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Thread: f(x).f(1/x)

  1. #1
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    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
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  2. #2
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    $\displaystyle \left\{\begin{array}{ll}f(x)f\left(\frac{1}{x}\rig ht)=f(x)+f\left(\frac{1}{x}\right)\\f(4)=65\end{ar ray}\right. \Longleftrightarrow 65f\left(\frac{1}{4}\right)=65+f\left(\frac{1}{4}\ right)\Longleftrightarrow f\left(\frac{1}{4}\right)=\frac{65}{64}$ (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).
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  3. #3
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    Quote Originally Posted by Curious_eager View Post
    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 $\displaystyle f(x) = 1+x^3$, which would make $\displaystyle f(6)=217$.
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  4. #4
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    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.
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  5. #5
    Senior Member pankaj's Avatar
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    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,$\displaystyle f(x)=1+x^3$
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