Question

For what values of x, if any, does `f(x) = secpix` have vertical asymptotes?

Answer 1

`x = ((2k+1) )/2, qquad k in mathbf(Z)`

Explanation:

for vertical asymptotes we look for a denominator of 0

here we have

`f = sec pi x = 1/ (cos pi x)`

the denom will be zero at
`pi x = pi/2, (3pi)/2 , ...., ((2k+1) pi)/2, qquad k in mathbf(Z)`

ie
`x = ((2k+1) )/2, qquad k in mathbf(Z)`