Connect with experts and get insightful answers on IDNLearn.com. Discover detailed and accurate answers to your questions from our knowledgeable and dedicated community members.
Sagot :
To determine the correct statement about the graph of the function [tex]\( f(x) = \log_2 x \)[/tex], let's analyze the graph step-by-step.
1. Vertical Asymptote:
- The logarithmic function [tex]\( \log_2 x \)[/tex] has a vertical asymptote where the argument of the logarithm (in this case, [tex]\( x \)[/tex]) approaches zero from the right.
- Therefore, the vertical asymptote for [tex]\( f(x) = \log_2 x \)[/tex] is at [tex]\( x = 0 \)[/tex].
2. Behavior of the Graph:
- As [tex]\( x \)[/tex] approaches 0 from the right ([tex]\( x \to 0^+ \)[/tex]), the value of [tex]\( \log_2 x \)[/tex] tends towards negative infinity. This confirms that the graph has a vertical asymptote at [tex]\( x = 0 \)[/tex].
- The function [tex]\( f(x) = \log_2 x \)[/tex] is defined only for positive [tex]\( x \)[/tex] values.
3. Sign of the Function on the Interval [tex]\( (0, 1) \)[/tex]:
- We next investigate the value of [tex]\( f(x) \)[/tex] on the interval [tex]\( (0, 1) \)[/tex]. We know that for [tex]\( x \)[/tex] in [tex]\( (0, 1) \)[/tex], [tex]\( 0 < x < 1 \)[/tex].
- For any [tex]\( x \)[/tex] between 0 and 1, [tex]\( \log_2 x \)[/tex] is negative because we are taking the logarithm base 2 of a number less than 1.
We can now address each of the statements given in the question:
A. The graph has an asymptote of [tex]\( y = 0 \)[/tex] and is increasing as [tex]\( x \)[/tex] approaches positive infinity. - This is incorrect. The asymptote is at [tex]\( x = 0 \)[/tex], not [tex]\( y = 0 \)[/tex].
B. The graph has an asymptote of [tex]\( x = 0 \)[/tex] and is negative over the interval [tex]\( (0, 1) \)[/tex]. - This statement is correct.
C. The graph has an asymptote of [tex]\( x = 0 \)[/tex] and is positive over the interval [tex]\( (0, 1) \)[/tex]. - This is incorrect. The function [tex]\( \log_2 x \)[/tex] is negative over the interval [tex]\( (0, 1) \)[/tex].
D. The graph has an asymptote of [tex]\( y = 0 \)[/tex] and is decreasing as [tex]\( x \)[/tex] approaches positive infinity. - This is incorrect. The asymptote is at [tex]\( x = 0 \)[/tex], not [tex]\( y = 0 \)[/tex], and the function is increasing as [tex]\( x \)[/tex] approaches positive infinity.
Based on this analysis, the correct answer is B. The graph has an asymptote of [tex]\( x=0 \)[/tex] and is negative over the interval [tex]\( (0,1) \)[/tex].
1. Vertical Asymptote:
- The logarithmic function [tex]\( \log_2 x \)[/tex] has a vertical asymptote where the argument of the logarithm (in this case, [tex]\( x \)[/tex]) approaches zero from the right.
- Therefore, the vertical asymptote for [tex]\( f(x) = \log_2 x \)[/tex] is at [tex]\( x = 0 \)[/tex].
2. Behavior of the Graph:
- As [tex]\( x \)[/tex] approaches 0 from the right ([tex]\( x \to 0^+ \)[/tex]), the value of [tex]\( \log_2 x \)[/tex] tends towards negative infinity. This confirms that the graph has a vertical asymptote at [tex]\( x = 0 \)[/tex].
- The function [tex]\( f(x) = \log_2 x \)[/tex] is defined only for positive [tex]\( x \)[/tex] values.
3. Sign of the Function on the Interval [tex]\( (0, 1) \)[/tex]:
- We next investigate the value of [tex]\( f(x) \)[/tex] on the interval [tex]\( (0, 1) \)[/tex]. We know that for [tex]\( x \)[/tex] in [tex]\( (0, 1) \)[/tex], [tex]\( 0 < x < 1 \)[/tex].
- For any [tex]\( x \)[/tex] between 0 and 1, [tex]\( \log_2 x \)[/tex] is negative because we are taking the logarithm base 2 of a number less than 1.
We can now address each of the statements given in the question:
A. The graph has an asymptote of [tex]\( y = 0 \)[/tex] and is increasing as [tex]\( x \)[/tex] approaches positive infinity. - This is incorrect. The asymptote is at [tex]\( x = 0 \)[/tex], not [tex]\( y = 0 \)[/tex].
B. The graph has an asymptote of [tex]\( x = 0 \)[/tex] and is negative over the interval [tex]\( (0, 1) \)[/tex]. - This statement is correct.
C. The graph has an asymptote of [tex]\( x = 0 \)[/tex] and is positive over the interval [tex]\( (0, 1) \)[/tex]. - This is incorrect. The function [tex]\( \log_2 x \)[/tex] is negative over the interval [tex]\( (0, 1) \)[/tex].
D. The graph has an asymptote of [tex]\( y = 0 \)[/tex] and is decreasing as [tex]\( x \)[/tex] approaches positive infinity. - This is incorrect. The asymptote is at [tex]\( x = 0 \)[/tex], not [tex]\( y = 0 \)[/tex], and the function is increasing as [tex]\( x \)[/tex] approaches positive infinity.
Based on this analysis, the correct answer is B. The graph has an asymptote of [tex]\( x=0 \)[/tex] and is negative over the interval [tex]\( (0,1) \)[/tex].
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.