Functions — IB Math AA HL AA HL Study Guide

For: IB Math AA HL candidates sitting IB Math: Analysis & Approaches HL.

Covers: domain, range, composite and inverse functions, core function types (quadratic, rational, exponential, log), graph transformations, algebraic and graphical inequality solving, and polynomial functions with the factor theorem.

You should already know: IGCSE / pre-DP math, comfort with proof and algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Math AA HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Are Functions?

A function is a well-defined relation that maps every input value from a set called the domain to exactly one output value in a set called the codomain, with the range being the set of all actual outputs produced by valid inputs. Functions are written with notation like $f(x)$, $g(t)$, or $h(\theta )$, where the variable in brackets is the input. Also called mappings or transformations, functions make up ~20% of marks across Papers 1, 2, and 3 of the AA HL exam, and act as the foundation for calculus, complex numbers, and statistical modeling later in the syllabus.

2. Domain, range, composite, inverse functions

Every function is defined by its domain (valid inputs) and range (attainable outputs) before any operation is applied.

• Domain rules: You cannot divide by 0, take the even root of a negative number, or take the logarithm of a non-positive value. For $f(x)=\frac{1}{\sqrt{4-x^2}}$, domain requires $4-x^2>0$, so $x\in (-2,2)$.
• Range calculation: Identify local maxima/minima and end behavior of the function. For the same $f(x)$, $\sqrt{4-x^2}\in (0,2]$, so the range is $\left[\frac{1}{2},\infty \right)$.

Composite functions

The composite function $(f\circ g)(x)=f(g(x))$ applies $g$ first, then $f$. Order matters: $(f\circ g)(x)$ is almost never equal to $(g\circ f)(x)$. The domain of $(f\circ g)(x)$ is all $x$ in the domain of $g$ where $g(x)$ is in the domain of $f$. For example, if $f(x)=\ln x$ and $g(x)=x-3$, $(f\circ g)(x)=\ln (x-3)$ has domain $x>3$, while $(g\circ f)(x)=\ln x-3$ has domain $x>0$.

Inverse functions

An inverse function $f^{-1}(x)$ exists only if $f$ is bijective (one-to-one, i.e., no two inputs map to the same output, and surjective, i.e., range equals codomain). The inverse swaps the domain and range of the original function, and satisfies $f(f^{-1}(x))=f^{-1}(f(x))=x$. To find the inverse:

1. Replace $f(x)$ with $y$
2. Swap $x$ and $y$
3. Rearrange to solve for $y$, then replace $y$ with $f^{-1}(x)$

Worked example: For $f(x)=2x+5$, swap $x=2y+5$ → $y=\frac{x-5}{2}$, so $f^{-1}(x)=\frac{x-5}{2}$. The graph of $f^{-1}(x)$ is a reflection of $f(x)$ over the line $y=x$.

3. Quadratic, rational, exponential, log functions

These four function types appear in 80% of function-focused exam questions:

• Quadratic functions: Form $f(x)=a{x}^2+bx+c$, $a\ne 0$. The vertex (turning point) is at $x=-\frac{b}{2a}$, and the discriminant $\Delta =b^2-4ac$ tells you the number of real roots: $\Delta >0$ gives 2 distinct real roots, $\Delta =0$ gives 1 repeated root, $\Delta <0$ gives no real roots. Completed square form $f(x)=a(x-h{)}^2+k$ explicitly shows the vertex at $(h,k)$ and axis of symmetry $x=h$.
• Rational functions: Form $f(x)=\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials, $Q(x)\ne 0$. Vertical asymptotes occur at roots of $Q(x)$ where $P(x)\ne 0$. Horizontal asymptotes exist if $\mathrm{deg}P\le \mathrm{deg}Q$: if $\mathrm{deg}P<\mathrm{deg}Q$, asymptote at $y=0$; if equal, asymptote at $y=\frac{\text{leading coefficient of }P}{\text{leading coefficient of }Q}$.
• Exponential functions: Form $f(x)=a\cdot b^x$, $b>0$, $b\ne 1$. If $b>1$, the function shows exponential growth; if $0<b<1$, it shows exponential decay. All exponential functions pass through $(0,a)$ and have a horizontal asymptote at $y=0$.
• Logarithmic functions: $f(x)={\log }_{b}x$ is the inverse of $b^x$. It has domain $x>0$, range all real numbers, passes through $(1,0)$, and has a vertical asymptote at $x=0$. The natural logarithm $\ln x={\log }_{e}x$, where $e\approx 2.718$ is the exponential constant.

Worked example: For $f(x)=3{\log }_2(x+2)-1$, the domain is $x>-2$, vertical asymptote at $x=-2$, and x-intercept at $x=2^{1/3}-2\approx -0.74$.

4. Transformations of graphs — reflection, stretch, translation

Graph transformations follow a fixed order: stretches and reflections first, then translations. All transformations modify either the input $x$ or the entire output $f(x)$:

Exam trap: Horizontal transformations are counterintuitive: $f(x-h)$ shifts the graph right by $h$ units, not left. Always rewrite transformations to isolate $x$ inside the function to avoid errors. Worked example: To transform $f(x)=\ln x$ to $g(x)=\ln (4-2x)$, rewrite as $g(x)=\ln (-2(x-2))$: reflect over the $y$-axis, stretch horizontally by factor $\frac{1}{2}$, then shift right by 2 units.

5. Solving inequalities graphically and algebraically

Examiners accept both methods, but you must show full working to earn full marks:

• Algebraic method: Rearrange the inequality to have 0 on one side, factor the expression, identify critical points (roots of the numerator and denominator), then test the sign of the expression in intervals around each critical point. Flip the inequality sign if you multiply or divide both sides by a negative number.
• Graphical method: Plot both sides of the inequality, find intersection points, then identify regions where one graph lies above or below the other. Exclude endpoints if the inequality is strict ($<$ or $>$), and include them if it is non-strict ($\le$ or $\ge$).

Worked example: Solve $\frac{x+1}{x-2}\ge 3$

1. Rearrange: $\frac{x+1}{x-2}-3\ge 0$ → $\frac{-2x+7}{x-2}\ge 0$ → $\frac{2x-7}{x-2}\le 0$
2. Critical points: $x=2$ (denominator root, undefined) and $x=3.5$ (numerator root)
3. Test intervals: $x<2$ (positive, invalid), $2<x<3.5$ (negative, valid), $x>3.5$ (positive, invalid)
4. Solution: $(2,3.5]$ (include $x=3.5$, exclude $x=2$)

6. Polynomial functions and the factor theorem

A polynomial of degree $n$ has the form $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_{1}x+a_0$, where $a_n\ne 0$ and all exponents are non-negative integers. The factor theorem states that for any polynomial $f(x)$, $(x-a)$ is a linear factor of $f(x)$ if and only if $f(a)=0$. This extends to factors of the form $(px-q)$: $(px-q)$ is a factor if and only if $f\left(\frac{q}{p}\right)=0$.

Worked example: Show $(x-2)$ is a factor of $f(x)=x^3-4x^2+x+6$, then factor completely.

1. Calculate $f(2)=8-16+2+6=0$, so $(x-2)$ is a factor.
2. Use synthetic division to divide $f(x)$ by $(x-2)$: the quotient is $x^2-2x-3$.
3. Factor the quadratic: $x^2-2x-3=(x-3)(x+1)$
4. Final factorization: $f(x)=(x-2)(x-3)(x+1)$, with roots at $x=-1,2,3$.

Exam tip: For higher-degree polynomials, always test integer factors of the constant term first, as these are the most common roots set by examiners.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to restrict the domain of a non-injective function before calculating its inverse, leading to a relation that is not a function. Why: Students assume all functions have inverses, but only strictly increasing/decreasing functions do. Correct move: Restrict the original function’s domain to an interval where it is one-to-one before solving for the inverse, and explicitly state the inverse’s domain (equal to the original function’s range).
• Wrong move: Applying translations before stretches/reflections, leading to incorrect transformed graphs. Why: Students treat horizontal and vertical transformations as interchangeable. Correct move: Follow BODMAS order for modifications to $x$: resolve coefficients and negatives first, then apply horizontal shifts.
• Wrong move: Cross-multiplying denominators when solving rational inequalities, without checking if the denominator is positive or negative. Why: Students treat inequalities like equations. Correct move: Always rearrange rational inequalities to have 0 on one side, so you only need to test sign changes at critical points.
• Wrong move: Using $f(-q/p)$ to test if $(px-q)$ is a factor of a polynomial. Why: Students misremember the sign for non-monic linear factors. Correct move: Set $px-q=0$ to get $x=q/p$, then test if $f(q/p)=0$ to confirm the factor.
• Wrong move: Including asymptote values in the range of a function. Why: Students plug in edge inputs without checking if outputs are attainable. Correct move: Evaluate the limit of the function as $x\to \pm \infty$ to confirm which end values are excluded from the range.

8. Practice Questions (IB Math AA HL Style)

Question 1

Let $f(x)=\sqrt{x+4}$, $x\ge -4$ and $g(x)=x^2-3$, $x\in \mathbb{R}$. (a) Find the domain and range of $(f\circ g)(x)$. (b) Find $f^{-1}(x)$, stating its domain and range. (c) Solve $f(g(x))=1$.

Solution

(a) $(f\circ g)(x)=f(x^2-3)=\sqrt{x^2-3+4}=\sqrt{x^2+1}$. The expression $x^2+1\ge 1$ for all real $x$, so domain is $x\in \mathbb{R}$. The range is $\sqrt{x^2+1}\ge 1$, so range $\left[1,\infty \right)$. (b) Let $y=\sqrt{x+4}$. Swap $x=\sqrt{y+4}$, square both sides: $x^2=y+4$ → $y=x^2-4$. The domain of $f^{-1}(x)$ is the range of $f(x)$: $x\ge 0$. The range of $f^{-1}(x)$ is the domain of $f(x)$: $y\ge -4$. So $f^{-1}(x)=x^2-4$, $x\ge 0$. (c) $\sqrt{x^2+1}=1$ → $x^2+1=1$ → $x^2=0$ → $x=0$. Verify: $f(g(0))=f(-3)=\sqrt{1}=1$, valid.

Question 2

(a) Describe the sequence of transformations mapping $y=e^x$ to $y=2e^{3-2x}-5$. (b) Solve $2e^{3-2x}-5\ge 3$ algebraically, giving your answer in exact form.

Solution

(a) Rewrite the transformed function: $y=2e^{-2(x-1.5)}-5$. The sequence is:

1. Reflect $y=e^x$ over the $y$-axis (for $-x$)
2. Stretch horizontally by scale factor $\frac{1}{2}$ (for coefficient 2 of $x$)
3. Stretch vertically by scale factor 2
4. Translate 1.5 units right and 5 units down. (b) Rearrange the inequality: $2e^{3-2x}\ge 8$ → $e^{3-2x}\ge 4$ Take natural log of both sides: $3-2x\ge \ln 4$ Rearrange: $-2x\ge 2\ln 2-3$ Multiply by -1, flip inequality: $2x\le 3-2\ln 2$ → $x\le \frac{3-2\ln 2}{2}$ Solution: $x\in \left(-\infty ,\frac{3-2\ln 2}{2}\right]$

Question 3

The polynomial $f(x)=2x^3+a{x}^2+bx+6$ has a factor $(x+2)$ and leaves a remainder of 12 when divided by $(x-1)$. (a) Find the values of $a$ and $b$. (b) Find all real roots of $f(x)=0$, giving exact values where appropriate.

Solution

(a) By the factor theorem: $f(-2)=0$ $2(-8)+a(4)+b(-2)+6=0$ → $-16+4a-2b+6=0$ → $2a-b=5$ (Equation 1) By the remainder theorem: $f(1)=12$ $2+a+b+6=12$ → $a+b=4$ (Equation 2) Add Equation 1 and 2: $3a=9$ → $a=3$. Substitute into Equation 2: $3+b=4$ → $b=1$. (b) $f(x)=2x^3+3x^2+x+6$. We know $(x+2)$ is a factor, so use synthetic division to get the quadratic quotient: $2x^2-x+3$. The discriminant of the quadratic is $\Delta =(-1{)}^2-4(2)(3)=1-24=-23<0$, so it has no real roots. The only real root is $x=-2$.

9. Quick Reference Cheatsheet

10. What's Next

Functions are the foundational building block for almost all remaining AA HL topics: you will apply the properties covered here to differentiate and integrate all core function types in the calculus unit, use polynomial functions for binomial theorem and complex number applications, and use exponential and log functions for real-world modeling in Paper 3 investigations. A strong grasp of graph transformations will also directly help you sketch derivative and integral graphs, a common 3-4 mark question on Paper 1.

To reinforce your understanding, work through official IBO past paper questions focused on functions, prioritizing multi-part 5-7 mark questions that combine subtopics like function composition, graph transformations, and inequality solving. If you get stuck on any question, or need clarification on any rule covered in this guide, you can ask Ollie, our AI tutor, for personalized explanations and extra practice problems at any time.