# 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:

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?

## Answers

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

A1.The compositions are:and

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

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

A2.The composition \(\normalsize{f \circ g}\) is\[\Large{(t \circ d)(x)=t(x+30)=1.15(x+30)=1.15x+34.5},\] \[\Large{(d \circ t)(x)=d(1.15x)=1.15x+30.}\]

A3.The compositions are:If taxes don’t apply to delivery costs, then the total cost is given by \(\normalsize{(d \circ t)(x)}.\)

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

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