Addition of two piecewise function

Question

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 $g(x)=y(x+2)+2y(x+1)$

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$$y(x+2)=\begin{cases} x+3\qquad &-2\leq x\leq-1 \\ -x\qquad & -1<x \leq0 \\ 0\qquad &\mathrm{elsewhere} \end{cases}$$

$$2y(x+1)=\begin{cases} 2x+4\qquad & -1\leq x\leq0\\ 2-2x\qquad & 0<x \leq1 \\ 0\qquad &\mathrm{elsewhere} \end{cases}$$

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 $-1+3=2$

Answer 1

\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 $g(-1)=4$.

Answer 2

In your calculation of $g(x)$, the first line, $x+3$, was obtained by adding $y(x+2)=x+3$ and $2y(x+1)=0$. The first of these two equations, $y(x+2)=x+3$, is valid for $x$ in the range $-2\leq x\leq-1$. The second, $2y(x+1)=0$, is valid when $x<-1$ and also when $x>1$. In particular, it is not valid when $x=-1$. So the $x+3$ line in your formula for $g(x)$ is not valid for $x=-1$ but only for $-2\leq x<-1$.

More generally, when you combine several formulas, each f which is valid for a certain range of $x$ values, then the combination will in general only be valid for those $x$ values that are in all the relevant ranges simultaneously.