Determine whether there exists a function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f\big(x^3+x\big)\le x\le\big(f(x)\big)^3+f(x)$$ for all $x\in\mathbb{R}$.

Source: Math Excalibur Volume 22 No. 4 Page 3 Problem 536 rephrased


1 Answer 1


Let $g(x) := x^3+x$. Then the problem asks for $f$ such that $f(g(x)) \leq x \leq g(f(x))$.

Since $g$ is bijective, choose $f = g^{-1}$, and the inequality becomes $x \leq x \leq x$, which is satisfied for all $x$.

EDIT: Since it was mentioned that there is

only one solution, here's a quick proof of that too:

Let $x$ be arbitrary, let $y := g^{-1}(x)$. We have $g(f(x)) \geq x$, but we also have $f(g(y)) \leq y$, so $g(f(x)) = g(f(g(y)) \leq g(y) = x$ by monotony of $g$. Hence, $g(f(x)) = x$, so $f(x) = g^{-1}(x)$.

  • 1
    $\begingroup$ Correct! And the only possible solution is the inverse of $x^3+x$! $\endgroup$ May 27, 2020 at 4:12

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