Is there a particular way to find a function using given conditions
If f(x) is a differentiable function satisfying $f^2 (x) + f^2 (y) + 2(xy – 1) = f^2 (x + y) $ , Also f(x) > 0 $f (\sqrt2)= 2$
I can only see a symmetry in $f^2 (x) + f^2 (y) + 2(xy)$.......$€(a+b)^2$ But I dont know how exactly should I approach such problems
Let $g(t)=f^2(t)$, which is also differentiable. Plug in $x=y=0$ to find $2g(0)-2=g(0)$, i.e., $$g(0)=2. $$ Now shift terms around $$ g(x+y)-g(x)=g(y)+2(xy-1)=g(y)-g(0)+2xy$$ and divide by $y$ (assuming $y\ne 0$): $$ \frac{g(x+y)-g(x)}y = \frac{g(y)-g(0)}y+2x.$$ Taking the limit as $y\to 0$, we recognize derivatives: $$ g'(x) = g'(0) +2x.$$ From this, we infer by integration $$g(x)=xg'(0)+x^2+C$$ for some constant $C$. We know $g(0)=2$ and $g(\sqrt 2)=4$, so must have $C=2$ and $g'(0)=0$. Hence ultimately, $$ f(x)=+\sqrt{g(x)}=\sqrt{x^2+2}.$$
Remark: We did not really need the given property $f(x)>0$.