What values can take expression

Question

What values can take the expression
$T=\frac{f(t)-f(0)}{f(t^2)+f(t)-2f(0)+2}\mspace{20mu},\mspace{15mu} $ where

$f(2x+y)-f(x+y)=2x , \mspace{20mu}x,y\in \mathbb{R}$

No idea what to do

Answer 1

When one substitutes the value $y= -x$ in the functional equation $f(2x+y)-f(x+y)=2x$ you get $f(x)-f(0)=2x$. So you instantly know that in $$T=\frac{f(t)-f(0)}{f(t^2)+f(t)-2f(0)+2}$$ the numerator $f(t)-f(0)$ must equal $2t$. And the denominator, $$f(t^2)+f(t)-2f(0)+2=(f(t^2)-f(0))+f(t)-f(0)+2$$ must reduce to $2t^2 + 2t+2$, so your fraction becomes $$T=\frac{2t}{2t^2+2t+2} = \frac t{t^2+t+1}.$$