# How do you find the domain & range for f(T)=sec(piT/4)?

Nov 27, 2015

Since $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$, and since you can't divide by zero, the domain is composed by all the points such that $\cos \left(\frac{T \pi}{4}\right) \setminus \ne 0$.

This means that:

• $\frac{T \pi}{4} \setminus \ne \setminus \frac{\pi}{2} + 2 k \setminus \pi$

• $\frac{T \pi}{4} \setminus \ne \frac{3 \setminus \pi}{2} + 2 k \setminus \pi$

Solving for $T$:

• $T \setminus \ne 2 + 4 k$

• $T \setminus \ne 6 + 4 k$.

As for the range, since the function has vertical asymptote, and the denominator changes sign as it crosses zero, we have that

${\lim}_{T \setminus \to {6}^{\setminus \pm}} \frac{1}{\cos} \left(\frac{T \setminus \pi}{4}\right) = \frac{1}{0} ^ \left\{\setminus \pm\right\} = \setminus \pm \setminus \infty$