`f(x) = (e^x)/(1 - e^x)` (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals...

`f(x) = (e^x)/(1 - e^x)` (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of concavity and the inflection points. (e) Use the information from parts (a)-(d) to sketch the graph of `f`.

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Textbook Question

Chapter 4, 4.3 - Problem 48 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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(a) vertical asymptotes can be where f tends to infinity at a finite point. Here the only possible point is where `1-e^x=0,` or x=0. The numerator not zero at this point so this is really vertical asymptote.

Horizontal asymptotes can be at `+-oo` if f has finite limit(s) there.

`f(x) = -1 + (1)/(1-e^x).`  `e^x -> +oo` when `x -> +oo` and `e^x -> 0` when `x -> -oo,` so

`f(x) -> -1` when `x -> +oo,` `f(x) -> 0` when `x -> -oo.`

y=-1 and y=0 are horizontal asymptotes.

(b) `f'(x) = (e^x)/(1-e^x)^2` >0 everywhere (except x=0 where it is undefined). So f(x) increases on `(-oo, 0)` and `(0, +oo)` .

(c) therefore there are no local minimums and maximums

(d) `f''(x) = e^x*[(1-e^x) - (-2)*e^x]/(1-e^x)^3 = e^x*(1+e^x)/(1-e^x)^3.`

this is >0 for x<0 and <0 for x>0, so f is concave upward on `(-oo, 0)` and is concave downward on `(0, +oo).` Probably we can call x=0 the inflection point.

(e) 

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