Functions: notation, domain, composition and inverse

What a function is

A function $f$ takes each input $x$ from a set called its domain and returns exactly one output $f(x)$. The set of outputs actually produced is the range. "Exactly one" is the whole definition: a rule that sends the input 4 to both 2 and $-2$ is not a function, which is why $\sqrt{4}$ means only $2$.

The letter inside the brackets is a placeholder. If $f(x) = x^2 - 3x$, then $f$ means "square the input, then subtract three times the input". So

$$f(a+1) = (a+1)^2 - 3(a+1) = a^2 - a - 2.$$

Every $x$ is replaced by the whole input, brackets included. Most notation errors come from replacing only part of it: writing $f(a+1)$ as $a^2 + 1 - 3a + 1$, or $f(2x)$ as $2f(x)$.

Finding the domain: three rules

When a question gives only a formula, the domain is every real number for which the formula makes sense. For the formulas in this course that comes down to three restrictions:

Expression	Requirement	Why
$\frac{P}{Q}$	$Q \neq 0$	Division by zero is undefined.
$\sqrt{E}$ (any even root)	$E \geq 0$	No real number squares to a negative.
$\log_a E$ or $\ln E$	$E > 0$	A positive base raised to any power is positive.

Odd roots such as $\sqrt[3]{E}$ impose nothing: $\sqrt[3]{-8} = -2$. When a formula has several pieces, write every condition and take the values that satisfy all of them at once. Watch for combinations: an even root in a denominator needs its radicand strictly positive, because $\sqrt{0} = 0$ would put a zero underneath.

Simplifying a formula never enlarges its domain. $\frac{x^2 - 1}{x - 1}$ equals $x + 1$ wherever both are defined, but the first is undefined at $x = 1$, so its domain is $x \neq 1$ and its graph is the line $y = x + 1$ with one point missing. Find the domain from the formula as given, before you cancel anything.

Composition: inside first

The composition $f(g(x))$, also written $(f \circ g)(x)$, means "apply $g$, then apply $f$ to the result". The function written nearer to $x$ acts first. In general $f(g(x)) \neq g(f(x))$, and the test likes to offer the other order as a wrong option.

To build $f(g(x))$ as a formula, take the formula for $f$ and put the whole of $g(x)$, in brackets, wherever $f$ has its variable. To evaluate $f(g(3))$ you do not need the formula at all: compute $g(3)$, then feed that number to $f$. Doing it both ways is a quick check on your algebra.

A composition also has a domain: $x$ must be in the domain of $g$, and $g(x)$ must be in the domain of $f$.

Inverse functions

The inverse $f^{-1}$ undoes $f$: if $f(a) = b$, then $f^{-1}(b) = a$. It exists only when $f$ is one-to-one, meaning no two inputs share an output. $f(x) = x^2$ on all reals is not one-to-one ($f(3) = f(-3)$), so it has no inverse there; restricted to $x \geq 0$ it has the inverse $\sqrt{x}$.

To find an inverse from a formula:

1. Write $y = f(x)$.
2. Solve for $x$ in terms of $y$.
3. Rename: the result is $f^{-1}(y)$; swap letters to write $f^{-1}(x)$.

The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$. On a graph, $f$ and $f^{-1}$ are mirror images in the line $y = x$, because every point $(a, b)$ becomes $(b, a)$.

The notation is a trap in itself: $f^{-1}(x)$ is the inverse function, not $\frac{1}{f(x)}$. When a question asks only for one value such as $f^{-1}(5)$, skip the formula: solve $f(x) = 5$.

Why this matters under time pressure

In multiple-choice function questions, wrong options tend to come from predictable slips: the other order of composition, a domain endpoint wrongly included, a reciprocal instead of an inverse. Before you choose, substitute one easy number into your answer and into the original; it takes seconds and catches most of these.