In Year 1, algebra was mostly about expressions: expanding, factorising, solving. In Year 2 the focus shifts to functions — rules that turn inputs into outputs — and to the tools that make later calculus possible.
The modulus function $|x|$ · graphs of $y = |\mathrm{f}(x)|$ and $y = \mathrm{f}(|x|)$
Domain and range · composite and inverse functions · combined transformations
| Word | Meaning | Example |
|---|---|---|
| Function | A mapping where every input has exactly one output. | $\mathrm{f}(x) = x^2$ |
| Domain | The set of allowed inputs. | $x \geqslant 0$ |
| Range | The set of outputs actually produced. | $\mathrm{f}(x) \geqslant 0$ |
| One-one | Different inputs always give different outputs. | $\mathrm{f}(x) = 2x + 1$ |
| Many-one | Two or more inputs share an output. | $\mathrm{f}(x) = x^2$, $x \in \mathbb{R}$ |
| Composite | One function applied to the output of another. | $\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x))$ |
| Inverse | The function that undoes $\mathrm{f}$. | $\mathrm{f}^{-1}(x)$ |
| Modulus | The size of a number, ignoring its sign. | $|-3| = 3$ |
| Rational expression | One polynomial divided by another. | $\dfrac{x+1}{x^2-4}$ |
| Improper | Numerator degree $\geqslant$ denominator degree. | $\dfrac{x^2}{x-1}$ |
| Partial fractions | A sum of simpler fractions equal to one fraction. | $\dfrac{2}{x-1} + \dfrac{3}{x+2}$ |
$\mathrm{f}(x) = |2x - 4|$, $x \in \mathbb{R}$, and $\mathrm{g}(x) = x + 1$, $x \in \mathbb{R}$. (a) State the range of $\mathrm{f}$. (b) Explain why $\mathrm{f}$ has no inverse. (c) Solve $\mathrm{fg}(x) = \mathrm{g}(x)$.
Every Year 2 part of content area B, and the page that covers it.
- BAlgebraic and Partial FractionsSimplifying rational expressions, algebraic division, and partial fractions with linear and repeated linear factors.
- BThe Modulus FunctionGraphs of $y = |ax + b|$ and $y = |\mathrm{f}(x)|$; solving modulus equations and inequalities.
- BComposite and Inverse FunctionsMappings, domain and range, one-one and many-one, composites, and inverses as reflections in $y = x$.
- BCombined Transformations$y = a\,\mathrm{f}(bx + c) + d$, the order of transformations, and $y = \mathrm{f}(|x|)$.
Function
One output for every input.
Inverse exists
Only for one-one functions.
Inverse graph
Reflection in $y = x$.
Composite
$\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x))$.
Modulus graph
$|\mathrm{f}(x)|$: reflect below-axis parts up.
$\mathrm{f}(|x|)$
Mirror the right half in the $y$-axis.
Partial fractions
Substitute roots, compare coefficients.
Improper
Divide first, then split.
Transformations
Inside: opposite, on $x$. Outside: as written, on $y$.
Ten questions drawing on the whole of this section. Work each one through before opening the solution.
Simplify $\dfrac{x^2 - x - 12}{x^2 - 16}$.
▶ Show solution
$\dfrac{(x - 4)(x + 3)}{(x - 4)(x + 4)} = \dfrac{x + 3}{x + 4}$
Express $\dfrac{3x + 7}{(x + 1)(x + 3)}$ in partial fractions.
▶ Show solution
$3x + 7 \equiv A(x + 3) + B(x + 1)$.
$x = -1$: $4 = 2A$, $A = 2$. $x = -3$: $-2 = -2B$, $B = 1$.
$\dfrac{2}{x + 1} + \dfrac{1}{x + 3}$
Solve $|4x - 3| = 9$.
▶ Show solution
$4x - 3 = 9 \Rightarrow x = 3$; $4x - 3 = -9 \Rightarrow x = -\tfrac32$.
Solve $|x + 1| \leqslant 4$.
▶ Show solution
$-4 \leqslant x + 1 \leqslant 4 \Rightarrow -5 \leqslant x \leqslant 3$.
$\mathrm{f}(x) = 3x + 2$ and $\mathrm{g}(x) = \dfrac{1}{x}$, $x \neq 0$. Find $\mathrm{fg}(x)$ and $\mathrm{gf}(x)$, stating any value that must be excluded from the domain of $\mathrm{gf}$.
▶ Show solution
$\mathrm{fg}(x) = 3 \times \dfrac{1}{x} + 2 = \dfrac{3}{x} + 2$, $x \neq 0$.
$\mathrm{gf}(x) = \dfrac{1}{3x + 2}$, which needs $3x + 2 \neq 0$, so $x \neq -\tfrac23$.
$\mathrm{f}(x) = \dfrac{x + 4}{2}$, $x \in \mathbb{R}$. Find $\mathrm{f}^{-1}(x)$ and verify that $\mathrm{f}\mathrm{f}^{-1}(x) = x$.
▶ Show solution
$y = \dfrac{x + 4}{2} \Rightarrow x = 2y - 4$, so $\mathrm{f}^{-1}(x) = 2x - 4$.
$\mathrm{f}\mathrm{f}^{-1}(x) = \dfrac{(2x - 4) + 4}{2} = \dfrac{2x}{2} = x$ ✓
The point $(2, 8)$ lies on $y = \mathrm{f}(x)$. Find its image on $y = \tfrac12\mathrm{f}(x + 3) - 1$.
▶ Show solution
$x + 3 = 2 \Rightarrow x = -1$; $y = \tfrac12(8) - 1 = 3$.
Image $(-1, 3)$.
Express $\dfrac{x^2 + 3x - 1}{x + 2}$ in the form $Ax + B + \dfrac{C}{x + 2}$.
▶ Show solution
$x^2 \div x = x$; subtract $x^2 + 2x$ to leave $x - 1$.
$x \div x = 1$; subtract $x + 2$ to leave $-3$.
$x + 1 - \dfrac{3}{x + 2}$, so $A = 1$, $B = 1$, $C = -3$. (Check: substituting $x = -2$ into the numerator gives $4 - 6 - 1 = -3$ ✓)
$\mathrm{f}(x) = (x + 1)^2 + 2$, $x \geqslant -1$. Find $\mathrm{f}^{-1}(x)$, state its domain, and describe how the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}^{-1}(x)$ are related.
▶ Show solution
Range of $\mathrm{f}$: $\mathrm{f}(x) \geqslant 2$.
$y = (x + 1)^2 + 2 \Rightarrow x + 1 = \sqrt{y - 2}$ (positive root since $x \geqslant -1$).
$\mathrm{f}^{-1}(x) = \sqrt{x - 2} - 1$, domain $x \geqslant 2$.
Each graph is the reflection of the other in the line $y = x$.
$\mathrm{f}(x) = \dfrac{5x - 1}{(x + 1)(2x - 1)}$, $x > \tfrac12$.
(a) Express $\mathrm{f}(x)$ in partial fractions. (b) Show that $\mathrm{f}(1) = 2$. (c) The function $\mathrm{g}$ is defined by $\mathrm{g}(x) = |x - 3|$, $x \in \mathbb{R}$. Find $\mathrm{gf}(1)$, and solve $\mathrm{g}(x) = 2x$.
▶ Show solution
(a) $5x - 1 \equiv A(2x - 1) + B(x + 1)$.
$x = -1$: $-6 = -3A$, $A = 2$. $x = \tfrac12$: $\tfrac32 = \tfrac32 B$, $B = 1$.
$\mathrm{f}(x) = \dfrac{2}{x + 1} + \dfrac{1}{2x - 1}$
(b) $\mathrm{f}(1) = \dfrac{2}{2} + \dfrac{1}{1} = 2$ ✓ (or directly, $\dfrac{4}{2 \times 1} = 2$).
(c) $\mathrm{gf}(1) = \mathrm{g}(2) = |2 - 3| = 1$.
$|x - 3| = 2x$: case $x - 3 = 2x$ gives $x = -3$, but then $2x = -6 < 0$: reject.
Case $3 - x = 2x$ gives $x = 1$; check $|1 - 3| = 2 = 2(1)$ ✓. So $x = 1$.