For what values of x, if any, does #f(x) = 1/((x-4)sin(pi+(3pi)/x) # have vertical asymptotes?

1 Answer
Oct 11, 2017

Answer:

A vertical asymptote occurs when #x=4, 0, and 3/(n-1), ninRR#

Explanation:

For there to be a vertical asymptote, the denominator bust be 0.

Here, we have #(x-4)sin(\pi+(3\pi)/x)# as the denominator. As long as one of these equals 0, the entire denominator will equal 0.

An asymptote will appear at #x=4# because if we put out #x# value in, we get #0sin(\pi+(3\pi)/4)=0#.

A second asymptote will appear at #x=0# because in the #sin# we have a fraction with #x# on the bottom, if this equals 0 then there can be no #y# value.

The final asymptotes will appear when #x=3/(n-1), ninRR#, this is because for #sin\theta# to equal 0, #\theta# must equal #n\pi, ninRR#.

So, we have #n\pi=\pi+(3\pi)/x#

If we rearrange to make #x# the subject, we get #x=(3\pi)/(n\pi-\pi)=3/(n-1)#

By putting this into #sin(\pi+(3\pi)/x)#, we get #sin(\pi+(3\pi)/(3/(n-1)))-=sin(\pi+(3\pi(n-1))/3-=sin(\pi+\pi(n-1))-=sin(\pi+n\pi-\pi)-=sin(n\pi)#