Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function such that for all $x$ we have $$f(f(x))+f(x)=x^4+3x^2+3,$$ prove that for all $x \in \mathbb{R}$, $f(-x)=f(x)$.

I noticed that $f$ cannot have any fixed points. If it had one, say $t$, we would have $$0=t^4+3t^2-2t+3=(t^2+1)^2+(t-1)^2+1$$ which is impossible. So, since $f$ is continuous, we either have $f(x)<x, \: \forall x \in \mathbb{R}$ or $f(x)>x, \: \forall x \in \mathbb{R}$. If the first were true, then we would get $$x^4+3x^2+3=f(f(x))+f(x)<f(x)+x<2x, \quad \forall x \in \mathbb{R}$$ which is absurd. So $f(x)>x, \: \forall x \in \mathbb{R}$.

Using this and the fact that $x^4+3x^2+3$ is strictly increasing on $[0, \infty)$ and strictly decreasing on $(-\infty,0]$, I managed to prove that $f$ is strictly increasing on $[0, \infty)$ and strictly decreasing on $(-\infty, 0]$. This is where I got stuck.

Edit: I think I made some progess. Suppose that there is $x_0$ such that $f(x_0)<0$. Then we get $0>f(x_0)>x_0$ and since $f$ is strictly decreasing on $(-\infty,0]$ it means that $f(0)<f(x_0)<0$, which contradicts $f(0)>0$. So $f(x) \geq 0, \: \forall x \in \mathbb{R}$

Now, suppose $f(x)>f(-x)>0$, for $x \neq 0$. Then $f(f(x))>f(f(-x))$ and summing these yields a contradiction with the hypothesis. The same happens if we suppose $0<f(x)<f(-x)$. So $f(x)=f(-x), \: \forall x \in \mathbb{R}$.

  • $\begingroup$ I've a problem with the use of "continous" here. Does it mean for you that $f(x)$ is "continously increasing or decending" ? Because in general it doesn't mean that. E.g. $e^{\sin\ln x} $ with $x>0$ (en.wikipedia.org/wiki/Continuous_function , section Construction of continuous functions last example) is also continous but neither increasing nor decending for the given value range. $\endgroup$
    – user90369
    Mar 22, 2018 at 14:03
  • $\begingroup$ No, it doesn't mean that it is continuously increasing or descending. Here the function is simply continuous. $\endgroup$
    – AndrewC
    Mar 24, 2018 at 14:11

1 Answer 1


Since you have proved that $f$ is strictly increasing on $[0, +\infty)$, from $$f(f(x)) + f(x) = x^4 + 3x^2 + 4 \quad (\forall x > 0)$$ it is not difficult to prove by contradiction that$$ \lim_{x \to +\infty} f(x) = +\infty. $$ Analogously,$$ \lim_{x \to -\infty} f(x) = +\infty. $$

Now for any $x > 0$, since $f$ is continuous, $f(x) > f(0)$, and $\lim\limits_{x \to -\infty} f(x) = +\infty$, there exists $y < 0$ such that $f(y) = f(x)$. Therefore,$$ x^4 + 3x^2 + 3 = f(f(x)) + f(x) = f(f(y)) + f(y) = y^4 + 3y^2 + 3. $$ Note that $x > 0 > y$ and$$ 0 = (x^4 + 3x^2 + 3) - (y^4 + 3y^2 + 3) = (x - y)(x + y)(x^2 + y^2 + 3), $$ therefore $y = -x$. Thus $f(x) = f(-x)$.


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