The inverse image of an interval Someone help me to find $f^{-1}(](x^2+1)-\varepsilon,(x^2+1)+\varepsilon[)$ where $$f(x)=\begin{cases} 0,~\text{if}~ x<0\\ x^2+1,~\text{if}~ x\geq0\end{cases}$$
I know that 
$f^{-1}(](x^2+1)-\varepsilon,(x^2+1)+\varepsilon[)=\{y\in \mathbb{R}, f(y)\in ](x^2+1)-\varepsilon, (x^2+1)-\varepsilon[\}$
let $y<0$ then $f(y)=0$ so $f(y)\in ](x^2+1)-\varepsilon, (x^2+1)-\varepsilon[ $ if $x^2+1-\varepsilon <0$ that is $ \varepsilon >x^2+1$
so if $\varepsilon >x^2+1$ $f^{-1}(y)=]-\infty,0[$ right ?
someone have an idea on how to do this ?

Hint:Based on the graph of $f(x)$
$${{f}^{-1}}\left( \left] a,b \right[ \right)=\left\{ \begin{align}
  & \phi \ \quad \quad \quad \quad \quad \quad \quad \quad ,if\ b\le 0\  \\ 
 & \left] -\infty ,0 \right[\quad \quad \quad \quad \quad \quad ,if\ b\le 1\ AND\ a<0 \\ 
 & \phi \quad \quad \quad \quad \quad \quad \quad \quad ,if\ b\le 1AND\ a\ge 0 \\ 
 & \left] \sqrt{a-1},\sqrt{b-1} \right[\quad \ \quad ,if\ a\ge 1 \\ 
 & \left] 0,\sqrt{b-1} \right[\quad \quad \quad \ \quad ,if\ b>1\ AND\ a\ge 0 \\ 
 & \left[ 0,\sqrt{b-1} \right[\quad \quad \quad \ \quad ,if\ b>1\ AND\ a<0 \\ 
\end{align} \right\}
$$
