Solve for the domain and range of f(x) if f(x) = (√((3x−5)/(x+3)))


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Question Number 207043 by necx122 last updated on 04/May/24

$S o l v e f o r t h e d o m a i n a n d r a n g e o f f (x) i f$ $f (x) = \sqrt{\frac{3 x - 5}{x + 3}}$
	Answered by Frix last updated on 04/May/24

	$1 . x + 3 \neq 0$ $x \neq - 3$ $2 . \frac{3 x - 5}{x + 3} ⩾ 0$ $(3 x - 5 < 0 \land x + 3 < 0) \lor (3 x - 5 ⩾ 0 \land x + 3 > 0)$ $x < - 3 \lor x ⩾ \frac{5}{3} [= domain]$ $f (x) = y = \sqrt{\frac{3 x - 5}{x + 3}} \Leftrightarrow x = - \frac{3 y^{2} + 5}{y^{2} - 3}$ $\Rightarrow f (x) ⩾ 0 \land f (x) \neq \sqrt{3} [= range]$
	Commented by necx122 last updated on 05/May/24

	$T h a n k y o u s o m u c h .$ $N o w, f o r g e t t i n g t h e v a l u e s o f t h e$ $i n e q u a l i t y \frac{3 x - 5}{x + 3} ⩾ 0$ $w h e n d o y o u d e c i d e t o u s e (3 x - 5 ⩾ 0 \land x + 3 > 0)$ $a n d (3 x - 5 < 0 \land x + 3 < 0) .$ $1) C a n t h e s e 2 i n e q u a l i t i e s s a t i s f y t h e$ $s a m e r a n g e o f v a l u e s ?$ $2) H o w d o w e c h o o s e w h i c h t o g o w i t h ?$ $3) w h a t s p e c i f i c r u l e s a r e t h e r e f o r$ $s o l v i n g r a t i o n a l i n e q u a l i t i e s ?$ $T h a n k y o u s o m u c h . I a n t i c i p a t e y o u r$ $r e s p o n s e .$
	Commented by mr W last updated on 05/May/24

	$\frac{3 x - 5}{x + 3} ⩾ 0 \Leftrightarrow (3 x - 5) (x + 3) ⩾ 0 \land x + 3 \neq 0$ ${i t}^{'} s e a s i e r t o t r e a t (3 x - 5) (x + 3) ⩾ 0 .$ $w e g e t d i r e c t l y x ⩽ - 3 \lor x ⩾ \frac{5}{3} .$ $s e e d i a g r a m b e l o w .$ $f o r y ⩾ 0 \Rightarrow x ⩽ x_{1} \lor x ⩾ x_{2}$ $f o r y ⩽ 0 \Rightarrow x_{1} ⩽ x ⩽ x_{2}$
	Commented by mr W last updated on 05/May/24

	Commented by Frix last updated on 05/May/24

	$\frac{g (x)}{h (x)} ⩾ 0 means that either both functions$ $are > 0 or both functions are < 0 .$ $[\frac{7}{4} > 0 and \frac{- 7}{- 4} is also > 0]$ $h (x) = 0 is allowed but h (x) = 0 leads to the$ $undefined \frac{(x)}{0} .$ $I wrote (g (x) ⩾ 0 \land h (x) > 0) \lor (g (x) < 0 \land h (x) < 0)$ $where the \lor means ‘ ‘ {or}^{″} so we get 2 regions$
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