For what values of x, if any, does #f(x) = 1/((3x-2)sin(pi+(2pi)/x) # have vertical asymptotes?

1 Answer
Mar 23, 2017

Answer:

#x={-2,2/3}#

Explanation:

Vertical symptotes occur where f(x) is undefined. This will occur where the function is over 0 in this case. We must therefore determine when the denominator equals 0. Whenever two things are multiplied together, such as in this case #3x-2# and #sin(pi+(2*pi)/x)#, the total will be equal to zero when either or both are zero so we must find when both equal 0. To do this, set each expression separately equal to 0 and solve for x.

Now we have the equations #3x-2=0# and #sin(pi+(2*pi)/x)=0#. The former is much easier to solve. Simply add two to both sides and divide both sides by three to get: #x=2/3#.

For the next equation we must first find where sine is equal to 0. This only occurs at the quadrantals of 0, #pi#, and #2*pi# within a range of 0 to #2*pi# for the angle. Since #2*pi# and 0 are the same angle we can disregard one. This means that sine is equal to 0 whenever its argument is 0 or #pi#. From this we can make two MORE equations from setting the argument equal to these two things.

This gives us: #pi+(2*pi)/x=0# and #pi+(2*pi)/x=pi#. For the second equation subtract #pi# from both sides and multiply both sides by x to get: #2*pi=0# which means that this case can never happen because #2pi# will NEVER equal 0 and we can disregard it. For the first equation factor out #2pi# from the left expression to get #2pi*(1/2+1/x)=0# then divide by #2pi# on both sides to get #1/2+1/x=0#. Subtract #1/2# from both sides and multiply by x on both sides to get #1=-x/2# then multiply by #-2# on both sides to finally get: #x=-2#.

Finally if we combine all of the x's we found we get that there are vertical asymptotes for this function at #x={-2,2/3}# within a range of 0 to #2pi#. Hope I helped!