Let-f-R-R-be-polynomial-function-satisfying-f-x-f-1-x-f-x-f-1-x-and-f-3-28-then-f-x-is-

Question Number 171435 by cortano1 last updated on 15/Jun/22

L e t f : R \to R b e p o l y n o m i a l

f u n c t i o n s a t i s f y i n g

f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x}) a n d

f (3) = 28, t h e n f (x) i s

Commented by infinityaction last updated on 15/Jun/22

n o w f u n c t i o n i s n o t d e f i n e R \to R

Commented by infinityaction last updated on 15/Jun/22

s i r i h a v e c h a n g e d t h e q u e s t i o n

Commented by mr W last updated on 15/Jun/22

f i s R \to R m e a n s :

f o r x \in R \Rightarrow y = f (x) \in R .

i t {d o e s n}^{'} t m e a n :

f o r x \in R \Rightarrow y = f (\frac{1}{x}) \in R .

Commented by infinityaction last updated on 15/Jun/22

y o u r q u e s t i o n s h o u l d b e

A p o l y n o m i a l f u n c t i o n f (x) s a t i s f y i n g

f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x}) a n d

f (3) = 28, t h e n f (x) i s

t h e n w e k n o w t h a t

f (x) = 1 \pm x^{n}

s o f (3) = 1 \pm 3^{n} = 28

1 + 3^{n} = 28

t h e n n = 3

f (x) = 1 + x^{3}

Commented by floor(10²Eta[1]) last updated on 15/Jun/22

f is not R \to R because we have \frac{1}{x} there

Commented by floor(10²Eta[1]) last updated on 15/Jun/22

x cant be zero \Rightarrow x \notin R

Commented by mr W last updated on 15/Jun/22

t h i s i s n o t t r u e s i r .

f (\frac{1}{x}) {d o e s n}^{'} t m e a n f i s n o t R \to R!

e x a m p l e :

f (x) = x

f i s R \to R .

f (\frac{1}{x}) = \frac{1}{x}

Answered by aleks041103 last updated on 15/Jun/22

g e n e r a l s o l u t i o n

f (x) = {\begin{cases} f_{1} (x), x > 0 \\ f_{2} (x), x < 0 \end{cases}

\frac{1}{f (x)} + \frac{1}{f (1 / x)} = 1

x = e^{t}

\frac{1}{f (e^{t})} = g (t)

\Rightarrow g (t) + g (- t) = 1

\Rightarrow g (t) = \frac{1}{2} + o (t)

\Rightarrow f (x) = \frac{1}{\frac{1}{2} + o (l n (x))}

\Rightarrow f (x) = {\begin{cases} {(\frac{1}{2} + o_{1} (l n (x)))}^{- 1}, x > 0 \\ {(\frac{1}{2} + o_{2} (l n (- x)))}^{- 1}, x < 0 \\ f_{0}, x = 0 \end{cases}

w h e r e o_{1, 2} (x) a r e o d d f u n c t i o n s, i . e .

o_{1, 2} (- x) = - o_{1, 2} (x)