# 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 ?

• There is no way to find $f(4)$ from just those conditions. Oct 8, 2019 at 16:17
• 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 Oct 8, 2019 at 16:20
• $(f(x)-1)(f(\frac{1}{x})-1) =1$
– user655800
Oct 8, 2019 at 16:24
• Are you assuming . $f(x)$ is a polynomial? Oct 8, 2019 at 16:37

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:

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

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 $$f(x)=\frac{g(x)}{g(x)-1}$$

• Kinda did the same thing but the answer isn't matching 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 $$f(x)=1+x^3.$$

• This is true if $f$ is known to be a polynomial. Oct 8, 2019 at 18:29
• 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? Oct 9, 2019 at 2:37
• I understood the starting statements but why have u taken g(X) = Kx^m can u please explain cause your answer is correct Oct 9, 2019 at 2:40
• @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. Oct 9, 2019 at 2:49
• Okay??? Sorry didn't get that Oct 10, 2019 at 3:06