Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 48

(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)