mathmath333
  • mathmath333
functions
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{if }\ f(x)\ \text{is a function satisfying} \hspace{.33em}\\~\\ & f(x)\cdot f(\frac{1}{x})=f(x)+f(\frac{1}{x}),\ \ f(4)=65 \hspace{.33em}\\~\\~\\~\\ & \normalsize \text{what will be the value of }\ \ f(6) \hspace{.33em}\\~\\ & a.)\ \ 37 \hspace{.33em}\\~\\ & b.)\ \ 217 \hspace{.33em}\\~\\ & c.)\ \ 64 \hspace{.33em}\\~\\ & d.)\ \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)
alekos
  • alekos
anyone got an idea on how to approach this one?
alekos
  • alekos
i think the answer is greater than 65 maybe (b)

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ParthKohli
  • ParthKohli
\(65\) 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.
anonymous
  • anonymous
looks like \(f(x)=x^n +1\) !
ParthKohli
  • ParthKohli
Yup, that's it. And we solve for \(n\) using \(f(4) = 65\)
anonymous
  • anonymous
Can we find all the functions satisfying the condition of this problem?\[f(x) f \left(\frac{1}{x} \right)=f(x)+f \left(\frac{1}{x} \right)\]
ParthKohli
  • ParthKohli
\[\left[f(x)-1\right]\left[f(1/x) - 1\right]=1\]
alekos
  • alekos
so you guys are saying that n=3 ?
ParthKohli
  • ParthKohli
So basically any function in the form \(h(x) = f(x) + 1\) where \(f(1/x) = 1/f(x)\), right?
anonymous
  • anonymous
you mean \(h(1/x)=1/h(x)\), right?
ParthKohli
  • ParthKohli
No, I mean that all functions \(h\) in the form \(h(x) = f(x) + 1\) satisfy this equation where \(f(1/x) = 1/f(x)\).
anonymous
  • anonymous
@alekos I'm not sure yet
alekos
  • alekos
n=3 seems to work for f(6) = 217 and satisfies the original equation for f(1/6)
mathmath333
  • mathmath333
is \(217\) the correct ans
welshfella
  • welshfella
6^3 + 1 = 217 yes
welshfella
  • welshfella
x^3 + 1 fits the original equation
mathmath333
  • mathmath333
in book 217 is given correct

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