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Math Help - Sketching y = x arccos(x)

  1. #1
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    Sketching y = x arccos(x)

    The question:
    Sketch the following curve, showing its main features:
    y = xcos^{-1}(x)

    I'm not sure how to attempt this. I know that the inverse of cosine is restricted to [-1, 1] in the domain, and must be restricted on the range to allow it to have an inverse, i.e. [0, \frac{\pi}{2}] I know the graph passes through the origin too. I'm stumped in regards to sketching it. I don't think I have all the information. Any assistance would be great!
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  2. #2
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    Quote Originally Posted by Glitch View Post
    The question:
    Sketch the following curve, showing its main features:
    y = xcos^{-1}(x)

    I'm not sure how to attempt this. I know that the inverse of cosine is restricted to [-1, 1] in the domain, and must be restricted on the range to allow it to have an inverse, i.e. [0, \frac{\pi}{2}] I know the graph passes through the origin too. I'm stumped in regards to sketching it. I don't think I have all the information. Any assistance would be great!
    The domain is [-1, 1]. Yes, (0, 0) is on the curve. I suggest you plot points eg. endpoints (the points where x = 1, -1), 1/2, -1/2 etc. and then try to fit a curve. Are you expected to find turning points? If so, calculus is required.
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  3. #3
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    Thanks. The question didn't mention anything about finding turning points. I did that anyway, and it only has one in the interval.
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  4. #4
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    Quote Originally Posted by Glitch View Post
    Thanks. The question didn't mention anything about finding turning points. I did that anyway, and it only has one in the interval.
    plot y = x Arccos[x] - Wolfram|Alpha
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  5. #5
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    It's an ugly thing, isn't it? :P
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  6. #6
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    Quote Originally Posted by Glitch View Post
    It's an ugly thing, isn't it? :P
    Everything is relative ....
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