# trig graph problem

• Jan 15th 2009, 02:09 PM
h4hv4hd4si4n
trig graph problem
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)?
• Jan 15th 2009, 02:18 PM
vincisonfire
Since $\displaystyle sec(x)=\frac{1}{cos(x)}$ the period is the same.
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.