Discover new knowledge and insights with IDNLearn.com's extensive Q&A platform. Ask your questions and receive comprehensive, trustworthy responses from our dedicated team of experts.
Sagot :
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].
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Find clear and concise answers at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.