Let $f(x)$ be a one-one, polynomial function such that $f(x)f(y)+2=f(x)+f(y)+f(xy) \ \forall \ x,y \in \mathbb R - \{0\}$, $f(1) \neq 1$, $f'(1)=3$. Find $f(x)$.

I tried to find the degree of the polynomial from the equation by using suitable substitution, but it didn't work. Also, I found that $f(1)=2$ and then I substituted $y=\dfrac{1}{x}$ to get $f(x)f(\dfrac{1}{x})= f(x) + f(\dfrac{1}{x})$. But I can't simplify further. Also, the answer given in my book is $f(x)=x^3+1$.

Any help will be appreciated.


To add the solution by @Rory Daulton: Set $y=0$ and you get $$f(x)f(0)+2=f(x)+2f(0)$$ With $f(0)=2$ you get $$f(x)=2f(0)-2$$ which contradicts $f^\prime(1)=3$. Thus $f(0)=1$.

EDIT 1: Take $\partial_x$ from both sides in $$f(x)f(y)+2=f(x)+f(y)+f(xy)$$ which yields $$f^\prime(x)f(y)=f^\prime(x)+yf^\prime(xy)$$ Set $x=1$:$$3f(y)=3+yf^\prime(y)$$ Thus $$3f(y)-3=yf^\prime(y)$$ From This ODE you can derive, that the degree of the polynomial is 3. Now you can solve this ODE...

  • $\begingroup$ How can we set $y = 0$? The functional relation is satisfied necessarily only for non-zero reals, right? $\endgroup$ – Anay Karnik Jan 19 '17 at 17:20
  • $\begingroup$ Oh sorry, just read Rory's solution and got it. $\endgroup$ – Anay Karnik Jan 19 '17 at 17:22

Here's something to get you started.

You know that $f(x)$ is a polynomial function, so it is also continuous. We also know that


if both $x$ and $y$ are zero. However, taking both $x$ and $y$ to approach zero, the equation is also true for either $x$ or $y$ or both being zero.

Substitute $y=0$. Substituting and solving yields


This could be true for all $x$ only if $f(0)=1$ or if $f(x)=\frac{2f(0)-2}{f(0)-1}$. This last would make $f(x)$ a constant polynomial, which is prohibited by $f'(1)=3$. Therefore, $f(0)=1$.

(After I finished my last edit for the last paragraph, I saw that @tampis had done basically the same using the previous version of my answer. He deserves due credit, but I did write this on my own!)

Similar substitution for $x=y=1$ gives you $f(1)=1$ or $f(1)=2$. The first is given to be false, so $f(1)=2$, as you already found.

  • $\begingroup$ Actually you were a little bit faster than me extending your answer... ;-) $\endgroup$ – Stephan Kulla May 19 '15 at 19:01
  • $\begingroup$ I'm not so sure, since this site lumps together edits done within five minutes. Anyway, +1 for you! $\endgroup$ – Rory Daulton May 19 '15 at 19:11

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.