How do you find the domain & range for #f(T)=sec(piT/4)#?
1 Answer
Nov 27, 2015
Since
This means that:

#(Tpi)/4 \ne \pi/2 + 2k\pi# 
#(Tpi)/4 \ne (3\pi)/2 + 2k\pi#
Solving for

#T \ne 2 + 4k# 
#T \ne 6 + 4k# .
As for the range, since the function has vertical asymptote, and the denominator changes sign as it crosses zero, we have that