The modulus function and its graphs and equations, composite and inverse functions, and resolving rational expressions into partial fractions.

What this dot point is asking

WJEC wants you to work with the modulus function $|x|$ (its graph, and solving modulus equations and inequalities), composite functions $fg(x)$ and inverse functions $f^{-1}(x)$ (including domains, ranges and the reflection in $y = x$), and to resolve a rational expression into partial fractions. Partial fractions in particular are a gateway skill for integration later in this unit.

The answer

The modulus function

The modulus (or absolute value) $|x|$ equals $x$ when $x \ge 0$ and $-x$ when $x < 0$, so it is never negative.

To solve $|f(x)| = k$ (with $k > 0$), solve $f(x) = k$ and $f(x) = -k$. For inequalities, $|x| < k$ means $-k < x < k$, while $|x| > k$ means $x < -k$ or $x > k$. A reliable alternative for modulus equations is to square both sides (since both are non-negative), turning $|2x - 1| = |x + 3|$ into $(2x-1)^2 = (x+3)^2$, then solving the resulting quadratic; this avoids missing a case.

Composite functions

A composite function applies one function then another. The notation $fg(x)$ means "do $g$ first, then $f$": $fg(x) = f(g(x))$.

Inverse functions

The inverse $f^{-1}$ undoes $f$, so $f^{-1}(f(x)) = x$. It exists only if $f$ is one-to-one (each output comes from exactly one input).

To find an inverse: write $y = f(x)$, swap $x$ and $y$, and solve for $y$. The domain of $f^{-1}$ is the range of $f$, and the graph of $f^{-1}$ is the reflection of $f$ in the line $y = x$.

Partial fractions

A proper rational expression (degree of numerator below degree of denominator) splits into a sum with one fraction per factor of the denominator.

Examples in context

Example 1. A modulus inequality. Solve $|2x - 3| < 5$. This gives $-5 < 2x - 3 < 5$, so $-2 < 2x < 8$ and $-1 < x < 4$. The single modulus inequality unfolds into a double inequality bounding $x$.

Example 2. Composite with a restriction. If $f(x) = \sqrt{x}$ and $g(x) = x - 4$, then $fg(x) = \sqrt{x - 4}$, which is defined only for $x \ge 4$ because the square root needs a non-negative input. The domain of the composite is set by the inner function feeding a valid value into the outer one.

Try this

Q1. Solve $|x - 2| = 6$. [2 marks]

• Cue. $x - 2 = 6$ or $x - 2 = -6$, so $x = 8$ or $x = -4$.

Q2. Given $f(x) = x^2$ and $g(x) = x + 1$, find $fg(x)$. [2 marks]

• Cue. Apply $g$ first: $fg(x) = (x + 1)^2$.

Q3. Find the inverse of $f(x) = \dfrac{1}{x - 2}$, $x \neq 2$. [3 marks]

• Cue. $y = \dfrac{1}{x-2}$, swap and solve: $x = \dfrac{1}{y-2}$, so $y = \dfrac{1}{x} + 2$, giving $f^{-1}(x) = \dfrac{1}{x} + 2$.