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Let $f:\mathbb{R}\to \mathbb{R}$ be such that $$f(x)=x^3+x^2+x+\{x\}$$ where $\{x\}$ denotes the fractional part of $x$. Whether $f$ is one-one or onto or both?

For one-one, we need to show that if $f(x)=f(y)$, then we should have $x=y$. For this, I am trying to show that $f$ is strictly increasing but unable to show it for the case when $x>y$ but $\{x\}<\{y\}$. Please help how to proceed?

For onto, since $f$ contains $x^3$, it is onto, which can be shown using intermediate value theorem.

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    $\begingroup$ I suggest that you plot the function on the range $[-1, 1]$ or $[0, 2]$. $\endgroup$
    – Martin R
    Commented Jun 15 at 9:48
  • $\begingroup$ "For onto, since $f$ contains $x^3$ , it is onto, which can be shown using intermediate value theorem." Be careful , you can apply the intermediate value theorem only for continous functions , but this function is not continous. $\endgroup$
    – Peter
    Commented Jun 15 at 10:00
  • $\begingroup$ Yes you are right. I. cannot apply that. $\endgroup$
    – PAMG
    Commented Jun 15 at 10:02

1 Answer 1

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Note that $f$ is continuous on $[-1,0)$, that you have $\lim_{x\to0^-}f(x)=1$ and that $f(-1)=-1$. Therefore, there is some $\alpha\in(-1,0)$ such that $f(\alpha)=0=f(0)$. So, $f$ is not injective.

But $f$ is indeed surjective. Take $y\in\Bbb R$. Let $n\in\Bbb Z$ be the largest integer such that $n^3+n^2+n\leqslant y$. Then $(n+1)^3+(n+1)^2+n+1>y$. The restriction of $f$ to $[n,n+1)$ is continuous, $f(n)=n^3+n^2+n\leqslant y$ and\begin{align}\lim_{x\to(n+1)^-}f(x)&=(n+1)^3+(n+1)^2+n+2\\&>(n+1)^3+(n+1)^2+n+1\\&>y,\end{align}and therefore there is some $\beta\in[n,n+1)$ such that $f(\beta)=y$.

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  • $\begingroup$ I don't understand how $\lim_{x\to 0^-}f(x)=1$. How it is coming? $\endgroup$
    – PAMG
    Commented Jun 15 at 10:33
  • $\begingroup$ Because $\lim_{x\to0^-}x^3+x^2+x=0$ and $\lim_{x\to0^-}\{x\}=1$. $\endgroup$ Commented Jun 15 at 10:39
  • $\begingroup$ Thank you for such a nice answer. $\endgroup$
    – PAMG
    Commented Jun 17 at 13:08

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