# Whether the given function is one-one or onto or bijective?

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.

• I suggest that you plot the function on the range $[-1, 1]$ or $[0, 2]$. Commented Jun 15 at 9:48
• "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. Commented Jun 15 at 10:00
• Yes you are right. I. cannot apply that.
– PAMG
Commented Jun 15 at 10:02

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 $$f(\beta)=y$$.
• I don't understand how $\lim_{x\to 0^-}f(x)=1$. How it is coming?
• Because $\lim_{x\to0^-}x^3+x^2+x=0$ and $\lim_{x\to0^-}\{x\}=1$. Commented Jun 15 at 10:39