Proving the range of a function involving absolute signs (using the double set inclusion technique) I have been trying to prove that the range of the function $f: \mathbb{R} \to \mathbb{R} $ given by $f(x)=\frac{|x+3|}{|x|+3}$ is the interval $[0,1]$ for some time now without success. I'm aware that in order to formulate a proof, I need to show that $f(\mathbb{R}) \subseteq [0,1]$, and $[0,1] \subseteq f(\mathbb{R})$ to show that $f(\mathbb{R})$ is indeed equal to $[0,1]$, however, I'm not sure how to continue from this point onward.
So, I have come up with rough work first, which is as follows: 

ROUGH WORK FOR ($f(\mathbb{R}) \subseteq [0,1]$)

We know that $0 \leq \frac{|x+3|}{|x|+3} \leq 1$. Hence, by doing some re-arragements, we can show that $0 \leq |x+3| \leq |x| + 3$ (since $|x| + 3 \ge 3$, we could multiply both sides). After multiplying, we basically get the triangle inequality, which has to always be true. Thus, this proves the first inclusion ($f(\mathbb{R}) \subseteq [0,1]$).

ROUGH WORK FOR ($[0,1] \subseteq f(\mathbb{R})$)

Now, to show the other inclusion $[0,1] \subseteq f(\mathbb{R})$, I'm not sure how to continue! I have tried using cases $x \ge 0$ and $x \leq 0$, but that didn't really help. More specifically, I tried showing that if $x \ge 0$, then $x = \frac{3(1-y)}{y-1}$ which always produces an $x$ in the image, however, since $y \in [0,1]$, we might get a division by zero. So, I'm not sure what can be done instead.
I am quite new to proofs, so any help would be immensely appreciated!
 A: Case 0 <= x.  f(x) = 1.  
Case -3 <= x < 0.
f(x) = (x + 3)/(-x + 3)
= (9 - x$^2$)/(3 - x)$^2$ >= 0
As x + 3 < 3 and 3 <= 3 - x, f(x) < 1/2.
Notice that in this case f takes every value in [0,1). 
Case x < -3.  f(x) = -(x + 3)/(-x + 3) >= 0
There remains to show f(x) <= 1. 
A: Your first part is correct. Since 
$$0\leq|x+3|\leq|x|+3 \implies 0\leq\frac{|x+3|}{|x|+3}\leq1$$
So $f(\Bbb R)\subset [0, 1]$
For the second part. If $x\geq0$, 
$$\frac{|x+3|}{|x|+3}=\frac{x+3}{x+3}=1$$
and if $x=-3$
$$\frac{|(-3)+3|}{|(-3)|+3}=\frac0{3+3}=0$$
Let's see if all the value in $[0, 1]$ could be obtain with $-3\leq x\leq0$.
If $-3\leq x\leq0$ then
$$\frac{|x+3|}{|x|+3}=\frac{x+3}{-x+3}$$
Let $y\in[0, 1]$, we need to find $x$ such that
$$y=\frac{x+3}{-x+3}$$
$$y(3-x)=x+3\implies 3y-3=xy+x\implies x=\frac{3y-3}{y+1}$$
It remains to show that 
$$-3\leq\frac{3y-3}{y+1}\leq 0$$
Since $y\in[0, 1]$, $y+1\geq0$ and $3y-3\leq0$, so
$$\frac{3y-3}{y+1}\leq0.$$
Since $y\in[0, 1]$, $y+1\leq2$ and $3y-3\geq-3$ then
$$\frac{3y-3}{y+1}\geq\frac{-3}2\geq-3.$$
