Algebra and functions

Q: What is the difference between the domain and the range of a function?

The domain is the set of all allowable input values (\(x\)), while the range is the set of all resulting output values (\(f(x)\) or \(y\)).

Q: Which types of mappings qualify as true functions, and which do not?

One-to-one and many-to-one mappings are functions; one-to-many mappings are NOT functions.

Q: In what order are functions applied in the composite function \(fg(x)\)?

Apply the inside function \(g(x)\) first, and then apply \(f\) to the result: \(f(g(x))\).

Q: What condition must a function satisfy to have an inverse function \(f^{-1}(x)\), and what is its graphical relationship to \(f(x)\)?

The function must be one-to-one ; graphically, \(y = f^{-1}(x)\) is the reflection of \(y = f(x)\) in the line \(y = x\).

Q: How is the graph of \(y = |f(x)|\) obtained from the graph of \(y = f(x)\)?

Reflect any part of the graph lying below the \(x\)-axis vertically upwards into the positive region, leaving parts above the \(x\)-axis unchanged.

Q: What makes an algebraic fraction improper, and what step must be taken before finding partial fractions?

An algebraic fraction is improper when the degree of the numerator is \(\ge\) the degree of the denominator; polynomial long division must be performed first.

Mathematics 2210 · CCEA A Level · 10 min read · Updated: 31 Aug 2026

Welcome to A2 Algebra and Functions

Welcome to one of the most fundamental chapters in your CCEA A2 Pure Mathematics journey! Think of algebra and functions as the toolkit of mathematics. Just as a master craftsperson needs sharp, reliable tools, you will use the skills learned here — such as working with composite functions, dissecting fractions, and graphing modulus equations — across almost every other A2 topic, especially calculus.

Don't worry if some of these ideas look a bit abstract at first. We will break every concept down into simple, bite-sized pieces with step-by-step methods and clear examples to build your confidence.

1. Understanding Functions, Domain, and Range

In AS Mathematics, you worked with equations like \(y = x^2 + 3\). At A2 level, we look more formally at functions. Think of a function as an input-output machine: you feed in a number \(x\), the machine follows a rule, and it outputs a single result \(f(x)\).

Key Terms: Domain and Range

Domain: The set of all allowable input values (all valid \(x\)-values).
Range: The set of all possible output values (all resulting \(f(x)\) or \(y\)-values).

Real-world analogy: Imagine a vending machine. The coins you are allowed to insert represent the domain. The snacks that can actually drop into the tray represent the range.

Types of Mappings

A mapping links elements from the domain to the range. For an expression to qualify as a true function, every input must produce exactly one output:

• One-to-One: Each \(x\)-value gives a unique \(y\)-value, and no two \(x\)-values give the same \(y\)-value (e.g., \(f(x) = 2x + 1\)).
• Many-to-One: Different \(x\)-values can produce the same \(y\)-value (e.g., \(f(x) = x^2\), since both \(x = 2\) and \(x = -2\) give \(y = 4\)). This is still a valid function!
• One-to-Many: A single \(x\)-value produces multiple \(y\)-values (e.g., \(y = \pm\sqrt{x}\)). This is NOT a function.

Did you know? You can test if a graph represents a function using the Vertical Line Test: if any vertical line drawn through the graph touches it more than once, it is not a function.

Composite Functions

A composite function is created when you apply one function to the result of another. We write this as \(fg(x)\) or \(f(g(x))\).

Crucial Rule: In \(fg(x)\), you apply the function closest to \(x\) first. Work from the inside out (apply \(g\) first, then apply \(f\) to the answer).

Step-by-Step Example:
Let \(f(x) = 2x + 3\) and \(g(x) = x^2 - 1\).
To find \(fg(4)\):
Step 1: Calculate the inside function: \(g(4) = (4)^2 - 1 = 15\).
Step 2: Put that output into \(f\): \(f(15) = 2(15) + 3 = 33\).
So, \(fg(4) = 33\).

To find the general expression \(fg(x)\):
Replace every \(x\) in \(f(x)\) with the entire expression \(g(x)\):
\(fg(x) = f(x^2 - 1) = 2(x^2 - 1) + 3 = 2x^2 - 2 + 3 = 2x^2 + 1\).

Inverse Functions

The inverse function, written as \(f^{-1}(x)\), completely reverses the action of \(f(x)\). If \(f(2) = 7\), then \(f^{-1}(7) = 2\).

Important Condition: A function must be one-to-one to have an inverse function. If it were many-to-one, the reverse process would be one-to-many, which is not allowed!

How to Find an Inverse Function Algebraically:
Step 1: Write the equation as \(y = f(x)\).
Step 2: Rearrange the equation to make \(x\) the subject.
Step 3: Swap \(x\) and \(y\), and write your answer as \(f^{-1}(x)\).

Example: Find \(f^{-1}(x)\) for \(f(x) = \frac{3x - 1}{2}\).
1. Set \(y = \frac{3x - 1}{2}\)
2. Multiply by \(2\): \(2y = 3x - 1\)
3. Add \(1\): \(2y + 1 = 3x\)
4. Divide by \(3\): \(x = \frac{2y + 1}{3}\)
5. Swap variables: \(f^{-1}(x) = \frac{2x + 1}{3}\)

Graphical Relationship:
The graph of \(y = f^{-1}(x)\) is the reflection of \(y = f(x)\) in the line \(y = x\).
Furthermore: Domain of \(f^{-1}(x)\) = Range of \(f(x)\), and Range of \(f^{-1}(x)\) = Domain of \(f(x)\).

Key Takeaway: Always remember the order of operations for composite functions (inside first) and note that finding an inverse is simply reversing the operations — geometrically, reflecting across \(y = x\).

2. The Modulus Function

The modulus (or absolute value) of a number is its non-negative size or distance from zero. It is written using vertical bars: \(|x|\).

• If \(x \ge 0\), then \(|x| = x\) (e.g., \(|5| = 5\))
• If \(x < 0\), then \(|x| = -x\) (e.g., \(|-5| = -(-5) = 5\))

Graphs of Modulus Functions

You need to know how to sketch two distinct types of modulus graphs:

1. Sketching \(y = |f(x)|\):
Step 1: Sketch the normal curve \(y = f(x)\) lightly in pencil.
Step 2: Any part of the graph that is below the \(x\)-axis (where \(y < 0\)) is reflected vertically upwards into the positive region.
Step 3: Keep the parts already above the \(x\)-axis unchanged.

2. Sketching \(y = f(|x|)\):
Step 1: Sketch the graph of \(y = f(x)\) for \(x \ge 0\) (the right-hand side of the \(y\)-axis).
Step 2: Delete any part of the graph on the left-hand side (\(x < 0\)).
Step 3: Reflect the right-hand side across the \(y\)-axis to form a symmetrical graph.

Solving Modulus Equations and Inequalities

When solving equations like \(|ax + b| = c\) or \(|ax + b| = |cx + d|\), we account for positive and negative possibilities.

Method 1: Consider Positive and Negative Cases
To solve \(|2x - 3| = 7\):
Case 1 (Positive branch): \(2x - 3 = 7 \implies 2x = 10 \implies x = 5\)
Case 2 (Negative branch): \(2x - 3 = -7 \implies 2x = -4 \implies x = -2\)
Always check your solutions in the original equation to ensure they are valid!

Method 2: Squaring Both Sides (Great for \(|f(x)| = |g(x)|\))
Since both sides are non-negative, squaring eliminates the modulus bars because \(|A|^2 = A^2\).
To solve \(|x + 2| = |2x - 1|\):
\((x + 2)^2 = (2x - 1)^2\)
\(x^2 + 4x + 4 = 4x^2 - 4x + 1\)
\(3x^2 - 8x - 3 = 0\)
\((3x + 1)(x - 3) = 0 \implies x = -\frac{1}{3}\) or \(x = 3\).

Solving Modulus Inequalities:
• \(|x| < a\) means \(-a < x < a\) (the value lies between \(-a\) and \(a\)).
• \(|x| > a\) means \(x < -a\) or \(x > a\) (the value lies on the outside wings).

Key Takeaway: The modulus always gives a positive output. Graphically, \(y = |f(x)|\) bounces off the \(x\)-axis, while solving modulus equations requires testing both the positive and negative cases.

3. Partial Fractions

In AS, you learned how to combine several fractions into one by finding a common denominator. In A2, partial fractions does the exact reverse: splitting a single complex algebraic fraction into a sum of simpler fractions!

Why is this useful? Splitting fractions makes complicated expressions much easier to integrate and expand using the Binomial Theorem.

Proper vs Improper Algebraic Fractions

Before splitting into partial fractions, look at the degree (the highest power of \(x\)) of the numerator and denominator:

• Proper Fraction: Degree of Numerator < Degree of Denominator (Ready to split!).
• Improper Fraction: Degree of Numerator \(\ge\) Degree of Denominator. You must carry out polynomial long division first!

Case 1: Distinct Linear Factors in the Denominator

Form: \(\frac{px + q}{(ax + b)(cx + d)} \equiv \frac{A}{ax + b} + \frac{B}{cx + d}\)

Worked Example: Express \(\frac{7x - 1}{(x - 1)(x + 2)}\) in partial fractions.
Step 1: Set up the identity:
\(\frac{7x - 1}{(x - 1)(x + 2)} \equiv \frac{A}{x - 1} + \frac{B}{x + 2}\)
Step 2: Multiply both sides by the common denominator \((x - 1)(x + 2)\):
\(7x - 1 \equiv A(x + 2) + B(x - 1)\)
Step 3: Choose clever values of \(x\) to eliminate terms (the substitution method):
Let \(x = 1\):
\(7(1) - 1 = A(1 + 2) + B(0) \implies 6 = 3A \implies A = 2\)
Let \(x = -2\):
\(7(-2) - 1 = A(0) + B(-2 - 1) \implies -15 = -3B \implies B = 5\)
Step 4: Write the final answer:
\(\frac{7x - 1}{(x - 1)(x + 2)} = \frac{2}{x - 1} + \frac{5}{x + 2}\)

Case 2: Repeated Linear Factors in the Denominator

When a denominator contains a squared factor like \((ax + b)^2\), you must include a term for both powers:

Form: \(\frac{px^2 + qx + r}{(ax + b)(cx + d)^2} \equiv \frac{A}{ax + b} + \frac{B}{cx + d} + \frac{C}{(cx + d)^2}\)

Top Exam Tip: A very common mistake is forgetting the single linear term \(\frac{B}{cx + d}\). Always include both powers up to the repeated power!

Worked Example: Express \(\frac{2x^2 + 7x - 2}{(x + 1)(x - 2)^2}\) in partial fractions.
Step 1: Set up the identity:
\(\frac{2x^2 + 7x - 2}{(x + 1)(x - 2)^2} \equiv \frac{A}{x + 1} + \frac{B}{x - 2} + \frac{C}{(x - 2)^2}\)
Step 2: Multiply across by \((x + 1)(x - 2)^2\):
\(2x^2 + 7x - 2 \equiv A(x - 2)^2 + B(x + 1)(x - 2) + C(x + 1)\)
Step 3: Substitute strategic values of \(x\):
Let \(x = 2\):
\(2(2)^2 + 7(2) - 2 = A(0) + B(0) + C(2 + 1) \implies 20 = 3C \implies C = \frac{20}{3}\) (or choose cleaner numbers in exam problems)
Let \(x = -1\):
\(2(-1)^2 + 7(-1) - 2 = A(-1 - 2)^2 + 0 + 0 \implies -7 = 9A \implies A = -\frac{7}{9}\)
Step 4: To find \(B\), equate coefficients of \(x^2\) on both sides:
\(2 = A + B \implies B = 2 - A = 2 - \left(-\frac{7}{9}\right) = \frac{25}{9}\).

Case 3: Improper Fractions

If the degree on top is greater than or equal to the degree on the bottom, divide first.

Example Form:
\(\frac{x^2 + 5x + 6}{x^2 - 1} = 1 + \frac{5x + 7}{x^2 - 1} = 1 + \frac{A}{x - 1} + \frac{B}{x + 1}\)

Key Takeaway: Check whether the fraction is proper or improper first. Use smart substitution values to find constants quickly, and equate coefficients if you run out of roots.

4. Transformations of Functions

You encountered basic graph transformations at AS level. At A2, you need to confidently combine multiple transformations and describe their exact order.

Summary of Single Transformations

• \(y = f(x) + a\): Translation vertically by vector \(\begin{pmatrix} 0 \\ a \end{pmatrix}\).
• \(y = f(x + a)\): Translation horizontally by vector \(\begin{pmatrix} -a \\ 0 \end{pmatrix}\) (moves left if \(a > 0\)).
• \(y = a f(x)\): Vertical stretch by scale factor \(a\) (parallel to \(y\)-axis).
• \(y = f(ax)\): Horizontal stretch by scale factor \(\frac{1}{a}\) (parallel to \(x\)-axis).
• \(y = -f(x)\): Reflection in the \(x\)-axis.
• \(y = f(-x)\): Reflection in the \(y\)-axis.

Memory Aid: Outside vs Inside

Outside the bracket affects \(y\) (does what you expect):
\(2f(x)\) doubles the \(y\)-coordinates; \(f(x) + 3\) shifts the graph up by \(3\).

Inside the bracket affects \(x\) (does the opposite / inverse):
\(f(2x)\) halves the \(x\)-coordinates; \(f(x + 3)\) shifts the graph left by \(3\).

Combining Transformations: Order Matters!

When applying multiple transformations, follow standard order of operations:

• For transformations affecting \(y\) (e.g., \(y = a f(x) + b\)):
Perform the stretch \(a\) first, then the vertical shift \(b\).

• For transformations affecting \(x\) (e.g., \(y = f(ax + b)\)):
Factorise the inside first: \(f(a(x + \frac{b}{a}))\).
This shows a horizontal shift by \(-\frac{b}{a}\) followed by a horizontal stretch of factor \(\frac{1}{a}\) (or stretch first, then shift by \(-b\) in the stretched system).

Key Takeaway: Keep "outside = vertical and normal" and "inside = horizontal and opposite" in mind, and always specify stretches with scale factors and translations with column vectors.

Chapter Review & Common Pitfalls

• Domain/Range Confusion: Always identify the input restrictions (like denominators \(\ne 0\) or square roots \(\ge 0\)) when finding domain.
• Inverse Function Existence: Remember that an inverse function \(f^{-1}(x)\) only exists if \(f(x)\) is one-to-one.
• Modulus Signs: When removing modulus brackets in inequalities, remember that \(|x| > k\) splits into two separate inequalities: \(x > k\) or \(x < -k\).
• Partial Fractions: Always check if long division is needed first (if the numerator power \(\ge\) denominator power). Never forget the single power term for repeated factors!

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the difference between the domain and the range of a function?

The domain is the set of all allowable input values (\(x\)), while the range is the set of all resulting output values (\(f(x)\) or \(y\)).

Which types of mappings qualify as true functions, and which do not?

One-to-one and many-to-one mappings are functions; one-to-many mappings are NOT functions.

In what order are functions applied in the composite function \(fg(x)\)?

Apply the inside function \(g(x)\) first, and then apply \(f\) to the result: \(f(g(x))\).

What condition must a function satisfy to have an inverse function \(f^{-1}(x)\), and what is its graphical relationship to \(f(x)\)?

The function must be one-to-one; graphically, \(y = f^{-1}(x)\) is the reflection of \(y = f(x)\) in the line \(y = x\).

How is the graph of \(y = |f(x)|\) obtained from the graph of \(y = f(x)\)?

Reflect any part of the graph lying below the \(x\)-axis vertically upwards into the positive region, leaving parts above the \(x\)-axis unchanged.

How is the graph of \(y = f(|x|)\) obtained from the graph of \(y = f(x)\)?

Delete the portion for \(x < 0\) and reflect the portion for \(x \ge 0\) across the \(y\)-axis.

How do the modulus inequalities \(|x| < a\) and \(|x| > a\) expand without modulus bars (where \(a > 0\))?

\(|x| < a\) becomes \(-a < x < a\), while \(|x| > a\) splits into \(x < -a\) or \(x > a\).

What makes an algebraic fraction improper, and what step must be taken before finding partial fractions?

An algebraic fraction is improper when the degree of the numerator is \(\ge\) the degree of the denominator; polynomial long division must be performed first.

What is the correct partial fraction decomposition form for distinct linear factors \(\frac{px + q}{(ax + b)(cx + d)}\)?

\(\frac{A}{ax + b} + \frac{B}{cx + d}\)

What is the correct partial fraction form when the denominator contains a repeated linear factor \((cx + d)^2\)?

\(\frac{px^2 + qx + r}{(ax + b)(cx + d)^2} \equiv \frac{A}{ax + b} + \frac{B}{cx + d} + \frac{C}{(cx + d)^2}\)

How do changes inside the bracket \(f(ax)\) versus outside the bracket \(af(x)\) transform a graph?

\(af(x)\) stretches the graph vertically by scale factor \(a\), whereas \(f(ax)\) stretches it horizontally by scale factor \(\frac{1}{a}\).

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