Question
A 2-column table with 6 rows. The first column is labeled t with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 12, m, 4, 0, negative 4, negative 2.
If the table of the function contains exactly two potential turning points, one with an input value of –1, which statement best describes all possible values of m?
m ≥ –12
–12 < m < 4
m ≤ 4
m ≥ 4 or m ≤ –12
If the table of the function contains exactly two potential turning points, one with an input value of –1, which statement best describes all possible values of m?
m ≥ –12
–12 < m < 4
m ≤ 4
m ≥ 4 or m ≤ –12
Answers
Answer
1. A 2-column table with 7 rows. The first column is labeled x with entries negative 5, negative 3, negative 1, 1, 3, 5, 7. The second column is labeled f of x with entries 8, 4, 0, negative 2, negative 2, 0, 4.
Which is a valid prediction about the continuous function f(x)?
f(x) ≥ 0 over the interval [5, ∞).
f(x) ≤ 0 over the interval [–1, ∞).
f(x) > 0 over the interval (–∞, 1).
f(x) < 0 over the interval (–∞. –1).
2. A 2-column table with 6 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2, 3. The second column is labeled f of x with entries 20, 0, negative 6, negative 4, 0, 0.
Which is an x-intercept of the continuous function in the table?
(–1, 0)
(0, –6)
(–6, 0)
(0, –1)
3. A 2-column table with 8 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2, 3, 4. The second column is labeled f of x with entries negative 2, 0, 2, 2, 0, negative 8, negative 10, negative 20.
Which could be the entire interval over which the function, f(x), is positive?
(–∞, 1)
(–2, 1)
(–∞, 0)
(1, 4)
4. Which table shows a function that is increasing only over the interval (–2, 1), and nowhere else?
A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 6, negative 3, negative 1, 1, 3, 6.
A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 2, negative 4, negative 1, 1, 4, 3.
A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 3, negative 5, negative 7, negative 6, 1, negative 1.
A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries 5, 7, 1, 0, negative 4, negative 2.
5. A 2-column table with 6 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2, 3. The second column is labeled f of x with entries negative 10, negative 8, negative 6, negative 4, negative 2, 0.
Which is an x-intercept of the continuous function in the table?
(0, –6)
(3, 0)
(–6, 0)
(0, 3)
6. A 2-column table with 6 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2, 3. The second column is labeled f of x with entries negative 8, 0, 0, negative 2, 0, 12.
Which lists all of the y-intercepts of the continuous function in the table?
(0, 0)
(–1, 0), (2, 0)
(–1, 0), (0, 0)
(–1, 0), (0, 0), (2, 0)
7. A 2-column table with 8 rows. The first column is labeled x with entries negative 6, negative 5, negative 4, negative 3, negative 2, negative 1, 0, 1. The second column is labeled f of x with entries 34, 3, negative 10, negative 11, negative 6, negative 1, negative 2, negative 15.
Using only the values given in the table for the function, f(x), what is the interval of x-values over which the function is increasing?
(–6, –3)
(–3, –1)
(–3, 0)
(–6, –5)
8. 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
9. 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 negative 10, 0, 0, negative 4, negative 6, 0.
Which is a y-intercept of the continuous function in the table?
(0, –6)
(–2, 0)
(–6, 0)
(0, –2)
10. A 2-column table with 7 rows. The first column is labeled x with entries negative 4, negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 6, negative 4, negative 1, negative 2, negative 5, negative 8, negative 16.
Which is a possible turning point for the continuous function f(x)?
(–3, –4)
(–2, –1)
(0, –5)
(1, –8)
Which is a valid prediction about the continuous function f(x)?
f(x) ≥ 0 over the interval [5, ∞).
f(x) ≤ 0 over the interval [–1, ∞).
f(x) > 0 over the interval (–∞, 1).
f(x) < 0 over the interval (–∞. –1).
2. A 2-column table with 6 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2, 3. The second column is labeled f of x with entries 20, 0, negative 6, negative 4, 0, 0.
Which is an x-intercept of the continuous function in the table?
(–1, 0)
(0, –6)
(–6, 0)
(0, –1)
3. A 2-column table with 8 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2, 3, 4. The second column is labeled f of x with entries negative 2, 0, 2, 2, 0, negative 8, negative 10, negative 20.
Which could be the entire interval over which the function, f(x), is positive?
(–∞, 1)
(–2, 1)
(–∞, 0)
(1, 4)
4. Which table shows a function that is increasing only over the interval (–2, 1), and nowhere else?
A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 6, negative 3, negative 1, 1, 3, 6.
A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 2, negative 4, negative 1, 1, 4, 3.
A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 3, negative 5, negative 7, negative 6, 1, negative 1.
A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries 5, 7, 1, 0, negative 4, negative 2.
5. A 2-column table with 6 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2, 3. The second column is labeled f of x with entries negative 10, negative 8, negative 6, negative 4, negative 2, 0.
Which is an x-intercept of the continuous function in the table?
(0, –6)
(3, 0)
(–6, 0)
(0, 3)
6. A 2-column table with 6 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2, 3. The second column is labeled f of x with entries negative 8, 0, 0, negative 2, 0, 12.
Which lists all of the y-intercepts of the continuous function in the table?
(0, 0)
(–1, 0), (2, 0)
(–1, 0), (0, 0)
(–1, 0), (0, 0), (2, 0)
7. A 2-column table with 8 rows. The first column is labeled x with entries negative 6, negative 5, negative 4, negative 3, negative 2, negative 1, 0, 1. The second column is labeled f of x with entries 34, 3, negative 10, negative 11, negative 6, negative 1, negative 2, negative 15.
Using only the values given in the table for the function, f(x), what is the interval of x-values over which the function is increasing?
(–6, –3)
(–3, –1)
(–3, 0)
(–6, –5)
8. 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
9. 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 negative 10, 0, 0, negative 4, negative 6, 0.
Which is a y-intercept of the continuous function in the table?
(0, –6)
(–2, 0)
(–6, 0)
(0, –2)
10. A 2-column table with 7 rows. The first column is labeled x with entries negative 4, negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 6, negative 4, negative 1, negative 2, negative 5, negative 8, negative 16.
Which is a possible turning point for the continuous function f(x)?
(–3, –4)
(–2, –1)
(0, –5)
(1, –8)
Answered by
GPT-4o
To find the possible values of \( m \) in the table for the function \( f(x) \), let's analyze the trend and conditions provided.
Given table of values:
\[
\begin{array}{|c|c|}
\hline
t & f(x) \\
\hline
-3 & -12 \\
-2 & m \\
-1 & 4 \\
0 & 0 \\
1 & -4 \\
2 & -2 \\
\hline
\end{array}
\]
We know that the function has exactly two potential turning points, and one turning point is at \( t = -1 \).
### Analyzing the intervals:
1. From \( t = -3 \) to \( t = -2 \):
- If \( f(t) \) changes from -12 to \( m \), the function's trend is either increasing if \( m > -12 \) or decreasing if \( m < -12 \).
2. From \( t = -2 \) to \( t = -1 \):
- If \( f(t) \) changes from \( m \) to 4, we need to ensure that this interval contains a turning point.
- Since \( t = -1 \) is a turning point, \( f(t) \) should be at a maximum or minimum at \( t = -1 \).
- Thus, the function must be increasing up to \( t = -1 \) (i.e., \( m < 4 \)) if \( t = -1 \) is a local maximum.
3. From \( t = -1 \) to \( t = 0 \):
- The change from \( f(-1) = 4 \) to \( f(0) = 0 \) shows that the function is decreasing.
4. From \( t = 0 \) to \( t = 1 \):
- The change from \( f(0) = 0 \) to \( f(1) = -4 \) shows that the function continues to decrease.
5. From \( t = 1 \) to \( t = 2 \):
- The function changes from \( f(1) = -4 \) to \( f(2) = -2 \), indicating an increase.
From this analysis, \( f(t) \) has two changes in direction (turning points), once at \( t = -1 \) and another one needed in the interval containing \( t = -2 \).
### Conditions for \( m \):
- \( m \) must be such that \( f(t) \) increases up to \( t = -1 \), meaning it must be less than 4 (local maximum reached at \( t = -1 \)) and ensured a change in trend from \( t = -3 \).
Thus, the appropriate condition that fits the scenario is:
\[ -12 < m < 4 \]
This interval ensures there are exactly two turning points in the function while satisfying the given constraints.
So, the best statement that describes all possible values of \( m \) is:
**\( -12 < m < 4 \)**.
Given table of values:
\[
\begin{array}{|c|c|}
\hline
t & f(x) \\
\hline
-3 & -12 \\
-2 & m \\
-1 & 4 \\
0 & 0 \\
1 & -4 \\
2 & -2 \\
\hline
\end{array}
\]
We know that the function has exactly two potential turning points, and one turning point is at \( t = -1 \).
### Analyzing the intervals:
1. From \( t = -3 \) to \( t = -2 \):
- If \( f(t) \) changes from -12 to \( m \), the function's trend is either increasing if \( m > -12 \) or decreasing if \( m < -12 \).
2. From \( t = -2 \) to \( t = -1 \):
- If \( f(t) \) changes from \( m \) to 4, we need to ensure that this interval contains a turning point.
- Since \( t = -1 \) is a turning point, \( f(t) \) should be at a maximum or minimum at \( t = -1 \).
- Thus, the function must be increasing up to \( t = -1 \) (i.e., \( m < 4 \)) if \( t = -1 \) is a local maximum.
3. From \( t = -1 \) to \( t = 0 \):
- The change from \( f(-1) = 4 \) to \( f(0) = 0 \) shows that the function is decreasing.
4. From \( t = 0 \) to \( t = 1 \):
- The change from \( f(0) = 0 \) to \( f(1) = -4 \) shows that the function continues to decrease.
5. From \( t = 1 \) to \( t = 2 \):
- The function changes from \( f(1) = -4 \) to \( f(2) = -2 \), indicating an increase.
From this analysis, \( f(t) \) has two changes in direction (turning points), once at \( t = -1 \) and another one needed in the interval containing \( t = -2 \).
### Conditions for \( m \):
- \( m \) must be such that \( f(t) \) increases up to \( t = -1 \), meaning it must be less than 4 (local maximum reached at \( t = -1 \)) and ensured a change in trend from \( t = -3 \).
Thus, the appropriate condition that fits the scenario is:
\[ -12 < m < 4 \]
This interval ensures there are exactly two turning points in the function while satisfying the given constraints.
So, the best statement that describes all possible values of \( m \) is:
**\( -12 < m < 4 \)**.
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