IDNLearn.com is the place where your questions are met with thoughtful and precise answers. Join our knowledgeable community and get detailed, reliable answers to all your questions.
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
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your questions deserve reliable answers. Thanks for visiting IDNLearn.com, and see you again soon for more helpful information.
