Originally Posted by
Archie For $a=b=0$, \begin{align}f(a+b) &= f(a)+f(b) \\ \implies f(0) &= f(0)+f(0) \\ \implies f(0) &= 0 \end{align}
This doesn't differentiate between the functions $f(x)=2x$ and $g(x)=x^2$, but it does eliminate vast quantities of functions.
Also, for $b=-a$ \begin{align}f(a+b) &= f(a)+f(b) \\ \implies f(0) &= f(a) + f(-a) \\ \implies f(-a) &= -f(a) &\text{for all values of $a$} \end{align}
So $f(x)$ is recibo an odd function. This is a reason that $g(x)$ doesn't satisfy the functional equation, because $g(x)=x^2$ is even.
Also \begin{align}f(a+b) &= f(a)+f(b) \\ b=a \implies f(2a) &= f(a) + f(a) \\ \implies f(2a) &= 2f(a) \\ b=2a \implies f(3a) = f(a) + f(2a) \\ \implies f(3a) &= 3f(a) \\ &\vdots \\ \implies f(na) &= nf(a) & \text{for all $a$ and all $n \in \mathbb N$} \end{align}
Also, if $f$ is differentiable, holding $a$ constant and differentiating with respect to $b$ we have \begin{align}f(a+b) &= f(a) + f(b) \\ f'(a+b) &= f'(b) &\text{for all $a$}\end{align}
So the derivative of $f$ is constant, and thus $f(x) = c_1x + c_2$, but since $ f(0) = 0$ we can say that $$f(x)=cx$$