Functional Equation (Probably needs a CAS)

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Homework Statement


If f'(x)>0 for all real positive x, where f:R+ ---> R and
f(x)+(1/x)=f-1(1/(f(x))),

f-1(1/(f(x)))>0 for all x>0. Find all the possible values of (i) f(2),(ii) f'(2) and (iii) Limit (x f(x)) as x ----->0 .


The Attempt at a Solution


Guessing from the last part, I intuitively, took f(x) = k/x. Using this, I get a quadratic in k which yields that k might be \frac{1 (+/-) \sqrt{5} }{2}. But this was just a guess. And there might still be other functions which satisfy that equation. I wonder if they could be manually solved for.
All that I need from you guys is to check out if all the possible solutions of this equation can be solved manually or not. If not, please let me know how can I get solutions to this equation using a Computer Algebra System. Actually, I have both Matlab R2008a and Mathematica 7 but I don't know how to use them to solve functional equations involving inverses. If anyone of you could provide me the syntax for these CAS on how to solve functional equations, I shall be very grateful.
Thanks.
Ritwik
 
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Hi Ritwik! Thanks for the PM! :smile:

Yes, f(x) = (1 ± √5)/2x certainly works …

I think you've done very well to get that!

Sorry, but I haven't a clue how to get any other solution :redface:

(and I don't know any computing)

 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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