Let $f$ : $\mathbb{R}$$\to$$\mathbb{R}$ be a differentiable function such that $f(x+h)$$-$$f(x)$ $=$ $h$$f'(x+{1\over2}h)$, for all real $x$ and $h$. Prove that $f$ is a polynomial of degree at most $2$.
At first, write it as ${f(x+h)-f(x)\over h}$$=$$f'(x+{1\over2}h)$. Put limit $h$$\to$$0$ both sides. It'll be $f'(x)$$=$$\lim_{h\to 0}$$f'(x+{1\over2}h)$. Now I am stucked to show that $f$ is a polynomial. Please help.
Hints: Putting $h=1$ helps you seeing that $f$ is 2 times diffenrentiable (and even more)
Use your expression with 2h instead of x: $f(x+2h)-f(x)=2hf'(x+h)$
Use x+h instead of x in the original expression : $f(x+2h)-f(x+h)=hf'(x+3h/2)$
Summing this expression with yours and comparing with the previous one leads to (putting y=2h) $$2f'(2y+x)=f'(y+x)+f'(3y+x)$$
Derive this expression two times with respect to y and set y=0 to find that $f^{(3)}=0$