Results 1 to 3 of 3

Math Help - inverse

  1. #1
    Senior Member
    Joined
    Feb 2010
    Posts
    456
    Thanks
    34

    inverse

    if f:[1,infinity)-[1,infinity) is defined by f(x)-2^(x(x-1)) then find inverse of f(x)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Certainly y = f(x) is strictly increasing in [1,+infty) and

    lim_ {x to +infty} f(x) = +infty

    This means that

    f : [1,+infty) -> [1,+infty)

    is bijective, as a consequence there exists f^{ -1 } .

    Take log in both sides of y = f(x) , solve the quadratic equation on x and choose the positive branch.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    There is some problem of 'interpretation', so I suppose that the function is...

    y= 2^[x (x-1)] (1)

    ... so that its inverse function is the solution(s) of the equation...

    x^2 - x - log2 y=0 (2)

    ... that are...

    x= 1/2 [1 +/- (1+ 4 log2 y)^(1/2)] (3)

    It has to be noted that (3) has two distinct brantches and is defined for y> 2^(- 1/4)...

    Kind regards

    chi sigma
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. left inverse equivalent to inverse
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: June 30th 2011, 03:58 PM
  2. Inverse tan and inverse of tanh
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 31st 2010, 06:20 AM
  3. secant inverse in terms of cosine inverse?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: August 30th 2010, 08:08 PM
  4. Replies: 1
    Last Post: April 9th 2010, 05:51 PM
  5. inverse trig values and finding inverse
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 6th 2009, 12:04 AM

Search Tags


/mathhelpforum @mathhelpforum