How do you find the domain and range of f(x) = 1/(2-e^x)f(x)=12ex?

1 Answer
Aug 3, 2015

Domain: (-oo, ln(2)) uu (ln(2), +oo)(,ln(2))(ln(2),+)
Range: (-oo, 0) uu (1/2, +oo)(,0)(12,+)

Explanation:

Right from the start, you can say that one value of xx 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)(2ex) must not be equal to zero.

The value of xx for which condition is not satisfied will be

2-e^x = 02ex=0

e^x = 2ex=2

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

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

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

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

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

1/(2-e^x) >012ex>0 for x < ln(2)x<ln(2) and 1/(2-e^x)<012ex<0 for x>ln(2)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]}