Show that the function f: R---(-1, 1) defined by f(x)= x/1+modulus x, x belongs to R is a bijective function

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Solution

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$f (x) = \frac{x}{1 + |x|} l e t f o r x 1 a n d x 2 f (x 1) = f (x 2) \frac{x 1}{1 + |x 1|} = \frac{x 2}{1 + |x 2|} x 1 + x 1 |x 2| = x 2 + x 2 |x 1| i f b o t h x 1, x 2 a r e b o t h p o s i t i v e o r n e g a t i v e t h e n x 1 \pm x 1 x 2 = x 2 \pm x 1 x 2 x 1 = x 2 f o r t h e y a r e o f a l t e r n a t e s i g n t h e n x 1 + x 1 x 2 = x 2 - x 1 x 2 o r x 1 - x 1 x 2 = x 2 + x 1 x 2 x 2 - x 1 = 2 x 1 x 2 o r x 1 - x 2 = 2 x 1 x 2 b u t a s t h e y a r e o f a l t e r n a t e s i g n i n e a c h c a s e L H S i s p o s t i v e a n d R H S n e g a t i v e s o n o s l u t i o n i n t h i s c a s e h e n c e x 1 = x f u n c t i o n i s o n e - o n e y = f (x) = \frac{x}{1 + |x|} = \frac{x}{1 \pm x} x = \frac{y}{1 \pm y} d o m a i n o f x = R - \{- 1, 1\} w h i c h i s s a m e a s r a n g e o f f (x) s o f u n c t i o n i s a l s o o n t o h e n c e b i j e c t i v e$

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