Question

For what values of x, if any, does `f(x) = 1/(xe^x-3)` have vertical asymptotes?

Answer 1

There is a vertical asymptote at the solution to `xe^x=3`

Explanation:

If there is a solution to `xe^x=3`, call it `c`, then there is a vertical asymptote at `x= c`, because the limit as `x rarr c` of `f(x)` will have the form `1/0`, so the function is either increasing or decreasing without bound as `x= c`.

It is clear that we can make `xe^x < 3` (for example, by making `x=0`).

We can also make `xe^x > 3` (for example, by making `x=10`)

We also know that `xe^x` is continuous on `(-oo,oo)`, so by the Intermediate Value Theorem, there is a solution to `xe^x=3`.

So we know that there is a vertical asymptote, and we can describe where it occurs.

Unfortunately there is not a nice expression for the solution to `xe^x=3`. Using numerical or technological methods we can get an approximation of `c ~~ 1.05`