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\begin{tabular}{|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & 0 & 1 & 2 & 3 \\
\hline
[tex]$f(x)$[/tex] & -50 & -25 & [tex]$-\frac{25}{2}$[/tex] & [tex]$-\frac{25}{4}$[/tex] \\
\hline
\end{tabular}


Sagot :

Let's solve the given problem step-by-step. We need to determine the values of [tex]\( f(x) \)[/tex] for each corresponding [tex]\( x \)[/tex] in the table.

The given values are:

[tex]\[ \begin{array}{|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 \\ \hline f(x) & -50 & -25 & -\frac{25}{2} & -\frac{25}{4} \\ \hline \end{array} \][/tex]

### Step-by-Step Calculation

1. For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -50 \][/tex]
So, the first entry in the solution is:
[tex]\[ (0, -50) \][/tex]

2. For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -25 \][/tex]
So, the second entry in the solution is:
[tex]\[ (1, -25) \][/tex]

3. For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = -\frac{25}{2} \][/tex]
Simplifying [tex]\( -\frac{25}{2} \)[/tex] to a decimal gives us:
[tex]\[ f(2) = -12.5 \][/tex]
So, the third entry in the solution is:
[tex]\[ (2, -12.5) \][/tex]

4. For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = -\frac{25}{4} \][/tex]
Simplifying [tex]\( -\frac{25}{4} \)[/tex] to a decimal gives us:
[tex]\[ f(3) = -6.25 \][/tex]
So, the fourth entry in the solution is:
[tex]\[ (3, -6.25) \][/tex]

### Final Table
Combining all these results, we form the resultant table:

[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 0 & -50 \\ 1 & -25 \\ 2 & -12.5 \\ 3 & -6.25 \\ \hline \end{array} \][/tex]

Thus, the solution can be summarized as the set of pairs [tex]\((x, f(x))\)[/tex]:

[tex]\[ [(0, -50), (1, -25), (2, -12.5), (3, -6.25)] \][/tex]

These pairs represent the values of [tex]\( f(x) \)[/tex] for the given [tex]\( x \)[/tex] values.
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