1. ## f(x).f(1/x)

If f(x).f(1/x) = f(x) + f(1/x) and f(4) = 65,

what will be the value of f(6) ?

Any help would be greatly appreciated.

Thanks,
Anshu

2. $\left\{\begin{array}{ll}f(x)f\left(\frac{1}{x}\rig ht)=f(x)+f\left(\frac{1}{x}\right)\\f(4)=65\end{ar ray}\right. \Longleftrightarrow 65f\left(\frac{1}{4}\right)=65+f\left(\frac{1}{4}\ right)\Longleftrightarrow f\left(\frac{1}{4}\right)=\frac{65}{64}$ (1)

Now if $f$ is a homogeneous function, then $f(\alpha x)=\alpha f(x)$. Thus $f(6)=f\left(\frac{24}{4}\right)=24f\left(\frac{1}{ 4}\right)$ (2)

Now use (1) and (2).

3. Originally Posted by Curious_eager
If f(x).f(1/x) = f(x) + f(1/x) and f(4) = 65,

what will be the value of f(6) ?
The answer could be almost anything, since the given information only determines the values of the function at x=4 and x=1/4. Probably the simplest function to satisfy the functional equation is $f(x) = 1+x^3$, which would make $f(6)=217$.

4. If it's a homogeneous function you don't even need to go through all the steps that I showed since you can just use the fact that $f(4)=65$ and $f(\alpha\cdot 4)=\alpha \cdot 65$

Try posting the question exactly as it was given.

5. The question actually is that :

If $f(x)$ is a polynomial function satisfying the condition that

$f(x)f(\frac{1}{x})=f(x)+f(\frac{1}{x})$

and $f(4)=65$.Then what is value of $f(6)$.

On assuming a poloynomial of degree $n$ and plugging it in the given condition and equating the coefficients we get

$
f(x)=1\pm x^n
$

Consequently, $f(x)=1+x^3$