For what values of x, if any, does #f(x) = tan((5pi)/4-x) # have vertical asymptotes?

1 Answer
Feb 27, 2018

Answer:

#f(x)# has vertical asymptotes at #....(3pi)/4,-pi/4,-(5pi)/4,...#

We can say that #f(x)# is undefined for any applicable #(3pi)/4+npi#, where #ninZZ#.

Explanation:

#tan(x)# is undefined at certain points:

#pi/2,(3pi)/2,(5pi)/2#, and so on.

Graphing #tan(x),# we find that, where #h# is a point where #tan(x)# is undefined:

#lim_(xrarrh^+)tan(x)=oo#, and #lim_(xrarrh^-)tan(x)=-oo#

We'll find the first three vertical asymptotes.

So here, #f(x) " DNE"# when #((5pi)/4-x)rarrh#.

Taking #h=pi/2#, we have:

#((5pi)/4-x)rarrpi/2#, which we can write as:

#(5pi)/4-x=pi/2#

#-x=pi/2-(5pi)/4#

#-x=-(3pi)/4#

#x=(3pi)/4#

Taking #h=(3pi)/2#, we have:

#(5pi)/4-x=(3pi)/2#

#-x=(3pi)/2-(5pi)/4#

#-x=pi/4#

#x=-pi/4#

For #h=(5pi)/2#, we have:

#(5pi)/4-x=(5pi)/2#

#-x=(5pi)/2-(5pi)/4#

#-x=(5pi)/4#

#x=-(5pi)/4#

This series continues to the left and right.