g(x) = 2 sec (3x - π) + 1
find:
(1) period
(2) domain
(3) range
(4) draw graph
Is the equation for the period is 2π/B (like it is for sine and cosine equations)?
Since $\displaystyle sec(x)=\frac{1}{cos(x)} $ the period is the same.
(Answer to your question is yes)
The domain is everywhere except where $\displaystyle cos(3x-\pi)=0 $ because division by 0 is not defined.
Range : What is the highest value of $\displaystyle cos(3x-\pi) $? What is then the minimum value of $\displaystyle \frac{1}{ cos(3x-\pi)} $?
Well as I told you $\displaystyle cos(3x-\pi)=0 $ is not possible.
I occurs when $\displaystyle 3x-\pi = \frac{(2k+1)\pi}{2} $, $\displaystyle k\in \mathbb Z $ In other words $\displaystyle 3x-\pi = \frac{\pi}{2},\frac{3\pi}{2},... $
It follows that $\displaystyle 3x = \frac{(2k+1)\pi}{2} + \pi =\frac{(2k+3)\pi}{2} $ and when $\displaystyle x = \frac{(2k+1)\pi}{6} $(we don't care because 1 and 3 gives the same set) $\displaystyle k\in \mathbb Z $ the function is not defined.