please help me y=(√(9−x^2 )) Q find domain and range of given real function i found the domain=[−3,3] but for range i followed this way y^2 =9−x^2 x^2 =9−y^2 ⇒x=(√(9−y^2 ))⇒y=[−3,3] but the answer is [0,3] after checking the graph y≥0. yes its true we have to take positive root but how can i write it


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Question Number 52610 by Kunal12588 last updated on 10/Jan/19

$p l e a s e h e l p m e$ $y = \sqrt{9 - x^{2}}$ $Q f i n d d o m a i n a n d r a n g e o f g i v e n r e a l f u n c t i o n$ $i f o u n d t h e d o m a i n = [- 3, 3]$ $b u t f o r r a n g e i f o l l o w e d t h i s w a y$ $y^{2} = 9 - x^{2}$ $x^{2} = 9 - y^{2} \Rightarrow x = \sqrt{9 - y^{2}} \Rightarrow y = [- 3, 3]$ $b u t t h e a n s w e r i s [0, 3]$ $a f t e r c h e c k i n g t h e g r a p h y ⩾ 0 . y e s i t s t r u e$ $w e h a v e t o t a k e p o s i t i v e r o o t b u t h o w c a n i$ $w r i t e i t$
Commented by Kunal12588 last updated on 10/Jan/19

$o f f c o u r s e s i r b u t t o m m o r o w i s m y t e s t i {c a n}^{'} t$ $d r a w g r a p h s t h e i r$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

$t o m y v i e w i t i s b e t t e r t o c o s u l t s g r a p h . . . a v a i l a b l e$ $i n p l a y s t o r e . . . s o m e q u e s t i o n s e a s i l y u n d e r s t o o d$ $f r o m g r a p h . . .$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

Commented by Kunal12588 last updated on 10/Jan/19

$t h a n k y o u s i r$
	Answered by kaivan.ahmadi last updated on 10/Jan/19

	$since y = \sqrt{9 - x^{2}} so y ⩾ 0 . it means \min (y) = 0$ $and found \max (y)$ $y is \max when 9 - x^{2} is \max$ $so \max (y) = \sqrt{9} = 3$
	Commented by Kunal12588 last updated on 10/Jan/19

	$t h a n k y o u s i r$
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