Question
A 2-column table with 6 rows. The first column is labeled x with entries negative 1, 0, 1, 2, 3, 4. The second column is labeled f of x with entries negative 9, negative 8, negative 7, 0, 19, 56.
What is the x-intercept of the continuous function shown in the table?
(0, –8)
(0, 2)
(–8, 0)
(2, 0)
What is the x-intercept of the continuous function shown in the table?
(0, –8)
(0, 2)
(–8, 0)
(2, 0)
Answers
Answer
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 18, 3, 0, 3, 6, 3.
Using only the values given in the table for the function f(x) = –x3 + 4x + 3, what is the largest interval of x-values where the function is increasing?
(,)
Using only the values given in the table for the function f(x) = –x3 + 4x + 3, what is the largest interval of x-values where the function is increasing?
(,)
Answer
A 2-column table with 7 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2, 3. The second column is labeled f of x with entries 15, 0, 3, 0, negative 3, 0, 15.
Predict which statements are true about the intervals of the continuous function. Check all that apply.
f(x) > 0 over the interval (−, 3).
f(x) ≤ 0 over the interval [0, 2].
f(x) < 0 over the interval (−1, 1).
f(x) > 0 over the interval (–2, 0).
f(x) ≥ 0 over the interval [2, ).
Predict which statements are true about the intervals of the continuous function. Check all that apply.
f(x) > 0 over the interval (−, 3).
f(x) ≤ 0 over the interval [0, 2].
f(x) < 0 over the interval (−1, 1).
f(x) > 0 over the interval (–2, 0).
f(x) ≥ 0 over the interval [2, ).
Answer
Nope try again must be more than one if your certain just pick another best answer if you really think its only one
Answer
f(x) > 0 over the interval (–2, 0). you forgot this!
- **\( f(x) \leq 0 \) over the interval \([0, 2]\)**.
- **\( f(x) \geq 0 \) over the interval \([2, \infty)\)**. correct though
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
.
- **\( f(x) \leq 0 \) over the interval \([0, 2]\)**.
- **\( f(x) \geq 0 \) over the interval \([2, \infty)\)**. correct though
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
.
Answer
There is no 1,2 for the second statement? its either 0,1 0,2 or -2,0
Answered by
GPT-4o mini
To find the x-intercept of the continuous function represented in the table, we need to look for the point where \( f(x) = 0 \).
From the entries given in your table:
- When \( x = 0 \), \( f(0) = -8 \)
- When \( x = 1 \), \( f(1) = -7 \)
- When \( x = 2 \), \( f(2) = 0 \)
The x-intercept occurs at \( x = 2 \) where \( f(2) = 0 \). Therefore, the x-intercept is at the point (2, 0).
Thus, the correct choice is:
**(2, 0)**
From the entries given in your table:
- When \( x = 0 \), \( f(0) = -8 \)
- When \( x = 1 \), \( f(1) = -7 \)
- When \( x = 2 \), \( f(2) = 0 \)
The x-intercept occurs at \( x = 2 \) where \( f(2) = 0 \). Therefore, the x-intercept is at the point (2, 0).
Thus, the correct choice is:
**(2, 0)**
Answered by
GPT-4o mini
To determine where the function \( f(x) = -x^3 + 4x + 3 \) is increasing using the given table of values, we will calculate the changes in \( f(x) \) for the intervals defined by the \( x \)-values in the table.
The \( x \) values and corresponding \( f(x) \) values from the table are:
- \( x = -3 \), \( f(-3) = 18 \)
- \( x = -2 \), \( f(-2) = 3 \)
- \( x = -1 \), \( f(-1) = 0 \)
- \( x = 0 \), \( f(0) = 3 \)
- \( x = 1 \), \( f(1) = 6 \)
- \( x = 2 \), \( f(2) = 3 \)
Now we will analyze the changes in the function values as \( x \) increases:
1. From \( x = -3 \) to \( x = -2 \):
- \( f(-3) = 18 \) to \( f(-2) = 3 \) → Decreasing
2. From \( x = -2 \) to \( x = -1 \):
- \( f(-2) = 3 \) to \( f(-1) = 0 \) → Decreasing
3. From \( x = -1 \) to \( x = 0 \):
- \( f(-1) = 0 \) to \( f(0) = 3 \) → Increasing
4. From \( x = 0 \) to \( x = 1 \):
- \( f(0) = 3 \) to \( f(1) = 6 \) → Increasing
5. From \( x = 1 \) to \( x = 2 \):
- \( f(1) = 6 \) to \( f(2) = 3 \) → Decreasing
From this analysis, we see that the function is increasing in the intervals \( (-1, 0) \) and \( (0, 1) \).
The largest interval where the function is increasing is from \( -1 \) to \( 1 \).
Thus, the answer is:
**(-1, 1)**
The \( x \) values and corresponding \( f(x) \) values from the table are:
- \( x = -3 \), \( f(-3) = 18 \)
- \( x = -2 \), \( f(-2) = 3 \)
- \( x = -1 \), \( f(-1) = 0 \)
- \( x = 0 \), \( f(0) = 3 \)
- \( x = 1 \), \( f(1) = 6 \)
- \( x = 2 \), \( f(2) = 3 \)
Now we will analyze the changes in the function values as \( x \) increases:
1. From \( x = -3 \) to \( x = -2 \):
- \( f(-3) = 18 \) to \( f(-2) = 3 \) → Decreasing
2. From \( x = -2 \) to \( x = -1 \):
- \( f(-2) = 3 \) to \( f(-1) = 0 \) → Decreasing
3. From \( x = -1 \) to \( x = 0 \):
- \( f(-1) = 0 \) to \( f(0) = 3 \) → Increasing
4. From \( x = 0 \) to \( x = 1 \):
- \( f(0) = 3 \) to \( f(1) = 6 \) → Increasing
5. From \( x = 1 \) to \( x = 2 \):
- \( f(1) = 6 \) to \( f(2) = 3 \) → Decreasing
From this analysis, we see that the function is increasing in the intervals \( (-1, 0) \) and \( (0, 1) \).
The largest interval where the function is increasing is from \( -1 \) to \( 1 \).
Thus, the answer is:
**(-1, 1)**
Answered by
GPT-4o mini
Let's analyze the values of \( f(x) \) from the table you provided:
- \( x = -3 \), \( f(-3) = 15 \)
- \( x = -2 \), \( f(-2) = 0 \)
- \( x = -1 \), \( f(-1) = 3 \)
- \( x = 0 \), \( f(0) = 0 \)
- \( x = 1 \), \( f(1) = -3 \)
- \( x = 2 \), \( f(2) = 0 \)
- \( x = 3 \), \( f(3) = 15 \)
Now, we can evaluate each statement based on these values:
1. **\( f(x) > 0 \) over the interval \( (-\infty, 3) \)**:
- From the table, we can see that \( f(x) > 0 \) for \( x = -3, -2, -1 \) (positive values) and \( x = 3 \) (also positive), but \( f(1) = -3 \) which means this value is not within the positive interval. Therefore, this statement is **false** because \( f(x) \) is not greater than 0 in the interval \( (-1, 1) \).
2. **\( f(x) \leq 0 \) over the interval \([0, 2]\)**:
- At \( x = 0\), \( f(0) = 0 \); at \( x = 1\), \( f(1) = -3\); at \( x = 2\), \( f(2) = 0\). Thus, \( f(x) \) is either 0 or negative from 0 to 2. Therefore, this statement is **true**.
3. **\( f(x) < 0 \) over the interval \((-1, 1)\)**:
- At \( x = -1 \), \( f(-1) = 3 \) (not less than 0) and at \( x = 1 \), \( f(1) = -3 \). Since \( f(-1) \) is positive, this statement is **false**.
4. **\( f(x) > 0 \) over the interval \((-2, 0)\)**:
- At \( x = -2\), \( f(-2) = 0\); at \( x = -1 \), \( f(-1) = 3\); and at \( x = 0\), \( f(0) = 0\). Therefore, \( f(x) \) is greater than 0 only at \( x = -1\), but it is 0 at the endpoints, making this statement **false**.
5. **\( f(x) \geq 0 \) over the interval \([2, \infty)\)**:
- At \( x = 2 \), \( f(2) = 0\) and at \( x = 3 \), \( f(3) = 15 \). Since there is no data provided for \( x > 3\), we can't definitively say about values beyond 3. Therefore, strictly speaking about the interval up to 3, at least \( f(2)\) is 0, thus it should include numbers \( \geq 0\) at least at the beginning, making this statement **false**.
In summary, the true statements are:
- **\( f(x) \leq 0 \) over the interval \([0, 2]\)**.
Thus, the only statement that is true is the second one.
- \( x = -3 \), \( f(-3) = 15 \)
- \( x = -2 \), \( f(-2) = 0 \)
- \( x = -1 \), \( f(-1) = 3 \)
- \( x = 0 \), \( f(0) = 0 \)
- \( x = 1 \), \( f(1) = -3 \)
- \( x = 2 \), \( f(2) = 0 \)
- \( x = 3 \), \( f(3) = 15 \)
Now, we can evaluate each statement based on these values:
1. **\( f(x) > 0 \) over the interval \( (-\infty, 3) \)**:
- From the table, we can see that \( f(x) > 0 \) for \( x = -3, -2, -1 \) (positive values) and \( x = 3 \) (also positive), but \( f(1) = -3 \) which means this value is not within the positive interval. Therefore, this statement is **false** because \( f(x) \) is not greater than 0 in the interval \( (-1, 1) \).
2. **\( f(x) \leq 0 \) over the interval \([0, 2]\)**:
- At \( x = 0\), \( f(0) = 0 \); at \( x = 1\), \( f(1) = -3\); at \( x = 2\), \( f(2) = 0\). Thus, \( f(x) \) is either 0 or negative from 0 to 2. Therefore, this statement is **true**.
3. **\( f(x) < 0 \) over the interval \((-1, 1)\)**:
- At \( x = -1 \), \( f(-1) = 3 \) (not less than 0) and at \( x = 1 \), \( f(1) = -3 \). Since \( f(-1) \) is positive, this statement is **false**.
4. **\( f(x) > 0 \) over the interval \((-2, 0)\)**:
- At \( x = -2\), \( f(-2) = 0\); at \( x = -1 \), \( f(-1) = 3\); and at \( x = 0\), \( f(0) = 0\). Therefore, \( f(x) \) is greater than 0 only at \( x = -1\), but it is 0 at the endpoints, making this statement **false**.
5. **\( f(x) \geq 0 \) over the interval \([2, \infty)\)**:
- At \( x = 2 \), \( f(2) = 0\) and at \( x = 3 \), \( f(3) = 15 \). Since there is no data provided for \( x > 3\), we can't definitively say about values beyond 3. Therefore, strictly speaking about the interval up to 3, at least \( f(2)\) is 0, thus it should include numbers \( \geq 0\) at least at the beginning, making this statement **false**.
In summary, the true statements are:
- **\( f(x) \leq 0 \) over the interval \([0, 2]\)**.
Thus, the only statement that is true is the second one.
Answered by
GPT-4o mini
Let's analyze the statements again based on the values provided in the table:
- \( x = -3 \), \( f(-3) = 15 \)
- \( x = -2 \), \( f(-2) = 0 \)
- \( x = -1 \), \( f(-1) = 3 \)
- \( x = 0 \), \( f(0) = 0 \)
- \( x = 1 \), \( f(1) = -3 \)
- \( x = 2 \), \( f(2) = 0 \)
- \( x = 3 \), \( f(3) = 15 \)
Now, let's evaluate each statement one by one:
1. **\( f(x) > 0 \) over the interval \( (-\infty, 3) \)**:
- From \( x = -3 \) to \( x = -2 \) (it is greater than 0), but it goes to negative between \( (-1, 1) \). So this statement is **false**.
2. **\( f(x) \leq 0 \) over the interval \([0, 2]\)**:
- At \( x = 0 \), \( f(0) = 0\); at \( x = 1 \), \( f(1) = -3\); at \( x = 2 \), \( f(2) = 0\). This statement is **true**.
3. **\( f(x) < 0 \) over the interval \((-1, 1)\)**:
- At \( x = -1 \) \( f(-1) = 3\) (not less than 0) and at \( x = 1 \), \( f(1) = -3\). Since \( f(-1) \) is positive, this statement is **false**.
4. **\( f(x) > 0 \) over the interval \((-2, 0)\)**:
- At \( x = -2 \), \( f(-2) = 0\) and at \( x = -1 \), \( f(-1) = 3\). Since \( f(-2) \) is equal to 0, this part of the interval does not satisfy \( f(x) > 0\). Therefore, this statement is **false**.
5. **\( f(x) \geq 0 \) over the interval \([2, \infty)\)**:
- At \( x = 2 \), \( f(2) = 0\); at \( x = 3 \), \( f(3) = 15\). The function is equal to 0 at \( x = 2 \) and positive for \( x = 3 \). We do not have values for \( x > 3 \), but within the interval from 2 to 3, it is valid that \( f(x) \geq 0\). So this statement can be viewed as **true** based on available data.
Thus, the true statements are:
- **\( f(x) \leq 0 \) over the interval \([0, 2]\)**.
- **\( f(x) \geq 0 \) over the interval \([2, \infty)\)**.
Thank you for your patience, and I hope this response satisfies your requirements!
- \( x = -3 \), \( f(-3) = 15 \)
- \( x = -2 \), \( f(-2) = 0 \)
- \( x = -1 \), \( f(-1) = 3 \)
- \( x = 0 \), \( f(0) = 0 \)
- \( x = 1 \), \( f(1) = -3 \)
- \( x = 2 \), \( f(2) = 0 \)
- \( x = 3 \), \( f(3) = 15 \)
Now, let's evaluate each statement one by one:
1. **\( f(x) > 0 \) over the interval \( (-\infty, 3) \)**:
- From \( x = -3 \) to \( x = -2 \) (it is greater than 0), but it goes to negative between \( (-1, 1) \). So this statement is **false**.
2. **\( f(x) \leq 0 \) over the interval \([0, 2]\)**:
- At \( x = 0 \), \( f(0) = 0\); at \( x = 1 \), \( f(1) = -3\); at \( x = 2 \), \( f(2) = 0\). This statement is **true**.
3. **\( f(x) < 0 \) over the interval \((-1, 1)\)**:
- At \( x = -1 \) \( f(-1) = 3\) (not less than 0) and at \( x = 1 \), \( f(1) = -3\). Since \( f(-1) \) is positive, this statement is **false**.
4. **\( f(x) > 0 \) over the interval \((-2, 0)\)**:
- At \( x = -2 \), \( f(-2) = 0\) and at \( x = -1 \), \( f(-1) = 3\). Since \( f(-2) \) is equal to 0, this part of the interval does not satisfy \( f(x) > 0\). Therefore, this statement is **false**.
5. **\( f(x) \geq 0 \) over the interval \([2, \infty)\)**:
- At \( x = 2 \), \( f(2) = 0\); at \( x = 3 \), \( f(3) = 15\). The function is equal to 0 at \( x = 2 \) and positive for \( x = 3 \). We do not have values for \( x > 3 \), but within the interval from 2 to 3, it is valid that \( f(x) \geq 0\). So this statement can be viewed as **true** based on available data.
Thus, the true statements are:
- **\( f(x) \leq 0 \) over the interval \([0, 2]\)**.
- **\( f(x) \geq 0 \) over the interval \([2, \infty)\)**.
Thank you for your patience, and I hope this response satisfies your requirements!
Answered by
GPT-4o mini
Let's analyze the new set of values in the table provided:
- \( x = -3 \), \( f(-3) = -16 \)
- \( x = -2 \), \( f(-2) = -1 \)
- \( x = -1 \), \( f(-1) = 2 \)
- \( x = 0 \), \( f(0) = -1 \)
- \( x = 1 \), \( f(1) = -4 \)
- \( x = 2 \), \( f(2) = -1 \)
### Finding Local Maximum and Minimum
1. **Local Maximum**: A local maximum occurs when the function changes from increasing to decreasing. We look for a point where \( f(x) \) seems to have a peak.
- Between \( x = -3 \) and \( x = -2 \):
- \( f(-3) = -16 \) to \( f(-2) = -1 \) (increasing)
- Between \( x = -2 \) and \( x = -1 \):
- \( f(-2) = -1 \) to \( f(-1) = 2 \) (increasing)
- Between \( x = -1 \) and \( x = 0 \):
- \( f(-1) = 2 \) (maximum) to \( f(0) = -1 \) (decreasing)
Thus, a local maximum occurs at \( x = -1 \).
2. **Local Minimum**: A local minimum occurs when the function changes from decreasing to increasing.
- Between \( x = -2 \) and \( x = -1 \):
- There's no local minimum here as it increases.
- Between \( x = -1 \) and \( x = 0 \):
- There is still a maximum at \( -1 \).
- Between \( x = 0 \) and \( x = 1 \):
- \( f(0) = -1 \) to \( f(1) = -4 \) (decreasing)
- Between \( x = 1 \) and \( x = 2 \):
- \( f(1) = -4 \) to \( f(2) = -1 \) (increasing)
Here, the function decreases to \( -4 \) at \( x = 1 \), then increases from there, so this indicates a local minimum occurs at \( x = 1 \).
### Conclusion
- A local maximum occurs over the interval \(\{ -1 \}\).
- A local minimum occurs over the interval \((1, 2)\) since the function rises after \( x = 1 \).
To summarize:
- Local Maximum: \( x = -1 \)
- Local Minimum: \( x \in (1, 2) \)
- \( x = -3 \), \( f(-3) = -16 \)
- \( x = -2 \), \( f(-2) = -1 \)
- \( x = -1 \), \( f(-1) = 2 \)
- \( x = 0 \), \( f(0) = -1 \)
- \( x = 1 \), \( f(1) = -4 \)
- \( x = 2 \), \( f(2) = -1 \)
### Finding Local Maximum and Minimum
1. **Local Maximum**: A local maximum occurs when the function changes from increasing to decreasing. We look for a point where \( f(x) \) seems to have a peak.
- Between \( x = -3 \) and \( x = -2 \):
- \( f(-3) = -16 \) to \( f(-2) = -1 \) (increasing)
- Between \( x = -2 \) and \( x = -1 \):
- \( f(-2) = -1 \) to \( f(-1) = 2 \) (increasing)
- Between \( x = -1 \) and \( x = 0 \):
- \( f(-1) = 2 \) (maximum) to \( f(0) = -1 \) (decreasing)
Thus, a local maximum occurs at \( x = -1 \).
2. **Local Minimum**: A local minimum occurs when the function changes from decreasing to increasing.
- Between \( x = -2 \) and \( x = -1 \):
- There's no local minimum here as it increases.
- Between \( x = -1 \) and \( x = 0 \):
- There is still a maximum at \( -1 \).
- Between \( x = 0 \) and \( x = 1 \):
- \( f(0) = -1 \) to \( f(1) = -4 \) (decreasing)
- Between \( x = 1 \) and \( x = 2 \):
- \( f(1) = -4 \) to \( f(2) = -1 \) (increasing)
Here, the function decreases to \( -4 \) at \( x = 1 \), then increases from there, so this indicates a local minimum occurs at \( x = 1 \).
### Conclusion
- A local maximum occurs over the interval \(\{ -1 \}\).
- A local minimum occurs over the interval \((1, 2)\) since the function rises after \( x = 1 \).
To summarize:
- Local Maximum: \( x = -1 \)
- Local Minimum: \( x \in (1, 2) \)
Answered by
GPT-4o mini
Let's analyze the function values from the table again, focusing specifically on identifying the local minimum correctly.
Here are the values given:
- \( x = -3 \), \( f(-3) = -16 \)
- \( x = -2 \), \( f(-2) = -1 \)
- \( x = -1 \), \( f(-1) = 2 \)
- \( x = 0 \), \( f(0) = -1 \)
- \( x = 1 \), \( f(1) = -4 \)
- \( x = 2 \), \( f(2) = -1 \)
### Analyzing for Local Maximum and Minimum:
1. **Identifying Local Maximum**:
- **Between \( x = -3 \) and \( x = -2 \)**: \( f(-3) = -16 \) to \( f(-2) = -1 \) (increasing)
- **Between \( x = -2 \) and \( x = -1 \)**: \( f(-2) = -1 \) to \( f(-1) = 2 \) (increasing)
- **Between \( x = -1 \) and \( x = 0 \)**: \( f(-1) = 2 \) (maximum) to \( f(0) = -1 \) (decreasing)
Thus, a **local maximum** occurs at \( x = -1\).
2. **Identifying Local Minimum**:
- **Between \( x = -2 \) and \( x = -1 \)**: No minimum here, as it increases.
- **Between \( x = -1 \) and \( x = 0 \)**: There’s a maximum at \( -1 \) transitioning to negative values.
- **Between \( x = 0 \) and \( x = 1 \)**: \( f(0) = -1 \) to \( f(1) = -4 \) (decreasing).
- **Between \( x = 1 \) and \( x = 2 \)**: \( f(1) = -4 \) to \( f(2) = -1 \) (increasing).
Thus, we see that the function decreases to \( -4 \) at \( x = 1 \) and then increases afterward, indicating that the local minimum occurs at \( x = 1 \).
To clarify with your intervals:
- A **local maximum** occurs over the interval \(\{-1\}\).
- A **local minimum** occurs between \( x = 0 \) and \( x = 1 \) when the function transitions from \( -1 \) to \( -4 \).
So, correcting for the interval choices based on your guidance:
- A local minimum occurs over the interval \([0, 1)\).
Thank you for your patience, and I hope this clarification meets your expectations!
Here are the values given:
- \( x = -3 \), \( f(-3) = -16 \)
- \( x = -2 \), \( f(-2) = -1 \)
- \( x = -1 \), \( f(-1) = 2 \)
- \( x = 0 \), \( f(0) = -1 \)
- \( x = 1 \), \( f(1) = -4 \)
- \( x = 2 \), \( f(2) = -1 \)
### Analyzing for Local Maximum and Minimum:
1. **Identifying Local Maximum**:
- **Between \( x = -3 \) and \( x = -2 \)**: \( f(-3) = -16 \) to \( f(-2) = -1 \) (increasing)
- **Between \( x = -2 \) and \( x = -1 \)**: \( f(-2) = -1 \) to \( f(-1) = 2 \) (increasing)
- **Between \( x = -1 \) and \( x = 0 \)**: \( f(-1) = 2 \) (maximum) to \( f(0) = -1 \) (decreasing)
Thus, a **local maximum** occurs at \( x = -1\).
2. **Identifying Local Minimum**:
- **Between \( x = -2 \) and \( x = -1 \)**: No minimum here, as it increases.
- **Between \( x = -1 \) and \( x = 0 \)**: There’s a maximum at \( -1 \) transitioning to negative values.
- **Between \( x = 0 \) and \( x = 1 \)**: \( f(0) = -1 \) to \( f(1) = -4 \) (decreasing).
- **Between \( x = 1 \) and \( x = 2 \)**: \( f(1) = -4 \) to \( f(2) = -1 \) (increasing).
Thus, we see that the function decreases to \( -4 \) at \( x = 1 \) and then increases afterward, indicating that the local minimum occurs at \( x = 1 \).
To clarify with your intervals:
- A **local maximum** occurs over the interval \(\{-1\}\).
- A **local minimum** occurs between \( x = 0 \) and \( x = 1 \) when the function transitions from \( -1 \) to \( -4 \).
So, correcting for the interval choices based on your guidance:
- A local minimum occurs over the interval \([0, 1)\).
Thank you for your patience, and I hope this clarification meets your expectations!
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