For what values of x, if any, does #f(x) = 1/((12x-9)sin(pi+(3pi)/x) # have vertical asymptotes?

1 Answer
Oct 1, 2016

Answer:

#x=0,3/k,kinZZ# (where #k# is an integer)

Explanation:

Note that for a rational function such as the given function #f#, there will be a rational function whenever its denominator equals #0#.

So, there is a vertical asymptote whenever #(12x-9)sin(pi+(3pi)/x)=0#.

We can split this into two parts:

#{(12x-9=0),(sin(pi+(3pi)/x)=0):}#

The first can be solved to show that #x=9/12=3/4#.

The second is a little more difficult: note that #sin(x)=0# when #x=0,pi,2pi#, and so on. This can be written as #x=kpi#, where #kinZZ#, which means #k# is an integer.

Thus:

#pi+(3pi)/x=kpi" "" "," " "kinZZ#

Subtracting #pi#, we see that the right hand side remains #kpi#, because some integer multiple of #pi# minus #pi# is still an integer multiple of #pi#.

#(3pi)/x=kpi" "" "," " " "kinZZ#

Rearranging:

#3pi=x(kpi)" "" "," "" "kinZZ#

#x=(3pi)/(kpi)=3/k" "" "," "" "kinZZ#

Note that the previous solution from #12x-9=0# gave #x=3/4#, which is included in the solution #x=3/k,kinZZ#.

Furthermore, since #(3pi)/x# is included within the sine function, there will be a vertical asymptote at #x=0#.