Adding and multiplying piecewise functions How do I add and multiply two piecewise functions?
$$
f(x)= 
\begin{cases}
x+3 &\text{if }x<2\\
\dfrac{x+13}{3} &\text{if }x>2
\end{cases}
$$
$$
g(x)= 
\begin{cases}
x-3 &\text{if }x<3\\
x-5 &\text{if }x>3
\end{cases}
$$
 A: Create an additional third case when $2<x<3$ and add/multiply the appropriate expressions for this in-between case.
A: It helps to think about the fact that you can only add and multiply stuff that actually exists (i.e. is defined):

We can only add and multiply these functions in places where they both exist at the same time, namely:
$x<2\\
2<x<3\\
x>3$
The final step is to check what the functions equal at each of these segments, then put it all together:
$$
f(x)\times g(x)= 
\begin{cases}
(x+3)\times(x-3) &\mbox{if } \quad x<2\\
(\frac{x+13}{3})\times(x-3) &\mbox{if } \quad 2<x<3\\
(\frac{x+13}{3})\times(x-5) &\mbox{if } \quad x>3
\end{cases}
$$
$$
f(x)+g(x)= 
\begin{cases}
(x+3)+(x-3) &\mbox{if } \quad x<2\\
(\frac{x+13}{3})+(x-3) &\mbox{if } \quad 2<x<3\\
(\frac{x+13}{3})+(x-5) &\mbox{if } \quad x>3
\end{cases}
$$
BONUS ROUND: What about when functions have different arguments, like $f(x)$ and g(y)? Since there is no $x$ in $g()$ and there is no $y$ in $f()$, we don't have to worry about one function being defined when the other one isn't. They live in different worlds (i.e. different domains) so we just have to add or multiply each case of one with each case of the other. For example:
$$
f(x)= 
\begin{cases}
x+3 &\mbox{if } \quad x<2\\
\frac{x+13}{3} &\mbox{if } \quad x>2
\end{cases}
$$
$$
g(y)= 
\begin{cases}
y-3 &\mbox{if } \quad y<3\\
y-5 &\mbox{if } \quad y>3
\end{cases}
$$
$$
f(x)\times g(y)= 
\begin{cases}
(x+3)\times(y-3) &\mbox{if } \quad x<2,y<3\\
(x+3)\times(y-5) &\mbox{if } \quad x<2,y>3\\
(\frac{x+13}{3})\times(y-3) &\mbox{if } \quad x>2,y<3\\
(\frac{x+13}{3})\times(y-5) &\mbox{if } \quad x>2,y>3
\end{cases}
$$
