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

 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

$$\:{please}\:{help}\:{me} \\ $$$${y}=\sqrt{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$$${Q}\:{find}\:{domain}\:{and}\:{range}\:{of}\:{given}\:{real}\:{function} \\ $$$${i}\:{found}\:{the}\:{domain}=\left[−\mathrm{3},\mathrm{3}\right] \\ $$$${but}\:{for}\:{range}\:{i}\:{followed}\:{this}\:{way} \\ $$$${y}^{\mathrm{2}} =\mathrm{9}−{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} =\mathrm{9}−{y}^{\mathrm{2}} \Rightarrow{x}=\sqrt{\mathrm{9}−{y}^{\mathrm{2}} }\Rightarrow{y}=\left[−\mathrm{3},\mathrm{3}\right] \\ $$$${but}\:{the}\:{answer}\:{is}\:\left[\mathrm{0},\mathrm{3}\right] \\ $$$${after}\:{checking}\:{the}\:{graph}\:\:{y}\geqslant\mathrm{0}.\:{yes}\:{its}\:{true} \\ $$$${we}\:{have}\:{to}\:{take}\:{positive}\:{root}\:{but}\:{how}\:{can}\:{i} \\ $$$${write}\:{it} \\ $$

Commented by Kunal12588 last updated on 10/Jan/19

offcourse sir but tommorow is my test i can′t  draw graphs their

$${offcourse}\:{sir}\:{but}\:{tommorow}\:{is}\:{my}\:{test}\:{i}\:{can}'{t} \\ $$$${draw}\:{graphs}\:{their} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

to my view it is better to cosults graph ...available  in playstore...some questions easily understood  from graph...

$${to}\:{my}\:{view}\:{it}\:{is}\:{better}\:{to}\:{cosults}\:{graph}\:...{available} \\ $$$${in}\:{playstore}...{some}\:{questions}\:{easily}\:{understood} \\ $$$${from}\:{graph}... \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

Commented by Kunal12588 last updated on 10/Jan/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by kaivan.ahmadi last updated on 10/Jan/19

since y=(√(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)=(√9)=3

$$\mathrm{since}\:\mathrm{y}=\sqrt{\mathrm{9}−\mathrm{x}^{\mathrm{2}} \:}\:\mathrm{so}\:\mathrm{y}\geqslant\mathrm{0}\:.\mathrm{it}\:\mathrm{means}\:\mathrm{min}\left(\mathrm{y}\right)=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{found}\:\mathrm{max}\left(\mathrm{y}\right) \\ $$$$\mathrm{y}\:\mathrm{is}\:\mathrm{max}\:\mathrm{when}\:\mathrm{9}−\mathrm{x}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{max}\: \\ $$$$\mathrm{so}\:\mathrm{max}\left(\mathrm{y}\right)=\sqrt{\mathrm{9}}=\mathrm{3} \\ $$

Commented by Kunal12588 last updated on 10/Jan/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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