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

1 Answer
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-3x=0

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

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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.