For what values of x, if any, does #f(x) = sec((-15pi)/8+2x) # have vertical asymptotes?

1 Answer
mason m
Oct 29, 2016

#x=(pi(19+8k))/16,kinZZ#

Explanation:

#f(x)=sec((-15pi)/8+2x)#

Rewriting using the definition of secant:

#f(x)=1/cos((-15pi)/8+2x)#

This function will have vertical asymptotes when its denominator equals zero, or when:

#cos((-15pi)/8+2x)=0#

Note that the function #cos(theta)=0# when #theta=-pi/2,pi/2,(3pi)/2#, which can be generalized as saying that #cos(theta)=0# for #theta=pi/2+kpi,kinZZ#. Note that #kinZZ# is the mathematical way of writing that #k# is an integer.

Thus, we see that there is a vertical asymptote in #f(x)# when:

#(-15pi)/8+2x=pi/2+kpi" "" "" "" ",kinZZ#

Adding #(15pi)/8# to both sides:

#2x=pi/2+(15pi)/8+kpi" "" "" "" ",kinZZ#

#2x=(19pi)/8+kpi" "" "" "" ",kinZZ#

#x=(19pi)/16+(kpi)/2" "" "" "" ",kinZZ#

#x=(pi(19+8k))/16" "" "" "" ",kinZZ#

So, there are an infinite number of times when there are vertical asymptotes. We can identify those close to the origin by trying #k=-1#, which gives an asymptote at #x=(11pi)/16#, or any other value of #k# for a different asymptote.

Related questions
See all questions in Infinite Limits and Vertical Asymptotes
Impact of this question
1413 views around the world
You can reuse this answer
Creative Commons License