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

1 Answer
May 20, 2016

Answer:

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.