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

1 Answer

Refer to explanation

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)