trig graph problem

h4hv4hd4si4n · #1 $Old$ January 15th, 2009, 04:09 PM

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)?

vincisonfire · #2 $Old$ January 15th, 2009, 04:18 PM

Since $sec(x)=\frac{1}{cos(x)}$ the period is the same.
(Answer to your question is yes)
The domain is everywhere except where $cos(3x-\pi)=0$ because division by 0 is not defined.
Range : What is the highest value of $cos(3x-\pi)$ ? What is then the minimum value of $\frac{1}{ cos(3x-\pi)}$ ?

h4hv4hd4si4n · #3 $Old$ January 15th, 2009, 04:19 PM

ok, then how do you figure out what the domain is?

vincisonfire · #4 $Old$ January 15th, 2009, 04:43 PM

Well as I told you $cos(3x-\pi)=0$ is not possible.
I occurs when $3x-\pi = \frac{(2k+1)\pi}{2}$ , $k\in \mathbb Z$ In other words $3x-\pi = \frac{\pi}{2},\frac{3\pi}{2},...$
It follows that $3x = \frac{(2k+1)\pi}{2} + \pi =\frac{(2k+3)\pi}{2}$ and when $x = \frac{(2k+1)\pi}{6}$ (we don't care because 1 and 3 gives the same set) $k\in \mathbb Z$ the function is not defined.

h4hv4hd4si4n · #5 $Old$ January 15th, 2009, 05:16 PM

Thanks for your help!