Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A 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]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
