Answer :
To determine which of the given functions has a range of [tex]\((-\infty, a]\)[/tex] and a domain of [tex]\([b, \infty)\)[/tex], we will analyze each function one by one.
Option A: [tex]\( f(x) = \sqrt[3]{x + b} - a \)[/tex]
1. Domain: Since the cube root function [tex]\(\sqrt[3]{x + b}\)[/tex] is defined for all real numbers, the domain of this function is [tex]\(\mathbb{R}\)[/tex], which doesn't match [tex]\([b, \infty)\)[/tex].
2. Range: The cube root function [tex]\(\sqrt[3]{x + b}\)[/tex] can take any real value, so [tex]\( f(x) \)[/tex] can take any real value shifted by [tex]\( -a \)[/tex]. Thus, the range is [tex]\(\mathbb{R}\)[/tex], which doesn't match [tex]\((-\infty, a]\)[/tex].
Option B: [tex]\( f(x) = -\sqrt{x - b} + a \)[/tex]
1. Domain: The square root function [tex]\(\sqrt{x - b}\)[/tex] is defined when [tex]\( x - b \geq 0 \)[/tex] or [tex]\( x \geq b \)[/tex]. Therefore, the domain is [tex]\([b, \infty)\)[/tex], which matches the given domain.
2. Range: The square root function [tex]\(\sqrt{x - b}\)[/tex] produces non-negative values, so [tex]\(-\sqrt{x - b}\)[/tex] produces non-positive values (ranging from 0 to [tex]\(-\infty\)[/tex]). When we add [tex]\(a\)[/tex] to [tex]\(-\sqrt{x - b}\)[/tex], the maximum value is [tex]\(a\)[/tex] and it can go down to [tex]\(-\infty\)[/tex], giving a range of [tex]\((-\infty, a]\)[/tex], which matches the given range.
Option C: [tex]\( f(x) = -\sqrt[3]{x + a} - b \)[/tex]
1. Domain: The cube root function [tex]\(\sqrt[3]{x + a}\)[/tex] is defined for all real numbers, so the domain of this function is [tex]\(\mathbb{R}\)[/tex], which doesn't match [tex]\([b, \infty)\)[/tex].
2. Range: The cube root function [tex]\(\sqrt[3]{x + a}\)[/tex] can take any real value, so [tex]\( f(x)\)[/tex] can take any real value shifted by [tex]\(-b\)[/tex]. Thus, the range is [tex]\(\mathbb{R}\)[/tex], which doesn't match [tex]\((-\infty, a]\)[/tex].
Option D: [tex]\( f(x) = \sqrt{x - a} + b \)[/tex]
1. Domain: The square root function [tex]\(\sqrt{x - a}\)[/tex] is defined when [tex]\(x - a \geq 0\)[/tex] or [tex]\(x \geq a\)[/tex]. Therefore, the domain is [tex]\([a, \infty)\)[/tex], which doesn't match [tex]\([b, \infty)\)[/tex].
2. Range: The square root function [tex]\(\sqrt{x - a}\)[/tex] produces non-negative values, so [tex]\( \sqrt{x - a} + b \)[/tex] produces values starting from [tex]\(b\)[/tex] to [tex]\(\infty\)[/tex], which doesn't match [tex]\((-\infty, a]\)[/tex].
Conclusion:
Among all the given options, the function [tex]\( f(x) = -\sqrt{x - b} + a \)[/tex] in Option B has the correct domain [tex]\([b, \infty)\)[/tex] and the correct range [tex]\((-\infty, a]\)[/tex].
Therefore, the correct answer is:
B. [tex]\( f(x) = -\sqrt{x - b} + a \)[/tex]
Option A: [tex]\( f(x) = \sqrt[3]{x + b} - a \)[/tex]
1. Domain: Since the cube root function [tex]\(\sqrt[3]{x + b}\)[/tex] is defined for all real numbers, the domain of this function is [tex]\(\mathbb{R}\)[/tex], which doesn't match [tex]\([b, \infty)\)[/tex].
2. Range: The cube root function [tex]\(\sqrt[3]{x + b}\)[/tex] can take any real value, so [tex]\( f(x) \)[/tex] can take any real value shifted by [tex]\( -a \)[/tex]. Thus, the range is [tex]\(\mathbb{R}\)[/tex], which doesn't match [tex]\((-\infty, a]\)[/tex].
Option B: [tex]\( f(x) = -\sqrt{x - b} + a \)[/tex]
1. Domain: The square root function [tex]\(\sqrt{x - b}\)[/tex] is defined when [tex]\( x - b \geq 0 \)[/tex] or [tex]\( x \geq b \)[/tex]. Therefore, the domain is [tex]\([b, \infty)\)[/tex], which matches the given domain.
2. Range: The square root function [tex]\(\sqrt{x - b}\)[/tex] produces non-negative values, so [tex]\(-\sqrt{x - b}\)[/tex] produces non-positive values (ranging from 0 to [tex]\(-\infty\)[/tex]). When we add [tex]\(a\)[/tex] to [tex]\(-\sqrt{x - b}\)[/tex], the maximum value is [tex]\(a\)[/tex] and it can go down to [tex]\(-\infty\)[/tex], giving a range of [tex]\((-\infty, a]\)[/tex], which matches the given range.
Option C: [tex]\( f(x) = -\sqrt[3]{x + a} - b \)[/tex]
1. Domain: The cube root function [tex]\(\sqrt[3]{x + a}\)[/tex] is defined for all real numbers, so the domain of this function is [tex]\(\mathbb{R}\)[/tex], which doesn't match [tex]\([b, \infty)\)[/tex].
2. Range: The cube root function [tex]\(\sqrt[3]{x + a}\)[/tex] can take any real value, so [tex]\( f(x)\)[/tex] can take any real value shifted by [tex]\(-b\)[/tex]. Thus, the range is [tex]\(\mathbb{R}\)[/tex], which doesn't match [tex]\((-\infty, a]\)[/tex].
Option D: [tex]\( f(x) = \sqrt{x - a} + b \)[/tex]
1. Domain: The square root function [tex]\(\sqrt{x - a}\)[/tex] is defined when [tex]\(x - a \geq 0\)[/tex] or [tex]\(x \geq a\)[/tex]. Therefore, the domain is [tex]\([a, \infty)\)[/tex], which doesn't match [tex]\([b, \infty)\)[/tex].
2. Range: The square root function [tex]\(\sqrt{x - a}\)[/tex] produces non-negative values, so [tex]\( \sqrt{x - a} + b \)[/tex] produces values starting from [tex]\(b\)[/tex] to [tex]\(\infty\)[/tex], which doesn't match [tex]\((-\infty, a]\)[/tex].
Conclusion:
Among all the given options, the function [tex]\( f(x) = -\sqrt{x - b} + a \)[/tex] in Option B has the correct domain [tex]\([b, \infty)\)[/tex] and the correct range [tex]\((-\infty, a]\)[/tex].
Therefore, the correct answer is:
B. [tex]\( f(x) = -\sqrt{x - b} + a \)[/tex]
