How do you find domain and range for $y=sqrt((4 - x²))$ ?

Question

57205 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-25T07:01:33

Refer to explanation

Since its a square root we have that

$4-x^2>=0=>(2-x)*(2+x)>=0$ which holds for $[-2,2]$ hence

$D(f)=[-2,2]$

The range is

$y=sqrt(4-x^2)=>y^2=4-x^2=>x^2=4-y^2>=0=>(2-y)(2+y)>=0$

Because $y>=0$ we have that $R(f)=(0,2)$

How do you find domain and range for y=sqrt((4 - x²))y=sqrt((4 - x²))?