Build the graph of a function with absolute value. The function is: 
And my idea of graphic (i did it using two graphs and deleting some parts)

Is it correct?
 A: Your choice of scale makes the graph a bit difficult to interpret, but it appears you correctly graphed $y = x^2 + 3$ when $|x| > 1$.  However, you incorrectly graphed $y = |x + 2| + |x - 1|$ in the interval $[-1, 1]$.  
Since $x - 1 \geq 0$ if $x \geq 1$ and $x - 1 < 0$ if $x < 1$, 
\begin{align*}
|x - 1| & = 
\begin{cases}
x - 1 & \text{if $x \geq 1$}\\
-(x - 1) & \text{if $x < 1$}
\end{cases}\\
& = 
\begin{cases}
x - 1 & \text{if $x \geq 1$}\\
-x + 1 & \text{if $x < 1$}
\end{cases}
\end{align*}
Since $x + 2 \geq 0$ if $x \geq -2$ and $x + 2 < 0$ if $x < -2$, 
\begin{align*}
|x + 2| & = 
\begin{cases}
x + 2 & \text{if $x \geq -2$}\\
-(x + 2) & \text{if $x < -2$}
\end{cases}\\
& = 
\begin{cases}
x - 1 & \text{if $x \geq -2$}\\
-x - 2 & \text{if $x < -2$}
\end{cases}
\end{align*}
Putting these rules together yields
\begin{align*}
|x + 2| + |x - 1| & = 
\begin{cases}
x + 2  + x - 1 & \text{if $x \geq 1$}\\
x + 2 - (x - 1) & \text{if $-2 \leq x < 1$}\\
-(x + 2) - (x - 1) & \text{if $x < -2$}
\end{cases}\\
& = 
\begin{cases}
2x + 1 & \text{if $x \geq 1$}\\
3 & \text{if $-2 \leq x < 1$}\\
-2x - 1 & \text{if $x < 1$}
\end{cases}
\end{align*} 
If $|x| \leq 1$, then $-1 \leq x \leq 1$.  If $-1 \leq x < 1$, then the rule for $-2 \leq x < 1$ applies, so $$y = |x + 2| + |x - 1| = 3$$  If $x = 1$, then the rule for $x \geq 1$ applies, so $$y = |x + 2| + |x - 1| = 2x + 1 = 2 \cdot 1 + 1 = 2 + 1 = 3$$  Thus, in the interval $[-1, 1]$, the graph of $y = |x + 2| + |x - 1|$ lies on the horizontal line $y = 3$.    
The graph of the function 
$$
y = 
\begin{cases}
x^2 + 3 & \text{if $|x| > 1$}\\
|x + 2| + |x - 1| & \text{if $|x| \leq 1$}
\end{cases}
$$ 
is sketched below.  While the grid lines make it somewhat difficult to tell, the circles at the points $(1, 3)$ and $(-1, 3)$ are filled since these points are on the graph while the circles at the points $(1, 4)$ and $(1, -4)$ are empty since these points are not on the graph.

A: The green parts are correct, but try to look at the blue part again. For example, we have
$$
f(1) = |1-1| + |1+2| = 3.
$$
