How to solve this functional equation given
$$ f(x)f(1/x) =f(x) + f(1/x) $$
and
$$  f(3) = 28 $$
then find the value of $f(4)$
the way i attempted the problem was that i found the value of $f(x)$ in terms of $f(1/x)$ which came put to be
$$ f(x) =\frac { f(\frac 1x)}{(f(\frac 1x) - 1)}$$
assuming that on putting$ x=3 \space \space \space , \text the \space \space RHS = 28$

So i assumed $f(\frac13)$ as $t$ and then solved for $t$ which gave me the value of $\frac {29}{27}$ so by judging this value ,  I  assumed 
$$t =\frac{(9x+2)}{9x}$$
 now putting this value of $t$ which i assumed to be $f(1/x)$ in RHS the function simplified to become $$ f(x) = 9x + 1 $$ so putting $x =3 \space $,  I absolutely get $f(3) = 28$ but putting $x =4$ it doesn't gives $65$ which is the answer. 

So can someone please help me which is the correct approach and method and tell where I went wrong ?
Give any $g:[1,\infty)\to\mathbb R\setminus\{1\}$ such that $g(1)=2,$ you can define:

$$f(x)=\begin{cases}g(x)&amp;x\geq 1\\ \frac{g(1/x)}{g(1/x)-1}&amp;x\in(0,1)\end{cases}$$

Then $f(x)$ satisfies your property. 

In particular, you can have any value of $f(4)=g(4)$ given the known value of $f(3).$



If $f$ is known to be a polynomial, then if you define $p(x)=f(x)-1$ then you have:

$$p(x)p(x^{-1})=1$$

Then you can show that $p(x)=\pm x^m$ for some natural number $m.$
Note to OP: Check that $f(x)=9x+1$ does not satisfy the original functional equation, which it has to.

Let $f(x)=A, f(1/x)=B; A-1=g(x), B-1=g(1/x)$, then $$AB=A+B \implies (A-1)(B-1)=1 \implies g(x)=\frac{1}{g(1/x)} \implies g(x)=K x^m, K=\pm 1$$,
So $$g(x)=\pm x^m \implies f(x)=1\pm x^m$$ the given condition $f(3)=28$, fixes $m=3$ and  $+$ sign here. Finally, we get $f(x)=1+x^3.$