How do you find the domain and range of #f(x) = 1/(2-e^x)#?

1 Answer
Aug 3, 2015

Answer:

Domain: #(-oo, ln(2)) uu (ln(2), +oo)#
Range: #(-oo, 0) uu (1/2, +oo)#

Explanation:

Right from the start, you can say that one value of #x# will be excluded from the domain of the function because of the restriction to the denominator of the function.

More epecifically, the expression #(2-e^x)# must not be equal to zero.

The value of #x# for which condition is not satisfied will be

#2-e^x = 0#

#e^x = 2#

#ln(e^x) = ln(2)#

#x * ln(e) = ln(2) => x = ln(2)#

The domain of the function will thus be #RR-{ln(2)}#, or #(-oo, ln(2)) uu (ln(2), +oo)#.

The range of the function will be affected by the fact that the graph has a vertical asymptote at #x=ln(2)#.

Now, because #e^x >0, (AA)x#, you get that

#1/(2-e^x) >0# for #x < ln(2)# and #1/(2-e^x)<0# for #x>ln(2)#.

That happens because you have

#lim_(x->-oo)e^x = 0#, for which you have #f(x) -> 1/2#

#lim_(x->oo)e^x = +oo#, for which you have #f(x) ->0#

The range of your function will thus be #(-oo, 0) uu (1/2, +oo)#

graph{1/(2-e^x) [-10, 10, -5, 5]}