Algebra of functions and composition

Doing arithmetic with functions

Two functions can be combined in the same way as two numbers. We do it point by point: at each $x$ we find $f(x)$ and $g(x)$, and then we add, subtract, multiply or divide the two numbers.

Example. Let $f(x) = \sqrt{x}$ and $g(x) = x - 4$. The domain of $f$ is $[0, \infty)$ and the domain of $g$ is $\mathbb{R}$. So the shared domain is $[0, \infty)$.

• $(f + g)(x) = \sqrt{x} + x - 4$, with domain $[0, \infty)$.
• $(fg)(x) = \sqrt{x}\,(x - 4)$, with domain $[0, \infty)$.
• $\left(\dfrac{f}{g}\right)(x) = \dfrac{\sqrt{x}}{x - 4}$. Also $g(x) \neq 0$, so $x \neq 4$. The domain is $[0, 4) \cup (4, \infty)$.

Doing one function after another

Think of converting a length. A machine $f$ changes metres to centimetres: $f(x) = 100x$. A second machine $g$ changes centimetres to millimetres: $g(x) = 10x$. Put them one after the other, and a length in metres becomes a length in millimetres. We can replace the two machines by one: $x \mapsto 1000x$.

This is a composite function. The composite of $f$ and $g$, with $f$ first, is written $g \circ f$ (read "$g$ after $f$"):

In the picture, $f(x) = 2x + 1$ and $g(x) = x^2$. For $x = 3$, we get $f(3) = 7$, and then $g(7) = 49$. So $(g \circ f)(3) = 49$.

The order matters. For the same functions,

At $x = 3$ the first is $49$ and the second is $19$. In general, $g \circ f \neq f \circ g$. Putting on your socks and then your shoes is not the same as shoes and then socks.

Set $b = 1$: the two lines meet everywhere, because multiplying by $1$ does nothing. For other $b$, the order of the two machines changes the answer.

The domain of a composite function

For $g(f(x))$ to make sense, two things must hold: $x$ must be in the domain of $f$, and then $f(x)$ must be in the domain of $g$.

Example. Let $f(x) = x - 4$ and $g(x) = \sqrt{x}$.

• $(g \circ f)(x) = \sqrt{x - 4}$. We need $x - 4 \ge 0$, so the domain is $[4, \infty)$.
• $(f \circ g)(x) = \sqrt{x} - 4$. We need only $x \ge 0$, so the domain is $[0, \infty)$.

The two functions are different in the rule and in the domain.

Three facts

• Composition is associative: $h \circ (g \circ f) = (h \circ g) \circ f$. So we can write $h \circ g \circ f$ without brackets.
• The identity function $I(x) = x$ changes nothing: $f \circ I = I \circ f = f$.
• It is usually not commutative: $g \circ f \neq f \circ g$.

Taking a function apart

A complicated function is often a composite of simple ones. To take it apart, ask: what do I do first to $x$, and what do I do last?

Example. $h(x) = \sqrt{x^2 + 1}$. First we square $x$ and add $1$, so $f(x) = x^2 + 1$. Last we take the root, so $g(x) = \sqrt{x}$. Then $h = g \circ f$.

Practise with new numbers

Try it in Python

Python can build a composite function from two functions. This program makes $g \circ f$ and $f \circ g$ and compares them.

Remember these

• $(f + g)(x)$, $(fg)(x)$ and so on are found point by point, on the domain that both functions share.
• For a quotient, remove the $x$ where the denominator function is zero.
• $(g \circ f)(x) = g(f(x))$: the function inside works first.
• In general, $g \circ f \neq f \circ g$.
• The domain of $g \circ f$: $x$ in the domain of $f$, with $f(x)$ in the domain of $g$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

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

2. Find the domain of $\dfrac{f}{g}$ when $f(x) = x + 1$ and $g(x) = x^2 - 1$.

3. Write $h(x) = (x - 5)^3$ as $g \circ f$.

4. $f(x) = x^2$ and $g(x) = \sqrt{x}$. Is $(f \circ g)(x) = x$ for every real $x$?