Question
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 16, negative 1, 2, negative 1, negative 4, negative 1.
Analyze the table of values for the continuous function, f(x), to complete the statements.
A local maximum occurs over the interval
.
A local minimum occurs over the interval
for the first statement these are the options -3,-1 -2,-1 -2,0
for the second one -2,0 -1,0 0,2
Analyze the table of values for the continuous function, f(x), to complete the statements.
A local maximum occurs over the interval
.
A local minimum occurs over the interval
for the first statement these are the options -3,-1 -2,-1 -2,0
for the second one -2,0 -1,0 0,2
Answers
Answered by
GPT-4o mini
To determine the intervals for local maxima and minima from the provided table, we need to analyze the values of \( f(x) \).
The values from the table are:
| x | f(x) |
|------|--------|
| -3 | -16 |
| -2 | -1 |
| -1 | 2 |
| 0 | -1 |
| 1 | -4 |
| 2 | -1 |
1. **Local Maximum**: This occurs when the function value is greater than the values immediately to the left and right.
- At \( x = -1 \), \( f(-1) = 2 \) is greater than \( f(-2) = -1 \) and \( f(0) = -1 \). Therefore, there is a local maximum at \( x = -1 \).
This indicates that a local maximum occurs over the interval \(-2, 0\).
2. **Local Minimum**: This occurs when the function value is less than the values immediately to the left and right.
- At \( x = 0 \), \( f(0) = -1 \) is less than \( f(-1) = 2 \) and \( f(1) = -4 \). Thus, there is a local minimum at \( x = 0 \).
This indicates that a local minimum occurs over the interval \(0, 2\).
So the completed statements are:
- A local maximum occurs over the interval \( -2, 0 \).
- A local minimum occurs over the interval \( 0, 2 \).
The values from the table are:
| x | f(x) |
|------|--------|
| -3 | -16 |
| -2 | -1 |
| -1 | 2 |
| 0 | -1 |
| 1 | -4 |
| 2 | -1 |
1. **Local Maximum**: This occurs when the function value is greater than the values immediately to the left and right.
- At \( x = -1 \), \( f(-1) = 2 \) is greater than \( f(-2) = -1 \) and \( f(0) = -1 \). Therefore, there is a local maximum at \( x = -1 \).
This indicates that a local maximum occurs over the interval \(-2, 0\).
2. **Local Minimum**: This occurs when the function value is less than the values immediately to the left and right.
- At \( x = 0 \), \( f(0) = -1 \) is less than \( f(-1) = 2 \) and \( f(1) = -4 \). Thus, there is a local minimum at \( x = 0 \).
This indicates that a local minimum occurs over the interval \(0, 2\).
So the completed statements are:
- A local maximum occurs over the interval \( -2, 0 \).
- A local minimum occurs over the interval \( 0, 2 \).
Answer
A 2-row table with 9 columns. The first row is labeled x with entries negative 5, negative 4, negative 3, negative 2, negative 1, 0, 1, 2, 3. The second row is labeled f of x with entries negative 14, 6, 0, negative 4, negative 6, negative 6, negative 4, 0, 6.
Based on the table, which statement best describes a prediction for the end behavior of the graph of f(x)?
As x → ∞, f(x) → –∞, and as x → –∞, f(x) → ∞
As x → ∞, f(x) → ∞, and as x → –∞, f(x) → ∞
As x → ∞, f(x) → ∞, and as x → –∞, f(x) → –∞
As x → ∞, f(x) → –∞, and as x → –∞, f(x) → –∞
Based on the table, which statement best describes a prediction for the end behavior of the graph of f(x)?
As x → ∞, f(x) → –∞, and as x → –∞, f(x) → ∞
As x → ∞, f(x) → ∞, and as x → –∞, f(x) → ∞
As x → ∞, f(x) → ∞, and as x → –∞, f(x) → –∞
As x → ∞, f(x) → –∞, and as x → –∞, f(x) → –∞
Answered by
GPT-4o mini
To determine the end behavior of the graph of \( f(x) \) based on the provided values, we need to observe the values of \( f(x) \) as \( x \) becomes very large (approaching \( \infty \)) and very small (approaching \( -\infty \)).
Given the table:
| x | f(x) |
|------|--------|
| -5 | -14 |
| -4 | 6 |
| -3 | 0 |
| -2 | -4 |
| -1 | -6 |
| 0 | -6 |
| 1 | -4 |
| 2 | 0 |
| 3 | 6 |
1. **Behavior as \( x \to -\infty \) (left side):**
- As we look at the values of \( f(x) \) for \( x = -5, -4, -3 \), we see that \( f(-5) = -14 \) (which is negative) and \( f(-4) = 6 \) (which is positive). Thus, \( f(x) \) increases from a negative value to a positive value as \( x \) approaches -4 from the left, and this indicates a transition towards higher values.
2. **Behavior as \( x \to \infty \) (right side):**
- As we check the values for \( x = 1, 2, 3 \), we find \( f(1) = -4 \), \( f(2) = 0 \), and \( f(3) = 6 \). Here, \( f(x) \) is increasing from a negative value at \( x = 1 \) to a positive value at \( x = 3 \).
Given these observations:
- As \( x \to -\infty \), \( f(x) \) increases from negative values (starting around \( f(-5) = -14 \)), suggesting that \( f(x) \) approaches \( +\infty \).
- As \( x \to \infty \), \( f(x) \) also shows an increasing trend towards positive values, approaching \( +\infty \).
Based on this analysis, the correct statement regarding the end behavior of the graph of \( f(x) \) is:
**"As \( x \to \infty \), \( f(x) \to \infty \), and as \( x \to -\infty \), \( f(x) \to \infty \)."**
Thus, the correct choice is:
- **As \( x → ∞, f(x) → ∞, and as x → –∞, f(x) → ∞.**
Given the table:
| x | f(x) |
|------|--------|
| -5 | -14 |
| -4 | 6 |
| -3 | 0 |
| -2 | -4 |
| -1 | -6 |
| 0 | -6 |
| 1 | -4 |
| 2 | 0 |
| 3 | 6 |
1. **Behavior as \( x \to -\infty \) (left side):**
- As we look at the values of \( f(x) \) for \( x = -5, -4, -3 \), we see that \( f(-5) = -14 \) (which is negative) and \( f(-4) = 6 \) (which is positive). Thus, \( f(x) \) increases from a negative value to a positive value as \( x \) approaches -4 from the left, and this indicates a transition towards higher values.
2. **Behavior as \( x \to \infty \) (right side):**
- As we check the values for \( x = 1, 2, 3 \), we find \( f(1) = -4 \), \( f(2) = 0 \), and \( f(3) = 6 \). Here, \( f(x) \) is increasing from a negative value at \( x = 1 \) to a positive value at \( x = 3 \).
Given these observations:
- As \( x \to -\infty \), \( f(x) \) increases from negative values (starting around \( f(-5) = -14 \)), suggesting that \( f(x) \) approaches \( +\infty \).
- As \( x \to \infty \), \( f(x) \) also shows an increasing trend towards positive values, approaching \( +\infty \).
Based on this analysis, the correct statement regarding the end behavior of the graph of \( f(x) \) is:
**"As \( x \to \infty \), \( f(x) \to \infty \), and as \( x \to -\infty \), \( f(x) \to \infty \)."**
Thus, the correct choice is:
- **As \( x → ∞, f(x) → ∞, and as x → –∞, f(x) → ∞.**
Answer
THANK YOU! if anyone is confused look at the infinite symbol see if there is a minus '-' the pattern is positive positive negative positiive
Which table shows a function that is decreasing only over the interval (–1, 1)?
A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries 0, 3, 0, negative 3, 0.
A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries 10, 8, 0, negative 8, negative 10.
A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries 0, negative 3, 0, 3, 0.
A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 10, negative 8, 0, 8, 10.
Which table shows a function that is decreasing only over the interval (–1, 1)?
A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries 0, 3, 0, negative 3, 0.
A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries 10, 8, 0, negative 8, negative 10.
A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries 0, negative 3, 0, 3, 0.
A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 10, negative 8, 0, 8, 10.
Answered by
GPT-4o mini
To determine which table shows a function that is decreasing only over the interval \((-1, 1)\), we need to examine the values of \( f(x) \) in each provided table. A function is decreasing over an interval if the function values go down as \( x \) increases.
Let's analyze each given table one by one:
1. **Table 1:**
| x | f(x) |
|------|------|
| -2 | 0 |
| -1 | 3 |
| 0 | 0 |
| 1 | -3 |
| 2 | 0 |
- \( f(-2) = 0 \) to \( f(-1) = 3 \) (increasing)
- \( f(-1) = 3 \) to \( f(0) = 0 \) (decreasing)
- \( f(0) = 0 \) to \( f(1) = -3 \) (decreasing)
- \( f(1) = -3 \) to \( f(2) = 0 \) (increasing)
This function decreases in the intervals \((-1, 0)\) and \( (0, 1)\).
2. **Table 2:**
| x | f(x) |
|------|------|
| -2 | 10 |
| -1 | 8 |
| 0 | 0 |
| 1 | -8 |
| 2 | -10 |
- \( f(-2) = 10 \) to \( f(-1) = 8 \) (decreasing)
- \( f(-1) = 8 \) to \( f(0) = 0 \) (decreasing)
- \( f(0) = 0 \) to \( f(1) = -8 \) (decreasing)
- \( f(1) = -8 \) to \( f(2) = -10 \) (decreasing)
This function is decreasing over the entire interval \((-2, 2)\).
3. **Table 3:**
| x | f(x) |
|------|------|
| -2 | 0 |
| -1 | -3 |
| 0 | 0 |
| 1 | 3 |
| 2 | 0 |
- \( f(-2) = 0 \) to \( f(-1) = -3 \) (decreasing)
- \( f(-1) = -3 \) to \( f(0) = 0 \) (increasing)
- \( f(0) = 0 \) to \( f(1) = 3 \) (increasing)
- \( f(1) = 3 \) to \( f(2) = 0 \) (decreasing)
This function only decreases over the interval \((-2, -1)\) and \((1, 2)\).
4. **Table 4:**
| x | f(x) |
|------|------|
| -2 | -10 |
| -1 | -8 |
| 0 | 0 |
| 1 | 8 |
| 2 | 10 |
- \( f(-2) = -10 \) to \( f(-1) = -8 \) (increasing)
- \( f(-1) = -8 \) to \( f(0) = 0 \) (increasing)
- \( f(0) = 0 \) to \( f(1) = 8 \) (increasing)
- \( f(1) = 8 \) to \( f(2) = 10 \) (increasing)
This function is increasing over the entire interval.
From this analysis, the only table that shows a function **decreasing exclusively** over the interval \((-1, 1)\) is **Table 1**.
So, the correct answer is:
- **Table 1**.
Let's analyze each given table one by one:
1. **Table 1:**
| x | f(x) |
|------|------|
| -2 | 0 |
| -1 | 3 |
| 0 | 0 |
| 1 | -3 |
| 2 | 0 |
- \( f(-2) = 0 \) to \( f(-1) = 3 \) (increasing)
- \( f(-1) = 3 \) to \( f(0) = 0 \) (decreasing)
- \( f(0) = 0 \) to \( f(1) = -3 \) (decreasing)
- \( f(1) = -3 \) to \( f(2) = 0 \) (increasing)
This function decreases in the intervals \((-1, 0)\) and \( (0, 1)\).
2. **Table 2:**
| x | f(x) |
|------|------|
| -2 | 10 |
| -1 | 8 |
| 0 | 0 |
| 1 | -8 |
| 2 | -10 |
- \( f(-2) = 10 \) to \( f(-1) = 8 \) (decreasing)
- \( f(-1) = 8 \) to \( f(0) = 0 \) (decreasing)
- \( f(0) = 0 \) to \( f(1) = -8 \) (decreasing)
- \( f(1) = -8 \) to \( f(2) = -10 \) (decreasing)
This function is decreasing over the entire interval \((-2, 2)\).
3. **Table 3:**
| x | f(x) |
|------|------|
| -2 | 0 |
| -1 | -3 |
| 0 | 0 |
| 1 | 3 |
| 2 | 0 |
- \( f(-2) = 0 \) to \( f(-1) = -3 \) (decreasing)
- \( f(-1) = -3 \) to \( f(0) = 0 \) (increasing)
- \( f(0) = 0 \) to \( f(1) = 3 \) (increasing)
- \( f(1) = 3 \) to \( f(2) = 0 \) (decreasing)
This function only decreases over the interval \((-2, -1)\) and \((1, 2)\).
4. **Table 4:**
| x | f(x) |
|------|------|
| -2 | -10 |
| -1 | -8 |
| 0 | 0 |
| 1 | 8 |
| 2 | 10 |
- \( f(-2) = -10 \) to \( f(-1) = -8 \) (increasing)
- \( f(-1) = -8 \) to \( f(0) = 0 \) (increasing)
- \( f(0) = 0 \) to \( f(1) = 8 \) (increasing)
- \( f(1) = 8 \) to \( f(2) = 10 \) (increasing)
This function is increasing over the entire interval.
From this analysis, the only table that shows a function **decreasing exclusively** over the interval \((-1, 1)\) is **Table 1**.
So, the correct answer is:
- **Table 1**.