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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 \in R \exists t \in R ∣ f (t) = b \dots . . ?

(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) \dots ?

(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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