Answer :
Let's consider the given piecewise function:
[tex]\[ f(x) = \left\{ \begin{array}{ll} x + 1 & \text{if } x < -2 \\ 2x - 3 & \text{if } x \geq -2 \end{array} \right. \][/tex]
We need to check the validity of the following statements:
1. The domain is [tex]\((- \infty, -2]\)[/tex]
2. The function is linear for all [tex]\( x \)[/tex]
3. The function is continuous at [tex]\( x = -2 \)[/tex]
4. The range includes only positive numbers
### Step 1: Checking the Domain
The domain of [tex]\( f(x) \)[/tex] is the set of all possible values of [tex]\( x \)[/tex] for which the function is defined. Looking at the piecewise definition:
- [tex]\( x + 1 \)[/tex] is defined for [tex]\( x < -2 \)[/tex]
- [tex]\( 2x - 3 \)[/tex] is defined for [tex]\( x \geq -2 \)[/tex]
Thus, the function is defined for all real numbers [tex]\( x \)[/tex].
Therefore, the correct domain is [tex]\( (-\infty, \infty) \)[/tex], not [tex]\( (-\infty, -2] \)[/tex].
The statement "The domain is [tex]\( (-\infty, -2] \)[/tex]" is False.
### Step 2: Checking Linearity
Both pieces of the function [tex]\( x + 1 \)[/tex] and [tex]\( 2x - 3 \)[/tex] are linear functions. They are in the form [tex]\( ax + b \)[/tex], where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are constants.
Hence, the entire function [tex]\( f(x) \)[/tex] is composed of linear functions for all [tex]\( x \)[/tex].
The statement "The function is linear for all [tex]\( x \)[/tex]" is True.
### Step 3: Checking Continuity at [tex]\( x = -2 \)[/tex]
To check for continuity at [tex]\( x = -2 \)[/tex], we need to verify if the limit of the function from the left side equals the limit from the right side and equals the function value at [tex]\( x = -2 \)[/tex].
- Left-hand limit: As [tex]\( x \)[/tex] approaches [tex]\(-2\)[/tex] from the left, we use [tex]\( f(x) = x + 1 \)[/tex].
[tex]\[ \lim_{x \to -2^-} (x + 1) = -2 + 1 = -1 \][/tex]
- Right-hand limit: As [tex]\( x \)[/tex] approaches [tex]\(-2\)[/tex] from the right, we use [tex]\( f(x) = 2x - 3 \)[/tex].
[tex]\[ \lim_{x \to -2^+} (2x - 3) = 2(-2) - 3 = -4 - 3 = -7 \][/tex]
- Function value at [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 2(-2) - 3 = -4 - 3 = -7 \][/tex]
The left-hand limit and the right-hand limit are not equal, hence the function is not continuous at [tex]\( x = -2 \)[/tex].
The statement "The function is continuous at [tex]\( x = -2 \)[/tex]" is False.
### Step 4: Checking the Range
We need to find the range of [tex]\( f(x) \)[/tex].
- For [tex]\( x < -2 \)[/tex], [tex]\( f(x) = x + 1 \)[/tex], which takes on all values for [tex]\( x < -1 \)[/tex].
- For [tex]\( x \geq -2 \)[/tex], [tex]\( f(x) = 2x - 3 \)[/tex]:
When [tex]\( x = -2 \)[/tex], [tex]\( f(-2) = -7\)[/tex], and as [tex]\( x \)[/tex] increases, [tex]\( 2x - 3 \)[/tex] covers all values greater than [tex]\(-7\)[/tex].
Thus, the range of the function includes all numbers less than [tex]\(-1\)[/tex] and all numbers greater than or equal to [tex]\(-7\)[/tex]. Hence, the range includes negative numbers and positive numbers as well.
The statement "The range includes only positive numbers" is False.
### Conclusion
After checking each statement step-by-step:
1. False: The domain is not [tex]\( (-\infty, -2] \)[/tex]; it's [tex]\( (-\infty, \infty) \)[/tex].
2. True: The function is linear for all [tex]\( x \)[/tex].
3. False: The function is not continuous at [tex]\( x = -2 \)[/tex].
4. False: The range includes both negative and positive numbers, not just positive.
So, the correct output corresponds to the statements:
[tex]\[ \text{(False, True, False, False)} \][/tex]
[tex]\[ f(x) = \left\{ \begin{array}{ll} x + 1 & \text{if } x < -2 \\ 2x - 3 & \text{if } x \geq -2 \end{array} \right. \][/tex]
We need to check the validity of the following statements:
1. The domain is [tex]\((- \infty, -2]\)[/tex]
2. The function is linear for all [tex]\( x \)[/tex]
3. The function is continuous at [tex]\( x = -2 \)[/tex]
4. The range includes only positive numbers
### Step 1: Checking the Domain
The domain of [tex]\( f(x) \)[/tex] is the set of all possible values of [tex]\( x \)[/tex] for which the function is defined. Looking at the piecewise definition:
- [tex]\( x + 1 \)[/tex] is defined for [tex]\( x < -2 \)[/tex]
- [tex]\( 2x - 3 \)[/tex] is defined for [tex]\( x \geq -2 \)[/tex]
Thus, the function is defined for all real numbers [tex]\( x \)[/tex].
Therefore, the correct domain is [tex]\( (-\infty, \infty) \)[/tex], not [tex]\( (-\infty, -2] \)[/tex].
The statement "The domain is [tex]\( (-\infty, -2] \)[/tex]" is False.
### Step 2: Checking Linearity
Both pieces of the function [tex]\( x + 1 \)[/tex] and [tex]\( 2x - 3 \)[/tex] are linear functions. They are in the form [tex]\( ax + b \)[/tex], where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are constants.
Hence, the entire function [tex]\( f(x) \)[/tex] is composed of linear functions for all [tex]\( x \)[/tex].
The statement "The function is linear for all [tex]\( x \)[/tex]" is True.
### Step 3: Checking Continuity at [tex]\( x = -2 \)[/tex]
To check for continuity at [tex]\( x = -2 \)[/tex], we need to verify if the limit of the function from the left side equals the limit from the right side and equals the function value at [tex]\( x = -2 \)[/tex].
- Left-hand limit: As [tex]\( x \)[/tex] approaches [tex]\(-2\)[/tex] from the left, we use [tex]\( f(x) = x + 1 \)[/tex].
[tex]\[ \lim_{x \to -2^-} (x + 1) = -2 + 1 = -1 \][/tex]
- Right-hand limit: As [tex]\( x \)[/tex] approaches [tex]\(-2\)[/tex] from the right, we use [tex]\( f(x) = 2x - 3 \)[/tex].
[tex]\[ \lim_{x \to -2^+} (2x - 3) = 2(-2) - 3 = -4 - 3 = -7 \][/tex]
- Function value at [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 2(-2) - 3 = -4 - 3 = -7 \][/tex]
The left-hand limit and the right-hand limit are not equal, hence the function is not continuous at [tex]\( x = -2 \)[/tex].
The statement "The function is continuous at [tex]\( x = -2 \)[/tex]" is False.
### Step 4: Checking the Range
We need to find the range of [tex]\( f(x) \)[/tex].
- For [tex]\( x < -2 \)[/tex], [tex]\( f(x) = x + 1 \)[/tex], which takes on all values for [tex]\( x < -1 \)[/tex].
- For [tex]\( x \geq -2 \)[/tex], [tex]\( f(x) = 2x - 3 \)[/tex]:
When [tex]\( x = -2 \)[/tex], [tex]\( f(-2) = -7\)[/tex], and as [tex]\( x \)[/tex] increases, [tex]\( 2x - 3 \)[/tex] covers all values greater than [tex]\(-7\)[/tex].
Thus, the range of the function includes all numbers less than [tex]\(-1\)[/tex] and all numbers greater than or equal to [tex]\(-7\)[/tex]. Hence, the range includes negative numbers and positive numbers as well.
The statement "The range includes only positive numbers" is False.
### Conclusion
After checking each statement step-by-step:
1. False: The domain is not [tex]\( (-\infty, -2] \)[/tex]; it's [tex]\( (-\infty, \infty) \)[/tex].
2. True: The function is linear for all [tex]\( x \)[/tex].
3. False: The function is not continuous at [tex]\( x = -2 \)[/tex].
4. False: The range includes both negative and positive numbers, not just positive.
So, the correct output corresponds to the statements:
[tex]\[ \text{(False, True, False, False)} \][/tex]
