Question
A local maximum of the function f(x) occurs for which x-value?
A 2-column table with 6 rows. The first column is labeled x with entries negative 4, negative 3, negative 2, negative 1, 0, 1. The second column is labeled f of x with entries 16, negative 2, 0, 6, 0, negative 2.
–4
–3
–2
–1
A 2-column table with 6 rows. The first column is labeled x with entries negative 4, negative 3, negative 2, negative 1, 0, 1. The second column is labeled f of x with entries 16, negative 2, 0, 6, 0, negative 2.
–4
–3
–2
–1
Answers
GPT-4o
To determine where a local maximum occurs for the function \( f(x) \) based on the values provided in the table, you need to compare the function values at each \( x \) and identify if there is an \( x \) value where the function reaches a peak compared to its immediate neighbors.
Here's the table again for clarity:
\[
\begin{array}{cc}
x & f(x) \\
\hline
-4 & 16 \\
-3 & -2 \\
-2 & 0 \\
-1 & 6 \\
0 & 0 \\
1 & -2 \\
\end{array}
\]
A local maximum is where the function value at a certain \( x \) is greater than the function values at the adjacent \( x \) values (its neighbors).
Let's evaluate each point:
- At \( x = -4 \), \( f(x) = 16 \):
- \( f(-4) = 16 \) is not compared to an earlier value (since -4 is the start of the range given) but it is greater than \( f(-3) = -2 \).
- At \( x = -3 \), \( f(x) = -2 \):
- \( f(-3) = -2 \) is less than both \( f(-4) = 16 \) and \( f(-2) = 0 \).
- At \( x = -2 \), \( f(x) = 0 \):
- \( f(-2) = 0 \) is greater than \( f(-3) = -2 \), but less than \( f(-1) = 6 \).
- At \( x = -1 \), \( f(x) = 6 \):
- \( f(-1) = 6 \) is greater than both \( f(-2) = 0 \) and \( f(0) = 0 \).
- At \( x = 0 \), \( f(x) = 0 \):
- \( f(0) = 0 \) is equal to \( f(-1) = 6 \), but greater than \( f(1) = -2 \).
- At \( x = 1 \), \( f(x) = -2 \):
- \( f(1) = -2 \) is less than both \( f(0) = 0 \).
Upon comparison, the local maximum occurs at \( x = -1 \), where \( f(-1) = 6 \) is greater than the function values at its immediate neighbors.
So, the local maximum of the function occurs at \( x = -1 \).
Here's the table again for clarity:
\[
\begin{array}{cc}
x & f(x) \\
\hline
-4 & 16 \\
-3 & -2 \\
-2 & 0 \\
-1 & 6 \\
0 & 0 \\
1 & -2 \\
\end{array}
\]
A local maximum is where the function value at a certain \( x \) is greater than the function values at the adjacent \( x \) values (its neighbors).
Let's evaluate each point:
- At \( x = -4 \), \( f(x) = 16 \):
- \( f(-4) = 16 \) is not compared to an earlier value (since -4 is the start of the range given) but it is greater than \( f(-3) = -2 \).
- At \( x = -3 \), \( f(x) = -2 \):
- \( f(-3) = -2 \) is less than both \( f(-4) = 16 \) and \( f(-2) = 0 \).
- At \( x = -2 \), \( f(x) = 0 \):
- \( f(-2) = 0 \) is greater than \( f(-3) = -2 \), but less than \( f(-1) = 6 \).
- At \( x = -1 \), \( f(x) = 6 \):
- \( f(-1) = 6 \) is greater than both \( f(-2) = 0 \) and \( f(0) = 0 \).
- At \( x = 0 \), \( f(x) = 0 \):
- \( f(0) = 0 \) is equal to \( f(-1) = 6 \), but greater than \( f(1) = -2 \).
- At \( x = 1 \), \( f(x) = -2 \):
- \( f(1) = -2 \) is less than both \( f(0) = 0 \).
Upon comparison, the local maximum occurs at \( x = -1 \), where \( f(-1) = 6 \) is greater than the function values at its immediate neighbors.
So, the local maximum of the function occurs at \( x = -1 \).