Question

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

Answer 1

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]}