For what values of x, if any, does #f(x) = 1/((4x+9)cos(pi/2+(4pi)/x) # have vertical asymptotes?

1 Answer

Answer:

Asymptote is vertical at

#x=-9/4, 4/k#

Where #k=0, \pm1, \pm2, \pm3, \ldots#

Explanation:

The given function:

#f(x)=1/{(4x+9)\cos(\pi/2+{4\pi}/x)}#

#f(x)=1/{(4x+9)\sin({4\pi}/x)}#

Above function will have vertical asymptotes when denominator becomes equal to zero i.e.

#(4x+9)\sin({4\pi}/x)=0#

#4x+9=0, \ \ or\ \ \sin({4\pi}/x)=0#

#x=-9/4, \ \ or\ \ \ {4\pi}/x=k\pi#

#x=-9/4\ \ \or \ \ \ x=4/k#

Where, #k# is any integer hence we get the set of points where asymptote is vertical

#x=-9/4, 4/k#

Where #k=0, \pm1, \pm2, \pm3, \ldots#