Addition of two piecewise function Suppose I have function 

$$y(x)=\begin{cases}
 x+1\qquad & 0\leq x\leq1 \\
 2-x\qquad &  1<x \leq2 \\
0\qquad & \mathrm{elsewhere}
\end{cases}$$
  I need to find the function $g(x)=y(x+2)+2y(x+1)$


$$y(x+2)=\begin{cases}
 x+3\qquad &-2\leq x\leq-1 \\
 -x\qquad &  -1<x \leq0 \\
0\qquad &\mathrm{elsewhere}
\end{cases}$$
$$2y(x+1)=\begin{cases}
 2x+4\qquad & -1\leq x\leq0\\
 2-2x\qquad &  0<x \leq1 \\
0\qquad &\mathrm{elsewhere}
\end{cases}$$
Now,
$$g(x)=y(x+2)+2y(x+1)=\begin{cases}
 x+3\qquad & -2\leq x\leq -1 \\
 x+4\qquad &  -1<x \leq0 \\
2-2x\qquad &0<x\leq1\\
0\qquad &\mathrm{elsewhere}
\end{cases}$$
Now the result looks Okay,
But the problem is the value of $y(x+2)$ at $x=-1$ is $2$ and $2y(x+2)$ at $x=-1$ is also $2$ so if I add both then $y(x+2)+2y(x+1)$ at $x=-1$ will be $2+2=4$,but the value of $g(x)$ at $x=-1$ is $-1+3=2$ 

\begin{eqnarray}
g_1(x)=y(x+2)=\begin{cases}
x+3&\text{ for }-2\le x\le -1\\
-x&\text{ for }-1< x\le0\\
0&\text{ otherwise}
\end{cases}
\end{eqnarray}
and
\begin{eqnarray}
g_2(x)=2y(x+1)=\begin{cases}
2x+4&\text{ for }-1\le x\le0\\
2-2x&\text{ for }0< x\le1\\
0&\text{ otherwise}
\end{cases}
\end{eqnarray}
Giving the result
\begin{eqnarray}
g(x)=\begin{cases}
x+3&\text{ for }-2\le x<-1\\
3x+7&\text{ for }x=-1\\
x+4&\text{ for }-1< x\le0\\
2-2x&\text{ for }0< x\le1\\
0&\text{ otherwise }
\end{cases}
\end{eqnarray}
Now we see that $g_1(-1)=2,\,g_2(-1)=2$ and $g(-1)=4$.
