Given certain conditions, prove a function G(x) is always equal to 4. Theres a question I've been having trouble with:
$G(x)$ is the function $|x+2|+|x-2|$. Show that if $-2<x<2$ then $G(x) = 4$.
Any help would be greatly appreciated. No calculus please :D

For problems with absolute values, break into cases. Remember,
$$|y|=\begin{cases}\hphantom{-}y & \text{if }y\geq 0,\\ -y & \text{if }y<0.\end{cases}$$
Thus,
$$\begin{align*}
|x+2|&=\begin{cases}\hphantom{-(}x+2 & \text{if }x+2\geq 0,\\ -(x+2) & \text{if }x+2<0\end{cases}\\\\\\
&=\begin{cases}\hphantom{-}x+2 & \text{if }x\geq -2,\\ -x-2 & \text{if }x<-2\end{cases}
\end{align*}$$
and
$$\begin{align*}
|x-2|&=\begin{cases}\hphantom{-(}x-2 & \text{if }x-2\geq 0,\\ -(x-2) & \text{if }x-2<0\end{cases}\\\\\\
&=\begin{cases}\hphantom{-}x-2 & \text{if }x\geq 2,\\ -x+2 & \text{if }x<2.\end{cases}
\end{align*}$$
Putting it all together,
$$\begin{align*}
G(x)&=|x+2|+|x-2|\\\\\\
&=\begin{cases}
\hphantom{-}x+2+|x-2| &\text{if }x\geq -2=\begin{cases}\hphantom{-}x+2+x-2 &\text{if }x\geq -2\textbf{ and }x\geq 2,\\
\hphantom{-}x+2-x+2 &\text{if }x\geq -2\textbf{ and }x<2
\end{cases}\\\\\\\\\\\\\\\\
-x-2+|x-2| &\text{if }x< -2=\begin{cases}-x-2+x-2 &\text{if }x<-2\textbf{ and }x\geq 2,\\
-x-2-x+2 &\text{if }x<-2\textbf{ and }x<2
\end{cases}
\end{cases}\\\\\\
&=\begin{cases}
\hphantom{-}2x & \text{if }x\geq -2\textbf{ and }x\geq 2,\\
\hphantom{-}4 & \text{if }x\geq -2\textbf{ and }x<2,\\
-4 & \text{if }x<-2 \textbf{ and }x\geq 2\; (\color{red}{\text{impossible}}),\\
-2x & \text{if }x<-2\textbf{ and }x<2
\end{cases}\\\\\\\\\\
&=\begin{cases}
\hphantom{-}2x & \text{if }x\geq 2,\\
\hphantom{-}4 & \text{if }-2\leq x<2,\\
-2x & \text{if }x<-2.
\end{cases}
\end{align*}$$
