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Math Help - 1 Trig Problem with inverse

  1. #1
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    1 Trig Problem with inverse

    1. Given: f(x)=arcsin(x/3)+5, and -3<=x<=3

    Unknown: f(x)^-1=?

    I got y= sin^-1(x-5)*5


    Any suggestions?
    Last edited by qbkr21; November 6th 2006 at 12:19 PM.
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  2. #2
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    I'm confused on your notation. What does f(x)^1 mean? Is that supposed to be the inverse, or f^{-1}(x)?
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  3. #3
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    Re:

    yes I am sorry
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  4. #4
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    Mmk. So to do this problem, just switch x and y.

    Original: y=\sin^{-1} \left(\frac{x}{3} \right) +5

    Switch x's and y's. x=\sin^{-1} \left(\frac{y}{3} \right) +5

    \sin^{-1} \left(\frac{y}{3} \right) = x-5

    Take the sine of both sides. Can you finish it?
    Last edited by Jameson; November 6th 2006 at 12:55 PM.
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  5. #5
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    Re:

    No I got to thant point, I can finish it.
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  6. #6
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    Re:

    What happens when you take a sin^-1 back across the equal's sign. Does it change back to just plain sin ?
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  7. #7
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    \sin^{-1} \left(\frac{y}{3} \right) = x-5

    Take the sine of both sides.

    \frac{y}{3}=\sin(x-5)

    f^{-1}(x)=3\sin(x-5)
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  8. #8
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    Re:

    Thanks Jameson, you are the Master of Math!!! And God's Gift to us all!!!
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