For what values of x, if any, does #f(x) = 1/((9x-5)sin(pi+(6pi)/x) # have vertical asymptotes?

Question

1455 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-21T08:50:02

#x = 5/9#

#x = 6/(2n-1) and 3/n , n in ZZ #

To compute the verticle asymptotes we must consider the values for what the function is undefined at, hence where denominator #=0#

#=> (9x-5)sin(pi + (6pi)/x )= 0 #

#=> 9x - 5 = 0 #

#=> color(red)(x = 5/9) # This is our first asymptote

To find the second....

#=> sin(pi + (6pi)/x ) = 0 #

#sin( pi + (6pi)/x) = sin(0) #

#------------#

Using our knowledge of general solutions...

If #sintheta = sin gamma #

# theta = {2pin + gamma , 2pin + pi - gamma }#

#------------#

#=> pi + (6pi)/x = {2pin , 2pin + pi } #

#=> (6pi)/x = {2pi n -pi , 2pin }#

#=> 1/x = { (2n-1)/6 , n/3 }#

#=> color(red)(x = {6/(2n-1) , 3/n }) #

#n in ZZ #

The graph... enter image source here

Interesting...