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

May 20, 2016

See below.

#### Explanation:

So in general, vertical asymptotes occur when the denominator in the function is zero, such that it's value doesn't exist,
Therefore, we need ${e}^{x} - 3 x = 0$

As it turns out, this is really quite hard to solve. I would use a graphical calculator or app to solve this.

But if we want to find solutions analytically then we could use the Taylor expansion for ${e}^{x}$ which is sum_{n=0}^inftyx^n/(n!). If we set this equal to $3 x$, the more terms you take in the Taylor series, the closer you will get to the actual solution.