y=sqrt(4-x^2)
I have a trouble finding range,
I used the inverse to find
worked out to be y=2 but
the answer is [0,2]
my question is how do I found 0 in [0,2]?
√x is defined only for x>=0
SO,√(4-x^2) has domain where
4-x^2 >= 0
x^2 <= 4
|x| <= 2
Using that, 0 <= y <= 2
Of course, you recognized this as the top half of a circle of radius 2 with center at (0,0), right?