f(xf(y)+x)=xy+f(x) f:R→R f(x)=?


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Question Number 198178 by universe last updated on 13/Oct/23

$f (x f (y) + x) = x y + f (x)$ $f : R \to R$ $f (x) = ?$
	Answered by Rasheed.Sindhi last updated on 13/Oct/23

	$L e t f (x) = a x + b$ $f (x f (y) + x) = x y + f (x)$ $\Rightarrow a (x f (y) + x) + b = x y + a x + b$ $\Rightarrow a x (a y + b) + a x = x y + a x$ $\Rightarrow a x (a y + b) = x y$ $y \to x$ $\Rightarrow a x (a x + b) = x^{2}$ $a^{2} x^{2} + b x = x^{2}$ $a^{2} = 1 \land b = 0$ $a = \pm 1 \land b = 0$ $f (x) = a x + b = (\pm 1) x + 0$ $\begin{matrix} \begin{matrix} f (x) = x ∣ f (x) = - x \end{matrix} \end{matrix}$ $V e r i f i c a t i o n :$ $(1) f (x) = x$ $f (x f (y) + x) = x y + f (x)$ $x f (y) + x = x y + f (x)$ $\Rightarrow x y + x = x y + x (✓)$ $(2) f (x) = - x$ $f (x f (y) + x) = x y + f (x)$ $\Rightarrow - x f (y) - x = x y - x$ $\Rightarrow - x f (y) - x = x y - x$ $\Rightarrow - x (- y) - x = x y - x$ $\Rightarrow x y - x = x y - x (✓)$
	Commented by mr W last updated on 13/Oct/23

	$f i n e!$ $b u t h o w i s y o u r a r g u m e n t t h a t f (x) i s$ $a l i n e a r f u n c t i o n ?$
	Commented by Rasheed.Sindhi last updated on 13/Oct/23

	$s i r, I^{'} v e n o a r g u m e n t! E v e n a b o u t$ $t h e f u n c t i o n t o b e p o l y n o m i a l! B u t {t h e r e}^{'} s$ $p o s s i b i l i t y o f b e i n g l i n e a r .$
	Answered by witcher3 last updated on 13/Oct/23
	$let b∈R ∃ t∈R∣f(t)=b.....? (x,y)=(1,y)⇒f(f(y)+1)=y+f(1) y=b−f(1) ⇒f(f(b−f(1)+1)=b,t=f(b−f(1))+1 f surjective if f(a)=f(b)⇒(a=b)...? (x,y)=(1,y) f(f(y)+1)=y+f(1)..y=a y=b⇒a=b ⇒ f is injective so f bijective f(f(y)+1)=y+f(1) for y=0⇒f(f(0)+1)=f(1) withe injectivity⇒f(0)+1=1⇒f(0)=0 ∀t∈R−{0}⇔f(t)≠0 by bijection“ (x,y)=(x,y^∗ ),suche f(y^∗ )=−1.y^∗ existe f surjective ⇔f(x.−1+x)=xy^∗ +f(x)=f(0)=0 f(x)=−xy^∗ f is linear f(x)=ax solution⇒(x,y)=(x,x) f(ax^2 +x)=a^2 x^2 +ax=x^2 +ax⇒a^2 =1 a∈{−1,1} ∀(x,y)∈R^2 ∣f(xf(y)+x)=xy+f(x)⇒f∈{f(x)=x,f(x)=−x}$
	$let b \in R \exists t \in R ∣ f (t) = b . . . . . ?$ $(x, y) = (1, y) \Rightarrow f (f (y) + 1) = y + f (1)$ $y = b - f (1)$ $\Rightarrow f (f (b - f (1) + 1) = b, t = f (b - f (1)) + 1$ $f surjective$ $if f (a) = f (b) \Rightarrow (a = b) . . . ?$ $(x, y) = (1, y)$ $f (f (y) + 1) = y + f (1) . . y = a y = b \Rightarrow a = b$ $\Rightarrow f is injective$ $so f bijective$ $f (f (y) + 1) = y + f (1) for y = 0 \Rightarrow f (f (0) + 1) = f (1)$ $withe injectivity \Rightarrow f (0) + 1 = 1 \Rightarrow f (0) = 0$ $\forall t \in R - {0} \Leftrightarrow f (t) \neq 0 by bijection ‘ ‘$ $(x, y) = (x, \overset{}{y}), suche f (\overset{}{y}) = - 1 . \overset{}{y} existe f surjective$ $\Leftrightarrow f (x . - 1 + x) = x \overset{}{y} + f (x) = f (0) = 0$ $f (x) = - x \overset{*}{y}$ $f is linear$ $f (x) = ax solution \Rightarrow (x, y) = (x, x)$ $f ({ax}^{2} + x) = a^{2} x^{2} + ax = x^{2} + ax \Rightarrow a^{2} = 1$ $a \in {- 1, 1}$ $\forall (x, y) \in R^{2} ∣ f (xf (y) + x) = xy + f (x) \Rightarrow f \in {f (x) = x, f (x) = - x}$
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