## mathmath333 one year ago functions

1. 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}}

2. alekos

anyone got an idea on how to approach this one?

3. alekos

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

4. 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.

5. anonymous

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

6. ParthKohli

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

7. anonymous

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

8. ParthKohli

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

9. alekos

so you guys are saying that n=3 ?

10. ParthKohli

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

11. anonymous

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

12. ParthKohli

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

13. anonymous

@alekos I'm not sure yet

14. alekos

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

15. mathmath333

is $$217$$ the correct ans

16. welshfella

6^3 + 1 = 217 yes

17. welshfella

x^3 + 1 fits the original equation

18. mathmath333

in book 217 is given correct