Let's analyze each statement one by one to understand how the graph of function [tex]\( g \)[/tex] compares to the graph of function [tex]\( f \)[/tex].
Given:
- Function [tex]\( f(x) = \ln x \)[/tex]
- Function [tex]\( g(x) = -5 \ln x \)[/tex]
1. The graphs of both functions have a vertical asymptote of [tex]\( x = 0 \)[/tex].
For [tex]\( f(x) = \ln x \)[/tex]:
- The natural logarithm function, [tex]\( \ln x \)[/tex], has a vertical asymptote at [tex]\( x = 0 \)[/tex].
For [tex]\( g(x) = -5 \ln x \)[/tex]:
- The expression [tex]\( \ln x \)[/tex] within [tex]\( g(x) \)[/tex] still implies a vertical asymptote at [tex]\( x = 0 \)[/tex].
Therefore, the graphs of both [tex]\( f \)[/tex] and [tex]\( g \)[/tex] have a vertical asymptote at [tex]\( x = 0 \)[/tex].
Answer: True
2. Unlike the graph of function [tex]\( f \)[/tex], the graph of function [tex]\( g \)[/tex] has a domain of [tex]\( \{x \mid -5
For [tex]\( f(x) = \ln x \)[/tex]:
- The domain of the natural logarithm function is [tex]\( x > 0 \)[/tex].
For [tex]\( g(x) = -5 \ln x \)[/tex]:
- The domain is dictated by the argument of the logarithm function, which is [tex]\( x > 0 \)[/tex].
Therefore, the domain of both functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] is [tex]\( x > 0 \)[/tex], and not [tex]\( -5 < x < \infty \)[/tex].
Answer: False
3. Unlike the graph of function [tex]\( f \)[/tex], the graph of function [tex]\( g \)[/tex] decreases as [tex]\( x \)[/tex] increases.
For [tex]\( f(x) = \ln x \)[/tex]:
- The function [tex]\( \ln x \)[/tex] increases as [tex]\( x \)[/tex] increases.
For [tex]\( g(x) = -5 \ln x \)[/tex]:
- The negative sign and the multiplication by 5 invert and stretch the function vertically.
- Thus, [tex]\( g(x) \)[/tex] decreases as [tex]\( x \)[/tex] increases.
Therefore, [tex]\( g \)[/tex] decreases as [tex]\( x \)[/tex] increases, whereas [tex]\( f \)[/tex] increases.
Answer: True
4. Unlike the graph of function [tex]\( f \)[/tex], the graph of function [tex]\( g \)[/tex] has a [tex]\( y \)[/tex]-intercept.
For [tex]\( f(x) = \ln x \)[/tex]:
- There is no [tex]\( y \)[/tex]-intercept because [tex]\( \ln x \)[/tex] is undefined at [tex]\( x = 0 \)[/tex].
For [tex]\( g(x) = -5 \ln x \)[/tex]:
- Similarly, there is no [tex]\( y \)[/tex]-intercept since [tex]\( \ln x \)[/tex] is undefined at [tex]\( x = 0 \)[/tex].
Therefore, neither function has a [tex]\( y \)[/tex]-intercept.
Answer: False
5. The graph of function [tex]\( g \)[/tex] is the graph of function [tex]\( f \)[/tex] reflected over the [tex]\( x \)[/tex]-axis and vertically stretched by a factor of 5.
For [tex]\( f(x) = \ln x \)[/tex]:
- The original function is [tex]\( \ln x \)[/tex].
For [tex]\( g(x) = -5 \ln x \)[/tex]:
- The negative sign indicates a reflection over the [tex]\( x \)[/tex]-axis.
- The multiplication by 5 indicates a vertical stretch by a factor of 5.
Therefore, the graph of [tex]\( g \)[/tex] is indeed the graph of [tex]\( f \)[/tex] reflected over the [tex]\( x \)[/tex]-axis and then stretched by a factor of 5.
Answer: True
Summarizing the correct answers:
- The graphs of both functions have a vertical asymptote of [tex]\( x = 0 \)[/tex]. (True)
- Unlike the graph of function [tex]\( f \)[/tex], the graph of function [tex]\( g \)[/tex] has a domain of [tex]\( \{x \mid -5 < x < \infty \} \)[/tex]. (False)
- Unlike the graph of function [tex]\( f \)[/tex], the graph of function [tex]\( g \)[/tex] decreases as [tex]\( x \)[/tex] increases. (True)
- Unlike the graph of function [tex]\( f \)[/tex], the graph of function [tex]\( g \)[/tex] has a [tex]\( y \)[/tex]-intercept. (False)
- The graph of function [tex]\( g \)[/tex] is the graph of function [tex]\( f \)[/tex] reflected over the [tex]\( x \)[/tex]-axis and vertically stretched by a factor of 5. (True)
