Question

functions

Answer 1

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

Answer 2

anyone got an idea on how to approach this one?

Answer 3

i think the answer is greater than 65 maybe (b)

Answer 4

\(65\) 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.

Answer 5

looks like \(f(x)=x^n +1\) !

Answer 6

Yup, that's it. And we solve for \(n\) using \(f(4) = 65\)

Answer 7

Can we find all the functions satisfying the condition of this problem?\[f(x) f \left(\frac{1}{x} \right)=f(x)+f \left(\frac{1}{x} \right)\]

Answer 8

\[\left[f(x)-1\right]\left[f(1/x) - 1\right]=1\]

Answer 9

so you guys are saying that n=3 ?

Answer 10

So basically any function in the form \(h(x) = f(x) + 1\) where \(f(1/x) = 1/f(x)\), right?

Answer 11

you mean \(h(1/x)=1/h(x)\), right?

Answer 12

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

Answer 13

@alekos I'm not sure yet

Answer 14

n=3 seems to work for f(6) = 217 and satisfies the original equation for f(1/6)

Answer 15

is \(217\) the correct ans

Answer 16

6^3 + 1 = 217
yes

Answer 17

x^3 + 1 fits the original equation

Answer 18

in book 217 is given correct