# Function Problem

• May 28th 2011, 08:10 AM
avik
Function Problem
Given f(x)=65 and f(x)f(1/x)=f(x)+f(1/x)... find the value of f(6)=???
• May 28th 2011, 08:21 AM
Plato
Quote:

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.
• May 28th 2011, 08:24 AM
avik
Yes I'm Sorry. :( That is f(4)=65.
• May 29th 2011, 12:52 AM
Opalg
There is not enough information here for the problem to have a unique solution. I suspect that the intended answer is that \$\displaystyle 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 \$\displaystyle f(x) = x^{g(x)}+1\$ is a solution, and f(6) can be a completely arbitrary number greater than 1.
• May 29th 2011, 09:31 AM
avik
Quote:

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 \$\displaystyle 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 \$\displaystyle 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(Clapping). Can you please explain how did you deduce \$\displaystyle f(x) = x^3+1\$ from the problem statement? please reply..
• May 29th 2011, 11:33 PM
Opalg
Quote:

Originally Posted by avik
Can you please explain how did you deduce \$\displaystyle f(x) = x^3+1\$ from the problem statement? please reply..

The equation \$\displaystyle f(x)f(1/x)=f(x)+f(1/x)\$ can be written as \$\displaystyle \bigl((f(x)-1\bigr)\bigl((f(1/x)-1\bigr) = 1.\$ So if \$\displaystyle g(x) = f(x)-1\$ then \$\displaystyle g(1/x) = 1/g(x)\$. There are very many functions having that property, but the most obvious solutions are \$\displaystyle g(x) = x^k\$ for some fixed k.

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