At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
To determine which function has the desired properties—a range of [tex]\((- \infty, a]\)[/tex] and a domain of [tex]\([b, \infty)\)[/tex], where [tex]\(a>0\)[/tex] and [tex]\(b>0\)[/tex]—let's analyze each option step by step.
### Option A: [tex]\( f(x) = -\sqrt{x - b} + a \)[/tex]
Domain:
- The expression inside the square root, [tex]\(x - b\)[/tex], must be non-negative.
- Therefore, [tex]\( x - b \geq 0 \Rightarrow x \geq b \)[/tex].
- The domain of [tex]\(f(x) = -\sqrt{x - b} + a\)[/tex] is [tex]\([b, \infty)\)[/tex].
Range:
- The square root function, [tex]\(\sqrt{x - b}\)[/tex], produces non-negative values.
- Hence, [tex]\(-\sqrt{x - b}\)[/tex] produces non-positive values (i.e., values [tex]\(\leq 0\)[/tex]).
- Adding [tex]\(a\)[/tex] to these non-positive values gives [tex]\(-\sqrt{x - b} + a \leq a\)[/tex].
- Therefore, the range of this function is [tex]\((-\infty, a]\)[/tex].
This function meets the criteria of having a domain [tex]\([b, \infty)\)[/tex] and a range [tex]\((-\infty, a]\)[/tex].
### Option B: [tex]\( f(x) = \sqrt[3]{x + b} - a \)[/tex]
Domain:
- The cube root function [tex]\(\sqrt[3]{x + b}\)[/tex] is defined for all real numbers.
- Therefore, the domain is [tex]\((-\infty, \infty)\)[/tex].
Range:
- The cube root function [tex]\(\sqrt[3]{x + b}\)[/tex] can take any real number.
- Subtracting [tex]\(a\)[/tex] shifts the entire range by [tex]\(a\)[/tex], meaning the range is still [tex]\((-\infty, \infty)\)[/tex].
This function does not meet the criteria.
### Option C: [tex]\( f(x) = \sqrt{x - a} + b \)[/tex]
Domain:
- The expression inside the square root, [tex]\(x - a\)[/tex], must be non-negative.
- Therefore, [tex]\(x - a \geq 0 \Rightarrow x \geq a \)[/tex].
- The domain of [tex]\(f(x) = \sqrt{x - a} + b\)[/tex] is [tex]\([a, \infty)\)[/tex].
Range:
- The square root function, [tex]\(\sqrt{x - a}\)[/tex], produces non-negative values.
- Adding [tex]\(b\)[/tex] to non-negative values produces values [tex]\(\geq b\)[/tex].
- Therefore, the range of this function is [tex]\([b, \infty)\)[/tex].
This function does not meet the criteria.
### Option D: [tex]\( f(x) = -\sqrt[3]{x + a} - b \)[/tex]
Domain:
- The cube root function [tex]\(\sqrt[3]{x + a}\)[/tex] is defined for all real numbers.
- Therefore, the domain is [tex]\((-\infty, \infty)\)[/tex].
Range:
- The cube root function [tex]\(\sqrt[3]{x + a}\)[/tex] can take any real number.
- Negating and subtracting [tex]\(b\)[/tex] still results in a range of [tex]\((-\infty, \infty)\)[/tex].
This function does not meet the criteria.
### Conclusion
The function that fits the given domain [tex]\([b, \infty)\)[/tex] and range [tex]\((-\infty, a]\)[/tex] is:
Option A: [tex]\(f(x) = -\sqrt{x - b} + a\)[/tex]
### Option A: [tex]\( f(x) = -\sqrt{x - b} + a \)[/tex]
Domain:
- The expression inside the square root, [tex]\(x - b\)[/tex], must be non-negative.
- Therefore, [tex]\( x - b \geq 0 \Rightarrow x \geq b \)[/tex].
- The domain of [tex]\(f(x) = -\sqrt{x - b} + a\)[/tex] is [tex]\([b, \infty)\)[/tex].
Range:
- The square root function, [tex]\(\sqrt{x - b}\)[/tex], produces non-negative values.
- Hence, [tex]\(-\sqrt{x - b}\)[/tex] produces non-positive values (i.e., values [tex]\(\leq 0\)[/tex]).
- Adding [tex]\(a\)[/tex] to these non-positive values gives [tex]\(-\sqrt{x - b} + a \leq a\)[/tex].
- Therefore, the range of this function is [tex]\((-\infty, a]\)[/tex].
This function meets the criteria of having a domain [tex]\([b, \infty)\)[/tex] and a range [tex]\((-\infty, a]\)[/tex].
### Option B: [tex]\( f(x) = \sqrt[3]{x + b} - a \)[/tex]
Domain:
- The cube root function [tex]\(\sqrt[3]{x + b}\)[/tex] is defined for all real numbers.
- Therefore, the domain is [tex]\((-\infty, \infty)\)[/tex].
Range:
- The cube root function [tex]\(\sqrt[3]{x + b}\)[/tex] can take any real number.
- Subtracting [tex]\(a\)[/tex] shifts the entire range by [tex]\(a\)[/tex], meaning the range is still [tex]\((-\infty, \infty)\)[/tex].
This function does not meet the criteria.
### Option C: [tex]\( f(x) = \sqrt{x - a} + b \)[/tex]
Domain:
- The expression inside the square root, [tex]\(x - a\)[/tex], must be non-negative.
- Therefore, [tex]\(x - a \geq 0 \Rightarrow x \geq a \)[/tex].
- The domain of [tex]\(f(x) = \sqrt{x - a} + b\)[/tex] is [tex]\([a, \infty)\)[/tex].
Range:
- The square root function, [tex]\(\sqrt{x - a}\)[/tex], produces non-negative values.
- Adding [tex]\(b\)[/tex] to non-negative values produces values [tex]\(\geq b\)[/tex].
- Therefore, the range of this function is [tex]\([b, \infty)\)[/tex].
This function does not meet the criteria.
### Option D: [tex]\( f(x) = -\sqrt[3]{x + a} - b \)[/tex]
Domain:
- The cube root function [tex]\(\sqrt[3]{x + a}\)[/tex] is defined for all real numbers.
- Therefore, the domain is [tex]\((-\infty, \infty)\)[/tex].
Range:
- The cube root function [tex]\(\sqrt[3]{x + a}\)[/tex] can take any real number.
- Negating and subtracting [tex]\(b\)[/tex] still results in a range of [tex]\((-\infty, \infty)\)[/tex].
This function does not meet the criteria.
### Conclusion
The function that fits the given domain [tex]\([b, \infty)\)[/tex] and range [tex]\((-\infty, a]\)[/tex] is:
Option A: [tex]\(f(x) = -\sqrt{x - b} + a\)[/tex]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.