I'm trying to solve this question using calculus, however am getting the incorrect answer. Can someone please point out my mistake.
Question: $$f(x)f(y) = f(x+y) + xy$$
My (incorrect) solution: $$f'(x) = \lim \frac{f(x+h) - f(x)}{h}$$ $$f'(x) = \lim \frac{f(x)f(h) - xh - f(x)}{h}$$ $$f'(x) = \lim [-x + f(x) ( \frac{f(h) - 1}{h})]$$ $$f'(x) = -x + f(x)(\lim\frac{f(h)-1}{h})$$ Assuming the limit converges, $$f'(x) = -x + kf(x)$$ I then solved this differential equation to get, $$f(x) = A + 1/k^2 + x/k$$ The correct answers are $$f(x) = 1 - x$$ or $$f(x) = 1 + x$$ which were obtained via more standard methods for solving these questions. However I cannot get the correct coefficents for my answer. Thanks for your help in advance.
$A$ and $k$ are not arbitrary. You know (both from the assumption about the limit, and from inspection of the functional equation) that $f(0) = 1$. Therefore, $A + 1/k^2 = 1$, so $f(x) = 1 + x/k$.
Now, plug this back into the functional equation
$$(1 + x/k)(1 + y/k) = 1 + (x+y)/k + xy)$$ $$1 + (x+y)/k + xy/k^2 = 1 + (x+y)/k + xy$$ $$xy/k^2 = xy$$
to observe that $k = \pm 1$.