Find the domain and range of a function The question is: find the domain and range of $f(x) = \sqrt{{(1-x)}/x}$.
For the domain, I took $x ≠ 0$ and $(1-x)/x ≥ 0$, which gave me $1 ≥ x$.
So the domain I found is $(-∞,0)∪(0,1]$. However, the proposed solution says that the domain is just $(0,1],$ without explaining, so I wonder why.
Also, any quick tip to find the range (without taking monotonicity, roots, limits to infinity etc.)? I tried inverting the function and the domain of the inverse is $\Bbb R$, which is clearly not the range of $f$.
Thanks in advance!

$f(x) = \sqrt{\frac {1-x}x}$ which is undefined if $x =0$ and if $\frac {1-x}x = \frac 1x - 1 < 0$.  $x\ne 0$ and $\frac 1x -1 < 0 \implies \frac 1x \ge  1$.   
If $x > 0$ then $\frac 1x \ge 1 \implies x \le 1$ and if $x < 0$ implies $x \ge 1$ which is a contradiction so $0 < x \le 1$ and domain is $(0,1]$.
[Alternatively $\frac 1x \ge 1 > 0$ so $\frac {\frac 1x}{\frac 1x} \ge \frac 1{\frac 1x}>\frac 0{\frac 1x} \implies 1 \ge x > 0$.]
As for range.
Now that we know $0 < x \le 1$ then $\frac 1x \ge 1$ and $\frac 1x - 1 \ge 0$ and $\frac {1-x}x = \frac 1x -1 \ge 0$ and $\sqrt{\frac {1-x}x} \ge 0$ with no restrictions.
That implies the range is $[0, \infty)$.
But just to make certain we should test if $\sqrt{\frac {1-x}x} = m$ is solvable for all $m \ge 0$.
$\sqrt{\frac {1-x}x} = m \ge 0 \implies$
$\frac{1-x}x =\frac 1x- 1= m^2 \implies$
$\frac 1x = m^2 + 1$.  So $m^2 + 1 \ge 1$ we can conclude
$x = \frac 1{m^2 + 1}$ and that exists for all possible $m\ge 0$
So range is all $[0, \infty)$
