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Question Number 49389 by behi83417@gmail.com last updated on 06/Dec/18

f o r x \neq 0, y \neq 0, xy \neq - 1, f (1) = \frac{1}{2}

f (x) + f (y) = f (x + y) + \frac{x + y}{1 + xy}

1 . f i n d : f (x), [i f p o s s i b l e]

2 . f i n d : f^{- 1} (1), [i f p o s s i b l e] .

Answered by kaivan.ahmadi last updated on 07/Dec/18

f (x) + f (x) = f (2 x) = \frac{2 x}{1 + x^{2}} \Rightarrow

2 f (x) = \frac{2 x}{1 + x^{2}} \Rightarrow f (x) = \frac{x}{1 + x^{2}}

Commented by Abdo msup. last updated on 08/Dec/18

f (x) = y \Leftrightarrow x = f^{- 1} (y)

f (x) = y \Leftrightarrow \frac{x}{1 + x^{2}} = y \Leftrightarrow x = (1 + x^{2}) y \Leftrightarrow y + {y x}^{2} - x = 0

\Leftrightarrow {y x}^{2} - x + y = 0

Δ = 1 - 4 y^{2} Δ ⩾ 0 \Leftrightarrow 4 y^{2} ⩽ 1 \Leftrightarrow∣ y ∣⩽ \frac{1}{2}

x_{1} = \frac{1 + \sqrt{1 - 4 y^{2}}}{2 y} a n d x_{2} = \frac{1 - \sqrt{1 - 4 y^{2}}}{2 y} \Rightarrow

f^{- 1} (y) = \frac{1 \bar{+} \sqrt{1 - 4 y^{2}}}{2 y} a n d w e s e e t h a t f^{- 1} (1) d o n t e x i s t

(1 \notin [- \frac{1}{2}, \frac{1}{2}])

Answered by kaivan.ahmadi last updated on 07/Dec/18

f^{- 1} (1) is not exist, since if (x, 1) \in f then

x^{2} - x + 1 = 0 that has no root