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Math Help - trig graph problem

  1. #1
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    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)?
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  2. #2
    Senior Member vincisonfire's Avatar
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    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)} ?
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  3. #3
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    ok, then how do you figure out what the domain is?
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  4. #4
    Senior Member vincisonfire's Avatar
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    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.
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  5. #5
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    Thanks for your help!
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