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Math Help - The inverse function

  1. #1
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    The inverse function

    Hey guys, was just wondering if you could help me solve an understanding issue. Was just curious if i had the function sin10x, with a restricted domain (obviously), would the inverse function be (1/10)sin-1x? I done this by simply swapping the x and y values and then rearranging the equation. If i am wrong please show me the correct working.
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by sanado View Post
    Hey guys, was just wondering if you could help me solve an understanding issue. Was just curious if i had the function sin10x, with a restricted domain (obviously), would the inverse function be (1/10)sin-1x? I done this by simply swapping the x and y values and then rearranging the equation. If i am wrong please show me the correct working.
    sin(10x) = y

    10x = arcsin(y)

    x = (0,1)(arcsin(y))

    Swap x and y

    y = (0,1)(arcsin(x))


    Note the domain of f equals the range of inverse f ; and the range of f equals the domain of inverse f. (y = f(x))
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  3. #3
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    Okay but when i sketch, y = .1arcsinx and y = sin10x, im not getting that inverse reflection look when insert y = x. Why is this?
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  4. #4
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    Quote Originally Posted by sanado View Post
    Okay but when i sketch, y = .1arcsinx and y = sin10x, im not getting that inverse reflection look when insert y = x. Why is this?
    Hmmmm, when I do this my hand hurts. Why is that?

    Tricky question to answer, isn't it? Without any other information ......

    Without any other information, I'd suggest that one or more of your sketch graphs are wrong.
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  5. #5
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    They look good to me....

    -Dan
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