To determine which of the given points could be a turning point for the continuous function [tex]\( f(x) \)[/tex], we need to analyze the behavior of the function around each point. A turning point is where the function changes its direction, which means the slope of the function (or the derivative) changes sign. In other words, at a turning point, the slope of [tex]\( f(x) \)[/tex] changes from positive to negative or vice versa.
Let us examine the function data step-by-step:
Given data points:
[tex]\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-4 & -6 \\
\hline
-3 & -4 \\
\hline
-2 & -1 \\
\hline
-1 & -2 \\
\hline
0 & -5 \\
\hline
1 & -8 \\
\hline
2 & -16 \\
\hline
\end{array}
\][/tex]
1. Compute the differences [tex]\( \Delta f(x) \)[/tex] between consecutive points to analyze the slope:
[tex]\[
\begin{aligned}
\Delta f(x_{-3}) &= f(-3) - f(-4) = -4 - (-6) = 2 \\
\Delta f(x_{-2}) &= f(-2) - f(-3) = -1 - (-4) = 3 \\
\Delta f(x_{-1}) &= f(-1) - f(-2) = -2 - (-1) = -1 \\
\Delta f(x_{0}) &= f(0) - f(-1) = -5 - (-2) = -3 \\
\Delta f(x_{1}) &= f(1) - f(0) = -8 - (-5) = -3 \\
\Delta f(x_{2}) &= f(2) - f(1) = -16 - (-8) = -8 \\
\end{aligned}
\][/tex]
2. Identify where the change in sign occurs between consecutive differences:
[tex]\[
\begin{aligned}
\Delta f(x_{-3}) \quad & 2 \\
\Delta f(x_{-2}) \quad & 3 \quad (\text{positive}) \\
\Delta f(x_{-1}) \quad & -1 \quad (\text{negative}) \\
\Delta f(x_{0}) \quad & -3 \\
\Delta f(x_{1}) \quad & -3 \\
\Delta f(x_{2}) \quad & -8 \\
\end{aligned}
\][/tex]
We see that [tex]\(\Delta f(x_{-2})\)[/tex] is positive, and [tex]\(\Delta f(x_{-1})\)[/tex] is negative. Hence, the change from positive to negative occurs between [tex]\( x = -2 \)[/tex] and [tex]\( x = -1 \)[/tex].
Given the candidates for possible turning points:
[tex]\[
(-3, -4), (-2, -1), (0, -5), (1, -8)
\][/tex]
Given that the slope change occurs around [tex]\( x = -2 \)[/tex], the possible turning point must be at [tex]\( x = -2 \)[/tex].
Thus, the possible turning point is:
[tex]\[
(-2, -1)
\][/tex]
