Finding the value of $f(2)$ if $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$, $f(0)=1$ and $f'(0)=-1$ Let $f:\mathbb{R}\to\mathbb{R}$ be a function satisfying the conditions
$$\begin{gather}
f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2} \tag{1}\\
f(0)=1 \tag{2}\\
f'(0)=-1 \tag{3}
\end{gather}$$
Find the value of $f(2)$ by proper explanation.
 A: To begin with, let us introduce the function $g(x)=f(x)+x-1$.
This function satisfies the midpoint property (1), and $g(0)=g'(0)=0$.
We claim that these three conditions imply that $g(x)=0$ for all $x\in\mathbb{R}$.
Since $g(0)=0$, it is easily seen that (1) implies $g(-x)=-g(x)$ for all $x$.
Next observe that $g(x)-g(y)=2g((x-y)/2)$ for all $x,y$.
It follows that $g$ is differentiable on $\mathbb{R}$ and that $g'(x)=g'(0)=0$ for all $x$.
Hence $g$ is constant and equal to $g(0)=0$.
So $f(x)=1-x$ for all $x$.
In particular, we find $f(2)=-1$.
A: First, an example of a function that satisfies all the given conditions: $g(x) = 1 - x$. Clearly, if we trust that the problem has only one answer, the answer is bound to be $g(2)=-1$.
Of course, this won't pass as a complete solution, because now we need to somehow prove that $f(2)=-1$ for an arbitrary $f$ that satisfies the three conditions. There is no guarantee (at least no obvious guarantee) that such an $f$ must be equal to $g$. And here is the main idea: it is quite easy to prove that $f(x)=1-x$ for any dyadic number $x \in \mathbb{R}$. It will automatically mean that $f(2)=-1$.
PS: and if you think just a bit more you can even prove that $f(x)=1-x$ for every $x \in \mathbb{R}$.
PPS: the answer is meant as a hint, but I can expand upon request.
UPDATE: expanding, although now with all these other answers there is little point to it.
The proposed solution would go something like this. Step one: let's explore what (1) tells us. Let's call $\alpha = f(0)$ and $\beta = f(1)-f(0)$. So, we have the equality $f(x) = \alpha + \beta x$ for $x = 0$ and $x = 1$. Using (1) it is almost trivial to expand this equality first to $\mathbb{Z}$, and then to the set of all dyadic numbers. So, for all dyadic numbers $x$ we have $f(x) = \alpha + \beta x$. Using (2) we now get that $\alpha = 1$, and using (3) we get $\beta = -1$.
What I said in PPS is a simple variation of the same logic.
This is the way I was thinking about the problem. If you strip it down to the minimum, you'll get a solution similar to WimC's answer.
A: From (1) and (2): $2^k\left(f(2^{1-k})-1\right)=f(2)-1$ for all $k \geq 0$.
From (3): $\lim_{k\to \infty}2^k\left(f(2^{1-k})-1\right)=-2$.
So $f(2)=-1$.
