Answer :
To determine the domain and range of the function [tex]\( f(x) = \sqrt{-x + 3} \)[/tex], we need to consider the properties of the square root function and the conditions under which the expression inside the square root is defined.
### Finding the Domain
The expression inside the square root, [tex]\(-x + 3\)[/tex], must be non-negative because the square root of a negative number is not defined in the set of real numbers. Therefore, we set up the inequality:
[tex]\[ -x + 3 \geq 0 \][/tex]
Solving this inequality for [tex]\(x\)[/tex]:
1. Subtract 3 from both sides:
[tex]\[ -x \geq -3 \][/tex]
2. Multiply both sides by -1 (and reverse the inequality sign):
[tex]\[ x \leq 3 \][/tex]
This means that [tex]\(x\)[/tex] can take any value less than or equal to 3. In interval notation, the domain is:
[tex]\[ x \in (-\infty, 3] \][/tex]
### Finding the Range
Next, we determine the range of the function [tex]\( f(x) \)[/tex]. Since [tex]\( f(x) \)[/tex] involves a square root, [tex]\( \sqrt{-x + 3} \)[/tex], the output (or range) consists of all non-negative values because the square root function always produces non-negative results.
The smallest value inside the square root occurs when [tex]\(-x + 3 = 0\)[/tex], which happens when [tex]\(x = 3\)[/tex]:
[tex]\[ f(3) = \sqrt{0} = 0 \][/tex]
When [tex]\( x < 3 \)[/tex], [tex]\(-x + 3\)[/tex] is positive, and as [tex]\( x \)[/tex] decreases (moving to the left of 3), the value [tex]\(-x + 3\)[/tex] increases, making the square root of it also increase.
Thus, as [tex]\( x \)[/tex] ranges from [tex]\(-\infty\)[/tex] to 3, [tex]\( f(x) \)[/tex] ranges from 0 to [tex]\(\infty\)[/tex]. This gives us the range:
[tex]\[ y \in [0, \infty) \][/tex]
To summarize:
- Domain: [tex]\( x \in (-\infty, 3] \)[/tex]
- Range: [tex]\( y \in [0, \infty) \)[/tex]
These intervals capture all the possible input values [tex]\(x\)[/tex] and the corresponding output values [tex]\(y\)[/tex] for the function [tex]\( f(x) = \sqrt{-x + 3} \)[/tex].
### Finding the Domain
The expression inside the square root, [tex]\(-x + 3\)[/tex], must be non-negative because the square root of a negative number is not defined in the set of real numbers. Therefore, we set up the inequality:
[tex]\[ -x + 3 \geq 0 \][/tex]
Solving this inequality for [tex]\(x\)[/tex]:
1. Subtract 3 from both sides:
[tex]\[ -x \geq -3 \][/tex]
2. Multiply both sides by -1 (and reverse the inequality sign):
[tex]\[ x \leq 3 \][/tex]
This means that [tex]\(x\)[/tex] can take any value less than or equal to 3. In interval notation, the domain is:
[tex]\[ x \in (-\infty, 3] \][/tex]
### Finding the Range
Next, we determine the range of the function [tex]\( f(x) \)[/tex]. Since [tex]\( f(x) \)[/tex] involves a square root, [tex]\( \sqrt{-x + 3} \)[/tex], the output (or range) consists of all non-negative values because the square root function always produces non-negative results.
The smallest value inside the square root occurs when [tex]\(-x + 3 = 0\)[/tex], which happens when [tex]\(x = 3\)[/tex]:
[tex]\[ f(3) = \sqrt{0} = 0 \][/tex]
When [tex]\( x < 3 \)[/tex], [tex]\(-x + 3\)[/tex] is positive, and as [tex]\( x \)[/tex] decreases (moving to the left of 3), the value [tex]\(-x + 3\)[/tex] increases, making the square root of it also increase.
Thus, as [tex]\( x \)[/tex] ranges from [tex]\(-\infty\)[/tex] to 3, [tex]\( f(x) \)[/tex] ranges from 0 to [tex]\(\infty\)[/tex]. This gives us the range:
[tex]\[ y \in [0, \infty) \][/tex]
To summarize:
- Domain: [tex]\( x \in (-\infty, 3] \)[/tex]
- Range: [tex]\( y \in [0, \infty) \)[/tex]
These intervals capture all the possible input values [tex]\(x\)[/tex] and the corresponding output values [tex]\(y\)[/tex] for the function [tex]\( f(x) = \sqrt{-x + 3} \)[/tex].
