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To determine which function has a range [tex]\((- \infty, a]\)[/tex] and a domain [tex]\([b, \infty)\)[/tex] where [tex]\(a > 0\)[/tex] and [tex]\(b > 0\)[/tex], let's analyze each option in detail:
Option A: [tex]\( f(x) = -\sqrt{x-b} + a \)[/tex]
1. Domain:
- The expression inside the square root must be non-negative for real numbers.
- Therefore, [tex]\(x - b \geq 0\)[/tex].
- Solving for [tex]\(x\)[/tex], we get [tex]\(x \geq b\)[/tex].
- Hence, the domain is [tex]\([b, \infty)\)[/tex].
2. Range:
- Since [tex]\(\sqrt{x-b}\)[/tex] is always non-negative, [tex]\(-\sqrt{x-b}\)[/tex] is always non-positive.
- Therefore, [tex]\(-\sqrt{x-b} \leq 0\)[/tex].
- Adding [tex]\(a\)[/tex] to the inequality, we get [tex]\(-\sqrt{x-b} + a \leq a\)[/tex].
- Hence, the range is [tex]\((-\infty, a]\)[/tex].
We see that this satisfies the conditions.
Option B: [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.
- Hence, the domain is [tex]\((-\infty, \infty)\)[/tex].
2. Range:
- The cube root function [tex]\(\sqrt[3]{x+a}\)[/tex] can take any real number.
- Therefore, [tex]\(-\sqrt[3]{x+a}\)[/tex] can also take any real number.
- Hence, [tex]\( -\sqrt[3]{x+a} - b \)[/tex] can take any real number as well.
- The range is [tex]\((-\infty, \infty)\)[/tex].
This does not satisfy the conditions.
Option C: [tex]\( f(x) = \sqrt[3]{x+b} - a \)[/tex]
1. Domain:
- The cube root function [tex]\(\sqrt[3]{x+b}\)[/tex] is defined for all real numbers.
- Hence, the domain is [tex]\((-\infty, \infty)\)[/tex].
2. Range:
- The cube root function [tex]\(\sqrt[3]{x+b}\)[/tex] can take any real number.
- Therefore, [tex]\(\sqrt[3]{x+b}\)[/tex] can also take any real number.
- Hence, [tex]\(\sqrt[3]{x+b} - a \)[/tex] can take any real number as well.
- The range is [tex]\((-\infty, \infty)\)[/tex].
This does not satisfy the conditions.
Option D: [tex]\( f(x) = \sqrt{x-a} + b \)[/tex]
1. Domain:
- The expression inside the square root must be non-negative for real numbers.
- Therefore, [tex]\(x - a \geq 0\)[/tex].
- Solving for [tex]\(x\)[/tex], we get [tex]\(x \geq a\)[/tex].
- Hence, the domain is [tex]\([a, \infty)\)[/tex].
2. Range:
- Since [tex]\(\sqrt{x-a}\)[/tex] is always non-negative, [tex]\(\sqrt{x-a}\)[/tex] is always non-negative.
- Therefore, [tex]\(\sqrt{x-a} \geq 0\)[/tex].
- Adding [tex]\(b\)[/tex] to the inequality, we get [tex]\(\sqrt{x-a} + b \geq b\)[/tex].
- Hence, the range is [tex]\([b, \infty)\)[/tex].
This does not satisfy the conditions.
After analyzing all options, we conclude that the correct function that has a range of [tex]\((-\infty, a]\)[/tex] and a domain of [tex]\([b, \infty)\)[/tex] is:
Option A: [tex]\( f(x) = -\sqrt{x-b} + a \)[/tex].
Thus, the correct answer is:
```
1
Option A: [tex]\( f(x) = -\sqrt{x-b} + a \)[/tex]
1. Domain:
- The expression inside the square root must be non-negative for real numbers.
- Therefore, [tex]\(x - b \geq 0\)[/tex].
- Solving for [tex]\(x\)[/tex], we get [tex]\(x \geq b\)[/tex].
- Hence, the domain is [tex]\([b, \infty)\)[/tex].
2. Range:
- Since [tex]\(\sqrt{x-b}\)[/tex] is always non-negative, [tex]\(-\sqrt{x-b}\)[/tex] is always non-positive.
- Therefore, [tex]\(-\sqrt{x-b} \leq 0\)[/tex].
- Adding [tex]\(a\)[/tex] to the inequality, we get [tex]\(-\sqrt{x-b} + a \leq a\)[/tex].
- Hence, the range is [tex]\((-\infty, a]\)[/tex].
We see that this satisfies the conditions.
Option B: [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.
- Hence, the domain is [tex]\((-\infty, \infty)\)[/tex].
2. Range:
- The cube root function [tex]\(\sqrt[3]{x+a}\)[/tex] can take any real number.
- Therefore, [tex]\(-\sqrt[3]{x+a}\)[/tex] can also take any real number.
- Hence, [tex]\( -\sqrt[3]{x+a} - b \)[/tex] can take any real number as well.
- The range is [tex]\((-\infty, \infty)\)[/tex].
This does not satisfy the conditions.
Option C: [tex]\( f(x) = \sqrt[3]{x+b} - a \)[/tex]
1. Domain:
- The cube root function [tex]\(\sqrt[3]{x+b}\)[/tex] is defined for all real numbers.
- Hence, the domain is [tex]\((-\infty, \infty)\)[/tex].
2. Range:
- The cube root function [tex]\(\sqrt[3]{x+b}\)[/tex] can take any real number.
- Therefore, [tex]\(\sqrt[3]{x+b}\)[/tex] can also take any real number.
- Hence, [tex]\(\sqrt[3]{x+b} - a \)[/tex] can take any real number as well.
- The range is [tex]\((-\infty, \infty)\)[/tex].
This does not satisfy the conditions.
Option D: [tex]\( f(x) = \sqrt{x-a} + b \)[/tex]
1. Domain:
- The expression inside the square root must be non-negative for real numbers.
- Therefore, [tex]\(x - a \geq 0\)[/tex].
- Solving for [tex]\(x\)[/tex], we get [tex]\(x \geq a\)[/tex].
- Hence, the domain is [tex]\([a, \infty)\)[/tex].
2. Range:
- Since [tex]\(\sqrt{x-a}\)[/tex] is always non-negative, [tex]\(\sqrt{x-a}\)[/tex] is always non-negative.
- Therefore, [tex]\(\sqrt{x-a} \geq 0\)[/tex].
- Adding [tex]\(b\)[/tex] to the inequality, we get [tex]\(\sqrt{x-a} + b \geq b\)[/tex].
- Hence, the range is [tex]\([b, \infty)\)[/tex].
This does not satisfy the conditions.
After analyzing all options, we conclude that the correct function that has a range of [tex]\((-\infty, a]\)[/tex] and a domain of [tex]\([b, \infty)\)[/tex] is:
Option A: [tex]\( f(x) = -\sqrt{x-b} + a \)[/tex].
Thus, the correct answer is:
```
1
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