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!
Best Answer
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, $f''(x) = 2x$.