Finding critical points of a function

Slightly more generally than the solution by @Sjoerd, you may look for roots of the reciprocal function - if you are interested in pole-like or root-like singularities:

f[x_] := (1 - 2 x)/(2 Sqrt[x - x^2])

Reduce[1/f[x] == 0, x]

(*
  ==> x == 0 || x == 1
*)

A simple method is to test for reality directly, as follows

Reduce[f'[x] ∈ Reals, x, Reals]

0 < x < 1

which does not give you the points where $f^\prime(x) \notin \mathbb{R}$. Alternatively, you can test the denominator directly for reality, as follows

Reduce[Denominator[f'[x]] ∈ Reals, x, Reals]

0 <= x <= 1

which taken with the previous result clearly shows that at $x = 0,1$, $f^\prime(x) \notin \mathbb{R}$. Explicitly searching for those points by combining the two equations, though, gives an odd result

Reduce[! (f'[x] ∈ Reals) 
      && (Denominator[f'[x]] ∈ Reals), x]

Im[x] != 0 && Re[x] == 1/2

despite the results being compatible when the search is expanded to include $x \in \mathbb{C}$

Reduce[f'[x] ∈ Reals, x]
Reduce[Denominator[f'[x]] ∈ Reals, x] // LogicalExpand // Simplify
Reduce[!%% && %, x, Reals]

0 < Re[x] < 1 && Im[x] == 0
(Im[x] == 0 && 0 <= Re[x] <= 1) || 2 Re[x] == 1
x == 0 || x == 1

Edit: in case anyone was wondering, looking for the region where $f^\prime(x) \notin \mathbb{R}$ directly

Reduce[! (f'[x] ∈ Reals), x]

gives the remarkably helpful answer

Im[(1 - 2 x)/Sqrt[x - x^2]] != 0

However, this does better

!Reduce[(f'[x] ∈ Reals), x] // LogicalExpand // Simplify

Re[x] <= 0 || Re[x] >= 1 || Im[x] != 0

Here's one approach, though it possibly answers a slightly different question:

Reduce[-Infinity < f'[x] < Infinity, x]

0 < Re[x] < 1 && Im[x] == 0

or

Reduce[-Infinity < f'[x] < Infinity, x, Reals]

0 < x < 1