Suppose that f: R R is differentiable and that for every a,b R there holds f (a + b) = f(a) + f(b).
Prove that f '(x) = f '(0) for all x R. What kind of function is f?
Start by noticing that f(0)=0. Indeed, putting a=b=0, one gets f(0)=2f(0), hence f(0)=0.
Now put a=x, b=h:
f(x+h)-f(x)=f(h)=f(h)-f(0)
Divide by h to get
(f(x+h)-f(x))/h=(f(h)-f(0))/h
Take the limit h->0 to finally get
f'(x)=f'(0).
This means that f'(x)=a is a constant: f(x)=cx+d, but d must be zero in order to the first relation to hold.
Hence f(x)=cx. Those are called linear functions.