WRITING A FUNCTION AS A COMPOSITION

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Consider the function $\,h(x) = 5(x-4)^3 - 7\,$.
This function $\,h\,$ does the following:

1. subtracts $\,4\,$
2. cubes the result
3. multiplies by $\,5\,$
4. 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\ $:   $f(x) = (x-4)^3$
• 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 $\,h\,$: $$ \cssId{s24}{(g\circ f\,)(x)} \cssId{s25}{= g(f(x))} \cssId{s26}{= g\bigl((x-4)^3\bigr)} \cssId{s27}{= 5(x-4)^3 - 7} \cssId{s28}{= h(x)} $$

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 $4$):   $f(x) = x-4$
  $g\,$ 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 $5$):   $f(x) = 5(x-4)^3$
  $g\,$ 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 $4$:   $a(x) = x-4$
• let $\ b\ $ cube:   $b(x) = x^3$
• let $\ c\ $ multiply by $5$ and subtract $7$:   $c(x) = 5x - 7$

Then,

$ \displaystyle \begin{align} \cssId{s47}{(c\circ b\circ a\,)(x)}\ &\cssId{s48}{= c(b(a(x)))}\cr &\cssId{s49}{= c(b(x-4))}\cr &\cssId{s50}{= c((x-4)^3)}\cr &\cssId{s51}{= 5(x-4)^3 - 7}\cr &\cssId{s52}{= 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’