$f:\mathbb R\to \mathbb R$ monotonically increasing.

Solve the system:

$$\begin{cases} f(x) + x = f(y) + y\\ x^2 + xy + y^2 = 12\end{cases}$$

Since $f$ is monotonically increasing and $f:\mathbb R\to \mathbb R$, isn't $f$ one-to-one?

But, I can't get it past that..

Can I replace $f(x)$ with $y$ so as to reduce the unknowns?


Since $f(x)+x$ is injective, as you already figured out, you can conclude that $x=y$. This makes the second equation trivial to solve.

You may not replace $f(x)$ with $y$, $y$ is a different, unrelated variable.

  • $\begingroup$ If I take x = y for granted, and continue with f(x) = f(y) f(x) + x = f(y) + y which is true based on the hypothesis, is it an acceptable solution? $\endgroup$
    – AQUATH
    Aug 2 '15 at 10:03
  • $\begingroup$ You already used the first equation for figuring out that x=y is the only option. You basically said $f(x)+x=f(y)+y \rightarrow f(x)=f(y)\rightarrow x=y $. $\endgroup$
    – orion
    Aug 2 '15 at 12:20

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.