Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By Plato

Math Help - f ^-1 (x) = (e^x - e^-x) / 2

  1. #1
    Junior Member
    Joined
    Jan 2013
    From
    Chicago
    Posts
    27

    f ^-1 (x) = (e^x - e^-x) / 2

    That is, the inverse of (e^x - e^-x) / 2. The answer is x + sqrt(x^2 + 1). If you start with x = (e^y - e^-y ) / 2, what intermediary steps lead to the answer?

    You can clear the denominator:
    2x = e^y - e^-y;

    you can re-write e^-y as 1/e^y:
    2x = e^y - 1/e^y;

    you can then clear that denominator:
    2xe^y = e^2y - 1;

    I'm basically out of ideas at this point. Hmm term on the right is a difference of two squares. I would appreciate any comments!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,963
    Thanks
    1784
    Awards
    1

    Re: f ^-1 (x) = (e^x - e^-x) / 2

    Quote Originally Posted by kjtruitt View Post
    That is, the inverse of (e^x - e^-x) / 2. The answer is x + sqrt(x^2 + 1). If you start with x = (e^y - e^-y ) / 2, what intermediary steps lead to the answer?
    You can clear the denominator:
    2x = e^y - e^-y;
    \\2x = e^y - e^{-y}\\2xe^y=e^{2y}-1\\e^{2y}-2xe^y-1=0\\\\\text{So }e^y=\frac{2x+\sqrt{4x^2+4}}{2}
    Thanks from HallsofIvy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,421
    Thanks
    1856

    Re: f ^-1 (x) = (e^x - e^-x) / 2

    When Plato go to e^{2y}- 2xe^y- 1= 0, he thought of that as a quadratic equation, (e^y)^2- 2x e^y- 1= 0 of the form au^2+ bu+ c= 0 with a= 1, u= e^y, b= -2x, and c= -1. The quadratic formula says that the solutions to au^2+ bu+ c= 0 are of the form u= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}. Setting a= 1, b= -2x, c= -1 that gives u= e^y= \frac{-(-2x)\pm\sqrt{(-2x)^2- 4(1)(-1)}}{2(1)}= \frac{2x\pm\sqrt{4x^2+ 4}}{2}.

    But notice that 4x^2+ 4> 4x^2 so that 2x- \sqrt{4x^2+ 4}< 0 and, or course, e^y is never negative so we can drop the "-" sign to get exactly what Plato gave.

    (It can be simplified a little more: 4x^2+ 4= 4(x^2+ 1) and \sqrt{4(x^2+ 1)}= 2\sqrt{x^2+ 1} so that \frac{2x+ \sqrt{4x^2+ 4}}{2}= x+ \sqrt{x^2+ 1}.)
    Last edited by HallsofIvy; July 18th 2013 at 03:38 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2013
    From
    Chicago
    Posts
    27

    Re: f ^-1 (x) = (e^x - e^-x) / 2

    Got it thanks
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum