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


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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
	$general solution f(x)= { ((f_1 (x),x>0)),((f_2 (x),x<0)) :} (1/(f(x))) +(1/(f(1/x)))=1 x=e^t (1/(f(e^t )))=g(t) ⇒g(t)+g(−t)=1 ⇒g(t)=(1/2)+o(t) ⇒f(x)=(1/((1/2)+o(ln(x)))) ⇒f(x)= { ((((1/2)+o_1 (ln(x)))^(−1) , x>0)),((((1/2)+o_2 (ln(−x)))^(−1) , x<0)),((f_0 , x=0)) :} where o_(1,2) (x) are odd functions, i.e. o_(1,2) (−x)=−o_(1,2) (x)$
	$g e n e r a l s o l u t i o n$ $f (x) = {\begin{cases} f_{1} (x), x > 0 \\ f_{2} (x), x < 0 \end{cases}$ $\frac{1}{f (x)} + \frac{1}{f (1 / x)} = 1$ $x = e^{t}$ $\frac{1}{f (e^{t})} = g (t)$ $\Rightarrow g (t) + g (- t) = 1$ $\Rightarrow g (t) = \frac{1}{2} + o (t)$ $\Rightarrow f (x) = \frac{1}{\frac{1}{2} + o (l n (x))}$ $\Rightarrow f (x) = {\begin{cases} {(\frac{1}{2} + o_{1} (l n (x)))}^{- 1}, x > 0 \\ {(\frac{1}{2} + o_{2} (l n (- x)))}^{- 1}, x < 0 \\ f_{0}, x = 0 \end{cases}$ $w h e r e o_{1, 2} (x) a r e o d d f u n c t i o n s, i . e .$ $o_{1, 2} (- x) = - o_{1, 2} (x)$
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