Equating Functions

f(x) = f(x^2)
f(x) - f(x^2) = 0

f(1/2) - f(1/4 ) = 0
f(1/4) - f(1/16) =0
. .
. . for n tending to infinity
f(1/n) - f(1/n^2) = 0
adding all the above equations
f(1/2) - f(0) = 0
f(1/2) = 1/2
2f(1/2) = 1

If $f(x)$ is a polynomial of degree $n$ , then $f(x^2)$ is a polynomial of degree $2n$ .

Therefore, no polynomial will satisfy $f(x) = f(x^2)$ .

No rational function will satisfy the equation, and no absolute value function will satisfy the equation too.

Therefore, the most likely answer is that $f(x)$ is a constant function. Given that $f(0) = \dfrac{1}{2}$ , we can say that

$f(x) = \dfrac{1}{2}$

Therefore, $2f(\dfrac{1}{2}) = 2(\dfrac{1}{2}) = \boxed{1}$