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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]
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