Answer :
To determine the domain and range of the region [tex]\( R = \{(x, y): y \leq x+2, y \geq 2-x, \text{ and } y \geq 0\} \)[/tex], we need to analyze the given inequalities step by step and understand how they intersect.
### Step 1: Analyzing the Inequalities
We are given three inequalities:
1. [tex]\( y \leq x + 2 \)[/tex]
2. [tex]\( y \geq 2 - x \)[/tex]
3. [tex]\( y \geq 0 \)[/tex]
### Step 2: Finding the Intersections
Let's find the points of intersection of these lines, as these will form the vertices of the region [tex]\( R \)[/tex].
#### Intersection of [tex]\( y = x + 2 \)[/tex] and [tex]\( y = 2 - x \)[/tex]:
Set [tex]\( x + 2 = 2 - x \)[/tex]:
[tex]\[ x + 2 = 2 - x \][/tex]
[tex]\[ 2x = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
So, substituting [tex]\( x = 0 \)[/tex] back into either equation gives:
[tex]\[ y = x + 2 = 0 + 2 = 2 \][/tex]
Therefore, one point of intersection is [tex]\((0, 2)\)[/tex].
#### Intersection of [tex]\( y = x + 2 \)[/tex] and [tex]\( y = 0 \)[/tex]:
Set [tex]\( x + 2 = 0 \)[/tex]:
[tex]\[ y = x + 2 \][/tex]
[tex]\[ 0 = x + 2 \][/tex]
[tex]\[ x = -2 \][/tex]
So, substituting [tex]\( x = -2 \)[/tex] back into the equation gives:
[tex]\[ y = 0 \][/tex]
Therefore, another point of intersection is [tex]\((-2, 0)\)[/tex].
#### Intersection of [tex]\( y = 2 - x \)[/tex] and [tex]\( y = 0 \)[/tex]:
Set [tex]\( 2 - x = 0 \)[/tex]:
[tex]\[ y = 2 - x \][/tex]
[tex]\[ 0 = 2 - x \][/tex]
[tex]\[ x = 2 \][/tex]
So, substituting [tex]\( x = 2 \)[/tex] back into the equation gives:
[tex]\[ y = 0 \][/tex]
Therefore, the third point of intersection is [tex]\((2, 0)\)[/tex].
### Step 3: Determining the Domain and Range
From the points of intersection, we know the vertices of the triangular region defined by these inequalities are [tex]\((0, 2)\)[/tex], [tex]\((-2, 0)\)[/tex], and [tex]\((2, 0)\)[/tex].
Domain:
The domain is all [tex]\( x \)[/tex]-values that are part of the region [tex]\( R \)[/tex].
Examining the [tex]\( x \)[/tex]-coordinates of the vertices:
- Lowest [tex]\( x \)[/tex]-value is [tex]\(-2\)[/tex]
- Highest [tex]\( x \)[/tex]-value is [tex]\(2\)[/tex]
So, the domain is:
[tex]\[ [-2, 2] \][/tex]
Range:
The range is all [tex]\( y \)[/tex]-values that are part of the region [tex]\( R \)[/tex].
Examining the [tex]\( y \)[/tex]-coordinates of the vertices and how [tex]\( y \)[/tex] varies within the region:
- Lowest [tex]\( y \)[/tex]-value is [tex]\(0\)[/tex]
- Highest [tex]\( y \)[/tex]-value is [tex]\(2\)[/tex]
So, the range is:
[tex]\[ [0, 2] \][/tex]
### Answer
The domain and range of the region [tex]\( R \)[/tex] are:
- Domain: [tex]\([-2, 2]\)[/tex]
- Range: [tex]\([0, 2]\)[/tex]
### Step 1: Analyzing the Inequalities
We are given three inequalities:
1. [tex]\( y \leq x + 2 \)[/tex]
2. [tex]\( y \geq 2 - x \)[/tex]
3. [tex]\( y \geq 0 \)[/tex]
### Step 2: Finding the Intersections
Let's find the points of intersection of these lines, as these will form the vertices of the region [tex]\( R \)[/tex].
#### Intersection of [tex]\( y = x + 2 \)[/tex] and [tex]\( y = 2 - x \)[/tex]:
Set [tex]\( x + 2 = 2 - x \)[/tex]:
[tex]\[ x + 2 = 2 - x \][/tex]
[tex]\[ 2x = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
So, substituting [tex]\( x = 0 \)[/tex] back into either equation gives:
[tex]\[ y = x + 2 = 0 + 2 = 2 \][/tex]
Therefore, one point of intersection is [tex]\((0, 2)\)[/tex].
#### Intersection of [tex]\( y = x + 2 \)[/tex] and [tex]\( y = 0 \)[/tex]:
Set [tex]\( x + 2 = 0 \)[/tex]:
[tex]\[ y = x + 2 \][/tex]
[tex]\[ 0 = x + 2 \][/tex]
[tex]\[ x = -2 \][/tex]
So, substituting [tex]\( x = -2 \)[/tex] back into the equation gives:
[tex]\[ y = 0 \][/tex]
Therefore, another point of intersection is [tex]\((-2, 0)\)[/tex].
#### Intersection of [tex]\( y = 2 - x \)[/tex] and [tex]\( y = 0 \)[/tex]:
Set [tex]\( 2 - x = 0 \)[/tex]:
[tex]\[ y = 2 - x \][/tex]
[tex]\[ 0 = 2 - x \][/tex]
[tex]\[ x = 2 \][/tex]
So, substituting [tex]\( x = 2 \)[/tex] back into the equation gives:
[tex]\[ y = 0 \][/tex]
Therefore, the third point of intersection is [tex]\((2, 0)\)[/tex].
### Step 3: Determining the Domain and Range
From the points of intersection, we know the vertices of the triangular region defined by these inequalities are [tex]\((0, 2)\)[/tex], [tex]\((-2, 0)\)[/tex], and [tex]\((2, 0)\)[/tex].
Domain:
The domain is all [tex]\( x \)[/tex]-values that are part of the region [tex]\( R \)[/tex].
Examining the [tex]\( x \)[/tex]-coordinates of the vertices:
- Lowest [tex]\( x \)[/tex]-value is [tex]\(-2\)[/tex]
- Highest [tex]\( x \)[/tex]-value is [tex]\(2\)[/tex]
So, the domain is:
[tex]\[ [-2, 2] \][/tex]
Range:
The range is all [tex]\( y \)[/tex]-values that are part of the region [tex]\( R \)[/tex].
Examining the [tex]\( y \)[/tex]-coordinates of the vertices and how [tex]\( y \)[/tex] varies within the region:
- Lowest [tex]\( y \)[/tex]-value is [tex]\(0\)[/tex]
- Highest [tex]\( y \)[/tex]-value is [tex]\(2\)[/tex]
So, the range is:
[tex]\[ [0, 2] \][/tex]
### Answer
The domain and range of the region [tex]\( R \)[/tex] are:
- Domain: [tex]\([-2, 2]\)[/tex]
- Range: [tex]\([0, 2]\)[/tex]
