Question

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

Answer 1

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-3x=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 `3x`, the more terms you take in the Taylor series, the closer you will get to the actual solution.