The language of functions (domain, range, composite and inverse functions), the modulus function and its graph, and solving modulus equations and inequalities.

What this dot point is asking

OCR wants you to use the language of functions (domain, range, one-to-one and many-to-one mappings), form composite functions in the correct order, find inverse functions and state their domain and range, understand that an inverse exists only for a one-to-one function (so a restricted domain may be needed), sketch the modulus function $y = |f(x)|$, and solve modulus equations and inequalities both algebraically and graphically.

The answer

Domain, range and mappings

A function maps each input in its domain to exactly one output; the set of outputs is the range. A function is one-to-one if no two inputs share an output, and many-to-one if some output comes from several inputs (like $x^2$). Only one-to-one functions have an inverse over their whole domain.

Composite functions

The composite $fg(x)$ means "do $g$ first, then $f$": $fg(x) = f(g(x))$. Order matters, so in general $fg \ne gf$. The domain of $fg$ is the set of inputs for which $g(x)$ is itself a valid input to $f$.

Inverse functions

The inverse $f^{-1}$ undoes $f$: if $f(a) = b$ then $f^{-1}(b) = a$. To find it, write $y = f(x)$, swap the roles and solve for the new output. The graph of $f^{-1}$ is the reflection of the graph of $f$ in the line $y = x$, and the domain and range swap.

The modulus function

The modulus $|x|$ is the distance of $x$ from zero, so it is never negative: $|x| = x$ for $x \ge 0$ and $|x| = -x$ for $x < 0$. The graph $y = |f(x)|$ takes the graph of $f$ and reflects any part below the $x$-axis up above it.

Examples in context

Restricting a domain for an inverse

A many-to-one function such as $f(x) = x^2$ has no inverse over all reals, because two inputs give the same output. Restricting the domain to $x \ge 0$ makes it one-to-one, and then $f^{-1}(x) = \sqrt{x}$.

Solving modulus problems

To solve $|f(x)| = g(x)$ algebraically, square both sides (valid because both sides are then squares) or split into the two cases $f(x) = g(x)$ and $f(x) = -g(x)$, checking each solution. A sketch confirms how many solutions there are and which region satisfies an inequality.

Try this

Q1. Given $f(x) = 2x + 5$, find $f^{-1}(x)$. [2 marks]

• Cue. $y = 2x + 5$ gives $x = \dfrac{y - 5}{2}$, so $f^{-1}(x) = \dfrac{x - 5}{2}$.

Q2. Solve $|3x - 2| = 7$. [2 marks]

• Cue. $3x - 2 = 7$ gives $x = 3$; $3x - 2 = -7$ gives $x = -\tfrac{5}{3}$.