A naive approach:
We know that there's "something special" about the points $x = -1, x = 1$. So let's do this:
We'll split the real number line into 5 parts:
Part 1: all numbers less than -1
Part 2: -1
Part 3: all numbers between -1 and 1
Part 4: 1
Part 5: all numbers greater than 1.
Now let's pick numbers in each part, to see what happens:
The 5 real numbers I will use are:
Part 1: -4
Part 2: -1 (no choice, here)
Part 3: 1/2
Part 4: 1 (again, no other choice)
Part 5: 6
Now we will look at $f(x_0)$ for $x_0$ being each one of these 5 numbers:
$f(-4) = 4\sqrt{1 - (-4)^2} = 4\sqrt{-15} = \text{bad}$ (undefined)
$f(-1) = (-1)\sqrt{1 - (-1)^2} = (-1)\sqrt{1 - 1} = (-1)\sqrt{0} = (-1)(0) = 0$ (OK!)
$f(\frac{1}{2}) = \frac{1}{2}\sqrt{1 - (\frac{1}{2})^2} = \frac{1}{2}\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{4}$ (OK!)
$f(1) = (1)\sqrt{1 - 1^2} = (1)(0) = 0$ (OK again!)
$f(6) = 6\sqrt{1 - 6^2} = 6\sqrt{-35} = ?$ (not so good).
So it looks as if what we want is parts 2,3, and 4, and NOT parts 1 and 5. This is:
$\{-1\} \cup (-1,1) \cup \{1\} = [-1,1]$
If you prefer, you can write this as:
$\text{dom}(f) = \{x \in \Bbb R: |x| \leq 1\}$
the absolute value of $x$, written $|x|$ is just another way of saying:
$\sqrt{x^2}$, if one understand square roots as always being non-negative. So if:
$x^2 \leq 1$, then
$\sqrt{x^2} \leq \sqrt{1} = 1$
so:
$|x| \leq 1$.