IDNLearn.com connects you with experts who provide accurate and reliable answers. Our community is ready to provide in-depth answers and practical solutions to any questions you may have.
Sagot :
To determine which function is increasing and has a domain of [tex]\((1, \infty)\)[/tex], we need to analyze the properties of each of the given functions.
Step 1: Determine the domain of each function
1. [tex]\( f(x) = \log (x-1) + 2 \)[/tex]
- The expression [tex]\(\log(x-1)\)[/tex] is defined when [tex]\(x-1 > 0\)[/tex], thus [tex]\(x > 1\)[/tex].
- Therefore, the domain is [tex]\((1, \infty)\)[/tex].
2. [tex]\( f(x) = -\log (x-1) + 2 \)[/tex]
- Similar to the first function, [tex]\(\log(x-1)\)[/tex] is defined when [tex]\(x-1 > 0\)[/tex], thus [tex]\(x > 1\)[/tex].
- Therefore, the domain is [tex]\((1, \infty)\)[/tex].
3. [tex]\( f(x) = \log (x-2) + 1 \)[/tex]
- The expression [tex]\(\log(x-2)\)[/tex] is defined when [tex]\(x-2 > 0\)[/tex], thus [tex]\(x > 2\)[/tex].
- Therefore, the domain is [tex]\((2, \infty)\)[/tex].
4. [tex]\( f(x) = -\log (x-2) + 1 \)[/tex]
- Similar to the third function, [tex]\(\log(x-2)\)[/tex] is defined when [tex]\(x-2 > 0\)[/tex], thus [tex]\(x > 2\)[/tex].
- Therefore, the domain is [tex]\((2, \infty)\)[/tex].
Step 2: Determine which function is increasing
To see if a function is increasing, we can look at the derivative or the behavior of the function itself:
1. [tex]\( f(x) = \log (x-1) + 2 \)[/tex]
- The derivative:
[tex]\[ f'(x) = \frac{d}{dx} [\log (x-1) + 2] = \frac{1}{x-1} \][/tex]
- For [tex]\(x > 1\)[/tex], [tex]\(\frac{1}{x-1} > 0\)[/tex], meaning [tex]\( f(x) \)[/tex] is increasing on [tex]\((1, \infty)\)[/tex].
2. [tex]\( f(x) = -\log (x-1) + 2 \)[/tex]
- The derivative:
[tex]\[ f'(x) = \frac{d}{dx} [-\log (x-1) + 2] = -\frac{1}{x-1} \][/tex]
- For [tex]\(x > 1\)[/tex], [tex]\(-\frac{1}{x-1} < 0\)[/tex], meaning [tex]\( f(x) \)[/tex] is decreasing on [tex]\((1, \infty)\)[/tex].
3. [tex]\( f(x) = \log (x-2) + 1 \)[/tex]
- The domain of this function is [tex]\((2, \infty)\)[/tex], so it is not defined on [tex]\((1, \infty)\)[/tex].
4. [tex]\( f(x) = -\log (x-2) + 1 \)[/tex]
- The domain of this function is [tex]\((2, \infty)\)[/tex], so it is not defined on [tex]\((1, \infty)\)[/tex].
Conclusion:
The function [tex]\( f(x) = \log (x-1) + 2 \)[/tex] is the only function that is increasing on [tex]\((1, \infty)\)[/tex] and has a domain of [tex]\((1, \infty)\)[/tex].
Therefore, the correct answer is:
A. [tex]\( f(x) = \log (x-1) + 2 \)[/tex]
Step 1: Determine the domain of each function
1. [tex]\( f(x) = \log (x-1) + 2 \)[/tex]
- The expression [tex]\(\log(x-1)\)[/tex] is defined when [tex]\(x-1 > 0\)[/tex], thus [tex]\(x > 1\)[/tex].
- Therefore, the domain is [tex]\((1, \infty)\)[/tex].
2. [tex]\( f(x) = -\log (x-1) + 2 \)[/tex]
- Similar to the first function, [tex]\(\log(x-1)\)[/tex] is defined when [tex]\(x-1 > 0\)[/tex], thus [tex]\(x > 1\)[/tex].
- Therefore, the domain is [tex]\((1, \infty)\)[/tex].
3. [tex]\( f(x) = \log (x-2) + 1 \)[/tex]
- The expression [tex]\(\log(x-2)\)[/tex] is defined when [tex]\(x-2 > 0\)[/tex], thus [tex]\(x > 2\)[/tex].
- Therefore, the domain is [tex]\((2, \infty)\)[/tex].
4. [tex]\( f(x) = -\log (x-2) + 1 \)[/tex]
- Similar to the third function, [tex]\(\log(x-2)\)[/tex] is defined when [tex]\(x-2 > 0\)[/tex], thus [tex]\(x > 2\)[/tex].
- Therefore, the domain is [tex]\((2, \infty)\)[/tex].
Step 2: Determine which function is increasing
To see if a function is increasing, we can look at the derivative or the behavior of the function itself:
1. [tex]\( f(x) = \log (x-1) + 2 \)[/tex]
- The derivative:
[tex]\[ f'(x) = \frac{d}{dx} [\log (x-1) + 2] = \frac{1}{x-1} \][/tex]
- For [tex]\(x > 1\)[/tex], [tex]\(\frac{1}{x-1} > 0\)[/tex], meaning [tex]\( f(x) \)[/tex] is increasing on [tex]\((1, \infty)\)[/tex].
2. [tex]\( f(x) = -\log (x-1) + 2 \)[/tex]
- The derivative:
[tex]\[ f'(x) = \frac{d}{dx} [-\log (x-1) + 2] = -\frac{1}{x-1} \][/tex]
- For [tex]\(x > 1\)[/tex], [tex]\(-\frac{1}{x-1} < 0\)[/tex], meaning [tex]\( f(x) \)[/tex] is decreasing on [tex]\((1, \infty)\)[/tex].
3. [tex]\( f(x) = \log (x-2) + 1 \)[/tex]
- The domain of this function is [tex]\((2, \infty)\)[/tex], so it is not defined on [tex]\((1, \infty)\)[/tex].
4. [tex]\( f(x) = -\log (x-2) + 1 \)[/tex]
- The domain of this function is [tex]\((2, \infty)\)[/tex], so it is not defined on [tex]\((1, \infty)\)[/tex].
Conclusion:
The function [tex]\( f(x) = \log (x-1) + 2 \)[/tex] is the only function that is increasing on [tex]\((1, \infty)\)[/tex] and has a domain of [tex]\((1, \infty)\)[/tex].
Therefore, the correct answer is:
A. [tex]\( f(x) = \log (x-1) + 2 \)[/tex]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For clear and precise answers, choose IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.
