range of function y = ((∣x−1∣)/(x+3))


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Question Number 85425 by john santu last updated on 22/Mar/20

$r a n g e o f f u n c t i o n$ $y = \frac{∣ x - 1 ∣}{x + 3}$
	Answered by john santu last updated on 22/Mar/20

	$f o r x ⩾ 1 \Rightarrow y = \frac{x - 1}{x + 3}$ $x = \frac{- 3 y - 1}{y - 1} = \frac{3 y + 1}{1 - y}$ $w e g e t : 0 ⩽ y < 1$ $f o r x < 1 \Rightarrow y = \frac{- x + 1}{x + 3}$ $x = \frac{- 3 y + 1}{y + 1}, w e g e t y < - 1 \lor y ⩾ 0$ $t h e r e f o r e r a n g e i s y < - 1 \lor y ⩾ 0$
	Commented by jagoll last updated on 22/Mar/20

	Commented by jagoll last updated on 22/Mar/20

	$this the graph$
	Answered by MJS last updated on 22/Mar/20
	$y= { ((((1−x)/(x+3))=−1+(4/(x+3)); x<1)),((((x−1)/(x+3))=1−(4/(x+3)); x≥1)) :} range= { ((R_1 =R\[−1; 0])),((R_2 =[0; 1[)) :} range: R=R_1 ∪R_2 =R\[−1;0[ or R=]−∞; −1[ ∪ [0; +∞[$
	$y = {\begin{cases} \frac{1 - x}{x + 3} = - 1 + \frac{4}{x + 3}; x < 1 \\ \frac{x - 1}{x + 3} = 1 - \frac{4}{x + 3}; x ⩾ 1 \end{cases}$ $range = {\begin{cases} R_{1} = R ∖ [- 1; 0] \\ R_{2} = [0; 1 [ \end{cases}$ $range : R = R_{1} \cup R_{2} = R ∖ [- 1; 0 [$ $or R =] - \infty; - 1 [\cup [0; + \infty [$
	Commented by john santu last updated on 22/Mar/20

	$m y a n s w e r c o r r e c t o r w r o n g ?$
	Commented by john santu last updated on 22/Mar/20

	$i {d o n}^{'} t u n d e r s t a n d t h e$ $n o t a t i o n] - \infty; - 1 [?$ $i t s a m e t o (- \infty, - 1) ? s i r$
	Commented by $@ty@m123 last updated on 22/Mar/20

	$Y e a h .$ $B o t h a r e s a m e .$
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