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

##### 1 Answer

Apr 30, 2017

#### Explanation:

Vertical asymptotes occur when the denominator equals

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#

This means that

#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

Note that the values found from the