Numbers can be ‘connected’ to get new numbers:
for example, $\,1+2=3\,$.
Indeed, addition ‘+’ is a connective for numbers.
Sets can be ‘connected’ to get new sets:
for example, $\,\{1,2,3\}\cap \{2\} = \{2\}\,$.
Indeed, the intersection operator ‘$\,\cap\,$’ is a connective for sets.
Functions can also be ‘connected’ to get new functions:
the most important way to do this, called function composition, is the subject of the next lesson.
In this lesson, we look at some other ways that functions can be combined to get new functions.
Consider the box at right. It shows how two functions $\,f\,$ and $\,g\,$ are ‘combined’ to get a new function, named $\,f+g\,$. (Don't be intimidated by the multisymbol function name ‘$\,f+g\,$’! It's a very natural name, as you'll see.) Here's how this new function $\,f+g\,$ works:

The idea is the same for the difference, product, and quotient functions:
function  definition  domain 
$f  g$  $(fg)(x) := f(x)  g(x)$  $\text{dom}(fg) = \text{dom}(f) \cap \text{dom}(g)$ 
$fg$  $(fg)(x) := f(x)g(x)$  $\text{dom}(fg) = \text{dom}(f) \cap \text{dom}(g)$ 
$\displaystyle\frac{f}{g}$  $\displaystyle\bigl(\frac{f}{g}\bigr)(x) := \frac{f(x)}{g(x)}$ 
$\text{dom}\bigl(\frac{f}{g}\bigr) = \text{dom}(f) \cap \text{dom}(g) \cap \{x\ \ g(x)\ne 0\}$ Since division by zero is not allowed, the domain of the quotient function is a bit more complicated. 
Let $\,f(x) = x^2\,$ and $\,g(x) = x+3\,$.
Find $\,f+g\,$, $\,fg\,$, $\,fg\,$, and $\,\frac{f}{g}\,$.
Solution:
$\displaystyle
\begin{align}
\cssId{s63}{(f+g)(x)}\ &\cssId{s64}{:= f(x) + g(x)}\cssId{s65}{ = x^2 + x + 3}\cr
\cr
\cssId{s66}{(fg)(x)}\ &\cssId{s67}{:= f(x)  g(x)}\cssId{s68}{ = x^2  (x + 3)}\cssId{s69}{ = x^2  x  3}\cr
\cr
\cssId{s70}{(fg)(x)}\ &\cssId{s71}{:= f(x)g(x)}\cssId{s72}{ = x^2(x+3)}\cssId{s73}{ = x^3 + 3x^2}\cr
\cr
\cssId{s74}{\bigl(\frac{f}{g}\bigr)(x)}\ &\cssId{s75}{:= \frac{f(x)}{g(x)}}\cssId{s76}{ = \frac{x^2}{x+3}}\cssId{s77}{\ \ \text{for}\ \ x\ne 3}
\end{align}
$
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
