Functions is the conceptual spine of IB Mathematics Analysis & Approaches HL. Every chapter that follows — calculus, complex numbers, statistics, sequences — assumes you can confidently sketch graphs, find inverses, classify transformations, and reason about domain and range. The HL extension over SL is large: oblique asymptotes via long division, odd/even classification, restricted-domain inverses, modulus transformations, rational inequalities via sign tables, and the $1/f(x)$ and $f(|x|)$ advanced graph transforms.
The single biggest source of dropped marks is sloppy notation around transformations — students mix up $f(x - a)$ vs $f(x + a)$, or invent the "rule" that $|f(x)| = f(|x|)$. This cheatsheet condenses every formula and trap from SL 2.1–2.11 and AHL 2.13–2.16 into one page you can revise from.
§1 — Lines & Function Basics SL 2.1–2.2, 2.5
Straight lines
Parallel: $m_1 = m_2$. Perpendicular: $m_1 m_2 = -1$, so $m_\perp = -\dfrac{1}{m}$ (negate AND reciprocate).
Function concepts
Domain: allowed inputs. Range: outputs.
Vertical Line Test (VLT): a graph defines a function iff every vertical line hits it at most once.
Inverse $f^{-1}$: exists iff $f$ is one-to-one. Domain $f^{-1}$ $=$ Range $f$. Graph: reflect in $y = x$. Compositions: $(f \circ f^{-1})(x) = x$. Method: write $y = f(x)$; swap $x \leftrightarrow y$; rearrange.
Composites & domain
$(f \circ g)(x) = f(g(x))$ — apply $g$ first, then $f$. In general $f \circ g \neq g \circ f$.
Domain of $f \circ g$: $x \in \text{dom}(g)$ with $g(x) \in \text{dom}(f)$.
Key graph checklist
Always find & label: $x$-intercepts ($f(x) = 0$); $y$-intercept ($f(0)$); VA (denominator $= 0$); HA ($x \to \pm \infty$); turning points; domain; range. Sketch = shape + key points labelled. Draw = accurate scale.
§2 — Quadratics & Rational Functions SL 2.6–2.8
Quadratic forms
- Standard: $ax^2 + bx + c$, $y$-intercept $(0, c)$.
- Factored: $a(x - p)(x - q)$, roots $p, q$.
- Vertex: $a(x - h)^2 + k$, vertex $(h, k)$ with $h = -\dfrac{b}{2a}$, $k = c - \dfrac{b^2}{4a}$.
Discriminant & quadratic formula
- $\Delta > 0$: two distinct real roots (crosses $x$-axis twice).
- $\Delta = 0$: one repeated root (touches $x$-axis).
- $\Delta < 0$: no real roots (no $x$-intercept).
Parameter conditions: set $\Delta > 0$, $\Delta = 0$, or $\Delta < 0$ and solve.
Rational $\dfrac{ax + b}{cx + d}$
Vertical asymptote: $x = -d/c$. Horizontal asymptote: $y = a/c$.
6-step sketch: 1) VA (denominator $= 0$); 2) HA (leading coefficient ratio); 3) $y$-int $f(0) = b/d$; 4) $x$-int (numerator $= 0$); 5) sign table for branches; 6) draw and label.
Self-inverse: $d = -a$. To prove: show $f(f(x)) = x$.
§3 — Exp/Log & Transformations SL 2.9, 2.11
$y = e^x$ and $y = \ln x$ are inverses — reflections of each other in the line $y = x$.$e^x$ vs $\ln x$
- $e^x$: domain $\mathbb{R}$, range $> 0$, HA $y = 0$, $y$-int $(0, 1)$.
- $\ln x$: domain $x > 0$, range $\mathbb{R}$, VA $x = 0$, $x$-int $(1, 0)$.
- Inverses of each other; reflect in $y = x$.
Log laws
- $\ln(xy) = \ln x + \ln y$
- $\ln(x/y) = \ln x - \ln y$
- $\ln(x^n) = n \ln x$
- $\log_a x = \dfrac{\ln x}{\ln a}$ (change of base)
- $e^{\ln x} = x$, $\ln(e^x) = x$, $a^x = e^{x \ln a}$
Hidden quadratic: for $e^{2x} - 5e^x + 4 = 0$, substitute $u = e^x$: $(u - 1)(u - 4) = 0$, so $x = 0$ or $\ln 4$.
SL transformations
| Transformation | Effect | Map |
|---|---|---|
| $y = f(x) + b$ | Translate up by $b$ | $(x, y) \to (x, y + b)$ |
| $y = f(x - a)$ | Translate right by $a$ | $(x, y) \to (x + a, y)$ |
| $y = p\,f(x)$ | Vertical stretch $\times p$ | $(x, y) \to (x, py)$ |
| $y = f(qx)$ | Horizontal stretch $\times \tfrac{1}{q}$ | $(x, y) \to (\tfrac{x}{q}, y)$ |
| $y = -f(x)$ | Reflect in $x$-axis | |
| $y = f(-x)$ | Reflect in $y$-axis |
Order for $p\,f(q(x - a)) + b$: 1) horizontal stretch $\tfrac{1}{q}$; 2) horizontal translate $+a$; 3) vertical stretch $p$; 4) vertical translate $+b$.
§4 — Oblique Asymptotes & Odd/Even AHL 2.13–2.14
Oblique asymptote
When $\deg(\text{numerator}) = \deg(\text{denominator}) + 1$ there is an oblique (slant) asymptote.
Method — long division:
$$\frac{P(x)}{dx + e} = \text{quotient} + \frac{R}{dx + e}.$$
The oblique asymptote is the quotient (the remainder $\to 0$ as $x \to \infty$).
Example: $\dfrac{x^2 - 14x + 24}{2(x + 3)} = \dfrac{x - 17}{2} + \dfrac{75/2}{x + 3}$. OA: $y = \tfrac{x}{2} - \tfrac{17}{2}$; VA: $x = -3$.
Odd & even functions
Even: $f(-x) = f(x)$; symmetric about the $y$-axis. e.g.\ $x^2$, $\cos x$, $|x|$, $\dfrac{1}{x^2 + 1}$.
Odd: $f(-x) = -f(x)$; $180^\circ$ symmetric about the origin, so $f(0) = 0$ always. e.g.\ $x^3$, $\sin x$, $\tan x$, $\dfrac{1}{x}$.
Proof: compute $f(-x)$, compare with $\pm f(x)$, state the conclusion.
Products: Even $\times$ Even $=$ Even; Odd $\times$ Odd $=$ Even; Even $\times$ Odd $=$ Odd.
Inverse on a restricted domain
- Pass the Horizontal Line Test on the restricted domain.
- State the restricted domain.
- Swap $x \leftrightarrow y$ and rearrange.
- Domain of $f^{-1}$ $=$ Range of the restricted $f$.
Self-inverse: $f(f(x)) = x$; graph symmetric in $y = x$. For $\dfrac{ax + b}{cx + d}$, self-inverse iff $d = -a$.
§5 — Inequalities & Modulus AHL 2.15–2.16
Inequalities — sign-table method
- Rearrange to $h(x) \geq 0$.
- Fully factorise $h(x)$.
- Critical values $=$ roots of numerator $+$ undefined points (zeros of denominator).
- Determine the sign of $h$ in each interval.
- State the solution set with the correct $\leq$ vs $<$.
Rational inequalities: $\dfrac{h}{k} \geq 0$ — exclude points where $k = 0$ from the solution.
GDC (Paper 2): graph both sides; find intersections.
Sign table template
| $x$ | $< r_1$ | $r_1$ | $(r_1, r_2)$ | $r_2$ | $> r_2$ |
|---|---|---|---|---|---|
| $(x - r_1)$ | $-$ | $0$ | $+$ | $+$ | $+$ |
| $(x - r_2)$ | $-$ | $-$ | $-$ | $0$ | $+$ |
| product | $+$ | $0$ | $-$ | $0$ | $+$ |
Product $\leq 0$: $r_1 \leq x \leq r_2$.
Modulus
$|x| = x$ if $x \geq 0$; $|x| = -x$ if $x < 0$.
Equations: $|f(x)| = k$ $\Rightarrow$ $f(x) = k$ or $f(x) = -k$ (require $k \geq 0$, else no solution).
Inequalities:
- $|f| < k \;\Leftrightarrow\; -k < f < k$
- $|f| > k \;\Leftrightarrow\; f > k$ or $f < -k$
Graph $y = |f(x)|$: reflect any negative parts upwards (result $\geq 0$).
Graph $y = f(|x|)$: keep the right half, then reflect right $\to$ left (result is even).
§6 — Advanced Transforms & Attack Plan AHL 2.16
$y = 1/f(x)$ rules
| Zero of $f$ | $\to$ VA of $1/f$ |
| VA of $f$ | $\to$ zero of $1/f$ |
| $f = \pm 1$ | Invariant point |
| $f$ local max $(a, b)$ | $1/f$ local min $(a, 1/b)$ |
| $f > 0$ | $1/f > 0$ (same sign) |
| $f \to \infty$ | $1/f \to 0$ |
$[f(x)]^2$ & $f(ax + b)$
$[f(x)]^2$: always $\geq 0$; zeros stay; negative parts reflect then stretch.
$f(ax + b)$ (AHL only): rewrite as $f\!\left(a\!\left(x + \tfrac{b}{a}\right)\right)$. Horizontal stretch $\times \tfrac{1}{a}$, then horizontal translate $-\tfrac{b}{a}$. Map: $(x, y) \to \left(\dfrac{x - b}{a}, y\right)$.
Attack plan
| See… | Do… |
|---|---|
| Find $f^{-1}$ | Swap; rearrange |
| Sketch rational | VA, HA, intercepts, branches |
| Describe transformation | Identify $p, q, a, b$ |
| Discriminant condition | Set $\Delta >/</= 0$; solve |
| Odd/even proof | Compute $f(-x)$ |
| Oblique asymptote | Long division; OA $=$ quotient |
| $|f(x)| = k$ | Two cases $\pm k$ |
| Rational inequality | Sign table |
| Self-inverse | Show $f(f(x)) = x$ |
| Restricted inverse | HLT; state new domain |
| $e^{2x} + \ldots$ | Substitute $u = e^x$ |
| $\log$ equation | Log laws; check domain |
| $1/f$ sketch | Zeros $\leftrightarrow$ VAs; invert TPs |
| $f(|x|)$ sketch | Right half + reflect to left |
Top exam traps
- $f^{-1}(x) \neq \dfrac{1}{f(x)}$ — inverse $\neq$ reciprocal.
- $f(x - a)$ shifts right, not left.
- $f(qx)$: stretch $= \tfrac{1}{q}$, not $q$.
- Inequality: NEVER multiply by an unknown-sign variable.
- Odd function: must have $f(0) = 0$.
- $|f(x)|$ is reflect-negative-parts-up, not "lift the whole graph".
- $f(|x|) \neq |f(x)|$ — completely different.
- Oblique asymptote $=$ quotient, not remainder.
- Rational inequality: exclude zeros of the denominator.
- Domain of $f^{-1}$ $=$ range of the restricted $f$.
- $\ln(a + b) \neq \ln a + \ln b$.
- Check $k \geq 0$ before splitting $|f| = k$.
Worked Example — Restricted Inverse & Composition
Question (HL Paper 1 style — 7 marks)
The function $f$ is defined by $f(x) = x^2 - 4x + 7$ for $x \geq 2$.
(a) Show that $f$ has an inverse on this domain. (b) Find $f^{-1}(x)$ and state its domain. (c) Find $(f^{-1} \circ f)(5)$.
Solution
- Complete the square: $f(x) = (x - 2)^2 + 3$. The vertex is at $(2, 3)$. (M1)
- For $x \geq 2$, $f$ is monotonically increasing (the right arm of the parabola). It passes the Horizontal Line Test, so $f$ has an inverse on $[2, \infty)$. (R1)
- Set $y = (x - 2)^2 + 3$ and swap: $x = (y - 2)^2 + 3$, so $(y - 2)^2 = x - 3$. (M1)
- Since $y \geq 2$, take the positive root: $y = 2 + \sqrt{x - 3}$. So $f^{-1}(x) = 2 + \sqrt{x - 3}$. (A1)
- Domain of $f^{-1}$ $=$ Range of $f$ $= [3, \infty)$, i.e.\ $x \geq 3$. (A1)
- For (c): $(f^{-1} \circ f)(5) = 5$ directly, since $f^{-1}$ undoes $f$ on the restricted domain (and $5 \geq 2$). (R1)(A1)
Examiner's note: Two common errors: (a) taking the negative square root and writing $y = 2 - \sqrt{x - 3}$ — that gives the inverse for the LEFT arm $x \leq 2$, not the right arm; (b) stating the domain of $f^{-1}$ as $\mathbb{R}$ instead of $[3, \infty)$. Always quote: "domain of $f^{-1}$ $=$ range of restricted $f$".