If $F(x)$ is a differentiable function that satisfies the above functional equation for all real x and y except -1. If $F(0) \ne 0$ and $F'(0) = 1$ then $F(x) = ? $

  • $\begingroup$ I assume "!=" means "not equals"? $\endgroup$ – Ethan Dlugie Apr 23 at 16:52
  • $\begingroup$ Yes. Sorry, I did not realise that it could be misinterpreted as factorial. Will edit. $\endgroup$ – Mamta Kumari Apr 23 at 17:06


Let $x=\tan x_1$ and $y=\tan y_1$ with $|x_1|<{\pi\over 2}$ and $|y_1|<{\pi\over 2}$. Then by defining $$g(x)=e^{f(\tan x)}$$ we have $$g(x_1+y_1)=g(x_1)+g(y_1)$$

| cite | improve this answer | |
  • $\begingroup$ I don't think that's working! But I did find the answer by exploiting the fact that the function is said to differentiable. F(x) turned out to be ((1+x)/(1-x))^(0.5) $\endgroup$ – Mamta Kumari Apr 23 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.