Finding range of a function using inequalities

[JC2000]

Homework Statement

If $f : R \Rightarrow R$and $g:R \Rightarrow R$ defined by $f(x) = |x|$ and $g(x)=(x-3)$ then ${g(f(x)) : -1.6 < x < 1.6}$

Relevant Equations

Given $f(x)$ and $g(x)$ for $-1.6 < x < 1.6$ we get $0\leq f(x)<1.6$
Now for $0\leq f(x)< 1$ $\implies$ $g(f(x))=-3$ since $-3\leq f(x)-3<-2$
Again for, $1\leq f(x)< 1.6$ $\implies$ $g(f(x))=-2$ since $-2 \leq f(x)-3 < -1.4$
Thus the required set is {-3,-2}

My attempt :

Given $f(x)$ and $g(x)$ for $-1.6 < x < 1.6$ we get $0\leq f(x)<1.6$
Thus, for $f(g(x))$ we get $-3 \leq g(f(x)) < -1.4$
Thus the required set should be the interval $[-3, -1.4)$?

My Questions :
1. What have I missed since my answer does not match the given solution.
2. In the given solution why is the $f(x)$ inequality broken into $0\leq f(x)< 1$ and $1\leq f(x)< 1.6$
3. After splitting the inequality I don't understand how the result is not an interval but two unique integers.

 

[PeroK]

Your answer is correct given the problem stated. The other answer may be the answer to a different question.

 

[WWGD]

My attempt :

Given $f(x)$ and $g(x)$ for $-1.6 < x < 1.6$ we get $0\leq f(x)<1.6$
Thus, for $f(g(x))$ we get $-3 \leq g(f(x)) < -1.4$
Thus the required set should be the interval $[-3, -1.4)$?

My Questions :
1. What have I missed since my answer does not match the given solution.
2. In the given solution why is the $f(x)$ inequality broken into $0\leq f(x)< 1$ and $1\leq f(x)< 1.6$
3. After splitting the inequality I don't understand how the result is not an interval but two unique integers.

You may want to use \mathbb r wrapped around in Latex for the Real numbers, instead of just R. It comes out like this $\mathbb R$. In some cases you may eliminate ambiguity.