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 ?

  • $\begingroup$ There is no way to find $f(4)$ from just those conditions. $\endgroup$ Oct 8, 2019 at 16:17
  • $\begingroup$ Literally there was just the functional equation given and the value of f(3) in the question and when I checked for it's solution the was nothing just the equivalent function in the solution which is equal to f(X) which isn't the equivalent function that I concluded $\endgroup$ Oct 8, 2019 at 16:20
  • 2
    $\begingroup$ $(f(x)-1)(f(\frac{1}{x})-1) =1$ $\endgroup$
    – user655800
    Oct 8, 2019 at 16:24
  • $\begingroup$ Are you assuming . $f(x)$ is a polynomial? $\endgroup$ Oct 8, 2019 at 16:37

3 Answers 3


Give any $g:[1,\infty)\to\mathbb R\setminus\{1\}$ such that $g(1)=2,$ you can define:

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

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:


Then you can show that $p(x)=\pm x^m$ for some natural number $m.$


HINT.-If you want $ab=a+b$ and $a=f(3)=28$, since $a=\dfrac{b}{b-1}$ and $b=\dfrac{a}{a-1}$ you can take an arbitrary function $g(x)$ as $g(x)=f(\frac1x)$ such that $g(3)=\dfrac{28}{27}$ and $g(1)=2$ and $g(x)\ne1$ for all $x$ so you get an example of solution defined by $$f(x)=\frac{g(x)}{g(x)-1}$$

  • $\begingroup$ Kinda did the same thing but the answer isn't matching $\endgroup$ Oct 9, 2019 at 2:36

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 $f(x)=1+x^3.$

  • $\begingroup$ This is true if $f$ is known to be a polynomial. $\endgroup$ Oct 8, 2019 at 18:29
  • $\begingroup$ There wasn't given that f is an invertible function so I think we can assume it to be polynomial as no additional info was given right? $\endgroup$ Oct 9, 2019 at 2:37
  • $\begingroup$ I understood the starting statements but why have u taken g(X) = Kx^m can u please explain cause your answer is correct $\endgroup$ Oct 9, 2019 at 2:40
  • $\begingroup$ @aditya prakash the $g(x)g(1/x)=1$ suggests that $g(x)$ can only be mono-nomial (one term) and the two constantants $K,m$ are allowed. You can check that we cannot add even a constant to $g(x)$. Thais means that even $g(x)=Kx^m+L, L\ne 0$ cannot be the solution. $\endgroup$
    – Z Ahmed
    Oct 9, 2019 at 2:49
  • $\begingroup$ Okay??? Sorry didn't get that $\endgroup$ Oct 10, 2019 at 3:06

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