Question
On a coordinate plane, a piecewise function has 2 lines. The first line has a closed circle at (negative 2, negative 2) and then goes up through (negative 4, 2) with an arrow instead of an endpoint. The second line has an open circle at (2, 1) and then goes up through (5, 4) with an arrow instead of an endpoint.Which values are within the domain of the function? Check all that apply.
x = –6x
x = –4
x = –2
x = 0
x = 2
x = 4
x = –6x
x = –4
x = –2
x = 0
x = 2
x = 4
Answers
Answer
include x = –4 and yes
On a coordinate plane, a piecewise function has 2 lines. The first line has a closed circle at (0, 3) and then goes down througacccch (negative 4, 1) with an arrow instead of an endpoint. The second line has an open circle at (0, 1) and then goes down through (4, negative 1) with an arrow instead of an endpoint.Consider the function given by the graph. What are these values?
f(–2 ) =___
f(0) =___
f(4) =___
On a coordinate plane, a piecewise function has 2 lines. The first line has a closed circle at (0, 3) and then goes down througacccch (negative 4, 1) with an arrow instead of an endpoint. The second line has an open circle at (0, 1) and then goes down through (4, negative 1) with an arrow instead of an endpoint.Consider the function given by the graph. What are these values?
f(–2 ) =___
f(0) =___
f(4) =___
Answer
2,3,-1
Evaluate f(3) for the piecewise function:
f(x)={3x/2 + 8, x < -6
{-3x - 2, -4 ≤ x ≤ 3
4x + 4, x > 3
Which value represents f(3)?
–11
8
12.5
16
Evaluate f(3) for the piecewise function:
f(x)={3x/2 + 8, x < -6
{-3x - 2, -4 ≤ x ≤ 3
4x + 4, x > 3
Which value represents f(3)?
–11
8
12.5
16
Answer
Which table represents the second piece of the function f(x) = StartLayout enlarged left-brace 1st Row 1st column negative 3.5 x + 0.5, 2nd column x less-than 1 2nd row 1st column 8 minus 2 x, 2nd column x greater-than-or-equal-to 1 EndLayout?A 2-column table has 3 rows. The first column is labeled x with entries negative 1, 0, 1. The second column is labeled f (x) with entries 4, 0.5, negative 3.A 2-column table has 3 rows. The first column is labeled x with entries negative 1, 0, 1. The second column is labeled f (x) with entries 10, 8, 6.A 2-column table has 3 rows. The first column is labeled x with entries 1, 2, 3. The second column is labeled f (x) with entries 6, 4, 2.A 2-column table has 3 rows. The first column is labeled x with entries negative 1, 2, 3. The second column is labeled f (x) with entries negative 3, negative 6.5, negative 10.
Answer
Defining a Piecewise FunctionOn a coordinate plane, a piecewise function has 2 lines. The first line has an open circle at (0, 1) and goes up through (negative 4, 3) with an arrow instead of an endpoint. The second line has a closed circle at (0, negative 2) and goes up through (2, 2) with an arrow instead of an endpoint.Which piecewise function is shown in the graph?f(x) = StartLayout enlarged left-brace 1st Row 1st column negative 0.5 x + 1, 2nd column x less-than 0 2nd row 1st column 2 x minus 2, 2nd column x greater-than-or-equal-to 0 EndLayoutf(x) = StartLayout enlarged left-brace 1st Row 1st column negative x + 1, 2nd column x less-than 0 2nd row 1st column 0.5 x minus 2, 2nd column x greater-than 0 EndLayoutf(x) = StartLayout enlarged left-brace 1st Row 1st column x minus 1, 2nd column x less-than-or-equal-to 0 2nd row 1st column 2 x minus 2, 2nd column x greater-than 0 EndLayoutf(x) = StartLayout enlarged left-brace 1st Row 1st column negative 0.5 x minus 1, 2nd column x less-than-or-equal-to 0 2nd row 1st column 2 x minus 2, 2nd column x greater-than-or-equal-to 0 EndLayout
Answer
Lucy needs to buy some organic apples, and her grocery store is having a sale on them. If she buys 3 or fewer pounds of apples, the price will be $1.50 per pound. If she buys more than 3 pounds of apples, the price is $1.10 per pound. What is the domain of the piecewise-defined function, where x represents the number of pounds of apples?
{x| x ≥ 0}
{x| x is a real number}
{x| 0 ≤ x ≤ 3}
{x| x ≥ 3}
{x| x ≥ 0}
{x| x is a real number}
{x| 0 ≤ x ≤ 3}
{x| x ≥ 3}
Answered by
GPT-4o mini
To determine the values within the domain of the piecewise function described, we need to identify the ranges of \( x \) for each line segment.
1. **First line segment**:
- It starts with a closed circle at \((-2, -2)\). This means that \(-2\) is included in the domain.
- The line goes up and extends toward \((-4, 2)\), which corresponds to \( x = -4 \) but this point is not included as there is an arrow suggesting the line extends beyond this point in the upward direction. Therefore, the left endpoint is \(-2\) and extends leftward to \(-\infty\).
- Thus, the first line segment represents the domain portion: \(-\infty < x \leq -2\).
2. **Second line segment**:
- It has an open circle at \( (2, 1) \), which means \( x = 2 \) is not included in the domain.
- The line goes up through \( (5, 4) \) and continues to extend upward, indicating that \( x \) can take all values starting just above \( 2 \) and going to \( 5 \) onward but not including \( 2 \).
- Thus, the second line segment represents the domain portion: \(2 < x < \infty\).
Now we can evaluate the provided options:
- \( x = -6 \) is within the interval \( (-\infty, -2] \): **Included.**
- \( x = -4 \) is *not* included in the first line segment; hence it is: **Not Included.**
- \( x = -2 \) is included as it is a closed circle at that point: **Included.**
- \( x = 0 \) is not included; it is not part of either interval: **Not Included.**
- \( x = 2 \) is not included due to the open circle: **Not Included.**
- \( x = 4 \) is within the second segment since \( 2 < 4 < \infty \): **Included.**
In summary, the values within the domain of the function are:
- \( x = -6 \)
- \( x = -2 \)
- \( x = 4 \)
1. **First line segment**:
- It starts with a closed circle at \((-2, -2)\). This means that \(-2\) is included in the domain.
- The line goes up and extends toward \((-4, 2)\), which corresponds to \( x = -4 \) but this point is not included as there is an arrow suggesting the line extends beyond this point in the upward direction. Therefore, the left endpoint is \(-2\) and extends leftward to \(-\infty\).
- Thus, the first line segment represents the domain portion: \(-\infty < x \leq -2\).
2. **Second line segment**:
- It has an open circle at \( (2, 1) \), which means \( x = 2 \) is not included in the domain.
- The line goes up through \( (5, 4) \) and continues to extend upward, indicating that \( x \) can take all values starting just above \( 2 \) and going to \( 5 \) onward but not including \( 2 \).
- Thus, the second line segment represents the domain portion: \(2 < x < \infty\).
Now we can evaluate the provided options:
- \( x = -6 \) is within the interval \( (-\infty, -2] \): **Included.**
- \( x = -4 \) is *not* included in the first line segment; hence it is: **Not Included.**
- \( x = -2 \) is included as it is a closed circle at that point: **Included.**
- \( x = 0 \) is not included; it is not part of either interval: **Not Included.**
- \( x = 2 \) is not included due to the open circle: **Not Included.**
- \( x = 4 \) is within the second segment since \( 2 < 4 < \infty \): **Included.**
In summary, the values within the domain of the function are:
- \( x = -6 \)
- \( x = -2 \)
- \( x = 4 \)
Answered by
GPT-4o mini
To determine the values of the function \( f(x) \) based on the given piecewise function description, let's analyze the two segments.
1. **First line segment**:
- There is a **closed circle** at \( (0, 3) \), which means \( f(0) = 3 \).
- The line goes down through \( (-4, 1) \) extending leftward, which indicates that the function describes part of a line that continues downwards from \( x = 0 \) towards \( x = -4 \) and includes the endpoint at \( (0, 3) \).
**Expression of the first line segment**:
- The slope can be calculated if we consider the two points \((0, 3)\) and \((-4, 1)\):
\[
m = \frac{3 - 1}{0 - (-4)} = \frac{2}{4} = \frac{1}{2}.
\]
The equation of the line starting from \( (0, 3) \) can be expressed as:
\[
y = -\frac{1}{2}x + 3.
\]
Now we can compute \( f(-2) \):
\[
f(-2) = -\frac{1}{2}(-2) + 3 = 1 + 3 = 4.
\]
2. **Second line segment**:
- There is an **open circle** at \( (0, 1) \) indicating that \( f(0) \) is not defined for this part.
- The line goes down through \( (4, -1) \) showing it continues downward all the way to the right.
**Expression of the second line segment**:
- The points \( (0, 1) \) and \( (4, -1) \) can also be used to determine its slope:
\[
m = \frac{-1 - 1}{4 - 0} = \frac{-2}{4} = -\frac{1}{2}.
\]
The equation can be determined to be:
\[
y = -\frac{1}{2}x + 1.
\]
Now we can compute \( f(4) \):
\[
f(4) = -\frac{1}{2}(4) + 1 = -2 + 1 = -1.
\]
### Summary of Evaluated Values:
- \( f(-2) = 4 \)
- \( f(0) \) is not defined, but since the closed dot on the left line at 3 is applicable, we can view it as:
- \( f(0) = 3 \) from the first segment or **undefined** from the second segment.
- \( f(4) = -1 \)
In conclusion:
- \( f(-2) = 4 \)
- \( f(0) = 3\) (based on closed circle context)
- \( f(4) = -1 \)
1. **First line segment**:
- There is a **closed circle** at \( (0, 3) \), which means \( f(0) = 3 \).
- The line goes down through \( (-4, 1) \) extending leftward, which indicates that the function describes part of a line that continues downwards from \( x = 0 \) towards \( x = -4 \) and includes the endpoint at \( (0, 3) \).
**Expression of the first line segment**:
- The slope can be calculated if we consider the two points \((0, 3)\) and \((-4, 1)\):
\[
m = \frac{3 - 1}{0 - (-4)} = \frac{2}{4} = \frac{1}{2}.
\]
The equation of the line starting from \( (0, 3) \) can be expressed as:
\[
y = -\frac{1}{2}x + 3.
\]
Now we can compute \( f(-2) \):
\[
f(-2) = -\frac{1}{2}(-2) + 3 = 1 + 3 = 4.
\]
2. **Second line segment**:
- There is an **open circle** at \( (0, 1) \) indicating that \( f(0) \) is not defined for this part.
- The line goes down through \( (4, -1) \) showing it continues downward all the way to the right.
**Expression of the second line segment**:
- The points \( (0, 1) \) and \( (4, -1) \) can also be used to determine its slope:
\[
m = \frac{-1 - 1}{4 - 0} = \frac{-2}{4} = -\frac{1}{2}.
\]
The equation can be determined to be:
\[
y = -\frac{1}{2}x + 1.
\]
Now we can compute \( f(4) \):
\[
f(4) = -\frac{1}{2}(4) + 1 = -2 + 1 = -1.
\]
### Summary of Evaluated Values:
- \( f(-2) = 4 \)
- \( f(0) \) is not defined, but since the closed dot on the left line at 3 is applicable, we can view it as:
- \( f(0) = 3 \) from the first segment or **undefined** from the second segment.
- \( f(4) = -1 \)
In conclusion:
- \( f(-2) = 4 \)
- \( f(0) = 3\) (based on closed circle context)
- \( f(4) = -1 \)
Answered by
GPT-4o mini
To evaluate \( f(3) \) for the given piecewise function:
\[
f(x) =
\begin{cases}
\frac{3x}{2} + 8 & \text{if } x < -6 \\
-3x - 2 & \text{if } -4 \leq x \leq 3 \\
4x + 4 & \text{if } x > 3
\end{cases}
\]
First, we need to identify which condition applies to \( x = 3 \).
Since \( 3 \) falls into the interval \( -4 \leq x \leq 3 \), we will use the second piece:
\[
f(x) = -3x - 2.
\]
Now we can substitute \( x = 3 \) into this expression:
\[
f(3) = -3(3) - 2.
\]
Calculating this gives:
\[
f(3) = -9 - 2 = -11.
\]
Thus, the value of \( f(3) \) is **–11**.
\[
f(x) =
\begin{cases}
\frac{3x}{2} + 8 & \text{if } x < -6 \\
-3x - 2 & \text{if } -4 \leq x \leq 3 \\
4x + 4 & \text{if } x > 3
\end{cases}
\]
First, we need to identify which condition applies to \( x = 3 \).
Since \( 3 \) falls into the interval \( -4 \leq x \leq 3 \), we will use the second piece:
\[
f(x) = -3x - 2.
\]
Now we can substitute \( x = 3 \) into this expression:
\[
f(3) = -3(3) - 2.
\]
Calculating this gives:
\[
f(3) = -9 - 2 = -11.
\]
Thus, the value of \( f(3) \) is **–11**.
Answered by
GPT-4o mini
To determine which table represents the second piece of the function
\[
f(x) =
\begin{cases}
-3.5x + 0.5 & \text{if } x < 1 \\
8 - 2x & \text{if } x \geq 1
\end{cases}
\]
we need to evaluate the second piece of the function, which is applicable for \( x \geq 1 \):
\[
f(x) = 8 - 2x.
\]
Now, we'll calculate \( f(x) \) for values in the relevant range (specifically \( x = 1, 2, 3 \)).
1. **For \( x = 1 \)**:
\[
f(1) = 8 - 2(1) = 8 - 2 = 6.
\]
2. **For \( x = 2 \)**:
\[
f(2) = 8 - 2(2) = 8 - 4 = 4.
\]
3. **For \( x = 3 \)**:
\[
f(3) = 8 - 2(3) = 8 - 6 = 2.
\]
Now, we can summarize the values we have computed:
- \( f(1) = 6 \)
- \( f(2) = 4 \)
- \( f(3) = 2 \)
Next, we compare these results to the options provided:
1. **Table 1**:
- \( x: [-1, 0, 1] \)
- \( f(x): [4, 0.5, -3] \) (Not matching)
2. **Table 2**:
- \( x: [-1, 0, 1] \)
- \( f(x): [10, 8, 6] \) (Not matching)
3. **Table 3**:
- \( x: [1, 2, 3] \)
- \( f(x): [6, 4, 2] \) (Matches our calculations)
4. **Table 4**:
- \( x: [-1, 2, 3] \)
- \( f(x): [-3, -6.5, -10] \) (Not matching)
### Conclusion:
The correct table that represents the second piece \( f(x) = 8 - 2x \) is **Table 3**:
| \( x \) | \( f(x) \) |
|---------|-----------|
| 1 | 6 |
| 2 | 4 |
| 3 | 2 |
\[
f(x) =
\begin{cases}
-3.5x + 0.5 & \text{if } x < 1 \\
8 - 2x & \text{if } x \geq 1
\end{cases}
\]
we need to evaluate the second piece of the function, which is applicable for \( x \geq 1 \):
\[
f(x) = 8 - 2x.
\]
Now, we'll calculate \( f(x) \) for values in the relevant range (specifically \( x = 1, 2, 3 \)).
1. **For \( x = 1 \)**:
\[
f(1) = 8 - 2(1) = 8 - 2 = 6.
\]
2. **For \( x = 2 \)**:
\[
f(2) = 8 - 2(2) = 8 - 4 = 4.
\]
3. **For \( x = 3 \)**:
\[
f(3) = 8 - 2(3) = 8 - 6 = 2.
\]
Now, we can summarize the values we have computed:
- \( f(1) = 6 \)
- \( f(2) = 4 \)
- \( f(3) = 2 \)
Next, we compare these results to the options provided:
1. **Table 1**:
- \( x: [-1, 0, 1] \)
- \( f(x): [4, 0.5, -3] \) (Not matching)
2. **Table 2**:
- \( x: [-1, 0, 1] \)
- \( f(x): [10, 8, 6] \) (Not matching)
3. **Table 3**:
- \( x: [1, 2, 3] \)
- \( f(x): [6, 4, 2] \) (Matches our calculations)
4. **Table 4**:
- \( x: [-1, 2, 3] \)
- \( f(x): [-3, -6.5, -10] \) (Not matching)
### Conclusion:
The correct table that represents the second piece \( f(x) = 8 - 2x \) is **Table 3**:
| \( x \) | \( f(x) \) |
|---------|-----------|
| 1 | 6 |
| 2 | 4 |
| 3 | 2 |
Answered by
GPT-4o mini
To determine which piecewise function corresponds to the given graph, we need to analyze the key features of the lines and their definitions according to your description:
1. **First line**:
- **Open circle at (0, 1)**: The function does not include the point at \( (0, 1) \) for \( x = 0\).
- **Goes up through (−4, 3)**: Indicates that there is a positive slope leading to \( (0, 1) \).
Since it goes upward from \((-4, 3)\) to \( (0, 1) \), we can determine the definition for this segment (where \( x < 0 \)).
To find the slope, we can use:
\[
\text{Slope} = \frac{1 - 3}{0 + 4} = \frac{-2}{4} = -0.5.
\]
The y-intercept can be derived from \( y = mx + b \): since the line is defined at \( (0, 1) \), we have:
\[
y = -0.5x + 1.
\]
2. **Second line**:
- **Closed circle at (0, -2)**: The function includes the point \( (0, -2) \).
- **Goes up through (2, 2)**: The line continues upward, and it has a positive slope.
Calculating the slope between \( (0, -2) \) and \( (2, 2)\):
\[
\text{Slope} = \frac{2 - (-2)}{2 - 0} = \frac{4}{2} = 2.
\]
Substituting for \( y \) when \( x = 0 \):
\[
y = 2x - 2.
\]
### Combining the Pieces:
Based on the analysis:
- For \( x < 0 \): \( f(x) = -0.5x + 1 \) (first line)
- For \( x \geq 0 \): \( f(x) = 2x - 2 \) (second line)
Thus, we represent this piecewise function as follows:
\[
f(x) =
\begin{cases}
-0.5x + 1 & \text{if } x < 0 \\
2x - 2 & \text{if } x \geq 0
\end{cases}
\]
### Checking the Options:
Now, let’s check which option matches our derived piecewise function:
1. **Option 1**:
\[
f(x) =
\begin{cases}
-0.5x + 1 & \text{if } x < 0 \\
2x - 2 & \text{if } x \geq 0
\end{cases}
\]
2. **Option 2**:
\[
f(x) =
\begin{cases}
-x + 1 & \text{if } x < 0 \\
0.5x - 2 & \text{if } x > 0
\end{cases}
\]
(Not matching)
3. **Option 3**:
\[
f(x) =
\begin{cases}
x - 1 & \text{if } x \leq 0 \\
2x - 2 & \text{if } x > 0
\end{cases}
\]
(Not matching)
4. **Option 4**:
\[
f(x) =
\begin{cases}
-0.5x - 1 & \text{if } x \leq 0 \\
2x - 2 & \text{if } x \geq 0
\end{cases}
\]
(Not matching)
### Conclusion:
The only matching option we have is **Option 1**, which correctly represents the piecewise function based on the graph provided. Therefore, the function is:
\[
f(x) =
\begin{cases}
-0.5x + 1 & \text{if } x < 0 \\
2x - 2 & \text{if } x \geq 0
\end{cases}
\]
1. **First line**:
- **Open circle at (0, 1)**: The function does not include the point at \( (0, 1) \) for \( x = 0\).
- **Goes up through (−4, 3)**: Indicates that there is a positive slope leading to \( (0, 1) \).
Since it goes upward from \((-4, 3)\) to \( (0, 1) \), we can determine the definition for this segment (where \( x < 0 \)).
To find the slope, we can use:
\[
\text{Slope} = \frac{1 - 3}{0 + 4} = \frac{-2}{4} = -0.5.
\]
The y-intercept can be derived from \( y = mx + b \): since the line is defined at \( (0, 1) \), we have:
\[
y = -0.5x + 1.
\]
2. **Second line**:
- **Closed circle at (0, -2)**: The function includes the point \( (0, -2) \).
- **Goes up through (2, 2)**: The line continues upward, and it has a positive slope.
Calculating the slope between \( (0, -2) \) and \( (2, 2)\):
\[
\text{Slope} = \frac{2 - (-2)}{2 - 0} = \frac{4}{2} = 2.
\]
Substituting for \( y \) when \( x = 0 \):
\[
y = 2x - 2.
\]
### Combining the Pieces:
Based on the analysis:
- For \( x < 0 \): \( f(x) = -0.5x + 1 \) (first line)
- For \( x \geq 0 \): \( f(x) = 2x - 2 \) (second line)
Thus, we represent this piecewise function as follows:
\[
f(x) =
\begin{cases}
-0.5x + 1 & \text{if } x < 0 \\
2x - 2 & \text{if } x \geq 0
\end{cases}
\]
### Checking the Options:
Now, let’s check which option matches our derived piecewise function:
1. **Option 1**:
\[
f(x) =
\begin{cases}
-0.5x + 1 & \text{if } x < 0 \\
2x - 2 & \text{if } x \geq 0
\end{cases}
\]
2. **Option 2**:
\[
f(x) =
\begin{cases}
-x + 1 & \text{if } x < 0 \\
0.5x - 2 & \text{if } x > 0
\end{cases}
\]
(Not matching)
3. **Option 3**:
\[
f(x) =
\begin{cases}
x - 1 & \text{if } x \leq 0 \\
2x - 2 & \text{if } x > 0
\end{cases}
\]
(Not matching)
4. **Option 4**:
\[
f(x) =
\begin{cases}
-0.5x - 1 & \text{if } x \leq 0 \\
2x - 2 & \text{if } x \geq 0
\end{cases}
\]
(Not matching)
### Conclusion:
The only matching option we have is **Option 1**, which correctly represents the piecewise function based on the graph provided. Therefore, the function is:
\[
f(x) =
\begin{cases}
-0.5x + 1 & \text{if } x < 0 \\
2x - 2 & \text{if } x \geq 0
\end{cases}
\]
Answered by
GPT-4o mini
In this situation, we are looking at the piecewise-defined function that describes the pricing of organic apples based on the number of pounds Lucy buys.
1. **Context of the Problem**:
- \( x \) represents the number of pounds of apples.
- The price structure is as follows:
- If Lucy buys **3 or fewer pounds**, the cost is \( \$1.50 \) per pound (i.e., \( 0 \leq x \leq 3 \)).
- If she buys **more than 3 pounds**, the cost is \( \$1.10 \) per pound (i.e., \( x > 3 \)).
2. **Determining the Domain**:
- Since \( x \) represents the number of pounds of apples, it cannot be negative. Therefore, \( x \) must be non-negative (\( x \geq 0\)).
- Additionally, there is no upper limit to the number of pounds Lucy can buy, meaning \( x \) can extend to any positive value.
Considering the above points, the complete domain includes all non-negative values of \( x\).
### Options Evaluation:
- **{x | x ≥ 0}**: This indicates all non-negative real numbers, which captures all possible values of \( x \) (0 pounds and more). **This is correct.**
- **{x | x is a real number}**: This includes all real numbers, both positive and negative. This is not appropriate since Lucy cannot buy a negative amount of apples. **Incorrect.**
- **{x | 0 ≤ x ≤ 3}**: This only describes the range for 3 or fewer pounds, not considering purchases greater than 3. **Incorrect.**
- **{x | x ≥ 3}**: This only describes purchases greater than 3 pounds, excluding the option of buying less than or equal to 3 pounds. **Incorrect.**
### Conclusion:
The correct representation of the domain of this piecewise-defined function is:
**{x | x ≥ 0}**.
1. **Context of the Problem**:
- \( x \) represents the number of pounds of apples.
- The price structure is as follows:
- If Lucy buys **3 or fewer pounds**, the cost is \( \$1.50 \) per pound (i.e., \( 0 \leq x \leq 3 \)).
- If she buys **more than 3 pounds**, the cost is \( \$1.10 \) per pound (i.e., \( x > 3 \)).
2. **Determining the Domain**:
- Since \( x \) represents the number of pounds of apples, it cannot be negative. Therefore, \( x \) must be non-negative (\( x \geq 0\)).
- Additionally, there is no upper limit to the number of pounds Lucy can buy, meaning \( x \) can extend to any positive value.
Considering the above points, the complete domain includes all non-negative values of \( x\).
### Options Evaluation:
- **{x | x ≥ 0}**: This indicates all non-negative real numbers, which captures all possible values of \( x \) (0 pounds and more). **This is correct.**
- **{x | x is a real number}**: This includes all real numbers, both positive and negative. This is not appropriate since Lucy cannot buy a negative amount of apples. **Incorrect.**
- **{x | 0 ≤ x ≤ 3}**: This only describes the range for 3 or fewer pounds, not considering purchases greater than 3. **Incorrect.**
- **{x | x ≥ 3}**: This only describes purchases greater than 3 pounds, excluding the option of buying less than or equal to 3 pounds. **Incorrect.**
### Conclusion:
The correct representation of the domain of this piecewise-defined function is:
**{x | x ≥ 0}**.
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