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The Inverse Function of f(x)=x^2+x

  1. Aug 21, 2011 #1
    I'm trying to work out the method of getting the inverse function of

    [itex] f(x) = x^2 + x [/itex]


    I already know the inverse but I would like to know the method used to obtain it. So far I have:

    Made f(x) = y:
    [itex]y = x^2+2[/itex]

    And then made y = x and x = y:
    [itex]x=y^2+y[/itex]

    And then I did this but I'm not sure if it's correct:
    [itex]x=y(y+1)[/itex]



    Apparently the solution is this
    [itex]x(y) = 1/2 (-1±sqrt(4 y+1))[/itex]

    But I need to know the steps to get that. Hope you can help. :)
     
  2. jcsd
  3. Aug 21, 2011 #2

    micromass

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    OK, let's leave from here. What you did afterwards is correct, but it won't help.
    You have

    [tex]y^2+y-x=0[/tex]

    this is a quadratic equation in y, so it can be solved with the quadratic formula. What does that give you?
     
  4. Aug 21, 2011 #3
    Oooh right so

    a = 1
    b = 1
    c = -x

    [itex]y=-1\pm \sqrt{(1+4x)}/2[/itex]

    I think

    Also sorry, I really fail with these itex tags.
     
    Last edited: Aug 21, 2011
  5. Aug 21, 2011 #4

    HallsofIvy

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    By the way, since a real valued function of a real variable is "single valued", that does not have a true "inverse". What you are saying is that you can divide it into two functions, on either side of the vertex of the parabola, one having inverse function [itex]f^{-1}(x)= (1/2)(-1+ \sqrt{4x+ 1})[/itex] and the other having inverse [itex]g^{-1}(x)= (1/2)(-1- \sqrt{4x+ 1})[/itex]
     
  6. Nov 29, 2011 #5
    try this:

    http://www.sosmath.com/algebra/quadraticeq/quadraformula/quadraformula.html

    y(x) = a(x^2)+bx+c

    0 = a(x^2)+bx+(c-y)

    ...

    x(y) = (1/2a)(-b+(b^2 - 4a(c-y))^.5)

    which is just the general quadratic with c replaced by (c-Y) so that x becomes a function of y. Abel showed this can't be done in general for polynomials with a finite number of arithmatic operations.

    I would like to know how to the math symbols in something other than text. Is that what badballer was refering to in "itex tags"?
     
  7. Feb 7, 2013 #6
    I do not know the formula, but know the answer. Your function x^2+x looks like not-complete standart quadratic equation
    a^2+2ab+b^2
    (where a^2=x^2, 2ab=x, but missing b^2),
    if it would be complete we could write it as
    (a+b)^2.
    To make it complete we add missing b^2,
    if our a=x and 2ab=x, then our b is 1/2.
    Our b^2 is 1/4, so we add it on both sides:
    y+1/4=x^2+x+1/4.
    y+1/4=(x+1/2)^2.
    Square root both sides:
    x+1/2=±sqrt(y+1/4)
    move 1/2 from left side to right
    x=±sqrt(y+1/4)-1/2.
    Now you can transform right side of your equation to what you have
    x=1/2(±2sqrt(y+1/4)-1).
    Put 2 under square root:
    x=1/2(±sqrt4(y+1/4)-1)
    Then just do simple math
     
  8. Feb 7, 2013 #7

    SammyS

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    You have just answered a thread that is more than 1 year old.
     
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