Introduction

Function composition builds a new function by plugging one function’s output into another. 

In simpler terms, if you had two functions, $f\left(x\right)$ and $g\left(x\right)$, then the composite function is $f\left(g\left(x\right)\right)$ (sometimes denoted as $f\circ g$; this is not multiplication; this is an open dot). 

This chaining of dependencies is useful for modeling. For example, suppose the function $T\left(d\right)$ gives the average daily temperature on day $d$, and $C\left(T\right)$ gives the cost to heat a house at temperature $T$. Then cost ($C$) as a function of days ($d$) is $C\left(T\left(d\right)\right)$. The cost depends on the temperature $T$, which depends on the days $d$.Combining the two steps into one function $C\left(T\left(d\right)\right)$ is the definition of forming the composition between function $C$ and function $T$ (or in other words, $C\circ T$

Although it may be hard to see the practical uses now, it is often very useful to see how a certain variable changes with respect to another. If you are taking AP Physics or AP Chemistry, you can see this pattern being used all the time. Though it may not be clear, every time you substitute an equation or variable in, you are, in some way, creating a composition of functions.

Definition and Notation

Formally, if $f$ and $g$ are functions, the composition $f\circ g$ is defined by $\left(\left(f\circ g\right)\left(x\right)\right)=f\left(g\left(x\right)\right)$.

We read this as “$f$ of $g$ of $x$” or “$f$ composed with $g$ at $x$.” The circle symbol ∘ means that we first apply $g$ to $x$, then apply $f$ to the result.

(Do NOT confuse this symbol with a multiplication dot. They do not always produce the same result.)

It can be difficult to understand just looking at the function notations themselves, so let’s look at an example. Say we had $f(x)=x^2$ and $g(x)=2x+5$. Then, the composition of the two functions would be defined as so:

$\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

Substituting $g\left(x\right)$ in, 

$f\left(g\left(x\right)\right)=f\left(2x+5\right)$

Then, plugging in $2x+5$ into our function $f$, $f\left(g\left(x\right)\right)={\left(2x+5\right)}^2$

Composition is a binary operation on functions (meaning that it takes two functions, uses composition, and produces another function), like addition or multiplication of functions, but note it is not the same as multiplying values: usually $f\left(g\left(x\right)\right)\ne f\left(x\right)\cdot g\left(x\right)$

Additionally, composition is also associative, meaning the order does not matter. 

What this means is that:$\left(\left(f\circ g\right)\circ h\right)\left(x\right)=\left(f\circ g\right)\left(h\left(x\right)\right)=f\left(g\left(h\left(x\right)\right)\right)$

and

$\left(f\circ \left(g\circ h\right)\right)\left(x\right)=f\left(g\circ h\right)\left(x\right)=f\left(\left(h\left(x\right)\right)\right)$

Evaluating Composite Functions

There are three ways to evaluate composite functions.

First off, though, to compute a composite value $(f\circ g)(x)$ at a specific $x$, you must always work from the inside outwards: first, evaluate $g\left(x\right)$, then evaluate $f$ at whatever value you get from your calculation of $g$ at $x$.

For example, if $f(x)=3x-1$ and $g\left(x\right)=x^2+4$, then

$\left(f\circ g\right)\left(2\right)=f\left(g\left(2\right)\right)=f\left(22+4\right)=f\left(8\right)=3\cdot 8-1=23$

Analytic/Formula Method

First, substitute $g\left(x\right)$ into $f$. Suppose $f\left(x\right)=x^2$ and $g(x)=2x+1$.

We first replace the $x$ in $f$ with $g\left(x\right)$ to get $f\left(g\left(x\right)\right)={\left(2x+1\right)}^2$. This gives us a single formula for the composite function, to which you can simply plug in whatever value of $x$ they ask you to find the composite function at. If we were asked to find the composite of $f$ and $g$ at $x=3$, for example, we would simply plug $x=3$ into our new composite.

$f\left(g\left(3\right)\right)={\left(2\left(3\right)+1\right)}^2={\left(6+1\right)}^2=7^2=49$

Numerical/Tabular Method

Here, we use the values of $g$ at $x$ as our inputs for the function $f$. Going back to our previous example, Suppose

$f\left(x\right)=x^2$ and $g(x)=2x+1$. We are still trying to find the composite of $f$ and $g$ at $x=3$.

Using this method, we would first find $g\left(3\right)$:

$g\left(3\right)=2\left(3\right)+1=6+1=7$

Then, we would plug in our output for $g\left(3\right)$, $7$, as our input for $f\left(x\right)$. 

$f\left(g\left(3\right)\right)=\left(g{\left(3\right)}^2\right)$

$f\left(7\right)=\left(7^2\right)=49$

The reason this method is sometimes also called the “tabular method” is because in later units (such as when you begin to analyze trig functions), you will often create a table of values for the parent trig function, and then apply the transformations in a separate column (this is a form of composite functions).

Graphical Method

This method is the most straightforward, but really only useful when you’re using a calculator to graph it. You simply graph the composition of $f$ and $g$ by plugging $f\left(g\left(x\right)\right)$ into a calculator. Then, you read off values from the graphs at specific values of $x$

Although, sometimes you may be provided separate graphs (or even a table of values) that describe $f$ and $g$. Say the graph of $g$ has a point $g\left(1\right)=3$, and the graph of $f$ has a point $f\left(3\right)=6$. Then, $\left(f\circ g\right)\left(1\right)=f\left(g\left(1\right)\right)=f\left(3\right)=6$

In summary, composing “processes” involves plugging the outputs of one function into another, whether by algebraic solution or by carrying intermediate values that are solved for, given graphically, or given numerical (usually through means of a table of a values).

Domain of a Composite Function

The domain of $\left(f\circ g\right)$ consists of those $x$ for which both steps make sense: $x$ must be in the domain of $g$ and $g\left(x\right)$ must be in the domain of $f$, so both $g\left(x\right)$ and $f\left(x\right)$ are defined.

In other words, the value of $x$ must be in the domain of $g$, and the range of $g$ must be included within the domain of $f$.

To save time on tests and/or quizzes on composite functions, you must first determine whether the domain will pose an issue.

Commutativity and Identity

Although it is associative, composition is not commutative, meaning (usually) $f\circ g\ne g\circ f$ (note, it is possible for $f\circ g=g\circ f$ in some cases; however, the chances are very unlikely).

The AP exam will often emphasize that order matters in composition.

Decomposing Functions

Sometimes a given function $h\left(x\right)$ can be rewritten as a composite $f\left(g\left(x\right)\right)$, which can simplify analysis. However, in the context of the AP exam, this is just free points.

Say the function $h\left(x\right)={\left(x+1\right)}^2$.

I could say that $h\left(x\right)=f\left(g\left(x\right)\right)$ when $f\left(x\right)=x^2$ and $g\left(x\right)=x+1$.

I could even say that $h\left(x\right)=f\left(g\left(x\right)\right)$ when $f\left(x\right)={\left(x+1\right)}^2$ and $g\left(x\right)=x$. Obviously, this example has no practical use. However, the point is, you can decompose functions into other functions by using the idea of composition of functions.

Transformations Viewed as Composition

Standard function transformations (shifts, stretches, reflections) can themselves be seen as compositions with simple functions. 

For shifts, say we have functions $f\left(x\right)={\left(x\right)}^2$ and $g\left(x\right)=x-1$. If we wanted to create a new function $h\left(x\right)$ where $h\left(x\right)=f\left(x\right)$ but shifted right by one, we could write $h\left(x\right)=f\left(g\left(x\right)\right)$.

Similarly, this can be applied to both stretches (dilations) and reflections. The same idea is applied.

Practice Questions

Solutions to Practice Problems

Question 1: $(f\circ g)(x)=\frac{1}{\sqrt{x+5}-3}$ Since we have a square root in the denominator, the value inside of the square root must be greater than or equal to $-5$. Additionally, since $\sqrt{4+5}-3=0,$ the domain can not include $x=4$, as that would result in the denominator being $0$. Therefore, our domain is $x\ge -5$, with $x\ne 4$.

Question 2: $(f\circ g)(x)=f(g(x))=f(x^2-4)=2(x^2-4)+1=2x^2-8+1=2x^2-7$

$(g\circ f)(x)=g(f(x))=g(2x+1)=(2x+1{)}^2-4=4x^2+4x+1-4=4x^2+4x-3$.

Question 3: One possible answer is

$g(x)=3x-5$

$f(x)=x^2+2$

$h(x)=(f\circ g)(x)=g(f(x))=(3x-5{)}^2-2.$

Question 4: $(f\circ g)(x)=f(g(x))=f(x-1)=\sqrt{x-1}$. The domain is $x\ge 1$.

$(g\circ f)(x)=g(f(x))=g(sqrt(x))=sqrt(x)-1$. The domain is $x\ge 0$.

Therefore, their domains are not equal.

Question 5: $g(f(x))=g(\sqrt{x-1})=(\sqrt{x-1}{)}^2-4=x-1-4$ for $x\ge 1=x-5$ for $x\ge 1$.

$f(g(f(x))=f(x-5)$ for $x\ge 1=\sqrt{x-5-1}$ for $x\ge 1=\sqrt{x-6}$.

Note that we can drop the original domain restriction that $x\ge 1$ because our final expression requires that $x\ge 6$, which is stronger than $x\ge 1$.

The domain needed is $x\ge 6$

Question 6: The domain of $g(x)$ is $x\ne -2$.

$f(g(x))=\frac{1}{\frac{1}{x+2}+1}$, so its domain is that $x\ne -2,-3$.

Question 7: Heating cost = $C(T(d))=C(30-0.5d)=1000-20(30-0.5d)=1000-600+10d=400+10d$.

$C(T(10))=500$ dollars.

Question 8: $(f\circ g)(x)=f(g(x))=g(x{)}^2$. If we want $(f\circ g)(x)=(x-3{)}^2,g(x)=(x-3)$. Since we are subtracting $3$ from the input, we are shifting the graph of $f(g(x))$ to the right by $3$, as it now has a zero at $x=3$ instead of $x=0$.