If $f$ is a function defined for all real $x$ and satisfies $f(x+y) = f(x) + f(y) + xy(x+y)$, do $f'$ and $f''$ exist?

Question

Let $f$ be a function defined for all real $x$ and satisfies $f(x+y) = f(x) + f(y) + xy(x+y)$. If $f'(0) =-1$, then prove that $f$ and $f'$ are differentiable for all $x$. Also find $f'(3)$ and $f'(4)$.

I have tried assuming $x=y=0$ and solving the equation but to no avail. I also tried differentiating both sides. I just can't get a thought on its solution. Please help!

Answer 1

Notice that $f(0+0) = f(0) +f(0)$ which implies that $f(0)=0$. Then, $$f'(0) = \lim_{h\to 0}\frac{f(h)-f(0)}{h} = \lim_{h\to 0} \frac{f(h)}{h}.$$

Using $$f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0}\frac{f(x)+f(h)+xh(x+h)-f(x)}{h}$$

So $$f'(x) = \lim_{h\rightarrow 0}\frac{f(h)}{h}+\lim_{h\rightarrow 0}\frac{xh(x+h)}{h} = f'(0) + x^2$$

So $$f'(x) =-1+x^2$$

Finally, $f''(x) = 2x$.

Answer 2

Let $$g(x):=f(x)-\frac{x^3}{3}$$ for all $x\in\mathbb{R}$. Then, $g$ satisfies Cauchy's functional equation $$g(x+y)=g(x)+g(y)$$ for all $x,y\in\mathbb{R}$ with the additional property that $g$ is continuous at $0$ (since $g'(0)=f'(0)=-1$). Thus, there exists $k\in\mathbb{R}$ such that $$g(x)=kx$$ for all $x\in\mathbb{R}$. Finally, as $k=g'(0)=-1$, we get $$f(x)=-x+\frac{x^3}{3}$$ for all $x\in\mathbb{R}$.