# Question 1b0ff

Sep 15, 2017

Domain is $- 1 \le x \le 1$, and range is $0 \le x \le 1$.

#### Explanation:

To find the domain, find where $\left(1 - {x}^{2}\right)$ is negative because if that is negative, the equation will come up with imaginary numbers.
First, you have to factor it.
$\left(1 - {x}^{2}\right) = - \left({x}^{2} - 1\right) = - \left(x - 1\right) \left(x + 1\right)$
Now, all we have to do is find when: $- \left(x - 1\right) \left(x + 1\right) < 0$
Multiplying both sides by $- 1$...
$\left(x - 1\right) \left(x + 1\right) > 0$. Now we just have to find when $\left(x - 1\right) \left(x + 1\right)$ 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 $x > - 1$.
$\left(x - 1\right) \left(x + 1\right)$ is only both negative or both positive when $x < - 1$, and when $x > 1$.

So, our domain is between those values, or $- 1 \le x \le 1$.

For range, we have to determine how big and small $\sqrt{1 - {x}^{2}}$ can be when $- 1 \le x \le 1$.

First of all, a square root cannot produce a negative number, so we know part of the range is $- 1 \le x$.

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 $a {x}^{2} + b x + c$ is $\left(- \frac{b}{2 a} , \frac{4 a c - {b}^{2}}{4 a}\right)$. 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 $0 \le x \le 1$.

So all in all, the domain is $- 1 \le x \le 1$ and the range is $0 \le x \le 1$. You could also graph the equation, but algebraically, this is how you would solve it.