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Sagot :
Let's analyze the predictions step by step based on the given function values.
### Given Data:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -3 & -15 \\ \hline -2 & -5 \\ \hline -1 & 0 \\ \hline 0 & 5 \\ \hline 1 & 0 \\ \hline 2 & -5 \\ \hline \end{array} \][/tex]
### Evaluate the Statements:
1. Statement: \( f(x) \leq 0 \) over the interval \( (-\infty, \infty) \)
- We need to check if \( f(x) \) is always less than or equal to zero for all \( x \).
- At \( x = 0 \), \( f(0) = 5 \), which is greater than 0.
- Therefore, this statement is False.
2. Statement: \( f(x) > 0 \) over the interval \( (-1, \infty) \)
- We need to check if \( f(x) \) is always greater than zero for \( x \) in the interval \( (-1, \infty) \).
- At \( x = -1 \), \( f(-1) = 0 \), which is not greater than 0.
- At \( x = 1 \), \( f(1) = 0 \), which is also not greater than 0.
- Therefore, this statement is False.
3. Statement: \( f(x) \geq 0 \) over the interval \( [-1, 1] \)
- We need to check if \( f(x) \) is always greater than or equal to zero for \( x \) in the interval \( [-1, 1] \).
- At \( x = -1 \), \( f(-1) = 0 \), which is equal to 0.
- At \( x = 0 \), \( f(0) = 5 \), which is greater than 0.
- At \( x = 1 \), \( f(1) = 0 \), which is equal to 0.
- Therefore, this statement is True.
4. Statement: \( f(x) < 0 \) over the interval \( (0, 2) \)
- We need to check if \( f(x) \) is always less than zero for \( x \) in the open interval \( (0, 2) \).
- Although \( f(2) = -5 \), we need to consider the open interval.
- As we observe, there are no given values between \(0\) and \(2\) to contradict this.
- Therefore, this statement is True.
### Conclusion:
The valid predictions about the continuous function \( f(x) \) are:
- \( f(x) \geq 0 \) over the interval \([-1,1]\)
- \( f(x) < 0 \) over the interval \((0,2)\)
Thus, our predictions are:
- \( f(x) \leq 0 \) over the interval \( (-\infty, \infty) \) is False
- \( f(x) > 0 \) over the interval \( (-1, \infty) \) is False
- \( f(x) \geq 0 \) over the interval \([-1,1]\) is True
- [tex]\( f(x) < 0 \)[/tex] over the interval [tex]\((0,2)\)[/tex] is True
### Given Data:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -3 & -15 \\ \hline -2 & -5 \\ \hline -1 & 0 \\ \hline 0 & 5 \\ \hline 1 & 0 \\ \hline 2 & -5 \\ \hline \end{array} \][/tex]
### Evaluate the Statements:
1. Statement: \( f(x) \leq 0 \) over the interval \( (-\infty, \infty) \)
- We need to check if \( f(x) \) is always less than or equal to zero for all \( x \).
- At \( x = 0 \), \( f(0) = 5 \), which is greater than 0.
- Therefore, this statement is False.
2. Statement: \( f(x) > 0 \) over the interval \( (-1, \infty) \)
- We need to check if \( f(x) \) is always greater than zero for \( x \) in the interval \( (-1, \infty) \).
- At \( x = -1 \), \( f(-1) = 0 \), which is not greater than 0.
- At \( x = 1 \), \( f(1) = 0 \), which is also not greater than 0.
- Therefore, this statement is False.
3. Statement: \( f(x) \geq 0 \) over the interval \( [-1, 1] \)
- We need to check if \( f(x) \) is always greater than or equal to zero for \( x \) in the interval \( [-1, 1] \).
- At \( x = -1 \), \( f(-1) = 0 \), which is equal to 0.
- At \( x = 0 \), \( f(0) = 5 \), which is greater than 0.
- At \( x = 1 \), \( f(1) = 0 \), which is equal to 0.
- Therefore, this statement is True.
4. Statement: \( f(x) < 0 \) over the interval \( (0, 2) \)
- We need to check if \( f(x) \) is always less than zero for \( x \) in the open interval \( (0, 2) \).
- Although \( f(2) = -5 \), we need to consider the open interval.
- As we observe, there are no given values between \(0\) and \(2\) to contradict this.
- Therefore, this statement is True.
### Conclusion:
The valid predictions about the continuous function \( f(x) \) are:
- \( f(x) \geq 0 \) over the interval \([-1,1]\)
- \( f(x) < 0 \) over the interval \((0,2)\)
Thus, our predictions are:
- \( f(x) \leq 0 \) over the interval \( (-\infty, \infty) \) is False
- \( f(x) > 0 \) over the interval \( (-1, \infty) \) is False
- \( f(x) \geq 0 \) over the interval \([-1,1]\) is True
- [tex]\( f(x) < 0 \)[/tex] over the interval [tex]\((0,2)\)[/tex] is True
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