If $f(\frac{x+y}{2})=\frac{f(x)-f(y)}{x-y}$ where $x$ and $y$ ar not equal, what can we say about the result? 
If $f(\frac{x+y}{2})=\frac{f(x)-f(y)}{x-y}$ where $x$ and $y$ ar not equal, what can we say about the result?

Can anyone explain how to go about this? Do we have to answer in terms of whether the function is differentiable and/or continuous? Any help would be appreciated thanks.
 A: Let $y=x+2k$. Then we may rewrite the equation for $f$ as
$$
\tag{1}
f(x+2k)=2kf(x+k)+f(x).
$$
Suppose $f(x_0)=a$ and $f(x_0+1)=b$, where $x_0$ is an arbitrary real number. Then we get from $(1)$ with $k=1$:
$$
f(x_0+2)=2f(x_0+1)+f(x_0)=2b+a,\quad f(x_0+3)=2f(x_0+2)+f(x_0+1)=5b+2a.
$$
On the other hand we have $f(x_0+1/2)=f(x_0+1)-f(x_0)=b-a$. Using now $(1)$ with $k=1/2$ gives:
$$
\begin{align}
&f(x_0+3/2)=f(x_0+1)+f(x_0+1/2)=2b-a,\cr 
&f(x_0+2)=f(x_0+3/2)+f(x_0+1)=3b-a,\cr
&f(x_0+5/2)=f(x_0+2)+f(x_0+3/2)=5b-2a,\cr
&f(x_0+3)=f(x_0+5/2)+f(x_0+2)=8b-3a.\cr
\end{align}
$$
The values for $f(x_0+2)$ and $f(x_0+3)$ are the same only if $2b+a=3b-a$ and $5b+2a=8b-3a$, that is if $a=b=0$. But $x_0$ was arbitrary, so we may conclude that $f(x)=0$ for every $x$.
A: If $f$ is supposed to be continuous, then for all $x \in \mathbb R$:$$f(x)=\lim\limits_{y \to x} f(\frac{x+y}{2})=\lim\limits_{y \to x} \frac{f(y)-f(x)}{y-x}=f^\prime(x)$$ so $f$ is differentiable and satisfies the differential equation $y^\prime=y$. Hence $f(x)=\lambda e^x$.
Is $f$ necessarily continuous... That is the question.
Using the comments from Henry and Ivan Neretin, $e^x$ cannot be solution. So if a solution exists, it cannot be continuous on $\mathbb R$.
A: If you replace y by x + a (y = x + a) then this equation can be easier to solve. You will get f(y) = f(x + a). Also if you change the counter by the following: f(x + a) = f(x) + f '(x) a + 0.5  f ''(x)(a)2, then it might become more clear.
