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Sagot :
Let's analyze the function [tex]\( y = \ln(x-2) + 3 \)[/tex] step by step to determine its domain, range, intervals of increase and decrease, and any asymptotes.
### Domain:
The natural logarithm function [tex]\( \ln(z) \)[/tex] is defined only for [tex]\( z > 0 \)[/tex]. Here, [tex]\( z = x-2 \)[/tex], so we need:
[tex]\[ x - 2 > 0 \][/tex]
[tex]\[ x > 2 \][/tex]
Thus, the domain of the function is:
[tex]\[ (2, \infty) \][/tex]
### Range:
The natural logarithm function [tex]\( \ln(z) \)[/tex] can take any real value from [tex]\(-\infty\)[/tex] to [tex]\(+\infty\)[/tex]. Since the function [tex]\( y = \ln(x-2) + 3 \)[/tex] merely shifts the entire graph of [tex]\( \ln(x-2) \)[/tex] upwards by 3 units, the range of [tex]\( y \)[/tex] will also span all real numbers. Therefore, the range is:
[tex]\[ (-\infty, \infty) \][/tex]
### Intervals of Increase:
To determine where the function is increasing or decreasing, we consider the derivative. The derivative of [tex]\( y = \ln(x-2) + 3 \)[/tex] is given by:
[tex]\[ y' = \frac{d}{dx}[\ln(x-2) + 3] = \frac{1}{x-2} \][/tex]
Since [tex]\( \frac{1}{x-2} > 0 \)[/tex] for all [tex]\( x > 2 \)[/tex], the function is always increasing for [tex]\( x > 2 \)[/tex]. Hence, the interval of increase is:
[tex]\[ (2, \infty) \][/tex]
### Intervals of Decrease:
Given that the function [tex]\( y = \ln(x-2) + 3 \)[/tex] is always increasing for [tex]\( x > 2 \)[/tex], there are no intervals of decrease.
### Asymptote(s):
Vertical asymptotes occur where the function approaches [tex]\(\pm \infty\)[/tex] as [tex]\( x \)[/tex] approaches a particular value. For the function [tex]\( y = \ln(x-2) + 3 \)[/tex], as [tex]\( x \)[/tex] approaches 2 from the right ([tex]\( x \to 2^+ \)[/tex]), [tex]\( \ln(x-2) \)[/tex] approaches [tex]\(-\infty \)[/tex]. Thus, the function has a vertical asymptote at:
[tex]\[ x = 2 \][/tex]
To summarize:
- Domain: [tex]\( (2, \infty) \)[/tex]
- Range: [tex]\( (-\infty, \infty) \)[/tex]
- Intervals of Increase: [tex]\( (2, \infty) \)[/tex]
- Intervals of Decrease: None
- Asymptote(s): Vertical asymptote at [tex]\( x = 2 \)[/tex]
These summaries align with the analysis done above.
### Domain:
The natural logarithm function [tex]\( \ln(z) \)[/tex] is defined only for [tex]\( z > 0 \)[/tex]. Here, [tex]\( z = x-2 \)[/tex], so we need:
[tex]\[ x - 2 > 0 \][/tex]
[tex]\[ x > 2 \][/tex]
Thus, the domain of the function is:
[tex]\[ (2, \infty) \][/tex]
### Range:
The natural logarithm function [tex]\( \ln(z) \)[/tex] can take any real value from [tex]\(-\infty\)[/tex] to [tex]\(+\infty\)[/tex]. Since the function [tex]\( y = \ln(x-2) + 3 \)[/tex] merely shifts the entire graph of [tex]\( \ln(x-2) \)[/tex] upwards by 3 units, the range of [tex]\( y \)[/tex] will also span all real numbers. Therefore, the range is:
[tex]\[ (-\infty, \infty) \][/tex]
### Intervals of Increase:
To determine where the function is increasing or decreasing, we consider the derivative. The derivative of [tex]\( y = \ln(x-2) + 3 \)[/tex] is given by:
[tex]\[ y' = \frac{d}{dx}[\ln(x-2) + 3] = \frac{1}{x-2} \][/tex]
Since [tex]\( \frac{1}{x-2} > 0 \)[/tex] for all [tex]\( x > 2 \)[/tex], the function is always increasing for [tex]\( x > 2 \)[/tex]. Hence, the interval of increase is:
[tex]\[ (2, \infty) \][/tex]
### Intervals of Decrease:
Given that the function [tex]\( y = \ln(x-2) + 3 \)[/tex] is always increasing for [tex]\( x > 2 \)[/tex], there are no intervals of decrease.
### Asymptote(s):
Vertical asymptotes occur where the function approaches [tex]\(\pm \infty\)[/tex] as [tex]\( x \)[/tex] approaches a particular value. For the function [tex]\( y = \ln(x-2) + 3 \)[/tex], as [tex]\( x \)[/tex] approaches 2 from the right ([tex]\( x \to 2^+ \)[/tex]), [tex]\( \ln(x-2) \)[/tex] approaches [tex]\(-\infty \)[/tex]. Thus, the function has a vertical asymptote at:
[tex]\[ x = 2 \][/tex]
To summarize:
- Domain: [tex]\( (2, \infty) \)[/tex]
- Range: [tex]\( (-\infty, \infty) \)[/tex]
- Intervals of Increase: [tex]\( (2, \infty) \)[/tex]
- Intervals of Decrease: None
- Asymptote(s): Vertical asymptote at [tex]\( x = 2 \)[/tex]
These summaries align with the analysis done above.
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