Question

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

Answer 1

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`