1. ## Function Problem

Given f(x)=65 and f(x)f(1/x)=f(x)+f(1/x)... find the value of f(6)=???

2. Originally Posted by avik
Given f(x)=65 and f(x)f(1/x)=f(x)+f(1/x)... find the value of f(6)=???
Surely that is a mistake (typo?) in what you have posted.

3. Yes I'm Sorry. That is f(4)=65.

4. There is not enough information here for the problem to have a unique solution. I suspect that the intended answer is that $f(x) = x^3+1$, and so f(6) = 217. But in fact if g(x) is an arbitrary function such that g(1/x) = g(x) and g(4)=3, then $f(x) = x^{g(x)}+1$ is a solution, and f(6) can be a completely arbitrary number greater than 1.

5. Originally Posted by Opalg
There is not enough information here for the problem to have a unique solution. I suspect that the intended answer is that $f(x) = x^3+1$, and so f(6) = 217. But in fact if g(x) is an arbitrary function such that g(1/x) = g(x) and g(4)=3, then $f(x) = x^{g(x)}+1$ is a solution, and f(6) can be a completely arbitrary number greater than 1.
This is what given in the problem. Anyways, that 217 answer is correct. Can you please explain how did you deduce $f(x) = x^3+1$ from the problem statement? please reply..

6. Originally Posted by avik
Can you please explain how did you deduce $f(x) = x^3+1$ from the problem statement? please reply..
The equation $f(x)f(1/x)=f(x)+f(1/x)$ can be written as $\bigl((f(x)-1\bigr)\bigl((f(1/x)-1\bigr) = 1.$ So if $g(x) = f(x)-1$ then $g(1/x) = 1/g(x)$. There are very many functions having that property, but the most obvious solutions are $g(x) = x^k$ for some fixed k.

If $f(4) = 65$ then $g(4) = 64 = 4^3$. That suggests that we should take k=3, giving $g(x) = x^3$ and hence $f(x) = x^3+1.$

But as I said in the previous comment, there is nothing unique about that solution. If there really was no further information given, then this is a very badly thought out problem.