For what values of x, if any, does #f(x) = sec((-15pi)/8+9x) # have vertical asymptotes?

Question

Emma Aldridge · Accepted Answer

#x = - (11pi)/72 + pi/8k# #" "# for #k# an integer. which may also be written as #x = (5pi)/72 + pi/8k# #" "# for #k an integer.

Recall #sec theta = 1/cos theta# The secant has a vertical asymptote where the cosine of the argument is #0#. The cosine is #0# where the argument is an odd multiple of #pi/2# The odd multiples of #pi/2# have form #(2k+1)pi/2#, which may also be written #pi/2 + pik# for #k# an integer. So this function has vertical asymptotes at the solutions to #(-15pi)/8 +9x = pi/2 + pik# #" "# for #k# an integer. #9x = (4pi)/8 -(15pi)/8 + pik# #" "# for #k# an integer. #9x = - (11pi)/8 + pik# #" "# for #k# an integer. #x = - (11pi)/72 + pi/8k# #" "# for #k# an integer. Alternatives We need not take #- (11pi)/72# as our base case. We can use any #x# on the list of solutions (any #x# that has the proper form). We got #9x = - (11pi)/8 + pik# #" "# for #k# an integer. Taking #k=2# we see that #9x = (5pi)/8# is a solution. So we could replace the solution with #9x = (5pi)/8 + pik# #" "# for #k# an integer. So, the solutions are: #x = (5pi)/72 + pi/8k# #" "# for #k# an integer.

Christopher Agresta · Answer

undefined The function f(x) = sec((-15pi)/8+9x) has vertical asymptotes when the value inside the secant function, (-15pi)/8+9x, equals odd multiples of pi/2.