Function involving real values

Question

Let $f$ be a function satisfying $f(x/2+y/2)=(f(x)+f(y))/2$, for all real $x$ and $y$. If $f'(0)$ exists and equal to $-1$ , then $f(2)$ equals: ?

I have tried this question by putting various values of $x$ and $y$ and I got $4$ equations and $4$ variables but still could not get the answer and I did not use $f'(0)$

Answer 1

Put $y=0$. $\forall x\in\mathbb{R},$

$$f(x)=\frac{f(x)+f(0)}{2}$$

$$f(x)=f(0)$$

$f$ is a constant function and $f'(x)=0\ne-1$ $\forall x\in\mathbb{R}$.

Such a function does not exist.

Answer 2

Let $a \ne b$.

Solve $x + y/2 = a; y+x/2 = b$ so $y = \frac {4b - 2a}3$ so $x = \frac {4a-2b}3; y = \frac {4b - 2a}3$.

So $f(a) = f (x + y/2) =\frac {f(x) + f(y)}2 = f(y + x/2) = f(b)$.

So the function is constant and $f'(x) = 0 \ne -1$.

So the problem has no solution.