1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Equation relating a function and its inverse
  1. Inverse functions (Replies: 3)

  2. Inverse function (Replies: 12)

  3. Inverse of a function (Replies: 6)

  4. Inverse function (Replies: 5)

Loading...