 # Consider the function: f:R->R f(x)=x/(|x|+1) I already showed that this function is continuous and injective. Maybe this could help? Maybe we can use the intermediate value theorem? The problem is that I don't have an interval [a,b] with a,b in R. I know that I can consider the limit, but we didn't proved the fact, that I can use the intermediate value theorem for limits. saucletbh 2022-09-17 Answered
Consider the function:
$f:\mathbb{R}\to \mathbb{R}$
$f\left(x\right)=\frac{x}{|x|+1}$
I already showed that this function is continuous and injective. Maybe this could help? Maybe we can use the intermediate value theorem? The problem is that I don't have an interval [a,b] with a,b $\in$ $\mathbb{R}$. I know that I can consider the limit, but we didn't proved the fact, that I can use the intermediate value theorem for limits.
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First show that $|f|<1$ (this should be pretty easy). Then take any $y\in \left(-1,1\right)$ and show that you can explicitly solve $f\left(x\right)=y$ for $x$, thereby showing that the image is $\left(-1,1\right)$ and nothing else. You might find it helpful to do $y\ge 0$ and $y<0$ separately.
###### Not exactly what you’re looking for? trkalo84
$|f\left(x\right)|=|\frac{x}{|x|+1}|=\frac{|x|}{|x|+1}<1$
thus as $f\left(x\right)$ is continuos and lim as $x\to ±\mathrm{\infty }=±1$
$-1