A question about domains and ranges: find the domain and range of the function $f(x)=\sqrt{1-x^2}$, without any digital help If $f(x)=\sqrt{1-x^2}$, then the domain of $f$ is $[-1,1]$. Respectively, the range of $f$ is $[0,1]$. 
How do I understand that the domain of $f$ is $[-1,1]$ and that the range of $f$ is $[0,1]$, without a graphing calculator? 
Let us suppose that  I don't know the given intervals from start. Instead, suppose that I know only that the function that I should find the domain and range of is $f(x)=\sqrt{1-x^2}$. The question will then be: 'Find the domain and range of the function $f(x)=\sqrt{1-x^2}$, without any digital help'

One way to determine the domain and range of $f(x) = \sqrt{1 - x^2}$ is to graph the function.  If $y = \sqrt{1 - x^2}$, then $y \geq 0$.  Moreover, \begin{align*}
y & = \sqrt{1 - x^2}\\
y^2 & = 1 - x^2\\
x^2 + y^2 & = 1
\end{align*}
which is the equation of the unit circle, that is, the equation of the circle with center at the origin and radius $1$.  
 
The restriction that $y \geq 0$ means that we obtain the upper semi-circle.

From the graph, we can see that $f(x) = \sqrt{1 - x^2}$ has domain $[-1, 1]$ and range $[0, 1]$.
If we did not know how to graph the function, we would start by observing that for $f(x) = \sqrt{1 - x^2}$ to be real-valued, $1 - x^2 \geq 0$.  Solving the inequality yields
\begin{align*}
1 - x^2 \geq 0\\
1 \geq x^2\\
1 \geq |x|\\
|x| \leq 1
\end{align*}
which implies that $-1 \leq x \leq 1$, from which we obtain the domain $[-1, 1]$.  
We further observe that if $x$ is in the domain, $1 - x^2 \leq 1$, with equality holding only at $0$.  Moreover, the square of a number in the domain is at most $1$, so $1 - x^2 \geq 1 - 1 = 0$. Since a polynomial function is continuous, $1 - x^2$ assumes every value in the interval $[0, 1]$, so we have $0 \leq 1 - x^2 \leq 1$. Taking square roots yields $0 \leq \sqrt{1 - x^2} \leq 1$, so the range is $[0, 1]$.
That said, it helps to recognize that the graph of $f(x) = \sqrt{r^2 - x^2}$ is the upper half of a circle with radius $r$ and center at the origin.
