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

Jan 28, 2016

There is a vertical asymptote at the solution to $x {e}^{x} = 3$

Explanation:

If there is a solution to $x {e}^{x} = 3$, call it $c$, then there is a vertical asymptote at $x = c$, because the limit as $x \rightarrow c$ of $f \left(x\right)$ will have the form $\frac{1}{0}$, so the function is either increasing or decreasing without bound as $x = c$.

It is clear that we can make $x {e}^{x} < 3$ (for example, by making $x = 0$).

We can also make $x {e}^{x} > 3$ (for example, by making $x = 10$)

We also know that $x {e}^{x}$ is continuous on $\left(- \infty , \infty\right)$, so by the Intermediate Value Theorem, there is a solution to $x {e}^{x} = 3$.

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 $x {e}^{x} = 3$. Using numerical or technological methods we can get an approximation of $c \approx 1.05$