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Using IFT to find inverse.

  1. Feb 4, 2012 #1
    How does one use IFT to find the inverse of a function? I thought it was something like [tex]\int \frac{dx}{df(x)}dx[/tex]. But that doesn't work with f(x)=x^2:[tex]\int \frac{dx}{2x}=\frac{1}{2} \log{x} \neq f^{-1}(x)[/tex].
  2. jcsd
  3. Feb 4, 2012 #2
    BTW, IFT = Inverse Function Theorem.

    The reason I asked is because I remember something like this from calc II and I was wanting to use it to find the inverse of a function.
  4. Feb 5, 2012 #3


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    You seem to be mixing up inverse with reciprocal. Example: the inverse of y=x2 is x = √y, not y= 1/x2.
  5. Feb 5, 2012 #4
    Thanks for the reply! I was beginning to wonder if I was so incorrigibly wrong that no one was even willing to try. :P

    The reciprocal is the multiplicative inverse, and multiplication is a function that forms a group over the rationals. The inverse of a function is the function created by mapping each element of the range to the element of the domain that maps to it (and it only exists for one to one functions). I'm generally advanced in math, I've just been slipping a little because I don't have any math classes this year.

    I got 1/(2x) from dx/(df(x)). df(x) = 2x dx.

    The theorem I'm referring to is stated here: https://en.wikipedia.org/wiki/Inverse_function_theorem#Statement_of_the_theorem. From that theorem, I should be able to integrate 1/f'(x) to find f inverse. Like this: [tex]f^{-1}(f(x))=\int \frac{dx}{f'(x)}[/tex]. However, this doesn't work with f(x)=x^2 (at least not for me, see my first post). I think I'm either integrating incorrectly or integrating in terms of the wrong variable. So, I guess my question should be: "How is IFT applied to find the inverse of x^2"?
  6. Feb 6, 2012 #5


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    f-1(f(x)) = x.
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