<s>[INST] 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'
 [/INST] Method 1:
Domain.  In the real numbers $\sqrt{w}$ exists if and only if $w \ge 0$, so the domain is all real $x$ so that $1-x^2 \ge 0$ or $x^2 \le 1$ or $-\sqrt 1 \le x \le \sqrt 1$ or $-1 \le x \le 1$.
So domain is $[-1,1]$.
Range:  $x$ may be anything so that $-1 \le x \le 1$.  That means $x^2$ may be anything so that $0 \le x^2 \le 1$ That means $-x^2$ may be anything $-1 \le -x^2 \le 0$.  Which means $1-x^2$ may be anything $0 \le 1-x^2 \le 1$ so $0 \le \sqrt{1- x^2} \le 1$.
So range is $[0,1]$.
Method 2:
The graph is $y = \sqrt{1-x^2} \ge 0$
So $y^2 = 1-x^2$
$y^2 + x^2 = 1$.  That's a graph of a unit circle centered at $(0,0)$
So $y = \sqrt{1 -x^2}$ a half circle restricted to $y \ge 0$.
In this half circle, $x$  goes from $-1$ to $1$ so the domain is $[-1,1]$.
In the full circle $y$ goes from $-1$ to $1$ but we are restricting to $y\ge 0$ so the range is $[0,1]$.
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