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

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- May 30th 2009, 05:17 AM #1

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- May 30th 2009, 05:55 AM #2
$\displaystyle \left\{\begin{array}{ll}f(x)f\left(\frac{1}{x}\rig ht)=f(x)+f\left(\frac{1}{x}\right)\\f(4)=65\end{ar ray}\right. \Longleftrightarrow 65f\left(\frac{1}{4}\right)=65+f\left(\frac{1}{4}\ right)\Longleftrightarrow f\left(\frac{1}{4}\right)=\frac{65}{64}$

**(1)**

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

Now use**(1)**and**(2)**.

- May 30th 2009, 08:56 AM #3
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 $\displaystyle f(x) = 1+x^3$, which would make $\displaystyle f(6)=217$.

- May 30th 2009, 04:36 PM #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 $\displaystyle f(4)=65$ and $\displaystyle f(\alpha\cdot 4)=\alpha \cdot 65$

Try posting the question exactly as it was given.

- May 30th 2009, 04:48 PM #5
The question actually is that :

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

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

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

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

$\displaystyle

f(x)=1\pm x^n

$

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