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Question on functions and trigonometry.

  1. Mar 7, 2010 #1
    Sorry, for overloading you guys, but I just can't get my head around these!

    Question 1
    1. The problem statement, all variables and given/known data

    Show that, when restricted to the domain [-1,1], g has an inverse function. Use a definition of a theorem, a graph will not be accepted.


    2. Relevant equations

    g(x) = x/(x^2 + 1)

    3. The attempt at a solution

    I tried solving this first by going

    y = x/(x^2 + 1)
    then making x the subject... so..

    yx^2 + y = x
    yx^2 - x + y = 0

    using quadratic formula, i end up with

    g^-1(x) = (1+- root(1 - 4x^2) ) / 2x

    But I don't get how to prove its an inverse function when x is between -1 and 1.. :(



    Question 2...
    1. The problem statement, all variables and given/known data

    Simplify sin(2tan^-1(x)), tan^-1 is arctan, and state where your simplification is valid.


    3. The attempt at a solution

    I made a right angled triangle, with adjacent length 1, opposite length x and hypotenuse length root(1 + x^2).

    But I don't know how to deal with that 2...



    Thanks (in advance) for helping :D
     
  2. jcsd
  3. Mar 7, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi mgnymph! :smile:

    (have a square-root: √ and a ± and try using the X2 tag just above the Reply box :wink:)
    That's inside-out1 :rolleyes:

    You mean x = (1 ± √(1 - 4y2))/2y.

    But anyway, wouldn't it be easier just to prove that g(x) is always increasing (by calculus or by a trig substitution)? :wink:
    Rewrite it …

    if y = tanx, then sin2y = … ? :smile:
     
  4. Mar 7, 2010 #3
    oh! Thanks for opening my eyes :D
     
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