Consider the following function. (If an answer does not exist, enter DNE.) f(x) = ln(7 − ln(x))

Question

Consider the following function. (If an answer does not exist, enter DNE.)

f(x) = ln(7 − ln(x))

(a) Find the interval where the function is increasing. (Enter your answer using interval notation.)

 

Find the interval where the function is decreasing. (Enter your answer using interval notation.)

 

(b) Find the local maximum and minimum values.

local maximum value
local minimum value

(c) 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) =

Answer 1

Domain: x > 0 and 7 - lnx > 0; lnx < 7; so, 0 < x < e⁷_;

_{(a) f'(x) = 1/(7 - lnx)·(- 1/x) = 1/[x(lnx - 7). So we have no critical points becousse x = 0 and lnx - 7 = 0; lnx = 7 x = e}⁷ not in domaiin. So, fr'(x) < 0 and interval (0, e⁷), which is all domain f(x) decreasing.

So, f(x) never increasing

(b) No local extrema

(c) f''(x) [x^-1(lnx - 7)^-1]' = -x^-2(lnx - 7)^-1 - x^-1(lnx - 7)^-2·1/x = - 1/[x²(lnx - 7)] - 1/[x²(lnx - 7)²] =

(- lnx + 7 - 1)/[x²(7 - lnx)²] = 0; - lnx + 6 = 0; lnx = 6 and x = e⁶; - lnx + 6 > 0; x < e⁶ but x > 0

So, on interval (0, e⁶) graph concave up; onterval (e⁶, e⁷) graph concave down; f(e⁶) = ln(7 - ln(e⁶)) =

ln(7 - 6) = 0. So, (e⁶, 0) is inflection point