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Sagot :
To determine characteristics about the continuous function that produced the given table of values, we need to analyze the changes in the [tex]$y$[/tex] values as [tex]$x$[/tex] increases. The table provides the following data points:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0.125 & -3 \\ \hline 0.5 & -1 \\ \hline 2 & 1 \\ \hline 8 & 3 \\ \hline 64 & 6 \\ \hline \end{array} \][/tex]
To find if the function has any [tex]$x$[/tex]-intercepts (where the function crosses the [tex]$x$[/tex]-axis, i.e., [tex]$y = 0$[/tex]), we look at the given [tex]$y$[/tex] values:
- At [tex]$x = 0.125$[/tex], [tex]$y = -3$[/tex]
- At [tex]$x = 0.5$[/tex], [tex]$y = -1$[/tex]
- At [tex]$x = 2$[/tex], [tex]$y = 1$[/tex]
- At [tex]$x = 8$[/tex], [tex]$y = 3$[/tex]
- At [tex]$x = 64$[/tex], [tex]$y = 6$[/tex]
A continuous function has an [tex]$x$[/tex]-intercept if the sign of [tex]$y$[/tex] changes between any two consecutive [tex]$x$[/tex] values. Let’s examine the sign changes between consecutive [tex]$y$[/tex] values:
1. From [tex]$x = 0.125$[/tex] ([tex]$y = -3$[/tex]) to [tex]$x = 0.5$[/tex] ([tex]$y = -1$[/tex])
- The sign of [tex]$y$[/tex] is negative in both cases. No sign change here.
2. From [tex]$x = 0.5$[/tex] ([tex]$y = -1$[/tex]) to [tex]$x = 2$[/tex] ([tex]$y = 1$[/tex])
- The sign of [tex]$y$[/tex] changes from negative to positive. This indicates the function must cross the [tex]$x$[/tex]-axis somewhere between [tex]$x = 0.5$[/tex] and [tex]$x = 2$[/tex].
3. We could continue checking, but once a change in sign is detected, it is sufficient to conclude that there is at least one [tex]$x$[/tex]-intercept.
Therefore, since there is a change in the sign of [tex]$y$[/tex] values between [tex]$x = 0.5$[/tex] and [tex]$x = 2$[/tex], we can confidently state that the function has at least one [tex]$x$[/tex]-intercept.
Hence, the correct answer is:
C. the function has at least one [tex]$x$[/tex]-intercept
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0.125 & -3 \\ \hline 0.5 & -1 \\ \hline 2 & 1 \\ \hline 8 & 3 \\ \hline 64 & 6 \\ \hline \end{array} \][/tex]
To find if the function has any [tex]$x$[/tex]-intercepts (where the function crosses the [tex]$x$[/tex]-axis, i.e., [tex]$y = 0$[/tex]), we look at the given [tex]$y$[/tex] values:
- At [tex]$x = 0.125$[/tex], [tex]$y = -3$[/tex]
- At [tex]$x = 0.5$[/tex], [tex]$y = -1$[/tex]
- At [tex]$x = 2$[/tex], [tex]$y = 1$[/tex]
- At [tex]$x = 8$[/tex], [tex]$y = 3$[/tex]
- At [tex]$x = 64$[/tex], [tex]$y = 6$[/tex]
A continuous function has an [tex]$x$[/tex]-intercept if the sign of [tex]$y$[/tex] changes between any two consecutive [tex]$x$[/tex] values. Let’s examine the sign changes between consecutive [tex]$y$[/tex] values:
1. From [tex]$x = 0.125$[/tex] ([tex]$y = -3$[/tex]) to [tex]$x = 0.5$[/tex] ([tex]$y = -1$[/tex])
- The sign of [tex]$y$[/tex] is negative in both cases. No sign change here.
2. From [tex]$x = 0.5$[/tex] ([tex]$y = -1$[/tex]) to [tex]$x = 2$[/tex] ([tex]$y = 1$[/tex])
- The sign of [tex]$y$[/tex] changes from negative to positive. This indicates the function must cross the [tex]$x$[/tex]-axis somewhere between [tex]$x = 0.5$[/tex] and [tex]$x = 2$[/tex].
3. We could continue checking, but once a change in sign is detected, it is sufficient to conclude that there is at least one [tex]$x$[/tex]-intercept.
Therefore, since there is a change in the sign of [tex]$y$[/tex] values between [tex]$x = 0.5$[/tex] and [tex]$x = 2$[/tex], we can confidently state that the function has at least one [tex]$x$[/tex]-intercept.
Hence, the correct answer is:
C. the function has at least one [tex]$x$[/tex]-intercept
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