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Grid with 2 points and 2 paths between them. One path has one segment labelled g circle f. Other path is made of two segments labelled f and g.
First f, then g.

Composing functions

When we view graphs as functions, taking a given value of \(\normalsize{x}\) and outputting a particular value of \(\normalsize{y}\) according to a rule \(\normalsize{y=f(x)}\), then an interesting new operation arises: composition.

In this step we look at composing functions rather generally, and give examples from the various kinds of linear, quadratic and inverse functions we have already studied.

Composing functions

If \(\normalsize{f}\) and \(\normalsize{g}\) are functions, then the composition \(\normalsize{g \circ f}\) is the function defined by:

\[\Large{(g \circ f)(x)=g(f(x))}.\]

It means that we first apply \(\normalsize{f}\) to \(\normalsize{x}\), and then apply \(\normalsize{g}\) to \(\normalsize{f(x)}\). This is one of those cases where it might be more natural to read right to left.

Composition of linear functions

Let’s do a few examples involving simple linear functions. If \(\normalsize{f(x)=x+3}\) and \(\normalsize{g(x)=5x-1}\), then

\[\Large{(g \circ f)(x)=g(f(x))=g(x+3)=5(x+3)-1=5x+14}.\]

Note that

\[\Large{(f \circ g)(x)=f(g(x))=f(5x-1)=(5x-1)+3=5x+2}\]

and that this is different from \(\normalsize{g \circ f}\).

We can also note that more generally the composition of two linear functions will be another linear function.

Q1 (E): If \(\normalsize{h(x)=x-1}\) and \(\normalsize{k(x)=3x+2}\), find \(\normalsize (h \circ k)(x)\) and \(\normalsize (k \circ h)(x)\).

However this is not so simple when we look at compositions of higher degree functions.

Composition of quadratic functions

If \(\normalsize{f(x)=x^2+1}\) and \(\normalsize{g(x)=3x^2-x}\) then

\[\Large{\begin{align}(f \circ g)(x)=f(g(x)) & =(3x^2-x)^2+1 \\ & =9x^4-6x^3+x^2+1.\end{align}}\]

In general we can see that the composition of two quadratic functions is going to be a quartic, or degree four, function.

However the composition of a quadratic function with a linear function will be a quadratic function.

Q2 (E): What is the composition \(\normalsize{f \circ g}\) of the functions \(\normalsize{f(x)=x^2+1}\) and \(\normalsize{g(x)=3x-4}\)?

Q3 (M): You buy a new kitchen table. Taxes are \(\normalsize{15\%}\) and delivery is \(\normalsize{\$30}\). If the cost of the table is \(\normalsize{x}\), then what is your total cost? In this question, there is a bit of an ambiguity, as we are not sure if taxes apply to delivery or not. Suppose we define functions \(\normalsize{t(x)=1.15x}\) and \(\normalsize{d(x)=x+30}\). Then what are \(\normalsize{(t \circ d)(x)}\) and \(\normalsize{(d \circ t)(x)}\)? If taxes don’t apply to delivery costs, then which composite represents your total cost?

Q4 (C): Compositions are not always defined, so one has to be careful. Let \(\normalsize{f(x)=-x^2}\) and \(\normalsize{g(x)=\sqrt{x}}\). For which values of \(\normalsize x\) are the compositions \(\normalsize{(f \circ g)(x)}\) and \(\normalsize{(g \circ f)(x)}\) defined?

Q5 (C): The morning operations \(\normalsize t\) for “brushing teeth” and \(\normalsize e\) for “eating breakfast” can be performed as \(\normalsize t \circ e\) or \(\normalsize e \circ t\). Daniel believes that the correct order of operations is \(\normalsize t \circ e\) because it keeps teeth clean for longer. Norman reckons that breakfast tastes better when we choose the order \(\normalsize e \circ t\). What do you think?


A1. The compositions are:

\[\Large (h \circ k)(x) = h(3x+2) = ((3x+2)-1) = 3x+1\]


\(\Large (k \circ h)(x) = k(x-1) = 3(x-1)+2= 3x-1\).

A2. The composition \(\normalsize{f \circ g}\) is

\[\Large{(f \circ g)(x)=f(3x-4)=(3x-4)^2+1=9x^2-24x+17.}\]

A3. The compositions are:

\[\Large{(t \circ d)(x)=t(x+30)=1.15(x+30)=1.15x+34.5},\] \[\Large{(d \circ t)(x)=d(1.15x)=1.15x+30.}\]

If taxes don’t apply to delivery costs, then the total cost is given by \(\normalsize{(d \circ t)(x)}.\)

A4. The composition \(\normalsize{(f \circ g)(x)}\) can be defined for all positive numbers \(\normalsize{x}\), since

\[\Large{(f \circ g)(x)=f(\sqrt{x})=-(\sqrt{x})^2=-x.}\]

However, the composition

\[\Large{(g \circ f)(x) = g(-x^2) = \sqrt{-x^2} }\]

can be defined only for \(\normalsize{x=0}\), since the value of the function \(\normalsize{f}\) is otherwise negative and we cannot take the square root of a negative number.

A5. Let’s take a vote in the discussion.

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This article is from the free online course:

Maths for Humans: Inverse Relations and Power Laws

UNSW Sydney