# Function Composition

**Function composition** refers to the pointwise application of one function to another, which produces a third function. When we compose the function $f$ with $g$, we obtain $f \circ g$. Sometimes, $f \circ g (x)$ is also denoted as $f \big( g(x) \big)$.

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## Existence of Composite

To ensure that this is well-defined, we must check that $f \circ g (x)$ makes sense. For example, if $f(x) = \sqrt{x}$ and $g(x) = x- 1$, then $f \circ g ( 0 )$ is not properly defined, because we cannot have $\sqrt{ -1 }$. As such, the condition that we have to check is that the codomain of $g(x)$ lies within the domain of $f(x)$. In a later section, we will look at the case when this condition is not satisfied.

For now, we will make the assumption that the range of $g(x)$ is equal to the domain of $f(x)$, which satisfies this condition. We consider the functions $f: Y \rightarrow Z$ and $g: X \rightarrow Y$, which allows us to define function $f \circ g : X \rightarrow Z$.

Let's look at the following example of function composition, where the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ are given by $f(x) = x^2$ and $g(x) = x+1.$ Let's see how we can find certain values of $f\big(g(x) \big)$ by plugging in $x = 1, 2, 3, 4, 5:$

This gives us a pictorial sense of what is happening. Algebraically, we evaluate that

$f \circ g (x) = f ( x+1) = ( x+1)^2 = x^2 + 2x + 1.$

We can do the same for $g \circ f (x)$ and show that

$g \circ f(x) = g \big( x^2\big) = x^2 + 1 .$

Observe that $f \circ g \neq g \circ f$, which is a common mistake made. We say that function composition is not commutative.

If $f = \{ (1,1), (2,3), (3,2), (4,4) \}$ and $g = \{ (1,2), (2,1), (3,4), (4,3) \}$, what is $f\circ g?$

We need to find $f \circ g (x)$ for $x = 1, 2, 3, 4$:

$\begin{aligned} f \circ g (1) &= f( 2) = 3 \\ f \circ g (2) &= f(1) = 1 \\ f \circ g (3) &= f( 4) = 4 \\ f \circ g(4) &= g(3) = 2. \end{aligned}$

Thus, $f \circ g = \{ (1,3), (2,1), (3,4), (4,2) \}$. $_\square$

Consider the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x) = x^2 + 2x$ and $g(x) = x+1$. What is $f \circ g (x)?$

We have

$f \circ g (x) = f ( x+1) = (x+1)^2 + 2(x+1) = x^2 + 2x + 1 + 2x + 2 = x^2 + 4x + 3.\ _\square$

The function $M(t)$ is the number of chairs that are produced each month if the factory operates $t$ hours each day. The function $P(c)$ is the monthly profit earned by selling $c$ chairs each month. What is the monthly profit earned by letting the factory operate 8 hours each day?

If the factory operates 8 hours each day, the amount of chairs produced is $M (8)$.

If $M(8)$ chairs were sold, then the monthly profit is $P \big( M ( 8) \big)$. $_\square$Note: More generally, we know that the monthly profit of operating $t$ hours each day is $P\big( M(t) \big)$.

Consider the functions $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ and $g: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ given by $f(x) = x+1$ and $g(x) = \frac{1}{x}$. What is $f \circ g (x)$ and $g \circ f(x)?$

We have

$\begin{aligned} f \circ g(x) &= f \left ( \frac{1}{x} \right) = \frac{1}{x} + 1 = \frac{ x+1} { x} \\ g \circ f(x) &= g (x+1) = \frac{ 1}{ x+1}. \ _\square \end{aligned}$

Bonus question:If $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ are defined by $f(x) = 2x^2 + 3$ and $g(x) = 3x - 2$, then find$\begin{aligned} 1) & f \circ g (x) \\ 2) & g \circ f (x) \\ 3) & f \circ f (0). \\ \end{aligned}$

We have $\begin{aligned} 1) f \circ g (x) & = f\big(g(x)\big) \\&= f(3x - 2) \\ & = 2(3x - 2)^2 + 3 \\ &= 2(9x^2 - 12x + 4) + 3 \\ & = 18x^2 - 24x + 11 \\\\ 2) g \circ f (x) & = g\big(f(x)\big)\\& = g(2x^2 + 3) \\ & = 3(2x^2 + 3) - 2 \\ & = 6x^2 + 7 \\\\ 3) f \circ f (0) & = f\big(f(0)\big)\\& = f\big(2(0) + 3\big) \\ & = f(3)\\& = 2(3)^2 + 3 \\ & = 21. \ _\square \\ \end{aligned}$

## Codomain of $g$ does not lie within Domain of $f$

As shown above with the example of $g(x) = x-1$ and $f(x) = \sqrt{x}$, if the codomain of $g$ does not lie within the domain of $f$, then the function $f\big(g(x)\big)$ might not make sense.

However, what we could do is to restrict the domain further such that the range of values of $g$ lies within the domain of $f.$ Simply put, we are getting rid of values which do not make sense:

Since $f(x)$ is only defined on the non-negative real numbers, the range of $g(x)$ has to be the non-negative real numbers, or $x - 1 \geq 0$. This implies that $x \geq 1$, which is the restricted domain that will allow us to define $f\big(g(x)\big)$.

Hence, $f \circ g : [ 1, \infty) \rightarrow \mathbb{R} , f \circ g(x) = \sqrt{x-1}$.

Consider the functions $f: [0,1] \rightarrow [1,2]$ given by $f(x) = x+1$, and $g: \mathbb{R} \rightarrow \mathbb{R}$ given by $g(x) = x - 1$. Define $f \circ g$ on a suitable domain.

The image of $g(x)$ is all real numbers, which is not a subset of the domain of $f(x)$. Hence, we have to restrict the domain of $g(x)$. We require that $0 \leq g(x) \leq 1$, or that $0 \leq x-1 \leq 1$, which implies $1 \leq x \leq 2$.

On this restricted domain, we have $f \circ g(x) = f ( x-1) = x-1 + 1 = x$.

Hence, $f \circ g : [1,2] \rightarrow [1,2], f \circ g (x) = x$. $_\square$

Consider the functions $f (x) = \frac{1}{x}, x \neq 0$ and $g(x) = x -1$. What is $f \circ g?$

In this case, since we are not given the domain, we will assume that it is the real numbers. Also, we know that $\frac{ 1}{x}$ is not defined for $x = 0$, so the domain of $f$ is $x \neq 0$. We have to restrict the domain of $g(x)$ such that $g(x) \neq 0$, or that $x - 1 \neq 0$. Hence, the domain to use is $x \neq 1$.

On this restricted domain, we have $f \circ g(x) = f( x- 1) = \frac{1}{x-1}$.

Hence, $f \circ g (x) = \frac{1}{x-1}, x \neq 1$. $_\square$

Consider the functions $f: [2,10] \rightarrow [1,3]$ given by $f(x) = \sqrt{x - 1}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ given by $g(x) =2x-3$. What is $f \circ g ?$

The image of $g(x)$ is all real numbers, which is not a subset of the domain of $f(x)$. Hence, we have to restrict the domain of $g(x)$. We require $2 \leq g(x) \leq 10,$ or equivalently $2 \leq 2x-3 \leq 10 ,$ which implies $\frac{5}{2} \leq x \leq \frac{13}{2} .$

On this restricted domain, we have $f \circ g(x) = f ( 2x-3) = \sqrt{2x-3-1} = \sqrt{2x-4} .$ Therefore, it follows that $f \circ g (x) = \sqrt{2x-4},$ where $\frac{5}{2} \leq x \leq \frac{13}{2} .$ $_\square$

Consider the functions $f(x) =\frac{1}{x^2-3x+2}$ and $g(x) =x+1$. What is $f \circ g ?$

The given function can be rewritten as

$f(x)=\frac{1}{x^2-3x+2}=\frac{1}{(x-1)(x-2)} . \qquad (1)$

From $(1),$ we know that $\frac{ 1}{x^2-3x+2}$ is not defined for $x = 1$ and $x=2$, so the domain of $f$ is $x \neq 1,2$. We have to restrict the domain of $g(x)$ such that $g(x) \neq 1,2 ,$ or equivalently $x + 1 \neq 1,2$. Hence, the domain to use is $x \neq 0,1$.

On this restricted domain, we have

$\begin{aligned} f \circ g(x) &= f( x+1)\\ &= \frac{1}{(x+1)^2-3(x+1)+2} \\ &= \frac{1}{x^2-x}. \end{aligned}$

Hence, $f \circ g (x) = \frac{1}{x^2-x},$ where $x \neq 0,1 .$ $_\square$

Consider the functions $f(x) =\frac{1}{\sqrt{x-4}}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ given by $g(x) =x^2-12$. What is $f \circ g ?$

The value of $\sqrt{x-4}$ is either positive or zero, and a denominator cannot be zero. Hence, the domain of $x$ for $f(x)$ is

$\begin{aligned} x-4 \geq 0 , x \neq 4 \implies x>4. \end{aligned}$

So, we must restrict the domain of $g(x)$ such that $g(x) > 4,$ or equivalently $x^2-12>4 ,$ which implies that the domain to use is $x <-4 \text{ or } x>4.$

On this restricted domain, we have

$\begin{aligned} f \circ g(x) &= f( x^2-12)\\ &= \frac{1}{\sqrt{x^2-12-4}} \\ &= \frac{1}{\sqrt{x^2-16}}. \end{aligned}$

Hence, $f \circ g (x) = \frac{1}{\sqrt{x^2-16}},$ where $x<-4 \text{ or } x>4 .$ $_\square$

Consider the functions $f(x) =\ln \big(\sqrt{x-4}-1\big)$ and $g(x) =\frac{1}{x}.$ What is $f \circ g ?$

The value of $\sqrt{x-4}-1$ is positive and the value of $\sqrt{x-4}$ is either positive or zero. Hence, it follows that the domain of $x$ for $f(x)$ is

$\begin{aligned} \sqrt{x-4}-1 > 0 \ \text{and} \ x-4 \geq 0 \implies x > 5. \end{aligned}$

So, the range of $g(x)$ is $g(x) > 5,$ or $\frac{1}{x} > 5 ,$ which implies that the domain to use is $x < \frac{1}{5}.$

On this restricted domain, we have

$\begin{aligned} f \circ g(x) = f \left ( \frac{1}{x} \right ) = \ln \left( \sqrt{\frac{1}{x}-4}-1 \right). \end{aligned}$

Hence, $f \circ g (x) = \ln \left ( \sqrt{\frac{1}{x}-4}-1 \right ),$ where $x< \frac{1}{5}.$ $_\square$

Consider $f(x) = x^2$ and $g(x) = 2^x$. Solve the equation $f \circ g (x) = g \circ f (x).$

We have

$\begin{aligned} f \circ g (x) &= f\big(g(x)\big) = f(2^x) = (2^x)^2 = 2^{2x}\\ g \circ f (x) &= g\big(f(x)\big) = g(x^2) = 2^{x^2}. \end{aligned}$

As

$\begin{aligned} f \circ g (x) & = g \circ f (x) \\ 2^{2x} & = 2^{x^2} \\ 2x & = x^2 \\ x^2 - 2x & = 0 \\ x(x - 2) & = 0, \end{aligned}$

we have $x = 0,2.\ _\square$

Bonus questions:

1) If $f(x) = 2x^2 + 3$ and $\ g(x) = 3x - 2,$ then find $g \circ f \circ f (3)$.

2) If $f(x) = 2, g(x) = x^2,$ and $h(x) = 2x,$ then find $h \circ g \circ f (x)$.

We have$\begin{aligned} 1)\ \ g \circ f \circ f (3) & = g\left[f\big(f(3)\big)\right] \\ & = g\big[f(18 + 3)\big] \\&= g\big[f(21)\big] \\ & = g\big(2(21)^2 + 2\big) \\&= g(885) \\ & = 3(885) - 2 \\ & = 2653 \\\\ 2)\ \ h \circ g \circ f (x) & = h\left[g\big(f(x)\big)\right] \\ & = h\big[g(2)\big] \\&= h[4] \\ & = 2(4) \\ & = 8.\ _\square \\ \end{aligned}$

## Iterated Function Composition

If the range of a function is a subset of the domain of a function, then we can compose this function with itself. If so, we use $f^2 (x)$ to denote $f \circ f (x)$. More generally, we say that $f^n (x)$ is $f$ composed with itself $n$ times, i.e.

$\underbrace{ f \circ f \circ \cdots \circ f }_{n \text{ times}}.$

A function which satisfies $f^2(x) = x$ is called an **involution**. This means that the function is its own inverse.

Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x) = x + 1.$ What is $f ^ n (x) ?$

We have

$\begin{aligned} f(x) &= x + 1 \\ f^2 (x) &= f( x + 1) = x + 1 + 1 = x + 2\\ f^3 (x) &= f(x + 2) = x+2 + 1 = x + 3. \end{aligned}$

This strongly suggests that $f^n (x) = x+ n$, which we can prove using induction. $_\square$

Consider the function $f(x) = \frac{x + 1}{x - 1}\ (x \neq \pm 1).$ Then find $f \circ f \circ f (x)$ and $f \circ f \circ f \circ f (x).$

Given $f \circ f (x) = \frac{x + 1}{x - 1},$ let us first find $f \circ f (x):$

$f \circ f = f\big(f(x)\big) = f \left(\dfrac{x + 1}{x - 1}\right) = \dfrac{\dfrac{x + 1}{x - 1} + 1}{\dfrac{x + 1}{x - 1} - 1} = \dfrac{\hspace{3mm} \dfrac{2x}{x- 1}\hspace{3mm} }{\dfrac{2}{x - 1}} = x.$

You can see that

$f(x)$ is an identity functionas $f \circ f (x) = x:$$\begin{aligned} f \circ f \circ f (x) & = f\big(\underbrace{f \circ f (x)}_{x}\big) \\ & = f(x) \\ & = \dfrac{x + 1}{x - 1} \\\\\\ f \circ f \circ f \circ f (x) & = f \big(f(\underbrace{f \circ f)}_{x}\big) \\ & = f \big(f(x)\big) \\ & = f \circ f (x) \\ & = x. \ _\square \end{aligned}$

In some questions where you have to find multiple composite functions, if you get $f \circ f (x)$ as Identity function $\big($which means $f \circ f (x) = x\big),$ you are lucky enough and you need not solve the whole problem. If $f \circ f (x) = x,$ then

$f^n(x) = \underbrace{ f \circ f \circ \cdots \circ f }_{n \text{ times}} \ (x) = \begin{cases} f(x) &&\text{when } n \text{ is an odd number} \\ x &&\text{when } n \text{ is an even number}. \end{cases}$

For example, take the above problem where we got that $f \circ f (x)$ is an Identity function. In $f \circ f \circ f (x),$ $f$ is repeated 3 times or it can be written as $f^3(x),$ where 3 is an odd number. So,

$f \circ f \circ f (x) = f(x) = \dfrac{x + 1}{x - 1}.$

In $f \circ f \circ f \circ f (x) = f^4(x),$ 4 is an even number, so

$f \circ f \circ f \circ f (x) = x.$

Bonus question: Consider the previous problem. If $f(x) = \frac{x - 1}{x + 1},$ then find $f^3(x)$ and $f^4(x).$

Given $f(x) = \frac{x - 1}{x + 1},$$f \circ f (x) = f \left( \frac{x - 1}{x + 1} \right) = \frac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1} = \frac{x - 1 - x - 1}{x - 1 + x + 1} = - \frac{1}{x}.$

Here, $f \circ f (x) \neq x,$ so we cannot apply any shortcut and must go through fundamentals:

$\begin{aligned} f^3 (x) & = f\big(f(x)\big) \\&= f \left(- \frac{1}{x}\right) \\ & = \frac{-\frac{1}{x} - 1}{-\frac{1}{x} + 1} \\ & = - \frac{x + 1}{x - 1} \\\\\\ f^4(x) & = f\big(f^3(x)\big) \\&= f \left( - \frac{x + 1}{x - 1} \right) \\\\ & = \frac{\frac{-x - 1}{x - 1} - 1}{\frac{-x - 1}{x - 1} + 1} \\\\ & = \frac{-x - 1 - x + 1}{-x - 1 + x - 1} \\\\ & = x.\ _\square \end{aligned}$

$$

Consider the function $g: \mathbb{R} ^ + \rightarrow \mathbb{R} ^+$ given by $g(x) = \frac{1}{x}$. What is $g^n (x)?$

We have

$\begin{aligned} g(x) &= \frac{1}{x} \\ g^2(x) &= g \left( \frac{1}{x} \right) = \frac{1}{\hspace{3mm} \frac{1}{x}\hspace{3mm} } = x \\ g^3 (x) &= g(x) = \frac{ 1}{x}. \end{aligned}$

Thus, we have

$g^n(x) = \begin{cases} \frac{1}{x} && n \text{ is odd } \\ x && n \text{ is even. } _\square \\ \end{cases}$

**Cite as:**Function Composition.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/functions-composition/