][/tex]
The value of [tex]\( g(-7) \)[/tex] is 93.

- Calculate [tex]\( g(8) \)[/tex]:
[tex]\[ g(8) = 2 \cdot 8^2 - 5 \][/tex]
The value of [tex]\( g(8) \)[/tex] is 123.

- Calculate [tex]\( g(9) \)[/tex]:
[tex]\[ g(9) = 2 \cdot 9^2 - 5 \][/tex]
The value of [tex]\( g(9) \)[/tex] is 157.

3. Fill in the table with our calculated values:

[tex]\[ \begin{array}{|c|c|c|} \hline x & f(x) & g(x) \\ \hline f(3) & 1 & - \\ \hline f(-5) & 6.33 & - \\ \hline g(-3) & - & 13 \\ \hline g(-7) & - & 93 \\ \hline g(8) & - & 123 \\ \hline g(9) & - & 157 \\ \hline \end{array} \][/tex]

So, the table should look like this:

[tex]\[ \begin{array}{|c|c|c|} \hline x & f(x) & g(x) \\ \hline f(3) & 1 & - \\ \hline f(-5) & 6.33 & - \\ \hline g(-3) & - & 13 \\ \hline g(-7) & - & 93 \\ \hline g(8) & - & 123 \\ \hline g(9) & - & 157 \\ \hline \end{array} \][/tex]

The correct values should be placed accordingly to ensure clarity in the table.