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mathmath333

  • one year ago

functions

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{if }\ f(x)\ \text{is a function satisfying} \hspace{.33em}\\~\\ & f(x)\cdot f(\frac{1}{x})=f(x)+f(\frac{1}{x}),\ \ f(4)=65 \hspace{.33em}\\~\\~\\~\\ & \normalsize \text{what will be the value of }\ \ f(6) \hspace{.33em}\\~\\ & a.)\ \ 37 \hspace{.33em}\\~\\ & b.)\ \ 217 \hspace{.33em}\\~\\ & c.)\ \ 64 \hspace{.33em}\\~\\ & d.)\ \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)

  2. alekos
    • one year ago
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    anyone got an idea on how to approach this one?

  3. alekos
    • one year ago
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    i think the answer is greater than 65 maybe (b)

  4. ParthKohli
    • one year ago
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    \(65\) is an interesting number in the sense that \(65 = \color{red}4^3 + 1\). Is the answer \(6^3+1 \) then? I'm not exactly sure. It does hint us about the form of the function.

  5. anonymous
    • one year ago
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    looks like \(f(x)=x^n +1\) !

  6. ParthKohli
    • one year ago
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    Yup, that's it. And we solve for \(n\) using \(f(4) = 65\)

  7. anonymous
    • one year ago
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    Can we find all the functions satisfying the condition of this problem?\[f(x) f \left(\frac{1}{x} \right)=f(x)+f \left(\frac{1}{x} \right)\]

  8. ParthKohli
    • one year ago
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    \[\left[f(x)-1\right]\left[f(1/x) - 1\right]=1\]

  9. alekos
    • one year ago
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    so you guys are saying that n=3 ?

  10. ParthKohli
    • one year ago
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    So basically any function in the form \(h(x) = f(x) + 1\) where \(f(1/x) = 1/f(x)\), right?

  11. anonymous
    • one year ago
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    you mean \(h(1/x)=1/h(x)\), right?

  12. ParthKohli
    • one year ago
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    No, I mean that all functions \(h\) in the form \(h(x) = f(x) + 1\) satisfy this equation where \(f(1/x) = 1/f(x)\).

  13. anonymous
    • one year ago
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    @alekos I'm not sure yet

  14. alekos
    • one year ago
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    n=3 seems to work for f(6) = 217 and satisfies the original equation for f(1/6)

  15. mathmath333
    • one year ago
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    is \(217\) the correct ans

  16. welshfella
    • one year ago
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    6^3 + 1 = 217 yes

  17. welshfella
    • one year ago
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    x^3 + 1 fits the original equation

  18. mathmath333
    • one year ago
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    in book 217 is given correct

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