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Help with finding implied domain and range of trig composite function watch

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    hi guys. im doing some trig and really need help with this question, ive done most except some trouble with the range. (if anythings wrong with the rest pls tell me) thanks!


    Find the domain and range of f(x)=tan(3arccos(x))

    (btw ran means range, dom means domain)

    let 3arccosx=h. dom(h)=dom(3arccosx)=ran(3cosx)= [-1,1]

    ran(h)=3*ran[arccosx]=3*[0,pi]=[0*3, 3*pi]=[0,3pi]

    Now for tanx:

    dom(tanx)=R\{...-3pi/2, -pi/2, pi/2, 3pi/2...}

    ran(tanx)=R


    Now to map one onto the other...

    so,

    since dom(h) is a subset of dom(tanx) ..... ALSO at f(0), 3arccos(0)=3pi/2=pi/2

    and tan(pi/2)=undefined so f is undefined at x=pi/2

    therefore dom(f)=[-1/1]\{0}


    range? :confused:
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    Are there any other points where \tan (3 \cos^{-1} x) are undefined between -1 and 1?
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    Remember, \cos \frac{\pi}{2} = \cos \frac{3\pi}{2} = \cos \frac{5 \pi}{2} = 0
    The range is correct.
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    (Original post by SimonM)
    Are there any other points where \tan (3 \cos^{-1} x) are undefined between -1 and 1?
    Not that I can figure. I used the graphic calc do guide my solutions, and f(0) is the only one i can see. There are afew other turning points though.

    The range is correct.
    I havent done the range yet, thats what i need help with...
    I've only done the component ranges but dunno how to find the final range of f.
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    Well, you wrote down where \tan x was undefined,

    \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5 \pi}{2}, \ldots above

    And you wrote down that the range of 3 \cos^{-1} x was [0, 3 \pi)

    Now which of those undefined values appears in the range of 3 \cos^{-1} x?

    (Sorry about the range bit, I saw "Range tan x" and assumed you'd got it)

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    4000 posts, what is wrong with me?!
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    pi/2 ? :confused:
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    Isn't 0<\frac{3 \pi}{2} < 3 \pi?
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    but the domain of f is [-1,1] so 3pi/2 doesnt count
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    Yes, but the RANGE of 3\cos^{-1} includes \frac{3 \pi}{2}
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    ok i went through the whole thing again and yes the range of 3arccos(x) does include 3pi/2 ...because its between 0 and 3pi.

    i dont understand how this relates to getting the range of tan(3arccos(x)) :confused:
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    Well, if you put in (0, \pi) into \tan x, which values do you get?
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    tan(0)=tan(pi)=0

    im not sure i follow..
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    I mean, what do you get if you put in all the values between 0 and \pi
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    values of range [0, ~1.5507]
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    if u write down the working out and answer i might be able to decipher the method/reasoning behind the solution. i dont think im getting anywhere, ive been stuck on it for hours now.
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    Plot \tan x for the domain I gave you (or the range which comes from 3 \cos^{-1} x)
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    yeh i get the standard tan graph..
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    So the range will be...
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    R?
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    Yes
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    (Original post by SimonM)
    Yes
    simply because it is the higher/greater range of the two composites (f and h) ?
 
 
 
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