For what values of x, if any, does #f(x) = 1/((5x+8)cos(pi/2-(12pi)/x) # have vertical asymptotes?

Question

William Abdalla · Accepted Answer

#x=-8/5#, #x=0#, and #x=-6/n# (see explanation)

For there to be a vertical asymptote, the denominator must equal to zero, and #0*"anything"=0#, so #(5x+8)=0, 5x=-8, x=-8/5#. Also, #cos((pi/2)-(12pi)/x)=0#. #cos^(-1)(0)=(pi)/2+n2pi#. Therefore, #pi/2-(12pi)/x=pi/2+n2pi#, #(12pi)/x=-n2pi#, #x=-(12pi)/(n2pi)#, #x=-6/n#. By rearranging we get #n=-6/x#. We can substitute this to get #pi/2-(12pi)/x=pi/2+2(-6/x)pi-=pi/2-(12pi)/x=pi/2-12/xpi#. For #n# to be an integer, #2n-=-12/x#, where #x# is any number which 12 is divisible by, to also give a whole number. For example, let's take #x=4, 12 " and " 24#: 4: #2n=-12/4=-3#, as #2n# is an integer, #pi/2-(12pi)/x# can equal #pi/2+2(-6/x)pi#, #pi/2-3pi=pi/2-3pi#, #cos(pi/2-3pi)=0#, #0(5(-3)+8)=0#. 12: #2n=-12/12=-1#, as #2n# is an integer, #pi/2-(12pi)/x# can equal #pi/2+2(-6/x)pi#, #pi/2-1pi=pi/2-1pi#, #cos(pi/2-pi)=0#, #0(5(-1)+8)=0#. 24: #2n=-12/24=-1/2#, as #2n# is not an integer, #pi/2-(12pi)/x# cannot equal #pi/2+2(-6/x)pi#, as only #cos(pi/2+2npi)# can equal 0, where #2n# forms an integer, but #-1/2# is not an integer. For example, #cos(pi/2)=0#, #cos(pi/2+2pi)=0#, and #cos(pi/2+4pi)=0#. Also, #x=0# is an asymptote as #(12pi)/0="undefined"#.

Emilia Aguilar · Answer

undefined The function f(x) has vertical asymptotes at values of x where the denominator of the function becomes zero. In this case, the denominator is (5x+8)cos(pi/2-(12pi)/x). To find the values of x that make the denominator zero, we need to solve the equation (5x+8)cos(pi/2-(12pi)/x) = 0. However, the cosine function is never equal to zero, so there are no values of x that make the denominator zero. Therefore, the function f(x) does not have any vertical asymptotes.