# Thread: Polynomial of Degree 2

1. ## Polynomial of Degree 2

My professor has this proof in his notes and I was wondering if anyone could help get me rolling...

Let $f :\mathbb{R} \rightarrow \mathbb{R}$ be differentiable. Suppose
$f(x + y) = f(x) + f(y) + 2xy$ for all $x,y\in \mathbb{R}$.

Show by completing the two parts below that $f$ is a polynomial of degree 2.
(1) Prove $f'(x) = f'(0)+2x$, for all $x\in \mathbb{R}$.
(2) Prove $f(x) = x^2+f'(0)x+f(0)$,for all $x\in \mathbb{R}$.

I think my biggest problem is coming up with the best way to represent $f'(x)$ which is obviously a big roadblock.

2. Originally Posted by zebra2147
My professor has this proof in his notes and I was wondering if anyone could help get me rolling...

Let $f :\mathbb{R} \rightarrow \mathbb{R}$ be differentiable. Suppose
$f(x + y) = f(x) + f(y) + 2xy$ for all $x,y\in \mathbb{R}$.

Show by completing the two parts below that $f$ is a polynomial of degree 2.
(1) Prove $f'(x) = f'(0)+2x$, for all $x\in \mathbb{R}$.
(2) Prove $f(x) = x^2+f'(0)x+f(0)$,for all $x\in \mathbb{R}$.

I think my biggest problem is coming up with the best way to represent $f'(x)$ which is obviously a big roadblock.
If you take the derivative with respect to x you get

$\displaystyle f'(x+y)=f'(x)+2y \implies f'(x)=f'(x+y)-2y$

This must hold for all $y \in \mathbb{R}$ so what choice of $y$ will give the result that you want?

3. $y=-x$??

So that we get $f'(x) = f'(0)+2x$?

4. Originally Posted by zebra2147
$y=-x$??

So that we get $f'(x) = f'(0)+2x$?
Yes that is the one

5. So is the information that you have helped me come up sufficient for the proof because it only has to work for all $x$ and not all $y$?

And do you have any hints for part 2?

Thanks!

6. Originally Posted by zebra2147
So is the information that you have helped me come up sufficient for the proof because it only has to work for all $x$ and not all $y$?

And do you have any hints for part 2?

Thanks!
Let $g(x)=f(x)-x^2-f'(0)x$. Note that $g'(x)=f'(x)-2x-f'(0)=0$ and since $g'(x)=0$ on an interval (in this case all of $\mathbb{R}$) we may conclude that $f(x)-x^2-f'(0)x=g(x)=c$ for some $c$. Noting that $g(0)=f(0)$ gives the desired results.