Finding period of function from the functional relation $f(x+T) = 1 + \left(2f(x)-f(x)^2\right)^{\frac12}$ I am supposed to find period of function following the below relation if exists, for domain belonging to all real numbers.
$$f(x+T) = 1 + \left(2f(x)-f(x)^2\right)^{\frac12}$$
I just took $1$ to other side and squared and then differentiated to get $f(x+T) = f(-x)$ which indicates that period is $2T$.
Query: I don't think differentiation is a good way, because it is not mentioned that $f(x)$ is differentiable. Can someone please help in clarifying the underlying ideas or suggesting a altogether different approach.

I think this shows it:
If you rearrange as you say:
$(f(x+T)-1)^2=2f(x)-f(x)^2$
$f(x+T)^2-2f(x+T)+1=2f(x)-f(x)^2$
$2f(x)−f(x)^2-f(x+T)^2+2f(x+T)=1$  (eq. 1)
Substituting $x+T$ for $x$:
$2f(x+T)−f(x+T)^2-f(x+2T)^2+2f(x+2T)=1$  (eq. 2)
Subtracting eq. 2 from eq. 1:
$2f(x)-f(x)^2-2f(x+2T)+f(x+2T)^2=0$
$2f(x)-f(x)^2=2f(x+2T)-f(x+2T)^2$
...so the period is $2T$.
