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Sagot :
To determine a possible turning point for the continuous function \( f(x) \), we look for points where the function changes direction. This typically occurs at local maxima or minima.
To identify such turning points, we should analyze the values of \( f(x) \) and check for changes in the direction (whether \( f(x) \) shifts from decreasing to increasing or from increasing to decreasing) by examining the neighboring points.
Given the table of points:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -6 & 8 \\ \hline -4 & 2 \\ \hline -2 & 0 \\ \hline 0 & -2 \\ \hline 2 & -1 \\ \hline 4 & 0 \\ \hline 6 & 4 \\ \hline \end{array} \][/tex]
Let us analyze:
1. For \( x = -6 \) with \( f(x) = 8 \):
- Previous: N/A
- Next: \( f(-4) = 2 \)
- Since 8 > 2, the function is decreasing.
2. For \( x = -4 \) with \( f(x) = 2 \):
- Previous: \( f(-6) = 8 \)
- Next: \( f(-2) = 0 \)
- Since 8 > 2 > 0, the function is continuously decreasing.
3. For \( x = -2 \) with \( f(x) = 0 \):
- Previous: \( f(-4) = 2 \)
- Next: \( f(0) = -2 \)
- Since 2 > 0 > -2, the function is continuously decreasing.
4. For \( x = 0 \) with \( f(x) = -2 \):
- Previous: \( f(-2) = 0 \)
- Next: \( f(2) = -1 \)
- Since 0 > -2 < -1, the function decreases to -2 and then increases past it.
Since the function changes direction at \( x = 0 \), it indicates a turning point.
5. For \( x = 2 \) with \( f(x) = -1 \):
- Previous: \( f(0) = -2 \)
- Next: \( f(4) = 0 \)
- Since -2 < -1 < 0, the function is continuously increasing.
6. For \( x = 4 \) with \( f(x) = 0 \):
- Previous: \( f(2) = -1 \)
- Next: \( f(6) = 4 \)
- Since -1 < 0 < 4, the function is continuously increasing.
7. For \( x = 6 \) with \( f(x) = 4 \):
- Previous: \( f(4) = 0 \)
- Next: N/A
- Since 0 < 4, the function is increasing.
Therefore, the possible turning point, where the function changes direction, is:
[tex]\[ (0, -2) \][/tex]
So, the answer is:
[tex]\((0,-2)\)[/tex]
To identify such turning points, we should analyze the values of \( f(x) \) and check for changes in the direction (whether \( f(x) \) shifts from decreasing to increasing or from increasing to decreasing) by examining the neighboring points.
Given the table of points:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -6 & 8 \\ \hline -4 & 2 \\ \hline -2 & 0 \\ \hline 0 & -2 \\ \hline 2 & -1 \\ \hline 4 & 0 \\ \hline 6 & 4 \\ \hline \end{array} \][/tex]
Let us analyze:
1. For \( x = -6 \) with \( f(x) = 8 \):
- Previous: N/A
- Next: \( f(-4) = 2 \)
- Since 8 > 2, the function is decreasing.
2. For \( x = -4 \) with \( f(x) = 2 \):
- Previous: \( f(-6) = 8 \)
- Next: \( f(-2) = 0 \)
- Since 8 > 2 > 0, the function is continuously decreasing.
3. For \( x = -2 \) with \( f(x) = 0 \):
- Previous: \( f(-4) = 2 \)
- Next: \( f(0) = -2 \)
- Since 2 > 0 > -2, the function is continuously decreasing.
4. For \( x = 0 \) with \( f(x) = -2 \):
- Previous: \( f(-2) = 0 \)
- Next: \( f(2) = -1 \)
- Since 0 > -2 < -1, the function decreases to -2 and then increases past it.
Since the function changes direction at \( x = 0 \), it indicates a turning point.
5. For \( x = 2 \) with \( f(x) = -1 \):
- Previous: \( f(0) = -2 \)
- Next: \( f(4) = 0 \)
- Since -2 < -1 < 0, the function is continuously increasing.
6. For \( x = 4 \) with \( f(x) = 0 \):
- Previous: \( f(2) = -1 \)
- Next: \( f(6) = 4 \)
- Since -1 < 0 < 4, the function is continuously increasing.
7. For \( x = 6 \) with \( f(x) = 4 \):
- Previous: \( f(4) = 0 \)
- Next: N/A
- Since 0 < 4, the function is increasing.
Therefore, the possible turning point, where the function changes direction, is:
[tex]\[ (0, -2) \][/tex]
So, the answer is:
[tex]\((0,-2)\)[/tex]
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