Composite of functions with absolute value range I have a really big problem with this next task, determing the range of composite g(f(x)).
$f(x)=-2x+4$
$g(x)=\lbrace{x^2-4};|x| \le 2\rbrace$
$g(x)=\lbrace{4x^2-x^4};|x|>2\rbrace$
The result is supposed to be:
$-1 \le x \le 3$ for the first part of g(x) with $|x| \le 2$ and $x<-1$ and $ x>3$ for the second part of g(x) with $|x| > 2$.
Thank you for your help
 A: By definition, $$\mathop{\mathrm{dom}} g \circ f = \{ x \in \mathop{\mathrm{dom}} f : f(x) \in \mathop{\mathrm{dom}} g\}.$$
Without an explicit restriction on $f$, assume its domain is the entire real line: $$\mathop{\mathrm{dom}} f = \mathbb{R}.$$ In other words, $x \in \mathop{\mathrm{dom}} f \iff x \in \mathbb{R}$.

In the case of $g(x) = x^2 - 4,\ |x| \leq 2$, $$\mathop{\mathrm{dom}} g = \{ x : |x| \leq 2 \}.$$ In other words, $x \in \mathop{\mathrm{dom}} g \iff |x| \leq 2$.
Therefore, $$\begin{align*}\mathop{\mathrm{dom}} g \circ f & = \{ x \in \mathop{\mathrm{dom}} f : f(x) \in \mathop{\mathrm{dom}} g \} \\ & = \{ x \in \mathbb{R} : |f(x)| \leq 2 \} \\ & = \{ x \in \mathbb{R} : |{-2x + 4}| \leq 2 \} \\ & = \{ x \in \mathbb{R} : -2 \leq -2x + 4 \leq 2 \} \\ & = \{ x \in \mathbb{R} : -6 \leq -2 x \leq -2 \} \\ & = \{ x \in \mathbb{R} : 3 \geq x \geq 1 \} \\ & = \{ x \in \mathbb{R} : 1 \leq x \leq 3 \} \\ & = (1,3).\end{align*}$$
Note $-1 \not \in \mathop{\mathrm{dom}} g \circ f$, contrary to the claim in your original post.

In the case of $g(x) = 4x^2 - x^4,\ |x| > 2$, $$\mathop{\mathrm{dom}} g = \{ x : |x| > 2 \}.$$ In other words, $x \in \mathop{\mathrm{dom}} g \iff |x| > 2$.
Therefore, $$\begin{align*}\mathop{\mathrm{dom}} g \circ f & = \{ x \in \mathop{\mathrm{dom}} f : f(x) \in \mathop{\mathrm{dom}} g \} \\ & = \{ x \in \mathbb{R} : |f(x)| > 2 \} \\ & = \{ x \in \mathbb{R} : |{-2x + 4}| > 2 \} \\ & = \{ x \in \mathbb{R} : -2x + 4 < -2 \text{ or } -2x + 4 > 2 \} \\ & = \{ x \in \mathbb{R} : x > 3 \text{ or } x < 1 \} \\ & = \{ x \in \mathbb{R} : x < 1 \text{ or } x > 3\} \\ & = (-\infty,1) \cup (3,+\infty). \end{align*}$$
Once more, I believe the answer you supplied in your original post is in error.

Note the range of $g$ never entered into our consideration; i.e., we could have found $\mathop{\mathrm{dom}} g \circ f$, e.g., in the first case, without knowing $g(x) = x^2 - 4$.
Furthermore, because the domain of $g$ in the first case, $|x| \leq 2$, is the complement of the domain of $g$ in the second case, $|x| > 2$, the domain of $g \circ f$ in the first case is the complement (relative to the range of $f$) of the domain of $g \circ f$ in the second case.
I include, for your consideration, an alternative definition of $\mathop{\mathrm{dom}} g \circ f$: $$\mathop{\mathrm{dom}} g \circ f = \mathop{\mathrm{ran}} f \cap \mathop{\mathrm{dom}} g.$$
