Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
