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Math Help - Tangent problem

  1. #1
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    Tangent problem

    f(x) = 3 + x^2 + tan(pi(x/2)), where -1 < x < 1.
    Find f^-1 (3).

    I got up to:
    f^-1 (x) = 3 + y^2 + tan(pi(y/2)), where -1 < y < 1
    then subing x= 3
    I get 0 = y^2 + tan(pi(y/2))
    And I'm stuck again!
    What should I do with the y within the tangent
    Any advice and hint is greatly appreciated ! Thanks !
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  2. #2
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    Hint: The easiest way for x^2 + \tan \left( \frac{\pi x}{2} \right) = 0 is for both terms to be zero in the first place.
    Last edited by mr fantastic; September 20th 2009 at 02:15 AM. Reason: Made the reply relevant to the OP
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  3. #3
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    Quote Originally Posted by letzdiscuss View Post
    f(x) = 3 + x^2 + tan(pi(x/2)), where -1 < x < 1.
    Find f^-1 (3).

    I got up to:
    f^-1 (x) = 3 + y^2 + tan(pi(y/2)), where -1 < y < 1
    then subing x= 3
    I get 0 = y^2 + tan(pi(y/2))
    And I'm stuck again!
    What should I do with the y within the tangent
    Any advice and hint is greatly appreciated ! Thanks !
    By inspection, y = 0 is a solution. You should prove that this is the only solution that satisfies -1 < x < 1.
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  4. #4
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    I'm sorry but how would I prove y = 0. Could you please give me some more hints?
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  5. #5
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    Quote Originally Posted by letzdiscuss View Post
    I'm sorry but how would I prove y = 0. Could you please give me some more hints?
    Try substituting y = 0 into 0 = y^2 + tan(pi(y/2)). Does it work?
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  6. #6
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    Yes it worked ! So my answer would be f^-1 (3) = 0.
    Thanks a lot !!!
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