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

Answer:

#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#.