Question

For what values of x, if any, does `f(x) = 1/((2x^2-8)cos(pi/2+(8pi)/x)` have vertical asymptotes?

Answer 1

`x=8/k` where `kinZZ`

Explanation:

Vertical asymptotes occur when the denominator equals `0`.

Focusing on the non-trigonometric part first:

`2x^2-8=0`

`x^2=4`

`x=pm2`

The other part is more difficult:

`cos(pi/2+(8pi)/x)=0`

`cos(x)=0` for `x=pi/2,(3pi)/2,(5pi)/2`, which can be expressed as `x=pi/2+kpi`, where `kinZZ` (which means that `k` is an integer).

This means that `cos(pi/2+(8pi)/x)=0` and `f(x)` is asymptotic when:

`pi/2+(8pi)/x=pi/2+kpi" "" "" ",kinZZ`

So:

`(8pi)/x=kpi" "" "" ",kinZZ`

`8/x=k" "" "" ",kinZZ`

`x=8/k" "" "" ",kinZZ`

So there are vertical asymptotes at `x=pm8/1,pm8/2,pm8/3,pm8/5,pm8/6...`

Note that the values found from the `2x^2-8` portion are included in this set, as a sort of special case where `k=pm4`.