Answer :
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]
