Question

Find the range of the real valued function √(9-x²)​

Answer 1

Answer:

Domain and range

$\sf \: f(x)=9−x2$

$\sf y= \sqrt{9 -x { }^{2} }$

$\sf \sqrt{9 - x {}^{2} \geqslant 0 }$

$\sf9 - x {}^{2 } \geqslant 0$

$\sf \: x { }^{2} \leqslant 9$

$\sf \: x≤−+3$

$\sf \:x ∈[−3,3]$

$\sf \: y {}^{2} =9−x2$

$\sf \: x {}^{2}= 9 - y {}^{2}$

$\sf \: x= \sqrt{9 - 2y {}^{2} }$

$\sf9 - y {}^{2} \geqslant 0$

$\sf \: y {}^{2} ≤9 \: y∈3,−3$

$\sf \: y=+ve$

$\sf∵y∈[0,3]$

Answer 2

Answer:

Find the range of the real valued function √(9-x²)