How can I find $f(3)$ if $f(2x)=f^2(x)-2f(x)-\frac{1}{2}$? If $f(2x)=f^2(x)-2f(x)-\frac{1}{2}$ and $f(1) = 2$ then find $f(3)$.
Can you give me any hint that I can start with?

$f(2x)=f^2(x)-2f(x)-\frac{1}{2}
$
and
$f(1) = 2
$.
I can see how to get
$f(2^n)$
and expressions for
$f(2^{-n})$,
but I don't see how to get
$f(3)$.
I show what I've got so far.
All fairly trivial.
$f(2x)+\frac32
=f^2(x)-2f(x)+1
=(f(x)-1)^2
$
so
$f(x)
=1\pm\sqrt{f(2x)+\frac32}
$. 
$x = 0
\implies f(0) = f^2(0)-2f(0)-\frac12
$
or
$f^2(0)-3f(0) = \frac12
$
or
$f^2(0)-3f(0)+9/4 
=\frac12+9/4
=\frac{11}{4}
$
or
$(f(0)-\frac32)^2 
=\frac{11}{4}
$
or
$f(0)
=\frac32\pm\frac{\sqrt{11}}{2}
$.
$x=1
\implies f(2) = -\frac12
$.
$x=\frac12
\implies f(\frac12) = 1\pm\sqrt{f(1)+\frac32}
=1\pm\sqrt{\frac72}
$.
And that's all.
