consider the following function. (if an answer does not exist, enter dne.)\n$f(x)=ln (3-ln (x))$\n(a) find…
consider the following function. (if an answer does not exist, enter dne.)\n$f(x)=ln (3-ln (x))$\n(a) find the vertical asymptote(s). (enter your answers as a comma - separated list.)\n$x=$$\nfind the horizontal asymptote(s). (enter your answers as a comma - separated list.)\n$y=$$\n(b) find the interval(s) of increase. (enter your answer using interval notation.)\n$$\nfind the interval(s) of decrease. (enter your answer using interval notation.)\n$$\n(c) find the local maximum and minimum values.\nlocal maximum value$$\nlocal minimum value$$\n(d) find the interval on which ( f ) is concave up. (enter your answer using interval notation.)\n$$\nfind the interval on which ( f ) is concave down. (enter your answer using interval notation.)\n$$\nfind the inflection point.\n( (x, y)=(quad) )
Answer
Explanation:
Step1: Find the domain
For (y = \ln(3-\ln x)), we need (x>0) and (3-\ln x>0). Solving (3-\ln x>0) gives (\ln x<3) or (x < e^{3}). So the domain is ((0,e^{3})).
Step2: Find vertical asymptote
As (x\rightarrow0^{+}), (\ln x\rightarrow-\infty), (3-\ln x\rightarrow\infty), (y = \ln(3-\ln x)\rightarrow\infty). As (x\rightarrow e^{3}), (\ln x = 3), (3-\ln x=0), (y=\ln(3 - \ln x)\rightarrow-\infty). So the vertical asymptote is (x = e^{3}).
Step3: Find horizontal asymptote
As (x\rightarrow0^{+}), (y\rightarrow\infty); as (x\rightarrow e^{3}), (y\rightarrow-\infty). There is no horizontal asymptote (DNE).
Step4: Find derivative
Using the chain - rule, (y^\prime=\frac{1}{3-\ln x}\cdot(-\frac{1}{x})=-\frac{1}{x(3 - \ln x)}).
Step5: Find intervals of increase/decrease
Set (y^\prime = 0), but (-\frac{1}{x(3 - \ln x)}=0) has no solution. For (y^\prime=-\frac{1}{x(3 - \ln x)}), in the domain ((0,e^{3})), (y^\prime<0) for all (x\in(0,e^{3})). So the function is decreasing on ((0,e^{3})) and there is no interval of increase (DNE).
Step6: Find local extrema
Since the function is always decreasing on its domain, there are no local maximum or minimum values (DNE for both).
Step7: Find second derivative
Using the quotient rule, (y^{\prime\prime}=\frac{-(3 - \ln x)+1}{x^{2}(3 - \ln x)^{2}}=\frac{\ln x - 2}{x^{2}(3 - \ln x)^{2}}).
Step8: Find concavity
Set (y^{\prime\prime}=0), then (\ln x-2 = 0), (x = e^{2}). When (0<x<e^{2}), (\ln x-2<0), (y^{\prime\prime}<0), function is concave down on ((0,e^{2})). When (e^{2}<x<e^{3}), (\ln x - 2>0), (y^{\prime\prime}>0), function is concave up on ((e^{2},e^{3})).
Step9: Find inflection point
When (x = e^{2}), (y=\ln(3-\ln(e^{2}))=\ln(3 - 2)=\ln1 = 0). So the inflection point is ((e^{2},0)).
Answer:
a. Vertical asymptote: (x = e^{3}); Horizontal asymptote: DNE b. Interval of increase: DNE; Interval of decrease: ((0,e^{3})) c. Local maximum value: DNE; Local minimum value: DNE d. Concave up: ((e^{2},e^{3})); Concave down: ((0,e^{2})); Inflection point: ((e^{2},0))