For what values of x, if any, does #f(x) = tan((pi)/4-9x) # have vertical asymptotes?

Question

Aurora Alvarado · Accepted Answer

Period: #pi/9#. For each period #(pi/12+k/9pi, 7/36pi+k/9pi)#, there are two terminal asymptotes x = end value, #k = 0, +-1, +-2, +-3, ...#

As #(pi/4-9x) to# (an odd multiple ) #(2k+1)# of #pi/2#,# f to +-oo#,

giving

#f to +-oo#, as #x to -(4k+1)/36pi, k = 0, +-1, +-2, +-3, ..#

So, in ine period #x in (pi/12, 7/36pi)#, we have two terminal

asymptotes

#x = pi/12 and x = 7/36pi#

The period for f is #pi/9=(7/36-1/12)pi#.

graph{y-tan(0.7854-9x)=0 [-5, 5, -2.5, 2.5]}

Evelyn Agard · Answer

undefined The function f(x) = tan((pi)/4-9x) has vertical asymptotes when the tangent function is undefined. The tangent function is undefined when the angle is equal to (2n+1)(pi)/2, where n is an integer.

Setting (pi)/4-9x equal to (2n+1)(pi)/2 and solving for x, we get:

(pi)/4-9x = (2n+1)(pi)/2
-9x = (2n+1)(pi)/2 - (pi)/4
x = [(2n+1)(pi)/2 - (pi)/4]/(-9)

Therefore, the function f(x) = tan((pi)/4-9x) has vertical asymptotes at x = [(2n+1)(pi)/2 - (pi)/4]/(-9), where n is an integer.