select all the correct answers.\nconsider function f and function g.\nf(x) = ln x\ng(x) = -5 ln x\nhow does…

select all the correct answers.\nconsider function f and function g.\nf(x) = ln x\ng(x) = -5 ln x\nhow does the graph of function g compare with the graph of function f?\n□ unlike the graph of function f, the graph of function g decreases as x increases.\n□ the graph of function g is the graph of function f reflected over the x - axis and vertically stretched by a factor of 5.\n□ unlike the graph of function f, the graph of function g has a y - intercept.\n□ the graphs of both functions have a vertical asymptote of x = 0.\n□ unlike the graph of function f, the graph of function g has a domain of {x|-5 < x < ∞}.
Answer
Explanation:
Step1: Analyze the slope of the functions
The derivative of $f(x)=\ln x$ is $f^\prime(x)=\frac{1}{x}>0$ for $x > 0$, so $f(x)$ is increasing. For $g(x)=- 5\ln x$, its derivative is $g^\prime(x)=-\frac{5}{x}<0$ for $x>0$, so $g(x)$ decreases as $x$ increases.
Step2: Analyze the transformation
The function $g(x)=-5\ln x=-1\times5\times\ln x$. The negative - sign reflects the graph of $y = \ln x$ over the $x$-axis and the coefficient 5 vertically stretches it by a factor of 5.
Step3: Analyze the $y$-intercept
The domain of $y = \ln x$ and $y=-5\ln x$ is $x>0$. Since $x = 0$ is not in the domain, neither function has a $y$-intercept.
Step4: Analyze the vertical asymptote
The domain of $f(x)=\ln x$ and $g(x)=-5\ln x$ is $(0,\infty)$. As $x\rightarrow0^{+}$, $\ln x\rightarrow-\infty$ and $-5\ln x\rightarrow\infty$. So the vertical asymptote of both functions is $x = 0$.
Step5: Analyze the domain
The domain of $f(x)=\ln x$ and $g(x)=-5\ln x$ is $x>0$ (i.e., $(0,\infty)$), not ${x|-5 < x<\infty}$.
Answer:
A. Unlike the graph of function $f$, the graph of function $g$ decreases as $x$ increases. B. The graph of function $g$ is the graph of function $f$ reflected over the $x$-axis and vertically stretched by a factor of 5. D. The graphs of both functions have a vertical asymptote of $x = 0$.