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Math Help - Inverse Function!!

  1. #1
    Newbie
    Joined
    Jan 2008
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    Inverse Function!!

    Hi, I am in the middle of solving this problem:
    Let f(x) = 3 +x^2 + tan (πx/2), where -1 < x < 1.
    a.) Find the(inverse) f^1 (3).
    b.) Find f(f^-1(5)).

    I know the steps on how to find the inverse equation, but I do not know how to solve for x, in order to get it alone, because of the tan[(pi)(x/2)]. I do not remember how to get x alone when it is in sin (x), tan (x), etc. Do you multiply everything by tan x in order to cancel it out?

    I'm pretty sure the second part requires a cancellation equation, but then again I'm stuck on the tan issue. Trigonemtric functions confuse me .

    Thank you so much for your help!!
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  2. #2
    Senior Member
    Joined
    Dec 2007
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    Melbourne
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    I do not know how to solve for x, in order to get it alone, because of the tan[(pi)(x/2)].
    This cannot be done. In this case there is no way of expressing f^-1(x) in terms of x. However, you do not need to solve for x to get an answer.

    It sounds like you have already got to x = 3 +y^2+\tan (\frac { \pi y}{2})
    Now substitute x = 3 to get

    3 = 3 +y^2+\tan (\frac {\pi y}{2})
    y^2+\tan (\frac {\pi y}{2}) = 0

    It is easy to see that y = 0 is a solution to this, just by recognising that tan (0) = 0.

    b.) Find f(f^-1(5)).
    You are expected to know that f(f^-1(x)) = x
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