Let's analyze the given function [tex]\( f(x) \)[/tex] for each interval based on the table provided:
[tex]\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-5 & 8 \\
\hline
-3 & 4 \\
\hline
-1 & 0 \\
\hline
1 & -2 \\
\hline
3 & -2 \\
\hline
5 & 0 \\
\hline
7 & 4 \\
\hline
\end{array}
\][/tex]
1. Checking [tex]\(f(x) \geq 0\)[/tex] over the interval [tex]\([5, \infty)\)[/tex]:
For [tex]\(x = 5\)[/tex], [tex]\(f(x) = 0\)[/tex] (which is [tex]\(\geq 0\)[/tex]).
For [tex]\(x = 7\)[/tex], [tex]\(f(x) = 4\)[/tex] (which is [tex]\(\geq 0\)[/tex]).
There are no values beyond 7 given, so based on these points, [tex]\(f(x) \geq 0\)[/tex] over the interval [tex]\([5, \infty)\)[/tex] seems valid.
2. Checking [tex]\(f(x) \leq 0\)[/tex] over the interval [tex]\([-1, \infty)\)[/tex]:
For [tex]\(x = -1\)[/tex], [tex]\(f(x) = 0\)[/tex] (which is [tex]\(\leq 0\)[/tex]).
For [tex]\(x = 1\)[/tex], [tex]\(f(x) = -2\)[/tex] (which is [tex]\(\leq 0\)[/tex]).
For [tex]\(x = 3\)[/tex], [tex]\(f(x) = -2\)[/tex] (which is [tex]\(\leq 0\)[/tex]).
For [tex]\(x = 5\)[/tex], [tex]\(f(x) = 0\)[/tex] (which is [tex]\(\leq 0\)[/tex]).
For [tex]\(x = 7\)[/tex], [tex]\(f(x) = 4\)[/tex] (which is not [tex]\(\leq 0\)[/tex]).
Since [tex]\(f(x)\)[/tex] is not [tex]\(\leq 0\)[/tex] at [tex]\(x = 7\)[/tex], this prediction is invalid.
3. Checking [tex]\(f(x) > 0\)[/tex] over the interval [tex]\((-\infty, 1)\)[/tex]:
For [tex]\(x = -5\)[/tex], [tex]\(f(x) = 8\)[/tex] (which is [tex]\(> 0\)[/tex]).
For [tex]\(x = -3\)[/tex], [tex]\(f(x) = 4\)[/tex] (which is [tex]\(> 0\)[/tex]).
For [tex]\(x = -1\)[/tex], [tex]\(f(x) = 0\)[/tex] (which is not [tex]\(> 0\)[/tex]).
Since [tex]\(f(x)\)[/tex] is not [tex]\(> 0\)[/tex] at [tex]\(x = -1\)[/tex], this prediction is invalid.
4. Checking [tex]\(f(x) < 0\)[/tex] over the interval [tex]\((-\infty, -1)\)[/tex]:
For [tex]\(x = -5\)[/tex], [tex]\(f(x) = 8\)[/tex] (which is not [tex]\(< 0\)[/tex]).
For [tex]\(x = -3\)[/tex], [tex]\(f(x) = 4\)[/tex] (which is not [tex]\(< 0\)[/tex]).
Since [tex]\(f(x)\)[/tex] is not [tex]\(< 0\)[/tex] for values in [tex]\((-\infty, -1)\)[/tex], this prediction is invalid.
After evaluating all intervals, the valid prediction about the continuous function [tex]\( f(x) \)[/tex] is:
[tex]\[ f(x) \geq 0 \text{ over the interval } [5, \infty). \][/tex]
