f(((1−x)/(1+x))) = x+2 then what is the domain and range of f^(−1) (x) ?


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Question Number 103457 by bobhans last updated on 15/Jul/20

$f (\frac{1 - x}{1 + x}) = x + 2 t h e n w h a t i s t h e d o m a i n$ $a n d r a n g e o f f^{- 1} (x) ?$
	Answered by Worm_Tail last updated on 15/Jul/20
	$f(((1−x)/(1+x)))=x+2 (((1−x)/(1+x)))=f^(−1) (x+2)_(x=x−2) ((3−x)/(−1+x))=f^(−1) (x) Domain=R−{1} 3−x=−f^(−1) (x)+xf^(−1) (x) x=((3+f^(−1) (x))/(1+f^(−1) (x))) Range=R−{−1}$
	$f (\frac{1 - x}{1 + x}) = x + 2$ $(\frac{1 - x}{1 + x}) = f^{- 1} {(x + 2)}_{x = x - 2}$ $\frac{3 - x}{- 1 + x} = f^{- 1} (x) D o m a i n = R - {1}$ $3 - x = - f^{- 1} (x) + {x f}^{- 1} (x)$ $x = \frac{3 + f^{- 1} (x)}{1 + f^{- 1} (x)} R a n g e = R - {- 1}$
	Answered by bemath last updated on 15/Jul/20

	$f^{- 1} (x + 2) = \frac{1 - x}{1 + x}$ $f^{- 1} (x) = \frac{1 - (x - 2)}{1 + x - 2} = \frac{3 - x}{x - 1} = \frac{- x + 3}{x - 1}$ $d o m a i n f^{- 1} (x) \Rightarrow x \in R \land x \neq 1$ $f o r f i n d R a n g e f^{- 1} (x)$ $\Leftrightarrow x = \frac{f^{- 1} (x) + 3}{f^{- 1} (x) + 1}; r a n g e$ $: y \in R \land y \neq - 1$
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