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Inverse functions?

  1. Apr 11, 2005 #1
    Inverse functions???

    I was just wondering...Isnot y=x the only function that is the inverse of itself??? How do you go on about saying it in a formal way??? I mean as response to your assignment question. where you are not just told to not only give fatcts but "Justify your claim with appropriate proofs!"???

    K. Cv.
  2. jcsd
  3. Apr 11, 2005 #2


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    There's a nice graphic-related proof.A function that is invertible on a domain is bijective,which means both injective & surjective.In terms of graphs,the graph of the function and its inverse (if it exists) are symmetrical wrt the graph of f(x)=x.If the two functions coincide,then they coincide with the identity function.

  4. Apr 11, 2005 #3
    Do you mean to say, y=x is the only function where [tex]f(x)=f^{-1}(x)[/tex]?

    Proof by counterexample:

  5. Apr 11, 2005 #4


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    Any function which is symmetrical from the line y=x is the inverse of itself, in that sense there will be an infinite number of them.

    Although perhaps very few simple functions, another easy one is [tex]y^2 + x^2 = r^2[/tex]
  6. Apr 11, 2005 #5


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    The identity function is invertible on all R...

  7. Apr 11, 2005 #6


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    how about y = -x, y = -1/x,....
  8. Apr 11, 2005 #7
    If your question was simply "prove of disprove: y = x is the only function that is its own inverse", you can just give a counterexample and be done. Is this how the question you were given was worded?
  9. Mar 1, 2007 #8
    precalculus question


    I have a similar question; I'm currently taking precalculus, and I have a question here that I'm stuck on:

    "Let f(x)=(ax+b)/(cx-a), (cx-a)?0. Show that f(x) is it's own inverse."

    I understand, graphically, that f(x) equals its inverse in this case, because they are reflections of each other over y=x. I've graphically solved this using technology and by-hand sketches. And I also understand that the restiction on the denominator come about because the equation would then be undefined (if it equalled 0). But I cannot figure out why this is, algebraically.

    Can anyone help me, or give me a hint?

    Thanks very much,
  10. Mar 1, 2007 #9
    Just compute f(f(x)).
  11. Mar 1, 2007 #10


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    Sam, just start with [tex]y = \frac{ax+b}{cx-a}[/tex] and then rearrage it to get "x" by itself on the LHS.

    Whats happening. Are you just getting stuck on the algebra to do this rearrangement?
  12. Mar 1, 2007 #11
    [tex]y^2 + x^2 = r^2[/tex] is not a function, because for any value of x < r, you can get two different values of y.
  13. Mar 1, 2007 #12
    A silly conjecture... An infinite number of counterexamples.
  14. Mar 2, 2007 #13

    Thanks to ZioX and uart for giving me the helpful direction.:smile: I think what I must have done before was just think too far ahead, and presume that it couldn't be simplified in that way. Thanks again! It turns out it actually ends up simplifying to just:


  15. Mar 3, 2007 #14
    what is the inverse function of sin x. I don't know if it is written correctly
  16. Mar 3, 2007 #15

    matt grime

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    arcsin, or sin^{-1}

    and if you think that it is avoiding the issue, then you need to rethink your view of what a function is.
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