Functions

At CSEC a function was a rule. At CAPE it is a rule together with a domain, and most of the marks in this topic come from taking the domain seriously.

Definition and notation

A function $f: A \to B$ assigns to each element of the domain $A$ exactly one element of the codomain $B$.

$$f: x \mapsto 2x + 3 \qquad \text{or} \qquad f(x) = 2x + 3$$

The range (or image) is the set of values actually produced. It is a subset of the codomain, and often a proper one.

Domain and range

When a domain is not stated, take the natural domain: every real value for which the rule makes sense. Two things break it:

Division by zero. For $f(x) = \dfrac{1}{x-3}$, the domain is all real $x$ except $x = 3$, written $\{x \in \mathbb{R} : x \ne 3\}$.

Square roots of negatives. For $f(x) = \sqrt{x-2}$, we need $x - 2 \ge 0$, so the domain is $x \ge 2$.

The range takes more thought:

• $f(x) = x^2$ on $\mathbb{R}$ has range $y \ge 0$, since a square is never negative.
• $f(x) = \sqrt{x-2}$ has range $y \ge 0$, since the root sign denotes the non-negative root.
• $f(x) = \dfrac{1}{x-3}$ has range $y \ne 0$, since a fraction with numerator 1 is never zero.

One-to-one and onto

A function is one-to-one (injective) if different inputs always give different outputs — no output is hit twice.

$f(x) = x^2$ on $\mathbb{R}$ is not one-to-one, because $f(2) = f(-2) = 4$.

$f(x) = 2x + 3$ is one-to-one: if $2a + 3 = 2b + 3$ then $a = b$.

A function is onto (surjective) if every element of the codomain is actually reached — the range equals the codomain.

$f: \mathbb{R} \to \mathbb{R}$ with $f(x) = x^2$ is not onto, because no real $x$ gives $-1$.

A function that is both is a bijection, and bijections are exactly the functions with inverses.

Composite functions

$fg(x)$ means apply $g$ first, then $f$: work from the inside out.

With $f(x) = 2x + 3$ and $g(x) = x^2$:

$$fg(x) = f(x^2) = 2x^2 + 3$$ $$gf(x) = g(2x+3) = (2x+3)^2 = 4x^2 + 12x + 9$$

These are different functions, as they usually are.

For $fg$ to exist, the range of $g$ must lie inside the domain of $f$. If $f(x) = \sqrt{x}$ and $g(x) = x - 5$, then $fg(x) = \sqrt{x-5}$ only makes sense for $x \ge 5$ — the composition has narrowed the domain.

Inverse functions

$f^{-1}$ undoes $f$: if $f(a) = b$ then $f^{-1}(b) = a$. Equivalently,

$$ff^{-1}(x) = x \qquad \text{and} \qquad f^{-1}f(x) = x$$

An inverse exists only if $f$ is one-to-one. If two inputs shared an output, the inverse would not know which to return.

Find $f^{-1}$ for $f(x) = 2x + 3$.

$$y = 2x + 3 \quad \Rightarrow \quad x = \frac{y-3}{2} \quad \Rightarrow \quad f^{-1}(x) = \frac{x-3}{2}$$

Check: $f(5) = 13$ and $f^{-1}(13) = \dfrac{10}{2} = 5$ ✓

The order reverses

$$(fg)^{-1} = g^{-1}f^{-1}$$

Undoing a sequence means undoing the steps in reverse: socks then shoes, but shoes off then socks off.

Domain and range swap

The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$.

Graphs and restrictions

The graph of $y = f^{-1}(x)$ is the graph of $y = f(x)$ reflected in the line $y = x$. Every point $(a,b)$ on one becomes $(b,a)$ on the other.

This is why $y = e^x$ and $y = \ln x$ are mirror images, and why one passes through $(0,1)$ while the other passes through $(1,0)$.

Restricting the domain

$f(x) = x^2$ on $\mathbb{R}$ has no inverse. Restrict the domain to $x \ge 0$ and it becomes one-to-one, so

$$f^{-1}(x) = \sqrt{x}, \qquad x \ge 0$$

The restriction is not a technicality — it is what makes the inverse exist, and it is why the calculator's $\sqrt{\phantom{x}}$ key returns only the positive root.