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)?

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- Jan 15th 2009, 02:09 PMh4hv4hd4si4ntrig 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 PMvincisonfire
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)} $? - Jan 15th 2009, 02:19 PMh4hv4hd4si4n
ok, then how do you figure out what the domain is?

- Jan 15th 2009, 02:43 PMvincisonfire
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. - Jan 15th 2009, 03:16 PMh4hv4hd4si4n
Thanks for your help!