When you're given a multistep task to perform, you may want to break it into pieces,
and assign different pieces to different people.
When you're given a function that does several things,
you may want to break it into ‘smaller’ functions that accomplish the same job!
EXAMPLE (breaking a composite function into pieces)
Consider the function $\,h(x) = 5(x4)^3  7\,$.
This function $\,h\,$ does the following:
 subtracts $\,4\,$
 cubes the result
 multiplies by $\,5\,$
 subtracts $\,7\,$
Break these four tasks into two pieces, as shown in the mapping diagram below:
 assign the first two tasks (subtract $\,4\,$, then cube) to $\,f\ $:
$f(x) = (x4)^3$
 assign the last two tasks (multiply by $\,5\,$, then subtract $\,7\,$) to $\,g\ $:
$g(x) = 5x  7$
With these assignments, the
composite function $\,g\circ f\ $ (where $\,f\,$ acts first, followed by $\,g\,$) accomplishes the same thing as $\,h\,$:
$$
(g\circ f\,)(x) = g(f(x)) = g\bigl((x4)^3\bigr) = 5(x4)^3  7 = h(x)
$$
You can break a task into pieces in different ways!
Of course, you can delegate the responsibilities in different ways:

$f\,$ takes the first step (subtract $4$): $f(x) = x4$
$g\,$ takes the other three steps (cube, multiply by $5$, subtract $7$): $g(x) = 5x^3  7$

$f\,$ takes the first three steps (subtract $4$, cube, multiply by $5$): $f(x) = 5(x4)^3$
$g\,$ takes the last (subtract $7$): $g(x) = x  7$
In both cases, be sure to check that $\,(g\circ f\,)(x) = h(x)\,$.
You can use more than two helpers!
Or, you can use more ‘helper’ functions.
For example:
 let $\ a\ $ subtract $4$: $a(x) = x4$
 let $\ b\ $ cube: $b(x) = x^3$
 let $\ c\ $ multiply by $5$ and subtract $7$: $c(x) = 5x  7$
Then,
$
\displaystyle
\begin{align}
(c\circ b\circ a\,)(x) &= c(b(a(x)))\cr
&= c(b(x4))\cr
&= c((x4)^3)\cr
&= 5(x4)^3  7\cr
&= h(x)
\end{align}
$
The exercises in this lesson give you practice with this process of writing
a function as a composition.
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
using a function box ‘backwards’