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
 A: 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,
$$\begin{align*}
G(x)&=|x+2|+|x-2|\\\\\\
&=\begin{cases}
\hphantom{-}x+2+|x-2| &\text{if }x\geq -2=\begin{cases}\hphantom{-}x+2+x-2 &\text{if }x\geq -2\textbf{ and }x\geq 2,\\
\hphantom{-}x+2-x+2 &\text{if }x\geq -2\textbf{ and }x<2
\end{cases}\\\\\\\\\\\\\\\\
-x-2+|x-2| &\text{if }x< -2=\begin{cases}-x-2+x-2 &\text{if }x<-2\textbf{ and }x\geq 2,\\
-x-2-x+2 &\text{if }x<-2\textbf{ and }x<2
\end{cases}
\end{cases}\\\\\\
&=\begin{cases}
\hphantom{-}2x & \text{if }x\geq -2\textbf{ and }x\geq 2,\\
\hphantom{-}4 & \text{if }x\geq -2\textbf{ and }x<2,\\
-4 & \text{if }x<-2 \textbf{ and }x\geq 2\; (\color{red}{\text{impossible}}),\\
-2x & \text{if }x<-2\textbf{ and }x<2
\end{cases}\\\\\\\\\\
&=\begin{cases}
\hphantom{-}2x & \text{if }x\geq 2,\\
\hphantom{-}4 & \text{if }-2\leq x<2,\\
-2x & \text{if }x<-2.
\end{cases}
\end{align*}$$
A: If $x>-2$, then $x+2>0$, thus $|x+2|=x+2$.
Likewise, if $x<2$, then $x-2<0$, thus $|x-2| = -(x-2) = 2-x$.
Thus, for $-2<x<2$ we have $|x+2|+|x-2| = x+2+2-x = 4$.
A: The graph of $x \mapsto |x + 2|$ is a vee-shape that touches the $x$ axis at $x = -2$ and then goes up with a gradient of 1 from there. The graph of $x \mapsto |x - 2|$ is a vee-shape that touches the $x$ axis at $x = 2$ and to the left of that has a gradient of $-1$. So when you add them together, the gradients (between $-2$ and $2$) will cancel out and you'll get a flat line, i.e. a constant function. It's now only necessary to work out how high up the flat line is – but just work out what it is for any value of $x$ ($x = 2$ is good) and the flat line must be at that height all the way across.
A: Draw a number line. In general, $|x-a|$ is the distance from $x$ to $a$. So $|x-2|$ is the distance from $x$ to $2$. And $|x+2|=|x-(-2)|$ is the distance from $x$ to $-2$. 
Thus $|x+2|+|x+2|$ is the sum of the distances from $x$ to $-2$ and to $2$.  If $x$ is anywhere in the interval $=2\le x\le 2$, the sum of its distances from $-2$ and $2$ is always equal to the distance between $-2$ and $2$, which is $4$. 

A more informal way of seeing it is to suppose that you are standing on the number line at $-2$, and start walking slowly to the right. At the beginning, the sum of your distances from $-2$ and $2$ is $4$. For every tiny step $s$ that you take, your distance from $-2$ increases by $s$. But your distance from $2$ decreases by $s$, so the sum of your distances from $-2$ and $2$ does not change. This continues until you reach $2$, at which point the sum of your distances from $-2$ and $2$ starts to increase.
