1. Consider the function below. f(x) = ln(3 - lnx) Consider the function below.
ID: 2860081 • Letter: 1
Question
1. Consider the function below. f(x) = ln(3 - lnx)
Consider the function below. F(x) = in(3 - in x) Find the vertical asymptote(s). (Enter your answers as a comma-separated lit. If an answer does not exist, enter DNE.) X= Find the horizontal asymptote(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) y= Find the interval where the function is decreasing. (Enter your answer using interval notation.) Find the local maximum and minimum values. (If an answer does not exist, enter DNE.) local maximum value local minimum value Find the interval where the function is concave up. (Enter your answer using interval notation.) Find the interval where the function is concave down. (Enter your answer using interval notation.) Find the inflection point. (x, y)= ()Explanation / Answer
(a)Values inside log must be > 0
From inner log, we get
x > 0
From outer log, we get
3 ln x > 0
ln x > 3
ln x < 3
x < e^3
Function is defined for 0 < x < e^3
and there are vertical asymptotes at x = 0 and x = e^3
(b)f'(x) = 1/(3lnx) * (1/x) = 1/(x(lnx3))
Function is decreasing when f'(x) < 0
Since x > 0, then f'(x) < 0 when (lnx3) < 0
lnx 3 < 0
lnx < 3
x < e^3
So function is decreasing on interval (0, e^3)
(c) f'(x) = 1/(x(lnx3)) = (x(lnx3))^(1)
f''(x) = 1(x(lnx3))^(2) * (1*(lnx3) + x*(1/x0))
f''(x) = (lnx2)/(x(lnx3))²
f''(x) = (2lnx)/(x(lnx3))²
Function is concave up when f''(x) > 0
Since denominator of f''(x) > 0, then we need to find when numerator > 0
2 lnx > 0
lnx > 2
lnx < 2
x < e^2
Function is concave up on interval (0, e^2)
(d) From (c) function is concave down on interval (e^2, e^3)
Point of inflection occurs when concavity changes sign, i.e. at x = e^2
y = ln(3 ln x) = ln(3 ln(e^3)) = ln(32) = ln(1) = 0
Point of inflection = (e^2, 0)
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