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Equation relating a function and its inverse

  1. Oct 25, 2012 #1
    so here's the question: if you have some equation relating a function, f(x), and its inverse, f-1(x), can you solve for the function?

    for example, solve for f(x):

    f(x)+f-1(x)=x^2

    how about:

    f(x)+f-1(x)=g(x)

    my math teacher (AP calc) was stumped on this one... any thoughts?
     
  2. jcsd
  3. Oct 25, 2012 #2

    rcgldr

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    By inverse, do you mean

    f -1(x) = 1 / f(x)

    or that

    if f(x) = y, then f -1(y) = x?
     
  4. Oct 25, 2012 #3
    by inverse i mean f(f-1(x))=x... e.g. ln(e^x)=x
     
  5. Oct 25, 2012 #4

    Mute

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    What if you differentiated the equation? Can you relate the derivative of the function and its inverse? (hint)
     
  6. Oct 26, 2012 #5
    I suppose you could differentiate both sides. Then you would get:

    f'(x)+f-1'(x)=2x

    f'(x)+1/(f'(f-1(x)))=2x

    but this equation still does not get rid of the f-1(x)... do you see something I don't?
     
  7. Oct 26, 2012 #6

    Mute

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    The way the equation is written on the wikipedia page somewhat obscures the point:

    $$\frac{df^{-1}(x)}{dx} = \frac{1}{\frac{df(x)}{dx}}$$

    (where the inverse exists)
     
  8. Oct 26, 2012 #7

    mathwonk

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    your question is a little strange, since f + f^-1 is only defined where both f and f^-1 are defined. but the domain of f^-1 equals the range of f, so they may not have any common domain at all.

    but lets say they do have a common domain interval, e.g. the whole real line.

    if you want a continuous solution there cannot be any.

    i.e. a continuous invertible function on the line is monotone, say increasing. but then also its inverse is increasing, hence also their sum,

    but x^2 is not increasing. so there do not exist any continuous, much less differentiable, f and f^-1 defined on all of R satisfying your equation.
     
  9. Oct 26, 2012 #8

    mathwonk

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    now lets take a finite interval, say [0,1]. then either f(0) = 0 and f(1) = 1 and f is increasing or else f(0) = 1 and f(1) = 0 and f is decreasing. lets assume increasing.

    then both f and f^-1 are non negative, and their sum equals x^2, hence both have values everywhere less than or equal to x^2.

    this is also impossible, since the graphs of f and f^-1 are symmetrical about the line y=x.

    Thus if f has values less than x^2, then f^-1 has values larger than x^2, contradiction.
     
  10. Oct 26, 2012 #9

    mathwonk

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    now lets take a half infinite interval like [a,infinity). then f(a) = a and also f^-1(a) = a, so their sum is 2a, not a^2, unless a=0 or 2. if a=0 i think we already did it above, if a = 2, then f^-1(2) = 2, and both f and f^-1 are increasing so again the graphs are both above the line y=x a contradiction.


    you might try to think of any examples at all of f and f^-1 where their sum is a polynomial function of x.


    or maybe you could look at examples of functions f and f^-1 with different domains, but where f+f^-1 = x^2, on some interval common to their domains. that would be harder to study.
     
    Last edited: Oct 26, 2012
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