Question #1b0ff

1 Answer
Sep 15, 2017

Domain is -1<= x <= 1, and range is 0<=x<=1.

Explanation:

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.