To find the domain, find where #(1-x^2)# is negative because if that is negative, the equation will come up with imaginary numbers.
First, you have to factor it.
#(1-x^2) = -(x^2-1) = -(x-1)(x+1)#
Now, all we have to do is find when: #-(x-1)(x+1) < 0#
Multiplying both sides by #-1#...
#(x-1)(x+1)>0#. Now we just have to find when #(x-1)(x+1)# is either both positive or both negative.
#x-1# is negative when #x<1#, and positive when #x>1#.
#x+1# is negative when #x<-1#, and positive when #x> -1#.
#(x-1)(x+1)# is only both negative or both positive when #x<-1#, and when #x>1#.
So, our domain is between those values, or #-1 <= x <= 1#.
For range, we have to determine how big and small #sqrt(1-x^2)# can be when #-1 <= x <= 1#.
First of all, a square root cannot produce a negative number, so we know part of the range is #-1<=x#.
At this point, when calculating it further, we only need the part inside the radical, or #1-x^2#. Just by looking at it, we can tell that the graph, without the domain restriction, has one absolute maximum point, or the vertex of the curve. The vertex of a parabola of equation #ax^2+bx+c# is #(-b/(2a),(4ac-b^2)/(4a))#. So, plugging in #a=-1#, #b=0#, and #c=1#, we get:
#(-0/(-2),((4(-1)(1)-(0))/((4)(-1))) = (0,(-4)/(-4)) = (0,1)#
The maximum of the equation #1-x^2# is #1#. We still have to square root it because we took the equation out of the square root, so...
#sqrt(1) = 1#. We now know that the range is #0<=x<=1#.
So all in all, the domain is #-1<=x<=1# and the range is #0<=x<=1#. You could also graph the equation, but algebraically, this is how you would solve it.
