For what values of x, if any, does #f(x) = 1/((12x-8)sin(pi+(3pi)/x) # have vertical asymptotes?

1 Answer
KillerBunny
May 2, 2018

The function has vertical asymptotes in the following points:

  • #x = 2/3#
  • All the points of the form #\frac{3}{2k-1}#
  • All the points of the form #\frac{3}{2k}#

Note that, since #k# can be any integer number, the last two points can be summed up in

  • All the points of the form #\frac{3}{n}#, with #n \in \mathbb{Z}#

Explanation:

A function usually has vertical asymptotes where it is not defined, either because of a zero denominator, or a zero argument in a logarithm.

In this case, we have to look for zeros of the denominator.

The denominator, in turn, is a multiplication of two factors, namely #(12x-8)# and #sin(\pi + \frac{3\pi}{x})#. A multiplication of two factor is zero if either of the two factors is zero, so let's examine them one by one.

#12x-8# equals zero if and only if...

#12x-8=0 \iff 12x = 8 \iff x = 8/12 = 2/3#

#sin(\pi + \frac{3\pi}{x})# equals zero if and only if the argument equals #2k\pi# or #\pi + 2k\pi#. Let's split the two cases again:

First case: the argument equals #2k\pi#

#\pi + \frac{3\pi}{x} = 2k\pi \iff \frac{3\pi}{x} = 2k\pi-\pi \iff \frac{x}{3\pi} = \frac{1}{2k\pi-\pi}#

#\iff x = \frac{3\pi}{2k\pi-\pi} = \frac{3}{2k-1}#

Second case: the argument equals #\pi+2k\pi#

#\pi + \frac{3\pi}{x} = \pi+2k\pi \iff \frac{3\pi}{x} = 2k\pi \iff \frac{x}{3\pi} = \frac{1}{2k\pi}#

#\iff x = \frac{3\pi}{2k\pi} = \frac{3}{2k}#

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