Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Sure, let's analyze the function [tex]\( y = \log_{1/2}(x + 9) + 4 \)[/tex]:
### Domain
For the logarithmic function [tex]\( \log_{1/2}(x + 9) \)[/tex] to be defined, the argument [tex]\( x + 9 \)[/tex] must be greater than 0:
[tex]\[ x + 9 > 0 \][/tex]
[tex]\[ x > -9 \][/tex]
Thus, the domain of the function is:
[tex]\[ (-9, \infty) \][/tex]
### Range
The range of a logarithmic function with any base is all real numbers because a logarithmic function can produce any real number output. Therefore:
[tex]\[ (-\infty, \infty) \][/tex]
### Intervals of Increase/Decrease
Since the base of the logarithm is [tex]\( \frac{1}{2} \)[/tex], which is less than 1, the function [tex]\( \log_{1/2}(x + 9) \)[/tex] is a decreasing function. Adding 4 does not change the nature of increase or decrease; it just shifts the entire function up by 4 units. Thus, the function decreases over its entire domain:
- There are no intervals where the function is increasing.
- The function is decreasing on the entire domain:
[tex]\[ (-9, \infty) \][/tex]
### Asymptote(s)
The vertical asymptote occurs where the argument of the logarithm approaches zero (i.e., [tex]\( x + 9 = 0 \)[/tex]):
[tex]\[ x = -9 \][/tex]
Thus, the vertical asymptote of the function is at:
[tex]\[ x = -9 \][/tex]
### Summary
- Domain: [tex]\( (-9, \infty) \)[/tex]
- Range: [tex]\( (-\infty, \infty) \)[/tex]
- Intervals of increase: None
- Intervals of decrease: [tex]\( (-9, \infty) \)[/tex]
- Asymptote(s): [tex]\( x = -9 \)[/tex]
### Domain
For the logarithmic function [tex]\( \log_{1/2}(x + 9) \)[/tex] to be defined, the argument [tex]\( x + 9 \)[/tex] must be greater than 0:
[tex]\[ x + 9 > 0 \][/tex]
[tex]\[ x > -9 \][/tex]
Thus, the domain of the function is:
[tex]\[ (-9, \infty) \][/tex]
### Range
The range of a logarithmic function with any base is all real numbers because a logarithmic function can produce any real number output. Therefore:
[tex]\[ (-\infty, \infty) \][/tex]
### Intervals of Increase/Decrease
Since the base of the logarithm is [tex]\( \frac{1}{2} \)[/tex], which is less than 1, the function [tex]\( \log_{1/2}(x + 9) \)[/tex] is a decreasing function. Adding 4 does not change the nature of increase or decrease; it just shifts the entire function up by 4 units. Thus, the function decreases over its entire domain:
- There are no intervals where the function is increasing.
- The function is decreasing on the entire domain:
[tex]\[ (-9, \infty) \][/tex]
### Asymptote(s)
The vertical asymptote occurs where the argument of the logarithm approaches zero (i.e., [tex]\( x + 9 = 0 \)[/tex]):
[tex]\[ x = -9 \][/tex]
Thus, the vertical asymptote of the function is at:
[tex]\[ x = -9 \][/tex]
### Summary
- Domain: [tex]\( (-9, \infty) \)[/tex]
- Range: [tex]\( (-\infty, \infty) \)[/tex]
- Intervals of increase: None
- Intervals of decrease: [tex]\( (-9, \infty) \)[/tex]
- Asymptote(s): [tex]\( x = -9 \)[/tex]
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.